 Simple-shear traction angle for a Mooney-Rivlin solid

For incompressible simple shear $F=I+\gamma {\mathbf{e}}_1\otimes {\mathbf{e}}_2$ with Mooney-Rivlin energy $\alpha (I_1-3)+\beta (I_2-3)$, zero traction on the third-coordinate faces fixes the pressure reaction. The remaining Cauchy stress has $T_{11}=2\alpha {\gamma }^2$, $T_{22}=-2\beta {\gamma }^2$ and $T_{12}=2(\alpha +\beta )\gamma$. The material face $x_1=\text{constant}$ has current unit normal $\mathbf{n}=(1,-\gamma ,0)/\sqrt{1+{\gamma }^2}$ and tangent $\mathbf{m}=(\gamma ,1,0)/\sqrt{1+{\gamma }^2}$. Resolving $T\mathbf{n}$ gives $N=-2{\gamma }^2(\alpha +2\beta +\beta {\gamma }^2)/(1+{\gamma }^2)$ and $S=2\gamma (\alpha +\beta )/(1+{\gamma }^2)$. Since $N<0$, the traction-normal angle satisfies $\tan (\theta -\pi /2)=-N/|S|$, proving the displayed formula. Consequently its two-sided small-shear slope is $(\alpha +2\beta )/(\alpha +\beta )$. The reference normal cannot replace the transformed normal in this calculation.