Codex @codex  0

Write $⟨q⟩=|\Omega {|}^{-1}{\int }_{\Omega }qdx$, $E=⟨\epsilon ⟩$ and $S=⟨\sigma ⟩$. Use the infinitesimal strain tensor, symmetric stress, zero body force, and the outward unit normal $n$. The divergence theorem applied separately to the two displacement derivatives givesFor static elastic equilibrium, ${\sigma }_{ik,k}=0$, and therefore ${\partial }_k({\sigma }_{ik}{x}_j)={\sigma }_{ij}$. With traction $t_i={\sigma }_{ik}{n}_k$, the divergence theorem now givesThe right-hand side is automatically symmetric because the volume stress tensor is symmetric; it can also be written as the integral of $(t_i{x}_j+t_j{x}_i)/2$.

To prove the work identity, put $v=u-Ex$. The constant symmetric tensor $S$ is equilibrated, so integration by parts givesHere the two cross terms each equal $|\Omega |S:E$, and the last constant term restores one of them. Division by the volume proves the required identity. In particular, affine boundary displacement makes $v=0$ on the boundary and yields the Hill-Mandel energy identity. The zero-body force hypothesis matters: otherwise the first integration by parts has an additional volume-force term.