User:Eml4500.f08.bottle.barnes/HW4

2-Bar Truss Eigenvectors
The eigenvalues of the global stiffness matrix are as follows:

$$\textbf{K}= \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5 \\ -0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5 \\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5  \end{bmatrix}\,$$

To find the eigenvectors corresponding to the eigenvalues of the global stiffness matrix, first the matrix of eigenvalues is defined in MATLAB:

k =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000   -2.5000    2.5000         0         0    2.5000   -2.5000    2.5000   -2.5000

Then, the following command is implemented:

EDU>> E = eig(k)

E =

-7.0650  -0.0000   -0.0000    0.0000    1.4401    7.1249

Plotting these values in yields the following graph: