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{align}

\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{align}$$ Using the symmetry of the stiffness tensor, we have,
 * $$\begin{align}

\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{align}$$ Therefore,

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