Past exam of the elasticity and plasticity course of the University of Oxford / 2011 / 1 / h / Solution

The radial equilibrium equation gives $d{t}_r/dr=(t_{\theta }-t_r)/r$. Set $C=a^2-A^2$, so that $R^2=r^2-C$. Substitution of the neo-Hookean solid stress difference eliminates both $q$ and $t_{\theta }$:This is a first-order ordinary differential equation for the radial stress, parametrized by the unknown inner radius. For inflation $C>0$, its right side is positive, so $t_r$ increases from the negative inner pressure to zero at the outer surface. The denominator stays positive because $r^2-C=R^2\ge A^2>0$.