University of Florida/Eml4507/s13.team3.GuzyR5

Problem 5.1

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

${\displaystyle m_{1}=3}$

${\displaystyle m_{2}=2}$

${\displaystyle k_{1}=10}$

${\displaystyle k_{2}=20}$

${\displaystyle k_{3}=15}$

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.

First plug given values into the stiffness matrix:

${\displaystyle K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}}$

${\displaystyle K={\begin{bmatrix}(10+20)&-20\\-20&(20+15)\\\end{bmatrix}}}$

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

Next plug stiffness matrix in equation below:

${\displaystyle [K-\gamma _{I}]x=0}$

Take the determinant:

${\displaystyle det{\begin{Bmatrix}30-\gamma &-20\\-20&35-\gamma \end{Bmatrix}}=[(30-\gamma )(35-\gamma )]-(-20)(-20)=0}$

Simplifying:

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

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

Using the Quadratic formula to solve:

${\displaystyle \gamma _{1}={\frac {65+{\sqrt {65^{2}-(4)(650)}}}{2}}}$

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

Final gamma values:

${\displaystyle \gamma _{1}=52.66}$

${\displaystyle \gamma _{2}=12.34}$

To find the eigenvectors, first we:

${\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-\gamma *{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}{\begin{bmatrix}]x_{1}\\x_{2}\\\end{bmatrix}}=0}$

To solve, lets set ${\displaystyle x_{1}}$ equal to 1:

${\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}$

${\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}+{\begin{bmatrix}-52.66&0\\0&-52.66\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}$

${\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30-52.66&-20\\-20&35-52.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}$

${\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}$

Multiplying out the matrices, we obtain:

${\displaystyle -20-17.66x_{2}=0}$

${\displaystyle x_{2}=1.1325}$

Using the same process we find ${\displaystyle x_{1}}$:

${\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0}$

${\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0}$

Multiplying out the matrices, we obtain:

${\displaystyle -22.66x_{1}-20=0}$

${\displaystyle x_{1}=0.8826}$