= Solution
\b[Compression and stretch viewed on rotated faces.] To isolate shear, take equal small principal strains $\epsilon_{xx}=-e$, $\epsilon_{yy}=+e$. A square with sides at $45^\circ$ to these directions becomes a rhombus: adjacent side lengths are unchanged to first order, but their right angle changes by $2e$. In axes rotated through $45^\circ$, the same strain tensor has zero diagonal entries and off-diagonal entries of magnitude $e$.
The <pure shear from opposite principal strains> therefore has engineering <shear strain> $\gamma=2e$, not $e$. Equal opposite principal stresses $(-S,+S)$ give purely tangential traction of magnitude $S$ on those rotated faces. Hence
$$
\boxed{\gamma=2e,\qquad \tau=S=G\gamma=2Ge.}
$$
If the compression and stretch are not equal and opposite, an additional dilatational strain remains; their deviatoric part still supplies shear.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1-c12-pure-shear.png]
{title=Original comparison of compression and transverse extension with the same stress resolved as shear on faces rotated through 45 degrees; deformation is exaggerated}
\b[The shaft's torsional stiffness.] At radius $r$, an end twist $\phi$ over length $L$ gives tangential displacement $r\phi$ and shear angle $r\phi/L$. The local <shear stress> is $Gr\phi/L$. An annular area $dA=2\pi r\,dr$ contributes torque $r\tau\,dA$, so
$$
C=\int_0^a r\frac{Gr\phi}{L}\,2\pi r\,dr
=\frac{2\pi G\phi}{L}\frac{a^4}{4}.
$$
Thus <torsion of a solid circular shaft> gives
$$
\boxed{C=\frac{\pi a^4G}{2L}\phi=\frac{GJ_p}{L}\phi,\qquad
J_p=\frac{\pi a^4}{2}.}
$$
Here $J_p$ is the <polar second moment of area>, in $\mathrm{m^4}$, not a mass moment of inertia. This $C$ is the transmitted torque magnitude; the shaft's restoring couple opposes the twist. The formula assumes small elastic shear and a circular uniform section.
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