Past exam of the solid mechanics course of the University of Oxford / 2011 / 2

In the classical smooth setting, a deformation in continuum mechanics is a sufficiently regular, injective map $y:\Omega \to {\mathbb{R}}^3$ with $\det Dy>0$. Injectivity excludes material interpenetration, and a positive determinant preserves orientation and local volume. We use $C^1$ regularity for these kinematics and additional differentiability where stress-divergence identities are applied.

Cauchy's stress theorem states that the contact force on an oriented surface element at a current point depends on the point and its unit normal, rather than on the shape of the chosen subbody. Thus the contact force on the boundary of a current subbody $v$ has the form ${\int }_{\partial v}t(z,n)da$. Static force balance, for a body force $b$ per unit current volume, isfor every subbody. A shrinking pillbox gives $t(z,-n)=-t(z,n)$; a shrinking tetrahedron then gives linear dependence on the normal, $t(z,n)=S(z)n$. Applying the divergence theorem and using arbitrary subbodies givesIf body force is specified per unit mass instead, replace $b$ by $\rho b$. Symmetry of $S$ follows additionally from conservation of angular momentum in the absence of body couples; it is not a consequence of force balance alone.

The surface element relation is Nanson's formula. The force on a corresponding pair of current and reference surface elements must agree:Therefore the first Piola-Kirchhoff stress tensor isIt measures force per reference area, with a current force direction and a reference normal direction.

In components, differentiate $T_{iA}=S_{ij}(y(x))(\mathrm{cof}F{)}_{jA}$. The Piola identity says ${\partial }_A(\mathrm{cof}Dy{)}_{jA}=0$. One proof writes the cofactor matrix as $\frac{1}{2}{\epsilon }_{jkl}{\epsilon }_{ABC}{y}_{k,B}{y}_{l,C}$; differentiating in $A$ produces symmetric second derivatives contracted against antisymmetric epsilon symbols, hence zero. The chain rule consequently givessince $F_{kA}(\mathrm{cof}F{)}_{jA}=J{\delta }_{kj}$. Thus, with the coordinate composition made explicit,The distinction between reference divergence and current divergence is essential to this identity.

For $R=|x|>0$, $DR=\theta$ and $D\theta =(I-\theta \otimes \theta )/R$. Differentiating the spherically symmetric deformation givesFor an orientation-preserving spherically symmetric deformation with positive radius, $r>0$ and $r^{\prime }>0$. The radial direction and its perpendicular plane are eigenspaces of this positive symmetric matrix, soAt a positive diagonal matrix $D$, every diagonal rotation with two minus signs commutes with $D$. The supplied equivariance relation therefore forces $\hat{T}(D)$ to be a diagonal matrix: each off-diagonal entry changes sign under one such rotation while $D$ stays fixed. Varying a positive diagonal entry shows that the diagonal entries of $\hat{T}(D)$ are ${\Phi }_{,j}$. Permutation symmetry of $\Phi$ implies ${\Phi }_{,2}={\Phi }_{,3}$ when ${\lambda }_2={\lambda }_3$.

Choose a rotation whose first column is $\theta$. Applying equivariance to $F=QD{Q}^T$ now giveswith the derivatives evaluated at $(r^{\prime },r/R,r/R)$. Thus the stress has one radial and one repeated tangential principal value, independently of the choice of tangent basis.

The Green-Lagrange strain tensor has eigenvalues $({\lambda }_j^2-1)/2$. Its Frobenius norm and trace are invariant under the orthogonal principal-axis change, soFor the homogeneous dilation $F=\rho I$, put $K=\mu +3\lambda /2$. Then $E=({\rho }^2-1)I/2$. Differentiating the energy through $E$ gives the second Piola-Kirchhoff stress tensor $2\mu E+\lambda \mathrm{tr}(E)I=K({\rho }^2-1)I$, and multiplication by $F$ gives the first Piola-Kirchhoff stress tensor. ThusThe normals of the reference and deformed spheres point in the same radial direction. As $\rho \downarrow 0$, the reference traction magnitude $|TN|$ tends to zero, while the current traction magnitude $|Sn|$ tends to infinity, provided $K>0$, as for physically stable Lamé parameters. The negative signs of the stress coefficients for $0<\rho <1$ indicate compression.

There is no contradiction: the current area is ${\rho }^2$ times the reference area, so the same force divided by the vanishing current area has the larger stress. Indeed $TN={\rho }^{2}Sn$. The limiting map at $\rho =0$ is not itself an admissible deformation in continuum mechanics; the statement concerns the approach through positive $\rho$. A degenerate choice $K=0$ would not have the asserted growth without bound.