User:Eas4200c.f08.WIKI.E/hw5

Mtg. 24 Monday 20 October 08
kinematic assumption drop down

4 zero strain components

5 zero stress components

$$\varepsilon -\sigma $$ relation in two forms (Tensorial 23-2 or Engineering 23-1b)

Rewrite $$\varepsilon -\sigma $$ relation: Full (not diagonal)

$$\begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}_{6x1}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix} \begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}_{6x1}$$

$$\varepsilon -\sigma $$ relation

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}= \begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix} \begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}$$

$$\mathbf{C}_{6x6}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix}$$

$$ \mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\mathbf{I}_{6x6}$$

$$\mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\begin{vmatrix} \mathbf{A_{3x3}A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B_{3x3}B}_{3x3}^{-1} \end{vmatrix}=\mathbf{I_{6x6}} $$

We are trying to prove that $$ \sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\gamma _{yz}=0$$

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}=

\begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix}\begin{Bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12} \end{Bmatrix}$$ $$\begin{matrix} \sigma _{11}=0\\ \sigma _{22}=0\\ \sigma _{33}=0\\ \sigma _{23}=2G\varepsilon _{23}\\ \sigma _{31}=2G\varepsilon _{31}\\ \sigma _{21}=2G\varepsilon _{12} \end{matrix}$$

Consider the 1D stress model