User:Eml4500.f08.jamama.justin/HW4

Note: Consider the case: $$\tilde{d}{_{4}}^{(e)}\neq 0$$

$$\tilde{d}{_{1}}^{(e)}=\tilde{d}{_{2}}^{(e)}=\tilde{d}{_{3}}^{(e)}=0$$

$$\tilde{f}^{(e)}=\tilde{k}^{(e)}\tilde{d}^{(e)}=0$$ 4X1  4X4  4X1   4X1

Interp of transverse DOF's.

Recall from Meeting 19: $$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$

similarly $$\tilde{f}^{(e)}=\tilde{T}^{(e)}f^{(e)}$$

Also: $$\tilde{k}^{(e)}\tilde{d}^{(e)}=\tilde{f}^{(e)}$$

$$\Rightarrow \hat{k}^{(e)}\tilde{T}^{(e)}d^{(e)}=\tilde{T}^{(e)}f^{(e)}$$

If $$\tilde{T}^{(e)}$$ is invertible then:

$$\left[ \tilde{T}^{(e)-1}\hat{k}^{(e)}\tilde{T}^{(e)}\right]d^{(e)}=f^{(e)}$$ (K matrix)

$$\tilde{T}^{(e)}$$=block diagonal matrix (P.19.3)

Consider a general block-diagram matrix:

$$A=\begin{bmatrix} D1 & &  &  &0 \\ & ..&  &  & \\  &  &  ..&  & \\  &  &  &  ..& \\  0&  &  &  & D5 \end{bmatrix}$$

Question: What is $$A^{-1}$$

Simpler example: Diagonal matrix

$$B=\begin{bmatrix} d_{11} & &  &  &0 \\ & d_{22}&  &  & \\ & &  d_{33}&  & \\ & &  &  d_{44}& \\ 0& &  &  &d_{mn} \end{bmatrix}= diag\left[d_{11},d_{22},d_{33}....d_{mn} \right]$$

assuming that $$d_{ij}\neq 0 $$ for i=1,....n then $$B^{-1}=diag\left[\frac{1}{d_{11}},\frac{1}{d_{22}},\frac{1}{d_{33}}....\frac{1}{d_{mn}} \right]$$

For a block diagonal matrix A:

$$A=diag\left[D_{1},....D_{5} \right]$$

$$A^{-1}=diag\left[D_{1}^{-1},....D_{5}^{-1} \right]$$

$$\tilde{T}^{(e)-1}=diag\left[R^{(e)-1},R^{(e)-1} \right]$$

P.19-2 $$R^{(e)T}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix}$$

$$R^{(e)T}R^{(e)}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}&l^{(e)} \end{bmatrix}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=I $$ (2x2 identity matrix) ... $$\Rightarrow R^{(e)-1}=R^{(e)T}$$

$$\Rightarrow \tilde{T}^{(e)-1}=diag\left[R^{(e)T},R^{(e)T} \right] =\left( diag\left[R^{(e)},R^{(e)} \right]\right)^{T}$$ ...  $$\tilde{T}^{(e)-1}=\tilde{T}^{(e)T}$$

so from earlier it can be shown that: $$\left[ \tilde{T}^{(e)T}\hat{k}^{(e)}\tilde{T}^{(e)}\right]d^{(e)}=f^{(e)}$$