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

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 $\alpha$.