= Solution
At small load the deformation is elastic: $\varepsilon_{\parallel}=\sigma/E$, and, using isotropic elasticity with $E=2G(1+\nu)$, $\nu=0.30$ gives lateral contraction $\varepsilon_\perp=-0.30\varepsilon_\parallel$. At maximum load,
$$
\boxed{\sigma_{\max}=\frac{12.6\,\mathrm{kN}}{\pi(2\,\mathrm{mm})^2}=1.003\,\mathrm{GPa}}.
$$
For the specified <slip system>, the tensile direction is proportional to $(-2,-3,5)$, the slip direction to $(1,-1,1)$ and the plane normal to $(1,0,-1)$. Hence the <Schmid factor> is
$$
m=\frac{|6|}{\sqrt{38}\sqrt3}\frac{|-7|}{\sqrt{38}\sqrt2}
=0.4512,\qquad \boxed{|\tau_{\max}|=0.452\,\mathrm{GPa}}.
$$
<Schmid's law> predicts glide when this reaches the <critical resolved shear stress>. Before that, unloading reverses the <strain>; after activation, glide produces permanent <strain> and may generate <work hardening> and additional slip. \b[The supplied data do not fix the yield load.] Neither the adjective “ultra-strong” nor $E/G$ supplies a critical shear <stress> or an absolute modulus. Other <slip systems> must also be checked before claiming which activates first. In particular, the specimen remains elastic through the maximum load if every relevant system stays below its activation threshold.
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