Structural Systems by L Stavridis

Structural systems: behaviour and design

Structural systems: behaviourand design

L. T. Stavridis

Published by Thomas Telford Limited, 40 Marsh Wall, London E14 9TP.

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First published 2010

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Structural systems: behaviour and design. Volume 1: Plane structural systems. Stavridis L.T. ISBN: 978-0-7277-4106-6

Structural systems: behaviour and design. Volume 2: Spatial structural systems, foundations and dynamics. Stavridis L.T.

ISBN: 978-0-7277-4107-3

Structural analysis with finite elements. Rugarli P. ISBN: 978-0-7277-4093-9

Steel—concrete composite buildings: Designing with eurocodes. Collings D. ISBN: 978-0-7277-4089-2

Designers guide to eurocode 1: Actions on bridges. EN 1991-2, EN 1991-1-1, -1-3 to -1-7 and EN 1990 Annex A2.

Calgaro J.-A., Tschumi M., Gulvanessian H. ISBN: 978-0-7277-3158-6

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ISBN: 978-0-7277-4105-9

# Thomas Telford Limited 2010

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‘To the memory of my parents’

Contents

Preface xi

Foreword xiv

1 Introductory concepts 11.1 Loads 11.2 The structural behaviour of basic materials 51.3 Behaviour of a reinforced concrete member under tension 131.4 Behaviour of a prestressed concrete member under tension 181.5 Numerical examples 221.6 The design process and its control 26

Reference 28

2 The use of equilibrium in finding the state of stress and deformation(statically determinate structures) 292.1 Introductory concepts 292.2 The handling of internal forces 412.3 Determining the deformations 652.4 Symmetric plane structures 842.5 Grid structures 87

3 The handling of deformations for determining the stress state in framedstructures (statically indeterminate structures) 933.1 Introduction 933.2 The force method 943.3 The deformation method 114

4 Simply supported beams 1394.1 Steel beams (reference material) 1394.2 Reinforced concrete beams 1504.3 Prestressed concrete beams 1634.4 Cantilever beams 1774.5 External prestressing 181

vii

4.6 Design control 184References 185

5 Continuous beams 1875.1 Introduction 1875.2 Steel beams 1885.3 Reinforced concrete beams 1955.4 Prestressed concrete beams 1955.5 Creep effects 1995.6 Composite beams 209

References 221

6 Frames 2236.1 Overview 2236.2 Single-storey, single-bay frames 2236.3 One-storey multibay frames 2356.4 Multi-storey frames 2406.5 Design of sections 2516.6 Plastic analysis and design 2576.7 Design and check of joints 266

References 269

7 The influence of deformations on the state of stress — elastic stability 2717.1 Overview 2717.2. Buckling of bars 2717.3 The influence of deformation on the response of beams

(second-order theory) 2777.4 The influence of deformation on the response of frames 2827.5 Lateral buckling of beams 2907.6 Plastic analysis 293

References 296

8 Arches 2978.1 Basic characteristics of structural behaviour 2978.2 Elastic stability — second-order theory 3048.3 The girder-stiffened arch system 3088.4 The tied-arch system 311

References 313

9 Cable structures 3159.1 Overview 3159.2 Cable—beam structures 3179.3 The freely suspended cable 3229.4 Prestressed cable nets 3259.5 Suspension bridges — the suspended girder 328

viii


9.6 Stiffening the suspension cable 3359.7 Cable-stayed bridges 339

References 349

10 Grids 35110.1 Overview 35110.2 Main characteristics of the structural behaviour of grids 35110.3 Layout and structural action of skew bridges 358

11 Plates 36111.1 The plate equation as a consequence of its load-bearing action 36111.2 Orthogonal plates 36911.3 Circular plates 38511.4 Skew plates 38611.5 Flat slabs 38911.6 Folded plates 399

References 402

12 Shells 40312.1 Introduction 40312.2 The membrane action as a basic design concept 40312.3 Cylindrical shells 40912.4 Dome shells 42312.5 Hyperbolic paraboloid shells 43612.6 Conoidal shells 451

References 453

13 Thin-walled beams 45513.1 General characteristics 45513.2 The basic assumption of a non-deformable cross-section 46013.3 Shear centre 46113.4 Warping of thin-walled beams and the stress state due to its

prevention 46213.5 The bimoment concept 47013.6 Two theorems of the bimoment 47213.7 Warping shear stresses 47313.8 The governing equation for torsion and its practical treatment 47413.9 Examples 477

References 479

14 Box girders 48114.1 General 48114.2 Rectilinear girders 48114.3 Curved girders 495

References 516

ix

Contents

15 Lateral response of multi-storey systems 51715.1 Introduction 51715.2 Formation of the system 51715.3 Lateral response 51915.4 Temperature effect 528

References 532

16 Dynamic behaviour of discrete mass systems 53316.1 Introduction 53316.2 Single-degree-of-freedom systems 53516.3 Multi-degree systems 55516.4 Approximate treatment of continuous systems 57716.5 Design for avoiding annoying vibrations 580

References 586

17 Supporting the structure on the ground 58717.1 Overview 58717.2 General mechanical characteristics of soils 58717.3 Shallow foundations 59217.4 Pile foundations 617

References 621

Bibliography 623

Index 625

x


‘‘Pepaideumnon gar estinepi tosout! to akrib& epizhtein

oson h tou prgmato& ’si& epidcetai’’Aristotlh&

‘‘Because it is the essence of educationto seek as much accuracy

as the nature of things allows’’Aristoteles

Preface

What a technically educated person (engineer, architect or constructor) understandstoday by the term ‘structural design’ is practically the same as what his fellow manmeant 500 — or even more — years ago, namely a procedure based on the applicationof a particular item of knowledge, and because of which a structure will ‘stand up’ andnot ‘fall down’, under whatever actions it may be exposed to during its life span.However, what changed over the years and took the so-called ‘structural theory’ out of

the realm of empiricism and intuition was the introduction of analysis in the assessmentof structural behaviour and its elevation to an applied scientific field. Of course, theevolution of computational methods together with the wide availability of computerfacilities played decisive roles in this.Structural mechanics is now a highly demanding subject, not only from the point of

view of its analytical treatment regarding structural behaviour but also with regard toits evaluation and practical application in structural design. Both of these directionshave quite distinct characteristics.The analytical approach always poses the question: Given a structural configuration and a

certain loading, what is the response and its corresponding deformations? This is a problemgoverned by strict analytical conditions and requirements, establishing in this way the scien-tific character of the subject matter, but sometimes creating the illusion that the processof analysis is an end in itself. Of course, the solution to this problem is nowadays ensured,due to powerful numerical methods and the wide availability of personal computers.On the other hand, the practical aspects focus on the application of load-carrying

behaviour in the conceptual and working design of structures, by posing the essentialquestion: Given the physical environment and the prevailing service requirements, whatstructural system made of the appropriate materials will meet the necessary load-bearingrequirements in an economical and aesthetically satisfactory way? This problem doesindeed constitute an end in itself.It is the first approach which has become established, over the years, in education for

many reasons, all of them hinging on the fact that it is this approach which ensures theappropriate ‘scientific’ profile. The second approach, although more realistic, seldomattracts the attention it deserves in the educational civil engineering curricula, a factthat the student unfortunately becomes aware of at a rather later stage, to his or herdisappointment.

xi

Indeed, the almost exclusive concentration on the computational aspect of structuralmechanics dramatically deprives the young and inexperienced engineer from thephysical perception of the load-carrying characteristics of a structural system, somethingwhich is not due solely to his or her lack of experience. The student concerned, beingused to the study of complicated computational scenarios, tends to lack the mentalclarity that permits the direct structural perception which is so often required eitherby the collaborating architect or the constructor on the building site.In this book, an approach to the understanding of load-carrying mechanisms and the

behaviour of a wide range of structural systems is presented, with subsequent applicationto relevant design decisions, which, relying in principle on physical comprehension, iscarried out through simple analytical reasoning. However, despite the prevailingnon-computer-oriented philosophy, it is recognised throughout this book that thecomputational procedure in the design office practice needs in every case to be under-pinned by the appropriate computer software. A necessary condition, though, for thesuccessful use of such software is the ability of the user to derive, even approximately,results based on a well-cultivated structural perception regarding the load-carryingaction and the subsequent preliminary design, which should in any case precede anycomputer modelling and the corresponding computation: this is exactly what the presentbook aims at.Thus, through the study of basic load-bearing actions and the behaviour of typical

systems, such as simply supported and continuous beams, frames, arches, cable structuresof any type, grids, plates, shells, rectilinear and curved thin-walled beams, andmulti-storey systems, an attempt has been made to bring insights into these load-carryingcharacteristics that are necessary for their design and are usually overshadowed by astrictly analytical examination. In this respect, particular attention has been given tothe use of reinforced and prestressed concrete, as well as to composite structures, inaddition to the ‘traditional’ consideration of steel as the most ‘convenient’ material.Due attention has also been given to the plastic analysis and design of skeletal structuresas well as to second-order theory and stability effects, with an emphasis on their practicaland efficient use.However, it has been deemed an absolute necessity to consider in advance and in

some detail the handling of plane skeletal statically determinate and indeterminatestructures in two respective chapters. Moreover, the dynamic behaviour of discrete-mass structures has been examined in the penultimate chapter, where not only theresponse of multi-storey systems to earthquake but also that of beams and plates tohuman-induced dynamic actions as well as to machine operation is discussed.Of course, complete consideration of structural design must also include the founda-

tion problem. Thus, the final chapter is dedicated to this issue, dealing primarily with thebehaviour and design of shallow foundations, but also with soil—structure interactionand, to a lesser extent, the pile foundations.It is hoped that, through the systematic examination of the above subjects, the basic

elements of structural perception are emphasised which permit the safe preliminarydesign and dimensioning of any structure, such as a bridge, building, roofing of a largespace etc. These elements constitute the basis of not only appropriate computer

xii


modelling but also of the deliberate acceptance — or not — of the numerical resultsproduced by software, which is a very important issue.This book follows a strictly progressive path in the presentation of various subjects.

Thus, later chapter build only on previously examined concepts, and some basicknowledge of elementary mechanics is considered a prerequisite.As is well known — since the first century BC, from the Roman architect Vitruvius — a

successful structural concept requires the satisfaction of four characteristic properties,namely technical safety, functionality, economy and aesthetic quality. Technical safetymeans that the available strength should be greater than the resulting response;functionality means, structurally, the control of different deformational characteristics,annoying vibrations included; economy means the successful selection of the structuraland foundation system, as well as of the appropriate construction method; and aestheticquality means the achievement of structural elegance.While the first two criteria are the subject of a knowledgeable technical analysis,

successfully meeting the last two requires, on the part of the engineer, ingenuity,creativity and aesthetic judgment — properties not acquired by studying the prevailingdesign characteristics of the structural systems but are nevertheless prerequisites.Thus, although the successful design of a structure is based on the technical andaesthetic talent of the engineer, the knowledge of the structural principles put forwardin this book are an essential basis for any such endeavour.This book is aimed at everyone engaged in the study of structural analysis and design,

either as a student, a practising engineer, or even as an architect seeking for a moreprofound structural understanding of bridge or building engineering. I hope that thisbooks proves useful and achieves its goals.Finally, I would like to express my gratitude to Thomas Telford Ltd for undertaking

the publication of this edition of this book and the excellent collaboration with theeditorial team, whose authoritative and knowledgeable direction under Matthew Lanehas led to an entirely satisfactory result.

L.T. StavridisAthens, June 2010

xiii

Preface

Foreword

In the process of designing, engineers have to consider the different alternatives forstructures, aiming generally towards an optimum of functional performance, resistance,durability and also towards a minimum of costs. In the dialogue with architects, theoverall design as well as the suitable structural elements are looked for. Hence, theengineer has to master the art of structural design, has to understand the behaviourand the functions of beams, frames, plates, shells etc. Engineering knowledge is morethan the ability to analyse a structure. It enables engineers to propose structural systemsand also to determine a first estimation of the overall and sectional dimensions needed.A computer aided numerical analysis can be applied only after the design is chosen. It is a

complementary tool, especially helpful to obtain results for dimensioning the structure andfor parameter variations. For checking and validating computer results it is needed to applyengineering judgement, plausibility considerations and e.g. simple equilibrium checks.An engineer is a good engineer when he or she — even on vacation — does not only

admire structures like bridges, towers, buildings, but when he or she is visualising theflow of forces, the structural elements for tension and those for compression, perhapseven imaging moment diagrams. This is rather easy where ‘form follows function’(L.H. Sullivan) as for instance in suspension bridges, yet difficult for more complexstructures such as thin shell structures.The book of Professor Stavridis Structural systems: behaviour and design follows the above

issues very consistently, making a remarkable contribution to this rather rarely encounteredbook genre. It is an excellent textbook, especially for Civil Engineering students and youngprofessionals. It covers the principal approaches to determine states of equilibrium and tocalculate the deformations. For statically indeterminate structures, alternative methodsare demonstrated using either forces or deformations as unknowns. In explaining the struc-tural behaviour of beams, frames, plates, shells, thin-walled elements, pre-stressed concrete,foundation elements and dynamically incited systems, basic concepts and directly applicableapproaches are presented that allow an estimation of crucial quantities for the design of avery wide range of structures, including bridges, buildings, roofing of larger spaces and foun-dations. Many graphic figures help to visualise the underlying mechanics. Thus, the mainobjective is very well achieved: to promote structural understanding as a complementarytool to computer analyses, to make students and readers fit for the art of designing.

Heinz DuddeckProfessor emeritus Dr.Ing., Dr.Ing. E.h.

Institute of Structural Analysis, Technical University of Braunschweig

xiv

Foreword

Methods of structural analysis have experienced an explosive growth during the last 40years. But it was the advent of powerful personal computers, along with the evolution ofnumerical tools (based mainly on the finite element method) and the parallel develop-ment of numerous reliable, comprehensive, commercially available computer software,that have enabled the engineer to tackle very complex structural systems. As aconsequence, in today’s design offices, analysis of even some rather simple systems isperformed (especially by the younger generation of engineers) with the use of suchcomputer codes. Classical as well modern methods of structural analysis (based on theprinciples of virtual work, compatibility of deformations, matrix analysis) are ratherrarely invoked in everyday practice. Yet, these theoretical tools often constitute themajor (if not the only) part of the curriculum in civil engineering schools.Several problems may arise from this state of affairs. First, the danger of the ‘black-box

syndrome’: when a sophisticated code is used without the analyst having the ability tocheck if the results are indeed reasonable and to spot any errors in the physical meaningof his/her implicit assumptions and on how these assumptions are materialised in themodel. Second, there is little if any training to help the young engineer develop adeeper understanding of how structural systems behave, let alone to sharpen his/herphysical intuition; such understanding and intuition are necessary especially in theconception and preliminary design stages. Indeed, conceptual clarity and physical insightare rarely mentioned as key objectives of structural analysis courses.The book by Professor Leonidas Stavridis offers a much needed addition to classical

computational structural analysis: a physical approach is developed in which a structuralsystem is decomposed into elements whose behaviour to the applied loads is easilycomputed ‘from the basics’. Starting in the first chapters with fundamental conceptsand applications, the step-by-step exposition becomes progressively more advanced.Structural analysis blends naturally with mechanics of materials — the latter includereinforced and pre-stressed concrete, steel and composites. The in-depth analysis ofstandard structural systems (such as simply and multi-supported beams, frames,arches, cabled beams) is followed by the exposition of some more advanced topicssuch as buckling, plates and shells, thin-walled and box girders, grids and curvedbeams, laterally-loaded multi-story frames and shear walls.It is amazing how the analysis of such complex systems is made so simple, clearly

understandable even to a non-specialist civil engineer, as the present writer. This isaccomplished to a large extent thanks to the numerous illustrative figures (sketches)

xv

which go far beyond the usual ‘formalistic’ figures of most available textbooks: they areimaginative, vivid, self-explanatory. What a difference they make when trying tocomprehend difficult topics! For instance, the chapter on ‘Shells’ contains 51 elaboratefigures, most of which comprise several sketches while a few of them are a whole pagelong. The 3D nature of cylindrical, spherical, paraboloid, and conical shells is elucidatedwith the help of ingeniously-selected isometric views and numerous cross-sections, sothat the reader feels that this is a rather simple subject.As an engineer with special interest in soil—foundation—structure interaction I was

particularly happy with the comprehensive treatment of foundations. Viewed mainlyfrom a structural engineer’s viewpoint, the pertinent chapter deals not only with someclassical deformation-settlement and stress-distribution problems, but also with theinterplay between foundation stiffness and structure distress.I believe this book will prove invaluable to both students and active engineers in helpingthem not only to absorb a huge volume of material but (more significantly) to cultivate‘engineering intuition’ and develop insight into the physics of structural analysis. Forstudents, in particular, all this will offer the motivation for further study and the desireto later apply in real-life projects both the material and the methodology developed inthe book.

George GazetasProfessor of Soil Mechanics and Foundation Engineering

National Technical University of Athens

xvi


1

Introductory concepts

1.1 LoadsLoads constitute the raison d’etre of structural systems, and therefore their examinationprecedes anything else. Structures are built and designed in such a way as to carry, safelyand in a functionally satisfactory way, only certain loads, and it is the responsibility of thedesigner to prescribe the loads which the structure is logically expected to be exposed toduring its life.The determination of loads raises a difficult problem. Because of the obvious need to

design structures by loading them in a manner agreed in advance for each category,structural loads are subject to regulations which typically differ from country tocountry. Thus, for example, the same bridge would be designed for different loads inthe USA than in the UK or in Japan.Before examining the causes of the existence of loads, a distinction between static and

dynamic loads should be made.A load P is considered to act statically when the time t1 needed for its full develop-

ment is definitely longer than the fundamental period T of the structure. This correspondspractically to the time it takes for a structure to perform a complete oscillation, when anarbitrary deflection from its equilibrium situation is imposed on the structure and thenleft free to oscillate (Figure 1.1).Thus a wind gust increasing from zero to its maximum in 3 s represents a static force

for a short, stiff building having a fundamental period of 0.5 s, whereas for a tall, flexiblebuilding having a fundamental period of 6 s the wind gust must be considered as adynamic loading. It is clear that the way in which a dynamic loading is handleddiffers radically from that of a static one, for the simple reason that, because of theinduced permanent motion, inertia forces are developed, which depend — at eachmoment — on the varying corresponding displacement of the structure. It is becauseof this complexity that it is always preferable, when feasible, to replace a dynamicloading with an equivalent static one.Loads, from consideration of their natural origin, can be categorised as below.

(1) Gravity loads (g¼ 9.81m/s2)Loads which consist of the weight of the permanent components of a structure arecharacterised as dead loads. They are determined according to the specific weight ofeach material (e.g. reinforced concrete¼ 25.0 kN/m3, steel¼ 78.5 kN/m3). Theseloads are clearly static ones.

Structural systems: behaviour and design Copyright Thomas Telford Limited # 2010978-0-7277-4105-9 All rights reserved

Loads due to the occupancy or the use, in general, of mainly horizontal surfaces arecharacterised as live loads. This use may arise from human activities (rooms in residencesor offices with equipment, people on pedestrian bridges or standing in stadiums, etc.) orfrom equipment installed in industrial spaces, or it may represent traffic loads due tomoving vehicles on road bridges, as well as trains on railway bridges.These loads are expressed either as surface loads (kN/m2), or as concentrated loads

(kN). They are termed ‘live’ because they may change their position in the structurepermanently. This fact always poses the question of the configuration of these liveloads, which then leads to consideration of the most unfavourable response in astructure.Live loads — especially moving traffic — are usually considered to act in a static

manner, even if, from their very nature, they act dynamically. This is accounted forby an additional percentage of the acting vehicle weight, known as the impact factor.However, these traffic loads also always induce horizontal braking forces, whichusually constitute about 5% of their corresponding weight.Normal human activities on specifically formed surfaces, as occur in rooms used for

gymnastics and in dance halls, as well as on pedestrian bridges, also have a dynamicinfluence.

(2) Loads arising from the soil that surrounds and/or supports the structure. Earth pressure is exerted on (mainly) vertical surfaces of the structure that come into

contact with the surrounding soil. Earth pressure basically arises from the fact that asoil volume resting freely on a horizontal level usually shows a more or less formedslope, the free sides of which are standing in a state of equilibrium (Figure 1.2).This physical slope angle generally depends on the nature of the soil; for example,for sands this angle is roughly equal to 308. Thus, when this sloped surface isbound, through a structural surface, to be retained at a different angle to the physicalone — usually vertically — the soil exerts on that surface a force that is stronger thegreater the divergence from the physical slope angle.

. This situation is in fact similar to that of the hydrostatic pressure, where the corre-sponding ‘physical slope angle’ obviously equals zero. This is why the earth pressure

2

P

tt1

t1 > T Static loadingt1 < T Dynamic loading

When the system is left free it willvibrate with the fundamental period T

P

P

Figure 1.1 Characterisation of load as a static or a dynamic load


(kN/m2) is determined as the product of the height of soil mass, times its specificweight, multiplied by a factor less than that applied for water, i.e. less than unity.Such additional pressures also act on a retaining wall surface, when live loads haveto be considered on the free horizontal soil surface (see Chapter 16).

. The soil, on which the structure is supported, may exhibit settlement due to causeswhich are either unrelated to the structure (e.g. groundwater lowering) or relatedto the deformability of the structure under loads acting on it. In the latter case,there is a substantial interaction between the foundation soil and the structure.Whatever the case, soil settlement definitely constitutes a loading on the structure,causing internal forces within it, unless the structure is statically determinate (aswill be examined in Chapters 2 and 3).

. During an earthquake the foundation soil exhibits horizontal and/or vertical vibra-tions, which are imposed on the foundations of the structure. As a result of thisdynamic loading, the whole structure is subjected to strong internal forces. As willbe examined in Chapter 16, it is the imposed accelerations on the foundation thatactually cause this state of stress in the structure; the ensuing soil settlement maybe an additional factor, as explained above.

(3) Loads arising from the aquatic environmentThe acting water may be either free, as in the case of a marine structure or a dam, orcontained in the surrounding soil. The main effect of water is the hydrostaticpressure, which acts on every surface embedded within a water-containing environment.In the case of strong water motion relative to the structure, additional hydrodynamicpressures come into play.If the structure is confined with its lower part in soil containing groundwater,

buoyancy forces develop, which are merely the upward hydrostatic pressure on thelower horizontal surface of the structure (Figure 1.3). Although the buoyancy forcedecreases the weight of the structure acting on the soil, the lower surface of the structureis acted on by the unreduced weight in the upward direction.

(4) Loads arising from climatic conditions. Wind forces represent basically a horizontal loading that is dynamic in character, as

explained above. The wind force experienced by a structure is directly related to

3

Retaining wall

Free slope surface

Underlying soil

Earthpressure

Slope angle

Figure 1.2 Earth pressure loading


the form of the structure and its height above the earth’s surface. A thoroughdetermination of wind forces is particularly difficult, as the forces may interact withthe ensuing structural deformations (e.g. in the case of cable-suspended bridges),leading to complex aerodynamic calculations. However, in most cases wind pressurescan be determined from the appropriate structural codes.

. Snow — in fact a gravity loading — is taken into account by means of a prescribedpressure on a roof, according to the appropriate structural codes. The prescribedpressure values take into account the geographic region as well as the altitude ofthe specific site.

. Temperature variation may be uniform (an increase or reduction) or linearlydistributed across the thickness of a structural element. It should be noted that theunavoidable shrinkage of concrete during its hardening may be simulated by auniform temperature fall of 208C. This causes a state of stress in the structure,which may not be negligible (see Chapter 15). A non-uniform temperature changealways appears in structural elements exposed to the atmosphere (space coveringroofs, bridges, etc.). It should be noted that, as in the case of soil settlement, atemperature variation in a statically determinate structure does not give rise to anyinternal forces at all (see Chapter 2).

(5) Special impact loadsImpact loads are dynamic in nature and are due to events such as an explosion, acollision of a body with the structure (e.g. a ship colliding with a bridge pier), etc.

4

The slab is acted on by the full (unreduced) weight of the structure

The soil is acted on by a reduced pressure because of buoyancy

Pressure on the slab

Pressure on the soil

Water table Water table

Water pressure Water pressure

Figure 1.3 Loads in an aquatic environment


1.2 The structural behaviour of basic materialsThe structural strength of a material is determined on the basis of the relationshipbetween the stress and the strain " experienced by an appropriate bar specimensubjected to pure tension or pure compression.Consider a specimen with a length L and a section A, subjected to a force F acting

along its longitudinal axis. Assuming a uniform distribution of the force on thesection, the concept of stress is introduced as

¼ F/A (kN/m2)

In this way the stress is uniformly distributed over the whole section A of the specimen(Figure 1.4). The force F concerns the whole specimen externally, while the stress characterises the internal state of stress of its longitudinal fibres. It may be consideredthat the application of the force F and the development of the stress over thewhole section A represent equivalent factors, given that F¼ A.The result of the application of the force F is an elongation or shortening of the

specimen, depending on whether F is applied as a tensile or as a compressive force.Thus the concept of strain " may be introduced as

"¼ /L (non-dimensional number)

The relationship between with " is obtained experimentally, by loading thespecimen progressively and at a constant rate, starting from a null loading level andcontinuously recording the pairs of magnitudes and " on a diagram over two ortho-gonal axes, as they are calculated on the basis of the measured values of F and .

1.2.1 SteelFigure 1.5(a) shows the experimentally obtained diagram for the behaviour of steel. Thediagram represents a test done on steel in tension, but applies equally to steel incompression. It is clear that, for each pair of values (, "), the force F and the

5

Deformed specimen

LL

δ

σ

F

F

A

A

σ = F/A ε = δ/L

Figure 1.4 The concept of stress


corresponding deformation can be determined as

F¼ A and ¼ " L (a)

It should be noted that the test is made by imposing on the specimen a specific defor-mation " rather than a stress . The rate of application d"/dt of " is approximately5 105/s. It can be seen from Figure 1.5(a) that during the test four distinct behavioursare observed.In region (1) the stress is proportional to the strain ", which means — as will be

shown later — that the corresponding force F is also proportional to the deformation. This fundamental statement, widely known as Hooke’s law, can be written as

¼ " E

The magnitude of E is derived geometrically as the slope of the straight segment (1). Itrepresents the most basic quantitative characteristic of the load-carrying behaviour ofthe material and is called the modulus of elasticity. It can be thought of as a measureof the resistance (stiffness) offered by the material against its unit elongation or short-ening. In the case of steel, the value of E is approximately 2.1 108 kN/m2.Substituting the above two equations (a) into the preceding equation gives the

relationship between the external force F and the deformation of the specimen:

F¼ (E A/L) The expression (E A/L) is called the axial stiffness of the specimen (kN/m). It gives theforce required (kN) to deform (either elongate or shorten) the specimen by 1m; in other

6

fy

fy

σ

εy εy

εy

ε: ‰

εpl

εtot

σ

εtot= εy+ εpl

Idealised σ–ε diagram for steel

(b)

2.0 5.0 8.0

E

Unloading

(a)

(1)

(3) (4)

(2)

0.10 ε

Figure 1.5 General characteristics of the behaviour of steel


words, it gives the resistance (stiffness) (kN) that the specimen offers when subjected toan axial deformation of 1m.The last equation may also be written in the form

¼ F (L/E A)

which shows how the axial deformation of the specimen can be calculated from theexternally applied force F. The expression (L/E A) is called the axial flexibility (ordeformability) (m/kN). It gives the axial deformation of the specimen (m) when anaxial force of 1 kN is applied. It is clear that these two concepts (and values) of ‘stiffness’and ‘flexibility’ are inversely related.The behaviour of the steel in region (1) is called linear behaviour. This is a very con-

venient material property because it is governed by an absolute proportionality betweenthe force F (or stress ) and the deformation (or strain "). Thus if the force F (or thestress ) is multiplied by a factor k, the deformation (or the strain ") is also multipliedby the same factor. However, this property is valid only up to a certain stress level fy, atwhich point region (2) of the behaviour of the steel begins (see Figure 1.5(a)).Up to the point where the stress reaches the value fy, if the load is removed from

the specimen the deformation (as well the strain ") will return to zero. In otherwords, up to this point the deformation is reversible, and for this reason the region upto fy is called the elastic region. For bars used in concrete reinforcement the value of fymay be considered equal to 420N/mm2, and the corresponding "y, i.e. ( fy/Es), isabout 2%. Representative values of fy for the two basic quality types of structuralsteel are 240N/mm2 and 360N/mm2.Once the stress has reached the value fy, the specimen can be deformed further

without an increase in the applied force (i.e. stress). The steel exhibits zero stiffness(the slope of the graph is equal to zero), which means it ‘yields’, and so the stress fy iscalled the yield stress. This state prevails up to a value of " of roughly 10%, where thesteel acquires a little stiffness again, as can be seen from the small slope at the beginningof region (3). To obtain a further increase in the stress ", a small increase in (i.e. in theforce F) is required, and this property of the steel is called hardening.After region (1), the steel no longer exhibits ‘elastic behaviour’. Removing the

loading from the specimen at any time will lead to a course parallel to the ‘elasticregion’, and so will not lead to zero deformation but to a ‘remaining’ deformation,represented by the strain "pl. Thus the specimen cannot revert to its initial length andwill be either a bit longer or a bit shorter, according to whether the stress imposed wasa tensile or compressive one. Therefore, it is said that the material has undergone aplastic deformation, i.e. a plastic strain "pl. It is clear that the total deformation of thespecimen is represented by the total strain "tot (see Figure 1.5(a)), which is given by

"tot¼ "yþ "pl

The material property of yielding, i.e. of deformability under the constant stress fy, is avery important structural property of steel, called the ductility, which confers beneficialeffects on the load-carrying capacity of structures. This holds not only for steel structuresbut also for reinforced or prestressed concrete structures, as will be explained later with

7


respect to the plastic behaviour of these materials. Now, regarding the very small slope ofregion (3), this minimal stiffness is considered to be practically zero, exactly as in theyield state, and consequently the very small increase in stress fy in this region issimply ignored. So, the state of yielding can be considered to extend up to values of "of about 10%, but for practical design purposes the value is taken as 8%. However,for the detailed control requirements in an investigational study, the aforementionedhardening property of steel must be taken into account.It can be see from region (4) in Figure 1.5(a) that to increase the deformation (i.e. of

strain ") further requires increasingly less force F (i.e. stress ), until the point where thespecimen fails. This region is of no practical interest for the design of structures.From the foregoing it can be concluded that, for practical purposes, the maximum

level of stress on the steel must not exceed the value fy. Thus in practice the behaviourof steel can be taken as the behaviour of a homogeneous, ideal, elastic—plastic material(see Figure 1.5(b)). Such behaviour characteristics are used in the development of theclassic methods of structural analysis.This idealised material behaviour is generally used in this book, but special attention

will be given to those peculiarities of concrete structures which do not necessarilyconform to this ideal behaviour.The stress—strain diagrams in Figure 1.6(a) are for naturally hard steel, the strength of

which is due entirely to its chemical composition. This quality of steel is used for bothstructural elements and reinforcing bars. However, when prestressed concrete is used, aprestressing steel of very high strength is needed (for reasons that will be explainedlater), and this is obtained through a cold forming procedure.The stress—strain diagram for high-strength steel is shown in Figure 1.6(c). Comparing

this with the graph for natural steel, it is immediately observed that there is no ‘yieldplateau’, which means that there is not a clearly recognisable yield stress in the waythat has been previously established. However, a yield stress fPy may be considered,which corresponds to the point where removing the load from the specimen will lead toa remaining plastic strain "pl of 2%. This yield stress may be considered as constantafter the value "Py 8% has been reached, as shown in the idealised diagram inFigure 1.6(b). For practical purposes the actual ultimate tensile strength of the prestressingsteel beyond fPy can thus be ignored.

1.2.2 ConcreteSteel exhibits practically the same stress—strain diagram whether the force applied is atension or compression force. However, the behaviour of concrete under tension istotally different from its behaviour under compression, as the compressive strength ofconcrete is much higher than its tensile strength. The stress—strain diagram for aspecimen of concrete under compression is shown in Figure 1.7.It is seen that up to a loading level roughly equal to one-third of the maximum

compressive strength fc the concrete exhibits linear elastic behaviour. The constantslope in this region represents the value of the modulus of elasticity of concrete Ec,which is roughly equal to 25 000 kN/m2.

8


9

Reinforcing steel

Structural steel

E

~E

~240

~400

σ: N/mm2

σ: N/mm2

σ: N/mm2

~1600

~1600

2‰

2‰

8‰ ε

8.0 10.0 ε: ‰

ε

εPy

Idealised σ–ε diagram for prestressing steel

(b)

(a)

(c)

fPy

Prestressing steel

Unloading

Figure 1.6 Characteristics of the behaviour of reinforcing and prestressing steel

1.5–2.5‰

σ: N/mm2

Linear behaviour

~fc/3

~20 to ~45fc

ε

Figure 1.7 Stress—strain diagram for concrete under compression


After the point fc/3 and up to the highest sustainable stress fc the relationshipbetween and " is no longer linear, and consequently Hooke’s law does not apply.The strain corresponding to the highest stress level fc lies roughly between 1.5% and2.5%. The stress fc represents the compressive strength of the specimen. Any furtherdeformation of the concrete requires an applied stress of less than fc. This fact makesconcrete in this last region of the stress—strain diagram, from a design point of view,effectively useless.An interesting fact observed experimentally is that increasing the speed of

imposing the deformation increases the maximum sustainable stress (i.e. strength) fc.It is found that if a stress " is imposed at a speed of 102 to 101/s — which correspondsto the rate during an earthquake — there is an increase in the compressive strength fcof 50%. It should be noted that no such increase in strength is observed in the caseof steel.As has been already pointed out, the tensile strength of concrete fct is much lower

than its compressive strength. The tensile strength can generally be estimated usingthe empirical relationship

fct ¼ 0:45 ffiffiffifc

pðN=mm2Þ

and is of the order of 2N/mm2.Obviously, the concept of a yield stress for concrete simply does not exist. Exceeding

the tensile strength of concrete leads automatically to cracking. In most concrete struc-tures the load under which cracking occurs is only a part of the loading that the structurehas to support under service conditions, i.e. under conditions of everyday use. Thus it isnecessary to place reinforcing bars in the concrete in order to overcome the tensilestresses within the structure. Despite this measure, the concrete can still crack, andthis poses different requirements on the design from those for a steel-only structure.The placement of reinforcing bars actually creates a new composite material, reinforcedconcrete, which is able to carry not only compressive but also tensile internal forces. Theparticularities of the behaviour of reinforced concrete under axial tension are dealt within Section 1.3.

1.2.2.1 CreepIf a compressive stress 0 within the elastic range is applied to a concrete specimen and ismaintained at a constant level for a long time, it will be seen that, after an initial elasticshortening, the specimen will continue to shorten as time progresses. This phenomenonis called creep and is an unavoidable property of concrete (Figure 1.8).This deformation will continue for a period of more than 1 year, and will finally take a

value that is larger than the initial elastic deformation. This extra deformation due tocreep depends on the initial elastic deformation "el and the creep factor ’. The valueof ’ depends on the age of the specimen at the time when the stress 0 was applied(accounted for by a factor k), the relative humidity of the environment (a dryenvironment enhances creep), and the ratio of water to cement weight.

10


The final creep strain "cr, i.e. the creep strain after an infinite time, can be written as(see Figure 1.8)

"cr ¼ ’1 "el ¼ ’1 0

Ec

where

’1 ¼ ’ k

Depending on the age of the concrete at the time when the stress 0 is applied, the factork takes different values, being 1.8, 1 and 0.02 for ages of 1 day, 28 days and 5 years,respectively.It must be pointed out that this deformation is not reversible because, after the

unloading of the specimen, i.e. after the removal of the stress 0 that caused "el, aremaining deformation is observed.Thus, the total strain "tot of the specimen at an infinite time is

"tot¼ "elþ "cr¼ (1þ’1) "elThe value of ’1 normally lies between 2.0 and 3.0. g

The basic problem with regard to creep is to determine the final strain "1 when an initialstress 0 is applied at time 0 (the ‘age’ of the concrete) and then varies (say, increases)gradually over time until its final value 1 (Figure 1.9). Systematic research over theyears has led to a practical and very functional conclusion known as Trost’s proposal(Menn, 1990) which, assuming that the principle of superposition of creep deformationsis generally valid, can be expressed as (see Figure 1.9)

"1 ¼ 0Ec

ð1þ ’1Þ þ 1 0

Ec

ð1þ ’1Þ

where the value of the coefficient (which depends on 0) is of the order of 0.85.The last relationship may be written in the form

"1 ¼ 0Einit

þ 1 0Edif

11

The compressive stress of concrete remains permanently constant

Immediatedeformation

Gradual (slow)deformation

σ0

σ0

σ0

εel = σ0/E

εcr = φ∞ · εelεtot = εel + εcr

εtot = σ0 · (1 + φ∞)/E = σ0/Einit

Figure 1.8 The creep behaviour of concrete


where

Einit ¼Ec

1þ ’1and Edif ¼

Ec

1þ ’1

This expression of "1 is of particular importance. First, it shows that the effect of creep isgenerally equivalent to a reduction in the modulus of elasticity Ec of concrete. For theinitially applied stress 0 a value of the modulus equal to Einit is considered, whereas forany further gradual increment in the stress — assumed as (1 — 0) — the value Edif

applies. In other words, the concrete responds instantaneously to an initially appliedstress through the value Ec, but over the course of time responds with the value Einit,on the condition that the applied stress remains constant. For any further responsedue to an additional gradual variation in the initially imposed stress, the value Edif

must be applied.In this way, an instantaneous elastic deformation w due to an initial stress results in

the value w1¼ w (1þ’1) by applying the modulus Einit, whereas the gradual appli-cation of an additional stress corresponding to an ‘instantaneously produced’ elasticdeformation w results in the value w,1¼w (1þ ’1) by applying the modulusEdif . Of course the total deformation that actually occurs is w1þw,1. g

When looking at the deformation properties of concrete, shrinkage must also beconsidered. During its curing and drying there is a shrinking of the concrete mass,and this is expressed as the corresponding shortening strain "s. For practical purposesthe value of "s is taken as 20 105. This is equivalent to the effect of a uniformtemperature fall of 208C, assuming a thermal coefficient of 105/8C, a value that isalso valid for steel.

1.2.2.2 RelaxationThe property of concrete relaxation has characteristics opposite to those of creep. Moreprecisely, if a shortening strain 0/Ec is imposed on a concrete specimen through anapplied compressive stress 0 and care is taken such that this deformation is kept

12

Additional deformation

ε = ∆σ/Edif

Influence of time duration

+

ε = σ0/Einit

Einit = E/(1 + ϕ∞)

Edif = E/(1 + µ · ϕ∞)

ε∞

Variation of the initially applied stress (Trost’s proposal)

Gradual (slow) additionalapplication of stress

Instantaneous applicationof stress

σ

σ∞

σ0

σ0

∆σ

tτ0 t∞

Figure 1.9 The creep behaviour of concrete with a variation in applied stress


constant (i.e. is blocked), then as time progresses there is a decrease in the concretestress (Figure 1.10). In other words, the stress required for the initially imposedspecimen shortening to remain constant decreases continuously over time.This reduction in stress under constant strain is called relaxation and is due to the

creep property of concrete. It can be explained as follows: if the initial compressivestress 0 is held constant, the specimen should shorten further over time, accordingto its creep behaviour. It is obvious that, in this hypothetical configuration, in orderto keep the initial shortening intact, an additional tensile stress has to be applied,which of course leads to a reduction in the initial compressive stress (see Figure 1.10).The amount of relaxation is easily determined on the basis of Trost’s relation (see

Section 1.2.2.1). It must simply hold that

"1 ¼ 0Ec

ð1þ ’1Þ þ 1 0Ec

ð1þ ’1Þ ¼ 0Ec

from which it is obtained that

1 ¼ 0 1 ’11þ ’1

In this way, for ’1¼ 2.0 and with ¼ 0.85, the final value 1 amounts to only 26% ofthe initially applied stress 0. This last equation plays a particular role in evaluating thestress in statically indeterminate concrete structures under an imposed deformation, aswill be explained in Section 5.5.1. g

It should be noted that the phenomenon of relaxation occurs also in prestressing steel,wherein there is a percentage loss of the initially applied stress under constant strain.This loss depends on the bar diameter and is roughly in the region of 5%.

1.3 Behaviour of a reinforced concrete member under tensionThe peculiarity of concrete as a structural material, as compared with steel, does not lieso much in the fact that the —" diagram is not linear after a certain stress level, but in

13

Immediatedeformation

σ0

Blockeddeformation

σ∞ < σ0

σ0

Freedeformation

Why the stress decreasesεel

Figure 1.10 The relaxation of concrete


its inability to afford a significant tensile resistance, a fact that makes the placementof reinforcing steel bars necessary, as already noted above. Thus, an understanding ofthe behaviour of a simple reinforced concrete member under axial tension is of basicimportance.Consider a straight concrete member (Figure 1.11) with a reinforcing bar placed at the

centre of its square section and sticking out a little at both ends of the member so that aself-equilibrating pair of tensile forces Z can be applied at its extremities. Let Ac and As

be the cross-sectional areas of the concrete and the steel member, respectively, and Ec

and Es the corresponding moduli of elasticity.Under the action of the forces Z the steel bar becomes elongated. It is of absolutely

basic importance that this steel elongation is imposed on the concrete too. This isachieved through the adhesion or bond forces between the steel and the concrete,which are more efficient if the surface of the steel bar is roughly formed (e.g. ribbed).The bond between the two materials ensures that all along the reinforcing bar "s¼ "c.Obviously, if there is a total absence of a bond then "c¼ 0, which means that theconcrete would be completely uninvolved in carrying the tensile load — somethingthat is undesirable, as will become clear.Before the material stresses near the ends of the member are examined, a section

near the middle of the specimen is considered, in order to determine the tensile stressof the concrete c and the steel s at this point. If Zc and Zs are the tensile forcescarried by the concrete and the steel, respectively, then considering the equilibriumof, say, the left-hand part as a free body (see Figure 1.11):

Z¼ Zcþ Zs¼Ac cþAs s¼Ac ("c Ec)þAs ("s Es)

14

Bond stresses

Bond stresses

Bond stresses

Steel stress

Concrete tensile steel

Equivalent section

Ac,i, Ec

Ac, Ec

Z

Z

Z

ZS

τS

ZS

ZC

ZC

Z

Zτ

σc (fct)

σsσc

Ib

Ib

Ib

As, Es

Z = Zs + Zc

Z = Zs + Zτ

Zτ = Zc

Zs < Z

Figure 1.11 The distribution of stresses due to bond forces


Given the identical strains of the two materials ("s¼ "c), then s¼c (Es/Ec) and, morespecifically, on the basis of the last equation:

s ¼ "s Es ¼Z

As þ Ac

Ec

Es

(a)

and

c ¼ "c Ec ¼Z

Ac þ As Es

Ec

¼ Z

Ac;i

(b)

From the above equation it may be deduced that the concrete behaves as if its sectionwere homogeneous and equal to

Ac;i ¼ Ac þ As Es

Ec

This cross-section Ab,i, which is somewhat larger than Ac, is called the equivalent cross-section (see Figure 1.11). The concept of the idealised cross-section, which will also beused in other cases in this book, allows the material — in this case the concrete — to beconsidered as ‘homogenous’, by considering a cross-section that is larger than its realdimensions.If the initial magnitude of the force Z is small enough, producing a stress c in the

concrete that is lower than its tensile strength fct — thus leaving the concrete uncracked— it is immediately concluded from equation (a) that s <Z/As and consequently,according to Figure 1.11, the tensile force on the reinforcing bar Zs¼As s at theconsidered point is always smaller than the external force Z. However, in order toexplain the longitudinal equilibrium of the reinforcing bar considered as a free body,it must be concluded, and indeed should be intuitively clear, that friction (oradhesion) stresses are developed along the bar on its lateral surface. These are betterknown as bond stresses s, and contribute to the overall equilibrium (see Figure 1.11).The distribution of these bond stresses along the reinforcing bar is shown qualitatively

in Figure 1.11 (Menn, 1990). It can be seen that these stresses develop over a restrictedlength lb and then disappear. This length lb, known as the bond or anchorage length, allowsthe introduction of the external force Z into the concrete in order to produce a uniformlydistributed state of stress. The anchorage length depends on the layout of the reinfor-cement and the quality of the concrete.The bond stresses act on the concrete in the opposite sense to the way they act on the

steel reinforcement, and they explain its equilibrium as a free body, given that no force isacting at its left-hand end, while in the location of the reinforcing section the tensileforce Zc¼Ac c is acting. The integral of the bond stresses acting over the length lbgives the bond force Z which, on the basis of equilibrium of the reinforcing bar, isZ ¼ Z — As s, while from the longitudinal equilibrium of the concrete part Z ¼ Zc

(see Figure 1.11).

15


Because of the distribution of the bond stresses, the steel stress initially has a highvalue, which falls rapidly to a lower value towards the end of the anchorage length,and thereafter remains constant. Conversely, the concrete stress is initially zero andincreases gradually up to a constant value c. Obviously, this pattern of stress distri-bution applies equally, in a symmetrical manner, to the right-hand end of themember. g

The case is now considered wherein the applied force Z increases gradually up to such avalue that the concrete stress c becomes equal to the tensile strength fct. This value ofthe external force Zr, the cracking force, results from the condition

fct ¼Zr

Ac;i

(c)

The concrete is now cracked in some place — not necessarily predeterminable — over thewhole height of its cross-section (Figure 1.12) and, as the steel bar has practically nocontact with the surrounding concrete at the crack, it develops the full cracking force Zr.The steel stress in the region of the crack is increased from the value given by

expression (a) up to the value sr¼ Zr/As. As shown in Figure 1.12, the introductionof the force Zr at the ends of the specimen through the developed bond stresses isapplied again, in an inverse manner, i.e. from ‘inside’ towards the ‘outside’, at bothsides of the crack. A minimal increase in the external force Zr will create furthercracks, as the concrete has already reached its tensile strength limit. The same stresspicture is repeated between two consecutive cracks, as if they were both ‘free’ edgeswith regard to the bond stresses and the development of the steel and concretestresses. It is clear that a new crack will occur each time that the maximum concretestress (max c¼ fct) is reached in a region.

16

Ib Ib

Zr Zr

Zr

Zr

Ib Ib Ib Ib

<2Ib

Next crack location

Crack Crack Crack Crack

Bond stressesCrack

Zr

Zr

σs Steel stress

Concrete tensile stressσc (fct) σc (fct)

fct

σc < fct

Figure 1.12 Cracking and stress distribution in a reinforced concrete member


If two cracks develop at a distance 2 lb apart, they will inevitably lead to a third crackat the middle of the distance between them (see Figure 1.12). As the maximum concretestress at the middle of the two new ‘halves’ is lower than the tensile strength fct, it can beconcluded that the occurrence of further cracking in these two ‘halves’ is impossible. Forthe same reason, if two consecutive cracks are closer together than 2 lb a new crackcannot form between them. Thus lb is the minimum possible distance between cracks,and there is therefore a maximum number of cracks that cannot be exceeded. Thiscrack distribution with a crack spacing of the order of lb is known as the stabilisedcrack pattern (Menn, 1990).It should be noted that the cracking of concrete is essential for the economical use of

reinforcement. Otherwise, the steel would be limited to a working stress s0 which, aspreviously determined, cannot exceed the value s0¼ fct (Es/Ec).However, this process of completion of the crack pattern is feasible only when at the

first crack the steel stress sr is smaller than the yield stress, i.e. if

sr¼ Zr/As< fsy

If this is not the case there will be an undesirable yielding of the reinforcement whereverthe first crack appears, together with an uncontrolled crack width. Thus maintaining thesteel stress at first cracking below the yield stress is an absolutely necessary condition forthe progressive formation of multiple cracks, thus avoiding the danger of the formationof a few excessively wide cracks. This condition leads to a minimum amount ofreinforcement required in a given section, independent of the tensile force acting onit. In practice, many, not always controllable, factors lead the concrete to experiencetensile stress, so that the minimum reinforcement should always be used in order toprotect the structure against such an undesirable situation.This minimum reinforcement As,min for a concrete section Ac can be directly deter-

mined from the last equation if the cracking force Zr is estimated first from equation(c) as: Zr¼ fct Ac. Then, practically:

As,min¼ ( fct/fsy) Ac

It should be emphasised that in the event of a change in the cross-section of the concreteAc along the examined element, the minimum reinforcement must be determined basedon the larger cross-section, as shown in Figure 1.13 for the case of a tensioned elementwith a hole in its interior. If region B is reinforced with a minimum reinforcement based

17

C B C

AB

AC

Minimum reinforcement according to cross-sectional area at C

Z Z

Figure 1.13 The minimum reinforcement required in the case of a change in the cross-section of theconcrete


on the section AB, then the cracking force of region C, which is equal to fct AC, wouldautomatically lead the reinforcement in region B to fail (i.e. yield), leading to theoccurrence of uncontrollably wide cracks. g

Something that is of particular interest in the design of a reinforced concrete structureis, of course, the crack width in the regular crack pattern configuration. It is known thatthe concrete that surrounds the steel protects it from corrosion because of the prevailingpH of the local micro-environment. The crack width that complies with maintainingthis situation is of the order of 0.2—0.3mm. If s is the spacing between the cracks inthe stabilised crack pattern (approximately equal to lb), and "sm represents theaverage strain on the reinforcing steel, then the crack width w, ignoring the elongationof the concrete, is

w¼ s "smAssuming that, approximately, "sm¼ (0.8 s)/Es (Menn, 1990):

w¼ 0.80 s lb/Es

It is clear that, in order to restrict the width of the cracks, the steel stress must be limitedaccordingly. g

Once the above described crack pattern has been attained, the cracked specimen can beloaded further. However, the cracked specimen has a greatly reduced stiffness comparedwith the uncracked specimen, as it is only the reinforcing bar that offers any resistance tothe external tensile force.It is evident from the foregoing that the specimen can be loaded until the value Zy,

when the steel stress Zy/As becomes equivalent to the yield stress (Figure 1.14).Beyond this level the specimen cannot be loaded further. Thus the ultimate tensileload of the specimen is

Zy¼As fsy

1.4 Behaviour of a prestressed concrete member under tensionThe case of a concrete member under tension may be used as an elementaryintroduction to the concept of prestressing.Consider a concrete member with a duct embedded along its central line, and assume

that the duct has sufficient bond with the concrete and contains a loose steel cable wirethat has no bond with the concrete. The cable has a cross-sectional area AP and is

18

The ultimate tensile load isindependent of the concrete section

As Zy = As · fsyZy

Figure 1.14 Ultimate tensile load


anchored at the left-hand end of the concrete member, while its other end hangs outfreely (Figure 1.15).Using a hydraulic jack a tensile force P is applied at the free end of the cable;

simultaneously, an equal and opposite force is exerted on the concrete. The cable isin equilibrium, with the applied force P at its free end and the equal and oppositeforce exerted on it by the concrete at its anchored end. The concrete member, onthe other hand, is in equilibrium due to the compressive anchor force acting at itsleft-hand end and the compressive force P acting through the jack at its other end.The two cooperating systems, i.e. the concrete and the steel, at first work totally

independently of each other, given that no bond forces exist between them. The steelcable undergoes a tension and an elongation under the jacking force P, while theconcrete undergoes a compression and a shortening under the action of the equal andopposite force DP (DP¼ P). When the externally applied jacking force reaches a desiredvalue P0, the cable is anchored to the right-hand free end of the member and thehydraulic jack is removed. A cement grout is then poured into the steel duct, giving acomplete bond between the steel and the cement. The concrete is now subjected to acompressive force P0, while an equal tensile force acts on the cable.After this technical intervention, the composite member can carry any tension force

Z, leaving a compressive stress on the concrete — i.e. without cracks — provided the forceZ is smaller than P0 (see Figure 1.15). As will be seen below, a value of Z slightly greaterthan P0 is needed so that the concrete will be under no stress at all, as would be the casein a reinforced concrete member in its unloaded state. The concrete member may carry afurther tensile load until a cracking load ZPr is reached by exploiting the tensile strengthof the concrete, exactly as described in Section 1.3 (Figure 1.16).The establishment of a bond between the cable wire and the concrete allows the

equations given in Section 1.3 to be applied directly:

P0

Ac

þ fct ¼ZPr

Ac;i

¼ ZPr

Ac þ AP Es

Ec

19

Z Z

Without bond

For Z < P0 the member continues to be compressed

The concrete is compressed with a stress P0/Ac DP = P0

Restoration of bond

P0 P0

DP

DP

Ac

Ap

Figure 1.15 Forces introduced by prestressing


This equation yields the value of ZPr, which is clearly greater than P0. The force Z whichleads the concrete to experience null stress may be determined by omitting from theleft-hand side of the equation the stress fct. It is clear that even this force is slightlygreater than P0.Consider now the cable stress, which has the initial value P0¼ P0/AP at the

prestressing level. On reaching the cracking force ZPr it shows an increase Pwhich, according to equation (a) in Section 1.3, is given by

P ¼ZPr

AP þ Ac Ec

Es

This increase is small in comparison to the initial stress P0. However, while theprevailing cable stress immediately before cracking is (P0þP), immediately aftercracking it is ZPr/AP¼ (P0þP)þ ( fct Ab)/AP. This abrupt change corresponds tothe previously existing concrete tensile force which, after the crack has occurred, istransferred to the cable itself. From this point on and for every further increase in thetensile force, only the cable section AP offers any resistance, and thus the memberhas a much lower stiffness when cracked than when uncracked. The member remainsfunctional regarding the maximum allowable crack width, provided that the furtherincrease in the steel stress does not exceed about 200N/mm2.It should be noted here that, in order to confront the cracking danger better,

reinforcing steel should additionally be used. Up to the moment the external tensileforce ZPr is reached, the reinforcing steel develops only a small tensile stress, as itbegins to be tensioned only after the vanishing of the compressive stress of theconcrete. After cracking occurs, the reinforcing steel, together with the prestressedcable, can be stressed further only up to a level of 200N/mm2. g

20

Z

ZPr

Z

ZPr/Ac,i = P0/Ac + fct

(Conclusion: ZPr > P0)

Without bond

Restoration of bond

Equivalent section

P0 P0

σc = P0/Ac Ac

Ac,i

Figure 1.16 Cracking force


Exactly as in the case of the reinforced concrete member, the ultimate tensile force Zy ofthe member with additional reinforcing steel corresponds to the yield stress of both thereinforcing and prestressing steel (Figure 1.17):

Zy¼AP fPyþAs fsy g

When designing a tensioned concrete member it must be kept in mind that reinforcingand prestressing steel have practically the same modulus of elasticity, but the yield stressof prestressing steel is about four times greater than that of reinforcing steel. Thus, forthe hypothetical case of a concrete tension member, using either only prestressedsteel or only reinforcing steel, and assuming that the concrete and the steel have thesame cross-sections in both cases, it can be concluded that (see Section 1.5):

. The cracking force is much greater for a prestressed member than for a simply rein-forced member. The elongation of the prestressed member is far less than if the forcewere taken up by the prestressing steel alone.

. The ultimate tension load resulting from the yield stress of the steel is about fourtimes greater for the prestressed member than for the reinforced one.

It is interesting to note that the cost ratio of prestressed to reinforcing steel for the samecross-sectional area is less than 4.

1.4.1 Loss of prestressAt this point it should be pointed out that, because of the creep and shrinkage ofconcrete, there will be some shortening of the member, and consequently of thetensioned cable itself, leading to a reduction in the prestressing force. This loss ofprestressing force can be determined on the basis of Trost’s relation, as described inSection 1.2.2.1.Due to the initial prestressing force P0, a uniform compressive stress 0¼P0/Ac is

applied to the member, whereas the developing prestressing loss P0 induces agradual stress variation 0¼ —P0/Ac in the member.The total shortening strain of the concrete "tot is due to the superposed effects of 0

and0 on one side, and the shrinkage strain on the other. According to Trost’s ‘consti-tutive relation’ (see Section 1.2.2.1),

"tot ¼0

Ec=ð1þ ’Þ þ0

Ec=ð1þ ’Þ þ "s

21

As · fsy

AP · fPy AP · fPy

As · fsy

As

AP

The ultimate tensile load is independentof the prestressing force P0

Figure 1.17 The limit tensile load of a prestressed concrete member


Now, the shortening " of the member due to creep and shrinkage will be

"¼ "tot "0

where "0¼0/Ec represents the instantaneous member shortening due to 0. As " isidentical to the shortening strain of the cable, the loss of prestress P0 will be

P0¼" EP AP

From the above equations, if the elastic shortening due to P0 is neglected, the pre-stressing loss is obtained as

P0 ¼ n 0Ac þ "s EcAc

1þ n ð1þ ’Þ

while the ratio of the loss to the initial prestressing force is

P0=P0 ¼ n ’þ "s ðEb=0Þ1þ n ð1þ ’Þ

where

n¼EP/Ec and ¼AP/Ac

It is clear that the force (P0 P0) should be taken as the effective prestressingforce for the member, and it is this force, rather than P0, which should not beexceeded by an external tensile load if the member is to be kept uncracked.Moreover, it is clear that the ultimate tension load Zy of the member remains unaffectedby the loss of prestress P0.

1.5 Numerical examples

1.5.1 Reinforced concreteConsider a reinforced concrete bar, 3m long and having a cross-section 30 30 cm(Ab¼ 0.09m2). The reinforcement area is As¼ 8.00 cm2, and Ec¼ 3 107 kN/m2 andEs¼ 2 108 kN/m2.The cracking load Zr of the bar can be determined according to equation (c) in

Section 1.3, assuming a tensile strength of the concrete fct of 200 kN/m2:

200 ¼ Zr

0:302 þ 8:00 104 6:67Thus Zr¼ 19.07 kN.This force causes a steel stress according to equation (a) of

s0 ¼19:07

8:00 104 þ 0:302 0:15¼ 1333:6 kN=m2

with a corresponding strain "s¼s/Es¼ 1333.0/(2 108)¼ 6.66 106. The barelongation is ¼ (6.66 106) 3.0m, i.e. practically negligible.

22


Up to this point the concrete remains uncracked, but immediately afterwards the barhas to be considered as cracked, and thus only the reinforcing area can offer a resistanceto the external force. This causes an abrupt change in the steel stress and an abruptchange in the axial stiffness of the bar. The steel stress becomes

sr ¼19:07

8:00 104¼ 23 837:0 kN=m2 (abrupt change)

The member can be further loaded and still remain functionally usable with an acceptedcrack width until the steel stress s reaches 200 000 kN/m2. That means that it can beloaded up to the value

Zs¼ 19.07þ (200 000.0 23 837.0) (8.00 104)¼ 160.00 kN

The member elongation is

¼ 200 000:0 1333:6

ð2 108Þ 3:0 ¼ 2:98 103 m (acceptable value)

The ultimate load in tension Zy corresponds to the steel yield stress ( fsy¼ 420000kN/m2),and is obtained as

Zy¼As fsy¼ (8.00 104) 420 000.0¼ 336.0 kN

The variation in the reinforcement stress with the increasing external force Z is shown inFigure 1.18. At a very early stage, i.e. at the cracking load, the abrupt change in stiffnesscan be observed, which means that the use of the bar as a tensioned element in afunctionally satisfactory way (small crack width, acceptable deformations) is veryrestricted.

23

σs: kN/m2

420 000

200 000

23 837

19.1 336.0160.0

Cracking Ultimate tensile loadFunctionally adequate

Z: kN

8.0 cm2

0.30 m

0.30 m

Figure 1.18 The variation in the steel stress in a reinforced section


1.5.2 Prestressed concreteThe same member as in Section 1.5.1 is considered, but this time the reinforcement (ofthe same sectional area) is prestressed steel with a yield stress fPy of 1 600 000 kN/m2.The reinforcement is prestressed with an initial stress equal to 70% of its yield stress,

i.e. with a force approximately equal to 900 kN ( 8.00 104 0.70 1 600 000). Thesame force is also applied to the concrete:

c ¼900:0

0:302¼ 10 000:0 kN=m2 (acceptable value)

P0 ¼900:0

8:00 104¼ 1 125 000:0 kN=m2

The member can be loaded, remaining at first in an uncracked condition, after therestoration of the bond, as set out in Section 1.3. In order to reach the cracking forceZPr, the null concrete stress must first be attained, and then the maximum tensilestress of the concrete. This means that the cracking force has to produce in the homoge-neous (uncracked) concrete member a ‘tensile stress’ equal to cþ fct, i.e. equal to(10 000þ 200) kN/m2. According to Section 1.4,

10 200:0 ¼ ZPr

0:302 þ 8:00 104 6:67Thus ZPr¼ 972.4 kN.The load that will overcome the concrete stress is then 972.4 10 000/102 000¼

953.3 kN. This value is somewhat higher than the value of 900 kN, as was previouslynoted would be the case.Due to ZPr, the initial stress P0 of the reinforcement increases by P:

P ¼972:4

8:00 104 þ 0:302 0:15¼ 68 000.0 kN/m2 (small in comparison to P0Þ

From this point on it is the reinforcement alone that carries the increasing externaltension force.It is understandable that, after cracking, there is an abrupt increase in the

reinforcement stress, because the tensile concrete force is transferred to the steel.This increase is equal to

200.0 0.302/8.00 104¼ 22 500.0 kN/m2

The member can be loaded further and remain functionally usable with an acceptedcrack width up to the point where the resulting increase in the steel stress P, beyond theone that prevailed immediately before cracking, reaches 200 000 kN/m2. This maximumusable load Zs can be obtained as

Zs¼ 972.4þ (200 000.0 22 500.0) (8.00 104)¼ 979.2þ 151.4¼ 1114.4 kN

Now consider the case when a mild steel reinforcement, say with the same cross-section As¼ 8.00 cm2, is also present. Initially this reinforcement is compressed, butunder the load of 953.3 kN it will also have null stress. Up to the cracking load it

24


develops a stress, according to equation (a) in Section 1.3, equal to

sr ¼972:4 953:3

2 8:00 104 þ 0:302 0:15¼ 1284:8 kN=m2 ðnegligibleÞ

After cracking has occurred, the effective section area consists of both the prestressedand the reinforcing steel. Further loading of the member can occur up to the point Zs

only, where the stress increase in both reinforcements of 200 000.0 kN/m2 will nothave been surpassed:

Zs ¼ 972:4þð200 000:022 500:0Þð8:00104Þþð200 000:01284:8Þð8:00104Þ

¼ 1273:4 kN

The ultimate load Zy corresponds to the steel yielding stress.In the case when there is only prestressing reinforcement:

Zy¼AP fPy¼ (8.00 104) 1 600 000.0¼ 1280.0 kN

In the case when there is both prestressing and reinforcing steel, it can be concluded thatit is the mild steel that will yield first. Therefore, finally:

Zy¼AP fPyþAs fsy¼ 1280.0þ (8.00 104) 420 000.0¼ 1280.0þ 336.0

¼ 1616.0 kN

The stress in the prestressing steel during the gradual increase in the external force Zis shown in Figure 1.19. The following conclusions may be established:

25

Pre

stre

ssin

gC

rack

ing

Ulti

mat

e lo

ad

Ult

imat

e lo

ad

Fun

ctio

nally

adeq

uate

Fu

nct

ion

ally

adeq

uat

e

σs: kN/m2

1 600 000

1 393 000

1 193 0001 125 000

0.30 m

0.30 m

0.30 m

0.30 m

8.0 cm2

8.0 cm2

8.0 cm2

Z: kN900 972 1114 1280

1273

1616

Figure 1.19 The variation in the stress in the prestressed reinforcement


. The capacity of a member to carry a tensile load in a functionally acceptable way is fargreater when prestressed steel is used rather than mild reinforcement.

. The ultimate resistance of a member against a tensile load for the same cross-sectionalcharacteristics is about 3.80 times greater for a prestressed concrete member than fora simply reinforced member.

The superior resistance of prestressed concrete is due to the high strength of theprestressing steel, under the condition of course that the quality of the concrete usedallows it to be compressed with a relatively higher compressive stress. The crackingload is directly dependent on this compressive stress, as discussed previously. Moreover,given that a certain loss of prestress of the order of 15% occurs, due mainly to timeeffects (creep), the initial prestressing of steel should be as high as possible. This naturallycannot be achieved within the allowable stress range of the reinforcing steel, and thus forprestressed concrete the use of high-strength steel is of primary importance.

1.6 The design process and its controlA structure is formed and designed in such a way that it will be able to carry the appro-priate loads in a safe and functionally satisfactory way.The process which, by taking into account the functional requirements of a project,

results in a concrete structural form, together with all necessary details, including thefoundation and possibly also the method of construction, is called design.The loads for which the structure is designed are determined, as mentioned

previously, according to specific codes. These loads cause, under the service conditionsof the structure, a certain state of stress and deformations, which can be deduced on thebasis of the geometrical, cross-sectional and construction characteristics of the examinedstructural form. The determination of this state of stress and deformation is the object ofthe structural analysis.Since the time when structures became subject to strict analysis (i.e. roughly since

1850), and up to about 40 years ago, the control of strength consisted mainly in thecontrol of stresses (i.e. controlling the state of stress that results from the serviceloads, according to the elastic analysis of the structure, such that it is less than, or atmost equal to, some allowable limits). These limits (stresses) were defined withinspecial structural codes as a percentage of the yield stress with respect to steel, or as apercentage of the compressive strength with respect to concrete. Although thismethod of design has functioned, generally speaking, satisfactorily for over 100 years,it was at some point, and after a number of incidents of damage to or collapse ofstructures, realised that the ‘determination’ of stresses is by no means a ‘sure’ issuedue to several uncontrollable or unverifiable factors. Consequently, a more consistentapproach to the subject of safety should now be followed.If P represents the maximum service loading for a structure (according to the appro-

priate structural codes) and a certain factor of safety (>1.0) is agreed in advance, thestructure should be safe against collapse for every load PþP less than P, while for aload P (or greater) the structure may reach a state of collapse.

26


According to the ‘old’ method, the so-called working stress design, if the load P wouldcause stresses equal to the magnitude of (ultimate strength/), it was assumed that aload P would lead to the ‘ultimate strength’ (i.e. to the collapse of the structure) and,moreover, that a load PþP < P should produce stresses that would be lower thanthis ‘ultimate strength’. Of course, according to the structural knowledge now available,both these assumptions are false.The concept of the factor of safety for a designed structure can be applied through the

theory of plastic analysis, which will be examined later in this book. However, the factorof safety can be practically ensured by means of the following consideration (seeChapters 5 and 6).Let R represent the ultimate resistance (strength) of a designed section in some part of

the structure, and let S represent a sectional force of that section corresponding to theload P, under the condition that the equilibrium requirements are satisfied. The‘sectional forces’ S account for the whole state of stress of the section under consid-eration, including its axial, bending, shearing and torsional stresses (this will be clarifiedin Chapter 2). However, until now, the examination of the state of stress has beenrestricted to the axial stress, and for the time being the axial force can be consideredas being representative of the ‘sectional force’ concept.If the structure is designed in such a way that the relation

SR (a)

is satisfied everywhere, then it can be proved that the structure will collapse at a loadgreater than or equal to P, which means that the existing factor of safety is greaterthan or equal to .In recent structural codes, the above relationship is given in the following form:

S SR/R (with S R¼ )

The factor S is intended to cover the uncertainty regarding the application of loads, aswell as all kinds of inaccuracy or inconsistency in the design of the structural model.Clearly, therefore, smaller factors are applied for permanent loads than for live loads(see e.g. Eurocode 1.3 for permanent loads and 1.5 for live loads).The factor R expresses the uncertainty with regard to the accepted strength of the

materials used (being greater for concrete than for steel). Thus, the control of technicalsafety lies in the verification that the ‘magnified’ internal forces during the ‘state ofservice’ are smaller than the reduced strengths of the designed sections.However, it should be emphasised that the dominant condition for safety remains the

initially formulated one, according to which a load of greater than or equal to P (butnot less) should lead to collapse. This is noted in particular because, in cases where thestructural safety is connected to the so-called buckling load — which is examined in detailin Chapter 7 — the mere satisfaction of the condition (a) above under simple equilibriumrequirements is not relevant.Of course, the control of a designed structure at the ultimate load level alone does not

suffice, as it does not ensure satisfactory behaviour of the structure in everyday use, a

27


factor that is equally important. Thus, there must be additional control of the responseof the structure to the service loads ‘P’, taking into consideration excessive deforma-tions, excessive crack widths in the case of concrete, or perhaps annoying vibrationsinduced by people using the structure or by machines, etc.Finally, it should be noted that in some special cases the design of a structure could be

based on some ‘allowable stresses’, which should not be exceeded.

ReferenceMenn C. (1990) Prestressed Concrete Bridges. Basel: Birkhauser Verlag.

28


2

The use of equilibrium in finding thestate of stress and deformation(statically determinate structures)

2.1 Introductory conceptsLoad-bearing structures consisting of longish members of, usually, constant cross-section, the largest dimension of which does not exceed about one-fifth of themember length, are called skeletal structures. These members, which are usually recti-linear but may also be curved, are assumed to be made of one of the materials examinedin Chapter 1. The skeletal structures consisting of these elements must be able to take upall the loads for which they are designed and transfer them safely to the ground throughappropriately designed supports, which together are called the foundation.Any such element is by its very nature a three-dimensional one but, in order to be

studied, it is idealised as the centroidal line of cross-sections all along the member(Figure 2.1), which is supposed to fulfil all its relevant bearing properties.In this chapter, all these lines that substitute as the elements of a structure are

considered as belonging to a certain plane, thus forming a plane skeletal system. Thissystem, if appropriately supported, may be loaded either in its own plane, in whichcase it is called a plane structure, or perpendicularly to it, in which case it is called agrid structure. This chapter deals mostly with plane structures.

2.1.1 ForcesThe concept of force is of basic importance. Each force is a vector quantity and is fullydetermined when the following items are known: (a) its application point, (b) its line ofaction (direction) with the respective sense and (c) its magnitude. It is understandablethat with the application point already known, the force is also determined if its projec-tions Px and Py on two arbitrary axes x and y (not necessarily orthogonal) are known. Theunit of force is the kilonewton (kN).The loads on plane structures are considered either as forces concentrated on specific

points or as distributed forces along a skeletal member (Figure 2.2). The latter aremeasured in kN/m, expressing how many kN are uniformly acting on a length of 1malong the element. Usually, when designing structures uniformly distributed loadswith a constant value of kN/m along a member are considered.


2.1.2 Conditions of equilibriumEquilibrium constitutes the most basic condition that must exist and prevail in astructure. It concerns the forces acting on the structure, as well as the forces that acton any part arbitrarily cut out of it.It should be made clear that the following three conditions are necessary and

sufficient for the equilibrium of any forces acting on a plane formation. This meansthat if equilibrium exists in a plane formation, the satisfaction of these three conditionswith regard to the acting forces is implied; and, conversely, the fulfilment of these threeconditions will ensure the equilibrium of any plane formation undergoing these forces.The three conditions are (Figure 2.3):

(1) The sum of the vectorial projections of all the acting forces on an arbitrary axisshould be zero, i.e. the forces should be mutually cancelled. This axis may be anarbitrarily chosen line with an arbitrary positive sense. The vectorial projectionsare signed with respect to the selected positive sense of the axis.

(2) The sum of the vectorial projections of all the acting forces on another arbitrary axisshould also be zero. This second axis should not be parallel to the first. All otherprevious remarks remain the same.

(3) The sum of all the moments of all the forces with respect to an arbitrary point oftheir plane should be zero. This presupposes for the plane considered that anarbitrary rotational positive sense should be selected (either clockwise or anti-clockwise). Thus the moment of each force with respect to the selected pointmay be analogously signed.

30

Foundation Foundation

Real member layout

Support Support

Centroidal axis Members are simulated by lines(model)

Figure 2.1 Idealisation of skeletal members

3.0 kN/m

2.0 kN/m4.0 kN/m

Px

Px

Py

Py

∆s

3.0 · ∆s

Figure 2.2 External forces on a member


The need to satisfy the above three conditions, which are expressed by the threeequations shown in Figure 2.3, has the following basic consequence. If a plane formationis in equilibrium under various forces and moments containing, at most, three unknownpieces of information, then these unknown data can be determined directly using thethree equations of equilibrium.Furthermore, the above equilibrium conditions lead to the following conclusions

(Figure 2.4):

. When a plane formation is in equilibrium under three concentrated forces, thesethree forces pass through the same point. This is explained by the fact that, otherwise,

31

Positive sense Positive sense

Arbitrary selection of axes

x y

OReference point

of moments

EquilibriumΣPx = 0ΣPy = 0ΣMo = 0

Figure 2.3 The three equilibrium conditions of a free formation

The three forces must pass through the same point

P3

P2

P1

P1 P2

P3

A

B

The two self-equilibrating forceslie on the direction AB

The acting forces at themember ends lie on their axes

Equilibrium Triangle of forces

Figure 2.4 The consequences of the three equilibrium conditions

Stress and deformation (statically determinate structures)

the moment of any one force with respect to the point of intersection of the other twocannot be balanced.

It should be noted that in the case when one force is completely known, but for theother two only their directions are known, then both their sense and magnitude may bequalitatively (and quantitatively) determined through the so-called triangle of forces, asshown in Figure 2.4. In the triangle of forces, all forces must follow the same rotatingsense. Thus the triangle can be used to quickly obtain an idea of the relativemagnitude and sense of the two forces, in contrast to the more time-consumingprocess of undertaking a purely analytical manipulation of the equilibrium equations.

. When a plane formation is in equilibrium under only two concentrated forces, thesetwo forces must act on the line connecting their points of application, and the twoforces must be equal and opposite in magnitude. This has a direct consequencewith regard to rectilinear members that are embedded in a plane structure in sucha way that the only forces they can receive are concentrated at their ends and withouta moment. Such (hinged) members develop an exclusively axial response (tensile orcompressive) (see Figure 2.4).

2.1.3 Principle of action and reactionThis is a fundamental principle of statics. When a body I comes into contact in any waywith a body II, then at the common point of contact two forces are developed: (a) body Ireceives from body II the force PI, and (b) body II receives from body I the force PII.According to the action and reaction principle, PI¼ —PII.

2.1.4 Support conditionsThere are three possible ways of supporting a plane skeletal structure at any point —simple, hinged and fixed support. Each way consists of a certain number of restrainedand freely developing displacements with regard to the supported point of the structure.The restrained displacements give rise to the development of forces that act as reactionson the structure.

. Simple support. In this case the displacement of a supported point is restrained (nulli-fied) in a specific direction a. This point may be displaced freely in the perpendiculardirection, and can rotate freely (Figure 2.5). Consequently, at the supported point the

32

Symbol Reaction in a specific direction aa

a

Simple support

Figure 2.5 Simple support


structure undergoes a reaction of only one concentrated force in the direction a, andno force in a perpendicular direction and no moment can be developed there.

. Hinged support. In this case the displacement of the supported point is restrained(nullified) in any direction, but the point may rotate freely (Figure 2.6). Con-sequently, at the supported point the structure may undergo as a reaction anyforce in any direction, while any moment is excluded.The notion of ‘hinged’ may also be used for the connection of two elements or

skeletal formations if, through this connection only, a force may be mutually trans-ferred (according to the principle of action and reaction). Obviously, this forcemay have an arbitrary direction (see Figure 2.6).

. Fixed support. In this case the supported point is restrained in any direction as wellas against rotation (Figure 2.7). As a direct consequence the structure may beacted upon at this point by a force of any direction as well as by a moment.The notion of ‘fixed’ is also used for the connection of two members or even

skeletal formations, when, between them, a force of any direction as well as amoment may be mutually transmitted. g

33

or

Symbol

Force with an arbitrary directionHinge

R

R

Figure 2.6 Hinged support or connection

Symbol

Fixed support Force with an arbitrary direction and moment

R

R

M

M

Figure 2.7 Fixed support or connection


As pointed out previously, a structure is in equilibrium under the acting external loads(forces and/or moments), as well as the reactions it receives from its supports. The systemof all these forces must satisfy the three equilibrium conditions listed above, thus being aself-equilibrated system (Figure 2.8). Furthermore, the supporting ground receives theequal and opposite forces of the reactions acting on the structure. How these forcesare carried by the soil mass itself is something which, although it may relate to thestress state of the structure, is examined as a soil—structure interaction (this is coveredmainly in Chapter 17).

2.1.5 Basic assumptionsAt this point it is appropriate to mention the basic assumptions that govern the analysisof skeletal structures, as considered in the present chapter and Chapter 3.

(1) As has been already mentioned, the length of the members is more than about fivetimes the maximum transverse direction of their cross-section.

(2) The linear relationship between stress and strain is valid: ¼E " (see Section 1.2).(3) The deformations of the structure are so small that the equilibrium, which obviously

exists at the deformed configuration, can be examined in the undeformed condi-tion. This important assumption, which is illustrated in Figure 2.9 and constitutesthe so-called first-order theory, is generally valid for the majority of skeletal structuresdesigned. However, for some structure types and in many cases of skeletal struc-tures, where under certain conditions this assumption has to be abandoned, adifferent approach must be taken to their analysis. These conditions are examinedextensively in Chapter 7 (see also Chapters 8 and 9).

A direct consequence of the two last assumptions is the so-called superposition principle.According to this principle, a structure receiving the total of loadingsA and B develops astate of stress and deformation, consisting of the sum of stress states and deformationsdue to the loadings A and B separately. Thus if a state of stress or deformation due to

34

Reactions on the structure

Actions on the ground

Equilibrium

Figure 2.8 The equilibrium of a structure


the loading A is considered, those states due to k A are obtained by multiplying theformer by the factor k.

2.1.6 Statically determinate formationIf, in a rigid skeletal formation under an arbitrary system of self-equilibrated forces, it ispossible to determine the forces and moments acting on the ends of an arbitrarily cut outpart of the formation on the basis only of the three equilibrium conditions, then theconsidered formation is called statically determinate.The most usual type of a statically determinate formation is the so-called ‘open’ or

‘tree’ skeletal system (Figure 2.10). Indeed, if the formation is cut at an arbitraryplace, then the equilibrium of either of the two pieces, which are now free bodies,will allow the determination of the three unknowns, and thus the specification of theforce and the moment at the point of the cut. It is clear that each cut out piece isconsidered ‘fixed’ to its adjacent one.A more complex type of statically determinate formation is the so-called three-hinged

connection between three formations (I, II and III in Figure 2.11). By considering themutually transferred forces (the action—reaction principle) at each hinge, threeunknown forces appear, namely RI—II, RII—III, RI—III. Each of these forces representstwo unknowns, e.g. the projections on two orthogonal axes x and y. The total of the

35

Undeformed

Undeformed

Deformed

Deformed

According to the first-order theory, the equilibriumis examined in the undeformed configuration

Figure 2.9 The influence of deformation on the equilibrium state

Figure 2.10 Statically determinate formation of the ‘open’ or ‘tree’ type


six unknowns is determined by setting out the system of three equilibrium equations forany two formations, say I and II. The unknowns thus determined ensure automaticallythe equilibrium of the third formation also.A special case of the three-hinged formation is the triangular truss formation of

rectilinear bars, which receives (self-equilibrated) forces (no moments) only at itsnodes (Figure 2.12). According to the above, the members of the truss formation willdevelop exclusively axial forces, and consequently there are three resultingunknowns. The equilibrium of each of the three nodes leads to the consecutive deter-mination of all three axial forces, by requiring each time the vanishing of the sum of the

36

RII–III

RII–III

RI–II

RI–II

RI–III

RI–III

I I

IIII

III

III

The equilibrium of I and II ensures also the equilibrium of III

Three-hinged formationLoading of the formation must be self-equilibrating

Figure 2.11 The three-hinged formation

(1)

(2) (3)

Self-equilibrating forces

Self-equilibratingforces

Equilibrium of projections on the selected axisdetermines an unknown force each time

Figure 2.12 The truss formation


projections of all the acting forces on an axis perpendicular to either of the unknownforces (see Figure 2.12). The addition of a further node linked to two other nodes ofthe existing formation, and so on, creates broader statically determinate formations,which can be analysed in the manner described above.If the formation is not a statically determinate one, it is called statically indeterminate.

For example, the closed system shown in Figure 2.13 is subjected to a self-equilibratingloading. Cutting the formation into two parts reveals five unknowns which cannot be‘covered’ by the three equilibrium equations for either part.

2.1.7 Statically determinate supportThere are two ways of supporting a plane formation (whether statically determinate orstatically indeterminate) such that the developed reactions can be determined usingonly the equilibrium conditions (equations): the simply supported manner and the fixedsupported manner.In the simply supported case (Figure 2.14), one point of the formation has a hinged

support while another has a simple support, restraining the displacement with respectto a direction a. Direction a is not allowed to pass through the hinged support; if itdoes so the structure will not be rigid, being a movable mechanism. Clearly, the

37

Two unknowns

Threeunknowns

Self-equilibrating loading The three equations of equilibrium are notsufficient for the determination of five unknowns

Statically indeterminate formation

Figure 2.13 A statically indeterminate formation

Specified direction: one unknownArbitrary direction: two unknowns

Figure 2.14 A simply supported formation


unknown reaction of the simple support may be determined directly if the momentequilibrium of all the acting forces is considered with respect to the hinged support,the corresponding moment of the hinge reaction obviously being null. The componentsof this reaction about two arbitrary axes are also determined directly by considering theequilibrium of the projections with respect to the perpendicular directions on either ofthese two axes.In the fixed support case, any point of the formation may be fixed (Figure 2.15). The

three unknowns of the reactions are determined by applying the three equilibriumconditions applied to the formation, which is considered as a free body, under theacting loads and reactions.When there are two plane formations, there is another manner of statically deter-

minate support: three-hinged support (Figure 2.16). The two plane formations aremutually connected at any one point with a hinge, while each of them is supportedon the ground also through a hinge. The three hinges are not allowed to be on thesame line, otherwise the structure is not rigid. In such a system there are threeunknown forces — the two reaction forces and the mutually transferred force throughthe common hinge — with unknown direction, which means there are six unknownsin total. As can be seen from Figure 2.16, these six unknowns can be determinedthrough the two groups of three equilibrium equations valid for each one of the twoformations.

38

Arbitrary direction: two unknowns

One unknown

Figure 2.15 A fixed formation

RG

RG

RARA

RBRB

A

G

B

Figure 2.16 Three-hinged support


At this point, the difference between formation and structure should be made clear. By‘formation’ is meant a free body that can be in equilibrium only under the action of a self-equilibrating system of forces. A ‘structure’ may be subjected to any forces and moments(i.e. loads) lying in its own plane, and it is the role of the supports to offer the appropriatereactions in order to establish the equilibrium.If a statically determinate formation is supported in one of the three ways described

above, the structure so created is called statically determinate (Figure 2.17). Clearly,the addition of further supports increases the number of unknowns, making it impossibleto determine them through the use of the equilibrium conditions alone. The structureformed in this case is called statically indeterminate (Figure 2.18). Moreover, it is obviousthat supporting a statically indeterminate formation with a statically determinatesupport leads also to a statically indeterminate structure.The distinction between statically determinate and statically indeterminate structures

is examined more thoroughly in the following section.

2.1.8 Compound structuresThe three types of statically determinate structure examined in the previous section areconsidered to be fundamental. The formation of any plane structure, however complex,is practically based on them. Indeed, by initially setting any one of the three types ofstructure, any point on it may be considered as the ‘ground’ to be used as the supportfor another of the three structural types previously examined. Each point of the new

39

Statically determinate support of statically determinate formationsleads to statically determinate structures

Figure 2.17 The transition from the statically determinate formation to a structure

A statically determinate formation supported indeterminatelyconstitutes a statically indeterminate structure

Figure 2.18 Statically indeterminate support of a statically determinate formation


formed system may be regarded as the ‘ground’ for supporting a further structure, andso forth. The final structure formed in this way is a statically determinate one, whichmeans that all its internal forces can be determined on the basis of the three equilibriumconditions alone.The way to ensure that this determination can be done directly is illustrated using the

example complex structure shown in Figure 2.19. First, the individual parts of thestructure are sought. These may be supported successively in any one of the threeways described in Section 2.1.7. Thus, part I is selected first, then part II is supportedon it (in a statically determinate manner), and finally part III is supported, again in astatically determinate way.Now if the ‘analysis’ of the structure follows the opposite path to that of ‘synthesis’, it

will proceed unimpeded. Indeed, the analysis of the last part (III) will directly yield somereactions, on the basis of which part II may be analysed. Finally, by applying to part I theequal and opposite reactions of parts III and II (which are already known), the wholeanalysis can be completed in a direct manner.It should at first be examined whether any composite structure can be formed in the

above manner. If it can, this means that the structure is statically determinate and rigid.If no composite structure can be formed, then it should be examined whether it ispossible for the structure to become statically determinate by appropriate modificationsto the continuity of some members or to the supports.

40

Analysis of structureOpposite path

Each time a simple structure is analysed All actions are known

R1

R1

R2

R2

R3

R3

Synthesis of structure

II

II

III

III

I

I

I: Fixed supported II: Simply supported III: Three-hinged supported

Figure 2.19 The formation and analysis of a composite statically determinate structure


Cutting the continuity at a point where two parts are monolithically (‘fixed’)connected, means that three magnitudes (i.e. unknowns) are removed. Of course,cutting off a hinged bar means the removal of only one magnitude, namely the axialforce (Figure 2.20).The ‘modification’ of a support also leads to the removal of some effective magnitudes.

For example, modifying a fixed support to become a simple support means the removal oftwo magnitudes, whereas modifying it to a hinge support means the removal of onemagnitude only. Moreover, removing a fixed, a hinged or a simple support leads tothe removal of three, two and one magnitude, respectively.The total number of magnitudes removed in the above way, so that the remaining

structure becomes statically determinate, gives the so-called degree of statical indeter-minacy of the initial plane structure considered (see Figure 2.20).The overwhelming majority of structures encountered in practice are statically

indeterminate. However, becoming familiar with handling statically determinate struc-tures is absolutely necessary for the clarification of the concepts that are essential forunderstanding the load-carrying action of structures.

2.2 The handling of internal forces

2.2.1 Sectional forcesIn order to ensure that any member of a plane skeletal structure has equilibrium as a freebody, there must be a suitable force and moment acting at its ends.Under the loading, the skeletal structure ACBD shown in Figure 2.21 develops the

vertical reactions RA¼ 24.0 kN and RB¼ 36.0 kN. The reaction RA can be determined

41

(–3)

(–3)

(–3)

(–1)

(–1)

(–1)Statically determinate structure

Three times statically indeterminate

Nine times statically indeterminate

Statically determinate structure

Figure 2.20 The degree of statical indeterminacy


by expressing the moment equilibrium of all external forces (uniform load, RA, RB), withrespect to point B (whereby the contribution of RB is null), while RB is obtained from theequilibrium of the projections of the external forces on a vertical axis after RA has beendetermined.Considering point m at a horizontal distance of 3.0m from A and examining the

equilibrium of part (Am) — chosen because it is ‘simpler’ than the part mCBD — thethree magnitudes, which represent the fixed connection between the parts Am andmCBD, are sought. What are specifically sought are the two components N and V ofthe force acting on section m, corresponding to the member axis and the perpendiculardirection to it, respectively, together with the acting moment M. The positive senses ofthe axes N and V as well as of the momentM are chosen arbitrarily. The determinationof N, V and M follows the three equilibrium conditions given in Section 2.1.2:

42

M = –49.50 kN mN = –2.85 kN V = –8.54 kN

5 kN/m

5 kN/m

5 kN/m

5 kN/m

5 kN/m

C

N

M

V

DB

A

m

m

m

m

m

2.0

3.0 4.0 2.0 m3.0

RB = 36 kN

RA = 24 kN

24 kN

24 kN

36 kN

36 kN

2.85 kN8.54 kN

49.50 kN m

49.50 kN m

49.50 kN m

9.00 kN

9.00 kN

The section is acted upon by one force and one moment

Figure 2.21 The internal forces required for the equilibrium


(1) Considering the moment equilibrium with respect to point m, the contributionsof Nand V vanish, and the moment M¼49.50 kNm is deduced directly, thenegative sign expressing the fact that it is applied in the opposite sense than hasbeen assumed.

(2) The equilibrium of the projections on directionN — the contribution of V being zero— allows direct determination of the component N¼2.85 kN, indicating therebythat it is also acting in the opposite sense.

(3) Finally, by considering the sum of the projections on direction V, the contribution ofN automatically vanishes and the component V¼8.54 kN may be determineddirectly. Its sense is again the opposite to what has been assumed.

It is clear that the equilibrium of part mCBD requires at the point m the application ofthe equal and opposite forcesM, V andN (see Figure 2.21). These three magnitudesM,V and N are called the bending moment, the shear force and the axial force, respectively,and constitute the sectional forces at point m. However, it must be made clear that sectionm receives, apart from the bending moment, a unique force, actually the resultant F of Vand N:

F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2:852 þ 8:542

p¼ 9:00 kN

This force, for the case examined, happens to be vertical, as the vertical equilibrium ofthe part Am implies (see Figure 2.21), and it is emphasised that the magnitudesN and Vsimply represent the components of this unique force F with respect to two arbitrarilychosen axes. However, these specific axes have been purposefully chosen along thedirection of the corresponding normal and shear stresses, as will be explained below. g

The action of a bending moment on a cross-section induces normal stresses , whichshould have such a distribution that they offer to the section precisely the actingmoment M (Figure 2.22). The resulting distribution is a linear one, has a zero valueat the centroidal axis of the section, and exhibits compressive and tensile stresses oneither side of it, thus offering a resultant compressive force on one part and a resultanttensile force on the other part of the section. It is understood that these two forces shouldbe equal and with opposite senses in order to offer the moment M when coupled.

43

Mz

D

Z

Compression

Centroidal axis

Diagram of normal stresses

Tension D = ZM = D · z

y

Figure 2.22 The development of bending stresses


It is found that the normal stresses developed at each fibre on the cross-section areproportional to its distance y of the fibre from the centroidal axis, as well as to the actingmoment M, while remaining constant across the width of the section. Moreover, thesestresses are inversely proportional to the moment of inertia I of the cross-section, andthus they obey the relation:

¼ M

I y

As discussed in Chapter 1, the axial force N produces uniformly distributed normalstresses (either compressive or tensile) according to the relation ¼N/A, where Arepresents the cross-sectional area.The coexistence of a bending moment and an axial force leads to a superposition of

the normal stresses (compressive and tensile) due to bending and those (compressiveor tensile) due to the axial force (Figure 2.23):

¼ N

AþM

I y

Moreover, acting of a shear force V on a section causes shearing stresses of the samesense, but not evenly distributed over the height of the section, which naturally sum overthe whole section to this force exactly (Figure 2.24).The developed shearing stress over the height of the section, which remains

constant across the width b of the section at the point considered, is given at a

44

Mz

Compression Compression Compression

Centroidal axis

TensionTension

DM = ZMM = D · zM = D · e1 + Z · e2

D – Z = N

N

DM D

ZZM

e2

e1

Figure 2.23 The stresses caused by the combined bending and axial force

Distribution of shearing stresses τ(rectangular section)

max τ = 1.50 · (V/A)Centroidal

axis

y

τ

V

Figure 2.24 The shearing stresses caused by the shear force


distance y from its centroidal axis by

¼ V Sb I

where S represents the so-called first moment of area of the portion of the section, lyingbeyond the place of the measured width b, with respect to the centroidal axis. g

It is appropriate now to examine how the positive sign of the above sectional forces isdefined.

. Axial force. The axial force N is considered ‘positive’ if it causes tensile stresses(Figure 2.25). In the example examined previously (see Figure 2.22), the axialforce at point m, which is equal to 2.85 kN, causes compressive stresses and is,consequently, ‘negative’.

. Shear force. The shear force V at a point m is considered positive if, together with theshear force of a neighbouring section, which for equilibrium reasons must act in theopposite sense, it constitutes a clockwise couple (Figure 2.26). Thus, in the previousexample, the shear force of 8.54 kN at point m is positive (see Figure 2.21).

. Bending moment. For the bending moment, signing is not physically important. How-ever, it is of primary importance to determine on each occasion on which side of thesection the tensile stresses (or compressive stresses) are developed (Figure 2.27). It isconcluded that in the above example the bending moment of 49.5 kNm causestension at the bottom fibres of section m. While the signing of the axial and shearforce does have a physical meaning, because it suggests the actual sense of theaxial and shearing stresses, respectively, on the section, the signing of the bendingmoment has no such meaning. However, if a so-called positive border is arbitrarilyestablished on the member, then the bending moment may be considered ‘positive’if the tensile stresses it produces are on this positive side; otherwise it is ‘negative’(see Figure 2.27).

45

Positive axial force

NN

NN

m

m

Figure 2.25 The convention for the positive axial force

Positive shear force

Clockwise couple m

mV

VVV

Figure 2.26 The convention for the positive shear force


2.2.2 Determination of sectional forcesThe explanation given in the previous section allows for the systematic determination ofthe sectional forces at every point on a skeletal model.Using the same example as above, in order to find the sectional forces at the

characteristic positions A, Cl, Cr, Bl, Br and D, the structure is cut at the respectiveposition each time, and that part is considered which seems more convenient fromthe aspect of equilibrium examination. The positive sectional forces N and V arethen placed on the section, the selected sense of M being indifferent (Figure 2.28).

. Position A. The tiny cut out part is shown in Figure 2.28, together with the knownexternal force and the unknown sectional forces. Following the determination ofeach force through the equilibrium of the projections on its own axis, it is foundthat N¼7.59 kN, V¼þ22.57 kN and M¼ 0.0 kNm.

. Position Cl. The selected part ACl is shown in Figure 2.28 together with its externalforces and the unknown sectional forces in their positive sense. It is found thatN¼þ1.90 kN, V¼5.69 kN and M¼þ54.00 kNm.

. Position Cr. The selected part ACr is shown in the Figure 2.28. It is found thatN¼ 0.0 kN, V¼6.0 kN and M¼þ54.00 kNm. Note that the values of N and Vhave been differentiated with respect to position Cl, whereas this is not the casefor the bending moment M.

. Position Bl. As the more convenient part for the examination of the equilibrium, partBlD (see Figure 2.28) is selected. It is found that N¼ 0.0 kN, V¼26.0 kN andM¼10.0 kNm.

. Position Br. The examined part is now BrD (see Figure 2.28). According to the positivesenses of the sectional forces, it is found that N¼ 0.0 kN, V¼þ10.0 kN andM¼10.0 kNm.

. Position D. The equilibrium of the tiny cut out part (see Figure 2.28) leads to theconclusion that all the sectional forces there are null. g

46

Tensioned fibres

49.50 kN m

49.50 kN m

(Moment = +49.50)

m

m

mTensioned fibres

Tensioned fibres

Positive border

Positive border

The selection of the positive border is arbitrary

The tensioned side is independent of the positive border

m

(Moment = –49.50)

Figure 2.27 The bending moment and tensioned fibres


The ultimate objective of this analysis is to construct three diagrams, giving for eachpoint of the member the value of the bending moment, the shear and the axial force.The diagrams are ‘read’ in the direction perpendicular to the member axis. Withregard to the axial force and the shear force diagrams, it does not matter on whichside of the member axis the positive values appear, but it is absolutely necessary thatthese sectional forces are signed. However, for the bending moments the conventionmust be followed that they must always appear on the tensioned side of the section.The values of M, V and N obtained in the example considered above are shown intheir appropriate positions in Figure 2.29. The procedure that must be followed in

47

N = –7.59 kNV = +22.57 kNM = 0.0 kN m

36 kN N = 0.0 kNV = –26.00 kNM = –10.0 kN m

N = 0.0 kNV = +10.00 kNM = –10.0 kN m

N = 0.0 kNV = 0.0 kNM = 0.0 kN m

N = +1.90 kNV = –5.69 kNM = +54.00 kN m

N = 0.0 kNV = –6.00 kNM = +54.00 kN m24 kN24 kN

5 kN/m 5 kN/m

5 kN/m

5 kN/m

5 kN/m

V VVM M M

N N

V

MN

V

M

A

A

N

V

MN

Nl

l

r

BD

C

C

r

C

DB D

B D

A A

24 kN

2.0

RB = 36 kN

l r l r

RA = 24 kN

3.0 3.0 2.0 m4.0

Figure 2.28 The systematic determination of sectional forces


order to round out these values to a corresponding continuous diagram is elucidated inSection 2.2.5.

2.2.3 Relationships between loading and sectional forcesIn a skeletal structure an element of lengths is cut out and its equilibrium is consideredunder the transverse load p(s) and the longitudinal load h(s). The parameter s is referredto the longitudinal sense of the element. This element is acted upon by the actionsM, Vand N at one end and by the (generally) modified actions MþM, VþV andNþN at the other (Figure 2.30).Applying the equilibrium conditions for this element, the following differential

relations are readily obtained:

dV

ds¼ pðsÞ dM

ds¼ VðsÞ dN

ds¼ hðsÞ

48

C B D

l r l r

5 kN/m

A RB = 36 kN

RA = 24 kN

[V ]

[M ]

[N ]

7.59

6.005.69

22.77

54.00 54.00

1.90

26.0

10.0

10.0

Figure 2.29 Diagrams showing the calculated sectional forces


On the basis of the first two, an equally important relationship is deduced:

d2M

ds2¼ pðsÞ

From the first relation it may be seen that the ‘bearing mechanism’ of the transverse loadp(s) on a beam is expressed through the variability (dV/ds) of the shear force. Moreover,the last relationship rewritten as

d

ds

dM

ds

¼ pðsÞ

expresses the same fact.The above equations are directly related to the diagrams of the sectional forces M, V

andN — as they are themselves actually functions referred to the axis of each member —and allow the following conclusions to be drawn:

. The slope of the tangent at every point on the diagram of V is identical to the value ofthe transverse distributed load acting at that point.

. The slope of the tangent at every point on the diagram ofM is identical to the value ofthe shear force V at that point. Consequently, wherever the shear force V is null, thediagram of M exhibits either a maximum or a minimum value.

. The slope of the tangent at every point on the diagram ofN is identical to the value ofthe longitudinal distributed load at this point.

Moreover, on the basis of the above, the following may also be concluded:

. A region of a rectilinear member which is not acted upon by a transverse distributedload has a constant V. In the same region M varies linearly.

. In a region of a rectilinear member having a constant transverse distributed load, theshear force V varies linearly. In the same regionM varies according to a parabolic law.

. A change in value in the transverse distributed load means a change in the slope inthe diagram for V (a higher transverse distributed load leads to a greater slope).

. A null longitudinal distributed load means constant N.

. A constant longitudinal distributed load means that the diagram forN varies linearly.

A concentrated transverse force at some point will cause a discontinuity in the Vdiagram, due to a different value of V on each side of the load (Figure 2.31), whereas

49

p(s)

h(s)

∆s

M M + ∆M

N N + ∆N

V V + ∆V

Figure 2.30 The equilibrium of an elementary rectilinear part


the value of bending momentM will be unaffected. However, the slope of tangent of themoment diagram M will be different on the two sides of the load, adapting to the corre-sponding values of the shear force.Finally, the application of a concentrated moment at a certain point will cause a

discontinuity in the bending diagram, with different values of M on either side;however, the corresponding values of V will remain the same.

2.2.4 The simply supported beamKnowing how the bending moment and shear force diagrams for a simply supportedbeam are drawn for some typical loadings is essential for the treatment of any planeskeletal structure involving sectional force diagrams.

2.2.4.1 A beam under a uniform loadBoth reactions are vertical and equal to half of the total vertical load (Figure 2.32). Thediagrams for V and M are shown in Figure 2.32.

50

∆s

M P M

Equilibrium: V r = V I – P

V l V r

Figure 2.31 The influence of a concentrated load

Equilibrium

Tangent

q

q · L/2 q · L/2

q · L/2

q · L/2

P · b/L P · a/L

P · b/L

P · a/L

P · a · b/L

P · a/LP · b/L

q · L2/8

q · L2/8

P

P

A B

L

L a b

[V ] [V ]

[M ][M ]

Figure 2.32 Sectional force diagrams for a simply supported beam


The following may be observed: the shear force diagram has a constant slope andbecomes zero at the midspan, where the maximum value of M occurs. The slope ofthe parabolic diagram for M follows the value of the shear force. The maximumbending moment has the value q L2/8 and causes tension of the lower fibres, as infact occurs over the entire length of the beam. Moreover, it is useful to note the wayin which the parabolic diagram for M is drawn, as shown in Figure 2.32.

2.2.4.2 A beam under a concentrated loadThe reactions and the diagrams for V and M are obtained as shown in Figure 2.32. Thedifferent shear forces on each side of the concentrated load satisfy the equilibrium ofthe tiny cut out segment as shown in Figure 2.32, and correspond to the differentslopes of the bending moment diagram.

2.2.4.3 An oblique simply supported beam under a uniform loadThe inclination angle of the beam axis is equal to a (Figure 2.33). The uniform load refersto the horizontal projection L. The beam receives thereby a constant transverse andlongitudinal distributed load. The specific values of these loads is not of interest. Bothreactions are vertical and equal to the half the total vertical load (q L), i.e. they areidentical to those of the ‘horizontal’ beam.

Diagrams V and N. Position A. From the equilibrium of the tiny cut out element (see Figure 2.33) it is

found that VA¼þ(q L/2) cos and NA¼(q L/2) sin.. Position B. From the equilibrium of the tiny cut out element it is found that

VB¼(q L/2) cos and NB¼þ(q L/2) sin.

The diagrams for V and N are linear due to the constant transverse and longitudinaldistributed loads, respectively.

Bending diagram MMidspan m. The equilibrium of the cut out part Am (see Figure 2.33) yields a bendingmoment equal to (q L2/8), tensioning the bottom fibres, i.e. exactly the same as forthe previous ‘horizontal’ beam. This means that at each point of the oblique beam thedeveloping bending moment is equal to that of the projected point on the ‘horizontal’beam. The parabolic diagram M referred to the oblique axis is drawn in the samemanner as shown previously (see Figure 2.33).

2.2.5 Drawing the bending moment diagram of a memberFollowing the procedure given in the preceding two sections, the moment diagram forany straight member, the bending moments at its ends having been already determined,may be drawn directly. Specifically, it is assumed that the end bending momentsM1 and

51


M2 of the skeletal member 1—2, which is subjected to a transverse loading in an arbitrarydirection, as shown in Figure 2.34, have already been determined. These moments aredrawn accordingly on a diagram [M] (see Figure 2.34).The transverse loading is represented indicatively by a distributed load. The length of

the projection 10—20 of the member in a direction perpendicular to the load is L. Asnoted in the previous section, the moment diagram M0 of the simply supported beam1—2 under the considered load is identical to that of the simply supported beam 10—20.

52

q

B

q · L/2

q · L/2

q ′q ′

Aα L′

L

Total load: q · L = q ′ · L′ (Determination of the load components is not necessary)

[V ]

(q · L/2) cos α

q · L/2 q · L/2

(q · L/2) cos α (q · L/2) sin α

(q · L/2) sin α

[N ]

[M ]

q

A

q · L2/8

q · L2/8

q · L2/8

L/2

L

V

VA B

N

N

q · L/2

m

The two moment diagrams areabsolutely equivalent

The moment in the middle is independentof the beam length

Tangent

Tangent

V: positive V: negativeN: compressive N: tensile

Figure 2.33 The load carrying response of an oblique simply supported beam


The question of how the bending diagram for the member 1—2 is drawn between thevaluesM1— andM2 can now be answered simply. It is drawn by ‘hanging’ the diagramM0

from the straight line connecting the two end valuesM1— andM2, to that side where thetensioned fibres of the beam 10—20 are. It is understood that diagram for M0, like valuesM1 and M2, is drawn perpendicularly to direction 1—2.Note that the shear forces at the ends of a straight member can always readily be

determined from the equilibrium requirements, once the end bending moments of themember are known.

2.2.5.1 Application to a cantileverThe cantilever, being a member that is fixed at one end, generally constitutes a basicelement in the formation of plane skeletal structures, and the understanding of itsstate of stress, as expressed through the corresponding sectional forces diagrams, hasits own importance.The loading of a horizontal cantilever of length L by a vertical uniformly distributed

load q causes at its fixed end the bending moment (q L2/2), which imposes tension inthe top fibres (Figure 2.35). Between this value and the zero value at the free end, thediagram of the simply supported beam is ‘hung’ with the characteristic value (q L2/8) in

53

The diagram may be ‘suspended’from any axis

1′ 2′

q · L2/8

q · L2/8

Tangent

M2

M0

M1

M0

2

1

2

2

1

1

2

1

V1

N1

N2V2

M1

M2N2

N1

V1

V2

q

q

[V] [N]

L

L

Figure 2.34 Sectional force diagrams for a member with known end actions


the middle. Clearly, the resulting diagram should exhibit a horizontal tangent at the freeend, given that the corresponding shear force is null.The same cantilever under a vertical concentrated force at its free end develops at its

fixed end the bending moment (P L), causing tension in the upper fibres. Because of theabsence of a transverse load, the bending moment values at the two ends are connectedby a straight line (see Section 2.2.3).Moreover, Figure 2.35 shows how the bending moment diagram for an ‘oblique’

cantilever is drawn, according to the above.

2.2.6 The three-hinged frameThe preceding sections were essentially devoted to the examination of beams. However,the formation of a skeletal structure that will ‘cover’ or ‘close’ a space leads tothe concept of a frame. In its simplest form a frame consists of a beam (called agirder) as the ‘horizontal’ element and two columns in the ‘vertical’ or ‘oblique’ sensewhich are rigidly connected to it. Clearly, in simply supported frames, like thoseshown in Figure 2.36, the bending response of the girder does not differ at all from

54

The fixed-end moment is independentof the oblique cantilever length

[M ]

[M ][M ]

[V] [V]

q · L2/2

q · L2/2

q · L2/2

q · Lq

q · L2/2

q · L

L L

q · L2/8

q · L2/8

L

q · L

P · L

P

P P

P · L

q

Figure 2.35 Sectional force diagrams for a cantilever


that of the simply supported beam. If, however, the structure is formed as a three-hingedone, as described in Section 2.1.7, this situation changes. The reactions are no longervertical and their horizontal components cause bending moments in the columns, aswell as at the ends of the girders, resulting in significantly reduced midspan moments(see Figure 2.36).Considering the three-hinged arch subjected to a vertical loading shown in Figure

2.37, it can be concluded from the ‘horizontal’ equilibrium requirement that thehorizontal components H of the reactions of the supports A and B are equal inmagnitude and of opposite sense.The moment equilibrium with respect to support B yields a relationship between the

components VA and H. One more relationship between these two components maybe obtained by considering the moment equilibrium of part AG with respect to pointG (see Figure 2.37). The remaining components VB, VG and HG can then be readilydetermined.In the case when the external loading acts in a perpendicular direction to the line

connecting the supports A and B (Figure 2.38), some interesting conclusions may bedrawn. First, on the basis of the previously described procedure, the vertical reactionsVA and VB work out to be identical to those of the simply supported beam AB.

55

Three-hinged frame

Bending moment of simply supported beam

The vertical reactions lead to bending moments identicalto those of a simply supported beam

Figure 2.36 The consequences of the three-hinged formation


The horizontal component H (thrust), may be expressed as

H ¼ MG

f

whereMG is the bending moment of the simply supported beamAB at the correspondingpoint G on the hinge (see Figure 2.38).The last equation has particular importance. It shows that the higher the frame, the

less its horizontal thrust, i.e. the more the reaction approaches the vertical. Conversely,if the frame is made lower, the horizontal thrust H increases and the inclination of thereaction becomes smaller.Another conclusion is that the component H is directed inwards inasmuch as the

bending moment MG of the simply supported beam at G causes tension in the lowerfibres. Otherwise (which is certainly unusual), the component H is directed outwards(see Figure 2.38). g

Finally, a graphical or ‘qualitative’ determination of the reactions and internal forces of athree-hinged arch is presented in Figure 2.39. When one of the two parts of the frame isunloaded, it is possible to determine qualitatively the direction and sense of both thereactions and the internal force in G. Given that the part AG is two-hinged and

56

The developed reactions at A and B are a single forceThe appearing components correspond to arbitrarily chosen directions

Determination of the components of the reaction at A

Moment equilibrium with respect to B Moment equilibrium with respect to G

H

H

H

H

H

H

VA

VA

VB

VB

VA

VG

B

B

A

A

A

G

G

G

Figure 2.37 Analysis of a three-hinged arch


unloaded, it is directly concluded (see Section 2.1.2) that the forces at A and G areacting along the line AG. Due to the fact that the whole structure is acted on bythree forces only (i.e. the two reactions and the resultant of the uniform load), itsequilibrium is governed by the corresponding triangle of forces. The acting forcesshould run in the same rotating sense (see Section 2.1.2). Thus the sense of directionof all the relevant forces is determined, and then, following the procedure given inSection 2.2.2, the sectional forces diagrams can be set out in a purely qualitativemanner (see Figure 2.39).

2.2.7 The funicular structureThe development of bending moments in a skeletal structure leads to a linear distribu-tion of stresses in a section, which means that the bearing capacity of all the longitudinalfibres of the member cannot be fully exploited. Obviously, the optimum utilisation of thematerial used is accomplished only when the fibres of each section are stressed uniformly,which clearly means the development of exclusively axial forces.Thus the question is raised about the form that a structure should have in order to

carry a specific vertical loading by developing exclusively axial stress, being itselfsupported by two hinges at its ends, spanning a distance L (Figure 2.40).

57

The horizontal thrusts are directed outwards

The vertical components are identical to the reactions of the simply supported beam

H

f

H

G

VA VB

VA VB

BA

A B

AB

A B

G

G

G

H = MG/f

MG

MG

[M ]

Figure 2.38 A three-hinged arch with a horizontal base line


This question may be answered directly from a physical point of view, if an absolutelyflexible medium, such as a cable, is fixed at the given points and is subjected tothe loading under consideration. The cable will take such a form that only tensileforces will be developed within it. In the case of a material that can sustaincompression rather than tension, the absolutely equivalent mirror image of the ‘cablesystem’ may be considered, and the loading sense may be reversed, thus causingexclusively compressive stresses (see Figure 2.40). In this way the familiar arch formmay be created.This form is called a funicular structure. It is understandable that at every point the

externally acting load is in equilibrium with the internal axial (compressive) forces. Inother words, the loads are transferred to the supports exclusively through axial(compressive) forces (see Figure 2.40).With regard to the geometrical form y(x) of such an arch, it is understandable (see

Section 2.2.6 and Figure 2.38) that, as the vertical components of the reactions are

58

Triangle of forces

Resultant ofdistributed load

G

G

G

A

A

P

P

B

B

RA

RG

RG

RB

RB

RB

RA

RA

[M ] [V ] [N ]

[M ] [V ] [N ]

Figure 2.39 Qualitative determination of the sectional forces in a three-hinged frame


equal to the reactions of the corresponding simply supported beam, the bending momentof the arch at a point with an ordinate y(x) will result from the corresponding bendingmoment M0(x) of the beam, reduced by the corresponding moment of the thrust H(Figure 2.41). Thus M(x)¼M0(x)H y(x).The requirement M(x)¼ 0 leads to the equation

yðxÞ ¼ M0ðxÞH

which states that the sought-after form is affine to the bending diagram of thecorresponding simply supported beam, by the undefined factor 1/H. This factor isfixed only when a third point of the arch is also defined, as for example the crown

59

Reversing the loads leads to a reversal of the responseDevelopment of exclusively compressive forces

Static configuration identical to the previous one

The cable develops exclusively axial tensile forcesbalancing the loads at each node

L

Figure 2.40 Conception of the funicular structure


height at a distance x0 from the support A. Then, with the term y(x0) already specified,

H ¼ M0ðx0Þy0ðx0Þ

and finally

yðxÞ ¼ yðx0ÞM0ðx0Þ

M0ðxÞ (see Figure 2.41)

The curve y(x) is also called the pressure line. For a constant uniformly distributed loadthe curve is a parabola (Figure 2.42). However, in order to take up a group of concen-trated loads, and according to the bending moment diagram M0, a polygonal funiculararch form is the appropriate one.At this point it should be noted that if the constant uniform load q stays perpendicular

to the axis of a very flexible medium such as a cable, then the cable takes the form of acircular segment with radius R, requiring a certain force P to be applied at its ends. This

60

H H

VAVB

VAVB

BA

H H

VAVB

BA

Pressure line: y(x) = M0(x) · [y0/M0(x0)] H = M0(x0)/y0

y0 (x = x0)

y(x)

x

x0

x

Once the ‘third’ point is specified the solution is ascertained

M = M0 – H · y = 0Infinite number of solutions: y(x) = M0(x)/H

y(x)

M0(x0)

M0(x)

Equal Equal

Figure 2.41 Formation of the funicular structure


force goes through the whole length of the cable as the only existing axial tensile force,and is given by P¼ q R (see Figure 2.42). Obviously, the same funicular form acts as acompression path for the same loading in the opposite sense. It should be pointed outthat, while in this funicular form the axial force is kept constant, in the examinedcase of a constant vertical uniform loading q the axial force in the funicular parabolicform keeps change constantly over the entire arch length, as shown in Figure 2.42.It is clear that the specific line pressure y(x) leads to an exclusively axial stress state

only for the specific loading under consideration. Any other loading will cause bending.Moreover, it is obvious that the slightest deviation from the funicular form under thegiven loading will equally lead to bending. Of course, the greater this deviation is, thegreater the degree of bending developed (see Section 6.2).

2.2.8 TrussesIn contrast to monolithic structures, which have rigid joints and develop bending andshear, trusses are load-bearing structures consisting of rectilinear two-hinged bars.They are loaded by concentrated forces at their nodes only, and consequently exclusivelyaxial forces develop in their members. The truss loading is applied to the nodes by meansof small beams supported on each node. It is these small beams that actually receivethe incoming loads, and then transmit them as concentrated forces to the truss joints(Figure 2.43). Despite the fact that the truss members are never connected by a pinjoint, this idealisation leads to safe results. Moreover, although the self-weight of thebars does not actually act directly on the nodes, it may, nevertheless, be safely consideredas directly acting on them. g

A statically determinate truss is analysed by first determining its reactions and thenexamining the equilibrium of each node as a free body, under the corresponding loadingand the unknown axial forces acting on the cut out bars. Using the two equations of

61

Variable axial force S with constant component H Constant compressive force

Circular arch

Constant tensile force

q · R

q

q

q

q · R

q · R q · R

S

H

HH

H = q · R

R ≈ L2/8 · f

q · L/2q · L/2

L

f

Figure 2.42 Typical forms of funicular arches


equilibrium for each node, a linear system of equations is obtained, which will lead to thedetermination of all the unknown axial forces.For the overwhelming majority of trusses it is also possible, by examining the equili-

brium at each node through use of the triangle of forces, to determine qualitatively thetwo unknown axial forces at each node, starting from the support joints where thereactions are already known. It is useful to note that in an unloaded node with threeconcurring bars, two of which are on the same line, the third bar is under no stress.Furthermore, if at a node there are four concurrent forces lying in only two differentdirections, then the two forces in each direction must be equal and opposite.Although the above described joint equilibrium method always leads to the full analysis

of a truss, it does not help one in understanding the bearing action of the truss itself, andthe determination of the axial force in specific bars is rather cumbersome. g

A truss consisting of an upper and a bottom chord, with vertical and diagonal barsbetween them, can instead be thought of as a monolithic structure, because both itsbending and shear action can be determined through its axial stress state, as describedbelow.If at a region of a simply supported beam, for example, a part is removed and substi-

tuted by three two-hinged bars, as shown in Figure 2.44, the resulting structure willcontinue to be statically determinate and rigid (see Sections 2.1.6 and 2.1.7). Cuttingthese three bars through a vertical section and considering, for example, the equilibriumof the left-hand part, it is understandable that the forces C and Z may be used assubstitutes for the compressive and tensile stresses of the section, respectively, andthus simultaneously give the corresponding bending moment at the position considered.Moreover, by examining the vertical equilibrium of the same part it is clear that thediagonal force D must provide the required positive shear force at the same place.This specific shear force may be offered only through a tensile force D, whereas forthe cross-symmetric position of the diagonal bar the positive shear force can beoffered only through a compressive force D (see Figure 2.44).Considering in this manner the region of interest, the stress state in a truss in that region

may always be determined using the sectional force diagrams for the correspondingmonolithic structure. g

For the truss shown in Figure 2.45, the axial forces O, U and D in the bars can bedetermined directly using the free-body equilibrium of the cut out left-hand part of

62

Figure 2.43 Indirect loading of a truss through joint loads


the truss. To determine the axial forceO, the moment equilibrium with respect to pointm2, where the forces U and D intersect, is considered, in order to get rid of their contri-bution. It is found that

O ¼ Mm2=hm2

(compressive)

To determine the axial force U, the moment equilibrium with respect to point m1 (theintersection of forcesO and D) is considered, in order to eliminate their involvement, asabove:

U ¼ Mm1=hm1

(tensile)

Finally, the axial forceD is determined through the requirement for vertical equilibrium,the contribution of force U being thereby eliminated:

D¼Vm/sinmþO/sin’

Thus the compressive axial force O ‘helps’ the tensile force D to ‘offer’ to theconsidered cut out left-hand part the required positive shear force Vm (see Figure2.45). Consequently, the value of D is lower than the value that would obtained ifthe upper chord was horizontal.

63

C

D

Z

C

D

Z

The three connecting bars can offer all the sectional forces

The inclination and the direction of the diagonal determine its response

Figure 2.44 The sectional forces occurring due to the truss action


Finally, the axial force of the vertical bar at node m1 may be obtained from the verticalequilibrium of the cut out node. g

As has been shown, an appropriately formed truss may be used as a substitute for anymonolithic structure, such as a beam, frame or arch, as it can provide all the necessarysectional bending and shear forces (Figure 2.46). The main structural advantage of trusses

64

Vm

Mm1Mm2

The inclined compressive force of the upper chordcontributes to take up the shear, therefore relieving the diagonal

D

D

U

U

O = Mm2/hm2

U = Mm1/hm1

D = (Vm – O sin ϕ)/sin αm

m2

m1hm2

hm1

O

O

ϕ

αm

Figure 2.45 Analysis of a truss considered as a monolithic beam

The axial forces of the truss offer all the sectional forcesof the corresponding monolithic structure

Three-hinged frame

Figure 2.46 A truss used as a substitute for a monolithic structure


is that, due to the exclusively axial response of their members, the cross-sectional area ofthe truss members can be exploited fully, and this naturally leads to material economy.

2.3 Determining the deformations

2.3.1 OverviewDeformations constitute the ‘visible’ part of the behaviour of structures, in the sense thatthe stresses and the sectional forces are ‘derived’ quantities, not being directly measur-able. Deformations may be ‘external’, such as the displacements and rotations developedat any point of a plane skeletal structure, or ‘internal’, the latter being better known as‘strains’ (Figure 2.47).When a plane skeletal structure is acted upon by a coplanar loading, and possibly also

a temperature variation, each point on the structure undergoes a displacement, which isrepresented by a vector and a rotation, either clockwise or anticlockwise, in the sameplane as the displacement.It should be emphasised that, during the deformation, the angles between any

monolithically connected parts (joints) — strictly speaking the angles between thetangents of the deformed lines at the joints — remain unchanged, meaning that therotation angles at the ends of all concurring members at a joint are equal. This doesnot happen when two members are connected by a hinge (see Figure 2.47).It should also be noted that, while the strains are deduced from the sectional forces

and from the temperature variation, this is not the case for the external deformations. g

The determination and general consideration of the external deformations, which willhereafter be referred to simply as ‘deformations’, has particular importance in thestudy of the behaviour of structures. The development of deformations in a structure,as a result of the incoming sectional forces in general, is inevitable and must be containedwithin prescribed limits. Excessive deformations, whether permanent or not, will weakena structure, and should always be controlled and, if necessary, appropriately settled.The determination of deformations is absolutely necessary in order to calculate andunderstand the response of statically indeterminate systems, which account for theoverwhelming majority of structures.

65

The relative rotation of neighbouring sections(‘internal deformation’)

The change in the relative angleat the hinge

The angle at each jointremains unchanged

Rotation

Displacement

Figure 2.47 Deformations of plane structures


2.3.2 Causes of deformationBefore dealing with the procedure for determining any deformation that occurs in askeletal plane structure, it is appropriate to examine the kind of strains that arecaused by the sectional forces and by temperature variations.

2.3.2.1 Bending momentA self-equilibrating pair of bending momentsM acting on the opposite, initially parallel,sides of a skeletal element of lengths causes a curving with a radius of curvature r and,consequently, a relative rotation’ of the two parallel sides (Figure 2.48). Such curvingmust occur because, otherwise, the shortening and the elongation of the extremefibres would change in an unacceptable way the right angles of the edges of the element.There is a basic ‘constitutive’ relationship between the curvature 1/r and the bendingmoment M:

1

r¼ M

EI

Given that: 1/r¼’/s, then

’ ¼ M

EIs

Obviously the curving of a straightmember will cause displacements transversely to its axis,and for skeletal structures this is the main reason for the development of deformations.The diagram of the bending moments allows it to be concluded directly how each

region is curved, given that the curvature always occurs on the tensioned side ofthe member (Figure 2.49). In areas where the moment is null, the correspondingregion has no curvature. To obtain a qualitative picture of the deformation for thewhole structure (see Figure 2.49), each member is first deformed separately, according

66

∆sr

∆ϕ

M M M M

Elongation

The change in the right angle is unacceptable (1/r) = M/EI∆ϕ = (M/EI ) · ∆s

Shortening

Figure 2.48 Bending deformation


to its moment diagram. The deformed members are then connected together,retaining the initial angles at each joint and considering the allowable deformationsat the supports.

2.3.2.2 Axial forceIf the opposite sides of an element are acted upon by two self-equilibrating axial forces,an elongation or a shortening l of the element will occur, depending on whetherthe element is tensioned or compressed respectively (Figure 2.50). The relationshipbetween the change in length and the axial force (Hooke’s law) is already knownfrom Section 1.2.1:

l ¼ N

EAs

Clearly, in structures bearing their loads exclusively through axial forces, such as trussesor structures having a funicular form, the finally developed deformation is due exclu-sively to the elongation or shortening of their elements.

67

[M][M]

At each point the curvature is proportional to the bending moment

Figure 2.49 A qualitative estimate of deformation on the basis of the bending moment diagram

Elongation ∆I = (N/EI) · ∆s Shortening

∆s ∆s

∆I/2 ∆I/2 ∆I/2 ∆I/2

NN N N

Figure 2.50 Axial deformation


2.3.2.3 Shear forceIf the opposite sides of the element are acted upon by a pair of shearing forces V, a certainmutual ‘sliding’ h of the two sides occurs. This results to the development of an ‘angleof shearing deformation’ , or shearing strain, which is given by ¼h/s (Figure 2.51).This may be written as

h ¼ V

GAs

where A is the cross-sectional area and G is the so-called shear modulus, which forconcrete has roughly the half value of its modulus of elasticity and for structural steelis 0.8 108 kN/m2.The contribution of the shearing deformation to the overall deformation of a skeletal

structure is generally negligible. It begins to play an appreciable role only in the case ofso-called deep beams, where the ratio of the member length to the section depth is lessthan 3. As stated at the beginning of this chapter, skeletal structures in general do notbelong to this category, and therefore the influence of shear forces on their deformationmay be disregarded.

2.3.2.4 Temperature variationIf an element is subjected to a uniform temperature T, which is different from thetemperature T0 that it was created under, then its length undergoes a change lTaccording to the equation

lT¼Ts aT s

where Ts¼TT0 and aT is the temperature coefficient, which has the same value forsteel and concrete up to 10—5/8C.An increase or decrease in the initial temperature of the element by Ts leads to an

elongation or shortening of the element, respectively. This change in length is notaccompanied by any axial stress at all (Figure 2.52).Usually the temperature change is not uniform across the depth h of the element,

which means that the the fibres at the two extremes of the section are at differenttemperatures. Thus, for example, in the cross-section shown in Figure 2.53, the

68

The contribution of the shear forces is neglected

∆h/2∆h/2

∆h/2

Vh

V

∆h/2

∆h = (V/GA) · ∆s

∆s

L/h > 3

L

Figure 2.51 Deformation due to the shearing force


temperature of the top fibre is 108C and that of the bottom fibre is þ408C, while theinitial temperature T0 is assumed to be þ58C. The temperature variation across theregion between the extreme fibres is assumed to be linear. Thus the middle fibre hasa temperature Tm¼þ158C, and consequently it undergoes a temperature increase ofTs¼ 15 5¼ 108C and the corresponding elongation.Any non-uniform temperature change can always be considered as a superposition of

a uniform temperature equal to the existing temperature of the middle fibre Tm and anon-uniform change T, which accounts for equal but opposite temperatures at thetop and bottom fibres, having the value (T/2). In the present case the value isT¼ 40þ 10¼ 508C (see Figure 2.53).An element that undergoes such a temperature variation (i.e. an equal shortening and

elongation of the two extreme fibres) develops a curvature with radius r, in the same

69

∆s ∆s

Temperature increase(elongation)

Temperature fall(shortening)

∆IT/2 ∆IT/2 ∆IT/2 ∆IT/2

∆IT = ∆Ts · αT · ∆s

Uniform temperature difference ∆Ts

Figure 2.52 Deformation due to a uniform change in temperature

Initial temperature: +5°C–10 +15

+15

–50/2

+15+40 +50/2

(∆T/2)

(∆T/2)

h

h

∆s

∆T ∆T

Temperature increase of the middle fibre: (15 – 5) = 10°C

Due to ∆T

Elongation Curvature in order to preservethe right angles

∆φ

Shortening

∆φ = (αT · ∆T/h) · ∆s

Figure 2.53 Deformation due to a non-uniform temperature change


manner as occurs in bending. However, without the presence of stress, the deformationis actually purely geometrical in nature. The consequent relative rotation’ of the twoopposite sides of the element is

’ ¼ T T

hs

Of course it is always valid that 1/r¼’/s.In conclusion, a non-uniform temperature change in an element, causes: (1) a

uniform elongation or shortening, depending on the temperature change Ts in themiddle fibre of the element; and (2) a curvature that is determined by the temperaturedifference T between the fibres located at the two extremes of the element.It should be noted that the imposition of any temperature variation on a statically

determinate structure does not cause reactions, and consequently does not affect thesectional force diagrams at all — if these in fact exist due to other loads. However, itdoes cause deformations.

2.3.3 The principle of virtual workThis principle allows the determination of any deformation in a skeletal structuresubjected to a certain loading, if the sectional forces diagrams (mainly M and N) arealready known. The principle of virtual work applies to any skeletal structural system,and links a freely selected (arbitrary), self-equilibrating loading, which consists ofconcentrated forces or/and moments, the so-called virtual loading, with those deforma-tions of the system under the real loading, which correspond to the specific loads ofthe virtual loading on the one hand, and the sectional forces diagrams due to the virtualand real loading on the other.Although this principle applies to any structure, whether determinate or indeter-

minate, and to any part of it, for simplicity it is illustrated below for a simple structure.When subjected to a uniform loading, the fixed structure shown in Figure 2.54

develops a state of stress, represented by the diagrams M and N, and a deformationdue to this stress state. This situation is called real loading. It is clear that at anypoint on the structure certain curvatures and strains are developed, and these arerepresented by the quantities ’real and lreal, respectively (see Section 2.3.2). Forthe same structure, another arbitrary (‘imaginary’) loading may be considered that istotally independent of the real one. This imaginary loading may consist of, forexample, the force P (2.0 kN) and the moment M (10.0 kNm), as shown in Figure2.54, which, together with the induced reactions RP (2.0 kN) and RM (4.0 kNm),constitute a self-equilibrating system of forces, called the virtual loading, as alsomentioned previously.In order to clarify how the above quantities are interconnected, first the structure

is considered under the virtual loading only, i.e. the self-equilibrating force systemPvirt that gives rise to the bending moment and the axial force diagrams Mvirt andNvirt, respectively. The real load is then superposed on the virtual loading, and thestructure deforms further. Obviously this additional deformation is identical to the

70


one caused by the real loading itself, represented by the quantities ’real and lreal

(see above). During this additional loading both the external (self-equilibrating)forces of the virtual loading and the corresponding internal (sectional) forces remainconstant and produce some work on the imposed (real) deformations (i.e. displacementsand rotations).Thus the external forces Pvirt of the virtual loading (reactions included) produce the

workXPvirt resp;real

on those real displacements (and rotations) that correspond to them and, moreover,the internal (sectional) forces Mvirt and Nvirt produce work on the corresponding realdeformations ’real and lreal, which can of course be expressed in terms of the real

71

Work of external forces of virtual loading on the corresponding real deformations: (2.0 · δ) + (10.0 · ϕ) + (4.0 · 0) + (2.0 · 0)

Equal to the work of internal forces of virtual loading on the corresponding real deformations:

Work of bending moments: –1.0 · 36.0 · 4.0 · 5.0/EI + [(1/3) · 36.0 · 6.0 · 3.0/EI – (1/3) · 6.0 · 9.0 · 3.0/EI ] = –558.0/EI

Work of axial forces: +1.0 · 24.0 · 2.0 · 5.0/EA = +240.0/EA (negligible)

8.0 kN/m 36.0

6.0

24.0

36.09.0

4.0 2.0

δϕ

5.0 m

M = 10.0 kN m

RM = 4.0 kN mRP = 2.0 kN

Virtual loading

Principle of virtual work

[M real] [N real]

3.0 mReal loading

P = 2.0 kN

[Mvirt] [Nvirt]

Figure 2.54 The principle of virtual work


sectional forces (see above). This work is given byðMvirt ’real þ

ðNvirt lreal ¼

ðMvirt M

real

EIdsþ

ðNvirt N

real

EAds

The principle of virtual work states that the above two works are equal, and consequently

XPvirt resp;real ¼

ðMvirt M

real

EIdsþ

ðNvirt N

real

EAds

For the case under consideration the left-hand side of the above equation can be writtenasX

Pvirt resp;real ¼ P þM ’þ RP 0þ RM 0

The deformations and ’ are shown in Figure 2.54, whereas the null terms represent thenull displacement and rotation of the fixed support in the real situation.In the above equation all the terms are known quantities except resp,real, i.e. the real

deformations ( and ’) involved. Thus, if on the left-hand side of the equation thesought-after deformation is the only unknown, it can be determined directly.However, for this to be possible the virtual loading must be chosen appropriately suchthat the work produced by it is restricted to this specific deformation only. It is thusobvious that, in order to determine a particular displacement, the virtual loadingshould consist of a corresponding force for that displacement plus the other forces(reactions) required for equilibrium, whereas the determination of a rotation requiresthe virtual loading to have a corresponding moment to this rotation, plus the otherforces that ensure equilibrium (Figure 2.55). It is, therefore, clearly convenient if the‘corresponding’ concentrated virtual action is chosen to be a unit force or moment. Ifthe calculated deformation has a negative sign, its sense is opposite to that of thevirtual load chosen.Thus, referring to Figure 2.55, the sought-after deformations are determined from the

equations

1:0 þ 3:0 0þ 1:0 0 ¼ðMvirt M

real

EI

and

1:0 ’þ 1:0 0 ¼ðMvirt M

real

EI

These integrals account for each separate member of the skeletal system; their evalu-ation is facilitated by using the values given in Table 2.1. The values given in thistable are for the integrals of the products of those functions that are commonly occurringvariations of the sectional forces diagrams M and N, referred over a length L, withconstant EI.It is clear that for the evaluation of the above integrals the bending moment diagrams

M must be appropriately signed, as indicated in Section 2.2.1. It should also be noted

72


that when a skeletal structure exhibits a bending response the contribution of the axialforces to its deformations is practically negligible compared with that of the bendingmoments. Thus usually only the bending moments need be taken into account in theintegrals. g

When, in addition to the above considered deformations there is a variation in tempera-ture, the internal forces due to the virtual loading will produce additional work on thecorresponding real deformations ’real and lreal (see Section 2.3.2):ð

Mvirt ’real þðNvirt lreal ¼

ðMvirt T T

hdsþ

ðNvirt T Ts ds

73

8.0 kN/m 8.0 kN/m

5.0 m 5.0 m

Sought-afterdeformation

Sought-afterdeformation

Real loading

ϕδ

3.0 m 3.0 m

1.0 kN m1.0 kN

[Mvirt]

Determination of δ

[Mvirt]

Determination of ϕ

Virtual loading1.0 kN 3.0 kN m 1.0 kN m

Figure 2.55 The choice of the virtual loading

Table 2.1 Characteristic integral values for the moment diagram products

M2 M2 M2

M1 1 M1 M2 L/EI 12 M1 M2 L/EI 2

3 M1 M2 L/EI

M112 M1 M2 L/EI 1

3 M1 M2 L/EI 13 M1 M2 L/EI

M112 M1 M2 L/EI 1

6 M1 M2 L/EI 13 M1 M2 L/EI


Because the above integrals contain a number of constant terms, the calculation, basedon the appropriate areas of the diagrams forMvirt andNvirt, is relatively simple. However,due attention must be paid to the sign of the work. The sign of the first integral is positiveor negative, depending on whether or not the curvature due toMvirt is of the same senseas the curvature caused by the existing T (see Section 2.3.2). The sign of the secondintegral (which is not negligible) is positive or negative depending on whether or not theaxial deformation due to Nvirt corresponds to that of the existing Ts. It is clear that ifthe above integrals are signed in the suggested manner, then all the quantities involvedmust be used with their absolute value. Thus the principle of virtual work may be writtenin its complete form asX

Pvirt resp;real ¼ðMvirt M

real

EIdsþ

ðNvirt N

real

EAds

þðMvirt T T

hdsþ

ðNvirt T Ts ds

It may be recalled here that the termMreal/EI expresses the curvature 1/r at every point(see Section 2.3.2) and, consequently, the first term on the right-hand side may also bewritten asð

Mvirt ð1=rÞ ds g

In the case of a truss the above relationship is simplified, and can be written as:XPvirt resp;real ¼

XNvirt N

real

EA lþ

XNvirt T Ts l

Finally, it is noted that if the shearing deformation has to be taken into account on thebasis of the existing shear forces, then the following term should be added to the right-hand side of the above equation:ð

Vvirt Vreal

G A ds

2.3.4 A wider use of the principle of virtual workThe conditions under which the principle of virtual work may be applied were describedin the previous section. However, it is also useful to examine the problem of finding thedeformations in statically indeterminate structures.As an example, the statically indeterminate frame shown in Figure 2.56 is considered.

Under the action of some loading, this frame develops the bending moment diagramshown in the figure. The horizontal displacement of the upper left-hand joint is tobe determined. This deformation can, in principle, be found by considering a virtualloading consisting of a unit horizontal force applied at this joint, and the ensuingreactions, and determining the corresponding bending moment diagram. The procedurefor determining the virtual work is then applied as set out in Section 2.3.3.

74


However, although this procedure is correct, it is practically impossible to carry out bysimple computational means. The self-equilibrating force system of virtual loading canbe chosen much more simply if the left column of the frame is thoroughly cut outand subjected to the same virtual load at the top, accompanied by the actionsrequired for equilibrium at its base (i.e. an equal and opposite force and an appropriatemoment). The bending diagram so obtained corresponds to that for a vertical cantilever(see Figure 2.56). The principle of virtual work may now be applied directly:

1 þ 1 0þ 12:0 0 ¼ðMvirt M

real

EIds

It is understood that the integral refers only to the selected members. g

In order to point out the absolute freedom to use the principle of virtual work, the leftupper column of the frame is considered alone (see Figure 2.56). The unit horizontal

75

Direct determination of δ

Virtual loadingReal loading

Virtual loadingReal loading

The principle of virtual workis always valid

15.05

15.05 15.05

12.0

12.0

1.0

8.98

7.37 7.37 1.0

1.06.0

1.0

4.604.60 6.0u

oδo

δu

ϕu

8.0 mInappropriate way of determining δ

δ

6.0 m

6.0 m

13.58

13.58

8.988.98

4.60

7.37

7.37 1.0

4.60

δ

Figure 2.56 A wider application of the principle of virtual work


force at the top of this column is appropriately equilibrated by a force and a moment at itsbase. The system of these three forces is a self-equilibrating one and clearly constitutes avirtual loading, and consequently the principle may be applied directly:

1 o 1 u 6:0 ’u ¼ðMvirt M

real

EIds

where o, u and ’u are the corresponding real deformations to the virtual forcesconsidered, i.e. the horizontal displacements at the points o and u, and the rotation atpoint u. The integral obviously concerns only the selected member. However, althoughthe above equation is absolutely correct, it is of no help at all in the determination of o,as this is not the only unknown in the equation.

2.3.5 The Betti—Maxwell theoremThis theorem is of particular importance in structural theory and is valid for any type ofstructure.Consider, for example, the structure shown in Figure 2.57. Now consider two separate

loadings: a concentrated load P in state I and a moment M in state II. Each of theseforces acts at a different point. In state I the structure develops a rotation ’ at thepoint where the moment is acting in state II, whereas in state II the structuredevelops a displacement at the point where the concentrated force is acting in state I.According to Betti’s theorem the (hypothetical) work produced by the force P

(remaining constant) on the ‘path’ is equal to the work produced by the moment M(remaining constant) on the ‘path’ ’. That is,

P ¼M ’Of course, instead of the moment M, there may be any other concentrated force in itsplace, i.e. instead of the pair (M, ’) another one (P, ) may be considered. Obviouslythe deformations considered in each instance should correspond to the type of forcesproducing work on their ‘path’. Thus it may be generally written:

(force I) (corresponding deformation II)¼ (force II) (corresponding deformation I)

76

Betti: P · δ = M · ϕMaxwell: if P = 1 and M = 1 then δ = ϕ

ϕ M

P δI II

Figure 2.57 The Betti—Maxwell theorem


The relationship above has particular importance if the forces P and M have unitvalue. Then it is:

¼’

Of course this relationship must be interpreted as an equality of work (1 ¼ 1 ’)rather than as an absurd equality between two quantities having different dimensions.Nevertheless, for unit forces it may be written as

(respective deformation to II)¼ (respective deformation to I)

The above relationship is also known as the reciprocal theorem or Maxwell’s theorem.

2.3.6 The beam equation and finding the deflection curveThe principle of virtual work can be used to determine the deformations at specificpoints of a structure. However, in order to find the deflection curve for one or moremembers of a skeletal structure, it is necessary to return to the basic equation for thebeam, governing its deflection w(x) under an applied distributed load p(x). As the curva-ture 1/r at a point of a beam with abscissa x is expressed to a very good approximation bythe value(dw2/dx2), its relationship to the bending moment (see Section 2.2.3) may bewritten as

dw2

dx2¼ M

EIðaÞ

Moreover, the equation for equilibrium between the bending moment and the load (seeSection 2.2.3) is

d2M

dx2¼ pðxÞ ðbÞ

By twice differentiating equation (a), the classical beam equation is obtained:

EIdw4

dx4¼ pðxÞ

The physical meaning of this equation is that, if the beam is ‘compelled’ to deform byw(x), it will offer a resistance qi equal to

qi ¼ EIdw4

dx4

This resistance, which is directed in the opposite sense to w, is of course in equilibriumwith the applied load p(x) causing this deformation, so that qi¼ p(x).Solving the above equation mathematically in order to determine the function w(x) is

cumbersome, and does not provide the required physical picture. Thus, in order to findthe deflection curve, instead of the purely mathematical approach, the followingprocedure (Mohr’s theorem) is preferred (Figure 2.58). It is clear that in the sameway that equation (b) can be used to find the diagram for M from the load p,

77


equation (a) can be used to determine w on the basis of M, i.e. as the bending momentdiagram that results from application of the load M/EI.Thus, in any rectilinear member of a skeletal structure having a diagram for M, the

deflection curve transverse to its axis and between its displaced ends may be drawn asthe moment diagram of the corresponding simply supported beam loaded with M/EI.Note that this loading must always be directed towards the tensioned fibres of thepart of the member under consideration (see Figure 2.58). It is also clear that thetransverse displacements at the ends of the member may be determined by appropriateapplication of the principle of virtual work (see Section 2.3.3).A further remark to be made is that equation (a) written in the form

d

dx

dw

dx

¼ M

EI

suggests an analogy with the equilibrium relation dV/dx¼p, and consequently theslope of the deflection curve may be found from the loading M/EI in the same waythe shear force can be found from the loading p. Thus the slopes of the non-displacedends of a beam with bending diagram [M] may be represented by the reactions of asimply supported beam loaded by the diagram for M/EI, directed as explained above.

78

1

1 2

2

1 2

1 2

Deflection curve

Developed deformation

Fictitious loading of a simply supported beam

Bending moment diagram M due to fictitious loading:deflection curve for unyielding ends

The diagram M drawn between the displaced endsleads to the final deflection curve

[M ]

[M ]

δ

δ

M/EI

M/EI

Figure 2.58 Finding the deflection curve using the bending moment diagram


2.3.7 Elastic support of structuresThe load-bearing structures are supported on the ground. Thus, at the points of supportthe ground takes up forces and moments that are equal and opposite to the reactions,and consequently the ground deforms. Obviously the role of the ground as a means ofsupport may be undertaken by any other appropriately formed structure. If the actionapplied by a structure on the body on which it is supported causes an elastic deformationexclusively in the sense of the action itself, then the corresponding support offered to thestructure is called elastic simple support or elastic clamped support, depending on whetherthe action is a force or a moment.

2.3.7.1 Elastic simple supportHere it is assumed that the ground at the point of application of a vertical concentratedforce P deforms by , according to the relationship

¼ fs PThe coefficient fs is called the coefficient of elastic support, and represents the settlement ofthe ground under a unit concentrated force. The elastic support is provided by a transla-tional spring. Clearly the settlement is directed in the same sense as P (Figure 2.59).The elastic simple support may be represented by any elastic body which, under aconcentrated load P, develops only a corresponding displacement , and no rotation,as would occur, for example, in an axially loaded bar or a simply supported beamloaded at its middle.

2.3.7.2 Elastic clamped supportHere it is assumed that the ground at the point of application of a momentM rotates by’according to the relationship

’ ¼ f’ M

The coefficient f’ is called the coefficient of elastic rotation, and represents the rotation ofthe ground under a unit moment. The elastic clamped support is provided by a rotationalspring. Clearly the rotation ’ has the same sense asM (Figure 2.60). The elastic clampedsupport may be represented by any elastic body that under a momentM develops only a

79

Ground

P PP

L

P

P

δ

δ δ

δ

δ = fs · P

fsfs

fs = L/(EA) fs = L3/(48 · EI)

L

Figure 2.59 Elastic simple support


rotation ’ and no displacement, as would occur, for example, at the simply supportedend of a beam or at the joint of an unmoveable frame. g

At this point it should be emphasised that a statically determinate structure, which issupported elastically in the above sense, develops exactly the same state of stress as ifit were unyieldingly supported, although it develops a different deformation.The cantilever shown in Figure 2.61, subjected to a uniform load p and elastically

clamped with a coefficient f’, develops an identical bending moment diagram to

80

fϕ = L/(6 · EI)

L

L

L

fϕ = L/(3 · EI)

GroundM

ϕ

ϕ

ϕ

ϕ

M M M

M

fϕ fϕϕ = fϕ · M

Figure 2.60 Elastic clamped support

Same diagram M as for unyielding support

Opposite sense than the rotation caused

Virtual loading

1.0

1.0 · L

1.0 · LP = 1

p · L2/2p · L2/2

p · L2/2

Ground

fϕ

fϕ

L

p

φ

Figure 2.61 Elastically clamped structure


the diagram produced for a fixed support. However, its deflection at the freeend differs. According to the principle of virtual work the bending moment may becalculated as

1 1 Lð Þ p L2

2

! f’

" #ðMvirt M

real

EIds

Note that, the moment 1 L, as an external force (reaction) of the chosen ‘virtual loading’,produces work on the corresponding ‘real’ rotation due to the action ( p L2/2) of the ‘realloading’ on the ground, but the two moments have the opposite sense.The fact that the reaction ( p L2/2) has an opposite sense to that of the developed

rotation at the support is important. Moreover, referring to the elastically supportedbeam shown in Figure 2.62, it is similarly noted that the reaction ( p L/2) acting onthe beam has an opposite sense to the corresponding support settlement at this point.It is understood that the latter remarks are generally valid, regardless of whether ornot the structure is statically determinate.

2.3.8 The concepts of flexibility and stiffnessIf in a structure a deformation caused by a certain force is expressed through the quantityf by the relationship Deformation¼ Force f, then it is obvious that, at the same point onthe structure, the force required in order to produce a similar deformation as the abovemay be expressed through the quantity k by the relationship Force¼Deformation k,where k¼ 1/f.The above remarks relate directly to the concepts of flexibility and stiffness, respec-

tively, which are merely a consequence of the deformability of structures. Thus theterm flexibility f means the deformation at some specific point of a structure caused bya unit force. The term stiffness k means the opposite; namely, the force that must beapplied at a certain point on a structure in order for it to develop a unit deformation.According to the above discussion of elastic simple and clamped supports, the

quantities fs and f’ introduced in Section 2.3.7 represent flexibility quantities,

81

Ground fs δ

fs δ

p · L/2

p · L/2

L

p p

Figure 2.62 A elastically simply supported structure


whereas the values 1/fs and 1/f’ represent stiffness quantities. Specifically, these lastvalues represent the force and moment required to act on the springs in order toproduce a unit displacement and rotation, respectively.The simply supported beam shown in Figure 2.63, under the concentrated load at the

middle, develops at this point a deflection equal to

¼ L3

48 EI P

The quantity (L3/48 EI) represents the flexibility of the beam with respect to thevertical deflection of its middle, which is the deformation itself as the force is unity(P¼ 1).If a structure is supported at the middle of the beam (see Figure 2.63), it can be

considered as elastically supported with a coefficient of elastic support fs equal to(L3/48 EI). Furthermore, given that P¼ (48 EI/L3) , the quantity (48 EI/L3)represents the stiffness of the beam at its middle with respect to the vertical deflection,and is the force required to produce the unit displacement ¼ 1. This specific stiffness ofthe beam is proportional to EI of its cross-section, and it decreases rapidly as the lengthof the beam increases.Moreover, in the axially loaded member (see Figure 2.63) the quantity EA/L

represents its axial stiffness, and is the compressive (or tensile) force required toproduce a unit shortening (or elongation). g

82

δ = P · L /(EA)

P = δ · (EA)/L

Flexibility

Stiffness

Flexibility

Stiffnessfs = L/(EA)

L

L

L

P

P

δ = 1

δ

δ

δ

P = stiffness

Offered resistance = stiffnessδ = P · L3/(48 · EI)

P = δ · (48 · EI/L)3

fs = L3/(48 · EI)

Figure 2.63 Flexibility and stiffness with respect to displacement


The same beam as above loaded with the concentrated momentM at its ends develops arotation angle ’ (Figure 2.64):

’ ¼ L

3 EI M

The quantity (L/3 EI) represents the flexibility of the beam with respect to its endrotation, being the rotation due to the unit moment M¼ 1.If a structure is monolithically connected to the end of the beam (see Figure 2.64), the

structure may be considered as elastically clamped with a coefficient of elastic clampedsupport f’ equal to (L/3 EI). Furthermore, given that M¼ (3 EI/L) ’, the quantity(3 EI/L) represents the stiffness of the beam with respect to the end rotation, and isthe moment required to produce the unit rotation ’¼ 1. This specific stiffness of thebeam is proportional to EI and inversely proportional to its length. g

From the two cases described above it can be seen that the stiffness represents theresistance (force) offered by the structure against a deformation (typically of unitvalue). Clearly the concept of stiffness is quite general, and it is not restricted to thecase of a simple beam. For example, the horizontal force that must be applied to thenode of the frame shown in Figure 2.65 in order to produce a unit horizontal displace-ment at that point is taken as the ‘lateral stiffness of the frame’ at that point.

83

Stiffness

Flexibility

M

L

L

Offered resistance = stiffness

M = stiffness

ϕ = 1

fφ = L/(3 · EI)

φ = M · L/(3 · EI)

ϕϕ

M = φ · (3 · EI/L)

Figure 2.64 Flexibility and stiffness with respect to rotation

P = stiffness δ = 1

Figure 2.65 Lateral stiffness


Finally, it should be noted that, while a change in the sectional properties EI of amember in a skeletal structure clearly affects its deformability, the state of stress ofthe statically determinate structure remains totally unaffected.

2.4 Symmetric plane structuresIn many cases structures are designed such that they possess an axis of symmetry withregard to their geometric and elastic properties (Figure 2.66). When such a structureis subjected to a symmetric or antisymmetric loading, its sectional forces, as well asdeformations, at the points of intersection with this axis must obey certain rules.An antisymmetric loading is one which, when the sense of all loads on each side of the

symmetry axis is inverted, results in a symmetric loading. An arbitrary loading of asymmetric structure may always be considered as a superposition of a symmetric andan antisymmetric part, as shown in Figure 2.67. The properties of both symmetric andantisymmetric loading are always very useful, and are described below.

2.4.1 Symmetric loadingA symmetrical loading with respect to a (vertical) axis of symmetry is visually directlyobvious. Symmetrical loading also includes any concentrated load P0 lying on thisaxis. All the sectional forces and the deformations are developed symmetrically to theaxis (Figure 2.68). Based on this fact the following conclusions may be deduced, andare valid for all points A lying on the intersection of the structure with the axis ofsymmetry.

(a) Deformations. The rotations and displacements perpendicular to the axis are null.. Displacements are developed only along the axis of symmetry.

(b) Internal forces. The (vertical) component along the axis of symmetry of the internal force acting on

the section is equal to P0/2, as may be concluded from the vertical equilibrium of thecut out joint around the axis of symmetry. In the case when P0¼ 0, this component

84

Symmetric loading Antisymmetric loading

M M

P

p p p p

P P P

M M

q q q q

Figure 2.66 Symmetric and antisymmetric loading


obviously vanishes. Note that this component does not generally represent the shear-ing force of the relevant section.

. Both the bending moment and the (horizontal) component perpendicular to the axisof symmetry of the internal force are normally developed.

The above stress and deformation state at point A allows the structure to be cut out ateach such point and to be provided there by a movable fixed support, as shown in Figure2.68. This kind of support permits movement parallel to the axis of symmetry, as well asthe development of both a bending moment and a horizontal reaction. However, thedevelopment of rotation, of displacement perpendicular to the axis and of a reactionalong the axis is not possible. In the case of a concentrated force P0 acting on thestructure, an applied force at point A equal to P0/2 must be considered. Thus it isconcluded that the whole structure works exactly the same as the above ‘half-model’.

85

Symmetric AntisymmetricArbitrary loading

q

q/2 q/2 q/2

P P/2

p/2 p/2 p/2 p/2

P/2 P/2 P/2

p

q/2

Figure 2.67 Splitting of an arbitrary loading into a symmetric and an antisymmetric part

Equivalent model

P/2

P0/2

P0/2P0/2

Always existingSymmetric stress state

Symmetricdeformation

A

A

A

A

Displacement = 0

Rotation = 0

P0P0

P P

δδ

ϕϕ

Figure 2.68 Properties arising from a symmetric loading


2.4.2 Antisymmetric loadingA concentrated force H0 may also count as an antisymmetric loading, perpendicular tothe axis, as may a concentrated momentM0, both acting at the intersection points of thestructure with the axis. All sectional forces and deformations are developed anti-symmetrically to the axis (Figure 2.69). Based on this fact the following conclusionsmay be deduced, and these are valid for all points A lying on the intersection of thestructure with the symmetry axis.

(a) Deformations. The displacements perpendicular to the axis are null.. The rotations and the displacements perpendicular to the axis are normally developed.

(b) Internal forces. The moment equilibrium of a joint cut out around the axis implies that the bending

moment is M0/2, and in the case when M0¼ 0 it obviously vanishes.. The (horizontal) component perpendicular to the axis of symmetry of the internal

force acting on the section is H0/2, as may be concluded from the horizontalequilibrium of the cut out joint around the axis of symmetry. In the case whenH0¼ 0, this component obviously vanishes. Note that this component does notgenerally represent the axial force of the section under consideration.

. The (vertical) component along the axis of symmetry of the internal force is normallydeveloped.

86

Antisymmetric deformation

Antisymmetric stress Always existing Equivalent model

Displacement = 0

A

A

A

(EI) N = 0

A

M0M0

M0/2

M0/2

H0/2 H0/2

H0/2

δ

δ

ϕϕ

H0 H0

(EI/2)

Figure 2.69 Properties arising from an antisymmetric loading


The above stress and deformation state at point A allows the structure to be cut out ateach such point and to be provided there with a simple support, as shown in Figure 2.69.This kind of support satisfies all the above requirements at point A. The presence of aconcentrated force or moment at joint A of the structure implies the action on thesupported point of the forces H0/2 and/or M0/2. Thus it is concluded that the wholestructure works in exactly the same way as the above ‘half-model’.It should be noted that in the case where members of the structure lie on the axis of

symmetry, in the ‘half model’ these members have to be considered to have EI/2.However, their bending and shear forces in the actual structure have a value doublethat in the ‘half model’. Moreover, the axial force in all such members vanishes. Thismay be explained if a value for the axial force of such a member is assumed and thenthe structure is ‘seen’ from the ‘back’, i.e. the same structure with an identicalloading but in the opposite sense. The fact that for the same member and the sameloading two opposite values of the axial force may be concluded proves that this axialforce simply cannot exist.

2.5 Grid structures

2.5.1 GeneralThe plane structures examined so far have been loaded and deformed within their ownplane. In this section the basic concepts concerning the stressing and deformation of gridstructures will be presented as they were defined in Section 2.1, i.e. the concepts as theyapply to plane structures that are loaded perpendicularly to their own plane (Figure2.70). In this respect it is appropriate for the grid to be referred to a horizontal planedefined by two arbitrary axes OX and OY.For any cut out part of a grid in equilibrium the following conditions should be

satisfied:XMx

! ¼ 0;X

My

! ¼ 0;X

Pz!¼ 0

87

The reactions at 1 and 4 togetherwith the external loading satisfy

the equilibrium conditions

Z

Y

O

X

1

2

4

3 1

2

4

3

EquilibriumΣ MOX = 0Σ MOY = 0Σ MOZ = 0

Figure 2.70 Equilibrium conditions for a loading perpendicular to a plane structure


This means that the sum of the moment vectors of all the forces acting on the examinedpart, with respect to both axisOX and axisOY, must be null. Furthermore, the sum of allacting vertical forces must also be zero.

2.5.2 Sectional forces and deformationsFor the better description of the sectional forces and deformations at a point on a gridmember, account should be taken of the vertical plane to the member at the pointconsidered. The intersection of this plane with the member reveals two free edges,both of which exhibit identical deformation vectors but equal and opposite vectors fortheir sectional forces.These magnitudes are referred to a local orthogonal system (x, y, z) that has its origin

at the centroid of the section (Figure 2.71). Its x axis coincides with the axis of themember, while the two other axes y and z lie on the edge plane, in the horizontal

88

V

V

Positive sectional forces

Plane Oxy coincides with the plane OXY Loading vertical to the plane OXY (grid)

MB

MB

MT

MT

w

z

y

x

O

Local system

ϕy

ϕx

Vz (V )

(MT)

(MB)

My

Mx

Sectional forces Deformations

The ‘screw’ convention

Z

O

Y

X

Figure 2.71 Sectional forces and the deformations developed


and the vertical senses, respectively. It is assumed that the section is symmetrical aboutthe horizontal y axis. The vectors with double arrows in Figure 2.71 are moments androtations that obey the ‘screw rule’, i.e. they express the sense in which a screw mustbe rotated in order to proceed in the direction indicated by the arrow.In a grid where each member is loaded within its corresponding plane (x, z) the

following internal forces are developed (see Figure 2.71):

. shear force Vz (i.e. V)

. bending moment My (i.e. MB)

. torsional moment Mx (i.e. MT).

The bending moment MB together with the shear force V induce bending within thevertical plane (x, z) of each member.Moreover, at each point on the grid, the deformations developed consist of a vertical

deflection w (with respect to the z axis) and a rotation represented by a horizontalvector. The projection ’y of this vector onto the y axis (i.e. perpendicular to themember axis) refers to the bending of the member (bending rotation), whereas theprojection ’x onto the x axis refers to its torsion (twisting angle).The above sectional forces are signed as follows (see Figure 2.71):

. The shear force is positive if, together with the shear force acting on the next cut outsection, it forms a clockwise couple.

. The bending moment MB is positive if it causes tension at the bottom fibres.

. The torsional moment MT is positive if its vector acts in a ‘tensile sense’ on thesection.

These three sectional forces can always be determined through the three previouslystated equilibrium conditions for a cut out part of a grid, provided that all the otherforces acting on it are known (Figure 2.72).For the determination of any deformation, the principle of virtual work is always

applicable, which, in order to allow for the contribution of torsional moments in

89

O

X

Z

Y

(Grid)

Equilibrium of cut out part

For the determination of sectional forces the numberof unknown magnitudes must not exceed 3

Figure 2.72 Determination of the sectional forces on a grid


carrying the vertical loads, is now written in the form:XPvirt resp;real ¼

ðMvirt

B MrealB

EIdsþ

ðMvirt

T MrealT

GITds

2.5.3 TorsionIt is already clear that the stressing factor that differentiates grids from plane structures isthe torsion. The basic property of the torsional behaviour of a member having a length Lis the relationship between the acting torsional moment MT at its free end with thecorresponding twisting angle ’T of the end section of this member (Figure 2.73):

’T ¼ MT L

GIT

where IT represents the torsional moment of inertia and G is the shear modulus of thematerial of which the member is composed (see Section 2.3.2).From the above equation it is concluded that the magnitude GIT/L expresses the

torsional stiffness of the member, being, according to Section 2.3.8, that torsionalmoment required to produce a unit twisting angle, or, in other words, the torsionalresistance offered by the member if subjected to a unit twisting angle ’T¼ 1.For an orthogonal section having dimensions b/t (b> t), IT is calculated according to

the expression IT¼ b t3/, the coefficient varying from 3.0 for oblong sections to 7.0for square sections.A torsional momentMT causes a rotating flow of shearing stresses on a full orthogonal

cross-section, diminishing towards the centre and having a resultant moment vectorwith respect to the centre of the section identical to MT (Figure 2.73). In the case oforthogonal sections, the maximum value max of these shearing stresses lies at themiddle of the longest side and is proportional to the stress magnitude MT/(b t2) by a

90

MT

MT

ϕT

ϕT

b > t

t

b

ϕT = MT · (L/GIT) τmax = (3 ~ 5) · MT/(b · t2)3 for b/t > 105 for b/t = 1

L

τmax

Figure 2.73 The torsional response of a beam having an orthogonal full section


factor of 3.0 for oblong sections and 5.0 for square sections. Note the inverse relationshipbetween max and the square of the thickness of the section.A commonly used type of section is the hollow box section (Figure 2.74). This

section has a constant wall thickness t, and takes up the torsional moment MT

through a rotating peripheral constant shearing stress , which, according to the Bredtformula, is given by

¼ MT

2 Fk twhere Fk is the area enclosed by the middle line of the walls, i.e. slightly less than the areaof the full section (see Figure 2.74).It may be concluded that each wall of the examined section of the rectilinear member

is subjected to a constant peripheral force flow vl¼ t¼MT/(2 Fk) acting per unitlength (kN/m), which thus causes a total shear force V over each wall (see Figure2.74). It is found that this shear flow v has a constant value for all walls, even if theirthicknesses are different.Of course the relationship between the twisting angle ’T and the torsional moment

MT is also valid here, the torsional moment of inertia IT now being equal to 4 t F2k=L,where L is the perimeter of the middle line of the section. Note that the quantity IT,and consequently the torsional stiffness GIT/L, is, for the same cross-sectional area,much greater in a hollow than in a full section.

2.5.4 Types of supportThe two following ways of supporting a point on a grid structure may practically be distin-guished (Figure 2.75).

. Simple support. This type of support simply prohibits the vertical movement. Thedeformations ’x and ’y are freely developed. The only reaction developed is a verticalforce R.

91

The shear force of each wall is proportional to its height

V = MT · b/(2 · Fk)

V = MT · h/(2 · Fk)

MTMT

ϕT

b

h

t

t

τ (constant)

Fk

Figure 2.74 The torsional response of a beam with a hollow box section


. Fixed support. With this type of support all possible deformations are constrained. Thedeveloped reactions are a vertical force R and the two moments Mx and My.

If the whole grid is supported in such a way that only three reaction forces are developed,the reactions can be determined using the three equilibrium conditions mentionedpreviously. If more than three reaction forces exist, the equilibrium conditions areinsufficient for the determination of these forces.

92

Simple support Fixed support

ϕx

ϕy

ϕx = 0ϕy = 0w = 0

R

R

My

Mx

Figure 2.75 Two ways of supporting a grid structure


3

The handling of deformations fordetermining the stress state in framedstructures (statically indeterminatestructures)

3.1 IntroductionAs pointed out in the previous chapter, the vast majority of plane frames are staticallyindeterminate structures. This means that it is not possible to determine their stressstate (i.e. the diagrams M, V and N) by using only the equilibrium equations, even iftheir reactions can be determined, as shown in Figure 3.1.In such a structure, one can always perform suitable modifications consisting of

either removing support elements or altering the continuity of the structure whereverneeded so that the system is changed to a statically determinate one (see Section2.1.8). Each modification corresponds to a specific number of forces that are auto-matically released (a maximum of three), which means that their development in thenew system is excluded. The statically determinate structure that results in this way iscalled the primary structure, and henceforth it constitutes the base on which theforces suppressed by the modifications will be determined. The number N of theseforces is the degree of redundancy of the initial structure (Figure 3.2).These forces are called statically redundant forces and may be considered as (unknown)

external actions applied to the primary (statically determinate) structure. It is clear thatat the points where the continuity in the statically redundant structure has beenreleased, the external actions may be considered to be acting on the two ends of thecorresponding section with equal and opposite sign (see Figure 3.2). It is obvious thatfor any arbitrary value of these forces and the given external loads, the primary structureis in a state of equilibrium and has a specific deformation.Thus in a statically indeterminate structure, which after suitable modifications

is transformed to a statically determinate system, an unlimited number of equi-librium states under the specific loading can be considered. This fact automaticallyraises the question of which stress state among all the possible stress states will reallybe developed in the structure. Assuming an elastic behaviour, one unique solutionexists for this problem. This solution can be found on the basis of the deformability ofthe primary structure, as the deformations which are developed at the modificationpoints are not compatible with those of the real structure (see Figure 3.2).


However, the compatibility is restored by the application of concrete external stati-cally redundant forces on the primary structure. Thus, if continuity is broken at somepoint in a monolithic member, both section ends should not only come in contactafter the final application of the statically redundant forces, but their relative rotationshould be zero, unless the break in continuity takes place at an internal hinge. It isclear that when a support is removed the developing shift (or rotation) should havethe value imposed by the real conditions. Clearly the number of statically redundantforces will always coincide with the number of deformation conditions that should besatisfied in this way, a fact that always leads to a unique solution.This way of dealing with the problem, which is described in further detail below, is

known as the force method.

3.2 The force method

3.2.1 Physical overview of the force methodThe procedure that may be applied for any statically indeterminate system involves thethree steps described below.

Step 1Specific modifications are made to the redundant system in order to transform it into ina statically determinate (and rigid) one, i.e. the primary structure. All considerationshereafter will concern this system only.

94

The three equilibrium equations are not sufficientfor the determination of the sectional forces

Statically determinate

Hinge removed

Figure 3.1 The inadequacy of the equilibrium equations for determining the stress state


The modifications have the following two consequences for the primary system:

. they exclude the development of some specific (redundant) forces X1, X2, . . . , etc.

. they allow the development of deformations at the specific points, corresponding tothe forces X1, X2, . . . , etc.

Therefore, the following deformations are calculated in the primary structure under thegiven external loads:

. deformation along the direction of X1

. deformation along the direction of X2.

Step 2The same deformations as above are calculated for the imposition of the external loadsX1, X2, . . . , etc. only. Each of these deformations results as a linear combination of X1,X2, . . . , etc.

95

Hyperstatic structure

There are various possibilities forselecting the primary system

Releasing of three redundant forces

Releasing of three redundant forcesRedundant forces

Development of:

• relative rotation• relative horizontal displacement• relative vertical displacement

Primary system Primary system

Primary system Primary system

Development of displacementDevelopment of rotation

Redundant forces

Requirement for restoring thecontinuity of deflection curve

Requirement forzero displacement

Requirement for zero rotation

Figure 3.2 The selection of the primary structure and the corresponding requirements

Stress state in framed structures (statically indeterminate structures)

Step 3The total deformations of the primary structure are expressed as a superposition of thosedetermined in Steps 1 and 2. It only remains, therefore, to enforce these deformations totake the values they must actually have. This automatically implies the determination ofa system of as many equations as there are unknowns X1, X2, . . . , etc. Thus the role ofthe statically redundant forces becomes explicit: the forces X1, X2, . . . , etc. should takesuch values that if applied as external forces to the primary structure together with theexternal loads they cause the specific deformations that occur in the real redundantsystem (compatibility of deformations).

It is clear that the force method consists exclusively in dealing with the deformationsthat are developed in the primary structure.It should be noted that the satisfaction of the two criteria of equilibrium and compatibility

of deformations is an absolutely necessary condition that a stress state in a staticallyindeterminate structure has to fulfil, within the premises of an assumed elastic behaviour. g

The procedure described above is now applied in the following two examples.

Example 1Consider the two-hinged frame shown in Figure 3.3. Of the various modifications thatcan be made to make this structure statically determinate, the interruption of thecontinuity of the girder by placing a hinge at node G is selected. The primary structurethus created is a three-hinged frame.

Step 1. With this modification, the possibility of the development of a bending momentX1 at node G automatically disappears, while the deformation corresponding to X1 isdeveloped, i.e. the relative rotation of the two ends at node G. The magnitude of X1

is therefore considered to be the statically redundant force.

Step 2. Now X1 is considered as an external individual loading on the primary structure,due to which a corresponding deformation along the direction of X1 is developed.

Step 3. Superposition of the deformations obtained in Steps 1 and 2 gives the totalrelative rotation at the ends of the hinge due to the external load X1, which willobviously depend on the value of X1. It is clear that the only acceptable value for X1

is the one that causes a deformation identical to the deformation of the real indeter-minate system, i.e. a zero relative rotation.

Example 2Consider the suspended beam shown in Figure 3.4. Of the various modifications that canbe made to make this structure statically determinate, the interruption of the continuityof bar CD and changing the fixed support A to a hinged one are selected.

Step 1. In the primary structure thus created:

. The development of an axial force X1 in the bar CD and of a bending moment X2 atnode A, is excluded.

96


. Deformations due to the external loads are developed in the directions of X1 and X2,i.e. a relative shift of the cut points in the bar CD and a rotation at end A respectively.The magnitudes of X1 and X2 are the statically redundant forces.

Step 2. The imposition of X1 and X2 only leads to a corresponding deformation in thedirection of X1 (a relative shift of the cut points) and a corresponding deformation inthe direction of X2 (a rotation of end A) will develop. Obviously these deformationswill depend on the values of X1 and X2.

Step 3. Superposition of the deformations obtained in Steps 1 and 2 gives the relativeshift and rotation that will result from the simultaneous action of the external loadsand any arbitrary values of the statically redundant forces X1 and X2.

However, it is obvious that only those values of X1 and X2 are sought after that willlead to the deformations that actually occur in the real redundant system, i.e. a zerorelative shift and a zero rotation.

3.2.2 The analytical application of the force methodAn analytical application of the force method is now presented using Example 2 inSection 3.2.1, as illustrated in Figure 3.5. As already explained, for the primary structure

97

Requirement for elimination of relative rotation

Primary structure Primary structure

Primary structure

Development ofrelative rotation

Development ofrelative rotation

X1

X1

G

G

Redundant structure

Figure 3.3 A redundant force as an external action


two deformations have to be determined when the given external loads as well as theloads X1 and X2 are applied: deformation in the direction of X1 and deformation inthe direction of X2.The first deformation is denoted by s

1 and the second one by s2. The deformation

s1 is considered positive when, corresponding to the deformation effect of the arbitrary

direction of X1, the two section ends are approaching each other. Deformation s2 is

considered positive when, corresponding to the deformation effect of the arbitrarydirection of X2, the end is rotating clockwise.The following loading states are considered:

. State 0: loading due to external loads.

. State 1: loading due to a unit force X1¼ 1.

. State 2: loading due to a unit moment X2¼ 1.

It is obvious that, according to the superposition principle, any deformation in the finalstate corresponding in the total action of the external loads and the individual actionsX1 and X2, is equal to the deformation value in State 0, plus X1 times its value in State 1,plus X2 times its value in State 2.Thus, for the relative shift s

1 of the section ends

1s ¼ (relative shift in State 0)þ X1 (relative shift in State 1)

þ X2 (relative shift in State 2)

98

Redundant structure

Primary structure

Requirement for null relative displacementRequirement for null rotation

C

BD

BD

A

C

A

Rotation Rotation Rotation

Relative displacement Relative displacement Relative displacement

X1

X1

X1

X1

X2

X2

Figure 3.4 Redundant forces as external actions


while for the rotation s2 of end A it will be

s2 ¼ (rotation in State 0)þ X1 (rotation in State 1)þ X2 (rotation in State 2)

At this point, a more appropriate notation is introduced for the deformations in order tosystemise their computation.Fij is used to denote the deformation of the (primary) structure, and this deformation

corresponds to the redundant force Xi when the structure is loaded only with Xj¼ 1, inother words the deformation corresponding to Xi in state j. As Fij corresponds to adeformation due to a unit load, it is called the flexibility coefficient (see Section 2.3.8).Fi0 is used to denote the deformation of the primary structure, corresponding to the

99

A = 0.002 m2

I = 0.00029 m4

E (homogeneous)

20 kN/m

20 kN/m

5.0 m

Redundant structure

Primary structure

10.0 5.0 m

∆s1

X1

X1

X2∆s

2

Step 1Dependent on the external loading

Step 2Independent of the external loading

Step 3 X1 = 299.70 kNX2 = –264.65 kN m

∆s1 = F10 + X1 · F11 + X2 · F12

∆s2 = F20 + X1 · F21 + X2 · F22

Requirement by redundant structure: ∆s1 = 0

Requirement by redundant structure: ∆s2 = 0

F10 = –5121.9/EIF20 = 2812.5/EI

F11 = 11.1/EI + 11.2/EAF21 = –4.97/EI

F12= –4.97/EIF22= 5.0/EI(F12 = F21)

· X1 · X2

562.501.49

1.0

[M0] [M1] [M2][0]

[1] [2]Rotation F20 Rotation F21 Rotation F22(virtual loading: [2])

1 kN

1 kN1 kN

Relativedisplacement F12

Relativedisplacement F11

Relativedisplacement F10 (virtual loading: [1])

20 kN/m

Figure 3.5 The analytical application of the method of forces


redundant force Xi when the structure is loaded only with the given external loading(State 0).Using the above defined notation and according to the aforementioned superposition,

it may be written for s1 and s

2:

s1¼ F10þX1 F11þX2 F12

(a)s

2¼ F20þX1 F21þX2 F22

The physical significance of the factors Fij is as follows:

. F10, the relative shift due to the external loads; F11, the relative shift due to X1¼ 1;F12, the relative shift due to X2¼ 1. (A positive sign corresponds to the deformationdue to X1.)

. F20, the rotation due to the external loads; F21, the rotation due to X1¼ 1; F22,the rotation due to X2¼ 1. (A positive sign corresponds to the deformation dueto X2.)

All the above deformations can be computed through the principle of virtual work. Forthe magnitudes F10, F11 and F12 the loading X1¼ 1 has to be used as the virtual one(State 1), whereas for the magnitudes F20, F21 and F22 the loading X2¼ 1 is used asthe virtual one (State 2). Thus:

1 F10 ¼ðM1

M0

EIds ¼ 5121:88

EI(kNm)

1 F11 ¼ðM1

M1

EIdsþ

ðN1

N1

EAds ¼ 11:10

EIþ 11:18

EA(kNm)

1 F12 ¼ðM1

M2

EIds ¼ 4:97

EI(kNm)

1 F20ðM2

M0

EIds ¼ 2812:5

EI(kNm)

1 F21 ¼ðM2

M1

EIds ¼ 4:97

EI(kNm)

1 F22 ¼ðM2

M2

EIds ¼ 5:0

EI(kNm)

It is observed that F12¼ F21, and this is justified according to the Betti—Maxwelltheorem (see Section 2.3.5). Thus, in general, Fij¼ Fji.Note that the above equations (a) allow the calculation of s

1, s2 for any specific

values of X1 and X2.The third step of the method of forces is now applied, i.e. the satisfaction of the

compatibility of the deformations. This means that s1 and s

2 are required to takethe values they have in the real structure, i.e. zero and zero, respectively. If the

100


left-hand side of equations (a) is known, then X1 and X2 can be determined from theresulting equations. From the solution of equations (a) it results that X1¼ 299.70 kNand X2¼264.65 kNm.Thus if X1 and X2 are replaced in the primary structure with their actual senses and

values then, together with the external loading, the sectional forces diagrams can bereadily obtained, being obviously identical to those of the actual redundant structure(Figure 3.6).

3.2.3 The effect of temperature changeThe imposition of a temperature change in a statically indeterminate structure withrespect to an initial temperature (either uniform across all fibres of the cross-section,or linearly variable along the depth of the section) causes a stress state; unlike in astatically determinate structure, where only deformations are caused and there is nostress at all (see Section 2.3.2). This is because under the applied temperature changethe selected primary structure is deformed (without being stressed), developing specificdeformations at the modified points of the structure. Therefore, specific redundant forcesmust be applied at the relevant points in order to re-establish the required deformationvalues, i.e. those occurring in the statically indeterminate structure (usually zero). Thestresses developed will be due to these redundant forces only.The redundant forces are determined in the same way as described previously. It is

clear that temperature effects occur only in State 0, i.e. in the deformations Fi0.For the structure under consideration and for the state of temperature depicted in

Figure 3.7, it is obtained (see Section 2.3.3):

1 F10 ¼ðN1 T Tsð Þ dsþ

ðM1

T T

hds

¼ 1:11:18 105 20 1

2 1:49 15:0 10

5 400:50

¼ 0:011 kNm

101

Redundant structure

10.0 m 5.0 m

5.0 m

20 kN/mA = 0.002 mI = 0.00029 mE (homogeneous)

[M ] [V ] [N]

122.97

57.0

268.0677.0343.0

62.50

264.65

250.0

34.99

299.70 kN20 kN/m

264.65 kN mPrimary structure

Figure 3.6 Sectional forces diagrams drawn on the basis of the determined redundant forces


1 F20 ¼ðN2 T Tsð Þ dsþ

ðM2

T T

hds

¼ 0þ 1

2 1:0 15:0 10

5 400:50

¼ 0:006 kNm

In the linear system of equations (a) given in the previous section, all the flexibilitycoefficients remain the same and, by requiring that s

1 ¼ s2 ¼ 0, their solution

yields X1¼ 40.32 kN and X2¼32.0 kNm.

3.2.4 The effect of support settlementThe imposition of specific shifts or rotations on the supports of indeterminate structuresalso creates stresses, unlike the case in statically determinate structures in which onlydeformations are caused, and no stresses (see Section 2.3.7). This can be explained asfollows (Figure 3.8).

102

5.0 m

Tinit = +20°C

Tinit = +20°C

T = 0°C

T = 0°C

T = +40°C

T = +40°C

Redundant structure

5.0 m10.0 m

A = 0.002 m2

I = 0.00029 m4

E = 2.1 · 108 kN/m2

∆s1

∆s2

Shortening

Curvature due to ∆T

No axial temperature variation

X1

X2

X1

Primary system

Figure 3.7 The application of the force method in the case of a temperature change

The redundant force causes stress

δ δ

δX1X1

First case Second case

Development of δ required Requirement for null rotation

Figure 3.8 The stress state due to support settlement


The consideration of a primary structure will show the imposed deformation either tocoincide with a redundant force or not. In the first case, it is obvious that the impositionof redundant forces on the primary structure will be required, these forces having suchvalues that the actual deformation state at the points of the supports is produced. Thestress state of the indeterminate structure is caused by these forces (see Figure 3.8).In the second case, where the imposed deformation does not coincide with some

selected redundant force, the primary structure under the imposition of the deformationconsidered at its supports will develop at the modified points certain deformations corre-sponding to the redundant forces, without of course being stressed. It is obvious that theimposition of redundant forces as an external loading will be required, so that theresulting deformations at these points are identical to the real ones. The stress stateof the redundant structure is caused by these forces (see Figure 3.8).Two examples, using the same structure considered above, are now given to demon-

strate the procedure used for the case where there are support settlements (Figure 3.9).In the first example (see Figure 3.9(a)), in addition to the uniform loading, the structureis subjected to an imposed rotation of the fixed support A by 0.01 rad in the clockwisesense. The coefficients ‘F’ in the previously established system of equations (a) (seeSection 3.2.2) obviously remain unaltered:

s1 ¼ F10 þ X1 F11 þ X2 F12

s2 ¼ F20 þ X1 F21 þ X2 F22

The redundants X1 and X2 result from the requirement to create through equations (a)such valuess

1 ands2 that are identical to the actually occurring ones, i.e.

s1 ¼ 0 and

s2 ¼ þ0:01. It is: 1¼ 376.50 kN, 2¼ 68.21 kN/m.In the second example, the same structure under the external loading is subjected to

both to an imposed rotation at point A of 0.01 rad clockwise and to a support settlementat point B of 0.02m downwards (see Figure 3.9(b)). The same superposition is now

103

The imposed rotation is not included in the primary system0.02 m

0.02 m

Primary systemPrimary systemX2 X2

X1 X1

X1 X1

(a) (b)

20 kN/m 20 kN/m

20 kN/m 20 kN/m

5.0 m

0.01 rad 0.01 rad

C C

A

A

A

A

B

B

B

B

D

D

D

D

C C

A = 0.002 m2

I = 0.00029 m4

E = 2.1 · 108 kN/m2

Redundant structure Redundant structure

10.0 m 5.0 m10.0 m 5.0 m

∆s1 ∆s

1

∆s2 ∆s

2

Figure 3.9 The application of the force method in the case of support settlements


made, by considering in State 0 the imposed settlement of support B of 0.02mdownwards, as it does not correspond to a statically redundant size. The role of X1

and X2 now is to cause a zero relative displacement at the cut point of bar CD anda þ0.01 rad rotation of end A, when applied to the primary structure subjected to theexternal loading and to the settlement of support B.Equations (a) again allow the calculation ofs

1 ands2 for any value of X1 and X2, but

the coefficients F10 and F20 are obviously different from those in Section 3.2.2, althoughall the other coefficients remain the same. According to the principle of virtual work:

1 F10 þ 0:298 0:02 ¼ðM1

M0

EIds ¼ 5121:88

EIðkNmÞ

1 F20 1

15 0:02 ¼

ðM2

M0

EIds ¼ 2812:5

EIðkNmÞ

The system of equations (a) leads to the determination of values for X1 and X2 that willcause s

1 ¼ 0 and s2 ¼ þ0:01 rad, i.e. X1¼ 412.34 kN and X2¼48.61 kNm.

3.2.5 The influence of EI in the stress stateIn the case where a statically redundant system is subjected only to external loads, thedeformations of the primary structure, F10, F20, . . . , etc., as well as Fij are inverselyproportional to the value EI for the members and hence, as examination of equations(a) in Section 3.2.2 shows, the stress state, being determined by the redundant forcesX1, X2, . . . , etc., definitely depends on the ratio of the individual values of EI to eachother and not on their absolute values. Thus, the frames (a) and (b) in Figure 3.10have exactly the same bending moment diagram, while frame (c) has a different one.However, in the case of a temperature change or an imposed support settlement, the

developing deformations in the fundamental structure are geometrical ones and do notdepend on EI. As a result — see equations (a) in Section 3.2.2 — the final stress state (i.e.X1, X2, . . . , etc.) depends on the absolute value of EI. Thus, in Figure 3.10, the higher EIvalues in frame (b) cause greater stresses, due to some imposed temperature change orsome support settlement, than those developed in frame (a). It is obvious that thepresence of external loads simultaneously with either a temperature change or animposed settlement, or even both, does not change the fact that X1, X2, . . . , etc.,ultimately depend on the absolute value of EI.

104

EI EI 3EI 3EI EI EI

(a) (b) (c)

3EI 9EIEI

Figure 3.10 The influence of member stiffness on the stress state


3.2.6 Checking the resultsGiven a statically indeterminate structure with an applied loading and the resultingsectional diagrams, there is always the question of whether or not these diagrams arecorrect.If suitable modifications are performed to the supports and/or inside the structure, so

that it becomes a statically determinate one, then the stress (and deformation) state ofthe redundant structure may be determined on the basis of the actual loading and theexternally imposed relevant redundant quantities, which may be taken immediatelyfrom the corresponding diagrams [M], [V] and [N].Now, the corresponding deformations at the points of modification in the primary

structure may be determined through the principle of virtual work, and the control ofthe validity of the sectional diagrams consists of checking whether the calculated defor-mations are identical to the real ones. If this is not the case, then the sectional diagramsare erroneous.Of course an equilibrium check must have been undertaken, i.e. a check to ensure

that the diagrams [M], [V] and [N], are in ‘equilibrium’ with the external loads (seeSection 2.2.5).

3.2.7 Analysis of elastically supported redundant structuresElastically supported redundant structures clearly develop a different stress state thanthe one developed by a redundant structure having immovable supports, as supportsettlements occur, and with a non-prescribed value.As discussed in Section 2.3.7, the settlement of an elastic support (displacement or

rotation) is due to the corresponding action on the ground, having always its direction,i.e. the direction opposite to that of the reaction on the structure. Thus the analysis ofelastically supported redundant structures does not present any peculiarity, and followsthe normal course of the analysis of statically indeterminate structures with supportsettlement (see Section 3.2.4). It need only be taken into account whether or not theselected redundant forces correspond to the elastic supports (Figure 3.11).In the case where the redundant size X1, for example, corresponds to an elastic

support, then the corresponding s1 actually takes the value (which will be required

to be developed in the primary structure):

s1 ¼ fs X1

It is clear that the modifications that will lead to the fundamental structure can alsoleave the flexible supports intact, and therefore the primary structure is simply a staticallydetermined system with elastic supports, and can be handled as described in Section2.3.7 (see Figure 3.11).

3.2.8 Qualitative handling of the method of forcesIn many cases the qualitative determination of the reactions and the sectionaldiagrams of statically redundant structures is possible without the need for any

105


particular numerical calculations. This is always desirable, not only in order to have adirect sense of the response of a structure, but also in order to be able to test anycomputed results.The strategy for making a qualitative analysis of a statically indeterminate structure

according to the force method does not differ in any way from the one used for its quanti-tative analysis (see Figures 3.13 and 3.14).After the choice of suitable modifications, the primary structure is considered

subjected to the external loading. The redundant forces are already located. Theprimary structure is deformed under the external loads. What is sought are thesenses of the corresponding deformations at the points of action, and these can befound once the whole deformation configuration of the primary structure has beendrawn. The drawing of the deformation configuration is accomplished on the basisof the corresponding bending moment diagram, as detailed in Section 2.3.2. Thebending of any part of a bar always occurs toward the side of tension fibres. Itshould not be overlooked that in any rigid joint the angles between the connectedmembers are maintained, regardless of any joint deformation. Only in hinged jointsdo adjacent members develop different rotations. As the specific deformations foundin this way will not generally comply with the actual ones of the redundantstructure, the suitable senses of the redundant forces may be easily concluded in

106

∆s1

∆s1 = F10 + X1 · F11

Requirement from redundant structure: ∆s1 = –X1 · fs

X1 = 57.9 kN

fϕ fϕ

fϕ fϕ

1 kN

· X1

20 kN/m

20 kN/m

EI = 42 000 kN/m2

20 kN/m

Displacement F11(virtual loading is state [1])

Displacement F10(virtual loading is state [1])[0] [1]

The settlement is not included6.0 m

Primary systemHyperstatic structure3.0 m

EI

Ground fs

fs

fϕ = 0.001 rad/kN mfs = 0.001 m/kN

X1

X1

Figure 3.11 The application of the force method in the case of elastic supports


order to restore these real deformations. Plotting the sectional diagrams is thenpossible.Figure 3.12 shows the application of the above technique for some typical structures,

such as the fixed-end beam, the continuous beam and the two-hinged frame.

107

Deduced senses of X1 and X2

Deduced senses of X1

Deduced deformation

Deduced deformation

Deduced deformation

DeduceddeformationDeduced sense of X1

Primary structureRedundantforce

Redundantforce

Redundantforce

Redundantforce

Redundantforce

Redundantforce

Redundantforce

Redundant structure

Primary structure

Redundant structure

Redundantstructure

Primarystructure

[M0]

Deduced sense of X1

Primary structure

[M0] Not developed

Redundant structure

Ttop > Tbottom

Ttop > Tbottom

[M0]

[M0]

[M ]

[M ]

[M ][M ]

X1 X2

X1

X1

X1

X1

X1

X1

X1

X1 X1

X1

X2

X2X1

X1 X1

X1 X1

X1 X1

Figure 3.12 Qualitative plot of the bending moment diagram


3.2.9 Consequences of the elastic support conceptThe beam AB in Figure 3.13 has both its ends A and B elastically clamped and acoefficient of elastic rotation f’ (see Section 2.3.7.2). The beam is subjected to a uniformload q.The determination of the redundant reactions X1 based on the primary structure of

the simply supported beam leads to the relationship

f’ X1 ¼ s1 ¼ F10 þ F11 X1

from which it results that

X1 ¼q L2=12

1þ 2 f’ EI=L

It is observed at first that for f’¼ 0, i.e. for a fully fixed support, the value of theredundant reaction is independent of EI and is equal to q L2/12, while as the term f’increases the reaction decreases, depending always on EI. For a very large f’ thevalue of X1 approaches zero and the beam behaves as if it were simply supported.

108

A B

A B

A B

Increasing of spring flexibility leads to decrease of end momentand vice versa

The loaded beam develops the same response in all three cases

LL1 L2

L

L

L1 L2

EI

EI

E1I1 E2I2

EIE1I1 E2I2

q

q

q

q

L1/(4E1I1) = fϕ L2/(4E2I2) = fϕ

L1/(3E1I1) = fϕ L2/(3E2I2) = fϕ

X1X1

X1X1

qL2/8

fϕ fϕ

Figure 3.13 The elastic support concept in a continuous beam


As explained in Chapter 2 (see Section 2.3.7), the existence of an elastic support can bemodelled by a connection to a simply supported beamwith lengthL and rigidity (EI) suchthat f’¼ L/3 (EI). Thus, the elastically clamped beam, behaves in precisely the sameway as the intermediate span of the continuous beam shown in Figure 3.13. It is alsounderstood (see Section 2.3.7) that an increase or a decrease in the flexibility f’ meansa decrease or increase, respectively, in the stiffness of the end beams, which is of courseexpressed as 3 (EI)/L, i.e. as the inverse of the flexibility f’.Therefore, according to the above behaviour of X1 with respect to the flexibility f’ and

using the more usual term of stiffness, it is concluded that the increase in the length L

leading to a reduction in stiffness causes a reduction in the support moment MA, whichis obviously identical to the redundant size X1 of the isolated beam. Of course, anincrease in the EI of the end beams leads to an increase in stiffness, and consequentlyan increase in the end moments MA. Moreover, it is concluded that an increase inthe EI of the beam AB leads, according to the above analysis, to a reduction in thesupport moment MA (see Figure 3.16). It is clear that an increase in, for example, themoment MA leads to a reduction in the moment in span AB, as the constantmoment diagram for the simply supported beam is always hanging by the points thatare associated with the moment MA.Clearly the existence of an elastic support can be related to the connection with a

simply supported beam or the connection with a fixed-hinged beam. In this case therotation deformability of the free end is smaller than the corresponding one of thesimply supported beam, according to the relationship

’ ¼ L

4 ðEIÞ M

and in this case f’ is computed from the expression L/4 (EI) (see Figure 3.13).It can be concluded from the above that the increase in stiffness of members

monolithically connected to the loaded beam always leads to an increase in themoment at the connection node, whereas a reduction in the stiffness of membersneighbouring the loaded beam always leads to a reduction in the support moment ofthe beam and, accordingly, to an increase in its span moment, as previously explained(Figure 3.14). Hence:

. A reduction in the length of the adjacent beams (i.e. an increase in their stiffness withrespect to the loaded beam AB), leads to an increase in the support momentsMA,MB

(with a simultaneous reduction in the span moment).. An increase in the stiffness of the loaded beam AB with respect to the adjacent beams

causes a reduction in the support momentsMA,MB and, consequently, to an increasein the span moment.

. A very long length (very low stiffness) of the adjacent beams causes the beam AB tobehave as a simply supported one, whereas a very short length (very high stiffness) ofthe adjacent beams causes the beam AB to behave as a fixed-end one.

It is obvious that the continuous beam shown in Figure 3.15 develops precisely the samebending response as the corresponding frame. (This happens because the frame does not

109


develop any shift at its nodes, like the continuous beam.) Accordingly, all the previousobservations regarding the continuous beam are also applicable to frames. Thus, anincrease in, for example, the height of the column leads to a reduction in the jointmoment MA, and vice versa. Because in the case of a frame the moment MA is dueexclusively to the horizontal reaction H, it is concluded that an increase in the heightof the column leads to a reduction in the horizontal reaction H, and vice versa (seeFigure 3.15).Finally, a reduction in the EI of the girder leads to an increase in the momentMA and,

accordingly, to a reduction in its span moment. Clearly, stiffening of the cross-section ofthe columns leads also to an increase in the moments MA, MB .

3.2.10 The influence of stiffness on the stress distributionIt is clear from the discussion in the previous section that, when a loaded member isconnected to other unloaded members, an increase in the stiffness of the unloadedmembers with respect to the loaded one leads to an increase in the end moments ofthe loaded member, and vice versa. This conclusion can be extended, in that a relativechange in member stiffness (bending or axial) in a structure results in an increase in thestresses in the strengthened members and to a reduction in the stresses in the less stiffmembers (Figure 3.16).Further to this property, which is attributed only to the static redundancy (the stress

state of statically determinate structures remains unaffected by any changes in the

110

Decrease in end moments

Increase in end moments

q

Relative increase (or decrease) in the stiffness of the adjacent memberscauses an increase (or decrease) in the end moments

q

q

q

q

A B

Figure 3.14 The effect of a variation in the stiffness of continuous beams


rigidity of their members), it may be concluded that, when a load (force or moment) istransferred to the supports through members that are compelled to develop commondeformations, the most rigid members carry the largest portion of the bendingresponse and the least rigid ones the smallest proportion. g

As a first example, a system of two beams connected by a vertical bar at their midpoints isexamined (Figure 3.17). Assume that the axial rigidity EA of the bar is very high, whichensures the same (vertical) shift at the midpoint of each beam. Considering as redundantthe axial force X1 of the binding bar, it is found that

X1 ¼P

1þ L2L1

3 ðEIÞ1ðEIÞ2

Expressing the stiffness of the two beams as k1¼ 48 (EI)1/L31 and k2¼ 48 (EI)2/L32, then

X1 ¼P

1þ k1k2

111

End moments decreaseSpan moment increases

Decrease in H

End moments increaseSpan moment decreases

Increase in H

End moments decreaseSpan moment increases

Decrease in H

End moments increaseSpan moment decreases

Increase in H

q

q

q

q

q

q

Relative increase (or decrease) in the stiffness of the adjacent memberscauses an increase (or decrease) in the end moments

q

H H

H H

BA

BA

BA

Figure 3.15 The influence of a variation in the stiffness of frames


This result has a particular importance as it shows the distribution of the load P on thetwo beams: The bottom beam is loaded with P2¼X1, while the top beam is loaded withthe force P1¼ PX1. It results that

P1 ¼ P k1k1 þ k2

and

P2 ¼ P k2k1 þ k2

If one of the two beams is turned by 908 in the horizontal plane, the above results andconclusions remain valid. Of course they concern the transfer of a vertical load P in twohorizontal directions, a subject that has a significant practical importance (seeChapter 10).If the actual axial stiffness ks¼EA/L of the binding bar is taken into account, the

above results are written as

X1 ¼P

1þ

112

Increasing EI causes a decreasein the tie force and consequentlya decrease in the joint moments

Increasing EI of the beam causes anincrease in its moments and a

decrease in the girder moments

Strengthening the support sections causes an increase in the support momentsand a decrease in the span moments

EI

EI

EI

Figure 3.16 The influence of the rigidity of the members on the bending moment distribution


with

P1 ¼ P 1

1þ

and

P2 ¼ P

1þ

where

¼ k2 1

k1þ 1

ks

As a second example, the joint between the connected bars in Figure 3.17 is considered.The moment M is applied to the joint; the joint is immovable. It is obvious that thesum of the moments M1, M2 and M3 should, for equilibrium reasons, be equal to M.Moreover, given that stiffness (2)> stiffness (1)> stiffness (3), then M2>M1>M3

(see Section 3.3.7).Finally, a fixed beam subjected to either a concentrated load N or a concentrated

torsional moment MT is examined. Both cases are characterised by the requirement

113

If stiffness (2) > stiffness (1)then beam (2) is loaded by >P/2

If stiffness (2) < stiffness (1)then beam (2) is loaded by <P/2

Beam (2) takes the largest part of the load

N1

N MT

N2 M1 M2

L L

a ba b

(3)

(2)(2)

(1)

(1)

(1) (1)

(2) (2)

M2 > M1 > M3

P

P P

M M

M3

M2

M1

X1

Figure 3.17 The distribution of a nodal load in statically redundant systems


that at the point of application of the load the axial displacement or the twist angle,respectively, must be common to the two parts of the beam. Thus, according to theforegoing, and to Sections 2.3.8 and 2.5.3, it is obtained that, respectively (seeFigure 3.17):

N1 ¼ N bL; N2 ¼ N a

L

and

M1 ¼ MT bL; M2 ¼ MT a

L

Note that the above results come out the same whether N or MT is thought of as thetransverse load P acting on a simply supported beam, with ‘N’ and ‘M’ being thereaction of the beam, respectively (see Figure 2.32).

3.3 The deformation method

3.3.1 IntroductionOn a plane structure subjected to a given loading, it is always possible to apply suitableexternal actions on its nodes such that all shifts and rotations developed at the nodes areeliminated. These actions (forces and moments) may be directly determined as follows.As each node, cut out as a free body, will receive at the member ends bending

moments, shear forces and possibly axial forces corresponding to the fixed state ofeach member, these external actions merely have to satisfy the equilibrium of thenode, and so are directly determinable. This is illustrated by the two examples shownin Figure 3.18.It is now understood that the stress state of the examined structure results by super-

posing the already known stress state of the ‘fixed’ structure onto the one resulting froma ‘nodal loading’, which consists merely of the equal and opposite forces to the aboveactions. It is then obvious that the nodal deformations of the structure are due exclu-sively to this last nodal loading. Thus, the problem of the initial structure is substantiallyreduced from the analysis point of view to one of a specific nodal loading, as the stressstate of the ‘fixed’ structure may be considered as ‘trivial’. g

The deformation method is based on the fact that the stress state of any member of astructure loaded only at its nodes is determined by the developing displacements androtations at its two ends. Of course, if the member is statically determinate, i.e. it iseither simply supported or a cantilever, it remains unaffected by the displacements orrotations of its supports. It should also be pointed out (something that will becommented on later as well) that the rotations that develop at the hinged ends ofmembers do not cause stresses.Given the above, the ‘active’ members in a structure (subjected to the determined

nodal loading according to the precedents), will fall into one of the following twocategories:

114


. monolithically connected at one end and hinged at the other (simply fixed)

. monolithically connected at both ends (fixed—fixed).

The deformation method consists of the following three steps:

(1) the introduction of specific rotations and vectors of nodal shifts of the structure asunknowns

(2) the expression of the sectional forces in the members on the basis of their nodaldeformations

(3) the determination of the unknown deformations from the equilibrium requirementsat each node.

It is obvious that the resulting stress state must be superposed on that of the fixedstructure.It is clear from step (2) the application of the deformation method is based on the

force method, which is essentially the only way to deal properly with the stress stateof a statically indeterminate structure. However, thanks to standardised results (aswill be explained later), it is possible to overcome this and apply the deformationmethod, which may eventually lead to a smaller number of unknowns than areencountered in the force method.Despite the fact that step (1) is the most critical, it is appropriate at this stage to

examine the simply fixed beam and the fixed—fixed beam, in order to obtain thestandardised results required in step (2).

115

6.0 kN/m

15.0 kN

6.0 kN/m

6.0 kN/m

6.0 kN/m15.0 kN 15.0 kN15.0 kN

12.5 kN m 12.5 kN m 12.5 kN m12.5 kN m

‘Fixed state structure’

‘Fixed state structure’

The nodal actions eliminatethe deformations of the nodes

Joint deformations

Joint deformations

10.0 kN10.0 kN

11.25 kN18.75 kN11.25 kN18.75 kN

10.42 kN m

10.42 kN m

4.0 kN/m

4.0 kN/m(5/8) · 6.0 · 5.0

6.0 · 5.02/84.0 · 5.0/2

4.0 · 5.02/12(3/8) · 6.0 · 5.0

5.0

m

5.0 m

6.0 · 5.0/26.0 · 5.02/12

The nodal actionseliminate the deformations

of the nodes

Figure 3.18 The concept of the fixed structure and the imposition of nodal actions


3.3.2 The simply fixed beam

3.3.2.1 The fixed stateThe redundant reaction X1 is determined for some external load, using the com-patibility requirement for the primary structure of a simply supported beam(Figure 3.19):

rotation of i¼ 0¼ rotation of i due to the external loadþ rotation of i due to X1

For a vertical load downwards,X1 causes tension in the top fibres, as may also be deducedfrom the developed curvature of the elastic line.A uniform load q causes the moment: X1¼ q L2/8. The bending moment diagram

is shown in Figure 3.19. The reactions at A and B are equal to 5 q L/8 and3 q L/8, where the dimension L refers to the length of the horizontal projection ofthe beam. g

Regarding now a temperature difference T between the top and bottom fibres (Figure3.20), the above compatibility requirement leads to the value X1¼ 3 (EI T T)/2h.It can be seen that the bending moment is independent of the length of the beam,

but the shear force is not.

116

The reactions and bending response are independent of beam inclination

[M ]

[M ]

Elastic line

α

q · L2/8

q · L2/8

q · L2/8

q · L2/8

5 · q · L/8

5 · q · L/8

3 · q · L/83 · q · L/8

L

L

q

q q

q

iii

k kk

X1

Figure 3.19 A simply fixed beam under a uniform load


3.3.2.2 Imposed rotation of the fixed endEnd i is subjected to an anticlockwise rotation ’, as shown in Figure 3.21. The redundantreaction X1 is determined from the compatibility requirement:

rotation of i¼’¼ rotation of i due to X1

The result is that X11¼ (3EI/L)’, causing tension at the top fibres. L is the length ofthe beam. The sense of this reaction may also be ascertained qualitatively by consideringthe curvature of the elastic line at end i. The bending moment diagram is shown inFigure 3.21.Clearly the reaction X1 is the moment required to rotate end i of the simply supported

beam ik by ’. Hence, the stiffness of the simply supported beam ik with respect to endrotation is 3EI/L.

117

[M ]3 · (EI · αT · ∆T) · cos α/(2hL)

The bending response is independent of the beam length

3 · (EI · αT · ∆T)/(2h)

3 · (EI · αT · ∆T)/(2h)

Ttop > Tbottom

Figure 3.20 A simply fixed beam under a temperature difference

(3 · EI/L2) · ϕ

(3 · EI/L2) · ϕ (3 · EI/L) · ϕ

(3 · EI/L) · ϕ

kk

L

ϕ

ϕ

Elastic line

i

i

[M ]Required moment to produce rotation

X1

Figure 3.21 A simply fixed beam subjected to a fixed end rotation


3.3.2.3 Imposed relative displacement of the beam endsEnd i is subjected to a transverse displacement downwards, as shown in Figure 3.22.The redundant reaction X1 is determined using the compatibility requirement:

rotation of i¼ 0¼ rotation of i due to displacement þ rotation of i due to X1

It results that X11¼ (3 EI/L2) , causing tension at the bottom fibres, as may also bededuced qualitatively from the corresponding curvature of the elastic line. The bendingmoment diagram is shown in Figure 3.22.From this qualitative approach it is obvious that the bending moments (and shear

forces) of the member will be the same as above if, instead of end i, end k is subjectedto a transverse displacement upwards, or, more generally, if a relative shift is imposed

118

Required force to produce displacement

[M ]Elastic line

Elastic line

Elastic line

Li

i

i

i

kkk

k

∆

∆

∆∆

The bending response is due onlyto the relative transverse shift

(3 · EI/L3) · ∆

(3 · EI/L3) · ∆

(3 · EI/L3) · ∆

(3 · EI/L3) · ∆

(3 · EI/L3) · ∆

(3 · EI/L2) · ∆

(3 · EI/L2) · ∆

X1

Figure 3.22 A simply fixed beam under an imposed displacement of the supports


transversely to the axis of the beam, provided, however, that the resulting elastic linecauses with its curvature at end i a tension on the same (bottom) fibres. It is concludedthat only the relative shift plays a role in the bending response, and not the individualshifts of the two member ends.It is also clear that the shear force at the support k is the transverse force required to

shift end k of the cantilever by . Hence, the cantilever stiffness with respect to thetransverse shift of its free end is 3 EI/L3.

3.3.3 The fixed-end beam

3.3.3.1 The fixed stateThe redundant reactions X1 and X2 are determined from the compatibility requirementin the primary structure of the simply supported beam (Figure 3.23):

rotation of i¼ 0¼ rotation of i due to external loadþ rotation of i due to X1

þ rotation of i due to X2

119

q q

q q

The reactions and bending response are independent of the beam inclination

Elastic line

q · L/2

q · L/2

q · L/2

q · L2/12

q · L2/12

q · L2/12

q · L2/12

q · L/2

q · L2/12

q · L2/12

(q · L/2) · cos α(q · L/2) · cos α

[V ]

[M ]

[V ]

[M ]

X1

X2

L

L

k k k

i i i

α

Figure 3.23 A fixed-end beam under uniform load


rotation of k¼ 0¼ rotation of k due to external loadþ rotation of k due to X1

þ rotation of k due to X2

For an external load in a downward direction it is obvious, from the curvature of theelastic line, that X1 and X2 are causing tension in the top fibres. A uniform load qcorresponds to the values X1¼X2¼ q L2/12. Note that the dimension L refers to thehorizontal projection length of the beam. g

Regarding now a temperature difference T between the top and bottom fibres, theabove compatibility requirements lead to the values (Figure 3.24)

X1¼X2¼EI T T/h

It can be seen that the value of the constant bending moment along the beam is inde-pendent of the beam length, and no shear forces develop.

3.3.3.2 Imposed end rotationEnd i is subjected to a anticlockwise rotation ’ (Figure 3.25). The redundant reactionsX1 and X2 are determined from the compatibility requirements:

rotation of i¼’¼ rotation of i due to X1þ rotation of i due to X2

rotation of k¼ 0¼ rotation of k due to X1þ rotation of k due to X2

It results that

X1 ¼4 EIL

’

and

X2 ¼2 EIL

’

as shown in Figure 3.25. The length of the beam is denoted by L. The resulting curva-tures at the ends of the elastic line of the beam automatically indicate the side of thetensioned fibres and, accordingly, the sense of the bending moments developed at the

120

No shear developed

The bending response is independent of the beam length

(EI · αT · ∆T )/h (EI · αT · ∆T )/h

[M ]Ttop > Tbottom

Figure 3.24 A fixed-end beam under a temperature difference


ends. It can be seen that the redundant reactions have the same sense, i.e. the one corre-sponding to the rotating end being double that of the other end. From the oppositecurvatures at the two ends, it is concluded that there will exist a point in the beamwhere the curvature, and thus the bending moment also, will be zero (inflection point).It is clear that the reaction X1 is the moment required to rotate the freely supported

end i of the simply fixed beam by ’, with a resulting moment diagram identical to theone above. Hence, the stiffness of the simply fixed beam ik with respect to rotation ofits freely supported end is 4 EI/L.

3.3.3.3 Imposed relative displacement of the beam endsEnd i is subjected to a transverse downward displacement (Figure 3.26). Theredundant reactions X1 and X2 are determined from the compatibility requirement:

rotation of i¼ 0¼ rotation of i due to displacement þ rotation of i due to X1

þ rotation of i due to X2

rotation of k¼ 0¼ rotation of k due to displacement þ rotation of k due to X1

þ rotation of k due to X2

It results that

X1 ¼ X2 ¼6 EIL2

The resulting curvatures at the ends of the beam automatically indicate the side of thetensioned fibres and, accordingly, the sense of the bending moments developed at the

121

Elastic line

Required moment to produce rotation

ϕ

ϕ

kk k

[M ]

i

i i

L

X1

X2

(6 · EI/L2) · φ (6 · EI/L2) · φ

(4 · EI/L) · φ

(2 · EI/L) · φ

(4 · EI/L) · φ

Figure 3.25 A fixed—fixed beam under an imposed end rotation


ends. It is observed that the redundant reactions have the same sense and are equal inmagnitude. As the curvatures at the two ends are opposite, it is concluded that at themidpoint of the beam the curvature, and hence the moment, will be zero (inflectionpoint).This qualitative analysis shows that the bending moments will be the same as those

above if, instead of end i, end k is subjected to a transverse displacement upwards, or,more generally, if a relative shift is imposed transversely to the axis of the beam,provided, however, that the resulting elastic line causes, with its curvature at end i,tension in the same (bottom) fibres. It is concluded that only the relative shift plays a role in the bending response, and not the individual shifts of the two memberends.

122

The bending response is due onlyto the relative transverse shift

Required force for producing displacement

∆

∆

∆∆

[V ]

[M ]

Elastic line

Elastic line

Elastic line

(12 · EI/L3) · ∆

(12 · EI/L3) · ∆

(12 · EI/L3) · ∆(12 · EI/L3) · ∆

i

k

(12 · EI/L3) · ∆

(6 · EI/L2) · ∆

L

(6 · EI/L2) · ∆

(6 · EI/L2) · ∆

(6 · EI/L2) · ∆

i

iii

k k k k

X1X2

Figure 3.26 A fixed-end beam under an imposed end displacement


It is also clear that the shear force at the support k is the transverse force required toshift the fixed end k of the beam by (without rotation). Hence, the fixed-end beamstiffness with respect to a relative transverse shift of its ends is equal to: 12 EI/L3.

3.3.4 Recognition of the unknown nodal deformations

3.3.4.1 Rotations and displacementsAt each rigid joint (and also at every point) of any plane structure a displacement vectorand rotation are developed. With respect to the deformation method, the rotation of anode can be considered as an unknown magnitude if a bending moment can bedeveloped in at least one of its adjacent members (Figure 3.27). The nodes k of thestructure shown in Figure 3.27 do not justify an unknown rotation, although themembers develop a rotation at their ends. The nodes m develop rotations that do notaffect the stress state of the two-hinged bar.With regard to the displacement vectors, these may be considered as unknowns at

each node, as they can be freely developed. However, in order to restrict theirnumber, the following justified assumption is made. The lengths of all bendingmembers (i.e. those that do not develop exclusively axial forces) are assumed toremain unchanged. This means that the projections of the displacement vectors ofthe two ends on the beam axis are identical, and thus a relative shift of the two endsalong the bar is excluded (Figure 3.28). Of course, it should not be concluded on the

123

Joint rotations leave thehinged bar unaffected

Unknown rotations are referred only to developing end moments

k

k

k

m m

Figure 3.27 Determination of the unknown nodal rotations

The displacement vectorshave equal projections

on the direction 2–3

1

2

4

3

Figure 3.28 Displacement vectors ensuring the length of the bars is unchanged


basis of this assumption that, according to Hooke’s law, the axial forces should beconsidered zero, as these are simply determined from the equilibrium of the nodes. Inthe vast majority of framed structures that carry their loading through a bendingmechanism, the effects of this assumption on the accuracy of the results are insignificant.Nevertheless, if a member develops exclusively axial forces, as a two-hinged bar withoutintermediate loading, then its stress state is determined on the basis of its change inlength, according to Hooke’s law (see Section 1.2.1).Thus, the number of unknown nodal displacements is considerably reduced. For

example, without assuming an unchanged length of the bars, the displacement components1 to 8 of the frame nodes shown in Figure 3.29 are all independent of each other(eight unknowns), while with the above assumption the two displacements 1 and 2

are sufficient to describe any deformed configuration of the nodes.

3.3.4.2 Basic consideration of the kinematics of membersFor the final determination of the unknown nodal displacements, according to thecriteria presented in the next section, it is essential first to become familiar with thekinematics of the hinged connection between three arbitrary points (Figure 3.30).The system of bars ABC shown in Figure 3.30 is immovable. Node B cannot be

displaced because this would mean a change in length of some bar. If node A is nowdisplaced by ~aa going to point A0 and node C is displaced by~cc going to point C0, thequestion arises of how node B will be displaced such that from its new position B0 thelengths B0A0 and B0C0 are equal to the lengths BA and BC, respectively.Based on the fundamental assumption of small deformations governing the linear

behaviour of structures, the replacement of circular arches by their tangent is allowed(see Figure 3.30). Therefore, two segments, equal and parallel to AB and CB, aredrawn from the points A0 and C0, respectively. The intersection of the vertical linesat the ends of these two segments will fix the new position B0 of point B.

124

∆6

∆5

∆8

∆7 ∆2

∆2

∆1

∆4

∆3 ∆1

Assumption of unchanged lengths

Figure 3.29 Restriction of the number of unknown nodal displacements


3.3.4.3 Selection of the unknown nodal displacementsIn order to determine the unknown node displacements in a plane skeletal structure,which may represent any arbitrary displacement configuration of the system, all rigidjoints are first changed to hinged ones, thus converting the whole structure to amechanism.A group of shift vectors is selected that satisfies the following two conditions

(Figure 3.31):

(1) If all the proposed possibilities of shifting (displacement vectors) are blocked, theshift of any node should be impossible.

(2) Releasing one by one and consecutively every displacement vector of the proposedgroup, while keeping all the others blocked, the displacement vector of each nodeshould be expressed kinematically in terms of the displacement of the released oneonly, according to the previous section (see Figure 3.31).

The process of recognising the unknown displacement vectors is a heuristic one. Thismeans that it is simply checked whether or not a proposed group ‘’ fulfils the criteria(1) and (2). If it does, then it is the right one. If not, then another group should besought.

3.3.5 The procedure for the deformation methodThe basic steps of analysis using the deformation method are described below, andapplied to the example structure shown in Figure 3.32.

125

a→

c→

E′

B′

C′

A′

A

C

B

ED

B′A′ = BAB′C′ = BC

DE = DE′

Displacement of point B Replaces circular segmentReplaces circular segment

with centre A′

Replaces circular segmentwith centre C′

Figure 3.30 Kinematic handling of a system with two hinged solid bars


Step 1: Recognition of the unknown rotations. All rotations of solid nodes are a prioriindependent of each other (see Section 3.3.4.1) (see Figure 3.32).

Step 2: Recognition of the unknown displacements. The structure is transformed into amechanism by replacing the rigid nodes with hinges and searching for a displacementgroup that satisfies the two criteria set out in Section 3.3.4.3 (see Figure 3.32).

Step 3: Determination of the stress state in the fixed structure and then of the required externalactions in order to ensure the nodal equilibrium. Automatic determination of the activenodal loads on the structure with equal and opposite actions. The nodal deformationsof the structure are exclusively due to this final nodal loading (see Figure 3.32).

Step 4: Determination of the developing moments and shear forces at the ends of each bar dueto the imposed rotations and shifts of the ends that have been considered as unknowns, based onthe assumed arbitrary senses of the unknown rotations and displacements (Figure 3.33). Inorder to facilitate the determination of the physical sense of the bending moments,the corresponding elastic line resulting for each member is drawn. After the physical

126

Based on 2–5–3

Fulfilment of the first condition

All nodes unshifted

Fulfilment of the second condition

Fulfilment of the first condition

All nodes unshifted

Fulfilment of the second condition

One unknown displacement

One unknown displacement

All nodes unshifted

Unshifted5

5 5

1

1 2 3 4 1 2 3 4

1

5

1 22

43 43

Based on 3–4–2

Based on 1–2–5 Based on 5–3–4

Figure 3.31 The method of determining the unknown nodal displacements


sense of the end moments and shear forces has been determined, the set up of the signedexpressions of the moments and shear forces at the ends of the members follows, on thebasis of superposition. The adoption of a specific code of positive sign moments is, ofcourse, necessary. In the example examined, the convention based on a predefinedpositive border is followed.With regard to the axial forces, as pointed out in Section 3.3.4.1, while their

development is recognised, they cannot be expressed through the displacementvectors of nodes, because of the assumption of unchanged lengths. They are determinedthrough the equilibrium of nodes.

127

3.0

m20.0 kN/m

20.0 kN/m

20.0 kN/mThis loading ‘fixes’ the structure The node deformations are due to this loading

41.67 kN m 41.67 kN m

50.0 kN

50.0 kN

110.0 kN110.0 kN

37.5 kN37.5 kN70.83 kN m

70.83 kN m

Step 2 2

43

1

2

43

3

35

43

1

2

435

1

1

Step 1Step 3

α

∆4/sin α

∆4

ϕ4ϕ3

∆4

[M ]

41.67 41.67

90.0

22.5

Based on 1–3–4

‘Fixed structure’

20.0 · 5.0/2 20.0 · 5.0/2

20.0 kN/m20.0 kN/m

20.0 kN/m

3.0 m 5.0 m

20.0 · 3.02/820.0 · 3.0

20.0 · 3.02/2

20.0 · 5.02/12

20.0 · 3.0 · 5/8

20.0 · 3.0 · 3/8

Figure 3.32 Steps 1 to 3 in the analysis of a framed structure


Step 5: Equilibrium equations. After the determination of signed end moments and shearforces at each member, in terms of the unknown shifts and rotations, the three equili-brium equations for each node of the structure are formulated (Figure 3.34). Theequation of equilibrium for the moments does not present a problem, but the axial

128

Step 4

Member 3–4

Member 1–3

Member 4–2

Conventional expressionsM42 = +(4EI/3.0) · ϕ4 – (6EI/3.02) · ∆4

M24 = –(2EI/3.0) · ϕ4 + (6EI/3.02) · ∆4

V42 = –(6EI/3.02) · ϕ4 + (12EI/3.03) · ∆4

Physical forces

Physical forces

Physical forces

(4EI/3.0) · ϕ4

(2EI/3.0) · ϕ4

(6EI/3.02) · ϕ4 (12EI/3.03) · ∆4

(6EI/3.02) · ∆4

(6EI/3.02) · ∆4

4

43

3

1 1

3

43

43

43

4

2 2

∆4

ϕ4

Conventional expressionsM34 = +(4EI/5.0) · ϕ3 + (2EI/5.0) · ϕ4 + (6EI/5.02) · ∆4/tan αM43 = –(2EI/5.0) · ϕ3 + (2EI/5.0) · ϕ4 + (6EI/5.02) · ∆4/tan αV34 = –(6EI/5.02) · ϕ3 – (6EI/5.02) · ϕ4 – (12EI/5.03) · ∆4/tan αV43 = V34

(12EI/5.03) · ∆4/tan α

(6EI/5.02) · ∆4/tan α

(6EI/5.02) · ∆4/tan α(6EI/5.02) · ϕ4

(6EI/5.02) · ϕ3

(2EI/5.0) · ϕ3

ϕ3

(4EI/5.0) · ϕ3

(4EI/5.0) · ϕ4(2EI/5.0) · ϕ4

∆4

∆4/sin α

∆4/tan α

ϕ4

ϕ3

Conventional expressionsM31 = –(3EI/4.24) · ϕ3 + (3EI/4.242) · ∆4/sin α V31 = –(3EI/4.242) · ϕ3 + (3EI/4.243) · ∆4/sin α

(3EI/4.24) · ϕ3

(3EI/4.242) · ϕ3(3EI/4.243) · ∆4/sin α

(3EI/4.242) · ∆4/sin α

∆4/sin α

Figure 3.33 Step 4 in the analysis of a framed structure


forces of bars may be additional unknowns in the equations for the equilibrium of theforces (see Figure 3.34). However, in many cases the determination of the unknownscan be done more rapidly if, after formulating the equilibrium equations for themoments at all nodes, suitable cut out parts of the structure are considered, in orderto supplement through their global equilibrium the number of missing equationsneeded (Figure 3.35).

Step 6: Determination of unknowns and calculation of final end moments and forces. Afterthe analysis of the system and the determination of the unknowns, the end moments ofthe members are calculated from the expressions found in Step 4. These moments aresuperposed on those of the fixed state according to Step 3, so that the final endmoments of the structure are determined (Figure 3.36). The shear forces are then calcu-lated from the equilibrium of each member. The axial forces are determined directlythrough the equilibrium of the nodes.

129

Step 5

N43 = N34

Unknowns involved: ϕ3, ϕ4, ∆4, N31, N34, N42

110.0 kN 50.0 kN

70.83 kN m

37.5 kN

43

41.67 kN m

+V43

+V42

+N42

+M42

+N43

+M43

Equilibrium of node 4(three equations)

Equilibrium of node 3(three equations)

+V31+N31

+N34

+V34

+M34+M31


O

110.0 kN

70.83 kN m37.5 kN

50.0 kN

41.67 kN m

+V42

+M42

+N42

+M31

+N31+V31

3 4

Solution: ϕ3 = –59.6/EI, ϕ4 = +44.0/EI, ∆4 = +266.8/EI

Moment equilibrium of part 3–4 w.r.t. O:

M31 + V31 · 7.07 – 70.83 – 110.0 · 5.0 – 37.5 · 5.0

– 41.67 – M42 + V42 · 5.0 = 0 (1)

Moment equilibrium of node 3:

M31 – M34 – 70.83 = 0 (2)

Moment equilibrium of node 4:

M43 – M42 – 41.67 = 0 (3)

Unknowns involved: ϕ3, ϕ4, ∆4



3.3.6 The effect of an elastic supportWithin the framework of the deformation method, it is appropriate to recognise theinfluence of an elastic support on the end reactions of a fixed beam subjected to auniform load (Figure 3.37).

Elastic support through a spring with rigidity ksIn order, at first, to eliminate the settlement of end A (see Section 3.3.1), the externalvertical force (3 p L/8) must be applied upwards, leaving the spring unstressed. It isclear then that the opposite force applied at end A downwards, will cause the developing

130

Step 6

[V ] [N ]

41.189.2

89.2124.8

82.4

148.5

72.7

72.7

27.3

60.0

1.3

1

1

2

2

3

3

4

4

[M ]

[M ][M ]

84.32

5.68

90.0

90.0

119.1

119.1

119.1

According to Step 4(ϕ, ∆)

According to Step 3(Fixed structure)

41.6741.6722.5

77.43

106.82

148.5

35.99

According to ϕ and ∆:M31 = +106.82 kN m M34 = +35.99 kN mM42 = –119.1 kN m M43 = –77.43 kN m M24 = +148.5 kN m



settlement. The real stress state obviously constitutes a superposition of these two states.Equilibrium of node A can be written as

3 EIL3

þ ks ¼ 3 p L8

and, given that the reaction RA of the beam at A is equal to the spring force, it resultsthat

RA ¼ ks ¼ 3 pL=8

1þ 3 EIL3 ks

¼ R0A

1þ 3 EIL3 ks

Thus the presence of the spring generally induces a reduction in the reaction R0A that is

developed at the unyielding support of the beam under any load, according to the aboveequation.

Flexible support through a rotational spring with rigidity kuIn order to eliminate the end rotation at A, the application of an external moment( p L2/12) is required, according to Section 3.3.3. The rotational spring is then unstressed.

131

p

p

p

p

(a) (b)

A

A A

B AB

EI EI

L Lks

ks

ks

ks

kϕ

kϕ

kϕ

kϕ

Stressless springStressless spring

Reaction at AEquilibrium

RA < 3 · p · L/8

Equilibrium

MA < p · L2/12

∆ = 0

∆

3 · p · L/8

3 · p · L/8

3 · p · L/8

p · L2/12

p · L2/12

p · L2/12

(4 · EI/L) · ϕ

ϕ = 0

ϕ

ϕ

ks · ∆ = RA kϕ · ϕ = MA

(3 · EI/L3) · ∆

Figure 3.37 The consequences of an elastic support of a beam


Application of an equal and opposite moment at end A will cause the development of arotation’ at this point. The real stress state constitutes a superposition of these two states.Moment equilibrium in node A requires that

4 EIL

’þ k’ ’ ¼ p L2

12

and, given that the reaction MA at support A is identical to the moment of the spring,

MA ¼ k’ ’ ¼ p L2=12

1þ 4 EIL k’

¼ M0A

1þ 4 EIL k’

Similarly, it is concluded that the rotational compliance of the support expressed by therotational spring induces a reduction in the reaction M0

A developed in the fixed supportof the beam under any load.

3.3.7 The distribution of a nodal actionThe determination of how a moment applied to a node is distributed to the adjacentmembers, assuming that no shift but only rotation is possible, is particularly useful.The qualitative conclusion drawn in Section 3.2.10 is now quantified using thedeformation method. Equilibrium of the examined node under the moment M showsthat the developing moments at the ends of the members have the opposite sense tothe moment above (Figure 3.38). Moreover, each Mi will be proportional to the rotation’ of the node, according to the expression

Mi¼ ki ’where ki is the rotational stiffness of the corresponding member, which is equal to 4 EI/Lior 3 EI/Li, depending on whether the opposite end of the member is fixed or hinged,

132

M

MNode equilibrium

Stressless member

(1)

(2) (3)

1

2

3

1

2

3

k1 = 3 · EI/L1

k2 = 4 · EI/L2

k3 = 4 · EI/L3

k1 · ϕ

k2 · ϕ k3 · ϕ

ϕϕϕ

kiMi = M ∑ k

Figure 3.38 The distribution of a nodal moment to the adjacent bars


respectively. It is obvious that if a bar with a hinged end is connected to the node, thenno moment is developed, and hence no end rotation in the member is developed.According to the above, nodal equilibrium for the applied moments requires:

M ¼ M1 þM2 þM3 þ . . . ¼ k1 ’þ k2 ’þ k3 ’þ . . . ¼ ’ X

ki

which results in

’ ¼ MPk

Thus, the moment developing in each member will be

Mi ¼kiPkM

The moment transmitted to the opposite fixed end is, of course, Mi/2, with the samesense as that resulting from the elastic line of the member.Thus, according to the last equation, the conclusion of Section 3.2.10 is justified, i.e.

the distribution of the nodal moment depends on the stiffness of the correspondingmember.This conclusion is also valid in cases where certain members are assigned under the

imposition of a force P to a common shift: The force P is distributed to the membersdepending on their rigidity. In the frame shown in Figure 3.39, due to excessive stiffnessof the girder, the nodes are displaced by , but are not rotated. The force P is distributedto the three columns as a shear force, proportionally to their shifting rigidity ks, asdetermined in Sections 3.3.2.3 and 3.3.3.3. Similarly to the above equation, thefollowing is obtained:

Vi ¼kðiÞsPks P

133

The joints are not rotated

H

H/2δ δ δ

P

P

P

M M/2

M

V V/4 2V

ks

2ks

ks/4

2M

[M ] [V ]

V = P/3.25 V/4

2V

P = V + V/4 + 2V

Figure 3.39 The distribution of a horizontal force on vertical columns


3.3.8 Qualitative handling of the deformation methodThe qualitative examination of a structure, according to the deformation method, doesnot differ substantially from its quantitative treatment, as described in the previoussections.The determination of the fixed structure and of the nodal loading that causes the

rotations and shifts at the nodes can be done as explained in Section 3.3.1. In a casewhere the structure does not develop nodal displacements, the senses of the nodalrotations are always identical to those of the nodal moments, thus permitting theelastic line for each member to be drawn directly, and the corresponding qualitativebending moment diagram plotted (Figure 3.40). The nodes that do not have a nodalloading are first considered fixed and the sense of their rotations deduced from thecorresponding acting moment transferred to them by the neighbouring rotating

134

[M ]

[M ]due to node

rotations

[M ]fixed structure

[V ]

Fixed moment 3–4 decreasesFixed moment 4–3 increasesFixed moment 4–5 decreases

Elastic line due to node external actions

External actions in order to fix the structure Opposite actions cause the node deformations

1 2

1 21 2

3 4 553 4

3 4 5

Figure 3.40 Qualitative determination of the moment diagram for immovable nodes


nodes. It should be remembered that the distribution of a nodal moment to the adjacentmembers occurs according to their rigidity (see Sections 3.3.7 and 3.2.10).The final moment diagram M will naturally result from superposing this last moment

diagram on the one for the fixed state. For the loaded members, the final diagram isessentially a ‘modified’ fixed-state one, which is usually not reversed by the superposedend moments. For the unloaded members the final diagram is deduced directly from thelast diagram, due to the incoming rotations. g

If the structure does develop a displacement, the sense of that displacement must bedetermined first. This can be done as follows (Figure 3.41). In a first stage, a suitableadditional support is added to the structure in order to prevent the development ofdisplacement, and the qualitative diagram of bending moments in the modified structure

135

Shift development Shift constraint

?

External actionsfor node fixation

[M0]Fixed structure

[M1]due to rotations

(Stronger moment at left-hand column)

Development ofnode rotations

Shifting goes to the right

Equilibrium

Stronger shear force at left-hand column(First stage)

[M0] + [M1] [V ]

Figure 3.41 Qualitative determination of the displacement sense at a node


is then plotted, according to the precedents. The acting shear forces on the nodes areobtained from this diagram of bending moments. Next, examination of the equilibriumof a suitably cut out part of the structure containing the imported support allows thedetermination of the sense of the reaction developed and, accordingly, of the developingdisplacement, as this must have the opposite sense to that of the reactions.After the determination of the displacement sense, it is clear that the final bending

state results by superposing the ‘first stage’ of the structure under the full loading withimmovable nodes on the ‘second stage’, where only the known developing displacementis imposed on the structure (Figure 3.42). This second stage is treated exactly as for aloaded structure with immovable joints (first stage), the only difference being that thefixed state due to the loads is now due to the ‘known’ displacement. The procedurefor the examined frame is shown in Figure 3.42.The bending state of the second stage qualitatively modifies the one in the first stage

only if the loading of the structure is along the direction of the developing displacement.Thus, for the frame shown in Figure 3.41 the form of moment diagram [M] after super-position with the one in Figure 3.42 remains qualitatively unchanged, except for thefixed support of the column where the existing moment is assumed to cause tension

136

[M1]

[M0] + [M1]

Second stage

[M0]Nodes fixed

Imposition of shiftNodes not rotated

Node equilibrium

Bending momentsof second stage

External node actionsfor release of rotations

Figure 3.42 Qualitative determination of the bending state in the second stage


137

Final moment diagram

[M ]

[M ]

[M ]

[M ]

[M ]

Equilibrium of fixed nodes Nodal actions for rotation release

Imposition of shift. Nodes fixed Bending state with fixed nodes

Nodal loading for rotation fixity Rotation development

Shift developmentShift preventionFixed structure

Secondstage

Firststage

Figure 3.43 Qualitative determination of the moment diagram for displaced nodes


on the external fibres, as concluded from the elastic line in Figure 3.42. However, in theexample shown in Figure 3.43, the bending state of the second stage is predominant.It is obvious that the above qualitative examination of redundant structures, through

the consideration of the deformations of their nodes, leads to a clear result only for simplestructures that have zero or, at most, one degree of displacement freedom. Nevertheless,such an examination can also be useful for more complex structures, in that numericalresults (particularly if obtained using a computer) can be predicted qualitatively orjudged a posteriori.

138


4

Simply supported beams

4.1 Steel beams (reference material)

4.1.1 Service conditions (elastic behaviour)The simply supported beam as a structure transferring loads to its two hinge supports hasthe biggest possible deviation from the corresponding funicular structure for the sametransverse loading (see Section 2.2.7) and, consequently, it develops for that reasonthe maximum bending response for that loading. Nevertheless, this ‘unfavourable’result is counterbalanced by the structural advantages arising from the geometriclinearity of the structure.

Loads acting transversely to the beam direction are carried to the supports throughthe axial direction, as shown in Figure 4.1, in a cross-section at the middle of thebeam. The forces D and Z represent the resultants of the compressive and the tensilestresses, respectively, and, due to equilibrium requirements, they are obviously equalin magnitude and opposite in direction. These forces produce the bending momentM¼D z at each specific point along the beam. As the distance z between the linesof action of the two forces is definitely smaller than the relatively small height h ofthe cross-section, it follows that the forces D and Z are going to be considerable, forexample significantly larger than the reactions. According to the well-knownrelationship given in Section 2.2.1 between stresses and moments, the normal stressdistribution reveals that a significant portion of the cross-section in the neighbourhoodof its centre is barely stressed, and therefore the section is not adequately exploited. Thisis the reason why bending is generally considered ‘undesirable’ or ‘unfavourable’.

The extreme values of the stresses appear at the top and bottom of the section, andare given by the expression

max¼M ( ymax/I)¼M/W

where W¼ I/ymax is the elastic section modulus for the appropriate top and bottom valuesof y (see Figure 4.1).

It should be noted that the above equation is based on the Bernoulli assumption(Euler—Bernoulli beam theory), which states that transverse lines perpendicular tothe mid-surface before deformation remain perpendicular to the deformed mid-surface. In other words, the strain distribution " through the thickness of the beam islinear (Figure 4.2). In practice, this means that, if a contour is drawn for an arbitrarycross-section before the deformation of the beam (application of the loads) in such a


way that its plane is perpendicular to the beam axis, this contour will maintain both itsflatness and its orthogonality to the beam axis also in the deformed configuration (seeFigure 4.2). g

In general, it is desirable to design a cross-section which can carry the largest possiblebending moment M for a given area and to preserve, of course, the values of the endstresses within acceptable limits. This means that the moment of inertia for the sectionunder consideration should be increased by amassing as much material as possibletowards the top and bottom of the section, and thus increasing to a maximum thedistance between the internal forces D and Z (Figure 4.3).

A question, however, arises: How thin may the part of the section become whichconnects the compression and the tension zones of the cross-section? To answer this,equilibrium in the longitudinal direction is considered separately for the upper andlower halves of the left-hand side of the beam. The left end of either part is not subjectedto any horizontal forces, whereas at the right end the applied forces are the compressiveforce D for the upper part and the tensile force Z for the lower part (Figure 4.4). Itbecomes apparent that, in order to satisfy the equilibrium condition in the horizontaldirection, it is required to have longitudinal forces T (kN/m) along the length of the

140

σ = M · y/l

Compressive stresses

Tensile stresses

D

M

Z

z

y

Neutral axis

Figure 4.1 Bending of a simply supported beam

Undeformed beam Deformed beam

εσh

Figure 4.2 Plane sections remain plane after deformation


two pieces with directions to the right for the upper half and to the left for the lower one.Development of such forces, i.e. shearing stresses, also implies the presence of equalshear stresses acting along the side surface in the vertical direction (see Figure 4.4).

It is easily seen that these shear stresses are due to the variation of the bendingmoments, i.e. due to the variation of forces D and Z, because of which the axialequilibrium of a specific part of the beam requires the development of horizontal andvertical shear stresses, as shown in Figure 4.4. This can be considered as the physicalinterpretation of the well-known equation dM/dx¼V (see Section 2.2.3). One shouldkeep in mind that the presence of shear stresses in the vertical direction may also bededuced from equilibrium in the vertical direction of the left half of the beam,because their integration over the thickness yields the shear force V.

141

Thickness?

D

Z

D

Z

M

Moving material towards the outer fibresdecreases the compressive and tensile forces

Greater bending-moment-carrying capacity

Figure 4.3 Moment-carrying mechanism for a cross-section

Shear stress distribution(orthogonal section)

Shear stress generation

(Section width b)

Identical longitudinal force Tand shear force V distribution

h[T ]

h/2

h/2

D D D + ∆D

Z Z Z + ∆Z

D · z = MT = τ · b

T = τ · b

τmax = 1.50 · (V/A)

τ τ

τ

τ

Figure 4.4 Mechanism for the development of shear stresses in a beam


These shear stresses are determined from the expression (see Section 2.2.1)

¼V S/b I

The distribution of shear stresses along the interface between the upper and lowerhalves of the beam and, consequently, of their horizontal resultant forces T followsthe distribution of the vertical shear forces V (see Figure 4.4). In addition, the shearstress distribution in the transverse section (or the depth of the beam) vanishes atthe top and bottom, and it appears to have a maximum value at the centre of thecross-section. Note that this maximum shear stress is bounded by the limit of failure,which is the value fy/

ffiffiffi3

p.

Consequently, in order to design a beam with a pair of flanges, one being undertension and the other under compression, so as to produce a bending moment, an appro-priate thickness for the connecting section (web) is required for the safe development ofhorizontal shear stresses between them.

This conclusion plays a very important role in the design of beams, as will be explainedlater.

The fact that the area around the mid-height of a beam is underworked compared withits top and bottom fibres has led to the subtraction of this material ‘surplus’ and thecreation of beams such as the one shown in Figure 4.5, aiming at a reduction in weight.

The flanges must be able to withstand the force M/z both in tension and incompression. This, however, is not enough, because the same force must also becarried in shear. That is, the resultant longitudinal shear force M/z must be held byall parts of the beam located between contiguous openings and crossed by the horizontalsection. In fact, this force varies along the length proportionally to the distribution of the

142

Identical T and V distribution

D

Z

D

z

Z

T T

Z = D = M/z

Figure 4.5 Shear transfer mechanism in a beam with openings


vertical shear force diagram (see Figure 4.5). This means that, in the case of a uniformdistribution for the openings, it is sufficient to check the regions closest to the twosupports. g

If the openings are quite wide, so that the beam can be modelled using the indeterminatestructure shown in Figure 4.6 (Vierendeel beam), then it should be taken into accountthat the constituent parts of the beam, namely the two-column frames, are particularlydeformable due to the shear forces developed by the beam (see Figure 4.6). This defor-mation of the frames is attributed only to bending, as shown in the figure. Thus, framesadjacent to the supports deform more than the rest. All of them, however, present thetypical behaviour of a two-column frame subjected to a force parallel to its girder, forwhich the moments vanish at the middle of the girder and close to the middle of thetwo columns (see Sections 2.4.2 and 6.2.1). Placement of hinges at these points removesthe difficulties of static indeterminacy and permits the direct evaluation of internal forcescarried by the members in the model (see Figure 4.6).

At this point, a remark is necessary. The deformation of the solid beam having a ratioof span length to depth greater than 3, as has been discussed in Section 2.3.2, is mostlycaused by the deformation of its elements due to bending moments, assuming that the

143

Addition of diagonal barssuppresses the shearing action

The deformations are drastically reduced

D1 = M1/z D2 = M2/z T = D2 · D1

D1

Z1

D2

Z2

T T

V1/2 V2/2

V1/2 V2/2

V1/2 V2/2

V1/2 V2/2

m

m mm

1 2

Constituent frame

z

Figure 4.6 Modelling of a beam with rectangular openings (Vierendeel beam)


corresponding shear forces make a, relatively, insignificant contribution. In the presentcase of a Vierendeel beam, shear deformation plays the primary role, as has already beenpointed out, and the resulting deformations are comparatively large. How much largercan be determined by drastically restricting the shear deformability, which can beeasily accomplished by adding diagonal members, as illustrated in Figure 4.6. Thus,the two-column frame with a diagonal bar becomes much less deformable, resulting ina vertical deflection at the middle of the new simply supported system which isalmost 10 times smaller than before. This last model has, of course, nothing to dowith the beam with openings under consideration.

It is obvious that a Vierendeel beam may also be designed to bridge a long span, with arelatively large available depth. g

It is common in practice to construct a composite beam consisting of two separate hori-zontal sections which are simply laid one onto the other without any adhesives orconnecting devices (unilateral contact). Assuming that between the two parts there isno friction, a comparison between such a composite beam and a homogeneous beamof the same length and with the same vertical load is unfavourable for the compositebeam (Figure 4.7).

If, for example, the two ‘half-beams’ had the same thickness, it would be wrong toconsider that the upper one provides the compressive force D and the lower one thetensile force Z for the formation of the bending moment, as in the corresponding caseof the homogeneous beam. The reason for this is that there is no shear interactionbetween these two forces. Thus, the developed normal stresses within each beamhave the distribution depicted in Figure 4.7, and every half-beam is stressed anddeformed independently of the other, with an effective moment of inertia correspondingto half the thickness of the homogeneous beam, e.g. for a rectangular cross-section theeffective moment of inertia is equal to I/8, where I applies to the entire section. In order,therefore, to restore this shear interaction, there must be an appropriate structuralformation so that each of the two halves can develop the horizontal shear forcepreviously mentioned and, indeed, in the correct direction (see Figure 4.7). Onlythen can the system of the two beams can be considered equivalent to the single

144

h/2

h/2 σ

σT T

h

The two beams work like a homogeneous one only if shear forces T are offered

Wrong structural pattern: impossible for forces T to develop Correct structural pattern: shear force T development

Figure 4.7 Coupling of two half-beams in bending requires their interaction in shear


homogeneous beam. In other words, only then can the moment of inertia I of thehomogeneous beam also be used for the composite beam. The structural formation inquestion requires a jagged interface between the upper and lower halves, which is verycommon for wooden beams, but for composite structures made out of steel and concreteit is realised in a different way, as will be studied extensively in the next chapter. g

In the discussion above, reference is always made to either the horizontal or verticalsection of a beam, for which the expressions of normal and shear stresses refer to aCartesian system of axes conveniently applied in the longitudinal direction of thebeam. This choice of reference axes is arbitrary. Instead of considering a transversesection at the middle of the beam, it may as well be a slant section for which theequilibrium requirements yield completely different normal (perpendicular to theplane of the section) and shear (in the plane of the section) stresses. To confirm this,let us consider an element with a rectangular area on the neutral axis of the beamwhose sides, as is already known, are subjected only to shear stresses (Figure 4.8). Itshould be pointed out that the directions of the shear stresses follow the equilibriumrequirements of Cauchy’s relation. If this rectangle is cut by an arbitrary straight line,two triangular shaped areas will be formed, and each of them will also have to be inequilibrium. On the inclined side, the developed normal and shear stress componentswill have values determined directly from the values of the shear stresses on the sidesof the rectangular element.

145

Tension

Tension

Compression

Principal stresses

(τ = 0) (τ = 0)

Compr

essio

n

σ2σ1

τ

τ

σ

Figure 4.8 Principal stresses and principal directions


A question can be posed: Is every single one of these numerous pairs of stresses (, )revealed by the randomly drawn sections realised in the structure? The answer is‘definitely no’. The structure is stressed only in the direction of a normal stress(without any shearing), which is the only one realised by the structure, and this byitself is able to maintain the equilibrium of the triangular element. It can be provedthat at every point in the structure there are only two such orthogonal directions,called principal directions (see Figure 4.8). The corresponding stresses (compressiveand tensile) are called principal stresses. Hence, the structure ‘recognises’ only theprincipal stresses, and although the (, ) pairs do not represent reality, they never-theless legitimately represent the principal stresses, and are used as suitable quantitiesfor the various structural checks.

Principal stresses have different values and directions at each point in the structure. Itcan be ascertained that a (homogeneous) simply supported beam subjected to auniformly distributed load develops the stress trajectories shown in Figure 4.9. It isnoted again that the principal directions are mutually orthogonal at every point. Solidlines indicate tensile stresses, while dashed lines indicate compressive stresses. It isobserved that at the end fibres the evaluated normal stresses coincide with the principalones, having compression at the top and tension at the bottom. In addition, because atthe neutral axis of the cross-section only shear stresses are computed, the predominantstate there is characterised by tensile and compressive principal stresses at a 458 angle.This may also be deduced by inspecting the deformation of a rectangular element whichis in equilibrium according to Cauchy’s relation (see Figure 4.8). Elongated diagonalsidentify the direction of the principal tensile stresses, while shortened diagonalsdenote the principal direction of compression. Understanding this approach for directdetermination of the principal directions of tension and compression is importantbecause it is used in many cases. g

146

q

R

Figure 4.9 Principal stress trajectories in a simply supported beam


In the case of steel beams, it should be noted that their deformability may play a moreimportant role in their design than the allowable state of stress. The deflection of asteel beam should generally not exceed 1/500 of its span.

At this point, it should be recalled that deformation due to bending, which in mostcases is solely responsible for the formation of the elastic curve of the beam(compared with the shear deformation), as was previously explained, is directlyrelated to the curvature (1/r), as this is used in the principle of virtual work (seeSection 2.3.3).

The curvature, which varies with position along the length of the beam, may, in anycase and independently of the material in use, be expressed as shown in Figure 4.10, withthe only precondition being the preservation of the plane nature of the section:

1

r¼

"upper þ "lower

h¼ "

h

The above equation yields, in the case of elastically behaving material, the well-knownexpression M/EI (see Section 2.3.2).

Finally, it can be shown that the deflection of a uniformly loaded simply supportedbeam of length l is 5 q l4/384 EI.

4.1.2 State of failure (plastic behaviour)According to the elastic behaviour consideration, the maximum allowable moment My

in a beam is that which causes at any of the two end fibres (top or bottom) the yield stressfy (Figure 4.11). This fibre will be the one located furthest away from the neutral axis, ata distance ymax. Thus, My¼ fy (I/ymax) and, in the case of a section exhibiting doublesymmetry,

My¼ fy [I/(h/2)]

147

r

h

Mεupper

εlower

∆ε

(1/r) = (∆ε/h) = M/EI

Figure 4.10 Correlation between curvature and a bending moment


In contrast to the case of a bar subjected to uniform tension, for which the developmentof the stress fy obviously means that all cross-sectional fibres yield simultaneously and,inevitably, failure of the bar under the axial force Ny¼A fy, in the case of beambending, it is possible to apply a greater moment, i.e. higher values than My, becausenot all the fibres of the cross-section will have yielded. Increasing the loading andrequiring the cross-sections to remain plain during the process — as can be experimen-tally verified — the rest of the fibres gradually yield either in tension or in compression.The final yield state is characterised by yielding of all the fibres in the location of themaximum bending moment. The distribution of stresses is uniform in the region oftension as well as in the region of compression. The fact that the tensile force shouldbe equal to the compressive force due to equilibrium requirements in the horizontaldirection helps to determine the plastic neutral axis at that location where the sectionis divided into portions of equal areas (above and below the line).

Now, the moment Mpl experienced by the beam is created by the couple of the tworesultant forces, and is equal to

Mpl¼ fy (A/2) zIt is obvious that for members possessing two planes of symmetry, the plastic neutral axisbisects the depth of the beam (axis of symmetry).

148

Sections of double symmetry

α = Mpl/My

α = 1.17α = 1.5 α = 1.7

h

Plastic neutral axis

A/2

A/2

My = 2 · fy · (l/h)

Mpl = fy · (A/2) · z

Mpl = fy · (A/2) · z

Mpl = pu · L2/8

My = py · L2/8

fy

fy

fy

z

y

Zpl

Dpl

My = fy · (1/ymax)

Zy

Dy

A

py

L

pu

Figure 4.11 Plastic behaviour of steel beams


According to the previous discussion, the ultimate bending moment, which causesfailure of the beam, can be higher than the value My if plastic analysis is adopted, or,in other words, the beam can withstand a greater bending moment before failing. Ameasure of this increase is the ratio a¼Mpl/My, which is called the shape factor, andtakes values between 1.14 and 1.70, as shown for the cases of sections with twoplanes of symmetry in Figure 4.11. g

By the time all the fibres in the critical section are ready to yield, two regions of totallength ly will have already developed on either side of the section, in which the bendingmoments Mpl begin to diminish at the critical position, reaching a minimum value of My

at the position where only the upper and lower fibres yield (Figure 4.12). The rest of thebeam continues to deform elastically, because there are no other sections with theirextreme fibres in a yield state.

The plastic region of length ly is characterised by a progressive increase of thecurvature from its ends inwards and up to the critical section, as implied by the equation

1

r¼

"upper þ "lower

h

149

Mpl = q · L2/8

q

Plastic hinge in a fixed-hinged beam:capacity for further increase of the load

Plastic hinge in simply supported beam:impossible to carry further loading

Elastic range

Elastic range Elastic range

Elastic rangeElastic range

ly

Plastic range

qu

qu

Mpl = qu · L2/8Mpl

MyMyMpl

Figure 4.12 The plastic hinge


In this region, although the stresses remain constant and equal to fy, the deformationsdue to yielding of the end fibres of the section increase continuously, resulting inconsiderably higher values for the curvature (1/r).

In the case of a simply supported beam, formation of such region is equivalent tofailure (collapse). However, because the behaviour along the plastic region is quitecomplicated, and in order to simplify the static analysis, the so-called plastic hinge wasdevised and implemented, as will be described next.

It is assumed that as long as the bending moment Mpl does not appear in the sectionwith the maximum moment, the whole structure behaves elastically, and the curvatureis expressed at any point of the beam as M/EI. Once the loading reaches the value qu (seeFigure 4.12) and the moment Mpl has developed in the critical section, at that exactpoint a hinge is considered which carries on its sides a couple of external momentsMpl that are equal in magnitude and opposite in direction. The structure is thuschanged, as it has acquired a hinge at that point and, although in the case of simplysupported beams it becomes a mechanism and thus cannot take any further loading,it remains in equilibrium under the load qu. This is justified by the presence of theexternal moments Mpl at the hinge, which would have also been there internally inthe absence of the hinge.

A characteristic property of the plastic hinge as it appears in the case of indeterminatestructures is the ability to develop a relative change in slope between the two memberson either side of the hinge with a further increase in the load, offering continuously aconstant resistance Mpl which acts as an external action (see Figure 4.12). Thereason for this is that, because the cross-section is in the yield state, the strains canincrease freely, as was previously explained. In this way, the curvature at thatlocation increases accordingly (see the equation above), without any correspondingrise in the stress values and hence in the resulting moment Mpl.

Of course, in the case of the simply supported beam, the modified structural systemwith the plastic hinge, which was previously described, fails to equilibrate even with aload slightly larger than the qu, and it inevitably fails.

Therefore, if the simply supported beam is loaded with a uniformly distributed load q(see Figure 4.12), the failure load qu may be evaluated by requiring qu l2/8¼Mpl, asqu¼Mpl/(l2/8).

It should be pointed out that the above concept assumes that the beam does not faildue to excessive shear stresses, whose limit value is fy/

ffiffiffi3

p, as previously mentioned.

4.2 Reinforced concrete beams

4.2.1 Service conditions

4.2.1.1 BendingIn a concrete beam the transverse loads are also carried to the supports in the axialdirection, through a compressive and a tensile force. The problem, however, is that,although the compressive force D can be supported by the concrete, the tensile forceZ may be sustained only up to a limiting value determined by the low and unreliable

150


tension strength of the concrete. Additionally, the two forces must be equal, i.e. D¼ Z,as explained earlier, and this implies that the bending moment which a section can safelywithstand is very small (Figure 4.13). To overcome this deficiency, rebars are placed atthe lowest possible position in the section so that there is perfect bonding with thesurrounding concrete, and thus they receive all the tension, which, as a result, hasthe desirable force Z. These bars actually work as described in Section 1.3, that is,within fractured material. The concrete is assumed to be cracked up to a certainheight of the cross-section, so that the remaining uncracked part of the thickness willbe able to develop the appropriate compressive force D required for the creation ofthe corresponding resultant bending moment M.

It should not be forgotten that an essential requirement for the tension rebars to playtheir role successfully is to have a sufficient anchorage length. g

The assumption that plane sections prior to loading continue to be plane in the beamunder load applies also in this case, which means that the distribution of the axial strains" is linear through the thickness of the uncracked (upper) region and that the value "s ofthe strain at the level of the steel bars is depicted on the extension of the same line (seeFigure 4.13). The cracked (lower) region of the concrete is ignored.

Assuming that Hooke’s law applies here, as was discussed in Section 1.2.2, the stressdistribution developed in the upper, uncracked region is going to be triangular, with aheight x and a base value c¼ "c Ec. Its zero value occurs exactly at the point wherethe cracked zone begins. The resultant compressive force for a constant width b ofthe section will be

D¼ b c x/2¼ b ("c Ec) x/2

acting at the centroid of the stress diagram, which is at a distance x/3 from the upperedge.

151

Uncracked

Cracked

Given: M, h, bUnknowns: As, x, εc, εs(practically z ~ 0.90 · h)

D0

Z0

Unreinforced concrete

Bending moment capacity M clearly limited

Sections remain planeD = Z

D

Z

M = Z · z

Mh

x

zAs

<fct

σc

εs = σs/Es

εc = σc/Ec

Figure 4.13 Bending mechanism for a reinforced concrete beam


On the other hand, if As is the total cross-sectional area of the reinforcing steel bars,located at a distance h from the free edge of the uncracked (compression) zone, thedeveloped tensile force becomes

Z¼As s¼As "s Es

The values "c, "s and x are not independent of each other. They are related due to therequirement for plane sections through the similar triangles of the figure:

x/(h x)¼ "c/"s

The equilibrium in the horizontal direction of either part obtained after ‘cutting’ thebeam at the cross-section, or in other words, the fact that the total axial force atthe cross-section is equal to zero, demands the compressive force to be equal to thetensile force. This is expressed as

b ("c Ec) x/2¼As "s Es

Additionally, the forces D and Z should produce the bending moment M of the section:

M¼ Z (h x/3)¼ (As "s Es) (h x/3)

The last three equations relate four quantities (As, x, "c and "s) which must bedetermined in order to obtain from this application of Hooke’s law the stresses c ands. It becomes apparent that if one of these quantities is chosen arbitrarily, then theremaining three can be calculated using the three equations (see Figure 4.13).

In the past, when the design was based on service conditions, the allowable stress limitof steel was exhausted (e.g. 240 N/mm2) in order to achieve a reduction in cost bychoosing the strain "s to have the value 240/2.1 105. Then, the other threeunknown quantities were determined according to this choice, in particular the cross-sectional area of the reinforcement bars and the compressive stress of concrete. Incase this stress turned out to be unreasonably large, reinforcement of an appropriatecross-sectional area A0

s was also incorporated for compression. It was placed as closeas possible to the free edge of the compression zone, so that its compressive force Ds

(Ds¼A0s 0

s) would have a noticeable contribution to the required total compressiveforce (DsþDc) of the section (equal in any case to Z), and thus permitting theconcrete to work at a lower and acceptable stress level. Apparently, this was anuneconomic solution, and that is why it was circumvented by increasing the height hof the section.

Today, the design of a member and, more specifically, the calculation of thereinforcement area As, is performed at the failure stage according to the plasticapproach, which will be described later. The aforementioned equations are only usedto determine stresses in the materials under service conditions (the working state),given, of course, As.

The lever arm z¼ (h x/3) of the internal axial forces D and Z of the sectionmay generally be considered to be about 0.90 h, which considerably simplifies thecalculations. g

152


The evaluation of the strains "b and "s permits the determination of the curvature 1/r atthe section in question (see Section 4.1.1), and, furthermore, its axial distribution alongthe beam, which may approximately follow that of the bending moments (Figure 4.14).

According to this, the deflection is evaluated directly from the principle of virtualwork (see Section 2.3.3), as the expression for the work done by the internal forcesdue to virtual loading

ÐMvirtual [(Mreal/EI) ds] is equivalent to

ÐMvirtual [(1/r) ds].

Obviously, due to the inevitable cracking, use of the moment of inertia I of theuncracked section would produce underestimated deformations.

A simple and approximate geometric evaluation of the transverse deflection is givenby the expression ¼ L2/(8 r), as shown in Figure 4.14.

4.2.1.2 ShearAs was explained in Section 4.1.1, the beam develops only principal stresses. Amongthese, the compressive stresses can be withstood by concrete without any problems,while the tensile stresses can be taken up to the very small value of about 2 N/mm2.Stresses quickly surpass this value, and thus concrete cracks under relatively smallloads. Higher values of tensile principal stresses appear in the horizontal direction atthe bottom fibres of the midspan and also close to the supports in a 458 direction(Figure 4.15). Tension in these directions causes cracks perpendicular to these direc-tions, a fact which can be confirmed experimentally. The vertical cracks near themidspan are called flexural cracks, while the inclined cracks near the supports arereferred to as diagonal tension cracks. The inclined cracks produce skewed solid(uncracked) areas between them, which are linked only by the upper uncrackedregion of the beam. It is clear that all these uncracked areas are under compression,and it is reasonable to provide reinforcement in all those directions along whichtension is expected. This explains the presence of longitudinal rebars at the bottomedge of the beam and also the need for diagonal reinforcement transverse to the cracks.

This picture of the cracked beam, together with the abovementioned reinforcement,led many years ago to the concept of a truss model for the beam. This truss may bestatically determinate, and consists of compression members, which correspond to thesolid parts of the beam, and tension members, which correspond to the variousrebars. Modelling the truss as a statically determinate structure and combining themany compression members into just a few is definitely a simplification, but it is

153

r

δ

(1/r)max

Approximate curvature distributionin the cracked beam

L

δ = L2/(8 · r)

Figure 4.14 Approximate distribution of curvatures along the beam


absolutely legitimate according to the so-called ‘static theorem’ of plastic analysis, whichwill be studied in Section 6.6.2.

As shown in Figure 4.15, there are two different layouts for the transverse reinforce-ment. In the first, the reinforcement is perpendicular to the axis of the beam, and in thesecond it forms a 458 angle with respect to the beam axis in the appropriate direction(i.e. 458 at the right end of the beam and 1358 at the left end). The first layoutconsists of separate reinforcement which is placed in the body of the section in theform of vertical stirrups. The second layout is usually realised through the lowerlongitudinal reinforcement, which bends upwards at specific points. Commonpractice, in the past, was a combination of these two layouts. Recently, though, the

154

z

z

2z

z

z

z

z

Cracks in concrete due to tension

(a)

(b)

(a)

(b)

i

i

ii

ii

i

i

ii

ii

Fc

Fd

Fc

Ft

Fc

Fd

Ft

Fc

Fw

Ft

Ft

Fw

Fw = V

MV

(Section i–i) (Section ii–ii)

Represents the number of stirrups (z/s)equally spaced within z (stirrup distance s)

The same shear force V requires lower stresses in ties and strutsfor layout (b) than for layout (a)

Fw = V/√2

Fd = V/√2

z/√2

Figure 4.15 Truss model of a reinforced concrete beam


first type of reinforcement with the vertical stirrups has dominated over the other,mainly because it is simple and convenient in construction, with the second type beinglimited to cases of high shear requirements, again using stirrups. In general, bendingup the bottom steel bars at an angle of 458 should be avoided, as it produces undesirablestress concentrations in the concrete in the neighbourhood of the corner points.

Next, how these two types of reinforcement are appraised through static con-siderations will be examined (see Walther and Miehlbradt, 1990).

The truss model (a) in Figure 4.15 has essentially grouped, on the one hand, the manycompression members formed after cracking into a specific number of 458 diagonal bars(struts), and, on the other hand, the numerous vertical stirrups into a number of verticalbars (ties). The height of the truss is considered equal to the lever arm z of the internalforces D and Z of the cross-section. In accordance with the truss-based model, all theloads of the beam are transferred to the joints, and thus every double-pinned bar ofthe truss develops only axial force (see Section 2.2.8).

Two different sections i—i and ii—ii are drawn somewhere in the left half of the beam.Analysis of the left-hand side part of the beam with respect to these sections revealsthat, in order to offer to the section the internal quantities M and V in the appropriatedirections necessary for equilibrium, the axial member forces should be directed asshown in Figure 4.15.

So, in order to produce the shear force V using section i—i,

Fd¼Vffiffiffi2

p

whereas section ii—ii yields

Fw¼V

It is clear that the diagonal strut represents a skewed solid part of the concrete beamhaving the transverse dimension z/

ffiffiffi2

p(i.e. the normal distance between consecutive

diagonal bars). The developed average compressive stress d, for a beam width bw, is

d ¼ Fd

bwðz=ffiffiffi2

pÞ¼ 2 V

bw z

Now, every vertical bar represents all the stirrups placed within the length z, and allthese vertical bars are equally spaced at a distance s. If the overall cross-sectional areaof every stirrup is Aw, then the total area of all stirrups will be

PAw¼Aw (z/s), and

the developed tensile stress will be computed as

w ¼ FwPAw

¼ V s

Aw z

Layout (b) with the inclined stirrups seems to be statically more favourable (seeFigure 4.15).

Every inclined strut denotes a skewed solid part of width 2 z/ffiffiffi2

p, which is twice as

long as that of the previous set-up. Correspondingly, every diagonal tension bar repre-sents stirrups placed within a transverse width equal again to 2 z/

ffiffiffi2

pand having

between them a constant horizontal distance s. Then, the distance between them s0

155


in the transverse direction becomes s0 ¼ s/ffiffiffi2

pand the total area

PAw of all the stirrups

represented by the bar isP

Aw¼ (Aw/s0) (2 z/ffiffiffi2

p).

Two different sections i—i and ii—ii (see Figure 4.15) reveal that, in order to obtain theshear force V,

Fd¼Vffiffiffi2

p(same as in the previous configuration)

and

Fw¼Vffiffiffi2

p

Thus, the compressive stress of the skewed solid parts becomes

d ¼ Fd

bw ð2 z=ffiffiffi2

pÞ¼ V

bw z(half of the value in the previous configuration)

while the tensile stress of the inclined stirrups is

w ¼ FwPAw

¼ V s

Aw z

1ffiffiffi2

p (reduced by 30%) g

Despite the fact that the system is stressed less using the configuration with the inclinedreinforcement and, consequently, the required steel bars have smaller sections, thevertical stirrup layout is always preferred, as was previously mentioned. The reason forthis is that the cost-reduction benefits due to the decreased weight of the requiredoblique reinforcement are negated by the cost of the labour necessary for its placement.

During the last 20 years, extensive experimental research on the issue of the verticalstirrup configuration has revealed that the angle of the cracks — and consequently theangle of the compressive members — is generally smaller than 458, especially in regions ofintense shear. This fact, as demonstrated by the analysis of the corresponding truss ofconfiguration (a) in Figure 4.15, causes an increase in the compressive stress in thesolid parts and a reduction in the tensile stress of the stirrups, as will be shown later.

In this case, every vertical bar of the truss represents all the stirrups placed within alength of z/tan. This is the horizontal projection of a crack which forms an angle withthe horizontal direction and crosses over the stirrups (Figure 4.16).

The force Fw developed in the stirrups gives the shear force V for the section ii—ii.More specifically,

Fw¼w (Aw/s) (z/tan)¼V

and, therefore,

w ¼ V s

Aw z tan

Furthermore, the diagonal bar represents the skewed solid part of the concrete, whichnow has the transverse dimension z cos. With reference to section i—i, the compressiveforce is

Fd¼V/sin

156


The mean compressive stress d for the beam width bw is

d ¼ Fd

bwðz cosÞ ¼V

bw z 1

cos sin

For an angle ¼ 458, these values are the same as those for layout (a) in Figure 4.15. Inpractice, the angle varies between 308 and 458. It is of note that the freedom of choicefor this angle is validated by the static theorem of plasticity, as will be explained inSection 6.6.2.

It is obvious that the above equations for d and w can be readily used for theadequacy test of the beam width bw, as well as of the stirrup area Aw.

A very important issue is the tension force developed by the reinforcement forbending due to shear in the neighbourhood of zero moments, as shown in Figure 4.16.Thus, this reinforcement does not stay inactive but instead provides the tensile force

Ft¼V/tan a

which requires, of course, an appropriate anchorage. g

The truss analogy for the beam can, and must, be taken into account in every case. So,for concentrated loads close to the supports it is clear that the loads are passed on to thesupports only through the adjacent diagonals and that the tensile force at the bottomedge is not caused by beam bending but is generated to ensure equilibrium of the supportjoint. The transverse members (stirrups) are obviously inactive (Figure 4.17).

4.2.1.3 TorsionThe association between principal stresses and shear stresses, along with the fact thatthese are experienced by the reinforcement, also has direct application to the torsion

157

z

Ft

Ft

Fd

Fd

Fc

Fc

α α

α

Fw Fw = V

Ft = V/tan α

Fd = V/sin α

V

V

Section ii–ii Number of stirrups z/(s · tan α)

Section i–iz/tan α

z · cos α

A = V

Figure 4.16 Variable-angle truss model


of concrete bars. Although this type of loading does not apply to simply supported beams,it is appropriate to study it in this chapter.

It is well known that in the case of torsion, higher values for shear stresses appear closeto the periphery of the member. This allows modelling of a solid section as hollow,having all the torque carried by the thin region around the periphery of the section,as discussed in Section 2.5.3. Of course, the higher levels of the obtained stress areon the safe side. Thus, the study of thin-walled tubes and hollow cross-sections has abroader application and importance.

Obviously, the shear flow around the thin wall which furnishes the torque at thesection produces the principal stresses — the only stresses that really exist — asdepicted in Figure 4.18. The tension principal stresses should be taken up by an appro-priate layout of the reinforcement. This is usually realised by closed rectangular stirrupsand longitudinal reinforcement, which offer the necessary oblique tensile force (seeFigure 4.18).

Allowing a service stress s for the steel results in a required section as per unit lengthfor the stirrups, which is equal to as¼ vl/s (cm2/m), in a total cross-sectional area Al forlongitudinal bars (uniformly distributed along the perimeter), equal to Al¼ (vl/s) L.It should be noted that the notation introduced in Section 2.5.3 has also been used here.

4.2.2 Ultimate state

4.2.2.1 BendingIt is assumed that the simply supported beam has longitudinal rebars for bending, alongwith transverse stirrups for shear. Following the stress state at the midsection of thebeam for a uniformly distributed loading, as it increases from an initial value (e.g. itsown weight) up to the value that produces the failure moment Mpl, the followingstages occur.

It is assumed that, under the initial load (dead load), the tension strength fct ofconcrete is not exceeded and the section remains uncracked. The internal forcesapplied to the section, i.e. the compressive force and the tensile forces of theconcrete as well as of the reinforcement, are determined by using a modified concretesection. In this transformed section, the area of steel has been replaced by an equivalentarea of concrete, which is obtained by multiplying the steel area by the ratio Es/Ec. This

158

A

Z

P

Load P taken up without ‘bending’ or ‘shear’

Figure 4.17 Transfer of a concentrated load through the truss mechanism


is justified by the fact that the concrete and the steel share the same strain at thatparticular fibre of the section, that is "s¼ "c or s/Es¼c/Ec, and consequently therelationship between stresses becomes s¼ (Es/Ec) c. So the tensile force (As s) ofsteel is transformed to the corresponding nominal tensile force ([As (Es/Ec)] c) ofthe concrete (Figure 4.19).

As the load increases, the concrete stress will reach its tensile strength c¼ fct after acertain time, while still in the uncracked state. The moment Mr causing this stress iscalled the cracking moment, and initiates cracks in the concrete. From this point on,or even with a slightly higher bending moment in the middle of the beam, thebending moment is considered to be furnished by a cracked section, as studiedpreviously. Once the concrete has cracked, the tension force of the steel increasesabruptly, just as occurs with the tension bar (see Section 1.3). It is assumed that thesteel section As is adequate to ensure that the new stress for the steel, which is deter-mined as explained in Section 4.2.2.1, does not reach its yield stress, which results in

159

Total area Alas (cm2/m)

MT

MTMT

Offered by stirrups

Offered by stirrups

Principal tensile direction

Principal compressive direction

Offered by longitudinal reinforcement

Offered bylongitudinal reinforcement



vl = MT/(2 · Fk)

vl = MT/(2 · Fk)

Figure 4.18 Thin-walled model for the torsional design of reinforced concrete members


an uncontrollable state of cracking, as mentioned in Section 1.3. So the initial phase ofcracking begins with the extreme compression stress of concrete being less than fc, andthe stress of steel being less than the yield strength fy (Figure 4.20). The diagram ofcompressive stresses is not necessarily linear, especially when it goes beyond the value

160

n = Es/Ec

As

Uncracked section

(n – 1) · As/2

Figure 4.19 Transformed concrete section

σc σsfc fy

2.0 3.5 εc: ‰ 2.0 5.0 εs: ‰Simplified σ–ε diagrams for concrete and steel

As

hM

εc < 2.0‰ εc < 3.5‰εc = 2.0‰ fc fc

εc < 3.5‰ fc fc

εc = 2.0‰ εc < 10‰εc > 2.0‰ fy

εs = 5.0‰ fy fy

fy

D D

AsAs

h

x 0.80 · x

z = h – 0.40 · x

Z = As · fy Z = As · fy

Figure 4.20 Stages of concrete beam bending


fc/2, according to the adopted stress—strain diagram of concrete (see Figure 4.20). Theinternal forces developed are governed by equilibrium requirements and by the alwaysapplicable linear distribution of the strains through the beam thickness, a fact thatallows their evaluation.

It should be recalled that the expression for the curvature, mentioned previously (seeSection 4.1.1), is valid for all cases. Accepting a distribution along the length of thebeam similar to that of the bending moment diagram enables the evaluation of themaximum deflection by applying the principle of virtual work (see Section 2.3.3 andFigure 4.14).

When the value of the bending moment becomes such that the tension strength fy ofthe steel reinforcement is required for equilibrium, the section is entering the yield state(see Figure 4.20). It should be pointed out that the selection of the cross-sectional beamdimensions and also of the reinforcement area should be such that the compressionstrength fc does not appear first.

As the stress fy first arises when c< fc, the corresponding strains will take values forelongation "s 2% and for shortening "c< 2%. An increase in load will lead to furtherincrease in the bending moment, causing, of course, a new distribution of the strains ",and, eventually, the concrete will develop the maximum possible compressive stress fcwhen "c has reached the value 2%. From that point on, due to the adopted —"diagram of concrete (see Figure 4.20), the extreme stress value of the compressionzone will remain constant, while the inner fibres will gradually begin to develop themaximum possible stress fc. The limit capacity of the section is determined accordingto the diagrams in Figure 4.20 by the maximum shortening strain "c of the concrete,which takes the value 3.5%, along with the maximum elongation "s of the steel,which may be considered to be equal to 10%. In the present case, it is expected toexhaust the elongation of steel. It should be pointed out that, with a value of "s

already equal to 5%, the bending moment produced and the beam loading will haveessentially reached their highest values.

A plastic hinge is assumed to be forming at the location where the section enters thisstage. This process is exactly the same as in the case of the homogeneous steel beamconsidered in Section 4.1.2. g

As previously mentioned, the sectional dimensions and reinforcement should be selectedso that steel will yield first for the predetermined moment of beam failure (see alsoSection 4.6). This happens for two basic reasons. First, failure comes without warningwhen the strength of concrete has been exceeded (brittle failure), while the graduallyincreasing curvature due to high values of "s (up to and above 10%) is clearly observablewhen yielding of steel has occurred. Second, this deformability is particularly desirable, asit allows the members adjacent to the plastic hinge to develop a larger relative rotationthere. This is not so important for statically determinate structures, but it is veryimportant for the indeterminate ones, as will be studied later with reference to theirplastic analysis.

The plastic moment Mpl developed by the section is determined by the tensile forceZ¼As fsy of the steel, the equal and opposite compressive force D of the concrete,

161


which is computed from the area of the compressive stress diagram, and the normaldistance z between the two forces (lever arm):

Mpl¼As fsy zThe location of the resultant compression force D, and thus of the lever arm z, can bevery accurately approximated assuming that the distribution of the compressive stresseshas a constant value fc up to a distance of 0.80 x from the outer compressed edge (seeFigure 4.20). Then, z¼ h 0.80 x/2, and the equation of the compressive and thetensile forces acting on the section yields

x ¼As fsy

0:80 fc b

It should be recalled that x denotes the distance between the zero-strain point of the "diagram and the outer compressed edge of the section, which is the height of thecompression zone.

In fact, for practical design purposes, the lever arm z can be set equal to 0.90 h, whichmakes the above equation an easy way to determine the steel section for a chosen Mpl. Amore accurate calculation may be achieved employing either the last two equations orusing appropriate computer programs.

The development of the bending moment Mpl presumes that the beam does not faildue to shear. This is going to happen only if the produced corresponding shear force isless than the computed shear resistance of failure, which, of course, depends on theexisting transverse reinforcement as well as on the dimensions of the section, as willbe described in the following section.

4.2.2.2 ShearShear is determined according to the statically determinate truss model, which entersthe failure state even if just one of its members fails. Thus, it is easy to deducecomputationally the state of shear resistance of the beam. This state is generatedeither because the concrete in the struts has reached its compression strength, orbecause the steel in the ties (stirrups) has exceeded its yield stress.

The compression strength of the struts fc, according to Section 4.2.1.2, is expressed bythe relationship

fc ¼Vd

bw z 1

cos sin

which may be used either to compute the maximum possible shear force Vd, or todetermine the minimum web thickness bw for a given ultimate shear Vd.

Similarly, the yield stress fsy of the stirrups is expressed, according to Section 4.2.1.2,as

fsy ¼Vd s

Aw z tan

162


which may be utilised either for the calculation of the maximum shear Vd or for thedetermination of a suitable configuration of the transverse reinforcement (Aw, s) for agiven shear Vd leading to failure.

Clearly, in order to compute the shear resistance of failure, the smallest of the abovetwo values should be considered.

4.3 Prestressed concrete beams

4.3.1 Service conditions

4.3.1.1 BendingThe aim of prestressing concrete, as discussed in the introductory study on tension bars(see Section 1.4), is to develop compressive stresses that prevail over the undesirabletensile stresses. However, in the case of flexible members such as beams, while thescope of prestress remains essentially the same, i.e. potential elimination of the tensilestresses and overall improvement in performance of structural concrete, the idea ofappropriately embedding a steel cable in the concrete is worth analysing here in asomewhat different way (Figure 4.21).

Every structure is subjected to external loads, which create the stress state for whichthe structure was originally designed. It would be of great benefit if there was a way ofapplying to this structure the exact opposite loads, so that their, a priori, unfavourableinfluence could be diminished to some degree. An ideal solution to this problem isprovided through a steel cable, which can receive any transverse loading along itslength (such as that applied to the structure) and be self-equilibrating — as a freebody under appropriate tensile forces at its ends. Under these loads, the cable takes aspecific shape — funicular — which, according to Section 2.2.8, replicates the momentdiagram of the corresponding simply supported beam and develops only tensile forces.It is clear that if a free cable is obliged to take the aforementioned shape and is subjectedto the specific tension forces at its two ends, then, in order to be in equilibrium, it mustbe acted on by the very same transverse loading along its length that led to the funicular

163

The cable gives back to the structurethe equal and opposite forces

Equilibrium of the cable

Figure 4.21 Static interaction between a cable and the structure in a prestressed beam


shape in question. If these transverse loads are provided to the cable by the structure, itis expected, due to the action—reaction law, that the cable will apply to the structure theequal and opposite transverse loads to those required for the cable to be in equilibrium asa free body. Consequently, the original question has been addressed. This approach isdepicted graphically in Figure 4.21.

Having discussed this general issue, a simply supported concrete beam is nowconsidered which has an orthogonal section and is subjected to its own weight, theuniformly distributed load g (Figure 4.22). The beam has embedded in it a parabolicallyshaped and flexible duct, which has sufficient bond with the body of the concrete. Itbegins and ends at the centroids of the two orthogonal end sections, and it contains aloose steel cable (tendon), with one end being anchored at one end of the beam (e.g.the right end), and the other freely hanging out of the other (i.e. left) end. The cableoffset from the centre-line at the midspan of the beam is considered to be f.

Through a hydraulic jack, a tensile force P is applied to the free (left) end of thetendon and, thus, an equal compressive force is experienced by the concrete due tothe equilibrium of the jack (see Figure 4.22). It is obvious that this tensile force atthe left end of the tendon is transferred to the right anchored end as a compressiveforce on the concrete, while the right end of the tendon receives as a reaction thesame tensile force P. The tendon, as a free and flexible funicular body having a parabolicprofile, is in equilibrium under the tension forces at its ends and the uniformlydistributed load u applied to it by the concrete. According to Section 2.2.7,

f ¼ ðu L2Þ=8

P

164

~P

g · L/2

~P

g · L/2

~P

g · L/2

~P

g · L/2

(g – u)

u

u

u

f f

g

g g

PP

PP

PP PP

P P

(g – u) ~P · e0 ~P · e0

The transverse load comes out smaller but thebending moment at the middle remains the same

Prestressing does not generate reactionsin statically determinate structures

g · L/2 g · L/2

e0 e0

Higher deviation forces due to larger f

Figure 4.22 Effective loads on a beam produced by a prestressing tendon


or

u ¼ 8 P f

L2

It has been assumed that there is no friction between the tendon and the duct, althoughin practice this is not true and, therefore, it results in a slightly smaller than P force at theright end.

The beam is acting on the steel tendon with forces which are equal to those requiredby the equilibrium conditions of the tendon. The reactions on the concrete beam are thetwo compressive (oblique) forces P at its ends and the uniform load u applied in theopposite direction. Obviously, all the aforementioned loads also form a self-equilibratingsystem acting on the beam.

Eventually, the beam is subjected to the following loads:

(1) A uniformly distributed load g u.(2) Two oblique concentrated forces on its end sections, each of which is decomposed

into a horizontal compressive and a vertical force. The two vertical componentsbalance the distributed loading u. In the present case, it is apparent that theprestressing does not create reactions at the supports.

In this way, the beam is under pure bending due to the load g u, and under uniaxialcompression due to the two horizontal component forces at the ends of the tendon,which can be taken to be approximately equal to P instead of the actual valueP cos, because the angle is quite small.

If one or both ends of the cable do not coincide with the centroid of the concretesection, i.e. there is an eccentricity e0, the eccentric compressive load P should bereplaced by its equivalent force P and moment P e0 at the centroid (see Figure 4.22). g

Clearly, the value of the load u should be as high as possible. According to the lastequation for u, this uniform load depends directly on the force P, as well as on thevalue of the deflection f. So, its maximum possible value is usually achieved bymoving the lowest point of the parabola closer to the bottom of the beam, taking intoconsideration, of course, the minimum allowable concrete cover outside of the conduit,which ranges between 7 and 10 cm.

The stress state at the midspan is determined by the axial force P applied to the centroidof the section and the moment MgþP¼ (g u) l2/8þ P e0, which generally producestension at the lower border of the beam, if the load g is greater than u (see Figure 4.22).

Superposition of the corresponding stress distributions is given graphically inFigure 4.23, and this can be checked directly through the following equations(negative stress denotes compression):

top ¼ P

A

Mgþ P

Wtop

bottom ¼ P

Aþ

Mgþ P

Wbottom

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Before applying the above equations for the evaluation of the stresses, it is useful to lookfirst at the physical interpretation of the various terms of the stress state (Figure 4.24).At any point in the beam, the axial force P (acting on the centroid) coexists with thebending moment MgþP (see Figure 4.22), and both of them together are equivalentto a single force of magnitude P acting at a distance c above the centroidal axis (forg> u), which is equal to c¼MgþP/P (see Figure 4.24). This coexistence is unequivocalbecause, on application of the prestress, the concrete beam rises slightly from theformwork due to the fact that the loads u begin to overcome the weight of the beam,and this occurs to its full extent once the formwork has been removed. Whether thiseccentric force P produces only compressive stresses in the beam depends on whether

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Central action of the prestressing force at the beam ends

Distribution of normal stresses for g < uDistribution of normal stresses for g > u

Due to transverse load (g – u)Due to transverse load (g – u)

P/A (Mg+P/W ) P/A (Mg+P/W )

Figure 4.23 Stress distribution in a beam for different values of the deviation forces

Bending moment due to prestressing is P · e

In statically determinate structures the compressive prestressing forceis applied always on the cable trace

(g = 0)P

f

P

P

P

P

PMg+P

(g = 0) (g = 0)

Mg+P = (g – u) · L2/8

MP = u · L2/8

c = Mg+P/P

c = MP/P = f

ee0 e0

Figure 4.24 Location of the compressive force on a beam cross-section


the point of action of P is located inside the so-called ‘core’ of the section. The borders ofthe core define the area within which a compression load may be applied without causingtension on the section, and are determined by the distances ktop and kbottom on eitherside of the centroidal axis, according to the expressions

ktop = Wbottom/A

kbottom = Wtop/A

Theoretically, when the weight g has not yet been applied (g¼ 0), the compressive forceP of the section is always located on the trace of the tendon in the section (obviouslyexterior to the core), because MP has the opposite direction to MgþP, and

c ¼ MP

P¼ u L2

8 P¼ f

This result applies not only to the midspan section but also to any section along thebeam, and, in fact, is independent of the tendon eccentricity at the ends of the beam(see Figure 4.24).

The conclusion is that, due to deviation forces u, the bending moment MP at anypoint in the beam is always equal to the moment of the compressive force P aboutthe centroidal axis (see Figure 4.24), thus

MP¼ P eThe above expression for MP helps to draw the following important conclusion: in order tocompensate for a moment diagram of the beam produced by gravity loads, the prestressedtendon must have such a profile that its eccentricities conform to the moment diagram.It should be recalled that, in order to achieve values for P which are smaller but which stillhave the greatest effect on the beam, the maximum possible eccentricity should accountfor the allowable concrete cover outside of the duct (7—10 cm).

In the case of, for example, a concentrated load acting on the beam, the prestressedtendon should have the profile shown in Figure 4.25. Clearly, the deviation force on theconcrete is simply the opposite of the concentrated force which causes the funicularshape of the cable. g

On the basis of the last conclusion, the eccentricities of the tendon are considered withrespect to the centre-line of the beam, which leads to another important conclusion, thatthe profile of the centre-line along the beam has the same influence on the developmentof the moment diagram due to prestress as does the tendon. Thus, the parabolic profile of

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Figure 4.25 Tendon profile that conforms with the bending moment diagram


the centre-line of the beam in Figure 4.26, for which the tendon is a straight line,produces the same deviation forces as those of a beam with constant cross-section anda parabolically shaped tendon.

Generally, in order to determine the forces imposed by the cable on a beam with non-horizontal or even a curved centre-line, it is recommended that the centre-line isconsidered as a straight line carrying along with it the tendon, and thus as a fictitiouscable having the same eccentricities as the real one. In this way, the prestress is dealtwith exactly as before, for the fictitious tendon (see Figure 4.26). g

It should be pointed out once again that prestressing a flexural statically determinatebeam which is subjected to a load g is equivalent to applying at every section anadditional compressive force P on the trace of the tendon. Therefore, every section issubjected to the compression D, the tension Z (both determined through the bendingmoment Mg), as well as the compressive force P (Figure 4.27). The resultant of thesethree forces is apparently equal to P, which is acting at a distance c¼MgþP/P fromthe centroidal axis.

Thus, re-examining the cross-section which is subjected initially only to theprestressing compressive force P (applied to the trace of the tendon), it emerges thatthe application of a bending moment M which produces tension at the bottom fibresraises the point of action of the compressive force P by a distance a¼M/P (Figure 4.28).

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The deviation forces u are the same as thoseof the fictitious cable

The deviation forces u are exactly the same

Fictitious cable

Figure 4.26 Influence of the beam center-line on the development of the deviation forces

Due to g Due to PD

ZP

P

e e

c = Mg+P/P

(D = Z ) Location of the resultant of D, Z, P

Figure 4.27 Prestress as a compressive force applied at the trace of the tendon


Indeed, according to the previous discussion, it is (see Figure 4.28)

a ¼ e þ c ¼ e þ M P e

P¼ M

P

This conclusion is practically very important. g

For a given section subjected to a particular bending moment Mg due to the self-weight g,there are three characteristic values of the prestress force P which satisfy the condition ofa tension-free stress development (Figure 4.29).

The first and lowest acceptable value of P1, coexisting with Mg, moves the point ofaction of the compressive force to the top border of the section core, according to theaforementioned discussion. Thus, the previous equation yields

P1 ¼Mg

ktop þ f

Every load in excess of the weight of the beam will raise the compressive force more, andwill immediately produce tensile stresses at the bottom fibres. This value of P1 producesdeviation forces u smaller than g, resulting in downward beam deflections significantlysmaller than those caused by g, when it acts alone on the beam.

The second characteristic value Pm of the prestressing force is the one that, due to Mg,moves the point of action of the compressive force P exactly to the centroid of the

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ea

P

P

P

MMc

e

a = M/P P · e

Initial position (g = 0)Initial position (g = 0)

Gravity loads g shift the compressiveprestressing force upwards

Figure 4.28 Relocation of the compressive prestressing force due to a bending moment

Characteristic values and locations of P to rule out tensile stressesdue to an applied bending moment M

ktop

kbottom

P1 P2

P2

Pm

Pm

P1

aa

a

M

P = M/a

P1 < Pm P2<

Figure 4.29 Critical locations of the compressive prestressing force


section (c¼ 0). Using the above equation, the load becomes

Pm ¼Mg

f

This value produces deviation forces u which cancel out the effect of the weight of thebeam. The beam is under pure compression and does not deflect at all.

The maximum acceptable value P2 is the one which, combined with the coexistentmoment Mg, relocates the point of action of the compressive force P to the bottomborder of the section core. Based on the above equation, the force is expressed as

P2 ¼Mg

f kbottom

The top fibre of the beam has zero stress, and the bottom fibre is under compression.Consequently, the section is capable of carrying the maximum possible additionalload, compared with the previous cases, causing at the bottom fibre as much tensionas the pre-existing compression due to P2, i.e. zero stress at the bottom fibre. Thedeviation forces u are significantly larger than the self-weight, and the beam deformsupwards. g

It is worth pointing out here that, besides friction, which has already been mentioned,there are also other factors contributing to the decrease in the initial prestressingforce from P0 to P. All the different reduction factors (friction, creep of concrete,relaxation of steel, etc.) combine to give the final value of the prestressing force P,which is approximately 0.85 P0.

Clearly, the values of the prestressing force P0 are directly proportional to the cross-section of the employed prestressing steel. Usually, this steel, with a total cross-sectionalarea AP, is strained during the prestress process with 70% of its yield stress fPy, andtherefore

P0¼ 0.70 AP fPy

4.3.1.2 Design of prestressed concrete beamsThe problem of designing a prestressed beam does not have a unique solution, and, first,the following can be stated.

A selection is sought for the concrete section and the area of the prestressed steel sothat the beam will be able to carry not only its own weight but also an additional live loadp, which produces by itself a particular bending moment Mp. The stresses developed inthe beam never exceed the maximum admissible compressive stresses under serviceconditions, while, at the same time, a specific safety factor is provided against failuredue to increased loads (see Section 4.6).

Basically, there are two design concepts. The first sets as a condition the completeabsence of tensile stresses in the section under service loads, referring either to thedead load or to a combination of the dead and live loads. In other words, for anybending moment M under service conditions for which Mg<M<Mgþp, the equivalent

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compressive force P on the section must be applied within its core. This managementof prestress is called full prestressing. The compressive prestressing force P will not tomove higher than the top border of the core, because of the moment Mgþp (seeFigure 4.29). Thus,

Mgþp/P< ktopþ f

and, consequently,

P > Mgþ p=ðktop þ f Þ

The advantage of the design in question is that the structure always remains uncrackedunder service loads, but in doing so it requires higher values of the prestress and, thus, anincreased area for the prestressing steel. The deviation forces u overcome the self-weight, and so the beam, due to its own weight, is subjected to a combined load inthe upward direction. This results in an upward deformation with a curvature whichis determined by the developed stress distribution, according to the well-knownexpression. However, this negative deflection increases considerably as time goes on,because of the creep of concrete (see Section 1.2), and it is often undesirably evident.

The second design concept, known as partial prestressing, aims at cutting down thevalue of the prestressing force. It ensures that the beam will not crack under its ownweight, but it permits cracking under live loads on the grounds that this is temporaryand that, once the loads have been removed, the cracks will close again due to thealways present compressive force P. There is an issue, however, about the width ofthe cracks, which has to be confined according to the discussion in Section 1.3, a factthat definitely requires the presence of more reinforcing steel. Economically, this isnot a disadvantage, because this reinforcement may be fully utilised in the developmentof the ultimate moment capacity of the section, as will be studied later. However, in thecase of full prestressing, the reinforcing steel required by regulation provides the sectionwith higher strength than necessary, making this approach even more uneconomic. g

Beam design under full prestressing is considered nowadays rather inefficient, and it isnot employed unless there are reasons to avoid cracking under service conditions. Inthis case, the following should be checked:

. tension does not develop anywhere

. the predefined limits of the maximum compressive stresses, according to the adoptedofficial codes and regulations, are not exceeded anywhere in the beam and this valueis close to 13 N/mm2. g

The stresses involved in the above checks are evaluated from the following equationsðMgþP < 0, Mp > 0):

top ¼ P

A

Mgþ P

Wtop

top ¼ P

A

Mgþ P

Wtop

Mp

Wtop

171


and also (compressive stresses are negative)

bottom ¼ P

Aþ

Mgþ P

Wbottom

bottom ¼ P

Aþ

Mgþ P

Wbottom

þMp

Wbottom

It should be noted that the application of the prestressing force P on the beam, as anaxial compression force, must be statically feasible. For example, it is obvious that abeam with firm supports develops zero axial force after applying the prestress. Thesupports receive all the prestressing force, and the beam receives only the deviationforces u (Figure 4.30).

Throughout the previous discussion, there was never a need for an assumption aboutthe bond between the cable and the ducts. This is taken care of, immediately after thefinal stressing, by pouring cement grout inside the steel duct which is embedded in theconcrete, until this is completely filled. As long as the compressive force of the concreteis located within the core under full prestresssing, the above equations are obviouslyvalid whether there is bond or not. In the case where the prestressing steel is bonded,the beam has a higher carrying capacity with respect to the required safety factor.The bonding is also an important issue, especially for the protection of the prestressingsteel against corrosion. g

It should be kept in mind that the design of the prestressed beam under service conditionsfor full prestressing can never be adequate by itself. As the loads are increased, cracksappear in some sections, strength conditions change radically, and the prestressing forcedoes not increase as fast as the bending moments. Therefore, it is imperative to thoroughlyexamine the actual degree of safety of the structure also for failure, where bonding betweenthe prestressing steel and concrete plays an important role.

The state of failure will be studied generally in Section 4.3.2 through partialprestressing, for which the prevalent state of cracking also calls for a check at theservice state.

4.3.1.3 ShearPrestressing has a favourable effect on shear, because the deviation forces reduce thedeveloped shear force. This is deduced from the inclined compressive force, which

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PP

The beam is not subjected to an axial compressionbut is acted on by the cable deviation forces

Figure 4.30 Axial deformation as a prerequisite for introducing the compressive force


acts at the trace of the cable on the section (Figure 4.31). If the cable has a slope atthat specific point and the shear force due to external loads (without deviation forces) isV, the effective shear Veff of the section is given by

Veff¼V P sin

The analysis of the shear effect on the basis of the truss model studied for the case ofreinforced concrete is always valid. Additionally, the favourable fact should be takeninto account that, due to the prestressing force and the prevailing longitudinalcompressive stresses, the angle of the struts formed by the principal compressivestresses is smaller than the value used for reinforced concrete, specifically being in therange 25—308. As a result, stress values for the stirrups are reduced and, thus, theirdesign becomes more economical (see Figure 4.16).

4.3.2 Partial prestressing — ultimate state

4.3.2.1 BendingAs mentioned in Section 4.3.1.2, the idea of designing a beam with less prestressing forcethan that used in full prestressing, ensuring at the same time a controlled crackingconfiguration under service loads, leads to a more logical and more economic solution.

In order to understand, under these conditions, the state of stress in the criticalsection (usually at midspan) of a beam, the consecutive loading phases increasing upto the ultimate state should be examined, by taking into account the prestressingsteel area AP, as well as the mild reinforcing steel As, as shown in Figure 4.32. Theexistence of bond between the concrete and the reinforcement is emphasised inparticular.

The initial loading stage, where only the prestressing force P is applied, is oftheoretical importance, as previously explained, given the partial or total mobilisationof the self-weight of the beam occurring in practice. However, it is important torecognise that in the pure state of prestressing, the compressive force D (¼ P) oneach section acts exactly on the centroid of the tendon, i.e. at that point where thetensile force of the tendon is also acting. The mild reinforcement, placed as low aspossible under the prestressed reinforcement, is obviously compressed.

Any bending moment M, which is afterwards applied to the section due to gravityloads, shifts the force D upwards at a distance a equal to a¼M/P. As the concrete

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The vertical component of the prestressing force Pdecreases the acting shear force VV

β

P

Figure 4.31 Reduction of the shear force due to prestressing


section is acted on only by the compressive force D, if this force lies within the core areaof the section, then only compressive stresses will develop. The maximum bendingmoment ensuring the exclusive compressive stress state of the section will cause ashift of the compressive force D to the upper core limit (see Section 4.3.1.1). Thismoment MD is called the decompression moment (see Figure 4.32). It is thus clear that

MD¼ P (ktopþ e)

The bending moment MD results in a triangle of compressive stresses and strains on thesection, with a zero value at the bottom edge. This means that, for each loadingexceeding this level, tensile stresses and corresponding cracks in the bottom regionwill develop, if the unreliable — at least for design purposes — tensile strength ofconcrete is disregarded. It can be seen that until the bending moment MD is reached,the tendon stress P may be regarded as essentially constant and equal to P/AP, whilethe mild reinforcement stress remains essentially equal to zero.

Every bending moment M greater than MD produces an additional elongation strain"s in the prestressed tendon, while the same elongation is also applied to the reinforcingsteel. In this way, the tendon undergoes an additional tensile stress P¼"P Es,whereas the mild reinforcement develops essentially the same stress. It is clear that,while the linear strain distribution diagram describes directly the prevailing state inthe concrete and the mild reinforcement As, regarding the tendon, it has significanceonly to what takes place after the application of the decompression moment MD (seeFigure 4.32).

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Compressive stressesin concrete

ktop

e

DD

DExternal moment

M = 0

MD

Mpl

Cracked region

MD = P · (ktop + e)

Decompression moment

Maximum allowable moment fortension-free stress developmentD = P

xa

M

εb fc

0.80 · x

AP

As

AP

As

zP zP

zszs

ZP

Zs

ZP

Zsεs

∆εP

σP = P/EP εP = σP/Es + ∆εPZP = AP · fPy

Zs = As · fsy

Figure 4.32 Stages of the bending response of the prestressed beam


Now, in this cracked situation, the compressive force D on the concrete, the tensiletendon force ZP and the tensile force Zs of the mild reinforcement must togetherrepresent the resultant axial and bending state of stress of the section. Once thetendon is bonded to the concrete, it can be considered to be an effective part of thesection, so that the resultant axial force acting on it equals zero:

D¼ ZPþ Zs (a)

where

ZP ¼ P þP AP

Zs ¼ ð"s EsÞ As

Moreover, the bending moment M (>MD) is equal to

M¼ ZP zPþ Zs zs (b)

The magnitude and position of the force D as the resultant of the compressive stressblock of the concrete may be determined from the shortening strain "c and theheight x of the uncracked concrete region (see Figure 4.32). It is clear that, due tothe valid assumption of a plane section, the magnitudes P and "s may also beexpressed in terms of "c and x.

If "c is still less than 2% (Bieger et al., 1993),

D ¼ "c ð6 "cÞ12

b x fc

a ¼ 8 "c

4 ð6 "cÞ x

On the basis of the prestressing force P and the cross-sectional areas of prestressing andreinforcing steel AP and As respectively, "c and x can be readily determined for eachbending moment M, from equations (a) and (b). Subsequently, the reinforcementsteel and tendon stresses are determined directly.

An increase in the bending moment leads, firstly, to the yielding of the steelreinforcement and, with a further increase in the moment M, under the constantstress fsy of the mild reinforcement, the tendon will, in turn, also reach the yieldingstress fPy. At this stage, the concrete develops the maximum compressive stress fc atthe top of the section, whereas its shortening strain "c, which is now greater than2%, cannot exceed 3.5%. The section will then develop its maximum bendingresistance Mpl:

Mpl¼AP fPy zPþAs fsy zs

At this stage, the magnitude and the location of the compressive concrete force may beestimated reliably by assuming a fictitious compressive stress block having a constantuniform stress of fc and extending from the top edge up to a depth of 0.80 x (seeFigure 4.32). In this way, the magnitudes D, zP and zs may be expressed as

D¼ fc b (0.80 x)

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zP¼ d cP 0.40 xzs¼ d cs 0.40 x

In the present state, the known forces ZP and Zs allow the determination of thecompressive concrete force D, the depth x of the compressed zone, and consequentlythe lever arms zP and zs may be readily deduced. It should be noted that the depth xof the compressive zone represents the distance of the neutral axis in the straindiagram up to the top of the section, and this must not be confused with the heightof the fictitious compressive stress diagram, which, of course, equals 0.80 x. g

It is clear that the previous consideration of the ultimate state concerns both full andpartial prestressing, and gives a unified treatment of the limit state of reinforced concretesections in general, either with mild or prestressed reinforcement. Actually, the onlydifference between them lies in the fact that, in the case of prestressing, a certainpre-elongation is stored in the high-strength prestressing steel, owing to which anadditional tensile force is available, which in turn means the development of anadditional equal compressive force on the section. This leads to an increase in thebending resistance, according to the corresponding moment of the resulting couple offorces.

It can be seen, then, that in the ultimate state the presence of prestressing is notconnected to the development of a bending moment as in the case of service conditions,but is reflected simply in the formation of the bending resistance Mpl of the section(Menn, 1990). However, as will be illustrated in the next chapter, this holds only forstatically determinate structures. g

According to the previous discussion, it is clear that the increase in the tensile force inthe prestressing steel from its initial value P — after the various losses have occurred — upto the value corresponding to the yielding stress fPy is feasible only because of the existingbond, which allows the cable to ‘keep pace’ with the ‘evolution’ of the linear straindiagram " over the height of the beam.

If the bond of the tendon is not established by grouting, as mentioned in Section4.3.1.2, the increase in the cable force will depend only on the global deformation ofthe beam, which is difficult to analyse. And given that the prestressing steel does notgenerally reach the yielding point in the design, it may be assumed, for safety reasons,that the tendon force will not be subjected to any increase beyond its initial value,thus resulting in a decreased bending resistance Mpl.

In many cases, the use of unbonded tendons may be — apart from avoiding thegrouting — preferable to bonded ones from a constructional point of view, because oftheir easy exchangeability and the possibility of re-stressing at a later time. Theseconstructional advantages may take precedence, in the design phase, over thedrawback of the clearly reduced ultimate bending resistance in comparison withbonded tendons. Although the first prestressed concrete structures of earlier timesused unbonded tendons, which were soon abandoned because of the clear structuraladvantages of bonded ones, the last two decades have seen the frequent use ofunbonded tendons (primarily in bridges), placed either ‘externally’ or ‘internally’ with

176


respect to the beam section. In such cases, however, according to what has beenpreviously discussed, the decompression moment MD, which is independent of theexistence of a bond, may represent a safe assessment of the ultimate bending resistanceof the section. g

As pointed out previously, when partial prestressing is adopted, particular attention hasto be paid to the service condition, where the cracking of the concrete must be takeninto account. An acceptable cracking width is ensured through the restriction of theadditional stress increase in both the tendons and the reinforcing steel to within200 N/mm2. The determination of these tensile stresses, and also of the compressivestresses of the concrete — which also have to be held within logical limits — may becarried out through the equations previously considered.

Finally, aspects that affect the necessary control of deflection under service conditionscan be assessed — according to Section 4.2.2.1 — on the basis of the relevant expressionof the curvature 1/r in Section 4.1.1 and by taking into account the previously obtainedtop and bottom strains of the section.

4.3.2.2 ShearEverything that has been mentioned in Section 4.2.2.2 regarding the shear ultimatestate in reinforced concrete is obviously also valid in the case of prestressing, where,according to Section 4.3.1.3, the struts are inclined at about 308. The relieving influenceof the cable prestressing force on the effective shear has to be also taken into account inthe ultimate state, as emphasised in this section. It is clear that, for safety reasons, theinclined compressive prestressing force after all losses should be taken into accountrather than the corresponding yielding force.

4.4 Cantilever beamsThe response of the cantilever beam to gravity loads is characterised by the fact that thebending moments as well as the shearing forces increase towards the fixed end. Thecantilever beam is a structural system that is particularly sensitive because it has asingle support where the maximum bending and shearing responses occur simulta-neously, and therefore a possible failure at this point will cause total collapse. Inaddition, the cantilever beam exhibits particularly strong deformability.

It is clear that the layout of internal forces due to gravity loads is the inverse of that forthe simply supported beam. The tensile stresses are developed in the top region, whereasthe bottom region is compressed.

The steadily increasing bending response towards the fixed end suggests a continuousdecrease in the structural height towards the free end, in order to achieve a betterexploitation of the various sections in bending and save material in this way, givenalso the favourable distribution of weight decreasing towards the free end. This struc-tural form has a beneficial effect on the shearing response of the beam, as is nowexplained.

177


The inclined compressive force in the lower region of the section contributes substan-tially, with its vertical component, to the total shearing force required for the equili-brium (Figure 4.33). Consequently, the developed shearing stresses on the verticalsection of the cantilever are less than those that would be developed if the sectionheight remained constant along the beam, as in this case the vertical component ofthe horizontal compressive force is simply missing. As a result, for a parabolicdecrease in the height of the beam towards the free ends, an approximately constantshear along the beam is achieved.

If an I-section is considered, then the inclined compressive force D equals M/z, and itsvertical component is (M/z) (z/x). In the case of a constant height section, theshearing stress is

¼ V

bw z

In the case of a variable section, the above stress is reduced to

¼ 1

bw z V M

zz

x

The resulting constant shearing response allows the adoption of a constant webthickness bw along the complete length of a cantilever beam. g

With regard to reinforced concrete beams, apart from the above introductory remarks,neither the bending nor the shearing load-carrying actions differ from what has beenpreviously discussed for simply supported beams. The statically determinate trussmodel of the beam takes the form shown in Figure 4.34, and consequently all the

178

The contribution of the vertical component of thecompressive force to the required V allows the beamshearing stresses to cover only the remaining force

V V

Approximately constant shearing stress Maximum shearing stress varies like V

(I-section)

D

Mz

∆z/∆x

DV

X

Figure 4.33 Favourable influence of variable beam height on the shearing response


relevant results in Section 4.2.1.2 are also valid in this case with respect to the state ofstress of either the inclined struts or the stirrups.

In the case of a variable beam height, it is clear that the previously discussedapproximately constant response in shear allows the use of stirrups of constant cross-sectional area. g

With regard to additional prestressed concrete beams, the following remarks apply:

(1) The deviation forces and the anchor forces of the selected tendon profile shouldproduce bending and shear moment diagrams which are of opposite sense tothose caused by the gravity loads, thus contributing to the effective reduction ofthese loads.

(2) In the case of a beam with constant cross-section, the layouts (a) and (b) inFigure 4.35 may be examined. Each layout produces quite different bendingmoments and shear diagrams, but nevertheless have the same bending values attheir ends (see Figure 4.35). It is, of course, obvious that even the inclined straightcable tendon profile (c) will cause a bending and shear response, which reduces thatdue to gravity loads.

It is clear that, in the considered cantilever system, the forces acting on the beam dueto the cable prestressing consist of the compressive force at the free end of the beam(inclined or not), as well as the deviation force of the tendon (see Figure 4.35). Ineach case, the static action of the prestressed tendon on a cross-section is equivalentto a compressive force equal to P applied at the centroid of the tendon with theappropriate inclination.

In contrast to the simply supported beam, in the cantilever system the prestressedtendon produces tensile bending stresses at the bottom fibres, whereas the gravityloads produce tension at the top. Assuming, now, that the beam is under the prestressingactions only, with the compressive force P acting on the corresponding trace of thetendon, a bending moment M due to a gravity load applied afterwards will cause ashift a of the force P downwards equal to a¼M/P (Figure 4.36).

It is also clear that the behaviour of the beam in response to a further increase in loadwill follow the same stages as described in previous sections, by simply inverting thetensile and compressed regions.

(3) For beams with variable height, in addition to the above mentioned shearingresponse, all that has been discussed in Section 4.3.1.1 regarding the handling of

179

Figure 4.34 Truss model for a reinforced concrete cantilever beam


the cable profile applies (see Figure 4.26). Thus, in the case, for example, where thecentroidal axis of the beam has a parabolic track, the straight cable shown inFigure 4.37 is equivalent to a fictitious parabolic one. This causes uniformupward deviation forces which oppose the gravity loads. Of course, the staticallymost favourable result is achieved when the cable profile, having exhausted allthe available eccentricity at the fixed end, follows a parabolic trace towards thecentroid of the free end cross-section, as shown in layouts (a) and (b).

180

(a) (b)

(c)

ff

Stronger shear reliefdue to prestressing

Higher bending resistancesdue to prestressing

u uP

P

MPMP

VP VP

Figure 4.35 Influence of the cable profile on the response of a cantilever beam

Initial location ofprestressing force

a = M/P

aM

P

P

Figure 4.36 Downward shifting of the compressive force due to the bending action of gravity loads


While in both layouts (a) and (b) the required prestressing force, in order to overcomethe bending moment at fixed end, is the same, layout (b) offers, for a given prestressingforce, greater ultimate bending resistance throughout the length of the cantilever, whilethe shear relief offered at the fixed end by layout (a) is greater. However, in the case ofcantilever-constructed girder bridges using the so-called ‘free cantilever method’, layout(b) is imposed for constructional reasons (see Section 5.5.2). g

Considering now the control of shear, it is clear that the effective shearing force Veff,according to the previous discussion, is equal to (see Figure 4.33)

Veff ¼ V M

zz

x VP

where VP is the vertical component of the cable action on the examined section(Figure 4.37). The actual shearing stress of the cross-section will then be

¼ Veff

bw z

4.5 External prestressingIn a beam, it is possible, through an appropriate external prestressing layout, to offerconcentrated loads opposing those due to gravity, according to the principle given inSection 4.3.1.1

181

Higher shearing force Higher bending resistances

Fictitious cable profile

Although the tendon is straightupward deviation forces are developed

(a) (b)

VP

P

Figure 4.37 Influence of the cable profile in a cantilevered beam of variable height


The layout of the structural system shown in Figure 4.38 offers the simply supportedbeam of span length L upwards concentrated forces through the two vertical struts,clearly reducing the bending response (g L2/8) of the beam if it were in isolation.The structural action of this statically indeterminate system under the permanentload g leads inevitably to a settlement at the strut joints, which may be eliminatedthrough appropriate prestressing of the three straight cables. More simply, theprestressing of the cables should be such that the developed strut forces are equal tothe reactions of a three-span continuous beam with rigid supports. Thus, the bendingresponse of the simply supported beam for the given loads will be identical to themuch more favourable one of a continuous beam (see Figure 4.38).

If the struts are ordered at the third points of the span, given that the reaction R atthe interior support of a continuous beam with three equal spans is R¼ 1.10 g L, thenthe prestressing force S, which must be introduced to the horizontal cable, can readily beobtained from the equilibrium at the nodes (Figure 4.39):

S ¼ 0:978 g L2

8 h

182

The appropriate prestressing force S ensures the exertion of force R

The deflections at the locations of the struts are eliminated and thebending response becomes identical to that of the continuous beam

The statically indeterminate structure develops deflections at the locations of the struts

Tension Tension

Tension

Compression Compression

R R

R R

S

L/3 L/3 L/3

h

Figure 4.38 The purpose of external prestressing


It is appropriate to confirm this last result by also remarking that the overall bendingmoment (g L2/8) of the system may consist of the bending moment of the continuousbeam at its middle section (0.025 q (L/3)2) and the corresponding moment (S h)exerted by the internal force S of the horizontal cable (see Figure 4.39):

g L2/8¼ 0.00278 g L2þ S h

Prestressing of the horizontal cable with the above force ensures, as pointed out, thefunction of the simply supported beam as a continuous beam over four rigid supports.An additional live load p will cause an increase of the axial force of all three cablesaccording to the behaviour of the actual statically indeterminate system, whichdepends on both the rigidity of the beam EI and the cross-sectional area As of thecable. g

In the ultimate limit state, under the load q, where q is the total service load of thestructure, it may be considered that the ultimate bending moment (( q) L2/8) isundertaken by the bending resistance of the beam Mpl and the moment which theyielding cable force contributes with respect to the centroid of the cross-section ofthe beam (see Figure 4.39):

( q) L2/8¼MplþAs fPy h

This equation may be used for the assessment of the cross-sectional area of the cable inthe preliminary design. g

The beam can be made of steel, reinforced concrete or prestressed concrete. The thirdchoice is made when the resulting ‘continuous’ concrete beam has a relatively long span(i.e. of the order of 15 m). Thus by providing the beam with a stronger ultimate bendingresistance Mpl, the required prestressing steel cross-sectional area As of the cable is

183

g

g

0.10 · g ·(L/3)2

0.025 · g ·(L/3)2

0.025 · g · (L/3)2

L/3

L/2 L/2

L/3 L/3

Service condition

γ · q

g · L/2 (γ · q) · L/2

g · L2/8h h

S

Mpl

AP fPy

(γ · q) · L2/8

Ultimate state

Figure 4.39 Structural action of the system


reduced, according to the last equilibrium equation. However, the choice of a high-strength steel for the cable is crucial.

It should be noted here that due to the presence of a significant compressive force inthe beam, a more realistic calculation of Mpl may rely on the factors discussed in Sections6.5.1 and 6.5.2, regarding the steel and reinforced concrete sections, respectively.

Finally, it should be pointed out that the optimal distance between the vertical struts,with regard the required prestressing force S of the cable in order to counterbalance thepermanent load of the beam, equals 0.44 L, instead of L/3. Thus,

S ¼ 0:92 g L2

8 h

4.6 Design controlAs has been pointed out in Section 1.6, the design control of a structure includes, inaddition to the control in service conditions of deformations, cracking, annoyingvibrations, etc., the control of ultimate strength. This is no longer the control ofthe stresses developed under service loads, but concerns checking that the specifiedworking loads q multiplied by a safety factor (depending on the code followed),which are approximately equal 1.80, are less than (or equal to) the correspondingultimate load which causes the failure of the structure. According to the so-called‘static theorem of plasticity theory’, which will be examined in particular in Chapters5 and 6, this condition is ensured — excluding failure by loss of elastic stability, or thepresence of second-order effects (see Chapter 7) — if the following equation, givenpreviously in Section 1.6, is satisfied:

SR (¼ S R)

The sense of the factors S and R, has been clarified in Section 1.6. The above equationapplies to any cross-section which can be regarded as critical and relates to the sectionalforces of the structure examined.

The term S represents the sectional force of the cross-section examined, and resultsfrom the corresponding sectional force diagram, which should be in equilibrium with theexternal loads q. This, with regard to the bending moments of a beam or frame structure,can be ensured by two things. First, the moment values at the joints must be in equilibrium,on the one hand, and the support reactions must ensure the global equilibrium of thestructure, on the other hand. Second, the bending moment diagram, which is drawnbetween the end values of a straight member, must be identical to the bending momentdiagram of the corresponding simply supported beam (see Section 2.2.5). While such abending moment diagram is compatible with the prevailing equilibrium of the structure,it may generally present a deviation from the corresponding diagram of the elasticsolution for the loads ( q), as stated in particular in Chapters 5 and 6. The shear forcediagram may obviously be determined from the adopted bending moment diagram.

The term R represents the corresponding ultimate resistance of the cross-section, e.g.if bending is considered it will be equal to Mpl.

184


Thus, while for the statically determinate structures the compatible moment diagramwith the overall equilibrium is unique, for statically indeterminate structures, where anunlimited number of adoptable bending moment diagrams can satisfy the equilibriumrequirements, the design of cross-sections has a large number of options, dependingon decisions made by the engineer.

Finally, the remark made in Section 4.3.2.1 is repeated here, namely that in the caseof a prestressed statically determinate structure, such as a simply supported beam or acantilever beam where prestressing does not cause any support reactions, thepresence of prestressing is reflected only in the ‘resistance term’ R, i.e. in thecorresponding determination of Mpl.

ReferencesBieger K.W., Lierse J., Roth J. (1993) Stahlbeton— und Spannbetontragwerke. Berlin: Springer Verlag.Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkauser Verlag.Walther R., Miehlbradt M. (1990) Dimensionnement des Structures en Beton. Lausanne: Presses

polytechniques et universitaires romandes.

185


5

Continuous beams

5.1 IntroductionThe form of the bending moment diagram of a continuous beam in which the tension isalternated between the top and bottom fibres between the regions of supports and spans,respectively (see Figure 3.14), suggests that such a beam behaves as consisting of inter-mediate simply supported beams resting on cantilevers, extending from the internalsupports up to the points where the bending moments become null. The length ofthese cantilevers constitutes a percentage of the corresponding span, which, forinternal spans having a constant depth, is of the order of 20%, depending on the typeof loading, as well as the span ratios (Figure 5.1).For beams of large spans, as are usually applied in bridges, it is statically purposeful to

follow a parabolic change in depth in all spans, with the maximum depth at the supportsand the minimum depth at the midspan sections. In this way there is an increase in themoments at the supports (see Section 3.2.10) and, according to the equilibriumrequirement (suspension of the moment diagram for the simply supported beambetween the values at the supports), a reduction in the midspan moments. Moreover,the cantilever lengths are increased compared with those of the corresponding beamwith constant depth (see Figure 5.1). Thus, in the case of a constant depth, thebending moment diagram is independent of the moment of inertia of the beam, asexplained in Section 3.2.5.The aforementioned change in depth leads to a proportional distribution of self-

weight that, for a support-to-midspan depth ratio of 3 (a common value for bridgeswith large spans), results in a significant increase (up to 35%) in the lengths of theconsidered ‘cantilevers’ of the internal spans. Therefore, the ‘simply supported beams’are limited to 30% of the span lengths, with far smaller bending moments than thoseof the ‘cantilevers’ (see Figure 5.1). The fact that the thus resulting support momentsdo not differ substantially from those of the cantilevers extending up to the midspangave rise to the construction of concrete bridges according to the so-called balancedcantilever method, namely the construction of progressively smaller prestressed cantileversegments from both sides of the two free-standing supports (piers) until the midspan,and thereafter the restoration of the beam continuity (see Section 5.5.2). This depthvariation in the ‘cantilevers’ contributes favourably to the shear response, as explainedin Section 4.4.

5.2 Steel beamsAs, mainly for reasons of economy, steel beams are nowadays designed using the ‘plastictheory’ rather than the ‘elastic’ one, it is advisable to be familiar with this means ofdesign, as applied to a statically indeterminate continuous beam, by considering firstthe behaviour of a fixed-ended beam and then that of a fixed simply supported beam.The bending (and shear) response at any span can be described using the basic charac-teristics for either beam type.

5.2.1 Fixed-ended beamA fixed-ended beam under a uniform load q develops the bending moment diagramshown in Figure 5.2. From the figure and the discussion above it is obvious that theremust be two cantilevers with length 0.21 L. The moment of the support q L2/12 is

188

L

~0.30 · L~0.35 · L ~0.35 · L

L

~0.60 · L~0.20 · L ~0.20 · L

Figure 5.1 Load-bearing action of a continuous beam


double the maximum moment of the midspan, which is of course the span moment ofthe ‘fictitious’ simply supported beam of length 0.58 L. It is assumed that the cross-section of the beam is symmetrical with respect to the horizontal axis of bending, andaccordingly it has the same bending resistance Mpl for moments causing tensioneither at the top fibres or at the bottom ones.For some load q1 the bending moment will first reach the plastic resistance value at the

supports. Thus q1 L2/12¼Mpl and q1¼ 12 Mpl/L2. For this load a plastic hinge will be

developed at both ends of the beam, as discussed in Section 4.1.2 (Figure 5.3). Obviously,the beam does not collapse under this load, and is still in a position to accept more loading.During the imposition of any additional load, the plastic hinges at the two ends rotateunder the constant moment Mpl, which can be considered as being externally applied.

189

q

0.21 · L 0.58 · L 0.21 · L

q · L2/12q · L2/8

Figure 5.2 Load-bearing action of a fixed-ended beam

Virtual loading for the determinationof rotation at the plastic hinge

q1

qu

Mpl

Mpl

Mpl

MplMpl

qu = 16 · Mpl/L2

qu · L2/8

11

11

Figure 5.3 Beam behaviour due to plastic hinge formation

Continuous beams

The diagram of the bending moment corresponding to each load q is obtained by hangingthe diagram for the simply supported beam q L2/8 from the constant points that corre-spond to Mpl. Clearly the beam will collapse when the load q takes the value qu atwhich a bending moment Mpl is developed at the midspan of the beam. In this case, anadditional plastic hinge will be created at the midspan (first-degree mechanism) and nofurther increase in load will be possible, as discussed in Section 4.1.2 for a simply supportedbeam. Thus it may be written that qu L2/8Mpl¼Mpl and, consequently, qu¼ 16 Mpl/L2 (see Figure 5.3).The problem now of designing a beam under a certain service load qser is thatMpl should

be defined such that the beam may sustain, or even better collapse at, a load qu¼ qser(see Section 4.6). According to the last equation the value sought for Mpl is given by

Mpl¼ ( qser) L2/16It should be pointed out that designing a beam for the ‘collapse’ load qser by followingthe elastic moment diagram means that a value of ( qser) L2/12 must be selected forMpl, which is of course higher than above value, and consequently represents anuneconomical solution. Precisely for this reason it was pointed out in Section 4.6 thatin the equation

SR (a)

the design value S does not necessarily correspond to the elastic solution for the load qser. Thus, in order to design a fixed-ended beam with length L, so that the ultimatebending resistance is exploited at the ends as well as at the midspan, one should considerthe bending moment diagram for the simply supported beam that ensures balance withthe applied load qser and then suitably shift its base until equal parts Mpl are cut atboth ends as well as the midspan (see Figure 5.3). Thus, it is required that

( qser) L2/8¼MplþMpl

This result coincides with the preceding one.It is important though to point out that, according to the static theorem of plastic

analysis (see Section 6.6.2), the satisfaction of the above inequality (a) at any pointon the beam ensures that the value qser is less than or equal to the ultimate load,provided that the value S belongs to a bending moment diagram which is inequilibrium with the external load qser. The confirmation that ( qser)< qu, can bemade when selecting both Mpl¼ ( qser) L2/16 and Mpl¼ ( qser) L2/12. It is, ofcourse, clear that while in the first case the ultimate load qu coincides with qser, inthe second case, according to the above equation, it results in

qu¼ 16 Mpl/L2¼ 1.33 ( qser)

It is purposeful at this point to find also the correspondingMpl that would result if thedesign was made on the basis of the allowable stress fy/. As the section at the beam endis the most unfavourable one, according to Section 4.1.2 My/¼ qser L2/12. AsMpl¼ a My, it follows that the selected section on which the load qser causes themaximum allowable stress fy/ should have a bending strength equal to

190


Mpl¼ a ( qser) L2/12. This value of the moment resistance leads to a collapse loadequal to qu¼ 16 Mpl/L

2¼ a 1.33 ( qser). For a¼ 1.15, which is appropriate forI-shaped cross-sections, qu¼ 1.53 ( qser). This result shows that the safety factor determined according to the method of permissible stresses is practically equal to 1.53 .Finally, another point of interest is the rotation developed in the plastic hinge at the

instant of collapse. This rotation, which must be developed in order to realise theaforementioned moment diagram, is limited by the plastic deformability (ductility) ofthe material, which for steel (as experimentally shown) is of the order of 0.1 rad. Thisrotation can be determined directly from the existing moment diagram for the instantof collapse by applying the principle of virtual work.A self-equilibrating system of unit moments, applied at the ends of the simply

supported beam, is selected as the virtual load. According to the principle of virtualwork, and taking into account the above (see Figure 5.3):

2 1 ’ ¼ðM M

EIds ¼

Mpl L3 EI ¼

2 a fy L3 E h (see Section 4.1.2)

For the values a¼ 1.15 and fy¼ 4.2 105 kN/m2, it results that ’¼ 0.77 10—3 (L/h). Itis concluded that for practical values of the ratio L/h, i.e. values of 25—30, the resultingvalue of ’ is feasible.The resulting value of the angle ’ for the elastic diagram of bending moments must be

equal to zero, as the diagram under consideration is supposed to satisfy, besides theequilibrium, the compatibility of the deformations. Such a requirement cannot, ofcourse, be satisfied for the adopted design diagram, as elastic behaviour has beenexceeded and plastic behaviour has already been established.

5.2.2 Fixed simply supported beamA fixed simply supported beam under a uniform load q develops the bending momentdiagram shown in Figure 5.4. It can be seen that there is a cantilever part of length

191

q

q · L2/8

q · L2/8

max M

0.25 · L 0.75 · L

0.375 · L

Figure 5.4 Load-bearing action of a fixed simply supported beam

Continuous beams

0.25 L, with a fixed end moment equal to q L2/8. The maximum span moment corre-sponding to the bending moment of a ‘fictitious’ simply supported beam of length 0.75 Loccurs at a distance 3 L/8 from the free end, and is equal to q L2/14.2.The beam at first develops a plastic hinge at the fixed end under the load q1, so that

Mpl¼ q1 L2/8, and it can be loaded further, up to the collapse under the total load qu.Under this load, and with Mpl applied externally at the henceforth ‘hinged’ end, thebeam develops its maximum span moment, which is also equal to Mpl (Figure 5.5). Itis found that the load qu that leads to collapse is qu¼ 12 Mpl/L

2.The same design procedure used for the fixed-ended beam can be equally applied to

the fixed simply supported beam also. On the basis of the bending moment diagram ofthe simply supported beam under an external load qser, a position of the rotatingbaseline around the simply supported end is sought at which the cut maximum valueof the span is equal to the one corresponding to the fixed end (see Figure 5.5). Theseequal values are obviously the plastic moment Mpl, and according to the above thisresults in Mpl¼ ( qser) L2/12.

192

q1

qu

L

Mpl

Mpl

Mpl

M

M

1

1

Virtual loading for the determinationof rotation at the plastic hinge

qu = 12 · Mpl/L2

qu · L2/8

5 · L/12

5 · L/12

(γ · qser) · L2/8

Figure 5.5 Ultimate strength and design of a fixed-ended beam


Again, the design of such a beam for a load capacity qser, made on the basis of thecorresponding maximum value of the elastic diagram involving Mpl¼ ( qser) L2/8,leads to a less economic solution, as according to the above expression the correspondingultimate load is

qu¼ 12 Mpl/L2¼ 1.50 ( qser)

Moreover, a design done using the method of permissible stresses under a ‘safety factor’0 imposes, according to Section 4.1.2, My/¼ qser L2/8, and consequentlyMpl¼ a ( qser) L2/8. This selected bending resistance (a¼ 1.15) leads to an ultimateload qu¼ 1.72 ( qser).The rotation developed in the plastic hinge is again found, according to the

principle of virtual work, on the basis of the adopted bending moment diagram (seeFigure 5.5):

1 ’ ¼ðM M

EIds ¼

Mpl L6 EI

a value which is identical to the one for the fixed-ended beam (see Section 5.2.1).

5.2.3 Continuous beamThe case of a continuous beam can now be examined. In general, given the load to becarried qser, the plastic moments Mpl at midspan and at the supports (Mm

pl, and Mlpl

and Mtpl, respectively) should be so determined that the resulting ‘suspended’

moment diagram M0 of the corresponding simply supported beam between any twosuccessive supports exhibits values that are less than or equal to the correspondingvalue of Mpl (Figure 5.6).This condition is ensured if the following equation is satisfied at each span (Menn,

1990):

M0ð qserÞ Mmpl þ

Mlpl þMr

pl

2ðaÞ

193

(γ · qser)

M lpl

M lpl

M rpl

M rpl

(γ · qser) · L2/8

M mpl

M mpl

Figure 5.6 Design of a continuous beam based on plastic behaviour

Continuous beams

For a beam of constant depth, it is purposeful to select a cross-section with doublesymmetry and having the same value of Mpl for the moments causing tension at thetop and at the bottom. The satisfaction of equation (a) for beams with roughly equalspans (min l/max l 0.8) and uniform load qser is ensured if for the external spans avalue Mpl¼ ( qser) L2/11 and for the internal spans a value Mpl¼ ( qser) L2/16are selected, as explained previously.To reiterate, according to the static theorem of plastic analysis, the design of the beam

(i.e. the determination of Mpl), based on the above equation of equilibrium (a), ensuresthat the load qser will be less than or equal to the ultimate load of the beam.It should also be emphasised that the adopted bending moment diagram, based on

equation (a), is only acceptable if the relative rotations of the plastic hinges in theinternal supports are within the permissible limits, according to the above. Theirdetermination is always made by direct use of the principle of virtual work, as shownin Figure 5.7. Assuming a uniformly distributed load qser, the developed (relative)rotation in the plastic hinge on any internal support is

’ ¼ ð qserÞ ðL31 þ L32Þ24 EI ðM1 þ 2 MÞ L1

6 EI ðM2 þ 2 MÞ L26 EI

Note again that, based on the elastic moment diagram, the aforementioned rotation iszero, given that the elastic solution is supposed to satisfy the compatibility requirementson both sides of the considered hinge at each support, and therefore requires the nulli-fication of the relative rotation, due to continuity of the deflection line.However, in cases of intensely different spans or rigidities, the relative rotation of the

plastic hinge at an internal support, determined on the basis of the correspondingadopted moment diagram, may be rather prohibitive, and the structure must bedesigned following the elastic solution, although always according to the strengthcondition SR.

194

Virtual loading for the determination of the relative rotation at the plastic hinge1 1

1 1

L1

M1 M2

L2

(γ · qser)

(γ · qser) · L12/8

(γ · qser) · L22/8

Figure 5.7 The required check of the relative rotation at the plastic hinge


5.3 Reinforced concrete beamsAlthough it has always been common practice to calculate the bending moments ofcontinuous beams for a specific loading on the basis of elastic analysis, it should be recog-nised that the variation in the moments along a continuous beam will cause cracking inthose parts where the cracking moment is exceeded, while all other regions will remainuncracked. The cracking moment is the bending moment that causes tensile stresses onthe homogeneous section equal to the tensile strength of concrete, which is of the orderof 200 kN/m2. However, given that cracking of the section causes a reduction in themoment of inertia, the variable rigidity along each span will obviously lead to amoment diagram that is different from the one resulting for the uncracked sectionsover the entire beam length. At the same time, factors irrelevant to the consideredloading, such as shrinkage, creep or possible thermal variation due to environmentaleffects, may lead to the tensile strength of the concrete being exceeded, and hence tocracking, before the anticipated loading to reach this point. For all these reasons aprecise follow-up of the stress state of the beam becomes practically impossible, andtherefore a plastic analysis will be more useful in such a case. This objective of theanalysis is to design and reinforce the beam such that the load qser is less than orequal to the ultimate load of the beam. This can be achieved, as stated previously,using the static theorem of plastic analysis (see Section 6.6.2), and by adopting sucha bending moment diagram that equation (a) in Section 5.2.3 holds true for each span.However, it should be emphasised that attention must be paid to ensure that the

deviation of the adopted moment diagram from the elastic solution does not exceed a‘reasonable’ limit. The strict criterion to determine whether the adopted momentdiagram leads to permissible results for design is the magnitude of the (relative)rotation that each plastic hinge is supposed to develop on the basis of the diagram.The maximum ‘allowable’ rotation value lies in the range 0.02—0.05 rad, dependingon the cross-sectional area of the existing reinforcement. This condition is generallysatisfied without special calculation if the deviation of the moment values at thesupports from the corresponding values in the elastic solution is not more than 20%.Of course, it is understood (always according to the static theorem of plastic analysis)

that the adoption of the elastic moment diagram for the load qser and the satisfaction atevery point of the relationship S(qser)R ensures the above objective, although notin the most economical way.Everything examined in Section 5.2 with regard to steel beams has an analogous

application in the case of reinforced concrete beams. Moreover, whatever applies tothe design of a simply supported beam and a cantilever with regard to reinforcedconcrete has a direct application in the determination of the reinforcements for acontinuous beam, as such a beam is composed of such analogous parts.

5.4 Prestressed concrete beams

5.4.1 Tendon design and structural performanceIn the case of prestressed concrete, the layout of tendons is also based on the synthesis ofcantilevers and simply supported beams, as explained previously. Basically, the tendon

195

Continuous beams

layout aims to provide the beam, via the induced deviation forces, with the oppositeloads to those due to gravity, which stress the beam. In order for the greatest possiblebending resistance to be offered, both in the regions of the supports for momentscausing tension in the top fibres and in the spans for moments causing tension in thebottom fibres, the tendons are arranged in such a way as to exhaust the correspondingplacement limits. The thus resulting reverse curvatures on the path of the tendon mustensure that the minimum applicable radius of curvature over the support for the specifictendon is not exceeded. It is understood that the equilibrium of the tendon as a free bodyimposes specific geometrical conditions on the positions of turning points and on thecurvatures of the tendons (Figure 5.8).As the continuous beam is a statically indeterminate structure, the bending moment

due to prestressing is no longer equal to the product of the prestressing force due to itseccentricity with respect to the centroidal axis. In other words, the compressive force onthe concrete section is not applied at the trace of the tendon, as is the case in the simplysupported beam or the cantilever. This is explained below using the example of acontinuous beam with two equal spans (Figure 5.9).For the analysis of the statically indeterminate beam under prestressing loads only, the

force method is applied and, in order to convert the structure into a statically deter-minate one, the intermediate support is removed by selecting the correspondingreaction as the redundant unknown. The simply supported beam is then deformedupwards under the deviation forces and, in order to restore the compatibility of deforma-tions with respect to the actual system, which requires an immovable intermediatesupport, a downward applied (redundant) force must be applied. Thus a triangulardiagram of bending moments MSP results, causing tension in the bottom fibres, and

196

Keeping the minimum radius of curvature

Common tangent

Self-equilibrating loading

uF uF

uS

uF

uS

LSLF

LSLF

uF · LF = uS · LS

Figure 5.8 Geometric requirements for the layout of tendons


this diagram is then superposed on the one for the ‘statically determinate’ prestressingbending moments, which are the product of the prestressing forces due to the existingeccentricity (see Figure 5.9). As a result, the compressive force on the concrete isshifted (as explained in Chapter 4, see Figures 4.28 and 4.36) upwards relative to thetrace of the cable, in the regions of both the midspan and the internal supports,where it is of course located outside the section.In this manner, the finally developed bending moments MP due to prestressing are

given by

MP¼M0þMSP

197


Deviation forces

Statically determinate beam Deformed statically determinate beam

P P

P P

Redundant force

[MSP]

[MP]

Redundant prestressing moments due tothe required concentrated (redundant) force

The compressive force on each cross-section shifts upwardsdue to the redundant prestressing moments

Bending moments due to self-equilibrating prestress loading

MP = M0 + MSP

Figure 5.9 Analysis of a prestressed statically indeterminate beam

Continuous beams

where M0 represents the statically determinate moments (¼ P e), and MSP are the so-called ‘statically indeterminate’ or ‘constraint’ moments, which are due to the action ofthe hyperstatic (redundant) force only.It should be noted that the development of the aforementioned redundant force

implies also the development of self-equilibrating reactions in the continuous beam,while in the simply supported system loaded with the self-equilibrating prestressingforces no such reactions are developed. It is clear that the moments MSP are causedby this system of self-equilibrating reactions. Of course, the fact that the developingreactions in the continuous beam are self-equilibrating arises immediately also fromthe fact that they are caused by a self-equilibrating loading such as prestressing.The momentsMP may be calculated directly from the statically indeterminate system

under the ‘external’ loads imposed by the prestressing, i.e. the deviation forces and thepossible moments at the two ends of the continuous beam due to eccentricity of theanchor forces (see Figure 5.9). The statically indeterminate moments will then be

MSP¼MPM0

It should be noted that as the loading is increased and the time of cracking due to amore intense bending response is approached, the redundant prestressing moments donot change significantly. However, when the ultimate state is approached and thecontinuous beam becomes in practical terms a number of simply supported partsconstituting a statically determinate system, the redundant prestressing momentsapproach vanishing. In any case, the above moments must be determined because, aswill be explained in the next section, they are taken into account in the control ofthe design.With regard to large-span beams, as occur in bridges, girders with variable depth

(following a parabolic law) are normally used, and consequently all that applies iswhat has been discussed already in Section 5.1. The bending moment diagram has afavourable shape and there is a positive influence on the shearing forces in theregions of the supports, which for practical purposes leads to a constant shear stressdesign value.

5.4.2 Control of the designThe adoption of a moment diagram that is in equilibrium with the external loads and hasbeen determined on the basis of equation (a) in Section 5.2.3 has already been examinedfor steel and for concrete beams.In the case of prestressed concrete, the statically redundant moments should be

superposed on the moments due to external loads. As explained previously (seeSection 4.6), the statically determinate prestressing moments do not count towardsthe loads, but simply contribute to the bending resistance. This is reasonable, as onlythe redundant prestressing moments modify the equilibrium state of the system, bychanging the reactions arising from the gravity loads.Despite the fact that the structure is designed for a supposed failure state where, as

previously stated, the statically redundant moments do not develop, the normal

198


values of the latter moments are nevertheless generally taken into considerationaccording to Section 5.4.1.Therefore, the strength relationship

M(qser)MR

can be written (Menn, 1990) as

M(qser)þMSPMR

It should be noted that this relationship is valid regardless of whether or not the depthvaries across the cross-section.The required bending resistance MR in various locations of the beam is achieved, as

for the simply supported beam, by considering a combination of prestressed tendons andreinforcing steel (see Section 4.3.2.1).It should be noted that the momentsMSP are ‘legitimately’ superposed with the values

from the diagram M(qser), multiplied by some factor P, due to the static theorem ofplasticity theory, as they arise from a self-equilibrating state (see Section 6.6.2). Ofcourse, at the ultimate limit state, where the system becomes statically determinate,these moments may be considered as zero (i.e. P¼ 0), as it in fact happens. g

As noted previously, at a preliminary stage the system is first designed based on ultimatestrength requirements, and this system is then checked in the serviceability state. It isobvious that the design principles laid out in Section 4.3 — particularly regarding partialprestress — may be directly applicable in the case of continuous beams.In the service condition, the satisfactory behaviour of the beam is first checked against

cracking. If only compressive stresses are developing in the concrete at every location,besides the ultimate strength, it should also be checked that the area of reinforcingsteel area is at least the minimum required by the appropriate structural code; this isusually expressed as a percentage of the section.The presence of tensile stresses in the concrete above the limit of 2N/mm2 will lead to

cracking (partial prestressing) and, accordingly, both the stresses in the reinforcing steeland the stress increase in the prestressed tendons must be checked. These stress valuesmust not exceed the limit of 200N/mm2.

5.5 Creep effectsThe presence of creep as a progressive cause of deformation raises immediately theproblem of the redistribution of the initial stresses in those systems where the state ofstress under service conditions is based on the compatibility of deformations (i.e. inthe statically indeterminate structures in general, and in this respect in continuousbeams as well). The problem arises of whether the system has been definitivelyformed or will undergo a change during the construction procedure. Although thisproblem is complex, thanks to the understanding of the creep behaviour of concrete(see Section 1.2.2) it may be dealt with in a direct manner (Menn, 1990), asdescribed below.

199

Continuous beams

5.5.1 Statically redundant systemsWhat may be said in general about any structure the members of which have the samecreep properties (i.e. structures that are made entirely concrete of the same specifica-tions and age) is that, although the creep causes an increase in deformations, it doesnot affect the state of stress (axial, bending and shear responses) caused by externalloading (excluding the cases of support settlement and temperature variation). Thereason for this lies simply in the fact that a uniform change in the elasticity modulusof the members of a statically redundant system (more specifically a reduction due tocreep) involves exactly what was stated in Section 3.2.5. Thus, if one takes intoaccount that creep implies a reduction in the elasticity modulus (see Section 1.2.2.1)and this reduction is uniform (for members that have the same creep properties),then, according to Section 3.2.5, no change takes place in the state of stress,although the deformation increases.The continuous beam can be used as a representative example of a statically

redundant system. For simplicity, a continuous beam of two spans under an arbitraryloading is considered (Figure 5.10). The statically determinate primary system consistsof the two adjacent simply supported beams, and thus the moment at the internalsupport B is the unknown redundant.

200

A B C

Initial diagram of bending responseFinal (redistributed) bending response due to creep

The right beam does notcreep at all

X1

X1

∆X1

F l11 F r

11

F l10 F r

10

X1 = 1

The bending diagram remains unchanged with timeif all members have the same creep properties

Figure 5.10 Creep effects on the state of stress of a continuous beam


With the load imposed on the continuous beam, the magnitude X1 is instantaneouslydeveloped. This value results from the compatibility condition

ðFl10 þ Fr10Þ þ X1 ðFl11 þ Fr11Þ ¼ 0

where F1 denotes the rotation at support B of the primary system due to the external loador due to X1¼ 1, for both the beam on the left (l) and the beam on the right (r). Thequantities (Fl10 þ Fr10) and (Fl11 þ Fr11) represent the corresponding rotation at end Bof the beams due to the external load and X1¼ 1, respectively (see Section 3.2.2).It is now assumed that X1 changes because of creep, so that after an adequate time it

reaches the value (X1þX1). The total value of the relative rotation at support B(¼ 0) is obtained by superposing the following magnitudes (see Sections 1.2.2.1 and3.2.2, where for simplicity ’1 is simply written as ’):

. the relative rotation due to external loads

Fl10 ð1þ ’Þ þ Fr10 ð1þ ’Þ. the relative rotation due to the imposition of X1

X1 ½Fl11 ð1þ ’Þ þ Fr11 ð1þ ’Þ. the relative rotation due to the gradual imposition of the change in X1, (i.e. of X1)

X1 ½Fl11 ð1þ ’Þ þ Fr11 ð1þ ’ÞIt is clear that multiplication of the ‘elastic’ deformation values by (1þ’) or (1þ ’)means that Einit or Edif, respectively, is adopted as the effective elasticity modulus ofconcrete, according to Section 1.2.2.1.Thus, the compatibility condition after some elapsed time takes the form

Fl10 ð1þ ’Þ þ Fr10 ð1þ ’Þ þ X1 ½Fl11 ð1þ ’Þ þ Fr11 ð1þ ’Þ

þX1 ½Fl11 ð1þ ’Þ þ Fr11 ð1þ ’Þ ¼ 0

Based on the aforementioned initial compatibility relation, this last equation can bewritten more simply as

X1 ½Fl11 ð1þ ’Þ þ Fr11 ð1þ ’Þ ¼ 0

from which it is immediately deduced that X1¼ 0.The conclusion is that X1¼ 0 is not valid if the creep is not identical in all the

members of the structural system. This is obvious in the ‘extreme’ case of the samebeam, when its right-hand part, being made of steel, does not creep at all(Figure 5.10). After some time has elapsed, for the primary system that was previouslyexamined the total value of relative rotation at support B (which is obviously also equalto zero) will be composed of superposing the following:

. the relative rotation due to the external loads

Fl10 ð1þ ’Þ þ Fr10

201

Continuous beams

. the relative rotation due to the imposition of X1

X1 ½Fl11 ð1þ ’Þ þ Fr11. the relative rotation due to the gradual imposition of the change in X1 (i.e. X1)

X1 ½Fl11 ð1þ ’Þ þ Fr11The compatibility condition is now written as

Fl10 ð1þ ’Þ þ Fr10 þ X1 ½Fl11 ð1þ ’Þ þ Fr11

þX1 ½Fl11 ð1þ ’Þ þ Fr11 ¼ 0

and, taking into account the initial condition, which naturally remains valid,

Fl10 ’þ X1 Fl11 ’þX1 ½Fl11 ð1þ ’Þ þ Fr11 ¼ 0

and therefore

X1 ¼ ’ ðFl10 þ X1 Fl11ÞFl11 ð1þ ’Þ þ Fr11

From the numerator in the above equation it may be observed that if the (absolute) valueof the bending moment at the internal support (i.e. X1) is smaller than that for the left-hand fixed simply supported beam, then X1 has the same sign as X1, i.e. the bendingmoment at the support increases. This becomes immediately clear in the case wherethe right-hand beam is unloaded. The system will then have an identical bendingaction to the frame ABC in Figure 5.11, where the column is made of steel. The reductionin the effective elasticity modulus of the loaded beam due to creep causes (see Section3.2.9) an increase in the bending moment at the support (see Figure 5.11).The last equation may also be applied in any other case where there is cooperation

between concrete (C) and steel (S) members (as shown in Section 6.2.1 also). In thiscase it is more convenient to write the above equation in the form

X1 ¼ ’ ðFC10 þ X1 FC11ÞFC11 ð1þ ’Þ þ FS11

g

202

Initial diagram of bending responseFinal (redistributed) bending responsedue to creep

A B

C

The bending behaviour of the frame is identicalto that of the corresponding continuous beam

The vertical member does not creepX1

∆X1

X1 ∆X1

Figure 5.11 Moment redistribution due to creep


It is known that the imposition of loading on a statically indeterminate structure causes astate of stress that depends only on the ratio of the rigidity EI of its members and not ontheir absolute values. However, if a deformation is imposed (support settlement ortemperature change), the absolute value of EI (i.e. the bending stiffness of eachmember) plays a predominant role. In such cases it is logical that the redundantforces change over time because of creep, as can be shown using the same example ofthe two-span continuous beam.In the continuous beam shown in Figure 5.12 a settlement is imposed on the

internal support. In the primary system of the simply supported beam, the reaction X1

is considered as the redundant force. The value acquired instantaneously by thereaction X1 (downwards) is determined from the compatibility equation

X1 F11¼ (i.e. X1¼ /F11)

where F11 is the displacement caused by the unit force X1¼ 1. This reaction decreaseswith time, and acquires a final value of (X1þX1). The reduction byX1 is determinedby satisfying the compatibility condition at the final time phase, as follows.The displacement of the reaction application point at the final time phase results from

the superposition of the following two effects: the deformation due to the application overa long time of the force X1; and the deformation due to the gradual imposition over thesame time interval of its variation X1. The first deformation is equal to X1 F11(1þ’),and the second deformation is equal to X1 F11(1þ ’). Thus,X1 F11(1þ’)þX1 F11(1þ ’)¼

203

Initial diagram of bending responseFinal bending response due to creep

∆X1

X1

δ

δ

Reduction in the instantaneously developed reaction

Deformation due to instantaneousimposition of settlement

Additional deformation due to creepwhich is eliminated by ∆X1

X1: Instantaneously developed reaction

Figure 5.12 Relieving creep effect on the state of stress due to an instantaneous settlement

Continuous beams

Combining this with the preceding relationship gives

X1 ¼ X1 ’

1þ ’and therefore the final value of the internal reaction (definitely decreased) is

ðX1 þX1Þ ¼ X1 1 ’

1þ ’

It is clear that the value in parentheses applies to both the bending moments and theshear forces of the beam. This reduction is very pronounced (see Figure 5.12). For typicalvalues of ’¼ 2.5 and ¼ 0.80, the final reaction is 16% of X1. Note here the absolutecorrespondence with the phenomenon of ‘relaxation’ (see Section 1.2.2.2). Indeed, theabove expression is identical to the one derived in Section 1.2.2.2 for the final stress ofconcrete.It is not unusual for the settlement to be imposed on the structure gradually, starting

from a zero value. This occurs in cohesive soils (e.g. clay), in which settlements developgradually (see Section 17.1.2). It is obvious that, in this case, no initial reaction X1 exists.However, a reaction is being gradually developed, acquiring its final valueX1 when theprocess of settlement growth is completed (Figure 5.13). If accepted that this processfollows the time function of creep deformation, then the above compatibility conditionmay be written as

X1 F11(1þ ’)¼

Considering again, as previously, the value X1¼ /F11, which corresponds to the initialreaction resulting from an instantaneous imposition of a settlement , it follows that

X1 ¼ X1 1

1þ ’In this case the developing reaction is 33% of X1.

204

Bending response due to instantaneous settlementDeveloped bending response due to creep

∆X1

δ

Gradually imposed settlement

Gradually developed reaction which finally causes δon the statically determinate beam

Figure 5.13 Relieving creep effect on the state of stress due to a gradually imposed settlement


The above results are generally valid for any type of imposed deformation and for anydegree of redundancy as long as the creep is identical in all the members in the structure.Thus, for an instantaneous imposition of a deformation, a reduction factor applied to theinstantaneously developing redundant forces is

1 ’

1þ ’whereas for a gradual imposition of a deformation the ratio of the developing sectionalforces to those corresponding to instantaneous imposition is equal to [1/(1þ ’)].It may be understandable that in all the above cases, the redistribution of stresses

caused by creep under service conditions are statically equivalent to superposing aself-equilibrating state of stress to the initial one and therefore does not affect theultimate state checks, according to the static theorem of plastic analysis (see Section6.6.2).

5.5.2 Change of structural systemFor the construction of continuous beams in bridge engineering with a certain number ofequal spans, it is customary to use precast girders for each span that are first treated assimply supported beams and then treated with their continuity restored in the regions ofthe supports by means of a monolithic connection (Figure 5.14). It is clear that at theinstant of connection the bending moment diagram for the system is composed of the

205

Initial bending diagramFinal bending diagram

g g

F10

F11

∆X1

∆X1 = 1

g · L2/8

2 · F10: Initial relative rotation (2 · F10) · ϕ: Increment of relative rotation in the free condition

Eliminated by the developed bending moment

Figure 5.14 Redistribution of the bending moments due to the change of structural system

Continuous beams

diagrams for the simply supported beams. However, as time passes and due to the creepproperties of concrete, bending moments are developed at the supports, and these tendto reach the values that correspond to a continuous system initially constructedmonolithically, with a consequent corresponding reduction in span moments.The reason for the development of this state of stress, as shown by the example of a

continuous beam with two equal spans in Figure 5.14, is that the initial rotation F10 atthe two ends of the internal supports will not remain constant over time in the freesystem, but will increase until the final time phase by the value F10 ’.This increase, however, is prevented by the restored continuity, through the gradually

developing moment X1. This moment considered alone causes at each end a rotationwith a final value X1 F11(1þ ’), where F11 is, as previously, the instantaneousrotation of each end due to X1¼ 1.The compatibility of deformations requires that the relative rotation at the internal

support is zero:

2 F10 ’þX1 2 F11(1þ ’)¼ 0 (see Figure 5.14)

Introducing the moment value X1¼2 F10/2 F11 that corresponds to the mono-lithically constructed system (due to the self-weight), it results that

X1 ¼ X1 ’

1þ ’For the values of ’ and already used in previous sections, it is found that the finallydeveloped moment at the support is 83% of the corresponding ‘elastic’ moment. g

A similar situation also exists when, in order to construct a bridge of a large span over adifficult natural obstacle (deep canyon, river, etc.) the two half parts of the bridge arepurposefully constructed as cantilevers, as mentioned earlier in this chapter, and thenconnected together (balanced cantilever method). In such a case, which is shown inFigure 5.15, it is clear that, as soon as continuity is restored at the connection, thediagram of bending moments for the system is composed of that for the two cantileveredparts. However, at the point of connection a bending moment develops in time, whichbrings tension to the bottom fibres and acquires a significant percentage of the momentcorresponding to the ‘monolithic’ system. It is obvious that, for the maintenance ofequilibrium, the moments at the supports are correspondingly decreased.The analysis is precisely the same as in the last problem (see Figure 5.15). The relative

rotation (2 F10) of the two ends in a hypothetical free state, up to the final instant,would increase by (2 F10 ’). This, however, is prevented by the gradual increase inmoment X1, which causes a relative rotation X1 (2 F11) (1þ ’). Thus, thecompatibility equation remains exactly the same as previously, and the result is identical.The ratioX1/X1 of the developing bending momentX1 to the span moment X1 thatwould be developed in a monolithic system is equal to ’/(1þ ’). Of course, since,until the two free ends approach each other, the concrete has already reached someage, the value of factor ’ is now smaller (see Section. 1.2.2.1). Therefore, a smallerpercentage of the span moment of the monolithic system develops, this being of theorder of about 60%. It is understandable that such a bending moment should be

206


taken up by an appropriate layout of tendons, placed on the bottom side (‘midspantendons’). g

The last problem considered here is again concerned with bridge engineering (Menn,1990). The structure considered is an a posteriori cast-in-place concrete deck plate onprecast prestressed concrete girders, already placed parallel to each other at equaldistances (Figure 5.16). Over the course of time, the initially unstressed plate willdevelop under its total self-weight a stress state due to creep of the beam. Obviously,if the beams are made of steel — i.e. they do not exhibit creep — this state of stresswill not be developed.The redistribution of stresses is due to fact that the top fibres of the beam will

continue to shorten even after the final casting of the top plate, and, therefore, thestresses will drift, with a proportional shortening — i.e. compression stress state — ofthe bottom fibres of the top plate as well. The active cross-section consists then ofthe beam and a certain ‘effective’ width of the top plate.

207

Bending diagram M just after the junctionBending diagram M just after a long time period

∆X

∆X

∆X: Developed bending moment after the junction

Eliminated by the deformational contributionof the developed bending moment

∆X · (2 · F11) · (1 + µ · ϕ)

2 · F10: Initial relative rotation2 · F10 · φ: Free increment of relative rotation

∆X = 1

∆X ∆X

F10

F11

Figure 5.15 Redistribution of the bending state of stress in a free cantilever construction

Continuous beams

At the moment of restoring of the full section, the total self-weight is taken only bythe beam. The beam is, therefore, subjected to the total self-weight and prestressingforces. In the characteristic midspan section, there is a compressive force NB

0 on itscentroid and a moment MB

0 that causes tension in the bottom fibres. As time passes,a compressive force NPl develops in the top plate, as already mentioned, which ofcourse will be accompanied by a corresponding change NB in the compressive forceof the beam and a change MB in its bending moment (see Figure 5.16).Due to the much smaller bending stiffness of the top plate compared with that of the

beam, it can be assumed that its developing bending MB will, in the course of time,become negligible.These last three unknowns are determined through the compatibility of deformation,

and through the equilibrium of the cross-section. Compatibility of deformation meansthat the variation in the strain "B in the extreme top fibre of the beam is identicalto the strain "Pl in the bottom fibre of the plate, while equilibrium of forces in thecross-section imposes that the variations in the sectional forces in the two partsconstitute a self-equilibrating system. Thus

"B0 ¼ NB0

EB ABþ MB

0

EB IB eB

and, given that both "B and "Pl are gradual changes,

"Pl ¼ NPl

EPl APl ð1þ Pl ’PlÞ

"B ¼ "B0 ’B þ NB

EB ABþ MB

EB IBeB

! ð1þ B ’BÞ

The differentiated values of and ’ are due to the fact that the two concrete parts havedifferent ages at the time when creep is considered to start.The previously stated conditions for the determination of the three unknowns NPl,

NB and MB are written as:

"B¼"Pl

208

a a

Apl ∆εPl

AB ∆εB

eB

The variations in the internal forcesare self-equilibrating

N0Pl = 0

N0B M0

B

∆NPl

∆M B∆NB

Figure 5.16 Redistribution of the internal forces in a concrete composite section


NPl¼NB

MB¼ a NPl

where a is the distance between the centroids of the plate and the beam cross-sections.As expected, it is found that the distribution of normal stresses on the full cross-

section tends towards the one that would be developed if the total cross-section wasuniformly cast (and prestressed) from the beginning.

5.6 Composite beams

5.6.1 Basic shear functionThe concept of using a composite beam to transfer a transverse load over two supportslies in the utilisation of a concrete zone of sufficient thickness to carry the compressiveforce, and of a steel beam to carry the tensile force, both forces arising from the bendingaction of the so-formed composite beam (Figure 5.17).As examined in Section 4.1.1, this can be accomplished only under the condition that

the horizontal — and consequently also the vertical — shearing cooperation betweenthe two parts is ensured. Thus, if the longitudinal equilibrium of the upper part of thebeam extending from the free end up to the midlength of the beam span is considered,

209

max M

max Z

M = 0

M = 0 max M

max D

[M ]

Figure 5.17 The equilibrium between internally developed actions

Continuous beams

a significant compressive force at the right-hand end will be ascertained, which can onlybe balanced by the development of shearing forces (stresses) on the top flange of thesteel beam. This is accomplished by attaching (welding) special steel connectors, inthe form of headed studs, to the top flange of the steel beam at the steel—concreteinterface. The connectors can transmit the horizontal forces, with sustainable stressesin the concrete flange (see Figure 5.17).Moreover, in the regions of the internal supports of a composite continuous beam, the

top concrete flange is under tension, and accordingly an appropriate reinforcement mustbe placed near the top surface of the flange. It is obvious that, in the same region, theshear connectors located between the support and the point where the momentvanishes are supposed to offer (and take) the tensile force developed by the reinforce-ment at the support section (see Figure 5.17).The aim of the subsequent elastic analysis is to determine: first, the normal stresses

developed in the cross-section; and, secondly, the horizontal shearing forces borne bythe shear connectors.

5.6.2 Bending behaviour of composite beamsIn order to determine the normal stresses due to bending, it is essential to consider anequivalent section made solely from steel, by dividing the concrete section area Ac bythe ratio n¼Es/Ec, also known as the ‘modular ratio’ (see Section 4.2.2.1). This isexplained by the fact that both a concrete cross-section Ac and a steel cross-section Ac/ndevelop the same force for the same imposed strain ", as Ac " Ec¼ (Ac/n) " Es

(Figure 5.18).IfAst, As andAc are the cross-sectional areas of the steel beam, the reinforcement and

the concrete, respectively, and if z are the corresponding centroidal distances from thebottom surface of the steel beam, then the distance zm of the centroid of the total cross-section will be

zm ¼ 1

Am

½Ast zst þ As zs þ ðAc=nÞ zc

210

As

Ast

Ac

Mh

z

σst

σc

TensionStresses due to the bending moment

n = Es/Ec

Compression

Ac/n: Statically equivalentsteel section

Figure 5.18 Composite cross-section


where

Am¼ [AstþAsþ (Ac/n)]

Moreover, if Ii0 is the moment of inertia of each part of the cross-section, by employingSteiner’s theorem the moment of inertia Im of the total cross-section with respect to thecentroidal axis will be

Im¼P

Ii0þP

Ai (zizm)2

It is obvious that the centroidal axis also constitutes the neutral axis of the total cross-section.Then, the stresses will result from the simple bending formula; for the concrete part

they should be divided by n. Thus, for a bending moment M causing tension at thebottom fibres

st ¼M

Im zm (tension)

s ¼M

Im ðzs zmÞ (compression)

c ¼1

nMIm

ðh zmÞ (compression)

5.6.3 Shear transferThe fact that composite beams are necessarily constructed in two stages is quiteimportant. First, the steel beam is set in place, and then the top concrete plate iscast. In the concrete-casting phase the steel beam is acting alone, simply carrying theweight of the concrete. In this phase it must be ensured that there is no noticeabledeformation of the beam under its own weight and that of the concrete, possibly byproviding suitable intermediate temporary supports. Only after the concrete hashardened are the intermediate supports removed, and any further loading can thenbe borne by the composite steel—concrete system. The developing stresses can thenbe calculated according to the above formulae.Naturally, an essential condition for the operation of the composite system is the

appropriate take-over of the horizontal shearing flow. This consists, as previouslyexplained, of the horizontal force vel per unit length being transmitted between the twomaterials at the steel—concrete interface, and is equal to the product of the consideredintersurface width (i.e. the width of the top flange) and the corresponding shearing stress:

vel ¼V ScIm n

The value Sc is the first moment of inertia of the concrete cross-section with respect tothe neutral axis of the composite cross-section previously determined, so that the value(Sc/n) corresponds to the assumed equivalent ‘steel section’.

211

Continuous beams

It should be checked, of course, that the above force, which varies along the length ofthe beam according to the corresponding value of the shear force V (Figure 5.19), can betaken over by the corresponding connectors located in the appropriate positions. In thecase where there is no need to use intermediate temporary supports, the aforementionedlongitudinal shearing forces will be developed only for those loads applied after theconcrete hardening.The resistance that a shear connector can offer is constituted by the shear resistance

VRs developed in its shaft and the local resistance VRc of the concrete, which receives byreaction the same force through the connector shaft, mainly through its specially formedhead, as previously mentioned. Thus (according to indicative regulatory requirements)

VRs¼ 0.7 fuD AD

where fuD is the tensile strength of the connector material and AD is the cross-section ofthe connector (¼ d2/4, where d is the diameter of the connector), and

VRc ¼ 0:35 d2 ffiffiffiffiffiffiffiffiffiffiffiffifc Ec

pNaturally, the minimum of these two VR values is the effective one.The strength of the stiffeners is checked using the condition that the total longitu-

dinal shear force V1 ¼Ðvel dx between points with zero and maximum bending

response (see Figure 5.17) is safely taken by the locally available number m ofconnectors:

V1=m VR=R ðR ¼ 1:1Þ

5.6.4 Construction stages — stress controlOnce the shearing cooperation between the two parts of the composite beam has beenestablished, it is necessary to understand the mechanism of the development of stress,

212

vel

The distribution of the longitudinal shearing forces is identical to that of the transverse ones

[V ]

Figure 5.19 The transfer of the longitudinal shearing forces


both during the construction phases and thereafter during service. As noted previously,the construction process plays an important role, and must take into account the creepproperties of the compressed concrete.When the finally formed composite beam is loaded with a permanent action, the

stresses developed due to this loading will change over time. As shown in Section1.2.2, the elasticity modulus of concrete decreases over time to the value Einit and,accordingly, the ratio n increases (see Section 5.6.2). In order to understand theimplication of this, three cases of constructing a simply supported beam are examined.

. A steel beam is without any intermediate supports, loaded by its self-weight and the weightof the still fresh concrete (Figure 5.20). Clearly, even after the concrete has hardenedand shear cooperation has been established, the concrete bears no compressivestresses and, provided that no additional load is applied, this stress state will notchange over time. If, however, an additional permanent load is applied immediatelyafter the hardening of the concrete, this will cause compressive stresses in theconcrete, which will decrease over time. After a long time the state of stress willbe equivalent to the superposition of the initial one (where only the steel beamwas stressed) and the additional one, which is of course calculated on the basis ofthe ratio n¼Es/Einit.

. The provisional support of the steel beam at two points. In the concrete-casting phase, thesteel beam acts as a three-span continuous beam under its self-weight, and the weightof the fresh concrete, developing reactions at the internal supports equal to A(Figure 5.21). After the concrete has hardened, the effective section of the beamchanges, but the reactions A do not change because the stiffness ratio is maintained.The removal, afterwards, of the intermediate supports is equivalent to the applicationof the forces A, with the opposite sense to the previous ones, on the restored simply

213

Concrete after hardening remains free of stresses

Figure 5.20 Restoration of the top flange of the beam with concrete

The compressive forces of concrete keep reducing with time

Reactions remain the same after hardening of concrete

Insertion of temporary supports

Removal of temporary supports

A A

A A

Figure 5.21 Provisional support for the restoration of the function of the top flange

Continuous beams

supported beam. This means that, according to the above, the resulting stresses(compressive in the concrete and tensile in the steel) will change over time, thefinal values being determined on the basis of the ratio n¼Es/Einit. Clearly, eachadditional permanent load applied after the hardening of the concrete will causestresses that can be calculated in the same way, whereas a transient live load willcause stresses that can be calculated on the basis of the modular ratio n¼Es/Ec.

. Providing the concrete with an initial compression force (i.e. prestressing). The steel beamis provided with two temporary supports at the points where the reactions A develop.Just before concreting, an equal upward displacement is imposed on each support. It isobvious that in this way the moments at the supports of the steel beam are increased,as are the corresponding reactions which are increased by AP (Figure 5.22). It isunderstandable that, after concrete casting and until concrete hardening, the totalreactions at the intermediate supports will be exactly the same and equal to(AþAP), where A includes also the beam reactions under the self-weight of theconcrete. No stresses have yet developed in the concrete. Next, the removal of theintermediate supports can be treated as previously, where the permanent forces(AþAP) are acting with opposite sense on the restored simply supported compositebeam (see Figure 5.22). In this way, an additional compressive stress is offered to theconcrete, which is irrelevant due to the permanent loads acting on the beam. Thethus caused stresses will again change over time and reach final values based onthe ratio n¼Es/Einit, with final compressive stresses in the concrete that are smallerthan the initial ones.

As it may be seen, in the last two cases, despite the various construction measures, noadditional final sectional forces are introduced, besides those caused by the self-weight g.Therefore, the normal stress diagram in each case is always equivalent to a bendingmoment (g L2/8) and a zero normal force.

214

Concrete free of stress also after hardening

Reactions unchangedafter concrete hardening

Due to self-weight

Due to imposition of displacements

Additional loading beyond self-weightAdditional compressive concrete stresses

Insertion of intermediate supports: imposition of upward displacements

Removal of intermediate supports

A A

AP AP

A + AP A + AP

Figure 5.22 Prestressing of the top flange via temporary intermediate supports


5.6.5 Temperature changeThere are two good reasons for considering the state of stress that develops in acomposite beam due to a temperature difference of its two constituent parts, i.e.concrete and steel. The first reason is that, due to the smaller thermal conductivity ofthe concrete, an exterior temperature change causes a short-term differentialtemperature between the steel and the concrete. The second reason is the concreteshrinkage, which is prevented to some degree by the steel and which, as mentionedin Section 1.2.2.1, is equivalent to an imposition of a fall in the temperature of theconcrete.The implications for the response of a composite beam of a temperature difference can

be assessed on the basis of equilibrium and compatibility considerations (Figure 5.23).For a fall in temperature of T in the concrete, for example, the concrete is left freeto shrink by "c¼ aT T, and a tensile force NT is then applied to restore thesection (Ac/n) of ‘concrete’ to its initial state:

NT¼ ("c Es) (Ac/n)

This force causes a corresponding tensile stress in the concrete, while leaving the steelpart unaffected.Given, however, that no external force is applied to the section, an equal force with

opposite sense (compressive) must be applied at the same point on the total (composite)section. This force, referred to the centre of gravity of the section, is accompanied by abending moment that causes tension in the bottom fibres, equal to

MT¼NT acwhere ac is the distance between the centre of gravity of the composite section and theconcrete part.The superposition of the two states contains the tensile force NT acting on the

concrete alone, and the compressive force NT and the moment MT acting on thecomposite section. The combined effect is the self-equilibrating stress diagram shownin Figure 5.23.

215

∆T N∆T

M∆T

ac

M∆TM∆T

N∆T

N∆T

ϕ ϕ

Simply supported beam: constant bending response, development of curvature

Equivalent actionsFree shortening Final stresses

+

+

–

Figure 5.23 The state of stress in a simply supported beam due to a temperature difference

Continuous beams

This moment (as well as the above stress diagram) is constant along the length of thebeam, and causes a constant curvature 1/r¼MT/(Es Im), as well as an equal rotation ’at both beam ends:

’ ¼ L

2 r ¼MT

2 Es Im L

(The same result is, of course, also obtainable via the principle of virtual work.) g

With regard to a fixed-ended beam — subjected only to this temperature change — theabove state of stress of a simply supported beam must have superposed on it two appro-priate equal and opposite moments applied at the beam ends, so that the above rotationsvanish. These moments are obviously equal to MT and, consequently, they eliminatethe curvature of the beam, leaving it straight (Figure 5.24). The state of stress of afixed-ended beam is thus composed of the superposition of the force NT acting onlyon the concrete part, and the compressive force NT acting on the composite section.The stresses that result are uniform in both the concrete and the steel section, andare given by, respectively,

c ¼NT

Ac

NT

n Am

¼ T T Es Ast

n Am

(tensile)

st ¼NT

Am

¼ T T Es Ac

n Am

(compressive)

Thus, with regard to the internal spans of a continuous beam, this last result constitutes asatisfactory approximation. g

With regard to a fixed simply supported beam, the support moment, which must cancelthe aforementioned rotation ’, is obtained according to the principle of virtual work,and is equal to 1.5 MT. This moment is superposed on the bending momentdiagram of the simply supported beam and leads to the diagram shown in Figure 5.25.The stresses that develop in the section are calculated through superposition of the

imposed axial forces NT, as above (see Section 5.6.2). Thus, for a moment Mcausing tension in the bottom fibres, the above stresses of the fixed-ended beam aresuperposed, for the concrete, on the compressive stress:

c ¼1

nMIm

ðh zmÞ

216

Fixed-ended beam: null bending response, null curvature

Additional actions required to eliminate the rotations

M∆T M∆T M∆T M∆T

Figure 5.24 The effect of a temperature change on a fixed-ended beam


For the steel they are superposed on the tensile stress:

st ¼M

Im zm

This state of stress can approximate to a satisfactory degree the one corresponding to theend spans of a continuous beam.

5.6.6 Continuous beam behaviour — prestressingAs pointed at the beginning of this chapter, in the case of a continuous beam and at theregion of internal supports, the concrete is under tension and needs to be reinforced.However, this does not prevent the concrete from cracking.The calculation of the stresses due to a bending moment causing tension in the top

fibres is performed, according to Section 5.6.2, by considering the total cross-sectionwithout the concrete part, i.e. Ac¼ 0. It is obvious that in this case the neutral axiswill be within the steel profile. The possible case of a cantilever is treated in the sameway.Checking of the shear connectors can again be done, as for the case of the simply

supported beam, on the basis of the relationship

Vl/mVR/R

where in this case the longitudinal shearing force Vl is equal to the maximum forcedeveloped in the reinforcement at the corresponding internal support position(Figure 5.26). g

Cracking under service conditions can be prevented if an analogous constructionprocedure to the one described in Section 5.6.4 is followed, in order to offer theconcrete an initial compressive force (prestressing). More specifically, in the exampleof the two-span beam shown in Figure 5.27, an upward shift is imposed on theinternal support before concrete casting. As a result, the beam develops, besides thebending moment due to the self-weight of the fresh concrete, an additional triangularbending diagram causing tension in the top fibres, resulting from the ‘statically

217

Additional moment for the elimination of rotation

M∆T

M∆T

M∆T1.5 · M∆T

0.5 · M∆T

Fixed simply supported beam: variable bending response

Figure 5.25 The effect of a temperature change on a fixed simply supported beam

Continuous beams

determinate’ simply supported beam loaded with an upward force equal to X0. Aftercasting, hardening of concrete and restoration of the composite section, while theconcrete remains free of stress, the same shift is imposed downwards on the internalsupport, so that a uniform level applies for all support points.It is clear that, in this second phase, bending moments opposite to those in the first

phase will be developed along the whole length of the beam, causing compression of theconcrete plate. These bending moments will be greater than those in the first phase, asthe composite section is now strengthened with respect to the previous one, where onlythe steel part existed (see Section 3.2.5). In fact the bending moments can be so largethat the resulting compression cancels the corresponding tension due to the additionalpermanent and live loads. Obviously, the resulting (triangular) bending momentdiagram is due to the downward reaction X1 (being greater than X0), acting on theprimary system of the simply supported beam (see Figure 5.27).The development of compressive stresses in the concrete plate will cause creep, which

leads directly to an undesired decrease in the plate compression stresses due to thedecrease in reaction X1. Note that, while the concrete plate is stressed only because

218

Total force Vl (=D) Total force Vl (=Z )

(M = 0) (max M) (min M ) (M = 0)

D Z = fs · As

Figure 5.26 The equilibrium of the internal forces in characteristic positions

Bending response due to imposition of displacement

Concrete free of stress before imposition ofdownward displacement

Self-weight of fresh concrete plus steel beam

Bending moment diagram causescompression of the concrete plate

δ δ

X0 Reaction developed

X1 Reaction developedX1 > X0

Reaction dueto self-weight

(X1 – X0): Reaction due to impositionof the two consecutive displacements

Figure 5.27 The introduction of prestressing in the top flange of a continuous beam


of X1, which decreases over time, the steel beam is stressed by the superposition of thecontinuous beam action under the total self-weight and the application of a downwardforce X1X0 on the simply supported system, which constitutes the additional reactionof the continuous beam due to the two imposed phases. However, this force decreasesover time following the decrease in X1 (see Figure 5.27).This decrease can be estimated using the analytic approach described in Section 5.5.1

via the force method, as this procedure is always adequate for dealing with creepproblems due to a redistribution of forces, by following the compatibility of deformations.Thus, the developed force X1 and its variation X1 are applied to the ‘staticallydeterminate’ simply supported beam for the whole time period examined (theoreticallyinfinite) (Figure 5.28). The resultant deformation due to the two actions must obviouslybe equal to . According to Section 1.2.2, the final shift due to X1¼ 1 is denoted by Finit11 ,which is obtained using the ratio n¼Es/Einit, while the final shift due to X1¼ 1 isdenoted by Fdif11 , and this is obtained using the ratio n¼Es/Edif . Thus (see Section 5.5.1)

¼ X1 Finit11 þX1 Fdif11

It should be noted that the developed reaction X1 is initially caused by the ‘instanta-neous’ imposition of downwards, or — in other words — the application of force X1

in the simply supported beam instantaneously causes the shift . The value of X1 isgiven by

X1¼ /F11

where F11 is obtained the basis of the modular ratio n¼Es/Ec. It is clear that

F11 < Fdif11 < Finit11

The two last equations lead to the result that the final value (X1þX1) of the reactionX1 is equal to

ðX1 þX1Þ ¼ X1 1 Finit11 F11

Fdif11

!(reduction)

Thus the reduction factor of the initial compressive stress of the plate (due toX1) is equal to

1 Finit11 F11

Fdif11

219

(t = ∞)

(t = ∞)

∆X1 = 1

X1 = 1

F1d1if

F i1n1it

Figure 5.28 The effect of creep on the development of the final stress state

Continuous beams

The final bending moment of the steel beam is obtained by superposing on its continuousbeam action under the gravity loads a concentrated downward force on the ‘simplysupported’ system (see Figure 5.27) that is equal to

X1 1 Finit11 F11

Fdif11

! X0

It is clear that the required prestressing of the concrete part should be calculated for theservice condition on the basis of the above reduction factor.

5.6.7 Plastic analysisAlthough elastic analysis is commonly used for the design of composite beams, it is usefulto consider some basic characteristics of their plastic design. First, it should be pointedout that the plastic design of composite structures is not affected at all by any stressredistribution that may occur under service conditions. g

With regard to the simply supported beam, exceeding the bending resistance Mpl of asection means collapse of the structure. The computation of the bending resistanceMpl requires the determination of the neutral axis position, so that the compressivelimit force developed in the concrete above the neutral axis is equal to the limittensile force in the steel below it. At this point the stresses are considered to havereached their maximum limit value in all fibres of the section, that is fc for concreteand fy for steel (Figure 5.29).Assuming that the neutral axis lies within the concrete section at a distance x from

the extreme fibre with an effective width beff , the compressive force in the section isequal to

D¼ fc beff x

220

beff

D

Z

Z

Support region

D

Span region

Mpl

Mpl

x

a

a

fy

fy

fsy

fc

fy

Figure 5.29 Plastic analysis of a composite section


The tensile force Z results from the entire steel sectional area As:

Z¼ fy As

The distance x is determined through the relationship D¼ Z, and hence, the position ofthe neutral axis is also verified. In the uncommon case where the resulting value of x ishigher than the thickness of the concrete plate, the procedure is repeated in the samemanner. The plastic moment Mpl is given by

Mpl¼ Z awhere a is the distance between the points of application of the forces D and Z.With regard to shear, the ultimate shear force Vpl is of course carried only by the cross-

section Aw of the steel web and, as described in Section 4.1.1, it is equal to

Vpl ¼fyffiffiffi3

p Aw

For a continuous beam — and analogously the cantilever — the computation of thebending resistance of the section at the internal supports where the concrete iscompletely cracked requires the determination of the neutral axis position in a purelysteel section. In this case, the neutral axis lies within the steel section, so that thesum of the tensile forces in the reinforcement and the corresponding top part of thebeam is equal to the compressive force of the bottom part. It is obvious that the limitstress fy prevails along the whole height of the section. The plastic momentMpl is calcu-lated accordingly (see Figure 5.29). g

With regard to the check of the shear connectors, the only difference from the elasticanalysis (see Section 5.6.3) lies in the determination of the total longitudinal shearforce Vl between points with zero and maximum bending stresses (i.e. at plastichinges). In regions where the bottom fibres are in tension due to bending, the shearforce Vl is equal to the tensile force of the steel beam fy Ast , provided that theneutral axis of the section lies within the concrete section. In the uncommon casewhere the neutral axis lies even lower, the force Vl is obviously equal to the compressiveforce of the concrete section fc Ac. Moreover, it is obvious that in regions where the topfibres are in tension due to bending, the force Vl is equal to the reinforcement tensileforce fsy As. The corresponding check of the available m shear connectors placed inthe examined region is performed through the requirement

V1m VR

ReferencesMenn C. (1990) Prestressed Concrete Bridges. Basel: Birkhauser Verlag.

221

Continuous beams

6

Frames

6.1 OverviewA frame is a skeleton structural system consisting of beams and columns designed to carryloads of various nature (gravity as well as horizontal loads) and to transfer them to theground. In this chapter, plane frameswill be considered, which are loaded only in their plane.Frames are either designed to cover or bridge a specific span, characterised as

‘one-storey frames’, where the girder of the frame may not always be a straightmember, or may constitute basic elements of a space system carrying superimposedfloors in a building, and therefore the girders placed on more than one level arenecessarily horizontal and straight. In this last case, the member system bearing thefloors is a space system, but in the overwhelming majority of buildings the wholestructural arrangement is such that the space system can be considered as a group ofmulti-storey plane frames connected with each other (see Chapter 15).

6.2 Single-storey, single-bay framesThese usually consist of a horizontal beam (girder) and two columns, which areconnected to the two ends of the beam and are based on the ground.

6.2.1 Design for vertical and horizontal loadsThe simplest form of such a structure is the post-and-beam frame, where the columns arefixed to the ground and the beam is simply supported on their tops. Thus, for verticalloads the beam acts as a simply supported beam, transferring its vertical reactions tothe column tops, without inducing bending moments in them (Figure 6.1).It is obvious that if two cantilevers are set up on the left and the right side, the

bending moment of the girder will be decreased, thus resulting in a statically morefavourable structure. If, indeed, the length of each cantilever is about 35% of thelength of the central span L then the span and support moments for a load q will beequal to q L2/16, which is half of those for the simply supported beam. In the casewhere a cantilever is placed only on one side, the length of the cantilever must beequal to 41% of the central span in order to have again equal span and supportmoments with a value of q L2/11.9.For a horizontal load W acting at the level of the girder, it is obvious that the beam

remains unstressed in both bending and shearing (see Figure 6.1), while the whole load


is carried equally by the two columns (provided that they have the same stiffness, seeSection 3.2.10), thus acting as two cantilevers with a horizontal load equal with W/2applied at their tops.Columns usually connect monolithically to the girder. Then, the frame is formed

connected either to the columns hinged to their foundation or fixed (Figure 6.2).The foundation is considered as a solid massive concrete element which is baseddirectly on the ground, being rigidly connected to it and aiming, through its largerdimensions, to transfer much smaller stresses to the ground than those developing atthe column base. The static conditions for their design will be examined in Chapter 17.In the case of a hinged connection at the foundation, the frame is either statically

determinate, if it has an additional hinge (usually at the midpoint of the girder), or ithas its girder continuous, in which case it is one-time statically indeterminate. In thecase where the columns are fixed to the foundation and the girder is continuous, thesystem is statically three-times indeterminate. g

Since it is useful to examine the bending of the above-considered frames, also in correlationwith the so-called pressure line, the following should be recalled (see Section 2.2.7).For any vertical load q(x) that must be taken up by a structure transferring the loads

only through axial forces, i.e. without bending (and shear), at two points on the groundlying on a horizontal line at a distance L and enclosing its projection, there exist infinitesolutions regarding the axis of the structure. All these solutions constitute affine curvesof the bending moment diagram M0(x) of a simply supported beam having length L andloaded by q(x) (Figure 6.3). It is obvious that q(x) can also include concentrated loads or

224

q q

q

L L 0.35 · L0.35 · L

L0.41 · L

hq · L2/8

q · L2/16

q · L2/11.9

Without bending

Without bendingW

V = +(W/2) V = +(W/2)

(W/2) · h (W/2) · h

Figure 6.1 Set-up of post-and-beam frames


even be composed only of such loads. Thus, if y(x) represents the curve which thestructure should follow, then

y(x)¼ k M(x)0

The choice of a value for k means the choice of a specific curve, i.e. a specific pressureline. This value for k is the inverse value 1/H of the horizontal component H of thereaction R of the system, which will of course have the direction of the tangent atthe beginning of the selected curve (see Figure 6.3).

225

Three-hinged frame Two-hinged frame Fixed frame

Figure 6.2 One-bay rigid frames

The selection of a third point establishes the pressure line and the value of k

BABA

The acting compressive force along the pressure line is not constantbut its horizontal component H has a constant value

RARB

A BH = 1/k

H Hy0

M0

x0

y

x

Pressure lines for the given loading

k

S1

S2

∆x

y(x) = k · M0

k = 1.20k = 1.0

k = 0.55

Figure 6.3 Characteristics and determination of a pressure line

Frames

The pressure line ensures at each point x the equilibrium between a correspondingvertical load q x and the two forces S(xx/2) and S(xþx/2) that act tangentiallyat the two ends of the segment x. It is emphasised that the aforementioned axial(compressive) forces of the pressure line do not have a constant value, but they dohave a constant horizontal component equal to H (see Figure 6.3).It is clear that with two support points, the determination of one more point y(x0)

lying on the curve is required in order to determine the curve completely. Then,k¼ y(x0)/M

0(x0), and y(x) is already defined, so that the structure that follows thiscurve will develop only axial forces. The structure will develop compressive forces ifthe point y(x0) lies above the line connecting the supports (AB), and will developtensile forces if it lies below the line. However, in this consideration the second caseis not of much interest (see Figure 6.3).Now, if a hinge is inserted at the position y(x0) and the structure does not follow

the specific curve but takes a different form while remaining a three-hinged frame, thenit will develop bending moments at each point, which are the product of the distance ofthis point from the pressure line and the corresponding axial force acting on it (Figure 6.4).The compressive sense of the axial forces of the pressure line will provide a particularly

useful result, i.e. it will indicate which fibres of the frame are under tension due tobending (see Figure 6.4). g

Regarding the portal frame in Figure 6.5, it can be seen that the placement of a hinge atthe midspan of the girder under a uniform load due to gravity causes tension exclusively atthe outside fibres. The form of the moment diagram is not necessarily favourable. Thegirders are bent with the same moment magnitude (q L2/8) that is developed in thesimply supported beam, and, moreover, the columns also bend with the same moment.

226

m n

m n

Rm

Rn

Rm

Rn

em

em

en

en

y0

x0A B

A B

RA

RB

RA

RB

Mm = Rm · em Mn = Rn · en

The bending moment at each point of the framedepends on its distance from the pressure line

Figure 6.4 Correlation between the pressure line and the developed bending


It is obvious that the form of the pitched frame significantly decreases its bending responsesince it is adapted more to the pressure line. Note that the horizontal component H of thereaction of the frame is inversely proportional to its height; the moment at the joint,however, is independent of the absolute value of the height (see Figure 6.5).Of course, for a horizontal load W acting on the three-hinged frame (Figure 6.5),

without taking into account the fact that the girder also participates in load bearing,the same bending moments are developed as those for the ‘cantilever’ shown inFigure 6.1. The developing moment at the column top is (W/2) h.If the girder is continuous, with regard always to hinged supports (Figure 6.6), the

beam develops, for a vertical uniform load q, moments at its ends that cause tensionon the top fibres. So, the maximum span moment is clearly decreased in comparisonwith that of the simply supported beam, since, as is known, this configuration isalways ‘suspended’ from the points of the end moments.If the stiffness ratio ¼ (IR/L)/(IS/h) of the girder and columns is introduced, the

expression for the exact value of the momentMB at the joint is (Salvadori and Levy, 1967)

MB ¼ q L2

12 3

3þ 2 For common stiffness ratios of the girder and columns (1.0<< 2.50), the bendingmoment diagram of the girder takes a zero value (placement of hinges) at distances

227

Pressure line

G

Pressure line

G

W

H = M0G/(h + f )

H = M0G/h

M0G

M0G/(1 + f /h)

f

h

h

Thrust inversely proportional to height

Independent of the height

W/2 W/2

W · h/2

Figure 6.5 Load-carrying action of a three-hinged frame

Frames

from the two ends in the range 0.13 L and 0.07 L, respectively. An intermediate value of0.10 L allows, through the equilibrium of the imaginary two-hinged extreme part, anapproximate computation of the horizontal component H (¼ 0.045 q L2/h) of thereaction and then of the moment at the top of the column (H h¼ 0.045 q L2). Inany case, the direction of the reaction does not differ substantially from the lineconnecting the hinges. Note that the pressure line passing from the supports as well asfrom the points of zero moments in the girder clearly shows the distribution of thebending response, according to the distance of each point of the frame from it, whilethe prevailing compression force has to increase towards the supports in order to alwaysmaintain a constant horizontal component.The bending response, due to a horizontal load at the girder level, does not differ

from the corresponding three-hinged frame with a hinge in the middle, given theantisymmetric character of the system (see Figure 6.6).In cases where it is not desirable for the ground to take up the horizontal reaction H, a

steel tension member (tie) is inserted in the frame, developing an axial force whichdepends on the ratio of its axial stiffness (Es As) to the bending stiffness (EI) of theframe (Figure 6.7). The structure has again one degree of statical redundancy. It canbe seen that an increase (or reduction) in Es As with respect to EI causes an increase(or reduction) in the tensile force of the member and, consequently, of the bending

228

W/2 W/2

W

H

L

Pressure line

H

h

W · h/2

[M ]

[M ]

q

q

0.045 · q · L2/h

IS

IR

q · L/2

~0.10 · L ~0.10 · L 0.10 · L

Equilibrium

~0.045 · q · L2

(~Direction of reaction)

0.40 · q · L

Figure 6.6 Load-carrying action of a two-hinged frame


moment at the column top, as well as of the axial compression of the girder (see Section3.2.10).In the case where the frame is made of concrete, a redistribution of internal forces

will of course take place due to creep. The concrete will present a decreasing E intime and, as a result, the tensile force in the steel member will increase, with thesame consequences for the frame as above.The change in the tensile force X1 of the tie is determined through precisely the same

expression that gives the change in the state of stress in the heterogeneous frameexample given in Section 5.5.1, which is

X1 ¼ ’ ðFB10 þ X1 FB

11ÞFB11 ð1þ ’Þ þ FSt

11

All the underlying deformation magnitudes arise according to the method of forces,introducing the force X1 of the tie as the statically redundant unknown.Of course, if the tie is prestressed with the force corresponding to the horizontal

reaction of the two-hinged frame, then no change in the whole state of stress willtake place, since the frame will behave as a statically redundant system with hingedsupports and uniform creep properties (see Figure 6.7 and Section 5.5.1). g

In the case of a fixed frame (Figure 6.8), the moment MB at the ends of the girder is alittle higher than the one corresponding to the two-hinged frame, that is (Salvadori andLevy, 1967),

MB ¼ q L2

12 2

2þ

The points of zero moments are now at larger distance from the ends, that is between0.15L and 0.09L, relatively to the values 1.0< a< 2.50. As an intermediate distance, itmay be considered roughly 0.12L. Moreover, since for a uniformly distributed verticalload the column top is simply rotating without shifting, it follows that the moment atthe top of the column will be double than the one at the base (see Section 3.3.3), andconsequently, the moment diagram of the column is zero at 1

3 of the height up fromthe base.

229

P P = H

EI

EsAs

Steel tie

Increase in the tie axial force due to creep Prestressing with the thrust force of the hinged frameleaves the bending moments unchanged

Bending response before creep

After creep

EI

Figure 6.7 Frame with a tension member — the creep effect

Frames

From the equilibrium of the extreme imaginary two-hinged part, it can again be realisedthat the direction of the force transferred through the bottom hinge does not significantlydiffer from the line connecting the hinges, which has, of course, a slope smaller than thatpreviously encountered in the two-hinged frame. Now, given the constant value (q L/2) ofthe vertical component of the force transferred through the bottom hinge, it can beconcluded that its horizontal component, which is obviously equal to the horizontalcomponent of the reaction of the frame, is higher than that of the two-hinged frame.From the equilibrium of the above extreme two-hinged part, H¼ 0.079 q L2/h, and,for the moment at the top of the column, H (2 h/3)¼ 0.053 q L2. The moment atthe fixed support is MV¼H (h/3)¼ 0.026 q L2. In this case, the correspondingpressure line can again be considered to be passing through the column hinges, as wellas the hinges in the girders, as shown in Figure 6.8.In the case of a fairly deformable soil, the foundation receiving this last moment

rotates, and, as a result, the point of zero moments — i.e. the imaginary hinge — shiftsdownwards, approaching the base of the column. The frame then behaves more like atwo-hinged one, causing a moment reduction at the end of the girder and, accordingly,an increase in the span moment.A horizontal load W acting at the level of the girder leads to a moment diagram, which

becomes zero at midspan due to antisymmetric conditions, as well as at points nearlythree-eighths of the column height from top. Thus, the top part of the structure acts asa three-hinged frame (see Figure 6.8). The horizontal components of the forces transferred

230

W/2

H

H HH

L

W/2

W

W/2

W/2

3 · h/8

3 · W · h/16

5 · W · h/16

~0.026 · q · L2

0.079 · q · L2/hIS

IR

Pressure line

q · L/2

0.38 · q · L

(~Direction of reaction)

~0.053 · q · L2

~0.12 · L ~0.12 · L 0.12 · L

Equilibrium

q

q

2 · h/3h

Figure 6.8 Load-carrying action of a fixed frame


through the column hinges, and consequently the reactions of the frame, are equal to W/2,while the moments at the column base are equal to (W/2) (5 h/8)¼ 5 W h/16, and themoments at the column top are equal to (W/2) (3 h/8)¼ 3 W h/16.In the case of a fairly deformable soil, the imaginable column hinges will be shifted

downwards, and, as a result, this latter moment at the top will increase, approachingthe value W h/2 of the two-hinged frame, while the moments at the fixed base of thecolumn will decrease, approaching the value (W h/2) of the two-hinged frame, whilethe moments at the fixed column base will decrease.

6.2.2 Lateral stiffnessIn addition to the developing state of stress in the frame, also of basic importance is theassessment of its lateral stiffness, that is, the resistance that the frame presents to theimposition of a specific (unit) horizontal displacement at the joint, or — to put itanother way — of its deformability (or flexibility), that is, of the produced displacementdue to a specific (unit) horizontal force at the joint. As explained in Section 2.3.8,stiffness and flexibility are mutually opposing concepts. If represents the displacementunder a force P and k is the stiffness, then, always, ¼ P/k.In the case of a frame with pinned supports, which is subjected to a horizontal force P

(Figure 6.9), a shift is developed according to the expression

¼ P h3

EIS 2þ 1=

12

corresponding to the following expression of stiffness k:

k ¼ EISh3

12

2þ 1=

It should be noted that in this shift the deformability of the columns plays a more signif-icant role than that of the girder, which enters the expression only via the ratio . Thestiffness kS of the columns in the case of a very stiff girder appears as the stiffness of reversedcantilevers, that is, kS¼ 3 EIS/h

3. Thus, the applied force P is opposed by the total stiffnessof the columns, that is, 2 kS¼ 6 EIS/h

3, and, therefore, ¼ P/kS¼P h3/6 EIS. Thisresult corresponds to an ‘infinite’ value of , and constitutes the maximum limit for k.It is clear that, by the reduction of the girder stiffness, this latter shift will be increaseddepending on the ratio .A fixed frame under a horizontal force P, according to the above, ‘exhibits’ its

imaginary column hinges at roughly five-eighths of the column height (see Figure6.9). Thus, the horizontal shift of the girder consists of the shift of the two-hingedframe with three-eighths of the height plus that of the top of the ‘cantilever’ bottomparts of the columns with height (5h/8), loaded with a force (P/2). Based on theabove, the resulting horizontal shift is roughly one-quarter of the previous one,

ffi P h3

EIS 2þ 1=

48

indicating a quadruple stiffness compared with the two-hinged frame (see Figure 6.9).

231

Frames

Once again, the deformability of the columns plays the main role. In the case of a verystiff girder, the ends of the columns do not rotate, and hence rigidity is kS¼ 12 EIS/h

3. Itfollows that

¼P/2 kS¼ P h3/24 EIS

a value that corresponds to ‘infinite’ , and constitutes the maximum limit for k. Theincreasing effect of the reduction in the girder stiffness is obvious.It is important to point out that the deformability of the frame, which is relatively

high, is, in practice, identified through the increase in the length d of its diagonal,and can be considerably limited by placing a hinged diagonal member. The state ofstress of the frame is, of course, mainly due to the horizontal displacement of itsnodes and, secondarily, to their rotation (see Figure 6.9).Applying the method of forces, the ‘statically redundant’ value of the tension member

force S results from the relative gap F10 of its ends due to the force P, and the relativeapproach F11 due to the force S¼ 1:

F10¼ (L/d)

232

Compressive member, ineffective because of bucklingActivated by inverting the loading sense

PP

PP

PP

Reduction in the horizontal displacement(increase in the horizontal stiffness)

For the same percentage of stiffness increment afour-fold cross-sectional area of tie is required

Elongation of the diagonal Four-fold stiffness

IR IR

IR IRδδ ~δ/4~δ/4

ISIS

ISIS

A

A

S/(3 ~ 5)Stiffness: k + ∆k

S S

Sh

~1/4 elongation

Stiffness: k

P/2 P/2

3 · h/8

Figure 6.9 Lateral stiffness of a two-column frame and its strengthening


F11¼ (L/d)2 (/P)þ d/ES A

S¼ F10/F11

It is obvious that the horizontal component Sh¼ S (L/d), being opposite to the loadingforce P, leads to a reduction in the horizontal displacement as well as in the stress state inthe frame. Based on the above relations, it follows that, for the hinged frame,

Sh ¼P

1þ ðd=LÞ2 k d

ES A

while for the fixed frame, the force Sh is computed on the basis of a four-fold higher valuefor k, i.e. (4 k).The frame displacement thus becomes ¼ (P Sh)/k, and, as a result, the existing

stiffness k increases by

k ¼ L

d

2 ES A

d

Obviously, this latter frame displacement is ¼ P/(kþk).The last relation permits the determination of a cross-section A for the truss member,

which reduces both the horizontal displacement as well as the stress state of the framedue to a load P to any desirable ratio %. Hence,

A ¼

100 k d

L

2 d

ES

while, for the fixed frame, the corresponding value is obviously four times higher.The new stiffness value is

k 100

100 g

Given that the above change in length of the diagonal is the same when elongated andshortened, it can be seen that a hinged member placed in the opposite direction andacting as a compression member has the same effect as above. However, due to thesmall section A with respect to the length of the compression member, this barbecomes immediately inactive due to buckling, as will be explained in Chapter 7.Since, therefore, the direction of the loading force P may be alternated (e.g. due towind or earthquake), it is common practice to place two crossing bars, so that for aspecific direction of the force P only the tension member is active, while the othermember is considered to be inactive (see Figure 6.9).Masonry embedded in a concrete or steel frame definitely causes a reduction in both

the deformation and stress state of the frame under a horizontal loading. The action ofthe wall can be simulated with a diagonal compression member having the samethickness as the wall and a width equal to approximately 1/4—1/5 of the length of thediagonal (Rosman, 1983).

233

Frames

Thus, for a fixed column frame with geometric characteristics L¼ 6.0m, h¼ 4.0m,IR¼ IS¼ 0.005m4, E¼ 30 000N/mm2 and wall thickness t¼ 15 cm, assuming thatEs¼ 5000N/mm2, then — based on the above — the ‘equivalent’ cross-sectional area isA¼ (d/5) 0.15¼ 0.216m2, which makes the frame stiffness four times stronger.It is assumed, of course, that the stress in the compression member,

¼ S

ð0:20 dÞ t¼ Sh

0:20 L t

has a bearable value with regard to its compressive strength.

6.2.3 Frames with inclined legsFrames with inclined legs are better than orthogonal frames for transferring gravity loadsat two support points on the ground, since they correspond more closely to the pressureline (Figure 6.10). These frames develop smaller bending moments, thus leading to more

234

g

p

p p/2

p/2

p/2

Symmetric part

p/2

Relatively small bending relief

The symmetric loading causes little bendingresponse, contrary to the antisymmetric one

The vertical component of the reaction reduces the bending of the leg

The small deviations from the pressure line reduce the bending

[M ]

The absence of horizontal reactionleads to intense bending

Monolithic extension from both sides

Antisymmetric part

W/2

W/2W

Pressure line

Figure 6.10 Behaviour of a frame with inclined legs


economic use of materials. In addition, their stressing due to horizontal loads at thegirder level is clearly favourable.The horizontal girder may be monolithically extended to both sides, so that a

continuous passage for transport loading (i.e. a bridge) is formed between two specificlocations (see Figure 6.10).In the case where the girder is extended from both sides through simply supported

beams, the sensitivity of the frame should be taken into account in view of bendingstresses towards half-loading (with live loads), as shown in Figure 6.10. However,with regard to bigger spans (up to 100m), the self-weight plays a more significant rolein this configuration than the live load and, consequently, the sensitivity is limitedthrough the monolithic connection of girders (see Figure 6.10).

6.3 One-storey multibay frames

6.3.1 Vertical loadsIn the case of a hinged connection of the girder to the columns being fixed at their base,the same characteristics appear exactly as in the frame with one span (see Figure 6.1). Itis obvious that a lateral load applied at the level of the girder will be taken up by theshear forces V at the top of the columns, distributed proportionally to their stiffness(3 EI/h3) and causing directly at the fixed support of the columns the moments V h(see Section 3.3.7 and Figure 3.42).These frames are usually formed with the girder monolithically connected to the

columns, and with spans that do not differ substantially between them. The columnsare considered to have the same height (Figure 6.11).In such a frame, a gravity load uniformly distributed over the whole length of the

girder does not cause a substantial horizontal displacement of the frame, and hencethe bending stresses in the columns depend only on the rotation of their top. Thisrotation can be neglected in practice for the internal columns, and thus neitherbending nor shear appears in them. As a result, the girder of each internal bay acts,basically, as a fixed-ended beam. However, the rotation at the two extreme jointscannot be neglected. This rotation causes a bending moment at the top of the twoexternal columns, as well as at the ends of the corresponding girders, with the samemagnitude, of course. The above moment MB, which causes tension of the externalfibres of the column and the girder, depends on the stiffness ratio ¼ (IR/L)/(IS/h),and for a uniform load q it is

MðhÞB ¼ q L2

12 1

1þ 1:33 (for a hinged connection)

and

MðfÞB ¼ q L2

12 1

1þ (for a fixed support on the ground)

Since the internal joints of the girder are considered not to rotate, it is obvious that themoment of the external joint under consideration will be transmitted with half its

235

Frames

magnitude to the interior joint, this time causing tension at the bottom fibres. Thismoment is superposed on the fixed simply supported moment q L2/8 of the extremespan, thus resulting in the final bending moment of the first internal joint (see Figure6.11). The difference between this moment and that from the fixed-ended beam ofthe neighbouring span should naturally be taken by the first internal column (jointequilibrium); however, it is small in most cases.In the case of a fixed frame, the above moment M

ðfÞB will also be transmitted with half

its magnitude to the column base, causing tension at its internal fibres (see Figure 6.11).Of course, at the external columns, the developed reactions have horizontal compo-

nents directed towards the interior of the frame, equal to MðhÞB /h or 1.50 (MðfÞ

B /h), whilethe column compressive forces result from the vertical equilibrium of each joint underthe corresponding shear actions of the girders.

6.3.2 Horizontal loadsWith respect to the lateral loading of the frame, a concentrated force P at the externaljoint of the girder is again considered. Any pre-existing different distribution of thehorizontal force along the girder will only affect its axial stress state and nothing else(Figure 6.12).Such a loading causes identical horizontal shifts and joint rotations in practice,

resulting in zero moments at the midspan of the girders, due to antisymmetricalrotation of their ends. The columns on the other side, which develop bendingmoments due mainly to their top displacement, in the case of fixed frames also, in

236

q

q

bL

IS

h IS

B B

B B

IR

IR

Without bending

Without bending

1.5 · MB(f)/h

MB(h)/h

1.5 · MB(f)/h

MB(f) MB

(f)

MB(h)/h

MB(h) MB

(h)

q · L2/12

q · L2/12q · L2/8 – MB(f)/2

q · L2/8 – MB(h)/2

Figure 6.11 Bending behaviour of a multibay frame under vertical loads


practice, exhibit zero moments at their midheight. However, at the external columns,the point of zero moments lies slightly higher than their midheight (compared withat three-eighths in single-bay frames), but this is not particularly important for theirpreliminary design. Of course, hinges may be inserted at the above zero-momentpoints, still leaving the frame statically indeterminate (see Figure 6.12).In a frame with columns hinged at their base, it can be considered that the n single-bay

frames that make up the structure carry the concentrated force P equally. Thus, eachsubframe takes a horizontal force P/n. The direct result is that the horizontal reaction(shear) at the hinge of the external columns is P/2 n, while in all the internal columnsit is equal to P/n (see Figure 6.12). The above considerations lead to the following results:

. As the vertical reactions corresponding to two adjacent subframes cancel each other,it may be concluded that the axial forces of all the internal columns are zero. Thus,the moment P h of the load with respect to the frame base is taken up only by the twovertical reactions of the external columns, which are at a distance b from each other,i.e. by axial forces P h/b (tensile and compressive).

. The shear force at the midspan hinge of each girder is P h/b, causing at each span enda moment of half the magnitude of that developing at the top of each internalcolumn. Indeed, the bending moment at the top of the internal columns is P h/n— double the corresponding value for the external columns — and the moment atthe ends of each girder is (P h/b) (L/2)¼ P h/2n. g

237

Fixed frameThe bending moments are halved

P/n P/n

P/n P/n

P/n

P/n P/n P/n

P/n P/n P/n P/n

P/2nP/2n

P/2n P/2n P/2n

P/2n P/2n

P/2n

h/2

h

P

P

P

N = 0

IS/2 IS/2

IRIR

P · h/b

P · h/b P · h/b P · h/b P · h/b

P · h/b

Without Nn: number of bays (P/n) · h

Figure 6.12 Bending behaviour of a multibay frame under horizontal loads

Frames

A frame with fixed columns at its base can be considered, according to the above, as aframe with hinged columns of height h/2, and, therefore, all the previously developedresults are, by analogy, valid. Thus, the shear forces in the columns remain the same,while the moments in the girders and the columns are halved with respect to those ofhinged frames.

6.3.3 Lateral stiffnessThe horizontal displacement of a multibay frame is identical to the shift of one of itssingle-bay subframes. The columns in each have half the moment of inertia of theactual value IS, while the moment of inertia of the girder is equal to the existingvalue IR (Figure 6.13).Thus, the displacement results from the expression for in Section 6.2.2 on the basis

of a horizontal force P/n, and a moment of inertia of the columns (IS/2):

¼ 2 P h3

n EIS 2þ 1=ð2 Þ

12(hinged columns)

¼ IR=L

IS=h

238

P

P

P IR δ

δ δ δ δ

δ δ δ δ

IR

ISIS

P/n P/n P/n P/n IRIR

IRIR

n: numberof bays

Hinged columns

IS/2 Stiffness = k* IS/2

IS/2 IS/2

Total stiffness: k = n · k*Fixed columns: stiffness = 4 · k*, caused δ/4

Total stiffness: k = n · k* + ∆kδ = P/k = (Peff – Sh)/k*

Sh

S S

Stiffness = k* + ∆k(with tie)

Stiffness = k*(without tie)

P – Peff

n – 1P – Peff

n – 1P – Peff

n – 1Peff

Figure 6.13 Lateral stiffness of a multibay frame and its strengthening


The stiffness k of each subframe is then

k ¼ EIS2 h3

12

2þ 1=2

while the stiffness k of the whole frame is k¼ n k. Note that for fixed columns the stiff-ness k is four times higher (see Figure 6.13), according to Section 6.2.2.Restriction of the horizontal shift can again be attained by inserting hinged diagonal

members in the frame. The placement of one such diagonal bar (with length d) in onlyone bay increases the stiffness k, according to Section 6.2.2, by

k ¼ L

d

2 ES A

d

and thus the stiffness of the whole frame becomes equal to n k þk.This ‘strengthened’ bay will now take up a higher horizontal force Peff (see Section

3.3.7), equal to

Peff ¼ P k þk

n k þk

and, according to Section 6.2.2, the horizontal component Sh opposing the action Peff

applied to the bay being examined will be

Sh ¼Peff

1þ d

L

2 ðk

þkÞ d

ES A

It should be recalled that the axial force S developing in the tension member is

S ¼ Sh d

L

The obviously reduced displacement of the whole frame now becomes

¼ P/(n k þk)

a value which may also result by simply considering the ‘strengthened’ opening accordingto Section 6.2.2:

¼ (Peff Sh)/k ¼ Peff/(k

þk)

Once again, it is possible to determine the required cross-section A of the tensionmember that reduces the horizontal displacement due to a load P to a desirablepercentage %. Thus,

A ¼

100 n k d

L

2 d

ES

while for the fixed frame the corresponding value is four times higher.

239

Frames

The new stiffness is obviously equal to

n k 100

100 g

It can be seen that multibay frames present a higher stiffness than single-bay ones, andthis means, of course, that they are more sensitive to the imposition of a constraint suchas temperature change or support settlement (see Section 3.2.5). With regard toconcrete frames, creep acts as a reducing factor for the stress state due to shrinkage.If such a constraint is applied within a small time interval (e.g. ground settlement ofnon-cohesive soil) the instantaneously resulting state of stress reduces after some timeby a percentage of the order of 100 ’/(1þ ’)%, whereas, if the constraint isimposed gradually (e.g. shrinkage, a temperature change), the final developed state ofstress is only a part (e.g. 100/(1þ ’)%) of that which corresponds to the finaldeformation, if applied instantaneously (see Section 5.5.1).However, the impending cracking will reduce the calculated stiffness for the

uncracked state, and, accordingly, the ‘elastic’ response of the structure to theimposed constraint is not as intense as initially calculated.

6.4 Multi-storey framesMulti-storey frames are a constitutive element of multi-storey buildings, offering as planesystems the possibility of taking up linear distributed loads at various levels, acting intheir own plane. The considered loads can be either gravity loads or horizontal loadsdue to wind or earthquake actions. In the case of horizontal loads, these may also becarried through a combined action of the frame and a shear wall, interconnected atvarious levels.The examined frames are considered as not having particular differentiation with

regard to the column sections along the storeys or the girder spans and the floorheights.Such a frame is a structure with a high degree of static redundancy, and its

analysis obviously requires the use of a special computer program. However, it isnecessary to cultivate a basic perception for the load-bearing action of the framebased on an approximate, yet structurally sound, approach that permits the preliminarydesign of girders and columns and which constitutes a solid basis for evaluating thecomputer results.

6.4.1 Vertical loadsThe gravity loads on the girders, as for the one-storey multibay frames examined above(see Section 6.3.1), are considered to cause negligible rotations at the joints of theinternal columns, and thus these are not stressed in bending but only axially. Therotations are, however, not negligible at the external columns, where they causebending moments. Thus, each girder may be considered as a continuous beam beingsubjected at its ends to bending moments causing tension at the top fibres (Figure 6.14).

240


For equal storey heights and column sections, as well as for the same loading of girdersin all floors, the rotations of the external nodes can be considered equal, inducing adirectly zero moment at the midheight of columns.This allows one to consider each girder to be connected at its external joint with two

columns hinged at their ends above and below, having height h/2, with the internal jointof the girder (approximately) fixed (see Figure 6.14). Using the girder to column stiffnessratio ¼ (IR/LR)/(IS/h), as introduced in Section 6.2.1, the moment MS at the bottom

241

h

h

h

h

h

h/2

h/2

h

h

IS

IS

IR

LR

L 0.21 · L 0.21 · L

IR

IR

MSMS

MS

MR

MR

M 0R

MR

MS

MS

Without bending Without bending

Figure 6.14 Bending of a multi-storey frame under vertical loads

Frames

end of the top column and at the top of the bottom column, according to this model, is

MS ¼1

2þ 0:66 Mð0ÞR

while the moment MR of the girder that balances them at the joint is

MR ¼ 2

2þ 0:66 Mð0ÞR

In these relations, Mð0ÞR is the fixed-end moment of the external fixed-ended span of the

girder.Note that considering the full height of columns with fixed ends above and below (see

Figure 6.14), and taking into account the effect of loading of the two girders, leads to theresult

MS ¼1:5

2:5þ M

ð0ÞR

which is unfavourable in comparison with the previous equation. However, consideringreinforced concrete frames, the use of this expression seems more logical, given theexpected reduction in girder stiffness relative to the columns, due to girder cracking.The fact that in the external columns at the ground level the column moment at their

top is double the moment at the fixed bottom and, accordingly, the moment becomeszero at a distance h/3 from the base, leads to the conclusion that the above resultsare always on the safe side.Considering now the internal spans of each girder, they may be considered as having

roughly fixed supports for the permanent loads, i.e. they may be treated as fixed-endedmembers with fictitious hinges placed at a distance from the supports equal to 21% of thespan length (see Figure 6.14). However, in view of the unfavourable layout of live loadsnecessary to obtain the highest moments in the spans, the imaginary intermediate‘simply supported’ beam should have a length of 80% of the span, instead of 58%with regard to the permanent loads. It is also obvious that when calculating thesupport moments due to live loads the corresponding ‘cantilevers’ must be consideredas having a length 21% of the span (see Section 5.1).

6.4.2 Horizontal loadsThe horizontal loads Pi are usually applied as concentrated forces at the levels of girders.This way, each floor can be considered as a multibay single-storey frame, which, throughthe shear forces of the columns of the upper floor, takes up the sum of all horizontal loadsacting above the considered girder level. Thus, according to Section 6.3.2, the midpointsof columns and girders have zero moment, and the shear forces of the columns aredetermined according to the above. Again, only the external columns develop asubstantial axial stress, as was also clarified in Section 6.3.2 (Figure 6.15).Each floor i receives, through the column shear forces of the upper floor, a horizontal

forceP

Pi — the summation of all the loads above this level — which is distributed to

242


the columns as explained earlier for single-storey frames. Each column with itscorresponding shear (

PPi/n) considered at the hinge position (i.e. at its midlength)

develops a moment Mst,i at the top and at the bottom equal to

Mst,i¼ (P

Pi/n) (h/2)The moments MR at the girder ends on the left and right side of the column are deter-mined by equilibrium at the corresponding node. For equal heights of columns and equalgirder spans, the moment of the girder between the ith and (iþ 1)th floor is

MR ¼ ðMst;i þ Mst;iþ 1Þ=2

that is

MR¼ (P

PiþP

Piþ 1) (h/4n)

It is clear that the bending stresses in the girders increase as long as their levelapproaches the base (see Figure 6.15).

243

ΣPi + 1/2n ΣPi + 1/2nΣPi + 1/n ΣPi + 1/n ΣPi + 1/n

P

P

ΣPi

ΣPi

Storey i + 1

Storey i

Storey i + 1

Storey i

Storey i

Without axialforces

Without axialforces

n: numberof bays

h

h

h

hTensile N Compressive N

Moments are increasing towards lower storeys

ΣPi /2n

MR

MR

ΣPi /2nΣPi /n ΣPi /n ΣPi /n

ΣPi /2n ΣPi /nΣPi /n ΣPi /n ΣPi /n

Figure 6.15 Bending of a multi-storey frame under horizontal loads

Frames

It is important to note that the column moments do not increase up to the foundationwith respect to height, i.e. much as would happen in a cantilever, although the multi-storey frame seems ‘macroscopically’ a tall structure (cantilever) fixed to the ground.The bending moment in each column depends only on the shear force that thecolumn transfers, as well as on its floor height. Thus, a single load applied at the toplevel causes in any column (through its antisymmetric part) roughly equal momentsin each floor up to its base. This kind of response is obviously due to the monolithicconnection between girders and columns, as Figure 6.16 illustrates, and it is a directconsequence of the ‘shear beam’ behaviour that such a frame exhibits, as will beexplained in the next section.Indeed, in the system shown in Figure 6.16, where the girders are connected to

the columns through hinges, the bending in the columns, due to the antisymmetricalpart of loading — for which the simply supported girders are unstressed — makesthem act as cantilevers fixed to the ground, developing moments much higher thanthe corresponding ones in a frame with monolithic joints.

6.4.3 Lateral stiffnessIn the case of multi-storey frames that are loaded with horizontal concentrated forces atthe level of the girders, the developing horizontal shift at the level of each girder is ofgreat interest.This shift is, of course, common to all the joints of each floor, and is basically

produced by the deformation of a typical bay panel taken from the ‘interior’ of theframe, as illustrated in Figure 6.17. This panel comprises the girder and the half-lengths of the two columns at its left and right ends, which are hinged at theirremote ends, according to the above. The hinged ends of the half-columns are actedon by the corresponding shear forces

PPiþ 1/n and

PPi/n, respectively (see Section

6.4.2).

244

P P

h

h H

h

(P/2) · h/2

‘Shear’ action ‘Bending’ action

Rigidly connected girders: column moments practically unchanged

(P/2) · H

Figure 6.16 Bending and shear of a multi-storey system under horizontal loads


It can be seen from Figure 6.17 that the relative horizontal shift w of the top andbottom hinged ends of the column places the relative horizontal shift between thetwo consecutive girder floors at the level under consideration. This shift contributesthe rotation ’R of the girder end (the same at both ends, hence the antisymmetricstress state) with the magnitude 2 ’R h/2, as well as the forces

PPiþ 1/n and

PPi/n

acting at the hinged ends of the cantilevers above and below with lengths h/2. Thus,

w ¼ 2 MR ðLR=2Þ3 EIR

h

2þP

Piþ 1

n ðh=2Þ

3

3 EISþP

Pi

n ðh=2Þ

3

3 EIS

¼P

Pi þP

Piþ 1

2 h2

12 n E LR

IRþ h

IS

The length LR obviously represents an average value of the girder spans.

245

h

h H GA

wp

xx

n: numberof bays

Storeyi + 1

Storey i

Storey i + 1

Shear deformation

Storey i

Increasing with the ‘shear force’,i.e. downwards

∆w ∆w

h/2

h/2

ϕR · h/2

ϕR · h/2

ϕR

ϕR

ΣPi + 1/n

ΣPi /n ΣPi /n

ΣPi + 1/n

IS

IR

Ast

N

Ast

Nn · LR

[M ](Shear beam)Storey drift proportional to bending moment

Figure 6.17 Deformation behaviour of a multi-storey frame under horizontal loads

Frames

The expression (P

PiþP

Piþ 1)/2 gives the shear force V of a fictitious ‘cantilever’beam at the position of the considered girder, and the shift w, according to the aboverelation, is proportional to this value (Franz and Schafer, 1988).It can be seen that the multi-storey frame acts like a beam, the elastic curve ‘w’ of

which is developed due to the shear and not the bending deformation of its elements.In such a shear beam, the slope dw/dx at every point is equal to the shear angle ,which is expressed on the basis of the shear force V at the point under consideration,and the effective cross-section A as ¼V/AG, that is

dw

dx¼ VðxÞ

A GðaÞ (see Section 2.3.2.3 and Figure 2.51)

Note that the deformation of this beam under a transverse load p is not governed by theequation dw4/dx4¼ p/EI (see Section 2.3.6) but by that resulting from differentiating thelatter relation, that is dw2/dx2¼ (dV/dx)/AG¼p/AG, conforming to the elastic curveof the cantilever beam shown in Figure 6.17.Staying with the shear beam concept, it can be alternatively stated that the deformations

of a shear beammay be obtained according to the following formulation of the principle ofvirtual work, where only the shear force terms come into play (see Section 2.3.3):

XPvir resp;real ¼

ðVvir Vreal

G Ads

In this way, a shear beam cantilever having a height H and subjected to a horizontalload p(x) develops at its top a shift , which can be determined by considering as virtualloading a unit force at the top. By applying the principle of virtual work, according to theabove relation,

1 ¼ 1

2 1:0 ð p HÞ H

G A¼ 1

G A p H2

2ðbÞ

which means that the deflection (or shift) at the top of a shear cantilever beam is propor-tional to the bending moment at its base.Returning now to the multi-storey frame being examined, since dw/dx¼w/h, an

analogous relation to the above (a) can also be established here. According to the above,

dw

dx¼ VðxÞ h

12 n E LR

IRþ h

IS

¼ kV VðxÞ

where

kV ¼ h

12 n E LR

IRþ h

IS

It is clear that kV now plays the role of the term 1/AG in relation (a).From the last relation the following conclusion is drawn:

wV ¼ kV

ðV dx ¼ kV MðxÞ ðcÞ

246


that is, the shift of the frame — as well as of any ‘shear beam’ — is proportional to thebending moment, exactly as was deduced previously using the virtual work principle.Of course, the last result re-establishes the above-mentioned role of the factor kV, incorrelation also to the above result (b).The effective difference between the two types of beam now becomes clearer. While

in the ‘bending beam’ the curvature (d2w/dx2) is proportional to the bending moment,according to the relation d2w/dx2¼ kM M(x), where kM¼ 1/EI (see Section 2.3.6), inthe case of the ‘shear beam’ it is the deflection that is proportional to the bendingmoment.After all the above analysis it should again be emphasised that, although the

multi-storey frame deforms under horizontal loads following the bending of itscolumns and girders, it exhibits a lateral stiffness behaviour, which may be simulatedby a shear beam with specific sectional characteristics described by the factor kV.The above relation (c) allows the calculation of the relative shift of each level with

respect to the top, which represents the origin for the coordinate x (see Figure 6.17).Thus, the shift wT

V at the top can first be calculated based directly on the momentdeveloping at the foundation (x¼H). For each inferior level A at a distance x fromthe top, the relative shift wA ¼ wT

V wAV results directly from applying the above

relation (c): wA¼ kV M(x). Thus, the distribution of shifts along the heightbecomes comprehensible, as shown in Figure 6.17.It is, of course, clear that in the case of hinged girders commented on in Section 6.4.2

(see Figure 6.16), the deformation of the two ‘bending beams’ gives a much higher shiftthan that resulting from the monolithic frame acting as a ‘shear beam’. g

With the increase in the frame total height with respect to its width, the axial forces ofthe external columns are also increased, so that, in combination with their axialdeformation (tensile and compressive, respectively), they cause a horizontal shift, inaddition to that caused by the bending of columns and girders.Since the external columns are in tension and compression, through the force

NðxÞ ¼ MðxÞn LR

the elongation or the shortening strain of these columns is

" ¼ N

EAst

Then, assuming the girders to be completely rigid, the frame can be simulated with a bentbeam, the developing curvature of which, according to Figure 6.17 (see Section 4.1.1and Figure 4.10), is

1

r¼ 2 "

n LR

¼ 2 N

n LR E Ast

¼ 2 MðxÞðn LRÞ2 E Ast

The displacement at the top wN is determined according to the principle of virtualwork, by considering as a virtual load a concentrated unit force at the top and

247

Frames

assuming a parabolic distribution of MðxÞ (see Section 2.3.3):

wN ¼ð

Mvir 1rdx ¼

H2 Mðx¼HÞ

3 ðn LRÞ2 E Ast

g

A restriction of the frame deformability, and of its stress state accordingly, can beachieved through the insertion of hinged diagonal bars, in the same sense as hasalready been examined for single-storey multibay frames (Figure 6.18).The developing diagonal tensile force S of a hinged bar with a cross-sectional area A in

a frame panel depends on the relative shiftw between the two levels. In this case, thetensile force may be expressed as S¼ F10/F11, through the developed gap F10 betweenthe ends of the cut tension bar and their relative approach F11 due to the force S¼ 1,according to the procedure of the method of forces followed in Section 3.2.2.As is shown in Figure 6.18, F10¼w cos!, where, according to the above,

w ¼ kV VðxÞ h

Considering now F11, this can be obtained in the same way as F10, that is

F11¼w1 (LR/d)þ d/(ES A)

248

Shortening of the diagonal due to S = 1

h

h

h

h

P

P

h

h

Elongation of the diagonal (S = 0)

(S = 0)

S = 1

∆w

∆w

w

Increasing with the ‘shear force’,i.e. downwards

SS

S

Sh

Sh

1/cos ω

1/cos ω∆w · cos ω

LR

Storey i

The girder is acted on by the difference of Sh which opposes P

Figure 6.18 Lateral stiffening of a multi-storey frame


where w1, according to Figure 6.18 and the previous analysis, may be calculated as

w1 ¼ kV 28 h

cos! kV ð28 1Þ h

cos!¼ kV h d

LR

Thus, for the horizontal component Sh¼ S (LR/d), the following expression isobtained:

ShðxÞ ¼ VðxÞ ðLR=dÞ2

1þ d

kV h Es A

From this expression, it is clear that the axial force of the tension bar is proportional tothe shear force V(x), as is also the case with w.In this way, the girder at each level receives the resultant of the forces Sh above and

below it, a force opposite to the horizontal loads P. The distribution of these ‘relieving’forces obviously follows that of the shear forces V, having a constant rate of change forequal loads P (see Figure 6.18).The role of masonry should be recalled here, in contributing to such ‘relieving’ actions

that lead to an effective increase in the lateral frame stiffness.

6.4.4 Mixed systemA mixed system is considered to be the coupling of a multi-storey ‘shear-acting’ framewith a cantilever ‘bending’ beam (such as a ‘shear wall’) with a large moment ofinertia, fixed at its base and connected to the frame at each level with a hinged —and hence axially stressed — bar. As such a system may represent a typical structuralconstituent of high-rise buildings, its behaviour under horizontal loads applied at eachlevel, generally equivalent to a distributed load ph (Figure 6.19), is of primary interest.The load ph is distributed between the two ‘members’ in such a way that a common

horizontal shift w is developed at each level. For the ‘bending-acting’ beam theimposition of a shift w(x) along its height requires the application of a distributedforce pb¼EI(dw4/dx4). Moreover, imposition of a shift w(x) along the height of the‘shear-acting’ beam requires the application of a distributed force ps which, accordingto the previous section, is expressed as ps¼(1/kV) (dw2/dx2). The coordinate xincreases from the bottom to the top.Thus, the following relation results:

ph ¼ pb þ ps ¼ EI dw4

dx4 1

kV dw2

dx2

The distribution pattern of the horizontal shift along the frame and the ‘bending’beam due to a uniform lateral load is quite different, as shown in Figure 6.19. Thedeformability of the ‘bending’ beam decreases in its lower regions and increases inits higher regions, in contrast to what occurs in the shear member. Thus the frame ‘issupported’ in its lower levels on the ‘bending’ beam, while in the higher levels it is

249

Frames

the ‘bending’ beam which is supported by the frame. This means that the hingedconnecting bars are in tension in the lower levels, whereas in the higher levels theyare compressed.Attempting to find an analytical solution to the above equation does not make much

sense, given the ability to analyse the system through an appropriate computer programfor framed structures. Nevertheless, of basic importance is an understanding of thestructural action of such a system. In order, therefore, to assess the developing stateof stress, the two members might be considered to be connected by a hinged bar onlyat their top. The corresponding compressive force X developed is then determined viathe method of forces.

250

ph

ph ph

ph

The force of the connecting bar results in a compressive force [M ]

EI EI

H

F10: Relative approach(negative)

F11: Relative gap(positive)

X = 1

Compression in connecting bars(increase towards the top)

Approximate consideration

Shearingdeformation

Tension in connecting bars(increases towards the bottom)

Bendingdeformation

Figure 6.19 Coupling of a multi-storey frame and a shear wall


If F11 is the relative gap between the top of the bending beam and the frame, due to amutual compressive force X¼ 1, then, according to Section 6.4.3,

F11 ¼1 H3

3 EIþ kV H (positive)

Compatibility requires that X¼F10/F11, where F10 is the relative displacement of thetops of the bending beam and the frame (negative), due to the external horizontal loads(see Figure 6.19).After the determination of the compression force in the bar, an estimate of the stress

state of the frame as well as of the cantilever beam can easily be made, according to theabove. This estimate is on the safe side.

6.5 Design of sectionsThe defining characteristic of a stress state of a frame, in contrast to a beam, is obviouslythe additional presence of axial forces — more so for the columns than for the girders. Ineach column coexist, besides the shear force, which is basically treated in the same wayas for the beams, an axial force and a bending moment that require a particular check.This check — and consequently the member design — refers to the cross-section resis-tance as well as to the stability of the whole, as will be examined in the next chapter.The cross-section check that will be considered next is performed — as is well known— utilising two aspects. Following the first aspect, the developing longitudinal normalstresses due to the existing pair of values (M, N) under the service condition shouldnot exceed some ‘allowable’ limits. According to the second aspect, the examinedsection should contribute — under the extreme stress state of its materials — the pairof values (Md, Nd) that correspond to service loads ‘magnified’ by S and will causethe section to ‘fail’:

Md¼ S MNd¼ S N

6.5.1 Steel sectionsThe developing longitudinal stresses at the extreme fibres of a steel cross-section under acompressive force N and a bending moment M, as shown in Figure 6.20, are calculatedusing the following relation that expresses superposition of axial and bending stresses:

¼ N þ M ¼ ðN=AÞ ðM=WÞThe above stresses should generally be lower than a value of about 160N/mm2.The stresses in the whole cross-section are distributed linearly between the above two

values, and the stress diagram creates the pair of values (M, N) (see Figure 6.20).As has been pointed out previously, the design of a cross-section is nowadays

performed at the ultimate state. In this state it is considered that all the fibres of thecross-section develop the yield stress fy (see Figure 4.11), and the section has dimensions

251

Frames

such that it can accommodate the pair of values (Md, Nd) — or some other ‘safer’ ones —through a suitable distribution of the stress diagram.In order to check whether this is possible, the following relation must be satisfied:

Nd

Npl

þ Md

1:18 Mpl

1

where Npl is the highest compression force that the cross-section can develop, i.e. fy A,while Mpl is the cross-section resistance in pure bending, as determined in Section 4.1.2.The above relation, considered as an equality, yields the moment MR, which, along

with a given axial force NR, causes cross-sectional failure, and vice versa. This isdepicted by the interaction diagram in Figure 6.20.If the above equation is applied for a specific cross-section, the axial force Nd is not

necessarily the limit value for the moment Md under consideration, but, generally,some other force N>Nd may be. The same procedure is applied if Nd is consideredas the basic sectional force, instead of Md.

6.5.2 Reinforced concrete sectionsThe presence of axial forces together with the bending moment requires the placementof reinforcement on both ‘active’ sides of the section. The reason for this is that columnbending results from, in particular, the lateral loads, which can be cyclical. This imposesa symmetrical layout on the reinforcement in the section.The determination of stresses in the concrete and the reinforcement for a given pair of

values (M, N) lies basically in finding an appropriate (linear) diagram of strains " in thecross-section. This means the determination of the following two values: the shorteningstrain "c of the concrete and the elongation (or even shortening) strain "s of thereinforcement (Figure 6.21). From these values, the stresses in the concrete and thereinforcement of the cross-section can be determined directly. Both "c and "s shouldhave such values that the computed tensile and compressive resultant forces arestatically equivalent to the pair of forces (M, N) in the cross-section.

252

NR

Npl

(= fy · A) Unsaferegion

Saferegion

(0, 0) 1.18 · MplMplσM

σMσN

Due to MDue to N

M

N

MR

Superposition Interaction curve

Figure 6.20 Check of a steel section in eccentric compression


Furthermore, in the case of reinforced concrete, the design is preferably implementedfor the ultimate state of the cross-section. This means that the concrete cross-sectionalarea Ac and the reinforcement cross-sectional area As are designed so that the pair ofvalues (Md, Nd) correspond to the highest possible stressing of concrete and steelreinforcement. For the concrete, the highest response in failure level is the developmentof a compressive strain equal to 3.5%, corresponding to a compressive resistance fc,depending on its quality, while for the steel it is the yield stress fy, roughly equal to420N/mm2, that corresponds to a strain of 2.0%. Since, in the ultimate state beingconsidered, the section remains plane too (Bernoulli), a linear distribution of strainsover the cross-section is sought, so that the resulting compressive and tensile forces‘cover’ the above pair of sectional forces (Md, Nd), on the basis of the correspondingstress—strain diagrams for concrete and steel.Keeping a constant strain of 3.5% at the outer fibre of the compression side, for each

selected strain line trimming out at the level of tensioned reinforcement, a value of "sbetween 2% and 10% (see Figure 6.21), corresponds to the ‘ultimate state’ pair (MR,

253

Statically equivalent to (N, M)

N

MZs

As As

As As

DsDc

DsDc

εs

εc

εc, εs: Determine D and Z Interaction curve

MRMpl

Npl

(0, 0)

εc = 3.5‰

2.0‰

εs

Zs NR (compressive)

NR

MR

Statically equivalent to (NR, MR)

Cracked region

Unsaferegion

Saferegion

Figure 6.21 Concrete section symmetrically reinforced under axial compression and bending

Frames

NR), so that for the given cross-section a particular curve can be drawn. This curve,called the interaction curve (between MR and NR), constitutes the means by whicheach reinforced concrete section can be identified. Thus, on the basis of the interactioncurve, a given value of the moment Md corresponds to an ‘ultimate’ axial force NR,which should be greater than Nd (see Figure 6.21).However, for preliminary design purposes and for a given pair of internal forces (Md,

Nd) corresponding to ‘factored’ service loads by S, the following consideration may befollowed instead, offering a more practical procedure.The internal forces are referred to the centroid of the section. Nevertheless the pair

(Md, Nd) can be substituted by a single compressive force Nd, applied at a distance fromthe centroid equal to c¼Md/Nd. Two cases will be examined.In the first case, if the shifted force Nd lies between the two extreme reinforcement

positions having a distance z, so that c< z/2, then, although the compressive force Nd

may lie outside the core of the section, mainly compressive stresses develop in thesection. In this case it may be assumed, in practice, that the cross-section reaches itsultimate state by superposition of the following two states: (a) a uniform distributionof ultimate compression stresses fc over the whole area affected by the compressiveforce, and (b) the development of a compressive and tensile yield stress fy at bothreinforcement layers which are acted on by the bending moment (Figure 6.22).Thus, for a specific section Ac with symmetric reinforcement As, that axial force NR

is sought, which, for a given c (meaning MR¼ c NR), causes the section to fail.Equivalency of the ‘internal’ forces of the section with the ‘external’ forces (MR, NR)requires, as shown in Figure 6.22, the moment resultants about the tension reinforce-ment to be equal, i.e.

fy As zþ fc Ac (z/2)¼NR (z/2)þMR

Setting as Npl the resistance of the cross-section in pure compression,

Npl¼ fy (AsþAs)þ fc Ac

254

MR = c · NR

MNR

NR

N

fc · Ac fy · As fy · As

fc fyfy

AsAsAsAs

fc

z

c

z z

Figure 6.22 Practical assessment of ultimate resistance for small eccentricity


and, consequently,

NR ¼ Npl 1

2 c

zþ 1

Thus, for the case where c< z/2, a practical check for a section is the examination ofwhether, for the given value of c (¼Md/Nd) the calculated value NR is equal to orhigher than the design axial force Nd.In the second case, where the shifted force Nd lies outside the cross-section, i.e. when

c> z/2, the neutral axis clearly limits the compression area. The section exhausts itsresistance with the development of yield stresses in the tension reinforcement and asuitable development of compressive force in the compression reinforcement and inthe compression area. Note that the total of these compression forces is higher thanthe force of the yielded tension reinforcement, in order for the corresponding‘external’ compression force of the section (Figure 6.23) to be offered.Again, as above, an axial force NR is sought which, acting on a given section with

symmetric reinforcement As under a given eccentricity c (meaning combined with themoment MR¼ c NR), causes the section to fail.Equivalency of the ‘internal’ forces of the section with the ‘external’ forces (MR, NR)

applied to the centroid of the section requires, as shown in Figure 6.23, the momentresultants about the compression reinforcement to be equal. Considering the distancebetween the concrete and steel compressive forces to be essentially negligible (seeFigure 6.23),

fy As z¼MRNR (z/2)and, hence,

NR ¼As fyc

z 1

2

255

z

c

z

AsAs

fyfy

AsAs

NR

MR = c · NR

NR

MR

NR

εs

εc

fy · As

fy · As

Dc

Figure 6.23 Practical assessment of ultimate resistance for large eccentricity

Frames

This relation can be used either for checking that for a given c based on the design values(c¼Md/Nd) the calculated NR is equal to or higher than the design axial force Nd, or forthe direct determination of the reinforcement area As based on the values Nd (¼NR)and c¼Md/Nd.

6.5.3 Prestressed concreteThe application of prestressing in frames concerns mainly the girders in order to countergravity loads. It also concerns, with regard to single-storey simply supported portalframes, the insertion of an externally prestressed tension member between thecolumn bases in order to avoid an outward thrust on the foundation, in case the soilis inadequate to offer the required resistance to this thrust (see Figure 6.7).Girder prestressing, basically, follows the logic of a simply supported prestressed beam

(or of a continuous beam), with a possible eccentricity of the anchored cable at theextreme joints, in order to compensate for the presence of end moments there(Figure 6.24).The issue of the active compression force in the girder should be noted. Due to the

stiffness of the columns, external application of a horizontal compression force P at

256

P P P P

The deviation forces remain unchanged

A decrease in girder compressive force More intense decrease in girder compressive force

Statically indeterminate Statically determinate

Resulting bending response: MPMP = M0 + MSP

(X1, X2)Redundant forces: induce MSP

They induce M0

X1 X1

X2

Figure 6.24 Prestressing action in the statically redundant frame


the extreme nodes of the frame does not imply the development of the same compressiveforce in the girder as well. As long as the columns are stiffer, the portion of thecompression force transferred to the girder is decreased. Therefore, although prestressingthe cable with a force P causes undiminished deviation forces on the girder based on P,the corresponding axial force in the girder resulting from the statically redundant frameaction is, nevertheless, clearly smaller than P (see Figure 6.24).The treatment of the statically ‘determinate’ and ‘redundant’ prestressing moments

takes place in the same way as examined for the continuous beam (see Section 5.4).More specifically, the frame under the external anchor forces, as well as the deviationforces, both calculated directly based on the force P, develops as a statically redundantsystem the bending moments MP, where (see Figure 6.24)

MP¼M0þMSP

The statically determined moments M0 expressed with respect to P (¼ P e) correspondto the statically determined primary system resulting through appropriate interventionsunder the support conditions (see Figure 6.24), since in this case the girder receives theentire compression force P. The redundant prestressing moments MSP result from theabove relation, and are treated as sectional forces in the strength-checking relation,as examined in Section 5.4.2.As explained in Section 5.4, the moments MSP are caused by the statically redundant

forces applied to the primary system under consideration, in order to restore the actualdeformation conditions at the support points (seee Figure 6.24). As was also pointed outin Section 5.4, these moments are essentially unaffected by the progressive cracking dueto load increment, since they depend on the relative stiffness of the frame membersrather than on their absolute values. Of course, in the phase of collapse where theframe has become a statically determined system (see Section 6.6), these momentsvanish. However, they may generally be considered to be multiplied by a varyingfactor that is normally equal to unity.

6.6 Plastic analysis and designAs in the design of beams, the main objective for frames and for any type of structure isto design the structure in such a way that the service loads factored by S will either leaddirectly to collapse or to a non-collapse state which, however, is as close as possible tocollapse.

6.6.1 General overview and plastic behaviourAny type of loading that is applied to a frame structure consists of distributed andconcentrated loads that can always be expressed in the form ‘ P’ associated with asingle parameter for each load (Figure 6.25). If the cross-sections of the structuralmembers are already designed with an ultimate resistance pair of values (MR, NR),then the parameter can always be determined so that the resulting loading inducescollapse of the structure.

257

Frames

The process of determination of the collapse load is substantially the same as thatfollowed for the cases of fixed-ended and fixed simply supported beams in Sections5.2.1 and 5.2.2, respectively.More specifically, with a gradual increase in the parameter starting from zero, the

first plastic hinge appears in a section. Next, the thus modified structure is consideredto have a hinge at the corresponding position, which receives at both its edges anexternally applied self-equilibrating pair of moments equal to Mpl, or at only one end

258

λ · P

λ · (P/2) 2.50 · EI

2 3 2 · Mpl 4

1 5

5.0 m

5.0 m 5.0 m

EI EIMpl Mpl

Mpl = 60.0 kN m

λ · (1.0/2)

λ · (1.0/2)

λ · (1.0/2)

λ · (1.0/2)

λ · 1.0

λ · 1.0

λ · 1.0

λ · 1.0

60.0

60.060.0

60.060.0

60.060.0

60.0

16.26

78.63

49.21

60.0

60.0

60.0

60.078.95116.35

60.0

60.0

60.0

60.0

20.0300.0320.0

60.037.40

25.652.8

Joint 4

120.0Joint 3

120.0

Joint 1

Collapse

102.82

Joint 5

λ = 45.43

λ = 54.85

λ = 61.00

66.59 62.76 40.93

λ = 64.00

420.0

120.0

480.0

50.53

170.53

140.95

Figure 6.25 Consecutive loading stages and sectional forces until total frame collapse


if this is a fixed support. Further loading increase — on the modified structure — leads tothe formation of a new plastic hinge at another location in the structure. The newlymodified structure is considered to have two hinges at the corresponding positions,loaded externally by the self-equilibrating moments Mpl.This procedure is continued until the developed plastic hinges make the structure

either statically determined or a first-degree mechanism. In the first case, it is under-stood that the next plastic hinge formation following the continuously consideredincrease in the parameter will lead to a first-degree mechanism, and consequentlyto collapse. In the second case, it is considered that the structure has already collapsed.This last value of will directly provide the collapse load as well as the prevailingmoment diagram at the instant of collapse in all cases (Duddeck, 1984). This diagramdoes not differ substantially in its qualitative characteristics from that correspondingto an elastic analysis.The whole process can be followed in the example given in Figure 6.25. g

At this point it is useful to follow the lateral deformation behaviour of a simple fixed (i.e.statically indeterminate) frame, as shown in Figure 6.26, subjected to a constant uniformload of 30 kN/m, under an ever-growing horizontal concentrated force P, by followingthe above incremental procedure. As is shown in Figure 6.26, until the moment ofcreation of the first plastic hinge, the behaviour is linearly elastic. Then, the implemen-tation of the first plastic hinge changes the system, which, under the constant externalaction of the plastic moment at node 3, may be further loaded, until the next plastichinge is formed at node 4. Again, the new system affords some resistance to theever-growing lateral load with ever-decreasing stiffness, with an additional plastichinge at node 1, until, by further increasing the horizontal load, the final plastic hingeis created near the midspan of the girder in the already statically determinate system,which implies the collapse of the structure. It is clear that the respective displacementsat the characteristic loading stages may be readily calculated on the basis of thecorresponding bending moment diagram and by using the principle of virtual workappropriately. The diagram relating the acting lateral load to the developing lateraldisplacement describes a typical non-linear behaviour of the so-called push-over responseof the frame under examination.Of course, it should be pointed out that, in order for the above procedure to

be valid, the ductility requirements must be met, which means, in practice, thatthe relative rotation of the first developed plastic hinge (e.g. at nodes 4 and 3 inFigures 6.25 and 6.26, respectively) should not exceed the corresponding values,given in Sections 5.2.1 and 5.3, for steel and concrete (0.1 rad and 0.02—0.05 rad,respectively). Should this maximum allowable value develop at an earlier phase, thecorresponding load and moment diagram should be considered to be the ultimatediagram. g

If, with the progressive increase in loading, a plastic hinge is formed in an axially-onlystressed tension bar with a cross-sectional area A, developing a yield force Ny¼ fy A,then this bar, not being able to develop a higher axial force, may be removed andsubstituted by the corresponding forces Ny applied at both its ends.

259

Frames

Similarly to the above, the relative displacement of the ends of a possible tension bar isassumed to not exceed, roughly, 15% of its length during the whole gradual increment ofthe loading.

6.6.2 DesignThe process described in the previous section considers the problem of finding thecollapse load provided that the design of the structure has already been undertaken.In practice, however, the problem of plastic design, as was stated at the very beginningof the discussion, is dimensioning a structure so that the service load magnified by thefactor S is smaller than or equal to the corresponding collapse load. This is ensured,according to the so-called static theorem of plastic analysis, provided that any moment

260

P

4.0 m

4.0 m 4.0 m

EIMplEI Mpl

30.0 kN/m30.0 kN/m

30.0 kN/m

30.0 kN/m

2 EI Mpl 3

1 4

2 3

1 4

2

1

2

1

2

14

(HE-B 220)Mpl = 206.0 kN m

(a) (b) (c)

(a)

(b)

(d)

(d)(c)

104 kN 112 kN 157 kN206 kN m

206 kN m

206 kN m

206 kN m

206 kN m

206 kN m

206 4450

66 80194

206 206

206

206

20694

206

94

206206

Joint 3 Joint 4 Joint 1

29 32 142 δ: mm63Collapse

178 kN

P: kN

178157

112104

Figure 6.26 Push-over behaviour of a portal frame


diagram being in equilibrium with the considered loading satisfies at any point thecondition

(M, N) (MR, NR)l

and that buckling of the structure is avoided. Of course, provided the structure is stati-cally redundant, an unlimited number of moment diagrams exist that are in equilibriumwith the given loading. The ‘elastic solution’ also belongs naturally to these diagrams.The ‘static theorem’ can also be formulated more widely as follows. If a structure is

due to a model that can be considered to be included in the structure, and which isdesigned on the basis of failure of some of its members through an adopted sectionalforce distribution satisfying everywhere the equilibrium requirements and not neces-sarily the compatibility conditions, then the collapse load for the model is less than(or at most equal to) the collapse load of the structure.This formulation of the static theorem is of particular value since it allows the ‘substi-

tution’ of continuous media with framed structures, as, for example, in Section 4.2.1.2(see also Section 6.7).According to the above, the design procedure is clearly a search procedure. An ideal

(economic) solution of the plastic design problem of structures lies in finding that designselection which, according to the above process, leads to collapse loads identical to theservice loads factored by S.The full achievement of this objective is feasible but not strictly necessary. In practice,

a bending moment diagram is sought that is in equilibrium with the magnified loads onthe one hand, and satisfies at all sections the requirements of the static theorem on theother. A diagram like this reflects the so-called moment redistribution with respect to thatof the elastic analysis.Such diagrams are acquired through inserting, at first, so many hinges at the nodes

and the fixed supports that the structure becomes statically determined, and thenimposing the ‘selected’ plastic moments externally at their edges, so that the staticallydetermined system can be directly solved under the given (factored) loads. If thedesign offers at any position resistances that are nowhere exceeded by the resultingcorresponding diagram, then, according to the ‘static theorem’, the collapse load isdefinitely higher than or equal to the imposed external load. It is obvious that this isalso still valid if — and this would indeed be desirable — the diagram exhausts in onlyone further location the corresponding moment resistance, thus implying an additionalhinge. Of course, the most economical solution possible is sought.Moreover, it should be understood at this point that superposing the elastic solution

(Mel, Nel) with a self-equilibrating stress state (M1, N1) without additional loads leads toa moment diagram M that is also in equilibrium with the existing external loads. Thus,the diagrams M¼ S MelþM1 and N¼ S NelþN1 can be used for application of the‘static theorem’, and so it may be stated that superposition of the self-equilibrating stressstate (M1, N1) does not affect the collapse load. An example of such a moment diagramM1 is the statically redundant moments MSP due to prestressing in a statically redundantsystem, as previously examined in Section 5.4. Another example is the internal forcesdeveloped in a statically redundant system due to support settlements.

261

Frames

Because of the above, it can be seen that the redistribution of moments due to creepin a statically redundant frame with different creep properties of its members (seeSection 5.5.1) does not affect the collapse load of the frame.It is therefore obvious that, for design, it is neither necessary nor useful to follow the

‘elastic’ moment diagram strictly but instead an alternative, modified, diagram that is inequilibrium with the (factored) external loads. The only restriction, as previouslypointed out, concerns the final relative rotation of the first created plastic hinge. g

Since the rotation capacity (ductility) of the plastic hinge created depends directly onthe capacity of the member to develop as large a curvature as possible, it is necessaryto pay particular attention to the compressed concrete members, as is explained next.From the expression for curvature

1

r¼ "c þ "s

h

and given that the presence of a compressive axial force limits the maximum ‘allowable’value "c to the order of 2% (instead of 3.5% for pure bending), it becomes clear that theability to develop a sufficient curvature for the requirement for plastic behaviour of a simul-taneously bent and axially compressed reinforced concrete element is perceptibly limited.In order to deal with this unfavourable characteristic, a dense arrangement of stirrups

(optionally set up in spiral form) may be provided. This results, as can be shown experi-mentally, in an increase in the compressive resistance of the concrete member and,accordingly, in an increase in the ultimate strain "c of up to three times, still dependingon the cross-sectional area of the arranged stirrups. This provision is known as concrete‘confinement’, and is usually applied to the extreme regions of each stressed element, bycombined compression and bending (Paulay and Priestley, 1992). g

In conclusion, any design decision leads, following the procedure in Section 6.6.1 for aspecific load arrangement, to a specific collapse load, which then corresponds to a particularmoment diagram that creates as many plastic hinges as required to make the structure afirst-degree mechanism. This resultant collapse load is either lower or higher, or possiblyequal to, the factored loads. In the first case, the selected design is unsafe. In the secondcase, the selected design is definitely safe, and economic as long as the deviation fromthe factored loads is smaller. The last case corresponds to the ideal solution.It is again pointed out that the development of sectional force diagrams due to

temperature differences, support settlement or moment redistribution due to creep,although affecting the serviceability state, leaves the collapse load unaffected, sincethey are self-equilibrating diagrams.

6.6.3 ExampleFor a practical understanding of the above issues, the metal frame in Figure 6.27 isexamined as an illustrative example. The loading shown in the figure is assumed todepict the factored loads for which the frame should be designed. The members ofthe frame are selected from the profile (HE-B) series. Obviously, the moment resistanceMpl is the side on which the fibres are under tension.

262


6.6.3.1 Elastic considerationAssuming constant rigidity (EI), the bending moment diagram is determined accordingto the statically redundant system. Considering the highest moments of each member asthe required Mpl, the profile sections HE-B 140 for the inclined column and HE-B 220for the remaining two members are selected. Since the diagram with the selectedsections in Figure 6.27 satisfies the requirements of the static theorem everywhere, itcan be concluded that the collapse loads of the frame are higher than the given loads.In order to confirm this last conclusion, the collapse load of the designed frame is

sought, according to the procedure described above. The considered loading consistsof the loads 0.333 (horizontal), 1 (vertical) and 1 (vertical), corresponding tothe factored loads. In order to expedite the procedure described in Section 6.6.2, athought experiment will now be undertaken.First, the elastic moment diagram, corresponding to the specific sectional properties of

the frame members, is determined (see Figure 6.27). Obviously, the differences in thisdiagram to the one corresponding to constant EI are not significant. Next, the existing

263

By superposing the two diagrams the ‘static theorem’is everywhere satisfied λu = 70.52 > 60.0

892.23

[Mr] [M1]

205.9

68.63

205.9

205.9205.9

205.9

205.9 9.98

1/3

1.01.04.0

2.0

343.17

127.264.69

120.049.01

166.47

166.47

[M] based on selectedcross-sections 0.61

0.40 0.58

0.12 0.80

0.80

(· λ)

[M ] based on constant EI

23.3 136.5HE-B 140: Mpl = 61.1 kN mHE-B 220: Mpl = 205.9 kN m

HE-B 140HE-B 220

HE-B 220

6.03.0 3.0 3.0

24.52

1 4

2 36.0 m

20 kN

60 kN 60 kN51.3

100.6444.5

163.4

163.4

Figure 6.27 Design check on the basis of the static theorem

Frames

ratio M/Mpl at each characteristic position in the frame is recorded. As the value of theparameter increases, it is logical to expect that the plastic hinge formation followsthe sequence of the corresponding ratios M/Mpl, starting from the one with the highestvalue and going to the lowest one. Now, so many plastic hinges (r in number) areconsidered that the next one to be developed makes the structure a first-degreemechanism. In a well-designed frame, the number r of these initial plastic hinges is equalto the number of the redundant degrees of freedom of the structure — in this case equalto 3 — and converts the structure to a statically determinate one while remaining stable.In any case, however, a modified structure with r hinges may be considered, with thecorresponding moments Mpl applied externally at their edges, thus leading to a bendingmoment diagram [Mr]. If [M1] is now the bending moment diagram of the samemodified structure, due to external loads corresponding to ¼ 1 only — in this case0.333, 1 and 1 — the value u of the parameter can be then determined so that thediagram [Mr]þu [M1] causes an additional plastic hinge, thus leading the so-formedfirst-degree mechanism to collapse.In Figure 6.27, the ‘priority indices’ M/Mpl for each characteristic section of the frame,

the diagram [Mr] and the diagram [M1] are shown. The next plastic hinge formation(which causes the frame to collapse) will occur at the top of the inclined column. Theultimate load factor u will result from the requirement 343.17 4.00 u¼ 61.1,that is, u¼ 70.52> 60.0 (18% higher), thus justifying the static theorem prediction.According to the above, one should also check the relative rotation at the final stage

of the first occurring hinge formation, i.e. that with the highest index M/Mpl, in this case3. This relative rotation’ can easily be obtained through the principle of virtual work,according to the relation

1 ’ ¼ð

Mvir Mreal

EI

whereMvir is themoment diagram of themodified structure with r hinges, due to the appli-cation of the external virtual moment 1 at the edges of hinge 3, and Mreal is the finallydeveloped moment diagram [Mr]þu [M1], which in this case is [Mr]þ 70.52 [M1](Mvir and Mreal are not shown in Figure 6.27). This results in ’¼ 0.094 rad, which isan acceptable value according to Section 5.2.1.It is noted again that in the case where the value of the relative rotation of the first

plastic hinge is not acceptable, the plastic hinge formation procedure should stop at anappropriate stage, followed by a corresponding reduction in the ultimate load.

6.6.3.2 Moment redistributionNow, a modification (redistribution) of the moments in the elastic diagram is attemptedin order to obtain a more economical solution. First, hinges are placed at the same pointsas described in the previous section, in order to convert the system into a staticallydeterminate one, on which the factored loads are applied. On applying an additionalarbitrary self-equilibrating pair of moments to the hinges, a particular momentdiagram results, which is in equilibrium with the external loads (Figure 6.28).

264


The goal is to consider moment values for the above external pairs such that, bydetermining Mpl for each member according to the moment developed, a moreeconomical solution compared with the ‘elastic’ solution is obtained. A value of140.0 kNm is obtained after trials (the use of classic analysis software is essential) forall the external moments. The corresponding moment diagram is given in Figure 6.28.According to this diagram, the section HE-B 100 is selected for the inclined columnand HE-B 200 for the remaining two members. The weight difference between the twosolutions is obvious and, since the requirements of the static theorem are satisfied, thedesigned structure will need higher loads than the factored ones in order to collapse,and hence it is safe.For confirmation of the latter conclusion, the same ‘shortened’ procedure as described

previously is followed. The ‘priority indices’ (M/Mpl) corresponding to each character-istic cross-section of the frame, the diagram [Mr] and the diagram [M1] are given inFigure 6.28. The next plastic hinge will occur again at the top of the inclined

265

By superposing the two diagrams the ‘static theorem’ is everywhere satisfiedλu = 60.20 > 60.00 (ideal)

693.76

160.1

160.1160.1

160.1

160.1

[Mr] [M1] ( · λ)

2.01/3

1.04.0

1.053.36

266.83

9.98

4.58116.86

53.21

128.3

164.3214.15

164.32

0.021.03

0.54 0.80 0.33

140.0

140.0

140.0

140.0

140.0

140.0

73.3

140.0

6.67

HE-B 100HE-B 200

HE-B 200

HE-B 100: Mpl = 26.0 kN mHE-B 200: Mpl = 160.1 kN m

60 kN

20 kN

60 kN

6.67

Figure 6.28 Plastic design based on a ‘preselected’ moment diagram

Frames

column. The factor u of the collapse load will result from the requirement that

266.83 4.00 u¼ 26.0

that is

u¼ 60.2> 60 (Figure 6.28)

It becomes clear that the economy of the solution is ascertained by this imperceptible(essentially nil) deviation of u from the ideal value of 60.0.It should be pointed out once more that the collapse loads in question are feasible only

under the condition of sufficient deformation capacity (plasticity) of the cross-sections,particularly of the cross-section that is first plastified. This can always be checked bycalculating the developed relative rotation of the first plastic hinge based on theprinciple of virtual work.Note that in the example the influence of the axial forces on the plastic moments was

not taken into account, as this is negligible in practice.

6.7 Design and check of jointsThe load-bearing action of a frame is based exclusively on the ability of its joints totransfer the sectional forces from one member to another. This process is not asobvious as perhaps the static analysis of a framed structure implies. The joint has toreceive reliably the compressive and tensile forces as well as the corresponding shearforces of the sections of the adjacent members, remaining essentially undeformed.Taking up these forces causes a particular stress state in the joint, as the developedprincipal stresses in its interior — which constitute the only invariable criterion of theelastic behaviour (see Section 4.1.1) — do not result directly from considering theframe as such. A complex state that is described by two-dimensional (or even three-dimensional) elasticity and not by one-dimensional elasticity (which is valid forframed structures) prevails in the joint.In steel frames, which are mainly formed by I-shaped cross-sections, where the com-

pressive and tensile forces in the cross-section are essentially taken up by the flanges,continuing the ‘internal’ flanges inside the joint is sufficient, as shown in Figure 6.29.Thus, the joint formation as a closed, very stiff frame ensures its undeformability underthe acting internal forces of the jointed members. However, due to the compressiveand tensile forces acting on the joint, a ‘third’ force acting at both the external and theinternal junctions of the joint needs to be provided in order to ensure local equilibrium.By limiting the examination in an external joint of a column and a girder, and withregard to moments causing tension at the outside fibres — as always happens forgravity loads — the system needs a compression force acting in the diagonal sense ofthe joint configuration, as shown in Figure 6.29. This compressive force cannotpossibly be offered by the joint panel corresponding to the web of the I-section,because of the risk of buckling and, for this reason, a profiled segment (stiffener) isplaced in the diagonal sense, in order to carry more reliably the arising compressionforce. In contrast, moments causing tension at the internal fibres, as can occur under

266


some forms of horizontal loading (e.g. earthquake), require the presence of a tensilediagonal force that can be provided by the node panel without any problem. g

Turning now to concrete frames, it is useful to examine the possibility of approaching thetwo-dimensional stress state in the joint through an appropriate truss action. This idea isbased on the fact that a two-dimensional stress state is composed of principal compres-sive and tensile stresses, as was mentioned in Section 4.1.1, and thus the internal forcesacting in the structure can generally be represented by the members of a statically deter-minate truss formation. As has also been pointed out in Section 4.2.1.2, the compressionmembers of this configuration correspond to the compressed uncracked concreteregions, whereas the tension members correspond to the reinforcement bars thatshould be placed in order to substitute for the cracked regions due to tension. Thechoice of a statically determinate configuration instead of a statically redundant one ispreferable, in order to have a clear view of the restored equilibrium, regardless of thecompatibility requirements. Indeed, ensuring equilibrium through a logically composed

267

Layout of stiffeningmember

Compression Tension

Compression Tension

Figure 6.29 Load transfer in a steel frame joint

Frames

model leads, according to Section 6.6.2, to a limit load smaller than the actual one (statictheorem of plastic analysis).This logic can also be applied in the design and check of joints (Figure 6.30). Again,

by limiting the examination to an external joint of a column and a girder, where themoments of connected members cause tension at the outside fibres, it is obvious fromFigure 6.30 that the balance of forces in the interior of the joint is ensured withoutproblem. This results because the diagonal region of the joint can offer the compressionbar needed for equilibrium at both the exterior and interior edges. However, in the casewhere the internal fibres of the jointed members are in tension, a diagonal tensionmember is required, as shown in Figure 6.30, and this, of course, means that thestirrups of both the horizontal and the vertical members have to continue to theinterior of the node, in order to receive the required diagonal tensile force (seeFigure 4.18). Similar truss configurations can also be formed in other cases of joints(see Figure 6.30).

268

Compression Tension

Compression Tension

Stirrup reinforcement

Stirrup reinforcement

Figure 6.30 Truss action in a concrete frame joint


ReferencesDuddeck H. (1984) Traglasttheorie der Stabwerke — Beton Kalender, Teil II. Berlin: Ernst.Franz G., Schafer K. (1988) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: Springer-Verlag.Paulay T., Priestley M.J.N. (1992) Seismic Design of Reinforced Concrete and MASONRY Buildings. New

York: John Wiley.Rosman R. (1983) Erdbeben-widerstandsfahiges Bauen. Berlin: Ernst.Salvadori M. and Levy M. (1967) Structural Design in Architecture. Englewood Cliffs, NJ: Prentice-Hall.

269

Frames

7

The influence of deformations on thestate of stress — elastic stability

7.1 OverviewThis chapter deals with problems that arise from the presence of high axial compressiveforces, due mainly to gravity loads, in beams and frames. These problems link to the factthat the equilibrium of all structures is realised physically, without exception, at theirdeformed state. Thus, under certain conditions, it can be the case that the analysisbased on the geometry of the undeformed structure, according to the usual practice( first-order analysis), leads to erroneous and unsafe sectional forces. It is understandablethat in such cases the design must necessarily take into account the influence ofdeformations on the state of stress of the structure; this is known as second-orderanalysis. The aim of this chapter is the practical understanding and the quantitativeassessment, using first-order analysis, of such a state of stress of beams and frames.

7.2. Buckling of barsTo understand the influence of the deflection on attaining the equilibrium in acompressed member, it is absolutely essential to first examine the centrally compressedbar, even though such a case is clearly a virtual one that cannot be encountered inreality.It is assumed that the centrally compressed straight bar is subjected to a pair of equal

and opposite forces P applied to its free ends (Figure 7.1). It is intuitively immediatelyunderstandable that, for a ‘small’ value of the force P, a light horizontal transverseaction will cause a transverse deformation, expressed through the deflection f, whichwill disappear once this action is removed. The understanding of this phenomenon iscrucial. It is clear that if a beam is obliged to curve due to a transverse action, then itwill develop an ‘internal’ bending resistance Mi¼EI/r, which should be in equilibriumwith the bending moment Me induced by the external loads (r is obviously a functionof f ).Of course Me¼ P fþM1, where M1 denotes the moment due to the transverse

action. Thus, if Mi¼Me, and given that Mi> P f, once the transverse action isremoved, the internal resistance takes precedence over the external moment and thebar reverts to its initial straight configuration. Assuming now that the force P is graduallyincreased, it would at some point reach a value Pcr at which a minimal transverse force


will suffice (and therefore a minimal moment M1 too) to produce a deflection f in thebar. Therefore, for any imposed curvature 1/r on the bar,

Mi¼ Pcr fi.e. the bar will equilibrate at any arbitrary position, without offering resistance to anychange in its curvature (see Figure 7.1). This automatically means that the bar willfail, given that under the action of the pair of sectional forces (N¼ Pcr and M¼EI/r¼ Pcr f ) the strength of the cross-section will be exhausted as soon as the curvature(i.e. the deflection f ) reaches the appropriate value (see Section 6.5).The load Pcr is called the buckling load and, although it is an absolutely theoretical

concept — given that in reality a perfectly central application of a compressive forcecannot occur and the bar cannot be perfectly rectilinear — it plays an important rolein dealing with practical problems concerning the assessment of the influence of defor-mations on the response, as will be explained below.The above relationship allows a rough approximation of Pcr by considering that

1/r 8 f/L2. However, the resulting value of 8 EI/L2 is about 20% less than the exactone. The reason for this deviation will be explained later. g

For the precise determination of the critical load Pcr a more general reconsideration ofthe situation than the one that has just been described is appropriate. The same beam asabove, subjected to equal and opposite compressive axial loads P, is considered, this timewith a transverse load p(x) as well (Figure 7.2). In order to be deflected by w(x), thebeam opposes a transverse resistance equal to qi¼EI (dw4/dx4) (see Section 2.3.6),and the fact that the beam is in equilibrium means that the external load p(x) is counter-balanced by the resistance qi, i.e. p(x)¼ qi.The presence of axial forces P throughout the curved beam signals the development of

transverse deviation forces pd, which add to the load p(x). According to Section 2.2.7,

272

P

P

P

P

Pcr

Pcr

Transverse action Mi = EI/r

Equilibrium: Mi = Me

Me = P · f + M1(M1) fEI

No transverse action is practically neededin order to cause a curvature

Figure 7.1 Buckling as the limit state of equilibrium


this force is given by

pd¼ P/r¼P (dw2/dx2)

which is obtained by adopting for the curvature the approximate expression (dw2/dx2)(a positive magnitude), instead of the strict one

1

r¼ w00

ð1þ w0Þ3=2

The equilibrium must therefore be pdþ p(x)¼ qi, which can be written as

EI dw4

dx4þ P dw

2

dx2¼ pðxÞ

This equation reflects the equilibrium of the compressed beam, in the presence of thetransverse load p(x). However, in the previously examined case the transverse loadp(x) is absent, so the equation is written as

EI dw4

dx4þ P dw

2

dx2¼ 0

This is the classical differential equation for buckling, which expresses, for the deformedstate of the bar, the equilibrium between the deviation forces P (dw2/dx2) and thetransverse bending resistance of the beam EI (dw4/dx4), both of which are mobilisedthrough the imposed deformation w(x).The solution of this equation has the form:

w ¼ C1 sin ðkxÞ þ C2 cos ðkxÞ þ C3 xþ C4

where

k ¼ffiffiffiffiffiffiffiffiffiffiP=EI

p273

P

P

x

wEI

P

P

P/r

p(x)

p(x) = 0 Equation of buckling

Equilibrium: p(x) + P/r = qi (loading resistance)

EI(d4w/dx4): transverse resistance qi

Figure 7.2 Equilibrium of a compressed beam in a deformed state due to bending

Influence of deformations on the state of stress (elastic stability)

The treatment of this solution with respect to the coefficients C is carried out by takinginto account the existing boundary conditions of the bar, which means the way in whichthe bar is supported, with the aim, of course, of determining the critical load Pcr. (Theimportance of the boundary conditions may be understood through the above roughcalculation of the critical load. The deviation of the value by 20% is due to the factthat at the ends of the bar the curvature was considered to be 8 f/L2 (and not zero asit ought to be, because of the null bending response prevailing there).This treatment results in the following expression for the buckling load:

Pcr ¼p2 EIs2k

The magnitude sk, which is known as the effective length, depends on the way in whichthe bar is supported, and it may be considered as the distance of those points of the defor-mation line of the bar which exhibit zero curvature, i.e. the inflection points. At thesepoints the bending moment of the bar is zero. Thus, for a bar hinged at both endssk¼ L, for a cantilever sk¼ 2 L, for a fixed simply supported bar sk¼ 0.70 L and for afixed-ended bar sk¼ 0.50 L (Figure 7.3). In all cases should be ensured that the pointof application of the compressive force P can be axially displaced, as shown in Figure7.3. In the case of a fixed-ended bar with one end free to move transversally, sk¼ L.As is apparent from the last equation, the critical load increases (i.e. the risk of bucklingreduces) as EI increases and sk decreases.It is also interesting to consider the critical axial stress cr, which is developed when

the buckling load is reached:

cr ¼PcrA

¼ p2 E2

274

PPPPP

sk = L

sk = Lsk = 2 · L

sk = 0.70 · Lsk = 0.50 · L

Slenderness λ = sk √A/I

Pcr = E · A · π2/λ2 = EI · π2/sk2

Figure 7.3 The influence of the support type on the value of the buckling load


where is a dimensionless quantity, the slenderness, which is the ratio of the effectivelength sk to the radius of inertia i of the cross-section:

i ¼ffiffiffiffiffiffiffiffiI=A

p¼ sk/i

The slenderness is the measure of dangerousness of the axial compressive response ofa bar — the greater the value of , the lower the compressive stresses that the bar cancarry, thus leaving the possibility of the cross-section taking on higher stressesunexploited. g

In the above expression for the buckling load Pcr a constant rigidity EI is assumed overthe entire bar length, which may not correspond to reality. For this reason, a moregeneral method for determining the critical load Pcr that is of particular importancepractically is examined below for the determination of the critical load Pcr. Thismethod may be applied in any case, and is known as Vianello’s method (Menn, 1990).It is assumed that the bar, which is axially loaded with the load P, has an initial

eccentricity equal to w0(x) (Figure 7.4). Under the bending moments M0¼ P w0(x)the resulting deflection w1 at a certain point of the bar can be calculated directly as afunction of P on the basis of the principle of virtual work (see Section 2.3.3). Themoments M1¼ P w1(x), in turn, also produce an additional deflection w2. Theprocedure, provided the load P is less than the critical one, converges, and the finaldeformation w is

w¼w0þw1þw2þ . . .

If the function w0(x), and consequently also w1(x), is affine to the buckling shape of thebar, then it can be proved that

w1 ¼P

Pcr w0 ðaÞ

Moreover,

w ¼ w0 1

1 P=PcrðbÞ

275

Due either to initial deformationor to transverse load

P P P Pw0 w0

w1 WFirst-order deflection due to compressive load Final deflection curve

If w0 = w1 then P = Pcr ξ = P/Pcr, w = w0/(1 – ξ)

Figure 7.4 The criterion for determining the critical compressive load


Equation (a) allows the direct determination of the critical load Pcr, as for P¼ Pcr it mustbe thatw1¼w0. More precisely, for a selected position x¼ x0, e.g. where the maximum woccurs, the deflection w1(x0) is calculated as a function of the load P and thecorresponding value w0(x0). Then, by equating the two deformations w1(x0) and w0(x0),the critical load Pcr is obtained directly (see Figure 7.4). Equation (b) allows the assessmentof the influence of the deformations on the response, as will be explained later.It should be noted that the concept of w0 is not necessarily linked to an initially

deflected beam exhibiting some sort of imperfection. It can also be simply consideredas the first-order deformation produced by an arbitrary transverse load acting on thebeam. The elastic deformation w0(x) thus developed necessarily satisfies the appropriateboundary conditions, and this ensures (as discussed previously) that the final result willapproximate the ‘accurate’ one as closely as possible.Thus, as considered above, the buckling load Pcr is the load which, for an initial

deformation w0(x) respecting the boundary conditions of the beam, produces throughfirst-order analysis the same deformation as the initial one at some selected point(usually the point having the maximum w). g

The following example (Figure 7.5) illustrates the procedure more precisely. In order toassess the buckling load of the fixed-ended beam, with one end freely movable as inFigure 7.5, the developed deflection w0(x) under a transverse load q is first determined.This load is calculated, on the basis of the corresponding bending diagram [M0], as thebending diagram of a simply supported beam loaded with [M0]/EI, according to Mohr’stheorem (see Section 2.3.6). The maximum deflection at midspan (x¼ L/2) isw0¼ 0.0026 q L4/EI.This deflection w0(x) is considered as the initial one, and the axial load P, after

developing the corresponding bending moment diagram [M], produces at the same

276

P

[M ]

q

EI

L

q · L2/12 q · L2/12

( · P · q · L4/EI )

[M0]

[M1]

0.125 · L0.125 · L1

0.125 · L

w0

0.00

03

0.00

11

0.00

18

0.00

24

0.00

26

0.00

24

0.00

18

0.00

11

0.00

03

First-order moment diagram due to the compressive load

Virtual loadingThe initial deformation due to the transverse load

satisfies the boundary conditions

The critical load causes a first-order deflection equal to the initial one

Figure 7.5 Estimation of the critical compressive load


point (x¼ L/2) the deflection w1. This deflection is calculated, according to the principleof virtual work, according to Figure 7.5, and is given by

w1 ¼ðL0

M M1

EIdx ¼ 0:025 P w0

L2

EI

The critical buckling load results from the requirement w1¼w0 at the specific pointconsidered. It is obtained as

Pcr ¼ 40:0 EIL2

This value is just 1% higher than the ‘accurate’ one.

7.3 The influence of deformation on the response of beams(second-order theory)

As is clear from the foregoing, the presence of axial forces in a beam de factomodifies thebending moment diagram and, consequently, the deformations too, which are anywaycaused by the bending response. The beam, which carries a transverse loading and issubjected to compressive axial loads P (Figure 7.6), develops a deflection line thatobeys the differential equation examined in Section 7.2. It is clear that both thedeformation developed and the resulting bending response of the beam are greaterthan those determined on the basis of first-order theory.Repeating once again the result of Section 7.2, if ¼ P/Pcr and W1 represents the

deformation of the beam according to a first-order result, then the final deformation

277

P P

P PP P

P P

W1

W1w

Final deflection curve First-order deflection curve

w0

w1

First-order deflection curveFirst-order deformation: W1 = w0 + w1

Initial deformation

Final deformation[second order]

ξ = P/Pcr

w = W1/(1 – ξ)

Increased moments relative to first-order values

Under constant axial load the superposition is valid

Figure 7.6 Estimation of the final deformation based on the value of the critical load


w of the beam will be

w ¼ W1 1

1

The presence of a compressive load imposes on the initially (to first order) developeddeformationW1 the magnifying factor 1/(1 ), which increases steadily as the compres-sive load approaches the value Pcr. Thus, for example, for a compressive force equal to50% of Pcr the value ofW1 is doubled, while for 80% of Pcr the value ofW1 is quintupled.It is also found that, as the compressive force remains constant, the deformation due to atransverse load is proportional to it, i.e. the principle of superposition is valid, but onlyunder the above condition.If M1 represents the first-order bending moments, the actually developed bending

moments M in the beam are

M¼M1þ P w g

In order to reinforce the understanding of the concept ofW1 in the above expression forthe final deformation w, a compressed beam subjected to both an initial deformation w0

and a transverse load (see Figure 7.6) is considered. Assuming that a deformation w1 isproduced to first-order by the load, then, by superposing the individual actions (which islegitimate, given that they are both referred to the same compressive force P), accordingto Section 7.2,

w ¼ w0 1

1 þ w1

1

1 ¼ ðw0 þ w1Þ

1

1 ¼ W1

1

1

i.e. W1 is the sum of the first-order products w0 and w1. g

The above analysis is now applied to three illustrative examples.

Example 1. A column with a fixed base and height L, in addition to a transverse load q, issubjected at its top to a horizontal force H, a moment M0 and a compressive load P(Figure 7.7). The horizontal deflection wA that is developed at the top of the columnis calculated on the basis of the corresponding first-order deflection WA

1 :

WA1 ¼ q L4

8 EIþH L3

3 EI M0 L2

2 EIBased on the buckling load

Pcr ¼p2 EI4 L2

it is obtained that

wA ¼ WA1 1

1 4 P L2

p2 EI

278


It is clear that the presence of the compressive force increases the deformability of thecolumn.The bending moment at the base of the column F is given by

MF ¼ MF1 þ P wA ¼ q L2

2þH LM0

!þ P wA

Example 2. The simply supported structure shown in Figure 7.8 is subjected at its endsto the loads P, and develops in the midspan a deflection wm, resulting from thecorresponding first-order deflection Wm

1 , by applying the magnifying factor 1/(1 ).Wm

1 is given by

Wm1 ¼ P e L2

8 EIand given that

Pcr ¼p2 EIL2

279

P P

m e

Wm

Final deformationW1

m

First-order deformation

Figure 7.8 Assessment of the final deformation

L q EI

H

F

A

wA

M0

P

Final deformation

First-order deformation

W1A

Figure 7.7 Assessment of the final deformation


it results that

wm ¼ P e L2

8 EI1

1 P L2

p2 EIThe bending moment in the middle of the beam is

Mm ¼ Mm1 þ P wm ¼ P eþ P wm

Example 3. The horizontal stiffness kR at the top of the column shown in Figure 7.9 isobviously influenced by the applied compressive load P. According to first-ordertheory, this stiffness is equal to kH¼ 3 EI/L3 (see Figure 3.24). The horizontal forceH¼ 1 produces, according to Example 1, a displacement at the top equal to

wA ¼ L3

3 EI 1

1 4 P L2

p2 EIand consequently the corresponding stiffness kR, i.e. the opposing resistance of thecantilever for an imposed unit displacement on its end, is 1/wA, which is written as

kR ¼ kH 1 4 P L2

p2 EI

¼ kH 1 P

Pcr

It can thus be seen how the presence of a compressive force leads to a reduction in thebeam’s stiffness. Indeed, by approaching the critical buckling value, this stiffness tends todisappear.The resulting bending moment at the base F due to a horizontal force H at the top is:

MF ¼ H Lþ P wAg

280

L

P

EI

A

F

wA

H = 1

kR = 1/wA

Final deformation

The stiffness decreases as the compressive loadapproaches the critical value

Figure 7.9 Assessment of the transverse stiffness of the cantilever


Thus it is clear that for a compressed beam any eccentricity in the compressive loaddistinguishes it from a centrally compressed beam, in that it develops a specificdeformation and bending response, which should be taken over by the correspondingcross-section. In a purely axially compressed bar no bending deformation occurs untilthe compressive load reaches its critical value, and thus the smallest disturbance isfatal for the structure.Therefore, the check of the strength capacity of a beam subjected to a compressive

service load P, having first-order deformations and bending moments W1 and M1,respectively, for a global safety factor applied to service loads, will be carriedthrough on the basis of an axial force

N¼ Pand a bending moment

M ¼ M1 þ P W1 1

1 PPcr

It should be checked, of course, that the pair (M, N) is ‘covered’ by the capacity of thesection, as described in Section 6.5. g

Having examined the compressed beam, it is useful now to consider a transverselyloaded beam under an additional tensile axial load (Figure 7.10). The presence oftensile forces P throughout the curved length of the beam means — analogously toSection 7.2 (see Figure 7.2) — the development of transverse deviation forces pd¼ P/rwhich, having the same sense as the bending resistance of the beam, oppose the loadp(x). The equilibrium requires now that p(x)¼ qiþ pd, and the correspondingequation takes the form

EI dw4

dx4 P dw

2

dx2¼ pðxÞ

281

P Pw

W1

Final deflection curve

ξ = P/Pcr

w = W1/(1 + ξ)

Reduced moments relative to the first-order ones

Valid superposition under a constant axial load

First-order deflection curve

Figure 7.10 Assessment of the final deformation under a tensile axial load


In this case it is understandable that both the developing deformation and the resultingbending response of the beam will be less than those determined on the basis of thefirst-order theory. The final deformation w is derived through an approach analogousto the one described previously, and is given by the relation

w ¼ W1 1

1þ ð ¼ P=PcrÞ

It is clear that the reduction coefficient 1/(1þ ) as applied to the first-order deformationW1 due to the transverse load, may take, depending on the acting tensile load, somesmall value.It is found here also that, under a constant tensile force, the deformation w due to a

transverse load is proportional to it, i.e. the principle of superposition is still valid.The developed bending moments M of the beam are

M¼M1 P w

where M1 are the corresponding first-order moments.The model of the tensioned beam has a direct application both in the analysis of

suspension bridges and in calculating the torsional response of thin-walled beams dueto prevented warping, as will be examined in Chapters 9 and 13, respectively.

7.4 The influence of deformation on the response of framesThe columns of frames are compressed members par excellence and, when theircompressive response increases significantly with respect to their slenderness, theirbehaviour as frame members leads generally to a reduction in the horizontal stiffnessof the frame (Figure 7.11). Of course, when this stiffness disappears, the appliedvertical loads have reached their critical values, which leads directly to collapse. Thepurpose of this section is to show how one can obtain a very good approximation ofthe frame response under the influence of occurring deformations, using only first-order analysis.

282

Vertical loads reduce the lateral resistance offered to horizontal loads

Figure 7.11 The influence of vertical loads on the behaviour in response to lateral loads


7.4.1 One-storey framesThe behaviour of a frame with significant axial forces on its columns against lateraldisplacements may illustrated using with the model shown in Figure 7.12 (Franz et al.,1991). The frame, which is loaded only by the vertical loads P1 and P2 at its joints,may be considered as equivalently consisting of two interconnected, geometricallyidentical systems. The first system, having its members connected through hinges attheir ends, has all the vertical loads applied at its nodes, being itself connected througha hinged bar to the unloaded actual frame (second system) at the level of the girder.First, the critical sum of vertical loads has to be determined that renders the system

elastically unstable, i.e. ready to collapse. An imposed small horizontal displacement causes an oblique position of the hinged columns, as shown in Figure 7.12. From theexisting equilibrium of the nodes, involving the vertical loads and the inclinedcompression forces of the columns, arise horizontal forces Ha that act on the monolithicframe. The frame obviously opposes a certain resistanceHp against the imposition of thedisplacement .From the figure it can be seen that the action Ha consists of the sum of the horizontal

components of the inclined compressive forces N in the columns. Given the smallinclination angles of the columns,

Ha ¼X

P tan ¼X

P tanð=hÞ X

P ð=hÞ ¼ X

ðP=hÞ

where h is the column height.

283

P1 P2

P1 P2

P1 P2

h

h

δ

δ

Ha Hp

ψ2

ψ1

Ha < Hp kH: horizontal stiffness

At the critical load the deviation force equals the resistance of the frameCritical load = kH · h (columns of equal height)

Total deviation force

Resistance of the frame

Figure 7.12 Fictitious load-carrying mechanism for the critical vertical load


If kH represents the lateral stiffness of the frame, then the resistance Hp will be

Hp¼ kH It should be noted that the resistance of the frame Hp does not depend on the verticalloads, whereas the action Ha does depend both on the vertical loads and on the angle ofinclination of the columns. For a relatively low value of vertical loadsHp>Ha and, afterremoving the displacement , the system returns to its initial undeformed configuration.However, if the loads take values that for any imposed are Hp¼Ha, the critical loadwill be reached and, consequently, collapse will occur. In this case, from the previousequations it can be deduced that the critical loads must satisfy the relationshipP

(P/h)cr¼ kH

and, if the columns have the same height h,

(P

P)cr¼ kH h g

Having considered how the above model works, it is useful to confirm the above resultsby examining the actual frame more directly (Figure 7.13). A horizontal displacement is again considered, which is imposed on the frame in Figure 7.13, which has a totalvertical load (

PP) acting on its nodes. The oblique action of the compressive force

N of the columns in their inclined position under the loads (P

P) is equivalent tothe action of a horizontal force Ha¼

PP (/h) on the undeformed configuration of

the frame. If this force suffices to exhaust the resistance of the frame Hp¼ kH , i.e.if Ha¼Hp, then the critical load (

PP)cr has been reached. If Ha<Hp, then obviously

the frame comes back to its initial configuration.

284

P1 P2 P3

P1 P2 P3

P1 P2 P3

δ δ δ

δ δ δ

Resistance of the frame

h

Hp Ha

Ha

kH: horizontal stiffnessAt the critical load the deviation force equals the resistance of the frame

Critical load = kH · h (columns of equal height)

The deviation force results from the horizontal components of the oblique column forces



Figure 7.13 Determination of the critical vertical load of the frame


Thus, it becomes apparent that the vertical loads acquire their critical value (P

P)crwhen the activated horizontal force Ha, due to an arbitrary deviation , causes the verysame , following a first-order analysis. g

It should now be understood that an initial (unintentional) constructional eccentricity0 in the frame, with respect to the undeformed geometrical configuration, will in thepresence of (significant) vertical loads P inevitably increase by (Figure 7.14). Thefinal equilibrium configuration of the frame with a total displacement ¼ 0þ,may be conceived as a consequence of a fictitious external horizontal force Ha actingat the level of the beam. According to the above, this is equal to

Ha ¼X

P 0 þ

h

This force imposes on the frame the additional displacement and is given by

Ha¼ kH According to the previous expression for (

PP)cr for columns of equal height, it is

obtained that

¼ 0 1P

Pð ÞcrPP

1

Thus, the final displacement ¼ 0þ is

¼ 0 1

1P

PPPð Þcr

It is found that the initial deviation is magnified, because of the presence of compressiveloads, by the same factor that is applied in the case of beams (see Section 7.3).The initial presence of a horizontal force H0 automatically means the existence of an

initial deviation 0 and, according to the above, the development of a further defor-mation leading to an additional response (Figure 7.15). The total response maybe conceived as being due to a fictitious horizontal force H which, caused by the

285

P1 P2

The additional displacement is dueto the fictitious horizontal loading

Ha

∆δ Additional displacementInitial displacement δ0

Figure 7.14 Increment in the initial displacement due to the influence of vertical loads


oblique position of the columns, is given by

H ¼ H0 þHa ¼ H0 þ

h

XP

This force obviously produces the total displacement (see Figure 7.15), which meansthat H¼ kH and, consequently,

¼ H0

kH P

P

h

This result may be easily verified as being identical to the previous one for columns of thesame height. If this is not the case, then

¼ H0

kH P P

h

286

P1 P2 P1 P2

Total displacementδ

Initial displacement δ0 ∆δ Additional displacement

H0 H

H H

Fictitious loading

H = H0 + Ha

Ha

The column shear forces balance the horizontal load

The fictitious loading causesthe total displacement

The column shear forces balance the fictitious load

... but they are reduced by the horizontal componentsof the inclined column forces

Figure 7.15 A fictitious horizontal force as an equivalent action on the frame


The above expression for H takes the following form after substituting from theequation above for columns of the same height:

H ¼ H0 1

1P

PPPð Þcr

i.e. the final bending response is derived from the first-order one due to the forces H0 byapplying the standard magnifying factor:

1

1P

PPPð Þcr

From the above results the activated additional force Ha can be written as

Ha ¼ H0 1P

Pð ÞcrPP

1

It should be made clear that this horizontal forceHa is fictitious, and its purpose is simplyto be able to represent the equilibrium in the deformed state (second-order theory) byusing first-order analysis, based on the (fictitious) horizontal load H¼H0þHa appliedon the undeformed frame. The column shear forces, of course, obviously equilibrate theforce H0 and not the force H. However, this is apparently in conflict with the fact thatthe increased bending response of the columns due to the forces H leads to increasedshear forces, which should equilibrate H rather than H0. The ‘paradox’ is removedif the inclined position of columns is taken into account, whereby the horizontalcomponents of the compressive axial forces do actually reduce the ‘increased’ shearforces due to the force H. Thus, the actual shear forces do indeed equilibrate thehorizontal force H0 (see Figure 7.15). g

Clearly, the resistance (stiffness) offered by a frame when a unit horizontal displacement(¼ 1) is imposed in the presence of loads

PP is equal to the force R that produces such

a displacement. According to the last results, for columns having the same height

R ¼ kH P

P

h

while for columns of unequal height

R ¼ kH X P

h

Thus, it has clearly been shown how the horizontal stiffness of the frame is reduced dueto the vertical loads (see Figure 7.11). This stiffness vanishes, of course, when the criticalload is reached.

287


7.4.2 Multi-storey frames

7.4.2.1 The influence of deformationThe behaviour of multi-storey frames is examined in a similar manner to that of one-storey frames. Clearly, an initially oblique position of the columns under the presenceof considerable axial forces, even with a small deviation from the vertical due either toa constructional imperfection or to the first-order effect of the application of horizontalforces, raises some additional moments, which increase this deviation even more. Thefinal deviations from the vertical cause the activation of ‘fictitious’ horizontal forces atthe joints of the frame, which may be considered as being applied to the undeformedframe, producing (by first-order means) the very same deformations that have mobilisedthem. The problem in multi-storey frames lies simply in the fact that at each storeylevel there is an unknown final horizontal drift, which necessarily has to be determined.The procedure is illustrated using the example of a three-storey frame (Figure 7.16)

(Franz et al., 1991). At each storey level the total vertical loadsP

P1,P

P2 andPP3 are acting, while the initial clockwise deviation angles of the columns from the

vertical are 10, 20 and 30, respectively. It is clear that these initial angles willincrease and reach their final values 1, 2 and 3. Consequently, the initial storeydrifts 10, 20 and 30 will finally attain the final values 1, 2 and 3, respectively. Itis understandable that in the final equilibrium state the inclined compressive forces ofthe columns, according to previous considerations (see Section 7.4.1), provide,through their horizontal components, deviation forces of the same sense, becausethose resulting on the lower side of the corresponding girder are always greater thanthose on the upper side, as can be seen in Figure 7.16.

288

The vertical loads cause an increase in theinitial deviations (possibly due to horizontal forces)

Additional fictitious forcesfor the first-order calculation

Σ(P1 + P2+ P3)

Ha,2

Ha,3

Ha,1

θ1

θ2

δ1

δ2

δ3

θ3

δ30

θ30

δ20

θ20

δ10

θ10

P3

P2

P1

h3

h2

h1

P3

(H0,3)

(H0,2)

(H0,1)

ΣP3

ΣP3

Σ(P2 + P3)

Σ(P2 + P3)

Figure 7.16 The activation of ‘fictitious’ horizontal forces


Thus the following deviation forces act at the girder level of each storey:

Ha,3¼ [3 P

P3]

Ha,2¼ [2 (P

P3þP

P2) 3 P

P3]

Ha,1¼ [1 (P

P3þP

P2þP

P1) 2 (P

P3þP

P2)]

The final deformation of the frame will consist of the initial deformation plus the oneproduced by the forces Ha,1, Ha,2 and Ha,3, following of course a first-order procedure.Introducing the flexibility coefficient ik as that displacement of level i which is

produced by a unit horizontal force acting at level k, as shown in Figure 7.17, it canbe written that (see Figure 7.16)

1 ¼ 10 þHa;1 11 þHa;2 12 þHa;3 13 ¼ 1 h12 ¼ 20 þHa;1 21 þHa;2 22 þHa;3 23 ¼ 1 h1 þ 2 h23 ¼ 30 þHa;1 31 þHa;2 32 þHa;3 33 ¼ 1 h1 þ 2 h2 þ 3 h3

Substituting the above expressions for Ha,1, Ha,2 and Ha,3, a linear system of threeequations with respect to the unknown deviations 1, 2 and 3 is obtained. Thus, theeffective actions Ha,1, Ha,2 and Ha,3 can be readily determined and the state of stressof the frame may be deduced according to classical first-order procedures.In this way, the final deformation and state of stress of the frame may be found as the

result of the coexistence of the initial deformations 10, 20 and 30 and the verticalloads. It is clear that if 10, 20 and 30 are due to the action of horizontal loads H0,1,H0,2 and H0,3 (see Figure 7.16) under first-order conditions, then the fictitiousforces Ha,1, Ha,2 and Ha,3 resulting from the above system have to be added to theinitial ones. Thus the final response of the frame will, under first-order consideration,correspond to the horizontal forces

H1¼H0,1þHa,1

H2¼H0,2þHa,2

H3¼H0,3þHa,3

289

1 Level 3

1 Level 1

1 Level 2

δ33

δ23

δ13

δ31

δ21

δ11

δ32

δ22

δ12

Flexibility coefficients

Figure 7.17 Description of the horizontal flexibility of the frame


7.4.2.2 Critical loadA factor is now sought which, by magnifying the vertical loads of the frame, makesthem critical for the elastic stability of the frame. Clearly, in this case the verticalloads should have such values that the activated forces Ha,1, Ha,2 and Ha,3 due to anarbitrary deviation configuration 1, 2 and 3 can produce these very same displace-ments, following a first-order procedure.By omitting the initial deviations, the equations given above (see Section 7.4.2.1)

may be rewritten as

1 ¼ Ha;1 11 þHa;2 12 þHa;3 13 ¼ 1 h12 ¼ Ha;1 21 þHa;2 22 þHa;3 23 ¼ 1 h1 þ 2 h23 ¼ Ha;1 31 þHa;2 32 þHa;3 33 ¼ 1 h1 þ 2 h2 þ 3 h3

where

Ha,3¼ [3 P

P3]

Ha,2¼ [2 (P

P3þP

P2) 3 P

P3]

Ha,1¼ [1 (P

P3þP

P2þP

P1) 2 (P

P3þP

P2)]

It is clear that the above relationships lead to a homogeneous system with respect to 1,2 and 3, i.e. a linear system in which all the constant terms are equal to zero. As it iswell known, such a system has a solution different from zero only if the determinant ofthe coefficients of the unknowns 1, 2 and 3 is equal to zero. This requirement leads toa third-degree algebraic equation with respect to , and it is obvious that the minimumreal value of satisfying this equation will represent the sought factor.

7.5 Lateral buckling of beamsThe resultant force of the compressive stresses on a beam subjected to gravity loadsrepresents a danger for the ‘stability’ of the upper part of the beam section, and conse-quently for the whole beam, given the reduced moment of inertia of this part (and of thebeam) about the vertical axis of symmetry. The situation is best visualised using theexample of a truss in which all the compression members have a buckling length sK(see Section 7.2) approximately equal to their actual length, whereas the whole upper(compressed) chord has as ‘buckling length’ equal to the whole span length for apossible deformation out of the truss plane. Under certain conditions the upper part(flange) of a symmetrical beam may bend out of the plane of symmetry, even if thebeam loads act in this plane (Figure 7.18). It is understandable that this deformationis inhibited by the torsional stiffness of the beam, which is obviously mobilised; while,on the other hand, the presence of the longitudinal internal tension force in thelower part of the beam plays a relieving role. This situation can become crucial forthe stability of the beam if the bending moment induced by the beam loads reaches acertain critical value. According to a classical result referring to a simply supportedrectangular beam, with ends restrained against torsion, and loaded in its vertical

290


symmetry plane, this valueMcr depends on the moment of inertia Iy of the beam sectionabout its vertical axis, the torsional moment of inertia IT and the beam length L (Gaylordand Gaylord, 1957):

Mcr ¼ LffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiG IT EIy

qwhere the dimensionless coefficient depends on the form of the bending diagramM andtakes the following values:

. for constant M, ¼ 1.0

. for parabolic M (uniform load), ¼ 1.13

. for triangular M (concentrated load at midspan), ¼ 1.35

. for a moment M applied at one end, ¼ 1.77.

The above expression is approximately valid also for I-beams having a section of doublesymmetry.It should be noted that the side on which the beam load is applied influences the

examined response, as may be seen in Figure 7.18. Applying the load on the top sideleads to a more adverse response than applying it on the bottom side, as in the lastcase a ‘relieving’ torsional moment is present throughout the beam length. g

The case is now examined where a simply supported beam of depth h with a single planeof symmetry (the vertical one) has an initial deformation which is characterised by thehorizontal deviations e0 and eu of the upper and lower part of the section (flange),respectively. The upper and lower part actually provide the bending resistance to thelateral response through their corresponding moments of inertia Io and Iu about the

291

Truss visualisation

Very high slenderness in the lateral direction

Through the lateral deformation the torsional resistance is mobilised

Applying the load at the bottom siderelieves the response of the beam

Figure 7.18 Visualisation of the lateral buckling


vertical axis of symmetry (Figure 7.19). Under the action of the compressive forceP¼M/h arising from an externally applied moment M, the deviation e0 will increaseand reach the value w. The beam will withstand this deviation and not lose its stability,and the value of w determines the response of the upper section under the compressiveforce P and the bending momentM¼ P w acting in the corresponding horizontal plane.The whole problem may be described by analogy to a fictitious compressed and

transversely loaded simply supported beam, such as the one examined in Section 7.2.This fictitious beam has an appropriate moment of inertia I together with anappropriate initial deformation e and is subjected to the same compressive force P asthat of the upper part of the actual beam (Mann, 1976).At first, the critical compressive force PK of the upper part (flange), which is fatal for

the stability of the considered beam (PK¼Mcr/h), may be calculated with satisfactoryaccuracy through the familiar expression for the buckling load:

PK ¼ p2 EI

L2

where

I ¼ Io ðc1 þ c2 Þ

292

Z

DM P

P

L

L

P

h

h

w

h

Z

D

M

w

w

I0

eu

eu

Iu

IT

e0

e0

Midspan section

Final deformation

Initial deformation

D = Z = P = M/h

e*EI*

Analogy with the initially deformed fictitious beam

Final deformation accordingto second-order theory

Figure 7.19 Towards the assessment of the final lateral deformation


and

c1 ¼1

2ffiffiffiffiffiffiffiffiffiffiffiffi1 Iu

Io

s

c2 ¼ffiffiffiffiffiffiffiffiffiffiffiffi1þ Iu

Io

s

¼ L

h

ffiffiffiffiffiffiffiffiffiffiffiffiG ITE Io

s

Now, after expressing the fictitious initial deformation e as

e ¼ e0 1þ P

Pþ PK 1

c2 2 c2 þ 1 c2 2 eu

eo

the final deformation w is found to be

w ¼ e 1

1 P

PK

It is remarkable that this result is identical to equation (b) in Section 7.2.It may be observed that in the case when Io¼ Iu, and where c1¼ 0 and c2¼

ffiffiffi2

p, the

last expression for PK coincides with the one given previously for the critical bendingmoment of a beam having a rectangular section (or one with double symmetry).It should be noted that, in the case of a reinforced concrete beam, the practically

negligible horizontal rigidity of its lower part, due to the existing cracked state, maybe taken into account by considering Iu¼ 0 in the above expressions. g

The problem of lateral stability is usually resolved by inserting a horizontal bracingconnecting the compressed longitudinal upper parts of two adjacent beams, or theupper chords of two adjacent trusses (Figure 7.20), in order to make a combinedelement, which exhibits a strong lateral rigidity in its compressed part. This combinedelement may serve as ‘lateral support’ for the remaining beams or trusses (seeFigure 7.20).

7.6 Plastic analysis

7.6.1 General remarksWhen a frame is to be dimensioned for horizontal forces accompanied by strongcompressive column forces on the basis of the ultimate limit state design (see Section6.6), it is necessary to take the influence of the deformations into account. In particular,the final displaced configuration of the frame must be assessed, in order to establishthose fictitious horizontal forces that will enable the analysis of the frame accordingto first-order considerations (see Section 7.4).

293


For a practical treatment of the problem, an appropriate factor is actually soughtwhich, applying to all loads on the frame, will lead the frame to collapse on the basisof a corresponding bending moment diagram, possibly also taking into account theinteraction of the internal forces M and N (see Section 6.5). Through this momentdiagram, the determination of the storey drift is also possible, using the principle ofvirtual work. The limit value of the horizontal force, which acts externally at eachgirder level, therefore includes the fictitious horizontal force corresponding to thedeformed (inclined) position of the columns (this force obviously not acting inreality). Thus the removal of this force from the limit horizontal force at each girderlevel allows the determination of the actual ultimate horizontal forces, in the contextof the deformed configuration, as well as of the load magnifying factor leading to thecollapse of the structure (Duddeck, 1984).The following example illustrates the basic points described above.

7.6.2 Numerical exampleThe steel frame shown in Figure 7.21 is made up of the cross-section HE-B 200 and isacted on by the vertical load P, as well as by the horizontal force 0.1 P. The ultimatevalue of the load P leading to failure is to be determined.The limit moment diagram (a) which leads the frame to collapse may be easily

deduced from the ‘elastic’ one (see Section 6.6). In particular, on the staticallydeterminate frame having a hinge in place of a fixed support, the plastic moment ofthe cross-section is applied externally. Then, the horizontal load is sought which willcause a plastic hinge also at the top of the column. This should balance the columnshear force, as may be deduced from the horizontal equilibrium of the girder:

0.1 P¼ 2 160.0/7.50¼ 42.66 kN

Thus the corresponding vertical load has the value P¼ 426.6 kN.

294

Bracing

Beam or truss

Beam or truss

Figure 7.20 Lateral bracing of beams or trusses


However, as it has been explained, this ultimate horizontal force will necessarilyinclude the fictitious horizontal force 426.6 (/7.50), where is the horizontal displace-ment of the frame. The latter is readily determined, according to the principle of virtualwork, on the basis of the bending diagram shown in Figure 7.21(a), considering as thevirtual loading a unit horizontal load acting on the statically determinate frame (seeFigure 7.21). It is obtained that (see Section 2.3.3)

1¼ (160.07.50/2320.07.50/6)7.5/(2.15700)þ(160.07.50/3)2.5/(2.15700)¼ 0.205m

Thus the fictitious horizontal force is equal to 426.6 (0.205/7.50)¼ 11.66 kN, andconsequently the actual ultimate force is 42.66 11.66¼ 31.0 kN and the ultimatevertical load is 31.0 10¼ 310 kN.It is also be interesting to check the ‘distance’ of this load from the critical buckling

load of the frame. As the (first-order) lateral stiffness of the frame is kH¼ 261.8 kN/m,the critical load will be (see Section 7.4) Pcr¼ 261.8 7.5¼ 1963.5 kN 310.0 kN. g

295

First-order ultimate horizontal force Second-order ultimate horizontal force

(Omission of the interaction (M, N))

Fictitious horizontal load

P

0.1 · P

0.1 · P 1 1

Equilibrium in the ultimate moment diagram

42.70 δ 7.50

7.50 m

7.50160.0

160.0

160.04.13

3.37

42.70 δ 31.0 kN

(a)

2.50 m Elastic diagram Virtual loadingUltimate moment diagram(first-order)

HE-B 200I = 5700 cm4

Mpl = 160 kN m

2 · 160.0/7.50 = 42.70

11.70 kN

Figure 7.21 Determination of the ultimate value of the horizontal load


However, it should be noted that the influence of the axial load on the ultimate bendingstrength has been omitted in the above. For the cross-section type used, HE-B, thefollowing relationship may be used for the interaction between M and N (see Section6.5.1):

M ¼ Mpl 1:1 1 N

Npl

Given that, in this case, Npl¼ 1601 kN, the previous equation for the determination ofP becomes

0.1 P¼ 2 160 1.1 (1 P/1601)/7.50

from which it is obtained that P¼ 364.0 kN, instead of the previously found value of426.6 kN.Now, the diagram shown in Figure 7.21(a), which will be used for the determination

of , instead of the plastic moment 160 kNm, has the moment 160 1.1 (1 364.0/1601)¼ 136.0 kNm, and on this basis it is obtained that

¼ 0.205 136.0/160.0¼ 0.174m

Consequently, the ultimate horizontal force is actually equal to 36.4 364.0 (0.174/7.50)¼ 27.95 kN and the ultimate vertical load is 27.95 10¼ 279.5 kN, which iscertainly reduced from the value of 310 kN.

ReferencesDuddeck H. (1984) Traglasttheorie der Stabwerke — Beton Kalender, Teil II. Berlin: Ernst.Franz G., Hampe E., Schafer K. (1991) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: Springer-

Verlag.Gaylord E.H., Gaylord C.N. (1957) Design of Steel Structures. New York: McGraw-Hill.Mann W. (1976) Kippnachweis und Kippaussteifung von schlanken Stahlbeton- und Spannbeton-

tragern. Beton- und Stahlbetonbau 2, 37—42.Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhauser Verlag.

296


8

Arches

8.1 Basic characteristics of structural behaviourArches represent that structural type which transfers gravity loads to two support pointsmainly through axial compressive forces, thus limiting the bending as much as possibleand offering aesthetically a particularly satisfactory result. This type of structure is usedeither for roofing relatively large areas or as a bridge over large obstacles (rivers or valleyswith abrupt bends) without intervening supports. In this last case, the live load isapplied to a horizontal beam lying above or below the arch, transferring the verticalloads directly to the arch through intermediate elements (usually vertical), acting assupports for the beam (Figure 8.1).The transfer of gravity loads without bending to the supports is achieved if the

arch axis precisely follows the pressure line corresponding to these loads. It is usefulto design the arch to always follow the pressure line corresponding to the permanentloads.It should be recalled that by the ‘pressure line’ between the support points of the arch,

we mean that line along which only compressive axial forces are transmitted thatbalance the externally applied gravity loads at any point, so that there is no need forthe structure to develop bending moments (and shear forces) in order to take upthese loads (see Sections 2.2.7 and 6.2.1).In this sense, for a specific vertical load g there exists an infinite number of such

‘pressure lines’, all of them governed by the equation

y(x)¼ k M0g(x)

which means that each diagram affine to the moment diagramM0g(x) of the corresponding

simply supported beam represents a pressure line. The corresponding developinghorizontal component H of the axial force at the ends of the pressure line is equal to1/k. The higher the arch (i.e. the higher the value of k) is, the smaller the horizontalthrust H (Figure 8.2). However, it should be noted, as in Section 6.1.1, that thecompressive axial force N along the pressure line increases towards the supports,although its horizontal component remains constant and equal to H everywhere.Then (see Figure 8.2), N¼H/cos.It is clear, in general, that the pressure line for a specific loading is specified only if

a third point is also found on the arch which the pressure line passes through(see Figure 8.2). For example, an arch having at its midspan the rise f (i.e.y(L/2)¼ f ) exhibits a value of k which, according to the last equation, is determined


as follows:

k¼ f/M0g(L/2)

The rise f of the arches is usually selected between one-fifth and one-tenth of the archspan.An arch with immovable supports designed in the above way essentially develops the

same axial compression forces, regardless of its statical form (three-hinged, two-hingedor fixed arch).Naturally, the above horizontal thrusts, as applied to the ground with an outward

direction, may cause a relative horizontal shift of the supports (Figure 8.3). The archis particularly sensitive to this relative shift, and is stressed in bending, depending onthe degree of static redundancy.Thus, while the three-hinged arch remains unaffected, the two-hinged arch and, in

particular, the fixed arch develop an increasing response. A reduction in f/L causes anotable increase in the sensitivity of the relative displacement (shift) of the supports,by developing bending moments that cause tension at the bottom fibres. However, in

298

Figure 8.1 The formation of typical arches

g g

g

N

N

H = 1/k H = 1/k

Mg0/f Mg

0/fy

x

f

(f /Mg0) · Mg

0(x)

Mg0(x)Mg

0α

Through the supports passes aninfinite number of pressure lines

The pressure line is determined once athird point on the arch is specified

Figure 8.2 Arch design following the pressure line


order for the outwards applied horizontal thrust to be taken up ‘internally’, it is possibleto insert a cable tie in the arch, prestressed with a force equal to the horizontal thrustfor immovable supports (see Figure 8.3). In that way, not only is the developmentof bending prevented but also, in the case of a concrete arch, a redistribution ofinternal forces due to creep is excluded for permanent loads, as explained previouslyin Section 6.2.1 (see Figure 6.7). It is obvious, then, that the reactions of the arch atthe supports consist only of two vertical forces, which is particularly favourable forthe foundation. g

It can now be seen that if an arch designed to coincide with the pressure line for thepermanent loads g is loaded with an additional live load p applied at an arbitraryposition, then the arch axis will no longer follow the pressure line corresponding tothe new loading (gþ p), so that bending will be developed (Figure 8.4). This ‘new’pressure line may be determined so that it passes through a specified intermediatepoint of the arch with zero bending moment.The magnitude of the bending moment results as the moment of the compression

force of the new pressure line with respect to the point of the arch being considered(see Figure 6.4). The fibres which will be tensioned by the bending are obviously deter-mined according to the relative position of the pressure line with respect to the point ofthe arch being considered (see Figure 8.4). g

299

The deformability of the soil causes a relative shift of the supportsand consequently a bending response due to the thrust decrease

Prestressing with the initial thrust forceensures the absence of bending

HH

Initial thrust acting on the ground

Figure 8.3 Effects due to the horizontal yielding of the supports, and the prevention of such yielding

gg + p

New loading

g Initial loading

The bending causes tension at the top fibres

Pressure line for the initial loadingPressure line for the new loading

Figure 8.4 Effects due to changes in the pressure line

Arches

At this point it should be noted that, particularly for materials with a limited tensilestrength (unreinforced concrete, masonry), bending is not prohibited in principle,provided that the compression force resultant is acting in the core of the cross-section. Thus, if the ‘new pressure line’ due to a live load lies within the core of allsections of the arch, then no tensile stresses are developed.This is examined more specifically in the following example. A two-hinged arch with

span L will bear the permanent uniform load g as well as the uniform live load p(Figure 8.5).Before the height of the arch is determined, it is clear that any arch configuration that

follows the moment diagram of the simply supported beam, that is, any second-degreeparabolic curve passing from the support points, essentially develops only axialcompressive stresses. Once the height of the arch f is determined, however, then the

300

Tension on the top fibresTension on the bottom fibres

p + g

p

f

g

h

g

g

Mg + p

Bending diagram valid also for the actual live load p

p · L2/64

p/2

L/2

p/2

p/2

p/2

p/2

p/2

M = 0

M

H = 0

H H = Mg0/f

Passing through the core of the sectionmeans bending without tensile stresses

Pressure line for (g + p)

The arch does not develop bending

L

Resultant compressive force

~H

~H

(M/H )

Mg0 = g · L2/8

(f/Mg0) · Mg

0(x)

Figure 8.5 Load-bearing function and stressing of an arch under an additional live load


equation of the pressure line, and hence the arch axis, will take the form

y(x)¼ (8 f/q L2) M0g(x)

while the horizontal componentH of the reaction will beH¼ q L2/(8 f ). It is clear thatthe axial force at the top of the arch also has the same value.The arch, of course, should also be examined for the live load p. It is obvious that

considering the live load over the whole length of the arch will proportionallyincrease the axial stressing without causing bending, because the form of the arch willstill be coincident with the pressure line that corresponds to the loading gþ p.However, as has been already mentioned for single-storey frames with inclined legs(see Section 6.2.3), the most unfavourable layout for the live load corresponds to theloading of only the half of the structure with the load p (see Figure 8.5).This loading comprises a symmetric part and an antisymmetric part, both being equal

to p/2. The symmetric loading increases the axial compression without causing bendingmoments. The antisymmetric loading causes zero axial stresses at the top of the arch,making, in consequence, the horizontal components of the support reactions, as wellas the bending moment at the crown, null (see Section 2.4.2). Moreover, accordingto the antisymmetric loading, the resulting vertical displacement at the crown is alsozero. The pressure line corresponding to the total load gþ p will pass, as statedpreviously, through the midpoint of the arch, where the bending moment is zero.Thus, the factored moment diagram of the simply supported beam can be appropriatelydetermined so that the pressure line passes through this point (see Figure 8.5). Theequation of this ‘new’ curve is

yðxÞ ¼ ½ f=M0gþ pðL=2Þ M0

gþ pðxÞ

Consideration of the pressure line for the loading gþ p, despite the fact that the arch wasformed for a pressure line corresponding to the load g, is important for an understandingof arch stressing. As also shown in Figure 8.5, the deviation of the arch axis from the‘new’ pressure line is clearly hints at the development of bending moments that causetension at the bottom fibres in the left half of the arch and at the top fibres in theright half. The absence of horizontal forces at the crown and at both ends of the archunder the antisymmetric loading p/2 allows one to conclude that the maximummoment developing in each half of the arch corresponds to

MF¼ ( p/2) (L/2)2/8¼ p L2/64

Of course, for a masonry arch, the already determined ‘new’ pressure line should not passoutside the core of each section (see Figure 8.5). Considering approximately (but never-theless conservatively) that at a quarter of the arch span the compression force due togþ p/2 is equal to H¼ (gþ p/2) L2/8 f, it is concluded that, for a cross-sectional depthequal to h,

h

6 M

H¼ p L2

64 8 fðg þ p=2Þ L2

¼ p f8 ðg þ p=2Þ

301

Arches

From the above condition and by introducing the non-dimensional ratio ¼ h/f, thefollowing limit live load value can be obtained:

p ¼ 4 g 3 2

Arch bending is favoured by the ‘free prestressing’ offered by the axial forces, so that theeventual reinforcement needed is more favourable than in a beam or even a frame girder,because of the presence of compression force.Considering now fixed arches, the antisymmetric loading p/2 induces bending

moments at the supports, which cause tension at the corresponding top and bottomfibres, as shown in Figure 8.6.This bending intensity is very little affected by a reduction or an increase in the ratio f/L,

being near the mean value of 0.970 ( p/2) (L/2)2/8¼ 0.015 pL2 (see Figure 8.6). gIt should be noted that, due to the existing slope of the arch axis, the distribution of self-weight over a horizontal projection is not uniform, being increased towards both ends ofthe arch (Figure 8.7). This fact, also depicted in Figure 8.2, leads to a pressure line that isdifferent to the second-degree parabolic one, and hence the arch with a parabolic axisdevelops some bending that is significantly weaker as the ratio f/L decreases. g

An arch with a large span develops high compressive forces under permanent loads,which cause elastic shortening of its axis, resulting in deviation from the pressureline, and hence bending, so that, strictly speaking, purely axial stressing of the arch is

302

Bending diagram valid also for the actual live load p

p/2

p/2 p

p/2

p/2

p/2

Parabola of second degree

H = 0

M = 0

M

L/2

M ≈ 0.970 · M0M0 = p · L2/64

Figure 8.6 Fixed arch behaviour due to antisymmetric loading

Realistic distribution of self-weight

g

Figure 8.7 Non-uniform distribution of the self-weight


only theoretically feasible, even under permanent loads (Figure 8.8). In the case of atwo-hinged arch, shortening of its length will require self-induced horizontal reactionsdirected outwards, in order to restore immovable supports in the statically determinedsystem, i.e. in the same sense as the horizontal reactions which are developed due toa horizontal outward yielding of the supports, as previously considered. This willnaturally result in a reduction in the horizontal thrust of the arch under the permanentloading, and consequently the corresponding pressure line must be considered to beshifted upwards from its initial position — i.e. the arch axis — in order to conform tothe decreased value of H. This upward deviation of the pressure line from the archaxis means, as can be seen in Figure 8.8, that the development of bending momentscauses tension at the internal fibres of the arch.The crown deflection wc of an arch developing a horizontal thrust H under a uniform

load g, due to the elastic shortening of the arch axis, may be roughly estimated from theexpression (Menn, 1990)

wc ¼H

E A L ½1þ 3ð f=LÞ24 ð f=LÞ

where A represents the cross-sectional area of the arch.In the case of a two-hinged arch, the bending momentMc induced at the crown by the

above deflection (Figure 8.9) may be assessed from the expression

MðhÞc ¼ 8 EI

L2 wc

where I represents the moment of inertia of the arch section.

303

g

g

HH

H – HN

H – HN

H – HN

HN

Decrease in horizontal thrust

Tension at the bottom fibres

(Reactions)

The pressure line coincides with the arch axis

The pressure line for H The pressure line for (H – HN)

Deformed axis

Upward shifting of pressure line due to thrust decrease

Figure 8.8 Cause of bending response due to shortening of the arch axis

Arches

The corresponding bending response in the case of a fixed arch Mc and Ms at thecrown and the fixed supports, respectively, may be approximately assessed as (Menn,1990)

MðfÞc ¼ 16 EI

L2 wc

MðfÞs ¼ 32 EI

L2 wc

At this point, it should be emphasised that in large concrete arches the presence of creep‘following’ the intense compressive stressing of the concrete always creates additionalshortening of the arch axis, which means deviations from the ideal pressure line and,hence, bending. This is also an appropriate point to mention that awareness of thecreep phenomenon itself happened about 100 years ago, when a particularly shallowthree-hinged concrete arch of 75m span with a rise-to-span ratio f/L equal to 15,designed by the then very young French engineer Eugene Freyssinet, showed somemonths after the completion of the bridge, and despite a successful decenteringprocedure, an unexpected sinking of 13 cm at the crown. This fact gave rise to theidea of an ‘unacceptable’ reduction in the modulus of elasticity of concrete with time,which allowed Freyssinet to conceive of and study the phenomenon of creep (seeSection 1.2.2.1).

8.2 Elastic stability — second-order theoryThe bearing capacity of an arch is based on the development of high compression forces,which means that particular attention must be paid to ensure its elastic stability(buckling), on the one hand, and to the fact that its response may be modified by thedevelopment of transverse deformations — according to second-order considerations —on the other, as for example in the case of the antisymmetric loading p/2 studied inthe previous section.Arch buckling leads, generally, to an antisymmetrical deformed configuration

(Figure 8.10). A conservative estimate of the critical compressive force Hcr (thrust)of the arch under a uniform load may result from the following expression(O’Connor, 1971):

Hcr ¼ C EIL2

304

Two-hinged arch Fixed arch

Mc(h)

Ms(f)

Mc(f)

Figure 8.9 Assessment of the bending response due to shortening of the arch axis


where the factor C depends directly on the ratio f/L as well as on the arch type. A fixedarch develops roughly double the critical compressive force of the two-hinged arch, andthis force increases as long as the ratio f/L decreases. The latter shows, for values of f/Lgreater than 0.2, practically the same critical thrust as the three-hinged arch. Indicativevalues of the factor C are given in Table 8.1.As mentioned previously, any loading q, the pressure line of which deviates from the

arch axis, causes bending. However, the transverse deformation caused by this bendingaugments the initial deviation further, consequently causing an additional bending, andso on, until the final convergence. Thus, the final bending depends on the finaltransverse deformation of the arch (second-order theory). This final bending obviouslycorresponds to the deviation of the deformed arch axis from the pressure line of loadingq, which can immediately be seen from an examination of the two-hinged arch example(see Figure 8.10).

305

Tension on the top fibresTension on the bottom fibres

qcr

p

g

e

w

w

e

Pressure line for (g + p)

Undeformed arch axis

Deformed arch axis

Buckling deformation

Arch axisHcr Hcr

Figure 8.10 Deformed state due to the application of a live load

Table 8.1 Values of factor C

f/L 0.1 0.2 0.3 0.4 0.5

Fixed-arch 75.2 63.2 47.7 35.1 24.8Two-hinged arch 35.0 28.8 20.0 13.6 9.2Three-hinged arch 27.8 24.8 20.0 13.6 9.2

Arches

It is obvious that, due to the total load gþ p, a downward shift w will take place at thepoints on the left half of the arch, due simply to the antisymmetrical part p/2 (seeFigure 8.10). The pressure line, however, corresponding to the total load gþ p, passesin the same left half above the undeformed arch axis at a distance e, and the total diver-gence from the actual arch axis thus comprises the sum of the above two deviations, i.e.wþ e, and is not due to w alone, as is the case for the rectilinear members according tothe second-order theory (see Section 7.3).Given that the pressure line gþ p passes in the examined left part above the deflected

arch axis, the resulting bending will cause tension at the bottom fibres, while in the rightpart, where it lies under the deflected axis, the top fibres will be tensioned (seeFigure 8.10).A calculation with regard to second-order conditions is generally necessary. Naturally,

the more the developing compressive force H diverges from the critical value Hcr forbuckling, the more the additional bending moment decreases. Considering a two-hinged arch again, it is possible to assess the final deflection w (occurring at thequarter point of the arch span), on the basis of the deviation e of the pressure linefrom the undeformed arch axis. Thus, according to the previous section (seeFigure 8.5),

e ¼ MF

H¼ p L2

64 8 fðg þ p=2Þ L2

¼ 1

ðg=pÞ þ 1=2 f8

The deflection w may be estimated (somewhat conservatively) through the equation(Franz et al., 1991)

w ¼ e H=Hcr

1H=Hcr

According to the above, the final value M of the ‘first-order’ bending momentMF¼ p L2/64 will be equal to

M ¼ MF eþ w

e¼ p L2

64 1

1H=Hcr

where an analogous ‘magnification factor’ is applied to the initial value of the bendingmoment MF, as also is the case for beams and frames, as discussed in Chapter 7.The compressive force H corresponds to the total load gþ p and — as previously

explained — is due to the symmetric part only (gþ p/2):

H ¼ ðg þ p=2Þ L2

8 f

The estimated values H and M may then represent a basis for the preliminary designchecking of the arch. However, using a global safety factor applied to service loads,the check should be made analogously to Section 7.3, for an axial force

Nd¼ H

306


and a bending moment

Md ¼ p L2

64 1

1 H=HK

g

At this point, the possibility of an out-of-plane buckling of a free-standing arch will beconsidered, in the sense already examined in Section 7.5, with regard to the lateralbuckling of beams. For arches, the degree of end fixity has a major influence on thisbuckling behaviour, the fixed arch having in this respect a definite advantage overthe two-hinged arch. Although in an out-of-plane deformation the torsional rigidityof the arch offers some help, this resistance can be neglected in the preliminarydesign and, instead, the moment of inertia of the section Iy about a vertical axis willbe considered as the main resistance factor.Thus, for a fixed arch, taking as the ‘effective length’ the half-length of the arch axis,

in analogy to the corresponding basic buckling formula according to Section 7.1, thefollowing expression may be used in order to approximately estimate the criticalcompressive axial force NK for the out-of-plane buckling:

NK ffi4 2 EIy

L2 ½1þ ð8=3Þ ð f=LÞ22¼ 39:5

½1þ ð8=3Þ ð f=LÞ22EIy

L2

In the case of two-hinged arches, as used in bridge construction, especially when theload-bearing roadway lies under the arch (see Section 8.4), in order to ensure (orenhance) the lateral stability of the two closely arranged arch ribs, a bracing mustbe arranged between them, consisting usually of horizontal ribs perpendicular to theplane of the arches and, obviously, high enough in order not to interfere with thetraffic (Figure 8.11). In this way, a ‘curved frame’ is formed, the buckling behaviourof which may be judged if considered to be plane, and investigated, as Figure 8.11suggests, according to the procedure described in Section 7.4.2.

307

Axial compressive force N

N

N

N

N

Figure 8.11 Lateral bracing of arches

Arches

8.3 The girder-stiffened arch systemAs mentioned at the beginning of this chapter, the transfer of traffic load from one pointto another through an arch imposes the requirement for the construction of a horizontal,or at least slightly inclined, deck (or girder). The deck transfers its loads to the archusually through vertical elements that provide, primarily, a direct support to the deck(see Figure 8.1).Initially, the structural function of the horizontal girder was limited simply to bearing

and transferring the live loads onto the arch via the vertical supporting members. Atsome point, however, it became clear that the whole structural action would be muchmore effective if the combined load bearing of the horizontal girder and the archcould be considered. This is explained below (Figure 8.12).The unfavourable arrangement of the live load in one-half of the structure, say the

left part, understandably causes a lowering of the left part of the arch and a raising ofthe right part. Unfavourable bending of the arch due to this deformation is con-siderably limited if the horizontal beam is made to follow it. However, in order for thebeam to deform in the same way, it is clear that it must receive downward forces in

308

The final arch loading tends to become uniform

The arch receives the opposite forces

The arch deformsand

the girder followsbut

it requires the application of forces

pg

EI G

EI A

Figure 8.12 Static operation of the girder arch system under a live load


the left part and upward forces in the right. These forces will obviously be providedby the arch, which will itself then receive by reaction the opposite forces, i.e.additional upward and downward forces will be applied to the left and right parts ofthe arch, respectively. As a result, the unfavourable unilateral loading of the archthat exists without participation by the horizontal girder tends to become uniformbecause of the correspondence of deformations, so that the bending of the arch isclearly limited (see Figure 8.12).It can be seen that the bending moment M of the arch, which, according to Section

8.1, is due only to the antisymmetric part ( p/2) of the live load, will now be distributedproportionally between the arch (MA) and the girder (MG), according to the bendingstiffness of the two elements (see Section 3.2.10), i.e.,

MG ¼ M EIG

EIG þ EIA(a)

MA ¼ M EIA

EIG þ EIA

In the extreme case where the girder stiffness EIG is much higher than the arch stiffnessEIA, bending will be experienced almost entirely by the beam, while in the opposite caseit will be experienced entirely by the arch. As neither case is aesthetically satisfactory,the solution EIGEIA is usually adopted, whereby the bending response is almostequally distributed. An estimate for the magnitude M of the distributed bending inthe examined case is equal to p L2/64 or p L2/113.6 in the two-hinged or fixed arch,respectively, according to Section 8.1 (Figure 8.13).If, now, the arch is formed according to the pressure line for permanent loads, it

can be seen that the girder will function essentially as a continuous beam onunyielding supports for these loads. However, as pointed out in Section 8.1, underpermanent loads, especially for long spans, a transverse deformation due to axialshortening of the arch is developed. The deflection wc at the crown (Figure 8.14),assessed in Section 8.1, will also be imposed on the girder, thus causing additionalbending stresses.

309

p

Two-hinged arch

Fixed arch~0.56 · M

~0.97 · M

M = p · L2/64

M

Figure 8.13 The effect of a live load on arch bending

Arches

Assuming fixed-end supports for the girder and the arch too, the beam bending maybe estimated for preliminary design purposes, through the expression

MG ¼ 24 EIG

L2 wc (see Section 3.3.3)

causing tensile stresses in the bottom fibres of the midspan and in the top fibres of the fixedsupports.It is obvious that the response of the girder as a continuous beam on unyielding

supports should be superposed to the above.The corresponding bending response of the fixed arch Mc and Ms at the crown and

the fixed supports, respectively, may be assessed as in Section 8.1 (Menn, 1990):

Mc ¼16 EIA

L2 wc Ms ¼

32 EIA

L2 wc

where IA represents the moment of inertia of the arch. g

The association between the girder and the arch increases perceptibly the buckling axialforce of the arch. A conservative estimate of the critical axial force Ncr of the fixed archis given by the expression (see Menn, 1990)

Ncr ¼2 EIc

ð0:7 L=2Þ2

where EIc¼ (EIGþEIA)/2 may be considered, or even its half value, allowing for possiblecracking.In the case of a concrete structure, the girder is usually prestressed in order to ensure

uncracked conditions in the serviceability state and thus a higher available stiffness EIG,as well as being able to deal more effectively with the developing alternating bendingmoments, depending on the left or right placement of the live load.Attention must also be paid here to the modified bending under the live load p, due to

the developing transverse deformations of the arch (second-order effect), which, ofcourse, also affects the girder, and is shared between them according to the aboveequations (a). The distributed bending moment M may now be roughly calculated,analogously to Section 8.2, following the expression

M ¼ MF 1

1N=Ncr

where MF represents the ‘first-order’ maximum span moment of the arch and N is theaxial force due to the permanent load and live load.

310

L

EI G

EI A wc

Figure 8.14 Check of the deflection of the arch top


8.4 The tied-arch systemIt is possible, and often desirable, to suspend the roadway from twin arch ribs from belowthrough hangers which transfer the applied loads on the deck upwards to the arch,instead of supporting the girder on the arch from above. The so-formed spatial bridgestructure may be considered as basically a plane arch system, such as the one depicted inFigure 8.1, where the deck is represented by the horizontal ‘tie’ element, which, beingsuspended from the arch, may also be treated structurally as a ‘girder’ stressed both axiallyand in bending. This type of bridge is used whenever the free space under the traffic deckis restricted, as is the case for low-level crossings, over rivers, etc., or even if the availablefoundation soil is not suitable for conventional arch foundations. As all the permanentloads are transferred to the arch through the hangers, the arch may be formed accordingto the pressure line of the loads, in the same way as discussed in Section 8.1. The relativelydense layout of hangers ensures that the arch receives the applied load on the ‘girder’ in arather uniform manner, so that the parabolic form of the arch is usually imposed. Giventhat the arch for permanent (uniform) loads is stiff, it can be seen that the hangers maybe appropriately stressed (prestressed) by the reaction forces they offer to the girder, andconsidered as a continuous beam over unyielding supports. Then, it is appropriate to alsoprestress the tie girder in order to provide the arch with the horizontal thrust required fora normal two-hinged arch. In this way, as has been explained in Section 8.1, no horizontalthrust is transferred to the foundation, which essentially receives vertical forces only.Although the deck may be made of concrete, which, if prestressed as mentioned, canwithstand the high tensile force the tie has to offer, it is more suitable for the arch, forboth constructional as well as for aesthetic reasons, to be made of steel.Although all the characteristics of the structural behaviour of the arch itself, the

stability issues included, as described Sections 8.1 and 8.2, are obviously also valid forthis type of arch system, it would nevertheless be useful to again examine thecarrying mechanism for the live load.As mentioned in Section 8.1, for the most unfavourable layout of the live load p,

namely the loading of only one half of the girder, given that the symmetric part ( p/2)causes no bending, it is the antisymmetric part ( p/2) that needs to be examined(Figure 8.15). Under this antisymmetric load the equivalent half-system acts approxi-mately like two simply supported beams connected through a tension element andsubjected to a load applied to the lower one. It is obvious from Figure 8.15 that theresulting bending moment M¼ p L2/64 has to be taken up from both the girder andthe arch. According to Section 3.2.10, if MA and MG represent the bending regionsof the girder and the arch, respectively, then they may be roughly assessed as

MA ¼ M 1

1þ MG ¼ M 1

1þ 1=

where

¼ 48 EG IG

ðL=2Þ3

ðL=2Þ3

48 EA IAþ f

2 EH AH

and the superscripts G, A and H refer to the girder, the arch and the hanger, respectively.

311

Arches

It is clear that both the girder and the arch have a more or less substantial bendingresponse to undergo, the reason for this being obviously the fact that both elementstake up the transferred forces only through bending.However, the situation changes if the truss-like layout of diagonal hangers is applied,

as in Figure 8.16. Then, the live load on the girder can be taken up in a much morefavourable way, because the whole system acts as a truss, which is structurally more

312

M = 0 M = 0

p/2

L/2

p/2

p/2

p/2

p/2

g

g

f

H

H = Mg0/f

EA · IA

E H · AH

E G · IG

M = p · L2/64

Figure 8.15 Load-bearing action of a tied arch for permanent and live loads

p

p/2

p/2

p/2

p/2

p/2

M = 0

f

H

H = Mg0/f

g g

The hanger forces are practicallytaken up only through bending

Possibility to take up the live loadsthrough a truss action

Figure 8.16 Favourable load-bearing action through a diagonal layout


effective than a bending element. Although the bending in both the girder and the archis not eliminated, it is nevertheless significantly reduced, to a value which is about 15%of that corresponding to vertical hangers, thus leading to an analogously smallerdeflection. Moreover, it is clear that the diagonal layout of hangers leaves the axialforces in the arch and the girder tie for the permanent loads (as well as the symmetricpart of live load) essentially unaffected.A further noticeable reduction in the deflection and in the bending moments in both

the girder and the arch is possible if a network of diagonal hangers is used, as shown inFigure 8.17 (Tveit, 1966).The hangers have a constant slope, and cross each other at least twice. In this case,

the axial forces both in the arch crown and the tie are slightly changed. As can be seenfrom Figure 8.17, the existing hanger tensile forces under permanent loads at themidspan section suggest that, for horizontal equilibrium, the crown compressive forceand the tie tensile force have to be increased and decreased by half of the resultanthorizontal force of the hangers (see Figure 8.17). It is clear that suitable computersoftware has to be used for a thorough analysis of the system.

ReferencesFranz G., Hampe E., Schafer K. (1991) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: Springer-

Verlag.Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhauser Verlag.O’Connor C. (1971) Design of Bridge Superstructure. New York: John Wiley.Tveit P. (1966) Design of network arches. Structural Engineer 44(7), 247—259.

313

q

q

f

R/2Mq0/f

Mq0/f R/2

R = q · f/tan2 ω

Figure 8.17 Network tied-arch layout

Arches

9

Cable structures

9.1 OverviewThe main constituent element of cable structures is the high-strength steel cables thatpresent particularly favourable load-bearing properties. The high tensile strength of steelcables, their main characteristic, allows them to take up large axial forces with smallcross-sections, and thus they are more economical in terms of price/offered strengththan structural steel. As the cable lengths used are very large with regard to theircross-section, the cables present an inherent flexibility, and thus do not offer resistanceto bending moments or compressive forces. Therefore, cables can only develop tension,and this allows the complete exploitation of the cable cross-section, as the danger ofbuckling does not exist. The aforementioned characteristics generally make cablestructures the most economic ones.Cable structures present unlimited possibilities for the creation of the most versatile

forms, and are particularly useful for economically covering large areas under roof forms,and as effective basic bearing members in bridges having very large spans.In this chapter the main load-bearing characteristics of various, mainly ‘plane’, forms

of cable structures are examined, including cable—beam roofing systems, suspended roofshaving relatively large spans, and suspension and cable-stayed bridges.The basic structural characteristic of cables is that they automatically acquire the

so-called ‘funicular form’ for every load pattern acting on them, whether the loadsare concentrated or distributed. This funicular form implies the development ofexclusively tensile forces at each point of the cable, which directly raises the problemof safely taking up the anchoring forces of the cable at both its ends. The solution ofthis problem is of great economic and aesthetic importance, and plays a decisive rolein the whole structural design (Figure 9.1).Of course, it should be pointed out that for a given plane loading there are two

families of funicular forms that can take up this loading exclusively by axial forces.The first form develops exclusively tensile forces and concerns the ‘cables’, while thesecond develops exclusively compressive forces and concerns the ‘arches’. It isobvious that the two form families are mirror images of each other.As became clear in Chapter 8, an arch is a structure having a funicular shape for some

system of coplanar loads, and for a different loading the same arch will no longer havethe same funicular structure, resulting in the development of bending. This does nothappen in the case of a cable, wherein the new system of loads will continue to causeexclusively tensile stresses. However, there is the serious disadvantage that the cable

Structural systems: behaviour and design Copyright Thomas Telford Limited # 2010

will automatically adopt a new funicular form because of a lack of bending stiffness(contrary to what happens in an arch), which means a more or less drastic change inthe form of the cable. A typical example of this problem is the application of winduplift forces on cables that have been appropriately formed for gravity loads (seeFigure 9.1). The upwardly applied wind loads may possibly exceed the existingpermanent loads, thus leading to the danger of an intense change in the cable form.A first measure for tackling this problem could be to increase the dead weight merely

by adding more weight along the length of the cable. However, this is an uneconomicsolution (Figure 9.2). The structurally proper solution is to ‘stiffen the cable’ bycovering the already suspended cable with concrete of constant cross-section andapplying additional tension to the cable itself as prestressing. The vertical upwarddeviation forces that are thus developed will set the so-formed inverse concrete archunder pure axial compression.This structural configuration, by retaining the form of the suspension cable practically

unaffected, allows the additional gravity loads to be borne as well as the upward loads.

316

Funicular form (tension)

Wind uplift forces may overtake the gravity loads

Cable sensitivity

Change in loading means change in the geometry

Taking up the same loading through an arch (compression)

Problem raised in taking up the anchor forces

A A

Solution I Solution IIEquilibriumat node A

High anchor force Less anchor force

Figure 9.1 The cable as a funicular form for carrying vertical loads


For the vertical gravity loads, the already prestressed arch, acting in an inverse way to itstypical function (see Chapter 8), receives an additional tension, which is suppressed bythe already existing compression forces, while the potential upward loads are taken upaccording to the typical arch action (see Figure 9.2). In both cases, the bending stiffnessof the arch ensures the non-deformability of the cable, even for additional asymmetricloads. This topic is treated more thoroughly in Section 9.6.

9.2 Cable—beam structuresA purely cable plane structure that acts as a single-span beam basically has the form of atruss, which, even before accepting the gravity loads that it is intended to receive, is inequilibrium, with all its members being under tension through prestressing. In such atruss, in which compression members cannot naturally exist, the upper and lowerchords will necessarily take a curved form, concave upward and downward, respectively.The reason for this is that the chords, each of which consists of a single cable, will makethe funicular form due to the tensile forces of the intermediate members that are appliedon them; these forces are directed downwards for the top chord and upwards for the

317

Typical arch action (compression)

Adding dead weight removes somehow the sensitivitywithout representing a structurally sound solution

Covering with a concrete thin layerand afterwards further tensioning (prestressing)

Deviation forces due to prestressing

Development of initial compressive stresses

Self-contained prestressing does not affect the anchorage forces

Arch working in tension

Additional loads

Figure 9.2 The need for stabilisation of the cable form

Cable structures

bottom chord (Figure 9.3). It is clear that, at each node, any existing external forces,along with the concurrent axial forces, must be in equilibrium.In a similar beam-like structure, the cables of the upper and bottom chord may alter-

natively be connected through vertical axially compressed struts, where the curvaturesof the two cables are obviously reversed (see Figure 9.3). In any case, the structureshould be subjected to such an initial tensile stress state through prestressing that theresulting compressive forces due to the imposition of external loads (gravity, wind,etc.) will be suppressed by the tensile forces introduced during prestressing. This willprevent the development of compressive forces anywhere within the structure, whichmight lead to slack members and, consequently, to failure of the whole system(Figure 9.4).It should be clearly understood that the design and behaviour of this type of cable

structure are based on the combined influence of the introduced prestressing in alltheir members and the mandatory consideration of equilibrium at the deformed state,i.e. at the final positions of the nodes. Thus a cable, which in the initial state ofprestressing is horizontal and symmetrically placed with respect to the vertical axis ofsymmetry of the structure, can, with suitable prestressing, offer in the case of anti-symmetrical loading the required (for equilibrium) vertical force along the axis,through the inclined position that the cable adopts with deformation.In order to illustrate this last condition better, the symmetrical system shown in

Figure 9.5 is considered. The system is composed of an upper and bottom cable andtwo vertical struts. It is assumed that the system is prestressed with the prestressingforces given in the figure. The reactions at the supports of the system are zero. Theapplication of an antisymmetrical loading causes antisymmetrical reactions which, dueto equilibrium at the support nodes, require the development of certain axial forces in

318

Without external loading, initial tensioning(prestressing) in all members

Compressed struts

Funicular form for the acting forces

Funicular form for the acting forces

Figure 9.3 Cable—beam formation


319

P

P

F

F

FF

FF

Equilibrium

Taking up of compressive forces is impossiblewithout prior prestressing of cables

Tension

Tension

Tension

Compression

Tension

Tension

Com

pres

sion

Com

pres

sion

Com

pres

sion

Com

pres

sion

Tension

TensionTension

Tension

Tension

TensionTension

Tension

Compression Compression

Figure 9.4 The importance of prestressing in the formation of cable structures

L

a

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Tension

Stressless

Stressless

Antisymmetric loading

Tension

P

P

P

P

F · (a/L)

F · (a/L)

F · (a/L)

F F

FF

F

F

F · (a/L)

F · (a/L) F · (a/L)Deformed configuration Equilibrium possible thanks to theinclined axial forces

Impossible vertical equilibriumCompression

Compression

Tension

Tension Tension

TensionTension

Tension

Tension

Com

pres

sion

Com

pres

sion

Figure 9.5 The need for prestressing in maintaining the equilibrium in the deformed configuration

Cable structures

the inclined cables. This, however, makes the maintenance of the overall equilibriumimpossible if the system is examined in the undeformed configuration. Indeed, thetensile forces of the horizontal cables resulting exclusively from the initial prestressing(the loading itself induces no forces) are unable to offer the required upwards verticalforce F (1 a/L) for equilibrium.It is clear that only after considering the deformed inclined position of the horizontal

cables, with their appropriate prestressing forces, can the required total vertical force inquestion be offered (see Figure 9.5). Of course, it is understood that the prestressingforce should be sufficient such that the equilibrium will be obtained with acceptablevalues of the nodal deformations. g

Usually, when determining the permanent loads in a symmetrical structure andfor preliminary design purposes, the deformed geometry may be ignored, by followingthe general procedure described below. However, for live loads with an arbitrarylayout, one must make an estimation using special software programs that take intoaccount the deformed geometry (position) of the nodes, when checking for equi-librium, through successive iterations. Note that, due to the large deformations andthe ‘geometric non-linearity’ involved, the superposition principle is not generallyvalid.The preliminary design is performed first for the permanent loads, thus determining a

functional and aesthetically acceptable geometrical form in combination with theadopted anchoring layout for the relevant tensile forces. It is clear that all the aboverequire the explicit determination of the introduced prestressing, which thus constitutesan essential parameter for the preliminary design.In any case, finalisation of the geometry in the initial prestressed state under the

permanent loads will mean that the equilibrium requirements are satisfied in theprecisely determined nodal positions, but this can be resolved only with the use of asuitable software program. However, for preliminary design purposes, the determinationof the required cable prestressing can be done as described below.As a first step, the tensile and compressive axial forces that are in equilibrium with the

considered external nodal permanent loads are determined. These axial forces, which,along with the external loads, keep each node in equilibrium, may be determined byanalysing the system as a plane frame with rigid joints (by means of a typical softwareprogram for framed structures) and by considering an extremely small bending stiffnessof the members, so that the developing bending moments and shear forces are limited topractically vanishing values.The sought-after self-equilibrating state of prestressing should be such that, if super-

posed on the results obtained in the first step, it will lead all the members to a tensilestress state. This prestressing state may be determined, as a second step, by imposinga uniform temperature decrease (or increase, if needed), e.g. of 1008C in a selectedgroup of bars, and then searching for the minimum possible factor for this adoptedtemperature, so that the superposition of the axial forces due to prestressing andthose due to the permanent loads ensures for all members of the structure themaintenance of a predetermined minimum safe amount of tension.

320


The application of the above technique is now illustrated in the following twoexamples:

(1) In this example (Figure 9.6) each concentrated load of 100 kN consists of apermanent load of 35 kN and a snow load of 65 kN.

(2) In this example (Figure 9.7) each concentrated load of 60 kN consists of apermanent load of 20 kN and a snow load of 40 kN.

321

100 kN100 kN 100 kN

100 kN

2.40

1.20

2.40

+184.5 +184.5+79.4 +79.4

–64.2

+78.0 +78.0+81.0 +81.0

+91.3

24.0 m

(λ = 10.0)+54.6 +54.6+33.0 +33.0

–153.1 –153.1

+28.5 +28.5

Prestressing:temperature reduction of 100°C

for all the diagonal members

Analysis as a skeleton frame

λ ·

+12.

3 +12.3

+0.393 +0.3

93

+23.4+23.4

+8.54 +8.54

–109

.4

–109.4

+56.0 +56.

0

–161.7–161.7

+6.73 +6.73

Figure 9.6 Determination of the required prestressing of cables: example 1

60 kN60 60 60 60 60

60

(λ = 10.0)

Prestressing:temperature increase of 100°C

for all the vertical members

Analysis as a skeleton frame

λ ·

–3.4

2

–3.1

5

–3.0

2

–2.9

8

–3.0

2

–3.1

5

–3.4

2

–29.

9

–29.

9

–29.

9

–29.

9

–29.

9

–29.

9

–29.

9

+32.8

+32.8+32.8

+32.8

+31.8

+31.8+31.8

+31.8+31.2

+31.2

+30.9

+30.9

+30.9

+30.9

+31.2

+31.2

–313.7

+310.6+310.6

–313.7

+304.1

–307.2–307.2

+304.1+300.2

–303.2

+298.3

–301.2

+298.3

–301.2

+300.2

–303.2

Figure 9.7 Determination of the required prestressing of cables: example 2

Cable structures

In both cases the resulting factor is equal to 10.0, as can easily be concluded from thebehaviour of the two extreme diagonals (Example 1, see Figure 9.6) and of the twointermediate members of the upper chord (Example 2, see Figure 9.7).The check for non-symmetric loads, i.e. the check for developing deformations and

for possible slack members (i.e. compression), is done, as explained above, by consideringthe equilibrium in the deformed state through the use of a special software program thattakes into account the geometric non-linearity of the system. In the two examplesexamined, this means the consideration of unilateral snow load. In Example 1 (seeFigure 9.6) the first two loads from the left are equal to 100 kN and the other twoloads are equal to 35 kN, while in Example 2 (see Figure 9.7) the first four loads fromthe left are equal to 60 kN and the other three loads are equal to 20 kN. It is clearthat, in each case, the cables retain the prestressing that was previously determined(with ¼ 10.0). It is checked that in both cases all cables are found to be under tension.

9.3 The freely suspended cableAccording to the previous section, a freely suspended cable of span L under a uniformpermanent load g (self-weight, etc.) takes a form affine to the bending moment diagramM0(x) of a simply supported beam under the load g. If Hg is the horizontal component ofthe anchoring force at both ends of the cable and f is the sag of the cable at its midlength,thenHg¼ g L2/8 f and, according to Section 2.2.7, the ordinates y(x) of the cable form are

yðxÞ ¼ M0g=Hg

i.e. a parabolic line (Figure 9.8).

322

f y

y

g

p

p

g

g

Funicular form

ηη = (Mp – Hp · y)/(Hg + Hp)

Mg0

Hg = Mg0/f Hg

Hp + Hg Hg + Hp

Mg

Mp

Figure 9.8 Determination of the funicular form under an additional load


For an additional load p the cable is deformed further, with ordinates , which bringsan additional horizontal component Hp to the cable force, and the cable takes a newfunicular form with ordinates ( yþ ) that are expressed, just as previously, throughthe equation

y þ ¼M0

g þ M0p

Hg þ Hp

Based on the above expression for y, the additional ordinates result as (see Figure 9.8)

¼M0

p Hp y

Hg þ Hp

In this expression for the force Hp is unknown. To determine Hp the relationship mustbe found, on the basis of Hooke’s law, between the geometric elongation of the cable dueto the ordinates , its elastic characteristics (i.e. the elasticity modulus Ec) and the cross-sectional are Ac. The following result may then be obtained (Timoshenko and Young,1965):

Hp

Ac Ec

Ls ¼g

Hg

ðL

0 dx 1

2ðL

0

d2

dx2 dx

As, in the present context, the interest is mainly in the covering of large areas and insuspension bridges, where the ratio ¼ f/L lies between 1/8 and 1/12, the aboveequation may be simplified, to a good approximation, as

Hp

Ac Ec

Ls ¼g

Hg

ðL

0 dx ðaÞ

where the length Ls is given by the expression

Ls ¼ L ð1þ 8 2ÞIt must be noted that the length Ls used in the above ‘constitutive equation’ of the cabledoes not represent the actual length of the cable Lc. This can be estimated using thefollowing equation:

Lc ¼ L 1þ 8

3 2

It is clear that from the last expression for that the integral in equation (a) above canbe determined with respect to the unknown value Hp. It is useful now to introduce,instead of Hp, the ratio

Z ¼Hp

Hg

as the unknown quantity, and the dimensionless parameters

¼ p

g

323

Cable structures

" ¼Hg

Ac Ec

where p is equal either to p or p/2 depending on whether the load p is extended overthe entire length of the cable or is located in its extreme half, respectively. Theformer live load layout leads to the minimum cable force, whereas the latter one leadsto the maximum deflection, at approximately a quarter of the span. The parameter "represents the strain of the cable at its lowest point due to the dead load g.Appropriate substitution in equation (a) leads to the following quadratic equation

with respect to Z:

Z2 þ1þ 16

3 2

1þ 8 2 1"

Z 16

3 2

1þ 8 2 1" ¼ 0

After the corresponding determination of Z (i.e. of Hp) the values of may be readilyobtained as shown above. It results that

ðL=2Þ ¼ f Z

Z þ 1

or

ðL=4Þ ¼ f 0:50 0:75 Z

Z þ 1

with respect to the valid layout of live load p as above. g

An approximate, but more practical, estimate of the increase f in the established sagf of the cable, due to a uniform load p along its whole length, can be obtained byconsidering its elongation Lc according to Hooke’s law (Timoshenko and Young,1965):

Lc ¼Hp L

Ac Ec

1þ 16

3 f2

L2

where

Hp ¼p L2

8 f

However, the changeLc may also be obtained from the above expression for the cablelength Lc due to the changef in its sag (i.e. as the first derivative with respect to f ), as

Lc ¼16

3 f

Lf

From these last two equations it results that

f ¼ p L2

8 Ac Ec

1þ 3

16 2

¼ p L

kc

324


where

kc ¼8 Ac Ec

L 1þ 3

16 2

expresses the stiffness of the cable, i.e. the uniformly distributed load ( p L) required tocause a unit vertical displacement f¼ 1 at its midpoint.

9.4 Prestressed cable netsAs explained in Section 9.2, the geometry of the system shown in Figure 9.9 is ensuredby the prestressing of its vertical members. The prestressing should be uniformlydistributed in order to obtain a parabolic form for both the upper and bottom cables.It is clear that if this distributed prestressing is equal to u (kN/m), it should be valid that

u ¼ H1

R1

¼ H2

R2

where H1 and H2 are the horizontal components of the axial force in each cable, and R1

and R2 are the corresponding radii of the curvatures.Nothing will change in the above-described developing forces if the following

progressive reformation of the system is followed. First, cable 1 is rotated by 908 andmoves vertically downwards until its midpoint coincides with the midpoint of cable 2,coming into contact underneath (Figure 9.10(a)). Next, the cables are considered tobe arranged parallel to one other at the same time and at small distances a in both direc-tions, while always maintaining the contact of cables 1 and 2. The resulting network canconstitute the cover for a specific area (Figure 9.10(b)). It is understood that thedistributed load u is applied downwards on cable 1 and upwards on cable 2, and thecables develop the forces H1 and H2, respectively, according to the above equation(Figure 9.10(c)).The curved surface created by cables 1 and 2 is a hyperbolic paraboloid and has

specific geometrical properties that have particular implications on the bearing

325

2

1

2

1

H1 H1

H2 H2

u

u

R1

R2

u = H1/R1 = H2/R2

Figure 9.9 The stress state of the prestressed cable system

Cable structures

behaviour of shells, as is examined extensively in Chapter 12. It should be noted that,in the constructional practice of such a prestressed cable network, only cables 2 areprestressed, and this automatically sets cables 1 in a prestressed state as well.It is clear that cables 1 and 2 must give up their forces to the outline of the considered

space. Therefore, in order for the structural members at the perimeter to be able to takeup these forces in an appropriate manner, they must be formed according to thefunicular form, thus developing analogously either compression or tension(Figure 9.11). In the first case the members will be composed of continuous archesconcave inwardly, while in the second case they will be composed of cables concaveoutwardly, both having a parabolic form, given that the loads received from thecables are practically uniform. Moreover, it is understood that the end points of

326

1

1

1

1

2

2

2

2

R2

R1

H1

H1

H1

H1

H2H2

H2H2

u

u

u

u

The state of prestressing in the cable net isdescribed by the condition u = H1/R1 = H2/R2

(a)

(b)

(c)

a

a

Figure 9.10 Creation of a prestressed cable network space system

Layout of compressed arch

12

Layout of tensioned cable

Figure 9.11 Formation of the bearing elements at the boundary


cables 1 will generally be higher than those of cables 2, and this certainly means ananalogous formation of the peripheral bearing element.The prestressing forces H1 and H2 must have such values that the imposition of

permanent and live loads always leaves both cable types under tensile stress. This ischecked through the following consideration (Figure 9.12). A uniformly distributedgravity load q (kN/m2) applied to the spatial system may be considered as beingdistributed to a load q1 (kN/m

2) along the direction of cables 1 and to a load q2 (kN/m)distributed along the direction of cables 2, causing additional tensile forces in cables 1and additional compressive forces in cables 2.Now, the requirement for loads q1 and q2, besides offering in total the load q, is to

cause the same vertical deformation in cables 1 and 2 at their midpoint, thusallowing the estimation of their value for preliminary design needs. This last requirementcan be written, according to Section 9.2, as

f ð1Þ ¼ ðq1 aÞ L1

kð1Þc

¼ f ð2Þ ¼ ðq2 aÞ L2

kð2Þc

that is

q1q2

¼ L2

L1

kð1Þc

kð2Þc

where

kð1Þc ¼ 8 Að1Þc Ec

L1 1þ 3

16 21

and

kð2Þc ¼ 8 Að2Þc Ec

L2 1þ 3

16 22

The last equation for q1 and q2, in combination with the requirement that q1þ q2¼ q,allows their direct determination.

327

2

1

(q2 · a)

(q1 · a)

Figure 9.12 Distribution of the vertical load in both directions

Cable structures

Cable 1 develops a total tensile force N1 that is equal to N1¼H1þ (q1 a) R1, whilethe total force N2 of cable 2 is equal to N2¼H2 (q2 a) R2. Of course, it should bechecked whether the force N1 can be taken by the cross-section of cable 1, as well aswhether the force N2 remains tensile with a sufficient margin of safety so that cable 2does not become slack.

9.5 Suspension bridges — the suspended girderFor bridging a span, an individual cable is, by itself, functionally useless. However, cablemay be used to the suspend a deck carrying traffic loads, in the form of a suspensionbridge, wherein it takes up the total permanent and live load acting on it and transfersthis load to the supporting towers and anchorage (Figure 9.13). The suspension isrealised through vertical tension elements (hangers) placed at small distances along thetwo longitudinal edges of the deck. The deck can be considered as a beam (a flexibleone), given its very long length with respect to its two other transverse dimensions

328

f

L

Anchorage Anchorage

Hangers Tower

Stiffening girder

Figure 9.13 Structural system of a suspension bridge


(depth and width), and it is obvious that the two cables used in the two created planes ofthe hangers may be considered as a single one in order to examine the load-bearingfunction of the system under vertical loads (see Figure 9.13).Usually a suspension bridge has a typical configuration of three spans — the main span

and two adjacent minor spans. This gives the optimum layout for bridging a largeobstacle by using two intermediate support points (the high bridge pylons). Thepylons offer a support for the suspension cables at their top, while at the deck levelthey allow a corresponding support for the girders from either side. Although thesegirders may be constructed continuously over the pylon’s supports, it is more usual forthem to be considered (and constructed) as simply supported ones (see Figure 9.13).The tension Hg on the cable is always adjusted with respect to the desirable sag f so

that, for the existing total permanent load g, the girder (deck) is absolutely horizontal.According to Section 9.2, Hg¼ g L2/8 f.The bending stress on the deck girder under the permanent loads can in practical

terms be considered as minimal due to the small distances between the hangers andthe care taken to ensure that they are in an absolutely horizontal position. However,an additional live load p applied on the girder tends to deform the cable, and it isclear that this additional deformation of the cable will also be entirely imposed tothe girder itself, given that the hangers, with their relatively small length and therelatively small additional axial force they have to carry, may be assumed to be non-deformable (Figure 9.14). The girder should be designed to limit through its rigiditythe cable deformation due to the live load p, on the one side, and to resist thebending imposed on it by this very same deformation , on the other side, as previouslyexplained.Thus the imposition of a live load p on the beam causes an additional deformation

both in the cable and in the beam, which results in an increase in the horizontalcomponent of the cable force by Hp, as well as an increase in the bending moment ofthe suspended beam (see Figure 9.14). To determine the magnitudes of these variables,the deformed geometry of the system should be taken into account. The aim of thefollowing analysis is precisely the essential comprehension and practical solution ofthis problem (see Stavridis, 2008).The girder is in equilibrium under the following loads (Figure 9.15): (1) the self-

weight g (permanent load) acting downwards, (2) a live load p applied over a certainlength and acting downwards, and (3) the actions qc(x) of the tension hangers actingupwards. Loads g and p can be considered as being uniformly distributed, but the loadqc(x) is non-uniform along the length of the girder. Applying a live load p on thegirder makes the cable deform vertically through the hangers and, because thehangers will not be uniformly tensioned, the cable will receive non-uniform forces(see Figure 9.14), which will lead to a new funicular form that is different to theprevious (parabolic) one. These same forces will obviously be applied with oppositesign on the girder, constituting precisely the actions qc(x), as shown in Figure 9.15.The distances between the hangers are so small that the above consideration of a

distributed applied load qc(x) on the cable as well as on the girder is justified. Thus, ifz(x) represents the ordinates of the cable position after application of the vertical

329

Cable structures

forces qc(x) (i.e. of its new funicular form) and H is the horizontal component of the totalaxial force of the cable, then, from the equilibrium of an elementary part of the cablealong the vertical direction (see Figure 9.14), the following basic equation results (seeSection 7.1):

qcðxÞ ¼ d2z

dx2 H

In this equation qc(x) is the downward force, and the term dz2/dx2 represents thecurvature (1/r) of the cable, which has a positive value. Thus the above equation is

330

H = Hg + Hp: horizontal component of axial force S

Vertical equilibrium

qc · ∆x = H · ∆(dz/dx)

S

Sqc

z(x)

qc

x

Hg: cable force due to permanent loadHp: increase of cable force due to live load

Loads on the cable

Undeformable hangers

η(x)

η(x)

p Additional live load

g Self-weight

x

z(x) = y + η y(x)

1/r : due to permanent load

Figure 9.14 Cable deformation and the acting forces

Forces acting on the girder

qc

gp

Figure 9.15 Actions applied to the girder


nothing more than the generally valid relationship q¼H (1/r) connecting thedistributed load on a cable with its tensile force and its curvature, which applieseverywhere a funicular form, be it in a prestressing cable (see Section 4.3.1.1), anarch (see Section 2.2.7) or even a shell (see Sections 12.2, or 12.3), etc.It is clear that, in this case (see Figure 9.14),

z(x)¼ y(x)þ (x)

and

H¼HgþHp

According to the above, the total load q(x) applied to the beam, considered as positivedownwards, is (see Figure 9.15)

q(x)¼qc(x)þ gþ p

This load is related to the girder deformation and the beam rigidity EI according toSection 2.3.6:

EI d4

dx4¼ qðxÞ

On the basis of the previous equations the same load may be written as

qðxÞ ¼ d2

dx2 ðHg þ HpÞ þ Hp

d2y

dx2þ p

Given that the term d2y/dx2 is equal to the negative curvature (1/r) of the cable underthe permanent loads g (i.e. is equal to 8 f/L2), the above differential equation for thesuspended girder takes the final form

EI d4

dx4 d2

dx2 ðHg þ HpÞ ¼ p

Hp

r

According to what was examined in Section 7.2, this equation represents the behaviourof a fictitious simply supported beam under a transversely applied load ( pHp/r) andsubjected to an axial tensile force (HgþHp), according to the second-order theory(Figure 9.16).The deflection curve (x) may be sufficiently satisfactorily approximated, according to

Section 7.2, by the equation

¼ W1 1

1þ

331

Fictitious beam:the produced deflection must satisfy the constitutive equation for the cable

p

Hg + Hp Hg + Hp

Hp/r

Figure 9.16 Actions on the fictitious simply supported beam

Cable structures

where W1 is the deflection curve of a simply supported beam under the transverselyapplied load ( pHp/r) and is equal to ¼ (HgþHp)/Pcr. Then, of course,Pcr¼2 EI/L2.It is therefore clear that, for a given load p having a specific application zone over

the suspended girder, the cable should develop such an additional force Hp thatthe deflection curve (x) resulting from the analysis of the fictitious tensioned beam(see Figure 9.16) satisfies the constitutive equation (a) for the cable deformation asformulated in Section 9.3:

Hp

Ac Ec

Ls ¼g

Hg

ðL

0 dx

Two cases of the application of the live load p are now examined according to the usualpreliminary design practice:

. Case 1: the live load is applied on the extreme half of the girder.

. Case 2: the live load is applied over the entire length of the girder’s length.

In the first case the aim is to determine the largest deformation of the cable and thegirder, and, consequently, the maximum bending moment. The second case leads tothe determination of the maximum cable force (Figure 9.17).In Case 1 the load p may be analysed as a symmetric load p/2 applied over the entire

length of the girder and as an antisymmetric load p/2, as shown in Figure 9.17. Then, thedeflection curve (x) of the fictitious tensioned beam may be considered as resultingfrom superposition of the deflection curves sym and ant corresponding to the symmetricload ( p/2Hp/r) and the antisymmetric load p/2, respectively (see Figure 9.17). Notethat the superposition principle is valid within the second-order theory, provided thatthe axial force remains constant for all superposed systems. Therefore, in this case theabove constitutive equation is written as

Hp

Ac Ec

Ls ¼g

Hg

ðL

0sym dx þ

ðL

0ant dx

As ant is an antisymmetric function, the second integral will be zero, and the equation issimplified to

Hp

Ac Ec

Ls ¼g

Hg

ðL

0sym dx

Obviously this last equation is also valid in Case 2, wherein the live load p is applied overthe entire whole length of the girder, and here the fictitious tensioned beam is loadedwith ( pHp/r).As mentioned previously, the deflection curve sym may be determined to a very good

approximation through the expression

sym ¼ W1 1

1þ

332


where W1 is the deflection curve of the simply supported beam under the load ( p Hp/r).The load p is equal to p/2 or to p, depending on whether one-half or the whole length ofthe girder is loaded.The moment diagram M0

1 for the simply supported beam is, of course, parabolic, with amaximum value ( p Hp/r) L2/8, and the deflection curve W1 is, according to Mohr’stheorem (see Section 2.3.6), equal to the moment diagram of the beam loaded by anexternal load (M0

1/EI):

W1ðxÞ ¼ð p Hp=rÞ L4

24 EI

x

L

4 2

x

L

3þ x

L

Now the ratio Z, as the unknown quantity instead of Hp, and the dimensionlessparameters , and ", as used in the previous section, will be used again. Moreover,the following dimensionless parameter G is introduced:

G ¼Hg L2

EI

333

First case: live load over the left half of the span

The antisymmetric loading is ineffective

Final loading of the fictitious beam

Second case: live load over the whole span

Fictitious beam

Fictitious beam

Hg + HpHg + Hp

Hg + HpHg + Hp

Hg + HpHg + Hp

p

Hp/r

p/2

p

Hp/r

Hp/r

p/2

Figure 9.17 Actions on the fictitious beam depending on the live load position

Cable structures

Substitution in the above constitutive equation for the cable leads, finally, to thefollowing quadratic equation with respect to the dimensionless ratio Z:

Z2 þp2

Gþ þ 1

Z ¼ 0

where

¼ 8 p2

15 " 2

1þ 8 2

The maximum cable force (HgþHp) thus occurs when the whole length of the girder isloaded with the live load, and the value of this force is obtained directly from the aboveequation by setting ¼ p/g.The maximum bending moment Mmax and the maximum deflection max of the

suspended girder are determined by loading the half-length of the girder. Thesemagnitudes can be evaluated for the purposes of preliminary analysis at the quarter ofthe span. By splitting the considered loading into a symmetric and an antisymmetricpart of equal halves, as explained previously, it is noted that, because of the anti-symmetric loading, no additional cable force develops. Moreover, the beam behaveslike a simply supported beam with a half-span and subjected to half the live load.Obviously, the additional cable force Hp is due to the loading of the whole girderwith the load p/2, and this force can be determined using the above equation bysetting ¼ p/(2g). Then, superposing the relevant magnitudes at the quarter and atthe middle of the span in the full- and the half-length beam, respectively, the followingexpressions are obtained:

Mmax ¼ Msym þ Mant

It is understandable that

Msym ¼ M0sym ðHg þ HpÞ sym

and

Mant ¼ M0ant ðHg þ HpÞ ant

Using the dimensionless parameters introduced so far, the following expressions areobtained:

Msym ¼ g L2

8

2 Z

0:75 0:0742 p

2 G ðZ þ 1Þp2 þ G ðZ þ 1Þ

Mant ¼g L2

8

0:125 0:013 p2 G ðZ þ 1Þ

4 p2 þ G ðZ þ 1Þ

and

sym ¼ 5 g L4

384 EI 0:7125

2 Z

p2

p2 þ G ðZ þ 1Þ

334


ant ¼5 g L4

384 EI 0:03125 4 p2

4 p2 þ G ðZ þ 1Þ

Note that in the above results the bending moments and the deflections are expressed interms of the corresponding quantities of the stiffening girder under the permanent loadg, considered as a simply supported beam of span L.Obviously, the maximum deformation of the cable will be:

max¼ symþ ant g

9.6 Stiffening the suspension cable

9.6.1 Conceptual analysisAs explained in Section 9.1, the automatic adoption of the funicular form for anyloading pattern acting on a cable constitutes an advantage in that it makes the cablethe most efficient load-bearing element from a strength exploitation point of view.However, it constitutes a weakness too, because of the excessive deformability of thecable under the application of an additional load with a different pattern from thepermanent one. It was shown in Section 9.5 how this sensitivity of the cable isovercome by the use of a horizontal girder which, for any additional (live) loading,restricts the deformability of the cable decisively, due to its appropriately chosenbending stiffness.It has already been mentioned in Section 9.1 (see Figure 9.2) how the stiffening of the

freely suspended cable may also be achieved by another means, namely by embeddingthe hanging cable in a relatively thin band of concrete, which is prestressed with anappropriate force so that the resulting inverted arch is under compressive stress. Thegravity loads to be applied later will cause tensile forces on the inverted arch. These,however, will be suppressed by the pre-existing compressive forces due to prestressing.More specifically, consider an orthogonal concrete cross-section with width b and

height d, and a total cross-sectional area of the cables of Ap. The cables are supposedto have no bond with the surrounding concrete. Assuming that, with concrete ribboncasting, the load transferred to the cable having the section Ap is g (the concreteweight plus the cable self-weight), the cable must be prestressed with a forceHg¼ g L2/(8 f ) in order to establish the predetermined sag f over the span to becovered L (Figure 9.18(a)).If, after hardening of the concrete, the cable is stressed further by an additional

prestressing force HV, then the inverted arch with section (b d) will receive, allover its length, from the cable a uniform upward deviation force uV¼HV/r(Figure 9.18(b)), and under this load a uniform compressive stress V will develop:

V ¼ uV L2

8 f 1

b d

this being obviously equal to HV/(b d).

335

Cable structures

For each additionally applied gravity load p that is smaller than uV the arch ribbon willcarry a total upward load equal to (uV p). That is, the arch will always be undercompression without bending, while possible upward loads due to wind pressure willbe carried normally by its operation as a typical arch. Of course, this assumption issimplified because the cable and the concrete band constitute a system that carries alladditional loads in common, and hence the need arises to determine the stressdistribution between the cable and the concrete section.Although this last assumption that the inverted arch can carry the gravity load p may

be practically plausible, in the case when the live load does not extend to the full lengthof the arch but only up to the midspan (a loading that is mostly unfavourable in manycases, as examined previously) the combined action of the arch and the cable must beconsidered, in order to estimate the bending response in the arch and the stiffeningeffect of the above layout.It may be understood that the concrete—cable system has essentially the same

characteristics as the suspended girder system considered in Section 9.5, in that therole of the girder as a stiffening element may be represented by the bending stiffnessof the arch. More specifically, the fact that the common deflection of the cable andthe arch is determined by both the axial rigidity of the cable and the bending rigidityof the arch itself, leads to the conclusion that this system behaves like a fictitioussuspension system with a separate cable and stiffening girder. Indeed, in such asystem, after an initial geometry for the cable sag under permanent loads has beenestablished through the introduction of an appropriate cable force, a later applied liveload to the stiffening beam produces identical deflections to the cable and the girderat each point, due to the presence of the vertical hangers. It may be concluded thatthe stress-ribbon system exhibits exactly the same characteristics as the suspension

336

L

uV = HV/r

g

Hg = g · r

HV

uV

Hg

HV

(a) Actions on the cable

f

(b) Actions on the inverted arch

Figure 9.18 Conceptual layout for the inverted arch


bridge and, consequently, this may be used as a fictitious model for the analysis of thesystem examined. Of course, in the fictitious system of a suspended beam with acable cross-section Ap and a beam cross-section (b d), the ‘permanent’ load ‘g’considered for the system must be one that establishes the sag f of the cable. In thepresent case this load is equal to the sum of the load g and the deviation load HV/rdue to prestressing. Indeed, the fictitious system of the suspended beam under thepermanent load w¼ (gþHV/r) develops in the cable an axial force (gþHV/r) L2/(8 f )¼HgþHV, which obviously results in the same value as for the initial cableforce in the actual system (Figure 9.19).In this way the results obtained in Section 9.5 for the stress state that develops in

suspension bridges due to an unfavourable layout of the live load can be applieddirectly also in the present case, in order to estimate the bending moment M of theinverted arch as well as the maximum deflection in the case where the live load isextended only up to the midspan position. Of course, when checking the stress statein the orthogonal section of the arch, the simultaneous action of a compressive forceN must also be taken into account. The value of this compressive force is given by

N ¼ ðuV p=2Þ L

8 ¼ HV p

2 L

8 It is obvious that, when covering areas with large spans, the ratio ¼ f/L may be evensmaller than that for suspension bridges.At this point, one may consider the the so-called stress-ribbon bridges (which are

mainly pedestrian bridges) in the same way as described above, where, in order toretain some maximum allowable slopes of the ‘inverted arch’ for pedestrian use, theratio f/L must be limited to the value 1/50 (see Stavridis, 2010).

9.6.2 Numerical exampleIn order to illustrate numerically the content of the previous section, the followingpurely indicative numerical example is presented.

337

Hw = w · r Hw = Hg + Hv

L

w = g + uvEI

r

f

Figure 9.19 Fictitious suspended beam

Cable structures

A hanging cable having a span of L¼ 60.0m, a sag f¼ 5.0m and a cross-sectionalarea Ap¼ 42 cm2 is to carry a permanent load g¼ 5.0 kN/m and a uniformly distributedlive load p¼ 3.0 kN/m. The cable has a radius of curvature equal to r¼ 60.02/(8 5.0)¼ 90.0m.

. According to Section 9.3, in order to find the maximum cable deflection thelive load is placed over the left half of the span. The relevant parameters arecalculated as:

¼ 3:0

2 5:0 ¼ 0:30

Hg ¼5:0 60:02

8 5:0 ¼ 450:0 kN

" ¼ 450:0

0:0042 2:1 108¼ 0:00051

Using the corresponding quadratic equation these lead to the value Z¼Hp/Hg¼ 0.294 and, according to the corresponding equation given in Section 9.3, it isobtained that

L

4¼ 5:0 0:50 0:60 0:75 0:294

0:294þ 1¼ 0:307m

. It is now assumed that the cable is stiffened by means of a steel beam having theprofile HE-B 450, with a moment of inertia equal to I¼ 79890 cm4 and a sectionmodulus W¼ 3550 cm3. According to Section 9.4, the relevant additional parameterG is

G ¼ 450:0 60:02

2:1 108 0:0007989¼ 9:656

and, for the same position of the live load, through the corresponding quadraticequation it is found that Z¼Hp/Hg¼ 0.290 (which is practically the same asbefore). According now to the corresponding equations given in Section 9.4, themaximum deflection is found to be noticeably reduced: max¼ symþ ant¼0.0016þ 0.0717¼ 0.073m.

With regard to the bending response of the beam, from the equations given inSection 9.4 it is found that Mmax¼ 135.0 kNm, which leads to a moderatemaximum bending stress of 135.0/0.00355¼ 38 028 kN/m2.

. Finally, the case is examined where the cable is embedded in a concrete ribbon 20 cmthick, according to Section 9.6.1. The weight of the concrete for a width of 1.0m,fits with the assumed permanent load, namely g¼ 0.20 25.0¼ 5.0 kN/m. Applyingnow an additional prestressing HV¼ 850.0 kN, the fictitious suspended beam, aspreviously explained, is considered, consisting of the very same 20 cm concretethick ribbon, and subjected to the fictitious permanent load w¼ 5.0þ 850.0/90.0¼ 14.44 kN/m.

338


Following the same procedure as above, the corresponding parameters arecalculated as

¼ 3:0

14:44¼ 0:208

" ¼ 450:0þ 850:0

0:0042 2:1 108¼ 0:00147

and

G ¼ ð450:0þ 850:0Þ 60:02

2:1 108 0:0006667¼ 33:427

For the same position of live load, through the corresponding quadratic equation fromSection 9.4, the value Z¼Hp/Hg is calculated on the basis of the value /2 as beingZ¼ 0.099. The maximum deflection is now found to be 0.12m and the maximumbending moment to be M¼ 19.72 kNm. The corresponding compressive force is,according to Section 9.6.1,

N ¼ 850:0 3:0

2 60:0

8 ð1=12Þ ¼ 715:0 kN

It is clear that the resultant compressive force, having an eccentricity of 19.72/715.0¼ 0.028m (<0.20/6), falls inside the core of the concrete section, leaving ituncracked.

9.7 Cable-stayed bridgesCable bridges with inclined straight cables have been used for many years to cover largespans, as an alternative solution to suspension bridges, and for at least 40 years havebeen the most commonly used solution for spans bigger than 150m. A cable bridge iscomposed of a long deck that is usually suspended by its two sides by sidelong cablesplaced at equal distances of about 5—8m. These cables usually lie in two distinctplanes, and are attached at the top to vertical pylons. The height of the pylons abovethe level of the deck is of the order of 1/5 of the main span of the bridge (Figure 9.20).The loads on the deck are transferred to the two planes of the cables through a simply

supported action — precisely as occurs in suspension bridges — and are then led throughthe cables to the top of the pylons (see Figure 9.20).In the plane model, the deck is simulated by a straight girder, the two planes of cables

being naturally incorporated into one, while the pylon is modelled by a vertical beamelement connected monolithically, or not, to the horizontal girder (see Figure 9.20).The deck transfers its loads to the cables through transverse bending, while in thelongitudinal direction it acts as required by the behaviour of the corresponding girderin the plane model.It is possible for the deck to be suspended from the pylons through only one plane of

cables along the length of its longitudinal axis, but in this case the torsional momentscaused by the eccentric loads on the deck must be taken up only by the torsional stiffness

339

Cable structures

of the deck itself. Therefore, the girder must be designed with an appropriate box sectionoffering an adequate torsional rigidity, as will be examined in Chapters 13 and 14.The cable-stayed bridge constitutes a characteristic example in structural design

where the state of stress due to the self-weight is determined on the basis of theerection procedure rather than on the model of the completed structure, which, ofcourse, will be considered for the assessment of the influence of live loads.More specifically, the bridge is normally built using the balanced-cantilever method,

with progressive construction of the parts between the hangers as ‘cantilevers’ from bothsides of each pylon, and finally the remaining central part of the main span is accordinglysupplemented. Each newly added segment is supported at one end at the free end of thepreviously constructed segment, while the other end is suspended from the top of thepylon (Figure 9.21).The basic static operation mechanism during the construction of such a bridge is

shown in Figure 9.21. Each added (statically determined) segment of the simplysupported type and the corresponding cable can bear either a vertical uniform load ora concentrated load at the extreme node. It is clear that the reaction of each newlyconstructed segment at its left end passes through the point of intersection of thesuspension cable and the resultant of the distributed load, while the force applied onthe right end has the direction of the cable. In this way, the added segment is undercompression.Naturally, this state applies also when erecting the first segment. With the addition of

the next part, the reaction force of the new segment is applied with opposite sign at theright end of the previous part. This reaction causes — as it results from the equilibrium inthe corresponding node — a tensile force in the corresponding cable and a compressionforce in the previously constructed segment (see Figure 9.21). Thus, with the addition of

340

Indicative pylon formation

Figure 9.20 Structural system of a cable-stayed bridge


each new part the tensile force of the immediately preceding cable is increased — but notthe tensile force of the cable before that — while the imported compression force is fullytransferred (i.e. it is added) to all the previous segments, so that, finally, the compressionforce of the deck’s girder acquires its maximum value near the pylon, and then decreases(always remaining compressive) towards the midspan. Moreover, it is understood thatthe tensile forces on the cables are increased when going away from the pylon, i.e. astheir angle of inclination is reduced (see Figure 9.21).It should be noted, of course, that the actual monolithic connection of the girder

segments (not the hinged one considered) does not practically alter, even quantitatively,the above conclusions.The presence of intense compression in the beam deck normally leads to the use of

concrete as the more economically suitable material to receive compression.At this point it must be pointed out that the consideration of a ‘complete’ statically

indeterminate system constructed in a ‘single phase’ does not attribute the progressiveincrease in the compression forces in the girder until the midspan. On the contrary,according to this model, the intermediate region appears to be in tension, as it iseasily understood if the self-equilibrating horizontal components of the cable forcesare applied as actions on the continuous girder (Figure 9.22).

341

The compressive stress of the girder accumulates toward thepylon, whereas the cable force correspondingly decreases

Taking up the self-weight of each added segment thegirder is compressed and the cables are tensioned

Each added segment receives from the next onean action augmenting both its own cable force

and its own compressive stress

Figure 9.21 Progressive construction of segments and the corresponding stress states

Cable structures

Naturally, for the assessment of the additional live load influence as well as of anyadditional permanent load, the complete system shown in Figure 9.22 should beessentially considered. For the analysis of this model, which has such a large degree ofstatical redundancy, it is essential to use an appropriate software program that willtake into account the geometric properties of the cables, the girder and the pylonitself.Remaining a while longer with the established stress due to the initial permanent

loads, it is pointed out that the restored monolithic system of the two cantilever-likehalves will generally present deformations downwards, which could create anundesirable visible curvature. Moreover, in the case when the girder is made ofconcrete, after connection of the two ‘cantilevers’ and the elapse of some time, ascreep appears the effective bending rigidity of the girder will be decreased, and thuslarger forces will be gradually transferred in the cables (see Section 3.2.10 andFigure 6.7). Thus, the additional elongation of the cable leads to an increase in thevertical shifts of the suspension points of the girder and to a subsequent increase incurvature over time, as well as to the development of an additional bending response.These negative characteristics practically disappear with the suitable prestressing of

each cable. The cable should be prestressed with such a force that its vertical componentcorresponds exactly to the reaction of the corresponding continuous beam on unyieldingsupports and under the acting permanent load (Figure 9.23).In this case, it is clear that the same level for all the nodes of the girder is ensured.

Precisely because the bending stress of the supported ‘statically redundant’ beam isnot altered due to creep, as time elapses (see Section 5.5.1) its reactions will not altereither, nor will the cable forces, and thus the suspension points will accordinglyremain immovable over time. It is obvious then, that the bending and shear stress ofthe girder, being constant over time and identical to that of the continuous beam onimmovable supports, will be definitely limited (see Figure 9.23). g

At this point, it must be noted that the axial stiffness of a ‘straight’ cable is considerablyaffected by the cable deflection under its self-weight, and this in turn depends on themagnitude of the imposed stress in the cable. This is equivalent to a fictitious reduction

342

Compressive axial forceTensile axial force

Figure 9.22 Tension due to the self-weight action on the whole system


in the elasticity modulus of the cable, which is expressed as (Leonhardt and Zellner,1980)

Ei ¼Ec

1þ 2 l2 Ec

12 3

The term l is the horizontal projection length of the cable and is the specific weight ofthe material. The inclination angle of the cable plays no role at all.It can be seen from the above equation that the influence of the cable stress is

important. For a cable with horizontal projection length 100m and ¼ 100N/mm2

(¼ 78.5 kN/m2, Ec¼ 2.0 108 kN/m2), the reduction in Ec is of the order of 50%,whereas for a cable stress ¼ 500N/mm2 there is no reduction. The importance ofusing high-strength steel, and thus allowing a high level of imposed prestressingforces, is evident, and at the same time it is explained why previous attempts at usingregular steel cables have failed. As the cables are designed for the serviceability state,with an allowable stress of about 45% of the ultimate one, for a usual quality ofprestressing steel with a yield stress/ultimate stress ratio of 1670/1860N/mm2, the fullnominal value for the elasticity modulus Ec may be used. g

343

The prestressing cable forces must offer thesupport reactions of the continuous beam

Concrete girder: state of stress and deformationunchanged over time due to creep

Bending diagram identical to that for the continuous beam

Ri

αi

g

s s

s s

g

g

Pi0

Pi0

Figure 9.23 Cable prestressing according to the reactions of the continuous beam

Cable structures

According to the above, the initial forces P0i of the cables are P0

i ¼Ri/sini, where Ri isthe corresponding reaction of the continuous beam (Ri g s) and i is the inclination ofthe corresponding cable.The application of the live load p causes, at first, an increase in the cable forces in

correspondence with the applied force p s, so that the final cable forces Pi can beestimated based on the initial statically determined model as Pi¼ (g sþ p s)/sini

(Figure 9.24). Hence, the required cross-section Ai for each cable will be Ai¼ Pi/adm, where adm is the admissible working stress for the cable. In practice, the cablesare grouped, and for each group of cables the same cross-sectional area is used.Clearly, the design bending moment of the girder depends mainly on its deformation

due to the application of the live load, as the bending moments due the permanent loadsg is only that of the continuous beam with equal spans of length s, i.e. with a maximumvalue 0.10 g s2. The above deformation (i.e. the support ‘settlements’ of thecontinuous beam) is related not only to the additional elongation of the cables due tothe live load p, but it is also particularly affected by the bending deformation of thepylon itself, which is directly related to its horizontal shift at the top (Figure 9.25).This is obviously caused by the resultant Fh of the horizontal components of thecable tensile forces acting on each side, and is higher only when the main (central)span is loaded with the live load p. Then, the left pylon is loaded to the right, whilethe right pylon is loaded to the left. Thus, for example, for the left pylon

Fh ¼Xright

s ðg þ pÞtani

Xleft

s g

tani

For the system illustrated in Figure 9.25, where a pylon is connected to a fixed supportthrough a backstay cable, it can be concluded that the horizontal shift of the top of thepylon is much more affected by the axial stiffness Ec Ac of the cable than by the bendingstiffness EI of the pylon itself. Indeed, the horizontal shift h of the pylon head resultsfrom the analysis of the statically indeterminate system:

h ¼ Fh d h3

3 d EI þ h3 cos2 Ec Ac

344

s s

Pi = (g · s + p · s)/sin αi

αi

g p Pi

Figure 9.24 Cable stressing due to the total load


It is obvious that the influence of EI of the pylon is much smaller than the influence ofEc Ac of the cable; therefore, by a suitable choice of a cross-section for the backstaycable, the horizontal shift of the pylon top is limited to such an extent that its effecton the beam deformation is practically negligible (see VSL, 1984). g

Provided that the deformation of the pylon is satisfactorily limited by a suitable selectionof the axial stiffness of the backstays, it can be considered that the deck girder acts withrespect to the live loads like a continuous beam elastically supported at the ends of thedeformable inclined cables. It is known that the inclined cables offer a variable resilienceof the supports along the length of the beam that can be described through springs andmust be taken into account (Figure 9.26).In order to calculate the stiffness of the spring that corresponds to cable i, a unit vertical

load may be applied to the corresponding node and the vertical shift of the node canbe determined accordingly. The developed cable force is (1/sini); the elongation liof the cable with respect to its length (h/sini) is li¼ h/(Ai Ei sin2i), while from

345

p

The bending response of the girder due to the live loadis increased through the deformation of the pylons

The difference of the two forces acts at the top of the pylon

Sum of horizontal components Sum of horizontal componentsFh

rightFhleft

FhrightFh

left

gpPi Pi

Pi · cos αi

αi

s s

Pi = s · g/sin αi Pi = s · (g + p)/sin αi

Maximum compressive force in the girder

EI h

Fh

Ec · Ac

α

d

The influence of the pylon bending stiffness on its deformationis much less than that of the axial stiffness of the cable

Figure 9.25 Pylon stressing

Cable structures

the geometry of the cable deformation (see Figure 9.26) the vertical shift vi of the nodeis vi ¼ li/sini.On the basis of the above, Ai¼ (g sþ p s)/(adm sini), and therefore the stiffness

ci of the spring is

ci ¼1

vi¼ sin2 i ðg þ pÞ s Ei

h admThus the girder may be considered as a beam on an elastic foundation (see Section17.3.3.1) with a modulus of subgrade reaction equal to ki¼ ci/s (kN/m2) (seeFigure 9.26). It is known that factor ki is variable and decreases towards the midspanregion.If a constant value km is considered as an approximation for the girder based on

the inclination of an intermediate cable, it results (Timoshenko, 1956) that theregion that should be loaded with the live load p, in order to cause the maximumbending moment at some point C in the main span, is extended to a length 2 asymmetrically placed with respect to point C (see Figure 9.26). This length is equal to2a¼ (/2) Ls, where

Ls ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4 EI=km

4p

is the so-called elastic length of the beam (see Section 17.3.3.1). The correspondingbending moment MC is assessed as approximately

MC ffi 0:20 p L2s

The analysis of a beam on an elastic support for any position of a live load p can clearly bedone with the aid of suitable software. g

346

h

A

p

v

1/sin α1

α

∆I

Deformed cable position

s s k = 1/(v · s)

Maximum bending response caused by the live load

The suspended girder acts like a beam on an elastic foundation

π · Ls/4 π · Ls/4C

Mc = 0.20 · ρ · Ls2

Figure 9.26 Girder function considered as a beam on an elastic foundation


As explained at the beginning of this section, the deck girder is under permanentcompression, introduced through the anchoring points of the cables, with high concen-trated loads in the midspan region, which are gradually reduced towards the pylons.Their values are s g/tani or s (gþ p)/tani, depending on whether a live load is, oris not applied, respectively, so that the largest compression force in the beam is equalto the horizontal force Fh applied on the pylon, according to the equation givenabove (see Figure 9.25). It is understood that this compression force raises, on theone hand, the problem of potential buckling of the beam, and, on the other, theinfluence of the bending deformation of the girder on the developing bendingmoments, in the sense of second-order theory (Figure 9.27).For the permanent loads the beam deformation is definitely limited by the cable forces

that are acting on the beam as ‘reinstatement forces’, not only for downward shifts —through the imposed elongation and the consecutive tension of the cables — but alsoto some degree for upward shifts. This means that the cables provide some stiffness incompression too, provided that the prestressing force is not entirely cancelled out (seeFigure 9.27). One must take into account the aforementioned effective elasticitymodulus, which for large spans may drastically decrease with the reduction in theexisting tensile force of the cable, with a consequent decrease in its ‘compressivestiffness’. However, the preliminary design of the girder based on the simultaneousaction of bending and compression seems to be secure, in general, from a possiblebuckling of the beam.Consider first the simple example of a suspended beam (Figure 9.28) under a

permanent load g¼ 100 kN/m. If the beam is designed to have a HE-B 800 profile

347

Prestressing: Rg/sin α Cable force: ~Rg + p/sin α

Deformed beam

α

g

g

g

p

g

p

Null deflection

Rg/tan α ~Rg + p/tan α

~Rg + p/tan α

Rg + pRg

Buckling danger?

P

P

D

D

Increase in the first-order bending momentdue to deformation of the beam

In as much as D < P, the prestressed cableoffers a ‘compressive’ stiffness (~E · A/L)

Figure 9.27 Influence of the cable prestressing force on the bending of the girder

Cable structures

based on the applied loads, the suspension system, with the particular prestressingcable, will have no problem at all in taking up this load. However, carrying the sameload under the externally applied cable forces becomes problematic due to buckling,either if the cable is taken away, or if the girder in the integrated suspension systemis designed as an HE-B 200 profile, which should of course be excluded by thepreliminary design.The same conclusion may also be drawn from the example of the suspended beam

shown in Figure 9.29 which, due to the introduced prestressing of cables, is predesigned,on the basis of its bending response as a continuous beam on immovable supports, as anHE-B 800 profile. It is shown that the axial loads on the beam in the suspension system,and for the particular prestressing forces involved, can lead to buckling only when thecross-section becomes an HE-B 360 profile, which is however excluded from the start.In any case, the control of elastic stability with regard to the existing safety margins

is always necessary, and must be performed for the case under the given permanentload, with the aid of a suitable software program that can also take into account thegeometrical and cross-sectional data for the pylon.Moreover, the pylons bear on their top the total load of the deck (permanent and live

load) and must be designed with sufficient safety for the critical load related to theirbuckling. It should be pointed out that the critical loads of both the pylon and thebeam significantly increase if the ends of the deck beam are formed as immovablesupports. On the other hand, the influence of the beam displacements due to the liveload on its final bending moment cannot be ignored (see Figure 9.27), as its compressiveaxial forces cause an increase in the first-order bending moments resulting from theconsidered model on elastic supports. It is understood that, for the assessment of thisinfluence, the use of a software program that will take into account the second-orderbehaviour is essential.

348

100.0 kN/m 100.0 kN/m

100.0 kN/m

1875 kN1875 kN

375 kN

375 kN

Prestress: 1912 kN

2.0 m

No buckling dangerHE-B 800

Ac = 25 cm2

10.0 m

It fails in bucklingbecause of lack of support by the cable

Figure 9.28 Consideration of the possible occurrence of buckling


ReferencesLeonhardt F., Zellner W. (1980) Cable stayed bridges. IABSE (International Association for Bridge and

Structural Engineering) Surveys, S-13/80.Stavridis L.T. (2008) A simplified analysis of the behavior of suspension bridges under live load.

Structural Engineering and Mechanics 30(5), 559—576.Stavridis L.T. (2010) Evaluation of static response in stress-ribbon concrete pedestrian bridges.

Structural Engineering and Mechanics 34(2), 213—229.Timoshenko S. (1956) Strength of Materials, Part II. Princeton, NJ: D Van Nostrand Company Inc.VSL (1984) VSL Stay Cables for Cable-stayed Bridges. Berne: VSL International Ltd.

349

Prestressing:

10.0 m

100 kN/m

100 kN/m

100 kN/m

10.0 10.0 10.0

50.0 m

No buckling dangerHE-B 800

Ac = 70 cm2

10.0 10.0

997

997

1010

1010

964

964

1134

1134

394

394

997 2020 2892 4536 1971

997 2020 2892 4536 1971

It fails in buckling because of lack of support by the cables

1410 2258 30494676

2010

Figure 9.29 Consideration of the possible occurrence of buckling

Cable structures

10

Grids

10.1 OverviewA grid is a plane formation of beams connected to each other with joints (basicallycrossing each other). The aim is to receive loads acting perpendicular to the gridplane and to transfer them in two or more directions. Despite the fact that the gridnodes are almost always formed as rigid, it is useful to consider the case where thecross-beams are simply supported by each other, thus transferring only an internalvertical force.In the case of rigid joints, the grid beams develop three types of sectional force

according to Section 2.5.1: bending moments, torsional moments and shear forces.The grid joints develop a vertical displacement and a rotation, represented by ahorizontal vector (Figure 10.1). In the case of nodes simply supported by each other,the beams develop only bending moments and shear forces. It should be recalled thatthe loads acting on any cut-out part of the grid must satisfy three equilibrium conditions,specifically equilibrium of the vertical forces, and the equilibrium of the vectorprojections of the moments with respect to two arbitrary horizontal axes of the gridplane (see Section 2.5).It is obvious that the grids constitute statically indeterminate structures with a large

degree of redundancy, and their analysis almost always requires the application ofsuitable computer software. Nevertheless, it is essential to examine some basic charac-teristics of the load-carrying behaviour of grids.

10.2 Main characteristics of the structural behaviour of gridsIn order to examine the main structural action of a grid, i.e. the transfer of vertical loadsin two horizontal directions, the simple model shown in Figure 10.2 is employed, where avertical load P is transferred in two directions via two cross-beams. The symmetryexisting in both directions involves the development of only one (vertical) displacementat the node. This displacement is common to both beams. However, the correspondingcurvature 1/r is different. The shorter beam will develop a greater curvature than thelonger beam and, according to the relation M¼EI/r, it will also develop a higherbending moment, provided that the stiffness EI of both beams is the same.As has been shown in a previous chapter (see Section 2.3.8), the bending stiffness or

‘rigidity’ of the simply supported beam with regard to its midpoint displacement is theforce required to cause a unit displacement equal to 48 EI/L3. The factor 48


becomes 192 for a beam with fixed ends. It thus appears that, for example, doublingthe length of the beam causes an eightfold decrease in the above stiffness, while it isproportional to EI.Analysing this simple system (using the method of forces or the method of

deformations), the important conclusion is found that if P1 and P2 are the forces

352

Node deformations

Sectional forces

Figure 10.1 Internal forces and displacements in grids

Beam 1 develops greater curvatureand hence a bigger bending response

2

2

1

P

1

P2

P1

Common displacement

δP1 (stiffness)1 =P2 (stiffness)2

Figure 10.2 Load distribution depending on member stiffness


carried by beams 1 and 2 at their midpoints, respectively, then

P1P2

¼ ðstiffnessÞ1ðstiffnessÞ2

(see Section 3.2.10)

It is easily concluded that if beam 2 has double the length of beam 1, then the loadcarried by beam 2 is one-ninth the applied load P, while the maximum moment of theshort beam 1 is four times greater than that of beam 2. If beam 2 is even longer, it isclear that its contribution to carrying the vertical load will be insignificant.It is, of course, clear that even in the case of beams with different lengths it is possible

to achieve a balanced load distribution based on the above equation, by designing thebeams in such a way that (stiffness)1¼ (stiffness)2. g

The conclusions resulting from the examination of the above simple model are also validfor regular grids that cover areas with an orthogonal outline. If the beams have the samecross-sectional characteristics, then those with a shorter length are more stressed, whilefor an aspect ratio larger than 1.50 the beams in the longitudinal direction are clearly‘underworking’ (Figure 10.3).The above relationship is obviously valid also in the case where the two cross-beams

are not at right angles (see Figure 10.3). On the basis of this fact, it can be seen that, tocover oblong areas (regardless of the aspect ratio), the adoption of a skew layout ofbeams ensures essentially the same stressing for each group of beams, as they havethe same length. Thus, the balanced load transfer through the two groups of beamsclearly implies smaller bending moments (i.e. smaller beam depths) and also smallerdeflections compared with the orthogonal layout.

353

The load is carried mainly by the short beam Only the beams in the short directionare practically stressed

The load is carried equally by the two beams Lower bending response of beamsSmaller deformations

P

P

Plan view

Figure 10.3 Influence of the grid layout on the load-carrying behaviour

Grids

Moreover, the presence of beams of short lengths and hence of high stiffness near thecorners provides fixed conditions for the relevant diagonal beams, which is favourablefor both the bending response and the deformation. g

The model examined above shows the contribution of crossed beams in transferring theload. An additional characteristic of the load-bearing action of grids is the possibility thatother neighbouring parallel beams also contribute to the transfer of a load that is appliedto the beam under consideration.More specifically, if a load is applied to the intermediate of three longitudinal beams

with a parallel layout, as shown in Figure 10.4, the two extreme beams will obviouslyremain unloaded, whereas the presence of a transverse beam connected to themtransfers the load also to the other two beams, thus relieving the intermediate beam.In order to examine the influence of the transverse beam, this beam may be consideredas a continuous beam of two spans resting on three flexible supports with spring stiffnessks¼ 48 EI1/L31, according to Section 2.3.7.Under the influence of the applied concentrated load, the intermediate longitudinal

beam deflects downwards by , imposing the same displacement on the central supportof the transverse beam too. At the extreme supports of the transverse beam, upwardreactions will develop, which will be applied with opposite sign to the longitudinalbeams (see Figure 10.4). From the analysis (see Chapter 3), the downward enddisplacement v is

v ¼

1þ ks L3

3 EI

354

(EI )1

(EI )1

(EI )1

L L

L

L

EI EI

EI

EI

Increasing EI tends to equalise the settlements

[M ]

[M ]

The spring settlements are proportional to theload carried by the longitudinal beams

They act with the opposite sense on the longitudinal beams

L1

ks ∆v v ksks

Transverse beam

Figure 10.4 The influence of a cross-beam on the transfer of a concentrated load


The stressing of each longitudinal beam depends exclusively on its midpoint displace-ment, whereby, the effect of EI according to the above equation becomes clear. Asmall value of EI causes a small v, whereas with an increase in EI the displacement vtends to approach the value . This means that the load is also transferred to thetwo extreme beams due to the stiffness of the transverse beams. The utilisation of avery stiff transverse beam ensures that, in practice, the load is equally distributedbetween the three parallel beams.In view of this conclusion, it becomes clear that in oblong orthogonal areas, where, as

was previously examined, only the beams in the shorter direction participate substan-tially in the distribution of loads, the longitudinal beams, despite the fact that theyare ineffective for the uniformly distributed permanent loads, nevertheless must beplaced in order to ensure, by their rigidity, the distribution of any concentrated loadto the neighbouring main beams too. g

In the above considerations it has been assumed that the cross-beams of the grid aresimply resting on each other by transferring only vertical support forces. In practice,the connections between beams are not designed this way, being usually rigid (ormonolithic), which means the development of additional torsional moments in thebeams. This fact does not alter the above conclusions at all, while it contributes tothe more favourable structural action of grids with respect to bending stresses anddeformations. This favourable contribution of torsion to the bearing capacity of gridsis explained immediately below.The symmetrical model in Figure 10.5 is considered, where the beams are mono-

lithically connected to each other. The cross-beam is simply supported at its endswhile the two longitudinal beams are supported so as to be able to take up torsionalmoments. Two equal forces P are applied to the internal nodes of the model.It is clear from the above that the longitudinal beams provide the transverse beam

with an elastic support at the corresponding nodes with a spring stiffness equal toks¼ 48 EI/L31. It is also clear that, besides the settlement of the support points, arotation ’ will also occur at these points in the transverse beam, which will be freelydeveloped if the longitudinal beams are simply supported, but not if these are connectedmonolithically. The reason for this is that in the last case the above rotation will benecessarily developed in the longitudinal beam as well, which will develop torsionalmoments (see Figure 10.5), because the extreme cross-sections cannot rotate, thusapplying a bending moment simultaneously to the transverse beam that resists thisrotation. It is clear that this moment will decrease the bending stress state anddeformation of the transverse beam accordingly (see Figure 10.5).More specifically, in order for the longitudinal beam to develop a unit rotation at its

considered internal node, it requires — or in other words, opposes the member imposingthis rotation — a specific moment k’, is called the ‘torsional stiffness’ (see Section 2.5.2).Thus, the imposition of a rotation ’ requires the application of a torsional moment(k’ ’), which is obviously also applied with opposite sign to the member imposingthis rotation, i.e. to the transverse beam. Therefore, the transverse beam does notonly have the translational springs ks in its two internal nodes but also the torsional

355

Grids

springs k’ that oppose the rotation of the nodes (Figure 10.6). According to Section2.5.3, as well as Section 3.2.10,

k’ ¼ 4 G ITL1

For the applied loads P and after direct application of the method of deformations (seeChapter 3) it is found that, for the particular layout shown in Figure 10.6, the developingvertical shift results from the relationship

¼ PL3

3 EI 1

1 3

4þ k’ L=EIþks L3

3 EI

From this equation one can draw the conclusion that the existence of torsional rigidity inthe longitudinal beams that is expressed by the presence of a rotational spring k’ causes areduction in the developing settlement of the supports of the transverse beam, in

356

Moment equilibriumat the joint

Transverse beam

The torsional moments on the joint reduce thebending response in the other direction

MT

MT

MT

MT

MT

MT

MT

MT

MT

MT

(GIT)

EI

(GIT)

(GIT)

(GIT)

(EI )1

(EI )1

MT = ϕ · kϕ

ϕ

ϕ

PL

P

P

L

2L

L1

MB(r)

MB(l)

MB(l)

MB(r)

Figure 10.5 Relieving effect of torsion in the bending and deformation states


contrast to the case where this torsional rigidity does not exist (k’¼ 0). In other words,the developing torsion of the longitudinal beams decreases the bending stress in both thetransverse and the longitudinal beams, given the fact that this reduction in also impliesdecreased corresponding curvatures (see Figure 10.2). The system with monolithicnodes thus acquires higher stiffness. This is, of course, at the expense of torsionalmoments, which means additional shearing stresses.It is clear that these conclusions are also applicable to larger grids: the existence of

monolithic nodes implies the development of torsional moments, which reduce thebending response of beams and increase the stiffness against vertical loads.The above ‘contribution’ offered by torsion — as well as its magnitude — depends

directly on the torsional rigidity of the beams, i.e. on the value of the torsionalconstant IT. Thus, in grids with beams of I cross-section, the developing torsionalmoments are much smaller than, for example, in grids of the same dimensions withhollow sections of the same moment of inertia, so that in the latter case the relievinginfluence of torsion on bending is more intense.It becomes clear that the structural action of grids actually constitutes an interaction

between torsional and bending rigidity, and is generally based on the ratio of thesevalues, that is, ¼GIT/EI. An increase in the ratio means an increased torsionalcontribution — and a corresponding bending restriction — in carrying the loads. g

At this point, it should be pointed out that the presence of torsion in the grids is notessential for the equilibrium, as is the case in all the models previously examined.This means that equilibrium in a grid node, as shown in Figure 10.7, can be ensured— besides the requirement for vertical equilibrium — by itself, in the presence ofbending moments. The presence of torsional moments acts simply as a helpful contri-bution for equilibrium to each direction, and is due to the requirements for compatibilityof deformations. This has a direct repercussion on the approach to the design of a gridwith respect to the ultimate state.In particular, for beams of reinforced concrete with orthogonal section the imminent

cracking will reduce their torsional much more than their bending stiffness, so that thetorsional moments are reduced whereas the bending moments are increased.

357

The presence of rotational springs limits δ

2L LL

EI

Transverse beam

[MB]

P P

kϕ kϕ

ks δ ksδ

Figure 10.6 Justification of the relieving effect of torsion

Grids

As an extreme situation, the torsional rigidity can be totally disregarded, implying thatthe highest possible bending, can, of course, be considered. In any case, according to thestatic theorem of plastic analysis (see Section 6.6.2) and provided that the conditions forequilibrium are maintained, the design involving the omission of torsional moments(and a corresponding increase in the bending moments) is, of course, on the safe side,and can be followed, at least for the aims of preliminary design, as the estimate of theeffective torsional stiffness due to cracking is rather uncertain.

10.3 Layout and structural action of skew bridgesFrom the previous section, it can now be understood how beams with relatively largespans and relatively high torsional rigidity can reduce their bending, because of thegrid action, through the bending stiffness of the transverse beams connected to them.This possibility is used in bridge design with the so-called skew layout, where the

existing transport line cuts sideways the axis of an existing obstacle (river, under-passing road, etc.) and results in an increase in the length of the structure comparedwith what would be the case if the two axes were crossing each other at a right angle.

358

Equilibrium may also exist without considering torsional stiffness

Joint equilibrium Joint equilibrium

Torsional stiffness considered Torsional stiffness disregarded

P

P P

P

P

P

[MB][MB]

[MT]

Figure 10.7 Load bearing with or without the development of torsion


The beam to be used to support each of the bridge beam ends should be arranged parallelto the axis of the obstacle, which is sideways with respect to the main beam (Figure 10.8).This support beam, which has a high bending stiffness, is simply supported on bearingsthat leave free (unhindered) the rotation about their connecting axis, so that notorsional stiffness is offered in practice. For the efficient structural action of thissystem, the main beam must carry significant torsional moments, and, as mentionedin Section 2.5.3 (see also Section 13.1), only hollow-type beam sections can meet thisrequirement.The above-described structural system is shown in Figure 10.8. The main beam with

length L is loaded with a uniformly distributed load q. At each node, the vector of thetransferred moment can only be perpendicular to the axis of the support beam, as theleast deviation from the perpendicular direction would create a torsional loading thatthe beam will not be able to carry. Considering this moment X as a statically redundantmagnitude, it is found from the requirement for compatibility of the developingrotations, by applying the force method (see Chapter 3), that

X ¼ q L2

8

2

3 ð1þ t g2 =Þ cosð ¼ GIT=EIÞ

It is clear that the moment X with its transverse component to the longitudinal axis ofthe main beam provides the relieving bending moment MB for the span, while with itscomponent MT on the axis it provides the torsional moment of the beam. This action

359

A

A

B

B

(Primary structure)

q

q

α

L

MT

MT

X

X

IT

Transverse beams rotate freely about their axisAxis

of obsta

cle

MBTorsional response is excluded

κT

MB(A) MB

(B)

Structure and sectional forces in plan view

Bending relief due to the skewness of transverse beams

q · L2/8

MB(A) = MB

(B) = X · cos α MT = X · sin α

Figure 10.8 Skew bridge layout

Grids

arises because of the angle of skewness , and, as long as this angle decreases, the greateris the bending relief in the main beam and the greater is its torsional response. Aspreviously pointed out, the latter increases as the ratio increases.With regard to concrete structures, it is understandable that prestressing, which is

used to limit the bending response of the main beams, also contributes substantiallyto guarantee their full torsional rigidity, which is necessary for the appropriate develop-ment of torsional moments.

360


11

Plates

Plates or slabs are flat monolithic structures that extend in two dimensions and have arelatively small, constant thickness. A plate is basically intended to take up distributedloads transversely to its plane; however, it presents a particularly high stiffness whenreceiving loads within its plane.The support of the plate — to where naturally its loads are transferred — is set along the

whole length or a part of its border, along the length of any line, or even at preselectedpoints on its plane (Figure 11.1).In this chapter, only plates receiving gravity loads are examined. Such plates are made

almost exclusively of concrete, and constitute the largest part of the entire concretevolume used for load-bearing construction in general.

11.1 The plate equation as a consequence of its load-bearing actionThe bearing action of plates can primarily be considered as arising from the bearingaction of grids, which was examined in Chapter 10. However, the fact that the plateconstitutes a continuous medium adds some particular bearing characteristics that areimportant for both the understanding of the load-carrying mechanism and for its design.The idea of modelling a plate as a dense grid of beams allows, first, the conclusion that

the bearing action of a plate is based on its bending in two directions and on the corre-sponding torsion. It is reasonable for the considered grid to be oriented in the directionsof a system (x, y), which is also common to the plate (Figure 11.2). The natural parallelof the plate with the grid enables the load-bearing action of the plate to be understood inanalogous terms to those for beam structures.With regard to the load resistance mechanisms, the load p(x, y) applied on a cut-out

square element of unit dimensions is equivalent to four ‘resistance’ forces, namely: theforces developed through the bending in the x and y directions, denoted as pb,x and pb,y,respectively; and the forces developed through torsion about the axes x and y, denotedby pt,x and pt,y respectively. Thus (Figure 11.3)

p¼ pb,xþ pb,yþ pt,xþ pt,y

To understand this better, the orthogonal plate shown in Figure 11.2 and the corre-sponding grid with its beams placed at unit distances are considered. The deformationpicture in the region of a considered point A shows that the deflections along thelength in each direction are increasing, whereas the slopes are decreasing towards the


centre of the plate, with a gradual increase in curvature in both the x and the y direc-tions. In the cut-out square element, the vector bending moments mx and my applied onthe corresponding sides are considered, with the moments on the opposite sidesdetermined, according to the law of differential increase, as mxþ (@mx/@x) 1 andmyþ (@my/@y) 1, respectively (see Figure 11.3). It is clear from the total picture ofdeformations that all the above moments are causing tension on the bottom fibres ofthe plate, this tension increasing toward the centre. This means that in the x and y direc-tions the element is acted on by the resultant moments (@mx/@x) 1 and (@my/@y) 1,respectively (see Figure 11.3).Considering first the x direction, the moment (@mx/@x) 1 requires, for equilibrium,

the development of shear forces qb,x¼ (@mx/@x) 1/1, which, decreasing toward thecentre, offer through their variation along x the corresponding ‘resistance force’ pb,xto the load p(x, y), i.e. pb,x¼ @2mx/@x

2.Now, if w(x, y) expresses the deflection of the plate, then the change in slope @w/@x at

point A along y is @2w/@x @y. Given, however, that this change in slope is of a torsionalnature along the y direction, expressing the relative twisting angle of the unit-lengthelement, a torsional moment mxy will develop in the y direction of the plate, accordingto relationship

mxy ¼ @2w

@x @yGIT (see Section 2.5.3)

362

p

The loading acts vertically to the plane of the plate

Figure 11.1 A plate as a continuous plane structure, loaded transversely

x

y

A

Transverse load p(x, y)

Figure 11.2 Deformed configuration of an orthogonal plate


As the relative twisting angle @2w/@x @y is decreasing along the y axis, the torsionalmoment mxy will also be decreasing, and its change (@mxy/@y) requires for equilibriumthe shear forces (@mxy/@y)/1, which, decreasing along the x direction, offer by theirvariation the corresponding ‘resistance force’ pt,x to the load p(x, y), which has thesame sign as pb,x and is equal to pt,x¼@2mxy/@x @y.In a similar manner, consideration of the y direction leads to the other two resistance

forces (with the same sign as the previous ones) associated with the correspondingparallel sides of the element, i.e. the ‘bending’ resistance force pb,y¼@2my/@y

2 andthe ‘torsional’ resistance force pt,y¼@2mxy/@x @y. It is obvious that the torsionalmoment mxy is the same in both directions.Thus, the above formulated equilibrium of load p with the four resistance forces can

be written on the basis of their corresponding expressions (see Figure 11.3):

@2mx

@x2þ 2

@2mxy

@x @yþ@2my

@y2¼ p

363

The pair of forces balancesthe vectorial sum of

the torsional moments

The variation in forcesoffers the corresponding

resistance to the load

The shearing pair balancesthe vectorial sum of

the bending moments

The variation in shearoffers the corresponding

resistance to the load

Tra

nsve

rse

load

p(x

, y)

Sen

se y

Sen

se x

Equilibrium

Equilibrium

∂my/∂y

Variation in y

Variation in y

mxy

mxy

my

mx

1 1

11

1

1

1

1

mxy + (∂mxy/∂x) · 1

(∂mxy/∂x) · 1

(∂mxy/∂y) · 1

∂mxy/∂x pt,y

pt,x

pb,x

∂2mxy/∂x ∂y

∂2mxy/∂x ∂y

∂2mx/∂x2

my + (∂my/∂y) · 1

pb,y

∂2my/∂y2

∂mxy/∂y

∂mx/∂x

mxy + (∂mxy/∂y) · 1

mx + (∂mx/∂x) · 1

Figure 11.3 Development of resistance forces due to the transverse load on the plate

Plates

In this way, the relieving role of the twisting moments mxy in the bending stress state ofthe plate becomes evident, either by the offer of the corresponding resistance forces pt,xand pt,y, or through the consideration of a bent strip of unit width (e.g. along thedirection x (or y)) receiving along its length the action of the resultant of momentsmxy on its two longitudinal edges (Figure 11.4).It must be emphasised that the moments mx, my and mxy are referred to a unit length.

The bending stress state is proportional to the corresponding curvature 1/r (see Section2.3.2), that is

mx ¼ @2w

@x2 EI

and

my ¼ @2w

@y2 EI

The expression formxy has been determined previously. In the present case, I¼ d3 1/12,IT¼ d3 1/6 (see Section 2.5.3) andG¼E/2. (Note that only half the torsional inertia ITis considered because the shear stresses receiving the torsional momentmxy are activatedonly through the thickness of the relevant section, and not through its width also, as isthe case in a typical cross-section.)The last expressions for the bending moments are not strictly correct, as they do not

contain the influence of the so-called Poisson’s ratio, which is related to the transverse defor-mation that generally accompanies an axial stress state. However, in any case the concretecracking limits this influence, and so it can be ignored for preliminary design purposes.The substitution of the expressions for the moments mx, my and mxy in the above

equilibrium equation leads to the classical equation for plates with respect to deflection w:

@4w

@x4þ 2 @4w

@x2 @y2þ @4w

@y4¼ p

K

where K is the plate bending stiffness, which is practically equal to Ed3/12.

364

Resulting actions due to the torsional momentsReduced bending response and deformation

x

y

Figure 11.4 Influence of the twisting moments on the bending stress state and deformation


In the past, this differential equation, together with the effective support conditionsalso formulated in terms of deflection w, was used for the analytical determination of thestress state in plates with orthogonal or circular boundaries. But this always constituted aparticularly difficult problem. The rapid spread of computers in recent years has allowedthe application of numerical methods that were developed long time ago, and hence,using suitable and accessible programs, the reliable analysis of plates is nowadayspossible, even when their geometry is highly complex. The need for a naturalinterpretation of the load-bearing behaviour of plates remains, nevertheless, essential,as does an appraisal of the characteristic particularities of certain typical constructionallayouts and their design. The present chapter will move within this frame. g

First, it may be pointed out that the reactions on a straight boundary do not coincidewith the shear forces q corresponding to the transverse bending mechanism, as occursin a grid, but are formed by the simultaneous presence of the twisting moments. Forexample, the twisting moments mxy that appear along the freely supported straightboundary of the orthogonal plate shown in Figure 11.5 can be represented, from a

365

Acting forces along the plate boundary due to the presence of torsional momentsThey are mutually balanced

Plate reactions are stronger than the corresponding shearing force

R

R R

R = 2 · mxy

Transverse load p(x, y)

1 1 1

1

1

mxy

mxy

mxy

mxymxy

mxymxy + (∂mxy/∂x) · 1

(∂mxy/∂x) · 1Effective difference

Not developed in thecase of a clamped edgeEquivalent actions

x

y

Figure 11.5 Development of reactions at the plate boundaries

Plates

natural point of view, only by corresponding pairs of forces with a unit lever arm, being ofcourse equal to mxy (see Figure 11.5). The variation in these forces, for example in the xdirection, involves the additional loading of this straight boundary of the plate by thedistributed load @mxy/@x (upwards), with the value of this load increasing towards theends of the boundary and resulting in a reaction equal to (qþ @mxy/@x). Thus itappears that the sum of the reaction forces is higher than the total load applied tothe plate. The reason for this paradox lies in the fact that at the four corners of theplate the above local ‘fictitious’ forces mxy have the same direction, and are added upin each corner to give a resultant downward force R, which cancels out the distributedload @mxy/@x. Thus the equilibrium between the external load, the distributed reactions(qþ @mxy/@x) and the corner forces R is finally ensured. It thus results that R¼ 2 mxy.Note that the force R is of the order of roughly 9% of the total uniform load of the plate,and the take up of this force requires a suitable anchorage of each corner of the plate, sothat this downward force can actually be offered.It should be pointed out that, in the case where the boundary of the plate is fixed, the

moments mxy are not developed along it. Because the value of the corresponding zeroslope is not altered in the other direction, the magnitude @2w/@x @y is thus equal tozero. Consequently, the reactions remain unaffected (see Figure 11.5). g

At this point it should be noted that, as the loads on the beams are transferred to thesupports via principal stresses (see Section 4.1.1), the loads on the plates are nottaken up along the arbitrary directions x and y (involving the development of thebending and twisting moments mx, my and mxy, respectively) but are taken upthrough the so-called principal moments. These principal moments are bendingmoments applied to each point of the plate in specific directions perpendicular toeach other.Thus, at each point of a plate there are two mutually perpendicular directions, each of

which is acted on exclusively by a moment vector parallel to the cut-out side, withoutany twisting action along it, which is the case in all other directions (Figure 11.6). Thisfact has particular importance as these directions reveal the paths that are followed foran exclusively load-bearing bending action. The two perpendicular directions at each

366

The plate behaves according to the principal moments developedAll bending moments are referred to the unit length

Principal moments

mxymxy

mxy

mxy

mymy

mxmx

m1 m2

y

x

Figure 11.6 The presence of principal moments


point, as well as the principal moments m1 and m2 themselves, are determined throughthe values of mx, my and mxy that prevail at the point considered, on the basis of equili-brium of the two triangular cut-out elements (see Figure 11.6).That this is the ‘correct’ way of perceiving the structural action of plates is confirmed

by the fact that the developing cracks in a concrete plate correspond precisely to theprincipal moments considered. Thus it is understandable that, regardless of the gridmodel used to approach the bearing action of a plate (see above), the essentialcharacteristic of its structural operation results from its monolithic continuity, whichis not valid for the grid, which exhibits a naturally structural discontinuity.The fact that a plate transfers its loads only by bending in the directions of the

principal moments (Figure 11.7) makes it possible to model it with a grid, theelements of which may be aligned as closely as possible to the principal directions,transferring to each other one vertical force (i.e. bending only, and no torsion). Thus,the obsession with the grid returns again, having once been utilised in an aestheticallyvery persuasive way by the Italian architect—engineer Pier Luigi Nervi for covering aspecific area using beams with negligible torsional stiffness (Figure 11.8) (Nervi,1956). g

Considering now, as previously, the equilibrium of the elements (see Figure 11.6), it isevident that the degree of divergence of the principal moment direction from thedirections x and y is related to the magnitude of the twisting moment mxy. Thus, forthe corner regions of the simply supported plate, where the biggest mxy occur and thebending moments mx and my are vanishing, it is possible to obtain a clear picture of theprincipal moments m1 and m2 developing there (Figure 11.9). Equilibrium of the twocorner elements (a) and (b) in the same region, with equal twisting moments mxy

acting at the vertical sides, implies that at the 458 cuts only the vector of principalbending moments m1 and m2 is acting and, in particular, m1¼m2¼mxy (see Figure 11.9).The resulting sense of m1 in Figure 11.9(a) shows that the top face is in tension, with

tensile stresses acting along the diagonal, and this is confirmed by the developing cracksin the oblique direction. Moreover, equilibrium of element (b) shows that the vector ofthe moment required for equilibrium on the skew side is also of a purely bending nature,

367

(a) (b)

Figure 11.7 Trajectories of the principal moments under (a) free and (b) fixed support conditions

Plates

368

Figure 11.8 Configuration of a grid following the trajectories of the principal moments

(a) (b)Principal moments

All bending moments are referred to the unit length

m1 = m2 = mxy

x

y

1

1 1

1 1√2

m1 m2mxy

mxy mxy

mxy

90° a

a

a

m1 = R · a/2 · a = R/2

R (= 2 · mxy)

1√2

Figure 11.9 Equilibrium in the corner region of a plate


representing the principal momentm2, causing cracks at the bottom face in the directionof vector m2.Note that the principal momentm1 relates directly to the corner force R, as this causes

a moment (R a) at a distance a from the diagonal, applied over a length 2 a. Thusm1¼ (R a)/(2 a)¼R/2, a result that is supported by the above conclusions (seeFigure 11.9).

11.2 Orthogonal platesOrthogonal plates are the most used form of plates in building construction and inbridges. In bridges, because of the concentrated wheel loads of vehicles, a specificload-bearing characteristic should be clarified as early as possible.The presence of a concentrated load causes curvature in both directions of the plate

and, consequently, a proportional bending response. The apparent paradox is that thisbending response is, in practice, independent of the dimensions of the plate(Figure 11.10). The reason for is that for a specific concentrated load P the totalmoment developing in each direction is proportional to the corresponding length,whereas for the corresponding unit width of application the bending moment comesout practically as a constant (see Figure 11.10).

11.2.1 Two-side supported slabs

11.2.1.1 Stress stateTwo-side supported plates are those that are supported on two opposite sides by eithersimple or fixed supports, while the other two sides are either free or are supported at sucha large distance apart that the transfer of loads in effect takes place within the shorterspan distance L.

369

Total bending moment: MDistribution width: L

M/L = m

Total bending moment: 2MDistribution width: 2L

2M/2L = m

The bending response remains constant

2L

L

2LL P P

2MM

Figure 11.10 The action of a concentrated load on a plate

Plates

These plates, when loaded with a uniform load p, act as beams and are substantiallystressed only in the supporting direction, with maximum span moments of m¼ p L2/8and m¼ p L2/24 for the simply supported and fixed cases, respectively.However, in the case of a concentrated load P (which of course is not referred to a

geometrical point, but is considered here as being distributed over a small square ofarea 0.0025 L2), the fact that the locally developing deflection causes curvature alsoin the transverse direction implies the development of moments in this direction too(Figure 11.11).The transfer of the concentrated load to the supports can also be considered similar to

the situation for a beam. The transfer is through an effective zone of width bm that is ofthe order of 75% of the span L for the simply supported plate and of the order of 45%of the span L for the fixed supported plate, assuming that the load is applied at themidspan. Thus, the statically developed total bending moment M, which at themidspan is P L/4 for the simply supported plate and P L/8 for the fixed supported one,is distributed over an effective width of 0.75 L and 0.45 L, respectively. Therefore,the developing moments can be estimated as m¼M/bm¼ P/3 and m¼M/bm¼ P/3.60,respectively. Special attention must be paid in the case where the concentrated loadis applied at the midpoint of the free edge. In this situation the effective widthis decreased by half, and the bending moments are doubled accordingly (seeFigure 11.11). Note that the transverse bending moments developed are about 70%of the longitudinal ones. Moreover, the total fixed support moment M¼ P L/8 in thefixed plate is distributed to the support over a larger width bm, which is of the orderof 70% of the opening L, and therefore the corresponding bending moment is estimatedas m¼M/bm¼ P/5.60. Thus, as noted initially, the bending moments due to concen-trated loads are independent of the span length, a fact that is of particular importancein bridge engineering, given the significant loads imposed by vehicle wheels.Finally, it is pointed out that for a linear load q (kN/m) extended over the span L of a

simply supported plate (Figure 11.12), the total developed bending moment q L2/8 isdistributed over a width bm¼ 1.35 L, so that the bending moment m is roughly equalto 0.10 q L, which is of course dependent on the length L.

370

The transverse bending intensity is ∼70%

Dou

bled

ben

ding

res

pons

e

Dou

bled

ben

ding

res

pons

eDistribution width Distribution width

0.75 · L 0.70 · L

0.45 · L

P P · L/4

P · L/8

P · L/8

P PP

Figure 11.11 The influence of a concentrated load on the stress state of a two-side supported plate


11.2.1.2 DesignPlates are generally designed and reinforced for bending rather than for shearing. This isbecause the developing principal tensile stresses in the regions of intense shear andnegligible bending are — for the applied design thicknesses — definitely smaller thanthe resistance ‘allowed’ in tension. With regard to simply supported plates, reinforce-ment is placed on the bottom region and is anchored at the ends. Thus, when theload is increased, with the accompanying development of the expected intensecracking, the compact compression zone of the concrete acts as a tied arch that takesup the loads without the need for stirrups (Figure 11.13).The design of the bending reinforcement as (cm

2/m) follows that examined in Section4.2.2.1, but in the present case the lever arm z of the internal forces is considered asequal to 0.95 h. Thus for a design moment md, it is obtained that as¼ R md/( fsy 0.95 h). gThe design of plates always includes the control of deflections in the service state. Forpermanent loads, for which one should not consider the uncracked section but perform amore detailed calculation instead, according to Section 4.2.1.1, the creep influenceintervenes, always adversely. The creep increases the deformation L due to the loadsby ’ L. Thus the total deflection may be estimated as L (1þ’). This should besmaller than an acceptable ratio of the opening L, usually L/250, while in the casewhere brick walls are supported on the plate this ratio must be smaller than L/500.These conditions are usually met for plate thicknesses of the order of d L/25.

PrestressingThe possible need for prestressing is treated as developed in Section 4.3. Prestressing ismainly applied in bridges with spans over 15m. The longest span that is economicallyacceptable is roughly between 35m and 40m. The maximum thickness of a slabbridge should not exceed 1.00—1.10m.

371

1.35 · L

Distribution width

q q · L2/8L

Figure 11.12 The influence of a linear load on the bending state of a two-side supported plate

Figure 11.13 The action of a concrete plate in the cracked state

Plates

In this respect, attention must be paid to the more intense bending response, aswell as the deformation, developed at the free edges, compared with that in themiddle region of the slab, due to vehicle loading (see Section 11.2.1.1). The useof special computer software is necessary to determine the stress state and thedeformations.To estimate the prestressing force two basic approaches may be followed, as described

in Section 4.3: full prestressing and partial prestressing. In full prestressing, the size ofthe prestressing force (i.e. the total section of the cables) is determined such thatunder the permanent and traffic loads the tensile stresses in the concrete arepractically eliminated, or at least tolerated if they are smaller than the resistance ofthe concrete in tension (2N/mm2). However, as explained in Section 4.3, thisapproach is uneconomical.In partial prestressing the prestressing force is limited to that size which allows only

the permanent loads to be taken up, without developing tension. The traffic loadsthat will be applied may lead to stresses higher than the tensile resistance of theconcrete, and cause cracking (of limited width), which, however, disappears aftertheir removal (see Section 4.3.2.1). Thus the partial prestressing approach is clearlymore economical.Prestressing is usually distributed uniformly over the slab width, i.e. the cables are

placed and anchored at equal distances from one other (Figure 11.14). For the fulltake-up of a permanent load g (kN/m2) over a span length L, the required prestressingforce Pg per metre is (see Section 4.3.1.1)

Pg ¼g L2

8 ðd=2 cÞwhere d is the slab thickness and c is the cable cover.Under the permanent loads and the prestressing force Pg no bending develops, only a

uniform compression across the plate thickness. As a reasonable consequence, this

372

Plan view

Figure 11.14 Loads are taken up through distributed prestressing


prestressing value usually offers, together with the necessary percentage of steel reinfor-cement (of the order of 1%), a bending resistance mR that is more than sufficient tosatisfy the design criterion (mgþmp)<mR, where mg and mp are the designmoments for the permanent and traffic loads, respectively. For this reason, and inorder to limit the ‘distance’ between the two members of inequality, thus reducingthe necessary cross-sectional area of cables, a percentage of the prestressing force Pgof the order of 75% is applied.In the case of a bridge, the bending response state along the free edges due to traffic

loads mrp is clearly more than that which occurs in the interior of the slab mm

p , asexplained previously. Thus, the above checking of the strength must be performednot only for the interior of the slab under the design moment Smd ¼ (mgþmm

p ), butalso for the extreme zones with the moment Srd¼ (mgþmr

p).At this point it should be noted that the force Pg is transferred as a compression force

in the slab only if the slab support allows movement in the direction of the force, so thatthe corresponding shortening can take place. If this is not the case, the slab cannotreceive axial forces (because the prestressing force is absorbed by the immovablesupports). Nevertheless, the slab is acted on by the full deviation forces, and thus,from a stress state point of view, the bending response, as previously considered, isstill valid (see Section 4.3.1.2). g

The saving in steel through partial prestressing may be even greater if the cables areconcentrated in two bands at the outer edge zones of the slab, over a width of theorder of 20% of the total width B of the slab, as shown in Figure 11.15 (Menn, 1990;Stavridis, 2001). In this case, due to prestressing, the plate develops a bendingmoment mP that is practically constant over the entire width of the plate, and can beeasily calculated on the basis of the deviation forces as follows. If the width of eachzone is ( B) and ! is the applied reduction factor for the prestressing force Pg (i.e.of the order of 50%), then the applied prestressing P per unit width of the zone isP¼! Pg/2 and the applied deviation forces uP on the zones are

uP ¼ 8 P ðd=2 cÞ=L2 ðkN=m2Þ

Thus the bending due to prestressing, which is practically constant across the entirewidth of the slab, is

mP¼ uP (2 B) L2/8B¼ 0.25 uP L2 (tension on the top fibres)

The above bending moment is composed of the so-called ‘statically determinate’ partm0P (tension on the top fibres), which is limited to the outer zones only and affects,as already stated, the bending resistance only in these regions, and the ‘staticallyredundant’ part mSP (see Section 5.4.1). Hence

mrSP ¼ mP m0P ¼ 0:25 uP L2 þ P ðd=2 cÞ

while

mmSP¼ 0.25 uP L2

373

Plates

As this last part is self-equilibrating, as explained in Section 5.4.1 (check:mr

SP 2 Bþ mmSP (B 2 B)¼ 0), it may be superposed on the bending moments

due to the external loads in the strength check, according to the equations

Srd þ mrSP < mr

R

374

Self-equilibratingresponse

Self-equilibratingresponse

mqrmq

r mqm

mqr = 514 kN m/m mq

m = –289 kN m/m

Sdr + m r

SP + mqr < mR

r

Sdm + mm

SP + mqm < mR

m

(r) (r)(m)

(r) (r)(m)

(r) (r)(m)

(r) (r)

(m)

(r) (r)(m)

λ · B λ · B λ · B λ · B

λ · B λ · B λ · B λ · B

qq

qq

mq0 = q · L2/8

mq = – 0.25 · q · λ · L2

Sdr = 4092 kN m/m Sd

m = 3825 kN m/m m rSP = 1425 kN m/m mm

SP = – 801 kN m/m

q = 11.15 kN/m2

Sdr Sd

rSdm m r

SP m rSPmm

SP

m rR = 6050 kN m/m

mRm = 2736 kN m/m

L = 24 mB = 15 mλ = 0.18

d = 0.95 m

m rSP m r

SP

uP

uP

uP

uP

mmSP = mP

Sdr Sd

m Sdr

uP = 30.92 kN/m2

(r):AP = 53 cm2/mAs = 19.3 cm2/m

(m):As = 57 cm2/m

P = 5936 kN/m mP = – 0.25 · uP · λ · L2

L

B

Figure 11.15 Taking up the load through prestressing of the edge zones


and

Smd þ mmSP < mR

for the outer and interior regions, respectively (see Figure 11.15).It may always be asserted that the moment mr

SP burdens the bending state, and themoment mm

SP relieves it. It is also known that the resistance moment mrR is formed by

both the prestressing and the steel reinforcement, while the resistance moment mmR is

formed only by the steel reinforcement. Of course, taking into account the favourableinfluence of the distributed compression force nP due to prestressing, a good approxi-mation is obtained: nP ¼ 2 P .As the above resistance check for the outer region of the slab is generally satisfied,

whereas that for the interior region — despite the relieving action of the momentmm

SP — is not, in order to ‘fix’ the situation, an additional self-equilibrating loading canalways be superposed, in view of the static theorem of the plasticity theory (seeSection 6.6.2). This loading consists, on the one hand, of a suitably selected uniformdownward load q (kN/m2), applied only at the two zones considered as being cut outfrom the slab, and, on the other hand, of the same load q applied upwards on the twozones of the full slab (see Figure 11.15). Thus, the outer region takes in addition thebending moment mr

q¼ q L2/8 0.25 q L2, while the interior one is relieved bythe bending moment mm

q ¼ 0.25 q L2. Superposition on the corresponding‘loading’ terms leads to the final satisfaction of both the above checks (see Figure 11.15).Despite the obvious need for more steel reinforcement in the interior region than is

required in the case of a uniform distribution of tendons over the whole width of theslab, the layout examined leads generally to a more economical solution. Obviously,the most favourable combination of the parameters and ! must be found by meansof trials.

11.2.2 Cantilever slabs

11.2.2.1 Stress stateThe introductory remarks given in Section 11.2 are valid also for cantilever slabs. Theuniform load p causes bending proportional to the square of the cantilevered length L,i.e. m¼ p L2/2. However, the concentrated load P applied at the free cantilever endcauses a total moment M¼ P L at the clamped edge, and this is distributed over alength roughly equal to 2 L, meaning that the load is transferred to the support at anangle of about 908. Thus, the bending moment is m¼ 0.50 P, i.e. it is independentof the length L (Figure 11.16). The moment acting along the longitudinal directiondue to the existing curvature in this sense causes, of course, tension on the bottomfibres, and may be considered roughly equal to the bending moment at the clamped side.It should be noted here that any stiffener or beam located along the free edge

somehow increases the distribution width (more favourable bending at the fixed end),while in case of a cantilever of variable thickness the distribution width is limited andthe bending moment increases (see Figure 11.16).

375

Plates

One should also note the resulting stress state in the region of a cantilever corner, asshown in Figure 11.17. For a uniform load p this stress is m¼ 1.50 p L2, while for aconcentrated load P on the external corner it is m¼ 1.65 P.

11.2.2.2 DesignCantilever slabs are designed, analogously to the description in Section 11.2.1.2, forbending rather than shearing, with the reinforcement placed at the top side. In a canti-lever there is no particular reserve of bearing resistance, because of its ‘statically deter-minate’ load-carrying action, and therefore particular attention must be paid to itsdesign, reinforcement, as well as its deformations, as even residential buildingsconstitute an area where higher live loads may be acting. Moreover, if a cantilever isconnected at its clamped end to relatively flexible slabs, its fixed end will be compliant(elastically rotating support), and this will increase the deformation of its free edge, incomparison with a completely fixed support (see Section 2.3.7).For the calculation of the deflections the same remarks as given for the two-side

supported plates in Section 11.2.1.2 are valid, whereby particular attention must bepaid to creep. For the deformations due to permanent loads, the principle of virtual

376

Plan view Section

Development of bending response alsoin the longitudinal sense

m ∼ 0.50 · PBottom fibres tensioned

m = 0.50 · PTop fibrestensioned

Width of load distribution

2 · L

Decrease in distribution width

Increase in distribution width

P

L

Figure 11.16 Cantilever action under a concentrated load

Uniform load p Concentrated load P

Top fibres tensioned Top fibres tensioned

m =

1.5

0 · p

· L2

m =

1.6

5 · p

m = 1.50 · p · L2 m = 1.65 · pL L

L L

P

Figure 11.17 Bending state of a cantilever corner under a uniform load and a concentrated load


work may be applied (see Section 4.2.1.1) by adopting the parabolic distribution of curva-tures shown in Figure 11.18. In any case, the expression L¼ (2 L)2/(8 r)¼ L2/(2 r)may be used as an acceptable estimate for the deformation L.The much stricter condition regarding the deformation at the free edge (< L/250),

which must be respected, can be met with a plate thickness of d> L/10. Finally, it isnoted that the deflection at the cantilever corner examined previously (seeFigure 11.17) for a uniform load is twice that of the adjacent cantilevers.Where there is a possible need for prestressing, the remarks made in Sections 4.4 and

11.2.1.2 are valid.

11.2.3 Four-side supported platesFour-side supported plates are the most commonly used type of slabs in buildingconstruction. They are supported perimetrically on beams or walls, and are usuallyarranged orthogonally, being connected to each other in a continuous, monolithic way.

11.2.3.1 Stress stateThe four-side supported slabs transfer their loads in both directions, differentiating theirload-carrying response depending on whether the load is uniformly distributed orconcentrated. In the case of a uniform load, the shorter direction also receives thelargest bending response, as has been explored for grids (see Figure 10.2). It should beclarified, however, that, with respect to a uniform load, the concept of ‘rigidity’ alongthe x or y sense concerns the distributed load px or py, respectively, which causes aunit deflection at the midpoint of a characteristic strip of unit width. This rigidity isexpressed as k (EI)/L4, where the factor k is equal to 76.8 or 384, depending onwhether the length L refers to a simply supported or to a fixed-end direction, respec-tively, and I¼ d3/12. Thus, it may be considered that the load p is ‘distributed’ to theloads px and py, i.e. p¼ pxþ py, while the requirement for common deflection of thetwo mutually orthogonal strips requires (according also to Section 10.2) that

pxðstiffnessÞx

¼py

ðstiffnessÞy

377

δL = (2 · L)2/(8 · r)

2 · LApproximate curvature distribution

in cracked cantilever

L

(1/r)max

δL

r

Figure 11.18 Cantilever deformation under a uniform load

Plates

that is

pxpy

¼ ðstiffnessÞxðstiffnessÞy

Thus, under conditions of uniform support, in the case where Ly¼ 2 Lx,px¼ 24 py¼ 16 py

As briefly pointed out in Section 11.2, the bending stress state due to a concentratedload is independent of the dimensions of the plate and is proportional to the load.Tables 11.1 and 11.2, which refer to Figure 11.19, show the bending moments of

perimetrically simply supported plates and fixed plates, for both their midpoints andtheir fixed supports, under a uniform and a concentrated load, for three characteristicaspect ratios. The case of an extremely oblong ground plan has already beenexamined in the case of two-side supported plates, but it is also included here forpurposes of comparison.It can be shown that the twisting moment mxy (developed only at the free support) is,

for the examined side ratios, roughly equal to 4% of the total load of the plate.With respect to a uniform load, it is observed that the momentmx at the midpoint of a

square plate is barely 30% of the corresponding moment for a two-side supported plate inthe case of simple supports, and 43% of the corresponding moment for the case of fixedsupports. However, these percentages are increased if the beneficial influence of thetwisting moments is ignored, which is equivalent to omitting the corresponding termin the differential equation for the plate (see Section 11.1). In the case of ribbed

378

Table 11.1 Bending moments of orthogonal plates under a uniform distributed load

Ly=Lx Perimetrically simplysupported plates Fixed plates

mx=p L2x my=mx mx=p L2x my=mx mex=p L2x mey=p L2x

1.0 0.037 1.0 0.018 1.0 0.052 0.0521.50 0.073 0.38 0.034 0.30 0.076 0.0571 0.125 0 0.042 0 0.083 0.057

Table 11.2 Bending moments of orthogonal plates under a concentrated load


mx=P my=mx mx=P my=mx mex=P mey=P

1.0 0.274 1.0 0.233 1.0 0.103 0.1031.50 0.311 0.83 0.257 0.88 0.127 0.0341 0.327 0.76 0.26 0.85 0.170


plates this influence is deliberately not taken into account, as will be examined later.Table 11.3 gives the corresponding data for the case of a uniform load where thetwisting moments are neglected. The developing bending moments are, of course,increased, but the ratio of the moments in the two directions remains substantiallythe same.An estimate of the reactions on the sides of the perimeter can be obtained by evalu-

ating the load acting on the regions created by the ‘bisectors’ of the corners withsupported sides and the midline of the two opposite longer sides (Figure 11.20). Inthe case where a fixed side is connected to a simply supported one, the ‘bisectors’ areset at an angle of 608 to the fixed side. g

In building construction, four-side supported plates are usually set in a continuousorthogonal arrangement, being supported on beams or, possibly, walls. Thus theirsupport along the length of the beams (or walls) can be considered as a simple one,and only a vertically distributed load is transferred, while the plates themselves aremutually monolithically connected. As the plates receive a permanent load g, as wellas a movable live load p, there is a need to determine the most adverse bendingmoments both at the centre of the plates and over their supports, i.e. in the commoncontact lines (Figure 11.21).

379

LxLx

(b)(a)

mx mx

my my mexmex

mey

mey

Figure 11.19 Characteristic locations of bending moments in orthogonal plates: (a) simple supports;(b) fixed supports

Table 11.3 Bending moments, with the twisting moments neglected, under a uniform load


mx=p L2x my=mx mx=p L2x my=mx mex=p L2x mey=p L2x

1.0 0.077 1.0 0.025 1.0 0.056 0.0561.50 0.128 0.38 0.043 0.31 0.085 0.0552.0 0.142 0.15 0.045 0.06 0.088 0.052

Plates

For this purpose, three distinct loading-layout cases may be considered, namely: (1)loading with ( gþ p/2), (2) loading with p/2, and (3) loading with ( gþ p).For each individual plate, the maximum bending moments in the centre can be

estimated by superposing the load in case (1), with fixed conditions all over theirperimeter, and the load in case (2), assuming simple support conditions over thesame support lines. The bending moments at each internal support (tension on thetop side) can be estimated from load case (3), with fixed conditions all around, andtaking always the mean of the values from each side (see Figure 11.21).

380

45

45

45

45 45

60

60

60 45

6045

60 6045

Figure 11.20 Distribution of the reactions at the sides

g + p/2 g + p/2

g + p/2 g + p/2

g + p g + p

g + p g + p

Span moments(bottom fibres tensioned)

Support moments(top fibres tensioned)

Fixed supports Fixed supports

Internal supports Simple support

Average of support moments

0.75 < Lx/Ly < 1.30

p/2 p/2

p/2 p/2

g, p g, p

g, p g, p

Figure 11.21 Evaluation of the bending moments in continuous plates


11.2.3.2 DesignThe reinforcement is determined on the basis of the bending response. The fact thatbending occurs in both directions in the centre of a slab implies that the reinforcementbars should be placed in two layers, one just over the other, in the x and y directionsrespectively. Obviously this results in a differentiated effective depth h (or lever armz) for the two senses. As is the case along the length of the supports, the bendingmoments develop only in the transverse direction; they are obviously covered by onlyone reinforcement layer, placed normally on the top side. To estimate the reinforce-ment, the equation for two-side supported plates (see Section 11.2.1.2) may beapplied for each direction.With regard now to the deformations, these can be determined through the

‘distributed loads’ px and py as has been explained previously:

px ¼ pL4x

L4x þ L4y

py ¼ pL4y

L4x þ L4y

The deformation w at the midpoint of a strip of unit width is

w ¼ 1

k px;y

12 L4x;yE d3

where k is determined according to Section 11.2.3.1. However, the estimated value of wdue to permanent loads must be increased by w ’ due to creep, and the final resultcannot exceed 1/250th of the corresponding span length. This condition is usuallymet for plate thicknesses d> L/30. g

In the rather rare case where, due to larger dimensions of the plate, the application ofprestressing is required, the permanent load g of the plate should be taken up by thedeviation forces of the cables. The required deviation forces can be ensured by auniform parallel arrangement of cables in one direction only, this direction being theshorter one. In this way the cables offer the largest possible deviation forces for thesame prestressing force, given that u¼ 8 P f/L2 (kN/m2) (Figure 11.22). It shouldalso be observed that, because the usually prevailing conditions do not allow a freeshift of the ‘prestressing front’, uniform compressive stresses due to prestressing withinthe plane of slab must be excluded, as explained previously (see Section 11.2.1.2),

11.2.4 Ribbed platesIf in a slab only the compression zone of the concrete is maintained, with the remainderof the material being supplemented in the form of parallel ribs arranged in both direc-tions, a statically more favourable system is acquired. This arrangement has, ofcourse, a higher structural depth, and hence smaller internal compressive and tensileforces, offering the possibility of using an even smaller thickness for the compressed

381

Plates

plate and less tension reinforcement (Figure 11.23). Moreover, the deformations of thisstructural system will obviously be smaller than those of a solid plate.If the plate is two-side supported, the parallel ribs are ordered in a dense layout only in

the load-bearing direction. However, one up to three transverse ribs still have to beplaced in the other direction (depending on the span), for reasons of wider distributionof the eventual live loads, as explained for grids (see Chapter 10).For an equal axial distance e between the ribs, under the self-weight G of a zone of

unit length and width e, and a live load p, the developing bending moment Mr on therib considered is Mr¼ (Gþ p e) L2/8, for a two-side supported plate. This momentis, of course, taken by the beam-like rib with depth d. The distance e does not haveto exceed 70 cm (see Figure 11.23).In the four-side supported plate the ribs are arranged, as already mentioned, along the

x and y directions. Provided that the same distance e (<70 cm) exists in both senses, theplate may be considered to behave as having a constant thickness. It should be notedthat, due to the relatively small thickness of the ribs, which is limited practically to athickness that allows easy placement of the reinforcement bars in one layer only,their torsional resistance is low, especially after the oncoming cracking. For thisreason, it is preferable, at least for preliminary design purposes, to ignore the beneficialinfluence of the twisting moments and apply what was discussed in Section 11.2.3.1. It isof course clear that the bending moment Mr of the ribs results by multiplying thecorresponding tabulated values by the distance e (see Figure 11.23). Moreover, theribs corresponding to the longer direction will develop the least bending response, asindicated by the ratio my/mx in Table 11.1. g

In a four-side supported plate, in order to achieve a preferred specific distribution of thebending response, or even for reasons associated with the construction process, it ispossible for the rigidities to vary in the two directions, thus creating an orthotropicplate (Figure 11.24). This is usually the case if the ribs are ordered along one directiononly, and consist of either concrete or of steel profiles, constituting thus a compositestructure (see Section 5.6). Of course, the equation for the load distribution considered

382

Layout A

With layout A a given total prestressing forcemay outbalance a greater load on the slab

Layout B

P

P

P

P

P P P

P PP

P

P

Figure 11.22 Layout for prestressing in a four-sided plate


383

Same bending response in the x- and y-sense

e

e

e

L

e

d

e

e e

The torsional momentsmay be ignored

Load-bearing ribs

It works like a solidfour-sided plate

(<70 cm)

Transverse rib

For a given dead load (volume)the developing internal forces in the ribbed plateare weaker due to the greater internal lever arm

Figure 11.23 The use of a rib arrangement in an orthogonal plate

Strong change in bending responsewith respect to the homogeneous plate Lx

Ly

Stiffness y Stiffness x

Figure 11.24 Stiffness differentiation in the two directions of a four-side supported plate

Plates

in Section 11.2.3.1 will be applied accordingly. For a total load p this equation is:

px ¼ p ðstiffnessÞxðstiffnessÞx þ ðstiffnessÞy

and

py ¼ p ðstiffnessÞy

ðstiffnessÞx þ ðstiffnessÞyTable 11.4 gives some specific typical results for four-side supported plates, with thebeneficial contribution of the twisting moments neglected (Stavridis, 1993). Theplates are considered perimetrically simply supported. It can be seen from the valuesgiven in the table how radically the distribution of bending moments can be influencedby the stiffness ratio adopted.The decks of steel bridges may also be included in this category of ‘orthotropic plates’.

Steel bridges are formed of two parallel main longitudinal beams with the distancebetween them being equal to the width of the deck. The beams are connected withtransverse beams placed at distances of the order of 5m, thus forming a grid ofbeams. The deck is formed by a steel plate with longitudinal stiffeners at distances ofthe order of 30 cm, thus creating a plate, the bending rigidity of which in the longitudinaldirection is much higher than the one in the transverse direction (i.e. rigidities ratio ofthe order of 20). The ribs are usually selected to have closed sections, so that with theexisting torsional stiffness the bending response will be limited, as explained previously.As the deck is designed to participate in receiving the intense compressive force of thetop flange of the main beams, its contribution to avoiding the lateral buckling of thesebeams may lead to an increase in its own bending rigidity (see Section 7.5). g

For oblong areas a skew layout of orthogonal ribs may be used, as discussed in Section10.2 (see Figure 10.3), in order to achieve an equal distribution of response in all theribs (Figure 11.25).Characteristic values of the bending moment Mr of a rib, with respect to the equal

distances e, are given in Table 11.5 for a 458 arrangement, for both a simply supportedand a fixed plate (Stavridis, 1993). Again, for the reasons already explained, thebeneficial influence of the torsional rigidity of the ribs has been ignored.

384

Table 11.4 Estimate of the bending moments in the two directions

Ly=Lx (stiffness)y/(stiffness)x

5 10 20

mx=p L2x my=mx mx=p L2x my=mx mx=p L2x my=mx

1.0 0.0225 5.76 0.0104 13.46 0.0038 37.71.50 0.0775 2.22 0.050 4.64 0.028 9.882.0 0.119 1.15 0.096 2.43 0.068 5.07


Clearly, the quantityMr/e corresponds to the moment m of a solid plate with constantthickness. Thus, by comparison with the corresponding values in Table 11.1, it is foundthat the bending moment of the ribs in the skew layout constitutes for the oblong areasonly 40—50% of the corresponding moment developed in the orthogonal arrangement.

11.3 Circular platesThe circular plates used for covering circular areas are either simply supported or fixed attheir perimeter. Under uniform load they develop a symmetrical stress state with respectto their centre, which is basically expressed by the principal moments mr in the radialdirection.For conditions of simple support, the highest span bending moment developed at the

centre of the plate can be practically approximated by considering a fictitious squareplate with sides 10% longer than the diameter of the circular plate (Figure 11.26).For a plate with diameter D under a uniform load p, mr¼ 0.047 p D2.For a fixed plate, the bending response at the centre of the plate is practically identical

to that of a fixed square plate that has sides of a length equal to the diameter of thecircle. As the cut-out corners of the square plate are substantially non-deformable,

385

Favourable distribution of the bending response in oblong rectangular areas

ee

ee

Lx Lx

Mr MrLy LyMr,xe

Mr,ye

Figure 11.25 The use of a skew layout of ribs in oblong areas

Table 11.5 Estimated bending moments for a skew layout of the ribs

Ly=Lx Simply supported plate Fixed plate

Mr=p e L2x Mr=p e L2x Mr;xe=p e L2x Mr;ye=p e L2x

1.0 0.035 0.021 0.040 0.0401.50 0.052 0.029 0.061 0.0462.00 0.069 0.034 0.072 0.044

Plates

this approximation is roughly valid also in the case of perimetrically simple supportconditions (see Figure 11.26). The bending moment at the fixed support is similarlylikened to that of a fixed plate with sides equal to 80% of the diameter of the circularplate (see Figure 11.26). Thus, at the centre of the fixed plate mr¼ 0.018 p D2,while at the clamped support mr¼0.031 p D2.

11.4 Skew platesPlates with a parallelogram outline and supported on two opposite sides are often used tocover skew transport crossings, and are called skew plates (Figure 11.27). The ‘skewness’becomes structurally more perceptible as the angle of the straight line that links thecorners of the obtuse angles with the supported sides becomes greater (seeFigure 11.27). The fact that the direction of this line is followed for the transferring

386

The moment at the centre is practically approachedby the circumscribed square plate

Identical fixed-end momentsfor a circular and square plate

1.10 · D 0.80 · Dmr mr

mr

DD

D

Figure 11.26 Assessment of the bending response in a circular plate

Traf

fic a

xis

Compression

Tension

Pressure distributionon the support

Bending response in obtuse anglescauses tension in the top fibres

Directions of principal moments

Direction of load transfer Top fibres tensioned

Bend

ing

of th

e fre

e ed

ge

Figure 11.27 Load-bearing action of a skew plate


of loads, it being the shortest (i.e. the stiffest) path to the supports, leads to theconclusion that the directions of the principal moments are also skew with respect tothe free edges of the plate, so that its bearing action results differ from those observedin an orthogonal plate. It may practically be considered that the directions of theprincipal moments that cause tension at the bottom face of the plate are found approxi-mately along the bisector of the obtuse angles. This results in a strongly unequaldistribution of the reactions as the ‘skewness’ increases, with a particular ‘accumulation’of forces at the obtuse corners, as there may possibly be a need to anchor the regions atthe acute corners, due to the developing downwards reactions.Moreover, an increasing ‘skewness’ leads to a longer free edge with larger deforma-

tions, while, simultaneously, the conceivable continuance of the free edge zone to theregion of the immovable supports at the obtuse corner creates there conditions offixity. Thus the principal moments that are perpendicular to the ‘bisectors’ of theobtuse corners cause tension on the top face of the plate. The great variation inbending moments along the free edge, i.e. from the ‘negative’ moments at thesupports to the ‘positive’ moments at the midlength, signifies the development oflarge shear forces (Q¼ dM/ds), and this confirms the ‘accumulation’ of reactions atthe obtuse corners of the plate (see Figure 11.27).A further characteristic of skew plates is that the skew placement of the directions of the

principal moments with respect to the plate outline under a uniform load is practicallyunmodified under a concentrated load. This is of particular importance because it practicallyallows the direct superposition of the principal moments due to both uniform and concen-trated loads. It should be noted, however, that for the final determination of the bendingmoments in skew plates, the use of suitable computer software is definitely required.A direct consequence of the skew placement of the principal moments with respect to

the free edge is that, apart from the free edge bending according to the developedcurvature, it is compelled to receive torsional moments also (Figure 11.28) by followingthe equilibrium of the cut edge zone. The internal face of this zone in Figure 11.28 hasbeen deliberately shaped as a ‘saw’ so that only the principal (bending) momentsconsidered are applied to it. As bending in the direction of the diagonal sense is

387

Obtuse angle

Principal moments Torsional loading ofthe free edge

Figure 11.28 Torsional response in the free edge zone

Plates

stronger than that in the transverse sense, the resultant ‘loading’ moment vector at eachpoint along the free edge of the plate permanently has the same sign, and consequentlycreates a torsional loading that must be taken up accordingly. Many skew bridges that donot have a sufficient number of stirrups in their edge zones for the uptake of thesetorsional moments (see Section 4.2.1.3) have developed cracks due to the abovetorsional response. g

For spans longer than 15m the slab should be prestressed. Moreover, in this case, as inorthogonal slabs, the sensitivity of the longitudinal edge zones with regard to theresponse and deformations must be taken into account, particularly for the appliedvehicle loads. The plate thickness should again not exceed 1.10m.The cables are placed parallel to the free sides, usually in a uniform distribution

(Figure 11.29). As examined in the case of orthogonal slabs, the full uptake of thepermanent load g by the prestressing force Pg per metre, for a skew span equal to L,according to the equation

Pg ¼g L2

8 ðd=2 cÞoffers, together with the obligatory steel reinforcement, much greater bending resistancethan required. For this reason, in order to follow a more economic design, only a percen-tage of the force Pg is applied, which, when augmented by the ‘skewness’ of the slab, mayeven be of the order of 50%.Note that in a skew plate the full take up of the load g by the cable deviation forces

resulting from the prestressing force Pg ensures an equal distribution of the supportreactions. The reason for this is that the deviation forces from the cables completelycancel out the load g of the plate, while the remaining equally distributed downwardresultants of the prestressing force acting directly on the supports (being totally equaland opposite to the deviation forces) cause uniformly distributed reactions.

388

Plan view d

L

L

Figure 11.29 Uptake of the loading by equally distributed prestressed cables


As in the case of orthogonal plates, there also exists for skew plates the possibility ofarranging the prestressing cables in the edge zones in a direction parallel to the freesides (see Section 11.2.1.2). In this way an even stronger restriction of the requiredprestressing force is achieved, so that, with an analogously increased percentage of steelreinforcement, the appropriate resistance check can be satisfied (Stavridis, 2001). Theprocedure for determining the cross-sectional area of the cables (i.e. the prestressingforce, by assuming the cable is initially stressed with 70% of its yield stress) follows preciselythe same course as that for the case of orthogonal plates (see Section 11.2.1.2). In this waya lower total cost may be achieved for the prestressed and steel reinforcement.

11.5 Flat slabs

11.5.1 State of stressThe load-bearing action of plates in transferring their loads in at least two directionsimplies that, normally, there should be immovable continuous supports at the ends ofthese directions to take up the loads transferred. These continuous supports areusually provided by beams of sufficient stiffness, which in turn are supported onvertical elements (columns) and, finally, transfer the loads to the bases of theseelements (foundations). It may also (more correctly) be considered that the platestransfer their loads to the horizontal girders of frames, which are formed by thesupporting beams and columns (Figure 11.30).It is intuitively known that a plate can alternatively be supported directly on columns

in a raster arrangement, without the use of beams, provided that the plate affords theappropriate rigidity. This arrangement permits the exploitation of the available netspace height over the entire area covered. As will be analysed next, such a plate,compared to a plate supported on beams, develops a higher bending response anddeformations on one side, as well as a particular stress state on the other, arising fromthe fact that the plate faces the danger of being punched by the columns. All theseconstitute, of course, disadvantages which are dealt with simply by increasing thethickness of the plate and by providing additional reinforcement, i.e. by takingmeasures that lead to a higher cost of construction, despite the simpler formwork

389

Figure 11.30 The transfer of the slab loads through the beams to the columns

Plates

required. In any case, the neater aesthetic result offered by this layout has led to arelatively wide application of these plates, particularly in cases with larger spans.In the orthogonal raster of columns considered (Figure 11.31), the natural tendency

of the plate to transfer its loads along the ‘most rigid’ (i.e. the shortest) path to itssupports appears clearly in the trajectories of the existing principal moments shown inFigure 11.32, where the regions between the columns are composed of ‘quasi-beams’on which the remaining internal ‘orthogonal panels’ are supported.The load-bearing action of such a system may first be considered as shown in

Figure 11.33, where each ‘horizontal’ row of columns constitutes, together with theimaginary strip of width Ly, a one-storey frame with equal spans Lx. In the same way,

390

The loads are transferred to the columnsthrough the more ‘stiff’ path

Figure 11.31 The transfer of the slab load to the columns without the presence of beams

Figure 11.32 Trajectories of the principal moments


the action of the plate in the y direction is taken into account. It is understood that theabsence of twisting moments along the length of the ‘free’ edges of this strip, due tosymmetry, allows it to be regarded as a free girder. It can be seen, of course, that eachspan of this frame receives the total load of q Lx Ly.The bending response of such a strip is substantially depicted by the form of the

moment diagram over the corresponding continuous beam with spans Lx and continuousuniform load q Ly (Figure 11.34). It is characterised by the corresponding span momentsMF (kNm) and support momentsMS (kNm). These moments, arising from the action ofeither the frame or the beam, are globally referred to the whole width of the strip and aredistributed between the ‘quasi-beam’ (i.e. an appropriate column strip) and theremainder of the strip. Thus the ‘column strip’, being more rigid than the remainderstrip, will receive the greater portion of the above moments (Franz and Schafer, 1988).

391

Bending action of the slab in the direction x–x: uptake of the total loadCorrespondingly also in the y–y direction: uptake of the total load

Lx Lx

One-storey frame

Lx

Lx Lx Lx

Ly

Ly

Ly Ly

Ly

Figure 11.33 Establishment of the action of the frame between the plate and the columns

Lx Lx Lx

Lx Lx Lx

MF

MF

MS

MS MS

Ly

Ly

Ly

Ly

0.40 · Ly

Ly

‘Quasi-beam’ strip

0.84 · mF

0.84 · mF

0.50 · mS

0.50 · mS

1.25 · mF 1.75 · mS

0.40

· L y

q · Ly · (Lx2)/8

The quasi-beam takes up 50% of the span moment MFand 70% of the support moment MS

mS = MS/ Ly mF = MF/ Ly

min (Lx, Ly) > 0.75 max (Lx, Ly)

Figure 11.34 Distribution of the bending moments of the strip

Plates

According to the German Standard DIN 1045, the width of the column strip may beconsidered equal to 40% of the width Ly, symmetrically arranged with respect to the axesof the column. Thus, with respect to the span momentsMF (kNm) for the column stripregion, instead of the ‘mean’ bending momentMF/Ly (kNm/m), 125% of this value maybe considered, while for the remainder strip width (0.60 Ly) 84% of the mean value isapplied (see Figure 11.34). This means that the moment MF is equally distributedbetween the quasi-beam and the remainder strip.Regarding the support moments MS (kNm), instead of the mean bending moment

MS/Ly (kNm/m), 175% of this value can be applied on the width of the quasi-beams,while in the remainder strip width (0.60 Ly) half of the above value is applied. Thus,the column strip region receives globally 70% of the moment MS, while the remainderstrip receives the other 30%.As the transfer of plate loads occurs also in the transverse y direction, the analogous

frame with span lengths Ly and strip width Lx should also be considered (Figure 11.35).This frame will again carry over each span the total load of the plate q Lx Ly, while themagnitudes of the momentsMF andMS will be distributed in the same manner as in theother direction.In this way, due to absence of continuous supports, the entire plate load is necessarily

taken into account in full, in both directions, in order to ensure equilibrium in each ofthe two directions. Thus the resulting bending response of the plate is clearly lessfavourable than the one developed when there is a continuous beam support. Ofcourse, each column practically receives the total load (q Lx Ly) of the correspondingcontributory area (see Figure 11.35).It should be clear that the above quantitative estimates concern the preliminary

design and are only valid if the existing aspect ratio of each panel is not less than0.75 (see Figure 11.34). It is obvious that for a complete determination of the state ofstress of a flat slab the use of appropriate computer software is required. g

392

Bending action of the slab in the sense y–yUptake of the total load

Tributary area ofcolumn load

Lx Lx Lx

Lx Lx

Ly

Ly

Ly

Ly

Figure 11.35 Establishment of the action of the frame in the other direction


As stated at the beginning of this discussion, a basic problem with flat slabs is the dangerof punching. The column loads are applied on the slab, practically as upward concen-trated loads, and, given their high value, particular attention is required for their safeintroduction into the slab. The region around a column presents a more intenseincrease in the shear forces than is the case in a continuous beam, as shown inFigure 11.36 for the example of a circular column.Consider a flat slab placed over a column raster with span lengths equal to L. Under a

uniform load equal to q the column load isN¼ q L2, and the shear force v (kN/m) of theplate at a radial distance x from the column centre will be

v ¼ N q p x2

2 p x ¼ q L2

2 p xx

2

!

The presence of x in the denominator indicates the particularly intense increase in theshear force, as the circular section shrinks to the outline of the column section.

11.5.2 Design for punching shearThe determination of reinforcement on the basis of the locally considered bendingmoments does not present any peculiarity in comparison to conventional slabs.However, as previously mentioned, designing for the bending moments only is notenough. The punching shear needs also to be taken into account, as examined below.As explained previously, the development of large shear forces, and hence shear

stresses in the column neighbourhood, induces skew tensile stresses that ultimatelycause a perimetric detachment of the slab from the remaining column region over asurface in the form of a cone (Figure 11.37). The failure does not occur at the perimeterof the column but at an angle of roughly 308 to the column face. The reason for this isthat in the immediate region where the the load is introduced the plate calculated shearstresses are hazardless, due to the high vertical compressive stresses, while at somedistance these relieving stresses practically vanish.

393

Strong increase in shear towards the column

v v(x) (kN/m)

N N

x

Figure 11.36 Variation in the shear force in the region of a column

Plates

For additional safety this checking is done at a distance of an angle of 458. In thisposition, the transferred ‘effective shear’ Veff from the slab to the body of the conethrough the corresponding failure inter-surface is obviously smaller than the load N ofthe column due to the presence of the uniform load applied over the column head. It is:

Veff¼N kwhere

k ¼ 1 cþ 2 hL

2" #

In this expression c is the dimension of the column and h is the mean static height of twosuccessive reinforcement layers of the slab in the region over the column head (seeFigure 11.37).Hence, the shear stress is directly determined as

¼ Veff

4 h ðcþ hÞThe ultimate value u of this stress is determined as a function of the ratio of the meanvalueAs (cm

2/m) of reinforcement in the two top layers to the ‘effective’ cross-section ofthe plate 100 h (cm2), the compression strength of concrete c and the yield stressof reinforcement fsy, on the basis of the equation (Herzog, 1996)

u ¼ c 1:6 ð fsy=cÞ

1þ 16 ð fsy=cÞ

The calculated ultimate force Nu of the column according to the above equations is

Nu¼ 4 u h (cþ h)/k

If Nu is greater than the design value N, then sufficient safety against punching isconsidered to exist. If Nu is smaller than the design value N, then vertical stirrups

394

Outline of action for Veff

h hc

h hc

h

hh

c

N

Failure surface

Effective shear: Veff = N · [1 – (c + 2h)2/L2)]

Figure 11.37 Region for checking the effective shear against punching


should be placed in the slab in order to deal with the ‘uncovered’ column force( NNu). The action of these stirrups, as shown in Figure 11.38, consists of offeringa relieving ‘suspension’ vertical force through the cracked conoidal surface, at an angle of308 and towards the loading force Veff, which is then ‘safely’ transferred as a com-pressive force to the remaining column head. Obviously the force (Nu k), representingthe transferable force from the slab through the lateral surface of the cone, should begreater than, or at most equal to, the loading force Veff, which is reduced by the stirrupforces, with a total cross-sectional area Aw that is cut by the failure surface, being underthe yield stress fsy. Thus

VeffAw fsy¼Nu k

and hence

Aw ¼ NNu

fsy k

It should be pointed out that the punching type of failure is a brittle one, i.e. it occurswithout previous warning, and its repercussions are particularly unpleasant anddangerous (much worse than, for example, failure of an individual beam or even apart of a plate). For this reason, when designing flat slabs one should avoid exhaustingthe ‘calculation limits’.

395

The relieving action of stirrup forces is basedon the fact that they are taken up

by the punching cone

VeffVeff

N

N

Acting shear: Veff – Aw · fsy

Layout of stirrups fordecreasing the acting shear

Aw · fsy Aw · fsy

Figure 11.38 Contribution of the stirrups to the uptake of the effective shear force

Plates

11.5.3 PrestressingPrestressing is particularly suitable for dealing with the bending response and deforma-tions of flat slabs, as well as for increasing their punching resistance.In order to counteract the bending influence of the load q on a flat slab over a column

raster arrangement, the initial idea consists of taking the force up through deviationforces ux,F and uy,F of the prestressed cables in each panel (Figure 11.39). It is reasonableto equally distribute this load to the two directions, i.e. ux,F¼ uy,F¼ 0.5 q, and thereforeto consider the corresponding sags f of the cables in both senses as practically equal. Thecorresponding prestressing forces per unit length will then be (see Section 4.3.1.1)

Px;F ¼0:5 q L2x;F

8 fand

Py;F ¼0:5 q L2y;F

8 fThe profile of each cable, transversally crossing a number of column strips, necessarilyfollows a path similar to that applied for a continuous beam (see Figure 11.39). Thispath passes towards the ends of each panel, through an inflection point, to anopposite curvature over the column strips, in both the x and y direction, and continuesin this way to the neighbouring panels (see Section 5.4.1).As explained in Section 5.4.1, the deviation forces of each cable between the column

axes constitute a self-equilibrating system, and thus each column strip is acted on by thedownward applied deviation loads: ux,S¼ ux,F (Lx,F/Lx,S) and uy,S¼ uy,F (Ly,F/Ly,S) in

396

Uptake of gravity loads q ux,S = ux,F · (Lx,F/Lx,S)uy,S = uy,F · (Ly,F/Ly,S)

ux,S

Lx,S

Lx,S

uy,S

Lx,F

Ly,F

Ly,S

Ly,S

uy,Fux,F

Because of these deviation forces,additional tendons are required

Prestressing should take up the total load in each direction: uneconomic solution

Figure 11.39 Taking up of loads through equally distributed prestressed cables


the x and y directions, respectively. These loads necessarily have to be taken up byprestressing, meaning that additional cables must be placed in the support strip, asshown in Figure 11.39. The distributed prestressing force of these cables per metrewidth in the corresponding column strip are obtained as

Px;S ¼uy;S L2x;F

8 fand

Py;S ¼ux;S L2y;F

8 fThus, the total prestressing force, for example in the x direction, is

Px;F Ly;F þ Px;s Ly;S ¼ðq Lx;F Ly;FÞ Lx;F

8 fThis clearly corresponds to the uptake of the total load q, not half of it, as was initiallyintended. Obviously, the same naturally also happens in the y direction and,consequently, the above cable arrangement does not represent an economic solution. g

A much more reasonable layout of tendons results by placing the cables in the columnstrips only, making sure that the load q is fully taken up (Figure 11.40). The total load isequally distributed to the deviation forces in the strips along the x and y senses, i.e.ux¼ 0.5 q Ly (kN/m) and uy¼ 0.5 q Lx (kN/m), respectively, leading to the totalprestressing force

Px ¼ux L2x;F8 f

397

Uptake of gravity loads q ux = 0.50 · q · Ly

uy = 0.50 · q · Lx

Ly

Lx

Ly,F

uy

uy

ux

ux

Lx,F

Figure 11.40 Taking up of loads through prestressing of the column strips

Plates

and

Py ¼uy L2y;F8 f

respectively, for each strip.In this way, the downwards acting deviation forces, being in equilibrium with those

in the panels, are loading the columns directly. Thus, on the one hand, they do notcreate additional stresses that should be dealt with, as in the previous case; and, onthe other hand, they relieve the punching stresses, as will be examined below. It isobvious that with this cable layout the bending moments in the panels are met usingsteel reinforcement.With regard to the column strips, it should be noted that the total developing bending

moment mP due to prestressing (i.e. due to the deviation forces) is separated into astatically determinate part m0 and a statically redundant part mSP (see Sections 5.4.1and 11.2.1.2). The bending moment mP should be determined using appropriatecomputer software. For the statically determinate part, m0¼ P e, where e is theeccentricity of the cable and the sign þ or depends on whether the cable is aboveor below the neutral axis, respectively. Thus, mSP¼mPm0. It should be remembered(see Section 5.4.2) that mSP must be added to the ‘loading side’ of the strength equation,whereas m0 is automatically taken into account in the determination of the bendingresistance of the plate section, whereby the always existing steel reinforcement mustalso be considered. g

Regarding now the punching shear, it is clear that, according to the above layout, thedeviation forces applied downwards are received directly by the columns, while thedeviation forces applied upwards reduce the effective shear (see Section 11.5.2)developed in the region of the punching cone (Figure 11.41). The deviation forcesare equivalent to the vertical components of the cable forces, and appear at theirintersection points with the punching cone lateral surface, and are equal toP

P sin. Angle is measured from the point where the cable intersects the failure

398

Relieving action of the deviation forces(ΣP · sin α)

ααPP

N

The effective shear is relieved

Figure 11.41 Relieving action of prestressing cables against the effective shear


surface and may be (unfavourably) considered equal to 458 (see Figure 11.41). Thus, thepunching check formulated in Section 11.5.2 may now be written, without the presenceof stirrups, as

VeffP

P sin<Nu kand finallyP

P sin> ( NNu) ka requirement that can easily be satisfied.

11.6 Folded platesAs pointed out at the beginning of this chapter, plates are not only able to carry loadstransversely to their plane through bending, but they can also offer much higher stiffnessagainst loads applied within their plane. Exploitation of this property in covering large(mainly orthogonal) areas is possible by connecting oblong plates along their longcommon sides, thus creating a folded form (Figure 11.42). This structural layoutobviously needs a much greater available height than does a single compact plate.Nevertheless, it achieves the covering of comparatively much larger spaces, with alimited plate thickness.A direct perception of the possibilities of a folded plate can be obtained if a sheet of

paper, supported on two opposite sides and deforming excessively because of its smallthickness, is folded as shown in Figure 11.43. Its deformation then becomes negligible,even if a small additional change is added.The load-bearing action of a folded plate is understandable from the typical layout

shown in Figure 11.42. It can be seen that this structure is composed of -shapedbeams, which, due to their large structural height and their much higher moment ofinertia compared to the plane plate, offer a considerably more rigid structure(Figure 11.44).The above remarks allow the conclusion that the folding edges may offer practically

immovable supports to the oblong plates that are formed by each plane surface. It is clear

399

Vertical diaphragms

Figure 11.42 Layout of a folded plate

Plates

that these plates act in bending as two-side supported slabs, as their length is muchgreater than their width. It should be noted, however, that the longitudinal edges are‘immovable’ only if frontal vertical diaphragms or appropriately formed frames withsignificant bending rigidity are supplied that are firmly connected to the ends of theplates. The structural action of a folded plate is examined below and is illustrated inFigure 11.45.First, imaginary supports are considered along each fold, these offering the same

vertical upward reactions as those of a corresponding continuous beam representingthe individual two-side supported plates. Obviously the bending response of the‘altered’ folded system will be identical to that of the continuous one, and thus maybe determined directly.Of course, the fact that the edges of a folded plate are actually free from external

forces implies that the above ‘reaction’ forces on each edge must be applied again, ina second phase, as loads of opposite sense, so that this state of stress may be superposedon the previous one (see Section 3.3.1). In this second phase, each edge load can beanalysed along the direction of the adjacent plates, and thus each plate is loadedwithin its plane by the resultant of the corresponding loads along its two longitudinalsides. These loads are taken up by the plate, which is acting now as a deep beam andtransfers the loads to the aforementioned frontal walls through the correspondingshear forces of its end sections. It is clear that the above treatment of the downwardedge loads refers to the taking up of these loads by the -shaped folded beams ofwhich the whole folded plate is composed.

400

Figure 11.43 Drastic improvement in the bearing capacity of a sheet of paper due to folding

Through folding a much higher stiffness is achievedby maintaining the same thickness

Figure 11.44 Drastic increase in the active moment of inertia due to folding


Thus a folded plate acts in a ‘combined’ way, i.e. as a combination of a system of rigiddeep beams in longitudinal bending together with the formed oblong slabs continuouslysupported in transverse bending (Figure 11.46).It should be pointed out that the above approach can be used only for preliminary

design purposes. For a more accurate determination of the response the use of asuitable finite-element program is necessary.As folded plates exploit advantageously the rigidity of their constituent slabs in their

own plane, then provided that the necessary space for the required folding height is

401

q

q

q · a q · a q · a

q · a q · a q · a

q · a

q · a q · a q · a q · a

q · a q · a

q · a q · a

L

a a a a

mspan ∼ q · a2/24Action of two-sided slabs

(unyielding supports)

Action of longitudinal beams

msupp ∼ q · a2/12

q · a/(2 · sin α) q · a/(2 · sin α)

Figure 11.45 Load-bearing action of a folded plate

Plates

available, they allow a wide variety of construction layouts, as surface loads can be takenup in vertical as well as in horizontal directions, as, for example, in retaining walls(Figure 11.47).

ReferencesFranz G., Schafer K. (1988) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: Springer-Verlag.Herzog M. (1996) Kurze baupraktische Festigkeitslehre. Dusseldorf: Werner Verlag.Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhauser Verlag.Nervi P.L. (1956) Structures. New York: McGraw-Hill.Stavridis L.T. (1993) Static and dynamic analysis of orthotropic rectangular plates [in German].

Stahlbau 62, 73—80.Stavridis L.T. (2001) Alternative layout for the prestressing of slab bridges. Journal of Bridge

Engineering ASCE 6(5), 324—332.

402

They have to be offered (and taken up)by the front diaphragms

(q · a/sin α) · L/2

(q · a/sin α) · L/2

q · a/sin α

L

Figure 11.46 Load-bearing action of a plate as a deep beam

Earth pressure

Figure 11.47 The possibility of taking up earth pressures


12

Shells

12.1 IntroductionCovering spaces of large dimensions which are about 15m long in both directions with aconcrete plate is inefficient in practice, and becomes impossible for larger dimensions.In the case of a beam, a single span (without involving prestress) has length limitationsthat are overcome by adopting a structure of the so-called funicular form (i.e. an arch).In exactly the same way, the flexural plate also has limited dimensions, which can bedealt with by adopting a monolithic type of thin structure, formed by an appropriatecurved surface of the same basic outline (Figure 12.1). This type of structure iscalled a ‘shell’.The contour of the shell boundary does not necessarily lie on a plane. It may be

represented by any properly shaped spatial curve, whose projection on the horizontalplane corresponds to the boundaries of the floor space which needs to be covered. Ofcourse, the shape of the shell is affected by its contour, and this shape, in turn, has amajor influence on its static behaviour. A typical shell may have a ratio of baselength to thickness of the order of 500.The morphological options for the problem of covering large spaces using shells are

essentially unlimited, and the aesthetic result is always of importance.

12.2 The membrane action as a basic design conceptA shell is always referred to by its middle surface, and for this reason it is necessaryto define some fundamental geometric attributes that play an important role in itsload-carrying behaviour.It is known that there is only one line passing from a point O of a surface that is

perpendicular to it. Two planes are considered now to pass from that line, havingarbitrary orientations and being perpendicular to each other. These two planes intersectthe tangent to the surface plane at point O along the straight axes x and y, and thesurface of the shell along the curves z¼ fx(x) and z¼ fy( y), where z is measured inthe normal direction, as shown in Figure 12.2.Each of these two curves has at the common point O a specific radius of curvature

and a curvature 1/Rx and 1/Ry, respectively. If the radii of curvature Rx and Ry arefound on the same side with respect to the surface, the surface is considered to be ofpositive (Gaussian) curvature, otherwise it is of negative (Gaussian) curvature.


Whether the surface has positive or negative curvature is very important for thestatic behaviour of the shell.At any point O of a surface, there is always a pair of mutually perpendicular planes

which produce curvatures corresponding, simultaneously, to the maximum andminimum values that they can achieve as the pair of planes rotates about the normalline at point O. These curvatures are called principal curvatures, as they are in thesame directions as the principal plane.Let a point O1 be considered on the curve fy at a small distance y from O, and

proceed with the intersection of the shell surface and a plane passing from point O1

that is parallel to the plane defined by the curve fx (see Figure 12.2). This intersectionis another curve, and if its tangent at point O1 forms an angle with the direction O1x,then the quantity /y, i.e. the rate of change of the angle , is called the ‘twist’ txy of

404

Figure 12.1 Hyperbolic paraboloid shell for covering a rectangular space

y

x

x

z

Surface twist = α/∆y = ∂2z/∂x ∂y

They belong to a plane parallel to Oxz

α

O

Perpendicular planes

Tangential plane at O

∆yRy

Rx

fy(y)fx(x)

O1

z = z(x, y)

Figure 12.2 Surface geometry


the surface at point O with respect to the coordinate system Oxy, and

txy ¼@2z

@x @ywhere z = z(x, y) is the function describing the surface of the shell.It becomes clear that the twist of the surface with respect to the directions of the

principal curvatures is always zero, since in that case the angle vanishes. g

It should be recalled here that in the case of an arch which carries the loads only throughcompressive internal forces, its shape is determined by considering the inverted shape ofa fictitious cable which is subjected to exactly the same loads. Exactly the sameprocedure may be applied in the case of a shell.More specifically, observation reveals that, if a membrane takes only tension and no

bending at all and is firmly supported along the contour of the shell, application to themembrane of the vertical loads which will be eventually received by the shell willgenerate a specific shape for the membrane (Figure 12.3). If this load is inverted andapplied to a shell having the same specific shape as the deformed membrane, theshell will be stressed exactly like the membrane, but with opposite sign, meaningsolely compression. Considering its mirror image with respect to the horizontal planeleads to the solution of the problem (Figure 12.4). This state of the shell is called themembrane state of stress, and plays a dominant role in design, since the describedapproach achieves uniform utilisation of the fibres of the section (see Section 2.2.7).Adoption of the specific directions x and y is necessary in order to consider the

membrane-carrying mechanism for a distributed load q acting along the z direction —the normal at point O of the surface — through the cut-off quadrilateral element ofthe shell, as shown in Figure 12.5.

405

Figure 12.3 Creation of a funicular membrane

Shells

The state of stress of the membrane consists of an axial membrane force N, which isnormal to the element side and tangential to the shell surface, and of a tangential shearforce S acting along the side. Both forces refer to a unit side length.Equilibrium of the quadrilateral element in the z direction reveals that the load p is

taken over by the membrane forces Nx, Ny and S, according to the following relation(see Figure 12.5):

p ¼ Nx

Rx

þNy

Ry

þ 2 S txy

The first two terms on the right-hand side of this equation are in direct correspondencewith the equilibrium equation of cables, as studied in Section 2.2.7 (see Figure 2.42).The third term refers to the vector resultant of the shear forces S along the z axis.It is clear that the shear force S does not contribute in carrying the load when the twist

of the surface at the point in question is zero, and, consequently, there is no reason for its

406

Figure 12.4 Shell formation by inverting the funicular membrane

Equilibrium:p = (Nx/Rx) + (Ny/Ry) + 2 · S · txy

Twist with respect to the system consideredinduces the development of shearing forces

Ny

Ny

Nx

Nx

Ry

SS

S Sy

z

x

p

Rx

Figure 12.5 Membrane equilibrium


development (S¼ 0). Obviously, this happens when the directions x and y correspond tothose of the principal curvatures. g

Although shells can exhibit unlimited morphologies, they can be placed into twocategories with respect to their basic geometric configuration: shells generated byrevolution and shells generated by translation (Figure 12.6). The first category refersto shells whose midsurface is generated by revolving a planar line, either straight orcurved, which is called the ‘generator’, about a constant axis. The shells of thesecond category are produced by sliding a planar line (straight or curved generator)on another constant line, referred to as the ‘directrix’, maintaining its plane alwaysparallel to itself. This distinction must be combined with that of positive or negativecurvature.For any distributed load p there is one and only one membrane state of stress (Nx, Ny

and S) which is in equilibrium with the load, and its determination can alwaysbe considered as a statically determinate problem. It should be pointed out that theuniqueness of the membrane solution is attributed to the existence of curvature. If,for example, a plate is subjected to in-plane loads, the developed membrane state

407

Generatrix(Positive curvature)

Generatrix(Negative curvature)

Translational shells(Straight-line generated surface: important constructional advantage)

(Positive curvature) (Negative curvature)

Shells of revolution

(Null curvature)

Generatrix GeneratrixGeneratrix

Axis ofrotation

Axis ofrotation

Axis ofrotation

Completely differentregarding structural action

Positive curvature Negative curvature

Figure 12.6 Different ways of generating shells

Shells

of stress cannot be determined simply through equilibrium considerations, becausethere are three unknown quantities, and there are only two available equilibriumequations for the plane case. This plane stress problem is addressed using the theoryof elasticity. g

The construction of a shell usually requires an arrangement of beams along itsboundaries, which should be designed to carry the opposite of the forces applied tothe shell at the boundary (Figure 12.7).The shell, by developing the membrane action under the applied loads and the forces

on its boundary required for equilibrium, exhibits some specific deformations. However,the same boundary forces of the shell are acting in the opposite direction on the beamsor, more generally, on the supporting structural elements, causing other deformationsthere. The question that arises is whether these two different types of deformationare compatible. If this is not the case, then the membrane state of stress for the shellis not possible, and re-establishment of the displacement compatibility requires one ofthe following two procedures.According to the first procedure, additional appropriate forces are applied to the shell

boundary and exactly opposite forces to the supporting structure, in order to make thefinal deformations equal. Application of such forces to the shell will cause bending,although bending is not acceptable for a shell whose natural state of stress is that of amembrane. Since the displacement compatibility is restored by introducing theseboundary forces, this means that the problem at hand turns into a ‘statically indeter-minate’ one, and the bending response will be dependent on the shell thickness. Ofcourse, the thickness does not affect the membrane action of the shell. In manycases, this bending of the shell occurs only in a small region within the neighbourhoodof the boundary.

408

Supporting beams

Restoration of the compatibility inducesthe development of local bending

Bending is avoided only by prestressing the beams

Boundary deformations

Are they compatible?

Beam deformations

Figure 12.7 The need for displacement compatibility at the boundary of the shell


In the second procedure, prestressing is introduced to the supporting structure, sothat its displacements are the same as those developed on the shell. This solution,whenever applicable, is statically preferable. It is also reasonable to apply prestress tothe shell instead of the supporting structure in order to achieve compatibility for thedisplacement between the shell boundary and the supports. g

The design of a shell, i.e. the selection of its shape and dimensions, can and must bebased on the membrane action, in order to utilise to the maximum its material.Nevertheless, in cases of marked incompatibility between the deformations of theshell boundary and the supporting system, the state of stress due to bending shouldalso be considered in the selection of dimensions, as was previously explained,although for shells with positive curvature this is confined within a small area close tothe boundary. However, a detailed computation of the stresses requires appropriatecomputer software.Obviously, given that shells covering large areas are subjected to high compressive

forces, particular effort should be made to avoid failure due to buckling, which maypossibly become the dominant factor in the determination of the thickness. Since thecritical compressive stress and the critical pressure applied to the shell are, in allcases, proportional to the modulus of elasticity E of the material, i.e. of the concrete,and given that there is always uncertainty about the implementation of the analyticalresults due to constructional defects, creep of concrete, etc., it is logical to adopt areduced value for E, in order to increase the desired safety factor.The basic characteristics of the load-carrying behaviour of the most common types of

shells are presented in the following sections.

12.3 Cylindrical shellsCylindrical shells of revolution are used either for storing liquids or granular materials(reservoirs or silos). Their longitudinal axis may either coincide with the verticaldirection, in which case they rest on a circular concrete base, or may be horizontal inorder to cover orthogonal areas, using only a part of their ring section.

12.3.1 Cylindrical shell under constant internal pressureThe perception of the load-carrying behaviour of a cylindrical shell which is supportedat its lower circular end and is loaded internally by constant pressure pr along its heightis of particular importance, as will also be seen later for other shells of revolution(Figure 12.8).The internal pressure pr, considered initially as a self-equilibrating system acting on

the free shell, produces the pure membrane tension N in the ring, while the straight-line generators, due to their infinite radius of curvature, do not develop any axialforces. All the foregoing may be directly concluded from the membrane equilibriumequation given in Section 12.2, which yields: N¼ pr R, where R represents theradius of the cylinder.

409

Shells

This tensile state of stress results in an angular elongation strain ", which depends onits thickness since "¼N/E d, and, consequently, its radial outward displacement wm

is equal to wm¼ " R. These last relations yield

pr ¼ wm E dR2

This expression reveals that the pressure pr is taken by the shell through membraneaction, just as in the case of an elastic spring base having stiffness k¼ (E d/R2) (seeSection 17.3.2.2). This stiffness is offered by the fictitious consecutive shell ringsthrough the whole height of the shell, and is proportional to their thickness d andinversely proportional to the square of the shell radius R. Obviously, wm representsthe deformation of the fictitious spring base (see Figure 12.8). g

It is now well understood that the fixed support around the circular bottom end of theshell is not compatible with the development of wm. As a consequence, bendingmoments and radial shear forces should act along the boundary, in order to make itsradial displacement vanish and to guarantee the zero slope of the generators at theclamped support (Figure 12.9).Otherwise, the shell may be considered to consist of vertical straight beams (having,

of course, a bending stiffness) and to be supported elastically by the horizontal shell ringsalong the shell height (Figure 12.9). These vertical beams may be considered as fixed orhinged at their base. In case they are not supported there, the constant pressure pr iscarried through the membrane action of the rings, much like springs with the afore-mentioned stiffness k. The fact that all along the vertical beams there is an imposedconstant displacement

wm ¼ ð pr R2Þ=ðE dÞ

410

wm wm

R

d

pr

pr

Nθ

Nθ

No vertical membrane forces are developedbesides the self-weight

The ring being the funicular structure for theinternal pressure develops only tensile forces Nθ

The ring behaves like a springand the shell as an elastic base

Radial displacements: circumferential elongationwm = εθ · R

pr = wm · (E · d/R2)

Nθ = pr · R

pr

Figure 12.8 Load-carrying behaviour of a free cylindrical shell under internal pressure


leads to the conclusion that these beams remain stress-free. If, however, their lowerend is supported, the developed transverse displacements will obviously change,resulting in the activation of their bending stiffness. Thus, the load pr is considered tobe carried by these beams as if these were resting on an elastic foundation (seeSection 17.3.3.1) with the modulus of the subgrade reaction k¼ (E d/R2) (seeFigure 12.9).The above consideration means that the load pr is carried by the membrane action of

the shell ring on the one hand and the bending resistance of the vertical beams on the

411

wm wm

Me MeBoundary forces

They restore the state of fixity at the lower boundary(Bending response)

Beam on an elastic foundation

EI(d4w/dx4) + (E · d/R2) · w = pr

prpr

wmMe

pr

pr

Mmax

k = E · d /R2

(E · d /R2)

The load is taken up by the cantilevers,which are elastically supported on the rings

(π/4) · c

(π/4) · c π

Figure 12.9 Load-carrying action of a cylindrical shell fixed at its base

Shells

other, both being expressed through the developed displacement w. So,

EI d4w

dx4þ E d

R2 w ¼ pr

In this way, the well-known equation of a beam of unit width resting on an elasticfoundation with the subgrade modulus k¼ (E d/R2) can be directly recognised (Bill-ington, 1965; see Section 17.3.3.1).If the lower shell boundary is fixed, as previously stated, the required uniformly

distributed radial forces and moments have to cancel not only the radial displacementwm due to membrane action but also the rotation of the generatrix at its base (seeFigure 12.9). It is clear that the bending moment resulting at the lower boundaryof the shell will be identical to the bending moment Me at the fixed end of thecorresponding beam on an elastic foundation, i.e. (Timoshenko, 1956)

Me ¼2

c2 EI pr

k

where

c ¼ffiffiffiffiffiffiffiffiffiffiffi4 EIk 1

4

r

Physically, the quantity c, which is called the ‘characteristic length’ (see Section 17.3.3.1),relates to the distance (c p=4) from the fixed end of the beam (equivalently from theend of the shell) over which the bending disturbance is extending (c ¼ 0:76

ffiffiffiffiffiffiffiffiffiR d

p),

while the rest of the beam remains essentially free of stress, due to the imposedconstant displacement wm (see Figure 12.9).It is interesting to show that the above expression for Me, after the necessary sub-

stitutions, yields the bending moment of a cantilever having a length c and carryingthe load pr, i.e.

Me ¼ pr c2=2 ¼ 0:29 pr R d

which, obviously, applies tension to the inner fibres of the cylinder (see Figure 12.9).It is useful to point out that the term pr/k in the initial expression forMe is equal to wm,

and thus may also be written as

Me ¼ 0:29 E d2 wm

R

In the case where the lower boundary of the shell is hinged, the maximum bendingmoment Mmax, which applies tension to the external fibres, occurs, by analogy to thebeam on an elastic foundation with one hinged end, at a distance c p=4 from thehinged end (Figure 12.9). This moment is given by (see Timoshenko, 1956)

Mmax ¼0:64

c2 EI pr

k

412


or

Mmax ¼ 0:092 E d2 wm

R¼ 0:092 pr R d

12.3.2 Cylindrical tanksThese are acted on by a permanent internal hydrostatic pressure pr perpendicular totheir surface and growing linearly with depth H, i.e. pr¼H , where is the specificweight of the liquid stored (Figure 12.10).The bending moment Me at the fixed boundary of the shell can be approximately

assessed according to the previous section, as the fixed-end moment of a cantileverhaving a length c and subjected to the uniform load (H c) , i.e.Me¼ (H c) c2/2

The maximum tensile force N occurs at a distance x¼ c /2 from the bottom, thusresulting in N¼ (H x) R (see Figure 12.10). g

413

pr

Nθ

Nθ

Nθ = pr · R(Increasing with the depth)

No vertical membrane forces are developed

The ring being the funicular structure for theinternal pressure develops only tensile forces

The load is taken up by the cantilevers,which are elastically supported on the rings

EI(d4w/dx4) + (E · d /R2) · w = pr

Boundary forces MeMe

wmwm

dR

H

pr

pr

Rotation of the generatrix

pr

Radial replacements: elongation of the peripherywm = εθ · R pr

Figure 12.10 Loading-carrying behaviour of a cylindrical liquid tank

Shells

Such a shell may also be loaded by the inward pressure of the possibly surrounding soil,while being empty inside (Figure 12.11). This pressure pE will correspond to the lateralearth pressure at rest, which also depends on the possibly acting live load p on the soilsurface (see Section 17.2.1). Thus,

pE¼ H 0.50þ p 0.50where is the specific weight of the soil andH is the depth, measured from the soil surface.It is obvious that the lateral pressure pE will produce a state of stress corresponding to

the previously considered hydrostatic pressure and constant internal pressure, acting inthe opposite sense. It is clear that in the last expression, the value 0.50 has been usedinstead of .However, the fact that the shell will be subjected to a ring compression raises the

problem of the risk of buckling. g

The critical external pressure pD of a closed cylindrical shell may be assessed from therelation (Pfluger, 1966)

pD ¼ 0:62 E RH d

R

2:50Any additional load acting along the longitudinal axis of the shell and producing acompressive stress x must be taken separately into account.The critical longitudinal compressive stress xD may be assessed as (Pfluger, 1963)

xD ¼ 0:48 E d=Rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ R

100 d

r

The coexistence of these two compressive states of stress can be regarded as secure, if thefollowing condition is satisfied (Flugge, 1960):

xxD

þ pEpD

1 g

414

p p

HpE pE

pE

Nθ

Nθ

Nθ = pE · R

The surrounding soil causes compressive ring forces

ρ · H · 0.50p · 0.50

Soil

Figure 12.11 Cylindrical shell under lateral soil pressure


The development of tensile ring stresses (N/d) in a cylindrical tank, may possibly beproblematic due to special requirements regarding leakage risk. Specifically, thesestresses should clearly be lower than the concrete tensile strength cz. Thus, a peripheralprestressing is often applied consisting either of circular tendons or of high-strength wirestightened along the perimeter over the whole height of the shell (Figure 12.12).If Z is the prestressing force of a cable circling the shell, referring to the unit length of

shell height and assuming its radius to be essentially equal to R, then the inward-actingdeviation forces uP¼ Z/R cause the compressive ring force NP¼ uP R, and, con-sequently, the developed compressive stress c in the shell equals (see Figure 12.12)

c¼NP/d¼ Z/d

The prestressing force, which entirely takes over the internal pressure pr (i.e. uP¼ pr),equals Zp¼ pr R. Prestressing with this force will lead to approximately zero stressesin the shell wall. However, in order to take additional tensile stresses into consideration,due to the influence of concrete shrinkage as well as of other secondary factors, aprecompression c of the order of 500—1000 kN/m2 needs to be provided through anadditional prestressing. The required additional prestressing force Zd for this purposeis, according to the above relation, equal to Zd¼c d. Thus, the applied prestressingforce Z in the circular cables — after losses — should be

Z¼ Zpþ Zd¼ pr Rþc dThe circular cables should be placed in the external region of the wall thickness, whichshould in principle not be thinner than 25 cm. Moreover, the distribution of prestressingforce Z over the height should in principle follow the corresponding distribution of thetensile forces N (see Figure 12.12).

12.3.3 SilosThese also have a circular section, and are used for storing granular material. Their ratioof height to diameter is considerably greater than that for liquid tanks, because theinternal pressure of the content in the case of tanks increases linearly with depth,

415

Prestressing cables

Prestressing cable

Z

Z

σc = NθP/d

NθP

pr

uPd

H

x ~ c · π/2

c = 0.76 · √R · d

Distribution of tensile forces Nθ

Figure 12.12 Peripheral prestressing of a cylindrical tank

Shells

whereas in silos containing granular material with an angle of internal friction ’ theinternal pressure increases linearly only up to the depth (Ciesielski et al., 1970)

z ¼ 2 R4 tanð0:6 ’Þ

and then remains constant to the bottom (Figure 12.13).The maximum internal pressure which can be developed is

ph ¼ z ¼ R2 tanð0:6 ’Þ

12.3.4 Barrel shellsA long, thin cylindrical shell with a closed circular cross-section placed horizontallyand supported only at its two extreme fronts in such a way that its profileremains undeformed (e.g. through an appropriate stiffener or a suitable circular wall)presents notable stiffness towards transverse loads, which allows it to bridge spansmultiple times longer than its diameter by developing exclusively membrane forces(Figure 12.14).Assuming that the shell is acted on only by its self-weight g and considering the

basic equation given in Section 12.2, which expresses how a load perpendicular to ashell surface can be taken over by its membrane forces, it can be deduced that, at thehighest and lowest points of the cross-section, the developed ring forces N — tensileor compressive — are N¼ g R. These ring forces vanish at the ends of the horizontaldiameter, given that the perpendicular component of g to the surface is zero there.Furthermore, it can be seen that the bending moment M0 over the span L, which isequal to

M0¼ (g 2Rp) L2/8

416

Angle of internal friction ϕ

ph

ph

z

(Depending only on the radius of the silo)

The pressure of the granular soilincreases only until a certain depth

R

Figure 12.13 Distribution of internal pressure in a silo containing granular material


can be provided through longitudinal membrane forces Nx, whose maximum value mayresult on the basis of an approximate value of moment of inertia, namely

I¼ d R3

Thus,

Nx¼ (M0 R/I) d¼ g L2/(4 R)This maximum value appears as compression at the upper edge of the profile and astension at the lower edge of the profile, while at the ends of the horizontal diameterNx becomes null (see Figure 12.14).It is now clear that longitudinal membrane shear forces are also developed along the

generatrices, arising from the longitudinal variation ofNx. Thus, given thatNx vanish atthe horizontal edges, by considering the horizontal equilibrium of the upper section cutout from, say, the right half of the closed shell, membrane shear forces Ns must be

417

Self-weight g

Compression

Tension

Tension

Compression

Nθ

Nθ

Nθ

Nθ

R

L

Thickness t

Ns

Ns Ns

Ns

Nx

Nx

NxNxNx

Nx

g · L2/(4 · R)(Maximum value)

Figure 12.14 Membrane load-carrying action of a closed beam-like cylindrical shell

Shells

developed in order to balance the compressive forces Nx acting over the cross-section atthe middle (see Figure 12.14). These shear forces obtain their maximum value at theedges of the shell, given that, acting along the periphery of the circular edges, theyrepresent the only possibility of equilibrating the total vertical load on the shell.According to Cauchy’s theorem, the longitudinal shearing forces at the edges havethe same value. This value is equal to Ns¼ 2 g (L/2). The above state of stress ofthe cylindrical shell is obviously a pure membrane state (see Figure 12.14). g

The form of a closed cylinder may not be suitable for covering an orthogonal area, unlesscut by a horizontal plane, and its resulting upper part is taken, again under the conditionthat its edges retain their cross-section, to be undeformable. A sheet of paper fashionedappropriately illustrates this directly (Figure 12.15)In order to now examine whether it is possible for a pure membrane state of stress to

be developed in such a structure, under the action of gravity loads, those membraneforces should at first be applied along its free edges which would be correspondinglydeveloped in a closed cylinder. If it is considered that the examined barrel constitutesa semicircle, given that the ring forces are at this location null, as previously stated,then the only forces which can be offered externally are the longitudinal shearingones Ns, which may be offered by a horizontal beam (Figure 12.16). However, such abeam, acted on by the opposite longitudinal shearing forces Ns, is tensioned, and itselongation is not compatible with the absence of strain at the longitudinal fibres ofthe shell, given that Nx¼ 0, as previously explained. It can be seen that for therestoration of compatibility, additional longitudinal shearing forces should be introducedat the free boundary of the shell — together, of course, with the corresponding oppositeforces on the beam — which will obviously cause tension in the lower regions of the shelland which, together with the unavoidable action of self-weight of the beam along thefree edge of the shell, will lead to a deviation from the pure membrane state, i.e. theywill cause (transverse) bending.If the intersection of the full cylinder with the horizontal level is made at a higher level

in order to maintain the membrane state, apart from the longitudinal shearing forces Ns

of the free boundary, the ring forces N should also be provided (see Figure 12.16).However, the gap in the longitudinal sense will now be greater, due to the presenceof the compressive force Nx at the boundary of the shell, and therefore the beamshould be able to provide, through its stiffness, the required inclined forces N. Thus,

418

As shown also in folded plates,a sheet of paper acquires stiffness through its form

Figure 12.15 Appearance of stiffness with the formation of a cylindrical cross-section


with the additional exertion on the free edge of the shell of the self-weight of the beam,the deviation from the membrane condition will be even more intense.Certainly, the shell and beam system may be designed by taking into account the

introduced bending state of stress for the restoration of the gap, as may be foundusing appropriate computer software. However, it may often be desirable to maintain,as far as possible, the membrane state of stress, and this can be done by prestressingthe edge beam.As shown in Figure 12.17, the compressive force of prestressing is able to eliminate the

gap between the two longitudinal edges (between the shell and the beam), while, on theother hand, both the ring forcesN and the self-weight of the beammay be taken over bythe cable deviation forces. g

The barrel shells can be divided into long and short ones, depending on the ratio of thelength L to the width b of their ground plan.

419

Self-weight g

Self-weight g

Causes additional bending to the shell

Causes additional bending to the shell

(Greater gap)

The restoring of the gapinduces bending

The restoring of the gapinduces bending

Edge shortening

Edge elongation

Edge elongation

NθNθ

Nθ

Ns

Ns

Ns

Ns

Ns

Zero elongation strain

Zero forces Nθ

Self-weight

Self-weight

Self-weight

Self-weight

Figure 12.16 Consequences of membrane action in a barrel shell

Shells

The long barrel shells have a ratio L/b greater than 2, and they may be designed at apreliminary stage as thin-walled beams, according to the ‘classical’ theory of beams. Onthe basis of the maximum bending moment M0, the compressive force Nx, occurring atthe top of the middle section, can be determined from the relation Nx¼M0 yo d/I(kN/m), whereas the tensile force Z, which will be taken over by a correspondingreinforcement, may be considered equal to Z¼M0/z, where z is the lever arm of theinternal longitudinal forces, estimated approximately as z¼ 0.90 h (Figure 12.18).The shearing forces Ns, which are applied to the end sections of the span and ensure

the equilibrium with the vertical loads, result from the vertical shear stresses (seeSection 2.2.1) fs¼ (V S)/(I b) as Ns¼ ( fs/cos) d. These forces show theirmaximum value at the centroidal axis of the section.Of course, apart from these forces, ring forces N are also developed, which have

their maximum value at the top of the arch. This value, for the preliminary designneeds, may be considered to be, according to the membrane state, N¼ g R (seeFigure 12.18).A possible additional support of the shell at an intermediate position implies a similar

treatment to the continuous beam.At this point, regarding the longitudinal bending response of the long barrel shell, an

alternative approach can be considered, based on the results of Section 12.3.1. Specifi-cally, by considering the vertical load on the barrel shell as playing, approximately, therole of constant internal pressure, but in the opposite sense to a closed cylindrical shell,direct use may be made of the results.By setting, approximately, g¼ pr, the following expressions for the bending moments

Me at a clamped-end support and the ‘span’ bending moment Mmax in the case of asimply supported shell (both per unit length) may be considered to be, respectively,

Me ¼ 0:29 g R d

420

The compressive prestressing forcecan eliminate the gap between the edges

The deviation forces of the tendonstake up the vertical loads of the beam

Self-weight

Self-weight

Self-weight g

Ns

Ns

Nθ

Nθ

Nθ

Figure 12.17 Prestressing the edge beam of a barrel shell


and

Mmax ¼ 0:092 g R dIt is clear that these values may be regarded only as approximate (see Section 12.3.1).As explained above, the deviation from the membrane state of stress also means, for

the shell, the development of transversal bending moments M causing tensioneventually at both the top and bottom fibres (see Figure 12.18). An approximateassessment of their values is given by the expression g (b/4)2/8.However, these transversal moments may be very limited if care is taken to preserve

the curved profile as undeformable, through the insertion of a number of transverse ribs,i.e. arches of significantly greater stiffness than that offered by the thickness of the cross-section of the shell only.North light shells should also be included in the long cylindrical shell category, as shown

in Figure 12.19, which, regarding lighting, show definite functional advantages comparedwith the full barrel shell, when they are used for covering industrial areas.These shells may be designed at a preliminary stage in the same way, namely as thin-

walled beams according to theory, taking into account the fact that their principal axesare skew. These beams develop a certain torsional response, as will be explained inChapter 13, but this is not of significance in their preliminary design. However, forthe validity of the above approach, it is important to ensure the undeformability of

421

Self-weight g

Reinforcement

Reinforcement

b

b

tα

As/2

Ns y0

h

Nθ

Mθ

b

b

b

bAs = M0/(0.90 · h · fs)

M0 = g · b · L2/8

L

As

Ns

Ns

Ns

Nx

Nx

Nθ

Figure 12.18 Long barrel shells as beams with an additional transversal response

Shells

the cross-section, e.g. with transversal ribs. Moreover, the ring forcesN are necessary inorder to maintain the equilibrium, and may be assessed as before.In short barrel shells having a ratio L/b less than 2, the distribution of longitudinal

membrane forces Nx may not be based on the theory of bending. Of course, the freeedges will develop tension, but tension may also occur in the top region. The ringforces N can be considered, as in the case of long shells, and the transversal bendingmoments may also be assessed here according to the last expression. However, for thefinal design of these shells, appropriate computer software should be used. g

The risk against buckling should in no case be neglected, given that this is the factorwhich, in preliminary design, determines the thickness of the shell. Here, particularattention must be paid to both longitudinal and ring compressive stresses.For the longitudinal (bending) compressive stresses, the following stress limit may be

considered as critical (Seide, 1981):

xD ¼ 0:58 E d=Rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ R

100 d

r

while for the transverse compression, the pressure pD can be considered to be the criticalloading, as established for the closed cylindrical shell in Section 12.3.2. Both criteriashould be examined. g

Cylindrical shells are particularly suitable for the aesthetically satisfactory covering ofsquare as well as triangular ground plans, by using a layout of intersected vaults.Figure 12.20 shows the covering of a square ground plan with four intersecting barrel

vaults, one based on each side. The arches created along the two diagonals receivethe practically constant ring forces N¼ g R from both their sides. The resulting ‘V’cross-section of arches gives them a clearly greater bending stiffness than that whichthe shell thickness alone can afford.

422

2

1

Self-weight g

g1 g2

g

Self-weight induces bendingabout both principal axes

Centre ofgravity

Development of torsion

Figure 12.19 Formation of north light shell beams for covering an orthogonal layout


The same logic is followed for covering a triangular ground plan, as depicted inFigure 12.21. The previous remarks are again valid.The preliminary design of these shells can be made on the basis of the above-

mentioned buckling criteria, as well as of the ring forces N according to the previouslyexposed membrane state. However, it bears repeating that, for ground plans of largedimensions, the buckling criterion is predominant.

12.4 Dome shells

12.4.1 Non-shallow shellsWith their middle surface being part of a spherical surface, dome shells are suitable forcovering not only circular areas but also various polygonal plans. The correspondingform is derived by cutting off the spherical surface with planes passing through everylinear segment of the periphery of the base (Figure 12.22). Of course, in the case of acircular base, the shell boundary will be a circle, but in all other cases it will consist ofconsecutive arch-like segments, representing the trace of the sphere on the previously

423

NθNθ

Figure 12.21 Intersecting barrel vaults for covering a triangular area

The arches take up the vertical components of Nθ

NθNθ

Figure 12.20 Intersecting barrel vaults for covering a square area (cross-vault)

Shells

mentioned planes. It is clear that these planes are not necessarily vertical, and a smalloutward slope leads to a more aesthetically appealing result.Spherical shells are particularly appropriate for carrying vertical distributed loads by

membrane action, something which is always aimed for via their appropriate design.However, the degree to which this may be accomplished depends on the support condi-tions at the boundary of the shell, as will be examined later.The rotational symmetry of a shell suggests the consideration of, for equilibrium

purposes, a surface element formed by two adjacent meridian and parallel curves(Figure 12.23). The membrane forces N’ and the membrane ring forces N act alongthe parallel and meridian curves and in a direction perpendicular to them. Moreover,if the vertical loading is also axisymmetric about the same axis as the shell, then it iseasily concluded that the membrane shearing forces along the edges of the element,either meridian or parallel, vanish.The membrane forcesN’ acting along a parallel curve can always be determined from

the equilibrium condition in the vertical direction of the typical cut-off element(Figure 12.24). If a represents the radius of the parallel circle, ’ is the angle betweena shell radius to the periphery and the vertical axis of symmetry, and V is equal to

424

Hexagonal planCircular plan

Figure 12.22 Design alternatives for covering areas with spherical shells

Vertical loading

Meridian curve

Parallel curve

Nθ

Nθ

Nϕ

Nϕ

Figure 12.23 Membrane forces in a spherical shell


the total load applied to the upper shell portion being examined, then

N’¼V/(2p a sin’)It is now assumed that the shell is subjected to a uniform load which is axisymmetricabout the vertical axis. Clearly, the value of N’ for an element at the top of the shellwhich is under uniform loading p — which may also represent the weight of the shellelement — cannot be determined from the last equation, but instead from themembrane equation of equilibrium in Section 12.2. So, application of this equation tothe four-sided shell element with its centre being located at the top of the shell yieldsthe relation 2 (N’/R)¼ p, since in the absence of twist it is S¼ 0. Equivalently,

N’¼ p (R/2)From the above two equations, it can easily be concluded that the last value of N’ isconstant over the entire shell, if a uniform constant load p is applied over the horizontalprojection of the shell (see Figure 12.24).

425

Nϕ Nϕ

Nϕ

Nϕ

Nϕ

Nϕ

Nϕ Nϕ

Nϕ Nϕ

ϕ ϕR

R

g

g

p

p

V

Ra

[Nϕ]

g · R/(1 + cos ϕ)

Nϕ = V/(2 · π · a · sin φ)

Nϕ = p · R/2

p · R/2

g · R/2

Top

Figure 12.24 Determination of the meridional membrane forces

Shells

It should be noted that, although the self-weight g of the shell is not uniformlydistributed over the horizontal projection of the shell, its value increasing towards theedges, it can nonetheless be considered uniform in the neighbourhood of the top ofthe shell, exactly as the load p. Therefore, the meridional forces N’ due to the self-weight g of the shell are expressed, according to the previous relation, as

N’ ¼ R g 1

1þ cos’

The ring force N forming a right angle with N’ is derived from the general membranerelation given in Section 12.2 by taking into account the corresponding principal curva-tures at the point under consideration. It should be noted that, whereas the meridionalcurve of radius R’¼R refers to a principal curvature at that point, the correspondingparallel circle of radius a, whose tangent at each point provides the line of action ofthe ring force N, does not represent the other principal curvature at the point underconsideration (see Figure 12.24). Instead, it corresponds to a circle of radius R¼ a/sin’¼R, which is a tangent to the parallel circle at that point, and N belongs toboth these planes as a common tangent to both circles (see Figure 12.23). Given thatS¼ 0, the following relation can be written:

N’

R’

þN

R

¼ pr

where pr is the component of the distributed load normal to the surface of the shell. So,the ring force N may be expressed as

N ¼ R pr N’

R

The developed ring force N, due only to the self-weight g of the shell, is evaluated fromthe expression (Billington, 1965)

N ¼ R g 1

1þ cos’ cos’

(positive values refer to tension)

In addition, for a uniformly distributed vertical load p,

N ¼ p R cosð2 ’Þ=2

The development of ring forces N can easily be understood by considering the equili-brium of a narrow ring cut off from the shell and subjected to the applied meridionalforces N’ along its upper and lower edges (Figure 12.25).It is well known that as long as the ring belongs to the upper shell region, the inward

horizontal component of N’ at the lower edge prevails over the outward one of theupper edge, resulting in a compression of the ring (see Section 2.2.7 and Figure 2.44).However, in the lower shell region, the outward component of N’ at the upper edgedominates, therefore leading to tensile ring forces N. The exact value of the angle ’for which the transition from positive to negative values of N takes place is derived

426


from the above expressions for N, as ’¼ 51.508 for the self-weight and ’¼ 458 for theuniform load p. g

In the case where the shell base is a horizontal circular plan, the meridional membraneforces N’ at its boundary also have to act on the element that supports the shell, but inthe opposite direction, i.e. outwards (Figure 12.26). This element may normally be acircular beam (ring) which, under the action of the horizontal components of N’, will

427

Nϕ

Nϕ

Nϕ Nϕ

Nϕ Nϕ

Nϕ Nϕ

ϕ ϕ

g: ϕ > 52° p: ϕ > 45°

g: ϕ < 52° p: ϕ < 45°

The ring is tensioned axially

The inward horizontal component prevailsThe ring is compressed axially

The outward horizontal component prevails

g

p

ϕ > 52°: Nθ tensileSelf-weight ϕ < 52°: Nθ compressive

ϕ > 45°: Nθ tensileUniform load ϕ < 45°: Nθ compressive

Figure 12.25 Behaviour of the ring membrane actions

The elongation of the ring and the shorteningof the boundary creates a gap

The boundary forces act like redundant quantities:they restore the gap

Shell

Ring

(∆H) (∆H)(∆ϕ) (∆ϕ)

H

HME

ME

Figure 12.26 Interaction between the shell and the ring beam

Shells

develop only tension without bending, since it represents the funicular form for such aradial uniform loading (see Section 2.2.7 and Figure 2.42). Consequently, its fibres tendto elongate, and do not conform to the deformation of the shell boundary as is implied byits ring forces N. If the ring beam lies over the limit of the approximately 528 angle, theshell boundary will shorten (N < 0), whereas under this limit it will elongate (N> 0).Hence, in both these cases there will be a gap in the deformation of the two systems,although in the latter the gap will be less. Even in the case of a hemispherical shell,where the horizontal component of N’ vanishes and the ring beam is not loadedwithin its plane, the existence of the tensile ring shell forces N leads again to a gapbetween the ring beam and the shell.Re-establishment of the deformation compatibility requires application of appropriate

additional forces on the shell boundary, which also act on the ring beam but in theopposite sense. These forces are suggested by those which the ring beam can sustainaccording to its stiffness, and these, obviously, are horizontally distributed radialforces H as well as radially distributed torsional loads ME (see Figure 12.26). Thismeans that the shell will be subjected to equal and opposite boundary loads, with thetorsional loads ME of the beam clearly representing bending moments for the shell.An analytical examination reveals that, due to the double curvature of the shell, thisdisturbance of the membrane state does not extend significantly into the body of theshell. Rather, it is restricted to its boundary region, and thus does not have an essentialinfluence on the preliminary design of the shell. Despite this, a thorough examination ofthe stress state and deformation of the boundary region is necessary in order to have abetter understanding of its response (Billington, 1965).

12.4.1.1 Shell boundaryGenerally, the shell boundary exhibits, under its loads, a radial horizontal displacementH and a rotation of angle ’ (see Figure 12.26).Due to its own weight g and under a membrane state of stress it is

H ¼ R2 gE d 1

1þ cos’ cos’

sin’

’ ¼ 2 R gE d sin’

It is well known that H is produced by the strain "¼N/E d of the horizontalboundary ring, which has the radius R sin’ and is subjected to the action of N.Consequently,

H¼ " (R sin’)

It should be noted that H is inwards when ’<51.508 and outwards when ’> 51.508.The development of ’ results in an increase of the end slope at the boundary ofthe shell.

428


According to the previous analysis, for a constant vertical load p uniformly distributedall over the shell,

H ¼ " R sin’ ¼ N

E d R sin ¼ R2 p2 E d cosð2 ’Þ sin’

Due to a uniform horizontal force H (kN/m) in the outward radial direction,

H ¼ 2 R2 sin2 ’E d c H

’ ¼ 2 R2 sin’E d c2

H

where c ¼ 0:76 ffiffiffiffiffiffiffiffiffiR d

p, according to Section 12.3.1. Here, H is directed outwards,

whereas the tangent slope at the edge is decreased due to ’.Due to a uniformly distributed bending moment ME applying tension to the inner

boundary fibres of the shell,

H ¼ 2 R2 sin’E d c2

ME

’ ¼ 4 R2

E d c3ME

whereH is directed outwards, whereas the tangent slope at the edge is decreased due to’.In the above expressions, it should be noted that the values of H and ’ are

reciprocally equal, according to the Betti—Maxwell theorem (see Section 2.3.5).

12.4.1.2 Ring beamIt is already known that a ring beam (section data: AR, IR) subjected to a distributedradial force H develops only an axial force NR¼H R and no bending at all, due toits funicular behaviour under that loading (see Section 2.2.7 and Figure 12.27).

429

NR = p · R NR

PLAN VIEW

No bending response

∆H

∆H = (R2/EA) · H

H H

R

Figure 12.27 Behaviour of a ring beam under internal pressure

Shells

The strain in the ring beam is "R¼NR/(E AR), the elongation of its circumference is"R 2R, and its radial displacement is

H ¼ "R R ¼ R2

E AR

H

On the other hand, the application of a torsional load ME uniformly distributed alongthe entire length of the periphery simply causes an equally directed rotation ’ of itssections,

’ ¼ R2

E IRME

and a constant bending momentM¼ME R, without creating any torsional moments, asshown in Figure 12.28. Under these conditions, the centroid of the beam section doesnot undergo a radial displacement H.In view of the above results, the evaluation of the magnitudes H and ME becomes

possible by imposing compatibility of the displacements at the shell boundary and thering beam, following a procedure identical to the force method, which was presentedin Chapter 3.However, it should be noted here that the displacements H have a much stronger

influence on the magnitudes H and ME than do the rotations ’. g

At this point it should be emphasised that the bending response of the shell edge isessentially identical to that of a cylindrical shell of the same thickness, whichcircumscribes the spherical shell and is tangential to it at their boundaries(Figure 12.29). The cylindrical shell in question obviously has the same radius as thespherical one, i.e. equal to R. This result, which can be confirmed through a moredetailed analysis, has clearly to do with the fact that the characteristic length cappears in the above expressions for the deformations of the spherical shell boundary.

430


Sections rotate in the direction of the loading

ME ME

R

No torsional response

M = ME · R M

PLAN VIEW

Figure 12.28 Behaviour of a ring beam under a distributed torsional load


This length was first introduced in the examination of the behaviour of cylindricalshells (see Section 12.3.1).In this way, and for ’> 208, the bending moment at the spherical shell boundary can

be approximately determined according to the results of Section 12.3.1, through theprojection (H/sin’) of the horizontal displacement H (due to its membrane state)on its radial direction. The value of the projection coincides with the radial displacementof the cylindrical shell that is tangentially equivalent.For a spherical shell with a fixed boundary, the moment is expressed as

ME ¼ 0:29 E d2 H

R sin’while for a hinged boundary it is

Mmax ¼ 0:092 E d2 H

R sin’H in the above equations takes the value which corresponds to the prevailing loading,which is either the self-weight or a uniform load.The resulting distributed thrust H (kN/m) acting on the shell boundary and directed

outwards can be determined directly from the corresponding equations of the previousdiscussion. g

As has already been pointed out, the resulting bending disturbance of the pure membranestate in a spherical shell is, in any case, restricted to a relatively narrow region by itsboundary, having a width of the order of magnitude of the characteristic length c.Prestressing of the ring beam can restore the incompatibility of the deformations.

This, additionally, solves the problem of taking on the tensile stresses in the case of aconcrete beam, and also restricts the dimensions of its section (Figure 12.30).

431

Estimated bending responseusing the cylindrical shell

lies on the safe side

Tangential cylindrical shell

d

d

∆H

∆H/sin ϕ

ϕ

RME

Figure 12.29 Identical bending behaviour at the boundary of a spherical and a cylindrical shell

Shells

By placing a circular tendon prestressed with a force P in a concrete ring beam, thering will be automatically subjected to an inward radial uniform pressure P/R. If thispressure is greater than the already existing pressure H, the ring will develop an axialcompressive force NR¼ (P/R) RH R¼ PH R, and consequently it will exhibitan inward radial displacement (see above):

H ¼ NR

E AR

R ¼ R2

E AR

P

RH

The prestressing force P can be selected in conjunction with the other structuralparameters, so that, according to the previous analysis, the resulting gap between thering beam and the shell boundary vanishes (see Figure 12.30).

12.4.2 Shallow shellsIn order to confine the covered volume, as well as for functional or aesthetic reasons, arestriction of the dome height is often necessary, and thus the shell is characterised as

432

Elongation of the ring and shorteningof the boundary create a gap

The ring receives an inward radial pressure (P/R – H)and the gap is eliminated

Deviation forceon the ring

P/R

R

P/R

H H

Prestressing the ring with force P

Nϕ Nϕ

NϕNϕ

PLAN VIEW

P/R

Figure 12.30 Prestressing of the ring beam


shallow. This implies a decrease in curvatures and, consequently, an increase of themembrane compression forces. In such a shell, where the height f is less than aboutone-fifth of the covered span a, the radius of curvature can be approximately expressedas

R ¼ f2 þ ða=2Þ2

2f

or simply R¼ (a/2)2/(2 f ), if f< a/10 (Figure 12.31).Such a shell can cover an orthogonal, usually square, area, where the resulting

boundary curves are sections of the spherical surface cut by vertical planes correspondingto the four sides of the shell base.Due to the self-weight g, the resulting meridional membrane forces N’ along any

‘parallel’ are, according to the relations given in Section 12.4.1,

N’ ¼ p a2 g2 p a sin’ ¼ g R

2

So, it can be concluded that this force remains constant essentially over the entireshell surface. On the basis of this result and according to the analysis in Section12.4.1, the ring force N, which is obviously compressive, is equal to the meridionalone: N¼ g (R/2).In the case of an orthogonal base, the above result may also be confirmed by

considering that the shallow shell consists of shallow arches which transfer their loadin both directions through compressive forces. These compressive forces N (per unitof width) are equal to

N ¼ g

2 a2

8 f ¼g R2

according to the previous approximate expression for the radius of curvature. g

433

a

a

f

f

d

N

N

N

N

Figure 12.31 Structural layout of a shallow shell

Shells

In such a shell, having essentially the form of a curved orthogonal plate, the bendingresponse should be taken into consideration even during the preliminary designphase, especially when its boundary is restrained from rotation. An estimation of thebending moments developed at the boundary can be made on the basis of an analogyto the ‘tangential’ cylindrical shell, as previously explained (see Section 12.2.1 andFigure 12.29). This bending moment is found to be

Me ¼ 0:29 p R d

and causes tension of the upper fibres. Note that, because of the shell shallowness, theself-weight can be considered essentially as a uniformly distributed load.This result lies, generally, on the safe side, i.e. the bending moment is over-

estimated. However, in order to estimate directly the influence on the bendingresponse of the ratios f/a and a/d using more accurate analytical methods (Stavridisand Armenakas, 1988), it is useful to provide Table 12.1, which gives shallowshells with square bases and fixed boundaries. The table values refer to the dimen-sionless magnitude Me=ð p d2Þ. In contrast to plates, the dependence of the bendingresponse on the shell thickness is clear here. Poisson’s ratio has been assumed tobe null.It should be noted that, in the case where the lateral transverse displacement of the

boundaries is not constrained — which means that no forces N’ can develop at theboundary — the values in Table 12.1 must be doubled. g

In the compressed dome, and particularly in the shallow dome, the danger of bucklingdeserves special attention. Of course, the existence of double curvature is a relievingfactor which increases the critical load. This can be estimated using the relation(Schmidt, 1961)

pD¼ 0.15 E (d/R)2

From this relation it is seen that an increase in the radius of curvature decreases theload-carrying capacity of the shell, whereas an increase in the shell thickness increasesthe critical load. One way to increase the critical load is to place transverse ribs alongthe mutually orthogonal directions, which increases the average shell thickness, as isoften seen in older structures. g

434

Table 12.1 Estimation of the bending moment (Me=p d2) at the fixed boundary of a shallow shell

a=d

f=d 100 150 200 250

0.2 4.20 5.20 5.90 6.500.1 12.80 16.70 19.90 22.800.05 33.50 43.80 53.20 62.00


As previously mentioned, a spherical dome can cover a triangular, square or polygonalarea. The sections of the shell surface on the corresponding planes passing through thesides of the polygonal base create arch-like openings with favourable lighting conditions(Figure 12.32).For the proper structural design of the shell boundary, it should be taken into account

that forces cannot be applied in a direction which is normal to the boundary andtangential to the shell, while shear forces should be applied along the boundary,taking up the whole vertical load on the shell, if equilibrium has to be maintainedthrough only a membrane state of stress.The fact that membrane forces N’ acting transversely to the arch-like boundary

cannot exist means that the vertical load in the boundary regions will be taken onlyby the membrane forces N, which is parallel to the arch-like edges (see Figure 12.32).If Rb represents the radius of curvature of the corresponding arch, then according tothe basic membrane relation given in Section 12.2

N¼ g Rb

Shear forces have to vanish along each axis of symmetry of the shell. So, in the arch-likeboundary segment with span length L and height f, they vary from zero at the crown to amaximum of S0 at its ends. Clearly, the integral sum of their vertical components over allthe boundary arches must equilibrate the total load g A of the shell, where A denotesthe area of the polygonal base. Assuming a linear distribution of the shear forces alongeach boundary arch, the approximate maximum value is directly obtained as

S0 ¼1

k 3 g A

4 f

435

Vertial load g

A

N N

S0

S0

S0

S0

S0

S0

H

H = S0 · L/4

Usually offered by a tie rod

Boundary arch

fZω

L

Figure 12.32 Load-carrying behaviour at the boundary of a shell with a polygonal base

Shells

where k is the total number of boundary arches (see Figure 12.32) (Salvadori and Levy,1967).These shearing forces can be directly applied to the shell by an arch of suitable section,

which may be adjusted to the shell boundary. Such an arch cannot transmit forces otherthan those lying in its own plane; therefore, it has to take on the shearing forces of theshell boundary along its axis and transfer them to the ends of the arch, which are also thesupport points of the shell. These forces are obviously directed towards the ends, andconsequently produce an outward horizontal thrust H. This thrust can be estimatedon the basis of the adopted linear distribution of shearing forces as H¼ S0L/4, andhas to be borne either by a fixed support or through a tie connecting the ends of eachboundary arch (see Figure 12.32).At each shell corner, the developed membrane shear forces S0 produce a tensile force

Z acting in a direction perpendicular to the angle and equal to

Z¼ S0/tan (!/2)

This tensile force has to be supported by appropriate reinforcement.

12.5 Hyperbolic paraboloid shells

12.5.1 OverviewThese shells belong to the translational shells category. Their midsurface is generated bya downwards curved parabola (generatrix), which ‘slides’ on another fixed parabolacurved upwards (guide), keeping its plane always parallel to itself. The created surfacehas the form of a ‘saddle’ and, obviously, has a negative curvature, in contrast to theshells examined previously (Figure 12.33).Such shells, apart from their impressive aesthetics resulting from the alternating

curvature, show a characteristic of particular constructional importance because oftheir negative Gaussian curvature, i.e. they may also be generated using straight lines.

436

ϕ1 · ϕ2 = constant ∆12

Generatrix

Generatrix

Directrix

Directrix

Plane of parallel reference Plane of parallel reference

∆12

ϕ1

ϕ2

Figure 12.33 Creation of hyperbolic paraboloid shells


Thus, if a straight line, regarded as a generatrix, moves in space with one of its edges,sliding over another fixed straight line, keeping itself parallel to a fixed vertical planeon the one hand and with a constant rate of variation of its slope over the travelleddistance by its sliding end on the other, then this line will produce the same surfaceas that previously described (see Figure 12.33). This will be explained in detail immedi-ately below. What will, however, be pointed out first is that an understanding of this‘dual’ nature of the hyperbolic paraboloid shell is of paramount importance for the com-prehension of its load-carrying potential, as well as for the design of these shells ingeneral, as they are able to cover orthogonal (and other) layouts with an unlimitedvariety of morphologies. In all cases, the possibility of creating the surface by usingexclusively straight-line segments is a crucial constructional advantage which givesthese shells a clear economic precedence.As an understanding of the geometry of this shell is necessary for understanding its

structural behaviour, it is appropriate first to examine it through the following analysis.

12.5.2 Perception of the geometryThe analytic description of the creation of the surface by the two parabolas is based onFigure 12.34.In the considered orthogonal system Oxyz the ‘generatrix’ parabola with a span and

rise equal to 2a and f1, respectively, has its plane parallel to the plane Oxz, while the‘guide’ parabola curved in the opposite direction lies in the plane Oyz, having a spanand rise (sag) equal to 2b and f2, respectively. The coordinate z of the created surfaceis determined from the relation

z ¼ x2

2R1

y2

2R2

where R1 and R2 are the radii of curvature of the two parabolas, assumed to be constants,and equal to R1¼ a2/2 f1 and R2¼ b2/2 f2, respectively. It is assumed that bothparabolas are ‘shallow’ in the sense described in Section 12.4.2.

437

Equation of surface: z = x2/2 · R1 – y2/2 · R2

2a

2b xx

z zyy

O O

ba

R2 f2

GeneratrixDirectrixR1

f1

Figure 12.34 Creation of a shell through a parabolic generatrix

Shells

It can be seen that the so-created surface covers a rectangular ground plan ofdimensions (2 a) (2 b). Every section of this surface with a vertical plane is aparabola, while a section with a horizontal plane produces a hyperbola.If a vertical plane is now considered whose trace on the horizontal planeOxy forms an

angle ! with the x axis such that tan2!¼R2/R1, then this plane intersects the shellalong a straight line. In addition, if the symmetric of this plane with respect to thex axis is considered, its intersection with the shell is also a straight line. In otherwords, the x axis bisects the angle 2 ! formed by the traces of the above two planesover the horizontal plane Oxy (Figure 12.35).If these two traces are considered as skew coordinate axes Ou and Ov, then it can be

proved that the coordinates z of the surface may also be obtained from the expression

z¼ k u vwhere k¼ 2/(R1þR2).It is clear that the intersection of the surface with a plane parallel to Ouz or parallel

to Ovz is a straight line. The same shell can be assumed to be created if a straight lineOa, initially coinciding with axis Ou, is transposed in space while remaining parallelto the plane Ouz, with its edge always lying on the axis Ov. During this linemovement, the ratio of ’, the change in the angle of its slope ’ relative to theplane Ouv, to the corresponding distance travelled by its edge O on the axis Ovshould be kept constant and equal to k (see Figure 12.35). It is clear that k representsthe constant twist of the surface relative to the coordinate system Ouv considered:

k ¼ @2z=@u @v g

The transition from one geometry to the other may also take place.

438

All vertical planes parallel to Ou and Ovintersect the curved surface along straight lines

a

a

a

ϕ2ϕ1

uTorsion: k = ∆ϕ/c

tan2 ω = R2/R1

2ω

c v

O

z

z

Surface equation: z = k · u · vTorsion of surface: k = 2/(R1 + R2)

The two created surfaces are identicalO

v

u

x

y

R1

R2

Surface equation: z = x2/2R1 – y2/2 · R2

ω

ω

Figure 12.35 Creation of a shell through a straight generatrix and the appearance of twist


Assume that the shell is intended to cover a parallelogram ground plan OACB withsides a, b enclosing an angle 2 !, while the skew axes Ou and Ov coincide with the twosides of the parallelogram (Figure12.36).Corresponding to the edge C with skew coordinates u¼ a, v¼ b, a point C0 is defined

with coordinate z¼ f and connected through a line segment to the two other edges Aand B. The hyperbolic paraboloid shell can be created if the segment OA movesparallel to the plane Ouz, inclined towards the lines OB and AC0. The coordinates zof the so-formed surface are z¼ k u v, where k¼ f/a b.It is clear that any vertical plane parallel either to the axis Ou or to Ov intersects this

surface along a straight line, thus making it possible for the surface to cover anyparallelogram ground plan.This same surface may now be depicted through a system of mutually orthogonal

parabolas with opposite curvatures in the following way (Figure 12.37).If the axis Ox is considered as the bisecting line of the angle 2 !, and the axis Oy is

drawn perpendicular to it, then each of the planes Oxz and Oyz intersects the surfacealong parabolas. The intersection with the plane Oxz shows an upward curvaturewith a radius

R1 ¼ 2 cos2 !=kwhile the intersection with the plane Oyz shows a downward curvature with a radius

R2 ¼ 2 sin2 !=kThus, the coordinates z of this same surface referred to the orthogonal systemOxy can beexpressed through the radii of curvature R1 and R2, as previously shown.From the above analysis, it is clear that a hyperbolic paraboloid shell with equal radii

of curvature R1¼R2¼R, being referred to the orthogonal axes Oxy, can also beconsidered as created from the straight generatrices representing its intersections with

439

u

z

b

v

CB

A

a

fC′

2ω

O

Equation of surface: z = k · u · vTorsion of surface: k = f/(a · b)

All vertical planes parallel to Ou and Ov intersect the curved surface along straight lines

Figure 12.36 Creation of a hyperbolic paraboloid surface over a skew ground plan

Shells

those vertical and mutually perpendicular planes Ouz and Ovz which bisect the rightangles of the axes Ox and Oy. The twist k of the surface corresponding to theequation z¼ k u v is, according to the above, equal to k¼ 1/R.In order to acquire a more direct understanding of the load-carrying behaviour, shells

with equal radii of curvature in both the x and y directions will now be considered.

12.5.3 Considerations of equilibriumIn the orthogonal layout shown in Figure 12.38, a2/2 f1¼ b2/2 f2¼R (see Section12.5.2). The shell, as can be seen from the figure, extends between the four parabolascorresponding to the sides of the orthogonal base.The twist txy of the surface with reference to the systemOxy is zero, as can be deduced

from the expression @2z=@x @y, which implies that the tangential shear forces S in thedirections x and y are also zero. The equilibrium of a curved quadrilateral element (seeFigure 12.38) shows that the load p may be carried by the compressive forces, as well asby the tensile forces Nx and Ny. This is clear from the fact that the load p causescompression of downward-curved parabolas, making them act as ‘arches’, while simulta-neously it applies tension to the upward-curved parabolas, making them behave as‘cables’. From the equilibrium equation in Section 12.2, it can be deduced that themembrane forces Nx and Ny have a common value equal to p R/2.However, in order for this membrane state in the shell to exist, the above axial

membrane forces should be appropriately introduced at the boundaries of the shell. If

440

u

z

b

v

CB

A

a

y

x

C′

O

All vertical planes parallel to Ox and Oy intersect the curved surface along straight lines

R2

R1

2ω

ωω

R2 = 2 · sin2 ω/kR1 = 2 · cos2 ω/k

Bisecting the angle 2ω

z = k · u · vk = f/(a · b)

The surface created from the two parabolasis identical with the initial one

Remark: In case of rhomboid layout (a = b) the axis Ox coincides with the line OC, but the two radii of curvature are not equal

Figure 12.37 Restoration of parabolic generatrices in a shell over a rhomboid ground plan


the shell is unyieldingly supported along its parabolic boundaries, then this is feasible.However, if the shell is supported on, for example, vertical parabolic arches, these arenot positioned to carry forces perpendicular to their plane, and thus the purelymembrane state of stress is impossible. The very fact, of course, that these arches areable to carry vertical loads means that vertical shear forces will necessarily be introducedin the shell, thus leading unavoidably to bending.Another support possibility is the existence of unyielding supports along the ends of

downward-curved (i.e. compressed) parabolic ‘arches’, while the edges of the upward-curved ones are free (Figure 12.39). Then, the load p has to be supported entirely bythe compressed parabolas, which will develop a compressive force Nx equal to p R,double the previous value (see Section 12.2).Of course, unyielding supports may also be considered only at the edges of

the tensioned parabolas, the sides of the other direction being left free. Then, theload p going through the tensioned parabolic trajectories with the axial forceNy¼ p R should be carried by prestressed tendons embedded in the concrete (seeSection 9.1). g

It is now of interest to examine whether the above-discussed case of a shell carryingits load p through compression paths (arches) can be realised without transferringtransversal loads to the edge-supporting arches, which has been previously shown as

441

Vertical load p

Cable action (p · R/2) Boundary arches

x

y

b

Arch action (p · R/2)

R

R

f2

f1

a

Arch action

Cable action

p · R/2

p · R/2

The membrane state with simultaneous arch and cable actionapplied on vertical boundary arches is problematic

Figure 12.38 Load-carrying action of a hyperbolic paraboloid shell with parabolic edges

Shells

problematic, while ensuring that a pure membrane state develops (see Franz andSchafer, 1988). This is achieved on the basis of the remark that the cuts of the shellwith vertical planes mutually perpendicular and with an angle of 458 to the axes xand y are straight lines, as mentioned previously (see Section 12.5.2).First, it is again assumed that the load p is carried by the axial compressive force

Nx¼ p R in one direction only, as in Figure 12.39, keeping the two other oppositeedges free from transverse loads. This state of stress will be superposed on anotherone, as explained in the following two cases.In the first case, where the ground plan is assumed to be square, a self-equilibrating

loading is further considered, as in Figure 12.40. This loading consists of a distributedload p R acting in the same sense perpendicularly to the opposite edges of thecompressed parabolas, being equilibrated by tangential shearing forces applied at thetwo other opposite sides.As is evident from Figure 12.40, the normal forces applied to the two boundaries may

be analysed in two perpendicular directions at 458 as tensile or compressive forces, whichare transferred directly to the neighbouring boundary where they are balanced by theshearing forces acting there. Thus, this additional loading is self-equilibrated withinthe shell itself, and therefore able to be superposed on the former state of stress. It isnow obvious that one boundary of the compressed parabolic paths will be free offorces, whereas on the opposite side an external compressive force 2 p R is requiredthrough an appropriate support (see Figure 12.40). It is also clear that the existingarches, at the other two opposite sides, are able to receive — and to transmit — thelongitudinal shearing forces without problem. It can also be seen that the requiredcompressive force Nx¼ p R is applied to the free edge, due to the simultaneoustensile forces inclined by 458 which meet on the boundary (see Figure 12.40), so thatthe overall compressive action of the parabolic shell arches is not disturbed.The second case, where one side in the ground plan is twice as long as the other, also

allows the development of a purely membrane state by considering the application of

442

Arch actionFree boundary

Free boundary

Unyielding boundary Unyielding boundary

Vertical load p

p · R

p · R

bb p · R

a

a

The vertical load is carried only through the arch action

Plan view

Arch action (p · R)

Figure 12.39 Feasible membrane equilibrium using arch action and free opposite boundaries


additional self-equilibrating loading tensile forces Nx¼ p R to the shorter sides of theshell (Figure 12.41).This load, which eliminates, of course, the initially obtained corresponding

compressive forces, may be considered as coexisting with the self-equilibratedshearing forces on each oblong side of the shell. The analysis of longitudinal tensileforces at both short sides in components at an angle of 458 makes the establishmentof the internal equilibrium possible. This is achieved due to the selected directionsbeing implemented in straight lines on the shell, which allows the direct transfer offorces between the oblong boundaries, where they are balanced by the shearing forcesacting there. These shearing forces can be imposed by the corresponding boundaryarches while the short sides remain free, as the tensile forces balancing there providethe compressive forces Nx¼ p R, leading to the compressive carrying action of thearches in the oblong sense (see Figure 12.41).

443

Release of one shell edge from externally applied normal forces

p · R p · R

p · R

p · R

p · R

p · R p · R

(Self-equilibrated loading)p · R

p · R

2 · p · R

a

Plan viewa

Thanks to the shell geometry compressive forces are introduced,although the edge remains free

They are offered bythe boundary arches

They are offered by the boundary arches

Tensio

n

0.70

7 · p

· R

0.70

7 · p

· R

Compr

essio

nTension

Compression

Compression

Tension

Tensio

n

Compr

essio

n

Figure 12.40 Feasible membrane equilibrium in a square layout with a free edge

Shells

12.5.4 Hypar shells with straight edgesTo cover orthogonal areas, hypar shells are usually formed with straight boundaries. Theshell is necessarily referred to a system of axes Oxy comprising two adjacent sides of theorthogonal base OABC (Figure 12.42). With the origin O kept fixed, the three otheredges can be moved vertically at any distance, as shown in Figure 12.42. The shell iscreated through the constant displacement of side OA0 in parallel to the plane Oxz,inclined towards the sides OB0 and A0C0, so as to result in position B0C0.

444

Release of two shell edges from the external action of normal forces

p · Rp · R p · R

p · R p · R

p · R p · R

p · R p · R

p · R p · R

p · R

p · R p · R

a

2 · a

(Self-equilibrated loading)

Plan view


Introduction of compressive forces,although the edge remains free


Tensio

n

Tensio

nCom

pres

sion

Tension

Tension

Compression

Tensio

n

Tensio

n

Compr

essio

n

Tension

Tension

Compression

Figure 12.41 Feasible membrane equilibrium in an oblong layout with two free edges

Edge beams

Torsion: k = (f3 – f1 – f2)/(a · b)

Stronger slopes

Vertical equilibrium: S = p/(2 · k)

S S

S S

(f3)

xz

a

p

yf2

f1

O

b

C′

CB′

B

A′

A

Direction y

Ny = 0

Directi

on x

N x = 0

Figure 12.42 Load-carrying shear mechanism of a hypar shell with straight edges


The twist k of the surface thus created is

k ¼ f3 f1 f2a b

The segments ‘f ’ are considered signed quantities with respect to the axis z. Accordingto Section 12.5.2 and given that 2 !¼ 908, both surface curvatures (i.e. ‘1/R’) are equalto k. These, understandably, appear along the skew directions at an angle 2 !/2¼ 458relative to the axis Ox.It should be noted that, in this case, the equation of the surface is not expressed by

z¼ k x y but by

z¼ k x yþ ( f1/a) xþ ( f2/b) y

which does not interfere at all with its geometry, as the twist of the surface according tothe expression @2z=@x @y remains equal to k.Considering now the equilibrium of an orthogonal shell element in the system Oxy

under the vertical load p, it can be seen that the only membrane mechanism throughwhich this load may be carried is the tangential shearing forces S acting along thesides of the element (see Figure 12.42). This happens because the shell elementpresents null curvatures about the two orthogonal axes (i.e. x and y), and, con-sequently, any developing internal forces Nx and Ny cannot offer a componentopposing the load, because they are collinear for any corresponding pair of oppositeelement sides. Thus, it follows that the above forces cannot be developed, so that thebasic equation for membrane equilibrium in Section 12.2 yields the followingconstant value for S:

S ¼ p

2k

Indeed, the difference in tilt at opposite sides of the element allows the forces S to offeran upward component, which equilibrates the vertical load p (see Figure 12.42).These forces Smust necessarily be applied at the straight edges of the shell too, which

is possible only if edge beams are arranged along each boundary.It should be noted here that the previous conclusion about zero membrane forces

Nx and Ny is strictly valid only if the load p acts perpendicularly to the shell surface.Since the acting load is vertical, it is clear that for small slopes of the surface theabove conclusion is essentially valid. ‘Small slopes’ is synonymous with the existenceof ‘small curvatures’, which in turn means that the shell should be ‘shallow’, and thisis the case if the edge beams have a slope not exceeding 188. For steeper slopes, thevertical load p has a non-negligible component in the direction of the straight genera-trices, leading to the development of axial forces in their direction and, consequently,the membrane shearing forces are no longer constant. However, for most cases ofroofing, the hypar shells used can be considered ‘shallow’.Of course, irrespective of the above impact of the equilibrium consideration for the

axes x and y, it should not be overlooked that, due to existing curvatures of 458 inboth oblique senses, axial compressive and tensile forces N that are equal in magnitude

445

Shells

are developed, which, according to Sections 12.5.2 and 12.5.3, are

N ¼ p R2

¼ p

2 ki.e. equal to the membrane shear forces (Figure 12.43).Thus, the shell always works as a group of compressed ‘arches’ and suspended ‘cables’

both developing the same axial force. These forces reaching the boundary may beconsidered as being applied to the vertical sides of unit length of the orthogonal tri-angular element, as shown in Figure 12.43, the hypotenuse of which is

ffiffiffi2

p. The

equilibrium of this element, given that the component of forces N perpendicular tothe boundary is zero, confirms the necessity for the membrane shearing force to actalong the boundary and to be equal to N. This is exactly the shearing force foundabove, running through the whole shell, which should, therefore, also be applied atthe boundary.The edge beams which transmit these shear forces to the shell are obviously subjected

to the opposite forces, and contribute their sum throughout the boundary length to eachsupport of the shell.As shown in Figure 12.44, the forces on the edge beams accumulate to their lowest

points, where they are taken up by appropriately designed columns that transfer thetotal load of the shell to them. The edge beams develop a progressively increasingcompression towards their low supports, and in all cases care must be taken to carrytheir self-weight, usually by providing a continuous support. g

The question arises, however, of whether the edge beams that will carry the aboveshearing forces can be designed free of intermediate support. In other words, is alayout of boundary beams which are supported (fixed) only at their correspondinglower ends and remain free all along their length, and, through the use of large canti-levers, provide an imposing aesthetic result, possible?

446

Compressive arch action (p · R/2)

Edge beams

f2

f1

S S

S S

O

x

p

a

Sb

y

p · R/2

p · R/2(1)

Equal radii of curvature R(f3)

Tensile cable action (p · R/2)

Plan Torsion: k = (f3 – f1 – f2)/(a · b) = 1/R

The compressive, tensile and shearing forces have the same value

S = p/(2 · k) = p · R/2

Direction y

Ny = 0

Directi

on x

N x = 0

(p · R/2) · √

2

Figure 12.43 Load-carrying mechanism through the arch and suspension action of a shell


The fact is that, although according to the membrane state these cantilevers do notcarry any part of the shell vertical loads but only longitudinal loads, they neverthelesshave a significant self-weight, which they are not able to carry as long they are actingas cantilevers. The question, therefore, which essentially arises is whether the shellcan participate in carrying the self-weight of the edge beams and, if so, with whatconsequences for its state of stress.The answer to this question is in principle positive, and is based on the fact that there

are straight generatrices between the edge beams which can act as cables. These ‘cables’can act against the tendency of the cantilevers to deform, by opposing tensile forcesalong the boundary, which relieve the cantilever from the response due to its self-weight. These forces are carried essentially unchanged from a given boundary edge tothe opposite one (Schlaich, 1970).As shown by the model in Figure 12.45 — which is provided with an optional support

at the high-point C to take into account a possible asymmetric loading — the system ofthe two adjacent ‘cantilevers’ OAB, which is subjected to the vertical loads of theirself-weight, is also acted on by the forces of the tensioned cables at every discretelevel. These cable forces Z counteract the tendency of the ‘cantilevers’ to tilt aboutthe axis AB, as they cause an opposite bending moment to M0 due to their self-weight, thus considerably relieving them. For simplicity, the structure is assumed tobe symmetrical about the vertical plane passing through the axis AB.An approximate estimation of the tensile cable forces can be made by assuming that

the cantilevers are elastically undeformable and behave like rigid bodies rotating freelyabout the hinge axis of their supportAB (Figure 12.46). It can be seen that the system ofcables actually consists of prestressed tendons anchored at the opposite edge beams.Thus, for each level i, the forces Zi are determined on the basis of the angle ! between

the cables considered in plan view, the difference zi of their ends, their projectionlength x on the vertical plane OC, the corresponding cable length Li, the distanceei of the corresponding resultant Ri of forces Zi at each node from the support axis AB

447

S S

S S

Shearing actions (per m)on the edge beams (equal)

The shearing actions are always summed up on the lowest corners

Figure 12.44 Forces acting on the edge beams of the straight boundaries

Shells

and, finally, the cross-sectional area Fi of the ‘cables’, which may at first represent thetributary shell strips between them.Thus (Schlaich, 1970),

Zi ¼ M0

Ki

2 cosð!=2Þ Xi

1

Ki eiwhere

Ki ¼Fi liL3i

ðx sin þzi cos Þ

and

x ¼ ðABÞ2 tanð!=2Þ

As mentioned previously, these tensile forces Zi should be taken up by prestressing(Figure 12.47). By introducing the appropriate prestressing forces along the straightgeneratrices, a purely membrane state in the shell may be feasible in practice, giventhat each compressive force applied at the shell boundary is not spread out but remainsin the ‘strip’ of the corresponding generatrix. In addition, the absence of curvature in thetendons essentially eliminates the friction losses. g

The synthesis of individual shells with a square ground plan in wider forms makes thecovering of larger areas with a small number of supports possible. With the appropriatejuxtaposition of shell parts, edge beams are needed only at the free boundaries inpractice, because at the ‘interior’ straight boundaries of each shell, the requiredshearing forces S for the equilibrium are provided mutually by the adjacent shells.

448

The cable forces relieve the cantilevers, and in this waystrongly decrease the bending action of the loads

Symmetry about the vertical plane AB

Carrying the weight of the boundary beam through the shell

Optional

B

C

A

O

i + 1

i + 1

i + 2

i + 2

i

i

Zi + 1

Zi + 1

Zi + 1

Zi + 1

Figure 12.45 Self-weight of the edge beams is taken up by ‘cable action’


449

The cable forces acting on part AOB

B

C

A

O

y

ii + 1

i + 2

i

i + 1

i + 2

i

i + 1

i + 2

Zi + 2Zi + 1∆zi

Ii

Gi

Gi + 1

Gi + 2

RiRi + 1

Ri + 2

Ziv = Ziu

Ziv

Ziu

A

OC

B

v

u

Ri

Zi

Zi

i + 2

i + 2

i + 2 i + 1

i + 1

i + 1

i

i + 2i + 1

i

i

iω/2ω

Rhomboid planRi = 2 · Zi · cos(ω/2)

∆x

The cable forces have to be equal in each directionbecause of symmetry about the vertical plane AB

The cable forces Z are determined from the conditionthat the moments of the R values about the axis AB

balance the moments of the G values

Figure 12.46 Assessment of the required tensile (prestressing) cable forces

The required tensile forces are provided through prestressing

O

A

C

B

Optional

Figure 12.47 Realisation of cable action through prestressing of a shell

Shells

The covering of a square ground plan using four individual shallow hypar shells is shownin Figure 12.48. Each shell develops a shearing state of stress in the sense of itsgeneratrices, which, as explained above, is caused by the simultaneous presence of theequal tensile and compressive forces of the ‘arch’ and ‘suspension’ action, respectively.It is concluded that the two imaginary cut strips along the top horizontal edges are

themselves in equilibrium under the oppositely acting shearing forces, as shown inFigure 12.48. Thus, the required shearing forces along the interior boundary of eachshell are provided by the adjacent shell. However, the required shearing forces at the‘external’ boundaries have to be provided by edge beams which, being acted on bythe opposite forces, have to give up their resultants to the corner columns. These resul-tants have a horizontal component acting along the diagonals of the ground plan,according to the arch action indicated in Figure 12.48 (see Section 8.1). The equilibriumof the column top necessitates the provision of a horizontal diagonal action in order toavoid an overstressing of both the columns and the edge beams, and this is accomplishedby providing diagonal ties that are prestressed accordingly. g

Despite the fact that the preliminary design of hypar shells may be based on theassumption that the loads are carried by membrane action, it should be noted thatvarious incompatibilities arising from this consideration may also affect the bendingaction of the shell. Thus, it is noted, among other things, that, for example, non-uniform loadings may cause bending in the shell, or even that the mechanismthrough which the self-weight of the straight edge beams is carried throughmembrane action may never be complete.As has been emphasised elsewhere, in order to determine the final state of stress of the

shell, the use of appropriate software is necessary, so that all details of its constructionalformation can be taken into account.

450

Edge beams

The horizontal acmes (beams) carry the self-equilibrating actions from the adjacent shells

Actions on the boundaries

Tie rod

Tie rod

Cable action

Arch action

Figure 12.48 Layout of four hypar shells to cover a square ground plan


12.5.5 Elastic stabilityThe presence of compressive forces in hypar shells constitutes a danger of bucklingwhich should be evaluated sufficiently during their preliminary design.In the case of shells with parabolic boundaries, it is clear that the presence of tensile

forces along the upward-curved parabolas clearly contributes to the stability of theembedded ‘arches’. Thus, the use, in this case, of the expression for the critical loadof a cylindrical shell having the same dimensions in plan as well as the same curvaturewill definitely lead to safe results (see Section 12.3.2).In the case of hypar shells with straight boundaries that are rigidly supported, having

ground plan dimensions a and b (a b), a height difference f and a thickness d, thecritical load pK may be assessed from the relation (Beles and Soare, 1972)

pK ¼ 0:40 E f2 d2

a2 b2If there are also edge beams having a moment of inertia equal to I, the critical load isgreater. This can be assessed from the relation

pK ¼ 2 E f2 d2

a2 b2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2

3 1þ 24 I

d3 a

s

12.6 Conoidal shellsAlong with the translational shells are also classified those shells which, in order to covera rectangular ground plan between two parabolic fronts over two opposite sides of therectangle, are formed through the segment of the other side as a generatrix by usingthe two parabolic edges as ‘guides’. This generatrix moves in parallel to a verticalplane, passing through the corresponding side (usually the oblong one), inclinedtowards the two parabolic guides, thus offering obvious advantages regarding theconstruction of the shell formwork (Figure 12.49). One of the two parabolic fronts isdeliberately made higher than the other, which could also simply be a straight. Thethus-formed conoidal shell can be supported on all four sides, i.e. at its parabolicedges through appropriately formed arches, and at the straight sides either throughfree-standing beams or rigid continuous supports.The preliminary design of such a shell can be based on considering it as an arch having a

height fm equal to the average height of the two front parabolas. If L is the span of theparabolic arch, the compressive force per unit metre of the shell due to the vertical load g is

Nd¼ g L2/(8 fm)In addition, the critical buckling load can be assessed on the basis of the average radiusof curvature R¼ L2/8 fm, from the corresponding relation for cylindrical shells (seeSection 12.3.2).In the case when the shell is not directly supported along its straight edges of length B,

it can be assumed that the above compressive ‘arch’ forces of the shell are carried by thedeep beams, which are formed along the edge region with an assumed structural heightof B/4 and act as simply supported beams (see Figure 12.49).

451

Shells

A not unusual use of conoidal shells is that of a shelter in the form of a cantilever,fixed at its parabolic edge and having all its three other straight edges free (Figure 12.50).Its structural function may be conceived as that of a cantilever of length B having a

parabolic cross-section, which presents an advantageous increase in its ‘moment ofinertia’ towards the fixed end, where the maximum tensile and compressive stressesare developed.

452

Boundary arch

Boundary arch

Tie rod

Tie rodL

B

N

N

NN

f2

f1

Average arch height: fm = (f1 + f2)/2 Beam action of the edge strip

Figure 12.49 Load-bearing action of conoidal shells

Thickness d

L

B

Compressive forces

Tensile forces

fNz

Mb

MbMtr

Mtr

Figure 12.50 Structural action of a conoidal shell as a cantilever


According to the above approach, the maximum tensile force Nz which results undera vertical load g is (see Kollar, 1984)

Nz ¼B2 g=20:265 f ¼ 1:90 B

2 gf

and is developed at the top of the supported parabolic edge.Beyond this ‘membrane’ load-carrying mechanism with regard to the cantilever

moment Me¼ g B2/2, a bending moment Mb of much less intensity also has to beconsidered, being assessed as Mb¼ (d/f )2 Me.Furthermore, a bending momentMtran in the transverse direction has to be taken into

account, which causes tension at the top fibres since it increases the transversalcurvature of the shell. Its value at the middle of span L may be assessed as

Mtran¼ 0.13 g L2

In the case when such shells are arranged monolithically connected in order to shelterlarger spaces (see Figure 12.50), the transversal moments are limited to about 25% of theabove value, with alternating signs across each field, causing a tension at the bottomfibres over the top of the connected ‘arches’.

ReferencesBeles A.A., Soare M.V. (1972) Berechnung von Schalentragwerken. Wiesbaden: Bauverlag.Billington D. (1965) Thin Shell Concrete Structures. New York: McGraw-Hill.Ciesielski R., Mitzel A., Stachurrski W., Suwalski J., Zmudzinski Z. (1970) Behalter, Bunker, Silos,

Schornsteine, Fernsehturme und Freileitungsmaste. Berlin: Wilhelm Ernst.Flugge W. (1960) Stresses in Shells. Berlin: Springer-Verlag.Franz G., Schafer K. (1988) Konstruktionslehre des Stahlbetons, Band II, A. Berlin: Springer-Verlag.Kollar L. (1984) Schalenkonstruktionen — Beton — Kalender 1984 — Teil II. Berlin: Wilhelm Ernst.Pfluger A. (1963) Zur praktischen Berechnung der axial gedruckten Kreiszylinderschale. Stahlbau 32,

161—164.Pfluger A. (1966) Zur praktischen Berechnung der axial gedruckten Kreiszylinderschale unter Mantel-

druck. Stahlbau 35, 249—252.Salvadori M. and Levy M. (1967) Structural design in Architecture. Englewood Cliffs, NJ: Prentice-Hall.Schlaich J. (1970) Zum Tragverhalten von Hyparschalen mit nicht unterstutzten Randtragern. Beton

u. Stahlbetonbau 3, 54—63.Schmidt H. (1961) Ergebnisse von Beulversuchen mit doppeltgekrummten Schalenmodellen aus

Aluminium. IASS Symposium Delft, pp. 159—181.Seide P. (1981) Stability of cylindrical reinforced concrete shells.American Concrete Institute ACI — SP

67, 43—62.Stavridis L.T., Armenakas A.E. (1988) Analysis of shallow shells with rectangular projection: analysis.

Journal of Engineering Mechanics ASCE, 114(6), 943—952.Timoshenko S. (1956) Strength of Materials, Part II. Princeton, NJ: Van Nostrand.

453

Shells

13

Thin-walled beams

13.1 General characteristicsThe concept of a thin-walled beam is applied in a wide variety of structural forms. Thesebeams are used mainly in bridges, but they find application in building construction aswell. The cross-section of a thin-walled beam is formed by using thin surface elements —mainly plane elements — that are monolithically connected to one other at theircommon edges. They thus form beam-type oblong members, resulting in a hugesaving in material compared with a corresponding solid cross-section of the sameconstructional height and, simultaneously, have better load-carrying properties. It isfor this reason that bridge superstructures, either suspended or not, are formed almostexclusively of thin-walled beams.As in beams having a solid cross-section, in thin-walled beams the dimension of

length L is clearly larger than the maximum transverse dimension B, with a length/width ratio (L/B) of more than 5, while the thickness t of the plane elements is thinenough that t/B 0.1 (Figure 13.1). Thus, the beam is formulated as a folded structureaccording to Section 11.6, and it can basically be analysed using suitable computersoftware. However, the main goal of the present chapter is the examination of thisstructural type as a linear beam element so that its load-bearing action can be appraisedand understood accordingly.The main difference between the thin-walled and the solid beam, and which causes

the biggest difficulties in analysing the thin-walled beam as a beam element, lies in theway the torsion is taken up. Not only shear but also longitudinal normal stressescontribute, a fact that is not foreseen in the so-called ‘classical technical theory’ ofsolid beams. The development of torsional stresses in thin-walled beams occurs bothin rectilinear bridges, due to the eccentricity of traffic loads with respect to the cross-sectional axis, and in curved bridges, where only the self-weight leads to a torsionalresponse.Thin-walled cross-sections are distinguished as ‘open’ and ‘closed’ sections, each

category having its own load-carrying characteristics. The closed cross-sections arealso called ‘box sections’ due to their closed form. g

The comparison between a closed and an open double-T section, as they are used inbridge design with respect to the evaluation of their structural behaviour, shows clearadvantages for the former, without excluding at all the effectiveness of the latter, in anumber of cases.


In a beam with a double-T section, the possibility of developing compressive forces inthe bottom region over the supports, either in a continuous system or in a cantilever(Figure 13.2), is certainly limited in comparison with a box section with the samedepth (see Figure 13.1), as the bottom plate of the box can participate fully in takingup the compressive force, something that the double-T section cannot do.This is automatically translated as a restriction of the bearable tensile force by the top

plate (in the case of concrete, through suitable reinforcement, prestressed or not — seeChapters 4 and 5), resulting in a limited corresponding bending resistance of the cross-section. g

As stated previously, in beams made of concrete, prestressing is necessary for spanslonger than 15—20m. Apart from the fact that the bending resistance of a closedsection is clearly greater, another ‘unfavourable’ characteristic of the open sectionappears in its use under service conditions, due to the combination of a smaller coredepth and a higher positioned centroidal axis, as explained below.Consider the midspan section of a simply supported prestressed beam. The

prestressing compressive force P that is initially applied to the cable trace is finallyshifted upwards to a distance a¼Mg/P, due to the self-weight bending moment Mg,as discussed in Section 4.3.1 (Figure 13.3). However, in an open section this positionis closer to the bottom core limit than is the case in a corresponding closed section.Hence the diagram of compressive stresses shows a greater difference betweenthe extreme fibres in the open cross-section than is the case in the correspondingclosed cross-section. Now, as already mentioned in Section 4.1.1, the curvature 1/r of

456

L/B > 5t/B < 0.1

Thin-walled sections

BB

L

Thickness t

Figure 13.1 The formation of thin-walled beams

Need for a bottom flange in order to take up the compressive forceDouble-T section: limited possibility

Figure 13.2 The need for a closed cross-section formation


the beam is expressed as |"o "u|/h, i.e. as ||/(E h). Thus the open section presentslarger deformations due to prestressing than does the closed one, and this differencebecomes two to three times larger under the influence of creep (see Section 1.2.2.1),which is, optically at least, annoying (see Figure 13.3). g

The last and perhaps most important difference between the two types of cross-section isrelated to torsion. A torsional moment MT applied to the open section is taken up,according to the classic technical theory, by the three oblong orthogonal partsthrough the development of shear stresses, the maximum values max of which occurat the external longitudinal sides (Figure 13.4), as explained in Section 2.5.3. Thusmax¼ (3 MT,i)/(B t2), where B is the length of each oblong side and MT,i is thatpart of MT which is carried by each part and offered naturally by the couple ofthe oppositely directed stresses . Note that the distribution of MT to the parts isproportional to their torsional inertia IT¼B t3/3 (see Section 3.2.10). However, dueto the small lever arm (roughly 2 t/3) of the above couple, the possibility of takingup a torsional moment through this mechanism is always limited (see Section 2.5.3).

457

h

h Centroidal axis

Centroidal axis

Core extent

ku

ko

kuMg/P

Mg/P

P Final position of compressive force

P Initial position of compressive force (g = 0)

P Initial position of compressive force (g = 0)

P Final position of compressive force

∆σ

∆σ

1/r = ∆σ/E · h

The larger stress difference causes larger permanent deformations

Figure 13.3 The influence of the type of section on the variation in stress due to prestressing acrossthe depth

Torsional moment is shared to the section wallsproportionally to their torsional rigidity

More favourable uptake of torsionthan in an open section

MT

MT

τ Tt t

τ

τ

v = τ · t = MT/(2 · Fk)

Smaller lever armslead to larger stresses

Larger lever armslead to smaller stresses

Higher torsional stiffness

Figure 13.4 The basic torsional behaviour of open and closed cross-sections

Thin-walled beams

Conversely, in a closed section, shear stresses of the same magnitude with a constantflow all over the profile have the ability to take up a much larger torsional moment,due to the relatively large lever arm they can offer. Thus, in each wall of the section,uniformly distributed shear stresses 1, 2, 3 and 4 are developed, each of which hasthe characteristic property of presenting a constant shear flow v¼ ( t), i.e. a constantshear force per unit length (see Figure 13.4). Thus v¼ 1 t1¼ 2 t2¼ 3 t3¼ 4 t4,while, according to the Bredt’s formula,

v ¼ MT

2 Fkwhere Fk is the area encompassed by the midline of the cross-section (see Section 2.5.3).Certainly, the torsional rigidity of a closed section is much greater than that of an

open section having the same cross-sectional area, as for constant thickness t, thetorsional inertia IT is equal to 4 t F2k=L, where L is the perimeter of the midline ofthe closed cross-section (see Section 2.5.3). This generally means that a closedsection has smaller torsional deformations than a corresponding open section. g

However, the possibilities of the open section to take up torsional moments are notexhausted by the classical shear stress mechanism mentioned previously, also knownas the St Venant mechanism. In addition to this mechanism there is another that isnot predicted by classical technical theory and which helps the St Venant mechanismin taking up the torsional moment. However, despite this, the ability of an opensection to carry torsional moments remains limited compared to the ability of theclosed section.In a last comparison of the two types of cross-section, the possibility of the partici-

pation of the two webs in receiving an eccentrically applied load will be examined,based on a gross estimate of the additional mechanism of torsional resistance mentionedabove for open sections, a detailed examination being reserved for later. Both types ofcross-section have a width b and a height h (Figure 13.5).Consider a concentrated loadQ applied at the midsection of a simply supported beam,

placed precisely on the left web of each section. The load is split into a symmetric and anantisymmetric load, as shown in Figure 13.5. The symmetric loading causes bending andshearing of the beam, whereas the antisymmetric load causes torsion.Examining first the beam with an open section located to the left of its midpoint, it

may be concluded that for the symmetric loading the shear force Q/2 acts throughthe downward forces Q/4 at each web, while the plate shear flows have the directionsshown in Figure 13.5 and vanish at the midpoint of the plate. In the antisymmetricloading, the torsional moment (Q b/4) is counterbalanced by a couple of opposingshear forces in the webs, which are equal to (Q b/4)/b¼Q/4, and the shear senses inthe plate that result are as shown in Figure 13.5. (At this point, it should be notedthat the directions of the shearing stresses generally correspond to those followed byan analogous hydraulic flow, regardless of whether they originate from shear ortorsion.) Of course, the presence of these shear forces implies bending of the left-hand web downwards and the right-hand web upwards, and clearly this opposite

458


bending of the two webs will cause such rotations at their ends that Bernoulli’s law ofplane sections is automatically ruled out. This is a rough description of the mobilisedbending mechanism of an open cross-section, which generally takes up a bigger shareof the torsional moment than does the St Venant mechanism. This leads to theconclusion that the left-hand web receives all the shear force Q/2, while the right-hand web remains unstressed (see Figure 13.5).The beam with a closed cross-section shows a better behaviour. Again, a cross-section

to the left of the beam midpoint is examined. In the symmetric loading the shear forceQ/2 again causes shear forces of Q/4 in the webs, and the shear stresses in the top andbottom slab result according to the above rule of ‘hydraulic flow’, as shown in Figure 13.5.In the antisymmetric loading, however, the torsional moment (Q b/4) causes aperipheral shear flow according to Bredt, which develops in each web a shear force(see Figure 13.5)

Q b4 ð2 b hÞ h ¼

Q

8

instead of Q/4, as in the open cross-section. Thus, the left-hand web receives a shearforce (3 Q/8), i.e. 75% of the shear force Q/2, while the right-hand web receives a

459

The right web remains practically uninvolved in uptaking the load

Participation also of the right web due to better torsional behaviour

Q

bh

Q/2

Q/2

Q/2

Q/2Q/2

Q · b/4

Q · b/4

Q · b/4Q/2

Q/2

Q · b/4Q/2

Q/4 Q/4

Q/4

Q/4

Q/4

Q/8

Q/8

Q/4

Figure 13.5 The uptake of an eccentric load by beams having an open and a closed cross-section

Thin-walled beams

shear forceQ/8, i.e. the remaining 25%. It is understandable that the superior behaviourof the closed cross-section in the distribution of an eccentrically applied load is owed to amore favourable uptake (with smaller shear forces) of the torsional moment (Q b/4)(see Figure 13.5).The basic principles governing the behaviour of thin-walled cross-sections will now be

examined in a more systematic way.

13.2 The basic assumption of a non-deformable cross-sectionA thin-walled beam with an open cross-section that is subjected to a torsional momentpresents a perceptible distortion in each cross-sectional plane (Figure 13.6). More speci-fically, the longitudinal fibres of the beam are deformed along their length so that nocross-section remains planar but undergoes warping. The reason for this will be explainedlater. However, the new outline of the cross-section, mapped over its initially singleplane, gives a form that is rotated with respect to its initial position, and does notpresent any deformation relative to its initial form (see Figure 13.6). This is the so-called assumption of a ‘non-deformable cross-section’. Despite the fact that thisassumption can be satisfied implicitly, the insertion of transverse diaphragms at certainreasonable distances over the length of the beam may be additionally required, which,given the negligible rigidity transversely to their plane, does not affect the deformabilityof the longitudinal fibres of the beam, i.e. its warping.The reason why a distortion of the cross-section occurs is that a torsional moment

applied to an open section is taken up not only by the local shear flow in each of itsconstituent thin-walled elements (as mentioned in Sections 2.5.3 and 13.1) but also,and mainly, by their bending within their plane, as shown for example in Figure 13.6.This involves, of course, a bending rotation of the sections of the elements withintheir plane, and the ‘synthesis’ of these bending rotations of the elements leads towarping of the cross-section. It is important to note that the presence of warpingshould not be confused with the requirement of non-deformability of the cross-section.For beams having a closed cross-section, the magnitude of warping is very limited. It

should be noted, however, that the assumption of the non-deformability of closed cross-sections that is adopted throughout this chapter may potentially be abandoned for

460

Plan view

Undeformable section

Deviation from the plane section (warping)

Figure 13.6 Warping and non-deformability of the cross-section


reasons of constructional practice reasons, with corresponding consequences on thestructural behaviour, as will be examined in Chapter 14.

13.3 Shear centreThe concept of a shear centre is of fundamental importance in describing the behaviourof thin-walled cross-sections in torsion. In the example shown in Figure 13.7, the thin-walled beam with a channel section is fixed at its left-hand end and is subjected to avertical distributed load, positioned over the web. At the fixed end, besides thebending moment, this load causes a shear flow over the cross-section of the beam.In this case, the determination of the senses of shear forces in the flanges requires

more than the analogy with a hydraulic flow (see above). To establish the longitudinal

461

No torsion developeda = MT/V

VV

VV

Twist

B

B

A

B

AA

L

S

p

p

p

Development of torsion

MT

Shear centreS

p · a

p

MT = p · a · L

The torsion modifies the wall shearing stresses

a

Figure 13.7 The shear centre in a thin-walled beam

Thin-walled beams

equilibrium of a cut-out part of the top and bottom flange that is acted on by differentlongitudinal normal forces requires a shear force on its internal face. The sense of theshear stresses on the plane of the cross-section is obtained, based on the above,according to Cauchy’s theorem (see Section 4.1.1).If, now, the beam is considered as a free body under the distributed uniform load, the

shear flow forces and the normal stresses on the fixed-end section, then the equilibriumis not satisfied because the shear forces of the two flanges produce an unbalancedmoment. It is clear that to establish the equilibrium a torsional moment must beapplied from the fixed end of the beam, with a vector directed towards the fixed end(see Figure 13.7). The twisting angle of each cross-section will be denoted by a vectorwith opposite sign.In order to exclude the above torsion, the plane of the distributed load must be

shifted to a suitable distance a on the outside, thus passing from the point S as itappears in Figure 13.7. With this loading arrangement, the three applied shearforces on the walls at the fixed cross-section will be in equilibrium with the loads,both in the vertical direction as well as with the developing moments with respect topoint S, so that no torsional moment need be imposed by the fixed end to maintainequilibrium.Thus, it is obvious that the beam does not twist. The point S is called the shear centre

of the cross-section, and it is the point to which torsion is always referred (seeFigure 13.7). Any load that passes through the shear centre does not cause torsionbut only a bending deformation, whereas the action of each load, acting at a distancee from it, is equivalent to that of the same load shifted to the shear centre plus a torsionalmoment. This moment, when the load p is a distributed one, constitutes a distributedtorsional moment mD¼ p e along the beam axis (see Figure 13.7), whereas when theforce P is concentrated it constitutes a concentrated torsional moment P e. It isobvious that the development of a torsional reaction on the beam will cause anadditional shear flow in the walls of the cross-section.Clearly, if a thin-walled section has one axis of symmetry, then its shear centre must

lie on this axis. It should also be pointed out that the shear centre is the point aboutwhich the cross-section rotates. In other words, it is the point of the cross-sectionthat under pure torsion remains immovable.The above means of ‘qualitatively’ searching for the shear centre (i.e. through the

development of shear flows) can, in principle, be applied to any cross-section. Ofcourse, the analytical determination of the shear centre (Zbirohowski-Koscia, 1967) iscumbersome and must be performed using suitable computer software. Softwareprograms of similar nature may also be employed to determine the shear flow distributionin the walls of a closed cross-section.

13.4 Warping of thin-walled beams and the stress state due to itsprevention

The uptake of a torsional moment by a thin-walled beam causes warping of itscross-section, as mentioned in Section 13.2. However, the different mechanisms of

462


taking up a torsional moment in open and closed sections involve some characteristicdifferences in the behaviour of thin-walled beams, depending on their type, as shownbelow.

13.4.1 Open sectionsThe beam shown in Figure 13.8 having an I cross-section is examined. This is consideredin free body equilibrium under two equal and opposite torsional moments, as shownbelow.It is understandable that the constant torsional moment at each cross-section of the

beam is taken up by the total of the shear stresses developed in each wall of the cross-section. These shear stresses vanish, of course, at the middle of each wall thickness, sothat the midplane of each wall is free from shear stresses (see Section 2.5.3).Under the influence of the torsional moments, each cross-section of the beam,

including the extreme ones, will rotate about its centre of symmetry (shear centre),which means that the section flanges will move in opposite directions. As shown inFigure 13.8, the assumption that the extreme cross-sections will retain their initialplane leads to a certain shear deformation of the beam flanges, which naturally is not

463

Acute angle

Acute angle

Right angle

Right angle

Right angle

Middle plane of section

Retaining of the plane section implies changingof the right angle, but this is impossible

The section develops exclusively shearing stresses

Retaining of the right angle impliesdeviation from the plane section, i.e. warping

Figure 13.8 Compulsory warping of a free beam under torsion

Thin-walled beams

possible, given the stressless state of the midplane of the section. Hence it is concludedthat the occurrence of the above shear stresses in the section walls for the uptake oftorsion involves the warping of the cross-section. It is clear that no other stress needsto develop in the element examined. g

The above stress state changes drastically if one end of the beam, say the left-hand one,is fixed while at the free end the previous torsional loading is maintained. It is under-stood that at the fixed end the cross-section cannot warp, and obviously remainsplanar (Figure 13.9).For the stress state prevailing at the extreme free cross-section, all that has been

mentioned previously is valid. The applied torsional moment MT causes a rotatingshear flow in each wall. The total developed torsional contribution of the walls accordingto St Venant isMT,S. Obviously, at the extreme cross-sectionMT¼MT,S, but this is notthe case for any other point in the beam.Because of the developing rotation of the cross-sections by , which naturally

varies along the length of the beam, the two section flanges are bent within their

464

Plan view

Top flange Bottom flangeB

BB

A

AA

MT

MT

MT

τω

τω

τω

τω

σω

σω

τS

Self-equilibrating σω

Offering of torsional moment MT,S(St Venant)

Offering of torsional moment MT,ω(warping prevention)

Figure 13.9 The development of the longitudinal stress state due to restrained warping


plane in opposite directions. For the particular sense of MT the top flange is bent tothe left and the bottom flange to the right. This bending of the flanges, i.e. the devel-oping curvature, is naturally accompanied by longitudinal stresses ! that vary alongthe length of beam, as the curvature also varies, being maximum at the fixed end andzero at the free end, where obviously normal stresses cannot be developed. It is clearthat this variation in longitudinal normal stresses ! leads, for equilibrium reasons, tothe development of shear stresses ! in both the top and the bottom flange (seeFigure 13.9).Thus, at each position of the top or the bottom flange, a bending moment is developed

corresponding to a linearly varying diagram of longitudinal normal stresses, as well as atransverse shear force in each flange for each position. It can be seen from Figure 13.9that, while the developing longitudinal normal stresses ! in the two flanges are self-equilibrating, i.e. they do not create an axial force in the cross-section, the shearforces in the two flanges offer a torsional moment MT,!. This moment, together withthe moment MT,S, constitutes the imposed torsional moment MT:

MT¼MT,SþMT,!

It is obvious that the contribution of MT,S to the total moment MT of the cross-sectiondecreases as one moves away from the free end, where of course it constitutes 100%.It becomes clear that both the stresses ! and the developing ! along the length of

the beam, which finally give rise to the contribution of MT,! in taking up the torsionalmomentMT, owe their existence to the prevention of the cross-sectional warping at thefixed end. In general, any prevention of warping of the cross-section in places wheretorsion develops causes the above stresses. It is clear that the higher this distortion,the bigger the stresses, and vice versa. A detailed examination of warping is, therefore,absolutely necessary, because the consequences of its prevention constitute the onlyreason why the technical theory of torsion (St Venant) needs to be revised. As amatter of fact, the latter predicts the sectional distortion, but not the stresses ! and! that appear due to its prevention.

13.4.2 Closed sectionsBefore examining warping in detail, it is important to point out that the above contri-bution of shear stresses ! to the total torsional moment MT applies only to opensections. The reason for this is that in closed sections the developing ! constitute aself-equilibrating system of stresses (in the same manner as the stresses !) and, con-sequently, do not contribute to the uptake of the torsional moment of the correspondingsection. Thus, in closed sections the developed shear flow, according to Bredt, is actuallythe only mechanism of taking up the applied torsional moment.This is now shown (Figure 13.10) for the same cantilever beam examined previously.

The applied torsional moment is considered to be introduced through a transversediaphragm that is assumed to be very stiff in its own plane. However, the diaphragmraises no resistance to transverse displacements, such as those that accompany thewarping of section. The diaphragm transmits (according to Bredt) a peripheral shear

465

Thin-walled beams

flow to the walls of the closed section (see Section 13.1). Thus the cross-section rotatesabout the shear centre due to the applied torsional moment by an angle that reduces tozero at the fixed support.To elucidate the deformation of the beam, the top plate is first considered cut out

from the webs, with the midpoint M of its left side simply pinned against thehorizontal displacement (see Figure 13.10). The plate is in equilibrium under theshear actions (according to Bredt) at both ends and the shear forces (according toCauchy) on its longitudinal sides. Under the above forces the plate is deformed inshear and the two end sides acquire a slope with respect to their initial (undeformed)

466

D

B

B

A

B

A

B

A

A B

A

C

M

M

σω

τω

τω

τω

τω

τωτω

τω

τω

τω

σω

σω

σω

σω

MTMTτS

τS

(Bredt)

Due to additionally imposedrotation of section

Lower warping stresses(self-equilibrating)

Due to Bredt

Reduced warping

(Compression)

(Tension)

The torsional moment is taken uponly through the Bredt stresses

Self-equilibrating τω

Figure 13.10 Restrained warping and the longitudinal stress state in a closed cross-section


positions, thus causing a deviation from the cross-sectional plane and, therefore,warping. Certainly, this deviation (warping) is reduced by the imposed movement ofthe right-hand side of plate ‘downwards’, with point M being fixed, due to the anti-clockwise rotation of the cross-section of the loaded end (see Figure 13.10). Thus it isclear that the developed warping in a closed cross-section is definitely smaller thanthat in an open one.It is clear that the existence of fixity does not allow the left-hand side to develop this

slope (i.e. to warp), and therefore in order for the plane section to maintain planar,longitudinal stresses ! must develop (see Figure 13.10) over the entire perimeter ofthe section, and the same applies for each individual wall of the cross-section. It isthus clear that each wall is subjected to bending. The developing longitudinal stressesare smaller than those of an analogous open section, because the warping of theclosed section is reduced. Here also the stresses ! constitute a self-equilibratingsystem, and the fact that they obviously vanish at the free end means that they keepreducing all along the length of the beam, thus giving rise to the development ofshear stresses !.Indeed, by cutting the top and bottom plates, as well as the two webs at their mid-

width along some length of the beam, the need to develop longitudinal shear stressesis apparent from the longitudinal equilibrium of the free part, which in turn (accordingto Cauchy’s theorem) produces corresponding shear stresses ! on the cross-sectionitself. However, as can be seen from Figure 13.10, the total of ! on the cross-sectionis self-equilibrating, and thus it does not contribute to the uptake of the torsionalmoment of the corresponding section, although it modifies to a small degree theexisting shear stresses according to Bredt. Hence, in closed sections,

MT¼MT,S

As to the length over which the stresses ! and ! extend, this is of the order of thecross-sectional width, given the self-equilibrating character of ! and the non-deformablity of the cross-section, according to the corresponding conclusion of elasticitytheory (the St Venant theorem).As the additional stresses ! and ! do not particularly modify the stress state

obtained by means of the technical theory, they may generally be omitted in thepreliminary design stage, provided that the ‘non-deformability’ of the cross-section isascertained.

13.4.3 Analysis of warpingIf each point of the cross-section outline is characterised by its distance s on it, measuredfrom a certain constant point, then the longitudinal displacement of a point s of thesection that is found on the abscissa x of the beam may be represented by the functionu(x, s), while (x) is the twisting angle at the specific position (x) of the beam(Figure 13.11).As is already clear from the above, the warping of a cross-section at position x results

from the new positions u(x, s) of its points (x, s), after the longitudinal deformation of the

467

Thin-walled beams

beam fibres. It is found that (Zbirohowski-Koscia, 1967)

uðx; sÞ ¼ d

dx !ðsÞ

where the ‘warping function’ !(s) is referenced at each point of cross-section outlineand is determined on the basis of the position of its shear centre. Thus each thin-walled cross-section has its own characteristic diagram of the warping function !(s).This diagram shows for each point of the cross-section how much this point will beshifted along the beam axis due to a unit change d/dx in the twisting angle of thebeam. The diagram !(s) (dimensions m2) always satisfies the equationð

!ðsÞ t ds ¼ 0

The determination of the warping diagram !(s) for a cross-section is a cumbersomeprocess and should be done using a specialised software program. The qualitativewarping diagrams of some characteristic cross-sections are shown in Figure 13.11. g

It should be noted that there are some thin-walled cross-sections that do not developany warping under the torsional response, and consequently they are never subjectedto the stresses ! and ! that arise due to the restraint of warping. The followingcross-sections belong to this category (Figure 13.12):

. cross-sections having a centre of symmetry — the shear centre coincides with thispoint

. open cross-sections having walls passing through a common point — the shear centrecoincides with this point

. all the closed cross-sections that can be circumscribed to a circle having a constantwall thickness (triangular, etc.)

468

Beam axis

s

x

Location of section [x = x1]u(x, s)

θ(x)

ω(s)

u(x1, s) = (dθ/dx)[x = x1] · ω(s)

The diagram ω(s) represents the section warping due to a unit torsional moment

Typical ω(s) diagrams

∫ ω(s) · t · ds = 0 Limited values of ‘ω’ in closed sections

Figure 13.11 Warping behaviour of cross-sections


. more generally, if in a closed cross-section the wall thicknesses are considered asvectors and the resultants of these vectors at each corner pass through a commonpoint, then the cross-section does not undergo warping.

13.4.4 Longitudinal stresses due to prevention of warpingIt is reiterated that the longitudinal normal stresses ! due to the restraint of warping arenot predicted by the classical technical theory of beams, as the only normal stresses thatappear therein refer either to bending moments or to axial forces. In each case thedeveloped stresses are statically equivalent to the occurring bending moment or axialforce. As the stresses ! are self-equilibrating, they do not have any static equivalenceto any force or moment, i.e. they do not contribute to the global equilibrium of thestructure. However, it has been shown that they may be expressed in the same way asthe normal stresses in technical theory, through a sectional magnitude M! called thebimoment (Figure 13.13).In absolute analogy with the technical theory of bending, it is found that for each

point of the cross-section the longitudinal warping stress can be expressed as (Vlasov,1961)

! ¼M!

I! !ðsÞ

where

I! ¼ðA!2 dA

The magnitude I! (dimensions m6) is called the warping constant, and is obtained byintegrating over each elementary part of the surface dA of the cross-section. It is under-stood that !(s) plays the role of ‘distances from the neutral axis’, as is the case in purebending. The similarity with pure bending is obvious. It is observed that, for a givenvalue of the bimoment, the distribution of ! is identical to the distribution of !(s).

469

S: shear centre t = constant

Sections that do not warp under a torsional moment

S

S S S

S S

t4

t4

t3

t3t2

t2

t1

t1

Figure 13.12 Cross-sections that do not develop warping

Thin-walled beams

The analytical determination of I! is very cumbersome and should be performed usingspecialised computer software.The bimoment itself (dimensions kNm2), the physical meaning of which is covered in

the following two sections, is defined, analogously to the bending moment in technicaltheory, through the expression

M! ¼ðA!ðsÞ !ðsÞ dA

If (x) is the function of the twisting angle along the length of the beam, then

M! ¼ EI! d2

dx2

As closed cross-sections exhibit a clearly smaller warping compared to open cross-sections, they clearly also have a much more ‘moderate’ warping function !(s), andconsequently their warping constant I! is comparatively smaller. Thus in closed cross-sections the normal stresses ! are limited.

13.5 The bimoment conceptIn a thin-walled beam the bimoment concept is directly related to the self-equilibratingsystem of the normal stresses that arise from the prevention of warping of the cross-section under a torsional moment. It is clear that preventing a constructional warpingof even a single section (e.g. the end of one beam) is sufficient to exert a restrainingeffect on the warping of all other sections of the beam. It should also be understoodthat this prevention of warping activates the bending rigidity of the section walls, and

470

Bimoment

Mω = ∫ σω(s) · ω(s) · dA = –(EIω) · (d2θ/dx2)

Iω = ∫ ω2 · dA Much less in closed than in open sections

The bimoment represents quantitatively the self-equilibrating longitudinal normal stresses

Beam axis

Mω(x)

θ(x)

MT

σω

σω

σω

x

σω(s) = (Mω/Iω) · ω(s)

Self-equilibrating σω

Distribution of ‘σ’ identical with that of ‘ω’

Figure 13.13 Size and significance of the bimoment


automatically causes longitudinal normal stresses that are not negligible, being in anycase obligatory in the inherent self-equilibrium along the longitudinal sense, giventhat no external axial force is present. It may be considered that these normal stressesrepresent the bimoment itself.The fact that a torsional moment generally causes bending, thus giving rise to the

concept of the bimoment, besides what was examined in Section 13.4.1, is illustratedin the model shown in Figure 13.14. The vertical beam with an open profile is fixedat its bottom end, while the top end is rigidly connected to a horizontal diaphragmthat is non-deformable within its own plane and has no transverse bending rigidity. Atorsional moment MT is applied on the diaphragm.The beam may initially be considered as being composed of mutually independent

vertical plane strips that exhibit a bending rigidity only within their plane and have atorsional rigidity according to St Venant. The forced movement of the horizontaldiaphragm under the torsional moment, consisting of a displacement and a rotation,causes a shear force and a torsional moment on each wall segment according to itsown bending and torsional rigidity, respectively. The horizontal diaphragm is in equili-brium under the externally applied torque, as well as the reacting shear forces andtorsional moments of the walls, which are acting in opposite senses. It is clear thatthese shear forces will amount to a total moment MT,! because, for equilibrium

471

MT

MTMT

Restoring the identical strains andstresses of the adjacent sides Summed up to MT,S

Statically equivalent to MT,ω

MT = MT,S + MT,ω

Self-equilibrating longitudinal stresses

Bimoment

Inducing bending in the strips

Activation of the bending and torsional stiffnessof the vertical strips

Figure 13.14 An example of the physical meaning of the bimoment

Thin-walled beams

reasons, it is impossible to offer any other force. MT,!, together with the summation ofthe St Venant moments MT,S, will then balance the moment MT:

MT¼MT,!þMT,S

The development of a shear force at the top of each wall segment causes bending,leading to axial normal stresses. These stresses are obviously self-equilibrating, i.e.they have no resultant along the longitudinal direction, as no axial force is acting onthe segment. This situation will not change even when a self-equilibrating system oflongitudinal shear forces (see Figure 13.14) is imposed along the free sides of the wallsegments in order to restore the incompatible deformations occurring there. This setof self-equilibrating axial stresses is actually equivalent to the existence of a bimoment.Returning now to the last equation, it is again pointed out that a torsional moment

MT is composed of a St Venant momentMT,S and a warping momentMT,!. It is under-standable that the relationship between these two contributions depends not only on thecross-section itself but also on the length of the beam. For example, as the lengthincreases, the ratio of the bending to the torsional rigidity for each wall segmentdecreases, and consequently the percentage contribution of MT,! with respect toMT,S in taking up the torsional moment MT decreases too.

13.6 Two theorems of the bimoment

Theorem 1A force P parallel to the beam axis and applied to a point on the cross-section havinga warping value ! causes a bimoment equal to M!¼ P ! (Figure 13.15). (Thecompressive loads are considered to be negative.)In the beam with an I cross-section, where three self-equilibrating forces are applied

(N¼ 0), a bimoment equal toM!¼ 2 (P/2) !1 is developed. This bimoment obviouslycauses axial stresses according to the basic equation given in Section 13.4.4, contrary towhat would be predicted using the technical theory.The equal and opposite compressive forces P on the beam in Figure 13.15, which can

be considered as prestressing forces, cause a positive bimoment M!¼ P !1, (P< 0,!1< 0) and, consequently, tensile normal stresses at the ends of the cross-section(!> 0), contrary to what would be concluded according to the technical theory.However, a uniform compressive stress p0 applied on the entire the cross-section doesnot cause a bimoment, because (see Section 13.4.3)

M! ¼ðp0 dA ! ¼ p0

ð! dA ¼ 0

and, therefore, the whole section is acted on by the above compressive stress.

Theorem 2A momentM applied on a plane parallel to the beam axis at a distance e from the shearcentre causes a bimoment M!¼M e. (The bimoment is considered positive if thevector of the applied moment M is directed towards the shear centre.)

472


According to the example shown in Figure 13.15, even though the two externallyapplied moments M on the beam are in equilibrium, they cause a bimoment

M!¼M (h/2)M (h/2)¼M hand accordingly they give rise to a longitudinal stress response.

13.7 Warping shear stressesAs explained previously, the warping shear stresses ! are required in order to ensurethe equilibrium, given the variation in the normal stresses ! along the length of thebeam. As stated, their distribution over the cross-section is not self-equilibrating, as isthe case for !, but leads to the warping torsional moment MT,!. It should be noted,however, that this particular effect occurs only in open sections. As shown in Section13.4.2, the stresses ! in closed sections constitute a self-equilibrating system, whichaffects the distribution of shear stresses according to Bredt. However, it does notcontribute to the uptake of the torsional moment MT, which is taken up exclusivelyby the mechanism of Bredt’s peripheral shear flow.

473

P P

P

P/2

P/2

h/2

h/2

S

M

M

ω1 ω1[ω]

[ω]ω1 Mω = 2 · (P/2) · ω1 Mω = P · ω1

The presence of bimoment means development of self-equilibrating stresses

Direction of moment vectorshowing away from S

Mω = M · h

Figure 13.15 The development of a bimoment in the absence of a torsional load

Thin-walled beams

It should be mentioned here that a warping moment MT,! is also present in closedcross-sections, and this leads to self-equilibrating shear stresses !, as explainedpreviously.The analytical determination of the distribution of stresses ! in the cross-section

(open or closed) based on the moment MT,! is generally cumbersome, and for thisreason specialised computer software must be used. The distribution of the warpingshear stresses ! in some typical cross-sections is shown in Figure 13.16.

13.8 The governing equation for torsion and its practical treatmentConsider a suitably supported rectilinear thin-walled beam that is subjected to adistributed torsional load mD, and, possibly, to a concentrated torque T as well. Thebeam is referenced to axes with an abscissa x, as shown in Figure 13.13.The characteristic deformation magnitude on the basis of which all the torsional

magnitudes of the beam are expressed is the rotation (x). Thus it is found that thewarping torsional moment MT,! is expressed as

MT;! ¼ EI! d3

dx3

while the torsional moment according to St Venant MT,S is

MT;S ¼ G ID d

dx

Hence the equation established in Section 13.4 (MT¼MT,!þMT,S) may be written as

MT ¼ G ID d

dx EI!

d3

dx3

and, given that the equilibrium relation is, in effect,

dMT

dx¼ mD

474

Indicative distribution of warping shearing stresses

MT MT MT MT

τωτω τω τω

Figure 13.16 Shear stresses due to the prevention of distortion


the last equation takes the form (Vlasov, 1961)

EI! d4

dx4G ID

d2

dx2¼ mD

It should be pointed out that this equation is valid for both open and closed cross-sections, despite the fact that in closed cross-sections the size ofMT,! leads, as previouslymentioned, to self-equilibrating shear stresses !. It is, however, possible that the lastresult may be also obtained from the equation

¼ T,!þ T,S

which is valid for both open and closed cross-sections.Moreover, as the bimoment M! in Section 13.4 is expressed as

M!¼EI! (d2/dx2)the following relationship between the bimoment and the warping torsional moment isobtained directly from the above:

MT;! ¼dM!

dx

This above equation is the fundamental relationship that governs the torsionalbehaviour of thin-walled cross-sections and, of course, in order to determine all theabove torsional magnitudes in terms of the function (x), the satisfaction of the pertinentrespective boundary conditions is required. Thus, the following boundary conditions aredistinguished:

(1) In the case of a fully fixed cross-section at position x

¼ 0

d/dx¼ 0

(2) In the case where twisting of the cross-section, but not its warping, is prevented,which means that in the considered cross-section no longitudinal warping stresses! can be developed, then ¼ 0 and M!¼ 0, and on the basis of the previousequations

¼ 0

d2/dx2¼ 0

(3) In the case of a completely free cross-sectionM!¼ 0 and MT¼ 0, and therefore onthe basis of previous equations

d2/dx2¼ 0

G ID d

dx EI!

d3

dx3¼ 0

The boundary conditions for other cases can be worked out based on the above. Thus,for example, when in the place of the internal support of a continuous beam instead

475

Thin-walled beams

two beams are linked, the arranged bearings, with or without a diaphragm placed abovethe support, prohibit the end rotation, and thus (1)¼ (2)¼ 0, while due to thecontinuity of normal stresses ! the equality of bimoments, M!(1)¼M!(2) must alsobe required at the same position. g

476

Table 13.1 Analogy between the magnitudes in the bending and the torsional response

Beam in bending ! Beam in torsion

Displacement w ! Rotation angle Tensile force H ! Torsional rigidity G ITMoment of inertia I ! Warping constant I!Uniform load p ! Uniform torsional moment mD

Concentrated transverse load P ! Concentrated torque TVertical component of the tensile force Hv ! Torsional moment according to St Venant MT,S

Bending moment M ! Bimoment M!

Shear force (vertical) Q ! Torsional moment MT

MT;!

! ¼ V! Hv

!(see Figure 13.17).

Prevention of rotation (θ = 0)Free warping (Mω = 0)

Imposition of rotation θFree warping (Mω = 0)

Prevention of rotation (θ = 0)Prevention of warping (Mω ≠ 0)

Imposition of rotation θPrevention of warping (Mω ≠ 0)

w = θ

w = θ

w = θ

Boundary conditions

V = HV + MT,ω

H

H H = G · ID

H

H

VV

V

MT,ω

MT,ω

HV = MT,SHV

HV

M = Mω

I = IωID IωT θ

mD

P = T

Torsional response Bending response

Analogy

Figure 13.17 The analogy between the torsional and the bending stress states of a tensioned beam


Although the solution of the above basic equation governing the torsional stress state ofa beam through the classical analytical treatment of differential equations is, in principle,feasible, it is very awkward and impractical, especially for more complex cases. However,it should be noted that the form of the equation in question is precisely the same as theequation for a tension beam under transverse loading according to the second-ordertheory, as already examined in Section 7.2 (Roik, 1983). This equation is

EI dw4

dx4H dw

2

dx2¼ pðxÞ

This observation is of crucial importance because it establishes a complete equivalencebetween the two problems, and this allows the torsional problem to be confronted usingthe conceptually much simpler problem of beam bending, for which many accessiblesoftware programs exist. This equivalence, and the corresponding dimensions, isillustrated in Table 13.1 and Figure 13.17. It is clear that all the boundary conditionsof the problem that are expressed through the specific requirements for the rotationangle , the bimoment M! or the torsional moment MT, may be attributed to thecorresponding terms for the beam in bending (see Figure 13.17). This result providesan overall and naturally more direct illustration of the torsional problem.

13.9 Examples

Example 1The aim is to determine the torsional stiffness of a cantilever of length 10.0m, i.e. todetermine the torsional moment required at its free end that will cause a rotation¼ 1 rad. The behaviour of both an open and a closed cross-section, having the sameexternal dimensions and the same cross-sectional area, is examined, considering ineach case the possibility of preventing or not the warping of the free end. It isassumed that the profile of the cross-section remains undeformable.Figure 13.18 shows the details of approaching the torsional behaviour through the

established analogy with the beam in bending based on the second-order theory (seeSection 13.8). The torsional stiffness is determined as the shear force that developsdue to an imposed support settlement on the bending model of the beam, after theassumed twisting angle at the free end.It can be seen that in the open section the prevention of warping at the free end increases

considerably the torsional rigidity, which is significantly higher than the value predictedaccording to the classic expression after St Venant (G ID/L). In the closed cross-sectionthese differences are much smaller and thus have a much lesser practical importance.

Example 2Consider the stress state of a simply supported concrete bridge, with a span of 40mand an open cross-section. The bridge is loaded at the midspan with an eccentricload of 1000 kN applied directly on the top of one web. Assume that the profile ofthe cross-section remains undeformable due to a suitable insertion of transversediaphragms (Figure 13.19).

477

Thin-walled beams

The load is analysed exactly as described in Section 13.1, i.e. by considering asymmetric and an antisymmetric load. For the sake of later comparison, it is notedthat the symmetric load, which consists of the two loads of 500 kN, causes at themidspan a maximum tensile stress of 6300 kN/m2. In this respect the interest is, ofcourse, focused on the antisymmetric loads of 500 kN (see Figure 13.19), a loadingthat causes a corresponding torsion. It is supposed that the layout of the bearings canoffer to the beam the required reactions for the development of the torsional momentof 1425 kNm at its ends. The analysis is done according to the analogy established inSection 13.8, and is shown in Figure 13.19.The resulting bendingmoment of 18 946kNmat themidspan represents the bimoment,

and this is used as the basis for determining the longitudinal warping stresses, as detailed inSection 13.4.4. Based on the corresponding diagram [!], the greatest tensile stress is4121 kN/m2, i.e. 65% of the tensile stress caused by bending. The contribution of MT,S,

478

2.50 m

2.50 m

t = 0.15

t = 0.15 E = 3.0 · 107 kN m G = 1.25 · 107 kN m

E = 3.0 · 107 kN/m2 G = 1.25 · 107 kN/m2

t = 0.30

1.50 m

1.50 m

BA

BA

BA

BA

St Venant torsional stiffnessG · ID/L = 32 375 kN m/rad

St Venant torsional stiffnessG · ID/L = 1.33 · 106 kN m/rad

Iω = 0.1318 m6

ID = 0.026 m4

Iω = 0.022 m6

ID = 1.064 m4

? MT

? MT

? MT

? MT

w = θ = 1 m

w = θ = 1 m

w = θ = 1 m

w = θ = 1 m

VB = 49 594 kN = MT

VB = 85 950 kN = MT

(∆T = –36 944°C)

N = G · ID = 1.33 · 107 kN

(∆T = –899.3°C)

N = G · ID = 323 750 kN

VB = 1.392 · 106 kN = MT

Imposition of rotation θ = 1 radPrevention of warping

Imposition of rotation θ = 1 radPrevention of warping

10.0 m

10.0 m

VB = 1.360 · 106 kN = MTImposition of rotation θ = 1 radFree warping

Imposition of rotation θ = 1 radFree warping

G · ID = 1.33 · 107 kN

G · ID = 323 750 kN

Figure 13.18 Determination of the torsional stiffness of a thin-walled beam


i.e. of the St Venant shear flow, in taking up the torsional moment MT is determinedaccording to Section 13.8. As shown in Figure 13.19, this contribution is zero at themidspan, while at the ends it amounts to 49%. The corresponding percentages for thewarping torsional moment MT,! are 100% and 51%, respectively.

ReferencesRoik K. (1983) Vorlesungen uber Stahlbau. Grundlagen. Berlin: Wilhelm Ernst.Vlasov V.Z. (1961) Thin Walled Elastic Beams. London: Oldbourne Press.Zbirohowski-Koscia K. (1967) Thin Walled Beams — From Theory to Practice. London: Crosby

Lockwood.

479

Midspan:Mω = 18 946 kN/m2

MT,S = 0MT,ω = 1425 kN m

End:MT,S = ϕ(GID) = 698 kN mMT,ω = 1425 – 698 = 727 kN m

5.26

1425

1425

14251425

2.77

[V ]

[σω][ω]

18 946 · 5.26/24.18 = 4121 kN/m2

Rotation at the end:ϕ = 0.23 · 10–3 rad

Rotation at midspan:ϕ = 0

18 946 kN m

[M ]

2850 kN

500 · 5.70 = 2850 kN m

G · ID = 3036 · 103 kN

[MT]

1000 kN

40.0 m

10.50 m

t = 0.25

5.70

Plan view

10.50

2.50

500 kN 500 kN

t = 0.50 m

ID = 0.253 m4

Iω = 24.18 m6

E = 3.0 · 107 kN/m2

G = 0.4 · E

Figure 13.19 Determination of the stress state in a thin-walled girder due to an eccentric load

Thin-walled beams

14

Box girders

14.1 GeneralBox girders, used mainly for bridges, are thin-walled beams having a closed section, and,as such, they obey everything in the previous chapter, on one condition: in order for thebasic equation as well as for all the related concepts (bimoment, etc.) concerningtorsional response to be valid, the undeformability of the box section must beensured. This condition, for beams with relatively small-section dimensions, can beguaranteed by a certain wall thickness, but for larger dimensions, stiff diaphragms arerequired over each support, as well as at two or three places along the span. If thesediaphragms are not desirable for constructional reasons, then the uptake of torsionalloads induces deformation of the section, and, consequently, an additional bendingresponse in the longitudinal as well as in the transverse sense is developed. In thischapter, rectilinear beams are first examined, followed by curved beams.

14.2 Rectilinear girders

14.2.1 General loading caseIn the typical box section shown in Figure 14.1, an eccentric layout of the traffic load isconsidered. This is specified in each case by the load standards followed.It is appropriate to consider those external actions applied to the nodes A and B of the

upper plate, which eliminate both their displacement and rotation. These ‘fixing actions’(see Section 3.3.1) consist of longitudinally distributed vertical loads qA and qB, as wellas longitudinally distributed moments mA and mB, which are determined directly fromthe fixed-end actions on the assumed clamped nodes A and B.Thus, following Section 3.3.1, it can be considered that the state of stress of the girder

results from the superposition of the fixed state I containing the loads with theappropriate nodal actions and of the state II containing only these nodal actions qAand mA, together with qB and mB, acting, however, in the opposite sense.The fixed state I, as a trivial state, is directly determinable, so that state II is

predominantly considered. This state — provided the section is symmetric — canalways be split into symmetric and antisymmetric parts. The symmetric part consistsof the symmetrically distributed loads qR/2 and moments mR/2 applied to the edges Aand B, respectively. The antisymmetric part consists of the mutually opposite loadsqS/2, as well as the moments in the same sense, mS/2 (see Figure 14.1).


The symmetric loading activates the longitudinal bending stiffness of the girder, andresults in the well-known sectional force diagrams for bending moments and shear.Moreover, it causes a transverse response, if the equilibrium of a segmental part of unitlength is considered. This girder strip is acted on by the resultant of the shear flowsapplied to each of its two faces (Figure 14.2). Given that the shear flow v is expressedas v¼V S/I, then dv/dx¼ (S/I) (dV/dx)¼ (S/I) qR so that the above resultant isexactly the differential shear flow dv acting on the strip, equal to dv¼ (S/I) qR. Itshould be noted that S represents the static moment of inertia of the cut-out part ofthe section being considered,with respect to its centroidal axis.Thus, the girder strip is in equilibrium under the symmetric loads qR/2, as well as

under mR/2, and the differential shear flow dv. The loads qR/2 and dv do not induce

482

A B

Fixed state

Nodalloading

Antisymmeticloading

Symmetricloading

mR/2 mR/2

qR/2 qR/2

mS/2 mS/2

qS/2 qS/2

qA qB

qA qB

mA mB

mA mB

b0

qR = qA + qB qS = qA – qB

mR = mA + mB mS = mA – mB

Figure 14.1 Derivation of nodal actions on a beam profile from the acting eccentric loading

qR/2

qR/2mR/2

mR/2dv

dv

∆x = 1

Differential shear flow(= resultant from the two faces)

qR

V V + dV

dv = (S/I) · qRdx

Flow directions follow the hydraulic analogy

Figure 14.2 Transverse response under symmetric loading


any bending at all, but only axial stresses. The loads mR/2 cause both bending and axialstresses (see Figure 14.2). This state of stress may be determined using classic computersoftware for plane frames. g

When speaking of the symmetric loading of a box girder, two points should be maderegarding haunched bridge girders either in a cantilevered system or, analogously, in acontinuous one over its internal supports — in addition to what has already beenmentioned in Sections 4.4 and 5.1 — mainly with respect to the free cantileveredmethod of construction.The first remark concerns the variation in the tensile axial forces Z in the top slab of a

cantilever, which is associated with the longitudinal shearing flow v, as shown inFigure 14.3. As explained above, the inclined compressive force of the bottom slabhelps with its vertical component, in offering to the cross-section the required resistanceto the induced vertical shear force, thus achieving, for a parabolic curving of the bottomslab, an approximately constant shear flow throughout the cantilever length. By cuttingout from the web an elementary segment of the top slab subjected to the axial tensileforces, as well as to the acting shear flow, and considering its horizontal equilibrium,

483

Transverse response

pu

D ≈ M/Z

D ≈ M/z

pu = D/r

τ · bw = ∆Z/∆xConstant shear flow τ · bw

1st

segment

etc.3rd2nd

3 2 1 1 2 3

3

2 3

1 2 3Acc

umul

ated

pres

tres

sing

forc

e

Tensile force Z

τ · bw

τ · bw

∆x

ZZ + ∆Z

V

Mz

Figure 14.3 Structural behaviour of a cantilevered haunched girder

Box girders

it is concluded that

v ¼ bw ¼ Z

x

where

bw ¼ 1

z V M

zz

x

(see Section 4.4)

The above relations are quite generally valid. In the case of a constant depth(z/x¼ 0), the tensile force Z exhibits a parabolic variation along the girder(z/x¼V/z). However, in the case of a variable depth leading to a constant flow v,the tensile force Z is linearly varied, which, with regard to the balanced cantileveredconstruction for continuous bridges, mentioned in Section 5.1 and suggested inFigure 14.3, means that an essentially constant amount of prestressing ‘dosage’ needsto be added at each consecutive construction stage, thus representing a constructionaladvantage. It should be pointed out that the prestressing reinforcement for the freecantilevered process (see also Section 5.5.2) is placed principally on the top slab ofthe girder section.The second remark regards the fact that the ever-changing inclination of the

compression force of the bottom slab, due to bending, creates an inward deviationpressure pu¼D/r (see Figure 14.3), which stresses the bottom slab in the transversesense, obviously counteracting its self-weight, but necessarily affecting the wholeclosed transverse segmental strip in bending, as shown in Figure 14.3. g

As the symmetric loading does not present any other peculiarities, the whole problem isshifted to the antisymmetric loading part. This acts as a distributed torsional load,mD¼ (qS/2) b0þmS (see Figure 14.1), resulting in a particular torsional momentdiagram (Figure 14.4). At each position of the beam, the torsional moment developedis taken up by the corresponding Bredt shear flow. Thus, for each wall of the sectionconsidered in its longitudinal sense, a shear force diagram can be assigned.If, now, the non-deformability of sections can be ensured through the layout of

transverse diaphragms, as mentioned previously, an additional longitudinal responseconsisting of axial and shearing stresses will take place, according to Chapter 13. Thefact that the section is closed leads to a limited response arising from the restrainingof the limited warping of the section.

484

The Bredt shear flow induces shear forces in each section wall

[MT] [MT]

Figure 14.4 Influence of the support layout on the distribution of the torsional response


However, the presence of the transverse diaphragms is generally not desirable forconstructional reasons. Thus, the girder section must be considered as deformable,the development of torsional response is no longer governed by the basic equation ofSection 13.8, and, consequently, the structure has to be considered, accordingly, as afolded system. In this case, as will be examined in the following section, the above-mentioned additional response to the shear flow, according to Bredt, consists of alongitudinal bending for each wall, as well as of a transverse response for the sectionprofile.

14.2.2 Response due to the deformability of the profile sectionThe imposed torsional loading (mD¼ (qS/2) b0þmS), due to the antisymmetric couple,causes deformation of the closed section, causing longitudinal bending of the sectionwalls (Figure 14.5), which is coupled with the resulting transverse bending of thesection profile itself (Schlaich and Scheef, 1982). It is again pointed out that thisresponse is in addition to the initial Bredt shear flow in the section walls.In order to now investigate the influence of the deformability of the section on the

response, the equilibrium of a cut-out girder strip of unit length is first consideredwith the antisymmetric loads qS/2 and mS/2 acting at the section edges A and B (seeFigure 14.5). It is clear that the segment is in equilibrium under the above external

485

∆v = (qS · b0 + 2 · mS)/(4 · Fk)∆x = 1

Differential shear flow(≈ resultant from the two faces)

∆vv

v ∆v

b0

v = MT/(2 · Fk) [kN/m]

Self-equilibrating system

mS/2

mS/2

qS/2qS/2

mS/2

qS/2

qS/2

mS/2

Bredt shear flow inducesshear forces in the section walls

Torsional moment diagram Equilibrium of section stripe

Antisymmeticloading

A B

Figure 14.5 Analysis of antisymmetric loading

Box girders

forces and the differential shear flowv, which is obtained as the resultant of the Bredtshear flows on the two faces of the strip considered.Given that

v ¼ MT

2 Fkthen

v

x¼ MT

x 1

2 Fk¼ mD

2 Fkand, consequently,

v ¼ qS b0 þ 2 mS

4 FkAccording to the section dimensions given in Figure 14.5, Fk¼ d (b0þ bu)/2.Obviously, the examined strip tends to deform under the forces qS/2, mS/2 and v.

This deformation consists essentially of a change in length of its diagonals, andinduces an additional state of stress for each section wall. So, it is considered appropriate— following Section 3.3 — to insert a hinged strut of infinite axial stiffness in the sense ofthe augmenting diagonal, i.e. a non-deformable one (Figure 14.6). With the insertion ofthis element throughout the length of the girder, the deformability of the section iseliminated, whereas the section itself develops a bending and axial state of stress, theformer being induced by the acting moments mS/2. It is clear that this diagonalelement develops a tensile axial force R (kN/m) distributed along the length of the

486

Cut out strip in equilibrium

qS/2qS/2

mS/2 mS/2

qS/2qS/2

mS/2 mS/2

∆v

Differential shear flow

R

RA = ∞

Opposite actions cause alongitudinal and transverse response

Additional response to the Bredt shearing one

The fictitious member istensioned with a force R

Figure 14.6 Additional girder response due to antisymmetric loading


girder. Applying now the opposite of the acting forces R on the corresponding longi-tudinal edges of the girder in absence of the diagonal strut, it is obvious that thesuperposition of the thus resulting state of stress with that of the blocked strip willgive the final response (see Figure 14.6). g

For the examination of this last ‘diagonal loading’ it is first considered that the forces Ract on a girder having hinged connections at the section edges instead of monolithicones (Figure 14.7). The forces R may be resolved equivalently at each edge into thetwo concurring walls; therefore, each of them can be considered as a longitudinalbeam loaded by the corresponding component, developing bending moments M0 andcorresponding normal stresses , according to the classical theory of bending.However, the resulting strains /E at the common edges of the walls are not equal —as they should be — and for this reason some distributed longitudinal forces have tobe additionally introduced along the edges of each wall, in order to establish thestrain compatibility at each edge (see Figure 14.7). It is clear that in this way the initiallydetermined normal stresses will be changed.By following the above analysis, the determination of the longitudinal axial stresses is

possible, e.g. for the left web of the ‘hinged’ box section in Figure 14.8 they can bedetermined through the classical bending formula on the basis of the aforementionedmoment M0. In this case, the moment of inertia I used is slightly larger than thenormal value Iw for the web by a factor kw, and the new ‘neutral axis’ lies at adistance y0 from the top fibre, which is somewhat less than its ‘normal value’, i.e. halfof the web height (Schlaich and Scheef, 1982).The moment M0 results from the loading of the left web with the corresponding

component Rw of the ‘diagonal force’ R (see Figure 14.7). It is found that Rw¼R bw/s.

487

R0

Rw

Ru

Rw

Rw Ro

Rw

Ru

RR

Beam span

Applied self-equilibrating shear forcesfor restoring the incompatibility of stresses

Figure 14.7 Taking up the diagonal loads through the longitudinal wall beams

Box girders

Thus,

0 ¼ M0

I y0

u ¼M0

I ðbw y0Þ

where

I ¼ Iw kw

Furthermore (see Figure 14.8),

kw ¼ 2 ½ð0 þ 2Þ ðu þ 2Þ 1ð1þ Þ ð3þ 3 þ 0 þ u Þ

y0 ¼1þ 2 þ u

3þ 3 þ 0 þ u bw

Thus, the longitudinal web beam obeys the following typical differential equation (seeSection 2.3.6):

EI d4w

dx4¼ Rw

488

Effective moment of inertia I* = Iw · kw

‘Neutral axis’

Beam span

σ0 = (M0/I*) · y0

σu = (M0/I*) · (bw – y0)

Inducing the bending moment M0

bw

bw

y0

Rw

Rw Rs

RIw tw

bu

b

b0

tu

t0Rw

σu

σ0β = (bu/b0)

αo = (t0 · b3)/(tw · bw · b0

2)αu = (tu · bu)/(tw · bw)

Figure 14.8 Bending response of the longitudinal section walls in a ‘hinged’ system


where Rw and I represent the initial distributed web loading and the equivalent moment

of inertia, respectively, as explained above. It is noted that w represents the in-planedeflection of the web, and it is due merely to the assumed deformability of the section(see Figure 14.8).Only now may the monolithic connection of the section walls be taken into account.The bending deformation of the hinged section profile, under the action of diagonal

forces R, has as a consequence an increase of its diagonal length by (Figure 14.9).However, this change cannot be realised without resistance, given that the transversestiffnessC of the closed monolithic section profile is automatically mobilised. This stiffnessis expressed through the relation r¼ C, as that diagonal pair of forces r required toproduce the deformation ¼ 1. The stiffness C clearly depends on the geometric andelastic data of the section, and may be determined using typical computer software forplane frames. Thus, the tendency of the hinged section profile to be deformed by iscounteracted by the resistance r of the monolithic closed frame, and in this way theweb is subjected to, in addition to its ‘initial’ loading Rw, the loading of the componentrw of the force r, obviously with opposite sense. Thus, rw¼ r (bw/s) (see Figure 14.9).Taking into account that the bending deformation w of the web can be expressed

through the diagonal deformation by the relation ¼w D (see Figure 14.9), where

D ¼ 2 bw s ð1þ Þ ð2þ 2 þ 2 2 þ 0 þ u 2Þ

3þ 3 þ 0 þ u

489

End diaphragmEnd diaphragm Rw

Rw

wK

w

w

w

bw

rw = K · w

The deflection w refers only to the profile deformationThe presence of a diaphragm means a corresponding support (w = 0)

(I* = Iw · kw)

E, I*

rr

rw

RR

δ

Deflections w induce an increase δin the diagonal δ = w · D

The frame opposes a resistance rto the increase in its diagonal

The web acts as if resting on an elastic base

r = δ · C rw = r · (bw/s) = K · w

Figure 14.9 Structural action of a section wall in the monolithic system

Box girders

it is found that the component rw is also expressed through w by the relation

rw ¼ w D C ðbw=sÞ ¼ w Kwhere

K ¼ D C ðbw=sÞThe differential equation for the web beam can now be adjusted to the restored rigidsection profile, and can be written as

EI d4w

dx4¼ Rw rw ¼ Rw w K

and, finally, takes the form

EI d4w

dx4þ K w ¼ Rw

This equation is recognised as the typical equation of a beam on an elastic foundationwith a modulus of subgrade reaction K (see Section 17.3.3.1). Indeed, the web is carriedby the elastic support offered by the profile resistance when undergoing diagonaldeformation (see Figure 14.9). g

It is, of course, clear that the presence of a single concentrated load on the web will leadto an applied concentrated moment and a corresponding shearing flow (Bredt) and,finally, to a concentrated load applied to the model of the elastically supported beam(Figure 14.10). If this eccentric load is the only one acting on the girder, then, in theabove equation, the distributed load Rw on the right-hand side does not exist. In thiscase, it is clear that the diagonal actions ‘R’ on the girder consist of a single pair ofequal and opposite forces. These forces, if a transverse diaphragm exists at thislocation, are taken up directly by it, without causing any ‘secondary’ stresses in thegirder, as would happen in an ‘undeformable section’. If such a transverse diaphragmdoes not exist, then the diagonal forces are taken up by the deformable section of thegirder itself, as has been previously examined in detail (see Figure 14.10) (Menn, 1990).Recapitulating, the deformation of the web under the component Rw of the diagonal

loading R induces the transverse stiffness of the closed profile. This stiffness arises as aload rw acting on the web in the opposite sense, being linearly related to its deflection,i.e. as occurs in a beam resting on an elastic foundation. It is clear that the thus-resultinglongitudinal bending response of the web will lead, through the developed axial stressesaccording to the above relations, to the edge stresses of the remaining section walls, as ateach edge the strains " and, consequently, the stresses " E are common. g

Regarding the boundary conditions that have to be fulfilled by the deflection w, it shouldbe pointed out that w specifically concerns only that deformation which arises from thedeformability of the section profile and not the overall deflection of the web beam, towhich the torsional twist angle of the section also contributes as a rigid body. Thus,at each position, where the profile remains undeformed because of the presence of atransverse diaphragm, the condition w¼ 0 has to be satisfied. This means that, at the

490


corresponding point of the elastically supported beammodel, an unyielding support mustbe inserted (see Figure 14.10). Moreover, at a fixed position of the girder, the followingcondition is imposed:

dw

dx¼ 0

Finally, the treatment of a girder support with no transverse diaphragm under thedeveloped support reactions arising from torsional response is examined (Figure 14.11).It should be noted that, in this case, no support needs to be provided for the model of

the elastically supported beam but, instead, the developed vertical reactions should beconsidered as antisymmetric loads. These exert a concentrated torsional moment and,according to the above, they lead to a concentrated force on the elastically supportedweb model. g

491

E, I*

R

R

Rw

Rw

P/2

P/2

P

P/2

P/2

P/2

P/2

Due to antisymmetric loading

[MT]

Equilibrium of girder strip

Figure 14.10 Behaviour of a box girder under an eccentric concentrated load

Box girders

Of course, the simultaneous transverse response of the section profile should not beoverlooked.Based on the deformation w of the web, the diagonal loading r of the closed rigid

profile, as shown in Figure 14.12, can be readily determined, thus allowing theevaluation of the transverse response of the section. According to the above,

r ¼ w K s

bw

14.2.3 Numerical exampleA simply supported orthogonal box girder with a 40.0m span, shown in Figure 14.13, isexamined. At the midspan and over the left web a concentrated load of 1000 kN is

492

Q Q

Q

Q

Q

R

R

RwRw

[MT]

Interior support

Development of torsional reaction

Equilibrium of girder strip

Figure 14.11 Behaviour at a support without a transverse diaphragm

Deflections w induce an increase δin the diagonal δ = w · D

Imposing an increase δ in the frame diagonalrequires action of the forces r

r = δ · C

Transverse response

w

w

r

r

δ

Figure 14.12 Transverse response due to profile deformation


considered, which, according to Figure 14.10, induces torsion through the antisym-metric couple of 500 kN.On the basis of the expressions in Section 14.2.2, the following cross-section data are

determined:

¼ 1, 0¼ 1.24, u¼ 0.79, kw¼ 1.002, y0¼ 1.18, D¼ 1.495, s¼ 6.69m

as well as

Iw¼ 0.504m4

Thus obtaining

I ¼ 0.504 1.002¼ 0.505m4

493

r = w · D · C (kN/m)

40.5 kN/m

Transverse response

σo(Tension)

σo(Compression)

σu(Compression)

σu(Tension)

Longitudinal web stresses

σo = 1201 kN/m2

σu = 1344 kN/m2

514 kN m[M]

w = 1.26 · 10–3 mK = 12 025 kN/m2

End diaphragm End diaphragmE, I*

w

250 kN

620 kN100 · 6.20 kN

Differential flow100 kN/m

620 kN

250 kN100 · 2.50 kN 250 kN

500 kN 500 kN

500 kN 500 kN

0.250.502.50 m

0.16

6.20 m1 kN/m

1 kN/m

Equilibrium of girder stripe

Diagonal increase 4.65 · 10–5 m2/kNC = 1/4.65 · 10–5 = 21 524 kN/m2

Figure 14.13 Antisymmetric loading of a girder caused by an eccentric load

Box girders

Moreover, the stiffness C (kN/m2) of the unit length of the girder strip with respectto a unit increase (1.0m) of the diagonal distance of its edges is determined asC¼ 21 524.0 kN/m2.From the equilibrium of the girder strip at midspan, the web is acted on by a

concentrated force of 250.0 kN. More specifically, the strip is in equilibrium underthe antisymmetric couple of 500 kN as well as the peripheral differential shear flow v,which, according to Section 14.2.2, is

v ¼ 500 2 6:204 2:50 6:20 ¼ 100 kN=m

The force Rw acting on the left web is

Rw¼ 500.0 v 2.50¼ 250.0 kN (see Figure 14.13)

The web, as explained previously, acts as a beam on an elastic foundation with amodulus of subgrade reaction equal to K¼ 1.495 21 524.0 2.50/6.69¼ 12 025 kN/m2.The corresponding bending moment diagram is obtained, as can be seen in

Figure 14.13, exhibiting a maximum value at midspan equal to 514 kNm. Moreover,the corresponding deflection is 1.26 10—3m.Thus, the normal stresses at the top and bottom fibres are obtained, from the relations

in the previous section, as

0¼ 514.0 1.18/0.505¼ 1201 kN/m2 (compression)

u¼ 514.0 (2.50 1.18)/0.505¼ 1344 kN/m2 (tension)

The above longitudinal stresses are solely due to the torsional response of the girder, bytaking into account the deformability of its cross-section. The complement of the stressimage for the whole section is shown in Figure 14.13.The development of the deflection w causes an increase in the diagonal length of the

profile, which requires the action of the corresponding diagonal forces r. These are,according to the above,

r¼ 1.26 10—3 1.495 21 524.0¼ 40.5 kN/m

On the basis of these forces, the corresponding bending, axial and shearing responses ofthe closed girder section can be readily deduced (see Figure 14.13).

14.2.4 SummaryAccording to the analysis above, the examination of the behaviour of rectilinear boxgirders under an eccentric loading consists of the following:

. The eccentric loading is reduced through the ‘trivial fixed state’, represented by thetransverse bending state [m0], into a non-symmetric nodal loading.

. The non-symmetric nodal loading is split into a symmetric and an antisymmetric part.

. The symmetric part causes a longitudinal bending and shearing response determinedby the corresponding sectional force diagrams [M] and [V], as well as a transverse

494


response concerning a transverse girder strip of unit length arising from a self-equilibrated loading consisting of the acting loads and the differential shear flow.The transverse response is represented by the corresponding bending moment,shear force and axial force diagrams [msym], [vsym] and [nsym], respectively, plottedalong the section profile.

. The antisymmetric part causes a torsional response along the girder, represented by atorsional diagram [MT], which enables the determination of the Bredt shear flow ateach position. This shear flow causes, in turn, on the one hand, shear of the sectionwalls, depicted by a the diagram [VT] for each wall, and, on the other hand, a long-itudinal ‘diagonal loading’ R, arising from the resulting differential shear flow of astrip. To this last response, the transverse loading arising from fixation of the sectionedges has to be added, represented by the diagrams [mf], [vf] and [nf].

. The ‘diagonal loading’ R causes a longitudinal bending and shear response, repre-sented by the sectional force diagrams [MR] and [VR], respectively, arising from theconsideration of a characteristic section wall (usually the web), as a beam resting onan elastic foundation. This beam is supported only at places where the deformation ofthe section profile is prevented.

. The above response is always accompanied by the transverse response of the sectionprofile, determined on the basis of its mobilised ‘diagonal stiffness’, and is representedby the corresponding sectional diagrams [mR], [vR] and [nR] along the profile peri-meter.

Thus, for the response of each wall, the following diagrams must be taken intoaccount:

. Longitudinal bending: [M]þ [MR].

. Longitudinal shear: [V]þ [VT]þ [VR].

. Transverse bending: [m0]þ [msym]þ [mf]þ [mR].

. Transverse axial response: [nsym]þ [nf]þ [nR].

. Transverse shear: [vsym]þ [vf]þ [vR].

The transverse shear is usually of secondary importance.

14.3 Curved girders

14.3.1 GeneralThe box girder is particularly suitable for bridges curved in plan, as these are subjected toa permanent torsional response caused even by non-eccentric loads, e.g. the self-weightof the girder.The skeletal model used for a curved girder is of the grid type (see Chapter 10), i.e. a

plane structure loaded transversely to its plane, and, as such, it develops three sectionalforces at each position, namely a vertical shear force V, a bending moment MB and atorsional moment MT (Figure 14.14).As will be shown immediately below, an equilibrium interrelation exists between

bending and torsional moments, and thus any redistribution of the bending response

495

Box girders

of the girder, for reasons associated with its plastic analysis (see Section 6.6.2), has,necessarily, to be accompanied by an adjustment of the corresponding torsionalmoments.

14.3.2 Determination of the bending and torsional response

14.3.2.1 Evaluation of the equilibrium equationsThe girder is assumed to have a constant radius of curvature equal to R. The loadingconsists of a vertical distributed load q passing through the shear centre of the cross-section, as well as of a distributed torsional moment mD (Figure 14.15). It should berecalled that, exactly as in the case for grids, any part of the beam must satisfy three

496

Vq

MB

MT

Figure 14.14 Curved beam model with its developed sectional forces

d2MB/ds2 = –q + (1/R) · (dMT/ds)

Analogy with determination of M:d2M/ds2 = –q

dMT/ds = –mD – MB/R

Analogy with determination of V:dV/ds = –q

MT + dMT

MB + dMB

MB + dMB

Bending equilibrium

Torsional equilibrium

PLAN VIEW

MB

MB

MT · ds = MB · dϕ

MT

ds

q

MD

dϕ

V + dV

V

The change in the bending moment vectormeans an additional torsional load mT = MB/R

Usually negligible

Figure 14.15 Evaluation of equations of equilibrium


conditions of equilibrium, i.e. those with respect to vertical forces, and those withrespect to the projections of moment vectors on two arbitrary horizontal axes.The equilibrium equations of an elementary segment of length ds forming an angle d’

(1/R¼ d’/ds), as shown in Figure 14.15, can be written in the form:

dV

ds¼ q (equilibrium of vertical forces)

dMT

ds¼ mD MB

R(equilibrium of moment vectors, on a tangential axis)

dMB

ds¼ V þMT

R(equilibrium of moment vectors, perpendicular to the axis)

It should be understood here, according to Figure 14.15, that the vectorial change in abending moment between two adjacent sections is equivalent to the action of adistributed torsional load. This can also be recognised from the presence of the termMB/R on the right-hand side of the second equation above. In other words, thepresence of a bending moment at the two edges of a curved beam segment meansthat, apart from mD, an additional distributed torsional load MB/R is acting on thebeam. In an analogous way, the vectorial change in the torsional moment MT isrelated to bending development, as can be seen from the third equation (Menn, 1990).The elimination of the shear force V from the above equations provides the two final

equations, which clearly describe the interdependence of the bending and torsionalmoments of the girder:

d2MB

ds2¼ q 1

R dMT

ds

ðaÞ

dMT

ds¼ MB

Rþ mD

ðbÞ

If the arc span length L is much smaller than the radius of curvature R (i.e. L/R< 0.3)and, moreover, the distributed torsional load mD is much smaller than q R, then, byexpressing MB as q L2/c — where c is of the order of about 10 — from equation (b) itmay be concluded that

1

R dMT

dsq

¼ L2

R2 cþ mD

R qwhich means that the term (1/R) (dMT/ds) on the right-hand side of equation (a) ismuch smaller than q.Thus, the first equation may be more simply written as

d2MB

ds2¼ q

i.e. resembling the equilibrium relation of a rectilinear beam between the bendingmoment and the load (see Section 2.2.3). This means that, under the above conditions,

497

Box girders

the bending moments of the curved girder may be approximated by the bendingmoments MB of a straight beam of span L equal to the arch length of the girder.Once the moments MB are determined in this way, then the right-hand side of

equation (b) takes a concrete value, and the equation itself refers directly to the equili-brium relation of a beam, dQ/ds¼ p, between shear forces and the distributed load.This means that the torsional moments MT may result as the shear forces of astraight beam of length L carrying the load MB/RþmD (see Figure 14.15).If the above conditions regarding the ratios L/R and mD/q R are not met and a better

approximation is needed, then, on the basis of the above initial values of MB, one canreturn to equation (a) and evaluate the term

q 1

R dMT

ds¼ q 1

R MB

Rþ mD

as a new loading of the beam, in order to obtain ‘improved’ values of the bendingmoments MB, and afterwards, through equation (b), to determine anew the torsionalmoments MT as shear forces for the loading

MB

Rþ mD

The repetition of the above procedure converges rapidly.

14.3.2.2 Summary. First, the ‘stretched’ girder is considered.. On the basis of vertical loads, the bending moment diagram MB is considered.. The ‘straight’ girder is loaded with the torsional load MB/R, according to the sense

dictated by the composition of the vectors MB, as applied to the correspondingelementary curved girder segment considered in plan view. The distribution of thistorsional loading is obviously identical to the corresponding bending diagram [MB].Any existing load mD must be added.

. The torsional loads MB/R are considered as distributed vertical downward loads, aslong as the bending moments MB cause tension at the bottom fibres; otherwise,they are directed upwards. The possibly present mD may also be considered as adistributed vertical load with the same direction as MB/R, if its sense coincideswith that of the torsional load MB/R; otherwise, it has the opposite direction.

. The torsional moment diagram [MT] is identical to the diagram of shear forces for theabove fictitious loading, but not with respect to its sign. g

It is clear that the above equations concern only the equilibrium, and, as they do nottake into account the compatibility of deformations in the case of redundantsupports, they do not lead to a strictly accurate solution, unless the ratio ¼EI/G ITis considered equal to zero. Nevertheless, their use for purposes of preliminary designis quite appropriate. It is noted that for a simply supported curved girder and for aratio L/R less than 0.3, which means a central angle not greater than 208, the

498


discrepancies from the ‘exact’ solution are less than 1%. Moreover, for a fixed-endedgirder, and thus for the approximately equal internal spans of a continuous girderwith L/R 0.3 and 5, the discrepancies are less than 1.5%. Furthermore, asmentioned initially, the above equations may have a direct utility when, for thepurposes of plastic design, the bending moments obtained from a computer solutionhave to be altered, in which case the redistribution of torsional moments mustnecessarily satisfy the equilibrium criterion.

14.3.2.3 Application examplesThe above procedure is applied to the following examples.In the first example (Figure 14.16) a simply supported girder allowing the take up of

torsional moments at its supports and a fixed-end girder are considered. Figure 14.16shows that the fixed-end girder develops a clearly lower torsional response, while itstorsional reactions are zero. This may be understood on the basis of the remark madein Section 2.3.6 that, as the beam is loaded by its bending moment diagram, its reactions

499

(Shear force diagrams due to transverse load)

[MT] [MT]

Reaction = 0 = torsional reaction(see Section 2.3.6)

Reaction = torsional reaction

MB/R

MB/R

MB/RMB/R

MB/RMB/R

Analogy between torsional and transverse load

Distribution of torsional load

Torsionalreaction

MB MBMB MB

MBMB

MBMB

[MB][MB]

q · L2/24

q · L2/12

q · L2/8

L

L

q (kN/m)

PLAN VIEW PLAN VIEW

q (kN/m)

Figure 14.16 Determination of the bending and torsional response of a curved girder

Box girders

(specifically the torsional ones) must correspond to the developed end rotation angles,which are obviously zero. In both cases examined, attention must be focused on thecorrect determination of the loading sense of the torsional loading MB/R, as shown inFigure 14.16.Regarding the internal span of a continuous beam, the consideration of a fixed-end

beam instead does not essentially alter the torsional response.In the second example (Figure 14.17), a beam is examined which can develop a

torsional reaction only at its left support, being in the first case supported on an inter-mediate point, whereas in the second case it acts as a cantilever.Despite the fact that the intermediate support cannot develop a torsional reaction,

it relieves the beam significantly with respect to the torsional response, since thetorsional reaction in the case of a cantilever is about eight times greater. Again, dueattention is paid to the direction of the torsional loading MB/R, according toFigure 14.17. g

500

Torsional moment diagram identical to the shear force one

Analogy between torsional and transverse load

(213 000/R) kN (1 713 000/R) kN

(213 000/R) kN m

(1 713 000/R) kN m

MB/R

MB/R

MB/R

MB/RMB/R

Distribution of torsional load

PLAN VIEW PLAN VIEW

g = 160 kN/m g = 160 kN/m

8000 kN m 128 000 kN m

[MB]

[MB]

14 000 kN m

MB/R

MB

MBMB

MBMB

MB

MB

MB

30.0 m 10.0 m40.0 m

Figure 14.17 Evaluation of bending and torsional response with cantilever action


It is now appropriate to examine how a girder behaves when subjected to a distributedtorsional moment mD along its whole length. First, it is obvious that a straight girder,where the ratio L/R is equal to zero, develops only torsional response and no bendingat all.However, as the arc length L over a certain chord, and consequently the ratio L/R,

increase, the torsional response is progressively reduced, being accompanied by ever-increasing bending. This can be confirmed using equations (a) and (b) in Section14.3.2, which are, of course, valid without making any assumption regarding thegeometry of the curved girder. However, if a girder in the form of a ring, i.e. exhibitingthe maximum possible value of the ratio L/R, namely equal to 2 p, is subjected to auniformly distributed torsional self-equilibrating loading moment mD, it develops notorsion at all but only constant bending, as was ascertained in Section 12.4.1 (seeFigure 12.28). Such a ring develops a constant twisting angle all along its perimeter.The above property of a ring makes possible the bridging of a span length 2 R through

a girder in the form of semicircle of radius R, suspended along its internal edge in order toexclusively develop a bending response without any torsion. This goal, however, isachieved under the condition that the girder is being subjected to an appropriatebending moment at both its supported ends, and this is in turn directly accomplishedif the girder is connected to two rectilinear beams of appropriate length, as shown inFigure 14.18 (Schlaich and Seidel, 1988).More specifically, the eccentrically suspended semicircular girder, under its self-

weight g, is subjected to a constant torsional load mD¼ g e, where e is the distance ofthe centroid to the suspension line (Figure 14.19). According to Section 12.4.1 (seeFigure 12.28), the equilibrium condition of the curved girder leads to the exclusivedevelopment of the bending response, if a bending moment MB¼mD R can beprovided at both its ends. This specific bending moment, which will be actingthroughout the curved girder, can be provided in practice by the fixed-supportmoment gS L2s /8 of a fixed simply supported beam. This is rigidly connected at the

501

Suspension forces

Figure 14.18 Layout of a semicircular cable-stayed bridge

Box girders

two girder ends, having an appropriate self-weight gS and length Ls, both determinedthrough the obvious requirement that mD R¼ gS L2s /8 (see Figure 14.19).

14.3.3 The response of cross-section wallsThe torsional response referring to the skeletal model of the girder as examined inSection 14.3.2 implies, beyond the Bredt peripheral flow, an additional response forthe box section walls. This response results from the way that gravity loads areintroduced to the girder.First, it should be noted that the acting compressive forces D and tensile forces Z in

the curved top and bottom flanges, respectively, cause distributed deviation forces q,according to the well-known relation

(distributed deviation force)¼ (axial force)/(radius of curvature)

It is clear from Figure 14.20 that the transversely distributed equal and opposite forcesqD and qZ, which the top and bottom slabs are obliged to carry, respectively, createa torsional load per unit of curved length, which is simply that resulting from thevectorial variation of bending moments, as examined in the previous section (Menn,1990):

qD ¼ D

R¼ MB

h R

qZ ¼Z

R¼ MB

h R

502

Centroid

Suspension points

g

mD = g · e

R

Ls

Lsgs gs

M = mD · R M = mD · R

gs · Ls2/8 gs · Ls

2/8

PLAN VIEW

Figure 14.19 Structural action of the system shown in Figure 14.18


and, given that D¼ Z,

qD h ¼ MB

R

Thus, it can be seen that, even without the action of an externally applied torsionalmoment mD, merely the existence of bending along a curved axis implies — as previouslyfound — the imposition of a distributed torsional load MB/R, which causes, of course, atorsional response, according to equation (b) in Section 14.3.2:

dMT

ds¼ MB

R

Of course, how the loads qD and qZ are introduced to the girder has to be examined,and, in particular, how and with what consequences a transverse sectional strip is

503

Additional response to the Bredt shear stress state

S = MB · s/(2 · b · h · R)

S

sMB/(2 · b · R) MB/(2 · h · R)

MB/(2 · h · R)

MB/(2 · h · R)

MB/(h · R)

MB/(h · R)

dv · b

Differential shear flowDeviation forces

dv · h

MB/(2 · h · R)

Charging of the interior web

The deviation forces createtorsional loading

The torsional loading is balanced bythe differential shear flow

1 1

h

h

b

MB /(2 · b · R)

Z

D

Z

D

R = ds/dϕ

Girder strip of unit length

qD = D/R = MB/(h · R)qZ = Z/R

qZ · dsσ(MB)

qD · dsds

dϕ dϕ

Figure 14.20 Introduction of a torsional response in a box girder walls

Box girders

self-equilibrated under the above loads. The cut-out strip of unit length, receiving theforces qD and qZ at its top and bottom sides, respectively, is in equilibrium with thedeveloped Bredt shear flow at both its faces (see Figure 14.20). The resultant of thesetwo flows is the so-called differential shear dv referred to the unit of transverse length(see Section 14.2.1). According to Bredt’s formula (see Section 2.5.2),

dv

ds¼ dMT

ds

1

2 b h ¼ MB

2 b h RThus, the strip, being in equilibrium as a plane structure, under the loads qD, qZ and dv,gives rise to the self-equilibrated diagonal loading of the profile

S ¼ MB

2 R ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

b2þ 1

h2

r

as shown in Figure 14.20, which causes longitudinal bending of the walls as well as trans-verse bending of the section profile, as examined in detail for the rectilinear girder.Although the analogy is not quite accurate, for preliminary design needs and for

limited curvatures, it can be considered that the left web wall takes the downwarduniform load Sw¼MB/2 R b, acting, as in the case of a rectilinear beam, like a beamresting on an elastic foundation with a ‘subgrade modulus’ K, on the basis of the stiffnessof the section profile against a diagonal length change. Of course, if the resultingresponse is high, transverse diaphragms should be provided. Then, the deviationforces of the top and bottom slabs that trigger the whole response will be transferredthrough corresponding horizontal bending to each diaphragm. This, in turn, willtransfer a concentrated torque to the section, by excluding at the same time its defor-mation that would otherwise take place.The above diagonal loading of the girder cross-section, when transverse diaphragms

are not present, makes it clear that — as shown in Figure 14.20 for the longitudinalresponse — the ‘internal’ web is burdened whereas the ‘external’ one is relieved.However, it is interesting to note that in the case of an open section, as shown inFigure 14.21, quite the opposite happens, as explained below.More specifically, the consideration of an elementary curved segment, as depicted in

Figure 14.21, shows that the action of bending moments MB is equivalent — aspreviously explained — to the application of a distributed torsional load mD¼MB/R. Itcan be seen that this torsional load can be taken up by the ‘opposite bending’ of thetwo webs, through the resolution in the continuously distributed vertical loads

q ¼MB/R bwhich obviously give an additional load to the external web while decreasing the load onthe internal one. If it is assumed that the existing vertical load q is equally distributedbetween the two webs, then their bending response will be obtained as

MB,ext¼ (1/2) MB(q)þMB(q)

MB,int¼ (1/2) MB(q)MB(q)

for the external and the internal web, respectively.

504


The compressive forces in the slab due to the bending momentsMB compel it to takeup ‘outward’ deviation forces, whereas the existence of compressive and — especially —tensile stresses in the webs create transverse distributed deviation forces which equili-brate those of the slab, causing at the same time transverse bending of the openframe of the section, as shown in Figure 14.21 (Menn, 1990).

14.3.4 The influence of prestressing in curved girdersAs was made clear in Chapter 4 on beams, the layout of prestressing tendons aims toprovide deviation forces that counteract the gravity loads for which the structure hasto be designed. In the case of curved girders, gravity loads indirectly create torsionalloads through the existing curvature, eventually being increased by their eccentriclayout on the bridge deck. Thus, it is useful to examine how the presence of prestressingin a curved box girder influences its torsional response and, moreover, to consider thepossibility of torsional relief through an appropriate tendon layout.Given that the tendons are arranged within the thickness of each web by using all of

its available depth, it is first pointed out that a cable, inclined with respect to the girderaxis, implies torsion (Figure 14.22). This is explained by the fact that the inclinedcompressive force on the section has a component in its own plane, which causes atorsional moment equal to the product of this component by its distance to the shearcentre of the section. The fact that, in the usual prestressing layouts, the prestressingdoes not induce torsion arises because the different torsional contributions of the

505

1

MB

MB

MB

MB mD = MB/R

D

D

Zex

Zex

Zin

b

ds

Change in the exterior web

q*

q* = MB/(R · b)

Transfer of the torsional loadto the webs

qD · ds

Deviation forces in web

dϕ

Deviation force in slab

qD = D/R

qZ,in = Zin/R

qZ,ex = Zex/R

Figure 14.21 Uptake of a torsional load by the webs of an open section

Box girders

cables cancel each other. However, an appropriate change in the cable inclinations mayinduce a torsional response that could relieve that arising from the loads.Of course, a cable lying in the top or bottom slab with an inclined position to the

girder axis may also cause a torsional moment with respect to the shear centrethrough the component of its compressive action on the section (see Figure 14.22).Moreover, in a curved beam, where the prestressing cable runs parallel to the shear

centre axis and is anchored at its two ends, no torsional loads are imposed, as theprestressing force does not offer anywhere a component on the corresponding sectionplane (see Figure 14.22).On the basis of the foregoing, it can be seen that in a curved statically determinate

girder an arbitrary tendon layout with a given prestressing force causes exactly thesame response (i.e. diagrams of sectional forces MB, Q and MT) as that developed inthe corresponding rectilinear (i.e. ‘stretched out’) girder. Why this happens is nowexplained.A statically determinate structure containing a prestressed tendon appropriately

anchored, if subjected only to the prestressing forces (anchor and deviation forces),

506

Application of torsional loadthrough deviation forces

The component of the compressive forcecauses torsion

If the compressive force acts perpendicularly to the section,no torsion is developed

P

PP

P

PS

P

Figure 14.22 Torsional response due to tendon inclination


does not develop any reaction at the points of support. Hence, any bending or torsionalmoment, or even shear force of a section, results from the prestressing force acting onit. The point of application and the direction of this force always, as explained inSection 4.3.1.1, correspond to the appropriate trace and local tangent of the tendon.If the component of this action on the section plane does not pass from the shearcentre, then a torsional moment is caused, as previously mentioned. Otherwise, notorsion is imposed. This situation remains unchanged if the initially consideredstraight girder becomes curved (Figure 14.23).Thus, while a rectilinear statically determinate girder, under, for example, its self-

weight, will develop torsion if it becomes curved, a prestressing tendon not producingtorsion in the rectilinear girder will continue to do so, even if the girder becomescurved. Accordingly, it may be concluded that in a curved statically determinategirder, if the prestressing deviation forces completely counterbalance the externalloads (e.g. self-weight), bending will be eliminated but torsion will not.It should, however, be noted at this point that the ‘bending moment’ MP due to

prestressing should not be perceived as creating (see Section 14.3.2) a torsional loadMP/R, just because the bending moment MP is not ‘genuine’, i.e. it is not directlyrequired for the equilibrium, as recalled in Figure 14.24.Of course, all the above is changed once the girder becomes statically indeterminate.

A curved continuous girder, for example, with a prestressing tendon that in the assumed‘stretched’ structure does not induce torsional moments develops a torsional response, asexplained below, and depicted in Figure 14.25.The continuous girder considered in Figure 14.25 has simple supports at both its ends

that are capable of developing torsional reactions, whereas it is simply supported in themiddle. The primary — statically determinate — structure obtained through removal ofthe intermediate support under the existing deviation and anchor forces does notdevelop any torsion, as already explained. However, the redundant vertical force thatmust be applied in order to eliminate the deflection occurring at the midpoint of the

507

Curving the beam induces torsionunder the applied loads

No torsiondeveloped

Curving the beamdoes not cause torsion

P P

Prestressing of astatically determinate beam

Figure 14.23 External loads and prestressing in a curved statically determinate beam

Box girders

primary structure does induce a torsional response. This externally imposed redundantforce, which differs very little from the corresponding reaction of the rectilinearcontinuous beam, provided that the values L/R as well as are small enough (seeSection 14.3.2.1), will induce a triangular bending moment diagram in the primary

508

Equilibrium requires the action of a moment No moment is required for equilibrium

g

u

P

P

Application of loads Application of prestressing

Figure 14.24 Prestressing in a statically determinate beam implies an internal moment

Torsional moment diagram for a prestressed curved girderdue exclusively to redundant prestressing moments [MSP]

[MT]

‘Shear force’ diagram

Torsional loading due to R Substitute transverse loading

Statically determinate structureIn the curved girder no torsion is developed

PLAN VIEW

PLAN VIEW

Deformation due to deviation forcesAn applied force R at the midspan is required

In the ‘stretched’ girder no torsion is developed

The force R on the curved girder causes torsion Bending diagram due to RStatically indeterminate prestressing moments

MSP/R MSP/R

[MSP]

R

R

R

Figure 14.25 Influence of prestressing on a curved redundant beam


structure. This represents the statically indeterminate prestressing moments [MSP], asexplained in Section 5.4.1. The bending diagram [MSP] applied to the curved staticallydeterminate girder will automatically produce the torsional loading [MSP/R]. Theimplied torsional response will then be obtained as the shear force diagram of thecorresponding rectilinear beam, due to the transverse triangular loading [MSP/R] (seeFigure 14.25). It is clear, then, that a prestressing tendon in a curved continuousgirder causes a torsional response solely due to the statically indeterminate prestressingmoments MSP of the corresponding rectilinear beam.Thus, according to the above and referring to Figure 14.26, the torsional moment

diagram [MT] of a curved statically indeterminate prestressed girder can be obtained

509

eP

P

P

R

R

R

f

u

Deviation forces

Bending diagram due to R(Statically indeterminate prestressing moments)

Torsional loading due to R

Substitute transverse loadingThe force R induces torsion

Deformation due to deformation forcesRequired force R = P · (f – e )/L

L

u

‘Shear force’ diagram

Torsional moment diagram for a prestressed curved girder due exclusively to [MSP]

[MT]

MSP/R

MSP/R

[MSP]P · (f – e )

MSP = MB – P · e

P · f

[MB]

Figure 14.26 Assessment of the influence of prestressing on a statically indeterminate curved girder

Box girders

on the basis of the bending moment diagram [MP] due to deviation and anchor forces(see Section 5.4.1), as follows. After the statically indeterminate moments (MSP¼MP P e) have been determined, where P and e are the prestressing force and thetendon eccentricity at each position, respectively, the torsional load [MSP/R] isapplied as a distributed vertical load on the statically determinate rectilinear girder.The resulting shear force diagram represents, then, the torsional moment diagram[MT] of the statically indeterminate curved girder.

14.3.5 Reducing the torsional response through prestressingThe distributed torsional action mD, or, more generally, the development of a torsionalmoment diagram [MT], in a box girder, may be reduced to a lesser degree throughprestressing (Menn, 1990).As has been stated previously, the presence of an inclination at some point on a

prestressing cable in a girder web principally implies a torsional action. If the cable isstraight, the torsional response developed is constant. If the cable is parabolic, theconstant deviation forces u induce a constant distributed torsional load: mD,P¼ u b/2(Figure 14.27).The arrangement of a straight cable in the top or bottom slab of the girder, in a

different direction to the girder axis, will also cause a constant torsional moment,while a parabolic cable, through its deviation forces, will impose a constant torsionalload mD,P¼ u hs, where hs is the distance from the shear centre (see Figure 14.27).It is clear, however, that arranging a parabolic prestressing cable in one web only

implies the imposition not only of a distributed torsional load but also of a distributedvertical load, which may be undesirable if only a torsional load mD,P is to be imposed.For this reason, another parabolic cable with opposite curvature should be arrangedin the other web (Figure 14.28). Analogously, in a horizontal cable layout, the girderslabs should be provided with cables of opposite curvature.The prestressing tendons can be specified separately in the first stage of design, to

counteract the bending and the torsional response of the girder. In a second stage,

510

P

u

uP

SPv

hs

bmD,P = u · b/2

mD,P = u · hs

Constant torsional moment: +Pv · b/2Constant torsional load due to thecorrresponding deviation forces

Figure 14.27 Creating torsional loading through separate prestressing cables


the synthesis of the two separate tracings into a single, final, one can be attempted, byretaining, if possible, the number of cables (i.e. the total prestressing force) required forbending. In this way, a differentiated tendon curvature can be obtained for both webs,which must ensure the required upward deviation forces, on the one hand, and offer therequired torsional counter-load, on the other.The above is illustrated by the example of the simply supported curved girder in

Figure 14.29. It becomes clear that the arrangement of a convex cable in the ‘outer’

511

u

uu

u

PP

P

P

PP

P

P

h

bmD,P = u · b mD,P = u · h

Figure 14.28 Basic layouts of prestressing cables for inducing torsional action

Significant decrease in torsion is accompaniedby an additional transverse response

Tendon profile in web 1

Tendon profile in web 2

Counteracting of torsion is achievedthrough a differentiated tendon profile

21

1

2

[MT]

[MT]

uexuin (uex – uin)/2

Differentiating the deviation forces

Prestressing the tendon does not induce any torsion

Aimed for torsional loading

g = 168 kN/m R = 50 m

2460 kN m MB/R

544 kN m/mPLAN VIEW

36.0 m

Figure 14.29 Relieving influence of the tendon layout on the torsional response

Box girders

512

Introduction of torsion

(uex – uin)/2

Transverse response

(Increase in response)


PRESTRESSING

SELF-WEIGHT

1 2

1 2

Figure 14.30 Influence of the prestressing layout on the transverse response of a girder

1

11

2

2

11

2

1 2

1 2

[MT]

[MT]

[MB]

[MT]

894 kN m

894 kN m 894 kN m

7173 kN m

Shearing of walls(Bredt)

928 kN m

MB/R

Introduction oftorsion

Differential shear flow(Bredt)

Transverseresponse

Span region

Support region

14 450 kN mg = 155 kN/m

R = 50.0 m

PLAN VIEW

26.0 m 26.0 m

Figure 14.31 Longitudinal and transverse response due to the self-weight of a girder


web and a concave cable in the ‘inner’ web will produce a torsional load counteractingthe torsional effect of the self-weight. Consequently, the final tendon profiles present astronger curvature for the outer web than for the inner one, both of them being, ofcourse, convex.If in the above case the deviation forces of the outer and inner web are uex and uin,

respectively (uex> uin), it can be seen that, while the symmetric total deviation force(uexþ uin)/2 counteracts the gravity loads, the resulting torsional load is due to theantisymmetric loading of each web with (uex uin)/2, thus being equal tomD,P¼ (uex uin) b/2. However, this torsional load will cause a correspondingperipheral shear flow, leading to an additional longitudinal and transverse response ofthe girder, according to the examination in Section 14.3.3 (Figure 14.30).

14.3.6 Summary examplesIt is appropriate at this point to recapitulate the complete torsional response of a single-span prestressed curved girder, according to the following four states of stress:

513


(uex – uin)/2


Transverseresponse

(Surcharging of section)1

1

1

2

1

1

2

1 2

2

1 2

[MT]

[MT]

[MT]

Transverse response uin uex


Web 1 Web 1

Web 2 Web 2

Deviation loads must conform in total to theinitial tendon profile

Aimed for torsional loading achieved througha differentiated tendon profile

PLAN VIEW

Figure 14.32 Influence of the differentiated tendon profile on the torsional response

Box girders

(a) The shear response of the section walls according to Bredt for gravity loads, due tothe existing curvature of the girder.

(b) The additional response due to the deformation of the section profile, on the basis ofthe differential shear flow in (a) and the deviation forces on the top and bottomslabs.

(c) The shear response of section walls according to Bredt, due to the torsional loadsinitiated by the eventually differentiated tendon profile in the webs.

(d) The additional response due to the deformation of the section profile, on the basis ofthe differentiated shear flow in (c) and the antisymmetric loading of the two webs bythe deviation forces of the tendons.

The application of the above in the case of the single-span girder examined, as inFigures 14.29 and 14.30, shows that the transverse response of the section due to thedifferentiated tendon profile in the webs increases that arising from the self-weight,resulting in greater longitudinal and transverse bending.It can be seen that in the case of a statically indeterminate girder the following states

of stress must, furthermore, be taken into account:

514



Transverseresponse

1

1

1

2

1

2

1 2

[MT]

[MT]


1

2

PLAN VIEW

The deviation forces must conform to thoseresulting from the differentiated tendon profile

Torsional moment diagram due to prestress

MSP/R

Statically indeterminate prestressing moments

[MSP]

[MT]

Figure 14.33 Influence of prestressing on the torsional response due to the static redundancy


515

Introduction of torsion Transverse response

(Relieving response)


PRESTRESSING

SELF-WEIGHT

1

1

2

2

1 2

Tendon layout in top slab

Tendon layout in bottom slab

Aimed for torsional loading

PLAN VIEW

Figure 14.34 Consequences of the arrangement of additional tendons in the top and bottom slabs

Tendon profile of bottom slab

Tendon profile of top slab

PLAN VIEW

Aimed for torsional relief

Torsional loading due to self-weight

u

u

Figure 14.35 Layout of additional tendons in the top and bottom slabs of a continuous girder

Box girders

(e) The shearing response of the section walls (Bredt), due to the torsional response onthe basis of the statically indeterminate prestressing moments MSP, according toSection 14.3.4.

(f ) The additional response due to the deformation of the section profile on the basis ofthe differential shear flow and the deviation forces of the top and bottom slabsoccurring in (e), according to Section 14.3.3, in the same way as in (b).

Figures 14.31—14.33 illustrate the treatment of a prestressed continuous curved girder,in order to reduce the torsional response due to the action of a permanent load as thegirder self-weight, following the above sequence:

. steps (a) and (b) are illustrated in Figure 14.31

. steps (c) and (d) are illustrated in Figure 14.32

. steps (e) and (f ) are illustrated in Figure 14.33. g

As previously explained, the torsional response due to permanent loads may clearly bereduced through the arrangement of additional tendons in the top and bottom slabs,which, by means of their deviation forces u, offer a relieving torsional momentmD,P¼ u h, as shown in Figure 14.34.The introduction of these horizontal deviation forces, together with the differential

shear flow they cause, lead, according to the foregoing, to a ‘diagonal’ state of stress,which drastically reduces that due to the permanent load (see Figure 14.34). It isclear that the above layout is much more effective for torsional reduction in acontinuous girder than the previously considered differentiation of the profiles of thetendons within the webs (Figure 14.35). However, this can be achieved only throughthe extra cost of additional tendons.The task of restricting the torsional response through prestressing must always be

undertaken for each case with regard to its technological—economic consequences,and taking account of the cost of the difficulties that will be encountered in therealisation of the prestressing layout.

ReferencesMenn C. (1990) Prestressed Concrete Bridges. Basel: Birkhauser Verlag.Schlaich J., Scheef H. (1982) Concrete Box-girder Bridges. Zurich: IABSE.Schlaich J., Seidel J. (1988) Die Fussgangerbrucke in Kelheim. Bauingenieur 63.

516


15

Lateral response of multi-storey systems

15.1 IntroductionIn this chapter the behaviour of a three-dimensional multi-storey building systemunder lateral forces is examined. These forces may result either through an imposedseismic ground motion, or as an incoming wind action, or even through a temperaturevariation which, as will be explained later, activates the lateral stiffness of the structuralsystem. The reason for making the consideration of temperature change an essentialdesign issue is not only because of the obvious environmental temperature influenceon the structure but also because of the fact that the shrinkage that the concreteslabs at each storey of the load-bearing system are subjected to, after the concrete hashardened, may be expressed through a temperature fall, thus causing a response ofthe whole system.

15.2 Formation of the systemMulti-storey systems are considered here as horizontal diaphragms (slabs) with very highrigidity within their own plane, and arranged vertically at a regular distance of a storeyheight and connected with each other as well as with the foundation ground throughvertical elements of rigidity, according to a particular layout plan. These verticalelements consist either of frames formed by the columns and the building girders, orof single columns or shear walls, or possibly of thin-walled elements having an openor closed section (Figure 15.1).It should be noted that the diaphragm concept used below is not necessarily restricted

to a solid concrete slab but can apply to any structural formation possessing a highstiffness in its own plane. The same naturally applies for all the vertical elements,which do not necessarily have to be made of concrete but can be formed of steelelements.Each of the horizontal diaphragms under the action of horizontal forces moves as a

rigid body within its own plane. This movement consists of a horizontal displacementas well as a rotation, and is obviously imposed unchanged on all vertical elements,causing a corresponding response (Figure 15.2).In the present examination, which, as previously pointed out, facilitates, in principle,

structural design rather than ‘accurate’ analysis, it may be assumed that all verticalelements, as the supporting skeleton of the diaphragms, constitute a set of elementsof purely plane stiffness (Stavridis, 1986).


By the term element of purely plane stiffness or simply plane element, it is meant that theelement is stressed only by that component of the diaphragm displacement which corre-sponds to its plane. Thus, for example, while the same displacement and rotation vectorof the diaphragm is applied to each girder of all frames in Figure 15.3, each frame is

518

Vertical frame

Vertical frame

Vertical frame

Vertical frame

Rigid diaphragm

Shear wall

Rigid diaphragm

Thin-walled element

Shear wall

Figure 15.1 Layout of horizontal diaphragms and vertical stiffness elements

P3

P2

P1

Diaphragm displacement3

2

1

Figure 15.2 Horizontal movement of the diaphragms


assumed to be stressed only by that component of the displacement vector corre-sponding to the specific frame plane.Although the bending stiffness of a frame with respect both to its own plane and

transversely to it can be described by two separate plane vertical elements, the clearpreponderance of the stiffness in its own plane allows the omission of the correspondingtransverse plane element. However, in a single column, two plane elements may beconsidered representative of its stiffness with respect to its two principal stiffnesssenses (Figure 15.4).Moreover, it may be understood that a vertical thin-walled element of either an open

or closed section may be represented by the plane elements constituting the wholesection. These plane elements are also able to offer the torsional stiffness of theelement, as any applied torsional load may be taken up by the plane state of stress ofthe constituent plane elements themselves (Figure 15.4).

15.3 Lateral response

15.3.1 Treatment of load-bearing actionThe basic problem may be stated in the following way.It is assumed that the diaphragm of the nth level is acted upon on its centroid by a

horizontal force P (Figure 15.5). The diaphragm, with reference to the orthogonal

519

Effective componentfor the plane element

Effective componentfor the plane element

Rigid diaphragm

Rigid diaphragm

Element ofplane stiffness

Element ofplane stiffness

Displacementvector of thediaphragm

Displacement vectorof the diaphragm

Figure 15.3 Activation of stiffness of the plane elements through the diaphragm


520

The two parallel walls cantake up the moment

The moment is taken up by thestiffness of the four plane elements

Three plane elements

Four plane elements

Plane element 1

Plane element 2The transverse stiffness

is ignored

Effective stiffness

Figure 15.4 Expressing the stiffness of vertical members through the elements of the plane stress

2

3

4

1

∆ω

∆x∆y

Px

y

O

Figure 15.5 Horizontal loading of a single diaphragm


axes Oxy having their origin O at the centroid of the diaphragm (i.e. its centre of mass),undergoes, through the action of the force P, horizontal movement as a rigid body,represented by the components of displacement x and y as well as by the rotation!, taken as positive if anticlockwise. This displacement is obviously resisted by thelateral stiffness of plane elements, as explained above.The lateral stiffness of each plane element is represented by the required horizontal

force S acting on a specific level n, in order to induce a corresponding unit displacement(1m) at that level. This force remains essentially the same, even if all the upper storeylevels are omitted, as shown in Figure 15.6. This fact, apart from its simplifying effect,has also a further convenient consequence, as will become clear later.The resistance of the plane elements to the diaphragm movement implies for each

element the action of a certain force at the corresponding level, resulting from theimposed displacements on them, while the same forces are returned in an oppositesense to the diaphragm itself. This must be in equilibrium under the above forces aswell as under the directly acting force P (Figure 15.7). Given that these ‘resistance’forces may be expressed — as will be shown later — in terms of the displacement compo-nents x, y and !, these components can be determined through the threeequations of equilibrium of the diaphragm. The above components thus allow theevaluation of the forces acting on each plane element at the specific level. However,it must be ensured that the diaphragm in question can safely take up the self-equili-brating coplanar forces that it is subjected to.The simultaneous loading of some or all of the diaphragms subjected to corresponding

forces P may be treated for each diaphragm separately, by neglecting all the above-lyinglevels of the plane elements involved, as has been previously indicated. Of course,the final response of each vertical element will be deduced from the total action ofthe corresponding forces determined at each level (Figure 15.8). These forces areused for the design of the plane elements and the assessment of their displacementsat the various levels, as well as for the response of the corresponding diaphragm.

521

The lateral stiffness S is practically unaffectedby the upper-lying levels

Stiffness S1 Stiffness S2 Stiffness S1 Stiffness S2

causesδ = 1 m

causesδ = 1 m

causesδ = 1 m

causesδ = 1 m

Figure 15.6 Lateral stiffness of a plane element


522

The acting forces remain the sameeven if the upper-lying levels are ignored

EQUILIBRIUM OF DIAPHRAGM

The diaphragm is stressed

Equal and opposite forces on the diaphragm

Acting forces on the plane elements(depending on ∆x, ∆y, ∆ω)

∆y ∆x ∆ω

P

Figure 15.7 Equilibrium of the loaded horizontal diaphragm

The forces at each level are due to theindividual action of ‘P ’ on the

corresponding diaphragm

4

3

2

1

P4

P3

P2

P1

Figure 15.8 Treatment of the loading of all diaphragms


15.3.2 Response of a plane elementEach plane element i is characterised by the angle i, described in an anticlockwise senseby the axis Ox with an arbitrarily determined positive direction on the trace of theelement in the ground plan. On this trace a characteristic point with coordinates xiand yi is selected (Figure 15.9).By considering a specific level i of an element where the displacement components

x, y and ! of the corresponding diagram are imposed, the corresponding dis-placement Di for the element may be determined, by assuming that the diaphragm, asalready pointed out, remains undeformable during its movement as a rigid body:

Di¼x cosiþy sin iþ! eiwhere ei is the (signed) distance of the trace of the element from the origin of thecoordinate axes O, being equal to

ei¼ xi sini yi cosi

In the level under consideration, the element i may be structurally represented by thecorresponding force S that must be applied in order to produce a unit displacement(1m), known also as the lateral stiffness. This magnitude can be determined by calcu-lating the displacement (m) due to a unit horizontal force (1 kN) applied to thatlevel (Figure 15.10). Then,

S¼ 1/ (kN/m)

As has already been pointed out in Section 15.3.1, neglect of the overlying levels doesnot affect the above displacement in practice, or, in consequence, the correspondingstiffness (see Figure 15.10).Thus, in order for the plane element i to contribute to the imposed diaphragm

displacement at the level under consideration, a horizontal force F¼ Si Di must be

523

∆y

∆x(negative)

xO O x

Element i Element i

Characteristic point(arbitrary)

βiyi

xi

ei

∆ω

y y

Di

Figure 15.9 Geometric and kinematic characteristics of the plane element


imposed on it, i.e.

F¼ Si (x cosiþy siniþ! ei)

15.3.3 General layoutIf the diaphragm of a specific level is subjected to a horizontal force P applied to itscentroid and represented by the components Px and Py, then it may be found fromits equilibrium, as previously considered, that the components x, y and ! of itsdisplacement must satisfy the following three-equation system:

K1 xþK2 yþK3 !¼ Px

K2 xþK4 yþK5 !¼ Py

K3 xþK5 yþK6 !¼ 0

where

K1¼P

Si cos2i, K2¼P

Si sin i cosi, K3¼P

Si ei cos iK4¼

PSi sin2i, K5¼

PSi ei sini, K6¼

PSi e2i

Once the components x, y and ! are known, the horizontal forces acting at thecorresponding levels for all plane elements can be readily determined, as described inSection 15.3.2 (Stavridis, 1986).The stiffness Si at any level may be thought of as a horizontal spring in the direction of

the corresponding element, which may be connected to the diaphragm at the position ofthe corresponding characteristic point of the element (Figure 15.11). In this way, thediaphragm acted upon by the force Pmaintains its equilibrium because of the developedspring forces acting on it, while the same forces are also acting on the correspondingplane elements.

524

Practically both displacements are equal so thatthe corresponding level stiffness remains unaffected

1 kN δ: m F D1 kN δ: m

StiffnessS = 1/δ

Requiredforce

Induceddisplacement

F = S · D

Figure 15.10 Determination and evaluation of the translational stiffness of a plane element


Alongside the above considerations, it is also useful to clarify the structural signifi-cance of the coefficients K1, K2, K3, K4, K5 and K6 in the above system of equations.This group of coefficients characterises the resistance offered by the system ofplane elements against the horizontal movement of the diaphragm. More specifically,the terms K1 xþK2 yþK3 !, K2 xþK4 yþK5 ! and K3 xþK5 yþK6 ! represent, respectively, the forces Px and Py, and the torsionalmomentM! that must be applied at the originO in order to cause a diaphragm displace-ment having the componentsx,y and!. In other words, the above terms representthe resistance offered by the diaphragm at point O if this is ‘obliged’ to be displaced byx,y and!. It is, of course, understood that specifying a ‘point of application’ for thetorsional moment M! does not make much sense, as the moment may be appliedanywhere in the diaphragm. However, the null value that is assigned to this momentin the above equation system is due to the fact that the displacement componentsx, y and ! are realised simply through the application of a single force P atpoint O, i.e. without imposing any moment M!. If for any reason the diaphragm isacted upon by an external moment M!, then this value — positive for an anticlockwisedirection — must be used instead of zero.From the above equations it can be deduced that the plane elements may be ‘shifted’

anywhere along their own trace, as only their corresponding angle i and their distance eifrom the origin of the coordinate axes matter. This may also be understood through theconcept of spring supports introduced above, which, owing to the absolute rigidity of thediaphragm, may be connected at any point along their line of action.

15.3.4 Orthogonal layoutThe vertical elements in building structures are usually arranged in mutually orthogonaldirections. In this case, the axes Ox and Oy are placed parallel to these two directions(Figure 15.12). Then, all the elements in the x sense have i¼ 0 and ei¼ yi, whereasall the elements in the y sense have i¼ 908 and ei¼ xi.Thus (see Section 15.3.3),

K1¼P

Sx, K2¼ 0, K3¼P

(Sx yi)K4¼

PSy, K5¼

P(Sy xi), K6¼

P(Sx y2i )þ

P(Sy x2i )

525

Characteristic pointof plane element

Ox

yS4

S2

S1

S3

Centroid

Plane elements (spring) may ‘slip’ on their trace

P∆ω

∆x∆y

Figure 15.11 Visualisation of the support of the loaded diaphragm through the corresponding springs


while x, y and ! are determined from the following expressions:

x ¼ PxK1

K3

K1

K3

K1

þ K5

K4

K23

K1

þ K25

K4

K6

; y ¼PyK4

K5

K4

K3

K1

þ K5

K4

K23

K1

þ K25

K4

K6

! ¼

K3

K1

Px þK5

K4

Py

K23

K1

þ K25

K4

K6

For the forces acting on the elements in the x direction at the level under consideration(see Section 15.3.2),

Fx¼ Sx (x! yi)whereas for the elements in the y direction the forces acting are

Fy¼ Sy (yþ! xi)The magnitudes of K3 and K5, i.e.

P(Sx yi) and

P(Sy xi), respectively, play an

important role. If both are zero, the diaphragm does not rotate at all (!¼ 0), andx and y are simply

x¼ Px/P

Sx, y¼ Py/P

Sy

This result may also be interpreted and evaluated according to the spring model of therigid diaphragm (Figure 15.13).A force Px acting on the diaphragm at point O in the x direction, such that it causes

no rotation but only a displacement , must lie on the line of action of the resultant ofthe spring forces (Sx ), which means that

P(Sx ) y¼ 0, i.e.

P(Sx y)¼ 0.

Moreover, for equilibrium of the diaphragm, Px¼P

(Sx )¼ (P

Sx), as hasbeen exactly obtained before. For this examination of the behaviour in the x sense, arigid beam may substitute for the diaphragm, as shown in Figure 15.13. Analogousresults are also obtained for the y sense.

526

PLAN VIEW

Centroid

S1

S2

S3

S5

S7

S6S4

O

P

y

x

Figure 15.12 Orthogonal layout of the plane elements


In this way, for an arbitrary displacement imposed in the x direction, the specificline of action of the resultant of the spring forces (Sx ) may be readily determined,while for another imposed arbitrary displacement in the y direction, the springforces (Sy ) allow the determination of the line of action of their own resultant(see Figure 15.13). It is obvious that the position of the above resultants is independentof the value , and hence for any force passing through the intersection of the aboveaction lines, the diaphragm will not rotate but only shift in the direction of the actingforce. This point, which is known as the centre of stiffness or the centre of resistance,may be paralleled to the shear centre of thin-walled sections (see Section 13.3).It is now clear that a zero value for the magnitudes K3 and K5, i.e. of

P(Sx yi) andP

(Sy xi), means that the pointO — representing in practice the centre of gravity of thevertical element compressive forces — coincides with the centre of stiffness.It should be made clear that all the foregoing equations are valid, independently of

whether the point of application of the load P on the diaphragm — which may be itscentroid — coincides or not with the centre of stiffness. However, it is clear that apossible application of a load P at a point different from the centre of stiffness meansthe action of a certain moment on the diaphragm, which must be balanced by thespring forces, i.e. the resistance forces offered by the plane elements. Moreover, thespring (plane element) layout must ensure the uptake of an arbitrary load P acting inany direction.The above remarks are illustrated in Figure 15.14.

527

PLAN VIEW

S5

S1

S6S4

S3

S2

S7

S6

S5

S7

S1 S2 S3 S4

PO

x

y

Centre of stiffness

Resultant

S5 · ∆

S6 · ∆

S7 · ∆

∆

∆

(S1 + S2) · ∆

S3 · ∆

S4 · ∆Res

ulta

nt

Rigid beam

The horizontal force P acts on the diaphragm’s centroid

If the force P does not passthrough the centre of stiffness

a diaphragm rotation is developed

Figure 15.13 Simulating the diaphragm behaviour through a spring-supported rigid beam


In case (a) the x component of the load P cannot be balanced by the spring forces, andconsequently the layout is unsafe. In case (b), although springs, i.e. plane elements, areavailable to the diaphragm for the uptake of any component of the load P, these areunable to provide a couple of forces, in order to balance the moment of the load withrespect to the centre of stiffness, and therefore the system may easily collapse.However, in case (c) the developed spring forces can balance any component of theacting load, as well as the moment of the load P with respect to the centre of stiffnessthrough the offered forces S1 and S3.It is now clear that, while the point of application of the load P is generally determined

by the diaphragm itself, the centre of stiffness is dependent on the selected layout of thesprings, i.e. the plane stiffness elements.It must be pointed out that selecting a layout of vertical elements — frames,

shear walls, etc. — that exhibits small or zero values of the magnitudesP

(Sx y)and

P(Sy x) has a definite advantage over a layout where the point of application

of load P, i.e. the centre of gravity of the diaphragm, is some distance from the centreof stiffness, thus causing a rotation ! of the diaphragm. This rotation is obviouslythe same as that occurring in each of the rigid beam models in Figure 15.13. As canbe seen from the above expressions, the rotation ! causes a non-uniform distri-bution in the response of the plane elements, depending on their position in theground plan (for a generalised treatment of the problem see Section 16.3.1.4).

15.4 Temperature effect

15.4.1 Treatment of load-bearing actionAny temperature variation in a diaphragm always causes a certain stress state on thevertical elements. The reason for this is that a dilatation or a shrinkage of a diaphragmalways imposes displacements at the corresponding level of the plane elements, whichare consequently stressed (Figure 15.15).More specifically, assuming that a diaphragm is subjected to a temperature increase of

T by retaining its centre of mass O absolutely fixed, it can be seen that some radialdisplacement will be imposed on all the vertical plane elements, which will lead to acertain shift along their trace (Figure 15.16).

528

(a) (b) (c)Centre of stiffness Centre of stiffness

Inapplicable layout Inapplicable layout Applicable layout

Centre of gravity Centre of gravity Centre of gravity

S1

S2

S1

S2S3

S1

S2S3 S3

P P P

S4

Figure 15.14 Criteria for the right or wrong layout of the plane stiffness elements


Due to this imposed displacement, the plane elements will offer resistance at theircorresponding level that will be acted on by the diaphragm, being obviously equal andopposite to that acting on the plane element itself. It is clear that the diaphragmunder the above forces will generally not be in equilibrium as a free body and, therefore,at the point of fixation O, a specific force and a moment have to be applied. The actionof these forces ensures the fixation of point O under the temperature variation of thediaphragm (Figure 15.16). However, given that, in reality, no external actions areapplied at point O, equal and opposite actions to those above must additionally beapplied at this point, in order to restore the real situation. Of course, the last actionson the ‘free’ system will cause displacement of the diaphragm with the componentsx, y and !, which in turn will cause an additional stress state on the planeelements, through the corresponding horizontal forces acting on their correspondinglevels (Figure 15.17).The final response of the vertical elements will result as a superposition of the two

above stress states (Stavridis, 1994), namely:

529

PLAN VIEW

δ ?

δ ?Temperature riseof the diaphragm

Figure 15.15 Displacements of the plane elements due to a temperature variation in the diaphragm

A B A B C D∆T

δ δ

(Rise of temperature)

Fixed centroid

Fixation actions

O

A

C

D

B

δ δ

δ(–PωT)

(–PxT)

(–PyT)

Plan view

FIRST STAGE

O

Figure 15.16 Temperature variation under fixation of the centroid of the diaphragm


. First stage. The state of stress resulting from the free dilatation of the diaphragm withits centroidO fixed through externally applied actions at this point (see Figure 15.16).This stress state is due to the imposed displacements on the specific level of the planeelements because of the radial dilatation of the diaphragm on the one hand and thetemperature variation itself imposed on the corresponding girder of the plane frameon the other hand.

. Second stage. The state of stress resulting in the — now — free system, through theimposition at the point O of the equal and opposite actions of the previous stage,which leads to the final displacement of the diaphragm, described by the componentsx, y and ! (see Figure 15.17).

15.4.2 General layoutAccording to the foregoing examination, and on the basis of Sections 15.2 and 15.3,it may be deduced that, for the specific level considered, the horizontal displacementDi of the ith plane element referred to a characteristic point (xi, yi) selected at themiddle of its trace may be expressed as follows, by conveying the influence of the twosuperposed states:

Di¼ [ i T T]þ [x cosiþy siniþ! ei]where i¼ xi cos iþ yi sini, and the temperature coefficient T¼ 10—5/8C.The temperature variation T is considered positive if the temperature of the

diaphragm increases, and vice versa.Thus, the force F acting on each plane element at the level examined is

F¼ Si Di¼ Si T i Tþ Si (x cos iþy sin iþ! ei)

530

(PωT)

(PxT)

(PyT)

Equal and oppositeto the fixation actions

SECOND STAGE

The induced element actions at the corresponding levelare superposed on those of the first stage

Plan view

Developed: ∆x, ∆y, ∆ω

Imposition of effective actions on the free diaphragm

O

Figure 15.17 Imposition of the opposite fixation actions to the free system


The determination of x, y and ! is by the solution of an analogously structuredsystem of equations as in the previous case of lateral loading:

K1 xþK2 yþK3 !¼ PTx

K2 xþK4 yþK5 !¼ PTy

K3 xþK5 yþK6 !¼ PT!

In the present case of temperature influence, however, the right-hand variablesrepresent — with opposite signs — the required external actions at the point O, tocomply with the aforementioned first stage:

PTx ¼T T P

(cosi Si i)PTy ¼T T

P(sini Si i)

PT! ¼T T P

(Si i ei)

15.4.3 Orthogonal layoutIn the usual case where the plane elements are arranged in an orthogonal layout, theabove expressions are simplified as follows.For the plane elements in the x sense, i¼ xi and ei¼ yi, whereas for the elements in

the y sense, i¼ yi and ei¼ xi.Then,

PTx ¼T T P

(Sx xi)PTy ¼T T

P(Sy yi)

and

PT! ¼T T P

(Sy xi yi Sx xi yi)while x, y and ! are determined as

x ¼ T T P

ðSx xiÞK1

K3

K1

K3

K1

þ K5

K4

K23

K1

þ K25

K4

K6

y ¼ T T P

ðSy yiÞK4

K5

K4

K3

K1

þ K5

K4

K23

K1

þ K25

K4

K6

! ¼ T T

K3

K1

X

ðSx xiÞ þK5

K4

X

ðSy yiÞ þX

Sx xi yi Sy xi yi

K23

K1

þ K25

K4

K6

531


According to the expression for Di in Section 15.4.2, the elements in the x and y sensesare subjected at the examined level to the following forces, respectively:

Fx¼ Sx Di¼ Sx (xi T Tþx! yi)Fy¼ Sy Di¼ Sy (yi T Tþyþ! xi)

It is clear, however, that the consideration of an applied temperature difference at morethan one diaphragm simultaneously may be treated by superposing the respective resultsfor each separate level, exactly as in the case of lateral loading examined in Section15.3.1. Thus, each plane element will be subjected to a corresponding force at eachspecific level where the temperature variation is applied.From the above equations, it may be concluded that, contrary to what happens in the

case of horizontal loading, the specific position of each plane element in the ground planplays its own role in the development of the response, so that the element may not ‘slip’along its axis.The symmetry in the ground plan with respect to either the x or y axis simplifies the

above expressions, as the coefficients K3 and K5 have null values.The existence of a double symmetry for the axes Ox and Oy for the stiffnesses Sx and

Sy, respectively, means that K3¼K5¼ 0, and leads to an important simplification:x¼y¼!¼ 0. Hence, the plane elements in the x and y senses are acted uponby the forces Fx¼ Sx xi T T and Fy¼ Sy yi T T, respectively.The forces Fx and Fy must be taken into account for all cases during the design of the

arrangement of the vertical elements — frames, shear walls — of a building structure.They meet precisely the criterion for the maximum building dimensions with regardto a monolithic construction, in order to withstand not only possible climatic variationassociated with temperature change but also the ever-present concrete shrinkage of thediaphragms (slabs), being equivalent to a temperature fall of about 258C. The typicallyconsidered ‘maximum length’ of 30m, where the ‘dilatation joints’ are normallyprovided, must clearly always be adjusted for, depending in each case on the planestiffness elements themselves, as well as on their specific layout in the ground plan.It is thus clear that the design of the vertical members of a multi-storey building

structure, apart from ensuring the safe uptake of any horizontal action either asseismic or wind loading, must also cover the ‘temperature variation’ of the diaphragms.

ReferencesStavridis L. (1986) Static and dynamic analysis of multistorey systems. Technika Chronika Scientific

Journal of the Technical Chamber of Greece 6(2), 187—219.Stavridis L. (1994) Beanspruchung von mehrstockigen Bauten infolge Temperaturanderung.

Bauingenieur 69(3), 117—122.

532


16

Dynamic behaviour of discretemass systems

16.1 IntroductionThe static loads, which all load-bearing structures have to take up, may be considered ascompletely independent of the structure itself, as far as its deformation is concerned.This uptake of static loads is realised through the stiffness of the structure, dependingon its geometric and material characteristics. The closed process (load action) !(uptake through stiffness) is accomplished through the process of the realisation ofequilibrium, i.e. of the immobility of the structure. In this static process the mass (m)of the structure obviously remains inactive. It is activated, however, as a factor ofgenerating loads according to Newton’s second law of motion (F¼m ), from themoment that the structure is somehow set in motion. There are many factors thatcan set the structure — which is fixed to the ground — in motion, such as wind forces,earthquake ground movements, ‘sudden’ load applications, and moving loads (e.g.vehicles or pedestrians on roads or footbridges, respectively), etc.

The problem created through the motion of the mass of the structure is that it impliesaccelerations () leading, according to d’Alembert’s principle, to inertial forces equal tom . The structure is thus stressed under these loads, which may clearly be treated asstatic ones in the usual manner. However, the problem lies in the fact that the aboveaccelerations, which vary in time, are unknown from their onset. In fact, they dependon the masses and the stiffness of the system, and to determine their variation inevery localised mass of the structure and at every instant of time constitutes adynamic problem that is complicated from the onset of the motion.

Thus, the whole dynamic problem may be described as follows. The load-bearingsystem consists of specific masses that are connected to each other as well as to theground through a massless structural web, which, apart from its own geometric layout,exhibits specific stiffness characteristics (Figure 16.1). When the structure is set inmotion, the inertia forces of the masses (m ) on the one hand and the elasticforces of the structural web on the other are activated, depending on its stiffnesscharacteristics and acting on the same masses. The solution to the problem of thedetermination of the accelerations ‘’ of the masses results from the requirement ofequilibrium for each mass under the action of the inertia as well as the elastic forces.However, the difficulty in determining the motion of the masses arises from the factthat, while the elastic forces are generally proportional to the displacements of the


masses, the inertia forces are proportional to the second derivative of these displace-ments with respect to time.

The structures constitute a priori continuous systems in the sense that their mass iscontinuously distributed and not concentrated at discrete locations. Nevertheless,lumping the mass on the basis of behavioural criteria is always attempted for selectedlocations, as the analytical treatment of the continuous mass distribution is mathemati-cally much more cumbersome. g

The possible motions of the discrete (lumped) masses generally determine the inertiaforces they are subjected to, and from this point of view the latter are distinguishedbetween translational and rotational inertia forces. In the present examination, which is,however, oriented to preliminary design needs, the inertia forces due the rotation ofthe masses will be neglected, except for the horizontal diaphragms of multi-storeybuildings undergoing a rotation !, as explained in Section 15.2.

Thus, in the plane frame shown in Figure 16.2, in order for a lateral excitation toapply, its masses are assumed — approximately — to be concentrated at each storey,only their horizontal displacement is recognised, and their rotations about an axisperpendicular to the plane of the frame, as well as their vertical displacements, areneglected as a source of inertia forces.

Similarly, in the continuous beam shown in Figure 16.2, where a transverse excitationis encountered, its mass is considered concentrated at specific points, and only thevertical displacement of these points is recognised, their rotations about their ownaxis being neglected as a source of inertia forces.

Finally, in the two-storey spatial system shown in Figure 16.2, where, a horizontalseismic excitation occurs, the mass is considered equally distributed in each horizontaldiaphragm, and only the horizontal displacement components (x, y), producinginertia forces in the x and y directions, respectively, as well as the rotation !, producingan inertia moment at each storey, are recognised.

534

Structural web(developing elastic forces)

Mass(developing inertia forces)

m1 m1 · γ1

m2m2 · γ2

γ1

γ2

Base motion (earthquake)

Figure 16.1 Masses in motion and the structural web


In this way, each system is characterised on the basis of its effective displacements,which are usually referred to as degrees of freedom. Thus, in the above examples, theframe, the continuous beam and the two-storey system have, respectively, three, fourand six degrees of freedom.

Despite the fact that load-bearing structures usually exhibit more than one degree offreedom, the study of the single-degree-of-freedom system is absolutely crucial in orderto understand all the essential characteristics and dynamic behaviour of multi-degreesystems. It is necessary, then, to begin an elementary study of the dynamic behaviourof a single mass with only one ‘active’ displacement, given also that this model appliesdirectly to the practical treatment of composite systems, particularly those with acontinuous mass distribution, as shown in the last two sections of this chapter.

16.2 Single-degree-of-freedom systems

16.2.1 Dynamic equilibriumA mass m connected to the free end of a cantilevered bar of length L will be used as anexample of a single-degree-of-freedom system. The bending stiffness of the bar is muchless than its axial stiffness, and therefore the only possibility for displacement of the massis transversely to the axis of the bar. Consequently, the stiffness k (kN/m) of the bar is

535

Structural web

Structural web

Structural web

Structural web

Mass

Mass

Mass

Mass

Effective motion possibilities(degrees of freedom)

Figure 16.2 Possible effective mass displacements (degrees of freedom)

Dynamic behaviour of discrete mass systems

considered with respect to the displacement of its free end (Figure 16.3), and is equal tok¼ 3 EI/L3.

It is assumed that the mass is acted upon by a horizontal force function F(t) and thatat a specific moment t1 it undergoes a displacement u(t1) in the same direction.

Moreover, it is assumed that the mass receives an additional resistance to its motiondue to what can be considered the ‘internal friction’ of the material of the bar, which is atfirst taken as proportional to the velocity of the mass _uuðtÞ, and thus equal to c _uuðtÞ. Thecoefficient c is called the damping coefficient, as by its nature it tends to dampen themotion of the mass.

Thus, at the time t1 the mass m, according to d’Alembert’s principle, may beconsidered in equilibrium under the following forces:

. the external force F(t1) in the same direction as u(t1)

. the elastic force k u(t1) having a sense opposite to that of u(t1)

. the d’Alembert’s inertia force m €uuðt1Þ with a sense opposite to that of u(t1)

. the damping force c _uuðt1Þ with a sense opposite to that of u(t1).

The static equilibrium of the mass requires (see Figure 16.3)

m €uuðt1Þ þ k uðt1Þ þ c _uuðt1Þ ¼ Fðt1ÞIt should be pointed out that the elastic force k u(t1) coincides with that needed to shiftthe mass m by u(t1), according to the stiffness of its structural web.

16.2.2 Free vibration

16.2.2.1 Undamped vibrationAs the dynamic characteristics of the system are really independent of the externalloading, their investigation is made easier if the force F(t1) is removed, and thus thefree vibration of the mass m is considered after an initial arbitrary displacement isimposed at the rest position. By temporarily neglecting the ‘damping’ factor, it may bewritten for any time t1 that

m €uuðt1Þ þ k uðt1Þ ¼ 0

536

Static equilibrium of mass

1

Stiffness ofstructural web

mm

k = 3 · EI/L3

u(t1)

F(t1)

c · u(t1)k · u(t1)

m · ü(t1)

k · u(t1)c · u(t1)

Figure 16.3 Dynamic equilibrium of a single mass in a cantilevered structural web


For the determination of the displacement function u, the kinematic conditions of massm at the start time are needed, i.e. its displacement u0¼ u(t¼ 0) and its velocity_uu0 ¼ _uu(t¼ 0). The solution of the above equation shows that the mass oscillatescontinuously in a periodic manner about the ‘rest’ position with a deviation u(t) thatvaries sinusoidally with respect to time (Figure 16.4).

The time needed for a full oscillation of the mass, after which the motion isrepeated in the exact same way, is called the natural period T (s). The inversemagnitude f of the period is called the frequency, and f¼ 1/T (1/s). The frequency frepresents the number of full oscillations that take place within a second, the unit ofthis quantity being the Hertz. It is common in vibration studies to also use the termcircular frequency instead of ‘frequency f ’, denoting the magnitude !, on the basis ofthe relation:

f¼!/2 p

The concept of frequency, or the natural frequency, of a system constitutes a basiccharacteristic of the system, directly influencing its dynamic behaviour towards the

537

An increase in the stiffness of the structural web or of the massimplies an increase or decrease in the eigenfrequency, respectively

ümax

ümax

ümax = A · (2π · f )2 = A · (k/m)

(u = 0)

AccelerationDeceleration


umax umax = A · (2π · f ) t: s(u = 0)

ümax

f = 1/T = 1.25 Hz 2π · f = √k/m

A A

A

A A A A

T = 0.80 s

u

umax = A

00.20 1.201.000.800.600.40

0 0.20 0.40 0.60 0.80 1.00 1.20 s

k · A k · A k · A k · A

m · ümax m · ümax m · ümax m · ümax

Equilibrium of mass in characteristic positions

Figure 16.4 Free vibration of a single mass


imposition of external forces. It is found that

! ¼ffiffiffiffik

m

r(rad/s)

and, consequently,

f ¼ 1

2 p ffiffiffiffik

m

r(1/s ¼ hertz)

It should be noted that the mass unit used for structures is generally the tonne (t), whichis equal to 1000 kg: 1 t¼ 1 kN s2/m, and, given a weight W in kN, its mass is

m¼W/9.81 t

As seen from the last equation, the larger the mass, the fewer the vibrations that occur in1 s, whereas the stiffer the connection of the mass with the ground — i.e. the structuralweb — the higher the frequency of the system. However, the presence of the square rootattenuates the effect of the above parameters.

Of course, the frequency f of the system is independent of both the initial deviation u0

and the initial velocity _uu0 of mass m.The mass m executes a steadily varying, oscillating motion (see Figure 16.4). Its

displacement u(t) is expressed as

uðtÞ ¼ _uu0

! sin! tþ u0 cos! t

Its maximum deviation A on both sides, also called the amplitude, is found to be

A ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu2

0 þ_uu0

!

2s

At these positions, u ¼ A, the mass has zero velocity and maximum acceleration (ordeceleration), €uumax ¼ A !2, whereas at the ‘rest position’ (u¼ 0) it has the maximumvelocity, _uumax ¼ A !.

Moreover, it should be pointed out that the elastic forces k u(t1) have the oppositesense to the corresponding d’Alembert forces m €uuðt1Þ, as is implied by the basicequation itself. More specifically, as the inertia force m €uuðt1Þ is opposite to theacceleration vector €uuðt1Þ, just before the maximum deviation A, the inertia force isopposite to the maximum deceleration developed €uumax, whereas just after this positionthe inertia force is opposite to the equal acceleration €uumax. This means that in bothcases the inertia force is directed to the right, i.e. in the opposite direction to theacting elastic force (k A) (see Figure 16.4) and, finally, from the equilibriumequation m €uumax¼ k umax¼ k A, it can be checked that €uumax ¼ A !2. g

It should be made clear that the selection of the cantilever-type single-mass system isonly indicative of a system with mass m and stiffness of structural web k, subjected toan external force F(t), with respect to its active stiffness k. A system consisting of, for

538


example, a mass connected to a spring, as shown in Figure 16.5, would be equallyrepresentative, provided, of course, that the self-weight of the mass m is ignored.

However, it should be recognised that while the cantilevered mass is offered as a moreillustrative model for the examination of systems consisting of a number of massesarranged in height, as in, for example, buildings or bridges, the mass—spring model ismore representative of the examination of vertical dynamic actions on existingstructures such as beams or plates, as in, for example, a machine permanently fixed inplace. The behaviour of the mass—spring model should thus be recognised as analogousto the cantilever system, as shown in Figure 16.6.

16.2.2.2 Damped vibrationConsideration of the damping coefficient c in the equation

m €uuðtÞ þ k uðtÞ þ c _uuðtÞ ¼ 0

does not alter the periodicity of the oscillation of the mass; however, it does result inthe continuous decrease of the amplitude in each vibration cycle. It should be notedthat the system vibrates with a somewhat increased period Td, which remains constant(Figure 16.7). It is found that the correspondingly decreased circular frequency !d isequal to

!d ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi!2

c

2 m

2s

At this point it is useful to clarify some facts concerning the damping, in order to betterunderstand its treatment. Damping is the loss of energy in each vibration cycle. Thisenergy is absorbed by the internal friction in the structure carrying the masses, i.e.the structural web, which gives stiffness to the system. Thus, the damping forcec _uuðtÞ is acting on the mass through the structural web, as a restraining force in the

539

c · u(t1)

c · u(t1)

k · u(t1)

k · u(t1)Stiffness ofstructural web

m

m

Static equilibrium of mass F(t1)

m · ü(t1)

1

ku(t1)

Figure 16.5 Dynamic equilibrium of a mass carried by a spring


same direction as the acting elastic force. Of course, the direct result of damping is, aspreviously mentioned, a continuous decrease in the vibration’s amplitude.

The description of damping as a force proportional to the velocity of the mass is notsatisfactory from a physical point of view. Nevertheless, it permits — as will be shownlater — a practical and effective quantitative description of the phenomenon, becauseof its convenient mathematical treatment.

Thus, it is found that if the damping coefficient c achieves the critical valueccr¼ 2 m !, then no vibration can be developed, and the mass moves from itsposition of maximum deviation directly to its resting point. Although such a situationis impossible in a structure, its consideration is nevertheless useful because it representsa reference level for the applied damping coefficient in each case.

First, the so-called damping ratio is defined, as the ratio of the damping coefficient toits critical value: ¼ c/ccr. On the basis of this equation,

¼ c/(2 m !)

and, consequently,

!d ¼ ! ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1 2Þ

q

540

ümax

ümax

ümax

(u = 0)



umax umax t: s(ü = 0)

ümax

A A

A

u

umax = A

00.20 1.201.000.800.600.40

0 0.20 0.40 0.60 0.80 1.00 1.20 s

A

A

k · A

k · A

k · A

k · A

m · ümax

m · ümax

m · ümax

m · ümax

Equilibrium of mass in characteristic positions

Figure 16.6 Free vibration of a mass—spring system


The ratio , despite the fact that it refers to the unrealistic concept of the coefficient c,has an essential physical meaning, something that may not only be deduced from the lastequation but also from the fact that between two consecutive amplitudes of the decayingvibration, with a ‘time distance’ Td,

¼ lnuðtÞ

u ðtþ TdÞ¼ c

2 m 2 p!¼ 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffi

1 2p

The above equation is of particular practical importance, because, from the measure-ment of the consecutive amplitudes of a damped vibration of a system exhibiting aconstant ratio, the value of may be directly deduced. As this value, for structuralsystems encountered in practice, does not usually exceed 10%, it may be consideredwith a good approximation to be

!d¼! (hence Td¼T)

on one side and

¼ 2 p on the other.

In real situations, however, for practical reasons as well as for accuracy, the amplitudesof two vibration cycles are measured as having a time distance rather equal to n T, in

541

The presence of damping ζ < 10% affects the oscillation period very little

Damping coefficient ζ = ln(λ)/2π λ = e2πζ

The ratio λ of two consecutive amplitudes is constant:(A1/A3) = (A2/A4) = λ

(Practically: f = √k/m /2π)

0 t : s

A

u

A · e–2π · f · t

A · e–2π · f · t

A2 A4

A3

A1

Td Td Td

Figure 16.7 Characteristics of damped vibration


which case (Figure 16.8)

n ¼ lnuðtÞ

uðtþ n TÞ ¼ n ¼ 2 p n

Hence,

¼ n2 p n

Finally, the expression of the displacement u(t) of a mass m, which, after being displacedby A, is left to vibrate freely with zero initial velocity, under a damping ratio (!d!), is

uðtÞ ¼ A

e ! t ð sin! tþ cos! tÞ

16.2.3 Forced vibrationThe aim of the above analysis of free vibration was to depict the basic dynamic charac-teristics of a single-degree-of-freedom system. An understanding of these characteristicsis necessary to understand the dynamic response of the system if the mass is acted on by aforce varying in time, i.e. exactly as was initially considered, according to the equation ofmotion (Muller and Keintzel, 1984):

m €uuðtÞ þ k uðtÞ þ c _uuðtÞ ¼ FðtÞwhere, after the introduction of the damping ratio ,

c¼ 2 m !It is useful to express the force F(t) in the form F(t)¼ F1 f(t), where f(t) describes thevariation of the force during the whole period of the time examined, while F1 representsthe maximum value attained by this force. For the design of a structure, the response of

542

(A1/A2) = (A2/A3) = (A3/A4) = ... = λζ = ln(A1/A4)/(2π · 3)

0 t : s

u

Td TdTd TdTd

3*Td (n = 3)A1

A2A3 A4

Figure 16.8 Practical determination of the damping ratio


the structural web connecting the mass with the ground is of primary interest. This responsedepends only on the displacement u. It is understandable that the maximum responsecorresponds to the maximum displacement umax, which is clearly larger than thatcorresponding to the ‘static’ loading of the system with the force F1, this being equal to

ustat¼ F1/k

Thus, if in a forced vibration the ratio DMF¼ u/ustat is defined as the dynamicmagnification factor, then its maximum value (DMF)max¼ umax/ustat is of definitedesign importance.

Of course, the response of the system under the loading F1 f(t) is fully described bythe exact solution of the above differential equation. This solution underlines the factthat the final response is the result of a superposition of the free vibration and theeffect of the force imposed on the system, being expressed by the so called Duhamelintegral, as shown below. Thus, the general solution of this equation is (Biggs, 1964)

uðtÞ ¼ e ! t u0 cos!d tþ_uu0 þ ! u0

!d

sin!d t

þ F1

m !d

ðt

0fðÞ e ! ðt Þ sin!d ðt Þ d

Obviously, direct use of this solution is not practical. However, the accessibility of specialsoftware dealing also with systems of multiple degrees of freedom allows the desiredresults to be obtained through a direct numerical approach rather than through theabove analytic approach. Nevertheless, the above expression is very useful, becauseit permits direct analysis of some typical cases and allows particularly importantconclusions to be obtained for the response of single-degree and, additionally, multi-degree systems. A few examples of such typical cases are listed below (see Figure 16.9):

. The ‘abrupt’ application of force F1, followed either by its equally ‘abrupt’ removalafter some period td, or by an immediate linearly varied decrease during the sameperiod td.

. The progressive application of the force up to the value F1 and, consecutively, itsdecrease down to zero over a total time period td, first in a linear fashion and thenin a parabolic manner.

. The progressive linearly increasing application of force F1, over a time period td,followed by maintenance of this acquired value for an unlimited time.

In all the above cases, the damping effect is ignored for simplicity.In the curves shown in Figure 16.9 the behaviour of the maximum value of the

dynamic magnification factor (DMF)max is depicted for each of the above cases, as afunction of the ratio of the characteristic time duration td to the natural period T ofthe vibrating system.

From curves (a), it can be seen that in the case of a constant applied force, when thetime reaches the value T/2, the factor (DMF)max takes its maximum value of 2, which isnot exceeded in any other case. This value is also valid for any td greater than T/2.

543


Otherwise, for a duration td less than about T/5, the factor (DMF)max is less than unity,i.e. the maximum deviation of the mass is even less than ustat¼ F1/k.

In contrast, the ‘immediately occurring’ linear decrease of the applied force has moremoderate results. Only after an unloading duration td of about 10 times the naturalperiod T does the factor (DMF)max begin to approach the value of 2.0, while for aduration less than T/3 the maximum deviation is less than the ‘static’ one ustat.

From curves (b), it can be seen that the maximum value of (DMF)max, i.e. themaximum mass deviation, occurs when the total loading—unloading durationapproaches 90% of the natural period T. This value equals 1.8 or 1.5 for a parabolicor linear force variation, respectively. For lower or greater durations td, the maximumvalue of (DMF)max is always lower.

Curve (c) represents the typical force application scheme that is encountered in manypractical cases. The more the duration td falls below T, the more the value of the factorincreases, never exceeding, though, the absolute maximum value of 2.0, no matter how‘abrupt’ the loading process is. However, for any duration td greater than T, themaximum value of factor DMF does not exceed the value 1.2. The above resultsconfirm the definition of the static loading, as given in the introduction to Chapter 1.

16.2.4 Periodic sinusoidal acting forceOf particular importance for the design of structures is the case when the acting forceF(t) is acting on the system in a periodic manner, according to the expressionF(t)¼ F0 sin t. The examination of the corresponding response permits one todraw useful conclusions about any kind of periodic excitation — machinery effects,human rhythmic activities, seismic action, etc. — as well as about measures requiredfor the attenuation of their effects (dynamic isolation).

544

td t

td t

td t

td t

td t

(DM

F) m

ax

2.4

2.0

1.6

1.2

1.0

0.8

0.4

0

(a)

(b)

(c)

0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25

td/T

F1

F1

F1

F1

F1

Figure 16.9 Behaviour of a single-mass system under a dynamically applied load


By applying the general solution of the previous section in the present case andignoring the effect of damping, the following expression is obtained:

uðtÞ ¼ U sinð! tþ ’Þ þ F0=k

1 ð=!Þ2 sin t

The first term on the right-hand side is essentially identical to the correspondingexpression for the general solution, where

U ¼ A and tan’ ¼ u0

_uu0=!

This term represents, as already mentioned, that part of the response correspondingto the free vibration of the system, whereas the second term is exclusively due to theexternally applied force.

Now, as consideration of even the least possible damping makes the contribution ofthe free vibration (i.e. the first term) soon vanish, the conclusion may directly be drawnthat the maximum value of the dynamic magnification factor (DMF)max will be

ðDMFÞmax ¼u

ustat

max

¼ 1

1 ð=!Þ2

It is clear that the ratio r ¼ =! plays a basic role. For values less than 0.20 and greaterthan 1.50, the dynamic influence is not important. More specifically, while for values of rtending to zero (static loading) the factor (DMF)max tends to 1, for a ‘high-frequency’excitation (r> 2.50) this factor tends towards 0. These correlations are clearly shownin Figure 16.10. However, as can be deduced from the equation above, when ¼ !,the factor (DMF)max is infinite. What is actually happening in this case is that ineach cycle the factor DMF is steadily increasing in time, according to the equation

(DMF)¼!¼ (sin! t! t cos! t)

On the basis of this equation:

ðDMFÞmax;¼! ¼umax

F0=k¼ 1

2 n p

where n is the ever-increasing number of cycles.The situation that occurs in the case when ¼ ! is called resonance, and is

something that should, of course, be avoided. However, as an infinite value of thefactor DMF corresponds to a system without any damping effect, it should pointedout that this does not happen in reality, as even a small amount of damping actionreduces the response significantly.

More precisely, for an existing damping factor equal to (Biggs, 1964),

ðDMFÞmax ¼1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

½1 ð=!Þ22 þ 4 2 ð=!Þ2q

where the restrictive contribution of to the response is clear (see Figure 16.10).

545


Thus, in the case of resonance ( ¼ !), (DMF)max; ¼ !¼ 1/(2), and, consequently,

umax ¼1

2 F0

k

However, until the system develops the above maximum deviation of mass, a largenumber of cycles is needed. g

Of particular interest is the assessment of the transmitted forces to the base, i.e. to thesupport of the structural web.

From the examination of the dynamic equilibrium of a system — see Section 16.2.1and also Figure 16.11 — the transmitted force Fg at the base of the structural web,consists of its elastic force (k u) plus the damping force (c _uu) acting in the same sense.

The maximum force transmitted to the base max Fg can be obtained according to theequation

max Fg ¼ F0

1þ 4 2 r2

ð1 r2Þ2 þ 4 2 r2

1=2

546

(DM

F) m

ax

Ω: Imposed frequencyω: System eigenfrequency

r = Ω/ω

In the case of resonance (Ω/ω): (DMF )max = 1/2ζ

0 0.50 1.00 1.50 2.00 2.50

5.00

4.50

4.00

3.50

3.00

2.50

2.00

1.50

1.00

0.50

0

ζ = 1.00

ζ = 0.50

ζ = 0.20

ζ = 0.10

ζ = 0

Figure 16.10 Behaviour of a system towards a periodic load action


The ratio T¼max Fg/F0 is called the transmission factor, and its dependence on the ratior ¼ =!, as well as on the damping factor, is depicted in Figure 16.12.

According to Figure 16.12, given the frequency of an imposed load, due to, forexample, an installed machine, by providing to the system — through appropriatemeans — a natural circular frequency !, which is less than about /2.50, the transmittedforce to the base is drastically limited.

16.2.5 Seismic excitationSeismic excitation consists of an imposed oscillating motion of the support of a structure.The cause of this motion is the abrupt (brittle) rupture in some region of an already

547

m m

m

m

u

u

F0 · sin Ωt

F0 · sin Ωt

m · ü

k · u

k · u

c · u

c · u

Fg Fg

m · ü(t1)

c · u(t1) k · u(t1)

c · u(t1)k · u(t1)

Transmission factorT = max Fg/F0

Figure 16.11 Transmission of dynamic actions at the base of a structural web

Ω: Imposed frequencyω: System eigenfrequency

Ω/ωEffective isolation with soft springsand low damping(Increased displacements)

0 1.00 √2 2.00 3.00

ζ = 0.25

ζ = 0.2

ζ = 0ζ = 0

Tra

nsm

issi

on fa

ctor

T

3.00

2.00

1.00

0

Figure 16.12 Transmissibility of dynamic actions at the base of a system


existing fault in the earth’s crust, as a result of accumulated stresses. This rupture isinstantaneously spread out along the fault, causing propagation of a vibration wavewithin the earth’s crust lasting for a period of about 15—30 s, and then coming to rest.

First, it is assumed that the imposed horizontal ground motion us(t) on the support ofthe structure may be expressed as us(t)¼ us,0 f(t), where us,0 represents the maximumvalue of the ground displacement in the region of the support. In the present case ofa single-mass system, the direction of us corresponds to the sense for that of the stiffnessk of the system (Figure 16.13).

It is clear that the elastic force acting on the mass m will be due to the deformation ofthe whole system, i.e. to the relative displacement of the mass to the ground, expressedas u us. Given that the mass m is not acted on by any external load, but is subjected tothe inertia force only, obviously with reference to its absolute displacement u, as well asto the elastic force, it can be written, if the damping force is ignored, that

m €uuðtÞ þ k ½uðtÞ usðtÞ ¼ 0

The absolute displacement u(t) of the mass m cannot describe the deformation of thestructure depending, as pointed out, exclusively on the relative displacement:

x(t)¼ u(t) us(t)

Then, substituting the last equation,

m ð€xxþ €uusÞ þ k x ¼ 0

It is clear that the response k x of the structure depends solely on the ground accel-eration €uusðtÞ (see Figure 16.13).

Thus, the characteristic of an earthquake, which is relevant for the design of astructure, is only the so-called accelerogram, which is recorded by special equipmenteach time. For a specific earthquake and within a major region, different accelerogramsmay be recorded for different locations, depending on the distance from the respective

548

Imposing a ground motion means essentially imposing an accelerogram,and this is turn means imposing an inertia force on a rigidly supported system

üs(t ) = a · fa(t )Accelerogram

(a: Maximum ground acceleration)

Fixed baseus

x = u – us

k · x

k · x

k · x

k · x

mm

u x

x

x

m · ü(t ) m · üs(t ) (m · a) · fa(t )m · x

Figure 16.13 Imposition of seismic ground motion and dynamic equilibrium


source of the event (hypocentre) as well as on the existing soil properties along the pathfollowed by the seismic waves to the place recorded.

The accelerogram €uusðtÞ may be expressed as

€uusðtÞ ¼ a fsðtÞ

where a represents the maximum occurring ground acceleration, while the function fa(t)describes its evolution in time. Now, the last equation may be written as

m €xxþ k x ¼ ðm aÞ faðtÞ

The form of this equation is exactly the same as that for the forced vibration of a single-degree-system on a fixed basis (see Figure 16.13), and, as the existing damping obviouslyaffects the relative velocity _xx of the mass m with respect to the ground, the equation maybe accordingly modified as follows:

m €xxþ k xþ c _xx ¼ ðm aÞ faðtÞIt should be noted that the right-hand term refers to a ‘fictitious force’ — representingalong with the force m €xx the total inertia force that the mass m is subjected to.

By assuming initial conditions corresponding to ‘rest’, i.e. x¼ 0 and _xx ¼ 0, the solutionof the last equation, analogously to Section 16.2.3, may be written as (Muller andKeintzel, 1984)

xðtÞ ¼ a

!d

ðt

0faðÞ e ! ðt Þ sin!d ðt Þ d

As previously explained (see Section 16.2.2.2), for values of up to 10%, the funda-mental circular frequency !d does not differ substantially from the natural frequencyof system ! without damping. In any case, however, for the design of a structure, it isthe maximum value Sd of the relative deviation that is of paramount interest, as itobviously leads to the maximum response:

Sd ¼ xmax ¼a

!ðt

0faðÞ e ! ðt Þ sin! ðt Þ d

max

It can be seen that at the time that the maximum deviation xmax occurs, the acceleration(or deceleration) €xx also takes its maximum value €xxmax, whereas the velocity _xx — andhence the damping force — vanishes. Thus, at this moment, the elastic force k xmax

acting on the mass m balances its inertia force, depending, of course, on its absoluteacceleration, and is equal to m (€xxmaxþ €uus).

Assuming that at this same moment the maximum ground acceleration is essentiallyattained, it can be written (Figure 16.14) that

m ð€xxþ €uusÞmax ¼ k xmax or ð€xxþ €uusÞmax ¼ !2 xmax

Thus, the maximum value Sa of the absolute acceleration of the mass is

Sa ¼ ð€xxþ €uusÞmax ¼ a ! ðt


max

549


This expression is strictly valid provided that the damping coefficient does not exceed10%. g

It is thus clear that the response of a single-degree system to a specific seismic excitation— i.e. with known characteristics regarding the time development fa(t) of the groundacceleration, as well as its maximum value a — depends only on the natural frequency,or natural period, of the system and the corresponding damping factor.

On the basis of the last equation and for a given accelerogram, the maximum absoluteacceleration Sa can therefore be plotted as a function of the natural period T, fordifferent values of the damping coefficient . Such a diagram represents the accelerationspectrum, and it permits, for any oscillator with a known natural period T, the directdetermination of its maximum absolute acceleration. It may be easily seen thatSa¼ Sd !2.

It is useful to point out here that, according to the above expression, Sa — as well as thespectral displacement Sd — is proportional to the maximum ground acceleration aaccompanying the specific accelerogram, so that if divided by a, it represents aspectrum corresponding to an accelerogram exhibiting the same evolution in time butwith respect to unit maximum acceleration.

Figure 16.15 shows an accelerogram recorded from the El Centro earthquake(California, USA, 1940), together with the acceleration spectrum (Sa/a) correspondingto unit ground acceleration, for two different damping coefficients.

It is can now be seen that for the practical evaluation of a specific earthquake beingdescribed by its accelerogram, the acceleration spectrum Sa — or the unit spectrum Sa/atogether with the ground acceleration a — represents the examined excitation quitesufficiently. Thus, for a specific system with a known natural period T, on the basis ofthe corresponding spectral value Sa, the maximum inertia force can be directlyobtained as m Sa. This force represents the maximum shear force Va developed at

550

Approximation: the maximum deviation coincideswith the maximum ground acceleration

üs(t) = a · fa(t)Accelerogram

xmax

xmaxm · xmax m · üs(t)

k · xmax

k · xmax

k · xmax

k · xmax

Maximum response

Inertia force

Maximum acceleration

m(xmax + a) = m · Sa

Strict consideration Acceptable approximation

Figure 16.14 Position of maximum response in a single-degree-of-freedom system


the base of the structure, called the base shear:

Va¼m Sa

It is clear that the above inertia force (m Sa) is identical to the maximum elastic force(k Sd) acting on the mass m (see Figure 16.15). Thus,

(m Sa)¼ k Sd¼ k Sa/!2

The use of the acceleration spectrum therefore allows the determination of themaximum response in the structural web through a static loading, which is, of course,of particular practical importance.

551

0 2 4 6 8 10 12 14 16 18 20

Gro

und

acce

lera

tion:

g

0.4

0.2

0.0

–0.2

–0.4

T: s

0 0.5 1.0 1.5 2.0

4.80

3.60

2.40

1.20

0

T: s

ζ = 0.02

ζ = 0.10

Sa/

a

Base shear

Static consideration

m · a

m · (Sa/a) · a

mm

m

m

m

Figure 16.15 Accelerogram and acceleration spectrum (El Centro, California)


What can be deduced from the spectrum of a specific earthquake is that there isalways a range of periods which is more sensitive towards the development of themaximum inertia force, or — in other words — of the maximum shear base. As can beseen in Figure 16.15, structures with a lower stiffness being reflected in values of anatural period T greater than, say, 1.0 sec, develop a clearly limited response.

The dynamic magnification factor (DMF)max, which as defined in Section 16.2.3 isthe ratio of the maximum relative displacement Sd to the ‘static’ displacement (m a)/k,is now examined:

ðDMFÞmax ¼Sd

ðm a=kÞ ¼!2 Sd

a¼ Sa

a

¼ ! ðt


max

Written differently:

Sd¼ (a/!2) (DMF)max

Sa¼ Sd !2¼ a (DMF)max

It should be noted that the dynamic factor (DMF)max, as can be seen from the aboveexpression, is identical to the acceleration spectral value (Sa/a). g

Assuming now, in an approximate approach, that the variation of the acceleration a intime is sinusoidal, i.e. fa(t)¼ sin t, then, given that the two last equations aregenerally valid, the previously found expression for the dynamic factor (see Section16.2.4) may be applied directly:

ðDMFÞ;max ¼1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

½1 ð=!Þ22 þ 4 2 ð=!Þ2q

Thus,

Sd; ¼ ða=!2Þ ðDMFÞ;max

and

Sa; ¼ a ðDMFÞ;max

The diagram in Figure 16.10 may therefore also be interpreted as an accelerationspectrum — with respect to unit ground acceleration — for any value of the ratio of thefrequency (taken as constant) to the natural frequency ! of the single-degree system.The magnitude of the assumed frequency of the maximum ground accelerationdepends on the geotechnical characteristics of the foundation soil. For stiff soils androcks, ranges from 20 to 60 rad/s, for medium stiff soils, from 12 to 50 rad/s, and forsoft or very soft soils, between 10 and 3 rad/s. g

Considering the influence of the foundation soil, it can be seen from Figure 16.16 that itsdeformability influences, in addition to the maximum developed ground acceleration,

552


the response of the structure itself, due to the variation of the effective inertia force ofthe mass. This force increases or decreases according to the region of the correspondingacceleration spectrum that the natural period of the structure belongs to, by assuming afixed base (see Figure 16.16).

In actual design practice, the so-called design spectrum, which contains the effects ofa large number of relevant accelerograms for the region examined, is used as anacceleration spectrum, also making a distinction between the various soil types, giventhe influence of the foundation soil deformability, as explained above.

Of course, the elaboration of the design spectral curves is a major issue for the regula-tions in a particular country. Examples of these curves are shown in Figure 16.17. Theinfluence of the soil type on the maximum effective acceleration Sa is obvious.

16.2.6 Influence of plastic behaviour on the seismic responseAccording to the previous section, the basic equation of motion of a single-degree-of-freedom system under a seismic excitation, with respect to a fixed base, can be

553

Increment of inertia force Decrement of inertia force

(Rigid support)

T: s

4.80

3.60

2.40

1.20

0

Sa/

a

0 0.5 1.0 1.5 2.0

Resilient ground

kϕkϕ

Imposing the same accelerogram on both systems causes in the case of resilient groundsmaller or greater accelerations (inertia forces) depending on the period T itself

1 1

m m

k = 3 · EI/L3 k = (3 · EI/L3)/p



Natural period T ρ = 1 + 3 · EI/(kϕ · L)

Smaller stiffnesshence greater period

Figure 16.16 Influence of the deformability of the soil on the seismic response


written as

m €xxþ k xþ c _xx ¼ ðm aÞ faðtÞThis equation expresses the static equilibrium between the externally acting inertiaforce (m €xxþm a fa(t)) together with the damping force on one side and the elasticforce developed by the structural web due to its deformation on the other. Thislast force (k x) increases for as long as the displacement x of the mass increases, i.e.as for as long as the deformation of the structural web increases, according to the existingstiffness factor k.

However, it is clear that a possible plastic hinge at the base of the cantilever, wherethe maximum bending response is developed, places an upper limit on the increase ofthe above-mentioned force, which remains constant afterwards, despite furtherincrease in the displacement x (Figure 16.18). More specifically, if L is the cantileverheight and Mpl represents the plastic bending strength of the structural web, then thelimit F up to which the force can be transmitted to the mass, i.e. the ‘elastic force’,may be increased, F¼Mpl/L. This, of course, is under the condition that the relevant

554

Sa/

g

T: s

RockStiff soilSoft soil

0.60

0.40

0.20

00 0.5 1.0 1.5 2.0

Figure 16.17 Design spectrum

Maximum acting force Maximum acting force(reduced)

The adoption of a plastic hinge limits the responseof a structural web until a predetermined valueElastic behaviour

L

k · xmax

k · xmax

xmax

Accelerogramüs(t) = a · fa(t)

(Inertia force)

m · (xmax + a) = m · Sa m · (Sa/q)

Mpl/L

Mpl/L

Mpl

Figure 16.18 Limitation of the inertia force due to plastification


section of the structural web affords the required ductility (rotational deformability), sothat the augmented displacement x can be realised.

It is clear that the limitation of the internal — elastic — force expressed by the term k xin the above equation restricts, accordingly, the level of the effective inertia force too.This means that, although the mass displacement x may increase, the above equilibriumequation can be satisfied with less internal force and, correspondingly, less inertia force,than required by a purely elastic behaviour of the structural web.

This fact may be taken into account through a corresponding reduction in the spectralacceleration value Sa, through the factor q, known as the behaviour factor. Thus, in thecorresponding equations in the previous section, the magnitude Sa/q is used instead ofSa. The selected value of the factor q is essentially the designer’s decision, dependingon the estimated — or even desired — ductility of the structural web. A value of qequal to 1.0 would indicate the acceptance of elastic behaviour throughout the webwithout any reduction in the inertia seismic forces. A steel structure may allow consid-eration of a behaviour factor up to 4.0. In reinforced concrete structures, with anappropriate layout of reinforcement in order to ascertain good deformability (ductility)of the ‘critical’ sections, a value of q of even 3.0 may be adopted, leading to acorresponding decrease in the acting seismic forces. However, it should be pointedout that in the design process the consideration (or not) of q greater than unity — i.e.the adoption (or not) of plastic hinges in the structural web to account for a givenmaximum ground acceleration — relies exclusively on the designer’s engineeringjudgement.

16.3 Multi-degree systemsAs multi-degree systems imply, they are systems that contain more than one discrete mass.These masses are ‘carried’ by a massless structural web exhibiting a clearly determinablebehaviour regarding its deformability and stiffness. Within this frame of behaviour themasses show displacements or rotations involving either negligible or non-negligibleinertia forces. These last displacements are considered as ‘effective’ for the correspondingmasses, and their number represents the so-called effective degrees of freedom of the system.Examples of such multi-degree systems have been mentioned in the introductory sectionof this chapter (see Figure 16.2).

16.3.1 The stiffness matrix

16.3.1.1 The concept of the stiffness matrixOf decisive importance for the description of the behaviour of a multi-degree system isthe so-called stiffness matrix, i.e. a set of magnitudes, describing the stiffness of its struc-tural web, which are determined as shown in the example of a three-storey plane modelin Figure 16.19. Here, it is assumed that the masses m1, m2 and m3 develop in ahorizontal excitation (e.g. earthquake) inertia forces with respect only to their horizontaldisplacements. Thus, the effective displacements of the system are the lateral storeyshifts 1, 2 and 3.

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The stiffness matrix consists in the horizontal juxtaposition of three groups of forcesthat are determined as follows (see Figure 16.19). The first group consists of those three— horizontal — forces (k11, k21, k31) that are required to be applied correspondingly to theeffective displacements in order to produce a unit shift in level 1 and zero shift in the twoothers (i.e. 1¼ 1, 2¼ 0, 3¼ 0).

Analogously, the second group consists of those forces (k12, k22, k32) needed to beacted on in the sense of the effective displacements, in order to produce a unit shiftin level 2 and zero shift in the two others (i.e. 1¼ 0, 2¼ 1, 3¼ 0). Finally, thethird group contains those forces (k13, k23, k33), which, acting along the effectivedisplacements, produce 1¼ 0, 2¼ 0, 3¼ 1.

Each one of the above groups comprises the reactions developed if, at first, fictitioussimple supports are inserted into the system, in order to restrain all its effectivedisplacements, and then the corresponding unit-support settlement (1 m) is imposed,as shown in Figure 16.19. In order to achieve this, the use of a computer is clearlynecessary.

Defining the positive senses of these forces as coinciding with those of the displace-ments (1, 2, 3), i.e. to the right, it is clear that the forces of each group willexhibit alternating signs.

556

m1

m3

m2

∆1

∆3

∆2

∆1 = +1

∆2 = +1

∆3 = +1

k11

k21

k31

k12

k22

k32

k13

k23

k33

Structural web

The stiffness matrix represents the stiffness of the structural webwith respect to the effective mass displacements

k11 k12 k13

k = k21 k22 k23

k31 k32 k33

Figure 16.19 The elements of a stiffness matrix


The structural web which carries the three masses may be represented by the stiffnessmatrix k by juxtaposing horizontally the above three force groups as previouslymentioned, according to the following formulation:

k ¼k11 k12 k13

k21 k22 k23

k31 k32 k33

264

375

It is useful to become familiar with matrix operations. However, it must be pointed outthat the use of matrices contributes very little, if at all, to the understanding of the staticor dynamic behaviour of any system — at least at this introductory level. Nevertheless,matrices are a necessary tool for the compact writing of equilibrium equations and therelevant equations, thus facilitating their methodical numerical treatment throughcomputer programs.

16.3.1.2 Matrix operationsIn the present context, of particular interest are either the so-called square matrices —having an equal number of rows and columns — or the so-called column matrices —consisting of the vertical juxtaposition of numerical elements.

A square matrix k of the order NN, has N rows and N columns (see Figure 16.19).The elements of the square matrix along the diagonal starting from the upper leftelement constitute the so-called diagonal elements. A column matrix a of the orderN 1 has N rows and 1 column.

Interchange of the rows and columns of a matrix — or vice versa — leads to thetranspose matrix, denoted by the letter T. Thus, while the matrix kT still has N rowsand N columns (i.e. NN), matrix aT has 1 row and N columns (i.e. 1N).

The basic physical concept of a matrix is directly related to the column matrix. This isthe recording of quantities in correspondence to specific magnitudes, e.g. the storeyshifts of the structure described previously, or the forces acting at each storey level.The appropriate horizontal juxtaposition of such column matrices leads to squarematrices, like the stiffness matrix which has been previously defined.

The sum of two column matrices is a matrix of the same order, and comprises the sumsof their corresponding terms. Although the addition also applies to two square matricesof the same order, the physical meaning of this operation is clearer and better under-stood in this context with respect to the column matrices.

The multiplication of two matrices is of basic importance, and it is accomplished asshown in Figure 16.20. The multiplication of two matrices is feasible only when thenumber of columns in the left matrix is equal to the number of rows in the rightmatrix. This may also be understood from the rule that the final order of the productmatrix is found by eliminating the ‘interior’ dimensions of the matrices — as shownbelow.

The product k a is written as (NN) (N 1), and is a matrix of order N 1,whereas product a k, which is written as (N 1) (NN), does not make sense. Of

557


course, product aT k does have a meaning, because it is written as (1N) (NN),which leads to a matrix of order 1 N, whereas it can be seen that product aT a,written as (1N) (N 1), leads to a matrix of order 1 1, i.e. a single number.

It is thus clear, also taking into account the schematic layout of Figure 16.20, that themultiplication of two matrices consists of the consecutive formation of matrix productsof the last type, (1N) (N 1), which, of course, represent a single number.

The square matrix, the diagonal elements of which consist of unity while all otherelements are null, is called the identity matrix, and is denoted by I. Consideration ofthe identity matrix allows reference to the basic concept of the so-called inversematrix of a square matrix A of the order NN, which is denoted by A1. Theinverse matrix is one that satisfies the equation

A A1¼ I

or (identically)

A1 A¼ I

Obviously, both matrix A1 and matrix I are of the order NN.The concept of the inverse matrix is directly connected to the solution of a linear

system of N equations with N unknowns. Such a system may be represented throughthe following matrix notation:

A X¼C

If matrix A consists of the coefficients of the unknowns, then column matrices X and Ccontain the unknowns and the ‘known terms’ of the right-hand side of the equations,respectively. Multiplying both sides by the inverse matrix A1 yields

A1 A X¼A1 Cthat is,

I X¼A1 C

558

Number of columns cT =number of rows a

Number of columns k =number of rows a

Number of columns k =number of rows b

= c1 · a1 + c2 · a2 + c3 · a3

Column matrix

P = k · ∆

MultiplicationcT · a

Multiplicationk · a

Multiplicationk · b

Columns

Rows

a1

a2

a3

a1

a2

a3

b11 b12 b13

b21 b22 b23

b31 b32 b33

k11 k12 k13

k21 k22 k23

k31 k32 k33

k11 k12 k13

k21 k22 k23

k31 k32 k33c1 c2 c3

Figure 16.20 Layout for the multiplication of matrices


and,

X¼A1 COf course, the inversion of matrix A requires a numerical procedure that is equallycumbersome as the solution of the linear system, so that the advantage in the use ofthe matrix notation lies only in the compact formulation and not in the computingeffort required, which in any case will be undertaken using a computer.

16.3.1.3 Correlation of loading and displacementsThe usefulness of the matrix notation in describing in a compact manner the behaviourof a system is illustrated using the stiffness matrix k (3 3) of the structural modelshown in Figure 16.21. From the physical definition of k in Section 16.3.1.1, it iseasily seen that the three forces (P1, P2, P3) required to be applied to the three storeylevels in order produce the corresponding displacements (1, 2, 3) can beobtained from the equation

P¼ k ðaÞwhere matrix P (3 1) represents the forces (P1, P2, P3), and matrix (3 1) representsthe displacements (1, 2, 3).

According to the introduced ‘matrix language’, it can be seen that if the loads P areknown and the displacements caused by them are sought, these can be obtained bymultiplying both sides of the above equation by the inverse k1 of the stiffness matrix k,i.e.

k1 P¼ k1 k so that

¼ k1 PClearly, this result represents nothing more than the solution of the linear system (a),without this compact matrix formulation helping the solution itself.

559

P = k · ∆

∆1

∆3

∆2P2

P1

P3

m2

m1

m3

Structural web

Figure 16.21 Correlation of loading and deformation in a multi-degree system


16.3.1.4 Multi-storey systemsAt this point it is useful — for later use too — to describe through the compact matrixmethodology the overall response of a multi-storey spatial system under horizontalloads, according to the previous examination in Chapter 15 (Figure 16.22).

From Chapter 15, we know that the structural web of a multi-storey spatial systemconsists of elements of plane stiffness. Each of the plane elements is represented by itsown stiffness matrixSi — as previously defined in Section 16.3.1.1 and shown in Figure 16.23.

The overall stiffness matrix K of the order 9 9, referring to the whole system, relatesthe loads Px and Py (order 3 1) to the developed displacements x, y and ! ateach diaphragm. Matrix K is constructed as follows:

K ¼K1 K2 K3

K2 K4 K5

K3 K5 K6

264

375

where

K1¼P

Si cos2i

K2¼P

Si sin i cosi

K3¼P

Si ei cos i

K4¼P

Si sin2i

K5¼P

Si ei sin i

K6¼P

Si e2i

all matrices being of the order 3 3 (Stavridis, 1986).

560

x

y

y

y

x

x

Structural web

Structural web

Structural web

m1

m2

m3

P1, ∆1

P2, ∆2

P3, ∆3

Figure 16.22 Layout of a multi-storey system


The correspondence with the relevant expressions in Section 15.3.3 — at least withreference to a single storey — is absolute.

Thus, the equation between loading and deformation of the system is written asfollows:

Px

Py

0

8><>:

9>=>; ¼

K1 K2 K3

K2 K4 K5

K3 K5 K6

264

375

x

y

!

8><>:

9>=>;

It should be noted that the matrices Px and x consist of the components in the x senseof the magnitudes (P1, P2, P3) and (1, 2, 3), respectively (see Figure 16.22). This isanalogous for matrices Py and y too. The matrix ! consists of the diaphragmrotations (see Section 15.3.1).

It is, of course, clear that, given a multi-storey system with N diaphragms, the aboveequation allows the determination of the storey displacements, through inversion ofmatrix K of the order (3N 3N), as examined above, through use of suitablecomputer software. However, in the context of the examination of dynamic behaviourfrom a preliminary design point of view, of prime interest is the orthogonal layout ofplane elements about the axes x and y. Then, the constituent matrices of the globalstiffness matrix K are expressed as follows (see Section 15.3.4):

K1¼P

Sx

K2¼ 0

K3¼P

(Sx y)K4¼

PSy

561

The lateral stiffness matrix is assembled from the reactionsdeveloped at the lateral (imaginable) supports of the plane element

∆1 = 1

∆3 = 1

∆2 = 1

S11 S12 S13

S21 S22 S23

S31 S32 S33

Figure 16.23 Determination of the matrix of lateral stiffness for a plane element


K5¼P

(Sy xi)K6¼

P(Sx y2

i )þP

(Sy x2i )

Thus, the above condensed equation may be split up as follows:

Px¼K1 xþK3 !

Py¼K4 yþK5 !

0¼K3 xþK5 yþK6 !

If the plane elements layout leads to nullification — even approximately — of the matricesK3 and K5, which should be the basic intention according to Section 15.3.4, then, fromthe last equation, ! ¼ 0, i.e. the diaphragms do not rotate, thus not subjecting planeelements to an additional response. Recall that the horizontal loads are acting on thecentroid of each diaphragm, to which the coordinates x and y are also referred. Bytaking into account the above,

Px¼K1 x¼ (P

Sx) x

and

Py¼K4 y¼ (P

Sy) y

The above equations mean that, in the examined case, the displacements in both the xand y senses are independent of each other. They are obtained by applying the loads Px

and Py on the two separate systems, respectively, which are formed through coupling ofall the respective plane elements in each sense, carrying at each level the correspondingmass (Figure 16.24). This coupling is accomplished at each level through hingedundeformable bars, as shown in the figure. The above conclusion is of particular practicalimportance for both the assessment of the dynamic characteristics of the multi-storey

562

∆1

∆3

∆2

∆1

∆3

∆2

P2

P1

P3

m2

m1

m3

m2

m1

m3

P2

P1

P3

In the case where the diaphragm rotation may be ignored,the multi-storey system acts in each sense like a plane structural web

P = (∑ sx) · ∆

Sense x–x

P = (∑ sy) · ∆

Sense y–y

Figure 16.24 Equivalent plane system in both directions


mass system and the determination of the equivalent static loading because of seismicexcitation, as will be examined later in this chapter.

16.3.2 Free vibration

16.3.2.1 Plane systemsThe study of multi-degree systems will be continued on an illustrative basis for thethree-degree model, without thereby limiting the general validity of any conclusionsfor systems with more degrees of freedom.

At this point, it is noted that the ‘rectilinear’ structural web in Figure 16.25 shouldnot necessarily be considered merely as a straight member carrying at some points themasses m1, m2 and m3 but as a structural system described by its stiffness matrix k.Consequently, the model may also be considered to represent a plane frame, such asthat shown in the figure, and having the same stiffness matrix k (see Figure 16.23).This clarification is necessary because the lateral shift behaviour of a vertical cantileveris quite different from that of the frame, as has already been pointed out in Section 6.4.3(see Figure 6.16).

First, it is assumed that the mass mi, acted on by an external force Fi(t) in the sense ofits own degree of freedom, i.e. horizontally, undergoes a displacement xi(t). The positivesense for all the forces and displacements is considered common (Figure 16.26).

Both the set of forces Fi(t) and the set of displacements xi(t) may be represented by thecolumn matrices F(t) and x, respectively. For ease of matrix manipulation only, thesystem of masses mi is represented by the square diagonal matrix m of order NN.In this matrix, the masses mi constitute the diagonal elements, with all the otherelements being zero. It can be written as

m¼ I m

563

P = k · ∆

The ‘rectilinear’ model is used merely torepresent a structural web having a stiffness matrix [k]

The two structural systems exhibit a totally different behaviour under lateral loads

P2

P1

P3

m2

m1

m3

P2

P1

P3

m2

m1

m3

Structural web

∆1

∆3

∆2

∆1

∆3

∆2

Figure 16.25 Simulation of a multi-degree frame with a rectilinear model


Just as in a single-degree system, a specific mass mi may be considered as being always inequilibrium under the following forces (see Figure 16.26):

(1) the external force Fi(t)(2) the total of the elastic forces Si transmitted from the structural web to the mass, as a

consequence of its deformational configuration x(t)(3) the inertia force mi €xxiðtÞ(4) the damping force ci _xxiðtÞ.As forces (2), (3) and (4) have the opposite sense to the force Fi(t), for the equilibrium ofeach mass mi (see Figure 16.26),

mi €xxiðtÞ þ ci _xxiðtÞ þ Si ¼ FiðtÞIt is clear that the deformational configuration x(t) for the specific time instant t may beproduced by a unique set of external forces depending on the stiffness of the structuralweb. According to the foregoing, this set of forces can be given by the expression k x(t)(column matrix N 1), and it can be seen that each of these forces equals the respectivesum Si of the elastic forces acting directly on the mass mi.

The simultaneous consideration of the equilibrium of all masses can now be expressedby the following matrix formulation:

m €xxþ c _xxþ k x ¼ FðtÞThe damping coefficients ci are represented by the square diagonal matrix c. g

As in the single-degree systems, it is expedient here to also seek the basic dynamic charac-teristics of the multi-degree system, through the examination of its undamped free vibration.

Free vibration is evoked by imposing an arbitrary displacement at a certain mass,according to its corresponding degree of freedom (e.g. a horizontal displacement3¼ ) and then leaving the system free to vibrate.

564

m1

m3

m2

Structural web S2 = S21 + S23

S2 balances the static force required todisplace the mass by x(t)

The whole of the static forces is presented by k · x(t)

Equilibrium of mass m2

x(t)

S21

S23

F2(t)m2 · x2(t)

c2 · x 2(t)

Figure 16.26 Dynamic equilibrium in a multi-degree system under external loading


In this case, the global equilibrium condition is written as

m €xxþ k x ¼ 0

By trying out the harmonic solution x¼’ sin! t, where the magnitude ! is, of course,unknown and, moreover, the column matrix ’ also consists of unknown quantities, andgiven that €xx¼!2 ’ sin! t, the above equation can be written as (Tedesco et al.,1999)

½m1 k !2 I ’ ¼ 0

This equation is known as the eigenvalue equation or frequency equation of the system, andrepresents a linear system of N equations with respect to the N unknowns ’. Accordingto the laws of linear algebra, in order for a non-zero solution to exist in a linear systemhaving all of its right-hand side members equal to zero, the determinant of the coeffi-cients of the unknowns must be equal to zero. This determinant, as can be concludedfrom the last equation, consists of a polynomial of Nth degree with respect to !2, forwhich a number of N values (roots) can be determined. It is, of course, obvious thatfor each eigenvalue !2 the corresponding solution of the above linear system does notsimply consist of a well-defined set of values ’ but rather of an infinite group ofvalues ’, where is an arbitrary number.

It is clear that because of the computational effort required for the determination ofthe above magnitudes, these can only be obtained through the use of appropriatecomputer software.

From a physical point of view, the above signifies that the N-degree system has Ndifferent eigenfrequencies !, and each one of them corresponds to a certain displacementconfiguration ’ for each degree of freedom, which is multiplied by an arbitrary factor,either positive or negative. This displacement configuration is called the eigenform ornatural mode (Figure 16.27). The physical meaning of any of the eigenforms ’ is that

565

m1 m1m1

m2

m2 m2

m3 m3 m3

M1 M2 M3

K1 = M1 · ω12 K2 = M2 · ω2

2 K3 = M3 · ω32

1st eigenform[ω1]

2nd eigenform[ω2]

3rd eigenform[ω3]

In a structural web deformed according to some eigenform and then left free,all the masses are vibrating in phase under the corresponding eigenfrequency

The whole system acts then as an ideal mass with an ideal stiffness

Figure 16.27 Eigenforms and dynamic characteristics of a multi-degree system


if it is somehow imposed on the system (i.e. through an appropriate application ofhorizontal forces) and the system is left free, then every mass of the system will vibratelike a single-mass oscillator under the same natural frequency !, which corresponds tothe selected eigenform.

An additional analytical property of the eigenforms ’ should be noted. For any pair ofdifferent eigenforms ’r and ’s the following is valid:

’Tr m ’s ¼ 0

’Tr k ’s ¼ 0

which is known as the orthogonality condition of the eigenforms ’.The fact that the whole system may vibrate at any specific frequency of the N

realisable eigenfrequencies !r exhibited for each mass amplitude (represented by theeigenform ’r) means that all masses are found at the same time either at the restposition or at their maximum deviation, and suggests the idea of considering an ideal(fictitious) mass Mr connected to an ideal (fictitious) stiffness Kr (see Figure 16.27).It is indeed found that for the corresponding eigenform ’r the above magnitudes maybe determined as

Mr ¼ ’Tr m ’r ¼

XN1

mr ’2r

Kr ¼ ’Tr k ’r

According, of course, to the characteristics of a typical single-degree system (Biggs, 1964):

!2r ¼ Kr=Mr

In this way, for a given multi-degree system, N single-mass oscillators may be considered,each with specific characteristics of mass, stiffness and frequency. The lowest frequency! is called the fundamental frequency, and has more physical importance than the others,as will be shown later.

More specifically, if an arbitrary horizontal displacement is imposed on any mass inany of the two systems shown in Figure 16.28, and the structure is then left free,each mass in the system will vibrate at a frequency that essentially coincides with thefundamental frequency. This fact, which may apply to any multi-degree system, leadsto an important result that allows the determination of the fundamental frequency ofthe system, as will now be shown.

If to every mass in a multi-degree system an arbitrary static force Fr of the same senseis applied, then the masses will be displaced by r (Figure 16.29), and, if the system isthen left free, a harmonic motion for every mass with a natural frequency ! will beimmediately established, whereby all masses take both their rest position and theirmaximum deviation r at the same time instant.

According to the characteristics of the single-degree systems, the maximum velocityof every mass at its rest position will be ! r, and, consequently, the kinetic energy ofthe masses as a whole will be

P12 mr (! r)

2.

566


Clearly, this kinetic energy should be equal to the total work done by the forces Fr,which is

P12 Fr r. Thus, the value of the fundamental frequency ! may be readily

obtained as (Biggs, 1964)

!2 ¼P

Fr rPmr 2

r

The computation is facilitated if the forces Fr are selected as equal (see Figure 16.29). g

The damping action is logically expressed through the damping coefficient with regardto each eigenform. Usually, the value of applied to the first eigenform corresponding to

567

Removal of the static force causes vibrationunder the fundamental natural period

F F

m1

m2

m3

m1

m2

m3

Figure 16.28 The fundamental frequency

∆1

∆3

∆2

F2

F1

F3

m2

m1

m3

It is more appropriate to select equal forces

Structural web

∆ = k–1 · F

Fundamental natural frequency

A = F1 · ∆1 + F2 · ∆2 + F3 · ∆3

B = m1 · ∆12 + m2 · ∆2

2 + m3 · ∆3

ω12 = A/B

Figure 16.29 Practical evaluation of the fundamental eigenfrequency


the fundamental frequency ! is that directly relating to the nature of the structural webmaterial, in accordance with Section 16.2.2.2. For each additional eigenform, the valueof may be considered to increase by 1%.

For the experimental determination of the damping coefficient, as well as of theeigenfrequency of each eigenform, the same procedure may be followed as in the caseof a single-degree system, i.e. by creating through appropriate forces the correspondingdisplacement configuration ’r and following the corresponding free vibration of thesystem (see Section 16.2.2.2).

The measured eigenfrequencies for each eigenform should not differ essentially fromthose obtained theoretically, as explained above, provided that the damping coefficientsdo not exceed 10%.

16.3.2.2 Multi-storey spatial systemsIn the case of a multi-storey building with horizontal diaphragms arranged vertically overeach other, and which are carried by the structural web consisting of vertical planeelements, as examined in Chapter 15, it becomes clear that the dynamic equilibrium ofthe diaphragm masses is related to the displacements x and y of each diaphragm, aswell as to its rotation !. While displacements x and y activate the inertia forcesm €xx and m €yy of the mass of each diaphragm, respectively, the rotation ! activatesthe rotational inertia moment ( €!!) on the basis of the rotational moment of inertia of the mass of the diaphragm with respect to its centroid.

For the case of an equally distributed mass in the diaphragm,

¼m (Ixþ Iy)

where Ix and Iy are the moments of inertia of the diaphragm area with respect to the axesx and y, respectively.

By taking into account the ‘expanded’ mass matrix (3N 3N),

mm ¼m 0 0

0 m 0

0 0

264

375

the eigenfrequency equation is analogous to that obtained previously (Stavridis, 1986):

½mm1 K !2 I ’ ¼ 0

The eigenfrequencies ! can be determined only through the use of a computer.For the case where the diaphragm rotations are negligible, even approximately,

according to Section 15.3.4 the fundamental eigenfrequency of the whole system maybe assessed with satisfactory accuracy on the basis of the expression for !2 given inSection 16.3.2.1, in both the x and y directions, according to the following procedure(Figure 16.30):

. A plane formation is created separately for the x and y senses, consisting of thecorresponding plane elements in each direction, as shown in Section 16.3.1.4.

568


. Unit loads are applied to the different levels of the structural web, for which therespective displacements are calculated. The fundamental eigenfrequency for theconsidered sense (x or y) is obtained by direct application of the last expression for!2 in Section 16.3.2.1.

Numerical exampleThe ground plan shown in Figure 16.31 applies for three consecutive storeys, eachhaving a height of 3.50 m and a slab thickness equal to 25 cm. All of the frame girdershave a width and depth equal to 30 cm and 80 cm, respectively, while all the columnshave a section of 50/30 cm. The wall elements have a thickness of 20 cm.

Considering as the effective mass m only that corresponding to the self-weight ofthe slab:

slab self-weight¼ 15.40 8.0 0.25 25.0¼ 770 kN

m¼ 770/9.81¼ 78.5 t

The distinction between the two plane systems is made as shown in Figure 16.30.

569

F

F

F

m2

m1

m3

F

F

F

m2

m1

m3

A = F · (∆1 + ∆2 + ∆3)

B = m1 · ∆12 + m2 · ∆2

2 + m3 · ∆32

(F = 1)

ω2 = A/B

∆ = F · [∑ (S)x]–1

Sense x–x

∆ = F · [∑ (S)y]–1

Sense y–y

∆1

∆3

∆2

∆1

∆3

∆2

Figure 16.30 Practical determination of the fundamental eigenfrequency of a multi-storey system

3.00

2.50

2.50

6.00

1.00

1.00

2.00 3.708.001.70

Figure 16.31 Plan view layout of vertical elements of stiffness


By applying horizontal unit loads at the corresponding levels of each plane system(F¼ 1 kN), the following displacements result:

. Plane system in the x direction:

1¼ 0.0477 mm, 2¼ 0.0298 mm, 3¼ 0.0117 mm

. Plane system in the y direction:

1¼ 0.0773 mm, 2¼ 0.0455 mm, 3¼ 0.0161 mm

According to the expression for !2 (see Section 16.3.2.1)

. For the x—x sense:

!2 ¼ 1 0:0477þ 1 0:0298þ 1 0:0117

78:5 ð0:04772 þ 0:02982 þ 0:01172Þ¼ 344:3 s2

!¼ 18.50/s

T¼ 2 p/!¼ 0.465 s

. For the y—y sense:

!2 ¼ 1 0:0773þ 1 0:0455þ 1 0:0161

78:5 ð0:07732 þ 0:04552 þ 0:01612Þ¼ 213:1 s2

!¼ 14.60/s

T¼ 2 p/!¼ 0.43 s

16.3.3 Forced vibrationThe response of a multi-degree system subjected to external forces is governed, asanalysed in Section 16.3.2.1, by means of a system of differential equations, whichhas been expressed in a condensed matrix form as (Muller and Keintzel, 1984)

m €xxþ c _xxþ k x ¼ FðtÞAs in the case of free vibration, it is also found here that the multi-degree systembehaves as a system of N discrete and independent single-mass systems, having afictitious mass M, as well as a fictitious stiffness K, according to the expressions inSection 16.3.2.1 (Figure 16.32). Each such system is governed by the equation

Mr €qqr þ 2 r Mr !r _qqr þ Kr qr ¼ ’Tr FðtÞ

which, for the usually encountered case where F(t)¼P f(t), i.e. when all loads exhibitthe same variation in time, may be rewritten as

Mr €qqr þ 2 r Mr !r _qqr þ Kr qr ¼ ’Tr P fðtÞ ¼

XN1

’ðrÞi Pi

fðtÞ

Although this equation can be treated analytically, as shown in Section 16.2.3, thenumeric solution obtained using suitable computer software is more practical, provided

570


that the time steps used in the solution are smaller than 1/10th of the correspondingeigenperiod Tr¼ 2p/!r.

It should be noted that the determination of the displacements qr(t) of the N fictitioussingle-degree systems allows the calculation of the displacements x(t) of the masses inthe multi-degree system, according to the equation x(t)¼ q(t), where is asquare matrix (NN), obtained from the juxtaposition of the N column matrices ’r

(see Figure 16.32). Moreover, the fact that each eigenform may be considered to bemultiplied by an arbitrary factor does not affect the resulting displacements x(t) atall, as this factor, according to the above equations, is eliminated.

16.3.4 Seismic excitation

16.3.4.1 Dynamic analysisThe governing factor for the response of a system to seismic excitation is, just as in thesingle-degree case (see Section 16.2.5), the accelerogram €uusðtÞ, expressed as

€uusðtÞ ¼ a faðtÞwhere a represents the maximum-occurring ground acceleration (Figure 16.33).

It is obvious that the inertia mass forces are caused by their absolute acceleration,whereas the elastic as well as the damping forces acting on the masses depend onlyon their relative displacements to the ground. Given that no external forces areacting on the system, the condensed equation for dynamic equilibrium according to

571

P2 · f(t)

P1 · f(t)

P3 · f(t)

m2

m1

m3

m2

m1

m3

m2

m1

m3

m2

m1

m3

x(t) x(t) x(t)

q1 q2 q3

K1 K2 K3

M1 M2 M3

The sum of x yields the final displacements and the responseof the structural web through the stiffness matrix [k]

1st eigenformx = q1 · 1

2nd eigenformx = q2 · 2

3rd eigenformx = q3 · 3

1T · P · f(t) 1

T · P · f(t)1T · P · f(t)

Figure 16.32 Dynamic response of a multi-degree system under external loads


the previous section is written as

m €xxþ I a faðtÞ þ c _xxþ k x ¼ 0

This equation can also be written as

m €xxþ c _xxþ k x ¼ m I a faðtÞwhere I represents a column matrix containing N unit elements.

Thus, the problem of seismic excitation is posed as a problem of forced vibrationaccording to the previous section, the multi-degree system being supported on fixedground and subjected to the external loads expressed by the right-hand side of thelast equation (see Figure 16.33).

According to the foregoing, the multi-degree system is ‘split’ into N single-massoscillators, each one corresponding to a respective eigenfrequency ! with the accom-panying eigenform ’. On this basis, the corresponding fictitious mass Mr may be deter-mined as well as the fictitious stiffness Kr. The equation of motion for the rtheigenfrequency is written as (Muller and Keintzel, 1984)

Mr €qqr þ 2 r Mr !r _qqr þ Kr qr ¼ ’Tr m I a faðtÞ

and, dividing by Mr,

€qqr þ 2 !r _qqr þ !2r qr ¼ ’T

r m I 1

Mr

a faðtÞ

This equation is identical to the equation of motion of the single-mass system inSection 16.2.5, which is repeated here:

€xxþ ð2 !Þ _xxþ !2 x ¼ a faðtÞ

572

m2 · a · fa(t )

m1 · a · fa(t )

m3 · a · fa(t )

m2

m1

m3

m2

m1

m3

üs(t ) = a · fa(t )Accelerogram

(a: Maximum ground acceleration)

Fixed support

Figure 16.33 Seismic loading of a multi-degree system


It is observed that the maximum ground acceleration a is now multiplied by the factor

r ¼ ’Tr m I

1

Mr

¼P

mi ’ðrÞiP

mi ½’ðrÞi 2

which is a pure number and expresses the degree to which the rth eigenform isparticipating in causing the response of the system (Figure 16.34), and is thuscalled the participation factor. Its value is less than unity, while the sum of all the Nparticipation factors r is logically equal to 1. Usually, the first (fundamental) eigenformexhibits the greatest participation factor, which gradually decreases for the highereigenfrequencies.

According to the results for the single-mass oscillator, it is possible to determinethe corresponding (qr)max through the acceleration spectrum for an earthquake, onthe basis of the corresponding value SðrÞa for the rth eigenfrequency, considering asmaximum ground acceleration the value ( r a) instead of a. Hence, according toSection 16.2.5,

ðqrÞmax ¼1

!2 ð€qqr þ €uusÞmax

¼ 1

!2 r a !

ðt0faðÞ e ! ðt Þ sin! ðt Þ d

max

that is,

ðqrÞmax ¼SðrÞa

!2r

r

573

m2 · a · fa(t )

m1 · a · fa(t )

m3 · a · fa(t )

Γ1 · M1 · a · fa(t ) Γ2 · M2 · a · fa(t ) Γ3 · M3 · a · fa(t )

[∑ mi · ϕi]Γi = [∑ mi · ϕi

2]

m2

m1

m3

K1 K2 K3

M1 M2 M3

qmax = Γ1 · [Sa(1)/ω1

2] qmax = Γ2 · [Sa(2)/ω2

2] qmax = Γ3 · [Sa(3)/ω3

2]

Superposition is not valid

1st eigenformxm

(1a)x = qmax · 1

P (1) = k · xm(1

a)x

2nd eigenformxm

(2a)x = qmax · 2

P (2) = k · xm(2

a)x

3rd eigenformxm

(3a)x = qmax · 3

P (3) = k · xm(3

a)x

Figure 16.34 Seismic analysis of a multi-degree system


Now, for each oscillator r, the mass displacements (xmax)r of the multi-degree systemmay be determined on the basis of the computed (qr)max (see Figure 16.34):

ðxmaxÞr ¼ ’r ðqrÞmax ¼ r SðrÞa

!2r

’r

Thus, a maximum deviation is assigned to each mass, corresponding to each of the Nfictitious oscillators. Of course, the actual displacement of a specific mass cannot bededuced as the sum of the above ‘component’ maximum deviations (xmax)r on thebasis of the N oscillators, because the corresponding (qr)max do not refer to the sametime instant. Nevertheless, according to an acceptable probabilistic consideration, thismass deviation may be obtained as the square root of the sum of the squares of theabove ‘components’.

Considering now the determination of the maximum response in the structural web, itcan be seen that for the realisation of the above displacements (xmax)r of the N masses,the action of those forces Pr is required, which are suggested by the stiffness of thestructural web itself according to the equation

Pr ¼ k ðxmaxÞr ¼ r SðrÞa

!2r

k ’r

The determination of the stiffness matrix k follows Section 16.3.1.1It is thus possible to determine the N systems of external forces P for the maximum

(real) deviations xmax corresponding to the N (fictitious) oscillators. However, as theseforces do not correspond to simultaneously developed displacements, their superpositionis not meaningful, and one has to use the previous treatment (the square root of thesum of their squares) in order to obtain an acceptable approximation for loading ofthe structural web.

16.3.4.2 Equivalent static loadsThe bulk of the calculations in Section 16.3.4.1 that take into account the contributionof all the eigenforms of the system in its seismic excitation are not especially manageable,particularly for preliminary design purposes. This inconvenience led to the search fora static loading acting on the masses of the system such that the resulting responseapproximates the ‘exact’ dynamic analysis. A practical suggestion to this end is thefact that, for a regular distribution of mass and stiffness in a structural web, it is thefirst eigenform that plays the governing role in the dynamic response of the web(Muller and Keintzel, 1984).

Thus, in seeking these equivalent static loads, it is assumed, as an approximation, thatthese may be represented by the inertia forces H to which the masses in the system aresubjected, if only the first eigenform (r¼ 1) is taken into account. According to Section16.3.4.1 and Figure 16.35 (see also Section 16.2.5),

H ¼ m ½ð€xxÞmax þ I a ¼ !2 m ðxÞð1Þmax ¼ !2 ðq1Þmax m ’

574


On the basis of the expression for (q1)max in Section 16.3.4.1,

H ¼ m ’ P

mi ’iPmi ’2

i

Sa

Looking now for a more convenient expression for these forces, their sum (P

H) overthe whole height of the structural web is considered. It is clear that this force constitutesthe sum of the actions to which the masses are subjected, transmitted directly to theground. It is thus found that

XH ¼

Pmi ’ið Þ2Pmi ’2

i

Sa

This expression reveals that the factorPmi ’ið Þ2Pmi ’2

i

¼ Xðmi ’iÞ

according to Section 16.2.5, represents a mass mm which, given that the value of for thefirst eigenform is near unity, may be considered as approximately equal to the sum M ofthe masses in the system, i.e.

mm ¼ Xðmi ’iÞ ffi

Xmi ¼ M

Thus,XH ¼ V ¼ M Sa

575

m2 · ω2 · xm

(1a)x

m1 · ω2 · xm

(1a)x

m3 · ω2 · xm

(1a)x

A · m2 · ϕ2

A · m1 · ϕ1

A · m3 · ϕ3

Ah · m2 · h2

Ah · m1 · h1

Ah · m3 · h3

m2

m1

m3

m2

m1

m3

m2

m1

m3

1st eigenformxm

(1a)x = qm

(1a)x · 1

Equivalent static loading

h3

h2

h1

Γ1 · [∑ mi · ϕ i] · Sa V = [∑ mi] · Sa

Base shearV = [∑ mi] · Sa

Base shear

∑ miA = ·Sa ∑ mi · ϕ i

∑ miAh = ·Sa ∑ mi · hi

Figure 16.35 Determination of equivalent static loading in a multi-storey system


This V force is called the base shear, and is used in the exact same sense as in Section16.2.5 (see Figure 16.35).

Now, the expression of the equivalent static loads H is written on the basis of theabove as

H ¼ m ’ 1Pmi ’i

P

mi ’ið Þ2Pmi ’2

i

Sa ¼ m ’ VPmi ’i

and, for each mass separately,

Hi ¼ V mi ’iPmi ’i

This simpler expression, though less accurate than the previous one, has neverthelessbeen adopted in most of the regulations (see Figure 6.35). However, in the search foran even simpler expression for the equivalent static loads for preliminary designpurposes, a further simplification is undertaken, namely the consideration of the firsteigenform as a straight line. Thus, by using the level heights hi instead of the eigenformcoefficients ’, the following final expression is obtained:

Hi ¼ V mi hiPmi hi

(see Figure 16.35)

More specifically, for a multi-storey building — as considered in Section 16.3.1.4 andfrom a dynamic viewpoint in Section 16.3.2.2 — which is subjected to a given groundacceleration, in order to determine the equivalent horizontal static load for eachlevel, the following procedure should be followed:

(1) On the basis of the fundamental eigenfrequency for both the x and y senses, calculatedaccording to Section 16.3.2.2, as well as of the maximum adopted ground acceleration,first the value Sð1Þa and then the base shear V are determined from the accelerationspectrum used separately for both senses in the two discrete plane systems.

(2) The corresponding equivalent static loads are obtained through direct application ofone of the two last-obtained expressions. In the case where the eigenform values ’i

are used, these may be considered to be the already determined displacements forthe equivalent plane system under horizontal unit loads, applied in order to assessits fundamental frequency, as described in Section 16.3.2.2. It is clear that for a uni-form distribution of masses, these static loads increase continuously for higher storeys.

(3) On the basis of the acting forces on each diaphragm of a storey, the response of thevertical — as well as the horizontal — elements of the multi-storey system may bereadily assessed, according to Sections 15.3.1 and 15.3.2.

Numerical exampleContinuing the example in Section 16.3.2.2, on the basis of the design spectrum inEurocode EC8 and for soil class B, both for the sense x—x with a fundamental periodT¼ 0.34 s and for the sense y—y with a fundamental period T¼ 0.43 s, a maximumspectral acceleration Sa¼ 2.50 is determined. Assuming a maximum ground acceleration

576


equal to 0.16g, the following total seismic force (base shear) is obtained:

V¼ 3 78.5 0.16 9.81 2.50¼ 924 kN

Considering an earthquake in the x direction, the equivalent static loads acting on thecentroid of each of the three diaphragms (slabs) are estimated on the basis of deviations’. According to Section 16.3.2.2:X

mi ’i ¼ 78:5 ð0:0000477þ 0:0000298þ 0:0000117Þ ¼ 0:007 t

Thus, the loads acting in the x direction are (see Figure 16.35)

P1¼ 924 78.5 0.0000477/0.007¼ 494 kN

P2¼ 924 78.5 0.0000298/0.007¼ 309 kN

P3¼ 924 78.5 0.0000117/0.007¼ 121 kN

Considering an earthquake in the y direction, the equivalent static loads are analogouslyestimated. According to Section 16.3.2.2:X

mi ’i ¼ 78:5 ð0:0000773þ 0:0000455þ 0:0000161Þ ¼ 0:011 t

The corresponding forces acting in the y direction are (see Figure 16.35)

P1¼ 924 78.5 0.0000773/0.011¼ 512 kN

P2¼ 924 78.5 0.0000455/0.011¼ 302 kN

P3¼ 924 78.5 0.0000161/0.011¼ 110 kN

i.e. essentially with the same distribution as the x sense.The assessment of the response of a multi-storey system under the above loads, for

both the x and y senses, may be deduced by following the procedure described inSection 15.3.1.

16.4 Approximate treatment of continuous systemsAs pointed out at the beginning of this chapter, the load-bearing structures have de factoa continuous mass distribution. However, the corresponding inertia forces that have tobe considered in order to express the equilibrium of each element leads to unmanageabledifferential equations, which excludes their use in structural design.

However, the discrete systems examined so far, although not corresponding absolutelyto physical reality, are nevertheless able to express the dynamic behaviour of anystructure, by considering its mass lumped in as many places of the structural web asdesired and by following the previously described procedures.

The total (continuous) mass m of a beam, for example, may be partitioned into fivelumped equidistant masses (m/5) distributed over the structural web, thus creating asystem with five degrees of freedom, if only those inertia forces, which correspond totransverse displacements, are counted (Figure 16.36).

577


In the same sense, the mass m of a plate may be divided in 25 equidistant masses (m/25),corresponding to a system with 25 degrees of freedom based only on the transversedeflections (Figure 16.36).

However, the analysis of the above discrete systems requires the use of appropriatecomputer programs. This does not directly meet the needs of a preliminary designwith respect to basic design parameters, such as, for instance, the fundamental frequencyof the system, or its maximum deformation, or even its maximum acceleration underspecific dynamic loads. Thus, it is appropriate — wherever possible and at least for theusual cases — to pursue the equivalent dynamic behaviour of the continuous systemsthrough single-mass systems exhibiting the same frequency as the fundamentalfrequency of the examined system. This idea is based on the fact that, in most practicalcases, the fundamental eigenfrequency of a system plays the governing role in itsdynamic behaviour, as pointed out in Section 16.3.4.2.

Figure 16.36 shows the ‘substitution’ of a beam and plate by a corresponding‘equivalent’ single-mass system, for which the mass, stiffness and equivalent dynamicloading are to be determined (Biggs, 1964).

The determination of the above parameters is based on the fact that, as has beendiscussed in Sections 16.3.2 and 16.3.3, the dynamic analysis of an N-degree offreedom system goes back to the analysis of N independent single-mass oscillators,

578

EQUIVALENT OSCILLATOR

Structural web

Structural web

mmm m m

m

m

mmm

P · f(t ) P · f(t )P · f(t )

P · f(t ) P · f(t ) P · f(t )

P · f(t )

δ = 1δ = 1

keff: the total load inducing δ = 1

keff: the total load inducing δ = 1

Ke = λP · keff Ke = λP · keff

Me = λM · ∑ m Me = λM · ∑ m

Fe = λP · ∑ PFe = λP · ∑ P

[ · f(t )] [ · f(t )]λM = (∑ ϕ2)/N

λP = (∑ ϕ)/N

Figure 16.36 Equivalent substitution of a continuous structure through a single-mass system


and, as explained above, it is the oscillator with the lowest frequency that can mostsuccessfully represent the whole system.

As described in Section 16.2.2, the mass Me of this oscillator can be expressed as

Me ¼ ’T1 m ’1 ¼

XNs¼ 1

ms ’2s ¼ M

XNs¼ 1

ms

where ’s are the deviation values of the first eigenform of the N-degree system and, in thecase of a uniformly distributed total mass M in the discrete model, the mass factor M is

M ¼PN

s¼ 1 ’2s

N(dimensionless number)

Thus, the mass Me of the equivalent oscillator is Me¼M M, expressed as a part(M< 1) of the total mass M of the system (see Figure 16.36).

Assume, now, that the discrete system is subjected to the dynamic loading

F(t)¼P f(t)

where matrix P consists of the specific loads acting on the N masses of the system.According to Section 16.3.3, the loading term of the single-mass ‘component’

oscillator with the lowest frequency (i.e. the ‘fundamental oscillator’) of the N-degreesystem is

F1 ¼ ’T1 P fðtÞ ¼

XNs¼ 1

Ps ’s

! fðtÞ

Considering that load Fe f(t) will act on the equivalent oscillator, then

Fe ¼XNs¼ 1

Ps ’s ¼ P XNs¼ 1

Ps (see Figure 16.36)

and, in the case of an equally distributed total load to the N masses of the discretesystem, the load factor P will be

P ¼PN

s¼ 1 ’s

N(dimensionless number)

It is clear that in the case of a uniform distribution of load p (kN/m) on a beam of lengthL,P

Ps¼ ( p L), whereas for a uniformly distributed load q (kN/m2) on an orthogonalslab with dimensions (a, b),

PPs¼ (q a b).

It should be noted that the basic aim of seeking an equivalent single-mass oscillator isto determine the kinematic behaviour (displacement, velocity, acceleration) of a specificcritical point of the initial system, and this, for preliminary design purposes in the case ofa single span beam or a plate, is obviously the central point of the structure. To this end,and extending the initially adopted notion of stiffness in Section 2.3.8, the concept ofthe effective stiffness keff is introduced as the total load which, under a given andprescribed distribution, causes unit displacement at the reference point of the structure

579


— i.e. at its midpoint (see Figure 16.36). It is, then, clear that the deflection at thatpoint, due to the total load

PPs, is, for the considered distribution (see Figure 16.37),

¼P

Ps

keff

For a uniform load distribution, in the case of a beam with length L, keff¼ 76.8 EI/L3,whereas in the case of a simply supported square plate with side L, keff¼ 271 EI/L2.

Thus, the sought-after stiffness Ke of the equivalent oscillator results from therequirement that the transverse displacement at the considered reference point ofthe continuous system, due to the static loads Ps, should be identical to the displacementof the equivalent oscillator under the load Fe¼P

PPs:

¼P

Ps

keff

¼ P P

Ps

Ke

and, hence,

Ke ¼ P keff (see Figure 16.36)

The fundamental frequency f1 of the continuous system, being equal to that of theequivalent oscillator, is thus obtained as

f1 ¼1

2p

ffiffiffiffiffiffiKe

Me

s¼ 1

2pffiffiffiffiffiffiP

M

sffiffiffiffiffiffikeff

M

r

where the factors P and M refer to a uniform distribution of load and mass, respectively.The two tables in Figure 16.37 give the values of the mass and load factors P and

M for various cases of beams with distributed or concentrated loads, as well as forperimetrically supported plates (Biggs, 1964).

16.5 Design for avoiding annoying vibrations

16.5.1 Human activitiesApart from the seismic disturbance of multi-storey systems, which has been examined inSection 16.3.4, checking for transverse vibrations of horizontal load-carrying elements —such as beams or plates in normal use by humans — constitutes a major design issue inseveral cases (Bachmann and Ammann, 1987). A pedestrian bridge, for example, or anoffice floor, or even a floor designed for sport or dance activities, should in no caseundergo vibrations which are felt by the people in a disturbing way. Such vibrationsoccur when the frequency with which the foot strikes the floor lies near the fundamentalfrequency of the load-carrying structure.

Within the framework of the preliminary design goals set out in this book, andwithout the need in the present context to dynamically analyse the human activity initself, footbridges should be designed with a fundamental frequency outside the range1.6—2.4 Hz and, if possible, not lower than 4.50 Hz. For floors intended for sportactivities, this frequency should not be less than 6.8 Hz, whereas for dance floors the

580


minimum value is 6.0 Hz. It is worth mentioning that, for the above uses, the maximumtolerable acceleration of the human body is about 5% of acceleration due to gravity g.Finally, in the case of office floors, where lower acceleration limits are required, a funda-mental frequency of at least 7.50 Hz should be targeted (Bachmann and Ammann,1995).

16.5.2 Vibrations induced by machinesThe running of machinery generally causes vibrations in the part of the structure inwhich it is installed, which may be felt by and annoy people standing near it. Theinstalled machines produce continuous periodical sinusoidal loading, characterised bythe maximum value of force F0 under a frequency f0, according to the data suppliedby the vendor.

As a measure of the acceptability level of accelerations felt by a person, the so-calledsensitivity factor K can be considered, which is calculated using the empirical formula(Bachmann and Ammann, 1987)

K ¼ d0:80 f2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ 0:032 f2p

where d is the maximum amplitude of vibration (mm) and f is the fundamentaleigenfrequency (Hz) of the structural part supporting the machine.

Machines are supported on a floor structure either directly or through appropriatesprings. The design goal is usually to limit the deflection max and/or the velocity vmax

of the supporting structural part at values tolerable by the machine, as well as to limit

581

1.0 0.31 0.45 271

0.9 0.33 0.47 248

0.8 0.35 0.49 228

0.7 0.37 0.51 216

0.6 0.39 0.53 212

0.5 0.41 0.55 216

1.0 0.21 0.33 870

0.9 0.23 0.34 798

0.8 0.25 0.36 757

0.7 0.27 0.38 744

0.6 0.29 0.41 778

0.5 0.31 0.43 866

(a) (b)

Type of loadingand beamLength L

Massfactor

λM

Loadfactor

λP

Effectivestiffness

keff

( · EI/L3)

Uniform loadon plate

a/b Massfactor

λM

Loadfactor

λP

Effectivestiffness

keff

( · Eh3/12 · a2)

0.5 0.637 76.80

0.5 1.0 48

0.479 0.595 185

0.479 1.0 107

0.396 0.523 384

0.396 1.0 192

Plate perimetricallysimply supported

Plate thickness: h

Plate perimetricallyfixed

Plate thickness: h

b

a

b

a

Figure 16.37 Equivalent substitute parameters of single-mass systems for beams and plates


the sensitivity factor K to under 2.0, so that the floor vibration is tolerable by peoplestanding nearby. g

In the case where the machine is directly supported on the floor, the whole system maybe described by a fictitious oscillator with a determinable mass Me and spring stiffnessKe, under an equivalent dynamic loading Fe (Bachmann and Ammann, 1987). Thisoscillator exhibits a damping coefficient identical to that of the supporting structureof the floor (Figure 16.38).

The mass Me of the equivalent oscillator is determined as the sum of the equivalentvibrating mass M M of the supporting structural part (usually a slab) having the massM and the mass m of the machine itself. So,

Me¼M Mþm

The spring stiffness Ke is expressed according to Section 16.4 through the effectivestiffness keff of the supporting part, multiplied by the factor P (see Figure 16.38).

If the fundamental eigenfrequency f1 of the supporting structure being calculated,according to Section 16.4,

f1 ¼1

2pffiffiffiffiffiffiP

M

sffiffiffiffiffiffikeff

M

r

is lower than the operation frequency f0 of the machine, instead of calculating thestiffness Ke of the oscillator as P keff, it is more useful to consider it equal to

Ke ¼ !20 Me ¼ ð2 p f0Þ2 Me

582

umax = Fe/(2 · ζ · Ke)

umax = umax · ω0

ümax = umax · ω02

Sensitivity factor K

Direct bearing of machine

m

M

F

F0

To be checked:

T0

t

ω0 = 2π/T0

Dynamic loading imposed by the machine

EQUIVALENT OSCILLATOR

max Fe = λP · F0

Me = λM · M + m

Ke = λP · keff Ke = ω02 · Me

Figure 16.38 Direct support of a machine in a continuous system


in order to take account of higher frequency values being applied to the supportingstructural part, which can lead to resonance.

Then, on the basis of the adopted damping coefficient for the structure, thedetermination of its maximum deflection umax — and consequently of the machineitself — is possible, according to Section 16.2.4, as

umax ¼1

2 Fe

Ke

¼ 1

2 P F0

Ke

which may be directly compared to the maximum max supplied by the vendor.Moreover, the maximum velocity and the maximum acceleration developed can bechecked, according to Section 16.2.2.1, as

_uumax ¼ umax ! ¼ ð2 p f0Þ umax

and

€uumax ¼ umax !2 ¼ ð2 p f0Þ2 umax

respectively (see Figure 16.38).However, of decisive importance for the acceptance (or not) of the dynamic

behaviour of the system is the value of the sensitivity factor K, according to the foregoingformula:

K ¼ umax

0:80 f20ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ 0:032 f20

qThis factor should not exceed a value of 2.0. g

If the above solution does not lead to acceptable results, then, in order to limit theresponse of the supporting structure, the machine should be placed on springs withappropriately selected stiffness. The two masses, of the machine and of the structure,are separated as shown in Figure 16.39, thus constituting a two-degree system consistingof an ‘upper’ mass mM supported through a spring having stiffness kM on the ‘lower’ massmS, being in turn supported through a spring of stiffness kS on a solid base.

The quantities mS and kS for the structure can be considered to be known, but thecharacteristics of the ‘upper’ system have to be determined, namely the machine massmM and the spring stiffness kM. The ‘machine mass’ is not necessarily the mass m onlybut can also include any appropriately attached mass mB — e.g. a concrete pedestal —required to fulfil the dynamic requirements (see Figure 16.39).

The eigenfrequencies f1 and f2 of the two-degree system may be directly determinedthrough the expression (Bachmann and Ammann, 1987)

f1=2 ¼1

8 p2 kM

mM

þ kM þ kS

mS

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikM

mM

þ kM þ kS

mS

2s

4 kM kS

mM mS

0@

1A

24

351=2

It is generally ensured whenever possible, through appropriately selected values, that thetwo frequencies are clearly separate, e.g. by making one a multiple of the other. This

583


allows the two systems to be decoupled, so that each mass can be considered asconnected to a fixed base through its spring. Then, the frequencies f1 and f2 obtainedfrom the above expression coincide — at least approximately — with the frequenciesfM and fS of the individual oscillators, respectively.

Thus, if the two requirements

fM ¼1

2 p

ffiffiffiffiffiffiffikM

mM

s¼

1

8 p2kM

mM

þ kM þ kS

mS


mM

þ kM þ kS

mS

2s

4 kM kS

mM mS

1=2

and

fS ¼1

2 p

ffiffiffiffiffiffikS

mS

s¼

1

8 p2kM

mM

þ kM þ kS

mS

þ


mM

þ kM þ kS

mS

2s

4 kM kS

mM mS

1=2

lead — approximately — to technically plausible values for the sought-after quantities mM

and kM, then the dynamic uncoupling of the two masses is feasible.Then, in the system [mM, kM], the maximum machine displacement as well as the

maximum transmitted force on the base of the machine are first determined, accordingto Section 16.2.4, and, on the basis of this force, the system [mS, kS] is afterwards

584

Two-degree system

If mass and spring stiffness lead to frequenciesapproximately equal to those of the two-degree systemthen the decoupling of the two oscillators is possible

Fg

kS

kS

kM

kM

mM

mM

mS

mS

To be checked:

umax

Factor K

Fgmax F0

Known

Unknown(to be determined)

mM = m + mB

(M)

(S)

(M)

(S)

To be determined

To be determined

m

mDynamic loading imposed by the machine

F

T0

t

F0

mS

kS

mS

kS

mB

kM

kM

Possibly required additionalattached mass mB

Figure 16.39 Support of a machine in a continuous system through springs


analysed with respect to its relevant dynamic characteristics of interest, which concernthe supporting structure itself.

More specifically, by considering for the machine component — according to theabove — its total mass (mM¼mþmB), as well as the stiffness kM of its correspondingspring (see Figure 16.39), the corresponding eigenfrequency is obtained as

fM ¼1

2p

ffiffiffiffiffiffiffikM

mM

s

The mass of the equivalent oscillator with a damping coefficient will be equal to

mS¼M MThe spring stiffness kS is determined through the effective stiffness keff, according toSection 16.4, as

kS ¼ P keff

The eigenfrequency fS, representing the fundamental eigenfrequency of the supportingstructure, is

fS ¼1

2p

ffiffiffiffiffiffikS

mS

s

If, now, the two systems are considered to be uncoupled, as previously explained, then,for the mass mM (machine), the maximum displacement umax as well as the maximumtransmitted force Fg to the supporting base can be readily determined, according toSection 16.2.4, through the following equations (¼ 0):

umax ¼ ustat 1

1 ð=!Þ2¼ F0

kM

1

1 ð f0=fMÞ2(must be <max)

and

Fg ¼ F0 1

1 ð f0=fMÞ2¼ umax kM

Due to decoupling, the above force may be considered to be acting directly on thesupporting structural system, with a cyclic frequency corresponding to fM. Thus, theresulting maximum displacement may also be expressed, according to Section 16.2.4, as

umax ¼Fg

kS

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1 fM=fSÞ2 þ ð2 fM=fSÞ2

qAs was previously pointed out, the decisive criterion for the acceptance of thevibrational characteristics of the supporting structure is the value of the correspondingsensitivity factor K, which may be directly calculated as

K ¼ umax

0:80 f2Sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ 0:032 f2S

q

585


ReferencesBachmann H., Ammann W. (1987) Vibrations in Structures — Induced by Man and Machines. Zurich:

IABSE.Bachmann H., Ammann W. (1995) Vibration Problems in Structures. Boston: Birkhauser Verlag.Biggs J.M. (1964) Introduction to Structural Dynamics. New York: McGraw-Hill.Muller F.P., Keintzel E. (1984) Erdbebensicherung von Hochbauten. Berlin: Ernst.Stavridis L. (1986) Static and dynamic analysis of multistory systems. Technika Chronika Scientific

Journal of the Technical Chamber of Greece 6(2), 187—219.Tedesco J.W., McDougal W.G., Ross C.A. (1999) Structural Dynamics — Theory and Applications.

California: Addison Wesley Longman.

586


17

Supporting the structure on the ground

17.1 OverviewAll structural systems can be considered as free bodies in equilibrium under the loads forwhich they have been designed and the forces acting on them from the ground on whichthey are supported. These forces (i.e. the ‘reactions’) are, of course, accompanied byother equal and opposite forces, which are acting on the soil. The transfer of the last-mentioned forces to the ground must be accomplished in an absolutely safe mannerregarding induced deformations and the soil stresses developed. For the successfultransfer of such a force to the ground, an appropriately formed contact region of thestructure with the soil is required. This constructional layout is called a foundation,and consists of a concrete structure firmly connected to the superstructure, so that bythe term ‘load-bearing structure’ one means the whole system of ‘superstructure’ plus‘foundation’ which is actually supported on the bearing ground. In this sense, thefoundation through its contact surface with the soil has a double role: it must bedesigned to withstand the self-equilibrating system of soil stresses and the actionsfrom the superstructure — equal and opposite to the developing ‘reactions’ — on theone hand, and to introduce such stresses in the soil mass that they can be safelytaken up by it, on the other (Figure 17.1). In this respect, another point comes intoquestion, namely the fact that the implied soil deformations — as also imposed on thestructure (the foundation included) — create an additional response to the otherwiserigidly supported structural system (see Figure 17.1). These deformations are, ofcourse, not known in advance, but they arise as a result of the soil—structure interaction,and in this respect the foundation design plays a decisive role.

Thus, in the design of a structure, the factor ‘supporting soil’ is — through thefoundation — always directly involved, influencing the whole structural concept.

The following development is not intended to cover the examination of themechanical properties of soils, nor the various constructional layouts of the foundationpossibilities. The aim is simply to point out particular basic structural characteristicsinvolved in the design of some typical forms of foundation systems, in order to assistthe understanding of their load-bearing behaviour.

17.2 General mechanical characteristics of soilsThere are two main categories into which the vast majority of soil types fall: the so-callednon-cohesive and the cohesive soils.


17.2.1 Non-cohesive soilsNon-cohesive soils (sand, gravel) constitute in dry conditions a mass of loose, uncon-nected grains, not smaller than 0.06 mm, mutually transmitting friction forcesthrough their small contact areas. These friction forces are proportional to the normalpressure acting on the contact surface, through the angle of internal friction ’ of thesoil examined, the maximum value being considered as its shearing strength. Itshould be pointed out that the shearing strength of soils represents the determiningfactor for their actual resistance, unlike solid materials, whose strength is instead repre-sented by normal tensile or compression stresses. So, ¼ tan’.

Water, which can penetrate the gaps between the grains, easily escapes underpressure. This is why settlement in these soils is developed within a short period oftime, once the loads are imposed. Non-cohesive soils, in a layer of medium to highdensity and of adequate thickness, generally constitute a good foundation ground. Itshould be noted, however, that in fine-grained, loose and water-saturated soil masses,the danger of loss of shearing strength exists (i.e. the loss of friction between thegrains), known as liquefaction, particularly in the case of intense cyclic loadinginduced by a strong earthquake.

The specific weight of these soils in dry conditions is of the order of 20 kN/m3.However, when the soil layer lies under the water level, the soil grains are underbuoyancy and the above value is reduced to 20—10¼ 10 kN/m3.

Non-cohesive soils in natural free embankments exhibit a slope that tends to coincidewith the angle ’ of internal friction (Figure 17.2). If, however, this embankment slope is

588

State of stress?

State of stress?

(Equilibrium)

(Equilibrium)

GroundDeveloping deformations

Ground

Foundation

The same deformations imposedFoundation

Structure

Structure

Figure 17.1 The system superstructure—foundation—soil interaction


prevented through a vertical front, e.g. a retaining wall against a certain soil volume,then this front is acted on over its entire height H by soil pressures that increase asthe internal angle of friction ’ decreases. The horizontal component H of thesestresses is proportional to the prevailing vertical pressure V in the considered depthh, due either to the soil self-weight (V¼ h), or to a vertical distributed load pacting on the free soil surface (V¼ p). Thus H¼K V, where the coefficient Krepresents the reducing effect of the internal friction (e.g. against water, where, ofcourse, K¼ 1), moving between a minimum value: Ka¼ tan2 (45’/2) and amaximum value K0¼ 1 sin’ (see Figure 17.2). The minimum value Ka correspondsto the case where slight resilience of the vertical front is allowed, referring to the so-called active pressure, whereas the maximum value K0 corresponds to an absolutelyunyielding front, and refers to the so-called pressure at rest. The resultant horizontalforce Eh acting on the vertical front due to the soil self-weight, under a triangularpressure distribution, is then Eh¼ H2 K/2, whereas that due to a surface load p itis Eh¼ p K H (uniform pressure distribution). For the combined action of the twoloadings, the above results are accordingly superposed (see Figure 17.2). Moreover, inFigure 17.3(a), the influence of a differentiated value of K in the case of a layered soilmass is shown.

If water is present in the soil mass, the specific weight of the soil is reduced to0 ¼ W, due to buoyancy, and this should, together with the full hydrostaticaction on the retaining structure, be taken into account (Figure 17.3(b)). It shouldbe noted that the angle of internal friction ’ remains essentially unaffected.

589

σH = K · σV = K · γ · H σH = K · σV = K · p + K · γ · H

p

p

Retaining frontRetaining front

Embankment slope ~ angle of internal friction

z

H

γ, ϕ

EhEh

ϕ

tan2(45 – ϕ/2) ø K ø 1 – sin φ

σV = γ · z + p

σH = K · σV

(K = 1 – sin ϕ)

Figure 17.2 Earth pressure in non-cohesive soils


17.2.2 Cohesive soilsCohesive soils (clay) constitute a consistent mass even in dry condition. They consist ofparticles of oblong or plate form with diameters ranging from 0.0002 mm up to 0.06 mm,having many more common contact points than non-cohesive soils. Given that thevoids between the particles are much larger than the dimensions of the grainsthemselves, any water present cannot escape as easily as in non-cohesive soils underan applied external pressure, because of the resistance offered by the many contactpoints of the grains. Thus, settlement develops very slowly and may continue forgreater periods of time (months or even years).

A basic characteristic of cohesive soils is, apart from certain friction forces betweengrains, the development of cohesion forces as a result of mutually acting electricalforces between the particles. It should be pointed out that friction forces are decreasedbecause of the existance of the water pressure u in the voids between the soil grains,called the pore pressure. The pore pressure, in the presence of a water table, is identicalto the hydrostatic pressure at the point considered.

Thus, the shearing strength basically consists of cohesion c (kN/m2), as well as of thefriction resistance 0 tan’. The stress 0 represents the actual normal pressure betweenthe solid grains, which, of course, results from the externally imposed normal pressure reduced by the existing pore pressure u. The angle of internal friction ’ generally hasmore moderate values than in non-cohesive soils. The cohesion c in water-saturatedsoils has a value of the order of 10—40 kN/m2. Thus,

¼ cþ ( u) tan’

590

Pressures on the retaining front

Pressures on the retaining front

h1

h2

h1

h2

1

2

γ, ϕ1

γ, ϕ2 < ϕ1

γ, ϕ

γ′ = γ – γW

K2 > K1

γW

The presence of water does not affectthe angle of internal friction

K · (γ · h1 + γ′ · h2) · γW · h2

(a)

(b)

K2 · γ · h1 K1 · γ · h1

K2 · (γ · h1 + γ · h2)

K1 · γ · h1

Figure 17.3 Earth pressures in a layered soil in the presence of a water table


The stress u is called the effective stress. It is clear that the shear strength of thecohesive soils depends, to a significant degree, on the presence of water in the pores,as well as on the cohesion c, which plays an important role, sometimes being moredecisive than the internal friction. It should be pointed out that the presence of watermay cause a decrease in the angle of internal friction ’, in contrast to whathappens in non-cohesive soils. Moreover, it is noted that in a cohesive saturated soil,the additionally imposed pressure on the soil mass will not imply a direct increase inthe contact pressure between the grains, as the extra pressure will be completelytaken up by the water mass, without thereby improving the shear strength of the soilat all.

The influx of water in cohesive soils generally affects their strength adversely, while apossible drying may give rise to pronounced deformations, accompanied by intensecracking, which will facilitate the further influx of water in the soil mass. In any case,however, under the same loading pressure the settlement of cohesive soils is greaterthan in the non-cohesive soils.

Regarding the pressures on vertical retaining walls, the presence of the cohesion cdefinitely has a relieving role (Figure 17.4). Thus, the total horizontal force on avertical front of height H consists of the force acting under the existing internalfriction and the relieving force due to cohesion:

Eh ¼ H2 K=2 2 c HffiffiffiffiK

p

It can be seen that the contribution of self-weight to the above force can be depicted by atriangular diagram, whereas the influence of cohesion is represented by a constant valueover the height. The treatment of coefficients K is the same as in the case of non-cohesive soils.

Moreover, in the above expression for Eh, the possible presence of a water table withinthe soil mass must be taken into account by the appropriate reduction of the specificweight due to buoyancy, this effect being equivalent to the fact that the porepressure u is equal to the exerted hydrostatic pressure.

591

Retaining front Retaining front

Internal friction(angle ϕ)

Cohesion c

Possible decreasein angle ϕ

Cohesion unaffected

γW

tan2 (45 – ϕ/2) ø K ø 1 – sin ϕ

2 · c · √K K · γ · H

Relievingeffect

Loadingeffect

2 · c · √K γW · H K · (γ – γW) · H

Relievingeffect

Loadingeffect

H

Figure 17.4 Earth pressure in cohesive soils


The above relation allows the estimation of the height H0 of an unsupported verticalcut, where Eh¼ 0 must hold (Figure 17.5):

H0 ¼ 4 c= ffiffiffiffiK

p

In the absence of internal friction (’¼ 0), where, of course, K¼ 1, the minimum heightof an unbraced vertical cut of a cohesive soil is

H0¼ 4 c/

17.3 Shallow foundationsIt is reasonable, in order to provide foundation to a structure, to always seek its safesupport at the shallowest level possible, since the cost of a foundation drasticallyincreases with its depth. This, of course, presumes that the bearing capacity of thesoil at the selected foundation level is adequate to permit the uptake of all the antici-pated loads on the structure. Such a foundation is called a shallow foundation, andcomprises the majority of foundations. However, if such a foundation is not feasible, aspecial structural layout must be provided so that the imposed loads are transmittedto lower soil layers with satisfactorily safe bearing capacity. The foundation is thencharacterised as a deep foundation. The cost of such a foundation constitutes a significantpercentage of the total cost of the structure.

What is presented below mainly concerns shallow foundations. However, some basicstructural characteristics of deep foundations will be examined in the final sections ofthis chapter.

17.3.1 The deformational behaviour of elastic soil under vertical loadsSoil may be considered as an elastic medium — possibly consisting of intermediate layers— being acted on by vertical loads on its free surface. Under these loads, soil behaveselastically. This assumption is supported by the results of geotechnical tests performedat the proposed site. Thus, each soil layer is characterised by its compression modulusEs, as well as by the corresponding Poisson ratio , determined by laboratory testsperformed during the geotechnical exploration of the site. The compression modulusEs is the ‘elasticity modulus’ of the soil material, and represents a quantity that

592

Cohesion c

Internal friction

Cohesion c

Internal frictionH0

Unsupported depth due tothe existing cohesion

4 · c

γ · √K

In case of the absence ofinternal friction, K = 1

In unsupported fronts the active pressure is suppressed by the cohesion

Figure 17.5 Effect of cohesion on the unbraced height of an open excavation


cannot be as precisely determined as in the case of, for example, steel. Its valuecorresponds to the geotechnical nature of the soil examined. Thus, non-cohesive soilsexhibit values from 30 000 up to 300 000 kN/m2, depending on the grain size and thedegree of consolidation (sand, 30 000—100 000 kN/m2; gravel, 70 000—300 000 kN/m2),whereas in cohesive clay soils the values of Es, clearly lower, are between 10 000 and40 000 kN/m2. The Poisson ratio has much narrower variation margins and asignificantly smaller influence on the deformation. Its order of magnitude for sand isabout 0.3, while for clay it is about 0.4.

Assuming that the examined soil mass, of theoretically infinite depth, has a uniformcompression modulus Es throughout and its free surface is loaded by a uniformlydistributed load p0 over a rectangular area (a b), then the settlement w developed ata distance x from its central point and in a parallel to the direction of side a, is(Figure 17.6)

w ¼ 1 2

Es 2p p0 a

where

¼ ð1 Þ ln

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1 Þ2 þ 2

qþ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1 Þ2 þ 2

q

þ ð1 þ Þ ln

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1 þ Þ2 þ 2

qþ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1 þ Þ2 þ 2

q

þ 2 ln

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1 Þ2 þ 2

qþ ð1 Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1 þ Þ2 þ 2

q ð1 þ Þ

and ¼ b=a, ¼ 2x/a (Stavridis, 1997). g

On the basis of the above result and in view of a later examination of soil—structureinteraction, a strip of length L and constant width b is considered, being divided into

593

Value of λ

ξ

α 0.0 2.0 6.0 10.0 14.0

1.0 7.04 2.06 1.00 0.39 0.28

0.50 9.60 3.69 1.31 0.78 0.55

0.20 13.19 6.72 3.00 1.88 1.34

p0

p0

a

PLAN VIEW

b

x x

w w

Elastic subspace(ES, v)

α = b/a

ξ = x/(a/2)

1 – v2

w = · p0 · a · λ ES · 2π

Figure 17.6 Settlement of the elastic subspace due to a uniformly loaded orthogonal area


a number of n equal segments of length (L/n), as shown in Figure 17.7, whereby themutual deformational influence of each of the above segments on each other is sought.

Applying a unit vertical load P¼ 1 consecutively on the midpoint of each segmentand assuming it to be uniformly distributed over that segment with an intensity p0,

p0 ¼ 1

ðL=nÞ b

The settlements at the midpoint of all the other segments can be readily determined onthe basis of the last relation. If the segments are numbered in ascending order from 1 ton, a number of n2 settlements fsr may be determined. fsr represents the settlement thatdevelops in segment number s, when the segment number r is loaded with P¼ 1. Thus,with a total number of segments equal to n¼ 3 (Figure 17.8), loading, for example,

594

f62

f62P = 1b

PLAN VIEW

1/(a · b)

L

a = L/6

Elastic subspace

61 2 3 4 5

Figure 17.7 Influence of a loaded area on the settlement of a point lying on the soil surface

(P1 = 1)

(P2 = 1)

(P3 = 1)

f21 f31f11

f22 f32f12

f23 f33f13w2 w3w1

2 31

P2 P3P1

PLAN VIEW

* P1

* P2

* P3

Figure 17.8 Determination of settlement due to specific concentrated loads


segment number 1, on the basis of the above relation, the quantities f11, f21 and f31 maybe directly determined. Similarly, by loading segment number 2, the quantities f12, f22

and f32 may be calculated, and by loading segment number 3, settlements f13, f23, f33

are analogously obtained.The above deformations allow the determination of the settlement at all midpoints, if

each segment is acted on by a certain load Pr. Thus, in this example (see Figure 17.8),the settlement w of each segment is, respectively,

w1¼ f11 P1þ f12 P2þ f13 P3

w2¼ f21 P1þ f22 P2þ f23 P3

w3¼ f31 P1þ f32 P2þ f33 P3

The extension to a greater number of segments is obvious:

w1¼ f11 P1þ f12 P2þ . . .þ f1n Pn

w2¼ f21 P1þ f22 P2þ . . .þ f2n Pn

. . .

wn¼ fn1 P1þ fn2 P2þ . . .þ fnn Pn

By assembling settlements ‘f ’ in a square matrix [F] (n n), which may be called theflexibility soil matrix, and further assembling settlements ‘wi’ and loads ‘Pi’ in thecolumn matrices (n 1) W and P, respectively, the above expressions may bewritten in the compact matrix form (see Section 16.3.1):

W¼ F P

The assumption of a constant compression modulus Es over the whole depth of a certainsoil mass is not always realistic. The case of consecutive vertically arranged horizontallayers, each with a compression modulus corresponding to the thickness of the layer,is not unusual. On the other hand, even in a ‘homogeneous’ soil mass, the compressionmodulus increases approximately linearly with depth. Although for preliminary designpurposes the consideration of an estimated ‘average’ compression modulus may besufficient, a more ‘correct’ determination of the settlement w is presented below forthe case of a layered soil profile (Stavridis, 2002).

As a basic tool, the following result is used, regarding the settlement w developed atany corner of an orthogonal area (a b) under a uniform load p0, lying over a soil layer ofthickness z, and a compression modulus Es. This layer is assumed to rest on an absolutelyrigid base (Figure 17.9):

wða; bÞ ¼ uðzÞ=Y

where

uðzÞ ¼ a p0

4 arctan

þ ln

þ !þ

!

þ ln

1

þ 1 !þ 1

! 1

595


and

Y ¼ Es

1 2

¼ b=a

¼ 2 z=a

! ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi1 þ 2

q

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 þ 2 þ 2

q

The settlement of a point S on the soil surface at a distance xs from the central point ofthe area (a b), in a parallel direction to side a, can be expressed as

ws¼ 2 [w(xsþ a/2, b/2)w(xs a/2, b/2)]

where the quantities w(. . . , . . .) inside the brackets refer to the corner settlements ofthe orthogonal areas, [(xsþ a/2) (b/2)] and [(xs a/2) (b/2)], respectively, accordingto the above expression for w(a, b) (see Figure 17.9).

By superposing the contribution of the compressibility of each soil layer on thedevelopment of the final settlement of point S (Figure 17.10), the following result isobtained:

ws ¼ uðz1Þ ð1=Y1 1=Y2Þ þ uðz2Þ ð1=Y2 1=Y3Þ

þ . . .þ uðzk 1Þ ð1=Yk 1 1=YkÞ þ uðzkÞ 1=Yk

It is clear that the formulation of the foregoing mutual interaction of the adjacentstrip segments, according to Figure 17.7, remains the same in the case of a stratifiedsoil.

596

ww

bb

b/2

ww

w

z

w

S

S

S

S

S

Elastic subspace(ES, v)

p0

p0

p0

p0

p0

a

a

PLAN VIEW

wS

xS

xS – a/2

xS + a/2

Settlement at position S

Figure 17.9 Settlement of an orthogonal area over a soil layer of finite thickness


17.3.2 Rectangular spread footingsIn order to provide a foundation for the base of a column element, where a specificvertical force V1, a horizontal force H and a moment M1 must be transmitted, anorthogonal concrete footing with a specific area and depth is usually constructed,normally having its larger base dimension parallel to the plane of action of themoment M1 (Figure 17.11). The reason why such a footing is necessary is obvious:the direct support by the soil of a column element, developing at its base the sectionalforces (V1, H, M1), is generally impossible because the soil cannot transmit to theelement those stresses required for the ‘realisation’ of the above sectional forces.Thus, for the proper design of a footing the following points have to be taken intoconsideration:

(A) The maximum developed contact stress with the soil, which should not exceed asafe value determined by the geotechnical study of the site, characterised as theallowable bearing pressure.

597

1

S

3

2

E1, v1

E3, v3

E2, v2

z1

z3

z2

Y = ES/1(1 – v2)

Contribution of layer 1

wS(z1, Y1)


wS(z3, Y3) – w(z2, Y3)


wS(z2, Y2) – w(z1, Y2)

The settlement of point S results from superposition of each layer contribution

Figure 17.10 Contribution of the existing layers to the soil surface settlement

Ground

V1M1

H

V1M1

H

d

σ2

σ2 σ1

σ1

Self-weight G

Actions on the footing(equilibrium)

Induced soil stresses

H < (V1 + G) · µ · 1.40Safety against slip

V1 + GM1 + H · d

H

Figure 17.11 Footing equilibrium and soil pressures


(B) The absolute magnitude of the maximum soil settlement developed, as well as theangle of rotation of the footing.

(C) The state of stress of the footing, which equilibrates as a free body under the actingsectional forces V1, H and M1, its self-weight G, as well as the acting contactpressures from the soil. The concrete footing must be reinforced and checkedappropriately so that it can take up all the above forces.

Of course, on the contact area between the footing and soil, a vertical force V¼V1þG,a horizontal force H as well as a moment M¼M1þH d are transmitted. It is underthese forces that the above points (A) and (B) should be checked.

The horizontal force H must be ‘covered’ by the ‘friction capacity’ (V ) between thesoil and foundation. A safe value for the friction coefficient is 0.50 for cohesive soilsand 0.60 for non-cohesive soils. However, by adopting a safety factor — usually takenas 1.40 — the following relation must be satisfied: H< (V ) (see Figure 17.11).Normally, the horizontal force H is not further involved in the design.

17.3.2.1 Soil pressures and settlementsIn a first attempt to estimate the developed soil pressures, the contact area of the footingis considered as a normal orthogonal section under the forces V and M. It is assumed thatforce V also includes the weight of the overlying soil mass. Force V is assumed to act onthe centroid of the footing area (Figure 17.12). It is, of course, clear that moment Mcauses a shift e of the force V along its plane of action, being equal to e¼M/V. Thus,the pair of forces (V, M) is statically equivalent to force V alone, shifted, however, bydistance e. As only compression contact pressures make sense over the foundationarea, it is appropriate to check whether or not the shifted force V lies within the coreof the corresponding section, i.e. whether it is e a/6. In that case, the maximumand the minimum contact (compression) stresses are, according to the bendingformula (see Section 2.2.1), respectively,

max ¼ V

Aþ M ða=2Þ

If

¼ V

a bþ 6 M

a2 b

min ¼ V

a b 6 M

a2 b

Of course, if the moment M is not acting, a uniform pressure ¼V/(a b) is developed.In the case where the shifted force V exceeds the core limits — something that should

generally be avoided — the above relations do not apply, since min becomes tensile.Then, the force V must be equilibrated by a triangular pressure diagram, having itsmaximum value on the nearest side to the force V, while its zero value to which it isextending will obviously lie within the foundation area (see Figure 17.12). As thedistance c of the force V to the near side is c¼ (a/2 e), the triangular diagram ofthe contact pressures extends over a length of 3 c, and the above maximum pressureresults from the equilibrium requirement:

12 max ð3 cÞ b ¼ V g

598


Although it is clear that all the above pressure distributions in each case satisfy thecorresponding equilibrium conditions for the magnitudes V and M, it is easily seenthat they lead to deformations which do not correspond to reality. More specifically,as the footing can be considered as essentially undeformable, the soil settlements,being identical with those of the foundation, should lie on a single plane exhibiting acertain rotation angle, because of the presence of the moment M. But even in the‘simple’ case of the central application of the force V, M¼ 0, it is clear that the resultinguniform pressure V/(a b) on the soil over a rectangular base does not lead — according toSection 17.3.1 — to a uniform settlement of all the points of the soil surface.

Thus, for the case of the central action of the force V, after dividing the foundationsurface into a certain number n of transverse strips of width a/n, the loads Pi for each striphave to be determined, which sum to the value V, on the one hand, and lead to the samesettlement (w) for all strips, on the other hand (Figure 17.13).

On the basis of the result of Section 17.3.1, the last requirement is expressed throughthe following equations:

w¼ f11 P1þ f12 P2þ . . .þ f1n Pn

599

The resultant of the soil pressuresis identical to the force V

σmax

σmin

σmax

3 · c

V lies outside the core(e > a/6)

The distribution is validonly if V falls within the core

(e < a/6)

e = M /V

e = M /V

V

c

e

Ve

The resultant of soil pressuresis identical to the force V

VM

V

M

VM

a

b

Figure 17.12 Influence of the location of the resultant force on the resulting soil pressures


w¼ f21 P1þ f22 P2þ . . .þ f2n Pn

. . .

w¼ fn1 P1þ fn2 P2þ . . .þ fnn Pn

In addition, the following equilibrium condition must be valid:

V¼ P1þ P2þ P3þ . . .þ Pn

The nþ 1 unknown quantities P1, P2, P3, . . . , Pn and w, can be determined through theabove linear system of equations.

By selecting a relatively small number of strips, e.g. n¼ 6, it can be ascertained that inan orthogonal footing under a central compression force V, forces P, instead of beingequal, clearly increase towards the edges of the footing (Figure 17.14).

600

P6P1 P2 P3 P4 P5

w

Grounda/6

The bottom face of the foundation

b

a

V

Those P values are sought which,having V as the resultant, induce ateach strip the same settlement w

Figure 17.13 More accurate determination of soil pressures and of uniform settlement

P6P1 P2 P3 P4 P5

2.0/6 m

2.00 m

2.00 m

V = 800 kN

Assumption of uniform distribution

Elastically deformable soil

Rigid footing

133.3 kN

193.

9 kN

193.

9 kN

105.

9

100.

0

100.

0

105.

9

Figure 17.14 Determination of soil pressures under a uniform settlement


The presence of the moment M introduces one more unknown, namely theinclination ’ of the resulting deformation plane of the soil surface (Figure 17.15).Thus, the above equations are modified as follows:

wþ 0 ’ (a/n)¼ f11 P1þ f12 P2þ . . .þ f1n Pn

wþ 1 ’ (a/n)¼ f21 P1þ f22 P2þ . . .þ f2n Pn

. . .

wþ (n 1) ’ (a/n)¼ fn1 P1þ fn2 P2þ . . .þ fnn Pn

Moreover, the static equivalence of magnitudes V and M with the forces P1, P2, P3, . . . ,Pn is required. Using the location of the force P1 as a reference point (see Figure 17.15),

V¼ P1þ P2þ P3þ . . .þ Pn

V (a/2 a/2nþM/V)¼ P1 (1 1) (a/n)þ P2 (2 1) (a/n)

þ P3 (3 1) (a/n)þ . . .þ Pn (n 1) (a/n)

It is clear that the nþ 2 unknowns P1, P2, P3, . . . , Pn, w and ’, can be determinedthrough the above nþ 2 linear equations. Obviously, for practical setting out andsolution of both the above linear systems, the use of appropriate computer software isessential. Moreover, it can be seen that the above equations are acceptable only forpositive (i.e. compressive) values of the forces ‘P’. Otherwise, an iterative processmust be followed. The developed soil pressures are obtained at each point byevaluating n P/a, whereas the maximum settlement is equal to wþ (n 1) ’ (a/n)(Stavridis, 2009).

17.3.2.2 Support on an elastic baseA less reliable but clearly more practical alternative to describing the deformationalbehaviour of the soil is offered by the concept of the modulus of subgrade reaction,represented by the coefficient k. It is based on the assumption that the settlement of

601

a/6

P6P1 P2 P3 P4 P5

V

Those P values are sought which,having as the resultants V and M,cause settlements on a single plane

M

Ground

ϕ

w

The bottom face of the footing

Figure 17.15 More strict determination of soil pressures and deformation for an eccentric loading


any region on the soil surface under a distributed load p is uniform over its entire extentand is equal to p/k. The coefficient k (kN/m3) represents the pressure (kN/m2) requiredto be applied to a soil surface in order to produce a settlement of 1 m, thus expressing thestiffness of the soil surface (Figure 17.16).

Although the concept of the subgrade modulus k is very convenient, being consistentwith the concept of elastic support (see Section 2.3.7), it does not constitute ameasurable soil property. More specifically, according to the above, the settlement of,for example, a rectangular surface under a pressure p is equal to p/k, i.e. it is independentof the surface itself. This is clearly not correct — as may easily be deduced fromFigure 17.6 — since, under the same pressure p, an increase in the loaded area alwaysleads to an increase in the corresponding settlement too (see Section 17.3.1). Thus,the determination of a subgrade modulus k has no meaning as a characteristic parameterof a given soil layer, as it depends directly on the specific loaded area. In addition, theassumption that all the points in the loaded area undergo the same settlement, whilethe surrounding region remains undeformed, is not realistic.

However, the computational convenience, together with the opportunity for thequalitative and quantitative estimation of the structural behaviour offered by the useof the parameter k, often overweigh, in practice, its serious ‘natural disadvantages’.

Certainly, a logical value of k for a specific orthogonal area may be obtained if thesettlement w, at its central point, under a pressure equal to 1 kN/m2 is calculatedaccording to Section 17.3.1, even by using the table in Figure 17.6. Of course, k¼ 1/w.However, for a square surface with a side equal to a, it is approximately

kEs/a (kN/m3)

In this way, the soil under each single footing having an area A, may be simulated by aspring with a stiffness ks equal to ks¼ k A, since, according to the above, a centrally

602

k (kN/m2)

Ground

w = 1.0 m

Neighbouring regionsremain undeformed

The loaded surfacesettles uniformly

The coefficient k does not constitute a soil constantas it depends on the loaded surface

Figure 17.16 Concept of subgrade modulus


acting load P on the footing causes a settlement w¼ (P/A)/k. Then, the whole structuremay be considered as including all the springs corresponding to its footings, being itselfnaturally supported on a rigid base (Figure 17.17). g

It can be seen that the induced rotation of a footing under the action of a moment Mdetermined — as usual — by assuming a fixed column base implies a reduction in thismoment, according to what has been discussed in Sections 3.2.9 and 3.3.6 regardingindeterminate structures. This footing rotation and the ensuing moment reductionneeds to be assessed, since it will lead to more favourable soil stresses.

Thus, whether the resilience of the footing can be described by a rotational springwith a specific elastic rotation flexibility f’, as defined in Section 2.3.7.2, i.e. as thedeveloped footing rotation under the action of a unit moment, will be examined.Obviously, such a moment should be accompanied by a vertical load V, in order toensure the development of contact pressures over the entire footing (Figure 17.18).

In the case of a rigid footing, transmitting the sectional forces V and M over its bearingarea A and, provided that only compressive pressures are developed over the entirefoundation surface — meaning that the shifted resultant V falls within the core of therectangular area — then, on the basis of a subgrade modulus k and according toSection 17.3.2.1, the settlements s1 and s2 at its two edges will be

s1 ¼ max

k¼ V

A kþ M ða=2Þ

If k

s2 ¼ min

k¼ V

A k M ða=2Þ

If kðIf ¼ b a3=12Þ

Consequently, the rotation ’ of the footing will be

’ ¼ s1 s2

a¼ M

If k

603

Each spring stiffness is proportional to thecorresponding footing area

Subgrade modulus k

Ground

A1 A3A2 k · A1 k · A2 k · A3

Figure 17.17 Simulation of the soil deformability through translational springs


This expression for ’ allows the determination of the coefficient of elastic rotation f’ —according to Section 2.3.7.2 — or, equivalently, of the stiffness coefficient with respect torotation, k’¼ 1/f’, simulating in this way the soil through a rotational spring (seeFigure 17.18):

f’ ¼ 1

If k

k’ ¼ If k

However, it should be pointed out that, while, on the basis of the previous more exactapproach to soil deformability (see Section 16.3.2.1) through the compression modulusEs, the load eccentricity may exceed the core limits of the foundation surface and yetmaintain the compression contact stresses everywhere, the adoption of a subgrademodulus k does not lead to such a consequence, as can be ascertained by means ofthe above expressions.

When the resultant force V acts outside the core, the settlement s1, according to theabove, is

s1 ¼ max

k¼ V

3 a=2 M

V

b k

604

V

V

a

b

M

M

Ground

kϕ = If · k

ϕ = M/(I · k)

If = b · a3/12

Soil simulation by a rotational springis only possible if the shifted resultantV falls within the core of the footing

Figure 17.18 Conditions for the soil simulation through a rotational spring


and the corresponding rotation ’ of footing will be

’ ¼ s1

3 c¼ V

9 a

2 M

V

2

b k

It is clear that rotation ’ is not proportional to the acting moment — depending also onthe value of the vertical force V — and, therefore, in this case the soil cannot be repre-sented by a rotational spring. g

On the basis of the resistance k’ offered by the rotational spring against the imposition ofa unit rotation, the reduction of the moment M previously mentioned may be estimatedas follows.

The rotational spring introduces the developed moment to the column, according tothe rotation of their common joint, which obviously rotates together with the spring. Ifan external moment is considered to act on this node so that its rotation vanishes, thenthis moment M must clearly be equal to that corresponding to a clamped column base(Figure 17.19). Thus, in order to make this moment disappear, since it does not actually

605

Ground

Ground

M

M

MM

[M ]

Diagram [M ] basedon fixed support

Diagram [M ]

External moment

Zero rotation

Removal ofexternal moment

Rotation developedStressless spring

Superposed onprevious [M ]

M foot

M foot

kϕ = If · k

Figure 17.19 Column—footing—ground interaction


exist, an equal and opposite moment M must be superposed, according to the processused in Section 3.3.1.

The above moment M causes a rotation ’ to both the rotational spring and thecolumn base, which may be determined from the equilibrium requirement of theircommon joint. Assuming the upper end of the column to be fixed (see Section 3.3.3.2):

’

f’þ 4 EI

H ’ ¼ M

Thus, the moment developed in the spring, i.e. the moment Mfoot taken up by thefooting and transmitted to the ground, is

Mfoot ¼’

f’¼ M

1 þ 4 EI

H f’

and is clearly reduced compared with M (see Figure 17.19).As can be seen, this result is identical to that based on Sections 3.2.10 and 3.3.7,

namely that the moment M is distributed in proportion to the ‘adjacent’ stiffnesses:

Mfoot ¼ M 1=f’

1

f’þ 4 EI

H

The moment Mfoot may be further reduced if beams are inserted that connect the lowerend of the examined column with the neighbouring ones (Figure 17.20).

In this case, too, the equilibrium of the joint of the rotational spring, which connectsthe column and the two connecting beams — with their far ends fixed — leads to thedistribution of the moment M proportionally to the stiffness of the three adjacentmembers (see Figure 17.20). The corresponding portion Mfoot assigned to the footingsurface is

Mfoot ¼ M 1=f’

1

f’þX 4 EI

L

¼ M 1

1 þ f’ X 4 EI

L

Generally it is useful to provide such ‘connecting ribs’ between the lower ends ofthe columns. However, these are indispensable in cases where an eccentric layoutof the footing with respect to the column is constructionally imposed, e.g. alongproperty lines where each column load V necessarily exhibits a certain eccentricity e(Figure 17.21).

Of course, the isolated footings normally follow a symmetric layout with respect to thecolumn section, in view of the possible shift of the column load V to either side, due tothe possible occurrence of an alternating moment action M, e.g. during an earthquake.

Returning now to the constructional eccentricity e of a footing, it can be seen that thecolumn moment M deduced by assuming its base to be fixed must be — in the mostadverse case — increased by V e and, consequently, the last expression for the

606


moment taken up by the foundation surface becomes

Mfoot ¼M þ V e

1 þ f’ X 4 EI

L

where, in the denominator, the column member and the connecting rib (i.e. twomembers) have to be taken into account. It is clear that in evaluating the contact

607

M

Diagram [M ]

External moment

Zero rotation

Removal ofexternal moment

Rotation developedStressless spring

kϕ = If · k

M

M

Ground

Ground

M

M foot

M foot

[M ]

Connecting beams

The presence of connecting beamsreduces the moment on the footing

Figure 17.20 The structural action of the connecting ribs

Ground

Ground

VConnecting beam

e

MAggravatingdiagram [M ] Mfoot

Mfoot

(M + V · e)

Figure 17.21 Treatment of an eccentric footing


pressures according to the pair (Mfoot, V), the force V is assumed to act centrally (seeFigure 17.21). g

It is noted that the coefficient f’ may be determined either through its previousexpression as f’¼ 1/(If k), which in the case of a square footing with side a, is f’¼ a/(If Es), or, more ‘reliably’, through the consideration of the foundation soil as anelastic medium, as shown in Section 17.3.2.1.

It should be pointed out here that the more resilient the foundation soil, the smaller isthe moment Mfoot acting on the ground. This is structurally convenient, because asmaller Es generally means a lower allowable soil pressure but also a greater reductionin the ‘effective moment’ acting on the foundation surface. However, in the case of alarge Es the coefficient f’ has a small value, and the reduction of the momentMþV e is also small. On the other hand, the capacity of the soil to safely take uphigher compression stresses is increased.

17.3.2.3 DimensioningThe footing is subjected as a free body to the soil pressures and to the transmittedsectional forces of the column base (equal and opposite to the corresponding reactions).Under the above forces, the footing acts like a slab loaded ‘upwards’. This loading causesbending, with the bottom face of the footing being in tension, whereas the compressedregion is limited in practice by the corresponding width of the column. The tensile force,clearly equal to the compressive force, is provided by reinforcement distributed over thewhole corresponding width of the footing (Figure 17.22). The dimensioning of the

608

d d

B

L

M

ASb

b

AS

Unreinforcedfooting

>60°

>45°

The tensile force is taken up by the concrete

No shear reinforcement required

M = B · (2σ1 + σ2) · L2/6

σ1σ2

Figure 17.22 Design of a concrete footing


reinforcement refers to a section having as the width b the width of the column and asthe depth d the depth of the footing itself (see Figure 17.22). It is clear that themaximum compressive force in the footing due to bending, as determined by thewidth b, must be kept below the corresponding allowable value.

However, if a tensile force required by the soil pressures, for their transference tothe column through the truss mechanism shown in Figure 17.22, can be provided bythe concrete itself, then the footing may be constructed without reinforcement. Thisis the case if the inclination angle of the corresponding strut in the above truss modelis at least 608 (see Figure 17.22).

On the other hand, if the reinforced footing has been formed so that the resultant of thesoil pressures outside the outline of the column can be directed to its edges under an anglegreater than 458, then the established truss mechanism, as shown in Figure 17.22, transfersthe soil pressures to the column without needing shearing reinforcement (Menn, 1990).

If the above conditions are not met, then the footing must be checked not only inbending but also in punching shear, following the procedure in Section 11.5.2.

17.3.3 Foundation beamsAs was pointed out in Section 17.3.2, the foundation of columns on single footingsshould take into account not only the maximum developed soil pressures but also thesettlements and rotations that are developed.

It is clear that the settlement of each single footing is generally also affected by itsneighbouring ones (see Section 17.3.1), except when the separation between them ismore than about triple their dimensions. In this case, the footings settle independentlyof each other. However, it should be kept in mind that the founded structure is — almostalways — a statically indeterminate one, and, as such, it is not sensitive to differentialsettlements of its supports. Of course, in the case of a relatively low allowable bearingpressure, the resulting footing dimensions could lead to footings lying either very nearto each other or even overlapping (Figure 17.23).

For all the above reasons, when designing the foundation of columns belonging to thesame vertical plane as in, for example, a frame, it is frequently deemed useful toconstruct a beam with an appropriate width to serve as a common foundation for the

609

In relatively deformable soil thefoundation beam layout is preferable

Figure 17.23 Incorporation of the foundation beam into the carrier system


group of columns under consideration. This beam, called the foundation beam, is loadedby the upward soil pressures and supported on its columns. Strictly speaking, thecorresponding — multi-storey — frame containing the columns and the foundationbeam is supported as a complete ‘closed’ structure on the ground (see Figure 17.23).

In such a frame, the tendency of a column to settle down as a single element isprevented by the bending stiffness of the foundation beam, which shares this settlementbetween the other columns, clearly limiting in this way the differential settlements.Thus, a foundation beam with a depth adequate to ensure the necessary bendingstiffness is clearly superior, as a foundation system, to that consisting of single spreadfootings. However, while the settlement behaviour of single footings is treated simplythrough appropriate spring supports — translational and/or rotational ones — supportedon a fixed base, the treatment of a foundation beam is more complicated, as the problemconsists of the determination of the soil pressures along the foundation beam itself,which implies examination of the interaction of the entire superstructure with thesoil. It is clear that this interaction affects not only the response of the foundationbeam but also that of the whole structure.

The basic concepts and procedures which allow the investigation of the interactionbetween the soil and the frame with its foundation beam will be examined below.

17.3.3.1 Soil simulation according to the Winkler modelThe behaviour of a foundation beam resting on an elastic base may be approachedthrough the concept of the subgrade modulus k, as defined in Section 17.3.2.2.According to this concept, assuming that a segment of a foundation beam undergoesa settlement w implies that the soil is acted on by a pressure psoil equal to psoil¼ k w,which, representing the resistance offered by the soil to the imposition of the abovesettlement, is obviously acting on the beam segment itself too. Expressing thispressure through the distributed load p (kN/m) along the foundation beam of widthb, psoil¼ p/b¼ k w and, consequently, p¼K w, where K¼ b k — expressed now inkN/m2 — represents the subgrade coefficient in the longitudinal direction of the beam(Figure 17.24).

Thus, the soil is simulated by a continuous distribution of springs, having the propertyof exhibiting for a uniform load p (kN/m) over some length a uniform settlement equal tow¼ p/K, whereas outside this region no deformation occurs.

The foundation beam having rigidity EI, under the distributed load q(x) is deflected byw(x). The resistance per unit length offered by the beam to this deformation equalsEI (dw4/dx4), while that offered by the soil equals K w, and, obviously, both resistancesbalance the externally applied load q(x). Hence (see Figure 17.24),

EI dw4

dx4þ K w ¼ qðxÞ

This differential equation, known as the Winkler equation, permits, through its closedsolution, the determination of analytic expressions for the response of a beam segmentembedded in an ‘elastic soil’ of the type examined, thus making possible the evaluation

610


of the response of the whole frame, including, of course, the foundation beam itself. Thisanalytic facility related to the adoption of the subgrade coefficient K makes this methodof evaluating the soil—structure interaction very convenient, particularly with the use ofan appropriate computer program.

The bending response of the foundation beam of a three-storey frame is shown inFigure 17.25, under a uniform girder loading of 80 kN/m. The outer and innercolumns have cross-sectional dimensions of 40/40 and 60/40 cm, respectively. Thegirders have a uniform section of 40/80 cm, whereas the section of the foundationbeam is 60/120 cm. The soil is assumed to have a compression modulusEs¼ 10 000 kN/m2, while its subgrade coefficient K, according to the foregoing, isconsidered equal to (see Section 17.3.2.2)

K¼ 0.60 k¼ 0.60 (10 000/0.60)¼ 10 000 kN/m2

The results are obtained by using appropriate computer software. g

The following points are again emphasised. The subgrade modulus k does not constitutein any case a soil constant. Ensuring consistency with its definition requires reference toa specific surface, which may be obvious for a footing of specific dimensions, but in thecase of the foundation beam is rather unclear. Moreover, loading a single foundationbeam with a uniform load causes a uniform settlement over the whole length of thebeam, meaning the absence of any bending along the beam, which obviously cannotbe valid. However, the concept of the subgrade coefficient k, although physically not

611

Resistance offered by the beam: EI · (d4w/dx4)Resistance offered by the soil: K · w

K · w

Equilibriumq(x) = EI · (d4w/dx4) + K · w

b

p

Beam cross-section

w(x)

q(x)

Beam

Surface of elastic base

x

For beam width b: K = b · k

K (kN/m2)w = p/K

∆x

Figure 17.24 Foundation beam on a Winkler foundation


consistent — as pointed out in Section 17.3.2.2 — nevertheless allows an approach to thedetermination of the soil pressures and to the response of the whole frame. In addition, itshould also be recognised that the development of a settlement at some location onthe soil surface implies soil deformation outside the considered region, because of thestiffness of the loaded beam, compensating in some way for the weakness of the soilmodel.

The adoption of a subgrade modulus k leads to the important concept of the so-calledelastic length Ls of the beam (see Sections 9.7 and 12.3.1):

Ls ¼ffiffiffiffiffiffiffiffiffiffiffi4 EI

b k

4

r

The concept of the elastic length applies directly in the case where the foundation beam— theoretically of an infinite length — is loaded by a concentrated load F, which mayrepresent a column load. It denotes that length over which the bending response ofthe beam extends from either side of the load F (Figure 17.26). Considering the soilpressure p as constant over the total double elastic length, i.e. p b¼ F/(2 Ls), andgiven that the bending moment at both its ends vanishes, it is found that thebending moment under the concentrated load resulting from either of the correspondingcantilevers represents a very good approximation to the bending moment of theelastically supported long foundation beam developed in practice at this location:

M ¼ ðF=2LsÞ L2s ¼ 0:25 F Ls

612

3.50

3.50

3.50

1.20

8.0 m 8.0 m

80 kN/m 80 kN/m

80 kN/m 80 kN/m

80 kN/m 80 kN/m

1517 kN m[M ]

54

1010

1105 26

7

267

1105

1010 54

Figure 17.25 Bending response of the foundation beam of a frame on an elastic base


Using the example given in Figure 17.25, the concentrated force F acting on themidpoint of the foundation beam is

F 3 (5 80.0 8.0/8) 2¼ 2400.0 kN

and its elastic length is equal to

Ls ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4 2:1 107 0:086=10 000

4q

¼ 5:18 m

Then, according to the above, the bending moment at the location of the concentratedload is M¼ 0.25 24 00.0 5.18¼ 3108 kN m. The discrepancy — on the safe side — fromthe previously calculated value of 1517 kN m is mainly due to the fact that the founda-tion beam is considered here to be free, i.e. not being prevented from deforming by thestiffness of the superstructure.

17.3.3.2 Soil simulation as an elastic mediumAccording to the soil model examined in Section 17.3.1, the structural frame issupported with its foundation beam on an infinitely extended elastic subspace, eitherhaving a constant compression modulus Es with a Poisson ratio , or consisting of acertain number of horizontal layers, with a corresponding Es for each layer.

According to precedents, a foundation beam of width b immediately poses theproblem of the determination of the distribution of soil pressures psoil(x) along itslength. These pressures expressed as a distributed load, p(x)¼ psoil(x) b (kN/m), mustfulfil the following two conditions:

(1) they must be in equilibrium with the vertical loads of the structure, the foundationbeam included

(2) the deformation of the foundation beam must match exactly the surface deforma-tion of the soil due to the load p(x).

613

F

F

M = p · Ls2/2

EI

[M ]

Elastic base

Coefficient K

Ls Ls

~p · Ls2/2

Elastic length

p = F/(2 · Ls)

All along the elastic length the pressure is assumed constant

Ls = (4 · EI/b · K)1/4

Figure 17.26 Physical meaning of the elastic length


It should be clear that both of these conditions are satisfied by the Winkler model, withthe only remark that, for the reasons mentioned earlier, the description of the soil surfacedeformation through the insertion of springs is not absolutely reliable. Thus, in order totake into account the soil behaviour as an elastic medium in a more consistent manner,the model shown in Figure 17.27 is considered.

The oblong orthogonal contact surface of a foundation beam of length L is dividedinto a number n of strips of equal width (L/n), at the midpoint of which the resultantof the corresponding soil pressures is acting. The whole structure is assumed to besimply supported at the above n midpoints. These rigid supports under the loads ofthe superstructure develop the reactions P0

i , which may be assembled into the columnmatrix P0 (n 1).

It can now be seen that, for a certain group of imposed settlements s1, s2, . . . , sn, thedeveloped reactions of the continuously supported structure R1, R2, . . . , Rn, willconstitute a self-equilibrating force system summing to zero. The fully developedreactions due to the structure loads and the settlements will be

V1 ¼ R1 þ P01;V2 ¼ R2 þ P0

2; . . . ;Vn ¼ Rn þ P0n

614

Working model of the examined frame

Plan view of the foundation beam

Sought-after imposed settlements

Developed self-equilibrating reactions

The induced soil settlements must coincidewith those imposed on the model

1 2 3 4 5 6 7 8 9

Ground

Ground

Si = (Pi0 + Ri)

R1 R2 R3 R8 R9

s1 s2 s3 s8 s9

P01 P0

2 P03 P0

8 P09

Figure 17.27 Frame foundation with a foundation beam on an elastic subspace


The above forces may be considered to be acting in the opposite sense also on theground, and, thus, the solution of the problem consists of finding the imposed settle-ments s1, s2, . . . , sn on the fictitiously supported frame such that the induced soil settle-ments under the loads V1, V2, . . . , Vn are identical to them (see Figure 17.27). This isachieved by the following procedure (Figure 17.28) (Stavridis, 2002).

Imposing on the first support 1 a unit settlement s1¼ 1, a set of self-equilibratingreactions (k11, k21, k31, . . . , kn1) is developed. Next, a unit settlement s2¼ 1 isimposed on support 2, leading to another set of self-equilibrating reactions (k12, k22,k32, . . . , kn2). Proceeding consecutively in the same way up to the last support n, aunit settlement sn¼ 1 is imposed equally, resulting in the self-equilibrating reactionsk1n, k2n, k3n, . . . , knn.

If the set of settlements s1, s2, . . . , sn is now imposed on the continuously supportedframe, the developed reactions R1, R2, . . . , Rn can clearly be expressed through thefollowing relations:

R1¼ s1 k11þ s2 k12þ s3 k13þ . . .þ sn k1n



Rn¼ s1 kn1þ s2 kn2þ s3 kn3þ . . .þ sn knn

Assembling the forces ‘kij’ into a square matrix Ksup (n n) and further assemblingsettlements ‘si’ and reactions ‘Ri’ in the (n 1) column matrices s and R, respectively,the above expressions may be written in matrix form (see Section 16.3.1):

R ¼ Ksup s

Also taking into account the reactions P0 due to the structure loading, as previouslymentioned, then the above expressions for the totally developed reactions ‘Vi’ in thefictitiously supported frame being assembled in the (n 1) column matrix V can beexpressed in matrix form as

V ¼ P0 þ R ¼ P0 þ Ksup s

According to the above, what remains is the requirement from the forces V1, V2, . . . , Vn,if applied in the opposite sense to the ground, to produce the soil settlements s1, s2, . . . ,sn (see Figure 17.28). For this purpose, the relation describing the deformationalbehaviour of the soil surface is recalled from Section 17.3.1:

W ¼ F Pwhere, now, W ¼ s and P ¼ V. This relation, after multiplying both sides by F

1,allows the following expression for the forces V:

V ¼ F1 s

By substituting the above relation, the following equation is finally obtained:

ðF1 þ KsupÞ s ¼ P0

615


By determining the settlements ‘si’ from this linear system, the forces ‘Vi’ acting on theground are readily obtained from the penultimate relation.

It is obvious that the whole procedure is feasible only by using an appropriatecomputer program for plane frame analysis.

616

Due to:s1, s2, … , sn

(Due to sn = 1)

s1, s2, … , sn

s1, s2, … , sn

Totalreactions:

Imposed settlements:

V1 V2 V3 Vn

V1 V2 V3 Vn

R1 R2 R3 Rn

Ground

REQUIREMENT

sn ·

s1 ·

s2 ·

k1n k2n k3nknn

(Due to s2 = 1)k12k22

k32

kn2

(Due to s1 = 1)k11

k21k31

kn1R1 R2 R3 Rn

P01 P0

2 P03 P0

n

Figure 17.28 Practical analysis of the soil—frame interaction


Numerical exampleFor reasons of direct comparison, the frame examined in Section 17.3.3.1 will again beconsidered, under the same loading and soil characteristics (Figure 17.29).

The model under consideration is supported at nine intermediate equidistant pointsof the foundation beam, as shown in Figure 17.29. According to the procedure describedabove, the following settlements are obtained:

s1¼ 0.0571 m, s2¼ 0.0562 m, s3¼ 0.0565 m, s4¼ 0.0577 m, s5¼ 0.0587 m,

s6¼ 0.0577 m, s7¼ 0.0565 m, s8¼ 0.0562 m, s9¼ 0.0571 m

This means that the state of stress and the deformation of the frame examined areidentical to those of the fictitiously supported model, under the existing loading andthe above support settlements.

The bending moment diagram of the foundation beam is shown in Figure 17.29. Bycomparing the results with those based on the adoption of the Winkler soil model, it isobserved that in the present case where the soil is considered as an elastic half-space, thespan moments have lower values, whereas the values of the bending response inthe region of the central column are greater.

17.4 Pile foundations

17.4.1 OverviewAs mentioned in Section 17.3, deep foundations transfer the loads of the superstructure,through weak unsuitable soils, to a lower stratum with sufficient bearing strength. The

617

1837 kN m

3.50

3.50

3.50

1.20

80 kN/m 80 kN/m

80 kN/m 80 kN/m

80 kN/m 80 kN/m

8.0 m 8.0 m

[M ]

5

815

752

127

127

752

815 5

1 2 9876543

Figure 17.29 Example of frame analysis, with the frame resting on an elastic subspace


usual way to accomplish this transfer is through the use of piles. Obviously, piles are usedwhen the subsurface conditions are not suitable for a shallow foundation. Piles arevertical or slightly inclined structural members made of concrete, up to 30 m long,having a circular cross-section and a diameter ranging from about 0.40 m up to1.50 m. The piles are arranged in a group of relatively dense layout, being connectedwith each other at their upper end through a thick slab called a pile cap, which isfurther connected to the superstructure. At their lower end, the piles have a slightlyenlarged base, resting on a soil layer of satisfactory bearing strength. In relatively rarecases, where this layer lies so deep that the length required from the piles wouldresult in a prohibitive total cost, the equilibrium of the loads transmitted to the pilesfrom the pile cap is mainly accomplished through skin friction forces — developedbetween the so-called friction pile — and the adjoining soil, through the whole lengthof the piles (Figure 17.30).

Thus, the foundation system consisting of the pile cap and the piles is in equi-librium under the action of the superstructure applied to the pile cap and the forcesacting on the lower end of piles from the sound stratum. It is clear that, in the caseof friction piles, the role of the acting soil force at the lower pile end is undertaken bythe total force of the lateral friction. Usually, there is a coexistence of friction andbearing forces in a rather uncertain proportion that does not affect the total axialforce, in practice, that is required to be taken up by the pile in order to fulfil theglobal equilibrium conditions.

The pile cap, as shown in Figure 17.30, is connected at its top to the supportedelement, which transmits the actions V, Mx, My (i.e. a vertical force and a momentwith an arbitrary horizontal vector), as well as a horizontal force H (Figure 17.31).

The piles are usually arranged in a symmetrical layout around the base of the foundedelement, and they are generally placed in a vertical position. With regard to thehorizontal force H, this is taken up either by the vertical piles themselves, or by batterpiles specially constructed for that purpose, with an angle of inclination not greaterthan 158.

618

Every pile has a maximum allowable load dependingon its dimensions as well as the soil strength

Ground Ground

Sound stratum Sound stratum

Uptake of loadthrough skin friction forces

Pile cap HM

V

Figure 17.30 Layout and load-bearing action of piles


17.4.2 Vertical loadsThe symmetrical layout of N piles around the point of application O of the load V impliesthe existence of the orthogonal axes of symmetry Ox and Oy. The components Mx andMy are referred to these two axes (see Figure 17.31).

For preliminary design purposes, it is sufficient to consider the magnitudes V, Mx, My

as being applied to a rigid section, consisting of the total layout of circular section areasof the piles. Given the coordinates xi, yi of the centres of the circular pile sections, withrespect to the centroid O of the section and assuming that all piles have a commonsection area A, the following sectional data are derived:

. Total area of section: As¼N A.

. Moment of inertia about the Ox axis: Ix¼A P

y2i .

. Moment of inertia about the Oy axis: Iy¼A P

x2i .

Thus, the consideration of biaxial bending, to which the total section is subjected, leadsto the determination of the axial force Pi for each pile separately (see Figure 17.31):

Pi

A¼ V

As

Mx yi

Ix

þMy xi

Iy

It is clear that the pile cap is in equilibrium under the loads V, Mx, My and the pile forcesPi, and must, therefore, be designed to safely take up the above actions. For preliminarydesign purposes, the eventual bending of the piles is ignored. Moreover, selecting alarge enough thickness for the pile cap, i.e. equal to at least half of the clear distancebetween the piles, the bending action of the cap is excluded and the load introducedfrom the column is transferred to the piles through a strut-and-tie model, normally a

619

Pile cap

O O

y y

x x

A AN = 3 N = 4

c.g. c.g.

c.g.: centre of gravity

Data of idealised sectionAS = N · AIx = A · ∑ yi

2

Iy = A · ∑ xi2

VV

My

Mx

My

Mx

Pi V Mx · yi My · xi = – +A AS Ix Iy

Figure 17.31 Loading of a pile system through a pile cap


three-dimensional one (Figure 17.32). In this model, which depicts the load transferthrough the most ‘stiff ’ — i.e. the shortest — paths, the struts represent the compressedconcrete between the column and the pile heads, whereas the ties are realised throughreinforcement bars, arranged within the narrow region of the pile circular section withgood anchor conditions, as shown in Figure 17.32, thus ensuring the equilibrium of theformed joints.

Designing in this way, that is, by taking up the design loads (i.e. service loadsmagnified by the corresponding safety factors) on the basis of a constructionallyfeasible equilibrium, ensures, according to the static theorem of the plastic analysis(see Section 6.6.2), that the collapse load is greater than that which the substitutestrut-and-tie model can carry.

17.4.3 Horizontal loadsThe uptake of a horizontal load by a vertical pile clearly implies a bending response of thepile. The pile acts as if laterally supported by an elastic medium described by a subgradesoil coefficient kh and substituted accordingly through a continuous distribution oflateral springs, as in the case of the foundation beam (Figure 17.33).

In non-cohesive soils, the coefficient kh can be considered to increase linearly to adepth t, according to the relation kh¼ nh t/D, where D is the pile diameter and nh isa quantity representing the lateral stiffness of the soil, ranging from 2 to 18 MN/m3,depending on the layer density.

In cohesive soils, the coefficient kh is considered to be constant, and is kh¼ nh/D,where nh is expressed in terms of cohesion c as nh¼ 160 c.

Clearly, the uptake of a horizontal force, or even a moment acting on top of a pile(transmitted through the pile cap), is accomplished through an appropriate distributionof lateral soil reactions along the length of the pile, implying either tension or

620

V Concrete struts

In the presence of the moment the force V is shifted(however within the core of idealised section)

and the space truss is formed accordingly

P P

PP

Tie reinforcement

Equilibrium at the joints

Figure 17.32 Load-bearing action of a pile cap


compression of the corresponding supporting springs (see Figure 17.33). In contrast tothe foundation beam, where the development of tensile spring forces is not normallyallowed, in the present case this is obviously meaningful, as any resulting spring forcerepresents compressive contact pressure between the soil and the pile.

It is clear that the above response may be determined using the same software as usedfor the analysis of foundation beams.

ReferencesMenn C. (1990) Prestressed Concrete Bridges. Basel: Birkhauser Verlag.Stavridis L. (1997) Tragweke auf elastischem Boden. Bauimgenieur 72(12) 565—569.Stavridis L. (2002) Simplified analysis of layered soil—structure interaction. Journal of Structural

Engineering (ASCE) 168(2), 224—230.Stavridis L. (2009) Rigid foundations resting on an elastic layered soil. Geotechnical and Geological

Engineering 27, 407—417.

621

H

H H

M

t

D

D

Distribution of subgrade coefficient

Compressedsprings

Tensionedsprings

Pressure distribution Bending moment distribution

kh = nh · t/DNon-cohesive soils

kh = 160 · c/DCohesive soils

Figure 17.33 Lateral response of piles


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624


Index

Page numbers in italics refer to figures separate from the corresponding text.

absolute displacement, 548acceleration spectra, 550—551, 551accelerograms, 548—551, 551action principles, 32active pressure, 589adhesion forces, 14allowable bearing pressure, 597allowable limits, 26anchorage lengths, 14, 15anchor forces, 509—510angles of internal friction, 588angles of skewness, 359—360antisymmetric loads, 302, 332—335, 481,

484—487aquatic environments, 3arbitrary loadings, 84, 85arches, 297—313axis shortening, 302—304conoidal shells, 452, 453elastic stability, 304—307formation, 297, 298girders, 308—310hyperbolic paraboloid shells, 440—442, 443, 444membrane action in shells, 405pressure lines, 297—298second-order theory, 304—307stiffness, 308—310straight edge hypar shells, 445—446, 450structural behaviour, 297—304suspension cables, 335—339tied-arch systems, 311—313

arch-like boundaries, 435axial deformations, 172axial flexibility, 7axial forces, 43, 44—45, 67, 226, 253—257box girders, 483—484, 486—487dome shells, 432hyperbolic paraboloid shells, 440—441

straight edge hypar shells, 445—446axial loads, 281axial stiffness, 6—7, 486—487axial stresses, 482—483, 487—489, 490

balanced cantilever methods, 206—207bar buckling, 271—277barrel shells, 416—423base shear, 550—551, 575basic material structural behaviour see material

structural behaviourbatter piles, 618beamsbox girders, 490—491, 491, 507, 508cable structures, 317—322, 331, 342, 343,

345—347cantilever beams, 177—181continuous beams, 187—221continuous systems, 577—580, 581cylindrical shells, 410—411, 418—421deflection curves, 77—78design control, 184—185discrete mass systems, 526, 527, 534, 535, 580,

581dome shells, 427—428, 429—432elastic mediums, 613—616elastic stability, 272—273, 277—282, 290—293,

294equilibrium, 50—52, 62—64, 77—78, 90—91external prestressing, 181—184formation, 455, 456frames, 223—269grid structures, 354—357ground structure supports, 609—617lateral buckling, 290—293multi-storey systems, 526, 527non-shallow shells, 429—432with openings, 142—144


beams (cont’d)plates, 396—397, 400, 401, 402prestressed concrete beams, 163—177rectangular openings, 143—144reinforced concrete, 150—163simply supported beams, 139—185soil simulations, 610—616statically indeterminate structures, 130, 131steel beams, 139—150, 188—194straight edge hypar shells, 446, 450suspension bridges, 331thin-walled beams, 455—479variable beam height, 178, 179—180, 181, 181Winkler model, 610—613

bearing pressure (allowable), 597bendingapplication examples, 499—502arches, 302—304, 309box girders, 482—483, 487—490, 495—503,

505—510conoidal shells, 453composite beams, 210—211continuous beams, 189—190, 196—198,

205—207curved girders, 496—502cylindrical shells, 410—413, 420—422deflection curves, 77—78dome shells, 429—431, 431, 434elastic stability, 272—273equilibrium, 43—46, 51—54, 66—67, 77—78, 89,

496—498frames, 253, 256, 257grid structures, 351—352, 355, 356ground structure supports, 611—613, 612hyperbolic paraboloid shells, 441mixed systems, 249—251multi-storey frames, 240—244, 246—247one-storey multibay frames, 235—238plates, 364—365, 370, 373—386, 391—392,

396—398simply supported beams, 139—140, 144—153,

158—170, 173—180single-storey, single-bay frames, 223—224, 226,

228, 234—235statically indeterminate structures, 107—110,

112, 118, 136—138suspension bridges, 329, 330

Betti—Maxwell theorem, 76—77, 429bimoment, 469—473, 470bonded tendons, 176—177bond forces, 14bond lengths, 14, 15

bond stresses, 14, 15boundary forces, 408box girders, 481—516cross-section walls, 502—505curved girders, 495—416prestressing, 505—511, 512, 513profile section deformability, 485—492rectilinear girders, 481—495torsion, 496—502, 510—511, 512, 513

bucklingcable-stayed bridges, 347—348, 349concepts, 27cylindrical shells, 422dome shells, 434elastic stability, 271—277, 290—293single-storey, single-bay frames, 233straight edge hypar shells, 451visualisation, 291

cable structures, 315—349box girders, 510, 511cable—beam structures, 317—322cable-stayed bridges, 339—348, 349, 501—502cylindrical shells, 415freely suspended cables, 322—325hyperbolic paraboloid shells, 440plates, 388—389, 396—397, 398—399prestressed cable nets, 325—328simply supported beams, 179—180, 181stiffening, 335—339straight edge hypar shells, 446—448, 448, 449suspension bridges, 328—335suspension cables, 335—339

conoidal shells, 451—453cantileversbox girders, 483—484conoidal shells, 452design, 376—377discrete mass systems, 535—536, 538—539equilibrium, 53—54mixed systems, 249—251multi-storey frames, 245—246plates, 375—377simply supported beams, 177—181straight edge hypar shells, 446—447, 448stresses, 375—376thin-walled beams, 465—466

cast-in-place concrete deck plates, 207—209Cauchy’s relation/theorem, 145—146, 418,

461—462centroid of frame section design, 254—255circular columns, 393

626


circular frequency, 537circular plans, 427—428circular plates, 365, 385, 386clamped supports, 79—81climatic conditions, 3—4closed beam-like cylindrical shells, 416—418closed cross-sectionsbox girders, 481—516general characteristics, 455—460non-deformable cross-sections, 460—461warping-based stresses, 462—463, 465—470

coefficients of elastic rotation, 79—80coefficients of elastic supports, 79cohesive soil types, 587, 590—592collapseconcepts, 26continuous beams, 190—191frames, 257—259, 261—262

column—footing—ground interactions, 605—606column matrices, 557columns, 223—269, 389—393, 397—398compatibility of deformations, 96composite beamsbehaviour, 217—220bending behaviour, 210—211construction stages, 212—214continuous beams, 209—221plastic analyses, 220—221prestressing, 217—220shear forces, 209—210, 211—212temperature, 215—217

compound structures, 39—41compressed beams, 214, 272—273compressed concrete, 8, 9compressionbox girders, 483—484, 502, 505—506, 507continuous beams, 214cylindrical shells, 414, 416—417, 419, 420, 422dome shells, 426, 432—433elastic stability, 275—277frames, 226, 253—257ground structure supports, 592—593hyperbolic paraboloid shells, 440—443, 443, 444membrane action in shells, 405, 409plates, 373, 381—382, 394simply supported beams, 165—172, 179, 180single-storey, single-bay frames, 226straight edge hypar shells, 445—446, 447,

450—451concentrated forces, 236—238concentrated loads, 2, 49—50, 51, 157, 158box girders, 490—491

grid structures, 354ground structure supports, 594—595plates, 369—370, 375—378, 387

concepts, 1—28bimoment, 469—472, 470control/design processes, 26—28discrete mass systems, 555—557elastic supports, 108—110equilibrium, 29—41, 81—84flexibility, 81—84loads, 1—5material structural behaviour, 5—22membrane action, 403—409prestressed concrete, 18—22, 24—26reinforced concrete, 10, 13—18, 22—23shells, 403—409statically indeterminate structures, 108—110stiffness, 81—84stiffness matrices, 555—557stress, 5

concretebeam bending stages, 159—161confinement, 262continuous beams, 195—199, 207—209,

213—214, 217—219creep, 10—12dome shells, 432frames, 229, 233, 240, 252—257, 262, 267—268grid structures, 357material structural behaviour, 8, 9, 10—18one-storey multibay frames, 240plates, 394rectangular spread footings, 608—609relaxation, 12—13simply supported beams, 150—179single-storey, single-bay frames, 229, 233suspension cables, 335—337two-side supported slabs, 371

concrete deck plates, 207—209constant internal pressure, 409—413constraint moments, 197—198continuous beams, 187—221, 396—397, 534, 535behaviour, 217—220bending behaviour, 210—211cable-stayed bridges, 342, 343composite beams, 209—221construction stages, 212—214creep, 199—209, 218—220design control, 198—199fixed-ended beams, 188—191fixed simply supported beams, 191—193plastic analyses, 220—221

627

Index

continuous beams (cont’d)prestressed concrete beams, 195—199prestressing, 217—220shear forces, 209—210, 211—212statically indeterminate structures, 106, 108,

109—110, 111statically redundant systems, 200—205steel beams, 188—194temperature change, 215—217tendon design/structural performance, 195—198

continuous girders, 507, 508continuous plane structures, 361, 362continuous plates, 380continuous supports, 389—399, 446, 447continuous systems, 577—580, 582—585, 584corner regions, 367, 368, 369coupling processes, 144—145, 249—251cracked states, 371cracking, 16—17, 19—21, 195, 199creepcontinuous beams, 199—209, 218—220frames, 261—262material structural behaviour, 10—12one-storey multibay frames, 240single-storey, single-bay frames, 229statically redundant systems, 200—205structural changes, 205—209

critical external pressure, 414critical loads, 275—279, 283—285, 290, 434, 451critical locations of compressive forces, 169—171critical longitudinal compressive stress, 414critical stress limits, 422cross-beams, 354cross-sectionsbimoment, 470—472box girders, 481—516conoidal shells, 452composite beams, 210cylindrical shells, 416—418, 421—422frames, 251—255non-deformable cross-sections, 460—461plates, 388—389shear centres, 461—462simply supported beams, 140, 141, 165—167single-storey, single-bay frames, 233—234straight edge hypar shells, 447—448suspension cables, 335—337thin-walled beams, 455—472warping-based stresses, 462—469

cross-vaults, 422, 423curvatures, 147, 153, 161dome shells, 426, 432—434

hyperbolic paraboloid shells, 436—437, 439—440membrane action in shells, 407—408shells, 403—404, 426, 432—434, 436—437straight edge hypar shells, 445—448

curved girders, 495—416bending, 496—502cross-section walls, 502—505prestressing, 505—511, 512, 513torsion, 496—502, 510—511, 512, 513

curved orthogonal plates, 433—434curved quadrilateral elements, 440, 441cylindrical shells, 409—423barrel shells, 416—423bending behaviour, 430, 431constant internal pressure, 409—413cylindrical tanks, 413—415silos, 415—416

damped vibration, 539—542deck plates, 207—209decompression moments, 173—174deep beams, 68, 400, 401, 402deep foundations, 592deflectionbox girders, 490—491curves, 77—78, 331—332plates, 376—377

deformable soil, 230—231deformation method, 352—353elastic supports, 130—132fixed-end beams, 119—123nodal action distribution, 132—133procedure, 125—129, 130qualitative handling, 105—107, 134—136, 107,

137, 138simply fixed beams, 116—119statically indeterminate structures, 114—138unknown nodal deformations, 123—125

deformationsarches, 304—306box girders, 489—490, 492cable—beam structures, 317—318, 319, 320causes, 66—70concepts, 29—41continuous beams, 199—209cylindrical shells, 410deformation method, 114—138, 352—353dome shells, 428, 431, 432equilibrium, 29—92force method, 94—114, 115frames, 93—138, 256, 257, 282—290grid structures, 87—92, 355, 356

628


ground structure supports, 592—596, 597,601—603

internal forces handling, 41—65lateral buckling, 290—293material structural behaviour, 7membrane action in shells, 408multi-degree systems, 559multi-storey frames, 244—249plastic analyses, 293—296plates, 361—362, 376—377, 381second-order theory, 277—282simply supported beams, 139—140, 143—144,

172single-storey, single-bay frames, 231—233statically indeterminate structures, 93—138stress states, 93—138, 271—296suspension bridges, 329—331symmetric plane structures, 84—87unknown nodal deformations, 123—125

degree of freedom systems, 534, 535—555degrees of statical indeterminacy, 41developing moment determination, 126—127, 128deviation forces, 165—166, 167—168box girders, 505, 509—510plates, 388, 396—399

deviation loads, 396—397diagonal elements, 557diagonal forces, 486—490diagonal loading, 503, 504diagonal tension cracks, 153diagonal tied-arch layouts, 312—313discrete mass systems, 533—586continuous systems, 577—580design spectra, 553, 554dynamic equilibrium, 535—536, 548, 563, 564equilibrium, 535—536, 548forced vibration, 542—544, 570—571free vibration, 536—542, 563—570human activities, 580—581loads, 543—545, 559, 572, 574—576machine induced vibrations, 581—585multi-degree systems, 555—577periodic sinusoidal acting forces, 544—547plane elements, 560, 561plastic behaviour, 553—555seismic excitation, 547—555, 571—577single-degree-of-freedom systems, 535—555stiffness matrices, 555—563

displacementcylindrical shells, 411, 412discrete mass systems, 534—538, 548, 550, 559dome shells, 432, 434

elastic stability, 285grid structures, 354—355member kinematics, 123—124, 125membrane action in shells, 408—409multi-storey systems, 528—529, 529one-storey multibay frames, 238—240selection, 125, 126statically indeterminate structures, 118—128,

135—138thin-walled beams, 466

distributed loads, 381distributed prestressing, 372DMF see dynamic magnification factordome shells, 423—436double-T cross-sections, 455—456downward acting deviation forces, 398downward applied deviation forces, 398—399downward applied deviation loads, 396—397downward applied forces, 196—198, 398—399downward-curved parabolas, 440—441, 442downward edge loads, 400downward shifting, 179, 180ductilitydiscrete mass systems, 554—555frames, 258, 259, 260, 262material structural behaviour, 7—8

Duhamel integral, 543dynamic equilibrium, 535—536, 548, 563, 564dynamic loads, 1, 2dynamic magnification factor (DMF), 543—544,

545—546, 552

earth pressure, 2—3cohesive soil types, 591non-cohesive soil types, 588—589, 590plates, 401—402

earthquakes, 3, 547—555, 571—577eccentric footings, 606, 607—608, 607eccentric loadsbox girders, 490—491ground structure supports, 601thin-walled beams, 458—460

edge beamscylindrical shells, 419, 420straight edge hypar shells, 446, 447, 447, 450

edge loads, 400edge-supporting arches, 441—442edge zonescylindrical shells, 418—419, 422hyperbolic paraboloid shells, 442, 443, 443plates, 373—375, 387—389

effective degrees of freedom, 555

629

Index

effective mass displacement, 534, 535effective shear, 394—395, 398—399effective stresses, 591eigenforms, 565—566eigenfrequencies, 565—567, 568, 569eigenvalue equations, 565EI value influence, 104elastically rotating supports, 376elastic bases, 601—608, 611—613, 612elastic componentscable-stayed bridges, 345—346frames, 261—264material structural behaviour, 7steel beams, 139—147

elasticity moduli, 6, 139, 213—214, 592—593elastic lengths, 612, 613elastic mediums, 613—616elastic soil, 592—596, 597elastic stability, 451arches, 304—307bar buckling, 271—277deformations, 271—296frames, 282—290lateral buckling, 290—293multi-storey frames, 288—290numerical examples, 294—296one-storey frames, 283—287plastic analyses, 293—296second-order theory, 277—282stress states, 271—296

elastic subspaces, 614—615, 617elastic supportsbox girders, 490—491, 491equilibrium, 79—81statically indeterminate structures, 105, 106,

108—110, 130—132element of purely plane stiffness, 518end rotation, 120—121equally distributed prestressed cable structures,

396—397equilibrium, 29—92action principles, 32antisymmetric loadings, 84, 85, 86—87basic assumptions, 34—35beam equations, 77—78bending moment diagrams, 51—54Betti—Maxwell theorem, 76—77box girders, 482—486, 495—498, 503—504, 507,

508cable—beam structures, 318, 319, 320—322compound structures, 39—41concepts, 29—41, 81—84

conditions of, 30—32continuous beams, 191, 198, 207—210, 218cylindrical shells, 417—418, 420deflection curves, 77—78deformations, 29—92discrete mass systems, 535—536, 548, 563, 564dome shells, 426, 427elastic stability, 271—273elastic supports, 79—81flexibility, 81—84forces, 29frames, 262funicular structures, 57—61grid structures, 87—92, 357—358, 358ground structure supports, 597—598hyperbolic paraboloid shells, 440—443, 444internal forces handling, 41—65loading forces, 48—50member bending moment diagrams, 51—54membrane action in shells, 407—408multi-degree systems, 563, 564multi-storey systems, 521, 522plates, 363—364, 373—375, 392, 396—397reaction principles, 32sectional forces, 41—50shell membranes, 405, 406simply supported beams, 50—51, 160, 161,

182—183single-degree-of-freedom systems, 535—536, 548single-storey, single-bay frames, 229—230statically determinate formations, 35—39statically indeterminate structures, 93, 94, 96,

126—129stiffness, 81—84straight edge hypar shells, 444, 445, 448stress, 29—92support conditions, 32—34, 37—39support types, 91—92symmetric plane structures, 84—87thin-walled beams, 461—463, 466—467,

470—474three-hinged frames, 54—57, 58torsion, 90—91trusses, 61—65virtual work, 70—76, 89—90

equivalency, 254—256continuous systems, 577—580, 581seismic excitation, 574—576stiffness matrices, 562—563thin-walled beams, 476, 476, 477

equivalent actions, 286, 287equivalent cross-sections, 14, 15

630


Euler—Bernoulli beam theory, 139—140external actions, 96, 97, 98, 126, 127external deformations, 65external fixed-ended spans, 241—242external forces, 29, 30, 70—71, 254—256external joints, 235—236external loads, 196, 197, 262, 564, 571externally-placed unbonded tendons, 176—177external nodes, 241—242external pressure, 414external prestressing, 181—184

factors of safety, 26failure states, 147—150fictitious beams, 331—334, 336—337‘fictitious’ horizontal forces, 286—289fictitious springs, 410final end force/moment calculations, 129, 130final stress states, 219—220finite thickness soil layers, 595—596first-order analyses/theories, 34, 35, 45, 271fixation actions, 529, 530fixed boundaries, 431, 434fixed-ended beams, 119—123, 188—193, 216fixed-end imposed rotation, 117fixed-end moment, 241—242, 413fixed-fixed beams, 120—121fixed states, 116, 117, 119—120fixed structurescontinuous beams, 191—193, 216—217frames, 229—234, 237—238statically indeterminate structures, 114, 115,

126, 127fixed supports, 33, 37, 38, 91—92, 367, 378, 379flat slab stressesplates, 389—399prestressing, 396—399punching shear design, 393—395

flexibilityelastic stability, 289equilibrium, 81—84material structural behaviour, 7soil matrices, 595see also stiffening/stiffness

flexible supports, 131—132flexural cracks, 153folded plates, 399—402forced vibration, 542—544, 570—571force method, 352—353, 359analytical application, 98—101EI value influence, 104elastic support, 105, 106, 108—110

physical overview, 94—97, 98qualitative methods, 105—107, 108, 110result checking, 105statically indeterminate structures, 94—115support settlement, 102—104temperature, 101—102

formation definition, 39foundation beamselastic mediums, 613—616ground structure supports, 609—617numerical examples, 617soil simulations, 610—616Winkler model, 610—613

foundations, 29flat slabs, 389foundation beams, 609—617ground structure supports, 592—621horizontal loads, 620—621lateral pile responses, 620—621layout, 618pile foundations, 617—621rectangular spread footings, 597—609vertical loads, 619—620

four-side supported plates, 377—381design, 381prestressing, 381, 382stresses, 377—380

frames, 223—269deformation method, 114—138design, 266—268elastic stability, 282—290equilibrium, 54—57, 58force method, 94—114, 115ground structure supports, 611—617horizontal loads, 223—231, 234—238,

242—244inclined legs, 234—235joint checks, 266—268lateral stiffness, 231—234, 238—240mixed systems, 249—251multi-degree systems, 563multi-storey frames, 240—251, 288—290one-storey frames, 235—240, 283—287plastic analyses, 257—266plates, 390—391, 392prestressed concrete, 256—257reinforced concrete, 252—256section design, 251—257single-storey, single-bay frames, 223—235steel sections, 251—252stress states, 93—138vertical loads, 223—231, 235—236, 240—242

631

Index

free cantilever constructions, 206—207free edge zonescylindrical shells, 418—419, 422hyperbolic paraboloid shells, 442, 443, 443plates, 372—373, 375, 387—389

free formations, 30, 31freely suspended cables, 322—325free support conditions, 367free vibrationdamped vibration, 539—542discrete mass systems, 536—542, 563—570multi-degree systems, 563—570multi-storey spatial systems, 568—569numerical examples, 569—570plane systems, 563—568single-degree-of-freedom systems, 536—542undamped vibration, 536—539

frequencies, 537—538, 565—568, 569friction piles, 618full prestressing, 372fundamental eigenfrequencies, 566—567fundamental frequencies, 566, 567fundamental periods (T), 1funicular membranes, 405, 406funicular structures, 57—61, 315—316, 322—325,

427—428

Gaussian curvatures, 403—404, 436—437generatrices, 436—439, 440, 451girdersarches, 308—310box girders, 481—516cable structures, 328—335, 345—347cross-section walls, 502—505curved girders, 495—416equilibrium, 54frames, 223—235prestressing, 505—511, 512, 513profile sections deformability, 485—492rectilinear girders, 481—495suspension bridges, 328—335torsion, 496—502, 510—511, 512, 513

gradually imposed settlements, 204—205granular material, 415—416gravity loads, 1—2multi-storey frames, 240, 241one-storey multibay frames, 235simply supported beams, 179, 180

grid structures, 87—92, 351—360plates, 367, 368skew bridges, 358—360structural behaviour, 351—358

ground structure supports, 587—621deformations, 592—596, 597elastic soil, 592—596, 597foundation beams, 609—617horizontal loads, 620—621pile foundations, 617—621rectangular spread footings, 597—609shallow foundations, 592—617vertical loads, 592—596, 597, 619—620

half-beams, 144—145half-loading, 234, 235hardeningsteel, 7

hertz, 537hinged boundaries, 431hingesequilibrium, 33frames, 258—262, 265—266multi-storey frames, 244—246one-storey multibay frames, 237—240single-storey, single-bay frames, 226—232

hollow box sections, 91Hooke’s law, 6horizontal circular plans, 427—428horizontal displacement, 238—240horizontal flexibility, 289horizontal forceselastic stability, 286—289single-storey, single-bay frames, 231statically indeterminate structures, 133

horizontal loads, 620—621elastic stability, 295multi-storey frames, 242—244one-storey multibay frames, 236—238single-storey, single-bay frames, 223—231,

234—235horizontally distributed radial forces, 427, 428horizontal yielding, 298—299hypar shells, 444—448, 449, 450, 451hyperbolic paraboloid shells, 403, 404, 436—451creation, 436—437equilibrium, 440—443, 444geometry, 437—440

hypocentres, 548—549

identity matrices, 558immovable nodes, 134—135impact factors, 2imposed relative displacement, 118—119, 121—123imposed rotation, 117, 120—121, 126—127, 128imposed settlements, 203—205

632


imposed shifts, 126—127, 128inclined legs, 234—235indirect loading, 61, 62inertia forces, 534, 551, 552, 554—555inertia moment, 399, 400, 452instantaneous settlements, 203—204interaction curves, 253—254intermediate longitudinal beams, 354—357intermediate supports, 213—214internal deformations, 65internal equilibrium, 443internal forces, 351, 352, 405, 420, 421continuous beams, 207—208, 217, 218equilibrium, 41—65, 70—71, 84—87frame design, 254—256single-storey, single-bay frames, 229

internal joints, 235—236internally developed actions, 209—210internally-placed unbonded tendons, 176—177internal pressure, 409—413, 415—416, 429internal supports, 194intersecting barrel vaults, 422—423inverse matrices, 558—559inverted arches, 335—336inward pressure, 414inward radial displacement, 432

joint check procedures, 266—268joint equilibrium methods, 61, 62, 62

large-span beams, 198lateral bracing, 307lateral buckling, 290—293lateral loads, 282lateral pile responses, 620—621lateral soil pressure, 414lateral stiffness, 560, 561multi-storey frames, 244—249one-storey multibay frames, 238—240single-storey, single-bay frames, 231—234

lateral transverse displacement, 434layered soil, 589, 590limit states, 271—272limit tensile loads, 20—21linear loads, 370, 371liquefaction, 588liquid tanks, 413live loads, 2, 379, 380arches, 300—302, 304—306, 308—309, 311—312cable—beam structures, 320cable-stayed bridges, 344suspension bridges, 329, 330, 332—335

loads, 1—5antisymmetric, 84—87, 302, 332—335aquatic environments, 3arches, 300—302, 304—306, 308—309, 311—313box girders, 481—488, 490—491, 502—504, 505,

509—510cable—beam structures, 320, 322cable-stayed bridges, 344—345conoidal shells, 451, 452, 453climatic conditions, 3—4continuous beams, 187—189, 191—193cylindrical shells, 409—413, 416—418, 420, 422discrete mass systems, 543—545, 559, 572,

574—576dome shells, 427, 428—430, 434—435elastic stability, 283, 295equilibrium, 48—50, 51, 52, 84—85frames, 257—259, 266—267freely suspended cables, 322—323gravity loads, 1—2grid structures, 351, 352, 353—354, 357, 358ground structure supports, 592—596, 597, 601,

619—621hyperbolic paraboloid shells, 440—443, 443membrane action in shells, 403, 405—408multi-degree systems, 559, 564, 571, 572,

574—576multi-storey systems, 519—525, 528—530orthogonal plates, 370, 371pile foundations, 619—621plates, 361—391, 396—402prestressed cable nets, 326—327rectilinear girders, 481—485simply supported beams, 164—165, 166single-degree-of-freedom systems, 543, 544, 545,

546single-storey, single-bay frames, 226—231soil supports/surroundings, 2—3special impact loads, 4straight edge hypar shells, 444, 444, 445—446,

447, 451supports, 2—3suspension bridges, 329, 330, 332—335symmetric plane structures, 84—85temperature, 528—530thin-walled beams, 458—460

long barrel shells, 419—420longitudinal beams, 354—357longitudinal bending moment, 370longitudinal forces, 212, 417—419, 420, 421, 422longitudinal stresses, 251—252, 414, 422,

469—470

633

Index

machine induced vibrations, 581—585masonry, 233, 249mass-spring systems, 538—539material structural behaviourconcepts, 5—22concrete, 8, 10—13numerical examples, 22—26prestressed concrete tension, 18—22reinforced concrete, 13—18, 22—23steel, 5—8, 9

maximum inertia forces, 551, 552maximum tensile forces, 413membrane action, 403—409membrane forcescylindrical shells, 417—419, 422dome shells, 424—428, 432—433, 435hyperbolic paraboloid shells, 440—441

membrane load-carrying mechanisms, 416—418,453

membrane states, 405—406, 421, 440—443meridional forces, 425—428, 433‘midspan tendons’, 206—207‘modular ratios’, 210, 213—214moduli of elasticity, 6, 139, 213—214moduli of subgrade reaction, 601—602momentbimoment, 470—473box girders, 487—488, 502—503, 505, 507—510calculations, 129, 130conoidal shells, 452, 453continuous beams, 191, 196—198, 202—203cylindrical shells, 410, 411, 412, 413, 420—422decompressive forces, 173—174developing moment determination, 126—127,

128discrete mass systems, 549—550dome shells, 429, 431, 434equilibrium, 51—54, 89fixed-end, 241—242frames, 236—237, 241—242, 261—262, 264—266grid structures, 358—360plates, 362—370, 375, 378—379, 382—388, 392,

398—400simply supported beams, 140, 141, 149, 167statically indeterminate structures, 126—127,

128, 134—135suspension bridges, 333thin-walled beams, 456—460, 469—474zero-moment points, 236—237, 241—242

monolithic structures, 361—402boundaries, 365—366cantilever slabs, 375—377

circular plates, 385, 386flat slabs, 389—399folded plates, 399—402four-side supported plates, 377—381load-bearing action, 361—369orthogonal plates, 361—362, 362, 365—366,

369—384, 385plates, 361—402prestressing, 371—375ribbed plates, 381—384, 385skew plates, 386—389two-side supported slabs, 369—375

monolithic systemsbox girders, 487, 489—490conoidal shells, 452, 453continuous beams, 206—207equilibrium, 62—63, 64one-storey multibay frames, 235, 236single-storey, single-bay frames, 224, 225, 234,

235see also monolithic structures

multibay frames, 235—240multi-degree systemsdiscrete mass systems, 555—577dynamic equilibrium, 564forced vibration, 570—571free vibration, 563—570seismic excitation, 571—577stiffness matrices, 555—563

multi-storey frames, 240—251elastic stability, 288—290horizontal loads, 242—244lateral stiffness, 244—249mixed systems, 249—251vertical loads, 240—242

multi-storey systemsformation, 517—519free vibration, 568—569lateral responses, 517—532layout, 524—525, 530—531, 560, 561loads, 519—515, 520, 522, 525, 528—530orthogonal layouts, 525—528, 531—532plane elements, 521, 523—526stiffness matrices, 560—563temperature, 528—532

natural frequency, 537—538natural modes, 565natural periods, 537negative Gaussian curvatures, 403—404, 436—437Nervi, Pier Luigi, 367, 368network tied-arch layouts, 313

634


nodal action distribution, 132—133nodal action impositions, 114, 115nodal deformationsmember kinematics, 124, 125selection, 125, 126statically indeterminate structures, 123—125

nodal load distributions, 111, 113nodal rotation, 123node-located displacement senses, 135—136, 138non-cohesive soil types, 587—589, 590non-deformable cross-sections, 460—461non-shallow shells, 423—432non-symmetric loads, 322non-uniform self-weight distributions, 302non-uniform temperature changes, 68—69north light shells, 421, 422

oblique simply supported beams, 51, 52oblong areas, 384, 385oblong layouts, 443, 444oblong plates, 399one-bay rigid frames, 224, 225one-storey frames, 223—240elastic stability, 283—287horizontal loads, 236—238lateral stiffness, 238—240vertical loads, 235—236

one-time statically indeterminate frames, 224open cross-sectionsgeneral characteristics, 455—460non-deformable cross-sections, 460warping-based stresses, 462—465, 468, 470

‘open’ skeletal systems, 35opposite fixation actions, 529, 530orthogonal areasdome shells, 433—434ground structure supports, 593, 595—596straight edge hypar shells, 444—448, 449, 450

orthogonal full sections, 90—91orthogonality conditions, 566orthogonal layoutscylindrical shells, 421, 422multi-storey systems, 525—528, 531—532

orthogonal parabolas, 440orthogonal plates, 369—384cantilever slabs, 375—377four-side supported plates, 377—379load-bearing action, 361—362, 362, 365—366ribbed plates, 381—384, 385two-side supported slabs, 369—375

orthogonal systems, 437—440orthotropic plates, 382—384

paraboloid shells, 403, 404, 436—451creation, 436—437equilibrium, 440—443, 444geometry, 437—440

partial prestressing, 171, 173—177, 199, 372—375participation factors, 573periodic sinusoidal acting forces, 544—547peripheral prestressing, 415peripheral shear flow, 465—466permanent loads, 372—373, 376—382, 380, 382arches, 311—312cable—beam structures, 320cable-stayed bridges, 342, 343suspension bridges, 329, 330, 331suspension cables, 336—337

physical slope angles, 2—3pile caps, 618, 619—620, 619pile foundationsground structure supports, 617—621horizontal loads, 620—621lateral pile responses, 620—621layout, 618vertical loads, 619—620

pinned supports, 231, 232plane elementsdiscrete mass systems, 560, 561multi-storey systems, 518—519, 521, 523—526stiffness matrices, 560

plane formations, 351—360plates, 367, 368skew bridges, 358—360structural behaviour, 351—358

plane sections, 139—140plane structures, 65, 84—87plates, 361, 362shells, 403

plane systemsfree vibration, 563—568stiffness matrices, 562—563

plastic analyses, 358continuous beams, 220—221design, 260—262elastic stability, 293—296examples, 262—266, 294—296frames, 257—266numerical examples, 294—296

plastic behaviour, 147—150, 193, 553—555plastic deformations, 7plastic hingescontinuous beams, 189—191, 194frames, 258—262, 265—266simply supported beams, 149, 150

635

Index

plasticity, 153—154plastic neutral axis, 148plates, 361—402boundaries, 365—366cantilever slabs, 375—377circular plates, 385, 386continuous systems, 580, 581equations, 361—369flat slabs, 389—399folded plates, 399—402four-side supported plates, 377—381load-bearing action, 361—369orthogonal plates, 361—362, 362, 365—366,

369—384, 385prestressing, 371—375ribbed plates, 381—384, 385skew plates, 386—389two-side supported slabs, 369—375

Poisson ratio, 592—593polygonal area coverage, 435polygonal bases, 435—436pore pressure, 590portal frames, 226—227, 259, 260positive Gaussian curvatures, 403—404post-and-beam frames, 223—224‘preselected’ moment diagrams, 264—266pressurecohesive soil types, 590, 591cylindrical shells, 409—416, 422dome shells, 429non-cohesive soil types, 588—589, 590rectangular spread footings, 597—598

pressure lines, 224—226, 234, 297—299prestressed concreteframes, 256—257loss of prestress, 21—22material structural behaviour, 18—22numerical examples, 24—26simply supported beams, 179

prestressed concrete beamscontinuous beams, 195—199design, 170—172, 198—199simply supported beams, 163—177tendon design/structural performance, 195—198

prestressingbox girders, 505—511, 512, 513cable—beam structures, 318, 319, 320—322cable nets, 325—328cable-stayed bridges, 342, 343, 346—347cable structures, 388—389, 396—397, 398—399continuous beams, 208, 214, 217—220curved girders, 505—511, 512, 513

cylindrical shells, 415, 419, 420dome shells, 431—432, 432flat slabs, 396—399four-side supported plates, 381, 382material structural behaviour, 8, 9membrane action in shells, 409straight edge hypar shells, 447, 448, 449two-side supported slabs, 371—375

principal curvatures, 404, 406—407principal directions/stresses, 145—146principal moment, 366—367, 368, 369, 385, 390‘priority indices’, 265—266provisional supports, 213—214punching shear, 393—395, 398—399pure membrane states, 431, 442—443push over responses, 259, 260pylon stressing, 344—345

qualitative methods, 105—107, 108, 110, 134—138quasi-beams, 391—392

radial directions, 385radial displacement, 432radial forces, 427, 428radii of curvature, 439—440reaction principles, 32real loading, 70—74rectangular ground plan coverage, 403, 404, 438rectangular spread footings, 597—609concrete footing design, 608—609dimensioning, 608—609elastic bases, 601—608soil pressures/settlements, 598—601

rectilinear bars, 36—37rectilinear girders, 481—495loads, 481—485numerical examples, 492—494profile section deformability, 485—492

rectilinear models, 563redundant forces, 96, 97, 98, 101, 116—118redundant structures, 105, 106redundant systems, 111, 113, 200—205, 507—508reinforced concrete, 10frame design, 252—256material structural behaviour, 10, 13—18numerical examples, 22—23simply supported beams, 158, 159, 178—179

reinforced concrete beamscontinuous beams, 195service conditions, 150—158simply supported beams, 150—163

relative displacement, 118—119, 121—123, 548

636


relative rotation, 194relaxation of concrete, 12—13relieving action, 398—399relieving influence, 511, 513relocation of compressive forces, 168—169‘resistance’ forces, 361—365, 363resistance moment, 375resonance, 545rhomboid ground plan coverage, 439, 440ribbed plates, 381—384, 385rib structures, 606, 607rigid beams, 526, 527rigidity, 109, 110, 112, 130—132, 351—352,

355—360, 377—378grid structures, 351—352, 355—360plates, 377—378statically indeterminate structures, 109, 110,

112, 130—132ring beams, 427—428, 429—432ring forces, 416—428, 433ring membrane actions, 426, 427ring stresses, 415ring tension, 409rotationcontinuous beams, 191, 194discrete mass systems, 554—555dome shells, 427, 428equilibrium, 79—80, 82—83flexibility, 82—83frames, 262grid structures, 355—356, 357inertia forces, 534multi-storey frames, 241—242one-storey multibay frames, 235plates, 376statically indeterminate structures, 117,

120—128stiffness, 82—83thin-walled beams, 464—467

rotational springsground structure supports, 603—604with rigidity, 131—132

‘safety factors’, 193second-order theory, 277—282, 304—307sectional forces, 41—54curved beams, 495, 496deformations, 88—90equilibrium, 41—54, 57—58, 62—63, 70—71,

88—90frames, 257—259statically indeterminate structures, 101

seismic excitationdiscrete mass systems, 547—555, 571—577dynamic analyses, 571—574equivalent static loads, 574—576multi-degree systems, 571—577numerical examples, 576—577single-degree-of-freedom systems, 547—553

self-equilibrating systems, 34, 70—71box girders, 503—504cable—beam structures, 320—322continuous beams, 191, 198, 207—209hyperbolic paraboloid shells, 442—443, 443plates, 373—375, 396—397thin-walled beams, 470—474

self-weightarches, 302cable-stayed bridges, 341—342continuous beams, 208, 213cylindrical shells, 418—419, 420dome shells, 426—427, 433, 434single-storey, single-bay frames, 234, 235straight edge hypar shells, 446—447, 447, 448

semicircular cable-stayed bridges, 501—502sensitivity factors, 581—582service conditionsmaterial structural behaviour, 10prestressed concrete beams, 163—173reinforced concrete beams, 150—158steel beams, 139—147

service states, 199, 371settlements, 3continuous beams, 203—205ground structure supports, 593—599, 597,

598—601statically indeterminate structures, 102—104

shallow foundationsfoundation beams, 609—617ground structure supports, 592—617rectangular spread footings, 597—609

shallow shells, 432—436‘shear-acting’ frames, 249—251shear base, 551, 552shear beam behaviour, 244, 246—247shear centres, 461—462shear flow, 465—466, 483—484, 490—491,

503—504shear forcesbox girders, 509—510continuous beams, 209—210, 211—212cylindrical shells, 410, 411, 417—419, 420dome shells, 435—436equilibrium, 68, 89

637

Index

shear forces (cont’d )membrane action in shells, 406—407one-storey multibay frames, 235, 237—238plates, 393simply supported beams, 144—145, 153—157,

162—163, 172—173, 177single-storey, single-bay frames, 223—224statically indeterminate structures, 126—127, 128straight edge hypar shells, 448, 450thin-walled beams, 457, 458, 461—462,

471—472shearing responses, 178shearing strain, 68shearing strength, 588shear mechanisms, 142—143, 444, 445, 445shear moduli, 68shear punching, 393—395, 398—399shear stresses, 44, 45, 140—141box girders, 484thin-walled beams, 458, 461—464, 473—474

shear transfer mechanisms, 142—143shells, 403—453barrel shells, 416—423canoidal shells, 451—453constant internal pressure, 409—413cylindrical shells, 409—423cylindrical tanks, 413—415design, 409dome shells, 423—436generation methods, 407hypar shells, 444—448, 449, 450hyperbolic paraboloid shells, 436—451membrane action, 403—409non-shallow shells, 423—432shallow shells, 432—436short barrel shells, 419silos, 415—416straight edge hypar shells, 444—448, 449, 450surface geometry, 404—405

simple supportsequilibrium, 32—33, 37—38, 79, 81, 91plates, 378, 379

simply fixed beams, 116—119simply supported beams, 139—185cable structures, 331cantilever beams, 177—181compression, 165—172, 179, 180continuous beams, 191—193, 215, 216, 217design control, 184—185equilibrium, 50—51, 52external prestressing, 181—184failure states, 147—150

partial prestressing, 173—177prestressed concrete, 179reinforced concrete, 150—163, 178—179steel beams, 139—150suspension bridges, 331ultimate states, 158—163, 173—177

single-degree-of-freedom systemsdesign spectra, 553, 554discrete mass systems, 535—555dynamic equilibrium, 535—536, 548forced vibration, 542—544free vibration, 536—542periodic sinusoidal acting forces, 544—547plastic behaviour, 553—555seismic excitation, 547—555

single-mass systemscontinuous systems, 577—580, 581free vibration, 358—359, 537

single-storey, single-bay frames, 223—235horizontal loads, 223—231, 234—235inclined legs, 234—235lateral stiffness, 231—234vertical loads, 223—231

sinusoidal acting forces, 544—547skeletal systems, 29, 30, 35skew bridges, 358—360, 384skew ground plans, 439skew layouts, 385skew plates, 386—389slabs see platesslenderness, 274—275snow, 4soils, 2—3, 230—231cylindrical shells, 414elastic mediums, 613—616elastic soil deformational behaviour, 592—596,

597foundation beams, 610—613mechanical characteristics, 587—592pressures, 598—601rectangular spread footings, 597—601seismic excitation, 552—553settlements, 598—601Winkler model, 610—613

space systems, 325—326special impact loads, 4spectral displacement, 550spherical shells see dome shellssplitting arbitrary loadings, 84, 85springsfictitious springs, 410ground structure supports, 602—604

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machine induced vibrations, 583—585multi-storey systems, 526, 527with rigidity, 130—132undamped vibration, 538—539

square ground plans, 422, 423dome shells, 433, 435straight edge hypar shells, 448, 450

square layouts, 442, 443square matrices, 557stabilisation of cable structures, 316—317stabilised crack patterns, 16—17statically determinate load-carrying action, 376statically determinate parts, 373, 398statically determinate structuresbox girders, 506—507, 508, 509—510continuous beams, 196—197, 217—218equilibrium, 35—40frames, 224

statically determined frames, 256—257statically determined problems, 407—408statically indeterminate prestressing moment,

197—198, 509—510statically indeterminate structuresbox girders, 507—509, 508, 509—510continuous beams, 196—198deformation method, 114—138EI value influence, 104elastic support, 105, 106, 108—110, 130—132equilibrium, 37, 39, 41fixed-end beams, 119—123force method, 94—114, 115frames, 224nodal action distribution, 132—133qualitative methods, 105—107, 108, 110,

134—138simply fixed beams, 116—119stress states, 93—138support settlement, 102—104temperature, 101—102, 116, 117, 120unknown nodal deformations, 123—125

statically redundant forces, 93, 95statically redundant parts, 373—375, 398statically redundant structures, 232—233, 256—257statically redundant systems, 200—205, 232—233static interactions, 163—164static loads, 1, 2, 574—576static theorem, 153—154, 260—261, 263—264, 358steel, 373—375, 384, 398concepts, 5—8, 9continuous beams, 216—217frames, 251—252, 266—267single-storey, single-bay frames, 233

steel beamscontinuous beams, 188—194, 213—214elastic behaviour, 139—147failure states, 147—150fixed-ended beams, 188—191plastic behaviour, 147—150service conditions, 139—147simply supported beams, 139—150, 191—193

stiffening/stiffnessarches, 308—310box girders, 486—487cable structures, 335—339cylindrical shells, 410, 418discrete mass systems, 560, 561elastic stability, 279—280equilibrium, 81—84frames, 256—257grid structures, 351—352material structural behaviour, 6—7multi-degree systems, 569—570multi-storey frames, 241—242, 244—249multi-storey systems, 517—521, 523, 524one-storey multibay frames, 238—240plates, 364—365, 367, 368, 382—384single-storey, single-bay frames, 228, 231—234statically indeterminate structures, 104, 106,

109—114suspension cables, 335—339

stiffness matricesconcepts, 555—557elements, 555—557multi-degree systems, 555—563multiplication, 557, 558multi-storey systems, 560—563operations, 557—559

stirrups, 394—395straight beams, 410—411straight boundary plates, 365—366straight edge beams, 450straight edge hypar shells, 444—448, 449, 450, 451straight-line generators, 409straight line segments, 436—437stresses, 5arches, 300—302box girders, 482—484, 487—489, 490cable structures, 325, 340—342, 344—345cantilever slabs, 375—376closed cross-sections, 462—463, 465—467cohesive soil types, 591concepts, 29—41continuous beams, 200, 203—207, 215—217,

219—220

639

Index

stresses (cont’d)cylindrical shells, 414—415, 421, 422deformations, 114—138, 271—296EI value influence, 104equilibrium, 29—92flat slabs, 389—393, 394—395force method, 94—114, 115four-side supported plates, 377—380frames, 93—138, 251—252, 282—290grid structures, 87—92hyperbolic paraboloid shells, 441internal forces handling, 41—65lateral buckling, 290—293material structural behaviour, 11, 12membrane action in shells, 405—406,

407—408open cross-sections, 462—465plastic analyses, 293—296plates, 365second-order theory, 277—282simply supported beams, 165—166stiffness, 110—114suspension bridges, 329, 330symmetric plane structures, 84—87thin-walled beams, 456—458, 461—474two-side supported slabs, 369—370, 371warping, 461—470

stress—strain diagrams, 6, 8, 9structural webs, 533—536, 534, 547structure definition, 39subgrade reaction moduli, 601—602supports, 2—3, 32—33, 32, 33arches, 298—299box girders, 490—491, 491, 492cable structures, 331cantilever beams, 177—181cohesive soil types, 587, 590—592continuous beams, 191—193, 213—216, 217deformations, 592—596, 597design control, 184—185elastic soil, 592—596, 597elastic stability, 274equilibrium, 32—34, 37—39, 50—52, 79—81,

91—92external prestressing, 181—184flat slabs, 389foundation beams, 609—617frames, 256, 257ground structures, 587—621horizontal loads, 620—621hyperbolic paraboloid shells, 440—442, 443internal supports, 194

membrane action in shells, 409multi-storey systems, 524, 525, 526, 527non-cohesive soil types, 587, 588—589, 590pile foundations, 617—621plates, 367, 376, 400prestressed concrete beams, 163—177rectangular spread footings, 597—609reinforced concrete beams, 150—163simply supported beams, 139—185statically indeterminate structures, 102—105,

106, 108—110, 130—132steel beams, 139—150suspension bridges, 331types, 32—33, 32, 33, 91—92vertical loads, 592—596, 597, 619—620

surface loads, 2suspension action, 445—446, 450suspension bridges, 328—335suspension cables, 335—339symmetric loads, 84—85, 332—335, 333, 481—484symmetric plane structures, 84—87

T see fundamental periodstemperature, 4, 528—532composite beams, 215—217continuous beams, 215—217equilibrium, 68—70statically indeterminate structures, 101—102,

116, 117, 120tendons, 195—198, 206—207, 505—507, 511, 513tensionbox girders, 483—484, 486—487, 502, 505cable-stayed bridges, 341—342conoidal shells, 452, 453continuous beams, 199, 216—217cylindrical shells, 413, 415—417elastic stability, 281equilibrium, 45, 46hyperbolic paraboloid shells, 442—443, 443plates, 381—382, 384prestressed concrete, 18—22simply supported beams, 153single-storey, single-bay frames, 229straight edge hypar shells, 445—448, 449,

450—451thin-walled beams, 455—479bimoment, 469—473, 470box girders, 481—516formation, 455, 456general characteristics, 455—460longitudinal stresses, 469—470non-deformable cross-sections, 460—461

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shear centres, 461—462shear stresses, 473—474torsion, 457—458, 462—477warping, 460—470

thin-walled models, 158, 159three equilibrium conditions of free formations,

30, 31three-hinged arch, 298, 305three-hinged connections, 35—38three-hinged frames, 54—58, 226—228, 230—231tied-arch systems, 311—313top flanges, 213—214, 217—219torsion, 89, 90—91, 157—158, 159application examples, 499—502bimoment, 470—472, 473box girders, 484—485, 495—507, 509—513curved girders, 496—502, 510—511, 512, 513dome shells, 427, 428, 430equations, 474—477equilibrium, 496—498grid structures, 355—360, 356, 357plates, 362—364, 367, 368, 387—388thin-walled beams, 457—460, 462—477warping-based stresses, 462—470

total frame collapse, 257—259total loads, 344, 397—398, 446, 447traffic loads, 372—373transformed concrete sections, 159, 160translational inertia forces, 534translational springs, 602—603translational stiffness, 523, 524transpose matrices, 557transversal loads, 441—442transverse beams, 354—357transverse bending moment, 370, 420—421, 422transverse compression, 422transverse diaphragms, 490—491, 492, 503, 504transverse displacement, 434transverse loads, 361, 362, 363—365transverse stiffness, 279—280‘tree’ skeletal systems, 35triangle of forces, 31, 32triangular bending moment, 196—198, 507—508triangular ground plans, 422, 423, 435triangular loading, 509triangular truss formations, 36—37Trost’s proposal, 11, 13, 21—22trusseselastic stability, 293, 294equilibrium, 36—37, 61—65frames, 267—268simply supported beams, 153—157, 158, 178—179

twist, 438see also torsion

twisting moment, 363—366, 378—379, 384two hinged solid bars, 124, 125two-hinge frames, 227—232two-side supported slabs, 369—375, 382, 399—400design, 371prestressing, 371—375stresses, 369—370, 371

two-span beams, 217—218two-storey spatial systems, 534, 535

ultimate bending moment, 149ultimate force, 394ultimate horizontal load values, 295ultimate states, 158—163, 173—177, 253—254ultimate strength, 192—193, 199ultimate tensile loads, 18, 21unbonded tendons, 176—177undamped vibration, 536—539uniform loadsbox girders, 504cylindrical shells, 413dome shells, 425, 426—428, 429, 434equilibrium, 50—51frames, 259, 260ground structure supports, 593one-storey multibay frames, 235plates, 370, 376—379single-storey, single-bay frames, 226—227statically indeterminate structures, 116,

119—120uniformly distributed forces/moment, 411, 412uniform settlement, 599, 600, 600uniform temperature changes, 68, 69unknown displacement, 126—127, 128unknown nodal deformations, 123—125, 125unknown rotation, 126—127, 128unyielding supports, 440—441, 442upward applied deviation forces, 398—399upward-curved parabolas, 451

variable-angle truss models, 157, 173variable beam heights, 178, 179—181vertical columns, 133vertical loadsbox girders, 504, 505, 509—510cable structures, 315—316conoidal shells, 453cylindrical shells, 420dome shells, 426, 427, 429elastic stability, 282, 283—285

641

Index

vertical loads (cont’d )ground structure supports, 592—596, 597,

619—620hyperbolic paraboloid shells, 441multi-storey frames, 240—241one-storey multibay frames, 235—236pile foundations, 619—620prestressed cable nets, 326—327single-storey, single-bay frames, 223—231straight edge hypar shells, 444, 445suspension bridges, 328—329

vertical shear forces, 141, 142vertical stiffness, 517, 518, 519, 520vertical straight beams, 410—411vibrationsannoying vibration avoidance, 580—585damped vibration, 539—542discrete mass systems, 547—553forced vibration, 542—544, 570—571free vibration, 536—542, 563—570human activities, 580—581machine induced, 581—585seismic excitation, 547—553, 571—577single-degree-of-freedom systems, 547—553undamped vibration, 536—539

Vierendeel beams, 143—144

virtual work, 376—377continuous beams, 191equilibrium, 70—76, 89—90multi-storey frames, 246, 247—248simply supported beams, 153, 161statically indeterminate structures, 104

warping, 461—467analyses, 467—469closed cross-sections, 462—463, 465—467constants, 469—470longitudinal stresses, 469—470open cross-sections, 462—465shear stresses, 473—474stresses, 461—470thin-walled beams, 460—470

water tables, 589, 590wind forces, 3—4Winkler model, 610—613working stress design, 27

yielding, 298—299yield states, 160, 161yield stresses, 7—8, 254—255, 394—395

zero-moment points, 236—237, 241—242

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Structural Systems by L Stavridis

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