User:EML4500.f08.RAMROD.E/HW4

HW: Eigenvectors for the zero Eigenvalues of K
Using Matlab, the eigenvalues for K were found (see HW 3) and the eigenvectors were plotted for the zero values. The eigenvectors were found to be:

$$E_1 = \begin{Bmatrix} -0.4494 & 0.6754 & 0.1682 & -0.3942 & 0.2812 & -0.2812 \end{Bmatrix}$$

$$E_2 = \begin{Bmatrix} 0.3877&-0.4281&0.3877&-0.4281&0.4163&-0.3995 \end{Bmatrix}$$

$$E_3 = \begin{Bmatrix} 0.5158&0.4825&0.5158&0.4825&-0.0116&-0.0449 \end{Bmatrix}$$

$$E_4 = \begin{Bmatrix} 0.0160&0.0244&0.0160&0.0244&0.7023&0.7107 \end{Bmatrix}$$

The plots are shown below.







These plots show the mode shapes corresponding to each of the zero eigenvalues and are representative of a linear combination of the four pure mode shapes (x-translation, y-translation, rotational, and mechanism).

For the eigenvalue/eigenvector problem:

$$Kv = \lambda v$$

We can let $$\begin{Bmatrix} u_1 & u_2 & u_3 & u_4 \end{Bmatrix}$$ be the pure eigenvetors corresponding to the four zero eigenvalues. Assuming this tells us that:

$$Ku_i = 0u_i$$ for i = 1,2,3,4.

To get the true eigenvector, w, for each zero eigenvalue, we take the sum of the four contributors, each multiplied by a constant $$\alpha _{1}$$, giving us:

$$ w = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 + \alpha_4 u_4$$

Because w consists of the combination of eigenvectors for zero eigenvalues, w too represents an eigenvector corresponding to a zero eigenvalue. That is that:

"$\bar{K}\bar{w} = K(\Sigma \alpha _iu_i) = \bar{0} = 0*/bar{w}$"