 Mean stress and strain boundary identities

Apply the divergence theorem to ${\partial }_k(x_i{\sigma }_{jk})$ and use momentum balance ${\sigma }_{jk,k}=\rho (a_j-b_j)$ and stress symmetry to obtain the displayed mean stress identity. Applying the same theorem to the infinitesimal strain tensor gives $V{\overline{e}}_{ij}=\frac{1}{2}{\int }_S(u_i{n}_j+u_j{n}_i)dS$. The acceleration term cannot be dropped during a transient.