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

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.