User:Eml4500.f08.delta 6.castrillon/3-bar truss MATLAB simulation

3-bar Truss Eigenvalues
We have two truss systems, one 3-bar and 5-bar. They are illustrated below:



Where $$E^{(1)}=E^{(2)}=E^{(3)}=1$$; $$A^{(1)}=A^{(2)}=A^{(3)}=1$$; $$L^{(1)}=L^{(2)}=L^{(3)}=1$$. It is possible to analyze the eigenvalues and eigen vectors according to $$\mathbf{K}v=\mathbf{\lambda} v$$ for the 3-bar truss case.

We construct the stffness matrix from each individual element stiffness matrix. It was found to be:
 * K=$$\begin{bmatrix}

0&0&0&0&0&0&0&0\\ 0&1&0&-1&0&0&0&0\\ 0&0&1&0&-1&0&0&0\\ 0&-1&0&2&0&0&0&-1\\ 0&0&-1&0&1&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&-1&0&0&0&1 \end{bmatrix}$$ This matrix was then inputed into MATLAB and its eigenvalues and eigenvectors were determined. The MATLAB results are as follow:

V = -0.0000   0.0000    0.0000   -0.4740    0.7103   -0.2504    0.4082    0.0000   -0.7071   -0.4674   -0.1769    0.2838   -0.0000   -0.7071         0   -0.2136   -0.3585   -0.5702   -0.8165   -0.0000    0.0000   -0.4674   -0.1769    0.2838    0.0000    0.7071    0.0000   -0.2136   -0.3585   -0.5702   -0.0000    0.0000   -0.0000   -0.0556    0.1402   -0.1014   -0.0000    0.0000    0.0000   -0.1604    0.3534   -0.1873    0.4082    0.0000    0.7071   -0.4674   -0.1769    0.2838 D = 3.0000        0         0         0         0         0         0    2.0000         0         0         0         0         0         0    1.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0   -0.0000