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,

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

We know that $\sigma _{11}+\sigma _{22}+\sigma _{33}=\sigma _{kk}$. Also a quantity that is $1$ when $i=j$ and $0$ when $i\neq j$ 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,

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

Elasticity