Past exam of the elasticity and plasticity course of the University of Oxford / 2009 / 1 / b / i / Solution

For a surface with unit normal $\mathbf{n}$, traction is $t_i={\tau }_{ij}{n}_j$. Thus ${\tau }_{ij}$ is the $i$th force-per-area component on a plane whose positive normal is ${\mathbf{e}}_j$. Diagonal entries are normal stresses and off-diagonal entries are shear stresses, with tensile normal stress positive in this convention.

The symmetric strain has six independent components and so does the symmetric stress. A general linear map between these two six-dimensional spaces therefore hasEquivalently choose the minor symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$. Components antisymmetric in the strain indices have no observable effect and are omitted from this representative.

A further distinction is important: if a strain energy density exists, $C_{ijkl}={\partial }^{2}W/(\partial e_{ij}\partial e_{kl})$ has the major symmetry $C_{ijkl}=C_{klij}$. Then the six-by-six map is symmetric and has $6\cdot 7/2=21$ independent coefficients. Thus 36 follows from the two stated tensor symmetries alone; 21 additionally uses the energy/reciprocity assumption, as in the variational setting later in the paper. These are the minor and major symmetries of linear elasticity.