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.

> #
> # Compute rate of deformation in laminar system
> #
> # Set up global base vectors
> #
> ex := vector([1,0,0]);
> ey := vector([0,1,0]);
> ez := vector([0,0,1]);
> #
> # Set up rotation matrix
> #
> ex_ehatx := dotprod(ex, ehat_x);
> ex_ehaty := dotprod(ex, ehat_y);
> ex_ehatz := dotprod(ex, ehat_z);
> ey_ehatx := dotprod(ey, ehat_x);
> ey_ehaty := dotprod(ey, ehat_y);
> ey_ehatz := dotprod(ey, ehat_z);
> ez_ehatx := dotprod(ez, ehat_x);
> ez_ehaty := dotprod(ez, ehat_y);
> ez_ehatz := dotprod(ez, ehat_z);
> Rlam := linalg[matrix](2,2,[ex_ehatx, ex_ehaty,
> ey_ehatx, ey_ehaty]);
> RlamT := transpose(Rlam);
> #
> # Compute rate of deformation in laminar system
> #
> Dlam := evalm(RlamT&*DefRate&*Rlam);