User:Eml4500.f08.gravy.sms/HW4

19.2
$$ \tilde{d}^{(e)}_2 = \overrightarrow{d}^{(e)}_{[1]} \cdot \overrightarrow{\tilde{j}} $$

$$ \tilde{d}^{(e)}_2 =(d^{(e)}_1 \overrightarrow{i} + d^{(e)}_2 \overrightarrow{j}) \cdot \overrightarrow{\tilde{j}} $$

$$ \tilde{d}^{(e)}_2 = d^{(e)}_1 (\overrightarrow{i} \cdot \overrightarrow{\tilde{j}}) + d^{(e)}_2(\overrightarrow{j} \cdot \overrightarrow{\tilde{j}}) $$

$$ \tilde{d}^{(e)}_2 = d^{(e)}_1 (-\sin \theta ^{(e)}) + d^{(e)}_2(\cos \theta^{(e)}) $$

Recall that $$m^{(e)} = \sin \theta ^{(e)}$$ and $$l^{(e)} = \cos \theta^{(e)}$$

$$ \tilde{d}^{(e)}_2 = d^{(e)}_1 (-m^{(e)}) + d^{(e)}_2(l^{(e)}) $$

$$ \tilde{d}^{(e)}_2 = \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} d^{(e)}_1\\ d^{(e)}_2 \end{Bmatrix} $$

21.3 Matlab
The following Matlab code was used to find the global stiffness matrix, K, for case (a), as well as the eigenvectors and eigenvalues corresponding to the global stiffness matrix.

It gave the following results. K is the global stiffness matrix, V is the eigenvector matrix, and D is the eigenvalue matrix with the diagonals as the eigenvalues:

Note that the first 5 eigenvalues are zero eigenvalues. The first 5 columns are the eigenvectors that correspond to the zero eigenvalues.

Below are the graphs of the eigenvectors corresponding to the the 5 zero eigenvalues in order.