University of Florida/Eml4507/s13.team3.GuzyR5

On my honor, I have neither given nor received unauthorized aid in doing this assignment. 

Given

$m_1=3$

$m_2=2$

$k_1=10$

$k_2=20$

$k_3=15$

Find

Solve by hand the gen. eigenvalue problem for the spring-mass-damper system on p.53-13, using the data for the masses in (2) p.53-13b, and the data for the stiffness coefficients in (4) p.53-13b.

Solution

First plug given values into the stiffness matrix:

$K=\left[\begin{array}{cc}(k_1+k_2)& -k_2\\ -k_2& (k_2+k_3)\end{array}\right]$

$K=\left[\begin{array}{cc}(10+20)& -20\\ -20& (20+15)\end{array}\right]$

$K=\left[\begin{array}{cc}30& -20\\ -20& 35\end{array}\right]$

Next plug stiffness matrix in equation below:

$[K-{\gamma }_I]x=0$

Take the determinant:

$det\left\{\begin{array}{cc}30-\gamma & -20\\ -20& 35-\gamma \end{array}\right\}=[(30-\gamma )(35-\gamma )]-(-20)(-20)=0$

Simplifying:

$1050-30\gamma -35\gamma +{\gamma }^2-400$

${\gamma }^2-65\gamma +650=0$

Using the Quadratic formula to solve:

${\gamma }_1=\frac{65+\sqrt{65^2-(4)(650)}}{2}$

${\gamma }_2=\frac{65-\sqrt{65^2-(4)(650)}}{2}$

Final gamma values:

${\gamma }_1=52.66$

${\gamma }_2=12.34$

To find the eigenvectors, first we:

$[K-{\gamma }_I]x=[\left[\begin{array}{cc}30& -20\\ -20& 35\end{array}\right]-\gamma *\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}]x_1\\ x_2\end{array}\right]=0$

To solve, lets set $x_1$ equal to 1:

$[K-{\gamma }_I]x=[\left[\begin{array}{cc}30& -20\\ -20& 35\end{array}\right]-52.66*\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]]\left[\begin{array}{c}1\\ x_2\end{array}\right]=0$

$[K-{\gamma }_I]x=[\left[\begin{array}{cc}30& -20\\ -20& 35\end{array}\right]+\left[\begin{array}{cc}-52.66& 0\\ 0& -52.66\end{array}\right]]\left[\begin{array}{c}1\\ x_2\end{array}\right]=0$

$[K-{\gamma }_I]x=[\left[\begin{array}{cc}30-52.66& -20\\ -20& 35-52.66\end{array}\right]\left[\begin{array}{c}1\\ x_2\end{array}\right]=0$

$[K-{\gamma }_I]x=[\left[\begin{array}{cc}-22.66& -20\\ -20& -17.66\end{array}\right]\left[\begin{array}{c}1\\ x_2\end{array}\right]=0$

Multiplying out the matrices, we obtain:

$-20-17.66x_2=0$

$x_2=1.1325$

Using the same process we find $x_1$:

$[K-{\gamma }_I]x=[\left[\begin{array}{cc}30& -20\\ -20& 35\end{array}\right]-52.66*\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]]\left[\begin{array}{c}{x}_1\\ 1\end{array}\right]=0$

$[K-{\gamma }_I]x=[\left[\begin{array}{cc}-22.66& -20\\ -20& -17.66\end{array}\right]\left[\begin{array}{c}{x}_1\\ 1\end{array}\right]=0$

Multiplying out the matrices, we obtain:

$-22.66x_1-20=0$

$x_1=0.8826$