ANALYTICAL AND FINITE ELEMENT BASED ...ufdcimages.uflib.ufl.edu/UF/E0/05/19/30/00001/...analytical model is first verified by comparing with results from finite element based micro-mechanics

ANALYTICAL AND FINITE ELEMENT BASED MICROMECHANICS FOR FAILURE THEORY OF COMPOSITES

By

SAI THARUN KOTIKALAPUDI

A THESIS PRESENTED TO THE GRADUATE SCHOOL

OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF

MASTER OF SCIENCE

UNIVERSITY OF FLORIDA

2017

© 2017 Sai Tharun Kotikalapudi

To Amma and Nanna for the incessant support and always believing in me

4

ACKNOWLEDGMENTS

I would like to express my gratitude to my thesis advisor Dr. Bhavani V. Sankar

for being a very supporting guide and mentor, and for giving me a chance to participate

in his research program critical to composite industry. He has consistently steered me in

the right direction and been a lamppost on this hazy road of exploration. I would also

like to thank Dr. Ashok V. Kumar for willing to be a member of the supervisory

committee and offer constructive criticism wherever needed.

I am grateful for my parents and Srinivas Namagiri for all their support and

encouragement. Without them this day would have never dawned. I am beholden to

Sumit Jagtap for his guidance on Abaqus as well as to Apoorva Walke and Maleeha Babar

for being a constant source of moral support. I would also like to acknowledge all my

teachers from University of Florida and SASTRA University, and my friends and family

for being there for me in the time of need.

5

TABLE OF CONTENTS page

ACKNOWLEDGMENTS .................................................................................................. 4

LIST OF TABLES ............................................................................................................ 6

LIST OF FIGURES .......................................................................................................... 8

LIST OF ABBREVIATIONS ........................................................................................... 11

ABSTRACT ................................................................................................................... 12

CHAPTER

1 INTRODUCTION .................................................................................................... 14

Literature Review .................................................................................................... 14

Research Scope ..................................................................................................... 15

2 ANALYTICAL EQUATIONS .................................................................................... 20

Introduction to the Three-Phase Model ................................................................... 20 Halpin Tsai Formulation for Composite Properties ................................................. 21 Longitudinal and Hydrostatic Stress Equations ....................................................... 23

Longitudinal Shear Stress in the x-y plane.............................................................. 27

Longitudinal Shear Stress in the x-z plane.............................................................. 32 Biaxial tension/compression in y-z plane ................................................................ 37 Transverse Shear Equations .................................................................................. 43

3 FINITE ELEMENT ANALYSIS AND COMPARISON .............................................. 50

Modelling and analysis of Hexagonal RVE ............................................................. 50 Comparison with analytical model .......................................................................... 61

4 ANALYTICAL MODEL RESULTS AND DISCUSSION ........................................... 65

Results for Kevlar/Epoxy ......................................................................................... 65

Carbon/Epoxy plots ................................................................................................ 69 Effects of Interface .................................................................................................. 73 Volume fraction analysis ......................................................................................... 81

Summary ................................................................................................................ 84

5 CONCLUSIONS AND FUTURE WORK ................................................................. 87

LIST OF REFERENCES ............................................................................................... 90

BIOGRAPHICAL SKETCH ............................................................................................ 92

6

LIST OF TABLES

Table page 2-1 Comparison of macro stresses with average micro stresses for longitudinal

shear stress ........................................................................................................ 22

2-2 Comparison of macro stresses with average micro stresses for normal Stress and in plane shear stress ................................................................................... 22

3-1 Properties of Kevlar/Epoxy used in the FEA ....................................................... 52

3-2 Coefficients of stiffness matrix obtained from unit strain analysis ....................... 56

3-3 Transverse strengths at various points for Kevlar/Epoxy (plane strain) .............. 60

3-4 Comparison of various transverse strengths for Kevlar/Epoxy (plane strain) ..... 63

3-5 Comparison of maximum principal stress for Kevlar/Epoxy (plane strain) .......... 63

3-6 Comparison of maximum von Mises stress for Kevlar/Epoxy (plane strain) ....... 63

3-7 Comparison of average of top 10% maximum principal stresses for Kevlar/Epoxy (plane strain) ................................................................................ 63

3-8 Comparison of average of top 10% von Mises stresses for Kevlar/Epoxy (plane strain)....................................................................................................... 63

3-9 Comparison of 10th percentile maximum principal stress for Kevlar/Epoxy (plane strain)....................................................................................................... 64

3-10 Comparison of 10th percentile maximum von Mises stress for Kevlar/Epoxy (plane strain)....................................................................................................... 64

4-1 Properties of Kevlar/Epoxy ................................................................................. 66

4-2 Strengths at various points for Kevlar/Epoxy (MPa) ........................................... 69

4-3 Properties of Carbon-T300/Epoxy-5208 ............................................................. 70

4-4 Predicted strengths of T300/5208/Carbon/Epoxy ............................................... 73

4-5 Comparison of strengths for Kevlar/epoxy including interface failure obtained using ADMM ....................................................................................................... 80

4-6 Comparison of strengths for Carbon/Epoxy including interface failure obtained using ADMM ........................................................................................ 81

7

4-7 Comparison of strengths for several composites with analytical model strengths ............................................................................................................. 85

4-8 %Difference of strengths for several composites relative to reference strengths ............................................................................................................. 85

8

LIST OF FIGURES

Figure page 1-1 Depiction of a RVE for the analytical model ....................................................... 16

1-2 Decomposition of macro stresses applied to an RVE of a fiber composite ......... 16

1-3 Macro stresses applied on the unit cell. (similar to 𝜏12 , 𝜏13 will be acting in the 13 plane and 𝜏23 will be acting in the 2-3 plane) .......................................... 17

1-4 Decomposition of applied state of macro stresses into five cases ...................... 18

2-1 Three-phase model ............................................................................................ 20

3-1 Representative volume element of a hexagonal unit cell .................................... 50

3-2 Coordinate system used in ABAQUS and principal coordinate system .............. 51

3-3 Sectional view and dimensions of the RVE ........................................................ 51

3-4 Meshed RVE, red bounded regions represent fiber and green unbounded region represents matrix ..................................................................................... 52

3-5 Element type used for meshing and analysis ..................................................... 53

3-6 Boundary conditions and loading in unit strain analysis (A) Direction 2 (B) direction 3 ........................................................................................................... 54

3-7 Schematic of the procedure followed to obtain Stiffness matrix.......................... 55

3-8 Initial and deformed hexagonal RVE under unit strain in A) 2nd Direction B) 3rd direction C) 2nd and 3rd direction..................................................................... 56

3-9 Schematic of procedure to plot a failure envelope in 2-3 plane .......................... 59

3-10 Failure envelopes of Kevlar/Epoxy in transverse direction obtained through unit strain analysis .............................................................................................. 60

3-11 Comparison of analytical and finite element model failure envelopes using maximum stress theory in the transverse plane (2-3 plane) ............................... 61

3-12 Comparison of analytical and finite element model failure envelopes using quadratic theory in the transverse plane (2-3 plane) .......................................... 62

4-1 Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy on 𝜎1 − 𝜎2 plane ................................................................................................................... 66

9

4-2 Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy in the 𝜎2 −𝜎3 plane .............................................................................................................. 67

4-3 Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy for longitudinal shear in the 1-2 or 1-3 plane ........................................................... 67

4-4 Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy subjected to both longitudinal and transverse shear stresses............................................. 68

4-5 Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy for shear in longitudinal direction and stress in fiber direction ............................................... 68

4-6 Comparison of MMN and QQN failure envelopes of Carbon/Epoxy in 1-2 plane ................................................................................................................... 70

4-7 Comparison of MMN and QQN failure envelopes of Carbon/Epoxy in 2-3 plane ................................................................................................................... 71

4-8 Comparison of MMN and QQN failure envelopes of Carbon/Epoxy for shear in longitudinal directions ..................................................................................... 71

4-9 Comparison of MMN and QQN failure envelopes of Carbon/Epoxy subjected to both longitudinal and transverse shear stresses............................................. 72

4-10 Comparison of MMN and QQN failure envelopes of Carbon/Epoxy for longitudinal shear and normal stress in fiber direction ........................................ 72

4-11 Interface effects on failure envelopes for Kevlar/Epoxy in 1-2 plane using maximum stress theory ...................................................................................... 74

4-12 Interface effects on failure envelopes for Kevlar/Epoxy in 1-2 plane using quadratic theory .................................................................................................. 75

4-13 Interface effects on failure envelopes for Kevlar/Epoxy in 2-3 plane using maximum stress theory ...................................................................................... 75

4-14 Interface effects on failure envelopes for Kevlar/Epoxy in 2-3 plane using quadratic theory .................................................................................................. 76

4-15 Interface effects on failure envelopes subjected to both longitudinal and transverse shear stresses using maximum stress theory ................................... 76

4-16 Interface effects on failure envelopes subjected to both longitudinal and transverse shear stresses using quadratic theory .............................................. 77

4-17 Interface effects on failure envelopes for Carbon/Epoxy in 1-2 plane using maximum stress theory ...................................................................................... 77

10

4-18 Interface effects on failure envelopes for Carbon/Epoxy in 1-2 plane using quadratic theory .................................................................................................. 78

4-19 Interface effects on failure envelopes for Carbon/Epoxy in 2-3 plane using maximum stress theory ...................................................................................... 78

4-20 Interface effects on failure envelopes for Carbon/Epoxy in 2-3 plane using quadratic theory .................................................................................................. 79

4-21 Interface effects on failure envelopes for Carbon/Epoxy for envelopes subjected to both longitudinal and transverse shear stresses ............................ 79

4-22 Interface effects on failure envelopes for Carbon/Epoxy longitudinal shear and normal stress in fiber direction using maximum stress theory ..................... 80

11

LIST OF ABBREVIATIONS

ADMM Analytical Direct Micromechanics Method

DMM

FEA

PBC

RVE

Direct Micromechanics Method

Finite Element Analysis

Periodic boundary Conditions

Representative Volume Element

12

Abstract of Thesis Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Master of Science

ANALYTICAL AND FINITE ELEMENT BASED MICROMECHANICS FOR FAILURE

THEORY OF COMPOSITES

By

Sai Tharun Kotikalapudi

December 2017

Chair: Bhavani V. Sankar Major: Mechanical Engineering

An analytical method using elasticity equations to predict the failure of a

unidirectional fiber-reinforced composite under multiaxial stress is presented. This

technique of calculating micro-stresses using elasticity equations and estimating the

strengths of a composite is based on the Direct Micromechanics Method (DMM).

Prediction of failure using phenomenological failure criteria such as Maximum Stress,

Maximum Strain and Tsai-Hill theories have been prevalent in the industry. However,

DMM has not been used in practice due to its prohibitive computational effort such as

the finite element analysis (FEA). The present method replaces the FEA in DMM by

analytical methods, thus drastically reducing the computational effort.

A micromechanical analysis of unidirectional fiber-reinforced composites is

performed using the three-phase model. A given state of macro-stress is applied to the

composite, and the micro-stresses in the fiber and matrix phases and along the fiber-

matrix interface are calculated. The micro-stresses in conjunction with failure theories

for the constituent phases are used to determine the integrity of the composite. The

analytical model is first verified by comparing with results from finite element based

micro-mechanics. Then, it is used to study the failure envelopes of various composites.

13

The effects of fiber-matrix interface on the strength of the composite is studied. The

results are compared with those available in the literature. It is found that the present

analytical Direct Micro-Mechanics (ADMM) predicts the strength of composites

reasonably well.

14

CHAPTER 1 INTRODUCTION

Literature Review

With the growing application of fiber composites, and with the tremendous

progress in low-cost manufacturing of composite structures, e.g., wind turbine blades,

there is a need to develop efficient predictive methodologies for the behavior of

composites. This should include probabilistic methods to aid in nondeterministic

optimization tools used in design. While methods to predict stiffness properties are well

established, methods to predict failure and fracture properties are still evolving.

Computational material science is the new field of study which attempts to use modern

computational analysis tools to perform multiscale analysis beginning from atomic scale

all the way up to structural scale.

While large scale computational methods are being advanced, there is always a

need for simple and efficient analytical methodologies. This is especially true for

strength prediction and failure behavior of composites. Currently available methods for

strength prediction either use numerical simulations such as finite element analysis [1]

or very simple methods such as mechanics of materials models (MoM) [2]. The former

can be expensive and time consuming, and the latter is only an approximate estimate to

be useful in practical design applications. The current study is aimed at developing an

analytical micromechanics method that is better than MoM models, but still not as

complicated as FEA based micromechanics. To this end we use the principles of Direct

Micromechanics Method [3] developed in 1990s in conjunction with the classical three-

phase elasticity model for unidirectional fiber composites [4].

15

Several failure theories for composite materials are available in the literature.

Majority of them use experimental results along with an empirical (phenomenological)

approach to plot the failure envelopes.

Contemporary failure theories, developed for unidirectional composites such as

Maximum Stress Theory, Maximum Strain Theory and Tsai-Hill Theory have been

thoroughly studied and implemented by various researchers and design engineers.

Direct micromechanics method (DMM) has a propitious approach to predict failure

strengths for an orthotropic composite material. First proposed by Sankar, it has been

widely used to analyze various phenomenological failure criteria, e.g., Marrey and

Sankar [5], Zhu et al. [6], Stamblewski et al. [7], karkakainen and Sankar. DMM

encompasses analytical techniques which is an alternative approach for physical testing

and experimental procedures. A micromechanical model is subjected to multiple macro

stresses which produce micro stresses in each element of the finite element model. The

micro stresses are used to devise a failure envelope considering various failure criteria

for fiber and matrix such as maximum principal stress theory and von Mises criterion.

The interface of fiber/matrix played a pivotal role in the failure of envelopes which was

also considered in DMM. Interfacial tensile stress and interfacial shear stress in the

composite are very sensitive properties that depend on various factors.

Research Scope

In this section, the research procedure followed will be discussed in detail. The

RVE for the analytical approach has been modeled as circular fiber surrounded by an

annular region of matrix. The description of analytical model is presented in Chapter 2.

16

Figure 1-1. Depiction of a RVE for the analytical model.

The applied macro stresses on the composite are denoted by the Cauchy stress

tensor [𝜎𝑥 𝜎𝑦 𝜎𝑧 𝜏𝑦𝑧 𝜏𝑥𝑦 𝜏𝑥𝑧], Note that throughout this study the principal material

coordinate system of the fiber composite will be denoted either by the standard 1-2-3

coordinates or x-y-z coordinates used in the commercial finite element software

ABAQUS. The two normal stresses 𝜎𝑦 and 𝜎𝑧 can be decomposed into two cases

hydrostatic stress state such that 𝜎𝑦 = 𝜎𝑧 = 𝜎𝐻 and a biaxial tension/compression

such that 𝜎𝑦 = −𝜎𝑧. The chart in Figure1-2 depicts various analysis of the stress cases

needed to complete the DMM.

Figure 1-2. Decomposition of macro stresses applied to an RVE of a fiber composite

Applied stress

Normal stress

Axail and hydrostaic

Biaxial tension and compresion

Shear stress

Transverse

Shear YZ

Longitudinal Shear XZ

Longitudinal Shear XY

Fiber

Matrix

Composite

17

Application of the stress field on the Representative Volume Element (RVE) of

the composite through individual cases generates macro strains. Every case has

distinct analytical equations for calculating the micro stresses in the fiber and matrix

phases. The load factors are calculated based on the type of failure criterion used either

maximum stress or some form of quadratic failure criterion, e.g. von Mises for isotropic

materials.

Figure 1-3. Macro stresses applied on the unit cell. (similar to 𝜏12 , 𝜏13 will be acting in the 13 plane and 𝜏23 will be acting in the 2-3 plane)

𝜏12

𝜏12

𝜏12

𝜏12

𝜎1

𝜎3

𝜎3

𝜎2 𝜎2

𝜎1

18

Figure 1-4. Decomposition of applied state of macro stresses into five cases A) Hydrostatic and longitudinal stress; B) Biaxial tension and compression; C) Shear in 2-3 plane; D) Shear in 1-2 plane; E) Shear in 1-3 plane

𝜏23

𝜏23 𝜏23

𝜎1

𝜎1

𝜎𝐻 𝜎𝐻

𝜎𝐻

𝜎𝐻 =(𝜎2 + 𝜎3)

2

(A) Case i

𝜎𝑇𝐶

−𝜎𝑇𝐶

−𝜎𝑇𝐶

𝜎𝑇𝐶 =(𝜎2 − 𝜎3)

2

(B) Case ii

(E) Case v

𝜏13

(C)Case iii

𝜏23

(D)Case iv

𝜏12

19

Figure 1-5. Schematic depiction of DMM followed to obtain failure envelopes

Figure 1-3 shows the six stresses acting on the unit cell of the composite which

are divided into five cases as shown in figure 1-4. For each case the micro stresses are

calculated at several locations using stress equations which is explained in detail in

chapter 2. Figure 1-5 portrays a schematic of the process followed in Direct

Micromechanics Method (DMM) to obtain the failure envelopes and strengths. Chapter

2 elaborates the stress equations used for micromechanical analysis and validation of

using energy methods. The analytical equations employed are further validated in

Chapter 3 through unit strain analysis in finite element analysis software ABAQUS.

Chapter 4 consists of the results obtained from the analytical model wherein a thorough

study is performed by considering two different materials i.e. isotropic and transversely

isotropic. A meticulous comparison on different strengths of composites with present

data is included in Chapter 4. A study is performed in Chapter 4 to understand the effect

of fiber volume fraction on the strength properties for few materials.

Macro stressesIsolating

stresses/formulating stress equations

Micro stresses

Eigen values/principal

stressesLoad Factors

Failure envelopes and strengths

20

CHAPTER 2 ANALYTICAL EQUATIONS

Introduction to the Three-Phase Model

In this section, the three-phase concentric cylinder composite assemblage model

is described. The three phase model proposed by Christensen [8] had been

successfully used in the past for predicting the elastic constants of fiber composites,

e.g. Flexible-resin/glass-fiber composite lamina [9]. In the present study we investigate

the use of three phase model to predict the strengths of unidirectional fiber composites.

Figure 2-1. Three-phase model

The model shown in Figure 2-1. consists of a single cylindrical inclusion (fiber)

embedded in a cylindrical annular region of matrix material. The composite cylinder is in

turn embedded in infinite medium properties of which are equal to that of the composite

material studied. The fiber and the composite are assumed transversely isotropic and

the matrix is isotropic. This enables us to use a polar coordinate system for the analysis.

Furthermore, the entire assemblage is in a state of generalized plane strain as the strain

2

1

3

r b

∞

a Fiber

Matrix

Composite

21

휀1 must be uniform and the same in all three phases. Thus, the problem becomes a

plane problem. Since we are using the model to calculate the micro-stresses in the fiber

and matrix for a give macro-stress state, the elastic constants of all three phases must

be available for the stress analysis. As a first step, the Rule of Mixtures and Halpin-Tsai

equations are used to estimate the elastic constants of the composite. In each analysis

energy equivalence verifies the validity of the input composite elastic constants as

explained in subsequent sections.

Halpin Tsai Formulation for Composite Properties

Halpin-Tsai equation is a widely used semi-empirical formulation for transverse

moduli 𝐸2, 𝐺12 of unidirectional fiber composites. The general form of Halpin-Tsai

equations for a property, say 𝑃, is as follows:

𝑃𝑐 = 𝑃𝑚 (1 + 𝜉𝜂𝑉𝑓

1 − 𝜂𝜈𝑓) (2-1)

Where,

𝜂 = ((𝑃𝑓/𝑃𝑚) − 1

(𝑃𝑓/𝑃𝑚) + 𝜉) (2-2)

𝑃𝑐: Property of the composite

𝑃𝑓: Property of the fiber

𝑃𝑚: Property of the matrix

𝜉: Curve fitting parameter

𝑉𝑓: Fiber volume fraction

The above formula was obtained using curve fitting the result for square array of

circular fibers. It is found that for 𝜉 = 2, an excellent fit is obtained for transverse

modulus 𝐸2. Whereas for shear modulus 𝐺12, the value 𝜉 = 1 was in excellent

22

agreement with the Adams and Doner [10] solution. In both cases a fiber volume

fraction of 𝑉𝑓 = 0.55 is used.

Since the analytical model in this paper has a circular fiber in an annular region

of matrix, the curve fitting parameter has been adjusted to formulate more accurate

predictions of moduli - 𝐸2, 𝐺12. The curve fitting parameter 𝜉 was estimated for the

present case by comparing the applied macro stresses in each case to the volume

average of the corresponding micro stresses. The modified values of 𝜉 are: 𝜉 = 1.16 for

transverse Young’s modulus 𝐸2, 𝜉 = 1 for transverse shear modulus 𝐺23 and 𝜉 = 1 for

longitudinal or axial shear modulus 𝐺12.

Table 2-1. Comparison of macro stresses with average micro stresses for longitudinal shear stress

case Fiber Matrix Average stresses

𝜏𝑥𝑧 𝜏𝑥𝑦 𝜏𝑥𝑧 𝜏𝑥𝑦 𝜏𝑥𝑧 𝜏𝑥𝑦

𝜏𝑥𝑧=1 1.235 0 0.647 0 1 0

𝜏𝑥𝑦=1 0 1.235 0 0.647 0 1

Table 2-2. Comparison of macro stresses with average micro stresses for normal stress

and in plane shear stress Case Fiber Matrix Average stresses

𝜎𝑥 𝜎𝑦 𝜎𝑧 𝜏𝑦𝑧 𝜎𝑥 𝜎𝑦 𝜎𝑧 𝜏𝑦𝑧 𝜎𝑥 𝜎𝑦 𝜎𝑧 𝜏𝑦𝑧

𝜎𝑥=1 1.612 0 0 0 0.081 0 0 0 1 0 0 0

𝜎𝑦=1 -0.18 1.166 -0.07 0 0.274 0.752 0.105 0 0 1 0 0

𝜎𝑧=1 -0.18 -0.07 1.166 0 0.274 0.105 0.752 0 0 0 1 0

𝜏𝑦𝑧=1 0 0 0 1.236 0 0 0 0.647 0 0 0 1

23

The adjusted curve fitting parameter 𝜉 for transverse modulus is 1.16,

considering the average stresses in the composite compared satisfactorily with the input

non-zero stress for each case as depicted in tables 2-1 and 2-2.

To summarize, longitudinal modulus 𝐸1 of the composite is calculated from rule

of mixtures:

𝐸1 = 𝐸1𝑓𝑉𝑓 + (1 − 𝑉𝑓)𝐸𝑚 (2-3)

Here, 𝐸1: Longitudinal modulus of the composite

𝐸1𝑓: Longitudinal modulus of the fiber

𝐸𝑚: Young’s modulus of matrix

𝑉𝑓: Fiber volume fraction

Transverse modulus 𝐸2, longitudinal shear modulus 𝐺12 and in-plane shear

modulus 𝐺23 are obtained from Halpin-Tsai equations.

Longitudinal and Hydrostatic Stress Equations

In this section, the displacement equations for the cases of longitudinal and

hydrostatic stresses are derived. Since we are using the composite cylinder model,

cylindrical coordinate system is used. The above two cases are also axis-symmetric.

Since the composite cylinder is under plane strain condition and an axisymmetric

model is assumed, we have only one non-zero displacement, which is the radial

displacement, 𝑢𝑟 . The displacement equation for all the three phases (fiber, matrix and

composite) is given below.

𝑢𝑟𝑖 = 𝐴𝑖𝑟 +𝐵𝑖

𝑟 (2-4)

24

Here, 𝑢𝑟𝑖 radial displacement in phase 𝑖 and 𝐴𝑖 , 𝐵𝑖 are constants to be

determined using various interface and boundary conditions.

The radial displacement equation for the fiber phase is shown below. Here, the

subscript 𝑓 denotes all the variables are pertaining to the fiber phase.

𝑢𝑟𝑓 = 𝐴𝑓𝑟 +𝐵𝑓

𝑟 (2-5)

One can deduce 𝐵𝑓 = 0 as the displacement at the center of the fiber is zero. The

strains are derived as follows:

휀𝑟 =𝜕𝑢𝑟

𝜕𝑟→ 휀𝑟𝑓 = 𝐴𝑓 (2-6)

휀𝜃 =1

𝑟

𝜕𝑢𝜃

𝜕𝜃+

𝑢𝑟

𝑟→ 휀𝜃𝑓 = 𝐴𝑓 (2-7)

The radial displacement equation for the matrix phase is derived below. The

subscript 𝑚 denotes all the variables are pertaining to the matrix phase.

𝑢𝑟𝑚 = 𝐴𝑚𝑟 +𝐵𝑚

𝑟 (2-8)

Following procedures as in fiber phase above, we get

휀𝑟𝑚 = 𝐴𝑚 −𝐵𝑚

𝑟2 (2-9)

휀𝜃𝑚 = 𝐴𝑚 +𝐵𝑚

𝑟2 (2-10)

The radial displacements in the composite phase are shown below. Here, the

subscript 𝑐 denotes all the variables are pertaining to the composite phase.

𝑢𝑟𝑐 = 𝐴𝑐𝑟 +𝐵𝑐

𝑟 (2-11)

휀𝑟𝑚 = 𝐴𝑐 −𝐵𝑐

𝑟2 (2-12)

휀𝜃𝑐 = 𝐴𝑐 +𝐵𝑐

𝑟2 (2-13)

25

Considering a composite unit cell, the strains along the longitudinal direction can

be equated as follows:

휀𝑥𝑓 = 휀𝑥𝑚 = 휀𝑥𝑐 = 휀0 (2-14)

Where, 휀0 is the longitudinal strain in the fiber direction applied to the composite.

Continuity Equations

Continuity of displacement and radial stresses must be ensured along the

fiber/matrix interface and matrix/composite interface. The interface continuity equations,

say between surfaces i and j, are given below:

𝑢𝑟𝑖(𝑟) = 𝑢𝑟𝑗(𝑟) (2-15)

𝜎𝑟𝑖 = 𝜎𝑟𝑗 (2-16)

From the equation of continuity of displacement along the fiber/matrix interface

𝑢𝑟𝑓(𝑎) = 𝑢𝑟𝑚(𝑎) (2-17)

𝐴𝑓𝑎2 − 𝐴𝑚𝑎2 − 𝐴𝑐 = 0 (2-18)

Here,

𝑎: Radius of the Fiber Phase

From the equation of continuity of radial stress along the fiber/matrix interface

𝜎𝑟𝑓(𝑎) = 𝜎𝑟𝑚(𝑎) (2-19)

The constitutive relation for transversely isotropic materials can be written as

{

𝜎𝑥

𝜎𝑟

𝜎𝜃

} = [𝐶11 𝐶12 𝐶12

𝐶12 𝐶22 𝐶23

𝐶12 𝐶23 𝐶22

] {

휀𝑥

휀𝑟

휀𝜃

} (2-20)

Hence, the stress continuity Equation (2-19) takes the form,

[𝐶12 𝐶22 𝐶23]𝑓 {

휀𝑥

휀𝑟

휀𝜃

}

𝑓

= [𝐶12 𝐶22 𝐶23]𝑚 {

휀𝑥

휀𝑟

휀𝜃

}

𝑚

26

𝐶12𝑓휀𝑥𝑓 + 𝐶22𝑚휀𝑟𝑓 + 𝐶23𝑓휀𝜃𝑓 = 𝐶12𝑚휀𝑥𝑚 + 𝐶22휀𝑟𝑓 + 𝐶23휀𝜃𝑚 (2-21)

For simplicity, we use the following notations

𝐶22𝑓 + 𝐶23𝑓 = 𝛼𝑓 (2-22)

𝐶22𝑓 − 𝐶23𝑓 = 𝛽𝑓 (2-23)

𝐶22𝑚 + 𝐶23𝑚 = 𝛼𝑚 (2-24)

𝐶22𝑚 − 𝐶23𝑚 = 𝛽𝑚 (2-25)

Then Equation (2-21) can be simplified as

𝐴𝑓𝛾𝑓 − 𝐴𝑚𝛾𝑚 +𝐵𝑚𝛿𝑚

𝑎2+ 휀𝑥(𝐶12𝑓 − 𝐶12𝑚) = 0 (2-26)

Similarly, considering continuity of displacement along the matrix/composite

interface we get

𝑢𝑚(𝑏) = 𝑢𝑐(𝑏) (2-27)

Equating displacements of matrix and composite phases

𝐴𝑚𝑏2 + 𝐵𝑚 − 𝐴𝐶𝑎2 − 𝐵𝐶 = 0 (2-28)

Here, 𝑏 is the radius of matrix phase

Similarly, continuity of radial stress along the matrix/composite interface yield

𝜎𝑟𝑚(𝑏) = 𝜎𝑟𝑐(𝑏) (2-29)

[𝐶12 𝐶22 𝐶23]𝑚 {

휀𝑥

휀𝑟

휀𝜃

}

𝑚

= [𝐶12 𝐶22 𝐶23]𝑐 {

휀𝑥

휀𝑟

휀𝜃

}

𝑐

Let

(𝐶22𝑐 + 𝐶23𝑐) = 𝛾𝑐 (2-30)

(𝐶22𝑐 − 𝐶23𝑐) = 𝛿𝑐 (2-31)

Then Equation (2-29) can be simplified as follows

27

𝐴𝑚𝛾𝑚 − 𝐴𝐶𝛾𝑐 −𝐴𝑚𝛿𝑚

𝑏2+

𝐵𝐶𝛿𝑐

𝑏2+ 휀𝑥(𝐶11𝑚 − 𝐶11𝑐) = 0 (2-32)

Boundary Conditions

At 𝑟 = ∞ the radial stress at the boundary of the composite 𝜎𝑟𝑐 = 𝜎𝐻

From the constitutive relation {𝜎} = [𝐶]{휀} we get

[𝐶12 𝐶22 𝐶23]𝑐 {

휀𝑥

휀𝑟

휀𝜃

}

𝑐

= 𝜎𝐻 (2-33)

𝜎𝐻 = 𝐴𝑐𝛾𝑐 + 휀𝑥𝐶12𝑐 (2-34)

Here, 𝜎𝐻 is the hydrostatic stress applied at the boundary of the composite

The constants 𝐴𝑓 , 𝐴𝑚, 𝐵𝑚, 𝐴𝑐, 𝐵𝑐 can be solved from Equations (2-18), (2-26), (2-

32) and (2-34). The hydrostatic stress 𝜎𝐻 is the remote stress applied to the composite.

The micro stresses in fiber phase and matrix phase for longitudinal and hydrostatic

stresses are calculated from the following equations:

𝜎𝑥𝑓 = 𝐶11𝑓휀𝑥𝑓 + 𝐴𝑓𝐶12𝑓 + 𝐴𝑓𝐶12𝑓 (2-35)

𝜎𝑟𝑓 = 𝐶12𝑓휀𝑥𝑓 + 𝐴𝑓𝐶22𝑓 + 𝐴𝑓𝐶23𝑓 (2-36)

𝜎𝜃𝑓 = 𝐶12𝑓휀𝑥𝑓 + 𝐴𝑓𝐶23𝑓 + 𝐴𝑓𝐶22𝑓 (2-37)

𝜎𝑥𝑚 = 𝐶11𝑚휀𝑥𝑚 + (𝐴𝑚 −𝐵𝑚

𝑟2) 𝐶12𝑚 + (𝐴𝑚 +

𝐵𝑚

𝑟2) 𝐶12𝑚 (2-38)

𝜎𝑟𝑚 = 𝐶12𝑚휀𝑥𝑚 + (𝐴𝑚 −𝐵𝑚

𝑟2) 𝐶22𝑚 + (𝐴𝑚 +

𝐵𝑚

𝑟2) 𝐶23𝑚 (2-39)

𝜎𝜃𝑚 = 𝐶12𝑚휀𝑥𝑚 + (𝐴𝑚 −𝐵𝑚

𝑟2) 𝐶23𝑚 + (𝐴𝑚 +

𝐵𝑚

𝑟2) 𝐶22𝑚 (2-40)

Longitudinal Shear Stress in the x-y plane

In this section, the equations for the case of longitudinal shear stress are derived.

Since we are using the composite cylinder model, cylindrical coordinate system is used.

28

The displacement equation for all the three Phases (fiber, matrix, and composite) is

given below:

𝑢𝑟𝑖 = 𝐶𝑖𝑥𝑐𝑜𝑠𝜃 (2-41)

𝑢𝜃𝑖 = −𝐶𝑖𝑥𝑠𝑖𝑛𝜃 (2-42)

𝑢𝑥𝑖 = (𝐴𝑖𝑟 +𝐵𝑖

𝑟) 𝑐𝑜𝑠𝜃 (2-43)

Here, 𝑢𝑟𝑖, 𝑢𝜃𝑖, 𝑢𝑥𝑖 are radial, angular, and axial displacements in the phase 𝑖

respectively. 𝐴𝑖, 𝐵𝑖, 𝐶𝑖 are constants varying with stresses in the phase 𝑖.

Strain Derivations

The six strain components [휀𝑥 휀𝑟 휀𝜃 𝛾𝜃𝑥 𝛾𝑟𝑥 𝛾𝑟𝜃] are derived in the shear model.

From the basic strain formulations for cylindrical system we get,


𝜕𝑟→ 휀𝑟 = 0 (2-44)

and

휀𝜃 =𝑢𝑟

𝑟+

1

𝑟

𝜕𝑢𝜃

𝜕𝜃→ 휀𝜃 = 0 (2-45)

and

휀𝑥 =𝜕𝑢𝑥

𝜕𝑥→ 휀𝑥 = 0 (2-46)

It is also observed that the transverse shear strain also vanishes:

𝛾𝑟𝜃 =𝜕𝑢𝜃

𝜕𝑟+

1

𝑟

𝜕𝑢𝜃

𝜕𝜃−

𝑢𝜃

𝑟= 0 (2-47)

Hence, only two shear strains will exist in the body, which are derived as follows:

𝛾𝜃𝑥 =𝜕𝑢𝜃

𝜕𝑥+

1

𝑟

𝜕𝑢𝑥

𝜕𝜃= − [𝐴 +

𝐵

𝑟2+ 𝐶] 𝑠𝑖𝑛𝜃 (2-48)

and

𝛾𝑟𝑥 =𝜕𝑢𝑟

𝜕𝑥+

𝜕𝑢𝑥

𝜕𝑟= [𝐴 −

𝐵

𝑟2+ 𝐶] 𝑐𝑜𝑠𝜃 (2-49)

29

Boundary Conditions

Since the displacement at point 𝑟 = 0 is finite, the constant pertaining to the fiber

phase 𝐵𝑓 must be zero. At 𝑟 = ∞, the longitudinal displacement 𝑢𝑥𝑐 in the composite

phase must be finite, hence 𝐴𝑐 = 0. The longitudinal shear strains in the composite

phase at 𝑟 = ∞ can be derived from Equations (2-48) and (2-49) are shown below:

𝛾𝜃𝑥 = −𝐶𝑐𝑠𝑖𝑛𝜃 (2-50)

𝛾𝑟𝑥 = 𝐶𝑐𝑐𝑜𝑠𝜃 (2-51)

The 3X3 rotation matrix for transformation of cylindrical coordinates to Cartesian

coordinates is

𝑅 = [𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 0

−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 00 0 1

]

Since the transverse shear strain is zero, the rotation matrix can be simplified to

a 2X2 matrix. Transforming the shear stresses from Equations (2-50) and (2-51) into

Cartesian form we get,

{𝛾𝑧𝑥

𝛾𝑥𝑦} = [

𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

] {𝛾𝜃𝑥

𝛾𝑟𝑥} (2-52)

{𝛾𝑧𝑥

𝛾𝑥𝑦} = {

0𝐶𝑐

} (2-53)

It can be observed that the shear strain 𝛾𝑥𝑦 is the only shear strain presiding in

this model and can be equated to 𝐶𝑐 (Constant in composite phase).

From the shear stress-strain relations formula, we deduce the following relation

𝐶𝑐 = 𝛾𝑦𝑥 =𝜏0

𝐺𝑥𝑦𝑐 (2-54)

30

Here, 𝐶𝑐: Constant 𝐶 in the composite phase 𝐺𝑥𝑦

𝑐 : Longitudinal shear modulus of the composite derived from Halpin-Tsai

Equation Continuity Equations


fiber/matrix interface and matrix/composite interface. The continuity equations are

shown below:

𝑢𝑟𝑖(𝑟𝑖) = 𝑢𝑟𝑗(𝑟𝑖) (2-55)

𝑢𝑥𝑖(𝑟𝑖) = 𝑢𝑥𝑗(𝑟𝑖) (2-56)

𝜏𝑟𝑥𝑖(𝑟𝑖) = 𝜏𝑟𝑥𝑗(𝑟𝑖) (2-57)

Here, 𝑢𝑟𝑖, 𝑢𝑟𝑗 ,𝑢𝑥𝑖, 𝑢𝑥𝑗 are the radial and axial displacements in consecutive

phases 𝑖, 𝑗 and 𝜏𝑟𝑥𝑖, 𝜏𝑟𝑥𝑗 are the radial shear strains in the corresponding phases.

From Equation (2-55) continuity of displacement along the fiber/matrix interface

at 𝑟 = 𝑎

𝑢𝑟𝑓(𝑎) = 𝑢𝑟𝑚(𝑎) → 𝐶𝑓𝑥𝑐𝑜𝑠𝜃 = 𝐶𝑚𝑥𝑐𝑜𝑠𝜃 (2-58)

Similarly, at 𝑟 = 𝑏

𝑢𝑚(𝑏) = 𝑢𝑐(𝑏) → 𝐶𝑚𝑥𝑐𝑜𝑠𝜃 = 𝐶𝑐𝑥𝑐𝑜𝑠𝜃 (2-59)

From Equations (2-58) and (2-59) we can infer that 𝐶𝑓 = 𝐶𝑚 = 𝐶𝑐

Let the constants 𝐶𝑓, 𝐶𝑚 and 𝐶 be equal to 𝐶:

𝐶𝑓 = 𝐶𝑚 = 𝐶𝑐 = 𝐶 (2-60)

Ensuring equal axial displacements at the fiber/matrix interface 𝑟 = 𝑎 we obtain

𝑢𝑥𝑓(𝑎) = 𝑢𝑥𝑚(𝑎) → 𝐴𝑓𝑎 + 𝐴𝑚𝑎 −𝐵𝑚

𝑎= 0 (2-61)

where, 𝐴𝑓 , 𝐴𝑚 and 𝐵𝑚 are constants in fiber and matrix phases

31

From Equation (2-57) continuity of shear stress along the fiber/matrix interface

𝜏𝑟𝑥𝑓(𝑎) = 𝜏𝑟𝑥𝑚(𝑎) → 𝐺𝑟𝑥𝑓

(𝐴𝑓 + 𝐶) − 𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑎2+ 𝐶) = 0 (2-62)

where, 𝐺𝑟𝑥𝑓

, 𝐺𝑟𝑥𝑚 are the longitudinal shear moduli of the fiber phase and matrix

phase respectively

Similarly, considering axial displacements at the matrix/composite interface we

get,

𝑢𝑥𝑚(𝑏) = 𝑢𝑥𝑐(𝑏) →𝐵𝑐

𝑏−𝐴𝑚𝑏 −

𝐵𝑚

𝑏= 0 (2-63)

Here, 𝐴𝑚, 𝐵𝑚𝐵𝑐 are constants in matrix phase and composite phase varying with

applied stresses. Similarly, continuity of shear stress along the matrix/composite

interface yields:

𝜏𝑟𝑥𝑚(𝑏) = 𝜏𝑟𝑥𝑐(𝑏) → 𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑏2+ 𝐶) − 𝐺𝑟𝑥

𝑐 (−𝐵𝑐

𝑏2+ 𝐶) = 0 (2-64)

where, 𝐺𝑟𝑥𝑚, 𝐺𝑟𝑥

𝑐 are the longitudinal shear moduli of the matrix phase and

composite phase respectively

The five constants (𝐴𝑓 , 𝐴𝑚, 𝐵𝑚, 𝐵𝑐, 𝐶) are solved from the following five equations:

𝐴𝑓 + 𝐴𝑚𝑎 −𝐵𝑚

𝑎= 0 (2-65)

𝐺𝑟𝑥𝑓

(𝐴𝑓 + 𝐶) − 𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑎2+ 𝐶) = 0 (2-66)

𝐵𝑐


𝐵𝑚

𝑏= 0 (2-67)

𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑏2+ 𝐶) − 𝐺𝑟𝑥

𝑐 (−𝐵𝑐

𝑏2+ 𝐶) = 0 (2-68)

𝐶 =𝜏𝑥𝑦

𝐺𝑥𝑦𝑐 (2-69)

32

The micro stresses in fiber and matrix phases for longitudinal shear XY case are

calculated from the following equations:

𝜏𝜃𝑥𝑓 = −𝐺𝑟𝑥𝑓

(𝐴𝑓 + 𝐶) 𝑠𝑖𝑛𝜃 (2-70)

𝜏𝑟𝑥𝑓 = 𝐺𝑟𝑥𝑓

(𝐴𝑓 + 𝐶) 𝑐𝑜𝑠𝜃 (2-71)

𝜏𝜃𝑥𝑚 = −𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑟2+ 𝐶) 𝑠𝑖𝑛𝜃 (2-72)

𝜏𝑟𝑥𝑚 = 𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑟2+ 𝐶) 𝑠𝑖𝑛𝜃 (2-73)

Longitudinal Shear Stress in the x-z plane

In this section, the displacement equations for the case of longitudinal shear

stress are derived. Since we are using the composite cylinder model, cylindrical

coordinate system is used.

The displacement equation for all the 3 Phases (Fiber, Matrix, and Composite) is

given below:

𝑢𝑟𝑖 = 𝐶𝑖𝑥𝑠𝑖𝑛𝜃 (2-74)

𝑢𝜃𝑖 = 𝐶𝑖𝑥𝑐𝑜𝑠𝜃 (2-75)

𝑢𝑥𝑖 = − (𝐴𝑖𝑟 +𝐵𝑖

𝑟) 𝑠𝑖𝑛𝜃 (2-76)

Here, 𝑢𝑟𝑖, 𝑢𝜃𝑖, 𝑢𝑥𝑖 are radial, angular, and axial displacements in the phase 𝑖

respectively. 𝐴𝑖, 𝐵𝑖, 𝐶𝑖 are constants varying with stresses in the phase 𝑖

Strain Derivations

The 6 fundamental strains [휀𝑥 휀𝑟 휀𝜃 𝛾𝜃𝑥 𝛾𝑟𝑥 𝛾𝑟𝜃] are derived in the shear model.

From the basic strain formulations for cylindrical system we get,


𝜕𝑟→ 휀𝑟 = 0 (2-77)

33

and

휀𝜃 =𝑢𝑟

𝑟+

1

𝑟

𝜕𝑢𝜃

𝜕𝜃→ 휀𝜃 = 0 (2-78)

and

휀𝑥 =𝜕𝑢𝑥

𝜕𝑥→ 휀𝑥 = 0 (2-79)

It can be observed that the longitudinal strain by the application of a discrete

shear stress in this model is zero.

Now the shear strains are calculated, since the shear is applied along the

longitudinal direction it is obvious that transverse shear is zero.

𝛾𝑟𝜃 =𝜕𝑢𝜃

𝜕𝑟+

1

𝑟

𝜕𝑢𝜃

𝜕𝜃−

𝑢𝜃

𝑟= 0 (2-80)

Apparently due to the nature of shear stress only two shear strains will exist in

the body, which are derived as follows:

𝛾𝜃𝑥 =𝜕𝑢𝜃

𝜕𝑥+

1

𝑟

𝜕𝑢𝑥

𝜕𝜃= − [𝐴 +

𝐵

𝑟2− 𝐶] 𝑐𝑜𝑠𝜃 (2-81)

𝛾𝑟𝑥 =𝜕𝑢𝑟

𝜕𝑥+

𝜕𝑢𝑥

𝜕𝑟= − [𝐴 −

𝐵

𝑟2− 𝐶] 𝑠𝑖𝑛𝜃 (2-82)

Boundary Conditions

Since the displacement at point 𝑟 = 0 is finite, the constant pertaining to the fiber

phase 𝐵𝑓 must be zero. At 𝑟 = ∞, the longitudinal displacement 𝑢𝑥𝑐 in the composite

phase must be finite, hence 𝐴𝑐 = 0. The longitudinal shear strains in the composite

phase at 𝑟 = ∞ can be derived from Equations (2-81) and (2-82) are shown below:

𝛾𝜃𝑥 = 𝐶𝑐𝑠𝑖𝑛𝜃 (2-83)

𝛾𝑟𝑥 = 𝐶𝑐𝑐𝑜𝑠𝜃 (2-84)

34

The 3X3 rotation matrix for transformation of cylindrical coordinates to Cartesian

coordinates is

𝑅 = [𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 0

−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 00 0 1

]

Since the transverse shear strain is zero the rotation matrix can be simplified to a

2X2 matrix. Transforming the shear stresses from Equations (2-83) and (2-84) into

Cartesian form we get,

{𝛾𝑧𝑥

𝛾𝑥𝑦} = [

𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

] {𝛾𝜃𝑥

𝛾𝑟𝑥} (2-85)

{𝛾𝑧𝑥

𝛾𝑥𝑦} = {

𝐶𝑐

0} (2-86)

It can be observed that the shear strain 𝛾𝑧𝑥 is the only shear strain presiding in

this model and can be equated to 𝐶𝑐 (Constant in composite phase). From basic shear

stress formula, we can deduce the following relation:

𝐶𝑐 = 𝛾𝑧𝑥 =𝜏𝑧𝑥

𝐺𝑟𝑥𝑐 (2-87)

Here,

𝐶𝑐: Constant C in the composite phase

𝐺𝑧𝑥𝑐 : Longitudinal shear modulus of the composite derived from Halpin-Tsai

equation

Continuity Equations


fiber/matrix interface and matrix/composite interface. The continuity equations are

shown below:

𝑢𝑟𝑖(𝑟𝑖) = 𝑢𝑟𝑗(𝑟𝑖) (2-88)

35

𝑢𝑥𝑖(𝑟𝑖) = 𝑢𝑥𝑗(𝑟𝑖) (2-89)

𝜏𝑟𝑥𝑖(𝑟𝑖) = 𝜏𝑟𝑥𝑗(𝑟𝑖) (2-90)

Here, 𝑢𝑟𝑖, 𝑢𝑟𝑗 ,𝑢𝑥𝑖, 𝑢𝑥𝑗 are the radial and axial displacements in consecutive

phases 𝑖, 𝑗 and 𝜏𝑟𝑥𝑖, 𝜏𝑟𝑥𝑗 are the radial shear strains in the corresponding phases. From

Equation (2-88) continuity of displacement along the fiber/matrix interface 𝑟 = 𝑎 we can

interpret

𝑢𝑟𝑓(𝑎) = 𝑢𝑟𝑚(𝑎) → 𝐶𝑓𝑥𝑠𝑖𝑛𝜃 = 𝐶𝑚𝑥𝑠𝑖𝑛𝜃 (2-91)

Similarly, at 𝑟 = 𝑏

𝑢𝑚(𝑏) = 𝑢𝑐(𝑏) → 𝐶𝑚𝑥𝑠𝑖𝑛𝜃 = 𝐶𝑐𝑥𝑠𝑖𝑛𝜃 (2-92)

From Equations (2-91) and (2-92) we can infer that 𝐶𝑓 = 𝐶𝑚 = 𝐶𝑐

For simplicity, the following assumption has been made and will be considered

for future derivations and equations

𝐶𝑓 = 𝐶𝑚 = 𝐶𝑐 = 𝐶 (2-93)

Ensuring equal axial displacements about the fiber/matrix interface at 𝑟 = 𝑎 we

get

𝑢𝑥𝑓(𝑎) = 𝑢𝑥𝑚(𝑎) → −𝐴𝑓𝑎 + 𝐴𝑚𝑎 +𝐵𝑚

𝑎= 0 (2-94)

Here, 𝐴𝑓 , 𝐴𝑚 and 𝐵𝑚 are constants in fiber and matrix phases varying with applied

stresses

From Equation (2-90) continuity of radial shear stress along the fiber/matrix

interface

𝜏𝑟𝑥𝑓(𝑎) = 𝜏𝑟𝑥𝑚(𝑎) → −𝐺𝑟𝑥𝑓

(𝐴𝑓 − 𝐶) + 𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑎2− 𝐶) = 0 (2-95)

36

Here, 𝐺𝑟𝑥𝑓

, 𝐺𝑟𝑥𝑚 are the longitudinal shear moduli of the fiber phase and matrix

phase respectively. Similarly, considering axial displacements at the matrix/composite

interface we get

𝑢𝑥𝑚(𝑏) = 𝑢𝑥𝑐(𝑏) →𝐵𝑐


𝐵𝑚

𝑏= 0 (2-96)

Here, 𝐴𝑚, 𝐵𝑚𝐵𝑐 are constants in matrix phase and composite phase varying with

applied stresses

Similarly, continuity of radial shear stress along the matrix/composite interface

yields

𝜏𝑟𝑥𝑚(𝑏) = 𝜏𝑟𝑥𝑐(𝑏) → −𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑏2− 𝐶) + 𝐺𝑟𝑥

𝑐 (−𝐵𝑐

𝑏2− 𝐶) = 0 (2-97)

Here,

𝐺𝑟𝑥𝑚: Longitudinal shear modulus of the matrix phase

𝐺𝑟𝑥𝑐 : Longitudinal shear modulus of the composite

The five constants (𝐴𝑓 , 𝐴𝑚, 𝐵𝑚, 𝐵𝑐, 𝐶) are solved from the following five equations:

−𝐴𝑓 + 𝐴𝑚𝑎 +𝐵𝑚

𝑎= 0 (2-98)

−𝐺𝑟𝑥𝑓

(𝐴𝑓 + 𝐶) + 𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑎2+ 𝐶) = 0 (2-99)

𝐵𝑐


𝐵𝑚

𝑏= 0 (2-100)

−𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑏2+ 𝐶) + 𝐺𝑟𝑥

𝑐 (−𝐵𝑐

𝑏2+ 𝐶) = 0 (2-101)

𝐶 =𝜏𝑥𝑦

𝐺𝑥𝑦𝑐 (2-102)

The micro stresses in each phase are calculated from the following equations

37

𝜏𝜃𝑥𝑓 = −𝐺𝑟𝑥𝑓

(𝐴𝑓 − 𝐶) 𝑐𝑜𝑠𝜃 (2-103)

𝜏𝑟𝑥𝑓 = −𝐺𝑟𝑥𝑓

(𝐴𝑓 − 𝐶) 𝑠𝑖𝑛𝜃 (2-104)

𝜏𝜃𝑥𝑚 = −𝐺𝑟𝑥𝑚 (𝐴𝑚 +

𝐵𝑚

𝑟2− 𝐶) 𝑐𝑜𝑠𝜃 (2-105)

𝜏𝑟𝑥𝑚 = −𝐺𝑟𝑥𝑚 (𝐴𝑚 −

𝐵𝑚

𝑟2− 𝐶) (2-106)

Biaxial tension/compression in y-z plane

In this section, the equations for the case of pure shear in the transverse plane

are derived. This two-dimensional problem can be dealt by assuming a suitable Airy

stress function ∅ which is shown below:

∅ = (𝐴𝑟2 + 𝐵𝑟4 +𝐶

𝑟2+ 𝐷) 𝑐𝑜𝑠2𝜃 (2-107)

Since plane strain is assumed, the three non-zero stresses pertaining to the 2-D

problem [ 𝜎𝑟 𝜎𝜃 𝜏𝑟𝜃] are derived from the Airy stress function as shown below. The

radial stress at any point in the model can be derived using the elasticity equation [11]

as follows

𝜎𝑟 =1

𝑟

𝜕∅

𝜕𝑟+

1

𝑟2

𝜕2∅

𝜕𝜃2 (2-108)

𝜎𝑟 = (2𝐴 + 4𝐵𝑟2 −2𝐶

𝑟4) 𝑐𝑜𝑠2𝜃 + (−4𝐴 − 4𝐵𝑟2 −

4𝐶

𝑟4−

4𝐷

𝑟2) 𝑐𝑜𝑠2𝜃

𝜎𝑟 = − (2𝐴 +6𝐶

𝑟4+

4𝐷

𝑟2) 𝑐𝑜𝑠2𝜃 (2-109)

The tangential stress derivation is shown below

𝜎𝜃 =𝜕2∅

𝜕𝑟2→ 𝜎𝜃 = (2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4) 𝑐𝑜𝑠2𝜃 (2-110)

The transverse shear in the plane is derived as follows

38

𝜏𝑟𝜃 = −𝜕

𝜕𝑟(

1

𝑟

𝜕∅

𝜕𝜃) (2-111)

𝜏𝑟𝜃 = −𝜕

𝜕𝑟(−

1

𝑟(𝐴𝑟2 + 𝐵𝑟4 +

𝐶

𝑟4+ 𝐷) 2𝑠𝑖𝑛2𝜃)

𝜏𝑟𝜃 = (2𝐴 + 6𝐵𝑟2 −6𝐶

𝑟4−

2𝐷

𝑟2) 𝑠𝑖𝑛2𝜃 (2-112)

The radial strain at any point is given by the following equations

휀𝑟 =1

𝐸′(𝜎𝑟 − 𝜈′𝜎𝜃) 𝑜𝑟 휀𝑟 =

𝜕𝑢𝑟

𝜕𝑟 (2-113)

Here, 𝐸′ is the plane strain modulus and 𝜈′ is the plane strain Poisson’s ratio

which is given by

𝜈′ = (𝜈

1 + 𝜈)

The radial displacement is calculated from the Equation (2-113) as follows

𝑢𝑟 = ∫(𝜎𝑟 − 𝜈′𝜎𝜃)𝑑𝑟 (2-114)

Substituting 𝜎𝜃 from Equation (2-110) in the above equation we get

𝑢𝑟 =1

𝐸′∫ ((−2𝐴 −

6𝐶

𝑟4−

4𝐷

𝑟2) 𝐶𝑜𝑠2𝜃 − 𝜈′ (2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4) 𝐶𝑜𝑠2𝜃) 𝑑𝑟

𝑢𝑟 =𝐶𝑜𝑠2𝜃

𝐸′∫ ((−2𝐴 −

6𝐶

𝑟4−

4𝐷

𝑟2) − 𝜈′ (2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4)) 𝑑𝑟 (2-115)


𝐸′((−2𝐴𝑟 +

2𝐶

𝑟3+

4𝐷

𝑟) − 𝜈′ (2𝐴𝑟 + 4𝐵𝑟3 −

2𝐶

𝑟3))

The radial displacement at any point on the surface can be obtained by the

following equation:


𝐸′((−2𝐴𝑟(1 + 𝜈′) − 4𝜈′𝐵𝑟3 +

2𝐶

𝑟3(1 + 𝜈′) +

4𝐷

𝑟)) (2-116)

39

The tangential strain at any point is given by the following equations:

휀𝜃 =1

𝐸′(𝜎𝜃 − 𝜈′𝜎𝑟) → 휀𝜃 =

𝑢𝑟

𝑟+

1

𝑟

𝜕𝑢𝜃

𝜕𝜃 (2-117)

Equation (2-117) can be modified to obtain the angular displacement gradient

with respect to 𝜃 as follows:

𝜕𝑢𝜃

𝜕𝜃= 𝑟 (

1

𝐸′(𝜎𝜃 − 𝜈′𝜎𝑟) −

𝑢𝑟

𝑟) (2-118)

Substituting 𝜎𝜃 and 𝜎𝑟 from Equations (2-110) and (2-109) we get the following

relations:

𝜕𝑢𝜃

𝜕𝜃= 𝑟 (

1

𝐸′((2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4) 𝐶𝑜𝑠2𝜃 + 𝜈′ (2𝐴 +

6𝐶

𝑟4+

4𝐷

𝑟2) 𝐶𝑜𝑠2𝜃) −

𝑢𝑟

𝑟)

𝜕𝑢𝜃

𝜕𝜃= (

𝐶𝑜𝑠2𝜃

𝐸′(2𝐴𝑟(1 + 𝜈′) + 12𝐵𝑟3 +

6𝐶

𝑟3(1 + 𝜈′) +

4𝐷𝜈′

𝑟)

−𝐶𝑜𝑠2𝜃

𝐸′((−2𝐴𝑟(1 + 𝜈′) − 4𝜈′𝐵𝑟3 +

2𝐶

𝑟3(1 + 𝜈′) +

4𝐷

𝑟)))

The obtained angular displacement gradient is as follows:

𝜕𝑢𝜃

𝜕𝜃= (

4𝐶𝑜𝑠2𝜃

𝐸′(𝐴𝑟(1 + 𝜈′) + 𝐵𝑟3(3 + 𝜈′) +

𝐶

𝑟3(1 + 𝜈′) +

𝐷

𝑟(𝜈′ − 1))) (2-119)

In the above equations 𝐴, 𝐵, 𝐶, 𝐷 are constants specific to each phase and vary

with applied stresses.

Since the stresses in fiber at 𝑟 = 0 are finite 𝐶𝑓 𝑎𝑛𝑑 𝐷𝑓 are equal to zero.

Continuity of 𝜎𝑟 , 𝜏𝑟𝜃, 𝑢𝑟 , 𝜕𝑢𝜃/𝜕𝜃 must be satisfied at 𝑟 = 𝑎 (fiber/matrix interface) and

𝑟 = 𝑏 (matrix/composite interface)

Continuity of radial stress along fiber/matrix interface yields the following

equation:

40

𝜎𝑟𝑓(𝑎) = 𝜎𝑟𝑚(𝑎) (2-120)

𝐴𝑓 − 𝐴𝑚 −3𝐶𝑚

𝑎4−

2𝐷𝑚

𝑎2= 0 (2-121)

Similarly, incorporating continuity of radial stress along the matrix/composite

interface we get,

𝜎𝑟𝑚(𝑏) = 𝜎𝑟𝑐(𝑏) (2-122)

−𝐴𝑚 −3𝐶𝑚

𝑏4−

2𝐷𝑚

𝑏2+ 𝐴𝑐 +

3𝐶𝑐

𝑏4+

2𝐷𝑐

𝑏2= 0 (2-123)

Continuity of radial displacement must be ensured along the fiber/matrix interface

𝑢𝑟𝑓(𝑎) = 𝑢𝑟𝑚(𝑎) (2-124)

1

𝐸𝑓′ (−2𝐴𝑓𝑎(1 + 𝜈𝑓

′ ) − 4𝜈𝑓′ 𝐵𝑓𝑎3)

−1

𝐸𝑚′

((−2𝐴𝑚𝑎(1 + 𝜈𝑚′ ) − 4𝜈𝑚

′ 𝐵𝑚𝑎3 +2𝐶𝑚

𝑎3(1 + 𝜗𝑚

′ ) +4𝐷𝑚

𝑎))

= 0 (2-125)

Similarly, considering the continuity of radial displacement along

matrix/composite interface we get the following equation:

1

𝐸𝑚′ ((−2𝐴𝑚𝑏(1 + 𝜈𝑚

′ ) − 4𝜈𝑚′ 𝐵𝑚𝑏3 +

2𝐶𝑚

𝑏3 (1 + 𝜈𝑚′ ) +

4𝐷𝑚

𝑏)) −

1

𝐸𝑐′ ((−2𝐴𝑐𝑏(1 + 𝜈𝑐) +

2𝐶𝑐

𝑏3(1 + 𝜈𝑐) +

4𝐷𝑐

𝑏)) = 0 (2-126)

Continuity of shear stress must be ensured along the fiber/matrix interface

𝜏𝑟𝜃𝑓(𝑎) = 𝜏𝑟𝜃𝑚(𝑎) (2-127)

(2𝐴𝑓 + 6𝐵𝑓𝑎2) − (2𝐴𝑚 + 6𝐵𝑚𝑟2 −6𝐶𝑚

𝑎4−

2𝐷𝑚

𝑎2) = 0 (2-128)

Similarly, considering the continuity of shear stress along matrix/composite

interface we get the following equation:

41

(2𝐴𝑚 + 6𝐵𝑚𝑏2 −6𝐶𝑚

𝑏4−

2𝐷𝑚

𝑏2) − (2𝐴𝑐 −

6𝐶𝑐

𝑏4−

2𝐷𝑐

𝑏2) = 0 (2-129)

Continuity of tangential displacement gradient must be ensured along the

fiber/matrix interface

𝜕𝑢𝜃𝑓

𝜕𝜃(𝑎) =

𝜕𝑢𝜃𝑚

𝜕𝜃(𝑎) (2-130)

1

𝐸𝑓′ (𝐴𝑓𝑎(1 + 𝜈𝑓

′ ) + 𝐵𝑓𝑎3(3 + 𝜈𝑓′ ))

− 1

𝐸𝑚′

(𝐴𝑚𝑎(1 + 𝜈𝑚′ ) + 𝐵𝑚𝑎3(3 + 𝜈𝑚

′ ) +𝐶𝑚

𝑎3(1 + 𝜈𝑚

′ ) +𝐷𝑚

𝑎(𝜈𝑚

′ − 1))

= 0 (2-131)



1

𝐸𝑚′

(𝐴𝑚𝑏(1 + 𝜈𝑚′ ) + 𝐵𝑚𝑏3(3 + 𝜈𝑚

′ ) +𝐶𝑚

𝑏3(1 + 𝜈𝑚

′ ) +𝐷𝑚

𝑏(𝜈𝑚

′ − 1))

−1

𝐸𝑐′

(𝐴𝑐𝑏(1 + 𝜈𝑐′) +

𝐶𝑐

𝑏3(1 + 𝜈𝑐

′) +𝐷𝑐

𝑏(𝜈𝑐

′ − 1))

= 0 (2-132)

Constants 𝐴𝑓 , 𝐵𝑓 , 𝐴𝑚, 𝐵𝑚, 𝐶𝑚𝐷𝑚, 𝐴𝑐, 𝐶𝑐 and 𝐷𝑐 can be found from the above

derived continuity equations. The stresses in each phase can be calculated from the

below mentioned equations:

Fiber Equations

𝜎𝑟𝑓 = −2𝐴𝑓𝐶𝑜𝑠2𝜃

𝜎𝜃𝑓 = (2𝐴𝑓 + 12𝐵𝑓𝑟2)𝐶𝑜𝑠2𝜃

𝜏𝑟𝜃𝑓 = (2𝐴𝑓 + 6𝐵𝑓𝑟2)𝑆𝑖𝑛2𝜃 (2-133)

42

𝑢𝑟𝑓 =𝐶𝑜𝑠2𝜃

𝐸𝑓′ (−2𝐴𝑓𝑟(1 + 𝜈𝑓

′ ) − 4𝜈𝑓𝐵𝑓𝑟3)

𝜕𝑢𝜃𝑓

𝜕𝜃=

4𝐶𝑜𝑠2𝜃

𝐸𝑓′ (𝐴𝑓𝑟(1 + 𝜈𝑓

′ ) + 𝐵𝑓𝑟3(3 + 𝜈𝑓′ ))

Matrix Equations

𝜎𝑟𝑚 = − (2𝐴𝑚 +6𝐶𝑚

𝑟4+

4𝐷𝑚

𝑟2) 𝐶𝑜𝑠2𝜃

𝜎𝜃𝑚 = (2𝐴𝑚 + 12𝐵𝑚𝑟2 +6𝐶𝑚


𝜏𝑟𝜃𝑚 = (2𝐴𝑚 + 6𝐵𝑚𝑟2 −6𝐶𝑚

𝑟4−

2𝐷𝑚

𝑟2) 𝑆𝑖𝑛2𝜃 (2-134)

𝑢𝑟𝑚 =𝐶𝑜𝑠2𝜃

𝐸𝑚′

((−2𝐴𝑚𝑟(1 + 𝜈𝑚′ ) − 4𝜈𝑚

′ 𝐵𝑚𝑟3 +2𝐶𝑚

𝑟3(1 + 𝜈𝑚

′ ) +4𝐷𝑚

𝑟))

𝜕𝑢𝜃𝑚

𝜕𝜃=

4𝐶𝑜𝑠2𝜃

𝐸𝑚′

(𝐴𝑚𝑟(1 + 𝜈𝑚′ ) + 𝐵𝑚𝑟3(3 + 𝜈𝑚

′ ) +𝐶𝑚

𝑟3(1 + 𝜈𝑚

′ ) +𝐷𝑚

𝑟(𝜈𝑚

′ − 1))

Composite Equations

𝜎𝑟𝑐 = − (2𝐴𝑐 +6𝐶𝑐

𝑟4+

4𝐷𝑐


𝜎𝜃𝑐 = (2𝐴𝑐 +6𝐶𝑐


𝜏𝑟𝜃𝑐 = (2𝐴𝑐 −6𝐶𝑐

𝑟4−

2𝐷𝑐

𝑟2) 𝑆𝑖𝑛2𝜃 (2-135)

𝑢𝑟𝑐 =𝐶𝑜𝑠2𝜃

𝐸𝑐′

((−2𝐴𝑐𝑟(1 + 𝜈𝑐′) +

2𝐶𝑐

𝑟3(1 + 𝜈𝑐

′) +4𝐷𝑐

𝑟))

𝜕𝑢𝜃𝑐

𝜕𝜃=

4𝐶𝑜𝑠2𝜃

𝐸𝑐′

(𝐴𝑐𝑟(1 + 𝜈𝑐′) +

𝐶𝑐

𝑟3(1 + 𝜈𝑐

′) +𝐷𝑐

𝑟(𝜈𝑐

′ − 1))

Substituting 𝑟 = ∞ in Equations (2-135) i.e. composite phase we get,

𝜎𝑟𝑐 = −(2𝐴𝑐)𝐶𝑜𝑠2𝜃

43

𝜎𝜃𝑐 = (2𝐴𝑐)𝐶𝑜𝑠2𝜃 (2-136)

𝜏𝑟𝜃𝑐 = (2𝐴𝑐)𝑆𝑖𝑛2𝜃

Transforming the stresses into Cartesian co-ordinates yields

[

𝜎𝑦

𝜎𝑧

𝜏𝑦𝑧

] = [𝐶𝑜𝑠2𝜃 𝑆𝑖𝑛2𝜃 − 𝑆𝑖𝑛2𝜃𝑆𝑖𝑛2𝜃 𝐶𝑜𝑠2𝜃 𝑆𝑖𝑛2𝜃𝐶𝑜𝑠𝜃𝑆𝑖𝑛𝜃 − 𝐶𝑜𝑠𝜃𝑆𝑖𝑛𝜃 𝐶𝑜𝑠2𝜃

] [−𝐶𝑜𝑠2𝜃𝐶𝑜𝑠2𝜃𝑆𝑖𝑛2𝜃

] 2𝐴𝑐

[

𝜎𝑦

𝜎𝑧

𝜏𝑦𝑧

] = [−2𝐴𝑐

+2𝐴𝑐

0] (2-137)

It can be observed in Equation (2-137), the shear in the transverse plane is zero

and the only non-zero stresses are 𝜎𝑦 and 𝜎𝑧 which are equal but are acting in opposite

directions which is equivalent to shear in 45 degrees.

Transverse Shear Equations

In this section, the stress equations for the case of transverse shear yz are

derived. This two-dimensional problem can be dealt by assuming a suitable stress

function [12] ∅ which is shown below

∅ = − (𝐴𝑟2 + 𝐵𝑟4 +𝐶

𝑟2+ 𝐷) 𝑠𝑖𝑛2𝜃 (2-138)

Since plane strain is assumed, the three fundamental stresses pertaining to a 2-

D problem [ 𝜎𝑟 𝜎𝜃 𝜏𝑟𝜃] are derived from the Airy stress function. The radial stress at any

point in the model can be derived using the elasticity equation as follows:

𝜎𝑟 =1

𝑟

𝜕∅

𝜕𝑟+

1

𝑟2

𝜕2∅

𝜕𝜃2 (2-139)

𝜎𝑟 = − (2𝐴 + 4𝐵𝑟2 −2𝐶

𝑟4) 𝑐𝑜𝑠2𝜃 + (4𝐴 + 4𝐵𝑟2 +

4𝐶

𝑟4+

4𝐷

𝑟2) 𝑠𝑖𝑛2𝜃

𝜎𝑟 = (2𝐴 +6𝐶

𝑟4+

4𝐷

𝑟2) 𝑠𝑖𝑛2𝜃 (2-140)

44

The tangential stress derivation is shown below:

𝜎𝜃 =𝜕2∅

𝜕𝑟2→ 𝜎𝜃 = − (2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4) 𝑠𝑖𝑛2𝜃 (2-141)

The transverse shear in the plane is derived as follows:

𝜏𝑟𝜃 = −𝜕

𝜕𝑟(

1

𝑟

𝜕∅

𝜕𝜃) (2-142)

𝜏𝑟𝜃 = −𝜕

𝜕𝑟(−

2

𝑟(𝐴𝑟2 + 𝐵𝑟4 +

𝐶

𝑟4+ 𝐷) 𝑐𝑜𝑠2𝜃)

𝜏𝑟𝜃 = (2𝐴 + 6𝐵𝑟2 −6𝐶

𝑟4−

2𝐷

𝑟2) 𝑐𝑜𝑠2𝜃 (2-143)

The radial strain at any point is given by the following equations:

휀𝑟 =1

𝐸′(𝜎𝑟 − 𝜗′𝜎𝜃) 𝑜𝑟 휀𝑟 =

𝜕𝑢𝑟

𝜕𝑟 (2-144)

Here, 𝐸′ is the plane strain modulus and 𝜈′ is the plane strain Poisson’s ratio.

The radial displacement is calculated from the Equation (2-144) as follows

𝑢𝑟 = ∫(𝜎𝑟 − 𝜈′𝜎𝜃)𝑑𝑟 (2-145)

Substituting 𝜎𝜃 from Equation (2-141) in the above equation we get,

𝑢𝑟 =1

𝐸′∫ ((+2𝐴 +

6𝐶

𝑟4+

4𝐷

𝑟2) 𝑠𝑖𝑛2𝜃 − 𝜈′ (− (2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4)) 𝑠𝑖𝑛2𝜃) 𝑑𝑟

𝑢𝑟 =𝑠𝑖𝑛2𝜃

𝐸′∫ ((+2𝐴 +

6𝐶

𝑟4+

4𝐷

𝑟2) + 𝜈′ (2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4)) 𝑑𝑟 (2-146)


𝐸′((2𝐴𝑟 −

2𝐶

𝑟3−

4𝐷

𝑟) + 𝜈′ (2𝐴𝑟 + 4𝐵𝑟3 −

2𝐶

𝑟3))

The radial displacement at any point on the surface can be obtained by the

following equation:

45


𝐸′((2𝐴𝑟(1 + 𝜈′) + 4𝜈′𝐵𝑟3 −

2𝐶

𝑟3(1 + 𝜈′) −

4𝐷

𝑟)) (2-147)

The tangential strain at any point is given by the following equations:

휀𝜃 =1

𝐸′(𝜎𝜃 − 𝜈′𝜎𝑟) → 휀𝜃 =

𝑢𝑟

𝑟+

1

𝑟

𝜕𝑢𝜃

𝜕𝜃 (2-148)

Equation (2-148) can be modified to obtain the angular displacement gradient

with respect to 𝜃 as follows:

𝜕𝑢𝜃

𝜕𝜃= 𝑟 (

1

𝐸′(𝜎𝜃 − 𝜈′𝜎𝑟) −

𝑢𝑟

𝑟) (2-149)

Substituting 𝜎𝜃 and 𝜎𝑟 from Equations (2-140) and (2-141) we get the following

relations:

𝜕𝑢𝜃

𝜕𝜃= 𝑟 (

1

𝐸′(− (2𝐴 + 12𝐵𝑟2 +

6𝐶

𝑟4) 𝑠𝑖𝑛2𝜃 − 𝜈′ (2𝐴 +

6𝐶

𝑟4+

4𝐷

𝑟2) 𝑠𝑖𝑛2𝜃) −

𝑢𝑟

𝑟)

𝜕𝑢𝜃

𝜕𝜃= (

𝑠𝑖𝑛2𝜃

𝐸′(−2𝐴𝑟(1 + 𝜈′) − 12𝐵𝑟3 −

6𝐶

𝑟3(1 + 𝜈′) −

4𝐷𝜗

𝑟)

−𝑠𝑖𝑛2𝜃

𝐸′((2𝐴𝑟(1 + 𝜈′) + 4𝜈′𝐵𝑟3 −

2𝐶

𝑟3(1 + 𝜈′) −

4𝐷

𝑟)))

The obtained angular displacement gradient is as follows:

𝜕𝑢𝜃

𝜕𝜃= (

4𝑠𝑖𝑛2𝜃

𝐸′(−𝐴𝑟(1 + 𝜈′) − 𝐵𝑟3(3 + 𝜈′) −

𝐶

𝑟3(1 + 𝜈′) +

𝐷

𝑟(1 − 𝜈′))) (2-150)

In the above equations 𝐴, 𝐵, 𝐶 and 𝐷 are constants specific to each phase and

vary with applied stresses.

Since the stresses in fiber at 𝑟 = 0 are finite 𝐶𝑓 𝑎𝑛𝑑 𝐷𝑓 are zero.

Continuity of 𝜎𝑟 , 𝜏𝑟𝜃, 𝑢𝑟 , 𝜕𝑢𝜃/𝜕𝜃 must be satisfied at 𝑟 = 𝑎 (fiber/matrix interface)

and 𝑟 = 𝑏 (matrix/composite interface)

46

Continuity of radial stress along fiber/matrix interface yields the following

equation:

𝜎𝑟𝑓(𝑎) = 𝜎𝑟𝑚(𝑎) (2-151)

2𝐴𝑓 − (2𝐴𝑚 +6𝐶𝑚

𝑎4+

4𝐷𝑚

𝑎2) = 0 (2-152)

Similarly, incorporating continuity of radial stress along the matrix/composite

interface we get,

𝜎𝑟𝑚(𝑏) = 𝜎𝑟𝑐(𝑏) (2-153)

(2𝐴𝑚 +6𝐶𝑚

𝑏4+

4𝐷𝑚

𝑏2) − (2𝐴𝑐 +

6𝐶𝑐

𝑏4+

4𝐷𝑐

𝑏2) = 0 (2-154)

Continuity of radial displacement must be ensured along the fiber/matrix interface

𝑢𝑟𝑓(𝑎) = 𝑢𝑟𝑚(𝑎) (2-155)

1

𝐸𝑓′ (2𝐴𝑓𝑎(1 + 𝜈𝑓) + 4𝜈𝑓𝐵𝑓𝑎3)

−1

𝐸𝑚′

((2𝐴𝑚𝑎(1 + 𝜈𝑚) + 4𝜈𝑚𝐵𝑚𝑎3 −2𝐶𝑚

𝑎3(1 + 𝜈𝑚) −

4𝐷𝑚

𝑎))

= 0 (2-156)

Similarly, considering the continuity of radial displacement along the

matrix/composite interface we get the following equation:

1

𝐸𝑚′

((2𝐴𝑚𝑏(1 + 𝜈𝑚) + 4𝜈𝑚𝐵𝑚𝑏3 −2𝐶𝑚

𝑏3(1 + 𝜈𝑚) −

4𝐷𝑚

𝑏))

−1

𝐸𝑐′

((2𝐴𝑐𝑏(1 + 𝜗𝑐) −2𝐶𝑐

𝑏3(1 + 𝜈𝑐) −

4𝐷𝑐

𝑏))

= 0 (2-157)

Continuity of shear stress must be ensured along the fiber/matrix interface

𝜏𝑟𝜃𝑓(𝑎) = 𝜏𝑟𝜃𝑚(𝑎) (2-158)

47

(2𝐴𝑓 + 6𝐵𝑓𝑎2) − (2𝐴𝑚 + 6𝐵𝑚𝑎2 −6𝐶𝑚

𝑎4−

2𝐷𝑚

𝑎2) = 0 (2-159)



(2𝐴𝑚 + 6𝐵𝑚𝑏2 −6𝐶𝑚

𝑏4−

2𝐷𝑚

𝑏2) − (2𝐴𝑐 −

6𝐶𝑐

𝑏4−

2𝐷𝑐

𝑏2) = 0 (2-160)

Continuity of tangential displacement gradient must be ensured along the

fiber/matrix interface

𝜕𝑢𝜃𝑓

𝜕𝜃(𝑎) =

𝜕𝑢𝜃𝑚

𝜕𝜃(𝑎) (2-161)

1

𝐸1𝑓(−𝐴𝑓𝑎(1 + 𝜗𝑓) − 𝐵𝑓𝑎3(3 + 𝜗𝑓))

−1

𝐸1𝑚(−𝐴𝑚𝑎(1 + 𝜈𝑚) − 𝐵𝑚𝑎3(3 + 𝜈𝑚) −

𝐶𝑚

𝑎3(1 + 𝜈𝑚) +

𝐷𝑚

𝑎(1 − 𝜈𝑚))

= 0 (2-162)



1

𝐸𝑚′

(−𝐴𝑚𝑏(1 + 𝜗𝑚′ ) − 𝐵𝑚𝑏3(3 + 𝜗𝑚

′ ) −𝐶𝑚

𝑏3(1 + 𝜗𝑚

′ ) +𝐷𝑚

𝑏(1 − 𝜗𝑚

′ ))

−1

𝐸𝑐′

(−𝐴𝑐𝑏(1 + 𝜗𝑐′) −

𝐶𝑐

𝑏3(1 + 𝜗𝑐

′) +𝐷𝑐

𝑏(1 − 𝜗𝑐

′))

= 0 (2-163)

Constants 𝐴𝑓 , 𝐵𝑓 , 𝐴𝑚, 𝐵𝑚, 𝐶𝑚𝐷𝑚, 𝐴𝑐, 𝐶𝑐 and 𝐷𝑐 can be found from the above

derived continuity equations. The stresses in each phase are calculated as follows

Fiber Equations

𝜎𝑟𝑓 = 2𝐴𝑓𝑠𝑖𝑛2𝜃

48

𝜎𝜃𝑓 = −(2𝐴𝑓 + 12𝐵𝑓𝑟2)𝑠𝑖𝑛2𝜃

𝜏𝑟𝜃𝑓 = (2𝐴𝑓 + 6𝐵𝑓𝑟2)𝑐𝑜𝑠2𝜃 (2-164)

𝑢𝑟𝑓 =𝑠𝑖𝑛2𝜃

𝐸𝑓′ (2𝐴𝑓𝑟(1 + 𝜈𝑓

′ ) + 4𝜈𝑓′ 𝐵𝑓𝑟3)

𝜕𝑢𝜃𝑓

𝜕𝜃=

4𝑠𝑖𝑛2𝜃

𝐸𝑓′ (−𝐴𝑓𝑟(1 + 𝜈𝑓

′ ) − 𝐵𝑓𝑟3(3 + 𝜈𝑓′ ))

Matrix Equations

𝜎𝑟𝑚 = (2𝐴𝑚 +6𝐶𝑚

𝑟4+

4𝐷𝑚


𝜎𝜃𝑚 = − (2𝐴𝑚 + 12𝐵𝑚𝑟2 +6𝐶𝑚


𝜏𝑟𝜃𝑚 = (2𝐴𝑚 + 6𝐵𝑚𝑟2 −6𝐶𝑚

𝑟4−

2𝐷𝑚

𝑟2) 𝑐𝑜𝑠2𝜃 (2-165)

𝑢𝑟𝑚 =𝑠𝑖𝑛2𝜃

𝐸𝑚′

((2𝐴𝑚𝑟(1 + 𝜈𝑚′ ) + 4𝜈𝑚

′ 𝐵𝑚𝑟3 −2𝐶𝑚

𝑟3(1 + 𝜈𝑚

′ ) −4𝐷𝑚

𝑟))

𝜕𝑢𝜃𝑚

𝜕𝜃=

4𝑠𝑖𝑛2𝜃

𝐸𝑚′

(−𝐴𝑚𝑟(1 + 𝜈𝑚′ ) − 𝐵𝑚𝑟3(3 + 𝜈𝑚

′ ) −𝐶𝑚

𝑟3(1 + 𝜈𝑚

′ ) +𝐷𝑚

𝑟(1 − 𝜈𝑚

′ ))

Composite Equations

𝜎𝑟𝑐 = (2𝐴𝑐 +6𝐶𝑐

𝑟4+

4𝐷𝑐


𝜎𝜃𝑐 = − (2𝐴𝑐 +6𝐶𝑐


𝜏𝑟𝜃𝑐 = (2𝐴𝑐 −6𝐶𝑐

𝑟4−

2𝐷𝑐

𝑟2) 𝑐𝑜𝑠2𝜃 (2-166)

𝑢𝑟𝑐 =𝑠𝑖𝑛2𝜃

𝐸𝑐′

((2𝐴𝑐𝑟(1 + 𝜈𝑐′) −

2𝐶𝑐

𝑟3(1 + 𝜈𝑐

′) −4𝐷𝑐

𝑟))

𝜕𝑢𝜃𝑐

𝜕𝜃=

4𝑠𝑖𝑛2𝜃

𝐸𝑐′

(−𝐴𝑐𝑟(1 + 𝜈𝑐′) −

𝐶𝑐

𝑟3(1 + 𝜈𝑐

′) +𝐷𝑐

𝑟(1 − 𝜈𝑐

′))

49

Substituting 𝑟 = ∞ in Equation (2-166) i.e. composite phase we get,

𝜎𝑟𝑐 = (2𝐴𝑐)𝑠𝑖𝑛2𝜃

𝜎𝜃𝑐 = −(2𝐴𝑐)𝑠𝑖𝑛2𝜃 (2-167)

𝜏𝑟𝜃𝑐 = (2𝐴𝑐)𝑐𝑜𝑠2𝜃

Transforming the stresses into Cartesian co-ordinates yields

[

𝜎𝑦

𝜎𝑧

𝜏𝑦𝑧

] = [𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛2𝜃 − 𝑠𝑖𝑛2𝜃𝑠𝑖𝑛2𝜃 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛2𝜃𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 𝑐𝑜𝑠2𝜃

] [𝑠𝑖𝑛2𝜃

−𝑠𝑖𝑛2𝜃𝑐𝑜𝑠2𝜃

] 2𝐴𝑐

We get,

[

𝜎𝑦

𝜎𝑧

𝜏𝑦𝑧

] = [00

2𝐴𝑐

] (2-168)

It can be observed in Equation (2-168), the shear in the transverse plane is the

only non- zero stress present in the plane whereas 𝜎𝑦 and 𝜎𝑧 are zero. So, the applied

transverse shear can be equated to 2𝐴𝑐.

The micro stresses in fiber and matrix from all the five-decomposed cases are

superimposed. Principal stresses are calculated from the resultant stresses in both the

phases. Depending on the failure criterion used, failure envelopes for multiaxial stresses

can be developed and various strengths can be predicted.

A plane strain case along the transverse plane of the composite is considered to

validate the analytical model using finite element analysis which will be discussed in

Chapter 3.

50

CHAPTER 3 FINITE ELEMENT ANALYSIS AND COMPARISON

Modelling and analysis of Hexagonal RVE

In this section, finite element modelling of a rectangular RVE (representative

volume element) of a composite with hexagonal unit-cell is discussed. The results will

be used in section 3.2 for comparing with analytical results. Only a planar hexagonal

array model is considered for the current analysis.

Figure 3-1. Representative volume element of a hexagonal unit cell

The benefit of using a rectangular RVE is that periodic boundary conditions can

be applied with least effort in the finite element analysis. Although models with

rectangular array of fibers is easier to analyze, the hexagonal unit-cell is a better

approximation of the three-phase model used in the present study. It is also a better

idealization of unidirectional fiber composites [13]

The commercial finite element software ABAQUS is used in the present study.

The coordinate system used in ABAQUS is depicted in figure 3-2. The conventional

principal material coordinate system 1-2-3 is designated as y-z-x in ABAQUS.

51

Figure 3-2. Coordinate system used in ABAQUS and principal coordinate system

A 2D deformable planar shell type element is chosen for the micromechanical

analysis. Due to symmetry, a quarter model was considered for the analysis. The

section sketch is depicted in the figure 3-3. For a fiber volume ratio 𝑉𝑓 of 0.6, the

dimensions of the rectangle are calculated as 15X8.66.

Figure 3-3. Sectional view and dimensions of the RVE

Multiple sections were created to obtain a uniform and regular mesh in both the

fiber and matrix phases. Fiber and matrix materials were assumed to be isotropic. The

following table 3-1 provides the list of properties used in the FEA.

52

Table 3-1. Properties of Kevlar/Epoxy used in the FEA

Property Fiber Matrix

Young’s modulus 130 GPa 3.5 GPa

Poisson’s ratio 0.3 0.35

Tensile strength 2.8 GPa 0.07 GPa

The meshing was fashioned differently for the fiber and matrix phases. Figure 3-4

highlights the different regions as well as the meshing pattern. The region enclosed in a

red border is the fiber and rest of the region is matrix. A 4-node bilinear plane strain

quadrilateral element (CPE4R) was chosen as an element type for both the phases.

More information regarding meshing can be observed in the figure 3-5. Reduced

integration was selected to reduce the computational time. A total of 756 elements were

created of which 144 belong to the matrix and 612 belong to the fiber. All the elements

in the fiber and matrix phase are quadrilaterals.

Figure 3-4. Meshed RVE, red bounded regions represent fiber and green unbounded region represents matrix

53

Figure 3-5. Element type used for meshing and analysis

Linear analysis was performed and micro stresses were calculated at the

centroid of each element. From the stress matrix, the maximum principal stress and von

Misses stresses were calculated in each element. Apart from the stresses, volume of

each element is also extracted to compute the macro stresses. Three different plane

strain analysis were performed on the RVE explicitly considering unit strain elongation

in y and z direction as well as unit strain elongation in both y and z directions together.

Figure 3-7 depicts a schematic which summarizes the whole procedure of obtaining the

coefficients of stiffness matrix 𝐶22, 𝐶33, 𝐶23, and 𝐶32 from the unit strain finite element

analysis.

54

Figure 3-6. Boundary conditions and loading in unit strain analysis (A) Direction 2 (B) direction 3

Since we are using plane strain analysis coefficients of stiffness matrix,

𝐶12 and 𝐶13 can also be obtained from the data. The coefficient 𝐶11 can be found using

the Rule of Mixtures

𝐶11 = 𝐶11𝑓𝑉𝑓 + 𝐶11𝑚(1 − 𝑉𝑓) (3-1)

where, 𝑉𝑓 = (𝑎

𝑏)

2

is the volume fraction of the fiber, which is assumed to be 0.6

for the analysis and individualistic stiffness coefficients 𝐶11𝑓 and 𝐶11𝑚 can be calculated

from the material properties of fiber and matrix.

The stiffness matrix 𝐶 can be populated by considering symmetry of [C].

𝐶 = [

𝐶11 𝐶12 𝐶13

𝐶12 𝐶22 𝐶23

𝐶13 𝐶23 𝐶33

]

l l

b

(A) (B)

b

55

Figure 3-7. Schematic of the procedure followed to obtain stiffness matrix

Unit strain analysis

Extract S11, S22, and S33 at the

centroid of each element

Multiply the stresses of each element

with their respective volumes

Summation of the products divided by

the total volume gives the macro

stress

Analysis along 2 and 3 directions yields 2nd and 3rd Columns of the

composite stiffness matrix respectively

56

Table 3-2. Coefficients of stiffness matrix obtained from unit strain analysis

Composite 𝐶22 𝐶33 𝐶23 𝐶32

Kevlar/Epoxy 1.63 1.63 0.730 0.730

E-glass/Epoxy 1.62 1.62 0.680 0.680

After the micromechanical analysis of the hexagonal RVE one can notice that

𝐶22 = 𝐶33 as well as 𝐶23 = 𝐶32. From table 3-2 it is apparent that analysis along one

direction of the transverse axis is sufficient to determine all the coefficients of the

stiffness matrix due to symmetry.

Figure 3-8 depicts deformed RVE from different unit strain analysis, the unit

strain difference can be discerned by overlapping the initial and the deformed bodies.

(A)

Figure 3-8. Initial and deformed hexagonal RVE under unit strain in A) 2nd Direction B) 3rd direction C) 2nd and 3rd direction

57

(B)

(C)

Figure 3-8 Continued

58

Unit strain analysis in both 2 and 3 directions, i.e. figure 3-8 C, is solely done for

comparison of maximum stresses with the analytical model and will be discussed later

in this chapter.

From the micro stresses obtained using the FEA unit strain analysis as well as

individual constituent material strengths and failure criteria, failure envelopes can be

plotted in the 2-3 plane. A flowchart explaining the procedures is given in Fig. 3-9.

The stiffness matrix for the case of Kevlar/epoxy composite has been found out to be

[𝐶] = [84.4 7.50 7.507.50 16.3 7.307.50 7.30 16.3

] 𝐺𝑝𝑎

Considering the tensile strengths mentioned in table 3-1 for fiber and matrix, we

determine the load factors in each element for maximum stress criterion and a quadratic

failure criterion. Identifying the maximum load factor from among the element load

factors will determine the strength of the composite for the given applied stress. Figure

3-10 shows the comparison of maximum principal stress theory and quadratic theory

(von Mises theory) in the transverse direction. The difference in the transverse strength

is attributed to the hexagonal RVE used for unit strain analysis. Figure 3-10 shows the

failure envelopes obtained for Kevlar/epoxy from unit strain analysis in 2 and 3

directions. Table 3-3 contains strengths calculated at different stress conditions. The

difference in the values of 𝑆𝑇+ in direction 2 and 3 can be attributed to the shape of the

hexagonal RVE. The strengths obtained are in a good agreement with strengths from

Zhu, Marrey, and Sankar [15] finite element model.

59

Figure 3-9. Schematic of procedure to plot a failure envelope in 2-3 plane

Inputs: Stiffness Matrix and Strengths of the materials

Calculating compliance matrix and composite properties

Apply a given set of macro-stresses on the composite and calculate corresponding macro-

strains

Calculating the micro stresses and principal stresses in fiber and

matrix

Calculating the load factor for each element using the strength

of the material

Finding out the maximum load factor and dividing it by the applied stress will result in the maximum stress that can be applied

60

A three-letter notation [15] has been used to refer the type of failure criterion

used for fiber and matrix in this section as well as in chapter 4. The first letter refers to

the failure criterion used for fiber and the second letter refers to matrix failure criterion.

The last letter denotes the type of interface failure incorporated and N denotes that the

interface was not considered in the micromechanical analysis. The letters M and Q refer

to maximum stress theory and quadratic failure theory, respectively. For example, QQN

means that quadratic failure criteria were used for both fiber and matrix and interface

failure was not considered.

Figure 3-10. Failure envelopes of Kevlar/Epoxy in transverse direction obtained through unit strain analysis

Table 3-3. Transverse strengths at various points for Kevlar/Epoxy (plane strain)

Failure Criteria 𝑆𝑇+ (Direction 2) 𝑆𝑇

+ (Direction 3) 𝜎2 = 𝜎3 𝜎2 = −𝜎3

MMN 54 38 58.2 28

QQN 71 59 126 34

61

Comparison with analytical model

In this section, comparison of results from finite element model and analytical

model are presented. The validation of the analytical model is done by comparing the

unit strain analysis results with finite element unit strain analysis.

Direct Micro-Mechanics method (DMM) is used in both models to obtain the micro

stresses which are used to calculate the load factors (inverse of factor of safety) to

develop the failure envelopes. ADMM refers to Analytical Direct Micromechanics

Method and FEA refers to Finite Element Analysis. Figure 3-11 shows the comparison

of maximum principal stress failure envelopes in the transverse plane. As one can

notice the envelopes compare fittingly.

Figure 3-11. Comparison of analytical and finite element model failure envelopes using

maximum stress theory in the transverse plane (2-3 plane)

Figure 3-12 portrays the comparison of the failure envelopes considering

quadratic failure theory which also compare favorably. Since the fiber is considered

isotropic, von Mises criteria is used.

62

Figure 3-12. Comparison of analytical and finite element model failure envelopes using

quadratic theory in the transverse plane (2-3 plane)

Table 3-4 shows the comparison of strengths at different locations. One can note

the values compare satisfactorily for (𝜎2 = −𝜎3) pure shear considering quadratic failure

for fiber and matrix. Tables 3-5 to 3-10 contains detailed comparison of various stresses

i.e. maximum principal stress, maximum von Mises stress. Since failure of an element

does not cause catastrophic failure of the composite, 10th percentile stress in each case

is also compared. An additional case for 휀2 = 휀3 = 1 is simulated specifically to compare

the stresses at various points.

In this case the results compared positively. Relatively noticeable difference can

be observed for the case of 휀3 = 1 which can be attributed to the shape of the

hexagonal RVE as discussed in the previous section.

63

Table 3-4. Comparison of various transverse strengths for Kevlar/Epoxy (plane strain)

Failure Criteria 𝑆𝑇+ (Direction 2) 𝑆𝑇

+ (Direction 3) 𝜎2 = 𝜎3 𝜎2 = −𝜎3

FEA DMM FEA DMM FEA DMM FEA DMM

MMN 54 53 38 53 58.2 64 28 44

QQN 71 75 59 75 126 137 34 39

Table 3-5. Comparison of maximum principal stress for Kevlar/Epoxy (plane strain)

Case Fiber Matrix % Difference

FEA DMM FEA DMM Fiber Matrix

휀2 = 1 20.6 20.2 19.4 20.6 2 -6.1

휀3 = 1 25 20.2 25.2 20.5 19 18.5

휀2 = 휀3 = 1 27 26.4 27 26.4 2 2.2

Table 3-6. Comparison of maximum von Mises stress for Kevlar/Epoxy (plane strain)



휀2 = 1 14.4 13.2 11.3 11.5 8.4 -1.4

휀3 = 1 15.1 13.2 14.4 11.4 12.7 20.5

휀2 = 휀3 = 1 10.5 10.6 12.3 11.9 -0.2 3.3

Table 3-7. Comparison of average of top 10% maximum principal stresses for

Kevlar/Epoxy (plane strain)



휀2 = 1 19.4 20.2 19.1 20.4 -3.8 -6.7

휀3 = 1 22.1 20.2 23.5 20.4 8.8 13.2

Table 3-8. Comparison of average of top 10% von Mises stresses for Kevlar/Epoxy

(plane strain)



휀2 = 1 12.3 13.2 11 11.1 -7.1 -0.7

휀3 = 1 14.4 13.2 12.8 11.1 8.5 13.0

64

Table 3-9. Comparison of 10th percentile maximum principal stress for Kevlar/Epoxy (plane strain)



휀2 = 1 19.2 20.2 18.9 20.2 -5.2 -6.9

휀3 = 1 21.2 20.2 22.8 20.2 4.7 11.4

휀2 = 휀3 = 1 26.6 26.0 26.6 26.0 2.3 2.3

Table 3-10. Comparison of 10th percentile maximum von Mises stress for Kevlar/Epoxy

(plane strain)



휀2 = 1 11.9 13.2 10.8 10.8 -10.9 0

휀3 = 1 13.9 13.2 12.0 10.8 5 10

휀2 = 휀3 = 1 10.4 10.5 12.2 12.0 -1 1.6

To summarize, reasonable comparison of the two models validates the analytical

model for further analysis and prediction of composite strengths currently used in

industry which will be discussed in Chapter 4.

65

CHAPTER 4 ANALYTICAL MODEL RESULTS AND DISCUSSION

In this section, the results obtained from Analytical Direct Micro-mechanics

(ADMM) will be discussed. The results are divided into four parts: (1) Failure envelopes

for composites with isotropic fibers, e.g., Kevlar/epoxy composite; (2) Failure envelopes

for composites with transversely isotropic fibers, e.g., carbon/epoxy composite; (3)

Interface effects on the above composites; (4) Volume fraction analysis for the above-

mentioned composites; and (5) Strengths for several composites used currently in

industries.

Results for Kevlar/Epoxy

Two failure criteria, maximum principal stress and quadratic criteria have been

compared in this chapter. The maximum principal stress theory is shown below

𝑆𝐿− < 𝜎1 < 𝑆𝐿

+

𝑆𝑇− < 𝜎2, 𝜎3 < 𝑆𝑇

+ (4-1)

𝜏12 < 𝑆𝐿𝑇

Where, 𝜎1, 𝜎2, 𝜎3 are stresses in the principal material directions.

Strengths 𝑆𝐿+, 𝑆𝑇

+, 𝑆𝐿𝑇 and 𝑆𝑇𝑇 have been tabulated at the end of each analysis.

𝑆𝑇𝑇 Refers to the shear strength in the 2-3 plane. Longitudinal compressive strength

𝑆𝐿− has not been studied since the failure due to compressive stress is due to buckling of

the fibers and is considered more of an instability phenomenon rather than failure of the

material. The notations such as MMN, QQY are defined as follows. The letters M and Q

refer to maximum stress theory and quadratic failure theory, respectively. For example,

QQN means that quadratic failure criteria was used for both fiber and matrix and

interface failure was not considered.

66

In this section, the effective strength properties of Kevlar/Epoxy will be studied

using analytical based micromechanics. The fiber and matrix materials were assumed

to be isotropic. Table 4-1 lists the constituent properties used

Table 4-1. Properties of Kevlar/Epoxy

Property Fiber Matrix

Young’s modulus 130 GPa 3.5 GPa

Poisson’s ratio 0.3 0.35

Tensile strength 2.8 Gpa 0.07 Gpa

Figure 4-1. Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy on 𝜎1 − 𝜎2

plane

67

Figure 4-2. Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy in the 𝜎2 − 𝜎3 plane

Figure 4-3. Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy for

longitudinal shear in the 1-2 or 1-3 plane

68

Figure 4-4. Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy subjected

to both longitudinal and transverse shear stresses

Figure 4-5. Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy for shear

in longitudinal direction and stress in fiber direction

69

Figures 4-1 through 4-5 shows the failure envelopes of Kevlar/Epoxy for various

combinations of normal and shear stresses. The strength properties of Kevlar/epoxy

have been extracted from the above plots and are tabulated in table 4-1. The strengths

obtained show good comparison with the properties obtained through finite element

micromechanics by Zhu, Marrey and Sankar [16].

Table 4-2. Strengths at various points for Kevlar/Epoxy (MPa)

Failure Criteria 𝑆𝐿+ 𝑆𝑇

+ 𝑆𝐿𝑇 𝑆𝑇𝑇

MMN 1573 53 56 44

QQN 1592 75 32 39

Carbon/Epoxy plots

In this section, the effective strength properties of carbon/epoxy will be studied

using analytical micromechanics. A transversely isotropic fiber is considered for the

current analysis. Table 4-3 lists the properties used in detail.

Since a transversely isotropic fiber failure is governed generally by Hashin type

failure theories [17], quadratic failure criterion for transversely isotropic composites is

incorporated for carbon with few modifications.

The quadratic failure can be summarized into four cases as shown below:

Tensile fiber mode

22 2

12 1311

21

L LTS S

(4-2)

Compressive fiber mode

11 LS (4-3)

Tensile matrix mode

70

2 2 2

22 33 23 22 33 12 23( )2 2 2

1 1 1( ) ( ) ( ) 1

T TT LTS S S

(4-1)

Compressive matrix mode

2

2

22 33 22 332

2 2 2

23 22 33 12 232 2

1 11 ( ) ( )

2 4

1 1( ) ( ) 1

T

T TT TT

TT LT

S

S S S

S S

(4-2)

Here 𝑆𝑇𝑇 refers to the transverse shear strength in the composite.

Table 4-3. Properties of Carbon-T300/Epoxy-5208

Property Fiber

Carbon T300

Matrix

Epoxy 5208

Young’s modulus 𝐸1 = 231 𝐺𝑃𝑎 𝐸2 = 13.7 𝐺𝑃𝑎 𝐸𝑚 = 3.5 𝐺𝑃𝑎

Poisson’s ratio 𝜗12 = 0.27 𝜗23 = 0.00125 𝜗𝑚 = 0.35

Tensile strength 𝑆𝐿+ = 2400 𝑀𝑃𝑎 𝑆𝑇

+ = 650 𝑀𝑃𝑎 𝑆 = 60 𝑀𝑃𝑎

Figure 4-6. Comparison of MMN and QQN failure envelopes of Carbon/Epoxy in 1-2

plane

71

Figure 4-7. Comparison of MMN and QQN failure envelopes of Carbon/Epoxy in 2-3

plane

Figure 4-8. Comparison of MMN and QQN failure envelopes of Carbon/Epoxy for shear

in longitudinal directions

72

Figure 4-9. Comparison of MMN and QQN failure envelopes of Carbon/Epoxy subjected

to both longitudinal and transverse shear stresses

Figure 4-10. Comparison of MMN and QQN failure envelopes of Carbon/Epoxy for

longitudinal shear and normal stress in fiber direction

Figures 4-6 to 4-10 shows the failure envelopes of Carbon/Epoxy for various

combinations of normal and shear stresses. The strength properties of Carbon/epoxy

73

have been extracted from the above plots and are tabulated in table 4-4. The strengths

obtained showed excellent comparison with the properties from principles of composite

material mechanics by Gibson [18].

Table 4-4. Predicted strengths of T300/5208/Carbon/Epoxy



MMN 1456 49 57 43

QQN 1456 62 33 31

Effects of Interface

In this section, effects of interface on different strength properties and envelopes

will be discussed. Fiber/ matrix interface plays a pivotal role in determining the strength

of the composites. Two failure criteria are used to study the effects of interface on

strengths. It is assumed that compressive stresses will not affect the interface.

Maximum stress criteria used is discussed as follows:

The normal stress acting at the interface is 𝜎𝑟 i.e. radial tensile stress acting at

the interface.

Interfacial tensile (𝜎𝑟 > 0)

𝜎𝑟 < 𝑆𝐼𝑇 (4-4)

Interfacial shear

√(𝜏𝑟𝜃2 + 𝜏𝑟𝑥

2 ) < 𝑆𝐼𝑆 (4-5)

Where 𝑆𝐼𝑇 is interfacial tensile strength and 𝑆𝐼𝑆 is interfacial shear strength

74

The quadratic interface theory is explained below:

𝜎𝑟2

𝑆𝐼𝑇2 +

(𝜏𝑟𝜃2 + 𝜏𝑟𝑥

2 )

𝑆𝐼𝑆2 = 1 (4-6)

Interface strength determination through experiments is inaccurate and difficult

as very limited data is available. Significant amount of work was done by Huges [19] on

carbon/Epoxy properties. Many factors were considered and there was no simple

conclusion found about the nature of Carbon/Epoxy interface. It was found out that the

interface has a very high stress concertation. The following plots demonstrates the

effect of interface on failure envelopes.

Figure 4-11. Interface effects on failure envelopes for Kevlar/Epoxy in 1-2 plane using

maximum stress theory

75


quadratic theory



76


quadratic theory

Figure 4-15. Interface effects on failure envelopes subjected to both longitudinal and

transverse shear stresses using maximum stress theory

77

Figure 4-16. Interface effects on failure envelopes subjected to both longitudinal and

transverse shear stresses using quadratic theory

Figure 4-17. Interface effects on failure envelopes for Carbon/Epoxy in 1-2 plane using


78


quadratic theory



79


quadratic theory

Figure 4-21. Interface effects on failure envelopes for Carbon/Epoxy for envelopes

subjected to both longitudinal and transverse shear stresses using maximum stress theory

80

Figure 4-22. Interface effects on failure envelopes for Carbon/Epoxy longitudinal shear

and normal stress in fiber direction using maximum stress theory

Figures 4-11 to 4-22 shows the failure envelopes of Carbon/Epoxy for various

combinations of normal and shear stresses. The strength properties of Kevlar/Epoxy

and Carbon/epoxy have been extracted from the above plots and are tabulated in tables

4-5 and 4-6. The strengths obtained showed acceptable comparison with the properties

obtained through finite element micromechanical analysis by Zhu, Marrey and Sankar

[20].

Table 4-5. Comparison of strengths for Kevlar/epoxy including interface failure obtained using ADMM



MMN 1573 53 56 44

QQN 1592 75 32 39

MMM 1573 23.4 22.55 20.38

QQQ 1592 23.4 18.15 20.38

81

Table 4-6. Comparison of strengths for Carbon/Epoxy including interface failure obtained using ADMM



MMN 1456 49 57 43

QQN 1456 62 33 31

MMM 1456 38.6 36 36.5

QQQ 1456 39 37 37

Volume fraction analysis

In this section, the variation of strengths with volume fraction will be discussed for

composites Kevlar/Epoxy and Carbon /Epoxy. Figures 4-23 to 4-28 depicts the variation

of 𝑆𝐿+, 𝑆𝑇

+, 𝑆𝐿𝑇 , 𝑆𝑇𝑇 with 𝑉𝑓.

Interface failure strengths are also considered and as explained in Chapter 3,

MMM refers to maximum stress theory for fiber, matrix and interface.

Figure 4-23. Variation of longitudinal shear strength with volume fraction in

Kevlar/Epoxy

82

Figure 4-24. Variation of Transverse tensile strength with volume fraction in

Kevlar/Epoxy

Figure 4-25. Variation of Transverse shear strength with volume fraction in Kevlar/Epoxy

83

Figure 4-26. Variation of longitudinal shear strength with volume fraction in

Carbon/Epoxy

Figure 4-27. Variation of transverse tensile strength with volume fraction in

Carbon/Epoxy

84

Figure 4-28. Variation of transverse shear strength with volume fraction in

Carbon/Epoxy

Summary

In this section, a detailed comparison of strengths obtained from the analytical

model with reference strengths extracted from Gibson [21] will be discussed.

Table 4-7 shows the comparison of properties obtained from maximum principal stress

theory and quadratic failure theory with the properties available in literature. von Mises

theory was used for an isotropic fiber and Hashin’s [22] criteria was used for

transversely isotropic fibers.

Table 4-8 shows the percentage difference with reference strengths. It can be

noticed that the model showed an excellent comparison with 𝑆𝐿+ in both MMN and QQN

except for boron/aluminum and E-glass/epoxy. In the case of boron/aluminum, yielding

plays a crucial role in determining the strengths which the present model will not

consider.

85

Table 4-7. Comparison of strengths for several composites with analytical model strengths

Reference MMN QQN

Composite 𝜗𝑓 𝑆𝐿+ 𝑆𝑇

+ 𝑆𝐿𝑇 𝑆𝐿+ 𝑆𝑇


+ 𝑆𝐿𝑇

Boron/5505 Boron/epoxy

0.5 1586 62.7 82.7 1593 46 65 1593 63 38

AS/3501 Carbon/epoxy

0.6 1448 48.3 62.1 1456 49 57 1456 62 33

T300/5208 Carbon/Epoxy

0.6 1488 44.8 62.1 1456 48 57 1456 62 33

IM7/8551-7 Carbon/Epoxy

0.6 2578 75.8 --- 2492 67 69 1492 86 42

AS4/APC2 carbon/PEEK

0.58 2060 78 157 2169 63 68 2169 81 41

B4/6061 Boron/Aluminum

0.5 1373 118 128 400 96 95 400 115 57

Kevalr49/epoxy aramid/epoxy

0.6 1379 27.6 60 1332 38 50 1332 45 31

Scocthply1002 E-glass/epoxy

0.45 1100 27.6 82.7 440 35 36 440 44 23

E-glass/470-36 E-glass/Vinyl ester

0.3 584 43 64 478.5 43 40 478.5 50 24

Table 4-8. %Difference of strengths for several composites relative to reference

strengths MMN QQN

Composite 𝜗𝑓 𝑆𝐿+ 𝑆𝑇


+ 𝑆𝐿𝑇

Boron/5505 Boron/epoxy

0.5 0 1 21 0 -31 54

AS/3501 Carbon/epoxy

0.6 0 -1 8 0 -28 47

T300/5208 Carbon/Epoxy

0.6 2 -7 8 2 -38 47

IM7/8551-7 Carbon/Epoxy

0.6 3 10 --- 3 -13 ---

AS4/APC2 carbon/PEEK

0.58 -5 18 57 -5 -5 74

B4/6061 Boron/Aluminum

0.5 67 9 26 67 -7 55

Kevalr49/epoxy aramid/epoxy

0.6 3 -38 17 3 -63 48

Scothply1002 E-glass/epoxy

0.45 60 -37 56 60 -59 72

E-glass/470-36 E-glass/Vinyl ester

0.3 18 0 38 18 -16 63

86

𝑆𝑇+ comparison has been acceptable for all the composites and maximum

principal stress theory is found out to be more suitable to calculate transverse tensile

strength. On the other hand, there is a lot of discrepancy for 𝑆𝐿𝑇 but maximum stress

theory proved to be best considering the cases of carbon fibers. Predicted strengths for

carbon-IM7/epoxy have also been presented in table 4-7. Overall maximum principal

stress theory overlooked quadratic theories for predicting strengths through Direct

Micromechanics Method (DMM).

87

CHAPTER 5 CONCLUSIONS AND FUTURE WORK

The three-phase composite model for unidirectional composites which is

originally proposed for homogenization is extended for a method called Analytical Direct

Micro-Mechanics (ADMM) to predict the strength properties of composites. In the three-

phase model the fiber is surrounded by an annular region of matrix and the fiber-matrix

assemblage is embedded in an infinite medium of composite. The elastic constants of

the composite are evaluated using a modified Halpin-Tsai type equations. In the ADMM

the micro-stresses are calculated in the fiber and matrix phases for a given macro-

stress applied to the composite. The micro-stresses in conjunction with failure theories

for fiber, matrix, and fiber/matrix interface are used to determine the failure of the

composites.

The ADMM is evaluated by comparing the results with that of finite element

based micromechanics. The ADMM results compare reasonably well with FEA based

DMM results for failure envelopes. Then the ADMM is extended to various composite

systems and compared with available results for strength values. The ADMM is able to

predict the strength reasonably well in majority of the cases. The strength in the fiber

direction compares well. The transverse strength properties are different in the 2 and 3

directions in the FEA model because of the hexagonal unit cell. However, the three-

phase model being axisymmetric predicts the same strength in the 2 and 3 directions,

which is closer to practical fiber composites. Since it is based on elasticity solution, it is

much faster than FEA based micro-mechanics. Due to its speed, ADMM can be used in

understanding the effects of variability of constituent properties on the composite

88

strength. The model will be very much suited for non-deterministic design of composite

structures.

Although present failure theories have been partially successful, many theories

have to be modified to keep up with the growing composite technology. The need for

unprecedented methods for predicting failure strengths is everlasting.

The normal strengths obtained from Direct Micromechanics Method (DMM) were

reasonable and are in good agreement with the properties of the composites currently

used in the industry. The modified Halpin-Tsai formulations played a productive role in

calculating the effective composite properties for the three-phase model. Although there

were slight discrepancies while comparing ADMM and FEA in principal direction 2 and

3, it still validates the analytical model since symmetry is not observed in a hexagonal

RVE. The unit strain finite element analysis of a hexagonal RVE validated the analytical

model. Interface strength has no effect on longitudinal tensile strength.

Future work. The newly developed Analytical Direct Micromechanics Method

(ADMM) has a lot of potential for future research. Since a preliminary model was

constructed, modifications can be done to consider buckling of fibers under

compression. The ADMM model with correction response surfaces can be a useful

analytical tool wherein a polynomial function of design variables is used to rectify the

difference from the experimental strength. It is also amenable to probabilistic

micromechanics, in which the effect of variability in fiber and matrix properties on the

stiffness and strength properties can be studied. The model can be extended to fracture

analysis (crack propagation in matrix and interface). Since there was a relatively

noticeable difference in longitudinal shear with properties available in literature, more

89

modifications to the model should be made. Thus, the ADMM paves the way for detailed

analysis of the interface.

90

LIST OF REFERENCES

[1] Nam Ho Kim., Course material for EML5526 Finite Element Analysis, University of Florida, Gainesville.

[2] Voigt, , “Theoretischestudien ber die elasticit tsverh ltnisse der krystalle”,

Abhandlungen der KöniglichenGesellschaft der Wissenschaften,Göttingen, Vol. 34, 1887, pp. 3-51.

[3] Zhu, H., B. V. Sankar, and R. V. Marrey. "Evaluation of failure criteria for fiber

composites using finite element micromechanics." Journal of composite materials 32.8 (1998): 766-782

[4] Christensen and K. H. Lo, “Solutions for effective properties in three phase sphere

and cyinder models.” (1979) [5] Marrey, Ramesh V., and Bhavani V. Sankar. "Micromechanical models for textile

structural composites." (1995) [6] Stamblewski, Christopher, Bhavani V. Sankar, and Dan Zenkert. "Analysis of three-

dimensional quadratic failure criteria for thick composites using the direct micromechanics method." Journal of composite materials 42.7 (2008): 635-654

[7] Karkkainen, Ryan L., Bhavani V. Sankar, and Jerome T. Tzeng. "Strength prediction

of multi-layer plain weave textile composites using the direct micromechanics method." Composites Part B: Engineering 38.7 (2007): 924-932

[8] Christensen and K. H. Lo, “Solutions for effective properties in three phase sphere

and cyinder models.” (1979) [9] Christensen. R. M and J. A. Rinde, “Transverse tensile characteristics of fiber

compsoites with flexible resin:theory and test results.” (1979) [10] Adams, D. F., Doner, D. R.,“Transverse normal loading of a

unidirectionalcomposite”, Journal of Composite Materials, Vol. 1, 1967, pp. 152-164. [11] Timoshenko.S and J.N. Goodier “theory of elasticity” (1951) [12] Timoshenko.S and J.N. Goodier “theory of elasticity” (1951) [13] Agarwal.B. D and L.J.Broutman “Analysis and performance of fiber compsites” John

Wiley & Sons, Inc (1990) [14] Zhu, H., B. V. Sankar, and R. V. Marrey. "Evaluation of failure criteria for fiber


91

[15] Zhu, H., B. V. Sankar, and R. V. Marrey. "Evaluation of failure criteria for fiber composites using finite element micromechanics." Journal of composite materials 32.8 (1998): 766-782

[16] Zhu, H., B. V. Sankar, and R. V. Marrey. "Evaluation of failure criteria for fiber


[17] Hashin, Zvi. "Failure criteria for unidirectional fiber composites." Journal of applied

mechanics 47.2 (1980): 329-334 [18] Gibson, Ronald F. Principles of composite material mechanics. CRC Press, 2011 [19] Hughes J.D.H, “The carbon fiber/epoxy interface – A review.” (1990) [20] Zhu, H., B. V. Sankar, and R. V. Marrey. "Evaluation of failure criteria for fiber


[21] Gibson, Ronald F. Principles of composite material mechanics. CRC Press, 2011 [22] Hashin, Zvi. "Failure criteria for unidirectional fiber composites." Journal of applied

mechanics 47.2 (1980): 329-334

92

BIOGRAPHICAL SKETCH

Sai Tharun Kotikalapudi was born in Hyderabad, India, in 1991. He grew up in

various southern states of India. He received a Bachelor of Technology in the field of

mechanical engineering from SASTRA University, India. From there he proceeded to

work for a year at Tata Consultancy Services Engineering and Infrastructure Services,

in Bangalore, India. He has worked under the guidance of personnel from various

internationally recognized companies such as Johnson & Johnson and Beckman

Coulter. His work involved developing and redesigning medical equipment in

SolidWorks. He defended his M.S. thesis in October 2017.

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ANALYTICAL AND FINITE ELEMENT BASED ...ufdcimages.uflib.ufl.edu/UF/E0/05/19/30/00001/...analytical model is first verified by comparing with results from finite element based micro-mechanics