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

\begin{bmatrix} \varepsilon_{11}\\\varepsilon_{22}\\\varepsilon_{33}\\\varepsilon_{23}\\\varepsilon_{31}\\\varepsilon_{12} \end{bmatrix} = \frac{1}{E} \begin{bmatrix} 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{bmatrix} \begin{bmatrix} \sigma_{11}\\\sigma_{22}\\\sigma_{33}\\\sigma_{23}\\\sigma_{31}\\\sigma_{12} \end{bmatrix} $$ Let us expand out the terms and put all of them in a similar form. Thus,
 * $$\begin{align}

E \varepsilon_{11} & = (1+\nu)\sigma_{11} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(1)\\ E \varepsilon_{22} & = (1+\nu)\sigma_{22} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(1)\\ E \varepsilon_{33} & = (1+\nu)\sigma_{33} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(1)\\ E \varepsilon_{23} & = (1+\nu)\sigma_{23} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(0)\\ E \varepsilon_{31} & = (1+\nu)\sigma_{31} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(0)\\ E \varepsilon_{12} & = (1+\nu)\sigma_{12} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(0) \end{align}$$ We know that $$\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 \varepsilon_{ij} = (1+\nu)\sigma_{ij} - \nu\sigma_{kk}\delta_{ij} $$ or,

\varepsilon_{ij} = \frac{(1+\nu)}{E}\sigma_{ij} - \frac{\nu}{E}\sigma_{kk}\delta_{ij} $$