Continuum mechanics/Thermoelasticity

Thermoelastic materials
A set of constitutive equations is required to close to system of balance laws. These are relations between appropriate kinematic quantities and stress measures that can be assigned a physical meaning.

Deformation gradient as the strain measure
In thermoelasticity we assume that the fundamental kinematic quantity is the deformation gradient ($$\boldsymbol{F}$$) which is given by

\boldsymbol{F} = \frac{\partial \mathbf{x}}{\partial \mathbf{X}} = \boldsymbol{\nabla}_{\circ} \mathbf{x} ~; \det\boldsymbol{F} > 0 ~. $$ A thermoelastic material is one in which the internal energy ($$e$$) is a function only of $$\boldsymbol{F}$$ and the specific entropy ($$\eta$$), that is

e = \bar{e}(\boldsymbol{F}, \eta) ~. $$ For a thermoelastic material, we can show that the entropy inequality can be written as At this stage, we make the following constitutive assumptions: 1) Like the internal energy, we assume that $$\boldsymbol{\sigma}$$ and $$T$$ are also functions only of $$\boldsymbol{F}$$ and $$\eta$$, i.e.,

\boldsymbol{\sigma} = \boldsymbol{\sigma}(\boldsymbol{F}, \eta) ~; T = T(\boldsymbol{F}, \eta) ~. $$

2) The heat flux $$\mathbf{q}$$ satisfies the thermal conductivity inequality and    if $$\mathbf{q}$$ is independent of $$\dot{\eta}$$ and $$\dot{\boldsymbol{F}}$$, we have

\mathbf{q}\cdot\boldsymbol{\nabla} T \le 0 \qquad\implies\qquad -(\boldsymbol{\kappa}\cdot\boldsymbol{\nabla} T)\cdot\boldsymbol{\nabla} T \le 0 \qquad\implies\qquad \boldsymbol{\kappa}\ge\mathbf{0} $$ i.e., the thermal conductivity $$\boldsymbol{\kappa}$$ is positive semidefinite. Therefore, the entropy inequality may be written as

\rho~\left(\frac{\partial \bar{e}}{\partial \eta} - T\right)~\dot{\eta} + \left(\rho~\frac{\partial \bar{e}}{\partial \boldsymbol{F}} - \boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}\right):\dot{\boldsymbol{F}} \le 0 ~. $$ Since $$\dot{\eta}$$ and $$\dot{\boldsymbol{F}}$$ are arbitrary, the entropy inequality will be satisfied if and only if

\frac{\partial \bar{e}}{\partial \eta} - T = 0 \implies T = \frac{\partial \bar{e}}{\partial \eta} \qquad\text{and}\qquad \rho~\frac{\partial \bar{e}}{\partial \boldsymbol{F}} - \boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T} = \mathbf{0} \implies \boldsymbol{\sigma} = \rho~\frac{\partial \bar{e}}{\partial \boldsymbol{F}}\cdot\boldsymbol{F}^T ~. $$ Therefore,

Given the above relations, the energy equation may expressed in terms of the specific entropy as

Effect of a rigid body rotation of the internal energy
If a thermoelastic body is subjected to a rigid body rotation $$\boldsymbol{Q}$$, then its internal energy should not change. After a rotation, the new deformation gradient ($$\hat{\boldsymbol{F}}$$) is given by

\hat{\boldsymbol{F}} = \boldsymbol{Q}\cdot\boldsymbol{F} ~. $$   Since the internal energy does not change, we must have

e = \bar{e}(\hat{\boldsymbol{F}}, \eta) = \bar{e}(\boldsymbol{F}, \eta) ~. $$   Now, from the polar decomposition theorem,  $$\boldsymbol{F} = \boldsymbol{R}\cdot\boldsymbol{U}$$ where $$\boldsymbol{R}$$ is the orthogonal rotation tensor (i.e., $$\boldsymbol{R}\cdot\boldsymbol{R}^T = \boldsymbol{R}^T\cdot\boldsymbol{R} = \boldsymbol{\mathit{1}}$$) and $$\boldsymbol{U}$$ is the symmetric right stretch tensor. Therefore,

\bar{e}(\boldsymbol{Q}\cdot\boldsymbol{R}\cdot\boldsymbol{U}, \eta) = \bar{e}(\boldsymbol{F}, \eta) ~. $$   We can choose any rotation $$\boldsymbol{Q}$$. In particular, if we choose    $$\boldsymbol{Q} = \boldsymbol{R}^T$$, we have

\bar{e}(\boldsymbol{R}^T\cdot\boldsymbol{R}\cdot\boldsymbol{U}, \eta) =         \bar{e}(\boldsymbol{\mathit{1}}\cdot\boldsymbol{U}, \eta) = \tilde{e}(\boldsymbol{U}, \eta)~. $$   Therefore,

\bar{e}(\boldsymbol{U}, \eta) = \bar{e}(\boldsymbol{F}, \eta) ~. $$   This means that the internal energy depends only on the stretch  $$\boldsymbol{U}$$ and not on the orientation of the body.

Other strain and stress measures
The internal energy depends on $$\boldsymbol{F}$$ only through the stretch $$\boldsymbol{U}$$. A strain measure that reflects this fact and also vanishes in the reference configuration is the Green strain Recall that the Cauchy stress is given by

\boldsymbol{\sigma} = \rho~\frac{\partial \bar{e}}{\partial \boldsymbol{F}}\cdot\boldsymbol{F}^T ~. $$ We can show that the Cauchy stress can be expressed in terms of the Green strain as Also, recall that the first Piola-Kirchhoff stress tensor is defined as

\boldsymbol{P} = J~(\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T})~\text{where}~ J = \det\boldsymbol{F} $$ Alternatively, we may use the nominal stress tensor

\boldsymbol{N} = J~(\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma}) $$ From the conservation of mass, we have $$\rho_0 = \rho~\det\boldsymbol{F}$$. Hence, The first P-K stress and the nominal stress are unsymmetric. Also recall that we can define a symmetric stress measure with respect to the reference configuration called the second Piola-Kirchhoff stress tensor ($$\boldsymbol{S}$$): In terms of the derivatives of the internal energy, we have

\boldsymbol{S} = \cfrac{\rho_0}{\rho}~\boldsymbol{F}^{-1}\cdot \left(\rho~\boldsymbol{F}\cdot\frac{\partial \bar{e}}{\partial \boldsymbol{E}}\cdot\boldsymbol{F}^T\right)\cdot\boldsymbol{F}^{-T} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}} $$ Therefore,

\boldsymbol{P} = \rho_0~\boldsymbol{F}\cdot\frac{\partial \bar{e}}{\partial \boldsymbol{E}} ~. $$ and

\boldsymbol{N} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}}\cdot\boldsymbol{F}^T ~. $$ That is,

Stress Power
The stress power per unit volume is given by $$\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v}$$. In terms of the stress measures in the reference configuration, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \left(\rho~\boldsymbol{F}\cdot\frac{\partial \bar{e}}{\partial \boldsymbol{E}}\cdot\boldsymbol{F}^T\right): (\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}) ~. $$ Using the identity $$\boldsymbol{A}:(\boldsymbol{B}\cdot\boldsymbol{C}) = (\boldsymbol{A}\cdot\boldsymbol{C}^T):\boldsymbol{B}$$, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \left[ \left(\rho~\boldsymbol{F}\cdot\frac{\partial \bar{e}}{\partial \boldsymbol{E}}\cdot\boldsymbol{F}^T\right)\cdot\boldsymbol{F}^{-T} \right] :\dot{\boldsymbol{F}} = \rho~\left(\boldsymbol{F}\cdot\frac{\partial \bar{e}}{\partial \boldsymbol{E}}\right):\dot{\boldsymbol{F}} = \cfrac{\rho}{\rho_0}~\boldsymbol{P}:\dot{\boldsymbol{F}} = \cfrac{\rho}{\rho_0}~\boldsymbol{N}^T:\dot{\boldsymbol{F}}~. $$ We can alternatively express the stress power in terms of $$\boldsymbol{S}$$ and $$\dot{\boldsymbol{E}}$$. Taking the material time derivative of $$\boldsymbol{E}$$ we have

\dot{\boldsymbol{E}} = \frac{1}{2}(\dot{\boldsymbol{F}^T}\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\dot{\boldsymbol{F}}) ~. $$ Therefore,

\boldsymbol{S}:\dot{\boldsymbol{E}} = \frac{1}{2}[\boldsymbol{S}:(\dot{\boldsymbol{F}^T}\cdot\boldsymbol{F}) + \boldsymbol{S}:(\boldsymbol{F}^T\cdot\dot{\boldsymbol{F}})]~. $$ Using the identities $$\boldsymbol{A}:(\boldsymbol{B}\cdot\boldsymbol{C}) = (\boldsymbol{A}\cdot\boldsymbol{C}^T):\boldsymbol{B} = (\boldsymbol{B}^T\cdot\boldsymbol{A}):\boldsymbol{C}$$ and $$\boldsymbol{A}:\boldsymbol{B} = \boldsymbol{A}^T:\boldsymbol{B}^T$$ and using the symmetry of $$\boldsymbol{S}$$, we have

\boldsymbol{S}:\dot{\boldsymbol{E}} = \frac{1}{2}[(\boldsymbol{S}\cdot\boldsymbol{F}^T):\dot{\boldsymbol{F}}^T + (\boldsymbol{F}\cdot\boldsymbol{S}):\dot{\boldsymbol{F}}] = \frac{1}{2}[(\boldsymbol{F}\cdot\boldsymbol{S}^T):\dot{\boldsymbol{F}} + (\boldsymbol{F}\cdot\boldsymbol{S}):\dot{\boldsymbol{F}}] = (\boldsymbol{F}\cdot\boldsymbol{S}):\dot{\boldsymbol{F}} ~. $$ Now, $$\boldsymbol{S} = \boldsymbol{F}^{-1}\cdot\boldsymbol{P}$$. Therefore, $$\boldsymbol{S}:\dot{\boldsymbol{E}} = \boldsymbol{N}^T:\dot{\boldsymbol{F}}$$. Hence, the stress power can be expressed as

If we split the velocity gradient into symmetric and skew parts using

\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{l} = \mathbf{d} + \boldsymbol{w} $$ where $$\mathbf{d}$$ is the rate of deformation tensor and $$\boldsymbol{w}$$ is the spin tensor, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\sigma}:\mathbf{d} + \boldsymbol{\sigma}:\boldsymbol{w} = \text{tr}~(\boldsymbol{\sigma}^T\cdot\mathbf{d}) + \text{tr}~(\boldsymbol{\sigma}^T\cdot\boldsymbol{w}) = \text{tr}~(\boldsymbol{\sigma}\cdot\mathbf{d}) + \text{tr}~(\boldsymbol{\sigma}\cdot\boldsymbol{w}) ~. $$ Since $$\boldsymbol{\sigma}$$ is symmetric and $$\boldsymbol{w}$$ is skew, we have $$\text{tr}~(\boldsymbol{\sigma}\cdot\boldsymbol{w}) = 0$$. Therefore, $$\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \text{tr}~(\boldsymbol{\sigma}\cdot\mathbf{d})$$. Hence, we may also express the stress power as

Helmholtz and Gibbs free energy
Recall that

\boldsymbol{S} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}} ~. $$ Therefore,

\frac{\partial \bar{e}}{\partial \boldsymbol{E}} = \cfrac{1}{\rho_0}~\boldsymbol{S} ~. $$ Also recall that

\frac{\partial \bar{e}}{\partial \eta} = T ~. $$ Now, the internal energy $$e = \bar{e}(\boldsymbol{E}, \eta)$$ is a function only of the Green strain and the specific entropy. Let us assume, that the above relations can be uniquely inverted locally at a material point so that we have

\boldsymbol{E} = \tilde{\boldsymbol{E}}(\boldsymbol{S}, T) \qquad\text{and}\qquad \eta = \tilde{\eta}(\boldsymbol{S}, T) ~. $$ Then the specific internal energy, the specific entropy, and the stress can also be expressed as functions of $$\boldsymbol{S}$$ and $$T$$, or $$\boldsymbol{E}$$ and $$T$$, i.e.,

e = \bar{e}(\boldsymbol{E}, \eta) = \tilde{e}(\boldsymbol{S}, T) = \hat{e}(\boldsymbol{E}, T)~; \qquad \eta = \tilde{\eta}(\boldsymbol{S},T) = \hat{\eta}(\boldsymbol{E},T) ~; \qquad\text{and}\qquad \boldsymbol{S} = \hat{\boldsymbol{S}}(\boldsymbol{E},T) $$ We can show that

\cfrac{d}{dt}(e - T~\eta) = - \dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} \qquad\text{or}\qquad \cfrac{d\psi}{dt} = - \dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} ~. $$ and

\cfrac{d}{dt}(e - T~\eta - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}) = - \dot{T}~\eta - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} \qquad\text{or}\qquad \cfrac{dg}{dt} = \dot{T}~\eta + \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} ~. $$ We define the  Helmholtz free energy as

We define the  Gibbs free energy as

The functions $$\hat{\psi}(\boldsymbol{E},T)$$ and $$\tilde{g}(\boldsymbol{S},T)$$ are unique. Using these definitions it can be shown that

\frac{\partial \hat{\psi}}{\partial \boldsymbol{E}} = \cfrac{1}{\rho_0}~\hat{\boldsymbol{S}}(\boldsymbol{E},T) ~; \frac{\partial \hat{\psi}}{\partial T} = -\hat{\eta}(\boldsymbol{E},T) ~; \frac{\partial \tilde{g}}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}~\tilde{\boldsymbol{E}}(\boldsymbol{S}, T) ~; \frac{\partial \tilde{g}}{\partial T} = \tilde{\eta}(\boldsymbol{S}, T)   $$ and

\frac{\partial \hat{\boldsymbol{S}}}{\partial T} = - \rho_0~\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} \qquad\text{and}\qquad \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} = \rho_0~\frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}} ~. $$

Specific Heats
The  specific heat at constant strain (or constant volume) is defined as

The  specific heat at constant stress (or constant pressure) is defined as

We can show that

C_v = T~\frac{\partial \hat{\eta}}{\partial T} = -T~\frac{\partial^2 \hat{\psi}}{\partial T^2} $$ and

C_p = T~\frac{\partial \tilde{\eta}}{\partial T} + \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} = T~\frac{\partial^2 \tilde{g}}{\partial T^2} + \boldsymbol{S}:\frac{\partial^2 \tilde{g}}{\partial \boldsymbol{S}\partial T}~. $$ Also the equation for the balance of energy can be expressed in terms of the specific heats as

where

\boldsymbol{\beta}_S := \frac{\partial \hat{\boldsymbol{S}}}{\partial T}    \qquad \text{and} \qquad \boldsymbol{\alpha}_E := \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. $$ The quantity $$\boldsymbol{\beta}_S$$ is called the  coefficient of thermal stress and the quantity $$\boldsymbol{\alpha}_E$$ is called the coefficient of thermal expansion.

The difference between $$C_p$$ and $$C_v$$ can be expressed as

C_p - C_v = \cfrac{1}{\rho_0}\left(\boldsymbol{S}-T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}\right): \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. $$ However, it is more common to express the above relation in terms of the elastic modulus tensor as

where the  fourth-order tensor of elastic moduli is defined as

\boldsymbol{\mathsf{C}} := \frac{\partial \hat{\boldsymbol{S}}}{\partial \tilde{\boldsymbol{E}}} = \rho_0~\frac{\partial^2 \hat{\psi}}{\partial \tilde{\boldsymbol{E}}\partial\tilde{\boldsymbol{E}}} ~. $$ For isotropic materials with a constant coefficient of thermal expansion that follow the St. Venant-Kirchhoff material model, we can show that

C_p - C_v = \cfrac{1}{\rho_0}\left[\alpha~\text{tr}{\boldsymbol{S}} + 9~\alpha^2~K~T\right]~. $$