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

Write the first strain invariant in components of the deformation gradient:Here lower-case and upper-case indices refer to the current configuration and reference configuration, respectively. Differentiating gives $\partial I_1/\partial F_{iI}=2F_{iI}$. The chain rule therefore givesSubstitution into the given hyperelastic material law, with its indicated matrix transpose, yieldsThe left Cauchy-Green deformation tensor is $\mathbf{B}=\mathbf{F}{\mathbf{F}}^T$. The scalar $p$ is the Lagrange multiplier enforcing incompressibility, rather than a function fixed by the strain energy density alone. The identity matrix $1$ makes this an isotropic constraint reaction in hyperelastic stress.

Use separate moving orthonormal bases in the reference and current cylindrical coordinate systems. For pure torsion of an incompressible cylinder, the deformation map isThe unit vectors depend on the angle: ${\partial }_{\theta }{\mathbf{e}}_r={\mathbf{e}}_{\theta }$ and ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-{\mathbf{e}}_r$. ConsequentlyIn particular, the last derivative includes the rotation of ${\mathbf{e}}_r$ along the cylinder. The material gradient has the reference coordinate scale factor $1/R$:The outer products specify the images of the three reference unit vectors. Thus, with current basis vectors indexing rows and reference basis vectors indexing columns,Its determinant is one, checking incompressibility. The radius is unchanged, so the current outer radius in part (d) is $a=A$. The formula at the coordinate axis is understood by continuity in Cartesian coordinates.

Multiplying the deformation gradient by its matrix transpose gives the left Cauchy-Green deformation tensor and its trace:The derivative of the strain energy density isFor brevity put $g(r)={\mu }_1+{\mu }_2(3+{\tau }^2{r}^2{)}^n$. Part (a) gives the Cauchy stress tensor, in the current cylindrical coordinate system orthonormal basis, asIn particular, the end-face shear stress is $T_{\theta z}=2g\tau r$. The undetermined pressure $p$ affects the normal stresses but does not affect this shear stress. No normal traction conditions have been specified to determine $p$.

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 $\alpha$.