Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 15

Problem 1: Part 15
Express the stress rate and the modified rate of deformation in the global coordinate system.

The modified laminar rate of deformation is

\begin{bmatrix} D_{xx} \\ D_{yy} \\ D_{xy} \end{bmatrix} = \begin{bmatrix} 2.4837 \\ -14.902 \\ -0.5 \end{bmatrix} $$

Alternatively, we can write

\mathbf{D}_{\text{lam}} = \begin{bmatrix} 2.4837 & -0.5 \\ -0.5 & -14.902 \end{bmatrix} ~. $$

The modified laminar stress rate is

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{xy} \end{bmatrix} = \begin{bmatrix} 100 & 60 & 0\\ 60 & 10 & 0 \\ 0 & 0& 30 \end{bmatrix} \begin{bmatrix} 2.4837 \\ -14.902 \\ -0.5 \end{bmatrix} = \begin{bmatrix} -645.76 \\ 0 \\ -15 \end{bmatrix} ~. $$

Alternatively, we can write

\cfrac{D}{Dt}(\boldsymbol{\sigma}_{\text{lam}}) = \begin{bmatrix} -645.76 & -15 \\ -15 & 0 \end{bmatrix} ~. $$

To get the global stress rate and rate of deformation, we have to rotate the components to the global basis using

\cfrac{D}{Dt}(\boldsymbol{\sigma}) = \mathbf{R}_{\text{lam}} \cfrac{D}{Dt}(\boldsymbol{\sigma}_{\text{lam}}) \mathbf{R}^T_{\text{lam}} ~; \qquad \mathbf{D} = \mathbf{R}_{\text{lam}} \mathbf{D}_{\text{lam}} \mathbf{R}^T_{\text{lam}} ~. $$

Computing these quantities gives us

\cfrac{D}{Dt}(\boldsymbol{\sigma}) = \begin{bmatrix} -337.88 & 322.88 \\ 322.88 & -307.88 \end{bmatrix} ~. $$ and

\mathbf{D} = \begin{bmatrix} -6.7092 & -8.6929 \\ -8.6829 & -5.7092 \end{bmatrix} ~. $$

The Maple code for the above computations is given below.