Past exam of the solid mechanics course of the University of Oxford / 2009 (C6.1a, DMAC 2466)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

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.

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.

The material symmetry group consists of invertible admissible reference transformations $H$ for which $W(FH)=W(F)$ for every admissible $F$. Its rotational part lies in $SO(3)$; a determinant-one reference convention restricts the full group to the corresponding special linear group. This right action is distinct from the spatial left action $W(QF)=W(F)$ of material frame indifference.

The proper cube-rotation group has $\boxed{24}$ elements. Choose the image of one outward face normal in six ways, then the image of an adjacent perpendicular normal in four ways; orientation fixes the third direction. No reflections are included. For an isotropic material every proper rotation belongs to the material symmetry group.

Combine material isotropy with the physical frame-indifference condition. The singular value decomposition may be written $F=R\mathrm{diag}(v_1,v_2,v_3)Q$ with $R,Q\in SO(3)$ because $\det F>0$. Left and right rotational invariance remove $R,Q$. Signed permutation rotations permit any permutation of the three diagonal stretches, soRight invariance alone, without the usual material frame indifference assumption, would not suffice to remove spatial orientation.

The strict Baker-Ericksen inequalities say that larger distinct principal stretches carry larger principal Cauchy stresses:A natural state of a hyperelastic material has zero stress. If two stretches differed, their equal zero stresses would contradict the strict inequality. Thus all three equal some $v>0$, and polar decomposition gives $\boxed{F=vQ}$, $Q\in SO(3)$. A weak non-strict inequality alone would not prove this conclusion, so its strict convention is explicit here.

For the radial deformation put $f(R)=r(R)/R$, $R=|\mathbf{x}|$, $\mathbf{\theta }=\mathbf{x}/R$. Since ${\partial }_{j}R={\theta }_j$,This symmetric tensor has radial eigenvalue $r^{\prime }$ and the repeated tangential eigenvalue $r/R$. Both are positive in the open annulus under the stated monotonicity/nonnegativity conditions. Therefore these are its principal stretches. The two cases below establish radial zero-energy rigidity by different arguments.

At every point the deformation gradient attains the differentiable energy's minimum, so $DW(D\mathbf{y})=0$ and it is a natural state. By the strict Baker-Ericksen argument just proved, all its principal stretches are equal. In the radial formula this givesThus $r(R)=vR$ for one constant $v>0$ on the connected radial interval, andThe constancy is global through the annulus, not a separately chosen dilation at each point.

The radial gradient $F=D\mathbf{y}$ is symmetric positive definite. If it belongs to a well $SO(3)U_i$, write $F=Q{U}_i$. Then $F^{T}F=U_i^2$, while positive symmetry makes $F$ itself the unique positive square root of an operator of $F^{T}F$. Hence $F=U_i$. Likewise a symmetric positive-definite matrix in $SO(3)$ must be $I$. ThereforeFor each radius, the tangential stretch $r(R)/R$ must belong to the finite set of eigenvalues of these matrices. The ratio is continuous on the connected radial interval, so it is one constant $v>0$. Thus $r(R)=vR$ and differentiation gives $r^{\prime }=v$ as well. HenceThis proof uses finiteness and positive definiteness of the wells, not isotropy or the Baker-Ericksen inequalities. It also covers $R_1=0$: the punctured three-dimensional ball remains connected, and no condition at the excluded center was used.