Basic Thermoelasticity

Transcription

1 Basic hermoelasticity Biswajit Banerjee November 15, 2006 Contents 1 Governing Equations Balance Laws he Clausius-Duhem Inequality Constitutive Relations Other strain an stress measures Stress Power Helmholtz an Gibbs free energy Specific Heats Appenix 9 1 Governing Equations he equations that govern the motion of a thermoelastic soli inclue the balance laws for mass, momentum, an energy. Kinematic equations an constitutive relations are neee to complete the system of equations. Physical restrictions on the form of the constitutive relations are impose by an entropy inequality that expresses the secon law of thermoynamics in mathematical form. he balance laws express the iea that the rate of change of a quantity (mass, momentum, energy) in a volume must arise from three causes: 1. the physical quantity itself flows through the surface that bouns the volume, 2. there is a source of the physical quantity on the surface of the volume, or/an, 3. there is a source of the physical quantity insie the volume. Let be the boy (an open subset of Eucliean space) an let be its surface (the bounary of ). 1

2 Let the motion of material points in the boy be escribe by the map x ϕ(x) x(x) (1) where X is the position of a point in the initial configuration an x is the location of the same point in the eforme configuration. he eformation graient (F ) is given by 1.1 Balance Laws F x X 0 x. (2) Let f(x, t) be a physical quantity that is flowing through the boy. Let g(x, t) be sources on the surface of the boy an let h(x, t) be sources insie the boy. Let n(x, t) be the outwar unit normal to the surface. Let v(x, t) be the velocity of the physical particles that carry the physical quantity that is flowing. Also, let the spee at which the bouning surface is moving be u n (in the irection n). hen, balance laws can be expresse in the general form ([1]) [ ] f(x, t) V f(x, t)[u n (x, t) v(x, t) n(x, t)] A + g(x, t) A + h(x, t) V. (3) t Note that the functions f(x, t), g(x, t), an h(x, t) can be scalar value, vector value, or tensor value - epening on the physical quantity that the balance equation eals with. It can be shown that the balance laws of mass, momentum, an energy can be written as (see Appenix): + v 0 v σ b 0 σ σ ė σ : ( v) + q s 0 Balance of Mass Balance of Linear Momentum Balance of Angular Momentum Balance of Energy. (4) In the above equations (x, t) is the mass ensity (current), is the material time erivative of, v(x, t) is the particle velocity, v is the material time erivative of v, σ(x, t) is the Cauchy stress tens b(x, t) is the boy force ensity, e(x, t) is the internal energy per unit mass, ė is the material time erivative of e, q(x, t) is the heat flux vect an s(x, t) is an energy source per unit mass. With respect to the reference configuration, the balance laws can be written as et(f ) ẍ 0 P 0 b 0 F P P F 0 ė P : F + 0 q 0 s 0 Balance of Mass Balance of Linear Momentum Balance of Angular Momentum Balance of Energy. (5) In the above, P is the first Piola-Kirchhoff stress tens an 0 is the mass ensity in the reference configuration. he first Piola-Kirchhoff stress tensor is relate to the Cauchy stress tensor by P et(f ) F 1 σ. (6) 2

3 We can alternatively efine the nominal stress tensor N which is the transpose of the first Piola-Kirchhoff stress tensor such that hen the balance laws become N : P et(f ) (F 1 σ) et(f ) σ F. (7) et(f ) ẍ 0 N 0 b 0 F N N F 0 ė N : F + 0 q 0 s 0 he graient an ivergence operators are efine such that Balance of Mass Balance of Linear Momentum Balance of Angular Momentum Balance of Energy. (8) v 3 i,j1 v i x j e i e j v i,j e i e j ; v 3 i1 v i x i v i,i ; S 3 i,j1 S ij x j e i σ ij,j e i. (9) where v is a vector fiel, BS is a secon-orer tensor fiel, an e i are the components of an orthonormal basis in the current configuration. Also, 0 v 3 i,j1 v i X j E i E j v i,j E i E j ; 0 v 3 i1 v i X i v i,i ; 0 S 3 i,j1 S ij X j E i S ij,j E i (10) where v is a vector fiel, BS is a secon-orer tensor fiel, an E i are the components of an orthonormal basis in the reference configuration. he contraction operation is efine as 1.2 he Clausius-Duhem Inequality A : B 3 A ij B ij A ij B ij. (11) i,j1 he Clausius-Duhem inequality can be use to express the secon law of thermoynamics for elastic-plastic materials. his inequality is a statement concerning the irreversibility of natural processes, especially when energy issipation is involve. Just like in the balance laws in the previous section, we assume that there is a flux of a quantity, a source of the quantity, an an internal ensity of the quantity per unit mass. he quantity of interest in this case is the entropy. hus, we assume that there is an entropy flux, an entropy source, an an internal entropy ensity per unit mass (η) in the region of interest. Let be such a region an let be its bounary. hen the secon law of thermoynamics states that the rate of increase of η in this region is greater than or equal to the sum of that supplie to (as a flux or from internal sources) an the change of the internal entropy ensity ue to material flowing in an out of the region. Let move with a velocity u n an let particles insie have velocities v. Let n be the unit outwar normal to the surface. Let be the ensity of matter in the region, q be the entropy flux at the surface, an r be the entropy source per unit mass. hen (see [1, 2]), the entropy inequality may be written as t ( η V ) η (u n v n) A + 3 q A + r V. (12)

4 he scalar entropy flux can be relate to the vector flux at the surface by the relation q ψ(x) n. Uner the assumption of incrementally isothermal conitions (see [3] for a etaile iscussion of the assumptions involve), we have ψ(x) q(x) ; r s where q is the heat flux vect s is a energy source per unit mass, an is the absolute temperature of a material point at x at time t. We then have the Clausius-Duhem inequality in integral form: ( ) q n η V η (u n v n) A t A + s V. (13) We can show that (see Appenix) the entropy inequality may be written in ifferential form as η ( ) q + s. (14) In terms of the Cauchy stress an the internal energy, the Clausius-Duhem inequality may be written as (see Appenix) 1.3 Constitutive Relations (ė η) σ : v q. (15) A set of constitutive equations is require to close to system of balance laws. For large eformation plasticity, we have to efine appropriate kinematic quantities an stress measures so that constitutive relations between them may have a physical meaning. Let the funamental kinematic quantity be the eformation graient (F ) which is given by F x X 0 x ; et F > 0. A thermoelastic material is one in which the internal energy (e) is a function only of F an the specific entropy (η), that is e ē(f, η). For a thermoelastic material, we can show that the entropy inequality can be written as (see Appenix) ( ) ē η At this stage, we make the following constitutive assumptions: ( η + ē ) F σ F : F + q 0. (16) 1. Like the internal energy, we assume that σ an are also functions only of F an η, i.e., σ σ(f, η) ; (F, η). 4

5 2. he heat flux q satisfies the thermal conuctivity inequality an if q is inepenent of η an F, we have q 0 (κ ) 0 κ 0 i.e., the thermal conuctivity κ is positive semiefinite. the entropy inequality may be written as ( ) ( ē η η + ē ) F σ F : F 0. Since η an F are arbitrary, the entropy inequality will be satisfie if an only if ē ē 0 η η an ē F σ F 0 σ ē F F. ē η an σ ē F F. (17) Given the above relations, the energy equation may expresse in terms of the specific entropy as (see Appenix) η q + s. (18) Effect of a rigi boy rotation of the internal energy: If a thermoelastic boy is subjecte to a rigi boy rotation Q, then its internal energy shoul not change. After a rotation, the new eformation graient ( ˆF ) is given by ˆF Q F. Since the internal energy oes not change, we must have e ē( ˆF, η) ē(f, η). Now, from the polar ecomposition theorem, F R U where R is the orthogonal rotation tensor (i.e., R R R R 1 ) an U is the symmetric right stretch tensor. ē(q R U, η) ē(f, η). Now, we can choose any rotation Q. In particular, if we choose Q R, we have ē(r R U, η) ē(1 U, η) ẽ(u, η). ē(u, η) ē(f, η). his means that the internal energy epens only on the stretch U an not on the orientation of the boy Other strain an stress measures he internal energy epens on F only through the stretch U. A strain measure that reflects this fact an also vanishes in the reference configuration is the Green strain E 1 2 (F F 1 ) 1 2 (U 2 1 ). (19) Recall that the Cauchy stress is given by σ ē F F. 5

6 We can show that the Cauchy stress can be expresse in terms of the Green strain as (see Appenix) Recall that the nominal stress tensor is efine as σ F ē E F. (20) N et F (σ F ). From the conservation of mass, we have 0 et F. Hence, N 0 σ F. (21) he nominal stress is unsymmetric. We can efine a symmetric stress measure with respect to the reference configuration call the secon Piola-Kirchhoff stress tensor (S): In terms of the erivatives of the internal energy, we have S : F 1 N P F 0 F 1 σ F. (22) S ( 0 F 1 F ē ) E F F ē 0 E an hat is, N 0 S 0 ( F ē ) E F F 0 F ē E. ē E an N 0 F ē E. (23) Stress Power he stress power per unit volume is given by σ : v. In terms of the stress measures in the reference configuration, we have ( σ : v F ē ) E F : ( F F 1 ). Using the ientity A : (B C) (A C ) : B, we have σ : v [( F ē ) ] E F F : F ( F ē ) : F N : F. E 0 We can alternatively express the stress power in terms of S an Ė. aking the material time erivative of E we have Ė 1 2 ( F F + F F ). S : Ė 1 2 [S : ( F F ) + S : (F F )]. 6

7 Using the ientities A : (B C) (A C ) : B (B A) : C an A : B A : B an using the symmetry of S, we have S : Ė 1 2 [(S F ) : F + (F S) : F ] 1 2 [(F S ) : F + (F S) : F ] (F S) : F. Now, S F 1 N. S : Ė N : F. Hence, the stress power can be expresse as If we split the velocity graient into symmetric an skew parts using σ : v N : F S : Ė. (24) v l + w where is the rate of eformation tensor an w is the spin tens we have σ : v σ : + σ : w tr(σ ) + tr(σ w) tr(σ ) + tr(σ w). Since σ is symmetric an w is skew, we have tr(σ w) 0. σ : v tr(σ ). Hence, we may also express the stress power as tr(σ ) tr(n F ) tr(s Ė). (25) Helmholtz an Gibbs free energy Recall that Also recall that S 0 ē E. ē E 1 0 S. ē η. Now, the internal energy e ē(e, η) is a function only of the Green strain an the specific entropy. Let us assume, that the above relations can be uniquely inverte locally at a material point so that we have E Ẽ(S, ) an η η(s, ). hen the specific internal energy, the specific entropy, an the stress can also be expresse as functions of S an, or E an, i.e., e ē(e, η) ẽ(s, ) ê(e, ) ; η η(s, ) ˆη(E, ) ; an S Ŝ(E, ) We can show that (see Appenix) (e η) η + 1 S : Ė t 0 or ψ t η S : Ė. (26) an t (e η 1 0 S : E) η 1 0 Ṡ : E or 7 g t η Ṡ : E. (27)

8 We efine the Helmholtz free energy as ψ ˆψ(E, ) : e η. (28) We efine the Gibbs free energy as g g(s, ) : e + η S : E. (29) he functions ˆψ(E, ) an g(s, ) are unique. Using these efinitions it can be showe that (see Appenix) ˆψ E 1 0 Ŝ(E, ) ; ˆψ ˆη(E, ) ; g S 1 0 Ẽ(S, ) ; g η(s, ) (30) an Ŝ 0 ˆη E an Ẽ 0 η S. (31) Specific Heats he specific heat at constant strain (or constant volume) is efine as C v : ê(e, ) he specific heat at constant stress (or constant pressure) is efine as We can show that (see Appenix) an C p : ẽ(s, ). (32). (33) C v ˆη 2 ˆψ 2 (34) C p η S : Ẽ 2 g 2 + S : 2 g S. (35) Also the equation for the balance of energy can be expresse in terms of the specific heats as (see Appenix) C v (κ ) + s + β S : Ė ) 0 (C p 10 S : α E (κ ) + s 0 α E : Ṡ (36) where β S : Ŝ an α E : Ẽ. (37) he quantity β S is calle the coefficient of thermal stress an the quantity α E is calle the coefficient of thermal expansion. 8

9 he ifference between C p an C v can be expresse as C p C v 1 0 ( ) S Ŝ : Ẽ. (38) However, it is more common to express the above relation in terms of the elastic moulus tensor as (see Appenix for proof) where the fourth-orer tensor of elastic mouli is efine as C p C v 1 0 S : α E + 0 α E : C : α E (39) C : Ŝ Ẽ 0 2 ˆψ ẼẼ. (40) For isotropic materials with a constant coefficient of thermal expansion that follow the St. Venant-Kirchhoff material moel, we can show that (see Appenix) C p C v 1 0 [ α tr(s) + 9 α 2 K ]. 2 Appenix 1. he integral is a function of the parameter t. Show that the erivative of F is given by F t t his relation is also known as the Leibniz rule. he following proof is taken from [4]. We have, Now, b(t) f(x, t) x a(t) b(t) F (t) f(x, t) x a(t) b(t) f(x, t) x + f[b(t), t] b(t) f[a(t), t] a(t). a(t) F t lim F (t + t) F (t). t 0 t F (t + t) F (t) 1 " b(t+ t) # b(t) f(x, t + t) x f(x, t) x t t a(t+ t) a(t) 1» b+ b b f(x, t + t) x f(x, t) x t a+ a a 1» a+ a b+ b b f(x, t + t) x + f(x, t + t) x f(x, t) x t a a a 1» a+ a b b+ b b f(x, t + t) x + f(x, t + t) x + f(x, t + t) x f(x, t) x t a a b a b f(x, t + t) f(x, t) x + 1 b+ b f(x, t + t) x 1 a+ a f(x, t + t) x. a t t b t a Since f(x, t) is essentially constant over the infinitesimal intervals a < x < a + a an b < x < b + b, we may write F (t + t) F (t) t b a f(x, t + t) f(x, t) t 9 x + f(b, t + t) b a f(a, t + t) t t.

10 aking the limit as t 0, we get lim t 0 " # F (t + t) F (t) t lim t 0 " # b f(x, t + t) f(x, t) x a t + lim t 0 " f(b, t + t) b t # lim t 0 " f(a, t + t) a # t F (t) t b(t) f(x, t) x + f[b(t), t] b(t) f[a(t), t] a(t). a(t) 2. Let (t) be a region in Eucliean space with bounary (t). Let x(t) be the positions of points in the region an let v(x, t) be the velocity fiel in the region. Let n(x, t) be the outwar unit normal to the bounary. Let f(x, t) be a vector fiel in the region (it may also be a scalar fiel). Show that t f V (t) f (t) V + (v n)f A. (t) his relation is also known as the Reynol s ransport heorem an is a generalization of the Leibniz rule. his proof is taken from [5] (also see [6]). Let 0 be reference configuration of the region (t). Let the motion an the eformation graient be given by x ϕ(x, t) ; F (X, t) 0 ϕ. Let J(X, t) et[f (x, t)]. hen, integrals in the current an the reference configurations are relate by (t) f(x, t) V f[ϕ(x, t), t] J(X, t) V ˆf(X, t) J(X, t) V. 0 0 he time erivative of an integral over a volume is efine as t f(x, t) V (t) lim t 0 1 t (t+ t) f(x, t + t) V f(x, t) V. (t) Converting into integrals over the reference configuration, we get t f(x, t) V (t) Since 0 is inepenent of time, we have t lim t 0 f(x, t) V (t) «1 ˆf(X, t + t) J(X, t + t) V ˆf(X, t) J(X, t) V. t 0 0 Now, the time erivative of et F is given by (see [6], p. 77) J(X, t) " # ˆf(X, t + t) J(X, t + t) ˆf(X, t) J(X, t) V t lim t [ˆf(X, t) J(X, t)] V 0 [ˆf(X, t)] J(X, t) + ˆf(X, t) [J(X, t)] «V (et F ) (et F )( v) J(X, t) v(ϕ(x, t), t) J(X, t) v(x, t). t f(x, t) V (t) «0 [ˆf(X, t)] J(X, t) + ˆf(X, t) J(X, t) v(x, t) V «0 [ˆf(X, t)] + ˆf(X, t) v(x, t) J(X, t) V ḟ(x, t) + f(x, t) v(x, t) V (t) where ḟ is the material time erivative of f. Now, the material erivative is given by f(x, t) ḟ(x, t) + [ f(x, t)] v(x, t). 10

11 Using the ientity we then have t f(x, t) V (t) t t «f(x, t) + [ f(x, t)] v(x, t) + f(x, t) v(x, t) V (t) f V (t) «f (t) + f v + f v V. (v w) v( w) + v w f V (t) «f + (f v) V. (t) Using the ivergence theorem an the ientity (a b) n (b n)a we have t f V (t) f (t) V + f (f v) n V (t) (t) V + (v n)f V. (t) 3. Show that the balance of mass can be expresse as: + v 0 where (x, t) is the current mass ensity, is the material time erivative of, an v(x, t) is the velocity of physical particles in the boy boune by the surface. Recall that the general equation for the balance of a physical quantity f(x, t) is given by» f(x, t) V f(x, t)[u n(x, t) v(x, t) n(x, t)] A + g(x, t) A + h(x, t) V. t o erive the equation for the balance of mass, we assume that the physical quantity of interest is the mass ensity (x, t). Since mass is neither create or estroye, the surface an interior sources are zero, i.e., g(x, t) h(x, t) 0. we have» (x, t) V (x, t)[u n(x, t) v(x, t) n(x, t)] A. t Let us assume that the volume is a control volume (i.e., it oes not change with time). hen the surface has a zero velocity (u n 0) an we get V (v n) A. Using the ivergence theorem we get Since is arbitrary, we must have Using the ientity we have Now, the material time erivative of is efine as v V v n A V ( v) V.» + ( v) V 0. + ( v) 0. (ϕ v) ϕ v + ϕ v + v + v 0. + v. + v 0. 11

12 4. Show that the balance of linear momentum can be expresse as: v σ b 0 where (x, t) is the mass ensity, v(x, t) is the velocity, σ(x, t) is the Cauchy stress, an b is the boy force ensity. Recall the general equation for the balance of a physical quantity» f(x, t) V f(x, t)[u n(x, t) v(x, t) n(x, t)] A + g(x, t) A + h(x, t) V. t In this case the physical quantity of interest is the momentum ensity, i.e., f(x, t) (x, t) v(x, t). he source of momentum flux at the surface is the surface traction, i.e., g(x, t) t. he source of momentum insie the boy is the boy force, i.e., h(x, t) (x, t) b(x, t). we have he surface tractions are relate to the Cauchy stress by Let us assume that is an arbitrary fixe control volume. hen,» v V v[u n v n] A + t A + b V. t t σ n.» v V v[u n v n] A + σ n A + b V. t Now, from the efinition of the tensor prouct we have (for all vectors a) ( v) V v (v n) A + σ n A + b V. (u v) a (a v) u. Using the ivergence theorem we have Since is arbitrary, we have Using the ientity we get Using the ientity we get From the efinition ( v) V (v v) n A + σ n A + b V. v V v n A ( v) V [ (v v)] V + σ V + b V» ( v) + [( v) v] σ b V 0. ( v) + [( v) v] σ b 0. (u v) ( v)u + ( u) v v + v + ( v)(v) + ( v) v σ b 0» + v v + v + ( v) v σ b 0 (ϕ v) ϕ v + v ( ϕ)» + v v + v + [ v + v ( )] v σ b 0 (u v) a (a v) u 12

13 we have Hence, [v ( )] v [v ( )] v.» + v v + v + v v + [v ( )] v σ b 0» + v + v v + v + v v σ b 0. he material time erivative of is efine as From the balance of mass, we have + v. [ + v] v + v + v v σ b 0. + v 0. he material time erivative of v is efine as Hence, v + v v σ b 0. v v + v v. v σ b Show that the balance of angular momentum can be expresse as: σ σ We assume that there are no surface couples on or boy couples in. Recall the general balance equation» f(x, t) V f(x, t)[u n(x, t) v(x, t) n(x, t)] A + g(x, t) A + h(x, t) V. t In this case, the physical quantity to be conserve the angular momentum ensity, i.e., f x ( v). he angular momentum source at the surface is then g x t an the angular momentum source insie the boy is h x ( b). he angular momentum an moments are calculate with respect to a fixe origin. Hence we have» x ( v) V [x ( v)][u n v n] A + x t A + x ( b) V. t Assuming that is a control volume, we have Using the efinition of a tensor prouct we can write» x ( v) V [x ( v)][v n] A + x t A + x ( b) V. [x ( v)][v n] [[x ( v)] v] n. Also, t σ n. herefore we have Using the ivergence theorem, we get» x ( v) V [[x ( v)] v] n A + x (σ n) A + x ( b) V.» x ( v) V [[x ( v)] v] V + x (σ n) A + x ( b) V. o convert the surface integral in the above equation into a volume integral, it is convenient to use inex notation. hus,» x (σ n) A i e ijk x j σ kl n l A A il n l A A n A 13

14 where [ ] i represents the i-th component of the vector. Using the ivergence theorem Differentiating, A il A n A A V V (e ijk x j σ kl ) V. x l x l»» σ kl σ kl A n A e ijk δ jl σ kl + e ijk x j V e ijk σ kj + e ijk x j V ˆeijk σ kj + e ijk x j [ σ] l V. x l x l Expresse in irect tensor notation, where E is the thir-orer permutation tensor.» i A n A h[e : σ ] i + [x ( σ)] i V x (σ n) A i he balance of angular momentum can then be written as i h[e : σ ] i + [x ( σ)] i V h i x (σ n) A E : σ + x ( σ) V.» h i x ( v) V [[x ( v)] v] V + E : σ + x ( σ) V + x ( b) V. Since is an arbitrary volume, we have Using the ientity, we get» x ( v) [[x ( v)] v] + E : σ + x ( σ) + x ( b)» x ( v) σ b [[x ( v)] v] + E : σ. (u v) ( v)u + ( u) v [[x ( v)] v] ( v)[x ( v)] + ( [x ( v)]) v. he secon term on the right can be further simplifie using inex notation as follows. [( [x ( v)]) v] i [( [ (x v)]) v] i herefore we can write he balance of angular momentum then takes the form x l ( e ijk x j v k ) v l» e ijk x j v k v l + x j v k v k v l + x j v l x l x l x l «(e ijk x j v k ) v l + (e ijk δ jl v k v l ) + e ijk x j x l [(x v)( v) + v v + x ( v v)] i [(x v)( v) + x ( v v)] i. [[x ( v)] v] ( v)(x v) + ( v)(x v) + x ( v v).» x ( v) σ b ( v)(x v) ( v)(x v) x ( v v) + E : σ v k x l v l x ( v) + v v σ b ( v)(x v) ( v)(x v) + E : σ x» v + v + v v σ b ( v)(x v) ( v)(x v) + E : σ 14

15 he material time erivative of v is efine as v v + v v. x [ v σ b] x v + ( v)(x v) ( v)(x v) + E : σ. Also, from the conservation of linear momentum Hence, he material time erivative of is efine as Hence, v σ b 0. 0 x v + ( v)(x v) + ( v)(x v) E : σ «+ v + v (x v) E : σ. + v. ( + v)(x v) E : σ 0. From the balance of mass In inex notation, Expaning out, we get Hence, + v 0. E : σ 0. e ijk σ kj 0. σ 12 σ 21 0 ; σ 23 σ 32 0 ; σ 31 σ σ σ 6. Show that the balance of energy can be expresse as: ė σ : ( v) + q s 0 where (x, t) is the mass ensity, e(x, t) is the internal energy per unit mass, σ(x, t) is the Cauchy stress, v(x, t) is the particle velocity, q is the heat flux vect an s is the rate at which energy is generate by sources insie the volume (per unit mass). Recall the general balance equation» f(x, t) V f(x, t)[u n(x, t) v(x, t) n(x, t)] A + g(x, t) A + h(x, t) V. t In this case, the physical quantity to be conserve the total energy ensity which is the sum of the internal energy ensity an the kinetic energy ensity, i.e., f e + 1/2 v v. he energy source at the surface is a sum of the rate of work one by the applie tractions an the rate of heat leaving the volume (per unit area), i.e, g v t q n where n is the outwar unit normal to the surface. he energy source insie the boy is the sum of the rate of work one by the boy forces an the rate of energy generate by internal sources, i.e., h v (b) + s. Hence we have» t v v V v v (u n v n) A + (v t q n) A + (v b + s) V. Let be a control volume that oes not change with time. hen we get» v v V v v (v n) A + (v t q n) A + (v b + s) V. Using the relation t σ n, the ientity v (σ n) (σ v) n, an invoking the symmetry of the stress tens we get» v v V v v (v n) A + (σ v q) n A + (v b + s) V. 15

16 We now apply the ivergence theorem to the surface integrals to get»» v v V v v v A + (σ v) A q A + (v b + s) V. Since is arbitrary, we have Expaning out the left han sie, we have»» v v v v v + (σ v) q + (v b + s).» v v e v v + «(v v) v v + e + v v. For the first term on the right han sie, we use the ientity (ϕ v) ϕ v + ϕ v to get» v v v» v v v + v v v v v v v v v v v v + v v v + v v v v + v v v + e v + 1 (v v) v 2 v + v v v + e v + ( v v) v v + v v v + e v + ( v v) v. For the secon term on the right we use the ientity (S v) S : v + ( S) v an the symmetry of the Cauchy stress tensor to get After collecting terms an rearranging, we get (σ v) σ : v + ( σ) v. «e + v + v + 12 «v v + v ««e + v v σ b v + + e v + σ : v + q s 0. Applying the balance of mass to the first term an the balance of linear momentum to the secon term, an using the material time erivative of the internal energy ė e + e v we get the final form of the balance of energy: 7. Show that the Clausius-Duhem inequality in integral form: ė σ : v + q s 0. «q n η V η (u n v n) A t A + s V. can be written in ifferential form as η q + s. Assume that is an arbitrary fixe control volume. hen u n 0 an the erivative can be take insie the integral to give q n ( η) V η (v n) A A + s V. Using the ivergence theorem, we get ( η) V ( η v) V 16 q s V + V.

17 Since is arbitrary, we must have q ( η) ( η v) + s. Expaning out η + η q (η) v η ( v) η + η q η v η v η ( v) ««η + v + v η + + η v + s + s q + s. Now, the material time erivatives of an η are given by + v ; η η + η v. ( + v) η + η q + s. From the conservation of mass + v 0. Hence, η q + s. 8. Show that the Clausius-Duhem inequality can be expresse in terms of the internal energy as η q + s (ė η) σ : v q Using the ientity (ϕ v) ϕ v + v ϕ in the Clausius-Duhem inequality, we get η q + s Now, using inex notation with respect to a Cartesian basis e j, or η 1 q q 1. + s. 1 ` 1 e j ` 2 e j 1 x j x j 2. Hence, Recall the balance of energy η 1 q q + s or η 1 ( q s) q. ė σ : v + q s 0 ė σ : v ( q s). Rearranging, η 1 1 q ( ė σ : v) + q η ė σ : v +. 2 (ė η) σ : v q 17.

18 9. For thermoelastic materials, the internal energy is a function only of the eformation graient an the temperature, i.e., e e(f, ). Show that, for thermoelastic materials, the Clausius-Duhem inequality can be expresse as Since e e(f, ), we have e F : F + e η «e η «η η σ : v q (ė η) σ : v q η + e «F σ F : F q. ė e F : F + e η η. or «e η Now, v l F F 1. using the ientity A : (B C) (A C ) : B, we have σ : v σ : ( F F 1 ) (σ F ) : F. η + e F : F σ : v q. Hence, «e η «e η η + e F : F (σ F ) : F q η + e «F σ F : F q. 10. Show that, for thermoelastic materials, the balance of energy ė σ : v + q s 0. can be expresse as η q + s. Since e e(f, ), we have Plug into energy equation to get Recall, Hence, e F : F + e η ė e F : F + e η η. η σ : v + q s 0. e e an η F σ F. (σ F ) : F + η σ : v + q s 0. Now, v l F F 1. using the ientity A : (B C) (A C ) : B, we have σ : v σ : ( F F 1 ) (σ F ) : F. Plugging into the energy equation, we have σ : v + η σ : v + q s 0 η q + s. 18

19 11. Show that, for thermoelastic materials, the Cauchy stress can be expresse in terms of the Green strain as Recall that the Cauchy stress is given by σ F e E F. σ e F F σ ij e Fkj F e F jk. ik F ik he Green strain E E(F ) E(U) an e e(f, η) e(u, η). Hence, using the chain rule, e F e E : E F e F ik e E lm E lm F ik. Now, E 1 2 (F F 1 ) E lm 1 2 (F lp Fpm δ lm) 1 2 (F pl F pm δ lm ). aking the erivative with respect to F, we get E F 1 F 2 F F + F F «F E lm 1 «Fpl F pm F pm + F pl. F ik 2 F ik F ik σ 1» e F 2 E : F F + F F «F σ ij 1» «e F 2 Fpl F pm F pm + F pl F jk. E lm F ik F ik Recall, A A A ij A kl δ ik δ jl an A A A ji A kl δ jk δ il. σ ij 1» e 2 `δpi δ lk F pm + F pl δ pi δ mk F jk 1» e E lm 2 (δ lk F im + F il δ mk ) F jk E lm σ ij 1» e 2 F im + E km e F il F jk σ 1 «e "F E lk 2 + F E " σ 1 «# e 2 F + e F. E E # e F E From the symmetry of the Cauchy stress, we have σ (F A) F an σ F (F A) F A F an σ σ A A. an we get «e e E E σ F e E F. 12. For thermoelastic materials, the specific internal energy is given by e ē(e, η) where E is the Green strain an η is the specific entropy. Show that (e η) η + 1 S : Ė t 0 an t (e η 1 S : E) η 1 Ṡ : E 0 0 where 0 is the initial ensity, is the absolute temperature, S is the 2n Piola-Kirchhoff stress, an a ot over a quantity inicates the material time erivative. 19

20 aking the material time erivative of the specific internal energy, we get ė ē E : Ė + ē η η. Now, for thermoelastic materials, ē η an S 0 ē E. ė 1 0 S : Ė + η. ė η 1 0 S : Ė. Now, ė ( η) + η 1 S : Ė t 0 ( η) η + η. t (e η) η + 1 S : Ė. t 0 Also, Hence, t 1 S : E 1 S : Ė + 1 Ṡ : E ė ( η) + η t t 1 S : E 1 Ṡ : E 0 0 t e η 1 0 S : E η 1 0 Ṡ : E. 13. For thermoelastic materials, show that the following relations hol: ψ E 1 0 Ŝ(E, ) ; ψ ˆη(E, ) ; where ψ(e, ) is the Helmholtz free energy an g(s, ) is the Gibbs free energy. Also show that Ŝ ˆη 0 E an g S 1 0 Ẽ(S, ) ; Ẽ η 0 S. g η(s, ) Recall that an ψ(e, ) e η ē(e, η) η. g(s, ) e + η S : E. (Note that we can choose any functional epenence that we like, because the quantities e, η, E are the actual quantities an not any particular functional relations). he erivatives are an Hence, ψ E 1 0 Ŝ(E, ) ; ψ E ē E 1 0 S ; g S 1 S 0 S : E 1 E ; 0 ψ ˆη(E, ) ; ψ η. g η. g S 1 0 Ẽ(S, ) ; g η(s, ) From the above, we have 2 ψ E 2 ψ E ˆη E 1 0 Ŝ. 20

21 an 2 g S 2 g S η S 1 Ẽ 0. Hence, Ŝ ˆη 0 E an Ẽ η 0 S. 14. For thermoelastic materials, show that the following relations hol: ê(e, ) ˆη 2 ˆψ 2 an Recall, ẽ(s, ) η + 1 S : Ẽ 0 2 g 2 + S : 2 g S. ˆψ(E, ) ψ e η ê(e, ) ˆη(E, ) an g(s, ) g e + η S : E ẽ(s, ) + η(s, ) S : Ẽ(S, ). an ẽ(s, ) ê(e, ) ˆψ ˆη + ˆη(E, ) + g η + η(s, ) S : Ẽ 0. Also, recall that ˆη(E, ) ˆψ ˆη 2 ˆψ 2, an Hence, an η(s, ) g Ẽ(S, ) 0 ẽ(s, ) g S ê(e, ) ˆη η 2 g 2, Ẽ 2 g 0 S. 2 ˆψ 2 η + 1 S : Ẽ 0 2 g 2 + S : 2 g S. 15. For thermoelastic materials, show that the balance of energy equation η q + s can be expresse as either or C v (κ ) + s + Ŝ 0 : Ė C p 1 S : Ẽ (κ ) + s Ẽ 0 0 : Ṡ where If the inepenent variables are E an, then C v ê(e, ) an C p ẽ(s, ). η ˆη(E, ) η ˆη E : Ė + ˆη 21.

22 On the other han, if we consier S an to be the inepenent variables η η(s, ) η η S : Ṡ + η. Since we have, either or ˆη E 1 Ŝ 0 ; ˆη Cv ; η S 1 Ẽ η ; an 0 1 η 1 Ŝ 0 : Ė + Cv η 1 0 Ẽ : Ṡ + 1 Cp 1 S : Ẽ. 0 Cp 1 S : Ẽ 0 he equation for balance of energy in terms of the specific entropy is Using the two forms of η, we get two forms of the energy equation: η q + s. Ŝ 0 : Ė + C v q + s an From Fourier s law of heat conuction an Rearranging, Ẽ 0 : Ṡ + C p 0 Ẽ S : Ẽ 0 q κ. q + s. Ŝ 0 : Ė + C v (κ ) + s : Ṡ + C p S : Ẽ 0 C v (κ ) + s. (κ ) + s + Ŝ 0 : Ė C p 1 S : Ẽ (κ ) + s Ẽ 0 0 : Ṡ. 16. For thermoelastic materials, show that the specific heats are relate by the relation C p C v 1 0 S Ŝ : Ẽ. Recall that an Also recall that C v : ê(e, ) C p : ẽ(s, ) ˆη η S : Ẽ. C p C v η S : Ẽ ˆη. η ˆη(E, ) η(s, ). 22

23 keeping S constant while ifferentiating, we have η ˆη E : E + ˆη. Noting that E Ẽ(S, ), an plugging back into the equation for the ifference between the two specific heats, we have C p C v ˆη E : Ẽ S : Ẽ. Recalling that we get C p C v 1 0 ˆη E 1 Ŝ 0 S Ŝ : Ẽ. 17. For thermoelastic materials, show that the specific heats can also be relate by the equations C p C v 1 S : E 0 + E 2 : ψ EE : E «1 S : E 0 + E S 0 : E : E «. Recall that Recall the chain rule which states that if S 0 ψ E 0 f(e(s, ), ). g(u, t) f(x(u, t), y(u, t)) then, if we keep u fixe, the partial erivative of g with respect to t is given by g f x x + f y y. In our case, Hence, we have u S, t, g(s, ) S, x(s, ) E(S, ), y(s, ), an f 0 f. S g(s, ) f(e(s, ), ) 0 f(e(s, ), ). aking the erivative with respect to keeping S constant, we have g S 6 f 0 4 E : E + f 7 5 Now, Again recall that, Plugging into the above, we get f ψ E 0 2 ψ EE : E + 0 f E : E + f. f E 2 ψ E 2 ψ EE an f E ψ E 1 0 S. «ψ : E E + 2 ψ E. 0 2 ψ EE : E S 1 0 S E : E S. «ψ. E 23

24 we get the following relation for S/ : S 2 ψ 0 EE : E S E : E. Recall that Plugging in the expressions for S/ we get: C p C v 1 S S «: E 0. C p C v 1 2 ψ S EE : E «: E 1 S + S 0 E : E «: E. C p C v 1 0 S : E + Using the ientity (A : B) : C C : (A : B), we have 2 ψ EE : E «: E 1 S : E 0 + S 0 E : E «: E. C p C v 1 S : E 0 + E 2 : ψ EE : E «1 S : E 0 + E S 0 : E : E «. 18. Consier an isotropic thermoelastic material that has a constant coefficient of thermal expansion an which follows the St-Venant Kirchhoff moel, i.e, α E α 1 an C λ µ I where α is the coefficient of thermal expansion an 3 λ 3 K 2 µ where K, µ are the bulk an shear mouli, respectively. Show that the specific heats relate by the equation C p C v 1 0 ˆα tr(s) + 9 α 2 K. Recall that, Plugging the expressions of α E an C into the above equation, we have C p C v 1 0 S : α E + 0 α E : C : α E. C p C v 1 0 S : (α 1 ) + 0 (α 1 ) : (λ µ I) : (α 1 ) α 0 tr(s) + α2 0 1 : (λ µ I) : 1 α 0 tr(s) + α2 0 1 : (λ tr(1 ) 1 + 2µ 1 ) α 0 tr(s) + α2 0 (3 λ tr(1 ) + 2µ tr(1 )) α 0 tr(s) + 3 α2 0 (3 λ + 2µ) α tr(s) + 9 α2 K. 0 0 C p C v 1 0 ˆα tr(s) + 9 α 2 K. 24

25 References [1]. W. Wright. he Physics an Mathematics of Aiabatic Shear Bans. Cambrige University Press, Cambrige, UK, [2] R. C. Batra. Elements of Continuum Mechanics. AIAA, Reston, VA., [3] G. A. Maugin. he hermomechanics of Nonlinear Irreversible Behaviors: An Introuction. Worl Scientific, Singapore, [4] M. D. Greenberg. Founations of Applie Mathematics. Prentice-Hall Inc., Englewoo Cliffs, New Jersey, [5]. Belytschko, W. K. Liu, an B. Moran. Nonlinear Finite Elements for Continua an Structures. John Wiley an Sons, Lt., New York, [6] M. E. Gurtin. An Introuction to Continuum Mechanics. Acaemic Press, New York,