User:Eml4500.f08.bottle.butler:Homework/4

Our goal is to find $$ \tilde{T}^{(e)}_{4X4} $$ that transforms the set of local element dof's $$ d^{(e)}_{4X1} $$ to another set of local element $$ \tilde{d}^{(e)}_{4X1}$$ such that $$ \tilde{T}^{(e)} $$ is invertible.

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From previous lectures we know that:

$$ \underbrace {\tilde{d}_{1}^{(e)}}_{q_{1}^{(e)}} = \begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

and $$ \tilde{d}_{2}^{(e)}= \vec{d}^{(e)}_{1}\circ \vec{\tilde{j}} $$ $$= -\sin \theta ^{(e)}d_{1}^{(e)}+\cos \theta ^{(e)}d_{2}^{(e)}$$

therefore:

$$ \tilde{d}_{2}^{(e)} = \begin{vmatrix} -m^{(e)} & l^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d_{2}^{(e)} \end{Bmatrix} $$

If we combine the two we get the following equation in matrix form: $$ \begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

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