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The polar decomposition in continuum mechanics and material isotropy remove the reference rotation from an elastic Cauchy stress tensor. Half-turns around principal stretch axes leave the diagonal left stretch tensor fixed, so material frame indifference forces the stress's off-diagonal entries to vanish in those axes. Rotations within repeated-stretch eigenspaces also force isotropic stress there. Consequently stress and the left Cauchy-Green deformation tensor are coaxial symmetric tensors.

Let $c=\cos (\alpha {\xi }_3)$, $s=\sin (\alpha {\xi }_3)$ and $r^2=x_1^2+x_2^2=({\xi }_1^2+{\xi }_2^2)/\lambda$. Differentiating the deformation givesFor the neo-Hookean solid, $W_{,A_{i\alpha }}=\mu A_{i\alpha }$. The constrained nominal stress tensor is therefore $N_{\alpha i}=\mu A_{i\alpha }+q(A^{-1}{)}_{\alpha i}$. The multiplier cannot be set arbitrarily to a constant; equilibrium and the traction-free curved surface determine it.

Push the nominal stress to the Cauchy stress tensor: since $J=1$, $\sigma =AN=\mu A{A}^T+qI$. In the current cylindrical orthonormal basis, the left Cauchy-Green deformation tensor has entriesThus ${\sigma }_{rr}=\mu /\lambda +q$, ${\sigma }_{\vartheta \vartheta }=\mu /\lambda +\mu {\alpha }^2{r}^2+q$, and ${\sigma }_{\vartheta 3}=\mu \alpha \lambda r$. Static radial equilibrium without body force givesThe current outer radius is $a=R/\sqrt{\lambda }$. Setting ${\sigma }_{rr}(a)=0$ integrates the reaction field:The other curved-surface traction components vanish because ${\sigma }_{r\vartheta }={\sigma }_{r3}=0$.

On the reference end ${\xi }_3=H$, the outward reference normal is $e_3$, and the nominal traction has components $t_i=N_{3i}$. The third row of $A^{-1}$ is $(0,0,{\lambda }^{-1})$. Hence the nominal end traction of a twisted neo-Hookean cylinder isThe current $x_1,x_2$ here are evaluated at ${\xi }_3=H$. These components are per unit reference area, not per unit current area.

For the moment, use current lever arms and reference-area traction. Its axial component isThe axial traction gives no axial moment. This torque also agrees with the reference-energy derivative at fixed axial stretch: the twist-dependent energy per reference length is $\mu {\alpha }^2\pi R^4/(4\lambda )$. It is important that $\alpha$ is twist per reference length, not twist per current length.

Use both material isotropy and material frame indifference. The polar decomposition in continuum mechanics writes $F=VR$, where $V$ is the left stretch tensor. Isotropy removes the reference rotation: $\sigma (F)=\sigma (V)$. Choose an orthonormal eigenbasis of $V$. A half-turn about any principal axis leaves $V$ fixed but reverses appropriate off-diagonal entries of $\sigma (V)$. Frame indifference therefore makes those entries zero. If a principal stretch is repeated, rotations in its eigenspace additionally force the stress to be isotropic there. Thus the principal stresses share the stretch axes: this proves coaxiality of isotropic elastic stress with $V$ and with $B=V^2=F{F}^T$.

For simple shear, the left Cauchy-Green deformation tensor isIts principal axes make an angle $\vartheta$ with the coordinate axes satisfying $2\cot (2\vartheta )=\gamma$ when $\gamma \ne 0$. If the two principal stresses are $s_1,s_2$, rotating their diagonal matrix back givesConsequently the universal simple-shear normal-stress identity isEquivalently, it follows immediately from the off-diagonal entry of $\sigma B-B\sigma =0$. At $\gamma =0$, the undeformed material has isotropic stress, so the identity holds there too. No particular elastic constitutive equation has been used.

The right Cauchy-Green tensor measures stretch on the reference configuration. Its eigenvalues are the same squared principal stretches as those of the left Cauchy-Green deformation tensor, but its eigenvectors are reference directions.