 Solution

ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2008/1/c/solution

Multiply equilibrium by the displacement and integrate. Integration by parts and the traction condition give

0 = \int_{B} u \cdot (\nabla \cdot T + ρ g) d V = \int_{\partial B} u \cdot σ d S - \int_{B} T : \nabla u d V + \int_{B} ρ u \cdot g d V .

The stress is symmetric, so it contracts only with the symmetric part of

\nabla u

, namely

e

. Hooke's law then gives

T : e = λ (tr e)^{2} + 2 μ e : e = 2 W

. Hence

\int_{B} ρ u \cdot g d V = \int_{B} 2 W d V - \int_{\partial B} u \cdot σ d S .

The identity is the linear-elastic work balance: twice the stored quadratic energy equals the body-force and boundary-traction work paired with the final displacement.

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