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# Continuum mechanics/Stress-strain relation for thermoelasticity: Wikis

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# Study guide

Up to date as of January 14, 2010

### From Wikiversity

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

Proof:

Recall that the Cauchy stress is given by

$\boldsymbol{\sigma} = \rho~\frac{\partial e}{\partial \boldsymbol{F}}\cdot\boldsymbol{F}^T \qquad \implies \qquad \sigma_{ij} = \rho~\frac{\partial e}{\partial F_{ik}}F^T_{kj} = \rho~\frac{\partial e}{\partial F_{ik}}F_{jk} ~.$

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

$\frac{\partial e}{\partial \boldsymbol{F}} = \frac{\partial e}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial \boldsymbol{F}} \qquad \implies \qquad \frac{\partial e}{\partial F_{ik}} = \frac{\partial e}{\partial E_{lm}}~\frac{\partial E_{lm}}{\partial F_{ik}} ~.$

Now,

$\boldsymbol{E} = \frac{1}{2}(\boldsymbol{F}^T\cdot\boldsymbol{F} - \boldsymbol{\mathit{1}}) \qquad \implies \qquad E_{lm} = \frac{1}{2}(F^T_{lp}~F_{pm} - \delta_{lm}) = \frac{1}{2}(F_{pl}~F_{pm} - \delta_{lm}) ~.$

Taking the derivative with respect to $\boldsymbol{F}$, we get

$\frac{\partial \boldsymbol{E}}{\partial \boldsymbol{F}} = \frac{1}{2}\left(\frac{\partial \boldsymbol{F}^T}{\partial \boldsymbol{F}}\cdot\boldsymbol{F} + \boldsymbol{F}^T\cdot\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{F}}\right) \qquad \implies \qquad \frac{\partial E_{lm}}{\partial F_{ik}} = \frac{1}{2}\left(\frac{\partial F_{pl}}{\partial F_{ik}}~F_{pm} + F_{pl}~\frac{\partial F_{pm}}{\partial F_{ik}}\right) ~.$

Therefore,

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

Recall,

$\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} \equiv \frac{\partial A_{ij}}{\partial A_{kl}} = \delta_{ik}~\delta_{jl} \qquad \text{and} \qquad \frac{\partial \boldsymbol{A}^T}{\partial \boldsymbol{A}} \equiv \frac{\partial A_{ji}}{\partial A_{kl}} = \delta_{jk}~\delta_{il} ~.$

Therefore,

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

or,

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

or,

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

From the symmetry of the Cauchy stress, we have

$\boldsymbol{\sigma} = (\boldsymbol{F}\cdot\boldsymbol{A})\cdot\boldsymbol{F}^T \qquad \text{and} \qquad \boldsymbol{\sigma}^T = \boldsymbol{F}\cdot(\boldsymbol{F}\cdot\boldsymbol{A})^T = \boldsymbol{F}\cdot\boldsymbol{A}^T\cdot\boldsymbol{F}^T \qquad \text{and} \qquad \boldsymbol{\sigma} = \boldsymbol{\sigma}^T \implies \boldsymbol{A} = \boldsymbol{A}^T ~.$

Therefore,

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

and we get

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