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

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.