 Solution

ID: 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} / d r = (t_{θ} - t_{r}) / r

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_{θ}

\frac{d t _{r}}{d r} = \frac{μ}{r} [\frac{r ^{2}}{r ^{2} - C} - \frac{r ^{2} - C}{r ^{2}}] = \frac{μ C ( 2 r ^{2} - C )}{r ^{3} ( r ^{2} - C )}, a < r < b .

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} \geq A^{2} > 0

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