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Year 4 of the mathematics course of the University of Oxford

Elasticity and plasticity course of the University of Oxford

Past exam of the elasticity and plasticity course of the University of Oxford

2010

The <traction> on the oriented solid element is
$$
\boxed{\boldsymbol\sigma=\tau\mathbf n,\qquad\sigma_i=\tau_{ij}n_j.}
$$
With constant Lamé coefficients and derivatives in the reference variables,
$$
\partial_j\tau_{ij}
=\lambda\partial_i(\partial_ku_k)
+\mu\partial_j(\partial_ju_i+\partial_iu_j)
=(\lambda+\mu)\partial_i(\nabla\cdot\mathbf u)+\mu\nabla^2u_i.
$$
Substitution into <balance of linear momentum> gives
$$
\boxed{\rho\mathbf u_{tt}=(\lambda+\mu)\nabla(\nabla\cdot\mathbf u)
+\mu\nabla^2\mathbf u+\rho\mathbf g.}
$$
To check the <vector> identity componentwise, contract the two alternating <tensors>:
$$
[\nabla\times(\nabla\times\mathbf u)]_i
=\epsilon_{ijk}\epsilon_{klm}\partial_j\partial_lu_m
=\partial_i\partial_ju_j-\partial_j\partial_ju_i.
$$
Rearranging proves $\nabla^2\mathbf u=\nabla(\nabla\cdot\mathbf u)-\nabla\times(\nabla\times\mathbf u)$. Therefore
$$
\boxed{\rho\mathbf u_{tt}=(\lambda+2\mu)\nabla(\nabla\cdot\mathbf u)
-\mu\nabla\times(\nabla\times\mathbf u)+\rho\mathbf g.}
$$

Solution

Past exam of the elasticity and plasticity course of the University of Oxford / 2010 / 1 / c / Solution

Past exam of the elasticity and plasticity course of the University of Oxford / 2010 / 1 / c / Solution

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