User:Eml4507.s13.team7.kermani

=Problem 1= On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Find
Solve the general eigenvalue problem by first converting it to the standard eigenvalue problem for the spring-mass-damper system given on p.53-13 of 'Fead.s13.sec.53b'.

Given
$$ M = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} $$

$$ K = \begin{pmatrix} 30 & -20 \\ -20 & 35 \end{pmatrix} $$

Solution
Converting the general eigenvalue problem to the standard eigenvalue problem and solving it General eigenvalue form: $$ Kx = \lambda Mx $$ Standard eigenvalue form: $$ K^* x^* = \lambda x^* $$ Where, $$ K^* = M^{-1/2} K M^{-1/2} $$ $$ x^* = M^{1/2} x $$ $$ K^* = \begin{pmatrix} 10 & -8.16 \\ -8.16 & 17.5 \end{pmatrix} $$ $$ M^{1/2} = \begin{pmatrix} \sqrt{3} & 0 \\ 0 & \sqrt{2} \end{pmatrix} $$ $$ M^{-1/2} = \begin{pmatrix} 1 / \sqrt{3} & 0 \\ 0 & 1 / \sqrt{2} \end{pmatrix} $$ $$ K^* x^* = \lambda x^* $$ $$ K^* M^{1/2} x = \lambda M^{1/2} x $$ $$ \lambda M^{1/2} x - K^* M^{1/2} x = 0 $$ $$ ( \lambda M^{1/2} - K^* M^{1/2})x = 0 $$ $$ \lambda M^{1/2} - K^* M^{1/2} = \begin{pmatrix} \sqrt{3} \lambda - 17.3 & 11.5 \\ 14.1 & \sqrt{2} \lambda - 24.7 \end{pmatrix} $$ $$ det( \lambda M^{1/2} - K^* M^{1/2}) = 0 $$ Calculating the determinant $$ (\sqrt{3} \lambda - 17.3)(\sqrt{2} \lambda - 24.7) - 162.15 = 0 $$ $$ 2.4 \lambda^2 - 67.2 \lambda + 427.3 = 0 $$ Therefore, the Eigenvalues are: $$ \lambda_1 = 18.2, \lambda_2 = 9.8 $$ The next step is finding the Eigenvectors $$ \lambda_1 = 18.2 $$ $$ ( \lambda_1 M^{1/2} - K^* M^{1/2})X_1 = \begin{pmatrix} 14.2 & 11.5 \\ 14.1 & 1 \end{pmatrix} \begin{pmatrix} x_{11} \\ x_{21} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

Letting $$ x_{21} = 1 $$ and calculating $$ x_{12} $$

$$ \begin{pmatrix} x_{11} \\ x_{21} \end{pmatrix} = \begin{pmatrix} -0.81 \\ 1 \end{pmatrix} $$ $$ \lambda_2 = 9.8 $$ $$ ( \lambda_2 M^{1/2} - K^* M^{1/2})X_1 = \begin{pmatrix} -0.326 & 11.5 \\ 14.1 & -10.8 \end{pmatrix} \begin{pmatrix} x_{12} \\ x_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

Letting $$ x_{12} = 1 $$ and calculating $$ x_{22} $$

$$ \begin{pmatrix} x_{12} \\ x_{22} \end{pmatrix} = \begin{pmatrix} 1 \\ 1.3 \end{pmatrix} $$

$$ \mathbf{|X_1 X_2|}=\begin{bmatrix} x_{11} & x_{12}\\ x_{21} & x_{22}\\ \end{bmatrix} = \begin{bmatrix} -0.81 & 1\\ 1 & 1.3\\ \end{bmatrix} $$ These results compare favorably with those from problem 5.6 using CALFEM. Mass Orthogonality: $$ x_i^T M_{xj} = 0 $$ The eigenvectors are therefore proven to be mass-orthogonal.