Assume D diagonalizable

Transcription

1 Chapter 5 Forced Response and Frequency Response 5. k k m c c m F F M ɺɺ Dɺ K = F( t) Assume D diagonalizable

2 If decoupled Decouple the system with the eigenvalues of ɺɺ y ζ iω i λ yɺ i y = V th so the i equation would be yi ζ iωi yi ωi yi = fi ( t) Responses to harmonic, periodic, or general forces as in Chapter Note that the modal force is a linear combination of many physical forces ɺɺ ɺ T Kɶ F

3 Modal equation ɺɺ y t yɺ t y t f t i ( ) ζ iωi i ( ) ωi i ( ) = i ( ) ζ iωit y ( t) = d e sin ω t φ ( ) i i di i t ζ ( ) sin ( ) iωit ζ iωiτ e fi t e ωdi t τ dτ ω di 3

4 An eample with three masses k k k 3 3 k 4 m m c c c 3 m 3 c 4 F F F 3 m =m =m 3 =Kg k =k =k 3 = k =3N/m C=.K M m k k k = m K = k k k k 3 3 m 3 k3 k3 k 4 4

5 Computation Using Matlab to calculate the eigenvectors and eigenvalues and hence the mode shapes and natural frequencies. V = ω =.94 ω =.73 ω =.6 3 5

6 5. Modal Analysis of Conservative Forced Systems Consider Mqɺɺ ( t) Kq( t) = f ( t), and let V be the modal matri: T T such that V MV = I, V KV = Λ n q( t) = y ( t) u = Vy( t) since u forms a basis i= i i i T T T V MVy ɺɺ ( t) V KVy( t) = V f ( t) ɺɺ ω = y i ( t) i y i ( t) fi ( t) With initial conditions y() V q(), y() V q() t = ɺ = y ( t) = f ( t τ )sinωτ dτ y ()cosω t i i i i i ωi ɺ ɺy i () sin ωit ω i 6

7 Forced Response Eample sin 3, qɺɺ t 3 3 q =. q() =, () = qɺ 6 8 = = and = 4 λ λ λ λ 7

8 Modal Transformation The modal equations then become: ɺɺ y ( t) y ( t) = sin 3 t, ɺɺ y ( t) 4y ( t) = sin 3 t. Note that the transformation decouples the equations of motion, but distributes the initial conditions and applied force. 8

9 Initial Conditions T.3 y() = V Mq() = T yɺ () = V Mqɺ () = 9

10 Time Response { 3.3 y ( t) = sin t cos t sin 3t y ( t) = sin t cos t sin 3t q( t) = Vy( t) 3.3 sin t cos t sin 3t / =.3.3. sin t cos t sin 3t sin t.3cos t sin 3t = 4 7.3sin t.3cos t.sin 3t

11 5.3 State Space Form I ui Az = λz, A = (3.6) zi = λ M K M D iui Alternately: z = q and z = qɺ z = z K zɺ K z M zɺ K D z A B = Az = λ Bz which is a symmetric eigenvalue problem

12 State Space Form Parameter of the linearized system M = K = D.. =.. state space Forms I = M K M D Az = λz A z ui = λ iui A =.... z.77 = z.77 =.77..

13 Response via state-space method assume c=. qɺɺ.. qɺ q = cos. t q.. q q 44 ɺɺ ɺ q() = qɺ = q = = q q qɺ qɺ qɺ T ɺɺ = cos.44 t.... () = 3

14 Discussion ɺɺ y ζ ω y ɺ Λ y i i = λ.λ = eig( A) i i = i i λ.3λ 3 = λ = i i λ = i i 4

15 Forced Response Response via State Space Method ɺ ( t) = A( t) f ( t) sx( s) = AX( s) F( s) () st where X( s) = ( t) e ds X( s) = si A () si A F( s) ( ) ( ) 5

16 5.4 Modal Analysis Let A = V T ΛV where Λ is the diagonal Jordon form of A. Let = Vz in the equation of motion: ɺ T T T Vz = AVz f V Vz = V AVz V f ɺ = T z Λz V f ɺ a decoupled vector equation T zɺ = λ z ( V f ) i=,,3 n scalar equations i i i i 6

17 Frequency response Function Amplitude (db) X (ω)/f (ω)) X (ω)/f (ω)) X 3 (ω)/f (ω)) Amplitude (db) 4 - X (ω)/f (ω)) X (ω)/f (ω)) X 3 (ω)/f (ω)) Frequency (ω) Frequency (ω) 7

18 5.5 Frequency Response Function jωt Substitute q( t) = ue in Mqɺɺ ( t) Kq( t) = f e ( ω ) jωt e = K M u f e jωt ( ω ) u = K M f α ( ω) = Receptance Matri Set each element of f = ecept the k th then α ik ω ( )= u i f k = The transfer function between a response measured at i and a force applied at k. jωt 8

19 9 Equation of Motion(/) m m T ɺ ɺ = 3 ) ( k k k V = 3 ) ( c c c F ɺ ɺ ɺ ɺ = 3 ) ( k k k m m V T L = = ɺ ɺ Using Largrange s method Kinetic Energy: Potential Energy: Rayleigh dissipation : Largrangian function:

20 Equation of Motion(/),,, i i i i d L L F Q i n dt = = ɺ ɺ, ) ( ) ( = k k k c c c m ɺ ɺ ɺɺ ) ( ) ( 3 3 = k k k c c c m ɺ ɺ ɺɺ = 3 3 k k k k k k c c c c c c m m ɺ ɺ ɺɺ ɺɺ Largrange s generation equation get the equation of motion the motion matri is

21 Natural Frequency = = i i k m = ɺɺ ɺɺ det( ) ω = K M 3 = ω 3 = = ω ω the natural frequency assume The characteristic equation we can get the eigenvalues the motion matri becomes

22 Eigenvector and normal mode Let ω = λ ( λi ) u = c K = u u ( λi ) u the mode shape K = u u The modal matri is T V MV = I V = the mode shape = c

23 V = Normal Mode Animation the maimum displacement of mode and mode =.77 mode mode 3

24 Normal Mode of Initial Displacement ( t) = State function of time response ( t) = A u sin( ω t φ) u A sin( ωt φ) I.C. () ɺ () = =.8 ω = u = ω = 3 u mass initial displacement = = mode mode ɺ ( t) = A u ω cos( ωt φ) u Aω cos( ωt φ) time response π π ( t) = sin( t ) sin( 3t ) π π ( t) = sin( t ) sin( 3t ) displacement time 4

25 Normal Mode of Initial Displacement ( t) =.8 mass initial displacement =.6.4 m Initial state displacement. -. m time - 5

26 Normal Mode of Initial velocity ɺ ( t) = State function of time response ( t) = A u sin( ω t φ) u A sin( ωt φ) ω cos( ωt φ) u Aω cos( ωt φ) () = ɺ() = ɺ ( t) = A u time response 3 ( t) = sin( t) sin( 6 3t) 3 ( t) = sin( t) sin( 6 3t) I.C. displacement ω = ω = 3 u u mass initial velocity = = = time 6 mode mode

27 ɺ ( t) = Normal Mode of Initial Velocity m m mass initial velocity = displacement time 7

28 Vibration response Mqɺɺ Dqɺ Kq = f due q = Vy we can rewrite the function y ζ ω y y = V T ɺɺ ɺ f i i Λ Use Laplace transform, the function become s ζ ω s y = V T I f i i Λ multiply V ( s I ζ ω s ) i i Λ Vy = V V T f ( I ζ ω ) i i Λ G( s) = V s s V T 8

29 Frequency Response Function V = Λ = c = c =. G( s) ( s I ) T Λ V ( G( s) = V s I ζ ω s ) i i Λ = V transferfunctions of the input and output transferfunctions of the input and output 8 6 V T 7 5 G(jw) Zero frequency ecitation.44 G(jw) 4 3 Zero frequency ecitation w w 9

30 Time Response without Damping(/3) ecitation force F( t) T = V f F = sin.44t = sin.44t ɺɺ y y =.77 sin.4t ɺɺ y Characteristic equation 3y =.77 sin.44t assume p = a cos.44t bsin. 44t 3

31 Time response without Damping(/3) due = Vy get equation of motion = sin.44t.77.77sin.44t.8 ( t) = ( t) sin. 44t = f (t) displacement time 3

32 Time Response without Damping(3/3) second m displacement. -. m displacement time time 3 3

33 Time Response with Damping(/3) ecitation force F( ) F = sin.44t = sin.44t characteristic equation of without damping ɺɺ y.yɺ y =.77 sin.44t ɺɺ y.3yɺ y =.77 sin.44t T t = V f 33

34 Time Response with Damping(/3) Let get equation of motion =.77 =.77 Vy [.693sin.44t.98cos. t].77[.6sin.44t.54cos. 44t] ( t) = sin.44t.98cos.44t.77.6sin.44t.54cos.44t [.693sin.44t.98cos. t].77[.6sin.44t.54cos. 44t] ( t) =

35 Time Response with Damping(3/3) second 6second displacement. -. displacement time time 35 35

36 System with Damping(/4) Proportional damping ζω ω ɺɺ y = ζω yɺ y ω ζ =%. ɺɺ y yɺ y =. 3 3 y ( t) e A cos( ω t) A sin( ω t) ωdi = ωi ζ ζωn = t i d i d i i 36

37 System with Damping(/4) ζ = % y y ω d =. ω =.t = ( t) e A cos( t) A sin( t) ζ = % ω 3 ω = = ( t) e A cos.7t A sin.7t.73t = 3 4 d q = Vy = e e.73t.t ( A 3 ( A cost A sint) cos.7t A4 sin.7t) 37

38 System with Damping(3/4) ( t) = Initial condition () = ɺ () = A =.77 A =.7 A 77 3 =. A 7 4 =. time response.t ( t) =.77e (.77 cost.7sin t).73t.77e (.77 cos.7t.7sin.7t).t ( t) =.77e (.77 cost.7sin t).73t.77e (.77 cos.7t.7sin.7t) 38

39 System with Damping(4/4).8 second m The system with damping.8 6second The system with damping.6.6 displacement.4. m displacement time time 39 39

40 System without Damping(/4) Proportional damping ζω ω ɺɺ y ζω yɺ ω y = ζ = = ɺɺy 3 y y ( t) e A cos( ω t) A sin( ω t) ζωn = t i d i d i ω = ω ζ d i i 4

41 System without Damping(/4) ω = y ( t) = A cos( t) A sin( t) ω = 3 y ( t) = A cos 3t A sin 3t q 3 4 = Vy.77 = ( A cost A.77 ( A3 cos 3t A 4 sint) sin 3t) ( A cost A sin t).77( A cos 3t A sin t) 3 ( A cost A sin t).77( A cos 3t A sin t) t) =.77 3 ( 4 ( t) =

42 System without Damping(3/4) ( t) = Initial condition () = ɺ () = A =.77 A = A =. 77 A = 3 4 time response ( t) =.5cost.5cos 3t ( t) =.5cost.5cos 3t 4

43 System without Damping(4/4).8 m mass intial displacement =.8 mass intial displacement =.6.4 m.6.4 displacement. -. displacement time time 43 43

44 System with Damping The system with damping The system with damping displacement. displacement time time 44 44

45 System without Damping mass intial displacement = mass intial displacement = displacement. -. displacement time time 45 45

46 5.6 Lumped-Parameter Models Mechanical Systems Mqɺɺ ( t) ( D G) qɺ ( t) ( K H ) q( t) = f ( t) M = M T = mass or inertia matri, usually symmetric D = D T = viscous damping matri (often denoted C) G = -G T = gyroscopic matri K = K T = stiffness matri H = -H T = circulatory matri (constraint damping) M can also be asymmetric but not so common 46

47 Sign of a Matri Consider the quadratic form, T A, a scalar. Note that A if I then: T A = 3 n Positive definite: If T A > for all nonzero real vectors, and T A = if and only if is zero. Positive semidefinite: If T A > for all nonzero real vectors, and (here T A could be zero for some nonzero ). Indefinite: If ( T A)( y T Ay) < for some pair of real vectors and y. 47

48 Conservative system: Classification of Systems Mqɺɺ Kq = f M positive definite, symmetric and K positive semidefinite symmetric Conservative Gyroscopic Systems: Mqɺɺ Gqɺ Kq = f Damped Non Gyroscopic Systems: G skew symmetric Circulatory Systems: Mqɺɺ ( K H ) q = f Mqɺɺ Dqɺ Kq = f D symmetric positive semidefinite H skew symmetric 48

49 Feedback Control Systems Mqɺɺ ( G D) qɺ ( K H ) q = K q K qɺ f p v control law Here K p is a position feedback gain matri and is a velocity feedback gain matri K v 49

50 State Space Formulation ɺ ( t) = ( t) Mɺ ( t) = ( D G) ɺ ( K H ) f ( t) (.9) where = q and = qɺ ɺ I ɺ ( t) = = ( ) ( ) ɺ M MD G M K H M A B f Which is of the form ɺ = A Bu 5

51 Eample of Gyroscopic Systems Summing forces and using only linear spring deflections yields: m c k k mω m m ɺɺ ɺ k mω q Ω q q = ( t) Here q =. Note that,, and are symmetric, and is skew symmetric y( t) M D K G 5

52 Matri Definiteness T M m = [ ] = ( ) > positive definite m m c T = c positive semi definite The quadratic form for a skew symmetric matri is always zero: T G = mω( ) = 5

53 Eample of Circulatory System: Pfluger s Rod Defines a follower force π 3 4 m EIπ η qɺɺ q = l 8 8 π 9 M is symmetric and positive definite, K is not symmetric but can be 4 written as: EI π π η η l η K H = EIπ 9 η ηπ 3 H 9 l K 53

54 Eample Feedback Control The open loop system: Fig.4 m c c c k k k m qɺɺ = c c qɺ k k q f Adding the control force: f = - g - g yields: m c c c k k k m qɺɺ = c c qɺ k g k g q An eample of position feedback. Note that the feedback destroys the symmetry but does add additional constants to tweak for improving performance. 54

55 Eperimental Models It is possible to eperimentally measure the transfer function of a structure resulting in: Several possible models could be fit to this transfer function. The plot suggest 3 DOF, maybe of the form: Fig.5 Physical: Modal: m c k m qɺɺ c qɺ k q = m 3 c 3 k 3 ζ ω ω rɺɺ ζ ω rɺ ω r = ζ 3ω 3 ω 3 55