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2.2 Thermoelasticity

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2.2 Thermoelasticity Section 2.2 2.2 Thermoelasticity In this section, thermoelasticity is considered. By definition, the constitutive relations for such a material depend only on the set of field variables {F,θ,Gradθ}. This general case will be considered toward the end of this section. Simpler cases, isothermal {}F , small- strain {}ε , one-dimensional {}ε , will be considered first. In what follows, upper case letters will be used to denote the energy potentials per unit volume. As before, lower case denotes per unit mass so, for example, U = ρu for internal energy. 2.2.1 Isothermal Small Strain Elasticity For isothermal deformations, the basic thermodynamic statement can be written as (see Eqn. 1.6.24) σ : ε& = ρψ& + ρφ (2.2.1) = Ψ& + Φ where Ψ& = ρψ& (rate of change of the free energy per unit volume) Φ = ρφ ≥ 0 (dissipation per unit volume) In isothermal elasticity1, Φ = 0, and so one need only specify a free energy function. One-Dimensional Analysis Consider a simple one-dimensional extension of material. The free energy is a function of kinematic state variables. The only state variable is the single strain ε and the only stress is σ and hence σε& = Ψ& (ε ) (2.2.2) i.e. dΨ(ε ) σε& = ε& (2.2.3) dε so that, since ε& is arbitrary, dΨ(ε ) σ = (2.2.4) dε 1 this is discussed more rigorously in §2.2.2 below Solid Mechanics Part IV 57 Kelly Section 2.2 which agrees with the general relation 1.6.33d. This is the stress-strain (constitutive) equation. In the context of elasticity theory, the Helmholtz free energy function is the elastic strain energy function (Eqn. 2.2.4 is the same as Part I, Eqn. 5.4.9). The classical linear elastic material corresponds to choosing a simple quadratic free energy function: 1 Ψ()ε = Eε 2 (2.2.5) 2 so σ = Eε , σ& = Eε& (2.2.6) A nonlinear example is ε 2 ε 3 ε 4 Ψ(ε ) = E ε + E + E + E 0 1 2 2 3 3 4 σ (ε ) = E + E ε + E ε 2 + E ε 3 0 1 2 3 (2.2.7) σ& = ()∂σ / ∂ε ε& 2 = ()E1 + 2E2ε + 3E3ε ε& One should ensure that (at least (1)) (1) σ (0) = 0 (zero strain at zero stress) which implies that E0 = 0 (2) ()dσ / dε ε =0 ≠ 0 (so one has a non-zero initial modulus) 2 2 (3) (d σ / dε )ε =0 = 0 (so that, with (2), the origin is a point of inflection, ensuring similar tensile and compressive strain behaviour) Note that Ψ has the dimensions of stress, so that the Ei also have the dimensions of stress. In the rate formulation: σ& = Dε& , D is the instantaneous elastic modulus (or stiffness) dσ / de or d 2 Ψ / dε 2 . D is a constant in the linear model, but is, in general, a function of 2 strain, as in the above example, where D = E1 + 2E2ε + 3E3ε . Three-dimensional Analysis Generalising the above to three dimensions, the free energy function is Ψ(ε ) Ψ(ε) (2.2.8) ij so that the basic thermomechanical identity becomes ∂Ψ dΨ σ ijε&ij = Ψ& (ε ij ) = ε&ij σ : ε& = Ψ& (ε) = : ε& (2.2.9) ∂ε ij dε Solid Mechanics Part IV 58 Kelly Section 2.2 Since this again holds for all strain rates one can deduce that ∂Ψ dΨ σ ij = σ = (2.2.10) ∂ε ij dε Again, for the linear elastic material, the free energy function is assumed to be a quadratic function of the strain: 1 1 Ψ = 2 Dijkl ε ijε kl Ψ = 2 ε : D : ε (2.2.11) so that ∂Ψ σ mn = ∂ε mn ∂ε 1 ij 1 ∂ε kl = 2 Dijkl ε kl + 2 Dijklε ij (2.2.12) ∂ε mn ∂ε mn 1 1 = 2 Dmnklε kl + 2 Dijmnε ij = Dmnklε kl provided the fourth order stiffness tensor D possesses the major symmetry: Dijkl = Dklij (2.2.13) (D satisfies the minor symmetries Dijkl = D jikl and Dijkl = Dijlk by virtue of the symmetry of the stress and strain.) Eqn. 2.1.12 is the stress-strain equation for an anisotropic linear elastic material. The rate form is formally σ& ij = Dijkl ε&kl (2.2.14) For a nonlinear material Ψ is no longer quadratic, and the instantaneous modulus tensor is a function of the strains. If one starts by proposing the rate form of the elastic law, with a given modulus function D()ε , then in general one cannot integrate Eqn. 2.2.14 to find a free energy function (although one can achieve this in one-dimension). Such a model is termed hypoelastic. A model in which Ψ does exist is called hyperelastic. Hyperelastic models automatically satisfy the laws of thermomechanics, hypoelastic ones do not. Note that the strain energy must be positive, Ψ > 0. From 2.2.11, this implies that Dijkl ε ijε kl > 0 ε : D : ε > 0 (2.2.15) for all strain tensors, that is, the stiffness tensor must be positive definite. Solid Mechanics Part IV 59 Kelly Section 2.2 Isotropic Elasticity The theory of isotropic elasticity with small strains is a special case of that for large strains discussed in Part III, §4.4. The theory discussed there applies here, only now with the left Cauchy Green strain replaced by the small strain tensor. Thus, the free energy function must be a function of a set of three invariants of the strain tensor ε , Ψ(E1 , E2 , E3 ) , which here are taken to be E1 = ε ii = trε 2 E2 = ε ijε ij = trε (2.2.16) 3 E3 = ε ijε jk ε ki = trε The stress-strain relation is hence given by: ∂Ψ σ = ij ∂ε ij (2.2.17) ∂Ψ ∂E ∂Ψ ∂E ∂Ψ ∂E = 1 + 2 + 3 ∂E1 ∂ε ij ∂E2 ∂ε ij ∂E3 ∂ε ij It is readily verified that {▲Problem 1} ∂E1 / ∂ε ij = δ ij ∂E1 / ∂ε = I ∂E2 / ∂ε ij = 2ε ij ∂E2 / ∂ε = 2ε (2.2.18) 2 ∂E3 / ∂ε ij = 3ε ik ε kj ∂E3 / ∂ε = 3ε so that ∂Ψ ∂Ψ ∂Ψ 2 σ = I + 2 ε + 3 ε (2.2.19) ∂E1 ∂E2 ∂E3 which is a quadratic polynomial in the strain tensor. If the free energy function does not depend on the third strain invariant E3 , the tensor relation is actually linear, though it should be appreciated that the coefficients ∂Ψ /,∂Ek k = 1,2,3, will in general depend in a nonlinear manner on the strain invariants. The classical linear model is recovered by choosing 1 2 1 2 2 Ψ = 2 λE1 + μE2 Ψ = 2 λ(trε) + μtrε (2.2.20) for then σ ij = λε kkδ ij + 2με ij σ = λ(trε)I + 2με (2.2.21) Solid Mechanics Part IV 60 Kelly Section 2.2 The Gibbs Function and the Complementary Energy From section 1.3, the Helmholtz free energy and the Gibbs energy functions are related by a Legendre transformation. Formally, consider again the one-dimensional stress-strain relation 2.2.4: dΨ(ε ) σ = (2.2.22) dε To invert this expression and express the strain as a function of stress, first partially integrate the equation with respect to strain to obtain σε = Ψ(ε ) + C(σ ) (2.2.23) where C(σ ) is an arbitrary function of integration, dependent only on stress. Differentiating with respect to σ leads to dC(σ ) ε = (2.2.24) dσ which is the inverse of the original equation. From Table 1.3.1, it is clear that the function C is the negative of the Gibbs energy function (per unit volume). In this context, C is the complementary energy discussed in Part I, §5.4. One says that the free energy is the dual of the complementary energy, and vice versa. For example, if Ψ = Kε 3 , σ = 3Kε 2 (2.2.25) then {▲Problem 2} 3 C(σ ) = 2K(σ / 3K) 2 . (2.2.26) This theory readily extends to the general three dimensional situation. The original stress strain equation is given by 2.2.10 so that σ ijε ij = Ψ(ε ij ) + C(σ ij ) σ : ε = Ψ(ε) + C(σ) (2.2.27) and the strains are now dC(σ ij ) dC(σ) ε ij = , ε = (2.2.28) dσ ij dσ 1 −1 For the linear material, Ψ = 2 ε : D : ε , ε = D : σ and the dual Complementary function 1 −1 is C = 2 σ : D : σ . The fact that D in the linear case is positive definite ensures that it is invertible. The inverse D−1 is called the elastic flexibility tensor C (see 2.2.11): Solid Mechanics Part IV 61 Kelly Section 2.2 1 1 C = 2 Cijklσ ijσ kl C = 2 σ : C : σ (2.2.29) 2.2.2 Large Strain Thermoelasticity In the general large strain thermoelastic case, the constitutive relations are of the form σ = σ(F,θ,∇θ ) q = q()F,θ,∇θ u = u()F,θ,∇θ (2.2.30) ψ =ψ ()F,θ,∇θ s = s()F,θ,∇θ Here, for convenience, the material gradient of 2.1.13 has been replaced with the spatial gradient ∇θ (through ∇θ = F −1 Gradθ ). The dissipation inequality 1.6.20 is 1 1 φ = −sθ& −ψ& + σ : d − q ⋅∇θ ≥ 0 (2.2.31) ρ ρθ Using the relation Part III, 3.7.25, σ : d = σF −T : F& , this can be expressed as −T 1 ρsθ& + ρψ& − σF : F& + q ⋅∇θ ≤ 0 (2.2.32) θ With ∂ψ ∂ψ ∂ψ ⋅ ψ& = : F& + θ& + ⋅∇θ (2.2.33) ∂F ∂θ ∂()∇θ one has ⎛ ∂ψ ⎞ ⎛ ∂ψ ⎞ ∂ψ ⋅ 1 ρ⎜ s + ⎟θ& − ⎜σF −T − ρ ⎟ : F& + ρ ⋅∇θ + q ⋅∇θ ≤ 0 (2.2.34) ⎝ ∂θ ⎠ ⎝ ∂F ⎠ ∂()∇θ θ For any given {}F,θ,∇θ , this inequality must hold for any arbitrary values of F& ,θ& and ⋅ ∇θ .
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