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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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