User:Eas4200c.f08.vqcrew.E/Homework5

Linear Stress-Strain Relations
The expressions giving the strain induced by a stress can be rewitten to allow for the deriviation of a second expression relating the stress induced for a given strain.
 * $$ \begin{bmatrix} \varepsilon_{11} \\ \varepsilon_{22} \\ \varepsilon_{33} \\ \varepsilon_{23} \\ \varepsilon_{13} \\ \varepsilon_{12} \end{bmatrix} = \begin{bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0 \\ \frac{-\nu}{E} & \frac{1}{E} & \frac{-\nu}{E} & 0 & 0 & 0 \\ \frac{-\nu}{E} & \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2G} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2G} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G} \end{bmatrix} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{13} \\ \sigma_{12} \end{bmatrix} $$

Becomes,


 * $$\left\{\varepsilon_{ij} \right\}_{6x1} = \begin{bmatrix}A_{3x3} & 0_{3x3} \\ 0_{3x3} & B_{3x3} \end{bmatrix}\left\{\sigma_{ij} \right\}_{6x1}$$

Where $$A_{3x3} = \begin{bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} \\ \frac{-\nu}{E} & \frac{1}{E} & \frac{-\nu}{E} \\ \frac{-\nu}{E} & \frac{-\nu}{E} & \frac{1}{E}\end{bmatrix}$$ and $$B_{3x3} = \begin{bmatrix}\frac{1}{2G} & 0 & 0 \\ 0 & \frac{1}{2G} & 0 \\ 0 & 0 & \frac{1}{2G}\end{bmatrix}$$

Solving for the stress tensor,


 * $$\left\{\sigma_{ij} \right\}_{6x1} = \begin{bmatrix}A^{-1}_{3x3} & 0_{3x3} \\ 0_{3x3} & B^{-1}_{3x3} \end{bmatrix}\left\{\varepsilon_{ij} \right\}_{6x1}$$

In order to verify this result, a new matrix is defined as


 * $$\begin{bmatrix}c\end{bmatrix}_{6x6} = \begin{bmatrix}A_{3x3} & 0_{3x3} \\ 0_{3x3} & B_{3x3} \end{bmatrix}$$ with inverse equal to $$\begin{bmatrix}c\end{bmatrix}^{-1}_{6x6} = \begin{bmatrix}A^{-1}_{3x3} & 0_{3x3} \\ 0_{3x3} & B^{-1}_{3x3} \end{bmatrix}$$

Due to the definition of the inverse, it is known that the multiplication of matrix c with its inverse will yield the identity matrix, I.


 * $$\begin{bmatrix}c\end{bmatrix}_{6x6}\begin{bmatrix}c\end{bmatrix}^{-1}_{6x6} = \begin{bmatrix}I\end{bmatrix}_{6x6}$$


 * $$\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}c\end{bmatrix}^{-1} = \begin{bmatrix}AA^{-1} & 0 \\ 0 & BB^{-1} \end{bmatrix} = \begin{bmatrix}I\end{bmatrix}$$

Now, the stress-strain relation can be found using the inverted matrices, A and B.


 * $$\left\{\sigma_{ij} \right\} = \begin{bmatrix}A^{-1} & 0_{3x3} \\ 0 & B^{-1} \end{bmatrix}\left\{\varepsilon_{ij} \right\} = \begin{bmatrix} \frac{E(v-1)}{(2v^2+v-1)} & \frac{-Ev}{(2v^2+v-1)} & \frac{-Ev}{(2v^2+v-1)} & 0 & 0 & 0\\ \frac{-Ev}{(2v^2+v-1)} & \frac{E(v-1)}{(2v^2+v-1)}  & \frac{-Ev}{(2v^2+v-1)} & 0 & 0 & 0\\ \frac{-Ev}{(2v^2+v-1)} & \frac{-Ev}{(2v^2+v-1)} & \frac{E(v-1)}{(2v^2+v-1)}  & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{2G} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G} \end{bmatrix}\begin{Bmatrix}0 \\ 0 \\ 0 \\ 0 \\ \varepsilon_{31} \\ \varepsilon_{12} \end{Bmatrix} \;$$