CHAPTER ONE INTRODUCTION - To Restore The Dignity … · i To derive a set of differential...

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CHAPTER ONE INTRODUCTION

1.1 PREAMBLE

A thin-walled structure is one which is m1ade from thin plates joined along their

edges. The plate thickness however is small compared to other cross sectional

dimensions which are in turn often small compared with the overall length of the

member or structure. Quantitatively, thin walled structures are structures in which the

ratio of the thickness t, to the two other linear dimensions (length l, and width w)

ranges within the limits ( t/l or t/w = 1/50 to 1/10), (Rekach 1979). Thus a thin walled

structure has two dimensions of the structural element much larger than the third

one, i.e., the thickness. The elements of a thin walled structure are plates and shells,

the later being plates having a curvature in one or two dimensions. Numerous

structural configurations require partial or even complete enclosure that can easily be

accomplished by plates. Thin walled structures have gained special importance and

notably increased application in recent years. The wide use of these thin walled

structures is due to their great carrying capability and reliability and to the economic

advantage they have over solid (column and beam) structures, (Heins 1975). Thin

walled structures which are designed with the nature of work in view, are lighter than

solid structures. Consequently their mechanical theory is of great importance in

engineering practice.

When two or more plates are joined together to form an open or closed structure

strength and rigidity are increased. For example, tanks, boilers, etc, are cylindrical

shell structures with increased strength and rigidity. Conical shell structures are also

common features in construction, mechanical engineering and aeronautical design.

Thin-walled structures are used extensively in steel and concrete bridges, ships, air

crafts, mining head frames and gantry frames. These are seen in the form of box

girders, plate girders, box columns and purlins (z and channel sections).

Generally, a thin walled structure can be of open cross section e.g. channel and

prismatic sections, or a closed cross section such as circular and rectangular

sections. Thin walled structures of closed sections are generally referred to as ‘box

structures’. Thus a girder structure cellular in section can be called box girder

structure. These find their uses in different fields of modern engineering such as box

culverts, and box girders in highway and bridge engineering.

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Thin walled structures of open cross section are subjected only to bending

stresses, no axial load or torsion, Heins (1975). Because of their thin wall

thicknesses, the shearing resistances are constant across the thickness of the plate.

On the other hand thin walled box structures may also be subjected to bending,

torsional and distortional stresses. Distortion alters the geometry of the cross section

and generates some additional stresses thereby reducing the bearing capability of

the box structural component.

Basically, a curved structural element has two interacting forces: bending and

torsion. Knowledge of this interaction leads to a successful design of the element.

The study of curved elements offers only one instance where torsional and bending

forces simultaneously occur. In other instances, e.g., straight girders (in either

bridges or buildings) may develop such forces and thus require appropriate analysis

and design. This situation arises where the load is eccentric with respect to the girder

axis or shear centre. The engineer is then forced to determine the shear centre,

resolve the forces into appropriate bending and torsional forces, Fig.1.1, and then

determine the stresses and deformations.

The torsional response of structural elements can be classified into two

categories: pure torsion and warping torsion or bi-moment. Most structural engineers

are familiar with the concept of pure torsion as this type of torsion is studied in

strength of materials courses. The second type of torsion, warping torsion and

distortion are probably new terms and phenomena which need to be fully

investigated in order to avoid their undesired effects particularly on box girder

bridges.

(a) Beam under eccentric load (b) Load transferred to shear

centre (c) Restoring moment at shear centre

= Sc

P

+ x

y

MZ

• x

y

• • sc c

x

z

P

y

ex xo

y

-

• • sc c c

Fig.1.1.1 Decomposition of eccentrically loaded beam

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The general theories described by Heins (1975) relative to straight and curved

girders assume thin walled sections which may be open or closed (box section).

Development of general equations was based on prismatic girders (solid

beams).Their solution then permits a proper engineering examination of box girder

elements when subjected to vertical torsional loads.

One major topic of interest presently challenging the engineer is box girder

bridges, straight and curved. Initially, design of box girder bridges was related to the

design of plate girder bridges. However, such design knowledge does not contain

important primary conditions of cross sectional deformations. Initially, box girders

could be examined as open sections to allow the application of general plate

theories. Then the section was modified to consider continuity and warping. All of

these investigations assumed no distortion of the cross section. The application of

cross sectional deformation equations formulated by Vlasov (1958) and Dabrowski

(1968), has opened a new way to analyze the torsional and distortional effects of

loads on such girders. This is the main thrust of this work.

1.2 AIMS AND OBJECTIVES OF THE STUDY

A complicated state of forces develops in eccentrically loaded straight girders

when they carry generalized loads. The forces that develop include bending

moments, shear forces, pure torsion, warping torsional moments and bi-moments.

Plane sections under torsion will generally warp as shown in Fig.1.2.1, i.e., will not

remain plane and Bernoulli’s beam theory is therefore violated. If restrained, these

out of plane deformations create additional normal and shear stresses which when

integrated over the cross section yield the bi-moment and warping torsional moment;

the warping stresses. Warping stresses can be as large as or even larger than the

bending stresses, and they can not be ignored. It usually has its maximum value at a

free edge of the cross section and plays a significant role in initiating local buckling

because free edges are the weak parts of a cross section.

The aims and objectives of this study are as follows.

i To derive a set of differential equations governing the behaviour of thin- walled box

girder bridge structures of arbitrary cross section on the basis of Vlasov’s theory.

ii To apply the obtained differential equations in the analysis of single cell and multi-

cell thin walled structures with mono–symmetric and doubly symmetric cross

sections.

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iii To apply generalized strain fields (modes) in computing the coefficients of the

differential equations in order to reduce the number of unknown displacement

functions.

iv To compare the results of analysis obtained using the derived equations with the

results obtained using other methods of analysis.

1.3 SIGNIFICANCE OF THE STUDY.

Structural components such as thin walled box girders are economical and

effective in resisting torsional and bending loads but due to thinness of the box walls,

loads applied to this type of structures give rise to warping and distortion of the cross

section. Consequently the geometry of the cross section is altered and some

additional stresses are generated thereby reducing the structural capability of the box

section.

Fortunately structural designers in steel are careful enough not to ignore the

effects of torsion on a structural member. Unfortunately however, the effects of

warping and distortion on a structural component are either poorly evaluated or

ignored in the analysis simply because of the rigorous mathematics involved in their

evaluation. There is therefore the need to develop a simple analytical model to

enable designers put into consideration the primary condition of cross sectional

deformations in the analysis and design of box girder structures. Thus a better and

more elaborate assessment of all the effects of loads on a thin walled box girder

bridge structure can be achieved by a consideration of the phenomena of warping

torsion and distortion in addition to bending and shear

1.4 SCOPE OF WORK

In this study, cellular box girders, with and without side cantilevers, are examined.

These include (a) single cell doubly symmetric section (b) multi cell doubly symmetric

Fig.1.2.1 Warping of closed thin-walled cross section

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section (c) single cell mono symmetric section and double cell mono symmetric

section.

The girders are assumed to be subjected to purely torsional / distortional loading,

therefore the bending action of the girders are neglected as the structure will undergo

torsional stresses, distortional stresses and distortional displacements only. However,

the lateral displacements of the box girders due to torsional loads are considered.

Since the basic equations of equilibrium form the fundamental principle in the

development of stress/strain relations, dynamic loads which are characteristic of

bridge loads are taken care of by static loads stipulated by appropriate codes of

practice.

Although the requirements for road way alignment may give rise to curved bridge

structure, the structural behaviour of such curved thin walled box girder bridges is

outside the scope of this work. Elastic analysis (not elasto-plastic analysis) is the

mode adopted for the work.

Various methods of analysis of thin walled box girders are outlined in the

literature review but the method of Vlasov, was adopted in this work because it

captures all peculiarities of cross sectional deformations

Finally it is noted that the behaviour of thin-walled structures is usually sensitive to

the nature and magnitude of initial imperfections which arise inevitably during

fabrication. The effects of such initial imperfections fall outside the scope of this work.

1.5 RESEARCH METHODOLOGY

The existence of distortional and warping stresses in a thin walled structure

whose transverse cross section is arbitrary and which contains several closed

contours was first substantiated by V.Z.Vlasov (1949), who also developed a theory

for their analysis. Research has shown that using Vlasov’s original displacement

fields for the analysis of thin walled closed structure leads to a large number of

displacement functions to be determined. Verbernov (1958) has shown that by using

generalize strain fields on Vlasov’s formulation, the number of unknown displacement

functions can be reduced to seven irrespective of the number of independent

displacements possessed by the structure.

In this work, Vlasov’s theory as modified by Verbernov was adopted. The

potential energy of the box girder bridge structure under the action of a distortional

load was obtained using the principle of minimum potential energy. By minimizing the

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energy functional using Euler-Lagrange technique the governing differential

equations of distortional equilibrium were obtained.

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CHAPTER TWO

LITERATURE REVIEW

2.1 REVIEW OF PAST WORK

The curvilinear nature of box girder bridges along with their complex deformation

patterns and stress fields have led designers to adopt approximate and conservative

methods for their analysis and design, Khaled et al (2002). Recent literatures, Hsu et

al (1995); Fan and Helwig (2002); Tetsuya et al (1995); Sennah and Kennedy (2003),

on straight and curved box girder bridges deal with analytical formulations to better

understand the behaviour of these complex structural systems. Few authors, Okil and

El-tawil (2004); Sennah and Kennedy (2003); have undertaken experimental studies

to investigate the accuracy of existing methods. These are discussed in this review of

past work.

Before the advent of Vlasov’s ‘theory of thin-walled beams’ the conventional

method of predicting warping and distortional stresses is by beam on elastic

foundation (BEF) analogy. This analogy ignores the effect of shear deformations and

takes no account of the cross sectional deformations which are likely to occur in a

thin walled box girder structure.

A modification of BEF analogy was developed by Hsu et al (1995) as a practical

approach to the distortional analysis of steel box girders. The equivalent beam on

elastic foundation (BEF) method as it is called is an enhancement of the BEF

method. It is adoptable to the analysis of closed (or quasi-closed) box girders and

provides a simplified procedure to account for deformation of the cross section, the

effect of rigid or flexible interior diaphragms and continuity over the supports

Fan and Helwig (2002) proposed a model of steel box girder bridge where

distortional stresses are controlled by internal cross frames that restrain the cross

section from distorting. They presented an analytical study on the distortional

behaviour of box girders with a trapezoidal cross sectional shape and obtained

distortional components from different torsional loads and used them to derive

expressions for the brace forces in the internal cross frames for quasi closed box

girders.

In a paper titled ‘Strength of thin walled box girders curved in plan’, (Yakubu et al

1995) used numerical methods to study the influence of local buckling and

distortional phenomenon on the ultimate strength of thin walled, welded steel box

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girder curved in plan. They proposed a modified stress strain curve to allow for local

buckling of plate components. In their non-linear flexure analysis, the effect of

distortion was considered through incorporating additional strain in the reference

stage of the incremental process.

Okeil and El-Tawil (2004) carried out detailed investigation on warping –related

stresses on 18 composite steel concrete box girder bridges. The differences between

stresses obtained taking warping into consideration and those calculated by ignoring

warping were used to evaluate the effect of warping. They concluded that warping

has little effect on both shear and normal stresses in all bridges.

Sennah et al (2003) used finite element model to carry out parametric study on

the shear distribution characteristics under dead and live loads in horizontally curved

composite multiple steel box girder bridges and used the results from tests on five

box girder bridge models to verify the finite element model. Based on the results from

parametric study simple empirical formulas for maximum shear (reactions) were

developed for the design office.

Osadebe and Mbajiogu (2006) employed the variational principles of cross

sectional deformation on the assumption of Vlasov’s theory and developed a fourth

order differential equation of distortional equilibrium for thin walled box girder

structures. Their formulation took into considerations shear deformations which were

reflected in the equation of equilibrium by second derivative term. Numerical analysis

of a single cell box girder subjected to distortional loading enabled them to evaluate

values of distortional displacement, distortional warping stresses and distortional

shear which they compared with BEF analogy results and concluded that the effect of

shear deformations can be substantial and should not be disregarded under

distortional loading.

Several investigators; Bazant and El-Nimeiri (1974), Zhang and Lyons (1984),

Boswell and Zhang (1984), Usuki (1986,1987),Waldron (1988), Paavola (1990),

Razaqpur and Lui (1991), Williams et al (1992), Fu and Hsu (1995), Tesar (1996),

have combined thin-walled beam theory of Vlasov and the finite element technique to

develop a thin walled box element for elastic analysis of straight and curved cellular

bridges.

Well over sixty literatures were reviewed. Various theories were postulated by

different authors examining methods of analysis, both classical and numerical. A few

others however carried out tests on prototype models to verify the authenticity of the

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theories. At the end of it all, it was concluded that Vlasov’s theory captures all

peculiarities of cross sectional deformation such as warping torsion distortion etc, and

that generalized strain fields can be used to reduce the number of differential

equations which poses difficulties in the application of Vlasov’s theory.

2.2 BOX GIRDER BRIDGE TYPES

The cross section of box girder bridges may take the form of single cell, multi cell

(cellular) or multi spine (multi box) profiles shown in Fig.2.2.1. Multi cell box girder

type provides greater torsional stiffness than multi spine due to the high efficiency of

the contiguous cells in resisting eccentric loadings.

Goodall (1971), Grant (1993) and Dean (1994), derived solutions to the torsional

stiffness and shear flow of regular multi cellular sections that are frequently used in

modern curved high way bridges. Fu and Yang (1996) extended the solution to the

torsional design of multi- box and multi-cellular reinforced concrete bridges. A

summary of research work done on load distribution and design of various box girder

cross sections is presented in the following sections.

2.2.1 Single cell box girder bridges

Trapezoidal steel box girder bridges are often used in curved bridges due to the

large torsional stiffness that result from the closed cross section. However the use of

single cell box girders in bridge construction is a common feature in curved bridges,

highway interchanges and ramps.

Fan and Helwig (2002), carried out an analytical study on the distortional

behaviour of single cell box girder bridge with a trapezoidal cross sectional shape.

Typical torsional loads on curved box girders were presented and distortional

components of the applied torsional loads were identified and used to derive

expressions for the brace forces in the internal cross frames for quasi-closed box

girders. Turkstra and Fan (1978), investigated the effect of warping on the

longitudinal normal stresses as well as on the transverse normal stresses in single

cell curved bridges.

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2.2.2 Multi-cell box girder bridges

In 1959, California design engineers who appreciated the large torsional rigidity of

closed cellular sections suggested to the American Association of State Highway and

Transportation Officials (AASHTO), a change in the load distribution factor (s / 5) then

used for bending moment in straight reinforced concrete box girder bridges, (Davis et

all (982). They recommended (s / 8) for one lane traffic and (s / 7) for two or more

traffic lanes, where s represents the cell width in feet. The National Standards of

Canada for design of highway bridges (CSA 1988) adopted the above specified

moment distribution factors for straight multicell bridge cross section even though

they do not give designers much information on the behaviour of the bridge or the

parameters influencing its response.

Trikha(1972) developed a set of design coefficients for two lane twin cell curved

girder reinforced concrete bridges as an aid to practical design of such bridges. The

effect of intermediate diaphragms was not considered in his study which he used

finite strip method to accomplish. Ho et all (1989) also used the finite strip method to

analyze straight simply supported, two-cell box girder and rectangular voided slab

bridges without intermediate diaphragms. Empirical expressions and design curves

were deduced for the ratio of the maximum longitudinal bending moment to the

equivalent beam moment.

2.2.3 Multiple spine box-girder bridges

Johanston and Mattock (1967) and Fountain and Mattock (1968) used a computer

program for the analysis of folded plate structures to study the lateral distribution of

loads in simple span composite multi-spine box girder bridges without intermediate

diaphragms. To verify their analysis and the authenticity of the computer program

they built and tested two-box and three-box bridge models under AASHTO truck

loading. The results were used to develop an expression for the live load bending

moment distribution factor for each box girder as a function of roadway width and

number of boxes. Their findings formed the bases for lateral distribution of loads for

(a) Single Cell (b) Multiple Spine (Multi-box) (c) Multi-cell (Cellular)

Fig.2.2.1 Box girder cross section types

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bending moment evaluation currently used by AASHTO (1998), OMTC (1983) and

CSA (1988), for the multi-spine box-girder bridges. However, the principle does not

account for the beneficial effect of cross bracing inside and between the boxes. More

over its application is limited to bridges with a number of spines equal to the number

of lanes. Therefore, AASHTO (1998) provides another expression for load distribution

for the ultimate limit state to obtain the live load-bending moment and shear force in

each box of the cross section.

Tung and Fountain (1970) presented an approximate method to determine the

torsional moments and angle of twist in curved box girders of single and multiple

spans. The box girders were assumed to be adequately stiffened by internal

transverse diaphragms so that both warping and distortional stresses were

considered negligible. Heins (1978) developed two expressions for the live load

longitudinal bending moment and torsional moment in a straight box girder, with two

modification factors for curvature. The expressions were developed from data provide

by Ho (1972) resulting from a thorough system analysis of 90 curved multi-spine box-

girder bridges. The work also included approximate formulas, for the box section

geometry, calculation of internal forces, and top chord and cross bracing

requirements.

Bakht and Jaeger (1985,1992) presented a method for analysis of a particular

case of multi-spine bridges having at least three spines, zero transverse bending

stiffness, with the load transfer between the various spines through transverse shear.

Based on these simplifications they proposed load distribution factor for bending

moment and shear. These formed the basis for live load distribution [OMTC 1972] for

multi-spine bridges. Salaheldin and Schmidt (1991) reported a detailed study, using

the finite element method, of the flange transverse membrane stresses of a simply

supported box girder reinforced with a rigid end diaphragm under symmetric

concentrated loads acting at the mid span. Empirical expressions were formulated to

predict the distribution and magnitude of the transverse flange stresses. Normandin

and Massicotte (1994) presented the result of a refined finite element analysis to

determine the distribution of live load patterns in multi-spine box girder bridges with

different characteristics and geometry. Results indicated that in some cases both the

AASHTO (1996) and OMTC (1992) distribution factors underestimate the live load

effects by a significant margin. Two empirical equations for longitudinal bending

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moment and shear in each box were developed for simply supported bridges when

the width of the steel section is of the same order of magnitude as the lane with.

Using the finite strip method, Cheung and Foo (1995) proposed expressions for

the ratio of the moment distribution factor of the curved composite multiple box-girder

bridge to that of the straight one. The disadvantage of this study is that it did not take

into account the beneficial effect of the diaphragms inside the box or cross bracings

between boxes. Furthermore, the effect of the number of boxes and dead load

distribution were not included. Foinquinos et all (1977), using the finite element

method assessed the influence of intermediate diaphragms on the live load

distribution of a straight three-spine box girder bridge, within the context of AASHTO

(1998).

2.3 BRIDGE LOADINGS AND RESULTANT STRESSES

Dead loads on a bridge structure are estimated based on the dimensions and

materials given in blue prints provided by the appropriate codes of practice. In USA,

live loads are considered according to AASHTO LRFD (1998) provisions. Following

the HL93-loading a uniform lane load of 9.3N/mm distributed over a 3.0 m width, is

considered in addition to a tandem load of two 110kN axles. Transversely, the loads

are positioned at the outermost possible location to generate the maximum torsional

effects, Fig. 2.3.1. Centrifugal forces (CE) are taken into account for curved segment

of the bridge. The tandem load is considered to act along the longitudinal direction of

the bridge axis, while the uniform lane load is positioned on each span separately.

(b) Tandem Load (one axle

shown)

600 b 1800

110KN

CE

1800

(a) Lane Load

3000

9.3 N/mm

Fig.2.3.1 Positioning of live loads for single-lane bridge

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Analysis of these load cases is used to generate envelopes for the live load force

distributions. Each set includes the bending moments (Mx, My) and bi-moment (Mw)

which causes warping, the shear forces (Vx and Vy). The torsional moment (T) acting

on the cross section has two components, pure torsion (Ts) and warping torsion (Tw).

Normal stresses (including the effect of warping) are calculated as follows, (Nakai

and Yoo (1988).

y ω ωxexact approx

x y ω ω

M M MMσ = y + x + ω =σ + ω

I I I I (2.3.1)

where,

Ix and Iy = the moments of inertia about x and y axes,

x and y = distances from the centroid of the cross section,

ωM = the bi-moment

ωI = the warping constant and

ω = warping function.

The sum of the first two terms in Eqn. 2.3.1 is from classical beam theory, which does

not account for warping and is hereby referred to as approxσ .

t approx , is the approximate shear calculated from classical beam theory by ignoring

the effects of warping, using the following equation, ( Okeil and El-Tawil 2004).

τ y xapprox x y

x y o

V V T= S (s)+ S (s)+

tI tI 2tA (2.3.2)

where

Vy and Vx = shear forces,

T = torsional moment acting on the cross section.

The other terms in the expression represent the geometric properties of the closed

cross section namely; box thickness (t), moments of area [ Sx and Sy], and enclosed

cross sectional area (Ao). The third term assumes that the entire torsional moment is

Saint Venant torsion.

If warping is considered the torsional moment s w

T τ τ= +

Where,

sτ = pure torsion

wτ = warping torsion

The shear stresses can be calculated as follows;

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τ ττ y s ωxexact x y ω

x y o ω

V V= S (s)+ S (s)+ + S (s)

tI tI 2tA tI (2.3.3)

wS = sectorial moment of area.

2.3.1 Bi-moment Effects

A bi-moment consists of two equal and opposite moments (couples) acting about the

same axis and separated from one another. Its value is the product of the moment

and the distance of separation.

The longitudinal stress at a point whose coordinates are (x, y) is given by;

y ωx

y x ω

M MMPσ = + y + x + ω

A I I I (2.4.1)

In Fig.2.3.2 the single load P can be treated as a combination of four sets of loads,

one of which represents axial loading, two of which represent bending about the axis

of symmetry of the cross section, and the last is a bi-moment. The first three sets of

loads will result in the deformation pattern familiar to engineers.

The bi-moment will result in distortions of the cross section parallel to the longitudinal

axis of the column and in a twisting of the column about its longitudinal axis [Fig.2.3.3

(b)]. In Fig 2.3.3(c) the web and flanges are separated from one another and a pair of

moments Pa/4, applied at the top of each flange. The flanges each bend as shown,

while the web remains straight. The integrity of the cross section can be restored by

twisting the web and each of the flanges through an angle q and since these

individual plates are very flexible the forces required to do this are likely to be small.

This pure twisting of each plate results in another set of stresses which are shear

stresses. They are called Saint Venant stressesst

τ , and the total torque required to

twist all of the plates in this way is the Saint Venant torque Mst. Thus it is seen that a

bi-moment can result in the twisting of a column and longitudinal stresses in the

flanges, and for the design engineer this can be very important consideration.

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Distortion of box girders occurs when the torsional loads are not distributed to the

plates of the cross section in proportion to a uniform Saint Venant shear flow (Fan

and Helwig 2002)

2.3.2 Pure tortional and distortional loads

Depending on the source of torsion, external torsional load on a box girder can be

modeled by either two opposing vertical or two opposing horizontal forces, as shown

(a) (c)

(d) (d)

(b)

Fig.2.3.2 Components of single load P resulting in warping of the profile.

(Saint Venants principle is not valid and Bernoulli’s hypothesis is not applicable for part (e))

(a ) I-column carrying (b) Distortional component of (c) Bimoment effects:

eccentric load P the eccentric load P (distortion and twisting)

Fig.2.3.3 A bi-moment applied to a column results in twisting and longitudinal stress

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in Fig 2.3.4 (a and b) respectively. Fig. 2.3.5 shows that torsional loads consisting of

opposing vertical forces usually result from gravity loads that are eccentric to the

centre line of the girder. The eccentric load can be modeled as a bending

component, Fig 2.3.5(b), superimposed on the torsional component, Fig 2.3.5(c).

The distortional behaviour of a box girder depends on the manner in which the

external torque is applied to the girder. A torsional load either a vertical couple or a

horizontal couple, can be modeled as a uniform torsional component superimposed

on a distortional component as shown in Figs. 2.3.6 and 2.3.7. The rectangular thin-

(a) Two opposing vertical force model

(c) Torsional Component

mT/2h mT/2h Tm

b

Tm

b

+ h mT/2b mT/2b

=

mT/2b

b mT/2b

mT/2h mT/2h (c) Distortional component (b) Pure torsional

component

(a) Torsional load

Fig.2.3.6 Torsional and distortional loads on rectangular box girder

due to vertical forces

b

(b) Bending Component

q

=

q

+

h

(a) Eccentric load

q

Concrete Slab

q

b

(b) Two opposing horizontal force model

e

Fig. 2.3.4 External torsional loads on box girder

w Girder

central

line

Fig.2.3.5 Decomposition of eccentric concrete load on box girder

2

w

wew

b ew

b

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walled box has a respective dept and width of h and b. Each couple can be modeled

by the uniform torsional components given in Fig.2.3.6(b) or Fig.2.3.7(b). Although

the boxes in Figs.2.3.6(a) or 2.3.7(a) are subjected to the same magnitude of torsion

(MT), the resulting distortional stresses are opposite in direction since the distortional

loads shown in Figs 2.3.6(c) and 2.3.7(c) are opposite. The pure torsional

component, Figs.2.3.6(b) and 2.3.7(b) generate a uniform Saint Venant shear flow

along the circumference of the box girder cross section and warping stresses due to

this torsional component are usually negligible. However, significant distortional

warping stresses may be induced due to the distortional components if the box is not

properly braced.

Vlasov (1961) was the first to study distortion of box girders while investigating the

torsional behaviour of thin-walled beams with closed cross section. Dabrowski (1968)

established a more rigorous theory when he developed the governing equations for

box girder distortion and provided solutions for several simple cases.

A fundamental issue for the distortional analysis of a box girder is the

determination of the distortional loads which are generally not explicitly applied to box

girders, but rather are included in the applied torsional loads as shown in Figs 2.3.6

and 2.3.7 for rectangular boxes. The pure torsional components of the applied

torsional loads [Figs 2.3.6(b) and 2.3.7(b)] do not cause the cross section to distort. A

distortional load consists of four force components on the plates of the box [Figs

2.3.6(c) and 2.3.7(c)]. These four components cause the cross section to distort,

however they do not induce torsion on the girder.

The distortional behaviour of box girders can be studied by separating the

distortional component from the applied torsional loads. The distortional components

are generally resisted by both in-plane and out of plane shears in the girder plates.

Fig. 2.3.7 Torsional and distortional loads on rectangular box girder due to horizontal forces

=

h

(a) Torsional

load

(b) Pure torsional

component (c) Distortional

component

mT/h

mT/h

mT/2h

mT/2h

mT/2b mT/2b mT/2b mT/2b

mT/2h

mT/2h

b

+

18

These two shear components result in different distortional stresses. Fig.2.3.8(a)

shows the typical distortional shape of the box girder that result in out-of-plane

bending of the plate components. Fig.2.3.8(b) shows the shears that develop in the

trough thickness direction as a result of the distortion. The distortional loads on the

flanges and webs are partially resisted by these through thickness shears that

develop in the plates. Out of plane bending stresses are induced with corresponding

moments shown in Fig.2.3.8(c).

Distortional loads are also partially resisted by the in-plane shears that develop on

the cross section of the individual plates, Fig.2.3.9. The large arrows represent the in-

plane shears that resist the distortional loads represented by the small arrows on the

girder plates. The individual plates will experience in-plane bending from these

shears, and longitudinal bending stresses may be induced on the cross section. The

longitudinal bending stresses are known as the distortional warping stresses.

Fig.2.3.9(b) shows a typical distribution of warping stresses in a trapezoidal box

girder.

(a) Typical distortional shape of box gird (b) Shears in trough thickness direction as a result of distortion (c) Corresponding moments due to out of plane bending

Fig.2.3.8 Out of plane distortional stresses in box girders

Fig. 2.3.9 In-plane distortional warping stresses in box girder

(a) In-Plane shears (b) Longitudinal bending stress

19

A pure torsional load consists of a shear component proportional to the Saint Venant

shear flow which intensity is constant along the perimeter of the box cross section,

Fan and Helwig (2000). Equilibrium and mechanics of thin-walled members can be

used to determine the intensity of the uniform shear flow f, as shown in Fig 2.3.10. A

torsional moment of T = f (2Ao) results from this uniform shear flow, where Ao is the

enclosed area of the box. Therefore, for an applied torsional moment T, the

equivalent pure torsional component is represented by uniform shear flow shown in

Fig 2.3.10, in which f = T/ 2Ao. If this distributed load along the perimeter of the cross

section is replaced with the shear loads on individual plates, the shear load on the ith

plate of the box is equal to biT / (2Ao), where bi is the width of the plate independent

of plate overhangs.

Figs 2.3.11 and 2.3.12 demonstrate that the applied torsional load is a function of the

pure torsional components and distortional components. Therefore if the equivalent

pure torsional components are known [Fig 2.3.11(b) or Fig 2.3.12(b)], they can be

subtracted from the applied torque [Fig.2.3.11 (a) or Fig.2.3.12(a)] to determine the

distortional components [Fig.2.3.11(c) or Fig. 2.3.12(c). For example, if the applied

torque consists of two opposing vertical forces P (kN/m), as shown in Fig.2.3.11 (a),

the resulting distributed torsional load is Pb (kNm/m). For a trapezoidal cross-section,

the vertical forces may be resolved into components in the web and in the flanges,

Fig.2.3.11(b). Thus,

2 tant

P P θ= (a)

/ cosw

P P θ= (b) (2.5.1)

0b

P = (c)

in which the subscript t, w, and b refer to the top plate, web plates and the bottom

plate respectively.

ƒ

Fig .2.3.10 Saint Venant shear flow in box girder

20

The intensity of the ‘shear flow’ of the equivalent pure torsional load is P x b / (2Ao).

Therefore, utilizing the expression, qi = bi T / 2Ao for the shear flow in the ith plate,

the component of the pure torsional load on the top plate is, from Fig. 2.3.11(b),

2

t,pureT

o o

Pb bq = b = P

2A 2A (2.5.2)

The distortional component on the girder plates can be derived by subtracting the

pure torsional components from the applied torsional load. For example, the

distortional component on the top flange of the box is given by;

2 2

t,dist t t,pureT

o o

Pb aq =P - q = 2Ptanφ - = - P

2A 2A (2.5.3)

in which the relation, 2Ao = h(a+b) and φtan = (b-a) / 2h , have been applied. Other

torsional and distortional components on the web and bottom plates of the box can

be similarly derived. The results are given in Table 2.1 where the expressions

represent the magnitude of the force components. The directions of pure torsional

and distortional components are shown in Fig. 2.3.11.

b

qt,dist

p

qw,dist

h

qw,Pure T

Pt

=

Pw

qt,pure T

=

p

p

h

a

p

θ

qw,Pure T

c

qb,Pure T

Pw

qw,dist

=

qb,dist

+

,t pureTq

Fig. 2.3.12 Pure torsional and distortional components in horizontal couple

,b pureTq

(a)

,wpureT

q

(b)

,wpureT

q

(c)

,t distq

,w distq

,w distq

,b distq

(a) (b) (c) (d)

Fig. 2.3.11 Pure torsional and distortional components in a vertical couple

+

21

Table 2.1 Pure torsional and distortional components of applied torsional

loads (Fan and Helwig 2002)

If the torsional load consists of two opposing horizontal forces acting on the top and

bottom plates of the box girder [Fig. 2.3.12(a)], the applied torque is Ph. The applied

torsional load consist of Pt = Pb= P and Pw = 0. The pure torsional and distortional

load components shown in Fig.2.3.12 (b and c) can be derived in a similar way

outlined above. The results are also presented in Table 2.1. The directions of the

force components are indicated in Fig. 2.3.12(b and c).

2.4 METHODS OF ANALYSIS OF THIN-WALLED BOX GIRDERS

During the time when ignorance superseded knowledge of the behaviour of box

girder beams, elementary beam theory (Bernoulli beam theory) was applied. The

usual assumption of beam bending theory is that plane sections remain plane over

the entire cross section of say a box girder, after bending. However it is easily

demonstrated that when certain thin-walled structures are twisted there is a so-called

warping of the cross section and Bernoulli hypothesis is violated. In other words,

warping brings about total violation of Bernoulli’s theory. Thinness of wall also leads

to distortion of box cross-section thereby also violating Bernoulli’s beam theory. All

these suggest that application of ordinary beam theory is not suitable for analysis of

Type / Magnitude

of load

Components in plate

members

Pure torsional

component

Distortional

component

Torsional moment

by vertical forces

(Pb)

Top plate, Pt =2Ptan θ

Bottom plate, Pb = 0

Web plate, Pw=P /cos θ

Pb2 / 2Ao

Pab / 2Ao

Pbc / 2Ao

Pa2 / Ao

Pab / 2Ao

Pac / 2Ao

Torsional moment

by horizontal

forces (Ph)

Top plate, Pt = P

Bottom plate, Pb = P

Web plate, Pw = 0

Pb / (a+b)

Pa / (a+ b)

Pc / (a+b)

Pa / (a+b)

Pb / (a+b)

Pc / (a+b)

22

box girder structures. This necessitated for a search for more accurate method.

Vlasov’s ‘theory of thin-walled beams’ (1951), translated into English language by

American National Science Foundation (USA NSF) and called ‘Thin-walled elastic

foundation (1961)’ was most accepted and adopted.

Some codes of practices, AASHTO (1996), CHBDC (2000), have recommended

several methods of analysis for only straight box girder bridges. These methods

include; (i) orthotropic plate theory, (ii) finite difference technique, (iii) grillage

analogy, (iv) folded plate, (v) finite strip and (vi) finite element techniques. Several

authors have applied these methods along with the thin-walled beam theory to

analyze straight and curved box girder bridges. These different approaches to box

girder bridge idealizations are further discussed in the following sections.

2.4.1 Orthotropic plate theory

In this method, the stiffness of the flanges and webs are lumped into an

orthotropic plate of equivalent stiffness and the stiffness of diaphragm is distributed

over the girder length. This method is suggested mainly for multi spine straight and

curved bridges. Bakht et al (1981) presented the various methods of calculating the

equivalent plate parameters which are necessary for analysis of straight cellular and

voided slab bridges. Cheung et al (1982) used the orthotropic plate method to

calculate the longitudinal moments and transverse shear in multi-spine box girder

bridges. The results were compared to those obtained from 3D analysis using the

finite-strip method to establish the limits of validity of the orthotropic plate method. It

was concluded that the orthotropic plate method gives accurate results provided that

the number of spines is not less than three.

2.4.2 Grillage analogy method

Canadian Highway Bridge Design Code (CHBDC 2000) limits the application of

grillage analogy method to voided slab and box-girder bridges in which the number of

cells or boxes is greater than two. In this method, the multi cellular super structure

was idealized as a grid assembly by Hambly and Pennels (1975). Similar idealization

was applied to curved multi-spine box girder bridges by Kissane and Beal (1975).

One difficulty in the grillage-analogy method lies in the representation of torsional

stiffness of closed cells. Satisfactory but approximate representation can be achieved

in modeling the torsional stiffness of a single closed cell by an equivalent I-beam

torsional stiffness,(Evans and Shanmugam 1984).

23

2.4.3 Folded plate method

The folded plate method utilizes the plane stress elasticity theory and the classical

two-way plate bending theory to determine the membrane stresses and slab moment

in each folded plate membrane The folded plate system consists of an assemblage of

longitudinal annular plate elements interconnected at joints along their longitudinal

edges and simply supported at the ends. No intermediate diaphragms are assumed.

Solution of simply supported straight and curved girders is obtained for any arbitrary

longitudinal load function by using direct stiffness harmonic method. The method has

been applied to cellular structures by Meyer and Scordelis (1977), Al-Rifaie and

Evans (1979) and Evans (1984). It was evident that the method was complicated and

time consuming.

2.4.4 Finite strip method

The finite strip method may be regarded as a special form of the displacement

formulation of the finite-element method. In principle, it employs the minimum total

potential energy theorem to develop the relationship between unknown nodal

displacement parameters and the applied load. In this method, the box girders and

plates are discretized into annular finite strips running from one end support to the

other and connected transversely along their edges by longitudinal nodal lines. The

displacement functions of the finite strips are assumed as a combination of

harmonics varying longitudinally and polynomials varying in the transverse direction.

2.4.5 Beam on elastic foundation (BEF) analogy

The response of a box cell to a loading that causes deformation of the cross

section can be represented by a differential equation identical in form to that for

beams on elastic foundations.

A direct analogy exists between the physical properties of the box cell and the beam

on elastic foundation. To develop an approximate theory of deformation of the cross

section one basic assumption regarding this in-plane motion is borrowed from the

theory of torsion of thin-walled beams of open section. It is assumed that the

distortions are accompanied by sufficient warping to annul the average shear strains

in the plates which form the cross section. The measure of distortion ω leads to

warping displacements proportional to ω' . Warping stresses depend on ω'' and

shear on ω''' . The forces per unit length which have their origin in warping and resist

24

deformation are then proportional to ω''' . These considerations suggest a governing

differential equation for ω in the form:

iv

ωωEI ω +kω =P (2.6.1)

in which

ωωI = warping constant

K = measure of deformation stiffness of a unit length of the box, and

P = the applied generalized distortional load per unit length

For distributed load P and beam deflection ω , the differential equation for the BEF

is

iv 4ω + 4β ω =P (2.6.2)

In which 4

b

Kβ =

4EI (2.6.3)

2.4.6 Finite element method

The finite element method which has been used extensively in structural

mechanics has also been used in the analysis of box girder bridge decks. Many

computer programs have been developed using the finite element procedure.

Modern computerized systems such as NASTRAN and BERSAFE for buckling and

stress analysis incorporating different types of finite elements have been developed

for elastic and in-elastic structures. Several finite element programs have also been

developed specifically for the analysis of box-girder bridges by Scordelis (1971).

There are some simplified approaches for box analysis such as those developed by

Sawko and Cope (1969) and Crisfield (1971) which retain the finite element method

of solution with reduced computer time.

2.5 SUMMARY OF LITERATURE SURVEY

Based on the numerous literatures consulted during the literature survey the

following observations and comments can be made.

1. Research work done on thin-walled box girder structures covers essentially three

types of cross section. (a) Single cell steel box girder structure, straight and curved.

(b) Twin cell reinforced concrete box girder, straight and curved. (c) Two-box and

three-box multiple spine box girder, steel and reinforced concrete

2. Literature on cellular steel box girders with three or more cells appears to be

scarce. Those available for two-celled box girder are in RC area. Thus, there appears

25

to be a dirt of information on the torsional-distortional behaviour of thin-walled box

girder bridge structure with three or more cells in cross section.

3. Specific areas covered by the available literature include:

(a) torsional stiffness and shear flow

(b) distortional design of multi-box and multi-cellular sections.

(c) distortional behaviour and brace forces

(d) effect of warping on longitudinal and transverse normal stresses

(e) design coefficients for loads and aids to practical design

(f) analysis of deformable sections / design curves

(g) load distribution

(h) approximate method to determine torsional moments- non deformable

sections

(i) bending moment expressions, shear and torsional moments for live loads

(j) bracing requirements

4. Available literature on torsional-distortional elastic behaviour of thin-walled

deformable box girder structure is surprisingly few. This could be attributed to the fact

that most authors assume that the use of intermediate stiffeners and diaphragms on

thin-walled closed and quasi-closed structures is an effective way of handling non-

uniform torsion and its attendant problems of warping and distortion.

5. It is believed that the use of triangular cells on trapezoidal cross section will

improve the torsional-distortional qualities of trapezoidal box girders and at the same

time make the use of diaphragms and intermediate stiffeners irrelevant. This is a

novel area for research.

2.6 THEORY OF THIN-WALLED BOX GIRDERS

2.6.1 Warping stresses and shear centre When a thin-walled girder is restrained from warping additional stresses arise in

the longitudinal and transverse directions. These stresses do not arise in the case of

uniform (Saint Venant ) torsion. Batch (1909) first noted these stresses by

considering a channel for which the centre of gravity S and the shear centre M do not

coincide. He observed that a transverse load P acting at the end of a cantilever

through S produced vertical deflection and rotation of the cross section about a

longitudinal axis. He also noticed that plane sections do not remain plane and that

warping out of the plane had occurred. By systematically changing the position of P

26

he found the position M (the shear center) and noted that when P acted through M,

Benoulli’s hypothesis was satisfied and the additional stresses also disappeared

Thus, in a thin-walled beam which is twisted only, there can be two sets of shear

stresses (and shear strains), that is, the Saint Venant shear stresses τ s and the

warping shear stress τω

.These are quite different from shear stresses associated

with bending.

There are no longitudinal stresses associated with Saint Venant (uniform ) torsion,

but there are with warping (non-uniform) torsion and they, together with the

associated warping shear stressesτω

, are called the warping stresses.

Every cross section has a shear centre M which is the point through which the

shear force must be applied if there is to be no twisting. When a cross section has

two planes of symmetry (doubly symmetric), the shear centre M and the centre of

gravity S coincide. If there is one plane of symmetry ( e.g. as there is for a channel)

M and S are located on the same plane and it is a relatively simple matter to

establish their coordinates. However when the cross section is less regular a

systematic method for locating M is required. The analysis in section 2.6.2 treats

general case of both open and closed profiles (mono symmetric and doubly

symmetric sections).

2.6.2 Saint Venant torsion of thin-walled cross section

The purpose of this section is to present back up theory for the analysis of beams

of various cross sections, i.e., open profiles, closed profiles and mixed profiles in

which part is open and the other part is closed.

Open Profile

The simplest example of open profile is a solid shaft of circular cross section.

Consider the case of a cylindrical shaft of radius R, length L, and applied torque M.

From the study of strength of materials the following are the expressions for the

maximum shear stress τmax , the maximum shear strain γ max ( i.e. at the outer

surface), and the angle of rotation φ of one end relative to the other end

τmax

p

MR=

J (2.6.1)

27

maxmax

G

τγ = (2.6.2)

p

MLφ =

GJ (2.6.3)

where, 2

pJ =πR /2 (2.6.4)

is the polar moment of inertia (for solid shaft) and GJp is the torsional stiffness of the shaft. Closed Profile

In the case of closed profile, Fig.2.6.1, it is seen that for longitudinal equilibrium of

the element A, τ τ1 1 2 2t dz = t dz i.e. T1 = T2 where T1 and T2 are shear flow, tτ=T

Thus the shear flow around a closed profile is constant. The moment of a force on an

element at C about an arbitrary point B is

BdM = Tρ ds (2.6.5)

By integrating around the cross section noting that

BdA =1/2ρ ds and ∫dA =1/2∫ Βρ ds = A , it is seen that;

M = 2TA (2.6.6) where A is the area enclosed by the profile.

The angle of twist is found by equating the internal and external energies.

ττ∫ ∫� �

21 L L tMφ = tγds = ds

2 2 2 G

Inserting eqn. (2.8.6), we obtain that

Bρ

B A

a

b C

•

1t

2t

2τ

1τ

dz

A

Fig.2.6.1 Closed Profile

28

∫

∫�

� 2

ds/tφ dφ 1 T M= = ds =

L dz 2GA t G 4A (2.6.7)

Comparing with eqn. (2.6.4) we see that for a closed profile with one cell the

equivalent property of the cross section to the polar moment of inertia is

∫�

2

p

4AJ =

ds/t (2.6.8)

This property is called the Saint Venant torsion constant or simply the torsion

constant.

Thin rectangular profile

Equation (2.6.5) can be used to derive the torsional property of a thin rectangle, (Fig

2.6.2), the element being treated as an open profile (Murray 1984).

Thus, the torsion constant of an isolated plate is

3

p

1J = at

3 (2.6.9)

τ 2

max

1M = at

3 (2.6.10)

For open cross section comprised of several thin rectangles joined together the

torsion constant is ∑ 3

p n n

1J =K a t

3 (2.6.11)

where K is a factor which makes allowance for small fillet. K is 1 for thin sections bent

from a sheet and 1.2 to 1.3 for rolled sections.

Closed Profile and one or more Open Profiles (Mixed Cross Section)

For mixed cross section the results obtained in closed profile and open coss

section of several thin rectangles joined together can be added. The total torque M is

the sum of the torques carried by the open profiles, ∑ openM and that carried by the

closed profile closedM . Hence,

maxτ a

t

Fig. 2.6.2 Rectangular profile

29

( ) ( ) ( ) ∑ ∑closed p popen open closed

dφM = M +M = G J + J

dz (2.6.12)

where Jp is the torsion constant of the entire cross section.

The torque carried by any single plate or the closed profile is obtained by simple

proportion.

( )∑ p open

open

p

JM =M

J ;

( )

p

p closedclosed

JM =M

J (2.6.13)

Having obtained the torque carried by, say, a plate of the cross section, the stress

can be found by applying the relevant formula.

Several Closed Profiles

When a cross section consists of several closed profiles joined together, Fig. 2.6.3,

equilibrium considerations at a joint such as A, show that the shear flow in the web is

;

web i i-1T = T - T

For the ith cell we have from eqn (2.6.6)

ii iM = 2A T

Hence, for the whole cross section

∑ i iM = 2A T (2.6.15)

The angle of twist is obtained from eqn (2.6.7), thus

∫ ∫ ∫ ∫B D

i-1 i i+1

ii i A C

dφ 1 Tds 1 ds ds ds= = -T +T - T

dz 2GA t 2GA t t t (2.6.16)

Since dφ

dz is the same for all cells, it is seen that a set of simultaneous equations in

the unknowns iTψ =

Gdφ/dz (2.6.17)

can be established and solved.

Substituting ψ value in eqns. (2.6.15) and (2.6.16) respectively the following

relationships are obtained.

∑ i i

dφM = G ψA

dz (2.6.18)

∫ ∫ ∫B D

i i-1 i i+1

A C

2A = -ψ ds/t +ψ ds/t -ψ ds/t (2.6.19)

i = 1

2

3

a

b

c

A B

D C

Fig. 2.6.3 Several closed profiles

2 3webT T T= −

2T

3T

30

Hence the torsion constant of a multi-celled profile is ∑p i iJ = 2 ψA .

When the profile consists of a single cell only then

∫�T 2A

ψ = =Gφ' ds/t

(2.6.20)

2.6.3 The out - of – plane displacement of profiles

Consider an element from a beam with open profile, Fig 2.6.4. The shear strain at

the middle surface is assumed to be zero. Hence,

( ) ( )γ∂ ∂

∂ ∂zs

u v= z,s + z,s = 0

s z (2.6.21)

It is also assumed that the shear strain normal to the surface, γ zn , is zero. By

integrating eqn.(2.6.21) between arbitrary starting point V and point i on the profile

distance s from V, Murray (1984) obtained the expression for the displacement in the

longitudinal, z direction as;

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫ ∫ ∫s

s s' ' '

B B B0 0

0

u z,s = ξ z - ε z cosα s ds -η z sinα s ds -φ z ρ s ds (2.6.22)

where, ( )zξ is a function of z only and arises from the integration, and the primes

indicate differentiation with respect to z.

Noting that ( )dx = dsCosα s and ( )dy = dsSinα s (2.6.23)

The following function is defined.

( ) ( )∫s

B B0

ω s = ρ s ds (2.6.24)

This term is called the warping function. It represents a sectorial area, the

magnitude depends on the location of the pole B and the point V in the profile from

which the integration is started (sectorial origin).

z , u

s , v

n , w dz ds

( )st

Fig.2.6.4 Element from a beam with 0pen profile

31

Substituting eqns.(2.6.23) and (2.6.24) into eqn. (2.6.22) the following expression

for u(z, s) is obtained.

' ' '

B B Bu(z,s) = ζ(z) - ε (z)x(s) -η (z)y(s) -φ (z)ω (s) (2.6.25)

where ζ(z) ≡ longitudinal extension, uniform over the cross section

≡'

Bε (z)x(s) rotation about the y axis

≡'

Bη (z)x(s) rotation about the x axis

≡'

Bφ (z)ω (s) the distortion or the warping of the cross section out of its plane

due to twisting.

Murray (1984) also considered the normal and tangential components of

displacement of a thin walled section rotating as a rigid body through an angle φ .He

obtained expressions for the components of the displacements as follows;

B B Bv(z,s) = ε cosα+η sinα+ρ φ (2.6.26)

B B Bω(z,s) = -ε sinα+η cosα+q φ (2.6.27)

where, the terms Βρ and Bq are the components of a radius vector Ai. He also

proved that Βρ is the perpendicular from the pole B to the tangent to the middle line

of the profile at the point i.

2.6.4 Von Karman and Christensen’s theory for closed profile

Consider an element from the ith cell of a box girder with closed profile .Von

Karman and Christensen , in their approximate theory for closed profile, observed

that the shear strain is not zero in this case but is given by the stress-strain

relationship expressed in terms of Saint Venant shear flow Ti in the ith cell.

Thus,

γ∂ ∂

∂ ∂zs

u(z,s) v(z,s) Ti(z)= + =

s z Gt(s) (2.6.28)

From eqns. (2.6.15) and (2.6.18) we have

ii '

T(z)ψ (z) =

Gφ (z) [for multi cell]

and

∫i

T(z) 2Aψ (z) = =

Gφ'(z) ds/t [for single cell] (2.6.29)

32

Hence,γ 'izs

ψ (z)= φ (z)

t(s) (2.6.30)

From eqns. (2.6.28), (2.6.30) and (2.6.26) we obtain

∫ ∫ ∫s s s

' ' ' iB B B

0 0 0

ψ (z)u(z,s) = ς(z) - ε (z) cosα(s)ds -η (z) sinα(s)ds -φ (z) ρ (s) - ds

t(s) (2.6.31)

This equation has the same form as eqn (2.6.22). Thus eqn. (2.6.22) is valid for both

open and closed profiles provided ( )sΒρ is correctly evaluated from eqn.(2.6.24) or

eqn.(2.6.26) and eqn.(2.6.27).

2.6.5 Warping function

From Fig 2.6.5 and eqn (2.6.24) it is seen that the warping function is an area whose

magnitude depends on the location of the pole, B and of point V in the profile from

which the integration is started (starting point).

Suppose that the beam is simply twisted about the axis through the pole B without

bending in either x or y directions and without longitudinal extension.

Under these circumstances,

' '

B Bζ(z) = ε (z) =η (z) = 0

Furthermore, if the angle of twist '( )zφ could be made equal to -1 radian per unit

length, eqn. (2.6.25) becomes

Bu(z,s) =ω (s) .

Thus it is seen that the warping function is the out of plane displacement of the cross

section when the beam is twisted one radian per unit length about its axis through the

pole without bending in either the x or y directions and without longitudinal extension.

B

A x

y ds

Middle line of

profile

i ½dωB(s)

ρB(s)

y

x

ωi(s) s B

Fig 2.6.5 The physical meaning of warping function

33

Warping function represents a scaled value of the warping or out of plane distortion

of the cross section. When they are plotted on the cross section in the direction of z

axis they give a pictorial representation of the warping.

34

CHAPTER THREE

ENERGY FORMULATIONS AND GOVERNING EQUATIONS OF EQUILIBRIUM

3.1 BACKGROUNG PRINCIPLES

The elastic strains and stresses as well as their corresponding strain relations for

a thin walled box girder can be obtained after the formulation of Vlasov’s

displacement functions. In this study generalized strain fields (modes) are employed

in order to reduce the number of unknown displacement functions in the equations of

equilibrium. Torsional- distortional analysis is accomplished by using energy

principles on the basis of minimum potential energy to obtain the energy functional of

thin walled box girder structures with arbitrary deformable and or non -deformable

cross sections. Using the principles of variational calculus the energy functional is

minimized with respect to its functional variables which are displacement functional

and their first and second order derivatives, to obtain the equations of equilibrium in

the form of a system of linear differential equations in displacement quantities.

The coefficients of these equations are determined using Morh’s integral for

displacement computations. By solving these linear differential equations the

displacement functions (longitudinal and transverse displacements) are obtained and

hence, normal and shear stresses are computed and the distortional bending

moments are also evaluated. These require the evaluation of cross sectional

parameters such as shear centre, torsion constant, warping constant and warping

function.

3.2 PRINCIPLE OF MINIMUM POTENTIAL ENERGY

Let us consider a structure in equilibrium whose deformed configuration is

characterized by the displacement field ui. Recall the principle of virtual work which

states that if a structure is in equilibrium and remains in equilibrium while it is subject

to a virtual distortion the external virtual work ( EδW ) done by the external forces

acting on the structure is equal to the internal virtual work (δU) done by the internal

stresses. If we consider a class of arbitrary displacements iu which are consistent

with all constraints imposed on the body, the iu will in general, differ from the actual

displacements by some amount, say iδu . That is

i i iu = u +δu (3.2.1)

35

where the variation ( iδu ) must vanish at those points in the structure where

constraints (supports) exist. Quite clearly the variation iδu is equivalent to the virtual

displacement ( iδu ). Likewise, the internal virtual work (δu ) may be interpreted as the

variation in the strain energy resulting from the variation in the displacement field,

(Touchert 1974). Thus we see that the internal virtual work may be regarded as the

first variation of the strain energy due to variation in strain components ije .

Similarly, the external virtual work ( EδW ) may be regarded as the work done by

the surface and body forces during a variation iδu in the displacement field. For most

types of external forces it is possible to define potential functions which, when

differentiated with respect to the displacement components, yield the corresponding

force components. For example, if i

G(u ) and ig(u )represent the potentials of the

surface forces iT and body forces if , respectively, then

∂

∂i

i

GT = -

u and

∂

∂i

i

gf = -

u (3.2.2)

When such potential functions exist, the external virtual work becomes

∂ ∂E EW = V (3.2.3)

where the potential of the external forces EV is given by

∫ ∫E

s v

V = Gds+ gdv (3.2.4)

Suppose for instance that the surface and body forces are functions of positions only;

i.e., they are independent of the deformation of the structure. Such external forces

are said to be conservative, and from eqn (3.2.2) it follows that

i iG = -Tu and i ig = -fu (3.2.5)

In this case the potential of the external force becomes

∫ ∫E i i i i

s v

V = - Tu ds - fu dv (3.2.6)

Thus, for a structure which possess a strain energy u and an external potential VE

the principles of virtual work equation may be written as

E Eδu -δW = δ(u+ V ) = 0 (3.2.7)

or δπ = 0 (3.2.8)

in which Eπ = u+ V , (3.2.9)

36

is the total potential energy of the structure.

Equation (3.2.8) is a mathematical statement of the principle of minimum potential

energy (pmpe) of a system. It may be stated as follows:

Of all displacement fields which satisfy the constraint conditions, the correct state is

that which makes the total potential energy of the structure a minimum.

It is important to note that while the principle of virtual work (pvw), is valid for both

elastic and inelastic structures subject to arbitrary loads, the pmpe is applicable only

to elastic structures (linear and non linear) acted upon by forces which are derivable

from potential functions.

3.3 PRINCIPLE OF VARIATIONAL CALCULUS

Many problems in engineering and physics are most naturally formulated in terms

of extremum principles. Ordinary extremum problems of the differential calculus

involve finding the extreme values (maxima or minima) of a function of one or more

independent variables. Recall that a function of a single variable u(x) will posses an

extremum at a certain point only if the first derivative of the function vanishes at that

point ( du

= 0dx

). Likewise a necessary condition for the existence of an extreme value

of a function of n variables u(xi . . . .,xn ), is that all its partial derivatives of first order

are zero (∂ ∂

∂ ∂i n

u u=..... = 0

x x ).

Let us now find, among a set of admissible functions, that function which maximizes

or minimizes a certain functional (i.e., a function of functions). Consider the problem

of determining the function u(x) which makes the following integral a minimum.

∫b

aI(u) = F(u,u',x)dx (3.3.1)

And which in addition, satisfies the prescribed end conditions, Fig 3.3.1.

u(a) = ua ; u(b) = ub (3.3.2)

The integrand F(u,u,x) is presumed to be a known function of 'u, u = du/dx, and .... x;

I is the functional to be minimized. The branch of mathematics concerned with

problems of this nature is known as the calculus of variation.

37

Let us hypothesize that the curve u(x) shown in Fig.3. 3.1 is the actual minimizing

function. In the calculus of variation we are interested in the behaviour of I(u) when

the curve u(x) is replaced by slightly different curve , say u(x) . A function u(x) in

the neighbourhood of u(x) can be represented in the form

u(x) = u(x)+ εη(x) (3.3.3)

where ε is a small parameter. The difference between u(x) and u(x) is called the

variation in u(x) and is denoted by

δu = u(x) -u(x) = εη(x) (3.3.4)

In other words the variation δu is a small arbitrary change in u from the value of u

which minimizes the integral Iu

Similarly, the difference between the slope of the minimizing curve u(x) and the

slope of the varied curve u(x) is called the variation in the slope and is denoted by

δu' : that is, δu' = u'(x) -u'(x) (3.3.5)

Comparing eqn.(3.3.5) with the expression obtained by differentiating eqn (3.3.4),

namely, (δu)' = u'(x) -u'(x) = εη'(x) (3.3.6)

it is seen that δu' = (δu)' = εη' (3.3.7)

Hence the process of variation and differentiation are permutable.

The difference I∆ between the minimum value of the integral I in eqn.(3.3.1) and

the value of I evaluated from the varied curve u(x) is given as ( Tauchert 1974),

∫ ∫ ∫b b b

a a a∆I = F(u,u',x)dx - F(u,u',x)dx = ∆Fdx (3.3.8)

For higher order variations ∆F is given by

21∆F = δF+ δ F+...

2! (3.3.9)

or ∫b

2

a

1∆I = (δF+ δ F+...)dx

2! (3.3.10)

The first and second order variations of the integral I are defined, respectively, as

a b

u

du

u(x)

_

( )u x

au

bu _

_

u

x

Figure 3.3.1 Boundary conditions of an integral

38

∫b

aδI = δFdx (3.3.11)

And ∫b

2 2

aδ I = δ Fdx (3.3.12)

Similarly the nth variation of I is defined by

∫b

n n

aδ I = δ Fdx (3.3.13)

Introducing these definitions into eqn. (3.3.10) gives

21∆I = δI+ δ I+...

2! (3.3.14)

A necessary condition for the integral I to have an extremum is that the first

variation Iδ vanishes. If the second variation 2δ I is positive definite, the extremum is a

maximum; if 2δ I is negative definite, it is a maximum.

3.3.1 Euler-Lagrange Equations

As demonstrated above, a necessary condition that the definite integral eqn.

(3.3.1) have an extremum under the boundary conditions eqn. (3.3.2) is

∂ ∂

∂ ∂∫b

'

'a

F FδI = ( δu+ δu )dx = 0

u u (3.3.15)

Noting that d

δu' = (δu)dx

, the second term in the integrand of eqn.(3.3.15) can be

integrated by parts giving:

∂ ∂ ∂

∂ ∂ ∂ ∫ ∫

bb b

I

I I Ia aa

F F d Fδudx = δu - δudx

u u dx u (3.3.16)

Therefore ∂ ∂ ∂

∂ ∂ ∂ ∫

bb

aa

F d F FδI = - δudx + δu = 0

u dx u' u' (3.3.17)

Since the variation δu = u(x) -u(x) vanishes at the end points x = a and x = b, the

integrated term in eqn (3.3.17) vanishes also. Furthermore since δu is arbitrary in the

range a < x < b, the bracketed term inside the integral must also vanish

independently; thus

∂ ∂

∂ ∂

F d F- = 0

u dx u' (3.3.18)

39

This differential equation is called the Euler-Lagrange equation. It represents a

necessary but not sufficient condition which the function u(x) must satisfy if it is to

yield an extremum for I(u). The Euler-Lagrange equation can be generalized to the

case in which the integrand F contains higher order derivatives of u. For example the

Euler-Lagrange equation for the integral

( )∫b

aI = F u,u',u'',x dx , is given by

∂ ∂ ∂

∂ ∂ ∂

2

2

F d F d F- + = 0

u dx u' dx u'' (3.3.19)

3.4 V.Z.VLASOV’S THEORY

3.4.1 Introduction

The theory of thin-walled structures whose transverse cross sections are arbitrary

and which contains several closed contours was developed by Vlasov (1958). A

prismatic frame is considered to consist of an infinite number of narrow strips having

the form of planer frames connected by absolutely rigid bars; the bars transmit

normal (Ns) and shearing (S) forces from frame to frame. This means that normal

and tangential stresses σ and τ , in a cross section (x = constant) are distributed

uniformly across the thickness δt of a strip frame. Normal (Ns), shearing (S), and

transverse (Q), forces and also bending moments (Ms) develop on areas of

longitudinal section (s = constant), Fig. 3.4.1. The position of a point on the middle

surface of a prismatic frame is determined by the coordinates x and s. Since the

plates are in plane stress state, the strain in them can be completely described if the

displacements, u(x,s) and v(x.s), of point c in the x-direction and along the tangent to

the contour of the transverse section, produced by the load P(x,s) and q(x,s) are

known.

o

o o o

o o

o

o o

o o o

o o

o o

o o o

o o

s

τ σ s

Q

40

3.4.2 Energy formulation of the equilibrium equations

The longitudinal warping and transverse (distortional) displacements given by Vlasov

(1958) are

ϕu(x,s) =U(x) (s) (3.4.1)

v(x,s) = V(x)ψ(s) (3.4.2)

where U(x) and V(x) are unknown functions governing the displacements in the

longitudinal and transverse directions respectively, and ϕ and ψ are generalized

warping and distortional strain modes respectively. These strain modes are known

functions of the profile coordinates, and are chosen in advance for any type of cross

section. The displacements may be represented in series form as;

ϕ∑m

i ii=1

u(x,s) = U (x) (s) (3.4.3)

∑n

k kk=1

v(x,s) = V (x)ψ (s) (3.4.4)

where, Ui(x) and Vk(x) are unknown functions which express the laws governing the

variation of the displacements along the length of the space frame.

ϕi(s) and kψ (s) are elementary displacements of the strip frame, respectively out of

the plane (m displacements) and in the plane (n displacements).

These displacements are chosen among all displacements possible, and are called

the generalised strain coordinates of a strip frame.

3.4.2.1 Vlasov’s stress – strain relations

Fig. 3.4.1 Vlasov’s model for stress and strain in a strip frame

41

From the theory of elasticity the strains in the longitudinal and transverse

directions are given by;

ϕ∂

∂∑m

i ii=1

u(x,s)= U'(x) (s)

x and

∂

∂∑

n

k kk=1

v(x,s)= V '(x)ψ (s)

x (3.4.5)

The expression for shear strain is γ∂ ∂

∂ ∂

u v(x,s) = +

s x

or ( ) ϕγ ∑ ∑m n

i i k ki=1 k=1

'(s)U (x)+ ψ (s)V '(x)x,s = (3.4.6)

Using the above displacement fields and basic stress-strain relationships of the

theory of elasticity the expressions for normal and shear stresses become:

ϕ∂

∂∑m

i ii=1

u(x,s)σ(x,s) = E = E (s)U'(x)

x (3.4.7)

γ ϕτ ∑ ∑m n

i i k ki=1 k=1

(x,s) = G (x,s) = G '(s)U (x)+ ψ (s)V '(x) (3.4.8)

The (m + n) functions sought for, iu (x) and kv (x) , are determined from (m + n)

equations for the strip frame, obtained by equating to zero the work done by external

and internal forces in (m + n) independent virtual displacements, Fig. 3.4.2.

Every virtual displacement is as a result of an infinitesimal variation experienced

by one of the generalized strain coordinates which determine the position of all joints

and bars of the frame. This application of the principle of virtual displacements is

called the method of variations.

Transverse bending moment generated in the box structure due to distortion is given

by; ( ) ∑n

k kk=1

M x,s = M (s)V (x) (3.4.9)

where Mk(s) = bending moment generated in the cross sectional frame of unit width

due to a unit distortion, V(x) = 1

42

3.4.2.2 Potential energy functional

The potential energy of a box structure under the action of a distortional load of

intensity q is given by:

Eπ =U+ W (3.4.10)

Where,

π = the total potential energy of the box structure,

U = Strain energy,

EW = External potential or work done by the external loads.

From strength of material, the strain energy of a structure is given by

τ

∫ ∫2 2 2

(s)L S

1 σ (x,s) (x,s) M (x,s)U = + t(s)+ dxds

2 E G EI (a)

and work done by external load is given by;

EW = qv(x,s)dxds= ϕ∑∫∫ h h

s x

q V (x) (s)dsdx = ∑∫ h h

x

q V dx (b)

Substituting expressions (a) and (b) into Eqn.( 3.4.10) we obtain that,

τ

∫ ∫

2 2 2

L S

1 σ (x,s) (x,s) M (x,s)π = + t(s)+ - 2qv(x,s) dxds

2 E G EI(s) (3.4.11)

where,

o 1

m = 6

n = 5

• • •

• • •

1 2 3

4 5 6

o o

o 2

3 o

4 5

(b) Double cell

(c) Triple triangular cell

•

• •

• • 2 1 3

4 5 o

o o

o 1

2

3 4

m = 5

n = 4

• •

• •

1 2

3 4 o

o o

o 1

2

3 4

(a) Single cell

m = 4

n = 4

Fig.3.4.2 Independent (m+n) virtual displacements

43

σ(x,s)= Normal stress

τ (x,s) = Shear stress

M(x,s) = Transverse distortional bending moment

q = Line load per unit area applied in the plane of the plate

3

(s) 2

t (s)I =

12(1- ν )= Moment of inertia

E = Modulus of elasticity

G = Shear modulus

ν = poisson ratio

t = thickness of plate

Substituting the expressions for σ(x,s) eqn (3.4.7), τ (x,s) eqn.(3.4.8), M(x,s)

eqn.(3.4.9) and v(x,s) eqn.(3.4.2) into eqn.(3.4.11) we obtain that:

ϕ ϕ∑ ∑i i j jπ =E (s)U'(x) * (s)U ''(x) * t(s)dsdx +

+ i j

ϕ ϕ ∑ ∑ ∑ ∑i k k j h hG '(s)U (x)+ ψ (s)V '(x) * '(s)U (x)+ ψ (s)V '(x) * t(s)dsdx +

+ ∑ ∑

n n

k k h hk=1 h=1

1M (s)V (x) * M (s)V (x) dsdx

EI - ∑∫ hh

x

q V dx

Simplifying further noting that t(s)ds = dA we obtain;

ϕ ϕ∑ ∑m m

i j i ji=1 j=1

1π = E U'(x)U '(x) * (s) (s) * dAdx

2+

+ ϕ ϕ∑ ∑m

j=1

m1 ' 'G U (x)U (x) (s) (s)dAdxi j i j2 i =1

+

+j

ϕ∑ ∑n n

j k kj=1 k=1

1G U (x)V '(x) * '(s)ψ (s) * dAdx

2 +

+ ϕ∑ ∑m m

i h i hh=1 i=1

1G U (x)U '(x) * '(s)ψ (s) * dAdx

2 +

44

+ ∑ ∑n n

k h k hh=1 k=1

1G V '(x)V '(x) * ψ (s)ψ (s) * dAdx

2 +

+ ∑ ∑n n

k hk h

k=1 k=1(s)

M (x)M (x)1* V (x)V (x) * dxds

2 EI

- ∑ h hq V dx (3.4.12)

Let,

ϕ ϕ∫ij ji i ja = a = (s) (s)dA (a)

ϕ ϕ∫ij ji i jb = b = '(s) '(s)dA (b)

ϕ∫kj jk k jc = c = '(s)ψ (s)dA (c)

ϕ∫ih hi i kc = c = '(s)ψ (s)dA (d)

∫kh hk k hr = r = ψ (s)ψ (s)dA; (e)

∫ k hkh hk

(s)

M (s)M (s)1s = s = ds

E EI (f)

∫h hq = qψ ds (g) (3.4.13)

Substituting eqns. (3.4.13) into eqn. (3.4.12) gives the potential energy functional:

∑ ij i j

1π = E a U'(x)U '(x)dx

2

+ ∑ ∑ij i j kj k j

1G b U (x)U (x)+ c U (x)V '(x)

2

+ ∑ ∑ih i h kh k h

1G c U (x)V '(x)+ r V '(x)V '(x) dx

2

+ ∑ hk k h

1E s V (x)V (x)dx

2 ∑- q V dx

h h (3.4.14)

3.4.3 Governing equations of equilibrium

The governing equations of distortional equilibrium are obtained by minimizing the

above functional eqn.(3.4.14), with respect to its functional variables u(x) and v(x)

using Euler Lagrange technique, eqns. (3.3.19)and (3.4.15).

∂ ∂ ∂ ∂

'

j j

π d π- = 0

U dx U (a)

45

'

∂ ∂

∂ ∂ h h

d- = 0

V dx V

π π (b) (3.4.15)

Carrying out the partial differentiation of eqn.(3.4.14) with respect to Uj and jU ' gives

∂ ∂∑ ∑ '

ij i kj k

j

= G b U (x)+ c V (x)U

π ,

∂

∂∑ '

ij i'

j

=E a U (x)U

π,

∂ ∂

''

ij i'

j

d=Ea U (x)

dx U

π

Therefore, ∂ ∂ ∂ ∂

'

j j

d- = 0

U dx U

π π

⇒ ∑ ∑ ∑' ''

ij i kj k ij iG b U (x)+ c V (x) -E a U (x) = 0

or ∑ ∑ ∑'' '

ij i ij i kj kE a U (x) -G b U (x) -G c V (x) = 0

Dividing through by G, and re-arranging we obtain;

∑ ∑ ∑m m n

k a U''(x)- b U (x)- c V '(x)=0ij i ij i kj k

i=1 i=1 k=1 (3.4.16a)

where υE

κ = = 2(1+ )G

Performing similar operations on eqn.(3.4.14) with respect to hV and hV ' we obtain

the second equation as follows.

∂

∂∑ ∑hk K h

h

=E s V (x) - qV

π

'

∂ ∂∑ ∑ih i kh k

h

= G c U (x)+ r V '(x)V

π

'

∂ ∂ ∑ ∑ih i kh k

h

d= G c U'(x)+ r V '(x)

dx V

π

'

∂ ∂∴

∂ ∂ h h

d- =

V dx V

π π - ∑ ∑ih i kh kG c U'(x)+ r V '(x) + ∑ ∑hk k hE s V (x) - q = 0

∑ ∑ih i kh kc U'(x)+ r V ''(x) - ∑ ∑hk k h

1κ s V (x)+ q = 0

G (3.4.16b)

Equations (3.4.16a) and (3.4.16b) are Vlasov’s differential equations of distortional

equilibrium for a box girder. The evaluation of the coefficients of these equations is

presented in the next chapter.

46

Equations (3.4.16a) and (3.4.16b) can be written in matrix notation. Thus,

''

''

''

11 12 13 1

21 22 23 2

31 32 33 3

a a a U

κ a a a U

a a a U

-

11 12 13 1

21 22 23 2

31 32 33 3

b b b U

b b b U

b b b U

-

'

'

'

'

1

11 12 13 14

2

21 22 23 24

3

31 32 33 34

4

Vc c c c

Vc c c c

Vc c c c

V

=0

(3.4.17a)

'

'

'

11 12 13

1

21 22 23

2

31 32 33

3

41 42 43

c c cU

c c cU

c c cU

c c c

-

111 12 13 14

221 22 23 24

331 32 33 34

441 42 43 44

Vs s s s

Vs s s sκ

Vs s s s

Vs s s s

+

1 ''

''

''

''

11 12 13 14

21 22 23 24 2

31 32 33 34 3

41 42 43 44 4

r r r r V

r r r r V

r r r r V

r r r r V

+

1

2

3

4

1

G= 0

q

q

q

q

(3.4.17b)

Expanding the matrix notation eqns. (3.4.17 (a) and (b)) we obtain:

11 1 12 2 13 3 11 1 12 2 13 3 11 1 12 2 13 3 14 4

'' '' '' ' ' ' 'κa U + κa U + κa U - b U - b U - b U - c V - c V - c V - c V = 0

21 1 22 2 23 3 21 1 22 2 23 3 21 1 22 2 23 3 24 4


1 32 2 33 3 31 1 32 2 33 3 31 1 32 2 33 3 34 4


31 (3.4.18a)

11 1 12 2 13 3 11 1 12 2 13 3 14 4 11 1 12 2 13 3 14 4

' ' ' '' '' '' ''c U + c U + c U - κs V - κs V - κs V - κs V + r V + r V + r V + r V = -q/G

21 1 22 2 23 3 21 1 22 2 23 3 24 4 21 1 22 2 23 3 24 4


31 1 32 2 33 3 31 1 32 2 33 3 34 4 31 1 32 2 33 3 34 4


41 1 42 2 43 3 41 1 42 2 43 3 44 4 41 1 42 2 43 3 44 4


47

CHAPTER FOUR

GENERATION OF THE STRAIN MODES AND EVALUATION OF THE COEFFICIENTS OF THE GOVERNING EQUATIONS OF EQUILIBRIUM

4.1 PREAMBLE

The governing differential equations of distortional equilibrium are given by eqns.

3.4.16(a) and 3.4.16(b).For ease of reference they are stated here below.

∑ ∑ ∑m m n

ji i ji i jk ki=1 i=1 k=1

k a U''(x) - b U (x) - c V '(x) = 0 4.1.1(a)

∑ ∑ ∑ ∑hi i kh k hk k h

1c U'(x)+ r V ''(x) - k s V (x)+ q = 0

G 4.1.1(b)

where, ji ij ji ij kj jk hi ih hk kh hk kha = a , b = b , c = c , c = c , r = r , s = s are called Vlasov’s coefficients

given by;

ϕ ϕ∫ij ji i ja = a = (s) (s)dA (a)

ϕ ϕ∫ij ji i jb = b = '(s) '(s)dA (b)

ϕ∫kj jk k jc = c = '(s)ψ (s)dA (c)

ϕ∫ih hi i kc = c = '(s)ψ (s)dA (d)

∫kh hk k hr = r = ψ (s)ψ (s)dA; (e)

∫ k hkh hk

(s)

M (s)M (s)1s = s = ds

E EI (f)

Before the solution of these differential equations can be obtained, the following

steps need to be taken.

(i) Appropriate choice of girder cross sections

(ii) Generation of strain modes for the girder sections

(iii) Evaluation of the coefficients of the differential equations for each girder cross

section

(iv) Development of the equations for various strain modes interactions

Steps (i) to (iii) are systematically considered and presented in this chapter while step

(iv) is considered in the subsequent chapter.

48

4.2 CHOICE OF BOX GIRDER CROSS SECTIONS

As stated in the aims and objectives of this study, thin walled bridge girders, both

mono-symmetric and doubly symmetric sections are examined in this report. From

literature survey it is observed that doubly symmetric sections are not common

features in current bridge design and construction, but they are prominently used in

box culverts and similar bridge structures of less importance. Consequently, the

following box girder sections, Figs. 4.2.1 and 4.2.2, are adopted in this work for the

purpose of torsional-distortional analysis. (a) Single cell mono symmetric section,

(b) double cell mono symmetric section, (c) single cell doubly symmetric section and

(d) multi cell doubly symmetric section,

A non symmetric section, Fig.4.2.3, is also chosen to highlight the effect of full

interaction of strain modes on the formulation of governing differential equations and

the analysis of thin walled box girders with such sections.

t

b

t

t

h

x

y

(a) Single Cell doubly symmetric

section

• C & S

Fig. 4.2.2 Doubly-symmetric box girder sections

b b b

h

t

t t t

x

y

C & S •

(b) Multi-Cell doubly symmetric Section

3050

2745 2745 1830 1830

915 915 3660 3660

203

203 203

203

(b) Double cell box girder section

Fig. 4.2.1 Mono-symmetric box girder sections

915 7320

2745 2745 3660

(a) Single cell box girder section

3050

915

t = 203

5490

100

49

4.3 STRAIN MODES AND VLASOV COEFFICIENTS

4.3.1 Introduction

From the energy formulation of the equilibrium equation in section 3.4.2 it was

noted that ϕ and ψ represent generalized warping and distortional strain modes

respectively and from eqns.( 3.4.3 and 3.4.4) ( )ϕi s and kψ (s) are elementary

displacements of the strip frame out of the plane (m displacements) and in the plane (

n displacements) respectively. It was also noted that these displacements are chosen

among all displacements possible and are called the generalized strain coordinates

of a strip frame. Thus, Vlasov’s coefficients of differential equations of equilibrium,

eqn.(3.4.13), which involve a combination of these elementary displacements and

their derivatives may be obtained by consideration of the box girder bridge cross

section as a strip frame and then applying unit displacement one after the other at

the nodal points of the frame in longitudinal direction, to determine the corresponding

out of plane displacement of every joint (regarded as fixed) on the frame. By applying

another set of unit displacements at the joints in n possible transverse directions, the

corresponding transverse (in-plane) displacements can also be obtained. The first

order derivatives of these displacement functions may be obtained by numerical

differentiation and used for computation of the coefficients with the aid of Morh’s

integral for displacement computations (appendix A).

4.3.2 Nodal displacement strain modes

Consider the single cell mono-symmetric section shown in Fig.4.3.1(a). The frame

has four degrees of freedom in the longitudinal direction and four in the transverse

direction. By applying unit displacement at the nodes in each of these directions the

50

longitudinal strain modes ( )ϕi s and transverse strain modes kψ (s) are obtained and

hence their first derivatives, Figs.4.3.1b and 4.3.2.

x

ψ1

x

ψ2

φ1(s) 1

1.0

1.0

x

y

φ2(s)

2

1.0

1.0

x

y

φ4(s)

4

1.0

x x

y

(a) Longitudinal Nodal Strain Modes

Fig.4.3.1 Longitudinal strain modes and their first derivatives

0.444

0.294

x

y

φI1(s)

0.294

0.444

x

y

φI2(s)

0.437

0.444

x

y

φI3(s) 0.444

0.437

x

y

φI4(s)

(b) Derivatives of Nodal Strain Modes

φ3(s)

3

1.0 y

x

51

Expanding the coefficients using eqn 3.4.13 we have;

(a) ϕ ϕ ϕ ϕ∑∫ji ij j i j ia = a = dA = dA

(b) ϕ ϕ ϕ ϕ∑∫ji ij i j i jb = b = ' 'dA = ' 'dA

(c) ϕ ϕ∑∫jk kj j k j kc = c = 'ψ dA = 'ψ dA (4.3.1)

(d) ϕ ϕ∑∫hi ih h i h ic = c = ψ 'dA = ψ 'dA

(e) ∑∫hk kh h k h kr = r = ψ ψ dA = ψ ψ dA

The coefficients aij, bij, ckj, cih, and rkh are determined using Morh’s integral chart.

4.3.3 Generalized strain modes

A consideration of the single cell mono-symmetric strip frame in Fig.4.2.1(a)

shows that it has four degrees of freedom in the longitudinal direction and four in the

transverse direction. From eqns.(3.4.3 and 3.4.4), where in this case m = 4 and n = 4,

it follows that we have sixteen displacement quantities to compute and hence,

sixteen differential equations of distortional equilibrium will be required.

For multi-celled profiles the number of degrees of freedom will increase and

hence the number of independent displacement quantities (m+n) will require 2(m x n)

differential equations to solve for the displacement quantities and this can be quite

cumbersome.

1.032

y 1.032

y

1.0

x

y

ψ3

1.0 y

ψ4

Fig. 4.3.2 Transverse strain modes

x

52

The application of Vlasovs generalized strain modes as modified by Varbanov

(1976) reduces the number of displacement quantities and hence the differential

equations of equilibrium required to solve for them to seven, irrespective of the

number of degrees of freedom possessed by the structure.

In the generalized strain modes, there are three strain fields in the longitudinal

direction, 1 2,ϕ ϕ and 3ϕ .Thus, from eqn. (3.4.3) we have,

ϕ ϕ ϕ1 1 2 2 3 3u(x,s) =U (x) (s)+U (x) (s)+U (x) (s)

or ϕ∑3

i ii=1

u(x,s) = U (x) (s) (4.3.2a)

In the transverse direction four strain modes are also recognized; 1 2 3ψ ,ψ ,ψ and 4ψ .

From eqn (3.4.4) we have;

1 1 2 2 3 3 4 4v(x,s) = V (x)ψ (s)+ V (x)ψ (s)+ V (x)ψ (s)+ V (x)ψ (s)

or ∑4

k kk=1

v(x,s) = V (x) ψ (s) (4.3.2b)

where 1ϕ = out of plane displacement parameter when the load is acting (vertically)

normal to the top flange of the girder, i.e., bending is about horizontal

axis.

2ϕ = Out of plane displacement parameter when the load is acting tangential

to the plane of the flanges, i.e., bending is about vertical axis.

3ϕ = Out of plane displacement parameter due to distortion of the cross

section, i.e., the warping function .

1ψ = In-plane displacement parameter due to the load giving rise to 1ϕ

2ψ = In-plane displacement parameter due to the load giving rise to 2ϕ

3ψ = In-plane displacement parameter due to the distortion of the cross

section, i.e., non uniform torsion.

4ψ = In-plane displacement function due to pure rotation or Saint Venant

torsion of the cross section.

4.3.4 Plotting of warping functions

In the preceding section, it was stated that 3ϕ , the out of plane displacement

parameter due to distortion of the cross section is obtained by plotting the warping

53

function of the cross section. The summary of the procedure for plotting the warping

function of any given cross section is presented below, followed by numerical

example calculations.

(i) Chose a coordinate axis (preferably along the axis of symmetry) and plot (a)

variation of x-ordinates along x-axis and (b) variation of y-ordinates along y-axis.

These give x diagram and y diagram as shown in Fig.4.3.3 (b) and (c) respectively

(ii) Choose a pole B and a starting point V. Evaluate and draw the warping function

with respect to pole B using eqns (2.6.24 and 2.6.29)

(iii) Evaluate the following section properties;

(a) F = Cross sectional area of profile

(b) Fx = first moment of area of section about y-axis

(c) Fy = first moment of area of section about x-axis

(d) ωF = First moment of sectorial area about pole B

(e) xxF = Second moment of area of profile about y-axis

(f) yyF = Second moment of area of profile about x-axis

(g) xyF = Product moment of area of profile, i.e., Fx.Fy.

(h) B

xωF = Sectorial product of area, Fx.FwB

(i) ByωF = Sectorial product of area, Fy.FwB

(j) B Bω ωF =Warping constant with respect to pole B

(iv) Locate the centroid s of the profile using the relations

__x

s

Fx =

F and

__y

s

Fy =

F (4.3.3)

The warping function with respect to the centroidal axis becomes

C B 0ω =ω -ω , where, 0 ωBω =F /F (4.3.4)

(v) Using the new (centroidal) axis plot x and y diagrams as in i(a) and (b) above.

(vi) Calculate the product integrals, Fxx, Fyy, Fxy, Fwx, Fwy, Fww, (based on the

centroidal axis), using the following transformation formulae:

54

__ _

__ _

__ _ _

__ _ _

__ _ _

__ _

2

xxxx x

2

yyyy y

xyxy x y

xωxω x ω

yωyω y ω

2

ωωωω ω

F =F -F /F

F =F -F /F

F =F -F F /F

F =F -F F /F

F =F -F F /F

F =F -F /F

(4.3.5)

(vii) Obtain the coordinates of the shear centre M from the relation

yω xx xω xy

m B 2

xx yy xy

F .F -F .Fx - x =

F .F -F (4.3.6)

and yω xy xω yy

M B 2

xx yy xy

F .F -F .Fy - y =

F .F -F (4.3.7)

(viii) The warping function Mω , with pole at shear centre M, is obtained from the

following transformation formulae (Murray 1984);

M C M B M Bω (s) =ω +(y - y )x - (x - x )y (4.3.8)

4.3.4.1 Warping functions for mono-symmetric box girder sections

(a) Single cell mono-symmetric box girder section

The warping functions ( Bω ) are given as follows:

∫B Bω (s) = ρ (s)ds for open sectional profile

∫B B

ψω (s) = ρ - ds

t for closed sectional profile

x

6.774

2.195

6.774

6.774

6.774

_

_ _

+

+

+

1

2

3

5 6

ωB

1

+ +

2

3 4

5

6 x

915 7320

2745 2745 3660

(a) Single cell box girder section

3050

915

t = 203

y

(b) x Diagram

y

+

1830

3660

4575

+

+ _

_

_ 4575

3660

1830

B,V 1 2

3

4

5 6 x

55

The computations of the warping functions are shown in Appendix One.

4.3.4.2 Warping function for multi cell doubly symmetric section

h=b

t t t

t

x C & S •

A3 A2 A1

Fig.4.3.3 : Single cell mono symmetric section and warping function

Fig.4.3.4 Double cell mono-symmetric section and warping function

(a) Double cell box girder

3050

2745 2745 1830 1830

915 915 3660

203

203

203

1889.4

1160.58

3660

y

x

(b) x Diagram

y

+

1830

3660

4575

+

+

_

_

_ 4575

3660

1830

B,V 1 2

3

4

5 6 x

(c) y Diagram

3050 3050 +

1

+ +

2

3 4

5

6 x

+

3050

y

B, V x

6.774

y 2.195

2.195

6.774

6.774

6.774

_

_ _

_

+

+

+

1

2

3

4

5 6

(d) Warping function ωB

ωB

56

For the doubly symmetric section in Fig.4.3.5 (a) the centroid C and the shear centre

S, coincide. Cross sectional area F = 10bt

Enclosed area A1 = A2 = A3 = b2

Fx = 0, symmetry about x-x axis

Fy = 0 , symmetry about y-y axis

∫ i

M M

ψω (s) = ρ - ds

t

For closed sections we have

∫ ∫ ∫B D

i i-1 i i+1

A C

ds ds ds2A = -ψ +ψ -ψ

t t t

For Cell 1 we obtain; 1 24ψ -ψ = 2bt (1)

Cell 2 ⇒ 1 2 3-ψ + 4ψ -ψ = 2bt (2)

Cell 3 ⇒ 2 3ψ + 4ψ = 2bt (3)

Solving eqns.(1, 2, & 3) above gives;

1 3

2

ψ =ψ = 5bt/7

ψ = 6bt/7

Using ( )∫M M iω = ρ -ψ /t ds we obtain,

+ +

+

+ +

+

_ _ _

_ _

_

2 3 8

4 7

tb

+

2 3 8

4 7

tb

+

2 3 8

4 7

tb

+

2 3 8

4 7

tb

+

2 1 3

4 7

tb

+

2 1 3

4 7

tb

+

(b) Warping function diagram

Fig.4.3.5 Doubly symmetric section and warping function diagram

x

y

57

For cell 1 and cell 3, ( )∫M Mω = ρ - 5bt/7 ds

Cell 2; ( )∫M Mω = ρ - 6bt/7 ds

Integrating from point V to point 1 we obtain;

M

b b 6btρ = - , ds = , ψ =

2 2 7

∴

2

1

1 3tω = -b +

4 7

From point 1 to point 2 we have;

M

b 5btρ = - , ds = b, ψ =

2 7

∴

2

2

3 8tω = -b +

4 7


M

b b 6btρ = ,ds = ,ψ = -

2 2 7

∴

2

4

1 3tω = b +

4 7


M

b 5btρ = , ds = b, ψ = -

2 7 ∴

2

3

3 8tω = b +

4 7

By reason of symmetry we observe that;

2

5 1

1 3tω = -ω = b +

4 7

2

6 2

3 8tω = -ω = b +

4 7

2

7 3

3 8tω = -ω = -b +

4 7

2

8 4

1 3tω = -ω = -b +

4 7.

The diagram of the warping function with the pole located at shear centre S is on

Fig.4.3.5 (b).

4.3.4.3 Warping function for non-symmetric section

3050

1.475

8.245

8.245

6.88

0 1

2 3

. c 1348

1702

x

200

5490

200

200

x

58

The warping function of the single cell non-symmetric profile (Fig.4.3.6a), based on

the centroidal axes is shown on Fig. 4.3.6b while that with respect to the shear centre

is on Fig. 4.3.13f

The computations of the warping functions are shown in Appendix One.

4.3.5 Strain mode diagrams

Consider a simply supported girder loaded as shown in Fig 4.3.7(a). If we assume

the normal beam theory, i.e., neutral axis remaining neutral before and after bending,

then the distortion of the cross section will be as shown in Fig.4,3.7(b) where, θ is

the distortion angle (rotation of the vertical axis). The displacement φ1 at any distance

R, from the centroid is given by 1ϕ = Rθ . If we assume a unit rotation of the vertical

(z) axis then 1ϕ = R , at any point on the cross section. Note that 1ϕ can be positive

or negative depending on the value of R, in the tension or compression zone of the

girder. Thus, 1ϕ is a property of the cross section obtained by plotting the

displacement of the members of the cross section when the vertical (z-z) axis is

rotated through a unit radian.

Similarly, if the load is acting in a horizontal (y- y) direction, normal to the x-z plane

in Fig.4.3.7(a), then the bending is in x-z plane and y axis is rotated through angle θ2

giving rise to 2ϕ , displacement out of plane. The values of 2ϕ are obtained for the

members of the cross section by plotting the displacement of the cross section when

y-axis is rotated through a unit radian.

The warping function 3ϕ ,of the beam cross section is obtained as detailed in

section 4.3.4. It has been explained that the warping function is the out of plane

displacement of the cross section when the beam is twisted about its axis through the

pole, one radian per unit length without bending in either x or y direction and without

longitudinal extension.

59

ψ1 and ψ2 are in-plane displacement of the cross section in x-z and x-y planes

respectively while ψ3 is the distortion of the cross section. They can be obtained by

numerical differentiation of 1ϕ , 2ϕ and 3ϕ diagrams respectively.

ψ4 is the displacement diagram of the beam cross section when the section is

rotated one radian in say, a clockwise direction, about its centroidal axis. Thus, ψ4 is

directly proportional to the perpendicular distance ( radius of rotation) from the

centroidal axis to the members of the cross section. It is assumed to be positive if the

member moves in the positive directions of the coordinate axis and negative

otherwise.

The generalized strain modes for (a) single cell mono-symmetric section and (b)

double cell mono-symmetric section are shown in Figs 4.3.8 and 4.3.9 respectively.

1

+

2

3

5

6 + 0.857

(d) Transverse strain mode in z-

direction

+ 0.857

+ y

z

4

'1 1

ϕ =ψψψψ

z

1.945 +

1

+ +

2

3 4

5 6

1.945

-1.105

1.945

1.105

-1.105

(c) Longitudinal Strain Mode Diagram

(Bending about o-y axis)

. y ϕ1111

1

+

2

5 6

4.575

3.66 3.66

+

- -

4.575

ϕ2222 1

+

0.514

2 5 6

0.514

1.0 _

-

'2 2

ϕ =ψψψψ

x

y

(a) Simply supported girder section

z +

-

θ

(b) Cross section distortion

y

Fig. 4.3.7 Simply Supported Girder and Cross Section Distortion

915 7320

2745 2745 3660

(a): Single cell box girder section

3050

915

t = 203

_

(b) Pure Rotation Diagram

. +

+

_

2.57

0

1.105

1ρ

2ρ

3ρ

4ρ

1.945 2.57

0

4

ψψψψ

y

60

(c)Longitudinal Strain Mode Diagram

(Bending About y-y Axis)

1.889 1.889

1.16 1.16 1.16

c

+

+ + +

- - -

-

1 2

3 4

5 6

y

z

c 0.857

0.857

+0.857

(d) Transverse stain mode in y-y direction

+ +

+

+0.857 +1.00

y

y

'1 1

ϕ =ψψψψ

Fig.4.3.8 Generalized strain modes for single cell mono-symmetric box girder frame

+

1 +

2

3 4

5

6

-0.159

(h) Distortion Diagram '

3 3( ) ( )s sϕ ψ=

+ -

-

-

y

z

-0.492

+1.162

-0.417

0.123

+1.162

+0.417

'3 3

ϕ =ψψψψ 0.966

+

1

2

4

5 6

+

z

+

-

- -

-

+ + + -

y

3

-0.966 -0.582

+0.582

+0.582

+0.901 -0.901

(g) Warping Function Diagram

+0.901

3Mω ϕ=

_ 1.00

+4.575 -3.66 +

- +3.66

3050

2745 2745 1830 1830

915 915 3660

203

203

203

(a) Double cell concrete box girder

1889.4

1160.58

3660

y

z

. +

+

_

_

1.0

0.514 0.514

0.514

1ρ

2ρ

3ρ

4ρ

(b) Pure rotation Diagram

4ψψψψ

y

61

Others are Fig.4.3.10 for single cell doubly symmetric section, Fig. 4.3.11 for multi-

cell doubly symmetric section and Fig.4.3.12 for single cell non- symmetric section.

Fig. 4.3.9: Generalized strain modes for double cell mono-symmetric box girder section

0.901 +

-

1 2

3

4

5 6

(g) Warping function, 3( )M

sω ω= diagram

0.901

+

+ +

-

-

-

-

0.901

0.966 0.582

0.582

0.582

0.966

3ϕ 0.417

0.492

+

+

1 2

4

5 6

(h) Distortion diagram

_

_ 0.417

0.159

3

'3 3

ϕ =ψψψψ

z

y

3.60

(b) Rotation diagram

⊕

⊕

-

-

4ψ

z

3m

7.2m

y

(a) Single cell doubly-symmetric section

200mm

200mm

3.60 1.50

1.50

y ⊕ ⊕

ϕ1 1

' =ψ ϕ1(s) y

- -

- 1.5

62

3.0m 3.0m 3.0m

y

3.0m 200

200

200

200

z

• 0,M

(a) Multi cell doubly symmetric section

1.50

4.50 4.50

(b) Pure rotation diagram

1.50

1.50

+ +

+ + -

- -

-

4ψ

y

1.0 1.0 1.0 1.0

+ +

+

y

1.50 - -

- - -

1.50 1.50 1.50 1.50

ϕ '1 1=ψ ϕ

1

(f) Transverse strain mode

in z-direction

z

y

⊕

⊕

⊕

(e) Longitudinal strain mode diagram

( bending about z-z axis)

2ϕ

y

z

ϕ2 2

' =ψ

1.0

1.0 3.60

3.60

-

-

-

3.60

3.60

y

⊕

z

− ⊕

ϕ3 M=ω

(g) Warping Function (h) Distortion diagram

z

y

Fig.4.3.10 Generalized strain modes for single cell doubly symmetric section

⊕

⊕

−

2.223

2.223 2.223

2.223

1.482

1.482

0.618

0.618

63

0 1

2 3

. C y

1.0 0.857

ϕ '1 1=ψ

0 1

2 3

. C y

1.348 1.348

+

+

-

-

-

1ϕ

(b) Rotation

0 1

2 3

. C y

z

+

+

_

_

1.944

1.348 2.414

1.702

4ψ

0 1

2 3

. C

2414 1246

1348

1702

(a) Cross section dimensions

5490

t = 100mm

(e) Longitudinal strain mode Diagram.

Bending about z-z axis

1 2 3 4

5 6 7 8

+

+

+ +

1.50 4.50 4.50

4.50 4.50

-

-

-

-

(f) Transverse strain mode in y-y direction

6 5 7

1 2 3 4

8

-

-

1.0

1.0

ϕ '2 2=ψ

ϕ2

8.808

8.808

_ _ _ +

+ +

+ + +

_ _

_

8.808 8.808

8.808 3.021 3.021

3.021 3.021

(g) Warping function 3 3ω ϕ=

Fig.4.3.11: Generalized strain modes for multi-cell doubly symmetric section

1.929

2.014 5.871

5.871 1.929

(h) Distortion '

3 3ϕ ψ= Diagram

2.014

ϕ '3 3=ψ

ϕ3

64

4.3.6 Tests for accuracy of strain modes

We define a function 0ϕ which is orthogonal to the strain modes 1 2,ϕ ϕ and 3ϕ

such that ϕ∫2

o dt = area of cross section. Since 0ϕ is orthogonal to 1 2,ϕ ϕ and 3ϕ it

follows that; ϕ ϕ∫ 0 idt = 0 , (for i =1,2,3) (4.3.10)

Equation (4.3.10) can therefore be used to verify the accuracy of the strain mode

diagrams. Additionally, the strain mode 1ϕ can be obtained from the relation,

1 1 0ϕ ϕ αϕ= + (4.3.11)

where,

1ϕ = auxiliary shape function

α = a factor

Substituting eqn (4.3.11) into eqn (4.3.10) gives, after simplification;

0 1

2 3

. C

0.608

(g) Warping function

0.421 0.430

0.633

3Mω ϕ= 028 y

z z

y

Fig.4.3.12 Generalized strain modes for single cell non symmetric section

0 1

2 3

. C y

z +

+

_

_

0.339

0.289

0.155

0.349

(h) Distortion Diagram

'3 3

ϕ ψ=

0 1

2 3

. C y

z

(e) Longitudinal strain mode:

bending in z-z direction

2ϕ

1.246

3.076

3.076 2.414

2.414

+

+

+

_

_

_

y

0 1

2 3

. C

z

(f) Transverse strain mode in z-z

direction

ϕ '2 2=ψ

1.00

0.514

1.0

_

_

+

65

ϕ ϕ

ϕ∫∫

0 1

2

0

dtα = -

dt (4.3.12)

With values of α obtained from eqn (4.3.12), 1ϕ can be computed from eqn (4.3.11)

and compared with 1ϕ diagram obtained using Morh’s integral.

The single cell mono-symmetric section is used to demonstrate the accuracy of

the strain mode diagrams. The cross sectional dimensions and the generalised

strain modes ( 1ϕ , 2ϕ and 3ϕ ) diagrams for this profile are shown in Fig.4.3.8. The

auxiliary shape function 1ϕ diagram, Fig.4.3.14(a), is obtained by arbitrary choice of

coordinate axis (different from the centroidal axis) and by drawing the displacement

diagram of the cross section when rotated through a unit radian about the x-x axis.

The 0ϕ diagram, Fig.4.3.14(b), is obtained by giving a unit displacement to each

member of the cross section in a positive direction.

Using Morh’s integral chart (appendix A), we have

ϕ∫2

0 dt =1.992 , 0 1ϕ ϕ∫ dt =1.839

From eqn(4.3.10) α = -0.923

From eqn (4.3.11) 1ϕ = 1ϕ -0.923 0ϕ (4.3.13)

A plot of eqn (4.3.13) gives exactly Fig.4.3.8 (b) thus, confirming that 1ϕ diagram is

correct.

From eqn. (4.3.10) we obtain

0 1ϕ ϕ∫ dt = (1.0)(-0.332)(9.15)(0.097)+½(1.0)(2.715-0.332)(3.557)(0.016)x2 +

+ (1.0)(2.718)(3.66)(0.016) = 0

By inspection it can also be seen that

1

+

2

3 4

5 6

+

+

+

1.0

1.0

1.0

1.0

(b) Orthogonal shape function 0ϕ

(Unit displacement diagram)

Fig.4.3.13 Auxiliary shape function and orthogonal shape function

1

+

2

3 4

5 6

+

+

3.05 3.05

(a) Auxiliary shape function 1ϕ

3.05

z

y

66

ϕ ϕ∫ 0 2dt = 0 and ϕ ϕ∫ 0 3dt = 0 are satisfied. Thus the accuracy of the strain mode

diagrams is confirmed.

4.4 COMPUTATION OF VLASOV’S COEFFICIENTS

As explained earlier, the coefficients ij ij kj iha , b , c , c and khr , of the differential

equations of equilibrium are computed with the aid of Morh’s integral chart.(Appendix

Three). Expanding eqn (3.4.13), the coefficients are obtained as shown in Table 4.1

The detailed calculations are shown in Appendix Two

Table 4.1a: Summary of Vlasov’s aij coefficients for various sections

Coefficient Elements

Mono-symmetric sections Doubly-symmetric Sections

Non-Symmetric Section

(a) Single Cell

(b) Double Cell

(c) Single Cell

(d) Multi- Cell

(e) Single Cell

11a 6.453

7.023

10.930

9.900

5.264

12 21a a=

0

0

0

0

1.536

13 31a a=

0

0

0

0

-0.001

22a 25.05 25.073 27.994 27.000 11.018

23 32a a=

-0.270

0.425

0

0

-0.038

33a 0.757 0.750 6.721 127.752

0.293

Table 4.1b: Summary of Vlasov’s bij coefficients for various sections


Mono-symmetric Sections Doubly-symmetric Sections

Non Symmetric Section

(a) Single Cell

(b) Double Cell

(c) Single Cell

(d) Multi-Cell

(e) Single Cell

11b 1.060 1.680 1.200 2.400 1.132

12 21b b=

0 0 0 0 0.314

0 0 0 0 -0.037

67

Table 4.1c: Summary of Vlasov’s Ckj coefficients for various

sections

Table 4.1d: Summary of Vlasov’s Cih coefficients for various

sections

13 31b b=

22b

2.982 2.982 2.880 3.600 2.018

23 32b b=

-1.066 -0.449 0 0 0.028

33b 1.407 1.533 3.736 55.727 0.244


Mono-symmetric Sections

Doubly-symmetric Sections


(a) Single Cell

(b) Double Cell

(c) Single Cell

(d) Multi-Cell

(e) Single Cell

11c 1.060

1.680

1.200

2.400

1.132

12 21c c=

0 0 0 0 0.314

13 31c c=

0 0 0 0 -0.037

22c 2.982 2.982 2.880 3.600 2.018

23 32c c=

-1.066 -0.449 0 0 0.028

33c 1.407 1.533 3.736 55.727 0.244

41c 0 0 0 0 0.148

42c -1.561 1.112 0 0 0.476

43c 1.265 1.295 3.732 24.759 0.262

Coefficient Mono-symmetric Doubly- Non Symmetric

68

Elements Sections symmetric Sections

Section

(a) Single Cell

(b) Double Cell

(c) Single Cell

(d) Multi-cell

(e) Single Cell

11c

1.060

1.680

1.200 2.400 1.132

21 12c c=

0 0 0 0 0.314

31 13c c=

0 0 0 0 -0.037

41 14c c=

0 0 0 0.148

22c

2.982 2.982 2.880 3.600 2.018

32 23c c=

-1.066 -0.449 0 0 0.028

33c

1.407 1.533 3.736 55.727 0.244

24c

-1.561 3.4041.112 0 0 0.476

34c

1.265 1.295 3.732 24.759 0.262


Mono-symmetric Sections Doubly-symmetric Sections


(a) Single Cell

(b) Double Cell

(d) Single Cell

(e) Multi-Cell

(f) Single Cell

11r

1.060

1.680

1.200

2.400 1.132

12 21r r= 0 0 0 0 0.314

13 31r r= 0 0 0 0 -0.037

14 41r r= 0 0 0 0 0.148

Table 4.1e: Summary of Vlasov’s rkh and hks coefficients for various

sections

69

4.4.1 The hks Coefficients

The coefficients hks given by eqn.(3.4.13(f)) depend on the bending deformation

of the strip characterized by kM (for k = 1, 2, 3, 4). To compute the coefficients we

need to construct the diagram of the bending moments due to the strain modes,

1 2 3 4ψ , ψ , ψ and ψ ,. Incidentally, 1ψ , 2ψ and 4ψ strain modes do not generate

distortional bending moments as they involve pure bending and pure rotation. Only

3ψ strain mode generates distortional bending moment which can be evaluated

using the distortion diagram for the relevant cross section. Consequently the relevant

expression for the coefficient becomes

∫ 3 3hk kh

ss

M (s)M (s)1s = s =

E EI, (4.4.1)

where 3M (s) is the distortional bending moment of the relevant cross section.

22r 2.982 2.982 2.880 3.600 2.018

23 32r r=

-1.066 -0.449 0 0 0.028

24 42r r=

-1.561 1.112 0 0 0.476

33r

1.407 1.538 3.736 57.727 0.244

34 43r r=

1.265 1.295 3.732 24.759 0.262

44r

14.616 14.485 22.032 35.100 10.358

=Shk

0.261x Is 0.723x Is 1.593x Is

65.065x Is

-

70

The hks coefficients for mono symmetric and doubly symmetric sections are shown in

Table 4.1e.

4.5 DISTORTIONAL BENDING MOMENT

The procedure for evaluation of distortional bending moment is not straight

forward, but attempt is made here to outline the procedure, backed up with numerical

calculations.

Step 1: Obtain the distortion diagram from numerical differentiation of the warping

function and indicate the distortional movement directions of the members of the

cross section with arrows, Fig.4.5.1 (a)

Step 2: Draw the base system for the frame (Fig.4.5.1 (b)) assuming all the joints are

fixed and use vectors, R1, R2,…Rn , to represent the magnitudes and directions of the

forces required to give unit translations to the joints in the direction of the forces. The

directions of the forces are guided by the directional movements of the members of

the cross section at distortion.

Step 3: Apply unit rotation to each of the joints, one after the other, and designate

the forces required to accomplish this by X1, X2, …Xn, Fig.4.5.1(b). Obtain the

bending moment diagram for the frame and by considering the equilibrium of each

joint determine the un-balanced moment at each joint. These are then considered as

coefficients ijr in the equilibrium equation.

Note that _

iM denotes the bending moment diagram due to unit rotation of joint i. A

system with n number of joints will have n bending moments and ( n x n) coefficients.

Step 4: Apply unit transverse displacement in the directions of R1 ,R2…Rn, one at a

time and draw the bending moment diagrams generated on the frame.

Step 5: Establish the equilibrium equations based on equality of internal work ij ijr X ,

and external work i iR∆ , eqn.(4.5.3).

Step 6: For each bending moment diagram due to unit joint translation determine the

un-balanced reaction ijR (coefficient) using static equilibrium of the joints.

Step 7: Multiply the coefficients ijR with the actual values of the distortion i

∆ ,

obtained from the distortion diagram to obtain values of ij iR ∆ and sum up to obtain,

for each joint the value of ∑ ij iR ∆ , Table 4.2.

71

Step 8: Substitute the values of ijr and∑ ij iR ∆ into the equilibrium equation and solve

for the unknown reactions, 1 2 nX , X , ...X .

Step 9: The final bending moment M due to distortion of the cross section is obtained

from the relation;

∑ ∑_ _

i Rii iM(s) = M X + M ∆ where i 3i∆ =ψ values. (4.5.1)

4.5.1 Evaluation of distortional bending moment for single cell mono-symmetric section

For uniform plate thickness top-flange bot-flange webI = I = I = I

Top Flange; 2

4EI/L = 0.546EI

2EI/L = 0.273EI

6EI/L = 0.112EI

Bottom Flange; 2

4EI/L =1.093EI

2EI/L = 0.546EI

6EI/L = 0.448EI

Web 2

4EI/L =1.125EI

2EI/L = 0.562EI

6EI/L = 0.474EI

+

1 2

3 4

5 6

-0.159

(a) Distortion Diagram 3ψ (s)

+ -

-

y

z

-0.492

-0.417

0.123

+0.417

915 915 7320

(b) Base System

3050

1

2

3

4

5 6

3660

X2

X1

X3

X4

R4

R3

R1

R2

Fig. 4.5.1 Distortion diagram and base system for evaluation of hks

2

4

1

3

0.582

0.546

_

1M

1.125

X 1

21r

31r

41r

11r

0.546EI

0.273EI

1.125EI

0.562EI

72

32r

42r

12r

0.273EI

0.546EI

1.125EI

0.562EI

22r 1 2

4

3

0.546EI

+0.273EI

1.125EI

0.562EI

_

2M

X 2

(b) Bending moment due to unit rotation of joint 2 and joints equilibrium diagram

1 2

4

3

0.562EI

_

3M

X 3

1.125EI

1.093EI

0.546EI

23r

0.56EI

33r

43r

13r

1.125EI

0.546EI

1.093EI

(c) Bending moment due to unit rotation of joint 3 and joints equilibrium diagram

24r

34r

44r

14r

0.546EI

1.093EI

0.562EI

1.125EI

1 2

4

3

0.562EI

_

4M

X 4

1.125EI

0.546EI

1.093EI

(d) Bending moment due to unit rotation of joint 4 and joints equilibrium diagram

Fig.4.5.2 Bending moments due to unit rotation of joints and joints equilibrium diagrams for single cell mono-symmetric section

1 2

3 4

R 1

∆1=1 ∆’

∆’

β

Unit Displacement of Joint 1

R 21

R 31 R 41

R 11

0.406EI 0.406EI

Joint Equilibrium

1 2

3 4

0.406EI

0.406EI 0.406EI

Moment Diagram

1M

R

73

(b) Unit Displacement of Joint 1 in R2 Direction, Moment Diagram and Joint Equilibrium

1 2

3 4

∆2=1

R2


in R 2 Direction

R 22

R 32

R 42

R 12

0.384EI

0.096EI

Joint Equilibrium

4

1 2

3

0.096EI

R 22

R12

R42

R32

0.096EI

0.384EI 0.384EI

Moment Diagram

4

2M

R

(c) Unit Displacement of Joint 3 in R3 Direction, Moment Diagram and Joint Equilibrium

R 23

R 33 R 43

R 13

0.406EI 0.406EI

Joint Equilibrium

1 2

3

4

∆3=1

R3


in R 3 Direction

R 23 1 2

3 4

0.406EI

0.406EI 0.406EI R 13

R 43 R 33

Moment Diagram

3M

R

(d) Unit Displacement of Joint 3 in R4 Direction, Moment Diagram and Joint Equilibrium

1 2

3 4

∆4=1


R4

R 24

R 34

R 44

R 14 0.096EI

0.384EI Joint Equilibrium

Moment Diagram

R 24 1 2

3

0.096EI

R14

R44

R34

0.096EI

0.384EI

0.384EI 4

4M

R

74

The bending moment diagrams due to unit joint rotations and the static joints equilibrium are shown in Fig.4.5.2. By considering the joint equilibrium diagrams for

_ _ _

1 2 3M , M , M and _

4M we obtain the following coefficients as reactions.

r11 = 1.671EI; r21 = 0.273EI; r31 = 0; r41 = 0.562EI r12 = 0.273EI; r22 = 1.671EI; r32 = 0.562EI; r42 = 0 r13 = 0; r23 = 0.562EI; r33 = 2.218EI; r43 = 0.546EI r14 = 0.562EI; r24 = 0; r34 = 0.546EI; r44 = 2.218EI From the joint equilibrium diagrams in Fig.4.5.3 we obtain the following coefficients. R11 = 0.406EI R21 = 0.406EI; R31 = 0.406EI; R41 = 0.406EI

R12 = - 0.096EI R22 = 0.096EI; R32 = -0.384EI; R42 = -0.384EI

R13 = 0.406EI R23 = 0.406EI; R33 = 0.406EI; R43 = 0.406EI

R14 = -0.096EI R24 = -0.096EI; R34 = -0.384EI; R44 = 0.384EI

Since the moment diagrams in Fig. 4.5.3 are based on unit translation of the joints in

the direction of the members’ displacements we now multiply these unit diagrams by

the actual displacements (distortion) of the members of the cross section obtained

from the distortion diagram, i.e.

Member 1 – 2 displaced ∆1 = 0.159m, left

Member 2 – 3 displaced ∆2 = 0.417m, down

Member 3 – 4 displaced ∆3 = 0.492m, right

Member 4 – 1 displaced ∆4 = 0.417m, up

Table 4.2 shows the summation of ij i

R ∆ for the single cell mono-symmetric section.

The equations of equilibrium is given by eqn (4.5.3) Equilibrium equations

11 1 12 2 13 3 14 4 1 1r X +r X +r X +r X +R ∆ = 0

21 1 22 2 23 3 24 4 2 2r X +r X +r X +r X +R ∆ = 0

31 1 32 2 33 3 34 4 3 3r X +r X +r X +r X +R ∆ = 0

41 1 42 2 43 3 44 4 4 4r X +r X +r X +r X +R ∆ = 0 (4.5.2)

75

Table 4.2 Summation of work done on the joints due to distortion of single cell mono symmetric section

Substituting the coefficients ijr and ∑ ij 1R ∆ into the equilibrium eqn.(4.5.2) and

solving gives; 1 2X = X = -0.1068 and 3 4X = X = 0.042 .

The final bending moment Fig.4.5.4(a) is obtained by summation of the products

of 1 2 3 4M , M , M and M diagrams and 1 2 3X , X , X & 4X respectively plus the summation of

the products of R1 R2 R3M , M , M & R4M diagrams and 1 2 3∆ , ∆ , ∆ & 4∆ respectively as per

eqn. (4.5.2) The distortional bending moment diagrams for single cell and double cell

mono-symmetric sections are shown in Fig.4.5.4 (a) and (b) respectively. From eqn.

(4.5.1) we obtain the hks coefficients for the mono-symmetric sections using Morh’s

integral chart (Appendix Three).

Joints

Work done ij iR ∆

1 2 3 4 R1j ∆1 R2j ∆2 R2j ∆3 R4j ∆4

1 0.406EI 0.406EI 0.406EI 0.406EI 2 -0.096EI -0.096EI -0.384EI -0.384EI 3 0.406EI 0.406EI 0.406EI 0.406EI 4 -0.096EI -0.096EI -0.384EI -0.384EI

∑ ij iR ∆ = 0.184EI 0.184EI -0.056EI -0.056EI

(a) Single Cell Mono-symmetric Section

1 2

3

0.167EI

0.251EI

0.251EI 0.251EI

0.167EI 0.167EI

4

3333M(S)M(S)M(S)M(S)

Fig.4.5.4 Bending moments due to distortion of mono-symmetric Sections

(b) Double Cell Mono-symmetric Section

0.324

-0.243

+

+

- - - 1 2 3

4

5 6

-

+

+ +

+ +

-

-

-

+

-0.560

0.280

0.280

0.324

0.229

0.229 -0.458 0.243


76

4.5.2 Evaluation of distortional bending moments for doubly symmetric

Sections

The distortion diagram and base system for evaluation of distortional bending

moment for multi-cell doubly symmetric section is shown in Fig.4.5.5. To determine

the distortional bending moment, we follow the same procedure outlined in section

4.5. For uniform plate thickness top-flange bot-flange webI = I = I = I

Flanges; ⇒4EI/h =1.333EI 2EI/h = 0.667EI , 26EI/h = 0.667EI

Webs ; ⇒4EI/b =1.333EI 2EI/b = 0.667EI , 26EI/b = 0.667EI

The bending moment diagrams due to unit joint rotations are shown in Fig.4.5.6.

By considering the joints equilibrium diagrams for 1 2 3M , M , M and 4M we obtain the

values of the coefficients as reactions. Thus,

From M1 diagram; 3111 21 81 41 51 61 71r = 2.667EI, r = 0.667EI, r = 0.667, r = r = r = r = r = 0

From M2 diagram; 12 22 32 72r = 0.667EI, r = 3.999EI, r = 0.667EI, r = 0.667, others are zero

From M3 diagram, 23 33 43 63r = 0.667EI, r = 3.99EI, r = 0.667EI, r = 0.667EI , others are

zero

From M4 diagram; 34 44 54r = 0.667EI, r = 2.667EI, r = 0.667EI,others are zero

From M5 diagram; 45 55 65r = 0.667EI, r = 2.667EI, r = 0.667EI, others are zero

Fig.4.5.5 Distortion diagram and base system for evaluation of hkS coefficient

(b) Base system

1.929

2.014 5.871

5.871 1.929

(a) Distortion Diagram

2.014 o

o o

o o

R 1 R 2 R3

R4

R5 R6

X1 X2 X3 X4

X5 X6 X7 X8

77

From M6 diagram; 36 56 66 76r = 0.667EI, r = 0.667EI, r = 3.999EI, r = 0.667EI, others are

zero.

From M7 diagram; 27 67 77 87r = 0.667EI, r = 0.667EI, r = 3.999EI, r = 0.667EI, others are

zero.

From M8 diagram; 18 78 88r = 0.667EI, r = 0.667EI, r = 2.667EI,others are zero.

(h) Bending moment due to unit rotation

7X

z

2 3 4

5 6 7 8

8888MMMM

z

1 2 3 4

5 6 7 8

1

0.667

0.667

0.667

7777MMMM

8X

1.333 1.333

0.667

0.667

z

6666MMMM

z

5555MMMM

0.667

0.667 0.667

0.667

0.667

1.333 1.333 5 5 6 6 7 7 8 8

(e) Bending moment due to unit rotation

of joint 5 (f) Bending moment due to unit rotation

of joint 6

y

z y

z 5X

6X

z z

1M 2222MMMM

(a) Bending moment due to unit rotation

of joint 1 (b) Bending moment due to unit rotation

of joint 2

1.333 0.667

0.667

1.333 0.667

1.333 0.667

0.667

1 2

y

z

y

z

1X 2X

z

3333MMMM

z

4444MMMM

(d) Bending moment due to unit rotation

of joint 4 (c) Bending moment due to unit rotation

of joint 3

1.333 1.333

0.667

0.667

0.667

0.667

0.667 3 4

y

z

y

z

4X 3X

78

1 2 3 4

5 6 7 8

1 2 3 4

5 6 7 8 4 o

0.667EI 0.667EI

R14 R24 R34

R44 RRRR4444MMMM

z

1 2 3 4

5 6 7 8

z

1 2 3 4

5 6 7 8

o

3

(e) Unit transverse displacement of joint

2

( Strain mode 3 )

0.667EI

0.667EI 0.667EI

0.667EI 0.667EI

(f) Bending moment diagram due to unit

displacement of joint 2 (Strain Mode 3)

R33

R23

R13

R83

R73

R63

0.667EI

R3R3R3R3MMMM

z

1 2 3 4

5 6 7 8

z

1 2 3 4

5 6 7 8

o

2

(c) Unit transverse displacement of joint

1

(Strain mode 2)

(d) Bending moment diagram due to unit


0.667EI

0.667EI

0.667EI

0.667EI

R12

R22

R72 R82

R2R2R2R2MMMM

z

1 2 3 4

5 6 7 8

z

1 2 3 4

5 6 7 8

1

0.667EI

0.667EI (a) Unit transverse displacement of joint

1 (Strain mode 1 ) (b) Bending moment due to unit

displacement of joint 1 (Strain mode 1)

o

0.667EI

0.667EI

R 11 R 21 R 31 R 41

R 51 R 61 R 71

R 81

R1R1R1R1MMMM

79

Similarly, a consideration of the equilibrium of joints for R1, R2, R3, R4, R5, and R6

moment diagrams gives;

From MR1 diagram; 11 21 31 41 51 61 71 81R =R =R =R =R =R =R =R = 0.667EI

From MR2 diagram; 12 22 72 82R =R =R =R = -0.667EI , 32 42 52 62R =R =R =R = 0

From MR3 diagram; 13 83R =R = 0.667EI , 33 63R =R = -0.667EI , 23 43 53 73R =R =R =R = 0

From MR4 diagram; 14 24 34 44 54 64 74 84R =R =R =R =R =R =R =R = 0.667EI

From MR5 diagram; 35 45 55 65R =R =R =R = -0.667EI , 15 25 75 85R =R =R =R = 0

From MR6 diagram; 26 76R =R = -0.667EI , 46 56R =R = 0.667EI , 16 36 66 86R =R =R =R = 0

Fig.4.5.7 Unit joint displacements and corresponding bending moment diagrams ( multi cell doubly symmetric section)

z

6

(k) Unit transverse displacement of joint

6

( Strain mode 6 )

1 2 3 4

5 7 8

z

1 2 3 4

5 6 7 8

(l) Bending moment diagram due to unit


0.667EI

0.667EI

0.667EI

0.667EI

0.667EI

R26

o

6

R36

R46

R56

R66

R76

R6R6R6R6MMMM

z

1 2 3 4

5 6 7 8

(j) Bending moment diagram due to unit


0.667EI

0.667EI

0.667EI

0.667EI

z

1 2 3 4

5 6 7 8

5

(i) Unit transverse displacement of joint

5

( Strain mode 5 )

o

R55

R45

R65

R35 R5R5R5R5MMMM

80

Since the moment diagrams in Fig. 4.5.7 are based on unit translation of the joints in

the direction of the joints displacements we now multiply these unit diagrams by the

actual displacements (distortion) of the joints obtained from the distortion diagram.

Table 4.3 shows the summation ofij i

R ∆ for the multi-cell doubly symmetric section.

The equations of equilibrium is given by eqn. (4.5.4)

Equilibrium equations

11 1 12 2 13 3 14 4 15 5 16 6 17 7 18 8 1r X +r X +r X +r X +r X +r X +r X +r X +R ∆ = 0







81 1 82 2 83 3 84 4 85 5 86 6 87 7 88 8 8r X +r X +r X +r X +r X +r X +r X +r X +R ∆ = 0 (4.5.4)

Substituting the coefficients ijr and ∑R ∆ij i into the equilibrium eqn. (4.5.4) and solving

we obtain; 1 4 5 8X = X = X = X = -0.1033 and 2 3 6 7X = X = X = X = 0.5166

The final bending moment Fig.4.5.8(a) is obtained by adding the products of

1 2 3M , M , M , 4M , 5 6 7 8M , M , M , M diagrams and 1 2 3X , X , X , 4X 5 6 7 8X , X , X , X

respectively to the products of R1 R2 R3M , M , M , R4M , R5 R6M , M diagrams and

1 2 3∆ , ∆ , ∆ , 4∆ , 5 6∆ , ∆ respectively as per eqn.(4.5.2).

Table 4.3 summation of work done on the joints due to distortion

Work done ( R ∆ij i )

Joint disp direction

Distorti0n ∆ R1j ∆1 R2j ∆2 R3j ∆3 R4j ∆4 R5j ∆5 R6j ∆6 R7j ∆7 R8j ∆8

1 1.929 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667

2 5.871 -0.667 -0.667 0 0 0 0 -0.667 -0.667

3 2.014 0.667 0.00 -0.667

0 0 -0.667 0 0.667

4 1.929 0.6670 0.667 0.667 0.667 0.667 0.667 0.667 0.667

5 5.871 0 0 - -0.667 -0.667 -0.667 0 0

81

The distortional bending moment diagrams for multi cell and single cell doubly symmetric sections are shown in Fig.4.5.8 (a) and (b) respectively. From eqn.(4.5.1)

we obtain the hks coefficients for the doubly symmetric sections using Morh’s integral

chart (appendix A).

4.6 INTERACTION OF STRAIN MODES

Vlasov’s coefficients of differential equations of equilibrium involve a combination

of elementary displacements. In the general strain mode there are three strain fields

in the longitudinal direction and four in the transverse direction. These are;

(i) Out of plane displacement due to vertical load characterized by 1ϕ

(ii) Out of plane displacement due to horizontal load normal to bridge axis,

characterized by 2ϕ

(iii) Out of plane displacement due to warping of the cross section, characterized by

3ϕ

(iv) In-plane displacement due to vertical load, characterized by 1ψ

0.667

6 2.014 0 -0.667 0 0.667 0.667 0 -0.667 0

∑R ∆ =ij i 0.000 -2.686 -2.686

0.000 0.000 -2.686 -2.686 0.000

Fig.4.5.8 Bending moment diagrams due to distortion of doubly-symmetric sections

_ _ +

+

+ + + +

_ _

_ 2.366EI

2.366EI

2.366EI -3.606EI -3.606EI

_

1.654EI

2.366EI

1.953EI

-3.606EI

2.366EI

-3.606EI

1.654EI

-2.366EI

1.953EI

+

_ _

+ + +

+

1.953EI 1.273EI


1.654EI 1.953EI

(a) Multi- cell section


(b) Single cell section

0.484EI

0.484EI

0.484EI

0.484EI

82

(v) In-plane displacement due to horizontal load normal to bridge axis, characterized

by 2ψ

(vi) In-plane displacement due to distortion of the cross section, characterized by 3ψ

(vii) In-plane displacement due to pure rotation, characterized by 4ψ

Some or all of these strain modes may be present in a given frame depending on the

geometry of the cross section and the nature of loading.

In section 2.5 we noted that torsional loads consisting of opposing vertical forces

result from gravity loads that are eccentric to the centre line of the girder and they

give rise to bending and torsion. We also noted that a torsional load can be modeled

as a uniform (Saint Venant) torsion and a distortional component. Therefore an

eccentric load on a bridge structure introduces interaction between bending, pure

torsion and distortion. The torsional- distortional interaction is of great interest

particularly in a thin walled box structure where the geometry of the cross section

comes into play.

From the computations of Vlasov coefficients in section 4.4 it is observed that

certain coefficients have zero values in the case of mono-symmetric and doubly-

symmetric sections, the number of cells not withstanding. These are

12 13 12 13 12 13 12a , a , b , b , c , c , r and 13r for mono-symmetric section. For doubly

symmetric sections we have 23 23 23 24 14 23 24a , b , c , c , c , r , r and 14r , in addition to the

zero coefficients for mono-symmetric sections. These coefficients with zero values

indicate that there are no interactions between the displacement parameters

represented by the coefficients. For example, the value of the coefficient c23 indicates

the degree of interaction between the in-plane displacement 2ψ due to load normal

to bridge axis and the in-plane displacement 3ψ due to distortion of the cross section.

Where there is no interaction between one strain mode and another the value of the

relevant coefficient is zero indicating that the analysis of the girder for such strain

modes can be done independent of each other.

A doubly symmetric section has only one interaction of strain modes, i.e.

torsional strain mode and distortional strain mode interaction. A mono-symmetric

section on the other hand, has three strain modes interactions. Torsion interacts with

distortion and each of these interacts with flexure about the non axis of symmetry.

Thus we have torsional-distortional interaction, flexural-torsional interaction, and

83

flexural-distortional interaction. The analysis of such sections requires solution of

three coupled differential equations of equilibrium.

A non symmetric section has multiple strain modes interactions: Each of torsion

and distortion interacts with flexure about both axes of non-symmetry in addition to

the interaction between themselves. Thus, we have torsiona-distortional interaction,

torsional-flexural interaction in major and minor axes and distortional-flexural

interaction in major and minor axes. These strain modes interactions are quite

inseparable that it is not possible to examine one strain modes interaction in isolation

of others.

84

(3.4.18b)

85

CHAPTER FIVE

FORMULATION OF DIFFERENTIAL EQUATIONS OF EQUILIBRIUM FOR VARIOUS BOX GIRDER SECTIONS

5.1 PREAMBLE

General expressions for the differential equations of distortional equilibrium for

box girders were derived in chapter three while the coefficients for various cross

sections (doubly symmetric, mono symmetric and non symmetric sections) were

evaluated in chapter four

In this chapter, specific expressions for different conditions of equilibrium are

obtained for doubly symmetric, mono-symmetric and non symmetric box girder

structures. These equilibrium conditions cover all possible strain modes interactions

such as;

(a) Flexural-torsional interaction.

(b) Flexural-distortional interaction.

(c) Torsional-distortional interaction.

(d) Flexural-torsional-distortional interaction.

5.2 DOUBLY SYMMETRIC SECTIONS

5.2.1 Flexural - torsional equilibrium

The relevant coefficients for flexural – torsional equilibrium for doubly symmetric

sections are 22 22 22 22 24 42 44a , b , c , r , r = r , and r . For doubly symmetric sections there

are no interactions between flexure and torsion (strain modes 2 and 4).

Consequently, the coefficients 24 42 24 42c , c , r , r are zero.

Substituting these relevant coefficients into the matrix notation eqn (3.4.17) we

obtain

22 22

''U U10 0 0 0 1

''κ 0 0 U - 0 0 U2 2

0 0 0 0 0 0'' UU 33

0 0

a b -

'

1

'

222 '

3

'

4

V0 0 0 0

V0 c 0 0 = 0

V0 0 0 0

V

(a)

and

22

'0 0 U1

0 0 'U

20 0 0'

U0 0 0 3

0

c-

V0 0 0 0 1

V0 0 0 0 2κ0 0 0 0 V

30 0 0 0 V

4

+

22

44

''V10 0 0''

V0 0 0 2''0 0 0 0 V3

0 0 0 ''V

4

0

r

r

0

=

1

2

3

4

q

q1+

qG

q

(b)

86

By multiplying out matrix equations (a) and (b) above we obtain the following three

equations.

22 2 22 2 22 2ka U '' -b U -c V ' = 0 (5.2.1)

222 2 22 2

qc U '+r V '' = -

G (5.2.2)

444 4

qr V '' = -

G (5.2.3)

From eqn.(5.2.2) we obtain;

22 22 2

22 22

r qU ' = - V '' -

c c G (5.2.4)

Differentiating eqn.(5.2.4) twice gives,

iv22

2 2

22

rU ''' = - V

c (5.2.5)

Differentiating eqn (5.2.1) once we obtain,

22 2 22 2 22 2κa U ''' -b U ' - c V '' = 0 (5.2.6)

Substituting eqns (5.2.4) and (5.2.5) into eqn.(5.2.6) we obtain,

22

iv22 22 222 2 22 2 22 22 2

22 22

r r qκa - V -b - V '' -b - - c V '' = 0

c c c G (5.2.7)

Noting that 22 22 22b = c = r and k = E/G , we obtain from eqn.(5.2.7);

iv 22

22

qV =

a E (5.2.8)

From eqn.(5.2.3) we obtain that

44

44

qV '' = -

r G (5.2.9)

Equations (5.2.8 and 5.2.9) are the differential equations of equilibrium for flexural-

torsional response of a doubly symmetrical box girder section.

5.2.2 Flexural - distortional equilibrium (doubly symmetric sections)

The relevant coefficients for flexural-distortional behaviour of doubly symmetric

sections are those involving strain modes 2 and 3. These are

87

, ,22 22 22 33 33 33a , b c , a , b , c , 33s and 33r . All other interacting coefficients,

23 23 23 23a , b , c , r , are zero because of symmetry in major and minor axis.

Substituting these coefficients into the matrix notation eqn (3.4.17) we obtain;

''' 11 1 ''' 2

22 2 22 2 22 ''' 3

33 33 33 3 33 '

4

V0 0 0 U 0 0 0 U 0 0 0 0

Vκ 0 a 0 U - 0 b 0 U - 0 c 0 0 = 0

V0 0 a U 0 0 b U 0 0 c 0

V

and

'

1

22 '

2

33 '

3

0 0 0U

0 c 0U -

0 0 cU

0 0 0

1

2

333

4

V0 0 0 0

V0 0 0 0k

V0 0 s 0

V0 0 0 0

+

''

1

''22 2

''33 3

''

4

0 0 0 0 V

0 r 0 0 V

0 0 r 0 V

0 0 0 0 V

1

2

3

4

q

q1+ = 0

qG

q

Expanding gives,

22 2 22 2 22 2κa U '' -b U -c V ' = 0 (5.2.9)

33 3 33 3 33 3κa U '' -b U -c V ' = 0 (5.2.10)

c 222 2 22 2

qU '+r V '' = -

G (5.2.11)

333 3 33 3 33 3

qc U '+r V '' - ks V = -

G (5.2.12)

From eqn. (5.2.11) we have,

22 2

22

qU ' = -V '' -

c G (5.2.13)

⇒ iv

2 2U ''' = -V (5.2.14)

Differentiating eqn. (5.2.9) once and substituting eqns. (5.2.13 & 5.2.14) we obtain

iv 222 2 22 2 22 22 2

22

qκa (-V ) -b (-V '') -b - -C v '' = 0

c G

⇒

iv 22 222 2 22 2 22 2

22

b q- κa V +b V ''+ - C v '' = 0

c G

⇒ =iv 222 2

q ka V

G or =iv 2

2

22

qV

ka G

Substituting E

k =G

we obtain iv 22

22

qV =

a E (5.2.15)


88

33 33 3 3

33 33

ks qU ' = -V '' + V -

c c G (5.2.16)

iv 333 3 3

33

ksU ''' = -V + V ''

c (5.2.17)

Differentiating eqns. (5.2.10) once we obtain;

33 3 33 3 33 3κa U ''' -b U ' - c V '' = 0 (5.2.18)

Substituting eqns.(5.2.16 & 5.2.17) into eqn.(5.2.18) we obtain

( ( )+ − −2

iv 33 33 22 33 333 3 3 33 3 3 33 33 3

33 33 33

a k s b ks qka V V '' -b V '') - V -b - c V '' = 0

c c c G

⇒ −2

iv 33 33 333 3 3 33 3

33

k a s qka V V ''+ks V =

c G

β⇒ −iv 2 4

3 3 3 3V α V ''+ 4 V = K

where, 2 333

33

ksα =

r

4 333

33

s4β =

a and 3

3

33

qK =

Ea

Thus, for flexural distortional equilibrium we have

iv 22

22

qV =

a E (5.2.19)

iv 2 4

3 3 3 3V -α V ''+ 4β V = K (5.2.20)

5.2.3 Torsional-distortional equilibrium (doubly symmetric section)

The relevant coefficients for torsional-distortional equilibrium are those involving

strain mode 4 ( for pure torsion), and strain mode 3 (for distortion). There will

however be additional coefficients due to the interaction between torsion and

distortion. These non-zero interacting coefficients modifying the displacement

quantities V3 (for distortion) and V4 (for torsion) are 34 43 34c ,c ,r , and 43r .Their values

are obtained as given in table 4.1. Thus, the relevant coefficients for torsional-

distortional behaviour of doubly symmetric sections are,

33 33 33 34 43 33 34 43 33a , b , c , c , c , r , r , r , s and 44r . Substituting these coefficients into

matrix notation eqn.(3.4.17) we obtain,

89

1

1 1

2

2 2

3

33 3 33 3 33 34

4

V '0 0 0 U '' 0 0 0 U 0 0 0 0

V 'k 0 0 0 U '' - 0 0 0 U - 0 0 0 0 = 0

V '0 0 a U '' 0 0 b U 0 0 c c

V '

1 1 1

1

2 2 2

2

33 33 34 3 3 333

3

43 43 44 4 4 4

0 0 0 0 0 0 0 V '' V q0 0 0 0U '

0 0 0 0 0 0 0 V '' V q0 0 0 0 1U ' + -k + = 0

0 0 c 0 0 r r V '' V q0 0 s 0 GU '

0 0 c 0 0 r r V '' V q0 0 0 0

Expanding the matrix notation we obtain the following equations.

33 3 33 3 33 3 34 4ka U '' -b U - c V ' - c V ' = 0 (5.2.21)

333 3 33 3 34 4 33 3

qc U '+r V ''+r V '' -ks V + = 0

G (5.2.22)

443 3 43 3 44 4

qc U '+r V ''+r V ''+ = 0

G (5.2.23)


44 43 3 4

43 43

r qU ' = -V '' - V '' -

c c G (5.2.24)

From eqn.(5.2.22) we obtain

34 33 33 3 4 3

33 33 33

r ks qU ' = -V '' - V ''+ V - = 0

c c c G (5.2.25)

Subtracting eqn. (5.2.25) from eqn.(5.2.24) gives

+

34 33 344 44 3

33 43 33 43 33

r ks qr q- V '' - V - = 0

c c c c c G (5.2.26)

Let γ

34 33 344 41 1 1

33 43 33 33 43

r ks qr q- =β ; = ; - =K

c c c c G c G

Eqn. (5.2.26) becomes

γ1 4 1 3 1β V '' - V +K = 0 (5.2.27)

Differentiating eqn. (5.2.21) once with respect to (x) we obtain;

33 3 33 3 33 3 34 4ka U ''' -b U ' - c V '' - C V '' = 0 (5.2.28)

Differentiating eqn. (5.2.24) twice gives

iv iv443 3 4

43

rU ''' = -V - V

c (5.2.29)

Substituting eqn. (5.2.24) and (5.2.29) into eqn. (5.2.28) we obtain,

90

iv iv44 44 433 3 33 4 33 3 33 4 33 33 3 34 4

43 43 43

r r qka (-V )+ka - V -b (-V '') -b - V '' -b - - c V '' - c V '' = 0

c c c G

⇒ iv iv33 44 33 44 33 433 3 4 33 3 4 33 3 34 4

43 43 43

ka r b r b q-ka V - V +b V ''+ V '' - c V '' - c V '' = -

c c c G

⇒ + −

iv iv33 44 33 44 33 433 3 4 34 4

43 43 43

ka r b r b qka V V + c V '' =

c c c G

⇒ +

iv iv 34 43 33 44 3344 43 4 4

43 33 43 33 43

c c -b r br qV V + V '' =

c ka c ka c G

⇒ iv iv

3 1 4 2 4 2V +α V +β V '' = K

where,

441

43

rα = ;

c

34 441

33 43

r rβ = - ;

c c 34 43 33 44

2

33 43

c c -b rβ = ;

ka c γ 33

1

33

ks= ;

c

3 41

33 43

q qK = -

c G c G; 33 334 4

2

33 43 33 43

b bq qK = =

ka c G a c E

Thus the set of differential equations governing torsional-distortional behavior of

doubly symmetric sections is

γ1 4 1 3 1

iv iv

3 1 4 4 2

β V '' - V +K = 0

V +α V +βV '' =K (5.2.30)

5.2.4 Flexural-torsional-distortional equilibrium (doubly symmetric sections)

The relevant coefficients for flexural – torsional – distortional equilibrium are those

involving strain modes 2, 3, and 4. These are 22 23 33 22a , a , a , b ,

23b , 22c , 23 22c , r , 33 33b , c , 24 42 34 43 34 43c = c , c = c , r = r , 24 42r = r , 44 33r and s . Because of

symmetry we note that there is no interaction between strain modes 2 and 3 and

between strain modes 2 and 4. Consequently, 23 23 23 23a = b = c = r = 0 and

24 42 24 42c = c = r = r = 0 .

Substituting these coefficients into matrix eqn. (3.4.17) we obtain,

1 1

22 2 22 2

33 3 33 3

0 0 0 U '' 0 0 0 U

κ 0 a 0 U '' - 0 b 0 U

0 0 a U '' 0 0 b U

-

1

2

22

3

33 34

4

V '0 0 0 0

V '0 c 0 0 = 0

V '0 0 c c

V '

91

1

22

2

33

3

43

0 0 0U '

0 c 0U '

0 0 cU '

0 0 c

-

1

2

33

3

4

0 0 0 0V

0 0 0 0V

κ 0 0 s 0V

0 0 0 0V

+

1

22

2

33 34

3

43 44

4

0 0 0 0V ''

0 r 0 0V ''

0 0 r rV ''

0 0 r rV ''

1

2

3

4

q

q1+ = 0

qG

q

Multiplying out we obtain

22 2 22 2 22 2ka U '' -b U - c V ' = 0 (5.2.31)

33 3 33 3 33 3 34 4ka U '' -b U - c V ' - c V ' = 0 (5.2.32)

0=222 2 22 2

qc U '+r V ''+

G (5.2.33)

333 3 33 3 33 3 34 4

qc U ' -ks V +r V ''+r V '' = -

G (5.2.34)

443 3 43 3 44 4

qc U '+r V '' +r V '' = -

G (5.2.35)

Since there are no interactions between flexure and other strain modes (torsion and

distortion), equations 5.2.31 to 5.2.35 can be grouped into two according to their

strain modes and handled independently. Thus, for flexural strain mode 2 we have

22 2 22 2 22 2ka U '' -b U - c V ' = 0 (5.2.31)

0=222 2 22 2

qc U '+r V ''+

G (5.2.33)

From eqn. (5.2.33), 22 2

22

qU ' = -V '' -

c G

Differentiating twice we obtain;

22 2

22

qU ' = -V '' -

c G and iv

2 2U ''' = -V (5.2.36)

Differentiating eqn. (5.2.31) once we obtain

22 2 22 2 22 2ka U ''' -b U ' - c V '' = 0 (5.2.37)

Substituting eqn. (5.2.36) into eqn. (5.2.37) we obtain

G

iv 222 2 22 2 22 22 2

22

qka (-V ) -b (-V '') -b (- ) - c V '' = 0

c

⇒ iv '' 222 2 22 2 22 2

q-ka V +b V '' - c V + = 0

G

⇒ iv 222 2

qka V =

G or iv 2

2

22

qV =

ka G (5.2.38)

92

For torsional and distortional strain modes 4 and 3 we have

33 3 33 3 33 3 34 4ka U '' -b U - c V ' - c V ' = 0 (5.2.32)

333 3 33 3 33 3 34 4

qc U ' -ks V +r V ''+r V '' = -

G (5.2.34)

443 3 43 3 44 4

qc U '+r V '' +r V '' = -

G (5.2.35)

These equations are the same as the expanded matrix equations 5.2.21, 5.2.22, and

5.2.23 in torsional-distortional equilibrium considerations. Therefore, for doubly

symmetric sections where there are no interactions between flexural strain mode and

other strain modes the governing equations of equilibrium are those obtained for

torsional-distortional equilibrium considerations. That is

γ1 4 1 3 1

iv iv

3 1 4 2 4 2

β V '' - V +K = 0

V +α V +β V '' =K (5.2.39)

where

441

43

rα = ;

c

34 441

33 43

r rβ = - ;

c c 34 43 33 44

2

33 43

c c -b rβ = ;

ka c γ 33

1

33

ks= ;

c

3 41

33 43

q qK = -

c G c G; 33 334 4

2

33 43 33 43

b bq qK = =

ka c G a c E

5.3 MONO-SYMMETRIC SECTIONS

5.3.1 Flexural - torsional equilibrium

The relevant coefficients for flexural – torsional equilibrium are those involving

strain modes 3 and 4, i.e., 22 22 22a , b , c , 24 42 22 24 42c = c ; r , r = r , and 44r . Substituting

these coefficients into the matrix eqn (3.4.17) we obtain

93

22 22

''U U10 0 0 0 1

''κ 0 0 U - 0 0 U2 2

0 0 0 0 0 0'' UU 33

0 0

a b -

'

1

'

222 24 '

3

'

4

V0 0 0 0

V0 c 0 c = 0

V0 0 0 0

V

and

22

42

'0 0 U1

0 0 'U

20 0 0'

U0 0 3

0

c

c

-

V0 0 0 0 1

V0 0 0 0 2κ0 0 0 0 V

30 0 0 0 V

4

+

22 24

42 44

''V10 0 0''

V0 0 2''0 0 0 0 V3

0 0 ''V

4

0

r r

r r

1

2

3

4

q

q1+ = 0

qG

q

. Expanding the above matrix equation we obtain

22 2 22 2 22 2 24 4ka U '' -b U - c V ' - c V ' = 0 (5.3.1)

222 2 22 2 24 4

qc U '+r V ''+r V '' = -

G (5.3.2)

442 2 42 2 44 4

qc U '+r V '' +r V '' = -

G (5.3.3)

From eqn. (5.3.3) we obtain;

44 42 2 4

42 42

iv iv442 2 4

42

r qU ' = -V '' - V '' -

c c G

rU ''' = -V - V

c

(5.3.4)


22 44 22 4 222 2 4 22 2 24 4

42 42

c r c q qc (-V '') - V '' - +r V ''+r V '' = -

c c G G

⇒

22 44 22 4 224 4

42 42

c r c q qr - V '' = -

c c G G

⇒

24 42 22 44 22 4 24

42 42

r c - c r c q qV '' = -

c c G G

⇒

22 4 42 24

42 24 22 44 42 24 22 44

c q c qV '' = -

c r - c r G c r - c r G

⇒ 4 1V '' =K (5.3.6)

where,

22 4 42 21

42 24 22 44 42 24 22 44

c q c qK = -


Differentiating eqn. (5.3.1) once we obtain,

94

22 2 22 2 22 2 24 4κa U ''' - b U ' - c V '' - c V '' = 0 (5.3.7)

Substituting eqns. (5.2.4) and (5.2.5) into eqn. (5.2.7) we obtain,

iv iv44 44 422 2 22 4 22 2 22 22 22 2 24 4

42 42 42

r r qκa (-V )+ka - V -b (-V '') -b - -b - - c V '' - c V '' = 0

c c c G

( )⇒ iv iv 422 42 2 22 44 4 22 44 24 42 4 22

q-κa c V +ka r V - b r - c c V '' = b

G

⇒1

iv iv

2 2 4 1 4 2α V +α V -β V '' =K (5.3.8)

Thus, the differential equations for flexural – torsional equilibrium are

1

4 1

iv iv

2 2 4 1 4 2

V '' =K

α V +α V -β V '' =K (5.3.9)

where,

( )

;

1 22 42 2 22 44 1 22 44 24 42

22 4 42 2 41 2 22

42 24 22 44 42 24 22 44

α =κa c ; α = ka r ; β = b r - c c ;

c q c q qK = - K = b

c r - c r G c r - c r G G

(5.3.10)

5.3.2 Flexural – distortional equilibrium (mono- symmetric Section)

The relevant coefficients are those involving strain modes 2 and 3. These are

22 23 33a , a ; a , 22,b 23 33 23 33 22 23 33 33b , b , c , c , r , r , r , s

Substituting these into matrix eqn. (3.4.17) we obtain

1 1

22 23 2 22 23 2

32 33 3 32 33 3

0 0 0 U '' 0 0 0 U

κ 0 a a U '' - 0 b b U

0 a a U '' 0 b b U

-

1

2

22 23

3

32 33

4

V '0 0 0 0

V '0 c c 0 = 0

V '0 c c 0

V '

1

22 23

2

32 33

3

U 'c c

U 'c c

U '

-

1

2

33

3

4

0 0 0 0V

0 0 0 0V

κ 0 0 s 0V

0 0 0 0V

+

1

22 23

2

32 33

3

4

0 0 0 0V ''

0 r r 0V ''

0 r r 0V ''

0 0 0 0V ''

1

2

3

4

q

q1+ = 0

qG

q


22 2 23 3 22 2 23 3 22 2 23 3ka U ''+ka U '' -b U -b U -c V ' - c V ' = 0 (5.3.11)

U32 2 33 3 32 2 33 3 32 2 33 3ka U ''+ka U '' -b U -b - c V ' - c V ' = 0 (5.3.12)

222 2 23 3 22 2 23 3

qc U '+c U '+r V ''+r V '' = -

G (5.3.13)

95

332 2 33 3 33 3 32 2 33 3

qc U '+c U ' -ks V +r V ''+r V '' = -

G (5.3.14)

From eqn. (5.3.13) we obtain that

23 23 22 3 2 3

22 22 22

iv iv23 232 3 2 3

22 22

c r qU ' = - U ' - V '' - V '' -

c c c G

c rU ''' = - U ''' - V - V

c c

(5.3.15)


32 23 32 23 32 2 33 32 2 3 33 3 33 3 32 2 33 3

22 22 22

c c c r c q q- U ' - c V '' - V '' - +c U ' -ks V +r V ''+r V '' = -

c c c G G

⇒

2 2

32 32 32 2 333 3 33 3 33 3

22 22 22

c c c q qc - U '+ r - V '' -ks V = -

c c c G G

⇒

2 2

33 22 32 33 23 32 32 2 33 3 33 3

22 22 22

c c - c c c -c c q qU '+ V '' -ks V = -

c c c G G

⇒

22 33 32 23 32 223 3 32 2 2

33 22 32 33 22 32 33 22 32 22

c ks c c qq cU ' = -V ''+ V '+ -

c c - c c c - c G c c -c c G

⇒

22 333 3 3 12

33 22 32

c ksU ' = -V '' + V +K

c c -c (5.3.17)

where

=

32 23 32 221 2 2

33 22 32 33 22 32 22

c c qq cK -

c c - c G c c -c c G

Substituting eqn. (5.3.17) into eqn. (5.3.15) gives

−

23 23 22 33 32 23 22 3 32 2

22 22 33 22 32 33 22 32 22

c c c ks c c qU ' = - (-V '') - V

c c c c - c c c -c c G

23 322

2

33 22 32 22

c q-c-

c c -c c G23 2

2 3

22 22

r q-V '' - V '' -

c c G

⇒

23 23 23 33 23 32 3 3 2 32 2

22 22 33 22 32 33 22 32

c r c ks c q U ' = V '' - V '' - V '' - V +

c c c c -c c c - c G

32 23 22

2

22 33 22 32 22

c c qq- -c G c c -c c G

Let

23 32 2

33 22 32

c qK =

c c -c G

32 23 22

2

22 33 22 32 22


96

∴

23 332 2 3 22

33 22 32

c ksU ' = -V '' - V +K

c c -c (5.3.18)

Differentiating eqn. (5.3.17 & 5.3.18) twice gives

iv 22 333 3 32

33 22 32

iv 23 332 2 32

33 22 32

c ksU ''' = -V + V

c c -c

c ksU ''' = -V - V

c c -c

(5.3.19)

Differentiating eqn. (5.3.11) once gives

22 2 23 3 22 2 23 3 22 2 23 3ka U '''+ka U ''' -b U ' -b U ' - c V '' - c V '' = 0 (5.3.20)

Substituting eqns. (5.3.17), ( 5.3.18) & (5.3.19) into eqn. (5.3.20) we obtain

( )

iv iv23 33 22 3322 2 22 3 23 3 23 32 2

33 22 32 33 22 32

c ks c kska (-V ) -ka V ''+ka -V +ka V ''

c c - c c c -c

23 33 22 3322 2 22 3 22 2 23 3 23 32 2

33 22 32 33 22 32

c ks c ks-b (-V '') -b - V -b K -b (-V '') -b V

c c -c c c - c

23 1 22 2 23 3+b K -c V '' - c V '' = 0

( ) ⇒

2

33 23 22 22 23iv iv

22 2 23 3 3 22 2 33 12

33 22 32

k s a c -a c-ka V -ka V + V '' -b K +b K = 0

c c -c

⇒ iv iv

1 2 2 3 1 3 3α V +α V -β V '' = +K (5.3.21)

( ); ; ;

2

33 23 22 22 23

1 22 2 23 1 3 23 1 22 22

33 22 32

k s a c -a cα = ka α = ka β = K = b K -b K

c c -c

=

32 23 32 221 2 2

33 22 32 33 22 32 22

c c qq cK -

c c - c G c c -c c G

23 32 2

33 22 32

c qK =

c c -c G

32 23 22

2

22 33 22 32 22



32 2 33 3 32 2 33 3 32 2 33 3ka U '''+ka U ''' -b U ' -b U ' - c V '' - c V '' = 0 (5.3.22)

Substituting eqns. (5.3.17), ( 5.3.18), & (5.3.19) into eqn. (5.3.22) we obtain

iv iv23 33 22 3332 2 32 3 33 3 33 32 2

33 22 32 33 22 32

c ks c kska (-V )+ka - V ''+ka (-V )+ka V ''

c c - c c c - c

23 33 22 3332 2 32 3 32 2 33 3 33 32 2

33 22 32 33 22 32

c ks c ks-b (-V '') -b - V -b K -b (V '') -b V

c c -c c c - c

97

33 1 32 2 33 3-b K -c V '' - c V '' = 0

( ) ⇒

2

33 33 22 32 23iv iv

32 2 33 3 32

33 22 32

k s a c -a c-ka V -ka V + V ''

c c - c

+( )

33 32 23 33 22

3 32 2 33 12

33 22 32

ks b c -b cV -b K -b K = 0

c c -c

γ⇒ iv iv

3 2 4 3 2 3 1 3 4α V +α V -β V '' - V = -K (5.3.23)

where, ( )

2

33 33 22 32 23

3 32 4 33 2 2

33 22 32

k s a c -a cα = ka ; α = ka ; β =

c c -c

( )

1γ

=

33 32 23 33 22

4 32 2 33 12

33 22 32

ks b c -b c; K = b K +b K

c c -c

Thus, the differential equations of equilibrium for flexural-distortional equilibrium of

mono-symmetric sections are:

γ

iv iv

1 2 2 3 1 3 3

iv iv

3 2 4 3 2 3 1 3 4

α V +α V -β V '' =K

α V +α V -β V '' - V = -K (5.3.24)

5.3.3 Torsional- distortional equilibrium (mono- symmetric Section)

The relevant coefficients for torsional-distortional equilibrium (strain modes 3 and

4), are a33, b33, c33, c34, r33, r34 = r 43, r44 and s33. Substituting these into the

matrix notation eqn. (3.4.17) we obtain;

1

2

33 3

0 0 0 U ''

κ 0 0 0 U ''

0 0 a U ''

-

1

2

33 3

0 0 0 U

0 0 0 U

0 0 b U

-

1

2

3

33 34

4

V '0 0 0 0

V '0 0 0 0

V '0 0 c c

V '

=0

33

43

0 0 0

0 0 0

0 0

0 0

c

c

1

2

3

U '

U '

U '

-

0 0 0

0 0 0

0 0

0 0

1

2

333

4

V0

V0κ

Vs 0

V0 0

+33 34

43 44

0 0 0 0

0 0 0 0

0 0

0 0

r r

r r

1

2

3

4

V ''

V ''

V ''

V ''

= −

1

2

3

4

1

G

qqqq


33 3 33 3 33 3 34 4ka U '' -b U -c V ' -c V ' = 0 (5.3.25)

333 3 33 3 33 3 34 4

qc U ' -ks V +r V ''+r V '' = -

G (5.3.26)

98

443 3 43 3 44 4

qc U '+r V '' +r V '' = -

G (5.3.27)

From eqn. (5.3.27) we obtain,

44 43 3 4

43 43

iv iv443 3 4

43

r qU ' = -V '' - V '' -

c c G

rU ''' = -V - V

c

(5.3.28)

Substituting eqns. (5.3.28 ) into eqn. (5.3.26) we have

( ) 4''q

G

33 34433 3 33 4 33 3 33 3 34 4

43 43

c qrc -v - c V '' - - ks V +r V ''+r V '' = -

c c G

⇒

33 44 3 33 434 4 33 3

43 43

c r q c qr - V '' -ks V = - +

c G c G

⇒

34 43 33 44 3 33 44 33 3

43 43

r c - c r q c qV '' -ks V = - +

c G c G

( )⇒ 3 434 43 33 44 4 43 33 3 43 43

q qr c - c r V '' - c ks V = -c +c

G G

γ⇒ 1 4 1 3 1β V '' - V =K (5.3.29)


33 3 33 3 33 3 34 4ka U ''' - b U ' - c V '' - c V '' = 0 (5.3.30)

Substituting eqns. (5.3.28) into eqn. (5.3.30) we have

iv iv44 44 433 3 33 4 33 3 33 4 33 33 3 34 4

43 43 43

r r -qka (-V )+ka - V -b (-V '') -b - V '' -b - c V '' - c V '' = 0

c c c G

( )

⇒

⇒

iv iv33 44 33 44 33 433 3 4 34 4

43 43 43

iv iv 33 433 43 3 33 44 4 33 44 43 34 4

ka r b r -b q-ka V - V + -c V '' =

c c c G

b qka c V +ka r V - b r - c c c V '' =

G

iv iv

3 2 4 2 4 2V +α V -β V '' =K (5.3.31)

Thus, the coupled differential equations of torsional-distortional equilibrium for mono-symmetric sections are:

γ1 4 1 3 1

iv iv

3 2 4 2 4 2

β V '' - V =K

V +α V -β V '' =K (5.3.32)

where, r

c

442

43

α = , 1 34 43 33 44β = r c -c r ; 33 44 34 432

33 43

b r -c cβ = ,

ka c

99

1γ 43 33= c ks ; 341 33 43

qqK = c -c

G G;

33 42

33 43

b qK =

ka c G

5.3.4 Flexural-torsional-distortional equilibrium (mono- symmetric Section)

The relevant coefficients for flexural-torsional-distortional equilibrium are those

involving strain modes 2, 3 and 4. These are; a32, a33, b32, b33, c32, c33, c34, c42, r22,

r32, r33, r34, r42, s33, and r44. Substituting these into the matrix eqn. (3.4.17), noting

that ij ji ij jia = a , b = b , etc, we obtain;

1

22 23 2

32 33 3

0 0 0 U ''

κ 0 a a U ''

0 a a U ''

-

1

22 23 2

32 33 3

0 0 0 U

0 b b U

0 b b U

-

1

2

22 23 24

3

32 33 34

4

V '0 0 0 0

V '0 c c c

V '0 c c c

V '

=0

1

22 23

2

32 33

3

42 43

0 0 0U '

0 c cU '

0 c cU '

0 c c

-

1

2

333

4

V0 0 0 0

V0 0 0 0κ

V0 0 s 0

V0 0 0 0

+

1

22 23 24 2

32 33 34 3

42 43 44 4

0 0 0 0 V ''

0 r r r V ''

0 r r r V ''

0 r r r V ''

1

2

3

4

1

G+ = 0

q

q

q

q

(5.3.33)

Multiplying out we obtain:

22 2 23 3 22 2 23 3 22 2 23 3 24 4ka U ''+ka U '' -b U -b U -c V ' - c V ' - c V ' = 0 (5.3.34)

32 2 33 3 32 2 33 3 32 2 33 3 34 4ka U ''+ka U '' -b U -b U -c V ' -c V ' - c V ' = 0 (5.3.35)

222 2 23 3 22 2 23 3 24 4

qc U '+c U '+r V ''+r V '' +r V '' = -

G (5.3.36)

332 2 33 3 33 3 32 2 33 3 34 4

qc U '+c U ' -ks V +r V ''+r V '' +r V '' = -

G (5.3.37)

442 2 43 3 42 2 43 3 44 4

qc U '+c U '+r V ''+r V ''+r V '' = -

G (5.3.38)

From eqn. (5.3.36) we obtain; 23 23 24 22 3 2 3 4

22 22 22 22

c r r qU ' = - U' - V '' - V '' - V '' -

c c c c G (5.3.39)


33 33 33 34 32 3 3 2 3 4

32 32 32 32 32

c ks r r qU ' = - U '+ V - V '' - V '' - V '' -

c c c c c G (5.3.40)

Equating eqns. (5.3.39 & 5.3.40) we obtain;

23 23 33 3324 23 2 3 4 3 3

22 22 22 22 32 32

c r c ksr qU' + V ''+ V '' + V'' + = U ' - V

c c c c G c c

100

33 34 32 3 4

32 32 32

r r q+V ''+ V ''+ V ''+

c c c G

⇒

23 33 33 23 34 33 324 23 3 4 3

22 32 32 22 32 22 32 32 22

c c r r r ks qr q- U ' = - V '' + - V'' - V + -

c c c c c c c c G c G

33 23 34 24

32 22 32 22 333 3 4 3

23 33 23 33 23 3332

22 32 22 32 22 32

r r r r- -

c c c c ksU ' = V ''+ V '' - V

c c c c c c- - c -

c c c c c c

3 2

23 33 23 3332 22

22 32 22 32

q q+ -

c c c cc G - c G -

c c c c

⇒ 3 1 3 2 4 3 3 1

iv iv

3 1 3 2 4 3 3

U ' =β V ''+β V '' -β V +K

U ''' =β V +β V -β V '' (5.3.41)

From eqn. (5.3.36) we obtain;

22 22 24 23 2 2 3 4

23 23 23 23

c r r qU ' = - U ' - V '' - V '' - V '' -

c c c c G (5.3.42)


32 33 32 34 33 2 3 2 3 4

33 33 33 33 33

c ks r r qU ' = - U' + V - V '' - V '' - V '' -

c c c c c G (5.3.43)

Equating eqns. (5.3.42) & (5.3.43) we obtain

= −32 3322 22 24 22 2 3 4 3 2 3

23 23 23 23 33 33

c Ksc r r qU '+ V ''+ V '' + V ''+ U ' = U' V

c c c c G c c

+ 32 34 32 3 4

33 33 33

r r qV '' + V ''+ V ''+

c c c G

⇒ − − − + −

32 32 34 33 322 22 24 22 2 4 3

23 33 33 23 33 23 33 33 23

c r r Ks qc r r q- U ' = V ''+ V '' V

c c c c c c c c G c G

32 3422 24

33 23 33 23 33 3 22 2 4 3

32 32 32 32 3222 22 22 22 2233 33 23

23 33 23 33 23 33 23 33 23 33

r rr r- -

c c c c ks q qU ' = V ''+ V '' - V + -

c c c c cc c c c c- - c - c G - c G -

c c c c c c c c c c

⇒ 2 1 2 2 4 3 3 2

iv iv

2 1 2 2 4 3 3

U ' = α V ''+α V '' -α V +K

U ''' = α V +α V -α V '' (5.3.44)

101

Substituting eqns. (5.3.41) & (5.3.44) into eqn. (5.3.38) we have;

42 1 2 42 2 4 43 3 3 42 2 43 1 3 43 2 4c α V ''+c α V '' - c α V +c K +c β V ''+c β V ''

443 3 3 43 1 42 2 43 3 44 4

q-c β V +c K +r V ''+r V ''+r V '' = -

G

( ) ( ) ( )⇒ 42 1 42 2 43 1 43 3 42 2 43 2 44 4c α +r V ''+ c β +r V ''+ c α +c β +r V ''

( ) 442 3 43 3 3 42 2 43 1

q- c α +c β V = - - c K -C K

G

β γ⇒ 4 2 5 3 6 4 1 3 3β V ''+β V ''+ V '' - V = -K (5.3.45)

Differentiating eqn. (5.3.34) we obtain

22 2 23 3 22 2 23 3 22 2 23 3 24 4ka U '''+ka U ''' - b U ' -b U ' - c V '' - c V '' - c V '' = 0 (5.3.46)

Substituting eqns. (5.3.41) & (5.3.44) into eqn. (5.3.46) we obtain;

iv iv iv iv

22 1 2 22 2 4 22 3 3 23 1 3 23 2 4ka α V +ka α V -ka α V ''+ka β V +ka β V - '

23 3 3 22 1 2 22 2 4 22 3 3 22 2 23 1 3 23 2 4-ka β V '' -b α V '' -b α V ''+b α V -b K -b β V '' -b β V ''

23 3 3 23 1 22 2 23 3 24 4+b β V -b K -c V '' - c V '' - c V '' = 0

β β⇒ iv iv iv

22 1 2 23 1 3 22 2 23 3 4 22 1 22 2ka α V +ka V +k(a α +a )V - (b α +c )V ''

23 3 23 1 22 3 23 3 22 2 24 23 2 4-(ka β +b β +kα α +c )V '' - (b α +c +b β )V ''

22 3 23 3 3 22 2 23 1+(b α +b β )V = b K +b K

β γ⇒ iv iv iv

4 2 5 3 6 4 7 2 8 3 9 4 2 3 4 α V +α V +α V -β V '' -β V '' - V ''+ V =K (5.3.47)


32 2 33 3 32 2 33 3 32 2 33 3 34 4ka U '''+ka U ''' -b U ' -b U ' - c V '' - c V '' - c V '' = 0 (5.3.48)

Substituting eqns .(5.3.46) & (5.3.44) into eqn. (5.3.48) we obtain

iv iv iv iv

32 1 2 32 2 4 32 3 3 33 1 3 33 2 4ka α V +ka α V -ka α V ''+ka β V +ka β V -

33 3 3 32 1 2 32 2 4 32 3 3 32 2 33 1 3 33 2 4-Ka β V '' -b α V '' -b α V ''+b α V -b K -b β V '' -b β V ''

33 3 3 33 1 32 2 33 3 34 4+b β V -b K -c V '' - c V '' - c V '' = 0

⇒ iv iv iv

32 1 2 33 1 3 32 2 33 2 4 32 1 32 2ka α V +ka β V +k(a α +a β )V - (b α +c )V ''

32 3 33 1 33 3 33 3 32 2 34 33 2 4-(ka α +b β +kα β +c )V '' - (b α +c +b β )V ''

32 3 33 3 3 32 2 33 1+(b α +b β )V = b K +b K

γ⇒ iv iv iv

7 2 8 3 9 4 10 2 11 3 12 4 3 3 5α V +α V +α V -β V '' -β V '' -β V ''+ V =K (5.3.49)

The differential equations for flexural- torsional- distortional equilibrium are;

102

4 2 5 3 6 4 1 3 3

iv iv iv

4 2 5 3 6 4 7 2 8 3 9 4 2 3 4

iv iv iv

7 2 8 3 9 4 10 2 11 3 12 4 3 3 5

β V ''+β V ''+β V '' - γ V = -K

α V +α V +α V -β V '' -β V '' -β V ''+ γ V =K

α V +α V +α V -β V '' -β V '' -β V ''+ γ V =K

(5.3.50)

32 3422 24

33 23 33 23 331 2 3

32 32 3222 22 2233

23 33 23 33 23 33

r rr r- -

c c c c Ksα = ; α = ; α ;

c c cc c c- - c -

c c c c c c

4 22 1 5 23 1 6 22 2 23 3α = ka α ; α = ka β ; α = k(a α +a β );

7 32 1 8 33 1 9 32 2 33 3α = ka α ; α = ka β ; α = +k(a α +a β );

33 23 34 24

32 22 32 22 331 2 3

23 33 23 33 23 3332

22 32 22 32 22 32

r r r r- -

c c c c ksβ = ; β = ; β =

c c c c c c- - c -

c c c c c c

( ) ( ) ( )4 42 1 42 5 43 1 43 6 42 2 43 2 44β = c α +r ; β = c β +r ; β = c α +c β +r

7 22 1 22β = (b α +c ) 8 23 3 23 1 22 3 23 9 22 2 24 23 2β = (ka β +b β +kα α +c ); β = -(b α +c +b β )

10 32 1 32β = (b α +c ) 11 32 3 33 1 33 3 33 12 32 2 34 33 2β = (ka α +b β +kα β +c ); β = (b α +c +b β )

( )1γ = 42 3 43 3c α +c β ; γ 2 22 3 23 3= (b α +b β ); γ 3 32 3 33 3= (b α +b β );

3 21

23 33 23 3332 22

22 32 22 32

q qK = -

c c c cc G - c G -

c c c c

; = −

3 22

32 3222 2233 23

23 33 23 33

q qK

c cc cc G - c G -

c c c c

;

43 42 2 43 1

qK = +c K +C K

G; 4 22 2 23 1K = b K +b K ; 5 32 2 33 1K = b K +b K

5.4 NON-SYMMETRIC SECTIONS

The governing equations of equilibrium are,

∑ ∑ ∑3 3 4

ji i ji i jk Ki=1 i=1 k=1

k a U''(x) - b U (x) - c V '(x) = 0 (5.4.1a)

∑ ∑ ∑3 4 4

hhi i hk k hk k

i=1 i=1 i=1

qc U' + r V '' - k s V + = 0

G (5.4.1b)

In matrix notation we have,

103

''

11 12 13 1

''

21 22 23 2

''

31 32 33 3

a a a U

κ a a a U

a a a U

-

11 12 13 1

21 22 23 2

31 32 33 3

b b b U

b b b U

b b b U

-

'

111 12 13 14 '

221 22 23 24 '

331 32 33 34 '

4

Vc c c c

Vc c c c

Vc c c c

V

= (5.4.2a)

11 12 13 '

1

21 22 23 '

2

31 32 33 '

3

41 42 43

c c cU

c c cU

c c cU

c c c

-

1

2

333

4

V0 0 0 0

V0 0 0 0κ

V0 0 s 0

V0 0 0 0

+

''11 12 13 14 1

''21 22 23 24 2

''31 32 33 34 3

''41 42 43 44 4

r r r r V

r r r r V

r r r r V

r r r r V

1

2

3

4

1=

G

q

q

q

q

(5.4.2b)

We note that in non-symmetric sections there are interactions between all the four

strain modes such that the elements of the matrix eqns. (5.4.2a) and (5.4.2b) are non

zero except matrix S, where only S33 has value as explained in section 4.5.

Expanding matrix eqns. (5.4.2a) and (5.4.2b) we obtain

'' '' '' ' ' ' '

11 1 12 2 13 3 11 1 12 2 13 3 11 1 12 2 13 3 14 4ka U +ka U +ka U -b U -b U -b U -c V -c V -c V -c V = 0 (5.4.3)

'' '' '' ' ' ' '


'' '' '' ' ' ' '


' ' ' '' '' '' '' 111 1 12 2 13 3 11 1 12 2 13 3 14 4

qc U +c U +c U +r V +r V +r V +r V = -

G (5.4.6)

' ' ' '' '' '' '' 221 1 22 2 23 3 21 1 22 2 23 3 24 4


G (5.4.7)

' ' ' '' '' '' '' 331 1 32 2 33 3 31 1 32 2 33 3 34 4 33 3

qc U +c U +c U +r V +r V +r V +r V -ks V = -

G (5.4.8)

' ' ' '' '' '' '' 441 1 42 2 43 3 41 1 42 2 43 3 44 4


G (5.4.9)

Multiplying eqn. (5.4.4) by 11 21(a /a ) and subtracting from eqn. (5.4.3) we obtain

( )'' ''

12 21 11 22 2 13 21 11 33 3 21 11 21 11 1 11 22 21 12 2(a a -a ka )U +(a a -ka a )U +(b a -a b )U + a b -a b U

( ) ( ) ( ) ( )' ' '

11 23 21 13 3 11 21 21 11 1 11 22 21 12 2 11 23 21 13 3+ a b -a b U + a c -a c V + a c -a c V + a c -a c V

( ) '

11 24 21 14 4+ a c -a c V = 0 (5.4.10)

⇒

'' 23 11 21 1322 11 21 12 12 21 11 221 2 2 3

11 21 21 11 11 21 21 11 11 21 21 11

b a -a bb a -a b a a -ka aU = - U - U - U

a b -a b a b -a b a b -a b

'' ' '13 21 11 33 21 11 11 21 22 11 12 213 1 2

11 21 21 11 11 21 21 11 11 21 21 11

a a -ka a c a -c a c a -c a- U - V - V


104

' '11 23 21 13 11 24 21 143 4

11 21 21 11 11 21 21 11

a c -a c a c -a c- V - V

a b -a b a b -a b (5.4.11)

Multiplying eqn. (5.4.5) by 11 31(a /a ) and subtracting from eqn .(5.4.3) we obtain

( )'' ''

12 31 11 32 2 13 31 11 33 3 11 31 31 11 1 11 32 31 12 2(ka a -ka a )U +(ka a -ka a )U +(a b -a b )U + a b -a b U

( ) ( ) ( ) ( )' ' '


( ) '

11 34 31 14 4+ a c -a c V = 0 (5.4.12)

⇒

''11 32 31 12 12 31 11 32 11 33 31 131 1 2 3

11 31 31 11 11 31 31 11 11 31 31 11

a b -a b ka a -ka a a b -a bU = - U - U - U


'' ' '13 31 11 33 11 31 31 11 11 32 31 123 1 2

11 31 31 11 11 31 31 11 11 31 31 11

ka a -ka a a c -a c a c -a c- U - V - V


' '11 33 31 13 11 34 31 143 4

11 31 31 11 11 31 31 11


a b -a b a b -a b (5.4.13)

Combining eqns. (5.4.11) and (5.4.13) we obtain

11 33 21 12 11 33 31 122

11 21 21 11 11 31 31 11

a b -a b a b -a b- U +

a b -a b a b -a b

''12 31 11 3212 21 11 222

11 21 21 11 11 31 31 11

ka a -a aa a -ka a- U +

a b -a b a b -a b

11 23 21 13 11 33 31 133

11 21 21 11 11 31 31 11


a b -a b a b -a b

''11 21 11 33 13 31 111 333

11 21 21 11 11 31 31 11

a a -ka a ka a -ka a- U +

a b -a b a b -a b

''11 31 31 1111 21 21 111

11 21 21 11 11 31 31 11

a c -a ca c -a c- V +

a b -a b a b -a b

'11 32 31 1211 22 21 122

11 21 21 11 11 31 31 11

a c -a ca c -a c- U +

a b -a b a b -a b

'11 23 21 13 11 33 31 133

11 21 21 11 11 31 31 11

a c -a c a c -a c- V +

a b -a b a b -a b

'11 34 31 1411 24 21 144

11 21 21 11 11 31 31 11

a c -a ca c -a c- V = 0

a b -a b a b -a b

⇒ '' '' ' ' ' '

1 2 2 2 3 3 4 3 5 1 6 2 7 3 8 4αU +α U +α U +α U +α V +α V +α V +α V = 0 (5.4.14)

where,

11 33 21 12 11 33 31 121

11 21 21 11 11 31 31 11

a b -a b a b -a bα = -

a b -a b a b -a b (a)

12 31 11 3212 21 11 222

11 21 21 11 11 31 31 11

ka a -a aa a -ka aα = -

a b -a b a b -a b (b)

11 23 21 13 11 33 31 133

11 21 21 11 11 31 31 11

a b -a b a b -a bα = -

a b -a b a b -a b (c)

105

11 21 11 33 13 31 111 334

11 21 21 11 11 31 31 11

a a -ka a ka a -ka aα = -

a b -a b a b -a b (d)

11 31 31 1111 21 21 115

11 21 21 11 11 31 31 11

a c -a ca c -a cα = -

a b -a b a b -a b (e)

11 32 31 1211 22 21 126

11 21 21 11 11 31 31 11


a b -a b a b -a b (f)

11 23 21 13 11 33 31 137

11 21 21 11 11 31 31 11

a c -a c a c -a cα = -

a b -a b a b -a b (g)

11 34 31 1411 24 21 148

11 21 21 11 11 31 31 11


a b -a b a b -a b (h) (5.4.15)

Multiplying eqn. (5.4.5) by 21 31(a /a )and subtracting from eqn. (5.4.4) we obtain

( )'' ''

12 31 21 32 2 13 31 21 33 3 21 31 31 21 1 21 32 31 22 2(ka a -ka a )U +(ka a -ka a )U +(a b -a b )U + a b -a b U

( ) ( ) ( ) ( )' ' '


( ) '

21 34 31 24 4+ a c -a c V = 0 (5.4.16)

⇒

''21 32 31 22 12 31 21 32 21 33 31 231 2 2 321 31 31 21 21 31 31 21 21 31 31 21a b -a b ka a -ka a a b - a bU = - U - U - U

a b -a b a b -a b a b - a b

'' ' '13 31 21 33 21 31 31 21 21 32 31 223 1 221 31 31 21 21 31 31 21 21 31 31 21ka a -ka a a c - a c a c - a c- U - V - V


' '21 33 31 23 21 34 31 243 4

21 31 31 21 21 31 31 21


a b -a b a b -a b (5.4.17)

Combining eqns. (5.4.11 and 5.4.17) we obtain

21 32 31 2211 22 21 122

11 21 21 11 21 31 31 21

a b -a ba b -a b- U +

a b -a b a b -a b

''12 31 21 3212 21 11 222

11 21 21 11 21 31 31 21

ka a -a aa a -ka a- U +

a b -a b a b -a b

11 23 21 13 21 33 31 233

11 21 21 11 21 31 31 21


a b -a b a b -a b

''13 21 11 33 13 31 21 333

11 21 21 11 21 31 31 21

a a -ka a ka a -ka a- U +

a b -a b a b -a b

''21 31 31 2111 21 21 111

11 21 21 11 21 31 31 21


a b -a b a b -a b

'21 32 31 2211 22 21 122

11 21 21 11 21 31 31 21


a b -a b a b -a b

'11 23 21 13 21 33 31 233

11 21 21 11 21 31 31 21

a c -a c a c -a c- V +

a b -a b a b -a b

'21 34 31 2411 24 21 144

11 21 21 11 21 31 31 21

a c -a ca c -a c- V = 0

a b -a b a b -a b

106

⇒ '' '' ' ' ' '

1 2 2 2 3 3 4 3 5 1 6 2 7 3 8 4ρU +ρ U +ρ U +ρ U +ρ V +ρ V +ρ V +ρ V = 0 (5.4.18)

where,

21 32 31 2211 22 21 121 11 21 21 11 21 31 31 21a b - a ba b - a bρ = -

a b - a b a b - a b (a)

12 31 21 3212 21 11 222 11 21 21 11 21 31 31 21ka a - a aa a -ka aρ = -

a b - a b a b - a b (b)

11 23 21 13 21 33 31 233 11 21 21 11 21 31 31 21a b - a b a b - a bρ = -

a b - a b a b - a b (c)

13 21 11 33 13 31 21 334 11 21 21 11 21 31 31 21a a -ka a ka a -ka aρ = -

a b - a b a b - a b (d)

5ρ

=

21 31 31 2111 21 21 1111 21 21 11 21 31 31 21a c - a ca c - a c-

a b - a b a b - a b (e)

21 32 31 2211 22 21 126 11 21 21 11 21 31 31 21a c - a ca c - a cρ = -

a b - a b a b - a b (f )

7ρ

11 23 21 13 21 33 31 2311 21 21 11 21 31 31 21a c - a c a c - a c= -

a b - a b a b - a b (g)

8ρ

=

21 34 31 2411 24 21 1411 21 21 11 21 31 31 21a c - a ca c - a c-

a b - a b a b - a b (h) (5.4.19)

Combining eqns. (5.4.13) and (5.4.17) we obtain

11 32 31 12 21 32 31 22 211 31 31 11 21 31 31 21a b - a b a b - a b- U +

a b - a b a b - a b

''12 31 11 32 12 31 21 32 211 31 31 11 21 31 31 21a a - a a ka a -ka a- U +

a b - a b a b - a b

11 33 31 13 21 33 31 23 311 31 31 11 21 31 31 21a b - a b a b - a b- U

a b - a b a b - a b+

''11 31 11 33 13 31 21 33 311 31 31 11 21 31 31 21a a -ka a ka a -ka a- U +

a b - a b a b - a b

'11 31 31 11 21 31 31 21 111 31 31 11 21 31 31 21a c - a c a c - a c- V +

a b - a b a b - a b

'11 32 32 12 21 32 31 22 211 31 31 11 21 32 31 21a c - a c a c - a c- V +

a b - a b a b - a b

'11 33 31 13 21 33 31 23 311 31 31 11 21 31 31 21a c - a c a c - a c- V +

a b - a b a b - a b

'11 34 31 14 21 34 31 24 411 31 31 11 21 31 31 21a c - a c a c - a c- V = 0

a b - a b a b - a b

⇒ '' '' ' ' ' '1 2 2 2 3 3 4 3 5 1 6 2 7 3 8 4η U +η U +η U +η U +η V +η V +η V +η V =0 (5.4.20)

where,

11 32 31 12 21 32 31 221 11 31 31 11 21 31 31 21a b - a b a b - a bη = -

a b - a b a b - a b (a)

107

12 31 11 32 12 31 21 321 11 31 31 11 21 31 31 21a a - a a ka a -ka aη = -

a b - a b a b - a b (b)

11 33 31 13 21 33 31 233 11 31 31 11 21 31 31 21a b - a b a b - a bη = -

a b - a b a b - a b (c)

11 31 11 33 13 31 21 334 11 31 31 11 21 31 31 21a a -ka a ka a -ka aη = -

a b - a b a b - a b (d)

5η

11 31 31 11 21 31 31 2111 31 31 11 21 31 31 21a c - a c a c - a c= -

a b - a b a b - a b (e)

6η

11 32 32 12 21 32 31 2211 31 31 11 21 32 31 21a c - a c a c - a c= -

a b - a b a b - a b (f)

7η

11 33 31 13 21 33 31 2311 31 31 11 21 31 31 21a c - a c a c - a c= -

a b - a b a b - a b (g)

8η

11 34 31 14 21 34 31 2411 31 31 11 21 31 31 21a c - a c a c - a c= -

a b - a b a b - a b (h) (5.4.21)

Multiplying eqn. (5.4.7) by 11 21(c /c )and subtracting from eqn. (5.4.6) we obtain

( )' ' '' ''12 21 11 22 2 13 21 11 23 3 11 21 21 11 1 12 21 22 11 2(c c - c c )U +(c c - c c )U +(r c - r c )V + r c - r c V +

+ ( )

'' '' 11 2 21 113 21 23 11 3 14 21 24 11 4 c q c qr c - r c V +(r c - r c )V = -

G G

⇒ ' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 1β U +β U +β V +β V +β V +β V =K (5.4.22)

where, ( )1 12 21 11 22β = c c - c c ; ( )2 13 21 11 23β = c c - c c ; ( )3 21 11 11 21β = c r - c r

( )4 21 12 11 2β = c r - c r ; ( )5 21 13 11 23β = c r - c r ( )6 21 14 11 24β = c r - c r (5.4.23)

Multiplying eqn. (5.4.9) by 31 41( / )c c and subtracting from eqn. (5.4.8) we obtain

( )' ' '' ''32 41 31 42 2 33 41 31 43 3 41 31 31 41 1 41 31 31 42 2(c c - c c )U +(c c - c c )U +(c r - c r )V + c r - c r V +

+ ( )

'' '' 31 4 41 341 33 31 44 3 41 34 31 44 4 33 3 c q c qc r - c r V +(c r - c r )V -ks V = -

G G

⇒ ' ' '' '' '' ''

1 2 2 3 3 1 4 2 5 3 6 4 7 3 2λ U +λ U +λ V + λ V +λ V +λ V - λ v =K (5.4.24)

where, ( )1 32 41 31 42λ = c c - c c ; ( )2 33 41 31 43λ = c c - c c ; ( )3 41 31 31 41λ = c r - c r

( )4 41 32 31 42λ = c r - c r ; ( )5 41 33 31 43λ = c r - c r ( )6 41 34 31 44λ = c r - c r (5.4.25)

Multiplying eqn. (5.4.8) by 11 31( / )c c and subtracting from eqn. (5.4.6) we obtain

108

( )' ' '' ''11 32 12 31 2 11 33 13 31 3 11 31 31 11 1 11 32 31 12 2(c c - c c )U +(c c - c c )U +(c r - c r )V + c r - c r V +

+ ( )

'' '' 31 1 11 311 33 31 14 3 11 34 31 14 4 11 33 3 c q c qc r - c r V +(c r - c r )V +kc s V = -

G G

⇒ ' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 7 3 3ν U +ν U +ν V +ν V +ν V +ν V +ν v =K (5.4.26)

Thus we have '' '' ' ' ' '1 2 2 2 3 3 4 3 5 1 6 2 7 3 8 4α U +α U +α U +α U +α V +α V +α V +α V =0 (5.4.14) '' '' ' ' ' '1 2 2 2 3 3 4 3 5 1 6 2 7 3 8 4ρ U +ρ U +ρ U +ρ U +ρ V +ρ V +ρ V +ρ V =0 (5.4.18) '' '' ' ' ' '1 2 2 2 3 3 4 3 5 1 6 2 7 3 8 4η U +η U +η U +η U +η V +η V +η V +η V =0 (5.4.20) ' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 1β U +β U +β V +β V +β V +β V =K (5.4.22) ' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 7 3 2λ U +λ U +λ V +λ V +λ V +λ V - λ v =K (5.4.24) ' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 7 3 3ν U +ν U +ν V +ν V +ν V +ν V +ν v =K (5.4.26)

Multiplying eqn. (5.4.18) by 1 1( / )α ρ and subtracting eqn. (5.4.14) from it, we obtain

( )'' '' '1 2 2 1 2 1 3 3 1 3 1 4 4 1 3 1 5 5 1 1(α ρ -α ρ )U +(α ρ -β ρ )U +(α ρ - α ρ )U + α ρ - α ρ V +

+ ( ) ( )' ' '1 6 6 1 2 1 7 7 1 3 1 8 8 1 4α ρ -α ρ V +(α ρ - α ρ )V + α ρ -α ρ V =0

⇒ '' '' ' ' ' '1 2 2 3 3 3 4 1 5 2 6 3 7 4μU +μ U +μ U +μ V +μ V +μ V +μ v = 0 (5.4.27)

where, 1 1 2 2 1μ =(α ρ -α ρ ); 2 1 3 3 1μ =(α ρ -β ρ ); 3 1 4 4 1μ =(α ρ -α ρ )

( )4 1 5 5 1µ = α ρ -α ρ ; ( )5 1 6 6 1μ = α ρ - α ρ 6 1 7 7 1μ =(α ρ - α ρ )

( )7 1 8 8 1μ = α ρ - α ρ (5.4.28)

Multiplying eqn. (5.4.20) by 1 1( / )α η and subtracting eqn. (5.4.14) from it, we obtain

( )'' '' '1 2 2 1 2 1 3 3 1 3 1 4 4 1 3 1 5 5 1 1(α η -α η )U +(α η -β η )U +(α η - α η )U + α η -α η V +

+ ( ) ( )' ' '1 6 6 1 2 1 7 7 1 3 1 8 8 1 4α η - α η V +(α η -α η )V + α η -α η V =0

⇒ '' '' ' ' ' '1 2 2 3 3 3 4 1 5 2 6 3 7 4ε U +ε U +ε U +ε V +ε V +ε V +ε v = 0 (5.4.29)

where, 1 1 2 2 1ε =(α η - α η ); 2 1 3 3 1ε =(α η - α η ); 3 1 4 4 1ε =(α η - α η )

( )4 1 5 5 1ε = α η - α η ; ( )5 1 6 6 1ε = α η - α η 6 1 7 7 1ε =(α η -α η )

( )7 1 8 8 1ε = α η - α η (5.4.30)

Multiplying eqn. (5.4.20) by 1 1( / )ρ η and subtracting eqn. (5.4.18) from it, we obtain

( )'' '' '1 2 2 1 2 1 3 3 1 3 1 4 4 1 3 1 5 5 1 1(ρ η -ρ η )U +(ρ η -ρ η )U +(ρ η -ρ η )U + ρ η -ρ η V +

109

+ ( ) ( )' ' '1 6 6 1 2 1 7 7 1 3 1 8 8 1 4ρ η -ρ η V +(ρ η -ρ η )V + ρ η -ρ η V =0

⇒ '' '' ' ' ' '1 2 2 3 3 3 4 1 5 2 6 3 7 4γ U +γ U +γ U +γ V +γ V +γ V +γ v = 0 (5.4.31)

where, 1 1 2 2 1γ =(ρ η -ρ η ); 2 1 3 3 1γ =(ρ η -ρ η ); 3 1 4 4 1γ =(ρ η -ρ η )

( )4 1 5 5 1γ = ρ η -ρ η ; ( )5 1 6 6 1γ = ρ η -ρ η 6 1 7 7 1γ =(ρ η -ρ η )

( )7 1 8 8 1γ = ρ η -ρ η (5.4.32)

Thus we have '' '' ' ' ' '1 2 2 3 3 3 4 1 5 2 6 3 7 4μU +μ U +μ U +μ V +μ V +μ V +μ v = 0 (5.4.27) '' '' ' ' ' '1 2 2 3 3 3 4 1 5 2 6 3 7 4ε U +ε U +ε U +ε V +ε V +ε V +ε v = 0 (5.4.29) '' '' ' ' ' '1 2 2 3 3 3 4 1 5 2 6 3 7 4γ U +γ U +γ U +γ V +γ V +γ V +γ v = 0 (5.4.31)

' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 1β U +β U +β V +β V +β V +β V =K (5.4.22) ' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 7 3 2λ U +λ U +λ V +λ V +λ V +λ V - λ v =K (5.4.24) ' ' '' '' '' ''1 2 2 3 3 1 4 2 5 3 6 4 7 3 3ν U +ν U +ν V +ν V +ν V +ν V +ν v =K (5.4.26)

Multiplying eqn. (5.4.22) by 1 1( / )λ β and subtracting eqn. (5.4.24) from it, we obtain

( )' '' '' ''1 2 2 1 3 1 3 3 1 1 1 4 4 1 2 1 5 5 1 3(λ β - λ β )U +(λ β - λ β )V +(λ β - λ β )V + λ β - λ β V +

( ) ( )''1 6 6 1 4 7 3 1 1 1 2+ λ β - λ β V +λ V = λ K -β K

⇒

' '' '' '' ''1 3 3 1 1 5 5 1 1 6 6 11 4 4 13 1 2 3 41 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1λ β - λ β λ β - λ β λ β - λ βλ β - λ βU = - V - V - V - V

λ β - λ β λ β - λ β λ β - λ β λ β - λ β

7 1 1 1 231 2 1 1 1 2 2 1λ λ K -β K- V +

λ β - λ β λ β - λ β

' '' '' '' ''3 1 1 2 2 3 3 4 4 5 3 4⇒U = -ω V -ω V -ω V -ω V -ω V +K (5.4.33)

where

1 3 3 11 1 2 2 1λ β - λ βω =

λ β - λ β;

1 4 4 12 1 2 2 1λ β - λ βω =

λ β - λ β;

1 5 5 13 1 2 2 1λ β - λ βω =

λ β - λ β

1 6 6 14 1 2 2 1λ β - λ βω =

λ β - λ β;

75 1 2 1 1λω =

λ β - λ β;

1 1 1 24 1 2 2 1λ K -β KK =

λ β - λ β

Multiplying eqn .(5.4.24) by 2 2( / )β λ and subtracting eqn. (5.4.22) from it, we obtain

( )' '' '' ''1 2 2 1 2 3 3 2 3 1 4 2 2 4 2 5 2 2 5 3(λ β - λ β )U +(λ β - λ β )V +(λ β - λ β )V + λ β - λ β V +

( ) ( ) ( )'' ''5 2 2 5 3 6 2 2 6 4 7 2 3 2 2 2 2+ λ β - λ β V + λ β - λ β V - λ β V = β K - λ K

110

⇒

' '' '' '' ''3 2 2 3 5 2 2 5 6 2 2 64 2 2 42 1 2 3 41 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1λ β - λ β λ β - λ β λ β - λ βλ β - λ βU = - V - V - V - V

λ β - λ β λ β - λ β λ β - λ β λ β - λ β

7 2 2 2 2 131 2 2 1 1 2 2 1λ β β K - λ K- V +

λ β - λ β λ β - λ β

' '' '' '' ''2 1 1 2 2 3 3 4 4 5 3 5⇒U = -π V -π V -π V -π V -π V +K (5.4.34)

Differentiating each of eqns. (5.4.27), ( 5.4.29) &( 5.4.31) once we obtain '' ' ''' '' '' '' ''1 2 2 3 3 3 4 1 5 2 6 3 7 4μU +μ U +μ U +μ V +μ V +μ V +μ v = 0 (5.4.35) ''' ' ''' '' '' '' ''1 2 2 3 3 3 4 1 5 2 6 3 7 4ε U +ε U +ε U +ε V +ε V +ε V +ε v = 0

(5.4.36) ''' ' ''' '' '' '' ''1 2 2 3 3 3 4 1 5 2 6 3 7 4γ U +γ U +γ U +γ V +γ V +γ V +γ v = 0 (5.4.37)

Differentiating eqn. ( 5.4.26) once we obtain ''''3333 = 0= 0= 0= 0

'' '' ''' ''' ''' '''1 2 2 3 3 1 4 2 5 3 6 4 7ν U +ν U +ν V +ν V +ν V +ν V +ν V (5.4.38)

But ' '' '' '' ''2 1 1 2 2 3 3 4 4 5 3 5U = -π V -π V -π V -π V -π V +K

'' ''' ''' ''' ''' '2 1 1 2 2 3 3 4 4 5 5U = -π V -π V -π V -π V -π V (5.4.39)

''' iv iv iv iv ''2 1 1 2 2 3 3 4 4 5 3U = -π V -π V -π V -π V -π V

Also,

' '' '' '' ''3 1 1 2 2 3 3 4 4 5 3 4U = -ω V -ω V -ω V -ω V -ω V +K

'' ''' ''' ''' ''' '3 1 1 2 2 3 3 4 4 5 3U = -ω V -ω V -ω V -ω V -ω V

''' iv iv iv iv ''3 1 1 2 2 3 3 4 4 5 3U = -ω V -ω V -ω V -ω V -ω V (5.4.40)

Substituting eqns. ( 5.4.39) and( 5.4.40) into eqn. (5.4.35) we obtain

( )iv IV iv iv ''1 1 1 2 2 3 3 4 4 5 3μ -π V -π V -π V +π V -π V + ( )'' '' '' ''2 1 1 2 2 3 3 4 4 5 3 4μ -ω V -ω V -ω V -ω V -ω V +K

( )iv iv iv iv ''3 1 1 2 2 3 3 4 4 5 3+μ -ω V -ω V -ω V -ω V -ω V '' '' '' ''4 1 5 2 6 3 7 4+μ V +μ V +μ V +μ V =0

⇒ ( ) ( ) ( ) ( )iv iv iv iv3 1 1 1 1 1 2 3 2 2 1 3 3 3 3 1 4 3 4 4- μ ω +μ π V - μ π +μ ω V - μ π +μ ω V - μ π +μ ω V

( ) ( ) ( )'' '' ''2 1 4 1 2 2 3 5 5 2 1 5 2 3 6 3- μ ω +μ V - μ ω +μ ω -μ V - μ π +μ ω -μ V

( ) ( )''3 4 7 4 2 5 3 4 2- μ ω -μ V - μ ω V +K μ =0

⇒⇒⇒⇒ iv iv iv iv '' '' '' ''1 1 2 2 3 3 4 4 5 1 6 2 7 3 8 4 9 3 2 4A V +A V +A V +A V +A V +A V +A V +A V +A V =μ K (5.4.41)

Also, substituting eqns. (5.4.39) and (5.4.40) into eqn. (5.4.36) we obtain

( )iv IV iv iv ''1 1 1 2 2 3 3 4 4 5 3ε -π V -π V -π V +π V -π V + ( )'' '' '' ''2 1 1 2 2 3 3 4 4 5 3 4+ε -ω V -ω V -ω V -ω V -ω V +K

( )iv iv iv iv ''3 1 1 2 2 3 3 4 4 5 3+ε -ω V -ω V -ω V -ω V -ω V '' '' '' ''4 1 5 2 6 3 7 4+ε V +ε V +ε V +ε V = 0

⇒ ( ) ( ) ( ) ( )iv iv iv iv1 1 3 1 1 1 2 3 2 2 1 3 3 3 3 1 4 3 4 4- ε π +ε ω V - ε π +ε ω V - ε π +ε ω V - ε π +ε ω V

( ) ( ) ( )'' '' ''2 1 4 1 2 2 5 2 1 5 2 3 6 3- ε ω +ε V - ε ω - ε V - ε π +ε ω - ε V

( ) ( )''2 4 7 4 2 5 3 2 4- ε ω - ε V - ε ω V +ε K =0

111

⇒⇒⇒⇒ iv iv iv iv '' '' '' ''1 1 2 2 3 3 4 4 5 1 6 2 7 3 8 4 9 3 2 4B V +B V +B V +B V +B V +B V +B V +B V +B V = ε K (5.4.42)

Substituting eqns. ( 5.4.39) and (5.4.40) into eqn. (5.4.37) we obtain

( )iv IV iv iv ''1 1 1 2 2 3 3 4 4 5 3γ -π V -π V -π V +π V -π V + ( )'' '' '' ''2 1 1 2 2 3 3 4 4 5 3 4γ -ω V -ω V -ω V -ω V -ω V +K

( )iv iv iv iv ''3 1 1 2 2 3 3 4 4 5 3+γ -ω V -ω V -ω V -ω V -ω V '' '' '' ''4 1 5 2 6 3 7 4+γ V +γ V +γ V +γ V =0

⇒ ( ) ( ) ( ) ( )iv iv iv iv1 1 3 1 1 1 2 3 2 2 1 3 3 3 3 1 4 3 4 4- γ π + γ ω V - γ π +γ ω V - γ π +γ ω V - γ π +γ ω V

( ) ( ) ( )'' '' ''2 1 4 1 2 2 5 2 1 5 2 3 6 3- γ ω +γ V - γ ω - γ V - γ π +γ ω -μ V

( ) ( )''2 4 7 4 2 5 3 2 4- γ ω - γ V - γ ω V +γ K =0

⇒⇒⇒⇒ iv iv iv iv '' '' '' ''1 1 2 2 3 3 4 4 5 1 6 2 7 3 8 4 9 3 2 4C V +C V +C V +C V +C V +C V +C V +C V +C V = γ K (5.4.43)

Substituting eqns. (5.4.39) and (5.4.40) into eqn. (5.4.38) we obtain

( )υ1111- +- +- +- +''' ''' ''' ''' '1 1 2 2 3 3 4 4 5 3π V +π V +π V +π V -π V ( )υ2222----

''' ''' ''' ''' '1 1 2 2 3 3 4 4 5 3ω V +ω V +ω V +ω V +ω V ''' ''' ''' ''' '3 1 4 2 5 3 6 4 7 3+υ V +υ V +υ V +υ V -υ V =0

(5.4.44)


( )iv iv iv iv ''

1 1 1 2 2 3 3 4 4 5 3-υ π V +π V +π V +π V -π V + ( )iv iv iv iv ''

2 1 1 2 2 3 3 4 4 5 3-υ ω V +ω V +ω V +ω V +ω V

iv iv iv iv ''

3 1 4 2 5 3 6 4 7 3-υ V -υ V -υ V -υ V -υ V = 0

⇒ ( ) ( ) ( )iv iv iv1 1 3 1 3 1 1 2 2 2 4 2 1 3 2 3 5 3υ π +υ ω -υ V + υ π +υ ω -υ V + υ π +υ ω -υ V

( ) ( )iv ''1 4 2 4 6 4 2 5 1 5 7 3+ υ π +υ ω -υ V + υ ω -υ π -υ V = 0

⇒⇒⇒⇒ iv iv iv iv ''

1 1 2 2 3 3 4 4 5 3D V +D V +D V +D V +D V = 0 (5.4.45)

Thus the coupled differential equations of equilibrium for non symmetric sections are

iv iv iv iv '' '' '' ''

1 1 2 2 3 3 4 4 5 1 6 2 7 3 8 4 9 3 2 4A V + A V + A V + A V + A V + A V + A V + A V + A V =µ K (5.4.41)

iv iv iv iv '' '' '' ''

1 1 2 2 3 3 4 4 5 1 6 2 7 3 8 4 9 3 2 4B V +B V +B V +B V +B V +B V +B V +B V +B V = ε K (5.4.42)

iv iv iv iv '' '' '' ''

1 1 2 2 3 3 4 4 5 1 6 2 7 3 8 4 9 3 2 4C V +C V +C V +C V +C V +C V +C V +C V +C V = γ K (5.4.43)

iv iv iv iv ''

1 1 2 2 3 3 4 4 5 3D V +D V +D V +D V +D V = 0 (5.4.45)

where

( )1 3 1 1 1A = µ ω +µ π ( )2 1 2 3 2A = µ π +µ ω

( )3 1 3 3 3A = µ π +µ ω ( )4 1 4 3 4A = µ π +µ ω

( )5 2 1 4A = µ ω +µ ( )6 2 3 3 5A = µ ω+µ ω -µ

( ) ( )7 1 5 2 3 6 8 3 4 7A = µ π +µ ω -µ A = µ ω -µ ( )9 2 5A = µ ω

112

( )

( )

( )

( )

1 1 1 3 1

2 1 2 3 2

3 1 3 3 3

4 1 4 3 4

B = ε π + ε ω

B = ε π + ε ω

B = ε π + ε ω

B = ε π + ε ω

( )

( )

( )

( )

5 2 1 4

6 2 2 5

7 1 5 2 3 6

8 2 4 7

B = ε ω + ε

B = ε ω - ε

B = ε π + ε ω - ε

B = ε ω - ε

( )9 2 5B = ε ω (5.4.47)

( )

( )

( )

( )

1 1 1 3 1

2 1 2 3 2

3 1 3 3 3

4 1 4 3 4

C = γ π +γ ω

C = γ π + γ ω

C = γ π + γ ω

C = γ π +γ ω

( )

( )

( )

( )

5 2 1 4

6 2 2 5

7 1 5 2 3 6

8 2 4 7

C = γ ω + γ

C = γ ω - γ

C = γ π + γ ω -µ

C = γ ω - γ

( )9 2 5C = γ ω (5.4.48)

( )

( )

( )

1 1 1 3 1 3

2 1 2 2 2 4

3 1 3 2 3 5

D = υ π +υ ω -υ



( )

( )4 1 4 2 4 6

5 2 5 1 5 7


D = υ ω -υ π -υ (5.4.49)

Table 5.1 Summary of Differential Equations of Equilibrium

Equilibriu

m

Condition

(a)

Doubly

symmetric

Sections

(b)

Mono-symmetric Sections

(c) Non-symmetric sections

Flexural-

torsional

equilibriu

m

qiv 2V =2

Ea22

qiv 4V =4

r G44

iv ivα V + α V - β V '' = K4 42 2 1 21

V '' = K4 1

iv iv iv iv '' ''A V + A V + A V + A V + A V + A V +4 4 51 1 2 2 3 3 1 6 2

'' ''+A V + A V + A V = µK7 4 43 8 9 3

iv iv iv iv '' ''B V + B V + B V + B V + B V + B V +4 4 51 1 2 2 3 3 1 6 2

'' ''+B V + B V + B V = ε K7 4 43 8 9 3 2

Flexural-

distortion

al

equilibriu

m

qiv 2V =2

Ea22

iv 2 4V - α V '' + 4β V = K3 3 3 3

γ

iv ivα V + α V - β V '' = K1 2 2 3 1 3 3

iv ivα V + α V - β V '' - V = -K4 43 2 3 2 3 1 3

113

Torsional

-

distortion

al

equilibriu

m

β V '' - γ V = -K41 1 3 1

iv ivV + α V + β V '' = K4 43 1 2 2

γβ V '' - V = K41 1 3 1

iv ivV + α V - β V '' = K4 43 2 2 2

iv iv iv iv '' ''C V + C V + C V + C V + C V + C V +4 4 51 1 2 2 3 3 1 6 2

'' ''+C V + C V + C V = τ K7 4 43 8 9 3 2

iv iv iv iv ''D V + D V + D V + D V + D V = 04 4 51 1 2 2 3 3 3

Flexural-

torsional-

distortion

al

equilibriu

m

γ

ivV = K2 0

β V '' - V = - K41 1 3 1

iv ivV - α V + β V '' = K4 43 1 2 2

β γβ V '' + β V '' + V '' - V = - K4 5 42 3 6 1 3 3

(2)γ

iv iv ivα V + α V + α V - β V '' - β4 5 4 72 3 6 2 8 3

-β V '' + V = K 4 49 2 3

(3)γ

iv iv ivα V +α V +α V -β V ''-β V ''7 42 8 3 9 10 2 11 3-β V ''+ V = K 4 512 3 3

114

CHAPTER SIX

TORSIONAL – DISTORTIONAL ANALYSIS OF BOX GIRDER SECTIONS 6.1 PREAMBLE

In this chapter the solutions of the differential equations of equilibrium derived in

chapter five are obtained and used to analyze the various cross sections of box

girder structures whose Vlasov’s coefficients were evaluated in chapter four.

Specifically, the following cross sections are considered.

(a) Single cell doubly symmetric section

(b) Multi cell doubly symmetric section

(c) Single cell mono-symmetric section

(d) Double cell mono-symmetric section

Each of these sections was analyzed for

(i) flexural- torsional interaction

(ii) flexural-distortional interaction

(iii) torsional-distortional interaction

(iv) flexural-torsional-distortional interaction.

It should be pointed out here that closed form solutions of the differential

equations require knowledge of the boundary conditions of the bridge girder. To this

effect, a three span simply supported bridge deck structure, Fig.6.2.1, was

considered. The torsional loads were obtained from AASHTO-LRFD(1998)

specifications. The dead load on the structure is not considered as they do not

contribute to torsional load (except where there is eccentricity in construction). The

flexural displacements concern strain mode 2 (horizontal bending) which has

interaction with strain modes 3 and 4. The vertical bending displacements concern

strain mode 1 which has no interaction whatsoever with other strain modes (except in

non symmetric sections) and is therefore not considered in the analysis.

Some of the governing equations of equilibrium were integrated by method of

trigonometric series with accelerated convergence, and others by Laplace transform

and traditional methods.

115

6.2 EVALUATION OF LOADS

Live loads are considered according to AASHTO-LRFD (AASHTO 1998),

following the HL-93 loading. Uniform lane load of 9.3N/mm distributed over a

3000mm width plus tandem load of two 110 KN axles. The loads are positioned at

the outermost possible location to generate the maximum torsional effects, Fig.6.2.2b

and Fig.6.2.4b. The tandem load is swept along the longitudinal direction of the

bridge axis, while the uniform lane load is positioned on each span separately. A

three span simply supported box girder bridge, Fig.6.2.1, is considered in the

analysis.

6.2.1 Doubly-symmetric sections

50m 50m 50m

Fig 6.2.1 Three Span Simply Supported Box Girder Bridge

e2 1500

600

100

110KN CL

(b) Tandem load (one axle shown)

Fig.6.2.2 Positioning of live loads for single lane bridge (AASHTO 1998)

3000mm 600mm

100mm

e1

CL

b

9.3KN/m

(a) Lane load

e2 2200m

P=138KN CL

b

h

(a) Torsional Load

P/2 CL

b

h

(b) Bending Component

P/2

116

6.2.2 Mono-symmetric sections

Fig.6.2.3 Components of bridge loads for doubly symmetric section

b

h=3m

(e) Pure Rotation

Tb,rot

Mq =

2h

Tt,rot

Mq =

2h

Tw,rot

Mq =

2b

Tw,rot

Mq =

2b

b

h=3m

(c) Torsionl Component

MT/b MT/b

b

h=3m

(d) Distortional

Component

Tw,dist

Mq =

2b

Tb,dist

Mq =

2h

Tw,dist

Mq =

2b

Tt,dist

Mq =

2h

1800

(a) Lane load (b) Tandem load (one axle shown)

600

b

110KN

3000 9.3 N/mm

CE

b 1800

Fig. 6.2.4 Positioning of Live Loads for Single Lane Bridge

2745 2745

2200 e 138KN

(C) Lane load + tandem load

2200 e 138KN

(a) Bridge eccentric Load

3.05m ⇒⇒⇒⇒

69KN 69KN

b

h = 3.05m

(b) Bending component (c) Torsional component

27.52KN

b

h = 3.05m ++++

Fig. 6.2.5 Components of Bridge Eccentric Load

117

Table 6.1 Distortional and rotational forces in the plates

S/No Sections Distortional and rotational forces (kN)

t,distq b,distq w,distq t,rotq b,rotq w,rotq

1 Single cell

doubly

symmetric

33.58 33.58 13.76 33.58 33.58 13.76

2 Multi-cell

doubly

symmetric

52.90 52.90 17.63 52.90 52.90 17.63

3 Single cell

mono-

symmetric

11.01 22.02 10.70 44.05 20.00 21.40

4 Doule cell

mono-

symmetric

11.01 22.02 10.70 44.05 20.00 21.40

t,distq = top flange plate distortional force, b,distq = bottom flange plate distortional force

w,distq = distortional force in the web plate, t,rotq = top flange plate rotational force

b,rotq = bottom flange plate rotational force, w,rotq = rotational force in the web plate

The distortional and rotational forces in the box girder plate elements are as shown in

Table 6.1. Details of the computations are shown in Appendix Four.

(c) Distortional component

(warping)

wDq wDq

tDq

bDq

Fig. 6.2.6 Components of the Torsional Load

(b) Pure torsional

component (rotation)

wRq

tRq

bRq

wRq

(a) Torsional load

27.52KN

b

h = 3.05m

27.52KN

⇒⇒⇒⇒ ++++

118

Table 6.2 Cross sectional dimensions and bridge loads

S/No Sections Dimensions Box girder loads

b (m) h (m) MT (kNm) 2q (kN) 3q (kN) 4q (kN)

1 Single cell

doubly

symmetric

7.32 3.05 201.48 0.00 620.45 1462.62

2 Multi-cell

doubly

symmetric

9.00 3.05 317.40 0.00 6495.23 2856.33

3 Single cell

mono-

symmetric

7.32 3.05 201.45 0.00 157.16 1446.51

4 Doule cell

mono-

symmetric

7.32 3.05 201.45 0.00 157.16 1446.51

MT = torsional moment, 2q = load causing bending about minor axis

3q = distortional load, 4q = rotational load

The computed box girder loadings are given in Table 6.2. Details of the computation

are given in Appendix Four

6.3 ANALYSIS OF DOUBLY SYMMETRIC SECTIONS

6.3.1 Flexural- torsional analysis: single cell doubly symmetric section

The governing differential equations for flexural-torsional equilibrium are

iv 22

22

qV =

Ea (6.3.1)

iv 44

44

qV = -

r G (6.3.2)

We note that for doubly symmetric section there is no interaction between flexure and

torsion. Consequently, eqns. (6.3.1 & 6.3.2) are independent of one another.

For single cell doubly symmetric section we have, from Table 4.1(a) and (e),

22 44a = 27.994, r = 22.032

For concrete we have 9 2 9 2G = 9.6X10 N/m ; E = 24x10 N/m , υ= 0.25

119

From analysis of bridge loads in Table 6.2, we obtained that q2 = 0KN;

4q = 1462.63KN

Thus, the solution to eqn. (6.3.1) is V2(x) = 0

Let 41

44

qK =

r G, Eqn. (6.3.2) can be written as

1 1v '' = -K (6.3.3)

The Laplace transform of the terms in eqn. (6.3.3) are,

{ }iv 4 3 2

1 1 0 1 2 3v = s v - s C - s C - sC -C

ℓ{ } 11

KK =

s

where 0 1 1 1 2 1 3 1C = v (0); C = v '(0); C = v ''(0); C = v '''(0)

Substituting into eqn. (6.3.3) we obtain;

2

4 0 3 1s v - sC -C -K /s = 0

2

4 1 0 1s v = +sC +C -K /s

2 3 14 2

sC +C -K /sv =

s

⇒ 1 11 0 2

C Kv = C + -

s s (6.3.4)

Taking the inverse transform of eqn (6.3.4) we obtain,

2

1 0 1 1

x xv = C +C -K

1! 2!

211 0 1

Kv = C +C x - x

2 (6.3.5)

where , 0 1 2 3C , C , C , C are constants determined by the boundary conditions.

Boundary conditions

⇒4 0V (0) = 0; C = 0 ; ⇒2

44 1

K LV (L) = 0; C L - = 2

2 or 4

1

K LC =

2

Substituting into eqn. (6.3.5) gives

2

1 14

K Lx K xV = -

2 2

For L = 50m we have

120

2

4 1 1V = 25K x -0.5K x ; (a)

Where, 3

-641 9

44

q 1462.63 *10K = - = = 6.9153 *10

r G 22.032* 9.6 *10

Substituting into (a) above we obtain the expression for variation of torsional

displacement along the length of the girder. Thus,

(6.3.6)

Table 6.3 Variation of torsional

displacement along the girder length

(single cell doubly symmetric section)

Distance x,

along the

length of

the girder

(m)

Torsional displacement ,

V4(x) x 10 -3(m)

0 0.000

5 -0.778

10 -1.400

15 -1.816

20 -2.075

25 -2.161

30 -2.075

35 -1.816

40 -1.400

45 -0.778

50 -0.000

-6 2 -4

4V (x) = 3.4577 *10 x -1.729 *10 x

121

6.3.2 Flexural-Torsional Analysis: Multi- cell doubly symmetric section

The governing differential equations for flexural-torsional equilibrium are the

same as for single cell doubly symmetric sections but the coefficients are different.

Thus we have,

iv 22

22

iv 44

44

qV =

Ea

qV = -

r G

(6.3.7)

22 44a = 27.994, r = 22.032; 9 2 9 2G = 9.6X10 N/m ; E = 24x10 N/m , υ= 0.25

From analysis of bridge loads we obtained that 2q = 0KN; 4q = 1462.63KN

Applying the same procedure as in single cell doubly symmetric sections we obtain

V2(x) = 0

2

4 2 2V (x) = 25K (x) -0.5K x

where , 3

-642 9

44

q 2856.33 *10K = - = - - 8.4768 *10

r G 35.100 * 9.6 *10

Therefore,

-4 -6 2

4V (x) = - 2.119 *10 x + 4.239 *10 x

Table 6.4 Variation of torsional

displacement along the girder length

(multi- cell doubly symmetric section)

Distance x,

along the

length of

the girder

(m)

Torsional displacement

V4(x) x 10 -3(m)

0 0.000

5 -0.954

10 -1.695

15 -2.225

20 -2.542

25 -2.648

122

30 -2.542

35 -1.225

40 -1.695

45 -0.954

50 -0.000

The variation of torsional displacement V4, along the length of the girder is shown in

Fig.6.3.2

6.3.3 Flexural-distortional analysis: single cell doubly symmetric section

The governing equations for flexural-distortional equilibrium for doubly symmetric

section are

( )

( )

a

b

iv 22

22

iv 2 4

3 3 3 3 3 3

qV =

Ea

V -α V ''+ 4β V =K

(6.3.8)

where

-3

33 22 33 33a = 6.721, a = 27.994, c = 3.736, s =1.062*10

υ9 2 9 2E = 24 *10 N/m , G = 9.6 *10 N/m , = 0.25

From analysis of bridge loading 2 3q = 0.0KN and q = 620KN

Consequently, V2 (x) = 0

Let the solution for eqn. (6.3.8b) be sought in the form; λx3V (x) = e

∴ λx3V '(x) = λe , 2 λx

3V ''(x) = λ e , iv 4 λx3V (x) = λ e (6.3.9)

Substituting eqn. (6.3.9) into eqn. (6.3.8) we obtain the auxiliary equation for

homogeneous condition as follows.

( )λx 4 2 2 4e λ -α λ + 4β = 0 ⇒ 4 2 2 4λ -α λ + 4β = 0

Let 2υ = λ ∴ 2 2 4υ -α υ+ 4β = 0

⇒2 4 4

1,2

α ± α -16β=

2υ ⇒

2 44

1

α αυ = + - 4β

2 4,

2 44

2

α αυ = - - 4β

2 4

∴2 4

4

1,2 1

α αλ = υ = ± + - 4β

2 4

123

2 4

4

3,4 2

α αλ = υ = ± - - 4β

2 4

where 2 333

33

ksα =

c, 4 33

33

s4β =

a

-32 -4333

33

ks 2.5x1.0625x10α = = = 7.11x10

c 3.736

-34 -433

33

s 1.0625x104β = = =1.581x10

a 6.721

-4 -7-4

1,2

7.11x10 5.0552x10λ = ± + -1.581x10

2 4

= -4 -7 -4± 3.555x10 + 1.2638x10 -1.581x10 4 43.555 10 1.5797 10x i x− −= ± +

Similarly, -4 -4

3,4λ = ± 3.555x10 - i 1.5797x10

Let -4a = 3.555x10 and -4 -2b = 1.5797x10 =1.2569x10

∴ ( )1

21λ = a+ib ( )

1

22λ = - a+ib ( )

1

23λ = a - ib ( )

1

24λ = - a - ib

Substituting in eqn. (6.3.8) we obtain

( ) ( ) ( ) ( )1 1 1 1

2 2 2 2a+ib - a+ib a-ib - a-ib

3 1 2 3 4V (x) = A e + A e + A e + A e

⇒ 3 1 2 3 4V (x) = A coshωxcosγx + A coshωxsinγx + A sinhωxcosγx + A sinhωxsinγx

where =ω real value of

1-4 2 23.555x10 +i1.2569x10

λ = imaginary value of 1

4 2 23.555 10 1.2569 10x i x− +

For λ = a+ib ,ω = Re a + ib , γ =Im a+ib

where -4a = 3.555x10 , -2b =1.2569x10

∴ a+ib =ω+iγ ⇒ ( )2 2 2a+ib = ω+iγ =ω +2iωγ - γ

( )2 2a+ib = ω - γ +2iωγ (6.3.12)

Equating equivalent terms on both sides of eqn. (6.3.10) we obtain

2 2ω - γ = a ……….(a) and 2ωγ = b ……..(b)

Squaring both sides of expressions (a) and (b) and adding up we have

( )2

2 2 2 2 2 2ω - γ + 4ω γ = a +b ⇒ 4 2 2 4 2 2ω - 2ω γ + γ + 4ω γ = a+b

124

( )⇒ 4 2 2 4 2 2ω +2ω γ +γ = a +b

( ) ( )⇒2

2 2 2 2ω +γ = a +b ∴ 2 2 2 2ω +γ = a +b ……….(c)

Adding expressions (a) and (c) we obtain

2 2 22ω = a+ a +b

⇒2 2a+ a +b

ω =2

( ) ( )

=

2 2-4 -4 -23.555x10 + 3.555x10 + 1.2569x10

2

=-4 -23.555x10 +1.2572x10

= 0.08042

Subtracting (a) from (c) we obtain

2 2 22γ = a +b -a

⇒2 2a +b -a

γ =2

( ) ( )

=

2 2-4 -2 -43.555x10 + 1.2569x10 -3.555x10

2

⇒ -36.10926x10 = 0.0782γ =

Thus, ω = 0.0804 , γ = 0.0782

3 homogeneous 1 2 3 4V (x) = A coshωxcosλx + A coshωxsinγx + A sinhωxcosγx + A sinhωxsinγx

Let the particular integral be sought in the form; 3V (x) = B (6.3.13)

This solution must satisfy eqn. (6.3.8b)

i.e., iv 2 4

3 3 3 3 3 3V -α V ''+ 4β V = K

From eqn. (6.3.13) we obtain iv

3 3V '' = 0, V = 0

Substituting into eqn. (6.3.8b) we obtain

4

44β B = K ∴ 3

4

KB =

4β ⇒ ( ) 3

3 4particular

KV x =

4β

where -123 33 39

33

q qK = = = 6.1995x10 q

a E 6.721x24x10

Boundary Conditions;

V(0) = 0; V(L) = 0; V''(0) = 0;

V''(L) = 0

33 1 2 3 4 4

KV (x) = A coshωxcosλx + A coshωxsinγx + A sinhωxcosγx + A sinhωxsinγx +

4β

125

[ ]3 1 2 3 4V ''(x) = 2 -A coshωxcosλx + A coshωxsinγx - A sinhωxcosγx + A sinhωxsinγx

∴ ⇒ 31 4

KV(0) = 0 : A = -

4β (a)

⇒''

3 4V (0) = 0 : 2A = 0 ⇒ 4A = 0 (b)

0 : ⇒ 31 2 3 4

KV(L) = -20.0315A -19.632A +20.019A + = 0

4β (c)

V''(L) = 0 : ⇒ 1 2 3-38.699A - 40.032A +38.724A = 0 (d)

Solving expressions (a), (b), (c), and (d) we obtain

31 4

KA = -

4β; ; 3

2 4

KA = 0.0256

4β 3

3 4

KA = 0.9649

4β; 4A = 0

Substituting values of 1 2 3 4A , A , A , & A into eqn. (6.3.14) we obtain

[ ]33 4

KV (x) = -coshωxcosλx +0.0256coshωxsinγx +1.0258sinhωxcosγx +1

4β

or [ ]33 4

KV (x) = 1.0258sinhωxcosγx - coshωxcosλx +0.0256coshωxsinγx +1

4β

where, 4 -44β =1.581x10 and -12

3 3K = 6.1995x10 q (6.3.15)

Substituting value of 3q into eqn. (6.3.15) we obtain

-12 33

4 -4

K 6.1995 *10 * 620.45 *10= = 0.02433

4β 1.581*10

∴ [ ]3V (x) = 0.02433 1.0258 * SinhωxCosγx -CoshωxCosγx +0.0256 * CoshωxSinγx +1

(6.3.16)

where, ω = 0.0804radian; γ = 0.0782radian

Table 6.5 Variation of distortional displacement along the girder length

(Single cell doubly symmetric section)

Distance (x)

from Left

Support (m)

1+ 1.0258 x

SinhwxCosvx -CoshwxCosvx 0.0256 x

CoshwxSinvx

Distortional displacement V3(x)

0 1.000 -1.0000 0.0000 0.000 x 103 (m)

5 1.3916 -1.0002 0.0106 9.78

10 1.6503 -0.9514 0.0242 17.39

15 1.6042 -0.7049 0.0429 22.92

126

20 1.0167 -0.0176 0.0665 25.93

25 -0.4090 1.4238 0.0902 26.88

30 -2.9725 3.9353 0.1028 25.93

35 -6.8345 7.6926 0.0843 22.93

40 -11.7639 124830 -0.0043 17.39

45 -16.7543 17.3327 -0.1759 9.79

50 -19.5351 20.0315 -0.4956 0.00

6.3.4 Flexural distortional analysis: multi- cell doubly symmetric section

The governing equations of flexural-distortional equilibrium for doubly symmetric

sections are

( )

( )

a

b

iv 22

22

iv 2 4

3 3 3 3 3 3

qV =

Ea

V -α V ''+ 4β V =K

(6.3.17)

where 2 433 33 33 3 3

33 33 33

ks s qα = ; 4β = ; K =

c a a E

For multi-cell doubly symmetric section we have;

c-2

33 33 33a =127.752, = 55.727, s = 4.338 *10

υ9 2 9 2E = 24 *10 N/m , G = 9.6 *10 N/m , = 0.25

From analysis of bridge loading 2 3q = 0.0KN and q = 6495.24KN

Consequently, V2 (x) = 0

Let the solution for eqn. (6.3.17b) be sought in the form; λx3V (x) = e (6.3.18)

∴ λx3V '(x) = λe , 2 λx

3V ''(x) = λ e , iv 4 λx3V (x) = λ e

Substituting derivatives of eqn. (6.3.18) into eqn. (6.3.17) we obtain the

characteristic equation for homogeneous condition as follows.

( )λx 4 2 2 4e λ -α λ + 4β = 0 ⇒ 4 2 2 4λ -α λ + 4β = 0

Let 2υ = λ ∴ 2 2 4υ -α υ+ 4β = 0

⇒2 4 4

1,2

α ± α -16β=

2υ ⇒

2 44

1

α αυ = + - 4β

2 4,

2 44

2

α αυ = - - 4β

2 4

127

∴2 4

4

1,2 1

α αλ = υ = ± + - 4β

2 4

2 4

4

3,4 2

α αλ = υ = ± - - 4β

2 4

where 2 333

33

ksα =

c, 4 33

33

s4β =

a

-32 -3333

33

ks 2.5 * 4.338 *10α = = =1.946 *10

c 55.727

-24 -433

33

s 4.338 *104β = = = 3.396 *10

a 127.752

-3 -6-4

1,2

1.946 *10 3.787 *10λ = ± + -3.396 *10

2 4

4 49.73*10 3.3865*10i− −= ± +

Similarly, -4 -4

3,4λ = ± 9.73 *10 - i 3.3865 *10

Let -4a = 9.73 *10 and -4 -2b = 3.3865 *10 =1.840210

∴ ( )1

21λ = a+ib ; ( )

1

22λ = - a+ib ; ( )

1

23λ = a - ib ; ( )

1

24λ = - a - ib

Substituting in eqn. (6.3.18) we obtain

( ) ( ) ( ) ( )1 1 1 1


3 1 2 3 4V (x) = A e + A e + A e + A e

⇒ 3 1 2 3 4V (x) = A coshωxcosγx + A coshωxsinγx + A sinhωxcosγx + A sinhωxsinγx


1-3 2 29.73 *10 +i1.8402*10

λ = imaginary value of

1-3 2 29.73 *10 +i1.8402*10

For λ = a+ib ,ω = Re a + ib , γ =Im a+ib

where -4a = 9.73 *10 and -2b =1.840210

∴ a+ib =ω+iγ ⇒ ( )2 2 2a+ib = ω+iγ =ω +2iωγ - γ

( )2 2a+ib = ω - γ +2iωγ (6.3.19)

Equating equivalent terms on both sides of eqn. (6.3.19) we obtain

2 2ω - γ = a ……….(a) and 2ωγ = b ……..(b)

128


( )2

2 2 2 2 2 2ω - γ + 4ω γ = a +b ⇒ 4 2 2 4 2 2ω - 2ω γ + γ + 4ω γ = a+b

( )⇒ 4 2 2 4 2 2ω +2ω γ +γ = a +b

( ) ( )⇒2

2 2 2 2ω +γ = a +b ∴ 2 2 2 2ω +γ = a +b ……….(c)


2 2 22ω = a+ a +b ⇒2 2a+ a +b

ω =2

= -3 -2ω 4.865 *10 +1.04076 *10 = 0.12358


2 2 22γ = a +b -a

⇒2 2a +b -a

γ =2

( ) ( )

=

-5 -4 -39.467 *10 + 3.386 *10 -9.73 *10

2

34.865*10−= −-21.04076 *10 = 0.074449

Thus, ω = 0.12358 , γ = 0.074449

3 homogeneous 1 2 3 4V (x) = A coshωxcosλx + A coshωxsinγx + A sinhωxcosγx + A sinhωxsinγx

Let the particular integral be sought in the form; 3V (x) =B (6.3.20)

This solution must satisfy eqn. (6.3.17b)

i.e, iv 2 4

3 3 3 3 3 3V -α V ''+ 4β V =K

From eqn. (6.3.20) we obtain iv

3 3V '' = 0, V = 0

Substituting into eqn. (6.3.17b) we obtain

4

44β B =K ∴ 3

4

KB =

4β ⇒ ( ) 3

3 4particular

KV x =

4β

where -133 33 39

33

q qK = = = 3.2615 *10 q

a E 127.752* 24 *10


V(0) = 0; V(L) = 0; V''(0) = 0; V''(L) = 0

33 1 2 3 4 4

KV (x) = A coshωxcosλx + A coshωxsinγx + A sinhωxcosγx + A sinhωxsinγx +

4β

(6.3.21)

129

∴ ⇒ 31 4

KV(0) = 0 : A = -

4β (a)

⇒''

3 4V (0) = 0 : 2A = 0 ⇒ 4A = 0 (b)

⇒ + 31 2 3 4 4

KV(L) = 0 : 240.8736A +15.6631A +240.744A +15.663A = 0

4β (c)

V''(L) = 0 : ⇒ 1 2 3 4-15.663A +240.744A -15.66314A +240.74665A = 0 (d)


31 4

KA = -

4β; -4 3

2 4

KA = - 2.34 *10

4β; 3

3 4

KA = 0.9964

4β; 4A = 0

Substituting values of 1 2 3 4A , A , A , & A into eqn. (6.3.21) we obtain

-43

3 4

KV (x) = -CoshωxCosλx - 2.34 *10 * CoshωxSinγx +1.0258SinhωxSosγx +1

4β

or [ ]33 4

KV (x) = 0.9964SinhωxCosγx -CoshωxCosλx +0.063576CoshωxSinγx +1

4β

where, 4 -44β = 3.396 *10 , -13

3 3K = 3.2615 *10 q and 3q = 6495.24KN

-13 3-33

4 -4

K 3.2615 *10 * 6495.24 *10= = 6.238 *10

4β 3.396 *10

∴ [ ]V (x) = 0.00624 0.9964 * SinhωxCosγx - CoshωxCosγx + 0.06358 * CoshωxSinγx +13

(6.3.22)

where, ω = 0.12358 radian ; γ = 0.074449 radian

Table 6.6 Variation of distortional displacement along the girder length

(Multi- cell doubly symmetric section)

Distance (x)

from Left

Support (m)

1+ 1.0258 x

SinhwxCosvx -CoshwxCosvx 0.0256 x

CoshwxSinvx

Distortional displacement V3(x)

0 1.000 -1.0000 0.0000 0.000 x 103 (m)

5 1.6556 -1.1970 -0.0001 2.863

10 2.5695 -1.8657 -0.0002 4.390

15 4.1016 -3.2694 -0.0003 5.195

20 6.855 -5.9609 -0.0013 5.579

25 11.915 -11.000 -0.0030 5.851

30 21.2729 -20.386 -0.0070 5.403

35 38.6125 -37.7616 -0.0143 5.220

130

40 70.7588 -70.0180 -0.0303 4.433

45 130.3628 -129.8341 -0.0632 2.900

50 240.877 -240.7468 -0.1302 0.000

6.3.5 Torsional - distortional analysis: single cell doubly symmetric section

The governing equations of equilibrium are

γ1 4 1 3 1β V '' - V = -K (6.3.23)

iv iv

3 1 4 2 4 2V +α V +β V '' =K (6.3.24)

γ

2

34 43 33 44 3344 441 1 2 1

43 33 43 33 43 33

r c -b r ksr rα = β = - ; β = ; =

c c c ka c c

3 33 441 2

33 43 33 43

q b qqK = - ; K =

c G c G a c E

-3

33 33 33 43 44 33a = 6.721, b = c = 3.736; c = 3.732; r = 22.032; s =1.062*10

υ 9 2 9 2= 0.25; E = 24 *10 N/m ; G = 9.6 *10 N/mm

From analysis of loads in section 6.2 we obtained that

3 4q = 620KN; q =1462.62KN .

The coefficients for the governing equations are evaluated using the expressions

above. Thus,

γ -4

1 1 1α = 5.904; β = -4.905; β = -1.0905; = 7.106 *10

Substituting the coefficients into eqns.(6.3.23 & 6.3.24) we obtain

-4

4 3 1 4.905V '' + 7.106 *10 V = K (6.3.25)

iv iv

3 4 4 2V +5.904V -1.0905V '' =K

The solution of these coupled differential equations is obtained using the method of

trigonometric series with accelerated convergence

Boundary conditions; 3 3 3 3V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0

4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0

We seek the solution in the form;

3 3 3

4 4 4

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.3.26)

131


-4 -4

4 3 4 14.905f ''+7.106 *10 f + 4.905g ''+7.106 *10 g =K

⇒ -4 -4

4 3 4 3 14.905g ''+7.106 *10 g = - 4.905f '' - 7.106 *10 f +K

⇒ -4

4 3 14.905g ''+7.106 *10 g =P (x)

where, -4

1 4 3 1P (x) = - 4.905f '' - 7.106 *10 f +K

Similarly,

iv iv iv iv

3 4 4 3 4 4 2f +5.904f -1.0905f ''+g +5.904g -1.0905g '' =K

⇒ iv iv iv iv

3 4 4 3 4 4 2g +5.904g -1.0905g '' = - f - 5.904f +1.0905f ''+K

⇒ iv iv

3 4 4 2g +5.904g -1.0905g '' =P (x)

where, iv iv

2 3 4 4 2P (x) = - f - 5.904g +1.0905g ''+K

∴ ( )

( )

a

b

-4

1 4 3 1

iv iv

2 3 4 4 2

P (x) = - 4.905f '' - 7.106 *10 f + K

P (x) = - f - 5.904g +1.0905g ''+ K (6.3.27)

We seek the complimentary function 3g (x)and 4g (x) in the form of sine series as

follows.

∞

∞

∑

∑

n3 3n

n=1

n4 4n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.3.28)

where nα = nπ

The auxiliary function can be sought in terms of algebraic function as follows.

2 3

3 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

4 0 1 2 3f (x) =B +B x +B x +B x (b) (6.3.29)

The constants 0 0 2 2A , B , A & B are easily defined using some of the boundary

conditions.

i.e. ⇒3 0V (0) = 0 A = 0

∑ n3 2 3 3n

α xV ''(x) = 2A +6A x - a sin

L

⇒3 2V ''(0) = 0 2A = 0 or 2A = 0

⇒4 0V (0) = 0 B = 0

∑ n4 2 3 4n

α xV ''(x) = 2B +6B x - a sin

L

132

⇒4 2V ''(0) = 0 2B = 0 or 2B = 0

Substituting values of 0, 0 1 1A B , A & B in eqn. (6.3.29) we obtain;

3

3 1 3f (x) = A x + A x (a)

3

4 1 3f (x) =B x +B x (b) (6.3.30)

The constants 1 3 1 3A , A , B & B are obtained using the remaining boundary

conditions. Thus, 3 3V (L) = f (L) = 0 ⇒ 3

1 3A L + A L = 0

3 3V ''(L) = f ''(L) = 0 ; ⇒ 36A L = 0 ∴ 3A = 0 and 1A = 0

4 4V (L) = f (L) = 0 ⇒ 3

1 3B L +B L = 0

4 4V ''(L) = f ''(L) = 0 ; 36B L = 0 ∴ B3 = 0 and B1= 0

Hence, 3 3

4 4

f (x) =P (x) = 0

f (x) =P (x) = 0 } ⇒

3 3

4 4

V (x) = g (x)

V (x) = g (x) (6.3.31)

Consequently eqn. (6.3.26) becomes

0=-4

4 14.905g ''+7.106 *10 g-K (a)

0=iv iv

3 4 4 2g +5.904g -1.0905g '' -K (b) (6.3.32)

But ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n4 4n

n=1

α xg (x) = a sin

L (b)

∞

∑ 2 n3 3n n

n=1

α xg ''(x) = - a α sin

L (c)

∞

∑iv 4 n3 3n n

n=1

α xg (x) = a α sin

L (d)

∞

∑ 2 n4 4n n

n=1


L (e)

∞

∑iv 4 n4 4n n

n=1


L (f) (6.3.33)

Substituting eqn. (6.3.33) into eqn. (6.3.32) we obtain;

∑ ∑

∑ ∑ ∑

2 -4 n4n n 3n 1

4 4 2 n3n n 4n n 4n n 2

α x-4.905 a α +7.106 *10 a sin =K

L

α xa α +5.904 a α +1.0905 a α sin =K

L

(6.3.34)

Equations (6.3.34) will always be satisfied if

133

( )α 2 -4

n 4n 3n 1-4.905 a +7.106 *10 a =K (6.3.35)

and ( )4 4 2

n 3n n n 4n 2α a + 5.904α +1.0905α a =K (6.3.36)

From eqn. (6.3.35), we have

-4

3n14n 2 2

n n

7.106 *10 aKa = - +

4.905α 4.905α (6.3.37)

For n = 1, ,2 -3 4 -5

n n n

π πα = = , α = 3.9478 *10 α =1.5585 *10

L 50

∴-4

-231 141 31 1-2 -2

7.106 *10 α Ka = - = 3.6697 *10 a - 51.642K

1.9364 *10 1.9364 *10

Substituting into eqn. (6.3.36) we have

( )-5 -3 -2

31 1 21.5585 *10 a + 4.397 *10 3.6697 *10 -51.642K =K

⇒

⇒

-4

31 2 1

31 2 1

1.7694 *10 a =K +0.22707K

a = 5651.63K +1283.32K

⇒ 41 2 1a = 207.397K - 4.623K

where = -53 41

33 43

q qK = - -2.2856 *10

c G c G

-633 42

33 43

b qK = = 8.634 *10

a c E

∴ -2

31a =1.946 *10 , -3

41a =1.896 *10

Substituting values of a31, a32 and eqn. (6.3.28) into eqn. (6.3.31) we obtain

-2

3

-3

4

πxV (x) =1.946 *10 Sin

50

πxV (x) =1.896 *10 Sin

50

Table 6.7 Variation of torsional and distortional

displacements along the length of the girder

(Single cell doubly symmetric section)

Distance (x)

from Left

Support (m)

πxSin

50

Distortional

displacement

V3(X)

X 10-3

(m)

Torsional

displacement

V4(X)

X 10-3

(m)

0 0.000 0.00 0.000

134

5 0.309 6.013 0.586

10 0.588 11.442 1.115

15 0.809 15.743 1.534

20 0.951 18.506 1.803

25 1.000 19.460 1.896

30 0.951 18.506 1.803

35 0.809 15.743 1.534

40 0.588 11.442 1.115

45 0.309 6.013 0.586

50 0.000 0.000 0.000

6.3.6 Torsional-distortional analysis: multi cell doubly symmetric section


γ1 4 1 3 1β V '' - V = -K (6.3.38)

iv iv

3 1 4 2 4 2V +α V +β V '' =K

; γ

2

34 43 33 44 3344 441 1 2 1

43 33 43 33 43 33

r c -b r ksr rα = β = - ; β = ; =

c c c ka c c

3 4 22 41 2

33 43 33 43

q q b qK = - ; K =

c G c G a c E

33 33 33 43 43a =127.752, b = c = 55.752; c = r = 3.732;

-2

44 33r = 35.100; s = 4.338 *10

υ 9 2 9 2= 0.25; E = 24 *10 N/m ; G = 9.6 *10 N/mm

From analysis of loads in section 6.2 we obtained that;

3 4q = 6495KN; q = 2856.33KN .

The coefficients for the governing equations are evaluated using the expressions

above. Thus,

γ -3

1 1 2 1α =1.418; β = - 0.973; β = - 0.1698; =1.946 *10

-7 -6

1 2K =1.2278 *10 ; K = 2.097 *10

Substituting the coefficients into eqns. (6.3.38) we obtain

135

-3

4 3 1

iv iv

3 4 4 2

-0.973V '' -1.946 *10 V = -K

V +1.418V -0.170V '' =K (6.3.39)

The solution of these coupled differential equations is obtained using the method of

trigonometric series with accelerated convergence


4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


3 3 3

4 4 4

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.3.40)

Substituting eqns. (6.3.39) into eqns. (6.3.40) we obtain

-3 -3

4 3 4 3 1-0.973f '' -1.946 *10 f -0.973g '' -1.946 *10 g = -K

iv iv iv iv

3 4 4 3 4 4 2f +1.418f -0.170f ''+g +1.418g -0.170g '' =K

⇒ -3

4 3 1-0.973f '' -1.946 *10 f =P (x)

⇒ iv iv

3 4 4 2f +1,418f -0.170f '' =P (x)

-3

1 4 1P (x) = -0.973g '' -1.946 *10 g+K (a)

iv iv

2 3 4 4 2P (x) = g +1.418g -0.170g '' -K (b) (6.3.41)


follows.

∞

∞

∑

∑

n3 3n

n=1

n4 4n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.3.42)

where nα = nπ


2 3

3 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

4 0 1 2 3f (x) =B +B x +B x +B x (b) (6.3.43)

The constants 0 0 1 1 2 2 3 3

A , B , A , B , A , B , A & B are easily defined using the boundary conditions.

From section 6.3.5 we note that for simply supported box girder the boundary

conditions are such that all the constants are zero.

Hence, 3 3

4 4

f (x) =P (x) = 0

f (x) =P (x) = 0 } ⇒

3 3

4 4

V (x) = g (x)

V (x) = g (x)


136

0

0

=

=

-3

4 3 1

iv iv

3 4 4 2

-0.973g '' -1.946 *10 g +K

g +1.418g -0.170g '' -K (6.3.45)

But ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n4 4n

n=1

α xg (x) = a sin

L (b)

∞

∑ 2 n3 3n n

n=1


L (c)

∞

∑iv 4 n3 3n n

n=1


L (d)

∑¥

2 n4 4n n

n=1


L (e)

∞

∑iv 4 n4 4n n

n=1


L (f) (6.3.46)


0.973

∑ ∑

∑ ∑ ∑

2 -3 n4n n 3n 1

4 4 2 n3n n 4n n 4n n 2

α xa α -1.946 *10 a sin = -K

L

α xa α +1.418 a α +0.170 a α sin =K

L

(6.3.47)


( )α 2 -3

n 4n 3n 10.973 a -1.946 *10 a = - K (6.3.48)

and ( )4 4 2

n 3n n n 4n 2α a + 1.418α +0.170α a =K (6.3.49)

From eqn. (6.3.48), we have

-3

14n 3n2 2

n n

K 1.946 *10a = - + a

0.973 α 0.973 α (6.3.50)

For n = 1, 2 -3 4 -5

1 1 1

π πα = = , α = 3.9478 *10 , α =1.5585 *10

L 50

∴-3 -7

-53141 31-3 -3

1.946 *10 α 1.2346 *10a = - = 0.5066a - 3.196 *10

0.973 * 3.9478 *10 0.973 * 3.9478 *10


( )-5 -4 -5 -6

31 311.5585 *10 a +6.9323 *10 0.5066a -3.196 *10 = 2.097 *10

⇒ -4 -8 -6

313.6677 *10 a = 2.2158 *10 +2.097 *10

137

3

45.779*10

3.667*10

−

−⇒ =

-6

31

2.119 *10a =

⇒ -3 -5 -3

41a = 0.5066 * 5.779 *10 -3.196 *10 = 2.896 *10


-3

3

-3

4

πxV (x) = 5.779 *10 Sin

50

πxV (x) = 2.896 *10 Sin

50



( Multi cell doubly symmetric section)

Distance (x)

from Left

Support (m)

πx

Sin50

Distortional

displacement

V3(X)

X 10-3

(m)

Torsional

displacement

V4(X)

X 10-3

(m)

0 0.000 0.000 0.000

5 0.309 1.786 0.895

10 0.588 3.398 1.703

15 0.809 4.675 2.343

20 0.951 5.496 2.754

25 1.000 5.779 2.896

30 0.951 5.496 2.754

35 0.809 4.675 2.343

40 0.588 3.398 1.703

45 0.309 1.786 0.895

50 0.000 0.000 0.000

6.3.7 Flexural-torsional-distortional analysis: single cell doubly symmetric

section

The differential equation governing flexural- torsional-distortional equilibrium for

doubly symmetric sections are

138

γ

iv

2 0

1 4 1 3 1

iv iv

3 1 4 2 4 2

V =K

β V '' - V = -K

V +α V +β V '' =K

(6.3.51)

The coefficients are as follows.

22 22 22 22a = 27.994, b = c = r = 2.880

33 33 33 43 43a = 6.721, b = c = 3.737; c = r = 3.732

-3

44 33r = 22.032; s =1.062*10

υ 9 2 9 2= 0.25; E = 24 *10 N/m ; G = 9.6 *10 N/mm

2 3 4q = 0.00KN, q = 620KN, q =1462.62KN

3444 441 1

43 33 43

rr rα = = 5.904; β = - = - 4.905

c c c

47.106*10γ −=2

43 33 44 332 1

33 43 33

c -b r ks β = = -1.0905 =

ka c c

-5 -63 33 441 2

33 43 33 43

q b qqK = - = - 2.2856 *10 ; K = = 8.634 *10

c G c G a c E

Substituting the coefficients into eqns. (6.3.51 ) we obtain

∴iv

2 2V = 0 V = 0

-4

4 3 14.905V ''+7.106 *10 V =K (6.3.52)

iv iv

3 4 4 2V +5.904V -1.0905V '' =K

These coupled equations are the same as eqns. (6.3.25) in torsional-distortional

analysis of section 6.3.5., and their coefficients are the same. Their solutions are

-2

3

-3

4

πxV (x) =1.946 *10 Sin

50

πxV (x) =1.896 *10 Sin

50

(6.3.53)

Therefore, flexural-torsional-distortional analysis of single cell doubly symmetrical

section results to torsional- distortional analysis of the same section because there

are no interactions between flexure and torsion or distortion in doubly symmetric

section.

139

Table 6.9 Variation of flexural, torsional and distortional

displacements along the length of the girder ( single cell doubly

symmetric section)

Distance (x)

from Left

Support (m)

πx

Sin50

Flexural

Displacement

V2(X) X 10-3

(m)

Distortional

Displacement

V3(X) X 10-3

(m)

Torsional

displacement

V4(X)

X 10-3

(m)

0 0.000 0.000 0.000 0.000

5 0.309 0.000 10.366 0.799

10 0.588 0.000 18.480 1.460

15 0.809 0.000 22.782 1.879

20 0.951 0.000 22.859 2.016

25 1.000 0.000 19.46 1.896

30 0.951 0.000 22.859 2.016

35 0.809 0.000 22.783 1.879

40 0.588 0.000 18.480 1.479

45 0.309 0.000 10.370 0.799

50 0.000 0.000 0.000 0.000

6.3.8 Flexural-torsional-distortional analysis: multi- cell doubly symmetric

section

The differential equation governing flexural- torsional-distortional equilibrium for

doubly symmetric sections are;

γ

∴iv

2 2

1 4 1 3 1

iv iv

3 1 4 2 4 2

V = 0; V (x) = 0

β V '' - V = -K

V +α V +β V '' =K

(6.3.54)

where,

; γ

2

34 43 33 44 3344 441 1 2 1

43 33 43 33 43 33

r c -b r ksr rα = β = - ; β = ; =

c c c ka c c

3 4 22 41 2

33 43 33 43

q q b qK = - ; K =

c G c G a c E

140

It has been established in section 6.3.7 that flexural –torsional-distortional analysis of

doubly symmetric sections does not differ from their torsional-distortional analysis.

Consequently, eqns. (6.3.54) have the same solutions as eqns. (6.3.38), i.e.

-3

3

-3

4

πxV (x) = 5.779 *10 Sin

50

πxV (x) = 2.896 *10 Sin

50

(6.3.55)

6.4 ANALYSIS OF MONO- SYMMETRIC SECTIONS 6.4.1 Flexural-torsional analysis: single cell mono symmetric section The governing equations of equilibrium are

iv iv

1 2 2 4 1 4 1

4 2

α V +α V - β V '' = - K

V '' =K (6.4.1)

where,

1 22 42 2 22 44α = ka c ; α = ka r

41 22 44 24 42 1 22

qβ = b r - c c ; K = b

G

42 2 22 42

42 24 22 44 42 24 22 44

c q c qK = - +


The relevant coefficients are

22 22 22 22

24 42 24 42

-4 -4

44 33

a = 25.05; b = c = r = 2.982

c = c = r = r = -2.515

r =14.616; s = 0.261* 6.667 *10 =1.740 *10

2 4

9 2 9 2

q = 0.00KN; q =1446.505KN

E = 24 *10 N/m ; G = 9.6 *10 N/m , k = 2.5

The coefficients of the governing equations are

1 2 1α = -157.502; α = 915.327; β = 49.910

-4 -5

1 2K = 4.4932*10 ; K = -1.206 *10


iv iv -4

2 4 4

-5

4

-157.502V +915.327V - 49.910V '' = 4.4932*10

V '' = -1.206x10 (6.4.2)

Integrating by method of trigonometric series with accelerated convergence we

have,

Interval= span L=50m. ∴ ≤ ≤0 x 50

141


4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


3 3 3

4 4 4

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.4.3)


iv iv iv iv -4

2 4 4 2 4 4

-5

4 4

-157.502f +915.327f - 49.910f '' -157.502g +915.327g - 49.910g '' = 4.4932*10

f ''+g '' = -1.206 *10

⇒iv iv

2 4 4 1

4 2

-157.502f +915.327f - 49.910f '' =P (x)

f '' =P (x)

iv iv -4

1 2 4 4

-5

2 4

P (x) = -157.502g +915.327g - 49.910g '' - 4.4932x10

P (x) = g ''+1.206 *10 (6.4.4)


follows.

∞

∞

∑

∑

n2 2n

n=1

n4 4n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.4.5)

where nα = nπ


2 3

3 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

4 0 1 2 3f (x) =B +B x +B x +B x (b) (6.4.6)

The constants 0 0 1 1 2 2 3 3A , B , A , B , A , B , A & B are easily defined using the boundary

conditions.

From section 6.3.5 we note that for simply supported span the boundary conditions

are such that all the constants are zero.

Hence, 2 1

4 2

f (x) =P (x) = 0

f (x) =P (x) = 0 } ⇒

2 2

4 4

V (x) = g (x)

V (x) = g (x) (6.4.7)


0=iv iv -4

2 4 4

-5

4

-157.502g +915.327g - 49.910g '' - 4.4932*10

g ''+1.206 *10 = 0 (6.4.8)

142

But ∞

∑ n2 2n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n4 4n

n=1

α xg (x) = a sin

L (b)

∞

∑ 2 n2 2n n

n=1


L (c)

∞

∑iv 4 n2 2n n

n=1


L (d) (6.4.9)

∑¥

'' 2 n4 4n n

n=1

α xg (x) = - a α sin

L (e)

∞

∑iv 4 n4 4n n

n=1


L (f)


∑ ∑ ∑

∑

4 4 2 -4nn 2n n 4n n 4n

2 -5nn 4n

α x-157.502 α a +915.327 α a + 49.910 α a sin = 4.4932*10

L

α x- α a Sin = -1.206 *10

L

(6.4.10)

Equations (6.4,10) will always be satisfied if

( ) ( )+4 4 2 -4

n 2n n n 4n-157.502α a 915.327α + 49.910α a = 4.4932*10 (6.4.11)

and ( )− 2 -5

n 4nα a = -1.206x10 ⇒-5

4n 2

n

1.206 *10a =

α (6.4.12)

For n = 1, 2 4

1 1 1α =π, α = 9.8696 α = 97.409

∴-5

-3

41 -3

1.206 *10a = = 3.055 *10

3.9478 *10

Substituting into eqn.(6.4.11) we have

( )

⇒

-3 -3 -4

21

-4-2

21 -3

-2.4547 *10 a +0.2113 3.055 *10 = 4.4932*10

-1.962*10a = = 7.993 *10

-2.4546 *10

Substituting values of a21, a41, and eqn. (6.4.5) into eqn. (6.4.7) we obtain

-3

2

-3

4

πxV (x) = 79.73 *10 Sin

50

πxV (x) = 3.055 *10 Sin

50

143

Table 6.10 Variation of flexural and torsional


(Single cell mono symmetric section)

Distance (x)

from Left

Support (m)

πxSin

50

Flexural

displacement

V2(x) x 10-3

(m)

Torsional

displacement

V4(x) x 10-3

(m)

0 0.000 0.000 0.000

5 0.309 24.637 0.944

10 0.588 46.881 1.796

15 0.809 64.502 2.471

20 0.951 75.823 2.905

25 1.000 79.73 3.055

30 0.951 75.823 2.905

35 0.809 64.502 2.471

40 0.588 46.881 1.796

45 0.309 24.637 0.944

50 0.000 0.000 0.000

6.4.2 Flexural-torsional analysis: double cell mono symmetric section The governing equations of equilibrium are

iv iv

1 2 2 4 1 4 1

4 2

α V +α V -β V '' =K

V '' =K (6.4.13)

where,

1 22 42 2 22 44α = ka c ; α = ka r

41 22 44 24 42 2 22

qβ = b r - c c ; K = b

G

42 2 22 41

42 24 22 44 42 24 22 44

c q c qK = - +


The relevant coefficients are

22 22 22 22

24 42 24 42

-4 -4

44 33

a = 25.073; b = c = r = 2.982

c = c = r = r =1.136

r =14.485; s = 0.723 * 6.667 *10 = 4.82*10

144

2 4

9 2 9 2

q = 0.0KN; q =1446.505KN

E = 24 *10 N/m ; G = 9.6 *10 N/m , k = 2.5

The coefficients of the governing equations are

1 2 1α = 71.209; α = 907.956; β = 41.903

;-4 -5

1 2K = 4.4932*10 K = -1.0709 *10 ;


iv iv -4

2 4 4

-5

4

71.209V +907.956V - 41.958V '' = 4.4932*10

V '' = -1.0723x10 (6.4.14)

Integrating by method of trigonometric series with accelerated convergence we have,

Interval = span, L = 50m. ∴ ≤ ≤0 x 50


4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


3 3 3

4 4 4

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.4.15)


iv iv iv iv -4

2 4 4 2 4 4

-5

4 4

71.209f +907.956f - 41.958f ''+71.209g +907.956g - 41.958V '' = 4.4932*10

f ''+g '' = -1.0723 *10

⇒iv iv

2 4 4 1

4 2

71.209f +907.956f - 41.958f '' =P (x)

f '' =P (x)

iv iv -4

1 2 4 4

-5

2 4

P (x) = 71.209g +907.956g - 41.958g '' - 4.4932x10

P (x) = g ''+1.0723 *10 (6.4.16)


follows. ∞

∑ n2 2n

n=1

α xg (x) = a sin

L;

∞

∑ n4 4n

n=1

α xg (x) = a sin

L (6.4.17)

where nα = nπ


2 3

3 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

4 0 1 2 3f (x) =B +B x +B x +B x (b) (6.4.18)

145


conditions.

From section 6.3.5 we note that for simply supported box girder the boundary

conditions are such that all the constants are zero.

Hence, 3 1

4 2

f (x) =P (x) = 0

f (x) =P (x) = 0 } ⇒

3 3

4 4

V (x) = g (x)

V (x) = g (x) (6.4.19)


iv iv -4

2 4 4

-5

4

71.209g +907.956g - 41.958g '' - 4.4932*10 = 0

g ''+1.0723 *10 = 0 (6.4.20)

But ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n4 4n

n=1

α xg (x) = a sin

L (b)

∞

∑ 2 n3 3n n

n=1


L (c)

∞

∑iv 4 n3 3n n

n=1


L (d) (6.4.21)

∑¥

'' 2 n4 4n n

n=1

α xg (x) = - a α sin

L (e)

∞

∑iv 4 n4 4n n

n=1


L (f)


71.209

∑ ∑ ∑

∑

4 4 2 -4nn 2n n 4n n 4n

2 -5nn 4n

α xα a +907.956 α a + 41.958 α a sin = 4.4932*10

L

α x- α a Sin = -1.0723 *10

L

(6.4.22)


( ) ( )+4 4 2 -4n 2n n n 4n71.209α a 907.956α + 41.903α a = 4.4932*10 (6.4.23)

and ( )− 2 -5

n 4nα a = -1.0723x10 ⇒-5

4n 2

n

1.0723 *10a =

α (6.4.24)

For n = 1, 2 -3 4 -5

1 1 1

πα = , α = 3.9478 *10 , α =1.5585 *10

50

∴-5

-3

41 -3

1.0723 *10a = = 2.716 *10

3.9478 *10

146


( )

⇒

-3 -3 -4

21

-5-2

21 -3

1.0942*10 a +0.1798 2.716 *10 = 4.4932*10

- 3.90168 *10a = = - 3.5658 *10

1.0942*10


-2

2

-3

4

πxV (x) = - 3.566 *10 Sin

50

πxV (x) = 2.716 *10 Sin

50

Table 6.11 Variation of flexural and torsional-


(Double cell mono symmetric section)

Distance (x)

from Left

Support (m)

πx

Sin50

Flexural

displacement

V2(x) x 10-3

(m)

Torsional

displacement

V4(x) x 10-3

(m)

0 0.000 0.000 0.000

5 0.309 11.019 0.839

10 0.588 20.968 1.597

15 0.809 28.849 2.197

20 0.951 33.913 2.583

25 1.000 35.660 2.716

30 0.951 33.913 2.583

35 0.809 28.849 2.197

40 0.588 20.968 1.597

45 0.309 11.019 0.839

50 0.000 0.000 0.000

6.4.3 Flexural-distortional analysis: single cell mono symmetric section The governing equations of equilibrium are

γ

iv iv

1 2 2 3 1 3 3

iv iv

3 2 4 3 2 3 1 3 4

α V +α V -β V '' =K

α V +α V -β V '' - V = - K (6.4.25)

147

The relevant coefficients are as follows.

22 23 32 33

22 22 22 33 33 33

a = 25.05; a = a = -0.270, a = 0.757

b = c = r = 2.982 b = c = r =1.407

23 32 23 32 23 32

-4 -4

44 33

b = b = c = c = r = r = -0.153

r =14.616; s = 0.261* 6.9712*10 =1.8195 *10

2 3 4

9 2 9 2

q = 0.0KN; q =196.46KN q =1446.505KN

E = 24 *10 N/m ; G = 9.6 *10 N/m , k = 2.5

The coefficients of the governing equations are as follows

1 22 2 23α = ka = 62.625; α = ka = -0.675

3 32 4 33α =Ka = -0.675; α =Ka =1.893

( ) 2

23 22 22 23 33 -4

1 2

33 22 32

a c -a c k sβ = = 8.252*10

c c - c

( )2

33 33 22 32 23 -4

2 2

33 22 32

k s a c -a cβ = = 6.04 *10

c c - c

( )γ 32 23 33 22 33 -4

1 2

33 22 32

a c -b c ks= = -4.5487 *10

c c - c

c

-53221 2

33 22 32

qcK = - = -1.4626 *10

c -c G

c

-723 32 2

33 22 32

c qK = - = -7.504 *10

c -c G

-6

3 23 1 22 2

-5

4 32 2 33 1

K = b K -b K = 4.47545 *10

K = b K +b K = -2.0463 *10


iv iv -4 -6

2 3 3

iv iv -4 -4 -5

2 3 3

62.625V -0.675V -8.2517 *10 V '' = 4.47545 *10

-0.675V +1.893V - 6.04 *10 V ''+ 4.5487 *10 = 2.0463 *10 (6.4.26)


Interval= span, L = 50m. ∴ ≤ ≤0 x 50


4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


2 2 2

3 3 3

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.4.27)

148


iv iv -4 iv iv

2 3 3 2 3

-4 -6

3

iv iv -4 -4

2 3 3

iv iv -4 -4 -5

2 3 3 3

62.625f -0.675f -8.2517 *10 f ''+62.625g -0.675g

-8.2517 *10 g '' = 4.47545 *10

-0.675f +1.893f - 6.04 *10 f ''+ 4.5487 *10 f

-0.675g +1.893g - 6.04 *10 g ''+ 4.5487 *10 g = 2.0463 *10

⇒ iv iv -4

2 3 3 1

iv iv -4 -4

2 3 3 2

62.625f -0.675f -8.2517 *10 f '' =P (x)

- 0.675f +1.893f - 6.04 *10 f '' + 4.5487 *10 f =P (x)

⇒iv iv -4 -6

1 2 3 3

iv iv -4 -4 -5

2 2 3 3 3

P (x) = 62.625g -0.675g -8.2517 *10 g '' - 4.47545 *10

P (x) = -0.675g +1.893g - 6.04 *10 g ''+ 4.5487 *10 g - 2.0463 *10 (6.4.28)


follows. ∞

∑ n2 2n

n=1

α xg (x) = a sin

L

∞

∑ n3 3n

n=1

α xg (x) = a sin

L (6.4.29)

where nα = nπ


2 3

2 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

3 0 1 2 3f (x) =B +B x +B x +B x (b) (6.4.30)


conditions. From section 6.3.5 we note that for simply supported span the boundary

conditions are such that all these constants are zero.

Hence, 2 1

3 2

f (x) =P (x) = 0

f (x) =P (x) = 0 } ⇒

2 2

3 3

V (x) = g (x)

V (x) = g (x) (6.4.31)


iv iv -4 -6

2 3 3

iv iv -4 -4 -5

2 3 3 3

62.625g -0.675g -8.2517 *10 g '' - 4.47545 *10 = 0

-0.675g +1.893g - 6.04 *10 g ''+ 4.5487 *10 g - 2.0463 *10 = 0 (6.4.32)

But ∞

∑ n2 2n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (b)

∞

∑ 2 n2 2n n

n=1


L (c)

∞

∑iv 4 n2 2n n

n=1


L (d) (6.4.33)

149

∞

∑ 2 n3 3n n

n=1


L (e)

∞

∑iv 4 n3 3n n

n=1


L (f)


∑ ∑ ∑

∑ ∑ ∑ ∑

4 4 -4 2 -6n2n n 3n n 3n n

4 -4 2 -4 n2n 3n n 3n n 3n

α x62.625 a α - 0.675 a α +8.2517 *10 a α Sin = 4.47545 *10

L

α x-0.675 a +1.893 a α +6.04 *10 a α + 4.5487 *10 a Sin

L

-5 = 2.0463 *10

(6.4.34)

Equations (6.4.34) will always be satisfied if;

4 4 -4 2 -6

n 2n n n 3n(62.625α )a +(-0.675α +8.2517 *10 α )a = 4.47545 *10 (6.4.35)

4 4 -4 2 -4 -5

n 2n n n 3n(-0.675α )a +(1.893α +6.04 *10 α + 4.5487 *10 )a = 2.0463 *10 (6.4.36)


( )4 -4 2 -6n n

2n 3n4 4

n n

-0.675α +8.2517 *10 α 4.4754 *10a = - a +

62.625α 62.625α (6.4.37)

For n = 1, 2 -3 4 -5

1 1 1

πα = , α = 3.948 *10 , α =1.5585 *10

50


-3 -3

21 3na = -7.4414 *10 a + 4.5858 *10


( )-5 -4 -3 -5 -5

31 31-1.05199 *10 -7.4414 *10 a + 4.4558 *10 + 4.8676 *10 a = 2.0463 *10

⇒ -4 -5 -8

3n4.8684 *10 a = 2.0463 *10 + 4.68746 *10

⇒-5

-2

31 -4

2.05099 *10a = = 4.213 *10

4.8684 *10

∴ -3

21 a = 4.272*10


-3

2

-2

3

πxV (x) = 4.272*10 Sin

50

πxV (x) = 4.213 *10 Sin

50

150

Table 6.12 Variation of flexural and distortional

displacements along the girder length (Single cell mono

symmetric section)

Distance (x)

from Left

Support (m)

πx

Sin50

V2(x)

X 10-3

V3(x)

X 10-3

0 0.000 0.000 0.000

5 0.309 1.320 13.018

10 0.588 2.512 24.772

15 0.809 3.456 34.083

20 0.951 4.063 40.065

25 1.000 4.272 42.130

30 0.951 4.063 40.065

35 0.809 3.456 34.083

40 0.588 2.512 24.772

45 0.309 1.320 13.018

50 0.000 0.000 0.000

6.4.4 Flexural-distortional analysis: double cell mono symmetric section


γ

iv iv

1 2 2 3 1 4 3

iv iv

3 2 4 3 2 3 1 3 4

α V +α V -β V '' =K

α V +α V -β V '' - V = -K (6.4.41)


22 23 32 33

22 22 22 33 33 33

a = 25.073; a = a = 0.425, a = 0.750

b = c = r = 2.982, b = c = r =1.533

23 32 23 32 23 32

-4 -4

33

b = b = c = c = r = r = -0.449

s = 0.261* 6.9712*10 =1.8195 *10

2 3 4

9 2 9 2

q = 0.0KN, q =196.46KN q =1446.505KN

E = 24 *10 N/m , G = 9.6 *10 N/m , k = 2.5


1 22

2 23

α = ka = 62.6825

α = ka =1.0625

151

3 32

4 33

α = Ka =1.0625

α = Ka =1.875

( ) 2

23 22 22 23 33

1 2

33 22 32

a c -a c k sβ = = 0.0283

c c - c

( )2

33 33 22 32 23 -3

2 2

33 22 32

k s a c - a cβ = =1.750 *10

c c - c

( )γ 32 23 33 22 33 -3

1 2

33 22 32

a c -b c ks= = -1.260 *10

c c - c

c

-53221 2

33 22 32

qcK = - = -1.115 *10

c -c G

c

-623 32 2

33 22 32

c qK = - = -1.684 *10

c -c G

-5

3 23 1 22 2

-5

4 32 2 33 1

K = b K -b K =1.003 *10

K = b K +b K = -1.634 *10


iv iv -5

2 3 3

iv iv -3 -3 -5

2 3 3 3

62.683V +1.0625V -0.0283V '' = -1.003 *10

1.0625V +1.875V -1.750 *10 V ''+1.26 *10 V =1.634 *10 (6.4.42)

Integrating by method of trigonometric series with accelerated convergence we

have,

Interval = span, L = 50m. ∴ ≤ ≤0 x 50


3 3 3 3V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


2 2 2

3 3 3

V (x) = f (x)+ g (x)

V (x) = f (x)+ g (x) (6.4.43)


( )a

iv iv iv iv

2 3 3 2 3

-5

3

iv iv -3 -3

2 3 3

iv iv -3

2 3

62.683f +1.0625f - 0.0283f ''+ 62.683g +1.0625g

-0.0283g '' = -1.003 *10

1.0625f +1.875f -1.750 *10 f ''+1.26 *10 f

+1.0625g +1.875g -1.70 *10 ( )b-3 -5

3 3g '' +1.26 *10 g = -1.634 *10

(6.4.44)

⇒ iv iv

2 3 3 1

iv iv -3 -3

2 3 3 2

62.683f +1.0625f -0.0283f '' =P (x)

1.0625f +1.875f -1.750 *10 f ''+1.260 *10 f =P (x)

152

where,

iv iv -5

1 2 3 3

iv iv -3 -3 -5

2 2 3 3 3

P (x) = 62.683g +1.0625g -0.0283g ''+1.003 *10

P (x) =1.0625g +1.875g -1.750 *10 g ''+1.260 *10 g -1.634 *10 (6.4.45)


follows.

∞

∞

∑

∑

n2 2n

n=1

n3 3n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.4.46)

where nα = nπ


2 3

2 0 1 2 3

2 3

3 0 1 2 3

f (x) = A + A x + A x + A x

f (x) =B +B x +B x +B x (6.4.47)


conditions.



Hence, 2 1

3 2

f (x) = P (x) = 0

f (x) = P (x) = 0 } ⇒

2 2

3 3

V (x) = g (x)

V (x) = g (x) (6.4.48)

Consequently eqn (6.4.45) becomes

iv iv -5

2 3 3

iv iv -3 -3 -5

2 3 3 3

62.683g +1.0625g -0.0283g ''+1.003 *10 = 0

1.0625g +1.875g -1.750 *10 g ''+1.260 *10 g -1.634 *10 = 0 (6.4.49)

But ∞

∑ n2 2n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (b)

∞

∑ 2 n2 2n n

n=1


L (c)

∞

∑iv 4 n2 2n n

n=1


L (d) (6.4.50)

∞

∑ 2 n3 3n n

n=1


L (e)

∞

∑iv 4 n3 3n n

n=1


L (f)

153


4

1.0625 ( )

1.0625 n

a

α

+

∑ ∑ ∑

∑ ∑ ∑ ∑

4 4 2 -5n2n n 3n n 3n n

4 -3 2 -3 n2n 3n n 3n n 3n

α x62.682 a α a α +0.0283 a α Sin = -1.003 *10

L

α xa +1.875 a α +1.75 *10 a α +1.260 *10 a Sin

L

( )b-5 =1.634 *10

(6.4.51)


4 4 2 -5

n 2n n n 3n(62.682α )a +(1.0625α +0.0283α )a = -1.003 *10 (6.4.52)

4 4 -3 2 -3 -5

n 2n n n 3n(1.0625α )a +(1.875α +1.75 *10 α +1.260 *10 )a =1.634 *10 (6.4.53)


( )4 2 -5n n

2n 3n4 4

n n

1.0625α +0.0283α 1.003 *10a = - a -

62.682α 62.682α (6.4.54)

For n = 1, 2 -3 4 -5

1 1 1

πα = , α = 3.948 *10 , α =1.5585 *10

50


-2

21 31a = -0.1313a +1.02672*10


( )-5 -2 -3 -5

31 311.656 *10 -0.1313a -1.0264 *1o +1.296 *10 a =1.634 *10

-3 -5

311.294 *10 a =1.616998 *10

⇒ -2

31

-2 -2

21

a = 1.2496 *10

a = - 0.1313 *1.2496 *10 +1.0267 *10

⇒ -3

21a = 8.626 *10

Substituting values of a21, a31 and eqn.(6.4.46) into eqn.(6.4.48) we obtain

-3

2

πxV (x) = 8.626 *10 Sin

50; -2

3

πxV (x) =1.250 *10 Sin

50

Table 6.13 Variation of flexural and distortional



Distance (x)

from Left

Support (m)

πx

SinL

V2(x)

X 10-3

V3(x)

X 10-3

0 0.000 0.000 0.000

154

6.4.5 Torsional-distortional analysis: single cell mono-symmetric section The governing equations of equilibrium are

γ1 4 1 3 1

iv iv

1 3 2 4 2 4 2

β V '' - V =K

α V +α V -β V '' =K (6.4.55)


33 33 33 33

34 43 34 43 44

a = 0.750; b = c = r =1.407

c = c = r = r =1.265; r =14.616

-4 -4

33s = 0.261* 6.9712*10 =1.8195 *10

2 3 4q = 0.0KN; q =196.46KN; q =1446.505KN

9 2 9 2E = 24 *10 N/m ; G = 9.6 *10 N/m ; k = 2.5

The coefficients for the governing equations are as follows

1 33 43

2 33 44

α = ka c = 2.372

α = ka r = 27.405

1 34 43 33 44β = r c - c r = -18.964

2 33 44 34 43β = b r - c c =18.964

9 2 9 2E = 24 *10 N/m , G = 9.6 *10 N/m , k = 2.5

1γ -4

43 33= c ks = 5.503 *10

-441 43 33

qK = - c +C =1.9163 *10

G G3333qqqq

5 0.309 2.665 3.863

10 0.588 5.072 7.350

15 0.809 6.978 10.113

20 0.951 8.203 11.888

25 1.000 8.626 12.500

30 0.951 8.203 11.888

35 0.809 6.978 10.113

40 0.588 5.072 7.350

45 0.309 2.665 3.863

50 0.000 0.000 0.000

155

-442 33

qK = b = 2.120 *10

G

Substituting the coefficients into eqns (6.4.55 ) we obtain

iv iv -6

3 4 4

iv -4 -4

4 3

2.371V +27.405V -18.963V '' = 2.120 *10

-18.964V -5.503 *10 V =1.9163 *10 (6.4.56)


Interval = span, L=50m. ∴ ≤ ≤0 x 50


4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


3 3 3

4 4 4

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.4.57)

Substituting eqn (6.4.57) into eqn (6.4.56) we obtain

iv iv iv iv -4

3 4 4 3 4 3

'' -4 '' -4 -4

4 3 4 3

2.371f +27.405f -18.964f '' +2.371g +27.405g -18.964g '' = 2.120 *10

-18.964f -5.503 *10 f -18.964g -5.503 *10 g =1.9163 *10

2.371⇒ iv iv

3 4 4 1

iv -4

4 3 2

f +27.405f -18.964f '' =P (x)

-18.964f -5.503 *10 f =P (x)

where, iv iv -4

1 3 4 4P (x) = 2.371g +27.405g -18.964g '' - 2.12*10 (a)

-4 -4

2 4 3P (x) = -18.964g '' - 5.503 *10 g -1.9163 *10 (b)

(6.4.58)


follows.

∞

∞

∑

∑

n3 3n

n=1

n4 4n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.4.59)

where nα = nπ


2 3

2 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

3 0 1 2 3f (x) =B +B x +B x +B x (b) (6.4.60)

156


conditions.



Hence, 3 3

4 4

f (x) =P (x) = 0

f (x) =P (x) = 0 } ⇒

3 3

4 4

V (x) = g (x)

V (x) = g (x) (6.4.61)


iv iv -4

3 4 4

-4 -4

4 3

2.371g +27.405g -18.964g '' - 2.12 *10 = 0

-18.964g '' - 5.503 *10 g -1.9163 *10 = 0 (6.4.62)

But ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (b)

∞

∑ 2 n3 3n n

n=1


L (c)

∞

∑iv 4 n3 3n n

n=1


L (d) (6.4.63)

∞

∑ 2 n4 4n n

n=1


L (e)

∞

∑iv 4 n4 4n n

n=1


L (f)


∑ ∑ ∑

∑ ∑

4 4 2 -4n3n n 4n n 4n n

2 -4 n4n n 3n

α x2.371 a α +27.405 a α +18.964 a α Sin = 2.120 *10

L

α x18.964 a α - 5.503 *10 a Sin =1.9163 *10 - 4

L

(6.4.64)


4 4 2 -4

n 3n n n 4n(2.371α )a +(27.405α +18.964α )a = 2.120 *10 (6.4.65)

2 -4 -4

n 4n 3n(18.964α )a - (5.503 *10 )a =1.9163 *10 (6.4.66)


( )−

2 -4n

3n 4n-4 -4

18.964α 1.9163 *10a = - a

5.503 *10 5.503 *10 (6.4.67)

157

⇒ 2

3n n 4na = 34461.203α a -0.3482

For n = 1, 2 -3 4 -5

1 1 1

πα = , α = 3.948 *10 , α =1.5585 *10

50


31 41a =136.053a -0.3482

Substituting into eqn.(6.4.66) we have

( )3.695 -5 -2 -4

41 412*10 136.053a -0.3482 +7.5297 *10 a = 2.120 *10

⇒ -2 -4 -5 -4

41

-3

41

8.0324 *10 a = 2.120 *10 +1.28667 *10 = 2.24667 *10

a = 2.7995 *10

( )⇒ -2

31a = 0.380879 -0.3482 = 3.268 *10


-2

3

-3

4

πxV (x) = 3.268 *10 Sin

50

πxV (x) = 2.80 *10 Sin

50



(Single cell mono symmetric section)

Distance (x)

from Left

Support (m)

πx

Sin50

V3(x)

X 10-3

V4(x)

X 10-3

0 0.000 0.000 0.000

5 0.309 10.098 0.865

10 0.588 19.216 1.646

15 0.809 26.438 2.265

20 0.951 31.079 2.663

25 1.000 32.680 2.800

30 0.951 31.079 2.663

35 0.809 26.438 2.265

40 0.588 19.216 1.646

45 0.309 10.098 0.865

158

50 0.000 0.000 0.000

6.4.6 Torsional-distortional analysis: double cell mono-symmetric section


γ1 4 1 3 1

iv iv

1 3 2 4 2 4 2

β V '' - V =K

α V +α V -β V '' =K (6.4.70)


33 33 33 33

34 43 34 43 44

a = 0.750; b = c = r =1.533

c = c = r = r =1.295; r =14.485

-4 -4

33s = 0.723 * 6.9712*10 = 5.04 *10

2 3 4

9 2 9 2

q = 0.0KN; q =196.46KN; q =1446.505KN

E = 24 *10 N/m ; G = 9.6 *10 N/m ; k = 2.5


1 33 43

2 33 44

α = ka c = 2.428

α = ka r = 27.160

1 34 43 33 44β = r c - c r = -20.528

2 33 44 34 43β = b r -c c = 20.528

1γ -3

43 33= c ks =1.632*10

-43 41 43 33

q qK = - c +C = 2.613 *10

G G

-442 33

qK = b = 2.87845 *10

G


iv iv -4

3 4 4

iv -3 -4

4 3

2.428V +27.16V - 20.528V '' = 2.87845 *10

-20.528V +1.632*10 V = 2.613 *10 (6.4.71)




4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


159

3 3 3

4 4 4

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.4.72)


( )

( )

a

b

iv iv iv iv -4

3 4 4 3 4 3

iv -3 iv -3 -4

4 3 4 3

2.428f +27.16f - 20.528f ''+2.428g +27.16g - 20.5284g '' = 2.87845 *10

-20.528f +1.632*10 f - 20.528g +1.632*10 g = 2.613 *10

⇒ iv iv

3 4 4 1

iv -3

4 3 2

2.428f +27.16f - 20.528f '' =P (x)

- 20.528f +1.632*10 f =P (x)

where,

iv iv -4

1 3 4 4

-3 -4

2 4 3

P (x) = 2.428g +27.16g - 20.528g '' - 2.87845 *10

P (x) = -20.528g +1.632*10 g - 2.613 *10 (6.4.73)


follows.

∞

∞

∑

∑

n3 3n

n=1

n4 4n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.4.74)

where nα = nπ


2 3

3 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

4 0 1 2 3f (x) =B +B x +B x +B x (b) (6.4.75)


conditions.

From section 6.3.5 we noted that for simply supported span the boundary conditions

are such that all the constants A0, B0, A1, B1, etc, are zero.

Hence, 3 3

4 4

f (x) =P (x) = 0

f (x) =P (x) = 0 } ⇒

3 3

4 4

V (x) = g (x)

V (x) = g (x) (6.4.76)


iv iv -4

3 4 4

-3 -4

4 3

2.428g +27.16g - 20.528g '' - 2.87845 *10 = 0

-20.528g ''+1.632*10 g - 2.613 *10 = 0 (6.4.77)

But ∞

∑ n3 3n

n=1

α xg (x) = a sin

L (a)

and ∞

∑ n4 4n

n=1

α xg (x) = a sin

L (b)

160

∞

∑ 2 n3 3n n

n=1


L (c)

∞

∑iv 4 n3 3n n

n=1


L (d) (6.4.78)

∞

∑ 2 n4 4n n

n=1


L (e)

∞

∑iv 4 n4 4n n

n=1


L (f)


220.528 α

−

∑ ∑ ∑

∑ ∑

4 4 2 -4n3n n 4n n 4n n

-3 -4n4n 3n

α x2.428 a α +27.16 a α +20.528 a α Sin = 2.87845 *10

L

α xa +1.632*10 a Sin = 2.613 *10

L

(6.4.79)


4 4 2 -4

n 3n n n 4n(2.428α )a +(27.16α +20.528α )a = 2.87845 *10 (6.4.80)

-3 -4

3n 4n(1.632*10 )a - (20.528)a = 2.613 *10 (6.4.81)


( )−

2 -4n

3n 4n-3 -3

20.528α 2.613 *10a = a

1.632*10 1.632*10

⇒ 2

3n n 4na =12578.43 α a -0.16011 (6.4.82)

For n = 1, 2 -3 4 -5

1 1 1

πα = , α = 3.948 *10 α =1.5585 *10

50


31 41a = 49.6596 a -0.16011


( )3.785 -5 -2 -4

41 41*10 49.6596 a -0.16011 +8.147 *10 a = 2.87845 *10

⇒

⇒

-2 -4 -6 -4

41

-3

41

8.3347 *10 a = 2.87845 *10 +6.0602*10 = 2.939 *10

a = 3.526 *10

( )∴ -2

31a = 0.1751-0.16011 = 1.500 *10

Substituting values of a31, a41 and eqn.(6.4.74) into eqn.(6.4.72) we obtain

161

-2

3

-3

4

πxV (x) =1.500 *10 Sin

50

πxV (x) = 3.526 *10 Sin

50




Distance (x)

from Left

Support (m)

πx

Sin50

V3(x)

X 10-3

V4(x)

X 10-3

0 0.000 0.000 0.000

5 0.309 4.635 1.090

10 0.588 8.820 2.073

15 0.809 12.135 2.852

20 0.951 14.265 3.353

25 1.000 15.000 3.526

30 0.951 14.265 3.353

35 0.809 12.135 2.852

40 0.588 8.820 2.073

45 0.309 4.635 1.090

50 0.000 0.000 0.000

6. 4.7 Flexural-torsional-distortional analysis: single cell mono-symmetric section The governing equations of equilibrium are

γ

γ

γ

4 2 5 3 6 4 1 3 3

iv iv iv iv

3 2 5 3 6 4 7 2 8 3 9 4 2 3 4

iv iv iv

7 2 8 3 9 4 10 2 11 3 12 4 3 3 5

β V ''+β V ''+β V '' - V = -K

α V +α V +α V -β V '' -β V '' -β V ''+ V =K

α V +α V +α V -β V '' -β V '' -β V ''+ V =K

(6.4.85)


22 23 33

22 22 22

23 23 23

a = 25.05, a = -0.270, a = 0.757,

b = c = r = 2.982

b = c = r = -0.153

162

24 42 24 42 2.515c c r r= = = = −

33 33 33 44

34 43 34 43

b = c = r =1.407, r =14.616

c = c = r = r =1.265

-4 -4

33s = 0.261* 6.9712*10 =1.8195 *10

2q 3 4

9 2 9 2

= 0.0KN, q =154.58KN, q =1446.505KN

E = 24 *10 N/m , G = 9.6 *10 N/m , k = 2.5

The coefficients of the governing equations are as follows.

32 22

33 231

3222

23 33

r r-

c c 19.38α = = = -1

cc -19.38-

c c

34 24

33 232

3222

23 33

r r-

c c -15.54α = = = 0.802

cc -19.38-

c c

-4-533

3

322233

23 33

ks 2.5 *1.74 *10α = = = -1.6679 *10

1.407 * (-19.38)cc- c

c c

33 23

221

23 33

22 32

r r-

c2 c -9.14477β = = = -1

c c 9.14477-

c c

34 24

32 222

23 33

22 32

r r-

c c -7.4246β = = = -0.812

c c 9.14477-

c c

-4-433

3

23 3332

22 32

ks 2.5 *1.74 *10β = = = -3.251*10

-0.153 * 9.14477c c- c

c c

;

( ) ( )4 42 1 42β = c α +r = -2.515 * -1 + -2.515 = 0 ; ( )5 43 1 43β = c β +r =1.265 * -1 +1265 = 0

( )6 42 2 43 2 44

7 22 1 22

β = c α +c β +r =11572

β = b α +c = 2.982* -1 +2.982 = 0

( )

( )

-4

8 23 3 23 1 22 3 23

9 22 2 33 2 24

β = ka β +b β +ka α +c = -7.89 *10

β = b α +b β +c = -1.266

163

( )

( )10 32 1 32

-4

11 32 33 3 33 1 33

β = b α +c = 0

β = ka +ka β +b β +c = -5.820 *10

( ) -4

12 32 2 33 2 34β = b α +b β +c = -1.90 *10

( )6 22 2 23 2

7 32 1

α = k a α +a β = 50.7734

α = ka α = 0.675

( )8 33 1

9 32 2 33 2

α = ka β = -1.893

α = k a α +a β = -2.078

( )

( )

γ

γ

-4

1 42 3 43 3

-9

2 22 3 23 3

= c α +c β = -3.532*10

= b α +b β = 4.8 *10 »0

( )3γ -4

32 3 33 3= b α +b β = -4.350 *10

-53 21

23 33 23 3332 22

22 32 22 32

q qK = - = -1.4625 *10

c c c c- c G - c G

c c c c

-73 22

32 3222 2233 23

23 33 23 33

q qK = - = -7.505 *10

c cc c- c G - c G

c c c c

-443 42 2 43 1

qK = +c K +c K = 1.7115 *10

G

( )

( )

-6

4 22 2 23 1

-5

5 32 2 33 1

K = b K +b K = -2.123 *10

K = b K +b K = -2.046 *10

Substituting the coefficients and the constants into eqn.(6.4.85) we obtain,

( )

( )

( )

a

b

c

-4 -4

4 3

iv iv iv -4 -6

2 3 4 3 4

iv iv iv -4 -4

2 3 4 3 4

-4 -5

3

11.572V ''+3.532*10 V = - 1.7115 *10

- 62.625V +0.675V +50.773V +7.89 *10 V +1.266V '' = -2.123 *10

0.675V -1.893V - 2.078V +5.820 *10 V ''+1.90 *10 V '' -

- 4.35 *10 V = - 2.046 *10

(6.4.86 )

Integrating by method of trigonometric series with accelerated convergence we have, Interval = span, L = 50m. ∴ ≤ ≤0 x 50


3 3 3 3V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0

4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


4 22 1

5 23 1

α = ka α = -62.625

α = ka β = 0.675

164

2 2 2

3 3 3

4 4 4

V (x) = f (x)+g (x)

V (x) = f (x)+g (x)

V (x) = f (x)+g (x)

(6.4.87)


-4 -4 -4

4 3 4 311.572f ''+3.532*10 f +11.572g ''+3.532*10 g = -1.7115 *10 (a)

iv iv iv -4 iv

2 3 4 3 4 2

iv iv -4 -6

3 4 3 4

-62.625f +0.675f +50.225f +7.89 *10 f +1.266f '' - 62.625g

+0.675g +50.225g +7.89 *10 g +1.266g '' = -2.123 *10 (b)

iv iv iv -4 -4

2 3 4 3 4

-4 iv iv iv -4 -4

3 2 3 4 3 4

-4 -5

3

0.675f -1.875f - 2.064f +5.768 *10 f ''+1.90 *10 f '' -

-4.35 *10 f +0.675g -1.875g - 2.064g +5.768 *10 g ''+1.90 *10 g '' -

-4.35 *10 g = -2.046 *10 (c)

⇒ -4

4 3 1 11.572f ''+3.532*10 f =P (x) (a)

iv iv iv -4

2 3 4 3 4 2-62.625f +0.675f +50.225f +7.89 *10 f +1.266f '' =P (x) (b)

iv iv iv -4 -4

2 3 4 3 4

-4

3 3

0.675f -1.875f - 2.064f +5.768 *10 f ''+1.90 *10 f '' -

-4.35 *10 f =P (x) (c)

where,

-4 -4

1 4 3P (x) =11.572g '' +3.532*10 g +1.7115 *10 (a)

iv iv iv -4 -6

2 2 3 4 3 4P (x) = -62.625g +0.675g +50.225g +7.89 *10 g +1.266g ''+2.123 *10 (b)

iv iv iv -4 -4

3 2 3 4 3 4

-4 -5

3

P (x) = 0.675g -1.875g - 2.064g +5.768 *10 g ''+1.90 *10 g '' -

-4.35 *10 g +2.046 *10 (c) (6.4.88)

We seek the complimentary function 2g (x) , g3(x) and g4(x) in the form of sine series

as follows.

∞

∞

∞

∑

∑

∑

n2 2n

n=1

n3 3n

n=1

n4 4n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.4.89)

where nα = nπ


2 3

2 0 1 2 3f (x) = A + A x + A x + A x (a)

165

2 3

3 0 1 2 3f (x) =B +B x +B x +B x (b) (6.4.90)

2 3

4 0 1 2 3f (x) = C +C x +C x + x (c)


conditions.

From section 6.3.5 we noted that for simply supported span the boundary conditions

are such that all the constants A0, B0, C0, etc, are zero.

Hence,

2 2

3 3

4 4

f (x) =P (x) = 0

f (x) =P (x) = 0

f (x) =P (x) = 0

} ⇒

2 2

3 3

4 4

V (x) = g (x)

V (x) = g (x)

V (x) = g (x)

(6.4.91)


-4 -4

4 311.572g ''+3.532*10 g +1.7115 *10 = 0 (a)

iv iv iv -4 -6

2 3 4 3 4-62.625g +0.675g +50.225g +7.89 *10 g +1.266g ''+2.123 *10 = 0 (b) (6.4.92)

iv iv iv -4 -4

2 3 4 3 4

-4 -5

3

0.675g -1.875g - 2.064g +5.768 *10 g ''+1.90 *10 g '' -

-4.35 *10 g +2.046 *10 = 0 (c)

But,

∞

∞

∞

∑

∑

∑

n2 2n

n=1

2 n2 2n n

n=1

iv 4 n2 2n n

n=1

α xg (x) = a sin

L


L


L

(a)

and

∞

∞

∞

∑

∑

∑

n3 3n

n=1

2 n3 3n n

n=1

iv 4 n3 3n n

n=1

α xg (x) = a sin

L


L


L

(b) (6.4.93)

166

Also,

∞

∞

∞

∑

∑

∑

4 n4 4n n

n=1

2 n4 4n n

n=1

iv 4 n4 4n n

n=1


L


L


L

(c)

Substituting eqn (6.4.93) into eqn (6.4.92) we obtain;

( )a∑ ∑

α x2 -4 -4n-11.572 a α + 3.532 *10 a Sin = -1.7115 *10n4n 3n L

∑ ∑ ∑ ∑ ∑

α x4 4 4 -4 2 n-62.625 a α + 0.675 a α + 50.773 a α + 7.89 * 10 a - 1.266 a α Sinn n n n4n 4n2n 3n 3n

L

( )b-6

= -2.123 * 10

(6.4.94)

∑ ∑ ∑ ∑ ∑

∑

α x4 4 4 -4 2 -4 2 n0.675 a α - 1.893 a α - 2.078 a α - 5.82 * 10 a α - 1.90 * 10 a α Sinn n n n n4n 4n2n 3n 3n

L

α xn-4 - 4.35*10 a Sin 3nL

( )c-5

= - 2.046 * 10


2 -4 -4

n 4n 3n(-11.572α )a +(3.532*10 )a = -1.7115 *10 (6.4.96)

( )4 4 -4 2 4 2 -6

n 2n n n 3n n n 4n(-62.625α )a +(0.675α +7.89 *10 α )a + 50.773α -1.266α a = -2.123 *10

(6.4.97)

( )

−

−

4 4 -4 2 -4

n 2n n n 3n

4 -4 2 -5

n n 4n

(0.675α )a - (1.893α +5.82*10 α + 4.35 *10 )a

2.078α +1.90 *10 α a = -2.046 *10 (6.4.98)

For n = 1, ,2 -3 4 -5

1 1 1

πα = , α = 3.948 *10 α =1.5585 *10

50

Substituting into eqn.(6.4.96, 6.4. 97, & 6.4.98) gives

( )

( )

( )

a

b

c

-2 -4 -4

41 31

-4 -5 -3 -6

21 31 41

-5 -4 -5 -5

21 31 41

-4.5684 *10 a +3.532*10 a = -1.7115 *10

-9.760 *10 a +1.3635 *10 a - 4.2066 *10 a = -2.123 *10

1.052*10 a - 4.658 *10 a -3.3136 *10 a = -2.046 *10

(6.4.99)

Solving eqns.[6.4.99 (a), (b), & (c)) we obtain

-2

21a = -3.287 *10 ; -2

31a = 4.282*10 ; -3

41a = 4.077 *10

Substituting values of a21, a31, a41 and eqn.(6.4.89) into eqn.(6.4.91) we obtain

167

-2

2

-2

3

-3

4

πx V (x) = -3.287 *10 Sin

50

πx V (x) = 4.282*10 Sin

50

πx V (x) = 4.077 *10 Sin

50

Table 6.16 Variation of flexural, torsional and distortional displacements along the length of the girder (Single cell mono symmetric section) Distance (x)

from Left

Support (m)

πx

Sin50

V2(x)

X 10-3

V3(x)

X 10-3

V4(x)

X 10-3

0 0.000 0.000 0.000 0.000

5 0.309 -10.156 13.231 1.260

10 0.588 -19.328 25.178 2.397

15 0.809 -26.592 34.641 3.298

20 0.951 -31.259 40.722 3.877

25 1.000 -32.870 42.820 4.077

30 0.951 -31.259 40.722 3.877

35 0.809 -26.592 34.641 3.298

40 0.588 -19.328 25.178 2.397

45 0.309 -10.156 13.231 1.260

50 0.000 0.000 0.000 0.000

6.4.8 Flexural-torsional-distortional analysis: double cell mono-symmetric section The governing equations of equilibrium are

γ

γ

γ

4 2 5 3 6 4 1 3 3

iv iv iv iv

3 2 5 3 6 4 7 2 8 3 9 4 2 3 4

iv iv iv

7 2 8 3 9 4 10 2 11 3 12 4 3 3 5

β V ''+β V ''+β V '' - V = -K

α V +α V +α V -β V '' -β V '' -β V ''+ V =K

α V +α V +α V -β V '' -β V '' -β V ''+ V =K

(6.4.101)


22 23 33

22 22 22

23 23 23

a = 25.073, a = 0.425, a = 0.750,

b = c = r = 2.982

b = c = r = -0.449

168

24 42 24 42 1.112c c r r= = = =

33 33 33 44

34 43 34 43

b = c = r =1.533, r =14.488

c = c = r = r =1.295

-4 -4

33s = 0.723 * 6.9712*10 = 5.040 *10

2 3 4

9 2 9 2

q = 0.00KN, q =154.58KN, q =1446.505KN

E = 24 *10 N/m , G = 9.6 *10 N/m , k = 2.5

The coefficients of the governing equations are as follows.

32 22

33 231

3222

23 33

r r-

c c 6.4385α = = = -1

cc -6.3485-

c c

34 24

33 232

3222

23 33

r r-

c c 3.321α = = = -0.523

cc -6.3485-

c c

-4-433

3

322233

23 33

ks 2.5 * 5.04 *10α = = = -1.295 *10

1.533 * (-6.3485)cc- c

c c

33 23

221

23 33

22 32

r r-

c2 c -3.2637β = = = -1

c c 3.2637-

c c

34 24

32 222

23 33

22 32

r r-

c c -3.257β = = = -0.998

c c 3.2637-

c c

-4333

23 3332

22 32

ksβ = = -8.5983 *10

c c- c

c c

( )

( )4 42 1 42

5 43 1 43

β = c α +r =1.112* -1 +1.112 = 0

β = c β +r =1.295 * -1 +1295 = 0

( )6 42 2 43 2 44

7 22 1 22

β = c α +c β +r =12.811

β = b α +c = 2.982* -1 +2.982 = 0

169

( )

( )

-3

8 23 3 23 1 22 3 23

9 22 2 33 2 24

β = ka β +b β +ka α +c = -9.031*10

β = b α +b β +c = -1.978

( )

( )10 32 1 32

-3

11 32 33 3 33 1 33

β = b α +c = 0

β = ka +ka β +b β +c = -1.7498 *10

( ) -4

12 32 2 33 2 34β = b α +b β +c = -1.07 *10

4 22 1α = ka α = -62.6825

5 23 1α = ka β = -1.063

( )6 22 2 23 2

7 32 1

α = k a α +a β = -33.8433

α = ka α = -1.063

( )8 33 1

9 32 2 33 2

α = ka β = -1.875

α = k a α +a β = -2.427

( )

( )

γ

γ

-3

1 42 3 43 3

-7

2 22 3 23 3

= c α +c β =1.2575 *10

= b α +b β =1.0533 *10

( )3γ -3

32 3 33 3= b α +b β = -1.260 *10

-53 21

23 33 23 3332 22

22 32 22 32

q qK = - = -1.3964 *10

c c c c- c G - c G

c c c c

-63 22

32 3222 2233 23

23 33 23 33

q qK = - = 4.09 *10

c cc c- c G - c G

c c c c

-443 42 2 43 1

qK = +c K +c K =1.7424 *10

G

( )

( )

-5

4 22 2 23 1

-5

5 32 2 33 1

K = b K +b K =1.8464 *10

K = b K +b K = 2.3197 *10

Substituting the coefficients and the constants into eqn.(6.4.101) we obtain,

( )

( )

a

b

-3 -4

4 3

iv iv iv -3

2 3 4 3 4

-7 -5

3

iv iv iv -3 -4

2 3 4 3 4

-4

12.611V '' -1.2573 *10 V = -1.400 *10

-62.6825V -1.063V -33.8433V +9.031*10 V ''+1.978V ''

-1.0533 *10 V =1.453 *10

-1.063V -1.875V - 2.427V +1.7498 *10 V ''+1.070 *10 V '' -

-1.260 *10 V ( )c-5

3 = -1.829 *10

(6.4.102)



Boundary conditions;

2 2 2 2V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0

170

3 3 3 3V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0

4 4 4 4V (0) = 0; V ''(0) = 0; V (L) = 0; V ''(L) = 0


2 2 2

3 3 3

V (x) = f (x)+g (x)

V (x) = f (x)+g (x) (6.4.103)

4 4 4V (x) = f (x)+g (x) (6.4.103)


-3 -3 -4

4 3 4 312.611f '' -1.2573 *10 f +12.611g '' -1.2573 *10 g = -1.400 *10 (a)

iv iv iv -3

2 3 4 3 4

-7 iv iv iv -3

3 2 3 4 3 4

-7 -5

3

-62.6825f -1.063f -33.8433f +9.031*10 f ''+1.978f ''

-1.0533 *10 f - 62.6825g -1.063g -33.8433g +9.031*10 g ''+1.978g ''

-1.0533 *10 g =1.453 *10 (b)

iv iv iv -3 -4

2 3 4 3 4

-4 iv iv iv -3 -4

3 2 3 4 3 4

-4 -5

3

-1.063f -1.875f - 2.427f +1.7498 *10 f ''+1.070 *10 f '' -

-1.260 *10 f -1.063g -1.875g - 2.427g +1.7498 *10 g ''+1.070 *10 g '' -

-1.260 *10 g = -1.829 *10 (c)

⇒ -3

4 3 12.611f '' -1.2573 *10 f =P(x) (a)

iv iv iv -3

2 3 4 3 4

-7

3 2

-62.6825f -1.063f -33.8433f +9.031*10 f ''+1.978f ''

-1.0533 *10 f =P (x) (b)

iv iv iv -3 -4

2 3 4 3 4

-4

3 3

-1.063f -1.875f - 2.427f +1.7498 *10 f '' +1.070 *10 f '' -

-1.260 *10 f =P (x) (c)

where,

-3 -4

1 4 3P (x) =12.611g '' -1.2573 *10 g +1.400 *10 (a)

iv iv iv -3

2 2 3 4 3 4

-7 -5

3

P (x) = -62.6825g -1.063g -33.8433g +9.031*10 g ''+1.978g ''

-1.0533 *10 g -1.453 *10 (b)(6.4.104)

iv iv iv -3 -4

3 2 3 4 3 4

-4 -5

3

P (x) = -1.063g -1.875g - 2.427g +1.7498 *10 g ''+1.070 *10 g '' -

-1.260 *10 g +1.829 *10 (c)

We seek the complimentary function 2g (x) , g3(x) and g4(x) in the form of sine series

as follows.

171

∞

∞

∞

∑

∑

∑

n2 2n

n=1

n3 3n

n=1

n4 4n

n=1

α xg (x) = a sin

L

α xg (x) = a sin

L

α xg (x) = a sin

L

(6.4.105)

where nα = nπ


2 3

2 0 1 2 3f (x) = A + A x + A x + A x (a)

2 3

3 0 1 2 3f (x) =B +B x +B x +B x (b) (6.4.106)

2 3

4 0 1 2 3f (x) = C +C x +C x +C x (c) (6.4.106)

The constants 0 0 0 1 1 1 2 2 2 3 3 3A , B , C , A , B , C , A , B , C , A , B ,C are easily defined using

the boundary conditions.

In section 6.3.5 we noted that for simply supported span the boundary conditions are

such that all the above constants are zero.

Hence,

2 2

3 3

4 4

f (x) =P (x) = 0

f (x) =P (x) = 0

f (x) =P (x) = 0

} ⇒

2 2

3 3

4 4

V (x) = g (x)

V (x) = g (x)

V (x) = g (x)

(6.4.107)


-3 -4

4 312.611g '' -1.2573 *10 g +1.400 *10 = 0 (a)

iv iv iv -3

2 3 4 3 4

-7 -5

3

-62.6825g -1.063g -33.8433g +9.031*10 g ''+1.978g ''

-1.0533 *10 g -1.453 *10 = 0 (b) (6.4.108)

iv iv iv -3 -4

2 3 4 3 4

-4 -5

3

-1.063g -1.875g - 2.427g +1.7498 *10 g ''+1.070 *10 g '' -

-1.260 *10 g +1.829 *10 = 0 (c)

But,

∞

∞

∞

∑

∑

∑

n2 2n

n=1

2 n2 2n n

n=1

iv 4 n2 2n n

n=1

α xg (x) = a sin

L


L


L

(a)

and

172

∞

∞

∞

∑

∑

∑

n3 3n

n=1

2 n3 3n n

n=1

iv 4 n3 3n n

n=1

α xg (x) = a sin

L


L


L

(b) (6.4.109)

Also,

;

∞

∞ ∞

∑

∑ ∑

4 n4 4n n

n=1

'' 2 iv 4n n4 4n n 4 4n n

n=1 n=1


L

α x α xg (x) = - a α sin g (x) = a α sin

L L

(c)


( )a∑ ∑

- 1.2573α x2 -3 -4n-12.611 a α *10 a Sin = -1.400 *10n4n 3n L

1.063 33.8433

( )b

∑ ∑ ∑ ∑ ∑− −

4 4 4 -3 2 2 -7-62.6825 a α a α a α - 9.031* 10 a α - 1.978 a α - 1.0533 * 10 a *n n n n4n 4n2n 3n 3n n 3n

α x -6nSin = -2.123 * 10

L

( )c

∑ ∑ ∑ ∑ ∑ ∑

4 4 4 -3 2 -4 2 -3-1.063 a α - 1.875 a α - 2.427 a α - 1.7498 * 10 a α - 1.07 * 10 a α - 1.260 * 10 a *n n n n n4n 4n2n 3n 3n 3n

α x -5nSin = - 2.046 * 10

L (6.4.110)


2 -3 -4

n 4n 3n(-12.611α )a - (1.2573 *10 )a = -1.400 *10 (6.4.111)

( )−

4 4 -3 2 -7

n 2n n n 3n

4 2 -5

n n 4n

(-62.6825α )a - (1.063α +9.031*10 α +1.0533 *10 )a

33.8433α +1.978α a = -1.453 *10 (6.4.112)

( )−

4 4 -3 2 -3

n 2n n n 3n

4 -4 2 -5

n n 4n

-(1.063α )a - (1.875α +1.7498 *10 α +1.260 *10 )a

2.427α +1.07 *10 α a = -1.829 *10 (6.4.113)

For n = 1, 2 -3 4 -5

1 1 1α =π, α = 3.948 *10 α =1.5585 *10

Substituting into eqn.(6.4.111, 6.4.112, & 6.4.113) gives

( )

( )

( )

a

b

c

-2 -3 -4

41 31

-4 -5 -3 -5

21 31 41

-5 -3 -5 -5

21 31 41

-4.9786 *10 a -1.2573 *10 a = -1.400 *10

-9.7691*10 a -5.2325 *10 a -8.3362*10 a =1.453 *10

-1.6567 *10 a -1.2961*10 a -3.8247 *10 a = -1.829 *10

(6.4.114)

Solving eqns.[6.4.114 (a), (b) & (c)] we obtain

, ,-2 -2 -3

21 31 41a = -3.557 *10 a =1.449 *10 a = 2.446 *10

173

Substituting values of a21, a31, a41, and eqn.(6.4.5) into eqn.(6.4.7) we obtain

∴ -2

2

-2

3

-3

4

πx V (x) = -3.557 *10 Sin

50

πx V (x) =1.449 *10 Sin

50

πx V (x) = 2.446 *10 Sin

50

Table 6.17 Variation of flexural, torsional & distortional



Distance (x)

from Left

Support (m)

πx

Sin50

V2(x)

X 10-3

V3(x)

X 10-3

V4(x) X 10-3

0 0.000 0.000 0.000 0.000

5 0.309 -10.991 4.477 0.756

10 0.588 -20.915 8.520 1.438

15 0.809 -28.776 11.722 1.979

20 0.951 -33.827 13.780 2.326

25 1.000 -35.570 14.490 2.446

30 0.951 -33.827 13.780 2.326

35 0.809 -28.776 11.722 1.979

40 0.588 -20.915 8.520 1.438

45 0.309 -10.991 4.477 0.756

50 0.000 0.000 0.000 0.000

174

6.5 DISTORTIONAL ANALYSIS OF A THIN WALLED BOX GIRDER: ( Numerical Example ) 6.5.1 Introduction

In this section, numerical analysis of a single cell doubly symmetric box girder

subjected to distortional loading is carried out using the present formulation. In order

to compare the results with those obtained by using other methods of analysis, e.g.,

beam on elastic foundation ( B.E.F) analogy, the example problem used by Osadebe

and Mbajiogu (2006) was chosen.

The box structure is a single cell doubly symmetric, simply supported box girder

with the following parameters.

Length, L = 10m; depth (height of section) h = 1.25m; width b = 2.50m; flange

thickness

tf = 0.200m; web thickness, tw = 0.250m. E = 213 x 106 KN/m2 ; v = 0.23. Distortional

load q = 50KN.

6.5.2 Strain mode diagrams and evaluation of Vlasov coefficients

Figure 6.5.1 shows the strain mode diagrams for the numerical example problem,

from where the Vlasov coefficients are evaluated using the procedures discussed in

chapter four. The summary of the coefficients is given on Table 6.16.

0.208

0.208 0.4166

0.4166

(b) Distortion

ψ3

0.2604

0.2604

0.2604 0.2604

0.2604

0.2604

(a) Warping function

x

y

ϕMω =3

(c) Bending moment due to distortion

1.079 EI

1.079 EI

1.079 EI 1.079 EI

1.079 EI

1.079 EI

M3

Fig. 6.5.1 Strain mode diagrams for numerical example problem

175

Table 6.18 Coefficients for numerical example problem

A ij = a ji b ij = b ji ckj = c jk r kh = r hk

11

12

13

22

23

33

a = 0.472

a = 0.000

a = -0.034

a =1.497

a = 0.000

a = 0.40

-3

33s = 2.07x10

11

12

13

22

23

33

b = 0.625

b = 0.000

b = 0.000

b =1.000

b = 0.000

b = 0.173

11

12

13

22

23

33

34

c = 0.635

c = 0.000

c = 0.000

c =1.000

c = 0.000

c = 0.173

c = 0.033

11

12

13

22

23

33

34

44

c = 0.635

c = 0.000

c = 0.000

c =1.000

c = 0.000

c = 0.173

c = 0.033

r =1.367

6.5.3 Differential equation for distortional equilibrium (numerical example)

From the formulation of flexural-distortional equilibrium equation for doubly

symmetric section in section 5.2.2 we noted that there was no interaction between

flexural strain mode 2 and distortional strain mode 3. Therefore, the two equations

obtained for flexure and distortion eqns.(5.2.19 and 5.2,20) are uncoupled and can

be solved independent of each other. The equation obtained for distortional

equilibrium for single cell doubly symmetric section is given by eqn. (5.2.20), i.e.,

iv 2 4

3 3 3 3V -α V ''+ 4β V =K (6.5.1)

where, 2 333

33

ksα =

r, 4 33

3

33

s4β =

a and 3

3

33

qK =

Ea

6.5.4 Distortional analysis ( numerical example problem) The governing equation for distortional equilibrium for the single cell doubly

symmetric section is

iv 2 4

3 3 3 3 3 3V -α V ''+ 4β V =K ( 6.5.1)

Where, 2 433 33 33 3 3

33 33 33

ks s qα = ; 4β = ; K =

c a a E

For the single cell doubly symmetric section we have;

-3

33 33 33 34 44 33a = 0.0396, c = r = 0.173, r = 0.1954, r =1.367, s = 2.07 *10

υ9 2 9 2E = 24 *10 N/m , G = 9.6 *10 N/m , = 0.25

176

3 3q = q x b = 50 x 2.5 =125KN

2 -2

3

4 -2

α = 3.111*10

4β = 5.227 *10

Let the solution be sought in the form; λx3V (x) = e (6.5.2)

∴ 2 λx3V (x) = λe , 2 2 λx

3V (x) = λ e , iv 4 λx3V (x) = λ e

Substituting eqn.(6.5.2) and its derivatives into eqn.(6.5.1) we obtain the auxiliary

equation for homogeneous condition as follows.

( )λx 4 2 2 4e λ -α λ + 4β = 0 ⇒ 4 2 2 4λ -α λ + 4β = 0

Let 2υ= λ ∴ 2 2 4υ -α υ+ 4β = 0

⇒2 4 4

1,2

α ± α -16β=

2υ ⇒

2 44

1

α αυ = + - 4β

2 4,

2 44

2

α αυ = - - 4β

2 4

∴2 4

4

1,2 1

α αλ = υ = ± + - 4β

2 4 ;

2 44

3,4 2

α αλ = υ = ± - - 4β

2 4

where 2 -233

33

ksα = = 3.11*10

c,

4 -233

33

s4β = = 5.227 *10

a

-2 -4-2

1,2

3.111*10 9.678 *10λ = ± + -5.227 *10

2 4

-2 -2= ± 1.555 *10 +i 5.2028 *10 = -2± 1.555 *10 +i2.281

Similarly, -23,4λ = ± 1.555 *10 - i * 0.2281

Let -2a =1.555 *10 and b = 0.2281

∴ ( )1

21λ = a+ib ( )

1

22λ = - a+ib ( )

1

23λ = a - ib ( )

1

24λ = - a - ib

Substituting into eqn.(6.5.2) we obtain

( ) ( ) ( ) ( )1 1 1 1


3 1 2 3 4V (x) = A e + A e + A e + A e

⇒3 1 2 3 4

V (x) = A coshωx cosφx + A coshωx sinφx + A sinhωx cosφx + A sinhωx sinφx (6.5.3)

177


1-2 21.555 *10 + i * 0.2281*

=φ imaginary value of

1-2 21.555 *10 + i * 0.2281*

For φ = a+ib ,ω =Re a+ib , φ =Im a+ib

∴ a+ib =ω+iφ ⇒ ( )2 2 2a+ib = ω+iφ =ω +2iωφ -φ

( )2 2a+ib = ω -φ +2iωφ

Equating equivalent terms we obtain

2 2ω -φ = a ……….(a) and 2ωφ = b ……..(b)


( )2

2 2 2 2 2 2ω -φ + 4ω φ = a +b ⇒ 4 2 2 4 2 2ω - 2ω φ +φ + 4ω φ = a+b

( )⇒ 4 2 2 4 2 2ω +2ω φ +φ = a +b

( ) ( )⇒2

2 2 2 2ω +φ = a +b ∴ 2 2 2 2ω +φ = a +b ……….(c)


2 2 22ω = a+ a +b

⇒2 2a+ a +b

ω =2

= -2 -4 -21.555 *10 2.418 *10 +5.2029 *10

+ = 0.349412 2


2 2 22φ = a +b -a

⇒2 2a +b -a

φ =2

( ) ( )

22-2 -21.555 *10 + 0.2282 -1.555 *10

= = 0.32642

Thus, ω = 0.34941, φ = 0.3264

3 homo. 1 2 3 4V (x) = A coshωx cosφx + A coshωx sinφx + A sinhωx cosφx + A sinhωx sinφx

Let the particular integral be sought in the form; 3V (x) =B (6.5.4)

This solution must satisfy eqn.(6.5.1)

i.e, iv 2 4

3 3 3 3V -α V ''+ 4β V =K

From eqn.(6.5.4) we obtain iv

3 3V '' = 0, V = 0

Substituting into eqn.(6.5.1) we obtain

178

4

44β B =K ∴ 3

4

KB =

4β ⇒ ( ) 3

3 4particular

KV x =

4β

where 3

-433 9

33

q 125 *10K = = =1.3152*10

a E 0.0396 * 24 *10


V(0) = 0; V(L) = 0; V''(0) = 0; V''(L) = 0

33 1 2 3 4 4

KV (x) = A coshωx cosφx + A coshωx sinφx + A sinhωx cosφx + A sinhωx sinφx +

4β

(6.5.5)

[ ]3 1 2 3 4V ''(x) = 2 -A sinhωx sinφx + A sinhωx cosφx - A coshωx sinφx + A coshωx cosφx

∴ ⇒ 31 4

KV(0) = 0 : A = -

4β (a)

⇒3 4V ''(0) = 0 : 2A = 0 ⇒ 4A = 0 (b)

⇒ 31 2 3 4 4

KV(L) = 0 : -16.3522A - 2.0117A -16.3221A - 2.008A + = 0

4β (c)

V''(L) = 0 : ⇒ 1 2 32.008A -16.3221A +2.0117A = 0 (d)


31 4

KA = -

4β; -3 3

2 4

KA = 7.873 *10

4β; 3

3 4

KA =1.0621

4β; 4A = 0

Substituting values of 1 2 3 4A , A , A , & A into eqn.(6.5.5) we obtain

-33

3 4

KV (x) = -Coshωx Cosφx +7.873 *10 * Coshωx Sinφx +1.062* Sinhωx Cosφx +1

4β

where, 4 -24β = 5.227 *10 , -4

3K =1.3153 *10 , ⇒ -33

4

K= 2.5163 *10

4β

ω = 0.3494, φ = 0.3264

∴

-33 -3

1.0621Sinhωx * Cosφx -Coshωx * CosφxV (x) = 2.5163 *10

+7.873 *10 * Coshωx * Sinφx +1

Table 6.19 Variation of distortional displacement along the girder

length:(numerical example problem)

Distance (x)

from Left

Support (m)

1+ 1.0621 *

sinhωx * cosφx

coshωx * cosφx- 7.873 x 10-3

*

coshωx * sinφx

Distortional displacement

V3(x)

179

0 1.000 -1.0000 0.0000 0.000 x 103 (m)

2 1.6388 -0.9964 0.0060 1.631

4 1.5287 -0.5626 0.0163 2.472

5 0.8193 0.1808 0.0232 2.575

6 -0.6085 1.5610 0.0301 2.473

8 -6.4690 7.0850 0.0327 1.632

10 -16.3340 16.3506 -0.01580. 0.000

6.5.5 Results From Other Methods of Analysis

Osadebe and Mbajiogu (2006) carried out a distortional analysis of the numerical

example problem using

(a) Vlasov’s theory with provision for shear deformations

(b) Beam on elastic foundation (BEF) analogy.

Table 6.18 shows the results they obtained for the two methods of analysis.

Substituting the actual values of q = q*b = 50 * 2.5 =125KN, and E = 24 *106 KN/m2

(for concrete), we obtain values in Table 6.19 which are comparable with the values

from the present formulation.

Table 6.20:Distortional Displacement V(x) (m) (Osadebe & Mbajiogu (2006))

X (m) 0 2 4 5 6 8 10

V(x)

(BEF)

0 31.654759

q

E 48.070156

q

E 50.074227

q

E 48.070217

q

E 31.654892

q

E

0

V(x)

Osadebe

&

Mbajiogu

0 32.479423

q

E 48.32075

q

E 49.86876

q

E 48.03229

q

E 32.479381

q

E

0

180

Table 6.21 Comparison of results from present formulation with

results from other methods of analysis (Numerical example)

X (m) Distortional displacement V3(x)

Present formulation x 10-3 (m)

Osadebe & Mbajiogus’ formulation x 10-3 (m)

BEF Analogy method x 10-3 (m)

0 0.000 0.000 0.000

2 1.631 1.691 1.649

4 2.472 2.516 2.505

5 2.575 2.597 2.608

6 2.473 2.516 2.503

8 1.632 1.691 1.649

10 0.000 0.000 0.000

181

CHAPTER SEVEN

7.0 DISCUSSION OF RESULTS, CONCLUSIONS AND RECOMMENDATIONS

7.1 General

In this work the torsional-distortional response of different cross sectional box

girder profiles were studied. They include (a) Single cell doubly symmetric section,

(b) multi cell doubly symmetric section, (c) single cell mono symmetric section and

double cell mono symmetric section.

Governing differential equations of equilibrium were derived for analysis of the

structure under the following modes of interaction.

(i) Flexural- torsional interaction

(II) Flexural distortional interaction

(iii) Torsional-distortional interaction

(iv) Flexural-torsional-distortional interaction

The results obtained from the analysis of a single cell doubly symmetric section,

numerical example problem, are also presented for the purpose of comparison.

7.2 Discussion of Results

7.2.1 Governing differential equations of equilibrium

The summary of the differential equations of equilibrium derived for each of these

modes of interaction is presented in Table 5.1, from where it can be observed that

the equilibrium equations for all doubly symmetric sections are the same for every

strain mode interaction. For example, the same equilibrium equation was obtained for

both single cell and multi cell doubly symmetric section for torsional-distortional

analysis. Similarly, the same equilibrium equations were derived for mono symmetric

sections for every strain modes interaction.

In doubly symmetric section it was observed that there was no interaction

between strain mode 2, (minor axis bending) and strain modes 3 and 4 (distortion

and torsion). Thus, the differential equations of equilibrium governing flexure (in

minor axis) is independent of distortion and torsion.

In mono symmetric sections, on the other hand, all the three strain modes interact

with each other and the equations of equilibrium governing any of the modes of

interaction were always coupled.

182

For doubly symmetric and mono symmetric sections the governing equations of

equilibrium are independent of strain mode 1 (major axis bending). In other words,

the major axis strain mode does not interact with any other strain modes in either

doubly symmetric or mono symmetric sections. It is for this reason that torsional

loads can be separated from the general loads on the bridge structure for the

purpose of torsional-distortional analysis.

For non-symmetric sections all the four strain modes interact in such a way that

the equilibrium conditions are inseparable. This means that a consideration of

torsional-distortional analysis of a non symmetric section can not be accomplished

without a consideration of all the strain modes interacting on the structure giving rise

to displacement in major axis V1(x), displacement in minor axis V2(x), distortion

V3(x) and torsion V4(x). It is for this reason that torsional-distortional analysis of non-

symmetric sections is most challenging, requiring a consideration of the general

loads on the box girder structure.

The governing differential equations of equilibrium presented in Table 5.1(a), (b)

and (c) are for analysis of deformable cross sections where distortion is not

prevented by the use of diaphragms or intermediate stiffeners, which is outside the

scope of this work.

7.2.2 Single cell doubly symmetric section

Fig. 7.2.1 shows the variation of torsional displacement along the length of the

girder while Fig.7.2.2 shows the variation of distortional displacement along the

length of the girder. Fig.7.2.1 is the result obtained for flexural-torsional analysis

while Fig.7.2.2 is for flexural distortional analysis. However, since there was no

interaction between flexure & torsion and between flexure & distortion in doubly

symmetric sections, both results represent the results for flexural analysis and

torsional analysis of single cell doubly symmetric section respectively. In the two

analysis , the flexural displacements are zero as a result of absence of component of

torsional load in minor axes plane and as a result of non interaction of strain mode 2

with other strain modes.

Fig.7.2.3 is the result for flexural-torsionl-distortional analysis of single cell doubly

symmetric section. It shows the variation of torsional displacement and distortional

displacement along the length of the girder, the flexural displacement being also zero

in this case.

183

0 5 10 15 20 25 30 35 40 45 500

20

40

Distance Along the Length of the Girder (m)

Dis

tort

ional D

ispla

cem

ent

(mm

)

0 5 10 15 20 25 30 35 40 45 500

1

2

3

4

5

Distance Along the Length of the Girder (m)Tors

ional D

ispla

cem

ent

(mm

)

Fig.7.2.1: Variation of torsional displacement along the length of the girder (Single cell doubly symmetric section)

Cross section

Fig.7.2.2: Variation of distortional displacement along the length of the girder (Single cell doubly symmetric section)

Cross section

184

0 5 10 15 20 25 30 35 40 45 500

10

20

30

40

50


Dis

pla

cem

ent

(mm

)



A closer look at Figs 7.2.1, 7.2.2 and 7.2.3 shows that a maximum (mid span)

torsional displacement of about 2mm was obtained for torsional analysis while 27mm

deplanation was obtained for distortional analysis. These values remain relatively

unchanged in the case of torsional distortional analysis as could be seen from

Fig.7.2.3. This shows that the effect of interaction between torsion and distortion is

negligible in single cell doubly symmetric section.

7.2.3 Multi-cell doubly symmetric section

Figures 7.2.4 and 7.2.5 show the variation of torsional deformation and distortional

deformation respectively, along the length of the girder, for multi-cell doubly

symmetric section. The mid span torsional deformation is 2.6mm, while the mid span

deplanation is 5.9mm.

The result of torsional distortional analysis for multi cell doubly symmetric section

is graphically presented in Fig 7.2.6. The maximum (mid san ) deformation was 6mm

while torsional deformation was 3mm. These values are very close to those obtained

by separate analysis for torsion and distorsion. Consequently, the equations of

equilibrium

Fig.7.2.3: Variation of torsional and distortional displacements along the length of the girder (Single cell doubly symmetric section)

Cross section

185

governing torsional-distortional response of doubly symmetric sections can be

0 5 10 15 20 25 30 35 40 45 500

1

2

3

4

5


Tors

ional dis

pla

cem

ent

(mm

)

0 5 10 15 20 25 30 35 40 45 500

2

4

6

8

10


Dis

tort

ional D

ispla

cem

ent

(mm

)

Fig.7.2.4: Variation of torsional displacement along the length of the girder (Multi- cell doubly symmetric section)

Cross section

Fig.7.2.5: Variation of distortional displacement along the length of the girder (Multi- cell doubly symmetric section)

Cross section

186

0 5 10 15 20 25 30 35 40 45 500

5

10

15


Dis

pla

cem

ent

(mm

)

Distortional Displacement

Tortional Displacement

decoupled to obtain two independent equations for distorsional displacement V3(x)

and torsional displacement V4(x). This implies that the weak interaction between

torsion and distortion can be ignored in all doubly symmetric sections, and that

analysis for torsion and distortion can be independently carried out.

7.2.4 Single cell mono symmetric section

The results for the analysis of single cell mono symmetric section are presented

as follows.

Flexural- torsional analysis; Fig. 7.2.7

Flexural-distortional analysis; Fig.7.2.8

Torsional-distortional analysis; Fig.7.2.9

Flexural-torsional-distortional analysis; Fig.7.2.10

For flexural torsional analysis, the mid span flexural displacement was 80mm

which dropped to 4mm in flexural distortional analysis. This shows that distortional

stresses tend to minimize flexural stresses. A comparison of maximum (mid span)

torsional values in Figs. 7.2.7 and 7.2.9 also shows that distortional stresses do not

encourage torsion in torsional- distortional analysis. However, when the three

modes of interaction

Fig.7.2.6: Variation of torsional and distortional displacements along the length of the girder (Multi- cell doubly symmetric section)

Cross section

187

0 5 10 15 20 25 30 35 40 45 500

20

40

60

80

100


Dis

pla

cem

ent

(mm

)

Flexural displacement


0 5 10 15 20 25 30 35 40 45 500

20

40

60

80


Dis

pla

cem

ent

(mm

) Flexural displacement


Fig.7.2.7: Variation of flexural and torsional displacements along the length of the girder (Single cell mono symmetric section)

Fig.7.2.8: Variation of flexural and distortional displacements along the length of the girder (Single cell mono symmetric section)

cross section

cross section

188

0 5 10 15 20 25 30 35 40 45 500

10

20

30

40

50

60


Dis

pla

cem

ent (m

m)



0 5 10 15 20 25 30 35 40 45 500

10

20

30

40

50

60

70


Dis

pla

cem

ent

(mm

)


Distorsional displacement


interplay, we notice from Fig.7.2.10 that distortional deformation (43mm) is higher

than torsional deformation (33mm) which in turn is higher than torsional displacement

(4mm).

Fig.7.2.9: Variation of torsional and distortional displacements along the length of the girder (Single cell mono symmetric section)

cross section

Fig.7.2.10: Variation of flexural, torsional and distortional displacements along the length of the girder (single cell mono symmetric section)

cross section

189

Generally we observe that distortional deformations (and hence stresses) were

consistently higher than higher than flexural and torsional stresses where ever either

or both of flexure and torsion interact with distortion. Also flexural deformations were

consistently higher than torsional deformations when ever both strain modes interact.

7.2.5 Double cell mono symmetric section

The results for the analysis of double cell mono symmetric section are presented

as follows. Fig. 7.2.11 is for flexural- torsional analysis, Fig.7.2.12 is for flexural

distortional analysis, Fig.7.2.13 is for torsional-distortional analysis, while Fig.7.2.14

is for flexural torsional- distortional analysis.

As in the case of single cell mono symmetric section, comparison of Figs. 7.2.11

and 7.2.12 shows that distortional stresses do not encourage deflection of the box

girder structure since mid span flexural deformation for flexural-torsional interaction

analysis was 33mm but for flexural- distortional interaction analysis it came down to

9mm.

When the three modes of interaction interplay we notice from Fig. 7.2.14 that

flexural deformations are significantly higher than distortional and torsional

deformations. This appears to represent the true picture of state of stress in bridge

structures where by more attention is given to flexural stresses than distortional and

torsional stresses in design.

0 5 10 15 20 25 30 35 40 45 500

10

20

30

40

50

60


Dis

pla

cem

ent (m

m)



Fig.7.2.11: Variation of flexural and torsional displacements along the length of the girder (Double cell mono symmetric section)

cross section

190

0 5 10 15 20 25 30 35 40 45 500

5

10

15

20

25

30

Distance Along the Length of the Girder(m)

Dis

pla

cem

ent

(mm

)


Flexural Displacement

0 5 10 15 20 25 30 35 40 45 500

5

10

15

20

25

30

35

40

45

50


Dis

pla

cem

ent (m

m)



Fig.7.2.12: Variation of flexural and distortional displacements along the length of the girder (Double cell mono symmetric section)

cross section

Fig.7.2.13: Variation of torsional and distortional displacements along the length of the girder (Double cell mono symmetric section)

cross section

191

0 5 10 15 20 25 30 35 40 45 500

10

20

30

40

50

60


Dis

pla

cem

ent

(mm

)


Distorsional displacement


7.3 COMPARISON OF RESULTS

Fig.7.3.1 shows the results obtained for distortional analysis of the numerical

example problem in section 6.5.1, using the present formulation. Table 6.19 shows

the results obtained from the present formulation and by Osadebe and Mbajiogu

(2006) for the same numerical example problem using (a) Vlasov’s theory with

modification for shear and (b) Beam on elastic foundation (BEF) analogy method.

The three results are graphically represented in Fig.7.3.2 from where it can be seen

that they are in agreement.

Fig.7.2.14: Variation of flexural, torsional and distortional displacements along the length of the girder (Double cell mono symmetric section)

cross section

192

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

Distance Along the Length of the Girder (mm)

Dis

tort

ional D

ispla

cem

ent

(mm

)

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5


Dis

tort

ional D

ispla

cem

ent

(mm

) Present formulation

Osadebe & Mbajiogu

BEF analogy

Fig. 7.3.1: Variation of distortional displacement along the length of the girder (numerical example problem)

cross section

Fig.7.3.2: Variation of distortional displacements along the length of the girder ( comparison of results)

cross section

193

7.4 CONCLUSIONS

In doubly symmetric profiles distortional deformations were found to be higher

than torsional deformations and the effect of interaction between torsion and

distortion is quite negligible. In fact, there was no interaction between (i) major axis

bending and distortion (ii) major axis bending and torsion (iii) minor axis bending and

distortion (iv) minor axis bending and torsion.

In mono symmetric section it was found that distortional stresses tent to reduce

flexural deformation more than torsional deformation. However, when the three strain

modes of flexure, torsion and distortion interplay, it was found that distortional

deformation was higher than flexural deformations only in single cell mono symmetric

section. In double cell mono symmetric profile flexural deformation was appreciably

higher than distortional and torsional deformations.

Application of the derived governing differential equations of equilibrium in the

analysis of doubly symmetric and mono symmetric sections gave good results.

Comparison of the results with those obtained using other methods of analysis show

that the derived equations of equilibrium are reliable.

For doubly symmetric and mono symmetric sections the governing equations of

equilibrium are independent of strain mode 1. In other words, the major axis strain

mode does not interact with any other strain modes in either doubly symmetric or

mono symmetric profiles. In non symmetric section all the four strain modes interact

in such a way that the equations of equilibrium are inseparable.

In real life structure with generalized loading, torsional- distortional interaction

analysis can not be separated from flexural- torsional- distortional interaction

analysis. Thus, in other to obtain reasonable results flexural–torsional-distortional

analysis were incorporated in the analysis.

As would be expected, multi cell sections offered greater resistance to flexural,

torsional and distortional deformations than single cell sections. This is because of

increase in the number of vertical members in multi cell sections contributing to the

rigidity of the structure. The introduction of middle member in double cell mono

symmetric cross section decreased torsional deformation and distortional

deformation by 40% and 67% respectively, while the lateral flexural deformation was

increased by 9%. The torsional-distortional deformations on mono symmetric

sections can not be compared with that of doubly symmetric sections because the

torsional moments are not the same even though the loadings are the same.

194

7.5 CONTRIBUTION TO KNOWLEDGE

The major contributions of this work to knowledge are as follows.

• For box girders possessing doubly symmetric sections, it was found that

neither torsion nor distortion interacts with flexure. Differential equations of

equilibrium were therefore derived for independent analysis of the structure in

flexure, torsion and distortion.

• For box girders possessing mono symmetric sections, it was ascertained that

both torsion and distortion interact with bending about the non symmetric axis

only. An independent equation for flexural analysis is therefore derived for the

single axis of symmetry.

• On the basis of (2) above differential equations of equilibrium for the following

cases were derived for box girders with mono symmetric section.

Case 1; flexural –torsional analysis (eqn.5.3.9),

Case 2; flexural-distortional analysis (eqn.5.3.24),

Case 3; torsional-distortional analysis (eqn.5.3.32),

Case 4; flexural-torsional-distortional analysis (eqn. 5.3.50 )

• For non symmetric sections, it was found that torsion, distortion and bending

interact and are therefore inseparable. Hence, a set of coupled differential

equations of equilibrium for generalized analysis of non symmetric sections in

flexure, torsion and distortion was derived.

• From all the above, the obtained differential equations of equilibrium can be

used to analyze box girders possessing doubly symmetric sections, mono

symmetric sections and non symmetric sections. Numerical examples

demonstrating the applicability of the obtained equations are given for box

girders possessing doubly symmetric section and mono symmetric section.

• Generalized equations for the flexural-torsional-distortional analysis of box

girder structures possessing non symmetric section, to the best of knowledge

of the author, is not available in technical literature. This makes this work

unique as it provides a holistic approach to torsional-distortional analysis of

thin walled box girder structures.

195

• The study has provided foundation for ramification of research work in elastic

analysis of thin walled structures made of concrete, steel, timber and other

structural materials. It only suffices to use the appropriate physical properties

(E, G, υ ) of a chosen material.

7.6 RECOMMENDATIONS FOR FURTHER WORK

Knowledge of higher engineering mathematics plays a vital role not only in the

formulation of the differential equations of equilibrium but also in the integration of the

derived equations to obtain solutions. In the case of non-symmetric sections, four 4th

order coupled differential equations of equilibrium were derived but not applied to

solution of problems. It is recommended that further work be carried out on non

symmetric sections so as to demonstrate the applicability of the derived equations

for flexural-torsional-distortional analysis of such profiles.

During the process of evaluation of warping functions it was found that material

thickness does not affect the warping function. However, the bending moments due

to distortion of cross sections vary with the thickness of the cross sectional members.

A parametric study on bridge deck arrangements is recommended. Such study would

high light the effect of cross sectional parameters on the distortional bending moment

of box girder sections and invariably on the torsional-distortional response of box

girder structures.

The present work considered only simply supported box girders in the analysis,

even though the derived governing differential equations are not limited to simply

supported bridge decks. The extension of this work to cover the analysis of bridge

decks with a combination of support conditions (rigid support, continuous support,

pinned support etc.) will bring the concept of this work to a logical conclusion.

196

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496.

Ontario Ministry of Transport and Communication OMTC (1992). Ontario

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highway bridge deign code, OHBDC, 2nd Ed., Downsview, Ontario.

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200

APPENDIX ONE

COMPUTATION OF WARPING FUNCTIONS

A1(a) Single cell mono-symmetric box girder section

Referring to Fig.A1 we obtain the following:

Cross sectional area F = 4.0446*106 mm2

Enclosed area A= 16,744,500mm2

Fy = 0, symmetry on y-axis

Fz = 2(3557 x 203 x 3050/2) + (3660 x 203 x 3050) = 4.648x10 9mm3

Integrating from point v to point 1, Fig.4.3.3b, we obtain; Bρ = 0, s = 3660, t = 203 mm

∫

22A 2x16,744500ψ = = = 375718mm

ds 7230 2x3557 3660+ +

t 203 203 203

6

1

44722ω = (0 - )(3660) = - 6.774x10

203mm2

From point 1 to point 2; 6

2ω = - 6.774x10

From point 1 to point 3;

o

Bρ = 3660sin59.04 = 3138.4mm

6 6 2

3

185083ω = (3138.4 - )(3556.88) - 6.744x10 = -2.194x10 mm

203

From point 3 to point 4; Bρ = 3050, s = 3660, t =203

∴66 2

4

185083ω = (3050 - )(3660) - 2.194x10 = 2.194x10 mm

203

By reason of symmetry we have,

915 7320

2745 2745 3660

FIG. A1: Single cell box girder section

3050

915

t = 203

201

6

5 1ω = -ω = 6.774x10 and 6 2

6 2ω = - ω = - 6.774X10 mm

The warping function B

ω with respect to the pole B for this profile is shown in

Fig.4.3.3 (d). The _

x and _

y diagrams are plotted in Figs. 4.3.3(b) and (c) respectively.

The moment of areas and product moments of areas are obtained using Morh’s

integral for displacement computations. Thus,

F = 4.045 x 106 mm

2 ; _

yF = 0

_

9

z

F = 4.468x10 ; __

13

yyF = 2.508x10 mm4

--

13

zzF =1.1388x10 mm4; --

yzF = 0

--

16

yωF = - 4.242x10 mm6, --

zωF = 0

--

19

ωωF = 7.253x10 mm6

The product integrals based on the centroidal axis are obtained from eqn.(4.3.5).

Thus,

113

yyF = 2.508x10 ; 12

zzF = 6.453x10

yzF = 0 ; 16

yωF = - 2.242x10 mm6

zωF = 0 ; 19

ωωF = 7.253X10 mm6

From eqn. (4.3.6) and (4.3.7) the coordinates of the shear centre are obtained as

below.

M B(y - y ) = 0 and M B(z - z ) =1692mm

From eqn. (4.3.8) the expression for the warping function is obtained:

M Bω (s) =ω +1692x (4.3.9)

The warping function for the single cell mono-symmetric section with respect to the

shear centre is shown in Fig. 4.3.8(g).

A1(b) Double cell mono-symmetric section

From Fig.A2 we obtain the following properties of the section.

Cross sectional area F =4,663,722 mm2

Enclosed area A1 = A2 = A3 = 8,372,250 mm2

Fx = 0, symmetry about y-y axis

Fy = (3660 x 203 x 3050) + (3050 x 203 x 1525)+2(3557x203x1525) = 5.4126 x

109mm3

-

202

∫B Bω (s) = ρ (s)ds for open section of the profile

∫B B

ψω (s) = ρ - ds

t for closed section of the profile

For closed sections we have, from eqn.(2.6.19)

∫ ∫ ∫B D

i i-1 i i+1

A C

ds ds ds2A = -ψ +ψ -ψ

t t t

For; Cell 1 we obtain; 6

1 259.59ψ -15.025ψ =16.7445x10 (1)

Cell 2 ⇒ 6

1 2 315.02ψ - 59.59ψ +35.57ψ = -16.7445x10 (2)

Solving eqns (1, & 2) above gives;

5

1 2ψ =ψ = 3.7572x10

For closed Cells 1 & 2, ( )∫B Bω = ρ -1850.84 ds

Integrating from point V to points 1(closed section) we obtain;

Bρ = 0, ds = 3660, t = 203, ∴ 6 2

1ω = -1850.84 * 3660 = - 6.774 *10 mm mm2

From point 1 to point 2 (open section) we have;

Bρ = 0, ds = 915, t = 203, ∴ 6 6

2ω = (0 * 915) - 6.774 *10 = - 6.774 *10 mm2

From point 1 to point 3 (closed section) we have;

Bρ = 3660 * sin59.036°= 3138.42, t = 203, ds = 3557,

( )∴ 6 6

3ω = -6.774 *10 + 3138.42 -1850.84)(3557 = -2.194 *10 mm2

By reason of symmetry we observe that;

6

4 3ω = -ω = 2.194 *10 mm2

and 6

5 1ω = -ω = 6.774 *10 mm2

3050

2745 2745 1830 1830

915 915 3660

203

203

203

1889.4

1160.58

3660

y

x

Fig. A2: Double cell mono symmetric box section

203

6

6 2ω = -ω = 6.774 *10 mm2

The section properties are evaluated using the same procedure used in section

4.3.4.1(a). The relevant values obtained are,

xy ωyF =F = 0

13

xxF = 2.5073 *10

16

ωxF = -4.242*10

Substituting these properties into eqns.(4.3.6 & 4.3.7) the coordinates of the shear

centre are obtained. Thus,

( ) 0M Bx x− = and ( )M By - y =1691.84mm

From eqn.(4.3.8) we obtain the expression for the warping function with respect to

the shear centre as,

M Bω =ω +1691.84x

where Bω is obtained from Fig.4.3.4(d).

The warping function Mω with respect to shear centre is on Fig.4.3.9 (g).

A1(c) Non-symmetric section

Fig.A3 shows the cross section of a single cell non symmetric section.

Cross sectional area F = 3.1514x106mm2

Enclosed area A= 13.954x106mm2

Fx = 4.2477x109mm3

Fy = 7.608x109mm3

∫B B

ψω (s) = ρ - ds

t for closed section of the profile

Fig.A3: Single cell non-symmetric section

1830

200

5490

200

200

3660

x

204

∫

22A 2x13,953,750x200ψ = = = 354230mm

ds 5490+3557+3600+3050

t

Integrating from point v to point 1; Bρ = 0, ds = 5490, t =100 mm

6

1

354230ω = (0 - )(5490) = -9.723x10

200mm2

From point 1 to point 2; Bρ = 5490sin59.036 = 4708mm

( ) 6

2ω = 4708 -1771 x3557 -9.723x10 = 6 20.724x10 mm

From point 2 to point 3;

Bρ = 3050, ds -3050, t =100

( )∴ 6

3ω = 3050 -1771 x3660+0.724x10 = 6 25.405x10 mm

The location of the centroidal axes of the section is given by,

9

x

6

F 7.608 *10x = = = 2414mm

F 3.1514 *10;

9y

6

F 4.2477 *10y = = =1348mm

F 3.1514 *10

The warping function with respect to the centroidal axes is given by,

C B 0ω =ω -ω , where Bω0

Fω =

F

The cross sectional parameters are evaluated following the procedures given in

section 4.3.4. The values obtained are as follows.

F = 3.151 *106 mm

2 , _

9

x

F = 7.608 *10

_

9

yF = 4.248 *10 , _

12

ωF = -4.647 *10

__

13

xx

F = 2.936 *10 mm4, 13

yyF =1.091*10 mm4

10

xyF = 4.359 *10 , 16

ωxF = -3.226 *10 mm6

15

ωyF = 7.202*10 ,

The product integrals based on the centroidal axis are obtained from eqn.(4.3.5).

Thus,

13

xxF = -1.099 *10 ; 12

yyF = 5.183 *10

12

xyF = -1.539 *10 ; 16

ωxF = -2.104 *10 mm6

16

ωyF =1.347 *10 ;

205

From eqn (4.3.6) and (4.3.7) the coordinates of the shear centre are obtained as

follows.

ωy xx ωx xy

M B 2

xx yy xy

F .F -F F(x - x ) = = 2119mm

F .F - (F )

and ωy xy ωx yy

M B 2

xx yy xy

F .F -F F(y - y ) = =1618mm

F .F - (F )

From eqn.(4.3.8) the warping function with respect to the shear center is obtained:

( ) ( )M C M B M Bω =ω + y - y x - x - x y

C=ω +1618x - 2119y

Bω 6

C B B

Fω =ω - =ω +1.475 *10

F

∴ 6

M Bω =ω +1.475 *10 +1618x - 2119y

The warping function with respect to the centroidal axes for this profile is shown in

Fig. 4.3.6(b) while that with respect to the shear centre is on Fig.4.3.12 (g).

206

APPENDIX TWO:

COMPUTATION OF VLASOV’S COEFFICIENTS

A2.1 Coefficients for single cell mono-symmetric section

ϕ ϕ∫a =a = (s) (s)dAij ji i j

' '( ). ( )b b s s dAi jij jisϕ ϕ= = ∫

11 1 1( ). ( ) 3.179s

a s s dAϕ ϕ= =∫ ' '

11 1 1( ). ( ) 0.522s

b s s dAϕ ϕ= =∫

12 21 1 2( ). ( ) 0.000s

a a s s dAϕ ϕ= = =∫ ' '

12 21 1 2( ). ( ) 0.000s

b b s s dAϕ ϕ= = =∫

13 31 1 2( ) ( ) 0.000s

a a s s dAϕ ϕ= = =∫ ' '

13 31 1 3( ). ( ) 0.000s


22 2 2( ) ( ) 12.341s

a s s dAϕ ϕ= =∫ ' '

22 2 2( ). ( ) 1.469s

b s s dAϕ ϕ= =∫

23 32 2 3( ). ( ) 0.133s

a a s s dAϕ ϕ= = = −∫ ' '

23 32 2 3( ). ( ) 0.525s

b b s s dAϕ ϕ= = = −∫

33 3 3( ). ( ) 0.373s

a s s dAϕ ϕ= =∫ ' '

33 3 3( ). ( ) 0.693s

b s s dAϕ ϕ= =∫

'( ). ( )c c s s dAjkj jk ks

ψ ϕ= = ∫ '( ). ( )i h

c c s s dAih hi s

ϕ ψ= = ∫

'

11 1 1( ). ( ) 0.522c s s dAψ ϕ= =∫ '

11 1 1( ). ( ) 0.522c s s dAψ ϕ= =∫

( ) '

12 21 1 2. ( ) 0.000s

c c s s dAψ ϕ= = =∫ ( ) '

12 21 1 2. ( ) 0.000s

c c s s dAψ ϕ= = =∫

'

13 31 1 3( ). ( ) 0.000s

c c s s dAψ ϕ= = =∫ '

13 31 1 3( ). ( ) 0.000s


'

22 2 2( ). ( ) 1.469s

c s s dAψ ϕ= =∫ '

22 2 2( ). ( ) 1.469s

c s s dAψ ϕ= =∫

'

23 32 2 3( ). ( ) 0.525s

c c s s dAψ ϕ= = = −∫ '

23 32 2 3( ). ( ) 0.525s

c c s s dAψ ϕ= = = −∫

'

33 3 3( ). ( ) 0.693s


33 3 3( ). ( ) 0.693s

c s s dAψ ϕ= =∫

'

41 4 1( ). ( ) 0.000s


41 4 1( ). ( ) 0.000s

c s s dAψ ϕ= =∫

'

42 4 2( ). ( ) 0.769S

c s s dAψ ϕ= = −∫ '

42 4 2( ). ( ) 0.769S

c s s dAψ ϕ= = −∫

207

'

43 4 3( ). ( ) 0.623s


43 4 3( ). ( ) 0.623s

c s s dAψ ϕ= =∫

( ). ( )r r s s dAkh hk k hs

ψ ψ= = ∫

11 1 1( ). ( ) 0.525s

r s s dAψ ψ= =∫

12 21 1 2( ). ( ) 0.000s

r r s s dAψ ψ= = =∫

13 31 1 3( ). ( ) 0.000s


14 41 1 4( ). ( ) 0.000s


22 2 2( ). ( ) 1.469s

r s s dAψ ψ= =∫

23 32 2 3( ). ( ) 0.525s

r r s s dAψ ψ= = = −∫

24 42 2 4( ). ( ) 0.769s

r r s s dAψ ψ= = = −∫

33 3 3( ). ( ) 0.693s

r s s dAψ ψ= =∫

34 43 3 4( ). ( ) 0.623s


44 4 4( ). ( ) 7.200s

r s s dAψ ψ= =∫

A2.2 Coefficients for double cell mono-symmetric section

( ) ( )a a s s dAij ji i j

ϕ ϕ= = ∫ ' '( ). ( )b b s s dAi jij jisϕ ϕ= = ∫

11 1 1( ). ( ) 7.023s

a s s dAϕ ϕ= =∫ ' '

11 1 1( ). ( ) 1.680s

b s s dAϕ ϕ= =∫

12 21 1 2( ). ( ) 0.000s

a a s s dAϕ ϕ= = =∫ ' '

12 21 1 2( ). ( ) 0.000s


13 31 1 2( ) ( ) 0.000s

a a s s dAϕ ϕ= = =∫ ' '

13 31 1 3( ). ( ) 0s


22 2 2( ). ( ) 25.073s

a s s dAϕ ϕ= =∫ ' '

22 2 2( ). ( ) 2.982s

b s s dAϕ ϕ= =∫

208

23 32 2 3( ) ( ) 0.425a a s s dAϕ ϕ= = =∫ ' '

23 32 2 3( ). ( ) 0.449s

b b s s dAϕ ϕ= = = −∫

33 3 3( ). ( ) 0.750s

a s s dAϕ ϕ= =∫ ' '

33 3 3( ). ( ) 1.533s

b s s dAϕ ϕ= =∫

'( ). ( )c c s s dAjkj jk ks

ψ ϕ= = ∫ '( ). ( )i h

c c s s dAih hi s

ϕ ψ= = ∫

'

11 1 1( ). ( ) 1.680s


11 1 1( ). ( ) 1.680s

c s s dAψ ϕ= =∫

( ) '

12 21 1 2. ( ) 0.000s

c c s s dAψ ϕ= = =∫ ( ) '

12 21 1 2. ( ) 0.000s


'

13 31 1 3( ). ( ) 0.000s

c c s s dAψ ϕ= = =∫ '

13 31 1 3( ). ( ) 0.000s


'

22 2 2( ). ( ) 2.982s


22 2 2( ). ( ) 2.982s

c s s dAψ ϕ= =∫

'

23 32 2 3( ). ( ) 0.449s

c c s s dAψ ϕ= = = −∫ '

23 32 2 3( ). ( ) 0.449s

c c s s dAψ ϕ= = = −∫

'

33 3 3( ). ( ) 1.533s


33 3 3( ). ( ) 1.533s

c s s dAψ ϕ= =∫

'

41 4 1( ). ( ) 0.000s


41 4 1( ). ( ) 0.000s

c s s dAψ ϕ= =∫

'

42 4 2( ). ( ) 3.404S


42 4 2( ). ( ) 3.404S

c s s dAψ ϕ= =∫

'

43 4 3( ). ( ) 0.886s


43 4 3( ). ( ) 0.886s

c s s dAψ ϕ= =∫


ψ ψ= = ∫

11 1 1( ). ( ) 1.680s

r s s dAψ ψ= =∫

12 21 1 2( ). ( ) 0.000s


13 31 1 3( ). ( ) 0.000s


14 41 1 4( ). ( ) 0.000s


22 2 2( ). ( ) 2.982s

r s s dAψ ψ= =∫

209

23 32 2 3( ). ( ) 0.449s

r r s s dAψ ψ= = = −∫

24 42 2 4( ). ( ) 3.404s


33 3 3( ). ( ) 1.533s

r s s dAψ ψ= =∫

34 43 3 4( ). ( ) 0.886s


44 4 4( ) ( ) 12.011r s s dAψ ψ= =∫

A2.3 Coefficients for single cell doubly symmetric section

( ) ( )a a s s dAij ji i j


11 1 1( ). ( ) 10.980s

a s s dAϕ ϕ= =∫ ' '

11 1 1( ). ( ) 1.200s

b s s dAϕ ϕ= =∫

12 21 1 2( ). ( ) 0.000s

a a s s dAϕ ϕ= = =∫ ' '

12 21 1 2( ). ( ) 0.000s


13 31 1 2( ) ( ) 0s

a a s s dAϕ ϕ= = =∫ ' '

13 31 1 3( ). ( ) 0.000s


22 2 2( ). ( ) 27.994s

a s s dAϕ ϕ= =∫ ' '

22 2 2( ). ( ) 2.880s

b s s dAϕ ϕ= =∫

23 32 2 3( ) ( ) 0.000s

a a s sϕ ϕ= = =∫ ' '

23 32 2 3( ). ( ) 0.000s


33 3 3( ). ( ) 7.709s

a s s dAϕ ϕ= =∫ ' '

33 3 3( ). ( ) 3.736s

b s s dAϕ ϕ= =∫

'( ). ( )c c s s dAjkj kj ks

ψ ϕ= = ∫ '( ). ( )i h

c c s s dAih hi s

ϕ ψ= = ∫

'

11 1 1( ). ( ) 1.200s


11 1 1( ). ( ) 1.200s

c s s dAϕ ψ= =∫

( ) '

12 21 1 2. ( ) 0.000s

c c s s dAψ ϕ= = =∫ '

12 21 1 2( ). ( ) 0.000s

c c s s dAϕ ψ= = =∫

'

13 31 1 3( ). ( ) 0.000s

c c s s dAψ ϕ= = =∫ '

13 31 1 3( ). ( ) 0.000s


'

22 2 2( ). ( ) 2.880s


14 41 1 4( ). ( ) 0.000s


210

'

23 32 2 3( ) ( ) 0.000s

c c s s dAψ ϕ= = =∫ '

22 2 2( ). ( ) 2.880s

c s s dAϕ ψ= =∫

'

33 3 3( ). ( ) 3.736s


23 32 2 3( ). ( ) 0.000s


'

41 4 1( ). ( ) 0.000s


24 42 2 4( ). ( ) 0.000s


'

42 4 2( ). ( ) 0.000S


33 3 3( ). ( ) 3.736s

c s s dAϕ ψ= =∫

'

43 4 3( ). ( ) 3.732s


34 43 3 4( ). ( ) 3.732s



ψ ψ= = ∫

11 1 1( ). ( ) 1.200s

r s s dAψ ψ= =∫

12 21 1 2( ). ( ) 0.000s


13 31 1 3( ). ( ) 0.000s


14 41 1 4( ). ( ) 0.000s


22 2 2( ). ( ) 2.880s

r s s dAψ ψ= =∫

23 32 2 3( ). ( ) 0.000s


24 42 2 4( ). ( ) 0.00s


33 3 3( ). ( ) 3.736s

r s s dAψ ψ= =∫

34 43 3 4( ). ( ) 3.732s


44 4 4( ). ( ) 22.032s

r s s dAψ ψ= =∫

A2.4 Coefficients for multi- cell doubly symmetric section

( ) ( )a a s s dAji ij i j


211

11 1 1( ). ( ) 9.900s

a s s dAϕ ϕ= =∫ ' '

11 1 1( ). ( ) 2.400s

b s s dAϕ ϕ= =∫

12 21 1 2( ). ( ) 0.000s

a a s s dAϕ ϕ= = =∫ ' '

12 21 1 2( ). ( ) 0.000s


13 31 1 2( ) ( ) 0.000s

a a s s dAϕ ϕ= = =∫ ' '

13 31 1 3( ). ( ) 0.000s


22 2 2( ). ( ) 27.000s

a s s dAϕ ϕ= =∫ ' '

22 2 2( ). ( ) 3.600s

b s s dAϕ ϕ= =∫

23 32 2 3( ) ( ) 0.000s

a a s sϕ ϕ= = =∫ ' '

23 32 2 3( ). ( ) 0.000s


33 3 3( ). ( ) 127.752s

a s s dAϕ ϕ= =∫ ' '

33 3 3( ). ( ) 55.727s

b s s dAϕ ϕ= =∫


ψ ϕ= = ∫ '( ). ( )i h

c c s s dAih hi s

ϕ ψ= = ∫

'

11 1 1( ). ( ) 2.400s


11 1 1( ). ( ) 2.400s

c s s dAψ ϕ= =∫

( ) '

12 21 1 2. ( ) 0.000s

c c s s dAψ ϕ= = =∫ ( ) '

12 21 1 2. ( ) 0.000s


'

13 31 1 3( ). ( ) 0.000s

c c s s dAψ ϕ= = =∫ '

13 31 1 3( ). ( ) 0.000s


'

22 2 2( ). ( ) 3.600s


14 4 1( ). ( ) 0.000s

c s s dAψ ϕ= =∫

'

23 32 2 3( ) ( ) 0.000s

c c s s dAψ ϕ= = =∫ '

22 2 2( ). ( ) 3.600s

c s s dAψ ϕ= =∫

'

33 3 3( ). ( ) 55.727s


23 32 2 3( ) ( ) 0.000s


'

41 4 1( ). ( ) 0.000s


42 4 2( ). ( ) 0.000S

c s s dAψ ϕ= =∫

'

42 4 2( ). ( ) 0.000S


33 3 3( ). ( ) 55.727s

c s s dAψ ϕ= =∫

'

43 4 3( ). ( ) 24.759s


34 4 3( ). ( ) 24.759s

c s s dAψ ϕ= =∫


ψ ψ= = ∫

11 1 1( ). ( ) 1.200s

r s s dAψ ψ= =∫

212

12 21 1 2( ). ( ) 0.000s


13 31 1 3( ). ( ) 0.000s


14 41 1 4( ). ( ) 0.000s


22 2 2( ). ( ) 3.600s

r s s dAψ ψ= =∫

23 32 2 3( ). ( ) 0.000s


∫24 42 2 4

s

r = r = ψ (s).ψ (s)dA = 0.00

∫33 3 3

s

r = ψ (s).ψ (s)dA = 57.727

∫34 43 3 4

s

r = r = ψ (s).ψ (s)dA = 24.759

∫44 4 4r = ψ (s)ψ (s)dA =35.100

A2.5 Coefficients for single cell non- symmetric section

( ) ( )a a s s dAji ij i j


11 1 1( ). ( ) 5.264s

a s s dAϕ ϕ= =∫ ' '

11 1 1( ). ( ) 1.132s

b s s dAϕ ϕ= =∫

12 21 1 2( ). ( ) 1.536s

a a s s dAϕ ϕ= = =∫ ' '

12 21 1 2( ). ( ) 0.314s


13 31 1 2( ) ( ) 0.001s

a a s s dAϕ ϕ= = = −∫ ' '

13 31 1 3( ). ( ) 0.037s

b b s s dAϕ ϕ= = = −∫

22 2 2( ). ( ) 11.018s

a s s dAϕ ϕ= =∫ ' '

22 2 2( ). ( ) 2.018s

b s s dAϕ ϕ= =∫

23 32 2 3( ) ( ) 0.038s

a a s sϕ ϕ= = = −∫ ' '

23 32 2 3( ). ( ) 0.028s


33 3 3( ). ( ) 0.293s

a s s dAϕ ϕ= =∫ ' '

33 3 3( ). ( ) 0.244s

b s s dAϕ ϕ= =∫


ψ ϕ= = ∫ '( ). ( )i h

c c s s dAih hi s

ϕ ψ= = ∫

'

11 1 1( ). ( ) 1.132s


11 1 1( ). ( ) 1.132s

c s s dAψ ϕ= =∫

213

( ) '

12 21 1 2. ( ) 0.314s

c c s s dAψ ϕ= = =∫ ( ) '

12 21 1 2. ( ) 0.314s


'

13 31 1 3( ). ( ) 0.037s

c c s s dAψ ϕ= = = −∫ '

13 31 1 3( ). ( ) 0.037s

c c s s dAψ ϕ= = = −∫

'

22 2 2( ). ( ) 2.018s


14 4 1( ). ( ) 0.148s

c s s dAψ ϕ= =∫

'

23 32 2 3( ) ( ) 0.028s

c c s s dAψ ϕ= = =∫ '

22 2 2( ). ( ) 2.018s

c s s dAψ ϕ= =∫

'

33 3 3( ). ( ) 0.244s


23 32 2 3( ) ( ) 0.028s


'

41 4 1( ). ( ) 0.148s


24 4 2( ). ( ) 0.476S

c s s dAψ ϕ= =∫

'

42 4 2( ). ( ) 0.476S


33 3 3( ). ( ) 0.244s

c s s dAψ ϕ= =∫

'

43 4 3( ). ( ) 0.262s


34 4 3( ). ( ) 0.262s

c s s dAψ ϕ= =∫


ψ ψ= = ∫

11 1 1( ). ( ) 1.132s

r s s dAψ ψ= =∫

12 21 1 2( ). ( ) 0.314s


13 31 1 3( ). ( ) 0.037s

r r s s dAψ ψ= = = −∫

14 41 1 4( ). ( ) 0.148s


22 2 2( ). ( ) 2.018s

r s s dAψ ψ= =∫

23 32 2 3( ). ( ) 0.028s


∫24 42 2 4

s

r = r = ψ (s).ψ (s)dA = 0.476

∫33 3 3

s

r = ψ (s).ψ (s)dA = 0.244 ∫34 43 3 4

s

r = r = ψ (s).ψ (s)dA = 0.262

∫44 4 4r = ψ (s)ψ (s)dA =10.358

214

APPENDIX THREE

MORH’S INTEGRAL FOR DISPLACEMENT COMPUTATION

Let 1f (x) = a+bx be any linear function and 2f (x) be any arbitrary curve or function as

shown in Fig.A1.

( )I = ∫ ∫1 2 2L Lf (x) * f (x)dx = a+bx f (x)dx

∫ ∫2 2L L

= a f (x)dx +b xf (x)dx

2 2= aA +bxA

where x is the centroid of area A2

( )I∴ 2 2 1= A a+bx = A xf (x)dx

2 1= A * ordinate of f (x) at x = x

Notes

(a) The function 1f (x) is a linear algebraic function

Example; 1 0 1 1 1 1 0f (x) = b +b x; f (x) = b x; f (x) = b

(b) The function 2f (x) is an arbitrary algebraic function

Example; 2 3

2 0 1 2f = a +a x +a x

(c) It is required that within the boundary, ≤ ≤0 x L , both functions must be

continuous. If however, there is discontinuity within the boundary, then L is broken up

into segments of piece-wise continuous functions.

common length L

1f (x) = a+bx

2f (x)

f (x) = a + bx1

x

• Area A2

Fig.A3: Functions f1(x) and f2(x)

x

y

215

Table A1: Morh’s Integral Chart - ∫L

1 20

f (x)f (x)dx

2a L

21a L

3

2 21

a +ab+b L3

acL

1acL

2 ( )

1c a+b L

2

1acL

2

1acL

3 ( )

1c 2a+b L

6

1acL

2

1acL

6 ( )

1c a+2b L

6

( )

1a c + d L

2 ( )

1a 2c + d L

6 ( ) ( )

1ac + a+b c + d +bd L

6

c

L

c

L

c

L

c L

d

a

L

1f (x)

2f (x)

a

L

a

L

b

216

APPENDIX FOUR

COMPUTATION OF DISTORTIONAL AND ROTATIONAL FORCES IN THE BOX

GIRDER PLATES

A4.1 Doubly symmetric section

Referring to Fig. 6.2.3a, the torsional load on the bridge girder is 138KN acting at

2.2m from the edge of the slab. and at (b/2 -1.46) m eccentricity. Thus, the torsional

moment MT on the girder is

T

bM =138 * -1.46

2KNm/m run. The torsional load is

decomposed into bending component Fig.6.2.3b, and torsional component

Fig.6.2.3c, which comprises of distortional component, Fig.6.2.3d and pure rotation,

Fig.6.2.3e.

A4.2 Single cell doubly symmetric section

For single cell mono-symmetric section, b = 7.32m, h = 3.00m, MT =

201.48KNm/m

From Fig.6.2.3 (d and e) we obtain the following.

t,dist Tq = M /2h = 33.58KN/m

b,dist t,distq = q = 33.58KN/m

w,dist Tq = M /2b =13.76KN/m

t,rot Tq = M /2h = 33.58KN/m

b,rot t,rot.q = q = 33.58KN/m

w,rot. Tq = M /2b =13.76KN/m

Work done on various plates of the cross section is given by

∫ i iq = qψds (A4.2.1)

where iψ is the transverse strain mode given in the generalized strain mode diagram

for single cell doubly symmetric section, Fig.4.3.10b.

e.g. for distortional strain mode 3:

3ψ = ±0.618 for bottom and top flanges

3ψ = ±1.482 for webs

For rotational strain mode 4:

4ψ = ±1.5 for the flanges and ± 3.6 for the webs

217

Substituting appropriate values of iψ and iq into eqn. (6.2.1) we obtain

( )3q = 2x33.58x0.618+2x13.76x1.482 x7.32 = 620.45KN

( )4q = 33.58x1.5x2+13.76x3.6x2 x7.32 =1462.62KN

A4.3 Multi-cell doubly symmetric section

For multi- cell doubly-symmetric section, b = 9.00m, h = 3.00m, MT =

317.40KNm/m

From Fig.6.2.2 we obtain the following.

t,dist Tq = M /2h = 52.90KN/m

b,dist t,distq = q = 52.90KN/m

w,dist Tq = M /2b =17.63KN/m

t,rot Tq = M /2h = 52.90KN/m

b,rot t,rot.q = q = 52.90KN/m

w,rot. Tq = M /2b =17.63KN/m

Work done on various plates of the cross section is given by eqn.(6.2.1)

Where iψ is the transverse strain mode given in the generalized strain mode diagram

for multi- cell doubly symmetric section, Fig. 4.3.11.

e. g., for distortional strain mode 3:

3ψ =1.929 for outer top and bottom flanges, 2.014 for middle top and bottom

flanges, 5.871 for the two outer webs and 2.014 for the two inner webs.

For rotational strain mode 4:

4ψ =1.5 0 for top and bottom flanges, 4.50 for the two outer webs and 1.50 for the

two inner webs.

Substituting appropriate values of iψ and iq into eqn. (6.2.1) we obtain

( )3q = 52.90x1.929x4+52.90x2.014x2+17.63x5.87x2 x9.00 = 6495.23KN

( )4q = 52.90x1.5x2+17.63x4.5x2 x9.00 = 2856.33KN

218

A4.4 Mono-symmetric sections

The torsional load on the bridge girder is 138KN, acting at 2.2m from the edge of the

slab, and at (b/2 -1.46) m eccentricity. Thus, the torsional moment MT on the girder is

T

bM =138 * -1.46

2KNm/m run. The torsional load is decomposed into bending

component Fig.6.2.5b, and torsional component Fig 6.2.5c, which comprises of

distortional component, Fig.6.2.6c and pure rotation, Fig.6.2.6b.

A4.5 Single cell mono-symmetric section

For single cell mono-symmetric section, b = 7.32m, h = 3.05m and hence, MT =

201.45KNm/m run.

The various components of pure torsional and distortional loads are evaluated using

Table 2.1.

( )

2

tR

Pbq = = 44.05KN/m

h a+b

( )bR

Pabq = = 20.0KN/m

h a+b

( )wR

Pbcq = = 21.40KN/m

h a+b

( )

2

tD

Paq = =11.01KN/m

h a+b

( )bD

Pabq = = 22.02KN/m

h a+b

( )wD

Pac= =10.702KN/m

h a+bq

Work done on the various components (plates) of the cross section is given by

∫i i iq = qψds (A4.5.1)

Values of iψ for various plates of the cross section are obtained from strain made

diagrams shown in Fig.4.3.8.

For bending strain mode 1: 1ψ = 0.857 for the webs and 0.0 for top and bottom

flanges.

∴ 1q = 69x0.857x2 =118.27KN/m

⇒ 1 q = qxb =118.27x9.15 =1082.17KN

For distortional strain mode 3: 3ψ = 0.159 for top flange, 0.417 for bottom flange and

0.123 for the webs.

∴

( ) ( ) ( )3 tD bD wD

3 3

q = q x1.105 + q x1.945 + 2 q x2.57

=11.01x0.159+ 22.02x0.492+10.702x0.417x2 = 21.47KN/m

q = q xb = 21.47x7.320 =157.16KN

219

For rotational strain mode 4: 4ψ =1.105 for top flange, 1.945 for bottom flange and

2.57 for the webs.

( ) ( ) ( )4 tD bD wD

4 4

q = q *1.105 + q *1.945 + 2 q * 2.57

= 44.05x1.105+ 20.02x1.945+ 21.4x2.57x2 =197.61

q = q xb =197.61x7.32 =1446.505KN

2q = 0 , since horizontal loads are not involved

A4.6 Double cell mono-symmetric section

For double cell mono-symmetric section, b = 7.32m, h = 3.05m and hence, MT =

201.45KNm/m run (same as in single cell).

Therefore 2q = 0.0KNm/m run,

3q =157.16KNm,

4q = 1446.505KNm, as in single cell mono-symmetric section

220

APPENDIX FIVE

LAPLACE SOLUTION OF TORSIONAL-DISTORTIONAL EQUATIONS OF

EQUILIBRIUM FOR MONO-SYMMETRIC SECTIONS

The Laplace transforms of the terms in eqn (6.3.8) are;

ℓ{ } ( ) ( ) ( )_

4 3 20 ' 0 '' 0 '''ivv s v s v s v sv v= − − − −

ℓ{ } ( ) ( )_

2'' 0 ' 0v s v sv v= − −

ℓ{ }_

v v=

ℓ{ } /K K s=

Substituting these expressions into eqn (6.3.8) we obtain

_ _ _4 3 2 2 2 2 2 4

0 1 2 3 0 1 4 /s v s C s C sC C s v sC C v K sα α α β− − − − − + + + = (A5.1)

where, 0 (0),C v= 1 '(0)C v= , ( )2 '' 0C v= , 3 '''(0)C v=

⇒ ( )_

4 2 2 44s s vα β− + = 3 2 2 2

0 1 0 2 3 1( ) /s C s C C C s C C K sα α+ + − + + − +

( )3 2 2 2_

0 1 2 0 3 1

4 2 2 4 4 2 2 4

1

4 4

s C s C C C s C CKv

s s s s s

α α

α β α β

+ + − + −= +

− + − + (A5.2)

Determination of partial factors for terms in eqn (A5.2)

We consider the denominator 4 2 2 44s sα β− +

Case I: 244 αβ ≥

If 4 24β α≥ then we have real and imaginary roots that can be expressed in the form

( )( )4 2 2 4 2 24s s s ps q s rs zα β− + = − + − + (A5.3)

where, , , ,p q r z are constants.

Multiplying out we have,

( ) ( )2 2 4 3 2 3 2 2s pz q s rs z s ps qs rs prs qrs zs pzs qz− + − + = − + − + − + − +

⇒ ( ) ( ) ( )4 2 2 2 4 3 24s s s p r s q pr z s pz qr s qzα β− + = − + + + + − + + (A5.4)

Comparing coefficients in eqn (6.2.16) we obtain that,

( ) 0p r− + = or p r= − (a)

221

2pr q z α+ + = − (b)

0pz qr+ = (c)

44qz β= ⇒ 44 /q zβ= (d)

Substituting (d) into (c) gives;

44 / 0pz r zβ+ =

∴ 44 / 0rz r zβ− + = [since p r= − ]

or ( )44 / 0r z zβ − = ( )4 24 0r zβ⇒ − = 4 24 zβ⇒ =

∴ 22z β=

From (d), 4 4 2 24 / 4 / 2 2q zβ β β β= = =

From (b); 2 2 22 2pr β β α+ + = − 2 2 24r β α⇒ − + = −

∴ 2 2 24r β α= +

2 24r β α⇒ = + ; 2 24p β α= − − ; 22q β= ; 22z β=

Thus, from eqn (6.2.15) we obtain,

( ) ( )4 2 2 4 2 2 2 2 2 2 2 24 4 2 4 2s s s s s sα β β α β β α β− + = + + + − + + (A5.5)

Let 2

2 2 214

2 4

αλ β α β= + = +

∴Eqn (A5.5) becomes,

( ) ( )4 2 2 4 2 2 2 24 2 2 2 2s s s s s sα β λ β λ β− + = + + − + (A5.6)

The partial fractions of the terms in eqn (A5.2) can be obtained using eqn (A5.6).

Thus,4 2 2 4 2 2 2 2

1

( 4 ) 2 2 2 2

E As B Cs D

s s s s s s s sα β λ β λ β

+ += + +

− + − + + + (A5.7)

⇒ ( ) ( )( ) ( ) ( )4 2 2 4 2 2 2 21 4 2 2 2E s s s As B s s s s Cs D s sα β λ β λ β= − + + + + + + + − +

4 2 2 4 4 3 3 2 2 2 21 4 2 2 2 2Es E s E As Bs A s s A s B sα β λ βλ β β= − + + + + + + +

4 3 3 2 2 2 22 2 2 2Cs Ds C s D s C s D sλ λ β β+ + − − + +

Comparing coefficients we obtain

0E A C+ + = (a)

( )2 0B D A Cλ+ + − = A C⇒ = (b)

( ) ( )2 22 2 0B D A C Eλ β α− + + − = (c)

222

( )22 0B Dβ + = B D⇒ = − (d)

41 4Eβ= 41/ 4E β⇒ = (e)

Solving eqns a,b,c,d and e above yields;

4

1

8A

β= − ;

2 2

4

2

8B

λ β

λβ

−= ;

4

1

8C

β= − ;

4 2

4

2

8D

β λ

λβ

−= ;

2

1

4E

β=

Substituting into eqn (A5.7) we obtain;

( ) ( )2 2 2 4 4 2 2 4

4 2 2 4 4 2 2 2 2

/ 8 2 /8 /8 2 /81 1 1.

( 4 ) 4 2 2 2 2

s s

s s s s s s s s

β λ β λβ β λ β λβ

α β β λ β λ β

− + − − − − = + +

− + − + + +

2 2 2 2

4 2 2 4 4 4 2 2 4 2 2

1 1 1 1 2 1 (2 ).

( 4 ) 4 8 2 2 8 2 2

s s

s s s s s s s s

λ λ β λ λ β

α β β λβ λ β λβ λ β

− + − − + −= + +

− + − + + +

4 4 2 2 2 2 2

1 1 1 2 1 1.

4 8 2 2 8 2 2

s

s s s s s

λ

β β λ β λβ λ β

−= − − − + − +

_

4 2 2 2 2 2

1 2 1 1

8 2 2 8 2 2

s

s s s s

λ

β λ β λβ λ β

+− + + + + +

(A5.8)

Considering the second term on the right hand side of eqn (A5.2) we obtain,

( )3 2 2 2

0 1 2 0 3 1

4 2 2 4 2 2 2 24 2 2 2 2

C s C s C C s C C Gs H Ks M

s s s s s s

α α

α β λ β λ β

+ + − + − + += +

− + − + + + (A5.9)

( ) ( ) ( )3 2 2 2 2 2

0 1 2 0 3 1 2 2C s C s C C s C C Gs H s sα α λ β⇒ + + − + − = + + + +

( ) ( )2 22 2Ks M s sλ β+ + − +

3 2 2 2 22 2 2 2Gs Hs G s H s G s Hλ λ β β= + + + + + +

3 2 2 2 22 2 2 2Ks Ms K s M s K s Mλ λ β β+ + − − + + (A5.10)

Comparing coefficients in eqn (A5.10) we have,

0 0G K C G C K+ = ⇒ = −

1( ) 2 ( )H M G K Cλ+ + − = (b)

2 2

2 02 ( ) 2 ( )G K H M C Cβ λ α+ + − = − (c)

2 2

3 1( ) ( ) / 2H M C Cα β+ = − (d)

223

Substituting (d) into (b) gives,

2 2

3 1 0 1( ) / 2 2 ( 2 )C C C K Cα β λ− + − = 2 2

3 1 0 1( ) / 2 2 4C C C C Kα β λ λ⇒ − + − =

2 2 2 2

3 1 0 3 1 0 11

2 2

4 2

8 2 4 8

C C C C C C CCK

α α λβ β

λβ λ λβ

− − + −∴ = + − =

⇒ ( )2 2 2

0 1 3

2

4 2

8

C C CK

λβ β α

λβ

− + += (A5.11)

From (a), ( )2 2 2 2

0 0 1 3

0 2

8 4 2

8

C C C CG C K

λβ λβ β α

λβ

− + + += − =

( )2 2 2

0 1 3

2

4 2

8

C C CG

λβ β α

λβ

+ + +⇒ = (A5.12)

( )2 2

1

2

2 2( )

8

CG K

β α

λβ

+− = ; ( )

2

0 3

2

4

4

C CG K

λβ

λβ

++ = (A5.13)

From (d), 2

3 1

22

C CH M

α

β

−= −

Substituting into expression (c) we obtain

2 22 20 3 3 1

2 02 2

42 2 2

4 2

C C C CM C C

λβ αβ λ α

λβ β

+ −+ − = −

2 2 220 3 3 1

2 02 2

44

2

C C C CM C C

λβ β λ α λλ α

λβ β

+ −⇒ + − = −

2 220 3 3 1

2 02

44

2

C C C CM C C

λ λ α λλ α

λ β

+ −⇒ + − = −

2 2 2

0 3 3 1 0 2

2 2

4

8 4 4

C C C C C CM

β λ λ α λ α

λ λβ λ

+ − −= + +

4 2 2 2 2 2 2 2

0 3 3 1 0 2

2 2

4 2 2 2 2

8

C C C C C Cλβ β λ α λ α λβ λβ

λ β

+ + − + −=

( ) ( )2 2 2 2 2 2 2 2

0 1 2 3

2 2

4 2 2 2 2

8

C C C CM

β λ α β λ α λ λβ β λ

λ β

+ − − + += (A5.14)

2

3 1

22

C CH M

α

β

−= −

( ) ( )2 2 2 2 2 2 2 2 2 2 2

3 1 0 1 2 3

2 2

4 4 4 2 2 2 2

8

C C C C C Cλ α λ β λ α β λ α λ β λ β λ

λ β

− − + + + − +=

224

( ) ( )2 2 2 2 2 2 2 2

3 2 1 0

2 2

2 2 2 4 2

8

C C C CH

λ β β λ λ α β λ β α λ

λ β

− + − − +⇒ = (A5.15)

Substituting eqns (A5.8) and (A5.9) into eqn (A5.2) while noting eqns (A5.11) to

(A5.15) we obtain;

3 3

4 4 2 2 2 2 2

1 2 1

4 8 2 2 8 2 2

K K s Kv

s s s s s

λ

β β λ β λβ λ β

−= − − − + − +

3 3

4 2 2 2 2 2

2 1

8 2 2 8 2 2

K Ks

s s s s

λ

β λ β λβ λ β

+− + + + + +

2 22 2

Gs

s sλ β+

− +

2 22 2

H

s sλ β+

− + 2 2 2 22 2 2 2

Ks M

s s s sλ β λ β+ +

+ + + + (A5.16)

Now, ( )22 2 2 22 2 2s s sλ β λ λ β− + = − − +

But 2 214

2λ β α= +

22 2

4

αλ β

⇒ = +

( )2

22 2 2 22 2 24

s s sα

λ β λ β β

∴ − + = − − + +

( )2

22 2 22 24

s s sα

λ β λ β

⇒ − + = − + −

( )2 2s λ ω= − − (A5.17)

where, 2

2 2

4

αω β

= −

(A5.18)

Similarly, ( )22 2 22 2s s sλ β λ ω+ + = + − (A5.19)

Noting that ( )2s sλ λ λ+ = + + and ( )2s sλ λ λ− = − − , eqn (A5.16) becomes,

( ) ( ) ( )3 3 3

2 2 24 4 22 2 2

1 1

4 8 8

K K Ksv

s s s s

λ λ

β β λβλ ω λ ω λ ω

−= − − −

− + − + − +

( ) ( ) ( )3 3

2 2 24 22 2 2

1

8 8

K Ks

s s s

λ λ

β λβλ ω λ ω λ ω

+− + +

+ + + + + +

( ) ( )

2 22 2

Gs H

s sλ ω λ ω+ +

− + − + ( ) ( )2 2 22

Ks M

ss λ ωλ ω+ +

+ ++ + (A5.20)

( ) ( ) ( )3 3 3 3

2 2 24 4 4 22 2 2

1 1

4 8 8 8

K K K Ksv

s s s s

λ λ

β β β λβλ ω λ ω λ ω

−⇒ = − + −

− + − + − +

225

( ) ( ) ( )

3 3 3

2 2 24 4 22 2 2

1

8 8 8

K K Ks

s s s

λ λ

β β λβλ ω λ ω λ ω

+− − +

+ + + + + +

( )

( ) ( )2 22 2

G s G

s s

λ λ

λ ω λ ω

−+ +

− + − + ( )

( )( )2 2 22

K sH

ss

λ

λ ωλ ω

++ +

+ +− +

( ) ( )2 2 22

K M

ss

λ

λ ωλ ω− +

+ ++ + (A5.21)

Taking the inverse transform of eqn (A5.21) we obtain;

3 3 3 3 3

4 4 4 2 4cos sin sin cos

4 8 8 8 8

x x x xK K K K Kv e x e x e x e xλ λ λ λλ

ω ω ω ωβ β β ω λβ ω β

− = − + − −

3 3

4 2sin sin cos sin

8 8

x x x xK K Ge x e x Ge x e xλ λ λ λλ λ

ω ω ω ωβ ω λβ ω ω

− −− + + +

sin cos sin sinx x x xH K Me x Ke x e x e x

λ λ λ λλω ω ω ω

ω ω ω− − −+ − + (A5.22)

( ) ( )3 3

4 4sin cos

4 8

x x x xK Kv x e e x e e

λ λ λ λλω ω

β β ω− −

⇒ = + − − +

( )3

4

1sin

8

x xKx e e

λ λλω

β ω ω−

+ −

( )cos x xx Ge Keλ λω −+ +

( ) ( ) 1sin sinx x x x

x Ge Ke x He Meλ λ λ λλ

ω ωω ω

− −+ − + + (A5.23)

3 3

4 4sinh .sinh cos .cosh

4 8

K Kv x x x x

λω λ ω λ

β β ω

⇒ = + −

-

3

2

1sinh .sinh

4

Kx xω λ

λβ ω

−

( ) ( )cos cosh sinh cosh sinhx G x x K x xω λ λ λ γ+ + + −

( ) ( )sin cosh sinh cosh sinhG H M K

x x x x xλ λ

ω λ λ λ λω ω

+ − + + + −

(A5.24)

3 3

4 4

1sin .sinh cos .cosh

4 4

K KV x x x x

λω λ ω λ

β ω λω β

⇒ = − − +

cos .cosh cos .sinhG x x G x xω λ ω λ+ + cos .cosh cos .sinhK x x K x xω λ ω λ+ −

sin .cosh sin .sinhG H G H

x x x xλ λ

ω λ ω λω ω

+ ++ +

226

sin .cosh sin .sinhM K M K

x x x xλ λ

ω λ ω λω ω

− −+ − (A5.25)

2

3 3

4 4

1sin .sinh cos .cosh

4 4

K KV x x x x

λω λ ω λ

β β ωλ

−⇒ = + −

( ) ( )cos .cosh cos .sinhG K x x G K x xω λ ω λ+ + + −

sin .coshG H M K

x xλ λ

ω λω

+ + − +

sin .sinhG H M K

x xλ λ

ω λω

+ − + +

(A5.26)

Addition of eqns.(A5.14) and (A5.15) gives

( )2

3 1

22

C CH M

α

β

−+ = (A5.27)

Subtracting eqn.(A5.13) from eqn.(A5.14) gives

( )2

2 3 0

2

2 2 2

4

C C CH M

λ λ α

λ

− − +− = (A5.28)

Substituting eqns.(A5.12),( A5.27) and (A5.28) into eqn.(A5.23) we obtain

2

3

4

11 sin .sinh cos .cosh

4

KV x x x x

λω λ ω λ

β ωλ

−= + −

2

0 3

2

4cos .cosh

4

C Cx x

λβω λ

λβ ω

++

( )2 2

1

2

2cos .sinh

4

Cx x

β αω λ

λβ ω

+ +

( )2 2

3 1

2

2 2sin .cosh

4

C Cx x

β αω λ

ωβ

+ − +

( ) ( )( )2 2 2 3 2 2 2

2 3 0

2 2

2 4 2 2sin .sinh

4

C C Cx x

λβ λ β λ β λβ αω λ

ωλ β

+ − + − + +

(A5.29)

⇒2

3 0 3

4 2

4cos .cosh

4 4

K C CV x x

λβω λ

β λβ ω

+= +

3

4cos .cosh

4

Kx xω λ

β−

( )2

3

4

1sin .sinh

4

Kx x

λω λ

ωλβ

− +

( )2 2

1

2

2cos .sinh

4

Cx x

β αω λ

ωλβ

+ +

( )2 2

3 1

2

2 2sin .cosh

4

C Cx x

β αω λ

ωβ

+ − +

227

( ) ( )( )2 2 2 3 2 2 2

2 3 0

2 2

2 4 2 2sin .sinh

4

C C Cx x

λβ λ β λ β λβ αω λ

ωλ β

+ − + − + +

(A5.30)

⇒2

3 0 3 33 4 2 4

4cos .cosh

4 4 4

K C C KV x x

λβω λ

β λβ ω β

+= + −

( )2 2

1

2

2cos .sinh

4

Cx x

β αω λ

ωλβ

+ +

( )2 2

3 1

2

2 2sin .

4

C Cx cos x

β αω λ

ωβ

+ − +

( ) ( ) ( )2 2 2 3 2 2 2 2

2 3 0 3

2 2 4

2 4 2 2 1sin .sinh

4 4

C C C Kx x

λβ λ β λ β λβ α λω λ

ωλ β ωλβ

+ − + − + − + +

(A5.31)

3 0 1 2cos .cosh cos .sinhV A A x x A x xω λ ω λ= + + + 3 4sin .cosh sin .sinhA x x A x xω λ ω λ+

(A5.32)

where

30 44

KA

β=

2

0 3 31 2 4

4

4 4

C C KA

λβ

λβ ω β

+= −

( )2 2

1

2 2

2

4

CA

β α

ωλβ

+ =

( )2 2

3 1

3 2

2 2

4

C CA

β α

ωβ

+ − =

(A5.33)

( ) ( ) ( )2 2 2 3 2 2 2 2

2 3 0 3

4 2 2 4

2 4 2 2 1

4 4

C C C KA

λβ λ β λ β λβ α λ

ωλ β ωλβ

+ − + − + − = +

Case II; 4 24 24 24 24444ββββ ≤ α≤ α≤ α≤ α

If 244 αβ ≤ , roots are real and the denominator of eqn (A5.2) can factorize as follows

442

22

24224 4422

4 βααα

βα +−

−

−=+− ssss

−−

−= 4

42

22 4

42β

ααs

2 4 2 42 4 2 44 * 4

2 4 2 4s s

α α α αβ β

= − + − − − −

⇒ 4 2 2 44s sα β− +2 4 2 4

2 4 2 44 * 42 4 2 4

s sα α α α

β β = − − − − + −

(A5.34)

228

Let 442

2 442

βαα

−−=u and 442

2 442

βαα

ω −+=

Eqn (A5.34) becomes

( )( )22224224 4 ωβα −−=+− susss

The partial factors of the first term in eqn (6.2.4) are as follows;

( ) 22224224 4

1

ωβα −

++

−

++=

+− s

EDs

us

CBs

s

A

sss

( )( ) ( ) ( ) ( ) ( )222222221 ussEDsssCBssusA −++−++−−=⇒ ωω (A5.35)

CHAPTER ONE INTRODUCTION - To Restore The Dignity … · i To derive a set of differential...

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Transcript of CHAPTER ONE INTRODUCTION - To Restore The Dignity … · i To derive a set of differential...