University of Florida/Eml4507/s13.team4ever.Wulterkens.R3

Given Equations
Given the two following equations, verify that they simplify to the bar element stiffness matrix $$\bar{k}^{(e)}$$.

$$ \mathbf 1) $$ $$\bar{k}^{(e)} = T^{(e)T} \hat{k}^{(e)} T^{(e)}$$

$$ \mathbf 2) $$ $$\bar{k}^{(e)} = \widetilde{T}^{{(e)}T} \, \widetilde{k}^{(e)} \, \widetilde{T}^{(e)}$$

Original Equation
$$\bar{k}^{(e)} = T^{(e)T} \hat{k}^{(e)} T^{(e)}$$

Given Equations
$$\hat{k}^{(e)} = k^{(e)}\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix} = \begin{bmatrix}k^{(e)} & -k^{(e)}\\ -k^{(e)} & k^{(e)}\end{bmatrix}$$

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

Plugging into original Equation
$$\bar{k}^{(e)} = \begin{bmatrix}l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix} \begin{bmatrix}k^{(e)} & -k^{(e)}\\ -k^{(e)} & k^{(e)}\end{bmatrix} \begin{bmatrix}l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix}$$

Solving
Multiple the first two equations. $$\bar{k}^{(e)} = \begin{bmatrix} l^{(e)} k^{(e)} & -l^{(e)} k^{(e)} \\ m^{(e)} k^{(e)}  & -m^{(e)} k^{(e)} \\ -l^{(e)} k^{(e)}  & l^{(e)} k^{(e)} \\ -m^{(e)} k^{(e)}  & m^{(e)} k^{(e)} \end{bmatrix} \begin{bmatrix}l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix}$$

Pull out the K term. $$\bar{k}^{(e)} = k^{(e)}\begin{bmatrix} l^{(e)} & -l^{(e)} \\ m^{(e)} & -m^{(e)} \\ -l^{(e)} & l^{(e)} \\ -m^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix}l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix}$$

Multiple the last two matrices to get the following equation: $$\bar{k}^{(e)} = k^{(e)} \begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\-l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$

This shows that the original equation can be simplified down to the bar element stiffness matrix.

Original Equation
$$k^{(e)} = \widetilde{T}^{{(e)}T} \, \widetilde{k}^{(e)} \, \widetilde{T}^{(e)}$$

Provided equations
$$\widetilde{T}^{(e)}=\begin{bmatrix} cos(\phi) & sin(\phi) & 0 & 0\\ -sin(\phi) & cos(\phi) & 0 & 0\\ 0 & 0 & cos(\phi) & sin(\phi)\\ 0 & 0 & -sin(\phi) & cos(\phi) \end{bmatrix}$$

$$\widetilde{k}^{(e)} = \frac{EA}{L}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Plugging into Original Equation
$$ k^{(e)} = \begin{bmatrix} cos(\phi) & -sin(\phi) & 0 & 0\\ sin(\phi) & cos(\phi) & 0 & 0\\ 0 & 0 & cos(\phi) & -sin(\phi)\\ 0 & 0 & sin(\phi) & cos(\phi) \end{bmatrix} \frac{EA}{L}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} cos(\phi) & sin(\phi) & 0 & 0\\ -sin(\phi) & cos(\phi) & 0 & 0\\ 0 & 0 & cos(\phi) & sin(\phi)\\ 0 & 0 & -sin(\phi) & cos(\phi) \end{bmatrix}$$

Solving
Begin by multiplying the first two matrices together. $$k^{(e)} =\frac{EA}{L}\begin{bmatrix} cos(\phi) & 0 & -cos(\phi)  & 0\\ sin(\phi) & 0 & -sin(\phi) & 0\\ -cos(\phi) & 0 & cos(\phi) & 0\\ -sin(\phi) & 0 & sin(\phi) & 0 \end{bmatrix}\begin{bmatrix} cos(\phi) & sin(\phi) & 0 & 0\\ -sin(\phi) & cos(\phi) & 0 & 0\\ 0 & 0 & cos(\phi) & sin(\phi)\\ 0 & 0 & -sin(\phi) & cos(\phi) \end{bmatrix}$$

Next, multiply the last two together. $$k^{(e)} =\frac{EA}{L}\begin{bmatrix} cos(\phi )^{2} & cos(\phi)sin(\phi ) & -cos(\phi )^{2} & -cos(\phi)sin(\phi )\\ cos(\phi)sin(\phi ) & sin(\phi )^{2} & -cos(\phi)sin(\phi ) & -sin(\phi )^{2}\\ -cos(\phi )^{2} & -cos(\phi)sin(\phi ) & cos(\phi )^{2} & cos(\phi)sin(\phi )\\ cos(\phi)sin(\phi ) & -sin(\phi )^{2} & -cos(\phi)sin(\phi ) & sin(\phi )^{2} \end{bmatrix}$$

This verifies our original expectation that the equation would simplify down to the bar element stiffness matrix.