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Finite element analysis forthe elastic stability of thinwalled open section columnsunder generalized loading

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FINITE ELEMENT ANALYSIS FOR THE

ELASTIC STABILITY OF THIN WALLED

OPEN SECTION COLUMNS UNDER GENERALIZED LOADING

t,

by

MASARACHIGE ARIYARATNA NANAYAKKARA

A Doctoral Thesis submitted in partial fulfIlment

of the requirements for the award of Doctor of Philosophy

of Loughborough University of Technology

December 1986 --

Supervisor: Dr PW Sharman,, MSc,, 'PhD., DCAe, CEng,, MRAeS

Department of Transport. Technology

(D by MA Nonoyakkara,, BSc. MTech

TABLE OF CONTEKTS

Paqe No

Synopsis ...... ooo. ooo. i Acknowledgements 00.000.0

CHAPTER 1: INTRODUCTION .0. Literature Survey ...... 0 2 Problem Formulation ... 3

CHAPTER 2: PHYSICAL DIMENSIONAL BOUNDARIES OF THIN WALLED BEAMS 0.. 0.0. *. *00 7 The Adopted Sign Convention .0.. 00 8 Torsional Behaviour of Non-Circular Bars ... 12

CHAPTER 3: WARPING THEORY OF THIN WALLED CROSS-SECTIONS 17

CHAPTER 4: BUCKLING ANALYSIS OF AN ELASTIC BEAM COLUMN 26 Introduction ...... 0.. 26 Potential Energy Formulation for Buckling in

the x-y and x-z Plane due to Axial load Fx 30 Potential Energy Formulation in Torsional Buckling due to Application of Axial Load Fx 33 Potential due to Applied Torque 46 Potential due to Warping Moment 54 Potential due to Lateral Bending and End Shear Loads .0....... -.. 0 57 Total Potential due to Buckling Deformations 66 Equations of Equilibrium 69

Buckling Analysis ... 70

Solution Procedure ... 000a00 71

CHAPTER 5: UTILIZATION OF FINITE ELEMENT METHOD FOR

BUCKLING ANALYSIS .... a. 090 72

TABLE OF CONTENTS

Page No

Synopsis 0.. 0.000. *00i Acknowledgements 00000000. *. 0 ii

CHAPTER 1: INTRODUCTION 01 Literature Survey 2 Problem Formulation ... 3

CHAPTER 2: PHYSICAL DIMENSIONAL BOUNDARIES OF THIN WALLED BEAMS 0.. 0.. 0a09*07 The Adopted Sign Convention ... 0008 Torsional Behaviour of Non-Circular Bars ... 12

CHAPTER 3: WARPING THEORY OF THIN WALLED CROSS-SECTIONS 17

CHAPTER 4: BUCKLING ANALYSIS OF AN ELASTIC BEAM COLUMN 26 Introduction .000*a000 26 Potential Energy Formulation for Buckling in the x-y and x-z Plane due to Axial load Fx 30 Potential Energy Formulation in Torsional Buckling due to Application of Axial Load Fx 33 Potential due to Applied Torque 46 Potential due to Warping Moment 54 Potential due to Lateral Bending and End Shear Loads. 000a*0000-. 00 57 Total Potential due to Buckling Deformations 66 Equations of Equilibrium 69 Buckling Analysis ... 70 Solution Procedure ... 71

CHAPTER 5: UTILIZATION OF FINITE ELEMENT METHOD FOR BUCKLING ANALYSIS ... 000000 72

Page No

Flexural Buckling of a Perfect Pin Ended Strut

Under Axial Compression 151

Examples in Pure Torsional Buckling 165

Torsional Buckling of a Flange Cruciform

Section Strut .00... 166

Elastic Stiffening and Large Deflection

Phenomenon of a Cantilever 0.178 Large Deflection Analysis of a Thin Walled

Channel Section Cantilever Beam Loaded at the

Centroid ..... 0.0.. 178

Large Displacement Analysis of a Shallow Truss 184

CHAPTER 11: DISCUSSION AND CONCLUSIONS ... 0.. 192

REFERENCES ............... 194

APPENDICES: Appendix 1 203

Appendix 2000 206

j

i

SYNOPSIS

The current interest in collapse characteristics brought about by

crashworthiness requirements ýas shown the need for a better

understanding and predictive capability for the thin walled open section structures. In general three possible modes exist in which a loaded thin walled open section column can buckle:

1. they can bend in the plane of one of the principal axes; 2. they can twist about the shear. centre; 3. or they can bend and twist simultaneously.

The following study was undertaken to investigate the general failure

of thin walled open section structures. A literature survey was conducted and it prevailed that a basic fundamental theoretical study was vital in describing the behaviour of thin walled structural members.

The following stages of theoretical study have been completed:

Formulation of the stiffness matrix to predict the generalised force-displacement relationships assuming the small displacement

theory in the linear elastic range. 2. Formulation of the geometric stiffness matrix to predict the

buckling criteria under generalised loading and end constraints in

the linear elastic range. 3. Formulation of the compound coordinate transformation matrix to

relate local and global displacements or forces.

4. Preparation of the associated finite element computer program to

solve general thin walled open sections structural problems.

ii

ACKNOWLEDGEMENTS

Drawing a dividing line between those who have assisted me in this

work during the past four years and those who have not, is an impractical, if not impossible, task. Since, as I infer, such a division does not exist, I am able to mention only a few of my

colleagues and friends; those whose influences have been a major

catalyst towards the successful completion of this project.

My supervisor, Dr Peter Sharman, instigated proceedings by giving me a broad introduction to the subject area. He was an immense help,

selflessly and patiently explaining the complex theories and

procedures involved, and effortlessly guiding me through the project. His constant supervision and encouragement is gratefully acknowledged. Thanks are also extended to my Director of Research in the Department

of Transport Technology.

I wish to thank Mr Joe Ogendo of the Department of Transport

Technology for his assistance in checking the mathematical validity

of numerous theories and debugging complex software modules. Our

endless, and often contentious philosophical discussions, were a

welcome deviation during the long hours of the night, somehow leaving

me miraculously refreshed mentally.

i also wish to thank Kapila, for checking the accuracy of integral

functions, Kuruppu and Ananda for helping me with the diagrams, and

Sana for his assistance in the proof reading of the manuscript. Such

are mundane tasks and the diligence with which they applied themselves, speak not only of their friendship but also of their

selflessness.

Nick Birkett and John Ascough, supervisors in my present job in the

Department of Mechanical Engineering, allowed me to work on this

thesis every Friday and I would like to thank them for their

iii

understanding and hope I will be able to reciprocate by justifying

their confidence in me.

I wish to acknowledge the Science and Engineering Research Council (Contract No RXA 313J) for funding the project.

At this point author, ý generally like to thank their typists. I cringe to be an exception! Thank you Janet. Also thank you Paul for the

endless cups of coffee and your profound suggestions on various aspects of life.

A very special thank you to my dearest wife, Amitha, for her invaluable love, companionship and also for giving me the highest

comfort which was a vital ingredient in performing this work.

Last, but in no way least, a big thank you to my daughters Sugandi

and Anushka for the nights and weekends spent away from me while spending time on the project.

1

CHAPTER 1

INTRODUCTION

Thin walled open section structural members are relatively easy to fabricate and erect. They are extensively used in the construction of ships, aircraft, automotive vehicles, civil engineering projects and in a large number of domestic and industrial applications. Due to their use these structural members are subjected to axial and transverse loads and moments producing direct axial and shear

stresses.

Under the action of these stresses three possible modes exist in which a loaded thin walled open section column can buckle: (1) they can bend in the plane of one of the principal axes, (2) they can twist about the shear centre axis, or (3) they can bend and twist simultaneously. For any given member depending on its length and geometry of its cross section, one of these three modes will be

critical. In the case of thin walled open sections the most obvious reason for the premature failure is that the torsional rigidity of an open section is lovi as it is approximately proportional to the cube of its wall thickness. Thus as a result of low torsional stiffness they

can therefore buckle by twisting at loads well below the flexural buqýkling loads.

Generally, at the buckling point, the behaviour basically falls into

two categories namely, just prior to buckling the stress throughout

the structure:

1. is in the linear elastic region, 2. is partially or completely in the plastic region (post yield).

In the case of the linear elastic condition, it is assumed that the

non-linearity associated with the deformations are purely due to its

geometric conditions. However in the case of plasticity the non-

2

linearity generally occurs with the combined effect of the geometry

and the state of strain in the material, which subsequently leads to

the total collapse of the structure.

LITERATURE SURVEY

Columns have been used for centuries as compression members in

buildings. In the early period, columns were designed empirically and their ultimate strength was determined entirely by the crushing

strength of the material similar to that of the fracture strength in

tension members. However, it was vaguely understood that the strength is somehow related to the column length and the cross-section. Van Musschenbrock(61 ) (1729) first recognised this and presented an empirical formula for column strength in terms of column length '1'.

The theo ry of el ast icfI exu ral bu ckl i ng of concent ri cal ly 1 oaded columns was first formulated by Euler. Wagner was the first to introduce open sections with the possibility of torsion. However, in his theory Wagner assumed arbitrarily, that the centre of rotation

coincides with the centroid which, in general, is not the case. The

exact solution for the torsional flexural buckling of thin-walled open

sections was first presented by Bleich(9), followed by Vlasov(62) ' Goodier and Timoshenko(58). These theories are readily available in

well known textbooks by Bleich(9) and Timoshenko and Gere(58). The

analysis of thin-walled structural systems was first approached using the method of transfer matrices, Vlasov (62)

, and the force

(flexibility) method, Bazant(6 ). These methods are however much less

versatile than the displacement (stiffness) method. The stiffness

matrix of a thin walled beam element, seems to have been first

presented by Krahula(36), for the case without the geometric (initial

stress) matrix. Kraiinovic(37) later included initial bending moments

and initial axial force in his derivation.

Barsoum and Gallagher(4) derived the stiffness and geometric stiffness

matrices with the inclusion of axial compression, biaxial bending and torsional effects with the aid of fundamental differential equations

3

derived by Bleich(9) and Timoshenko and Gere(58 ). Raiasekaran(54

appears to be the first to derive the same stiffness and geometric

stiffness matrices of Barsoum and Gallagher( 4) by considering the

second order terms in the strain equation while maintaining the thin-

walled open section beam assumptions of Vlasov(62). Rajasekaran's

work(53) was well recognised and attempts soon followed by Bazant and El-Nimeri( 6 ), Ehouney and Kirby(17) and Kasemest and Nishino and Lee(33) to develop the concept further. Bazant and El-Nimeri(6 )

suggested the development of a curved element, with the inclusion of the geometric stiffness matrix, such that initially curved members may be analysed while maintaining the geometric compatibility at nodes,

whose angles of intersections are less than 1800.

Recently Gaafar and Tidbury(19 ) observed experimentally that a centroidally end-loaded cantilevered channel section in large displacement work, failed always at a predetermined distance from the fixed end. Further they also observed that the torque could vary considerably as the rotational displacement gets larger along the

axial direction in an approximately linear manner. This was a useful finding as previous investigators assumed that the torsional loading

remained unaltered in their analyses as the structure deformed

progressively.

PROBLEM FORMULATION

Considerable attention was paid to the two analytical techniques

presented by Barsoum and Gallagher (A ) and Raiasekaran(52). Let us first consider the technique developed by RajasekaraO3. He used Vla sov, S(62) assumptions for thin walled open section prismatic

columns and obtained the expressions for the corresponding axial and

shear strains. The second order terms were included in the strain

equation in order to represent the geometric non-linearity. Finally

the strain equation was substituted into the virtual work expression

and by considering the state of equilibrium, the compatibility,

sectional properties, the associated material properties and Vlasov's

hypothesis, Rajasekaran(52) established the loading and reaction

4

vectors. This suggests that Rajasekaran was now left with very little freedom to allow the external loading vector to vary between the nodes of an element, as Geafar and Tidbury(19) suggested earlier. See al so equation 12.45 of ref. (19).

Unlike Rajasekaran, the method developed by Barsoum and Gallagher(4)

considered that the total potential was made up of two constituents. This method utilised the total strain energy equation which was developed by Bleich( 9) and the differential equation of lateral and torsional buckling of Timoshenko(58) to form the total potential for the columns under consideration. Initially the potential due to each loading component is formulated separately and simultaneously. By the

use of the theory of superposition the total potential for the entire column is established.

As mentioned previously, Gaafar and Tidbury(19) observed an unusual failure of beams, due to the increase of to'rsional and thus also resulted warping loadings, due to large angular displacements. The reason for this is that Vlasov assumed that the applied torsional load remained unaltered during loading (see equation 3 of ref. 19). This

assumption is incorrect in large displacement analysis. It is

suggested that the external torsional and warping loadings, may therefore be approximated as linear functions between the nodes.

In comparison to the two techniques described above, the method developed by Barsoum and Gallagher(, 4) provides flexibility and better

facilities to introduce variable external loads and therefore appears to be easier to adopt than the alternative method developed by

Rajasekaran.

Unfortunately, Barsoum and Gallagher( 4) made several errors and

misjudgements in their analysis, and they are as follows:

From equation (4.30.0) of Chapter 4, it can be seen that when the

axial force Fx is applied to an arbitrary point on the cross section, the resultant bending in two principal planes and the

5

torsion about the longitudinal axis are coupled. However, the

result and the state of equilibrium cannot be treated as if the

columns were loaded by two uncoupled separate bending moments and a torque acting simultaneously. Regardless of this fact Barsoum

and Gallagher( ) allowed the force Fx to act arbitrarily on the

cross section and a solution was formulated as if the force Fx was acting at the shear centre, thereby making bending and torsion

actions independent of each other as shown in equation (4.32.0).

2. The warping moment generated by the axial load acting at the point of application on the cross section was ignored.

3. Any external warping moment loading on the system was not cons1dered.

4. It was also assumed that the torsional and also the warping moment loads remained equal in magnitude at the two nodes of the column during loading. This is an invalid assumption, according to the experiments conducted by Gaafar and Tidbury(19). This should also become evident by referring to equations (4.53.0) and (4.55.0).

Furthermore none of. the researchers considered examining the nature of the displacements and associated forces when a beam element is placed arbitrarily orientated in space. By referring to the published work it is evident that the work carried out up to now was mainly devoted to

.I cases where the local axial direction of the element was conveniently parallel to anyone of the global axes. In order to analyse large

assembTies of elements, it is crucial to investigate columns when they

are loaded arbitrarily in space.

The purpose of this thesis is to investigate the fundamental effect of large displacements and to predict the failure mode of columns

arbitrarily placed in space (i. e. buckling loads and their associated

modes") with the inclusion . of arbitrary torsional and warping moment loads into the generalized loading vector.

6

Because of the versatile nature of the finite element method it is desirable to develop a relevant procedure for solving large displacement problems of beams and frames, so that a structure with arbitrary geometry, complex loading and boundary conditions can be treated. Although the finite element formulation can be based on either assumed stress or displacement fields, most often the displacement based finite element formulation is applied in practice. Hence in this thesis a displacement model is used to arrive at the force displacement relationship for a beam-column element, by considering the principle of total potential energy and also the application of calculus of variations to obtain the equations of equilibrium and the state of stability. The conventional stiffness formulation for small deflections is modified by the effect of axial shortening due to (a) direct compression, (b) lateral bending, (c) flexural -torsion. This is incorporated in a geometric (incremental)

stiffness formulation. The non-linear load displacement path is traced by a linearized mid-point tangent predictor procedure together with coordinate transformation at every increment in the load.

6

7

CHAPTER 2

PHYSICAL DIMENSIONAL BOUNDARIES OF THIK WALLED BEAMS

Prior to the study of the behaviour of beam-columns, the basic

governing equations for biaxial bending and torsion of thin-walled

elastic beam-columns are established first. The dimensional bounds (i. e. aspect ratios of -ýand! ), assumptions and hypotheses of which LL the generalized theory are based and formulated, are as follows: 1. Material is elastic and homogeneous.

2. The column is long and prismatic, typically:

D 0.1 L

B 0.1 L

3. The cross-section is thin-walled, typically:

t 0.1 D

t 0.1 B

4. The shear deformation is neglected. 5. No distortion in cross section can exist.

(2.1.0)

(2.2.0)

8

The adopted sign convention:

1

x. ei

Z Fig. 2-ZI

Z. @Z

cz Fig. 2.2.2 Fig. 2.2.3 Fig. 2.2.4

FIGURE 2.2.0

k.

The analytical formulation presented here does not strictly follow the

conventional right handed rule, shown in Figure 2.2.0. In Figure 2.2.2

the rotation on the x-y plane, has been reversed in order to keep the

analysis in it as simple as possible. In doing so, the same displacement function can be used to describe the behaviour of x-y,

y-z and z-x planes and also the torsion warping characteristic along the x-axi s.

In order to describe the sign convention fully, let us begin with the

following example, as illustrated in Figure 2.3.0. The beam is simply supported and carries a uniformly distributed load qy, applied in the

negative y-direction.

9

The boundary conditions for the problem in Figure 2.3.0 are:

V(X=O) = ý2V (X=O) = V(X=, ). =

d 2V

. (X= k) =0 dX2 dx7

From Macaulay's principle, the deflection curve for the problem under consideration is given by,

qx23_3 -? 4-E1 (2tx xz

y

x

FIGURE 2.3.0

By differentiating,

v qy 2- U3 3 - (6. tx . 24EI

af x (Z -

Vi= -q 2if (I - 2x)

EI

(2.3.0)

(2.4.0)

(2.5.0)

(2.6.0)

(2.7.0)

10

The five functions derived so far are now plotted as shown in Figure

2.4.0:

deflection curve

8 z

-I . -X I

I. P. "t lhe etastic cu rye

F Ig. 2.4.2

moment -Fig.

2-4.3

sbear force

Fig. 2-4-4

pa

load cý

Fig. (2.4.1)

Fig. (2.4.2)

Fig. (2.4.3)

Fig. (2.4.4)

Fig. (2.4.5)

FIGURE 2.4.0

11

Let us consider a similar sign convention for the case of pure torsion. The angle of twist Ox and its derivatives for a thin-walled

open cross section resemble the same set of equations (see reference (35)) as the one derived for bending, provided the system is subjected to a similar set of boundary conditions as shown below,

0. (x=O) = e"(X=O) = 0., (X=k) = o"(X=z) =0 xx

In this analogy, the torsional moment Mx corresponds to the transverse

shear forces QY or Qz while the warping moment Mxx corresponds to the bending moment in the simple beam.

By using equations (2.4.0), (2.5.0), (2.6.0), (2.7.0) and Figures (2.4.1), (2.4.2), (2.4.3), (2.4.4) and (2.4.5) the following chart is tabulated:

Subjected to lateral load Qy

Subjected to torsional load mx

Subjected to lateral load Qz

Displacement y Ox z Slope 0

z 1 -Y -el x ey = _Z# Bending moment Mz = EI V

yy z Mx x Ere x M y = EI 0' zz y Warping moment = EIyyy" = EIzzz"

Shearing forces QY M, z Mx Mx'x Qz = MI y and Torsional -EI yy z -Ere'" x = -EI zz y moments -EI yyy .I = -EIZZZIII Load' qy -, Q; m x

I -M x qz = _Q Iz

= EI 0111 yy z = Ere, "' x = El e"I zz y = EIyyy"18 = EIZZZ1111

TABLE 2. T. 0

12

TORSIONAL BEHAVIOUR OF NON-CIRCULAR BARS

Let us begin the analysis by examining the torsional behaviour of prismatic bars. In the case of torsion of prismatic bars not circular in cross section, the argument that a plane cross section remains plane during deformation is no longer valid. To illustrate the argument, consider the behaviour of elements A, B, C and D of the bar

shown in the foTlowing example:

x

y

Fig. 2-S. 1

c Fig. 2.5.2

Fig. 2-S-3

FIGURE 2.5.0

For simplicity, elements A and D are shown with right angled corners. As no shearing stresses exist on these elements, any component of shearing stress on element A, for example, would require stresses to be developed on the outside surface of the bar, which is generally not admissible.

13

Elements B and C, however, can develop shearing stresses, so long as they are parallel to the boundary line ad. Again, this is because

shearing stresses directed normal to any boundary, are inadmissible on the outside surface of the bar (Figure 2.5.2). It follows that when the sides of elements B and C acquire a shearing strainy, elements A

and D must undergo rigid body rotations for the sake of continuity. Hence the outside corners of elements A and B are displaced out of the

plane of the cross sections, as indicated in Figure 2.5.3. Out of

plane displacements such as those discussed above are the warping displacements. Other than the fact that the cross section warps,

perhaps one of the most obvious characteristics of the behaviour, is

the absence of normal stresses. No external forces or bending moments

are present, and since no end constraints exist, the only stress

components needed to provide the equilibrium of any transverse segment are shearing stresses in the cross sectional plane.

x

FIGURE 2.6.0

From the three components of the shearing stresses, only T XY and -r xz can result in a twisting moment. The component Tyz is zero which can

be verified by examining plane areas parallel to the bar axis.

14

Thus it is concluded for the non-circular prismatic bar in torsion

that,

ax =0y= az =T yz = 0, Ex =cy= cz = yyz =0 (2.8.0)

The fact that 'ryz is zero implies that cross sections do not distort

in their own planes.

In other words, the angle between any two lines on a cross section is

not changed during the deformation of the bar. This means that deformations of an element such as that indicated in Figure 2.5.0, do

not exist and that a point in the zy plane merely rotates about a centre of twist.

2

FIGURE 2.7.0

Application to thin-walled beam cross-section:

FIGURE 2.8.0

15

In thin-walled sections (closed or open), the warping displacement

relative to the origin can be expressed as (see page53 of reference-48)

f Txs ds - 2eu) (2.9.1) G

! -U + an (2.9.2)

xs as ax

where: u is warping in the x-direction n is the warping in the tangential (s)- direction

is the twist per unit length is the warping function

T xs is the shear stress in the s-direction

However, in the case of an open section it is necessary to pay closer attention to the definition of u. The shearing stress 'rxs in the case

of an open section, varies linearly throughout the wall thickness and is zero at the centre line of the wall. This means that S cannot be

measured along the centre line, or else the integral is zero. If, on the other hand, the maximum value of rxs is used and S is measured

along paths on the outside and then the inside periphery to two points

on opposite sides of the wall, the integral would represent the

difference between the warping displacement of these points, which is

a very small quantity for thin walled sections. It appears that for

open sections the integral in equation 2.9.1 is negligible in

comparison with the term 20w. Physically it is true that open

sections are much more flexible than the closed sections. In fact,

the torsional stiffness of a closed section is often an order of

magnitude greater than that of an open section of similar dimensions.

Thus the influence of shearing strain yxs is negligible, and the

warping displacements are due almost exclusively to the twist of the

section.

16

20w (2.10.0)

This follows that,

Lu + "I as ax

More elaborate analysis is found in reference (48). The physical interpretation to the last expression is as follows. The tangent to

the middle line at every point on the outline of the cross section

remains perpendicular to the associated longitudinal fibre after

torsion. This means that the bending and torsional stresses in the

beam can be assumed to be uncoupled, if the beam is subjected to

siumultaneous bending and torsional loadings (i. e. the behaviour of an

open sectioned beam is the same as that of a solid beam).

17

CHAPTER 3

WARPING THEORY FOR THIN WALLED CROSS-SECTIONS

y-

z

k,

where C(0,0) is the centroid O(ey, ez) is the shear centre D is an arbitrary point, conveniently placed to

measure distance S

a is the sliding radius

FIGURE 3.1.0

In Figure 3.1.0, point 'C' represents the origin of the coordinate axes and 0 represents the point of rotation of a typical thin-walled

cross-section. The objective is to establish the displacement of a point on the middle surface of the cross-section, in terms of UY09 Uzo

and UXC which are the transverse displacements of the shear centre, and the axial displacement of the origin respectively. The kinematics

of the cross-section are illustrated in Figure 3.2.0

18

Y-

VY

where: N is the initial position N' is the intermediate position N" is the final position UY0 is the translation of 0 in y direction Uzo is the translation of 0 in z direction

FIGURE 3.2.0

The displacement of an arbitrary point on the plane of cross-section

can be obtained by decomposing its displacements to a biaxial

translation of the shear centre followed by some rotational displacement of the cross-section about the shear centre. This can be illustrated as follows:

The rigid body translational displacement of point N, UY and UZ is

given by,

UY = UYO -a cos$ (1 - cos ex)- a* si nO sin Ox (3.1.1)

Uz = Uzo +a cosB sin ex -a sina 0- cos ex)

19

See Figure 3.2.0. Also by referring to Figure 3.1.0 it can be seen that,

a cosa =y- ey

a sinB =z- ez

(3.2.1)

(3.2.2)

By substituting equations (3.2.1) and (3.2.2), in equations (3.1.1)

and (3.1.2) yields,

Uy = UyO - (z-ez) sin ex - (y-ey)(1 - cos ex)

Uz = Uzo + (y-ey) sin Ox - (z-ez)(i - cos Ox) (3.3.2)

However, for small magnitudes of twisting rotations,

sin ex .% ex

Cos ex =1 (3.4.2)

Thus it follows that equations (3.3.1) and (3.3.2) are now simplified to,

uy=u YO - (z-ez) ex

Uz = Uzo + (y-ey) ex (3.4.2)

20

Equations (3.4.1) and (3.4.2) can now be used to describe the transverse displacements Uy and liz of an arbitary point.

To establish the expression for the axial displacement Uxt of an arbitrary point, it is essential to investigate the tangential displacement of point N in the direction of the mid-surface contour. This may be written as,

ut = Uzo cosa - Uyo sina + Poex (3.5.0)

See Figure 3.1.0 and Figure 3.2.0. I

Using the assumption that no shear deformation exists in the mid plane of the cross-section:

zu t zu x =x + Ds =0 (3.6.0)

Substitution of equation (3.5.0) in equation (3.6.0):

au x -- -Uzo cosci + U' sina -p (3.7.0) as yo 0x

in which primes denote differentiation with respect to x. Equation (3.7.0) is now:

[UXIS= _ul f cosa ds + U' fs sinct ds - Gx' fsP, ds (3.8.0)

D Zo D YO D

By referring to Figure (3.1.0) it can also be seen that,

21

.! LV-

= -sin ds

dz = cosa (3.9.2) TS

k, By substituting equations (3.9.1) and (3.9.2) into equation (3.8.0):

ux(X, Y, z) =u wo el (3.10.0) XD - (Z-ZD)Uzo - (Y-YD)Uyo - DS x

where w0 is defined as the sectorial coordinate of point S based on DS shear centre 0 and the reference point D Thus,

wfP ds DS ýD0

0 which gives rise to warping, when In general it is the term wDS

structures are subjected to torsional loadings. The assumption that the plane section remains plane during biaxial bending and rotation of the plane section about the shear centre due to torsional loadings, is based on the principle of rigid body translation and rotation. Thus,

although point C is not on the contour, the longitudinal displacement

at C may be obtained by visualizing a connection to it from any point

on the contour and by substituting the coordinates in equation (3.10.0) to those of the origin. Thus,

uu, + (3.11.0) XC ý XD + ZD Uzo YD U; o - 'obc x

By combining equations (3.10.0) and (3.11.0), the expression for the

axial displacement equation (3.12.0) is found,

22

Ux(X, Y, Z) =U. ZUZIO -y, 00A

xc . Uýo - O)Dc - 'ODS)OX (3.12.0)

At this point Rajasekaran(51) made the following statement. 'At a suitably selected reference point D on the outline of the cross- section, the quantity w0 could vanish'. This assumption is incorrect Dc and it can be seen by investigating the sectorial properties of the following exampi, e,

h

1 Z-

i7. r1-- ý-ý4

- %

-- & C. 6

b2t

I b3tlh2 2ht+bt, -IT-* FTT MT,

FIGURE 3.3.0

In the example shown above in Figure (3.3.0), the sectorial coordinate of the centroid is given by _dh . Furthermore, point D is not an 2' arbitrary point as Rajasekaran suggested, it is a unique point of any given section and its position can only be found by the unique method illustrated in reference (66) Section 2.2.1.

By differentiating equation (3.12.0) twice with respect to x and

neglecting all higher order terms, the express. ion for the total axial

strain can be found, and this is shown in equation (3.13.0)

23

11 U0 E=U, c-y0-Zu- (wo w s)ell (3.13.0) x UY M Dc Dx

where: C is the total axial strain at a point on the cross-section defined by (y, z)

U is the axial strain due to axial load only UY0 is the curvature about the z-axis

It UZO is the curvature about the y-axis e 11 is the warping curvature about the x-axis x

Since in the assumptions in linear elasticity only uniaxial stress is

significant,

a !:: Ec (3.14.0)

where a is the stress resulting due to strain E is the Young's modulus of elasticity

By considering the state of equilibrium under the loading condition, stress and stress resultants can be found as follows:

fy =I Iryx dA (3.15.1)

fz =f -r zx dA (3.15.2) A

fx =f adA (3.15.3) A

my =f aZ dA (3.15.4) A

mx f- ay'dA (3.15.5) A

mw =fa wo sA D (3.15.6)

(3.15.0)

24

t.

where fy is the shear force in the y-direction fz is the shear force in the z-direction fX is the direct force in the axial direction

my is the bending moment about the Y-axis mz is the bending moment about the z-axis MW is the warping moment about the x-axis T yx is the shear stress in the y-direction

T zx is the shear stress in the z-direction A is the area of the entire cross-section

By combining the relationship developed in equations (3.14.0) and (3.15.0), the following matrix representation can be developed

fx

ni y

mz

m

f EtdA f EtzdA f EtydA fE0 AAA t(woDc-&)Ds)dA ux, z

f Etz2dA f EtyzdA AA

f Ety2dA f AA

Symmetric

Et (w 0--1,0 .

bs)zdA -u" Dc zo

0 Et(wDoc-wDS)ydA -u" yo

Et(wo -wo )2 OH Dc Ds -x

By selecting coordinate axes y and z to be the principal axes, C and 0

as the centroid and the shear centre and also point D for the

principal radius, the off-diagonal terms in equation (3.16.0) will vanish. Thus equation (3.16.0) finally. reduces to,

fx f EtdA) u'xc A

mf Etz2dA) uz'O' (3.17.2) A

(3.17.0)

mz Ety2dA) U; O' (3.17.3)

f Et(woc - wOS)2 dA) e" ADDx

(3.17.4)

25

Thus, by considering' equation (3.17.0) shown above, the reaction forces required for a thin-walled cross-section under the generalized load condition can be found.

26

CHAPTER 4

BUCKLING ANALYSIS OF AN ELASTIC BEAM COLUMN

INTRODUCTION

A typical configuration displaced of a thin-walled cross-section is illustrated in Figure 3.2.0 in Chapter 3. These displacements are generally obtained by independent or simultaneous application of axial and transverse moments and torsional and warping forces. As was explained in the previous chapter, during loading the shear centre is

subjected to translational displacements, followed by rotation of the

entire cross section about the shear centre. This suggests that it

will make the analysis far simpler if all displacements and associated forces are referred to a set of axes which, in general are parallel to the principal. axes and pass through the shear centre. However, referring forces to an axis system at the shear centre generally generates additional moments and warping moments. These secondary forces can easily be accounted for by a suitable transformation

matrix, as described in equation (8.14.0).

I- With the assumptions and definitions made so far, the idealized

element under consideration can be described as illustrated in Figure (4.1.0)

-MZ -oz

FIGURE 4.1.0

N. M.. .

27

An orthogonal coordinate system xy, z was chosen, so that the y- and z-axes coincided with the principal axes of the cross-section and the

axial direction x coincided with the undeformed centroidal axis Cl-C2.

Let v, w denote the displ acements of the centroidal axes in the y- and z-directions respectively. ý is the angle of twist, and u is the

axial displacement. The quantities u, v, w and 0 are all assumed to be

of a small order so that the square of each quantity is negligible when compared with their individual magnitudes.

The assumption that no shear deformation can exist in the middle surface (i. e. the fourth assumption in Chapter 2), leads to the definition of the angular displacements, in terms of the first derivatives of the transverse displacements (i. e. exactly the same as in the case of a solid beam under bending loading).

dV Tx-

dw y 27X

The element shown in Figure (4.1.0) is loaded by an axial force FxI'

which acts at the shear centre, whose coordinates yo, zo are measured from the centroid, the end shear forces and end moments, Qy1I Qy2,

QZ1, Qz2, Myll My2l Mzj and Mz2 acting along axes parallel to the

principal axes through the shear centre. The torque Mx, and Mx2 and the warping moments Mxxi and Mxx2 are acting along an axis through the

shear centre.

The total potential energy 7r p for the structural system illustrated in

Figure (4.1.0) is given by,

7r p=u-v (4.2.0)

28

where U is the strain energy stored during the deformation of the structure, and

V is the potential (work done) by applied loads moving in the direction of associated degrees of freedom.

The method of calculating strain energy during deformation of structures is well known and the corresponding expression for the 6 problem illustrated in Figure (4.1.0) is as shown in equation (4.3.0).

The origin and proof of this expression can be found on pageJ58of ref e rence 4.

V, - 2 Wi, 2 112 t2 2]dx (4.3.0) Uf CE Iz + EI + Erex + GJOX + EAU 2

jt y

prime denotes the second derivative duz ( idfrxl) E is the elastic modulus G is the shear modulus IYPIZ are second moments of area of the cross-section about its

principal axes y and z respectively r is the torsion warping constant of the section i is the St Venant torsion constant, and A is the cross-sectional area of the section concerned.

where, single prime denotes the first derivative (d) and double

prime denotes the second derivative d2 TX_ ( dufxl)

E is the elastic modulus G is the shear modulus IYPIZ are second moments of area of the cross-section about its

principal axes y and z respectively r is the torsion warping constant of the section i is the St Venant torsion constant, and A is the cross-sectional area of the section concerned.

The quantities may be identified as:

EIzV,, 2 = the strain energy due to bending in the y-x plane

EI y W,, 2 the strain energy due to bending in the z-x plane

Erex,, 2 the strain energy due to warping about the x-axis

EAU, 2 the strain energy due to axial loading.

The potential of the applied loads in buckling analysis consists of two constituents. The first part V, is the work done by the applied loads in the direction of associated degrees of freedom (i. e. pre- buckling deformations) and thus can be shown as in equations (4.4.0)

29

Therefore

V= FX(Ul - U2) + QylVl + QY2V2 + QzlWl + QZ2W2 + Myloyl + MY2ýY2

m (4.4.0) Z1 zi 1 MZ20Z2 + Mxl'Oxl + Mx26X2 + Mxxlexl + Mxx2ex2

See Figure (4.1.0).

where Fx, UP U2 are the axial load and k, the displacements at nodes 1

and 2

are the transverse shear forces and the associated Qyl, QY2, VlIV2 t displacements in the y-direction at nodes 1 and 2

Qzl, Qz2, Wl, W2 are the transverse shear forces and the associated displacements in thez-direction at nodes 1 and 2

are the moments and associated angular My 1* My 2, Oyls Oy 2 displacements in the y-direction at nodes 1 and 2 Mz1IMz2I Ozll Oz2 are the moments and associated angular displacements in the z-direction at nodes I and 2 Mxl, Mx2, Oxl, 8x2 are the torsional moments and associated angular displacements in the x-direction at nodes 1 and 2. Mxxl, Mxx2lax'llox'2 are the warping moments and the associated twist gradient in the x-direction at nodes 1 and 2.

The utilization of the potential energy concept in the establishment of relationships for instability analysis presumes that pre-buckling deformations have occurred and are followed by buckling deformations.

However, the buckling deformations are mainly due to flexural and torsional action only, and hence there must be a proportion of work done by the applied loadings in their associated buckling deformations

and are given by V in equation (4.5.0).

flu (4.5.0)

The potential in the buckling deformations is formed by the following constituents:

30

V, is the bending in the x-y plane due to axial load Fx V2 is the bending In the x-z plane due to axial load Fx V3 is the twist about the x-axis due to axial load (torsional buckling)

V4 is the bending in the x-y plane due to transverse forces Qyj and Qy2

V5 is the bending in the x-z plane due to transverse forces Qzj and Qz2

V6 is the bendi ng in the x-y plane due to moments Mzj and Mz2

V is the bending in the x-z plane due to moments MI and M2 7yy V8 is the bending in the x-y and x-z planes due to torsional moments

MX, and Mx2

V9 is the bending in the x-y and x-z planes due to warping moments Mxxj and Mxx2*

Let us now begin by analysing each component independently.

Potential energy formulation for buckling in the x-y and x-z planes due to axial load Fx:

Y. V I

FIGURE 4.2.0

31

Consider the column loaded by an axial thrust Fx, as shown in Figure (4.2.0). The bend in the beam cau ses Fx to create a bending action as well as the compressive stress. This is a geometric non-linearity rather than a non-linearity caused by the stress-strain relationship for the mate rial, when the yield point is exceeded.

To apply the energy method to the problem, it is necessary to consider the work done by Fx in travel 11 ng through the di stance A, as wel 1 as Fx in travelling through the distance U, calculated by direct

compression. Thus the total potential energy of the system can now be

expressed as follows:

V EI

Z (Vii)2 dx - FXA - FXU (4.6.0) 02

where EI

V,, 2 is the strain energy in bending in the x-y plane 2 Fx is the potential energy due to buckling deformation, and FxU is the potential energy due to pre-buckling deformation.

Assuming that the distance A generated by the bending is far greater than the axial shortening U, then the final length of the bowed beam,

will be, Z. Therefore,

Y. = ds

and

Z- A= f dx (4.7.2)

By considering a small length ds in the bowed beam,

dS2 = dV2 + dX2

32

i e. ds = El + (Lv)']i dx dx

and by using the binomial expansion and neglecting the higher order terms, equation (4.8.1) is simplified to,

ds = El + _I (0) 21dx (4.8.2) 2 dx

Substitution of equation (4.8.2) into equation (4.7.2) and also by integrating:

.1f (dV)2 dx (4.9.0) 2 dx

Now by substituting equation (4.9.0) in equation (4.6.0), the

potential due to application of axial load Fx, causing bending in the

x-y plane is obtained, thus,

t V=1f EI V"2dx fF V'2dx -FU (4.10.0)

-2 0z2oxx

Similarly, for bending in the x-z plane, the corresponding equation is

given by,

W'-2dx -1x W'2dx -FU Vf El yi Fx x 2o20

where FxU is the potential of the pre-bukcling deformation and k*k 1 FXfV'2dx, .1f FxW12dx are the potential of the buckling

02 o* Lformations in x-y and x-z planes respectively.

33

potential. Energy Formulation in torsional buckling due to application

of axial load, Fx:

Y9V

iJ

-�C

FIGURE 4.3.0

Let the column shown in Figure (4.3.0) be loaded by an axial thrust

Fx. The differential equation governing the equilibrium is well

established (see page 2-of reference58) and is given by,

EI d4y

- Fx d2y = dX4 dX2 (4.12.0)

Y-11*' . . "%..

It.

dp

FIGURE 4.4.0

34

Now consider the cruciform sectioned column under uniform compression, as shown in Figure (4.4.0). The column has four identical flanges of width b and thickness t and also y and z are the axes of symmetry of the cross-section. Under the compression, the load Fx, causes torsional buckling as shown in Figure (4.4.0). During deformation,

the axis of the bar remains straight, wh_ile each flange buckles by

rotating about the x-axis.

For the purpose of determining the compressive force, which produces torsional buckling, it is necessary to consider the deflections of the

flanges during buckling. However, the deflection criterion of the

flange is identical to the deflections shown in the pin ended column in Figure (4.3.0).

Therefore, the deflection curve of the strut shown in Figure (4.4.0)

and the corresponding bending stress can be found by assuming that the

strut is loaded by a fictitious lateral load of intensity,

F d2V (4.13.0) A dx2

Let us now consider an element mn in the form of a thin strip of length dx, located at distance, p, from the x-axis and having cross-

sectional area tdp. - Owing to torsional buckling, the deflection of this element in the y-direction is,

v=0 (4.14.0)

The compressive force acting on the rotated ends of the element mn is

otdp, where a=Fx denotes the initial compressive stress. These W- compressive forces are statically equivalent to a lateral load of,

35

(atdp) ý! V- (4.15.0) dx2

which can be written in the form of,

cr t pdp d2o

A, (4.16.0) 7X-2

Therefore , the moment about the x-axis of the fictitious lateral load

acting on the element mn is given by,

cr d2 ý dx tp2dp (4.17.0) dX2

In the general case of a column of thin-walled open cross-section, buckling failure usually occurs by a combination of torsion and bendi ng. In order to investigate this type of buckling, consider the

assymmetrical cross-section shown in Figure (4.5.0):

C

Fig. 4-5.1

rl; 442

FIGURE 4.5.0

36

The y- and z-axes are the principal centroidal axes of the cross- section and eys ez are the coordinates of the shear centre 0. During buckling, the cross-section will undergo translational and rotational displacements. The translational displacements are defined by the deflections V and W in the y- and z-directions of the shear centre 0. Therefore the final deflection of the centroid C during buckling,

according to equations (3.4.1) and (3.4.2) in Chapter 3 is given by,

yc =V+ ezBx (40' 18.0)

zc =W- eyex (4.18.2)

If the only load acting on the column is a central thrust Fxq as in

the case of a pin ended column, then the bending moments with respect to the principal axes at any cross-section are,

Mz = -Fx (V + ezex) (4.19.0)

My= -Fx (W - eyex) (4.19.2)

Therefore, the differential equation for the deflection curve of the

shear centre is given by,

2 EI y=- Fx (W - eyox) (4.20.1)

(4.20.0)

-I d2W Ez--- Fx (V + ezex) (4.20.2) dX2

To obtain the equation for the angle of twist ex, the following

argument can be made. Consider a longitudinal strip of cross-section tds defined by coordinates (y, z) in the plane of the cross-section.

37

The components of its deflection in the y- and z-directions during buckling as stated previously in Chapter 3, are

UY =V- (z - ez) ex (4.21.1) (4.21.0)

Uz =w+ (y -eyx (4.21.2) il

Taking the second derivative of these expressions with respect to x and again considering an element of length dx, the compressive forces

Itds acting on slightly rotated ends of the element, produce forces in the y- and z-directions with intensities of,

-(cy tds ) d2 [V - (z - ez)exl (4.22.1) ly ý TX2 (4.22.0)

Fz = -((jtds) d2 [W - (y - ey) ex] (4.22.2) dX2

By taking moments about the shear centre axis of the above forces, the torque per unit length of the bar is given by,

dmx = -(crtds)(z - ez)[d 2V-

(z - ez) d2e

+(6tds)(y -e dx 2 dX2 y

[d2W + (y _ ey) d2e

(4.23.0) 2 dx dx-

Integrating over the entire cross-sectional area A and observing that,

cy f US = FX9 fz tds =f ytds =0 A (4.24.0)

f y2tds = Izo f z2tds =I AAy

38

and defining Io = Iz +Iy+ A(ez2 +ey 2).

The equation (4.23.0) is simplified to,

d2V d2W 3+ LO

Fx d2eX

(4.25.0) Mx f dmx = Fx Eez ---7 - ey A dx d7 A -dx--T-

where Io is the polar moment of area, referred to the shear centre.

in the previous section, the buckling of columns, was subjected only to centrally applied compressive loads. The following analysis intends

to discuss columns subjected to the action of bending couples MY and Mz at the ends, combined with the central compression Fx . It is also

assumed in this analysis, that the effect of Fx on bending stresses

are negligible. Therefore the normal stress at any point in the bar

is assumed to be independent of x and is given by the following

equation:

cyx !

-X - ý_Zy_

- 2-

(4.26.0) A Iz Iy

where y and z are the centroidal principal axes of the cross-section. Furthermore, the initial deflection of the bar due to the couples My

and MZ will be considered as very small in comparison to the geometry

of the column. As before, due to buckling deformation, the components

of deflection of any longitudinal fibre of the bar can be defined by

coordinates y and z. Thus,

uy =v- (z - ez) ex

Uz =w+ (y - ey) ex

39

f

11-11

Ns. z

FIGURE 4.6.0

Following the same argument used in the previous section it can also be said that the intensities of the fictitious lateral loads and distributed torque are given by,

q= -(cytds ) d2 [V - (z-ez)Ox] (4.27.1) yý -X2

qy = -(otds) d2 [W + (y-ey)8x3 (4.27.2) ; -X2

(4.27.0)

dmx =-f (atds)(z-ez) d2 [V : (z-ez)a X] +

A dx2

tds) (y-e y) d2 [W + (y-e y )G x] dx2 (4.27.3)

Substituting equations (4.26.0), (4-27.1) and (4.27.2) into equation (4.27.3) and performing the integration, the following relationship is

obtained,

40

dmx= f (-(a tds) (z-ez) d2

[V+(z-ez)Oxl+(atds)(y-e y)

d2 W-(y-ey)Bxl

A dX2 dX2

2 d26 x d2W d20

-(atds)(z-ez)ýz4 + (z-ez) ]+(atds) (y-ey) - (y-e ) x]) x2

y dX2 dx2 X2 dx2

=f (_( FX

tl Myz mzy

}tds(z-ez)[d2V + (z-e d2e F-Mm

tds(y-ey) AAII dX2 Z)---!

]+'--A iyZ

yz dX2 yIz

2W d2e

(y-e ) --=! ]) dX2 y dx2

22 2) =f EiýF-X-(z-ez)+ýý(zez-z2)+lz (ezy-zy)lLzv-+{ýx-(e -2ze +z AA1yIZ dX2 AZZ

m2 3) + jL(zez -2z2ez+z y

m d20X Fm + (yez2-2zyez+yz2), (e -y)+. ýZ(zey-zy)

17 dx2 {7ý YIy

mz (yey_y2) JALW iz dX2

F2 2)+ mY (ze 2-2yze +zy2 +.

m z(y. 2-2y2

d2e x x (e -2ye +y e +yl) yyIyyyTzYy dX2

(4.28.0)

Noting that,

f US = A, f Ads f Ytds =0 AAA

f y2tds = Izq f z2tds y AA

41

and defining Io = Iz +Iy+A (ez2 +ey 2)

where Io is the polar moment of area referred to the shear centre. Thus equation (4.28.0) is simplified to,

d2V d2w [. 1 dmx= {e Fx-My 1- {(eyFx+Mz }- + {F I +M _(f z3dA+f zy2dA) -2ezl Z dX2 dX2 x0YIyAA

d2e

Mz f y3dA +f yz2dA) - 2eyl x (4.29.0) zAA

dX2

Let us now define the following quantities where,

in which,

s0=A0+ ezýj + eyO2 (4.30.1)

-L (f z3dA +f zy2dA) - 2zo (4.30.2) Iy AA

ý2 I (f y3dA +f yz2dA) - 2yo (4.30.3)

zAA

Therefore, equation (4.29.0) can now be rewritten as,

dm = {, e F -My) d2V

- {eyF +M It-w+U 10+M

ý1+MzB21 d2e

X xzx ý72 xz dX2 )c TA y dX2

(4.30.0)

Now consider the open section element shown in Figure (4.7.0) where force Fx is applied at a point on the cross-section whose coordinates

are (ay, az) from the centroid.

42

Z

cy. %1 ey

FX

FIGURE 4.7.0

The system shown above is equivalent to:

My= Fx. azs Mz = Fx. ay

and a force Fx at the centroid. Applying the values of, My and MZ in

equation (4.27.1) and (4.27.2) would result,

d26

qy = Jx ý2w

- Fx (ay - ey) 2x (4.31.1)

dx2 dx

qz Fx d2V + Fx (az - ez)

d2ex (4.31.2)

dx2 dX2

d2V d2W 10 d2e

x dmx = -Fx (az-ez) + Fx(ay-ey). =--- + Fx + azRl +a

dX2 y 02) dX2

(4.31.3)

This equation becomes very simple if the force Fx acts along the shear

centre axis i. e. az = ez and a. = ey.

Then equations (4.31.1), (4.31.2) and (4.31.3) are simplified to

43

d2W qz = -F Xý -X2 (4.32.1)

(4.32.0)

qy = -Fx d2V (4.32.2)

and dX2 d2o

dmx = -Fx (LO + ezol + ey 02) 1-12ý- (4.32.3) A dX2

As Timoshenko(58) pointed out, with the assumptions made previously, equations (4.32.1), (4.32.2) and (4.32.3) become totally independent

of one another. In this instance, lateral buckling in the two

principal planes and torsional buckling may occur independently. The first two equations give the usual Euler conditions providing critical buckling loads under bending conditions, while the third equation gives the critical load corresponding to pure torsional buckling of the column.

Furthermore, for any other location of the axial load on the cross- section, the nature of buckling criterion is not independent and generally occurs by the combined effect of bending and torsion.

Barsoum and Gallagher(4 ) had probably realised the consequence of placing the axial load at points other than the shear centre. In

spite of warnings given by Timoshenkc)(58) they analysed the problem of arbitrary axial load applications, as though the axial force was acting at the shear centre. This is incorrect with respect to equation (4.31.3). Rajasekaran also ignored this effect and placed the axial force at the centroid. Except for Bleich( 9),

no other authors produced correct results for problems with wide flanges. (For details see ref. (9 )).

One of the main objectives in this analysis was to provide the facility for placing the axial force anywhere on the cross-sectional plane. The additional bending moment and the warping moments (i. e. Mxx

= Fxw p) generated are carefully taken into account in a suitable transformation matrix and which is explained in equation (8.14.0) In Chapter 8.

44

Potential due to axial load:

fir

FIGURE 4.8.0

Consider the thin-walled beam column shown in Figure (4.8.0) subjected to an axial thrust Fx- As a result, under this load, the column sections produced translational and rotational displacement (see

equations (4.21.1) and (4.21.2)). Let us assume the angle of twist at distance x, to be ex. Thus, at distance (x-ft), the twist = ex+Aexo Therefore the work done by the torque mx in increasing the twist from

ex to(e+, &B) is given by,

R= mXSOX dx (4.33.0)

but the Increment of angular displacement6Ox can be expressed as,

6ex = do

X dx (4.34.0) dx

45

Therefore the work done SW,

6W = mx (f dx) dx dx

but from equation (4.32.3),

J d29

mx Fx (eyß, + ezß2 + -7Z-) - dx2'-

Substituting the value of mX in equation (4.32.3):

+ 10 d26 dox

sw= Li Fx (eyß ,+e-Mf- dxldx AA )dX2 dx

(4.35.0)

t.

(4.36.0)

Thus the total potential energy in torsional buckling is given by,

Z d20 da

Wf E(Fx(eyß, + ezß2 + )--- x� - dx]dx (4.37.0)

0 dx7 dx

but from equation (4.30.1), So = ey6l + ezO +10 27

Thus, 9 d2e do

x W CIT S --ý) f dxldx (4.38.0) X0 FX2 dx

Since, d [( dox)2j

= 2( do

X) d2e

X (4.39.1) 3T dx dx dX2

Therefore, .1f (dOx) 2=j d2o

x do

x dx (4.39.2) 2 dx dxT- -Tx-

46

The integral function on the right hand side of equation (4.39.1) is

the same as the integral function of equation (4.39.2). Therefore,. by

substituting equation (4.39.2) in equation (4.38.0):

t do Wf Fx So ( 2dx (4.40.0)

20 dx

Equation (4.40.0) now provides the necessary relationship in

calculating the potential for columns under axial thrust, providing the column is at pure torsional buckling.

Since equations (4.32.0) are independent of one another, thus

uncoupled lateral buckling in two principal planes and torsional buckling may occur independently.

The three modes of uncoupled buckling due to application of an axial load Fx can now be fully described by the equation (4.32.0). The

potential comprises three quantities and by using the result of

equations (4.10.0) and (4.11.0), are as follows:

1. Due to uncoupled. lateral buckling about the y-axis and moment

equivalent tol f Fx V2 dx. 2 o,,

2. Due to uncoupled lateral bucklings about the z axis and moment I equivalent to f Fx W'2 dx.

2o

3. Due to purely torsional buckling about the x-axis and moment

equivalent tol fýXSO (dfl2dx. 2o dx

Potential due to applied torque:

Application of a torsional load to a column with constrained ends will generally produce an axial stress field. This phenomenon has already been explained in the introduction to non-uniform torsion. Due to the resulting axial stress field, the column may reach the unstable

47

condition and will produce large displacements about the x-axis and also in the x-y and x-z planes. Since these deflections are generally

coupled with large angles of twist, the final collapse will always take place in the form of lateral torsional buckling.

k.

c

FIGURE 4.9.0

Consider the configuration of a displaced column which is loaded by a torsional load about the column axis. Let us also assume that, prior to loading, the axial direction of the column coincided with the

global x-direction. To define the orientation of the column in space fully, let us define an orthogonal coordinate system in which the axis E

represents the local x-axis while the n and C axes represent the

instantaneous positions of the principal axes.

However, the displaced configuration in Figure (4.9.0) is solely due

to the applied torque, and hence the relationship between the moving

and the fixed coordinate systems can be expressed as:

48

Cos(ýX) cos(EY) cos(cz) x

cos(nx) cos(ny) cos(nz) y

cos(cx) cos(cy) cos(cz) z

(4.41.0)

where the quantities- in the square matrix are the direction cosines between the two axis systems. Let us also denote the small rotation angles of E, n, C axes about the x-, y- and z axes by exp ey and 0z

respectively as shown in the following table.

The axis pair under consideration The small angle

X-t

Y-n Z-ý

TABLE 4. T. 0

Then, by inserting the values in Table (4. T. 0) in equation (4.41.0)

and performing the right handed rule of vector multiplication, the

following relationship can be found

cose y cose z sinoz siney x

Tj -si no z cosex cose z sinex y

siney -sinex cose x cosey z (4.42.0)

Now let us consider an arbitarily chosen point P, in the x-y and x-z planes, as shown in Figures (4.10.1) and (4.10.2).

49

I LL

Fig. 4.10.1

z

Fig. 4.10.2 FIGURE 4.10.0

By referring to Figures (4.10.1) and (4.102) the slopes of the column to the x-axis, with the inclusion of axial shortenings are,

tane y= -ve (4.43.1)

tanez = -wl (4.43.2)

By considering bending only and ignoring possible axial shortening, equations (4.43.1) and (4.43.2) become:

since, cose =I A+ tan2e

tanaz (4.44.1) 1 +F-

0

we tane y =- (4.44.2) I+c

0

and sine = tane .9 47 tan2e

Then the square matrix [R] in equation (4.41.0), now yields to

50

[RI (l+E:

0) Wi

4(1+F, 0

)2+(WI)J. 4(1+F 0

)2+(VI)i 4(1+C 0

)2+(V1)2) ý(l+FE 0

)2+(W1)2)

A(I+e 0

)2+(VI))

wI

-Ve

4(l 0

)2+(WI )2)

cosex. (I+cO) Sinex 4(1+c

0 )2+(Vf)i

- Sinex Cosex. (I+cO)

Al +c 0

)2+(Wl)i

(4.45.0ý

If we assume the deformations are small in magnitude compared to the

original dimensions of the column, with respect to unity, the quantity

vo, tends to zero. For small angles of ex, sin ex tends to ex and cosex tends to 1. Thus the square matrix [R] now simplifies to:

1

ER] -V' 1 ex

wo -ex 1

Substituting equation (4.46.0) in equation (4.41.0) yields,

v

16x

wl -6 x

(4.46.0)

(4.47.0)

(Note, for orientations only, the absolute position of (X, Y, Z) to (x, y, z) is immaterial).

51

Since the curvature of the deflected column has the same orientation

as the local axis system, the relationship between the local and

global curvatures is:

0j1v4 -i 0 'x

OT, -vl i ex W" (4.48.0)

cý, ý -

wl --ex 1- vil

where 6" is the local warping curvature t is the global warping curvature

is the local bending curvature in the n plane W" is the global bending curvature in the y plane 6, is the local bending curvature in the plane V is the global bending curvature In the z plane

Thus, by applying matrix multiplication to equation (4.48.0),

et, = ex, + VIW" - W'Vll (4.49.0)

By differentiating both sides again, with respect to E, x:

oil =e Is + vlwill, - wivill (4.50.0) x

According to the sign convention adopted, warping curvature T is

negative, (see page 11). Then, the potential due to the gradually

applied torque Mx is given by*

14T = -. 1 Mx f (V'W" - W'V")dx (4.51.0) 21

52

By referring to equation (4.51.0), it can be seen that it is the same as the resul t shown in ref. (4). Barsoum and Gal 1 agher , emphasised that equation (4.51.0) was strictly valid only to the buckling of closed circular shafts. By referring to Figure (4.9.0)

and also equation (4.51.0), it can be seen quite clearly that the only assumption made in developing the theory was that the column under torsional load will produce warping displacements at any cross-section and also by restraining those warping displacements it will produce an axial stress field, thereby exposing the column to a possible instability. It was also shown in Chapter 2, in equation (2.5.0),

that a closed circular shaft can never produce warping displacements

under torsional loadings since their cross sections are axisymmetric. Thus the statement made in ref. (4 ) is contrary to the physical behaviour of thin-walled structures in general.

Recently, Gaafar and Tidbury(19) observed experimentally that at large

angles of twist, the torsional load along the column varied considerably and approximated this with a linear function. By

referring to Figure (4.9.0) and equation (4.49.0) it is fairly obvious that because at large angles of twist, the translational and

rotational displacements are significant. The shape of the column

under consideration, is now bent, and hence the final equilibrium

position for any cross-sectional plane is now affected by the

components of axial, flexural and rotational forces rather than the

applied torsional load on its own, as was the case at the beginning of loading.

Thus at significant angles of twist the torsional load Mx, as suggested by Gaafar and Tidbury is given by,

Mx ý Mx 1(1- 'ýý) XX Mx 2 (4.52.0) tX

where Mxj and Mx2 are the torsional loads at ends 1 and 2 respectively, andtlength of the beam column between nodes 1 and 2.

53

Referring back to the original problem, the potential due to applied torque is,

N= mJox (4.52.1)

where 6W is the potential generated in applying external torque Mx at an elemental length 6x

6ex is the difference of the angle of twist, at the two ends.

de But sex = -a-XX.

dx

also from equation (4.49.0),

do x= exi + VIWII - WIVII (4.52.2) -jx-

The term, ex' in the equation shown above, was considered in equation (4.4.0), as contributory deformation due to torsional load in pre- buckling displacements. Thus for the buckling deformations only, equation (4.52.1) is simplified to,

do x= Mal - WIVII (4.53.0) , u-x

Now, by applying equation (4.53.0) in equation (4.52.1),

wft do

X dx (4.54.0) 20

Mx -Ux

f EMxl (1 - '2ý) + MX23(V'W" - W'V") dx (4.55.0) 2o I

54

Thus, equation (4.55.0) can now be used to estimate the potential due

to applied torque.

Potential due to warping moment:

By referring to Table (2. T. 0), the torque exerted on the column due to

application of the warping moment can be expressed as,

Mx = -E r d"x

(4.56.1) dx2

d20 Mxx = Er

dX2 x (4.56.2)

where Mxx is the applied warping moment Mx is the torque generated.

By recalling equation (4.54.0), the potential generated due to the

applied torque is,

6w =M da

X dx (4.54.0) X dx

By applying equation (4.56.1) in equation (4.54.0) yields,

6w =- dM

XX de

x dx dx» -dx-

dM Xx dx)

do X

-Tx-- dx

SW 6m - do

x (4.57.0) xx -d-x

55

Thus, the potential due to the gradually applied warping moment, Mxx is,

w=- .1fI mxx do

X dx (4.58.0) 2o dx

This theory can also be explained qualitatively with the aid of the following example:

x

FIGURE 4.11.1

The base of the H section column shown in Figure (4.11.0) is fixed

while the upper end is free and subjected to a torsional load T in the

positive direction.

A warping moment will be generated due to the restrained condition at the lower end and consequently the two flanges will be distorted as

shown in Figure (4.11.0). This specific example and its phenomenon is_.

found well described in reference (59).

, --P-Z Fig, 4.11.1

56

Now consider the state of equilibrium of an element at a distance x from the base in Figure (4.11.0).

The displacement of the web on the y-z plane at distance x= el. The displacement of the web on the y-z plane at distance x+ dx = e2,

Within the distance dx the angular deflection of the web is

Since el =b Ix

e2 =b (ex +6ex) k"

therefore, e2- el

b de

x --Tx- Tx-

But the work done by the bending moment bM acting on el'ement 6x, to

produce a rotational displacement ft, at a distance x is given by,

6w 6m Ch ux-

= 6m b da

x ix-

Since there are two webs involved in warping, the final work done, 6W,

of the element 6x is,

6W =2 6M b

According to reference (66), the warping moment for an H section is defined as 6Mxx = 26Mb

do 6w = 6mxx

dx x

57

Thus, the total work done along the entire length of the column is,

-1 f ý'

M. do

X dx 20 dx

which is the same as the result shown in equation (4.58.0).

Since the warping moment is proportional to the applied torque, from

equation (4.52.0), the applied warping moment is,

'2 Mxx 1 (1 - i)+

'it Mxx 2

(4.59.0) Mxx '

and also from equation (4.50.0), the warping curvature is,

d2e x=

ell + v#wiit- WIVIII 7x-i -x

Now by applying equation (4.59.0) in equation (4.50.0), as before,

without the pre-buckling warping displacement O'ý; the final

expression for the potential due to the warping load becomes

f [Mxxl (1 - Z) + -ý MXX23 (V'W"'- W'V ... )dx (4.60.0)

0

Potential due to lateral bending and end shear loads:

Two possible modes of instability were discussed previously in

equations (4.27.1), (4.27.2) and (4.29.0). They are briefly:

1. Bending in the plane, of one of the principal axes 2. Twisting followed by bending, in the lowest plane of rigidity.

58

So far, the buckling mode mentioned in category 1 has been discussed in detail in equations (4.10.0) and (4.11.0) and the buckling mode in category 2 due to application of axial load Fx, was discussed in equation (4.32.3).

However, the application of forces QY Qz and moments My, Mz could also cause instability, and, depending upon their end constraints and loading condition, the associated mode of 6ckling could be either the bending in the plane of one of the principal axes or the twisting followed by bending, in the lowest plane of rigidity.

Let us now consider the stability of the beam column shown in Figure (4.14.0). The potential of the forces xan be obtained by considering the equilibrium of the beam column with two cantilevers mounted back to back as shown in Figure (4.12.1). The appropriate signs used in the two cantilevers are also shown in Figure (4.12.0) and (4.13.0),

according, to the chosen right handed sign convention (se Table (2. T. 0) .

Let us assume that the instability had occurred at the beam column. The associated lateral and rotational deformations which correspond to the two cantilevers under consideration are as shown by Figures (4.15.1), (4.15.2) and Figures (4.16.1), (4.16.2) respectively.

The potential of a differential element dx, due to application of the transverse force QY and moment MZ is as follows:

y

x

FIGURE 4.12.1

59

Fig. 4.12.2 L--

T-X

Fig. 4.12.3

Fig. 4.12.4

moment

Fig. 4.12.5

FIGURE 4.12.0

60

Ck y

F ig. 4.13.3

Moment ýI

Fig. 4.13.1

Fig. 4.13.2

Fig. 4.13.3

Fig. 4.13.4

FIGURE 4.13.0

61

(ly, (IY2

y

c Mzl I;

--- Mki

FIGURE 4.14.0

Yj

N2

4 Plan

dx . -v! ýx A

$00,

x Plan

e: ý%

Elevation

u

L$2

d`2

Elevation 2

plan

Fig. 4.15.1

Fig. 4.15.2

Fig. 4.15.3

Fig. 4.15.4

FIGURE 4.15.0

62

For the cantilever beam shown in Figure (4.14.0) the appropriate sign convention can be obtained by referring to Figure (2.4.0) and Table (2. T. 0). Corresponding results are illustrated in Figures (4.12.0)

and (4.13.0). Thus, the curvature (ý 2 ý-) for the cantilever shown in

Figure (4.14.0) is negative. dX2

From Pigure (4.15.2), the angle of rotation due to the applied load

Qy2 is I

zx = Lx - V"dx (4.61.0) R

where 6zx is the angle rotated by an element of length dx at a distance x from the origin in the positive direction. , Now,

considering triangle ABC, (see Figure (4.35-4),

dV2 = tan Ozx = ezx (4.62.0)

L-x

for small angles Ozxo

Therefore, by substituting equation (4.62.0) in equation (4.61.0):

dV2 = -V" (. Z-x) dx (4.63.0)

Figure (4.15.3) illustrates the case of the deflected element (shown

in Figure (4.15.2)) twisted by an angle ex* Due to this rotation

point 2' has now moved to 2".

Thus,

dV2 VI'8x (L-X) dx (4.64.0)

63

and this is shown in Figure (4.15.3) and (4.15.4).

Since it was assumed that plane sections remain plane during bending, the rotation of a beam on the x-z plane can be expressed as:

de' dw 2 (4.65.0) Z2 x T- ý-V OX dx . t-x

Since the displacement dV2 and d82 are known, the potential (work done) can now be calculated.

Thus, dVQ2 ý» - QY2 V" ex (£-x)dx (4.66.0)

dV M2 ý MZ2 V" ex dx (4.67.0)

where dVQ2 and dVm2 are the potentials due to application of Qy2 and Mz2 respectively.

Thus, the total work done in deforming the entire length VS. ) is given by 9

V2 'x - QY2 f Vlg Ox (£-x)dx + MZ2 f T' exdx (4.68.0) P. Z

in order to account for unequal end moments, it is necessary to consider the case where point 2 is fixed and point 1 is free as shown in Figure (4.16.1). Therefore, using the same argument, the work done by Qyj and Mzj is given by,

V, Qyj fV Ox x dx - Mzj f V" ex dx (4.69.0)

64

I

Hz

Elevation

Fig. 4.16.

pun

Etevation

Fig. 4.16.2

Plan

FIGURE 4.16.0

65

Assuming all the loads are gradually applied (i. e. QY1, Qy2, Mzj and Mz2 were initially set to zero), the work done in the general motion of the element is given by,

Vfinal '" (Vl + V2) 2

Qyj fV Ox xdx -" eY2 f V" ex(. t-x)dx - 't

2

mzj fV 6xdx +"f MZ2 V" exdx (4.70.0) x2t

Now, applying the result shown in equation (4.70.0) to the case of lateral bending in the x-z plane, the following result can be directly

obtained. See Figure (4.17.0) and equation (4.71.0).

a z11 Z2

GLEE -- 1112---,

FIGURE 4.17.0

Vfinal '"' (Vl + V2) 2

=-1 Qzl f Wil 0 xx dx - -' QZ2 f W" ex(£-X)dx - 2x 1 My, f W" ex dx + .1f MY2 W" 6x dx (4.71.0) 2x21

66

Total potential due to buckling defomations:

Since all the constituents of potential in buckling deformations are established, equation (4.5.2) can now be written in its explicit form

as follows:

V=1 fIFXV'2dx + .1fI Fx W'2dx +f Jt FXSO(OX')2dx Ve xxdx 220- .1 Qy f 020

-1 "e (. Z-x)dx -1M vile Z, f V'G xdx +if MZ2 xdx QY2 fVx20

00

xdx Q x)dx -1 My, f W"exdx Qzl f w"ex Z2 f W"ex('-

20020

fm dx -ý Y2 V"ex f EMX1(-1-2ý) + 2ý MX21(V'W"-W'V")dx 200k

1 W'V"ldx (4.72.0) Ein xxm xxl

(1 j) + -i xx23 (V'W" 1

Now, by combining equations (4.3.0), (4.4.0) and (4.72.0), the final

form of the total potential ITp, can be established. However, the total potential of a conservative system can be expressed as:

np =U-V (4.73.0)

where, U is the strain energy stored in the structure V is the potential due to pre-buckling and buckling

deformations.

Therefore, by combining equations (4.3.0), (4.4.0) and (4.72.0) in their explicit form, equation (4.74.0) is obtained:

67

Ilp = -lZ'f't[EjZVo12 Wo, 2 1,2 12 1 21dx Jt F Ve2dx + El + Ere + Gjex + EAU 0yx20x

f Fx W' 2dx - .1J FXS,, ýex, )2dx + .1 Qyl'f V"exxdx+.! Qy2f

z V"Ox'(L-x)dx

02o202o

+ .1mz V"exdx Mz2f V"Oxdx +1J W"exxdx 2 lo iQzl 0

1 -x)xdx + .1 My, 'j

k W"ex dx - -if Myy"exdx + -j QZ2

- Wex(i

22o

1. W'V")dx + -f EMA (1 x) + 2ý Mx23(V'W"- 20ZZ

+ -. 1t MXX23(V'W"! W'V". )dx-Fx(ul+u2)

- QylVl - QY2V2 - QzlWl - QZ2W2 - myleyl - MY20Y2 -Mzlozl - Mz2ez2

m (4.74.0) X1 OX1 - Mx2 Ox2 -M xl OX1 - Mxx2ox'2

Substitution of equations (5.7.0), (5.11.0) and (5.12.0), into

equation (4.74.0):

Z

zä[{fll 11 Ilp m [.! f, EEIZ LV6 v»lLf"jdx]{8'}+EI LW8yJ[{fwlLf"Jdxl{e 1

20vZywy

ErL0 0X'J[{foxlLfexJdxl{exl + GJLG eýEff'exiLfe2dxlfexl + x Oký x6 x-

EALuJ[Ifu)LfýJdxl{jjll- jF LVOZ4[{fv')LfJ dx]{V' 0xv ez

f FXLWB. J [{f; )Lfw'Jdx]{w 0y0y

68

f2F Sj. OXG'J [{fex}LfeJ dx]{ox} +1L QZILVOJ [{VIU j Xdx]{ex} xx el 2f0- 20xZvx ex

1101ze

xfM LVOJ [{f"}LfeJx dxl{exl - 2f

QZýVO J [{f ")Lf, J (. E-x)dxl {, X, ) +- 0zv. x0yzvx

M LVOZJ fUV )Lf. i xdx]{ox} + Ilk f dxl {O [{f" vj

f y2

x., +-2 yx ei 2x 00x

1Z ")Lf J (2-x)dxllex, ) +i Z fM LI-IOJ [lf; lLf, J dx]{ox) - 2f

qy2L140YJ JUW 60 zl y x -2 x 0* 0x0

f Mzý Wo J [{fý)Lf, J dxl{ X, 1 +'2 Im l(1-2ý)+ -ý M 2]LVOJ [ifv')Lf "Jdxl{w x 0ex0xZxxZw0y

I ]LI40J [jfýjLf"Jdxl{v I 2 [Mxl MX2

v D+2y

0z

1L -X

x IL'VGJ [{f')Lfý'9 dxllw l- 1f im xx «i f [Mxxl (1 P +i Mxx2 Zvay xxl(' t)+ Z

Mxx2)aýýl 0y0

[{fw')Lf"'jdxl{v )-LUJ{f I-LVeJ {Q I-LWoJ {Qzl - v0xI tiy- Y M; zz

- Le eIJ lý1X) xxM xx

(4.75.0)

Equation (4.75.0) can now be rewritten in its very compact form:

p= 21 L djT [KI (-dj- - 17 Ld JT EN] fd} 4 dJT M (4.76.0)

where LdJT =Lu v woxeyezeýjT

69

{ F) = Fx QY Qz mx MY mz mxx

where the first term in equation (4.76.0) is the strain energy stored in the structure by deforming, and the second and third terms are the

potentials of the applied loads in buckling and pre-buckling deformations. Furthermore the square matrices EKI and [N] in equation (4.76.0) are generally known as the system stiffness and geometric

stiffness matrices.

Equations of Equilibrium:

It has been shown previously that the total potential of a system can be

expressed as:

11 p=-! Ldj E KI ( dl - 17 Ldj E NI 1d1 -LdJT f FI

27

where LdY is the appropriate d. o. f. CKI is the system stiffness matrix CNI is the geometric stiffness matrix {Fj is the applied loading vector acting on the system.

See equation (4.76.0).

Using calculus of variations, it can be shown that for a stable equi Ii bri um:

70

r 311 d

Illp ad 2

.

3d

See reference (47).

Thus, the above equation of total potential is simplified to:

fF) = [Klld) - [Nlfd) (4.77.0)

Buckling Analysis:

The equation of total potential (4.76.0) would not be valid for

structural systems which are not in equilibrium. However, stable equilibrium conditions can be achieved by deforming the overall structure under applied loads in which case the work done is stored simultaneously as the strain energy in the structure and the

relationship remains valid up to the point of buckling.

Mathematically this phenomenon can be explained as the function n(total potential) reaching the neutral condition. Thus, for the

neutral condition,

L21I =0 (4.78.0)

3, &2

Since, 3211 _ det l[K] - [NII (4.79.0)

BA2

71

Therefore for buckling Det I[K] - [NII =0

(4.80.0)

Solution_Procedure:

The solution to equation (4.80.0) can be found if matrix [N] can be factorized with a single parameter x. A close examination of the terms in in matrix [N] indicates that the special condition mentioned above may be achieved as follows,

a) by applying axial load only b) by applying My only c) by applying MZ only.

Under these circumstances, the standard eigenvalue approach may be

used in extracting buckling coefficient x, along with associated

modes.

However, in the case of a generalized problem, matrix [N], consists of terms associated with the generalized degrees of freedom, as shown in

equation (4.48.0). Under these circumstances, equation (4.78.0)

becomes non-linear, and hence the straightforward eigenvalue analysis becomes impractical. An alternative approach is proposed, which is

commonly known as the mid-point tangent technique in which the loads

are predicted along with the associated displacement history.

72

CHAPTER 5

UTILIZATION OF FINITE ELEMENT METHOD FOR BUCKLING ANALYSIS

The finite elem! ent method is nothing more than a procedure in which a generally continuous problem is discretized by a series of equivalent

elements, providing the compatibility and the continuity of the nodal functions and the element boundaries are maintained. This can usually be achieved by selecting suitable admissible displacement functions

for each element under consideration, providing the chosen functions

and their derivatives can be used in predicting all rigid body modes for the elements and appropriate displacements.

Let us begin now by applying this technique to the problem in equation (4.78.0). With respect to Table (2. T. 0), it is seen that the solution

used for the problem in torsion is very similar to the solution used for the problems of bending in the x-y and x-z planes. Consider the

problem of bending in the x-y plane. The nodal information V1, ezj,

V21 ez2 are used in deriving the displacement (shape) function and this is shown in equation (5.1.2). Similar functions are developed for

axial, torsion, torsion warping and bending in the x-z plane and they

are shown by equations (5.1.1), (5.1.4), (5.1.5) and (5.1.3)

respectively.

U0 (1 - ")Ul + U2 = LFUJ JUI

v V=f lVl + f2 OZ1 + f3V2 + f40Z2 = LfvJ{e z

(5.1.2)

fW= Lf,, Jiw 1 11 '+ f2eyl + f3W2 + f4eY2 0y

73

ex f0+f0.1 1 xl 2 xl + f39 x2 + feX'2 = Lf XJ {ex,

01=fJ01'''= Lf J{O x1 xl +f xl + f' 0 x2 +P (5.1.5) 23 fe x2 ex 6

where, suffices 1 and 2 denote the corresponding nodal displacements

for every degree of freedom at nodes 1 and 2 respectively.

Y, v

"Z

a X. u

KZ

z2

Fig. 5.1.1

FIGURE 5.1.0

Y. V X. u

z Z, W

Fig. 5.1.2

Figure (5.1.1) illustrates the general displacements of a beam column

-under axial, end shear and moment loads. With respect to the sign

convention described in Figure (2.4.0) and Table (2. T. 0) the

generalized displacements at node 1 and node 2 can be expressed as

Node 1 Node 2

Linear, V, Linear, V2

Rotational, e rdV Rotational, rdV X= 1, Z1 7 X=O z2 [ý-Xvl

74

By using matrix notation, the displacement vector A can be defined as,

V,

Z1 V2

z2

As in the case of an axial member it is possible to choose a suitable polynomial to describe the displacement field A. However, the degree

of freedom (d. o. f) for this particular problem is four.

Thus a cubic polynomial is adequate to describe the entire displacement field effectively. Hence:

V= alx3 + a2 x2 + a3x + a4 (5.2.0)

where a,, a2, a3 and a4 are some unknown constants. By differentiating

equation (5.2.0) with respect to x, I

dV = 3alx2 + 2a2x + a3 (5.3.0)

ax-

as described previously,

e dV X=o = -a3 zi [ý-Xvl

and rdV L2 - a3 [i-xvlx= t, 2 -3a, - 2a2L z2 '

75

Similarly, substituting the boundary condition, x=0 and x in

equation (5.2.0) yields,

V, = a4

anq, V2 = alt3 + a2 Z2 + a3z + a4

.0. V1 a4-

ez, = -a3

V2 = alt3 + a2l 2+ aTý + a4

Oz2 2" -3alt2 - 2a2l - a3

This can be rewritten as:

vi 0001a,

ezl 00 -1 0 a2 (5.4.0)

V2 Z3 x a3

Oz2 -3t2 -2z -1 0 a4

The inverse of the square matrix in equation (5.4.0) is as follows:

76

a, a -2 V,

a2 31 2k2 3Z k2 ez, (5.5.0)

a3 0 _. t3 00V2

a4 j3 000 ez2

flow by substituting the result of equation (5.5.0) in equation (5.2.0) :

2 -2 V,

Lx 3 X2 x lj -3k ZE2 31 12 OZI

(5.6.0) 30-.

t 300V2

E3 000 eZ2

which can be rewritten as:

v=f {Ay) (5.7.0)

where, LfJ = Lf, f2 f3 f4 jT (5.8.0)

f 1-3 (2ý X

)2+23 (5.8.1)

f2 -x 1)2 (5.8.2)

f3 = -2(. 1)3 + 3(x)2 X

(5.8.3)

f4 = -X((. 1)2 X - X

(5.8.4)

77

The sign convention and the cubic polynomial used in describing the displacement for bending in the x-y plane is similar for both bending in the z-x plane and the torsion warping problem about the x axis. Therefore the corresponding displacement functions in terms of nodal values are given by:

21 -2 1 W1

-3z -2Z2 U _Z2 0 W OLx 3 x2 x jj II

yl (5.9.0)

g3 0 t3 00 W2

E3 000 eY2

and 2

-3t ex =LX3 x2 x lj 1

£3 0

-2 exl

-Z2 a -2Z2 U exl

(5.10.0) 00 OX2

00 01 OX2

Similarly, the equations corresponding to equations (5.9.0) and (5.10.0) also given by:

W =LfJIAZI

ex =UJUaýI (5.12.0)

where values of Lf-J in equations (5.11.0) and (5.12.0) are identical to values of LfJ in equation (5.6.0).

78

From the previous section, the shape function which describes the displacements in the x-y plane is as follows:

V=LfJ {A y) (5.7.0)

where, LfJ= Lf 1f2 f3 f4-J T k.

fl =1- 3g)2 + 2(X)3 ZZ

f2 z -x ( g)_ 1) 2 91

f3 = -2 (. 1)3 + 3(X)2 .ZL

f4 ý -x «. 1)2 - (X )) 2t

and

Y Vi OZ1 V2 e z2

By differentiating equation (5.7.0) with respect to x:

V=Lf 'J U (5.13.0)

I where, V= Lf 1 f2l f, fit T 3 4J

and f1= §-X + ý-X2 (5.13.1)

z2 t3

fý = (- 3x2 +U- 1) (5.13.2) 7

f3' = ý-x + §-X (5.13.3) p3 jt2

f4 = IX2

- X + (5.13.4)

t2

79

Similarly by differentiating equation (5.13.0) with respect to x

V =Lf"J {Ayl

where, Lf"J 11 fit fit fiojT Lfl 234

and f, 6 + 12x

z2 y3

fit 2 6x + 4)

fit 12x +6 3 g, 3 t2

fit 4 6x +2 7

Jz

(5.14.0)

(5.14.1)

(5.14.2)

(5.14.3)

(5.14.4)

Finally by differentiating equation (5.14.0) with respect to x:

I where, Lf "-

and ei.

f oil 2

3

4

Vill =Lf "D IAyI (5.15.0)

Lei' f Ill fl, I f-! jT 234

12 (5.15.1) 3

(5.15.2) ;2

. 12 (5.15.3)

. t3

-fL (5.15.4) 0

The sign convention used for bending in the x-y, y-z planes and the torsion warping problem about the x axis are identical to one another

80

(see Table 2. T. 0). Consequently, the same relationship must exist for the shape functions and their derivatives.

By substitution of the shape functions and their derivatives into the

corresponding terms of equation (4.75.0) and integrating within the

element boundaries the associated potentials can be calculated.

Thus, the corresponding term for bending in the x-y piane in equation (4.75.0) :

11 Elz LVO Z-J1{-f'V'}Lfýjdx {ve I z

(5.16.0)

-L LVO ZJ (12X-6. t)" L (12X-6t)-(6Px-4t2)-(12X-6t)-(6tx-2. t2)jdxlv 1

Wcz

(RX-42)

1-(12X-6t)

C(6tx-20 (5.16.1)

Hence the potential becomes:

E IZLVj e Z, V2 0 z2j L2 Synnetric V1 3

-6 4 F Z1 T -

-12 6 12 - v 2 T 3 T2

y3

-6 264 Oz2 L z2 z2 I

(5.16.2)

By a similar process the following contributions may be found:

81

zw f EI LW eyJ[{f")Lf; Jdxl 1,1

ywy

- HyLW1 0 yl W2 0

y2 J 12 [.

t3

-6 2

-12

jt 3

-6 2

Symmetric Wi

yl

6 12 w 2 t23

264 y2

12 ex} yf EILGXG; JE(f0"lLf"xJdxl{äxi

(5.17.0)

(5.17.1)

(5.18.0)

ErLO 0'6 12 Symmetric 1 x2 Ox2j X1 X1 .X Z3

-6 4 Ix, i k (5.18.1)

-12 6 12 x2

, t3 z3

-6 264 2X

OX2

Similarly, for torsion about the x axis the corresponding term in

equation (4.75.0):

.tý Ox 1.1 GJLOxOP [{fexlLfexJdxl {O 11 (5.19.0) 2ox

GJ LOX, ex'l Ox2 Ox'2J Symmetric OX1 5z

-1 2Z exii 10 15 (5.19.1)

-6 16 Ox2 5z 10 5z

-1 1 21 x2

!I 10 30 10 115]

82

Similarly, the term which corresponds to the direct action due to axial force in equation (4.75.0) is:

I f EALUJ[ffu'lLfýJdxl {u) (5.20.0) 0

EA LU1 U2J p. -1 u (5.20.1)

11u2

The integrated terms calculated so far can now be used in preparing the conventional elastic stiffness matrix [K] and this is found in

chart (5.1) of Chapter 5.

Similarly, the term which corresponds to buckling in the x-y plane due

to axial force Fx in equation (4.75.0) is:

1z f FXL V6ZJ [Ifv'l Lfpdxl le I 2o z

Fx LV, ezl V2 GZ2-J -6 Symmetric Vi 5z

1 E) Zj 10 15

6 -1 -6 v2 5z 10 5x

6 z2 L 10 30 10 15 _j

(5.21.0)

Likewise the term which corresponds to buckling in the x-z plane due to axial force Fx in equation (4.75.0) is:

11w 2.0f

FXL W eyJ V f; j L%Jdx] j01 (5.22.0)

83

Fx LW1 eyl W2 OY2J ': -6 Symmetric W, 5. t

1 -2-t TO 1-5 yl

6 -1 -6 w TO 5z 2

-2-E ey2 16 jo- 10 15

LI

(5.22.1)

The term which corresponds to torsional buckling about the x axis due

to axial force Fx in equation (4.75.0) is:

k f Fx SoLe&J [ffex}LfeýJdxl 10*1

2ox

FxSoL-exl6x'lox2ox'2J -6 Symmetric exl 5z

1 -2t ax'l 10 15

6 -1 -6 i ex2 E t 10 51

1k -1 -2Z e x, 2 10 30 10 15 L

(5.23.0)

(5.23.1)

The term which corresponds to lateral torsional buckling due to shear force Qzj in equation (4.75.0):

2 Qzl Lv ezJ {f")LfexJx dxl 0 -10 vfx

0 x-i

(5.24.0)

84

qzl LVj Ozi V2 ez2-J -1 10

5

10

Lý101-

0 10 10 Gxl,

-X2 -2 0 37 -ýT- ex'

0 -11 -Z 10 To- e x2

. t2 -gL _k2

- - 1 e 2 30 1 o Tg x

(5.24.1)

The solution shown above cannot be used directly in the geometric stiffness matrix [N], since the matrix [N] should be symmetric by definition. However, by making the following changes, the symmetry is

achieved while maintaining the absolute value unaffected and this is

shown by equation (5.24.2):

I J1 QzlLVlezlV2oz2exlexIox2ex2

I

The term which corresponds to

force Qz2 in equation (4.75.0)

0 2U -2-U

1 2,2 0 TU - -0 - TU

10- 11 z TU 2U -M

I t2 9 R, jt2

IzII 2U 10 - -20 -M 2-0 17U 2-U -M

Y2 0

-6-0 0 zu

1x 11 9z 2-U - TU -M- 2U

0 t2 (5.24.2) 2U I ateral torsional buckling due to shear is:

0 1 fzQZ2 LVOZ-J {Fvl&l LFOXJ (. Z-x)dx 2

, x, 0

xi (5.25.0)

vI

ezI

v2

OZ2 ex,

X'l

x2

6 X'2

QZ2 LVl6zlV2oz2oxloxlex2ex2j 11 Z

ro

L

vi

6 zl

v2

0z2

exl

a.,

1 72-0

0

0 2-0 10 TU - -0

11 9z 11 1 2U -; FO -W TU

k t2 0 M- -2U -M 20 7-0 - IU - 10

Y2 00 L zu

and the term which corresponds to lateral torsional buckling due to

moments My,, My2 in equation (4.75.0) is:

1f (myl - MY2) LVOZJ[{f"1 Lfexjdxl x (5.26.0)

(Myl-My2)LVlezlV2oz2oxle'xlex2o)'

1

(5

X2

x2

. 25.1)

6 16 1- Tri 2u TUT w 11 29,1 Z *zu - 3u -, 29 uu

6 16 1 TU - 2u -TU -2u

2 11 2x

6 11 ro 29.

, zu - 3u*, » -Z m

Tri - 2u -IM- -Z 9.1

and the term which corresponds to lateral torsional buckling due to shear force Qyj, in equation (4.75.0) is:

V. 1 6 zl

v2

z2 0

xl

0, xi

11 0 x2 1

[0 x- 2j

(5.26.1

11 Qyj LWay-l If " )LfexJ x dx (5.27.0) 0w0 XT

86

Qyl Lwl eyl w2oy2oxlexlex2ox

k. 11 9k 11 z 20 2u -2-6 1-0 .t-

t2 ko

k - 0

m - To- TU o 6U

0 _0

Z t Ti 7u 0

gt 2u Z2

- -20 g m

t2 w lyl

11

20 Z zu

1 - -2-0 0 WZ

k Tij 0 L k2

Tu -ff 8

- Y, , ex1

ex, 1 ex2

8 6x2

7- (5.27.1) The term which corresponds to lateral torsional buckling due to shear force Qy2 in equation (4.75.0) is:

11 f QY2 Lw6yJ 0

QY2 Lwleylw26y2oxloxlex2eý

EIfWI LfoxJ (. Z-x)dxl x (5.28.0) 6 x-

- 2u «zu «ZU 0- w 1 gt £2 A, z2

- - 0 6 0 yl

2u - 0 w 2

0 2u - y2 11 9x 11 L 6 - 2-0 ww TU x t2 t 0 'x'i

2u m- 2u - 1-0 x2 0

L £2 0 t2

60 ex2

(5.28.1)

and the term which corresponds to lateral torsional buckling due to Mz1, Mz2 in equation (4.75.0) is:

f (Mzl - MZ2)Lwoy-JI{fwl Lf exJdxl ex (5.29.0)

x'

01

= (M 81 zl-Mz2)4w16Y1W20y20x 1 Oxl 6X20X2

6 16 -

1 v io ict 20

yl 6 16 1

Tul 2u- Hig. Tu- w2 1 t- 11

--- - 2t

20 0 20 30 y2

-6 11 -

6 1 A Tui 2u Tui 1

mi- 21

- -Z 1

M- Z

ur xl 6 1 6 11

x2

20 ei- - 1

M- 21

- MT- 8 0X2 j

(5.29.1)

The first term which contributes to lateral torsional buckling, due to torsional load Mxj in equation (4.75.0) is:

Mxj LTO zJ EI fv'l Lf" J (1 - -I) dxl fw1 (5.30.0) w 9. ey

= Mx, LVp ZlV2 ez2Wl IýlW2 3 4 3 1 TO. t 2 TOi - -fOiL2 lo v1

7 13 7 4 - zl Mot ur 20x 3ö

3 4 3 2

1 -To-

v 2 3 7

- 3

- 1 -

0z2 2ri n mi t W

3 7 3 3 n in2 7ZUZ- - "Rit-2 w1

-4 13 4 7 1-ö-£ - ur 'Rfz- Zo yl

3 7 3 3 -1 0)-Z2 - rot w2

1 'TU-

4 - 3u

1 lo

1 30 0 y2

(5.30.1)

88

The first term which contributes to lateral torsional buckling due to Mx2 in equation (4.75.0) is:

I m LVOZJE[fv'l Lf Iiii dxl w1 (5.31.0) To Y2 w16y

Mx2 LViOzi V2'z2 ý416ylW2 0y

-3 3 3 7 TO-Z2 -2U- TO=I -20e

1 4 Tal - -75, 1 -01 75'

33 3 7 'TUT2 - 201 T -0 .

t2 -FT- t 47 4 13 IT - 7-0 M, 6-0

3 1 3 4 Toy2 - lot 1 oy, 2 To-

3 1 3 7 ME - -70 201 -60

3 1 3 4 "MT W t To- - -I"u

7 4 7 13 Mi ýx ME -ff

(5.31.1)

The second term which contributes to lateral torsional buckling due to torsional load Mxj in equation (4.7.50) is:

k f Mxj LWeyJ Ufw') Lfv'! J (i - X)dxl Iv (5.32.0)

r0z

Mxl LWP X1 W2ey2VlezlV2 e ZZJI

-3 -7 3 3

JOZ2 20t JOZ2 20Z 13 - 4 7

lot 60 lot 60

-1- 7 3 3

lot2 201 IOZ2 201 .1 4 1 1 1-0 TO T-0 30

-3 4 3

1 OZ2 lot 1012

7 -

13 -

7 - ý Ot ý0 ý Oz 3 4 3

ýOZ2- Tot ýOt2

3 -7

3 20t 60 20t

0

30

10

(5.32.1)

v1

ez

V2

ez 2

wI

ey,

W2

OY2

Wi

eyl

w2

OY2

v1

ez

v2

OZ2

89

The second term which contributes to lateral torsional buckling due to torsional load Mx2 in equation (4.75.0) is:

-1t 'ILf 11 1X dxl{v 1 (5.33.0) f MX2 LWeyJ[{fw V- 10 z

Mx2L-Wleyl W2ey2Vlezl V2

3 3 7 10Z2 'MY-£

1 4 Tr9, -'3 0

3 7 1Z 02 10Z2 2 Ö9.

4 7 4 13 i 0- u-o -f-u- -w

3,134 To-k2 -10JE - -fot2,10 w M

eyl

W2

ey2

Vi

ezl V2

ez2

3 13 7 lot 'jo -2 -6 60 3 13 4

lot Tot -i UP lo 7 47 13

2u-t - -ir - Mok - -6u-

(5.33.1)

The first term which contributes to lateral torsional buckling due to

warping moment Mxxl in equation (4.75.0) is:

-f MxxLV ezJ E(f v1 Lf W". ý] (1 - -X, ) dx] {w 1* (5.34.0)

2o0y

6363 Mxxl LV16Z1V26Z2W1BY1W2GY2-j 3 -1) T 35-2, -5-D-y -T. -7 -N - A' ýA, &. Z-

7-2 ZT 2-Z2 4£

6 3 6 3 29.7 - -2F -2-P 2Z2

7ii -47 -i Z-2 4£ -6 161 ý7 -2-kn- 1 -ý -V 2.0 3131

2 Y, 2 TTZ ý-o 4t 6161

43 ý-e -k3 2. t2 (5.34.1) 3= 131 2Z2 4z - 20 4. t

Vi

e, l v2 Oz2

Wi

0 Y11

W2

6y 2i

90

The first term which contributes to lateral torsional buckling due to warping moment Mxx2 in equation (4.75.0) is:

mw XX2 LVOJ f V') Lf dx j{1 (5.35.0) Wtey

=M xx2 L V, 0 ZlV2 ez2Wl eylW2 6 3 6

21 - - 27 7

6 3 6 2Z -3

6 1 6 2. z 3 T-

jtZ - -2- 3 3 2

3 1 3 1 N 7 7- 1 YET 71 6 1 6 1

Zill YET YET 3 1 3

Y7 7-1 ME -4 T

3

1 47£

I

v

ez

v

OZ21

wi

ey,

W2

eY2

(5.35.1)

The second term which contributes to lateral torsional buckling due tc

warping moment Mxxl in equation (4.75.0) is:

v -f Mxxj LWeyJ[[ fw') Lf"! J (1 - . 1) d*x] 1, ) (5.36.0)

20vz

Mxxl Up y1W e6 -3 6

2ey2V1ez1V2 zý 71-3

Z. t2 -2F F 71-2

636 213 2Tf TU

7- I 6 6 1

3 NET -2-E7 -3 3 1

. t2 t2 4L -6 6 1 'ý j -L3 -2 TT -f FT 212 (5.36.1) -3 1 3 1 T-

jt 2 - 7F TI-2 U

3w 2X2

1 --4-t t ey,

3 «2 w2

GY2

v1

6 zi

v2

6 z�l

91

The second term which contributes to lateral torsional buckling due to

warping moment Mxx2 in equation (4.75.0) is:

1 Jt VI f MXX2 LWOý [Ifwl) UP 2L dxlfe (5.37.0) z

=M xX2 LWP ylW2oy2Vlo ZlV20 6 3 6 3

2 Wi

27T -Ti- ey,

6 zu

3 -7 LT

6 3 w 2

1 -7 =Z

1 -4 -X 7Z7 - Tz- ey 2

-6 -L3 6

"Z 1 v 1

3 Yk- r -

3 Tz- 7 az 1

6 77 -

6 TY--S v 2

3 22

1 3 ez

(5.37.1)

Equations (5.16.2), (5.17.1), (5.18.1), (5.19.1). (5.20.1) (5.21.1),

(5.22.1), (5.23.1), (5.24.1), (5.25.1), (5.26.1), (5.27.1), (5.28.1),

(5.29.1), (5.30.1), (5.31.1), (5.32.1), (5.33.1), (5.34.1), (5.35.1),

(5.36.1) a nd (5.37.1) are now in their explicit form representing individual potentials and are used in preparing the matrix [N] for the thin-wall ed element shown in Figure (4.1.6). The final form of the

geometric stiffness matrix is illustrated in Chart (5.2) of Chapter 5.

92

The Linear Stiffness Matrix:

a Ul bv

wi SYMMETRIC I d ex I

ef eyl

-9 k ezi

-h ex I -a au2

-b 9bv2 ecw2

-d hd ex 2 -e ef OY2

-9 9k eZ2

-h kh X'2

where, a=1 EA Z*

b= 12 EI .i z» Z

12 C

9.3 Y

d L2 EC + -ý Gk z3 w 5t

e=-6 EI k2 Y

f= AEI

.ZY

zi -6 EI Z2Z

h J- EC Gk £2 w 10

i=2 EI 2Y

i= -ý EI Z

k EC Lo Gk 2'w- «3 -o

EC +2 w iu tGk

CHART (5.1 )

93

The Geometric Stiffness Matrix:

a

a

-c -d So a

-e fgb

fhib

jk Sof 11ms0b

-a c

-a d

op -SOa

e' ft

f hl y

-n' k' So f

e -f n

-f -h -k qr -Sof uvw

vIu si

wo sis sou

where, a=6F 31 x

2 1,9

11 Q6 2U Qzl - 'W z2 -M (Myl -my2)

IQII+6 (M M yl + 'M Qy2 TUTY, A -Mz2)

Sa=6

SYMMETRIC

a

a

-0 -P soa

-e ' -f xb

-f -h' -V b

n' -k' -Sof ill m" Sob

u

w

ex,

GY,

ozi

exi

U2

V2

w2

ex2

GY2

ez 2 ex 2

371 e 7rjt M

x2 + ME mA+I Ot Mx2

+4m+3 ' uo A YZ-7 (Mxx2+Mxxl)

2U (Mxx2-Mxxl)

ei =7 tle 3m4M 7rt x2 + 72N A+ TU x2

+1m+3 (M +M WA YJF xx2 xxl)

+-1 2. t'r (mxx27 Mx

x

94

IF TU

19 11 TU Qyl +M LQy2 + 2U (Mzl-Mz2)

-1m+4m+3m+7m+1 (M 3 T5lx2 TO-t x1 Tol x2%. -2rt xl YLT xx2-Mxxl) - Ni (mxxemxxl)

hl= 1M4m+3m+7M+1 (M m3 (M JU-t A' 'RY-t x2 M- x1 TO-L x2 -2 t-- 7 xx2- xxl) +Tiz xx2+Mxxl )

i= -L Q+90 11 (Myl -My2) 10 zl TU z2 +Z

9A+zQ 11 (M To- A IU z2+ 2U yl -MY2)

I+I "Fo 'Qz2 -2-0 (myl-my2)

11 k= -ro Q y2 + TOY (Mzl-Mz2)

QI 0- yl +M (Mzl-Mz2l

0 Tij x

z2 Q t2 2x (m Wo yl - TU QY2 30 zl-MZ2)

t2 Q z2 2x (m TU yl -ZU Qy2 3u zl -Mz2)

m= k2 Q t2 Q- 2t (m -m ro- z1M z2 -3U yl y2)

t2 Q jt2 Q 2t (m IM zl -M z2 - '30 yl -My2)

95

Sob S2Y, F ox

yl -M z2 cm

y2)

Q1 (M -M ) 0 ZI - NO

yl y2

1Q6 TJ zl 'Z z2 + 10. t

(Myl-My2)

= 11

+1 +- 6 (M P 2U Qy 1 TOY Qy 2 IN zl-Mz2ý

zz1 q -T Qyl -' M Qy2 - IN (Mzl-Mz2)

Q1 To- QZ1 , 2U z2 - 'M (myl-my2)

(M m -6 -0 QzI' TOY YI - y2)

-ý! Q1 60 z2 + TOT (Myl-My23

+i+I (M m 20 Qyl '70 Qy 2 2U zl- z2)

1F 30

SuS. -- tF

00Zx

v4m7M7m -ý mII (M 'Z x2 - Wo A- -0 x2 - 30 A- '4-tCMxx2-Mxxl) TF xx2-Mxxl

I

96

7m+4m4m+7m+1m ZU x2 MA+M x2 TOY xI xx2-Mxxl) +U (Mxx2-Mxxl)

;2 ,Q -FO yl +W (Mzl-Mz2)

x 9z Q1 11 (M Fof yl - To- Qy2 YOT zl-mz2)

y=-ZQZQ1 (M 1-M 2) zl ' IU z2 + '20 yy

CHART (5.2)

97

CHAPTER 6

THE MID POINT TANGENT METHOD

In Chapter 4, the difficulty in adopting the uZe of a straightforward eigenvalue approach to solve buckling problems which are subjected to

generalized loadings and end constraints was shown. However, if the

structure is loaded incrementally, the non-linear path may be traced by a suitable method which solves the conditions for equilibrium at every load step. One such method is known as the mid point tangent

method.

The solution procedure is as follows. At the first loading increment,

the geometric stiffness matrix [N] makes no contribution to the state of equilibrium of equation (4.78.0). This is mainly due to the fact

that the displacements are assumed to be small. Thus, equation (4.78.0) is simplified to:

{FI = IKI (A) (6.1.0)

Initially, the terms in the stiffness matrix [K] are calculated from the parameters of the problem under consideration (i. e. length, cross sectional properties etc). Then, by solving equation (6.1.0) for a specific loading incremental vector {&F), the unknown displacement

vector {SA) and the associated reaction force vector {AR) are calculated.

Prior to the next loading increment, the new geometry of the structure is obtained by combining appropriately, the displacement vector{6A) with the. undeformed geometry of the structure. This enables us to form the new stiffness matrix. Similarly the new geometry, along with the reaction force vector {ARI , is also used in formulating the geometric

98

stiffness matrix [N]. Subsequently, the two matrices (K] and [N] are inserted in equation (4.7i. 0) to examine the equilibrium contribution for the next loading increment{&F"

Mathematically, for the ith iteration, the method described above can be expressed as:

IAF)i = [Kli-l {Ali - [NIi-1 fAli (6.2.0)

where (AF)i is the ith loading increment *

[K]i_l is the stiffness matrix whose calculation is based on the structural geometry and the (i_, )th step

[Nli_l is the geometric stiffness matrix obtained by using the

accumulated reaction forces up to the (i-1) increment

and {Ali is the displacement vector corresponding to the ith loading increment.

However, equation (6.2.0) does not accurately describe the equilibrium

condition at the ith loading increment. This is mainly due to the fact that both [K] and [N] are calculated with the aid of the column length*and orientation of the (i-1) increment. However, a better

approximation can be made if the column lengths and the orientations

are estimated, based upon the average values of the ith and i-1th

steps, i. e.

{Ali = 0{, &Ij-l (6.3.0)

where, a is the ratio of the load increment at step i and that at step i-1.

The displacements at the ith increment can now be estimated by using the average stiffness values between the ith and i_lth increments.

99

With respect to reference05) it can be seen that a better estimation for the stiffness and the geometric stiffness matrices is found by predicting the structural geometry half way through the i_1th and ith increments. This approach of modifying the structural stiffness halfway between two increments is known as the mid-point tangent technique.

The above method must be i mplemented wi th caref ul ly chosen I oadi ng steps, and the collapse or the unstable condition of the structure is observed by comparing the displacements calculated at the ith increment to what was calculated at the (i_, )th increment. Complete discussion of the results obtained from the method is found in Chapter 10.

Implementation of the Mid-point Tangent Technique:

The non-linear behaviour of the structure is now determined by solving a sequence of linear problems. The total load is applied as a sequence of sufficiently small increments such that during loading the structure is assumed to respond linearly. This can be understood with the aid of the diagram shown in Figure (6.1.0).

Z'

rl .M e3

Lood[P)

Fig. 6.1.0

FIGURE 6.1.0

100

The step by step approach required in implementing the mid-point tangent technique is as follows:

1. Assuming the load increment, &pl, is in the linear elastic region. the equation,

{Apl) = [KI 0 {äul) (6.4.0)

is solved.

2. By predicting the new structural geometry half way through the

next increment, the nodal coordinates are updated. Thus,

(coord)ll= (coord), + '&Ul '6P2 (6.5.0)

2 Ap,

3. By recording the reaction forces calculated in the first increment

and inserting them in the geometric stiffness matrix

appropriately, the equation,

'AP21 = [K], 'AU21 - [N]l (AU21 (6.6.0)

is solved.

4. By predicting the new structural geometry half way through the following increment, the nodal coordinates are updated. Thus,

(coord)2 'ý (coord), + AU2 (6.7.0)

(coord)2'ý (coord)2 + AU 2 LP3

(6.8.0) 2 'ýP2

101

or in general

(coord)i = (coord)l_, + äul (6.9.0)

(coord)i, = (coord)i + Aul Pi+I -7- -Api

41

Equation (6.10.0) is used to modify the structural geometry and by

solving equation (6.11.0), the required displacements at the (i+, )th loading increment are calculated

{Api+ll= [K]i {Aui+ll - [N]i tAui+ll

The procedure is repeated until the desired convergence of the load displacement history is obtained. However, this method does not iterate within an increment for equilibrium. Thus the solution is dependent on sufficiently small increments.

102

CHAPTER 7

ORIEKTATION OF THE LOCAL AXES SYSTEM TO THE GLOBAL AXIS SYSTEM

The basic beam element is a three dimensional object. Thus, to define a beam element fully in space, a minimum of three nodes are required. In Figure (7.2.0) the axial direction and the plane of cross-section may be obtained by defining the coordinates of node 1, node 2 and node 3.

Under a generalized loading condition a thin wall beam will have seven degrees of freedom per node when referred to the local axis system (see equation (4.78.0)) or nine degrees of freedom per node if a global axis system is chosen (see equation (8.14.0). The solution procedure utilized in the mid-point tangent method (see equation (6.11.0)) presumes an overall definition of the updated coordinates of nodes 1,2 and 3 prior to the application of the next loading

i ncrement. The method of obtaining the updated positions for nodes 1 and 2, has already been established inChapter 5and in this chapter proposes the method to establish the position of the third node.

Let us recall the theory developed in Chapter 3 for predicting the deformation of thin walled cross sections under generalized loads.

ý-j )-Slnpl

FIGURE 7.1.0 Clen P-cos( P-1 u

103

In Figure (7.1.0) point C is the origin of the coordinate axes systems at node 1. Point 0 is the shear centre and N is an arbitrarily chosen point on the cross section. Due to loading, the cross section will undergo translational and rotational displacements. The magnitudes of these displacements may be estimated by inserting appropriate conditions into equations (3.4.1) and (3.4.2):

k,

Uno - (c - e. ),,

Uco + (n -e 11 )0 t (3.4.. 2)

(Note: equations (3-4.1) and (3.4.2) represent the combined translational rotational displacement of an arbitrary point defined with respect to axes nand c).

.. a ol EN

r2' PodeZ(

1.2

1

C'

FIGURE 7.2.0

104

Let us examine the kinematic behaviour of thin walled beams in space. In Figure (7.2.0), the displacement of point N can be decomposed into its components. The final position of point N' is obtained by the

axi al di spl acement &x followed by translational displacements Ur. N, and UnN" pa ral I el to the ý and n axes respect i vely. Note al so that the

effect of axial twist 6C is automatically accounted for in equations (3.4.1) and (3.4.2). The aim of the exercise is to predict the

position of N", which is an arbitrary point on the plane normal to Cj- C2 in which C2 is the new position of node 2 relative to node 1. Thus, if point N is conveniently chosen on the C axis, by connecting C2 and N", the new position of the principal axis is predicted. Simultaneously N" becomes the new third node.

The coordinates of node 1 with respect to (Cnc) system is given by,

U, 2 = Ax (7.1.0)

Substituting n= O, c =0 in equations (3.4.1), (3.4.2)

Unl, 2 ý UTIO +er. e& (7.2.0)

u-e (7.3.0) 0,2 U; o

As explained previously, let us place point N" conveniently on the C

axis. Then the coordinates of N, prior to any form of deformation has taken place, are given by,

EN = JE

=0

cN=I

105

where, Lis the length of the element before deformation. By applying the above coordinates into equations (3.4.1) and (3.4.2):

Zm, =k+ Ax (7.4.0)

UnN, = Urlo - (Z -epZ (7.5.0)

CN, ý- li v-en0C (7.6.0)

The vectorial movement of N" with respect to the axis system

at node 1 is illustrated in Figure (7.3.0)

I*, &,. W'"CO

FIGURE 7.3.0

106

A The unit vector n, in the direction of Cl C2 is defined as, ru

+ IAX

n Unl, 2

UC1,2

where, ol 2+ U2 +u2 x nl 2 cl 2)

Applying-vector algebra to the triangle C2N"N',

4-

C2N' = C2N" + N"N'

C-ýNll + (C-, ýN' .; ); 22 ru

N' ZY, + Ax - (L + AX; N" (I + AX) but C2 C2

I 'UnN' - U711 21n-U nl, 2

Uý N' - UO 2JC- UC1,2

4

by substituting C2"N', C2 - N" and in equation (7.8.0),

I +A xc0 Z+ AX L+Ax

UnNI n+ UnN I- Uril 20 Unl, 2 -aA Unl, 2

UCN' UW - Ucl 2 UU 2 UU, 2 Lj

(7.7.0)

k.

(7.8.0)

107

or mr Z0 £+, ä x £+A x k+AX

9 TI Lý N' - Un 12*% Unl, 2,1 2 Unl 2 -. UilN'

UCNI - UC, 2 UZ1,2 UU 2 u08

%ý -i 1 (7.9.0)

Note that U. 1,2, Uý1,2, U,, N, and UcN, are obtained by inserting the corresponding coordinates into equations (3.4.1) and (3.4.2).

Thus, by substitution of Ura, 2, Uia, 21 UrJVI LýN, along with a2 into

equation (7.9.0), the vector

can be calculated.

C The vector n now provides the new location of the third node for

the next load Cj increment on the local C axis.

By considering the global coordinates of node 1 (point Cj) and the direction cosines used in the local-global transformation matrix (equation 8.14.0) used in the previous load increment, the new position of node 3 with respect to the global system is established and this is illustrated as shown in equation (7.10.0)

xc xa Xi

y [TG1]T [TG2]T, n' + [TG1]T Ya' + Yl (7.10.0)

z za IiI Zi

108

I x

where, he vector y is the equivalent representation of the local

vector C1. z

For details of equation (7.10.0), see proof of equation (9.16.0) in Chapter 9.

Assembly of Finite Element Equations:

The element stiffness and the geometric stiffness matrices shown in

equation (4.78.0) have been evaluated with respect to nodal displacement, referred to the local coordinate system &, n and However, in practice, structures are constructed with a variety of shapes and lengths in arbitrary orientations. Nevertheless,. if a global axis system is chosen to measure nodal forces and displacements, then the solution becomes a relatively simple task.

---- 11 t

x

12

FIGURE 7.4.0

109

Figure (7.4.0) describes a typical arrangement of a structural assembly, in which the orientation of the two local axes (Cl, n1p ý1) and (& 2* ý2, C2) are defined with respect to the global axis (x, y, Z). Initially, the element and the geometric stiffness matrices for the system shown above are calculated with respect to their local axis system. However, the combined effect of both elements cannot be

analysed unless both elements are referred to the common set of axes (x, y, z). Let us assume that a relationship between the two axis systems can be obtained with a suitable transformation matrix [T], as shown below.

Thus "ElC [T] I UEIG

or LU T =LUEjT [T]T (7.12.0) EJLI G

where 1 UE) L is the displacement referred to the local axis system 1 UE) G is the displacement referred to the global axis system [T] is the transformation matrix

Since the energy in deforming the structure under applied loads is independent of both local and global axis systems, then the work done in both systems is:

TT 1 LUEJL IF G) (7.13.0) 2LI=

lL UEJG IF 2

but from equation (4.78.0)

{FLI = [K]{UE'L - [N]'UE'L

or [[K] - [NIIIUE)L

110

Now substituting. {FLI in equation (7.13.0)

TT LUEJL HKI - [NII 'UE'L ý UE'G {FG' (7.14.0)

Substituting equation (6.12.0) forL UE T

in equation (7.13.0) yields JL

T IF (7.15.0) LUEJG [T]T EEK] - [N]l 'UE'L "ýLUEJG G)

T By comparing the multiplications of LUEJG on both sides of the

equation (7.15.0) yields,

[T]T EEKI - ENII 'UE'L 'x IFGI (7.16.0)

from equation (6.11.0), (UE)L "ý ETIIUE)G

.0. IFG' [T]T [EK] - END (T] IUEIG

or {FGI [KIG {UE)G (7.17.0)

where, [KIG z [T]T [EK] - ENII ETI (7.18.0)

The term [KIG in equation (7.18.0), provides the necessary transformation between the local and global systems.

ill

CHAPTER 8

FORMULATION OF THE LOCAL GLOBAL TRANSFORMATION MATRIX

In Chapter 7 the relationship between the local and global coordinate systems was discussed. In Figure (8.1.0) a special case is illustrated, in which the local axial direction& of the element is parallel to the global axis x'. During loading of the structure, the axial load was placed at 'A' while the transverse forces and moments were applied at point B. The location of points A and B were referred with respect to the global system as shown in Figure (8.1.0).

C( Yc'.. c

CG

FIGURE 8.1.0

II-1 -1 where, (yc, zc), (y ,z) are the coordinates of the centroid and the

shear centre of the cross section, measured with respect to the global axis system. Also (e

Tj eC) is the position of the shear centre referred to the local coordinate system, situated at the centroid.

112

As established in equations (3.4.1), (3.4.2) and (3.12.0), the displacement of an arbitrary point with respect to the local axis system is given by:

UT, = Uno - (ý

-e le

Z (3.4.1)

Uý = Uýo + (Ti - e�)e 9 ý> (3.4.2)

uz = Ug - (ý - e, )e-0 - n (n - e, )8� - (wOs - wo DoPý (3.12.0)

By differentiating equations (3.4.1) and (3.4.2) with respect to C

6n =GTlo + (n -en) eý

0; = Gco + (c - e, ) e (8.2.0)

If the loading is applied at points A and B, then a rigid connection

must exist between the load points and the cross-section.

As shown below, in predicting deformation let us use the global system

instead of the local system. Thus, the displacement in the global y' direction becomes:

y, = UYI 0- (Z' - z')OX# (8.3.0)

by applying the coordinates of point B, (i. e. the point at which the

transverse loads and moments are applied)

113

Uy iB- Uv, 0- (bz o- il ) Ox, (8.4.0)

Since the local and global axis systems are parallel to each other,

UyU rio (8.5.0)

UY'B Uno - (bze - ; P)Oxi (8.6.0)

U r, 0 = Uy ,- (r - bz i) Ox t (8.7.0)

From now on let us denote the displacement in global Y' as uyi instead

Of uy'A' Similarly, the displacement in the local ý direction is,

UCO = Uzi + (Y' - b; )Oxl (8.8.0)

and equations (7.1.0) and (7.2.0) become,

OC 0= Oz ,- (P - bz') Ox 1 (8.9.0)

Orp = Gy, + (7 - bý)Oxe (8.10.0)

Finally, the axial displacement UCO (i. e. the point where the axial load is applied) is,

114

Ur =Ux, - . (z '-e rc z-)ey- - (yc - ey-)ez' - (%CcG-%ACG)OX

(8.11.0)

where wDCcg and L)DAcG are the sectorial coordinates of the centroid and point A, referred to the global origin cG and the reference point D on the cross section.

ý11 Let us concentrate on the general case, where the two axis systems are independent to one another as shown in Figure (8.2.0). Furthermore

the basic parameters shown in Figure (8.2.0) are the same as that illustrated in Figure (8.1.0) except in this situation the axial direction of the local system is not parallel to that of the global system.

-n -- gw, ) 9 RL

1aj

tvyji

Poi nt Coordinates with respect to

x, y, z system

(ex » ey, ez) B (bx, by, bz)

c (xcp YC9 ZC) 0 (x-. Y». i)

FIGURE 8.2.0

115

By using the direction cosines, the relationship between the (x, y, z) and the (x', y', z') system is found to be:

x Cosaxlx COSOIXI y Cosaxlz x

Y, Cosa y Ox Cosa yIy Cosa y lz y (8.12.0)

zi Cosaz, x cosazi y Cosazlz z

Since'the axis system (x', y', z) was drawn parallel to the( E, Ti c) system,

Cosa&x Cosa EY Cosacz x

TI Cosanx Cosa ny

Cosanz y (8.13.0)

Cosacx Cosa cy Cosa cz z

(Note that equation (8.13.0) only describes the relationship of direction cosines and not the absolute value of coordinates).

By substituting the result of equation (8.13.0) in equations (7.7.0),

(8.8.0). (8.9.0)9 (8.10.0) and (8.11.0), the final form of the transformation matrix is established and this is shown in equation (8.14.0). The square matrix (M) in equation (8.14.0) represents the transformation of displacements at one node only. However, the

generalized beam element always consists of two nodes. Thus, the

overall transformation of the displacements at nodes 1 and 2 is:

Al' [Tlll 101 Aj

A2 101 [Tlll 'ý2

A

116

CD CD cc,

LIP &IP fs

UP 6 N

ts %A C) Ln in Iýp (D

0 0 0

0

>1 OA 0

LIP LO ts 6 >1

0 en C) CD eD 4A CA W. Cý (D

+

0 0 0 ts

0 + V3 177 u L. P

t th

ýt, : 0 cl u V)

x 0 LO

15 cs x 4 P4 U

Cp U) in LIP C) ýp C 0

0 0 0 d 15 ts u + w

2 Ln

0 Ln 0

W)

0 N +

. u u u Q)

N N + x C

.0 .0 15 >% u CA

+ + 0

ts LIP

0 N N N >, >I ts W

u

V) (A ts

Lo ts

G7 I'LP C) (D C) LP C,

ts 0 u +

x 0 u

0 tn

11 CA Vb . .

2ý

0 0 0 0 0 X rlý u u u u u W

+ 0

>1 + + x c

u 0 u

cli LIP

t5 UP

15 >ý >ý >1 x x ri tn

x (u

in 0

U) 0

W C5

kp Cl C) Cý kf

C5 c

ts 0 u

0 CA in Ln V) 0 0 0 0 0 x

u

x 0 (A

.0 -0

x x k t4 Up

cl o

d x x X x U

tA tA LIP Isc

ýr CD CD CD

;; 2; - + 6 0 0 0

tA - I-P >2 LM

LA 0 0 0 14 '3 tn C5 x 0 v 0 u 0 u u L. P C, 0 (A e

ts cs u 0 tA .A 4A 0 0 U- U- LL. u u I>w

I Lr C) N I r4 + + I I

ts C) C) Ln In 'A + + x u u u

>, >. , 15 a

LP zs

C 15

4A 0

4A 0

I I I

Ln Vw u x , N 0 0 C,

Z

>1 >1 >1 u u Ix M LIP

'3ýp C> C) C> C. ) CD C) >, 1>*

4) Ln 0

LM 0

L4 0 6

ul IA + u 0 0 0 + +

x - 0 u u x x Lp CD co -

- -

- -

L. P C c5 (-) 01 . 3c r -C 15 ts tA 3 ut 0

ko x C, ) 4D

0 u

0 u

Q I (5 Cý

cs C5 6 C) (D C) C X 4. j Cý La LLj ,M 46 tA Ix Ix a) 3 - - - CO L. )

SO 0

o w C, ..

It .........

LIP 0 C7 L-F LIP C) C) CD CD OD

117

where the terms in [T11] are given by the square matrix (9x9) in

equation (8.14.0).

The vectors A1

and consist of local and global displacements at 2L A2 G

nodes 1 and 2 respectively.

An equivalent form of the transformation matrix for forces can be

found by considering the energy at both local and global levels. Thi s is explained in the following analysis.

Since the energy remaining is fixed at both local and global levels,

then,

-F1TfF (8.16.0) LU I=i LUEJG E 2EL {E}L IG

is valid where, 'UE) are the displacements

IFG) are the forces

and L and G are the identities for local and global systems.

From equation (7.11.0)

IUEIL ý ETIIUEIG orLUCff =LUEjT [T]T

Substituting equation (7.11.0) in equation (8.16.0) yields,

LU jT ET]T {F = LUEjT {F (8.17.0) E C, EIL G ÜG

By considering the terms to the right of LU jT in both sides of the EG equation (8.17.0),

118

IFOG ý [T]T {FE'L (8.18.0)

The equation shown above provides the necessary transformation of the forces from the local to the global axes system.

Application of the solution in equation (8.18.0) to a specific

probI em:

cf. T. 21

C(. c-Yc rc, A (by. by. bl

z CG

FIGURE 8.3.0

Assumptions: 1. Local and global axis systems are parallel to each other

2. Transverse loads are applied at point B on the cross section

3. Axial forces are applied at A on the cross section.

From assumption (1), the angles,

a 17 =(I ny =Ctýz = 00saCx =alnx =a nz =Yry =YLY = goo

Thus, the square matrix in equation (8.14.0) is simplified to,

119

A, = i', B, = bz, C, = F, D, = by, E, = ezs F, = ey, H, = zc, H2 = Yc

G00 1 v' ODC - wDA -y (e - zc) - Y(yc -ey

(For definitions of A,, Bj, Cj, Dj, Ej, F, and G, see subtitles of equation (B. 14.0).

Substitution of these values in the corresponding elements of the

square matrix in equation (8.14.0) yields,

UEO 1 0 0 0 (e zc) Z- (ey-yc) (A c-w 0 -(ez-zc)--Z(yc

D DA-Y -e y0C

Uno 0 1, 0 (bz-Tz) 0 0 0 0C

UCO 0 0 1 (y-by) 0 0 0 0C

0 Co 0 0 0 1 0 0 0 0C

no 0 0 0 0 1 0 (7-b y 0C

Co 0 0 0 0 0 1 (b - Z-Z) 00

81 to 0 0 0 0 0 0 1 00

0 0 0 0 0 0 0 0 00

0 J

0 L

0 0 0 0 0 0 00

(8.19.0)

Substitution of the transpose of the above square matrix in equation (8.18.0 ) yiel ds the correspo nding transformation matrix for the

forces. This is shown in equation (8.20.0).

uy

uz

x

0 y

0z

0f

120

Fx

QY

Qz

mx

MY

mz

y

z

L=

00 0 0 0 0 0 0 F

0 10 0 0 0 0 0 -0 Qn

0 01 0 0 0 0 0 0 Q

0 (bz-z-)(Y--by) 1 0 0 0 0 0 m

(ez-zc) 00 0 1 0 0 0 0 m

(ey-yc) 00 0 0 1 0 0 0 m

,w00 -Y(ez-zc)-Y(yc-e ) DC-u)DA y 00 0 (T-by) (bz-z-) 1 0 0 m

0 00 0 0 0 0 0 0 0

0 00 0 0 0 0 0 0 0 L

(8.20.0)

The elements of the square matrix in equation (8.20.0) is in agreement with the result shown in reference (26), except for the term (wDC- 0

wDA-y(ez-zc)-z(yc-ey)), which is the generation of the warping moment due to application of the axial load. Many authors in the past seem to have ignored the effect of the warping moment due to application of axial load and transverse moments. Furthermore, what is shown in

equations (8.19.0) and (8.20.0) is a special case of the more generalized condition shown in equation (8.14.0). Up to the time of

writing this thesis, no information was found regarding the existence

of a generalized transformation matrix.

121

Calculation of Sectorial Coordinates Referred to any Arbitrary Point Other than the Shear Centre:

The sectorial coordinate can be described as an entity which always represents the amount of warping displacement along the periphery of the cross-section. However, to estimate the value of the sectorial coordinates at a particular point, the following information must be

provided:

a) Linear coordinates referred to the centroidal principal axis b) The shear centre is the point of rotation and also the origin of

the sliding radius C) The principal radius is found by referring to a specific point on

the cross-section whereby any sectorial coordinate referred from this point would become principal sectorial coordinates.

The sectorial coordinate values in equation (3.12.0) and in the transformation matrix in equat . ion (8.14.0) are referred to the global reference point. A relationship must therefore be established such that the sectorial coordinates referred to any other arbitrary point can be conveniently transformed from its known principal sectorial coordinate.

Let us consider the diagram shown in Figure (8.4.0). The local and global axes are placed parallel to each other and they are displaced

as in Figure (8.1.0). The sign convention used in defining sectorial

coordinates is the same as the right handed convention illustrated in Table (2. T. 0).

In Figure (8.4.0) B is an arbitrarily chosen point on the cross section. Two normals are projected to the tangent at point B, from

the local and global origin whose instantaneous lengths are defined as PO and P respectively.

By definition, the sectorial coordinate with respect to the local axis

system, is

122

z

k,

FIGURE 8.4.0

wo = fs Po dS 0

where D is the principal reference point and OD is the principal

radius.

Similarly, the sectorial coordinate with respect to the global axis

system, is

s fP dS

123

From Figure (8.4.0)

dwPdS 00 , ellý 220

-1 (n+d )(C+dC + .1 (2C+dc)(-dn) nC 2

-1 [riý+ýdn+ndc- &dn- nq (ndý -ýdn) (8.21.0)

However, the two linear coordinate systems are related by the following relationship as follows, I

n=Y, - and C= z' - Tl (8.22.0)

0 Therefore, the sectorial coordinate wDB is:

fs2 wDB

00 dS =1 (y' - y- 1) dz' - (7 -P) dy'l

0= fs {y'dz' - 7'dz' - z'dy' DB 0

s f(y'dz'

- Z'dy)ds- Y'(Z' - ZD) + Y'(Y' - YD) 0

WO = fs (y'dz' - z'dy')ds - y'z' + li (8.23.0) DB 0

Y'ZD + 7'y' - Y'YD

Now, apply the relationship found in equation (8.23.0) to the problem in Figure (8.1.0), where the axial load is acting arbitrarily, but normal to the plane of cross section.

124

Equation (8.23.0) can be re-written as,

w0*wcG_, yl I +_1 ,-I- -is DB ý"* DB zY ZD + "Y Yý (8.24.0)

By applying this result in equation (8.14.0) to the centroid and to

point A- k.

0 CG

ww DC ý DC - Y' Zc "7z lycl +y& ZD' -Z1 YD (8.25.0)

wo w CG

e'z +z el (8.26.0) DB 'ý DB yy+ Yz'D - z. yý

The definitions for (eý, ez), (yr, zc'. ) are found in Figure (8.1.0).

Therefore, by subtracting equation (8.26.0) from equation (8.25.0) and rearranging:

cGcG00 (tÜDC - tÜDB ) : -- ('ODC - wDB) +7 (zc-e; ) -

-z' (y'-e' ) cy (8.27.0)

Apply the results developed in equation (8.27.0) to the more general problem shown in Figure (8.5.0)

x

FIGURE 8.5.0

125

The relationship between the (x, y, z) system and the (x', y', z') system, as before, is given by:

x Cosa X'x

y cosa yx

z Cosa zx

See also Figure (8.2.0).

cosaxt y cosaxlz

coscl yIy Cosa yIz

cosazl y coscxzlz

x

y (8.28.0)

zi

Since the (x', y', z') system is parallel to the (e, n, system, then it follows that,

X1 cosa&x Cosa EY cosacz x

Y, Cosanx Cosa Tly cosanz 4y (8.29.0)

ZI cosa; x Cosa CY cosacz z

(See also the footnote for equation (8.13.0).

Substituting the result of equation (8.29.0) in equation (8.27.0):

cGcG0 (O'DC -ODA MwDOC -wDA)+(Tcosclnx + 37cosa

ny + zcosanzmxccosaýý+Yccosa CY +

+ zccosa ,)- (ex cosacx + ey cosar ýz y+ ez cosacz

- (7cosax+ YCOSa a+

Fcosacz){(XccosaTlx + YCcOSarLy +zC COS (ITIZ )

- (exCOSanx + eycosa rly + ez cosctinz)l (8.30.0)

126

CG CG This result (wDC - 'ODA ) can now be used in equation (8.11.0) to

calculate the term G1 of the transformation matrix shown in equation (8.14.0).

k.

127

CHAPTER 9

EVALUATION OF THE DIRECTION COSINES AND LINEAR DIMENSIONS REQUIRED IN THE LOCAL GLOBAL TRANSFORMATION MATRIX

In Chapter 8 the development of the local global transformation matrix and its use in large structural assemblies was obtained. To compute the transformation matrix in equation (8.14.0) one must provide the direction cosines between the local and global axis systems. In

practice the choice of structural sections is virtually unlimited and also in complex asemblies, one may decide to join specific points on the cross-sections of the structural elements which are not necessarily the centroidal points, but are chosen mainly. from the

physical design constraints. Under these circumstances one may conveniently choose a point such as 'P' in Figure (9.1.0) to define

the global nodal geometry. Simultaneously the local axis is also

placed at P so that the cross-sectional geometry can be defined

locally.

y

P(xl. yl. z

x

CL(x2

-

X

C2

yy c

FIGURE 9.1.0

128

The axial nodes of the structural element shown in Figure (9.1.0) are P(xl, yl, zl) and Q(x2, Y2, z2)- The third node R(x3, y3, z3) is placed on the local z' axis so that x', y' z' are also orthogonal.

The axis X, Y, Z is drawn parallel to the x, y, z system and point P

such that, k,

xx Xi

y+Y, ZI Zi

The position and orientation of the centroid and the principal axis was calculated and placed at point C(xa, ya, za) with respect to the local axis system x', y', z'. A fourth set of axes parallel to x', y 1, za was also drawn at the centroid, and is defined by X', Y', ZI. From the conditions mentioned so far, the following relationships can be obtained:

x Cosctx x y Cosa yx z Cosol zx

[TG1

or x V=

z [TG1]T

Cosa xy Cosax azx Cosa yy Cosa y lz y (9.2.0)

Cosa zy Cosazlz- z

x

Y (9.3.0)

z

4Y (9.4.0)

zo

129

This can only be X written since matrix ETG11 is orthogonal. By

substituting for [Yj in equation (9.1.0): z

xx Xi

y [TGll y yj (9.5.0)

zo z Zi

Similarly, by considering the axis systems (X', Y, Z') and (c , TI, C), the following can be written:

COSCV1 cosa, yl

CosaTxl cosarlyt

COSE cxl Cosa Cyl

This can be re-written as:

x

[TG21 Y

ZI

or

X1 Y', [TG2]T

ZI

cosa&ZI xi

Cos arlz 1 4 yI

cosaýzl Z

(9.6.0)

(9.7.0)

(9.8.0)

This can only be achieved since matrix [TG2] is orthogonal. The position of the centroid is also defined with respect to the axis

system (x, y, z'). Therefore,

130

II xI X4 xal

4yI= IV +y a' ý (9.9.0)

zI ZI j za'

X1 By substituting for Y'. in equation (9.7.0),

ZI

E X, ýa ETG2] 'y Yy'a (9.10.0)

czI ial

x Now, substituting yl in equation (9.10.0):

zi

x XI xa [TG21

( [TGll jy Ull Ya

z Zi za

or r

X-Xi xa n ETG2][TGll Y-Y, - [TG21 , yal (9.12.0)

c Z-Zi za

x By substituting Y, in equation (9.4.0) into equation (9.1.0):

z

Jx

y-= [TG1]T

z

Xi

+- 'Yl Zi

(9.13.0)

131

and by substituting equation (q. 8.0) in equation (9.13.0):

xx

xa Xi

.m= [TGlIT

fyi

+ Ya'ý + "yl (9.14.0)

. zý zi za' Zi

Substituting equation (9.8.0) in equation (9.14.0):

xa Xi y ETG1]T . [TG2]T + ya +, Yl (9.15.0)

zz z al I

By rearranging the above equation, it can be written as,

xT -T

xa Xi yý = [TGll CTG21 n+ [TGll , y; + yl, (9.16.0)

.z ýc za Zi

Calculation of the terms in the square matrix [TG11:

By referring to Figure (9.1.0),

PR , x3 - xl) ý (X3-xl) + (Y3-Y1) + (Z3-Z1) k

y3 - yl (9.17.0)

z3 - zl

132

PQ X2 - X1 (x2-xl) '+ (y2-YI) j+ (z2-zl) nu ru

Y2 - yl (9.18.0) (Z2

- Z1

By taking the vector product between PR and P, vector Pý is obtained. fýj IQ, Thus,

k,

PR x PQ = PN = (X rýI x, ) Z

(Z Z, ) lb ry tu 3- (Y3-Y1) 3-

(X2-xl) (Y2-Y1) (Z2-Z1)

Evaluation of the determinant in the expression above yields,

PN = ((y3-yl)(z2-zl) - (y2-yl)(z3-zl))i 1ý0 FV

((x3-xl)(z2-zl)-(x2-xl)(z3-zl))j ru

+ ((x3-xl)(y2-yl) - (x2-xl)(y3-yl))k (9.19.0) N

more precisely equivalent to,

PN = ali + bjý + clk N rv ru

where the absolute values of ap b, and c, are given in equation (8.19.0).

Now, by taking the scalar products of PQ, PN and Pý with Ahe unit

vectors in the directions of X, Y and Z respectively, the direction NNN

cosines required in matrix [TG11, are calculated. This can be

and z to described as follows. Let us define the unit vectors in x, be i, and k.

ru rV

133

Then, the direction cosines are:

Cosax iX (x2-xl)

Cosaxly (y2-yl)J (9.20.0)

COSaxo k (z2-zl)k

By solving equation (9.20.0):

cosaxlx (x2-xl)/Al

Cosa X, Y (y2-yl)/Al (9.21.0)

Cosot xaz (z2-zl)/Al

where Al = ((x2-xl) 2+ (y2-yl )2 + (z2-zl )2) 1

I Similarly, COSa yIx and cosaz Ix are calculated.

COSa y ly cosazty cosa y 8z cosazoZ

The matrix [TG1] is now fully defined.

It was pointed out that both the principal axes n, c and axes y', z' are on the same plane as that of the cross-section, and thus, the

orientation between the two sets of axes can be found by satisfying the condition shown below:

tan 2a 21y lZI

- (9.22.0)

yIyI-I ZIZI

134

Equation (9.22.0) gives the condition required for the principal axes, in which Iy oz, is the product moment of area and I Y, y i, Izoz, are the second moment of area about y' and z' axes.

�I ---'I

FIGURE 9.2.0

Similarly, axes Y', Z' are parallel to axes n-C and also axis XI is

overlapping with axis &, then,

aCX, ý 00 a=a=a= ax= goo cyl cze nXI cI

aTly, = ct, aCZ, = cc, (%TZ= Cy, = go+ 1

90-a' '('

See Figures (8.1.0) and (8.2.0).

Thus applying the conditions above in to equation (8.6.0) yields to,

[TG21 100 0 cosa sina (9.23.0)

0 -sina cosa

By making a close examination of equation (9.12.0), it can be seen

135

that for the orientation of the two axes systems,

x ETG2][TGll y (9.24.0)

z

(Note that the relationship in equation (9.24.0) is only valid for the

calculation of angles between the axis only and for the coordinate transformation one must strictly use equation (9.12.0)).

Let us define, [TG31 = ETG2][TG11.

Since matrices [TG21 and [TG1] are fully known, matrix [TG3] can be

calculated, and the terms in matrix ETG3] can now be used as a direction cosine required in the global/local transformation matrix of

equation (8.14.0).

Calculation of the positions of the shear centre with respect to the

global axis system:

4. pfp. q. pe

WOO

FIGURE 9.3.0

In Figure (9.3.0), point '0' is the position of the shear centre (e 719

eC), defined by the principal axes T, and C. Let us also define another point P(p

n, PP. used in predicting shear centre coordinates as expressed by the equation shown below,

136

an z-- pn + wpos cdA (9.25.1)

nn

apf wpos ndA (9.25.2) A

where wpos ýdA and Wpos ndA are the linear sectorial moments about the nand ý axes.

By inserting (aT,, aý) in equation (9.16.0) the position of the shear centre with respect to the global axis can now be calculated. By

simultaneous application of A(b y a, bz9) and B(e y e, ezi) in equation (9.10.0) followed by equation (9.16.0), the position of the axial and transverse load applications, with respect to the global axis, can also be calculated (see Figure (8.2.0)).

The coordinates of the shear centre, axial and transverse force

applications with respect to the global axis system, and also the direction cosines between the principal and global axis are used in

the local/global transformation matrix of equation (8.14.0). The

application of the theory into a programming concept is illustrated in the Appendix I.

137

CHAPTER 10

SOLUTION TO PRACTICAL PROBLEMS

The basic theory for buckling of thin-walled structural members has

been extensively discussed in the earlier chapters and some practical

problems to which alternative solutions are available in the literature are discussed in this chapter.

Solutions for Small Displacement Theory:

Let us concentrate on a beam element with the cross-section shown in

Figure 10.1 to investigate the behaviour for:

1. bending on the x-y plane 2. bending on the x-z plane 3. torsion and torsion warping about the axial direction.

y&z 237

I F or F Yl

I S. C

-2.0 Op

='2000.0 F or xy

5y or 6. x x

N ote: - all dimensions are in mm thickness 2.0 mm all over

FIGURE 10.1

-ý 40.0 ý- View on section X-X

The deflection caused by bending due to load F at the free end of the

cantilever was initially calculated using the Euler-Bernoulli theory

leading to equation 10.1.0. The results were compared for single

element solutions as shown in Table 1O. T. 1:

138

Fy Y. ' ' 10.1.1 6y 3 EIZZ

(10.1.0)

6z = Fy Z' 10.1.2 3 EI

yy

/

The cases for torsion and-torsion warping depend upon the boundary

conditions used in the differential equation (10.2). The solution to St Venant torsion is obtained with free warping throughout the beam

while for the case of the combination of torsion warping and St Venant, the solution is found by introducing the condition of warping restrained at the fixed end. The generalized solution for the torsion

problems shown in equation (10.2) is presented in equation (10.3):

I it I Mx = Tsqv + Tw GJ Oý -E rox

Hence,

(10.2)

ex mx

(Cl + C2 cOsh 'x + C3 s'nh Xx + Ax) (10.3)

X3 Er

where, X= v/ GJ Er

St Venant torsion problem:

1= 303. 137

Fx 6 x ý--Lx

ox

CO Sc c CD

At x 0, ICI=.

u

12201 ex =0

Note: - 0 if =0 all dimensions are in mm x

thickness 1.6 mm all wer FIGURE 10.2 View on section Y-X

139

L- 0

C) CD

LLJ

cli C%j C) CD

0 1-4 -4 x x

u 4 j N -4 %D f c = a 6E tD r- Xý >Jý 0) CV) wo-

Cý C; C LL.

co I C-j-

0--4-3 C) - 4-J, a -4 4j U (U X u (U E -4

tD all CD 4- '0 Ms ; (V C; C

0

LLI a LL.

C: 01 C-i 0 -ý 4J

- 4J C 4-3 U Q) X

U a) E

0) L- cu C) CV) 4- "0 W cli

Cl cl) >,

4J 4-A a =5 (U 0 LO kD E J3 C) CD 0 -4 -4 E X x

-. 0 f-I

-0 CU NE -zr q: r C: L- CY) CV) 0 to ; u C W 4-

0) 0 LO LO

E .0 CD C) 0 (a - r-q r-. 4

E. X x tv -d' .r

CY) (Y) LO LO 0 (a ; u C; C (V 4- Ln 0

a 0

r . - 4j c? ) U CD (1)

U L-- L. . - CD 010 ; LL. C N

r_ 0

r . - 4-) wt U C)

(1) cu -4 x

CD ;

LL. CD C

Cl

LLI

-i

in

140

The boundary condition shown in Figure (10.2) yields C, 0 C2 !- C3=0 in equation (10.3). Thus equation (10.3) is simplified to,

MX -x1

_mx

X2 Er GJ

Therefore at x=k, the twist at the free end,

BXZ= mg.

Z

GJ

Torsion warping problem:

z

1= 31903.0

x Lx

Z

137

ex ý-

ý*-=

C. " sc C3,

Cl 220

k.

(10.10.4)

At x=

N ote. - all dimensions are in mm

thickness 1.6 mm all ever

6x=0-

. 6", = FIGURE 10.3

View on section X-x

The boundary condition shown in Figure (10.3) yields C, r- - C2 r-

-Tanho,. Z) and C3 ý -1 in equation (10.2). Thus equation (10.2) is

simplified to,

141

ex = mx

- tanh Xg, + tanh -Xt. cosh Xx - si nh Xx + Xx) X3Er

At x=;,, the twist at the free end:

t' m 6

r, £ =» 'y- (- tanh X9, + XY, ) (10.10.5) X3Er

As in the previous case, the results obtained by the F. E. method for a single element are shown in Table (10. T. 2).

Generalized transformation matrix:

The transformation matrix developed in this thesis has the following features:

1. It transfers deformations and forces between the local and global axis systems,

2. It takes into account the effects of offset loading and moments.

Two tests were conducted independently to examine the accuracy of the bending, torsion and torsion warping properties of the transformation

matrix developed in Chapter 9.

Bending Test:

Node 1 of the cantilever shown in Figýre (10.4) was constrained fully

at the global origin while maintaining the axial direction parallel to the global x axis. The structure was loaded at the free end, in the

global z direction,

The cantilever in Figure 10.4 was then shifted to a new position as shown in Figure 10.5 and the constraints applied at node 1 to the same degrees of freedom as in the previous test. An equivalent loading

142

system can be found such that the problem illustrated in Figure (10.4)

and Figure (10.5) are identical.

z2 Ow. Ou

: 303.0

k, 2L.

x 1.0

Note: - y2, yx F=6121

all dimensions are in mm F

View on section x-y

thickness 1.6 mm all ever y

yA

IGURE 10. node2

F FIGURE 10.4 node2

4

Fz

nodel X X

nodel is fixed on this plane.

z FIGURE 10.5

The equivalent system:

FX(N) Fy(n) Fz(N) Resultant M

25.0 25.0 -50.0 61.237 in the local z- direction (i. e. same as in the previous case)

143

tý

CD CD

CD C: )

cli

X. ý r-t C) 41 00 x

Lr) r- L. ii 0 Ln

LC)

4- 4J

E:

0

cli

4J -- 00 x U Ck 4-) r- %t to x= Vý r1- x (D - -4 Ln

- LU 0 CD LC) vi

CD C) r=

X E- CD CD CM Ln

r- CD CD

x x

c2 C: 3

E A ý4 CD CD m CY)

.: t CD CD 12 E- cli

E --e

LC) Lr)

C%i E x

CM E cli Co Co

CD CD cli -4

E X WE Co Co

C: ) C: )

0

S- 4-) o

4-)

4J

0 r_

(A CL S- S-

4-) o (0 cn ý- ýr

cli

CD

LLJ

144

The displacements predicted for both tests were compared as shown in Table (10. T. 3) and found to be identical.

'1 1 1<

all dimensions are in mm

1= 303-0

xx 2z --- -X

I

Note-. -

sc I

Cb

22-0

View on section X-X thickness 1.6 mm all Gver

FIGURE 10.7 FIGURE 10.6

A similar test was conducted for torsion and torsion warping problems using the channel cross-section beam, shown in Figure (10.7). The

reason for this that channel section beams produce warping displacements, whereas the cruciform section beam does not. As

before, the two cantilever systems shown in Figures (10.6) and (10.7)

were tested. The agreement was found to be identical to within the

accuracy of the machine and can be seen in Table (10.17A).

The non-linear theory:

Generally, the buckling of a thin-walled structural member may be

induced by:

a) Bending caused on either of the principal planes due to

application of the axial load, commonly known as Euler buckling

145

:!. Test Cantilever placed coincided Cantilever placed in the

Description with the global axes'system space and loaded at the loaded at the free end with free end in x, y and z a transverse force in the directions

z direction

Node I x coord 0.0 25.0 y coord 0.0 25.0 z coord 0.0 25.0

Node 2 x coord 303.0 200.0 y coord 0.0 200.0 z coord 0.0 200.0

I(mm) 303.0 303.0 IYAMM 4) 0.534 x 104 0.534 x 104

IZZ(MM4) 0.534 x 104 0.534 x 104 E (N/MM2) 0.208 x 106 0.208 x 106 F (N) 0.0 25.0 The components of forces h dF x 2 are c osen Fx. F an

a t Fy(N) 0.0 25.0 'a Sulb a asn"er th t he resultant is always in the

Fz(N) 612.0 _5D. O local positive z-direction

dx(mm) 0.0 0.209 dy(mm) 0.0 0.209 dz(mm) 0.512 -0.418 Rotx(rad) 0.0 -0.179 x 10-2

Roty(rad) -0.253 x 10-2 0.179 x 102

Rotz(rad) 0.0 0.0

Resultant 0.512 mm in the positive 0.512 mm linear local z direction displacement at node 2 (mm)

Resultant -0.253 x 10-2 radians in -0.253 x 10-2 radians rotational the x-z plane displacement at node 2 (rad)

TABLE (IO. T. 3)

146

Test Cantilever placed coincided Cantilever placed In the Description with the global axes system space and loaded at the

loaded at the free end with free-end 'with moments in a torsional moment in the x, y and z directions

positive x direction

Node 1 x coord 0.0 25.0 y coord 0.0 25.0 z coord 0.0 25.0

Node 2 x coord 303.0 200.0 y coord 0.0 200.0 z coord 0.0 200.0

z (mm) 303.0 303.0 (MM4) I 0.454 x 105 0.454 X 10

Yy I zz (mm4) 0.711 x 104 0.711 X 10

(MM6) 0.241 x 107 0.241 x 107

E (N/mm2) 0.208 x 106 0.2081 106

G (N/MM2) 0.832 x 105 0.832 x 103. Mx (Nmm) 520.0 0 The components of momeau

.I and Nz are chosen In 300

,%

my (Nmm) 0.0 c . h manner that the 300.0 resultant moment is almys

mz (Nmm) 0.0 in the local positive X- 300-0 directim

dx (mm) 0.0 0.677 X 10-13 dy (mm) 0.0 -0.574 X 10-13

dz (mm) 0.0 -0.974 x 10-14

Rotx (rad) 0.557 x 10-2 0.321 x 10-2

Roty (rad) 0.0 0.321 X 10-2

Rotz (rad) 0.0 0.321 X 10-2

Resultant 0.0 0.8929 x 10-14 linear displacement at node 2 (rad)

Resultant 0.557 x 10-2 radians 0.557 x 10-2 radians rotational about the x axis displacement at node 2 (rad)

TABLE (IO. T. 4)

147

b) Torsional buckling of the cross-sectional plane due to application of the axial load

C) Lateral torsional buckling of structural members at the plane of the least second moment of area when loaded arbitrarily on the cross-section.

In Chapter 6 it was also mentioned that there are two basic approaches to buckling analysis as follows

1. Eigen value approach 2. Large displacement approach.

In the first approach the geometric stiffness matrix in equation (6.2.0) is factorized with a single multiplier and by the use of the Eigen value method the force required for the buckling condition is

evaluated. Mathematically, this is defined as the condition for the singularity of the assembly stiffness matrix, i. e.

det ([k] - [N]) =0

where, k is the linear stiffness matrix N is the geometric stiffness matrix.

However, the terms in the geometric stiffness matrix cannot be

represented by a single multiplier (i. e. all reaction forces in the geometric stiffness matrix are distributed independently). Thus, the the Eigen value approach is not applicable.

The two classes of nonlinear problems under consideration are:

1. Bifurcation type 2. Large displacement type (i. e. buckling associated with the

displacement history).

4

146

The computer program developed in this thesis is based on the large displacement approach. Thus, any bifurcation type of problem must be

converted to the large displacement type. The conversion is generally done by applying the following techniques:

a) By introducing geometric imperfections (e. g. an Euler buckling

problem becomes an imperfect structure) b) By applying a perturbation force.

In general, the latter method is easier than having to introduce any

geometric imperfection to the structure.

Buckling of an imperfect strut: -

XI

Noie,

A dimensions are in mrft thickness 2.0 mm att over

z4W. C(ioad inuement

z 23.7

s

F. i

2.0 co

View on seEtion )(-X

FIGURE 10.8

4

The strut illustrated in Figure (10.8) is slightly bowed with a

maximum imperfection of 40 mm at the mid-span. Atypical cross-section

along the span is also shown in Figure (10.8). The structure was then

constrained as shown in the table in Figure (10.9). The theoretical

centre deflection at the mid-span may also be calculated using

equation 10.7:

b1

(i -,; ) "P

(10.7)

14q

I Euler buckling of an imperfect pin ended strut under axial compression

1; 160000

u 140000

120000

100000

30000

$0000

0000

E a u 20000

a

it. Ir 20

tr --T vww an lw"m X-V

NO%,

an d.. Ros; ons we in Mck"51 2.0 ON all Mw

t. I

Legend L element solution A Graphl 10element solution X GraP"2

Exact solution. (3 GrcPK3 so 100 50 200 250 300 350 400L 450 Soo 550 Soo

Deflection in the Y direction of the rnid spcn mrm

Cmawl. in C=dltlý.

Nwe No& Ex Uf CZ It= RDTY R= WX Vr UZ

11 0100011

10 "1

11 00aII

Mnstralm C-dltiý.

Nade NO: Dx DY DZ RMX RM lUrz NX Vf VZ

1 1 " 1 1 0 0 1 . 1 . , . . 3 0 0 0 1

5 . 0 0 0 1

1 0 0 0 1 1 0

0 0 0 . 0 10 0 0 1 1 11 1 1 1 1

Notat Any dýVve or r .. Is fl-d. IV I Any dgme of rr.. d- Is from. Iwo

FIGURE 10.9.

150

k.

L- co V-4 CD

IA z c

LU

2:

L) :30 CD

; a) 4-) C U L) C) CC)

4- (1)

u cc 0x C) CD

.0 (1) a 9 0 L

r- (1) -ý co co

(a u 4-3 to U L- :3

-ý 0ý 41 4- 0

Ict -It CD

-cr E E X X

qzr Co

X x

-d' mt

LO LO

CD C) cli 14

E E x x

CC) CC) C: ) C)

LU Clt Cý

4- 00

(L) Cý CD E

0 CL)

0'

LU -i Co

151

where, b, is the magnitude of the initial out of straightness measured at the mid-span P is the compressive force acting on the structural member Pe is the Euler critical buckling load for a pin ended strut

(i. e. Pe = 7r2 EI)

z2

The results obtained for 4 and 10 finite elements at the mid-span are shown in Figure (10.9 ). Finally, the trends shown in Figure (10.. 9) are as follows,

Critical buckling Critical buckling Critical buckling The load (N) calcula- load (N) using 4 load (N) using 10 % ted from eq. 10.7 elements elements Error

170000 180000 5.8 175000 2.9

Flexural buckling of a perfect pin ended strut under axial compression:

Two examples were considered: 1. Cruciform section - see Figure 10.10 2. Flanged cruciform section - see Figure 10.11.

The geometry of the cruciform section strut under consideration is illustrated in Figure (10.10). The solution obtained for 10 finite

elements with a perturbation lateral force varying from 0.001N to LON is shown in Figure (10.12).

The magnitude of the critical buckling load calculated for each perturbation force is given in Table (10. T. 5). By referring to Figure (10.1-7), it can be seen that, as the magnitude of the perturbation force decreases the corresponding load displacement curve changes from

a large displacement to a bifurcation type. It is also evident that

152

when the perturbation force is 0.001N the accuracy of the solution obtained by the finite element method was almost identical to that of the exact solution.

ý2 20-0 NI 10ad increment)

t.

I19. yM MO cm I

C3, C, i q 1.0

* ýp

I

Note: - View an section X-X all dimensions are in mm thickness 1.0 mm all ever

FIGURE 10.10

Let us consider next, the flexural behaviour of a cruciform section under axial compression when flanges are introduced to its radiating members as shown in Figure (10.11):

FrND-ON(toad increment) x

M 400 u -20.0

View on ---ai0n X-x

140te. - ali dimensions are in mm thickness 1.0 mm all ever

FIGURE 10.11

I bJ

Flexural Buckling of a pin ended strut.

go

700.,

.2

600.

500.

CA c 0 a 400

300

200

F- 100

0

10 element solution

lbod kutowntl

Not&- all dipwaisms are In as Ikkknas 1.8 an sit over

a

I 7soj

Legend fy a 0-001 N A Croph 1

ty u 0-1 N X Croph 2 fý a O-SN 0 Grcph 3 fy a I-ON 0 Creph 4

Exact solubon it Croph 5

jo tic 260 2. i 0 360 350 400 450 500 Deflection in the Y direcHon at the rnid span mrm

CýOla Cnditiom.

sk. de X.:

10 I,

SK VY IDZ It= MM X= ICK W IMZ

1, 0 1 1 . 1 1

0 0 1 1 1 . 1 1 0 0 11 MWAS Am dv" or f, .4 Is rhad. 6ýud by I

Any O-vw cr tmrft. Is trw. dwuA by 8

FIGURE 10-12

154

Flexural Buckling of a pin ended strut.

700 20 element solution

9; X Fj-20 ON I lead Imromeme j C 600

.2

a Sao

MAY

.R 12

400 2"

300

0

20D

Legend Note. -

fymO-OIN A GrapKI all dimemia-mWe Is 00 loo fyzO-C5N X '-rGPh2 E thickness 140MM11 Iwar

0 U fyw O-5N C Graph3

Exact solution 0 GrcPh4 0 --r---I ;a1; 0 ýo 200 2ýO 300 3; 0 , Go 450 Soo

Deflection in the Y direct ion of the mid span mm.

QXWLMIX QwAlttý.

N-de bb: EK Dt Im X= vorf F= IM W ut

11 2

50

0

1 12 0 13 0 1.0 15 0 1.1 17 0 Is 19 20 21 0

wvw,. My dev a or f is nud. d=MW by I AM oevý or ry" Sa rra. lhmoud b, 0

FIGURE 10.13

155

Comparison 10&20 elements, Flexural Buckling of a pin ended strut.

r 600 0

56D

400

300

2bO

M too E 0 L)

0-13 0

k,

yri on spcticfl X-X

Wate. - &H climensions'are in an thickness 1.0 am all over

0.4 0.6 0.8 Deflection in the Y direction at the mid span mm.

C. Swals cDmiti-.

W-de N.: Ix vi mz Yam any X= wx VY NZ

0 0 0 0 0 0

0 l' 0 0 0 0 0 0 . 0 0 0 0

: 0 0 0 0 . : . 0 0 0 0 0 l'

ounmin Qdjtiý.

N-de Not DX Dt M IR= MMY lem2 IIX W IJZ

I I I I 1 11 11 0 11 2 - . . . . . . ,1 3 a 0 0 11 0 0 11 11 0 0 0 . : . .

0 0 : : 0 0 . . : . . 11 1: 0 0 0 0 0 11 0 11 11 . 0 . . . . . II 12 0 0 a : 1 0 11 13 0 0

: : 0

5 16 0 0 17 . . . .

: . . 11

18 0 0 0 0 0 0 0 11 11 . " . . . I . 11 20 0 0 0 0 0 21 0 1 1 1 0

Notes Any 6.1 a of V Is find. demoted by I Arw dev se or I, -- Is frýss. OwyLd by 0

FIGURE 10-14

IOCNIIcad increswall

Legend fy s 0.0001 NA Graph 1 fy v 0.0001 NX Graph 2

156

Effect of axial load increment on Flexural Buckling .

70

q

6DO

50D x D 4D

z

400

300

200

0- too E 0 L)

10 element solution

y" en sertion X-X

1(ad k. a. r)

Note, all dimensions are in no Legend

thickness 11-8 an all ever Fx-5-ON. fy&0-001N A Graph 1

Fx=10-ON. fyvO-00IN X Graph 2

0qI -- 0 0ý 1 1.5 2 2. S Deflection in the Y direction of the mid span mm.

C-91tralft OwAltiom.

Node NOS lox Dr m JtDTX FDTY Ro7z wx wy VZ

0 0 0 0 0

0 ' : 0 0 ' 0 ' 0 - . i 0 0 0 0 l' 0 0 0 0 0 0

10 0 0 . , 0 l' l' 0 1 1 1 0 0 0 1 1 Notes ADY d. 91" of rroedou Is fixed. 6. noted by I

AM dev- Or freedom to fr". denoted by 0

FIGURE 10-15

157

Effect of axial load increment onnexural Buckling . 20 element solurion

C

£00

i6 -5 Soo a

400

0 0 En c

300

0

200

M loo E 0 u

0

t,

qý? vi-i on wetio" X-X

F111. aJ h'u, mw$I

Naftý all dwem*. Dm are is as

thiCkr"S 1-0 U& 811 &AW

p

Legend Fxz 5-ON. lywO. 00IN A Graph 1 FxslO. ON . fyzO. 001 NX Graph 2

0.5 1 LS 2 Deflection in the Y direction at the rnid span mrn.

Camtmin cDndjtiom.

W. de Ws ix Vf 12 le= Rmy ranz Vx vy liz

0 0 a 0 0 0 0 1 0 : 0 0 I 0 . 1-I I . :

0 0 0 1 ;

a 0 0 . 1 1 12 a 0 0 is 0 0 . IN a 0 0 0 1 1 is 0 . . , 1 16 0 0 0 0 0 1 17 0 11 0 0 0 1

0 0 0 0 1 1 20 . . 1 1

21 0 0 0 1 1

Notes ArW &Cý 01' f* welow Is fised. denoted bs, i Any d. V em . freedom IV frv*, dwotd by 0

FIGURE 10.16

158

C- Ol Lr) 0

1r; ui

En

ý he Lý C) C) u C ý U: 0

C; C D .0 (1) 4J

CD C: )

UU LO to

m rý-

C\j cli

4- (v CX

4J

-le 0- CC) CC) um :3x

C\i cli

C) C) 0 ý (1) -ý qzr

CC) co cli C"i ý 0-

4-x 4- 0

to Co C)

E E x

N %: r

C%3 cli cli cli

to CD (D

E E x x

-tr -tt cli c"I C%j cli

C) Co

x

co co C) C)

LLJ cli cl-i

4- 41 0a

0) (D CD

E -4 cli

C;

Fi CD

LU -i

h-

-C,

159

The example shown above was modelled with 10 finite elements and with the constraint conditions as in the previous example. The structure

was then loaded with-increments of 30ON and also with a lateral

perturbation force 0 This was repeated for a range of perturbation forces (0.01N to 0.05N) and the results are shown in Figure (10.17).

As in the previous example the approximate bifurcation condition was achieved at a ptrturbation force of LON. The results are compared

with the exact solution and shown in Table (10. T. 6).

To investigate the effect of a finer discretisation the structural

member shown in Figure (10.31) was re-modelled with 20 elements. However, the results show that there is no significant improvement

over that of the 10 element model. By referring to Figures (10.12*. 3)

and (10A2.3), it is also noticed a rapid change of the curve for

buckling force around 300N. This is because as the load approaches

towards the buckling load the determinant of the assembled stiffness

matrix approaches near zero, thereby producing large displacements.

However, since the loading increment is finite in magnitude, it is

impossible to attain the singularity condition of the assembled

stiffness matrix. Similarly, for load increments beyond the

fictitious buckling load, the A. S. M. can still be non-singular and

will produce displacements which correspond to the shift in origin. This can be seen in Figures (10.12. ) and (10.13). The results obtained for 10 and 20 elements were compared and are shown in Figure (1014).

The effect of the magnitude of the axial load increment on the final

result was examined according to the combinations shown in Table

(10. T. 7).

Test No No of Elements Magnitude of the Magnitude of. the . . axial force (N) lateral force (N)

1 10 5.0 0.7 2 10 10.0 0.5 3 20 5.0 0.5 4 20 10.0 0.5

TABLE 10. T. 7

160

10 element solution

q x

25000-

13

-. 2 20000

Ov c

U a

10000

5000

E 0 u

t,

PW -F-t-z"

Nar, L-

all diffienssong are in an thickness 1.0 an all 4wer

0-f 0

W. " NO:

2 3

S

7 U

ID 11

nexural Buckling of a pin ended sfrut.

Legend fy a 0-01 NA Crcp'h ly a 0-05N X Cra Ph 2 fy a O. SN C3 Graph 3 fy a I-ON 0 Graph 4

Excrt solution it Graph 5

1234 D&action in the Y direction ot the rn7id spon mrm

chwu*IM ýitsý.

mx ur ir Z= wm JE= IAK vr liz

1 1 " 1 : 0 1 . 1: . 0 :

:: : :

0 0 4 00

ti. t., Any dcrue c nwd- is irl-d. d-aud by I Am #-Vý rmý is rrweý d--ad by 0

FIGURE 10-17

6

161

nexural Buckling of a pin ended strut.

20 etement sotution 30000ý

X 25000

, 10

X 20000

C', too 0

15000 IM

S .0

0 1z e 10000

Vw am Wflan. -I-K Legend fy a 0.001 NA Graph

5000 wateý a dk-M; Gn* we In 0,4

fy a 0.65 NX Gr0ph2 M E *019"S 1.0 DeaR @. Of fy! n O-SN 0 Greph3 0 fy aI ON 0 G-ph 4

Exact solution 11 GraphS 1 0 0.5 2. 5 3

Deflection in f he Y direction cf the mid span mm.

ammmin cmutiý.

v, - bb: ox ve m i= mny n= ux vy vz

I I I I 1 0 0 0 1 1 2 0 0 0 0 0 0 0 1 1 3 0 a 0 0 0 0. 0 1 1 41 0 a 0 0 a a 0 1 1 5 0 0 0 0 0 0 0 1 .1 & 0 0 0 0 a 0 0 1 1 1 0 0 0 0 0 0 0 1 1 a 0 0 a 0 0 0 0 1 3 9 0 0 0 0 a 0 0 1 1 10 0 0 0 0 0 a 0 1 1 It 0 0 a 0 0 0 0 1 1 12 0 0 0 0 0 0 0 1 1 13 0 0 0 0 0 0 0 1 1 14 0 0 0 0 a a 0 1 3

0 0 0 0 0 0 0 1 2 0 0 a 0 0 0 0 1 1

17 0 0 0 0 0 0 a I I Is 0 0 0 0 0 0 0 1 2 19 0 a 0 a 0 0 0 1 1 20 a 0 a 0 0 0 0 1 1 21 0 0 0 1 1

W. to. kw daffý or fr. ". is rimw. dewtd by I Any deCr" of rr.. d- Is fro*, dow%&d bY 0

FIGURE 10.1 B

162

300

Cý

25000,

. c3 x a 20000 fu x in c

15000

10000

500C 0-

ComDarison 10&20 elements, nexural Duckling of a pin ended s1rut.

wofeý . 11 diniersions arc In as

thickness 1.0 small NO

O-f 0 0. " 0. *6 O. a i 1.2 1.4 1.6

Deflection in the Y direction at the mid spon mm.

Cmau-W- onditlam.

Node Noz DX Uf EZ RMX *MY am %x VY IJZ

I I I 1 0 0 0 1 1 a 0 0 0 0 0 0 1 1 0 0 0 a 0 0 a I I 0 a 0 0 0 0 0 1 1 0 0 0 a 0 0 0 1 1 0 0 0 a a 0 0 1 1 0 0 0 a a 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 a I I a 0 0 0 0 0 0 1 1 0 1 1 1 a 0 0 1 1

CbrALralm Cmgltjý.

.. te" .M Uf VZ Jc= mm'r R= VX VY la

I I I I 1 0 0 0 1 1 2 a 0 0 0 0 0 a 1 1 3 .0 0 a a 0 0 0 1 1

0 0 0 0 a 0 0 1 1 0 a 0 0 0 a 0 1 .1 6 a 0 0 0 0 '0 0 1 1

7 0 0 0 0 a 0 0 1 1 a a 0 0 0 0 a 0 1 1 9 0 a 0 a a 0 0 1 1 10 a 0 0 0 0 0 0 1 1 11 0 0 0 0 0 0 0 1 1 12 a a 0 a 0 0 a 1 1 13 0 0 0 0 a 0 0 1 1 IN 0 0 0 a 0 0 a I I Is 0 0 0 0 a 0 0 1 1 16 0 0 0 a 0 0 0 1 1 17 0 0 0 a 0 0 0 1 1 Is 0 0 0 0 a a 0 1 1 Is 0 0 0 0 0 0 0 1 1 20 0 0 0 0 0 a 0 1 1 21 0 1 1 1 a 0 0 1 1

Wtat MY dc- or fre-dom is timed. d---ted by I AM d. C- of fl-w- is rr". d-1.4 by 0

FIGURE 10 . 19

F.. bOO ON I Icad I-crt-l"I I

*I-

40.0 200

Y

Legend fy*O. 5N A Graph

fy *0 .5NX 'Graph 2

Exact Solution 0 Graph 3

163

Effect of axial load increment onnexural Buckling

10 element solution 30000 -

q ýII oad Imr. -W) X C I :2 25000-

k-

U

-6 X 0 20000-

-200

CM C 0 a

CD View an wun X-X 10000-

sit dimensions or* In she thickness 1.0 on all ever

44 91

CL 5000. Legend E Fxz300. ON. fysO-OSN A Graph 0

Fx a 600.0 N. fy a 0.05 NX Graph 2

0- 0.00 -r- I

0.05 O. io 015 020 0.25 O. io

Deflection in the Y direction of the rrdd spon mm.

Cbnslr. in Cmdittý.

N. de N. - EK IN DZ g= FMN x= IAX wy VZ

0 1 1 3 0 : 0 1 , 0 0 1 1

. . . 1 I a 0- 0 0

: I

o 0 0 1 1 7 l 11 0 1 1

0 0 0 0 0 0 : I I

. . . 0 0 0 1 1 10 11 a 0 0 0 0 0 1 1 11 . I I 1 0 0 0 1 1

Now Any derý or freedom Is fixed, denoted bY kw deg cr I'modow Is free. d--Aad by 0

FIGURE 10.20

164

Effect of axial food increment on Flexural Buckling .. 20 element solution

C

C 0

U I-

"0

C IC U S

-c C) C 0 0 C) C

U 0

U 0

S > S S S

C- E 0

U

k,

II @ad lmm^Wl

20.0

=4- 1.0

Vow on SKtion X-X

Legend , Fx a 100.0 N. fya0.05 NA Groph 1 Fx a600.0 N. fy a 0.05 NX Graph 2

Note, 611 dimensions at@ in as thickness 1.0 am all ever

0.0 0.1 0.2 0.3 0.4 4 Deflection in the Y direction at the mid span mm.

CýMmlm cbwi%j-.

w.. de »ic. 3 ZK Vt ZIZ RM RMY x= IM VY UZ

0 0 0 0 0 0 0 C 0 C C C 0 C 0 0 C 0 0 I

10 11 12 13 14 Is 16 17

20 21

limta: Arv deorw af freed= Is flaW. denotad by I Am dcvý or tr*"m to frý. dowLed by 0

FIGURE 10,21

165

The results obtained for the runs shown in Table (10. T. 7) can be seen in Figures (10.15) and (10.16).

Examples in pure torsional buckling:

Basically there exists two classes of torsional buckling problems: 1. St Venant torsion 2. Combined effects of St Venant torsion and torsion due to

restrained warping.

Let us begin with St Venant torsion by considering the behaviour of a cruciform section strut under axial compression. The geometry of the

strut is shown in Figure (10i-22). The strut was loaded by an axial force and a range of perturbation torsional moments acting simultaneously at the mid-span as shown in Figure (10.22). The result

obtained for 10 and 20 element assemblies with an axial load increment

of 10ON and perturbation torques varying from 0.01 Nmm to 0.5 Nmm are shown in Figures (10.23) and (10024). The axial load increment was then changed to MON and the perturbation torques were varied from 0.01 Nmm to 0.5 Nmm. The results obtained for these runs can also be

seen 'in Figures (10.23) and (10.24). These results are compared with the exact solution shown in equation (10.8). As in the case of flexural buckling, near bifurcation behaviour occurred for smaller magnitudes of perturbation torques. The accuracy of the agreements found was almost perfect as shown in Table (10.8).

.L rff 2 Er + GJ) Pcr ý lp %-F-

where Pcr is the compressive force under pure torsional buckling

A is the area of the cross-section Ip is the polar second moment of area r is the torsion warping constant E is the Young's elastic modulus G is the modulus of rigidity in shear

166

j is the St Venlent torsion constant is the length of the strut Fi=100-ONI[oad incrementi

20-0

1.0

View on section X-X

Note: - all dimensions are in min thickness 1.0 mm all over

FIGURE 10.22

Torsional buckling of a flanged cruciform section strut:

The flanged cruciform section strut under investigation is shown in Figure (10.28jand has been modelled with 10 and 20 elements. For each mesh the strut was constrained and loaded in the same way as in the

previous examples. The magnitude of the increment of the axial force

was kept at 10ON and the results obtained for both models can be seen in Figures (10.28) and (10.29).

By referring to equation 10.8 it is apparent that the buckling force

under St Venant torsion is a function only of the torsional buckling

rigidity . 21 and is independent of the column length. However, in the

case wherIePthe warping constant is non-zero, the buckling force will ' U2 Er increase by

p 9, and therefore the mode of buckling is somewhat similar to that of flexural buckling. Under these circumstances the magnitude of the perturbation torque was kept at a larger magnitude than in the previous section.

167

C) C)

C) CD LLI

C-C

Wu 4-3 CD C)

00. - 01 cli 4-) 4- '0 C) C)

U

LLJ U LL.

C-) _0

to = r- -a 00

CD

00 cli 4-)4- 0 C) C)

(n r-_ r- - V) %D to

x

U .0 LO LO

C) CD 01 -4 -4

E x x 0i C"i Cl) m CC) co

4.0 t" C) C)

CNJ -4 1-4 E x x

UJ E LO Lf) C) CD cli C*j

to C) Co Fý E- ; ;

C C

CC

CD

x

NE -cr E M ce)

LO 0

LO

CD CD

x

L. C) LC)

E ko kc)

c%j C-i 0

0 r_ Co 0W

W

Co

4

Co

168

k,

x 16000.

14000.

0

0

sono U 0

6000

V >

4000

V

IL

IF I 0.00 O. az

Torsional buckling of a pin ended strut.

10 element solution

x L LýL*

700 (, i

Legend

mx v 0.01 Nmm A Graph 1

Nvteýsjj dimensions are In min mx a 0.05 Nmm X Graph 2 mxxO. lNmm 0 Graph 3

thitkness 1.0 me all Over Mx aa-5 Nffn 0 Graph 4

Exact solution 11. Graph 5

0.04 0.06 0.08 0.10 0.12 0.14 Twist in the axial direction at the mid span radL

0.16 0.18

cbmwalft C-dltjý.

W. ft lb: UC tFr Elt It= ZYff RDM VX WY WZ

0 0 0 1 1 : 0 0 :: : 0 1 1

.

0 . a 0 0 1 1 40 0 a 0 1 1

: 0 0 0 1 1 10 . . . . . . 1 1 Is 0 1 1 A0 0 0 1 1 bbis: hv iko wr fr-d-- is fixed. der4ted by I

kw dw - at f".. 6m Is free. denoted bi 0

FIGURE 10.23

169

Torsional buckling of a oin ended strut.

20 element solution

16000-

14000-

12000.

loooo.

V) c 8000 U a ei e 6000 0

4000

92. E 2ooo

0.0

ý-Wfl INI load increment I

iLv On wtion V-X

Ncteý a ti dimensions are in as thicknus 1.0 so aLl over

4., o 6., 0 8.0 10.0 12.0 14.0 Twist in the axial direction at the mid span rod.

QW. UVIN ýtjý.

Node Nog DX cy Im RMX F= Im ux W UZ

3 1 1 . 1

7 1 1 1

10 11 12 13 0 0 a 0 11 1. 0 0 : : 0 11 Is . .

: : 0 1 1

16 0 0., 0 0 17 . . I : 0 0 1 . . 20 0 00 0 1 l 1 1 21 . II I

: : e 1 1

W. W9 A. 1 d or r--d. L, ri- &-Ud by I My wr" or rrtdm to fý. d., owd by 0

Legend rnx z 0.001 Nmm A Graph 1

mx a 0.005 Nmm X Graph 2

mX a 0.015Nmm C3 Graph 3

Exact solution 0 Graph 4

16.0 18.0 -icf'

FIGURE 10.24

170

cý x 16000

14000

12000

0)

cn BODO

6000

20DO

Comparison 10&20 elementsTorsional, Buckling of a pin ended strut. 18000

t-50,0 N1 t4wd Increment)

Note, all dwsmium are In me #utkftss tO mia all &W

04- 0.0 2.0 4.0 6.0 9.0 10.0

Twist in the mial direction of the rnid span rocL

Cmwwsin onwiti-A.

Nod. No:

1 2 3

10 11

EIK MM RM MMY gg= UX VY %IZ

I 1 0 0 11 11 0 . 0 . . II 0 0 0 0 0 0 0 11 0 11 0 0 0 0 11 0 . 0 0 0 0 a 11 : 0 0 1 1 . 0 0 0 0 0 1 1 0 0 11 11 11 . I I 1 0 0 0 11

Comtrals Qxidi%lý.

Wo6e Nos IX Vf M R= RM ADIZ IN WY 1-7

I I I 1 0 0 0 1 1 a 0 0 0. 0 0 1 1

0 1 0 1 .1 0 0 0 0 1 1

7 0 I 1 I I

I a 0 0 1 1 10 0 I , I I 11

0 0 a

0 . 0

0 : . 1 1 12 11 : : : 0 0 0 1 1 1 3 . . 0 1 0 0 0 0 : 15 l . 1. o 11 0 0 1 1 17 11 0 : . . : . 1 1 1. . . . . . . 1 1 11 0 11 11 11 0 0 0 1 1 20 . I . 0 1 1 21

0

. 0

I 0

I 1 0 0 0 1 1

W*ws Any derý st fýdvm Is flu. 4. danated by I AIW ". gý ct treed- in free. denwýd by 0

Legend

mx 0-001 Nmrn A Graph 1 mx 0.001 Nmm X Graph 2

0 14.0 s1O

FIGURE 10-25

171

18000 q x 16000 c 0 u

0 x 12000

10000

0

8000

0

Li 6000 0

W 4000

CL E 2000 0 u

Effect of axial load hcrement on Torsional Buckling . 10 element solution

x

q Wo

4-- V. j on setion X-X

gor ag d-amiem are In an umknesil 1.0 an alt @wer

I load tftnmeatj

Legend Fx a 50.0 N. mx a 0.001 Nmm A Graph 1 FxzlOO-ON. mxmO-OOI Nmm X Graph 2

06 1 0.0 5.0 do Iýc 2ýO 2;. o Ust in the cLxocd direction at the "M span rod.

Cms'Lraill CwAlU-.

)bd. No: IIX pf EZ x= NM NMZ %IX W IM

I I I I I 2 3 0 0 0

0

6 0 c 7 . o a 0 0 0 11 I - 0 . 0 10 0 0 a 0 9 1 1 11 - 1 1 1 : : 0 1 1

Now. AnY der, ** or rfv-&* Is flmd. dwLed kW I kv devve or rrweom to rý. 4-md bw 9

k

FIOURE 10.26

172

Effect of axial load increment on Torsional Buckling

20 element solution 18000

q x 16000- c F licad Wcrtp*ntl 2 . z

14000-

x a 12000-

200

1 0 10000-

. 1 0) i 0 6000 Z4-

Ywj on wction X-X 0

wy Nate" > all dimensions are In as

,; i 4000 0 thickness 1.0 an all ever E 0- Legend E 2000 0 Fxx5O-ON. mxaO-OlNmm A Graph u

ov Fx&300.0

.

N. mx a 0.01 Nmm X Graph 2

IT000 0.002 CA4 0.46 0.0108 0.010 0.0012 Twist in the axial direction at the rnid spon rod.

C-Auýft II-Itlem.

W, ' Ibc IX VI M R: YrX AMT 11= WX VY VZ

3 0 0 0 0 9 0 0 1 1 0 0 a a 0 0 1 1 0 a 0 9 0 0 1. 1

0 a 0 0 0 0 0 1 1 0 a 9 a a a 0 1 1 0 0 0 a 9 0 a I I

9 0 0 0 0 0 0 a I I 10 0 0 0 0 v 0 0 1 1 11 0 a 0 0 a 0 0 1 1 12 0 a 0 0 a a 0 1 1 13 0 0 0 a 0 0 0 1 1 14 a 0 0 0 0 0 0 1 1 Is 0 a 0 a a 0 0 1 1 16 0 0 0 0 9 0 0 1 1 17 0 0 0 0 0 a 0 1 1 le 9 0 0 .0 0 0 0 1 1 19 a 0 0 0 9 0 a 1 1 20 0 0 0 a a 0 0 1 1 21 0 1 1 1 a 0 0 1 1

WWAt AM 6evve or fnmdm Is fl=d, dmated by I Any d. 0 - or TF--d- Is 1ý. dowtod bY 0

FIGURE 10-27

173

Torsional buckling of a pin ended strut,

10 element solution

C; x c u 20000

.2 x 0

a :E 15000- en 3. c 0

9 . 40.0 10000- 20-0

.2 .0

5000- ALO-0 vi. on SKtion %-X

M

0 L)

Note- d 11

to, U dimensions are In As

Not ' u n

th tck us 1.0 on all over

VD0 0NI load Incromtot 1

A*7

Q Legend

mx r. 3-5 Nmm A Graph 1

MX a 6.5 NMM X Graph 2

mx a 12.5 Nmm Cl Graph 3 IJIAI- mx w25.0 Nmm 0 Graph 4

Exact solution It Graph 5

0 F;; IIII 0.00 0.02 0.04 0.; 6 0.68 0.10 0.12 0.14 0.16 0.

Twist in the axial direction at the mid span rod.

cb. Kumin ýluý

)WO Not IX Vr Im R= IM 1= WK WY WZ

I , .

0 0 0 . 0 -. 0 11

Wtas Any dto'ft or treed= ts tismd. d.. wted by I Arw dný or fs to free. d-ted bi 0

4 FIGURE 10.2 8

174

Torsional buckling of a pin ended strut.

20 element solution 250001

200DD- 100 1RI tud Increment I ri J/F,

-00

kq

200

10000-1 1.0

v Vý an %Ktion X-X Z Legend

0) Fx x 100 .0N& mx a1 .5 Nmm A Graph 1 Watt,

all doinsions are in ma Fx a 100 -ON 9 mx a35 Nm X Graph 2 E 0 thickness 10 Min all ow Fx z1OO. ON & mxs4. ONmm 0 Grapn 3 I

I'ItAlr

Fx z1OO. O N9 mx a 6.0 Nmm 0 Graph 4 If

0.00o 0. ý05 1). ý10 0.015 0.020 0. ý25 0-ý30 0.035 0.040 TWist in the axial direction at the rniid span rod.

C-WLMIN n-WWO.

w0de No: Dr Ez R-x im a= ux vy wz

I I I I 1 0 2 0 0 a 0 0 3 0 0 0 0 0 Al 0 0 a a a 5 0 0 0 0 0 6 0 0 0 a 0 7 0 0 0 a 0 a a 0 a 0 40 9 0 0 0 0 0 10 0 0 a a 0 11 0 0 0 0 0 12 a a 1) a a 13 0 0 0 0 0 ift 0 0 a 0 0 15 0 0 a 0 0 16 0 0 0 0 0 17 0 0 0 a 0 la 0 0 a 0 0 19 0 0 0 0 0 20 0 0 0 0 0 21 0 1 1 1 0

mtes Any deg or frvW-0 Is ri=d. dýted by I ArW OW m or fr.. dm to true. dcrated by 0

FIGURE 10-. 29

175

25000

.0 x 0 20000 Ju X

0 0 15000

10000

0.

Comparison 10&20 elements, Torsional Buckling of a pin ended strut.

k,

Vim on saftion Y-X

FItoad Im«c'

wcneý aU dimensions are in an ftichnus 1.0 as &U aw Legend

Fx z100-0 & mx a 3-5 Nmm. 10 elements A Graph 1 Fx m100-0 & nix a 3-5Nmm. 20 elements x Graph 2

0 ,I n 0.0 0 . 0.005 0.010 0.015 0-60 0.5

Twist in the axiol direction of the rnid spon rod.

Cmwwajh ýItiý.

W-do No: EX of m RM R= F= ux vy la

I I I I 0 1 1 10 2 - . - . ..: 11 3 0 0 0 000 11

0 . 0 0 S II 0 0- 0 0 0 . . . 11 7 0 0 0 11 00011

0a00 00

10 0 0 0 000011 11 . I I 1 11 .. .11

nitri1n cnd1tto,.

Node )bz ZX DY EZ ROIX RM MM VX VY la

I I I I 1 0 2 0 0 0 0 a 0 0 0 11 0 0 0 0 . : 0 " a 0 1. 1 0 11 0 0 11 7

: : : 0 0 1 1 : . : . I I 10 0 0 : 0 11 0 . : : : . 12 0 0 0 0 0 1 1 13 a 0 0 a 0 1 1 1 4 0 1 , 1 1 1 5 . 0 1 1

1 , 17 0 15 0 0 0 0 0 1 1 1. 0 0 0 1 0 1 1 20 : 0 0 "1 0 0 21 1 . 0 0

Note: Any dv-e or fýdm is fixd, d by I AM degý of fýdm In free, dýted by 0

FIGURE 10-130

176

Effect of axial load increment on Torsional Buckling

10 element solution 30000-

0 25000-

20000- 7 4)

X

0) C 0 Aso 0 15000-

--1 200

L)

1.0 10000-

0 Vý on Srr"Oft X-x

In W 49 diavalong are In as SDOO-

thOnts to on all ever Legend E 0 Fx a 100.0 N, mX z 3.5 Nmm L Graph

FX z150-0 N. mx a 3.5 Nmm X Graph 2

0.000 0.005 0.010 0.015 0.020 0.65 0.030 Twist in the axial direction of the rriml span rO&

C. mLrala Cvzdltiý.

Node Pb: Dt pf ZZ IIM F= Ic= %lx VY IIZ

I I 1 0 0 0 11 0 0 a 0 1 1

3 a 0 0 0 1 1

1 1 1

0 1 1

: 0 11 10 0 0 a 0 0 0 11 . I I 1 0 0 0 11

bb%at Am ämý er rived.. In fl.. d. dýmiad by 1 ftu u-c - er fr.. a- tu frft. dwwud bi 0

FlOuRE1 0.31

177

Effect of axial food increment on Torsional Duckling .

30000

9; x c 25000

.0 2000D

0 15000

U D

10000

5000

CD=U, atn cbrdjtiý.

20 element solution

)dmk bbs EX DY CE x= MM R=z lix VY UZ

I I I I I a 0 0 11 2 0 0 0 a 0 0 0 3 0 0 0 0 0 0 0 IN 0 0 0 0 0 0 0 11

0 0 0 0 0 a 0 1 1 0 0 0 0 0 0 0 . 11 7 0 0 0 0 0 0 11

0 o 0 0 a a II a 0 0 0 0 0 11 10 0 0 0 0 0 0 0 11

11 0 0 0 a 0 0 0 11 12 0 0 0 0 0 0 0 11 13 0 0 0 a 0 0 0 11 1% 0 0 0 0 0 0 0 11 Is 0 0 a 0 0 0 0 11 16 0 0 0 0 0 0 0 11 17 0 0 a 0 0 0 0 11 Is 0 0 0 0 0 0 0 11 It 0 0 9 0 0 0 0 11 20 a 0 D 0 0 0 0 11 21 0 1 1 1 0 0 0 11

NOW* &W dcfý Or fr"dom Is filed* dwoted by I AfW d=ý Or 1r-cdm IN rme. d. -Wd by 0

F

Awn 0

201

wvftý A do-mism are in a. lkkbms tf as all ow Legend

Fxm 100-0 N. mx x3_5 Nmm L Graph 1 Fx a 150-0 N. mic =3-5 Nmm X Graph 2

FIGURE 10.32

0.000 0.005 0.010 0.015 0.020 0.025 Twist in the axial direction at the r6id span rod.

178

-The results obtained for the case of free torsional buckling is shown in Figures (10.28) and (10.29). It can be seen that up to the value of the axial thrust of 9000N, the behaviour of the strut is almost linear. Beyond this value the trend becomes non-linear, leading to buckling. As before, the analysis was extended to lower increments of the axial load and the results obtained are shown in Table (10. T. 9).

6

Elastic Stiffening and Large Deflection Phenomenon of a Cantilever:

The channel section cantilever shown in Figure (10.1) is constrained

at its fixed end except for the warping degrees of freedom. It is

also constrained for the lateral degrees of freedom at all nodes for

the two loading conditions in the x-y and x-z planes, and loaded with a transverse force acting on the shear centre at its free end. The

6 Da2

results of the normalized deflection and the normalized load ('ý ) El

at the free end for both the x-y and x-z planes are shown in Figures

(10.31), (10.34) and (10.35). These results are. compared with the

exact solution obtained by D. C. Drucker (see reference 8) and also

with a finite element solution by T. Y. Yang (see reference 64). For

single element modes, the agreement was quite satisfactory, and using 10 elements the results show an almost perfect agreement with the

exact solution.

Large deflection analysis of a thin walled channel section cantilever beam loaded at the centroid:

The arrangement illustrated in Figure (10.36) is similar to that of

the previous example, except that the transverse force at the free end

acts through the centroid. The warping displacement is constrained at

the root. No significant axial twist was computed when the force was

applied along the y axis. This is because the line joining the shear

centre and the centroid is overlapped by the direction of the

transverse force.

179

0 CD C) ZA o

CD (D LJ

00 .ý a) .ý

U 4-3 CD CD L- = o

00ý --t 4J 4- 0

W CV)

to r_ 41 UU

o- 4-3 -1& U 42) L- :3

0 . -(Ijlo V) u0 CD C) L- L- = o 00 4-3 -, It

4-) 4- 0) 1-4 E m CY)

dia C: 0 LLJ

LI: 4-3 -be U

LO U') C) C>

CD E X X cli C%j cr) (Y) CC) co

Cý CD -. 1 1-4 X X

LAJ E CO co C) CD C%j cli

co co C) (D -4 -4

ý x x r- E 1-4

E -Kr CV)

cc CD Cý 10i ME m m

E cn CV) cn

CD Cl -4 1-1 x x

NE 4"t Kr E cli C%i

C-i CIO

tD to CD C) X x

lc: r Wýr C\j cli cli C%i

C) CD

to to 4 1-4

4- 0 r_

(L) C) C) EE cli

LLJ

ýz 8 Uj

-j ca

180

Elastic' stiffening and large deflection phenomenon of a cantilever

C 0.7- 0Y

j rX Y J-

I

as- t

C 04-

.2 ! -Y a; 0.3- Note,

all diffiens*. Onz are in MA w

0.2- thickness 2.0 imisall Dwf V. Pv on wrt" x- Legend

1 element solution A Graph E 0 10 element solution 0 Graph 2

0.1- B element solution by T. Y. Yang [3 Graph 3

Exact solution by K. E. BisShopp 0 Graph 4 of III

01246ia2 Normalized load ociing at the tip of the cantilever. Fy-l

El

cm. ft"Ill cmditf".

W-de Not EX DY M RM RDTY F= NX VY IJZ 1111111011 2001100011

O-dru'ala CN4ittom.

N-de Ne: IX Llt CZ x= RM x= NX wy IJZ 1 1 1 1 1 1 1 0 11 2 0 0 1 1 1 0 0 11 3 0 0 1 1 1 0 0 11

0 0 1 1 1 0 0 11 0 0 1 1 1 0 0 11 0 0 1 1 1 0 0 11 0 0 1 1 1 0 0 11

IN 0 0 1 1 1 0 0 11 9 0 0 1 1 1 0 0 11 10 0 0 1 1 1 0 0 11 11 0 0 1 1 1 0 0 11

3dote3 AnY &. a of f,.. dm In fixd. d.. Ytad by I AnY 6*cý ct rrwti-- is fý. d-ated by 0

FIGURE 10.

181

Elastic stiffening and large deflection phenomenon of a cantilever

0 U

0

. 9- 0.6-

237 21

k- -Y

0 0.4-

X-X

a)

.N Legend

all d'Mens'. Ons am In IM 0.2- E thichnns 2.0 an all O. W

I element solution L Graph

0 10 element Solution C Graph 2

8 element solution by T. Y. Yong D Graph 3 Exact solution by r, E. Bisshopp 0 Graph 4

ii456i 10 Normalized load acting of the Up of the cantilever.

ons%min ateditigug.

Nods Not lix Dr Ve F= R= It= WX Vf WZ

.0 11 11

carommID omdilucm.

bwe N32 Ex DY rz l= PM i= vx VY uz 1 1 1 1 1 1 1 0 1 1 2 0 1 0 1 0 1 0 1 1 3 0 1 0 1 0 1 0 1 1

1 0 1 0 1 0 1 1

6 0 7 0 1 0 1 0 1 0 1 1 6 0 1 0 1 0 1 0 1 1 9 0 1 0 1 0 1 0 1 1 10 0 1 0 1 0 1 0 1 1 11 0 1 0 1 0 1 0 1 1

Notez »w dww or f im fixed. demt4d by 1 kW 9tWýft er rýOM JS frý. 0. »t" bi 0

t.

FIGURE 10.34

182

Comparison of elastic stiffening on XY&XZ planes with published work. 10 element solution comparison

0.9- Lal -

0.8- 2 C a C,

0.7 - y&Z

U

. W,

rx x a 6

orb I LX . 9-

0.5- 23

C

.2 0.4-

F OrF

U 9-Y WOO, 0.3- A jtj IrLaam; ens are in as ET

Wiarass 2.0 - a" Gver VWJ M Vche" X-X Legend

'5 E

0.2- 10 elemenj solution loading in z direction A Graph

0 10 element solution loading in y direction 0 Graph 2

Z 0.1- 8 element solution by T. Y. Yang a Graph 3

Exact solution by K. E. Bisshopp 0 Graph 4

a_ 1 fIII 23456ii1,

Normalized load acting at the lip of the cantilever. lFyw FzI - 12 - E I

COMMIM

Wo6e IbS LIK DY Ez It= 932T R= lix VY ijz I 2

I 0

I 1

I 0

I 1

I 0

1 1

0 0

1 1

1 1 3 0 1 0 1 0 1 0

0 1 1

I 1 . 0

: 1 1 1 .

00 11 11 00 1 1 0 1 0 1 0 1 1 . 1 . 1 0 1 1

Wwt4: Any derres or f to Is fi-d. 6. mtad bY I ArW decre or fý*m Is f"w. 6um%ed W0

FIGM 10.35

183

r-X y Lx

J-x

4n -1

1.6

S! - N ote. - all dimens; ons are in mm --122.0 thickness 2.0 mm all over View on smtion X-X

FIGURE 10.36

A transverse force of 5. ON was applied along the z direction through the centroid and a significant twist along the span was observed as shown in Table (10.17.10).

ZI

137

C C, SC

C C, d,

224) 1

View on sertion X-X

700.0

x ý-Ex F Note: - 2 all dimensions are in Ma FIGURE 10.37 thickness 1-6 mm all wer

In this situation, where displacements are small, the twist produced due to the offset transverse force is in agreement with that of the

exact solution. The structure was then loaded with 5. ON load

increments. The axial twists produced were compared with experimental

and the continuum solutions produced by Gaafar and Tidbury (see

reference 19).

184

The cantilever was remodelled with meshes of 9 and 18 elements and the results obtained are shown in Figures (10.35). (10.39) and (10.40). A discrepancy was noted for transverse forces greater than 20ON and the maximum errors are shown in Table (IO. T. 11).

No of Elements Distance from the The error when the load fixed end, x (mm) at the free end is 700N, %

9 360.0 33.0 9 560.0 -30.0 9 700.0 16.5

18 360.0 0.0 18 560.0 -8.16 18 700.0 20.8

TABLE 1O. T. 11 Note: If the angle of twist predicted by the computer program is greater than the experimental observations, then the error is positive.

Large displacement analysis of a shallow truss:

The structure shown in Figure (10.4ý) has been analysed by Williams(64)

and is used in several publications as a benchmark. When loaded at

the apex the behaviour is similar to a toggle, and represents a very

severe test to any non-linear solution technique.

1 0753

-! LaL-. 41 Fyu aS tb I Increment)

12-p4

View on section X-X

Note: - all, dimensions are in inches Sig 10-0201a 1-41SIdeg)

thickness 0.243 in all ever

FIGURE 10.42

185

Test Description

k, Node 1 x oord y coord z coord

Node 2 x coord y coord z coord

0.0 0.0 0.0

a

114>- *

9. (Irn)

Iyy (MM4) Izz (MM4)

r (MM6) E (N/MM2)

G (N/MM2) Fx (N) Fy ( N) Fz (N) dx (MM) dy (MM)

dz (MM)

Rotx (rad)

Roty (rad)

Rotz (rad)

700.0 0.0 0.0

700.0 0.454xlo5 0.711X, 04 0.241X107 0.208X, 06 0.832405 0.0 0.0

10.0 0.0 0.0 0.211 0.65640-2

-0.43240-3 0.0

IL

0.0 0.0 0.0 0.0 0.0 0.0

700.0 700.0 0.0 0.0 0.0 0.0

700.0 0.454405 0.711X, 04 0.241407 0.208xiO6 0.832405 0.0 0.0

-10.0 0.0 0.0

-0.211 -0.65640-2 0.43240-3 0.0

700.0 0.454105 0.711X, 04 0.241X, 07 0.208xlo6 0.832405 0.0

10.0 0.0 0.0 0.733 0.0 0.0 0.0

-0.166xlo-2

0.0 0.0 0.0

700.0 0.0 0.0

700.0 0.454105 0.711X, 04 0.241X, 07 0.208406 0.832405 0.0

-10.0 0.0 0.0

-0.773 0.0 0.0 0.0 0.16640-2

TABLE (IO. T. 10)

186

Comparison of axial Wst of 9&18 elements at x=360 with published work

2 700 V

600

500

0

300

p 200

C

12 100

-X . .-I&.

- SC C :3y

7000

X Vol

X , o,,

NOL-

all domarisions are in as

thickma 1.6 so all off Legend

Experimental results of Goafar 8, Tidbury A Graph 1

Exact solution of Goolar & Tidbury X Graph 2

10 finite element solution 0 Graph 3

20 finite element solution 0 Graph 4

iiS67 Twist in the oxial direction at x=360. deg.

0 0

CNWLMID 0.41ticam.

ND60 lb: Ix Vf M M= MY It= WX wy UZ I I I I I I 1 0 11 2 0 1 0 0 0 1 0 11 3 0 0 0 0 1 0 11 4 0 11 0 a

0 0 1 0 0 0 1 0 11

0 a 0 1 11

Q=tross Cw4itims.

WO& Wo: IX IN EZ It= RMY R= SIX WY IIZ I I I I I I 1 0 11 2 a 1 0 0 0 1 0 11 3 0 1 a 0 0 1 0 11 4 0 1 0 1) 0 1 0 11 5 0 1 0 0 0 1 0 11 6 0 1 0 0 0 1 0 11 7 0 1 0 0 a 1 0 11 a 0 1 0 0 0 1 a II 9 0 1 0 0 0 1 a II

a 1 0 0 0 1 0 11 o f a 0 0 1 0 11

12 a 1 0 0 0 1 0 11 13 0 1 o o 0 1 0 11 "I : 1 0 0 c 1 0 1 1 15 1 . . o 1 . 1 1 16 0 1 0 0 11 1 1 1 17 . I . . . I

: 1 1

1: 0 1 c 1 , 1 , 1 1 , 1 1 11 0 1 o . 0 1 0 1 1

21 0 1 0 0 0 1 a II Note: Any degme or fýdm Is fixed. demted bY I

Am decree or fret6m Is free. desawd by 0

FIGURE 10-3b

187

Comparison of axial twist of 9&18 elements at x=ý560 With published work

300-

700- W

0) 600-

W SIC

500- 19201

- _ny

70"

X6 0X

0) 300-

doe"Siam are j.

ý 200- bms 1-6 on au Wer Legend

Experimnial restits of GaafCr Z 7idbUry A Graph C a E-ct sdLdian of G"afor & 7' 'dt"y X Graph 2 1'- 100- 10 finite element solution D Graph 3 20 finite eleffent soU; on 0 Graph 4

0246 Twist in the axial direction cif x--560. deg. lex)

Dwowlis fk-ftlý.

Nods lb: Dc Vt M Itarx X= M= VX WY UZ I I I I I I 1 0 11 2 0 1 a a D 1 0 11 a 0 1 a 0 a 1 0 11 4 a 1 0 a a 1 0 11 5 0 1 0 0 0 1 0 11 6 0 1 0 9 a I a 11 7 D 1 0 0 a 1 0 1 a 0 1 a a a 1 0 1 9 0 1 0 0 0 1 0 11 10 0 1 a 0 0 1 0 11 11 0

.1 0 0 0 1 a Ii

Mmumift C. Aitiý.

Node No: IX DY M R= MY R= %m VY %IZ I I I I I I 1 0 1 1 2 0 1 0 0 0 1 0 1 1 3 0 1 a 0 0 1 0 1 1 4 0 1 a 10 0 3 0 1 1 5 0 1 0 0 0 1 0 1 1 6 0 1 0 0 0 1 0 1 1 7 0 1 0 0 0 1 a I I a 0 1 0 0 0 1 0 1 1 9 0 1 0 0 0 1 0 1 1 10 0 1 0 0 0 1 0 1 1 is 0 f 0 0 0 1 0 1 1 12 0 1 0 0 0 1 0 1 1 13 0 1 a 0 0 1 a 1 1 14 0 1 0 0 a 1 0 1 1 Is 0 1 0 0 a 1 0 1 1 16 0 1 0 0 0 1 0 1 1 17 0 1 0 0 0 1 0 1 1 le 0 1 0 1) 0 1 0 1 1 19 0 1 0 0 a 1 0 1 1 20 0 1 0 0 0 1 0 1 1 21 0 1 0 9 0 1 0 1 1

una: Any 4kc-t or rýdm is fixed. dýLed by I ArW dep" or fmcdom to fma. dýLtd by 0

FIGURE 10.39

188

Comparison of axial Wist of 9&18 elements of x--700 with published work

BOD- z

6OD- oh

0 Sao- view on st(fiam v

z

400- 700.0 0) u

0 x

W 2 300-

udft- F doonzions are in No

200- 6 on All &W Le end > g

arifor & Tidbury L Graph ExPedmentcd results Of 6 0

o, '- Exact suLdian Of Goafar & 'ndbury X Graph 2 too-

10 ffite element soUion 0 Graph 3

20 frite element sokdion 0 Graph 4

0 0 2

Twist in the axial direction at x--70D. deg. lex)

omwz"» O-diti-.

H-do lb2 IX IN CZ It= a= Jl= WX vy IJZ I I I I I I 1 0 1 1 2 0 1 0 0 0 1 0 1 1 3 1 0 D 0 1 0 1 1

1 0 1 1 1 . 1 1 1 0 1 1

7 1 . 1 . 1 0 0 0 1 0 1

0 1 a a 0 1 0 1 10

ammtmin aw41ti..

Noft Not CDC vr ta RM z= it= lix wy %m I 2

I 0

I 1

I 0

I 0

I 0

1 1

0 0

1 1

1 1

3 0 1 0 0 0 1 0 1 0 1

1 0 0

0 9

a 0

1 1

0 0

1 1 1 6 1 0 0 a 1 0 1 1 7 0 1 0 0 a 1 0 1 1 a .0 1 0 0 a I a I I 9 0 1 a 0 0 1 0 1 1 10 0 1 0 0 0 1 0 1 1 11 o f o 0 0 1 0 1 1 12 0 1 0 0 0 1 0 1 1 13 0 1 0 0 0 1 0 1 1 14 0 1 0 0 0 1 0 1 1

is a 1 9 0 0 1 0 1 1 16 a 1 0 0 0 1 0 1 17 0 1 0 0 0 1 0 1 to 0 1 0 a 0 1 0 1 1 19 0 1 0 0 a 1 0 1 1 20 0 1 0 0 0 1 0 1 21 0 1 0 0 0 1 9 1

Wlte: AMY dtCr" Of rr"Gos to tb-d. G. ML*d by I Any dcme or fr"d- Is t'".. dowted by 0

FIGURE 10.40

Elastic stiffening and large deflection phenomenon of a cantilever

1000

9- -D C

I,

C, -C

a

-5 0 I. C C, U C 0 C,

0

C) I, I- C,

II, C 0

goo

60C

401

20

i-"

Cb, anivis 0., dltlý.

N. ds Nos EX DY Im il= MY lit= UX WY %M I I I I I I 1 0 1 1

2 0 1 0 0 0 1 0 1 1 3 0 1 0 0 0 1 a I I Al 0 1 0 0 0 1 0 1 1 5 0 1 0 0 0 1 0 1 1 6 0 1 0 0 0 1 a I I 7 0 1 0 0 a 1 0 1 1

0 1 0 0 a I a I I 0 1 0 0 0 1 0 1 1

10 0 1 0 0 0 1 a I I 11 1 0 1

Cmstr. 10 Omwitions.

)we Not rDc IN m R= icily a= wx wy L7 I 2

I 0

I 1

I 0

I 0

I 0

1 1

0 0

1 1

1 1 3

A a 0

I 1

a 0

0 0

0 0

1 1

0 0

1 1

1 1 5

a 0 a

1 1

0 0

0 0

0 0

1 1

0 0

1 1

1 1 7 0 1 0 a 0 1 0

a 9

0 0

1 1

0 0

0 0

0 0

1 1

a 0 1 1 so

Is 0 0 1

f- : : 00 1.

12 0 1 0 0 0 1 13 0 1 0 0 0 1 0 1 1 Sol 15

0 0

1 1

a 0

0 0

0 0

1 1

0 0

1 1

1 1

16 0 1 0 0 0 1 0 1 1 17 0 1 0 0 0 1 0 1 1 is 19

0 0

1 1

0 0

1) 0

a 0

1 1

0 0

1 1

1 1 20 0 1 0 0 0 1 0 1 1 2t a 1 0 0 0 1 0 1 1

N-%e: MY dtfyve of r wed Is fixed. d-01.4 b. v I Arv degrve of rm. dm is I'T". dý%. d b, 0

21

12201

V. fv on section Y-W

an Graph 1

FIGURE 10.41

In-plone defledion ct ihe free end mrrL 116Z 1

190

The structure shown in Figure (10.42) was model led with two meshes of

10 and 20 elements with an initial load increment of 1.0 lb. The load

displacement history predicted is shown in Figure (10.43) and compared

with the theoretical and experimental results of Williams. To improve

the accuracy, the problem under consideration was remodelled with the

load increment halved (0.5 lb). No significant change in the result

was found and in both tests a maximum error of approximately 22% was

recorded at a maximum load of 70 lbs.

191

Large deflection of a eigidly fixed ioggle*looded transversely at apex

60- 0759

AM

yn

Vwv on 1016DA X. X

40- y OSM14031 fdl

.0 'LCIL. 12ft Z;

30- X

*0 20- SIL di-enslars are is iswbes all"MI-1-41SUI-11

thickness 0 243 in all wer Legend

Experimental results of Williams A Graph

10 0 Exact solution of WiffiamS. X Graph 2

20 We element solution 0 Graph 3

It) finite element solution 0 Graph 4 ol I 0.0 0.1 0.2 0.3 CA 0.5 0.6

Deflection in Y direction at the apex inches

cwwtmin cbndfuý.

N. de wog EX DY tz RDIX MMY x= VX VY %IZ

0 0 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 1

41 0 0 1 0 1 0 0 1 1 5 0 0 1 0 1 0 0 1 1 6 0 0 1 0 1 0 0 1 1 7 0 0 1 0 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 9 0 0 1 0 1 0 0 1 1 10 0 0 1 0 1 0 0 1 1 11 1 1 1 1 1 1 1 1 1

Oxwtrain QwditJons.

Nods lb: rx Dy in RM It= it= wx wy vz I I II I I I I 1 1 2 0 01 . 1 0 0 1 1 3 0 01 0 1 . 0 1 1

0 01 0 1 0 a I I 11 11 1 0 1 0 0 1 1 0 01 1 1 . . 1 1

7 a 01 0 1 0 0 1 1 0 01 0 1 0 1 0 . 1

10 . .I a I a a 11 0 aI a 1 0 .0 1 1

0 01 0 1 0 0 1

0 03 0 1 0 0 1 1 11 1 0 1 11 0 1 1

Is 01 0 1 . . I I 0 01 0 1 0 0 1 1 0 1 . 0 1 1 0 0 ' l 1 1 , 1, 1 1 0 . I l 1 0 1 1

2. 0 01 0 s a a 1 1 21 1 1. I I I I I I I

Holes AW derrvt of freedam to fixed, devoted by I Any derme or rr.. dm is rr.. dtd b, c

FIGUPE 10.43

19L

CHAPTER 11

DISCUSSION AND CONCLUSIONS

In the majority of the literature published in the past, two independent reference axis systems are considered in deriving the

governing differential equations of thin walled open section structural members.

1. For displacements related to axial and flexural loads an axis system parallel to the principal axis located at the centroid

2. For displacements related to torsional and warping moments an axis system parallel to the principal axis located at the shear

centre.

This situation leads to some confusion as, in practice, the Joint

construction normally dictates the position of the loading, and also in many cases, the centroid and the shear centre are not coincident. Furthermore for the sake of simplicity, generally the authors in the

past only considered cases where the local elemental axial direction

was parallel to one of the global axes. Unfortunately however, this

option certainly restricted the classes of problem that are likely

to be solved by the established theory. However in this thesis, the

situation has been overcome by formulating a transformation matrix,

as shown in equation (8.14.0), while taking into account any of the

secondary forces that are likely to develop due to offset loadings.

The work conducted by Gaafar and Tidbury (19) was carefully formulated into a finite element approach and included in the total

potential formulation as shown in eqijation (4.72.0). Without this

modification, solving large displacement structural problems would

not have not been possible and instead the analysis would have been

rather limited to just bifurcation type of problems. Briefly the

refinements and devel6pmerits that are introduced into the present theory of the thin walled beam structural members are as follows:

193

1. The coupling effects of the resultant bending in two principal planes, torsion and torsion warping, due to an arbitrarily placed axial load Fx on the cross-sectional plane. For details see equations (4.31.3), (4.32.3) and (8.14.0);

2. Inclusion of unequal torsion and warping moment loads as suggested by Gaafar and Tidbury (19) to the generalised finite element theory of buckling. For details see equation (4.72.0);

3. Method of transformation of sectorial properties from the local cross- sectional plane to the global axis system in the space. For details see equations (8.14.0) and (8.30.0);

4. Development of local-global transformation matrix to encounter secondary forces and moments developed due to application of offset forces and moments. For details see Figure 8.2.0 and equation (8.14.0);

5. Theoretical prediction of the displaced position of the third node and the local axis system relative to the global axis system due to previous load increments. For details see ejuation (7.9.0);

A series of tests were carried out to examine the accuracy of the cýTputer predictions made by the following parts of the theory with the published work:

a) the linear elastic theory b) the local-global transformation matrix c) the non-linear elastic theory.

The linear theory of bending for a channel section beam was tested for both principal pl anes as illustrated in Figure 10.1. The local

axis system of the structure was purposely placed coincident with the global axis system so that no contribution from the transformation matrix was assumed to be made to the analysis. The

194

critical buckling load predicted for a single element was compared with the Euler-Bernoulli theory and found to be exact as shown in Table 10. T. 1. Under a similar geometric condition the structure was next tested for St Venant torsion and torsion warping problems, as shown by Figures 10.2 and 10.3. The results obtained for both cases were compared with the exact solution shown in equation (10.3) and found to be exact as shown in Table IO. T. 2.

To examine the behaviour of the transformation matrix under bending

only, a cruciform sectioned beam was placed in the space (note under torsion cruciform sections do not warp) as shown by Figures 10.4 and 10.5. At first the beam was loaded in the z direction for the case

where the lor-al and the global axes are coincident as shown in

Figure 10.4. The beam was next placed in the space, all degrees of freedom were constrained at node 1 and loaded at node 2 in the X, y

and z directions so that the resultant force at node 2 still

amounted to 61.237N as shown in Figure 10.5. The displacements

produced for both test cases were compared, as shown in Table 10. T. 3

and found to be exact. Therefore from this test it can be seen that

the bending feature of the transformation matrix appears to function

satisfactorily. A very similar test was carried out for torsion and torsion warping effects of a channel section cantilever, as shown in

Figures 10.6 and 10.7. As before initially the structure was placed

coincident with the global axis system, constrained at node 1 and loaded with a torsional moment at node 2. Then the beam was placed ' Iz in the space, constrained at node 1, and loaded with moments Mx, My

and Mz as shown in Figure 10.7. The results produced for both cases

were compared and found to be exact as shown in Table 10. T. 4.

To examine the performance of the transformation matrix for offset loading, the channel section cantilever was next loaded at the

centroid along the y and z directions as shown in Figures 10.36 and

10.37. By intuition it can be seen that there should not be any

torsional displacement developed when loaded at the centroid in the

y direction. However for the case when loaded along the z

direction, the induced torsional moment amounts to Fzee y and thus

195

must produce a significant torsional displacement. The direction of the loads was also reversed to observe the sign and the accuracy of the results generated and was found to be very satisfactory with the expected results, as shown in Table 10. T. 10. However the exact solution for this particular problem is rather complicated as

pointed out by Gaafar and Tidbury (19) and thus the agreement of the

finite element result to that of the exact solution is found by

comparison with the experimental results shown in Figures 10.38,

10.39 and 10.40.

The theory developed for non-linear analysis in Chapter 5 was tested

for several classes of problems and these are as follows:

The channel section column illustrated in Figure 10.8 has a

maximum imperfection of 40 mm at the mid-span and varies

gradually over a span of 2000 mm. The column was tested for

flexural buckling under an axial load with an increment of 400. OON. The computed results of buckling load for 4 and 10

element meshes are shown in Figure 10.9. The errors estimated for both the meshes are listed in the table shown on page 151.

The solution produced for the 10 element model was found in

fairly good agreement with a maximum error of 2.9%, that of the

exact solution in equation (10.7).

2,, a) A perfect cruciform section strut of 400 mm span was tested

for flexural buckling under an axial load, as shown in

Figure 10.10. The results were obtained for 10 and 20

element meshes with a lateral perturbation force ranging

from 0.001N to LON as shown in Figures 10.12 and 10.13.

With the 10 element solution the predicted buckling load

became the same as the Euler load when the perturbation force was 0.001N, whereas for the 20 element solution, the

critical buckling load became exact with the Euler load when

the perturbation force was 0.01N (i. e. ten times greater

than that of the 10 element solution). By referring to

Figure 10.15 it can also be seen that as the load increment

196

is reduced by 50% the prediction rapidly became much closer to the exact solution.

b) Generally any strut under an axial load produces some degree of cross-sectional rotation about the shear centre and this has been described in detail in Chapter 4 (page 33). Therefore in order to investigate the flexural buckling of a flanged cruciform strut it is essential to

prevent any cross-sectional rotation caused by an axial load, as shown in Figure 10.17. However flanged cruciform sections under axial rotation produce warping displacements and therefore care must be taken not to restrain the warping displacements, as shown in Figure 10.17. However due to relatively large flexural stiffness, the magnitude of the perturbation force used for the flanged cruciform section strut was greater than that for the straightforward cruciform section strut. As before the strut was tested for 10 and 20 element meshes with a range of lateral perturbation forces and axial load increments. For the 10 element solution with 30ON load increment the predicted buckling load was found to be within 7.79% that of the Euler load, whereas for the 20 element mesh under similar loading

conditions, a critical buckling load was found within 5.35% that of the exact solution as shown in Table 10. T. 6.

3. The exact solution for torsional buckling under axial load is

shown in equation (10.8). In this test two types of structural sections were used:

a) a cruciform section with zero torsion warping constant b) a flanged cruciform section with 0.341 x 108mm6 of torsion

, warping constant

a) As before the cruciform section strut was tested for 10 and 20 element ; eshds. The structure was also tested for a range of axial load increments ranging between 50 to NON

with a perturbation torque acting at the mid-span ranging

197

between 0.001 Nmm to 0.5 Nmm. The results obtained for these tests are shown in Figures 10.23,10.24,10.25,10.26 and 10.27. The results were compared with the exact solution and found to be in perfect agreement, as shown in Table IO. T. 8.

b) The tests carried out for the torsional buckling of the flanged cruciform section are shown in Figures 10.28,10.29, 10.30,10.31 and 10.32. The axial load increment was kept

at WON or 150N. As in the case of flexural buckling due to

relatively large torsional rigidity (see equation 10.8), the

magnitude of the perturbation torque was kept between 1.0 NmM to 25.0 Nmm. The results predicted were compared with the exact shown in equation (10.8) and found to be in

perfect agreement as shown in table IO. T. 9-

4. The problems discussed so far have only dealt with flexural and torsional buckling of struts due to application of an axial load. In the following example, the criterion of elastic stiffening, rather than its "inverse" i. e. buckling, of a cantilever was investigated. The cantilever beam shown in Figure 10.33 was constrained laterally for all out-of-plane displacements, as shown in Figure 10.33. The structure was loaded transversely at the tip and the normalized in plane tip displacement and the applied load was plotted for meshes with 1

and 10 elements. The tests were repeated for both the y and z directions as shown in Figures 10.33 and 10.34. The results produced by both tests were compared with the exact solution produced by Bishop (8) and the non-coupling finite element solutions produced by Yang (65) and found to be in close agreement.

The cantilever was then loaded at the centroid in the z direction and all in plane displacements constrained as shown in Figure 10.38. The axial twist produced due to increasing transverse load acting at the centroid, was examined at distances of 360 mm, 560 min and 700 mm from the fixed end, as

shown in Figures 10.38,10.39 and 10.40. By referring to Figure 10.38 it can also be seen that for the axial twist at 360 mm from the fixed node the displacements produced by the theory

developed tend to show a better agreement to the experimental results than the exact solution produced by Gaafar and Tidbury (19). However at 560 mm for transverse load around 800N, the theory developed in the thesis tends to produce much greater torsional stiffening than the exact solution developed by Gaafar

and Tidbury. A maximum error of 8.16% was also estimated against the experimental results at a force magnitude of 8ooN. The torsional displacement produced for the 20 element mesh at 700 mm showed the highest early torsional stiffening. However

with the 10 element mesh the theory developed in the thesis

produced much less stiffer torsional displacements than the

results due to theory and the experiments conducted by Gaafar

and Tfdbur .A maximum error of 20% was estimated for torsional d1splacement at BOON against the experimental results.

5. Finally a highly nonlinear structure of a shallow truss was chosen to complete the analysis. By referring to the geometry constrained and the loading condition shown for the structure in Figure 10.43 it can be seen that this particular test should examine the theory for flexural as well as the inherited toggle- like behaviour of the structure under consideration. The

A predicted results for the test showed a general trend similar to the results produced by Williams (64). However, as in the

previous case, some early stiffening was also noticed for loads beyond 2S lb. By using a much refined mesh no significant change was made to the results and a maximum error of 22% was estimated at a maximum loading of 70 lb.

With regard to the discussion made so far, it can be concluded that for most problems the results of the linear analysis were found to be In Perfect agreem4; nt with that of the theory developed. However, the accuracy of the computer predfctions to the non-linear problems -5ýGws a favourable agreement with their exact solutions and, as it

199

was noticed for more complex problems like:

1. Large rotational twist of a cantilever when loaded transversely

at the centroid 2. The shallow truss with toggle effect loaded vertically down at

the apex

Some early stiffening and discrepancies were observed.

The theory and the computer program developed in the thesis are believed to have a much wider scope for solving general assemblies

of space frameworks with mixed thin walled beam structural members than in the past. To solve such problems by using the program developed, however, needs more verification of the results that it

predicts. The satisfactory results produced by the program for the

test problems dealt with so far do not necessarily provide a justification for its general use. Therefore for wider applications it is suggested that the program be thoroughly tested against

analytical and/or experimental results for a range of typical highly

nonlinear industrial problems prior to its possible general use. Once the program is thoroughly tested it can be used in:

1. Linear analysis of assemblies of beam elements in space frames

2. Nonlinear analysis of assemblies of beam elements in space *-I frames.

I

200

REFERENCES

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.4

2. Anderson, J. M. and Trahair, N. S. Stability of Monosymmetric Beams and Cantilevers, Journal of the Structural Division, ASCE, Vol. 98, No. ST1, Proc. Paper 8646, Jan, pp 269-286 (1972).

3. Austin, W. J., Yegian, S. and Tung, T. P. Lateral Buckling of the Elastically End Restrained Beams, Proceedings of the ASCE, Vol. 81, Sept. No. 673, April, pp 1-25, (1955).

4. Barsoum, S. Roshdy and Gallagher, Richard H. Finite Element Analysis of Torsional and Torsional Flexural Stability Problems, Int. Journal for Numerical Methods in Engineering, Vol. 2,335-352 (1970).

5. Bazant, Z. P. Non-uniform Torsion of Thin-walled Bars of Variable Cross-section,

International Association for Bridge and Structural Engineering,

Publication Vol. 25, pp 245-267 (1965).

6. Bazant, P. Zdenek and EI-Nimeiri Mahjoab. Large Deflection Spatial Buckling of Thin-walled Beams and Frames,

Journal of Engineering Mechanics Division, ASCE, 1973, pp 1259- 1279.

7. Biot, M. A. Mechanics of incremental Deformation, John Wiley and Son (1965).

-Zol

8. Bishop, K. E. and Drucker, D. C. Large Deflection of Cantilever Beams, Quart. Appl. Maths. 3,272 (1945).

9. Bleich, Friedrich and Bleich H. Hans. Buckling Strength of Metal Structures, McGraw-Hill Book Co.,

1952. k.

10. Bradford, A. Mark and Trahair, S. Nicholas. Distortional Buckling of I-Beams, Journal of the Structural Division, ASCE, 1981, pp 355-371. -

11. Chajer, Alexander and Winter, George. Torsional Flexural Buckling of Thin-walled Members. Journal of the Structural Division, ASCE, 1965, pp 103-124.

12. Chen-Wai-Fah and Atuto Toshio. Theory of Beam Col umns, Vol umes 1 and 2, McGraw-Hi 11 Book Co., 1977.

-13. Chen, W. F. and Atsuta, T. Interaction Equations for Biaxially Loaded Sections, Journal of the Structural Division, ASCE, Vol. 98, No. ST5, Proc. Paper 8902, pp 1035-1052, May (1972).

14. Clark, J. W. and Hill, H. N.

Lateral Buckling of Beams, Journal of the Structural Division,

ASCE, Vol. 86, No. ST7, Proc. Paper 2559, July pp 175-196,

(1960).

15. Culver, C. G. Exact Solution of the Biaxial Bending Equations, Journal of the

Structural Division, ASCE Vol. 92, No. ST2, Proc. Paper 4772, pp 63-83, April (1966a).

-202-

16. Dux, F. Peter and Kitiparuchai, Sritawat. Elastic Buckling of Laterally Continuous I-Beam, Journal of the Structural Division ASCE, 1982, pp 2099-2117.

17. Ehouney, M. Mohammed and Kirby B. Jeffrey. Warping Restraint in Three-dimensional Frames. Journal of the Structural Division ASCE, 1981, pp 1643-1656.

18. Fukumoto, Yuhshi and Hathori, Ryoji. Lateral Buckling Tests on Welded Continuous Beams, Journal of the Structural Division, 1982, pp 2245-2263.

19. Gaafar, M. L. and Tidbury, G. Large Deflections Analysis of a Thin-walled Channel Section Cantilever Beam, Int. Journal Mechanical Science, 1980, Vol. 22,

pp 755-766.

20. Galambos, T. V. Structural Members and Frames, Prentice Hall , Inc., Englewood Cliffs, New Jersey, (1968).

21. Galambos, T. V. and Fukumoto, Y. Inelastic Lateral -Torsional Buckling of Beam-Columns, Journal of the Structural Division, ASCE Vol. 92, No. ST2, Proc. Paper 4770, April, pp 41-60 (1966).

22. Gallagher, H. Richard. Finite Element Analysis Fundamentals, Prentice-Hall Inc., 1975.

23. Ghoborah, A. Ahmed and Tro K. Wai.

Overall and Local Buckling of Channel Columns, Journal of the Engineering Mechanics Division, ASCE, 1969, pp 447-461.

24. Gjelswik, Atle. The Theory of Thin-walled Bars, John Wiley and Son, '1981.

913

25. Hancock, J. Gregory. Interaction Buckling in I-Section Columns, Journal of the Structural Division, ASCE, 1981, pp 165-179.

26. Hancock, J. Gregory. Nonlinear Analysis of Thin-walled I-Section in Bending, Research

Report, University of Sydney, School of Civil Engineering.

ký 27. Hollinger, A. Bruce and Manglesdorf, C. P.

Inelastic Lateral Torsional Buckling of Beams, Journal of the Structural Division, ASCE, 1981, pp 1551-1566.

28. Horne, M. R. The Flexural Torsional Buckling of Members of Symmetrical I-

Sections Under Combined Thrust and Unequal Terminal Moments,

Quarterly Journal of Mechanics and Appli ed Mathematics, Vol. 7,

Part 4, pp 410-426 (1954).

29. Horne, M. R. The Plastic Design of Columns, British Constructional Steelwork

Association, Publication No. 23, London, (1964).

30. Johnston, B. G. Lateral Buckling of I-Section Columns with Eccentric Loads in

Plane of Web, Journal of Applied Mechanics, pp A-176 (1941).

31. Johnston, B. G. Buckling Behaviour above the Tangent Modules Load, Journal of the

Engineering Mechanics Division, ASCE. Vol. 87, No. EM6, Proc.

Paper 3019, pp 79-99, December (1961).

32. Johnston, B. G. The Structural Stability Research Council Guide to Stability

Design Criteria for Metal Structures, Jonh Wiley, 3rd edn, (1976).

ýzp A-

33. Kasement Chesda, Nishino Fumio and Lee-Seng-Lip. Inelastic Stability of Beams Under Biaxial Bending, Journal of the Engineering Mechanics Division, 1974,965-989.

34. Kennedy, B. John and Madugula, K. S. Murity. Buckling of Angles, Journal of the Structural Division, ASCE, 1982, pp 1967-1977.

35. Kollbrunner, C. F. and Basler, K. Torsion in Structures, Springer-Verlag, Berlin, Heisdelberg, 1969.

36. Krahula, J. L. Analysis of Bent and Twisted Bar Using the Finite Element Method, AIAA, Journal 5 (1967).

37. Krajacinovic-Dusan, C. Consistent Discrete Element Technique for Thin-walled Assemblies, Int. Journal Solid Structures, 1969, Vol. 3, pp 639-662.

38. Livesley, R. K. Matrix Methods of Structural Analysis, Pergamon Press, Oxford (1964).

39. Martin, H. C. Finite Elements and the Analysis of Geometrically Nonlinear

Problems, Proceedings of Japan-US Seminar on Recent Advances in Matrix Methods of Structural Analysis and Design, Tokyo, Japan (1969).

40. Masur, F. Ernest. Large Deflection Spatial Buckling of Thin-walled Beams and Frames, Journal of the Engineering Mechanics ASCE, pp 82-87.

20ra

41. Mei , C. Coupled Vibration of Thin-walled Beams of Open Section Using the Finite Element Method, International Journal of Mechanical Sciences, Vol. 12, pp 883-891, (1970).

42. Murray, D. W. and Rajasekaran, S. A Technique for Formulating Beam Equations, Journal of the Engineering Mechanics Division-, ASCE, Vol. 101, No. EM10, Proc. Paper 11613, October, pp 561-573 (1975).

43. Nanayakkara, M. A. and Sharman, P. W. Formulation of Stiffness Matrices for Elastic Buckling of Thin-

walled Open Section Prismatic Columns, Internal Report, Department of Transport Technology, Loughborough University of Technology, TT 8402 (1984).

44. Nethercot, D. A. Recent Progress in the Application of the Finite Element Method to Problems of the Lateral Buckling of Beams, Proceedings, Conference on Finite Element Methods in Civil Engineering, Engineering Institute of Canada, Montreal, June, pp 367 (1972a).

45. Nethercot, D. A. The Solution of Inelastic Lateral Stability Problems by the Finite Element Method, Proceedings, 4th Australian Conference of Mechanics of Structures and Materials, Brisbane, pp 183-190, (1973d).

46. Nitzsche, N. and Miller, R. Robert. Torsion and Flexure of Curved Thin-walled Beams or Tubes, Journal

of the Engineering Mechanics Division, ASCE, 1972, pp 867-889.

47. Oden, J. T. Calculation of Geometric Stiffness Matrix for Complex Structures, AIAA Journal , Vol . 4, No. 8,1966, pp 1480-1482.

ZoG

48. Oden, J. T. Mechanics of Elastic Structures, McGraw-Hill Book Co., 1967.

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50. Powell, G. H. and Klingner, R. t'

Elastic Lateral Buckling of Steel Beams, Journal of the Structural Division, ASCE, Vol. 96, No. ST9, Proc. Paper 7555, September, pp 1919-1932, (1976).

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of Open Section, Proceedings, First International Conference on Finite Element Methods in Engineering, University of New South Wales, August 28-30, Sydney, Australian (1974).

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International Symposium on Computer Methods for Partial

Differential Equations, Leish University (1975).

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Z. 07

56. Seide, Paul. Accuracy of Some Numerical Methods for Column Buckling, Journal of the Engineering Divisions.

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See also a French translation by P. Massuet, Essai de Physique,

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2-09

64. Williams, F. W. An Approach to the Non-linear Behaviour of the Members of a Rigid Jointed Plane Frame Work with Finite Deflections, (Quarterly

Journal Mech. and Applied Math., Vol XVII, Pt. 4,1964,432-469).

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66. Zbriohowski-Koscia, K. Thin-walled Beams from Theory to Practice, Crosby Lockwood and Son Ltd., 19657.

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(1951).

zoct

APPENDIX I

Structural programming of the finite element technique is well established and therefore no attempt has been made in this thesis to discuss each stage and techniques that are fairly common to any ordinary finite element program. Instead it is intended to emphasise the unusual aspect of the parts of the program which are unique in

automatic data preparation for the thin walled beam column element.

As mentioned previously in Chapter 6, the non-linear behaviour is determined by solving a sequence of linear problems. The load is also

applied with a sequence of sufficiently small increments so that during loading the structure is assumed to respond linearly. For each load increment the displacement increment is computed by solving

equation 4.78.0. At each increment of loading the geometry of the

structure is also modified and updated by using equation 6.10.0, which basically implements the midpoint tangent technique. (For details of the method see Chapter 6).

Similarly a subsequent increment in load was also applied and the

process is repeated until the desired large displacement or the

collapse criterion is achieved and this is illustrated in the

following flow chart.

e2 (0

START

Read for each element the coordinate of the nodal points the node number and the cross-sectional geometry

Calculation of semi bandwidth

Read structural constraints and the loading vector

Setting up element stiffness matrix

For each element call subroutine 'GLOLOI and calculate,

a) The direction cosines of the local principal axis to the global reference axis system.

b) Calculations of shear centre and centroidal coordinates with respect to the global axis system.

C) Calculation of coordinates of the axial load, transverse loads and moments applications with respect to the global axis system.

d) Calculation of sectorial coordinate of the axial load application.

Subroutine 'GLOLO' internally calls up subroutine 'SPRO' which calculates the following for each cross-section of the element a) Position of the centroid and the shear centre b) Position of the principal axis and the second moment area

corresponds to each principal axis c) St Venent torsion constant, torsion warping constant and also

for each segmental area of the section, i. the sectorial coordinate

ii. linear sectorial moment of area and the statical moment of area about the two principal axis system

d) Calculation of Timoshenko torsional buckling constants

Formulation of element stiffness matrix subroutine BFW3D if the number of the run is greater than 1 then the formulation of the

geometric stiffness matrix subroutine BTW4D

211

Assembly of the stiffness matrix.

I Constraining of assembled matrix-

Call subroutine 'Reduce' Call subroutine 'Backsub'

resulting, establishment of nodal deformations.

Calculation of the global nodal forces.

Calculation of the new position of the third node is obtained by calling subroutine 'Third P'.

Accumulation of the forces in the global direction and convert them to corresponding components in the local direction which is to be used in the geometric stiffness

matrix in the next loading increment,

Accumulation of the displacement in the global direction.

Prepare next loading increment, predict new geometry and repeat the process from setting up element stiffness matrix.

If the buckling is achieved, the assembled stiffness matrix becomes singular or the displacement produced becomes

excessive. However in examples like the elastic problem of a cantilever they become progressively stiffer.

sl? -

APPENDIX 2 I

THE STRUCTURE OF SUBROUTINE 'SPRO'

Let us consider a general cross-section of a thin walled beam, as illustrated in Figurte A. M. In general, the geometry of the cros. s- section can be complicated by branching as in FigureA2.1. The section is divided into a number of manageable sizes of trapezoidal segments as shown. The origin of the axis which defines the geometry is

also arbitrary. C I

Let us estimate the position of the centroid and the segmental

properties for a typical segment shown in Figure A. M.

-d FIGURE A. 2.2

213

I

The geometry of the element mentioned is as follows:

Node 1 Node 2 Coordinates (nj, Cl) (r'2, C2) Thickness ti t2

and i denotes the ith segment length of the element =L distance to the local centroid from node 1 ý' nLJO

Therefore from basic geometry,

L The area of the segment = Ai =ý (tI, i + t2j) (A. 2.1) 7-

Li C2t2, i+tl, i) Distance to the local centroid from node 1= "L, i '-- -U'

(A. 2.2)

The second moment of area about rIL = InL

= Li

(t3, i+t2 2,

+t 3,

-2F8 2 2, itýi+t2, itl 110 (A. 2.3)

The second moment of area about ýL ICL 3

L22 i (t i+4t, (A. 2.4)

'K 2 it2, i+tl, i)

The product moment of area I %ýL

=0

Let us also define the inclination of the segments to the n axis as, 0

ý2, i Thus, tan6 = (A. 2.5) T'2, i Tl J

'214

However, by using the parallel axis theorem,

The second moment of area about the n axis

-2

n7 Ci ' (A. 2.6) I (ITIL + ICL) +1 (lTIL - ICL) Cos20 +AiCi

Similarly, the second moment of area about the ý axis

OTIL + '; L) ("nL - l; L) COS2 e+ Ai (A. 2.7) 2

The product moment of area = Ai

where, Tli + "L, i cOso

(A. 2.8)

sinO

Thus to obtain the properties for the entire cross-section by summing the segmental properties yields,

nA n

(A. 2.9.1) Ai

n AiTi

n (A. 2.9.2)

Ai

where, (TT) is the coordinates of the centroid of the entire cross- section from the arbitrarily chosen axes n and C.

'ZI

n is the number of finite segments that the cross-section is composed of. Thus, for the entire cross-section,

(Ai (T, 2

A-2-- 72 (n, - 2n ni +n+ llnLi

2 A 27n- Ai -ni +Ai -TL-2 +n

n n

2 Ai ýi n

n A )2

1 n Ai TIi2 n

Ai +-+ InLi JA Ai

n I Ai

2 2f

n

n A, T, 2 n

Ai)n2 + n

InLi) Y Ai A. 2.10.1)

j=l

Similarly, I n

AiT, 2 n Ai) T2 +

n IcLi (A. 2.10.2)

nyl

The product moment of area I nc

n Ai(-ni - -n)(-Ci +

n jI =l nLC L j=l i

n but 10 (see equation (A. 2.10.1)).

Therefore,

n Ai (ni ci - ni cin + n-C)

nc j=l

nnn AiTlici Ain Ai -Ci + Ai

21&

nnnn nAj Ai

Ai iii Ci

nn

+i

Ji A Ji

nAi Ai -niTi - (Ai) (A. 2.11)

Calculation of Sectorial properties:

The sectorial coordinates of a point with respect to a pole and a reference point is defined as the net area traced by the sliding radius, as it moves away from the reference point to the point of interest as shown in Figure (A. 2.3)

c

Fig. &. 2.3

where, D is a reference point placed arbitrarily on the outline of the

section C is a pole point placed arbitrarily on the (4-0 plane P is a point of interest defined by (Ti, c).

If pole C is replaced by the shear centre and CD is the principal radius, then the sectorial coordinate at point 'P' becomes unique and defined as the principal sectorial coordinate. A good detailed

analysis of prinaipal sectorial coordinates is found. on pages 20 and 21 of reference (66).

'alT

The position of the shear centre with respect to the (n. C) axis is given by,

icc Af

'OCDP CdA -I nc Af wCDP ndA

an = bT, +AII- 12

A

nn CC nc (A. 2.12)

inn f wCDP ndA I nc

f wCDP cdA bcAI112A

nn CC TIC

Sectorial coordinate with respect to an arbitrary pole and reference point:

e2 112

dA IL

ol " IL

FIGURE A. 2.4

The sectorial coordinate anywhere between the two nodes can be found

by approximating a linear variation of the nodal values as shown in

Figure (A. 2.4).

T'L

Therefore the sectorial coordinate w(nL) w wl + (4)2, i 10) L

The finite area,

'al 91 .,, "--, 7- -"- "- --"*'

dA(TIL) =dI- tj"i) + tl, i nL' lgtl, i + (t2, i

(t 2,1 nL +dnL

L

-t Tli

+ (t 2'i-tl") dnL dA('IL) = d"L Etl, i + (t2, i 2L

(A. 2.13)

, By ignoring second order terms,

Ti (110 = Eti, l + (ti, 2 - ti, l) "ý] dT'L (A. 2.14) L

Therefore the linear sectorial moment of area,

fudA =f( ol ,i+

(w2,, i )! -) d IIL A0- ul, i) 'LLL)(tl, i + (t2, i - tl, i nL

fwdA = L[wl, itl, i +w l(t2, i - tl, i) +

A2

(A. 2.15) 3

The linear sectorial moment of area about the axis is given by,

fCw dA Thus, A

f9 wdA= L nL n f (ýl, i + (C2, i + (w 2, i - ol, i)-j-) A0LL

(tl, i + (t2A tl, i)2L)dT, L

Zlot

DT i +t l, ilcl, i Dwi+DY iwl, i)

2

DT {yj ipw. +DY )Ayjýw. t DY-Dw-DT - ilol 'i + .1 -1. Lil (A. 2.16) 34

Similarly the linear sectorial moment of area about the c axis is

given by f.,?, wdA A

A fnwdA

= L[nl, iwl, itl. i nl. i(wl, iDTi+tl , iDwi)+Dniwl, itl, i

in, Dw-DT +Dni(wl, iDTi+tl Do. ) 1 11 1i

3

where, w= WCDP

DTi ý t2, i - tl, i

D=- wi, 1 �1

Dni " T'2, i - nl, i

Dq ` C2, i - Cl, i

2

Dn, Dw I DT.

(A. 2.17) 4 -1]

By substituting I nrl -, kc - ýC -in wdA and

iý wdA for the entire

cross-section in equations (A. 2.12.1) and ( . 2.12.2) the position of

the shear centre (a. a. ) can be calculated.

Calculation of principal sectorial coordinate:

To calculate the principal sectorial coordinates, the following

conditions must be met:

. 2zo

1. Shear centre as the pole point 2. Principal radius, which is defined by the principal reference

point chosen on the outline of the section and the shear centre.

Since the absolute position of the principal reference point is not relevant at this stage, we calculate the sectorial coordinate. In

, general this is done by using the shear centre as the pole point and also an arbitrary point on the outline as the reference point.

Thus by definition the condition that is required to satisfy principal sectorial coordinates Yields,

ýw dA

wo =Aý dA

(A. 2.18)

where wo is the sectorial ' coordinate of the principal reference point

on the outline of the section, providing that the shear centre is

chosen as the pole point and any other arbitrary point is as the

reference point.

Therefore the principal sectorial coordinate at the ith node of the

section is given by,

w P'i ý Wi - loo

Also, by definition the torsion warping constant r is given by,

n-W )2 Yf dA (, üp� 0 i=l A

2 2) dA =if (wp� - 2wp, 1 wo + wo i=l A n22n

= y' f (wp»i dA - 2wo if wdA + Wo j Ai (A. 2.19) i=l A i=l A i=l

, *Z? 4

where ý wdA is the principal linear sectorial moment of area and can n be calculated by substitutiný the principal sectorial coordinate into

equation (A. 2.15), wt Ai is the net area of the complete section.

Calculation of the torsion warping constant:

As before the distribution of the principal sectorial coordinate of the ith segment,

w (w2, i - wl, 1) "Li

(A. 2.20) "L. i Li

Similarly the distribution of the thickness of the segment,

ti *x tl, i + (t2, i - tl, i) "L

-i Li

(See also equation (A. 2.14)).

Therefore the torsion warping constant r, for the complete section,

nL 12 w2 dA =Ifw, 2 dA L i=l A0

"-ldn T11

Li

Cwl, iDTi+2Dui tl. i) 2Dwiwl, DT +Dw? t 2iii Li{wl, i tl, i + wl, i 23-

Dw*? DT +ý ii (A. 2.21)

4

22: L

where, as before, DTj = t2j - tl, i

'62, i- '01 ,i

Calculation of Timoshenko torsional buckling constant:

The phenomenon of torsional lateral buckling under axial load was described in Chapter 4 and the corresponding equations which govern the buckling constants are as follows:

L (f Z3 dA +fz y2 dA) - 2zo (A. 2.21.1) IyA

02 = _L (f y3 dA +fY Z2 dA) - 2yo (A. 2.21.1) IzAA

I S=0+e (A. 2.21.3)

0 -W z+ ey02

where, Iy and Iz are the second moments of area about the principal

axes, Io is the polar moment of area at the shear centre defined by,

io = lz +Iy+A (ez2 + ey2) (A. 2.21.3.1)

where, ey and ez are the coordinates of the shear centre with respect

to the principal axes system.

The position of the shear centre and the terms Iyj, Iz are already

established and by substituting into equation A2.21.3.1, the polar

moment of area at the shear centre is tabulated. The method of

estimating terms a, and 02 is as follows,

Z23

�1'

-'q

FIGURE A. 2.5

As before let us begin by assuming a linear variation of coordinates along the element as shown in Figure (A. 2.5). Thus the coordinate of an arbitrary point on the element can be expressed as,

ni, l + ('120 - nl, i) T'L, i (A. 2.22.1) Li

ý .) T'L, i (A. 2.22.2)

Let us first establish the term ý C3 dA. From equation (A. 2.22.1) the c coordinate of an arbitrary point is

given by,

Cl, i + (C2, i - Cl, i) T'L, i (A. 2.23) Li

ý 1ý By substituting equation (A. 2.23), for Ci in the expression above,

f C, 3 = jCl,, + (C2, i - Cl, i) L 1)3 dA (A. 2.24)

IZ2-11-

Also by substituting equation (A. 2.13) for dA in equation (A. 2.24)

yields,

nL - ýL, i f c13 dA = Lf'Cl, i + (C2, i - C"') L3

(ti, l + (t2, i-tl, i) L )dnL, i ii

n2 2 +3 itl,, D;. Lai + 3Cj, iDýitj, i(_Li)2

Li1Li Li

3 n' 3 DCi tij

ýLpi )3+ Cl, iDti

Li

2 nL i3 2 'ILgi)2 + Ul iD ýiDtj( wý 1,4) 3Cl, i DCiDti(

Li: p Li

3 TI ,- )41 dILi DýIiNi

L

C, 3 33222 dA = Clpitl, iLi +Z Cl, itl, iDCiLi + Cl, iDCitliLi

33 Dc tl, iLi + Dti Cl, i Li

4i

23213 -DT-L- + Cl, i DCiDtiLl + -5 DCI DtiLl ý1, i DC, ii -Z

(A. 2.2 5)

Finally the results of the equation shown above now provides the fC3dA

for a finite segment shown in Figure (A. 2.1). Thus for the entire

section,

, Z75

3n3+3222 ýl dA (Cl, itl, iLi Cl, itl, iDCiLi + Cl, iDCitl, iLi 2

3 Dc'i tl, iLi + zl DtiCl ,i

Li 4

232 Cl, i DCiDtiLi +W Cl, i DCi DTiLj + -ý' DCi3DtiLi)

(A. 2.26)

where, Dti = t2j - tl, i

Ni ý ý2, i - ýlj

A similar expression was also developed for the term j T13 dA:

3, (n3 222 ni dA 1', tl. iLi +-! nl, itl, iDniLi + nliDnitl, iLi 2

33 Dni tl, iLi + .1 Dti nl9i Li 42

223 Ill, i Dni DtiLi +1 nl, i Dni DtiLi + .1 Dni Dti Li)

45 (A. 2.27)

Let us now concentrate on the term C T12 dA. As before the corresponding term for the ith

2 segment f Ci Tli dA

-"L i Tl si )2. dA f 'Cl, i + (C2, i-ýlj) '11TI19i + (T'2. i-Tll, i)-17 Li I

{ti, l + (t2, i _ tl,, )r'L, i Id"L, i ri-

2Z

2n Lj C

Lf{ nl, iC,, iti'l + 2Ti-l, iDii 'i

i

2i Dýi "L, i Dri 1 (-Ei-- ipi 191 + ni --Ei- ti, i

+ 2n,, i Dni D; i (nLi)2 tl,, + MiDrl 2 "L, i)3 t -1-i i

+ 2n -Z1, i DiliDti( 711. i ýIj Dti -L i 1;, l -, i-

"L i2 rl L" DTI Dti (--ý 11 DCi Dti

Li9

T'LJ 3+ Dn 2

MiDtj ( nL, i)4}

dn +2 nl, i Dni Mi Dtj (iL Lj Fi i (A. 2.28)

After integrating:

n2 Jýj i dA {nl, iC,, iti,, Li + nj, jDnjLjCj, jtj, j

+2 Dný L- ýjjt n.,, i Dci Litl'i II l'i 2

22 nl, i Dni DCiLitl, i + Wl Ni Dni Lit,

,i+ nl, iC,, i DtiLj

3

2 + Cl, i DniDtiL + Wl Cl, iDni DtiLi 14

3

T12 ,i

DCi DtiLi + .1 nl, i Dni DCiDtiLi 2

+-l Dn2i Ni Dti Li) (A. 2.30) 5

12-7ý

By substituting equations (A. 2.26), (A. 2.27), (A. 2.28) with the

coordinates of the shear section into equations (A. 2.21.1) and (A. 2.21.2) the parameters a, and are established. Finally, further

substitution of, 01, ý2 and the polar moment of area about the shear

centre from equation (A. 2.21.3.1) into equation (A. 2.21.3) yields the Timoshenkian torsional buckling constant.

Calculation of the St Venent torsion constant:

By referring to page 3 of reference (66) the generalized expression of the torsion constant for the section shown in Figure (A. 2.1) is as f ol I ows:

3 - Liti (A. 2.31) .3

where, Li is the length of the ith segment ti ý (ti, 2 + ti,, )/2 is the average thickness of the ith

segment n designates the total number of segments of which the

section is comprised.