Example 5
edit
Given:
An isotropic material with Young's modulus 
and Poisson's ration 
.
Find:
The compliance matrix of the material in terms of the Young's modulus and Poisson's ratio.
Solution
edit
The strain is related to the stress via the compliance matrix by
the equation
For an isotropic material
![{\displaystyle \varepsilon _{ij}={\frac {1}{E}}\left[(1+\nu )\sigma _{ij}-\nu \sigma _{kk}\delta _{ij}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e454b334db32ee670a27126821b68f64e05c9f4d)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc4c87c8d2923510983dda0e364e7b791d3316d8)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d5f2f50f39efc4dfab6c537feaea0aa8c253970)
Converting into matrix notation,
We may also write the above equation as
where 
is the shear modulus.
