Continuum mechanics/Stress-strain relation for thermoelasticity

Relation between Cauchy stress and Green strain Show that, for thermoelastic materials, the Cauchy stress can be expressed in terms of the Green strain as $\mathit{\sigma }=\rho 𝑭\cdot \frac{\partial e}{\partial 𝑬}\cdot {𝑭}^T.$

Proof:

Recall that the Cauchy stress is given by

$\mathit{\sigma }=\rho \frac{\partial e}{\partial 𝑭}\cdot {𝑭}^T⟹{\sigma }_{ij}=\rho \frac{\partial e}{\partial F_{ik}}{F}_{kj}^T=\rho \frac{\partial e}{\partial F_{ik}}{F}_{jk}.$

The Green strain $𝑬=𝑬(𝑭)=𝑬(𝑼)$ and $e=e(𝑭,\eta )=e(𝑼,\eta )$. Hence, using the chain rule,

$\frac{\partial e}{\partial 𝑭}=\frac{\partial e}{\partial 𝑬}:\frac{\partial 𝑬}{\partial 𝑭}⟹\frac{\partial e}{\partial F_{ik}}=\frac{\partial e}{\partial E_{lm}}\frac{\partial E_{lm}}{\partial F_{ik}}.$

Now,

$𝑬=\frac{1}{2}({𝑭}^T\cdot 𝑭-1)⟹E_{lm}=\frac{1}{2}(F_{lp}^T{F}_{pm}-{\delta }_{lm})=\frac{1}{2}(F_{pl}{F}_{pm}-{\delta }_{lm}).$

Taking the derivative with respect to $𝑭$, we get

$\frac{\partial 𝑬}{\partial 𝑭}=\frac{1}{2}(\frac{\partial {𝑭}^T}{\partial 𝑭}\cdot 𝑭+{𝑭}^T\cdot \frac{\partial 𝑭}{\partial 𝑭})⟹\frac{\partial E_{lm}}{\partial F_{ik}}=\frac{1}{2}(\frac{\partial F_{pl}}{\partial F_{ik}}{F}_{pm}+F_{pl}\frac{\partial F_{pm}}{\partial F_{ik}}).$

Therefore,

$\mathit{\sigma }=\frac{1}{2}\rho [\frac{\partial e}{\partial 𝑬}:(\frac{\partial {𝑭}^T}{\partial 𝑭}\cdot 𝑭+{𝑭}^T\cdot \frac{\partial 𝑭}{\partial 𝑭})]\cdot {𝑭}^T⟹{\sigma }_{ij}=\frac{1}{2}\rho \left[\frac{\partial e}{\partial E_{lm}}(\frac{\partial F_{pl}}{\partial F_{ik}}{F}_{pm}+F_{pl}\frac{\partial F_{pm}}{\partial F_{ik}})\right]F_{jk}.$

Recall,

$\frac{\partial 𝑨}{\partial 𝑨}\equiv \frac{\partial A_{ij}}{\partial A_{kl}}={\delta }_{ik}{\delta }_{jl}\text{and}\frac{\partial {𝑨}^T}{\partial 𝑨}\equiv \frac{\partial A_{ji}}{\partial A_{kl}}={\delta }_{jk}{\delta }_{il}.$

Therefore,

${\sigma }_{ij}=\frac{1}{2}\rho \left[\frac{\partial e}{\partial E_{lm}}({\delta }_{pi}{\delta }_{lk}{F}_{pm}+F_{pl}{\delta }_{pi}{\delta }_{mk})\right]F_{jk}=\frac{1}{2}\rho \left[\frac{\partial e}{\partial E_{lm}}({\delta }_{lk}{F}_{im}+F_{il}{\delta }_{mk})\right]F_{jk}$

or,

${\sigma }_{ij}=\frac{1}{2}\rho [\frac{\partial e}{\partial E_{km}}{F}_{im}+\frac{\partial e}{\partial E_{lk}}{F}_{il}]F_{jk}⟹\mathit{\sigma }=\frac{1}{2}\rho [𝑭\cdot {\left(\frac{\partial e}{\partial 𝑬}\right)}^T+𝑭\cdot \frac{\partial e}{\partial 𝑬}]\cdot {𝑭}^T$

or,

$\mathit{\sigma }=\frac{1}{2}\rho 𝑭\cdot [{\left(\frac{\partial e}{\partial 𝑬}\right)}^T+\frac{\partial e}{\partial 𝑬}]\cdot {𝑭}^T.$

From the symmetry of the Cauchy stress, we have

$\mathit{\sigma }=(𝑭\cdot 𝑨)\cdot {𝑭}^T\text{and}{\mathit{\sigma }}^T=𝑭\cdot (𝑭\cdot 𝑨{)}^T=𝑭\cdot {𝑨}^T\cdot {𝑭}^T\text{and}\mathit{\sigma }={\mathit{\sigma }}^T⟹𝑨={𝑨}^T.$

Therefore,

$\frac{\partial e}{\partial 𝑬}={\left(\frac{\partial e}{\partial 𝑬}\right)}^T$

and we get

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

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