Introduction to 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.