User:EML4507.s13.team4ever.Bonner/Bonnerreport5p1

Problem R5.1
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Description
We are to solve, by hand, the generalized eigenvalue for the mass-spring-damper system on p. 53-13 using the following data for the masses and stiffness coefficients:

$$ m_1 = 3 $$

$$m_2 = 2 $$

$$k_1 = 10 $$

$$k_2 = 20 $$

$$k_3 = 15 $$

We are then to compare this result with those obtained using CALFEM. Finally, we are to verify the mass-orthogonality of the eigenvectors.

Given:
We are given that for a generalized eigenvalue problem:

$$ K \phi = \lambda M \phi $$

We also know that:

$$ M= \begin{bmatrix} m_1 & 0\\ 0 & m_2\\ \end{bmatrix} $$ and $$ K= \begin{bmatrix} k_1 + k_2 & -k_2\\ -k_2 & k_2 + k_3\\ \end{bmatrix} $$

Therefore, we have:

$$ M= \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} $$ and $$ K= \begin{bmatrix} 30 & -20\\ -20 & 35\\ \end{bmatrix} $$

Solution:
The first step would be to find an eigenvalue, $$ \lambda $$ that satifies:

$$ K \phi = \lambda M \phi $$

To do this, we must find an eigenvalue that satifies:

$$ det(K - \lambda M) = 0 $$

We then have:

$$ det( \begin{bmatrix} 30 - 3 \lambda & -20\\ -20 & 35 - 2 \lambda\\ \end{bmatrix} = 0 $$)

This gives us

$$ (30 - 3 \lambda)(35 - 2 \lambda) + 400 = 0 $$

$$ 6 \lambda^2 - 165 \lambda + 650 = 0 $$

Solving this equation using the quadratic equation gives us eigenvalues of:

$$ \lambda_{1,2} = 13.75 \pm 8.985 $$

$$ \lambda_1 = 4.765$$

$$ \lambda_2 = 22.735$$

We will first take the lowest eigenvalue. We'll now try to find the corresponding eigenvector. Plugging this eigenvalue into the equaiton

$$ K \phi = \lambda M \phi $$

$$ K \phi - \lambda M \phi = 0 $$

$$ (K - \lambda M) \phi = 0 $$

$$ (K - 4.765 M) \phi = 0 $$

We need to find a value of $$ \phi $$ that satisfies the above equation.

$$ \begin{bmatrix} 30 - (3)(4.765) & -20\\ -20 & 35 - (2)(4.765)\\ \end{bmatrix}\begin{bmatrix} \phi_1\\ \phi_2\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} $$

$$ \begin{bmatrix} 15.705 & -20\\ -20 & 25.47\\ \end{bmatrix}\begin{bmatrix} \phi_1\\ \phi_2\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} $$

Augmenting the right hand matrix onto the left hand matrix and putting this in row reduced echelon form, we obtain,

$$ \begin{bmatrix} 1 & -1.273 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $$

This means that the corresponding eigenvector is:

$$ \phi_1 = \begin{bmatrix} 1.273\\ 1\\ \end{bmatrix} $$

We will now take the higher eigenvalue. We'll now try to find the corresponding eigenvector. Plugging this eigenvalue into the equaiton

$$ K \phi = \lambda M \phi $$

$$ K \phi - \lambda M \phi = 0 $$

$$ (K - \lambda M) \phi = 0 $$

$$ (K - 22.735 M) \phi = 0 $$

We need to find a value of $$ \phi $$ that satisfies the above equation.

$$ \begin{bmatrix} 30 - (3)(22.735) & -20\\ -20 & 35 - (2)(22.735)\\ \end{bmatrix}\begin{bmatrix} \phi_1\\ \phi_2\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} $$

$$ \begin{bmatrix} -38.205 & -20\\ -20 & -10.47\\ \end{bmatrix}\begin{bmatrix} \phi_1\\ \phi_2\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix} $$

Augmenting the right hand matrix onto the left hand matrix and putting this in row reduced echelon form, we obtain,

$$ \begin{bmatrix} 1 & 0.5234 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $$

This means that the corresponding eigenvector is:

$$ \phi_2 = \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} $$

We will now verify the mass orthogonality of the eigenvectors. To do this, we need to show that

$$ \phi^T_i M \phi_j = 0 $$ for i not equal to j. For our case, we have:

$$ \phi^T_1 = \begin{bmatrix} 1.273 & 1 \\ \end{bmatrix} $$

$$ M = \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} $$ and

$$ \phi_2 = \begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} $$

We have then that

$$ \phi^T_i M \phi_j = \begin{bmatrix} 1.273 & 1 \\ \end{bmatrix}\begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix}\begin{bmatrix} -0.5234\\ 1\\ \end{bmatrix} = 0 $$

Thus verifying the mass-orthogonality of the eigenvectors.

Using our own code from Report 3, we can compare this with CALFEM: