User:Eml4500.f08.qwiki.bishop/Lecture 20

Note

 * Consider case: $$\tilde{d}^{(e)}_{4}\neq 0$$
 * $$\tilde{d}^{(e)}_{1}=\tilde{d}^{(e)}_{2}=\tilde{d}^{(3)}_{4}=0$$
 * $$\tilde{f}^{(e)}=\tilde{k}^{(e)}\tilde{d}^{(e)}$$=0
 * (4x1)=(4x4)(4x1)=(4x1)
 * The zero matrix represents the fourth column of the elemental stiffness matrix, k

T matrix application to the elemental Force matrix
\tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m_{(e)}&l^{(e)} \end{bmatrix}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{e}_{2}\\ \end{Bmatrix}=\textbf{R}^{(e)}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{(e)}_{2}\\ \end{Bmatrix}$$ \tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2}\\ \tilde{f}^{(e)}_{3}\\ \tilde{f}^{(e)}_{4} \end{Bmatrix}=\begin{bmatrix} \mathbf{R}^{(e)}_{2x2} & \mathbf{0}_{2x2}\\ \mathbf{0}_{2x2} & \mathbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{(e)}_{2}\\ {f}^{(e)}_{3}\\ {f}^{(e)}_{4} \end{Bmatrix}\Rightarrow \tilde{\mathbf{f}}^{(e)}_{4x1}=\tilde{\mathbf{T}}^{(e)}_{4x4}\mathbf{f}^{(e)}_{4x1}$$
 * p.19-3, $$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$
 * Similarly, $$\tilde{f}^{(e)}=\tilde{T}^{(e)}f^{(e)}$$
 * $$\begin{Bmatrix}
 * $$\begin{Bmatrix}
 * The set of R and zero matrices can be represented by the T matrix.
 * If we look at the force-displacement relationship, we can apply the T matrix conversion between global and axial coordiantes.
 * $$\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{d}}^{(e)}=\tilde{\mathbf{f}}^{(e)}\Rightarrow \tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}\mathbf{d}^{(e)}=\tilde{\mathbf{T}}^{(e)}\mathbf{f}^{(e)}$$
 * But if the T matrix can be inverted, then:
 * $$[\tilde{\mathbf{T}}^{(e) -1}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

General block diagonal matrix
d_{11}&0 &0 \\ 0& d_{22}&0 \\ 0& 0 &d_{33} \end{bmatrix}=diag[d_{11}, d_{22}, d_{33}]$$ l^{(e)} & -m^{(e)} \\ m_{(e)}&l^{(e)} \\ \end{bmatrix}$$ 1 & 0 \\ 0&1 \\ \end{bmatrix}=\mathbf{I}$$
 * $$\tilde{\mathbf{T}}^{(e)}$$ is a block diagonal matrix.
 * To denote a block diagonal matrix in shorthand use the following notation:
 * $$\mathbf{B}=\begin{bmatrix}
 * The inverse of B is:
 * $$\mathbf{B^{(-1)}}=diag[\frac{1}{d_{11}},\frac{1}{d_{22}}, \frac{1}{d_{33}}]=diag[d_{11}^{(-1)}, d_{22}^{(-1)}, d_{33}^{(-1)}]$$
 * Assuming that no value of d is zero
 * In applying this to the T-matrix:
 * $$\tilde{\mathbf{T}}^{(e)-1}=diag[\mathbf{R}^{(e)-1},\mathbf{R}^{(e)-1}]$$
 * In reference to p.19-2:
 * $$\mathbf{R}^{(e)T}=\begin{bmatrix}
 * In order to prove that the transpose of R is equivalent to the inverse of R:
 * $$\mathbf{R}^{(e)T}\mathbf{R}^{(e)}=\begin{bmatrix}
 * So, $$\mathbf{R}^{(e)T}=\mathbf{R}^{(e)-1}$$
 * The inverse of the T-matrix can be written as:
 * $$\tilde{\mathbf{T}}^{(e)-1}=diag[\mathbf{R}^{(e)T},\mathbf{R}^{(e)T}]=\tilde{\mathbf{T}}^{(e)T}$$

Force-Displacement Relationship with T

 * $$[\tilde{\mathbf{T}}^{(e)T}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$
 * To show that $$[\tilde{\mathbf{T}}^{(e)T}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]=\mathbf{k}^{(e)}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l^{(e)} &-m^{(e)} & 0 &0 \\ m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&-m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}\begin{bmatrix} 1 & 0 &-1 &0 \\  0& 0 & 0 & 0\\  -1&0  &1  &0 \\  0&  0&  0& 0 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & 0 &0 \\ -m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l &0 &-l  & 0\\ m& 0& -m & 0\\ -l & 0& l & 0\\ -m& 0 & m&0 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & 0 &0 \\ -m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l^2 &lm &-l^2  & -ml\\ ml& m^2& -ml & -m^2\\ -l^2 & -lm& l^2 & lm\\ -ml& -m^2 & ml&m^2 \end{bmatrix}$$
 * So, the relationship between the elemental k matrix in global coordinates and the k matrix in axial coordinates using the T matrix is confirmed.