Past exam of the solid mechanics course of the University of Oxford / 2009 / 2 / 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 $\mathbf{n}da=J{F}^{-T}\mathbf{N}dA$. Equating the actual surface force in both descriptions gives $T_R=JT{F}^{-T}$, and thereforeFor incompressibility $\boxed{J=1}$. The strain-energy derivative only fixes the constitutive stress up to a constraint reaction in hyperelastic stress: $T_R=DW-p\mathrm{cof}F$, 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 $B=F{F}^T$, its eigenvalues are $v_i^2$, so isotropy yieldsFor the prescribed simple shear,The $x_3$ faces retain normal $\pm {\mathbf{e}}_3$, and zero traction there gives $p=2(\alpha -\beta )$. ThusThis 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 areResolve the current traction $\mathbf{t}=T\mathbf{n}$ into these directions:Since $N<0$ for nonzero $\gamma$, the angle lies above $\pi /2$, andUsing $\arctan z\sim z$ at zero givesUsing 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.