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CONTACT MECHANICS OF LAYERED COMPOSITES
UNDER AXISYMMETRIC INDENTATION

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ZHENWEN WANG

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CONTACT MECHANICS OF LAYERED COMPOSITES
UNDER AXISYMMETRIC INDENTATION

By

Zhenwen Wang

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSIPHY

Department of Mechanical Engineering

2005

ABSTRACT

CONTACT MECHANICS OF LAYERED COMPOSITES
UNDER AXISYMMETRIC INDENTATION

By
Zhenwen Wang

The analytical studies on contact mechanics have been limited to problems of
either half space or single layer due to mathematical difficulty, though layered
composites subjected to indentation are commonly encountered in industrial
applications. The studies of indentation of layered composites, however, are
based on numerical approaches. This dissertation provides a theoretical method
for the contact mechanics of layered composites subjected to axially symmetric
indentation. A new function is introduced in this dissertation to reduce the
complexity of the mathematical process, and mathematical solutions are
provided for all the problems investigated in this research. However, the
mathematical solution for the final integration could only be obtained for the point
loading condition. A numerical method was used to evaluate the final results for
other loading conditions, such as uniform stress, flat indentation and spherical
indentation. In this dissertation, the effects of material property, layer thickness,
boundary condition, loading condition, and lamination on contact mechanics were
investigated for the cases of a half space, a single layer bonded to a rigid base, a
single layer bonded to an elastic half space and two-layered composites bonded
to a rigid base. The dissertation also investigated the frictional effect at the

contact interface. Both shear slip and normal separation theories were

incorporated into the mathematical formulation, allowing the study of debonding
at the interface between layers. New transformed shear slip and normal
separation coefficients are proposed to study the imperfect bonding interfaces
with a finite length. Contact mechanics models have been proposed based on
numerical results. These models provide insight into the relationships among
total load, maximum displacement, contact radius, layer thickness and material

properties, and guidelines for engineering applications.

ACKNOWLEDGEMENTS

The author is indebted to many people during the research and development of
this dissertation. First of all, I own many thanks to my adviser, Dr. Dahsin Liu, for
his encouragement through the years to keep from giving up. Without Dr. Liu's
tireless work with me over the weekends to accommodate my regular working
schedule, I would never have been able to turn in this dissertation. Furthermore,
the suggestions and comments from the professors in my guidance committee
are invaluable and much appreciated. I also own a debt to my wife, Xinlan, who
encouraged me to come back to the program after a year-long interruption, for
her patience, understanding and continuous support through the program to
achieve my life time goal. Finally, thanks goes to my two lovely children, six year
old Claire and three old Collin, who were understanding when father had to go to

work on the weekends in the past few years.

TABLE OF CONTENTS

List of Tables ....................................................................................................... vii
List of Figures ..................................................................................................... viii
List of Figures ..................................................................................................... viii
Chapter One Introduction ................................................................................... 1
1.1 Introduction ..................................................................................................... 1
1.1.1 History ......................................................................................................................................... l
1.1.2 Summary of contact mechanics research .................................................................. 1
1.1.3 Contact mechanics in composite material ................................................................. 6
1.1.4 Approach to layered composites .................................................................................... 8
1.2 Object of this study ......................................................................................... 9
1.3 State of the problem ..................................................................................... 11
1.4 Contents of this dissertation ......................................................................... 11
Chapter Two Development of Theoretical Analysis ............................................ 15
2.1 Elasticity Analysis Based on Cylindrical Coordinate System ........................ 15
2.2 Hankel Transformation and Bessel Functions .............................................. 20
2.3 Displacement Solutions ................................................................................ 22
2.4 Stress Solutions ............................................................................................ 23
2.5 Exact Solution for Half Space under Point Load ........................................... 29
2.6 Traction Boundary Conditions for Contact Problems .................................... 35
Chapter Three Numerical Solution and Verifications ....................................... 38
3.1 Elastic Layer on Rigid Base .......................................................................... 38
3.2 Elastic Layer on Elastic Base ....................................................................... 42
3.3 Two-Layer Composite on Rigid Base ........................................................... 46
3.4 Validations .................................................................................................... 50
3.4.1 Numerical integration vs. theoretical solution ........................................................ 51
3.4.2 Validations among cases ................................................................................................. 53
(i) Comparisons between case (1) and case (2) .......................................... 54
(ii) Comparisons between case (1) and case (3) ......................................... 57
(iii) Comparisons between case (2) and case (4) ........................................ 59
Chapter Four Basic Applications ..................................................................... 63 .
4.1 Effect of Material Property ............................................................................ 63
4.2 Effect of Thickness ....................................................................................... 67
4.3 Effect of Base Condition ............................................................................... 76
4.4 Effect of Loading Condition ........................................................................... 84
4.5 Effect of Lamination ...................................................................................... 93
4.6 Cylinder Shape lndenter ............................................................................... 98

Chapter Five Contact Theories ....................................................................... 102

5.1 Half space ................................................................................................... 102
5.2 Single layer on rigid base ........................................................................... 103
5.3 Single layer on elastic base ........................................................................ 106
5.4 Two layer composite on rigid base ............................................................. 110
Chapter Six Advanced Study ............................................................................ 114
6.1 Shear Loading ............................................................................................ 114
6.2 Contact Problems with Frictions ................................................................. 118
6.3 Bonding Theory .......................................................................................... 122
6.3.1 Entire Imperfect Interface .............................................................................................. 124
6.3.1.1 Verifications ................................................................................... 126
6.3.1.2 Variation of shear slip coefficient Kr .............................................. 128
6.3.1.3 Variation of normal separation coefficient Kz ................................ 131
6.3.2 Imperfect bonding with finite length .......................................................................... 134
6.3.2.1 Verifications ................................................................................... 138
6.3.2.2 Variation of shear slip coefficient kr ............................................... 142
6.3.2.3 Variation of normal separation coefficient kz ................................. 145
Chapter Seven Summary and Conclusions ................................................... 148
7.1 Summary of the Research .......................................................................... 148
7.2 Conclusions ................................................................................................ 148
7.3 Suggestions and Future Study ................................................................... 154
Appendix A Mathmatica ® C for General Derivation ....................................... 157
Appendix B Mathematica® Code for Half Space ............................................ 159
Appendix C Mathematica® Code for Point Loading ....................................... 161
Appendix D Mathematica® Code for Two-layer Half Space ........................... 163
Appendix E Mathematica® Code for Two-layer on Rigid ................................ 166
Appendix F Mathematica® Code for Shear Loading ...................................... 169
Appendix G C++ Source Codes for Numerical Calculations ........................... 172
Appendix G1 C++ source code for half space ..................................................... 173
Appendix G2 C++ source code for single layer bonded to rigid base ............. 181
Appendix G3 C++ source code for single bonded to elastic half space .......... 189
Appendix G4 C++ source code for two-layer bonded to rigid base ................. 201
Appendix G5 C++ code for single layer bonded to an elastic half space
subjected to shear loading ...................................................................................... 204
Appendix G6 C++ source code for single layer bonded to an elastic half space
with shear slip and normal separation theories ................................................... 218
Bibliography ...................................................................................................... 232

vi

List of Tables

Table 1 Summary of axisymmetric indentation interface stress distribution and its

Hankel transforms ....................................................................................... 37
Table 2 Material properties for steel, aluminum and nylon. ................................ 64
Table 3 Material properties used for thickness study ............. I ............................. 67
Table 4 Material properties for friction study ..................................................... 119
Table 5 Material properties for friction study ..................................................... 126
Table 6 Material properties for friction study ..................................................... 128
Table 7 Material properties for friction study ..................................................... 131
Table 8 Some values for different imperfect bonding length ............................. 137

Table 9 Material properties for friction study ..................................................... 139

vii

LIST OF FIGURES

Figure 1 Boundary condition of the indentation .................................................... 9
Figure 2 Stresses acting on an element of a solid revolution. ............................ 16
Figure 3 Half space with axisymmetric loading p(r). ........................................... 29
Figure 4 Single layer situated on rigid base: ...................................................... 38
Figure 5 Elastic layer situated on elastic base. ................................................... 42
Figure 6 Two-layer composite on rigid base. ...................................................... 46
Figure 7 Gaussian point integration algorithm. ................................................... 52
Figure 8 Numerical solution vs. exact solution for elastic half space. ............. 53
Figure 9 Comparison of normal stresses between case (1) and case (2) ........... 55
Figure 10 Comparison of shear stress between case 1 and case 2. .................. 55

Figure 11 Comparison of radial displacements between case (1) and case (2). 56

Figure 12 Comparison of normal displacements between case (1) and case (2).

.................................................................................................................... 56
Figure 13 Comparison of normal stresses between case (1 ) and case (3) ......... 57
Figure 14 Comparison of shear stresses between case (1) and case (3) ........... 58

Figure 15 Comparison of radial displacements between case (1) and case (3). 58

Figure 16 Comparison of normal displacements between case (1) and case (3).

oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

Figure 17 Comparison of normal stresses between case (2) and case (4) ......... 60
Figure 18 Comparison of shear stresses between case (2) and case (4) ........... 60
Figure 19 Comparison of radial displacements between case (2) and case (4). 61
Figure 20 Comparison of radial displacements between case (2) and case (4). 61

Figure 21 Normal stress distributions based on different materials. ................... 65

viii

Figure 22 Shear stress distributions based on different materials. ..................... 65

Figure 23 Radial displacement distributions based on different materials. ......... 66
Figure 24 Normal displacement distributions based on different materials. ........ 66
Figure 25 Normal stresses at the contact surface. ............................................. 68
Figure 26 Normal stresses at 0.25 mm below the contact surface. .................... 68
Figure 27 Normal stresses at 0.5 mm below the contact surface. ...................... 69
Figure 28 Normal stresses at 1.0 mm below the contact surface. ...................... 69
Figure 29 Shear stress distribution at the contact surface. ................................. 71
Figure 30 Shear stress distribution at 0.25 mm below the contact surface ......... 72
Figure 31 Shear stress distribution at 0.5 mm below the contact surface ........... 72
Figure 32 Shear stress distribution at 1.0 mm below the contact surface ........... 73
Figure 33 Radial displacement distributions at the contact surface. ................... 73

Figure 34 Radial displacement distributions at 0.5 mm below contact surface... 74
Figure 35 Radial displacement distributions at 1.0 mm below contact surface... 74
Figure 36 Normal displacement distributions at the contact surface ................... 75
Figure 37 Normal displacement distributions at 0.5 mm below contact surface. 75

Figure 38 Normal displacement distributions at 1.0 mm below contact surface. 76

Figure 39 Normal stress distributions at the contact surface. ............................. 77

Figure 40 Normal stress distributions at z=0.5 mm from contact surface. .......... 78

Figure 41 Normal stress distributions at z=1.0 mm from contact surface. .......... 78

Figure 42 Shear stress distributions at contact surface. ..................................... 79

Figure 43 Shear stress distributions at z=0.5 mm from contact surface. ............ 80

Figure 44 Shear stress distributions at z=1.0 mm from contact surface. ............ 80
ix

 

Figure 45 Radial displacement distributions at the contact surface. ................... 81
Figure 46 Radial displacement distributions at z=0.5 mm for contact surface. 81
Figure 47 Radial displacement distributions at z=1.0 mm from contact surface. 82
Figure 48 Normal displacement distributions at the contact surface ................... 82

Figure 49 Normal displacement distributions at z=0.5 mm from contact surface.
.................................................................................................................... 83

Figure 50 Normal displacement distributions at z=1.0 from contact surface ....... 83

Figure 51 Normal stress distributions at contact surface. ................................... 85
Figure 52 Normal stress distributions at z=0.5 mm from contact surface. .......... 85
Figure 53 Normal stress distributions at z=1.0 mm from contact surface. .......... 86
Figure 54 Normal stress distributions at z=2.0 mm from contact surface. .......... 86
Figure 55 Shear stress distributions at contact surface. ..................................... 87
Figure 56 Shear stress distributions at z=0.5 mm from contact surface. ............ 87
Figure 57 Shear stress distributions at z=1.0 mm from contact surface. ............ 88
Figure 58 Shear stress distributions at z=2.0 mm from contact surface. ............ 88
Figure 59 Radial displacement distributions at the contact interface. ................. 89

Figure 60 Normal displacement distributions at z=0.5 mm from contact surface.
.................................................................................................................... 89

Figure 61 Radial displacement distributions at z=1.0 mm from contact surface. 90
Figure 62 Radial displacement distributions at z=2.0 mm from contact surface. 90
Figure 63 Normal displacement distributions at the contact surface ................... 91

Figure 64 Normal displacement distributions at z=0.5 mm from contact surface.
.................................................................................................................... 91

Figure 65 Normal diplacement distributions at z=1.0 mm from contact surface .92

Figure 66 Normal displacement distributions at z=2.0 mm from contact surface.
.................................................................................................................... 92

 

Figure 67 Normal stresses at the bonding interface for nylon/aluminum. ........... 94
Figure 68 Shear stresses at the bonding interface for nylon/aluminum .............. 94
Figure 69 Radial displacements at the bonding interface for nylon/aluminum. 95
Figure 70 Normal displacements at the bonding interface for nylon/aluminum. .95
Figure 71 Normal stresses at the bonding interface for aluminum/nylon. ........... 96
Figure 72 Shear stresses at the bonding interface for aluminum/nylon. ............. 97
Figure 73 Radial displacements at the bonding interface for aluminum/nylon. 97

Figure 74 Normal displacements at the bonding interface for aluminum/nylon. .98

Figure 75 Force vs. deflection for single layer on rigid base. ........................... 104
Figure 76 Stiffness vs. Young’s modulus for single layer on rigid base. ........... 105
Figure 77 Relationship between stiffness and h/a. ........................................... 106
Figure 78 Force vs. deflection for single layer on elastic base. ....................... 107
Figure 79 Stiffness vs. the ratio of Young’s moduli. .......................................... 108
Figure 80 Normalized stiffness vs, the ratio of Young’s moduli. ....................... 108
Figure 81 Stiffness vs thickness changes ......................................................... 109
Figure 82 Force vs. deflection for two-layer composite on rigid base. .............. 111
Figure 83 Relationship between stiffness and the ratio of Young’s moduli. ...... 112
Figure 84 Stiffness vs. thickness ratio of the two layers. .................................. 113
Figure 85 Two-layer half space with shear loading ........................................... 115
Figure 86 Normal stresses at bonding interface between layer 1 and 2 ........... 119
Figure 87 Shear stresses at bonding interface between layer 1 and 2 ............. 120

Figure 88 Radial stresses at the bonding interface between layer 1 and layer 2
.................................................................................................................. 120

Figure 89 Displacements at bonding interface of layer 1 and 2 ........................ 121

xi

 

 

Figure 90 Displacements at bonding interface of layer 1 and 2 ........................ 121

Figure 91 Shear slip theory of two layer half space .......................................... 123
Figure 92 Normal stress verification ................................................................. 126
Figure 93 Shear stress verification .............. . .................................................... 127
Figure 94 Shear displacement verification ........................................................ 127
Figure 95 Normal displacement verification ...................................................... 128
Figure 96 Normal stresses vs. shear slip coefficient kr ..................................... 129
Figure 97 Shear stresses vs. shear slip coefficient kr ....................................... 130
Figure 98 Radial displacement Ur vs. shear slip coefficient kr .......................... 130
Figure 99 Normal displacement Uz vs. shear slip coefficient kr ........................ 131
Figure 100 Normal stresses vs. normal separation coefficient kz ..................... 132
Figure 101 Shear stress vs. normal separation coefficient kz .......................... 133
Figure 102 Shear displacement Ur vs. normal separation coefficient kz .......... 133
Figure 103 Normal displacement Uz vs. normal separation coefficient kz ........ 134
Figure 104 Two-layer composite of imperfect bonding with finite length .......... 134
Figure 105 Shear slip coefficient kr (é) distribution. .......................................... 136
Figure 106 Shear slip coefficient kz(f,) distribution ............................................ 136
Figure 107 Shear slip coefficient vs. imperfect bonding length ......................... 138
Figure 108 Normal separation coefficient vs. imperfect bonding length ........... 138
Figure 109 Normal stress for all three cases. ................................................... 140
Figure 110 Shear stress for all three cases. ..................................................... 140
Figure 111 Displacement Ur for all three cases. ............................................... 141
Figure 112 Displacement Uz for all three cases. .............................................. 141

xii

 

Figure 113 Normal stress vs. shear slip coefficient kr. ..................................... 143

Figure 114 Shear stress vs. shear slip coefficient kr. ....................................... 143
Figure 115 Radial displacements Ur vs shear slip coefficient kr ....................... 144
Figure 116 Normal displacement Uz vs shear slip coefficient kr ....................... 144
Figure 117 Normal stresses vs. normal separation coefficient kz. .................... 145
Figure 118 Shear stresses vs. normal separation coefficient kz. ...................... 146
Figure 119 Shear displacements vs normal separation coefficient kz. ............. 146
Figure 120 Normal displacements vs. normal separation coefficient kz. .......... 147
xiii

 

CHAPTER ONE INTRODUCTION

1.1 Introduction

1.1.1 History

Contact mechanics is a subject that has been discussed for more than a century.
Early in 1882, the first academic paper, “On the Contact of Elastic Solids”, was
published by Heinrich Hertz, With the increasing demand in the engineering
development in railway and marine reduction gears, rolling contact of bearing
industry and construction engineering, extensive efforts were made to the contact
mechanics study, along with other engineering mechanics theory. Hertz’s theory
was restricted to frictionless surfaces and perfect elastic bodies. Over the last
half century, research in contact mechanics has focused on the removal of these
restrictions for the pure contact problem. At the same time, development of the
theories of plasticity and linear visco-elasticity paved the road to investigate the
stresses and deformations at the contact of inelastic bodies. With the advances
of composite manufacturing technology and Its extensive usage in the aerospace
and automotive industry, contact mechanics was applied to composite study in

the past few decades.

1.1.2 Summary of contact mechanics research

The Boussinesq-Flamant problem was the very first contact problem that was
approached with theoretical mechanics. It was defined as a half space loaded by
a point load. Flamant (1892) obtained a solution by way of a three dimensional

solution for normal loading on a straight boundary. Boussinesq (1892) extended

 

 

the solution to the case of an inclined force, which can be visualized as a

combination of normal and tangential loadings.

Dean, Parsons and Sneddon (1944) presented a theoretical approach to semi-
infinite elastic solid with defined stress applied at the interior of the solid. The
direction of the loading conditions is perpendicular to the surface of the semi-
infinite solid. In order to simplify the mathematic work, the authors had to assume
the material is incompressible and the loading is uniform stress. They also
investigated the compressible elastic solid with numerical method. They found
that by using a single numerical factor, a fair accuracy can be achieved for the

general case from the equivalent incompressible medium.

During the course of investigation of the distribution of stress and displacement in
elastic solids, it used to be carried out through guessing appropriate combination
of solutions, which satisfy the prescribed boundary conditions in any special
case. Love (1939) used this cumbersome method in the case of a conical punch.
With some experienced engineers’ pioneer work on rigid indention, Harding and
Sneddon (1945) noticed that a systematic application of integral transforms into
the rigid indentation of semi-infinite elastic solid will reduce the problem into a
pair of integral equations, which fall into the classic mathematical cases studied
by Titchmarsh (1937) and Busbridge (1938). In the Harding and Sneddon study,
they established the foundation for the mathematical solution for axial symmetric
indentation problem. Though the development is limited to the semi-infinite
elastic solid, the authors pointed out the possibility of extending the method to

the solution for the stress distribution in a plate with finite thickness. Sneddon

 

 

(1951) presented a solution for a finite thickness plate perfectly bonded to a rigid
support. He used a symmetrical approach, mirrored the material and the loading
along the bonding interface with a rigid support. Therefore, by solving a finite
plate with doubled thickness and symmetrical loading at both top and bottom
surfaces, it leaded to the solution equivalent to the single thickness plate
perfectly bonded to a rigid base. Later, Sneddon used the theory developed by
Harding and Sneddon (1945) to successfully obtain solutions for the Boussinesq

problem with flat-ended cylindrical (1946) and rigid cone indenters (1948).

With more and more theoretical development driven from the industry, more and
more research was focused on sliding contact. After the half-space point solution
by Boussinesq and Cerutti, it seemed that integration of the point loading
condition would lead to all other loading conditions. However, this direct
approach normally leads to a series of intractable integrals with mathematical
difficulties. Green (1949) introduced a method for the stress analysis of normally
loaded half space, which can be extended to the shear loading of the half space.
Based on the Green study, Hamilton and Goodman (1966) developed the
equations for the complete stress field due to a circular contact region carrying
Hertzian normal pressure and a proportional distributed shearing traction. The
shearing traction is represented by the normal pressure multiplied by the
frictional coefficient. A complex harmonic function was used in their investigation.
The solution for the stresses inside the indented material was developed to
illustrate the potential failure point. Stress boundary conditions were specified for

the half space solid in their research.

Based on some Russian pioneer work, Meijers (1968) developed a symbolic
solution for the contact problem of a rigid cylinder pressed on an elastic layer
connected perfectly to a rigid base. Hertz contact boundary condition was
assumed in his development with no friction between the cylinder and the layer
for the generalized plain stress problem. Meijers focused on developing a
symbolic solution for both thin and thick layer conditions. He used a basic integral
equation for the pressure distribution at the contact interface presented by
Aleksandrove (1962) and Koiter (1963), by introducing a complex kernel function
for the contact stress and displacement and expanding the kernel function into a
power series. The first three coefficients of the power serials were evaluated by
using numerical integration for different Poisson ratios. With this approximating
work, Meijers was able to develop a general solution for both thin and thick single
layer indentations. The details of the numerical processing are quite
cumbersome and it is very difficult to extend the approach to layered composite

due to the complexity of mathematics.

Pao et al (1971) studied the upper and lower bounds of the maximum stress of
an elastic layer bonded to a half space elastic foundation by a rigid cylinder. The
symmetrical loading technique from Sneddon’s (1951) was used for his
calculation. A combination of half space and single layer solutions from
Sneddon’s development was used in this investigation. Pao used a segment
approach to divide pressure distribution into many small intervals. By assembling
a matrix from each interval, he was able to come up with a series of functions to

gain a numerical solution. Different boundary conditions, frictionless and rigid

bonding, between the layer and rigid base were investigated. The stress and
displacement component was evaluated to demonstrate the effects from position

ratio, friction between layer and the rigid support, and layer thickness.

Ting (1966) presented symbolic and numerical solution of the contact stresses
between a rigid indenter and a viscoelastic half-space, which extended the
research from elastic media to viscoelastic media. In his approach, he introduced
a method which can solve an arbitrary axisymmetric indention on viscoelastic half
space. A dynamic rebound of a rigid sphere on viscoelastic half space was
studied by Hunter (1960). A numerical method was used to evaluate the result by

both authors.

Dhaliwal (1970a) developed a solution for the single layer pressed against an
elastic foundation. A different numerical method was used to develop the
solution, which achieves good agreement. He also developed a symbolic solution
for an arbitrary profile (1970b). However, no numerical result was presented in

his further investigation.

Gupta et al (1973 and 1974) studied stress distributions in plane strain of a
layered elastic solid subjected to arbitrary normal and shear boundary loading.
Fourier transform of Airy stress function was used in his investigation. His
analysis result shows that internal stress distributions as obtained by elliptical
contact pressure are valid for most cases. He also pointed out that the half width
and maximum pressure on the surface may be quite different from Hertzian

solutions in order to solve the problem with his approach.

 

 

Matthewson (1981) presented a theory for the indentation of a soft coating by a
rigid body. The theory is based on the assumption that the coating is perfectly
bonded to the substrate, and behaves linearly elastically. The displacement of
the contact boundary condition is approximately approached by a finite power
series. The approximation used quantities averaged through the coating
thickness. This assumption differs from many other publications, where an exact
displacement boundary condition was met, and stress on the boundary problem

was approached approximately.

Jaffar (1993) used Chebyshev series method to study the surface deformation of
a bonded elastic layer indented by a rigid cylinder. His result shows close
agreement with the exact solution for the half plane problem with the derived

asymptotic formula for a thick layer.

Giannakopoulos (1997a, 1997b) investigated the indentation of solids with
gradients in elastic properties. A power law material model was used in his
approach. Due to the difficulties, commercial finite element code was used to

study different indentation case.

1.1.3 Contact mechanics in composite material
There are a lot of studies and publications in laminate composites study. Pagano '
(1970) presented a theoretical solution for laminate composites. Swanson (2004)
studied the Hertzian contact loading of orthotropic materials. He used both
procedures, developed by Willis (1966) and Pagano (1970) separately, and

further developed a detailed numerical method to achieve the results for laminate

composites. This research is to focus on the layered composite. Each layer has
homogeneous material property. Literature on laminated composites is not

reviewed in details.

Layered composite indentation was very challenging in contact mechanics. In the
application, a soft coating on a hard substrate is very efficient for protection. In
automotive safety design, a soft layer on a hard supporting structure provides the
necessary stiffness to meet the design requirements and also provide cushion to
minimize the injury of vehicle occupant. To the author’s knowledge, few
publications are available in this subject. The major theoretical study was limited
to indentation of single layer bonded to an elastic half space. The majority of the
industry currently uses finite element approach with commercial code, such as
LS-DYNA®, Abaqus®, PAMCRASH®, MADAMO® for dynamic research,

Abaqus®, Nastran®, ANSYS®, etc. for static study.

Shield and Bogy (1989) studied a multiple region contact solution for a flat
indenter on a single layer bonded to a half space. The flat indenter has sharp
corners, which has singularity. Since it is a multi-region contact problem, each

region has its own solutions.

Jergensen et al (1998) studied spherical indentation of composite laminates with
controlled gradients in elastic anisotropy. His study provides an insight of
indentation of the indentation problem on orthotropic media. There is no known
analytical method for the problem he was trying to solve. Finite element approach

was used in his investigation.

 

 

Lu and Liu (1992) presented an lnterlayer Shear Slip Theory for delaminating
analysis, which was extended from a layer-wise laminate theory, the previous
development by Lee and Liu et al (1991 ). The continuity condition of the
interlaminar shear stresses on the composite interfaces was satisfied, which is
only valid for analysis of laminates with shearing mode (mode It and mode III). In
order to be able to handle the normal mode (mode I), Liu et al (1994) extended
the lnterlayer Shear Slip Theory by introducing linear normal separation theory.
Combining both linear shear ship theory and linear normal separation theory, the
theory became robust to solve complex loading conditions. In this research, both
shear slip theory and normal separation theory was implemented into the

development as part of the advanced study.

There is tremendous literature regarding the classic topic of contact mechanics.
The books authored by Gladwell (1980) and Johnson (1985) provided rich
reference literature on contact mechanics. Only some critical publications are
reviewed here, which represent major progresses of the research and theory

development.

1.1.4 Approach to layered composites

In order to solve the contact problem, simplifying boundary conditions was critical
for mathematical success. For a rigid indenter, the contour of the indenter shape
defines the deforming shape contact interface. Specifically, the displacement at
the contact interface can be defined, and the normal tractions outside the contact
area disappear. It seems that the proper boundary conditions have to be mixed

displacement and traction. In order to solve this boundary value problem which

involves biharmonic equations, certain transformation is needed. With the mixed
boundary condition, mathematical difficulty for the theoretical approach to the

contact problem rises.

V

 

 

 

Figure 1 Boundary condition of the indentation

In this research, the focus is on contact mechanics of the axially symmetric
indentation. The general setup of the problems is illustrated in Figure 1. Hankel
transformation is used in order to solve the biharmonic equation. The details of

Hankel transformation are discussed in Chapter Two.

1.2 Object of this study
In automotive safety development, head injury is based on the acceleration level.
In crash accident, the human head contacts the vehicle interior trim. The human
head is a complex structure, which can be simplified as a rigid skull and a

deformable skin. In recent development, the dummy head was optimized as a

 

 

vinyl skin and polyurethane skull structure, which represent the human structure
more closely. In these developments, due to the complexity of the mathematical
modeling, there is no thorough theoretical study of the structure. All the research
work done in the last three decades was based on experiments and experience,
though a lot of papers were published. Most of the research in the industry was

done on experiments and finite element analysis.

The vehicle interior trim normally is constructed by sandwich composite with a
metal supporting base. The sandwiched composite is made of foam and plastic
materials, such as Ensolite®, rubber foam, urethane, polyurethane,
thermoplastics, etc. There were a lot of studies on the structure design, but few
literatures from contact mechanics study can be found on indentation study of
layered composite. The author’s early experimental study (Liu, Dang and Wang,
1996 and1997) of foam composite under low velocity impact was the initial drive
that leads into the research. It was realized that the theoretical studies for the
contact mechanics of layered composites are needed to better understand the
performance of the composite materials. This dissertation develops a theoretical
approach to this problem to provide some design guidance for the composite

structures.

From the literature review, most of the research is limited to two dimensional,
semi-infinite and single layer materials. To the author’s knowledge, the results of
indentation to the layered composites have not been seen. The objective of this
study is to investigate the layered composites subjected to indentation by an

axisymmetric indenter. A new technique is proposed and significantly reduces

10

the complexity of the mathematics, which provides a unique way to look into
stress and displacement distribution. The final goal is to develop a model for the
composite indentation that can be used as guidance for the design. Also shear
slip and normal separation theory are incorporated into this technique to study
the bonding interface. An imperfect bonding with finite length was studied with a
step function for the shear slip coefficient, which provides insight into the local

debonding of the composite material.

1.3 State of the problem

This dissertation will study the layered composite indented by an axisymmetric
indenter. An elastic theoretical study of the indentation was carried out in this
study. Studied cases include a half space, one layer bonded to rigid base, one
layer bonded to an elastic half space, and two layer bonded to rigid base. The
shear slip theory was incorporated into the above development and investigated

in this dissertation as well.

1.4 Contents of this dissertation
This dissertation includes seven chapters. Chapter One reviews the studies done

in the past, and states the subjects that are to be investigated in this study.

Chapter Two discusses the basic theory developed for this study. In this chapter,
we develop the basic theoretical derivation, discuss the Bessel function used in
the study, and the Hankel transform that is the core part of the mathematical

base for this study. We also discuss the exact solution of some simple cases.

11

Lastly, we discuss the details of the approach with a stress boundary condition

for the contact problem.

Chapter Three discusses the detail of a numerical approach to the final solution
for complicated cases. Mathematica® was used in this approach to reduce the
mathematical work significantly. Microsoft® C++ programs were developed for
the numerical calculation in the final result for this investigation. Verifications
were carried out among these derivations. For example, the numerical result is
verified against the theoretical solution for point loading. By increasing the
thickness of the first layer, a two-layer case would be simplified to a one-layer

case, etc.

Chapter Four discusses some basic applications. Six different cases are

investigated in this chapter:

Case 1, material study: with half space and point loading, we investigated
aluminum, steel, nylon material to understand the stress and displacement for

different material under the same loading condition.

Case 2, thickness study: two layer half spaces are used to investigate the
thickness effect. The composite consists of an aluminum layer bonded perfectly
with a very thick nylon base. Spherical indentation is investigated during the

study.

12

Case 3, boundary condition study: in this case, a two layer half space with a
combination of aluminum/steel, aluminum/nylon, aluminum/rigid are investigated,

using spherical loading.

Case 4, loading conditions: an aluminum half space with different loading
conditions, point loading, uniform pressure, flat indentation, and spherical shape

indenter are studied.

Case 5, lamination study: two different cases are investigated — material order
and thickness ratio. For material order study, the thickness ratio is set to
constant, and the composites of Nylon/Aluminum and Aluminum/Nylon were
investigated with spherical indentation. For thickness study, with a fixed material
order, both Nylon/Aluminum and Aluminum/Nylon, the thickness ratios are

presented in this chapter.

Case 6, Shear slip theory: This was studied for two-layer composted indented

against a rigid base.

Chapter Five presents models with the techniques developed in the previous
chapters. The relationship of force, deflection, Young’s modulus, thickness, and
contact radius is incorporated into simple models. These models provide simple
and valuable information as guidance to design and applications for layered

composite.

Chapter Six is presents some insight into the advanced study. The pure shear

traction at the indentation interface is presented. With friction coefficient to relate

13

normal traction and shear traction, the solution to a more general contact
problem can be obtained. Moreover, the shear slip theory and normal separation
theory are introduced into the technique developed in the previous chapters. With
the shear slip coefficient being set to a step function with Hankel transform, we

investigate the limited imperfect bonding for the composite under indentation.

Chapter Seven summarizes the results and concluded the study of this
dissertation. Some insights into the theory are discussed and suggestions to

future study were presented.

Lastly, bibliographies and appendices provide references and the information
that can’t be covered in detail in the main section of this dissertation due to

limited space.

14

CHAPTER TWO DEVELOPMENT OF THEORETICAL ANALYSIS

In this chapter, we develop a new technique that will be used subsequently to
solve contact problems consisting of the layered composites. Validations of the
technique are presented as they are necessary for the justification of later
applications. For the simple case of point loading, the new technique leads to the
same solution as otherwise presented by Sneddon (1953). For other loading
cases, such as uniform pressure, flat contact and spherical punch, the stress and
displacement solutions are in integral forms and explicit expressions are not

found. Therefore, numerical methods are sought to investigate these cases.

2.1 Elasticity Analysis Based on Cylindrical Coordinate System

Based on the free-body diagrams shown in Figure 2, we can establish the
following equilibrium equations (Timoshenko and Goodier, 1970), which are also
the governing equations for cylindrical bodies under static loading. Note that both

equations are independent of i9 and the body force is not considered in the

 

formulation.

60’ +61” +07 —06 =0 (2-1a)
0r 6r r

Qflfigfizzo (2-1b)
6r 82 r

15

 

 

 

 

Figure 2 Stresses acting on an element of a solid revolution.

Since linear elastic analysis for isotropic bodies is of interest in this research, the

following Hooke’s law is imposed

a, = $0. -v<og +0.11 (Ma)
80 = gag — v<o-r + a. )1 (2—2b)

16

6. =%[0r -v<ar +0911 (2-2c)

where E is Young’s modulus and v is Poisson’s ratio. Based on the cylindrical

coordinate system, we can have the following linear strain-displacement

relations.

3,. = 9‘1 (2-3a)
6r
,

£2 __. 9‘s (2-3c)
62
(3a,. auz

7rz —6z —6r ( I

where a and y are strains and u is displacement. The subscripts r, 6? and 2

represent the coordinates in the radial, circumferential and thickness directions,
respectively. There are four strain components while only two displacement
components. The relations between Cartesian and the cylindrical coordinate

systems are expressed as follows (Timoshenko 1970)
ox = 0",. cos2 9+ag sin2 6—2rrg sin6cost9 (2-48)

0y = 0', sin2 6 + 0'6 cos2 9 + 27,9 sin 6cos€ (2-4b)

17

rxy = (a, —0'g)sinficost9+ z'r6;(cos2 6—sin2 6) (2-4C)

Substitute the above relations into the comp ability equations in the Cartesian

coordinate systems, we have,

2

VZO'r-i(0'r-0'g)+—l——a——(0'r+0'g+0'z)=0 (2-5a)
,2 l+v r2

V209+—2—(0'r—0'g)+—1—l—a—(0’r+0'g+0'z)=0 (2-5b)
2 1+vr 6r

r

For convenience, it is possible to introduce a stress function ¢ (Love, 1927). The

stress components are related to the stress function in the following manners:

or = A~V2¢z _ 2(/l '1' #)¢rrz (2'63)
0. = (32 + 4mv2¢z — 2e + #szz (2—6b)
2 2
09 = ’1V ¢z " :(l + .u)¢zr (2-66)
6
rrz = (A + 2mgvz¢ — 2(2 + mm mm
2 62 1 a 52 . ,
where V = — + —— + —— 1s Laplace operator, and 2 and y are Lame 5

6r2 r 5" 622
constants. They are associated with Young’s modulus E and Poisson ratio v by

the following equations

18

l = ___E’v___ (2-7a)
(1 + v)(l — 2v)

E
y ' 2(1+ v) (Nb)

 

It is noted that Equations (2-6) satisfy the equilibrium equations automatically. If
we substitute Equations (2-6) into the compatibility equations, i.e. Equations (2-

5), we find the following biharmonic equation

v4¢ = o (2-8)

This equation combines equilibrium equations, linear strain-displacement
relations and compatibility equations. It is the ultimate governing equation for

linear elastic analysis for isotropic solids under static loading.

Similar to the stress components, the displacement components can also be
expressed in terms of the stress function by combining Equations (2-1), (2-3) and

(2-6). The following two equations are useful for later applications

 

 

A +
u, =— ”ti... <2-9a)
y
”z = 1:12#V2¢- A+lu¢zz (2-9b)

19

2.2 Hankel Transformation and Bessel Functions
Before we get into the details of solving the above biharmonic equation, we
should briefly discuss the basic theories of Hankel transformation and Bessel

functions, as they are useful for developing the solutions of contact problems.

When a function K (a,x) is a known function of two variables a and x, and the

integral

7(a) = jf(x)K(a,x)dx (2-10a)
0

is convergent, then Equation (2-10a) defines a function of the variable a. This
function is called the integral transform of function fix) by the kernel K (a,x) .

Examples of some commonly used kernel functions are:

Laplace transform K (a,x) = e‘”

fax

Fourier transform K (a, x) = e—

Mellin transform K (a,x) = xo‘”l

If the kernel function K (a,x) = va(xa), then we have

a)

flu) = jxf(x)./,,(xa)dx (2-10b)
O

20

 

 

f(x)= jaf<a>Jv(xa)da <2-10c)
0

where Jv(xa) denotes a Bessel function of order v of the first kind, i.e.

x)=n20_( 1) n(x/2)2n+v

n!F(n+v+1)

 

where F(n +1) = n! is the Gamma function when n is an integer.

The function f(a) defined by Equation (2-10b) is called the v’h -order Hankel

transform of function f(x). Equation (2-10c) is called the Hankel inversion

theorem.
If we let x=r and 01:6, then Equation (2-10b) can be rewritten as follows

7(5) = jrf(r)Jo(r§)dr (2-10d)
0

The Hankel transformation has the following property

w

jai— j—{—)Jo(r:)dr - -4: flé) (2-11)
0

V4¢ can be expressed as V2(V2¢). If we take the Hankel transform of Equation

(2-8) and use the relationship of Equation (2-11), then we have

21

er W0(§r)dr = (—z——: ) er0(r§)dr (2-12a)
0 dz 0

oo 2 oo
IrV2(V2¢)Jo(§r)dr = (5—2—42) er2¢/o(r<§)dr (2-12b)
0 Z 0

Substituting Equation (2-12a) into (2-12b), we have

m

2 00
JVV4¢10(§’)dr = (if - :212 IrWo(r§)dr = 0 (2'13)
0 dz 0

00

Let G(§,z) = Ir¢/0(r§)dr , i.e. the Hankel transform of ¢(r,z), then
0

d2 22
(—2--6 ) G(§.ZI=0 (2-14)

dz

The integration of Equation (2-14) is elementary and its solution can be

expressed as
G(éj,z) = (A + Ede—Z: + (C + Dz)ez§ (2-15)

where A, B, C, and D are functions of 5. Now we can develop stress and

displacement expressions based on Equation (2-15).

2.3 Displacement Solutions

If we multiply both sides of Equation (2-9a) by rJ0'(r§) and integrate over r from

0 to co, we have the following equation after integration by parts

22

J‘ru,.11(r§)dr =
0

1+2: fig
dz

lnverting this result by using the Hankel inversion theorem, i.e. Equation (2-10c),

we obtain the following expression

 

(X)

1+ dG

ur= “ jéz—Jlrred: (2-16a)
it 0 dz

Similarly, from Equation (2-9b) we have

00 2
d G 2+2
jruzJo(r:)dr-= 2 — ”.520
0 dz

 

lnverting this result by using the Hankel inversion theorem, i.e. Equation (2—10c),

 

we have
u. = lei-1" 26— ’1’” ” €20)J0(r~f)d§ (2-16b)
0 dz 1”

2.4 Stress Solutions
In a similar manner, we can find the expressions for stress components.

Multiplying both sides of Equation (2-6a) by rJ0(r§), Integrating them over r and

inverting the equation by using the Hankel inversion theorem, i.e. Equation (2-

10c), we find

23

 

°° 3
0. =j51u+md G

d 3 —(3l+4#)§2%g]Jo(r€)dé (MM
0 Z

In a similar fashion, multiplying both sides of Equation (2-6d) by U 1 (r5),

integrating them over r and inverting the equation by using the Hankel inversion

theorem, Equation (2-100), we find

00 2
rrz = 1521/1 d f +(2 + 2/1)§ZG]J1(r§)d§ mm
0

 

dz

Obtaining the expressions for a, and 0'6 is not as simple as the components we

just obtained. If we add Equation (2-63) to Equation (2-60), then we have

2
6
or +09 =[2(/I+#)-a—2—-2#V2]¢z
Z

Multiplying both sides of this equation by rJ0(§r), and integrating them with

respect to r from 0 to 00, we obtain

 

°° 3 2

d G 0’ d0
jrrar +ae)Jo(ré)dr=2(/1+m 3 —2#(—2 —:2)—
0 dz dz dz

lnverting the above equation by using the Hankel inversion theorem, i.e.

Equation (2-100), we obtain

°° d3G d
or +09 = 16(1—7+#§2—Q)J0(r§)d¢f (2-16e)
0 d dz

Z

24

Now let us rewrite the equilibrium equation, Equation (2-1a), in the following form

62-020,) = r(0',. +09)—r2 E
r

Substituting Equation (2-16d) and (2-16e) into the above equation, it yields

2dG

—(r 0r)=2r1§(1——+#§d7)Jo(r€)d§-r2j§[1;§;2+é(1+2#)—]Jo(r§)dcf

Integrating the foregoing equation with respect to0 r, interchanging the order of

the integrations and using the integrals

eroogur = 111 (r6) and
0 5

jr2J11r6idr = 2%J1(r5)‘ —’-Jo<r6>.
6
0 5
we obtain
=j61/1— + 6 (J1 + 33,t1)—I~/o(ré‘)d§-2(li + —"—) j62 i’g—Ji (r6>d6 (2-16f)
dz3 r 0 dz
From Equations (2-16e) and (2-16f), we have

‘62 d6

°° 3G
09 = ,1 j§(-:Z—3G- )—Jo(r r6>d6+ 2” + —”—) 0j62; dG J (r 61d6 (2469)
0

25

As discussed earlier, A, B, C and D of Equation (2-14) are functions of 5.

Without losing any mathematical generality, we can let

(1+v)(l—2v)

E62 [(A+Bz)e‘25 +(C+Dz)e25];7(6) (2-17)

6*(692) :

 

where 5(5) = er(r)J0(r§)dr is a function of 5, and J0 and J, are Bessel
0

functions of its first kind. Substituting Equation (2-17) into Equations (2-16), we

have

0; = j6{1A6 + 8(1— 2v + z-6IIe‘i‘f +1—C6 + 00 — 2v — 2611er ifi<6>Jo<r6>d6
O

(2-18a)

rrz = j{1A6 — B<2v — 2611625 +166 + D<2v + 26>]er 16(6)J1(r6)d6
0

(2-18b)

or = j6{1—A6 + B<1+ 2v — 26m”: + 1C6 + D<1+ 2v + 26%:5 }fi(<§)Jo(r6)d§
0

— j {ti—A6 + 8(1— 26)]e'25 + [C6 + 00+ 2611er }%17(6)J1 (r6166
0

(2-18c)

26

09 = j2v6(Be‘25 + De“ )17(§)Jo(r§)d§
0

§(§)J1(r§)d§

‘r|v-—

+ j{[(—A6 + 3(1- 26)]e'24‘ + [C6 + D(1+ 26)]e2‘5}
O
(2-18d)

u, = jg—i'gEu—M + 3(1— 26)]e‘25 + [C6 + D<1+ z6>1ezfm6>J1<r6>d6
0

(2-18e)

u2 = ojl%fl’—){1A6 + 8(2 — 4v + 26m”: + 1C6 + 0(1 + z6)1ezfn3(6)Jo(r6)d6
0
(2-18f)
The details can be found from appendix A. For simplicity, let us define
a; = [A6 + 30- 2v + 2.6)]e‘24‘ + [—C6 + D(1— 2v — 26)]ez‘5 (2-19a)

r; = [A6 — B(2v — 26H?“ + [C6 + D(2v + 2.6)]ez‘f (2-191))

u;=1::"{[—A6 + 30- 26)]63‘24r + [C6 + 0(1 + 26)]e24} (2-190)

 

u’ = —1—2—"{[A6 + 3(2 — 41/ + 26)]?25 +[C6 + D(1+ 2.6)]e24’ } (2-19d)

Z

 

Since ,u = , then Equations (2-190) and (2-19d) become

E
2(1 +v)

27

2w: = [-Aé‘ + 30- 26)]e‘zf + [C6 + D(1+ 26)]ezs‘ (2—19e)

— 2w: = [A6 + 3(2 -4v + Zak—24‘ + [06 + D0 + 26ml? (2-19f)

Accordingly, the stress and displacement components can be expressed by the

following relations.

07. = Edméflwfidé (2-20a)
rm = ETLfléNoUfldcf (2-201,)
u,- = °(:jui625(6)J1(r6)d6 (2-20c)
u. = Lju2m61Jo (r6)d6 (2-20c)

The use of Hankel transformation greatly simplifies the expressions of stress and
displacement components and has significant benefit of reducing mathematical
complexity in imposing boundary conditions. In the following sections, we will
apply these results to a few special contact problems of the layered composites

consisting of perfect bonding interfaces.

28

J W
I. “.
—LLII

 

2.5 Exact Solution for Half Space under Point Load
The solution for a half space subjected to a point load was developed by
Sneddon (1953). The same problem is solved below to validate the new

technique.

 

”IV

 

Figure 3 Half space with axisymmetric loading p(r).
Since both stress and displacement components have to be zero when 2
approaches infinity, C and D must vanish in Equation (2.17), i.e.

(l+v)(1—2v)

2 (A + lane—25 5(6) (221)
1‘35

G*(~§,Z)=

 

Equations (2-18) can then be simplified as

29

 

 

a. = j6{1A6 + Bo - 2v + z6>1}e‘“‘ 17(6)Jo(r6‘)d€ (2-22a)
0

6.2 = 11A6 — B<2v — z6)1e‘256(6>Jo(r6)d6 <2-22b)
0

a, = 161-A6 + B<1+ 2v - z6)1e'29‘;7(6)Jo(r6)d6
0

(2-22c)
— Zia—:46 + 8(1— z6)1e‘2‘=‘ £6<6>J1 (r61d6
0, + 019 = 361-146 + B<1+ 4v — 261163f 17(§)Jo(ré)d€ (2-22d)
u, = GEE—£33146 + 8(1- z6)1e‘zffi(6)J1(r6>d6 (2-22e)
uz = gig-Vlmwa—m 26>1e‘2513(6)Jo(r6>d6 (2-220

Similarly, Equations (2-19a), (2-19b), (2-19e) and (2-19f) can be simplified as

a; =[A6+ B(1-2v+z§)]e‘z9‘ (2-23a)
r; = [A6 — B(2v — 26)]e‘24’ (2-23b)
2m: =[—A6+B(1—26)]e‘z§ (2-23c)

30

— 2w: = [A6 + 3(2 ~4v + Zak-29‘ (2-23d)

With the use of Equation (2-17), the boundary conditions can also be simplified.
This provides us with a great advantage of solving the contact problems of

layered composite, i.e.

oz = :1602 17(6)Jo(r¢f)d§ (2-25a>
rrz = :IT;17(6)Jo(r€)dé (2-251,)
ur = Es‘u:§(€)J1(ré)d6 (2-25c)
uz =§u;l_7(§)~10(r§)d§ (2-25d)

The boundary conditions for the contact problem defined in Figure 3 are

az|z=0 = p(r) when r < a (2-26a)
az|z=0 = 0 when r > a (2-26b)
1,212: 0 = 0 (2-26c)

31

If the indenter has a smooth edge, the normal stress component vanishes at the
edge. However, if the indenter has a sharp edge, there is singularity around the

edge.

From Equations (2-26) and (2-25), with Hankel inversion theorem, Equation (2-

10c), we have

a:

0, =1 (2-27a)

z=O

 

a:
TI’Z

: 0 (2-27b)
z=0

 

With the above boundary condition and Equations (2-23a) and (2-23b), we have
A; + 80— 2v) =1 (2-28a)
A if — 2vB = 0 (2-28b)

Solving the above equations, we have

Therefore, function G defined in Equation (2-21) becomes

C(62) = (H ”5):; 2”) (2v + z6)e‘zf 13(6)

 

32

Substituting the solutions A and B into Equations (2-22), we can find all stress

and displacement components in the half space. Details of the analysis are given

in Appendix B.

02 = 1(1 +26)€e‘z‘5fi(6)Jo(rcf)d6 (2-29a)
0

rm = 1262625 17(§)J1(rcf)d§ (2.291))
0

a) = 1(1- z6)6e‘2517(6)Jo (r6)d6 +§ j(—1 + 2v + Z§)e“25fi(é)~/1(ré)d6
0 0
(2-29c)

019 = 121/662‘ .5(6)Jo(r6)d6 —} j (—1 + 2v + z6)e‘zi 13(6)J1 (r6)d6
0 0

 

 

(2-29d)

u) = 10+”1 gzv'ziie-zé p(6)J1(r6)d6 (2-29e)
0

u. = I (I + v)”; 2" — “5) e‘ziz—2(6)Jo(r6)d6 (2-29f)

0

For near point loading, the normal stress distribution on the contact surface found

by Sneddon (1951) is

33

f.)__3_ (2 30a)
27! _3_
(a2 + r2)2

p(r) = (-

The Hankel transform of p(r) is

(I)

6(6): jrp(r)Jo(r6)dr=(—f—)e““5 (2-301))
0 71'

Substituting Equation (2-30b) into Equations (2-29), we have all stress and

displacement components. Details of analysis can be found in Appendix C.

 

 

 

2 2
az:_(_21:)(3z+a)(z+a) +ar (2-31a)
7r __
["2 +(a+z)2]2
Trz = 4%) "3rz(a+z) (2-32b)
[r2 +(a+z)2]§
0'6? =—(—2£—)[-l-(l-2v)+ (—I+2v)(a+z) + 2vr(a+z)—rz3] (2-330)
70' r

r[r2 +(a+z)2]5 [r2 +(a+z)2]5

a',=—P—[l(1—2v)+ (-l+2v)(a+z) _ (a+22)r +rz[2(a+z)2—r2]]
2717' r 5

_ _3_ _
r[r2+(a+z)2]2 [r2+(a+z)2]2 [r2+(a+z)2]2
(2-34d)

z (l+v)P

1
a” 22:5-

[ (l—2v)(a+z) + rz 3] (2-356)
r

(—l+2v)+
r[r2 +(a+z)2]§ [r2 +(a+z)2]§

34

u =(1+V)P (5-4v)az+(3—2v)22+2(1_V))(02+r2)

z 27$ 1 (2-36f)

 

[ 3
[r2 +(a+z)2]2

These solutions are identical to the solutions presented by Sneddon (1951 ).

2.6 Traction Boundary Conditions for Contact Problems

The most challenging part of using the new technique to solve general contact
problems is the integral involved in Equations (2-29). A transformation technique
will be necessary to reduce the complexity of integration. However, the type of
boundary condition is also of a major concern as the inversion of the transform
will become very difficult, if not impossible, if mixed boundary conditions are
imposed. In other words, either displacement or traction should be used in each
contact problem to reduce the complexity of the dual integrations. Moreover, in
reality, all contacts occur with limited contact surfaces and the displacements are
unknown in the non-contact areas adjacent to the contact surfaces, even though
we can assume there is no deformation at areas far away from the contact areas.

Therefore, specifying displacements is not feasible in solving contact problems.

On the contrary, since there is no traction on the surfaces of non-contact areas, it
automatically leads to specifying traction boundary conditions for studying
contact problems. The focus perhaps should be placed on how to define the
tractions on the contact surfaces. As the tractions on the contact surfaces
depend on geometry and surface condition of the contact problem, there is no
easy rule to guess the traction on the contact surface. However, many pioneers

have defined tractions with their experience and knowledge in mathematics and

35

physics. The traction boundary conditions have been refined to be close to reality
for some commonly used indenters, such as those with point, flat, and spherical

noses.

Johnson (1985) summarized the pressure distribution for contact problems with a
circular region. For a circular contact region with a radius a, it is required to find
displacements at a surface point and stresses at an internal point due to the

pressure distributed over the circular region. Solutions in closed form can be

found for axisymmetric pressure distributions in the form of p = p0(l — r2 / a2 )" .

When n = 0, it is a uniform pressure. When n = -%, it is for a flat punch. When

n = %, it represents a spherical indention condition. Combining with a point load,

5(r)

, we have defined the
2n*r

which can be represented by a function of p(r) = —

 

pressure distributions for indenters with regular shapes. We will use Hankel

transformation in the following chapters. For point loading, Sneddon (1951) used

p(r) = -i 0 instead of p(r) = — am

for the ointloadin a roach,
22(a2+,2)3/2 27m 9 9 pp

 

 

because it lead to singularity solution at the contact point with infinite stress when

using function p(r) = — 250;) . With revised stress distribution
If r

 

function p(r) = - —}—)- a

, it eliminates the singularity with more realistic

 

resun.

36

The contact traction distributions and associated Hankel transforms for several

contact problems are summarized in Table 1.

Table 1 Summary of axisymmetric indentation interface stress distribution and its

Hankel transforms

 

Hankel Transform

 

 

 

 

 

 

 

 

 

 

lndenter Contact Interface Stress 5(5) =
Profile Distribution Total Load 00
jrp(r)Jo(r6)dr
0
Near p a P
Point P(") = "— -P 5(5) = __e-0~§
Loading 2” (02 + ”2):”2 2n
= — h 5
”mm“ M m w en 0 rsa 2 ‘(6) — 3301 (a6)
Pressure ‘ ”a P0 p ‘ 6 1
p(r) = 0 when r>a
r2 "I"
p(r) = -Po(1- —) 2
Flat a 2 2 _ poasin(a§)
Punch " 27661 PO 12(5): ’—
when OSrSa é:
Ip(r) = 0 when r>a
r2 l
p(r)=-Ivo(l-—2)2 _ p0 ..
Spherical a __ 3,112 mg) ="—3—
lndenter 3 P0 05
When OSrSa [05 (308(06) _ sin(a§)]
P(r) = 0 when r>a
Shape p(r) p0 W e f' I? _ ”(r22 _ r12)p0 P(€) 5
Uniform
Pressure p(r) = 0 when r<r1 & r>r2 [r2J1(r26) - nJ1(r16)]

 

37

 

CHAPTER THREE NUMERICAL SOLUTION AND VERIFICATIONS

In this chapter, we use the theory developed in Chapter Two to obtain solutions
for several contact problems, eg. a single-layer material situated on a rigid base,
a single-layer material situated on an elastic base and a two-layer composite
situated on a rigid base. A numerical scheme is developed to calculate the
integrations which cannot be executed analytically. Validations of the numerical

scheme are carried out through the investigations.

3.1 Elastic Layer on Rigid Base

A single elastic layer situated on a rigid half base is one of the most common
contact problems in the real world, e.g. soft thin films and protection coatings on
metals. In this section, we will develop a generic solution for this contact problem.
For any loading condition as shown in Figure 4, we assume there is no friction on

the contact surface and the bonding between the elastic layer and the rigid base

 

 

 

 

is perfect.
p(r)
l I 1) I l 4,
Elastic Layer I4—2a-0—J r
/////9 /// /
Rigid Base
v 2

Figure 4 Single layer situated on rigid base.

38

Based on the above assumptions, we have the following boundary conditions:

at z = 0
02 = p(r) (3-18)
6,, = 0 (3-1b)
at z = h
142 = 0 (3-1c)
u, = () (3-1d)

Combining Equation (2-18a) with the boundary condition Equation (3—1a), we

have

ja?6j(6)Jo(r6)d6 = p(r)
0

If we apply Hankel’s inverse transformation to the above equation, we obtain the

following relationship

a; :1 (Ma)

where a; is defined in Equation (2-17a). Similarly, combining Equations (2-18b)

and (3-1b), we have

39

erzr(6)Jo(r6)d6 = 0
0

After applying Hankel’s inverse transformation to it, we have
1,2 = 0 (3-2b)

where 2'; is defined in Equation (2-17b). Moreover, combining Equations (2-18d)

and (3-1c), we have

ju;fi(§)Jo(r6)d§ = 0
0

Using Hankel’s inverse transformation, we have
“2 = 0 (3-20)

where u; is defined in Equation (2-17d).

As the fourth exercise, combining Equations (2-18c) and (3-1d), we have

(I)

jui6fi(6)J1(r6)d6 = 0
0

After using Hankel's inverse transformation, we have

u, = o (3-2d)

4O

where u: is defined in Equation (2-17c).

As given in Chapter Two, we have defined

(1 +v)(1—2v)

G( , )=
62 E52

 

[(A + Bz)e_z‘5 + (C + one“ 115(6)

If we use Equations (2-17), Equations (3-2) can be rewritten as follows.

1+B(2—v)+D(1—2v)+A6—C6=0 (3-4a)
(A + C); - 2(3 — D)v = 0 (3-4b)
Age—h?” -C.6e”5 (1 + v)— Be‘h‘f (1 — kg) — Dehé (1 +116) = 0 (3-40)
Age—’75 + Céehé: + 36"“ (2 -4v+ h6)+ Dehé‘ (—2 +4v+ kg) = o (3-4d)

The variables A, B, C and D can be found from solving the foregoing equations,

i.e.

= 28”; (—2+4(1 -e2”5 )v2 41262 +v(5—3e2h5 + 2115))

 

 

 

A (3-5a)
6(3+e"”f (3—4v)—4v+2e2hi(s—12v+sv2 +2h24‘2))
B : e2“ (—1 + e2” (—3 + 4v) + 2126) (3%)
3 +e4”9‘(3-4v)-4v+ 2e2h5(5 -12v+8v2 + 211262)
2((3 — 4v)v+ e2” (2 + 4v2 + 12262 + v(-5 + 216») (3-5c)

— 6(3+e4”5 (3 —4v)—4v+2e2"5 (5-12v+8v2 + 212262))

41

3 —4v+e2”5(1 + 2kg)

(3-5d)
3+e4”5(3—4v)—4v+ 2e2”5(5 -12v+8v2 + 2h262)

B:—

 

3.2 Elastic Layer on Elastic Base
Shown in Figure 5 is an elastic layer situated on a semi-infinite elastic base. It
can be considered as an extension from the previous case which is an elastic

layer situated on a rigid base.

 

 

 

p(r)
ll . ll _
L 0 I T
r
Layer1 28“—’
h
Layer2
V Z

 

Figure 5 Elastic layer situated on elastic base.

To simplify the boundary conditions, there is no friction at the contact surface and
layer 1 is perfectly bonded to layer 2. The thickness of layer 1 is h and the
contact radius is a. If we use superscript to represent the layer, the boundary

conditions for this contact problem can be summarized as follows

atz=0

42

09’ =p(r) (3-6a)

:2.) = o (3-6b)
at z = h

09’ =02?) (3-6c)

A? = r)? (3—6d)

11(1) : “(2) (3-6e)

.9) = .9) (s-er)

The function of G(<§,z)for layer 1 and layer 2 can be written, respectively, as

 

 

61(532) = (1 + ”2):; 2'“ )1(A1 + Bide—“5 + (Cl + Dlz)e251i(6) we
I
02(62) = (”’2’“ g2’2)1(A2 + 322)e-z€ +(C2 + 022)” ]17(5) (3-7b)

524‘

For layer 2, the stress at a point far from the origin vanishes. Therefore, 02 and

02 shall be zero and function GZ(§,z) becomes

(1+v2)(l—2v2)

2 (A2 + 1322):”; 17(6) (3-7c)

02(622) =

 

529‘

43

With the use of Equations (2-18) and Hankel’s inverse transformation, the

boundary conditions, i.e. Equations (36), become

at z = 0
0:“) =1 (3-8a)
6;“) = o (3-8b)
at z = h
0:0) = (5(2) (MC)
6332‘” = 5,52) (3—8d)
14:“) = 1):”) (3-8e)
1):“) z u:(2) (Hf)

Substituting Equations (2-17) into the foregoing equations and defining 02 =0 and

02:0, we have

1+Bl(1-2v1)+01(1—2v1)+Alg—Clg=0 (3-9a)

—281v1+2D1v1+ A1§+C1§ = 0 (3-9b)

44

Ale-”15 6- Aze‘hl'56 — Clehléé- 3167’": (—1 + 2v] 4116) (3-9c)
—32e‘h15(1—2v2 +h16)— Dleh15(—1+2v1+h16)= 0

Ale‘hl4‘6- Aye—”156 + Clehléé + Bze—hlg (2v1 — r216) (3-9d)
— ale-”1592122 + 111;) + ole-”15am + 1116) = 0

_ Ale—”150 + v1) + Clehl‘fa + v1) + Age—h1§(1+v2)+ Ble‘h1‘5(1+v1)(1—h16)

 

 

 

 

El 51 E2 E19"
+ Bze’h15(1+ v2)(-1 +h16) + Dleh1‘5(1+ v1)(-l + 1116) = 0
E25 515
(3-9d)

Ale‘h1§(1+ v1) + Cleh1§(l+ v1) _ Aze‘hl‘5(1 + v2)

 

 

 

E1 El E2
+ 325’": (1 + v2)(-2+4v2 -h16) + 8184115 (1 + v1)(2—4V1 + 1116‘) (3-9e)
E16 E15
+ Dieh1§(1+V1)(—2+4V1 + 1116) = 0
E16

Solving the six equations simultaneously, we can identify A1, 8,, C1, D1, A2 and
82. With the use of Mathematica ®, the computational work can be significantly
reduced since the formulae can be manipulated for the convenience of
programming and the potential errors can be reduced to minimum. Details of the

computational scheme can be found in Appendix D.

45

3.3 Two-Layer Composite on Rigid Base

The formulation of a two-layer composite situated on a rigid half base will allow
us to evaluate the effect of interfacial bonding condition on composite
performance. In this investigation, we assume there is no friction on the contact
surface and there are perfect bonding conditions between layer 1 and layer 2,

and between layer 2 and the rigid half base.

p(r)

H.

r
Layer 1 26"—> m

 

 

Layer2 IE2
/////////////////

Rigid Base

 

172

Figure 6 Two-layer composite on rigid base.

The boundary conditions for this problem are as follows

at z = O
0 £1) = P0) (3-10a)
IS) = 0 (3-10b)

46

atz=h1

0'9) = of,” (3-100)

6);) = z)? ‘ (3-10d)

u,(.1) = u,(.2) (3-10e)

119) = u?) (3-10f)
at z = h] + h2

14,12) = 0 (3-109)

115,2) = o (3-10h)

The function of G(.§,z) for layer 1 and 2 can be written, respectively, as

 

 

G1 (6.2) = (1+ ”2):; 2’“ ’ [(A1 + Ewe—if + (C1 + D1z)ez‘v‘ 117(6) (Ma)
1
02(62) = (1 + ”2)" 7121qu + 3220f“; + (C2 + Dzz)e2"='117(6)(3-11b)

526

Combining Equations (2-18) and (3-10), and using Hankel’s inverse

transformation, we have

atz=0

47

0:“) :1 (3-123)

7:)” = o (3-12b)
at z = h]

0;“) = e29) (3-12c)

1;“) = 632‘” (3—12d)

11:“) = 15(2) (3-12e)

u:(l) = 15(2) (3-12f)
at z = h] + h2

11:”) = o (3129)

11;”) = o (3-12h)

Substituting Equations (2-17) into the above boundary conditions, we Obtain the

following eight equations

l+Bl(l-2V1)+D1(1—2v1)+Alé—Clg=0 (3-133)

231v1+201v1+ A1§+C1§ = 0 (3-13D)

48

Ale‘h156 - Aze“”1§6 -— Cle“"1‘5 6 + Czehl‘56 + Bze‘h15(—1 + 2v2 — I216)
+ 31947150 - 2v1+h1§)- Dleh1§(—1 + 2v] + hlg‘) (3-13c)
+Dzeh1‘5(-1+2v2 +h14‘) = 0

Ale‘hlé‘g- Aze'hl‘v‘g + Cle'h15 6 —C2e”156 + Bze_h1§(2v2 —h16)

(3-13d)
+ ale-”15(-2vl + 1116) + Dlehlg (21)] + i116) + Dzeh15(—2v2 — r1116): 0

_ Ale—hl‘ffl +111) + Clehl‘fa + v1) + Aye—”150 + v2)
51 El 52
__ Czeh15(l+ v2) + Ble‘h1§(1+ v1)(l 4116) + Bze'h15(1+ v2)(-1 + 1216)
52 515 526
+ Dleh1§(l+v1)(l + I216) _ Dzeh1§(1+ v2)(1 +1116) z
1514‘ 526

 

 

 

(3-13e)

0

 

 

Ale—”150 +121) + Clehléa +v1) _ Aze’h15(1+v2) __ Czehléfl +122)
El El 52 52
+ Bze’hl‘f (1 + v1 )(—2 + 4121 — 1116) + ale-”15 (1 + v1)(2 — 4v] + 1116)
525 526
+ Dlehlgfl +v1)(—2 +4121 + 1216) _ Dzeh15(1+ v2)(-2 +4v2 + I116) =
515 526

 

 

 

(3-13f)

O

 

 

49

_ Aze-(hl ”'2 ”(1 + v2) _ 3267“" “’2 )5 (I + V2 )(2 - 4V2 + (116 + I726)

 

 

 

 

 

 

E E
2 25 (3-139)
_ C2e(hl ”2); (1 + v2) _ 0262“” ”2’5 (1 + v2)(—2+4v2 + h]: + I126) _ 0
E2 526
_ Aze‘IhI +h2)‘5(1 + v2) + 326“" “’2 “5(1 +122 )(1 4116 - (226)
E
52 2‘5 (3-13h)
+ C2e(h1+h2)5(1+ v2) + Dze(”1+"2)5(1 + v2)(1 +1116 H126) _ 0
E2 525

Solving the eight equations simultaneously, we can obtain solutions for A1, B1,
C1, D1, A2, B2, C2 and D2. In order to simplify the mathematical work, it is an
advantage to use Mathematica®. Details of the procedures are given in Appendix

E.

3.4 Validations

To the author’s best knowledge, there is no analytical solution for the stress and
displacement components of the contact problems except the case investigated
in Chapter Two, i.e. a half elastic space loaded by a point force. Hence,
numerical solutions are sought to evaluate the stress and displacement
components for the three contact problems mentioned in this chapter. In this
research, Gaussian integration method is used for the stress and displacement
integrations. With the use of Mathematica to simplify the formulations, C++
programs are developed to evaluate the integrations. The following validations

are then carried out:

50

1. Validate the numerical solution for the point-loaded half space problem
with comparison to the analytical solution.

2. Validate the numerical solution for the problem of an elastic layer on a
rigid half base with comparison to the analytical solution for the point-
loaded half space problem.

3. Validate the numerical solution for the problem of an elastic layer on a
rigid half base with comparison to the analytical solution for the point-
loaded half space problem by largely increasing the thickness of the
elastic layer.

4. Validate the numerical solution for the problem of the two-layer composite
on a rigid half base with comparison to that of an elastic layer on a rigid
half base by largely increasing the thickness of layer 2.

5. Validate the numerical solution for the problem of the two-layer composite
on a rigid half base with comparison to that of an elastic layer on a rigid
half base by using identical material properties for the two layers.

3.4.1 Numerical integration vs. theoretical solution

In Chapter Two, we derive the theoretical solution for the elastic half space under
point load. The range of the integration is from 0 to 0° for the parameter 5.
However, this kind of integration is not possible in the numerical process.
Gaussian integration is thus used to evaluate the integration and to determine

the integration upper limit of parameter 9‘ of Equations (2-25).

51

f(§)

 

 

 

64 points 64 points 64 points
A A [_A_\
Y \ ......
1 1 I 1 g
0 a1 32 ...... a19 a20
6

Figure 7 Gaussian point integration algorithm.

The numerical integration uses 64 Gaussian points. The integration interval is
divided into many small sections, typically 20 sections as shown in Figure 7, to
achieve better convergence and accuracy. The parameter I; is normally chosen

to be 40 for a unit contact radius.

52

 

’’’’’’’’’

' __o-

  
 

-(').04 AW
1 -0.06 ——z=0 exact
ii“ -0.08 —----z=1 exact
g -------- z=2 exact
i3 '0'10 --—-z=3exact

1: z=0 numerical

o z=1 numerical)
A z=2 numerical i
_o z=3 numerical i

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0
rla

Figure 8 Numerical solution vs. exact solution for elastic half space.

As shown in Figure 8, the numerical integration and exact solution matches very
well. The Gaussian integration points and the upper limit of g for the integrations
are therefore used subsequently to evaluate more complex contact problems in

this research.

3.4.2 Validations among cases

Solutions for the following contact cases are obtained: (1) a half space, (2) an
elastic layer on a rigid half base, (3) an elastic layer on an elastic half base and
(4) a two-layer composite on a rigid half base. Among these solutions, some
cases should share the same solutions if the thicknesses and material properties

of layers are well chosen. For example, case (2) should become case (1) if the

53

thickness of the elastic layer in case (2) approaches infinity. Similarly, case (3)
should reduce to case (1 ), if the elastic layer and the base of case (3) has the
same material properties as the half space of case (1 ). Similar comparisons can
be applied to cases (3) and (4) and others. Only the aforementioned

comparisons will be presented in this research.

(i) Comparisons between case (1) and case (2)

If we increase the thickness of the elastic layer of case (2) to infinity, we should
arrive at the same result of case (1 ). For a spherical indentation, if the layer
thickness is 30 mm, Young’s modulus is 7GPa, Poisson’s ratio is 0.33 and the
contact radius is 1 mm, the comparisons of stress and displacement components
between case (1) and case (2) are shown in Figure 9 to Figure 12. Case (2)

reduce to case (1) very well.

54

    
   

r—B- 2:0 ha—Ifispac;

 

 

40") ' +z=1 half space 1 i
E 40.0 +z=2 half space 1
E. —e—z=3 half space . l
6 430.0 ‘ —-)I(—z=4 half space 1 i
+z=0 one layer on rigid
40-0 ,. —°—z=1 one layer on rigid I
+z=2 one layer on rigid j
'50‘0 ‘ —0—z=3 one layer on rigid
1 . .
-60.0 , f—z=4 one layer on rrgid i y
0.0 1.0 2.0 3.0 4.0 5.0 6.0
r (mm)

 

 

 

 

 

 

 

1.0 ~ ~ ]
0.0
'1‘0 ‘ -B—z=0 half space 71 1
A -2.0 +z=1 half space
‘ g +z=2 half space '
g -3 0 - -e—z=3 half space ’
g +z=4 half space i
'4-0 +z=0 one layer on rigid 1
—0—z=1 one layer on rigid I
-5.0 1 . . ‘
+z=2 one layer on ngrd I
-6.0 -0-z=3 one layer on rigid i
___:z=4 one layer on rigid I
-7.0 . . 1
0.0 1.0 2.0 3.0 4.0 5.0 6.0

r (mm)

Figure 10 Comparison of shear stress between case 1 and case 2.

55

 

 
  

—B— z=0 one layer
—e—z=1 one layer
( +z=2 one layer
1- —e—z=3 one layer
" +z=4 one layer

....... i +z=0 one layer on rigid I

  

 

.7 . -
1.2E-02

1.0E-02

8.0E-03

Uz (mm)

4.0E-O3 -

2.0E-03

1 0.0E+OO - 1
0.0 1.0 2.0

Figure 12 Comparison of normal displacements between case (1) and case (2).

6.0E-03 A

 
  
   
   

 

//////

nnnnnnn

nnnnn

r‘ .
n"-
"r‘

a--- "n
fihHHS-Q--.'.c "u

--
:::::::-“’c..... ~-~
....,,,.: ....
-..

..

 

_ -a—z=0 one layer 1 1
[ —e—z=1 one layer I i
‘ +z=2 one layer 1 1
I -e—z=3 one layer 1 1
1 +z=4 one layer .
i —-—z=0 one layer on rigid ; l

—o—z=1 one layer on rigid {
' -—z=2 one layer on rigid {
I —o—z=3 one layer on rigid 1|
w :-z=4-91)e layer on rieig

  

-‘.

 

 

56

-, ‘ -o—z=1 one layer on rigid .
1' —t—z=2 one layer on rigid i
" 1 —0—z=3 one layer on rigid 3
:1 A_,_E=4£D? ILL/er QMIQE .'

i

(ii) Comparisons between case (1) and case (3)

The case consisting of an elastic layer situated on an elastic half base, i.e. case
(3), can be reduced to the half space case, i.e. case (1 ). The numerical results
are given in Figure 13 to Figure 16 based on the foregoing material properties

and thickness.

 

1 —13—z=Oonelayer ‘1
—e—z=1 one layer I
I +z=20nelayer I
1 —e—z=3onelayer ‘
_ +z=40nelayer

—o—z=0twolayer I
I —o—z=1twolayer I;
—-—z=2twolayer I
—-z=3twolayer I
—z=4twolayer 1

 

I 0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

1 r (mm)

 

Figure 13 Comparison of normal stresses between case (1) and case (3).

57

 

 

     

' —e—z=1onelayer
+z=20nelayer
1 —e—z=3onelayer
+z=40nelayer !
—-—z=01wolayer 1 1
—+—z=1twolayer I
1 —~—z=2twolayer I,
' +z=3Molayer I“ (
I —z=4twolayer J1

l

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 14 Comparison of shear stresses between case (1) and case (3).

8.0E-04 ,

 

 

 

 

1 mm 1.12.11

1 —e—z=1 one layer ‘I

4.05414 1 I + z=2 one layer 1

—e—z=3 one layer

- "-"l' -, ,_ 1 +z=4 one layer 1’

’5‘ 0.0E+00 I —o—z=0 two layer I

g 1" 2:, -o—z=1twolayer|I

a. ;. (151‘? if. h '—.—z=21wo|ayer 'I
3 H 3'5 " '

4015414 N,» 1 _._z=3 two layer I,

,_ ' 1.-,-z=,4;!1veleyer I

-8.0E-04 « U. 1

‘.‘ 2'1 I

-1.2E-03 1 1 1 1 I

0.0 1.0 2.0 3.0 4.0 5.0

Um) I _ I

Figure 15 Comparison of radial displacements between case (1) and case (3).

58

—e:z_=06ne‘laye? 1
1 —e—z=1 one layer ’1
I1 +z=20nelayer1|

l

  
 
 
 
 
 
  
 
 

I —e—z=30nelayer I
+z=40nelayer

I —-—z=0twolayer I

—o—z=1twolayer 1 1

: —-—z=2twolayer >

I —o—z=3twolayer i

  

 

 

4.015433 .

I If—z=4twolayer 1 ,
‘ 1
I 2.0E-03 1 1
1 1 1
1 0.0E+00 1 1 1 A 1'
0.0 1.0 2.0 4.0 5 o '

Figure 16 Comparison of normal displacements between case (1) and case (3).
(iii) Comparisons between case (2) and case (4)

If the two layers of case (4) have identical material properties, it will hold the
same results as case (2). Details of the results are shown in Figure 17 to Figure

20.

59

  
   
 

—a—z=0 one layer I, -

'10'0 +z=1 one layer 1 1
I 13400 ‘ +z=2 one layer ‘ 1
5 -e—z=3 one layer 1 ‘
1 :300 +z=4 one layer I ‘ ‘
I b ' —-—z=0two layer I
I -40.0 -o—z=1 two layer 1

1 —-—FZIwolayer )
'50'0 * —o—z=3twolayer 1 l
‘ -—z=4twola r I

«60.0 . 1 2-- W - ye” ,

0.0 1.0 2.0 3.0 4.0 5.0 6.0 1

r (mm)

 

 

 

 

1.0 1
1
0 0 _ 1'3: :2: :13: 3:: 2:: 3:: 2:: :2: 22:: 22:77:.” ‘" "

1 —1.0 1
7; -2.0 - ‘ —a— z=0 one layer I I
g . —e— z=1 one layer l

I 4,-3.0 1 —a—z=2 one layer I
E —e— z=3 one layer 1
b .4_o + z=4 one layer 1

1 1 --— z=0 two layer ‘

-5.0 —o—z=1 two layer 1
1 + z=2 two layer '
—6.0 —¢— z=3 two layer I
1 -—z=4 two layer 1
-7.0 . T i ,
0.0 1.0 2.0 3.0 4.0 5.0 6.0

r (mm)

Figure 18 Comparison of shear stresses between case (2) and case (4).

60

 

        
 

-a— z=0 one layer I
—o—z=1 one layer I

—A— z=2 one layer I
f —e— z=3 one layer
‘ +z=4 one layer I
: —-— z=0 two layer I
- —o—z=1 two layer
—~— z=2 two layer
_._ z=3 two layer I
I —z=4 two layer I I

 

 

.0 3.0 4.0 5.0 .
r (mm) I,

I 0.0 1.0 2

I 1.0E-02 , ,, a V, a,” tea-:-

I I —e—z=0 one layer I

7 ' —o—z=1onelayer II

I +z=20nelayer II I
—e—z=3 one layer II

I +z=4 one layer I I

 

   
 
 
 
 
 

I

I 6.0E-03 I —o—Z=0 two layer I I

I? I —~—z=1twolayer II I

I —-—z=2two layer I

IE. 40E-031 I —o—z=3twolayer I I
s l

-— z=4 two @ng .I

  

 

 

I

ZOE-03 l I

I W _. _. A_ I I
I -2.0E-03 I I
I 0.0 1.0 2.0 3.0 4.0 5.0 ‘
I r(mm) I

Figure 20 Comparison of radial displacements between case (2) and case (4).

61

In the foregoing comparisons, we have simplified the complex cases to simple
cases. We have validated the numerical solutions with the analytical solution for
the half space problem, the problem of an elastic layer on a rigid half base with
the problem of half space, the problem of an elastic layer on an elastic half base
with the problem of half space, and the problem of two-layer composite on a rigid
half base with the problem of an elastic layer on a rigid half base. These
calculations serve as cross-validations for the derivation and programming of the
numerical scheme. With all these exercises, we have successfully developed and
validated the numerical scheme. In the next chapter, we will use the numerical

scheme to study more applications.

62

CHAPTER FOUR BASIC APPLICATIONS

We have validated the solution formulae and numerical scheme for studying
contact problems in the previous chapter. In this chapter, we use the numerical
scheme to study some basic contact problems, such as effects due to material
property, thickness, boundary condition, loading condition and lamination of
layers. Material property has strong effect on both stress and displacement
distributions. We study it first. Finite thickness, especially thin layer, can alter the
stress and displacement distributions in a half space. It is investigated secondly.
Base condition and loading condition are another two fundamental elements for
mechanics analysis. They are also critical to the response of materials subjected
to contact forces. They are examined subsequently. The last part of this chapter
is focused on the lamination effect of composite layers. We confine our studies to
composites with perfect bonding interfaces. Only the lamination with different

material properties and thicknesses are of interest.

4.1 Effect of Material Property

Three different materials, such as steel, aluminum and nylon, are used in the
following investigations. They represent hard, intermediate hard and soft
materials. For simplicity, we choose to investigate the half space problem loaded
by a point force. The material, loading condition and boundary condition are
summarized in Table 2. We use relatively small force and contact radius in the
studies to avoid the possible numerical overflow in the C++ program. These

values can be scaled by 100 times to 1,000,000 times without changing the

63

numerical values during calculations. For the rest of the chapters, we continue to

use Newton (N) for force and millimeter (mm) for length.

Table 2 Material properties for steel, aluminum and nylon.

 

 

 

 

 

 

Steel Aluminum Nylon
Total Load P 100 N 100 N 100 N
lndentor Shape near-point load near-point load near-point load
Young’s Modulus 200,000 MPa 7,000 MPa 2,500 MPa
Poisson Ratio 0.29 0.33 0.40
Contact Radius 1.0 mm 1.0 mm 1.0 mm

 

 

 

 

 

 

Shown in Figures 21 and 22 are the elastic stress distributions. As expected,
they are identical for all different materials. These results can also be attributed
to the equilibrium equations, Equations (2-2), which are not dependent on any
material property. The displacement distributions, however, are dependent on the
material property. They are higher for softer materials, which have lower Young’s

modulus, as illustrated in Figures 23 and 24.

64

  

 

 

I 0.0
I -2.0 I.L—9_:z=—0STIT _'I
I I +z=1STL II
I '40 +Z=2 STL I
I -e-z=3$TL - I
I -6.0 - I +z=4 STI. I
A +z=0 AL . I I
I g -8.0 I —o—z=1 AL I I
I ‘u' 100 I +z=2 AL I I
I b - . I —o—z=3 AL
. +z=4 AL
I 42.0 I _....z=0 Nylon I
I {14.0 I I —o—Z=1 Nylon I | I
‘ +Z=2 Nylon I I
I -16.0 .I I _._z=3Nylon I I
' —l—Z=4 Nylon I ;
-18.0 i flPM ' I
I

0.0 1.0 2.0 3.0 4.0 5.0 6.0

-B— z=0 STL I
+z=1 STL
+ z=2 STL I
—9— z=3 STL I
+ z=4 STL
—a— z=0 AL I
—o— z=1 AL I
—-¢— z=2 AL I
—o— z=3 AL I
+ z=4 AL I I
—0— z=0 Nylon I
—+—Z=1 NYIOD I
+ z=2 NYIOl'l
—o—z=3 Nylon I I

:fl'fiflj I

 

 

5.0 6.0 I
I

 

Figure 22 Shear stress distributions based on different materials.

65

-5.0E-04 -

Ur (mm)

I 10504 a

-9.0E-04 '

 

 

I -1.1E-03 It #Ifii I fly I
I 0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

 

 

125-02 -- . I
l I

1.05-02 a

8.0E-03

6.0E-03 I

Uz (mm)

4.0E-03 ,

 

I 2.05-03 I

 

 

Figure 24 Normal displacement distributions based on different materials.

66

4.2 Effect of Thickness

In this section, we study the thickness effect on the stress and displacement
distributions in contact problems. An elastic layer situated on an elastic half base
and subjected to a spherical indenter is selected for investigation. The elastic
layer is made of aluminum while the elastic half base is made of nylon. The
thickness of the top layer, aluminum, varies from 1 mm to 8 mm. Table 3

summarizes the contact problems.

Table 3 Material properties used for thickness study

 

 

 

 

 

 

 

Layer 1: Aluminum Layer 2: Nylon
Total Load 100 N 100 N
lndenter Shape spherical spherical
Young’s Modulus 7.000 MPa 2,500 MPa
Poisson Ratio 0.33 0.40
Thickness 1, 2, 4 and 8 mm infinity
Contact Radius 1.0 mm 1.0 mm

 

 

 

 

 

Using the analytical formulae and numerical scheme developed in Chapter
Three, we calculated the stress and displacement distributions and present them

in the following diagrams.

67

   

0'2 (MPaI

+z=0(h=2)I I I

i

+Z=0(h=4)II I

-5o.o I ,
.+§££h:8_)II I

-6o.oI L -2- A, H». _ #I__fi _____ I I
0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

I

 

 

   
 

+z=0.25(h=1)I
+z=o.25(h=2)I
I—o—z=o.25(h=4)I ; I
L+F£§<E§fl II I

Oz (MPa)

 

7 *’I

0.0 2.0 4.0 60

Figure 26 Normal stresses at 0.25 mm below the contact surface.

68

   
 

02 (MPa)
lb
0
O

It¥s—z=0.5(h=1)II
I+z=0.5(h=2)II
. I+FO.5(h=4)II
40-0 ' +2-05(h-8)I

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0

   
 

 

 

5.0 I, i 2, # —v I

0.0 I ._ I

,.v .~2222' 9...._..~ I

-5.0 r222 2222 I

is 40.0 I I

E. I I

"‘ 45.0 I I-

-20.0 I+Z=1 (h=2) I

.250 ’ -‘°‘Z:1(h=4)II
-30.0 .2 ,, . 2 ,2

0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 28 Normal stresses at 1.0 mm below the contact surface.

69

The normal stress distributions at the contact surface between the indenter and
the top elastic layer are identical for all different thicknesses of the top layer.
Apparently, the contact stress distribution is of a local phenomenon and is not
affected by the thickness of the layer underneath it. As the top elastic layer
becomes thicker, the normal stress distribution at the same depth below the
contact surface becomes higher. This is because the elastic half base, which is
made of nylon, has lower stiffness than the top elastic layer, which is made of
aluminum, and tends to release the normal stress distribution. This phenomenon
can also be identified from Figures 26 and 27 in which the normal stress
distributions for h= 2, 4 and 8 mm are almost identical for z= 0.25 and 0.5 mm
and from Figure 28 in which the normal stress distributions for h= 4 and 8 mm

are almost identical for z= 1mm.

The shear stress distributions vanish identically on the contact surface for all
cases, as shown in Figure 29, due to the free shear traction boundary condition.
When the top elastic layer becomes thicker, the shear stress distribution at the
same depth becomes lower. This phenomenon is opposite to that of normal
stress distribution. Moreover, at the interface between the top layer and the
bottom half space, the shear stress distribution for h= 1mm reduces significantly,

as shown in Figure 32.

The radial displacement distributions, shown in Figures 33 to 35, change from
negative at the contact surface to positive at z= 1mm. Whether it is positive or
negative, the thinnest top layer always has the largest deformation. The normal

displacementdistributions, shown in Figures 36 to 38, are related to the shape of

70

the spherical indenter. As the top layer becomes thicker, the vertical

displacement becomes smaller.

I 1.05-11 ; , ~ , I

’ » —9-z=0 (h=1)I

I+z=0(h=2)I '

5-0E'12 i I+z=0(h=4)

E Iii—,FQthB). ’

' E 0,0E+OO W i

I u .

b I I
-5.0E-12-I I
I |

-1.0E-11 - .

0.0 1.0 2.0 3.0 4.0 5.0 6.0

r (mm)

Figure 29 Shear stress distribution at the contact surface.

71

   
    

I

I—e—z-025 (h-1)I I

+z=0.25(h=2) I

+z=o.25(h=4)III

I —1t—Z'0.25 (h=8)I

-2000 .. - -2- ,2 __ 2__F___. I
0.0 1.0 2.0 3.0 4.0 5.0 5.0 I

r (mm) I

I
I
I I
I

+z=0.5(h=1)II I
+z=0.5(h=2)II I
+z=o.5(h=4)II I
j:- z=0. 5 (h=8)I I I
I 0.0 1.0 2.0 3.0 4.0 5.0 6.0 _
r(mm) I

 

Figure 31 Shear stress distribution at 0.5 mm below the contact surface.

72

 

an (MP3)

—:—z=1 (h=1)I I
. . +z=1(h=2)I I
I '8'0 " . .‘ +z=1(h=4)I I
I +z=1 (h-8) I I

-10.0 ~: -:— : :~ : :— : I

0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 32 Shear stress distribution at 1.0 mm below the contact surface.

 

 

 

 

 

F F — W—m—“fi ' ” W '
I 0.05+00 - :::—— —: :I I
I 33” AAAAAAAAAAAAAAA I I
I I -' . ‘ w "' " -- v“i“-:‘-:‘-::‘-:::: I I
I 20504 I - I
I E I . ’ “““““““““ I I
I E 405-04 5 I I
I 3 ' " .. I+z=O(h=1) I <
‘ -6_0E-04 I :::=:==.‘""..-.'- I+FO (h=2) I I
I I I+z—0 (h __BI)II I
43.0504 - - : :I :V. I - . »

I I
0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

I r(mm) I

Figure 33 Radial displacement distributions at the contact surface.

73

 

 

l 1.0E-04 l e» f f f — _+__;‘~m if
. z=0.5(h=1)"‘

l l
I l +z=0.5(h=2)‘ l
'» l+z=°5 (h=4)
0.0E+oo .tFO-Lmell l

Ur (mm)

Figure 34 Radial displacement distributions at 0.5 mm below contact surface.

l 8.0E-04 . . ____ ‘1} l

143—z=1 (h=1)i l
I—a—z=1(h=2)‘ w
'—o—z=1(h=4)!.' l

 

l 6.0E-O4 !

  
 
  
  

9

.. 4.05-04; flail L
5 2.0E-O4l '1

4 0-0E+00 . I t V I I V ‘ ‘ ‘ ‘ : : '9” »/‘"”‘“““:- fif17‘?‘T‘=‘:l.=-3.z- -;-:. i ‘

 

 

-2.oeao4l ___.. , w
l 0.0 1.0 2.0 3.0 4.0 5.0 6.0 l

Figure 35 Radial displacement distributions at 1.0 mm below contact surface.

74

7, ,, , ,7 W # i ,1

-e—poapn‘

fs—zmawan‘
—o—Z=0 (h=4)l
fidbffiufifih

 
 
   
 

6.0E-03 ‘

4oEo3“"“

.
. .‘
s :
-
. 3.
. ..
. .
- L“
5‘ “
| b. .,
.- n‘
.__ -‘
.-
-_..
.
O.-

Uz (mm)

  

l

0.0E+00 » , ,i - -_ ,7 , M __ . ;

0o 10 20 so 4o so 6o ‘

’ r(mm) l

14;;QEEEBi
Ls—ppsopm'l
-—o— z=0.5 (h=4)l % ‘
:%:PfléflEQ;i l

  
  

 

‘ 0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 37 Normal displacement distributions at 0.5 mm below contact surface.

75

8.0E-03 v — 7* f * ~ vw=.‘-:f:- =
r —a—z=1 (h=1)' l
3+z=1 (h=2)l !
l—o—z=1(h=4) ,
—-:—z?1 (h=8)fll l

   

 

 

 

Figure 38 Normal displacement distributions at 1.0 mm below contact surface.

4.3 Effect of Base Condition

In this section, we study the effect of different material combinations on the
stress and displacement distributions. The elastic layer situated on an elastic half
base with a spherical indenter is chosen for investigations while steel, aluminum
and nylon, given in Table 2, are used for material combinations. We compare the
results of aluminum indenting on a soft, hard and rigid base. The numerical

results are presented in the following diagrams.

76

10.0 ~ ~ » , ~ l

0.0
l 400
-20.0 «

    

Oz (MPa)

-30.0

:9— z=0 AL/Nylon, .
+ z=0 AL/STL l , l
l + z=0 AL/Rigid j i I

l 40.0 -
-50.0 -

 

 

 

450.0 , . . .
0.0 1.0 2.0 3.0 4.0 -5.0 6.0 l

r (mm) ‘

Figure 39 Normal stress distributions at the contact surface.

Shown in Figure 39, the normal stress distributions at the contact surface for all
three combinations are exactly the same, indicating again the local phenomenon
at the contact surface. As the stiffness of the half base increases, the normal

stress increases, as shown in Figures 40 and 41.

77

   

02 (MPa)
it»
.o
O

 

4;— z=0.5 AIL/Nylon} ‘ l
a + z=0.5 AL/STL ‘ ‘
A+ z=0.5 AL/Rigidj , l

 

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0 l

 

 

02 (MPa)

 

-30.0 1‘- wz—=A1AL/Nyiori'
l l .
+z=1ALISTL l‘ l

+51 ALjRigid ‘,

 

 

i 40.0 r
l 0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 41 Normal stress distributions at z=1.0 mm from contact surface.

78

The shear stress distributions vanish at the contact surface, as shown in Figure
42, due to the free shear traction boundary condition. When the stiffness of the
half base increases, the shear stress distribution decreases for the depth z= 0.5
mm, as shown in Figure 43. However, it increases for the depth z= 1 mm, as
depicted in Figure 44. This is because a harder base can sustain a higher stress
than a softer base. The displacement distributions, both radial and normal
components, as depicted in Figures 45 to 50, can also be interpreted based on

this argument.

 

 

_ F- _ . ___—____ _ _ - ________,___T
l 1.55-11 . _- .2

‘ ~ l
| 1.05-11 l

A 5.05-12 I

. a: l

l .

‘ g ODE-+00 — l

1 5 ~ l
. -5.0£-12 . W
l —9—F0 AL/Nytonif

-1 05-11 l+z=0 AL/STL H ‘

:- FQAHRiQid_§ .

-1.5E-11 : 1 . l

0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

Figure 42 Shear stress distributions at contact surface.

79

  
  

 

:a—z=0.5 ALlNyion. .
—A— z=0.5 AUSTL 1 g
l + F05Afl3‘9fl '

 

 

l 0.0 1.0 2.0 3.0 4.0 5.0 6.0
r (mm)

O’rz (MPa)

 

I—B-z=1AL/Ny|on‘é

 

 

‘ ! +Z=1 AUSTL I
+§1M91<L§
-8.0 . . , l

l 0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 44 Shear stress distributions at z=1.0 mm from contact surface.

80

l
l

2.0E-04
0.0E+00

-2.0E-04 ‘ '

  
 

Ur (mm)

 

405-04
l—a—z=0 AL/Nylonl
505'“ ” ' _ + z=0 AL/STL ‘;
g—x— z=0 AL/Rigid l

43.0504 . » ”fl“.

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0
r(mm) l

Figure 45 Radial displacement distributions at the contact surface.

. 8.0E—05 , ,- 3W , ,

I _ [uh-a— z=0.5 Ail/won , l
6.0E-O5 + z=0.5 AL/STL

'+_z=0.5 AL/Rigid ,

 

 

   

 

4.0E-05

205-05 «

Ur (mm)

0.0E+00

AA
AAAAAAAAAAAAAAAAAA

-2.0E-05

 

 

-4.0E-05 , . .
l 0.0 1.0 2.0 3.0 4.0 5.0 6.0 l

r (mm) '.

Figure 46 Radial displacement distributions at z=0.5 mm for contact surface.

81

 

   

8.0E-O4 " i __k W..

l +z=1ALlNylonl Q
60E-04 7 .-_-.»_--, 1—1§—Z=1AL/STL ‘ f
l L+z=1ALIRigid1 5
l E 4.0E-04

E
l t
* D ZOE-04
N 20504

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0
‘ r (mm)

    
  
 

 

 

 

5,- _ ,_. if- *‘fi
8.0E-03 1- -, w»- l
+z-OALINylon. 1
‘ 6.0E-03 I—A—Z-‘O ALISTL l l
+z=0 AURigid l

‘ l
l 5 405-03 - I l
5 l ,
5' 20503 ~ ,
l .
l l
0,0E+00 MW ' l
l ,
405-03 , . . . 3 l

0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 48 Normal displacement distributions at the contact surface.

82

l 305-03 5.5 _

  
 
 
 

 

 

3 +iuz=V0.5 AUNyton V

' l+§95 AA_L/_R:ig_i_<_[ l
l

l g 405-03 » l

l g ‘==:-- l

l 5 205-03 -. l

[ 0,0E+00 W 1

l

-2.05-03 ~~ . , 4 l

0.0 1.0 2.0 3.0 4.0 5.0 6.0 8

Figure 49 Normal displacement distributions at z=0.5 mm from contact surface.

 

   
 
 

? 3.05-03- —— ,~ 1

l I—a—F1AL/Nylon‘
l+z=1AUSTL l l

l 605-03» =,_—*—m_z_=1AL/Rigi_d_l l

l a '

, 5, 405-03

l N

l D

l 2.05-03 ‘

l

 

 

‘ 0.0E+OO ::.-.'---- - .,. _ . : j .
0.0 1.0 2.0 3.0 4.0 5.0 6.0 E

Figure 50 Normal displacement distributions at z=1.0 from contact surface.

83

4.4 Effect of Loading Condition

In this section, we study the effect of loading condition on stress and
displacement distributions. The half space problem (made of aluminum) is
chosen along with various load conditions, eg. near-point load, uniform stress,
flat punch and spherical loading, for the investigations. All cases have an

identical total force although the geometry of the indenters is different.

At the contact surface, we see numerical oscillations for all cases except the one
subjected to a near-point load, which has exact solution. The spherical indenter
has the highest maximum normal stress followed by flat punch and then by
uniform stress. The near-point load gives the highest maximum normal stress.
Besides, it should be noted that the maximum normal stress is located away from

the origin except for the near-point load case.

84

 

 

 

 

 

 

 

 

, 200.0 — ,
l l
l 0.0 ,
, -200.0 - '
l E
‘ E -400.0 3 ,. l
g M __ ._ . l
l -600.0 ‘ + z=0 pomt l l
, l + z=0 uniform l '1 l
l -800.0 i . + z=0 flat l
‘ i--2_=0_sphefica|_si l
) -10000 4 . . . ;
0.0 1.0 2.0 3.0 4.0 5.0 6.0 i
l r (mm) ;
Figure 51 Normal stress distributions at contact surface.
I i v ’i’** l
‘ 50.0 , ,
l
‘ l
l
‘ 73‘ l
o.
g l
l 3 l —9— F05 point l l
* 100 0 _ 4+ z=0.5 uniform l l
' . 9 + z=0.5 flat 1 5 ,
fi , —-— z=0.5 spherical 3 ' l
l -1500 , . . . 4: l
0.0 1.0 2.0 3.0 4.0 5.0 6.0 l

Figure 52 Normal stress distributions at z=0.5 mm from contact surface.

85

  
 
  
 

 

-5.0 l

l -10.0

. r; -15.0 i

l a

I g -20.0 1 . ._ _ j

g -250 1 l—B—Z=1 DUI-m l I 1
+z=1 uniform

'30'0 —o—z=1 flat
'35'0 w air—"z=1 spherical] *
-40.0

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0

r (mm)

 

 

0.0
-2.0
l A -4.0 l
l E
.5. '60 _ ________,
l g -80 ,-a—z=2 point :
1 ' ‘ +z=2 mifoml
40.0 ,4—Z=2flat ‘. ‘
l«an—z=2 spherical," ‘
-12.0 . -

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0
r (mm)

Figure 54 Normal stress distributions at z=2.0 mm from contact surface.

86

 

 

 

 

 

  
 

 

 

 

 

 

l 0.0 1.0 2.0 3.0 4.0 5.0 6.0

l 0.1

l 3 l

* E 0.0 W . '

l . - 55 l
‘3 —a—z=0 pomt l .

l -0.1 l+z=0 uniform l . l

l +Z=0 flat ‘ - l
l—n—z=0 sphericall .

l 02 w ' *7 l

l 0.0 1.0 2.0 3.0 4.0 5.0 6.0 l

r(mm) !

5 5 5 _ _ 555 v 5 5,5l

Figure 55 Shear stress distributions at contact surface.

l ’7 7 i 7 7 + Al

‘ 5.0 ., - 1

0.0 l l

-5.0 I ll

1; -10.0 ~ ? l

“- l l

E -15.0 l i:
E II , !

° -20.0 - ,—a—z=0.5 point l' ‘

‘ 25.0 _. ‘—a—z=0.5 uniform ll ‘
l 30 0 , + z=0.5 flat '5 l

l ‘ ' “ - l:;=0.5 sphericalli ,’

-35.0 - . . , l i :

l

Figure 56 Shear stress distributions at z=0.5 mm from contact surface.

87

   
 
   
 

 

psi—z=1 point 7
l+z=1 uniform l7
y-o—z=1 flat l
‘—-—z=1sp_hericalj‘ l
-12.0 i . 5 . , , l
0.0 1.0 2.0 3.0 4.0 5.0 6.0

l r (mm)

 

 

 

O’rz (MPa)

Ca— z=2 point
l+z=2 uniform i‘ l
i —o— z=2 flat , l

i -+- z=2 spherical l

l
l '3.5 i . l I l ‘
' 0.0 1.0. 2.0 3.0 4.0 5.0 6.0 l

 

 

 

Figure 58 Shear stress distributions at 2:20 mm from contact surface.

88

0.0E+00

   

 

 

 

l
l
1 40504 < l
|
l E l
1 5E, -2.05-04 - .
! I- ._ . _ 555 5 l
i :J 1 —a— z=0 point i ‘ l
3.05-04 : + z=0 uniform i i l
1 ——-— z=0 spherical i . ,
-4.0E-04 . 1— ' .
0.0 1.0 2.0 3.0 4.0 5.0 6.0 l

r (mm) l

 

 
 
   
 
  

1.5504 - ’

—a— z=0.5 point . l

; 1+z=0.5 uniform l l
l 1‘0“” ‘ +z=0.5 flat l l
l E ‘--z=0-5spheri¢al_l l
E, 50505 - 1
5 l
0.05+00 l

l

-5.05-05 i

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 60 Normal displacement distributions at z=0.5 mm from contact surface.

89

9.0E-05 1 555_5 5

 
  

 

 

l 1.1-.51 point I 1
7.0E-05 ens—z=1 uniform

I I+z=1flat I i
I 7; 5.0505- -:Z_=1$phericaL l
l IE? . l
* a 305-05- I
l l
I 10505 I
l 4.0505 .- , . . I
00 10 20 30 40 50 60 ?

l r(mm) l
_ 5____ _ 5

Figure 61 Radial displacement distributions at z=1.0 mm from contact surface.

 

I 5.0E-05 . " l.

  
  

 

+z=2 point “I I

I 4.05-05 +z=2 uniform ll "
I l—o—z=2 flat I i
l ‘g 3.05-05 ,—»—z=2spherical. 1
, .5, 1
i 5 2.05-05 » I
l 1.05-05 ~ " . l
l

0.05+00

 

 

Figure 62 Radial displacement distributions at 2:20 mm from contact surface.

90

5, 55-5 - I
I —a— z=0 point 7
+ z=0 uniform I

305-03 I + 50 flat ' 1

‘_--_z_=9_s59£ei£c.aU

Uz (mm)
M
O
rn
5:
OJ

 

 

I 0.05+00

 

 

., _ ._,_ _ ._ _ 5 I

l—a— z=0.5 point I . I
‘ + z=0.5 uniform l I I

8.0E-04 -. I —o— z=0.5 flat I I
I I l —-7 z=0.5 spherical Ij - I
l

Uz (mm)

l
I
i
|
i

 

I
|
1
l
I

 

 

, 0.0E+00 , - I
0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 64 Normal displacement distributions at z=0.5 mm from contact surface.

91

I—a—z=1 point
+z=1 uniform 1
—o—Z=1 flat
“5"‘51595'Efi93'3

  
  

 

 

0.0E+00 * 1 1 ,
0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

r (mm) 1

  
 
 

4.0E-04 . ___5__5
—a— z=2 point .
25329 . + z=2 uniform I.
3.0E-04 ' ’ '
—o— z=2 flat

» +Z=2 §Ph§§9ah 1
2.0504 - ‘ 1
l

|
l

U: (mm)

1.0E-04

 

 

0.0E+OO
0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 66 Normal displacement distributions at z=2.0 mm from contact surface.

92

4.5 Effect of Lamination

In this section, we study the effect of lamination on the stress and displacement
distributions. The two-layer composite situated on a rigid half base and subjected
to a spherical indenter is chosen for investigation. Two cases are studied: the
ratio of Young’s moduli and the ratio of thicknesses between the two layers. For
the ratio of Young’s moduli, we use nylon/aluminum (the ratio E1/EZ= 0.0356)
and aluminum/nylon combinations. For the ratio of thickness, we base it on
nylon/aluminum. More specifically, we fix the thickness of the top nylon layer and
change the thickness of the bottom aluminum layer so the relative location of

interest to the loading surface can be maintained the same.

From the calculations, we can see the stress distributions do change with the
thickness ratio, as shown in Figures 67 and 68. The displacement distributions,
however, change with the thickness ratio. As the thickness ratio between the first
layer and the second layer decreases, the radial displacement distributions

increases slightly while the normal displacement decreases.

93

0.0 -

  
 
  

 

 

-5.0
-100 -
f E -15.0
1 g x
~ -200 1 5 #- '1 1
1 b I—a—z=1 (111/112:2)
; -25.0 - +z=1 (111/112:1) 1 I
. 1 ,
-30.0 . —e—z=1 (h1/hZ-0.5)
_:’F—Z=515£‘1’h2=5052.5).
-35.0 5 1
0.0 1.0 2.0 3.0 4.0 5.0 6.0 1
r(mm) I

 

 

0.0
1
-2.0 » ‘
. A I
I g -4.0
, .5. 1
E '6 O .5. ,5 ~- - 555 I
b - +z=1 (111/112:2)
+z=1 (h1/h2=1) I
'80 1 +z=1 (111/112:0.5) I
1 tFlihflM-Zfilj ‘
-10.0 - ~ 1 1
0.0 1.0 2.0 3.0 4.0 5.0 6.0 1

r (mm)

Figure 68 Shear stresses at the bonding interface for nylon/aluminum

94

1.2E-04 5 5 5, -- .

  
  
 
  
  

 

 

 

I 1451111111241 1 1
1.0504» 1+F1(h1/112?1) I 1
I1 1+z=1 (111/112:0.5) I
’5 8.0E-05 1 116.211- (h1/h2=0.25)1
I
1 E, 6.0E-05-
'5 .1
4.0505 4
20505 I
0.05+00 I

0.0 1.0 2.0 3.0 4.0 5.0 6.0 1

Figure 69 Radial displacements at the bonding interface for nylon/aluminum.

I , , 5
1‘ 6.0E-04

 

'A"—1a_'/z'=_1(h1mz=2) 11
1+z=1 (h1/h2=1)
+z=1(h1/hZ=0.5) I

  
  
  

5.0E-04

 

 

4.0E-O4 ..
1 A 1+z=1_(h1/h2=0.25)Ig
E 3.0E-04 i
I 3 2.0504 - :I
I 10504
0.05+00 - “"=‘:: 5????“ : f f .. . .
-1.0E-04 . 5 5'
0.0 1.0 2.0 3.0 4.0 5.0 6.0
r(mm)

Figure 70 Normal displacements at the bonding interface for nylon/aluminum.

95

Similarly, we investigate the stacking order of aluminum/nylon, E1/E2= 28.1. As
the thickness ratio decreases, the normal stress decreases as shown in Figure
70. However the shear stress increases as shown in Figure 71, and, both radial

and normal displacement distributions increase.

 

  
  

 

2.0 i m- . mam. _
0.0 -
-2.0 ~
1; -4.0
O.
I g -6.0 ,., I
I 6‘

-8.0 . —e— zi=_ir(h1/l12=_2)n __ II I

' +z=1 (h1/h2=1)

 

 

-1o.o ..
420 ‘-e—z=1(h1/h2=0.5) II
' ' I—x—z=1(h1/h2=0.25)li
-14.o - * . “ . I
0.0 1.0 2.0 3.0 4.0 5.0 6.0

r (mm)

Figure 71 Normal stresses at the bonding interface for aluminum/nylon.

96

AAAAA
vvvvvv

---------

   

Orz (MPa)

£— z=1(h1/h2=2) If
I +z=1 (h1/h2=1) .
I—e—z=1 (h1/h2=0.5) I I

I+ z=_1 (h1/h2=0.25u I I

 

 

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

r (mm)

Figure 72 Shear stresses at the bonding interface for aluminum/nylon.

 

 

 

  

 

 

Ii , i i i, _

I 6.0E-0 ,mi , - , - i: .

I __ -B-F1 (h1/h2=2) II I

I 50504 A ' I+F1(h1m2=1) I: I

I—e—z=1(h1/l12=0.5) I

I 4 0504 f - Late—z=1 (h1/h2=0.25) ; I

I '5‘ ' I

E. 305434 s f" I

I 5 /[“n

D 2“” I

I 2.05-04 - I

I I: I

105-04 I I I I

"='f::;::; I

I 0.0E+00 . I " . ' I
I 0.0 1.0 2.0 3.0 4.0 5.0 6.0

Figure 73 Radial displacements at the bonding interface for aluminum/nylon.

97

. —a- z=1d(h1lh2=Zi)

  

 

 

4.0E-O3 = +z=1 (h1/hZ=1)
+z=1 (h1/l12=0.5)
E 3-05'03 ”ex +z=1(h1/h2=0.25)I
I “é‘ * ._ A— -s-“_.
I E, 205-03
| N
I D ::I:'-
1.05-03 - N
0.0E+OO s
-1.0E-O3 , I
I 0.0 1.0 2.0 3.0 4.0 5.0 6.0 I

Figure 74 Normal displacements at the bonding interface for aluminum/nylon.

4.6 Cylinder Shape lndenter

We can also study the cylinder shape indenter for each case with the
development in Chapter 3. Unfortunately, only the uniform stress in a cylinder
shape has a mathematical solution for the Hankel transform on stress on
boundary condition p(r). The others cases need numerical solution for the
surface. Here we will use the half space to demonstrate the solution. The stress

and displacement distributions for aluminum material are shown in the following
figures. We have set r1=2.0, r2 =3.0, a=1.0 for the calculation. The other cases

are straightfon/vard and not discussed here.

98

p(r)

 

 

 

 

 

 

 

 

 

r (mm)

Figure 76 Normal stress distributions for cylinder shape uniform stress

99

2.0

  
 

1.5 +50
+z=0.5 I I
I 1'0 “ I+Z=1.0 I .
I a“? 0,5 - I—O—z=1.5 I I
o. -' - !
E I " I I
I b -0.5 - I
-1.0 I
-1.5 ~ I
-2.0

 

 

0.0 2.0 4.0 6.0 8.0 10.0 I

r (mm) I

Figure 77 Shear stress of distributions for cylinder shape uniform stress

2.05-05 ~s #—
105-05 -

0.0E+OO

 

 

A -1.0E-05 - ,, -

E ‘J + z=0 I

I a 4.0505 I —-— z=0-5 I
, I + z=1.0
‘ -3.0E-05 I —0— z=1.5

I -*- F20 I

-4.0E-O5 ‘ I

0.0 2.0 4.0 6.0 8.0 10.0 I
r (mm) I

Figure 78 Radial displacement for cylinder shape uniform stress

100

3.0E-04 -.

  
 
 
 

 

 

I+Z=O I l I
2.5E-04 ’ I—I—Z=0.5 I

I I I

+Z=1.0

‘ 2.0E-04 ~ I
I I—o—z=1.5! I I
~ ’g 1.5E-04 ,- +Z=£~9I I I
I g I I
I 5' 1.0E-04 .
I I

I .
, 5.0E-05 . I

I
0.0E+OO I I I
0.0 2.0 4.0 6.0 8.0 10.0 I

r (mm)

Figure 79 Normal displacement for cylinder shape uniform stress

101

CHAPTER FIVE CONTACT THEORIES

In this chapter, we will summarize some important results obtained in the
previous chapters by presenting the relationship between the applied load and
the maximum deflection for each case. These relationships, also called contact

theories, will be useful in engineering practices.

5.1 Half space

For the case of half space, there is an explicit expression between the load and
the deflection and the maximum deflection at the location at z=0 and r=0. If we
substitute z=0 and r=0 into Equation (2-36f) for a point load, we have the

following relationship

a:
P: ” ‘; E*Uz (5-1a)
(1-v )

 

This equation is the contact theory for a point load exerted on a half space and

has also been obtained by other researchers.

For other loading conditions, we don’t have explicit relationship from the previous
derivations. The relationships between the loads and the deflections, however,
are implicit and complicated except at point (0,0). These relationships, for point

(0,0), can be obtained by using Mathematica® and are given below

(#02

2 sz (5-1b)
(1--v )

uniform pressure P =

102

a

 

 

flat indentation P = 2 E *Uz (5-1c)
(1 ‘V )
spherical indentation P = 4" 5* U2 (5-1d)
3(1 — v2)

For all half space cases, the individual load and the deflection relationship is
linear at point (0,0). However, the coefficients of them vary with the shape of the
indenter. We may use the same manner to evaluate the deflection at point (0,0)
for cases other than half space. However, due to the complexity of the integrals
involved in the derivations, we are not able to extract the relationships
analytically. Therefore, a numerical method should be used to determine the
relationships between the load and the deflection and between the stiffness and
the ratio of layer thickness. In the following sections, we will use the spherical
indenter as an example to develop these correlations for the cases of a single
layer on a rigid base, a single layer on an elastic base and a two-layer composite

on a rigid base.

5.2 Single layer on rigid base

Firstly, we focus on the co—relationship between the load and the deflection.
Since our derivations have been based on the linear elastic theory, the load and
the deflection should have a linear relationship. Aluminum was chosen for this
study, which had a Young’s modulus of 70,000 MPa, a Poisson's ratio of 0.33, a

material thickness of 1 mm. The following relationship can be achieved:

103

 

 

 

 

1200 -
1000 y = 299936x + 0.0002
R2=1
800
.2”:
8 600 —
'5
u.
400 ~
200 <
0 i . I
0 0.001 0.002 0.003 0.004
Deflection (mm)

 

 

Figure 80 Force vs. deflection for single layer on rigid base.

From the force-deflection diagram shown in Figure 80 and the associated least-

squares fitting line, the load and the deflection, as expected, have a perfectly

linear relationship since the square regression R2 = 1. If we define the stiffness
as the load divided by the deflection, we can study the relationship between the

stiffness and the material, i.e. Young’s modulus.

104

15805 ' y = 2.10211x + 0.023 . !
1.4E+05 R2 =1 ‘
1.2E+05 I I
1.0E+05 . I
8.0E+04 I

6.0E+04

Ple (Nlmm)

I 4.0r—:+04 I
2.0E+04 , I

I 0.0E+00 . I I I
0.05+00 2.0E+04 4.0E+04 6.0E+04 8.0E+04 I

l
Young's Modulus E (MPa) I

 

 

 

 

Figure 81 Stiffness vs. Young’s modulus for single layer on rigid base.

Figure 81 shows that the stiffness and the Young’s modulus has a linear
relationship. The layer thickness can be normalized by the contact radius which

is directly related to the radius of the spherical indenter. We will investigate the

relationship between U—PE and the ratio of thickness and contact radius, h/a.
Z

105

30.0

 

 

25.0 5"
20.0 , y=3.8149x'°‘6389

9': R2=0.908

a 15.0

I a
10.0 «
5.0 -
0.0 . . i I
0 l 2 3 4 5 6
hla

Figure 82 Relationship between stiffness and h/a.

The exponential fit seems to give the best fit to the data points given in Figure 82.
Therefore, the case of a single layer on a rigid base can be represented by the

following equation

j—0.6389

P=3.8149*E*U2*I1’- (5-2)

a

5.3 Single layer on elastic base

We have developed a contact relationship for a single layer situated on a rigid
support. In reality, most of the base is not really rigid, but elastic. In this section,
we will use the same numerical approach to develop the contact model for the

case of a single layer on an elastic base.

106

 

 

1200

1000

800

600

Force (N)

400

200

 

y = 133105x + 0.0002
R’- = 1

0.002

 

I

0.004 0.006 0.008
Deflection (mm)

 

Figure 83 Force vs. deflection for single layer on elastic base.

107

 

Figure 85 Normalized stiffness vs, the ratio of Young’s moduli.

Stiffness PIU (Nlmm)

90001

 

 

 

‘ o 51:70000 I
80001 -
If” a 51=20000
70001 *+ A 51=30000I
60001 :28 x E1=40000 I
50001 "W + E59200
40001 A
30001 - 000°m°°°m°°m°°°m
20001 oooooooooooooooooooo
10001
1
0.0 5.0 10.0 15.0 20.0 25.0
52/51
Figure 84 Stiffness vs. the ratio of Young’s moduli.
8.0
7.0 9
I
6.0 I
A 5.0
N
l” I
g. 4'0 y=1.4666x’°'9058 I
‘1 3.0 R2=0.9977 I

2.0 «
1.0 ~

0.0

 

 

0.0 5.0 10.0 15.0 20.0
E2IE1

108

From Figure 83, we can conclude that the stiffness increases with the Young’s
modulus of the top layer. We further normalize the data points by dividing them
with the Young’s modulus of the bottom layer, i.e. E2. The results are shown in
Figure 85. The normalized curves imply that there is a linear relationship
between the stiffness of PM and the Young’s modulus of the top layer with fixed
a=1 and h=4. A simple contact model can be proposed to reflect the correlation

given in Figure 85:

P=1.4666*U*E2*[%
1

-0.9058
I (5-3)

We can see, as the ratio between the Young’s moduli increases to 10, the curve

levels off, implying the supporting base is approaching rigid.

1.2 ~ y= 0.7936x-02658 I I
1 R2 = 0.9769

 

Stiffness PIUIE2

 

Figure 86 Stiffness vs thickness changes

109

Figure 86 shows that the stiffness is also associated with the thickness. As the
thickness increases, the stiffness is approaching to a half space result. With
known materials, such as E1=70000 and E2=200000 for this case, the following

model can approximately represent the contact relation:

—0.2658
] (54>

P=o.7936*u*52*[—‘
a

Combining equations 5.3 and 5.4, we have

E —0.4529 h
P=1.0788*U*E * —2 —‘- (5-5)
2 E

—0.1329
. J

5.4 Two layer composite on rigid base

We use a similar approach to study the case of two-layer composite situated on a
rigid base. With the numerical scheme developed earlier, we can identify the
effect of the layer materials, if no simple explicit analytical relationship can be

idenfified.

110

 

1200 *

 

 

 

1000 y = 133105x + 0.0002
R2=1
800
2
§ 600
O
I.L
400
200
0 fig
0 0.002 0.004 0.006 0.008

Deflection (mm)

 

 

 

Figure 87 Force vs. deflection for two-layer composite on rigid base.

111

 

 

y = 1 .6383x'0'854
R2 = 0.9951

PIUIEZ

 

 

0.0 5.0 10.0 15.0 20.0

 

 

 

Figure 88 Relationship between stiffness and the ratio of Young’s moduli.

From the calculated data points, a contact model is presented as follows:

—0.854
51] (5-6)

P=l.6383*U*E2 *[
51

Similarly, the stiffness and the thickness ratio are calculated and plotted in Figure

89. The relationship is

P=0.9423*U*Ez *[ifl—

—0.3962
2 ]

Combining Equations 5.6 and 5.6, we have

112

E —0.427 It —0.l83l
P=l.2425*U*E2*[—2—) [A] (5-8)

Stiffness (PIUIEZ)

51 112

 

2.5
2.0 ,
1.5 y = 0.9423X-08962
R2 = 0.9705
1.0
0.5
0.0 1 1

 

 

0.0 1.0 2.0 3.0 4.0 5.0 1
h1/h2 :

Figure 89 Stiffness vs. thickness ratio of the two layers.

113

CHAPTER SIX ADVANCED STUDY

In this chapter we will apply the techniques developed in previous chapters to
study cases involving contact interface and bonding interface. First of all, we will
derive a general formulation for cases with shear loading condition. Combining
normal and shear loading conditions, we will be able to study frictional effect at
indentation surfaces. We will also incorporate a shear slip condition into
interfaces of layered materials to simulate cracks in the layers and to investigate

the effect of the interfacial bonding on the performance of the layered materials.

6.1 Shear Loading

All the equations derived for normal loading in previous chapters can be
extended to shear loading. Here, the two-layer half space case is presented to
demonstrate the extension. it then is used to study the friction effect at the

contact surface.

114

9(r)

 

Layer2

 

Figure 90 Two-layer half space with shear loading.

For simplicity, there is no normal loading at the contact surface and the two
layers are perfectly bonded. The thickness of layer 1 is h and the contact radius
is a. If we use superscripts (1) and (2) to represent layer 1 and layer 2,
respectively, the boundary conditions for this contact problem can be

summarized as follows:

2 = 0
020) = 0 (6-1a)
ti? = q<r> <6-1b)
z = h
0;” = a?) (649

115

r2) = r2” (6—1d)
ug) = 11:2) (6-1e)

ugh =ug2> (6-1f)

The function of G(§,z) for layer 1 and layer 2 can be written, respectively, as

 

0105.2) = (1+ ”EX; 2‘ 2‘“ ) [(A1 + 812)!“ + (C1 + 012x229” 1cm) (6-2a)
l
Gztm = (1* ”2)“ ' ”Nu/12 + 3206—25 + (62 + 022W 151:) (6-2b)

 

5252

For layer 2, the stress at a point far away from the origin vanishes. Therefore, C2

and Dz shall be zero and function 02022) becomes

021:,z>= (”VZXIQZVZ’W + hoe-Zing) , (6-2c)

526

 

With the use of Equations (2-18) and Hankel’s inverse transformation, the

boundary conditions, i.e. Equations (6-1), become

0:“) = 0 (6-3a)

.116

7:9) :1 (6-3b)

z = h
0:0) : 0:;(2) (MC)
4,“) = 62‘” (6-3d)
21:“) = 15(2) (6-3e)
u:(1) : u:(2) (5-30

Similarly, if we use Equation (2-19) and replace 13(5) with (7(6) in the stress

function of 605,2), we then have the following equations:
31(1—2v,)+01(1—2v1)+Alg—Clg=0 (6-4a)

-]-231vl+201v1 +A1§+Cl§=0 (64D)

A1§*A2§*C162h1§§-31(-1+2V1‘h1§)-32(1-2V2 +’714‘) (6-4c)
—Die2”15(—1+2vl +1115) = 0

_ 73/116 _ _ _
A16 Azé-Cle 3: 32(2vz h1§>+31(2V1+/71§)

(6-4d)
- 01e2h15(2v1 + 121:) = 0

117

 

 

 

_ A1(l+v1) + C1(1+vl)e2”lcf + A2(l+v2) + Bl(l+v1)(l-h1§)
El 51 52 E15

 

 

 

 

 

(6-5e)
+ 32(1 + V2)(‘1 + 1114:) + 010 + V] )(1 +hl‘kahlé: ___ 0
E26 E15
Al(1+vl) + C1(1+V1)€2h1§ _ A2(1+V2) + 32(1+Vl)(‘2+4"2 -h11§)
E1 E1 E2 E1; (6'50
+ Bl(l+v1)(2-4v1 44114") + 01(1 +V1)("‘2+4’V1 +hlg)82hlé : 0
E16 £15

From Equations (6-5), we can determine A1, B1, C1, D], A2, and 32. Details of

the procedures are given in Appendix F. Based on the above equations, we can

evaluate the internal stress in contact conditions involving shear loading.

6.2 Contact Problems with Frictions
In the previous section, we developed a technique to investigate contact
problems involving shear loading. There is a relation between normal and shear

loading in Coulomb friction or sliding friction. It can be defined as
61(5) = f * p(é) (6-6)

where p(g) and q(§) are normal and shear stresses, respectively, and f is the
friction coefficient on the contact surface. In most cases, we consider f a

constant between the two materials contacting each other. We can always
separate the contact force into normal and shear components. The two force
components can be identified independently as shown in Chapter Four and

section 1 of this chapter. Based on linear elasticity, the results from the normal

118

and the shear loadings can be superimposed to represent the total solution. A

C++ program is written to evaluate the case mentioned above. When f = 0, it

represents a smooth contact interface. The cases of f equals 0, 0.05, 0.10 and

0.20 and 0.40 are examined. The results contributed by the shear force only are

presented in the following diagrams. The material properties are listed in table 4.

Table 4 Material properties for friction study

 

 

 

 

 

 

 

 

 

 

 

 

 

   

 

lndenter Shape Young’s Poisson .
and Total Force Modulus Ratio Thickness
Matef'a' 1 Spherical 7.0E+4 MP3 0.33 1 mm
(Aluminum) P=100 N
Material 2 -
43,68.) (compreSSIon) 2.0E+5 MPa 0.29 Half space
0.5
0.0 1 ..........................
-O.5
E
-1.0 -
E4 5
g ..
6 -2.0 I
-2.5 l
-3.0 ‘ +'zé17E0.05)l
-3,5 ‘-&—Z=1 (f=0.1) I
+z=1 (f=0.2L':
-4.0 ’i‘ h
0.0 1.0 2.0 3.0 4.0 5.0 6.0
r(mm)

 

 

Figure 91 Normal stresses at bonding interface between layer 1 and 2

119

 

    
    

* +£— z=1 (f=0.05)l
*—a—z=1 (f=0.1) I l
7*.Z_=1_ “70:21-1

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0 I
r (mm)

 

.-rn” _________________________
'pi ,.
v'. "

 

lie—3:1 Riki?
fat—z=1 (f=0.1) 1
7.6633368);

 

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0
r (mm)

Figure 93 Radial stresses at the bonding interface between layer 1 and layer 2

120

I... . .
0.0E+OO '7,

 

 

 

.szf=:::‘
L -1.0E-06 ’ .A. ‘cacc'
. ‘:::.... I
-2.0E-06 -
x , 114‘“ I
h 3-3.0E—06 n‘.issé;;;§3g.}éé5*‘fiiéiafl I
l .5. I
5 54.0506 « I
} l
-5.0E-06 I M . I
+z=1(f=0.05)l I
-6.0E-06 . +z=1(f=0.1) I I
‘ I ‘—x—z=1(f=0.2) l l
JOE—06 , . , . .

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0 I
r(mm) l

 

I I—s—2¥1(1=0.05)l l
' '—-e—z=1(f=0.1)I I

Ian—z=1 (1:02)

    
 

 

 

 

1 1.0E-05 1
‘ E:: | l
. 5.0E-06 - . I
‘ “::=::; ..... 1' ~-- . I I
I 0.0E+00 . ; i
0.0 1.0 2.0 3.0 4.0 5.0 6.0 1
r(mm) 1

Figure 95 Displacements at bonding interface of layer 1 and 2

121

From the results shown in the foregoing diagrams, the stresses and

displacements both increase as the friction coefficient increases.

6.3 Bonding Theory

Liu et al (1994) used a bonding theory to analyze imperfect composite laminates.
The theory was based on a relationship between the mismatch of displacements
on the opposite sides of an interface and the shear stress on the laminate

interface. The theory is defined as follows:
ul" w?) = krarz (6-7a)
149) — u 12> = kzaz (6-7b)

where u?) represent the displacement along j direction in layer i, and k, and

k2 represent the shear slip coefficient along rdirection and the normal
separation coefficient alongz direction, respectively, at the interface. When k,
and k2 approach zero, it presents a perfect bonding condition. The interface can

consist of an imperfect bonding zone and a perfect bonding zone. A distance d

from r=0 is defined to represent the imperfect bonding zone.

122

p(r)

 

 

H I011 l .
Layer1L—28—* h r
l
Layer2

 

Figure 96 Shear slip theory of two layer half space

In order the solve this problem, the boundary condition in Equations (3-6) are

modified as follows

2 = 0
09) = p(r) (6'83)
.19 = 0 (6-8b)
z==h
09) = a?) (6-8c)
4;) = ,g.) (6-8d)
14$” —u§2) =krorz <6-8e)

123

u?) — u?) = kz oz (6-8f)

6.3.1 Entire Imperfect Interface
When d is infinite, k, and k: are finite, the boundary conditions, i.e. Equations (6-

8), can be expressed as follows with the use of Equations (2-18) and Hankel’s

inverse transformation:

2 = 0
6;“) =1 (6-9a)
1,25” = o (6-9b)
z = h
6:“) = 62(2) (6-9c)
r22") = r13” (6-9d)
fl” -1132) =60; <6-9e)
11;“) w?” =kzcr§ (6-90

When we substitute Equations (2-19) into the above functions, we have the

following six equations:

1+Bl(l-2v])+D](l-2v])+Alé-C1§=O (6-108)

124

_281VI+2D1V1+ A1§+C1§ = 0

Ala: - A2; — C1e2h15§ — BI(—1 + 2v] -/11é)+ 32(1— 2122 + 111.5)
— DleZhl‘fG-l + 2v] + I216): 0

A16 - A25 + Clezmé + 32(2vz 7 h1§)+ B1(-ZV1+ h1éf)
+ 01e2h15(2v1+h1§)= o

_ A1(1+v1) + CIe2h15(1 +121) + 31(1 +v1)(l 411;)
151 51 E16
+ Dle2h1‘5(1+v1)(1+h15)+A2(1+v2 4%,)
E15 E2
(1+v2)(—1+h1:)) = 0
529

 

 

 

 

+ 32 (21(er - hlkré: +

A](1+v1) + CIe2h15(1+vI) + B](1+v1)(2—4v1 +1115)

 

 

 

 

El El E15
+ 01e2h1‘5(1+v1)(—2+4v1 +1115) +A2(-l+v2 4:26)
1:315 E2
+ 320-162 + 2km — 121ng + (1+ V2)” + 4V2 415)) = 0

525

(6-10b)

(6-10c)

(6-10d)

(6-10e)

(6-10f)

From Equations (6-10), the problem is solved with help of Mathematica®. A C++

program is also written to obtain the numerical results.

125

6.3.1.1 Verifications

First of all, if both k, and k2 equal zero, we have the same result as perfect

bonding condition as shown in the following diagrams. The material properties

are listed in table 5.

Table 5 Material properties for friction stud)!

 

 

 

 

 

 

 

 

 

 

 

  
 
 

lndenter Shape Young’s Poisson Th' k
and Total Force Modulus Ratio '6 ness
Material 1 .
(aluminum) gamete]! 7.0E+4 MPa 0.30 1 mm
Material 2 .
(nylon) (compreSSIon) 2.5E+3 MPa 0.29 Half space
0.0 ' — — —— _ __ _ w .z _
-1.0 . I
-2.0 I
E -3.0 I
E .
6 -4.0
-5.0 ,-__- ___ ,_ 7 7
+z=1 (kr=0, kz=0)
-6.0 . .
" - —9—z=1 (perfect bonding) .I
-7.0 1 - - 7. - ~ ’7 " ;. f 7 :z ‘i;:"i;t.
0.0 2.0 4.0 6.0
r(mm)

 

 

 

 

Figure 97 Normal stress verification

126

 

 

 

   

 

 

 

 

 

Ur (M Pa)

-0.2 ~
-0.4 1'
-0.6 j ‘
E 1
1: -1.0 I
b I
-1.2 1 I
'1-4 "1 —o—z=1 (kr=0, kz=0) I I
l l
'1-5 ‘ —e—z=1 (perfect bonding) 3 .
-1.8 as I , ---- if” - ’ ‘f—VI
0.0 1.0 2.0 3.0 4.0 5.0 6.0
r(mm)
Figure 98 Shear stress verification
0.0 - __

 

0.0 'I — -— —~ I
I _ +2=1(kr=0,kz=0) I ‘
0.0 1 I
~ —e—z=1 (perfect bonding) I
0.0 VI--- . -. __ 7. : 77:77:- I 7:1
0.0 1.0 2.0 3.0 4.0 5.0 6.0

r (mm)

 

Figure 99 Shear displacement verification

127

 

 

 

Uz (M Pa)

0.0 I

0.0 1

0.0 i
0.0

 

1.0 2.0

 

+ F1 (kl=0, kz=0)

—e—z=1 (perfect bonding) i

 

.7, f’

3.0 4.0
r (mm)

5.0

1‘.
I.

6.0

 

 

Figure 100 Normal displacement verification

6.3.1.2 Variation of shear slip coefficient Kr

If we let kz equal zero, we can study the stress and displacement as a function of

kr. Table 6 lists the material properties used in the calculations.

Table 6 Material properties for friction study

 

 

 

 

 

 

 

 

 

lndenter Shape Young’s Poisson .
and Total Force Modulus Ratio Th'c'mess
Material 1 .
- Spherlcal 7.0E+4 MPa 0.30 1 mm
(11:01.1?
(nylon) (compressron) 2.5E+3 MPa 0.29 Half space

 

From the following results, we can see that when kr is very small (less than

0.00001), the stress and displacement is close to the perfect bonding. Figure 101

and figure 102 show that kr can affect the shear stress more significantly than the

128

 

normal stress. Similarly, Figure 103 and 104 show that kr has more effect on the

shear displacement than the normal displacement.

For kr between 0.0001 and 0.01, some sinusoidal oscillations can be found in the
numerical results. These values of kr represent changes of interface from perfect
bonding to complete debonding. The sinusoidal oscillations is due to the Bessel

function in the numerical evaluation. When the integral range for 5 increases, the

magnitude of the vibration decrease and approaches stable as shown in Figure
102. Since the Bessel function converges very slowly, to achieve a perfect

numerical result would not be realistic.

    

 

 

I 0.0 I
-1.0 ~ I
A-2.0 1 ’ 7 M II I
. m —a—z=1(kr=0) I 1
$73“ +z=1 (kr=0.00001) ‘.
I ~40 1 +z=1(kr=0.00005) I ‘
j °_50 , +z=1 (k=0.0001) . I
I ' —e-z=1 (kr=0.0002) I
76-0 . —o—z=1 (kr=0.001)
-10 ,, +F1 (kr=0.01) I I
+z=1 (kr=0.1) I '
‘8.0 l 1 1 I

 

0.0 1.0 2.0 3.0 4.0 5.0 6.0?
r(mm)

Figure 101 Normal stresses vs. shear slip coefficient kr

129

   
 

  

 

 

 

 

 

 

 

0.0
I .10
I320 l
E30 " —a—‘ z=1 kFO I l
E +z=1 kr=0.00001 I ,
I 04.0 ~ +z=1 kr=0.00005 . l
—e—z=1 kr=0.0001 .

-5-07 , _ —e—z=1 kr=0.0002 I l
-- +z=1 kr=0.001) ‘ w
I 45-01 ~ —o—z=1 kr=0.01) I l
‘ 70 —)|(-Z=1 kr=9.1) . .

l
0.0 1.0 2.0 3.0 4.0 5.0 6.0 I
I r (mm) I

 

 

—B— z=1 (kr=0)

 

1-05-03 7 +z=1(kr=0.00001) I
i +z=1 (kr=0.00005) I
8.0E-04 1 ...... —e— z=1 (kr=0.0002)

I ' ~ I“ + z=1 (kr=0.001)
6.0E-04 ,7 +z=1 (kr=0.01)
I 4'5"???" .....,,_II‘_II-II‘IIIIII ‘¢""~‘.::.:"-'.i 2:1 (”301)

 

l

I

I7 . l
: E 4.0E'04 fl 1:”, “Hi, I
E, 4 w
5 205-04I I
I l
I 0.0E+00 I
I -2.05-04 -I I
l

' 4.05-04 5— - -7- -_ _ .7 7. . __ I
I 0.0 1.0 2.0 3.0 4.0 5.0 6.0 I
l

I r (mm)

Figure 103 Radial displacement Ur vs. shear slip coefficient kr

130

1.0E-02 * FF ~ * - —~- WA ~#—~
I —e—z=1(kr=0)

   
 
 
 
 

 

905-03 5;?“ +2: :1(kr=0.00001I I I
8.0E-O3 +z=1(kr-000005 I
A 7.0E-03 .......... ‘ :§:1I$888%2I I
I E 6.0E-03 '- .- :z_1(g_-8.o1) I
I}: 5.05-03 n.“ 2- ( - .) I I
I3 4.0E-03 " . w . I;¢\::"--- I
3.0E-03I ‘ f ,
2.05-03 I I
1.0E-03 I I
0.0E+00 I. , — . . W . we a I I
0.0 1.0 2.0 3.0 4.0 5.0 6.0'

r (mm) , _ I

Figure 104 Normal displacement Uz vs. shear slip coefficient kr

6.3.1.3 Variation of normal separation coefficient Kz

We have discussed the stress and displacement variation when kz equals zero.
Similarly, we will study the changes of stress and displacement with respect to kz
when kr equals zero. The results are presented in the following diagrams and the

material properties are listed in table 7.

Table 7 Material properties for friction study

 

 

 

 

 

 

 

 

 

lndenter Shape Young’s Poisson ,
and Total Force Modulus Ratio Thickness
Material 1 .
ialuminum) imam 7.0E+4 MPa 0.33 1 mm
Material 2 .
(nonn) (tenSIon) 2.553 MPa 0.29 Half space

 

 

Figures 105 and 106 show that the bonding condition is near perfect when kz is

very small, i.e. less than 0.00001. When kz increases, the normal stress at the

131

bonding interface decreases significantly; however, the shear stress is not as
much affected by the change of kz as the normal stress. Similarly, the normal
displacement at the bonding interface also decreases more significantly than the

shear displacement as shown in Figure 108.

, +z=1(kz=0)
+z=1(kz=0.00001) I
I —e—z=1(kz=0.00005) I
, + z=1 (kz=0.0001) I
I —e—z=1(kz=0.0002) II I
I —o—Z=1 (kz=0.001) I I
. .I +z=1(kz=0.01) I
I +z=1(kz=0.1) IN I
- ._ I
I

 

 

 

I 0.0 I I I 1' , , I
0.0 1.0 2.0 3.0 4.0 5.0 6.0

r (mm)

 

Figure 105 Normal stresses vs. normal separation coefficient kz

132

II
II

 

 

+Z=1ékz=m
160 I +z=1 kz=0.00001)
' +z=1 Ikz=0.00005) I
1 40 I I —e—z=1 kz=0.0001)
- r —e—z=1Ikz=0.0002)
, +z=1 kz=0.001)
1 1.20 I I +2§1Iszo.01) I
I @100 kz—0.1) -0 I
0' . l .. 3
£0.80 ;
, u
I 00.60 -
I 0.40 I 7- I
I 0.20 I
I ~ I
I 0.00 W W -—W
0.0 1.0 2.0 3.0 4.0 5.0 6.0 I
r(mm) .

Figure 106 Shear stress vs. normal-separation coefficient kz

 

0.0E+00

 

 

 

 

-4.0E-04 ,
, ‘. —9—z=1 kz=0)
I"-8 0E-04 +Z=1 kz=0.00001
1 E - kz=0.00005
Iv ¢ o :Fl rig-33??
, h _ _ I “ """"""" z= - .
'3 ”E 03 +2=1Ikz=0.01)
I —e—z=1 kz=0.1)
' -1.6E-03 I
I I “.'\‘.““-‘:-‘;:::3:32;ch

4.05-03 , - W I I I

0.0 1.0 2.0 3.0 4. 5.0 6.0

 

Figure 107 Shear displacement Urvs. normal separation coefficient kz

133

    

—a—z=1 Ikz=0) I

  

|

IA _ _ I +z=1 kz=0.00001)
: E 2'0E 02 +z=1 kz=0.00005)
I: I +FI F8883?) -
I +F Z= . ‘
=3 '3-05'02 . Ikz=0.01)

I i —e—z=1 kz=0.1)

I 405-02 I IIIIIII I
I -5.0E-02 - IIIIIIII I - — -----

0.0 2.0 4.0 6.0

r(mm) I

 

 

 

I_.-‘__.+ ___

Figure 108 Normal displacement Uz vs. normal separation coefficient kz

6.3.2 Imperfect bonding with finite length

II IN =
Layer1 L'Za —’ T r

 

 

 

 

Figure 109 Two-layer composite of imperfect bonding with finite length
For a finite imperfect interface with a radius d, we will study the following problem

to demonstrate some fundamental phenomena. The problem involves a step

134

function. When r S d, kI.(r) = k,. and when r > d, k,(r) = 0 where d is the

half crack length as shown in Figure 104 and i represents either r or z. The

nonlinear shear slip coefficient and normal separation coefficient are defined as

follows:
15(5) = Irk,<r)JI(r:)dr (6-106.)
5(5) = Irkz(r)Jo(r€)dr (6-10b)

Apply equation (6-10) to the above condition, we have

 

 

16(5) = ’2’: w. (d§)Ho(aé‘) -JI(dr:)H. (de (5-123)
. k
19(5) = :2 Mali) (6-12b)

where HI.(a/;) is the Struve function, for i=0 and 1,

3 5
Z Z

2
H =— — + —.-- 6-12c
0(2) ”[2 12-32 1232.52 1 ( )

 

2 z2 z4 z6
Hz=— — + I I —
'() ”[12-3 12-33-5 123-.527

 

...] (6-12d)

The k: and k; distributions are shown in Figures 110 and 111.

135

Kr (i)

0.30
0.25
0.20
0.15

0.10 -

0.05
0.00
-0.05

 

kr=1.0 when r510
kFOD when r>1.0

 

 

15

Figure 110 Shear slip coefficient kr (g) distribution.

0.50
0.40

0.60

1

0.30 ~

0.20
0.10
0.00
-0.10

Figure 111 Shear slip coefficient kz(§) distribution.

 

kz=1.0 when r510
kz=0.0 when r>1.0

20

 

 

136

10
f.

15

We can’t see the direct relationship between the coefficients and the length of the
imperfect bonding, though significant efforts were made to find proper functions
for the shear slip and normal separation coefficients. We plot the coefficients
against 5 and also summarize some critical values from the curves in table 8.
Finding better functions for the shear slip and normal separation coefficients still

remains for future study.

Table 8 Some values for different imperfect bonding length

 

 

 

 

 

 

 

 

 

 

d max kr* «f @743?” { @k1rs; time max kz* f @klfigme
1.0 0.2543 2.4516 5.8843 0.5 3.8317
2.0 1.0174 1.2254 2.9422 2.0 1.9159
3.0 1.9614 0.8172 2.2891 4.5 1.2772
4.0 4.0695 0.6129 1.4711 8.0 0.9579
5.0 6.3585 0.4903 1.1769 12.5 0.7663
6.0 9.1563 0.4086 1.3473 18.0 0.6386
8.0 16.2778 0.3065 0.7355 32.0 0.4790
10.0 25.4341 0.2452 0.5884 50.0 0.3832

 

 

 

 

 

 

137

 

. i
l: l
l l l
. l
20 . ‘ d-8
F i -
l /
i i
15 i
ii . l
. 5 i‘-
10 "
.' \
l ' ‘i
l. . .
‘ "'
x .‘ it u x
f l . i". .
/ _q 9".
Mr. ‘ S ‘4. ‘
'. ; A ,4. . g‘ Y . ‘, xv irr‘frx ._ _ ”u __ H . , 2,,
H ' ‘ l" ’ ‘1"‘4w" 2'.‘ */J 35%.)7773‘11 x.’:‘.;..,"r;13-"'nr"\ ‘rrur ' .P‘.;.',}:Jf;\'.' x l.’ Li~
‘ 1 ' .iflr' 7.. .
l 2 3 4 5 6
_5 L

<5

Kz* 50 (1:10

Figure 113 Normal separation coefficient vs. imperfect bonding length

6.3.2.1 Verifications

Verifications for the technique outlined above include the following special

1) perfect bonding on the entire interface as presented earlier

138

2) any d value with both kr and kz equal to zero, i.e. perfect bonding on the entire

interface

3) d equals zero with random kr and kz. i.e. perfect bonding on the entire

interface

The results are shown together in the following diagrams. The material properties

are listed in table 9.

Table 9 Material properties for friction study

 

lndenter Shape

Young’s

Poisson

 

 

 

 

 

 

 

 

and Total Force Modulus Ratio Thickness
Matef'a' 1 Spherical 7.0E+4 MPa 0.33 1 mm
(aluminum) _
M t _ | 2 P-100 N
a ena (compression) 2.5E+3 MPa 0.29 Half space
(nylon)

 

139

 

1 0.0 ,,,,,,
i -1.0
l -2.0
i $43.0
, .2,
1 64.0 ,v. # __ 7__ 2. __
l _50 - —~—z=1.0(kr=0,kz=0) I
l —o-Z=1.0(d=0) l
I .60 l —e—z=1.0(perfectbonding) l
l -7.0 ' , T5”. ‘ “ ;
, 0.0 2.0 3.0 4.0 5.0
i r(mm)
Figure 114 Normal stress for all three cases.
0.0 1 ’
-0.2 :
-o.4 i
i a -o.e ;
. o.
i E -0.8
l E
l b ‘1-0* “2 .__ __ __ __
; -12 . l —a—z=1.0(kr=0,kz=0) l
14 i +z=1.0(d=0) l
. _ _ . i ‘
i . —°—z=1.0(perfectbonding) !
-1.6 5 fl .._ f -_ ~ 2222.
0.0 2.0 4.0

 

  

 

    

 

  
 

 

6.0

 

2 ' (mm) _

6.0

Figure 115 Shear stress for all three cases.

140

U r (mm)

8.0E-03

U2(mm)

7.0E-04

6.0E-04
5.0E-04
4.0E-04

3.0E-04 ~

2.0E-04

1.0E-04 *

0.0E+00

Figure 116 Displacement Ur for all three cases.

,#_L_ ,___1. 'A 7‘]

—+— z=1.0 (kr=0,kz=0)
+ z=1.0 (d=0)

 

 

0.0

7.0E-03
6.0E-03
5.0E-03
4.0E-03
3.0E-03

2.0E-03 *
1.0E-03 -

0.0E+00

Figure 117 Displacement Uz for all three cases.

 

  
 
 

.0. z=1.0 (perfect bonding) j

1.0 2.0 3.0 4.0 5.0
r (mm)

6.0

 

; +z=1.0 (d=0)

 

 

—o— z=1.0 (perfect bonding

7+F1.bagr;o,kpor 1

) l

 

0.0

1.0 2.0 3.0 4.0 5.0
_ r (mm)

141

6.0

6.3.2.2 Variation of shear slip coefficient kr

Similarly, we will investigate how kr affect the stresses and displacements when
kz equals zero. Using the C++ program established earlier and incorporating kr,
we have the following numerical results. The material properties are listed in‘

table 8 before.

We can see that kr doesn’t poses significant effect on the normal stress and
displacement, as shown in Figure 118 and 121. However, kr affects the shear
stress and shear displacement significantly. Similar to section 3.1.2, there are
some sinusoidal oscillations when kr is between 0.0005 and 0.01. Increasing the
integral range doesn’t not eliminate the oscillations due to the slow convergence
of the Bessel function in the integration. Beyond this range, the results are close
to perfect bonding when kr is very small and the shear stress is very close to

zero when the interface is completely debonded.

142

   

 

,+ z=1 (kr=0) i ,

 

 

l

‘ a -2.0 ‘—a—z=1(kr=0.0001) ;

~ g -30 ,+z=1(kr=0.0005)

7; 4.0 —o—z=1(kr=0.0008)! j

‘ D _5.0 o~z=1(kF0.0009) ,

45.0 r i+z=1(kr=0.001) l _

. +z=1(kr=0.01) ,

'7'0 F —o—z=1(kr=0.1) l _
so. * ' 2.224 .
0.0 5.0 10.0 15.0 1

r(mm)

Figure 118 Normal stress vs. shear slip coefficient kr.

   
 

 

l
—9—z=1 (RFD) i
. +z=1 (kr=0.0001) ‘

45— z=1 (kn-10.0005)
‘ +z=1 (kr=0.0008)
l e z=1 (kr=0.0009)
* +z=1 060.001)

+ z=1 (kr=0.01)

—+—z=1 (kr=0.1) l:

 

 

 

 

-5.0 . F . .
0.0 2.0 4.0 6.0 8.0 10.0 12.0

__ T (m).

Figure 119 Shear stress vs. shear slip coefficient kr.

143

 

 

 

 

Ur (Miss)—

8.0E-04 ——. - - ______ ._ l

  
 
  
 
 
  

 

 

 

 

+¥1 (MEET *7 '

7.05-04 - I+z=1 (kr=0.0001) .
6.0504 g ' .a—z=1(kr=0.0005) f
i +z=1 (kr=0.0008)
5.05-04 - ' 49— z=1 (kr=0.0009) ,
- w +z=1 (kr=0.001) ? 1
4'0E 04 i +z=1 (kr=0.01) ‘ i

30504 - _, l_._z=1 (kr=0.1) J '
2.05-04 l ' i
1.0E-04 l
0.0E+00 . r 5‘ -

 

 

0.0 2.0 4.0 6.0 8.0 10.0 12.0 i
r(mm) i

Figure 120 Radial displacements Ur vs shear slip coefficient kr.

 

 

9.0E-03
8.0E-03
7.0E-03

 

Ke— z=1 (kr=0)

+ z=1 (kr=0.0001)
‘ ~9— z=1 (kr=0.0005)
. + z=1 (kr=0.0008)

 
    
  
 
 

 

 

 

 

6.0E-03 - e z=1 (kr=0.0009)
5 0503 . +z=1 (kr=0.001)

' . +z=1 (kr=0.01)
4.05-03 - g—l—z=1 (kr=0.1) ,'
3.0E-03 - l
2.05-03 4 !
1.05-03 - """"""""" ‘
0.0E+00 ~ . . , .

0.0 2.0 4.0 6.0 8.0 10.0 12.0

Figure 121 Normal displacement Uz vs shear slip coefficient kr.

144

6.3.2.3 Variation of normal separation coefficient kz

If we set kr to zero, we can investigate the variations of stress and displacement
against kz. The results are presented in the following diagrams. It seems kz has
a significant effect on both normal stress and normal displacement as shown in
Figures 122 and 125. The influence of kz to the shear stress and shear

displacement is, however, not significant.

 

 

I+z=1 (kz=0) i
rla—z=1 (kz=0.0001) j
A~—I—z=1(kz=0.0005) l

 
  
 
    
    

 

‘ a; 30 - i.+z=1(kz=0.0009)'! l
N ,+Z-_-1 (kz=0.001) l i
b 2.0 ‘ '

j+z=-1(kz=0.01) ‘ l
L:o—z=1 (kz=0.1) '

.,. ‘\

 

 

 

l . - “5"“? 1

Figure 122 Normal stresses vs. normal separation coefficient kz.

145

 

  
 
     
  
  
 

 

 

 

 

i 1.4 , —9—z=1(kz=0)
l +z=1(kz=0.0001) -
12 - ~~B—z=1(kz=0.0005)
i ’ —o—z=1(kz=0.0008)
. €10 - -—& z=1(kz=0.0009)
l a +z=1(kz=0.001) ,1:
i €08 l +z=1(kz=0.01) i
. —+——z=1 (kz=0.1) g
: D 0.6 - L‘ ----- ~— ' ‘;
l 0.4 . 7 ‘
0.2 f‘ .. .
0.0 l r l T i ‘

0.0 2.0 4.0 6.0 8.0 10.0 12.0
‘ r (mm) i

Figure 123 Shear stresses vs. normal separation coefficient kz.

 

 

 

 

 

 

, 0.0E+00 (
-2.0E-04 '5
'4-05'04 E? +z=1 (RFD)
‘ 43.05-04 g, - -' +z=1 (kz=0.0001) i
0.“? -8 0504 # K _ " ' 4F z=1 (kz=0.0005) ‘
l g ' * +F1 (kz=0.0008)
1 5 -1.0E-03 —O z=1 (kz=0.0009) I
. _1. 2503 . +z=1 (kz=0.001) l
. i +z=1 (kz=0.01) l
‘1-4E‘O3i --+—z=1 (kz=0.1)
-1.6E—03 - ‘ i
-1.8E-03 l ~+~ ~ -~ ~ -- e 2 - l

0.0 2.0 4.0 6.0 8.0 10.0

Figure 124 Shear displacements vs normal separation coefficient ikz.

146

0.0E+00 '

    
 

-5.0E-03
405-02
'1-554’2‘ *:2=1'(‘szO)W ,
,+z=1 (kz=0.0001)l,

TE
11
i E, -2.0E-02 , .
, 5. -0 z=1 (kz=0.0005)

 

-2.5E-02 .. l—o—z=1(kz=0.0008)l
ho z=1(kz=0.0009)'

'3-054’2 :+z=1(kz=0.001) i

l 3.55-02 * +z=1 (kz=0.01) .
40E 02 --*-Z=1 09:01) ."

 

0.0 2.0 4.0 6.0 8.0 10.0 12.0
r (mm)

 

 

 

Figure 125 Normaldisplacements vs. normal separation coefficient kz.

147

CHAPTER SEVEN SUMMARY AND CONCLUSIONS

7.1 Summary of the Research

Indentation of layered composites were studied in this research. A new technique
is presented to solve the layered composites under different indentation
conditions. No theoretical solution can be derived from the final solutions of the
stress and displacement except for a very simple case, since the solution carries
integration from zero to infinity. Therefore a numerical method is used to evaluate
these integrations. Cases of half space, single layer bonded to the rigid base,
single layer bonded to an elastic half space, and two-layer bonded to rigid base
are investigated in this research. Based on the research, mathematical models
are presented for each different case. The mathematical models provide simple
guidelines for the layered composite design. In the advanced study section, the
shear loading at the contact interface was studied. With co-relationship between
shear stress and normal stress by the friction coefficient, friction effect can be
studied at the contact interface. Finally, shear slip and normal separation theories
were incorporated into the technique developed previously. This study provides

some insight into the debonding for the layered composite.

7.2 Conclusions
With the theoretical and mathematical development, we are able to study a few

parameters that have an effect on the indentation.

148

A half space was used to study the material effect. We validated that material
properties have no influence to the stress distribution, while the softer material

intends to have higher displacement.

An elastic layer perfectly bonded to the elastic half base and indented with
spherical indenter was used to study the thickness effect to the stress and
displacement. We found that the normal stress distributions at the contact
surface were identical for layers with different thicknesses. Apparently, the
contact stress distribution is a local phenomenon and is not affected by the
thickness of the layer underneath it. As the top elastic layer becomes thicker, the
normal stress distribution at the same depth below the contact surface becomes
higher. This is because the elastic half base, which is made of nylon, has lower
stiffness than the top elastic layer, which is made of aluminum, and tends to
release the normal stress distribution. The shear stress distributions vanish
identically on the contact surface for all cases due to the free shear traction
boundary condition. When the top elastic layer becomes thicker, the shear stress
distribution at the same depth becomes lower. This phenomenon is opposite to
that of normal stress distribution. The radial displacement distributions change
from negative to positive at the contact surface. Whether it is positive or negative,
the thinnest top layer always has the largest deformation. The normal
displacement distributions are related to the shape of the spherical indenter. As

the top layer becomes thicker, the vertical displacement becomes less.

The same case, an elastic layer perfectly bonded to elastic half space indented

by a spherical indenter, was used to investigate the boundary condition. We

149

found that the normal stress distributions at the contact surface for all three
combinations are exactly the same, indicating again the local phenomenon at the
contact surface. As the stiffness of the half base increases, the normal stress
increases. The shear stress distributions vanish at the contact surface due to the
free shear traction boundary condition. When the stiffness of the half base
increases, the shear stress distribution decreases for the depth z= 0.5 mm, as
discussed in Chapter Four. However, it increases for the depth 2= 1 mm. This is
because a harder base can sustain a higher stress than a softer base. The
displacement distributions, both radial and normal components have the similar

relationship.

We studied the effect of loading condition on stress and displacement
distributions. The half space problem (made of aluminum) is chosen along with
various load conditions, eg. near-point load, uniform stress, flat punch and
spherical loading, for the investigations. All cases have an identical total force
although the geometry of the indenters is different. At the contact surface, we see
numerical oscillations for all cases except the one subjected to a near-point load,
which has an exact solution. The spherical indenter has the highest maximum
normal stress followed by flat punch and then by uniform stress. The near-point
load gives the highest maximum normal stress. It should be noted that the
maximum normal stress is located away from the origin except for the near-point

load case.

The effect of lamination on the stress and displacement distributions was studied.

The two-layer composite situated on a rigid half base and subjected to a

150

spherical indenter was chosen for investigations. Two cases were studied: the
ratio of Young’s moduli and the ratio of thicknesses between the two layers. For
the ratio of Young’s moduli, we used nylon/aluminum (the ratio E1/E2= 0.0356)
and aluminum/nylon combinations. For the ratio of thickness, we based it on
nylon/aluminum. More specifically, we fixed the thickness of the top nylon layer
and changed the thickness of the bottom aluminum layer so the relative location
of interest to the loading surface can be maintained. From the calculations, we
can see the stress distributions do change with the thickness ratio. The
displacement distributions, however, change with the thickness ratio. As the
thickness ratio between the first layer and the second layer decreases, the radial
displacement distributions increase slightly while the normal displacement
decreases. Similarly, we investigate the stacking order of aluminum/nylon,
E1/E2= 28.1. As the thickness ratio decreases, the normal stress decreases.
However the shear stress increases, and both radial and normal displacement

distributions increase.

Furthermore, we developed some simple indentation models for the indentation
condition developed previously. For half space, we came up with an exact
theoretical solution and the following relationships among loading force, material

properties, indenter shape, contact area and maximum deflection:

 

a:
Near point loading: P = 7! C; E * Uz
(l-v )
. (Jr * a)2
Uniform pressure: P = —TE * Uz
(l-v )

151

a

 

 

Flat indentation: P = 2 E * Uz
(l-v )
Spherical indentation: P = 4a 2 E *Uz
3(l-v )

From the numerical solution, we also developed simple models for the layered

composites for a spherical indenter shape. The following models are presented:

Single layer indented against a rigid base:

P=3.8149"‘E"‘Uz"‘(fl

—O.6389
a)

Single layer indented against an elastic base:

E —0.4529 ,1 —0,]329
P =1.0788*U * 152 *[ij {—1)
E] a

Two layered composites indented against a rigid base:

E —0.427 It —0.183l
P=1.2425*U*E2 {—2) [A]
El 112

As advanced study, we investigated the shear loading at the contact interface.
This allowed the author to investigate the friction effect on the internal stress and
displacement. We found that both stress and displacement increase with the

increase of the friction.

The shear slip and normal separation theories were incorporated into the early

development of this research. The results show that when kr is very small (less

152

than 0.00001), the stress and displacement is close to the perfect bounding. The
shear slip coefficient kr can affect the shear stress more significantly than the
normal stress and has more effect on the shear displacement than the normal
displacement. When the shear slip coefficient kr was between 0.0001 and 0.01,
some sinusoidal oscillations were observed in the numerical results. These
values of kr represent changes of interface from perfect bonding to complete
debonding. The sinusoidal oscillations were caused by the slow convergence of

the Bessel function in the numerical integrations.

For the effect of normal separation coefficient, we found that the bonding
condition is near perfect when k2 is very small, i.e. less than 0.00001. When kz
increases, the normal stress at the bonding interface decreases significantly;
however, the shear stress is not as much affected by the change of k2 as the
normal stress. Similarly, the normal displacement at the bonding interface also

decreases more significantly than the shear displacement.

Moreover, we extended the shear slip and normal separation theories to the finite
imperfect bonding interface with the introduction of a step function for the shear
slip and normal separation coefficients. We investigated the effect from both
coefficients. We found that shear slip coefficient kr doesn’t pose a significant
effect on the normal stress and displacement. However, kr affects the shear
stress and shear displacement significantly. Some sinusoidal oscillations were
observed when kr was between 0.0005 and 0.01. Beyond this range, the results
were close to perfect bonding when kr was very small and the shear stress was

very close to zero when the interface was completely debonded. Similarly, we

153

found that the normal separation coefficient kz has a significant effect on both
normal stress and normal displacement, while the influence of kz on the shear

stress and shear displacement is not significant.

7.3 Suggestions and Future Study

Contact mechanics involves mathematical analysis under some mechanics laws.
The development in this dissertation was aimed at filling some voids that were
not addressed in the literature. Sneddon’s (1951) work was used as the base in
the research development, i.e. using Hankel transformation to solve the
biharmonic equations. With the use of computer software packages such as
Mathematica® and Microsoft Visual Studio ®, the goal to tackle the sophisticated
contact mechanics problems became possible, though tremendous efforts in
mathematical manipulation were still necessary. However, Hankel transformation
did not allow the technique to be used for other composites, such as anisotropic
and/or inhomogeneous fiber composites. The technique was limited to

axisymmetric problems.

This research in this dissertation touched some fundamental issues. Many
related issues are still up for grasp. In this research, we only developed contact
models involving a spherical indenter. Other types of indenters should also be
investigated. Furthermore, in order to mathematically solve the imperfect bonding
with finite length, specially transformed shear slip and normal separation
relationships were used. It would be worthwhile to explore other relationships that

allow theoretical solutions to the imperfect bonding problems.

154

Finally, a theoretical study for specially mixed boundary conditions would prove
to be very useful in advancing the contact mechanics. It would allow the true
boundary condition to be exactly matched instead of the semi-guessing involved

in the current practice.

155

APPENDICES

156

Appendix A Mathmatica ® C for General Derivation

H=EO/(2*(l+v))

A=EO*v/((l+v)*(1-2*v))

G0=(l+v)(1-2*v)/(E0*§‘2)((A+B*z)Exp[-
§*z]+(C+D*z)Exp[§*z])*ph
Gl=Co11ect[Simplify[D[GO,z]],{Exp[-§*z],Exp[§*z]}]
G2:Co11ect[Simplify[o[cl,z]],{Exp[-§*z],Exp[§*z]}]
G3=Collect[Simplify[D[G2,z]],{Exp[-§*z],Exp[§*z]}]
oz=Collect[Simplify[§*((A+2*u)G3-
(3*A+4*u)*§‘2*Gl)*BesselJ[0,§*r]],{Exp[-§*z],Exp[§*z]}]
trz=Collect[Simplify[€‘2*(A*G2+(A+2*u)*§‘2*G0)*Besse1J[1,§*r
]],{Exp[-§*z],Exp[§*z]}]
or1=Collect[Simplify[§*(A*G3+(A+2*u)*§*2*Gl)*Besse1J[o,§*r]]
.{Exp[-§*z],Exp[§*z]}]
or2=Collect[Simplify[2(1+0)/r*§‘2*G1*BesselJ[l,§*r]],{Exp[-
§*z],Exp[§*z]}]
oel=cO11ect[Simplify[A*§*(G3—§‘2*G1)*Besse1J[0,§*r]],{Exp[-
§*z] ,Exp [5*2] }]
092=Collect[Simplify[2(A+u)/r*§‘2*Gl*BesselJ[l,§*r]],{Exp[-

§*ZJ,EXpl€*ZJ}]

ur=Collect[Simplify[(1+0)/u*§‘2*Gl*BesselJ[l,§*r]],{Exp[-

6*21 .EXp[§*21 }]

157

uz=Collect[Simplify[§*(G2(A+2*u)/u*§‘2*G0)*BesselJ[0,§*r]],

{EXp[-€*z],EXp[§*z]}l

158

Appendix B Mathematica® Code for Half Space

Half space - General Derivation

138: (1+ v) (1-2*v) / (HHE‘Z) Lwrwp (r) BasaelJ[0, Er] dlr

C0=0

D0=0

u=E0/(2*(l+v))

A=E0*v/((l+v)*(l-2*v))
G0=(A+B*z)Exp[-§*z]+(C+D*z)Exp[§*z]

G1=Collect[D[G0,z], {Exp[-§*z],Exp[§*z]}]
G2=Collect[D[Gl,z], {Exp[-§*z],Exp[§*z]}]
GB=Collect[D[G2,z], {Exp[-§*z],Exp[§*z]}]
ozit=Simplify[§((A+2*u)G3-(3*A+4*u)*§*2*G1)*BesselJ[0,§*r]]
trzit=Simplify[§‘2(A*G2+(A+2*u)*§‘2*G0)*BesselJ[l,§*r]]
orit=Simplify[§(A*G3+(A+2*u)*5‘2*G1)*BesselJ[0,§*r]-
2(A+u)/r*§‘2*Gl*BesselJ[l,§*r]]

oeit=Simplify[A*§(G3-
§‘2*Gl)*BesselJ[0,§*r]+2(A+u)/r*§‘2*G1*BesselJ[l,§*r]]
urit=Simplify[(A+u)/u*§‘2*Gl*BesselJ[l,§*r]]

uzit=Simplify[§(G2—(A+2*u)/u*§‘2*G0)*BesselJ[0,§*r]]
oz = foazita: pI-Id§
0

:12 = farmitar pl-IdE
O

or: Jonit*wd§
0

159

oe= fmwit*plid§
0

ur=Jmurit1~pHd§
0

112 = Inuit. pHd§
0

160

Appendix C Mathematica® Code for Point Loading

Half space - point loading

C0=0

D0=0

A0=2*v/§

BO=1

Ph=(-P/2/N)EXpl-a*€]

u=E0/(2*(l+v))

A=E0*v/((l+v)*(1—2*v))
G0=(1+v)(1-2*v)/(Eo*§*2)((A0+Bo*z)Exp[—
§*z]+(C0+D0*z)Exp[§*z])*ph

G1=Simplify[D[G0,z]]

G2=Simplify[D[G1,z]]

G3=Simplify[D[G2,z]]
ozit=Simplify[§((A+2*u)G3-(3*A+4*u)*§‘2*Gl)*BesselJ[0,§*r]]
trzit=Simplify[§‘2(A*G2+(A+2*u)*é‘2*G0)*BesselJ[l,§*r]]
orit=SimplifY[€(A*G3+(A+2*u)*§‘2*Gl)*BesselJ[0,§*r]-
2(A+u)/r*§‘2*Gl*BesselJ[l,§*r]]

oeit=Simplify[A*§(G3—
§*2*G1)*Besse1J[0,§*r]+2(A+p)/r*§‘2*G1*Besse1J[1,§*r]]
urit=Simplify[(A+u)/u*§‘2*Gl*BesselJ[1,§*r]]

uzit=Simplify[§(Gz-(A+2*u)/u*§‘2*G0)*Besse1J[0,§*r]]

<mz=~fwozithE
0

161

1212 = Lmtmtdg
or = Lmoritdlg
as = Lmoeitdé‘
ur = Lauritdlg'

112 = .J-mUZitdlg
O

162

Appendix D Mathematica® Code for Two-layer Half Space

2 layer half space, perfect bonding with z=0 at top surface,

layer 1 thickness =hl, layer 2 thickness=h2

ul=El/(2*(l+v1))

u2=E2/(2*(l+v2))

Al=El*vl/((1+v1)*(1-2*v1))

AZ=E2*v2/((l+v2)*(l-2*v2))

note: ph was removed from GiO to simplify calculation
G10=(1+v1)(1-2*v1)/(E1*§*2)((A1+Bl*z)Exp[-

§*z] + (C1+D1*z) Exp [§*z])
Gzo=(1+v2)(1-2*v2)/(E2*§*2)(A2+Bz*z)Exp[-§*z]
G11=Collect[Simplify[D[G10,z]],{Exp[—z é], Exp[z §]}]
G12=Collect[Simplify[D[G11,z]],{Exp[-z 5], Exp[z §]}]
G13=Collect[Simplify[D[G12,z]],{Exp[-z é], Exp[z §]}]
G21=Collect[Simplify[D[G20,z]],{Exp[-z é], Exp[z {1}]
G22=Collect[Simplify[D[G21,z]],{Exp[-z é], Exp[z §]}]
G23=Collect[Simplify[D[G22,z]],{Exp[—z é], Exp[z §]}]
Note: §*BesselJ[i,€*r] was removed from oz, trz, ur, uz to
simplify calculation
ole=Collect[Simplify[((Al+2*ul)G13-
(3*A1+4*ul)*§*2*611)],{A1,Bl,c1,Dl}l/. zaO
ozll=Collect[Simplify[((Al+2*ul)G13-

(3*A1+4*u1)*§*2*G11)],{A1,Bl,c1,01}]/. zehl

163

ozZl=Collect[Simplify[((12+2*02)G23-
(3*A2+4*p2)*§*2*G21)],{A2,Bz}]/. zshl
trle=Collect[Simplify[§(Al*G12+(Al+2*ul)*§‘2*G10)],{A1,B1,C
l,Dl}]/.za0
trzll:C011ect[Simplify[€(11*G12+(Al+2*p1)*§*2*G10)],{A1,Bl,c
1,D1}]/.z4h1
trz21=Collect[Simplify[§(22*G22+(12+2*u2)*§A2*G20)],{A2,BZ}]
/.zahl
ur11=Collect[Simplifyl(Al+ul)/ul*Gll*§],{Al,Bl,Cl,D1}]/.
zahl

ur21=Collect[Simplify[(22+u2)/02*G21*§],{A2,BZ}]/. zahl
uzll:Co11ect[Simplifylel2-(Al+2*ul)/ul*§*2*G10],
{Al,Bl,Cl,Dl}]/. zahl
u221=cO11ect[Simplifyiszz-(A2+2*02)/02*§*2*Gzo], {A2,Bz}]/.
zahl

eql=Collect[Simplify[ole+1],{A1,Bl,Cl,Dl,A2,BZ}]
eq2=Collect[Simplify[tr210],{A1,B1,C1,D1,A2,BZ}]
eq3=Collect[Simplify[ozll-oz21],{A1,B1,C1,D1,A2,B2}]
eq4=Collect[Simplify[trzll—rrzZl],{A1,Bl,Cl,Dl,A2,BZ}]
eq5=Collect[Simplify[urll-ur21],{Al,Bl,Cl,Dl,A2,BZ}]
eq6=Collect[Simplify[u221—uzll],{A1,B1,C1,D1,A2,BZ}]
Solve[{eq3:0,eq4:0},{A1,Cl}]

Simplify[%]

Solve[{eql:0,eq2:0},{A1,Cl}]

Simplify[%]

164

Solve [{eql==0, eq2::0, eq3==0, eq4==0, eq5==0, eq6==0,eq7==0, eq8==0} ,{

A1,B1,C1,D1,A2,B2}]

Simplify[%]

165

Appendix E Mathematica® Code for Two-layer on Rigid

Two-layer on rigid, perfect bonding woth z=0 at top surface,

layer 1 thickness =h1, layer 2 thickness=h2

ul=El/(2*(l+vl))

u2=E2/(2*(1+v2))

A1=E1*v1/((l+vl)*(1-2*vl))

Az=E2*v2/((1+v2)*(1-2*v2))

note: ph was removed from GiO to simplify calculation
G10=(1+v1)(1-2*v1)/(El*§‘2)((A1+Bl*z)Exp[-
§*z]+(Cl+Dl*z)Exp[§*z])
020=(1+v2)(1—2*v2)/(Ez*§*2)((A2+Bz*z)Exp[—
§*z]+(C2+D2*z)Exp[§*z])
G11=Collect[Simplify[D[G10,z]],{Exp[-z é], Exp[z §]}]
G12=Collect[Simplify[D[Gll,z]],{Expi—z g], Exp[z §]}]
G13=Collect[Simplify[D[G12,z]],{Exp[-z é], Exp[z §]}]
G21=Collect[Simplify[D[G20,z]i,{Expl-z é], Exp[z §]}]
G22=Collect[Simplify[D[G21,z]],{Exp[-z 5], Exp[z §]}]
G23=Collect[Simplify[D[G22,z]],{Exp[-z é], Exp[z {1}]
Note: §*BesselJ[i,§*r] was removed from oz, trz, ur, uz to
simplify calculation
ole=Collect[Simplify[((Al+2*ul)G13-

(3*A1+4*u1)*§*2*G11)],{A1,Bl,c1,01}]/. 240

166

ozll=Collect[Simplify[((Al+2*ul)G13-
(3*A1+4*u1)*§*2*G11)],{A1,Bl,c1,D1}]/. zehl
0221=Collect[Simplify[((22+2*02)G23-
(3*A2+4*p2)*§‘2*621)],{A2,82,c2,02}]/. zahl
0222=Collect[Simplify[((AZ+2*02)G23-
(3*12+4*02)*§‘2*G21)],{A2,BZ,C2,D2}]/.ze(h1+h2)
trle=Collect[Simplify[§(Al*G12+(Al+2*u1)*§‘2*G10)],{A1,B1,C
l,Dl}]/.ze0
trzll=Collect[Simplify[§(Al*G12+(Al+2*ul)*€62*G10)],{A1,Bl,C
l,Dl}]/.zeh1
tr221=cO11ect[Simplify[§(A2*G22+(12+2*u2)*§*2*G20)],{A2,Bz,c
2,D2}]/.zah1
tr222=cO11ect[Simplify[§(12*622+(A2+2*02)*§*2*G20)],{A2,Bz,c
2,D2}]/.ze(hl+h2)
ur11=Collect[Simplify[(21+ul)/ul*Gll*§],{Al,Bl,Cl,Dl}]/.
zahl
ur21=Co11ect[Simplifyl(12+u2)/02*621*§],{A2,Bz,c2,02}]/.
zahl
ur22=Collect[Simplify[(22+u2)/02*G21*§],{A2,BZ,C2,D2}]/.za(
h1+h2)

uzll=Collect[Simplify[G12-(Al+2*ul)/ul*§‘2*G10],
{Al,Bl,Cl,Dl}]/. zahl
u221=Collect[Simplify[G22-(22+2*02)/02*§A2*G20],

{A2,Bz,c2,02}]/. zahl

167

u222=Collect[Simplify[G22-(22+2*02)/H2*§A2*G20],
{A2,BZ,C2,D2}]/.ze(hl+h2)
eq1=Collect[Simplify[oz10+l],{A1,B1,C1,D1,A2,BZ,C2,D2}]
eq2=Collect[Simplify[trle],{A1,Bl,Cl,D1,A2,BZ,C2,D2}]
eq3=Collect[Simplify[ozll-ozZl],{A1,B1,C1,D1,A2,B2,C2,D2}]
eq4=Collect[Simplify[trzll-trz21],{A1,B1,C1,D1,A2,BZ,C2,D2}]
eq5=Collect[Simplify[urll-ur21],{A1,B1,Cl,Dl,A2,B2,C2,D2}]
eq6=Collect[Simplify[u221-uzll],{A1,Bl,Cl,Dl,A2,BZ,C2,D2}]
eq7=Collect[Simplify[u222],{A1,Bl,C1,D1,A2,B2,C2,D2}]
eq8=Collect[Simplify[ur22],{Al,Bl,C1,D1,A2,B2,C2,D2}]

Solve [{eq7::0,eq8=:0}, {A2,C2}]

Simplify[%]

Solve[{eqlxo,eq2:0,eq3:0,eq4:0},{A1,B1,C1,D1}]

Simplify[%]

Solve[{eql:0,eq2:0},{A1,Cl}]

Simplify[%]

Solve [{eql==0, eq2==0, eq3==0, eq4==0, eq5==0, eq6==0, eq7==0, eq8==0} ,{

A1,Bl,C1,D1,A2,B2,C2,D2}]

168

Appendix F Mathematica® Code for Shear Loading

Two layer half space, shear stress q(r) on top surface, with z=0 at top surface,

layer 1 thickness =h1, layer 2 thickness=h2

u1=E1/(2*(1+v1))

u2=E2/(2*(l+v2))

Al=E1*v1/((1+vl)*(1-2*vl))

22=E2*V2/ ( (1+v2) * (l-2*v2))

note: qh was removed from Gi0 to simplify calculation
G10=(1+v1)(1-2*v1)/(El*§*2)((A1+Bi*z)Exp[-
§*z]+(Cl+D1*z)Exp[§*z])
Gzo=(1+v2)(1-2*v2)/(E2*§*2)(A2+B2*z)Exp[-§*z]
G11=Collect[Simplify[D[GlO,z]],{Exp[-z é], Exp[z €]}]
G12=Collect[Simplify[D[Gll,z]],{Exp[-z é], Exp[z 5]}]
G13=Collect[Simplify[D[G12,z]],{Exp[-z g], Exp[z §]}]
G21=Collect[Simplify[D[G20,z]],{Expi-z é], Exp[z 6]}1
G22=Collect[Simplify[D[G21,z]],{Exp[-z é], Exp[z §]}]
G23=Collect[Simplify[D[G22,z]],{Expl-z é], Exp[z 6]}]
Note: €*BesselJ[i,§*r] was removed from 02, trz, ur, uz to
simplify calculation
ole=Collect[Simplify[((Al+2*ul)613-
(3*Al+4*u1)*§*2*G11)],{A1,Bl,c1,D1}]/. 240
ozll=Collect[Simplify[((Al+2*ul)Gl3-

(3*A1+4*p1)*§‘2*G11)],{A1,Bi,c1,Dl}]/. zahl

169

ozZl=Collect[Simplify[((22+2*02)G23-
(3*Az+4*02)*§*2*G21)],{A2,Bz}]/. zehl
trle:Co11ect[simplify[€(A1*G12+(A1+2*u1)*§*2*G10)],{A1,Bl,c
1,Dl}]/.za0
trzll:Co11ect[Simplify[§(xl*G12+(A1+2*u1)*§*2*G10)],{A1,Bl,c
l,D1}]/.zahl
:r221=cO11ect[Simplify[€(A2*G22+(A2+2*u2)*§*2*G20)],{A2,Bz,c
2,D2}]/.zahl
urll=Collect[Simplify[(21+ul)/ul*Gll*§],{Al,Bl,Cl,Dl}]/.
zahl
ur21=Co11ect[Simplifyl(12+u2)/02*G21*§],{A2,Bz,c2,02}l/.
zahl

uzll=Collect[Simplify[G12-(Al+2*ul)/ul*§‘2*GlO],
{Al,Bl,Cl,D1}]/. zahl
u221=Collect[Simplify[G22-(A2+2*02)/02*§‘2*G20],
{A2,Bz,c2,02}]/. zahl
eql=Collect[Simplify[ozlo],{Al,Bl,Cl,Dl,A2,BZ}]
eq2=Collect[Simplify[rrle—l],{A1,Bl,C1,Dl,A2,B2}]
eq3=Collect[Simplify[ozll-oz21],{A1,Bl,C1,D1,A2,B2}]
eq4=Collect[Simplify[trzll-tr221],{A1,Bl,Cl,Dl,A2,BZ}]
eqS=Collect[Simplify[urll-ur21],{A1,B1,C1,D1,A2,BZ}]
eq6=Collect[Simplify[uzZl—uzll],{Al,Bl,C1,Dl,A2,BZ}]
Solve[{eq1:0,eq2:0},{A1,Cl}]

Simplify[%]

Solve[{eq3:0,eq4:0},{A2,BZ}]

170

Simplify[%]
Solve[{eq5:0,eq6:0},{A2,BZ}]

Simplify[%]

Solve [{eq1::0,eq2::0,eq3::0,eq4::0,eq5::0,eq6::0,eq7::0,eq8::0} ,{
Al,Bl,Cl,Dl,A2,B2}]

Simplify[%]

171

Appendix G C++ Source Codes for Numerical Calculations

172

Appendix G1 C++ source code for half space

// 3D half space

// All units are in SI unit
#include<iostream.h>
#include<fstream.h>
#include <iosfwd>
#include<iomanip.h>
#include<io.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>
//time calculation headfile
#include <time.h>

using namespace std;

main(void)

{
”Running time calculation start
time_t BeginSec, EndSec, Total; // calculate running time
time(&BeginSec);

//**** Setup input and output file
char
lnFile1[80],
OutFile[80];
printf("\n Please enter the input file name: ");
gets(lnFile1);
printf("\n Please enter the output file name: ");
gets(OutFile);

//****Error message of opening files
constchar
ErrorMsg[]="\n***main: unable to open file: ";
fstream
lnMatl(lnFile1,ios::in),
OutComp(OutFile,ios::out);
if (lnMatl.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnFile1<<"\n\n";
exit(-1);

else if(OutComp.fail()){
cerr<<ErrorMsg<<OutFile<<"\n\n";
exit(-1 );
}

173

//Read Gaussian point and weight fucntion x[],w[]
constchar
lnGaussPoint[]="gaussian64.txt";
fstream lnGauss;
lnGauss.open(lnGaussPoint,ios::in);
if (lnGauss.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<|nGaussPoint<<"\n\n";
exit(-1);
}
double x[64],w[64];
int i, j, k;
for(i=0;i<64;i++){
x[i]=0.0;
w[i]=0.0;

while (llnGauss.eof()){
lnGauss>>i;
lnGauss>>x[i];
lnGauss>>w[i];

lnGauss.close();
cout<<setprecision(6);

”generate symmetrucal Gaussian points
//1) copy 0-31 to 32-64 for (0,1 )
for(i=63;i>31;i—){

x[i]=x[i-32];

w[i]=w[i-32];

}
l/2)establish 0-31 (-1,0) backward from 32-64
for(i=0;i<32;i++){
x[i]=-x[63-i];
w[i]=w[63-i];
}

l/****Read imput file
double E=0,Nu=0,hc=0.0,a=0.0,p0=0.0,P=0.0;
int indenter=0;
double r___max=0; // max value of z and r
double a0=0, b0=0; l/integration range
int Nz=0,Nr=0,Nab=0; ”number of points for z and r
while(!lnMatl.eof()){
lnMatl>>indenter; // lndenter Shape
lnMatl.ignore(80,'\n');
lnMatl>>E; // Young’s modulus

174

lnMatl.ignore(80,'\n');

lnMatl>>Nu; // Possion ratio
lnMatl.ignore(80,’\n');
lnMatl>>hc; // layer thickness

lnMatl.ignore(80,'\n');
lnMatl>>a; ”contact radius
lnMatl.ignore(80,'\n');

lnMatl>>a0; // integration lower limit
lnMatl.ignore(80,'\n');

InMatl>>b0; // integration upper limit
lnMatl.ignore(80,'\n');

lnMatl>>Nab; // number of integration zones in [a0,b0]
lnMatl.ignore(80,'\n');

lnMatl>>r_max; // ----9

lnMatl.ignore(80,'\n');

lnMatl>>Nz; //number of points for z -------10
lnMatl.ignore(80,'\n');

lnMatl>>Nr; // number of points for r --—----11
lnMatl.ignore(80,'\n');

lnMatl>>P;

lnMatl.ignore(80,'\n');

l/calculate p0
if (indenter==1){
p0=P/(3.14159*pow(a,2));//***uniform pressure

else if (indente ==2){ //***flat punch - uniform displacement
p0=P/(2*3.14159*pow(a,2));

else if (indenter==3){
p0=3*P/(2*3.14159*pow(a,2)); //***Spherical lndenter
}

lnMatl.close();

OutComp<<"Half Space Indentation"
<<"\n\nlndenter shape="<<indenter
<<" (0- point loading 1- uniform pressure 2-flat 3-sphere)"
<<"\n\nYoung' s Modulus: E="<<E
<<"(Steel 200E+3MPa, Aluminim 70E+3MPa, Nylon 2.5E+3MPa)"
<<"\nPossion Ratio:Nu="<<Nu
<<"(Steel 0.29, Aluminum 0.33, Nylon 0.40)"
<<"\nCalculated to Depth:hc="<<hc
<<"\nContact radius:a="<<a
<<"\nLower limit:a0="<<a0
<<"\nUpper limit:b0="<<b0
<<"\nNumber of integration zones:Nab="<<Nab
<<"\nr_max="<<r_max

175

<<"\nCalculated points of z:Nz="<<Nz
<<"\nCalculated points of r:Nr="<<Nr
<<"\nContact interface stress constant:p0="<<p0
<<"\nTotal applied load:P="<<P;
//the following code defines calculation points (z,r)
double z[20]={0},r[200]={0};
for(i=0;i<Nz;i++){
z[i+1]=z[i]+hc/Nz;
cout<<"\n"<<z[i];
}
for(i=0;i<Nr;i++){
r[i+1]=r[i]+r_max/Nr;

//****Algrorithm to calculate stress and displacement
//Jn: Bessel function, library _jn(x)

double Sigma_z[11][151]={0,0}, Tau_rz[11][151]={0,0},
Sigma_r[11][151]={0,0};

double Sigma_theta[11][151]={0,0}, U_r[11][151]={0,0},U_z[11][151]={0,0};
double y0=0; //point in [a0,b0] respect to Gaussian point at x[k]
double fc[6]={0};
double *f_integrant(int indenter,double E,double Nu,double a,
double z,double r,double xi, double p0, double P,double fc[]);

l/********Multiple integration region - code start ********
double Sigma_z_m[50]={0},Tau_rz_m[50]={0},Sigma_r_m[50]={0};
double Sigma_theta_m[50]={0},U_r_m[50]={0},U_z_m[50]={0};
double b[50]={0};
int m=0;
for(m=0;m<Nab;m++){
b[m+1]=b[m]+(b0-a0)/Nab;
}

for(i=0; i<=Nz; i++){
for(j=0;j<Nr;j++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){
y0=(b[m+1]-b[m])*x[k]l2+(b[m+1]+b[m])l2;
f_integrant(indenter,E,Nu,a,z[i],r[j],y0,p0,P,fc);
Sigma_z__m[m]=Sigma_z_m[m]+w[k]*(b[m+1 ]-b[m])l2.0*fc[0];
Tau_rz_m[m]=Tau_rz_m[m]+w[k]*(b[m+1]-b[m])l2.0*fc[1];
Sigma_r_m[m]=Sigma_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[2];
Sigma_theta_m[m]=Sigma_theta_m[m]+w[k]*(b[m+1]-
b[m])l2.0*fc[3];
U_r_m[m]=U_r_m[m]+w[k]*(b[m+1 ]-b[m])/2.0*fc[4];

176

U_z_m[m]=U_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[5];

}

}

for(m=0;m<Nab;m++){
Sigma_z[i][j]=Sigma_z[i][i]+Sigma_z_m[m];
Tau_rz[i][j]=Tau_rz[i][i]+Tau_rz_m[m];
Sigma_r[i][i]=Sigma_r[i][i]+Sigma_r_m[m];
Sigma_theta[i][j]=Sigma_theta[i][j]+Sigma_theta_m[m];
U_r[illl]=U_rlillil+U_r_m[ml;
U_Zlil[ll=U_Zlil[il+U_Z_mlml;

}

for(m=0;m<Nab;m++){
Sigma_z_m[m]=0.0;
Tau_rz_m[m]=0.0;
Sigma_r_m[m]=0.0;
Sigma_theta_m[m]=0.0;
U_r_m[m]=0.0;
U_z_m[m]=0.0;

}

}
}

”Running time calculation end

time(&EndSec);

Total = EndSec - BeginSec;

OutComp<<"\nTotal Running Time="<<Total<<"seconds";

//********Multiple integration region - code end ******

/l======Output results======
OutComp<<"\n\nSigma_z[ij]"<<"\nr[i]";
for(k=0;k<=Nz;k++){
OutComp<<";Z="<<Z[k]§

}
for(i=0;i<Nr;i++){ .
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<Sigma_z[j][i];
}
}

l/********************

OutComp<<"\n Tau_rz[ij]"<<"\nr[i]";

for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

}

for(i=0;i<Nr;i++){

177

OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){

OutComp<<";"<<TaU_rZ[i][i];
}

}

I ********************

OutComp<<"\n Sigma_r[ij]"<<"\nr[ij";

for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<Sigma_r[j][i];
}

}

l/********************

OutComp<<"\n Sigma_theta[ij]"<<"\nr[i]";

for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<Sigma_theta[j][i];
}

}

//********************

OutComp<<"\n U_r[ij]"<<"\nr[i]";

for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

}
for(i=0;i<Nr;i++){ ‘
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<U_r[j][i];
}

}

/ ********************

OutComp<<"\n U_z[ij]"<<"\nr[i]";
for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

178

 

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<U_z[j][i];
}
}

OutComp.close();
cout<<"\n";
return 0;

}

//uniform pressure fucntion-half space
”calculate f(x) at gaussian point in domain (a0,b0)
/lx (Xi) Gaussian point
double *f_integrant(int indenter,double E,double Nu,double a,double z,double
r,double xi, double p0,double P,double fc[]){

double pH=0.0; //Hankel transform of fucntion p

double Mu=0.0, Lambda=0.0;

Mu=El(2*(1+Nu));

Lambda=E*Nu/((1+Nu)*(1-2*Nu));

if (indenter==0){

pH=-P/(2*3.14159)*exp(-a*xi); //point loading

if (indenter==1){ .
pH=(-a*p0/xi*_j1(a*xi)); //***uniform pressure

else if (indenter==2){ //***flat punch - uniform displacement
pH=(-p0*a*sin(a*xi)/xi);

else if (indenter==3){
pH=(p0)*(a*xi*cos(a*xi)-sin(a*xi))/(a*pow(xi,3));
//***Spherical lndenter

}

//**** Half speace

double A=0.0,B=0.0;

double GO=0.0,G1=0.0,GZ=0.0,G3=0.0;
A=2*Nu/xi;

B=1;

G0=(1+Nu)*(1-2*Nu)/(E*pow(xi,2))*(A+B*z)*exp(-xi*z);

G1 =(1+Nu)*(1-2*Nu)/(E*pow(xi,2))*exp(-z*xi)*(B-A*xi-B*z*xi);

G2=(1+Nu)*(1-2*Nu)/(E*pow(xi,2))*exp(-z*xi)*(-
2*B*xi+A*pow(xi,2)+B*z*pow(xi,2));

G3=(1+Nu)*(1-2*Nu)/(E*pow(xi,2))*exp(-z*xi)*(3*B*pow(xi,2)

179

-A*pow(xi,3)—B*Z*DOW(Xi.3))3

fc[0]=xi*((Lambda+2*Mu)*G3-(3*Lambda

+4*Mu)*pow(xi,2)*G1 )*pH*_j0(xi*r); //Sigma__z

fc[1]=pow(xi,2)*(Lambda*G2+(Lambda

+2*Mu)*pow(xi,2)*GO)*pH*_j1(xi*r); //Tau_rz

if(r==0.0){

fc[2]=xi*(Lambda*G3+(Lambda+2*Mu)*pow(xi,2)*G1)*pH*_j0(xi*r)
-2*(Lambda+Mu)*pow(xi,2)*G1*pH*(xi/2); l/Sigma_r

fc[3]=Lambda*xi*(G3-pow(xi,2)*G1)*pH*_j0(xi*r)
+2*(Lambda+Mu)*pow(xi,2)*G1*pH*(xi/2); l/Sigma_theta

}
else{
fc[2]=xi*(Lambda*G3+(Lambda+2*Mu)*pow(xi,2)*G1)*pH*_j0(xi*r)
-2*(Lambda+Mu)/r*pow(xi,2)*G1*pH*_j1(xi*r); //Sigma_r
fc[3]=Lambda*xi*(G3—pow(xi,2)*G1 )*pH*_j0(xi*r)
+2*(Lambda+Mu)lr*pow(xi,2)*G1*pH*_j1 (xi*r); l/Sigma_theta
}
fc[4]=(Lambda+Mu)/Mu*pow(xi,2)*G1*pH*_j1(xi*r); /IU_r
fc[5]=xi*(GZ-(Lambda+2*Mu)/Mu*pow(xi,2)*G0)*pH*_j0(xi*r); //U_z
return 0;

180

 

Appendix GZ C++ source code for single layer bonded to rigid base

// 3D single layer against rigid
II All units are in SI unit
#include <iostream.h>
#include<fstream.h>
#include<iosfwd>
#include<iomanip.h>
#include<io.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>
#include<time.h>
using namespace std;

double *f_integrant(int indenter,double E,double Nu,double h,double a,double
z,double r,double xi,double p0,double P,double fc[]);

main(void)
{
time_t BeginSec, EndSec, Total; // calculate running time
time(&BeginSec);
//**** Setup input and output file
char
lnFile1[80],
OutFile[80];

printf(”\n Please enter the input file name: ");
gets(lnFile1);

printf("\n Please enter the output file name: ");
gets(OutFile);

I/****Error message of opening files

constchar
ErrorMsg[]="\n***main: unable to open file: ";
fstream
lnMatl(lnFile1,ios::in),
OutComp(OutFile,ios::out);
if (lnMatl.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnFile1<<"\n\n";

exit(-1 );

}
else if(OutComp.fail()){
cerr<<ErrorMsg<<OutFiIe<<"\n\n";

exit(-1 );
}

181

l/Read Gaussian point and weight fucntion x[],w[]
constchar
InGaussPoint[]="gaussian64.txt";
fstream lnGauss;
lnGauss.open(lnGaussPoint,ios::in);

if (lnGauss.fail()){ //detect if the file can be open

cerr<<ErrorMsg<<lnGaussPoint<<"\n\n";
exit(-1 );
}

double x[64],w[64];

int i, j, k;

for(i=0;i<64;i++){
x[i]=0.0;
w[i]=0.0;

while (llnGauss.eof()){
lnGauss>>i; .
lnGauss>>x[i];
lnGauss>>w[i];

lnGauss.close();

cout<<setprecision(6);

for(i=63;i>31;i-){
x[i]=x[i-32];
w[i]=w[i-32];

}

for(i=0;i<32;i++){
x[i]=-x[63-i];
w[i]=w[63-i];

}

//****Read imput file

double E=0,Nu=0,h=0.0,hc=0,a=0.0,p0=0.0,P=0.0;

int indenter=0;

double r_max=0; // max value of z and r
double a0=0, b0=0; ”integration range

int Nz=0,Nr=0,Nab=0; ”number of points for z and r
while(!lnMatl.eof()){

lnMatl>>indenter; // lndenter Shape
lnMatl.ignore(80,'\n');

lnMatl>>E; ”Young's modulus
lnMatl.ignore(80,'\n');

lnMatl>>Nu; //Possion ratio

lnMatl.ignore(80,'\n');

182

}

lnMatl>>h; // layer thickness
lnMatl.ignore(80,'\n');
lnMatl>>hc; // calculated layer thickness

lnMatl.ignore(80,'\n');
lnMatl>>a;
lnMatl.ignore(80,'\n');

//contact radius

lnMatl>>a0;

lnMatl.ignore(80,'\n'); // integration lower limit

lnMatl>>b0; // intergation upper limit
lnMatl.ignore(80,'\n');

lnMatl>>Nab; /l number of integration zones in [a0,b0]

lnMatl.ignore(80,'\n');
lnMatl>>r_max;
lnMatl.ignore(80,'\n’);

”calculation zone of max r

lnMatl>>Nz; //number of points for z
lnMatl.ignore(80,'\n');
lnMatl>>Nr; // number of points for r

lnMatl.ignore(80,'\n');
lnMatl>>P;
lnMatl.ignore(80,'\n');

lnMatl.close();

//*****calculate p0
if (indenter==1){

p0=P/(3.14159*pow(a,2));

else if (indenter==2){
p0=F>/(2*3.14159*pow(a.2)):

else if (indenter==3){

}

p0=3*P/(2*3-14159*P°W(3:2))§

// total load P

l/***uniform pressure

//***flat punch - uniform displacement

//***Spherical lndenter

OutComp<<"Single Layer Indentation"
<<"\n\nlndenter shape="<<indenter
<<" (0-point loading, 1-uniform pressure, 2-flat, 3-sphere)"

<<"\nh/a="<<h/a

<<"\n\nYoung' s Modulus. E="<<E

<<"(Steel 200E+3MPa, Aluminum 70E+3MPa, Nylon 2. 5E+3)"
<<"\nPossion Ratio. Nu="<<Nu

<<"(Steel 0.29, Aluminum 0.33, Nylon 0.40)"

<<"\nLayer thickness:h="<<h

<<"\nCalculated layer thickness:hc="<<hc

<<"\nContact radius:a="<<a

<<"\nLower limit:a0="<<a0

<<"\nUpper Iimit:b0="<<b0

183

<<"\nNumber of integration zones:Nab="<<Nab
<<"\nr_max="<<r_max

<<"\nCalculated points of z:Nz="<<Nz
<<"\nCalculated points_of r:Nr="<<Nr
<<"\nContact interface stress constant:p0="<<p0
<<"\nTotal applied load:P="<<P;

// use 1/10 a as contact radius for near point case
”the following code defines calculation points (z,r)
double z[20]={0},r[200]={0};
for(i=0;i<Nz;i++){

z[i+1]=z[i]+hc/Nz;

cout<<"\n"<<z[i];
}
for(i=0;i<=Nr;i++){

r[i+1]=r[i]+r_max/Nr;

double Sigma_z[11][151]={0,0}, Tau_rz[11][151]={0,0},
Sigma_r[11][151]={0,0};

double Sigma_theta[11][151]={0,0}, U_r[11][151]={0,0},U_z[11][151]={0,0};

double y0=0; //point in [a0,b0] respect to Gaussian point at x[k]

double fc[6]={0};

//********Multiple integration region - code start ********
double Sigma_z_m[50]={0},Tau_rz_m[50]={0},Sigma_r_m[50]={0};
double Sigma_theta_m[50]={0},U_r_m[50]={0},U_z_m[50]={0};
double b[50]={0};
int m=0;
for(m=0;m<Nab;m++){
b[m+1]=b[m]+(b0—a0)/Nab;
}

for(i=0; i<=Nz; i++){
for(i=0;i<Nr;i++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){
y0=(b[m+1]-b[m])*x[k]l2+(b[m+1]+b[m])l2;
f_integrant(indenter,E,Nu,h,a,z[i],rjj],y0,p0,P,fc);
Sigma_z_m[m]=Sigma_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[0];
Tau_rz_m[m]=Tau_rz_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[1];
Sigma_r_m[m]=Sigma_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[2];
Sigma_theta_m[m]=Sigma_theta_m[m]+w[k]*(b[m+1j
-b[m])/2.0*fc[3];
U_r_m[m]=U_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[4];
U_z_m[m]=U_z_m[m]+w[k]*(b[m+1 ]-b[m])/2.0*fc[5];

184

}

for(m=0;m<Nab;m++){
Sigma_z[i][j]=Sigma_z[i][j]+Sigma_z_m[m];
Tau_rz[i][i]=Tau_rz[i][j]+Tau_rz_m[m];
Sigma_r[i][i]=Sigma_r[i][j]+Sigma_r_m[m];
Sigma_theta[i][j]=Sigma_theta[i][j]+Sigma_theta_m[m];
U_r[illi]=U_r[i]ljl+U_r_mlml;
U.Z[il[il=U_Zlilli]+U_Z_m[m]:

}

for(m=0;m<Nab;m++){
Sigma_z_m[m]=0.0;
Tau_rz_m[m]=0.0;
Sigma_r_m[m]=0.0;
Sigma_theta_m[m]=0.0;
U_r_m[m]=0.0;
U_z_m[m]=0.0;

 

}

//********Multiple integration region - code end ******
time(&EndSec);

Total = EndSec - BeginSec;
OutComp<<"\n\nTotal_Running_Time="<<Total<<"seconds";

//======Output results======
OutComp<<"\n\nSigma_z[ij]"<<"\nr[i]";
for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<Sigma_z[j][i];
}

l/********************

OutComp<<"\nTau_rz[ij]"<<"\nr[i]";

for(k=0;k<=Nz;k++){ .
OutComp<<";z="<<z[k];

}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointIios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<Tau_rz[j][i];
}

185

I ********************

OutComp<<"\nSigma_r[ij]"<<"\nr[i]";

for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<Sigma_r[j][i];
}

}

//********************

OutComp<<"\nSigma_theta[ij]"<<"\nr[i]";

for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<Sigma_theta[j][i];
}

I ********************

OutComp<<"\nU_r[ij]"<<"\nr[i]";

for(k=0;k<=Nz;k++){
OutComp<<";z="<<z[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<";"<<U_r[j][i];
}

}

ll********************
OutComp<<"\nU_z[ij]"<<"\nr[i]";
for(k=0;k<=Nz;k++){

OutComp<<";z="<<z[k];
}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz;j++){
OutComp<<”;"<<U_z[j][i];
}

186

}

}
OutComp.close();

cout<<"\n";

return 0;

l/calculate f(x) at gaussian point in domain (a0,b0)

//x (Xi) Gaussian point

double *f_integrant(int indenter,double E,double Nu,double h,double a,double
z,double r,double xi,double p0,double P,double fc[])

{

double ph=0.0; //Hankel transform of fucntion p
double Mu=0.0, Lambda=0.0;
Mu=E/(2*(1+Nu));
Lambda=E*Nu/((1+Nu)*(1-2*Nu));
if (indenter==0){
ph=P/(2*3.14159)*exp(-a*xi); //point loading
}

if (indenter==1){
ph=a*p0/xi*_j1(a*xi); //***uniform pressure

else if (indenter==2){ //***flat punch - uniform displacement
ph=a*p0*sin(a*xi)/xi;

else if (indenter==3){
ph=-p0*(a*xi*cos(a*xl)-sin(a*xi))/(a*pow(xi,3)); //***Spherical
}

//**** Single Layer on Rigid
double A=0,B=0,C=0,D=0, AM=0.0;
double G0=0.0,G1=0.0,GZ=0.0,G3=0.0;

//Perfect bonding between layer and rigid base
AM=3+exp(4*h*xi)*(3-4*Nu)-4*Nu+2*exp(2*h*xi)*(5-12*Nu+8*pow(Nu,2)
+2*pow(h,2)*pow(xi,2));
A=2*exp(2*h*xi)*(—2+4*(-1+exp(2*h*xi))*pow(Nu,2)-pow(h,2)*pow(xi,2)
+Nu*(5-3*exp(2*h*xi)+2*h*xi))/(AM*xi);
B=exp(2*h*xi)*(—1+exp(2*h*xi)*(-3+4*Nu)+2*h*xi)lAM;
C=2*((3-4*Nu)*Nu+exp(2*h*xi)*(2+4*pow(Nu,2)+pow(h,2)*pow(xi,2)
+Nu*(-5+2*h*xi)))/(AM*xi);
D=-(3-4*Nu+exp(2*h*xi)*(1+2*h*xi))/AM;

GO=(1+Nu)*(1-2*Nu)/(E*pow(xi,2))*((A+B*z)*exp(-xi*z)
+(C+D*z)*exp(xi*z));
G1 =(1 +Nu)*(1-2*Nu)/(E*pow(xi,2))*(exp(-z*xi)*(B-A*xi-B*z*xi)
+exp(z*xi)*(D+C*xi+D*z*xi));
GZ=(1+Nu)*(1-2*Nu)/(E*pow(xi,2))*(exp(-z*xi)*(-2*B*xi+A*pow(xi,2)

187

+B*z*pow(xi,2))+exp(z*xi)*(2*D*xi+C*pow(xi,2)
+D*z*pow(xi,2)));
G3=(1+Nu)*(1-2*Nu)/(E*pow(xi,2))*(exp(-z*xi)*(3*B*pow(xi,2)
-A*pow(xi,3)-B*z*pow(xi,3))+exp(z*xi)*(3*D*pow(xi,2)
+C*pow(xi,3)+D*z*pow(xi,3)));
fc[0]=xi*((Lambda+2*Mu)*G3-(3*Lambda
+4*Mu)*pow(xi,2)*G1 )*ph*_i0(xi*r); //Sigma_z
fc[1]=pow(xi,2)*(Lambda*G2+(Lambda+2*Mu)*pow(xi,2)*GO)*ph*_j1(xi*r);
lfTau rz

if(r==0.0){

fc[2]=xi*(Lambda*G3+(Lambda+2*Mu)*pow(xi,2)*G1 )*ph*_j0(xi*r)
-2*(Lambda+Mu)*pow(xi,2)*G1*ph*(xi/2); //Sigma_r
fc[3]=Lambda*xi*(GB-pow(xi,2)*G1)*ph*_j0(xi*r) +2*(Lambda
+Mu)*pow(xi,2)*G1*ph*(xi/2); //Sigma_theta
}
else{
fc[2]=xi*(Lambda*G3+(Lambda+2*Mu)*pow(xi,2)*G1 )*ph*_j0(xi*r)
-2*(Lambda+Mu)/r*pow(xi,2)*G1*ph*_j1(xi*r); //Sigma_r
fc[3]=Lambda*xi*(G3-pow(xi,2)*G1 )*ph*_j0(xi*r)
+2*(Lambda+Mu)/r*pow(xi,2)*G1*ph*_j1(xi*r); //Sigma_theta

}
fc[4]=(Lambda+Mu)/Mu*pow(xi,2)*G1*ph*_j1(xi*r); //U_r

fc[5]=xi*(G2-(Lambda+2*Mu)/Mu*pow(xi,2)*G0)*ph*_j0(xi*r); //U_z
return 0;

188

Appendix G3 C++ source code for single bonded to elastic half space

// 30 two half space, split at z=0, internal load p(r) at z=0, downward as positive
for p(r)

II All units are in SI unit
#include<iostream.h>
#include<fstream.h>
#include<iomanip.h>
#include<io.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>
I/time calculation headfile
#include <time.h>

double *f_integrant(int indenter,double E1 ,double E2,double Nu1,double
Nu2,double h1,double h2,
double a, double z,double r,double xi,double p0,double P,double fcfl);

main(void)
{
time_t BeginSec, EndSec, Total; // calculate running time
time(&BeginSec);
//**** Setup input and output file
char
lnFile1[80],
OutFile[80];

printf("\n Please enter the input file name: ");
gets(lnFile1);

printf("\n Please enter the output file name: ");
gets(OutFile);

I/****Error message of opening files
constchar
ErrorMsg[]="\n***main: unable to open file: ";
fstream
lnMatl(lnFile1,ios::in),
OutComp(OutFile,ios::out);
if (lnMatl.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnFile1<<"\n\n";
exit(-1);

}

else if(OutComp.fail()){
cerr<<ErrorMsg<<OutFile<<"\n\n";
exit(—1 );

}

189

//Read Gaussian point and weight fucntion x[],w[]
constchar
lnGaussPointfl="gaussian64.txt";
fstream lnGauss;
lnGauss.open(lnGaussPoint,ios::in);
if (lnGauss.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnGaussPoint<<"\n\n";
exit(-1);

}

double x[64],w[64];

int i, j, k;

for(i=0;i<64;i++){
x[i]=0.0;
w[i]=0.0;

while (!lnGauss.eof()){
lnGauss>>i;
lnGauss>>x[i];
lnGauss>>w[i];

lnGauss.close();
cout<<setprecision(6);
l/generate symmetrucal Gaussian points
//1) copy 0-31 to 32-64 for (0,1)
for(i=63;i>31;i-){

x[i]=x[i-32];

w[i]=w[i-32];

}
//2)establish 0-31 (-1,0) backward from 32-64
for(i=0;i<32;i++){

x[i]=-x[63-i];

w[i]=w[63-i];

}

//****Read imput file

double E1=0,E2=0,Nu1=0,Nu2=0,h1=0.0,h2=0.0,a=0.0,p0=0.0,P=0.0;
int indenter=0;

double r_max=0; // max value of r

double a0=0, b0=0; ”integration range

int Nz1=0,N22=0,Nr=0,Nab=0; //number of points for z and r
while(!lnMatl.eof()){

lnMatl>>indenter; // layer thickness
lnMatl.ignore(80,'\n');

lnMatl>>E1; //Young's modulus
lnMatl.ignore(80,'\n');

lnMatl>>E2; ”Young's modulus

190

lnMatl.ignore(80,'\n');
lnMatl>>Nu1;
lnMatl.ignore(80,'\n');
lnMatl>>Nu2;
lnMatl.ignore(80,’\n');
lnMatl>>h1;
lnMatl.ignore(80,'\n');
lnMatl>>h2;
lnMatl.ignore(80,'\n');
lnMatl>>a;
lnMatl.ignore(80,'\n');
lnMatl>>a0;
lnMatl.ignore(80,'\n');
lnMatl>>b0;
lnMatl.ignore(80,'\n');
lnMatl>>Nab;
lnMatl.ignore(80,'\n');
lnMatl>>r_max;
lnMatl.ignore(80,'\n');
lnMatl>>Nz1;
lnMatl.ignore(80,'\n');
lnMatl>>N22;
lnMatl.ignore(80,'\n');
lnMatl>>Nr;
lnMatl.ignore(80,'\n');
lnMatl>>p0;
lnMatl.ignore(80,'\n');
lnMatl>>P;
lnMatl.ignore(80,'\n');

//Possion ratio
//Possion ratio

// layer thickness

// layer thickness
l/contact radius

// integration lower limit
// integration upper limit
// number of integration zones in [a0,b0]
ll max r value

//number of points for 2
//number of points for z
// number of points for r
// stress constant

// total force

lnMatl.close();

//*****calculate p0
if (indenter==1){
p0=P/(3.14159*pow(a,2));/l***uniform pressure

else if (indenter==2){ //***flat punch - uniform displacement
p0=P/(2.0*3.14159*pow(a,2));

else if (indenter==3){
p0=3.0*P/(2*3.14159*pow(a,2)); //***Spherical lndenter
}

OutComp<<"Two Half Space Perfectly Bonded at z=0";
OutComp<<"\n\nlndenter shape="<<indenter
<<" (0-point Ioading,1-uniform pressure,2-flat,3-spherical)"

191

<<"\nh1/a="<<h1/a
<<"\nh2/a="<<h2/a
<<"\nh1/h2="<<h1/h2
<<"\nE1/E2="<<E1/E2
<<"\n\nYoung's Modulus:E1 ="<<E1
<<"\nYoung’s Modulus:E2="<<E2
<<"\nPossion Ratio:Nu1="<<Nu1
<<"\nPossion_Ratio:Nu2="<<Nu2
<<"\nLayer thickness:h1="<<h1
<<"\n Layer thickness:h2="<<h2
<<"\nContact radius:a="<<a
<<"\nLower limit:a0="<<aO
<<"\nUpper limit:b0="<<b0
<<"\nNumber of integration zones:Nab="<<Nab
<<"\nr max="<<r__max
<<"\nCalculated points along z(layer1):Nz1="<<Nz1
<<"\nCalculated points along z(layer2):N22="<<N22
<<"\nCalculated points of r:Nr="<<Nr
<<"\nContact interface pressure constatant:p0="<<p0
<<"\nTotal applied load at contact interface:P="<<P;
//the following code defines calculation points (z,r)
double z1[20]={0}, 22[20]={0},r[200]={0};
”2 represent x axis, r represent y axis
z1[0]=-h1;
for(i=0;i<Nz1;l++){
z1[i+1]=z1[i]+h1/Nz1;

}

for(i=0;i<N22;i++){
22[i+1]=z2[i]+h2/N22;

} .

for(i=0;i<Nr;i++){
r[i+1]=r[i]+r_max/Nr;

}

double Sigma1_z[11][151]={0,0}, Tau1_rz[11][151]={0,0},
Sigma1_r[11][151]={0,0},Sigma1_theta[11][151]={0,0};

double Sigma2_z[11][151]={0,0}, Tau2_rz[11][151]={0,0},
Sigma2_r[11][151]={0,0},Sigma2_theta[11][151]={0,0};

double U1_z[11][151]={0,0},U1_r[11][151]={0,0};

double U2_z[11][151]={0,0},U2_r[11][151]={0,0};

double y0=0; //point in [aO,b0] respect to Gaussian point at x[k]

double fc[12]={0};

//********Multiple integration region - code start ********

double Sigma1_z_m[50]={0},Tau1_rz_m[50]={0},
Sigma1_r_m[50]={0},Sigma1_theta_m[50]={0};

192

double Sigma2_z_m[50]={0},Tau2_rz_m[50]={0},
Sigma2_r_m[50]={0},Sigma2_theta_m[50]={0};

double U1_z_m[50]={0},U1_r_m[50]={0},
U2_z_m[50]={0},U2_r_m[50]={0};

double b[50]={0};

int m=0;

for(m=0;m<Nab;m++){
b[m+1]=b[m]+(b0-a0)/Nab;

//**** Layer 1 ****
for(i=0; i<=Nz1; i++){
for(j=0;j<Nr;j++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){
y0=(b[m+1]-b[m])*x[k]l2+(b[m+1]+b[m])l2;
f_integrant(indenter,E1,E2,Nu1,Nu2,h1,h2,a,z1[i],r[j],y0,p0,P,fc);
Sigma1_z_m[m]=Sigma1_z_m[m]+w[k]*(b[m+1]-b[m])l2.0*fc[0];
Sigma1_r_m[m]=Sigma1_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[2];
Sigma1_theta_m[m]=Sigma1_theta_m[m]+w[k]*(b[m+1]
-b[m])/2.0*fc[4];

Tau 1 _rz_m[m]=Tau 1 _rz_m[m]+w[k]*(b[m+1 ]-b[m])/2.0*fc[6];
U1_z_m[m]=U1_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[8];
U1_r_m[m]=U1_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[10];

} }

for(m=0;m<Nab;m++){
Sigma1_z[i][j]=Sigma1_z[i][j]+Sigma1_z_m[m];
Sigma1_r[i][j]=Sigma1_r[i][j]+Sigma1_r_m[m];
Sigma1_theta[i][j]=Sigma1_theta[i][j]+Sigma1_theta_m[m];
Tau1_rz[i][j]=Tau1_rz[i][j]+Tau1_rz_m[m];
U1_z[i][j]=U1_z[i][j]+U1_z_m[m];
U1_r[i][j]=U1_r[i][j]+U1__r__m[m];

}

for(m=0;m<Nab;m++){
Sigma1_z_m[m]=0.0;
Sigma1_r_m[m]=0.0;
Sigma1_theta__m[m]=0;
Tau1_rz_m[m]=0.0;
U1_z_m[m]=0.0;
U1_r_m[m]=0.0;

}

}
}

//**** Layer 2 *****

193

 

for(i=0; i<=N22; i++){
for(j=0;j<Nr;j++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){
y0=(b[m+1]-b[m])*x[k]/2+(b[m+1]+b[m])l2;
f_integrant(indenter,E1,E2,Nu1,Nu2,h1,h2,a,22[i],r[j],y0,p0,P,fc);
Sigma2_z_m[m]=Sigma2_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[1];
Sigma2_r_m[m]=Sigma2_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[3];
Sigma2_theta_m[m]=Sigma2_theta_m[m]+w[k]*(b[m+1]
-b[m])/2.0*fc[5];
Tau2_rz_m[m]=Tau2_rz_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[7];
U2_z_m[m]=U2_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[9];
U2_r_m[m]=U2_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[1 1];
}

}

for(m=0;m<Nab;m++){
Sigma2_z[i][i]=SigmaZ_z[i][j]+Sigma2_z_m[m];
Sigma2_r[i][j]=Sigma2_r[i][j]+Sigma2_r_m[m];
Sigma2_theta[i][j]=Sigma2_theta[i][j]+Sigma2_theta_m[m];
Tau2_rz[i][i]=Tau2_rz[i][j]+Tau2_rz_m[m];
U2_z[i][j]=U2_z[i][j]+U2_z_m[m];
U2_r[i][j]=U2_r[i][j]+U2_r_m[m];

}

for(m=0;m<Nab;m++){
Sigma2_z_m[m]=0.0;
Sigma2_r_m[m]=0.0;
Sigma2_theta_m[m]=0;
Tau2_rz__m[m]=0.0;
U2_z_m[m]=0.0;
U2_r_m[m]=0.0;

}

}

//********Multip|e integration region - code end ******
time(&EndSec);

Total = EndSec - BeginSec;
OutComp<<”\n\nTotal_Running_Time="<<Total<<"seconds";

//****‘k*****************************************

l/* *
//* Output results - Layer 1 *
ll'k *-

/ *i'i'*******************************************

194

 

OutComp<<"\n\nSigma1_z[ij]"<<"\nr[i]";
for(k=0;k<=NZ1ik++){
OutComp<<";z="<<z1 [k];

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Sigma1_z[i][i];
}

}

l ***************i****

OutComp<<"\nTau1_rz[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1[k];

}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Tau1_rz[j][i];
} .

}

ll********************

OutComp<<"\nSigma1_r[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Sigma1_r[j][i];
}

}

I ********************

OutComp<<"\nSigma1_theta[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Sigma1_thetafi][i];
}

195

I ********************

OutComp<<"\nU1_r[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<U1_r[j][i];
}

}

//**'k***************** .

OutComp<<"\nU1_z[i,j]"<<"\nr[i]";

for(k=0;k<=Nz1 ;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<U1_z[j][i];

/ ********************************************************************
//*

*

//* Output results Layer 2

//*

*

//**********************************************i*********************
OutComp<<"\n\nSigm32_z[ij]"<<"\nr[i]";
for(k=0;k<=N22;k++){

OutComp<<";z="<<22[k];
}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Sigma2_z[j][i];
}

}

/ ********************

OutComp<<"\nTau2_rz[ij]"<<"\nr[i]";
for(k=0;k<=N22;k++){

196

OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Tau2_rz[j][i];
}

}

//********************

OutComp<<"\nSigma2_r[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Sigma2_r[j][i];
}

}

/ ********************

OutComp<<"\nSigma2_theta[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Sigma2_theta[i][i];
}

}

//********************

OutComp<<"\nU2_r[ij]"<<"\nr[i]";
for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<U2_r[j][i];
}

}

l/********************

197

OutComp<<"\nU2_z[ij]"<<"\nr[i]";
for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=NzZ;j++){
OutComp<<";"<<U2_z[j][i];
}

}

OutComp.close();
cout<<"\n";

return 0;

//3DZlayer half space calculation
llcalculate f(x) at gaussian point in domain (a0,b0)
//x (Xi) Gaussian point

double *f_integrant(int indenter,double E1 ,double E2,double Nu1,

double Nu2,double h1,double h2, double a, double 2,
double r,double xi,double p0,double P,double fc[]){
double ph=0.0; //Hankel transform of fucntion p
if (indenter==0){
ph=P/(2*3.14159)*exp(-a*xi); //point loading

}
if (indenter==1){
ph=a*p0/xi*_j1(a*xi); //***uniform pressure

else if (indenter==2){ //***flat punch - uniform displacement
ph=a*p0/xi*sin(a*xi);
}

else if (indenter==3){
ph=-p0*(a*xi*cos(a*xi)-sin(a*xi))/(a*pow(xi.3)); //***Spherical

}

//**** 3D 2 layer on rigid, z=0 on contact surface - top surface

double C1=0,D1=0,A2=0,B2=0;

D1=E1*(1+Nu2)/(-E2*(-3+Nu1+4*pow(Nu1,2))+E1*(1+Nu2));

B2=-E2*(1+Nu1)/(E2*(1+Nu1)-E1*(-3+Nu2+4*pow(Nu2,2)));

C1=-(1+82-D1+4*D1*Nu1)/(2*xi);

A2=-(1+B2-D1-4*BZ*Nu2)/(2*xi);

double G10=0,G11=0,G12=0,G13=0,GZO=0,G21=0,(322=0,GZ3=0;

198

G10=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*(C1+D1*z)*exp(xi*z);
GZO=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(A2+BZ*z)*exp(-xi*z);

G11=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*(D1+C1*xi+D1*z*xi)*exp(xi*z);
GZ1=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(BZ-A2*xi-BZ*z*xi)*exp(-xi*z);
G12=(1+Nu1)*(1-
2*Nu1 )/(E1*pow(xi,2))*(2*D1+C1*xi+D1*z*xi)*xi*exp(xi*z);
G22=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(-2*BZ+A2*xi
+BZ*z*xi)*xi*exp(-xi*z);
G13=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*(3*D1+C1*xi
+D1*z*xi)*pow(xi,2)*exp(xi*z);
G23=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(3*BZ-A2*xi
-BZ*z*xi)*pow(xi,2)*exp(-xi*z);
double Lambda1=0,Lambda2=0,Mu1=0,Mu2=0;
Mu1=E1/(2*(1+Nu1));
Mu2=E2/(2*(1+Nu2));
Lambda1=(E1*Nu1)/((1+Nu1)*(1-2*Nu1));
Lambda2=(E2*Nu2)/((1+Nu2)*(1-2*Nu2));
//===for Sigma_z ===
fc[0]=xi*((Lambda1+2*Mu1 )*G13-(3*Lambda1
+4*Mu1)*pow(xi,2)*G1 1 )*ph*_j0(xi*r);
fc[1]=xi*((Lambda2+2*Mu2)*G23-(3*Lambda2
+4*Mu2)*pow(xi,2)*G21)*ph*_j0(xi*r);
if (r==0.0){
//===for Sigma_r ===
fc[2]=xi*(Lambda1*G13+(Lambda1
+2*Mu1 )*pow(xi,2)*G1 1 )*ph*_j0(xi*r)
-2*(Lambda1+Mu1)*pow(xi,2)*G1 1*ph*(xi/2);
fc[3]=xi*(Lambda2*G23+(Lambda2
+2*Mu2)*pow(xi,2)*G21 )*ph*_j0(xi*r)
-2*(Lambda2+Mu2)*pow(xi,2)*GZ1*ph*(xi/2);
//=== Sigma_theta ===
fc[4]=Lambda1*xi*(G13-pow(xi,2)*G1 1 )*ph*_j0(xi*r)
+2*(Lambda1+Mu1)*pow(xi,2)*G1 1*ph*(xi/2);
fc[5]=Lambda2*xi*(G23-pow(xi,2)*GZ1 )*ph*_j0(xi*r)
+2*(Lambda2+Mu2)*pow(xi,2)*G21*ph*(xi/2);

else{

//===for Sigma_r ===

fc[2]=xi*(Lambda1*G13+(Lambda1
+2*Mu1)*pow(xi,2)*G1 1 )*ph*_j0(xi*r)
-2*(Lambda1+Mu1)/r*pow(xi,2)*G11*ph*_j1(xi*r);

fc[3]=xi*(Lambda2*G23+(Lambda2
+2*Mu2)*pow(xi,2)*G21)*ph*_j0(xi*r)
-2*(Lambda2+Mu2)/r*pow(xi,2)*G21*ph*_j1(xi*r);

//=== Sigma_theta ===

199

fc[4]=Lambda1*xi*(G13-pow(xi,2)*G1 1 )*ph*_j0(xi*r)
+2*(Lambda1+Mu1)/r*pow(xi,2)*G1 1*ph*_j1(xi*r);
fc[5]=Lambda2*xi*(G23-pow(xi,2)*G21)*ph*_j0(xi*r)
+2*(Lambda2+Mu2)/r*pow(xi,2)*GZ1*ph*_j1(xi*r);
}

//===for Tau_rz ===

fc[6]=pow(xi,2)*(Lambda1*G12+(Lambda1
+2*Mu1 )*pow(xi,2)*G10)*ph*_j1(xi*r);

fc[7]=pow(xi,2)*(LambdaZ*G22+(Lambda2
+2*Mu2)*pow(xi,2)*G20)*ph*_j1(xi*r);

//===for U_z ===
fc[8]=xi*(G12-(Lambda1+2*Mu1 )/Mu1*pow(xi,2)*G10)*ph*_j0(xi*r);
fc[9]=xi*(622-(Lambda2+2*Mu2)/Mu2*pow(xi,2)*G20)*ph*_j0(xi*r);

//===for U_r ===
fc[10]=(Lambda1+Mu1)/Mu1*pow(xi,2)*G1 1*ph*_j1(xi*r);
fc[1 1]=(Lambda2+Mu2)/Mu2*pow(xi,2)*G21*ph*_j1(xi*r);
return 0;

200

Appendix G4 C++ source code for two-layer bonded to rigid base
Replace the function *f_integrant in G3 with the following code, the rest are the
same.

llcalculate f(x) at gaussian point in domain (a0,b0)
l/x (Xi) Gaussian point

double *f_integrant(int indenter,double E1 ,double E2,double Nu1,
double Nu2,double h1,double h2, double a, double z,double r,
double xi,double p0,double P,double fc[]){

double ph=0.0; l/Hankel transform of fucntion p

if (indenter==0){
ph=P/(2*3.14159)*exp(-a*xi); //point loading

if (indenter==1){
ph=a*p0/xi*_j1(a*xi); //***uniform pressure

else if (indenter==2){ //***flat punch - uniform displacement
ph=a*p0*sin(a*xi)/xi;

else if (indenter==3){
ph=-p0*(a*xi*cos(a*xi)-sin(a*xi))/(a*pow(xi,3)); //***Spherical lndenter
}

//**** 30 2 layer on rigid, z=0 on contact surface - top surface
double BM21=0,BM22=0,BM23=0,BM24=0,BM251=0,BM252=0,
BM253=0,BM254=0,BM255=0,BM256=0,BM25=0,BM2=0;

BM21=3-4*Nu1+exp(4*h1*xi)*(3—4*Nu1)+2*exp(2*h1*xi)*(5—12*Nu1
+8*pow(Nu1 ,2)+2*pow(h1 ,2)*pow(xi,2));
BM22=3-4*Nu2+exp(4*h2*xi)*(3-4*Nu2)+2*exp(2*h2*xi)*(5—12*Nu2
+8*pow(Nu2,2)+2*pow(h2,2)*pow(xi,2));
BM23=1+exp(4*h1*xi)—2*exp(2*h1*xi)*(1+2*pow(h1 ,2)*pow(xi,2));
BM24=pow((3-4*Nu2),2)*(1+exp(4*h2*xi))-2*exp(2*h2*xi)*(9-24*Nu2
+16*pow(Nu2,2)+2*pow(h2,2)*pow(xi,2));
BM251=(exp(4*h1*xi)+exp(4*h2*xi))*(—3+2*Nu1+2*Nu2)*(-3+4*Nu2);
BM252=(—3+4*Nu2)*(5-6*Nu2+Nu1*(-6+8*Nu2))*(1+exp(4*(h1+h2)*xi));
BM253=-2*(exp(2*h1*xi)+exp(2*(h1+2*h2)*xi))*(3-10*Nu2
+8*pow(Nu2,2))*(—1+2*Nu1+2*pow(h1,2)*pow(xi,2));
BM254=-2*(exp(2*h2*xi)+exp(2*(2*h1+h2)*xi))*(~1+2*Nu1 )*(3-10*Nu2
+8*pow(Nu2,2)+2*pow(h2,2)*pow(xi,2));
BM255=-3+6*pow(h1 ,2)*pow(xi,2)-16*h1*h2*pow(xi,2)—2*pow(h2,2)*pow(xi,2)
+4*pow(h1 ,2)*pow(h2,2)*pow(xi,4)
+8*pow(Nu2,2)*(-1+2*pow(h1 ,2)*pow(xi,2));

201

 

BM256=Nu2*(10-20*pow(h1 ,2)*pow(xi,2)+16*h1*h2*pow(xi,2))
+2*Nu1*(3+8*pow(Nu2,2)+8*h1*h2*pow(xi,2)
+2*pow(h2,2)*pow(xi,2)-2*Nu2*(5+4*h1*h2*pow(xi,2)));

BM25=BM251+BM252+BM253+BM254+4*exp(2*(h1+h2)*xi)*(BM255+BM256);

BM2=pow(E2,2)*pow((1+Nu1),2)*BM21*BM22
+pow(E1,2)*pow((1+Nu2),2)*BM23*BM24
-2*E1*E2*(1+Nu1)*(1+Nu2)*BM25;

double BN21=0,BN22=0,BN23=0,BN24=0,BN2=0,
DN21=0,DN22=0,DN23=0,DN24=0,DN2=0,BZ=0,DZ=0;

BN21=1-2*h1*xi-6*h2*xi+8*h2*Nu1*xi+exp(2*(h1+h2)*xi)*(—3+4*Nu1)*(-3+4*Nu2);

BN22=exp(2*h2*xi)*(-3+4*Nu2)*(-1+2*h1*xi)
-exp(2*h1*xi)*(-3+4*Nu1+2*h2*xi*(1+2*h1*xi));

BN23=3-4*Nu2-6*h1*xi-2*h2*xi+8*h1*Nu2*xi+exp(2*(h1+h2)*xi)*(3-4*Nu2);

BN24=-exp(2*h2*xi)*(-3+4*Nu2)*(-1+2*h1*xi)
+exp(2*h1*xi)*(-3+4*Nu2+2*h2*xi*(1+2*h1*xi));

BN2=4*exp(2*(h1+h2)*xi)*E2*(-1+pow(Nu2,2))*(E2*(1+Nu1)*(BN21+BN22)
+E1*(1+Nu2)*(BN23+BN24));

DN21=-(—3+4*Nu1 )*(-3+4*Nu2)-exp(2*(h1+h2)*xi)*(1+2*h1*xi
+6*h2*xi-8*h2*Nu1*xi);

DN22=exp(2*h1*xi)*(-3+4*Nu2)*(1 +2*h1*xi)+exp(2*h2*xi)*(-3+4*Nu1
+2*h2*xi*(-1+2*h1*xi));

DN23=3-4*Nu2+exp(2*(h1+h2)*xi)*(3+6*h1*xi+2*h2*xi-4*Nu2*(1+2*h1*xi));

DN24=exp(2*h1*xi)*(-3+4*Nu2)*(1+2*h1*xi)+exp(2*h2*xi)*(-3
+4*Nu2+2*h2*xi*(-1+2*h1*xi));

DN2=-4*E2*(-1+pow(Nu2,2))*(E2*(1+Nu1)*(DN21+DN22)
-E1*(1+Nu2)*(DN23+DN24));

BZ=BN2IBM2;

DZ=DN2IBM2;

double A2=0,C2=0,A1=0,B1=0,C1=0,D1=0;
A2=(-D2*exp(2*(h1+h2)*xi)*(-3+4*Nu2)+BZ*(-1+4*Nu2-2*(h1+h2)*xi))/(2*xi);
CZ=(82*exp(-2*(h1+h2)*xi)*(-3+4*Nu2)+D2*(1-4*Nu2-2*(h1+h2)*xi))l(2*xi);
B1=(BZ*(-E2*(1+Nu1)+E1*(-3+Nu2+4*pow(Nu2,2)))+exp(2*h1*xi)*(E2*(1+Nu1)
-E1*(1+Nu2))*(2*CZ*xi+D2*(-1+4*Nu2+2*h1*xi)))/(4*E2*(-1+pow(Nu1,2)));
D1=exp(-2*h1*xi)*(D2*exp(2*h1*xi)*(-E2*(1+Nu1 )+E1*(-3+Nu2+4*pow(Nu2,2)))
+2*A2*(-E2*(1+Nu1)+E1*(1+Nu2))*xi+BZ*(E2*(1+Nu1)-E1*(1+Nu2))*(-1
+4*Nu2-2*h1*xi))/(4*E2*(-1+pow(Nu1 ,2»);
A1=(BZ-D1*exp(2*h1*xi)+D2*exp(2*h1*xi)-4*BZ*Nu2+2*A2*xi+2*BZ*h1*xi
+B1*(-1+4*Nu1-2*h1*xi))/(2*xi);
C1=(D1-DZ+B1*exp(-2*h1*xi)-BZ*exp(-2*h1*xi)-4*D1*Nu1+4*D2*Nu2
+2*CZ*xi-2*D1*h1*xi+2*D2*h1*xi)/(2*xi);

double G10=0,G11=0,G12=0,G13=0,G20=0,G21=0,622=0,G23=0;
G10=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((A1+B1*z)*exp(-xi*z)

202

 

 

+(C1+D1*z)*exp(xi*z));
GZO=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*((A2+BZ*z)*exp(—xi*z)

+(C2+D2*z)*exp(xi*z));
G1 1=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((B1-A1*xi-B1*z*xi)*exp(-xi*z)

+(D1+C1*xi+D1*z*xi)*exp(xi*z));
621=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*((B2-A2*xi-BZ*z*xi)*exp(-xi*z)

+(D2+CZ*xi+D2*z*xi)*exp(xi*z));
G12=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((-2*B1+A1*xi+B1*z*xi)*xi*exp(—xi*z)

+(2*D1+C1*xi+D1*z*xi)*xi*exp(xi*z));
G22=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*((-2*82+A2*xi+82*z*xi)*xi*exp(-xi*z)

+(2*DZ+CZ*xi+D2*z*xi)*xi*exp(xi*z)); '
G13=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((3*B1-A1*xi-B1*z*xi)*pow(xi,2)*exp(-
xi*z)

+(3*D1+C1*xi+D1*z*xi)*pow(xi,2)*exp(xi*z));
G23=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*((3*BZ—A2*xi-82*z*xi)*pow(xi,2)*exp(-
xi*z)

+(3*DZ+CZ*xi+D2*z*xi)*pow(xi,2)*exp(xi*z));

double Lambda1=0,Lambda2=0,Mu1=0,Mu2=0;
Mu1=E1/(2*(1+Nu1));
Mu2=E2/(2*(1+Nu2));
Lambda1=(E1*Nu1)/((1+Nu1)*(1-2*Nu1));
Lambda2=(E2*Nu2)/((1+Nu2)*(1-2*Nu2»;
//===for Sigma_z ===
fc[0]=xi*((Lambda1+2*Mu1)*G13-
3*Lambda1+4*Mu1)*pow(xi,2)*G1 1 )*ph*_j0(xi*r);
fc[1]=xi*((Lambda2+2*Mu2)*G23-
3*Lambda2+4*Mu2)*pow(xi,2)*621 )*ph*_j0(xi*r);

if (r==0.0){
//===for Sigma_r ===
fc[2]=xi*(Lambda1*G13+(Lambda1+2*Mu1)*pow(xi,2)*G1 1 )*ph*_j0(xi*r)
-2*(Lambda1+Mu1)*pow(xi,2)*G1 1*ph*(xi/2);
fc[3]=xi*(Lambda2*G23+(Lambda2+2*Mu2)*pow(xi,2)*G21 )*ph*_j0(xi*r)
-2*(Lambda2+Mu2)*pow(xi,2)*621*ph*(xi/2);
//=== Sigma_theta ===
fc[4]=Lambda1*xi*(G13-pow(xi,2)*G1 1 )*ph*_j0(xi*r)
+2*(Lambda1+Mu1)*pow(xi,2)*G1 1*ph*(xi/2);
fc[5]=Lambda2*xi*(G23-pow(xi,2)*621 )*ph*_j0(xi*r)
+2*(Lambda2+Mu2)*pow(xi,2)*G21*ph*(xi/2);

else{
//===for Sigma_r ===

fc[2]=xi*(Lambda1*G13+(Lambda1+2*Mu1)*pow(xi,2)*G1 1 )*ph*_j0(xi*r)
-2*(Lambda1+Mu1)/r*pow(xi,2)*G1 1*ph*_j1(xi*r);

203

fc[3]=xi*(Lambda2*GZ3+(Lambda2+2*Mu2)*pow(xi,2)*G21 )*ph*_j0(xi*r)
-2*(Lambda2+Mu2)/r*pow(xi,2)*G21*ph*_j1(xi*r);
//=== Sigma_theta ===
fc[4]=Lambda1*xi*(G13—pow(xi,2)*G1 1 )*ph*_j0(xi*r)
+2*(Lambda1+Mu1)/r*pow(xi,2)*G1 1*ph*_j1(xi*r);
fc[5]=Lambda2*xi*(G23—pow(xi,2)*G21 )*ph*_j0(xi*r)
+2*(Lambda2+Mu2)/r*pow(xi,2)*GZ1*ph*_j1(xi*r);

//===for Tau_rz ===

fc[6]=pow(xi,2)*(Lambda1*G12
+(Lambda1+2*Mu1)*pow(xi,2)*G10)*ph*_j1(xi*r);

fc[7]=pow(xi,2)*(Lambda2*G22
+(Lambda2+2*Mu2)*pow(xi,2)*GZO)*ph*_j1(xi*r);

//===for U_z ===
fc[8]=xi*(G12-(Lambda1+2*Mu1 )/Mu1*pow(xi,2)*G10)*ph*_j0(xi*r);
fc[9]=xi*(622-(Lambda2+2*Mu2)/Mu2*pow(xi,2)*G20)*ph*_j0(xi*r);

//===for U_r ===
fc[10]=(Lambda1+Mu1)/Mu1*pow(xi,2)*G1 1*ph*_j1(xi*r);
fc[1 1]=(Lambda2+Mu2)/Mu2*pow(xi,2)*G21*ph*_j1(xi*r);

return 0;

Appendix GS C++ code for single layer bonded to an elastic half space
subjected to shear loading

// 30 two layer half space with shear boundary condition on top surface
// All units are in SI unit
#include<iostream.h>
#include<fstream.h>
#include <iosfwd>
#include<iomanip.h>
#include<io.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>

lltime calculation headfile
#include <time.h>

using namespace std;

double *f_integrant(int indenter,double E1 ,double E2,double Nu1,

204

double Nu2,double h1, double a, double z,double r,double xi,
double p0,double P,double f,double fc[]);

main(void)
{
time_t BeginSec, EndSec, Total; // calculate running time
time(&BeginSec);
//**** Setup input and output file
char
lnFilel [80],
OutFile[80];

printf("\n Please enter the input file name: ");
gets(lnFile1);

printf("\n Please enter the output file name: ");
gets(OutFile);

//****Error message of opening files
constchar
ErrorMsg[]="\n***main: unable to open file: ";
fstream
lnMatl(lnFile1,ios::in),
OutComp(OutFile,ios::out);

if (lnMatl.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnFile1<<"\n\n";
exit(-1);

}

else if(OutComp.fail()){
cerr<<ErrorMsg<<OutFile<<"\n\n";
exit(-1 );

}

//Read Gaussian point and weight fucntion x[],w[]
constchar
lnGaussPoint[]="gaussian64.txt";
fstream lnGauss;
InGauss.open(lnGaussPoint,ios::in);

if (lnGauss.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnGaussPoint<<"\n\n";
exit(-1);
}
double x[64],w[64];
int i, j, k;
for(i=0;i<64;i++){
x[i]=0.0;
w[i]=0.0;

while (!lnGauss.eof()){

205

 

lnGauss>>i;
lnGauss>>x[i];
lnGauss>>w[i];

lnGauss.close();
cout<<setprecision(6);

//******generate symmetrucal Gaussian points
//1) copy 0-31 to 32-64 for (0,1)
for(i=63;i>31;i-){

x[i]=x[i-32];

w[i]=w[i-32];

}
//2)establish 0-31 (-1,0) backward from 32-64
for(i=0;i<32;i++){
x[i]=-x[63-i];
w[i]=w[63-i];
}

//****Read imput file

double E1=0,E2=0,Nu1=0,Nu2=0,h1=0.0,hc1=0.0,h02=0.0,a=0.0,
p0=0.0,P=0.0,f=0;

int indenter=0;

double r_max=0; // max value of r

double a0=0, b0=0; Ilintegration range

int Nz1=0,N22=0,Nr=0,Nab=0; I/number of points for z and r

while(!lnMatl.eof()){

lnMatl>>indenter; // layer thickness
lnMatl.ignore(80,'\n');

lnMatl>>E1; ”Young's modulus
lnMatl.ignore(80,'\n');

lnMatl>>E2; //Young's modulus
lnMatl.ignore(80,'\n');

lnMatl>>Nu1; l/Possion ratio
lnMatl.ignore(80,'\n');

lnMatl>>Nu2; //Possion ratio
lnMatl.ignore(80,'\n');

lnMatl>>h1; // layer thickness
lnMatl.ignore(80,'\n');

lnMatl>>hc1; II calculated thickness in layer one
lnMatl.ignore(80,'\n');

lnMatl>>hc2; // calculated thickness in layer two
lnMatl.ignore(80,'\n');

lnMatl>>a; l/contact radius
lnMatl.ignore(80,'\n');

lnMatl>>a0; // integration lower limit

206

 

lnMatl.ignore(80,'\n');

lnMatl>>b0; // integration upper limit
lnMatl.ignore(80,'\n');

lnMatl>>Nab; // number of integration zones in [a0,b0]
lnMatl.ignore(80,'\n');

lnMatl>>r_max; // max r value
lnMatl.ignore(80,'\n');

lnMatl>>Nz1; ”number of points for z
lnMatl.ignore(80,'\n');

lnMatl>>N22; //number of points for z
lnMatl.ignore(80,'\n');

lnMatl>>Nr; // number of points for r
lnMatl.ignore(80,'\n');

lnMatl>>P; // total force
lnMatl.ignore(80,'\n');

lnMatl>>f; // friction coefficient

lnMatl.ignore(80,'\n');

lnMatl.close();

//*****calculate p0
if (indenter==1){
p0=P/(3.14159*pow(a,2)); //***uniform pressure

else if (indenter==2){ //***flat punch - uniform displacement
p0=P/(2.0*3.14159*pow(a,2));
}

else if (indenter==3){
p0=3.0*P/(2*3.14159*pow(a,2)); //***Spherical lndenter
}

OutComp<<"Two Layer Composites Half Space with Shear Force at
Boundary Condition";
OutComp<<"\n\nlndenter shape="<<indenter
<<" (0-point loading,1-uniform pressure,2-flat,3-spherical)"
<<"\nh1/a="<<h1/a
<<"\nE1/E2="<<E1/E2
<<"\n\nYoung's Modulus:E1="<<E1
<<"(Steel 200E+3MPa, Aluminum 70E+3MPA, Nylon 2.5E+3MPa)"
<<"\nYoung's Modulus:E2="<<E2
<<"\nPossion Ratio:Nu1="<<Nu1
<<"(Steel 0.29, Aluminum 0.33, Nylon 0.40)"
<<"\nPossion_Ratio:Nu2="<<Nu2
<<"\nLayer one thickness:h1="<<h1
<<"\nCalculated layer one thickness:hc1="<<hc1

207

<<"\nCalculated layer two thickness:h02="<<hc2
<<"\nContact radius:a="<<a
<<"\nLower limit:a0="<<aO
<<"\nUpper limit:b0="<<b0
<<"\nNumber of integration zones:Nab="<<Nab
<<"\nr max="<<r_max
<<"\nCalculated points along z(layer1):Nz1="<<Nz1
<<"\nCalculated points along z(layer2):N22="<<N22
<<"\nCalculated points of r:Nr="<<Nr
<<"\nContact interface pressure constatant:p0="<<p0
<<"\nTotal applied load at contact interface:P=”<<P
<<"\nFriction coefficient:f="<<f;
llthe following code defines calculation points (z,r)
double z1[20]={0}, z2[20]={0},r[200]={0}; //2 represent x, r represent y
for(i=0;i<Nz1;i++){
z1[i+1]=zl[i]+hc1/Nz1;

}

22[0]=h1;

for(i=0;i<N22;i++){
22[i+1]=z2[i]+hc2/N22;

}

for(i=0;i<Nr;i++){
r[i+1]=r[i]+r_max/Nr;
}

double Sigma1_z[11][151]={0,0}, Tau1_rz[11][151]={0,0},
Sigma1_r[11][151]={0,0},Sigma1_theta[11][151]={0,0};

double Sigma2_z[11][151]={0,0}, Tau2_rz[11][151]={0,0},
Sigma2_r[11][151]={0,0},Sigma2_theta[11][151]={0,0};

double U1_z[11][151]={0,0},U1_r[11][151]={0,0};

double U2_z[11][151]={0,0},U2_r[11][151]={0,0};

double y0=0; //point in [a0,b0] respect to Gaussian point at x[k]

double fc[12]={0};

//********Multiple integration region — code start ********
double Sigma1_z_m[50]={0},Tau1_rz_m[50]={0},
Sigma1_r_m[50]={0},Sigma1_theta_m[50]={0};
double Sigma2_z_m[50]={0},Tau2_rz__m[50]={0},
Sigma2_r_m[50]={0},Sigma2_theta_m[50]={0};
double U1_z_m[50]={0},U1_r_m[50]={0},U2_z_m[50]={0},U2_r_m[50]={0};
double b[50]={0}; ‘
int m=0;
for(m=0;m<Nab;m++){
b[m+1]=b[m]+(b0-a0)/Nab;
}

208

//**** Layer 1 ****
for(i=0; i<=Nz1; i++){
for(j=0;j<Nr;j++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){
y0=(b[m+1]-b[m])*x[k]/2+(b[m+1]+b[m])l2;
f_integrant(indenter,E1,E2,Nu1,Nu2,h1,a,z1[i],r[j],y0,p0,P,f,fc);
Sigma1_z_m[m]=Sigma1_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[0];
Sigma1_r_m[m]=Sigma1_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[2];
Sigma1_theta_m[m]=Sigma1_theta_m[m]+w[k]*(b[m+1]
-b[m])/2.0*fc[4];
Tau 1 _rz_m[m]=Tau 1 _rz_m[m]+w[k]*(b[m+1 ]-b[m])/2.0*fc[6];
U1_z_m[m]=U1_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[8];
U1_r_m[m]=U1_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[10];
} }
for(m=0;m<Nab;m++){
Sigma1_z[i][j]=Sigma1_z[i][j]+Sigma1_z_m[m];
Sigma1_r[i][j]=Sigma1_r[i]_[j]+Sigma1_r_m[m];
Sigma1_theta[i][j]=Sigma1_theta[i][j]+Sigma1_theta_m[m];
Tau1_rz[i][j]=Tau1_rz[i][j]+Tau1_rz_m[m];
U1_z[i][i]=U1_z[i][i]+U1_z_m[m];
U1_r[i][i]=U1_r[i][j]+U1_r_m[m];
}
for(m=0;m<Nab;m++){
Sigma1_z_m[m]=0.0;
Sigma1_r_m[m]=0.0;
Sigma1_theta_m[m]=0;
Tau1_rz_m[m]=0.0;
U1_z_m[m]=0.0;
U1_r_m[m]=0.0;

//**** Layer 2 *****
for(i=0; i<=N22; i++){
for(j=0;j<Nr;j++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){

y0=(b[m+1]-b[m])*x[k]/2+(b[m+1]+b[m])l2;
f_integrant(indenter,E1,E2,Nu1,Nu2,h1,a,22[i],r[j],y0,p0,P,f,fc);
Sigma2_z_m[m]=Sigma2_z_m[m]+w[k]*(b[m+1]-b[m])l2.0*fc[1];
Sig ma2_r_m[m]=Sigma2_r_m[m] +w[k]*( b[m+1 ]-b[m])/2.0*fc[3];

209

 

Sigma2__theta_m[m]=Sigma2__theta_m[m]+w[k]*(b[m+1]
-b[m])/2.0*fc[5];
Ta u2_rz_m[m]=Tau2_rz_m[m] +w[k]*(b[m+1 ]-b[m])/2.0*fc[7];
U2_z_m[m]=U2_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[9];
U2_r_m[m]=U2_r_m[m]+w[k]*(b[m+1]-b[m])l2.0*fc[1 1];
}

}

for(m=0;m<Nab;m++){
Sigma2_z[i][j]=Sigma2_z[i][j]+Sigma2_z_m[m];
Sigma2_r[i][i]=Sigma2_r[i][i]+Sigma2_r_m[m];
Sigma2_theta[i][j]=Sigma2_theta[i][i]+Sigma2_theta_m[m];
Tau2_rz[i][j]=Tau2_rz[i][j]+Tau2_rz_m[m];
U2_z[i][j]=U2_z[i][j]+U2_z_m[m];
U2_r[i][j]=U2_r[i][j]+U2_r_m[m];

}

for(m=0;m<Nab;m++){
Sigma2_z_m[m]=0.0;
Sigma2_r_m[m]=0.0;
Sigma2_theta_m[m]=0;
Tau2_rz_m[m]=0.0;
U2_z_m[m]=0.0;
U2_r_m[m]=0.0;

//********Multiple integration region - code end ******

time(&EndSec);
Total = EndSec - BeginSec;
OutComp<<"\n\nTotal_Running_Time="<<Total<<"seconds";

//*******************************************
//*

*

//* Output results - Layer 1
//*

*

I ************************i******************

OutComp<<"\n\nSigma1_z[ij]"<<"\nr[i]";
for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){

210

 

 

 

OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){

OutComp<<";"<<Sigma1_z[i][i];
}

}

//********************

OutComp<<"\nTau1_rz[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<zl [k];

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Tau1_rz[j][i];
}

}

/ ********************

OutComp<<"\nSigma1_r[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nzl ;j++){
OutComp<<";"<<Sigma1_r[j][i];
}

}

I/********************

OutComp<<"\nSigma1_theta[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Sigma1_theta[j][i];
}

}

//********************

OutComp<<"\nU1_r[ij]"<<"\nr[i]";
for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

211

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<U1_r[j][i];
}

}

I ********************

OUtC0mp<<"\n U 1_Z[i ’j]"<<"\n r[ilu;
for(k=0;k<=Nz1;k+-I-){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<U1_z[j][i];

//*******************************************

//*

*

//* Output results Layer 2 *
//*

*

I *-******************************************

OutComp<<"\n\nSigma2_z[ij]"<<"\nr[i]";
for(k=0;k<=N22;k++){

OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Sigma2_z[j][i];
}

}

//********************

OutComp<<"\nTau2_rz[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

for(i=0;i<Nr;i++){

OutComp<<setiosflags(ios::showpoint]ios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){

212

 

OutComp<<";"<<Tau2_rZ[ll[i];

}

l/********************

OutComp<<"\nSigma2_r[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Sigma2_r[j][i];
}

}

I ********************

OutComp<<"\nSigma2_theta[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=NzZ;j++){
OutComp<<";"<<Sigma2_theta[j][i];
}

}

ll********************

OutComp<<"\nU2_r[ij]"<<"\nr[i]";

for(k=0;k<=NzZ;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<U2_r[i][i];
}

}

/ ********************

OutComp<<"\nU2_z[ij]"<<"\nr[i]";
for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){

213

OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){

OutComp<<";"<<U2_Z[i][i]I
}

}

OutComp.close();
cout<<"\n";
return 0;

}

l/3D2layer half space calculation
llcalculate f(x) at gaussian point in domain (a0,b0)
l/x (Xi) Gaussian point

double *f_integrant(int indenter,double E1,double E2,double Nu1,
double Nu2,double h1,double a, double z,double r,double xi,
double p0,double P,double f,double fc[]){

double ph=0, qh=0.0; ”Hankel transform of fucntion p
if (indenter==0){
ph=P/(2*3.14159)*exp(-a*xi); //point loading
qh=ph*f;

}

if (indenter==1){
ph=a*p0/xi*_j1(a*xi); //***uniform pressure
qh=ph*f;

}

else if (indente ==2){ //***flat punch - uniform displacement
ph=a*p0/xi*sin(a*xi);
qh=ph*f;

else if (indenter==3){
ph=-p0*(a*xi*cos(a*xi)-sin(a*xi))/(a*pow(xi.3)); //***Spherical
qh=ph*f;

}

//**** 3D 2 layer half space, z=0 on contact surface - top surface

double AM1=0,AM2=0,AM3=0,AM=0;

AM1 =1 +exp(4*h1*xi)-2*exp(2*h1*xi)*(1+2*pow(h1 ,2)*pow(xi,2));

AM2=-3+2*(Nu1+Nu2)+exp(4*h1*xi)*(5—6*Nu2+Nu1*(-6+8*Nu2))
-2*exp(2*h1*xi)*(—1+2*Nu2)*(-1+2*Nu1+2*pow(h1 ,2)*pow(xi,2));

AM3=3+exp(4*h1*xi)*(3-4*Nu1)-4*Nu1+2*exp(2*h1*xi)*(5-12*Nu1
+8*pow(Nu1 ,2)+2*pow(h1 ,2)*pow(xi,2));

AM=pow(E1,2)*pow((1+Nu2),2)*(-3+4*Nu2)*AM1-

214

2*E1*E2*(1+Nu1)*(1+Nu2)*AM2-pow(E2,2)*pow((1+Nu1),2)*AM3;

double BN11=0,BN12=0,BN1=0,B1=0;

BN11=-E2*(1+Nu1)+E1*(-3+Nu2+4*pow(Nu2,2));

BN12=E1*(1+Nu2)*(-1+exp(2*h1*xi)-2*h1*xi)—E2*(1+Nu1)*(-1
+exp(2*h1*xi)*(-3+4*Nu1)-2*h1*xi);

BN1 =—exp(2*h1*xi)*BN1 1*BN12;

B1=BN1/AM;

double DN11=0,DN12=0,DN1=0,D1=0;

DN1 1=E1*(-3+Nu2+4*pow(Nu2,2))*(1+exp(2*h1*xi)*(—1+2*h1*xi));
DN12=E2*(1+Nu1)*(-3+4*Nu1+exp(2*h1*xi)*(—1+2*h1*xi));
DN1=(-E2*(1+Nu1)+E1*(1+Nu2))*(DN11-DN12);

D1=DN1lAM;

double AN1 1=0,AN12=0,AN13=0,AN1 =0,A1 =0;

AN1 1=(-3+4*Nu2)*(1+exp(2*h1*xi)*(-1+2*Nu1)+2*h1*xi
+2*pow(h1,2)*pow(xi,2)-2*Nu1*(1+2*h1*xi));

AN12=5+8*pow(Nu1,2)+exp(2*h1*xi)*(3-10*Nu1+8*pow(Nu1,2))
+2*h1*xi+2*pow(h1 ,2)*pow(xi,2)-2*Nu1*(7+2*h1*xi);

AN13=exp(2*h1*xi)*(—1+2*Nu1)*(5-6*Nu2+Nu1*(-6+8*Nu2))
+(-1+2*Nu2)*(-1+h1*(2-4*Nu1)*xi+2*pow(h1,2)*pow(xi,2));

AN1=exp(2*h1*xi)*(pow(E1,2)*pow((1+Nu2),2)*AN1 1+pow(E2,2)*pow((1
+Nu1),2)*AN12-2*E1*E2*(1+Nu1)*(1+Nu2)*AN13);

A1 =AN1l(-xi*AM);

double CN11=0,CN12=0,CN13=0,CN1=0,C1=0;

CN1 1=(-1+2*Nu1)*(-3+2*Nu1+2*Nu2)+exp(2*h1*xi)*(-1+2*Nu2)*(-1
+2*h1*(-1+2*Nu1)*xi+2*pow(h1,2)*pow(xi,2));

CN12=3~10*Nu1+8*pow(Nu1 ,2)+exp(2*h1*xi)*(5+8*pow(Nu1 ,2)
-2*h1*xi+2*pow(h1 ,2)*pow(xi,2)+2*Nu1*(—7+2*h1*xi));

CN13=(-3+4*Nu2)*(-1+2*Nu1+exp(2*h1*xi)*(1-2*h1*xi
+2*pow(h1,2)*pow(xi,2)+Nu1*(-2+4*h1*xi)));

CN1=-2*E1*E2*(1+Nu1)*(1+Nu2)*CN11
+pow(E2,2)*pow((1+Nu1),2)*CN12
+pow(E1,2)*pow((1+Nu2),2)*CN13;

C1 =CN1/(-AM*xi);

double AN21=0,AN22=0,AN2=0,A2=0;

AN21=-1-h1*xi-pow(h1 ,2)*pow(xi,2)+exp(2*h1*xi)*(-1
+2*Nu2)*(-1+h1*xi)+2*Nu2*(1+h1*xi);

AN22=-1+Nu1+Nu2-h1*xi+2*h1*Nu2*xi-pow(h1 ,2)*pow(xi,2)
+exp(2*h1*xi)*(-1+Nu1+3*Nu2-4*Nu1*Nu2-h1*xi+2*h1*Nu1*xi);

AN2=8*exp(2*h1*xi)*E2*(-1+pow(Nu1 ,2))*(-E1*(1+Nu2)*AN21
+E2*(1+Nu1)*AN22);

A2=AN2/(-xi*AM);

215

double BN21=0,BN22=0,BN2=0,BZ=0;

BN21=-1+exp(2*h1*xi)—2*h1*xi;

BN22=-1+exp(2*h1*xi)*(-3+4*Nu1 )-2*h1*xi;

BN2=4*exp(2*h1*xi)*E2*(-1+pow(Nu1,2))*(E1*(1+Nu2)*BN21
-E2*(1+Nu1)*BN22);

BZ=BN2IAM;

double G10=0,G11=0,G12=0,G13=0,G20=0,G21=0,622=0,G23=0;
G10=(1+Nu1 )*(1-2*Nu1)/(E1*pow(xi,2))*((A1+B1*z)*exp(-xi*z)
+(C1+D1*z)*exp(xi*z));
G20=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(A2+BZ*z)*exp(-xi*z);
G1 1=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((B1-A1*xi-B1*z*xi)*exp(-xi*z)
+(D1+C1*xi+D1*z*xi)*exp(xi*z));
GZ1=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(BZ-A2*xi-82*z*xi)*exp(-xi*z);
G12=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((-2*B1+A1*xi
+B1*z*xi)*xi*exp(-xi*z)+(2*D1+C1 *xi+D1*z*xi)*xi*exp(xi*z));
622=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(-2*BZ+A2*xi
+82*z*xi)*xi*exp(-xi*z);
G13=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((3*B1-A1*xi-
B1 *z*xi)*pow(xi,2)*exp(-xi*z)+(3*D1 +C1 *xi
+D1*z*xi)*pow(xi,2)*exp(xi*z));
GZ3=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(3*BZ-A2*xi
-BZ*z*xi)*pow(xi,2)*exp(-xi*z);

double Lambda1=0,Lambda2=0,Mu1=0,Mu2=0;
Mu1=E1/(2*(1+Nu1));

Mu2=E2/(2*(1+Nu2));
Lambda1=(E1*Nu1)/((1+Nu1)*(1-2*Nu1));
Lambda2=(E2*Nu2)/((1+Nu2)*(1-2*Nu2»;

//===for Sigma_z ===

fc[0]=xi*((Lambda1+2*Mu1)*G13—
(3*Lambda1+4*Mu1)*pow(xi,2)*G1 1 )*qh*_j0(xi*r);

fc[1]=xi*((Lambda2+2*Mu2)*G23-
(3*Lambda2+4*Mu2)*pow(xi,2)*621)*qh*_j0(xi*r);

if (r==0.0){
//===for Sigma_r ===
fc[2]=xi*(Lambda1*G13+(Lambda1+2*Mu1)*pow(xi,2)*G1 1 )*qh*_j0(xi*r)
-2*(Lambda1+Mu1 )*pow(xi,2)*G1 1*qh*(xi/2);
fc[3]=xi*(Lambda2*G23+(Lambda2+2*Mu2)*pow(xi,2)*621 )*qh*_j0(xi*r)
—2*(LambdaZ+Mu2)*pow(xi,2)*G21*qh*(xi/2);

//=== Sigma_theta ===
fc[4]=Lambda1*xi*(G13-pow(xi,2)*G1 1 )*qh*_j0(xi*r)

216

+2*(Lambda1+Mu1)*pow(xi,2)*G1 1*qh*(xi/2);
fc[5]=Lambda2*xi*(G23-pow(xi,2)*G21)*qh*_j0(xi*r)

+2*(Lambda2+Mu2)*pow(xi,2)*621*qh*(xi/2);
}

else{
//===for Sigma_r ===

fc[2]=xi*(Lambda1*G13+(Lambda1+2*Mu1)*pow(xi,2)*G1 1 )*qh*_j0(xi*r)
-2*(Lambda1+Mu1)/r*pow(xi,2)*G1 1*qh*_j1(xi*r);

fc[3]=xi*(Lambda2*G23+(Lambda2+2*Mu2)*pow(xi,2)*G21)*qh*_j0(xi*r)
-*(Lambda2+Mu2)/r*pow(xi,2)*621*qh*_j1(xi*r);

//=== Sigma_theta ===
fc[4]=Lambda1*xi*(G13-pow(xi,2)*G1 1 )*qh*_j0(xi*r)
+2*(Lambda1+Mu1)/r*pow(xi,2)*G1 1*qh*_j1(xi*r);
fc[5]=Lambda2*xi*(G23-pow(xi,2)*G21)*qh*_j0(xi*r)
+2*(Lambda2+Mu2)/r*pow(xi,2)*621*qh*_j1(xi*r);
}

//===for Tau_rz ===

fc[6]=pow(xi,2)*(Lambda1*G12+(Lambda1
+2*Mu1 )*pow(xi,2)*G10)*qh*_j1(xi*r);

fc[7]=pow(xi,2)*(LambdaZ*G22+(Lambda2
+2*Mu2)*pow(xi,2)*620)*qh*_j1(xi*r);

I ===for U_z ===

fc[8]=xi*(G12-(Lambda1+2*Mu1 )/Mu1*pow(xi,2)*G10)*qh*_j0(xi*r);
fc[9]=xi*(622-(Lambda2+2*Mu2)/Mu2*pow(xi,2)*GZO)*qh*_j0(xi*r);
//===for U_r ===
fc[10]=(Lambda1+Mu1)/Mu1*pow(xi,2)*G11*qh*_j1(xi*r);

fc[1 1]=(Lambda2+Mu2)/Mu2*pow(xi,2)*GZ1*qh*_j1(xi*r);

return 0;

217

Appendix GG C++ source code for single layer bonded to an elastic half
space with shear slip and normal separation theories

l/ 30 two layer half space, shear slip theory at bonding interface and within d
II All units are in SI unit
#include<iostream.h>
#include<fstream.h>
#include <iosfwd>
#include<iomanip.h>
#include<io.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>

lltime calculation headfile
#include <time.h>

//using namespace std;
double *f_integrant(int indenter,double E1 ,double E2,double Nu1,

double Nu2,double h1,double h2,double a, double z,double r,
double xi,double p0,double P,double kr,double kz,double d,double fc[]);

main(void)
{
time_t BeginSec, EndSec, Total; // calculate running time
time(&BeginSec);
//**** Setup input and output file
char
lnFile1[80],
OutFile[80];

printf("\n Please enter the input file name: ");
gets(lnFile1);

printf("\n Please enter the output file name: ");
gets(OutFile);

//****Error message of opening files
constchar
ErrorMsg[]="\n***main: unable to open file: ";
fstream
lnMatl(lnFile1,ios::in),
OutComp(OutFile,ios::out);

if (lnMatl.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnFile1<<"\n\n";

exit(-1);

}

218

else if(OutComp.fail()){
cerr<<ErrorMsg<<OutFile<<"\n\n";
exit(—1 );

}

//Read Gaussian point and weight fucntion x[],w[]
constchar
lnGaussPoint[]="gaussian64.txt";
fstream lnGauss;
lnGauss.open(lnGaussPoint,ios::in);

if (lnGauss.fail()){ //detect if the file can be open
cerr<<ErrorMsg<<lnGaussPoint<<"\n\n";
exit(-1);
}
double x[64],w[64];
int i, j, k;
for(i=0;i<64;i++){
x[i]=0.0;
w[i]=0.0;
while (!lnGauss.eof()){
lnGauss>>i;
lnGauss>>x[i];
lnGauss>>w[i];

lnGauss.close();
cout<<setprecision(6);

llgenerate symmetrucal Gaussian points
//1) copy 0-31 to 32-64 for (0,1)
for(i=63;i>31;i-—){

x[i]=x[i-32];

w[i]=w[i-32];
}
II2)establish 0-31 (-1,0) backward from 32-64
for(i=0;i<32;i++){

x[i]=-x[63-i];

w[i]=w[63-i];
}

//****Read imput file

double E1=0,E2=0,Nu1=0,Nu2=0,h1=0.0,h2=0.0,a=0.0,p0=0.0,P=0.0;
int indenter=0;

double r_max=0; // max value of r

double a0=0, b0=0; ”integration range

int N21 =0,N22=0,Nr=0,Nab=0; //number of points for z and r

219

double kr=0.0, kz=0.0, d=0.0;
//kr (k1)-coefficient Ur, kz(k2) - coefficient for U2
while(!lnMatl.eof()){

lnMatl>>indenter;
lnMatl.ignore(80,'\n');
lnMatl>>E1;
lnMatl.ignore(80,'\n');
lnMatl>>E2;
lnMatl.ignore(80,'\n');
lnMatl>>Nu1;
lnMatl.ignore(80,'\n');
lnMatl>>Nu2;
lnMatl.ignore(80,'\n');
InMatl>>h1 ;
lnMatl.ignore(80,'\n');
lnMatl>>h2;
lnMatl.ignore(80,'\n');
lnMatl>>a;
lnMatl.ignore(80,'\n');
lnMatl>>a0;
lnMatl.ignore(80,'\n');
lnMatl>>b0;
lnMatl.ignore(80,'\n');
lnMatl>>Nab;
lnMatl.ignore(80,'\n');
lnMatl>>r_max;
lnMatl.ignore(80,'\n');
lnMatl>>Nz1;
lnMatl.ignore(80,'\n');
lnMatl>>N22;
lnMatl.ignore(80,'\n');
lnMatl>>Nr;
lnMatl.ignore(80,'\n');
lnMatl>>P;
lnMatl.ignore(80,'\n');
lnMatl>>kr;
lnMatl.ignore(80,'\n');
lnMatl>>kz;
lnMatl.ignore(80,'\n');
lnMatl>>d;
lnMatl.ignore(80,'\n');

// layer thickness
”Young's modulus
”Young's modulus
llPossion ratio
//Possion ratio

// layer thickness
// layer thickness

llcontact radius

// intergation upper limit

// number of integration zones in [a0,b0]
// max r value

//number of points for 2

//number of points for z

// number of points for r

// total force

llcoefficient along r axis

llcoefficient along 2 axis

//crack radius

lnMatl.close();

//*****calculate p0
if (indenter==1){

220

p0=P/(3.14159*pow(a,2)); //***uniform pressure

else if (indenter==2){ //***flat punch - uniform displacement
p0=P/(2.0*3.14159*pow(a,2));

else if (indenter==3){
p0=3.0*P/(2*3.14159*pow(a,2)); //***Spherical lndenter

OutComp<<"Two Layer Composites Half Space with Shear Slip Theory
within aa at bonding interface";

OutComp<<"\n\nlndenter shape="<<indenter
<<" (0-point loading,1-uniform pressure,2—flat,3-spherical)"
<<"\nh1/a="<<h1/a
<<"\nh2/a="<<h2la
<<"\nh1/h2="<<h1/h2
<<"\nE1/E2="<<E1/E2
<<"\n\nYoung's Modulus:E1="<<E1
<<"\nYoung's Modulus:E2="<<E2
<<"\nPossion Ratio:Nu1="<<Nu1
<<"\nPossion_Ratio:Nu2="<<Nu2
<<"\n Layer thickness:h1="<<h1
<<"\nLayer thickness:h2="<<h2
<<"\nContact radius:a="<<a
<<"\nLower limit:a0="<<aO
<<"\nUpper Iimit:b0="<<b0
<<"\nNumber of integration zones:Nab="<<Nab
<<"\nr max="<<r_max
<<"\nCalculated points along z(layer1):Nz1="<<Nz1
<<"\nCalculated points along z(layer2):N22="<<N22
<<"\nCalculated points of r:NF"<<Nr
<<"\nContact interface pressure constatant:p0="<<p0
<<"\nTotal applied load at contact interface:P="<<P
<<"\nShear slip theory, coefficient for Ur:kr(k1)="<<kr
<<"\nShear slip theory, coefficient for Uz:kz(k2)="<<kz
<<"\nShear slip crack length:d="<<d;

llthe following code defines calculation points (z,r)
double z1[20]={0}, 22[20]={0},r[200]={0}; llz represent x, r represent y axis
for(i=0;i<Nz1;i++){

z1[i+1]=z1[i]+h1/Nz1;

}
22[0]=h1;

for(i=0;i<N22;i++){
22[i+1]=22[i]+h2lN22;
}

221

for(i=0;i<Nr;i++){
r[i+1]=r[i]+r_max/Nr;
}

double Sigma1_z[11][151]={0,0}, Tau1_rz[11][151]={0,0},
Sigma1_r[11][151]={0,0},Sigma1_theta[11][151]={0,0};

double Sigm32_z[11][151]={0,0}, Tau2_rz[11][151]={0,0},
Sigma2_r[11][151]={0,0},Sigma2_theta[11][151]={0,0};

double U1_z[11][151]={0,0},U1_r[11][151]={0,0};

double U2_z[11][151]={0,0},U2_r[11][151]={0,0};

double y0=0; //point in [a0,b0] respect to Gaussian point at x[k]

double fc[12]={0};

//********Multiple integration region - code start ********
double Sigma1_z_m[50]={0},Tau1_rz_m[50]={0},
Sigma1_r_m[50]={0},Sigma1_theta_m[50]={0};
double Sigma2_z_m[50]={0},Tau2_rz_m[50]={0},
Sigma2_r_m[50]={0},Sigma2_theta_m[50]={0};
double U1_z_m[50]={0},U1_r_m[50]={0},U2_z_m[50]={0},U2_r_m[50]={0};
double b[50]={0};
double k1 =0,k2=0;
int m=0;
for(m=0;m<Nab;m++){
b[m+1]=b[m]+(b0-a0)/Nab;
}

//**** Layer 1 ****
for(i=0; i<=Nz1; i++){
for(i=0;i<Nr;i++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){
y0=(b[m+1]-b[m])*x[k]/2+(b[m+1]+b[m])12;
f_integrant(indenter,E1,E2,Nu1,Nu2,h1,h2,a,z1[i],r[j],
y0,p0,P,kr,kz,d,fc);
Sigmal_z_m[m]=Sigma1_z__m[m]+w[k]*(b[m+1]—b[m])/2.0*fc[0];
Sigma1_r_m[m]=Sigma1_r_m[m]+w[k]*(b[m+1]-b[m])l2.0*fc[2];
Sigma1_theta_m[m]=Sigma1_theta_m[m]+w[k]*(b[m+1]-
b[m])l2.0*fc[4];
Tau1_rz_m[m]=Tau1_rz_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[6];
U1_z_m[m]=U1_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[8];
U1_r_m[m]=U1_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[10];
}
}

for(m=0;m<Nab;m++){
Sigma1_z[i][i]=Sigma1_z[i][i]+Sigma1_z_m[m];
Sigma1_r[i][j]=Sigma1_r[i][j]+Sigma1_r_m[m];

222

Sigma1_theta[i][j]=Sigma1_theta[i][i]+Sigma1_theta_m[m];
Tau1_rz[i][j]=Tau1_rz[i][j]+Tau1_rz_m[m];
U1_z[i][j]=U1_z[i][j]+U1_z_m[m];
U1_r[i][j]=U1_r[i][j]+U1_r_m[m];

}

for(m=0;m<Nab;m++){
Sigma1_z_m[m]=0.0;
Sigma1_r_m[m]=0.0;
Sigma1_theta_m[m]=0;
Tau1_rz_m[m]=0.0;
U1_z_m[m]=0.0;
U1_r_m[m]=0.0;

lltiti Layer 2 *****
for(i=0; i<=N22; i++){
for(i=0;i<Nr;i++){
for(m=0;m<Nab;m++){
for(k=0;k<64;k++){
y0=(b[m+1]-b[m])*x[k]/2+(b[m+1]+b[m])l2;
f_integrant(indenter,E1,E2,Nu1,Nu2,h1,h2,a,22[i],r[j],
y0,p0,P,kr,kz,d,fc);
Sigma2__z_m[m]=Sigma2_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[1];
Sigma2_r_m[m]=Sigma2_r_m[m]+w[k]*(b[m+1]-b[m])l2.0*fc[3];
Sigma2_theta_m[m]=Sigma2_theta_m[m]+w[k]*(b[m+1]-
b[m])/2.0*fc[5];
Tau2_rz_m[m]=Tau2_rz_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[7];
U2_z_m[m]=U2_z_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[9];
U2_r_m[m]=U2_r_m[m]+w[k]*(b[m+1]-b[m])/2.0*fc[1 1];
} }
for(m=0;m<Nab;m++){
Sigma2_z[i][j]=Sigma2_z[i][j]+Sigma2_z_m[m];
Sigma2_r[i][i]=Sigma2_r[i][j]+Sigma2_r_m[m];
Sigma2_theta[i][j]=Sigma2_theta[i][j]+Sigma2_theta_m[m];
Tau2_rz[i][j]=Tau2_rz[i][j]+Tau2_rz_m[m];
U2_z[i][j]=U2_z[i][j]+U2_z_m[m];
U2_r[i][j]=U2_r[i][j]+U2_r_m[m];
}
for(m=0;m<Nab;m++){
Sigma2_z_m[m]=0.0;
Sigma2_r_m[m]=0.0;
Sigma2_theta_m[m]=0;

223

Tau2_rz_m[m]=0.0;
U2_z_m[m]=0.0;
U2_r_m[m]=0.0;

//********Multiple integration region - code end ******

time(&EndSec);
Total = EndSec - BeginSec;
OutComp<<"\n\nTotaI_Running_Time="<<Tota|<<"seconds";

//*'k*****************************************
//*
*

//* Output results - Layer 1
//*

/ *******************************************

OutComp<<"\n\nSigma1_z[ij]"<<"\nr[i]";
for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Sigma1_z[j][i];
}

}

/ ********************

OutComp<<"\nTau1_rz[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Tau1_rz[j][i];
}

}

l/********************

OutComp<<"\nSigma1_r[ij]"<<"\nr[i]";
for(k=0;k<=Nz1;k++){

224

OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Sigma1_r[j][i];
}

}

//********************

OutComp<<"\nSigma1_theta[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<Sigma1__theta[j][i];
}

}

I ********************

OutComp<<"\nU1_r[ij]"<<"\nr[i]";

for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<U1_r[j][i];
}

}

ll********************

OutComp<<"\nU1_z[i,j]"<<"\nr[i]";
for(k=0;k<=Nz1;k++){
OutComp<<";z="<<z1 [k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz1;j++){
OutComp<<";"<<U1_z[j][i];
}

}

I *******************************************

225

//*

//* Output results Layer 2 *
//*

I/******************************************‘k
OutComp<<"\n\nSigma2_z[ij]"<<"\nr[i]";

for(k=0;k<=Nz2;k++){
OutComp<<";z="<<22[k];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Sigma2_z[j][i];
}

}

I ********************

OutComp<<"\nTau2_rz[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Tau2_rz[j][i];
}

}

//********************

OutComp<<"\nSigma2_r[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<z2lk];

}
for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<Sigma2_r[j][i];
}

}

I ********************

OutComp<<"\nSigma2_theta[ij]"<<"\nr[i]";

for(k=0;k<=N22;k++){
OutComp<<";z="<<22[k];

}

226

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointllos::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){ ’
OutComp<<";"<<Sigma2_theta[j][i];
}

}

I ********************
OutComp<<"\nU2_r[ij]"<<"\nr[i]";
for(k=0;k<=N22;k++){

OutComp<<";z="<<22[k];
}

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=Nz2;j++){
OutComp<<";"<<U2_r[j][i];
}

}

l/********************
OutComp<<"\nU2_z[ij]"<<"\nr[i]";
for(k=0;k<=Nz2;k++){
OutComp<<";z="<<22[k];

for(i=0;i<Nr;i++){
OutComp<<setiosflags(ios::showpointlios::scientific)<<"\n"<<r[i];
for(j=0;j<=N22;j++){
OutComp<<";"<<U2_z[j][i];
}

}

OutComp.close();
cout<<"\n";

return 0;

ll3D2layer half space calculation
llcalculate f(x) at gaussian point in domain (a0,b0)
/lx (Xi) Gaussian point

double *f_integrant(int indenter,double E1 ,double E2,double Nu1,

double Nu2,double h1,double h2,double a, double z,double r,
double xi,double p0,double P,double kr,double kz,double d,double fc[]){

double ph=0.0; //Hankel transform of fucntion p

227

 

/I=========shear slip theory———-=--——=
double k1=0,k2=0;
double Struve0=0.0,Struve1=0.0,serial=1 .0;
//-------Calculate the StruveH function -—-------
intkk=0;
for(kk=0;kk<10;kk++){
serial=serial*pow((2*kk+1 ),2);
Struve0=Struve0+pow(-1 ,kk)*pow(d*xi,2*kk)/serial;
Struve1=Struve1 +pow(-1 ,kk)*pow(d*xi,2*kk+1 )/(serial*(2*kk+3));
}
//

 

k1 =pow(d,2)*kr*(_j1 (d*xi)*Struve0-_j0(d*xi)*Struve1 );
k2=kz*d/xi*_j1(d*xi);

”===:=========:=======================
if (indenter==0){

ph=P/(2*3.14159)*exp(-a*xi); //point loading
}

if (indenter==1){
ph=a*p0/xi*_j1(a*xi); //***uniform pressure

else if (indenter==2){ //***flat punch — uniform displacement
ph=a*p0/xi*sin(a*xi);
}

else if (indenter==3){
ph=p0*(a*xi*cos(a*xi)-sin(a*xi))/(a*pow(xi.3)); //***Spherical
}

//**** 3D 2 layer on rigid, z=0 on contact surface - top surface
double BM1=0,BM2=0,BM3=0,BM41=0,BM42=0,BM43=0,BM4=0,BM=0;
BM1=-3-2*Nu2+5*pow(Nu2,2)+4*pow(Nu2,3)
+2*E2*(1-pow(Nu2,2))*(k1-k2)*xi+pow(E2,2)*k1*k2*pow(xi,2);
BM2=1+exp(4*h1*xi)«2*exp(2*h1*xi)*(1 +2*pow(h1 ,2)*pow(xi,2));
BM3=(3-4*Nu1)*(1+exp(4*h1*xi))+2*exp(2*h1*xi)*(5-12*Nu1
+8*pow(Nu1 ,2)+2*pow(h1 ,2)*pow(xi,2));
BM41=-3+2*Nu1-Nu2+2*Nu1*Nu2+2*pow(Nu2,2)+E2*xi*(k1-k2)*(1-Nu1);
BM42=5-Nu2-6*pow(Nu2,2)+Nu1*(-6+2*Nu2+8*pow(Nu2,2))
-E2*xi*(k1-k2)*(1-Nu1);
BM43=(-1+Nu2+2*pow(Nu2,2))*(1-2*Nu1-2*pow(h1,2)*pow(xi,2))
-2*E2*h1*pow(xi,2)*(k1+k2)*(1-Nu1);
BM4=BM41+exp(4*h1*xi)*BM42+2*exp(2*h1*xi)*BM43;
BM=pow(E1,2)*BM1*BM2-pow(EZ,2)*pow((1+Nu1),2)*BM3—
2*E1*E2*(1+Nu1)*BM4;

double AN21=0,AN221=0,AN222=0,AN22=0,AN2=0,A2=0;
AN21=1-2*Nu1+2*Nu2-4*h1*Nu2*xi+2*pow(h1,2)*pow(xi,2)

228

+exp(2*h1*xi)*(-1+6*Nu2-2*h1*xi+Nu1*(2-8*Nu2+4*h1*xi));
AN221=(-1-2*pow(h1,2)*pow(xi,2)+4*h1*Nu2*xi)*(1+Nu2)
+2*E2*k1*Nu2*xi-E2*h1*pow(xi,2)*((k1+k2)+(k1-k2)*(2*Nu2-h1*xi));
AN222=1-4*h1*pow(Nu2,2)*xi+E2*pow(h1,2)*(k1+k2)*pow(xi,3)
+h1*xi*(2+E2*(k1+k2)*xi)-Nu2*(-1+2*E2*k1*xi
+2*h1*xi*(1+E2*(k1+k2)*xi));
AN22=AN221+exp(2*h1*xi)*AN222;
AN2=4*exp(2*h1*xi)*E2*(-1+pow(Nu1,2))*(E2*(1+Nu1)*AN21+E1*AN22);
A2=-AN2/(xi*BM);

double BN21=0,BN22=0,BN23=0,BN2=0,BZ=0;

BN21=-1+exp(2*h1*xi)*(-3+4*Nu1 )+2*h1*xi;

BN22=1+Nu2-2*h1*xi-2*h1*Nu2*xi-E2*xi*(k1-h1*k1*xi+h1*k2*xi);

BN23=exp(2*h1*xi)*(-1-Nu2+E2*xi*(k1+h1*k1*xi+h1*k2*xi));

BN2=4*exp(2*h1*xi)*E2*(-1+pow(Nu1,2))*(E2*(1+Nu1)*BN21
+E1*(BN22+BN23));

BZ=BN2IBM;

double BN11=0,BN12=0,BN13=0,BN14=0,BN15=0,BN1=0,B1=0;
BN1 1=-1+exp(2*h1'xi)*(-3+4*Nu1)+2*h1*xi;
BN12=-1+exp(2*h1*xi)+2*h1*xi;
BN13=-3-2*Nu2+4*pow(Nu2,3)+2*E2*(k1-k2)*xi
+pow(E2,2)*k1*k2*pow(xi,2)+pow(Nu2,2)*(5+2*E2*(-k1+k2)*xi);
BN14=(1-2*h1*xi)*(1-Nu2-2*pow(Nu2,2))+E2*xi*(k1+k2)*(1-Nu1);
BN15=5-Nu2-6*pow(Nu2,2)-E2*xi*(k1-k2)+Nu1*(-6+2*Nu2
+8*pow(Nu2,2)+E2*(k1-k2)*xi);
BN1=exp(2*h1*xi)*(pow(E2,2)*pow((1+Nu1),2)*BN11
+pow(E1,2)*BN12*BN13-
2*E1*E2*(1+Nu1)*(BN14+exp(2*h1*xi)*BN15));
B1=-BN1IBM;

double DN11=0,DN12=0,DN13=0,DN14=0,DN15=0,DN1=0,D1=0;
DN1 1=3-4*Nu1+exp(2*h1*xi)*(1+2*h1*xi);
DN12=-1+exp(2*h1*xi)*(1+2*h1*xi);
DN13=-3-2*Nu2+4*pow(Nu2,3)+2*E2*(k1-k2)*xi

+pow(E2,2)*k1*k2*pow(xi,2)+pow(Nu2,2)*(5+2*E2*(—k1+k2)*xi);
DN14=-3-Nu2+2*pow(Nu2,2)+E2*xi*(k1-k2)+Nu1*(2+2*Nu2

+E2*(-k1+k2)*xi);
DN15=(-1-2*h1*xi)*(1-Nu2-2*pow(Nu2,2))+E2*xi*(k1+k2)*(1-Nu1);
DN1=pow(E2,2)*pow((1+Nu1),2)*DN11+pow(E1,2)*DN12*DN13

+2*E1*E2*(1+Nu1)*(DN14-exp(2*h1*xi)*DN15);
D1=DN1IBM;

double A1 =0,C1=0;

A1=(BZ-D1*exp(2*h1*xi)-4*B2*Nu2+2*A2*xi+2*BZ*h1*xi
+B1*(-1+4*Nu1-2*h1*xi))/(2*xi);

229

C1=((B1-82)*exp(-2*h1*xi)+D1*(1-4*Nu1-2*h1*xi))/(2*xi);
double G10=0,G11=0,G12=0,G13=0,G20=0,GZ1=0,G22=0,GZ3=0;

G10=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((A1+B1*z)*exp(-xi*z)
+(C1+D1*z)*exp(xi*z));

GZO=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(A2+BZ*z)*exp(-xi*z);

G1 1=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((B1-A1*xi-B1*z*xi)*exp(—xi*z)
+(D1+C1*xi+D1*z*xi)*exp(xi*z));

G21 =(1 +Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(BZ-A2*xi-82*z*xi)*exp(-xi*z);

G12=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((-2*B1+A1*xi
+B1*z*xi)*xi*exp(-xi*z)+(2*D1+C1*xi+D1*z*xi)*xi*exp(xi*z));

622=(1+Nu2)*(1-2*Nu2)l(E2*pow(xi,2))*(-2*BZ+A2*xi
+BZ*z*xi)*xi*exp(—xi*z);

G13=(1+Nu1)*(1-2*Nu1)/(E1*pow(xi,2))*((3*B1-A1*xi-
*z*xi)*pow(xi,2)*exp(-xi*z)
+(3*D1+C1*xi+D1*z*xi)*pow(xi,2)*exp(xi*z));

G23=(1+Nu2)*(1-2*Nu2)/(E2*pow(xi,2))*(3*82-A2*xi-
*z*xi)*pow(xi,2)*exp(—xi*z);

double Lambda1=0,Lambda2=0,Mu1=0,Mu2=0;
Mu1=E1/(2*(1+Nu1));

Mu2=E2/(2*(1+Nu2));
Lambda1=(E1*Nu1)/((1+Nu1)*(1-2*Nu1»;
Lambda2=(E2*Nu2)/((1+Nu2)*(1-2*Nu2»;

//===for Sigma_z ===
fc[0]=xi*((Lambda1+2*Mu1)*G13—
*Lambda1+4*Mu1 )*pow(xi,2)*G1 1 )*ph*_j0(xi*r);
fc[1]=xi*((Lambda2+2*Mu2)*G23—
*Lambda2+4*Mu2)*pow(xi,2)*GZ1)*ph*_j0(xi*r);

if (r==0.0){
//===for Sigma_r ===

fc[2]=xi*(Lambda1*G13+(Lambda1+2*Mu1)*pow(xi,2)*G1 1 )*ph*_j0(xi*r)
-2*(Lambda1+Mu1)*pow(xi,2)*G1 1*ph*(xi/2);
fc[3]=xi*(Lambda2*G23+(Lambda2+2*Mu2)*pow(xi,2)*G21 )*ph*_j0(xi*r)
-2*(Lambda2+Mu2)*pow(xi,2)*GZ1*ph*(xi/2);
//=== Sigma_theta ===
fc[4]=Lambda1*xi*(G13-pow(xi,2)*G1 1 )*_j0(xi*r)
+2*(Lambda1+Mu1)*pow(xi,2)*G1 1*ph*(xi/2);
fc[5]=Lambda2*xi*(G23-pow(xi,2)*621 )*_j0(xi*r)
+2*(Lambda2+Mu2)*pow(xi,2)*621*ph*(xi/2);
}

else{

230

/l===for Sigma_r ===

fc[2]=xi*(Lambda1*G13+(Lambda1
+2*Mu1)*pow(xi,2)*G1 1 )*ph*_j0(xi*r)
-*(Lambda1+Mu1)/r*pow(xi,2)*G11*ph*_j1(xi*r);

fc[3]=xi*(Lambda2*G23+(Lambda2+2*Mu2)*pow(xi,2)*G21)*ph*_j0(xi*r)
-*(Lambda2+Mu2)/r*pow(xi,2)*621*ph*_j1(xi*r);

//=== Sigma_theta ===
fc[4]=Lambda1*xi*(G13-pow(xi,2)*G1 1 )*_j0(xi*r)
+2*(Lambda1+Mu1)/r*pow(xi,2)*G1 1*ph*_j1(xi*r);
fc[5]=Lambda2*xi*(G23-pow(xi,2)*G21 )*_j0(xi*r)
+2*(Lambd32+Mu2)/r*pow(xi,2)*G21*ph*_j1(xi*r);
}

//===for Tau_rz ===

fc[6]=pow(xi,2)*(Lambda1*G12+(Lambda1
+2*Mu1 )*pow(xi,2)*G10)*ph*_j1 (xi*r);

fc[7]=pow(xi,2)*(Lambda2*G22+(Lambda2
+2*Mu2)*pow(xi,2)*620)*ph*_j1(xi*r);

I/===for U_z ===
fc[8]=xi*(G12-(Lambda1+2*Mu1 )lMu1*pow(xi,2)*G10)*ph*_j0(xi*r);
fc[9]=xi*(622-(Lambda2+2*Mu2)/Mu2*pow(xi,2)*GZO)*ph*_j0(xi*r);

//===for U_r ===
fc[10]=(Lambda1+Mu1)lMu1*pow(xi,2)*G11*ph*_j1(xi*r);
fc[1 1]=(Lambda2+Mu2)/Mu2*pow(xi,2)*G21*ph*_j1(xi*r);
return 0;

231

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