 Solution

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

e_{R} = X / R

, so

X \otimes X / R^{2} = e_{R} \otimes e_{R}

. The identity matrix is the sum of the three basis outer products. Substitution in the given deformation gradient gives

F = f^{'} (R) e_{R} \otimes e_{R} + \frac{f ( R )}{R} (e_{Θ} \otimes e_{Θ} + e_{Φ} \otimes e_{Φ}) = diag (f^{'} (R), \frac{f ( R )}{R}, \frac{f ( R )}{R}) .

The reference and current radial directions coincide for this spherically symmetric deformation. The positive principal stretches are therefore

λ_{r} = f^{'} (R)

and

λ_{θ} = λ_{ϕ} = f (R) / R

f^{'} (R) \frac{f ( R ) ^{2}}{R ^{2}} = 1, \frac{d}{d R} f (R)^{3} = 3 R^{2} .

Integrating and using

f (A) = a

gives

r = f (R) = (R^{3} + a^{3} - A^{3})^{1/3}, b^{3} = B^{3} + a^{3} - A^{3} .

In particular

b^{3} - a^{3} = B^{3} - A^{3}

, expressing conservation of shell volume. Writing

λ = r / R

also gives

λ_{r} = λ^{- 2}

and

λ_{θ} = λ_{ϕ} = λ

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