Elasticity/Energy methods example 1

Given:

${\displaystyle \Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )+\Delta {\boldsymbol {\sigma }}(\mathbf {x} )]-\Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )]=\int _{\mathcal {R}}[U^{c}({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})-U^{c}({\boldsymbol {\sigma }})-{\boldsymbol {\varepsilon }}:\Delta {\boldsymbol {\sigma }}]~dV}$

Show:

${\displaystyle \Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )+\Delta {\boldsymbol {\sigma }}(\mathbf {x} )]-\Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )]=\int _{\mathcal {R}}U^{c}(\Delta {\boldsymbol {\sigma }})~dV}$

For a linear elastic material, the complementary strain energy density is given by

${\displaystyle U^{c}({\boldsymbol {\sigma }})={\frac {1}{2}}{\boldsymbol {\sigma }}:{\text{S}}:{\boldsymbol {\sigma }}}$

where ${\displaystyle {\text{S}}}$ is the compliance tensor.

Therefore,

${\displaystyle U^{c}({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})={\frac {1}{2}}({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }}):{\text{S}}:({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})={\frac {1}{2}}(\sigma _{ij}+\Delta \sigma _{ij})S_{ijkl}(\sigma _{kl}+\Delta \sigma _{kl})}$

or (using the symmetry of the compliance tensor),

${\displaystyle {\begin{aligned}U^{c}({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})&={\frac {1}{2}}\left[\sigma _{ij}\sigma _{kl}+\sigma _{ij}\Delta \sigma _{kl}+\sigma _{kl}\Delta \sigma _{ij}+\Delta \sigma _{ij}\Delta \sigma _{kl}\right]S_{ijkl}\\&={\frac {1}{2}}\left[\sigma _{ij}S_{ijkl}\sigma _{kl}+\sigma _{ij}S_{ijkl}\Delta \sigma _{kl}+\sigma _{kl}S_{ijkl}\Delta \sigma _{ij}+\Delta \sigma _{ij}S_{ijkl}\Delta \sigma _{kl}\right]\\&={\frac {1}{2}}\left[\sigma _{ij}S_{ijkl}\sigma _{kl}+\varepsilon _{kl}\Delta \sigma _{kl}+\varepsilon _{ij}\Delta \sigma _{ij}+\Delta \sigma _{ij}S_{ijkl}\Delta \sigma _{kl}\right]\\&={\frac {1}{2}}\left[\sigma _{ij}S_{ijkl}\sigma _{kl}+2\varepsilon _{kl}\Delta \sigma _{kl}+\Delta \sigma _{ij}S_{ijkl}\Delta \sigma _{kl}\right]\\&={\frac {1}{2}}{\boldsymbol {\sigma }}:{\text{S}}:{\boldsymbol {\sigma }}+{\boldsymbol {\varepsilon }}:\Delta {\boldsymbol {\sigma }}+{\frac {1}{2}}\Delta {\boldsymbol {\sigma }}:{\text{S}}:\Delta {\boldsymbol {\sigma }}\\&=U^{c}({\boldsymbol {\sigma }})+{\boldsymbol {\varepsilon }}:\Delta {\boldsymbol {\sigma }}+U^{c}(\Delta {\boldsymbol {\sigma }})\end{aligned}}}$

Therefore,

${\displaystyle U^{c}({\boldsymbol {\sigma }}+\Delta {\boldsymbol {\sigma }})=U^{c}({\boldsymbol {\sigma }})+{\boldsymbol {\varepsilon }}:\Delta {\boldsymbol {\sigma }}+U^{c}(\Delta {\boldsymbol {\sigma }})}$

Plugging into the given equation

${\displaystyle \Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )+\Delta {\boldsymbol {\sigma }}(\mathbf {x} )]-\Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )]=\int _{\mathcal {R}}[U^{c}({\boldsymbol {\sigma }})+{\boldsymbol {\varepsilon }}:\Delta {\boldsymbol {\sigma }}+U^{c}(\Delta {\boldsymbol {\sigma }})-U^{c}({\boldsymbol {\sigma }})-{\boldsymbol {\varepsilon }}:\Delta {\boldsymbol {\sigma }}]~dV}$

or,

${\displaystyle {\Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )+\Delta {\boldsymbol {\sigma }}(\mathbf {x} )]-\Pi ^{c}[{\boldsymbol {\sigma }}(\mathbf {x} )]=\int _{\mathcal {R}}U^{c}(\Delta {\boldsymbol {\sigma }})~dV}}$

Hence shown.