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

The displacement follows a material point from its reference position to its current position:The deformation gradient is the Jacobian matrix $F=\partial \mathbf{x}/\partial \mathbf{X}=I+{\nabla }_X\mathbf{u}$, and the volume Jacobian isFor an orientation-preserving physical deformation, $J>0$ measures the local volume expansion factor. Incompressibility requires $J=1$; conservation of mass gives $\rho (\mathbf{x},t)J={\rho }_0(\mathbf{X})$. For small displacement gradients, $J=1+{\nabla }_X\cdot \mathbf{u}+O(|{\nabla }_X\mathbf{u}{|}^2)$. If a spatial representation is desired, the same displacement is $\mathbf{u}(\mathbf{x},t)=\mathbf{x}-\mathbf{X}(\mathbf{x},t)$.

A Cartesian Cartesian second-rank tensor has components transforming under a constant orthogonal change of axes asHere ${\mathbf{X}}^{\prime }=P\mathbf{X}$ and ${\mathbf{u}}^{\prime }=P\mathbf{u}$. By the chain rule and orthogonality,Thus the displacement gradient is a Cartesian second-rank tensor. Its symmetric part transforms in the same way:Hence the infinitesimal strain tensor is also second-rank. This is a passive coordinate rotation, under which both reference and spatial coordinates rotate; it does not assert finite-deformation frame indifference of the linearized strain formula.

The traction on the oriented solid element isWith constant Lamé coefficients and derivatives in the reference variables,Substitution into balance of linear momentum givesTo check the vector identity componentwise, contract the two alternating tensors:Rearranging proves ${\nabla }^2\mathbf{u}=\nabla (\nabla \cdot \mathbf{u})-\nabla \times (\nabla \times \mathbf{u})$. Therefore