AN ABSTRACT OF A THESIS APPLICATION OF GROEBNER …equations, 3) and Groebner Basis is employed to decouple the algebraic equations to find analytic solutions in terms of the material

AN ABSTRACT OF A THESIS

APPLICATION OF GROEBNER BASES TO

GEOMETRICALLY NONLINEAR ANALYSIS OF

AXISYMMETRIC CIRCULAR ISOTROPIC PLATES

Timothy M. Harrell

Master of Science in Civil Engineering

This thesis demonstrates a new application of Groebner basis by finding an analytical

solution to geometrically nonlinear axisymmetric isotropic circular plates. Because technology is

becoming capable of creating materials that can perform materially in the linear elastic range

while experiencing large deformation geometrically, more accurate models must be used to

ensure the model will result in realistic representations of the structure. As a result, the governing

equations have a highly nonlinear and coupled nature. Many of these nonlinear problems are

solved numerically. Since analytic solutions are unavailable or limited to only a few simplified

cases, their analysis has remained a challenging problem in the engineering community.

On the other hand, with the increasing computing capability in recent years, the

application of Groebner basis can be seen in many areas of mathematics and science. However,

its use in engineering mechanics has not been utilized to its full potential. The focus of this thesis

is to introduce this methodology as a powerful and feasible tool in the analysis of geometrically

nonlinear plate problems to find the closed form solutions for displacement, stress, moment, and

transverse shearing force in the three cases defined in Chapter 4.

The procedure to determine the closed form solutions developed in the current study can

be summarized as follows: 1) the von Kármán plate theory is used to generate nonlinear

governing equations, 2) the method of minimum total potential energy combined with the Ritz

methodology converts the governing equations into a system of nonlinear and coupled algebraic

equations, 3) and Groebner Basis is employed to decouple the algebraic equations to find analytic

solutions in terms of the material and geometric parameters of the plate. Maple 13 is used to

compute the Groebner basis. Some examples of Maple worksheets and ANSYS log files for the

current study are documented in the thesis.

The results of the present analysis indicate that nonlinear effects for the plates subjected

to larger deformation are significant for predicting the deflections and stresses in the plates and

necessary compared to those based on the linear assumptions. The analysis presented in the thesis

further shows the potential of the Groebner basis methodology combined with the methods of

Ritz, Galerkin, and similar approximation methods of weighted residuals which may provide a

useful procedure of analysis to other nonlinear problems and a basis of preliminary design in

engineering practice.

APPLICATION OF GROEBNER BASES TO

GEOMETRICALLY NONLINEAR ANALYSIS OF

AXISYMMETRIC CIRCULAR ISOTROPIC PLATES

______________________

A Thesis

Presented to

the Faculty of the College of Graduate School

Tennessee Technological University

by

Timothy M. Harrell

______________________

In Partial Fulfillment

of the Requirements of the Degree

MASTER OF SCIENCE

Civil Engineering

______________________

August 2014

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iii

DEDICATION

This thesis is dedicated to my late grandfather who kindness and compassion is something I

will always look up to. Also, to my parents and wife who have always shown faith in me.

They inspire me to make myself better.

iv

ACKNOWLEDGEMENTS

I would like to thank Dr. Jane Y. Liu for being my advisor and contributing so much

time and patience during this process. I would also like to thank Dr. John Peddieson and Dr.

Guillermo Ramirez for serving on my committee. I would like to thank the Department of

Civil and Environmental Engineering for funding my graduate degree. Last but not least, I

would like to thank my wife and family for the encouragement throughout my time in

graduate school.

v

TABLE OF CONTENTS

Page

LIST OF TABLES ........................................................................................................................ viii

LIST OF FIGURES ........................................................................................................................ ix

NOMENCLATURE ..................................................................................................................... xiii

CHAPTER 1 - INTRODUCTION ................................................................................................... 1

1.1 Motivation and Overview .......................................................................................................... 1

1.2 Scope of Work ........................................................................................................................... 3

CHAPTER 2 - LITERATURE REVIEW ........................................................................................ 5

2.1 Plates .......................................................................................................................................... 5

2.2 Groebner Basis ........................................................................................................................... 8

CHAPTER 3 - THEORETICAL BACKGROUND ...................................................................... 10

3.1 Review of Thin Plate Theory ................................................................................................... 10

3.1.1 Kirchoff Plate Theory ................................................................................................... 12

3.1.2 Transformation of Coordinate System .......................................................................... 15

3.1.3 Geometrically Nonlinear Plate Theory ......................................................................... 18

3.2 Principle of Minimum Total Potential Energy ......................................................................... 21

3.2.1 Total Potential Energy Functional ................................................................................ 22

3.2.2 The Ritz Method ........................................................................................................... 23

3.3 Groebner Basis ......................................................................................................................... 24

3.3.1 Preliminary Concepts from Abstract Algebra ............................................................... 25

3.3.2 Monomials and Polynomials ......................................................................................... 25

vi

Page

3.3.3 Monomial Ideals and Dickson’s Lemma ...................................................................... 26

3.3.4 Polynomial Ideals and Hilbert’s Basis Theorem ........................................................... 27

3.3.5 Definition of a Groebner Basis ..................................................................................... 28

3.3.6 Affine Variety: the link between the Groebner bases and the original generating set

functions ................................................................................................................................. 29

CHAPTER 4 - IMPLEMENTATION AND METHOD VALIDATION ...................................... 32

4.1 Case Studies ............................................................................................................................. 32

4.1.1 Case 1: Fully Clamped Circular Plate ........................................................................... 32

4.1.2 Case 2: Simply Supported Immovable Circular Plate ................................................... 33

4.1.3 Case 3: Simply Supported Immovable Circular Plate with Overhang .......................... 34

4.2 Shape Functions ....................................................................................................................... 36

4.3 Trial Functions for Ritz Method .............................................................................................. 37

4.4 Computer implementation ....................................................................................................... 39

4.4.1 Maple 13 ....................................................................................................................... 40

4.4.2 ANSYS 13 .................................................................................................................... 41

4.5 Validation of Current Study ..................................................................................................... 45

4.5.1 Validation of Shape Functions ...................................................................................... 45

4.5.2 Validation with Load .................................................................................................... 52

4.5.3 Current Study Solution versus Two Coefficient Ritz Solution ..................................... 55

CHAPTER 5 - RESULTS AND DISCUSSION ............................................................................ 58

5.1 Results for Fully Clamped Circular Plate (Case 1) .................................................................. 58

vii

Page

5.2 Results for Simply Supported Immovable Edge Circular Plate (Case 2) ................................ 67

5.3 Results for Simply Supported Immovable Edge Circular Plate with Overhang (Case 3) ........ 76

CHAPTER 6 - CONCULSION ..................................................................................................... 87

REFERENCES .............................................................................................................................. 89

APPENDICES ............................................................................................................................... 96

APPENDIX A - DEFINITIONS OF SOME PRELIMINARY CONCEPTS ........................ 97

APPENDIX B - EXAMPLES OF AFFINE VARIETIES ................................................... 102

APPENDIX C - COMPARISION OF EXACT LINEAR SOLUTION FROM THE

LINEAR THEORY TO THE SOLUTION FOUND FROM THE RITZ METHOD FOR

THE CIRCULAR PLATE WITH OVERHANG (CASE 3) ............................................... 108

APPENDIX D - MAPLE FILE WITH CURRENT STUDY SOLUTION FOR

GEOMETRICALLY NONLINEAR CIRCULAR PLATE WITH FIXED EDGE (CASE

1) .......................................................................................................................................... 118

APPENDIX E - ANSYS LOG FILE CASE 3 IMMOVABLE SUPPORTED PLATE

WITH OVERHANG ........................................................................................................... 135

APPENDIX F - RITZ METHOD COEFFICIENTS EQUATIONS AND GROEBNER

BASIS EQUATIONS FROM CASE 2 AND CASE 3 ........................................................ 140

APPENDIX G - STRESS IN THETA DIRECTION AND TRANSVERSE SHEAR

PROFILE FOR EACH CASE ............................................................................................. 144

VITA ............................................................................................................................................ 151

viii

LIST OF TABLES

Page

Table 4.1: Case 1: Dimensionless ��/� versus Dimensionless Load �/� and % Error in

��/� (�/ = 1/100 and = 0.3) ........................................................................... 52


��/� (�/ = 1/100 and = 0.3) ........................................................................... 53


��/� (�/ = 1/100, �/ = 1.5, and = 0.3) ..................................................... 54


��/� (�/ = 1/100 and = 0.3) ........................................................................... 55


��/� (�/ = 1/100 and = 0.3) .......................................................................... 56


��/� (�/ = 1/100, �/ = 1.5, and = 0.3) ..................................................... 57

ix

LIST OF FIGURES

Page

Figure 3.1: Geometry of a Rectangular Thin Plate and Coordinate System .................................. 11

Figure 3.2: Rectangular Plate Variables and Sign Convention ...................................................... 11

Figure 3.3: Kirchoff Kinematics in ��-plane ................................................................................. 13

Figure 3.4: Geometry of Circular Thin Plate and Coordinate System ........................................... 15

Figure 3.5: Sign Convention for the Circular Plate Variables ....................................................... 16

Figure 3.6: In-plane Kinematics for Large Plate Deformation ...................................................... 19

Figure 3.7: Pictorial Description of a Field, Ring, and Polynomial Ring ...................................... 26

Figure 3.8: Pictorial Description of an Ideal .................................................................................. 27

Figure 4.1: Case 1: Fully Clamped Circular Plate ......................................................................... 33

Figure 4.2: Case 2: Simply Supported Immovable Circular Plate ................................................. 34

Figure 4.3: Case 3: Simply Supported Immovable Circular Plate with Overhang ........................ 35

Figure 4.4: Boundary Conditions of Finite Element Quarter Plate Model for Each Case Study ... 42

Figure 4.5: Convergence Study for Case 1 (� = 5��, ν = 0.3 , � = 29� + 6��, =10��, and � = 0.1��) ............................................................................................ 43

Figure 4.6: Mesh of Finite Element Model .................................................................................... 44

Figure 4.7: Case 1: Transverse Displacement versus Radial Distance (� = 5��, ν = 0.3 ,

� = 29� + 6��, and �/ = 1/100) ..................................................................... 46


� = 29� + 6��, and �/ = 1/100) ..................................................................... 47


� = 29� + 6��, �/ = 1.5, and �/ = 1/100) .................................................. 48

Figure 4.10: Case 1: Radial Stress versus the Radial Distance of the Plate (� = 5��, ν = 0.3 , � = 29� + 6�� , and �/ = 1/100) ..................................................... 49

x

Page

Figure 4.11: Case 2: Radial Stress versus the Radial Distance of the Plate (� = 5��, ν = 0.3 , � = 29� + 6�� , and �/ = 1/100) ..................................................... 50

Figure 4.12: Case 3: Radial Stress versus the Radial Distance of the Plate (� = 5��, ν = 0.3 , � = 29� + 6�� , �/ = 1.5 and �/ = 1/100) ................................... 51

Figure 5.1: Case 1: Dimensionless Displacement versus Dimensionless Load varying

Poisson’s Ratio, .................................................................................................... 62

Figure 5.2: Case 1: Dimensionless Transverse Displacement versus Dimensionless Load

varying Thickness Ratio, �/ (! = 0.3) ................................................................. 63

Figure 5.3: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse

Displacement varying Radial Distance "/ ( = 0.3 and � = �/2) ....................... 64

Figure 5.4: Case 1: Dimensionless Moment versus Dimensionless Transverse Displacement

( = 0.3) ................................................................................................................. 65

Figure 5.5: Case 1: Moment Profile (�#/(��#) = 500/29 and = 0.3) ................................... 66

Figure 5.6: Case 1: Stress Profile through the Thickness (�#/(��#) = 500/29 and = 0.3) .. 67

Figure 5.7: Case 2: Dimensionless Transverse Displacement versus Dimensionless load

varying Poisson’s ratio, ........................................................................................ 71


varying Thickness Ratio, �/ ( = 0.3) ................................................................. 72

Figure 5.9: Case 2: Dimensionless Radial Stress versus Dimensionless Maximum Transverse

Displacement varying Radial Distance "/ ( = 0.3 and � = �/2) ....................... 73


( = 0.3) ................................................................................................................. 74

Figure 5.11: Case 2: Moment Profile (�#/(��#) = 500/29 and = 0.3) ................................. 75

xi

Page

Figure 5.12: Case 2: Stress profile through the Thickness (�#/(��#) = 500/29 and =0.3) .......................................................................................................................... 76

Figure 5.13: Case 3: Dimensionless Displacement versus Dimensionless Load varying

Poisson’s Ratio, (�/ = 1.5) ................................................................................ 81


varying Thickness Ratio, �/ ( = 0.3 and �/ = 1.5) .......................................... 82

Figure 5.15: Case 3: Non-dimensional radial stress versus non-dimensional maximum

transverse displacement varying radial distance "/ ( = 0.3, �/ = 1.5, and

� = �/2) .................................................................................................................... 83


( = 0.3 and �/ = 1.5) .......................................................................................... 84

Figure 5.17: Case 3: Moment Profile versus Radial Distance varying �/ (�#/(��#) =500/29 and = 0.3) ................................................................................................ 85

Figure 5.18: Case 3: Stress Profile through the Thickness (�#/(��#) = 500/29, �/ = 1.5,

and = 0.3) .............................................................................................................. 86

Figure B.1: The Affine Variety, %(& − �(), in ℝ( ...................................................................... 103

Figure B.2: The Affine Variety, %(�( + &( + �( − 25), in ℝ* .................................................. 104

Figure B.3: The Affine Variety, %(�* − &* − �*), in ℝ* ........................................................... 105

Figure B.4: The Affine Variety, %(& − �*, & − �) = {(0,0), (−1,−1), (1,1)} in ℝ( ................ 106

Figure B.5: The Affine Variety %(�( + &( + �( − 25, �) = �( + &( − 25 in the �& plane. ..... 107

Figure G.1: Case 1: Transverse Shearing Profile (�#/(��#) = 500/29 and = 0.3). ............ 145

Figure G.2: Case 1: Stress in Theta Direction Profile through the Thickness (�#/(��#) =500/29, �. = 0.3). ........................................................................................... 146

Figure G.3: Case 2: Transverse Shearing Profile (�#/(��#) = 500/29 and = 0.3). ............ 147

xii

Figure G.4: Case 2: Stress in Theta Direction Profile through the Thickness (�#/(��#) =500/29, �. = 0.3). ........................................................................................... 148

Figure G.5: Case 3: Stress in Theta Direction Profile through the Thickness (�#/(��#) =500/29, �/ = 1.5, �. = 0.3) ......................................................................... 149

Figure G.4: Case 3: Stress in Theta Direction Profile through the Thickness (�#/(��#) =500/29, �/ = 1.5, �. = 0.3). ....................................................................... 150

xiii

NOMENCLATURE

/ Variable to Simplify Expression

0 Variable to Simplify Expression


0( Variable to Simplify Expression


234 Shear Strain in "5 plane

2�6 Shear Strain in �& plane

2�7 Shear Strain in �� plane

267 Shear Strain in &� plane

8 Variational Operator (Chapter 3)

8 Variable to Simplify Expression (Chapter 5)

944 Normal Strain in 5 direction

933 Normal Strain in " direction

9�� Normal Strain in � direction

966 Normal Strain in & direction

977 Normal Strain in � direction

5 Theta Component Polar Coordinates

xiv

: Expression Simplification Variable

Π Total Strain Energy

<44 Normal Stress in 5 direction

<33 Normal Stress in " direction

<�� Normal Stress in � direction

<66 Normal Stress in & direction

<77 Normal Stress in � direction

=�6 Shear Stress in �& plane

=�7 Shear Stress in �� plane

=67 Shear Stress in &� plane

Poisson’s Ratio

>?@ Shape Function for in-plane displacement

>A@ Shape Function for in-plane displacement

Radius from center to support

� Radius from center to overhang free edge

B Plate Bending Rigidity

� Modulus of Elasticity

C Shear Modulus

xv

DE Leading Coefficient

DF Leading Monomial

DG Leading Terms in a Polynomial

F�� Moment in the � direction

F66 Moment in the y direction

F�6 Moment in the �& direction

F33 Moment in the " direction

F44 Moment in the 5 direction

H33 Normal force in " direction

H44 Normal force in 5 direction

� Transverse uniform distributed load

I44 Transverse Shearing Force in the 5 direction

I33 Transverse Shearing Force in the " direction

I�� Transverse Shearing Force in the � direction

I66 Transverse Shearing Force in the & direction

" Radial Component Polar Coordinates

J In-plane Displacement Component in � direction

J3 In-plane Displacement

xvi

J3KL In-plane Displacement in overhang region

M Total Strain Energy

MN Strain Energy from bending

M� Strain Energy from membrane

! In-plane Displacement Component in & direction

M Potential Energy

� Transverse Displacement

�KL Transverse Displacement in overhang region for Case 3

ℂ The Set of Complex numbers

ℚ The Set of all Rational numbers

ℝ The Set of all Real numbers

ℤ The Set of all Integers

ℤR The Set of all Positive Integers

1

CHAPTER 1

INTRODUCTION

The goal of this thesis is to apply the Groebner basis methodology to generate analytical

solutions in plate problems as a relative new alternative to the commonly used numerical methods

such as the finite element methods to the engineering community. In this thesis, it will be

accomplished by exploring a method in which Groebner basis is used as a tool of solving a

system of nonlinear algebraic equations to find a purely analytical solution for an axisymmetric,

geometrically nonlinear circular isotropic plate. Although Groebner basis has been found useful

in the mathematics community, there have not been many applications found for applied

engineering problems; see section 2.2 for examples.

1.1 Motivation and Overview

Today, industries are trying to become more cost competitive. They are trying to achieve

this through a variety of methods such as weight savings. One way to try and find weight savings

is by using more advanced materials that has highly nonlinear and coupled nature. The

geometrically nonlinear behavior in the structure components is one of them, such as the material

of the component is in the linear elastic range even when subjected to large displacements. When

structures are subjected to large displacements more accurate models must be used. These more

accurate models have a higher degree of difficulty in analyzing. However, they yield a more

accurate answer and ultimately result in a much lighter weight structure than developing a design

using traditional linear assumptions. Comparisons have shown that the linear equations can

produce a large inaccuracy in the analysis at high load levels [1]. Much of the research when

dealing with large displacements is analyzed by using the finite element method or some other

2

numerical method. A quick literature search will show hundreds of papers with different

numerical methods to find a solution. These methods can be very computer and user intensive

requiring a large amount of parameters to be run to determine the best design. Unfortunately,

there are very few analytical equations that describe the geometrically nonlinear plates and most

all are merely solutions for the maximum stress or deflection at a certain point. None have shown

a purely symbolic closed form solution.

For geometrically nonlinear analysis of plates, the analytical solutions tend to be shied

away from do to its complexity. The complexity in finding a solution comes from including the

addition of the von Kármán equations for strain. In traditional linear assumptions, the in-plane

displacement is considered negligible. However, when subjected to large deflections the

assumption that the in-plane displacement is negligible is no longer true, so the addition of the

von Kármán strains account for the membrane strains due to in-plane loading. This makes the

plate problem more complex by having to account for the in-plane and out-of-plane displacement.

Through the use of the Ritz method approximate the displacement function can be used to

minimize the total potential energy in a system and the result is a system of algebraic equations.

Groebner basis is used to take this system of equations developed by the Ritz method and

decouple the multivariate equations into a solvable system of equations. This tool, Groebner

Basis, is not utilized very often in the engineering community and this thesis will demonstrate

how powerful it can be to help solve complex system of equations. There are many engineering

problems that can be helped by the use of Groebner basis. And now, many symbolic computation

software packages have created algorithms to easily compute Groebner basis making it accessible

to many engineers. Therefore, it is vital to try and make the existence of this tool known to the

community.

3

1.2 Scope of Work

The primary objective of this study is to develop a general methodology of using

Groebner bases in conjunction with the energy method for solving geometrically nonlinear

isotropic circular plates analytically. As an example, three case studies defined in Chapter 4 are

analyzed using the proposed methodology to obtain the closed form solutions; and evaluate the

performance of using this method for the geometrically linear and nonlinear behaviors of the

plates by comparing the solutions from this method to the numerical solutions from a commercial

finite element software package, ANSYS.

This thesis is organized into six chapters as follows. In Chapter 2, a brief literature review

of plate theories and Groebner bases. The plate section provides a brief history and many

examples of plate analysis. Also, included are references of plate solutions based on different

methods such as finite strip and finite element. The Groebner basis section discusses some

historic background and some limited applications found in different engineering fields such as

ones in railway interlocking systems, buckling problems, and robotics.

In Chapter 3, the two most common plate theories, Kirchoff (classical plate theory) and

von Kármán (lager strain theory) theories, are reviewed briefly. The total potential energy is

discussed and presented in the polar coordinate system for an axisymmetric plate problem and

followed by a review of the concept of Ritz method. Finally, a detailed review of Groebner basis

is provided to give the reader a comprehensive background on the subject.

In Chapter 4, three case studies are defined in Section 4.1 used as examples for the

validation of the method. The trial functions used for the Ritz approximation are presented for

each case. Also, the computer software packages used in the study are presented. In the final

section, a comparison of the results between ANSYS and the current study is presented with

detailed discussions. The comparisons show good agreement between the two solutions.

4

Chapter 5 mainly demonstrates the results of the parametric studies for each case. First

the closed form solutions for the transverse displacement, moment, shear force, and stress

function are calculated and presented from each case study. Then based on the closed form

solutions, a set of plots from parametric study in terms of the material and geometric properties

for each case are generated and presented to show the effects of the plate performance under

different parameters.

Finally, the conclusion and recommended future work are given in Chapter 6.

5

CHAPTER 2

LITERATURE REVIEW

The current topic being discussed is using Groebner Basis to solve a geometrically

nonlinear plate problem. The majority of literature on the subject of geometrically nonlinear

plates indicates that they have been solved by using numerical methods such as finite element

analysis. The availabilities of analytical solutions for geometrically nonlinear plate problems are

very few due to the fact that the coupling of the plate equations and the von Karman constraints

makes the problem highly nonlinear. The methodology developed in this study is not found

currently in the body of knowledge and has shown itself to match proven research. The following

sections will summarize the works found on plate analysis and Groebner basis.

2.1 Plates

Plate analysis is currently used to determine adequacy of design across the civil, marine,

aerospace, and many other engineering disciplines. Plates are very important structural

components used in dam walls, slabs in floors of buildings, hulls of ships, aircraft fuselages, and

much of the propulsion system in rockets. Before advanced mathematics was developed, much of

the work was handed down by so called "rules of thumb" from generations of early engineers.

One of the first "rules of thumb" passed down was for designing floor slabs in buildings [2]. The

designers decided the strength of the plate by how thick it needed to be and the designer used this

method for many years until mathematicians tried to describe this physical problem and thickness

is the first described variable that was associated with modern plate assessment.

The first mathematical models of plates were developed by Sophie Germain who

correctly described the differential equation that is the foundation of plate theory. Navier was the

6

first to apply Bernoulli's beam hypothesis to plates and correctly described the plate constant and

the first differential equation subject to transverse loads by using Fourier series to force a

solution. This solution only worked for simply supported rectangular plates, and could never be

used for different boundary conditions. Then Kirchhoff, who is considered the founder of

extended plate theory [3], was able to develop the first complete theory of plates with

supplementary boundary conditions. This development led to the boundary conditions being able

to be described as functions of displacement and their derivatives. One of his most significant

discoveries is that in-plane displacement cannot be neglected when the displacement becomes

large. Much of the linear solutions can be found in classic books such as Timoshenko's Theory of

Plates and Shells [4].

Researchers today tend to solve the large displacement of plates through numerical

methods. There is a great deal of literature on this subject, an eight-node hexahedral finite

element analysis of geometrically nonlinear plate in static and dynamic loading conditions was

shown in a paper by Duarte Filho and Awruch's [5]. A smoothed finite element method by Cui,

et al [6] was used to describe both linear and geometrically nonlinear plates using bilinear

quadrilateral elements. A boundary element mesh free plate analysis by Tiago and Pimenta [7]

was proposed and had good accuracy when the displacement and stress field were relatively

smooth. There are also examples of using finite strip method to solve plates with large

deflections by Shahidi et al [8]. Narayana and Krishnamoorthy developed a shell element that has

large-scale capabilities due to the stiffness matrices being relatively small [9]. Other elements

have been developed to deal with problems that have a geometric and material nonlinearity such

as with Bathe and Bolourch's [10] shell element that could solve large deflections of a square

plate due to a pressure load. These are just a few of the examples, however numerical results are

only good for one set of parameters and can also be prone to errors.

7

Babuska and Li evaluated the swings between a theoretical solution compared to a

computational one [11]. They noted the importance of boundary conditions and how they can

give significant inaccuracies especially in areas around the boundary. Wan's paper talks about

stress boundary conditions and their importance to accurately describing plate bending [12] and

develops boundary conditions that can be used for high order plate analysis. A paper by Reddy

and Averill [13] discusses some refined theories of composite plates and the aspects that make

Kirchoff's assumptions not valid for modeling composite plates.

Analytical solutions are much more vigorous and may not lead to some of the troubles

including edge concerns and singularities. However, exact solutions for geometrically nonlinear

plate analysis are very hard to obtain, and can only be found for problems of the simplest

geometry and simplest boundary conditions as in a circular plate on an elastic foundation [14] and

a rectangular plate subject to a concentrated load with different boundary conditions [15]. Chia's

book on nonlinear analysis of plates discusses many different solutions [16], however the

formulas developed are only good for the maximum deflection or the maximum stress. Large

deflections of circular plates with clamped and simply supported edges are developed in Chia's

book by assuming the in-plane deflection as third degree polynomial and solving the governing

equations.

Some literature have transformed the plate differential equation to develop analytical

solutions for plates with differing degrees of nonlinearity since there is no method that can solve

ever nonlinear differential equation exactly. The Airy stress function is used in Thomas and

Bilbao's paper [17] to produce semi-analytical results for geometrically nonlinear flexural

vibrations of plates. Lagrange equations are used in the paper by Haterbouch and Benamar [18]

to study simply supported thin isotropic plates. Amabili and Carra [19] study the thermal effects

on a simply supported plate. A hybrid method derived by Yeh et al [20] uses the finite difference

method and the differential transformation method to solve a clamped orthotropic rectangular

8

plate with large deflections. The Galerkin Method was used to solve a rectangular plate problem

with three clamped edges and one simply supported edge [21].

The use of Rayleigh-Ritz method is used extensively in flexural vibrations of plates. In

Reddy's book [22] the Ritz method is used to solve many examples of plate problems. Dickinson

and Blasio used the Ritz method to solve natural frequency and plate buckling examples of

isotropic and orthotropic rectangular plates [23]. Singh and Chakraverty developed orthogonal

polynomials to study natural frequencies in the transverse direction and develop mode shapes

[24]. Liew developed many solutions to vibrations of Mindlin plates using the Ritz method [25].

2.2 Groebner Basis

Groebner bases for ideals in polynomial ring were introduced in 1965 by Buchberger and

named after his dissertation advisor. The related work of “standard bases” for ideals in power

series ring was developed in 1964 by Hironaka independently. However, Buchberger was the

first to provide a useful algorithm for the determination of Groebner bases in his doctoral

dissertation [26], which has been implemented in many mathematical symbolic computational

software packages. This is a main reason that the Groebner basis methodology now becomes

feasible for many science and engineering applications.

Much of the literature for Groebner basis is being utilized in a different fashion than

computational mechanics. The core use for Groebner basis is to solve nonlinear algebraic

equations. However, researchers have applied the use of Groebner basis to solve difficult

mathematical proofs [27] in nonlinear geometry, robotic kinematics [28], and railway

interlocking systems [29]. At Johannes Kepler University in Linz, Austria, the Research Institute

for Symbolic Computation is continually exploring new applications for Groebner basis.

Boege et al demonstrate some examples and show current limitation by Groebner basis,

as well as the importance of polynomial ordering effects on the polynomials returned from

9

Groebner basis [30]. Borisevich et al discusses different algorithms to solving Groebner basis

and show solvability criteria and uniqueness criteria [31]. Since its inductance into symbolic

computation software packages it is being used to solve problems analytically [32] which is more

desirable to have than numerical solutions.

Ioakimidis is leading the way in applying Groebner basis to solve different physics and

mechanics problems. He has shown how to solve a particle moving along the circumference of

the circle of a radius[33], elasticity problems [34], a truss problem [35] and determining critical

buckling loads [36]. Also at Tennessee Technological University, research is being done to apply

Groebner basis to mechanics problems such as Liu's paper on nonlinear cables [37] and

Vandervort's thesis on geometrically nonlinear rectangular plates [38].

The small amount of research indicates that Groebner basis is not known among a large

majority of the engineering community. This may be due to the fact that before the widespread

use of technology, finding a Groebner basis could be a long task to complete. However, now

many symbolic computational software packages such as Maple and Mathematica come with

procedures to have the Groebner basis calculated. These procedures will be used to help solve

the nonlinear plate problem.

10

CHAPTER 3

THEORETICAL BACKGROUND

This chapter provides theoretical background for the present study. The chapter begins

with a brief review of two widely used plate theories, the Kirchoff plate theory (classical) and

geometrically nonlinear plate theory. The Kirchoff theory is presented in Cartesian and

cylindrical coordinate systems in section 3.1.1 and 3.1.2 respectively. The geometrically

nonlinear plate theory is then reviewed briefly in section 3.1.3. Following that, the current study

is to analyze the plate bending problem using the energy method, in section 3.2 the principle of

total potential energy and the Ritz method are also briefly reviewed. Lastly in this chapter, the

method of Groebner basis is reviewed in detail.

3.1 Review of Thin Plate Theory

In mechanics, the plate is a three dimensional structure defined by the thickness being

much less than the other two dimensions. Consider a rectangular plate in the �&-plane with a

uniform thickness, � and a mid-plane area, S with side lengths and �. Let J(�, &, �) and

!(�, &, �) be the in-plane displacement and �(�, &, �) be the transverse displacement of a point

(�, &) on the mid-surface of the plate. In general, a plate is considered to be thin when the

thickness to the length ratio, T� is roughly between 1/50 to 1/500. In Figure 3.1, the geometry of a

rectangular thin plate is depicted and the sign convention of the plate variables in the rectangular

coordinate system is shown in Figure 3.2.

11

Figure 3.1: Geometry of a Rectangular Thin Plate and Coordinate System

Figure 3.2: Rectangular Plate Variables and Sign Convention

12

3.1.1 Kirchoff Plate Theory

The kinematics of Kirchoff plates are derived using the following assumptions.

• Thin plate assumption: Since the thickness, � is much smaller in comparison to

the length in the �&-plane, the elongation in the � direction is much smaller than

in the � and & directions. That means the strain, 977 is negligible. Therefore,

977 = 0 or UAU7 = 0 (3.1).

This leads to the conclusion that the transverse displacement only varies with � and &,

� = �(�, &) (3.2).

• Plane sections remain plane and normal to the middle plane assumption: The

plane sections before bending remain plane after bending; the normals to the

middle plane before bending remain normal to this plane after bending as

depicted in Figure 3.3.

This implies that the shearing strains, 267 and 2�7 are small enough to be negligible.

Taking 2�7 = 0 and 267 = 0, we have

V�V& + V!V� = 0

V�V� + VJV� = 0

(3.3)

13

which leads to the in-plane displacements J and ! as follows

J = −� V�V�

! = −� V�V&

(3.4).

Figure 3.3: Kirchoff Kinematics in ��-plane

Due to all of the assumptions above, the following assumption can be concluded.

• Middle plane unstretched assumption: The middle plane of the plate remains

unstretched even though the middle plane becomes curved after bending.

Therefore, the resulting kinematic equations based on Kirchhoff assumptions are shown

in equation (3.5).

14

WXYXZ9��9662�6 [X\

X] =WXXYXXZ ∂J∂�∂!∂&∂J∂& + ∂!∂�[XX

\XX] = −�

WXXYXXZ V(�V�(V(�V&(2 V(�V�V&[XX

\XX]

(3.5)

From equation (3.5), the constitutive equations for linear elastic isotropic material in

terms of transverse displacement, � are shown in equation (3.6).

_<��<66=�6` = −z �(1 − ()b

1 0 1 00 0 (1 − )2 cWXXYXXZ V(�V�(V(�V&(2 V(�V�V&[XX

\XX]

(3.6)

where � is the modulus of elasticity and is the Poisson’s ratio. The bending moments can be

determined by integrating over the thickness presented in equation (3.7).

dF��F66F�6e = f _<��<66=�6` �.�

T(gT( = D

WXXYXXZV

(�V�( + V(�V&(V(�V&( + V(�V�((1 − )2 V(�V�V&[XX\XX]

(3.7)

where B is the plate bending rigidity, B = iTj1((1gkl). In the same manner, the transverse shearing forces can be described in the terms of the

bending moments illustrated in equation (3.8).

mI��I66n = WXY

XZ∂F��∂� + ∂F�6∂&∂F66∂& + ∂F�6∂� [X\X]

(3.8)

15

3.1.2 Transformation of Coordinate System

Since the circular plate will be examined in all case studies, it is convenient for the

analysis to use cylindrical coordinates instead of rectangular coordinates. In this section, the

governing equations based on the Kirchoff theory will be transformed from rectangular

coordinates to cylindrical coordinates. Let’s consider a plate in the "5-plane with a uniform

thickness, � and a middle plane area, S with radius . Let J3(", 5, �) and J4(", 5, �) be the in-

plane displacement and �(", 5) be the transverse displacement of a point (", 5) on the mid-

surface of the plate. Figure 3.4 shows the geometry of the circular thin plate and the sign

convention of the circular plate variables in the cylindrical coordinate system are illustrated in

Figure 3.5.

Figure 3.4: Geometry of Circular Thin Plate and Coordinate System

16

Figure 3.5: Sign Convention for the Circular Plate Variables

The relations between polar and rectangular coordinates are

"( = �( + &(, 5 = "o�� p6�q (3.9)

� = "or�(5), & = "��(5)

(3.10)

from which

U3U� = or�(5), U3U6 = ��(5) (3.11)

U4U� = − stu(4)3 , U4U6 = vKs(4)3 (3.12)

the transverse loading and deflection are considered as a function of " and 5, that is, � = �(", 5) and � = �(", 5). Based on the chain rule the first derivatives,

UAU� and UAU6 can be determined in

terms of UAU3 and

UAU4 shown in Equation (3.13).

17

WYZV�V�V�V&[\

] = bcos(5) −1" sin(5)sin(5) 1" cos(5) c |V�V"V�V5} (3.13)

Performing the same transformation, the second derivatives in terms of " and 5 are given in

Equation (3.14).

WXXYXXZ V(�V�(V(�V&(V(�V�V&[XX

\XX] =

~�� or�((5) 1"( ��((5) −2" or�(5)��(5)��((5) 1"( or�((5) 2" or�(5)��(5)or�(5)��(5) 1" or�(5)��(5) 1" or�((5) − 1" ��((5)��

��

WXXYXXZV(�V"(V(�V5(V(�V"V5[XX

\XX]

+~��

1" ��((5) 2"( or�(5)��(5)1" or�((5) 2"( or�(5)��(5)1" or�(5)��(5) 1"( ��((5) − 1"( or�((5)��|V�V"V�V5}

(3.14)

For the axisymmetric circular plate, the transverse loading and deflection become a

function of only " that is to say, � = �(") and � = �("). The kinematic equations in cylindrical

coordinates based on the Kirchoff theory for the axisymmetric case are reduced to

WXYXZ933944234 [X\

X]

��= −�

WXYXZV(�V"(1" V�V"0 [X\

X]

(3.15).

Based on Equation (3.15) the stress-strain relations will be

�<33<44�� = �(1 − () �1 1� |−�V(�V"(−� 1" V�V"} (3.16).

18

Similarly to the Cartesian bending moment equations, the moment can be found by

integrating over the thickness.

mF33F44n = f �<33<44� �.�

T(gT( =−BWY

ZV(�V"( + 1" V�V"1" V�V" + V(�V"([\]

(3.17)

The transverse shearing forces can be described in the terms of displacement and is shown in

Equation (3.18).

mI33I44n = d ∂∂" �V(�V"( + 1" V�V"�0 e

(3.18)

3.1.3 Geometrically Nonlinear Plate Theory

In the previous section, the Kirchoff plate theory was discussed, which is a linear theory

based on the assumption that the plates will be subject to small displacements. In cases in which

deformations are no longer small in comparison with the thickness of the plate but are still small

compared to the other dimensions, the small displacement assumption is no longer valid for these

cases. Figure 3.6 shows the in-plane kinematics for large plate deformation.

19

Figure 3.6: In-plane Kinematics for Large Plate Deformation

The strain is calculated by taking the change in length and dividing it by the original

length in Equation (3.19).

ε�� = A�B�� − ABAB

(3.19)

where S� is the original length from point S to point �, S′�′′ is the deformed length from point

S′ to point �′′. The strain can be calculated as follow

ε�� = �dx( + p∂w∂x dxq(�1( − dxdx

(3.20).

manipulating Equation (3.20) yields the following equation

20

ε�� = �1 + �∂w∂x�(�

1( − 1

(3.21).

then using a Taylor expansion, the strain in the � direction is reduced to

ε�� = 12 �∂w∂x�(

(3.22).

only the first term of the Taylor series is used for this derivation.

Similarly to the calculation of 9�� the other strains 966 and 2�6 are derived in the same

manner. The strains shown in Equation (3.23) are only due to large transverse displacements.

WXYXZ9��9662�6[X\

X]�Ku�á3�áu

=WXXYXXZ

VJV� + 12 �V�V��(V!V& + 12 �V�V&�(12 �V�V�� V�V&�¡[XX\XX]

(3.23)

transforming these strains into cylindrical coordinates for an axisymmetric plate, yields

WXYXZ933944234 [X\

X]

�Ku�á3�áu=

WXYXZVJ3V" + 12 �V�V" �(J3"0 [X\

X] (3.24).

Notice that in equation (3.24) the in-plane displacement (J3) is included in the kinematic

equation for large deformation known as von Kármán strains which will be used in the

formulation of the strain energy functional for this study.

As a result of the large deformation kinematic equation, in plane forces are present in the

plate. The in-plane forces can be calculated by integrating over the thickness. Equation (3.25)

21

shows the in-plane forces based on the large deformation kinematic equations for axisymmetric

circular plates. Because the plate is considered axisymmetric, H34 is equal to zero.

mH33H44n = f �<33<44� .�T(

gT( = �ℎ(1 − () WXYXZ VJ3V" + 12 �V�V" �( + J3"J3" + �VJ3V" + 12 �V�V" �(�[X\

X] (3.25)

3.2 Principle of Minimum Total Potential Energy

The principle of virtual work states that if the force system including internal and external

forces is in equilibrium the total amount of work done by all the forces moving through an

arbitrary virtual displacements has to be zero. The principle of minimum total potential energy is

a special case of the principle of virtual work that deals with elastic bodies. It shows that the

displacements in the elastic body, in the absence of any energy loss, will deform to a position

where the total potential energy will be minimized. The principle can be expressed in equation

(3.26) equivalently.

12f<t£9t£.Ω¥ −f¦t8Jt.Ω

¥ = 0 (3.26)

where <t£ is the stress tensor components, 9t£ is the strain tensor components, ¦t is the external

loading, 8Jt is the arbitrary virtual displacement in the elastic body, and Ω is the domain. Let the

first integral in equation (3.26) be the total strain energy and will now be denoted by M and the

second integral in equation (3.26) is the potential energy by external forces and will now be

denoted by §. Let Π be the total potential energy which is described in equation (3.27).

Π = M + § (3.27)

The variation of the total potential energy is in the form of

8Π = 8(M + §) (3.28)

22

based on the principle of minimum of total potential energy, the first variation of the total

potential energy must be equal to zero.

8Π = 0 (3.29)

3.2.1 Total Potential Energy Functional

The following section will show the derivation of the total strain energy for the

geometrically nonlinear isotropic axisymmetric circular plate. Given that the deflections are

geometrically nonlinear, membrane (in-plane) effects must be taken into account. Total strain

energy will be split into two components: the bending strain energy, MN and the membrane strain

energy, M�. Therefore, the total strain energy is

M = MN + M� (3.30).

In cases for the axisymmetric circular plate the total strain energy M can be formed in cylindrical

coordinate system as shown in Equation (3.30).

12f f (<33933 + <44944).Ω.�

¥T(

gT( (3.31)

The bending strain energy, MN can be calculated by substituting equations (3.15) and

(3.16) into equation (3.31). The resulting strain energy in terms of transverse displacement is

shown in equation (3.32).

MN = ¨Bf ©" �V(�V"( �( + 1" �V�V" �( + 2 V(�V"( V�V" ª ."«¬ (3.32)

The membrane strain energy, M� can be calculated by substituting equation (3.24) into

equation (3.31) and the linear elastic constitutive equations. The resulting strain energy in terms

of in-plane and transverse displacements (von Kármán strains) is shown in equation (3.33).

23

M� = 12¨Bf _VJ3V" + 12 �V�V" �(®( + pJ3" q( + 2 J3" VJ3V" + 12 �V�V" �(®` "."«¬ (3.33)

The potential energy due to the transverse load, � and transverse displacement can be

described as following

V = f ��"."«¬ (3.34).

The total potential energy functional is shown in equation (3.35).

Π = f ©¨B " �V(�V"( �( + 1" �V�V" �( + 2 V(�V"( V�V"¡«¬

+ 12¨B _�VJ3V" + 12 �V�V" �(�( + pJ3" q(+ 2 J3" �VJ3V" + 12 �V�V" �(�° " − "��ª ."

(3.35)

3.2.2 The Ritz Method

The Ritz method is a procedure to obtain the displacement components by assuming

displacement functions in the form of linear combinations. For example, the assumed

displacement functions for axisymmetric circular thin plate is shown in equation (3.36).

J3 = J3¬ +±St>?@�t²1

� = �¬ +±�t>A@u

t²1

(3.36)

where J3¬ and �¬ are displacement functions chosen to take the value of predescribed

displacement components on the displacement boundary conditions; >?@ and >A@ are

displacement shape functions that satisfy the essential boundary conditions; St and �t are ³ + �

24

independent coefficients for which to be determined. It is noticed that the displacement variation

are realized by the variation of the independent coefficients.

Based on the principle of minimum total potential energy, 8Π = 0 with the Ritz

approximation, the functional will be converted to a ³ + � system of algebra equations by

variation of coefficients

VΠVSt = 0

VΠV�t = 0 (3.37).

Based on the geometric nonlinear circular plate problem, the system of algebra equation

will be coupled and nonlinear. Therefore, Groebner can be used to decouple the system of

algebra equations to be solvable analytically.

3.3 Groebner Basis

The Groebner basis was introduced by B. Buchberger in 1965 [26]. With the increasing

capability of symbolic computation in recent years, Buchberger’s Algorithm to compute a

Groebner basis has been implemented into different computer algebraic systems like Maple,

Mathematica, AXIOM, CoCoCA, Macaulay, etc., which made Groebner basis become a feasible

tool for many scientific and engineering applications. The present study is to apply a

combination of Groebner basis and Ritz methodologies to the plate bending problems. In this

section some basic concepts and definitions related to Groebner basis will be reviewed, and the

details of the mathematical background and underlying proofs can be found in books such as Cox

et. al [39]. In particular, we are interested in solving nonlinear systems of equations using

Groebner bases.

25

3.3.1 Preliminary Concepts from Abstract Algebra

To get a basic understanding of what a Groebner Basis is, some basic concepts from

abstract algebra such as fields, commutative rings, ideals and monomial orderings are presented

in Appendix A. Fields, commutative rings, ideals are all forms of algebraic structures. An

algebraic structure is a set which impose operators that meet certain axioms. A formal definition

of a field is presented in Appendix A-1. Field are commonly represented by the letter ´. Some

well known fields are the set of real numbers, ℝ, the set of rational number, ℚ, and the complex

set of numbers, ℂ. Another important algebraic structure to understand is a commutative ring. A

formal definition of a commutative ring is presented in Appendix A-2. A commonly known

commutative ring is the set of integers, ℤ.

3.3.2 Monomials and Polynomials

A monomial in � variables �1, �(, … , �u is a product of the form �1¶·�(¶l …�u¶¸ where

all the exponents are positive integers. The total degree of a monomial is calculated by adding

the / terms (/TKT = /1 + /( +⋯/u).

A polynomial in � variables �1, �(, … , �u is a finite linear combination of monomials

with coefficients from the field ´. Let ¦ be a polynomial, then the polynomial can be described

as

¦ = ±¶�¶ ,¶ ∈ ´¶

(3.38).

where the sum is over a finite number of �-tuples / = (/1, … , /u). The set of all polynomials in

�1, … , �u with coefficients from ´ is denoted as ´»�1, … , �u¼. Figure 3.7 shows pictorial relations

among field, commutative ring, and polynomial ring.

26

Figure 3.7: Pictorial Description of a Field, Ring, and Polynomial Ring

3.3.3 Monomial Ideals and Dickson’s Lemma

A general definition of an ideal is presented in Appendix A-3. A monomial ideal in the

polynomial ring ´»�1, … , �u¼ is considered to be a monomial ideal if there is a subset S ⊂ ℤR

such that ¾ ⊂ ´»�1, … , �u¼ consists of all polynomials which are finite sums of the form:

±ℎ¶�¶ , �ℎ�"�ℎ¶ ∈ ´»�1, … , �u¼¶∈¿ (3.39).

In this case, we write ¾ =< �¶: / ∈ S >.

Dickson’s Lemma introduces an important idea that states that an ideal, which is an

infinite set, can be defined by a finite set of generating monomials. The lemma states that a

ℝ ℚ ℂ

Examples:

Definition:

See Appendix A-2

Examples:

´

ℤ

��.��"³��´»�1, … , �u¼ Definition:

See Appendix A-1 ´:¦��Ã.´�Ã�³��:¦1, … , ¦�,…

27

monomial ¾ =< �¶: / ∈ S ⊂ ℤR >⊆ ´»�1, �(, … , �u¼ can be written in the form ¾ =<�¶(1), … �¶(u) > where �¶(1), … �¶(u) ∈ S ⊂ ℤR and that monomial ideals have a finite basis.

3.3.4 Polynomial Ideals and Hilbert’s Basis Theorem

Let ¦1, … , ¦u be polynomials in the polynomial ring ´»�1, … , �u¼. Then a polynomial

ideal can be defined as:

<¦1, … , ¦s >= ±ℎt¦tu

t²1 ∶ ℎ1, ℎ(, … , ℎu ∈ ´»�1, … , �u¼ (3.40).

as a result, ¾ is finitely generated if there exist ¦1, … , ¦s ∈ ´»�1, … , �u¼ such that ¾ =< ¦1, … , ¦s >

and ¦1, … , ¦s is a basis of ¾. Figure 3.8 shows a pictorial relationship between a polynomial ring

and a polynomial ideal.

Figure 3.8: Pictorial Description of an Ideal

´»�1, … , �u¼ Definition:

See Appendix A-3 0¦1 + ¦( ¦1 ∙ ¦* ¦*

28

Hilbert’s Basis Theorem states that every ideal ¾ ∈ ´»�1, … , �u¼ has a finite generating

set of polynomials. That is ¾ =< ¦1, … , ¦s > for some ¦1, … , ¦s ∈ ¾. It indicates that any

polynomial ideal ¾ can be generated by different bases.

3.3.5 Definition of a Groebner Basis

A finite set of polynomials, ¦1, … , ¦s ∈ ´»�1, … , �u¼ generates an ideal, ¾ =<¦1, … , ¦s >in ´»�1, … , �u¼. The set ¦1, … , ¦s is known as a basis of ¾. This means that every

ideal, ¾ in ´»�1, … , �u¼ can be generated by different finite set of polynomials. The Groebner

Basis is a type of generating set where if the monomial order is fixed then a finite generating set

Ç1, … , ÇT of an ideal ¾ is said to be a Groebner Basis if:

< DG(Ç1), … , DG(ÇT) >=< DG(¾) > (3.41).

There exists a theorem called “Buchberger’s S-pair criterion” which shows that a basis of

an ideal ¾ in ´»�1, … , �u¼ is a Groebner basis. The theorem states that a basis C = {Ç1, … , ÇT} for

¾ is a Groebner basis if and only if for all pairs when � ≠ É, the remainder on the division of

Ê(Çt , Ç£) by C with a fixed monomial order is zero. The proof of this theorem can be found in

Buchberger's thesis, which was translated and reprinted into English in this reference [26]. The

proof proposes that to compute a Groebner basis for an ideal ¾ generated by Ë = {¦1, … , ¦s} one

must extend the basis Ë to a Groebner basis by successively adding nonzero remainders to Ë.

Buchberger’s algorithm is presented in Appendix A-5.

The generated Groebner basis will typically have unnecessary elements, and once further

constraints are applied these unnecessary elements will be eliminated resulting in a minimal

Groebner basis. The minimal Groebner basis, C, for an ideal ¾ is defined as:

DE(�) = 1∀� ∈ C ∀� ∈ C, DG(�) ∉< DG(C − �) >

(3.42).

29

Although in a given ideal in ´»�1, … , �u¼ there can still remain multiple minimal Groebner bases.

Therefore, further constrains can lead to a reduced Groebner basis which is unique. A reduced

Groebner basis for an ideal ¾ is a Groebner basis C such that:

DE(�) = 1∀� ∈ C ∀� ∈ C, no monomial of � belongs to < DG(C − �) >

(3.43).

There are many monomial ordering schemes for generating different Groebner Bases,

some typically ordering schemes found in symbolic computation software packages include

lexicographic order, graded lexicographic order, and graded reverse lexicographic order. Some

examples of these ordering schemes are provided in Appendix A-4.

Every polynomial ideal had finite generating sets and for a fixed monomial order every

ideal ¾ in ´»�1, … , �u¼, other than {0}, has unique reduced Groebner basis, and this basis has

proven to have many uses in multiple fields. For example, this thesis will use the Groebner basis

to help solve a nonlinear system of polynomial equations. More applications were shown in

Chapter 2.

3.3.6 Affine Variety: the link between the Groebner bases and the original generating set

functions

In particular, we are interested in solving nonlinear systems of equations using Groebner

bases. To understand the connection between Groebner bases and algebraic equations, it is

necessary to grasp the concept of affine space and affine varieties. It has the ability to link

polynomials from algebraic objects to geometric objects. For a given field ´ and a positive

integer �, then the affine space is defined as

ú = {(1, … , u): 1, … , uÎ´} (3.44).

30

namely, an �-dimensional affine space, ú is simply a set of the collection of all the points of �-

tuples from field ´. For example, ´ = ℝ, then ´( would be a collection of all the points (�, &) in

a two dimensional real number space, ℝ( in linear algebra.

Let ´ be a field, and a set of finite polynomials ¦1, … , ¦s ∈ ´»�1, … , �u¼ the affine

varieties defined by ¦1, … , ¦s are a set of points

§(¦1, … , ¦s) = {(1, … , u) ∈ ú|¦t(t , … , u) = 0}∀1 ≤ � ≤ � (3.45).

where § denotes the affine variety. Therefore, an affine variety §(¦1, … ¦s) ∈ ú is a set of

solutions or points which satisfies {¦1(�1, … , �u), … , ¦s(�1, … , �u)} = 0. Several examples of

affine variety are given in Appendix B.

Based on the definitions of the ideal ¾ =< ¦1, … , ¦s >, generated by ¦1, … , ¦s, it is easy to

show that the affine variety defined by the ideal < ¦1, … , ¦s > is its generating set:

§(¦1, … , ¦s) = §(< ¦1, … , ¦s >) (3.46).

On the other hand, if the finite set Ç1, … , ÇT is a basis of the ideal generated by ¦1, … , ¦s, then

¾ =< ¦1, … , ¦s >=< Ç1, … , ÇT > (3.47).

Therefore,

§(¾) = §(< ¦1, … , ¦s >) = §(< Ç1, … , Çu >) (3.48).

In particular, if the set Ç1, … , ÇT is the reduced Groebner basis for the ideal < ¦1, … , ¦s >, finally,

we have

§(¾) = §(¦1, … , ¦s) = §(Ç1, … , ÇT) (3.49).

31

It is the key that the affine variety of the Groebner basis will be the same as the original

system of equations. Therefore, the Groebner basis can be used to solve the original system of

equations.

32

CHAPTER 4

IMPLEMENTATION AND METHOD VALIDATION

In this chapter, the validation and implementation are presented in detail. The current

study is tested by examining three cases of geometrically nonlinear circular plates defined by

different boundary conditions. Section 4.1 introduces each of the three case studies and their

boundary conditions. The shape functions used for each case are provided in section 4.2 and the

trial functions for each case are shown in section 4.3. In section 4.4, the implementation of the

two computer software packages, ANSYS and Maple used for the current study are discussed.

Then in Section 4.5, two types of comparisons are presented. One is a comparison of the results

from the current study to the results generated by the finite element simulation and the results of

the exact solution using the linear theory, and the other is a comparison of the results from the

current study using only one coefficient in the trial function for the transverse displacement to the

results of using two coefficients in the trial functions for the Ritz approximation. The

comparisons serve as a validation for the current study.

4.1 Case Studies

4.1.1 Case 1: Fully Clamped Circular Plate

Case 1 is a fully clamped circular plate with the support located at the edge of the plate,

, and subjected to a uniform distributed transverse load of intensity, �, on its upper surface as

shown in Figure 4.1. The boundary conditions are expressed in equation (4.1).

4.1.2 Case 2: Simply Supported Immovable Circular Plate

edge of the plate,

upper surface


edge of the plate,

upper surface


edge of the plate,

upper surface

Displacement Boundary Conditions:


Case 2 is a simply supported immovable circular plate with the support located at the

edge of the plate,

upper surface




edge of the plate,

upper surface as shown in




edge of the plate,

as shown in

Fixed


Force



edge of the plate, , as shown in

Fixed


Force



and subjected to a uniform distrib

as shown in

Fixed

Figure


Force Boundary Conditions:




as shown in Figure

Figure


Boundary Conditions:




Figure

Figure 4.1






Figure 4

1: Case 1: Fully Clamped Circular Plate






4.2. The boundary conditi

: Case 1: Fully Clamped Circular Plate






. The boundary conditi















33


Displacement Boundary Conditions: J�V�V"

Boundary Conditions: None





33


J3|�|3V�V" |None






3²�3²�|3²

None



and subjected to a uniform distributed transverse load of intensity,



� == 0

²� =



uted transverse load of intensity,

. The boundary conditions are expressed in equation (4


0;0

= 0




ons are expressed in equation (4


; J;





J3|3V�V"





3²¬

V�V" |3





= 0

3²¬




0

= 0




0











uted transverse load of intensity, �, on its

ons are expressed in equation (4.2).

(4


, on its

.2).

4.1)


, on its

)


, on its

Simply Supported Immovable

4.1.3 Cas

distance,

uniform dist

Figure 4

rℎ


4.1.3 Cas

distance,

uniform dist

Figure 4

ℎ signifies the functions that describe the displ


4.1.3 Cas

distance,

uniform dist

Figure 4

signifies the functions that describe the displ



4.1.3 Case 3: Simply Supported Immovable Circular Plate with Overhang

Case 3 is a simply supported immovable plate with the support located at the radial

distance, uniform dist

Figure 4.3. The boundary conditions for Case 3 are shown in equation




e 3: Simply Supported Immovable Circular Plate with Overhang


, with an overhang that has a free edge at the radial distance,

uniform distributed load of intensity,

.3. The boundary conditions for Case 3 are shown in equation



Figure


Force



with an overhang that has a free edge at the radial distance,

ributed load of intensity,




Figure


Force








Figure 4


Force








4.2: Case 2: Simply Supported










: Case 2: Simply Supported


































ributed load of intensity, �.3. The boundary conditions for Case 3 are shown in equation








�, on its upper surface within the support









, on its upper surface within the support



34



Boundary Conditions: Fe 3: Simply Supported Immovable Circular Plate with Overhang






34


J3|�|V�V"F3 Ò|e 3: Simply Supported Immovable Circular Plate with Overhang





signifies the functions that describe the displacement and force in the overhang region

: Case 2: Simply Supported Immovable Circular Plate|3²�3²�V� |3²Ò|3²�






acement and force in the overhang region

Immovable Circular Plate

� =� =²¬ =

Ò � =e 3: Simply Supported Immovable Circular Plate with Overhang






Immovable Circular Plate= 0= 0

= 0Ò = 0







Immovable Circular Plate; J

0







Immovable Circular PlateJ3|3








3²¬








¬ =





.3. The boundary conditions for Case 3 are shown in equation (4.3


Immovable Circular Plate0



with an overhang that has a free edge at the radial distance, �, and subjected to a


3) where the subscript




, and subjected to a


where the subscript




, on its upper surface within the support as shown in

where the subscript




as shown in

where the subscript




as shown in

where the subscript


(4



as shown in

where the subscript


4.2)



as shown in

where the subscript


)



as shown in

where the subscript



support to the free edge;

edge;

transverse


<support to the free edge;

edge;

transverse


" ≤support to the free edge;

edge; Ftransverse


≤ �. Therefore,


F3KLtransverse shearing force from the edge of the support to the free edge.

Figure


. Therefore,


L is the moment from the edge of the support to the free edge; and

shearing force from the edge of the support to the free edge.

Figure


. Therefore,


is the moment from the edge of the support to the free edge; and


Figure 4.


. Therefore,




.3: Case 3: Simply Supported


. Therefore, �support to the free edge;





�KLsupport to the free edge; J3K





L can be described as the transverse displacement from the edge of the

3KL is the in





can be described as the transverse displacement from the edge of the

is the in





is the in





is the in-plane displacement from the edge of the support to the free





plane displacement from the edge of the support to the free









35





: Case 3: Simply Supported Immovable Circular Plate with Overhang

35





Immovable Circular Plate with Overhang

















































is the moment from the edge of the support to the free edge; and I3K




3KL is the



is the



is the



36

Displacement Boundary Conditions: J3|3²� = 0; J3|3²¬ = 0

J3KL|3²� = J3|3²�

�|3²� = 0

�KL|3²� = �|3²�

ÒV�V" Ó3²¬ = 0 V�V" Ó3²� = V�KLV" Ó3²�

Force Boundary Conditions: F3 Ò|3²� = F3KL Ò|3²�

F3KL Ò|3²N = 0

I3KL Ò|3²N = 0

(4.3)

4.2 Shape Functions

The exact linear solution of the transverse displacement from the linear theory for each

case can be found by solving the governing equation in (4.4).

B .." 1" .." �" .�(")." �¡ = � (4.4)

In equation (4.4), B is the plate bending rigidity and � is a uniform distributed transverse load.

For Case 1 and 2, the shape of the transverse displacement from the exact linear solution

is used as a shape function in the assumed displacement functions. For case 3, the exact linear

solution for the transverse displacement is used as a shape function. The exact linear solution for

Case 3 is derived in Appendix C. The in-plane shape functions are assumed so that the functions

satisfied the boundary conditions. The shape functions used for each case are presented in

equations (4.5) through (4.7).

37

Displacement Shape Functions for Case 1

>A = �1 − p"q(�(

>? = " �1 − p"q(�

(4.5)


>A = �1 − p"q(� �5 + 1 + − p"q(� >? = " p1 − "q

(4.6)


>A = ��"#64B + �(Õ(( − 1) − 2�(( + 1)Ö"(32B�(( + 1)− �#Õ−2(( − 1) − 3�(( + 1)Ö"(64�(( + 1)B �

>A×Ø = ��#( − 1)"#32B�((1 + ) + �#Õ((1 − ) + 2ln()�(( + 1)Ö32B�(( + 1) − �#ln(")16B �

>? = " p1 − "q

>?×Ø = p1 − "q

(4.7)

4.3 Trial Functions for Ritz Method

There are two sets of trial functions chosen for each case. One set includes only one

unknown coefficient for transverse displacement and two unknown coefficients for the in-plane

displacements. The other set has two unknown coefficients for transverse displacement and three

unknown coefficients for in-plane displacement. The in-plane displacement has one more

unknown coefficient for accuracy. From here on, the trial functions with one unknown

38

coefficient for the transverse displacement will be known as the “current study” which is shown

in equations (4.8) – (4.10).

Current Study Trial Functions for Case 1

� = �1 − p"q(�( (S¬) J3 = " �1 − p"q(� ��¬ + p"q( �1�

(4.8)

Current Study Trial Functions for Case 2

� = �1 − p"q(� �5 + 1 + − p"q(� (S¬) J3 = " p1 − "q ��¬ + p"q( �1�

(4.9)

Current Study Trial Functions Case 3

� = ��"#64B + �(Õ(( − 1) − 2�(( + 1)Ö"(32B�(( + 1)− �#Õ−2(( − 1) − 3�(( + 1)Ö"(64�(( + 1)B � (S¬) �KL = ��#( − 1)"#32B�((1 + ) + �#Õ((1 − ) + 2ln()�(( + 1)Ö32B�(( + 1) − �#ln(")16B � (S¬)

J3 = " p1 − "q ��¬ + p"q( �1�

J3KL = p1 − "q ��¬ + p"q( �1�

(4.10)

The trial functions for the transverse displacement with two coefficients is used to

examine the accuracy of the current study and the comparison can be found in section 4.5.3. The

trial functions with two coefficients for each case are shown in equations (4.11) – (4.13).

39

Two Coefficient Trial Functions for Case 1

� = �1 − p"q(�( (S¬ + S1 p"q() J3 = " �1 − p"q(� ��¬ + p"q( �1 + p"q# �(�

(4.11)

Two Coefficient Trial Functions for Case 2

� = �1 − p"q(� �5 + 1 + − p"q(� �S¬ + S1 p"q(�

J3 = " p1 − "q ��¬ + p"q( �1+p"q# �(�

(4.12)

Two Coefficient Trial Functions Case 3

� = ��"#64B + �(Õ(( − 1) − 2�(( + 1)Ö"(32B�(( + 1)− �#Õ−2(( − 1) − 3�(( + 1)Ö"(64�(( + 1)B � �S¬ + S1 p"q(�

�KL = ��#( − 1)"#32B�((1 + ) + �#Õ((1 − ) + 2 ln() �(( + 1)Ö32B�(( + 1) − �# ln(")16B � �S¬+ S1 p"q(�

J3 = " p1 − "q ��¬ + p"q( �1 + p"q# �(�

J3KL = p1 − "q ��¬ + p"q( �1 + p"q# �(�

(4.13)

4.4 Computer implementation

The main part of this study is to explore the application of Groebner basis methodology

in an analysis of nonlinear circular plate problems. The Maple software package is used in each

case study to implement the Groebner basis calculation which converts a set of coupled

polynomial algebraic equations into an equivalent set of uncoupled polynomial algebraic

40

equations. The methodology will provide analytical solutions to the problems. The following

section describes how to find the analytical solutions for the current study with Maple.

4.4.1 Maple 13

An example of a working Maple file to solve Case 1 using the methodology presented

can be found in Appendix D. The first line initialize the worksheet with the command

“with(Groebner)” that allows the subroutines associated with Groebner basis to be open. The

next line is used to display the total potential energy function. The capital letters associated with

the differentials, “Diff”, and integrations, “Int” tell the program not to evaluate, which provides a

convent way to check the equations to reduce errors. The detailed explanations and comments for

the process can be found in Appendix D as well.

The Maple 13 command “Basis” is used to find the reduced Groebner basis for a fixed

monomial order. If an elimination order is specified with “lexdeg”, the “Basis” converts the set

of coupled polynomial equations into a set of uncoupled ones in which the first equation contains

only one variable. If the first equation can be solved analytically, then the subsequent equations

can be solved by substituted back into the remaining equations to find all other unknowns. After

the unknown coefficients are solved, they can be substituted back into the displacement functions.

One of the advantages of finding the solution symbolically in Maple is that the fully symbolic

solutions make the parametric studies much easier than the numerical solutions.

The example presented in Appendix D shows how the Ritz method was implemented in

conjunction with the trial functions for case 1 in equation (4.8), to determine the approximate

result of the geometrically nonlinear circular plate analysis. ��S0, ��0, �.��1 are from

taking the derivatives of the total potential energy function (3.36) with respect to the unknown

Ritz coefficients. Equation (4.14) is the resulting system of equations which are coupled and

nonlinear.

41

��S0 = − ¨315(( ( − 1) (384��S¬* + 84��S¬(�1 + �1 − �¬ + 3 �¬)+ 560��*S¬ + 105�#( ( − 1)) ��0 = − 2¨��15( ( − 1) p(3 − 1)S¬( + 5(�1 + 2�¬)q

��1 = − 2¨��15( ( − 1) p(1 + )S¬( + 4�1 + 5�¬q

(4.14)

The application of Groebner basis methodology can be seen when using the “Basis” function on

��S0, ��0, �.��1 from the Ritz method. The output equations called reduced Groebner

bases in (4.15) contain 3 equations which are much easier to determine the unknown coefficients

since the first equation, C�»1¼ is uncoupled with only one unknown, S¬, which can be solved

symbolically. Since C�»2¼ and C�»3¼ are in terms of S¬ and only one other unknown, �¬ or �1,

the unknowns can be solved symbolically as well.

C�»1¼ = 525�#(1 − () − 2800��*S¬ + 32��(14 ( − 21 − 39)S¬*

C�»2¼ = 15�¬ + (7 − 9)S¬(

C�»3¼ = 3�1 + (3 − )S¬(

(4.15)

In summary, this is the way Groebner basis is utilized in the current study. The Ritz

method produces a system of equations that are coupled and nonlinear, such as equation (4.14).

The calculation of Groebner basis generates a system of equations that are uncoupled and

solvable, such as equation (4.15). Case 2 and 3 are presented in Appendix F.

4.4.2 ANSYS 13

ANSYS 13 (Mechanical APDL) [40] is used to check the validity of the solutions found

by the current study. Shell93 elements are used for the simulation since they have large

deflection capabilities. A quarter model of the circular plate is used to complete all the analyses.

The quarter plate model has symmetric boundary conditions applied to the two inside edges and

42

boundary conditions applied based on which case is being analyzed. Figure 4.4 shows the

boundary conditions applied to the model.

Figure 4.4: Boundary Conditions of Finite Element Quarter Plate Model for Each Case Study

where ÊÜF stands for symmetric boundary condition, Ë¾Ý�B stands for fixed boundary

condition, and ÊÊ stands for simply supported immovable boundary condition. Before

performing all of the analyses, a convergence study was conducted to determine the amount of

elements needed. A steel circular plate with a radius of 10 inches and a thickness of 0.1 inches

was selected to run the convergence study. Figure 4.5 presents the results from the convergence

study for Case 1:

Case 1 Case 2 Case 3

43

Figure 4.5: Convergence Study for Case 1 (� = 5��, ν = 0.3 , � = 29� + 6��, = 10��, and � = 0.1��)

where �� is the maximum displacement at the center of the plate. Adding more than 500

elements to the finite element model did not change the results with any significance. Therefore,

583 elements are used in the quarter plate model to conduct the validations and comparisons for

Case 1 and 2. A plot of the element mesh density for Case 1 and 2 is shown in Figure 4.6.

� �� (��oℎ

��)

4

Total Number of Elements in Finite Element Model

the simply supported immovable circular plate with overhang (Case 3) log file is presented in

Appendix E. The first portion of the

/PREP7 command. The preprocessor section is where the element data, material properties,

mesh, and geometry are entered. The ET command selects the type of element to be used in the

analysis, which

material data was inputted using the MPDATA commands. The plate geometry was generated by

using the CLY4 command to create a quarter segment of a circle. The next set of comman

used to mesh the plate. ANSYS has an automesh option which allowed a mesh to be generated

automatically in the model. The boundary conditions were applied by using the DL commands to

apply the in

Afterwards, the pressure load was applied to the area inside of the support on the plate. The





analysis, which





apply the in






analysis, which





apply the in


Log files were created for each of the three cases presented in this thesis. An example of





analysis, which





apply the in







analysis, which





apply the in-plane, transverse, and symmetric boundary condition







analysis, which is SHELL93 as indicated by the third parameter of the ET command. Then, the





plane, transverse, and symmetric boundary condition







is SHELL93 as indicated by the third parameter of the ET command. Then, the































Figure













Figure













Figure 4.6













6: Mesh of Finite Element Model













Mesh of Finite Element Model



Appendix E. The first portion of the log file is to develop and start the preprocessor with a













log file is to develop and start the preprocessor with a










44














44


























































































plane, transverse, and symmetric boundary conditions as seen in Appendix E.












s as seen in Appendix E.

























































using the CLY4 command to create a quarter segment of a circle. The next set of commands was












ds was












ds was












ds was





45

analysis type is setup to solve using large displacement assumptions. The "SOLVE" command

runs the ANSYS solver to complete the finite element simulation.

The last portion of the log file is post processing. All the analyses ran for this validation

outputted contour plots for in-plane displacements, transverse displacements, and stresses. In

addition, the /OUTPUT command generated text files which was used to gather results for in-

plane displacement, transverse displacement and stresses on the nodes. Also, the element and

nodal data were outputted so that they could be used to plot the ANSYS solution with the current

study.

4.5 Validation of Current Study

ANSYS, a commonly used finite element software package is utilized as a tool to

validate the results from the current study. In the following sections, the validation study will

show that the current procedure is of great value in generating closed form solutions compared to

the numerical solutions from ANSYS.

4.5.1 Validation of Shape Functions

The first check is to verify the shape functions. The shape functions are verified by

plotting the displacement with respect to the radial distance to insure the assumed shape functions

are accurate assumptions. The transverse displacement from the current study is shown on the

same figure as the ANSYS solution and the exact solution from the linear theory which allowed

for comparison of results. Figure 4.7 through Figure 4.9 show the transverse displacement for

each case. “Linear Theory” in all table and figures represents the exact solution based on the

linear theory. The load condition was given a magnitude in which the circular plate would be

subjected to geometrically nonlinear displacements. The same load and material properties were

used for each case and the magnitude of the transverse displacement for Case 1, 2 and 3 are

respectively 1.39�, 1.9� and 1.9�. The results in Figure 4.7 – 4.9 show that the transverse

46

displacement of the current study with one Ritz coefficient agrees with the ANSYS solution with

a fine mesh. The ANSYS solution for Case 1 and 2 had 583 elements and for Case 3 had 1154

elements due to the addition of the overhang which created additional elements. As expected

when the plate is subjected to large displacement the linear solution does not agree with the

current study nor the ANSYS results, the maximum displacement at the center of the plate for the

exact linear solution is over two times larger for Case 1, six times larger for Case 2 and five times

larger for Case 3 than the results from the current study and ANSYS solution. Therefore, the

exact linear solution assumptions are no longer accurate for those cases. However, in Appendix

D on page 126 when the plate is subject to small displacements pAÞßàT < 0.5q the transverse

displacement for the exact linear solution agree with the current study and the ANSYS solution.

Figure 4.7: Case 1: Transverse Displacement versus Radial Distance (� = 5��, ν = 0.3 , � = 29� + 6��, and �/ = 1/100)"/

� ��/�

ANSYS

Current Study

Linear Theory

47

Figure 4.8: Case 2: Transverse Displacement versus Radial Distance (� = 5��, ν = 0.3 , � = 29� + 6��, and �/ = 1/100)

"/

� ��/�

ANSYS

Current Study

Linear Theory

48

Figure 4.9: Case 3: Transverse Displacement versus Radial Distance (� = 5��, ν = 0.3 , � = 29� + 6��, �/ = 1.5, and �/ = 1/100)

Figures 4.10 – 4.12 show the radial stress, <33, with respect to the radial distance of the

plate for Case 1, Case 2 and Case 3, respectively. The same loading conditions used for the

transverse displacement are used for the radial stress figures as well. The results in Figures 4.10

– 4.12 show that the radial stress of the current study with one Ritz coefficient agrees with the

ANSYS solution. Similarly to the transverse displacement solution when the plate is subjected to

large displacement, the exact linear solution does not agree with the current study nor the ANSYS

results, the radial stress at the center of the plate for the exact linear solution is over 1.5 times

larger for Case 1, three times larger for Case 2 and 2.5 times larger for Case 3 than the current

study and ANSYS result. Therefore, the exact linear solution assumptions are no longer accurate

"/

� ��/�

ANSYS

Current Study

Linear Theory

49

for stress as well. However, in Appendix D on page 128 when the plate is subject to small

displacements pAÞßàT < 0.5q the radial stress for the exact linear solution agree with the current

study and the ANSYS solution.

Figure 4.10: Case 1: Radial Stress versus the Radial Distance of the Plate (� = 5��, ν = 0.3 , � = 29� + 6�� , and �/ = 1/100)

ANSYS

Current Study

Linear Theory

< 33/<33 Þß

à

"/

50

Figure 4.11: Case 2: Radial Stress versus the Radial Distance of the Plate (� = 5��, ν = 0.3 , � = 29� + 6�� , and �/ = 1/100)

< 33/<33 Þß

à

"/

ANSYS

Current Study

Linear Theory

51

Figure 4.12: Case 3: Radial Stress versus the Radial Distance of the Plate (� = 5��, ν = 0.3 , � = 29� + 6�� , �/ = 1.5 and �/ = 1/100)

Case 3 has one anomaly at the free edge where the radial stress should be equal to zero.

This is due to the fact that we did not enforce the natural boundary condition at that point.

However, it is acceptable because the energy method is minimized over an integration and not at

every single point. In summary, Figures 4.7 to 4.12 show that the transverse displacement and

radial stress results from the current study with one coefficient matched the results of the ANSYS

solution accurately. Therefore, the assumed shape functions used for the current study are

deemed acceptable because they provided an accurate match against the ANSYS solution for

each case.

< 33/<33 Þß

à

"/

ANSYS

Current Study

Linear Theory

52

4.5.2 Validation with Load

The next check for the current study is to check the accuracy compared to the ANSYS

and exact linear solution with different loading. In the Tables 4.1 to 4.3, the non-dimensional

transverse displacement is given for the current study, the ANSYS solution, and the exact linear

solution. The displacement conditions are based on the parameters that would make the current

study AÞßàT equal to a certain value for the current study. The same parameters are then used in

the ANSYS and exact linear solutions to develop a difference between the current study, ANSYS

and the exact linear solution. Then, the percentage error is shown to give the accuracy of the

current study to ANSYS and the current study to the exact linear solution. Table 4.1 shows the

comparison of results for Case 1, Table 4.2 shows the comparison results for Case 2 and Table

4.3 shows the comparison results for Case 3.

Table 4.1: Case 1: Dimensionless AÞßàT versus Dimensionless Load

ái and % Error in AÞßàT

(�/ = 1/100 and = 0.3)

âã(äågæ) çèéêë % Error in

çèéêë

CS LT ANSYS %ãííîí = ïð − ñòñò %ãííîí = ïð − óôðõðóôðõð

3.299 0.5 0.563 0.496 11.1% 0.8%

4.154 0.6 0.709 0.594 15.3% 1.0%

5.114 0.7 0.873 0.691 19.7% 1.2%

6.199 0.8 1.058 0.790 24.4% 1.3%

7.425 0.9 1.267 0.883 28.9% 1.8%

8.811 1.0 1.503 0.979 33.5% 2.0%

10.373 1.1 1.770 1.075 37.9% 2.3%

12.130 1.2 2.070 1.171 42.0% 2.4%

14.100 1.3 2.406 1.270 46.0% 2.3%

16.300 1.4 2.781 1.357 50.0% 3.0%

18.747 1.5 3.199 1.451 53.1% 3.3%

CS = Current Study, LT = Exact Solution from Linear Theory, and ANSYS = ANSYS Solution

53



(�/ = 1/100 and = 0.3)


çèéêë


3.299 0.5 0.732 0.496 31.7% 0.78%

4.154 0.6 1.001 0.594 40.1% 0.97%

5.114 0.7 1.338 0.695 47.7% 0.74%

6.199 0.8 1.752 0.786 54.3% 1.71%

7.425 0.9 2.255 0.883 60.1% 1.80%

8.811 1.0 2.858 0.975 65.0% 2.46%

10.373 1.1 3.573 1.070 69.2% 2.74%

12.130 1.2 4.411 1.166 72.8% 2.78%

14.100 1.3 5.382 1.254 75.9% 3.52%

16.300 1.4 6.499 1.347 78.5% 3.77%

18.747 1.5 7.771 1.442 80.7% 3.89%


54



(�/ = 1/100, �/ = 1.5, and = 0.3)


çèéêë


3.299 0.5 0.618 0.481 19.05% 3.76%

4.154 0.6 0.844 0.581 28.90% 3.10%

5.114 0.7 1.126 0.683 37.83% 2.48%

6.199 0.8 1.473 0.784 45.70% 1.95%

7.425 0.9 1.895 0.885 52.51% 1.72%

8.811 1.0 2.401 0.985 58.35% 1.52%

10.373 1.1 3.000 1.086 63.33% 1.26%

12.130 1.2 3.701 1.182 67.58% 1.50%

14.100 1.3 4.515 1.279 71.21% 1.58%

16.300 1.4 5.450 1.377 74.31% 1.66%

18.747 1.5 6.515 1.474 76.97% 1.71%


The results from Table 4.1 – 4.3 show that the current study with only one Ritz

coefficient matches the ANSYS solution. The largest error between the current study and the

ANSYS solution is less than 4% when the maximum transverse displacement is 0.5 < AÞßàT ≤1.5. This proves that the results presented are accurate. When comparing to the current study to

the linear solution, it is shown that the linear solution has a large amount of error. The maximum

error seen when the maximum transverse displacement is 0.5 < AÞßàT ≤ 1.5 is 80.7%. Therefore,

it can also be shown that the exact linear solution will be inaccurate when the plate is subjected to

large displacement.

55

4.5.3 Current Study Solution versus Two Coefficient Ritz Solution

The final validation study is to determine the difference in accuracy achieved by adding

another coefficient to the assumed displacement functions. Equations (4.11) to (4.14) were used

for the two coefficient Ritz solutions. Tables 4.4 to 4.6 show the percentage difference between

the two coefficient solution versus ANSYS and the two coefficient solution versus the current

study with one coefficient for all three cases. The displacement conditions were based on the

parameters that would make the current study AÞßàT equal to a certain value. The same parameters

were used in the ANSYS and two coefficient Ritz solution to develop a difference between the

current study and the two different solutions. Table 4.4 shows the comparison results for Case 1,

Table 4.5 shows the comparison results for Case 2 and Table 4.6 shows the comparison results for

Case 3.



(�/ = 1/100 and = 0.3)


çèéêë

TCS CS ANSYS %ãííîí = òïð − ïðòïð %ãííîí = òïð − óôðõðóôðõð

3.299 0.493 0.5 0.496 1.52% 0.70%

4.154 0.589 0.6 0.594 1.93% 0.90%

5.114 0.684 0.7 0.691 2.31% 0.98%

6.199 0.779 0.8 0.790 2.65% 1.35%

7.425 0.874 0.9 0.883 2.96% 1.00%

8.811 0.969 1.0 0.979 3.24% 1.06%

10.373 1.063 1.1 1.075 3.50% 1.13%

12.130 1.157 1.2 1.171 3.75% 1.22%

14.100 1.250 1.3 1.270 3.98% 1.56%

16.300 1.343 1.4 1.357 4.22% 1.01%

18.747 1.436 1.5 1.451 4.45% 1.03%

CS = Current Study, TCS = Two Coefficient Ritz Method Solution, and ANSYS = ANSYS Solution

56



(�/ = 1/100 and = 0.3) âã(äågæ)

çèéêë % Error in çèéêë


1.053 0.497 0.5 0.496 0.51% 0.27%

1.440 0.596 0.6 0.594 0.70% 0.27%

1.923 0.694 0.7 0.695 0.92% 0.16%

2.518 0.791 0.8 0.786 1.15% 0.57%

3.241 0.887 0.9 0.883 1.41% 0.42%

4.109 0.984 1.0 0.975 1.67% 0.83%

5.137 1.079 1.1 1.070 1.96% 0.83%

6.341 1.174 1.2 1.166 2.24% 0.60%

7.737 1.268 1.3 1.254 2.54% 1.06%

9.342 1.361 1.4 1.347 2.84% 1.04%

11.171 1.454 1.5 1.442 3.14% 0.87%


57



(�/ = 1/100, �/ = 1.5, and = 0.3)


çèéêë


1.041 0.473 0.5 0.481 5.72% 1.74%

1.422 0.575 0.6 0.581 4.41% 1.17%

1.897 0.677 0.7 0.683 3.42% 0.85%

2.482 0.779 0.8 0.784 2.67% 0.67%

3.193 0.881 0.9 0.885 2.11% 0.36%

4.045 0.983 1.0 0.985 1.68% 0.14%

5.054 1.085 1.1 1.086 1.35% 0.07%

6.236 1.187 1.2 1.182 1.09% 0.43%

7.607 1.289 1.3 1.279 0.89% 0.71%

9.181 1.390 1.4 1.377 0.72% 0.95%

10.976 1.491 1.5 1.474 0.60% 1.13%


As expected, adding the 2nd unknown coefficient provided more accurate results;

however, adding the 2nd unknown coefficient made the solution not solvable analytically because

the computed Groebner basis system of equations had a polynomial of degree order 9 which is

unable to be solved analytically. This is one of the limitations of this method, since it is shown in

the Abel-Ruffini impossibility theorem that there is no closed form solution for a polynomial of

degree order 5 or higher. The current study with one Ritz coefficient is still deemed acceptable,

because even though the answer (~4% different) is slightly conservative it is much more accurate

than the exact linear solution (~80% different) and there is still a closed form solution.

58

CHAPTER 5

RESULTS AND DISCUSSION

This chapter presents the results of each case study provided in section 4.1. The sections

in this chapter are organized by each individual case. And each section begins by presenting the

closed form solutions for the transverse displacement, stresses in the radial and theta directions,

moments, and shearing force generated from the current study. Once the closed form solutions

for the transverse and in-plane displacements functions are determined, the moments, shearing

force, and stresses can easily be calculated symbolically in terms of the structural parameters such

as material properties and plate geometry, which are also presented in each section. In addition,

the results from the parametric study for each case are plotted in Figures 5.1 to 5.15 and some

additional plots are in Appendix G.

5.1 Results for Fully Clamped Circular Plate (Case 1)

The closed form solution for the transverse displacement, � in terms of structural

parameters such as plate geometry and material properties is presented in equation (5.1). The

maximum displacement �� at the center of the plate is shown in equation (5.1) as well. The

additional variables /, 0, :, 2 and 8 in equations (5.1) to (5.6) are used to simplify the

expressions of transverse displacement, moments, shearing force, and stresses which are provided

in equation (5.7).

� = �1 − "((�(ö5251*/1*12Etβ + 2(525)(*Et*3α1* û

�Ò|3²¬ = �� = ö5251*/1*12Etβ + 2(525)(*Et*3α1* û (5.1)

59

Equation (5.2) presents the closed form solution of the radial moment, F33 the moment at

the support F33 Ò|3²�, and the moment at the center F33 Ò|3²¬. The maximum moment is at the

edge and is used in Figure 5.4 to develop a profile for the moment F33.

F33 = −Bü8"(# : − 4(1 + ) �1 − "((�( :ý

F33 Ò|3²¬ = B �4(1 + )( :�

F33 Ò|3²� = F33_�� = −B 8( :

(5.2)

Equation (5.3) presents the closed form solution of the moment in the 5 direction, F44

the moment at the support F44 Ò|3²�, and the moment at the center F44 Ò|3²¬.

F44 = −Bü 8"(# : − 4(1 + ) �1 − "((�( :ý

F44 Ò|3²¬ = F44_�� = B �4(1 + )( :�

F44 Ò|3²� = −B 8 ( :

(5.3)

Equation (5.4) presents the closed form solution of the transverse shearing force, I3 and

the transverse shearing force at the edge I3 Ò|3²�.

I3 = 32"# :

I3 Ò|3²� = Q�� = 32* :

(5.4)

60

In Equation (5.5) the closed form solution for the radial stress <33 and the stress in the

theta direction <44 are expressed in two terms. The terms with � are due to bending effects and

the rest is affected by in-plane displacement.

<33 = �1 − ( ��−�ü8"(# : − 4(1 + ) �1 − "((�( :ý − 2"(* �8 + 2 "((�

+ 2"(* �1 − "((� 2 + (1 + ) �8 + 2 �"((�� 1 − "((�+ 8:("( �1 − "((�(# �

�

<44 = �1 − (��−�ü8 "(# : − 4(1 + ) �1 − "((�( :ý + 8 "( �1 − "((�(# :(

+ (1 + ) �1 − "((� �8 + 2 �"((�� + 2 "(* �2 − 8 − 22"(( ��

(5.5)

The stresses at the support <33 Ò|3²� and at the center of the plate <33 Ò|3²¬ are presented in

equation (5.6). In equation (5.6), the maximum stress is at the edge when a uniform loading is

applied. This maximum stress is used to non-dimensionalize the stresses at the top, bottom and

middle plane shown in Figure 5.6.

<33 Ò|3²¬ = �1 − ( �4�(1 + )( : − (1 + )8 �

<33 Ò|3²� = <�� = �1 − ( �−8�( : − 2(2 + 8) � (5.6)

61

The additional variables /, 0, :, 2 and 8 appeared in equations (5.1) – (5.6) above are

given as follows.

/ = �(�(0( ö−27�#(1 − ()+�−268800��(0 + 729�((1 − ()(û

0 = −39 − 21 + 14 (

: = ö5251*/1*12Etβ + 2(525)(*Et*3α1* û

2 = :( − 33

8 = :( 9 − 7 15

(5.7)

Figure 5.1 shows how a non-dimensional displacement AÞßàT versus load

á�Ti varies with

Poisson’s ratio for Case 1. In the figure, the exact solutions from the linear theory, hereon

denoted as “Linear Theory” in all figures, are plotted with the solutions from the current study.

The results show that Poisson’s ratio does not have much effect on the maximum displacement

for the given load for both linear and nonlinear solutions. However, it is clearly shown that the

prediction based on the linear theory is only accurate when AÞßàT is less than 0.5, which is

expected.

62

Figure 5.1: Case 1: Dimensionless Displacement versus Dimensionless Load varying Poisson’s

Ratio,

Figure 5.2 presents a parametric study how the geometry properties of the plate affect the

displacement for Case 1. Notice that the thickness of the plate has a large effect on the maximum

displacement at the center. The plot indicates that as T� increases, the dimensionless displacement,

AÞßàT decreases. Additionally, the differences between the solutions of linear and nonlinear are

reduced for a given load.

� ��/�

�#/(�#�)

Current Study

Linear Theory

Figure

and moment at the center and at the edge of the plate respectively for Case 1.

that the displacement has a large effect on the stresses and moments. Again, the stresses and

moments calculated based on linear and nonlinear theory have no difference only for the

displacements

Figure




displacements

� ��/�

Figure 5




displacements

� ��/�

5.2: Case 1: Dimensionless Transverse Displac

Figure 5.3 and 5.4 present




displacements

: Case 1: Dimensionless Transverse Displac





displacements






displacements AÞßàT






Þßà <






< 0






0.5. Therefore, the nonl






. Therefore, the nonl


Thickness Ratio,







Thickness Ratio,

Figure 5.3 and 5.4 present a parametric study of the displacement






Thickness Ratio,

a parametric study of the displacement






Thickness Ratio,







Thickness Ratio,






63


Thickness Ratio,





. Therefore, the nonlinear analysis for these cases is necessary.

�/

63


Thickness Ratio, �/





inear analysis for these cases is necessary.

/� (10E

: Case 1: Dimensionless Transverse Displacement versus Dimensionless Load varying (!






(10E

ement versus Dimensionless Load varying ! =






(10E-6)

ement versus Dimensionless Load varying = 0.3






6)

Current Study

Lin

ement versus Dimensionless Load varying 3)






Current Study

Linear Theory

ement versus Dimensionless Load varying






Current Study

ear Theory







Current Study

ear Theory


a parametric study of the displacement A





Current Study

ear Theory


AÞßàTand moment at the center and at the edge of the plate respectively for Case 1.





Þßà versus the stress

and moment at the center and at the edge of the plate respectively for Case 1. The figures show





versus the stress

The figures show





versus the stress

The figures show





versus the stress

The figures show





versus the stress

The figures show




versus the stress

The figures show



versus the stress

The figures show



Figure

< 33(/(��(

)

Figure

< 33(/(��(

)

Figure 5..3: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse : Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse

Displacement varying Radial Distance

: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse


















Current Study

Linear Theory

64



�

Current Study

Linear Theory

64



��

Current Study

Linear Theory


Displacement varying Radial Distance "/��/�

Current Study

Linear Theory

: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse / (

�

Current Study

Linear Theory


( =: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse = 0: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse 0.3 and


and


and � =: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse = �/: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse /2)


: Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse : Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse : Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse : Case 1: Dimensionless Radial Stress versus Dimensionless Maximum Transverse

moment

middle plane along the radius of the plate can be easily plotted for a given load shown in

5.5 and 5.6. The given load is indicated in the caption of Figure

Figure

moment



F 33( /(��

# )

Figure

moment



F 33( /(��

# )

Figure 5

Since the results for the stress and moments are expressed symbolically, the profile of the

moments in the radial



F 33( /(��

# )

5.4: Case 1: Dimensionless Moment versus Dimensionless Transverse Displacement


s in the radial



: Case 1: Dimensionless Moment versus Dimensionless Transverse Displacement


s in the radial





s in the radial





s in the radial "middle plane along the radius of the plate can be easily plotted for a given load shown in




" and





and 5middle plane along the radius of the plate can be easily plotted for a given load shown in




5 direction and the profile of the stresses at the top, bottom, and





direction and the profile of the stresses at the top, bottom, and














( Since the results for the stress and moments are expressed symbolically, the profile of the




65

: Case 1: Dimensionless Moment versus Dimensionless Transverse Displacement = 0Since the results for the stress and moments are expressed symbolically, the profile of the




�

65

: Case 1: Dimensionless Moment versus Dimensionless Transverse Displacement 0.3)





��: Case 1: Dimensionless Moment versus Dimensionless Transverse Displacement

)





��/t






/t: Case 1: Dimensionless Moment versus Dimensionless Transverse Displacement









5.5 and 5.6. The given load is indicated in the caption of Figure 5.5 and 5.6.





5.5 and 5.6.

Current Study

Linear Theory





5.5 and 5.6.

Current Study

Linear Theory





5.5 and 5.6.

Current Study

Linear Theory





5.5 and 5.6.

Current Study

Linear Theory





Current Study

Linear Theory












middle plane along the radius of the plate can be easily plotted for a given load shown in Figure




Figure




Figure



Figure

66

Figure 5.5: Case 1: Moment Profile (�#/(��#) = 500/29 and = 0.3)

F/F ��

"/

F33/F33_�� F44/F44_��

67

Figure 5.6: Case 1: Stress Profile through the Thickness (�#/(��#) = 500/29 and = 0.3)

5.2 Results for Simply Supported Immovable Edge Circular Plate (Case 2)

The closed form solution transverse displacement function, � in terms of structural

parameters such as plate geometry and material properties is presented in equation (5.8). The

maximum displacement �� at the center of the plate is shown in equation (5.8) as well. The

additional variables /, 0, :, 2 and 8 found in equation (5.8) – (5.14) are used to simplify the

expressions of the transverse displacement, moments, shear force, and radial stress and are given

in equation (5.15).

"/

< 33/<��

� = − �2 � = + �2 � = 0

68

� = �1 − "((��5 + 1 + − "((� :

�Ò|3²¬ = �� = �5 + 1 + � : (5.8)

Equation (5.9) presents the closed form solution for radial moment function, F33 and the

moment at the center, F33 Ò|3²¬. Since the plate is simply supported, the moment at the edge,

F33 Ò|3²� is equal to zero. The maximum moment is at the center and is used in Figure 5.11 to

develop a profile for the moment F33.

F33 = B �4( − ")( + ")(3 + )# :�

F33 Ò|3²¬ = F33_�� = B �4(3 + )( :�

(5.9)

Equation (5.10) presents the closed form solution of the moment in the 5 direction, F44,

the moment at the center F44 Ò|3²¬ and the moment at the edge F44 Ò|3²�.

F44 = B �4((( − 3"() + 3( − "()# :�

F44 Ò|3²¬ = F44_�� = B �4(3 + )( :�

F44 Ò|3²� = B 8(1 − )( :

(5.10)

Equation (5.11) presents the closed form solution of the transverse shearing force, I3 and

the transverse shearing force at the edge I3 Ò|3²�.

69

I3 = 32"# :

I3 Ò|3²� = Q�� = 32* :

(5.11)

In Equation (5.12) the closed form solution for radial stress <33 and the stress in the theta

direction <44 are expressed in two terms. The terms with � are due to bending effects and the rest

is affected by in-plane displacement.

<33 = �1 − ( ö−� 8"(# : − 2(1 + )( : �5 + 1 + − 2"(( + 1�¡+ (1 + ) p1 − "q �8 + "(( 2� − "( �8 + "(( 2� + 2"(* 2 p1 − "q+ 12# −2": �5 + 1 + − 2"(( + 1�¡(¡

<44 = �1 − ( ö−� 8 "(# : − 2(1 + )( : �5 + 1 + − 2"(( + 1�¡+ (1 + ) p1 − "q �8 + "(( 2� − "( �8 + "(( 2� + 2 "(* 2 p1 − "q+ 2# −2": �5 + 1 + − 2"(( + 1�¡(¡

(5.12)

The stress at the support <33 Ò|3²� and at the center of the plate <33 Ò|3²¬ are presented in

equation (5.13). In equation (5.13), the maximum stress is at the edge when a uniform load is

applied. This maximum stress is used to non-dimensionalize the stresses at the top, bottom and

middle plane shown in Figure 5.10.

70

<33 Ò|3²¬ = <�� = �1 − ( �−� −2(1 + )( �5 + 1 + + 1�¡+ (1 + ) 8�

<33 Ò|3²� = �1 − ( ö−� 8( : − 2(1 + )( : �5 + 1 + − 1�¡− 1 (8 + 2) + 12# −2: �5 + 1 + − 1�¡(¡

(5.13)

The additional variables /, 0, :, 2 and 8 appeared in equations (5.8) – (5.13) above are

given as follows.

/ = �(�(0(ü9�#( ( − 1)( + 7)+ − 30 ( + 7)(( + 1)((4765967360�(�( + 7)(1 + )− 27�(0( − 1)()�1(û

0 = 6696449 � + 99521302 � + 478692103 # + 67588516 *− 6799490129 ( − 20445519770 − 14008216855

: = 279255901* öα1*(1 + ν)4Etβ + 8(279255901*)E(t#(1 + ν)(ν + 7))4Etα1* û

2 = 64:((24 * − 43 ( − 1181 − 58)5181(1 + )(

8 = −4:((17387 * + 105785 ( − 2951 − 645749)108801(1 + )(

(5.14)

71

Figure 5.7 shows how non-dimensional displacement, AÞßàT versus load

á�Ti varies with

Poisson’s ratio for Case 2. In the figure, the exact solutions from the linear theory are plotted

with the solutions from the current study. The results show that Poisson’s ratio does not have

much effect on the maximum displacement for a given load for both linear and nonlinear

solutions. However, it is clearly shown that the predictions based on the linear theory are only

accurate when AÞßàT is less than 0.5, which is expected.

Figure 5.7: Case 2: Dimensionless Transverse Displacement versus Dimensionless load varying

Poisson’s ratio,

�#/(�#�)

Current Study

Linear Theory

� ��/t


displacement at the center. The plot indicates that as

Areduced for a given load.

Figure



AÞßàTreduced for a given load.

Figure

� ��/�



Þßà decreases. In addition, the differences between the solutions of linear and nonlinear are


Figure 5

� ��/�

Figure



decreases. In addition, the differences between the solutions of linear and nonlinear are


5.8:

Figure





: Case

Figure 5.8 presents a param





Case

.8 presents a param





Case 2: Dimensionless Transverse Dis

.8 presents a param





Dimensionless Transverse Dis

.8 presents a param






.8 presents a param





.8 presents a param





Thickness Ratio,

.8 presents a parametric study how the geometry properties of the plate affect the





Thickness Ratio,

etric study how the geometry properties of the plate affect the





Thickness Ratio,






Thickness Ratio,






Thickness Ratio,

72






Thickness Ratio,

�/

Current Study

Linear Theory

72





Dimensionless Transverse Displacement versus Dimensionless Load varying

Thickness Ratio, �//� (10E

Current Study

Linear Theory



displacement at the center. The plot indicates that as T� increases, the dimensionless


placement versus Dimensionless Load varying ( (10E

Current Study

Linear Theory



increases, the dimensionless


placement versus Dimensionless Load varying =(10E-6)

Current Study

Linear Theory





placement versus Dimensionless Load varying = 0.36)

Current Study

Linear Theory





placement versus Dimensionless Load varying 3)

Current Study

Linear Theory





placement versus Dimensionless Load varying























increases, the dimensionless displacement,





displacement,





displacement,





displacement,





displacement,





displacement,


2. Notice that the exact solution from the linear theory stress result at the edge is ze

to the fact that the exact linear solution cannot account for in

due to bending. This is another condition where the linear solution cannot accurately predict the

solution.




solution.

Figure

< 33(/(��(

)2. Notice that the exact solution from the linear theory stress result at the edge is ze



solution.

Figure

< 33(/(��(

)Figure




solution.

Figure 5.

Figure




.9: Case 2:

Figure 5.9 shows how the stress is affected when the displacement is increased for Case




: Case 2:


.9 shows how the stress is affected when the displacement is increased for Case




: Case 2:












Dimensionless Radial Stress versus Dimensionless Maximum Transverse








Current Study

Linear Theory







Current Study

Linear Theory







Current Study

Linear Theory







Current Study

Linear Theory







Current Study

Linear Theory







Current Study

Linear Theory

73







�

73







��






Displacement varying Radial Distance "/��/�





Dimensionless Radial Stress versus Dimensionless Maximum Transverse / (

�



to the fact that the exact linear solution cannot account for in-



( =



-plane stress and the stress is only


Dimensionless Radial Stress versus Dimensionless Maximum Transverse = 0



plane stress and the stress is only


Dimensionless Radial Stress versus Dimensionless Maximum Transverse 0.3 and






and






and � =





Dimensionless Radial Stress versus Dimensionless Maximum Transverse = �/





Dimensionless Radial Stress versus Dimensionless Maximum Transverse /2)


2. Notice that the exact solution from the linear theory stress result at the edge is zero, this is due





ro, this is due





ro, this is due





ro, this is due





ro, this is due





ro, this is due



the center for Case 2. This figure shows that displacement has a large effect on moment. Again

it can be shown, the moment from when dis

become



become

Figure

F 33( /(��

# )the center for Case 2. This figure shows that displacement has a large effect on moment. Again


becomes

Figure

F 33( /(��

# )Figure



s increasing

Figure 5.

F 33( /(��

# )Figure



increasing

.10: Case 2: Dimensionless Moment versus Dimensionless Transverse Displ

Figure 5.10 presents a parametric study of the effect that displacement has on moment at



increasing

: Case 2: Dimensionless Moment versus Dimensionless Transverse Displ

.10 presents a parametric study of the effect that displacement has on moment at



increasingly





y inaccurate the larger the deflection becomes





inaccurate the larger the deflection becomes
































(

74



it can be shown, the moment from when displacement is small


: Case 2: Dimensionless Moment versus Dimensionless Transverse Displ= 0�

74



placement is small


: Case 2: Dimensionless Moment versus Dimensionless Transverse Displ0.3)

��



placement is small



)

��/t



placement is small



t

Current Study

Linear Theory



placement is small



Current Study

Linear Theory



placement is small

inaccurate the larger the deflection becomes.


Current Study

Linear Theory



placement is small 0.


Current Study

Linear Theory



0.5


Current Study

Linear Theory



> A




AÞßàT




Þßà is accurate and




is accurate and




is accurate and




is accurate and

acement



is accurate and

acement



is accurate and

acement



is accurate and

75


moments in the radial " and 5 direction and the profile of the stresses at the top, bottom, and

middle plane along the radius of the plate can be easily plotted for a given load shown in Figure

5.11 and 5.12. The given load is indicated in the caption of Figure 5.11 and 5.12.

Figure 5.11: Case 2: Moment Profile (�#/(��#) = 500/29 and = 0.3)

F/F ��

"/

F33/F33_�� F44/F44_��

76

Figure 5.12: Case 2: Stress profile through the Thickness (�#/(��#) = 500/29 and = 0.3)

5.3 Results for Simply Supported Immovable Edge Circular Plate with Overhang (Case 3)

The closed form solution for the transverse displacement, � and �KL in terms of

structural parameters such as plate geometry and material properties is presented in equation

(5.15). The maximum displacement �� at the center of the plate is shown in equation (5.15) as

well. The additional variables /, 01, 0(, :, 2 and 8 in equations (5.15) to (5.21) are used to

simplify the expressions of the transverse displacement, moment and radial stress and are given in

equation (5.22).

"/

< 33/<��

� = + �2 � = − �2 � = 0

77

� = ��"#64B + �(Õ(( − 1) − 2�(( + 1)Ö"(32B�(( + 1)− �#Õ−2(( − 1) − 3�(( + 1)Ö"(64�(( + 1)B � :

�KL = ��#( − 1)"#32B�((1 + ) + �#Õ((1 − ) + 2ln()�(( + 1)Ö32B�(( + 1) − �#ln(")16B � :

�Ò|3²¬ = �� = −�#Õ−2(( − 1) − 3�(( + 1)Ö"(64�(( + 1)B :

(5.15)

Equation (5.16) presents the closed form solution of the radial moment, F33 the moment

for the overhang region F33KL and the moment at the center F33 Ò|3²¬. The maximum moment is

at the center and is used in Figure 5.17 to develop a profile for the moment F33.

F33 = − �:16�( Õ(3 + )"(�( − 2(1 + )(�( + ( − 1)#Ö

F33KL = �#:16"(�( (�( − "()( − 1) F33 Ò|3²¬ = F33_�� = − �:16�( (−2(1 + )(�( + ( − 1)#)

(5.16)

Equation (5.17) presents the closed form solution of the moment in the 5 direction, F44

and F44KL, the moment at the center F44 Ò|3²¬, and the moment at the edge F44 Ò|3²�.

F44 = − �:16�( (−2( �( + 3"(�( + # + "(�( − # − 2(�() F44KL = :"(�( ("( + �( − "( − �() F44 Ò|3²¬ = F44_�� = − �:16�( (−2( �( + # − # − 2(�()

F44 Ò|3²« = − �:16�( (( �( + # − (�( − #) (5.17)

78

Equation (5.18) presents the closed form solution of the transverse shearing force I3, the

transverse shearing force in the overhang region I3KL, and the transverse shearing force at the

edge I3 Ò|3²�.

I3 = −6"��*� :( ( − 1) I3KL = 0

I3 Ò|3²� = Q�� = −6��*� :( ( − 1) (5.18)

Equation (5.19) shows the closed form solution for the radial stress <33, <33KL and

Equation (5.20) the stress in the theta direction <44, <44KL. The terms with � are due to bending

effects and the rest is affected by in-plane displacement.

<33 = �1 − ( ö−� 3(3 + )4��* (1 − ()�"(+ 3(1 − ()(�4�(��* Õ(( − 1) − 2�(( + 1)Ö� :+ 1 + p1 − "q �8 + 2 "((� − "( �8 + 2 "((� + 2"(* p1 − "q 2+ 12 3�"*4��* (1 − ()+ 3(�(1 − ()"4��(�*(1 + ) Õ(( − 1) − 2�(( + 1)Ö¡( :(¡

<33KL = �1 − (ü−� �3( − 1)(1 − ()�#4�(��* + 3(1 − ()�#4��*"( � : − 1 �8 + 2 "((�+ 2"( p1 − "q 2 + 12�3( − 1)(1 − ()�#"4�((1 + )��* − 3(1 − ()�#4��*" �( :(+ p1 − "q �8 + 2 "((�" û

(5.19)

79

<44 = �1 − ( ö−� 3(3ν + 1)4��3 (1 − 2)�"2+ 3(1 − 2)2�4�2��3 p2( − 1) − 2�2( + 1)q¡ :+ 1 + p1 − "q �8 + 2 "22� − "2 �8 + 2 "22� + 2ν"23 p1 − "q 2+ ν2 3�"34��3 (1 − 2)+ 32�(1 − 2)"4��2�3(1 + ) p2( − 1) − 2�2( + 1)q¡2 :2¡

< KL = �1 − (ü−��3( − 1)(1 − 2)�44�2��3 + 3(ν − 1)(1 − 2)�44��3"2 � :− ν�8 + 2 "22� + 2ν"2 p1 − "q 2+ ν2�3( − 1)(1 − 2)�4"4�2(1 + )��3 − 3(1 − 2)�44��3" �2 :2+ p1 − "q �8 + 2 "22�" û

(5.20)

The maximum stress which happens at the center of the plate <33 Ò|3²¬ is presented in

equation (5.21).

<33 Ò|3²¬ = <�� = �1 − ( ö−� ��(Õ(( − 1) − 2�(( + 1)Ö16B�( � : + (1 + ) (8)û (5.21)

The additional variables /, 01, 0(, :, 2 and 8 appeared in equations (5.14) – (5.21) above

are given as follows.

80

/ = 27��( − 1) − 36#��(( + 1)+ ((19289340( − 7)( − 1)�((¬− 12859560(1 + )(13 − 85)( − 1)��(�(1+ 34992(1 + )(17327 ( − 88500 − 106070)( − 1)��(�#1�− 43740(26921 ( − 126732 − 154613)(1 + )(( − 1)��(��1#− 81( − 1)#(−9�(�(1771167 ( − 7683330 − 9561185)(1 + )* + 250880�(�( − 1)()1(+ 9720( − 1)*(−�(�(77947 ( − 311674 − 396213)(1 + )*+ 12544�(�( − 1)()(1 + )�(1¬− 1296�#( − 1)((−��((143359 ( − 528770 − 687425)(1 + )# + 15680�(�(15 + 17)( − 1)()(1 + )+ 81285120�(��(5 + 7)(1 + )(( − 1)*�− 6773760��(�(3 + 5)(15 + 17)( ( − 1)(#+ 13547520�(��1¬( − 1)(3 + 5)((1 + )*(− 752640�(��1((3 + 5)*(1 + )*)/( − 1)(01)1(

01 = 26460( − 7)( − 1)# − 52920(1 + )(3 − 19)( − 1)*�(�+ 144(1 + )(2509 ( − 11860 − 14450)( − 1)(�##− 12( − 1)(30797 ( − 128700 − 161705)(1 + )(��(+ (143359 ( − 528770 − 687425)(1 + )*�

0( = 8�14701*� ( − 1)(�(�#Õ3( − 1)(# − 6�(( ( − 1)(+ �#(3 + 5)(1 + )Ö(01)

: = 8(1470)1*�(�( �0( + (�(�(( − 1)#01(/)(*�3#�( − 1)(01(�(�(( − 1)#01(/)1*

2 = − 3:(�(3584�#�(�� ( − 1)(�(−78( *�( + 23 *�# + 42# * − 210# (+ 54(�( ( + 197�# ( + 294# + 78( �( + 325 �# − 126#− 54(�( + 151�#) 8 = − :(�(3584�#�(�� ( − 1)(�(−474( *�( + 437 *�# + 126# * − 630# (+ 1362(�( ( − 289�# ( + 882# + 474( �( − 1889 �#− 378# − 1362(�( − 1163�#)

(5.22)

81

Figure 5.13 shows how a non-dimensional displacement AÞßàT versus loads

á�Ti varies

with Poisson’s ratio for Case 3. In the figure, the exact solutions from the linear theory are

plotted with the solutions from the current study. The results show that Poisson’s ratio does not

have much effect on the maximum displacement for given load for both linear and nonlinear

solutions. However, it is clearly shown that the prediction based on the linear theory is only

accurate when AÞßàT is less than 0.5, which is expected.

Figure 5.13: Case 3: Dimensionless Displacement versus Dimensionless Load varying Poisson’s

Ratio, (�/ = 1.5)

Current Study

Linear Theory

� ��/�

�#/(�#�)

displacement for Case 3. Notice that the thickness of the plate has a lar


Areduced for a g

Figure



AÞßàTreduced for a g

Figure

� ��/�



Þßà decreases. Additionally, the differences between the solutions of linear and nonlinear are

reduced for a g

Figure 5

� ��/�

Figure



decreases. Additionally, the differences between the solutions of linear and nonlinear are

reduced for a g

5.14

� ��/�

Figure




reduced for a g

14: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying

Figure 5.14 presents a parametric study





: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying

.14 presents a parametric study




iven load.






iven load.






iven load.


Thickness Ratio,






Thickness Ratio,






Thickness Ratio,






Thickness Ratio,






Thickness Ratio,






Thickness Ratio, �





: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying �/

�

82

.14 presents a parametric study of





( �/�

82

f how the geometry properties of the plate affect




: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying = 0

Current Study

Linear Theory

� (10E

how the geometry properties of the plate affect


displacement at the center. The plot indicates that as T� increases, the dimensionless displacement,


: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying 0.3

Current Study

Linear Theory

(10E






and

Current Study

Linear Theory

(10E-6)






and �/

Current Study

Linear Theory





: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying / =

Current Study

Linear Theory





: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying = 1





: Case 3: Dimensionless Transverse Displacement versus Dimensionless Load varying 1.5)


ge effect on the maximum











































3. To determine the difference in the stress result, the stress from the linear assumption is

determined at the same load level deflection for the geometrically nonlinear deflection, i.e. the

load to take the

solution. Again, the stress from when displacement is small

increasing

F



load to take the


increasing

Figure

< 33(/(��(

)3. To determine the difference in the stress result, the stress from the linear assumption is


load to take the


increasing

igure 5

< 33(/(��(

)Figure



load to take the


increasingly

5.15

displacement varying radial distance

Figure



load to take the



15: Case 3: Non


Figure 5.15 shows how the stress is affecte



load to take the current study solution is also the load used to determine the stress in the linear



Case 3: Non


.15 shows how the stress is affecte



current study solution is also the load used to determine the stress in the linear



Case 3: Non








Case 3: Non








Case 3: Non-dimensional radial stress versus non








dimensional radial stress versus non


















Current Study

Linear Theory









Current Study

Linear Theory









Current Study

Linear Theory









Current Study

Linear Theory

83








displacement varying radial distance "/

Current Study

Linear Theory

�

83







dimensional radial stress versus non/ (

��

d when the displacement is increased for Case






dimensional radial stress versus non =��/�







dimensional radial stress versus non= 0.3�





solution. Again, the stress from when displacement is small 0.

dimensional radial stress versus non-3, �





.5 >

-dimensional maximum transverse �/





> AÞßà

dimensional maximum transverse =





AÞßàT

dimensional maximum transverse 1.5





Þßà is accurate and become

dimensional maximum transverse 5, and





is accurate and become

dimensional maximum transverse

, and �






dimensional maximum transverse � =






dimensional maximum transverse �/2






dimensional maximum transverse 2)

















is accurate and becomes






the center and at the support for Case 3. This figure shows that displacement has a large effect on

moment. Again, the moment from when displacement is small

become



become

Figure

F 33( /(��

# )the center and at the support for Case 3. This figure shows that displacement has a large effect on


becomes

Figure

F 33( /(��

# )Figure



s increasing

Figure 5.

Figure



ncreasing

.16: Case 3: Dimensionless Moment versus Dimensionless Transverse Displacement

Figure 5.16 presents a parametric stud



ncreasing


.16 presents a parametric stud



ncreasingly



























(






( =





: Case 3: Dimensionless Moment versus Dimensionless Transverse Displacement = 0.





: Case 3: Dimensionless Moment versus Dimensionless Transverse Displacement .3 and

Current Study

Linear Theory

84






and

Current Study

Linear Theory

84

y of the effect that displacement has on moment at





and �/"/

Current Study

Linear Theory





: Case 3: Dimensionless Moment versus Dimensionless Transverse Displacement =/

Current Study

Linear Theory





: Case 3: Dimensionless Moment versus Dimensionless Transverse Displacement = 1.5

Current Study

Linear Theory





: Case 3: Dimensionless Moment versus Dimensionless Transverse Displacement 5)








moment. Again, the moment from when displacement is small 0.




0.5




> A




AÞßàT




Þßà is accurate and




is accurate and




is accurate and




is accurate and




is accurate and




is accurate and




is accurate and



5.17 and

Figure



.17 and

Figure

F/F ��



.17 and

Figure 5

F/F ��

Since the results for the stre



.17 and 5.18. The given load is indicated in the caption of Figure

5.17




.18. The given load is indicated in the caption of Figure

17: Case 3: Moment Profile versus Radial Distance varying





: Case 3: Moment Profile versus Radial Distance varying







moments in the radial "middle plane along the radius of the plate can be easily plotted for a given load shown in




" and





and 5middle plane along the radius of the plate can be easily plotted for a given load shown in




5 direction and the profile of the stresses at the top, bottom, and









ss and moments are expressed symbolically, the profile of the










and






and

85






and =

85





: Case 3: Moment Profile versus Radial Distance varying = 0.

"





: Case 3: Moment Profile versus Radial Distance varying .3)

"/



















.18. The given load is indicated in the caption of Figure 5.17 and





.17 and

: Case 3: Moment Profile versus Radial Distance varying �/




.17 and

(�

F33F44




.17 and 5.18.

�#/

33/F44/




.18.

/(�

F33/F44




��#)

33_��44_��




) =

��



middle plane along the radius of the plate can be easily plotted for a given load shown in Figur

500



Figur

500/29



Figure

29



e

86

Figure 5.18 is a stress profile that shows the top, bottom and middle plane stress along the

plate for a certain load case. <�� is determined from equation (5.20).

Figure 5.18: Case 3: Stress Profile through the Thickness (�#/(��#) = 500/29, �/ = 1.5, and = 0.3)

� = − �2 � = + �2 � = 0

"/

< 33/<��

87

CHAPTER 6

CONCULSION

Groebner basis is utilized in this thesis to demonstrate the usefulness in solving nonlinear

systems of equations to develop more analytical solutions. Analytical solutions are preferred

because the parametric study is much easier to perform over a numerical solution. We would like

the engineering community to become more familiar with this method and therefore, a

background of Groebner basis is provided in Chapter 3. The current study solves geometrically

nonlinear circular isotropic plates subject to uniform distributed load. The current study uses the

total potential energy of the plate in polar coordinates with the Ritz method to develop system of

algebraic equations. The resulting system of equations is coupled and nonlinear. Therefore,

Groebner basis is used to decouple the nonlinear algebraic equations and make it possible to solve

analytically. Purely symbolic solutions for displacement, stress and moments for three cases

defined in Chapter 4 were able to be found and are shown in Chapter 5.

The solutions of the current study were verified through the use of a finite element

simulation. The comparison between ANSYS and the current study showed that the two

solutions had good agreement with an error less than 4%. The results of the validation are

presented in Chapter 4. Included in the validation the exact solution from the linear theory was

compared to the ANSYS solution and the current study solution. The linear theory solution

demonstrated to be not accurate for displacements greater than half of the thickness which was

expected. The percent error between the linear theory solution and the current study is over 80%.

Therefore, it is shown the current study provided significant accuracy in the nonlinear range.

88

To provide accuracy using the Ritz method, typically additional unknown coefficients

can be added to the trial functions. The addition of a coefficient to the transverse displacement

did demonstrate a limitation of using Groebner basis to help solve system of equations, because

the resulting Groebner basis had a polynomial of degree order 9. Therefore, the system could not

be solved analytically because there is no closed form solution for a polynomial of degree order

five or more. However, the two coefficient solutions were solved numerically and shown to only

to provide 2% of accuracy increase over the one coefficient solution. When compared to

providing 80% accuracy increase provided by the one coefficient solution to the linear solution,

the one coefficient solution is acceptable.

There are many nonlinear problems especially in computational mechanics that can be

helped by Groebner basis to be solved analytically. In future studies, this thesis work can be

extended to problems with different boundary conditions, varied thickness, or thermal and

dynamic loading. In conclusion, the Groebner basis methodology is of great value in generating

analytical solutions combined with the methods of Ritz, Galerkin, and similar approximation

methods of weighted residuals which can produce correspondingly low order systems of

polynomial algebraic equations.

89

REFERENCES

90

[1] S. Qaqish, "Comparison between linear and nonlinear structural analysis and the use of very

small increments in solving nonlinear structural analysis," Computers & Structures, vol. 33,

no. 4, pp. 923-928, 1989.

[2] R. Szilard, Theories and Applications of Plate Analysis. United States: John Wiley & Sons,

Inc., 2004.

[3] R. Dugas, History of Mechanics. New York, NY, United States: Dover Publications, 1988.

[4] S. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells. New York:

McGraw-Hill, 1959.

[5] L. Duarte Filho and A. Awruch, "Geometricall nonlinear static and dynamic analysis of

shells and plates using eight-node hexahedral element with one-point quadrature," Finite

Elements in Analysis and Design, vol. 40, pp. 1297-1315, 2004.

[6] X. Cui, et al., "A Smoothed Finite Element Method (SFEM) for Linear and Geometrically

Nonlinear Analysis of Plates and Shells," Computer Modeling in Engineering and Science,

vol. 28, no. 2, pp. 109-125, 2008.

[7] C. Tiago and P. Pimenta, "An EFG method for the nonlinear analysis of plates undergoing

arbitraily large deformations," Engineering Analysis with Boundary Elements, vol. 32, pp.

494-511, 2008.

[8] A. Shahidi, M. Mahzoon, M. Saadatpour, and M. Azhari, "Very Large Deformation but Smal

Strain Analysis of Plates and Folded Plates by Finite Strip Method," Advances in Structural

Engineering, vol. 8, no. 6, pp. 547-560, 2005.

91

[9] G. Narayana and C. S. Krishnamoorthy, "A Simplified formulation of a plate/shell element

for geometrically nonlinear analysis of moderately large deflections with small rotations,"

Finite Elemnts in Analysis and Design, vol. 4, pp. 333-350, 1989.

[10] K. Bathe and Bolourchi, "A Geometric and Material Nonlinear Plate and Shell Element,"

Computers & Structures, vol. 11, pp. 23-48, 1980.

[11] I. Babuska and L. Li, "The Problem of Plate Modeling: Theoretical and Computational

Results," Computer Methods in Applied Mecahnics and Engineering, vol. 100, pp. 249-273,

1992.

[12] F. Wan, "Stress Boundary Conditions for Plate Bending," International Journal of Solids

and Structures, vol. 40, pp. 4107-4123, 2003.

[13] J. Reddy and R. Averill, "Advances in the Modeling of Laminated Plates," Computing

Systems in Engineering, vol. 2, no. 5/6, pp. 541-555, 1991.

[14] X. Zheng and Y. Zhou, "Exact solution of Nonlinear Circular Plate on Elastic Foundation,"

Journal of engineering mechanics, vol. 114, pp. 1303-1316, 1988.

[15] K. Yeh and X. Zheng, "An analytical formula of the exact solution to Von Kármán's

equations of a circular plate under a concentrated load ," International Journal of Non-

Linear Mechanics, vol. 24, no. 6, pp. 551-560, 1989.

[16] C. Chia, Nonlinear Analysis of Plates. New York, United States: McGraw-Hill International

Book Co, 1980.

[17] O. Thomas and S. Bilbao, "Geometrically nonlinear flexural vibrations of plates: In-plane

boundary conditions and some symmetry properties," Journal of Sound and Vibration, vol.

92

315, pp. 569-590, 2008.

[18] M. Haterbouch and R. Benamar, "Geometrically Nonlinear Free Vibrations of Simply

Supported Isotropic Thin Circular Plates," Journal of Sound and Vibration, vol. 280, pp.

903-924, 2005.

[19] M. Amabili and S. Carra, "Thermal Effects on Geometrically Nonlinear Vibrations of

Rectangular Plates with Fixed Edges," Journal of Sound and Vibration, vol. 321, pp. 936-

954, 321.

[20] Y. Yeh, M. Jang, and C. Wang, "Nonlinear Analysis of Large Deflections of Clamped

Orthotropic Rectangular Plate Using Hybrid Method," in Applied Simulation and Modelling,

Rhodes, Greece, 2006, pp. 15-20.

[21] J. Yang, L. Sun, and C. Lei, "Geometrically nonlinear analysis of Orthotropic Rectangular

Thin Plates with Three Edges Clamped and One Edge Simple Supported," China Acedemic

Journal of Engineering Mechanics, vol. 19, pp. 39-43, Jun. 2002.

[22] J. Reddy, Theory and Analysis of Elastic Plates and Shells. Boca Raton, FL, United States:

CRC Press, 2008.

[23] S. Dickinson and A. Blasio, "On the Use of Orthogonal Polynomials in the Rayleigh-Ritz

Method for the Study of the Flexural Vibration and Buckling of Isotropic and Orthotropic

Rectangular Plates," Journal of Sound and Vibration, vol. 108, no. 1, pp. 51-62, 1986.

[24] B. Singh and S. Chakraverty, "Transverse Vibrations of Completely-free Ellipitc and

Circular Plates Using Orthogonal Polynomials in the Rayleigh-Ritz Method," International

journal of mechanical sciences, vol. 33, no. 9, pp. 741-751, 1991.

93

[25] K. Liew, Vibration of Mindlin plates: programming the p-version Ritz Method. Oxford,

United Kingdom: El Sevier Science Inc, 1998.

[26] B. Buchberger, "Bruno Buchberger's PhD thesis 1965: An algorithm for finding the bsis

element of the residue class ring of a zero dimensional polynomial idel," Journal of Symbolic

Computation , vol. 41, pp. 475-511, 2006.

[27] B. Buchberger and F. Winkler, "Groebner Bases and Applications," Cambridge University

Press London Mathmatical Society Lecture Note Series, 251, 1998.

[28] B. Buchberger, "Applications of Groebner Bases in Non-Linear Computational Geometry,"

in Proceedings of the International Symposium on Trends in Computer Algebra, London,

UK, 1987, pp. 1-29.

[29] E. Roanes-Lozano and E. Roanes-Macias, "Some Applications of Groebner Bases,"

Computing in Science and Engineering, vol. 6, pp. 56-60, 2004.

[30] V. Borisevich, V. Potemkin, S. Strunkov, and H. Wood, "Global Methods for Solving

Systems of Nonlinear Algebraic Equations," Computers and Mathematics with Applications,

vol. 40, pp. 1015-1025, 2000.

[31] W. Boege, R. Gebauer, and H. Kredel, "Some examples for solving systems of algebraic

equations by calculatin Groebner bases," Journal of Symbolic Computation, vol. 2, no. 1, pp.

83-98, 1986.

[32] N. Ioakimidis, "Mathematica-based formula verification in applied mechanics," in First

National Conference on Recent Adcances in Mechanical Engineering, Patras, Greece, 2001,

pp. 1-6.

94

[33] N. Ioakimidis and E. Anastasselou, "Application of Groebner Bases to Problems of

Movement of a Particle," Computers & Mathematics with Applications, vol. 27, no. 3, pp.

51-57, 1994.

[34] N. Ioakimidis and E. Anastasselou, "Computer-based manipulation of systems of equations

in elasticity problems with Groebner bases," Computer Methods in Applied Mechanics and

Engineering, vol. 110, pp. 103-111, 1993.

[35] N. Ioakimidis and E. Anastasselou, "Groebner Bases in truss problems with Maple,"

Computer & Structures, vol. 52, pp. 1093-1096, 1994.

[36] N. Ioakimidis, "Determination of Critical Buckling Loads with Groebner Bases," Computers

& Structures, vol. 55, no. 3, pp. 433-440, 1995.

[37] Y. Liu, G. Buchanan, and J. Peddieson, "Application of Groebner Basis Methodology to

Nonlinear Static Cable Analysis," Journal of Offshore Mechanics and Arctic Engineering,

vol. 135, no. 4, Jul. 2013.

[38] R. Vandervort and Y. Liu, "Geometrically Nonlinear Analysis of Rectangular Plates with the

Groebner Bases," in Proc. 44 the Annual Technical Meeting Society of Engineering Science

2007, Texas A&M University, College Station, Texas, October 21-24, 2007.

[39] J. L. a. D. O. D Cox, Ideals, Varieties, and Algorithms: An Introduction to Computational

Algebraic Geometry and Commutative Algebra, 3rd ed. New York, NY, USA: Springer,

2007.

[40] ANSYS. ANSYS Mechanical APDL, Release 13.0.

95

[41] N. Bencharif and S. Ng, "Linear and Nonlinear Deflection Analysis of Thick Rectangular

Plates - I. Theoretical Derivation," Computers & Structures, vol. 50, no. 6, pp. 757-761,

1994.

[42] R. Liu, "Nonlinear Bending of Circular Snadwich Plates," Applied Mathematics and

Mechanics, vol. 2, no. 2, pp. 189-208, Aug. 1981.

[43] E. Efraim and M. Eisenberger, "Exact vibration analysis of variable thickness thick annular

isotropic and FGM plates," Journal of Sound and Vibration, vol. 299, pp. 720-738, 2007.

96

APPENDICES

97

APPENDIX A

DEFINITIONS OF SOME PRELIMINARY CONCEPTS

98

A-1. Field:

A field consists of a set ´ in which two operators are imposed ∙ and +. A list of the properties

is presented below:

(i) ∙ (� + o) = ∙ � + ∙ o∀, �, o ∈ ´ (distributive)

(ii) + � = � + �. ∙ � = � ∙ ∀, �, o ∈ ´ (commutative)

(iii) ( + �) + o = + (� + o)�. ∙ (� ∙ o) = ( ∙ �) ∙ o∀, �, o ∈ ´

(associative)

(iv) 0,1 ∈ ´�Joℎ�ℎ� + 0 = ∙ 1 = ∀ ∈ ´ (identities)

(v) C�!�� ∈ ´�ℎ�"�� ∈ ´�Joℎ�ℎ� + � = 0 (additive inverse)

(vi) C�!�� ∈ ´, ≠ 0, �ℎ�"��o ∈ ´�Joℎ�ℎ� ∙ o = 1 (multiplicative inverse)

A-2. Commutative Ring:

A commutative ring consists of a set � in which two operators are imposed ∙ and +. A list of

the properties is presented below:

(i) ∙ (� + o) = ∙ � + ∙ o∀, �, o ∈ � (distributive)

(ii) + � = � + �. ∙ � = � ∙ ∀, �, o ∈ � (commutative)

(iii) ( + �) + o = + (� + o)�. ∙ (� ∙ o) = ( ∙ �) ∙ o∀, �, o ∈ �

(associative)

(iv) 0,1 ∈ ��Joℎ�ℎ� + 0 = ∙ 1 = ∀ ∈ � (identities)

(v) C�!�� ∈ ��ℎ�"�� ∈ ��Joℎ�ℎ� + � = 0 (additive inverse)

99

A-3. Ideal:

An ideal, ¾ is a subset of a commutative ring � that satisfies the following properties:

(i) 0 ∈ ¾ (ii) ¾¦¦, Ç ∈ ¾�ℎ��¦ ∙ Ç ∈ ¾ (iii) ¾¦¦ ∈ ¾�.ℎ ∈ ��ℎ��ℎ¦ ∈ ¾

A-4. Monomial Ordering:

A monomial ordering on ´»�1, … , �u¼ is any relation, >, on ℤR, or any relation on the set of

monomials �¶, where / ∈ ℤR, satisfying the following:

(i) > is a total ordering on ℤR

(ii) ¾¦/ > 0 and 2 ∈ ℤR, then / + 2 > 0 + 2

(iii) Every non-empty subset of ℤRhas a smallest element under >

Using a function ¦ = �#&(�� + ��&*�� as an example with the ordering � > & > �, then

the following orders can be determined:

Lexicographic Order:

If / and 0 are n-tuples consisting of the powers of a monomial then / >�� 0 if when / − 0 the

left most nonzero entry of the vector difference is positive.

�#&(�� → / = (4,2,7) ��&*�� → 0 = (5,3,5) / − 0 = (−1,−1,2) 0 >�� / therefore ¦ = ��&*�� + �#&(��

(A.1)

100

Graded Lexicographic Order:

If / and 0 are n-tuples consisting of the powers of a monomial, then / >�3�� 0 if the sum of the

n-tuples for / is greater than the sum of the n-tuples for 0, |/| = ∑ /tut²1 > |0| = ∑ 0tut²1 . In

cases where |/| = |0|, lexicographic order is used to determine the order; / >�� 0.

�#&(�� → / = |(4,2,7)| = 13 ��&*�� → 0 = |(5,3,5)| = 13 0 >�3�� /, because |/| = |0| and / − 0 = (−1,−1,2) therefore ¦ = ��&*�� + �#&(��

(A.2)

Graded Reverse Lexicographic Order:

If / and 0 are n-tuples consisting of the powers of a monomial, then / >�3�� 0 if the sum of

the n-tuples for / is greater than the sum of the n-tuples for 0, |/| = ∑ /tut²1 > |0| = ∑ 0tut²1 .

In cases where |/| = |0|, if when / − 0 the right most nonzero entry of the vector difference is

negative.

�#&(�� → / = |(4,2,7)| = 13 ��&*z� → 0 = |(5,3,5)| = 13 / >�3�� 0, because |/| = |0| and / − 0 = (−1,−1,2) therefore ¦ = �#&(�� + �&*��

(A.3)

101

A-5. Buchberger’s Algorithm:

Let I =< f1, … , f� >≠ 0 be a polynomial ideal. The Groebner basis G for I cab be constructed in

a finite number of steps using the following algorithm:

INPUT: F = (f1, … , f�) OUTPUT: A Groebner basis G = (g1, … , g�) for I, with F ⊂ G

G ≔ F

REPEAT

G� ≔ G

FOR each pair {p, q}, p ≠ q in G′ DO

S ≔ S(p, q) !""

IF S ≠ 0 THEN G ≔ G ∪ {S} UNTIL C = C′

where S(p, q) !"" is notation for the remainder on the division of S(p, q) by the ordered t-tuple G��.

102

APPENDIX B

EXAMPLES OF AFFINE VARIETIES

103

Example 1:

The affine variety, %(¦) defined by ¦ = & − �( is shown in Figure B.1, which is a collection of

all the points that satisfy ¦ = 0 in ℝ(. For this example, the solution is a parabolic line.

Figure B.1: The Affine Variety, %(& − �(), in ℝ(

%(& − �()

& − �( = 0

�

&

104

Example 2:

The affine variety, %(¦) defined by ¦ = �( + &( + �( − 25 is shown in Figure B.2, which is a

collection of all the points that satisfys ¦ = 0 in ℝ*. And for this example, the affine variety is

the surface of a sphere with the radius of 5.

Figure B.2: The Affine Variety, %(�( + &( + �( − 25), in ℝ*

%(�( + &( + �( − 25)

�( + &( + �( − 25 = 0

�&

�

105

Example 3:

Figure B.3 shows the affine variety the affine variety, %(¦) defined by ¦ = �* − &* − �*, which

is a collection of all the points that satisfy ¦ = 0 in ℝ*.

Figure B.3: The Affine Variety, %(�* − &* − �*), in ℝ*

%(�* − &* − �*)

�* − &* − �* = 0

�&

�

106

Example 4:

Figure B.4 illustrates the affine variety, %(¦1, ¦() defined by ¦1 = & − �* and ¦( = & − �, which

is a collection of all the points that satisfy ¦1 = 0 and ¦( = 0 in ℝ(. The affine variety, %(& −�*, & − �), equals {(0,0), (−1,−1), (1,1)} which corresponds to the three intersection points of

the polynomials ¦1 and ¦(.

Figure B.4: The Affine Variety, %(& − �*, & − �) = {(0,0), (−1,−1), (1,1)}, in ℝ(

%(& − �*, & − �)

& − � = 0

& − �* = 0

�

&

107

Example 5:

In Figure B.5, the affine variety, %(¦1, ¦() defined by ¦1 = x( + y( + z( − 25 and ¦( = �, which

is a collection of all the points that sarisfy ¦1 = 0 and ¦( = 0. %(x( + y( + z( − 25, �) equals

�( + &( − 25 which is a circle in the � = 0 plane.

Figure B.5: The Affine Variety %(�( + &( + �( − 25, �) = �( + &( − 25 in the �& plane.

%(�( + &( + �( − 25, �)

x( + y( + z( − 25 = 0 � = 0

�&

�

108

APPENDIX C

COMPARISION OF EXACT LINEAR SOLUTION FROM THE LINEAR THEORY TO

THE SOLUTION FOUND FROM THE RITZ METHOD FOR THE CIRCULAR PLATE

WITH OVERHANG (CASE 3)

109

> restart:with(DETools):with(plots):with(linalg): #This command will restart the maple worksheet and start the

packages DETools, plots and linalg

Governing Equation of Linear Deflection for Axisymmetric

Bending of Circular Plate We are going to start out this Maple worksheet by solving the exact solution for Case 3

Immovable Simply Supported with Overhang

> #Showing the Governing equation symbolically D11/r*Diff(r*Diff(Diff(r*Diff(w(r),r),r)/r,r),r)=q;

The governing equation is split into two differential equations. The first with a uniform

distributed load, q and the second with no load.

> #Computing the differentials GE0:=D11/r*diff(r*diff(diff(r*diff(w(r),r),r)/r,r),r)=q:

GE1:=D11/r*diff(r*diff(diff(r*diff(w(r),r),r)/r,r),r)=0:

> def:=w(r);

Solving both differential equation

The first governing equation is solve GE0.

> #Solving the first differential. def2:=subs(dsolve(GE0),def):

w0:=def2;

Finding constants based on boundary conditions, since there are 4 unknown constants there are

four boundary conditions we can apply to figure out those constants.

> ##Boundary Condition Equation 1 subs(r=0,w0);

=

D11

d

d

r

r

d

d

r

d

d

r

r

d

d

r( )w r

r

rq

:= def ( )w r

:= w0 + + + − + q r4

64 D11

_C1 r2

2_C2 ( )ln r

1

2_C3 r2 ( )ln r

_C3 r2

4_C4

+ _C2 ( )ln 0 _C4

110

> #w(0) cannot be infinite, so C2 cannot exists due to ln(r) w0:=subs(_C2=0,w0);

> #Computing the transverse shearing force Q0[rr]:=simplify(-D11*diff((diff(w0,r,r)+(1/r)*diff(w0,r)),r)):

expand(Q0[rr]);

> #Because Q0[rr] cannot be infinite at r=0, C3 must equal zero w0:=subs(_C3=0,w0);

We have determined two constants to be zero C2 and C3.

> ##Boundary Condition Equation 2 eq1:=subs(r=a,w0)=0;

> diff(w0,r); subs(r=0,diff(w0,r))=0;

> #Last equation did not any value, so now compute the moment M0[rr]:=-D11*(diff(w0,r$2)+nu/r*diff(w0,r));

Since displacements are based on each other the second differential equation is solved to be able

to apply the boundary conditions at r=a

> #Solving second governing equation defo:=w(r):

defo2:=subs(dsolve(GE1),defo):

:= w0 + + − + q r4

64 D11

_C1 r2

2

1

2_C3 r2 ( )ln r

_C3 r2

4_C4

− − r q

2

2 _C3 D11

r

:= w0 + + q r4

64 D11

_C1 r2

2_C4

:= eq1 = + + q a4

64 D11

_C1 a2

2_C4 0

+ r3 q

16 D11_C1 r

= 0 0

:= M0rr

−D11

+ + 3 r2 q

16 D11_C1

ν

+

r3 q

16 D11_C1 r

r

111

w1:=_C5*r^2+_C6*r^2*ln(r)+_C7+_C8*ln(r);

> #Computing the transverse shearing force Q0[rr]:=simplify(-D11*diff((diff(w0,r,r)+(1/r)*diff(w0,r)),r));

Q1[rr]:=simplify(-D11*diff((diff(w1,r,r)+(1/r)*diff(w1,r)),r));

Knowing that the transverse shearing force at the free edge must be equal to zero the next

boundary condition can be applied.

> ##Boundary Condition Equation 3 simplify(subs(r=b,Q1[rr])=0);

> #Therefore _C6 must be zero w1:=subs(_C6=0, w1);

The transverse displacement at r=a must be zero, because of the support at that location

> ##Boundary Condition Equation 4 eq2:=subs(r=a,w1)=0;

The moment at r=a must equal for both moment functions

The moment at r=b must equal zero

And the slopes at r=a must be equal for both transverse displacement functions

> ##Boundary Condition Equation 5,6,7 M1[rr]:=-D11*(diff(w1,r$2)+nu/r*diff(w1,r));

eq3:=simplify(subs(r=a,M0[rr])=subs(r=a,M1[rr]));

eq4:=simplify(subs(r=b,M1[rr])=0);

eq5:=subs(r=a,diff(w0,r))=subs(r=a,diff(w1,r));

:= w1 + + + _C5 r2 _C6 r2 ( )ln r _C7 _C8 ( )ln r

:= Q0rr

−r q

2

:= Q1rr

−4 D11 _C6

r

= −4 D11 _C6

b0

:= w1 + + _C5 r2 _C7 _C8 ( )ln r

:= eq2 = + + _C5 a2 _C7 _C8 ( )ln a 0

112

We have now found all but 5 non zero constants.

> w0;w1;

Fortunately, we have 5 equations and therefore can determine those last 5 constants.

> ##Solving the constants and substituting them back into the ##transverse displacement function

w0:=subs(solve({eq1,eq2,eq3,eq4,eq5},[_C1,_C4,_C5,_C7,_C8])[1],w

0);

w1:=subs(solve({eq1,eq2,eq3,eq4,eq5},[_C1,_C4,_C5,_C7,_C8])[1],w

1);

Now we have solved the exact solution for the simply supported circular plate with an overhang

:= M1rr

−D11

− + 2 _C5_C8

r2

ν

+ 2 _C5 r

_C8

r

r

eq3 − − − − 3 a2 q

16_C1 D11

ν a2 q

16ν _C1 D11 = :=

−D11 ( ) − + + 2 _C5 a2 _C8 2 ν _C5 a2

ν _C8

a2

:= eq4 = −D11 ( ) − + + 2 _C5 b2 _C8 2 ν _C5 b2

ν _C8

b20

:= eq5 = + a3 q

16 D11_C1 a + 2 _C5 a

_C8

a

+ + q r4

64 D11

_C1 r2

2_C4

+ + _C5 r2 _C7 _C8 ( )ln r

w0q r4

64 D11

a2 q ( )− + − − a2 a2ν 2 b2 2 ν b2 r2

32 b2 ( ) + 1 ν D11 + :=

q a4 ( )− + − − 2 a2 2 a2ν 3 b2 3 ν b2

64 b2 ( ) + 1 ν D11 −

w1 :=

+ − ( )− + 1 ν q a4 r2

32 b2 ( ) + 1 ν D11

1

32

q a4 ( ) − + + a2 a2ν 2 ( )ln a b2 2 ( )ln a b2

ν

b2 ( ) + 1 ν D11

1

16

q a4 ( )ln r

D11

113

> #Plotting the results from the solution plot1:=plot(subs(a=10,b=15,D11=10000,q=10,nu=0.3,w0),r=0..10,lab

els=[`r`,`w`],legend=`Linear Theory w`):

plot2:=plot(subs(a=10,b=15,D11=10000,q=10,nu=0.3,w1),r=10..15,la

bels=[`r`,`w`],legend=`Linear Theory woh`):

display(plot1,plot2,implicitplot(y=0,x=0..15,y=-0.05..0.05,

color=black,legend=Àxis Line`),axes=boxed,title=`Transverse

Displacement for Linear Solution (Case

3)`,font=[TIMES,ROMAN,15],titlefont=[TIMES,ROMAN,15]);

> ##Plotting the moment M0[rr]:=simplify(-D11*(diff(w0,r$2)+nu/r*diff(w0,r)));

M1[rr]:=simplify(-D11*(diff(w1,r$2)+nu/r*diff(w1,r)));

plot4:=plot(subs(a=10,b=15,D11=10000,q=10,nu=0.3,M0[rr]),r=0..10

):

plot6:=plot(subs(a=10,b=15,D11=10000,q=10,nu=0.3,M1[rr]),r=10..1

5):

114


color=black),axes=boxed,title=`Radial Moment for Linear Solution

(Case 3)`,font=[TIMES,ROMAN,15],titlefont=[TIMES,ROMAN,15]);

>

Ritz Method Analysis To insure our results will give us the correct assumption in the Ritz solution it is computed using

the exact solution.

> #Symbolically showing the Total Potential Energy TPE:=Int(Pi*D11*(r*Diff(wr0(r),r,r)^2+1/r*Diff(wr0(r),r)^2+2*nu*

Diff(wr0(r),r)*Diff(wr0(r),r,r))-

:= M0rr

−( ) + − + − − r2 b2

ν a4ν 2 a2

ν b2 3 r2 b2 a4 2 a2 b2 q

16 b2

:= M1rr

( )− + + − r2ν ν b2 r2 b2 a4 q

16 r2 b2

115

2*Pi*q*wr0(r)*r,r=0..a)+Int(Pi*D11*(r*Diff(wr1(r),r,r)^2+1/r*Dif

f(wr1(r),r)^2+2*nu*Diff(wr1(r),r)*Diff(wr1(r),r,r)),r=a..b);

> ##Inputting the assumed displacement functions wr0:=r->(1/64/D11*q*r^4+1/32*a^2*q*(-a^2+a^2*nu-2*b^2-

2*nu*b^2)/b^2/(1+nu)/D11*r^2-1/64*q*a^4*(-2*a^2+2*a^2*nu-3*b^2-

3*nu*b^2)/b^2/(1+nu)/D11)*a11;

wr1:=r->(1/32/b^2/(1+nu)*(-1+nu)/D11*q*a^4*r^2+1/32*q*a^4*(a^2-

a^2*nu+2*ln(a)*b^2+2*ln(a)*b^2*nu)/b^2/(1+nu)/D11-

1/16*1/D11*q*a^4*ln(r))*a11;

> ##Computing the Total Potential Energy TPE:=int(Pi*D11*(r*diff(wr0(r),r,r)^2+1/r*diff(wr0(r),r)^2+2*nu*

diff(wr0(r),r)*diff(wr0(r),r,r))-

2*Pi*q*wr0(r)*r,r=0..a)+int(Pi*D11*(r*diff(wr1(r),r,r)^2+1/r*dif

f(wr1(r),r)^2+2*nu*diff(wr1(r),r)*diff(wr1(r),r,r)),r=a..b)

assuming a>0,b>a;

TPE

⌠

⌡

0

a

π D11

+ + r

d

d2

r2( )wr0 r

2

d

d

r( )wr0 r

2

r2 ν

d

d

r( )wr0 r

d

d2

r2( )wr0 r :=

2 π q ( )wr0 r r − rd

d

⌠

⌡

a

b

π D11

+ + r

d

d2

r2( )wr1 r

2

d

d

r( )wr1 r

2

r2 ν

d

d

r( )wr1 r

d

d2

r2( )wr1 r r +

wr0 r1

64

q r4

D11

1

32

a2 q ( )− + − − a2 a2ν 2 b2 2 ν b2 r2

b2 ( ) + 1 ν D11 +

→ :=

q a4 ( )− + − − 2 a2 2 a2ν 3 b2 3 ν b2

64 b2 ( ) + 1 ν D11 −

a11

wr1 r

→ :=

+ − 1

32

( )− + 1 ν q a4 r2

b2 ( ) + 1 ν D11

1

32

q a4 ( ) − + + a2 a2ν 2 ( )ln a b2 2 ( )ln a b2

ν

b2 ( ) + 1 ν D11

1

16

q a4 ( )ln r

D11

a11

116

> ##Setting the equation for the unknow coefficient eq1:=diff(TPE,a11)=0;

> #Solving the unknown coefficient a11:=solve(eq1,a11);

Since the unknown coefficient is 1. We know that the assumed shape function will yield

the same exact solution as is predicted from the linear theory.

> simplify(subs(r=0,w0)); simplify(wr0(0));

Showing the resulting functions from the Ritz solution and the exact solution from the

linear theory. Notice that they match exactly.

> plot7:=plot(subs(a=10,b=15,D11=10000,q=10,nu=0.3,wr0(r)), r=0..10,style=point,symbol=diamond,color=blue,numpoints=5,legend

=`Linear Ritz w`):

plot8:=plot(subs(a=10,b=15,D11=10000,q=10,nu=0.3,wr1(r)),r=10..1

5,style=point,symbol=diamond,color=blue,numpoints=5,

legend=`Linear Ritz woh`):

display(plot1,plot2,plot7,plot8,implicitplot(y=0,x=0..15,y=-

0.05..0.05, color=black,legend=Àxis

TPE π q2 a11 a2 12 a6ν b2 16 a4 b4

ν 6 a11 a8ν 6 a11 a6 b2 3 a11 a8

ν2

− − + + ( :=

5 a11 a4 b4 6 a11 a6ν

2 b2 8 a11 a4 b4ν 3 a11 a4

ν2 b4 12 a6 b2 16 a4 b4

+ − + + − −

3 a11 a8 + ) 768 D11 ( ) + 1 ν b4( )

a11 2 a6 q2π ( ) − + + − + − a4 b2 2 a4

ν b2 b2ν

2 a4 b6ν

2 b6 2 b4 a2ν 2 a2

ν2 b4

256 D11 ( ) + 1 ν b6 −

eq1 π q2 a2 12 a6ν b2 16 a4 b4

ν 6 a11 a8ν 6 a11 a6 b2 3 a11 a8

ν2

− − + + ( :=

5 a11 a4 b4 6 a11 a6ν

2 b2 8 a11 a4 b4ν 3 a11 a4

ν2 b4 12 a6 b2 16 a4 b4

+ − + + − −

3 a11 a8 + ) 768 D11 ( ) + 1 ν b4( ) π q2 a11 a2

+

( )− + + + − + + + 6 a8ν 6 a6 b2 3 a8

ν2 5 a4 b4 6 a6

ν2 b2 8 a4 b4

ν 3 a4ν

2 b4 3 a8 (

768 D11 ( ) + 1 ν b4 )

a11 a6 q2π ( ) − + + − + − a4 b2 2 a4

ν b2 b2ν

2 a4 b6ν

2 b6 2 b4 a2ν 2 a2

ν2 b4

128 D11 ( ) + 1 ν b6 − 0 =

:= a11 1

−q a4 ( )− + − − 2 a2 2 a2

ν 3 b2 3 ν b2

64 b2 ( ) + 1 ν D11

−q a4 ( )− + − − 2 a2 2 a2

ν 3 b2 3 ν b2

64 b2 ( ) + 1 ν D11

117

Line`),axes=boxed,title=`Transverse Displacement for Linear

Solution & One parameter Ritz Solution

(Case 3)`,font=[TIMES,ROMAN,15],titlefont=[TIMES,ROMAN,15]);

118

APPENDIX D

MAPLE FILE WITH CURRENT STUDY SOLUTION FOR GEOMETRICALLY

NONLINEAR CIRCULAR PLATE WITH FIXED EDGE (CASE 1)

119

> ##This restarts the Maple worksheet, plus it also opens the ##Groebner and plots packages so the associated commands can be

##used.

restart:with(Groebner):with(plots):

> ###Parameters for graphs#### param_q5:=[E=29000000,a=10,nu=0.3,q=5,t=0.1]:

param_q1:=[E=29000000,a=10,nu=0.3,q=0.25,t=0.1]:

param_q01:=[E=29000000,a=10,nu=0.3,q=0.1,t=0.1]:

FEA Solution Results

Several Solutions were run as a check and are inputted below as arrays.

These results were used for large displacement with the following parameters:

Ran Geometrically Nonlinear in ANSYS 13.0

Material Properties: E=29000000, nu=0.3

Geometry Properties: a=10,t=0.1

Loading: q=5

> ###In-plane displacement### u_r_ANSYS_q5:=[0.0,0.0],[0.1880,2.4235E-5],[0.3760,4.8342E-

5],[0.564,7.2194E-5],[0.752,9.5665E-5],[0.9441,1.1912E-

4],[1.1362,1.4192E-4],[1.3462,1.6592E-4],[1.5561,1.8879E-

4],[1.7484,2.0862E-4],[1.94070,2.2723E-4],[2.133,2.4452E-

4],[2.3253,2.6036E-4],[2.5176,2.7463E-4],[2.7099,2.8723E-

4],[2.9023,2.9806E-4],[3.0946,3.0701E-4],[3.2869,3.140E-

4],[3.4792,3.1895E-4],[3.6715,3.2179E-4],[3.8638,3.2248E-

4],[4.0561,3.2096E-4],[4.2484,3.1722E-4],[4.4407,3.1126E-

4],[4.633,3.0308E-4],[4.8253,2.9271E-4],[5.0176,2.8023E-

4],[5.2099,2.6571E-4],[5.4023,2.4925E-4],[5.5946,2.3101E-

4],[5.7869,2.1117E-4],[5.9792,1.8992E-4],[6.1715,1.6749E-

4],[6.3638,1.4417E-4],[6.5561,1.2026E-4],[6.7484,9.6085E-

5],[6.9407,7.2027E-5],[7.133,4.8477E-5],[7.3253,2.5858E-

5],[7.5176,4.6001E-6],[7.7099,-1.4843E-5],[7.90230,-3.2041E-

5],[8.09460,-4.653301E-5],[8.2869,-5.7935E-5],[8.47920,-6.5875E-

5],[8.6715,-7.0074E-5],[8.8638,-7.0316E-5],[9.0559,-6.6555E-

5],[9.248,-5.885401E-5],[9.436,-4.779400E-5],[9.624,-3.3721E-

5],[9.81200,-1.7439E-5],[10.0,0.0]:

###Out-of-plane displacement###

w_r_ANSYS_q5:=[0.0,0.13942],[0.188,0.13934],[0.376,0.13911],[0.5

64,0.13872],[0.752,0.13819],[0.9441,0.13748],[1.1362,0.13661],[1

.3462,0.13548],[1.5561,0.13416],[1.7484,0.13278],[1.9407,0.13125

],[2.133,0.12956],[2.3253,0.12772],[2.5176,0.12572],[2.7099,0.12

356],[2.9023,0.12126],[3.0946,0.11881],[3.2869,0.11621],[3.4792,

0.11346],[3.6715,0.11058],[3.8638,0.10755],[4.0561,0.1044],[4.24

84,0.10111],[4.4407,9.7698E-2],[4.633,9.4167E-

2],[4.8253,9.0522E-2],[5.0176,8.6771E-2],[5.2099,8.292E-

2],[5.4023,7.8977E-2],[5.5946,7.495401E-2],[5.7869,7.0859E-

120

2],[5.9792,6.6704E-2],[6.1715,6.2501E-2],[6.3638,5.8263E-

2],[6.5561,5.4005E-2],[6.7484,4.9743E-2],[6.9407,4.5494E-

2],[7.133,4.1276E-2],[7.3253,3.7111E-2],[7.5176,3.3019E-

2],[7.7099,2.9025E-2],[7.9023,2.5152E-2],[8.0946,2.1431E-

2],[8.2869,1.7889E-2],[8.4792,1.4558E-2],[8.6715,1.1471E-

2],[8.8638,8.6644E-3],[9.0559,6.1774E-3],[9.248,4.0479E-

3],[9.436,2.3516E-3],[9.624,1.0814E-3],[9.812,2.7993E-

4],[10.0,0.0]:

> ###Stress at the bottom of the plate### FEA_sigma_bot_q5:=[0,-3669],[0.376,-3667.6],[0.752,-

3663.7],[1.1362,-3653.4],[1.5561,-3633.2],[1.9407,-

3599],[2.3253,-3544.9],[2.7099,-3461.4],[3.0946,-

3338.3],[3.4792,-3164.5],[3.8638,-2925.3],[4.2484,-

2604.6],[4.633,-2185.2],[5.0176,-1645.3],[5.4023,-

963.96],[5.7869,-

115.88],[6.1715,925.06],[6.5561,2186.9],[6.9407,3700.3],[7.3253,

5496.1],[7.7099,7606.8],[8.0946,10063],[8.4792,12901],[8.8638,16

152],[9.248,19863],[9.624,23953],[10,28580]:

###Stress at the top of the plate###

FEA_sigma_top_q5:=[0,14363],[0.376,14352],[0.752,14319],[1.1362,

14259],[1.5561,14162],[1.9407,14037],[2.3253,13872],[2.7099,1365

8],[3.0946,13387],[3.4792,13046],[3.8638,12623],[4.2484,12102],[

4.633,11466],[5.0176,10696],[5.4023,9772],[5.7869,8670.9],[6.171

5,7368.8],[6.5561,5840.4],[6.9407,4058.4],[7.3253,1994.9],[7.709

9,-379.03],[8.0946,-3090.7],[8.4792,-6171.8],[8.8638,-

9652],[9.248,-13575],[9.624,-17855],[10,-22656]:

These results were used for analyzing the limit of small displacement with the following

parameters:




Loading: q=0.25

> ###In-plane displacement###

u_r_ANSYS_q1:=[0,0],[0.188,2.7124E-07],[0.376,5.4023E-

07],[0.564,8.0472E-07],[0.752,1.0625E-06],[0.9441,1.3168E-

06],[1.1362,1.5596E-06],[1.3462,1.8095E-06],[1.5561,2.0405E-

06],[1.7484,2.2336E-06],[1.9407,2.4071E-06],[2.133,2.5598E-

06],[2.3253,2.6901E-06],[2.5176,2.7969E-06],[2.7099,2.8795E-

06],[2.9023,2.9370E-06],[3.0946,2.9691E-06],[3.2869,2.9755E-

06],[3.4792,2.9563E-06],[3.6715,2.9118E-06],[3.8638,2.8426E-

06],[4.0561,2.7496E-06],[4.2484,2.6338E-06],[4.4407,2.4967E-

06],[4.633,2.3397E-06],[4.8253,2.1649E-06],[5.0176,1.9743E-

06],[5.2099,1.7702E-06],[5.4023,1.5550E-06],[5.5946,1.3317E-

06],[5.7869,1.1031E-06],[5.9792,8.7213E-07],[6.1715,6.4193E-

07],[6.3638,4.1571E-07],[6.5561,1.9669E-07],[6.7484,-1.1885E-

08],[6.9407,-2.0683E-07],[7.133,-3.8507E-07],[7.3253,-5.4364E-

121

07],[7.5176,-6.7985E-07],[7.7099,-7.9117E-07],[7.9023,-8.7557E-

07],[8.0946,-9.3116E-07],[8.2869,-9.5683E-07],[8.4792,-9.5180E-

07],[8.6715,-9.1613E-07],[8.8638,-8.5040E-07],[9.0559,-7.5639E-

07],[9.248,-6.3637E-07],[9.436,-4.9721E-07],[9.624,-3.4082E-

07],[9.812,-1.7289E-07],[10,0]:


w_r_ANSYS_q1:=[0,0.014547],[0.188,0.014537],[0.376,0.014507],[0.

564,0.014455],[0.752,0.014384],[0.9441,0.01429],[1.1362,0.014176

],[1.3462,0.014027],[1.5561,0.013854],[1.7484,0.013675],[1.9407,

0.013476],[2.133,0.013258],[2.3253,0.013022],[2.5176,0.012768],[

2.7099,0.012496],[2.9023,0.012208],[3.0946,0.011903],[3.2869,0.0

11583],[3.4792,0.011249],[3.6715,0.010901],[3.8638,0.01054],[4.0

561,0.010167],[4.2484,0.0097829],[4.4407,0.0093892],[4.633,0.008

9865],[4.8253,0.0085762],[5.0176,0.0081593],[5.2099,0.007737],[5

.4023,0.0073105],[5.5946,0.0068814],[5.7869,0.0064509],[5.9792,0

.0060204],[6.1715,0.0055913],[6.3638,0.0051653],[6.5561,0.004743

7],[6.7484,0.0043283],[6.9407,0.0039207],[7.133,0.0035226],[7.32

53,0.0031357],[7.5176,0.0027619],[7.7099,0.002403],[7.9023,0.002

0607],[8.0946,0.0017374],[8.2869,0.0014348],[8.4792,0.001155],[8

.6715,0.00090016],[8.8638,0.00067236],[9.0559,0.0047398],[9.248,

0.00030703],[9.436,0.00017633],[9.624,0.000080113],[9.812,0.0000

020453],[10,0]:

> ###Stress at the bottom of the plate### FEA_sigma_top_q1:=[0,1892.4],[0.376,1261.5],[0.752,1257.2],[1.13

62,1244.2],[1.5561,1222],[1.9407,1187.3],[2.3253,1146],[2.7099,1

095.6],[3.0946,1036.1],[3.4792,967.34],[3.8638,889.56],[4.2484,8

02.57],[4.633,706.35],[5.0176,601.07],[5.4023,486.43],[5.7869,36

2.63],[6.1715,229.61],[6.5561,87.371],[6.9407,-64.1],[7.3253,-

224.88],[7.7099,-394.92],[8.0946,-574.28],[8.4792,-

762.75],[8.8638,-960.49],[9.248,-1167.3],[9.624,-1383.8],[10,-

1603.5]:


FEA_sigma_bot_q1:=[0,-1834],[0.376,-1141.7],[0.752,-

1137.6],[1.1362,-1125.1],[1.5561,-1103.8],[1.9407,-

1070.4],[2.3253,-1030.6],[2.7099,-982.15],[3.0946,-

924.77],[3.4792,-858.48],[3.8638,-783.35],[4.2484,-

699.23],[4.633,-606.03],[5.0176,-503.91],[5.4023,-

392.52],[5.7869,-272.03],[6.1715,-142.35],[6.5561,-

3.4267],[6.9407,144.78],[7.3253,302.4],[7.7099,469.39],[8.0946,6

45.85],[8.4792,831.6],[8.8638,1026.8],[9.248,1231.4],[9.624,1445

.8],[10,1663.6]:

122

These results were used for analyzing the limit of small displacement with the following

parameters:




Loading: q=0.1

> ###In-plane displacement### u_r_ANSYS_q01:=[0,0],[0.188,0.000000044265],[0.376,0.00000008815

9],[0.564,0.00000013132],[0.752,0.00000017337],[0.9441,0.0000002

1485],[1.1362,0.00000025445],[1.3462,0.00000029518],[1.5561,0.00

000033282],[1.7484,0.00000036425],[1.9407,0.0000003925],[2.133,0

.00000041731],[2.3253,0.00000043847],[2.5176,0.00000045579],[2.7

099,0.00000046913],[2.9023,0.00000047838],[3.0946,0.00000048346]

,[3.2869,0.00000048435],[3.4792,0.00000048106],[3.6715,0.0000004

7365],[3.8638,0.00000046221],[4.0561,0.00000044688],[4.2484,0.00

000042785],[4.4407,0.00000040535],[4.633,0.00000037963],[4.8253,

0.00000035101],[5.0176,0.00000031983],[5.2099,0.00000028647],[5.

4023,0.00000025134],[5.5946,0.0000002149],[5.7869,0.00000017762]

,[5.9792,0.00000013998],[6.1715,0.00000010249],[6.3638,0.0000000

65682],[6.5561,0.00000003007],[6.7484,-

0.0000000038189],[6.9407,-0.000000035464],[7.133,-

0.00000006437],[7.3253,-0.000000090055],[7.5176,-

0.00000011208],[7.7099,-0.00000013005],[7.9023,-

0.00000014363],[8.0946,-0.00000015252],[8.2869,-

0.00000015654],[8.4792,-0.00000015557],[8.6715,-

0.00000014962],[8.8638,-0.0000001388],[9.0559,-

0.00000012339],[9.248,-0.00000010376],[9.436,-

0.000000081043],[9.624,-0.000000055537],[9.812,-

0.000000028167],[10,0]:


w_r_ANSYS_q01:=[0,0.0058748],[0.188,0.0058707],[0.376,0.0058582]

,[0.564,0.0058375],[0.752,0.0058086],[0.9441,0.0057707],[1.1362,

0.0057242],[1.3462,0.005664],[1.5561,0.005594],[1.7484,0.0055214

],[1.9407,0.005441],[2.133,0.0053528],[2.3253,0.0052572],[2.5176

,0.0051542],[2.7099,0.0050443],[2.9023,0.0049275],[3.0946,0.0048

043],[3.2869,0.0046748],[3.4792,0.0045395],[3.6715,0.0043987],[3

.8638,0.0042527],[4.0561,0.0041019],[4.2484,0.0039467],[4.4407,0

.0037875],[4.633,0.0036248],[4.8253,0.003459],[5.0176,0.0032905]

,[5.2099,0.0031199],[5.4023,0.0029476],[5.5946,0.0027743],[5.786

9,0.0026005],[5.9792,0.0024267],[6.1715,0.0022535],[6.3638,0.002

0815],[6.5561,0.0019114],[6.7484,0.0017439],[6.9407,0.0015794],[

7.133,0.0014189],[7.3253,0.0012629],[7.5176,0.0011122],[7.7099,0

.00096758],[7.9023,0.00082965],[8.0946,0.00069939],[8.2869,0.000

57751],[8.4792,0.00046485],[8.6715,0.00036223],[8.8638,0.0002705

3],[9.0559,0.00019068],[9.248,0.0001235],[9.436,0.000070919],[9.

624,0.000032217],[9.812,0.0000082239],[10,0]:

123

> ###Stress at the bottom of the plate### FEA_sigma_top_q01:=[0,496.51],[0.376,494.74],[0.752,489.51],[1.1

362,480.51],[1.5561,466.46],[1.9407,449.75],[2.3253,429.4],[2.70

99,405.35],[3.0946,377.61],[3.4792,346.24],[3.8638,311.18],[4.24

84,272.44],[4.633,230.07],[5.0176,183.98],[5.4023,134.24],[5.786

9,80.841],[6.1715,23.78],[6.5561,-36.943],[6.9407,-

101.36],[7.3253,-169.44],[7.7099,-241.22],[8.0946,-

316.6],[8.4792,-395.64],[8.8638,-478.28],[9.248,-

564.75],[9.624,-652.4],[10,-744.35]:


FEA_sigma_bot_q01:=[0,-476.96],[0.376,-475.22],[0.752,-

470.07],[1.1362,-461.21],[1.5561,-447.39],[1.9407,-

430.93],[2.3253,-410.89],[2.7099,-387.19],[3.0946,-

359.86],[3.4792,-328.92],[3.8638,-294.33],[4.2484,-

256.08],[4.633,-214.24],[5.0176,-168.68],[5.4023,-

119.48],[5.7869,-66.623],[6.1715,-

10.103],[6.5561,50.088],[6.9407,113.99],[7.3253,181.57],[7.7099,

252.88],[8.0946,327.81],[8.4792,406.45],[8.8638,488.71],[9.248,5

74.83],[9.624,662.19],[10,753.87]:

End of Finite Element Results

Exact Solution from Linear Theory

Inputting the exact solution found from the linear theory as a comparison to the thesis solution.

> w_linear:=r->1/64/D11*q*r^4-1/32*a^2*q/D11*r^2+1/64/D11*q*a^4;

1-parameter solution (Current Study)

Assigning polynomials to the out of plane and in plane displacement functions:

> ##Step 1. Input Assumed Displacement Functions:

w:=r->(1-r^2/a^2)^2*(A0);

u:=r->(r/a)*(1-(r/a)^2)*(B0+B1*(r/a)^2);

:= w_linear → r − + 1

64

q r4

D11

1

32

a2 q r2

D11

q a4

64 D11

:= w → r

− 1

r2

a2

2

A0

:= u → r

r

− 1

r2

a2

+ B0

B1 r2

a2

a

124

Integrating the Total Potential Energy Equation:

> ##Step 2. Input Total Potential Energy Equation:

TPE1:=int(Pi*D11*(r*diff(w(r),r,r)^2+1/r*diff(w(r),r)^2+2*nu*dif

f(w(r),r)*diff(w(r),r,r))+Pi*E*t/(1-

nu^2)*((diff(u(r),r)+1/2*(diff(w(r),r)^2))^2*r+u(r)^2/r+2*nu*u(r

)*(diff(u(r),r)+1/2*diff(w(r),r)^2))

-2*Pi*q*w(r)*r,r=0..a):

D11:=E*t^3/12/(1-nu^2):

Taking the derivative of each coefficient.

This will result in a set of nonlinear equations that can be solved by using Groebner Basis.

> ##Step 3. Make Equations for each unknown taking the

##derivative with respect to that unknown

eqA0:=diff(TPE1,A0):

eqB0:=diff(TPE1,B0):

eqB1:=diff(TPE1,B1):

Developing a set to run a Groebner Basis on.

> ##Step 4. Make a set of the unknown equations

WL:=[eqA0,eqB0,eqB1]:

Running the Basis function in Maple will run a Groebner Basis on the WL Set

> ##Step 5. Run the Groebner Basis Algorithm

GB:=Basis(WL,lexdeg([B1,B0],[A0])):

The map function will show what variables are in each of the equations found in first the "WL"

set and second the "GB" set

> ##Step 5a. (Optional) Check original set and Groebner Basis set

##to insure the system of equations are decoupled

map(indets,WL);

map(indets,GB);

Notice that in the "GB" set equation 1 b111 and b121 are removed and only a111 remains.

Notice that in the "GB" set equation 2 b121 is removed and a111, b111 remain.

Notice that in the "GB" set equation 3 b111 is removed and a111, b121 remain.

[ ], ,{ }, , , , , , ,A0 B0 B1 E a ν q t { }, , , , , ,A0 B0 B1 E a ν t { }, , , , , ,A0 B0 B1 E a ν t

[ ], ,{ }, , , , ,A0 E a ν q t { }, , ,A0 B0 a ν { }, , ,A0 B1 a ν

125

Now that the equations are decoupled the a111 can be found in the first equation and substituted

into the next equations to find b111 and b121.

> ##Step 6. Solve the system of equations

A00:=solve(GB[1],A0):

A00:=simplify(remove(has,[A00],I)):

A0:=A00[1]:

B00:=solve(GB[2],B0):

B00:=simplify(remove(has,[B00],I)):

B0:=B00[1]:

B10:=solve(GB[3],B1):

B10:=simplify(remove(has,[B10],I)):

B1:=B10[1]:

> ############################################################### ###############################################################

###### ######

###### SOLUTION IS SOLVED ######

###### NOW PLOTTING RESULTS ######

###### ######

###############################################################

###############################################################

>

>

Results

The thesis solution is compared to the finite element model and the exact solution

> ##Inputting Radial Stress Functions which are used in the graphs below for stress

##The first 3 equations are for the thesis solution

epsilon[rr]:=-

z*(diff(w(r),r,r))+diff(u(r),r)+1/2*(diff(w(r),r)^2):

epsilon[tt]:=-z*diff(w(r),r)/r+u(r)/r:

sigma[rr]:=E/(1-nu^2)*(epsilon[rr]+nu*epsilon[tt]):

> ##The next 3 equations are for the exact solution epsilon_linear[rr]:=-(z)*(diff(w_linear(r),r,r)):

epsilon_linear[tt]:=-(z)*1/r*diff(w_linear(r),r):

sigma_linear[rr]:=E/(1-

nu^2)*(epsilon_linear[rr]+nu*epsilon_linear[tt]):

>

SMALL DISPLACEMENTS (q=0.1, wmax_exact/t = .06)

We are going to start off by checking the solution for a small displacement exact where the exact

solution from the linear theory at the center of the plate (wmax_exact) over the thickness is much

greater than 0.5

126

> ##Plotting the transverse displacement

plot1:=plot([subs(param_q01,w_linear(r))],r=0..10,linestyle=dash

dot,color=grey,legend=["Linear Theory"],labels=[`r`,`w`]):

plot2:=plot([subs(param_q01,w(r))],r=0..10,legend=["Current

Study"],labels=[`r`,`w`]):

plot3:=plot([w_r_ANSYS_q01],style=point,symbolsize=15,symbol=dia

mond,color=green,legend=["ANSYS"],labels=[`r`,`w`]):

> display([plot1,plot2,plot3],axes=boxed,title=`Transverse Displacement for Linear Solution, Current Study & ANSYS

w_linear/t=.06 (Small Displacement Example) (Case


> ##Plotting the in-plane displacement

##Only finite element and thesis solution plotted because

##in-plane displacement for exact solution at the middle plane

##is zero

plot5:=plot([subs(param_q01,u(r))],r=0..10,legend=["Current

Study"]):

plot6:=plot([u_r_ANSYS_q01],style=point,symbolsize=15,symbol=dia

mond,color=green,legend=["ANSYS"]):

127

> display(plot5,plot6,implicitplot(y=0,x=0..10,y=-0.05..0.05, color=black,legend=Àxis Line`),axes=boxed,title=Ìn-Plane

Displacement for Linear Solution, Current Study and ANSYS



> ##Plotting Top, Bottom & Middle Plane Radial Stress

sig_plot_nl_top01:=plot(subs(param_q01,subs(z=t/2,sigma[rr])),r=

0..10, legend=`Current Study z=Top`):

sig_plot_nl_bot01:=plot(subs(param_q01,subs(z=-

t/2,sigma[rr])),r=0..10,legend=`Current Study z=Bottom`):

sig_plot_nl_mid01:=plot(subs(param_q01,subs(z=0,sigma[rr])),r=0.

.10,legend=`Current Study z=Midplane`):

sig_plot_FEA_top01:=plot([FEA_sigma_top_q01],style=point,symbols

ize=15,symbol=diamond,color=green,legend=["ANSYS z=Top"]):

sig_plot_FEA_bot01:=plot([FEA_sigma_bot_q01],style=point,symbols

ize=15,symbol=diamond,color=green,legend=["ANSYS z=Bottom"]):

sig_plot_l_top01:=plot(subs(param_q01,subs(z=t/2,sigma_linear[rr

])),r=0..10,color=grey,linestyle=dashdot,legend=`Linear Theory

z=Top`):

128

sig_plot_l_bot01:=plot(subs(param_q01,subs(z=-

t/2,sigma_linear[rr])),r=0..10,color=grey,linestyle=dashdot,lege

nd=`Linear Theory z=Bottom`):

display(sig_plot_nl_top01,sig_plot_nl_bot01,sig_plot_nl_mid01,si

g_plot_l_top01,sig_plot_l_bot01,sig_plot_FEA_top01,sig_plot_FEA_

bot01,implicitplot(y=0,x=0..10,y=-0.05..0.05,

color=black,legend=Àxis Line`),axes=boxed,title=Ìn-Plane




The results for the small displacement show that the results for displacement and stress match

closely.

129

LIMIT OF SMALL DISPLACEMENT (q=0.25, wmax_exact/t

= .5)

Checking the thesis results compared to finite element model and exact solution at the threshold

between small and large displacement

> plot1:=plot([subs(param_q1,w_linear(r))],r=0..10,color=grey, linestyle=dashdot,legend=["Linear Theory"]):


Study"]):

plot3:=plot([w_r_ANSYS_q1],style=point,symbolsize=15,symbol=diam

ond,color=green,legend=["ANSYS"]):

> display([plot1,plot2,plot3],axes=boxed,title=`Transverse


w_linear/t=.5 (Limit of Small Displacement) (Case


> ##Plotting the in-plane displacement ##Only finite element and thesis solution plotted because

##in-plane displacement for exact solution at the middle plane

##is zero


Study"]):

130

plot6:=plot([u_r_ANSYS_q1],style=point,symbolsize=15,symbol=diam



color=black,legend=Àxis Line`),axes=boxed,title=Ìn-plane





sig_plot_nl_top1:=plot(subs(param_q1,subs(z=t/2,sigma[rr])),r=0.

.10,legend=`Current Study z=Top`):



sig_plot_nl_mid1:=plot(subs(param_q1,subs(z=0,sigma[rr])),r=0..1

0,legend=`Current Study z=Midplane`):

sig_plot_FEA_top1:=plot([FEA_sigma_top_q1],style=point,symbolsiz

e=15,symbol=diamond,color=green,legend=["ANSYS z=Top"]):

sig_plot_FEA_bot1:=plot([FEA_sigma_bot_q1],style=point,symbolsiz

e=15,symbol=diamond,color=green,legend=["ANSYS z=Bottom"]):

sig_plot_l_top1:=plot(subs(param_q1,subs(z=t/2,sigma_linear[rr])

),r=0..10,color=grey,linestyle=dashdot,symbol=diamond,legend=`Li

near Theory z=Top`):

131


t/2,sigma_linear[rr])),r=0..10,color=grey,linestyle=dashdot,symb

ol=diamond,legend=`Linear Theory z=Bottom`):

display(sig_plot_nl_top1,sig_plot_nl_bot1,sig_plot_nl_mid1,sig_p

lot_l_top1,sig_plot_l_bot1,sig_plot_FEA_top1,sig_plot_FEA_bot1,i

mplicitplot(y=0,x=0..10,y=-0.05..0.05, color=black,legend=Àxis

Line`),axes=boxed,title=`Radial Stress for Linear Solution,

Current Study and ANSYS



The results at the limit of small displacement show that the results for displacement and stress

match closely.

LARGE DISPLACEMENTS q=5, wmax_exact/t = 3

Checking the thesis results compared to finite element model and exact solution at a large

displacement where wmax_exact/t >> 0.5

> plot1:=plot([subs(param_q5,w_linear(r))],r=0..10,color=grey, linestyle=dashdot,legend=["Linear Theory"]):

132


Study"]):

plot3:=plot([w_r_ANSYS_q5],style=point,symbolsize=15,symbol=diam


> display([plot1,plot2,plot3],axes=boxed,title=`Transverse


w_linear/t=3 (Large Displacement) (Case


> ##Plotting the in-plane displacement ##Only finite element and thesis solution plotted because in-

plane displacement for exact solution at the middle plane is

zero


Study"]):

plot6:=plot([u_r_ANSYS_q5],style=point,symbolsize=15,symbol=diam



color=black,legend=Àxis Line`),axes=boxed,title=Ìn-plane


133




sig_plot_nl_top1:=plot(subs(param_q5,subs(z=t/2,sigma[rr])),r=0.

.10,legend=`Current Study z=Top`):



sig_plot_nl_mid1:=plot(subs(param_q5,subs(z=0,sigma[rr])),r=0..1

0,legend=`Current Study z=Midplane`):

sig_plot_FEA_top1:=plot([FEA_sigma_top_q5],style=point,symbolsiz

e=15,symbol=diamond,color=green,legend=["ANSYS z=Top"]):

sig_plot_FEA_bot1:=plot([FEA_sigma_bot_q5],style=point,symbolsiz

e=15,symbol=diamond,color=green,legend=["ANSYS z=Bottom"]):

sig_plot_l_top1:=plot(subs(param_q5,subs(z=t/2,sigma_linear[rr])

),r=0..10,color=grey,linestyle=dashdot,symbol=diamond,legend=`Li

near Theory z=Top`):


t/2,sigma_linear[rr])),r=0..10,color=grey,linestyle=dashdot,symb

ol=diamond,legend=`Linear Theory z=Bottom`):

134

display(sig_plot_nl_top1,sig_plot_nl_bot1,sig_plot_nl_mid1,sig_p

lot_l_top1,sig_plot_l_bot1,sig_plot_FEA_top1,sig_plot_FEA_bot1,i

mplicitplot(y=0,x=0..10,y=-0.05..0.05, color=black,legend=Àxis

Line`),axes=boxed,title=`Radial Stress for Linear Solution,

Current Study and ANSYS



The results for large displacement show that the results for displacement and stress do not match

closely.

135

APPENDIX E

ANSYS LOG FILE CASE 3 IMMOVABLE SUPPORTED PLATE WITH OVERHANG

136

/BATCH

/FILNAME,loadstudy--case3-dist1,0 !Create a filename

!

!Change 'D:\Office Thesis--CPT\...\loadstudy--case3-!dist1' to a folder location on your

!computer. This will be the location you find the finite element model

!and all resulting solutions

!

/CWD,'D:\Office Thesis--CPT\Case 3--Overhang\Distributed Load\ANSYS\Results\loadstudy--

case3-dist1' !Sets current working directory

/PREP7 !Start Preprocessor

ET,1,SHELL93 !Selecting the element

R,1,0.1,0.1,0.1,0.1, , ,

KEYW,PR_STRUC,1 !Setting Structural Analysis

!!!!!!!!!Material Data!!!!!!!!!!!

MPTEMP,,,,,,,,

MPTEMP,1,0

MPDATA,EX,1,,29000000 !Elastic Modulus (using Steel)

MPDATA,PRXY,1,,0.3 !Poisson's Ratio

MPTEMP,,,,,,,,

MPTEMP,1,0

MPDE,EX,1

MPDE,PRXY,1

MPDATA,EX,1,,29000000

MPDATA,PRXY,1,,0.3

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!Creating Quarter Circle Geometry!!!!!!

CYL4,0,0,10,0,10,90

CYL4,0,0,10,0,15,90 !These two lines create the quarter circle segment from 0 to 90

FLST,2,2,5,ORDE,2

FITEM,2,1

FITEM,2,-2

AGLUE,P51X

FLST,2,6,4,ORDE,4

FITEM,2,1

FITEM,2,-4

FITEM,2,8

FITEM,2,-9 !These lines are selections of lines created

LGLUE,P51X

MSHAPE,0,2D

MSHKEY,0

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!Meshing the geometry!!!!!!

FLST,5,2,5,ORDE,2

FITEM,5,1

FITEM,5,3

CM,_Y,AREA

137

ASEL, , , ,P51X

CM,_Y1,AREA

CHKMSH,'AREA'

CMSEL,S,_Y

SMRT,2 !Setting Automesh 10 is most coarse mesh

! 1 is most fine mesh

AMESH,ALL !Meshing everything in the model,

!the only object in the model

!is the circular plate

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

EPLOT !Plotting the mesh

/UI,COPY,SAVE,PNG,GRAPH,COLOR,REVERSE,PORTRAIT,YES !Saving the plot to a

picture file (PNG)

FINISH

!!!!!!!!!Boundary Conditions!!!!!!!!!!!!

/SOL

FLST,2,4,4,ORDE,4 !Selecting lines to have boundary conditions applied

FITEM,2,2

FITEM,2,-3

FITEM,2,8

FITEM,2,-9

DL,P51X, ,SYMM !Applying symmetric boundary conditions

FLST,2,1,4,ORDE,1

FITEM,2,1

/GO

DL,P51X, ,UX,0.0 !Applying the immovable boundary constraints

FLST,2,1,4,ORDE,1

FITEM,2,1

!*

/GO

DL,P51X, ,UY,0 !Applying the immovable boundary constraints

FLST,2,1,4,ORDE,1

FITEM,2,1

!*

/GO

DL,P51X, ,UZ,0 !Applying the out of plane boundary constraints

FLST,2,1,5,ORDE,1

FITEM,2,1

/GO

SFA,P51X,1,PRES, 0.100000 !Applying Pressure Load

!!!!!!!!!!!!!!Analysis Type!!!!!!!!!!!!!!

NLGEOM,1 !NLGEOM,1 indicates to run for large displacements

NROPT,AUTO, ,

STAOPT,DEFA

138

LUMPM,0

EQSLV, , ,0, ,DELE

PRECISION,0

MSAVE,0

PCGOPT,0, ,AUTO, , ,AUTO

PIVCHECK,1

SSTIF

PSTRES

TOFFST,0,

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

SOLVE !Running the ANSYS solver

!!!!!!!!!!!!!!!!!!!!!!Post processing!!!!!!!!!!!!!!!!!!!!

FLST,5,2,4,ORDE,2

FITEM,5,3

FITEM,5,8

LSEL,R, , ,P51X

CM,CM_1,LINE

CMSEL,A,CM_1

CMSEL,S,CM_1

NSLL,S,1

CM,CM_2,NODE

/POST1

/EFACET,1

PLNSOL, U,X, 0,1.0 !Contour plot of in-plane displacement

/UI,COPY,SAVE,PNG,GRAPH,COLOR,REVERSE,PORTRAIT,YES

/EFACET,1

PLNSOL, U,Z, 0,1.0 !Contour plot of out-of-plane displacement


/VIEW,1,,1

/ANG,1

/REP,FAST


PLESOL, S,X, 0,1.0 !Contour plot of stress

/VIEW,1,,,1

/ANG,1

/REP,FAST


/output,ELE_X,txt

elist,all,,,0,0 !Outputting elements to text file "ELE_X.txt"

/out

CMSEL,A,CM_2

/output,U_X,txt !Outputting in-plane displacements to text file "U_X.txt"

PRNSOL,U,X

/out

139

/output,U_Z,txt !Outputting out-of-plane displacements to text file "U_Z.txt"

PRNSOL,U,Z

/out

/output,U_Z,txt

PRNSOL,U,Z

/out

/output,Stress,txt

PRNSOL,S,COMP !Outputting stress results to text file "Stress.txt"

/out

/output,NODES,txt

NLIST,ALL, , , ,NODE,NODE,NODE

/out !Outputting node geometry values to text file "Nodes.txt"

SAVE,loadstudy--case3-dist1,db

140

APPENDIX F

RITZ METHOD COEFFICIENTS EQUATIONS AND GROEBNER BASIS EQUATIONS

FROM CASE 2 AND CASE 3

141

Case 2 Ritz approximation equations:

��S0 = − ¨3465(( − 1)( + 1)� (−4��S¬(−1540�(( * + 10 ( + 24 + 22)+ 20317�¬ + 8034�1+ 1155� #( � + 10 # + 23 * + 12 ( − 17 − 22)+ 4��S¬(10780�( + 1056S¬((494 + 939)− 11 �¬(3243 + 622 ) + 2 *�1(71 ( + 887 + 3498)+ 11 *�¬(41 ( + 405 + 1138) + 14!�1(86 − 871)− 8085�#) ��0 = − ¨��630( # + 2 * − 2 − 1) p4 (S¬((41 + 323) + 126�1(1 + )(+ 315�¬(1 + )( + 4S¬((451 − 1847)Ö ��1 = − ¨��13860( # + 2 * − 2 − 1) p16 (S¬((71 + 745) + 2145�1(1 + )(+ 2772�¬(1 + )( + 208S¬((149 − 309)Ö

(F.1)

Case 2 Computed Groebner Basis equations:

C�»1¼ = −12665155�#( + 7)( − 1)( + 1)#− 670214160��*S¬( + 7)( + 1)* + 16��S¬*(6696449 �+ 99521302 � + 478692103 # + 67588516 * − 6799490129 (− 20445519770 − 14008216855) C�»2¼ = 108801�¬(1 + )( + S¬((69548 * + 423140 ( − 11804 − 2582996) C�»3¼ = 5181�1(1 + )( − S¬((−1536 * + 2752 ( + 75584 + 3712)

(F.2)

142

Case 3 Ritz approximation equations:

��S0 = ¨��((672��(�� *�1S¬�# + 1008��(�� *�¬S¬�#+ 1472�(��S¬�1 *� − 47251#S¬*�(�( �− 236251#S¬*�(�( # + 236251#S¬*�(�( ( + 47251#S¬*�(�(+ 2673�(S¬*� � − 2673�(S¬*� # − 2673�(S¬*� (+ 2673�(S¬*� − 4480��(� − 3360(��(�+ 2392�(��S¬�¬ *� − 7296*�(��S¬�¬��− 4464*�(��S¬�1�� − 4456�(��S¬�¬� − 2432�(��S¬�1�+ 512�(��S¬�1� ( − 2016��(��1S¬�#− 3360��(��1S¬�# ( − 3024��(��¬S¬�#− 5040��(��¬S¬�# ( − 1968*�(��S¬�1 *��− 3072*�(��S¬�1 *�� − 238141¬�(S¬*�� (+ 238141¬�(S¬*�� # − 79381¬�(S¬*�� + 90721(S¬*�(�#− 90721(S¬*�(�# ( − 90721(S¬*�(�# # + 90721(S¬*�(�# �+ 141751��(S¬* # − 189001��(S¬* * + 141751��(S¬* (− 56701��(S¬* + 189001#S¬*�( ��( + 5346�(S¬* ��+ 9451��(S¬* � − 56701��(S¬* � + 9451��(S¬*− 189001#S¬*�( �( − 10692�(S¬* *� + 5346�(S¬* �− 18144 �1(S¬*�(�# + 36288 *1(S¬*�(�# − 18144 1(S¬*�(�#+ 79381¬�(S¬*�� + 3360(S¬ ��(� − 3360(S¬ *��(�− 3360(S¬ (��(� + 2800S¬��(� + 3360(S¬��(�+ 1680#S¬�#�(� − 8960 ��(� − 4480 (��(�+ 3360( (��(� − 1680#S¬ (�#�(� + 1680#S¬ *�#�(�+ 1680S¬ *��(� − 1680#S¬ �#�(� + 7280S¬ ��(�+ 6160S¬�J(��(� + 4464*�(�� (S¬�1��+ 7296*�(�� (S¬�¬�� − 6520�(��S¬�¬ �+ 328�(��S¬�¬ (� + 1968*�(��S¬�1 ��+ 3072*�(��S¬�¬ �� + 4704��(�� 1S¬�#+ 7056��(�� ¬S¬�# − 3392�(��S¬�1 �)( − 1)/(17920( + 1)��11�*) ��0 = − ¨4480 (−384% ��(S¬(�( + 299� ��(S¬(�# + 12611 ��(S¬(− 88211�(S¬( # − 557��(S¬(�# # + 1680%�(S¬(�( #− 1824%�(S¬( *�( − 598��(S¬^2 *�# + 226811 *�(S¬(− 277211�(S¬( ( + 1114��(S¬(�# ( − 768%�(S¬(�( (+ 2208%�(S¬( �( + 299��(S¬( �# + 163811 �(S¬(− 557��(S¬(�# + 1344�(��1�# − 37811�(S¬(+ 2240�(��¬�# − 912%�(S¬(�()/(��#( ( − 1)) ��1 = −¨/6720 ∗ (−369% ��(S¬(�( + 276� ��(S¬(�# + 12611 ��(S¬(− 88211�(S¬( # − 456��(S¬(�# # + 1575%�(S¬(�( #− 1674%�(S¬( *�( − 552��(S¬( *�# + 226811 *�(S¬(− 277211�(S¬( ( + 912��(S¬(�# ( − 738%�(S¬(�( (+ 2043%�(S¬( �( + 276��(S¬( �# + 163811 �(S¬(− 456��(S¬(�# + 1568�(��1�# − 37811�(S¬(+ 2016�(��¬�# − 837%�(S¬(�()/(��#( ( − 1))

(F.3)

143

Case 3 Computed Groebner Basis equations:

C�»1¼ = 16405201#S¬*�(�( � + 50274001#S¬*�(�( # − 56624401#S¬*�(�( (− 10054801#S¬*�(�( − 385411�(S¬*� �+ 83397�(S¬*� # + 989439�(S¬*� ( − 687425�(S¬*�+ 2007040��(� + 1505280(��(� + 57948841¬�(S¬*�� (− 57683881¬�(S¬*�� # + 19139641¬�(S¬*�� − 20808001(S¬*�(�# + 13232161(S¬*�(�# (+ 35493121(S¬*�(�# # − 27917281(S¬*�(�# �− 33075001��(S¬* # + 41013001��(S¬* *− 29370601��(S¬* ( + 11377801��(S¬* − 48157201#S¬*�( ��( − 1587601#A¬*�(ν��(+ 11351881¬�(S¬* �� − 3695641¬�(S¬*ν��− 1502913�(A¬*ν�� + 143359�(S¬* ��− 3439801��(S¬*ν� + 15082201��(S¬*ν� + 264601��(A¬*ν�− 1852201��(S¬* + 7938001#S¬*�(ν*�( + 41806801#A¬*�(ν�(− 11616841¬�(S¬*ν*�� + 3960601¬�(S¬*ν��+ 2575749�(A¬*ν*� − 1216195�(S¬*ν�+ 361296ν�1(A¬*�(�# + 3765312ν�1(A¬*�(�#− 8661168ν*1(A¬*�(�# + 4534560ν1(A¬*�(�#− 19404601¬�(S¬*�� + 1505280(S¬ν(��(�− 1254400A¬��(� − 1505280(S¬��(�− 752640#S¬�#�(� + 2007040ν��(� − 1505280( ��(�− 752640#S¬ν(�#�(� + 1505280#A¬ν�#�(�− 2007040A¬ν��(� − 752640A¬ν(��(�

C�»2¼ = 12611 ��(S¬( − 88211�(S¬( # + 226811 *�(S¬( − 277211�(S¬( (+ 163811 �(S¬( − 37811�(S¬( − 474% ��(S¬(�(+ 2310%�(S¬(�( # − 2724%�(S¬( *�( − 948%�(S¬(�( (+ 3198%�(S¬( �( − 1362%�(S¬(�( + 437� ��(S¬(�#− 1163��(S¬(�# # − 874��(S¬( *�# + 2326��(S¬(�# (+ 437��(S¬( �# − 1163��(S¬(�# + 3584�(��¬�#

C�»3¼ = 12611 ��(S¬( − 88211�(S¬( # + 226811 *�(S¬( − 277211�(S¬( (+ 163811 �(S¬( − 37811�(S¬( − 234% ��(S¬(�(+ 630%�(S¬(�( # − 324%�(S¬( *�( − 468%�(S¬(�( (+ 558%�(S¬( �( − 162%�(S¬(�( + 69� ��(S¬(�#+ 453��(S¬(�# # − 138��(S¬( *�# − 906��(S¬(�# (+ 69��(S¬( �# + 453��(S¬(�# + 3584�(��1�#

`

(F.4)

144

APPENDIX G

STRESS IN THETA DIRECTION AND TRANSVERSE SHEAR PROFILE FOR EACH

CASE

145

Case 1 Stress in Theta Direction:

Figure G.1: Case 1: Stress in Theta Direction Profile through the Thickness (�#/(��#) =500/29 and = 0.3)

< 44/<��

"/

� = − �2 � = + �2 � = 0

146

Case 1 Transverse Shearing Force:

Figure G.2: Case 1: Transverse Shearing Profile (�#/(��#) = 500/29 and = 0.3)

"/

I 3/I��

147


Figure G.3: Case 2: Stress in Theta Direction Profile through the Thickness (�#/(��#) =500/29 and = 0.3)

� = − �2 � = + �2 � = 0

"/

< 44/<��

148


Figure G.4: Case 2: Transverse Shearing Profile (�#/(��#) = 500/29 and = 0.3)

I 3/I��

"/

149


Figure G.5: Case 3: Stress in Theta Direction Profile through the Thickness (�#/(��#) =500/29, �/ = 1.5, and = 0.3)

< 44/<��

"/

� = − �2 � = + �2 � = 0

150


Figure G.6: Case 3: Transverse Shearing Profile (�#/(��#) = 500/29, �/ = 1.5, and =0.3)

"/

I 3/I��

151

VITA

Timothy M. Harrell was born in Philadelphia, Pennsylvania on November 20, 1986. He

attended elementary school at St. Bernard’s Catholic School. In 1998, he and his family moved to

Winston-Salem, North Carolina, where he began middle school. In the middle of 8th grade year,

he and his family moved to Hendersonville, Tennessee where he attended and graduated from

Ellis Middle School. He attended Hendersonville High School in June 2005, where he received

many honors including the President’s award as well as participating in many extracurricular

activities including being the president of Future Business Leaders of America (FBLA). The

following August he started at Tennessee Technological University and in December 2010

graduated cum laude and received a Bachelors of Science degree in Civil Engineering with a

concentration in structural mechanics. He then enrolled at Tennessee Technological University

for the Master’s program and graduated in August 2014 with a Masters of Science in Civil

Engineering with a focus in structural mechanics.

Documents

AN ABSTRACT OF A THESIS APPLICATION OF GROEBNER …equations, 3) and Groebner Basis is employed to decouple the algebraic equations to find analytic solutions in terms of the material