Solution

If the strain energy density is given by equation (1), then (for linear elastic materials) the stress and strain can be related using

${\displaystyle {\text{(3)}}\qquad \sigma _{ij}={\frac {\partial U({\boldsymbol {\varepsilon }})}{\partial \varepsilon _{ij}}}}$

We will show that equation (2) is equivalent to equation (3). We start off with equation (2) and work backward.

${\displaystyle {\text{(4)}}\qquad U({\boldsymbol {\varepsilon }})={\frac {1}{2}}\sigma _{ij}\varepsilon _{ij}}$

For linear elastic materials,

${\displaystyle {\text{(5)}}\qquad \sigma _{ij}=C_{ijkl}\varepsilon _{kl}}$

Substituting equation (5) into equation (4), we get,

${\displaystyle {\text{(6)}}\qquad U({\boldsymbol {\varepsilon }})={\frac {1}{2}}C_{ijkl}\varepsilon _{kl}\varepsilon _{ij}}$

Recall that, for a second order tensor ${\displaystyle {\boldsymbol {A}}\,}$,

${\displaystyle {\frac {\partial A_{ij}}{\partial A_{kl}}}=\delta _{ik}\delta _{jl}}$

and that for a fourth order rensor ${\displaystyle {\mathsf {C}}\,}$ (substitution rule),

${\displaystyle C_{ijkl}\delta _{ir}=C_{rjkl}\,}$

Differentiating equation (6) with respect to ${\displaystyle \varepsilon _{rs}\,}$, we have,

${\displaystyle {\begin{aligned}{\frac {\partial U({\boldsymbol {\varepsilon }})}{\partial \varepsilon _{rs}}}&={\frac {1}{2}}C_{ijkl}\varepsilon _{kl}\delta _{ir}\delta _{js}+{\frac {1}{2}}C_{ijkl}\varepsilon _{ij}\delta _{kr}\delta _{ls}\\&={\frac {1}{2}}C_{rskl}\varepsilon _{kl}+{\frac {1}{2}}C_{ijrs}\varepsilon _{ij}\end{aligned}}}$

Using the symmetry of the stiffness tensor, we have,

${\displaystyle {\begin{aligned}{\frac {\partial U({\boldsymbol {\varepsilon }})}{\partial \varepsilon _{rs}}}&={\frac {1}{2}}C_{rskl}\varepsilon _{kl}+{\frac {1}{2}}C_{rsij}\varepsilon _{ij}\\&={\frac {1}{2}}\sigma _{rs}+{\frac {1}{2}}\sigma _{rs}\\&=\sigma _{rs}\end{aligned}}}$

Therefore,

${\displaystyle \sigma _{ij}={\frac {\partial U({\boldsymbol {\varepsilon }})}{\partial \varepsilon _{ij}}}}$

which is the same as equation (3). Hence shown.