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Use orthonormal cylindrical bases in the reference and deformed configurations. The radial coordinate remains $\rho$, the angular coordinate becomes $\theta =\Theta +\alpha {\xi }_3$, and the axial coordinate is unchanged. Removing the rotation of the cylindrical basis, the deformation gradient isIt is a local simple shear with unit determinant. The squared principal stretches are the eigenvalues ofOne principal stretch is one. The remaining squared principal stretches have product one and sum $2+s^2$. Consequently they can be written ${\lambda }^2$ and ${\lambda }^{-2}$, whereFor negative $s$ their ordering reverses, which is immaterial to the strain energy density of an isotropic material.

Let the reference length be $L$. By material isotropy, the total elastic energy and end rotation areFor a quasistatic change, incompressibility makes the constraint reaction do no work. The velocity on an end section is its rigid rotational velocity; its traction power is the resultant torque times the angular velocity. Thus conservation of energy, or equivalently the torque and axial force of a twisted incompressible cylinder work identity, gives $Md{\Theta }_{\mathrm{end}}=dE$ andThe radius $a$ is fixed in this derivative. This argument computes the required end torque along the prescribed deformation family; proving that a suitable end-only traction distribution sustains that family is unnecessary here.

For a neo-Hookean solid, the identity $\lambda -{\lambda }^{-1}=s$ gives ${\lambda }^2+{\lambda }^{-2}-2=s^2$. ThereforeThis is the signed torque in the twist direction; its magnitude is $\pi \mu a^4|\alpha |/2$. For the positive twist convention the signed value equals the magnitude.

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.