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With the right Cauchy-Green deformation tensor $C=F^{T}F$, the Green-Lagrange strain tensor measures the change in the material metric. Since ${\dot{E}}^{(2)}=F^{T}DF$, its work-conjugate stress and strain partner is the second Piola-Kirchhoff stress tensor $F^{-1}\tau F^{-T}$. The formula follows by cycling factors in $T:{\dot{E}}^{(2)}$, not by identifying nominal and symmetric material stress.

For $\gamma >0$, direct multiplication gives $|FN{|}^2-|N{|}^2=\gamma N_2(2N_1+\gamma N_2)$. Thus length-preserving line directions form the union of two planes in direction space: $N_2=0$ and $2N_1+\gamma N_2=0$. Their intersection is the third coordinate axis. This concerns segment length rather than pointwise fixed lines; a segment can move and turn while keeping its length.

The two nontrivial principal stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, with $v_{+}{v}_{-}=1$. The maximizing reference direction is proportional to $(1,v_{+},0)$. Consequently it tends to $(1,1,0)$ as $\gamma$ tends to zero from above. At zero itself all stretches equal one, so every direction maximizes length; the limit selects a direction from a degenerate endpoint. This follows from the largest eigenvector of the right Cauchy-Green deformation tensor.

With the right Cauchy-Green deformation tensor $C=F^{T}F$ and the second Piola-Kirchhoff stress tensor $S$, this mixed material stress is generally nonsymmetric. In the material-first nominal stress tensor convention, $PF=SC=M^T$. Thus replacing $P{F}^{-T}$ by $PF$ changes the stress measure. The relation follows from $\mathcal{P}=FS$, without any assumption of material isotropy.

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.

For an incompressible neo-Hookean solid, a convenient strain energy density is $W=\mu (\mathrm{tr}C-3)/2$, where $C$ is the right Cauchy-Green deformation tensor. An additive constant in $W$ does not affect the stress. The shear modulus $\mu$ fixes the small-strain stiffness.

A material fibre segment of reference length $L$ and unit direction $l$ has reference vector $Ll$. The uniform deformation gradient maps it to $LAl$, whose current length isTherefore its length is preserved if and only ifThis is the inextensible fibre constraint. The quadratic form uses the right Cauchy-Green deformation tensor; a subsequent rigid rotation does not change it, consistently with the polar factorization $A=RU$.

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.