= Solution
{c}
We first derive the <stress> and energy facts used in both parts. The Cauchy <traction> $s=Tn$ is <force> per unit area in the deformed configuration. The first Piola-Kirchhoff <traction> $s_R=T_RN$ is <force> per unit reference area. Their <force> vectors are measured in the same spatial coordinates, but the area measures and the normals used to specify the surface are different.
Parametrize a reference surface by $x=x(u,v)$. Its oriented area vector is $(x_u\times x_v)\,du\,dv$. Under deformation, the two tangents become $Fx_u,Fx_v$. The cross-product identity
$$
(Fa)\times(Fb)=(\operatorname{cof}F)(a\times b)=JF^{-T}(a\times b)
$$
therefore proves <Nanson's formula> $n\,da=JF^{-T}N\,dA$. Equality of the physical <force> on corresponding surface patches now gives
$$
T_RN\,dA=Tn\,da=JT F^{-T}N\,dA.
$$
Since this holds for every reference <normal>,
$$
\boxed{T_R=JT F^{-T},\qquad T=J^{-1}T_RF^T.}
$$
The <first Piola-Kirchhoff stress tensor> is generally not <symmetric>, whereas the <Cauchy stress tensor> is <symmetric> in the absence of couple stresses.
For the stated energy on orientation-preserving gradients, <differentiation> gives
$$
T_R=DW(F)=\mu\bigl(F-J^{-2}\operatorname{cof}F\bigr)
=\mu(F-J^{-1}F^{-T}).
$$
A stress-free state satisfies $FF^T=J^{-1}I$. Taking <determinants> yields $J^2=J^{-3}$, so $J=1$ and $FF^T=I$: precisely $F\in SO(3)$. These states are also the global energy <minimizers>. If $v_i>0$ are the <singular values>, the <arithmetic-geometric mean inequality> gives $|F|^2\geq3J^{2/3}$, with equality only when the <singular values> are equal. The function $3J^{2/3}/2+J^{-1}$ decreases up to $J=1$ and increases thereafter, with <minimum> $5/2$. Thus $W\geq5\mu/2$, with equality precisely at <rotations>. A constant shift of the energy can set this <minimum> to zero without changing the <stress>.
For the given <simple shear>,
$$
F=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix},\qquad
F^{-T}=\begin{pmatrix}1&0&0\\-1&1&0\\0&0&1\end{pmatrix},\qquad
T_R=\mu\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}.
$$
For a <unit normal> $N$, $|T_RN|^2=\mu^2(N_1^2+N_2^2)=\mu^2(1-N_3^2)$. Hence \b[every reference <plane> whose <normal> lies in the $e_1,e_2$ <plane> is maximizing]:
$$
\boxed{N=(\cos\alpha,\sin\alpha,0),\quad
x_1\cos\alpha+x_2\sin\alpha=0,\qquad\max|s_R|=\mu.}
$$
There is a family of maximizing planes, rather than one distinguished <normal>.
Back to article page
