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Behaviour of Thin Walled Stiffened Plates under In-Plane Axial Loads B.Tech Project Report by Abhijeet Singh (Y4008) Manish Kumar (Y4210) supervised by Dr. Ashwini Kumar Department of Civil Engineering Indian Institute of Technology Kanpur April, 2008

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Page 1: manishkumar.orgmanishkumar.org/research/Docs/BTP Report.pdfAbstract Unlike other slender structural members like columns, plates have the capability to resist significant amount of

Behaviour of Thin Walled Stiffened Plates under In-Plane Axial Loads

B.Tech Project Report

by Abhijeet Singh (Y4008) Manish Kumar (Y4210)

supervised by Dr. Ashwini Kumar

Department of Civil Engineering

Indian Institute of Technology Kanpur

April, 2008

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Certificate

It is certified that the work contained in this report titled “Behaviour of Thin Walled Stiffened Plates under In-Plane Axial Loads” is the original work done by Abhijeet Singh (Y4008) & Manish Kumar (Y4210), and has been carried out under my supervision.

Prof. Ashwini Kumar Supervisor Department of Civil Engineering Indian Institute of Technology Kanpur

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Abstract

Unlike other slender structural members like columns, plates have the capability to resist significant amount of compressive axial load after they start to buckle. Advantage can be taken of this special characteristic of plates while determining the allowable load-deflection criteria for the purpose of design of structures. Some established approximate methods of analysis have been used in this project to analyze orthotropic plates for post buckling behaviour. The approach can be easily extended to the analysis of structures where the properties are different in two orthogonal directions because of provision of stiffeners.

Firstly effective width of a stiffened plate has been found out considering it to be an orthotropic plate and then the result has been compared with the codal provisions. Then expression for critical load for buckling of a stiffened plate has been derived.

Post buckling strength of orthotropic has been considered and expression for effective width of the loading in post buckled state has been found out by two methods: 1) Stress Method 2) Successive approximation method. Load versus deflection curve has been drawn to see the buckling mode and post buckling behavior of the plate. Then comparison has been done between the effective width found out by the two methods. Finally a conclusion has been drawn based on results and graphs obtained.

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Acknowledgement

We are immensely grateful to our guide, Dr. Ashwini Kumar for the guidance he provided to us during this project. Dr. Kumar has always been ready to help us in any difficulty we faced and explained the correct approach to the problem. He has not only helped us by giving extremely helpful comments and suggestions regarding the work but also by providing the required material to study and have a better understanding of the subject.

We would also like to thank all the faculty and staff members of the Department of Civil Engineering, IIT Kanpur who made us skilled enough to work on this project by providing a highly fruitful training at this institute.

Abhijeet Singh

Manish Kumar

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Table of Contents

Certificate Acknowledgement Table of Contents Chapter 1: Introduction

1.1 IS Code Provisions ……………………………………………………………………1 1.2 Developing the differential equation………………………………………………….2 1.3 Effective thickness of a stiffened plate ……………….................................................8 1.4 Comparison………………………………………………………………………......10 1.5 Corrugated Plate …………………………………………………………………….11

Chapter 2: Post Buckling Strength: Stress Solution

2.1 Effective width concept in thin walled steel section…………………………………13 2.1.1 Post Buckling Strength………………………………………………………...13

2.2 Stress Solution……………………………………………………………………….14 2.2.1 Compatibility…………………………………………………………………..14

2.3 Post buckling behavior of axially compressed plates………………………………..16 2.3.1 Boundary conditions…………………………………………………………...16

2.4 Using Galerkin method for finding f ………………………………………………..18 2.5 Finding Effective width: Stress Method…………………………………………….20 Chapter 3: Post Buckling Strength: Method of Successive Approximations

3.1 Introduction……………………………………………………………………….…21 3.2 Problem background…………………………………………………………………21 3.3 Compressive load problem…………………………………………………………..26

3.3.1 Boundary conditions…………………………………………………….……..26 3.3.2 Homogeneous solution…………………………………………………………31 3.3.3 Particular solution……………………………………………………………...33

3.4 Calculation of effective width……………………………………………………….36 3.5 Non-dimensionalization of expressions………………………………….…………..37 3.6 Graphical Analysis…………………………………………………………………...40

3.6.1 Load deflection curve…………………………………………………………40 3.6.2 Comparison of results of effective width by two methods……………………42

3.7 Conclusion…………………………………………………………………………...44

References…………………………………….……………………………………………….45  

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Chapter 1

Introduction

1.1 IS Code Provisions

IS: 801-1975 suggests an expression to calculate the effective thickness of a stiffened plate. If the intermediate stiffeners are spaced so closely that the flat width ratio between stiffeners does not exceed (w/t)lim , all the stiffeners may considered effective. Only for the purposes of calculating the flat width ratio of entire multiple-stiffened element, such element shall be considered as replaced by an element without intermediate stiffeners whose w is the whole width between webs or from wed to edge stiffener, and whose equivalent thickness heff is determined as follows:

heff = 312

wI s

(1.1)

Where Is = moment of inertia of the full area of the multiple stiffened element, including the intermediate stiffeners, about its own centroidal axis

(w/t)lim is given by

“Maximum allowable overall flat width ratio w/h disregarding intermediate stiffeners and taking h as the actual thickness of the element shall be as follows:

For stiffened compression element with both longitudinal edges connected to other stiffened elements= 500”

Fig 1.1: Buckled shape

A closer examination of the behavior of the element in compression reveals that the given formula by code needs to be reviewed. Introduction of Intermediate stiffeners transfer the plate into an orthotropic plate having much higher flexural rigidities. Plate is being analyzed considering different flexural rigidities in two directions.

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1.2 Developing the differential equation

The equilibrium equation governing the buckling of a thin plate is given by

2

2

xM x

∂∂

- 2yx

M xy

∂∂

∂ 2

+ 2

2

yM y

∂+ 2

2

xwN x ∂

∂ + 2

2

ywN y ∂

∂ + yx

wN xy ∂∂∂ 2

2 = 0 (1.2)

Fig: 1.2: Rectangular plate under axial load

We have the following relationships

][1 yyx

yx

xxx

Eενε

ννσ +

−= (1.3a)

][1 xxy

yx

yyy

Eενε

ννσ +

−=

(1.3b)

xyxyxy G γσ = (1.3c)

The strains xε , yε and xyγ can be expressed in terms of displacements as

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xwzu∂∂

−= ywzv∂∂

−= (1.4a)

xu

x ∂∂

=ε yv

y ∂∂

=ε yv

xu

xy ∂∂

+∂∂

=γ (1.4b)

Substituting the values of u and v in strain equations and then strain values in stress train

relationships we get 2

2

xwzx ∂

∂−=ε 2

2

ywzy ∂

∂−=ε

yxwzxy ∂∂

∂−=

2

2γ (1.5a)

Therefore

][

1 2

2

2

2

yw

xwzE

yyx

xxx ∂

∂+

∂∂

−−= ν

ννσ (1.6a)

][1 2

2

2

2

xw

ywz

Ex

yx

yyy ∂

∂+

∂∂

−= ν

ννσ (1.6b)

yxwzGxyxy ∂∂

∂−=

2

2τ (1.6c)

Consider rectangular plate axially compressed in one direction and simply supported along all the edges.

For this case of plate having thickness “h” and given dimensions, we have

zdzM

h

hxxx ∫

+

=2/

2/

σ

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−= 2

2

2

23

)1(12 yw

xwzh

Ey

yx

x ννν

(1.7)

or

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xM = - xD [ 2

2

xw

∂∂ + yν 2

2

yw

∂∂

] (1.8a)

yM = - yD [ 2

2

yw

∂∂

+ xν 2

2

xw

∂∂ ] (1.8b)

Similarly

zdzM xyhhxy σ2/

2/+−∫=

xywhGxy ∂

∂−=

23

61 (1.8c)

or

xywDM xyxy ∂

∂−=

2

(1.8d)

where )1(12

3

yx

xx

hED

νν−= (1.9a)

)1(12

3

yx

yy

hED

νν−= (1.9b)

3

6h

GD xy

xy = (1.9c)

are called flexural rigidities.

Now substituting the values of moments in governing differential equations

2

2

xM x

∂∂

- 2yx

M xy

∂∂

∂ 2

+ 2

2

yM y

∂+ 2

2

xwN x ∂

∂ + 2

2

ywN y ∂

∂ + yx

wN xy ∂∂∂ 2

2 = 0

2

2

xM x

∂∂

= - xD2

2

x∂∂ [ 2

2

xw

∂∂ + yν 2

2

yw

∂∂

]

2

2

xM x

∂∂

= - xD [ 4

4

xw

∂∂ + yν 22

4

yxw∂∂

∂] (1.10a)

Similarly

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2

2

yM y

∂= - yD [ 4

4

yw

∂∂

+ xν 22

4

yxw

∂∂

] (1.10b)

And

yxM xy

∂∂

∂ 2

= - xyD 22

4

yxw

∂∂

(1.10c)

Substituting for all the expressions

- xD [ 4

4

xw

∂∂ + yν 22

4

yxw∂∂

∂] -2 xyD 22

4

yxw

∂∂ - yD [ 4

4

yw

∂∂

+ xν 22

4

yxw

∂∂

] + 2

2

xwN x ∂

∂ + 2

2

ywN y ∂

∂ +

yxwN xy ∂∂

∂ 2

2 = 0

or

2

2

xwN x ∂

∂ + 2

2

ywN y ∂

∂+

yxwN xy ∂∂

∂ 2

2 = xD [ 4

4

xw

∂∂ + yν 22

4

yxw∂∂

∂] + 2 xyD 22

4

yxw

∂∂

+ yD [ 4

4

yw

∂∂

+

xν 22

4

yxw

∂∂

] (1.11)

In this case we are considering rectangular plate axially compressed in one direction and simply supported along all the edges.

Boundary Conditions:

w = 0 for x = 0, a (1.12a)

w = 0 for y = 0, b (1.12b)

xM = 0, yM = 0 at edges

Hence

2

2

xw

∂∂ + yν 2

2

yw

∂∂

= 0 for x= 0, a (1.13a)

2

2

yw

∂∂

+ xν 2

2

xw

∂∂ = 0 for y= 0, b (1.13b)

Also since curvature of plate will be zero at the boundaries,

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2

2

yw

∂∂

= 0 for x = 0, a (1.14a)

2

2

xw

∂∂

= 0 for y = 0, b (1.14b)

Substituting from (1.14a) in (1.13a) & from (1.14b) in (1.13b)

2

2

xw

∂∂ = 0 for x = 0, a (1.15a)

2

2

yw

∂∂

= 0 for y= 0, b (1.15b)

Hence from all the conditions derived, it can be said that all the four edges of the plate will be undeflected as well as curvature along an axis perpendicular to edge will be zero at the edge.

As yN and xyN are zero, the final equation can also be written as

02 2

2

22

4

4

4

22

4

22

4

4

4

=∂∂

+⎥⎦

⎤⎢⎣

⎡∂∂

∂+

∂∂

+∂∂

∂+⎥

⎤⎢⎣

⎡∂∂

∂+

∂∂

xwN

yxw

ywD

yxwD

yxw

xwD xxyxyyx νν (1.16)

Let the solution be

∑∑∞

=

=

=1n 1

mnA(b

ynSina

xmSinx,y) wm

ππ for m = 1,2,3…..∞ ; n = 1,2,3…….∞ (1.17)

Substituting in our governing differential equation

( )( ) 0

21 122242224222444

422244

=⎥⎥⎦

⎢⎢⎣

−++

++∑∑∞

=

=n m xyxy

yxmn b

ynSina

xmSinmnmDnmnD

nmmDA ππ

πμπλπλνπλ

πλνπ (1.18)

where xNa 22 =μ and aspect ratio of plate ba

As Amn = 0 gives trivial solution, Amn ≠ 0 gives

( ) ( ) 02 22242224222444422244 =−++++ πμπλπλνπλπλνπ mnmDnmnDnmmD xyxyyx (1.19)

or

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( ) 2222

222

44

2

2222

2

2

2 naD

nm

na

Dnma

DN xyxyyxx λπλνλπλνπ

+⎟⎟⎠

⎞⎜⎜⎝

⎛+++= (1.20)

As n increases, Nx increases. Hence for the lowest value of Nx, n must assume one as the numerical value. This implies that the plate buckles with one half sine wave along the y-direction. The number of half sine waves along x-direction that correspond to minimum value of Nx can be obtained by taking derivative of Nx w.r.t. m, with n = 1.

0=∂∂

mN x => ( ) 022 3

4

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+

mDmD yx

λ

4

x

y

DD

bam = (1.21)

Substituting the value of m,

( ) ( )[ ]xyxyyxyxcrx DDDDD

bN 222

2

+++= ννπ (1.22)

For non-integer values of λ , the buckling load is higher than that for integer values. For such cases, equation (1.20) can be written as

ψπ2

2

bDN xx = (1.23)

where ψ =buckling load coefficient

After putting the value of n=1, ψ can be written as

( )

( )⎥⎦⎤

⎢⎣

⎡++−

+++=x

xx cc

ccc

mcm

νν

νλλ

ψ21

142

2

2

2

2

2

(1.24)

where x

y

x

y

EE

cνν

== = ratio of properties in two orthogonal directions.

Hence the buckling load coefficient ψ can be plotted for integer values of m and the minimum value for each aspect ratio can be taken as the governing buckling load coefficient.

Here min value of ψ is plotted against aspect ratioλ for 25.0=xν and 2.0=c .

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Fig1.3: variation of buckling load coefficientψ with apect ratio λ for 25.0=xν and 2.0=c

1.3 Effective thickness of a stiffened plate

Now consider a stiffened plate with given dimensions as shown in the figure.

formula for effective thickness of this stiffened plate heff is derived.

Fig 1.4: Stiffened Plate Let I1 be the moment of inertia of plate and I2 be the moment of inertia of stiffener on two sides with portion of plate between them.

( ) ( )[ ]xyxyyxyxcrx DDDDDb

N 222

2

+++= ννπ (1.25)

)1(12

3

ν+=

EhDxy = )1(

1

ν+wEI (1.26a)

0

0.5

1

1.5

2

2.5

3

3.5

4

0 2 4 6 8

ψ min vs λ

ψ min

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where 1I = 12

3wh

)1(12 2

3

ν−=

EhDx = )1( 2

1

ν−wEI

(1.26b)

)1(12 2

3

ν−=

EhDy + w

EI 2

where 12

3

2tHI =

yD = )1( 2

1

ν−wEI

+ w

EI 2 (1.26c)

)1(2 2

1

ν−=+

wEIDD yx +

wEI 2

= ( )⎥⎦

⎤⎢⎣

⎡−+

−2

1

22

1 12)1(

νν I

Iw

EI (1.27)

yx DD = ( )2

1

22

1 11)1(

νν

−+− I

Iw

EI

Substituting in the expression for ( )crxN

( )⎥⎥⎦

⎢⎢⎣

⎡−+⎟⎟

⎞⎜⎜⎝

⎛−++⎟⎟

⎞⎜⎜⎝

⎛−+⎟⎟

⎞⎜⎜⎝

⎛−

= )1(2)1(2)1(12)1(12

2

1

22

1

22

3

2

2

ννννν

πII

IIEh

bN crx

Comparing with the expression for isotropic plate,

⎥⎥⎦

⎢⎢⎣

⎡+−+⎟⎟

⎞⎜⎜⎝

⎛−+=⎟⎟

⎞⎜⎜⎝

⎛1)1(

2)1(1

21 2

1

22

1

2

3

νννII

II

hheff (1.28)

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1.4 Comparison

Closely spaced stiffened plate has been considered here and then treating that as an orthotropic plate code gives the following formula.

heff = 312

wI s

Where Is = moment of inertia of the full area of the multiple stiffened element, including the intermediate stiffeners, about its own centroidal axis.

For our case Is = I2 + I1

The above equation can be written as (heff)3= 12Is/w

Also we have I1 = wh3/12

So (heff/h)3 = 1+ I2/ I1

Also as derived earlier ⎥⎥⎦

⎢⎢⎣

⎡+−+⎟⎟

⎞⎜⎜⎝

⎛−+=⎟⎟

⎞⎜⎜⎝

⎛1)1(

2)1(1

21 2

1

22

1

2

3

νννII

II

hheff

This formula is compared with the one derived earlier for stiffened plate. Typical values have been taken to calculate ratio. Although two formulae can not be compared directly as there are restrictions on code formula under which it can be applied.

The curve of heff/h vs H/h for code formula is as shown on the next page.

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Fig1.5: heff/h vs. H/h for code formula

As we can see two values are close when H/h (I2/ I1) is low but diverges as we increase the ratio H/h(I2/ I1)

1.5 Corrugated Plate

A stiffened orthotropic plate can be modeled as a corrugated plate. Approximate flexural rigidities have been considered in different directions and then final formula for effective width of corrugated plate has been calculated.

Fig 1.6: Corrugated Plate

0

0.5

1

1.5

2

2.5

3

0 5 10 15

exact

code

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Modeling as a corrugated plate and taking flexural rigidity different in X and Y direction

We have

Ds

bDy 2≈

DsDx /12bhI3≈

DDxy )1(sb ν−≈

DsbDyx νν ≈

DsbDD xyyx ≈+ν

Now substituting these values in the equation

( ) ( )[ ]xyxyyxyxcrx DDDDDb

N 222

2

+++= ννπ

And simplifying, following result is derived

31

21

3

32 ⎥

⎢⎢

⎡⎟⎠

⎞⎜⎝

⎛+≈sh

Is

bh

h seq

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Chapter 2

Post Buckling Strength: Stress Solution

2.1 Effective width concept in thin walled steel section

The theoretical critical load for plate is not necessarily is a satisfactory basis for design, since the ultimate strength can be much greater than the critical buckling load. A plate loaded in uniaxial compression will undergo stress redistribution as well as develop transverse tensile membrane stresses after buckling that provide post buckling support. Thus additional load often be applied without structural damage.

2.1.1 Postbuckling Strength

Postbuckling strength in plates is mainly due to the redistribution of axial compressive stresses. Local buckling causes a loss of stiffness and a redistribution of stresses. Uniform edge compression in the longitudinal direction results in nonuniform stress distribution after buckling and buckled plate derives almost all of its stiffness from the longitudinal edge supports.

The fact that much of the load carried by the region of the plate in the close vicinity

of the edges suggests the use of “effective width concept”. The maximum strength of plates can be estimated by the use of effective width concept.

This concept makes a simplifying assumption that maximum edge stress acts uniformly over two “strips” of the plate and the central region remains unstressed. Thus only a fraction of the width is considered to be effective in resisting the applied compression.

Fig 2.1: Postbuckling Stress Distribution

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2.2 Stress Solution

The equilibrium equation in the z-direction is obtained as

( ) 0222

2

2

2

2

4

4

22

4

4

4

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂+

∂∂

+∂∂

−∂∂

+∂∂

∂+++

∂∂

yxwN

ywN

xwN

ywD

yxwDDD

xwD xyyxyxyxyyxx νν

(2.1)

2.2.1 Compatibility

The compatibility relations are obtained as

20

0 21

⎟⎠⎞

⎜⎝⎛∂∂

+∂∂

=xw

xu

xε (2.2a)

2

00 2

1⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=yw

yv

yε (2.2b)

and yw

xw

yv

xu

xy ∂∂

∂∂

+∂∂

+∂∂

= 000γ (2.2c)

Where 0xε , 0yε and 0xyγ are middle surface strains in x and y directions and in shear respectively. 0u and

0v are displacements at the middle surface and w is the displacement in transverse direction.

Strains can be related to forces as

( )yxxx

x NNhE

νε −=1

0 (2.3a)

( )xyyy

y NNhE

νε −=1

0 (2.3b)

xy

xyxy G

Nh1

0 =γ (2.3c)

Differentiating (2.2a) twice w.r.t. y, (2.2b) twice w.r.t. x and (2.2c) successively w.r.t. x and y and combining the results

2

2

2

2220

2

20

2

20

2

yw

xw

yxw

yxxyxyyx

∂∂

∂∂

−⎥⎦

⎤⎢⎣

⎡∂∂

∂=

∂∂

∂−

∂+

∂∂ γεε

(2.4)

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15  

This is deformation compatibility equation. To reduce the number of equations that must be solved, a stress function is introduced. Let the in-plane forces be defined in terms of a function ( )yxF , as

2

2

yFhN x ∂

∂= (2.5a)

2

2

xFhN y ∂

∂= (2.5b)

yx

FhN xy ∂∂∂

−=2

(2.5c)

Using equations (2.3a) to (2.3c)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

= 2

2

2

2

01

xF

yF

E xx

x νε (2.6a)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

= 2

2

2

2

01

yF

xF

E yy

y νε (2.6b)

yx

FGxy

xy ∂∂∂

−=2

01γ (2.6c)

substituting from (2.6a), (2.6b) and (2.6c) into (2.4) and from (2.5a), (2.5b) and (2.5c) into(2.1) gives

2

2

2

222

22

4

22

4

4

4

22

4

4

4 11yw

xw

yxw

yxF

Gh

yxF

xF

EyxF

yF

E xyy

yx

x ∂∂

∂∂

−⎥⎦

⎤⎢⎣

⎡∂∂

∂=

∂∂∂

+⎥⎦

⎤⎢⎣

⎡∂∂

∂−

∂∂

+⎥⎦

⎤⎢⎣

⎡∂∂

∂−

∂∂ νν

or 2

2

2

222

4

4

22

4

4

4 11yw

xw

yxw

yF

EyxF

EEGh

xF

E xy

y

x

x

xyy ∂∂

∂∂

−⎥⎦

⎤⎢⎣

⎡∂∂

∂=

∂∂

+∂∂

⎥⎥⎦

⎢⎢⎣

⎡−−+

∂∂ νν

(2.7a)

and

( ) 02222

2

2

2

2

2

2

2

2

4

4

22

4

4

4

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂∂

∂−

∂∂

∂∂

+∂∂

∂∂

−∂∂

+∂∂

∂+++

∂∂

yxw

yxFh

yw

xFh

xw

yFh

ywD

yxwDDD

xwD yxyxyyxx νν

(2.7b)

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16  

2.3 Post buckling behavior of axially compressed plates

Fig2.2: Simply supported plate compressed in x-direction

2.3.1 Boundary conditions

Transverse boundary conditions corresponding to simply supported edges are

02

2

=∂∂

=xww at ax ,0=

02

2

=∂∂

=yww at by ,0=

Thus

Average value of applied compressive stress ∫−=a

xxa dyNah 0

1σ (2.8)

Assuming lateral deflection function as b

yna

xmfw ππ sinsin= (2.9)

and substituting in the equation (2.7a)

2

2

2

222

4

4

22

4

4

4 11yw

xw

yxw

yF

EyxF

EEGh

xF

E xy

y

x

x

xyy ∂∂

∂∂

−⎥⎦

⎤⎢⎣

⎡∂∂

∂=

∂∂

+∂∂

⎥⎥⎦

⎢⎢⎣

⎡−−+

∂∂ νν

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17  

which reduces to

⎥⎦⎤

⎢⎣⎡ +=

∂∂

+∂∂

⎥⎥⎦

⎢⎢⎣

⎡−−+

∂∂

byn

axm

banmf

yF

EyxF

EEGh

xF

E xy

y

x

x

xyy

πππνν 2cos2cos2

1122

4222

4

4

22

4

4

4

(2.10)

Solution of the equation consists of a homogeneous part and a particular part i.e.

ph FFF +=

For homogeneous part of solution, RHS has to be set equal to zero. But this will be equivalent to letting0=w . Therefore the homogeneous solution corresponds to the in-plane stress distribution in the plate

just prior to buckling. But at that instant xN will be uniform and yN and xyN will be zero. Therefore a

homogeneous solution of the equation will be

2AyFh = (2.11)

as hN xax σ−=

we get

2

2yF xa

−= (2.12)

Now corresponding to the RHS, the particular solution can be written as

b

ynCa

xmBFpππ 2cos2cos += (2.13)

Substituting back in the differential equation and comparing the coefficients of the terms a

xmπ2cos and

bynπ2cos ,values of B and C are obtained as

22

222

32 bmnafE

B y=

and 22

222

32 anmbfE

C x= (2.14)

using which complete solution of F is obtained as

2

2cos32

2cos32

2

22

222

22

222 yb

ynanmbfE

axm

bmnafE

F xaxy σππ−+= (2.15)

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18  

2.4 Using Galerkin method for finding f

b

yna

xmfw ππ sinsin=

The Galerkin equation is

( ) ( ) 0,0 0

=∫ ∫ dxdyyxgfQa a

(2.16)

Where on arranging the terms Q(f) is obtained as

( )( )

byn

axm

amhf

bnyCos

amhfE

amxCos

bnhfE

banmDDD

bnD

amD

f

fQ

xaxy

xyxyyxyx

ππ

πσππππ

ννπ

sinsin2

82

8

2

2

22

4

443

4

443

22

22

4

4

4

44

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

+⎟⎟⎠

⎞⎜⎜⎝

⎛++++

=

(2.17)

Substituting the value of ( )fQ in Galerkin equation, taking ( )

byn

axmfyxg ππ sinsin, = and integrating

with proper limits the following is obtained

( )⎥⎥⎦

⎢⎢⎣

⎡++

⎥⎥⎦

⎢⎢⎣

⎡++++= 4

4

4

4222

22

22

4

4

4

4222

162

bnE

amEaf

banmDDD

bnD

amD

ham yx

xyxyyxyx

xaπννπσ

which can also be expressed as

⎥⎥⎦

⎢⎢⎣

⎡++= 4

4

4

4222

16 bnE

amEaf yx

crxaπσσ

(2.18)

where

( )⎥⎥⎦

⎢⎢⎣

⎡++++= 22

22

4

4

4

4222

2banmDDD

bnD

amD

ham

xyxyyxyx

cr ννπσ (2.19)

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19  

Now as

2

2

yF

xx ∂∂

−=σ

substituting the value of F , xxσ is obtained as

xax

xx bnyCos

afmE

σππσ +=

28 2

222

(2.20)

Also we have

⎥⎥⎦

⎢⎢⎣

⎡+

−=

4

4

4

4

2

2

22 )(28

bnE

amE

ma

f

yx

crxx σσπ

so substituting the value in the equation (2.20), xxσ is obtained as

bnyCos

EE

bn

ma

x

y

crxxxaxx

πσσσσ 2

1

)(2

4

4

4

4

⎥⎦

⎤⎢⎣

⎡+

−+=

Now let Ma

m=

π

and N

bn

Then the equation becomes

bnyCos

EE

MN

x

y

crxxxaxx

πσσσσ 2

1

)(2

4

4

⎥⎦

⎤⎢⎣

⎡+

−+= (2.21)

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20  

2.5 Finding Effective width: Stress Method

∫=b

xxe dyb0

max σσ (2.22)

maxσ =

⎥⎦

⎤⎢⎣

⎡+

−+=

=

x

y

crxxxayxx

EE

bn

ma

4

4

4

40

1

)(2 σσσσ

now,

∫∫ +=b

xa

b

xx dybnyCosbdy

00

2(........) πσσ

so by xa

b

xx σσ =∂∫0

hence

( )ησσ

σ

σ

+−

+=

1)(2 crxa

xa

xae

bb

(2.23)

where x

y

EE

bn

ma

4

4

4

4

=η (2.24)

Now we have crxx

xa

PP

=σσ

so

( )η+

−+

=

1

)1(2cr

cr

cre

PP

PP

PP

bb

(2.25)

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21  

Chapter 3

Post Buckling Strength: Method of Successive Approximations

3.1 Introduction

In the method of successive approximations, the set of Von Karman large deflection equations for the plates are converted from three nonlinear partial differential equations into infinite number of linear partial differential equations by expanding the displacement terms into a power series form of a parameter. The first few equations of the infinite set of equations are obtained as small deflection equations which give the solution for pre-buckling stage. Further solution of more equations gives approximate solution for post-buckling range.

The study of post-buckling behaviour of a simply supported orthotropic plate subjected to longitudinal compression is presented here.

3.2 Problem background

For a plate with no lateral loads, Von Karman equations for large deflection can be written as

0=∂

∂+

∂∂

yN

xN xyx (3.1a)

0=∂

∂+

yN

xN yxy (3.1b)

( ) 0222

2

2

2

2

4

4

22

4

4

4

=∂∂

∂−

∂∂

−∂∂

−∂∂

+∂∂

∂+++

∂∂

yxwN

ywN

xwN

ywD

yxwDDD

xwD xyyxyxyxyyxx νν (3.1c)

The strain force relations are the following

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

y

yy

x

xx E

NEN

hνε 1

(3.2a)

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

x

xx

y

yy E

NEN

hνε 1 (3.2b)

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22  

xy

xyxy G

Nh1

=γ (3.2c)

From equation (2a), (2b) & (2c), forces can be expressed in terms of strains as

yx

xx

hEN

νν−=

1)( yyx ενε + (3.3a)

yx

yy

hEN

νν−=

1)( xxy ενε + (3.3b)

xyxyxy hGN γ= (3.3c)

Strain-displacement equations are

2

21

⎟⎠⎞

⎜⎝⎛∂∂

+∂∂

=xw

xu

xε (3.4a)

2

21

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=yw

yv

yε (3.4b)

yw

xw

xv

yu

xy ∂∂

∂∂

+∂∂

+∂∂

=γ (3.4c)

substituting these relations we get the force-displacement relations as

⎟⎟

⎜⎜

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+∂∂

+∂∂

−=

22

21

1 yw

xw

yv

xuhE

N yyyx

xx νν

νν (3.5a)

⎟⎟

⎜⎜

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

−=

22

21

1 xw

yw

xu

yvhE

N xxyx

yy νν

νν (3.5b)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+∂∂

+∂∂

=yw

xw

xv

yuhGN xyxy (3.5c)

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23  

Now assuming that u , v & w may be expanded in a power series in terms of an arbitrary parameterα , they can be expressed as

∑∞

=

=2,0n

nnuu α (3.6a)

∑∞

=

=2,0n

nnvv α (3.6b)

∑∞

=

=3,1n

nnww α (3.6c)

Here u and v are assumed to start from a term having zero power of α whereas w is assumed to start with non-zero power ofα . The reason is that just prior to buckling, the in-plane deflections u and v are expected to have finite value whereas the transverse deflection w is expected to be zero until the plate buckles.

The plate can buckle in either direction but ( )yxw , should be independent of that except for a sign. Hence for + ve to –ve values ofα , w should just only change its sign and therefore only odd powers of α are assumed in the power series expansion of w .

Opposite to this, the in-plane deflections u and v should be independent of the direction of buckling (and hence of the sign of α ). So only even powers of α are assumed in the power series expansion of u and v .

Substituting these in equations (5a) to (5c)

nmnmy

nm

yx

x

m n

nny

n

n yx

xx y

wy

wx

wx

whEyv

xuhE

N +∞

=

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+∂∂

∂∂

⎟⎟⎠

⎞⎜⎜⎝

−+⎟⎟

⎞⎜⎜⎝

⎛∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

−= ∑ ∑∑ αν

νναν

νν 121

1 3,1 3,12,0

nmnmx

nm

yx

y

m n

nnx

n

n yx

yy x

wx

wy

wy

whEx

uyvhE

N +∞

=

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+∂∂

∂∂

⎟⎟⎠

⎞⎜⎜⎝

−+⎟⎟

⎞⎜⎜⎝

⎛∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

−= ∑ ∑∑ αν

νναν

νν 121

1 3,1 3,12,0

∑ ∑∑∞

=

=

+∞

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=3,1 3,12,0 m n

nmnmxy

n

nnnxyxy y

wx

whG

xv

yu

hGN αα

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24  

Expressing these equations as sum of two terms as

∑ ∑∑∞

=

=

+∞

=

+=3,1 3,1

),(

2,0

)(

m n

nmnmx

n

nnxx NNN αα (3.7a)

∑ ∑∑∞

=

=

+∞

=

+=3,1 3,1

),(

2,0

)(

m n

nmnmy

n

nnyy NNN αα (3.7b)

∑ ∑∑∞

=

=

+∞

=

+=3,1 3,1

),(

2,0

)(

m n

nmnmxy

n

nnxyxy NNN αα (3.7c)

where

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

−=

yv

xuhE

N ny

n

yx

xnx ν

νν1)( (3.8a)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+∂∂

∂∂

⎟⎟⎠

⎞⎜⎜⎝

−=

yw

yw

xw

xwhE

N nmy

nm

yx

xnmx ν

νν121),( (3.8b)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

−=

xu

yvhE

N nx

n

yx

yny ν

νν1)( (3.8c)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+∂∂

∂∂

⎟⎟⎠

⎞⎜⎜⎝

−=

xw

xw

yw

ywhE

N nmx

nm

yx

ynmy ν

νν121),( (3.8d)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=xv

yu

hGN nnxy

nxy

)( (3.8e)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=y

wx

whGN nm

xynm

xy),( (3.8f)

Now as stated earlier in this chapter, for a plate with no lateral loads, Von Karman large deflection equations can be written as

0=∂

∂+

∂∂

yN

xN xyx

0=∂

∂+

yN

xN yxy

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25  

( ) 0222

2

2

2

2

4

4

22

4

4

4

=∂∂

∂−

∂∂

−∂∂

−∂∂

+∂∂

∂+++

∂∂

yxwN

ywN

xwN

ywD

yxwDDD

xwD xyyxyxyxyyxx νν

Substituting the power series expansions of all the terms in these equations, three equations are obtained each of which equates a power series of α to zero. For the equation to hold true for all values ofα , coefficients of all the exponents of α in each of the three equations should be equal to zero.

Equating the coefficients of 00 =α ; the following relations are obtained

0

0

)0()0(

)0()0(

=∂

∂+

=∂

∂+

∂∂

yN

xN

yN

xN

yxy

xyx

} (3.9a)

and equating the coefficients of 01 =α ;

( ) 022 12

)0(2

12

)0(2

12

)0(4

14

221

4

41

4

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂+

∂∂

+∂∂

−∂∂

+∂∂

∂+++

∂∂

yxwN

ywN

xwN

ywD

yxwDDD

xwD xyyxyxyxyyxx νν

(3.9b) Similarly equating the co-efficient of 02 =α

0

0

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

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

=∂

∂+

∂+

∂+

=∂

∂+

∂+

∂∂

+∂

xN

xN

yN

yN

yN

yN

xN

xN

xyxyyy

xyxyxx

} (3.9c)

and equating the co-efficient of 03 =α the next equation is obtained as

( )

[ ] [ ] [ ]yx

wNNywNN

xwNN

yxw

Nyw

Nxw

Nyw

Dyx

wDDD

xw

D

xyxyyyxx

xyyxyxyxyyxx

∂∂∂

++∂∂

++∂∂

+

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂+

∂∂

+∂∂

−∂∂

+∂∂

∂+++

∂∂

12

)1,1()2(2

12

)1,1()2(2

12

)1,1()2(

32

)0(2

32

)0(2

32

)0(4

34

223

4

43

4

2νν

(3.9d)

Similarly equating higher coefficients of α to zero, more equations can be obtained.

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3.3 Compressive load problem

The plate under compressive loading along x- direction has the length a , width b and thickness h. The origin lies at one corner of the plate as shown in figure. The edges are all simply supported with compressive load of xN per unit width.

Fig: 3.1- Loading condition on the plate

3.3.1 Boundary Conditions

Zero transverse deflection at the edges

( ) ( ) ( ) ( ) 0,0,,,0 ==== bxwxwyawyw

Zero moment at edges

0),(

2

2

)0,(2

2

),(2

2

),0(2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

bxxyay xw

xw

xw

xw

Constant displacement along an edge

0),()0,(),(),0(

=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

bxxyay xv

xv

yu

yu

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27  

Zero shear stress at edges

0),(),0(),()0,(

=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

yaybxx xv

xv

yu

yu

Loaded edges

( ) PdyNb

x −=∫0

at ax ,0=

Unloaded edges

( ) PdxNa

y −=∫0

at by ,0=

The load is equal to or greater than the buckling load. Substituting the relations from equation (8a) & (8c) into the condition for loaded edge, the following is obtained

n

n

nPP α∑∞

=

=2,0

)( (3.10)

where

For loaded edges

( )dyNPb

x∫−=0

)0()0( at ax ,0= (3.11a)

( )dyNNPb

xx∫ +−=0

)1,1()2()2( at ax ,0= (3.11b)

( )dyNNPb

xx∫ +−=0

)3,1()4()4( 2 at ax ,0= (3.11c)

(as )1,3()3,1(xx NN = )

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28  

Similarly for unloaded edges

( ) 00

)0( =∫ dxNa

y at by ,0= (3.12a)

( ) 00

)1,1()2( =+∫ dxNNa

yy at by ,0= (3.12b)

( ) 02

0

)3,1()4( =+∫ dxNNa

yy at by ,0= (3.12c)

Now using equations (1a) and (1b)

0)0()0(

=∂

∂+

∂∂

yN

xN xyx

and 0)0()0(

=∂

∂+

yN

xN yxy

Substituting the relations from equations (3.8a), (3.8c) and (3.8e)

01

0000 =⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

∂∂

+⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

−∂∂

xv

yu

hGyy

vx

uhEx xyy

yx

x ννν

and 01

0000 =⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

−∂∂

+⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

∂∂

xu

yvhE

yxv

yu

hGx x

yx

yxy ν

νν

rewriting these

011

02

20

2

20

2

=∂∂

∂⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

− yxv

GE

yu

GxuE

xyyx

yxxy

yx

x

ννν

νν (3.13a)

0

110

2

20

2

20

2

=∂∂

∂⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

−+

∂∂

yxu

GE

yvE

xv

G xyyx

xy

yx

yxy νν

ννν

(3.13b)

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29  

Solutions of these equations satisfying the boundary conditions will be of the form

⎟⎠⎞

⎜⎝⎛ −=

200axUu

⎟⎠⎞

⎜⎝⎛ −=

200byVv

Using the loading boundary conditions the values of 0U and 0V are obtained as

hbEPU

x

)0(

0 −=

and hbE

PV

x

x)0(

=

so that

⎟⎠⎞

⎜⎝⎛ −−=

2

)0(

0ax

hbEPu

x

(3.14a)

⎟⎠⎞

⎜⎝⎛ −=

2

)0(

0by

hbEP

vx

xν (3.14b)

and therefore

0)0()0(

)0()0(

==

−=

xyy

x

NNb

PN (3.15)

Now 1w can be obtained from the equation (3.9b) which has the solution of the form

b

yna

xmWw ππ sinsin11 = (3.16)

NyMxW sinsin1=

which satisfies the boundary conditions. Putting this solution in the equation (3.9b) and substituting the results of the equation (3.15)

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30  

( )

0sinsinsinsin

sinsin2sinsin

12

22)0(

14

44

122

422

14

44

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

++++

byn

axmW

am

bP

byn

axmW

bnD

byn

axmW

banmDDD

byn

axmW

amD

y

xyxyyxx

ππππππ

πππννπππ

which gives the value of )0(P as

( )

2

22

4

44

22

422

4

44

)0(

2

am

bnD

banmDDD

amDb

Pyxyxyyxx

π

ππννπ⎥⎦

⎤⎢⎣

⎡++++

= (3.17)

Now using equation (3.9c)

0

)1(2)1(

1122

21

2122

=⎥⎦

⎤⎢⎣

⎡∂∂

∂∂

∂∂

+⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

∂∂

+⎥⎥⎦

⎢⎢⎣

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

−∂∂

+⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−∂∂

yw

xwhG

yyv

xuhG

y

yw

xwhE

xyv

xuhE

x

xyyxy

yyx

xy

yx

x

ν

ννν

ννν

or

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−−=

∂∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

21

211

21

12

12

12

122

22

2

22

2

)1()1()1(

xw

xw

yxw

yw

G

yxw

yw

xw

xwE

yxv

GE

yu

GxuhE

xy

yyx

xxy

yx

yxxy

yx

x ννννν

ννν

Substituting the value of 1w and simplifying the equation becomes

⎥⎥⎦

⎢⎢⎣

⎥⎥⎦

⎢⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

+−⎟

⎟⎠

⎞⎜⎜⎝

=∂∂

∂⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

NyCosNGENM

MENMM

MxSinw

yxv

GE

yu

GxuhE

xyyx

xy

yx

xy

xyyx

yxxy

yx

x

2)1(2

)()1(2

)(2

2

)1()1(

222222

1

22

22

2

22

2

ννν

ννν

ννν

νν

(3.18a)

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31  

Similarly after simplifying, the second equation of (3.9c) becomes

⎥⎥⎦

⎢⎢⎣

⎥⎥⎦

⎢⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

−+

−⎟⎟⎠

⎞⎜⎜⎝

−−

=∂∂

∂⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

−+

∂∂

MxMGEMN

NEMNN

NySinw

yxu

GE

yu

GyvhE

xv

G

xyyx

yxx

yx

yx

xyyx

xyxy

yx

xxy

2cos)1(2

)()1(2

)(2

2

)1()1(

222222

1

22

22

2

22

2

22

2

νννν

ννν

ννν

νν

(3.18b)

Now the total solution of the equation is given by

hp uuu 222 +=

hp vvv 222 +=

Total solution = Particular solution + homogeneous solution

3.3.2 Homogeneous solution

As the homogeneous part of the equation (3.18a) and (3.18b) in 2u and 2v is same as that in ou and

ov in equation (3.13a) and (3.13b), the homogeneous solution will of the same form.

Let ⎟⎠⎞

⎜⎝⎛ −=

222axUu

& ⎟⎠⎞

⎜⎝⎛ −=

222axVv

Now ⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−=

yv

xuhE

N yyx

xx

22)2(

)1(ν

νν

[ ]22)1(

VUhE

yyx

x ννν

+−

=

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

−=

21

21)1,1(

)1(2 yw

xwhE

N yyx

xx ν

νν

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32  

Substituting the value of SinMxSinNyWw 11 =

[ ]21

21

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

SinMxCosNyNwCosMxSinNyMwhE

N yyx

xx ν

νν+

−=

similarly

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−=

yu

xvhE

N xyx

yy

22)2(

)1(ν

νν

[ ]22)1(

UVhE

yyx

y ννν

+−

=

[ ]21

21

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

CosMxSinNyMwSinMxCosNyNwhE

N xyx

yy ν

νν+

−=

Using the relation from equation (3.11b)

( ) dyNNP

ax

b

xx,00

)1,1()2()2(

=∫ +−= for loaded edges

[ ] NySinWM

hEVU

hENN

yx

xy

yx

xaxxx

221

222,0

)1,1()2(

)1(2)1( ννν

νν −++

−=+

=

Similarly from equation (3.12b)

( ) 0

,00

)1,1()2( =+−=

∫ dxNNby

a

yy

for unloaded edges

[ ] MxSinWN

hEUV

hENN

yx

yx

yx

y

byyy22

12

22,0

)1,1()2(

)1(2)1( ννν

νν −++

−=+

=

Using the integrals ∫ =b bNydySin0

2

2

& ∫ =a aMxdxSin0

2

2

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33  

The following is obtained

For loaded edge

[ ] bWMhE

bVUhE

Pyx

xy

yx

x 21

222

)2(

)1(4)1( ννν

νν −++

−=−

For unloaded edge

[ ] aWN

hEaUV

hE

yx

yx

yx

y 21

222 )1(4)1(

0νν

ννν −

++−

=

Solving these two equations

( )( ) ⎥

⎥⎦

⎢⎢⎣

−+−=

41

21

22)2(

2WNM

hbEPU

yx

y

x ννν

( )( ) ⎥

⎥⎦

⎢⎢⎣

−−

+=41

21

22)2(

2WNM

hbEP

Vyx

x

x

x

νννν

3.3.3 Particular Solution

The particular solution will be of the form

[ ]NyBCosAMxSinu p 222 +=

[ ]MxDCosCNySinv p 222 +=

Substituting in two equations and comparing the co-efficient of terms

( )M

NMWA y222

1

16ν−

−= (3.19a)

(Comparing the co-efficient of MxSin2 )

( )N

MNWC x222

1

16ν−

−= (3.19b)

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34  

(Comparing the co-efficient of NySin2 )

Again comparing coefficient of NyMxCosSin 22

( ) ( )( )( ) ( ) ⎥

⎥⎦

⎢⎢⎣

⎡+

−−

+=

⎥⎥⎦

⎢⎢⎣

⎥⎥⎦

⎢⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

−+

⎥⎥⎦

⎢⎢⎣

⎡+

−2

22212

2

11811MNG

ENMMWGE

MNDGNEM

B xyyx

x

yx

yxy

yx

yxxy

yx

x

ννννν

ννν

νν

comparing coefficient of MxNyCosSin 22

( ) ( )( )( ) ( ) ⎥

⎥⎦

⎢⎢⎣

⎡+

−−+

=⎥⎥⎦

⎢⎢⎣

⎡+

−+

⎥⎥⎦

⎢⎢⎣

⎥⎥⎦

⎢⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

−NMG

EMNNWGM

ENDG

EMNB xy

yx

y

yx

xxy

yx

yxy

yx

xy 2222

122

11811 ννννν

ννννν

Solving above two equations B and C are obtained as

16

21MWB = (3.19c)

16

21NWD = (3.19d)

Therefore the complete expressions for 2u and 2v can be written as

( )( )

( )⎥⎥⎦

⎢⎢⎣

⎡+

−−+⎟

⎠⎞

⎜⎝⎛ −⎥⎥⎦

⎢⎢⎣

−+−= NyCosMW

MNMWMxSinaxWNM

hbEPu y

yx

y

x

21616

2241

21

2221

21

22)2(

2

νννν

(3.20a)

and

( )( )

( )⎥⎦

⎤⎢⎣

⎡+

−−+⎟

⎠⎞

⎜⎝⎛ −⎥⎥⎦

⎢⎢⎣

−−

+= MxCosNWN

MNWNySinaxWNMhbE

Pv x

yx

x

x

x 21616

2241

21

2221

21

22)2(

νννν

(3.20b)

where

amM π

= and b

nN π=

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35  

Now from equation (3.9d)

( )

[ ] [ ] [ ]yx

wNNywNN

xwNN

yxw

Nyw

Nxw

Nyw

Dyx

wDDD

xw

D

xyxyyyxx

xyyxyxyxyyxx

∂∂∂

++∂∂

++∂∂

+

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂+

∂∂

+∂∂

−∂∂

+∂∂

∂+++

∂∂

12

)1,1()2(2

12

)1,1()2(2

12

)1,1()2(

32

)0(2

32

)0(2

32

)0(4

34

223

4

43

4

2νν

(3.21)

R.H.S. of the equation (3.21) is the sum of three terms

R.H.S = Term 1 + Term 2 + Term 3

Analyzing term by term

Term 1 21

221

2122

21

)1( xw

yw

xw

yv

xuhE

yyyx

x

∂∂

⎥⎥⎦

⎢⎢⎣

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−= νν

νν

substituting the value of 122 &, wvu in Term 1 and simplifying we get co-efficient of SinMxSinNy in Term1 as

( ) ( )[ ][ ] 12

221 2)(21

WMDCNVBAMUE

C yyx

x +++++−

−= ννν

Similarly co-efficient of SinMxSinNy in Term 2

( ) ( ) [ ][ ] 12

222 )(221

WNBAMUDCNVE

C xyx

y +++++−

−= ννν

and co-efficient of SinMxSinNy in Term 3

( )[ ]3113 8 MNWWDNBNhMNGC xy −+=

Hence co-efficient of SinMxSinNy in Term 2 in the R.H.S of equation

321 CCCC ++=

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36  

So the equation becomes

( ) +++= SinMxSinNyCCCSHL 321.. A term free of SinMxSinNy

The homogeneous part of equation (3.21) is same as that in (3.9b). Hence SinMxSinNy will also be a homogeneous solution to the equation (3.21). But a term of SinMxSinNy appears on the right-hand-side. Therefore no solution to the equation (3.21) will be possible that satisfies the boundary conditions unless co-efficient of SinMxSinNy on R.H.S =0

0321 =++∴ CCC

Substituting the values of coefficients and solving this gives the value of 21W as

( ) ( )⎥⎥⎦

⎢⎢⎣

+−+

−=

)1()1()2(

822

44

2)2(2

1

yx

xyyxyx

yx

yx NMEENEME

MbhPW

νννν

νννν

(3.22)

3.4 Calculation of effective width

Axial shortening is given by

),(),0( yauyu −=Δ

),(),(),( 20 yxuyxuyxu +=

Now ⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−=−

22),(),0( 0000

aUaUyauyu

= aU 0−

hp uuu 222 +=

0),(),0( 22 =− yauyu pp

aUyauyu hh 222 ),(),0( −=−

Hence the Total shortening

),(),0( yauyu −=Δ

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37  

aUaU 20 −−=Δ

Substituting the values of U0 and U2

( )( ) aWNM

hbEPa

hbEP

yx

y

xx ⎥⎥⎦

⎢⎢⎣

−++=Δ

41

21

22)2(2

)0(

ννν

α

( ) ( )( ) 41

21

222)2(2)0( WNMhbE

PPa yx

y

x ννναα

−+

+=

Δ

( )( ) 41

21

222 WNMhbE

Pa yx

y

x νννα

−+=

Δ

Effective width is given by ha

EPbx

e Δ=

so

( )( ) 41

21

222 WNMhbE

PhbE

P

bb

yx

y

x

xe

νννα

−+

= (3.23)

3.5 Non dimensionalization of expressions

The relations obtained in this chapter are non-dimensionalised in this section so that all the expressions can be evaluated using just the aspect ratio of plate and the ratio of properties in two orthogonal directions.

From equation (3.22)

( )( )( ) ( )[ ]2244

2)2(2

1 218

NMEENEMEM

bhPW

xyyxyxyx

yx

νννννν

+−+−

−=

Let cEE

x

y

x

y ==νν

& λ=ba

(3.24)

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38  

Substituting a

mM π= &

bnN π

=

The equation reduces to

( )

( )( )[ ]2224442

22

2

)2(2

1 2218

λνλνλν

π nmccnmcacm

hEPW

xx

x

x −+−

−= (3.25)

Let shortening at the start of buckling i.e. at )0(PP = be 0Δ=Δ

Then ( ) ( )yauyu ,,00 −=Δ

and ⎟⎠⎞

⎜⎝⎛ −==

200axUuu

therefore ( ) ( )yauyu ,,0 000 −=Δ

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−=

22 00aUaU

aU 0−=

hence hbEaP

x

)0(

0 =Δ

or hbE

Pa x

)0(0 =

Δ

Now from previous analysis

( )( ) 41

21

222 WNMhbE

Pa yx

y

x νννα

−+=

Δ

dividing the two equations

( )( ) )0(

21

222

)0(0 41 P

hbEWNMPP x

yx

y

νννα

−+=

ΔΔ

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39  

after substituting

)2(

)0(2

PPP −

=α & ( )

( )( )[ ]2224442

22

2

)2(2

1 2218

λνλνλν

π nmccnmcacm

hEPW

xx

x

x −+−

−=

this reduces to

( )( )( )[ ]

( ))0(

)0(

2224442

2222

)0(0 22

2P

PPnmccnmc

ncmmPP

xx

x −−+−

−+=

ΔΔ

λνλνλν

Now Let

( )

( )( )[ ] Knmccnmc

ncmm

xx

x =−+−

−2224442

2222

222

λνλνλν

(3.26)

which results in

( ))0(

)0(

)0(0 P

PPKPP −

+=ΔΔ

or

KK

KPP

++

ΔΔ

+=

111

0)0(

and

( )( ) 41

21

222 WNMhbE

PhbE

P

bb

yx

y

x

xe

νννα

−+

=

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40  

which reduces to

( ) K

PPK

PP

bb

cr

cre

−+=

1 (3.27)

crPP =)0(Q

3.6 Graphical analysis

3.6.1 Load deflection curve

First Graph is drawn between the crPP

Vs 0ΔΔ

Recalling the expression ( )

)0(

)0(

)0(0 P

PPKPP −

+=ΔΔ

Where ( )

( )( )[ ]2224442

2222

222

λνλνλν

nmccnmcncmm

Kxx

x

−+−

−=

Graph is plotted for m=1 and m=2 and other values fix as shown.

m n νx c λ K η 1 1 0.25 0.2 1 0.83151 0.2 2 1 0.25 0.5 2 0.647399 0.5

Where cEE

x

y

x

y ==νν

λ=ba

and x

y

EE

bn

ma

4

4

4

4

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41  

Fig 3.2: Load vs. Deflection curve for c=.2 and λ=1

 

Fig 3.3: Load vs. Deflection curve for c=.5 and λ=1

As we can see, the graph crPP

Vs 0ΔΔ

is linear, which is expected since we haven’t introduced any second

order term of the deflection although we have taken non linearity in deformation into account. Now the curves have been shown for aspect ratio λ=1 that is a square plate.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 0.5 1 1.5 2 2.5

P/P0(m=1)

P/P0(m=2)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 0.5 1 1.5 2 2.5

P/P0(m=1)

P/P0(m=2)

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42  

3.6.2 Comparison of results of bbe by two methods:

bbe has been derived by two methods.

3.6.2a) Successive approximation Method

( ) KPPK

PP

bb

cr

cre

−+=

1

where ( )

( )( )[ ]2224442

2222

222

λνλνλν

nmccnmcncmm

Kxx

x

−+−

−=

3.6.2b) Stress Method

( )η+

−+

=

1

)1(2cr

cr

cre

PP

PP

PP

bb

where x

y

EE

bn

ma

4

4

4

4

Graph is plotted for m=1 and m=2 and other values fix as shown.

m n νx c λ K η 1 1 0.25 0.2 1 0.83151 0.21 1 0.25 0.5 1 0.647399 0.5

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43  

Fig 3.4: Effective width vs. deflection curve of orthotropic plate for c=0.2 and λ=1

Fig 3.5: Effective width vs. deflection curve of orthotropic plate for c=0.5 and λ=1

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2 2.5

be/b(m=1)

be/b stress(m=1)

be/b(m=2)

be/b stress(m=2)

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2 2.5

be/b(m=1)

be/b stress(m=1)

be/b(m=2)

be/b stress(m=2)

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44  

3.7 Conclusion

From effective width vs. deflection curves of the two methods: 1) Stress Method and 2) successive approximation method, results have been compared and it can been seen that the values by two methods agree to an extent. Little deviation can be attributed to fact that the two methods are approximate methods. Also as the value of m increases or number of half sine curves in which the plate buckle increases, the effective width decreases.

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45  

References

1. Stein M. , Loads and Deformations of Buckled Rectangular Plates. NASA TR R-40, 1959

2. Indian Standard Code of Practice for Use of Cold-Formed Light gauge Steel Structural Members In General Building Construction (First Revision). IS : 801-1975

3. Timoshenko S. P., Woinowsky-Krieger S., Theory of Plates and Shells. Second Edition.

4. Iyengar NGR., Elastic Stability of Structural Elements.

5. Chajes A., Principles of Structural Stability Theory.