lateral-torsional buckling of non-prismatic thin-walled ... · ... and the obtained results are...

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The Twelfth East Asia-Pacific Conference on Structural Engineering and Construction

Lateral-Torsional Buckling of Non-Prismatic Thin-Walled Beams with Non-Symmetric Cross Section

B.ASGARIANa, M.SOLTANIb

aAssociate Professor, K.N.Toosi university of Technology, Tehran, IranbPhD Student in Structural Engineering, K.N.Toosi university of Technology, Tehran, Iran

Abstract

A numerical method to evaluate the lateral buckling load of non-prismatic thin-walled straight beams with non-symmetric cross-section is proposed. For this purpose, the potential energy considering the effects of initial stresses, the elastic strain energy, and the work of external forces for non-prismatic thin-walled beam with non-symmetric cross-section are derived. Then equilibrium equations and boundary conditions are obtained from the total potential energy. Power series approximation is used to solve the fourth –order differential equations of non-prismatic thin-walled beam with variable geometric parameters and with generalized end conditions. Based on power series expansions of displacement components explicit expressions for displacement parameters are derived. And also, it is assumed that the functions which describe the beam's variable parameters such as: flexural rigidity and loads can be expanded in to power series form. Finally, the lateral buckling loads are determined by solving eigenvalue problem. Several numerical examples of non-prismatic thin-walled beams have been presented to verify the accuracy and validity of this method, and the obtained results are compared with the finite element solutions using Ansys software and other available numerical or analytical approaches. The method can be applied for the elastic critical buckling load of prismatic members as well as non prismatic members.

© 2011 Published by Elsevier Ltd. Selection Keywords: Lateral-Torsional buckling; Non-prismatic beams; Power series method; Total potential energy

1. Introduction

The use of variable cross section beam has been increasing in the steel construction industry. This is because of their ability to increase stability of structure, and sometimes to satisfy architectural and functional requirements in many engineering structures. So, stability analysis of non-prismatic beam with general end conditions is of interest to many researchers [5-15]. Closed form solution of differential equation governing the bending of the non-prismatic beams [1-4] with variable coefficients is often

1877–7058 © 2011 Published by Elsevier Ltd.doi:10.1016/j.proeng.2011.07.208

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1654 B. ASGARIAN and M. SOLTANI / Procedia Engineering 14 (2011) 1653–1664

difficult. In many cases with general variation in the cross-section and general boundary conditions, closed form solution is impossible. Many of previous studies have been concerned with uniform cross section beams (e.g. [5, 6 and 7]) and only a few studies are on tapered beams (e.g. [8-15] Among the recently investigations, the most important ones are the studies of Polyzois and Raftoyiannis [11] and Polyzois and Qing [10], which use shell elements to model the web-tapered I beams.

2. Derivation of Equilibrium Equations

A thin-walled beam with non-symmetric cross-section subjected transverse forces acting in the Z and Y direction that cause an additional bending couples yM and zM is considered (Fig.1 (a)). The material of non-prismatic thin-walled beam with non-symmetric cross-section is homogenous and isotropic. It is also assumed that the beam’s geometrical and material parameters are symmetric relative to axis x. The length of the thin-walled beam is lager compared to the dimension of non-symmetric cross-section. Therefore, the displacements are small, and the shear deformation is negligible. (Fig. 1 (b)) shows displacement parameters of thin-walled beams defined at the non-symmetric cross-section in which x and y; z are the centroidal and principal axes. U, V, W and are the four displacement parameters and rotation of O.

Fig. 1: (a) Thin-walled beam with non-symmetric cross-section, (b) Co-ordinate system and notation for displacement parameters ofthe non-prismatic beam

The three translation displacement components of O can be expressed as:

xtxzy

xtxwz

xtxvytxutzyxU ),(

),(),(),(

),(),,,( (1)

),(),(),,,( txztxvtzyxV (2)

B. ASGARIAN and M. SOLTANI / Procedia Engineering 14 (2011) 1653–1664 1655

),(),(),,,( txyztxwtzyxW (3)

In which U represents the axial displacement at the centroid. Displacement components W and V represent lateral and vertical displacement (indirection y and z). The warping function is defined using Saint Venant’s torsion theory on the cross section. The two components of vertical and lateral displacement at centroid can be replaced by the displacements at the shear center C as follows:

),(),(),( txztxvtxv CC (4)

),(),(),( txytxwtxw CC (5)

Then, the stress components for isotropic materials are:

,0,,, 332213121111 231312 GGE (6)

And the most general case of initial stresses associated with the bending moments My and Mz, and shear forces Vy and Vz are considered as:

AV

AV

IzM

IyM zy

z

z

y

y 013

012

011 ,, (7)

Using Eq. (1) and (4)-(6) the elastic strain energy of a beam can be written as:

l

A

l

A ijijE dAdxGGEdAdx0

213

212

211

0 2

1

2

1(8)

Where

0

,,

,,,,

2

22

AAAA

y

yC

z

zCAzA

AyAzAyA

dydzydydzzdydzyzdydz

II

zII

yzdydzIdydzI

dydzyIdydzzIdydzyIdydzA

, (9)

The substitution of last expression into Eq. (8) leads to the following equation:

dxGJyyIzzIwyIvzIwyIvzIyI

zIIyIzIyIzIwIvIuAE

CCzCCyCCzCCyCCzCCyCz

CyCzCyCzCyCzCyC

l

E

222

22222222222222

0

444422

442

1

(10)

Using non-linear strain and initial stresses, the potential energy due to effects of the initial stresses can be defined:


l

A

l

A ijijijG dAdxdAdxe0

*13

013

*12

012

*11

011

0

0

2

1)( (11)

Substituting the strain-displacement relations into Eq. (11), and integration over the cross-section, the following equation can be expressed:

dXMMzMzMyMyMvMwM zzyyCzCz

l

CyCyCzCyG22

0 2

1

2

1(12)

where

CAy

y ydAzyyI

2)(1 22 , C

Azz zdAzyy

I2)(

1 22 (13)

Let),,(),,,( zyxPzyxP zy , denotes the distributed forces in structural domain in y and z direction

respectively, then the work of external forces can be expressed as:

dxMMwqvqdxdydzWPVPWl

ttCzCyV zy 0

2

21ˆ (14)

where

,

,)()(ˆ,)(,)(

A yzt

A CyCztA zzA yy

dydzzPyPM

dydzzzPyyPMdydzPxqdydzPxq

The total potential energy of a thin-walled element of length l under consideration can be defined as:

GE -W (15)

By variation of Eq. (15) with respect to CC wvu ,, and the equilibrium equations for non-prismatic thin-walled beam with non-symmetric cross-section are derived as

0dxduEA

dxd

(16)

qMdxd

dxdzEI

dxdzEI

dxd

dxvdEI

dxd

yzCyCyC

y 2

2

2

2

2

2

2

2

2

2

2 (17)

zyCzCzC

z qMdxd

dxdyEI

dxdyEI

dxd

dxwdEI

dxd

2

2

2

2

2

2

2

2

2

2

2 (18)


MMdxdMM

dxdzMzMyMyM

dxvd

Mdx

wdMxIzzIyy(

dxdE

dxvd

IzE

dxwd

Iy(Edx

vdIz(

dxdE

dxwd

Iy(dxdEx)IzIy(E

dxd)IzIy(

dxdE

dxd(GJ)

dxd

dxdIzIyIIE

dxd

ttzzyyCzCzCyCy

Cz

CyyCCzCC

CyC

CzC

CyC

CzCyCzC

yCzCyCzC

ˆ

)()2)(

))2)2)(

4

2

2

2

2

2

2

2

2

2

2

2

222

222

222

2

2

(19)

All variables in the last expression such as moment of inertia of the beam’s cross section Iz(x) and Iy(x) and I can be presented in power series form as follows:

0000

0 000

()()()(

)()()()

i

iiyy

i

iiZZ

i

ii

i

iCC

i i

iCC

ii

i

iiyy

i

iizz

xMx)MxMxMxJxJxyxy

xzxzxIxIxIxI xI(xI

i

i

,,,,

,(20)

By neglecting Eq. (16) which is corresponding to axial displacement, substituting Eq. (20) into Eq. (17)-(19) and introducing a new non-dimensional variable

Lx

, Eq. (17)-(19) can be written as:

LqLLMddL

ddLzIj

ddE

LzIjjddE

dvd

LIEddE

i

iiiy

i

iiiz

i

jijiCiy

i j

jijiCiy

C

i

iiiy

j

j

0

4

02

22

0

12

2

0 0

22

2

2

2

02

2

)()1(2

)()1)(2(

1

2

(21)

0

4

02

22

0

12

2

0 0

22

2

2

2

02

2

)()1(2

)()1)(2(

1

2

i

iiiz

i

iiiy

i

jijiCiz

i j

jijiCiz

C

i

iiiz

LqLLMddL

ddLyIj

ddE

LyIjjddE

dwd

LIEddE

j

j

(22)

1 1 1 1

2 22

2 20 0

2

0 0 0

4 ( 1)( 1)j m j m

i i i ii i

i i

i j m i j mC C zi C C yi

i j m

d d d dE I L L G J Ld d d dx

d dE j m y y I z z I Ld d


2 2 2 2

1 1

4

0 0 0

2 21 1

2 20 0 0 0

( 1)( 2)( 1)( 2) ( )

2 ( 1) ( 1)

j m j m

j j

i j m i j mC C zi C C yi

i j m

i j i j i j i jC Czi C yi C

i j i j

E j j m m y y I z z I L

d w d vdE j I y L j I z Ld d d

(23)

2

2

1 2 1 2

22

20 0

22

20 0

3

0 0 0

( 1)( 2)

( 1)( 2)

2 ( 1)( 1)( 2) (

j

j

j m j m

i j i j Czi C

i j

i j i j Cyi C

i j

i j m i j mzi C C yi C C

i j m

d wE j j I y Ld

d vE j j I z Ld

dE j m m I y y I z z Ld

2

2 22

0 0

)

( 1)( 2) ( 1)( 2) ( )j j

i j i jyi C yi C

i jL j j M y i i M y L

2

2

0

22

2

0

2

0 0

22

2 )()2)(1()2)(1(2

dvdLML

dwdLML

LzMiizMjjL

C

i

iiiz

C

i

iiiy

i j

jijiCizCiz jj

2

0 0 0 0

ˆ ( )i i

i j i j i i i iyi y j z j z j t t

i j i i

d dL M M L M L M Ld d

If the general solution of (Eq. (21)-(23)) is presented by power series of the form

0

)(i

iiC bv (24)

0i

iiC cw (25)

di

ii

0

)( (26)

Furthermore, introducing new variables:


,,,,

,,,,i

izzi

iyyi

iZzi

CCi

CC

iiyy

iii

iii

iiyy

iizz

LqqLMMLMMLyyLzz

LqqLIILJJLIILII

iiiiiii

iii

*****

*****

(27)

And substituting Eq. (24)-(27) into Eq. (21)-(23) the following equation can be found:

qLdMddLdzIkj

ddE

dzIjjddEbjjI

ddE

i

iy

i

ji

jjz

i j k

kjikCy

i j k

kjikCy

i j

jijy

iiji

jii

0

*4

0 0

*2

22

0 0 01

**2

2

0 0 0

**2

2

0 02

*2

2

1

2

)1)(1(2

)1)(2()2)(1(

(28)

qLdMddLdzIkj

ddE

dyIjjddEcjjI

ddE

i

iz

i

ji

jjy

i j k

kjikCz

i j k

kjikCz

i j

jijz

iiji

jii

0

*4

0 0

*2

22

0 0 01

**2

2

0 0 0

**2

2

0 02

*2

2

1

2

)1)(1(2

)1)(2()2)(1(

(29)

0 0 02

**2

**

0 0 02

**2

**

0 0 0 01

******

0 0 0 01

******

0 01

*2

0 02

*2

2

22

11

2222

1111

)2)(1)(2)(1(

)2)(1)(1(2

)1)(2)(1)(2)(1(

)1)(1)(1(4

)1()2)(1(

i j k

kjikCykCz

i j k

kjikCykCz

i j m k

kmjikCCyCCz

i j m k

kmjikCCyCCz

i

ki

kki

i k

kiki

bzIcyIkkjj

bzIcyIkkjddE

dyzIyyIkmmjjE

dyzIyyIkmjddE

dJkddGLdIkk

ddE

jiji

jiji

mjimji

mjimji

(30)


0 0 0

****2

0 0 0

****2

00 0 0

******

22

22

2121

)2)(1()2)(1(

)2)(1()2)(1(

)2)(1)(1(2

i j k

kjikCC

i j k

kjikCC

k

kk

i j m

mjiCCyCCz

dzMiizMjjL

dyMiiyMjjL

dzzIyyImmjdd

jizjiz

jiyjiy

mjimji

2 * *2 2

0 0 0 0

2 * * * *1

0 0 0

4 * 4 *

0 0 0

( 1)( 2) ( 1)( 2)

( 1)

ˆ

yi zi

yi y zi z

i i

i j i jj j

i j i j

i j kj j k

i j k

i i jt t j

i i j

L j j M c j j M b

dL k M M dd

L M L M d

To satisfy Eq. (28) to (30) for all values of , one must have the following recurrence formula about 444 ,, kkk dcb :

0

2*

4 4*1

2* *

2 20 0

2* *

1 30 0

1( 4)( 3)( 2)( 1)

( 4)( 3)( 2)( 1)

( 2)( 1)( 2)( 1)

2 (

i

i

i

k

k y k iiy

jk

y C j i k jj i

jk

y C j i k jj i

b E I b k i k i k kEI k k k k

E I z d j i j i k j k j

E I z d j

22 4 *

20

1)( 3)( 2)( 1)

2 1 0,1,2,.......i k

k*y k i y

i

i k j k j k j

L M d (k )(k ) L q for k

(31)

0

2*

4 4*1

2* *

2 20 0

1( 4)( 3)( 2)( 1)

( 4)( 3)( 2)( 1)

( 2)( 1)( 2)( 1)

i

i

k

k z k iiz

jk

z C j i k jj i

c E I c k i k i k kEI k k k k

E I y d j i j i k j k j


2* *

1 30 0

2 ( 1)( 3)( 2)( 1)i

jk

z C j i k jj i

E I y d j i k j k j k j

22 4 *

20

2 1 0,1,2,.......i k

k*z k i z

iL M d (k )(k ) L q for k (32)

0

1 1 1 1

2

2*

4 4*1

1* * * * * *

20 0 0

* *

1( 4)( 3)( 2)( 1)

( 4)( 3)( 2)( 1)

4 ( 1)( 1)( 2)( 1)

i

i j i m j i j i m j

i j i m

k

k k ii

jk m

z C C y C C k mm j i

z C C

d E d I k i k i k kEI k k k k

E I y y I z z d j i m j k m k

E I y y2 2 2

1 1

2 2

* * * *

0 0 0

1

1* * * *

3 30 0

* * * *2 2

1 2 1 2 1

2 ( 1)( 3)( 2)( 1)

(

j i j i m j

i j i i j i

i j i i j i

jk m

y C Cm j i

k m

jk

z C k j y C k jj i

y C k j z C k j

I z z

d j i j i m j m j k m

E I y c I z b j i k j k j k

I z b I y c0 0

1)( 2)( 1)( 2)jk

j ij i j i k j k j

1 3 1 3

2

1* * * * * *

0 0 0

2 * 2 *2 2

0 0

2 * *

2 ( 1)( 3)( 2)( 1)

( 2)( 1) ( 2)( 1)

( 1)( 2

i j i m j i j i m j

i i

i j i

jk m

z C C y C C k mm j i

k k

y k i z k ii i

y C

I y y I z z d j i m j m j m

L M c k i k i L M b k i k i

L M y j i j i2

1* *

0 0

) ( 1)( 2)i j i

jk

y C k jj i

M y i i d

2 2

12 * * * *

0 0

12 * * * *

1 1 20 0

( 1)( 2) ( 1)( 2)

( 1)( 1)

i j i i j i

yi y zi z

jk

z C z C k jj i

jk

j j k jj i

L M z j i j i M z i i d

L k k j M M d

14 * 4 * 2 *

20 0

ˆ ( 1)( 2) 01 2 3 4 k k

ti k i tk i k ii i

L M d L M L GJ d k k i for k , , , , ,....... (33)


with the recurrence formula, the fundamental solution of Eq. (17) to (19) could be obtained unambiguously in term of the six constants ( 3210 ,,, cccc ), ( 3210 ,,, bbbb ) and ( 3210 ,,, dddd ) which can be determined by imposing the natural boundary conditions. The general solution of Eq. (17) to (19) can be expressed in the following form:

7362514033221100 CCCCCCCCC vdvdvdvdvbvbvbvbv (34)

7362514033221100 CCCCCCCCC wdwdwdwdwcwcwcwcw (35)

Where ),,, (i ,wv CiCi 7....,210 are the fundamental solutions of Eq. (28) to (29). Knowing that the first four coefficients ( 3210 ,,, cccc ), ( 3210 ,,, bbbb ) and ( 3210 ,,, dddd ) are functions of the displacements DOF, then all the coefficients ,..)6,5,4(kck , ,..)6,5,4(kbk , and

,..)6,5,4(kdk are also functions of the displacements DOF. Thus the displacement perpendicular to the beam’s axis ( )(Cv and )(Cw ) and the angle of lateral rotation of the cross section ( )( ) can be obtained as a function of the displacement DOF.

3. Numerical Example

In order to check the validity and accuracy of the elastic buckling load computed by present method, a comparative study was made with the variable analytical solution of certain types of non- prismatic thin-walled beam such as tapered beams. This example deals with the linear buckling of web- tapered cantilevers with equal or unequal uniform flange, the web height is made to vary linearly along the length so that, at the free end of cantilever beams, the height is reduced to half, as shown in figure. (2). The lengths of beams vary between 4.0m to 10.0m and the tip load is acting at the mid- line of top flange.

Table 1 and 2 provide the values of crP (the linear critical load) obtained by present study, one dimensional method proposed by Andrade and Camotim [13], finite element method by means of Ansys software, and the relative errors which are given by the expression 100/)( FEM

crFEM

crcr PPPfor tapered cantilever beam displayed in Fig.2. The beam has been modeled with SHELL63 of Ansys[16]. Shell63 has both bending and membrance capabilities. Both in-plane and normal loads are permitted. The element has 6 degrees of freedom at each node, translation in the nodal x, y and z directions and rotations about the nodal x, y, and z axes.

Fig. 2: Tapered I-section cantilever beam: geometry, material and loading properties.


Table 1: Web tapered cantilevers with equal flanges (Fig.2): elastic critical loads and relative errors.

)(mL )(KNPcr

(Present study)

)(KNP FEMcr

(Using Ansys Software)

)(KNP IDcr [14] (%)

4.0 44.45 42.96 47.42 3.468343

6.0 23.09 22.40 23.45 3.080357

8.0 14.07 13.897 14.21 1.244873

10.0 9.24 9.17 9.41 0.763359

Table 2: Web tapered cantilevers with unequal flanges (Fig.2): elastic critical loads and relative errors.

)(mL )(KNPcr

(Present study)

)(KNP FEMcr

(Using Ansys Software)

)(KNP IDcr [13] (%)

4.0 26.92 25.7 27.98 4.747082

6.0 15.09 14.88 15.47 1.41129

8.0 9.642 9.58 9.83 0.647182

10.0 6.74 6.72 6.73 0.297619

4. Conclusions

For the lateral buckling analysis of thin-walled beam with non-symmetric cross-section, the equilibrium equations have been derived by using energy method. The boundary conditions have also been defined. Finally, power series approximation is used to solve the fourth–order differential equations of non-prismatic thin-walled beams. On the basis of the preceding discussion the following conclusions can be stated:The total potential principle is used to derive the equilibrium equations of non-prismatic thin-walled beams. Exact numerical solution for differential equation of non-prismatic thin-walled beams with non-symmetric cross-section under general loading, variable moment of inertia, and general boundary conditions has been obtained using power series method. The proposed method can be applied in various forms of non-prismatic and prismatic members. As demonstrated in the numerical example section, the results obtained using the proposed computations are in close agreement to those obtained by other analytical and numerical solutions.


References

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Engineering Mechanics 2005; 131(6):586–97. [14] Bradford MA, Cuk PE. Elastic buckling of tapered monosymmetric Ibeams. Journal of Structural Engineering

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