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

$\mathbf {D} _{\text{lam}}=\mathbf {R} _{\text{lam}}^{T}\mathbf {D} \mathbf {R} _{\text{lam}}={\begin{bmatrix}2.4837&-0.5\\-0.5&0\end{bmatrix}}~.$

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);

Nonlinear finite elements/Homework 6/Solutions/Problem 1