User:Eml4500.f08.delta 6.guzman/Lecture Notes 10/10

$$f^{(e)}_{4x1} = k_{4x4} ^{(e)} * d_{4x1}^{(e)}$$ Note: Consider the case where $$ d_{4} \neq 0 $$ then $$ d_1^{(e)} = d_2^{(e)} = d_3^{(e)} = 0 $$

you will get the following relationship $$ f_{4x1} ^{(e)} = k_{4x4} ^{(e)} * d_{4x1} ^{(e)} = 0_{4x1} $$ where the zero matrix is the 4th column of matrix $$ k^{(e)} $$ (interpretation of the transverse dofs)

Recall from 19-3 that $$ d^{(e)} = T^{(e)} * d^{(e)} $$ similarly there is the same relationship for f where $$f^{(e)} = T^{(e)} * f^{(e)} $$

Also: $$ k^{(e)} = T^{(e)} * d^{(e)} $$

The lesson here is that we should use the transverse dofs instead of just using the inverse dofs therefore finding the relationship for the global matrix $$ k=d*f $$ $$k^{(e)} *T^{(e)} * d^{(e)}= T^{(e)} * f^{(e)}$$

If $$ T^{(e)}$$ is inversible, then the matrix that we are looking for is:

$$f^{(e)} = d^{(e)} * \begin{bmatrix} T^{(e)^{-1}} * k^{(e)} * T^{(e)} \end{bmatrix}$$ where $$T^{(e)}$$ is the block diagram matrix.

Considering a general block-diagram matrix A:

$$ A=

\begin{bmatrix}

D_{1} & 0 & 0 & 0\\

0 & . & 0 & 0\\

0 & 0 & . & 0\\

0 & 0 & 0 & D_{s}

\end{bmatrix} $$

so what will be the inverse of matrix A? This can be answered by observing the simple example of the diagonal matrix B:

$$ B=

\begin{bmatrix}

d_{11} & 0 & 0 & 0\\

0 & d_{22} & 0 & 0\\

0 & 0 & . & 0\\

0 & 0 & 0 & d_{mm}

\end{bmatrix} $$

$$ B= Diag \begin{bmatrix} d_{11}, d_{22}, ... , d_{mm} \end{bmatrix} $$

$$ B^{-1} = Diag\begin{bmatrix} \frac{1}{d_{11}}, \frac{1}{d_{22}}, ... , \frac{1}{d_{mm}} \end{bmatrix} $$

Assuming $$ d_{ii} \neq 0 $$  for i=1,…n

For a block diagonal matrix A

$$ A = Diag\begin{bmatrix} D_{11}, D_{22}, ... , D_{s} \end{bmatrix} $$

$$ A^{-1} = Diag\begin{bmatrix} D_{11}^{-1}, D_{22}^{-1}, ... , D_{s}^{-1} \end{bmatrix} $$

$$T^{(e)^{-1}} = Diag \begin{bmatrix} R^{(e)^{-1}}, ... , R^{(e)^{-1}} \end{bmatrix}$$

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

Since $$l^{2}*m^{2} = 1 $$ the matrix becomes where $$ I_{2x2} $$ is the identity matrix.

$$R_{2x2}^{(e)^{T}}*R_{2x2}^{(e)} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}_{2x2} = I_{2x2} $$

$$ R^{(e)^{T}}= R^{(e)^{-1}} $$ $$ T^{(e)^{-1}} = Diag\begin{bmatrix} R^{(e)^{T}}, R^{(e)^{T}} \end{bmatrix}$$ where $$ T^{(e)} = \begin{pmatrix} Diag \begin{bmatrix} R^{(e)}, R^{(e)} \end{bmatrix} \end{pmatrix}^{T} $$ $$ T^{(e)^{-1}} = T^{(e)^{T}} $$ $$ \begin{bmatrix} T^{(e)^{-T}} * k^{(e)}* T^{(e)} \end{bmatrix} *d^{(e)}= f^{(e)} $$ where $$ k^{e} = \begin{bmatrix} T^{(e)^{-T}} * k^{(e)}* T^{(e)} \end{bmatrix} $$