ENGINEERING MATHEMATICS 4 (BDA 34003)

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1 ENGINEERING MATHEMATICS 4 (BDA 34003) Lecture Module 7: Eigenvalue and Eigenvector Waluyo Adi Siswanto Universiti Tun Hussein Onn Malaysia This work is licensed under a Creative Commons Attribution 3.0 License.

2 Topics Eigenvalue and eigenvector in Engineering Characteristics equation Power Method Shifted Power Method Inverse Power Method Lecture Module 7 BDA

3 Learning Outcomes Students have the knowlegde of eigenvalue and eigenvector in engineering applications Students will be able to determine natural frequencies of structural engineering problems Students will be able to decide the appropriate method for a particular structural engineering problem Lecture Module 7 BDA

4 Eigenvalue and Eigenvector in Engineering The bridge was shaken by the wind. This external shaking frequency coincided or very close to the natural frequency of the bridge. Since the external frequency met with the natural frequency, the displacement amplitudes were then amplified (resonance). The bridge was collapse because of this resonance. Lecture Module 7 BDA

5 Eigenvalue and Eigenvector Magnification of amplitude at resonance Vibration on its Natural frequency External excitation Resonance result Therefore you have to know the natural frequency, so you know dangerous external citations that will damage the structure because of resonance phenomenon Lecture Module 7 BDA

6 Eigenvalue and Eigenvector Natural Frequencies = characteristics Every system has its own natural frequencies. Therefore natural frequencies shows the characteristics (German = eigen) of the system. The equation representing the characteristics is then called characteristics equation Every system has its own characteristics f n =0 Hz mode = f n =0 Hz mode={,} T f n2 =25 Hz mode 2={, 0.8} T Lecture Module 7 BDA

7 Eigenvalue and Eigenvector Natural Frequencies and modeshapes Natural frequencies are obtained from the eigenvalues Mode shapes are obtained from the eigenvectors Eigenvalue no Eigenvector no f n =0 Hz mode={,} T Eigenvalue no 2 f n2 =25 Hz Eigenvector no 2 mode 2={, 0.8} T Lecture Module 7 BDA

8 Eigenvalue and Eigenvector Characteristics equation From this simple vibration illustration, we will see the characteristics equation m x + (k + k 2 ) x k 2 x 2 =0 m 2 x 2 + k 2 ( x 2 x )=0 m k k 2 x x + (k + k 2 ) x m k 2 x m 2 =0 x 2 + k 2 (x m 2 x )=0 2 m 2 x 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {x x + [ 0 ]{ 0 x x 2} =0 Lecture Module 7 BDA

9 Characteristics equation m k x [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {x x + [ 0 ]{ 0 x x 2} =0 sin representing x =V sin(ω t) x 2 =V 2 sin(ω t) fluctuation x = V ω 2 sin(ωt) x 2 = V 2 ω 2 sin(ωt) m 2 k 2 x 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {V V + [ ω2 0 2]{ V 2} 0 ω V =0 Lecture Module 7 BDA

10 Characteristics equation m k k 2 x [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 2} {V V + [ ω2 0 2]{ V 2} 0 ω V =0 2} {V V [ ω2 0 λ 0 ω 2]{ V V 2} =0 m 2 x 2 _ [ A] {V } λ[ I ] {V } = 0 Standard eigenvalue-eigenvector problem Lecture Module 7 BDA

11 Characteristics equation _ [ A] {V } λ[ I ] {V } = 0 [[ A] λ [ I ]]{V }=0 Since {V } 0 [ A] λ [ I ] =0 In mathematics is called Characteristics equation λ {V } eigenvalue eigenvector Lecture Module 7 BDA34003

12 Example 7- k =20 k =0 m = m 2 = m k k 2 x By using characteristics equation: Find the eigenvalues of the system Find the natural (resonance) frequencies of the system Find the eigenvectors of the system Draw the mode shape corresponding to to each natural frequency m 2 x 2 Lecture Module 7 BDA

13 k =20 k 2 =0 m = m 2 = and λ=ω 2 k m k 2 m 2 x x 2 [(k + k 2) k 2 m m ] k 2 k 2 m 2 m 2 [ 30 2} {V V [ ω2 0 2]{ V 2} 0 ω V = ]{ V V 2} [ λ 0 0 λ]{ V V 2} =0 [(30 λ) 0 0 (0 λ)]{ V 2} V =0 (30 λ) 0 0 (0 λ) =0 Characteristics equation Lecture Module 7 BDA

14 The characteristics equation can be expanded: (30 λ) 0 0 (0 λ) =0 m k x (30 λ)(0 λ) ( 0)( 0)= λ 30 λ+ λ 2 00=0 λ 2 40 λ+ 200=0 k 2 λ = λ 2 =34.42 Eigenvalue Eigenvalue 2 m 2 x 2 In SMath you can write (to solve a polynomial): Lecture Module 7 BDA

15 λ = λ 2 =34.42 Eigenvalue Eigenvalue 2 m k x Angular natural frequency: ω = rad/s ω 2 =5.843 rad/s k 2 Natural frequency: f = Hz m 2 x 2 f 2 =0.93 Hz Lecture Module 7 BDA

16 λ = Eigenvalue k m k 2 m 2 x x 2 [(30 λ) 0 0 (0 λ)]{ V 2} V =0 Normalizing V = (30 λ )V =0V 2 V 2 =2.442 Eigenvector { V V 2} = { 2.442} Lecture Module 7 BDA

17 λ 2 =34.42 Eigenvalue 2 k m k 2 m 2 x x 2 [(30 λ) 0 0 (0 λ)]{ V 2} V =0 Normalizing V = (30 λ )V =0V 2 V 2 = Eigenvector 2 { V V 2} = { 0.442} Lecture Module 7 BDA

18 Physical interpretation of eigenvectors: Eigenvector { V V 2} = { 2.442} Eigenvector 2 { V V 2} = { 0.442} k m x k 2 m 2 x Lecture Module 7 BDA

19 Example 7-2 k =20 k =0 m = m 2 = m k k 2 x Use FreeMat Find the eigenvalues of the system Find the natural (resonance) frequencies of the system Find the eigenvectors of the system Draw the mode shape corresponding to to each natural frequency m 2 x 2 Lecture Module 7 BDA

20 λ = λ 2 =34.42 Eigenvalue Eigenvalue 2 Angular natural frequency: ω = ω 2 =5.843 rad/s rad/s Natural frequency: f = f 2 =0.93 Hz Hz Lecture Module 7 BDA

21 λ = Eigenvalue { V V 2} = { } { V V 2} = { 2.442} Lecture Module 7 BDA

22 λ =34.42 Eigenvalue 2 { V V 2} = { } { V V 2} = { 0.442} Lecture Module 7 BDA

23 Example 7-3 x x 2 x 3 k k k m m m Develop the characteristics equation of the system above Find the eigenvalues of the system Find the natural (resonance) frequencies of the system Find the eigenvectors of the system Draw the mode shape corresponding to to each natural frequency (Use FreeMat to verify the eigenvalues and eigenvectors) Lecture Module 7 BDA

24 x x 2 x 3 k k k m m m [ 2k k 0 k 2k k x ]{x 3}+[ m m 0 ẍ 2 0 k k x 0 0 m]{ẍ ẍ 3}=0 x =V sin(ω t) x 2 =V 2 sin(ω t) x 3 =V 3 sin(ω t) [ 2k 2 k 0 ω 0 0 k 2k k V 0 k k ]{V 2 0 m ω V 3} [m 2]{V 2 0 V m ω V 3}=0 Lecture Module 7 BDA

25 [ V 0 ]{V 3} [ 2 mω 0 0 k mω k V m ω k ]{V V 2 V 3}=0 λ= mω2 k [ 2 λ 0 2 λ V 2 0 λ]{v V 3}=0 The characteristics equation 2 λ 0 2 λ 0 λ =0 Lecture Module 7 BDA

26 2 λ 0 2 λ 0 λ =0 (2 λ)[(2 λ)( λ) ( )( )] ( )[( )( λ) 0]=0 λ λ 2 6 λ+ =0 λ =0.98 Eigenvalue λ 2 =.555 Eigenvalue 2 λ 2 =3.247 Eigenvalue 3 Lecture Module 7 BDA

27 λ =0.98 Eigenvalue λ 2 =.555 Eigenvalue 2 λ 2 =3.247 Eigenvalue 3 natural frequency λ= mω2 k ω =0.98 k / m ω 2 =.555 k / m ω 3 =3.247 k / m rad/s rad/s rad/s f = π k / m f 2 = π k / m Hz Hz f 3 = π k / m Hz Lecture Module 7 BDA

28 λ =0.98 Eigenvalue [ 2 λ 0 2 λ V 2 normalized V = 0 λ]{v V 3}=0 (2 λ)v 2 V 3 =0 ( )V V 2 + ( λ)v 3 =0.809V 2 V 3 = V V 3 =0 V 2 =.8023 V 3 = Eigenvector {V V 2 V 3}= { } Lecture Module 7 BDA

29 λ 2 =.555 Eigenvalue 2 [ 2 λ 0 2 λ V 2 normalized V = 0 λ]{v V 3}=0 (2 λ)v 2 V 3 =0 ( )V V 2 + ( λ)v 3 = V 2 V 3 = V V 3 =0 V 2 =0.445 V 3 = {V V 3}={ Eigenvector V 0.809} Lecture Module 7 BDA

30 λ 3 =3.247 Eigenvalue 3 [ 2 λ 0 2 λ V 2 normalized V = 0 λ]{v V 3}=0 (2 λ)v 2 V 3 =0 ( )V V 2 + ( λ)v 3 =0.247V 2 V 3 = V V 3 =0 V 2 =.2469 V 3 = {V V 3}={ } Eigenvector V Lecture Module 7 BDA

31 x x 2 x 3 k k k m m m Mode Mode 2 {V V 2 V 3}= { } {V V 3}={ 2 V } {V V 2 V 3}= { } Mode Lecture Module 7 BDA

32 If you normalise the eigenvectors: Eigenvector Eigenvector 2 Eigenvector 3 {V V 2 V 3}= { } {V V 3}={ 2 V } {V V 2 V 3}= { } Lecture Module 7 BDA

33 This equation _ [ A] {V } λ[ I ] {V } = 0 Power Method can be rewritten [ A]{V }=λ {V } So that the eigenvalue can be calculated from the eigenvector In iterative manner {V } k+ [ A]{V }k = λ k+ This method will find the highest eigenvalue and its corresponding eigenvector This iteration requires an initial trial eigenvector {V } k In Engineering, this method is called Matrix Iteration Method Lecture Module 7 BDA

34 Example 7-4 Find the eigenvalue and the corresponding eigenvector in Example 7-3 by implementing Power Method (Matrix Iteration Method) Use the initial eigenvector {V }=( ) T Stop the iteration when λ k+ λ k < Lecture Module 7 BDA

35 The system has been identified: [ V 0 ]{V 3} [ 2 mω 0 0 k mω k V m ω k ]{V V 2 V 3}=0 Therefore [ A]=[ ] Lecture Module 7 BDA

36 Lecture Module 7 BDA

37 Continue. until convergence satisfied Lecture Module 7 BDA

38 This equation Shifted Power Method _ [ A] {V } λ[ I ] {V } = 0 [ A]{V }=λ {V } [ A]{V } s[ I ]{V }=λ{v } s {V } can be shifted with a scalar value s [ A s[ I ]]{V }=(λ s){v } [ A s ]{V }=λ s {V } So that the eigenvalue can be calculated from the eigenvector In iterative manner {V } k+ = [ A s]{v } k λ s k+ λ=λ s + s This method will find the lowest eigenvalue and its corresponding eigenvector In some vibration books [ A s ] is called the deflated matrix Lecture Module 7 BDA

39 Example 7-5 Continue Example 7-4 to find the lowest eigenvalue Lecture Module 7 BDA

40 Lecture Module 7 BDA

41 Continue. until convergence satisfied Lecture Module 7 BDA

42 This equation Inverse Power Method _ [ A] {V } λ[ I ] {V } = 0 [ A]{V }=λ {V } can be modified by multiplying [ A] [ A][ A] {V }=λ [ A] {V } {V }=λ[a] {V } So that the eigenvalue can be calculated from the eigenvector In iterative manner {V } k+ = [ A] {V } k λ k+ This method will find the lowest eigenvalue and its corresponding eigenvector λ= λ k+ This iteration requires an initial trial eigenvector {V } k Standard Matrix Iteration Method In Engineering, this method is called Standard Because it gives the lowest eigenvalue, which is the standard information in vibration. Lecture Module 7 BDA

43 Example 7-6 Do similar problem Example 7-4 by using the Standard Matrix Iteration (Inverse Power Method) Lecture Module 7 BDA

44 Lecture Module 7 BDA

45 Continue. until convergence satisfied Lecture Module 7 BDA

46 Student Activity Lecture Module 7 BDA

47 Problem 7-A x x 2 x 3 k k k m m m k=2 m=2 Derive the dynamics equation of motion of the system Find the natural frequencies and the corresponding mode shapes by using - characteristics equation - starndard iteration method - verify your result using freemat Lecture Module 7 BDA

48 Lecture Module 7 BDA