Past exam of the physics B course of the University of Cambridge / 2018 / Paper 1 / Section C / C12 / Solution

Compression and stretch viewed on rotated faces. To isolate shear, take equal small principal strains ${ϵ}_{xx}=-e$, ${ϵ}_{yy}=+e$. 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. HenceIf the compression and stretch are not equal and opposite, an additional dilatational strain remains; their deviatoric part still supplies shear.

The shaft's torsional stiffness. At radius $r$, an end twist $\varphi$ over length $L$ gives tangential displacement $r\varphi$ and shear angle $r\varphi /L$. The local shear stress is $Gr\varphi /L$. An annular area $dA=2\pi rdr$ contributes torque $r\tau dA$, soThus torsion of a solid circular shaft givesHere $J_p$ 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.