Elasticity/Constitutive example 2

Example 1

Convert the stress-strain relation for isotropic materials (in matrix form) into an equation in index notation. Show all the steps in the process.

Solution

The stress-strain relation is

$\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ {\epsilon }_{33}\\ {\epsilon }_{23}\\ {\epsilon }_{31}\\ {\epsilon }_{12}\end{array}\right]=\frac{1}{E}\left[\begin{array}{cccccc}1& -\nu & -\nu & 0& 0& 0\\ -\nu & 1& -\nu & 0& 0& 0\\ -\nu & -\nu & 1& 0& 0& 0\\ 0& 0& 0& 1+\nu & 0& 0\\ 0& 0& 0& 0& 1+\nu & 0\\ 0& 0& 0& 0& 0& 1+\nu \end{array}\right]\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{33}\\ {\sigma }_{23}\\ {\sigma }_{31}\\ {\sigma }_{12}\end{array}\right]$

Let us expand out the terms and put all of them in a similar form. Thus,

$\begin{array}{cc}E{\epsilon }_{11}& =(1+\nu ){\sigma }_{11}-\nu ({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})(1)\\ E{\epsilon }_{22}& =(1+\nu ){\sigma }_{22}-\nu ({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})(1)\\ E{\epsilon }_{33}& =(1+\nu ){\sigma }_{33}-\nu ({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})(1)\\ E{\epsilon }_{23}& =(1+\nu ){\sigma }_{23}-\nu ({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})(0)\\ E{\epsilon }_{31}& =(1+\nu ){\sigma }_{31}-\nu ({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})(0)\\ E{\epsilon }_{12}& =(1+\nu ){\sigma }_{12}-\nu ({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})(0)\end{array}$

We know that ${\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33}={\sigma }_{kk}$. Also a quantity that is $1$ when $i=j$ and $0$ when $i\ne j$ can be represented by the Kronecker $\delta$. Therefore, we can write the above equations as

$E{\epsilon }_{ij}=(1+\nu ){\sigma }_{ij}-\nu {\sigma }_{kk}{\delta }_{ij}$

or,

${\epsilon }_{ij}=\frac{(1+\nu )}{E}{\sigma }_{ij}-\frac{\nu }{E}{\sigma }_{kk}{\delta }_{ij}$