Past exam of the solid mechanics course of the University of Oxford / 2009 / 1

The matrix exponential is the absolutely convergent series $e^{tA}={\sum }_{n=0}^{\infty }{t}^n{A}^n/n!$, with derivative $A{e}^{tA}$ and inverse $e^{-tA}$. If $K^T=-K$, then $(e^{tK}{)}^T{e}^{tK}=e^{-tK}{e}^{tK}=I$. Its determinant is continuous in $t$, is never zero, and starts at one, so it is one throughout: $e^{tK}\in SO(3)$. Conversely, differentiating $(e^{tK}{)}^T{e}^{tK}=I$ at zero gives $K^T+K=0$. ThusThe quantifier matters: membership at one isolated time alone would not imply skewness.

Write $P(F)=DW(F)$ and $J=\det F>0$. For any skew matrix $K$, differentiate frame indifference along $e^{tK}F$:A matrix orthogonal to every skew matrix is symmetric. Therefore $P(F)F^T$ and the Cauchy stress tensor $T=J^{-1}P(F)F^T$ are symmetric.

Let $\mathcal{C}=D^{2}W(I)$ and assume $P(I)=0$. Differentiating symmetry of $P(F)F^T$ at $I$ in direction $H$ shows $\mathcal{C}[H]$ is symmetric for every $H$. Hence $c_{ijkl}=c_{jikl}$. As a Hessian, $\mathcal{C}$ also has pair-interchange symmetry $c_{ijkl}=c_{klij}$. Combining the two gives $c_{ijkl}=c_{ijlk}$. Equivalently $\mathcal{C}$ annihilates infinitesimal rotations. This proves both requested minor symmetries and shows exactly where the stress-free assumption enters. It is stress-free linearization of an objective energy.

For the particular energy, ${\sum }_i{v}_i^2=\mathrm{tr}(F^{T}F)$ and ${\prod }_i{v}_i=J$, soAt $I$ this is $[4\alpha +h^{\prime }(1)]I$. Therefore the reference is stress-free exactly when $\boxed{h^{\prime }(1)=-4\alpha }$. To differentiate again, use $\mathrm{cof}F=J{F}^{-T}$. The determinant and inverse derivatives at $I$ givewhich also proves the supplied hint. It follows thatFor a symmetric $e$ this becomesNo additional positivity of the bulk modulus follows from the stated smoothness of $h$ alone; this calculation identifies the moduli rather than imposing extra material assumptions.