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An incompressible Mooney-Rivlin solid has strain energy density $W=C_1(I_1-3)+C_2(I_2-3)$, with right Cauchy-Green deformation tensor $C=F^{T}F$, $I_1=\mathrm{tr}C$ and $I_2=\mathrm{tr}{C}^{-1}$ when $\det C=1$. It extends the neo-Hookean solid by a second invariant. A convention $W={\mu }_1(I_1-3)/2-{\mu }_2(I_2-3)/2$ means $C_1={\mu }_1/2$, $C_2=-{\mu }_2/2$; the minus sign must be retained when comparing material constants.

Use $z={\lambda }_z>0$. In cylindrical orthonormal bases the principal stretches are $r_{,R}$, $r/R$ and $z$. Incompressibility givesIntegrate from the inner surface to obtain the inflation and extension of an incompressible tube kinematicsIn particular $b^2=a^2+(b_0^2-a_0^2)/z$. The radial principal stretch is $1/(z\lambda )$, so the reference-volume strain energy density is $\hat{W}(\lambda ,z)$ with the three arguments $(1/(z\lambda ),\lambda ,z)$.

For reference length $L_0$, the stored energy is $L_0\mathcal{E}$, whereLet $N_{\mathrm{wall}}$ denote the total axial resultant acting on the material annulus. The work of the inner pressure on the cylindrical wall is $2\pi az{L}_{0}p\dot{a}$, and axial work is $N_{\mathrm{wall}}{L}_0\dot{z}$. Reversible virtual work therefore saysThe independent derivatives of the circumferential stretch areEquating the coefficients of $\dot{a}$ and $\dot{z}$ provesand the corresponding axial resultant isHere ${\hat{W}}_{,z}$ holds $\lambda$ fixed. If $N$ means the independently applied load on pressurized closed ends, the pressure also does axial end-cap work $\pi a^{2}p{L}_0\dot{z}$. Equivalently its complete work is $pd(\pi a^{2}z{L}_0)/dt$. In that convention,If $N$ includes the end-pressure resultant, it is $N_{\mathrm{wall}}$ itself. Stating this distinction is necessary because the ends are not specified; the pressure formula is the same in both conventions.

For the specified Mooney-Rivlin solid convention, differentiating the constrained strain energy density givesThusPut $c=z{a}^2-a_0^2$. If $c\ne 0$, $z{\lambda }^2-1=c/R^2$ yields $dR/R=-z\lambda d\lambda /(z{\lambda }^2-1)$. The factor cancels becauseThe limits are ${\lambda }_a=a/a_0$ and ${\lambda }_b=b/b_0$, so direct integration givesFor $c=0$, all circumferential stretches equal $z^{-1/2}$ and the original integrand vanishes, so this formula still holds, with $p=0$. This also verifies its continuous limiting value without dividing by zero.

Use reference cylindrical coordinates $(R,\Theta ,Z)$ and current coordinates $(r,\theta ,z)$. The prescribed map has $r={\lambda }^{-1/2}R$, $\theta =\Theta +\alpha Z$ and $z=\lambda Z$. In the corresponding orthonormal cylindrical bases, differentiate physical line elements to obtain the deformation gradientThe radial direction is an eigenvector of the right Cauchy-Green deformation tensor with eigenvalue ${\lambda }^{-1}$. Its other block isIts trace is $s={\lambda }^{-1}(1+{\alpha }^2{R}^2)+{\lambda }^2$ and its determinant is $\lambda$. Thus its eigenvalues solve $z^2-sz+\lambda =0$. The principal stretches are the positive square roots of these eigenvalues:Their product is one, independently checking incompressibility.

The strain energy density $W$ is per reference volume, so stored energy per initial cylinder height is $\mathcal{E}(\alpha ,\lambda )=2\pi {\int }_0^{A}W({\lambda }_1,{\lambda }_2,{\lambda }_3)RdR$. The end rotation is $\alpha H$ and the current height is $\lambda H$, giving applied power per initial height $M\dot{\alpha }+N\dot{\lambda }$. For quasistatic reversible loading, equate this with $\dot{\mathcal{E}}$ for independent changes of $\alpha ,\lambda$. This proves the torque and axial force of a twisted incompressible cylinder formulasIn the second derivative hold $\alpha$, the twist per initial length, fixed. Holding twist per current length instead would mix the generalized forces.

For the Mooney-Rivlin solid, compute invariants directly from $F$ rather than differentiating individual eigenvalue square roots:The second identity follows from $I_2=\mathrm{tr}(F^{T}F{)}^{-1}$ when $J=1$. Retain the paper's minus sign on its ${\mu }_2$ term. Integrating $R$ and $R^3$ givesConsequentlyAs checks, untwisted unit stretch gives $M=N=0$, and setting ${\mu }_2=0$ recovers the neo-Hookean solid formulas. A convention with a positive second-invariant coefficient corresponds to replacing this ${\mu }_2$ by its negative.