 Solution

ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2010/1/c/solution

The traction on the oriented solid element is

σ = τ n, σ_{i} = τ_{ij} n_{j} .

With constant Lamé coefficients and derivatives in the reference variables,

\partial_{j} τ_{ij} = λ \partial_{i} (\partial_{k} u_{k}) + μ \partial_{j} (\partial_{j} u_{i} + \partial_{i} u_{j}) = (λ + μ) \partial_{i} (\nabla \cdot u) + μ \nabla^{2} u_{i} .

ρ u_{tt} = (λ + μ) \nabla (\nabla \cdot u) + μ \nabla^{2} u + ρ g .

To check the vector identity componentwise, contract the two alternating tensors:

[\nabla \times (\nabla \times u)]_{i} = ϵ_{ijk} ϵ_{k l m} \partial_{j} \partial_{l} u_{m} = \partial_{i} \partial_{j} u_{j} - \partial_{j} \partial_{j} u_{i} .

Rearranging proves

\nabla^{2} u = \nabla (\nabla \cdot u) - \nabla \times (\nabla \times u)

. Therefore

ρ u_{tt} = (λ + 2 μ) \nabla (\nabla \cdot u) - μ \nabla \times (\nabla \times u) + ρ g .

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