Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 12

Problem 1: Part 12
Rotate the rate of deformation so that its components are with respect to the laminar coordinate system.

The global base vectors are

\mathbf{e}_x = \begin{bmatrix} 1 \\ 0 \end{bmatrix}; ~\qquad \mathbf{e}_y = \begin{bmatrix} 0 \\ 1 \end{bmatrix}~. $$

Therefore, the rotation matrix is

\mathbf{R}_{\text{lam}} = \begin{bmatrix} \mathbf{e}_x\bullet\widehat{\mathbf{e}_x} & \mathbf{e}_x\bullet\widehat{\mathbf{e}_y} \\ \mathbf{e}_y\bullet\widehat{\mathbf{e}_x} & \mathbf{e}_y\bullet\widehat{\mathbf{e}_y} \end{bmatrix} = \begin{bmatrix} 0.7071 &0.7071 \\ -0.7071 &0.7071 \end{bmatrix} ~. $$

Therefore, the components of the rate of deformation tensor with respect to the laminar coordinate system are

\mathbf{D}_{\text{lam}} = \mathbf{R}^T_{\text{lam}}\mathbf{D}\mathbf{R}_{\text{lam}} = \begin{bmatrix} 2.4837 & -0.5 \\ -0.5 &0 \end{bmatrix} ~. $$ The Maple script used to compute the above is shown below.