User:Eml4500.f08.ateam.mcnally/Week 6

Further Evaluation and Understanding of the Zero Eigen Values
Meeting 21 on Monday October 13th 2008 was a review and further understanding of a previously assigned homework problem dealing with the Eigen values of the Global K Matrix for a two bar truss system. The solution to the Eigen value problem actually yields four zero Eigen which corresponds to rigid body motion. The four zero values correspond to the global dof constraints: two translations, one rotation, and one mechanism which relates the idea of zero stored energy in the truss system if you remove both supports from the two bar truss system.

The next section is a plot of the Eigen vectors which correspond to the zero Eigen values of the two bar truss system.

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

The corresponding Eigen values in a column vector are as follows:

eig(K)

ans =

-0.0000  -0.0000    0.0000    0.0000    1.4706   10.0294

The command [U,D]=eig(K) produces a matrix U whose columns are the eigenvectors of K and a diagonal matrix D with the eigenvalues of K on its diagonal

[U,D]=eig(K)

U =

-0.1118   0.5043   -0.0000   -0.5931    0.6174   -0.0139   -0.0814   -0.8634    0.0000   -0.3476    0.3565   -0.0080   -0.4628    0.0089    0.0000   -0.4803   -0.5409    0.5123    0.5266   -0.0053   -0.0000   -0.5429   -0.4330   -0.4904   -0.4947    0.0071   -0.7071    0.0313   -0.0765   -0.4984    0.4947   -0.0071   -0.7071   -0.0313    0.0765    0.4984

D =

-0.0000        0         0         0         0         0         0   -0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    1.4706         0         0         0         0         0         0   10.0294

From these matrices (U and D) the four Eigenvectors corresponding to the zero Eigen values can be pulled from matrix U and are as follows:

E1 =

-0.1118  -0.0814   -0.4628    0.5266   -0.4947    0.4947

E2 =

0.5043  -0.8634    0.0089   -0.0053    0.0071   -0.0071

E3 =

0        0         0         0   -0.7071   -0.7071

E4 =

-0.5931  -0.3476   -0.4803   -0.5429    0.0313   -0.0313





The results of the plots are the Mode Shapes which are a linear combination of the pure mode shapes. (pure rigid body motion two translations, one rotation, and one pure mechanism). The results explained above can also be represented mathematically with the following arguments.

$$Kv = \lambda v$$

Let $$\begin{Bmatrix} U_1 & U_2 & U_3 & U_4 \end{Bmatrix}$$ be the pure eigenvetors corresponding to the four zero eigenvalues. $$K*U_i = 0*U_i$$ for i = 1,...,4.

To get W for each zero eigenvalue, we take the sum of the four contributors, each multiplied by a constant $$\alpha _{1}$$, which gives us the following relation:

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

W consists of the combination of eigenvectors for zero eigenvalues, therefore W also represents an eigenvector corresponding to a zero eigenvalue.