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

${ \rho~\left(\frac{\partial \bar{e}}{\partial \eta} - T\right)~\dot{\eta} + \left(\rho~\frac{\partial \bar{e}}{\partial \boldsymbol{F}} - \boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}\right):\dot{\boldsymbol{F}} + \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} \le 0 ~. }$

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,

${ T = \frac{\partial \bar{e}}{\partial \eta} } \qquad\text{and}\qquad { \boldsymbol{\sigma} = \rho~\frac{\partial \bar{e}}{\partial \boldsymbol{F}}\cdot\boldsymbol{F}^T ~. }$

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

${ \rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. }$

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

${ \boldsymbol{E} = \frac{1}{2}(\boldsymbol{F}^T\cdot\boldsymbol{F} - \boldsymbol{\mathit{1}}) = \frac{1}{2}(\boldsymbol{U}^2 - \boldsymbol{\mathit{1}}) ~. }$

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

${ \boldsymbol{\sigma} = \rho~\boldsymbol{F}\cdot\frac{\partial \bar{e}}{\partial \boldsymbol{E}}\cdot\boldsymbol{F}^T ~. }$

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,

${ \boldsymbol{P} = \cfrac{\rho_0}{\rho}~\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}~~\text{and}~~ \boldsymbol{N} = \cfrac{\rho_0}{\rho}~\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma} }$

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 ($\boldsymbol{S}$):

${ \boldsymbol{S} := \boldsymbol{F}^{-1}\cdot\boldsymbol{P} = \boldsymbol{N}\cdot\boldsymbol{F}^{-T} = \cfrac{\rho_0}{\rho}~\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}~. }$

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,

${ \boldsymbol{S} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}} } ~;~~ { \boldsymbol{P} = \rho_0~\boldsymbol{F}\cdot\frac{\partial \bar{e}}{\partial \boldsymbol{E}} }~;~~ \boldsymbol{N} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}}\cdot\boldsymbol{F}^T$

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

${ \frac{\rho_0}{\rho}~\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{P}:\dot{\boldsymbol{F}} = \boldsymbol{N^T}:\dot{\boldsymbol{F}} = \boldsymbol{S}:\dot{\boldsymbol{E}} ~. }$

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

${ \frac{\rho_0}{\rho}~\text{tr}~(\boldsymbol{\sigma}\cdot\mathbf{d}) = \text{tr}~(\boldsymbol{P}^T\cdot\dot{\boldsymbol{F}}) = \text{tr}~(\boldsymbol{N}\cdot\dot{\boldsymbol{F}}) =\text{tr}~(\boldsymbol{S}\cdot\dot{\boldsymbol{E}}) ~. }$

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

${ \psi = \hat{\psi}(\boldsymbol{E}, T) := e - T~\eta ~. }$

We define the Gibbs free energy as

${ g = \tilde{g}(\boldsymbol{S}, T) := - e + T~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E} ~. }$

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

${ C_v := \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} ~. }$

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

${ C_p := \frac{\partial \tilde{e}(\boldsymbol{S}, T)}{\partial T} ~. }$

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

${ \begin{align} \rho~C_v~\dot{T} & = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s +\cfrac{\rho}{\rho_0}~T~\boldsymbol{\beta}_S:\dot{\boldsymbol{E}} \\ \rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{\alpha}_E\right) ~\dot{T} & = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\boldsymbol{\alpha}_E:\dot{\boldsymbol{S}} \end{align} }$

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

${ C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{\alpha}_E + \cfrac{T}{\rho_0}~\boldsymbol{\alpha}_E:\boldsymbol{\mathsf{C}}:\boldsymbol{\alpha}_E }$

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

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

References

1. T. W. Wright. The Physics and Mathematics of Adiabatic Shear Bands. Cambridge University Press, Cambridge, UK, 2002.
2. R. C. Batra. Elements of Continuum Mechanics. AIAA, Reston, VA., 2006.
3. G. A. Maugin. The Thermomechanics of Nonlinear Irreversible Behaviors: An Introduction. World Scientific, Singapore, 1999.