User:Eml4500.f08.team.guan/fixed108

 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.

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. 