Past exam of the elasticity and plasticity course of the University of Oxford / 2008 / 1

The infinitesimal strain is $e_{ij}=(u_{i,j}+u_{j,i})/2$. A constant translation has zero gradient. For the remaining term, $u_i={ϵ}_{ikl}{\omega }_k{x}_l$ has $u_{i,j}={ϵ}_{ikj}{\omega }_k$, a skew-symmetric matrix. ThereforePhysically $\mathbf{a}$ is a translation and $\mathbf{\omega }\times \mathbf{x}$ is an infinitesimal rigid rotation, with axis along $\mathbf{\omega }$ and rotation angle $|\mathbf{\omega }|$. It is the linearized rotation displacement, not the exact displacement of a general finite rotation. This distinction matters because the infinitesimal strain omits quadratic displacement-gradient terms.

Choose zero energy in the unstrained state. The required strain energy density isIts variation is $\delta W=(\lambda \mathrm{tr}(e)I+2\mu e):\delta e$, giving the prescribed stress as the tensor gradient. This variational notation handles the symmetric off-diagonal components without a spurious factor of two.

Decompose $e=e^{\mathrm{dev}}+(\mathrm{tr}e)I/3$ into orthogonal deviatoric and spherical parts. ThenThe two squares vanish together only for $e=0$. Thus $\mu >0$ and $3\lambda +2\mu >0$ are sufficient for positive definiteness. They are necessary by testing a nonzero trace-free strain and a pure dilation separately. Equivalently both shear modulus $\mu$ and bulk modulus $K=\lambda +2\mu /3$ must be positive.

Multiply equilibrium by the displacement and integrate. Integration by parts and the traction condition giveThe stress is symmetric, so it contracts only with the symmetric part of $\nabla \mathbf{u}$, namely $e$. Hooke's law then gives $T:e=\lambda (\mathrm{tr}e{)}^2+2\mu e:e=2W$. HenceThe 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.

Apply energy uniqueness for a linear elastic traction problem. Subtract two solutions with the same loading, writing their displacement difference as $\mathbf{v}$. Linearity makes its body force and boundary traction zero. The preceding energy identity therefore yields ${\int }_{B}2W(e(\mathbf{v}))dV=0$. Positive definiteness and continuity imply $e(\mathbf{v})=0$ everywhere. By the permitted zero-strain characterization,Thus the solution is unique modulo infinitesimal rigid translations and rotations, on each connected body. Conversely adding such a displacement changes neither strain nor stress, so these are exactly the unconstrained modes. This is conditional uniqueness; existence of a pure-traction solution additionally requires balanced resultant force and torque.