Mechanics and Mechanisms of Fracture

Mechanics andMechanisms of

Fracture:An Introduction

Alan F. Liu

ASM InternationalMaterials Park, Ohio 44073-0002

www.asminternational.org

2005 ASM International. All Rights Reserved.Mechanics and Mechanisms of Fracture: An Introduction (#06954G)


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Library of Congress Cataloging-in-Publication DataLiu, A.F.

Mechanics and mechanisms of fracture: an introduction / A.F. Liu.p. cm.

Includes bibliographical references and index.1. Fracture mechanics. I. Title.

TA409.L58 2005620.1126dc22 2005042107

ISBN: 0-87170-802-7SAN: 204-7586

ASM InternationalMaterials Park, OH 44073-0002


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iii

Contents

Preface ........................................................................................................................v

Chapter 1 Solid Mechanics of Homogeneous Materials ..................................................11.1 Key Types of Mechanical Behavior ......................................................................31.2 Stress and Strain ...............................................................................................71.3 Principal Stresses and Principal Strains ................................................................ 101.4 Equivalent Stress and Equivalent Strain ............................................................... 151.5 Stress Analysis of Monolithic Load-Carrying Members .......................................... 151.6 Stress Analysis Using Finite Element Methods ..................................................... 261.7 Local Stress Distribution at a Geometric Discontinuity ........................................... 331.8 Stress Analysis of Cracks .................................................................................. 39

Chapter 2 Deformation and Fracture Mechanisms and Static Strength of Metals ........... 472.1 Elastic and Plastic Behavior .............................................................................. 472.2 Yield Criteria .................................................................................................. 572.3 Fracture Criteria .............................................................................................. 592.4 Fracture Mechanisms and Appearances ............................................................... 622.5 Fracture Strengths ............................................................................................ 842.6 Residual Stresses ............................................................................................. 962.7 Material Toughness .........................................................................................1002.8 Deformation and Fracture under Sustained Loads .................................................102

Chapter 3 Fatigue Strength of Metals ........................................................................1213.1 Mechanical Behavior under Cyclic Loads ...........................................................1213.2 Microscopic and Macroscopic Aspects of Fatigue and Crack Propagation .................1273.3 Fatigue Life, Crack Initiation, Crack Growth, and Total Life ..................................1363.4 Innite-Life (Stress-Based) Fatigue Strength .......................................................1383.5 Finite-Life (Strain-Based) Fatigue Strength .........................................................1513.6 Some Practical Fatigue Design Considerations .....................................................158

Chapter 4 Static and Dynamic Fracture Toughness of Metals ......................................1694.1 Linear Elastic Fracture Mechanics .....................................................................1704.2 Plane-Strain Fracture Toughness: Static KIC .........................................................1774.3 Dynamic KIC .................................................................................................1874.4 Plane-Stress Fracture Toughness, KC ..................................................................2004.5 Fracture under Mixed Modes 1 and 2 .................................................................208

Chapter 5 Damage Tolerance of Metals .....................................................................2155.1 Stress-Intensity Factor and Damage Tolerance Analysis .........................................2155.2 Determination of Stress-Intensity Factors for Nonstandard Congurations ................2295.3 Fatigue Crack Growth in Room-Temperature Air .................................................236



Chapter 6 Nonlinear Fracture Mechanics ..................................................................2816.1 Elastic-Plastic Fracture Mechanics .....................................................................2816.2 Time-Dependent Fracture Mechanics .................................................................286

Chapter 7 Mechanical Behavior of Nonmetallic Materials ...........................................2977.1 Ceramics and Glasses ......................................................................................2977.2 Polymers .......................................................................................................3007.3 Fractography .................................................................................................315

Chapter 8 Mechanics of Fiber-Reinforced Composites ................................................3198.1 Types of Composites .......................................................................................3208.2 Coding System ...............................................................................................3218.3 Stresses and Strains in Composite Laminates .......................................................3238.4 Failure Mechanisms of Composites ....................................................................3288.5 Fracture Mechanics for Fibrous Composites ........................................................3368.6 Damage Tolerance of Composites ......................................................................3408.7 Some Practical Issues ......................................................................................343

Appendix 1 Lattice Structure and Deformation Mechanisms in MetallicSingle Crystals ......................................................................................357

A1.1 Crystal Lattice Structures .................................................................................357A1.2 Lattice Response under Stress ...........................................................................362A1.3 Mechanical Twinning ......................................................................................368A1.4 Stacking Faults ..............................................................................................370

Appendix 2 Close-Form Representation of Tangential Stress Distribution atCircular and Elliptical Holes ...................................................................373

A2.1 Innite Sheet .................................................................................................373A2.2 Finite-Width Sheet ..........................................................................................374A2.3 Saffs Close-Form Equation .............................................................................374A2.4 Comparison between Saffs Close-Form Equation and the Analytical Solutions .........377

Appendix 3 Nonarbitrary Crack Size Concept for Fatigue Crack Initiation ..................379

Appendix 4 Fatigue Spectrum Editing .......................................................................383A4.1 Spectrum Simplication: RACETRAK Method .......................................................384A4.2 Cycle Counting: Range-Pair Method ..................................................................385

Appendix 5 Stress Severity Factor .............................................................................391

Appendix 6 Specimen Orientation and Fracture Plane Identication ............................395

Appendix 7 Mechanical Properties Data for Selected Aluminum Alloys ........................397A7.1 Conventional and High-Strength Aluminum Alloys ..............................................397A7.2 P/M Aluminum ..............................................................................................398A7.3 Aluminum-Lithium Alloys ...............................................................................398A7.4 Aluminum Casting Alloys ................................................................................398

Appendix 8 Mechanical Properties Data for Selected Titanium Alloys ..........................411

Appendix 9 Mechanical Properties Data for Selected Titanium Aluminides ...................415

Appendix 10 Mechanical Properties Data for Selected Steels ........................................423

Appendix 11 Conversion Table .................................................................................429

Appendix 12 Glossary of Terms, Symbols, and Abbreviations ......................................431

Index ........................................................................................................................443

iv



vPreface

Failure of a machine part (or structural part) could be attributed to many known or unknownfactors. The most fundamental one is due to static overload of a structural member. The term overloadmight mean the structural member is subjected to a load that exceeds the ultimate strength of thematerial. Overload might also mean the applied load is higher than the anticipated design load of thestructure, the value of which is normally set to lower than the materials ultimate strength. Inciden-tally, structural parts often fail at load levels much lower than the design load. An initial defect (ordamage) in the part, cyclic load-induced cracking, or time-dependent/environmentally-assistedcrack-ing all can cause a machine part to fail prematurely. The studies of the problems and the remediesfor handling the failure modes in the rst two types of failures mentioned are called fracture me-chanics and fatigue; high-temperature creep, stress-corrosion, corrosion-fatigue, or hydrogen-embrittlement are the sources for the rest.

Fracture mechanics is the study of the inuence of loading, crack size, and structural geometry onthe fracture resistance of materials containing natural aws and cracks. When applied to design, theobjective of the fracture mechanics analysis is to limit the operating stress level so that a preexistingcrack would not grow to a critical size during the service life of the structure. Fatigue is the studyof the effects of repeated loading on a machine part that was initially defect free, and how thoseloads (in combination with other factors) may shorten its anticipated life. Understanding the mech-anisms of creep and stress-corrosion/hydrogen-embrittlement is important when a machine part isintended for use in an extreme environment.

Both the metallurgical and mechanical elements in each type of these failure modes will be dis-cussed. In the metallurgical eld, the model that offers explanations to a phenomenon is often casespecic. Experience gained may or may not be applicable to another case, however similar it mayseem. Revised or updated concepts or solutions appear in the literature from time to time. It isimpossible to discuss all the concepts and methods in a single book. Only those relevant to theintended specic themes are included here. In contrast, the mechanical causes of failure can beanalyzed by means of solid mechanics. Nevertheless, the solution to a problem is valid only for theset-up and assumptions as intended; thus it is also somewhat case specic.

The main focus of this book is to explicate how materials respond to applied forces, and relatethem to design analysis, material evaluation, and failure prevention. Metals occupy the main part ofthe book; also included are non-metallic materials such as ceramics, plastics, and ber-reinforcedpolymer-matrix composites. Both the fundamental and practical concepts of fracture are describedin terms of stress analysis and the mechanical behavior of materials. The metallurgical aspects ofdeformation and fracture in metals also are discussed. The rst two chapters of this book can beregarded as the fundamentals of stress analysis and mechanical behavior of materials. These chaptersprovide necessary knowledge for the understanding and appreciation of the contents in chapters thatfollow.

With regard to structural design and analysis, a strong emphasis is placed on showing how fracturemode is inuenced by the state of stress in the part. The stress analysis section in Chapter 1 servesas a crash course (or a refresher course) in the strength of materials and prepares the reader with thebasic analytical tools for the remainder of the book. A progressive approach is taken to show theeffect of structural geometry and loading conditions to the resulting stressesrst to show the stressesin undamaged structural members; next to show the stresses in structures that contain a geometric



discontinuity; then nally, to show the stress eld in the vicinity of a crack. Stress analysis of cracksrequires the use of a new analytical tool called fracture mechanics. This is discussed in Chapter 1for primary elastic with small scale yielding at the crack tip, and in Chapter 6 for large scale yieldingat the crack tip. Application of fracture mechanics to design/analysis and the related tasks is presentedin Chapters 4 through 6. Numerous examples are given throughout this book to illustrate the elasticand plastic behavior of materials at a stress raiser, and how the static, fatigue, and residual strengthsof a machine part might be affected by it. Finally, the structural analysis methods, as well as thedamage tolerant aspects of ber-reinforced composites, are discussed in Chapter 8.

Strangely, in the last couple years, 101 has become a household term. It penetrates into ourhomes by way of television, radio, and newspaper. Politicians, entertainers, the media, and talk showhosts and their guests, would freely use Whatever 101 to link to the topic or event that is beingdiscussed. Therefore this book could, in that sense, qualify to be titled Mechanics and MechanicalBehavior of Materials 101.

To reiterate: This book is about how machine (or structural) parts fail, why one piece fails in acertain way and another piece fails differently; and will provide engineering tools for analyzing thesefailures, and ultimately, preventing failure. This book can be used as a desktop reference book or asa self-study book, and can be used by engineering students and practicing engineers with or withoutsome prior training in solid mechanics and/or mechanical metallurgy.

I want to sincerely thank the reviewers at ASM for their constructive comments. A special thanksto Steve Lampman for his effort in integrating handbook content in the area of brittle and ductilefractures, and the mechanism of intergranular fracture, into Sections 2.4.3 and 2.4.4. Last but notleast, I wish to acknowledge the excellent work that Kathy Dragolich put in the coordination of thisbook.

A.F. LiuWest Hills, California

vi



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CHAPTER 1

Solid Mechanics ofHomogeneous Materials

Table 1.1 Bond energies for various materials

Bond energy

Bond type Material kJ/mol kcal/molIonic NaCl 640 153

MgO 1000 239Covalent Si 450 108

C (diamond) 713 170Metallic Hg 68 16

Al 324 77Fe 406 97W 849 203

van der Waals Ar 7.8 1.8Cl2 31 7.4

Hydrogen NH3 35 8.4H2O 51 12.2

Fig. 1.1 General characteristics of major classes of engineering materials

ENGINEERING MATERIALS can be con-veniently grouped into ve categories: metalsand alloys, intermetallics, ceramics and glasses,polymers (plastics), and composites. General ad-vantages of these materials are summarized inFig. 1.1 along with the predominant type ofbonding and spatial arrangement of their con-stituents. The bond type is related to the sharingof electrons due to differences in the number ofelectrons in the outer electron shells of individ-ual atoms. This results in different types ofbonding and bond strength (Table 1.1), includ-ing strong primary bonds (covalent and ionicbonds), intermediate-strength metallic bonds,and weaker secondary bonds (van der Waalbonds and hydrogen bonds). The type of bond-ing is an essential factor that inuences thephysical and mechanical properties of a sub-stance, and makes each class of materials

unique. The exception may be composites,which are not homogeneous materials. A com-posite is a kind of engineered material that con-sists of particulate llers or strong/stiff bers ina soft matrix.

Mechanics and Mechanisms of Fracture: An IntroductionAlan F. Liu, p1-45 DOI:10.1361/memf2005p001

Copyright 2005 ASM International All rights reserved. www.asminternational.org

2 / Mechanics and Mechanisms of Fracture: An Introduction

The bonds between atoms also occur in vari-ous spatial orientations. The spatial arrangementof atoms may be random (amorphous), or theatoms may be periodically arranged either in atwo-dimensional array (as in the case of a graph-ite sheet structure) or in a three-dimensional ar-ray of a crystal lattice. Over still greater dis-tances, spatial arrangements may be completelyamorphous, a single crystal, or polycrystalline(as is the case in most metals and alloys, wheremany individual crystalline grains are rotatedwith respect to one another). Microstructuresmay also be mixtures of crystalline and non-crystalline regions or have regions of orientedstructures (as in the case of aligned carbon back-bone chains in polymers).

Mechanical behavior depends on both thetype of bonding and the long-range spatial ar-rangement of constituents in a material. Metalsrepresent the majority of the pure elements andform the basis for the majority of the structuralmaterials. The metallic bonds typically formcrystalline grains, and the most common crystalstructures in most engineering metals includeface-centered cubic (fcc), body-centered cubic(bcc), or hexagonal close-packed (hcp) struc-tures. Deformation and fracture mechanisms ofmetals are described in Appendix 1.

Ceramics and glasses, which have strongionic-covalent chemical bonds, are very strong,stiff, and brittle. Crystal structures of ceramicmaterials are often more complex than those ofmetallic materials. High packing density cubicstructures do exist in some ceramic materials(rock salt and cesium chloride structures), butthe presence of more than one type of atom to-gether with ionic bonding causes differences incritical stresses for ductile (slip) and brittle(cleavage) behavior compared to metallic ma-terials. Qualitatively, the critical stresses for slipare so high relative to those required to propa-gate cleavage that these materials are consideredto be inherently brittle.

The physical and mechanical properties of in-termetallics fall in between ceramics and metals,but their bonding system is a combination of me-tallic, ionic, and/or covalent types. Intermetallicsare also crystalline, and so the mechanisms ofdeformation and fracture can be discussedwithin the same framework as metals (e.g., therole of antishear twinning, microtwinning,etc.*). Themicrostructure of thesematerials con-

trols microscale fracture mechanisms and frac-ture site initiation. In this book, intermetallics,such as TiAl, Ti3Al, NiAl, and Ni3Al, are treatedthe same as metals because they are used forhigh-temperature applications the same way ashigh-temperature superalloys.

Polymers, on the other hand, are long-chainmolecules that may bond together by the weakvan der Waals force, the stronger hydrogenbond, or the much stronger covalent bond. Thepolymer itself is a long chain of monomer units(typically made from compounds of carbon, butalso from inorganic chemicals, such as silicatesand silicones) that form a chain from strong co-valent bonds between the monomer units. Theselong polymer chains may also bond together invarious ways. For example, weak van der Waalsbonds occur between polymer chains in the caseof thermoplastic polymers, while stronger hy-drogen or covalent bonds occur between thepolymer chains in cross-linked thermosettingplastics.

Crystallinity is seldom complete in a poly-meric material, although some limited crystallin-ity can occur in polymers. Crystallinity occurswhen the polymer chains arrange themselvesinto an orderly structure. In general, simplepolymers (with little or no side branching) crys-tallize very easily. Crystallization is inhibited inheavily cross-linked (thermoset) polymers andin polymers containing bulky side groups. Theability of polymeric materials to form crystallinesolids depends in part on the complexity of thependant side groups of atoms attached to the co-valently bonded carbon atom backbone. Poly-meric materials containing mixtures of crystal-line and noncrystalline regions show decreasedductility and increased strength and stiffness asthe degree of crystallinity increases. Conversely,polymers exhibit low strength, low stiffness, andsusceptibility to creep when the temperature isabove the glass-transition temperature (i.e.,when the polymer structure becomes amorphouswith pronounced viscoelastic behavior).

The aim of this book is to introduce the vari-ous types of mechanical behavior/properties ofthese materials and provide analytical tools toassess the strength and life of a structural designin terms of mechanical properties. The rst taskof designing a mechanical/structural componentis to select a group of materials that can with-stand the in-service environment. For instance,among all the materials, one type of alloy isknown to have good resistance to corrosion; an-other type exhibits excellent high-temperature

*See Appendix 1 for a description of micromechanisms inthe deformation and fracture of metals.

Chapter 1: Solid Mechanics of Homogeneous Materials / 3

Fig. 1.2 Cylindrical bar subjected to axial load. (a) Linearelongation along the loading direction. (b) Free body

diagram. (c) Linear relationship between force and elastic defor-mation (or stress and elastic strain)

performance; a third group is specially devel-oped for low-temperature application, and so on.Depending on the service environment, as wellas the stiffness, strength, and life requirements,there is a little leeway left for material selection.Structural mechanics is an essential tool for de-sign. It helps to understand what leads a materialto respond to stresses in a certain way. An op-timum design is the one that turns the initiallyunfavorable failure mode to a favorable one.This can be done by cleverly combining stressanalysis and material selection. Much of thisbook will focus on various aspects of stress anal-ysis and related topics.

1.1 Key Types of Mechanical Behavior

Most engineering materials of concern in thisbook can be regarded as deformable solids. Theyrespond to applied loads in three different ways:elastic, plastic, and viscoelastic. Elastic andplastic behaviors deal with deformation (whenthe body is subjected to external force). Visco-elastic behavior actually is a combination ofelastic and viscous behavior. A material havingthis property is considered to combine the fea-tures of a perfectly elastic solid and a perfectuid. To honor those who rst introduced thethree types of ideal substances, the perfectlyelastic solid is commonly called the Hookesolid; the perfectly plastic solid is known as theSt. Venant solid. The perfect uid, which is con-sidered a viscous material, is called the Newto-nian liquid.

Consider a cylindrical bar or rod xed at oneend and with an external force P pulling at theother end, as shown in Fig. 1.2(a). The two gagemarks placed on the surface of the rod in itsunstrained state will move apart as the length ofthe rod is elongated under the applied load. Thedeformation d, i.e., the change in gage length tothe original gage length ratio, is called the av-erage strain and is represented by:

e d/L (Eq 1.1a)0

As shown in Fig. 1.2(b), the external load P isbalanced by the internal resisting force r A,where r is the stress normal to the cutting planeand A is the cross-sectional area of the rod.Therefore, in equilibrium:

r P/A (Eq 1.1b)

If we gradually increase the strain of a Hookesolid, the stress is also increased proportionally.The relation of traction to strain is a linear lineas shown in Fig. 1.2(c). This behavior is com-monly known as Hookes law. However, it neednot be a perfectly straight line; the denition ofelasticity only requires that on gradually dimin-ishing strain the same curve is retraced, and thaton complete release of the stress no permanentdeformation remains.

For the St. Venant solid, we assume the forceper unit of area that causes an extension or con-traction must reach a certain nite value for theinitiation of permanent distortion. When thischaracteristic load is reached, the substancestarts to deform permanently, or to yield, andcontinues so at this load. We further assume thatthe change in shape during increasing load issmall. This case is then represented by drawinga horizontal straight line in the diagram in whichthe permanent extensions are plotted as abscis-sas and the forces per unit area as ordinates, asshown in Fig. 1.3(a). This behavior is commonlyknown as rigidperfectly plastic.


Fig. 1.3 Schematic idealized ow curves (a) Rigidperfectly plastic material. (b) Perfectly plastic material with an elastic region. (c)Piecewise linear (strain-hardening) material. (d) Typical of engineering alloys

Very few substances exhibit only one form ofdeformation. They usually behave in combina-tion of any two of the three, and in a few casesall three. For example, viscoelasticity is a com-bination of viscosity and elasticity. However, themost common type of substance combines elas-ticity and plasticity. The theory of plasticity con-siders three types of substances that involveplastic ow, presented in idealized forms as:rigid ideal plastic material (Fig. 1.3a), ideal plas-tic material with an elastic region (Fig. 1.3b),and piecewise linear (strain-hardenable) mate-rial (Fig. 1.3c). As the assumption in materialbehavior changes from (a) toward (c), the com-plexity in mathematical representation of eachof these models increases.

The curve in Fig. 1.3(b) represents an elasticperfectly plastic material. It exhibits elastic be-havior up to the yield stress, then plastic owfollows without further increase in stress. Uponrelease of stress (at some point), that part of thetotal deformation is regained. When a body isstressed, and lets the strain increase indenitely,two things may happen. If the material behavior

is plastic, after the yield stress is reached it de-forms endlessly. If the material is elastic andbrittle, it will break at a stress level correspond-ing to its breaking (fracture) stress. A morerealistic representation of an elastic-plastic ma-terial is presented in Fig. 1.3(c). This stress-de-formation diagram shows that at rst the mate-rial deforms elastically. After the materialreaches its yield point, increase in strain contin-ues when the material is subjected to higherstresses. Still, Fig. 1.3(c) represents only a sim-plied version of a real material. A typicalstress-strain curve of most metallics does nothave a break between the elastic and the plasticregions; the transition is rather smooth (Fig.1.3d). The inelastic portion of the curve may bevery long or very short, depending on the duc-tility of the material. It may even become nil ifthe material is very brittle. Ductility and brittle-ness are not just the inherent characteristics of amaterial. Depending on many factors, such asthe ambient environment (including tempera-ture) and the active stress system, a normallyductile material may become (or act) brittle, and


vice versa. These factors are discussed at greatlength throughout this book.

In Chapter 2, a more detailed account is givenon the stress-strain curve, the information ob-tained from a tensile stress-strain curve, andwhat that information means in design analysisand failure prevention. Before presenting thestress-strain relations or getting into the char-acteristic details of the stress-strain curve, it isnecessary to introduce some fundamental elasticconstants that link the stresses and strains. Firstnotice the initial elastic behavior of the tensilestress-strain curve in Fig. 1.3(d). The slope ofthe straight line OQ that relates e and r is calledthe modulus of elasticity in tension, or simplythe Youngs modulus, designated by the symbolE. In the case of compression, E is the modulusof elasticity in compression. Within the elasticlimit, the modulus of elasticity in compressionis the same as in tension. The linear relationshipis known as Hookes law. This law states thatfor a homogeneous and isotropic material, stressis proportional to strain as long as the stress doesnot exceed a limiting value. If the material issubjected to shear deformation, as in torsion, theratio of the shearing stress to the shearing strainwithin the elastic range of stress is called themodulus of rigidity (or modulus of elasticity inshear), designated by the symbol G. The unitsof E and G are the same as stress. The strain andstress are related as follows:

r e E (Eq 1.1c)

for tension or compression, and

s c G (Eq 1.1d)

for torsion, with s and c being the torsionalstress and strain, respectively.

When a piece of structural member (or testbar) is loaded in tension, the axial elongation isaccompanied by a lateral contraction. The ab-solute value of the ratio of the lateral strain tothe axial strain is known as the Poisson ratio,and is designated by the symbol m (or l as it alsoappears in the literature). For a perfectly isotro-pic material, m 0.25. For some metals m canbe as high as 0.33. For many ductile metals, m 0.3. Within the elastic limit, the Poisson ratioin compression is the same as in tension. Phys-ically, the material expands in the lateral direc-tion during compression. The volume change ina material during deformation is given by:

DV (e e e )Vx y z1 2m

(r r r ) V (Eq 1.2a)x y z Ewhere V is the original volume and DV is thevolume change. For many applications/models,constancy of volume is assumed. In other words,the numerical sum of strains in a three-dimen-sional solid is zero. That is:*

e e e 0 (Eq 1.2b)x y z

It has been shown that the elastic constants E,G, and m are not independent but are related by:

EG (Eq 1.3)

2(1 m)

This equation is derived on the basis of the two-dimensional, pure shear condition. The magni-tude of the stress in one direction is equal, andopposite, of stress in the other direction. The twostresses are acted normal to each other.

It is also shown that in the plastic region mincreases with strain. When the plastic strain isvery large, the value of m approaches a limitingvalue of 0.5 for an incompressible material (Ref1.2). Substituting 0.5 for m in Eq 1.2(a), DV 0. Therefore, Eq 1.2(a) is reduced to Eq 1.2(b)in the large plastic region. The intermediate val-ues of m is given by:

cot m m (m m) (Eq 1.4)

cot

Here, is a constant angle, associated with Eand the elastic strain e. The angle is associ-ated with the total strain (the elastic strain e plusthe plastic strain e). The value of the angle starts with and becomes smaller as the strainincreases. When

m m r 0 m r m

Here, m is the elastic Poissons ratio, and m isa limiting value, which is 0.5 (see Fig. 1.3d).

To describe the deformation behavior of a vis-cous substance, we rst need to describe the vis-

*Unless otherwise noted, derivation and proof of all theequations presented in this and subsequent sections areavailable in Ref 1.1 to 1.3, among other textbooks on solidmechanics.


Fig. 1.5 Schematic of viscous deformationFig. 1.4 Viscous ow between two parallel plates. Source:

Ref 1.2

cosity behavior of a Newtonian liquid. Considera viscous substance, such as a heavy lubricatingoil, sandwiched between two closely spaced par-allel plates as shown in Fig. 1.4. One plate ismoved relative to the second one in a directionparallel to the plates. It will be found that theforce per unit area of the contact surface s isproportional to the relative sliding velocity u andinversely proportional to the distance of the twoplates. Thus, a uid exerts an internal resistanceagainst this motion. This shearing stress is ex-pressed by:

s l u/a (Eq 1.5a)

where a is the distance between the two plates.The material constant l is called the coefcientof viscosity. Equation 1.5(a) can be expressed ina more general manner by letting rectangular co-ordinates X in the direction of motion and Y inthe direction perpendicular to the plates:

s l (u/y) (Eq 1.5b)

Now we can plot the tangential force per unitarea s versus u/a, or u/y. The quantity u/a, oru/y, is called rate of shear. A typical plot ofthis behavior is shown in Fig. 1.5.

Now consider a viscous substance in the formof a rod (Fig. 1.2), but hang the rod verticallyand attach a dead weight at its bottom end. Therod will be seen to stretch continuously veryslowly at a constant velocity that is proportionalto the attached weight. Let u be the displacementand v be the velocity in a point x at time t, as-suming that u 0 at t 0. The permanent unitstrain ex will increase proportionally with thetime t with constant ex/t, where the subscriptx denotes the direction of deformation parallelto the axial (length) direction of the rod. There-fore:

v u/t, e u/x, e /t v/xx x(Eq 1.5c)

Figures 1.2 and 1.5 are analogies, as are Eq1.1(d) and 1.4(b). Thus, we can write an equa-tion for the viscous substance in analogy to Eq1.1(c) as:

r f (v/x) f (e /t) f ex x x(Eq 1.5d)

where the proportionality factor f may be calledthe coefcient of viscosity for tension. It is ap-parent that that f is in analogy to E, l to G, and

to ex. Since E is related to G, f is related toexl in the same way. If a value of m 0.5 issubstituted into Eq 1.2, E 3G; likewise, for aviscous substance, f 3l. That is, the coef-cient of viscosity for tension is equal to threetimes the coefcient of viscosity for shear.

As shown in more detail in this chapter, thestress and strain system in a three-dimensionalelastic solid consists six components of stresses(three tension, three shear) and six componentsof strains. Also, because the stresses are thesame in an elastic solid and a viscous substance,we know that the rate of deformation in a vis-cous material consists of six quantities. Thismeans that both materials have the same formof equations. One group of the equations can betransformed into the other group if the quantitiesappearing in the stress-strain relations (ex, ey, ez,cyz, czx, cxy, E, and G) are replaced for quantitiesof strain-rate relations in a viscous material ( x,ey, z, yz, zx, xy, f, and l). An example of suche e c c c

substitution has just been shown using the anal-ogies between Eq 1.1(c) and 1.5(d). Similarly,we can transform Eq 1.2 to suit a viscous ma-terial as:

e e e 0 (Eq 1.5e)x y z

The strength of a crystalline solid, such asmetal or ceramic, is deformation dependent. The


Fig. 1.6 Normal and shear stresses acting on a plane inclinedto the loading direction

strengths of viscoelastic polymers are deforma-tion rate dependent. Crystalline solids may be-come deformation rate sensitive when stressedat high temperature for a period. Time-depen-dent creep is an important phenomenon in met-als and ceramics at high temperatures. The vis-coelastic behavior of many polymers at roomtemperature is well known, and hence its timedependence is a rule rather than an exception.This type of transformation (from strain tostrain-rate behavior) is discussed later in the sec-tions for creep, creep crack growth, and poly-mers.

1.2 Stress and Strain

1.2.1 Normal and Shear Stresses

Consider a cylindrical bar xed at one end andwith an external force P pulling at the other end,as shown in Fig. 1.2(a). If X is the loading di-rection, the stress over the circular cross sectionthat is perpendicular to the axial load P will be:

r P/A (Eq 1.6a)x

where A is cross-sectional area. Now considerthe cross section pq in Fig. 1.6, which is inclinedto the axis by an angle u. As shown in Fig. 1.6,there is a normal stress component rn and a

shear stress component s acting on the pq plane.The normal stress component is:

r P/A (P cos u)/(A/cos u)n2 P cos u/A (Eq 1.6b)

The shear stress component is:

s P sin u cos u/A (Eq 1.6c)

or

s P sin 2u/2A (Eq 1.6d)

Note that this group of equations can also beused in the case of compression. Tensile stressis assumed positive and compression stress neg-ative (see Fig. 1.7a and b). Therefore, we mayonly change the signs for rx and rn to indicatecompression. The sign convention for the shear-ing stress would be that shown in Fig. 1.7(c) and(d). Thus, positive sign for shearing stress istaken when they form a couple in the clockwisedirection and negative sign for the opposite di-rection. Similarly, applying this sign conventionto a biaxial stress case, the shearing stress s inFig. 1.7(e) would be positive whereas s1 wouldbe negative.

In another situation, when an out-of-planetraction force is applied at an arbitrary anglewith respect to a plane, as shown in Fig. 1.8, theapplied force is resolved into two components:one normal to the plane, another acting on theplane. The stress corresponding to the force act-ing normal to the plane is given by:

Pr cos h (Eq 1.6e)

A

The shear stress lying in the plane along the lineOC, directly projected from P, will be:

Fig. 1.7 Sign conventions for normal and shear stresses


Fig. 1.9 Components of stressesFig. 1.8 Resolution of traction force into stress components

Ps sin h (Eq 1.6f )

A

This shear stress can be resolved into two com-ponents, along the X and Y directions, respec-tively.

PX direction: s sin h sin (Eq 1.6g)

A

PY direction: s sin h cos (Eq 1.6h)

A

Thus, in general, a given plane may have onenormal stress and two shear stresses actingon it.

Now consider a cube (Fig. 1.9), its size is in-nitesimally small. For equilibrium calculations,assume the traction is applied through a point atthe centroid of the cube. As shown in Fig. 1.9,one normal stress and two shearing stresses acton each face of the cube. The notations of thesestresses are as depicted: r for the normal stressesand s for the shear stresses. The subscript for rdenotes the direction of the stress. The rst sub-script for s indicates the plane on which theshear stress lies and the axis to which this planeis normal. The second subscript for s denotesthe direction. Two subscripts for the normalstresses follow the same notation system. Forexample, ry or ryy is the stress perpendicular tothe plane normal to the Y-axis, acting in the Y-direction. In the one-subscript system, theplane normal to the Y-axis is implied. Thecomponent syx is the shear stress lying on theplane normal to the Y-axis, acting in the X-direction.

Again, as shown in Fig. 1.9, one normal andtwo shearing stresses act on each face of the

cube. The notations and magnitude of each pairof components (acting on the back-to-backfaces) are equal. However, each pair of thesestresses acts on opposite directions. Thus, thefollowing nine stress components are parallel tothe three coordinate axes:

r s sx xy xzs r syx y yz s s rzx zy z

Their positive directions are shown as solid ar-rows in Fig. 1.9, for the three faces of the ele-ment the external normals of which point in thedirections of the positive X, Y, and Z axes.

From the condition of equilibrium with re-spect to moments of the forces transmittedthrough the six faces of the cube, the six com-ponents of the shearing stresses must satisfy theequalities:

s s , s s , s s (Eq 1.7)xy yx yz zy zx xz

Thus, the state of stress at a point is completelydescribed by six components: three normalstresses and three shear stresses (Ref 1.2, 1.3).

Many engineering problems are two dimen-sional rather than three dimensional. For exam-ple, stresses may occur on the X-Y plane of athin plate when forces are applied at the bound-ary, parallel to the plane of the plate. The stresscomponents rz, sxz, and syz are zero on both platefaces. For now, we can also assume that the Z-directional stresses are zero within the plate.Thus, the state of stress is then specied by rx,ry, and sxy only, and is called the plane stresscondition. However, as will be shown later, the


Fig. 1.11 Shear strainFig. 1.10 Stresses on an oblique plane in plane stress

strain in the Z-direction is a nonzero quantity.Also, the plane stress assumption requires thenonzero stresses to be only functions of x and y.

Consider an element lying on the X-Y planesubjected to stresses rx and ry, and shear stresssxy(syx). Figure 1.10 shows a part of this ele-ment with an oblique plane intersecting the Xand Y axes. The oblique plane can be at anyorientation with respect to the X and Y axes. InFig. 1.10, it is shown at an angle with the Y-axis. The stresses acting on the oblique planewould be r and s, as shown. These stresses aregiven by:

1 1r (r r ) (r r ) cos (2)x y x y2 2

s sin (2) (Eq 1.8a)xy

1s (r r ) sin (2) s cos (2)x y xy2

(Eq 1.8b)

The characteristics of these equations can besummarized as: The maximum and minimum values of nor-

mal stress on the oblique plane through thecenter of the element occur when the shearstress is zero.

The maximum and minimum values of bothnormal stress and shear stress occur at anglesthat are 90 apart.

The maximum shear stress occurs at an anglehalfway between the maximum and mini-mum normal stresses.

The variation of normal stress and shearstress occurs in the form of a sine wave, witha period of 180.

1.2.2 Linear and Shear Strains

The circular bar shown in Fig. 1.2(a) will de-form linearly corresponding to the applied axialload. The average linear strain, which is a di-

mensionless quantity, is the ratio of the changein length to the original length (Eq 1.1a). Thisis regarded as the linear strain. In a two-dimen-sional problem, consider a rectangle on the X-Yplane (Fig. 1.11). When the rectangle is sub-jected to in-plane loading, the shear stresses willdeform the rectangle as indicated by the dottedlines. This shearing action is called the simpleshear. The angular change in a right angle isknown as shear strain. As illustrated in Fig.1.11, the angle at point A, which is originally aright angle, is reduced by a small amount h. Theshear strain c is equal to the displacement a di-vided by the distance between the planes h. Theratio a/h is also the tangent of the angle throughwhich the element has been rotated. That is:

c a/h tan h (Eq 1.9a)

For the small angles usually involved, the tan-gent of the angle and the angle (in radius) areequal. Therefore, shear strains are often ex-pressed as angles of rotation:

c h (Eq 1.9b)

This representation of shear strain is also appli-cable to the three-dimensional problem, becausethe X-Y plane can be considered one face of thecube.

The entire strain system for the three-dimen-sional case is the same as the stress system. Thatis, six components of strain are sufcient to de-ne the strains at a point. The components ofstrains corresponding to the component ofstresses are, similarly: ex, ey, ez, cxy, cyz, and czx.

1.2.3 Stress-Strain Relations

In a one-dimensional problem, and referringto an experimentally established phenomenon


(Fig. 1.2, 1.3), elastic stress and elastic strain arerelated simply by:

e r /E (Eq 1.10a)x xwhere E is the modulus of elasticity in tension.Extension of the element in the X-direction isaccompanied by lateral contractions in the Y andZ directions. Therefore:

e e m e m r /E (Eq 1.10b)y z x xwhere m is Poissons ratio. Equations 1.10(a) and1.10(b) can be used also for simple compression.Within the elastic limit, the modulus of elasticityand Poissons ratio in compression are the sameas in tension. Physically, the material expands inthe lateral direction during compression.

For a three-dimensional problem, that is,when the stress system includes rx, ry, and rz:

1e [r m(r r )]x x y zE

1e [r m(r r )]y y z xE

1e [r m(r r )] (Eq 1.11a)z z x yE

In many practical problems, all displacementscan be considered to be limited to the X-Y plane,so that strains in the Z-direction can be neglectedin the analysis. Examples for such a system area very thick plate, or constraints applied to re-strict the plastic ows in the Z-direction. In anal-ogy to plane stress, plane strain is mathemati-cally dened as ez cxz cyz 0, whichensures that sxz syz 0 and rz m(rx ry). Therefore, similar to plane stress we needconsider only the same three stress components:rx, ry, and sxy. The strain-stress relationships forthe condition of plane strain thus become:

1 me [(1 m)r mr ]x x yE

1 me [(1 m)r mr ] (Eq 1.11b)y y xE

The modulus of rigidity (G) has been denedas the ratio of the shearing stress to the shearingstrain. According to Timoshenko and Goodier(Ref 1.3), the relations between the shearingstrain components and shearing stress compo-nents then are:

c s /G, c s /G, c s /Gxy xy yz yz zx zx(Eq 1.12)

The elongations (Eq 1.11a) and the distortions(Eq 1.12) are independent of each other. Hencethe general case of strain, produced by three nor-mal and three shearing components of stress, canbe obtained by superposition. Addition of thethree equations in Eq 1.11(a) results in:

1 2me e e (r r r )x y z x y zE

(Eq 1.13)

The left-hand term of Eq 1.13 is called the vol-ume strain, designated by the symbol K. UsingEq 1.13 and solving Eq 1.11(a) for rx, ry, andrz, we nd the stress in terms of strain:

mE Er K ex x(1 m)(1 2m) 1 m

mE Er K ey y(1 m)(1 2m) 1 m

mE Er K ez z(1 m)(1 2m) 1 m

(Eq 1.14)

For the case of plane stress (rz 0), this groupof equations reduces to:

Er (e me ) (Eq 1.15a)x x y21 m

Er (e me ) (Eq 1.15b)y y x21 m

1.3 Principal Stresses andPrincipal Strains

In the vast majority of practical applications,a two-or three-dimensional state of stress existsrather than the simple one-dimensional or uni-axial type shown in Fig. 1.12(a). The 2D and 3Dstates of stress are shown in Fig. 1.12(b) and (c).These latter systems, called state of combinedstress, greatly affect the strength and ductility

Fig. 1.12 State of stress


of materials. In general, each plane passingthrough a point of the body will be acted on byboth normal and shearing stresses. The problemis, however, greatly simplied if the planes cut-ting out the element of material are so rotatedthat only normal stresses appear on the faces ofthe element.

This has been done in Fig. 1.12. In otherwords, the stresses shown in Fig. 1.12 coincidewith the X, Y, or Z axis. It is these normalstresses acting on planes on which no shearingstresses exist that are of greatest signicance instress analysis problems and in the relation ofstress to mechanical properties. Those normalstresses acting on planes containing no shearingstresses are called principal stresses. At a pointin a stressed body, there are three principalstresses acting on mutually perpendicularplanes. They are designated by r1, r2, and r3,where the subscripts refer to the directions of thestresses. Customarily, these symbols also areused for labeling the magnitudes of the principalstresses. The subscript 1 indicates the largestprincipal stress, and the subscripts 2 and 3 in-dicate the intermediate and the smallest, respec-tively. The graphical method described below isthe simplest way to derive a set of equations forthe principal stresses in terms of the normalstresses and shearing stresses of the Cartesiancoordinate. This method is best known as theMohrs circle. The next section describes themethod with a two-dimensional plate (biaxialstress) problem, demonstrating how to constructa Mohrs circle to determine r1, r2, and smaxprovided that rx, ry, and sxy are given.

1.3.1 Mohrs Circle for Stresses

Consider rx and ry acting normal to the edgesof a rectangular plane element as shown in Fig.1.13(a). Shearing stresses also act around the

edges of this element so that, by denition, noneof these stresses is a principal stress. To nd themagnitudes and the directions of the principalstresses r1 and r2, construct a Mohrs circle asfollows. Referring to Fig. 1.13(b), select a scalefor stresses and measuring normal componentsalong the horizontal axis and shearing compo-nents along the vertical axis. Then, take point Oas the origin and plot the positive (tensile)stresses to the right, negative (compressive)stresses to the left. The rst step is to set thestresses rx and ry to scale at points E and E1.The second step is to let ED and E1D1 equal thepositive and negative values of sxy. The thirdstep is to connect DD1 and use DD1 as the di-ameter of the Mohrs circle. The intersection ofthis diameter with the X-axis gives the center Cof the circle, so that the circle can be readilyconstructed. The magnitudes of the principalstresses are dened by points A and B becausethe shear stresses are zero at these points. Themaximum shearing stress is equal to the radiusof the circle. Using the Mohrs circle, the for-mulas for calculating r1, r2, and smax can beobtained:

r rx yr OA 1 22r rx y 2 s (Eq 1.16a)xy 2

r rx yr OB 2 22r rx y 2 s (Eq 1.16b)xy 2

r r1 2s max 22r rx y 2 s (Eq 1.16c)xy 2

Fig. 1.13 Two-dimensional Mohrs circle. Note that s in the gure is the same as sxy.


The directions of the principal stresses canalso be obtained from the Mohrs circle. The an-gle DCA, labeled 2u, is the double angle be-tween one of the principal axes with respect tothe axis of reference. The X-axis is usually takenas the reference axis. For the calculation of thenumerical value of the angle u we have, fromFig. 1.13:

DE sxy|tan 2u| (Eq 1.17)(r r )/2x yCEThis equation denes two mutually perpendic-ular directions for which the shear stress is zero(the shear stresses along u and 90 u arezero). These directions are called the principaldirections of stress. The normal stresses cor-responding to these directions are the principalstresses. The sign convention for u is explainedas follows: When we constructed theMohrs cir-cle, we took the s-axis positive in the upwarddirection and considered shearing stresses aspositive when they gave a couple in the clock-wise direction. So, we set sxy upward at point E.However, by denition the lower half-circle ofFig. 1.13(b) represents the stress variation for allvalues of 2u between 0 and 180 (or, simply, 0 u p/2). The upper half-circle gives stressfor 0 u p/2.* Therefore, the angle be-tween the X-axis and r1 in this case (Fig. 1.13a)is negative; that is, u moves clockwise from rx.

In summary, when we know rx, ry, and sxywe can use Eq 1.16 and 1.17, or construct aMohrs circle, to determine the magnitude of theprincipal and the maximum shear stresses, andthe direction of these stresses. On the other hand,if we know the quantities of the latter, we canconstruct a Mohrs circle in a reverse manner todetermine rx, ry, and sxy. The 2D Mohrs circlecan be drawn the same way for the one-dimen-sional stress system. In that case, rx is the sameas r1. The 1D Mohrs circle would then be usedfor determining rn and s on oblique planes re-spective to the loading direction.

1.3.2 Mohrs Circle for Strains

Instead of using computational methods,stresses on the surface of a structural piece canbe determined on the basis of experimentallymeasured strains. The strain gage method is the

*As stated in Ref 1.3, this rule is used only in the construc-tion of Mohrs circle. Otherwise, the rule given in Fig. 1.10still holds.

most popular experimental technique. However,data reduction is not as simple as one mightthink. Rather than detailing the strain gage tech-nique, this section touches on a few key pointsregarding conversion of measured strains to ac-tual strains in a multidirectional strain eld.

A Mohrs circle for strains can be constructedusing the same method for constructing aMohrs circle for stresses. In a two-dimensionalplane problem, the Mohrs circle for strains islargely useful for determining principal strains,or strains along any desired orientation, from ex-perimentally measured strains. Then thesestrains can be converted to stresses through thestress-strain relationships described earlier.

In general, the state of strain is completelydetermined if ex, ey, and cxy can be measured.However, strain gages can make only directreadings of linear strain. Shear strains must bedetermined indirectly. Thus, three independentmeasurements of linear strain in different direc-tions are needed in order to calculate the mag-nitudes and directions of the two principalstresses. Strain gages are usually made up as aset in the form of a rosette. The most commontype consists of three gages positioned at relativedirections. The one with its legs placed 45 apartfrom one another is called the rectangular ro-sette. Figure 1.14(a) shows a general arrange-ment of a three-gage rosette. The three axes ofits legs (A, B, and C) are placed at arbitrary an-gles in relation to a pair of reference axes, OXand OY (90 apart). If corresponding linearstrains ea, eb, and ec are measured in their re-spective directions, one can calculate the linearand shearing strains, ex, ey, and cxy, correspond-ing to the OX and OY axes of reference. Thevalues of ex, ey, and cxy are calculated in termsof the measured strains ea, eb, and ec from thefollowing set of simultaneous equations (Ref1.4):

2 2e e cos h e sin h c sin h cos ha x a y a xy a a2 2e e cos h e sin h c sin h cos hb x b y b xy b b2 2e e cos h e sin h c sin h cos hc x c y c xy c c

(Eq 1.18)

The principal strains would then be calculatedfrom the expressions:

e ex ye 1 22e e 1x y 2 e (Eq 1.19a)xy 2 4


e ex ye 2 22e e 1x y 2 e (Eq 1.19b)xy 2 4

By changing designation subscripts x and y to1 and 2, respectively (Eq 1.15), the magnitudesof the principal stresses r1 and r2 can be cal-culated. The directions can be determined fromthe following relationship:

cxytan(2u) (Eq 1.20)e ex y

A Mohrs circle for strains can be constructedthe same way as that in Fig. 1.13. This time, thelinear strains are plotted as the abscissa and cxy/2 is plotted as the ordinate.

The following demonstrates how to determinee1 and e2 (as well as r1 and r2) using the directlymeasured strains ea, eb, and ec. Because thedemonstration must be case specic, only an ex-ample for the rectangular rosette with threegages is given. In Fig. 1.14(b), the three legs forthe rectangular rosette are OA, OB, and OC, 45apart between OA and OB, and between OB andOC. If the OA axis of the rosette (Fig. 1.14b) istaken as the reference, then OA is consideredcoincident with OX. For the arrangement of thestrain gage axes in the rectangular rosette withha 0, hb 45, and hc 90, Eq 1.18 re-duces to:

e e e e (Eq 1.21a)x a y c

and

e e /2 e /2 c /2 (Eq 1.21b)b a c xy

or

c 2e (e e ) (Eq 1.21c)xy b a c

Subsequently, Eq 1.19(a) and (b) can be ex-pressed as:

e ea ce 1 21 2 2 (e e ) [2e (e e )] a c b a c2

(Eq 1.22a)

e ea ce 2 21 2 2 (e e ) [2e (e e )] a c b a c2

(Eq 1.22b)

Comparing Eq 1.19(a) and (b) to Eq 1.22(a) and(b), and considering the fact that OA is perpen-dicular to OC and OB lies in the middle andforms 45 angles with the other gages, a Mohrscircle is constructed as shown in Fig. 1.15. Thedimensions A, B, C, D, and E are dened as:

e ea cA (Eq 1.23a)2

1 2 2B (e e ) [2e (e e )] a c b a c2(Eq 1.23b)

e ea cC (Eq 1.23c)2

D e (e e )/2 [2e (e e )]/2b a c b a c(Eq 1.23d)

Fig. 1.14 Rosette diagram. (a) Generalized conguration. (b) Rectangular rosette. A, B, and C are strain gage positions; 1 and 2 areprincipal stress directions


Fig. 1.15 Mohrs circle for the rectangular rosette with three observations of strain. Source: Ref 1.4

E e e (Eq 1.23e)a c

Substituting Eq 1.21(a) and (c) into Eq 1.20yields the expression:

2e (e e )b a ctan (2u) (Eq 1.24)e ea c

which yields two values of u. Each value relatesto a principal stress axis. That is, u1 correspondsto r1, and u2 corresponds to r2. The followingrules make up the sign convention for determin-ing u1 and u2: When eb (ea ec)/2, u1 lies between 0

and90, measured positive in the counter-clockwise direction from the positive OAaxis of the strain rosette to the positive O1axis, which corresponds to the direction ofr1.

When eb (ea ec)/2, u1 lies between 0and 90.

When eb (ea ec)/2 and ea ec, ea e1 and u1 0.

When eb (ea ec)/2 and ea ec, ea e2 and u1 90.

u2 90 u1.The proofs of these rules are given in Ref 1.4.

The principal stress can now be determined.Actually, in most cases, one does not need toknow the numerical values of the principalstrains; therefore, a little time and effort can besaved by using the directly measured strains.The expressions for r1, r2, and smax have beengiven by Murray and Stein (Ref 1.4) as:

e e 1a cr E 1 2(1 m) 2(1 m)2 2 (e e ) [2e (e e )] a c b a c

(Eq 1.25a)e e 1a cr E 2 2(1 m) 2(1 m)

2 2 (e e ) [2e (e e )] a c b a c (Eq 1.25b)


Es max 2(1 m)

2 2 (e e ) [2e (e e )] a c b a c(Eq 1.26)

More equations like these, for the other types ofrosette congurations, are available in the liter-ature (Ref 1.4, 1.5).

1.4 Equivalent Stress andEquivalent Strain

In design and analysis, extensive use is madeof test data developed from laboratory couponsunder uniaxial loading condition. Structural/machine components operate under multiaxialstress conditions. It is often helpful to convertdata derived under one state of stress to anotherstate of stress. This can be accomplished by useof the so-called tensile equivalent stresses andstrains. When a structural piece is subjected toa set of combined loads, biaxial or triaxial, itmust have a stress level of which the equivalentstress will take into account the effect of eachstress component. The term equivalent simplymeans the combined stress is equivalent to thesituation of simple tension so that the severityof the stress system can be evaluated by meansof the materials tensile test data. Among themany proposed methods, the von Mises rela-tionship is most appealing from the standpointof continuum calculation and experimental datatreatment, and is almost universally used by en-gineers. The von Mises relationship for equiva-lent stress is represented by (Ref 1.2):

r 1 2 2 2[(r r ) (r r ) (r r ) ]1 2 2 3 3 12

(Eq 1.27a)

The stress components in Eq 1.27(a) are prin-cipal stresses. For the condition of biaxialstresses (e.g., r3 0) Eq 1.27(a) is reduced to:

2 2r r r r r (Eq 1.27b) 1 1 2 2

In simple tension, the value of in these equa-rtions reduces to r1. Alternatively, these equa-tions can also be expressed using the rectangularcoordinate system:

r 1 2 2 2[(r r ) (r r ) (r r )x y y z z x2

2 2 2 6(s s s )]xy yz zx(Eq 1.28a)

and

2 2 2r r r r r 3s (Eq 1.28b) x x y y xy

The equivalent strain is given by:

1e

32 2 2 2[(e e ) (e e ) (e e ) ] 1 2 2 3 3 1

(Eq 1.29a)

and

1e

3

2 2 22 (e e ) (e e ) (e e )x y y z z x 3 2 2 2 (c c c )xy yz zx 2(Eq 1.29b)

When the axial components are absent, as inthe state of pure shear (in-plane loading or tor-sion), the equivalent stress and strain become:

r 3s (Eq 1.30a)

and

e c/ 3 (Eq 1.30b)

1.5 Stress Analysis of MonolithicLoad-Carrying Members

The force system acting at a given cross sec-tion of a structural member can be resolved into: A force normal to the section and acting

through the centroid, producing axial load-ing

A couple lying in a plane normal to the planeof the cross section, producing exure orbending

A force lying in the plane of the cross section(perpendicular to the longitudinal axis of themember), producing cross shear


Fig. 1.16 Axial compression causing buckling

A couple lying in the plane of the cross sec-tion, producing torsion

Because material failure (or fracture) is gov-erned by tensile and shear properties, any typeof loading should be resolved to tensile andshear components.

This book is not a textbook that covers allaspects of structural analysis. For handbook so-lutions, a library of stress formulas is availablein Ref 1.6. This section presents a few simpleyet typical cases. The sections that follow pres-ent some advanced structural analysis concepts.Because stress analysis procedures are case spe-cic, generalized solutions are rare. Therefore,several key steps that lead to the nal solutionwill be shown as required. Otherwise, derivationof all the equations in this section can be foundin Ref 1.2, 1.3, 1.7, and 1.8, as well as othertextbooks that cover strength of materials.

1.5.1 Axial Loading

The distribution of stress across any trans-verse section of an axially loaded member isusually assumed uniform. For a monolithicmember, say a rod, subjected to tension loads atboth ends (Fig. 1.2a), the stress is assumed uni-formly distributed over the cross section. If thematerial is homogeneous and if the rod has aconstant cross section, all the adjacent cross sec-tions remain plane and parallel under load. Thecross-sectional stress is simply the applied loaddivided by the cross-sectional area:

r P/A (Eq 1.31)

If the rod is long and slender, and is subjectedto axial compression instead of tension, the loadmay be sufciently large to cause buckling ofthe member, as shown in Fig. 1.16(a). Afterbuckling occurs, two adjacent plane sectionssuch as AA and BB, although they remain plane,do not remain parallel. These planes actually ex-perience a combination of axial load and bend-ing load as schematically illustrated in Fig.1.16(b). The critical stress (the critical P/A valuein this case) for a rod of length L is given by thewell-known Euler equation:

2p Er K (Eq 1.32)cr 2(L/k)

Here K is a factor accounting for loading con-dition. Columns of differing boundary condi-

tions have different values of K. Typical valuesfor K range from 0.25 to 4, K 0.25 for oneend free and one end xed, K 1 for both endsfree, K 4 for both ends xed. The parameterk is the radius of gyration with respect to theaxis about which buckling will occur, given intextbooks of statics as:

k I/A (Eq 1.33)

where I is moment of inertia given in text-books that cover strength of materials. It is seenthat for a given material the value of the criticalstress depends on the magnitude of the ratio L/k, which is called the slender ratio. Eulers equa-tion for columns is not valid for small values ofslender ratio because of its inherent limitationthat stresses must be below the proportionallimit. A number of empirical and semiempiricalformulas have been developed for a wide rangeof slender ratios. They can be found in textbooksthat cover strength of materials.

1.5.2 Simple Bending

The stress analysis procedures covered in thischapter deal with deformation in the elasticrange only, with one exception: bending in theplastic range, which is included due to its im-portance in structural design/analysis.

Elastic Bending. A beam or a rod can be bentby a couple of forces/moment applied at the endsof the beam, or by lateral forces acting on thebeam (Fig. 1.17a). The axial stresses acting nor-mal to the cross section of the beam resultingfrom the bending moment are schematically


Fig. 1.17 Bending of a beam. (a) A couple of forces appliedat the ends of the beam, or lateral force acting on

the supports at the ends of the beam. (b) Axial stress distributionin a beam subjected to bending moment M

shown in Fig. 1.17(b). The conventional formulafor the bending stress is:

r M y/I (Eq 1.34)x Z

where the subscript x indicates that the stress isin the X-direction, M is the bending moment (tobe dened later), y is the distance from the neu-tral axis of the beam, and Iz is the cross-sectionalmoment of inertia about the Z-axis. Note that thegeometry of the cross section shown in Fig.1.17(b) is rectangular; its neutral axis is in themiddle of the cross section. The resulting axialstresses along the depth direction are tension onone side and compression on the other side aboutthe neutral axis, and are symmetrically balanced.For a nonsymmetric cross section, the neutralaxis would be shifted and the axial stress distri-bution would be unbalanced.

The idealized loading conditions required bythe bending formula are essentially realized bythe center portion of a beam supported at theends and carrying two equal and symmetricallyplaced loads. However, Eq 1.34 is universallyapplicable to any loading condition that pro-duces bending, as long as the resulting stressesdo not exceed the materials proportional limit.

Equation 1.34 is also based on several assump-tions:

Staticsa. The resultant of the external forces is a couple

that lies in, or is perpendicular to, a plane ofsymmetry of the cross section.

b. The beam is in equilibrium.Geometryc. The longitudinal axis of the beam is straight.d. The beam has a constant cross section

throughout its length.e. A plane section that is normal to the longi-

tudinal axis of the beam before the beam isbent remains plane after the beam is bent.

f. The beam bends without twisting.Material Propertiesg. The beam material is homogeneous and iso-

tropic.h. The stresses do not exceed the proportional

limit of the material.The most common types of bending problems

encountered by engineers are the three-pointbend and the four-point bend congurationsshown in Fig. 1.18 and 1.19, respectively. Thesegures simply show that a simple supportedbeam is subject to a single force or a pair offorces acting opposite to the supporting forces.The bending moment M and the shearing forceV acting at a given location of the beam can bedetermined as follows.

Consider, for example, a simply supportedbeam with a single concentrated load P (see Fig.1.18). To calculate the reaction forces R1 and R2at the supports, choose the point of support at

Fig. 1.18 Three-point bending


Fig. 1.19 Four-point bending

the right as the pivot point. All bending mo-ments about that point should be balanced.Therefore:

R l P b (Eq 1.35a)1

Thus

R Pb/l (Eq 1.35b)1

and

R Pa/l (Eq 1.35c)2

Taking a cross section mn to the left of P, it canbe concluded that at such a cross section:

V R Pb/l (Eq 1.36)1

and

M R x Pbx/l (Eq 1.37)1

These equations imply that the shearing forcesremain constant at locations to the left of theapplied force. The bending moment varies inproportion to the distance x. The distributions ofV and M are shown in Fig. 1.18, where V and Mare represented by the straight lines ac and a1c1,respectively. The diagram for the shearing forceis called the shear diagram, and the diagram forthe bending moment is known as the momentdiagram. For a cross section to the right of theapplied load:

V R P Pa/l (Eq 1.38)1

and

PbxM R x P(x a) P(x a)1 l

(Eq 1.39)

with x always being the distance from the leftend of the beam. The shearing force for this por-tion of the beam remains constant and negative.Note that the area on the left side of the shearingforce diagram is balanced by the area on theright side.

Following this procedure, we can easily de-termine the shear and the moment diagrams forthe four-point bend conguration (Fig. 1.19).This procedure can be applied to any other load-ing condition that involves only four variables:number of Ps, the magnitude for each P, the po-

sition of each P, and the span between the sup-ports. Note that the moment diagram in Fig. 1.19is a unique feature of the four-point bendingcondition. It always results in pure bending be-tween the two applied forces. Pure bendingmeans there is a constant bending moment act-ing on that segment of the bar. In this case, whenthe two forces are symmetrically placed with re-spect to the center of the bar, R1 P and R2 P. A constant bending moment exists in betweenthe two forces; its value is (P a). No shearingforce is associated with pure bending.

For bending of a cantilever beam, the samemethod described previously is used to constructthe shearing force and bending moment dia-grams (Fig. 1.20). Measuring x from the left endof the beam and considering the portion to theleft of the load P2(0 x a):

V P and M P x (Eq 1.40)1 1

The minus signs in these expressions follow therule depicted in Fig. 1.21(a) and (b) and 1.22(a)and (b). For the right portion of the beam(a x l):

V P P and M P x P (x a)1 2 1 2(Eq 1.41)

The corresponding diagrams of shearing forceand bending moment are shown in Fig. 1.20(b)and (c).


Fig. 1.20 Bending of a cantilever beam

Fig. 1.22 Sign conventions for shearing forceFig. 1.21 Sign conventions for bending moment

Plastic Bending. The bending formula forsymmetrical bending (Eq 1.34) was derived onthe basis of the assumption that stress and strainare proportional. If the proportional limit of thematerial is exceeded, assumption (h) is no longervalid. However, the relationship between thestress and the moment in a beam can be deter-mined if we assume that the bending stress-strain characteristic of the beam is similar to thestress-strain characteristics in tension (or com-pression). Assumptions (a) through (g) can stillbe considered valid.

When the material is loaded into the nonlinearrange, the relationship between stress and nor-mal strain can be represented by the generalequation:

r f (e) (Eq 1.42)

A possible stress-strain relationship that is pre-sumably determined from a tension and a com-pression test is illustrated in Fig. 1.23. Since ex-perimental evidence shows that a plane sectionbefore bending remains plane after bending,even after plastic action is taking place, the unitstrain on a cross section of a beam is propor-tional to the distance from the neutral axis. Fig-ure 1.24 presents a schematic illustration.Hence,

e /e y/c (Eq 1.43)y 0

where ey is the unit strain at the point a distancey from the neutral axis, and e0 is the unit strainat a reference point located at a distance c fromthe neutral axis. In other words, the strain dis-tribution across the beam is linear, as shown inFig. 1.24(a). This relationship is valid for bothelastic and plastic ranges. The stress distributioncorresponding to the strain distribution of Fig.1.24(a) is shown in Fig. 1.24(b), which is ob-tained by combining Fig. 1.23 and 1.24(a). It isseen that while the strain distribution is alwayslinear, the stress distribution is nonlinear. How-ever, nonlinearity starts from the outermost -bers of the beam, where the bending stress is thehighest at a given load level. While the appliedload level is not high enough to cause yieldingacross the entire cross section of the beam, therestill is a linear region of stress in the middle ofthe beam. This can be called a stage of elastic-plastic bending. The distribution of bendingstresses then become totally nonlinear when thestress-strain relationship is fully plastic.


Fig. 1.25 Plastic bending stress-strain distribution in a beamwith arbitrary cross section and thickness t. The

stresses and strains are actually distributed along the length of thebeam, as in Fig. 1.24.

Fig. 1.24 Distribution of strain (a) and stress (b) in a beamabove the proportional limit

Fig. 1.23 Stress-strain curve r f (e) for bending. Subscriptsc and t indicate compression and tension, respec-

tively.

Fig. 1.26 Lengthwise view of a beam subjected to plasticbending (with a rectangular shape for simplicity).

e is the distance between the neutral axis and the centroid (itsvalue can be zero).

Here are a few key steps for obtaining anequation for the stresses in plastic bending. Be-gin with a beam that has a cross section of anarbitrary shape and thickness t (Fig. 1.25). As-

sume that the radius of curvature of the neutralsurface produced by the moments M is equal toq (Fig. 1.26). In this case, the unit elongation ofa ber a distance y from the neutral surface is:

e g/q (Eq 1.44)

Using gc and gt to denote the distances fromthe upper and lower beam surfaces respectiveto the neutral axis, the elongation in the out-most bers is:

e g /q, e g /q, g g tt t c c t c(Eq 1.45)

where the subscripts t and c stand for tensionand compression, respectively. It is seen that theelongation or contraction of any ber is readilyobtained, provided we know the position of theneutral axis, say gt or gc, and the radius of cur-vature, q . Since assumptions (a) and (b) are stillvalid, these two quantities can be found from thetwo equations of statics:

r dA 0 (Eq 1.46)and

r g dA M (Eq 1.47)From Fig. 1.25, dA b dg, therefore:

r b dg q r b de (Eq 1.48) t t

and

2r b g dg q r b e de M t t

(Eq 1.49)


where the integral over thickness t is integratingfrom gc to gt, or from ec to et whereverapplicable.

Simplifying the problem for this demonstra-tion, the cross section becomes rectangular;thus, b is no longer a variable, but is independentof g. Therefore, the cross section A is equal tob t. Inserting Eq 1.42 into Eq 1.46 and 1.47yields:

q b f (e) de 0 (Eq 1.50a)t

i.e.,

f (e) de 0 (Eq 1.50b)t

and

2q b f (e) e de M (Eq 1.50c)t

From Fig. 1.26,

h l/q (Eq 1.51)

where l is the length of the bar for which theangle h was observed. From Eq 1.45, which re-lates e, g, and q:

1 (e e )/t (Eq 1.52)t cq

and

l(e e )t ch (Eq 1.53)t

Substituting Eq 1.52 into Eq 1.50(c) gives a for-mula for the bending moment as a function ofthe positive and negative strains:

2t bM f (e) e de (Eq 1.54)2 (e e ) tt c

Now, further assume that we have tensile andcompression stress-strain curves for the materialin question. That is, the stress-strain relationshipof Eq 1.42 is presented as those shown in Fig.1.23. Introducing the areas under the respectivestress-strain curves for compression and for ten-sion, and on the basis of Eq 1.50(b), yields:

A A (Eq 1.55)c t

which gives:

r de r de (Eq 1.56)c c t t

After some manipulations, using Eq 1.50(b) to1.56, we nally obtain the Herbert equation (Ref1.2).

1 d r rt c2 2(Mh ) bt (Eq 1.57)h dh r rt c

For a material with the same stress-strain curvefor tension and compression:

r r , e e , de de (Eq 1.58)c t c t c t

Eq 1.57 becomes:

1 d rt2 2(Mh ) bt (Eq 1.59a)h dh 2

or

2 dMr 2M h (Eq 1.59b)t 2 bt dhwhere M is the bending moment applied at bothends of the bar. In the case of four-point bending(Fig. 1.19), M P a. If h or q can be measuredby running an experiment, and a diagram for Mversus h (Fig. 1.27) is constructed, Eq 1.56(b)can be interpreted by a graphical solution as:

2r (2PC PB) (Eq 1.59c)t 2bt

Fig. 1.27 Typical M vs. h diagram in pure bending. Source:Ref 1.2


for any given h. Substituting Eq 1.58 into Eq1.53, we obtain the corresponding tensile strainas:

te h (Eq 1.60)t 2l

where l is the length of the bar for which theangle h was observed, and t is the thickness ofthe beam. The subscript t indicates tension.

Cross Shear. It has been shown that duringsimple bending, a bending stress rx acts normalto the Y-Z plane of a cross section (see Fig. 1.17).In addition to rx (produced by the bending mo-ment), a shearing stress sxy (produced by theshear force V, which is in a way also producedby the bending moment) also acts on the Y-Zplane. This shearing stress is called the crossshear. The cross-shearing force V inducesshearing stresses on the vertical transverse planeon which it acts and, in addition, on horizontalplanes (see Fig. 1.28(a) and (b)). The shear stressthat acts on the horizontal plane is call the lon-gitudinal shear. At any point in the member, themagnitudes of the shear stresses on the two per-pendicular planes are equal, that is, syx sxy.

Let us begin with a simple conguration: arectangular cross section. The equation for thatis:

2V h 2s s y (Eq 1.61)xy yx 1 2I 4zwhere V is the total vertical shear from the sheardiagram, Iz is the moment of inertia about the Z-

axis, b is beam width, h is beam height, and y1is the vertical distance starting from the neutralaxis toward the top or bottom of the beam. Thisequation imposes that the shear stress is maxi-mum at the midheight of the beam and zerostress at both the top and bottom of the beam.

Because each cross-sectional geometry (e.g.,an I-beam, a bar with a circular cross section) isunique, all the geometric variables in Eq 1.61must be reformulated each time. No one singlesolution can cover all situations. To complicatethe matter further, twisting may occur when theprincipal plane of the beam is not a plane ofsymmetryfor example, a channel, an angle, ora Z-section. It would require some effort of anal-ysis to obtain a solution that is equivalently asimple bending without torsion. This type ofanalysis is beyond the scope of this book. Read-ers are referred to textbooks on solid mechanicsor strength of materials.

1.5.3 Simple Torsion

Let a shaft of length L and radius a be sub-jected to pure torsion by torque. The torqueforces MT are applied at the shaft ends, as shownin Fig. 1.29. A straight line MA drawn on thesurface of the shaft and parallel to the longitu-dinal axis when the shaft is unstrained will be-come a helix when the shaft is twisted. Point Aon the circumference will move to point B. Themovement of A relative to M (i.e., the arc AB)will be proportional to L. Now consider a smallelement that consists of points M and N on thesurface of the shaft, as shown in the enlargedview at the left in Fig. 1.29. This element de-

Fig. 1.28 Shearing stresses in a beam


Fig. 1.29 Simple torsion

forms as shown by the dotted lines, and the shearstrain on the outer perimeter is:

c x/L AB/L h a/L (Eq 1.62)a

where h is the angle of twist (in radian), and ais the maximum radius of the shaft ( d/2). Theshear stress is:

s G c, s G h a/L (Eq 1.63)a

The shearing strain and shearing stress at anypoint of the cross section is equal to:

c h r/L, s G h r/L (Eq 1.64)

where G is the modulus of rigidity, and r is theradius at a point on the cross section. If the mag-nitude of the twisting moment is known, theshearing stress can be expressed as:

s M r/J (Eq 1.65)T

where MT is the moment of twist (torque), andJ is the polar moment of inertia of a plane area.

The above equations apply to elastic defor-mation only. Beyond the torsional yield strength,the shear stress over a cross section of the bar isno longer a linear function of the distance fromthe axis. The procedure for obtaining a solutionis very similar to that for plastic bending thatwas demonstrated earlier. For a work-hardeningmaterial, the maximum shear stress at the outerber of the circular shaft (i.e., at r a) is:

1 dMTs 3M ha T3 2pa dh (Eq 1.66a)

where h is the angle of twist per unit length,that is, h h/L, or

1s (3AP CP) (Eq 1.66b)a 32pa

by interpretation of the torque-twist diagram(i.e., Fig. 1.30). The corresponding shear strainat the outer ber of the bar is:

c h a (Eq 1.67)a

1.5.4 Pressurized Tubes

In the case of pressurized tubes, pipes, or ves-sels (Fig. 1.31), the initial elastic hoop stress rh,the radial stress rr, and the stress in the axialdirection rx are as follows:

2 2 2P r (r r )i 0r (Eq 1.68a)h 2 2 2r (r r )0 i2 2 2P r (r r )i 0r (Eq 1.68b)r 2 2 2r (r r )0 i

Fig. 1.30 Torque-twist diagram


Fig. 1.31 Cross-sectional view of a pressurized thick-wallcylinder

2P r ir (Eq 1.68c)x 2 2r r0 i

where P is the internal pressure, r is the radialdistance, and the subscripts i and o stand forinner and outer, respectively. This set of equa-tions is for a thick-wall cylinder; rr is always acompressive stress and rh a tensile stress.

The maximum shearing stress acts at the innersurface of the cylinder:

2r r P rh r 0s

Mechanics and Mechanisms of Fracture

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