Elasticity/Equilibrium example 1

Example 1
 Given:

Euler's second law for the conservation of angular momentum
 * $$ \text{(1)} \qquad

\int_{\partial B} e_{ijk}~x_j~n_l~\sigma_{lk}~dS + \int_B \rho~e_{ijk}~x_j~b_k~dV = \frac{d}{dt}\left(\int_B \rho~e_{ijk}~x_j~v_k~dV \right) $$ The divergence theorem
 * $$ \text{(2)} \qquad

\int_{\partial B} n_i~\sigma_{ij}~dS = \int_B \frac{\partial \sigma_{ij}}{\partial x_i}~dV $$ The equilibrium equation (Cauchy's first law)
 * $$ \text{(3)} \qquad

\frac{\partial \sigma_{ij}}{\partial x_i} + \rho~b_j = \frac{d}{dt}\left(\rho~v_j\right) $$

 Show:


 * $$ \text{(4)} \qquad

\sigma_{ij} = \sigma_{ji} $$

Solution
Let us first look at the first term of equation~(1) and apply the divergence theorem (2). Thus,
 * $$\begin{align}

\int_{\partial B} e_{ijk}~x_j~n_l~\sigma_{lk}~dS & = \int_{\partial B} n_l~(e_{ijk}~x_j~\sigma_{lk})~dS \\ & = \int_{B} \frac{\partial{(e_{ijk}~x_j~\sigma_{lk})}}{\partial x_l}~dV \\ & = \int_{B} \left(e_{ijk}~\frac{\partial x_j}{\partial x_l}~\sigma_{lk} +              e_{ijk}~x_j~\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV\\ & = \int_{B} \left(e_{ijk}~\delta_{jl}~\sigma_{lk} +              e_{ijk}~x_j\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV\\ & = \int_{B} \left(e_{ilk}~\sigma_{lk} +              e_{ijk}~x_j\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV \end{align}$$ Plugging this back into equation~(1) gives

\int_{B} \left(e_{ilk}~\sigma_{lk} + e_{ijk}~x_j\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV + \int_B \rho~e_{ijk}~x_j~b_k~dV = \frac{d}{dt}\left(\int_B \rho~e_{ijk}~x_j~v_k~dV \right) $$ Therefore, bringing all terms to the left hand side,
 * $$\text{(5)} \qquad

\int_{B} \left[e_{ilk}~\sigma_{lk} + e_{ijk}~x_j\left(\frac{\partial\sigma_{lk}}{\partial x_l}  + \rho~b_k  - \frac{d}{dt}\left(\rho~v_k\right)\right)\right]~dV = 0 $$ Using the equilibrium equations~(3), equation~(5) reduces to
 * $$\text{(6)} \qquad

\int_{B} e_{ilk}~\sigma_{lk}~dV = 0 $$ Since this holds for any $$B$$, we have
 * $$\text{(7)} \qquad

e_{ilk}~\sigma_{lk} = 0 $$ If you work this expression out, you will see that $$\sigma_{ij} = \sigma_{ji}$$. Hence, the stress tensor is symmetric.