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MULTIPLICATIVE DECOMPOSITION OF DEFORMATION GRADIENT IN FINITE-STRAIN THERMOELASTICITY UDOH, PAUL JAMES B.Sc (Ilorin), M.Sc. (Ilorin). A THESIS SUBMITTED TO THE DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILORIN, ILORIN, IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE AWARD OF THE DEGREE OF DOCTOR OF PHILOSOPHY (Ph.D.) IN MATHEMATICS. JULY, 2008 ii CERTIFICATION This is to certify that the work in this thesis was carried out by UDOH, PAUL JAMES (Matric. No. 82/4551) in the Department of Mathematics, Faculty of Science,University of Ilorin, Ilorin, Nigeria. …………………………. Prof.. J. S. Sadiku (Supervisor) Department of Mathematics University of Ilorin, Ilorin, Nigeria ………. ………………… Date …………………………………. Prof. O. M. Bamigbola Head, Department of Mathematics University of Ilorin, Ilorin. Nigeria . …………………….. Date …………………………………… Dr. M. O. Oyesanya (External Examiner) :……………………… Date iii DEDICATION To my Late Auntie, Rev. Sister Veronica Akpan (M.M.M.) and my family. iv iv ACKNOWLEDGEMENT I appreciate the works of the Almighty God on me for good health of mind and body. In no small measure, I thank my Supervisor Prof. J. S. Sadiku for having the love and patience to read through this work and for his valuable contributions, suggestions, moral and financial supports. I wish to acknowledge with thanks the support and contributions of the Head of Mathematics Department, Prof. O. M. Bamigbola, for his untiring encouragement during the course of this research work. I also wish to acknowledge with special thanks and appreciation the role of Prof. J. A. Gbadeyan in the course of completing this thesis. I acknowledge the support and encouragement of Drs. O. A. Taiwo, R. B. Adeniyi, J. O. Omolehin, Babalola and lecturers in Mathematics Department of the University of Ilorin. I appreciate the University of Uyo for sponsoring me on study leave to undertake this programme and Mr. Enefiok Udoh who assisted me immensely. My sincere thanks go to Profs. R. W. Ogden of Glassgow University, UK; L. A. Taber, J. D. Humphrey of Michigan State University, USA, Marjid Mirzaei of University of Iran; K.Y. Volokh, University of Baltimore; S. R. Seremet, University of Moldova; C. O. Horgan, University of Virginia ,USA ; G. Saccomandi and A. Guillon, University of Lecce, Italy; V. A. Lubarda, University of Yugoslavia and all those who have contributed in one way or the other to this thesis . v Lastly, I acknowledge with gratitude, my lovely wife, Obonganwan (Mrs.) Bassey P. J. Udoh and my children, my parents, brothers and sisters, my cousins and friends. I acknowledge my noble colleagues in the programme for their kindness. vi ABSTRACT This thesis is devoted to finite-strain thermoelasticity capable of admitting large deformations. The relevant constitutive formulation is usually obtained by introducing the concept of thermodynamics into the well-known constitutive relations of linear elasticity. Consequently, the corresponding field equations are obtained. Multiplicative decomposition of total deformation gradient is one of the theories that can be used to obtain these constitutive formulations with the corresponding field equations. This theory exists for three material configurations, namely, the initial (undeformed), the current (deformed) and the intermediate configurations. Two cases of the theory had been presented for elastoplasticity and growth, where the total deformation gradient F is decomposed into Fe . Fp for elastoplasticity, and Fe . Fg for growth where Fe , Fp and Fg are elastic , plastic and growth parts respectively. The multiplicative theory for elastoplasticity and growth had earlier been presented and found very useful in experimental and other theoretical situations. In this context, the concept of multiplicative decomposition of total deformation gradient is introduced in this thesis for the theory of thermoelasticity, where the total deformation gradient F is now decomposed into the product of elastic and thermal parts as Fe . F . Then the associated strain, stress, energy and entropy were derived using three configurations as opposed to two configurations for the classical theory. The specific and latent heats are discussed and the comparison with vii the results of the classical thermoelasticity is given. We extend the analysis to growth theory under thermoelastic influence and finally, we apply thermoelasticity to approximate stress and strain of some known energy functions. Our results showed that while the free energy in the classical theory is determined by the temperature field and the Lagrangian strain, it is determined with the introduction of the multiplicative decomposition theory by the thermal stretch ratio and the elastic strain of the intermediate configuration. While the stress and entropy in classical theory are determined by the rate of change of the free energy with respect to Lagrangian strain and temperature respectively, the same stress in multiplicative theory is determined by the product of thermal stretch ratio, the Lagrangian strain and the temperature and the entropy is determined from the product of thermal stretch ratio and the bulk modulus. viii CONTENTS Title i Certification ii Dedication iii Acknowledgement iv Abstract vi Contents viii CHAPTER ONE: GENERAL INTRODUCTION 1 1.1Background to the Study 1 1.2 Objectives of the Study 9 1.3 Organization of the Thesis 9 1.4 Mathematical Preliminaries and Definition of Relevant Terms. 10 1.4.1 Governing Equations 10 1.4.2 Geometry of Deformation 11 1.4.3 Basic Derivations 11 1.4.4 Balance of Mass 13 1.4.5 Conservation of Mass 15 1.4.6 Balance of Linear Momentum 15 1.4.7 Balance of Angular Momentum 19 1.4.8 Balance of Energy 24 1.4.9 Clausius- Duhem Inequality 27 ix 1.4.10 Constitutive Relations 28 1.4.11 Other Strain and Stress Measures 30 1.4.12 The Stress Power 34 CHAPTER TWO: FINITE-STRAIN THERMOELASTICITY BASED ON MULTIPLICATIVE DECOMPOSITION OF DEFORMATION GRADIENT 2.1 Preamble 37 37 2.2 Thermoelastic Analysis Based on Multiplicative Decomposition of Deformation Gradient 38 2.3 Decomposition of Lagrangian Strain 39 2.4 Free Energy and Constitutive Expressions 45 2.5 Analysis of Stress Response 47 2.6 Entropy Expressions 51 2.7 Some Known Results in Classical Theory of Thermoerlasticity 56 2.8 Comparison of Field Equations and Related Expressions in Classical Theory with Those of the Multiplicative Decomposition. 58 CHAPTER THREE: THERMOELASTIC ANALYSIS BASED ON MULTIPLICATIVE DECOMPOSITION FOR GROWTH. 61 3.1 Preamble 61 3.2 Toy – Tissue Model 62 3.2.1 Basic Assumptions 65 x 3.3 Mechanics of Growth 65 3.3.1 Basic Kinematics 66 3.4 Mass Balance 69 3.4.1 Mass Balance Equation 72 3.4.2 Balance Equations for Momentum 73 3.4.3 Energy Balance and Constitutive Equations 74 3.5 Development of Thremal Residual Stress due to Thermal Changes. 77 3.5.1 The Thin Circular Tube 82 3.5.2 Instantaneous Point Source in an Infinite Body 85 3.5.3 Decomposition of the Free Energy and Evolution of Residual Stresses with Growth 90 3.5.4 Growth Conditions 98 CHAPTER FOUR: DEVELOPMENT of STRESS EXPRESSIONS FOR SOME. KNOWN ENERGY FUNCTIONS BASED ON MULTIPLICATIVE DECOMPOSITION 101 4.1 Preamble 101 4.2 Basic Equations 101 4.3 Explicit Representation of Strain - Energy 113 4.4 Representation of the Free Energy. 115 4.5 Approximating Response Coefficients of Some Known Energy Functions using Thermoelasticity Theory 119 xi 4.6 Some Thermoelastic Results Based on Gent Model 122 4.7 Application of Thermoelasticity to Axial Shear Problem 126 CHAPTER FIVE: GENERAL CONCLUSION 140 5.1 Discussion on Results 140 5.2 Thesis’ Contribution to Knowledge 142 5.3 Conclusion 143 5.4 Recommendation 144 REFERENCES 145 . CHAPTER ONE GENERAL INTRODUCTION 1.1 Background to the Study Elasticity is one of the aspects of continuum mechanics. Unlike in Physics, where materials are considered from the molecular (or atomic) point of view, here materials are viewed as manifold of particles. That is, any material is believed to consist of continuous particles. The constitutive theory of finite- strain thermoelasticity is a classical and well developed topic of the non-linear continuum mechanics, (eg.Trusdell and Noll [61], Nowaki [41]). The formulation of the general theory is given in the thermodynamic framework by introducing the Helmholtz free energy as a function of the finite strain and temperature, and by exploring the conservation of energy and the second law of xii thermodynamics. These yield the constitutive expressions for the stress and entropy, and the condition for the positive-definiteness of the heat conductivity tensor. Only two configurations of the material sample are considered in this approach at an arbitrary instant of deformation; that is, the initial unstressed configuration at the uniform reference temperature and the deformed configuration characterized by nonuniform stress and temperature fields. Some fundamental kinetic aspects of finite deformation elastoplasticity theory within the framework of multiplicative decomposition of deformation gradients had been presented by Lubarda [29]. In this work, there exists three material configurations, namely, the initial (undeformed), the current (deformed), and the intermediate configurations. The intermediate configurations is obtained from the current (deformed) material configurations by elastic destressing of the latter to zero stress. It differs from the initial configuration by residual deformation. It was also shown that the total elastoplastic deformation gradient can be decomposed into its elastic and plastic parts. The same theory of multiplicative decomposition gained prominence in growth theory of biomechanics where the total deformation gradient is again decomposed into the product of elastic and growth parts [59]. In the same way, we now introduce the theory of multiplicative decomposition of total deformation gradient into thermoelasticity. This is based on three configurations, the third configuration is introduced by a conceptual, isothermal destressing of the current configuration to zero stress state (ZSS). The total xiii deformation gradient is then decomposed into the product of purely elastic and thermal parts. The resulting decomposition is then used to build the constitutive analysis. One of the objectives of this research work is to elaborate on this topic, and to compare the results obtained by two different approaches. In the classical formulation of thermoelasticity, some new points are added in the consideration of the latent heats at finite strain and the derivation of the constitutive equations. Particular accent is given to the quadratic dependence of the free energy on the finite Lagrangian strain. The classical theory without the multiplicative decomposition of the deformation gradient is a well known concept. But in this work, we incorporate thermodynamic theory of thermoelasticity and show our results explicitly for elastically and thermally isotropic materials, with an extension to transversely isotropic and orthotropic materials. A physically appealing representation of the free energy is introduced and employed to derive the stress response and entropy expressions. The relationship between the specific and latent heats at constant elastic and total strain are also presented and discussed, as well as the general connection between the two constitutive formulations. Indeed, in non-linear elasticity, explicit constitutive equations are usually obtained using a phenomenological approach based on heuristic considerations, Earman and Mark [9]. In this contribution, we assume the well known results of the classical theory with two configurations of the material sample at an arbitrary instant of deformation, namely, the initial unstressed configuration B0 at uniform reference xiv temperature 0 and the deformed or current configuration Bt, characterized by nonuniform stress and temperature, . With this assumption of the classical theory of finite strain thermoelasticity, we introduce the thermoelastic analysis based on the multiplicative decomposition of the deformation gradient. This theory exists for three material configurations, namely, the initial (undeformed), the current (deformed) and the intermediate configurations. Two cases of the theory had been presented for elastoplasticity and growth, where the total deformation gradient F is decomposed into Fe . Fp for elastoplasticity, and Fe .Fg for growth and where Fe , Fp and Fg are elastic , plastic and growth parts respectively. The multiplicative decomposition of the deformation gradient is an alternative approach to develop the constitutive theory of thermoelastic material response and is based on the introduction of an intermediate configuration B0, which is obtained from the current configuration Bt by isothermal elastic unloading to zero stress (ZSS). The isothermal elastic deformation gradient from B0 to Bt is denoted by Fe, and the thermal deformation gradient from B0 to B by F. The total deformation gradient F, which maps an infinitesimal material element dX from the initial configuration to dx = F.dX in the current configuration, can then be decomposed as F = Fe.F .(1.1) An analogous decomposition of the elastoplastic deformation gradient in its elastic and plastic part was introduced by Lee, E.H.[27]. Earlier contributions toward the introduction of the intermediate configuration in the constitutive analysis of different xv materials include Chadwick [8] and Khul & Steinmann [24]. For the inhomogeneous deformation and temperature fields, only F is a true deformation gradient, whose components are the partial derivatives x X . In contrast, the mappings from B0 to Bt and from B0 to B are not, in general, continuous one-to-one mappings, so that Fe and F are not defined as the gradient of their respective mapping (which may not exist), but as the point function (local deformation gradient). Various geometric and kinematic aspects of the incompatibility of the intermediate configurations are discussed in [17]. The decomposition in Equation (1.1) is not unique because arbitrary rigidbody rotation can be superposed to B preserving it unstressed. However, we shall specify F, and thus the decomposition (1.1) is unique in each considered case or type of the material anisotropy. For example, for transversely isotropic material with the axis of isotropy parallel to the unit vector n0 in the configuration B0, we specify F by F = (-) a0 x a0 + I, .(1.2) where = () is the stretch ratio due to thermal expansion in the direction n0, while = () is the thermal stretch ratio in any direction within the plane of isotropy (orthogonal to n0). An extension of the representation (1.2) to orthotropic material is easily accomplished by decomposition of the Lagrangian strain as: E = ET . Ee. F. + E where (1.3) xvi Ee = ½ (FeT. Fe-I), E = ½ (FT.F-I) , ET are the elastic, thermal strain tensors and strain transpose respectively. In this thesis also, we extend multiplicative decomposition theory to growth of soft biological tissues. From mechanics perspective, as pointed out by Skalak [55], volumetric growth is analogous to thermal expansion. In linear elastic problems, growth (and thermal) stresses can be superposed on the mechanical stress field, but in non-linear problems, another approach must be used. The fundamental idea is to refer the strain measures in the constitutive (stress- strain) equations of each material element to its current ZSS configuration, which changes as the element grows. In this thesis, we adopt the definition used by many researchers that consider growth as a change in mass and geometry. This distinguishes growth from other remodelling, which is often regarded as rearrangement of the microstructure in the tissue but is not considered here. Our concern therefore is to apply the multiplicative decomposition of the deformation gradient to describe the mechanics of growth. The principal notion to be borne in mind while developing a continuum formulation for growth is that one is presented with a system that is open with respect to mass. Scalar mass sources and vectorial mass fluxes must be considered, (Epstein & Maugin [11]). A mass source was the first to be introduced in the context of biological growth. The mass flux is a more recent addition by Epstein and Maugin [11], Kuhl and Steinmann [24] also xvii incorporated the mass flux. Ogden [44] incorporated the mass balance, and linear momentum in his formulations. So far, no thermal terms have been introduced in the continuum formulation for growth. But in our work in this thesis, we introduced the thermal terms and suggest that the stress-strain relations should be analogous to thermoelasticity where the role of the temperature is played by the mass density; the increase of the mass density result in the volume expansion of the tissue given as: W P F ( o ) E (1.3) where, W is the strain-energy of non-growing material, E = (FTF – I)/2 is the Green strain tensor and I is the identity tensor, ρ , ρ0 are densities, T is a symmetric tensor of growth modulli, are related to the material volume expansion for the increasing mass density. The first term on the right-hand side of equation (1.3) is that of the classical hyperlasticity without growth. The first qualitative notion of the analogy between growth and thermal expansion is due to Skalak [55 ]. Finally, we introduce the constitutive equation for mass flux in the simplest Fickean form: = - (-0) or in thermoelastic form: (1.4) xviii = - (-0) (1.5) where is the mass conductivity of the material and is the gradient operator with respect to referential coordinates. The similarity between equations (1.4) and (1.5) is obvious after replacing the mass density increment by the temperature increament, mass flux by heat flux, mass conductivity by thermal conductivity. In this case, equation (1.4) is nothing but the thermoelastic generalization of Hookes law and equations (1.4-1.5) are just the Fourier law of heat conduction. The thermoelastic analogy allows for a better understanding of parameters of the growth process. For example, the vector of mass flux is analogous to the vector of heat flux. Besides the incorporation of a referential mass source, volumetric growth is addressed by means of a multiplicative decomposition of the overall deformation gradient into elastic and a growth distortion. Menzel,[36]. We have been able to obtain some closed form solution for the growth equation by utilizing deformation gradient approach. This research work accounts for the effects of growth on stress but not the effects of stress on growth. Rodriguez et al, [50] formulated a continuum theory that accounts for the coupling between stress and finite growth. In nonlinear elasticity, theories appropriate to small but finite deformation have been introduced by Murnaghan [35], Rivlin [51] and Signorini [54]. The scheme used by these authors was to approximate the strain-energy density function by polynomials in the appropriate invariants. In this way, a particular material is then xix characterized by the constant coefficients of the polynomial rather than by functions. Because the field equations are not approximated the solutions based on this method are approximate solutions for the general hyperelastic material and are exact for special classes of materials. The Rivlin-Signorini method has also been used by Saccomandi [52] in the study of elastic dielectrics and by Martin and Carlson [33] for elastic heat conductors. 1.2 Objectives of the Study: (i) To determine the field equations in thermoelastic forms using multiplicative decomposition of the deformation gradient. (ii) Comparing the strain, stress and entropy expressions of the multiplicative decomposition of deformation gradient with the existing ones earlier obtained for the classical theory with two configurations. (iii) To apply thermoelasticity based on the theory of multiplicative decomposition of the deformation gradient to growth of soft biological tissues. (iv) To obtain approximate stresses and strains of some known strain-energy functions based on multiplicative decomposition of the deformation gradient using the theory of thermoelasticity . 1.3. Organization of the Thesis xx This thesis consists of five chapters. The first chapter dealt with the general introduction, the objectives of the study and the derivation of basic equations. In chapter two, the multiplicative decomposition of the deformation gradient was introduced into the theory of finite-strain thermoelasticity with coupled heat equations. The free energy was also introduced and constitutive expressions for the stress and entropy were derived. Chapter three dealt with the study of growth of soft biological tissues based on the theory of multiplicative decomposition of the deformation gradient. We formulated the basic kinematics of growth in continuum settings. Relevant energy balance and constitutive equations were derived and growth conditions specified. An illustration is given using a toy-tissue model to explain the regular and point-mass supply to the system. The theory is applied to a circular cylindrical tube subjected to extension and inflation and internal pressure when the wall thickness changes as a result of persistent high pressure. In chapter four, we applied thermoelasticity theory to approximate material response functions of some known strain –energy functions based on multiplicative decomposition of deformation gradient. Some thermoelastic models were used as standard references. In chapter five, we stated the concluding remarks and possible extensions to the work done in this thesis. 1.4 Mathematical Preliminaries and Definition of Relevant Terms xxi 1.4.1 Governing Equations The equations that govern the motion of a thermoelatic solid include the balance laws for mass, momentum and energy. Kinematic equations and constitutive relations are needed to complete the system of equations. Physical restrictions on the form of the constitutive relations are imposed by an entropy inequality that expresses the second law of thermodynamics in mathematical form. The balance laws express the idea that the rate of change of quantity (mass, momentum, energy) in a volume must arise from three causes, namely, 1. The physical quantity itself flows through the surface that bounds the volume. 2. There is a source of the physical quantity inside the volume. 3. There is a source of the physical quantity outside the volume. 1.4.2 Geometry of deformation Let B be the body (an open subset of Euclidean space) and let ∂B be its surface (the boundary of B). Let the motion of material points in the body be described by the map x ( X ) x( X ) (1.6) where X is the position of a point in the initial configuration and x is the location of the same point in the deformed configuration. The deformation gradient (F) is given by xxii F x 0 x dX (1.7) 1.4.3 Basic Derivations This section presents the basic balance laws controlling thermomechanical response of simple continua. We emphasize that these balance laws are valid for all simple continuum, so they are valid for a wide class of materials which include: thermoelastic, elastic-plastic solids, etc. The equations that characterize the response of a particular material are called constitutive equations. In this thesis, our attention is focused on thermoelasticity with finite strain and their constitutive equations. Let f(x ,t) be a physical quantity that is flowing through the body, g(x ,t) be sources on the surface of the body and let h(x, t) be sources inside the body. Let n(x ,t) be the outward unit normal to the surface ∂B and v(x, t) be the velocity of the physical particles that carry the physical quantity that is flowing. Also, let the speed at which the boundary surface ∂B is moving be un (in the direction n). Following the work of Gent[14], the balance laws can be expressed in the form: d f ( x , t )dV B f ( x , t ) un ( x , t ) v( x , t ) .n ( x , t ) dA dt B B g( x , t )dA B h( x , t )dV (1.8) where the functions f(x,t), g(x,t) and h(x,t) can be scalar-valued, vector-valued,or tensor-valued, depending on the nature of the physical quantity that the balance xxiii equation deals with. We now state and show the balance laws of mass, momentum and energy as follows: + .v = 0 Balance of mass (1.9) v - . - b = 0 Balance of linear momentum (1.10) = T Balance of Angular momentum e - . (v) + .q - s = 0 Balance of energy (1.11) (1.12) where (x ,t) is the mass density (current), is the material time derivative of , v(x ,t) is the velocity of the particle, v is the material time derivative of v, (x ,t) is the Cauchy stress, b(x ,t) is the body force density, e(x ,t) is the internal energy per unit mass, e is the material time derivative of e, q(x ,t) the heat flux vector and s(x, t) is an energy source per unit mass. 1.4.4 Balance of mass Theorem 1.1: The balance of mass of a material can be expressed as . v 0 (1.13) where (x,t) is the current mass density, is the material time derivative of , and v(x,t) is the velocity of physical particles in the body B bounded by the surface ∂B. Proof: We recall that the general equation for the balance of a physical quantity f(x,t) is given by xxiv d f ( x , t )dV B f ( x , t )un ,( x , t ) v ( x , t ) .n( x , t )dA dt B B g( x , t )dA B h( x , t )d V (1.14) To derive the equation for the balance of mass, we assume that the physical quantity of interest is the mass density ρ(x,t). Since mass is neither created nor destroyed in a closed system, the surface and interior sources are zero ie g(x,t) = h(x,t) = 0. Therefore, we have, d ( x , t )dV B ( x , t )un ( x , t ) v( x , t ).n( x ,t )dA dt B (1.15) Let us assume that the volume B is a controlled volume (i.e., it does not change with time). Then the surface ∂B has zero velocity (un=0), and we get B dV ( v .n )dA t B (1.16) Using divergence theorem, B .vdv B ( v .n )dA (1.17) we get B dV B .( v ) dV t or (1.18) .( v ) dV 0 B t Since B is arbitrary, we have .( v ) 0 t (1.19) xxv Using the identity, .( v ) .v .v (1.20) we have .v .v 0 t (1.21) Now, the material time derivative of ρ is defined as .v t (1.22) Therefore equation (1.21) becomes .v 0 (1.23) and hence the proof. 1.4.5 Conservation of mass The conservation of mass requires the total mass of the region B to remain constant, i.e., B dv B0 0 dv (1.24) where B0 is the region in the reference configuration associated with B, and 0 is the mass density in the reference configuration. Using the relation, dv = JdV (1.25) where, dv and dV are volume elements from current and initial configurations resp. It follows that the integral over B0 can be converted to an integral over B to obtain xxvi 1 dV 0 B 0 J (1.26) For arbitrary B, the local form of the conservation of mass becomes ρ = 0 J-1 (1.27) 1.4.6 Balance of Linear Momentum We now show that the balance of linear momentum can be expressed as: v . b 0 (1.28) or div + ρb = v (1.29) where ρ(x,t) is the mass density, v(x,t) is the velocity, (x,t) is the Cauchy stress and ρb is the body force density. Theorem 1.2 : Equation (1.28) can be obtained using the physical quantity of the momentum density. Proof: From Eqn (1.14), the physical quantity of interest is the momentum density, that is, f(x,t) = ρ(x,t) v(x,t). The source of momentum flux at the surface is the surface traction i.e., g(x,t) = T. The source of momentum inside the body is the body force h(x,t) = ρ(x,t) b(x,t). Therefore, we have, xxvii d vdV B vun v .ndA B t dA B bdV dt B (1.30) The surface traction are related to the Cauchy stress by T = .n Therefore, (1.31) d v dv B v ( un v . n )dA B .n dA B b dv ) dt B (1.32) Assuming that B is an arbitrary fixed control volume, then un = 0 and so equation (1.32) gives B ( v ) dv B v( v .n ) dA B .n dA B b dV dt (1.33) Now from the definition of the tensor product, we have (for all vectors a) (uv). a = (a.v)u (1.34) Therefore, equation (1.33) becomes B ( v ) dV B ( v v ).n dA B .ndA B b dV t (1.35) Using divergence theorem, B .vdV B v .ndA (1.36) We have B or ( v ) dV B . ( v v )dV B .dV B bdV t B ( v ) .( v ) v . t Since B is arbitrary, we have b dV 0 (1.37) xxviii ( v ) .[( v ) v ] . b 0. t (1.38) Using the identity. .( u v ) ( . v ) u ( u ).v (1.39) we get v v ( .v )( v ) ( v ).v . b 0. t t or v t .v v t ( v ).v . b 0 (1.40) Using the identity ( v ) v v ( ) (1.41) we get v . v v [ . v v ( )]. v . b 0 t t (1.42) From the definition (1.4.29), we have v ( ).v v .( )v (1.43) Hence v t .v v t v .v v .( )v . b 0 v or .v .v v v .v . b 0 t t (1.44) xxix The material time derivative of ρ is defined as .v t (1.45) Therefore, equation (1.44) on using equation (1.45) leads to .v v v v . v . b 0 t (1.46) From the balance of mass equation, we have .v 0 (1.47) Hence, equation (1.46) on application of (1.47) gives v v .v . b 0. t (1.48) The material time derivative of v is defined as v v v . v t (1.49) Hence, equation (1.47) finally gives: v . b 0 or div b v (1.50) 1.4.7 Balance of Angular Momentum Theorem 3: The balance of angular momentum can be expressed as T Proof: (1.51) xxx We assume that there are no surface couples on B or body couples in B. Using our general balance equation (1.9) we note in this case that the physical quantity to be conserved is the angular momentum density, ie f = x (ρv). The angular momentum source at the surface is then g = x t and the angular momentum source inside the body is h = x (ρb). The angular momentum and the moment are calculated with respect to a fixed origin. Hence we have: d x ( v ) dV B X ( v )un v .ndA dt B B X t dA B X ( b ) dV (1.52) Assuming that B is a controlled volume, we have B X ( v ) dV B X ( v ) v .ndA t (1.53) B X t dA B X ( b ) d V Using the definition of a tensor product, we can write X ( v ) v .n [ X ( v ) v ] . n (1.54) Using, t = . n, we have x v dV x v v .n dA B B t x .ndA x b dv B On using divergence theorem, we get B (1.55) xxxi x v dv .x v v x .ndA x b dv B B B B t (1.56) It is most convenient to use index notation to convert surface integrals into volume integrals, thus: B x .n dA B eijk x j kl nl dA B Ail nl dA B A. ndA i (1.57) where x .n dA represents the i-th component of the vector. B i Using the divergence theorem on equation Aie dv e ijk x j kl dv B x e B x i A.ndA . Adv B (1.58) B (1.59) Differentiating equation (1.59), we have kl A. ndA e ijk jl kl e ijk x j dV x e B B kl e ijk kj e ijk x j dV x l B e ijk kj e ijk x j . l dV B Expressing in direct tensor form, we have, T A . ndA : i ( x ( . ) ,i dV B B where is a third – order permutation tensor. Therefore (1.60) xxxii T B x . n dA B : i x . dV i i (1.61) or T x .n dA : x . dV B (1.62) B The balance of angular momentum can be written as x v dV .x v v dV B B t : T x . dV x b dV B (1.63) B Since B is an arbitrary volume, we have: x v . x v v : T x . x b t or x v . t b .x v v : T (1.64) Using the identity, .u v .v u u.v (1.65) we get, . x v v .v x v x v .v The second term on the RHS can be further simplified using index notation as follows: (1.66) xxxiii x v .v i x v .v i eijk x j vk ve xe x j v e ijk x j v k v e v k v e x j k v e x e x e x e v e ijk x j v k v e e ijk je v k v e e ijk x j k v e x e x e x v .v v v x v .v i x v .v x v .v i (1.67) Therefore, we can rewrite equation (1.66) as: .x v v .v x v .v x v x v .v The balance of the angular momentum then takes the form: x v . b .v x v .v x v t T x v .v : or x v v .v . t b .v x v .v x v : T v x v v .v . b t t .v x v .v x v : T Using equation (1.49) we have, (1.68) (1.69) xxxiv X v . b X v ( .v ) ( X v ) t ( .v ) ( X v ) : T (1.70) Also from equation (1.50), equation (1.69) becomes v ( .v )( X v ) ( .v )( X v ) : T t .v .v ) ( X v ) : T t 0 X (1.71) From equation (1.22), we have .v ( X v ) : T 0 (1.72) Using equation (1.47), we get : T 0 (1.73) In index form, equation (1.73) gives eij ,k kj 0 (1.74) Expanding out we have the following results 12 21 0 ; 23 32 0 ; 31 13 0 (1.75) which shows that 12 21 , 23 32 , 31 31 Hence, T 1.4.8 Balance of Energy Theorem 1.4. The balance of energy equation can be expressed as (1.76) (1.77) xxxv e : ( v ) .q s 0 , (1.78) where ( x , t ) is the mass density, e(x,t) is the internal energy per unit mass, (x,t) is the Cauchy stress, v(x,t) is the particle velocity, q is the heat flux vector, and s is the rate at which the energy is generated inside the volume (per unit mass) Proof: We recall the general balance equation (1.14) as given below, d f ( x , t ) dv B f ( x , t )U n ,( x , t ) v( x , t ) .n( x , t )dA dt B B g( x , t )dA B h( x , t )d v (1.79) In this case, the physical quantity to be considered is the total energy density which is the sum of the internal energy density and kinetic energy density, that is, ƒ=e + ½ v .v . The energy source at the surface is a sum of the rate of work done by the applied tractions and the rate of heat leading the volume (per unit area), thus, g = v.t - q.n, where n is the outward unit normal to the surface. The energy source inside the body is the sum of the rate of work done by the body forces and the rate of energy generated by internal sources, that is, h = v. ( b ) s . Hence we have 1 ( e v .v )( un v .n ) dA d 1 2 ( e v .v ) dv B dt B 2 B ( v .t q .n )dA B ( v .b s )d v (1.80) xxxvi Let B be a control volume that does not change with time, then we get 1 1 B ( e v .v ) dV B ( e v .v )( v .n ) dA t 2 2 B ( v .t q .n )dA B ( v .b s )dV (1.81) Using the relation t = .n, the identity v. (.n) = (T.v). n, and invoking the symmetry of the stress tensor, we get B 1 1 ( e v .v ) dv B ( e v .v ) ( v .n ) dA B ( .v q ).ndA t 2 2 B ( v .b s )dV (1.82) Applying the divergence theorem to the surface integrals (1.82), we get B 1 1 ( e v .v ) dV B . ( e v .v ) v dA .( .v ) dA t 2 2 B .qdA B ( v .b s ) dV (1.83) Since B is arbitrary we have 1 1 ( e v . v ) dV . ( e v .v ) v ( .v ) .q ( v .b s ). t 2 2 (1.84) Expanding out the left hand side, we have 1 1 e 1 ( e v .v ) ( e v .v ) ( v .v ) t 2 2 t t 2 t 1 e v ( e v .v ) .v t 2 t t For the first term on the right hand side of (1.4.76), we use the identity (1.85) xxxvii .( v ) .v .v (1.86) and have, 1 1 1 . ( e v .v )v ( e v .v ) .v ( e v .v ) .v 2 2 2 1 1 1 ( e v .v ) .v ( e v .v ) .v ( e v .v ) v 2 2 2 1 1 ( e v .v ) .v ( e v .v ) .v e .v 2 2 1 ( v .v ).v 2 1 1 ( e v .v ) .v ( e v .v ) .v e .v 2 2 T ( v .v ).v (1.87) 1 1 v .v ) .v ( e v .v ) .v ve .v 2 2 ( v .v ).v ( e For the second term on the RHS of equation (1.82), we use the identity .(STv) = S: v + (.S).v (1.88) and the symmetry of the Cauchy stress tensor gives .(.v) = : v + (.).v (1.89) After collecting terms and rearranging, we get 1 v .v .v e v .v .v . b .v 2 t t e e .v : v .q s 0 t (1.90) Applying the balance of mass to the first term and the balance of linear momentum to the second term, and using material time derivative of the internal energy, xxxviii e e e .v t (1.91) We get the final form of the balance of energy as e : v .q s 0 (1.92) 1.4.9 Clausius-Duhem Inequality The Clausius-Duhem inequality can be used to express the second law of thermodynamics for elastic materials. This inequality is a statement concerning the irreversibility of natural processes, especially when energy dissipation is involved. Just like in the balance laws, we assume that there is a flux of a quantity, a source of the quantity, and an internal density of the quantity per unit mass. The quantity of interest in this case is the entropy. Thus, we assume that there is an entropy flux, an entropy source, and the internal entropy density per unit mass () in the region of interest. Let B be such a region and let ∂B be its boundary. Then using the second law of thermodynamics, we have d dv B ( un v .n )dA B q dA B .v dv dt B where is the internal entropy per mass, (1.93) q is the entropy flux at the surface, r is the entropy source per unit mass. The scalar entropy flux can be related to the vector flux at the surface by the relation q ( x ).n For isothermal condition, (1.94) xxxix ( x ) q( x ) ,r s (1.95) where qis the heat flux vector, s is an energy source per unit mass, and is the absolute temperature of a material point at x at time t. It is possible to show that the Clausuis-Duhem inequality in terms of (i) Integral form as d B dV B ( Un v .n )dA B q .n dA B s dV dt (1.96) (ii) Cauchy stress and internal energy ( e ) : v q . (1.97) 1.4.10 Constitutive Relations A set of constitutive equations is required so close to system of balance laws. For large deformation elasticity, we define appropriate kinematic quantities and stress measures so that constitutive relations between them may have a physical meaning. Let the fundamental kinematic quantity be the deformation gradient (F) which is given by F x 0 x ; X det F 0 (1.98) A thermoelastic material is one in which the internal energy (e) is a function only of F and the specific entropy (), that is e e ( F , ). (1.99) xl Theorem 1.5 : For a thermoelastic material, the entropy inequality satisfy Clasusius Duhem inequality and so e e q . .F T : F 0. F (1.100) Here, we make some constitutive assumptions: (1) Like the internal energy, and are also functions of F and , thus = (F,), = (F,), (1.101) ((2).The heat flux q satisfies the thermal conductivity inequality and if q is independent of and F , we have; q. 0 - (K.). 0 K 0. (1.102) Therefore, the entropy inequality may be written as e e .F T : F 0 F (1.103) Since and F are arbitrary, the entropy inequality will be satisfied if and only if e e 0 and e e T .F T 0 .F F F Therefore, (1.104) xli e and e . FT F (1.105) Hence, the energy equation may be expressed in terms of the specific entropy as or .q s div q s (1.106) 1.10 Other Strain and Stress Measures The internal energy depends on F only through the stretch U, the symmetric right stretch tensor. A strain measure that reflects this fact and also vanishes in the reference configuration is the Green strain. E = ½ (FT.F-I) = ½ (U2-I) (1.107) The Cauchy stress is given by: = e . FT F (1.108) It is possible to show that the Cauchy stress can be expressed in terms of the Green strain as σ = F. e T .F E (1.109) and the proof is stated below From equation (1.108), we can write in index notation ij = e T e FiK F jk F Fik (1.110) We define the Green strain tensor E = E(F) = E (U) and e = e (F,) = e (U,) (1.111) xlii Using the chain rule, we have, e e E e e Em . . F E F Fik Em Fik (1.112) Now, E = ½ (FT.F-I) Elm = ½ ( FTp F pm m ) 1 ( F p F pm m ) 2 (1.113) Taking the derivative with respect to F, we get Fpm E 1 F T F Em 1 Fpm .F F T . Fpm Fp F 2 F F Fik 2 Fik Fik Therefore, 1 2 ij e E FT F T .F . .F F T . F F F pm 1 e F p Fijk F pm F p 2 Eem Fik Fik Aji A Aij AT We recall that; ik j and jk i A Ak A Ak Therefore, ij = 1 e ( pi lk Fpm Fpl pi mk F jk 2 Elm 1 e ( lk Fim Fil mk F jk 2 Elm (1.114) xliii or, ij = 1 2 e ( Fim Fil Ekm 1 2 ij or ij e E ) F jk F T F T . F . .F F T . F F F pm 1 e F p Fijk F pm F p 2 Eem Fik Fik 1 e e ( Fim Fil F jk 2 E Elk 1 2 e T e T F . F . . F E E e T e 1 FT or F . 2 E E e T e 1 .F T F . 2 E E From the symmetry of the Cauchy stress, we have = (F.A). FT and T = F.(F.A)T = F.AT.FT And = T AT = A Therefore, e ∂e/∂E = E T (1.115) xliv and we get, F . e T .F E (1.116) The nominal stress tensor is defined as N = det F (.F-T) (1.117) From the conservation of mass, we have 0 = det F (1.118) 0 .F T (1.119) Hence, N= The nominal stress is unsymmetric. We can define a symmetric stress measure with respect to the reference conjugation called the second Piola-Kirchoff stress tensor (S): as: S F 1 .N .F T 0 1 F . .F T (1.120) In terms of the derivatives of the internal energy, we have S 0 1 e T T e F F. .F .F 0 F . E E (1.121) and N That is, 0 e e ( F . .F T ).F T 0 F . E E (1.22) xlv S 0 e E and N 0 F . e E (1.123) 1.4.12 The Stress Power The stress power per unit volume is given by : v (1.124) In terms of the stress measures in the reference configuration, we have e : ( F .F 1 ) : v F . E (1.125) Using the identity A : ( B .C ) ( A.C T ) : B , we have : v F . e T T .F .F : F E e : F = F E = N : F 0 (1.126) Alternatively, we can express the stress power in terms of S and E . Taking the material time derivative of E, we have, 1 E ( F T .F F T .F ) 2 (1.127) Therefore, 1 1 S : E S : ( F T .F F T .F S : ( F T .F ) 2 2 Using identities (1.128) xlvi A : ( B .C ) ( A.C T ) : B ( B T .A ) : C , (1.129) A : B AT : B T and from the symmetry of S, we have 1 S : E S .F T : F T ( F .S ) : F 2 = 1 F .S T : F ( F .S ) : F 2 = ( F .S ) : F (1.130) S F 1 .N (1.131) Now, Therefore, S : E N .F (1.132) Hence, the stress power can be expressed as : v N : F S : E (1.133) If we split the velocity gradient into symmetric and skew parts using v l d w (1.134) where d is the rate of deformation tensor and w is the spin tensor. Hence, we have, : v : d .w tr T .d tr T .w tr .d tr .w (1.135) Since σ is symmetric and w is skew, we can set tr .w = 0 Therefore, : v = tr .d Hence, we can express the stress power as, (1.136) xlvii tr( .d ) tr( N T .F ) tr( S .E ) CHAPTER TWO (1.137) xlviii FINITE –STRAIN THERMOELASTICITY BASED ON MULTIPLICATIVE DECOMPOSITION OF DEFORMATION GRADIENT 2.1 Preamble In this chapter, mathematical formulations are based on the framework of thermodynamics. We specify the initial configuration B0 and the current configuration as Bt. Associated with the configurations are their respective temperatures denoted by θ0 and θ. The Helmholtz free-energy is introduced as Ψ = u- θη = e – θη, where u or e is the specific internal energy (per unit mass), and η is the specific entropy. A second-order tensor of the latent heat ℓE and the specific heat CE, both at constant temperature are introduced to obtain coupled heat equation. A thermoelastic analysis based on multiplicative decomposition with three configurations is shown where the intermediate configuration, Bθ, at non uniform temperature θ is obtained from the deformed configuration Bt by isothermal destressing to zero stress state (ZSS). The deformation gradient from initial to deformed configuration F is decomposed into elastic part Fe and the thermal part Fθ such that F = Fe . Fθ (2.1) Constitutive expressions are derived for the stress, entropy and temperature. A couple energy equation is finally obtained with specified components. xlix 2.2 Thermoelastic Analysis Based On the Multiplicative Decomposition. An alternative approach to develop the constitutive theory of thermoelastic material response is based on the introduction of an intermediate configuration B0 which is obtained from the current configuration Bt by isothermal elastic unloading to zero stress (ZSS). (See fig. 2.1 below). B ,σ Figure 2.1 Multiplicative Decomposition of Deformation Gradient The intermediate configuration Bθ at non- uniform temperature θ is obtained from the deformed configuration by isothermal destressing to zero stress. The deformation gradient from initial to deformed configuration F is decomposed into elastic part Fe, and thermal part Fθ, such that F = F e . Fθ The isothermal elastic deformation gradient from Bθ to Bt is denoted by Fe, and the thermal deformation gradient from B0 to Bθ by Fθ. The total deformation l gradient F, which maps an infinitesimal material element dX from the initial configuration to dx = F.dX in the current configuration, can be decomposed as F = Fe.Fθ (2.2) The decomposition (2.2) is not unique because arbitrary rigid-body rotation can be superposed to Bθ preserving it unstressed. However, we shall be able to specify Fθ and thus the decomposition (2.2), uniquely in each considered case or type of the material anisotropy. For example, for transversely isotropic material with the axis of isotropy parallel to the unit vector n0 in the configuration B0, we specify Fθ by Fθ = (ξ – υ) n0 n0 + υI (2.3) where = (θ) is the stretch ratio due to thermal expansion in the direction n0, while υ = υ(θ) is the thermal stretch ratio in any direction within the plane of isotropy (orthogonal to n0 ) and I, is the unit tensor. Equation (2.3), is in one preferred direction. This representation can be extended to orthotropic materials. 2.3 Decomposition of Lagrangian Strain In this section, following [29], we decompose the finite Lagrangian Strain so as to extend the representation in equation (2.3) to orthotropic material. The decomposition takes the form E = ET . Ee . Fθ + Eθ (2.4) where Ee, E are elastic and thermal strain tensors respectively and Fθ is as earlier defined. We define these elastic and thermal strain tensors as li Ee = ½ (FeT . Fe – I) Eθ = ½ (F T F – I) and (2.5) From Equation (2.4), the elastic strain can be expressed as: Ee = Fθ –T.(E-Eθ). Fθ –1 (2.6) That is, E e FT .E .F1 FT .E .F1 (2.7) The material time derivative of Ee is 1 E e FT .E .F 1 FT . F T .F F T .F .F 1 F T .E .F 1 2 1 FT .E .F 1 FT FT .F FT .F .F 1 FT E .F 1 2 FT .Ê .F1 D Ee .L LT E (2.8) where L F .F1 (2.9) and Dθ is the symmetric part. We now restrict our analysis to isotropic material for which the thermal part of the deformation gradient is Fθ = (θ)I (2.10) where I is a unit tensor. From Equation (2.6) Ee = and Eθ = 1 2 (E - Eθ) 1 2 (υ -1)I 2 From Equation (2.11), using (2.12), we have (2.11) .(2.12) lii Ee = 1 2 [E – ½ (υ2-1)I]. (2.13) By generalizing mechanical theory, we define the thermal strain according to [28] by Eθ = α (θ-θ0)I .(2.14) And by replacing the total strain E in the constitutive equation for the stress by the quantity E - Eθ. Equating Equation (2.12) to Equation (2.14), we have (υ2-1)I = 2α (θ-θ0)I .(2.15) Consequences of Equation (2.15) (i) For isothermal condition, (θ=θ0) and so Equation (2.15) gives υ2 = 1, i.e., ( ) 1 This result is consistent with the condition that υ = υ (θ) is the thermal stretch ratio in any direction within the plane of symmetry of isotropy as shown in the diagram below. θ -1 1 liii Figure 2.2 The Symmetry of the Thermal Stretch Ratio. The symmetry of the thermal stretch showing the range of the temperature from -1 to +1. (ii) Upon thermal expansion (θ >θ0), from the initial temperature θ0 to the current temperature θ, an infinitesimal volume element dv0 from the configuration B0 becomes dVθ = (det Fθ) dV0 in the configuration Bθ such that d dV d det F dV0 dt dt (2.16) From equation (2.9), and upon using Equation (2.10), we have L F .F1 I (2.17) where 1 I 0 0 0 1 0 0 0 1 (2.18) is a unit tensor. Using equation (2.10), the time derivative gives dF d d F . V ( ) I dt d dt (2.19) liv Therefore, L 1 d I d (2.20) From Equation (2.16) d dV d det F dV0 dt dt d d det F dV0 (det F ) dV0 dt dt d d det F dV0 sin ce dV0 0 dt dt (2.21) Using the result d det F det F tr F 1 F dt therefore, (2.22) d det F det F tr F 1F det F tr L dt where, L vi vi Lij tr L div v xj xj (2.23) (2.24) Consequently, d d v det F tr L dv dt (2.25) Thus, div v measures the rate at which the volume changes during motion, (iii) Since F is nonsingular, by convention, the volume element must be positive, so that J det F > 0 . (2.26) And by definition dv = J dV (2.27) lv since J is a measure of the change in volume under deformation, and if the deformation is isochoric (no change in volume),then J det F > 0 (2.28) In this case, equation (2.23) becomes d ( dV ) det F tr L dV dt Thus , d ( dV ) tr L dV dt (2.29) where dV0 is the initial elemental volume. Substituting Equation (2.20) into Equation (2.29), we have 1 d 3 d d ( dV ) tr .I dV dV dt d d therefore , d dV 3 d dV dt d (2.30) . (2.31) The temperature-dependent coefficient of linear thermal expansion is defined as (see [18,19] ), , so that d dV 3 dV . dt (2.32) From Equation (2.31), the coefficient of volumetric thermal expansion is equal to 3α. Comparing equation (2.30) and (2.31), we have lvi 1 dv d (2.33) Equation (2.33) establishes a differential connection between the thermal stretch ratio and the coefficient of thermal expansion and upon integration, the equation gives exp d 0 (2.34) In view of equation (2.7) and (2.32), the rate elastic strain can now be written as E e 1 2 E I 2 E (2.35) 2.4 Free Energy and Constitutive Expressions In the framework of finite-strain thermoelasticity based on multiplicative decomposition, we define the Helmholtz free energy as: e E e , ( ) (2.36) where ψe is an isotropic function of the elastic strain Ee and temperature θ. Taking time derivative of equation (2.36), we have, e E e e . . . E e t t dt (2.37) e e : Ee E e (2.38) Substituting Equation (2.38) into (2.35) leads to e e 1 : 2 E ( I 2 E ) E e ( ) lvii 1 e e e :E 2 :( 1 2 E ) 2 Ee Ee (2.39) The comparison of Equation (2.39) with the classical form 1 0 T : E (2.40) then clearly yields the constitutive relations T 0 e 2 E e (2.41) and the entropy is given as: e e : I 2 E E e (2.42) We note the identity 1+2E = υ2 (I+2Ee) (2.43) Using the relationship according to [29], ρ0 = (det Fe) ρθ = υ3 ρθ (2.44) between the densities ρ0 in the configuration B0 and ρθ in the configuration Bθ. Equation of the stress from equation (2.41) can be written as : T Te , Te e E e Using equation (2.41), we have from equation (2.45) (2.45) lviii T 0 e . det Fe 0 E E e det Fe F1 .Te . FT (2.46) Equation (2.46) is a more general relationship for T. 2.5 Analysis of Stress Response. An appealing feature of the thermoelastic constitutive formulation based on the multiplicative decomposition is that the function ψe (Ee,θ) can be taken as one of the well-known strain energy function of the isothermal finite-strain elasticity (eg Ogden [20], Holzapfel [21] ), except that the coefficients of the strain-dependent terms are now functions of the temperature. Theorem 2.1 Suppose that ψe is a quadratic function of the strain components, such that ρ0ψe = ½ λ(θ) (trEe)2 + μ(θ) Ee: Ee (2.46) where λ(θ) and μ(θ) are temperature-dependent Lame moduli and Te = λ(θ) (trEe)I + 2 μ(θ) Ee = Λ0 (θ): Ee Then, T 1 ( ) ( trE )I 2 ( )E 3 ( )k ( )I ( ) 2 (2.47) (2.48) Proof: We define the elastic modulli tensor as Λe(θ) = λ(θ)II + 2μ(θ) I I Substituting Equations (2.23) and (2.48) into Equation (2.45) gives (2.49) lix T = υ Te, where, Te = ρ0 e Ee , and so T = υρθ e E e T = υ{λ(θ) tr (Ee) I + 2 μ(θ)Ee = υ{λ(θ) II + 2 μ(θ)I I}: Ee λ(θ) tr (Ee) I + 2 μ(θ)Ee = λ(θ) II: Ee+2μ(θ)I I: Ee. that is , or, 1 1 T = λθ tr 2 ( E E ) I 2 FeT .Fe I 2 ( )I I : FT .E E .F1 2 II : 1 T Fe .F I 2 ( )( tr Ee 2 ( ) Ee ( ( ) 2 ( ))I : Ee 2 1 ( )( tr Ee 2 ( ) Ee ( ( ) 2 ( ))I : Ee ( ) But from equation (2.7) Ee = Fθ –T.(E-Eθ). Fθ –1 and so after rearrangement, we have, T 1 ( trE )I 2 E 1 1 k I 2 (2.50) where k(θ) = λ(θ) + 2μ(θ) Consequences of Equation (2.50) (i) Equation (2.50) is an exact expression for the thermoelastic stress response associated with the quadratic representation of ψe in terms of the finite- strain Ee lx (ii) If the Lame modulli, λ(θ) and k(θ) are taken to be temperature independent, and if the approximation υ(θ) ≈ 1 + α0(θ-θ0) (2.52) according to [52 ], (α0 being the coefficient of linear thermal expansion at θ = θ0), and hence we can rewrite equation (2.50) as T 1 3 2 ( )( trE )I 2 ( )E ( 1 0 ( 0 ) 1 ] k ( ) I 1 0 ( 0 ) 2 ( ) ( trE ) I 2 ( ) E 3 0 0 k ( ) I (2.53) which reduces to classical formulation equation. (iii) When E and T are interpreted as the infinitesimal strain and the Cauchy stress and since we assume linear themoelasticity, the strain tensor is defined by: ij 1 ( ui , j u j ,i ) 2 (2.54) and the equilibrium equations are i j ,i f j 0 (2.54)1 We apply multiplicative decomposition theory to equation (2.54) so that we split the total strain tensor into elastic (mechanical) and thermal parts as: i j ei j i j (2.54)2 lxi For the elastic stresses, we define ei j 1 i j k k E E e i j 1 2G i j and G k k i j 1 i j E 2 ( 1 ) (2.54)3 In order to obtain an expression for the thermal strains, we may write dST ( 1 ) d S0 ( d ST d S0 ) d S0 (2.54)4 where dS0 is the initial length and dST is the length as a result of a temperature change and is the coefficient of the thermal expansion. Accordingly, the thermal strains are: ij i j (2.54)5 Adding equations (2.54)3 and (2.54)5 to obtain the total strain, we have i j ei j i j 1 2G k k i j 1 i j i j (2.54)6 Equation (2.54)6 coincides with Duhamel-Neumann expression for isotropic linear thermoelasticity. Hence equation (2.46) can be recast in terms of the total strain as 0 e 1 1 3 1 3 ( ) ( trE ) 2 ( )E : E k ( ) trE 2 1 2 2 4 This confirms the result of equation (2.50) through T 0 e E (2.55) lxii (iv). Due to small strain assumption on Equation (2.53), tr E turns out to be approximately equivalent to the relation in volume change, that is, trE V V (2.55) and the coefficient of volumetric thermal expansion α0 ≈ 3α (2.56) The stress tensor of Equation (2.53) can now be written as T = λ(θ) (trE)I + 2μ(θ)E – m (θ-θ0)I where (2.57) m = 3 αk(θ) (2.58) Consequences of (2.57) (a) When the strain vanishes, (E = 0), then from equation (2.57) T = m( θ-θ0)I (2.59) which is the pressure in an isotropic body. (b) When the stress vanishes (T = 0), E m( 0 )I ( ) ( trE )I 2 2 = λ (θ-θ0) I therefore, E = p(θ-θ0)I (2.60) where p= - m/2μ. That is, when the stress vanishes, the strain equals omnidirectional dilatation or hydrostatic compression in an isotropic body. 2.6 Entropy Expressions The entropy expression of equation (2.42) can be recast using the stress expression of equation (2.41) as: lxiii T 0 e 2 E e (2.61) So that the new entropy expression 1 0 2 0 e e .. :( 1 2E ) 2 E e e T : 1 2 E 0 By evaluating the term e (2.62) in the case of the quadratic strain-energy representation as of equation (2.46) , and using equations (2.38), (2.42),we have 1 2 0 e 3Te : E e (2.63) Thus, 3 d 1 dT e Te : E e 3 e : E e 2 d 2 d Ee Ee 2 0 3 1 T e or . 0 2 ( )T : Ee 3 e : Ee . 2 E e E e 2 3 1 1 2 ( )T : 2 E E 3 ( Te / ) : E e 2 2 1 Te 3 1 ( )T : E 2 1 I . 3 : Ee 2 2 2 Ee 3 1 1 ( )T : E 2 1 trT 3 ( Te / ) : Ee 2 2 2 The temperature gradient of the stress tensor is (2.64) lxiv or, d Te E e e : Ee d (2.65) d Te e : E e Ee d (2.66) By [29.] T e : E 2 1I 2 1 1 (2.67) Since T = υTe (2.68) Te T Ee E T 3 k I ... (2.69) Using the result of Classical Theory (see [28]), we arrive at Te T Ee e E 3 I T : E 21 1 trT 1 : E 2 1 2 2 (2.70) Substituting (2.70) into (2.66) leads to: 3 1 e E e T : E 2 1 trT 2 2 0 1 T 1 2 1 2 : E 1 I ) T : E 1 trT . 2 t E 2 2 2T : E 1 1 T 2 2 trT 2 2 E 1 : E 2 1 2 I Substituting equation (2.71) into equation (2.62), the entropy equation becomes (2.71) lxv 1 20 2 trT 1 T 2 0 E 1 : E 2 1 2 I (2.72) Since trT 3k trE 3 2 1 2 I (2.73) Equation (2.72) can be rearranged as 1 20 I 1 2 : E 1 2 E 3 kI T (2.74) Using the expression for latent heat, the entropy is 1 1 3 1 E kI : E 2 1 2 0 2 I . (2.75) Also, equation (2.75) can be denoted by expressing the elastic strain-energy ψe (see [49.] ) as 1 2 0 e T : E 1 2 1 trT 4 (2.76) The partial differentiation then gives: 1 1 2 1 2 e E : E 1 trLE trT 2 4 20 E however, when ,we recover equation (2.75) E , The latent heat E can be calculated from equation (2.72) as, E 1 0 T E (2.77) lxvi E 1 0 T 3 kI 1 de 1 : E 2 1 d 2 I (2.78) Consequences of Equation (2.78) If the elastic moduli are independent of temperature, and the components of the stress T are much smaller, the equation reduces to E 3 0 k I (2.79) (ii) Using Equation (2.79) in equation (2.75), we have I 1 1 3 3 1 kI kI : E 2 1 2 0 0 2 0 3 2 0 1 : E 1 2 0 3 kI 2 0 3k 3 0 trE 2 1 0 2 (2.80) Remark: The function ψθ is given (see [54.] ) by 1 C E0 9 2 0 k 0 0 2 0 0 2 C E0 0 9 2 0 k 0 0 0 0 Hence Equation (2.81) gives . . (2.81) (2.82) lxvii 0 3k 3 2 9 2 C E trE 1 k 0 0 2 0 0 0 0 . (2.83) Using Equation (2.72) in equation (2.83),we have 1 0 C E0 trT 0 9 0 02 k 0 0 (2.84) Since the entropy is not quadratic dependent, it gives 3 0 0 k0 trE C E0 0 0 . (2.85) Equation (2.85) is in perfect agreement with equation of the classical form of thermoelasticity,(see [54]) 2.7 Some Known Results in Classical Theory of Thermoelasticity 2.7.1 The Free Energy, Stress and Entropy The Helmholtz free energy per unit mass is given as Ψ = u – θη (2.85) And on differentiation we have u which gives 1 0 T : E (2.87) (2.88) We assume that the stress T linearly depends on the finite strain E, and if the specific and latent heats depend linearly on the temperature θ ,then lxviii C E C E0 c ( 0 ), E 0 . 0 E (2. 88)1 where c is a constant while C E0 and 0E are the specific and latent heats respectively. It follows readily that the free energy is given as: 0 0 1 0 : : ( E E ) ( 0E E ) 0 C E0 C 0 0 ln 0 1 c 0 2 2 (2.88)2 where, 0 is a fully symmetric fourth-order tensor of isothermal elastic modulli at the temperature θ = θ0. For isothermal condition, θ = θ0, and so the equation (2.88)2 gives: 1 20 0 : : E E (2.88)3 or 2 0 0 : : E E where : : double trace product . The corresponding stress and entropy expressions are respectively, T 0 : E 0 0 ( 0 ) 0 E (2.88)4 and 1 0 0E : E C E0 c 0 ln c( 0 ) 0 (2.88)5 lxix T 0 Also, E (2.89) (2.90) 2.7.2 The Latent and Specific Heats of the Classical Theory The latent and the specific heats are given as: T 2 E 0 E 1 CE 2 . 2 (2.91) (2.92) and so E : E C E (2.93) So the latent heat tensor in equation (2.93) can also be expressed as F . E .F T 1 (2.93)1 where σ is the Cauchy stress and is the temperature. 2.8. Comparison of Field Equations and Related Expressions in Classical Theory with Those in the Multiplicative Decomposition The specific entropy can be expressed as either of the two functions, ˆ E , E e , (2.94) Consequently, without loss of generality, we may express equation (2.94) as: d e : dE C E d E e : d Ee C E e d The two latent heat tensors at fixed temperature are given as: (2.95) lxx E ˆ , E Ee ˆ Ee (2.96) while the two specific heat tensors at constant elastic strain are also given as: CE ˆ ˆ , CE e (2.97) The specific heats at constant elastic strain CEe represents the amount of heat required to increase temperature of the unit mass by dθ, while the latent heat E e is a second order tensor representing the amount of heat associated with a change of the corresponding strain component dEeij. Similar interpretations hold for CE and E . Therefore, there is a relationship in terms of specific heats in both the Classical Theory and the Multiplicative Decomposition. ( see equations (2.92) and (2.97)). In multiplicative theory, the latent heat (see equation 2.79) is given as: Ee 3 0 k I (2.98) while in Classical Theory (see equation (2.91)) is given as E 1 0 T 2 E (2.99) This shows that the latent heat in equation (2.98) is determined by thermal stretch ratio, the coefficient of thermal expansion, the temperature and the bulk modulus. Equations (2.89) and (2.90) again are the equations for stress and entropy respectively, and can be used to determine the stress and entropy in Classical Theory lxxi whereas in Multiplicative Theory, the stress is a product of the square of the thermal stretch ratio and thermal stress. The entropy expression in Multiplicative Theory as given in equation (2.62) can be written in the form: 1 0 2 0 e e .. :( 1 2E ) 2 E e e T : 1 2 E 0 Clearly, it is to easy construct (2.100) which can be fitted to experimental data which is nonexistent in Classical Theory. So the Multiplicative Theory has the advantage of incorporating many field quantities in the determination of field equations. For details on equation (2.100) see equations (2.61) – (2.75). lxxii CHAPTER THREE 3.1 THERMOELASTIC ANALYSIS BASED ON MULTIPLICATIVE DECOMPOSITION FOR GROWTH Preambles In this chapter, we summaries a general continuum mechanical theory that takes account of growth in material capable of large deformations, with particular reference to soft biological tissues such as the artery. The type of growth considered here is “volumetric growth”. Here we adopt the definition used by many researchers that consider growth “as a change of mass and geometry.” We use the toy – tissue model to illustrate the concepts of volumetric growth and the analogy between growth and thermoelasticity and illustrate that volumetric growth is analogous to thermal expansion. We focus on the mechanics of growth, basic kinematics, balance equations of mass, linear and angular momentum and energy and all these are reviewed for volumetric and surface sources. General constitutive equations that include the effect of residual stresses are also described. The theory of multiplicative decomposition of deformation gradient is applied and again the intermediate configuration is studied for residual stresses. In this configuration, the equation of motion is decomposed into that involving cylindrical coordinates associated with a thin wall cylinder and that involving spherical lxxiii coordinates associated with instantaneous point source when the body is regarded as infinite. 3.2 Toy – tissue Model Material always occupies some volume. In saying “material points”, we mean a very small volume. Such small volumes are considered on the mesoscale of growth deformation process. By considering a simple toy-tissue model presented in figure 3.1, a reasonable insight into the tissue growth can be gained. In the model, the regular initial tissue can be seen on the top of the figure. This is the collection of the regularly packed balls. The balls are interpreted as the tissue elementary components---cells, molecules of the extracellular matrix, etc. The balls are arranged in a regular network for the sake of simplicity and clarity. They can be organized more chaotically, this does not affect the subsequent qualitative analysis. lxxiv Material Source Figure3 Toy-tissue model: regular (top), point mass supply (bottom). ( Adopted from Volokh (2005) and modified for use) We assume that a number of new balls are supplied pointwise as it is shown in Figure 3.1. This result is considered as an injection. Usually, the new material is created in real tissues in a more complicated manner following a chain of biochemical transformation [63]. However, the finally produced new material still appears lxxv pointwise from the existing cells. This kind of model can be constructed physically. The result of this thought – experiment is represented in the figure and it can be described as follows: (a) the number of the balls in the toy-tissue increases with the supply of the new ones. (b) the new balls are concentrated at the edge of the tube and they do not spread uniformly over the tissue. (c) the new balls cannot be accommodated at the point of their supply---the edge of the tube, the packing of the new balls get denser around the edge of the tube. (d) the more balls are injected, the less room remains for the new ones. (e) the new balls press the old ones. (f) the new balls tend to expand the area occupied by the tissue when overall ball arrangement reaches the tissue surface. We now translate these six qualitative features of the toy-tissue microscopic behaviour into the language of macroscopic theory accordingly as: (A) Mass of the tissue grows. (B) Mass growth is not uniform, i.e., the mass density changes from one point to another. (C) There is diffusion of mass. (D) The diffusion is restricted by the existing tissue structure and its mass density, the denser the tissue is, the less material it can accommodate. (E) Growth is accompanied by stress. lxxvi (F) The expansion of the tissue is volumetric i.e., it is analogous to thermal espansion of structural material. 3.2.1 BASIC ASSUMPTIONS We assume that material points are everywhere dense during the volumetric mass growth, so that in any small neighbourhood around the particle, there are always points that existed before growth. This assumption enables us to treat the problem of volumetric mass growth by using the usual continuum mechanics concepts, such as deformation gradient and strain tensors. The deformation gradient in the biomechanics theory of volumetric mass growth is due to both, the mass growth and the deformation caused by externally applied and growth-induced stresses. It is crucial to note that material particles change within the material point due to division and diffusion [Volokh, (2005)]. This means that referential mass density changes during deformation process, that is ρ ≠ constant, and mass is not conserved. The violation of the mass conservation is inherent in the open systems exchanging material with their environment. 3.3 MECHANICS OF GROWTH In modeling the mechanical behaviour of soft tissue, two rather different approaches present themselves: (i) the knowledge of the mechanics of the constituents of the microstructure of the tissue, together with knowledge of their interactions so as to develop a theoretical framework that would best describe the mechanics of the whole tissue, lxxvii (ii) the macroscopic description of the tissue as a whole, and how various features of the tissue evolve under changes in its mechanical environment. We refer to such models as phenomenological models, and they are based on the notion of cause and effect and have the advantage of not requiring the detailed knowledge of the internal structure of the tissue. 3.3.1 Basic Kinematics In this section, we present a diagram for volumetric growth and describe the mathematical formulation for it. Reference configuration density ρo Βo Βt F Loaded grown configuration density ρo Fe Fr Βr Unloaded, Residual-stressed configuration density ρr=ρ Figure 3.2, Schematic representation of configurations resulting from deformation and growth, starting from the fixed configuration В0. The current (loaded) configuration is Вt and Вr is the configuration ( evolving with growth) obtained from Вt by removal of loads.(Adopted and modified from Ogden & Quillou, (2006)) The density at X in B0 is denoted ρ0. We also assume that material points of the body at time t existed at the beginning of the growth process so that they can be mapped contiguously from the reference configuration to the current configuration. The deformation from B0 to Bt is described by a one-to-one mapping χ such that x ( X , t ) . The deformation gradient from B0 to Bt is then defined by lxxviii F Grad x . In standard notation, J det F 0. Let the deformation gradient from the (evolving) relaxed configuration B0 to Bt be represented by Fe. For convenience, we denote by ( r ) the specialization of χ to Br and we write X ( r ) ( r )( X ,t ) (3.1) The deformation gradient Fr Grad ( r ) from B0 to Bt then contains information about the growth as well as information on the rearrangement of the material that ensure geometric compatibility of the growth and is responsible for the appearance of additional residual stresses. We note as in Chapter Two that F Fe .Fr (3.2) which is analogous to F Fe .Fg (3.3) Here we adopt Fr for uniformity in notation. and emphasize that since Br is residually-stressed configuration. each of Fe, Fr and F is a deformation gradient and no “fictitious” stress-free configuration is assumed. The approach we adopt here is different from that initiated by Rodriguez et at [50], which involves a (systematric) ‘growth’ tensor that is not in general the gradient of a deformation. Let J r det Fr (3.4) Then, since Fr is isochoric, we have J Jr (3.5) F LF , (3.6) Again let where L is the velocity gradient, gradv, v being the material velocity in Bt Similarly, let Fr Lr Fr , (3.7) lxxix F e Le Fe , and (3.8) Here and subsequently a superposed dot signifies the material time derivative. We define the velocity gradient as: L = grad v or Lij (3.9) vi xj (3.10) using the identity ( see [44]) Grad u ( grad u ) F (3.11) equation (3.9), gives Grad v ( grad v )F LF Since v x (3.12) we also have Grad x Grad x F t (3.13) Hence we have the important connection F LF , Here v (3.14) is the material velocity in Bt (det F ) (det F )tr( F 1 F ) t (3.15) Using the result above on equation (3.14), we have or (det F ) (det F ) tr ( L ) t (3.16) J Jtr( L ) J div V (3.17) Since J det F , tr( L ) Lii vi divV xi Thus divV measures the rate at which the volume changes during the motion. Result For an isochoric deformation, (3.18) lxxx J 1 J 0 (3.19) Hence div V 0 (3.20) From Equation (3.14), we define Fr Lr Fr and F e Le Fe (3.21) Then L Le Fr Lr Fe1 (3.22) By incompressibility, trLe 0 (3.23) hence trL trLr (3.24) where tr denotes the trace of a second-orders tenor. 3.4 Mass Balance Definition.3.1 Sigma-algebra ( σ-algebra) : A set A of a subset of a set X is called a σ-algebra if (i) X is in A (ii) Y A Y c A (ii) Yn A , n 1 ,2 ,3 ,......... n 1Yn A Definition 3.2. Measurable space : A map : X Y , where ( X,A ), ( Y,A ) are measurable spaces ( measurable ) if 1 ( B ) A , for all B В Definition 3.3. Borel set : A Borel set ( B-set ), (or B- measurable set ) in a topological space is an element in the smallest σ-algebra which contains all compact sets. A B-set can be obtained from not more a countable number of operations of union and intersection of the open sets. We now utilize the results of the above definitions to cast and prove the following theorem. lxxxi Theorem 3.1. Suppose there is a function m defined over a set of bodies, such that for an arbitrary body, B , m( B ) 0 (3.25) then m is a Borel set of B Proof: Since m is assumed to be a well-defined function over the set of bodies, then m is a measure defined on measure space B ( for arbitrary B). Let X B, and X is finite, then B has the following properties: (i) the empty set is in B (ii) for any X1, X2, in B, X1 B, and X2 B (iii) for any collection of a countable number of sets in B, the union and intersection of a countable number of sets are also in B. Hence the sets in B are measurable and consequently are Borel sets. Since m(B) 0, then m is assumed to be additive. Using the properties of definition 3.1, we can write m ( B B ) m ( B ) m ( B ) (3.26) m ( B B ) m ( B ) m ( B ) (3.27) for arbitrary disjoint bodies B and B , wherein B B is the usual notation of set theory. Finally, m is assumed to be continuous with respect to the volume of the body. Therefore, m( B ) 0 as the volume tends to zero. lxxxii We assume that m is continuous with respect to the volume of the body. Hence it follows that m( B ) is intrinsic to the body B , independent of the motion of B . Thus d m( B ) 0 dt for each body. (3.28) In other words, m(B) is independent of the configuration B. Consequently, equation (3.25) is an objective scaler, called the conservation of mass hence the proof. Consequences of the Theorem 3.1 (i) The continuity requirement ensures that for each configuration Bi of B , there exists a scalar field defined over B at time t such that m( B ) B ( x , t )d V (3.29) where dV is the volume element appropriate to B . We call ρ mass density of the material (ii) If we assume that ρ is smooth, we have that ( x ,t ) 0 (3.30) (iii) Since m( B ) is independent of the configuration В we can write B ( x , t )dv B0 0 ( X )dV (3.31) for an arbitrary choice of reference configuration B0 for B , where ρ0 is the mass density of B in the configuration B0 and dV is the volume element for B 0 both independent of t (iv) Applying equation (3.31) to an arbitrary body, the continuity of ρ and the equation lxxxiii dv (det F )dV JdV (3.32) it follows that densities ρ and ρ0 are related by J 1 0 (3.33) And since J det F 0 (3.34) and by incompressibility, J det F 1 det F J 0 / 0 (3.35) where F is the deformation gradient from B0 to Bt (v) Since J 0 , it follows that 0 if and only if 0 0 (vi) Equation (3.33) is the local form of the principle of conservation of mass 3.4.1 Mass Balance Equation. To account for volumetric sources of mass and surface flux of mass, we let denote the bulk mass source per unit mass in the deformed configuration Bt and m the surface mass influx per unit area of Bt , the boundary of Bt . Using the Eulerian local form of the mass balance for open system, [11] we have div v div m (3.36) Using the relation J tr (L) = J div V, we have tr Lr div m (3.37) In the absence of mass sources, equation.(3.36) reduces to the standard conservation of mass equation, and can also be described as local form of mass conservation div v Since 0 J J div v (3.38) (3.39) lxxxiv Substituting for div v from (3.38) then gives J J that is , 0 ( J ) 0 J t (3.40) is a constant for any material particle. (3.41) In the reference configuration, J 1 , so that, J r or J 1 r (3.42) where r is the mass density in the reference configuration. Equation (3.42) is another form of the mass conservation equation 3.4.2 Balance equations for momentum For growth process, the standard linear momentum balance equation must be modified. As with mass balance, there are two contributions to linear momentum production, one is volumetric and the other is surface contribution. Let b denote the bulk source of momentum production which includes body forces and also possibly other volumetric sources. The surface source of momentum production consists of a contribution due to surface traction, Epstein and Maugin [11] treated these terms explicitly, but have for simplicity and convenience , we embody such sources into one volume and one surface term. Thus, the integral form takes the form d T Bt d Bt b d Bt n da dt (3.43) where is the Cauchy stress tensor. Using mass balance and transport formula, we have: v b div (3.44) lxxxv For consistency of notation, equation (3.44) can be written as and b div (3.45) b b div m v , where b is the effective body force and is the effective Cauchy stress, for alternative form of the equation [24]. Since is symmetric, we have that T (3.46) 3.4.3. Energy balance and constitutive equations We define u to be the internal energy per unit mass λ the volumetric growth energy supply (per unit mass), g the growth energy surface influx vector (per unit area of Bt ) h the heat supply, (per unit mass and q the influx vector (per unit area of Bt ). Then the local form of energy balance equation can be written as u t r ( L ) r div g h div q (3.47) where the overbars are similar to that used for b and , We now consider the Helmholtz free energy defined by u (3.48) where 0 , is the absolute temperature and is the specific entropy per mass Equation (3.47) may then be rewritten as tr ( L ) r div g h div q where the last four terms represent contributions from thermal affects. Using rate of entropy in Bt and second law of thermodynamics, we can rewrite equation (3.49) as (3.49) lxxxvi tr ( L ) div g p h div q 0 (3.50) The inequality in equation (3.50) is quite general and does not include my indication of the variables on which depends. We suppose that depends separately on the density and its gradient. This is analogous to thermoelasticity where the temperature gradient is included with the temperature. We embody this information by the structural tensor 0 ,which may be a vector, a tensor or collection of such objects. Furthermore, will in general depend on the residual stress within B0 and we denote it by 0 (which may be zero). For the growth phase, it is convenient to use a ‘pullback’ ( or Lagrangian ) version of the instead of itself. This is denoted L and defined by L = J, so that the mass balance equation (3.36) can now be written as L J ( div m ) (3.51) where we define . Considering elasticity and growth, has the dependence F , L , , 0 , 0 , . (3.52) Since we are considering volumetric growth here and no structural changes (no remodelling), 0 changes only through the effect of F. Consequently, Grad L From eqn (3.52) and the inequality (3.50) and with Grad , equation (3.50) as (see [45]). (3.53) we can write lxxxvii tr F L . r div g h F L (3.54) div q 0. For incompressible material elastic response, the constraint tr Le 0 (3.55) holds. But L may also be treated as arbitrary since L r , being not on its own constreained by the specification of and m in equation (3.36). Thus, during growth, the inequality in (3.54) must hold in particular, for arbitrary L. Hence we have F F (3.56) In structure, equation (3.56) has the standard form of conventional (thermoelasticity) in the absence of growth. We emphasis, however that the stress depends on growth and for suitable specifications of constitutive laws for and m , the growth will depend on the stress, so our formulation has the advantage of providing two way coupling. Now, L is not arbitrary since it appears in mass balance equation. We can write the residual inequality in (3.54) as L . div g h div q 0 L (3.57) The solution to the residual inequality depends on the specification of and g . We can represent (3.57) as lxxxviii L . G TH 0 L (3.58) where G , TH represent the growth and thermal structures respectively. In the sequel, we analyze the thermal changes and its effects. 3.5 Development of Thermal Residual Stresses due to Thermal Changes In this section, we consider a long cylinder under internal and external pressure, Pi and P0, and internal and external wall temperature, Ti and T0 at the inner and outer radii ri and ro. The artery is modeled as circular cylindrical tube. As the temperature is raised or lowered, with respect to the reference temperature, the body expands or contracts and thermal stresses (or residual stresses) may arise. Usually, the solution of the thermoelasticity problem begins with finding the time and position dependent temperature field. We may write the general form of the heat conduction equation as S k a ,ii ; a . t Cv Cv (3.59) In which is the temperature field, S is the strength of the heat sources in the field, is the density, Cv is the specific heat and k is internal conductivity tensor. Since in the absence of growth, the deformation is isochoric implies for a stationary field with no heat sources, we have , ii = 0 (3.60) Hence equation (3.59) gives S = t CV (t ) = S t + f ( t) CV ( 3.61) lxxxix At rest when t = 0, that is, (t1) = (t) = S t CV (3.62) S S S t1, (t2)= t2, ……, (tn) = tn Cv CV CV (3.63) Using cylindrical coordinates, we have 2 = 2 1 1 2 + + =0 r z 2 r r r r 2 2 (3.64) Since the tube is long, we neglect the end effects and assume a plane condition and eliminate the z dependence. Also, because the problem is axis symmetric, we eliminate dependence. Hence 2 = 1 r =0 r r t (3.65) Since = (r), r2 = x2 + y2 + z2 (3.66). r x x . 1 ( r ). . 1 ( r ) x r x r r 2 1 1 x r 1 x r (r ) 2 . ( r ) ( r ). 2 x r r x r x 1 1 x2 x2 = ( r ) 3 ( r ) 2 ( r ) r r r . (3.67) xc Similarly, 2 1 1 y2 1 y 2 11 ( r ) ( r ) (r ) y 2 r r3 r2 (3.68) 2 1 1 z2 1 z 2 11 ( r ) ( r ) (r ) z2 r r3 r2 (3.69) Therefore, 3 1 1 ( r ) 3 ( r )( x 2 y 2 z 2 ) 2 ( r )( x 2 y 2 z 2 ) r r r 2 = = 3 1 2 ( r ) ( r ) ( r ) = ( r ) ( r ) r r r Hence, 2 = 1 2 r ( r ) ( r ). = 0 rr r r This implies that (3.72) Solving equation (3.71) using thermal boundary conditions, we have 1 r rr r i .e . r (3.71) 2 ( r ) ( r ). = 0 r 2 That is, ( r ) ( r ). = 0 r (3.70) 0 A (r ) r A B r xci Consequently, A Hence, Ti Ti T0 , ri ln r0 B Ti T0 Ti ln ri . ri ln r0 T0 Ti r ln i r r ln i r0 (3.73) Remark : Equation (3.73) is the fundamental model for determining the effect of temperature during growth when the parameters are specified as experimental data. For simplicity and convenience, and without loss of generality, we now use the linear elastic strain instead of finite strain and with its usual definition, we state ias ei j 1 ( ui , j u j ,i ) 2 (3.74) The equilibrium equation still remains ij, j f i 0 (3.75) Due to our formulation based on multiplicative deformation, we have F = Fe . Fg = Fe . Fr (3.76) We can split the strain into elastic (mechanical) and thermal parts as: ei j eimj eig j eim j eiT j (3.77) It is convenient to work in terms of two independent stretches (or axes). Following the work of [30], we define eimj = 1 i j K K i j E E (3.78) xcii and G= E 2(1 ) (3.79) where is the Poisson’s ratio, E is the Young’s modulus, ij is the Kronecher delta. Using equation (3.79) in equation (3.78), we have eim j = 1 ij KK ij 2G 1 (3.80) In order to obtain an expression for the growth strain, we may write dST = (1 + αθ) dSo (3.81) where dSo is the initial length of a line element and dST is the length as a result of a temperature change in θ, and α is the coefficient of thermal expansion. Equation (3.81) results from the inflation and the extension in the tube. Consequently, we define the growth strain as eiT j ij (3.82) Adding equations (3.81) and (3.82) we have e i j eimj eiT j 1 i j i j i j i j 2G 1 (3.83) Equation (3.83) is the so called Duhamel-Neuman constitutive equation. Contracting on i and j indices, we have ei i 1 3 1 l l 3 2G 1 (3.84) xciii = 1 2 1 l l 3 2G 1 l l 2G (1 ) 1 2 e l l 3 (3.85) (3.86) Substituting equation (3.86) into (3.83), we have ei j = 1 i j 2G el l 3 ii j i j 2G 1 2 1 i j el l i 2G 1 2 j 1 i 1 2 j Hence i j 2G ei j ( 1 ) e l l l l 2G i 1 2 1 2 j (3.87) The above equation is the generalized from for Navier‘s displacement and as is given in terms of displacement as u i, j i 1 1 2( 1 ) ui , i j f i ,i 1 2 G ( 1 2 ) (3.88) 3.5.1 The Thin Circular Tube. Next we assume that the temperature field is symmetrical about the center of the tube and does not vary over the thickness, so we will find the thermal stresses when we regard the artery as a thin cylinder. The equilibrium equation and boundary conditions together still remains. xciv d rr rr 0 , rr P on r a , rr 0 on r b. dr r (3.89) Our pseudo-compatibility equation in thermo elastic form is err 1 ( rr ) E (3.90) e 1 ( rr ) E (3.91) In terms of the stresses can be written as rr E 1 2 E 1 2 err e ( 1 ) (3.92) e err ( 1 ) (3.93) Substituting equations (3.92) and (3.93) into the equilibrium equation (3.89) , we have r d E err e ( 1 ) 2 dr 1 =r E err e 1 e err 1 0 1 2 d err e 1 err e 1 r d dr dr (3.94) In terms of displacements, equation (3.94) can be rewritten as d 2 ur 1 dur ur d 2 1 2 dr r dr r dr (3.94)1 The above equation is nothing but the simplified form of the generalized Navier equation xcv for the current problem whose general solution for the displacements, radial and circumferential stresses are respectively, 1r C ur 1 r dr C1 r 2 ra r For rr E And E 1 r E r dr 2 r a 1 2 1 r rdr E r2 a (3.95) 1 C1 1 C 2 1 r 2 1 C1 1 C 2 1 r 2 (3.96) (3.97) For a soild disc, a = 0. such that r = a = 0, and lim r 0 1r r dr 0 r 0 (3.98) At the outer edge of the tube, rr 0 and hence C2 = 0. and at r = b C1 1 b r dr b2 1 (3.99) 0 b 1 r rr E 2 r dr 2 r dr r a b 0 E 1 b 1r rdr 2 rdr 2 b 0 r a (3.100) (3.101) At r = r1, rr P1 and r = r0, rr P0 1 b 1 r1 that is, E 2 r1 dr1 2 r1 dr1 P1 r1 0 b a (3.102) 1 b 1 r0 r dr r0 d r0 P0 0 0 2 2 r0 0 b a (3.103) E xcvi 3.5.2 Instantaneous Point Source in an Infinite Body We now consider a quantity of heat Q deposited at time t = 0 at a point Xi = 0 inside the tube. The heat will gradually spread into the body and every point will experience an increase and later decrease in temperature with time. Eventually, the body will return to its initial temperature after a long period of time. With the theory of multiplicative decomposition theory guiding the formulations in this thesis, shows that this relaxing to its initial state after the thermal loads are lifted does not imply the body returns to undeformed state but to intermediate configuration with some residual stresses. In order to obtain resulting thermal residual stresses, we must find time and position dependent temperature field. Since the spread of a point source in an infinite body is a 3 Dimensional event and is the same in all directions, the problem can be treated using spherical coordinates. At a point source, the proper form of the equation is of heat conduction is given as: 1 2 r a 2 t r r r (3.104) We define the Laplace transform as ˆ ( s ) ( t ) e s t d t (3.105) 0 so that, ˆ L s t (3.106) Therefore the Laplace transform of equation (3.104) is a d 2 d r s ˆ L 2 r dr d r = a d 2 dˆ r r 2 dr d r and the general solution to the equation (3.107) (see [31]) is given as (3.107) xcvii A ̂ e r s r a s r a B e r (3.108) To ensure that ̂ to remains bounded, we set B = 0, and so s r a A ̂ e r (3.108)1 Since Q the total amount of heat deposited in the body just the same as the fomous expression Q = m c ( t2 - t1 ), where m represents the mass, and c the specific heat. To find the constant A, we define Q as in [29] Q = 4 r 2 CV dr (3.109) 0 To ensure that heat is supplied instantaneously, Q must be chosen as a Heaviside step function. Next we take the Laplaces’ transform of the above equation (3.109) and we proceed as follows: 1 Q = 4 r 2 CV ˆ dr s 0 = 4 Cv A e s a r r dr 0 = 4 CV A Hence, A= a s (3.110) Q (3.111) 4 Cv a Finally, the time and position dependent temperature field for our problem is : ˆ Q e 4 CV a r s r a (3.112) The classical inverse transform of equation (3.112) gives as Q 4 a t 2 CV 3 e r2 4 at (3.113) xcviii Using the displacement filed equation of the generalised form of Navier’s equation: U i , jj 1 1 2 1 U i , jj f i i 1 2 G 1 2 (3.114) the general solution to equation (3.114) is g ui uih uip (3.115) where u hi is the complementary solution and u ip is the particular solution. We may assume that the particular solution can be derived from the thermoelastic potential Ф. Hence uip = Ф, i (3.115) Using equation (3.115) in (3.114) in the absence of body force, we have 1 , ,j j 1 2 Ф jj = 2 Ф = j j 2(1 ) 1 2 ,I 0 1 1 (3.116) (3.117) Assuming a non-stationary source-free temperature distribution, a solution for the above equation can be written as Ф= t 1 . dt Ф 0 + t Ф1 1 0 (3.118) where 2 Ф 1 =0 and Ф 0 represents the initial thermoelastic potential. For the homogenous part of the problem, since there are no body forces or surface traction, we may assume u hi =0 using equation (3.113) in (3.118), we have: Ф= 1 Q a 3 1 4 a 2 CV t 0 r2 4 at 3 t2 d t . Ф0 + Ф 1 t (3.119) xcix For ease of formulation, we introduce the following r2 V= V 4at 2 at r 2 3 r2 T= t2 4 aV 2 dt r3 3 2 (3.120) 8a V 3 r2 dV 2 aV3 Substituting equation (3.120) into (3.119), we have 1 Q a Ф= 3 1 4 a 2 CV = 1 Q a 3 1 4 a 2 CV 1 Q a = 3 1 4 a 2 CV = 1 Q a 3 1 4 a 2 CV r e V 2 at r2 dV . Ф 0 + t Ф 1 3 2 a V 2 r3 0 3 8a2 V3 3 4a 2 r r e V . 2 dV + Ф0 + t Ф 1 a 2 at 1 4 a2 2 e v dv +Ф 0 + t Ф 1 r r 2 at 2 0 e V2 dV r 2 V + Ф0 + t Ф 1 (3.121) 2 at To evaluate e V dV 2 0 Let u = V2 and 2VdV = du d V dV du 1 2V 2u 2 (3.122) Therefore, 1 V e d V = 0 2 1 2 u u e du 20 (3.123) c Equation (3.123) compares with the Gamma function u x1 u (3.124) e du That is x - 1= - ½ x 1 2 u Thus, u e du 0 2 v hence, e d V 0 1 2 1 2 (3.125) 1 . 2 (3.126) Therefore, r 1 Q 2 at v 2 2 1 Ф= e dV . Ф0 + t Ф 1 1 4 r CV 0 2 1 Q = 1 1 4 r CV = = r 2 1 Q 1 erf 1 4 r CV 1 Q 1 4 r CV 2 at e 0 v 2 d v Ф0 + t Ф 1 Ф0 + t Ф 1 2 at r erf c ( erf r 2 at ) Ф0 + t Ф 1 (3.127) r where erf is the error function defined as: erf r 2 at = 2 2 at v e dv 2 o and erf c is the error function complement. After a long period of time, the displacement and stresses return to zero and the relaxed configuration now posses residual potential energy. So Ф 0 and Ф 1 can now be redefined as Ф0r and Ф1r, where (r) denotes residual potential ci To determine Ф 0 r , Ф 1 r we assume Ф 1r = 0 Ф0r = - Ф Фr 1 1 4 r CV as t (3.128) r 1 erf 1 4 r CV 2 at (3.129) Consequently, the displacement, strain and stress fields can be determined as follows: ui u hi u pi Ф , j e ij Ф , jj 1 2 ij 2G ,ij ,kk ij (3.130) 1 1 2 ij (3.131) 3.5.3 Decomposition of the Free Energy We now decompose the free energy into two parts, namely, the isotropic and anisotropic parts as follows 0 0 iso 0 aniso (3.132) where 0 iso is the isotropic contribution of the matrix material. This material is endowed with cells called collagen fibres which are embedded in the blood stream,(see [19]). It is the collagen fibres that give the material response its anisotropic character and the associated energy function is 0 aniso . At low pressure, the main contribution to o is provided by 0 iso and whereas at high pressure 0 aniso become predominant. This loading and unloading generates the residual stresses. We also consider that the internal pressure is increased. In our own case, our internal pressure takes the form of a thermal potential. So the artery now experiences cii hypertensive thermal pressure, so that growth is initiated. At the start of the growth process, Br is identical to the unloaded configuration Bo associated with no growth and has the residual stress distribution equation. As time passes and growth takes place, Br evolves and changes in the residual stress distribution are generated. During growth, the relevant governing equation is div b 0 (3.133) From our previous equation, we have. L J div m (3.134) For simplicity, let div m = 0, so that we have L J (3.135) We still hold on to the equilibrium equation (3.89) with boundary condition rr P on r = a, rr 0 on r = b (3.136) Since we have set m = 0, the mass balance equation in the form of equation (3.36) and (3.51) together reduce to L J tr Lr L (3.137) Because of the symmetry, the radial part of the deformation has the form r = r(R), Z is uniform and r = r (R), = r/R (3.138) where the prime indicates differentiation with respect to R. The free energy is specialized to r , , L L (3.139) ciii The principal stresses are given by: rr r , , zz z r z (3.140) But here the residual stress distribution is not included separately The equilibrium equation and boundary conditions together are d rr 1 ( rr ) 0 dr r rr P , on r = a, rr 0 on r = b (3.141) The above equation can now be written as d 1 1 rr rr dr r r (3.142) Equation (3.142) is a linear differential equation whose solution is rr c r (3.143) At rr P on r = a, we have from the above equation c P a (3.144) And at rr 0 on r = b, c b (3.145) Using equation (3.144) in (3.143), we have c c P a r rr Also rr c c a b (3.146) (3.147) civ c a Therefore, rr r b a b c for P a r b a (3.148) (3.149) ab When b = a, P vanishes which suggests that the pressure is zero at the boundary of the tube, and hence rr c rb But r a 1 r 2 So that 1 r (3.150) ar R 2 A 2 1 2 (3.151) r 2 a2 2 R A2 (3.152) And for r = r(R), dr dr dR r 2 a2 . r ( R ) dr dR dr 2R 2R dr therefore, r ( R ) 2 2 r a d r Hence r(R) = r 2 2 R dr dr a 2 dr (3.153) (3.154) (3.155) To determine the principal stretches in equation (3.140), we construct the matrix cv r Z r r r r r r Z r Z Z Z Z Z Z (3.156) Specializing equation (3.140) subject to equation (3.132), we have (i) for isotropy r , , z (ii) for anisotropy, 0 (3.157) , , z , L , L , (3.158) Computing the matrix entries in equation (3.156) following equation (3.140), we have 2 2 r i R r a 2R r 0 0 Z 0 0 r 0 0 Z 0 r 2 a2 2R i R 0 (3.159) For simplicity, we define the following in Cartesian coordinates, x xr, , z , y yr, , z , z zr, , z , r x 2 y 2 z 2 x r cos , y r sin Therefore x r . r x r x x 1 2 (3.160) (3.161) (3.162) Following the steps in (3.162) we have r 2 a 2 r 2 a2 r 0 r R r r R 2 R x x r 2 R (3.163) cvi r 2 a 2 R 0 2 since z is uniform. (3.164) Z z (3.165) We note that includes dependence on growth directly through L and L and indirectly through the total stretches r , , z . The specialization of the decomposition F = Fe.Fr. (3.166) for the present situation is r e r re rr , , where the superscript (e) signifies the elastic part of the deformation which satisfies the incompressibility condition re e ze 1 (3.167) The residual (growth) part of the deformation is indicated by the superscript (r). Suppose that Rr and Zr denote the axial coordinates in the configuration Br and that Ar < Rr < Br, and 0 < Zr < Lr, then the elastic part of the deformation (from Br to Bt) is described by the equation r 2 a 2 ze Rr2 Ar2 , z ze Z r From the above equation, we can determine Rr as a function of Ar. From equation (3.168), Rr zr r 2 a 2 A2 dR rzr dr R rzr dr R dR 1 1 2 (3.168) cvii That is, r R R r zr r R R r Br 1 r z 1 dR = r Ar Therefore, r(R) = r 21 B 1 r z 2 r Ar2 r z 1 Br R2 2 Ar (3.169) Now, B r2 2 r 1 ( R ) r zr A r2 Br Ar 2r ( R) rzr 2r ( R ) ( r r z Ar2 (3.170) ) Br2 (3.171) 3.5.4 Evolution of Residual Stresses with Growth At any stage of growth, the equilibrium equation must be satisfied in the configuration Br and the associated boundary conditions are (r ) d RR 1 (r ) (r ) =0 on Rr = Ar, RR ( r ) 0 in Ar < Rr < Br , RR dRr Rr (3.172) (r ) Where RR and ( r ) denote the radial and circumferential residual stresses in Br at time t of the growth process. The specialization of the equation given below as div r o , ( r ) Fr Fr , L , , 0 , p( r ) I F (3.173) to the present situation with ˆ ( r ) , (z r ) , L , L , (r r ) , ( r ) , (z r ) , L , L , ˆ . is (r ) (r ) RR ( r ) ˆ ( r ) Integrating equation (3.172) following our previous methods, (3.174) (3.175) cviii we have , Br Br Ar Ar (r ) d RR dRr Br Br ˆ (r ) (r ) dR RR dR ( r ) Ar When rr( r ) 0 , we have , ( r ) Ar That is, (0r ) ˆ ( r ) (3.176) ˆ dRr 0 ( r ) Rr Br ln 0 Ar (3.177) Equation (3.176) provides a programme for the evaluation of growth and the resulting development of residual stress in the considered problem. A detailed analysis based on the coupled equations discussed above is beyond the scope of this thesis. We assume that during growth, the circumferential stress in Bt has non-uniform, time-dependent form c ( r , t ) (3.178) The specialization of , L , L. provide for growth law. This then gives the . combination tr L , which allows changes in density at fixed volume. Here we prescribe the value of , instead of the growth stretch rates. Furthermore, we set it to be proportional to the gradient of the circumferential stress. Thus, L L , r (3.179) where is a constant. Growth stops when is uniform. For simplicity of illustration, we now assume that is linear in r, so that c r , t c0 c1 t r (3.180) cix where co is a constant and c1 depends on time only. This is not unreasonable as a first approximation for a wall thickness that is not too large. If we choose co to have a uniform value in the absence of growth as given in our previous equation a0 P0 = 0 b0 a0 (3.181) then c1 can be interpreted as measuring the deviation in the circumferential stress distribution through the wall from its initial uniform value due to the hypertensive thermal pressure P. From the integration of the equilibrium equation (3.137) and the application of equations (3.136) - (3.144), we have, P b a c0 a c1 ( b a ) 2 c0 c1 2 P 2 b a2 2a 1 a b (3.182) 3.5.5 Growth Conditions (i) Growth stops when c1, regarded as a function of time vanishes that is, a 2 c0 2 P 2 2 b a or and a 2 P 2 2 b a 2 c0 1 0 a b 1 a b P ba c0 a (3.183) Equation (3.183) represents the final steady state condition with constant and uniform circumferential stress co. (ii) Let a = a1(> aO) be the internal radius of the artery arising from the (instantaneous) increase in pressure from PO to P. Another assumption based on cx experimental findings is now adopted. When the artery experiences high blood pressure, but no change in its blood flow rate, its lumen radius is essentially unchanged (see [47], [59], [60], [65]). (iii) If the inner radius a1 in the current configuration Bt remains constant during the whole growth process, equation (3.181) with a = a1 shows how the thickness of the artery has increased as a result of high blood pressure. (iv) In particular, in the special case c0 = σ0 for which the homeostatic circum ferential stress is unchanged by the pressure increase, we obtain from equation (3.183) b a1 P a1 c0 (3.184) when a = a0 , P = P0 , we have b0 a0 P a P 0 c0 0 0 a0 c0 b a0 (3.185) Using equation (3.185) in (3.184), we have b a1 P a1 a0 P0 or b a1 a1 P a0 P0 b 0 b 0 a0 a0 (3.186) (3.187) The thickness is then proportional to the pressure and also to the inner radius. (v) If the uniform circumferential stress is increased as a result of growth, then the right hand side of equation (3.187) can be scaled by a factor increase is proportionately less. That is 0 c0 and the thickness cxi b a1 0 a1 P b0 a0 c0 a0 P0 (3.188) Finally, we remark that the calculation of the residual stress resulting from growth requires the specification of . CHAPTER FOUR cxii DEVELOPMENT OF STRESS EXPRESSIONS FOR SOME KNOWN ENERGY FUNCTIONS BASED ON MULTIPLICATIVE DECOMPOSITION 4.1 Preamble In this chapter, stresses and strains of some known strain-energy functions based on multiplicative decomposition of the deformation gradient are obtained using the theory of thermoelasticity. In the heart of the chapter, we present a method of approximating response coefficients using rational function approximation instead of polynomial functions. Here, we generalize our results to isotropic and anisotropic thermoelasticity by using the more general class of rational functions to approximate the material response coefficients. 4.2. Basic Equations of Thermoelasticity In the framework of the thermoelasticity of elastic materials, it is usual to assume the following set of constitutive equations for the Cauchy stress tensor T, the specific entropy ,the internal energy e, the temperature , and the heat flux vector q: T = T(F, , ), q = q(F, , ), ˆ ˆ F , , , e eF , , . (4.1) where is the absolute temperature, (see [5 ], [42 ]). We introduce the Helmholtz free energy per unit mass as: = u - (4.2) Clausius-Duhem inequality, which preserves the assumption that the local entropy production is non-negative, is given by cxiii 1 ê ˆ tr TD q . 0 .(4.3) where is the mass density in the deformed configuration and D is the rate of deformation tensor, the superposed dot denotes the material time derivative. For inequality (4.3) to hold for all processes only the response function, q, may depend on , and we obtain our earlier results in Chapter Two as follows: (i) The Cauchy stress is given by T F F (ii) The entropy relation is given by ˆ (iii) The response function q obeys the heat conduction inequality T q (F, , ). 0 F (4.4) (4.5) (4.6) x , is the deformation gradient, X is a position vector in reference X configuration and x is the corresponding vector in the deformed configuration. The right Cauchy-Green strain tensor is given by C = FTF and for an isotropic hyperelastic material the principal invariants are defined as: I1 = tr C, I2 1 2 tr C tr C ’ 2 2 I3 = det C (4.7) In case of isotropy, it is convenient to use the left Cauchy-Green strain tensor B = FFT, which has the same principal strain invariants as C. On applying the principle of material frame indifference and assuming that the material is isotropic, ( see [5 ], [44 ]), equation (4.4) becomes cxiv T 2 0 I 2 I3 I 3 I 2 1 I B I3 B I I 1 2 (4.8) and the constitutive equation for the heat flux can be written as q = (x0I + x1B + x2B2) 4.9) where again, I1 = tr B, I 2 1 2 I tr B 2 , I3 = det B 2 1 (4.10) and i = i (I1, I2, I3, , ., .B, .B2), i = 1,2,3 (4.11) For incompressible material, I3 = 1 (4.12) and equation(3.8) is replaced by T pI 2 1 B2 B I 1 I 2 (4.13) where p denotes an arbitrary hydrostatic pressure. In thermoelastic settings, following the work of [52], we introduce our incompressibility constraint as J I3 1 g( ) (4.14) where g() 0 for all and is such that g(0) = 1, where 0 denotes a constant reference temperature. This means that a change of mass density is possible with a change of temperature because J det F 0 g( ) 0 and therefore (4.15). When equation (4.15) holds, the stress response is still given by equation (4.13). cxv Equation(4.12) is not compatible with the peculiar thermomechanical behaviours exhibited by elastomers under elastic deformations [52 ], whereas the modified constraint (4.15) does allow one to describe these thermoelastic effects to a considerable extent. Definition I: G(T) is said to be an isotropic tensor function of T if G(Q T QT ) = QG(T)QT (4.16) Definition 2: Let be a scalar function defined over a space of (positive definite) symmetric tensors. Then its value (T) at T is said to be a scalar invariant of T if (QTQT) = (T) (4.17) for all proper orthogonal Q. Such a Q is called an isotropic scalar function. Remark An isotropic scalar-valued function of T is also called a scalar invariant of T. Consequences of Definition 2. We verify that principal invariants I1, I2, I3 of T are scalar invariants according to equation (4.2): (a) I1 = tr(T). Since Q is orthogonal, tr(QTQT) = tr(QTQT) = tr(IT) (b) For I3 = det(T)) = tr (T) = det(QTQT) = (detQ)(detT)(det QT) ( 4.18) (4.19) Therefore the stored energy density of an isotropic material can also be expressed as a function of the principal invariants. Hence we can write W = (I1, I2, I3) (4.20) cxvi We call W simply the stored energy per unit volume of the reference configuration. Since in isotropic case, the only constitutive relation between the tensor C and second Piola - Kirchoff (P-K) strain tensor is S2 W C (4.21) then, S 2 I 1 I 2 I 3 W 2 C I 1 C I 2 C I 3 C (4.22) Calculating the derivatives of the principal invariants with respect to right Cauchy stress Ci j, I 1 I I , 1 1J C C1 J To calculate I 2 C (4.23) , we apply the product rule to equation (4.7), that is, II det C trC 1 det C trC 1 . C C C = det CtrC 1 C 1 det C But trC 1 C 1 Hence, But, 11 KL C 1 trC 1 C KL trC 1 KL C 1 C1 J C1 J C KL C 1 LM KM (4.24) KL (4.25) (4.26) cxvii Differentiating this identity with respect to CIJ, we have, 1 K JL C 1 C 1 LM 0 LM C KL C1 J (4.27) Multiplying this equation by (C-1)KN, gives C 1 C1 J KL C C 1 1 1K (4.28) JL (4.30) That is, . = = I 3 det C C 1 . In other words, C I 3 I 3 C 1 C (4.31) Hence equation (4.29) follows from equations (4.27) and (4.28). From equations (4.25), (4.29), (4.30) follow the following results for the second Piola-Kirchoff stress tensor, SYMBOL 102 \f "Symbol" \s 122 are scalar invariants of T. Proof: (a) If equation (4.35) holds, with 0, 1, 2 scalar invariants of T, then G(T) is clearly isotropic. (b) If G(T) is isotropic, we must show that it necessarily has the representation (4.9). We know that G(T) is coaxial with T and its cxviii principal values are scalar invariants of T. Let 1, 2, 3 and g1, g2, g3 be the principal values of T and G(T) respectively and consider the equations 0 + 1i + 2 i2 = gi , (i = 1, 2, 3) (4.36) for the three unknowns 0, 1, 2 . Assuming that i and gi are given and that the i are distinct, it follows that 0, 1, 2 are determined uniquely in terms of i and gi which are themselves scalar invariants of T. Thus 0, 1, 2 are scalar invariants of T, and G(T) and T are coaxial. Hence the theorem follows. Consequences of Theorem 4.1 (i) For an isotropic elastic solid the Cauchy stress is given by = G(T) (4.37) and from equation (4.29), T = T (4.38) cxix Explicitly, we now have = G(T) = 0I + 1T + 2T2 (4.39) where 0, 1, 2 are invariants of the left stretch tensor T. According to Theorem 3.I, each of 0, 1, 2 may be written as a function of the principal invariants of T, namely, I1(T) = 1 + 2 + 3 (4.40) I2(T) = 23 + 31 + 12 (4.41) I3(T) = 123 (4 .42) these being symmetric in the principal stretches 123 (ii) Equally, we may regard 0, 1, 2 as symmetric functions of 1,2,3 and write the principal components of equation (4.28) as ti = 0 + 1i + 2 i2 = gi , (i = 1, 2, 3) (4.43) where ti are the principal Cauchy stresses. Introducing the response function such that = ti = g(i, j, k) (4.44) where (i, j, k) is a permutation of (1, 2, 3) and g is defined on (0, )3. Thus g(i, j, k) = 0 + 1i + 2 i2 (4.45) and this embodies the symmetry g(i, j, k) = g(i, k, j) (4.46) implied by the symmetry of 0, 1, 2 as functions of 1, 2, 3 . Consequently, if F denotes the deformation gradient, then F= F T = 2. Hence = 0I + 1T + 2T2 (4.47) cxx From the principle of objectivity, thermoelastic body depends only on X, C and . That is, .