= Solution
{c}
Only the $k=0$ term of the <polynomial> contributes to the limit:
$$
\frac{M(\tau)}\tau
=\frac{A^4}{2}(\mu_1+3^n\mu_2)+O(\tau^2).
$$
Thus the requested <torsional stiffness per unit twist> is
$$
\boxed{\alpha=\frac{A^4}{2}(\mu_1+3^n\mu_2).}
$$
The small-<strain> <shear modulus> is $G=2\hat w'(3)=2(\mu_1+3^n\mu_2)$. Therefore $\alpha=GA^4/4$, consistently with the usual circular-shaft <torque> relation after restoring the $2\pi$ factor. If the prescribed end rotation is the <angle> $\varphi=\tau L$, the physical <torsional stiffness> is $dN/d\varphi=2\pi\alpha/L$ at zero twist. This also explains why $L$ does not enter the requested $\alpha$.
Back to article page
