= Energy release from an antiplane square-root coefficient
{title2=$G=\pi\mu k^2/4$}
Use antiplane energy $\frac\mu2\int|\nabla u|^2$ 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>.
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