Elasticity/Energy methods example 1

Example 1

Given:

${\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)+\mathrm{\Delta }\mathit{\sigma }(𝐱)]-{\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)]=\underset{ℛ}{\int }[U^c(\mathit{\sigma }+\mathrm{\Delta }\mathit{\sigma })-U^c(\mathit{\sigma })-\mathit{\epsilon }:\mathrm{\Delta }\mathit{\sigma }]dV$

Show:

${\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)+\mathrm{\Delta }\mathit{\sigma }(𝐱)]-{\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)]=\underset{ℛ}{\int }{U}^c(\mathrm{\Delta }\mathit{\sigma })dV$

Solution

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

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

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

Therefore,

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

or (using the symmetry of the compliance tensor),

$\begin{array}{cc}{U}^c(\mathit{\sigma }+\mathrm{\Delta }\mathit{\sigma })& =\frac{1}{2}[{\sigma }_{ij}{\sigma }_{kl}+{\sigma }_{ij}\mathrm{\Delta }{\sigma }_{kl}+{\sigma }_{kl}\mathrm{\Delta }{\sigma }_{ij}+\mathrm{\Delta }{\sigma }_{ij}\mathrm{\Delta }{\sigma }_{kl}]S_{ijkl}\\ & =\frac{1}{2}[{\sigma }_{ij}{S}_{ijkl}{\sigma }_{kl}+{\sigma }_{ij}{S}_{ijkl}\mathrm{\Delta }{\sigma }_{kl}+{\sigma }_{kl}{S}_{ijkl}\mathrm{\Delta }{\sigma }_{ij}+\mathrm{\Delta }{\sigma }_{ij}{S}_{ijkl}\mathrm{\Delta }{\sigma }_{kl}]\\ & =\frac{1}{2}[{\sigma }_{ij}{S}_{ijkl}{\sigma }_{kl}+{\epsilon }_{kl}\mathrm{\Delta }{\sigma }_{kl}+{\epsilon }_{ij}\mathrm{\Delta }{\sigma }_{ij}+\mathrm{\Delta }{\sigma }_{ij}{S}_{ijkl}\mathrm{\Delta }{\sigma }_{kl}]\\ & =\frac{1}{2}[{\sigma }_{ij}{S}_{ijkl}{\sigma }_{kl}+2{\epsilon }_{kl}\mathrm{\Delta }{\sigma }_{kl}+\mathrm{\Delta }{\sigma }_{ij}{S}_{ijkl}\mathrm{\Delta }{\sigma }_{kl}]\\ & =\frac{1}{2}\mathit{\sigma }:\text{S}:\mathit{\sigma }+\mathit{\epsilon }:\mathrm{\Delta }\mathit{\sigma }+\frac{1}{2}\mathrm{\Delta }\mathit{\sigma }:\text{S}:\mathrm{\Delta }\mathit{\sigma }\\ & =U^c(\mathit{\sigma })+\mathit{\epsilon }:\mathrm{\Delta }\mathit{\sigma }+U^c(\mathrm{\Delta }\mathit{\sigma })\end{array}$

Therefore,

$U^c(\mathit{\sigma }+\mathrm{\Delta }\mathit{\sigma })=U^c(\mathit{\sigma })+\mathit{\epsilon }:\mathrm{\Delta }\mathit{\sigma }+U^c(\mathrm{\Delta }\mathit{\sigma })$

Plugging into the given equation

${\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)+\mathrm{\Delta }\mathit{\sigma }(𝐱)]-{\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)]=\underset{ℛ}{\int }[U^c(\mathit{\sigma })+\mathit{\epsilon }:\mathrm{\Delta }\mathit{\sigma }+U^c(\mathrm{\Delta }\mathit{\sigma })-U^c(\mathit{\sigma })-\mathit{\epsilon }:\mathrm{\Delta }\mathit{\sigma }]dV$

or,

${\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)+\mathrm{\Delta }\mathit{\sigma }(𝐱)]-{\mathrm{\Pi }}^c[\mathit{\sigma }(𝐱)]=\underset{ℛ}{\int }{U}^c(\mathrm{\Delta }\mathit{\sigma })dV$

Hence shown.