Solid mechanics course of the University of Oxford (C5.1, C6.1a, C6.1)

Original solutions to past examinations of the solid mechanics course of the University of Oxford.

Solutions to all three questions of the solid mechanics course of the University of Oxford Part C examination, sat on 8 June 2023. Source: original released PDF, inclusive bundle pages 313–316; paper pages 1–4. All eighteen terminal parts are covered, including every question despite the best-two assessment rule. The torque normalization in Question 1 follows the printed integral; the signed pressure in Question 2 is positive for compression.

Write the first strain invariant in components of the deformation gradient:Here lower-case and upper-case indices refer to the current configuration and reference configuration, respectively. Differentiating gives $\partial I_1/\partial F_{iI}=2F_{iI}$. The chain rule therefore givesSubstitution into the given hyperelastic material law, with its indicated matrix transpose, yieldsThe left Cauchy-Green deformation tensor is $\mathbf{B}=\mathbf{F}{\mathbf{F}}^T$. The scalar $p$ is the Lagrange multiplier enforcing incompressibility, rather than a function fixed by the strain energy density alone. The identity matrix $1$ makes this an isotropic constraint reaction in hyperelastic stress.

Use separate moving orthonormal bases in the reference and current cylindrical coordinate systems. For pure torsion of an incompressible cylinder, the deformation map isThe unit vectors depend on the angle: ${\partial }_{\theta }{\mathbf{e}}_r={\mathbf{e}}_{\theta }$ and ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-{\mathbf{e}}_r$. ConsequentlyIn particular, the last derivative includes the rotation of ${\mathbf{e}}_r$ along the cylinder. The material gradient has the reference coordinate scale factor $1/R$:The outer products specify the images of the three reference unit vectors. Thus, with current basis vectors indexing rows and reference basis vectors indexing columns,Its determinant is one, checking incompressibility. The radius is unchanged, so the current outer radius in part (d) is $a=A$. The formula at the coordinate axis is understood by continuity in Cartesian coordinates.

Multiplying the deformation gradient by its matrix transpose gives the left Cauchy-Green deformation tensor and its trace:The derivative of the strain energy density isFor brevity put $g(r)={\mu }_1+{\mu }_2(3+{\tau }^2{r}^2{)}^n$. Part (a) gives the Cauchy stress tensor, in the current cylindrical coordinate system orthonormal basis, asIn particular, the end-face shear stress is $T_{\theta z}=2g\tau r$. The undetermined pressure $p$ affects the normal stresses but does not affect this shear stress. No normal traction conditions have been specified to determine $p$.

Using $a=A$ and the end-face shear stress from part (c), the printed torque normalization givesThe binomial theorem expands the remaining integral into powers of $r$:Integrating each term gives an expression valid also at $\tau =0$:For example, $n=1$ gives $M=({\mu }_1+3{\mu }_2)A^4\tau /2+{\mu }_2{A}^6{\tau }^3/3$. The torque is an odd polynomial in the twist per unit reference length, as expected when reversing the twist reverses the end-face shear stress.

For a physical end-face torque $N$, integration over the azimuth normally givesThe original PDF explicitly defines $M$ without this $2\pi$, so the boxed answer retains its convention. Since $\mathbf{T}$ is the Cauchy stress tensor, its shear component here is spatial $T_{\theta z}$; the printed capital $Z$ in $T_{\theta Z}$ does not require a change to a nominal stress tensor.

Only the $k=0$ term of the polynomial contributes to the limit:Thus the requested torsional stiffness per unit twist isThe small-strain shear modulus is $G=2{\hat{w}}^{\prime }(3)=2({\mu }_1+3^n{\mu }_2)$. Therefore $\alpha =G{A}^4/4$, consistently with the usual circular-shaft torque relation after restoring the $2\pi$ factor. If the prescribed end rotation is the angle $\phi =\tau L$, the physical torsional stiffness is $dN/d\phi =2\pi \alpha /L$ at zero twist. This also explains why $L$ does not enter the requested $\alpha$.

The radial unit vector in spherical coordinates is ${\mathbf{e}}_R=\mathbf{X}/R$, so $\mathbf{X}\otimes \mathbf{X}/R^2={\mathbf{e}}_R\otimes {\mathbf{e}}_R$. The identity matrix is the sum of the three basis outer products. Substitution in the given deformation gradient givesThe reference and current radial directions coincide for this spherically symmetric deformation. The positive principal stretches are therefore ${\lambda }_r=f^{\prime }(R)$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=f(R)/R$. Incompressibility imposesIntegrating and using $f(A)=a$ givesIn particular $b^3-a^3=B^3-A^3$, expressing conservation of shell volume. Writing $\lambda =r/R$ also gives ${\lambda }_r={\lambda }^{-2}$ and ${\lambda }_{\theta }={\lambda }_{\varphi }=\lambda$.

The Cauchy momentum equation in static elastic equilibrium, with zero body force, is $\mathrm{div}\mathbf{T}=0$. The Cauchy stress tensor is symmetric. Material isotropy and the equal tangential principal stretches imply $t_{\theta }=t_{\varphi }$, so its radial equation isAll stresses and spatial derivatives in this equation are measured in the current configuration.

Let $W_r,W_{\theta },W_{\varphi }$ denote derivatives of the strain energy density with respect to its three principal stretches. The principal stress formula for incompressible hyperelasticity isIt follows by applying $\mathbf{T}=\mathbf{F}({\partial }_{\mathbf{F}}W{)}^T-p1$ in principal axes. Equal tangential principal stretches and material isotropy give $W_{\theta }=W_{\varphi }$ at $({\lambda }^{-2},\lambda ,\lambda )$. Differentiating the reduced strain energy density by the chain rule givesThe unknown pressure cancels in this difference. Substituting in part (i), and then using the supplied derivative of $\lambda$, givesConsequently the desired ordinary differential equation isFor completeness, the supplied derivative follows from $R^3=r^3-(a^3-A^3)$: $dR/dr=r^2/R^2={\lambda }^2$, and differentiating $\lambda =r/R$ gives $(1-{\lambda }^3)/R$. At the undeformed value $\lambda =1$ the division must be interpreted by a limit; the spherically symmetric deformation with $a=A$ has $\lambda \equiv 1$ and should be handled directly by the radial equilibrium equation.

Take the inner surface to be free of traction, as appropriate to an empty cavity with no surface tension. At the outer surface the imposed pressure is compressive when positive. The outward normal vectors are $-{\mathbf{e}}_r$ at $r=a$ and $+{\mathbf{e}}_r$ at $r=b$, givingIntegrating the ordinary differential equation from part (ii) between the two surfaces givesThe outer stretch follows from incompressibility:Hence an expression entirely in terms of the inner stretch and the reference radii isEquivalently $P=-{\int }_{{\lambda }_b}^{{\lambda }_a}{h}^{\prime }(\lambda )/({\lambda }^3-1)d\lambda$. For an expanding shell, $1<{\lambda }_b<{\lambda }_a$; when $h^{\prime }>0$ on this range, $P<0$, correctly describing an outward tensile loading. If an inner pressure $P_i$ were present, the same derivation would give $P-P_i$ on the left. Thus the empty-cavity condition specifies the pressure reference needed for the requested answer.

First let $A\downarrow 0$ at fixed positive cavity radius $a$ and fixed reference outer radius $B$. Thenand the pressure sustaining a finite cavity isThe onset of cavitation is the additional limit $a\downarrow 0$. The outer stretch then tends to one, and the signed critical pressure isWhen this improper integral is positive and finite before the minus sign, the tensile loading magnitude is $-P_{\mathrm{crit}}>0$. Thus the cavity opens under tension, even though the load variable is named pressure. The critical value is independent of the original sphere radius $B$.

For $A=0$ and $a>0$, the deformation map is smooth away from $R=0$ but singular at the original centre: $r=(R^3+a^3{)}^{1/3}$ approaches $a$, rather than zero, as $R\downarrow 0$. This is the elastic cavitation mechanism; it cannot be represented by a smooth deformation map through the centre.

The neo-Hookean solid corresponds to $\alpha =1$, with positive shear modulus $\mu$. Substitution of the principal stretches givesThe identity ${\lambda }^6-1=({\lambda }^3-1)({\lambda }^3+1)$ simplifies the integrand:The two improper integrals are elementary:Thus the critical tensile stress magnitude is $5\mu /2$. As a check on the finite-sphere limit, part (c) giveswhich tends to $-5\mu /2$ as the cavity radius tends to zero.

For general $\alpha >0$, put $S(\lambda )={\lambda }^{-4}+2{\lambda }^2$. The reduced strain energy density and its derivative areThe same factorization as in part (i) yieldsThis form removes the apparent singularity at $\lambda =1$: its limit is $4\mu \alpha 3^{\alpha -1}$, finite for every fixed $\alpha >0$. Thus only the upper endpoint can obstruct convergence of the improper integral.

As $\lambda \to \infty$, $S(\lambda )\sim 2{\lambda }^2$, givingFor $\mu >0$, the comparison test with a power integral implies convergence exactly when $2\alpha -4<-1$. Thereforeis precisely the range with finite critical pressure, the cavitation criterion for power-law strain energy. At $\alpha =3/2$ the divergence is logarithmic; for larger $\alpha$ it is a power divergence. These materials require an unbounded tensile loading magnitude in this ideal model. The qualification $\mu >0$ excludes the degenerate zero-energy material.

The first equation is the local balance of linear momentum. Use $\mathsf{M},\mathsf{L},\mathsf{T}$ for the dimensions of mass, length, and time, to distinguish dimensional time from the Cauchy stress tensor $\mathbf{T}$.

Here $t$ has dimension $\mathsf{T}$. The hypothesis ${\partial }_t\mathbf{b}=0$ means that the prescribed spatial body force field is the same for every member of the static family and the resulting motion.

Restore the particle label suppressed in the question, writing the static family as ${\mathbf{x}}_0(\mathbf{X};\tau )$. At each fixed $\tau$, its Cauchy stress tensor satisfiesThis dynamic continuation of a static incompressible elastic deformation sets $\tau =t$ to form a motion. At every time its deformation gradient is the same as that of the corresponding static member. For an elastic material the elastic part of the Cauchy stress tensor depends on that deformation gradient, while incompressibility leaves an arbitrary scalar constraint reaction in hyperelastic stress. Therefore writeHere $1$ is the identity matrix. Subtracting static balance of linear momentum from dynamic balance of linear momentum givesSince the homogeneous incompressible material has constant mass density, define $\xi =(p-p_0)/\rho$. NecessarilyConversely, suppose this acceleration potential exists. Choose $p=p_0+\rho \xi$. ThenThe added isotropic stress is symmetric, so angular momentum conservation remains satisfied. This proves both directions for the field equations and the elastic constitutive equation. The surface tractions for the motion are those of the adjusted stress; identical prescribed static tractions would impose an additional boundary restriction on $\xi$.

The pressure adjustment from part (i) can be writtenIndeed, since $\mathrm{div}(q1)=\nabla q$, the difference of the two Cauchy momentum equations isThus $q+\rho \xi$ is constant in space on each connected component. The function $C(t)$ is fixed by any prescribed pressure or surface traction datum. In terms of the usual compressive pressure multiplier, $p=p_0-q=p_0+\rho \xi -C(t)$. The plus sign in $q1$ therefore does not identify $q$ with the compressive multiplier $p$ itself.

For a twice continuously differentiable scalar potential, mixed second partial derivatives commute. Taking the spatial curl of the equation from part (i) givesFor example, its first component is $-{\partial }_2{\partial }_3\xi +{\partial }_3{\partial }_2\xi =0$. The curl acts on the spatial representation of the material acceleration, not directly on its expression in reference coordinates.

This is a necessary local condition. It is sufficient locally, and sufficient globally on a simply connected domain under the usual smoothness assumptions. On a domain with holes, a global scalar potential additionally requires every closed-loop line integral of the material acceleration to vanish. The homogeneous motion in part (c) admits an explicit global scalar potential, so this topological qualification causes no obstruction there.

Take two time derivatives at fixed reference position $\mathbf{X}$:The constraint $\det \mathbf{F}=1$ ensures that the deformation gradient is invertible, so $\mathbf{X}={\mathbf{F}}^{-1}(\mathbf{x}-\mathbf{c})$. Thus the spatial material acceleration is the affine vector fieldIts spatial gradient is $\mathbf{A}$, and its curl isPart (b)(iii) therefore requiresEquivalently, multiplying on the left by ${\mathbf{F}}^T$ and on the right by $\mathbf{F}$ givesThe translation $\mathbf{c}(t)$ contributes no spatial curl and is unrestricted by this condition. A useful equivalent first integral isthe antisymmetric matrix in parentheses must be constant, and need not vanish. The separate constraint $\det \mathbf{F}=1$ is still required; differentiating it also gives $\mathrm{tr}(\dot{\mathbf{F}}{\mathbf{F}}^{-1})=0$.

The condition is understood in the setting of part (b), where a static family already exists under the prescribed body force. For a homogeneous elastic deformation gradient, the elastic stress $\mathbf{S}(\mathbf{F})$ is spatially constant. Static balance of linear momentum consequently gives $\nabla p_0=\rho \mathbf{b}$, ensuring that $\mathbf{b}$ has a scalar potential. If the static-family assumption were dropped and an arbitrary time-independent body force allowed, the general necessary condition would instead be $\nabla \times \ddot{\mathbf{x}}=\nabla \times \mathbf{b}$.

For this homogeneous motion, suppose the matrix $\mathbf{A}=\ddot{\mathbf{F}}{\mathbf{F}}^{-1}$ is symmetric. Put $\mathbf{y}=\mathbf{x}-\mathbf{c}(t)$. The acceleration potential of a homogeneous motion iswhere $g(t)$ is arbitrary. When taking the spatial gradient, $t$, $\mathbf{c}$, $\ddot{\mathbf{c}}$, and $\mathbf{A}$ are fixed. Symmetry givesThis verifies the potential directly and also proves sufficiency of the symmetric matrix condition within the static-family setting. Its dimension is ${\mathsf{L}}^2{\mathsf{T}}^{-2}$, so the pressure correction $-\rho \xi$ has the dimension of stress. Combining with part (b)(ii), the corresponding dynamic Cauchy stress tensor iswhere $\tilde{C}(t)=C(t)-\rho g(t)$. Substituting its divergence into the Cauchy momentum equation reproduces $\rho [\mathbf{Ay}+\ddot{\mathbf{c}}]$, as required.

Source release: Oxford examination archive release thread, Part C bundle pages 426–429. The cover identifies Solid Mechanics, Tuesday 7 June 2011, 9.30am–11.00am.

Answer availability checked on 28 October 2026: bounded searches of official Oxford sources by exact year, title, module and formal code found no advertised exact-paper answers. The 2020–21 Solid Mechanics course page lists an exam compilation covering 2011–2020, but does not advertise it as solutions; its linked resource could not be fetched in the bounded check. No later-year release policy was used to infer 2011 availability.

A homogeneous hyperelastic material has a strain energy density per unit reference volume $W(F)$ depending on the deformation gradient, with no explicit dependence on the reference configuration. For orientation-preserving gradients $F\in G{L}^{+}(3)$, its first Piola-Kirchhoff stress tensor is the derivative:Equivalently the infinitesimal stored-energy change is $DW(F)[H]=T:H$. Reference and spatial indices are distinguished here by upper- and lowercase index letters.

The material symmetry group may be written $G_W=\{H\in G{L}^{+}(3):W(FH)=W(F)\text{ for every }F\in G{L}^{+}(3)\}$. A volume-preserving relabelling convention restricts the candidate $H$ to $SL(3)$; the material isotropy criterion is the same. The material has material isotropy when $SO(3)\subseteq G_W$, meaning $W(FQ)=W(F)$ for every proper rotation $Q$. If only rotational symmetries are included in the definition of the group, material isotropy means that group equals $SO(3)$. This right-rotation invariance is distinct from material frame indifference, the left-rotation invariance $W(QF)=W(F)$.

The polar decomposition in continuum mechanics gives the unique factorizationswhere $U,V$ are positive-definite symmetric matrices. Their common positive eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$ are the principal stretches, equivalently the singular values of $F$. They describe the local stretch magnitudes after separating out a rigid rotation.

Define the invariants by the characteristic polynomialBecause $B=F{F}^T$ has positive eigenvalues ${\lambda }_j^2$, its invariants are the elementary symmetric functions in those eigenvalues:The Frobenius norm gives $|F{|}^2=\mathrm{tr}(F^{T}F)={\sum }_j{\lambda }_j^2$. Write $J=\det F>0$. Since $\mathrm{cof}F=J{F}^{-T}$, its singular values are $J/{\lambda }_j$, soFinally $\det B=\det F\det F^T=J^2$. ThereforeThese identities also give $I_B=\mathrm{tr}B$ and $I{I}_B=((\mathrm{tr}B{)}^2-\mathrm{tr}(B^2))/2$.

Take a singular value decomposition $F=Q_{1}D{Q}_2^T$, with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$ positive and $Q_1,Q_2\in SO(3)$. Such a choice is possible because $\det F>0$. Left material frame indifference removes $Q_1$, and right material isotropy removes $Q_2$, giving $W(F)=W(D)$.

Moreover $W(D)$ is unchanged when the positive diagonal entries are permuted. Every permutation can be implemented by a signed permutation matrix of determinant one; the signs disappear in conjugating a diagonal matrix. Thus the energy depends only on the unordered principal stretches.

The three invariants determine precisely this unordered triple: the roots of $t^3-I_B{t}^2+I{I}_{B}t-II{I}_B$ are ${\lambda }_1^2,{\lambda }_2^2,{\lambda }_3^2$, and their positive square roots give the stretches. Consequently, on the set of admissible invariant triples, define $h(I_B,I{I}_B,II{I}_B)$ to be $W(D)$. The preceding argument proves that this is well defined and thatNot every positive triple is realizable by three positive roots. Since the requested statement only asserts existence of a function on $(0,\infty {)}^3$, extend $h$ arbitrarily on the nonrealizable triples; no choice there affects the formula for a deformation in continuum mechanics. Smooth invariant representations, where needed for differentiating the constitutive equation, are understood locally on the realizable set.

Retain the material frame indifference assumption used in part (c), in addition to material isotropy. Put $p=u_{,1}$, $q=u_{,2}$ and $s=p^2+q^2$. For this antiplane shear,The invariant identities give $I_B=I{I}_B=3+s$ and $II{I}_B=1$. Hence the restriction of the strain energy density isAntiplane energy does not determine the full stress: the restriction gives one scalar energy along antiplane gradients. It does not determine the full stress derivative with respect to every entry of $F$.

Indeed, using the isotropic hyperelastic stress invariant representation and differentiating the energy, with $h_j$ denoting the corresponding partial derivatives, givesUsing $S=J^{-1}T{F}^T$, the general objective isotropic Cauchy stress tensor isAll derivatives are evaluated at the current invariant triple. The extra terms generally remain nonzero on antiplane gradients, so the printed assertion that the full stress must be $\beta B$ does not follow.

A stress-free, objective isotropic counterexample is the compressible neo-Hookean solid $W(F)=\frac{\mu }{2}(|F{|}^2-3)-\mu \log J$, with $\mu >0$. Its derivative is $T=\mu F-\mu F^{-T}$, so on antiplane gradientsTake $u(x_1,x_2)=x_1$. Then $B_{22}=1$ but $S_{22}=0$, whereas $B_{13}=1$ and $S_{13}=\mu$. The diagonal entry would force $\beta =0$, contradicting the shear entry. Thus this is not a matter of choosing the right scalar coefficient.