Introduction to Elasticity/Constitutive example 5

Example 5

Given:

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

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

\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

\varepsilon_{ij} = \frac{1}{E} \left[(1+\nu)\sigma_{ij} - \nu\sigma_{kk}\delta_{ij}\right]

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}

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}

Converting into matrix notation,

\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

\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 $\mu$ is the shear modulus.

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