User:Eml4500.f08.ramrod.c/10-8-08

Goal - We want to find T tha ttransfer the set of local elemental dofs (degrees of freedom), d, to another set of local elemental dofs, d, such that T is invertible.



$$\tilde{\underline{d}}^{(e)}_{4x1} = \tilde{\underline{T}}^{(e)}_{4x4}\;\tilde{\underline{d}}^{(e)}_{4x1}$$

Equation (1): $$\tilde{d}_1^{(e)}=\begin{bmatrix} l^{(e)}&m^{(e)} \end{bmatrix} = \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

$$\tilde{d}_2^{(e)} = \overrightarrow{d}_1^{(e)} \overrightarrow{\tilde{j}}$$

$$\tilde{d}_2^{(e)} = -\sin \theta^{(e)}d_1^{(e)} + \cos \theta^{(e)}d_2^{(e)}$$

Equation (2): $$\tilde{d}_2^{(e)} = \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

Put Equations (1) and (2) together in matrix form:

$$\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}& l^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

$$\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)}\\ \tilde{d}_4^{(e)} \end{Bmatrix} = \begin{bmatrix} \underline{R}_{2x2}^{(e)} & \underline{O}_{2x2}\\ \underline{O}_{2x2}& \underline{R}_{2x2}^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)} \end{Bmatrix}$$



$$\tilde{\underline{f}}^{(e)} = k^{(e)}\begin{vmatrix} 1 &0 &-1  &0 \\ 0 &0  & 0 &0 \\  -1& 0 & 1 & 0\\ 0 & 0 &0  &0 \end{vmatrix} \tilde{d}^{(e)}$$

$$\tilde{\underline{f}}_{4x1}^{(e)} = \tilde{\underline{k}}_{4x4}^{(e)}\;\tilde{\underline{d}}_{4x1}^{(e)}$$