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Part 3 Chapter 13 Eigenvalues PowerPoints organized by Prof. Steve Chapra, Tufts University All images copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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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.

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Dynamics of Three Coupled Bungee Jumpers in Time Is there an underlying pattern???

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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 system so that the solution would be meaningful???

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Mathematics (a 11 – ) x 1 + a 12 x 2 + a 13 x 3 = 0 a 21 x 1 + (a 22 – ) x 2 + a 23 x 3 = 0 a 31 x 1 + a 32 x 2 + (a 33 – ) x 3 = 0 [A] – {x} = 0 [ ] What about a homogeneous system like: or in matrix form For this case, there could be a value of that makes the equations equal zero. This is called an eigenvalue.

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Graphical Depiction of Eigenvalues

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Physical Background: Oscillations or Vibrations of Mass- Spring Systems

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m2m2 d 2 x 2 dt 2 = – k(x 2 – x 1 ) – kx 2 m1m1 d 2 x 1 dt 2 = – k x 1 + k(x 2 – x 1 ) m2m2 d 2 x 2 dt 2 – k (x 1 – 2x 2 ) = 0 m1m1 d 2 x 1 dt 2 – k (– 2x 1 + x 2 ) = 0 Collect terms: Model With Force Balances (AKA: F = ma )

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x i = X i sin ( t) where TpTp Differentiate twice: x i ” = – X i 2 sin ( t) Substitute back into system and collect terms Assume a Sinusoidal Solution

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X 1 – kk m1m1 X 2 k m1m1 X 2 kk m2m2 X 1 + k m2m2 Given: m 1 = m 2 = 40 kg; k = 200 N/m (10 – 2 ) X 1 – 5 X 2 = 0 – 5 X 1 + (10 – 2 ) X 2 = 0 This is now a homogeneous system where the eigenvalue represents the square of the fundamental frequency.

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Evaluate the determinant to yield a polynomial Solution: The Polynomial Method 10 – 2 5 X 1 0 – 5 10 – 2 X 2 0 = = ( 2 ) 2 2 10 – 2 5 – 5 10 – 2 The two roots of this "characteristic polynomial" are the system's eigenvalues: 2 = 15 5 or 3.873 Hz 2.36 Hz

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INTERPRETATION 2 = 5 /s 2 = 2.236 /s T p = 2 /2.236 = 2.81 s (10 – 2 ) X 1 – 5 X 2 = 0 – 5 X 1 + (10 – 2 ) X 2 = 0 (10 – ) X 1 – 5 X 2 = 0 – 5 X 1 + (10 – ) X 2 = 0 X 1 – 5 X 2 = 0 – 5 X 1 + X 2 = 0 X 1 = X 2 V = –0.7071 (10 – 1 ) X 1 – 5 X 2 = 0 – 5 X 1 + (10 – 1 ) X 2 = 0 – X 1 – 5 X 2 = 0 – 5 X 1 – X 2 = 0 X 1 = –X 2 V = –0.7071 0.7071 2 = 15 /s 2 = 3.873 /s T p = 2 /3.373 = 1.62 s

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Principle Modes of Vibration T p = 1.62 T p = 2.81

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Iterative method to compute the largest eigenvalue and its associated eigenvector. The Power Method [[A] [I]]]{x} = 0 [A]{x} = {x} Simple Algorithm: function [eval, evect] = powereig(A,es,maxit) n=length(A); evect=ones(n,1);eval=1;iter=0;ea=100; %initialize while(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 = maxit,break,end end

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Example: The Power Method 40 20 0 20 40 20 0 20 40 111111 = 20 0 20 = 20 101101 First iteration: 40 20 0 20 40 20 0 20 40 101101 = 40 20 40 = 40 111111 Second iteration: | a | = 40 20 40 100% = 50%

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Example: The Power Method Third iteration: | a | = 80 40 80 100% = 150% 40 20 0 20 40 20 0 20 40 = = 80 0.75 1 0.75 111111 60 80 60 Fourth iteration: | a | = 70 80) 70 100% = 214% 40 20 0 20 40 20 0 20 40 = = 70 0.71429 1 0.71429 50 75 50 0.75 1 0.75

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Example: The Power Method Fifth iteration: | a | = 68.51714 70 70 100% = 2.08% 40 20 0 20 40 20 0 20 40 = = 68.51714 0.71429 1 0.71429 48.51714 68.51714 48.51714 0.71429 1 0.71429 Note that the smallest eigenvalue and its associated eigenvector can be determined by applying the power method to the inverse of A The process can be continued to determine the largest eigenvalue (= 68.284) with the associated eigenvector [ 0.7071 1 0.7071]

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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 0.7071 lambda = 5 0 0 15

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Dynamics of Three Story Building

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Principle Modes of Vibration

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