A.- .. ._. ‘_____. _‘r -

   
     

  

LISRARY

Michigan State
University

This is to certify that the

thesis entitled

PRE- AND POST-BUCKLING BEHAVIOR
OF PLATES OF VARIABLE STIFFNESS
USING FINITE DIFFERENCES

presented by

Mohammad Ali Barkhordari

has been accepted towards fulfillment
of the requirements for

__£h.._D_.__degree in Weering

( . ’D
a . _ ) , . ,
“21/ Aft (we a (34;..[4'1

Major professor <

Date December 24L 1980

 

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PRE-AND POST-BUCKLING BEHAVIOR
OF PLATES OF VARIABLE STIFFNESS
USING FINITE DIFFERENCES

By
Mohammad Ali Barkhordari

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
DOCTOR OF PHILOSOPHY

Department of Civil and Sanitary Engineering

1980

 

 

ABSTRACT
PRE-AND POST-BUCKLING BEHAVIOR

OF PLATES OF VARIABLE STIFFNESS
USING FINITE DIFFERENCES

By

M. Ali Barkhordari

The Von Karman large deflection equation is applied to plates
of variable stiffness. Equilibrium equations and the in-plane
compatibility equation are derived. The ordinary finite difference
technique is employed to solve the nonlinear coupled partial differental
e(Nations. Iva different methods of formulation are considered:

a) In terms of lateral displacement w and a stress function,

5) In terms of displacement components u, v and w.

Stiffness variation can be implemented in two different ways, either
by varYiug the thickness of the plate with constant E, or by taking

a uniform thickness plate of variable E. Both types of stiffness
variation are considered. The nature of in-plane displacements on

the boundary is a significant factor in postbuckling. This effect

is examined by considering plates with different in-plane displacement
boundary conditions.

Several problems with different stiffness variation and
bOundary conditions are solved. The applicable computer program is
utilized to carry out the numerical solutions. In each case the

problem is investigated for different stages of loading as follows:

 

 

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a). Membrane solution analyzes the behavior of in—plane
forces and displacements for undeflected plates,
b) Stability analysis investigates the buckling and effect
of stiffness variation on critical loads and buckling
modes,
c) Postbuckling discusses the behavior of various aspects
of the problem due to edge loads or displacements higher
than critical values.
For clarity, the results are always accompanied by graphical illustrations
of membrane and bending stress as well as displacement components.
The accuracy of the solution is evaluated by comparison of the results
obtained with results from past studies and exact results, where
these results are available. The influence of the grid-spacing on
the accuracy of the results is investigated by taking successively
finer grid-spacings. The numerical results are analyzed and the effect
of stiffness variation on different aspects of the problem discussed.
One objective is to design a plate with stiffness variation
such that it be optimum in some respect. Some possible cases of
optimization are discussed and,as examples,some problems related to
buckling are solved. The results indicate that a considerable weight

and/or material savings can be achieved by using an efficient stiffness

variation pattern.

 

 

 

 

 

 

ACKNOWLEDGMENTS

The author wishes to express his most sincere gratitude to
Ids major Professor, Dr. W. A. Bradley, Professor of civil engineering
for his encouragement and constant help and guidance in the author's
academic development and preparation of this dissertation.

Thanks are also expressed to the other members of the guidance
committee - Dr. N. Altiero, Dr. J. L. Lubkin and Dr. R. K. Wen -
for their guidance and encouragement. The author also owes his
appreciation to Mrs. Clara Hanna for typing of this dissertation.

Special appreciation is also due his wife and son for their patience

and understanding.

ii

 

TABLES OF CONTENTS

Page
LIST OF TABLES-o.oooooooooo0.0000000000000000. v

LIST OF FIGURESOOOCOOOOOOOIOOOOOOOOOOOOOOO0.00 Vi

Chapter
I INTRODUCTION.................................. 1
1.1 General Remarks.......................... 1
1.2 Previous Developements................... 3
1.3 Present Investigation.............. ..... . 8
1.4 Notations................................ 10
II THEORETICAL DERIVATIONS 13
2.1 General.................................. 13
2.1.1 Nonlinear Equilibrium Equations... 14
2.1.2 Relation Between Stress Resultants
and Displacements................. 18
2.2 Formulation in Terms of Stress Function
and w.................................... 19
2.3 Formulation in Terms of Displacements
u, v and w............................... 21
2.4 Finite Difference Approximations......... 22
2.4.1 Principles of Finite Differences.. 22
2.4.2 Finite Difference Approximation
to Method Discussed in Section
(2.2)............................. 26
2.4.3 Finite Difference Approximation of
Method Discussed‘fin Section (2.3). 37
2.5 Boundary Conditions...................... 45
2.5.1 Some Examples of Practical B.C.... 48
2.6 Summary.................................. 48
2.6.1 Membrane Solution................. 49
2.6.2 Lateral Loading................... 49
2.6.3 Stability Analysis................ 50
2.6.4 Postbuckling...................... 51
III APPLICATION...................................' 52
3.1 Force Boundary Conditions................ 56

3.1.1 membrane Solution................. 58
3.1.1.1 Analysis of Results From

Membrane Solution......... 70

3.1.2 Buckling.......................... 77

iii

 

 

Chapter Page

3 2.1 General Buckling.............. 81
3.2.2 Analysis of the Results. ...... 88
3.1.3 Postbuckling. ...................... 93
3. .3.1 General Procedure. ............ 93
3. 3. 2 Numerical Solution............ 96
a) Square Plate with R a 1/10. 96
b) Square Plate with Uniform
Stiffness, R = 1........... 104
c ,d) Square Plate with R = 1/2
and R = 10. ......... ..... 110
3.1.3.3 Analysis of Postbuckling
Results....................... 119
3.1.3.3.1 Simply-Supported Edges...... 119
a) Uniform Stiffness Plate.... 119
b) Variable Stiffness, R = 1/10 124
c) Variable Stiffness, R 8 1/2 126
d) Variable Stiffness, R = 10.. 126
3.1.3.3.2 Clamped Edges................ 129
3.1.4 Optimization........................ 147
3.1.5 Summary............................. 154
3.2 Displacement Boundary Condition........... 156
3.2.1 Membrane Solution.................. 159
3. 2. 2 Buckling Solution.................. 164
3.2.2.1 Convergence Check and
Comparison.................... 164

.1.
l.

Idld

3.2.2.2 Optimization Analysis......... 169
3.2.2.3 Analysis of Buckling Mbdes.... 172

3. 2.3 Postbuckling....................... 175
3.2.3.1 Uniform Thickness Plate,

 

 

Simply Supported.............. 175
3.2.3.2 Simply Supported Square Plate

of Variable Thickness......... 181
3.2.3.3 Uniform Thickness Plate,

Clamped Boundary.............. 189
3.2.3.4 Clamped Plate with Variable

Thickness..................... 194

3.3 Comparison of Two Methods................. 202

IV CONCLUSIONooooooooooooocoo-000......ooooooooso. 204
4.1 The Problem Summary....................... 204
4. 2 Conclusions............................... 206
4.3 Recommendations........................... 208

BIBLIOGRAPE‘IY.oooooooooo'oo0.000000000000000ooooo 210

APPENDICES.°O............°.°.'...Ooooooooooooooo 214

iv

 

 

 

 

 

 

Table

3.1

3.2

3.3

3.4
3.5

3.6

3.7

3.8

3.9

3.10

3.11

3.12

3.13

3.14

LIST OF TABLES

Stress Function and Stress Resultant Ratios
Square Plate with R = 1/10.. ....................

Stress Function and Membrane Forces, R = 1/2....

Stress Function and In-plane Resultant Ratios,
R = 1 ...........................................

Stress Function and Membrane Forces, R = 10 .....
Convergence of the Solution, R = 1/10... ........

Eigenvalues of Simply-Supported Square Plate,
R 8 1/10, with Different Grid Spacings.... ......

Comparison of First Eigenvalues of Different
solutions (R=I)OOOOOOOOOOOCOOOOOOOOOOOOOOOO .....

Critical Loads of Simply-Supported Square Plate
R81 ...... 0.0... ........ .0 OOOOOOOOOOOOOOOOOOOOO

Critical Load of Clamped Square Plate Under
Bi-axial Uniform Load, R = 1............ ........

Boundary Displacement for Square Plate of
Figure 3.62................ .....................

Critical Displacements for a Simply-Supported
Plate using Different Mesh Sizes, Uniform
Stiffness Plate ..... ........ ....................

Critical Displacements for a Clamped Square
Plate using Different Mesh Sizes, Uniform
Stiffness .......... . ........... . ................

Convergence of Postbuckling Solution with Mesh
Size. Simply-Supported Plate...................

Convergence of Postbuckling Solution with Mesh
Size. Clamped PlateOOOIOOO.COOOOOOOOIOOOOOOOOOO

Page

65

67

68
68

69

80

80

82

91

158

165

166

176

189

 

 

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LIST OF FIGURES

Figure Page
2.1 Rectangular Flat Plate .......................... 15
2.2 Plate Element dxdy in Undeformed Configuration. . 15
2.3 Schematic Illustration of Internal Forces and
Moments on the Element of Middle Surface in
Deformed Configuration ........ . ................. 17
2-4 Function f(x) . . ................................. 23
2-5 Nodal Pattern..l........ ......................... 24
2.5 Two Dimensional Operator for fxy ................ 25
2.7 Nodal Arrangement. . . . . . . . . ...................... 27
2.8 Difference Operator for Left Hand Side of
Equilibrium Equation (2.14) ..................... 28
2.9 Difference Operator for Right Hand Side of
Equation (2.14) . . ............................... 30
2.10 Difference Operator for Left Hand Side of
Compatibility Equation (2.15) ................... 33
2.11 Operators for x—Equilibrium Equation (2.16) ..... 39
2.12“ Operators for y-Equilibrium Equation (2.17) ..... 41
3.1 Geometry and Stiffness Variation of Square
Plate-coo ooooooooooooooo Oooooooo ooooooooooooooo o 54
3-2 Stiffness Variation....... ...................... 57
3.3 Geometrical Plan and (9 Function ..... . ........... 61
3.4 Stiffness Ratio at Nodes and Intermediate
Nodes, Square Plate, R = 1/10 ......... . ......... 64
3.5 Convergence of Membrane Solution.. .............. 70

Force Distribution for Undeflected Square

3.6
Plate, R - 1...........

Figure Page

3.7 Contours of In-plane Displacement,

Undeflected Square Plate, R = 1 ................ 72
3.8 Distribution of In-plane Force and Displace-

ment, Square Plate, R = 1/10 ................... 73
3.9 Distribution of In-plane Force and Displace-

ment, Square Plate, R = 1/2 .................... 75
3.10 Distribution of In-plane Force and Displace-

ment, Square Plate, R = 10 ..................... 76
3.11 Convergence of Eigenvalue, R = 1/10 ............ 79
3-12 Modes of Buckling of Square Plate, R = 1/10,

Simply-Supported ....................... . ....... 84
3-13 Modes of Buckling of Square Plate, R = 1/10,

Clamped........‘. ............................... 84
3.14 Modes of Buckling of Square Plate, R = 1/2,

Simply-Supported ........ . ...................... 85
3.15 Modes of Buckling of Square Plate, R = 1/2,

Clamped. . . . . ................................... 85
3.16 Modesof Buckling of Square Plate, R = 1,

Simply-Supported..............................
3.17 Modes of Buckling of Square Plate, R = 1,

Clamped..0.O.C00.0.0...OOOIOOOOOOUOOOCOOO0.0..
3.18 Modes of Buckling of Square Plate, R = 10,

Simply-Supported” . ...................... 87
3.19 Modes of Buckling of Square Plate, R = 10,

Clamped........_ ........................ . ....... 87
3.20 Deflected Shape for s—s and Clamped Plate ...... 90
3.21 Flow Chart of Iterative Procedure .............. 94
3.22 Plots of w, U, Nx and M x’ Square Plate, 8-8,

R8 llloooooooooooo o oooooooo oo ooooooo 000000000. 100
3.23 Plots of Stress Components, Square Plate, s-s,

Rs 1/10....... .......... ..... ......... o ....... 101
3 24 Contours of In-plane Force and Profiles of

N ’ 5-3, R = 1/10.. .......................... 102

x

 

 

3.25

3.26

3.27

L28

L29
L30

3.31

L32

L33

L34

L35

L36

3.37

IL38
L39
IL40

3.41

3.42

3.43

In-Plane Displacement, Square Plate, s-s,
R=1/10 ooooooooooooooooooooooo oooooo oooooooooooo

Plots of w, U, Nx and M x’ Square Plate, s-s,
R = 1 ......................................... ...

Plots of Stress Components, Square Plate, s-s,
R=1000 OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

Contours of In-Plane Force and Profiles of N x’
s—s, R = 1 .......................................

In-Plane Displacement, Square Plate, s—s, R = 1..
Square Plate Under Uniform Load, s-s, R = 1 ......

Plots of w, U, Nx and M x’ Square Plate, s-s,
R = 1/2 ................................ . .........

Plots of Stress Components, Square Plate, s-s,
Ra 1/20 oooooooo o oooooooooooooooooooooooooooooooo

Contours of In-Plane Force and Profiles of N x’
S-S’R=1/20 ..... 00...... ..... O... ....... O ......

In-Plane Displacement, Square Plate, s-s, R = 1/2.

Plots of w, U, Nx and M x’ Square Plate, s-s,
R=100 0...... 000000 O ..... O OOOOOOOOOOOOOOOOOOOO

Plots of Stress Components, Square Plate, s-s,
R=100000000000 OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

Contours of In—Plane Force and Profiles of N x’
8-3 , =10 oooooooooooo o ooooooooooooooooooooooooo

In-Plane Displacement, Square Plate, s-s, R = 10.
Convergence of the Solution with Mesh Size .......

Deflected Shape for Various R Values... ..........

Plots of w, U, Nx and Mx’ Square Plate, Clamped,
R=1/10ooooooo ooooo ooooo oooooooooooooooooooooooo

Plots of Stress Components, Square Plate,
Clamped, R = 1/10. .................... . ..........

Contours of In-Plane Force and Profiles of N x’
Clamped, R = 1/10 ooooo o ooooooooooooooooooo o ......

 

105

107
108

109
111

112

114

117
118

120

132

133

 

‘ A

 

3.44

3.45

3.46

3.47

3.48

3.49

3.50

3.51

3.52

3.53

3.54

3.55

3.56

3.57

3.58

3.59

3.60

3.61

In-Plane Displacement, Square Plate, Clamped,
R - 1/10 ...................... . .................. 134

Plots of w, U, Nx and Mx , Square Plate, Clamped,
RF 1.......... ................................... 135

Plots of Stress Components, Square Plate,
Clamped, R= l ........ . ............ .. ........... 136

Contours of In-Plane Force and Profiles of N x’
Clamped, R . 1 ...................... . .......... 137

In—Plane Displacement, Square Plate, Clamped,
R. 1... .......... ...... ..... . ........... . ..... 138
R-l/z ooooo oooo oooooooooooooooooooo oooo ooooooo 139

Plots of Stress Components, Square Plate, Clamped
R'l/Zooo ..... ........ ............ . ........... 140

Contours of In-Plane Force and Profiles of N ,
Clamped, up 1/2.............................¥.. 141

In-Plane Displacement, Square Plate, Clamped,
R.1/2....OO.......0...O....0.000....0.0...... 142

Plots of w, U, Nx and M x’ Square Plate, Clamped,

R a 10... ............. . ............. . .......... 143
Plots of Stress Components, Square Plate,

Clamped, R a 10... ............. . ...... . ........ 144
Contours of In—Plane Force and Profiles of

Nx’ Clamped, R = 10......... ................... 145
In-Plane Displacement, Square Plate, Clamped,
R810. ....... .0.............I. ....... . ........ 146
Thickness Variation... ..... ..... ..... .... ...... 148

Critical Load vs. RT, Square Plate, Simply—
SupportedOOOOOOO0..............I..O0.00.00 ..... 152

Critical Load vs. RT, Square Plate, Clamped.... 153

Central Deflection for Different R Values,
8-8 and Clampedooooooooo ooooooo ooo ooooooooooooo 155

Plan.......0... ....... OOOOOOOOOOOOOI0.0000 ...... 156

 

 

3.62 Node Arrangement, Square Plate, h = a/8 ........ 157

3.63 Contours of Membrane Force, Principal Stress
and U-displacement, Undeflected Plate, RT = 1.. 160

3.64 Contours of Membrane Force, Principal Stress
and U-displacement, Undeflected Plate, RT=1/4... 162

3.65 Contours of Membrane Force, Principal Stress
and U-displacement, Undeflected Plate, RT = 2... 163

 

3.66 Convergence of Buckling Solution. ............... 165
3.67 Convergence of Buckling Solution ....... . ........ 166
3.68 Plan ................................... . ........ 168
3.69 Variation of Critical Displacement Versus RT.... 170
3.70 Variation of Critical Displacement Versus RT.... 171
3.71 Buckling Modes of Simply-Supported Square Plate,
RT!l..................... ....... . 0000000000000 174
3.72 Buckling Modes of Clamped, Square Plate, RT = 1.. 174
3.73 Convergence of Center Deflection, s-s, Square
Plate... ............................. .. ........ . 176
3.74 Plots of Stress Components, Square Plate,
S-S, RT=loooo ooooooo ooooooooooo ooooooo o ooooooo 178
3.75 In-Plane Displacement, Square Plate, s—s,
RT:1..0............O........... ............... 179
3.76 Square Plate Under Uniform Lateral Load......... 180
3.77 Central Deflection Versus Edge Displacement for
Different RT Values, Simply-Supported Square
Plate........ ...... ........O.............. ..... O 182
3.78 Plots of Stress Components, Square Plate,
Simply-Supported, RT = 1/4................. ..... 183
3.79 Plots of Stress Components, Square Plate, Simply-
Supported, RT = 2. ...... ................ ........ 185
IL80 Principal Stress Versus Edge Displacement for
Different RT Values, Simply-Supported Square
Plate...... .............................. .. ..... 187

 

 

 

In

II)

3.81

3.82

3.91

A3

A4

BI

32

InéPlane Displacement, Square Plate, Simply-

Supported, RT = 1/4 ............................. 188
In-Plane Displacement, Square Plate, Simply-

Supported, RT = 2 ............................... 188
Convergence of Solution ......................... 190

Central Deflection Versus Edge Displacement for
Different RT Values, Clamped Edges .............. 190

Plots of Stress Components, Square Plate, Clamped,
RT = 1 .......................................... 192

In-Plane Displacement, Square Plates, Clamped,
RT = 1.... ..... . ................................ 193

Plots of Stress Components, Square Plate, Clamped,
RT = 1/4 ........................................ 196

Plots of Stress Components, Square Plate, Clamped,
RT = 2 ........................................... 197

Max. Principal Stress Versus Edge Displacement
for Different RT Values, Square Plate, Clamped... 199

In—Plane Displacement, Square Plate, Clamped,
RT = 1/4 ........................................ 201

In-Plane Displacement, Square Plate, Clamped,
RT 8 2 ..... . .................................... 201

Difference Operator for Left Hand Side of Equi-
librium equation (2.24), a = 1.................. 215

Difference Operator for Left Hand Side of Equation
(2.14), a - 1, Uniform Stiffness P1ate.......... 216

VFinite Difference Operator for Left Hand Side of
Compatibility Equation (2.15), a = l............ 217

Finite Difference Approximation of Equation (2.29),

a.l.........l...0...............0.0.......0.0. 218

Nbde Arrangement for Force Boundary Condition, Square
Plate,h-8/8.................................. 220

Nede Arrangement for Force Boundary Condition, Square
Plate,h.all-6.00.00...000......000.0.0.0.0.000 221

xi

 

 

 

 

0

B3 Node Arrangement for Displacement Boundary
condition,ha8/12....0.0.....0.0.0.......O.I.0. 222

B4 Nbde Arrangement for Displacement Boundary

Conditon, h = a/l6............................... 223
C1 Nede Numbers..................................... 227
C2 Arrangement of Vector K......................... 228

C3 Arrangement of Nodes in Vector KK............... 231

xii

 

CHAPTER I
INTRODUCTION

1.1 GENERAL REMARKS

The widespread use of plate elements in many engineering
structures such as buildings, bridges, pavements, missiles, containers,
ship structures and space structures has made plate analysis the
subject of scientific investigation for more than 200 years. Because
of their two dimensional action, the mechanical behavior of plates
under thrust loads is completely different from beam elements. In
contrast to beam elements, in which buckling is usually associated
with collapse of the structure, the buckling of a plate is not an
end point in the serviceability of the structure.

The capability of a plate to carry load after buckling
is an interesting subject which has motivated many investigators
to study posbuckling behavior of plates, especially in connection
with weight-sensitive space applications. Most of the plate analyses
involve-uniform stiffness plates. However, elastic plates of variable
stiffness are used in many engineering structures such as aircraft
wings, turbine disks, etc. The need to conserve material and/or

minimize weight motivates the designers to make optimum.use of

the material.

 

 

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From the structural point of view, knowledge of critical
buckling loads is of great importance. To make an optimum design
with respect to some variables, an extensive analysis of the variable-
stiffness plate is necessary. The failure strength of a thin plate
can exceed the buckling strength appreciably. In many cases, the
structure is not sensitive to large deflection. Thus, it is of
technical importance to consider the postbuckling behavior of plates
(especially the variable stiffness plate) in order to optimize
the design.

Although a considerable amount of work has been done in
the area of variable stiffness plates, most studies have achieved
solutions by analytical methods which are restricted to some specific
geometry and boundary conditions. (See Section 1.2)

The purpose herein is to investigate the behavior of a'variable
stiffness plate so that a full history of the stress and strain
components of plates with different stiffness variation can be
presented. Such a history will help give the designer a better
understanding of the behavior of the plate and the effect of stiff-
ness variation on various aspects of the problem so that a more-
nearly optimum design may be achieved. Two different types of
variation in stiffness are possible: one with uniform thickness and
varying E, such as reinforced concrete or fiber-reinforced plastic.

The other has variable thickness and constant E. Both cases are

considered and analyzed.

 

 

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1.2 PREVIOUS DEVELOPMENTS

 

The study of plate theory began in the l7601s. Euler (17)
presented the first mathematical approach to plate studies in
1766.

In 1815, Sophie Germain (22) presented a fairly satisfactory
fundamental equation for the flexural vibrations to the French
Institut as the result of her investigation during the 1809 to 1815
period. Within the same period (in 1811), Lagrange arrived at his
equation, which is known, therefore, as Lagrange's equation for the
flexure and the vibration of plates. Kirchhoff (1824-1887) is
considered the founder of the extended plate theory which takes into
account combined bending and stretching. In 1910, Von Karman
introduced a set of differential equations valid for plates subject
to large deflection. These equations are referred to in the liter-
ature as the large deflection equations.-

The development of the modern aircraft industry directed
the attention of many scientists and researchers toward the study
of plate vibration, plates subject to in-plane loads and postbuckling
behavior of plates. The earliest solution of a flat plate stability
problem apparently was given by Bryan (lO)in 1891.

The ability of a plate to carry additional load after
buckling was apparently discovered in the late 1920's through
experimental studies made in connection with the design of air-
planes. In 1929, wagner (49) studied a shear web and based on his

findings, established a criterion for postbuckling strength of

the web. In 1942, Levy (29) presented solutions to the plates with

 

 

 

 

l.2 PREVIOUS DEVELOPMENTS

The study of plate theory began in the 176013. Euler (17)
presented the first mathematical approach to plate studies in
1766.

In 1815, Sophie Germain (22) presented a fairly satisfactory
fundamental equation for the flexural vibrations to the French

Institut as the result of her investigation during the 1809 to 1815

 

period. Within the same period (in 1811), Lagrange arrived at his

equation, which is known, therefore, as Lagrange's equation for the

 

flexure and the vibration of plates. Kirchhoff (1824-1887) is
considered the founder of the extended plate theory which takes into
account combined bending and stretching. In 1910, Von Karman
introduced a set of differential equations valid for plates subject
to large deflection. These equations are referred to in the liter-
ature as the large deflection equations.-

The development of the modern aircraft industry directed
the attention of many scientists and researchers toward the study
of plate vibration, plates subject to in-plane loads and postbuckling
behavior of plates. The earliest solution of a flat plate stability
problem apparently was given by Bryan (lO)in 1891.

The ability of a plate to carry additional load after
buckling was apparently discovered in the late 1920's through
experimental studies made in connection with the design of air-
planes. In 1929, wagner (49) studied a shear web and based on his
findings, established a criterion for postbuckling strength of

the web. In 1942, Levy (29) Presented solutions to the plates with

_———_-—‘-

large deflections under combined edge compression and lateral
loading. His investigation was based on analytical solutions using
Fourier series. He also considered postbuckling analysis of plates.
In 1970, Supple (42) analyzed a rectangular plate with constant
in-plane compressive loads on opposite edges using the out-of—plane
deflection, w, and the Airy stress function as variables.

A considerable amount of work has been done on plate
analysis by different methods; among these are solutions of the
equilibrium equations by series expansion, energy methods , and Vlasov's
method (46). Until relatively recent times, however, the investigations
have centered on analytical solutions, which are in most cases limited
to relatively simple geometry, load, and boundary conditions.
In many particular cases where these conditions are more complex,

the analysis via the classical route becomes increasingly difficult

and is often impossible. In such cases, the use of an approximate
approach becomes more practical due to the flexibility and quick
results.

Near the end of World War II, the invention of digital
computers,with their capability of processing large numerical
problems, caused rapid development of various numerical techniques.
Of these, the finite element, finite difference and boundary integral
methods are of most general use.

Although previous analysis of variable stiffness plates
has been limited, there has been a considerable amount of work

done on uniform stiffness plates using finite element techniques,

and dealing with stability and postbuckling of plates. The finite

 

element method waSintroduced by Turner, Clough, Martin, and Topp
(45) in 1956. Argyris (A) and Zienkiewicz (53) have made numerous
contributions in this field. Gallagher (20) and Hartz (24) also
have made great contributions in improving the method and including
nonlinear terms. A series of studies considering postbuckling
behavior of plates was made in the 1970's (13, 19, 52) using the
finite element technique. Murray and Wilson (31) have conducted
research on postbuckling of plates,considering various aspect

ratios and applying the finite element method. Other significant

 

contributions include papers by Conner (l4) and Yang (52). They
applied the finite elementnethods to solve postbuckling plate problems.
The finite difference method is also one of the general
numerical solution techniques which has been frequently used. The
finite difference method was first used by N. J. Neilsen (33) for
analysis of plates in 1920.
The first finite difference solution of the large deflection

of plates is due to Kaiser (26). More recently, Basu and Chapman

 

(5) contributed to this study. Kaiser also carried out some ex-
perimental tests which verified the theoretical results. Both
aforementioned investigations were formulated in terms of lateral
deflection, w, and a stress function. The finite difference ap-
proach to large deflection of plates was also used by Brown and
Harvey(9) who have studied large deflection of plates subject to
lateral pressure combined with different ranges of edge loadings.
More recently a new method of solving finite difference equations—

namely,the dynamic relaxation method was described by Otter (35). The

————_—l

 

basis of the method is to add dynamic terms to the equations. The
addition of dynamic terms such as acceleration and viscous damping
makes the problem analogous to a vibration problem. The damping
coefficients are taken corresponding to critical damping resulting
in a motion which dies out quickly. Thus, the solution to the
static problem is obtained. Rushton (38,39,40) has published papers
applying the dynamic relaxation method to large deflection of plates
subject to lateral load and to postbuckling of plates under in-plane
loads. Since the method is an alternative technique to solution
of the finite difference equations, it has the advantage that
variations in stiffness can be included. Rushton has stated that,
with appropriate time increment and damping coefficient, a solution
can be obtained with no difficulties.

The Boundary integral equation (BIE) method has also proved
to be successful in solving plate bending problems. Jaswon (25)
and Haiti (30) introduced the direct method of solution and recently
.Altiero and Sikarskie (3) presented the indirect method of solution
fifliich proved to be more efficient. In 1980, Wu (50) modified the

“method by moving the integration contours outside the real boundaries.

Tile plate of interest is embedded in a fictitious plate for which
Ifle Green's function is readily known. Fictitious forces and moments
ire then applied outside the real boundary and the solution can be
>btained by finding the magnitude of these fictitious loads such
111st the original boundary conditions are satisfied. The method

r213 proved to be very efficient in general plate bending problems.

 

 

 

Particularly pertinent to this study is a paper by Prabhakara
.). In his paper, he considered postbuckling of orthotropic
xtes. Recently (in 1980), Kennedy and Prabhakara (27) have
1died the postbuckling behavior of orthotropic skew plates and

:ained solutions to some problems using a series expansion method.

 

 

 

L.3 PRESENT INVESTIGATION

 

In the study of thin plates subject to lateral and edge
loading, especially in the postbuckling range where the deflections
are not small, the Kirchhoff theory (which neglects stretching and
shearing in the middle surface) can not yield satisfactory results.
In this case the Von Karman large deflection equation can be employed

to obtain more accurate results.

 

In Chapter II, a brief review of the theoretical background
is given and the derivation of the compatibility and equilibrium

equations is first presented. Next, by applying the ordinary finite

 

difference method, the required operators are derived and the pro-
cedures for solution of different problems are briefly discussed.

Two different alternative methods of formulation are considered:

a) in terms of lateral displacement, w, and a stress function,

b) in terms of the displacement components, u, v, and w.

Yhe solution procedures for both methods are also discussed. A few
imamples of practical boundary conditions are listed and theoretical

ialations for each boundary condition are mentioned.

Chapter III includes numerical solutions and analysis
f the results. A computer program and the required subroutines
iseee computer program in Appendix C) have been developed to facilitate
P1>lication of procedures discussed in Chapter II.

In order to provide a more complete view of the variable
tiliffness plate and it's behavior relative to the uniform stiff-
3538 plate, several different types of variation in stiffness are

)rlsidered. Uniform stiffness plate results are given for comparison.

_-——!QM

 

For clarity, in the procedure presented, the results are always
accompanied by graphical illustrations of membrane and bending stress
as well as displacement components. The behavior of those graphs
and their relations with applied load is discussed.

The accuracy of the solutions is evaluated by comparison
of the results obtained with results from past studies and exact
results, where these results are available.

Convergence of the solutions is examined by using different
umsh sizes with extrapolation.

Results obtained for the effect of stiffness variation on
in-plane forces, bending moments, in-plane displacements and lateral
deflection, provide a good source of information for optimization
in each case. Although the optimization procedure is straight-forward,
the stability optimization with respect to amount of material used
is presented as an example. Two computer programs are provided,
One for force boundary conditions and the other for displacement
‘50undary conditions. Both programs are listed in appendix (C).

It was found that convergence was easily obtained for the range of
lxaading less than the second critical load because the assumed
Single-wave buckled shape is the only possible pattern of stable
equilibrium other than the flat plate. For loading beyond the
E3€=cond critical load, due to different possible equilibrium states,
time problem does not converge easily. For solution beyond that

range a proper deflection shape must be enforced, as appropriate

:EKDr the physical conditions of the problem.

 

 

10

NOTATIONS

The symbols are properly identified when first introduced;

for the reader's convenience, symbols are tabulated here.

Side length of square plate

a
c1,c2,.. Constants
Et3
D =-———-—7f- Flexural rigidity of the plate
12(1-v )
D Stiffness of a uniform stiffness, unit thickness .}
0 plate
Dr Reference stiffness = stiffness at center of the plate
E Young's modulus
F Force function
h,k' Mesh intervals in x and y
K = Et Membrane rigidity
Kr Reference membrane stiffness = Et at center of the
plate
K3 Membrane rigidity of a uniform stiffness, unit
thickness plate
*1 ,M.,M Bending and twisting moments per unit width
1‘ y xy
of plate

 

“ 2
(D1 - M ; M ) = (Dat )(Mx; M.; M ) Dimensionless moments per
1 y unit width

be Applied edge force per unit width of plate

In-plane stress resultants per unit width of plate

2
5e
(1W - N*; N* ) = (%—)(N ; N ; N ) Dimensionless membrane forces
0 x y xy per unit width of plate

(IV - ". " = . .
N , N ) (Nx’ Ny’ ny)/N Membrane force ratios

 

11

(N;; N'; N; ) = (Nx; Ny; ny)/Ncr Membrane force ratios

 

 

Y
q Lateral distributed load per unit area of plate
6': qaa/Doti Dimensionless lateral load per unit width of plate
Q Transverse shear per unit width of plate
_ edge stiffness .
- central stiffness Stiffness ratio
_ edge thickness . .
- central thickness Thickness ratio
t Plate thickness
ti Unit thickness
T Temperature
15v,w Displacement components in x,y, and 2 directions
3 Volume
uo Edge displacement
(U; V) = (u; v) a/t2 Dimensionless displacement
i .
U a
U/Ucr
9 KO
U a ”if Dimensionless displacement
U0 Dimensionless edge displacement
[1* = U
cr
W a 55—— Dimensionless lateral deflection
i
":37,z Cartesian coordinates

(I(; Y) = (x; y)/a Dimensionless coordinates

0'- ==£~ Grid size ratio
81 = 51- Membrane stiffness ratio
Kr
oflr Coefficient of thermal expansion
6 Di
1 § -D—— Flexural stiffness ratio

 

 

12

.1,A2,... Eigenvalues

Ll,A2,... Eigenvectors

,s ,8 Strain components

{ Y KY

,0 Components of normal stress
i Y

2
'. ' = a . o 0
3x, 0y) (D t.)(ox’ 0y) Dimen51onless stress components

 

,T ,... Shear stress components
Ky xz

= t@' Stress function

Airy's stress function

 

= cp/Na2 Dimensionless stress function
i] Coefficient matrix for m
1w], [bw] Coefficient matrices for w

:Aw]; [Bw]) - ([aw]; [bw])xh4 Coefficient matrices for w

 

.u], [Bu] Coefficient matrices for u
v], [Bv] Coefficient matrices for v
ul], [AuZ] Coefficient matrices for u

V1], [Av2] Coefficient matrices for v

 

 

 

CHAPTER II

THEORETICAL DERIVATIONS

 

2.1 General

In this chapter, the equilibrium and compatibility equations

of the plate based on the theory of elasticity are first derived. Then,

 

the finite difference approximationsto these equations are developed.

These will be used to facilitate numerical solutions of those equa-

tions,for which,in most of the cases, closed form solutions,if not

impossible,are very tedious.
Thin plate theory is applied and homogeneous, isotropic

material is assumed.

Depending on the boundary conditions, two different approaches

are possible. Here, both approaches will be diScussed.

Geometrical and material nonlinearity can arise in plate

Prxiblems. In this study only geometrical nonlinearity will be con-

Sid ered.
Figure (2.1) shows the geometry and orientation of a plate

it: the cartesian coordinate system. The x-y plane lies in the middle

Filaine of the plate and z is normal to the middle plane.

Internal forces and moments acting on the edges of a dx by dy

pl«ate element, as shown in Figure (2.2), are related to the internal

Stresses by the equations:

13

   

14

t/Z t/z
Nx = f oxdz N = f 0 dz
-t/2 y ’t/z y
t/ t/
2 2
N 8 f T dz N = f I dz (2.1)
t/2 t/z
Q =f r dz Q =[ r dz
x -t/2 XZ y -t/2 yz
’ t/z t/z
Mx = f oxzdz My = f o zdz
't/Z -t/2
t/2 t/
2
M =f ‘I zdz M =f r zdz
xy -t/2 xy yx -t/2 yx

where Nx’ N , N , Nyx = in-plane normal and shearing stress resultants.
Qx’ Qy = transverse shearing stress resultants.
M , M = bending moments.

X Y

M , M = twisting moments.
KY YX

2.1.1 NONLINEAR EQUILIBRIUM EQUATIONS
In the literature, nonlinear behavior is commonly classified

as either

1) Material nonlinearity

2) Geometric nonlinearity

Material nonlinearity may arise in case of time-dependent
lfilteria]. or materials with nonlinear stress-strain relations (plastic,
tlastoplastic, viscoe'lastic, etc.) .

Geometric nonlinearity is usually associated with large

iJBplacements. It may also occur for small displacement if the

 

 

 

15

 

 

Figure 2.1 Rectangular flat plate

 

 

Figure 2.2 Plate element dxdy in undeformed configuration

 

16

behavior is such that variation in the applied load alters the

distributions of displacement.

In this report only geometric nonlinearity is considered

and the material is assumed linear elastic, isotropic,

To determine equilibrium equations applicable to moderately

large deformations, they must be derived using slightly deformed

configurations. Figure (2.3) represents stress resultants and

internal moments for an element dx by dy in the deformed con-

figuration. B and B are rotations in the xz
x 3’ 3N

x
-—-dx etc.
3x

and yz planes

 

respectively, and N: denotes Nx +

Summation of forces in the x-direction gives:

8N
__§

-Nxdy + (Nx + 3x

3N
_ +.__Z§_ a
dx)dy Nyxdx + (Nyx 3y dy)dx 0

which simplifies to

 

 

aux 3N x
—3—}—{ +—X-ay =0 (2.2)
Similarly, summation of forces in the y-direction leads to:
EN EN
is); +—]—3y =0. (2.3)
Fromsummation of forces in the z-direction, we obtain
3Q 3Q 2 2 2
- x-—1=q+N3—1"—-+N a"’+2N3—"—- (2.4)
EX 3y x ax2 x ay2 Ky 3x3y

SLnnmation of moments about

x and y axes will result in:

 

and homogeneous.

 

17

 

3M 3M
Qy 3 3y + 3x
BMX 6M x
Q = + —X— (205)
x 3x 3y

 

Figure 2.3. Schematic illustration of internal forces and moments

on the element of middle surface in deformed configuration.

 

18

2.1.2 Relation between stress resultants and displacements.

 

From Hooke's

where:

For moderately large

law, we have

 

 

N = (e + v e )

x l-v2 x y

Ny — Et (5 + v ex)
1-v

xv YX2(l+v) xy

_ Bu 22.2
8x - 3x + l/2(8x)
3v 3w 2
=-——+12_
6y 8y / (3y )
_ Bu 3v 3w 3w
Y .___ .__. .___

xy ‘ ay 3x 3x 3y

displacements, the relations between moments

and lateral displacement are:

 

32w 32
M = -D(—— +v—3)
x 2 2
9x 8y
2
M = 4,49% + 1%)
y 3y 3x
M = M =-1)(1-\»)32w
xy yx 3x3y
Et3
where D = —-—-—2- is the flexural rigidity of the plate.
12(l-v )

 

(2.6)

(2.7)

(2.8)

 

 

19

2.2 Formulationimxterms of stress function and w

 

By substitution of Equation (2.8) into (2.5) and (2.4),
we obtain the equilibrium equations in the z-direction in terms of

membrane resultants and lateral displacement w:

 

2 2 32D 32w 32D 32w 32D 32w
V (W w) ‘ (1“) ‘2—3 ‘2 3x3 3x3 2 2
8x 3y y y 3y 3x
2 2 2
_ 8 w 3 w 3 w

3x By

The compatibility equation for mid—plane strains is:

 

 

2 2 ' 2

3 Ex 3 e a Y 32w 2 82W 32W

8y2 + 2 - Bxay = (axay) - 2 2 (2'10)
3x 3x By

and from (2.6), theastrainsin terms of membrane forces are

0)
ll

1
Et (Nx-vNy)

x
e = JL—-(N -vN ) ' (2 11)
y Et . y x '
- 5.131.
ny Et ny

Now, we define a stress function, Q, similar to Airy's

stress function, so that:

2
N .152
x 2
8y
32
N -——‘§ (2.12)
y 3x
2
N -12..

 

 

 

 

 

 

 

 

\
l
I

 

20

The Airy's stress functions is defined as

2.52.
O a
x 2
3y
32 '
0y = ——‘§L (2.13)
8x
- - 239:
Txy 3x3y

are N = t ox = t-BJEE etc.

X By
is is suitable for a uniform thickness plate. However, in the case of
:iable thickness, if we define m', as (2.13), it will complicate

a formulation. For example, substitution in equilibrium equations

.2), would result in

‘33—:32 2 J" 2 3:3 ‘ta 2=0
3y axay y axay axay

[ch in the case of uniform thickness, leads to 0 = 0.
For our purpose the definitions of (2.12) will be used.
)stitution of (2.12) into (2.9), will result in the equilibrium

Jation,in terms of m and w, as

 

 

2 2 32D 32w 32D 82w 82D 32w
V (nv ‘0 - (I'V) 7 7 "'2 axay axay + —-2' "-2
3x 3y 3y 8x
_ 329 32w+ azgz 3w 322 32w
- Q+ .__2+ 2 3y2 2 Bxay axay ° (2'14)
3y 3x 8x

substitution of (2.12) into (2.11) and then into (2.10), we

tain the compatibility equation in terms of m and w:

 

 

 

 

 

 

21
2 2
3 a 2(l+v)
ayz [t yy xx aXZE 3X y fipxy
Z 2 2
3x3y2 3 2 3y2
32
where cp .- __52 , etc.
xx 3x2

2;; Formulation in terms of 3 displacements u,v,gand w

Substituting equation (2.7) into (2.6) and then into (2.2)
and (2.3),along with substitution of (2.8) into (2.5) and then into
(2.4),results in 3 equilibrium equations in terms of u,v, and w.

The equation of equilibrium in the x. direction (2.2)

 

 

 

becomes
2 2
Et 1:... 32V wit-k v(32v + 32w 3):) l-v Et 3 u 8 v
2
l-v2 3x2 3x2 3:: My axay ay 2 1w 3y2 3an

2 2
+§ifl+flu +_}___9EE)_ _3_1_1_+_];(_3_w >24. \)g_( E)... %(_:_W_)2

 

axay 3y ax 3y2 1_\)2 ax ax 2 3x
+ 1"“2 30%) fl+fl+flflJ .. 0. (2.16)
2(1_\, ) 3y 3y 3)! 3}! 3y

Similarly, the equation of equilibrium in the y-direction (2.3),

 

 

 

 

 

becomes
Et 32v 32w aw azu 32w aw l-v Et 32v azu
2 2 + 2 a—y ”(any + axay E) 2 2 2 + axay
1"v 33' 3y l-v ax
82w aw aw 32w 1 3(Et) 3v 1 3w 2
+axa E+T—2’+ 2 7+2?) +
y 3,, 1_\, 3y 5* Y
Bu v 3w 2 l-v 3(Et) 8V 311 3W W
+—-— -— —— -— —— I
v5 2(3x) :] +2(1-v2) 3x [3x + 3y + 3:: 3y] 0 (2.17)

 

22

Interchanging u and x with v and y respectively in
equation (2.17) results in equation (2.16).

The equilibrium equation in the z-direction (out of plane)
can be expressed in terms of 3 displacements by substituting (2.7)
into (2.6) and then the results into (2.9). We obtain:

2 2 2 2 2

 

 

 

2
2 2 3 D 8 w 3 D 8 w 8 D 8 w
V(DVw)-(l—v)—--—-- +——-

3x2 ay2 axay 3x3y 3y2 3x2

2
Et Bu 13w2 av v3w2 3w Et 3v 13w2
=q+—— —-+—(—-—) +v—+—(-— ]—+—[—+-(—)
1_\’2 [3x 2 8x 3y 2 3y 3x2 1-v2 8y 2 3y
+ Bu +3(__3_3)2 32w + Et(l-v)fl1_+_ay_+_3l§3 32w (2 18)

3x 2 3x 3y2 l-v2 3y 3x 3x 3y axay °

NOte: Since in this approach we are working with displacements,
compatibility need not be checked.
.L-é FINITE DIFFERENCE APPROXIMATION
So far we have derived the necessary equations for analysis
of tflle plate, but solving these coupled nonlinear partial differential
equations analytically may be difficult.
Here we employ finite difference techniques to transform
the differential equations into ordinary algebraic equations in terms
of values of the functions themselves at certain specified points.
'ggizl;‘flgrinciple of finite differences
The derivation of finite difference expressions is based on a
Tayl‘DIT series expansion. We expand the function at some
successive grid points, truncate higher order terms, and

. 801.
"ee for desired derivatives, we can obtain approximate expressions

 

 

 

'23

for first, second, or higher order derivatives in terms of values

of function at the discrete points.

Y

f’,. f(x)

 

 

 

 

 

m;2 m-l m m+1 m+2

AL»

Figure 2.4. Function f(x).

19 one dimensional cases we obtain the following approximation of

the derivatives of the function:

. ._1_ - - l 2 m

f'(x)m 2(Ax) (f1n+1 fm—l) 6(Ax) fm +....

n _1__ _ _ l— 2 .V

f (x)m a 2 (Em+1 2£m + fm_l) 12(Ax) f; + (2.19)
(AX)

' ._l;___ - _ _ l_ 2 v

f"(x)m 3 (fm+2 2 fIn+l + me-l fm-Z) 4(Ax) fm +...
2(Ax)

iv
1

f(x)In - 837; (fm+2 -4 fm+l + 6 fm - 4 fm_1 + fm-Z)

l 2 vi
-g(Ax) fm + ....

 

 

 

 

24

 

In practice, we truncate the terms following the parentheses.

These represent the error in the approximation. We will refer to these

error terms later in the discussion of accuracy.

x
NN
O
I
l
'N
Y NW:|3— ——O———ONE
I, I
I l l
| I
wwo———‘"O———Oo——-(l)§-— 05:
l I I
I l
I ., !
Sw3——_¢_)_..._ SE
l
|
555 b
Figure 2.5

To determine the finite difference approximation for two dimensional

problems, we consider Figure (2.5), and the fact that similar re-

lations for the approximations to the derivatives in the X-direction

hold also in y-direction.

Thus,we will be able to derive expressions for any order
of derivatives in x and y and combinations of x and y derivatives.
For example, if we consider grid points of Figure (2.5) ,

the derivatives with respect to x and y at point 0 are

l
fx- 2h (fE-f

)

W

 

 

fy g'2k (f8 - fN)

f = i— (f -2f + f )

xx 2 E 0 W
h

fyy = 7 (fS-ZfO-l- fN)
h

f _ 1

xy ' 4hk (fSE fsw' fNE+ wa)

or if £ = a then

fxy = F (fSE-fSW-fNE + wa) etc.

Often, these formulas for derivatives are represented

geometrically by stencil patterns, such as in Figure 2.6.

--—®

igure 2.6. Two dimensional operator for fx

(9-- -®- -j®
772- CD" (ID-- (:9
C9-- {'9

(2.20)

26

2.4.2 FINITE DIFFERENCE APPROXIMATION OF METHOD DISCUSSED IN SECTION(2.2)

In section (2.2), we derived equations of equilibrium and
compatibility as well as the components of internal forces and dis-
placements, in terms of lateral displacement, w, and stress function,
Q. In this section, we will discuss numerical solution of those
equations using finite difference techniques.

To find a solution to a plate problem, we must satisfy
both equilibrium in the z direction, and compatibility.

a) Equilibrium Ermation I

For this purpose, we will apply relations (2.20) to equilibrium
equation (2.14), and the results will be represented in two dimensional
Operator form. Introducing
Dr *3 flexural stiffness of the plate at some reference point

(center of the plate in this case),

U

- __i_
61 ‘ D
r

A finite difference operator for the left hand side of the equilibrium

6.and6
c

e‘l'aation (2.14) is given in Figure 2.8 where Ga, 6,), (1

refer to midpoints a, b, c and d of Figure (2.7) as explained in

reference ( 8).

 

 

 

27

 

 

 

 

 

 

 

 

 

 

 

 

Figure 2.7

Note: All terms in the operator of Figure (2.8) are coefficients of w.

If the grid spacing is the same in the x and y directions (a = l),
oPet'ator (2.8) will be simplified to the operator given in appendix
(A. 1) .

In case of a uniform stiffness plate, where 6, =- D— = 1
for all points, the operator reduces to the usual finite difference

OI’erator for V4w, as given in appendix (A.2).

 

‘

28

 

.A¢H.~v coauaauu assessaaauo ac cuss can; some now “compose ouaououuan .o.~ masses

oauuu Paoonqom I > . MI 5

 

 

 

 

 

 

 

u
6.33 coo—6:: I lab I «a
a m
o 6
u e
unwound: mucououou I a
d
:aisai 2;? 32-5.. 2353....
d
a are N was $3.3 $33 $8.. necréfiua

 

 

 

 

 

 

 

 

 

u 6
2.? in? 3 €ud~a~+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v a I u A I a
2 2. 3? dues. z m 3 u 2 0+ 3? Gag. NI
u . S :7 . 2+ 3.3+ 3.? . A: .a: 3:: .. an
A a+ avau +H Auap~+sa+~v as 9+ N u

Seize: Iain??? 222333

6 I z o I 0 I

a? 33% A 0+ 21533 $3 2., same A

20 a

 

 

 

,oc

.o-o

 

 

 

 

 

29

For the right hand side, if cp values are known, we can
derive the finite difference operator as coefficients of w, as

shown in Figure (2 .9 .a) , where

32 .. (FE-2'90 + W

 

 

 

 

 

 

<9 = —%
xx 3x h2
2

fie .. (phi-2'90 + CPS _ 0‘ (IN-2'90 + c"3) (2.21)
‘Pyy ' ‘ 2 ‘ 2

3y k h

a 3259 .. CPSE + (PNw'cpsw'cPNE g c‘(CPSE + ‘PNw'q’sw'q'Nfl

.2ny axay 4hk 2

4h

For a = l, we get the expression shown in Figure 2.9(b).
flLUTION OF THE EQUILIBRIUM EQUATION
To solve the equilibrium equation for w, we must have either

t _
he cp values, or the in plane forces (cpxx, cpy and cpxy).

Y
Substitution of these values into the operator of Figure
2.9 (a) or (b) will result in a known Operator at each node.
By applying the operator at each node, we will be able to
5011:: a matrix of coefficents of w, which along with the qi vector
will form the right hand side of equation (2.14) as

{q} + wa] {w}
where {q} is the lateral load vector and [bw] is the coefficient matrix

containing constants. Similarly, application of the operator
of Figure (2.8) will result in the formation of matrix DranHw} in
the left hand side of equation (2.14). [aw] is a constant coefficient
matrix.

Therefore, we can represent the equilibrium equation in

numerical form as :

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

30
_ _ ._ 2 _ _
' I ‘ny °‘ ‘9... I + 9‘2" ‘ny
I I I
I I I
I I -I
l ("W L' " 'I -mpyy + 0‘ 'pxx) I' — " (PYY
2
h I I I
I I l
4 I I
2
+%'ny I-“ 0' cpxx ___ -%CPXY
(a)
1 ' _ __ _ _1 -
’FW’SEWNw'q’NE‘IIIF '7 ‘PE'Z‘I’oJ'q’w EWSEWNW (ENE-(98W)
I I I
I I I
I I I
-2(<p +cpw+<9 +<Iz )
1 - — — — N s — — — -2 ,
;4 (PN 2Cp°+<ps + 5‘9. (PN <9°+<PS
I I I
I I I
I I I
1 —— -2 F—fI-(cp +<P -<2 -IP I
ISWSEWNW-CPNE-CPSW 'I cPE C99% 8 ‘SE NW sw "NE
(b)
FIG 2.9 DIFFERENCE OPERFITOR FOR RIGHT HFINU SIDE OF
EQUILIBRIUM EQUFITION 2.14
‘ Er- — -——

 

 

 

 

 

 

 

31
Dran]{w} = {q} + [bw]{w}. (2.22)

In complete matrix form we can write

 

 

 

 

 

 

 

 

 

 

 

Dran]{w} = [IJIQI + [bw]{w} . (2.23)
Both [aw] and [bw] havea factor of —1Z; so, we can rewrite
h
DrEAwJIW} = [IJIqh4} + [Bw]{w} (2.24)
4 4
where [Aw] = [anh , [Bw] = h [bw]
or
1 {qh4}
{[Aw] - D—- [Bw]}{w} = D (2.25)
r r
We will discuss solution of this equation later.
5) Compatibility Eguations.
To approximate the right hand side of the compatibility
ecluation (2.15), we apply relations (2.20) to get
32w 3 w3E + wW-WSW-WNE
axay 4hk
32w wE-Zwo + wW
2 = 2 (2.26)
3:: h
32w - wN-Zwo + wS
ayz k2

In order to obtain a finite difference operator representing
the left hand side of equation (2.15), we will differentiate and

ngroup terms, resulting in

32

 

2
Int. __2_[8_(__Et) a -—(v2<p) + aggt) 3%(‘72‘9J’H'Ei6 3 (Et) _ 2 ,23(Et)) )

 

 

 

Et Et 3x 3x2 (1302‘ ex
2 2
(J -v JIJ- [<31 3 2(39- 2 2<3§EtI>2 ML; -v a—%>1 (2.27)
3x2 3y2 3y (Et) y 3y 3x
+ zuwflagig 3(Et) 2 _ _1_ 3 203133329 }.

3y (Et)2 Et Bxay ax3y

Now by applying relations (2.20) and adding all contributions
at each node, we will obtain the finite difference operator of
Figure (2-10):

‘where

Kr: Et = in—plane rigidity of the plate at center.

K1 = in-plane rigidity at point i.

If a --£-- 1, this operator will be simplified to the one
given in Appendix (A.03)
An alternate approximation to the compatibility equation

(Z-GIS) in finite difference form can be developed.

Denoting
1
Fl - K ((Pyy “V (9“)
F=-1-(<p wt?) (228)
2 K xx yy °
1
F3 - K wxy

.313 coauusvo 23233322923 no 030 can: 30.— 202 232.2020 02.0.20qu .AVH.~ was»:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

33

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

H
M a
I I 2 A .
A A V .r 2 a
V2 z2I22 .2 .2
AA. I22I22+A2VA>+AVIA2 I A2IA2 2m.V.2I 12 2+. 2AI 22V ? A22I22I22¢22V 35% +
. A . .2
A....|.2.IM..AII.A....2.. . ..A 21:2,..- 2.... III. .- .2... .223? -
2I2 I2 NII. 2I2 .I 22II2INI.
A 2 2+22 2VN2 2A2A A 2 2VAM$AVA2+AA£AVA2 2.2 A 2+ 2 22 2VA2 + 2A..A
2 2 . z . 2 ,
2 2 . I2 . . I2VI2INIA2+2AI2V3+
AAI 28W» IA 2+ 2A .2VA3+ 2+ 2AI22V AaA? AaVA+ AA22I22VAIAIIA2I I AA A2,. A 2
22A 22+. 2 2 V A22IA2V .MIA +AA22+. 2AI22VI Iml + 2....
AA 2I 22V IA 1 2AI 2VI 22- IleflalIlVI IA22+.2AI2 2VA ASIAVA+ A 22I22 A
A22I22VAA£AV+AA£AV.22I A 2$A _A.2+2V.2 A22I22VAA2+AVIAA2+AV.2...I . .2
'M I.
AA. I22I22+A2VA>+AYAI+ AA22I22 2.2V .mIA 2+A2 2+. 2AI 222V 21 A22I22Iu2+a2VA>+AVAa
.2A A
I AiAVAI I I .45... +
HAzuI22VA3MI22VH_ NA=I 22VII IA: 2+ 22 222V 3+ HA22I2 22VAH2 I2 2VHA «a
A 22+ 22.22-22VI_A. I 2.A_.A A2 I22VA Aa+AV A_.IA AAo+AV _. .2..- A2 2-2 2+2 2- 28h. +.2A_.A
A N

 

 

 

 

 

 

 

 

 

NO

 

 

 

 

34

Equation (2.15) can be written as

 

 

2 2 2
a F 3 F a F 2 2 2

1 2 3 3 w 2 3 w 3 w
8y?" 3x2 8x8y axay 8x2 3y2

Now we can approximate derivatives of F1, F2, F3 as

a2Fl F1 -2F'l + F1N
2 S 0 etc. , where Fl , according to (2.27),can be

ayz k2 S
approximated as

cp-ZCP +4) (I? -2¢ +2)
[53 S 41-» SE 3 SW ,etc.

1
F = _—
K k2 h2

l
S S

For a = £ = 1, this approximation results in the operator

given in Appendix (AJI)
W TO THE COMPATIBILITY EQUATION
To be able to solve the compatibility equation, we must have

Values of w at nodal points. Then, we are able to compute the

right hand side at each node using expressions (2.26).

To determine the left hand side, we apply the operator of
Figure (2.10) or the one in Appendix (A.3) or (A.4) at each node.

By adding all contributions, a coefficient matrix will be

formed, Thus, we have:
[AJICPI = {w} (2.30)

where the w vector is known, and solutions of this system of

e
quations results in the cp values at prescribed nodes.

 

 

35

c) Approximation to other equations

After solving equilibrium and compatibility equations, we
may be interested in calculating in-plane forces and displacements
as well as bending stresses. Finite difference approximation to
some of these equations will be discussed below.
Insplane forces

By definition (2.12) we have:

- L22 2 28.20 + ‘PN

 

 

 

 

N
x ayZ k2
322 cI’E'ZCI’O + qu
N = 2 = 2 (2.31)
y 3x h
__ 32g) __. cPSI: + (PNW-‘%I3 ‘PNE
xy axay 4hk

infiplane Displacements

Comparing equations (2.7) and (2.11)

-22122

l
E:x - 3x 2(3x) 2 Et<Nx-vNy)

OI’

Bu 1

1 3w 2
_- (Nx-VNY) - 2 (E

3x Et

'rhe right hand side of this equation is known. By approximating the

Zleft hand side in finite difference form we have

36

HR "Uw

 

l 1 3w 2
2h Et(Nx-VNy)- 2(3x)
or
1 1 3w 2
uE-uw - ZhIEt(Nx-vNy)- 2(3x) ] (i)
similarly, for v displacement, we obtain (2.32)
_ _1_ £312 ..
vS-VN - ZRIEt(Ny va) 2(ay) J (11)

If we apply the operator of the left hand side of equations
(2.32) (i) and (ii) at each node and add all contributions, we can

form two coefficient matrices for u and v as

[Au]{u}

{Bu} (2.33)

[Av]{v}

{Bv}. (2.34)
Where [Au], [Bu], [Av] and [Bv] consist of constants only.

Solving these systems of equations, we obtain u and v, the
displacements at each node.

Note:

In order to get compatible displacements, we need to have
some points at which u, v or both are known. Points of this nature
can be found on boundaries or lines of symmetry.

BENDING MOMENTS

Having the solution for w displacement, we use approximations

(2-20) to compute bending moments according to equations (2.8) .

 

_‘_—. _

 

 

 

2 .
w
Mx = -1)(——8 ‘2’ + a g) - -D[ E g +v 3 2° N]
3x 3y h k
2 2 w -2w +w w -2w +w
s o w o w
M = -D(§-—‘"2- + a ‘2') = -n[—-—-—§-— + II—E—T—J (2.35)
y 3y ax k h
2 w +w -w -w
a w SE NW sw NE
Mm- -D(l-v)ax3y D(l-v)( 4hk )

where D is already defined.

2.4.3 FINITE DIFFERENCE APPROXIMATION OF METHOD DISCUSSED IN SECTION

(2.3)

 

In this method, 3 equilibrium equations in the x, y and z
directions, as derived in (2.16), (2.17) and (2.18) respectively,
flmst be solved. In this section, the finite difference approximation
Of each equation will be derived.
glgEquilibrium,in x-direction

Considering the equilibrium equation (2.16) and regrouping
Variables results in

82u + K(l-v) azu + @5333 + l-v 8K Bu]
2 2 2 32 32 2 $73—37
3x By

[K

---J + (i)

2
[Kgl-I-v23v +v§§§x+l-v_3_1g3v
2 Bxay 8x 3y 2 8y 3x

32w fl K(l+v) 32w _a_w_ + K(l-v) 32w 1%:

{K +
ax2 8x 2 3x8y 8y 2 ayZ 8x
13K awz 312 l-vglgflfl_
23'£[(ax) +"(332)14' 2 Syaxay} 0

‘where K = Et and both sides of equation (2.16) have been multiplied

by (l-vz)
Now, the difference approximations,(2.20), will be applied

and the! results will be represented as finite difference operators.

 

 

 

38

The operator representing the u contribution to this equation
is given in Figure (2.11-a). Similarly, difference approximation
to 'v results in operator Figure (2.ll-b).

The contribution of w to this equation can be approximated

 

 

 

         

 

 

as
wE -2w'+w’ w -w w +w w -w
xw function = K (E 0 w)( E w)+(liy')K02( SE NW WSW w—NEH S N)
0 2 2h 2 O 2 2h
h 4h
w -2w +W’N wE - w -w
l-v 2 s o _1_( 2:w 2 s N 2
K - wS -w
l-v 2 S KN N
+<—2 )a< 2h )[(wE :‘W 211)]

Therefore, the equilibrium in the x—direction can be schematically

shown as.

(u-operator) u + (v-operator) v + xw function = O (2.36)

Note: In this case, because of the absence of body force in the

x—direction, the right hand side of the equation is always zero; thus

 

the equation is first divided by ( 12).

l-v
b) Equilibrium ingy-direction
The same procedure will be followed to derive operators
representing finite difference approximations to equilibrium in the
y—direction.

Equilibrium equation (2.17) can be rewritten (after regrouping

 

variables u, v and w,

                

 

:Znas

 

 

|I-‘

 

 

39

 

 

 

KS-KN

 

 

 

 

 

 

 

 

 

—

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

—

 

 

 

 

_

 

 

 

 

 

 

 

0 ‘I’ — "az (Liz-'2'“,-
4I if
I I
I I
Ko- KE4KW - --Ko[-2-a2(l-v)]
l IT
I I
I I
K -K
2 1- ° N
O — — “0‘ {—23) (Kg!- )
a) U-UPERHTUR
1+V ____._
ij9 K. ”(KE‘KW)
I I
I I
I I
l-V __ _.__
‘(THKs'KN’ °
1 I
I I
I I
_li! K —-—-— v
2 I . (KE'Kw)

 

 

 

 

 

 

b) V—OPERHTOR

FIG 2.11 OPERRTORS FUR X-EQUILIBRIUI‘I

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

EQ.(2.16)

40

2

 

 

 

 

(l+v) 3 u l—v 3K Bu 8K Bu
.___—2|. _—
[K 2 axayJr 2 32 3y " ay 3x]+
2 2
K(l-v) a v 3 v l-v 3K 3v 3K 3v
I 2 2+K"—2+T“;—I; 3‘3“]
3x 3y y y
{K 32w‘§3'+ K(1+v) 32w .§HL+ K(l-v) 82w‘§!.+ (2 37)
3y2 8y 2 Bxay 3x 2 3x2 3y
1 3K 3w 2 3w 2 l—v 3K 3w 3W' _
2 3y{(3y) + v (3x) 1 + 2 axiax 3y} - 0

By finite difference approximation, we obtain the operator of
Figure (2.12-a) to represent the u-contribution and (2.12-b) as the
v-operator.

Terms including w can be approximated as:

a2K0(l+v) W KO (l-v) w -2w +w w -w

SEWNw'wsw'wNE WE 'ww

 

 

 

 

YV function = 2 [( 4h2 )( 2h )}+ 2 [( E 3 W)( 2h N)]+
- h
-2w +w w -w . - w -w w -w
3 ws 0 N s N 1- KISKIII E w s N
K0a[( h, )< 2h>1+<—2¥>(——§-fi—>a< 2h>(2h )+
K -K 2 w 2w w -w

S N a S N 2 v E W 2
a(——§H—)D§‘G‘jfir§ +‘5I 2h ) J

 

Finally, the equilibrium equation in the y-direction can

{be represented in the following operator scheme:

(u-operator) u + (v-operator) v + yw function = 0 (2.38)

‘22~§Eugilibrium in z-direction
The equation of equilibrium in the out-of-plane direction

1
s 1Jltroduced in Equation (2.18).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

41

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

_ 1+v
L‘1—§”-> K. — - -<-l-I-"-><KII-KII? - - -<—2-> Kc
I I I
I I I
I I I
_g_ l;v-(KS-KN) '- — O " ‘ V<KS-KN)
4h2
I I I
I I I
I I I
1+ l-v .13
- TV) KO —- -I (T) (KB-Kw) — — ( 2 ) KO
a) U-OPERRTOR
K —
l O — —a2(K°- 84K“) — - 0
I I I
I I I
I I I
2 $<42~KI+KII> - — 202.222) — $3422.22
I I I
I I I
, I I I
K -

 

 

 

 

 

 

 

 

b) V-OPERRTOR

FIG 2.12 OPERFITORS FUR Y-EQUILIBRIUI’I EQ.(2.17)

P

 

 

42

Approximation to the left hand side of this equation has

already been explained in Section (2.3.2-a).

As for the right hand side, we approximate the derivatives

of displacements to get
-v 2 w -w

 

 

 

 

(RHS) = q + -—K—21II(—-u-§—huw) +-12— (32-21% + 2 (vfih N) + V—g‘ (3h “>21
<-5::9-+:-) + [III— S ——N-h> +£33- (%)2+ 23%;“) +3 <———- 23h ”)23
22(——$-::-w§9-:) + (I-II) [AIS—u: :1”) + (3%?) + 2(WE;:W)(WE:N)J

a(WSE+WN:;:sw’wNE)} (2.39)

Finally, equilibrium in the z-direction can be represented

[Aw]{w} = RHS (2.40)

where matrix [Aw] consists of contants.

<1) Solution_procedure using this method

 

In order to perform numerical analysis in computer programs,
we arrange the equations in suchaway that the equations can be

represented in matrix form.

To find a solution by this method, we must satisfy all three

equilibrium equations. These contain three unknowns, u , v, and w.

There is no simple technique providing a direct solution to these

coupled nonlinear equations; thus, we employ an iterative technique

to solve them.

If w is known (or assumed),the first two equations of

e
quilibrium will become two uncoupled equations in u and v

¥ __

43

and can be solved as follows.

Applying the u—operator corresponding to equilibrimn in x
at each node and adding the contributions, and doing the same to the

v-operator, we get a matrix representation. of equation (2.16) as

[AulJNxN{u}le + [Avl]NxN {VIle = -{xw Function}le (2.41)

Repeating the same procedure for y-equilibrium equation (2.17) we
obtain

I:Au2]NxN{u}le + [Av2]NxN{v}le = -{yw Function}le (2.42)
Where [Aul], [Avl], [Au2] and [Av2] are constant coefficient matrices.

Both equations are coupled in u and v, and each contains

N equations in 2N unknowns.

One way of approaching this problem is to try to solve the
equations by iteration until reaching a solution that satisfies
both equations.

An easier approach can be employed if we realize that,
although the equations are coupled in u and v, there are no
In"Lited terms containing both u and v.

Therefore, we can combine the two, to get

Aul Avl u xw Function
-------- ' ' ---------- (2.43)

Au2 Av2 v yw Function
J

which is not only more efficient in computer programing but leads

t
o a. unique solution for the u and v displacements at specified

11 .
Ode points,based on an assumed (or known) w.

 

 

44

To solve the z-equilibrium equation, we substitute values
of u, v and w, in the right hand side (2.39), and solve equation

(2.40) for new values of w. The iteration will continue until the

new values of u, v and w, are equal to or very close to old ones.

A practical use of this method, including the details, will be dem-

onstrated in chapter 3.

POST SOLUTION DETAILS

After a solution is found for the three displacements

11, v and w, any components of stress and strain can be computed.

Average strain

Average strain at each node can be found by the finite
difference approximation of equations (2.7).

“E'uw 1 WE'ww 2
5 3 2h +‘2'(2h )

X
vs"'N 22 wS'WN 2
6y “—23-— +7 (‘73—) (“4)
uS-uN V 'VW W -‘Ww Ink-“W
E + II E M 3 N)

 

ny=°‘(2h )+ 2h (2h 2h

EEEEEQEane Forces
Substitution of equations (2.7) into (2.6) and approximating

the derivatives by finite differences, will result in

 

 

 

 

 

 

 

 

u- w-w v-v zw-w
N 2 K Eu” 2.2112 L! m 8 N2
X l_v2[2h +2(2h)+w(2h)+2(2h)]

v-v 2w-w - w-w
N a_BK SN g_ 5 N2 “E“w 3 Ew2 (2.45)
y 122M 2h>+2 (22 > +II(———2h )+2<2h >1
III-u V-‘V W'W W‘W
N aRSI-v) S N EW Ew S N
20 2) [a(-——2h )+(2h )+a( 2h )( 2h )1
-\)

 

45

BENDING MOMENTS

The bending moment approximation is given in equations (2.35).

2. 5 BOUNDARY CONDITIONS
In the small deflection theory of.plates, we consider only

out of plane (or flexural) boundary conditions because the effect

of in-plane displacements on the boundary is negligible. However,

-they become the chief factor in large deflections behavior and in

the postbuckling range. Thus, we discuss in-plane boundary conditions

as well as out of plane conditions.

The flexural boundary conditions, as commonly discussed in

elementary plate theory, are:

a) Simply-supported boundary

w=0

32
M =—D (.__w_ +
x 2
8x

2
v 2%) = 0 (on boundaries parallel to y)

3y

b ) Fixed boundary :

w=0

3w (n, normal to the boundary)

_20

an

c) Free boundary :

d) Others, such as elastic support, or partially fixed support, etc.

 

 

46

For the in-plane boundary conditions, there is a variety of

possible combinations that may occur in the postbuckling range.
On each edge, either 11 or v displacements can either be unrestrained

or have some specified values; also, there could be restrictions on

derivatives of either u or v, or both.

In terms of in-plane boundary conditions, we can classify

the problem in three major types.

1) - Force boundarL conditions (i.e. in—plane forces are specified

on the boundary).

If applied forces Nx’ Ny and N are known, we can use

to choose values of the Q functions on the

A practical

relations (2.12)
boundary points so that they satisfy boundary conditions.
example of this nature is discussed in chaper 3.

2; Displacement boundary condition Some possible cases are:

u and/or v are specified on the boundary,in which case the

a)

vallies of displacements would be '
x

 

assrlgned to boundary points.

 

 

 

b) Edge remains straight and parallel to y. (u-displacement is

constant all along the x = 0 edge).

e) Edge remains straight with no shear force along the edge; in this

Case, from equations (2.7) and (2.6), we have along edge x 2 0,

47

.__E_t__ 22 a_v 3222 .
ny 2(l+v) [3y + 3x + 3x 3y] 0 (2°46)
On supported edges (fixed or hinged), %§;- 0, thus
.. A! 2 22 . (1)
3x 3y

If the x-edge (x = constant) is straight with u constant, then
Bu 8v
-53; = 0; therefore, equation (1) results in 3;-= 0 along the edge-
d) Other conditions include possible restrictions on u, v or
their derivatives which lead to particular relations between
displacements or their difference approximations. For

example,the edge can be subjected to thermal expansion

(see section 3.2) such that %§" ex - constant along edge y = 0.

1). - Mixed boundary conditions. i.e.,case 1 applies to part of the

boundary, while the rest of the boundary is defined by case 2.

The computer program developed can solve either case 1 or 2.

Therefore, to handle a problem with mixed boundary conditions, we
can solve the problem by trial and error, as follows.

i) Assign some fictitious displacement values to the points
at which forces are specified, and solve the problem as
one with displacement boundary conditions.

ii) Compute forces at the boundary points.

iii) Compare with actual forces at the points.

iv) Correct previous fictitious displacements in such a way

that the solution is improved.

v) Repeat steps (ii) to (iv) until the computed forces are

equal to or close enough to the actual ones.

 

 

 

48

2.5.1 SOME EXAMPLES OF PRACTICAL B.C.

 

subject

1.

2.

Following is a list of some practical examples of plates

to various loading and boundary conditions.

Window glass can undergo large-deflection under lateral
wind pressure; the out-of-plane boundary condition is in

most cases simply-supported or sometimes built-in. Either
case may be accompanied by:

a) in—plane displacement possible.

b) in-plane displacement restricted.

Plates on stringers forcing the plate edges to remain straight,
as in many ship and aircraft sections surrounded by stringers.
Mechanical and instrumental plate elements subject to tens
perature change will be subjected to tension or compression
on some or all edges due to temperature change in surrounding
elements. Various combinations of boundary conditions are
possible.

The webs of structural steel profiles used in construction can
be categorized as plates subject to in-plane shear and normal

forces along the edges.

6 SUMMARY

\—

We will summarize the theoretical formulations discussed

11‘ ‘211apter 2, and mention procedures of solving some problems.

.Among several types of problems which can be solved numerically

based on the finite difference approximations shown, and using the

computer program which has been written, are:

49

2.6.1 MEMBRANE SOLUTION

For flat plate with w = 0 everywhere, the equilibrium in the
z-direction (normal to the plane of the plate) is trivial; to
find a solution for in-plane resultants and displacements,

a) In the case of force boundary condition, we solve compat-

ibility equation (2.30) with the w vector equal to zero.

[AJICPI = 0 (2-47)

The solution results in Cp values at discrete nodes, which can be
used in equations (2.31), (2.32), and (2.33) to find in—plane forces
and displacements.

b) If the displacements are specified on the boundaries, we

solve equation (2.43). Considering w - 0 everywhere,

we have

...... <..- =2.» (2.48)

 

 

 

 

 

for which the solution results in displacement values at the nodes.
Application of equation (2.45) then leads to the membrane resultants.
2-\6.2 LATERAL LOADING
a) Force boundary conditions.
1) Small displacement.In this case, the (9 values and
in—plane resultants are known from (2.5.1), so we can
solve the equilibrium equation (2.25) to obtain w.

Then (2.33), (2.34) and (2.35), can be applied to

‘_—;_g____ A

50

compute in-plane resultants, displacements and bending

moment 3 .

ii) Large deflection.

Compatibility equation (2.30)-and equilibrium equation
(2.24) could be solved iteratively, and the force resul-
tants and displacements can be calculated as discussed in a).
b) Displacement boundary conditions
i) For small deflections, we can solve equation (2.48),
neglecting the effect of w on in—plane solutions; then,
solve the z-equilibrium equation (2.40) for w, by ignoring
wbterms in the RHS.
ii) Large displacement problem - this requires an iterative
solution of equations (2.40) and (2.43) as discussed
in 2.3.3 (d).
L623 STABILITY ANALYSIS
a) Force boundary conditions.
The in-plane forces and m values are known from part
2'5-1 -a; then,we can use the equilibrium equation (2.25). If
q " 0, this will result in a characteristic matrix, the eigenvalues
of Which lead to the critical forces and the eigenvectors represent
the buckling modes.
1)) Displacement boundary condition.
Using in-plane resultants and displacements obtained from
(2""5-1.b) and forming the R.H.S. as a coefficient matrix for w,

with _ 0
q ~ , results in the characteristic matrix equation

51

{[Aw] - [Bw]} {w} = 0, (2-49)

for which the eigenvalues and eigenvectors lead to critical

boundary displacements and the mode shapes, respectively.

2.6.4 POSTBUCKLING

Since after buckling the plate takes on a state of stable

equilibrium, we can analyze the plate as a regular large deflection

C388 .

a) Force boundary condition.

We solve the equilibrium equation (2.24) and the

compatibility equation (2.30) iteratively.

b) Displacement boundary condtions.
In this case we employ the iteration technique to solve
the z-equilibrium equation (2.40) and the in-plane

equilibrium equation (2.48).

CHAPTER III

APPLICATION AND RESULTS

In this chapter, the theory and the methods developed in the

preceding chapters are applied to a variety of problems. A computer

program has been developed which is applicable to rectangular plates

with different boundary conditions and variation in stiffness. The

objective is to illustrate the application of the method to plates with
several types of variations in stiffness, as well as to the uniform stiff-
ness plates. Since solutions to the uniform stiffness plate are

known, it provides a good measure for “verifying the accuracy of the

SOlution procedure. For the plates considered, the solution is

Obtained for a few problems for all successive steps of loading from
zero load up to secondary buckling and the results are analyzed.

Convergence of the solution is checked and accuracy of the results

13 examined via comparison with known results when possible. Some

oPtilnzization problems are presented at the end of each section.

Chapter III is divided into two sections. Section 3.1

deals with force boundary conditions . In section 3.2, plates with

displacement boundary conditions are considered.

A square plate with a symmetrical variation in stiffness is

considered. Stiffness is symmetric with respect to both centerlines

a
nd diagonals as shown in Figure 3.1; thus only a quadrant of the
Pl

ate will be considered. The variation in stiffness is such that

52

53

in quadrant (I) of the plate (see Figure 3.1) the stiffness is a

function of x only. For example,in section 3.1 a parabolic variation

in stiffness is considered; the flexural stiffness can be

represented by a parabolic equation:

D(x) - DrER + 4(1-R)(§)21 (3.1)
where D is stiffness at point x
Dr is stiffness at center of the plate
R is ratio of edge stiffness to center stiffness
a is length of each edge of square plate
Ekate:
: Et3
Since bending stiffness, D =-—————-—- , and membrane stiff-
12(l-v )

ness, K - Et, are both present in the plate equations, it will make a

difference whether the stiffness variation is due to a variation in

13 (or in t.

a)

For the variation in E, with t constant, the D

variation and K variation will have the same pattern.

K1

Let Bi - KE" the ratio of membrane stiffness at point
r

i to membrane stiffenss at reference point, and let
D
6 = -2- be the bending stiffness ratio at corresponding

1 D
r

points. Then,

81 Bit/Err
a 2 3 3 2 1 °r 81 ' 51
i E t /E t

i r

 

54

 

 

 

 

 

 

 

 

 

 

 

 

‘X
\_ /
\ /
\ /
\ /
\ /
\ /
/
3 ,
A - - - A o
\
/ \
/ \
I \
/ \
/ \
/ \
1’.
A a A]
Y
I D
edge
Dcenter

Stiffness variation at section A-A

F
igure 3.1. Geometry and stiffness variation of square plate .

b) For the variation in

s =3
1 K
1'
D
51:3—
r

i-

55

t, with E constant,

Et

_123
Et t
r I'

Eti
12(1-02)

Et3
r

12(1-v2)

Both cases can be considered without any major difficulties.

In section 3.1 (excluding 3.1.4) case (a) is assumed,

and in 3.1.4 and the entire section 3.2, the variable thick-

ness case (b) is considered.

In terms of boundary conditions, two separate classes of

PrOblems are considered :

l - Force boundary condition

2 - Displacement boundary conditon

In order to avoid computational difficulties,the following

not1‘-d:l.mensional variables are introduced and frequently used in the

anal3'sis .

t'-
t/ti

== 3
D a

DI

1{/a

‘Y " 37/a

w 22 W/ti

where t

= unit thickness.

flexural stiffness of a uniform stiffness, unit thickness plate.

Inembrane rigidity of a uniform stiffness, unit thickness plate.

 

 

U = ua/ti 56
K0
'-
U “N"

2 2
V va/ti

N'; N '; N' = - -
( x y xy) (Nx, Ny, ny)/Ncr

* 2 k 2

N. N' N )=(a_.)(N. N' N )
( x’ y’ xy Do x’ y’ xy
N- N; N = N; N; N N
(x. y xy> ()2( y WM
—0 -o - = a o o
(Mx, My, MU) (Dot1)(Mx’ My, Mxy)

a2
0' = ( )o
Doti
‘6'= qa4/D t
o i

<9 cp/Nz

3.1 Force Boundary Condition

The previously mentioned plate case (a) is considered sub-
ject to a compressive normal in-plane force 'N' on all edges
(at):restrainton.in-plane displacements).The solution is obtained
for: three successive ranges of loading (membrane, buckling, and
POStbuckling), for different variations in stiffness, and for both
sinlPly'supported and fixed edges. The effect of a transverse load
is also examined.

The behavior of the plate within each range is observed.
For each range the in-plane stress resultants, the in-plane dis-
plaCements and the out-of-plane displacement are calculated and
Plots are given. The results of solution for variable stiffness
plates are compared with those for the uniform stiffness plate.

S
ection 3.1 deals with different phases of the behavior, as follows:
3.1.1 Membrane solution

3.1.2 Stability analysis

3.1.3 Postbuckling behavior

_w

 

 

.o.|

‘7

~,
,
‘L:~

 

 

o.
\.
. I
i.

‘

v‘.‘ ,

b
V“:

57

In order to have a common base in different cases of stiff-

ness variation for comparison of the results, the reference stiff-

ness, Dr’ is taken such that the volume under the stiffness curve

be constant for all cases. (i.e.,mean stiffness is constant)

The stiffness variation in this section is

 

x 2
D(x) - DrER + 4(1-R) (E) J a/2
+——-—1
as introduced in equation X
(3.1). The volume under
a/ 2
the curve over a quadrant II.

of the plate (see Figure
3/2
3.2) is vol.- 2] 9. D(x)dx

 

Substituting for 2 and

{r
90‘) we obtain De].- [D
D(x r

a/
vol . zpf 2(-a~x)[R+4(1-RX£)2]dx.
r o 2 a
Integrating produces Figure 3.2

2
V°1 =- ————a Dr (1 + SR
24 )'

This volume is to be constant for all cases; thus, the variation of

Dr W1th R must have the form,

D a constant
1: 1+5R
D

'I'
he base stiffness is chosen to be 63
o

= l for a uniform stiffness

plate. and the Dr for different stiffness ratios are tabulated below:

1/10
1/2

10

3.1 .1 Membrane Solution

58

1.7142857

.117647058

In this part, a flat plate with no initial deflection is

considered and the solution obtained.

Since w is zero, the solution

can be obtained by solving the compatibility equation (2.47) only.

Boundary Conditions

To determine the value of the stress function, <9, along

the boundary, recall Equations (2.12); along the x-edges, we have:

32
NX III J2 s-N (1)
3y
N 3.2};
KY “Exay a 0 (11)

From (1) we have

a
3 (~33) = -N

int eg: at ing

a
$5 =' ~Ny+cl +f1(x)

2
(P ‘3 lNZ + c1

b
“t from (11)

y + y fl(x) + f2(x) + c2

59

3.39:. 29.
By (3x) 0 + 3x constant along the edge.

df1(x) + df2(x)
dx dx

 

.-. _nggy

= constant
8x

From the above we can conclude:

2
=-N — _—
¢ -%-- + cly + c2

similarly, along y-edges, we will get:

-N 2
¢ = g + c3x + c4

 

The constants must be chosen such that the given boundary
conditions are satisfied. Since the second derivatives of Q
determine the resultants, and, in this case, the plate is symmetrical
with respect to its centerlines, ¢ can be chosen such that it will
be symmetric about both centerlines. Thus, arbitrarily choosing

w - 0 at the corners leads to:

along edge yi= 0

since at corner x = 0, w = 0, and at corner x = a, m — 0

or

¢ =«g-(ax - x2) (3.2)

60

figfi -
3x 2 (a 2x)
A‘E=N_a =
8x 2 at x O
fla-flé. =
8x 2 at x a
similarly, along edge x = 0 we get:
N 2
<P=§(aY-y) (3.3)
.352
Since 3x is constant all along this edge, and from Equation
(3.2) above,is equal to 4§§- at the corner, ‘%¥'=-%% all
along the x - 0 edge; therefore (x = 0, %¥-= §§> and (x = a,%¥ = - 5%);

similarly from equation (3.3) (y = 0, g; = if) and
(y =
Now, we are able to compute (p values at each boundary

node. Figure 3.3 (a) shows the geometry of the plate and location

of node points.

The <p values at nodes A,B, and C according to Equation (3.3) are:

 

2
N a a 2 _ Na
QA ='§'[a(§9 - (59 3 --EF-
= 5t (3) - (3)2] = 3&2
‘PB 2 a4 4 32
°9C=0

The value of Q at some imaginary exterior points such as
a and b 2h; Figure (3.3.5) are needed because when the difference

operator is applied at the first interior point, such as (2),

61

 

 

 

 

 

 

 

 

 

h
*‘x—l
B A B C

c C
b B 3 2 3 B
a A 2 l 2 A

B 2 3 B

C B _ A B C

y a) Geometry and node labels

 

 

(b) @ function

Figure 3.3. Geometrical plan and m function.

I‘:

 

62

the value of Q at the first exterior point, a, is required. This

can be determined from the boundary condtions. We already determined

that along edge x = 0, the slope g;- is equal to -%?.
Among the -EQ' approximations which may be used are:

3x

1) Two point difference,using known slope at edge and QA:

(as) £13.23
3x A h
or
= - if = _ Ea
cpa 09A h(3x)A cpA h(2)
- _ £12
(pa-CPA h(2) (3.4)
a Na2
For h =-z-, ma = ?A --—§- , where '?A is already determined.

2) Two point,using known slope at A and the central difference

operator:
(3(2) anZ-wa N_a.=CPZ-q’a
3x 2h ’ 2 2h
A
or
(Pa = (92 ' h(Na) (3.5)
2
a _ _ F1...
For case b = Z", @a ‘ 92 4

Both of these approximations were tested for the plate of constant
stiffness. Approximation (2) led to values of Nx - Ny - N at all

points in the plate but approximations (1) did not.

Al

63

Approximation (2) was used for the first derivative in

this case. In this example the values at exterior points are:
= -2123
cPa C?2 4
2
_ Na
‘Pb’q’3‘T

Next, the finite difference operator Figure (A.4 ) which
represents the compatibility equation (2.15) was applied at each
interior node. The right hand side of equation is zero, since

w = 0.

3.1.1.a The Square Plate With h = %-, R = 1/10

Stiffness of the plate at each node and intermediate nodes
is shown in Figure (3.4). 7

Application of operator to all nodes results in the following
systgn of linear equations

P

1 r W
34.802542 -57.451238 10.341004 ¢1W -1.5384615

-14.3628095 93.0717725 -58.708963 ¢2>=' 5.9879807 Na

A
V
N

 

 

 

 

 

2.5852510 -58.708963 122.8929425 Q3

L a ‘

£11.4951922)

64

 

 

 

 

 

 

 

 

 

 

 

 

10 9 8 7
9 5 5 4 s 5
II II II II
a 3 ,2 5 5
II I I II
7 4 4 1 4
II I I II
5 3 2 3 5
II II . II 11
5 5 4 5
K
Node No %—--'EE
1 i
1 l.
2 3.07692
3 3.07692
l\/j 4 10'
5 10.
6- 10.
I 1.649484
II 6.4

Figure 3.4. Stiffness ratio at nodes and intermediate nodes

square plate R = 1/10.

65

solving the equations, we get:

The membrane resultants can be computed using Equation (2.12)

 

 

For example:

N

x(3)

”.28204873

< 62 ? < .23348206

.199145001

 

\

 

3y2 (3) h2

1.136927044N

Na2

= .233482-2(.199l45)+3/32

 

a 2
(29

<Na2> =

similarly, stress resultants at each node are calculated;the results

for nodes shown, are given in Table 3.1.

Table 3.1 Stress function and stress resultant ratios at each.node.

 

 

 

 

 

 

 

 

 

 

 

 

 

Square plate with R = 1/10, v = .25, h = a/4.
Stress function
Node Cp/Na2 Nx/N Ny/N ny/N
1 .28204873 1.55413 1.55413 0.0
2 .23348206 1.09879 .95865 0.0
3 .100145001 1.13693 1.13693 .03205
4 .12500 1.0 .52857 0.0
5 .0937500 1.0 .67636 0.0
6 0 1.0 1.00 0.0
6
5 3
4 1

66

Equilibriumis satisfied along any section of the plate.

For example, along x = .25a, using the block approximation:

P = (Nx)5(%)(2) + (Nx)3(%)(2) + (Nx)2(-:‘-) = -.9999999 Na 3 -Na

and along x = .5a:
p = (Nx)4(%)(2) + (Nx)2(%)(2) + (Nx)l(%) = -.9999999 Na

which are both very close to in-plane resultant at the edge of the

block along x = 0:

P = -N x a = -Na

3.1.1.b The Same Problem With R - .5

The same square plate is considered except the ratio of

g edge stiffness a 5

stiffness 13’ R center stiffness

Resulting values of the stress function m, and the calculated

in-plane stress resultants at the nodes shown below, are listed in

Table 3.2.

 

 

 

 

67

Table 3.2 Stress function and membrane forces

(R = .5, h = a/4, v = .25)

 

 

Stress function
P°int gnga2 Nx/N Ny/N ny/N
1 .26057639 1.19625 1.19625 0.0
2 .22319361 1.04536 .97297 0.0
3 .19052623 1.02574 1.02574 9.01058
4 .125000 1.000 .8578 0.0
5 _.0937500 1.000 .90316 0.0
6 0.00000 1.000 1.000 0.00

 

 

 

 

 

 

 

As in the preceding problem, equilibrium is satisfied along all

sections of the plate.

3.1.l.c Uniform Stiffness Plate (R = l)*

It is anticipated that as the stiffness of the plate approaches
uniformity, the solution will converge to the known results for uni-
form stiffness plate. Thus, this case is considered as a measure
of verification. The results shown in Table 3.3, are as expected.

The membrane resultant ratios are equal to 1.0 everywhere and

shear stress is zero; these are exact values.

*Both E and thickness are uniform all over the plate.

68

Table 3.3 Stress function and in-plane resultant ratios.

(R - 1, h =-%)

 

 

 

 

 

 

Point gglNaZ Nx/ N Ny/ N ny/ N
1 .2500000 1.00 1.00 0.0
2 .21875000 1.00 1.00 0.0
3 .1975000 1.00 1.00 0.0
4 .125000 1.00 1.00 0.0
5 .0937500 1.00 1.00 0.0
6 .0000 1.00 1.00 0.0

 

 

 

L1.1.d The Same Problem as in 3.1.1.a With R = 10

In contrast to the previous problems, in this case the

stiffness is increasing from center to edges and at the edges it

Results are shown in Table 3.4.

18 ten times stiffer than at the center.

 

 

 

 

 

 

 

Equilibrium is satisfied along all sections. 2
4 2 1
Table 3.4 Stress function and membrane force ratios, square plate
R " 1.0, h a % 9 V a '25

\
._Efleggpc g/Na2 Nx/N Ny/N ny/N

.1 .21667651 .32022 .32022 0.0

2 .20666961 .78443 1.14660 0.0

3 .18215623 1.02229 1.02229 -.03332

4 .125000 1.00 1.38657 0.0

5 .09375000 1.00 1.171 0.0
L11-ji .000 1.00 1.00 0.0

 

 

 

 

 

 

69

Improvement of the solutions

Since this is an approximate method, it is desirable to study

the convergence of the solution with increasing numbers of node points

(decreasing grid spacing). In this section the same

problems are solved using finer grid spacings (h = g at h = i— .

Solutions to all four problems are obtained. The node arrangements

are shown in appendices 3.1 and 3.2.

The convergence of the solution for the case R = 1/10

is shown in Table 3.5 and illustrated in Figure 3.8.

Table 3.5 Convergence of the solution, square plate R . 1/10, v = .25

 

 

 

 

 

Grid ‘65 -values by extra) N , at extrapolation 3 pt.
spacing at node 1 olation noée 1 results * extrap-
h/a olation
% .28204873 1.55413
.27472129 1.59060
% .27655315 1.58149 1.59026
.27465677 1.59028
T15 . 27513087 1 . 58809
L\ _L

 

 

 

 

 

* Iiiczhardson's extrapolation.

70

 

 

 

2 .
cp/Na Nx/N
.28 «-
«1.65
T 1
.275 ~ 1.6
NX
41.55
.27 5
4 8 l6 a/h

Figure 3.5 Convergence of membrane solution

§;J;Ll:l. Analysis of Results from Membrane Solution.
i) Uniform stiffness plate (R = 1)

Compared to theoretical values, in this case the finite
difference solution gives accurate results. The
difference operators agree exactly with the usual
difference operator for uniform stiffness plate.
Since the m function is parabolic in this case, the
difference approximationsto the second and higher

derivatives of' m are exact, and the solution is exact,

a
even for h =-Z. The in-plane stress resultants are uniform

 

71

over all the plate and there is no shear stress, as
expected. Distribution of Nx/N is shown in Figure
(3.6). The displacements vary linearly from the symmetric
centerlines (see Figure 3.7), and the strain is constant,
which agrees with elasticity theory. The solution
obtained with the grid-spacing, h =-% is exact, as are

solutions with finer grid-spacings.

 

 

(L.
niece

N/N=1
X

 

 

 

 

 

 

 

 

 

(a) Plan ' (b) Contours of Nx/N

F — r“ w r-

 

 

 

 

 

 

 

 

 

 

9-9 8-3 C-C 0-0 E-E
N
(c) Profiles of-i?
2 1 -
SCHLE

FIEUINe 3.6. Force distribution for undeflected square plate R = 1,

Figure 3.7. Contours of in-plane displacement (U' =

ii)

72

 

 

 

 

 

 

 

.375 .28 .1875.094 0.
K

N) for

 

undeflected square plate, R = 1, v = .25.

Square plate with R = 1/10.

In this case, because of variation in stiffness, the
stresses vary over the plate. Thus,the m function

is not a smooth, parabolic one as it was in the uniform
stiffness case and the solution would be an approximate
one. Solutions with finer grid spacing were compared

(see Table 3.5) and they show fairly good convergence.

Study of in-plane resultants in Figure (3.8) shows the

expected behavior,with a shifting of the load toward

the stiffer parts of the plate.

Equilibrium is satisfied along any arbitrary section
of the plate. Shear stress is zero at points of symmetry

but at nonsymmetric points, because of the load shifting

process, there is a small shear force created, as expected..

73

 

 

 

 

 

 

 

 

 

 

 

D...—
m

a) Plan b)

 

 

 

 

 

—

 

 

L.

8-9 8-8 C-C

c) Profiles of Nx/N

Contours of Nx/N

1

 

[3-0

 

 

 

 

 

.5.4 .3 .2 .1

d) U'-displacement

.75

'1.0

\ ‘2
\ ~——
\ 1.
. P
1 (7115.0.

25

 

F
igul‘e 3.8. Distribution of in-plane force (Nx/N) and displacement

K
(U' = U 3;) square plate, R = 1/10.

74

In-plane displacement patterns are illustrated in

Figure(3-8 d)-It is observed that the displacement

normal to the edge is increasing as we approach the corners,
at which stiffness is less than at the center. This behavior
seems reasonable since there is no displacement restraint

along the edges.

iii) Square plate with R = 1/2

iv)

This problem can provide a good check on solutions,
because it lies between two previous cases. As
we go from the R = 1/10 case to the uniform stiffness
case (R = 1), we would expect the solutions to approach
the uniform stiffness results. Investigation of the
results in Figure(3.9) along with the results found by different
grid-spacings and comparing with cases (i) and (ii), indicates:
a) The convergence as the grid spacing becomes finer
b) Results for the stresses and displacements trend
toward those for the uniform thickness case as R is
changed from 1/10 to 1/2.
Square plate with R = 10.
Solution for this case shows convergence as the grid
spacing is decreased, and it also agreeswith previous
results in that:
a) Load in the plate shifts toward the edges which

are stiffer as illustrated in Figure (3.10).

.b) Displacement normal to the edge is a little larger

at center of the edge and decreases toward the

corner; (see Figure 3.10 d).

75

 

 

' J
1.0

1.10
Q l 1 1 F

b) Contours of Ng/N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a) Plan
7"
_J
R-H B-B C-C 0-0 E-E

c) Profiles of Nx/N

 

 

 

 

 

 

. .__J
.35.3 .2 .1

d) U'-disp1acement

Figure 3.9. Distribution of in-plane force (Nk
K
(U' = U 7;) square plate, R = 1/2.

/N) and displacement

 

fr

II
‘1.
‘ V
5116 3
I

76

 

 

I I 1
\\
\ 1.25

‘~

1.0
..75
l l ///’/"-—~‘;:::EEEE .50

9 édd 4'—

R
b) Contours of Nx/N

 

 

 

 

 

 

 

 

 

a) Plan

 

 

 

 

 

 

 

 

 

 

 

_J _J

8-9 8-8 8-8 . 00-0 E—E
c) Profiles of Nx/N
2 13-1-8
SCHL

 

 

 

 

d) U'displacement

Figure 3.10. Distribution of in-plane force (Ng/N) and displacement
K

(U' = U'jf) square plate R = 10.

77

3.1.2 Buckling

 

To study buckling and determine the critical value of the

applied compressive force, NC , the equilibrium Equation (2.14)
r

is employed. The left hand side of this equation is approximated
by the operator Figure (2.8) and the right hand side is represented
by the difference operator Figure (2.9). The stress function,
9, values obtained in the membrane solution for the plate with

no lateral displacement are used in the equilibrium equation. This
is acceptable,because in the stability analysis we are seeking

bifurcation of an initially undeflected plate.

Boundary conditions

 

In the case of buckling and postbuckling, the out-of-plane
displacement, w, has a considerable effect on the solution and
out-of-plane boundary conditions must be considered. In the simply

supported case, along edge x = 0, we have:

 

 

i) w = 0
82w 32w
11) M --D(—-+ v —) = 0
x 2 2
3x 8y
But --—5 = 0, so that (ii) becomes ——§-= 0. Using a difference
3y 3x
approximation,
32w _ W1+1 ' 2"1 + w1-1
8x2 h2

we will get at point B (Figure 3.3),

since wB

In the case of fixed support, the conditions are:

i) w B 0

11) gg=o

using a central difference approximation for slope we have:

8x 2h b

 

3.1.2.a Buckling of Plate with R = 1/10.

Let us consider problem a) again (R

supported edges and h 31%.

= 1/10) with simply

In the right hand side of equation (2.14) the m values have

already been obtained for the initially flat plate.

Applying operator Figure (2.8) at each point, and substituting

Q values in the right hand side, the following eigenvalue problem

is obtained.

P

 

 

 

 

1.071875 -5.5375 6.5725 w3

 

_J 1 J b
2
or: calling A = 32-
D
r
r
14.9375-.388533ZA ’ -20.52500+ .38853321 4.2875 1

 

L 1.071875+3004006l -5.5375+ .1421158A

-5.13125+.06867411 10.86875- .25717881 -5.5375+ .11983.GK

F' )
14.9375 -2o.52500 4.28751rw 3.885332 -.3885332

-5.l3125 10.86875 -5.5375 <w2>=r -.0686741 .2571788
-.0040060 -.l421158

 

6 . 57 25- . 284231?

0 ‘1 w

.1198306

.2842316

W

W

l

2

3

79
Solving leads to these eigenvalues,

f
5.20748

20.42753

>4
II
A

46.71190

 

\

01'

5.20749

N =1—‘2' = 20.42753

Nln

46.71190

The corresponding eigenvectors are:

11_ .12. .12.
1.0 1.0 -.69810
.84441 .36007 1.0

.63179 -.57565 .03122

To check convergence, the same problem was solved with finer grid

a a
spacings, (h -'5) and (h 16).

The eigenvalues shown in Table 3.6 were obtained.

The con-

vergence is fairly good and extrapolation flnproves the results

 

 

further.
A1 6.
5. ~—————___11 ._1
4._
'r. . r
4 8 16

Figure 3.11. Convergence of eigenvalue, R = 1/10.

a/h

80

Table 3.6 Eigenvalues of s-s square plate, R = 1/10,with different

grid spacings.

 

h/a A1 2 pt. extrapolation 3 pt. extrapolation

 

1

1/4 5.20748
4.727158
1/8 4.84724 4.855814
4.847773

1/16 4,84764

 

 

 

 

 

 

The first mode shapesin the three solutions are very close to each other.

3.1.2.b Buckling of Uniform Stiffness Plate

As before, a plate with uniform stiffness is considered;
the solution can be used for an accuracy test since the exact solution
is known. The first eigenvalue obtained using different grid

spacings is tabulated below and compared with the exact solution.

Table 3.7 Comparison of first eigenvalues of different solution (R - 1)

 

h/a Al 21 exact difference 2 Pt. extrqr- 3 Pt. extrap-
olation Aolation

 

1/4 18.74517 19.7392088 5%
19.734057

1/8 19.48684 1.2

N

19.739197
19.738873

1/16 19.67587 .3%

 

 

 

 

 

 

 

 

81

It can be seen that the results are very close to the exact values

given in Reference (44);

Na 2

N = 2V or, A =-——— = 2N = 19.7392088

The solution is satisfactorily converging to the exact value.
Three point extrapolation results in an error of less than 10—6.
Table (3.8) shows the critical loads obtained for two dif-
ferent loading cases and gives comparison with previous work as
well as exact values. It can be seen that even with 8 x 8 nodes
the results are satisfactorily within engineering accuracy. Mbre
accurate results can be obtained by extrapolation. In Table 3.8,
the results given by Clough and Felippa are obtained by the finite
element method and Dawe used the "discrete element displacement

method", which in principle is the same as finite element method.

Eigenvalue problems for other cases (R = 10, and R = 1/2)
were solved, and the convergence was examined with increasingly

finer grids. Details will be discussed later.

3.1.2.1 General Buckling

 

So far, the problem was considered symmetric with respect
to both centerlines and the diagonals. Therefore, the solutions

are limited to symmetrical modes of buckling only and nonsymmetrical

82

cowumaoamuuxo usaoalm mo oasmom
um

 

 

 

 

 

 

 

 

 

 

 

«smoooo.~
ommm.a oH\H
mmmmmmmm.a
.N wmm.a omqqnm.a w\H
mumma.~ scammoumaoo
mwm.H mmmmm.a «\H Hoaxmlam enemas:
oH\H
.c Hmo.e mmm.m om.m m\a cofimmouaaoo
oNH.¢ m~m.m «\H Hmfixmasa auouwsa
coquaaom Ado mmmwaom soHumHo nuanmom m\:
Hmowmmmao can nwsoao AnHV mama Iamuuxm hm uawmmum ouam 5mm: ommu mafivooq
. nu: Ho
mmm. u 9 H u an .mml I «z .mumaa mumavm wouuonmsm manawm mo mumoa Hmowufiuo w.m NHANH

 

83

modes are absent. This means that 12 in this solution is not
the eigenvalue corresponding to the second mode, but it represnts
the eigenvalue corresponding to the second symmetric mode of buckling.
In this particular case the 2nd symmetric mode corresponds to the
5th general buckling mode.

To obtain more accurate results, the plate was considered
without imposing any symmetry. Thus, all possible degrees of
freedom were allowed, within the restrictions imposed by the choice
of grid spacing. The solution for each case was Obtained and the

buckling modes and corresponding critical loads are studied in the

next section.

3.1.2.143 General Buckling_of Square Plate with R = 1/10.

1) Simply-supported boundaries.
Problem a is solved for general buckling (no
symmetry imposed) with h = a/8 and assuming simple
support along all edges. The first few modes of buckling
and the corresponding eigenvalues are shown in Figure
(3.12)

ii) Clamped edge.
The same problem as in (i), but with all edges clamped,
was solved. The first six modes of buckling and the

corresponding eigenvalues are shown in Figure (3.13).

84

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-— +
+
+ .—
* * 3 44 * 44
N1 = 19.39 N2 9. N3 39.
+ _ + _
+ +
- +
* 61,61 N* 75 15 N* 76 52
N4 . 5 . 6 .

Figure 3.12. Modes of buckling of square plate, R = 1/10,simp1y supported.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+
+
- +
* 42 24 N* - 66 60 N* — 66 60
N1 . 2 . 3 .
+
+ - + _ +
_ + - + _
* 91 60 N* 103 32 * 3
N4 . 5 . N6 . 118.28

Figure 3.13. Modes of buckling of square plate, R = 1/10, clamped,
(N* - Na2/Do).

cur

85

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+
+ + -
* 20 04 N* 46 73 N* 46 73
N1 . 2 . 3 .
+
+ - +
— + +
N* 74 79 N* - 88 78 N* - 89 28
4 O 5 _ O 6 - 0

Figure 3.14. Modes of buckling of square plate, R = l/2,simply supported.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

- +
+ +

N* 49 56 N* 82 0 *

1 . . 2 . 3 N3 - 82.03

+ +

+ —

"' " +

* 114 20 N* 12 o *
N4 . 5 7.3 N6 = 137.84

Figure 3.15- Moges of buckling of square plate, R = 1/2, clamped,
(N a NaZ/Do).

86

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

' *
N: = 19.48684 N: = 47.23375 N3 = 47.23375
+ — + +
- +
+
* * *
N4 = 74.98066 N5 = 88.75994 N6 . 88.75994

Figure 3.16. Modes of buckling of square plate, R = 1,simply supported.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+
+ + -
N* 49 567 N* 82 131 N* - 82 131
l O 2 . 3 .- O
+ - + +
- + +
* * *
N4 = 112.693 N5 = 124.778 N6 = 137.127

Figure 3.17. Moges of buckling of square plate, R = 1, clamped,
(N a NaZ/Do).

 

 

 

 

*
N

l

= 18.42

 

 

 

 

 

 

* 70 04
N4 .

Figure 3.18. Modes of buckling of square plate R = 10,simp1y supported.

 

 

 

 

* 52 09
Nl .

 

 

 

 

 

 

N* 101 78
4 .

Figure 3.19. Modes of buckling of square plate,R =

(N* = Na2/Do)

87

 

 

 

 

 

 

 

 

 

+

N* - 47 00
2 " o
+
+

N* - 83 01
5 - .

iéz‘é'.

 

 

 

 

 

 

 

 

+
N* - 81 37
2 - o
+
+
N* 11 02
5 3.

 

 

 

 

*
N = 47.00

 

 

 

N* 90 24
6 .

 

 

 

 

 

 

+
* 8
N3 - 1.37
+
N* 1 3 16
6 3 .

10, clamped,

 

 

 

 

A‘s _
.5 . «a
o .
a.
d— 44“

88

3.1.2.1(b,c,d)

 

The plates with R = 1/2, R = l and R = 10 were solved
for both simply supported and clamped boundaries; the buckling
modes and critical loads are illustrated in Figure (3.14)

through (3.19).

3.1.2.2 Analysis of the Results
A. Simply-supported edges:
1) Uniform stiffness plate.

As discussed earlier, the critical load obtained for
this problem agrees very well with the exact solution.
Mbdes of buckling based on theoretical solutions are
combinations<mfone or more half-sine waves in each
direction. The first mode shape is one half-sine wave

in each direction, x and y.

The first mode obtained by the difference solution was
examined and the deflected plate after buckling was
found to consist exactly of one half—sine waves in both
directions. First mode deflected shapes are shown on
Figure (3.20).
‘The second and third modes had two half-sine waves
in one direction and one half-sine in the perpendicular
direction.
ii) Square plate with R = l/10.
If we investigate mode shapes and critical loads and

compare with the uniform stiffness case, the following

iii)

89

properties are observed:

1. First mode of buckling is not a half-sine shape,

but it is flatter at the center where the plate is
stiffer and more curvature appears near the edges
where the stiffness decreases. This phenomenon could
be predicted, based on the nature of the plate.

In the higher modes, in contrast to the uniform stiff-
ness case, in which the modes are formed by two or
more half sine waves in one or two directions, there
is a flat region around the stiff center of the plate

and larger bending near the edges. See Figure (3.20).

. The second eigenvalue in the uniform stiffness case

is about 2.42 times the first one, while in the present
case the ratio of second to first eigenvalue is

2.03. This can be interpreted as follows:

The first mode shape has greater curvature in the
central region, while the second mode has zero cur-
vature in the center. Therefore, it is expected that

plates with more flexible edges correspond to lower

second eignevalue.

Case R = 1/2.
Since this problem lies between the two previous cases,
mode shapes are something in between which supports the

aforementioned ideas. The ratio of second to first eigen-

90

values is 2.33, which lies between 2.44 for case (i)

and 2.03 for case (ii).

 

 

 

 

iv) Case R = 10.
In this case, the variation of stiffness is reversed,
with the plate being stiffer near the edge. Investigation
of the mode shapes indicates more bending in the less stiff
lcentral region and flat curvature near edges where the
plate is stiffer, supporting the ideas introduced above.
c:
C)
("3'-1 Simply-supported edges
—-—-—Clamped edges, R = 1
E:
X/0
90.00 0.25 Oi-SO 01.75 11.00
ca‘
c>
°. 10
‘7‘ 1
1/2
1/10

Figure 3.20.

Deflected shape for s-s and clamped plate.
(N = 2.40 Ncr)

91

B. Clamped Boundaries.

1) Uniform stiffness plate.

The first eigenvalue is found to be 49.56763 which

is very close to the exact value, obtained by Levy (28),

of 5.0378

2 = 49.71319.

Table 3.9 shows convergence and accuracy of the results,

and gives comparison with some previous works and the exact value.

Extrapolation of the results shows an accuracy of about .22.

Table 3.9. Critical load ‘N* = N

2
a

or NZD

bi-axial uniform load, R = l, v = .316.

of clamped square plate under

 

 

 

 

 

 

 

 

Grid-spacing» A Extrapolation Levy (28) Clough & Classical
(hlg) 1 Felippa Solution
1
‘2 5.625
.% 5.02225 5.037 5.399 5.31
5.29921
1
IE’ 5.22997

 

The first buckling mode agrees almost exactly with the theoretical

mode shape (1- cos 22E5, for m = 1.

ii) Case R = 1/10.

111)

See dashed curve in Figure (3.20).

Similar to the s-s case, in the central region the plate

remains flat and sharper curvature occurs near the edges.

Case R = 1/2.

As anticipated, results obtained for this problem lie

between cases (i) and (ii) supporting the validity of

the solutions.

 

92

iv) Case R = 10.'
As before, sharp curvature is observed in the central

region of the plate due to smaller stiffness of that

region.

afzer
be in
Theref
be an;

sewn:

and t

the e

93

3.1.3 POSTBUCKLING

3.1.3.1 General Procedure

 

As discussed in the preceding chapter, a plate will stabilize
after first buckling. Immediately after buckling, the plate would 4
be in a state 0f Stable equilibrium with moderately large deflection.
Therefore, the large deflection theory discussed in chapter 2 can
be employed to study postbuckling behavior of the plate up to the

secondary buckling point.

In this section, the procedure followed will be discussed
and the results will be analyzed.
For a solution to the large deflection behavior of a plate,
the equilibrium and compatibility conditions must be satisfied;
both equations are coupled in w and m. In this case, in addition
to the variation in stiffness, geometrical nonlinearity caused by
large deflection will also be involved.
To solve these coupled, nonlinear equations, an iterative
technique is employed. A schematic flow chart of the procedure
is given in Figure (3.21). The steps of the procedure in Figure (3.21)
are as follows:
Step 1 - Solve the equilibrium equation (2.14)
The left hand side is approximated by the operator of Figure
(2.8),which will be applied at each node to form the matrix
[Aw]. To calculate the right hand side of this equation,
initial values for m and w are needed. w is assumed

consistent with the first buckling mode shape, and the

94

 

 

Apply load N, (N > Ncr)

 

 

 

 

Input @ and w based on previous

’ solutions

 

 

 

 

 

Solve equilibrium equation
(2.14)

 

 

 

 

 

 

 

 

A A Get new w

 

 

Solve compatibility equation
(2.15)

 

 

 

 

 

 

 

Get new m

‘ no ////:::{:\\\\\g Yes

+1
convergenc

 

 

 

 

Increase Load

 

 

 

 

N = N + AN
Yes N<N I
= max

No

 

 

 

Stop

 

Figure 3.21. Flow chart of iterative procedure.

95

m values calculated for the undeflected plate are used.
Second derivatives of m and w can be approximated by
finite differences to obtain the known vector {01}.

Then the linear system of equations, [Aw]{w} = {Q1}, is
solved and the new values of wi calculated.

Step 2 - Solve compatibility equation

To solve compatibility equation (2.15), the difference
operator,Figure (2.10),which represents equation (2.15),
is applied at each node and matrix [A] formed.

The right hand side of this equation includes derivatives
of w only. The values of {w} computed in step 1

are used along with the finite difference approximation
to the derivatives to form the right hand side vector,
{B}. The linear system of equations [A]{¢l} = {B}

is solved, resulting in new values for the stress function,
@1-

Step 3 - Taking the new values of w' and $1, steps 1 and 2 are.repeated.

1
Step 4 - Convergence
Aftereach iteration, the new values are compared with
the old ones. If they are close enough,the iteration will
be stopped and the last computed values of m and w
will be accepted as the final solution for the given
load. If the new values are not satisfactory, the iteration
procedure will continue. The convergence was accelerated by
using the convergence—inducing technique using 3 successive

values, as discussed in reference (53) and it proved to

be very useful.

96

Step 5 - The applied load is increased by a small amount,

AN, and steps 1 to 4 repeated again to get the solution

corresponding to load N + AN.

The above iterative procedure was continued until the

applied load approached the second critical load. Near

the second critical load, the equilibrium approaches in-

stability and it was observed that the solution would not

converge.

More details of the numerical solutions are discussed in

the following sections.

3.1.3.2 Numerical Solutions

 

In this section some problems will be solved and the results

analyzed.

8)

Square plate with R = 1/10, h = a/8

D
A square plate with R = BEQBE-—- = 1/10
center
subject to an in-plane load, N, on all four edges,

2.
8

considered and later the problems were solved with the

was considered. A grid spacing of h - was first

grid spacing of h a %E' for more accuracy.

The first critical load was found in (3.1.2 a) to be

N = 4.84724 2%- and the m values are known from Table
a

(3.1). A deflected shape, w, was assumed based on the

first mode shape obtained from the buckling solution.

97

D
Initially, N = 5.1—; was assumed; this is slightly
a

above the critical load. The transverse load, q, was

taken to be zero.

The iteration was continued until the new values of m

and w differed from the previous values by not more

IA

w -w
than a predetermined tolerance of .SZ; (I-éf—-| .005).

The last values of m and w were accepted as converged
values and based on them the in-plane forces and
displacements, as well as the bending moments and

.bending stresses, were computed.

In the very first step, convergence depends greatly

on first input values; in the choice of these, ex-
perience will play a major role. In the following steps,
convergence was achieved after about ten to twenty

iterations - depending on the size of load increment.
D

_29
a

which is ten percent of initial load. Variation with

In this case, the load increments were taken as .5

load of the lateral deflection, w; inéplane displacement, u:
membrane force, Nx and bending moment, Mx’

are plotted in Figure (3.22).

At the end of each step, having m and w at each
node, we can calculate in-plane stress resultants from

the difference approximations of equations (2.31) and

98

the bending moments from the difference approximations

of equations (2.35).

So far, we have enough information at each node to
illustrate the behavior of in-plane forces, bending
moments and principal stress; they are plotted along
the centerline of the plate in Figure (3.23), for three
different load levels. The variation of maximum
principal stress with load increase is also plotted

in 3.23 d.

N
Contours of the in-plane force ratio, 3%, at the three

different levels of loading are shown on Figure (3.24),
and the distribution along various sections of the plate
is shown on Figure (3.24 e). The variation of the
in-plane force at the plate center with load increase

can be seen on Figure (3.22 c).

It should be noted that Ny can be determined from Nx

plots by considering symmetry.

To investigate in-plane displacements, equations (2.33)

and (2.34) can be used.

We have already determined Nx and Ny, and the derivatives

of w can be easily obtained by difference approximations;

we are thus able to compute ux ='§E and v --§!

3x y 3y
at each node. Then (RE). 8 (bu) and (2!) - (bv)
8x 1 1 8y 1 1’
where the right hand sides (bu) and (bv) are computed

constant values. Applying the difference operator to

the left hand sides will give:

99

[Au] {u} {bu}

[Av] {v} {bv}

In order to obtain a solution to these equations, we

need to know some boundary values of u and v.

In this case, we will take advantage of symmetry. Since
the geometry and the loading is symmetric, we can conclude

the following (see Figure 3.3):

1) Because of symmetry in x, the u displacements
along the axis x = 3’ will be zero.
ii) Because of symmetry in y, the v displacements
. along axis y = %- will be zero.
iii) Symmetry about diagonals implies that u = v along

the diagonal, x = y, at nodes 1, 3, and 6 in Figure (3.4).

Taking advantage of these assumptions enables us to solve
a system of linear equations and obtain the u and v

values for each node.

The patterns of u displacements at each load step
and the contours of equal displacements are given in
Figure (3.25). The v displacement behavior can be

easily visualized considering symmetry.

The variation of in-plane displacement with load at some

specific points is illustrated in Figure (3.22 b).

100

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3. 3.
3* 8‘
8 a
#1 SJ
3 g A B
5 0+
.2 ‘2
3 3
3‘ 8‘
A
O
8 I 9 B 9
61 1 NO initial 2'4
, deflection
, ——— with initial
8 , imperfections
0 l 0
91.00 0100 {.00 2140 3.20 L11.00 10.00 20.00 30.00 40.00
CENTRHL w/ t1 U
a) Central deflection b) U-displacement
8 8
J“ 81
D
0:8 0:9
“—15-: 1.1.10.
1— u—‘l‘
Z 2
LLJ tu
tJo 13°
F—Q F—Q
05.0.; "‘81
Hz u: I
I
D I
w o
.1 =: '
- H I , , , I
' .00 15.00 , 30.00 45.00 50,00 115,00 . 30.00 35:00
N N
c) Nx at center d) Mx at center

Figure 3.22. Plots of w,U,Nx, and Mx; square plate, s-s, R = 1/10

IN-PLSNE STRESS RESULTBNT RSTIS (N )

PRINCIPSL STRESS (CT)

NOTRTION FOR CURVE IDENTIFICRTION.

101

ux
----- NY

 

/

 

 

SENDING HOHENTS(H)

 

 

 

o) NX HND NY SLONO HXIS Yza/Z

 

 

 

450.
c
b
a:
300.. 3
C
.—
a
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33
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150.. E
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0 ' T ' 0125' ' T 0.50
(X/al

c) PRINCIPRL STRESS flLONO 9x15 v=a/2
n) "21.0

  
   

---BX
J
80. C
30. B

 

r1
0.28
(X/OJ

b) atuutuo nontnrs atouo 9x13 7:0/2

480‘

180-

 

I l r

1.0

I I

T

 

4.0

T
2.0
I

N
d)PRINCIPRL STRESS VS. flPPLIEO LORD

(UNOEFLECTEO) a) "21.10 ‘0) "£2.00 .

FIGURE 3.23. PLOTS 0F STRESS COMPONENTS-SOURRE PLRTE.SIHPLY SUPPORTEooRzllloo

102

 

 

 

i/
W,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.l. l.
1.25 -5
1.5
a) uu055150150 (~51.1 5) 0005150 (uéz.25)
a 6' l c ECL
%2 7 '
1
0.
<;:-——‘-ln
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1 f U; I
n 8. 0 t 5
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1 I
n—n 8-8 0-0 0-0 5-5
2 1 -1 -
sanE

5) 515555 RESULTRNT 55051155 («22.251

FIGURE 3.24. CONTOURS OF IN-PLSNE FORCEINX) RNO PROFILES OF NX RT SPECIFIED
SECTIONS FOR LORD 2.25 TII‘IES THE CRITICRL LORO.SIHPLY SUPPORTE05R=IIIO

103

To compare the behavior of an imperfect plate, a small
initial deflection was imposed by applying a small
transverse load, q, at node 1 only. The plot of w

vs load is shown in Figure (3.22 a).

Convergence of the solution was checked by solving the
problem using two different grid spacings (h/a = 1/8
and h/a = 1/16) and the results proved to be very close.
For example, corresponding to N = 1.5Ncr the central
deflection was found to be 3.335299 for h =‘%- and
3.3735 for h 8 fig- 3 difference of only 1%. For more
accurate results in all succeeding analyses, solutions

a
with h 16 are given.

 

 

 

 

 

 

 

 

 

\

 

 

 

 

.5.'4.’3.2 .1 0 4.3.2.1. 0 50. 710. 0.
a) Undeflected plate b) Buckled (N/Ncr=l.15) c) Buckled (N/Ncr-2.0)

Figure 3.25. In-plane displacement (U - ua/ti), square plate, simply-
supported, R = 1/10.

b)

104

Square plate with uniform stiffness, R = 1.
As before, this problem, for which solutions have al—
ready been published, is undertaken in order to investigate

the accuracy of the difference method.

Justas in problem (a), the iteration procedure was fol-

Dr
lowed, starting with the load N ' 20'—§’5 WhiCh is
a

g
2 O

a
load increment was taken as 2.5, which is about 12.5%

slightly above the critical load, Ncr = 19.48 The
of the initial load and the tolerance range was taken
as .52. Convergence was usually achieved after

7 or 8 iterations.

Graphs of the w and u displacements versus load are
given in Figure (3.26).

Plots of the membrane forces, bending moments, principal
stress along the centerline (y ..%), and principal

stress versus load are shown in Figure (3.27).

For further illustration, the contours of in-plane
N
force ratio, —E- and its distribution along various

N
sections of the plate are shown in Figure (3.28).
Figures (3.26 c) and (3.26 d) show the variation with
load of membrane force and bending moments at the center

of the plate. In-plane displacement, u, is illustrated

by the contours in Figure (3.29).

105

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D
9 A B
8.1
8
6.
'
O
9
3.
3 / 8,
:1 / 5;
/ A
’ L
8 I g B
H I No 1 itia 2-
° , deflection
, -—-"-With initial
3 . deflection 8
50.00 0100 1150 2:40 —5'.20 ”0.00 (0.00 £0.00 $0.00 70.00
CENTRHL w/t1 U
a) Central deflection b) U-displacement
O
m 9
.q O!
1m
c:
033 of?
“452 11.10.
F— ,_v
2 2
l” m
U c.)
O a
hJQ F_o
GO. (ID...
x' 01
:3 c: ,/
7 ‘3 -'
I 7 7— l ' T 1
0.00 15.00 N' 30.00 45.00 c13.00 15.00 , 50.00 45.00
N .
c) Nx at center d) Mx at center

Figure 3.26. Plots of w,U,Nx, and Mx’ square plate, s-s, R = 1

s

2-

{.I

Oahzl hZChJDCUK amflzbn EZCJLIIN

‘bu

mmwzhn Jam—Ulntk

1136

 

 

 

\
'———-—ID(

" a \ ----- NY
l3: ‘ \

53 \

E: . \

‘5 \

EC \

25 3. \

‘_l

a; \

u: \

“ \

I?! a“ \\

m “~._~

.5 ‘ _ ~“}Y§_ A
U

x

C

.J

I.

I

55

 

a) NX 8ND NY RLONO RXIS Y=G/2

 

 

 

320,
240-
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33 a:
In In
°= es
5;:00- a,
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C
if a.
a-c . z;
0
3
a:
A
o I I I oI28 I T I aka
(X/o)

 

SENDING HOHENTS(N)

c) PRINCIPHL STRESS HLONG RXIS Yzalz

NDTHTION FOR CURVE IDENTIFICHTION.

FIGURE 3.27. PLOTS 0F STRESS CORPONENTS.SOUHRE PLRTE.SIflPLY SUPPORTEO.R=1

n) «21.0 tuuoerLecreo)

  

I

l

10..

’-"
—
"

 

0&5
(X/a)

b) aeuoxuo nonsurs nLouo nxxs v=q12

 

 

:20T
240.
100-
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0.0 1.0 2.0 3.0 4.0
NI

alraxNCIPnL sracss vs. flPPLIED LORD
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107

 

 

 

 

 

 

 

 

fix= 1 '1.
Hr
altUNOEFLECTED ("21.1 b) auchso (£21.25)

”MEI

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c) auchco (N22.40) d) PLnu
._7
n-n B—B c-c 0-0 5-5
2 1 - -
SCRLE

e) STRESS RESULTRNT PROFILES (NE2.40)

FISURE 3.28. CONTOURS 0F IN-PLRNE FORCEINX) 9N0 PROFILES 0F NX RT'SPECIFIEO
SECLIONS FOR LORD 2.40 TIHES THE CRITICRL LORD.SIHPLY SUPPORTED.R=1 .

108

 

 

 

 

 

 

 

 

 

 

 

 

 

 

— ‘

 

1.0 .5 0. 1.51.0 .25 0 15.1015. 0 o
a) Undeflected plate b) Buckled (ulN = 1.15) c) Buckled (N/N = 2 30)
cr cr °

Figure 3.29. In-plane displacement (U = ua/ti), square plate, simply-
supported, R = 1.

109

Comparison of the results

 

In order to compare the results obtained by the iterative
difference method with some currently available results,
a square plate under uniform lateral load, q, was solved,
and the results compared with those of Kaiser [26] and

Basu [5].

Figure (3.30) demonstrates the very close agreement with

the previous works.

 

 

c:
c?
‘r7
:3
c?
m“
c:
c:
3“? A Kaiser
D Basu
--Present result
c:
c? .
c:
c:
' 1
c33.00 1'0.00 2'0.020 $0.00 40.00
qalI/Doti x10

Figure 3.30. Square plate under uniform load, boundary free of

any in-plane force, simply supported v 8 .316.

110

c,d) R = 1/2 and R = 10.
Problems c and d were also solved similarly and

corresponding graphs are illustrated in Figures (3.31)

to (3.38).

 

111

 

 

 

 

 

 

 

 

 

 

 

O D
9 9
31 8*
A B
s s
O O
9 9
{fr 81
*2 *2
8 8
6. 6.
u- N
o A
g 9
'0‘ a: B D
8. 8.
93.00 0100 {.00 2140 a‘.20 “0.00 {0.00 $.00 03.00 40.00
CENTRRL w/t1 U
a) Central deflection b) U-displacement
s 3
.51 81
0
ad: 059
U '4 was
._0 ._1’
Z
w E
U U
a o
#49 b—Q
Gas 020.
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[2 I}:
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U) a /
.2 s ,
' 1 I v ‘ v
0-00 115-0." N. ”-00 45-00 cb.00 1%.00 , 30.00 3500
c) Nx at center d) Mx at center
Figure 3.31. Plots of w,U,Nx, and Mx’ square plate, s-s, R = 1/2

112

 

 

 

 

 

 

 

 

l\ -————Nx

; '5. \ """ NY 40..
—- \
E: \
t; 4 x ,_ -
i \ l5
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2

A
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a) NX 9ND NY RLONG RXIS Y=O/2

 

 

 

b) SENDING HOHENTS RLONG RXIS 7:0/2

 

 

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(X/a) N

c) PRINCIPRL STRESS HLONG RXIS Yzc/Z
NOTHTION FOR CURVE IDENTIFICRTION.

leRINCIPRL STRESS VS. HPPLIED LORD

a) uét.0 tunnerLecrenn a) «£1.10 0) ué2.40 .

FIGURE 3.32. PLOTS OF STRESS COHPONENTS.SDURRE PLRTE.SIHPLY SUPPORTED.R=1/2 .

113

 

 

.95
l

W

 

 

 

 

3

 

 

I
a) unnerLecrso (Hf-.1.) b) 00010.50 (N=1.10)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

l.
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c: BUCKLED (NI-1.35) d) M...
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e) smass ncsuunur PROFILES (4422.35) 8an

FIGURE 3.33. CONTOURS OF IN-PLRNE FORCEINX) RND PROFILES OF NX RT SPECIFIED

SECTIONS FOR LORD 2.35 TIRES THE CRITICRL LOflooSlflPLY surrourao.n=1/2.

114

 

 

 

 

 

 

 

 

 

 

 

 

 

1 e

 

 

 

.4 .1 o :1 2.1..5 o 10'3'2. 00
a) Undeflected plate b) Buckled (N/Ncr = 1.25) c) Buckled (N/Ncr‘=2.20)

Figure 3.34. In-plane displacement (U = ua/ti), square plate, simply-
supported, R = 1/2.

115

 

 

 

 

 

 

 

 

 

 

 

 

 

8 8
2'1 81
A B
s a
gs 3'1
8. a:
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2. 'z
8 8:
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94 a. B ’
o n
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°0.00 0300 1100 2'.40 3120 °0.00 10.00 €0.00 30.00 4'0.00
CENTRAL w/ t u
a) Central deflection _ b) U-displacement
:1
O O
"2 c5
'01 (01
c:
0:8 0::
EC)" B171
2 / 2
DJ ILJ
U U
o o
v-Q h-Q
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‘3 =2 f
'0.00 1300 , 311.00 4‘5-00 cb.00 1's.00 , SE00 4%.00
N N
c) Nx at center d) Mx at center

Figure 3.35. Plots of w,U,Nx, and Mx’ square plate, s-s, R = 10

116

 

 

 

 

 

 

 

 

 

 

T
\ -—Nx
l2 5‘ \ -----
o \
E . \
2 \ ..
h:
h- \ -v
:12: \ 1‘?-
1— 3.. z
5’ \ s
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1: \
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a) ‘s‘ \ H
U ‘s. O
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U
2
C
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I
E.
o I I r I I I I I
~ 0.25 0.50
(X/a)
a) NX 9ND NY HLONG RXIS Y=a/2 b) SENDING HOHENTS RLONG RXIS Yza/Z
.1 _
2‘01: 240d
‘2 ‘2
w :2
3 1001 m 160-4
K D:
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= a
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1x10) 14'

C) PRINCIPRL STRESS RLONG RXIS Y:a/2 leRINCIPRL STRESS VS. RPPLIED LORD
NOTRTION FOR CURVE IDENTIFICRTION. R) N;1.D IUNDEFLECTED) S) NE1.ID Cl NE2.4D .

FIGURE 3.36. PLOTS OF STRESS COHPONENTS.SOUHRE PLRTE.SIHPLY SUPPORTED.R=ID .

 

 

 

 

 

117
\1.25 LE1 5
l.
l.
.5

 

 

 

 

a) uuosrLecreo tn':1.1 In women (ML-1.101

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I0
I | J I i
I [L n d d d Efi
c1 00011.50 (ML-2.001 d, pm.
_1
9—9 8—8 C‘C D‘D E-E
2 1 - -
e) STRESS RESULTRNT PROFILES 1né2.001 SCH”:

FIGURE 3.37. CONTOURS OF IN-PLRNE FORCEINX) FIND PROFILES OF NX RT SPECIFIED
SECTIONS FOR LORD 2.00 TIRES THE CRITICRL LORD.SIHPLY SUPPORTED.R=ID.

118

 

 

 

 

 

 

 

 

 

 

 

 

.5 3 3.130 2 1. .5 o
a) Undeflected plate b) Buckled (N/Ncr = 1.25) c) Buckled (N/Ncr = 2.20)

Figure 3.38. In-plane displacement (U = ua/ti), square plate, simply-
supported, R - 10. -

119

3.1.3.3 ANALYSIS OF POSTBUCKLING RESULTS

3.1.3.3.1 SIMPLY SUPPORTED EDGES

 

a) Uniform stiffness plate:

Lateral Deflection

 

The plot of w-displacement versus edge load,Figure
(3.26aOshows a rapid increase in w immediately

after buckling, continuing in a smooth manner.

Consider the plate under combined loading (bending

— .__L.
Kb Aw center’

the slope of the curve is the instantaneous stiffness.

and membrane) and introduce

Before buckling, w at the center of plate is
increasing almost linearly. As the critical load is
approached, the stiffness, Rfi, begins to decrease
rapidly. Finally, after passing the buckling range,
it increases again and remains almost constant but less

than the initial stiffness.

This reaffirms that, in contrast to beams, the plate

has the capability of carrying load after buckling, but
with a lower equivalent bending stiffness. This be-
havior is in agreement with previous research and results

published.

120

Convergence of the Solution

 

To check variation of the solution with grid spacing, the
problem was solved with different mesh sizes and the w

results for each solution are shown in Figure (3.39).

Study of the plots leads us to the conclusion that:
i) the solution converges with decreasing mesh size.
a

ii) solution obtained with grid spacingh = 8 is reasonably

accurate for engineering purposes.

Investigation of membrane forces, bending moments, and
in-plane displacements also indicates convergence, but the
rate of convergence in bending moments is much greater

than the others. This is attributed to the fact that the

w function is very smooth.

 

 

 

= a/8

= a/12

-— r" h = a/l6
O
“3
04
D
c.’

‘0 .00 1'00 2'.00 3 .00 41.00

Figure 3.39. Convergence of the solution with mesh size.

121

In-Plane Stress Resultants

 

In the uniform stiffness plate, because of symmetry in
loading (hydrostatic), Nx and N& are equal to N every-

where before buckling.

After buckling, there is a change in distribution of the
in-plane resultants as illustrated in Figure (3.26),which

shows a tensile stress resultant at the center of the plate.
This can be explained as follows:} due to buckling, the

plate will deflectailarge amount - especially in this case with
free in-plane displacement and no restraint on lateral slope.
In this deflected position, the plate is analogous or similar

to a shallow spherical shell with horizontal load applied

all around the edges. _4,/”"'——_—_———““\\;

7 ‘

The variation of membrane resultants

agrees,qualitatively, with that of a shell. Shell analysis
shows that, due to this type of loading, N¢ (which is analogous
to Ng) along the axis, y = g, has its maximum near the edge,
decreases towards the center and sometimes becomes tensile

at points far from the edge as discussed in reference ( 7).

N6, the membrane resultant in the direction of the parallels,
will be the major load carrying element and will have values
much greater than N¢, the meridional resultant, near the
edge. In this case, Ny along the axis y 8 a/2 is the

counterpart of N in a shell and shows similar behavior.

8

Equilibrium was examined along different arbitrary sections

122

and the results were very good.

Referring to the paper by Kennedy and Prabhakara [27],
although the conditions are different, their results show the
existence of a tensile force at the center of a plate under
edge compression:1 Figure (3.26 c) illustrates the variation
of the membrane resultant Nx at the central point. It
shows that before buckling, in a flat plate or the small
deflection case, the compressive in-plane resultant is in-
creasing in proportion to the applied load. After buckling,
however, it will start to decrease (sharply in a flat plate
and smoothly in the small imperfection case) with an increase
in the edge load and subsequently the deformation behavior

of the plate is similar to that of a shallow shell.

Bending

Since the bending moment is a function of curvature, immedi-
ately after buckling, when considerable deflection occurs,
bending moments will be developed within the plate. The
Figure (3.27 b) shows the variation of moment along the

centerline, y = a/2, corresponding to different loading stages.

At the beginning, maximum Mi is at center, but as the
deflection increases and a flat region is formed around the

center, the location of Mg maximum shifts toward the edges.

k

1This phenomenon was studied by W.A. Bradley, (author's adviser),
using a frame-work approximation (or analogy) and the behavior proved
to be qualitatively in agreement with the above result.

Note

123

Principal Stress

 

Due to dependence of principal stress on both in-plane
forces and bending moments, it is varying with load both

in distribution and magnitude. Since maximum stress is one
of the major keys to design problems, the maximum principal
stress is plotted versus load and the point at which the
maximum is located is shown on Figure (3.27 c). It is

seen that due to edge loads appreciably above

critical load, the maximum principal stress is located

at the middle point of the edges in the postbuckling range,

while in membrane analysis it is uniform all over the plate.

The principal stress is a function of both membrane and
bending stresses and these two are functions of thickness
in different degree. Thus, the principal stress would vary
with reference thickness and the values given here are
valid only for this case, (i.e.,unit thickness plate with

E varying).

In—Plane Displacements

 

Study of graphs (Figure 3.29) shows that before buckling,
displacements are linear as expected (constant strain).
However, after buckling, geometrical nonlinearity caused

by w introduces some changes in their patterns. As
illustrated in (3.29), along edge x - 0, the u displacement

(normal to the edge) is increasing from the corner towards

b)

124

the center, which is consistent with lateral bulging. The

v displacement along the x = O edge (same as the u
displacement along the y a 0 edge) shows larger compressive
strain at center and smaller strain near corner; this is
caused by the considerable compressive membrane force in

this direction near the middle of the edge..

The in-plane displacement is illustrated in (3.26). It
shows that:
Before buckling, the edge displacement is linear

with respect to applied load.

After buckling, the rate of in-plane displacement u,

will increase at the corners. Define-A—N-————- - K the

m ’
corner

equivalent ineplane stiffness; it will decrease with in-
crease in deformation. Also, there will be much greater
reduction in IE; at the middle of the edge (x = 0, y = a/Z),
than at the corner.

Variable stiffness plate R = 1/10

Lateral Deflection

 

According to graph (3.40), in this case the lateral de-
flection w, follows the same pattern as in the uniform
stiffness case. However, when the lateral deflection gets
moderately large, a wide flat region occurs at the center
where the plate is stiffer and there is sharper curvature

near the edges, because of the smaller stiffness.

125

Membrane Resultants

Study of Figure (3.22, 23, 24) shows similar behavior in

membrane resultant as for case (a) the uniform stiffness plate.

Bending
Figure (3.23.b) shows bending moment behavior similar to

that of the uniform stiffness plate. The only noticeable
difference is that the location of the maximum Mk moment

is closer to the edge because of the greater curvature

near the edge, although the moment is a function of stiffness

as well as curvature.

Principal Stress

 

The maximum principal stress is found to be located at a
point close to the edge in the postbuckling range; whereas,
before buckling and during the first stages of postbuckling,

the maximum stress was at the center.

In—Plane Displacements

 

The graphs of Figure (3.25) show that along the edges, the
displacement normal to edge is almost constant, as in the
uniform stiffness case. As we move towards the center,
however, we observe a change in the pattern which can be
explained on the basis of the force distribution and the
stiffness. Along sections parallel to y, near the center

x z a/2 and moving toward the edges, the reduction in stiffness

C)

d)

126

is greater than the reduction in the membrane forces.

Therefore, there is less displacement at the center and

larger displacement near the edge.
As a quantitative example, as seen in Figure (3.24.a), along
a

a a
the section, x a-Z, at point (x = 23 y = 29’ Nx = 1.12N,

and the in-plane stiffness Et 8 ’325(ET)r° At point

. %3 y a 0) on the edge, Nx = .64 while Et 3 .1(Et)r.
Thus:
N
e x 3 1.12N g_ a)

N .
3.446 Etr at p01nt (

x = i? .325031)r 4 ’ 2

x Et .1 Et Et

N
e = —- = —- = 6.4 —- at point (2 , 0) on the edge
r r 4

(Note: Poisson's ratio will not cause a major change.)

After buckling, there is also a change in displacement normal

to the edge, similar to that explained in case (a).

R-1/2
This case is between cases (a) and (b); thus, we expect
the results to be so. All results obtained do lie between
the two previous cases and exhibit behavior already
explained.

R = 10

Lateral Deflection
The graph of Figure (3.35a) shows the general behavior of the
center deflection to be similar to that of the uniform stiff-

ness case; for the lateral deflection along the center line,

127

Figure (3.40) shows an expected difference compared to previous
cases, in line with the stiffness variation. There is greater
curvature in the central region in contrast to case (b), which

has a flat region in the center.

Membrane Resultants

 

The overall effect of large deflection after buckling is

similar to that in previous cases.

Principal Stress
The maximum principal stress always occurs at the middle

point of the edges.

Bending

Studying graph (3.36 b), we observe that M.x maximum, occurs at
a point closer to the center, where considerable curvature

takes place.

In contrast to previous cases, My reaches its maximum not
at the center but at some midpoint. This is because

the stiffness is minimum there and My is a function of

both D and curvature, although the curvature in y-direction

is maximum at the center.

128

In-Plane Displacement

Before buckling, in contrast to case ( b), we observe a
smaller u—displacement in the vicinity of the centerline
(y I g). This is consistent with the in-plane force dis-
tribution and the stiffness distribution as calculated in

case (b).

After buckling, the in-plane displacements due to N are
increased by the effect of large lateral deflection; also,
the region having more closely-spaced contours is moved

away from the center toward the edge.

 

 

 

O

C?

“’1

3

c, X/0

:0 .00 01.25 01.50 0J.75 500
- 10
- 1

S - 1/2

«7 - 1/10

Figure 3.40. Deflected shape for various R values.

Simply-supported edge; N/N a 2.40.
cr

129

3.1.3.3.2 CLAMPED EDGES

The same stiffness variations discussed for plates with
simply-supported edge are considered in this section, but
for plates with clamped edges. The corresponding plots are
given in Figures (3.41) to (3.56), which show the displace-
ments and stress resultants. The behavior is largely
similar to that for the simply-supported plate; notable

differences are mentioned in the following.

In-plane Stress Resultant

Overall behavior of the in-plane forces is similar, but in
the simply-supported case they decrease more rapidly at the
center. For example, compare Figure (3.26c) with (3.45c)
for the uniform stiffness plate. The plate with simple
support shows a much greater drop in Nx at the center

corresponding to edge load of two times the critical load.

Bending;Mbments

Bending moments exhibit the greatest differences between
clamped and simply-supported boundary conditions. Their
magnitudes and distributions are appreciably different
as a comparison of Figures 3.22 d and 3.23 b with

Figures 3.41 d and 3.42 b, for example, shows.

130

Ineplane Displacements

The in-plane movement of the plate in the postbuckling range
follows almost the same patterns for the simply-supported
and clamped cases. However, the plate with clamped edges
undergoes smaller displacement along the edges. Compare

Figures (3.25) and (3.44).

131

 

 

 

 

 

 

 

 

 

 

 

 

g 8
g 3.
' 8
8‘ -‘ B
8 3.
o'. "H
*2” *2“
8 a
21 81
A
8 a
8‘ 3‘ B ‘1
8 3.
a) . no 1'. on 2'. no 3'. on ‘2 00 ch 0 00 ST. DD 1'. .00 2" «DO SE a DO
CENTRRL w/ti U
a) Central deflection b) U-displacement
D
g 0.
531 81
D
ad? 0:9
w .4 IUD-1
hv‘ P-w
Z 2
U IJJ
U U
o 8
F- “'-
c:‘3_ cue.
XD (0 I
12 It I
o I
o D I
.1 9 1
'0.00 40.00 , 00.00 120.00 c11.00 4'0.00 , 50.00 1120.00
N N
c) Nx at center d) Mx at center

Figure 3.41. Plots of w,U,Nx and Mx’ square plate, clamped, R = 1/10

132

120.

 

 

IN-PLHNE STRESS RESULTHNT Rnrto (i 1

 

 

 

 

-120J

a) NX RND NY HLONO HXIS Y:a/2

D) SENDING HOHENTS RLONO flXIS Y:°/2

 

 

 

 

- 1

780.4 750‘

C
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(X/a) N'

C) PRINCIPHL STRESS RLONO BXIS Y=O/2 leRINCIPBL STRESS VS. 8?? [50 L080

NOTRTION FOR cunve Incurtrrcnrxou. n) N21.o (UNDEFLECTED) a) «21.10 C) "$2.50 .

FIGURE 3.42. PLOTS OF STRESS COHPONENTS.SOURRE PLRTE.CLHHPED. R=l/10 .

133

 

 

 

 

 

 

 

 

 

 

1.0
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(fl 1.5 J/I‘,.aaazrr—A ‘
a) UNOEFLECTEO (N;1.) b) auaneo (u;1.3oz

WHEEL

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r?)

 

 

 

 

c) aucheo (hé2.4on d) PLRN

 

 

 

 

 

 

 

O) STRESS RESULTRNT PROFILES (N;2.40)

FIGURE 3.43. CONTOURS OF IN-FLHNE FORCEINX) RNO PROFILES OF NX RT SPECIFIED
SECTIONS FOR LORD 2.40 TIHES THE CRITICHL LORO.CLRHPEO. R=1/IO .

134

 

 

 

 

 

 

 

 

r——-—"

 

 

 

 

 

 

.5“ .1 o

 

\( n

a) Undeflected plate b) Buckled (N/Ncr = 1.25) c) Buckled (N/Ncr .:2,25)

Figure 3.44. In-plane displacement (U = ua/ti), square plate, clamped
edges, R - 1/10.

135

 

 

 

 

 

 

 

 

 

 

 

 

8 8
6 6 A B
8* fij
. 8
°‘ SJ
8‘ ..
O O
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=2 s
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8* ‘3‘ B"
8. 8.
“13.00 1100 2200 3100 4100 “13.00 0100 $0.00 €4.00 €2.00
CENTRRL w/t1 U
a) Central deflection b) U-displacement
O
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0.00 40.00 N‘ 00.00 120.00 °0.00 40.00 , 00.00 120.00
N
c) Nx at center d) Mx at center

I?igure 3.45. Plots of w,U,Nx, and Mx’ square plate, clamped, R = 1

136

 

 

 

‘--—-NX
A 3“ _____ NY
'5 \
c, d \
t: \
a: x —.
z :3
"5 s
5' a:
co :3
u: As:
2.“ a
a 1g
53 u:
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O) NX RNO NY RLONO HXIS Y=al2

 

 

 

 

 

b) nanotuo nonsurs atom nxxs 7:0/2

 

 

0001 000.

000.. 000.
b 2

00 ' an '
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I
tX/a) N

d)PRINCIPRL STRESS VS. HPPLIEO L030

0) PRINCIPRL STRESS HLONO RXIS Y=°V2

NOTRTION FOR CURVE IOENTIFICRTION. H) NEI.O (UNOEFLECTEO) S) N;1.10 Cl Né2.SO .

FIGURE 3.46. PLOTS OF STRESS COHPONENTS.SGUHRE PLHTE.CLRHPEO. R=1 .

 

137

 

 

[

r—H
ooxooor-a

 

 

 

 

/

I
a) UNOEFLECTED (~21.) b) aucxtao (N=1.201

Ma

 

 

 

 

 

 

(Lg

 

t—‘i—‘NNW
O MOLDOUIOJ

 

 

 

 

 

 

 

 

 

 

(fl c. ||

nfd

c) aucheo (ué2.90) d) PLRN

D.—_...
m

 

 

 

 

 

e) srness RESULTRNT PROFILES (Né2.90)

FIGURE 3.4“]. CONTOURS OF IN-PLRNE FORCEINX) RNO PROFILES OF NX RT SPECIFIED
SECTIONS FOR LORD 2.90 TIHES THE CRITICRL LORD.CLRHPED. R=1 .

138

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1. .5 o. 2. 1.5 1. .5 o. 20. o
a) Undeflected plate b) Buckled (N/Ncr = 1.10) c) Buckled (N/Ncr =2.90)

Figure 3.48. In-plane displacement (U = ua/ti), square plate, clamped
edges, R = l.

139

 

 

 

 

 

 

 

 

 

 

 

 

8 8
é a
S“ 9‘:
A B
D
s O.
.8 .8
D O
O D
8‘ {2‘
I: ‘2
O D
s s
21 SW
D D
° e B .
31 3‘ I
3. 3
°0~00 1'.00 2300 3100 4'.00 P0.00 3100 £0.00 €4.00 $.00
CENTRRL w/ti U
a) Central deflection b) U-displacement
' o
D D
D. a
on 8T
0 D
Go MC:
uJ 0.. mo.
hi" y-0
z z
m w
Q U
o 8
Ho I—O
<58- 88
l2 D:
O
D D
.5 9‘ /
l T j— I 1 1
'0.00 40.00 , 00.00 120.00 °0.00 40.00 , 80.00 120.00
N N
c) Nx at center d) Mx at center

Figure 3.49. Plots of w,U,Nx, and Mx’ square plate, clamped, R = 1/2

140

 

 

 

 

 

 

 

 

 

 

 

 

___Ex
A 3.1 -—---NY
IE
3
.—
c a
m: |=
.— up
(D
g z
.—
" 1%
a a
N” Z
“ a
a a
O
E a
C) D
U
z
55 0
fi'
3.
-t‘ 'IOOd
a) NX HND NY RLONG RXIS Y=°/2 b) SENDING NONENTS RLDNG RXIS Y=°/2
' 7
000. 0004
‘2 "‘ .2
U)
a ‘00- g ‘0 .1
25 £5
0: C 0:
_l .1
8 ‘ § ‘
8
E 200.. a E Md
0 ‘ 0
' r ' 0125 ' ' T 0.150 0.0 ‘ {.0 T 210 ' 330 ' 4.!)
(X/a) N'

NOTHTION FOR CURVE IDENTIFICRTION.

c) PRINCIPHL STRESS HLONG RXIS Y=°/2

n) N21.0 (UNDEFLECTEO)

leRINCIPflL STRESS vs. RPPLIEO L000
0) «21.10 0) "£2.40 .

FIGURE 3.50. PLOTS OF STRESS COHPONENTS.SOUHRE PLRTE.CLHHPED. R=1/2 .

141

 

 

Fag—ops
.00 o
r—‘N

 

 

 

x)?

 

 

 

 

 

 

 

 

 

 

 

 

a) UNDEFLECTED ("£1.) b1 aucuteo (Nat-30)
R a (i C Eq-
‘\\::::1 3'? I I
(gm
1.5
1.
.5
rd 0.

 

 

c) auaneo (uéz.so) d) PLRN

 

 

 

 

 

 

 

o SCRLE
e) STRESS RESULTRNT PROFILES IN=2.SO)

FIGURE 3.51. CONTOURS OF IN-PLRNE FORCEINX) RND PROFILES OF NX RT SPECIFIED
SECTIONS FOR LORD 2.50 TIHES THE CRITICRL LORooCLRHPED. R=I/2 .

142

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.4 .1 0 3. 1. O 20 O

a) Undeflected plate b) Buckled (N/Ncr = 1.30) c) Buckled (N/Ncr=-2.50)

Figure 3.52. In-plane displacement (U a ua/ti), square plate, clamped
edges, R - 1/2.

 

143

 

 

 

 

 

 

 

 

 

Figure 3.53.

 

 

8 8.
,8. 8;.
A B
o 8
9 6
81 3‘
f5 D
9 9
1 ID.
* 8 ‘ p
:2 :z
8 8
.3: g.
s s A
6. 6. .
N N
8 8.
cb.00 1100 2100 3100 4100 cb.00 0100 13.00 21.00 51.00
' CENTRRL w/ti U
a) Central deflection b) U-displacement
O
8 =2
«1 81
c:
353 m9
h€"q Efig‘
z
m E
U L)
c:
*8 1-‘5’
.c‘éq $3..
5: u:
on
Q o
7 . =2 /’———
‘ jfi 1 1
0.00 40.00 N‘ 00.00 120.00 °0.00 40.00 N, 00.00 1120.00
c) Nx at center d) Mx at center

Plots of w,U,Nx, and Mx’ square plate, clamped, R = 10

IN-PLRNE STRESS RESULTRNT RRTIO (N I

mucxrm. STRESS 10')

144

---JNY

I

SENDING HOHENTSIH)

 

 

 

-I-

 

a) NX RND NY RLONG R/IS Y:°/2

 

ODD-1
b
8
4100... g
4..
U)
_l
"‘ I
D...
H
U
2
n
200.. E
D A
'1: T I 0'15 1 l 1 D ho
(X/a)

C) pRINCIPRL STRESS RLONG RXIS V=OIZ

 

 

200..

 

b) SENDINP HOHENTS RLONG RXIS Y=9/2

 

I

 

I I I

I

N
leRINCIPRL STRESS VS. RPP'IED LORD

110 3J0 ' 4.5

NOTRTION FOR CURVE IDENTIFICRTION. R) Né140 (UNOEFLECTED) S) N21410 Cl N;2.4O .

FIGURE 3.54. PLOTS OF STRESS COHPONENTS.SDURRE PLRTE.CLRHPED. R=10 .

145

 

 

 

 

 

 

1.25
1.0
1.0 .
' .75
fl '75 ' F15
7 .50

 

 

 

0) UNOEFLECTEO (N;1.) b) auanto («£1.1s1

R i (i EEG-

l l

 

 

Ot—IHNNW
U‘IOLflOUIO

 

 

 

 

 

 

1W

 

 

 

c1 aucheo 1N&2.401 d) PLAN

 

 

 

 

 

 

 

 

[ l
r/_\1

 

e) STRESS RESULTRHT PROFILES 1u22.401

FIGURE 3.55. CONTOURS OF IN-PLRNE FORCEINX) RND PROFILES OF NX RT SPECIFIED
SECTIONS FOR LORD 2.40 TIHES THE CRITICRL LORO.CLRHPEO. R=IO .

146

 

 

 

 

 

 

 

 

 

 

 

 

1 \

1.5 l. .5 0 2. 1. 0 10 g 0

 

 

a) Undeflected plate b) Buckled (N/Ncr = 1.10) c) Buckled (N/Ncr=-2.40)

Figure 3.56. In—plane displacement (U = ua/ti), square plate, clamped
edges, R = 10.

147

3.1.4 OPTIMIZATION

 

The main purpose of this study is to investigate the ap-
plication of the difference method to the variable stiffness plate.
The goal is to design a plate with a stiffness variation such
that it be optimum in.some respect.

For example, we may be concerned with weight or the amount
of material used.as is the case in most space structures, aircraft,
etc. In these cases, we are seeking the form of variation in thick-
ness of the plate for which the plate will be most efficient from
the point of view of either stress or displacement. Following
are some examples:

a) In case of a flat plate (membrane solution), we may

want to use a certain volume of material to construct
a plate which will result in minimum in-plane displace-
ment at some edge point, or in a minimum stress at some
point. It might also be desired that due to nonuniform
load, the stress be almost equal everywhere, or the

maximum stress be minimized.

b) In stability analysis, we may want to utilize a constant
amount of material to achieve maximum critical load,
or to minimize the center deflections immediately

after buckling.

c) In the postbuckling range, there is the possibility
of many different forms of optimization. For example,

corresponding to some combination of lateral and edge

148

loads within some range, we may wish to maintain an almost
uniform principal stress to prevent local yielding. Also,

it may be desirable to minimize the central deflection.

OPTIMIZATION EXAMPLE 1

 

Consider the square plate of

 

 

 

 

 

I. a 4:%
Figure (3.57) with thickness
varying from the edge toward
the center as shown. We __— ____. . x ____ ____
want to find the optimal &
slope (ratio of £5) so that
the critical load under bi- 1——->x
axial compression téI Itc
. t(x)

be maximum for a
Figure 3.57.

constant amount of material.

For a square plate of thickness, t, and sides, a, the total

volume of material used is
?'= azt-

For a variable thickness plate, if the variation in thick-
ness is linear, the thickness t(x) at point x is

t-t

c e
t(x) te + a/2 x

where

tC = thickness at center

149
te = thickness along the edges

I:
Let '32 = RT= Ratio of edge thickness to thickness at
c

center. Then:

_ £15
t(x) - tCERT + (l-RT) a

and the volume is

 

a
- 5. 2x 2x
v 4 f0 (a-Zx)tc(RI+-:r--RT:;)dx
azt
_._ 2 l + ZRT _ c
v - 4a tCG-jf§- )- 3 (l + ZRT). (i)

If the volume is limited to the original value, then

t —7I;2L__— ; for '3 = azti (ti = unit thickness)
c a (1+2RT)

3t1
t:c a (1+2RT) (11)

To maintain a constant volume, the center thickness must

vary with ratio RT, according to equation (ii).

For example:

 

RI tc/t1 v
2

.l 2.5 a T1

_1_ 2 11

2 2

1. 1. "

2 .6 II

150

It is evident that as the ratio, RT, and the center thickness,
tc, change, the in-plane and flexural stiffness of the

plate at each node will change.

A subprogram is PrOVidEd to compute center thickness and

stiffness at each node in each trial for a given RT.

The buckling problems were solved for a family of given
ratios, and the variation of critical load with ratio RT,
is shown in Figure (3.58). In this problem the plate is

simply supported along all edges and v = .316.

Graph (3.58 a) indicates that the maximum critical edge
load for a given volume of material occurs with RT : .2._
To find a more accurate value, the trial is continued with
finer intervals between RT = .15 and RT - .25. The larger
scale graph (3.58 b) shows that the'maximum

critical load is (N* )

cr imun = 25.66, corresponding

nl n
m

to RT= = .22. Thus, consider a simply supported square

C

plate under bi-axial compression.
From the stability point of view,the minimum material can

be used if RT ‘ .22 for the linear thickness variation

25.66-19.48
19.48

over the buckling load for a uniform thickness plate of the

introduced in Figure (3.57). An increase of - 31.6%

same volume results.

151

11935
i) The result obtained above is not the absolute optimum.

For the given amount of material, a still greater critical
edge force could no doubt be obtained with a thickness
variation other than linear. This form was used for
practical simplicity.

ii) For other types of loading or boundary conditions,
appropriate forms could be proposed and analyzed by

trial.

EXAMPLE 2
The plate in this example is the same as in example 1,
but with clamped edges; results are shown in Figure (3.59) .
In this case, the maximum critical load is found for an RT
of approximately 0.8.
Figure (3.59 b) is obtained by taking finer intervals
(AR)?8 .01) between .6 and .8. This graph indicates the maximum
critical load is (Nzr)max. = 473.962, corresponding to RT =.71.

51.20-49.56
49.56

over the plate of uniform thickness. In the case of fixed support,

The results show an increase of = 3.3%

from the stability point of view, it is not worthwhile constructing
-a plate of variable thickness. The effect of variation in thickness
on other aspects of the problem such as internal forces, bending

moments and lateral deflection will be analyzed later in this chapter.

152

28.00 32.00

4.00

25.1

20.00

 

c115.00

.00 0.40 0180 11120 1160 2100 2140

a)

26.80

2§.SO

 

 

.12 0116 0120 0124 0128
RT

c26.40

b)

Figure 3.58. Critical load vs RT,for a simply supported square

plate under Bi-axial compression, N.

153

9-00 47.00 55.00

31.00

 

c23.00

.20 1100 1180 2160 3140 4120 43100
RT

53.80

53.40
1

53.20

 

 

.50 0165 0170 0175 0.00
b) RT

c153.00

t
Figure 3.59. Critical load VS RT= 7?-, for square plate of constant
C

volume under Bi-axial compression;clamped edges.

154

3.1.5 SUMMARY

 

In Section 3.1 a method was developed for force boundary
condition problems, and the related computer program was applied to
a variable stiffness plate and the results were discussed extensively.

It seems necessary to emphasize that the plates discussed
in Section 3.1.1 through 3.1.3 have uniform thickness with varying
E, so that patterns of membrane and bending stresses follow exactly
the pattern of in-plane forces and bending moments respectively.
Solutions to similar problems are not available in the literature;
however, comparison with the uniform stiffness plate,as a special
case, supports the accuracy of the solutions. Convergence
of the solutions with an increasing number of nodes strengthens con-
fidence in the method.

In Section 3.1.4 the weight-saving advantage of a variable
thickness plate,from the stability point of view, was demonstrated as
an example. However, one can apply the optimization to any possible
aspects of stress or strain as desired.

Figure (3.60) shows variation of central deflection with load
for all cases. The graphs show that the plates with less stiff
edges undergo larger lateral deflection because,in the postbuckling
range,the main portion of the in-plane load is carried by portions

of the plate near the edges.

155

 

0.00

 

400

I

‘b.00 0100 110 .40 0120

D 2
CENTRHL w t 1
a) s-s

100.00
A

00.00

094m

1

40.00

20.00

 

 

cp.00

0100 4100
i

.00 1100 2100
CENTRRL w / t
b) Clamped

F
iiiure 3.60. Central deflection for different R values, s-s and

clamped edges.

156

3.2 DISPLACEMENT BOUNDARY CONDITION

In this section, problems with specified in-plane displace—
ments along the boundaries will be considered.

A.computer program hasbeen developed to solve this type of
problem based on the procedure discussed in section 2.3.

Plates with uniform thickness are analyzed, as well as variable
thickness plates. The results are compared with previous works or
exact solutions when they are available. As an example, optimization
of the thickness variation, from the stability point of view, is

also considered.

a - Geometry and Loading;Conditions

For an example, let us take a plate under uniform edge
displacement due to thermal load, and examine the membrane, buckling,
and postbuckling behavior of plates of different thickness variation,
‘with both simply-supported and clamped 'boundaries. Figure (3.61)

shows a plate, surrounded by a

 

 

 

 

 

 

 

 

g 0 .4

rigid frame that undergoes a r‘ ’T
////////j//Z I "

temperature change of either / /

/ /
expansion or contraction. g a

/

/

The strain in the frame will a 5

M /
impose a state of displace- 5 j

/ /
ment on the plate edges. //////[///[]/ I
The strain in the frame is

 

s: = 011(A'1‘)
Figure 3.61. Plan

157

where aT= coefficient of thermal expansion of the frame and

AT a temperature change.

Egg
1) It is assumed that the effects of the reaction forces
of the plate edges are negligible, so that the stresses
created within the frame have negligible effect on the
strain and displacements of the frame.

ii) Because of symmetry, we analyse only a quadrant of

 

 

 

 

the plate.

31 __32___33___34 35 36
r' I” T-_'T—_‘
: 1 I I l i

301__2 22| 2l 24 25
I
1
|

29L__29 13 14 15 lb
1
l

28i..__12 17- 7 8 9
|
|

27 I 18
r—-- 11 6 3 4
l
I

264 17. 10 5 2 1 _

 

 

 

 

 

 

 

Figure 3.62. Node arrangement, square plate, h = a/8

Since, because of symmetry, the centerlines of the
plate will not move, displacements enforced along the
edges will be as follows:

1 - Along edge x - 0, u displacement would be constant

0.

and equal to - ETaAT.

Table 3.10 .

2 — Because of constant strain (in frame),

158

the v-displace-

ment along x-edge will 'be linear with respectto

y, so that v =0TAT(y - g).

The boundary displacements are tabulated in Table

(3.10).

Boundary displacementfbr square plate of Figure 3.62

 

 

Point u/-JamTAT v/- aaTATfi

__7r—_ “3F“
17 1. 0.
18 1. .25
19 1. .5
20 1. .75
21 1. l.
22 .75 1.
23 .5 1.
24 .25 1.
25 0. 1.

Based on the given data,

analysed step-by step in the following pages.

the problem is solved and the results

159

3.2.1 MEMBRANE SOLUTION

 

In the solution for in-plane forces and displacements,
assuming zero w—displacement, the in—plane equilibrium equations
can be approximated by finite differences to obtain a set of linear
equations in u and v. (See equation (2.43)).

The equations are solved and in-plane forces and displace-

ments lead us to following conclusions.

a) In case of a uniform thickness plate, the equilibrium
equations (2.16) and (2.17) are linear differential
equations, and in this symmetric case the solution leads
to exact values of membrane stress resultants and dis-

placements, even with very coarse mesh sizes.

The theoretical solution predicts constant strain in both

 

" directions.
-uo -2uo a
.y = 8.. = .72 = T W (81> : 8)

and the in-plane stress resultants are

 

 

 

 

 

 

 

-2uo _ 110
N = N a Etz [e + ve ] = Et [(1+v)(————)] = ZEt --
y X l-V x y 1_v2 a (l-v) a
* 32 -2Eti no 82 (1-V2)12 “ca
3 N —— = __ = _ _—
Nx x Do (l-v) a. ( 3 ) (1+v)(24) t2
Eti i
uoa
or, calling -—§-= UO
‘1
*
Nx
U0 = -24(l+v) *
N
and for v = .316, x = 31.58

C.‘

O

160

Similarly,
N:
I = -31.58
N = 0
X?
and
a 2uO a
u = u(x) = 8(3 -x) = T (-2- -x)
a 2v0 a
v = v<y> = e<§ -y) = —a— (5 -y)

These values agree exactly with the solution obtained with
the computer program listed in Appendix:C. Contours of the membrane
force, N* , the principal stress and U-displacement are shown in Figure

(3.63). N& and v-displacement can be obtained considering symmetry.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N; = 31.58 o'= 31.53
U0 U0
1. 25 50 25 o
0) 11151101111111: 18011c1:.11’;‘{/Uo b1 911111019111. STRESS o'/U° c) U-OISPLRCEHENT

FIGURE 3.63.CONTOURS OF HEHSRRNE FORCE,PRINCIPRL STRESS,RNO U-DISPLRCEHENT
UNDEFLECTED SDURRE PLRTE. RT=I.

161

SOLUTION OF PLATES WITH VARIABLE THICKNESS

To observe the behavior of a variable thickness plate,two

different problems were solved. One with RT = .25, and the other had

RT - 2, and the results studied. Parameter RT =

edge thickness
center thickness

is the thickness ratio.

b)

Square plate with RT 8 1/4

Figure (3.64) illustrates membrane forces, principal

stresses and in-plane displacements;from the figures, we can

conclude the following.

i)

ii)

iii)

Figure (3.64 a) shows shifting of the load carrying
toward the center where the plate is thicker. The
shifting of the load becomes greater as we move
toward the center, and it is nearly uniform along
the edge.

Although there are larger in-plane stress resultants
in the central region, because the thickness is
large,the stress is smaller. Figure (3.64 b) shows
the principal stress within the plate. It is
observed that the minimum stress exists at the

center and it becomes larger toward the thin edges.

In Figure (3.64 c), the wide-spacing of contours of in-
plane displacement U in the central region, indicates
small strain corresponding to smaller stress in that

region.

 

 

20

25

 

,30

[1..

 

 

 

01 1151101111112 FORCEM; /Uo

162

 

 

50 40

b1 PRINCIPRL STRESS 0'71]o

Q—SO
4O

30
.25

r—jZO

 

 

 

 

 

l. .75 .50 .25

c1 U-D ISPLRCEHENT

FIGURE 364. CONTOURS OF HEHBRRNE FORCE.PRINCIPRL STRESS.RNO U-DISPLRCEI‘IENT

UNDEFLECTED SDURRE PLRTE. RT=I/4

c) Square plate with RT - 2.

The same problem is considered except with

RT . edge thickness

center thickness

are shown in Figure (3.65).

- 2. and the results obtained

following conclusions.

This leads to the

1) Supporting discussion ofthe preceding section, less

load is carried by the thin central region, and as

we move toward the edges, more load is transmitted.

11) Figure (3.65 b)

region.

shows larger stresses in the central

iii) Contours of in-plane displacements are consistent

with part (iix i.e.,larger strain occurs in the central

region,due to larger stresses.

Clearly, the out—of-plane support condition has

no effect on the membrane solution of the plate.

 

163

It should be noted that Figures (3.63), (3.64) and (3.65)
correspond to plates with RT = 1, 1/4 and 2, respectively,

but with thickness such that total material volume is the

same in each case.

 

 

 

 

 

 

 

 

 

 

 

//’-_______.——~*—135 I‘I““‘30
30 35
A 25 , f 40 _
1 . .75 .5 .25

 

 

01 1151101111111: F°R°E"‘;/Uo b1 PRINCIPHL STRESS o"/U° c1 u-msrmceneur

FIGURE 3.65.CONTOURS OF HEHBRRNE FORCE.PRINCIPRL STRESS.RND U-DISPLRCEHENT
UNDEFLECTED SDURRE PLRTE. RT=2.

164

3.2.2 BUCKLING SOLUTION

 

As is discussed in Section (3.1), the buckling solution
is based on the eigensolution of the out-of-plane equilibrium
equation (2.40) (h1the right hand side of this equation,the effect
of w on in-plane forces is considered to be zero. Thus, in the matrix
[Bw] of Equation (2.49), the in—plane forces obtained from the

membrane solution of the unbent plate will be used. The eigenproblem

 

will be solved, giving the critical edge displacements and the
corresponding buckled mode shapes.

Following-are solutions to some stability problems.

3.2.2.1 Convergence Check and Comparison

In order to check the convergence and consequently the
accuracy of the buckling solution, a uniform thickness plate
(problem 3.2 a) was solved using different mesh sizes. The critical
loads obtained in each solution are shown in Table (3.11) for
s-s boundaries. Also,the values of critical displacements are
compared to exact values.

Convergence of the solution is graphically illustrated in
Figure (3.66) for s-s boundaries.

Table (3.12) and graph (3.67) show the convergence pattern
of critical displacement for the case of boundaries fixed against
out-of-plane displacement.

The convergence tables show that the result obtained using
h 8:- in the simply supported case and h - {12- in the clamped

case are sufficiently accurate for engineering purposes.

 

 

165

More accurate results can be predicted by extrapolation.

The critical edge displacement obtained from extrapolation of the

last 3 lines has an accuracy of £H)0016 in the simply-supported

case and .0013 in case of clamped edges, compared to exact results.

Table 3.11 .

using different mesh sizes.

Critical displacements for a simply-supported plate
Uniform stiffness plate.

 

 

 

 

 

 

 

 

 

 

 

v = .316.
Mesh size Present solution Exact(1) Difference Two point 3 point
(h/a) Uo % extrap- extrap-
ér olation olation
1- 5935 5
4 .
.6248
§ .61698 1.2 .624972
.62497 .62495
1
"1‘2— “62141 5 .62498
.624975
1
16' .62297 .3
ID
0
61
. . . . Exact
O
(D
o.‘ //_——‘
‘2
3
O"
8
“11.00 4100 0100 12.00 {0.00 a/ h

Figure 3.66.

Convergence of buckling solution.

 

 

 

 

166

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Table 3. 12 . Critical displacements for a clamped square plate
using different mesh sizes. Uniform stiffness. v = .316.
Mesh size Present Solution Exact (1) Difference 2 point 3 point
(h/a) Ufi Z extrap- extrap-
r 'olation olation
% 1.34062 19.2
1.6456
_1_
8 1.56939 5.4 1.65620
1.6593 1.6550
12 1.61697 2.5 1.65709
1.6566
1. 1.6343 1.5
16
2
£1
Exact
ID
"3.
2
3
ID
‘3 .
“0.00 4100 0100 12.00 16.00 a/11
Figure 3.67. Convergence of buckling solution.
(1)The exact value of critical displacement is derived on page 167.

 

 

 

 

 

buc‘:
pla
all

for

bu

V
t:

 

 

 

167

Since it is assumed that the plate is perfectly flat before
buckling, the in-plane displacements are linear within the entire
plate (constant strain), and the membrane forces will be the same at
all points. Thus, critical displacements can be related to edge

forces as

along x-edges

1
ex Et [Nx - vNy]

but in this case

 

-u
cr _ g
(Ex)cr a/2 ’ and Nx - Ny Ncr
therefore
-a
ucr 2Et [l-VJNcr
anDr
For s-s square plate under bi-axial compression Ncr = --§—-.
a
[see Table 3.8]
Thus
u :3 l2 D— (1-\,) = i112..— .
cr a Et 12a(l+v) ’
u §__= .822467 .
or t2 1+V
.822467
Uo = -I:;- or for v = .316, UO = .62497
cr cr
nzDr
similarly, considering Ncr = 5.31 2 for clamped boundaries
a

[see Table 3.9 ],the critical edge displacement is

Uo = 1.6593 for v = .316
cr

 

"|\I

.
l
:2:—

I"

U'fi

 

 

 

 

 

 

168

EXAMPLE 1
Tofurther check the accuracy of the method, the solution
to problem (9.13) of Reference (44 ) is examined. In this example,

along the edges x = 0, and x = a, the

 

u-displacement is prevented. In the

y-direction, constant strain, a,
is assumed. Therefore, v = 6(3 - y) and
assuming that centerlines of the plate

coincide with axes of symmetry

during deformation, (6 a 353-. /

--_-——————

 

 

 

No u-displacement is
Y

allowed alon ed es = 0,
g g y Figure 3.68. Plan

and y = a, and with regard
to out of plane displacements, all edges are taken as simply supported.
The boundary data.are tabulated below. For the node arrange-

ment, see Figure ( 3.62).

 

Node u v w
17 0 0 0
l8 0 .25vo 0
19 0 .5 v0 0
20 0 .375vo 0
21 0 v0 0
22 0 v0 0
23 0 v0 0
24 0 v 0

25 0 v 0

 

 

169

Using the data for input, the buckling solution was obtained.

The critical uniaxial edge displacement was found to be
2
v0 = 1.23033 Ea— .
cr

This is to be compared to Timoshenko's results (44), obtained by

au1 energy method. Remember that he was taking the total edge strain to be
2v
ey, while in our case the compressive strain e = ——9

9
cr a
2

e . 2v = 2.46066 L.
cr __.°cr 2
a a 2

. _ h
Timoshenko found ecr - .632 —a—2 but he used a plate

 

of sides 2a=§'and thickness h. By substituting t. for h, and

-' 2 2
éfora in Timoshenko's result it becomes e = .632 h— - .632 t—-—-— =
2 2 cr 2 - 2

1: «'8 (a/2)
2'528122 . The results differ by 2.6%, fairly close

a
for such a coarse mesh size (h = 3).

3.2.2.2 Optimization Analysis

Using the same data as in problem 3.2.a, and solving for
the eigenvalues, two series of solutions are obtained for simply-
supported and for fixed out-of-plane boundary conditions—considering
different thickness variations. The variation in thickness is
taken to be linear, as in Section 3.1.4. The variation in thickness
is to be optimized in order to give the greatest edge displacement
at buckling for a given volume of material.

In the case of the simply-supported edge, Figure (3.69 a)

shows that the optimum variation corresponds to

a edge thickness
center thickness

 

RT = .15, with an increase in buckling displacement

 

170

0.80 0.90

0.70

 

 

O

9

a.

C3

“3

c13.00 0140 0160 12120 1160 2100 572140

a) Simply supported

2.00

1-75

1

.50

fl

 

 

ID
‘1‘
0
°.
'b.00 0140 0160 1120 1160 2100 2140
RT
b) Clamped

Figure 3.69. Variation of critical displacement versus RT.

 

 

171

0.65 0.70

0.60

D

0.55

 

 

.00 4100 0100 .00 16.00 20.00 21.00

cp.60

a) Simply supported

1-40 1.60

.20

D...‘

1.00

 

 

 

.00 4100 6100 .00 16.00 20.00 21.00

cp.60

12
RT
b) Clamped

Figure 3.70. Variation of critical displacement versus RT.

 

 

172

of '8119igiiél6 = 33% with respect to uniform thickness plate

of the same volume.

 

Figure (3.69 b) illustrates the variation of critical displace-
ment with respect to thickness variation for a clamped square plate.

The maximum critical displacement can be achieved if the edge thickness

1.589-1.569
1.569

over the critical edge displacement for the plate of uniform thickness.

is .85 times the center thickness with an increase of a 1.2%
To check the buckling behavior of the plate, within a wider

range of variation in thickness, the critical displacement for

different RT values up to 20 is obtained and plotted in Figure (3J0 );

the plot shows a decrease in critical displacement all the way up

‘to RT = 20,

Note: Examination of mode shapes shows that as the center of the
plate gets thinner, beyond some point, the buckling mode associated
with the lowest critical displacement is not a single concave shape. The
Ibuckling mode corresponding to the lowest eigenvalue for a clamped plate
with RT > 2.2 consists of more than one buckled wave. These results are
not shown in this thesis.
3.2.2.3 Analysis of Buckling_Modes

In this section, the shape of buckling modes will be reviewed
and compared with the exact shape in those cases where the exact
solutionsare available.

Up to this point, all solutions were based on the assumption
of symmetry about both axes of the plate. Thus, only a quarter of the

plate was considered.

 

173

Therefore, the nonsymmetric modes are missing. To obtain

all modes of buckling, a solution is obtained by considering every

node as an independent degree of freedom, from the out-of-plane displace-

ment point of view, while symmetry and anti-symmetry in u and v

are assumed, as before. Solution for the simply-supported plate

of uniform thickness shows the first mode shape to consist of half-sine

waves in both directions, and the second mode to consist of two

half-sine waves in one direction and one in the perpendicular direction.
The first few mode shapes for the simply-supported plate

are shown in Figure (3.71) and for clamped edges in Figure (3.72) .

 

174

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+
+ + -
0* - 619 U* - 1 51 0* - 1 51
1 ' ' 2 ‘ ' 3 ‘ °
+
+ —
+ — +
- +
0* - 2 458 0* - 2 75 * - 2 95
5 - . 6 - . U6 - .

Figure 3.71. Buckling modes of simply supported square plate, RT= 1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0* - 1 56 * 2 58 * 2 58
1 - . 02 — . U3 - .
.1.
+
+ —
- + +
0* - 3 55 0* - 3 94 * - 4 3o
4 — . 5 - . 06 - .

Figure 3.72. Buckling modes of clamped square plate, RT = 1.

 

175

3.2.3 POSTBUCKLING

 

In this section, a study is made of the postbuckling behavior

of uniform and variable thickness plates under uniform bi—axial edge

displacement compressed beyond the critical displacement. Convergence

of the solution is examined by solving the problem with different

mesh sizes. Variation of in-plane forces as well as in-plane displace—

ments is also studied.

3.2.3.1 Uniform Thickness Plate, s-s Boundaries

 

A simply-supported uniform thickness plate is compressed

beyond the critical displacement. and the solution is obtained.

Following are some results from the solution.

8)

anvergence of the solution

 

The problem is solved,successively taking

h/a = 1/4, 1/8, 1/12 and 1/16, and the results

are compared. Table (3;E3) shows the central deflection
of the plate due to edge displacements of 1.26 E;—, which

is almost two times the critical displacement.

Table 3.13 .

176

Convergence of postbuckling solution with mesh size

 

 

Mesh size IN-center Difference % Extrapolation N:-center Mgécenter
(h/a)
%- .96795 -118.99 100.80
9.7 .853016
%- .88175 -13o.17 98.55
1.5 .8643366
%5- .86869 -132.98 98.27
.1 .8701166
%g" .86976 -136.74 98.88

 

 

 

 

 

 

 

Review of Table (3.13 ) indicates that the solution obtained

by only 8 x 8 nodes is satisfactorily close to the converged solution,

keeping in mind that the accuracy of the iterative solution is set

to be one percent.

is small.

Figure 3.73.

O
O

.11

0.96
v

l

l

 

 

dues

I

4.00 0100

12.00 10.00

The results can be improved by extrapolation.

a/h

The difference for mesh sizes finer than 8 x 8

Convergence of center-deflection, s-s square plate.

(U = 2 U
c

r)

 

177

 

 

 

 

 

 

 

 

.’—--~ —_ Nx
_God \\\ ----NY
“ \
g; \\
'2' ‘ \x ..
E ‘ '5
.1 \ a!
ELJN. \ 12
u: \ g;
a: \ c1
8 \ =
In . \ <3
:5. \ E
0) In-.. ------ \ g
g—zo. “~- 1. A
a." \”“-~--h .4 -
\
if x
. x
\
\
“~11;
‘—12 r T’ 1 I
0.25 0.50
(X/OI
0) NX RNO NY RLONG RXIS Y=0/2
90.
70 ..
b' '3
a) 'I .6
a: a:
E g
“' SO 3 r-
00 j 40
.1 .1
a: 4:
0. a.
0-0 _. H
c0 c0
12 12
H H
a: 4:
°- 00. ‘-
2A
10
I f I o :25 I I I 0 .ko
(X/O)

c1 Pmucmu. STRESS 610110 111110 1:012

 
  

.___.fix

12. ""’"Y
03
J

 

I
0.50

I I I I I r

0 .26
(X/al

5) SENDING HOHENTS RLONG RXIS Y=°/2

 

 

so.
10.
00.
-I
00..
d
10
0.0 '110 1 2'0 ' 030 ' 470
0

019911401991. STRESS v0. U

NOTRTION FOR CURVE IDENTIFICRTION. R) UEI.O IUNDEFLECTED) S) 031.20 C) 022.40 .

FIGURE 3.74. PLOTS OF STRESS COHPONENTS.SDURRE PLRTE.SIHPLY SUPPORTEO.RT=1

b)

e)

178

Stress Analysis

The distribution of the membrane forces will change

as the lateral deflection becomes larger.

Also,the bending moments will vary with variation of the

deflected shape of the plate.

Graphs of IFigures (3.74 a) and (3.74 b) show the variation of

center-line membrane forces and bending moments as the

edge compression varies. Study of these graphs leads

to the following conclusions.

1) As the center of the plate deflects transversely,

ii)

iii)

iv)

the in-plane load shifts with more of the load
being carried by the portion of the plate near the
edges.

The bending moment, which is maximum at the center,
is increasing with the increase in deflection w.
Along the centerline parallel to x, the membrane
force Ny is increasing toward the edge, while Nx
is almost constant.

Bending moments Mx and My both are maximum

at the center and dimimish to zero along the edges,

as expected.

Infplane displacements

 

The distribution of in-plane displacements in the plate

is shown in Figure.(3. 75) at different stages of loading.

179

Analysis of these leads to the following conclusions:
1) Before buckling, the plate is perfectly flat.
Since there is no w effect, in-plane displacements
are linear, as expected.

ii) After buckling, the plate will undergo more contraction
near the edges and the central region carries less
compressive load; less inrplane displacement
occurs there.

iii) As the load increases this phenomenon becomes more
visible, so that at large edge displacements, the
curves show very little u and v displacement

in the central regions.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.625 .468 .312 .156 0 J5 50 .25 .1. 0 1.5 1. .5 .25 0
a) Undeflected (U/Ucr=1'0) b) Buckled (u/Ucr=1'20) c) Buckled (U/Ucr=2.40)

Figure 3.75. In-plane displacement (U = ua/ti), square plate,
simply-supported, RT = 1.

 

180

d) simply—supportedJ square_plate, loaded by uniform

normal pressure

 

To check the reliability of the method by comparing with
previous results, a problem similar to one presented by

Levy is considered. In Reference (29), Levy has solved

a square plate under a uniform normal pressure, with zero
in-plane displacement along the edges. iLevy applies the large

deflection equations and uses the series expansion method.

In this section, a solution is obtained for the same
problem to compare with Levy's. The plot of (3.76 )
shows the variation of w with lateral load as deter-

mined by Levy and by the difference method.

 

 

 

c:
C)
‘¢_

Present result

c:
c3
c6“ A Levy's result
:3
c:

2.3“,-
c:
c?
c:
c?

I
°0.00 10.00 20.020 30.00 70.00
x10
4
_£Ei__
D t
o 1

Figure 3.76. Square plate under uniform lateral load;no in-plane

displacements on boundary;v = .316.

181

3.2.3.2 Simply Supported Square Plate of Variable Thickness

To study the variable thickness plate, a solution was obtained

edge thickness
center thickness

 

for two different cases. In the first, RT = = %3

for which the membrane stiffness at the edge is l-that at the center,

4
while the flexural stiffness at the edge is %z-of the center stiffness.
In the second, RT - 2, with the edge flexural stiffness being-g-of

the central stiffness. These two opposite variations are chosen
so that the results, along with those for the uniform thickness
plate, would give an idea about the effects of variation in thick-
ness.‘ The discussion follows.

Figure (3.77 ) shows the central deflection of the
plate with variable thickness and also the result for a uniform
thickness plate. Comparison of the three plots leads to the conclusion
that, corresponding to the same edge compression, less deflection
occurs in the plate with RT 8 %-and the plate with RT - 2 under-

goes larger deflection. It should be noted that all these plates

contain the same volume.

This behavior is expected because, in a simply-supported

plate, more bending is occurring in the central region. Thus, plates

with thicker central region will experience less deflection and

plates with thin central regions are more likely to have greater

curvature and deflect large amounts.

182

2.00

1020 1060
Ran/4

0.80

0.40

 

 

—l

.00 0'.50 14.00 {.50 2.00

cp.00

Figure 3.77. Central deflection versus edge displacement for different

a)

b)

RT values, simply supported square plate.
Solution Procedure

In (3.2.3.1 a) it was shown that a grid spacing of
h =‘%g-will be accurate enough for engineering design

use. Both problems were solved with a grid spacing

of h =-%g- in determining the results plotted.

Stress Analysis

Distribution of in-plane forces, bending moments and

principal stress,along axes of the plate, are shown
1

in Figure (3.78 ) for RT = 2"

The Graph of Nx and Ny shows a decrease in membrane

forces with an increase in edge displacement in the central

183

 

 

  
  

 

 

 
 

 

 

 

 

 

 

 

I" ....... ux fix
42. x “x “““ "Y :0.
*A
'2 ‘ .. ‘
C’ h:
.— no
.1 {'3
033-24- 2 20d
01 01
K s
33 1:
U .1 o -
a. ‘z
.— H
°° s
u: u1
E"‘~ ‘9 10..
.4
“r
‘2
H A
-8 I 1 I t 1 r 1 i 0 1 1 l l I I I 4
0.28 0.50 0.28 0.80
(X/O) (Kid)
0) NX HNO NY HLGNG 9X18 Y=alé bl SENDING flOflENTS HLONO HXIS Y=G/2
ISO- 100.
120.. 120..
a: ' a: 7
a: in
01 In
a. 1m
'5 00.4 s; .0.-
4 .4
c: is
9.: 2:
c1 ' La ‘
.2. E.
m: a:
a. ‘0. 0. ‘0.-
0 0 a
' ' ' 0125 ' ' ' 0.h.0 .0 l 1'.0 ' ¢.'0 ' sfa ' 411
(11101 F

c) PRINCIPSL STRESS RUONO flXlS Y=°/2 leRINCIFHL STRESS VS. U
NOTRTION FOR CURVE IOENTIFICRTION. B) 021.0 (UNOEFLECTEO) B) UEI.ZS C) 052.75 .

FIGURE 3.78. PLOTS OF STRESS COHPONENTS.SOURRE PLHTE.SIHFLY SUPPORTEO.RT=1/4.

184
region, while in a region close to the edge, Ny is large.

As discussed before, the central region will carry less
load because of the transverse deflection and the load

will be shifted toward edges.

The moments are maximum at the center and vanish at

the edge, but fairly large moments occur at almost

halfway from center to edges. This is due to the relatively
larger thickness at center. Because most of the bending will
occur in the outer region, creating large curvature and

resulting in large moments.

Since we are concerned with the state of stress within

the plate and not necessarily membrane force or bending
moments individually, in Figure (3.78c) , the variation

of principal stress along the axis of plate is plotted

The curve shows maximum stress occuring approximately at a
point x =-%- on this axis. It should be noted that this

stress is also the absolute maximum for the entire plate.

Figure (3.79 ) show the same variables for a plate

with RT = 2.

In this case. because of the thicker edge region, beyond
buckling, the membrane load is sharply shifted toward

the edges.

185

 

 

 

 

 

 

 

 

 

 

 

 

‘----‘\ ——NX
-SO.¢ \ -----NY
\\
A \
*2 J \\
- \
.— \
Ei’i \ '5
_1 \ 0‘)
s x g
0.1 . \ 1.1
K M.-.‘ \ g
0) ‘ ~‘s \ a
33'2“ “x \ a
0: ‘~‘ \ 2
B “’5- “‘ \ 3
1.1 “ ---~.‘"--~. “Q E
z x --~-.-'=Ԥ m
a: m
.1 x‘
*‘52‘ ‘
E.
-‘ I r r I I I I I f r I r j
0.25 0.50 0.25 0.50
(X/O) (X/O)
o) NX RNO NY RLONG 8X13 Y:°/2 b) SENDING HOHENTS RLONG flXIS Y=Ol2
90.4 c 90.
J ..
S 7°~ g 10..
a: 00
a: o:
:2 + s -
I- I—
a: 00
.1 50.1 .1 00..
a: a:
4‘: * e:
u u
s. . s -
a: a:
a. 0..
30... 30.
A
V __// d
10 10
' I ' 0128 f ' ' 0.k0 0.0 ' 1'.0 ' 210 ' 3.11 ' 4.0
(X10) U

c) PRINCIPRL STRESS SLONG RXIS Y=°/2 leRINCIFflL STRESS VS. U

NOTFITION FOR CURVE IDENTIFICRTION. fl) 17:1.0 (UNOEFLECTEO) S) 0:1.25 C) 17:2.20 .

FIGURE 3.79. PLOTS OF STRESS COHPONENTsoSOURRE PLRTE.STHPLY SUPPORTEO.RT=2 .

186

The bending moments have a completely different pattern
than for the case RT =-%, because the more flexible
central region results in maximum curvature and maximum
moments in this central region. Also, because of smaller
thickness and smaller flexural rigidity, the maximum
principal stress is always at the center point. In
Figure (3.80 )shows the variation of the maximum principal
stress with postbuckling edge compression for the
different variation in thickness. It can

be seen that the least stress occurs for the uniform
thickness plate,and the plate with RT a-l- is subject to

4

largest stress at points away from center.

c)Inep1ane Displacements

 

Contours of the in-plane displacement u, in Figure (3.81 )
for RT 8 %-, and Figure (3.82 ) for RT - 2., show

a decreasing displacement in the central region due to

w deflection and corresponding decrease in membrane
forces. In the case of RT- %, we observe very small
displacements in the central region, while the more

closely-spaced contours in the vicinity of the edge

indicate very large strain in this region.

Comparison with the uniform thickness plate shows it

to be between these two variations, as expected.

187

100.00 125.00

78.00

 

PRINC IPHL STRESS 0'

 

 

Q
Do .00 0'.so 1100 1'.50 21.00

Figure 3.80. Principal stress versus edge displacement for different

RT values, simply-supported square plate.

188

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/5/25/

a) Undeflected (U/Ucr-l.) b) Buckled (U/Ucr-1.SO) c) Buckled (U/Ucr-2.50)

W

O 2. .50 .25 O

.8 .6 .4 .2 0

Figure 3.81. In-plane displacement (U=ua/t:), square plate, simply-supported,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1
RT 2'.
.54 .4 .27 .13 0 .7 .6 A1 2. 0 l. .75 .50 .25 0

a) Undeflected (U/Ucr= 1.) b) Buckled (U/Ucr-1.30) c) Buckled (UfUcr-1.85)

Figure 3.82. In-plane displacement (U - ua/ti), square plate, simply-
supported, RT = 2.

189

3.2.3.3 Uniform Thickness Plate, Clamped Boundaries
The problem studied in 3.2.3.1 was solved with clamped
boundaries, and the results of the solution are as follows.

a) Convergence of the solution.

Table<314) presents results for different mesh sizes corresponding

to a boundary displacement of (i°:59 = 1.93 times the

 

critical displacement.)

Table 3.14. Convergence of postbuckling solution with grid spacing:

 

 

 

3153.21.13 w-center Difference Extrapolation 1*? -center H -center
hla % x x
%. 1,9657 , 314.73 295.45
34.4 1.295366
-% 1.4632 _ 277.99 263.23
4.6 1.376613
-%5 1.39826 289.56 262.96
2.2 1.357113
'%5 1.3674 285.00 260.576

 

 

 

 

 

 

Considering that the accuracy test of the iterative solution
was set at 12 in successive trials, the convergence as illustrated
in graph (3. 83) is good.

Plot of central deflection with edge compression is shown

in Figure (3.84).

 

190

 

 

 

 

8
“"1
g
:4
3
S
:1
B
“0.00 4100 0'.00 1'2.00 16.00 a/h
Figure 3.83. Convergence of solution.
‘
g. .7) 7'
m 4: 4.?
a .2
:3 «‘5
m-
[D
c?
N‘
2'
D
'9
1.0
0'
a-
D
c?
I
c0.00 0.50 Lino 1'.50 2.00

Figure 3.84. Central deflection versus edge displacement for

different RT values, square plate, clamped edges.

191

b) Stress Analysis

 

Distribution of membrane forces, bending moments, and

principal stress along the axis of the plate, are plotted

in Figure (3. 85). These figures show:

i)

ii)

iii)

Larger membrane force occurs at the edge and the
central membrane forces decrease with an increase

in postbuckling edge compression.

Bending moment, Mx’ is positive at the center and
along the edge it is negative with an absolute
value larger than the central moment for the larger
deflection.

Principal stress is almost equal at center and edge
in the early postbuckling stage, increasing on the

edge with larger deflection.

c) Ineplane Displacements

 

Contours of in-plane displacement, u, shown in Figure

(3.86 ) indicate the following.

i)

ii)

111)

Before buckling, in the flat plate, in-plane dis-

placements are linear everywhere as expected.

Immediately after buckling, as the deflection starts to

increase, the u—displacement tends to decrease in the
central region.
The pattern of displacement is qualitatively similar

to that for the simply supported case.

192

 

 

 

 

 

 

 

’1’, ~‘\\ .___—_Nx
-100‘II \\ ----- NY
A \
\
1: J \
" \
SE \ -
5— w
.J 0‘)
D I-
O) 2
1.1 1.1
K I:
O
0) t
0)
01 c:
0: 2
h- 0.0
0) O
2
01 01
2 111
S
.J
0..
I
2
H
-20 T’ f 1’ I I I
0.25 0.00
(X/a)
0) NX 0N0 NY RLONG RXIS Y:°/2
fi
200.
b b
so no
8 C 8
0: 1:
h- F-
0) a:
A .J
C: e a:
L 0.
H 0..
L.) U
2 2
H H
a 120.. a:
Q. 0.
.1
A
‘0 ' 4T ' 0125 ' ' ' 0.00
(X/O)

C) PRINCIPRL STRESS RLONG RXIS Y=0V2

30.1

 

 

 

-80 .1

b1 SENDING noncurs RLONO 0x13 7:012

ZOO-J

200 .4

150 .1

40

 

I I I I

0.0 1.0 210

u
leRINCIPRL STRESS vs. U

 

310 ' 4.0

"(ITRTION FOR CURVE IDENTIFICRTPON. B] 021.0 (UNOEFLECTED) B) 051.20 C) 052.25 .

FIGURE 3.85. PLOTS OF STRESS COflPONENTS.SOUflRE PLHTE.CLRHPED. RT=! .

193

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.56 1.25 .33 .41 0 2. 1. f o 3. 2. 1. .5 o
a) Undeflected (U/chfllj b) Buckled (U/Ucr=l'20) c) Buckled (U/Ucr32.2)

Figure 3.86. In-plane displacement (U = ua/ti), square plate, clamped
edges, RT - l.

194

3.2.3.4 Clamped Plate with Variable Thickness

Again,the plates discussed in Section 3.2.3.2 (RT = 2%, and
RT = 2)are solved with clamped boundaries along all edges.

For comparison,the variation of w with edge compression
is plotted in.Figure 3.84 for plates with RT =-l- and RT = 2,

4
along with uniform thickness plate.

It can be seen that in the case of the clamped plate, due to
edge displacement appreciably greater than critical displacement,

the plate with RT =-% undergoes less deflection than either of the

plates with RT = 1 and RT = 2. This is similar to the case of
simple support, but in the early stages of postbuckling, the plate

with RT = 1 has the smaller lateral displacement.

195

a) Stress Analysis

 

Figure €3.87) illustrates the distribution of membrane

forces, bending moments, and principal stresses for' the

1
plate with RT = 2:. Analysis leads to the following

conclusions:

1)

ii)

iii)

iv)

While the membrane forces are decreasing with in-
crease in edge displacement at the center, these forces
increase sharply on the edges.

Bending moments are maximum.at the center, and the
negative moments along the edge are small because

of small flexural rigidity.

Principal stress is maximum on the edge and in-
creases with an increase in edge displacement.
Maximum principal stress occurs at the center of

the edge.

Study of membrane forces, bending moments, and principal

stresses for' the plate with RT = 2, in Figure (3.88), results

in the following observations.

1)

Ny decreases in the central region as the edge
displacement increases,while it increases sharply

near the edge.

ii) Nx also decreases at the center and decreases along

the edge.

196

 

1 ‘1
L
O
i

.1.

O

P
-0500100 000601810)

 

IN-PLRNE STRESS RESULTRNT

 

 

~10

 

 

 

 

 

 

 

I I I I I r I I
0.28 0.80
(X/Ol
0) NX RNO NY HLONG RXIS Y=0/2 b) SENDINC HOHENTS RLONG RXIS Y=°/2
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‘6 b
a: 7 a: 7
33 33
15 55
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.1 _J
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r ' ' 01.28 ' ' ' 0.150 0.0 ' 1r0 ‘ 210 ' 330 r 4.11
(XI 1 U

:1 901001901 STRESS 01000 0x18 7:012 01901001901 STRESS vs. 0
00101100 900 cuave 10001191001100. 01 051.0 1000091501001 01‘Ue1.30 01 Ue2.30 .

FIGURE 3.87. PLOTS OF STRESS COHPONENTS.SOURRE PLRTE.CLRHPED. RT=1/4 .

197'

   

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NOTRTION FOR CURVE-IOENTIFICRTION. R) UEI.O (UNOEFLECTEO) B) 051.30 C) U22.40 .

FIGURE 3.88. PLOTS OF STRESS COHPONENTS.SOURRE PLRTE.CLRHPED. RT=2 .

198

iii) Bending moments increase in magnitude at the
center and edges as the edge displacement increases,
as expected. The negative moments at the edges are
much larger, because the flexural rigidity is larger there.
iv) The graph of the principal stress shows an increase everywhere
as the edge displacement increases,but it is alwaysaa
maximum at the center of the plate.
v) Figure (3. 89) shows the variation of maximum
principal stress with edge compression, 0 , for the
three different RT values and for plates of constant

volume.

It can be seen that the plate with RT = %—
always undergoes larger stress. Although the
uniform thickness plate is less highly stressed
than the case RT = 2 for smaller edge displacements,

in the higher range of edge compression it is more highly

stressed than the plate with RT = 2.

NOte: It should be noted thatfor RT = l, and RT =~£,
the maximum.principal stress is located at the center of the
edge while for case of RT = 2, the location is .at the center

of the plate.

199

{x

62.50

J

'810‘
37.60 69.00

J

25.00

J

RT=2

CIPHL STRESS cr

PRIN
250

j

 

 

I I I fi
000 1000 ‘ 2000 3000 4000

ALDO

Figure 3.89. Max. Principal stress versus edge displacement for

different RT values, square plate, clamped boundaries.

200

b) Ineplane Displacements
Contours of in-plane displacement, U, are plotted in
Figure (3.90 ) for plate with RT = %-, and in Figure
(3.91 ) for plate of RT = 2. These contours show the

following:

1) Before buckling, the contours are exactly the same as
those for simple support, because in the undeflected
position the out-of-plane boundary condition has no

effect on the solution.

ii) After buckling, as usual, the in-plane displacement
in the central region is smaller compared to undeflected
case (i.e.,the contours are expanding at the center
and compacted contours are located away from center

toward the edge depending on the RT values.)

iii) In the case RT = %- closer contours are located
in the vicinity of the edge, while for the plate
RT = 2, because of stiff edges these concentrated
contours are seen to be close to the center.

iv) The behavior of the uniform thickness plate falls between

the cases RT =-%- and RT = 2.

201

/ W

1.25 1. .5 .25 o 1.5 1. .50 o 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

a) Undeflected (U/Ucral.) b) Buckled (U/Ucr=l.35) c) Buckled (U/Ucr=2.30)

Figure 3.90. In-plane displacement (U = ua/ti), square plate, clamped,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

RT = 1/4.
1.40 1.05 .70 .35 0 1.50 1. .50 0 3. 2. 1. 0

a) Undeflected (UflUCEI1.) b) Buckled (UflJcr-l.25) c) Buckled (UYUcr-2.15)

Figure 3.91. In-plane displacement (U - ua/ti), square plate, clamped,
RT = 2.

202

3.3 COMPARISON OF TWO METHODS

 

In order to compare results from the two methods discussed-

formulation in terms of the stress function or in terms of displacements-

two problems are solved by both methods and compared. Their results

can also be used as a measure of the accuracy of either of the methods.

a) Simply Supported

 

i) A simply-supported uniform stiffness plate, with

ii)

iii)

no restraint on in-plane boundary displacement and
loaded beyond the critical load was solved using the method
discussed in Section (2.2) (in terms of m and w).
The solution includes in-plane displacements, u

and v, on the boundary.

The problem is solved using the method of Section (2.3)
(in terms of three displacements, u, v and w),

by applying ‘the boundary displacements obtained in (i)
as boundary conditions.

If both methods are correct, we expect (1) and

(ii) to result in the same solutions, and.they did

turn out to be very close. The central deflection (w/t)
is .9613 in (i) and .9736 in (ii), witha difference

of 1.3% . This is very good agreement. Other
components of stress and displacement are also

very close.

b)

203

Clamped Edges

 

A uniform stiffness plate with clamped boundaries was

also examined in the same way as discussed in part (a).

The central deflection (w/t) was found to be 1.67121,

using the method of Section 2.2 (i.e.,formulation in

terms of stress function). It was 1.68578 using the formulation
in terms of displacements, with a difference of only .82.

The results are considered to be very good.

CHAPTER IV
CONCLUSION

4.1 THE PROBLEM SUMMARY

In the preceding chapters, the behavior of variable stiffness
plates was studied in the prebuckling and postbuckling range. In-
stability criteria were also examined. The work used the ordinary
finite difference technique. No results for similar variable stiffness
plates are available in the literature to confirm the validity of the
solutions. However, a uniform stiffness plate was included in each
case to serve as a control problem, and the results obtained by the
difference method were compared with those of analytical solutions
and other published results.-

Since the main purpose of varying stiffness is to optimize
the plate with respect to some design variables, some optimization
examples were presented in the buckling analyses.

In order to provide a better perspective on the change in
behavior of plates due to stiffness variation, the different problems
considered were assumed to contain a constant mean stiffness
or a constant amount of material (see Section 3.1.4). Two different
approaches were discussed:

1 Formulating in terms of stress function, Q, and the lateral

deflection, w.

204

205

2. Formulating in terms of three displacement components,
u, v and w.
First the applicable difference operators were worked out; then
the solution procedures were presented in detail. Finally, an
increasing number of nodes were used to check the convergence of
the solutions and also to provide guidance in selecting an
appropriate mesh size to obtain the desired accuracy.

The behavior of different stress and displacement components
was illustrated in suitable graphs and the results were analyzed.

In Section (3.1.1), uniform thickness plates with different
variations in E were considered,and the behavior of in-plane forces
and displacements was analyzed in Section (3.1.1.1). The buckling
of those plates was considered in Section (3.1.2.1) and the effect of
variation in stiffness on stability criteria was discussed in Section
(3.1.2.2),In Section (3.1.3), postbuckling behavior of those plates
‘was examined and the results were analyzed in Section (3.1.3.3).
Sections (3.1.4) and (3.2.2.2) deal with optimization of
variation in thickness from the stability point of view. The remainder
of Section 3.2. shows the effect of variation in thickness (with
constant B) on displacement, forces and moments as well as buckling

behavior.

206

4.2 CONCLUSION

 

The comparison of problems solved in Section 3.1 (with free
in-plane movement on the boundaries) with those of Section 3.2
(restricted in-plane displacement on the boundaries), shows the
great effect of in-plane boundary restraints in the postbuckling state.
Study of figures (3.60), (3.77) and (3.84), leads to
the conclusion that the behavior of central deflection due to vari-
ation in stiffness is not only dependent on the out of plane boundary
conditions but also greatly affected by in-plane boundary conditions.
Figures (3.60), (3.77) and (3.84) show that the trend of
central deflection of plates with different variation in stiffness
is not the same for all postbuckling ranges. For example, in Figure
(3.77), we observe that corresponding to the same load, a simply
supported plate with RT = l undergoes larger deflection than a plate
with RT =-% . Study of Figure (3.84) shows that the same plates
with clamped boundaries exhibit different behavior. (Although, for highly
compressed edges, less deflection is observed for the plate with
RT = %3 when the edge compression is only slightly above critical
displacement, the larger deflection corresponds to plate with RT = %).
The method and corresponding computer program utilized for
force boundary condition (Section 2.2) has been shown to give more
accurate results than that developed for the displacement boundary
condition (Section 2.3). This is due to the difference in the order of
derivatives involved in the formulation. In the force boundary

condition formulation, only second and higher order derivatives

of the two functions, 0 and w, are involved in the equilibrium and

207

compatibility equations (equation (2.14) and (2.15). The displace-

ment formulation, however, involves first and higher order derivatives

of the three functions, u, v, and w. Hence accumulative errors

could be expectedfiln difference approximation equations (2.19), the first
error term in the first order derivative includes the third derivative

of the function, and the error term in the second derivative includes

the fourth derivative of the function, etc).

Investigation of convergence indicates that for engineering

purposes, grid spacings for h =-% in the force formulation (see

Figure 3.39) and h =-%§ in the displacement formulation (see Table
3.13) are reasonably accurate; more accurate results can be obtained
by using finer grids and applying Richardson's extrapolation to the

results. Comparison of the results with known values supports the

reliability of the solutions.

208

4.3 RECOMMENDATIONS

The objective of this work was to examine application of the

two formulations in general, and to investigate the behavior of plates

with different boundary conditions and stiffness variations. Re-

finements and extensions of the method which are possible and de-

sirable include:

1.

Application of improved difference methods involving

more accurate approximations to the derivatives.

Inclusion of more nonlinear terms in the strain components
and a study of their effect on the results.

The difference operators can be revised to make them
applicable to orthotropic plates, and the computer

program improved so that it can be applicable to orthotropic
and nonhomogeneous materials.

Due to the absence of experimental sources to guarantee

the accuracy and practicability of the results,and re-
cognizing the advantages in the use of variable stiffness
plates, an experimental study of such plates from the
stability point of view and in the postbuckling range
should be very useful.

The computer programs developed here were mainly aimed

to solve particular problems. Although they are

more general than needed for the problems solved here,for
applications to loading and geometry different from the
ones presented here, the programs should be used with
caution and appropriate changes made. Also, the efficiency

of the programs can be improved.

209

Application of other methods such as the finite element

method using:

a) Elements with variable stiffness within the element.

b) Constant stiffness within an element but variation
of stiffness from element to element,

The boundary integral method might also be considered.

Consideration of the same cases and comparing numerical

results, convergence, computer cost, etc., would be of

interest.

B IBLOGRAPHY

10.

11.

12.

BIBLIOGRAPHY

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APPENDICES

APPENDIX A

APPENDIX A

Finite Difference Operators

214

215

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APPENDIX B

APPENDIX B
The plan and node arrangement on the portion of plate considered
in each case is shown. It should be noted that the second rows of
exterior nodes are auxiliary nodes for defining the K vector at edge
nodes only, and those nodes do not participate in any calculations. Thus,

'tlie node number for them could be any number or repetition of previous

Ones.

219

220

 

 

 

 

 

 

 

 

 

 

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Figure Bl.

condition square plate, h = a/8.

1

Node arrangement for force boundary

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I 51l 42 34 27 21 \Q.
I \
1 ' {921,-
56_ ._ _ __ ._ ‘ A
I 56" 41 33 26 20 15 {2;},
1 ' \*
I I \
5 _ _ __ \
67 ‘491' 40 32 25 19 14 10
' 1
S61. __ _ _1_ _ _ \
I 48. 39 31 24 18 13 9 6
l 1
\
wL——¢——
1 47l 38 3o 23 17 12 8 5 3
l 1
56'- _ . i _ _
46 37 29 22 16 ll 7 4 2
Figure B2. Node arrangement for force boundary condition, square plate,

_§_
h"l o

 

 

222

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.5. .4. Q. .6 .5 ,O Q. A. .1
,o_ ,o_ .4 1. 9. 1.
_ _
.5fi-.l.45+.1-A8 .5 04 0.
,6 ,6. A. 1. 9. 1. no 95 9.

_ _

_ _
<fi1.l I9fi-1.l
,o_ ,0. AI .4 1. A. 9: ,0 .5

.4 1. 9. 1.
_ _
11-._-.-I
5
,6 .1 ,6 1. 7. Q. 7. 1. .U

_ ,6 A. 1. .7. 1. 1. 1. .1

_ _

111111
.5 AU .5 7. 1. no 0. R. 7.
,0 ,o_ .4 1. 9. a; .i 1. 1.

_

_

_

51:94-10 1 o m a 7 e
,6 .5_ A. 1. a. 7. 7. 7.

_

_ _

.1.! IRS 1.1.1. 7. 1. 0.

"Q .5 .4 A. A. .4 .3 .8 my

_ _ _ _ _ _ _ 3_

_I _ _ _ _ _ _ _
.317110T15T11T1111111 11-
,fi .5 .5 :5 .4 7. 1. no

_ _ _ 5_ 2 5 5

_ _ _ _ _ _ _ _
_IIIFIIPIIPIIFIIFIILIILIIg
.5 .5 .5 .5 .5 .5 .5 .5 .5
,o ,o ,6 ,6 .6 ,o ,0 ,0

Node arrangement for displacement boundary condition solution,

Figure B3.

223

lOJ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

101 101 101 101 101 101 101 101 101 101 .
F—’T’_—I___I—-_I_-_I-_"I_—T_“ F——l_—_I'
I I I I I 1 1 | |
I I ' I I I ' I I
I I I I I |
I l I I
I I I I l
101 L _9QI_ 73 74 75 76 77 7 79' an 31
I i '
101 L 89' 72 57 58 59 60 61 62 63 64
. I
101 ' _831_ _71 55 43 44 4 46 47 4 49
r' I
l
1011 87: 7o 55 42 31 32 33 34 35 36
' I
I
101 L_ 8_6:__ _69 54 41 3o 21 22 23 24 25
' I
I
101 ,_ __ 85:... _68 53 4o 29 2 13 14 15 16
I I
101! 84I 67 52 ‘ 39 28 19 12 7 8 9
" I
I I
101I_ _ 8__31_ _66 51 38 27 18 11 6 3 4
I I
I I
0.1 L 821 65 50 37 26 17 1o 5 2 1
Figure B4. Node arrangement for displacement boundary condition solution,

h = a/16.

APPENDIX C
COMPUTER PROGRAMS
C.l DESCRIPTION OF THE PROGRAMS

A listing of the computer programs is presented in this
appendix. In the following a brief description of the main programs
and subroutines is given.

There are two programs, PLATE 1 and PLATE 2, corresponding
to Sections 3.1 and 3.2, respectively. Program PLATE 1 consists
of a main program and seven subroutines. Main programs direct the
flow of the computations by calling the appropriate subroutines,
in addition to reading data and performing minor calculations.
The subroutine FEOPRT computes values of the operator for the compat—
ibility equation (2.30) as well as the contribution of each node
considering the node number and the boundary condition. The sub-
routine AMATRX forms the coefficient matrix [A] by adding the
contribution of nodes. It also forms the vector {B}, the right-
hand side of compatibility equation (2.15). Formation of the operator
for the right-hand side of equilibrium equation (2.22) is accomplished
by subroutine WOPRT. The subroutine BWMAT forms the right-hand side
of equilibrium equation (2.22) as matrix [Bw]. The subroutine

AXLOAD is provided to calculate stress components and the subroutine

I

224

 

225

DISPLMT computes in-plane displacements. Subroutine WRITE controls
the arrangement of the large matrices in the write-outs.

Program PLATE 2 consists of a main program and subroutines
AMATRX, WOPRT and WRITE, in addition to six other subroutines as
follows:
The subroutine UVOPRT computes numerical values for the u and v
operators in equations (2.36) and (2.38). The subroutine AUVMAT
forms the coefficient matrices [Aul], [AuZ], [Avl] and [Av2]
which are assembled as in equation (2.43) by subroutine ASSMBL.
In subroutine WFUNCT, the derivatives of w are calculated and vectors
[Buvl] and [Buv2], the right-hand side of equation (2.9), are
formed. The subroutine KKVECT stores a nine node operator based
on a 13-node operator and subroutine RSHW is designed to compute
the vector {Bw} , the right-hand side of the equilibrium equation
in the z-direction. The subroutine BWBUCKL forms the right-hand side

of equation (2.24) for the eigensolution.

 

226

C.2 VARIABLES USED IN THE PROGRAMS

The variable names used in the Programs are defined below

in the order they appear in the Program:

PROGRAM PLATE 1

 

NSOLN Number of problems solved in one run;

ITYPE Variable controlling solution type. If EQ. 1,
membrane solution only. If EQ. 2, solve also
the eigenvalue problem. If EQ. 3, postbuckling
solution (skip eigensolution). If EQ. 4,
solve all steps;

RATIOI R or RT values (stiffness or thickness ratios);

RINCR R or RT increment for optimization;

Ll Number of trials for optimization;

N Number of real nOdes;

NPR Number of interior nodes;

NOUT Number of imaginary exterior nodes;

NA Number of intermediate nodes;

POS Poisson's ratio;

H Grid spacing;

E Modulus of elasticity;

T Reference thickness;

DF(I) Degree of freedom. If EQ. 1, interior node.

If EQ. 0, boundary node.

node.

If EQ. -l, exterior

 

227

IBC(I) Variable defining the out-of-plane boundary
condition. If EQ. l, clamped. If EQ. 2,

simply supported;

NSYM(I) Variable defining in-plane displacement symmetry.

If EQ. 1, node of antisymmetry for u. If
EQ. 2, pole of antisymmetry for v. If EQ.3,
pole of antisymmetry for both u and v. If

EQ. 4, node of diagon of symmetry (u ).

= v
south west

If EQ. 5, u-displacement zero (on boundary)

 

If EQ. 6, v = 0. (on boundary);

K
DEL(I)==%f- t E;-= Membrane stiffness ratio;
1 i

DELA(I) = Membrane stiffness ratio for intermediate nodes;
FEE(I) = The difference between stress function at
exterior nodes, and first interior nodes (i.e.

in Figure (Cl), $9 + (FEE)

Q18 ' 18‘

LP(I) = Interior node corresponding to each exterior

node in Figure Cl, node 9 is LP(18);

 

 

 

 

 

 

 

 

 

 

 

Figure Cl. Node numbers.

K(I,J)

TC

RK

DR
FE(I)
A(I,J)

AW(I,J)

BW(I,J)

B(I)

PA(I,J)

EIGVAL(I)

NTRY

228

= Vector defining the nodes participating in
operators at each node, ordering from top node

downward and rightward as shown on Figure C2;
1

 

 

 

10 ll 12

 

13

 

 

 

 

 

 

Figure C2. Arrangement of vector K.
= Central thickness (calculated);
a Reference membrane stiffness;

= Reference flexural stiffness;

3 Stress function Q;

= Coefficient matrix for compatibility equation;

= Coefficient matrix for left hand side of
equilibrium equation;

= Coefficient matrix for right hand side of
equilibrium equation;

= Vector in right hand side of compatibility
equation;

3 Auxiliary matrix;

a Eigenvalues (output);

= Number of loading steps required in iterative

procedure;

NITR

FORC

FORICR

DIF = e

Q(I) =

QINCR(I) = AQ

SUBROUTINE FEOPRT

 

R(I)

no.7) =

SUBROUTINE AXLOAD

 

XN =

YN =

SIGMAX

SIGMAY

SIGMANX

SIGMANY =

PRSTRES

SUBROUTINE DISPLMT

 

U(I). V(I)

229

Max. number of iterations allowed;
In-plane load N per unit width;
Load increment, AN;

Tolerance of convergence check;
Nodal lateral load vector;

= Lateral load increment;

Operator for compatibility equation;

Matrix defining contribution of each node based

on the position of the node;

Nx/N = Force ratio in x-direction;

Ny/N = Force ratio in y—direction;

ny/N = In-plane shear force ratio;

Mi = Bending moment in x-direction;

M? = Bending moment in y-direction;

Mky = Twisting moment;

Bending stress at extreme fibers in x-direction;
Bending stress at extreme fibers in y-direction;
In-plane stress in x-direction;

In-plane stress in y-direction;

Mbhr's circle radius

Maximum principal stress, ignoring transverse

shears;

In-plane displacements;

 

PROGRAM PLATE 2

 

AUl(I,J); AVl(I,J)

AU2(I,J); AV2(I,J)

AUV(I,J)

BUV(I)

UINCR

VINCR

31(1); B2(I)

SUBROUTINE UVOPRT

 

RUl(I,J); RVl(I,J)

RU2(I,J); RV2(I,J)

DU(I,J); DV(I,J)

SUBROUTINE AUVMAT

 

KK(I,J)

230

Sub-matrices representing contributions
of u and v in x-equilibrium equation;
Sub-matrices representing contributions
of u and v in y-equilibrium equation;
Coefficient matrix obtained by assembling
[Avl], [AVl], [AUZJ and CAVZ];

Right hand side of in-plane equilibrium
equations;

Increment of edge-displacement u;
Increment of edge-displacement v;
Sub-matrices on the right-hand side of

in-plane equilibrium equations;

Operators representing u and v con-
tributions in x-equilibrium equation;
Operators representing u and v con-
tributions in y-equilibrium equation;
Contributions of u and v at each
node considering node number, symmetry

and boundary conditions;

Vector containing nodes contributing to
operators of u and v contribution as

shown on Figure C3;

 

 

 

 

 

 

231
l 2 3
4 5 6
7 9

 

Figure C3. Arrangement of nodes in vector KK.

SUBROUTINE WFUNCT

 

BWl(I), BW2(I) = Contributions of w on the right-hand
side of x and y-equilibrium equations,

respectively;

232

C.3 COMPUTER PROGRAM

PROGRAM PLATE1(INPUTQOUTPUToTAPE5=INPUTQTAPE6=OUTPUT)

not...eagtgtttscgctc.cits...got-totstoosoootetogtstt«attests.
THIS PROGRAM USES FINITE DIFFERENCE TO ANALYZE PRE-BUCKLING
BUCKLING AND POSTBUCKLING OF VARIABLE STIFFNESS PLATES.
SUITABLE FOR FORCE BOUNDARY CONDITIONS (NO RESTRICTION ON
DISPLACEMENTS ALONG THE BOUNDARIES}.

itii...tfitfifiittttfififitiiit...OOQOQQtDIOQQQQOOOOOQOIQGI’QQQQQOOI

annnnnna

DIMENSION UKAREACSO)9ALFRC50).ALFI(50)OBETA(50)oITER(50)
DIMENSION EIGVAL(SO)QZ(50950)Q LP(65)90(50)9P8(50)9PA(50.50)
DIMENSION PBl(50)9ABU(50050)9 BUI‘50,001(50)QH3(50950)
DIMENSION FElCGS). FESCSOOTS)OFEE(65)9U1(65)
COMMON/ll POSOBBOCC'HONONPRODROTORK
CONHONIZIKI13950)"H(‘95°)QDEL(60’ODELA‘SO’ONSYNISO)
COMMON/3!OF(75)QO(13913).IBC(60)9R(13)oFE(75)
COMMON/4/A(50950)98(50)
COMMON/SIAU(50950)QBU(50950 )9RH(13)oDELTCéO)oDELTA(50)QH(65)
INTEGER AMcOF
II
c........READ INPUT DATA.
fl
READ (59') NSOLN
DO 99 JJ=19NSOLN
READ (59*) ITYP
READCSoI) RATIOIQRINCRQLI
READCSQI) NQNPRQNOUTQNAQPOSQH'EOT
NT=NONOUT
NPl=NOI
NP2=NPR01
READCSQ')(DF(I)9I=10NT)
READ (50*) (IBC(J)9J=NP29N)
READ(59*) (NSYM(I)0I=19N)
REAO(500)(DEL(J)9J=19N)
READC59*)(DELA(I)QI=10NA)
READCSQ‘) (FEE(I’9I3NPZQNT)
READ(59" (LP(I)QI=NP19NT)
DO 300 L=N929NT
FE(L)=FEE(L)
300 CONTINUE
BB=loOPOS
CC=1.-POS
DO 401 I310N
DELTCI):1./DEL(I)
001 CONTINUE
DO 402 J=IQNA
DELTACJ)=Io/DELA(J’
902 CONTINUE
URITECGQSOI)
URITE(69502)
URITEC69503) NoNPRoNOUToNAcPOSoHQDRoT
HRITEC69504)
URITEC60505)(DF(I’OI319NT)
URITEC695II)
DO 3 I=lvN
READCSQQ) LQCKCJQI’9J31913’0(AM(J0I)0J=IQ4)
URITECéQA) IQCKCJoIioJ=1013)Q(AM(J9I)9J=100)
3 CONTINUE
DO 999 LL=19L1
RK=EtT
DR=EOTfifi3IIIZofiBB'CC)

 

233

URITEI60506)
URITEC60507) (DEL(J).J=1.N)
URITEI69508)
URITEI60507) (DELA(J’OJ310NA’
URITEC69998) LLORATIOQRKODROT
998 FORMAT(I///.0TRIAL NO '.I3.* R=toF5.3o* ET CENTER=*9F8.3* D CENT
SER=*9F8.39* CENTRAL THICKNESS: *9F8.5)
DO 65 I=loNPR
FE(I)=B(I)=OI(I)=0.0
DO 66 J=19NPR
Z(IOJT=0.0
A(I.J)=AU(I.J)=BU(I.J):0.0
66 CONTINUE
65 CONTINUE
DO 10 M=loNPR
In
C........CALCULATE OPERATOR FOR COMPATIBILITY EQUATION AT EACH NODE.
I.
CALL FEOPRTIM)
I
C........FORM'COEFFICIENT MATRIX A IN LEFT HAND SIDE OF COMPATIBILITY EQUATION.
M
CALL AMATRXCM)
10 CONTINUE
HRITE‘60213)
DO 53 J=loNPR
HRITEI69214) BCJ)
PBCJ)=B(J)
DO 51 I=19NPR
PAIIQJD=AII9JT
51 CONTINUE
53 CONTINUE
DO 27 J=NP29NT
IF(FEE(J) .NE. 0.) GO TO 28
27 CONTINUE
GO TO 67
I
C........SOLVE COMPATIBILITY EOUATION.FOR LARGE MATRICES APPROPRIATE BANDED
C MATRICES AND PROPER SOLUTION ROUTINES MUST BE USED.
I
28 CALL LEOTIFI AQIQNPRosooBcavUKAREAQIER)
DO 54 I=19NPR
FE(I):B(I)
54 CONTINUE
67 URITEI69216)
DO 56 J=NPIONT
FE(J)=FEE(J)OFE(LP(J))
56 CONTINUE
DO 55 J=19NT
URITE‘60217) J9FE‘J)
FE1(J)=FE(J)
UCJ)=0.0
55 CONTINUE
FORC1=10
I
C........COMPUTE IN-PLANE FORCESOBENDING MOMENTS AND PRINCIPAL STRESSES.
m
CALL AXLOAD(FE19FORC19FEE)
M
C........COMPUTE IN'PLANE DISPLACEMENTSQU AND V.
I'll

234

CALL DISPLMTCFEI)
IFIITYP .EQ. 1) GO TO 99
DO 19 I=IONPR
DO 2 J=10NPR
A(IQJ)=0.0
2 CONTINUE
19 CONTINUE
DO 403 M=1.NPR
M
C........CALCULATE OPERATOR FOR LEFT HAND SIDE OF EQUILIBRIUM EQUATION.
I.
CALL HOPRT(M)
M
C........COMPUTE COEFFICIENT MATRIXQAU. FOR EQUILIBRIUM EQUATION.
M
CALL AMATRXCM)
m
C........FORM MATRIXQBUQ IN RIGHT HAND SIDE OF EQUILIBRIUM EQUATION.
M
CALL BUMATIM)
403 CONTINUE
DO 404 IzloNPR
DO 405 J=10NPR
AUIIOJ1=AII9J1
405 CONTINUE
404 CONTINUE
IFCITYP .EQ. 3) GO TO 18
M
C........COMPUTE EIGENVALUES AND EIGENVECTORS.
M
CALL EQZQFCA QSOOBHQSOONPROZQSO’
CALL EQZTFCA 950.BU950oNPR.EPSA.EPSB.Z.50.IER1
CALL EQZVFIA 950QBHQSOONPRCEPSAQEPSBOALFRQALFIQBETAQZQ501
URITEC6925)
DO 41 I=loNPR
EIGVAL(I)=ALFR(IDIBETACI)
URITEC6026) ALFR(I).ALFI(I198ETA(I).EIGVALII)
41 CONTINUE
18 CONTINUE
M
C........READ INITIAL VALUES FOR POSTBUCKLING TRIALS.
M
READI5911) NTRYQNITRQFORCQFORICRvDIF
READ (59*) (0(I19I=10NPR1
READI5991 QINCR
REAO(59*1 (FEIII10I319NPR1
READ(5.*) (UCI).I=19NPR)
NN=0
URITEI69224)
URITEIBOSZI)
URITEI69522)NTRYQNITRQFORCQFORICRODIFQH
URITEC60523)
URITEI60524) (QII)QI=1vNPR)
DO 5 M310NTRY
FORC1=IFORC*NN*FORICRD*DR
DO 6 I=loNPR
QII)=Q(I)¢QINCR
P81(I1=PB(I)iFORC1
6 CONTINUE
DO 20 I=NP2 9N
FEIIII=FEECI19FORC1

235

FE(I)=FE1(I)
20 CONTINUE
NN=NN¢I
HH=0
MHH=1
URITE (6.224)
URITE (6.221) NNvFORCI
DO 7 L=10NITR
MM1:MM-1
nnzznn-z
IF(MM-3.MMM) 226.2259226
225 MMMzMMMOI
m
C........APPLY CONVERGENCE-INOUCING TECHNIQUE.
m I
DO 60 I=1.NPR
IF((FE3IMM.I)-FE3(MM1.I)10(FE3CMM1.II-FESCMMZQI)).GE. 0.160 TO 60
FE!I1=IFE3¢MMI.I)fit2-FE3IMM01)fiFE3KMM2cI))l(2.'FE3(MM1.I)-FE3(MM.
OI)-FE3(MM2.I))
U(I)=(US(MM1.I)002-03(MM.I)IU3(MM2.I))I(2.'U3(MMIOI)-H3(MM.I)-H3(
OMMZQI’) _
60 CONTINUE
GO TO 61
226 00 49 I=IvN
FE(I)=FE1(I)
49 CONTINUE
61 MMzMMOI
OO 58 I=19NPR
DO 59 leONPR
BUII.J)=0.0
59 CONTINUE
58 CONTINUE
n
C........CALCULATE RIGHT HAND SIDE OF EQUILIBRIUM EQUATION.
a

 

008 I=19NPR
CALL BUMATCI)
01(I,=OOO
8 CONTINUE
DO 50 I=19NPR
009 J=IQNPR
ABU(I.J)=AU(I.J)
A(IQJ13PAII0J)
QI(I)=QI(I)¢BH(I.J)9U(J)
9 CONTINUE
01(I1=(01(IIOQCITthfi41/OR
UICI1=UIII
50 CONTINUE
m
C........SOLVE EQUILIBRIUM EQUATION.GET NEH U.
I“.
CALL LEQTIFCABU.loNPR050001989UKAREAoIERI
DO 12 I=19NPR
HII)=Q1(I)
12 CONTINUE
OO 14 J=10NPR
C........COMPUTE RIGHT HAND SIDE OF COMPATIBILITY EQUATION BASED ON NEH VALUES.
6:0 .0
UFAC1=(H(K(12.J1)OUIKCZQJID-UIKI10.J)1-U(K(4.J)))I4.
UFAC2=UCKIBQJ1IOHCKI6OJ11-2.4U(K(79J)1
HFAC3=UIKC39JT)OH(K(11.J1)-2.0H(K(7.J)1

 

14
m

236

G=GOHFAC1"2-HFAC2*UFAC3
BUIIJT=GORKOPBIIJI
CONTINUE

C........SOLVE COMPATIBILITY EQUATION.GET NEH FEE VALUES.

15

17
6

CALL LEQT1FI A91QNPRQSDQBUIQBQUKAREAQIER1
DO 15 I=IONPR

FEIII1=BUICIT

CONTINUE

DO 17 I=19N

FE3IMM.I)=FE1(I)

U3IMMQII=UII1

CONTINUE

C........CHECK CONVERGENCE.

16

7
72
71
64

63
a

DO 16 J=10NPR
IFIABS¢IH1(J)-U(J))lUIJTD .GT. DIF) GO TO 7
CONTINUE

GO TO 72

CONTINUE

URITEC69222) MM

DO 71 I=10N

HRITEC6913) FE1(I).U(I)
CONTINUE

DO 63 J=NPI.NT
FEIIJ)=FEE(J)OFORC10FE1(LP(J))
CONTINUE

C........CALCULATE IN-PLANE FORCES AND DISPLACEMENTS.

5
999
99
1

q
11

13
25

26
211

212
213
214
216
217
218
220

221
222
224
501
502

CALL AXLOAD(FE19FORC10FEE)

CALL DISPLMTCFEIT

CONTINUE

CONTINUE

CONTINUE

FORMATI4I5.F5.3.3F10.5)

FORMATI18I4)

FORMATIZIS.4F10.5)

FORMATC10X.F15.805X9F15.8)

FORMATI/II/.10X.IALFA REALI.5X.0ALFA IMAGO.8X.*BETA4910X.0EIG

4ENVALUESOI

FORMAT(I910X.F10.5.5X.F10.5.5X.F10.5.5X.F12.4)
FORMATI1H1.20X.*ELEMENTS 0F COEFFICIENT MATRIX-COLUMN509I20' TO 0

99121

FORMATI/95X910F10.5)

FORMATI/IOBXQ‘B VECTOROII)

FORMATISX9F12.B)

FORMATI'O'vIOXo'FE VALUES AT EACH NOOE’)
FORMATI/95X91298X9F12.8)

FORMATIIII/.20X9'ELEMENTS 0F MATRIX AU.)
FORMAT(III/020X99EIGENVECTORS CORRESPONDING TO EACH EIGENVALUE-

OVECTORS'QIZQ'TO *OIZI

FORMATI/l/I.50X.*LOAO POINT NO POIJOSXQPLO‘D:.OF802)
FORMATIII.0ITERATION NO *0I202X.‘ FE *010X9'U!)
FORMATIIII/.30X.9AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA'T
FORMAT(//.20XQ'OOOOO GEOMETRICAL INPUT DATD 000000)

FORMATI/IQ'NO OF NODES'.2X.*NO OF INTERNAL NOOESOQZXQ'NO OF EXTERI

00R NODES‘.2X.*NO OF INTERMEDIAT NODES492X0'POISON S RATIO'.2X9
4*GRID SPACING REF STUFNESS THICKNESS')

 

 

innnnna

a
C
n

237

503 FORMATII93X9I5012X.15.15X0I5015X9I5.20X.F5.3.10X.3F10.5)

504 FORMATI/l/920X9*DEGREES OF FREEDOM FOR EACH NODE 41=INTERIOR0 0:80
*UNDARY POINT.-1=EXTERIOR NODE...)

505 FORMATII95X940I3)

506 FORMATCIl/920X9'1/BETA=DR/Dfi)

507 FORMAT(5(5X910F10.5/))

508 FORMATI/l/Q20X041/BETA FOR INTERMEDIATE POINTS.)

511 FORMAT!///.20X.0CORRESPONDING POINTS PARTICIPATING IN EACH NODE O
OPERATOR'I) '

521 FORMATI/I/05X04NO OF LOAD POINT5402X04NO OF ITERATION*.2X.*FORCE*9
OZXQOFORCE INCREMENTOQZthDIFFERENCEi92X.*H=AIN*)

522 FORMATII012X9I5912X9I505X9F9.308X9F6.3910X9F6.395X9F6.31

523 FORMATIIICZOXO'EXTERNAL TRANSVERS LOAD.)

524 FORMAT(5(5X010F6.3/))

END

SUBROUTINE FEOPRTIM)

9.0...it.099......Ctfittfiittttttttttttttttitficttfitttt99.0.0...
THIS ROUTINE CALCULATES THE VALUES OF OPERATOR FOR COMPATIBILITY
EQUATION AND DETERMINES THE CONTRIBUTION OF EACH NODE BASED ON
NODE NUMBERS (MATRIX D 1

it.ti.Q.fit..40....0..0......ttttflfififififitfifiififitttifit...0.0.0...

COMMON/ll POSOBBQCCOHONQNPR'DRQT
COMMON/ZIKI1305070AMI4050100ELC60)ODELAC50)9NSYMI50)
COMMON/3/DFI75IQDI1301330IBCI6019R(13’9FEI75)
COMMON/4’AI50050’QB(50)
COMMON/5’AHI50950)QBUC50050 ’9RHC13)oDELTC60)oDELTA(SOIoUI65)
INTEGER AMQDF
R(I)=DEL(K(39M))
R(2)=‘(DEL(KI39M)1+DELIKI69M)1IOPOSOZ..BB.DELA(AMI3OM)1
R(3)=-2.*CC¢(DELCKC7OM)I‘DELIKI39M1’1'2.*BB'(DELAIAM(39M)IODELAIAM
9(49M111
R(473‘POS'IDELIKI39M))ODELIK(89M)I’02..BB'DELAIAM(4.M)I
R(5)=DEL(KI60M))
RI6I='Z.'CC*IDEL(KI79M)I’DELIKI6OM))1'2.'BB*(DELA(AM(29MII’DELAIAM
0(3QMIII
R(7,=DELIK(39HI)‘DELIKIGOH’,.DELIKIBOM)I’DELIKIIIOH),.BOPCCMDELIK
9679M)102.'BB*IDELA(AMI10M)10DELA(AM (20M)I’DELAIAMI39M1)*DELACAM(
O49MIIJ
R68’=-ZOPCCPIDELIKI79H)IPOELIKIBQM’I)‘20'88'IDELB‘AHIIOM’,.DELAIAH
9644M111
R(9)=DEL(K(80M)1
R11013-(DELIKI6QM))ODELIKIIIQMIIIOPOS02.OBB*DELA(AM(29M)3
R(II)=‘2.'CC*(DEL(KITOM1I‘DELIKI119M1)I-2.*BB.(DELA(AM(19M))‘DELA
0(AMI24MII)
R(IZ)=-IDEL(K(80M1IODELIKIIIQMII)‘POSOZ.'BB*DELAIAM(IQM)1
R613):DEL(K(119M11

OOOOOOOOCONSTRUCT D MATRIX AT EACH POINT.

DO 72 J=1913
OO 71 I=1ol3
D(IQJ)=0.0
71 CONTINUE
72 CONTINUE

 

238

D(7.7)=1.0

IFIDF(K(19M)) .NE. -1 .AND. DF(KC59M)) .NE. -1) 0(292)=1.0
IF‘DF(K(19M)) .NE. -1 .AND. DF(KI99M)) .NE. '1) D(494):1.0
IFIDF(K(59M)) .NE. -1 .AND. DF(KI139M)) .NE.-1) 0(10910)=1.0
IF(DF(K(99M)) .NE. '1 .AND. DF(KI139M)) .NE.-1) D(12912)=1.0
IFIDFIKCI9M))) 10091019102

100 0(791)=1.0
BIM)=B(M)-R(2)'FE(K(29M))-R(3)*FE(K(39M))-R(4)*FE(K(49M))-R(1)*FE
9(KC19M))
GO TO 40
102 0(191)=1.0
101 0(393)=1.0
IFIDFIKI19M)) .EQ. 0) BIM)=B(M)'R(1)0FE(K(19M))
40 IFIDF(K(59M))) 11091119112
110 0(795)=1.0
BIM)=BIM)-R(6)*FE(K(69M))-R(10)*FEIK(109M))-R(5)0FE(K(59M))
IFIDF(K(19M)) .NE.-1) B(M)=B(M)-R¢2)*FE(K(29M))
GO TO 41
112 0(595)=1.0
111 0(696)=1.0
IFIDFIKI59M)) .EQ. 0) BIM)=B(M)'R(5)*FEIK(59M))
41 IFIDFIKI99M)) ) 12091219122
120 0(799)=1.0
BIM)=B(M)-R(8)'FEIK(89M))-R(9)*FE(K(99M))-R(12)*FE(K(129M))
IFIDFIKII9M)) .NE. ’1) BIM)=BIM)-R(4)0FEIK(49M))
GO TO 42
122 D(999):1.0
121 0(898)=1.0
IFIDFIKI99M)) .EQ. 0) BIM)=B(M)-R(9)*FEIK(99M))
42 IF(DF(K(139M)) ) 13091319132
130 0(7913)=1.0
BCM)=8(M)-R(11)tFE(K(119M))-R(13)0FE(K(139M))
IFIDFIKIS9M)) .NE. '1) BIM)=BIM)'R(10)'FEIKI109M))
IFIDFIKI99M)) .NE. '1) BIM)=BIM)'RI12)*FE(K(129M))
GO TO 43
132 0(13913)=1.0
131 O(11911):1.0
IFIDFIKI139M)) .EQ. 0) 8(M)=8(M)-R(13)*FE(K(139M))
43 RETURN
END
M
M
M
m
M
SUBROUTINE AMATRXIM)
M
C .............................................................
C........CALCULATE COEFFICIENT MATRIX FOR COMPATIBILITY EQUATION.
C .....t..............................t........................
M
COMMON/1’ POS9BB9CC9H9N9NPR9DR9T
COMMON/2’KII3950)9AM(4950)9DEL(60)9DELA(50).NSYM¢50)
COMMON/3/DFI75)9D(13913)9IBCI60)9R(13)9FE(75)
COMMON/4/A(50950)9B(50)
DO 51 J: 1913
DO 52 L31913
IF(K(L9M) .GT. NPR) GO TO 52
A(M9KIL9M))=A(M9K(L9M))¢DIL9J)OR(J)
52 CONTINUE
51 CONTINUE

annnna

I
C...
M

03

71

72

100
1001

1002

239

RETURN
END

SUBROUTINE HOPRT(M)

9.....9.....ect....9.........t.....tt............e..t.9not...
THIS ROUTINE CALCULATES OPERATOR FOR EQUILIBRIUM EQUATION AND
CONTRIBUTION OF EACH NODE BASED ON NODE NUMBERS.

O‘COOOQCQOOOQQOOQOQOOOifiiiitittififiififitfitifittitiifiiiflQflttfiitii

COMMON/1’ POS9BB9CC9H9N9NPR9DR9T
COMMON/2/K(13950)9AM(4950)9DELI60)90ELA(50)
COMMON/3IDFI75)90(13913)9IBC(60)9R(13)9FE(75)
COMMON/5’AHC50950)980(50950 )9RHC13)90ELT(60)90ELTA(50)
INTEGER AM90F

.....CALCULATE OPERATOR AT EACH NODE USING FIGURE A1.

RUC1)=DELT(K¢3.M)) ,
Ru121=2.tcc-DELTA<AH:5.H))opos-IDELTIKI6.HI)ooELTIK13.H)I)
RUI3)=-2.t(889(DELT(K(7.M))ODELT(K(3.H)))OCC*(DELTA(AM(39M))9
OOELTAIAM¢49M))))
RU(4)=2.OCCOOELTA(AM(4.M)DOPOSOIOELT¢K(3.M))ODELT(K(8.M))3
RH(5)=OELT(K(69M)) .
89(6)=-2.ttaettoELT(Kt7.HI)ooELTIK(6.H)1)oCCccotLTAIAntzgn)1+
oDELTAIAn¢3.N))))
RH(7)=8.988005LT(K(7.M))OOELT(K(8.M))ODELT(K(6.M))OOELT(K(3.M))0
ODELT(K(119M))+2.*CC~(0ELTA(AM(19M))OOELTA(AM(29M))OOELTA(AM(3.M))
oooELTAIAHI4.M)I)
RH(8)=-2.t(88*(OELT(K(7.M))ODELTTKteoM)1)9CC9(0ELTA(AM¢1.M))9
oDELTAIAH¢4.n)1)I
RU(9)=OELT(K(8.M))
RH!101:2.9CC90ELTACAM(2.M))OPOSO(OELT(K(11.M))ODELT(K(6.M)))
RHC11)=-2.*(839(DELT(K(7.M))OOELT(K(119M)3)OCC9(0ELTA(AM(19M))9
ootLTAIAn(2.H))))
RH(12)=2.¢CC*DELTA(AM(19M))OPOSt(DELT(K(8.M)DODELT(K(119M))I
RU(13)=OELT(K(11.M))

CONSTRUCT 0 MATRIX AT EACH POINT

DO 72 J=1913

DO 71 I=1913

DII9J)=0.0

CONTINUE

R(J)=RU(J)

CONTINUE

06797I=100

IFIDFIKI19M)) .NE. -1 .AND. DF(KI59M)) .NE. '1) 0(292)=1.0
IFIDFIK119M1) .NE. '1 .AND. DF(KI99M)) .NE. ‘1) 0(494)=1.0
IFCDFIK159M)) .NE. '1 .AND. DF(KII39M)) .NE.-1) 0(10910)=I.O
IFIDFIKC99M)) .NE. ‘1 .AND. DF(KI139M)) .NE.-1) 0(12912)=1.0
IFIDFIKI19M))) 10091019102

IFCIBCIKI39M))-2) 10019100291002

0(791)=1.0

GO TO 40

0(791)='1.0

GO TO 40

 

102
101

110
1101

1102

112
111

120
1201

1202

122
121

130
1301

1302
132

131
43

306003

240

0(191)=1.0

D(393):1.0

IFIDFIKI59M))) 11091119112
IFIIBC(K(69M))-2) 11019110291102
0(795)=1.0

GO TO 41

0(795)=-1.0

GO TO 41

0(595)=1.0

0(696)=1.0

IFIDFIKI99M)) ) 12091219122
IFCIBC(K(89M))-2) 12019120291202
0(799)=1.0

GO TO 42

01799)=-1.0

GO TO 42

0(999)=1.0

0(898)=1.0

IFIDFIKI139M)) )13091319132
IFIIBCCKI119M))-2) 13019130291302
01701313100

GO TO 43

0(7913)=-1.0

GO TO 43

0113913)=1.0

0(11911)=1.0

RETURN

END

SUBROUTINE BUMATCJ)

socogeotaootst.0.....00090900.....ocoocs.ttosottosot.0......t
THIS ROUTINE IS UTILIIZED TO CALCULATE RIGHT HAND SIDE OF
EQUILIBRIUM EQUATION.

Otfiififittfiiflfi.09....9..QOIIQCQQOitfiitiiiifitflitfiititifitfififitltifii

COMMON/1’ POS9BBOCC9M9N9NPR9OR9T
COMMON/ZIK(13950)9AMI4950)9DEL(60)9OELA(50)9NSYMI50)
COMMON/310F175).0(13913)9IBC(60)9R(13)9FE(75)
COMMON/5/AUI50950)980(50950 )9RUII3)9DELTI60)9DELTA(50)
FEFAC1=FE(K(119J))-2.*FE(K(79J))4FEIKC39J))
FEFAC2=FEIKIB9J))-2.9FE(K(79J))0FEIKI69J))
FEFAC33-(FEIK1129J))OFE(K(29J))-FEIK(109J))‘FE(K(49J)))IB.
IF(K(29J) .GT. NPR) GO TO 1
BUIJ9KI29J))=BU(J9K(29J))OFEFAC3

IF(K(39J) .GT. NPR) GO TO 2
BUIJ9KI39J))=BU(J9K(39J))OFEFAC2

IFIKI49J) .GT. NPR) GO TO 3
BUCJ9K¢49J))=BH(J9K(49J))-FEFAC3

IF(K(69J) .GT. NPR) GO TO 4
BHIJ9KC69J))=BU(J9K(69J))OFEFACI

IFIK(79J) .GT. NPR) GO TO 5
BH!J9K(79J))=BU(J9K(79J))-2.*FEFAC1-2.OFEFAC2
IFIKIB9J) .GT. NPR) GO TO 6
BU¢J9K189J))=BH(J9K(89J))OFEFACl

IF(K(109J).GT. NPR) GO TO 7
BHIJ9KI109J))=BH(J9K(109J))-FEFAC3

 

annnna

241

7 IF(K(119J).GT. NPR) GO TO 8
BH(J9K(119J))=BH(J9K(119J))OFEFAC2
8 IF(K(129J).GT. NPR) GO TO 9
BU(J9K(129d))=BU(J9K(129J))OFEFAC3
9 RETURN
END

SUBROUTINE AXLOA0(FE19FORC19FEE)

........t...................t....................c...c.......
THIS ROUTINE COMPUTES IN-PLANE FORCES9BENDING MOMENTS AND
PRINCIPAL STRESSES AT EACH NODE.

i.0..O.C....QOQOQOCOCOOCCOOGOOGOOOOOOOOOGOO0.00.0.9..tfitifififii

DIMENSION FE1(75)9FEE(6S)
COMMON/II POS9BB9CC9H9N9NPR9DR9T
COMMONIZ/K(13950)9AM(4950)90EL(60)90ELA(50)9NSYM(50)
COMMON/3/0FC75)90(13913)9IBC(60)9R(13)9FE(75)
COMMON/SIAH(50950).BU(50950 )9RH(13)90ELT(60)90ELTA(50).H(65)
NP2=NPR91
MM=N-1
URITE (6921)

21 FORMAT(II//910X9*STRESS RESULTANT RATIO AT EACH NODE 4)
DO 1 M=NP29N
UF=10
IFIIBCIM) .EQ. 2, UF3-10
IF(NSYM(M)-6) 35936937

35 U(K(69M))=UF*U(K(B9M))
U(K(109M))=UF*H(K(129M))
GO TO 1

37 U(K(69M))=HF00(K(89M))
U(K(29M))=UF00(K(12.M))

36 H(K(39M))=UFOH(K(119M))
U(K(49M)):UFOU(K(129M))

1 CONTINUE

URITE(6922)

22 FORMAT(III/92X.0N00E*95X94NX/N4.6X.ONY/Nfi95XotNXY/NO. 9X90MX09
*8X9tMY'97X9'MXY'911X9'MXT/21‘95X9'NX/T'910X9'MYT/21'95X9'NY/T PRI
ONCIPAL STRESS.)

00 10 I=19N

IF(NSYM(I) .EQ. 7)FE1(K(29I))=FE1(K(1291))02.*FEE(K(69I))iFORCI

XN=-(FE1(K(1191))9FE1(K(391))-2.0FE1(K(79I)))/(H992*FORC1)

YN=-(FE1(K(69I))OFE1(K(B9I))-2.tFE1(K(79I)))ICHst2-FORC1)

XYN=¢(FE1(K(29I))OFE1(K(129I))-FE1(K(491))-FE1(K(109I)))/(4.'H'*2
9*FORC1)

HXX:(U(K(6.I))+H(K(89I))-2.OU(K(79I)))IH992

UYY=(U(K(119I))9H(K(39I))-2.'H(K(79I)))IH902

UXY=(U(K(4.I))OU(K(109I))-U(K(129I))-H(K(29I)))/(4.'Ht*2)

0P=0ROOELT(I)

XM=-0Pt(UXX¢POSOUYY)

YM=-0P*(UYYOPOS'UXX)

XYM=0P4CC*HXY

SIGMAX=XM96.IT992

SIGMAY:YM06.IT**2

SIGMANx=XNiFORC1lT

SIGMANYzYNtFORCIIT

XSIGMAX=ABS(SIGMAX)9ABS(SIGMANX)

 

 

242

YSIGMAX3ABS(SIGMAY)‘ABS(SIGMANY)
RAD=SQRT(((XSIGMAX-YSIGMAX)/2.)4420(XYN.FORC1/(2.*T))042)
PRSTRES =(XSIGMAXOYSIGMAX)/2.ORAD
URITE(6923) I9XN9YN9XYN9XM9YMOXYMOSIGMAXOSIGMANX9SIGMAY9SIGMANY
$9PRSTRES
23 F0RMAT(I/92X91292X93F10.594X93F10.392(2X9F10.39F12.3)9F12.3)
10 CONTINUE
RETURN
END '

3 3533 3

SUBROUTINE DISPLMT(FE1)

OtfififiitfiifiifitfififififififlfiCQQIQOIQOI.Oiifiitiiittfififitfiiflififi.t.00...

THIS ROUTINE COMPUTES IN-PLANE DISPLACEMENTS9U AND V AS
DESCRIBED IN SECTION 3.1.3.2.

fitt...Q...Ofiifitfitfifitit.Itfiiififitifiifittfifitiifii0......0.0.0....

annnna

DIMENSION AU(50950)9AV(50950)98U(50)9BV(50)9FE1(75).HKAREA(50)
DIMENSION NU(50).NV(50)9AUU(50950)9AVV(50950)988U(50)oBBVISO)
COMMON/ll POS9BB9CC9H9N9NPR9DR9T9RK
COMMON/2/K(13950)9AM(4950)90EL(60)90ELA(50)9NSYM(50)
COMMON/3/OF(75)9O(13913)9IBC(60)9R(13).FE(7S)
COMMON/SIAUCSO950)980(50950 )9RH(13)9OELT(60)90ELTA(50).U(65)
INTEGER OF
00 1 I=I9N
DO 2 J=19N
AU(I9J)=AV(I9J)=0.0

2 CONTINUE
BBU(I)=BBV(I)=0.0

1 CONTINUE

CocooooooIMPLY OUT'OF'PLANE BOUNDARY CONDITION.

NP2=NPR41
DO 22 M=NP29N
HF=10
IFCIBC(M) .EQ. 2) HF:-1.
IF(NSYM(M)-6) 35936937

35 U(K(69M))=UF0H(K(B9M))
H(K(39M))=HF'U(K(119M))
GO TO 22

37 H(K(69M))=UF*U(K(89M))
U(K(29M))=UF4U(K(129M))

36 U(K(39M))=HFOU(K(119M))

22 CONTINUE
N1=N2=0
DO 3 M=19N
UX=(U(K(89M))-U(K(69M)))/(2.*H)
UY=(H(K(39M))-U(K(119M)))/(2.*H)
XN=(FE1(K(119M))4FE1(K(39M))-2.4FE1(K(79M)))/(H0'2)
YN=(FE1(K(69M))OFEI(K(89M))-2..FE1(K(7.M)))I(Hfi*2)
C1=.5*HX**2
C2=.S'UY**2

C........CONSIOERING SYMMETRY AND BOUNDARY NOOES 9FORM COEFFICIENT MATRICES.

IF(NSYM(M) .NE. 0) GO TO 6

25 N1=N1¢1
N2=N2+1
NU(N1)=M
NV(N2):M

 

 

243

IF(NSYM(K(89M)) .EQ. 1 .OR. NSYM(K(89M)) .EQ. 4) GO TO 18
AU(N19K(79M))=AU(N19K(79M))-1./H
AU(N19K(B9M)):AU(N19K(89M))+1./H
GO TO 19
18 AU(N19K(79M)):AU(N19K(79M))-1./H
19 AV(N29K(79M))=AV(N29K(79M))¢1./H
AV(N29K(119M))=AV(N29K(119M))-1./H
BU(N1)=DEL (M)*(XN-POS*YN)IRK-C1
BV(N2)=DEL (M )*(YN-POS*XN)/RK-C2
GO TO 3
6 GO 70(79893915925925925) NSYM(M)
7 N2=N241
NV(N2)=M
IF(DF(K(39M)) .EQ.-1) GO TO 20
AV(N29K(39M))=AV(N29K(39M))O.5/H
AV(N29K(119M))=AV(N29K(119M))-.5/H
GO TO 21
20 AV(N29K(79M))=AV(N29K(79M))41./H
AV(N29K(119M))=AV(N29K(119M)1'1./H
21 BV(N2)=DEL (M )*(YN°POS*XN)/RK-C2
GO TO 3
B N1=N141
NU(N1)=M
AU(N19K(89M))=AU(N19K(89M))¢1./H
AU(N19K(79M))=AU(N19K(79M))-1./H
BU(N1)=0EL (M)*(XN-POS'YN)/RK-C1
GO TO 3
15 N2=N2¢1
NV(N2)=M
AV(N29K(79M))=AV(N29K(79M))91./H
AV(N29K(119M))=AV(N29K(119M))-1./H
BV(N2)=DEL (M )9(YN-POS*XN)/RK-C2
3 CONTINUE
DO 5 I=I9N1
DO 4 J=19N1
AUU(I.J)=AU(I9NU(J))
4 CONTINUE
5 CONTINUE
DO 17 I=l9N2
DO 14 J=19N2
AVV(I9J)=AV(I9NV(J))
14 CONTINUE
17 CONTINUE

COOOOOOOOSOLVE FOR V‘DISPLSCEHENT.

CALL LEQT1F(AVV919N29509BV989HKAREA9IER)
DO 10 J=19N2
BBV(NV(J))=BV(J)
10 CONTINUE
NN=0
00 23 M=19N
IF(NSYM(M) .EQ. 1 .OR. NSYM(M) .EQ.3) GO TO 23
IF(NSYM(M) .EQ. 4) GO TO 24
NN=NN+1
IF(NSYM(K(89M)) .NE. 4) GO TO 23
BU(NN)=BU(NN)+BBV(K(89M))IH
GO TO 23
24 BBU(M)='BBV(M)
23 CONTINUE

caucuses-SOLVE FOR U‘OISPLSCEMENTO

CALL LEQT1F(AUU919N19509BU989UKAREA9IER)
DO 9 I=19N1

 

 

m
C

244

BBU(NU(I))=BU(I)
9 CONTINUE
BBU(N)=-BBV(N)
HRITE (6911)
11 FORMAT(////910X9*NODE U-DISPLACEMENT
DO 16 I=19N
URITE(6912) I9BBU(I)9BBV(I)
12 FORMAT(11X91293X9F12.698X9F12.6)
16 CONTINUE
RETURN
END

SUBROUTINE URITE(NPR9A)

V-OISPLACEMENTfi9/l)

OfififiitttItitt...titiittfifittfit.tfifitfittfifitifiiii0......Qititiitt

C........THIS ROUTINE ARRANGES THE URITE- OUT OF LARGE MATRICES.

C.O...Ottitifitittfifit.tfififififii.Ittitfiitiifiitifittfifiififit.Qtttttti

DIMENSION A(50950)
NMATzNPR/IO‘I

DO 57 11:19NMAT
Ill=109(11-1)91

112:10411

IF(IIZ .GT. NPR) 112=NPR
IF(111 .GT. NPR) GO TO 57
URITE(693) 1119112

3 FORMAT(III/.SOX'Q--------------- COLUMNS ..12.. ‘0 ..12.. --------

$-------‘.//)

00 52 I=l9NPR

HRITE(694) (A(I9J)9J=1119112)
4 FORMAT(/95X910F12.5)
52 CONTINUE
57 CONTINUE

RETURN

END

 

 

245

PROGRAM PLATE2(INPUT9OUTPUT9TAPE5=INPUT9TAPE6=OUTPUT)

Qfifittfitiiifiifittifitfifitifififiifiit.Otififiititfitfiififififit...fittifitflfifi

THIS PROGRAM IS UTILIZED TO SOLVE PRE-BUCKLING9BUCKLING AND
POSTBUCKLING OF VARIABLE STIFFNESS PLATES SUITABLE FOR
DISPLACEMENT BOUNDARY CONDITIONS.

THIS PROGRAM USES THE GIVEN SUBROUTINES IN ADDITION TO THREE
ROUTINES (AMATRX9UOPRT AND URITE) AS GIVEN IN PROGRAM PLATE1.

0..tfiifittttfi.00....ifiifiitiifiiifititfififiiit....OOOOOOOOQIOOOQOQI

annnnnnna

DIMENSION 0(81)9BU(81)9A(1289128)9B(128)9UINCR(81)9VINCR(81)
DIMENSION AHH(64964)9UH(64925)9HKAREA(128)
DIMENSION EIGVAL(64)9ALFR(64).ALFI(64)9BETA(64)92(64964)
COMMON/IIPOS9BB9CC9K(13981)9DEL(81)9DF(101)9H9NSYM(81)9DR9RK
COMMON/2/0U(999)90V(9.9).KK(9964).RU1(9)9RU2(9)9RV1(9)9RV2(9)
COMMONI3/BI(65)982(65)9U(81)9V(81)9BU1(64)9BH2(64)9H(100)
COMMON/4] AU1(64964)9AU2(64964)9AV1(64964)9AV2(64964)9AUV(1289128)
COMMON/5i BUV1(64)9BUV2(64)9BUV(128)
COMMON/6/DELT(81)9DELTA(8 )9AM(4981)9DELA(8 )9IBC(81)
COMMON/7IRU(13)90U(13913)9AH(64964)9BUB(64964)
COMMON/8/XNX9YNY9XNY
INTEGER AM90F
m
COOOOCOOORE‘O INPUT DATA.
m
READ (59*)NSOLN
DO 99 NO=19NSOLN
REA0(59*) ITYP
READ(59*) RAT1019RINCR9L1
READ (591) N9NPR9NOUT9NA9POS9ALFA9H9E9T
NP2=NPR¢1
NT=N0NOUT
READ (59') (IBC(I)9I=NP29N)
READ (59') (DF(I)9I=19NT)
READ(59*) (NSYM(I)9I=19N)
READ (59*) (U(I)9I=NP29N).
READ (59*) (V(I)9I=NP29N)
00 100 I=19N
READ(5.4) L.(K(J9I)9J=1913) 9(AM(J.I)9J=194)
HRITE(6.4) I9(K(J9I).J=1913) .(AM(J9I) 9J=194)
100 CONTINUE
BB=1.¢POS
CC=1.°POS
URITE(69501)
READ(59*T (U(J)9J=19N)
READ (59*) (Q(I)9I=19N)
URITE(69511)
URITE(69502)
URITE(69503) N9NPR9NOUT9NA9POS9H9DR9T
URITE(69504)
URITE(69505)(DF(I)9I=19NT)
N2=24NPR
CALL KKVECT(NPR)
DO 999 LL=19L1
CALL DESIGN(RATIOI9RINCR9N9LL9TC9RATIO9NA)
T=TC
OR=E9T993/(12.9BBtCC)
RK=E9T
URITE(69506)
URITE(69507) (DEL(J)9J=19N)
HRITE(69508)

246

URITE(6.507) (OELA(J)9J=19NA)
HRITEC69998) LL9RAT109RK90R9T
998 FORMAT(////.*TRIAL NO 0.13.9 R=9.F5.3.* ET CENTER=9.F8.3* O CENT
SER=O.F8.3.* CENTRAL THICKNESS: '9F8.5)
DO 103 I=l9NPR
81(1)=82(I)=0.0
00 104 J=19NPR
AUl(I9J)=AU2(19J)=AV1(I.J)=AV2(I.J)=AU(I.J)=BHB(I.J)=0.0
104 CONTINUE
103 CONTINUE
Ill
CocooooooCALCULATE OPERATORS AND FOR" THE MATRICES FDR IN'PLANE
C EQUILIBRIUM EQUTIONS.
m
00 110 M=19NPR
CALL UVOPRT(M)
CALL AUVMAT(M.NPR)

M
C........COMPUTE U-OPERATOR FOR OUT-OF-PLANE EQUILIBRIUM EQUATION.
m
CALL UOPRT(M)
M
C........CALCULATE CONTRIBUTION OF H IN RIGHT HAND SIDE OF IN-PLANE
C EQUILIBRIUM EQUATIONS.
u
CALL UFUNCT(ALFA9M)
m
C........FORM COEFFICIENT MATRIX FOR OUT-OF-PLANE EQUILIBRIUM EQUATION.
m
CALL AMATRX(M.NPR)
110 CONTINUE
m
C........ASSEMBLE U AND V COEFFICIENT MATRICES.
m
CALL ASSMBL(NPR)
URITE(69265)
URITE(69240)
DO 115 I=19N2
B(I):BUV(I)
115 CONTINUE
DO 117 I=I9N2
DO 116 J=19N2
A(I9J)=AUV(I9J)
116 CONTINUE
117 CONTINUE

(I

C........SOLVE IN-PLANE EQUILIBRIUM EQUATIONS.

C NOTE. FOR LARGE MATRICES PROPER BANDED MATRICES MUST BE
C FORMED AND APPROPRIATE SOLUTION ROUTINES USED.

m
CALL LEOT1F(A919N291289B989HKAREA9IER)
DO 150 I=I9NPR
UCI):3(I)
NPV=I*NPR
V(I)=B(NPV)

150 CONTINUE

URITE(69201)
DO 156 M:19NPR
ITRENO=1

m

C........CALCULATE RIGHT HAND SIDE OF OUT-OF-PLANE EQUILIBRIUM EQUATION.

 

 

247

M
CALL RHSH(M9Q9ALFA9BU9NPR9N9ITREND)'
n
C........FORM COEFFICIENT MATRIX IN RIGHT HAND SIDE OF OUT-OF-PLANE
C EQUILIBRIUM EQUATION.
m
CALL BUBUCKL(M9NPR)
00 154 I=I9NPR
AUU(M9I)=AU(M9I)
154 CONTINUE
156 CONTINUE
IF(ITYP .EQ. 1) GO TO 99
IF(ITYP .EQ. 3) GO TO 119
00 131 I=I9NPR
DO 132 J=19NPR
AUCI9J)=AU¢I9J)-DR/Htta
132 CONTINUE
131 CONTINUE
HRITE(69261)

 

C........SOLVE EIGENVALUE PROBLEM.
fll

 

CALL EQZQF(AU9649BUB9649NPR92964)
CALL EQZTF(AU9649BUB9649NPR9EPSA9EPSB9Z9649IER)
CALL EQZVF(A00649BUB.64.NPR9EPSA9EPSB9ALFR.ALFI9BETA9Z964)
URITE(6925) ‘
DO 41 I=19NPR
EIGVAL(I)=ALFR(I)/BETA(I)
HRITE(6926) ALFR(1)9ALFI(I)98ETA(I)9EIGVAL(I)

41 CONTINUE

999 CONTINUE '

119 IF(ITYP .EQ. 2) GO TO 99

C........REAO INITIAL VALUES FOR POSTBUCKLING TRIALS.

REA0(59*) (U(I)9I=19NPR) '
READ (59*) (UINCR(I)9I=NP29N) 9DIF9NTRY9NITR
READ (59*) (VINCR(I)9I=NP29N)
REA0(59') QINCR
URITE(69224)
HRITE(69521)
URITE(69522)NTRY9NITR9FORC9FORICR9DIF9H
HRITE(69523)
URITE(69524) (Q(I)9I=19NPR)
HRITE(69525) QINCR
DO 111 II=19NTRY
IF(II .EQ. 1) GO TO 118
00 121 I=I9NPR
0(I)=a(I)OOINCR
121 CONTINUE
m
Cooooo...INCREASE EDGE DISPLACEMENTS.
m
00 112 L=NP29N
U(L)=U(L)'(1.O(II-1)*UINCR(L))I(1.¢(II-2)*UINCR(L))
V(L)=V(L)*(1.+(II-1)*VINCR(L))/(1.9(II-2)*VINCR(L))
112 CONTINUE
118 URITE(69262) 1190(1)
JJJ=1
DO 120 JJ=19NITR
ITREND=0

 

248

00 114 I=19NPR

81(I)=82(1)=0.0

DO 2 J=19NPR
AU1(I9J)=AU2(I9J)=AV1(I9J)=AV2(I9J)=0.0

2 CONTINUE
m
C........FORM COEFFICIENT MATRICES IN RIGHT HAND SIDE OF IN-PLANE
C EQUILIBRIUM EQUATIONS.
M

CALL UVOPRT(I)
CALL AUVMAT(I9NPR)

m
C........CALCULATE RIGHT HAND SIDE OF IN'PLANE EQUILIBRIUM EQUATIONS.
III
CALL UFUNCT(ALFA91)
114 CONTINUE
m
C........ASSEMBLE U AND V COEFFICIENT MATRICES.
HI
CALL ASSMBLCNPR)
DO 42 I=19NP
00 43 J=19N2
A(19J)=AUV¢19J)
43 CONTINUE
42 CONTINUE
Ill
C........SOLVE IN-PLANE EQUILIBRIUM EQUATIONS.
I!
CALL LEQTIFTA.19N2.128.8UV.89UKAREA.IER)
DO 44 1:1.NPR
L:I+NPR
U(1)=BUV(I)
V(1)=BUV(L)
44 CONTINUE
OO 68 J=19NPR

n
C........CALCULATE RIGHT HAND SIDE OF IN-PLANE EQUILIBRIUM EQUATIONS.
m
CALL RHSU(J9Q9ALFA9BU9NPR9N9ITREN0)
68 CONTINUE
DO 45 J=19NPR
DO 46 L=19NPR
AU(J9L)=AUU(J9L)
46 CONTINUE
45 CONTINUE
M
C........SOLVE OUT-OF-PLANN EQUILIBRIUM EQUATION.
m
CALL LEQT1F(AU 919NPR9649BU989HKAREA9IER)
DO 151 I=19NPR
U(I)=HH(I9JJ)=BU(I)
151 CONTINUE
JJ1=JJ-1
JJ2=JJ-2
IF(JJ .EQ. I) GO TO 120
m
C........CHECK CONVERGENCE.
m
00 152 J=19NPR
IF(UH(J9(JJ-1)) .EQ. 0.) GO TO 152
IF(ABS((HU(J9JJ)-HH(J9(JJ-1)))IUU(J9(JJ-1))) .GT. 01F) GO TO 160

249

152 CONTINUE
GO TO 1111
160 IF(JJ-S'JJJIIZ091589120
158 JJJ=JJJ91
DO 159 I=14NPR
m
C........APPLY CONVERGENCE -INDUCING TECHNIQUE.
fl'l
IF((UUIIQJJI-UUIIoJJ1))*(UU(IQJJII-UUCIQJJZII .GT. 0.) GO TO 159
HIII=CHUCI0JJ1)*.2-UU(IQJJI’HUCI0JJ2))/(2.*UH(IQJJ1I-UU(IQJJI-HUI
SIQJJZII
159 CONTINUE
120 CONTINUE
1111 ITREND:1
URITEC60263) JJ
URITEC69251)
DO 153 L310N
URITEC692SZIUCL)oVIL)oU(L)
153 CONTINUE
URITEC60201)
DO 688 J=IoN
M
C........CALCULATE IN-PLANE FORCESOBENDING MOMENTS AND PRINCIPAL STRESSES.
I
CALL RHSUIJQQQALFAQBUQNPRQNQITREND)
688 CONTINUE
111 CONTINUE
99 CONTINUE
1 FORMAT(4I592F5o39F15.1202F10o3)
4 FORMATI18I4)
5 FDRHAT(//oI599F7.445X99F7.4)
6 FORMAT (I95X09F3.0010X09F3.0)
25 FORMATI/I/IQIOXQ'ALFA REAL'05XQ*ALFA IMAG'QBXQ'BETA'910X9'EIG
*ENVALUES') ‘
26 FORMATIIO10X9F10.505X9F10.595X9F10.595X9F10.5)

201 FORMAT!III/44XQ‘NODE'Q7Xo‘NX'oBXo‘NY'oTXo'NXY'oI6X9'MX'08X9'MY*
$97X94HXY*97X9*MAX STRESS IN X'oZXQ'MAX STRESS IN Y PRINCIPL STRES
SS.)

220 FORMAT!III/92OXo'EIGENVECTORS CORRESPONDING TO EACH EIGENVALUE')

224 FORMATII/IIOSOXO'AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA')

240 FORMATI/I/IvSXv'ELEMENTS OF SUV-VECTOR.)

250 FORMAT(/910X9F10.5)

251 FORMATI/I/IQBX0*U-DISPLACEMEMENT*09X9'V'DISPLACEHENT*912Xv'U-DISPL
OACEMENT')

252 FORMAT(/98X0F10.5015XgF10.5015X.F10.5)

261 FORMATI/l/IQZOX04OCIHS)9' MATRIX BU *940(1HS)/I)

262 FORMAT(///920X940(IHS)Q' LOADING STEP NO .0139. Q='9F10.2Q40(1H$))

263 FORMATII/I/9'----'----¢‘-------- ITERATION NOth3.*-------------t)

265 FORMAT(////.20Xo40(1H$)9* MATRIX AUV '940(1HS)/I)

501 FORMATCIIQZOXQ‘OOOOO GEOMETRICAL INPUT DATD OOOOO’)

502 FORMATI/Io'NO OF NODES'OZXQ'NO OF INTERNAL NODES'.2X0*NO OF EXTERI
OOR NODES'QZXQ‘NO OF INTERMEDIAT NODES'02X9*POISON S RATIO'QZXQ
O'GRIO SPACING REF STUFNESS THICKNESS.)

503 FORMATI/QSXQISQIZXQI5915X9I5915X015120XvF5.3Q10X03F10.5)

504 FORMATI/I/cZOXo‘DEGREES OF FREEDOM FOR EACH NODE ¢1=INTERIOR¢ 0=BO
OUNDARY POINT9-1=EXTERIOR NODEOoc)

505 FORMATI/95X94013)

506 FORMATII/IQZOXQ'1/BETA=DR/D*I

507 FORMAT(5(5X010F10.5/))

508 FORMATC/I/oZOXo'I/BETA FOR INTERMEDIATE POINTS.)

511 FORMATIIIIQZOXQ'CORRESPONDING POINTS PARTICIPATING IN EACH NODE O

250

OPERATOROII

521 FORMATIIIIQSXQ‘NO OF LOAD POINTS*Q2XO'NO OF ITERATION‘OZXO*FORCE'O
92XO‘FORCE INCREHENT‘I2XQ'DIFFERENCE'92X9'H3A/N'I

522 FORMAT(’912X9I5012X9I595XQF9o308XQF603910X.F6.305XQF603I

523 FORMAT(//920X0’EXTERNAL TRANSVERS LOAD.)

524 FORMATI5I5X010F1202/II

525 FORMAT‘I/920X9.'-"-'-LATERAL LOAD IHCREHENT=*0F8.200'----------*)

STOP
END
M
M
M
M
M
SUBROUTINE UVOPRT(M)
M
C Ottitiittittttitttttitttttttttttfifittitfitiittittttttttttittttt
C THIS ROUTINE CALCULATES OPERATORS FOR U AND V FOR IN'PLANE
C EQUILIBRIUM EQUATION AND THE CONTRIBUTION OF EACH NODE.
C cottooctactttctt0.00.99.90.09.atoOttaoatonoot.coatcgttttccttc
M

COMMON/IIPOSOBBICCOK(13,81)OOELIBIIOOFIIOI)OHONSYHIBIIQDRORK
COMMON/ZIDUC949)oDV‘9o9)9KKI9964)9RU1(9)QRU2(9IQRV1(9)0RV2(9)
COHHON’3/BIIGSIQBZI65)OU‘BI)QVIOIIQBUIION’OBHZIONIQUIIOO’
INTEGER OF
DO 1 I:IO9
RUIIII=RUZIII=RVIIII=RVZII)=Ooo
DO 2 J=109
OUIIQJI=DVIIOJI=OQO
2 CONTINUE
I CONTINUE
X1=IolDELIKK(50HII
x2310/OELIKKIOON)I'Io/OELIKKI4OHI’
stlo/DEL‘KKIBQHII'Io/DELIKKIZOHI’
M
CQQOOOOQQCALCULATE OPERATOR FOR U1 IN X' EQUILIBRIUM
M
RUII2’305‘CC'IXI'Q25*X5)
RU1(4)=X1'X2/4.
RUI‘5’=(POS'50’.XI
RUI‘6’=XI.X2’~0
RUI‘8)305.CC.(XI.X3/4o’
M
COOQOOOQOCALCULATE OPERATOR FOR U2 IN Y’EOUILIBRIUH
M
RU2I1)=RU2(9I=0.5'88*X1
RU2I2)=‘05'CC'X2
RU2(3’=RU2(7)='RU2(I)
RU2I4)='pOSRXO
RUZ‘6’='RU2(4)
RU2‘8I3-RU2(2)
M
CocooooooCALCULATE OPERATOR FOR VI IN X-EDUILIBRIUH
M
RVIIIIzRVII9I=RUZIII
RVIIZI:-POS'X2
RV1I3I=RV1(7I=-RV1(II
RVII4I3‘05‘CC‘X3
RVII6I3'RVII4’
RVI‘B):-RV1(2I

g...

 

251

CocooooooCALCULATE OPERATOR FOR V2 IN T'EOUILIBRIUM

33033 33333

nnn

20

21

22

23

24

RV2I2I=X1'.25*X3

RV2I4I=CCRIOS 'XI'OI25‘X2)

RV2ISI=IPOS'30)4XI

RV2(6)=CCt(.StX10.125tX2)

RV2I8I=XI.X3/4o

OO 3 J=Io9

IF(NSYM(M) .NE. 0) GO TO 21

DUIJOJ’:DVIJQJI=100

GO TO 3

IF(NSYM(M)-2) 22023024

DUI103,3DU(496)=DU(799I=‘100
DUIIOII=DUI2Q2I=DUI4O4I=DU(595)=DU(707I=DU(808)=100
DVIJOJI=100

GO TO 3

DV‘107,3DVI2!8I=DV(309)=-1.0
DVIIQII=DVI292I=DVI393I=DVI4O4I:DV(505)=DV(646)3100
OUCJQJI=1.0

GO TO 3

DUI193)=OUC4Q6I=DU(799)='1.0
DVI197)=DVI2ORI=DV(399)=-IQO
DUIIQII=DU(242I=DU(404I=DU(595)=DUI7’7I=DU(808)3100
DV‘IQII=DVIZQ2I=DVISgOI=DVI494I=DVISQSI=DVI6OGI=100
CONTINUE

RETURN

ENC

SUBROUTINE AUVMAT(M9NPR)

CALCULATES SUB-MATRICES FOR IN-PLANE EQUILIBRIUM EQUATIONS.

COMMON/2/DU(999)QDV¢999TvKKI9v64D9RU1(9)0RU2(9)oRVI(9)9RV2(9)
COMMON/3/81(65)982(65)oUCBI)oVCBI)vBU1(64).BUZ(64).U(100)
COMMON/4i AU1(64964).AU2(64.64).AVII64964)9AV2(64.64D.AUVI1289128)
.fififfifitififl It'.....*...if...*.............fii.iiO.Ci....OOOQC'OQ
CALCULATE A-MATRIX FOR U IN X EQUILIBRIUM

.fitffiflfiifltfi iiiiifiifiififiQ......tfifiififltfifitfiti.itifi0.......O.....*.
DO 2 L=Io9

IF(KKILoM) .GT. NPR) GO TO 3
DO 1 I=199
AU1(MQKK(L9M))=AU1(M9KK(L9M))+DU(L9I)tRU1(I)
AU2(M9KK(L0M))=AU2(M9KK(L0M))+DU(L.I)*RU2(I)I4.
AVIIMQKK(L9H))=AVI(M0KK(L9M))ODVIL.I)*RV1(I)/4.
AV2(M9KK(L0H))=AV2(H0KK(L9M)I*OV(L1I)*RV2(I)
CONTINUE
GO TO 2
DO 4 J=1.9
BI(M)=BIIM)-DU(L9J)'RU1(JI'UIKKILvM))-OV(L.J)*RV1(J)'V(KK(L9M)’14.
BZ(M)=BZ(M)-DU(L.J)¢RU2(J)OU(KK(L9M))I4.-DV(L9J)tRV2(J)OV(KK(L.M))
CONTINUE

CONTINUE

RETURN

END

annna

annna

252

SUBROUTINE ASSMBLINPR)

itfiitfi...tfittittfitifitfiifiiiitfifiOiiittflfitfiitififiitfifiifiiQtttttiifi

ASSEMBLE MATRICES FOR IN-PLANE EQUILIBRIUM EQUATIONS.

Q.OtifitflfiiiiiififitiififitiitififtflIfifiiittfltfitii.fifiiifififitfitfiiii...

COMMON/4’ AUIC64964).AU2(64064).AVI(64964I9AV2(64.64).AUVCIZBQIZBI
COMMON/SI BUVII64).BUV2(64)QBUV(128)

- DO 1 I=IQNPR

NPI=I+NPR

auvcx)=euv1(x)

auvcupx»:auv2xt: -

DO 2 J=IQNPR ‘
NPR2:NPR¢J '
auv¢t.u)=au1¢1.a)

AUV(IoNPR2)=AV1(IoJ)

auvcupx.u)=nu2¢1.a)

nuv¢~p1.~paz)=av2¢1.4) LL
co~71~uc

convruu:

RETURN

sun

SUBROUTINE HFUNCTIALFAQM)

fififififiiifitfififititfltfiiflfifltfit...ititfifittfiiiitfitiifltittfiifitttttiflfi

CALCULATE RIGHT HAND SIDE OF IN-PLANE EQUILIBRIUM EQUATIONS.

OiflfiifiifitfiiifififififitttttfiiOffififfifitfitifiOiffifitififiififitifiiOitfiiflti.

COHHON/l/PO?.BB.CC.K(13.81).OEL(81).DF(101)oH.NSYM(81).DR.RK
COMMON/SIBITGS).82(65).U(81).Vtal).801(64).BH2T64).U(100)_
COMMON/SI BUV1(64).8UV2(64).BUVI128)
UX=.5*(U(K(8.H))-U(K(6.H)))IH
HY:.5«ALFA~(U(K(11.H))-U(K¢3.M)))IH
UXX:¢U(K(8.N))¢H(K(6.H))-2.tH(K(7.M)))IH0-2
UVY:ALFA¢*2*(U(K(3.H))90(K(11.H))-2.'U(K(7.H)))IHotz
UXY=.250ALFA*(U(K(2.H))-U(K(4.H))-H(K(10.H))¢U(K(12.H)))/Htt2
X1=1.IOEL(K(7.H))

x2=.5.(1./DEL(K(89M))-1.IOEL(K(6.H)T)lH
x3=.5tALFAt¢1.IDEL(K(11.H))-1./DEL(K(3.H)))IH
BUI(H)=x1-uxx.uxo.5cx1o(aacuxv-uvocc-uvvtux)+.5.x2-(UXo-zopos-
SHYO'ZTOCCOX3*UX¢UYO.5 '
802(H)=.5*x1tBBtHXY-UXO.SOXIOCCOHXX*UYO.5-Cch2*HXtUY0.50x3-(HY-t2
soPOSOUX**2)¢x1tHYYtHY

BUV1(H):BI(N)-801(H)tht2

BUVZCH)=BZIMT-8UZTN)th02

RETURN

END

253

SUBROUTINE KKVECTINPR)
m

C tittttttttittittttttttttttititttfitttitflttttOitttttiiittttttti
Co.......STDRES A 3 BY 3 STENCIL PATTERN FROM USUAL 13 NODE STENCIL.
C ti...tQttfiltfiiitfitttttttfitttttOtifitttitiittttittttltOttitttt.

M

COMMON/I/POSQBBQCCQKI13081,ODELCBIIQDFI101)cHoNSYMI81)9DR9RK
COMMON/2/DU(999)QDV(919)0KKI9964IQRU1(9)QRU2(9)oRVICB).RV2(9)
DO 1 J31 .NPR

KKIIQJ)=K(20J)

KKCZQJI=KC3QJI

KKC3QJI=KC49JI

KKI4QJI=K(64J)

KKK59J1=KTTQJI

KK‘69JI=K(89J)

KKITQJI=KIIOQJI

KKCBQJI=KCIIOJ)

KK‘9QJ)=K(129JI

1 CONTINUE

RETURN

END
M
M
M
M
M

SUBROUTINE RHSUCMOQOALFAvBUQNPRQNQITREND’
M
C titt'ttttttttttttttitttttittttOifititttittfitiitttttttti.cit...
C THIS ROUTINE CALCULATES RIGHT HAND SIDE OF OUT-OF-PLANE
C EQUILIBRIUM EQUATION AS HELL AS IN-PLANE FDRCESQBENDING
C MOMENTS AND PRINCIPAL STRESSES.
C tifitttfitttfittottfitfitfifititttttttfittttttfittttttfitttt.t000......
M

DIMENSION Q(BI)QBU(81)
COMMON/l/POSQBBQCCQKCI3981IoOELCBI)QDFIIOIIvHoNSYMCBIIQORoRK
COMMON/3/BIC6SIQBZIBSIOUCBIIQVCBIIoBU1(64)oBH2(64)QU(100)
COMMON/6/DELT(81)cDELTA(8 )9AM(4981)QDELA(8 ’vIBCCBII
COMMON/BIXNXQYNYQXNY
IF(NSYM(M) .NE. 0) GO TO 104
UX305.IU(K(80MI)‘UIKIGGMIII/H
VY=ALFA'.5'(V(K(IIQMII-V(K(30M!))IH
GO TO 20

104 GO TO (11.12913014915916vITI NSYM(M)

11 UX=-U(K(69M))/H
VY=ALFA'.5*(V(K(11¢M))-V(K(39M)I)IH
GO TO 20

12 UX=.5*(U(K(80M))-U(K(69M)IIIH
VY3'ALFA'V(K(39MIIIH
GO TO 20

13 UX='U(K(69M))IH
VY=-ALFA'V(K(39M))IH

20 UY=ALFA*.50(U(K(IIQMII-U(K(3gM)))lH
VX3.5.(V(K(MOM)I‘VIKI6QMI)I/H

14 GO TO 25

15 UX3(U(K(80M)I‘UIK(70MIII/H
VX:(V(K(89M))-V(K(70M)))IH
UY=ALFA'.5'(U(K(IIQMII-UCKC39MII)IH
VY=ALFA*.5*(V‘KIIIQMII'VIKI3QMII)IH
NP2=NPR¢1
IF(M .EQ. NP2) VY:ALFA*IVIK(70M)I'VIKC39M))I/H

 

 

22

16

17

21

25

202

254

IF(IBCCH) .E0. 2) 60 TO 22
H(K(2.M))=U(K(49M))
H(K(6.M))=H(K(8.MT)
u¢x¢1o.n»::u¢x¢12.H))
so TO 25
HIKIIOoM))=-U(K(129M))
UIKI2.M))=-H(K(4.M))
u:x¢5.n»:=-utx¢s.n))
so TO 25
UY:ALFA*(U(K(II.M)T-UIKITQM)T)lH
VY=ALFAtCV¢K(11.M))-V(K(T.M))IIH
UX=.5‘(U(K(8.M)T-U(K(6.M))D/H
IF(M .Ea. u) UX=(U(K(70M))-U(K(6.M)))lH
VX=.5'(V(K(8.M))-V(K(6.M))T/H
UIKI4.M))=H(K¢12.M))
IF(IBCIH) .EO.'2) H(K(4.M))=-U(K(129M)I
GO TO 25
UX=(U(K(89M))-U(K(7.M)T)IH
VX=(V(K(8.M))-V(K(7.M)))/H
UY=ALFA0(U(K(II.M))-U¢K(79M))I/H
VY=ALFAtCVIKIIIQMI)-V(K(7.M)))lH
IF(IBCCH) .E0. 2) GO TO 21
H(K(6.H))=U(K¢8.M))
U(K(3.M))=U(K(119M))
ucxcz.n))=H(K¢4.H))=H(K(12.H))
so to 25 ‘
u¢K(6.M))=-9(K¢3.n))

TU(K(30M11=-U(K(119M1)

UTK(2.H))=U(K(4.H))=-U(K(12.H))
ux=.5t(H(K(8.H))-U(K(6.H)))IH
UY:.5¢ALFA0(H(K(11.H))-2(K(3.H)))lH
UXX=(U(K(8.M)IOUIKISoM))-2.*H(K(7.MJ)D/HttZ
UYY=ALFAt22t(H(K(3.H)DoutK(llon))-2.-U(K(7.H)))lHrtz
UXY:.25*ALFAc(H(K(2.H))-H(K(4.H))-H(K(10.HD)¢H(K(12.M)))lHt-Z
x1=1./DEL(K(1.HT)
XNx=x1tRKt(UXO.5*Ux0-29POSOVY¢.5*POScUY¢fiZTICBBtCC)
YNY=x1-RK¢(VYO.5*HY¢*20POS'UX+POS*.5fiUXtt2)/(88tCC)
XNY=.59X1'RKOIUYOVXOUXOUYTIBB
IF(ITREND .E0. 0) GO TO 2
DP=DRaDELT(H)
XH=-DPt(HxxoPos-UYY)
YH:-OP¢(HYYOPOS'HXX)
XYM=OP*CC*UXY
T2=12.tBBfiCCtDP/(x1.RK)
T:SORT(T2)
SIGHAX=6otxan2
SIGMAY=6.0YHIT2
SIGMANX=XNXIT

SIGHANY=YNYIT
XSIGMAX=ABSISIGMAXIOABSCSIGMANX)
YSIGHAX=ABS(SIGHAY)+ABS(SIGHANY)
RAD=SQRT(I(XSIGMAX-YSIGMAXIIZ.ItOZOIXNY/(ZotT)3'02)
PRSTRES=(XSIGHAX¢YSIGMAXTI2.+RAO
HRITEtsgzoz) H.XNX.YNY.XNY.xnoYH.XYH.XSIGHAX.YSIGHAX.PRSTRES
FORMAT!l/osx.IZ.3X.3F10.4.5X.3F10.4.10x.2F12.4.5X.F12.4)
BUCMT=O(M)¢XNX*HXXOYNYtHYY¢2.4XNYtHXY
BUIM)=BU(M)*H*¢4IDR «
RETURN
END

255

M

M

M

SUBROUTINE BHBUCKLIHONPR’

C it!!!it...tttfititittttttttttttttfittifitfitctttitttttttQtiitttti
C........FORM COEFFICIENT MATRIX IN RIGHT HAND SIDE OF Z-EQUILIBRIUM EQUATION.
C 11.5.20....92....ottttttttatgttttcggoootttoctttcttottc..tttctgo.

M
COMMON/IIPOSoBBoCC9K(13981)oDEL(81)QDF(101)oHoNSYM(81)oDRgRK
COMMON/TIRUII3)QOUII3QI3)5AHI64964)98UB(64064)
COMMON/BIXNXvYNYoXNY
IF(K‘20") .GT. NPR) GO TO 2
BUBIMQKIZQM))=BHB(M.K(20M))OXNY/(2.*Hii2)

2 IF(KISQM) .GT. NPR) GO TO 3
BUB(M0K(39M))=BUB(M9K(35M))OYNY/H**2
3 IF(KI4gM) .GT. NPR) GO TO 4

BUBIM'KC4'M))33UBIM0KI4QM))-XNY/(2..H'02)

IF(KI69M) .GT. NPR) GO TO 5

BUBIMOKC60M))=BHB(M9K(69M))OXNXIH002

5 IF(KITQM) .GT. NPR) GO TO 6

BHBIMQK‘TQM))=BUB(M9K(70M))-2.*(XNXOYNY)/Ht*2

6 IF(KCBQM) .GT. NPR) GO TO 7

BUBIMQKCBQM))=BHB(M9K(89M))+XNXIH4*2

7 IF(KIIO'M) .GT. NPR) 50 TO 3

BUBIMOK(100M))=BUB(MOK(100M))-XNY/(2.*H*fi2)

IF(KIIIQM) .GT. NPR) 50 TO 9

BHBCMQKIIIQM))=BUB(M9K(110M))OYNY/Hti2

9 IF(KIIZQM) .GT. NPR) GO TO 10
BUBIMQKCIZQM))=BHB(M5K(129M))OXNY/(ZofiH9*2)
10 RETURN
END

O

 

 

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"HIM!"