Introduction to 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

${\displaystyle {\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\\varepsilon _{23}\\\varepsilon _{31}\\\varepsilon _{12}\end{bmatrix}}={\frac {1}{E}}{\begin{bmatrix}1&-\nu &-\nu &0&0&0\\-\nu &1&-\nu &0&0&0\\-\nu &-\nu &1&0&0&0\\0&0&0&1+\nu &0&0\\0&0&0&0&1+\nu &0\\0&0&0&0&0&1+\nu \\\end{bmatrix}}{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{31}\\\sigma _{12}\end{bmatrix}}}$

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

{\displaystyle {\begin{aligned}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{aligned}}}

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

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

or,

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