Past exam of the solid mechanics course of the University of Oxford / 2023 / 3

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.