part 3 chapter 13 eigenvalues {contains corrections to the texbook} powerpoints organized by prof....
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Part 3Chapter 13
Eigenvalues
{contains corrections to the Texbook}
PowerPoints organized by Prof. Steve Chapra, Tufts UniversityAll images copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter Objectives
• Understanding the mathematical definition of eigenvalues and eigenvectors.
• Understanding the physical interpretation of eigenvalues and eigenvectors within the context of engineering systems that vibrate or oscillate.
• Knowing how to implement the polynomial method.• Knowing how to implement the power method to
evaluate the largest and smallest eigenvalues and their respective eigenvectors.
• Knowing how to use and interpret MATLAB’s eig function.
Representing Linear Algebra
• Matrices provide a concise notation for representing and solving simultaneous linear equations:
a11x1 a12x2 a13x3 b1
a21x1 a22x2 a23x3 b2
a31x1 a32x2 a33x3 b3
a11 a12 a13
a21 a22 a23
a31 a32 a33
x1
x2
x3
b1
b2
b3
[A]{x} {b}
Dynamics of Three Coupled Bungee Jumpers in Time
Is there an underlying pattern???
Mathematics
[A] {x} = {b}
[A] {x} = 0
Up until now, heterogeneous systems:
What about homogeneous systems:
Trivial solution:
{x} = 0
Is there another way of formulating the systemso that the solution would be meaningful???
Mathematics
(a11 – l) x1 + a12 x2 + a13 x3 = 0 a21 x1 + (a22 – l) x2 + a23 x3 = 0 a31 x1 + a32 x2 + (a33 – l) x3 = 0
[A] – l [ I ] {x} = 0[ ]
homogeneous system:
or in matrix form
For this case, there could be a value of thatmakes the equations equal zero. This is calledan eigenvalue.
+ a11a22
+ × {correct textbook}
{correct textbook}
Graphical Depiction of Eigenvalues
Physical Background:Oscillations or Vibrations of Mass-
Spring Systems
m2
d 2x2
dt 2 = – k(x2 – x1) – kx2
m1
d 2x1
dt 2 = – k x1 + k(x2 – x1)
m2
d 2x2
dt 2– k (x1 – 2x2) = 0
m1
d 2x1
dt 2– k (– 2x1 + x2) = 0
Collect terms:
Model With Force Balances(AKA: F = ma)
xi = Xi sin ( w t) where w = 2pTp
Differentiate twice:
xi″ = – Xi w2 sin ( w t)
Substitute back into system and collect terms
Assume a Sinusoidal Solution
- w 2X1 – 2k
m1
X2 =0k
m1
- w 2X2 =02k
m2
X1 +k
m2
Given: m1 = m2 = 40 kg; k = 200 N/m
(10 – w 2) X1 – 5 X2 = 0
– 5 X1 + (10 – w 2) X2 = 0
This is now a homogeneous system where the eigenvalue represents the square of the
fundamental frequency.
Evaluate the determinant to yield a polynomial
Solution: The Polynomial Method
10 – w 2 -5 X1 0 – 5 10 – w 2 X2 0
=
= (w 2)2 -20w 2+7510 – w 2 -5 – 5 10 – w 2
The two roots of this "characteristic polynomial" are the system's eigenvalues (λ = w2)
w2 = w =3.873 rad/s2.36 rad/s
15 rad2/s2
5 rad2/s2
0.616 Hz0.356 Hz
f=
INTERPRETATIONw2 = 5 /s2
w = 2.236 /sTp = 2p/2.236 = 2.81 s
(10 – w2) X1 – 5 X2 = 0 – 5 X1 + (10 – w2) X2 = 0
(10 – 5) X1 – 5 X2 = 0 – 5 X1 + (10 – 5) X2 = 0
5 X1 – 5 X2 = 0– 5 X1 + 5 X2 = 0
X1 = X2
V =–0.7071–0.7071
(10 – 15) X1 – 5 X2 = 0 – 5 X1 + (10 – 15) X2 = 0
–5 X1 – 5 X2 = 0– 5 X1 – 5 X2 = 0
X1 = –X2
V =–0.70710.7071
w2 = 15 /s2
w = 3.873 /sTp = 2p/3.373 = 1.62 s
Principle Modes of Vibration
Tp = 1.62Tp = 1/f
Tp = 2.81Tp = 1/f
Iterative method to compute the largest eigenvalue and its associated eigenvector.
The Power Method
[[A] -l[I]]]{x} =0
[A]{x} =l{x}Simple Algorithm:
function [eval, evect] = powereig(A,es,maxit)n=length(A);evect=ones(n,1);eval=1;iter=0;ea=100; %initializewhile(1) evalold=eval; %save old eigenvalue value evect=A*evect; %determine eigenvector as [A]*{x) eval=max(abs(evect)); %determine new eigenvalue evect=evect./eval; %normalize eigenvector to eigenvalue iter=iter+1; if eval~=0, ea = abs((eval-evalold)/eval)*100; end if ea<=es | iter >= maxit,break,endend
Example
40 -20 0-20 40 -20 0 -20 40
111
=200
20
= 20101
First iteration:
40 -20 0-20 40 -20 0 -20 40
101
=40-4040
= 401-11
Second iteration:
|ea| = 40 -20
40 100% = 50%
Example: The Power MethodThird iteration:
|ea| = -80 -40
-80 100% = 150%
40 -20 0-20 40 -20 0 -20 40
= = -80-0.75
1-0.75
1-11
60-80 60
Fourth iteration:
|ea| = 70 -(-80)
70 100% = 214%
40 -20 0-20 40 -20 0 -20 40
= = 70-0.71429
1-0.71429
-50 70-50
-0.751
-0.75
Example: The Power MethodFifth iteration:
|ea | = 68.51714 - 70
70 100% = 2.08%
40 -20 0-20 40 -20 0 -20 40
= = 68.51714-0.708331
-0.70833
-48.51714 68.51714-48.51714
-0.714291
-0.71429
Note that the smallest eigenvalue and its associatedeigenvector can be determined by applying the
power method to the inverse of A
The process can be continued to determine the largesteigenvalue (= 68.284) with the associated eigenvector[-0.7071 1 -0.7071]
Determining Eigenvalues & Eigenvectors with MATLAB
>> A = [10 -5;-5 10]
A = 10 -5 -5 10
>> [v,lambda] = eig(A)
v = -0.7071 -0.7071 -0.7071 0.7071lambda = 5 0 0 15
Dynamics of Three Story Building
Principle Modes of Vibration