Elasticity/Constitutive example 3

Example 3

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

(1) U (ε) = \int_{0}^{ε} σ : d ε .

Show: For linear elastic deformations and small strains,

(2) U (ε) = \frac{1}{2} σ : ε .

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

(3) σ_{i j} = \frac{\partial U (ε)}{\partial ε_{i j}}

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

(4) U (ε) = \frac{1}{2} σ_{i j} ε_{i j}

For linear elastic materials,

(5) σ_{i j} = C_{i j k l} ε_{k l}

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

(6) U (ε) = \frac{1}{2} C_{i j k l} ε_{k l} ε_{i j}

Recall that, for a second order tensor $𝑨$ ,

\frac{\partial A_{i j}}{\partial A_{k l}} = δ_{i k} δ_{j l}

and that for a fourth order rensor $𝖢$ (substitution rule),

C_{i j k l} δ_{i r} = C_{r j k l}

Differentiating equation (6) with respect to $ε_{r s}$ , we have,

\begin{matrix} \frac{\partial U (ε)}{\partial ε_{r s}} & = \frac{1}{2} C_{i j k l} ε_{k l} δ_{i r} δ_{j s} + \frac{1}{2} C_{i j k l} ε_{i j} δ_{k r} δ_{l s} \\ = \frac{1}{2} C_{r s k l} ε_{k l} + \frac{1}{2} C_{i j r s} ε_{i j} \end{matrix}

Using the symmetry of the stiffness tensor, we have,

\begin{matrix} \frac{\partial U (ε)}{\partial ε_{r s}} & = \frac{1}{2} C_{r s k l} ε_{k l} + \frac{1}{2} C_{r s i j} ε_{i j} \\ = \frac{1}{2} σ_{r s} + \frac{1}{2} σ_{r s} \\ = σ_{r s} \end{matrix}

Therefore,

σ_{i j} = \frac{\partial U (ε)}{\partial ε_{i j}}

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