Past exam of the solid mechanics course of the University of Oxford / 2008 (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.

By the spectral theorem a symmetric positive-definite matrix $C$ has $C=Q\mathrm{diag}(c_i)Q^{\mathsf{T}}$, with $Q$ orthogonal and $c_i>0$. Define $S=Q\mathrm{diag}(\sqrt{c_i})Q^{\mathsf{T}}$. It is symmetric positive-definite and $S^2=C$. For uniqueness, any symmetric positive-definite square root $T$ commutes with $C=T^2$, hence preserves each eigenspace of $C$. On its $c$-eigenspace, $T$ is symmetric positive-definite and all its eigenvalues have square $c$, so all equal $\sqrt{c}$. Therefore $T=S$, including repeated eigenspaces.

For $\det F>0$, putThen $R^{\mathsf{T}}R=I$ and $\det R=\det F/\det U>0$, so $R\in SO(3)$; also $F=RU=VR$. Any other right polar decomposition has $F^{\mathsf{T}}F=U^2$, so uniqueness of the positive root fixes $U$ and then $R$. The left factor is consequently fixed too.

The principal stretches $v_i$ are the positive eigenvalues of $U$, equivalently the singular values of $F$. For a unit vector $N={\sum }_i{c}_i{N}_i$ in an orthonormal eigenbasis of $U$,Thus $\boxed{\underset{|N|=1}{\max }|FN|=v_{\max }}$, attained precisely by the unit vectors in the eigenspace of $U$ belonging to $v_{\max }$. With a simple largest eigenvalue these are its two opposite unit eigenvectors; degeneracy allows every unit combination within that eigenspace.

For the simple shear,A line with nonzero direction $N$ keeps its length exactly whenHence $\boxed{N_2=0\text{or}2N_1+\gamma N_2=0}$, with $N_3$ arbitrary. Every segment in the cube parallel to one of these directions preserves length; there is no restriction on its starting point other than remaining in the cube.

The two in-plane stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, and the third is one. A maximizing direction for $\gamma >0$ is $N\parallel (1,v_{+},0)$, since the $C$ eigenvalue equation gives $N_2/N_1=(v_{+}^2-1)/\gamma =v_{+}$. Therefore $\boxed{N\to \pm (1,1,0)/\sqrt{2}}$ as $\gamma \downarrow 0$. At exactly zero shear all directions tie; the limit refers to the selected maximizing direction for positive shear. These are the length-preserving directions and principal directions of simple shear.

Cauchy's stress hypothesis says that traction across a cut depends locally on its point, time and oriented normal, rather than on the global shape of the cut. Here all areas and normals are referential: $E$ is a fixed material region in reference coordinates, ${\rho }_R$ its mass density per reference volume, $v$ the current particle velocity, $b$ body force per unit mass, and $s_R$ force per unit reference area. The time derivative on the left is total material momentum rate; the two right terms are contact and body forces.

A shrinking pillbox gives $s_R(-N)=-s_R(N)$. In a shrinking tetrahedron the volume contributions scale as its linear size cubed, whereas face forces scale as size squared. Their leading balance expresses traction on an arbitrary face as the normal-component-weighted sum of tractions on three coordinate faces. Defining those three tractions as tensor columns gives $\boxed{s_R(N)=T_{R}N}$. Applying the divergence theorem to momentum balance and localizing over arbitrary $E$ then yieldsThe referential tensor is the first Piola-Kirchhoff stress and need not be symmetric; it should not be silently identified with the spatial Cauchy stress.

Incompressibility is $\boxed{\det F=1}$. For a homogeneous hyperelastic material the referential constitutive relation becomeswhere the undetermined scalar pressure $p$ enforces the constraint. Homogeneity removes explicit reference-position dependence from $W$, not the possible spatial variation of that pressure.

For the neo-Hookean energy, $DW=2\alpha F$. For the positive diagonal ansatz the nominal principal stresses are $t_i=2\alpha v_i-p/v_i$. Since $F$ is constant, equilibrium requires $\mathrm{cof}F\nabla p=0$, so $p$ is constant. Stress-free endfaces give $\boxed{p=2\alpha v_3^2>0}$. The side normals include $(0,-1,0)$ and $\pm (\sqrt{3}/2,1/2,0)$, so their prescribed tractions force $\boxed{t_1=t_2=T}$.

For fixed positive $p$, the scalar function $2\alpha v-p/v$ is strictly increasing on $v>0$. Thus equality of the two nominal stresses implies $v_1=v_2=v$. Incompressibility gives $v_3=v^{-2}$, and substitution givesThe load function has derivative $2\alpha (1+5v^{-6})>0$, tends to minus infinity as $v\downarrow 0$ and to infinity as $v\to \infty$. It therefore determines exactly one positive $v$ for every specified load, with $v\ge 1$ for outward tensile $T\ge 0$. The resulting constant stress satisfies equilibrium and every prescribed face traction, proving existence and uniqueness within the requested positive diagonal class. It does not claim uniqueness among all possible nonhomogeneous deformations. This is the equibiaxial dead loading of a neo-Hookean prism.

Put $n=x/R$ and $q=r(R)/R$. Differentiating $y=qx$ uses $\nabla R=n$ and $\nabla q=(r^{\prime }-q)n/R$. HenceIts radial eigenvalue is $r^{\prime }$ and its two tangential eigenvalues are $q$. They are positive by hypothesis, so they are also the principal stretches:For an isotropic stored energy the nominal stress has the same radial/tangential projectors,where symmetry of $\Phi$ gives equal tangential derivatives. Since $\mathrm{Div}(n\otimes n)=2n/R$ and the scalar coefficients depend only on $R$,Thus equilibrium is exactly $\boxed{d{\Phi }_1/dR=2({\Phi }_2-{\Phi }_1)/R}$.

For the particular energy define $\delta =r^{\prime }{q}^2=\det F>0$. Its two derivatives areAlso $q^{\prime }=(r^{\prime }-q)/R$. Differentiating the first formula and substituting the equilibrium equation givesThe last terms cancel. Strict positivity of $h^{\prime \prime }$ and $q$ yields $\boxed{{\delta }^{\prime }=0}$, so $r^2{r}^{\prime }=D{R}^2$ for one positive constant $D$. Integration gives $r^3=D{R}^3+C$.

The displacement data mean $r(1)={\lambda }_1$ and $r(2)=2{\lambda }_2$, not $r(2)={\lambda }_2$. Solving the two scalar equations givesThe given inequality makes $D>0$. Since $r^3$ increases from ${\lambda }_1^3>0$ to $8{\lambda }_2^3$, it is positive on the whole shell, and $r^{\prime }=D{R}^2/r^2>0$. Thus the formula is smooth and admissible and satisfies equilibrium. Conversely every admissible radial equilibrium had to have this constant determinant and these two integration constants, proving uniqueness. This is a constant-Jacobian radial equilibrium for a stretch-sum energy.

On each half-space the deformation is affine. Continuity on the separating plane makes the two affine constants equal and gives $(A-B)t=0$ for every tangent vector $t\perp N$. Decomposing any $x$ into its tangential part and $(x\cdot N)N$ givesThis is the Hadamard jump condition. Since $B=A-a\otimes N=A(I-A^{-1}a\otimes N)$, the matrix determinant lemma givesBoth determinants are positive, hence $\boxed{A^{-1}a\cdot N<1}$.

Write $a_1={\eta }_1$, $a_2={\eta }_2$, $S=a_1^2+a_2^2$ and $\Delta =a_2^2-a_1^2\ne 0$. The two proposed angles have $c=2a_1{a}_2/S$ and $s=\kappa \Delta /S$, and $c^2+s^2=1$. Therefore the displayed rotations lie in $SO(3)$. Direct multiplication givesThis is a rank-one connection for each sign. With a unit normal one may choose $N=(1,\kappa ,0)/\sqrt{2}$ and jump vector $a=\sqrt{2}\Delta (-a_2,\kappa a_1,0)/S$. The two rotations are distinct because $\Delta \ne 0$; the permitted at-most-two result therefore proves these are all solutions. Flipping both $a$ and $N$ changes no connection.

It remains to test the material plane. Because $y(0)=0$, the affine constants vanish. Let $t=(-p_2,p_1,0)$, so the plane is spanned by $t$ and $e_3$. Both deformation gradients send $e_3$ to $a_1{e}_3$. If the plane is not the twin plane, it cuts the two half-spaces; its two image halves are coplanar exactly when the in-plane vectors $Bt$ and $At$ are parallel. Put $d=(-a_2,\kappa a_1)$ and $n=(1,\kappa )$ in their two-dimensional sections. ThenIts prefactor is nonzero, so coplanarity requires $p_1=\kappa p_2$, precisely $p\parallel (1,\kappa ,0)\parallel N$. Conversely, on that twin plane $(A-B)x=0$, so the two affine maps agree and its image is a plane. Thus $\boxed{\Pi \text{ remains plane}⟺p\parallel N}$. All other allowed planes acquire a kink along their intersection with the twin. This is the planarity test across a cubic-to-tetragonal twin.