Introduction to Elasticity/Constitutive example 5
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[edit] Example 5
Given:
An isotropic material with Young's modulus E and Poisson's ration ν.
Find:
The compliance matrix of the material in terms of the Young's modulus and Poisson's ratio.
[edit] Solution
The strain is related to the stress via the compliance matrix by the equation
For an isotropic material
![\varepsilon_{ij} = \frac{1}{E}
\left[(1+\nu)\sigma_{ij} - \nu\sigma_{kk}\delta_{ij}\right]](http://upload.wikimedia.org/math/8/7/7/8778b1fb70b2ee1cba2c09315b0f0012.png)
Therefore,
![\begin{align}
\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{align}](http://upload.wikimedia.org/math/f/4/e/f4ed3e24574c25ccd17d1e654a129d29.png)
In engineering notation,
![\begin{align}
\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{align}](http://upload.wikimedia.org/math/4/c/2/4c208e30ebe69a2ad845b29c944833a8.png)
Converting into matrix notation,
We may also write the above equation as
where μ is the shear modulus.
