User:Eml4500.f08.team.guan/1013

10/13 Homework Assignment
Finding the Eigenvalues of a Three Bar Truss in a Square Configuration (Case A), and finding the eigenvalues of A Four Bar Truss in a Square Configuration with the Fourth Truss as a Diagonal Member (Case B). Plot the eigenvectors of Case A.

Matlab Code for Case A:

The below picture represents the plot produced by Matlab showing the Case A truss. 

For this case, the code found the the global stiffness matrix resulted as follows. $$ K= \begin{bmatrix} 0 & 0 & 0 &  0 & 0 &  0 & 0 &  0\\ 0 &  6 & 0 &  -6 & 0 &  0 & 0 &  0\\ 0 &  0 & 6 &  0 & -6 &  0 & 0 &  0 \\ 0 &  -6 & 0 &  6 & 0 &  0 & 0 &  0\\ 0 &  0 & -6 &  0 & 6 &  0 & 0 &  0\\ 0 &  0 & 0 &  0 & 0 &  6 & 0 &  -6\\ 0 &  0 & 0 &  0 & 0 &  0 & 0 &  0\\ 0 &  0 & 0 &  0 & 0 &  -6 & 0 &  6\\ \end{bmatrix}

$$ This is a singular matrix meaning that the determinant is zero and thus it is not invertible. The eigenvalues are as follows. $$\lambda$$=[0,0,0,0,0,12,12,12].

Matlab Code for Case B:

The below picture represents the plot produced by Matlab showing the Case B truss. 

For this case, the code found the the global stiffness matrix resulted as follows. $$ K= \begin{bmatrix} 2.1213  &    2.1213& 0&0 &     -2.1213&  -2.1213  &          0  &          0\\       2.1213 &      8.1213& 0     &      -6   &   -2.1213 & -2.1213   &         0   &         0\\ 0& 0      &    6&  0   &        -6    &     0   &         0 &           0\\ 0  &         -6&  0      &      6   &         0   &      0    &        0  &          0\\      -2.1213 &     -2.1213  &         -6  &          0    &   8.1213&    2.1213   &         0    &        0\\      -2.1213&      -2.1213    &        0  &          0   &    2.1213  &     8.1213  &          0  &         -6\\            0  &          0    &        0   &         0     &       0&         0   &         0    &        0\\            0   &         0     &       0     &       0      &      0    &       -6   &         0  &          6\\ \end{bmatrix}$$

This is also a singular matrix meaning that the determinant is zero and thus it is not invertible. The eigenvalues were as follows. $$\lambda $$=[0,0,0,0,3.8183,12,12,16.667].