# Introduction to Elasticity/Constitutive example 5

## Example 5

Given:

An isotropic material with Young's modulus ${\displaystyle E}$ and Poisson's ration ${\displaystyle \nu }$.

Find:

The compliance matrix of the material in terms of the Young's modulus and Poisson's ratio.

### Solution

The strain is related to the stress via the compliance matrix by the equation

${\displaystyle {\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{bmatrix}}={\begin{bmatrix}S_{11}&S_{12}&S_{13}&S_{14}&S_{15}&S_{16}\\S_{21}&S_{22}&S_{23}&S_{24}&S_{25}&S_{26}\\S_{31}&S_{32}&S_{33}&S_{34}&S_{35}&S_{36}\\S_{41}&S_{42}&S_{43}&S_{44}&S_{45}&S_{46}\\S_{51}&S_{52}&S_{53}&S_{54}&S_{55}&S_{56}\\S_{61}&S_{62}&S_{63}&S_{64}&S_{65}&S_{66}\end{bmatrix}}{\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}}$

For an isotropic material

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

Therefore,

{\displaystyle {\begin{aligned}\varepsilon _{11}&={\frac {1}{E}}\left[\sigma _{11}-\nu \sigma _{22}-\nu \sigma _{33}\right]\\\varepsilon _{22}&={\frac {1}{E}}\left[\sigma _{22}-\nu \sigma _{11}-\nu \sigma _{33}\right]\\\varepsilon _{33}&={\frac {1}{E}}\left[\sigma _{33}-\nu \sigma _{11}-\nu \sigma _{22}\right]\\\varepsilon _{23}&={\frac {1}{E}}\left[(1+\nu )\sigma _{23}\right]\\\varepsilon _{31}&={\frac {1}{E}}\left[(1+\nu )\sigma _{31}\right]\\\varepsilon _{12}&={\frac {1}{E}}\left[(1+\nu )\sigma _{12}\right]\end{aligned}}}

In engineering notation,

{\displaystyle {\begin{aligned}\varepsilon _{1}&={\frac {1}{E}}\left[\sigma _{1}-\nu \sigma _{2}-\nu \sigma _{3}\right]\\\varepsilon _{2}&={\frac {1}{E}}\left[\sigma _{2}-\nu \sigma _{1}-\nu \sigma _{3}\right]\\\varepsilon _{3}&={\frac {1}{E}}\left[\sigma _{3}-\nu \sigma _{1}-\nu \sigma _{2}\right]\\\varepsilon _{4}&={\frac {1}{E}}\left[2(1+\nu )\sigma _{4}\right]\\\varepsilon _{5}&={\frac {1}{E}}\left[2(1+\nu )\sigma _{5}\right]\\\varepsilon _{6}&={\frac {1}{E}}\left[2(1+\nu )\sigma _{6}\right]\end{aligned}}}

Converting into matrix notation,

${\displaystyle {\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{bmatrix}}={\frac {1}{E}}{\begin{bmatrix}1&-\nu &-\nu &0&0&0\\-\nu &1&-\nu &0&0&0\\-\nu &-\nu &1&0&0&0\\0&0&0&2(1+\nu )&0&0\\0&0&0&0&2(1+\nu )&0\\0&0&0&0&0&2(1+\nu )\end{bmatrix}}{\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}}$

We may also write the above equation as

${\displaystyle {\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{bmatrix}}={\begin{bmatrix}1/E&-\nu /E&-\nu /E&0&0&0\\-\nu /E&1/E&-\nu /E&0&0&0\\-\nu /E&-\nu /E&1/E&0&0&0\\0&0&0&1/\mu &0&0\\0&0&0&0&1/\mu &0\\0&0&0&0&0&1/\mu \end{bmatrix}}{\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}}$

where ${\displaystyle \mu }$ is the shear modulus.