 Energy release from an antiplane square-root coefficient

Use antiplane energy $\frac{\mu }{2}\int |\nabla u{|}^2$ and normalize the tip singularity as $u=k\sqrt{r}\sin (\theta /2)$ plus a less singular term. The energy release rate is the small-circle flux $\int [\mathcal{W}{n}_x-\mu u_r{u}_x],r,d\theta$. Here $\mathcal{W}=\mu k^2/(8r)$, $u_r=k\sin (\theta /2)/(2\sqrt{r})$ and $u_x=-k\sin (\theta /2)/(2\sqrt{r})$. Integrating over $(-\pi ,\pi )$ gives the displayed coefficient. The full three-dimensional antiplane stress has squared Frobenius norm $2{\mu }^2|\nabla u{|}^2$, so its leading norm ahead of the tip is $\mu |k|/\sqrt{2r}$. These constants depend on the normalization of the stress intensity factor.