Past exam of the solid mechanics course of the University of Oxford / 2023 / 2

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.