 Solution

ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2009/1/b/ii/solution

Differentiating the displacement gives

E = α diag (1, - ν, - ν), tr E = α (1 - 2 ν) .

τ_{xx} = α [λ (1 - 2 ν) + 2 μ], τ_{yy} = τ_{zz} = α [λ (1 - 2 ν) - 2 μν] .

The two transverse traction-free conditions set the latter entries to zero. Hence

λ = 2 ν (λ + μ)

and

ν = \frac{λ}{2 ( λ + μ )}, τ = diag (E α, 0, 0), E = \frac{μ ( 3 λ + 2 μ )}{λ + μ} .

Substitution shows

E = 2 μ (1 + ν)

; solving for the two Lamé parameters gives

μ = \frac{E}{2 ( 1 + ν )}, λ = \frac{ν E}{( 1 + ν ) ( 1 - 2 ν )} .

To determine the allowed interval without assuming signs not given, put

K = λ + 2 μ /3 > 0

. Then

ν = (3 K - 2 μ) / [2 (3 K + μ)]

, so

1 + ν = \frac{9 K}{2 ( 3 K + μ )} > 0, \frac{1}{2} - ν = \frac{3 μ}{2 ( 3 K + μ )} > 0.

Therefore

- 1 < ν < 1/2

. The endpoints would require a degenerate elastic modulus, so they are excluded by the strict material inequalities.

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