# Introduction to Elasticity/Constitutive example 3

## Example 3

Given: The strain energy density for a material undergoing small strain

${\text{(1)}}\qquad U({\boldsymbol {\varepsilon }})=\int _{0}^{\boldsymbol {\varepsilon }}{\boldsymbol {\sigma }}:d{\boldsymbol {\varepsilon }}~.$ Show: For linear elastic deformations and small strains,

${\text{(2)}}\qquad U({\boldsymbol {\varepsilon }})={\frac {1}{2}}{\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}~.$ ### Solution

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

${\text{(3)}}\qquad \sigma _{ij}={\frac {\partial U({\boldsymbol {\varepsilon }})}{\partial \varepsilon _{ij}}}$ We will show that equation (2) is equivalent to equation (3). We start off with equation (2) and work backward.

${\text{(4)}}\qquad U({\boldsymbol {\varepsilon }})={\frac {1}{2}}\sigma _{ij}\varepsilon _{ij}$ For linear elastic materials,

${\text{(5)}}\qquad \sigma _{ij}=C_{ijkl}\varepsilon _{kl}$ Substituting equation (5) into equation (4), we get,

${\text{(6)}}\qquad U({\boldsymbol {\varepsilon }})={\frac {1}{2}}C_{ijkl}\varepsilon _{kl}\varepsilon _{ij}$ Recall that, for a second order tensor ${\boldsymbol {A}}\,$ ,

${\frac {\partial A_{ij}}{\partial A_{kl}}}=\delta _{ik}\delta _{jl}$ and that for a fourth order rensor ${\mathsf {C}}\,$ (substitution rule),

$C_{ijkl}\delta _{ir}=C_{rjkl}\,$ Differentiating equation (6) with respect to $\varepsilon _{rs}\,$ , we have,

{\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,

{\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,

$\sigma _{ij}={\frac {\partial U({\boldsymbol {\varepsilon }})}{\partial \varepsilon _{ij}}}$ which is the same as equation (3). Hence shown.