 Solution

ID: past-exam-of-the-solid-mechanics-course-of-the-university-of-oxford/2023/2/d/i/solution

α = 1

, with positive shear modulus

μ

. Substitution of the principal stretches gives

h (λ) = \frac{μ}{2} (λ^{- 4} + 2 λ^{2} - 3), h^{'} (λ) = 2 μ (λ - λ^{- 5}) .

The identity

λ^{6} - 1 = (λ^{3} - 1) (λ^{3} + 1)

simplifies the integrand:

\frac{h ^{'} ( λ )}{λ ^{3} - 1} = 2 μ (λ^{- 2} + λ^{- 5}) .

The two improper integrals are elementary:

P_{crit} = - 2 μ (\int_{1}^{\infty} λ^{- 2} d λ + \int_{1}^{\infty} λ^{- 5} d λ) = - 2 μ (1 + \frac{1}{4}) = - \frac{5}{2} μ .

Thus the critical tensile stress magnitude is

5 μ /2

. As a check on the finite-sphere limit, part (c) gives

P (a) = - 2 μ (λ_{b}^{- 1} + \frac{1}{4} λ_{b}^{- 4}), λ_{b} = (1 + a^{3} / B^{3})^{1/3},

which tends to

- 5 μ /2

as the cavity radius tends to zero.

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