= Solution
The first <Piola-Kirchhoff stress tensor> $T_R$ maps a reference normal to force per reference area. The <Cauchy stress tensor> $T$ maps a current normal to force per current area. <Nanson's formula> gives $\mathbf n\,da=JF^{-T}\mathbf N\,dA$. Equating the actual surface force in both descriptions gives $T_R=JTF^{-T}$, and therefore
$$
\boxed{T=J^{-1}T_RF^T.}
$$
For <incompressibility> $\boxed{J=1}$. The strain-energy derivative only fixes the constitutive stress up to a <constraint reaction in hyperelastic stress>: $T_R=DW-p\operatorname{cof}F$, hence $T=DW F^T-pI$. The scalar $p$ is a <Lagrange multiplier> determined by equilibrium and boundary <tractions>, rather than by the isochoric energy alone.
In <principal stretch> axes, the extra <principal stress> is $v_i\partial W/\partial v_i=2\alpha v_i^2-2\beta v_i^{-2}$, up to that arbitrary spherical reaction. With $B=FF^T$, its <eigenvalues> are $v_i^2$, so isotropy yields
$$
\boxed{T^*=2\alpha B-2\beta B^{-1},\qquad T=T^*-pI.}
$$
For the prescribed <simple shear>,
$$
F=\begin{pmatrix}1&\gamma&0\\0&1&0\\0&0&1\end{pmatrix},\quad
B=\begin{pmatrix}1+\gamma^2&\gamma&0\\\gamma&1&0\\0&0&1\end{pmatrix},\quad
B^{-1}=\begin{pmatrix}1&-\gamma&0\\-\gamma&1+\gamma^2&0\\0&0&1\end{pmatrix}.
$$
The $x_3$ faces retain normal $\pm\mathbf e_3$, and zero <traction> there gives $p=2(\alpha-\beta)$. Thus
$$
\boxed{T=2\gamma\begin{pmatrix}\alpha\gamma&\alpha+\beta&0\\\alpha+\beta&-\beta\gamma&0\\0&0&0\end{pmatrix}.}
$$
This constant stress satisfies the interior equilibrium equation, and the remaining face forces maintain the deformation.
To compute the angle, the material face $x_1=1$ becomes the spatial plane $y_1-\gamma y_2=1$. Its outward <unit normal> and an in-plane unit tangent are
$$
\mathbf n=\frac{(1,-\gamma,0)}{\sqrt{1+\gamma^2}},\qquad\mathbf m=\frac{(\gamma,1,0)}{\sqrt{1+\gamma^2}}.
$$
Resolve the current <traction> $\mathbf t=T\mathbf n$ into these directions:
$$
N=\mathbf n\cdot T\mathbf n=-\frac{2\gamma^2(\alpha+2\beta+\beta\gamma^2)}{1+\gamma^2},\qquad
S=\mathbf m\cdot T\mathbf n=\frac{2\gamma(\alpha+\beta)}{1+\gamma^2}.
$$
Since $N<0$ for nonzero $\gamma$, the angle lies above $\pi/2$, and
$$
\boxed{\theta-\frac\pi2=\arctan\left(\frac{|\gamma|(\alpha+2\beta+\beta\gamma^2)}{\alpha+\beta}\right).}
$$
Using $\arctan z\sim z$ at zero gives
$$
\boxed{\lim_{\gamma\to0}\frac{\theta-\pi/2}{|\gamma|}=\frac{\alpha+2\beta}{\alpha+\beta}.}
$$
Using the undeformed normal $\mathbf e_1$ would give the wrong coefficient. The transformed <unit normal> is essential to this <simple-shear traction angle for a Mooney-Rivlin solid>.
Back to article page
