ANALYTICAL AND FINITE ELEMENT BASED MICROMECHANICS FOR FAILURE THEORY OF COMPOSITES

Transcription

1 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

2 2017 Sai Tharun Kotikalapudi

3 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. 4

5 TABLE OF CONTENTS page ACKNOWLEDGMENTS... 4 LIST OF TABLES... 6 LIST OF FIGURES... 8 LIST OF ABBREVIATIONS ABSTRACT CHAPTER 1 INTRODUCTION Literature Review Research Scope ANALYTICAL EQUATIONS Introduction to the Three-Phase Model Halpin Tsai Formulation for Composite Properties Longitudinal and Hydrostatic Stress Equations Longitudinal Shear Stress in the x-y plane Longitudinal Shear Stress in the x-z plane Biaxial tension/compression in y-z plane Transverse Shear Equations FINITE ELEMENT ANALYSIS AND COMPARISON Modelling and analysis of Hexagonal RVE Comparison with analytical model ANALYTICAL MODEL RESULTS AND DISCUSSION Results for Kevlar/Epoxy Carbon/Epoxy plots Effects of Interface Volume fraction analysis Summary CONCLUSIONS AND FUTURE WORK LIST OF REFERENCES BIOGRAPHICAL SKETCH

6 LIST OF TABLES Table page 2-1 Comparison of macro stresses with average micro stresses for longitudinal shear stress Comparison of macro stresses with average micro stresses for normal Stress and in plane shear stress Properties of Kevlar/Epoxy used in the FEA Coefficients of stiffness matrix obtained from unit strain analysis Transverse strengths at various points for Kevlar/Epoxy (plane strain) Comparison of various transverse strengths for Kevlar/Epoxy (plane strain) Comparison of maximum principal stress for Kevlar/Epoxy (plane strain) Comparison of maximum von Mises stress for Kevlar/Epoxy (plane strain) Comparison of average of top 10% maximum principal stresses for Kevlar/Epoxy (plane strain) Comparison of average of top 10% von Mises stresses for Kevlar/Epoxy (plane strain) Comparison of 10 th percentile maximum principal stress for Kevlar/Epoxy (plane strain) Comparison of 10 th percentile maximum von Mises stress for Kevlar/Epoxy (plane strain) Properties of Kevlar/Epoxy Strengths at various points for Kevlar/Epoxy (MPa) Properties of Carbon-T300/Epoxy Predicted strengths of T300/5208/Carbon/Epoxy Comparison of strengths for Kevlar/epoxy including interface failure obtained using ADMM Comparison of strengths for Carbon/Epoxy including interface failure obtained using ADMM

7 4-7 Comparison of strengths for several composites with analytical model strengths %Difference of strengths for several composites relative to reference strengths

8 LIST OF FIGURES Figure page 1-1 Depiction of a RVE for the analytical model Decomposition of macro stresses applied to an RVE of a fiber composite 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) Decomposition of applied state of macro stresses into five cases Three-phase model Representative volume element of a hexagonal unit cell Coordinate system used in ABAQUS and principal coordinate system Sectional view and dimensions of the RVE Meshed RVE, red bounded regions represent fiber and green unbounded region represents matrix Element type used for meshing and analysis Boundary conditions and loading in unit strain analysis (A) Direction 2 (B) direction Schematic of the procedure followed to obtain Stiffness matrix Initial and deformed hexagonal RVE under unit strain in A) 2 nd Direction B) 3 rd direction C) 2 nd and 3 rd direction Schematic of procedure to plot a failure envelope in 2-3 plane Failure envelopes of Kevlar/Epoxy in transverse direction obtained through unit strain analysis Comparison of analytical and finite element model failure envelopes using maximum stress theory in the transverse plane (2-3 plane) Comparison of analytical and finite element model failure envelopes using quadratic theory in the transverse plane (2-3 plane) Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy on σ1 σ2 plane

9 4-2 Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy in the σ2 σ3 plane Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy for longitudinal shear in the 1-2 or 1-3 plane Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy subjected to both longitudinal and transverse shear stresses Comparison of MMN and QQN failure envelopes of Kevlar/Epoxy for shear in longitudinal direction and stress in fiber direction Comparison of MMN and QQN failure envelopes of Carbon/Epoxy in 1-2 plane Comparison of MMN and QQN failure envelopes of Carbon/Epoxy in 2-3 plane Comparison of MMN and QQN failure envelopes of Carbon/Epoxy for shear in longitudinal directions Comparison of MMN and QQN failure envelopes of Carbon/Epoxy subjected to both longitudinal and transverse shear stresses Comparison of MMN and QQN failure envelopes of Carbon/Epoxy for longitudinal shear and normal stress in fiber direction Interface effects on failure envelopes for Kevlar/Epoxy in 1-2 plane using maximum stress theory Interface effects on failure envelopes for Kevlar/Epoxy in 1-2 plane using quadratic theory Interface effects on failure envelopes for Kevlar/Epoxy in 2-3 plane using maximum stress theory Interface effects on failure envelopes for Kevlar/Epoxy in 2-3 plane using quadratic theory Interface effects on failure envelopes subjected to both longitudinal and transverse shear stresses using maximum stress theory Interface effects on failure envelopes subjected to both longitudinal and transverse shear stresses using quadratic theory Interface effects on failure envelopes for Carbon/Epoxy in 1-2 plane using maximum stress theory

10 4-18 Interface effects on failure envelopes for Carbon/Epoxy in 1-2 plane using quadratic theory Interface effects on failure envelopes for Carbon/Epoxy in 2-3 plane using maximum stress theory Interface effects on failure envelopes for Carbon/Epoxy in 2-3 plane using quadratic theory Interface effects on failure envelopes for Carbon/Epoxy for envelopes subjected to both longitudinal and transverse shear stresses Interface effects on failure envelopes for Carbon/Epoxy longitudinal shear and normal stress in fiber direction using maximum stress theory

11 LIST OF ABBREVIATIONS ADMM DMM FEA PBC RVE Analytical Direct Micromechanics Method Direct Micromechanics Method Finite Element Analysis Periodic boundary Conditions Representative Volume Element 11

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 Chair: Bhavani V. Sankar Major: Mechanical Engineering By Sai Tharun Kotikalapudi December 2017 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 fibermatrix 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. 12

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. 13

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 threephase elasticity model for unidirectional fiber composites [4]. 14

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. 15

16 Fiber Matrix Composite 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 [σ x σ y σ z τ yz τ xy τ xz ], Note that throughout this study the principal material coordinate system of the fiber composite will be denoted either by the standard coordinates or x-y-z coordinates used in the commercial finite element software ABAQUS. The two normal stresses σ y and σ z can be decomposed into two cases hydrostatic stress state such that σ y = σ z = σ H and a biaxial tension/compression such that σ y = σ z. The chart in Figure1-2 depicts various analysis of the stress cases needed to complete the DMM. Applied stress Normal stress Shear stress Axail and hydrostaic Biaxial tension and compresion Transverse Shear YZ Longitudinal Shear XZ Longitudinal Shear XY Figure 1-2. Decomposition of macro stresses applied to an RVE of a fiber composite 16

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. σ 1 τ 12 σ 3 τ 12 τ 12 σ 2 σ 2 σ 3 σ 1 τ 12 Figure 1-3. Macro stresses applied on the unit cell. (similar to τ 12, τ 13 will be acting in the 13 plane and τ 23 will be acting in the 2-3 plane) 17

18 σ 1 σ H = (σ 2 + σ 3 ) 2 σ H σ H σ H σ TC σ 1 (A) Case i τ 23 σ TC σ TC = (σ 2 σ 3 ) 2 τ 23 τ 23 τ 23 σ TC (B) Case ii (C)Case iii τ 12 τ 13 (D)Case iv (E) Case v 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 18

19 Macro stresses Isolating stresses/formulating stress equations Micro stresses Eigen values/principal stresses Load Factors Failure envelopes and strengths 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. 19

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. 2 r a b 3 Fiber Matrix Composite 1 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 20

21 ε 1 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 E 2, G 12 of unidirectional fiber composites. The general form of Halpin-Tsai equations for a property, say P, is as follows: P c = P m ( 1 + ξηv f 1 ην f ) (2-1) Where, η = ( (P f/p m ) 1 (P f /P m ) + ξ ) (2-2) P c : Property of the composite P f : Property of the fiber P m : Property of the matrix ξ: Curve fitting parameter V f : 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 E 2. Whereas for shear modulus G 12, the value ξ = 1 was in excellent 21

22 agreement with the Adams and Doner [10] solution. In both cases a fiber volume fraction of V f = 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 - E 2, G 12. 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 E 2, ξ = 1 for transverse shear modulus G 23 and ξ = 1 for longitudinal or axial shear modulus G 12. Table 2-1. Comparison of macro stresses with average micro stresses for longitudinal shear stress case Fiber Matrix Average stresses τ xz τ xy τ xz τ xy τ xz τ xy τ xz = τ xy = Table 2-2. Comparison of macro stresses with average micro stresses for normal stress and in plane shear stress Case Fiber Matrix Average stresses σ x σ y σ z τ yz σ x σ y σ z τ yz σ x σ y σ z τ yz σ x = σ y = σ z = τ yz =

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 E 1 of the composite is calculated from rule of mixtures: E 1 = E 1f V f + (1 V f )E m (2-3) Here, E 1 : Longitudinal modulus of the composite E 1f : Longitudinal modulus of the fiber E m : Young s modulus of matrix V f : Fiber volume fraction Transverse modulus E 2, longitudinal shear modulus G 12 and in-plane shear modulus G 23 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, u r. The displacement equation for all the three phases (fiber, matrix and composite) is given below. u ri = A i r + B i r (2-4) 23

24 Here, u ri radial displacement in phase i and A i, B i 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. u rf = A f r + B f r (2-5) One can deduce B f = 0 as the displacement at the center of the fiber is zero. The strains are derived as follows: ε r = u r r ε rf = A f (2-6) ε θ = 1 r u θ θ + u r r ε θf = A f (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. u rm = A m r + B m r Following procedures as in fiber phase above, we get (2-8) ε rm = A m B m r 2 (2-9) ε θm = A m + B m r 2 (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. u rc = A c r + B c r (2-11) ε rm = A c B c r 2 (2-12) ε θc = A c + B c r 2 (2-13) 24

25 Considering a composite unit cell, the strains along the longitudinal direction can be equated as follows: ε xf = ε xm = ε xc = ε 0 (2-14) Where, ε 0 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: u ri (r) = u rj (r) (2-15) σ ri = σ rj (2-16) From the equation of continuity of displacement along the fiber/matrix interface u rf (a) = u rm (a) (2-17) A f a 2 A m a 2 A c = 0 (2-18) Here, a: Radius of the Fiber Phase From the equation of continuity of radial stress along the fiber/matrix interface σ rf (a) = σ rm (a) (2-19) The constitutive relation for transversely isotropic materials can be written as σ x C 11 C 12 C 12 ε x { σ r } = [ C 12 C 22 C 23 ] { ε r } (2-20) σ θ C 12 C 23 C 22 ε θ Hence, the stress continuity Equation (2-19) takes the form, ε x [C 12 C 22 C 23 ] f { ε r ε θ }f = [C 12 C 22 C 23 ] m { ε x ε r ε θ }m 25

26 C 12f ε xf + C 22m ε rf + C 23f ε θf = C 12m ε xm + C 22 ε rf + C 23 ε θm (2-21) For simplicity, we use the following notations C 22f + C 23f = α f (2-22) C 22f C 23f = β f (2-23) C 22m + C 23m = α m (2-24) C 22m C 23m = β m (2-25) Then Equation (2-21) can be simplified as A f γ f A m γ m + B mδ m a 2 + ε x (C 12f C 12m ) = 0 (2-26) Similarly, considering continuity of displacement along the matrix/composite interface we get u m (b) = u c (b) (2-27) Equating displacements of matrix and composite phases Here, b is the radius of matrix phase A m b 2 + B m A C a 2 B C = 0 (2-28) Similarly, continuity of radial stress along the matrix/composite interface yield Let ε x [C 12 C 22 C 23 ] m { ε r ε θ }m σ rm (b) = σ rc (b) (2-29) ε x = [C 12 C 22 C 23 ] c { ε r ε θ }c (C 22c + C 23c ) = γ c (2-30) (C 22c C 23c ) = δ c (2-31) Then Equation (2-29) can be simplified as follows 26

27 A m γ m A C γ c A mδ m b 2 + B Cδ c b 2 + ε x(c 11m C 11c ) = 0 (2-32) Boundary Conditions At r = the radial stress at the boundary of the composite σ rc = σ H From the constitutive relation {σ} = [C]{ε} we get ε x [C 12 C 22 C 23 ] c { ε r = σ H (2-33) ε θ }c σ H = A c γ c + ε x C 12c (2-34) Here, σ H is the hydrostatic stress applied at the boundary of the composite The constants A f, A m, B m, A c, B c can be solved from Equations (2-18), (2-26), (2-32) and (2-34). The hydrostatic stress σ H 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: σ xf = C 11f ε xf + A f C 12f + A f C 12f (2-35) σ rf = C 12f ε xf + A f C 22f + A f C 23f (2-36) σ θf = C 12f ε xf + A f C 23f + A f C 22f (2-37) σ xm = C 11m ε xm + (A m B m r 2 ) C 12m + (A m + B m r 2 ) C 12m (2-38) σ rm = C 12m ε xm + (A m B m r 2 ) C 22m + (A m + B m r 2 ) C 23m (2-39) σ θm = C 12m ε xm + (A m B m r 2 ) C 23m + (A m + B m r 2 ) C 22m (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. 27

28 The displacement equation for all the three Phases (fiber, matrix, and composite) is given below: u ri = C i xcosθ (2-41) u θi = C i xsinθ (2-42) u xi = (A i r + B i ) cosθ (2-43) r Here, u ri, u θi, u xi are radial, angular, and axial displacements in the phase i respectively. A i, B i, C i are constants varying with stresses in the phase i. Strain Derivations The six strain components [ε x ε r ε θ γ θx γ rx γ rθ ] are derived in the shear model. From the basic strain formulations for cylindrical system we get, and and ε r = u r r ε r = 0 (2-44) ε θ = u r r + 1 u θ r θ ε θ = 0 (2-45) ε x = u x x ε x = 0 (2-46) It is also observed that the transverse shear strain also vanishes: γ rθ = u θ r + 1 u θ r θ u θ r = 0 (2-47) Hence, only two shear strains will exist in the body, which are derived as follows: and γ θx = u θ x + 1 u x r θ = [A + B + C] sinθ (2-48) r2 γ rx = u r x + u x r = [A B + C] cosθ (2-49) r2 28

29 Boundary Conditions Since the displacement at point r = 0 is finite, the constant pertaining to the fiber phase B f must be zero. At r =, the longitudinal displacement u xc in the composite phase must be finite, hence A c = 0. The longitudinal shear strains in the composite phase at r = can be derived from Equations (2-48) and (2-49) are shown below: γ θx = C c sinθ (2-50) γ rx = C c cosθ (2-51) The 3X3 rotation matrix for transformation of cylindrical coordinates to Cartesian coordinates is cosθ sinθ 0 R = [ sinθ cosθ 0] 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, { γ zx γ } = [ cosθ xy sinθ sinθ cosθ ] {γ θx γ } (2-52) rx { γ zx γ xy } = { 0 C c } (2-53) It can be observed that the shear strain γ xy is the only shear strain presiding in this model and can be equated to C c (Constant in composite phase). From the shear stress-strain relations formula, we deduce the following relation C c = γ yx = τ 0 c G (2-54) xy 29

30 Here, C c : Constant C in the composite phase G c xy : Longitudinal shear modulus of the composite derived from Halpin-Tsai Equation Continuity Equations Continuity of displacement and radial stresses must be ensured along the fiber/matrix interface and matrix/composite interface. The continuity equations are shown below: u ri (r i ) = u rj (r i ) (2-55) u xi (r i ) = u xj (r i ) (2-56) τ rxi (r i ) = τ rxj (r i ) (2-57) Here, u ri, u rj,u xi, u xj are the radial and axial displacements in consecutive phases i, j and τ rxi, τ rxj are the radial shear strains in the corresponding phases. at r = a From Equation (2-55) continuity of displacement along the fiber/matrix interface Similarly, at r = b u rf (a) = u rm (a) C f xcosθ = C m xcosθ (2-58) u m (b) = u c (b) C m xcosθ = C c xcosθ (2-59) From Equations (2-58) and (2-59) we can infer that C f = C m = C c Let the constants C f, C m and C be equal to C: C f = C m = C c = C (2-60) Ensuring equal axial displacements at the fiber/matrix interface r = a we obtain u xf (a) = u xm (a) A f a + A m a B m a = 0 (2-61) where, A f, A m and B m are constants in fiber and matrix phases 30

31 From Equation (2-57) continuity of shear stress along the fiber/matrix interface τ rxf (a) = τ rxm (a) G f rx (A f + C) G m rx (A m B m + C) = 0 (2-62) a2 where, G f m rx, G rx are the longitudinal shear moduli of the fiber phase and matrix phase respectively get, Similarly, considering axial displacements at the matrix/composite interface we u xm (b) = u xc (b) B c b A mb B m b = 0 (2-63) Here, A m, B m B c are constants in matrix phase and composite phase varying with applied stresses. Similarly, continuity of shear stress along the matrix/composite interface yields: τ rxm (b) = τ rxc (b) G m rx (A m B m b 2 + C) G rx c ( B c + C) = 0 (2-64) b2 where, G m c rx, G rx are the longitudinal shear moduli of the matrix phase and composite phase respectively The five constants (A f, A m, B m, B c, C) are solved from the following five equations: A f + A m a B m a = 0 (2-65) f (A f + C) G m rx (A m B m + C) = 0 (2-66) a2 G rx B c b A mb B m b = 0 (2-67) G m rx (A m B m b 2 + C) G rx c ( B c + C) = 0 (2-68) b2 C = τ xy c G (2-69) xy 31

32 The micro stresses in fiber and matrix phases for longitudinal shear XY case are calculated from the following equations: τ θxf = G f rx (A f + C) sinθ (2-70) τ rxf = G f rx (A f + C) cosθ (2-71) τ θxm = G m rx (A m B m + C) sinθ (2-72) r2 τ rxm = G m rx (A m B m + C) sinθ (2-73) r2 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. given below: The displacement equation for all the 3 Phases (Fiber, Matrix, and Composite) is u ri = C i xsinθ (2-74) u θi = C i xcosθ (2-75) u xi = (A i r + B i ) sinθ (2-76) r Here, u ri, u θi, u xi are radial, angular, and axial displacements in the phase i respectively. A i, B i, C i are constants varying with stresses in the phase i Strain Derivations The 6 fundamental strains [ε x ε r ε θ γ θx γ rx γ rθ ] are derived in the shear model. From the basic strain formulations for cylindrical system we get, ε r = u r r ε r = 0 (2-77) 32

33 and ε θ = u r r + 1 u θ r θ ε θ = 0 (2-78) and ε x = u x x ε x = 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. γ rθ = u θ r + 1 u θ r θ u θ 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: γ θx = u θ x + 1 u x r θ = [A + B C] cosθ (2-81) r2 γ rx = u r x + u x r = [A B C] sinθ (2-82) r2 Boundary Conditions Since the displacement at point r = 0 is finite, the constant pertaining to the fiber phase B f must be zero. At r =, the longitudinal displacement u xc in the composite phase must be finite, hence A c = 0. The longitudinal shear strains in the composite phase at r = can be derived from Equations (2-81) and (2-82) are shown below: γ θx = C c sinθ (2-83) γ rx = C c cosθ (2-84) 33

34 The 3X3 rotation matrix for transformation of cylindrical coordinates to Cartesian coordinates is cosθ sinθ 0 R = [ sinθ cosθ 0] 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, { γ zx γ } = [ cosθ xy sinθ sinθ cosθ ] {γ θx γ } (2-85) rx { γ zx γ xy } = { C c 0 } (2-86) It can be observed that the shear strain γ zx is the only shear strain presiding in this model and can be equated to C c (Constant in composite phase). From basic shear stress formula, we can deduce the following relation: equation Here, C c : Constant C in the composite phase C c = γ zx = τ zx c G (2-87) rx G c zx : Longitudinal shear modulus of the composite derived from Halpin-Tsai Continuity Equations Continuity of displacement and radial stresses must be ensured along the fiber/matrix interface and matrix/composite interface. The continuity equations are shown below: u ri (r i ) = u rj (r i ) (2-88) 34

35 u xi (r i ) = u xj (r i ) (2-89) τ rxi (r i ) = τ rxj (r i ) (2-90) Here, u ri, u rj,u xi, u xj are the radial and axial displacements in consecutive phases i, j and τ rxi, τ rxj 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 u rf (a) = u rm (a) C f xsinθ = C m xsinθ (2-91) Similarly, at r = b u m (b) = u c (b) C m xsinθ = C c xsinθ (2-92) From Equations (2-91) and (2-92) we can infer that C f = C m = C c For simplicity, the following assumption has been made and will be considered for future derivations and equations get stresses interface C f = C m = C c = C (2-93) Ensuring equal axial displacements about the fiber/matrix interface at r = a we u xf (a) = u xm (a) A f a + A m a + B m a = 0 (2-94) Here, A f, A m and B m are constants in fiber and matrix phases varying with applied From Equation (2-90) continuity of radial shear stress along the fiber/matrix τ rxf (a) = τ rxm (a) G f rx (A f C) + G m rx (A m B m C) = 0 (2-95) a2 35

36 Here, G f m rx, G rx are the longitudinal shear moduli of the fiber phase and matrix phase respectively. Similarly, considering axial displacements at the matrix/composite interface we get u xm (b) = u xc (b) B c b A mb B m b = 0 (2-96) Here, A m, B m B c are constants in matrix phase and composite phase varying with applied stresses yields Similarly, continuity of radial shear stress along the matrix/composite interface τ rxm (b) = τ rxc (b) G m rx (A m B m b 2 C) + G rx c Here, G rx m : Longitudinal shear modulus of the matrix phase G c rx : Longitudinal shear modulus of the composite ( B c C) = 0 (2-97) b2 The five constants (A f, A m, B m, B c, C) are solved from the following five equations: A f + A m a + B m a = 0 (2-98) G rx f (A f + C) + G m rx (A m B m + C) = 0 (2-99) a2 B c b A mb B m b = 0 (2-100) G m rx (A m B m b 2 + C) + G rx c ( B c + C) = 0 (2-101) b2 C = τ xy c G (2-102) xy The micro stresses in each phase are calculated from the following equations 36

37 τ θxf = G f rx (A f C) cosθ (2-103) τ rxf = G f rx (A f C) sinθ (2-104) τ θxm = G m rx (A m + B m C) cosθ (2-105) r2 τ rxm = G m rx (A m B m C) (2-106) r2 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: = (Ar 2 + Br 4 + C + D) cos2θ (2-107) r2 Since plane strain is assumed, the three non-zero stresses pertaining to the 2-D problem [ σ r σ θ τ rθ ] 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 σ r = 1 r r r 2 θ 2 (2-108) σ r = (2A + 4Br 2 2C r 4 ) cos2θ + ( 4A 4Br2 4C r 4 4D r 2 ) cos2θ σ r = (2A + 6C r 4 + 4D ) cos2θ (2-109) r2 The tangential stress derivation is shown below σ θ = 2 r 2 σ θ = (2A + 12Br 2 + 6C ) cos2θ (2-110) r4 The transverse shear in the plane is derived as follows 37

38 τ rθ = r (1 r θ ) (2-111) τ rθ = r ( 1 r (Ar2 + Br 4 + C + D) 2sin2θ) r4 τ rθ = (2A + 6Br 2 6C r 4 2D ) sin2θ (2-112) r2 The radial strain at any point is given by the following equations ε r = 1 E (σ r ν σ θ ) or ε r = u r 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 u r = (σ r ν σ θ )dr (2-114) Substituting σ θ from Equation (2-110) in the above equation we get u r = 1 6C (( 2A E r 4 4D r 2 ) Cos2θ ν (2A + 12Br 2 + 6C ) Cos2θ) dr r4 u r = Cos2θ E (( 2A 6C r 4 4D r 2 ) ν (2A + 12Br 2 + 6C )) dr (2-115) r4 u r = Cos2θ E (( 2Ar + 2C r 3 + 4D r ) ν (2Ar + 4Br 3 2C r 3 )) The radial displacement at any point on the surface can be obtained by the following equation: u r = Cos2θ E (( 2Ar(1 + ν ) 4ν Br 3 + 2C r 3 (1 + ν ) + 4D r )) (2-116) 38

39 The tangential strain at any point is given by the following equations: ε θ = 1 E (σ θ ν σ r ) ε θ = u r r + 1 u θ r θ (2-117) Equation (2-117) can be modified to obtain the angular displacement gradient with respect to θ as follows: relations: u θ θ = r ( 1 E (σ θ ν σ r ) u r r ) (2-118) Substituting σ θ and σ r from Equations (2-110) and (2-109) we get the following u θ θ = r ( 1 E ((2A + 12Br2 + 6C r 4 ) Cos2θ + ν (2A + 6C r 4 + 4D r 2 ) Cos2θ) u r r ) u θ θ = (Cos2θ E (2Ar(1 + ν ) + 12Br 3 + 6C r 3 (1 + ν ) + 4Dν r ) Cos2θ E (( 2Ar(1 + ν ) 4ν Br 3 + 2C r 3 (1 + ν ) + 4D r ))) The obtained angular displacement gradient is as follows: u θ θ = (4Cos2θ E (Ar(1 + ν ) + Br 3 (3 + ν ) + C r 3 (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 C f and D f are equal to zero. Continuity of σ r, τ rθ, u r, u θ / θ must be satisfied at r = a (fiber/matrix interface) and r = b (matrix/composite interface) equation: Continuity of radial stress along fiber/matrix interface yields the following 39

40 σ rf (a) = σ rm (a) (2-120) A f A m 3C m a 4 2D m a 2 = 0 (2-121) Similarly, incorporating continuity of radial stress along the matrix/composite interface we get, A m 3C m b 4 σ rm (b) = σ rc (b) (2-122) 2D m b 2 + A c + 3C c b 4 + 2D c b 2 = 0 (2-123) Continuity of radial displacement must be ensured along the fiber/matrix interface u rf (a) = u rm (a) (2-124) 1 E f ( 2A fa(1 + ν f ) 4ν f B f a 3 ) 1 E m (( 2A m a(1 + ν m ) 4ν m B m a 3 + 2C m a 3 (1 + θ m ) + 4D m a )) = 0 (2-125) Similarly, considering the continuity of radial displacement along matrix/composite interface we get the following equation: 1 E m (( 2A m b(1 + ν m ) 4ν m B m b 3 + 2C m (1 + ν b 3 m ) + 4D m )) 1 b E (( 2A c b(1 + ν c ) + c 2C c (1 + ν b c) + 4D c 3 b )) = 0 (2-126) Continuity of shear stress must be ensured along the fiber/matrix interface τ rθf (a) = τ rθm (a) (2-127) (2A f + 6B f a 2 ) (2A m + 6B m r 2 6C m a 4 2D m a 2 ) = 0 (2-128) Similarly, considering the continuity of shear stress along matrix/composite interface we get the following equation: 40

41 (2A m + 6B m b 2 6C m b 4 2D m b 2 ) (2A c 6C c b 4 2D c b 2 ) = 0 (2-129) Continuity of tangential displacement gradient must be ensured along the fiber/matrix interface u θf θ (a) = u θm θ 1 E f (A fa(1 + ν f ) + B f a 3 (3 + ν f )) (a) (2-130) 1 E m (A m a(1 + ν m ) + B m a 3 (3 + ν m ) + C m a 3 (1 + ν m ) + D m a (ν m 1)) = 0 (2-131) Similarly, considering the continuity of shear stress along matrix/composite interface we get the following equation: 1 E m (A m b(1 + ν m ) + B m b 3 (3 + ν m ) + C m b 3 (1 + ν m ) + D m b (ν m 1)) 1 E c (A cb(1 + ν c ) + C c b 3 (1 + ν c ) + D c b (ν c 1)) = 0 (2-132) Constants A f, B f, A m, B m, C m D m, A c, C c and D c can be found from the above derived continuity equations. The stresses in each phase can be calculated from the below mentioned equations: Fiber Equations σ rf = 2A f Cos2θ σ θf = (2A f + 12B f r 2 )Cos2θ τ rθf = (2A f + 6B f r 2 )Sin2θ (2-133) 41

42 u rf = Cos2θ E f ( 2A f r(1 + ν f ) 4ν f B f r 3 ) u θf θ = 4Cos2θ E f (A f r(1 + ν f ) + B f r 3 (3 + ν f )) Matrix Equations σ rm = (2A m + 6C m r 4 + 4D m r 2 ) Cos2θ σ θm = (2A m + 12B m r 2 + 6C m r 4 ) Cos2θ τ rθm = (2A m + 6B m r 2 6C m r 4 2D m ) Sin2θ (2-134) r2 u rm = Cos2θ (( 2A E m r(1 + ν m ) 4ν m B m r 3 + 2C m m r 3 (1 + ν m ) + 4D m r )) u θm θ = 4Cos2θ (A E m r(1 + ν m ) + B m r 3 (3 + ν m ) + C m m r 3 (1 + ν m ) + D m r (ν m 1)) Composite Equations σ rc = (2A c + 6C c r 4 + 4D c r 2 ) Cos2θ σ θc = (2A c + 6C c r 4 ) Cos2θ τ rθc = (2A c 6C c r 4 2D c ) Sin2θ (2-135) r2 u rc = Cos2θ (( 2A E c r(1 + ν c ) + 2C c c r 3 (1 + ν c ) + 4D c r )) u θc θ = 4Cos2θ (A E c r(1 + ν c ) + C c c r 3 (1 + ν c ) + D c r (ν c 1)) Substituting r = in Equations (2-135) i.e. composite phase we get, σ rc = (2A c )Cos2θ 42

43 σ θc = (2A c )Cos2θ (2-136) τ rθc = (2A c )Sin2θ Transforming the stresses into Cartesian co-ordinates yields σ y Cos 2 θ Sin 2 θ Sin2θ Cos2θ [ σ z ] = [ Sin 2 θ Cos 2 θ Sin2θ] [ Cos2θ ] 2A c τ yz CosθSinθ CosθSinθ Cos2θ Sin2θ σ y 2A c [ σ z ] = [ +2A c ] (2-137) τ yz 0 It can be observed in Equation (2-137), the shear in the transverse plane is zero and the only non-zero stresses are σ y and σ z 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 = (Ar 2 + Br 4 + C + D) sin2θ (2-138) r2 Since plane strain is assumed, the three fundamental stresses pertaining to a 2- D problem [ σ r σ θ τ rθ ] 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: σ r = 1 r r r 2 θ 2 (2-139) σ r = (2A + 4Br 2 2C r 4 ) cos2θ + (4A + 4Br2 + 4C r 4 + 4D r 2 ) sin2θ σ r = (2A + 6C r 4 + 4D ) sin2θ (2-140) r2 43

44 The tangential stress derivation is shown below: σ θ = 2 r 2 σ θ = (2A + 12Br 2 + 6C ) sin2θ (2-141) r4 The transverse shear in the plane is derived as follows: τ rθ = r (1 r τ rθ = r ( 2 r (Ar2 + Br 4 + C + D) cos2θ) r4 θ ) (2-142) τ rθ = (2A + 6Br 2 6C r 4 2D ) cos2θ (2-143) r2 The radial strain at any point is given by the following equations: ε r = 1 E (σ r θ σ θ ) or ε r = u r 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 u r = (σ r ν σ θ )dr (2-145) Substituting σ θ from Equation (2-141) in the above equation we get, u r = 1 6C ((+2A + E r 4 + 4D r 2 ) sin2θ ν ( (2A + 12Br 2 + 6C )) sin2θ) dr r4 u r = sin2θ E ((+2A + 6C r 4 + 4D r 2 ) + ν (2A + 12Br 2 + 6C )) dr (2-146) r4 u r = sin2θ E ((2Ar 2C r 3 4D r ) + ν (2Ar + 4Br 3 2C r 3 )) The radial displacement at any point on the surface can be obtained by the following equation: 44

45 u r = sin2θ E ((2Ar(1 + ν ) + 4ν Br 3 2C r 3 (1 + ν ) 4D r The tangential strain at any point is given by the following equations: )) (2-147) ε θ = 1 E (σ θ ν σ r ) ε θ = u r r + 1 u θ r θ (2-148) Equation (2-148) can be modified to obtain the angular displacement gradient with respect to θ as follows: relations: u θ θ = r ( 1 E (σ θ ν σ r ) u r r ) (2-149) Substituting σ θ and σ r from Equations (2-140) and (2-141) we get the following u θ θ = r ( 1 E ( (2A + 12Br2 + 6C r 4 ) sin2θ ν (2A + 6C r 4 + 4D r 2 ) sin2θ) u r r ) u θ θ = (sin2θ E ( 2Ar(1 + ν ) 12Br 3 6C r 3 (1 + ν ) 4Dθ r ) sin2θ E ((2Ar(1 + ν ) + 4ν Br 3 2C r 3 (1 + ν ) 4D r ))) The obtained angular displacement gradient is as follows: u θ θ = (4sin2θ E ( Ar(1 + ν ) Br 3 (3 + ν ) C r 3 (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 C f and D f are zero. Continuity of σ r, τ rθ, u r, u θ / θ must be satisfied at r = a (fiber/matrix interface) and r = b (matrix/composite interface) 45

46 equation: Continuity of radial stress along fiber/matrix interface yields the following σ rf (a) = σ rm (a) (2-151) 2A f (2A m + 6C m a 4 + 4D m a 2 ) = 0 (2-152) Similarly, incorporating continuity of radial stress along the matrix/composite interface we get, σ rm (b) = σ rc (b) (2-153) (2A m + 6C m b 4 + 4D m b 2 ) (2A c + 6C c b 4 + 4D c b 2 ) = 0 (2-154) Continuity of radial displacement must be ensured along the fiber/matrix interface u rf (a) = u rm (a) (2-155) 1 E f (2A fa(1 + ν f ) + 4ν f B f a 3 ) 1 E m ((2A m a(1 + ν m ) + 4ν m B m a 3 2C m a 3 (1 + ν m) 4D m a )) = 0 (2-156) Similarly, considering the continuity of radial displacement along the matrix/composite interface we get the following equation: 1 E m ((2A m b(1 + ν m ) + 4ν m B m b 3 2C m b 3 (1 + ν m) 4D m b )) 1 E c ((2A cb(1 + θ c ) 2C c b 3 (1 + ν c) 4D c b )) = 0 (2-157) Continuity of shear stress must be ensured along the fiber/matrix interface τ rθf (a) = τ rθm (a) (2-158) 46

47 (2A f + 6B f a 2 ) (2A m + 6B m a 2 6C m a 4 2D m a 2 ) = 0 (2-159) Similarly, considering the continuity of shear stress along matrix/composite interface we get the following equation: (2A m + 6B m b 2 6C m b 4 2D m b 2 ) (2A c 6C c b 4 2D c b 2 ) = 0 (2-160) Continuity of tangential displacement gradient must be ensured along the fiber/matrix interface u θf θ (a) = u θm θ 1 E 1f ( A f a(1 + θ f ) B f a 3 (3 + θ f )) (a) (2-161) 1 E 1m ( A m a(1 + ν m ) B m a 3 (3 + ν m ) C m a 3 (1 + ν m) + D m a (1 ν m)) = 0 (2-162) Similarly, considering the continuity of shear stress along matrix/composite interface we get the following equation: 1 E m ( A m b(1 + θ m ) B m b 3 (3 + θ m ) C m b 3 (1 + θ m ) + D m b (1 θ m )) 1 E c ( A cb(1 + θ c ) C c b 3 (1 + θ c ) + D c b (1 θ c )) = 0 (2-163) Constants A f, B f, A m, B m, C m D m, A c, C c and D c can be found from the above derived continuity equations. The stresses in each phase are calculated as follows Fiber Equations σ rf = 2A f sin2θ 47

48 σ θf = (2A f + 12B f r 2 )sin2θ τ rθf = (2A f + 6B f r 2 )cos2θ (2-164) u rf = sin2θ E f (2A f r(1 + ν f ) + 4ν f B f r 3 ) u θf θ = 4sin2θ E f ( A f r(1 + ν f ) B f r 3 (3 + ν f )) u θm θ Matrix Equations σ rm = (2A m + 6C m r 4 + 4D m r 2 ) sin2θ σ θm = (2A m + 12B m r 2 + 6C m r 4 ) sin2θ τ rθm = (2A m + 6B m r 2 6C m r 4 2D m ) cos2θ (2-165) r2 u rm = sin2θ ((2A E m r(1 + ν m ) + 4ν m B m r 3 2C m m r 3 (1 + ν m ) 4D m r )) = 4sin2θ ( A E m r(1 + ν m ) B m r 3 (3 + ν m ) C m m r 3 (1 + ν m ) + D m r (1 ν m )) Composite Equations σ rc = (2A c + 6C c r 4 + 4D c r 2 ) sin2θ σ θc = (2A c + 6C c r 4 ) sin2θ τ rθc = (2A c 6C c r 4 2D c ) cos2θ (2-166) r2 u rc = sin2θ ((2A E c r(1 + ν c ) 2C c c r 3 (1 + ν c ) 4D c r )) u θc θ = 4sin2θ E c ( A c r(1 + ν c ) C c r 3 (1 + ν c ) + D c r (1 ν c )) 48

49 Substituting r = in Equation (2-166) i.e. composite phase we get, σ rc = (2A c )sin2θ σ θc = (2A c )sin2θ (2-167) τ rθc = (2A c )cos2θ Transforming the stresses into Cartesian co-ordinates yields We get, σ y cos 2 θ sin 2 θ sin2θ sin2θ [ σ z ] = [ sin 2 θ cos 2 θ sin2θ ] [ sin2θ] 2A c τ yz cosθsinθ cosθsinθ cos2θ cos2θ σ y 0 [ σ z ] = [ 0 ] (2-168) τ yz 2A c 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 σ y and σ z are zero. So, the applied transverse shear can be equated to 2A c. 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. 49

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 is designated as y-z-x in ABAQUS. 50

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 V f 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. 51

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 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 52

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 C 22, C 33, C 23, and C 32 from the unit strain finite element analysis. 53

54 b b l (A) l (B) 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, C 12 and C 13 can also be obtained from the data. The coefficient C 11 can be found using the Rule of Mixtures C 11 = C 11f V f + C 11m (1 V f ) (3-1) where, V f = ( a b )2 is the volume fraction of the fiber, which is assumed to be 0.6 for the analysis and individualistic stiffness coefficients C 11f and C 11m can be calculated from the material properties of fiber and matrix. The stiffness matrix C can be populated by considering symmetry of [C]. C 11 C 12 C 13 C = [ C 12 C 22 C 23 ] C 13 C 23 C 33 54

55 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 2 nd and 3 rd Columns of the composite stiffness matrix respectively Figure 3-7. Schematic of the procedure followed to obtain stiffness matrix 55

56 Table 3-2. Coefficients of stiffness matrix obtained from unit strain analysis Composite C 22 C 33 C 23 C 32 Kevlar/Epoxy E-glass/Epoxy After the micromechanical analysis of the hexagonal RVE one can notice that C 22 = C 33 as well as C 23 = C 32. 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. Figure 3-8. Initial and deformed hexagonal RVE under unit strain in A) 2 nd Direction B) 3 rd direction C) 2 nd and 3 rd direction (A) 56

57 (B) (C) Figure 3-8 Continued 57