User:Eml4500.f08.team.Bengtson/teamHW4

 HW Redo: The derivation for the transverse displacement degrees of freedom is added. The figure of the elemental subject to axial and transverse degrees of freedom and forces is also added.

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Eml4500.f08.team.foskey.ckf 00:30, 3 November 2008 (UTC)

Transforming Local Degrees of Freedom
Goal: Want to find $$ \tilde{\mathbf{T}}^{(e)}$$ that frames fom the set of local (element) definions $$\boldsymbol{d}^{(e)}$$ to another set of local (element) defs $$\tilde{\boldsymbol{d}}^{(e)}$$ that $$ \tilde{\mathbf{T}}^{(e)}$$ is invertible.

$$\tilde{\boldsymbol{d}}^{(e)}= \tilde{\mathbf{T}}^{(e)}\boldsymbol{d}^{(e)}$$ 

This image shows that one can show displacements can be made in the x and y axis but the axises can also be aligned along the length of the beam. This causes the x~ axis displacement to be either tension or compression.

$$\tilde{\boldsymbol{d}}^{(e)}_{1}=\boldsymbol{q}^{(e)}_{1}$$ (comp of $$\tilde{\boldsymbol{d}}^{(e)}_{[1]} $$ along the $$\jmath $$ also known as $$\tilde{y}$$ axis

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

$$\tilde{\boldsymbol{d}^{(e)}_{2}=\boldsymbol{d}}^{(e)}_{[1]}\jmath$$

$$=-\sin \theta ^{e}d_{1}^{(e)}+ \cos \theta ^{e}d_{2}^{(e)}$$

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

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

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

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

$$ \begin{Bmatrix} \tilde{d}^{(e)}_{1}\\ \tilde{d}^{(e)}_{2}\\ \tilde{d}^{(e)}_{3}\\ \tilde{d}^{(e)}_{4} \end{Bmatrix} =\begin{bmatrix} \boldsymbol{R}^{(e)} & \boldsymbol{0}\\ \boldsymbol{0} & \boldsymbol{R}^{(e)} \end{bmatrix} \begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix} $$

$$\boldsymbol{R}^{(e)}$$ is a 2x2 matrix $$\boldsymbol{0}^{(e)}$$ is a 2x2 0 matrix.

$$\tilde{\boldsymbol{f}}^{(e)} = K^{(e)}\begin{bmatrix} 1 & 0& -1 &0 \\ 0 &0  & 0 & 0\\ -1 &0  & 1 & 0\\ 0 & 0 & 0 &0 \end{bmatrix} \tilde{\boldsymbol{d}}^{(e)} $$

This means only the forces that act on the beam are the ones along beam. This means the force either acts a compression or tension.

Transverse forces do not stretch the springs. 