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

$\left[\begin{array}{c}{\epsilon }_1\\ {\epsilon }_2\\ {\epsilon }_3\\ {\epsilon }_4\\ {\epsilon }_5\\ {\epsilon }_6\end{array}\right]=\left[\begin{array}{cccccc}{S}_{11}& S_{12}& S_{13}& S_{14}& S_{15}& S_{16}\\ S_{21}& S_{22}& S_{23}& S_{24}& S_{25}& S_{26}\\ S_{31}& S_{32}& S_{33}& S_{34}& S_{35}& S_{36}\\ S_{41}& S_{42}& S_{43}& S_{44}& S_{45}& S_{46}\\ S_{51}& S_{52}& S_{53}& S_{54}& S_{55}& S_{56}\\ S_{61}& S_{62}& S_{63}& S_{64}& S_{65}& S_{66}\end{array}\right]\left[\begin{array}{c}{\sigma }_1\\ {\sigma }_2\\ {\sigma }_3\\ {\sigma }_4\\ {\sigma }_5\\ {\sigma }_6\end{array}\right]$

For an isotropic material

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

Therefore,

$\begin{array}{cc}{\epsilon }_{11}& =\frac{1}{E}[{\sigma }_{11}-\nu {\sigma }_{22}-\nu {\sigma }_{33}]\\ {\epsilon }_{22}& =\frac{1}{E}[{\sigma }_{22}-\nu {\sigma }_{11}-\nu {\sigma }_{33}]\\ {\epsilon }_{33}& =\frac{1}{E}[{\sigma }_{33}-\nu {\sigma }_{11}-\nu {\sigma }_{22}]\\ {\epsilon }_{23}& =\frac{1}{E}[(1+\nu ){\sigma }_{23}]\\ {\epsilon }_{31}& =\frac{1}{E}[(1+\nu ){\sigma }_{31}]\\ {\epsilon }_{12}& =\frac{1}{E}[(1+\nu ){\sigma }_{12}]\end{array}$

In engineering notation,

$\begin{array}{cc}{\epsilon }_1& =\frac{1}{E}[{\sigma }_1-\nu {\sigma }_2-\nu {\sigma }_3]\\ {\epsilon }_2& =\frac{1}{E}[{\sigma }_2-\nu {\sigma }_1-\nu {\sigma }_3]\\ {\epsilon }_3& =\frac{1}{E}[{\sigma }_3-\nu {\sigma }_1-\nu {\sigma }_2]\\ {\epsilon }_4& =\frac{1}{E}[2(1+\nu ){\sigma }_4]\\ {\epsilon }_5& =\frac{1}{E}[2(1+\nu ){\sigma }_5]\\ {\epsilon }_6& =\frac{1}{E}[2(1+\nu ){\sigma }_6]\end{array}$

Converting into matrix notation,

$\left[\begin{array}{c}{\epsilon }_1\\ {\epsilon }_2\\ {\epsilon }_3\\ {\epsilon }_4\\ {\epsilon }_5\\ {\epsilon }_6\end{array}\right]=\frac{1}{E}\left[\begin{array}{cccccc}1& -\nu & -\nu & 0& 0& 0\\ -\nu & 1& -\nu & 0& 0& 0\\ -\nu & -\nu & 1& 0& 0& 0\\ 0& 0& 0& 2(1+\nu )& 0& 0\\ 0& 0& 0& 0& 2(1+\nu )& 0\\ 0& 0& 0& 0& 0& 2(1+\nu )\end{array}\right]\left[\begin{array}{c}{\sigma }_1\\ {\sigma }_2\\ {\sigma }_3\\ {\sigma }_4\\ {\sigma }_5\\ {\sigma }_6\end{array}\right]$

We may also write the above equation as

$\left[\begin{array}{c}{\epsilon }_1\\ {\epsilon }_2\\ {\epsilon }_3\\ {\epsilon }_4\\ {\epsilon }_5\\ {\epsilon }_6\end{array}\right]=\left[\begin{array}{cccccc}1/E& -\nu /E& -\nu /E& 0& 0& 0\\ -\nu /E& 1/E& -\nu /E& 0& 0& 0\\ -\nu /E& -\nu /E& 1/E& 0& 0& 0\\ 0& 0& 0& 1/\mu & 0& 0\\ 0& 0& 0& 0& 1/\mu & 0\\ 0& 0& 0& 0& 0& 1/\mu \end{array}\right]\left[\begin{array}{c}{\sigma }_1\\ {\sigma }_2\\ {\sigma }_3\\ {\sigma }_4\\ {\sigma }_5\\ {\sigma }_6\end{array}\right]$

where $\mu$ is the shear modulus.