Pure shear

The deformation gradient for pure shear is given by:

${\displaystyle F=\left[\begin{array}{ccc}1& \gamma & 0\\ \gamma & 1& 0\\ 0& 0& 1\end{array}\right]}$

Note that this gives a Green-Lagrange strain of:

${\displaystyle E=\frac{1}{2}\left[\begin{array}{ccc}{\gamma }^2& 2\gamma & 0\\ 2\gamma & {\gamma }^2& 0\\ 0& 0& 0\end{array}\right]}$

Here there is no rotation occurring, which can be seen from the equal off-diagonal components of the strain tensor. The linear approximation to the Green-Lagrange strain shows that the small strain tensor is:

${\displaystyle ϵ=\frac{1}{2}\left[\begin{array}{ccc}0& 2\gamma & 0\\ 2\gamma & 0& 0\\ 0& 0& 0\end{array}\right]}$

which has only shearing components.