# 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
    $\mathbf {q} \cdot {\boldsymbol {\nabla }}T\leq 0\qquad \implies \qquad -({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla }}T)\cdot {\boldsymbol {\nabla }}T\leq 0\qquad \implies \qquad {\boldsymbol {\kappa }}\geq \mathbf {0}$
i.e., the thermal conductivity ${\boldsymbol {\kappa }}$ is positive semidefinite.
Therefore, the entropy inequality may be written as
    $\rho ~\left({\frac {\partial {\bar {e}}}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}\leq 0~.$
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
    $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.$