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
![{\displaystyle {\boldsymbol {\sigma }}=\rho ~{\boldsymbol {F}}\cdot {\frac {\partial e}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e365222d92611dbeb330ebe46b0b1d07648920d)
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Proof:
Recall that the Cauchy stress is given by
![{\displaystyle {\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_{kj}^{T}=\rho ~{\frac {\partial e}{\partial F_{ik}}}F_{jk}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb3905f057298e098babe0e3d6bb61c3c9d09c91)
The Green strain
and
. Hence, using the chain rule,
![{\displaystyle {\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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/454acee8a4cb6184994dc288f3f0acff63c3f701)
Now,
![{\displaystyle {\boldsymbol {E}}={\frac {1}{2}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})\qquad \implies \qquad E_{lm}={\frac {1}{2}}(F_{lp}^{T}~F_{pm}-\delta _{lm})={\frac {1}{2}}(F_{pl}~F_{pm}-\delta _{lm})~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de253877f271d387268278ace7916088d534dcbd)
Taking the derivative with respect to
, we get
![{\displaystyle {\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)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8525ab50ef9148253b55ca6f348a552a2a9af9fe)
Therefore,
![{\displaystyle {\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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a06b9baf5b1d812e964fd026b251f56bdddb8c87)
Recall,
![{\displaystyle {\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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ced0432a67947e8fc3003f9aaad2b3724dd81f23)
Therefore,
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/169dd125d3830d61445485f1838ef1e494fc716c)
or,
![{\displaystyle \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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a32ed5e79da6a66cb9a5b3a9eb98b2e802f65c02)
or,
![{\displaystyle {\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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/904a753aaf2024ad1e653e59121c8335e88ccf1d)
From the symmetry of the Cauchy stress, we have
![{\displaystyle {\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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73d3edf3cadeaeab6443de104162e2a591c23dd1)
Therefore,
![{\displaystyle {\frac {\partial e}{\partial {\boldsymbol {E}}}}=\left({\frac {\partial e}{\partial {\boldsymbol {E}}}}\right)^{T}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50f467a6557043da8d9d74d3338248938b500919)
and we get
![{\displaystyle {{\boldsymbol {\sigma }}=~\rho ~{\boldsymbol {F}}\cdot {\frac {\partial e}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd195aa3d26743cc2cfb9e7ecbd44b76f48fc281)