User:Eml4500.f08.ateam.boggs.t/Lecture 19

The goal is to find $$ \tilde{T}^{(e)}$$, a 4x4 matrix that transforms the set of local element degrees of freedom $$ d^{(e)} $$ a 4x1 matrix to another set of local, element, degrees of freedom $$ \tilde{d}^{(e)}$$ such that $$ \tilde{T}^{(e)}$$ is invertible.

$$ \tilde{d}^{(e)}$$ is the rotated degrees of freedom matrix corresponding to the $$\tilde{x}$$ coordinate system. The relationship is described below,


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

The components that make up the $$\tilde{d}^{(e)}$$ matrix are as follows,


 * $$ \tilde{d_{1}}^{(e)} =

\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$



Here $$\tilde{d_{1}}^{(e)}$$ directly corresponds to the axial displacement $$q_{1}^{(e)}$$. And,


 * $$ \tilde{d_{2}}^{(e)} = \overrightarrow{d_{[1]}}^{(e)} \cdot \overrightarrow{\tilde{j}}$$

where $$ \tilde{d_{2}}^{(e)}$$ is a component of $$\overrightarrow{d_{[1]}}^{(e)}$$ along the unit vector $$\overrightarrow{\tilde{j}}$$, i.e., along the $$\tilde{y}$$-axis.

Here,


 * $$ \overrightarrow{i} \cdot \overrightarrow{\tilde{j}} = \cos{(\theta + \frac{\pi}{2})} = -\sin\theta$$


 * $$ \overrightarrow{j} \cdot \overrightarrow{\tilde{j}} = \cos\theta $$

This becomes,


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

Where,


 * $$-\sin\theta^{(e)} = -m^{(e)}$$
 * $$ \cos\theta^{(e)} = l^{(e)}$$

So,



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

Combining these two equations equates to



\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} $$

Here,



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

Looking at the entire element e and combining the local degrees of freedom one gets,



\begin{bmatrix} \tilde{d_{1}}^{(e)}\\ \tilde{d_{2}}^{(e)}\\ \tilde{d_{3}}^{(e)}\\ \tilde{d_{4}}^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{1}^{(e)} \end{bmatrix} $$

The first matrix is the local element degrees of freedom corresponding to the $$\tilde{x}$$ coordinate system, the second is the transformation matrix, and the third is the matrix of the local element degrees of freedom.

The transformation matrix can be broken down as follows,





\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{bmatrix} R^{(e)} & 0\\ 0 & R^{(e)} \end{bmatrix} $$

If the transverse displacement is taken away, the force-displacement relationship will look the same as in previous lectures.


 * $$ k^{(e)}

\begin{bmatrix} 1 & -1\\     -1 & 1 \end{bmatrix} \begin{bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{bmatrix} = \begin{bmatrix} p_{1}^{(e)}\\ p_{2}^{(e)} \end{bmatrix} $$

Looking at the force-displacement relationship,



\tilde{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{d}^{(e)} $$

Here the axial displacements are present and the transverse displacements are taken to be zero because if one acts along the transverse the axial displacements are zero and the spring has no effect.

This now becomes


 * $$ \tilde{f}^{(e)} = \tilde{k}^{(e)} \tilde{d}^{(e)} $$

In this equation force is represented by a 4x1 matrix, the spring stiffness is represented by a 4x4 matrix, and the displcement is a 4x1 matrix.