Past exam of the solid mechanics course of the University of Oxford / 2023 / 1 / d

Using $a=A$ and the end-face shear stress from part (c), the printed torque normalization givesThe binomial theorem expands the remaining integral into powers of $r$:Integrating each term gives an expression valid also at $\tau =0$:For example, $n=1$ gives $M=({\mu }_1+3{\mu }_2)A^4\tau /2+{\mu }_2{A}^6{\tau }^3/3$. The torque is an odd polynomial in the twist per unit reference length, as expected when reversing the twist reverses the end-face shear stress.

For a physical end-face torque $N$, integration over the azimuth normally givesThe original PDF explicitly defines $M$ without this $2\pi$, so the boxed answer retains its convention. Since $\mathbf{T}$ is the Cauchy stress tensor, its shear component here is spatial $T_{\theta z}$; the printed capital $Z$ in $T_{\theta Z}$ does not require a change to a nominal stress tensor.

Only the $k=0$ term of the polynomial contributes to the limit:Thus the requested torsional stiffness per unit twist isThe small-strain shear modulus is $G=2{\hat{w}}^{\prime }(3)=2({\mu }_1+3^n{\mu }_2)$. Therefore $\alpha =G{A}^4/4$, consistently with the usual circular-shaft torque relation after restoring the $2\pi$ factor. If the prescribed end rotation is the angle $\phi =\tau L$, the physical torsional stiffness is $dN/d\phi =2\pi \alpha /L$ at zero twist. This also explains why $L$ does not enter the requested $\alpha$.