Ch. 1: Introduction of Mechanical Vibrations Modeling

Transcription

1 1.0 Outline That You Should Know Newton s Second Law Equations of Motion Equilibrium, Stability 1.0 Outline

2 1.1 That You Should Know Vibration is the repetitive motion of the system relative to a stationary frame of reference or nominal position. Principles of Motion Vibration Modeling Math Vibration Analysis *** design the system to have a particular response *** 1.1 That You Should Know

3 Spring-Mass Model Mechanical Energy = Potential + Kinetic From the energy point of view, vibration is caused by the exchange of potential and kinetic energy. When all energy goes into PE, the motion stops. When all energy goes into KE, max velocity happens. Spring stores potential energy by its deformation (kx /). Mass stores kinetic energy by its motion (mv /). 1.1 That You Should Know

4 Springs and Masses connection as the way to model the vibrating system. 1.1 That You Should Know

5 Real-life system may not see the springs/masses connection explicitly! You have to devise the simple model smartly, suitable to the requirements. 1.1 That You Should Know

6 Various forms of springs: linear, traverse, torsion spring Gravity force can make up the spring! Because work done by the gravity force is a kind of PE. When the altitude change, PE change KE change. 1.1 That You Should Know

7 Spring Stiffness (N/m or Nm/rad) Spring has the characteristic that force is the function of deformation: F = k(x). If k is constant, the spring is linear. Practically it is not constant. k is generally slope of the F-x curve, and is known as the stiffness. 1.1 That You Should Know

8 Stiffness is the property of the object which depends mainly on its shape and material properties, e.g. E. Equivalent massless spring constants 1.1 That You Should Know

9 Equivalent massless spring constants (cont.) 1.1 That You Should Know

13 Spring connection: series or parallel Forces add up Disp. add up The calculated stiffness value, k eff, is used in the modeling system. k eff 1.1 That You Should Know

14 Ex That You Should Know

15 Ex. Derive the equivalent spring constant for the system in the figure. 1.1 That You Should Know

16 let the system be deviated by torque τ F k = at that point Δ τ τ at distance l = a, Feq = and Δ= δ kl= a= a aδ τ b τa at distance l = b, F = and Δ= δ k = = k b a b δ τ kb kb = kl= a= δ a a 1 kkb 1 kl= ais in series with k1 keq = = ka 1 + kb k k eq l= b l= a That You Should Know

17 Perturb the system by force F and observe the displacement F keq = but F = k1 ( x aθ ) x k ( x aθ ) need to find relation between x and θ x 1 keq = ( θ) MO = 0 kbθ b k1 x a a= 0 kax kkb θ = = ka kb ka kb 1 1 k eq x 1.1 That You Should Know

18 Ex. 3 The system shown in the figure consists of two gears A and B mounted on uniform circular shafts of equal Stiffness GJ/L; the gears are capable of rolling on each other without slip. Derive an expression for the Equivalent spring constant of the system for the radii ratio R a / R b = n. 1.1 That You Should Know

19 Apply torque M at gear A, then gear A rotates θ. Shaft exerts resistant moment M and gear B exerts reaction force F. In turns, at gear B, opposite reaction force F happens and shaft exerts moment M to resist rotation θ. B A B A k eq@a M = need to write M and θa as functions of given parameters. θ A M = 0 M = MA + FRA and MB = FRB GJ GJ by spring stiffness, MA = θa and MB = θb L L RA θb by geometry, n = = R θ GJ M = θa + L A B GJ θb L R R M GJ keq@a = = + θ L B A A ( 1 n ) GJ = + L ( 1 n ) θ A 1.1 That You Should Know

20 θ B M A = = n θa M B trans torsional spring at gear B is connected to gear A k B A B = = = gear A sees stiffness at gear B = n k B θb L n θa eq@a M GJ 1 M springs are connected in parallel, equivalent stiffness at gear A k = k GJ L ( n) A + nkb = That You Should Know

21 Ex. 4 The two gears of the system have mass polar moment of inertia I A and I B. Derive an expression for the equivalent mass polar moment of inertia for the radii ratio R A / R B = n. 1.1 That You Should Know

22 Ignore the torsional spring and write the equation of motion at gear A M = I θ M FRA = I AθA to eliminate F, consider motion at gear B FRB = I Bθ B θb RA Employ the geometric constraints: = n = θa RB IBn θa M RA = I AθA M = ( IA + n IB) θa R A I = I + n I eq@a A B 1.1 That You Should Know

23 Mass, Inertia To store / release kinetic energy 1.1 That You Should Know

24 Damper viscous coefficient (Ns/m or Nms/rad) To dissipate energy fd = F = cx 1.1 That You Should Know

25 1.1 That You Should Know

30 1. Newton s Second Law Many ways to derive the equations of motions Newton-Euler is one, which is suitable for situations system in planar motion force and motion have constant direction system is simple 1. Newton s Second Law

31 Relative Velocity v v v v A B A/B rel = velocity of the particle A = velocity of the particle B = velocity of A relative to B A/B rel v = v + ω r + v A B A/B rel = ω r + v velocity of A as seen from the observer = x i+ y j = velocity of A as seen from an observer ω r = at anywhere fixed to the rotating x-y axes A/B ( vrel observed from nonrotating x-y) ( v observed from rotating x-y) rel If we use nonrotating axes, there will be no ω r term. This makes v rel = v A/B A/B, which means the velocity seen by the observer is the velocity of A relative to B. If B coincides with A, r = 0. This makes v = v, A/B rel A/B which means the velocity seen by the observer is the velocity of A relative to B, even the observer is rotating. 1. Newton s Second Law

32 Relative Acceleration Differentiating the relative velocity equation results in the relative acceleration equation: a = a + ω r + ω r + v A B A/B A/B rel Recall the vector differentiation relation, r = v + ω r v = a + ω v A/B rel A/B rel rel rel dω dω dω ω = + ω ω, = dt xy dt XY dt xy angular acceleration observed in fixed frame = that observed in the rotating frame Note: r = xi + yj v = x i + y j a = xi + yj A/B rel rel ( vrel and arel = velocity and acceleration seen by the rotating observer) a a ω r ω ( ω r ) ω v a = A B A/B A/B rel rel ω V 1. Newton s Second Law

33 Newton formula for translational motion of the rigid body F = ma G G F = the resultant of the external forces = forces applied to the rigid body m = total mass a = the acceleration of the center of mass G FBD 1. Newton s Second Law

34 Euler formula for planar rotational motion of the rigid body M Gz = I G ω I G ω = M Gz = moment of the external forces about CM (G) in z direction = centroidal mass moment of inertia about z axis through CM (G) rate of change of the angular velocity FBD 1. Newton s Second Law

35 Pure rotation stationary point in the body M Oz = I O ω I O ω = M Gz = moment of the external forces about O in z direction = mass moment of inertia about z axis through axis of rotation (O) rate of change of the angular velocity I = I + md O G FBD With this formula, the constraint force is eliminated 1. Newton s Second Law

36 1.3 Equations of Motion 1.3 Equations of Motion

37 1.3 Equations of Motion

40 Ex. 5 Simple 1 DOF m-k system 1.3 Equations of Motion

42 Ex. 6 Simple DOF m-k system 1.3 Equations of Motion

45 Ex. 7 Simple 1 DOF pendulum 1.3 Equations of Motion

46 Ex. 8 Compound DOF pendulum A uniform rigid bar of total mass m and length L, suspended at point O by a string of length L 1, is acted upon by the horizontal force F. Use the angular displacement θ 1 and θ to define the position, velocity, and acceleration of the mass center C in terms of body axes and then derive the EOM for the translation of C and the rotation about C. 1.3 Equations of Motion

48 L Fsθ + mgcθ Tcθ 1= m L 1θ1sθ 1 L 1θ1 cθ 1 θ L Fcθ mgsθ+ Tsθ 1= m L 1θ1cθ 1+ L 1θ1sθ 1+ θ L L ml F cθ T sθ 1 = θ 1 eliminate the constraint force T 1.3 Equations of Motion

49 Ex. 9 Washing machine 1.3 Equations of Motion

50 Ex. 9 Washing machine 1.3 Equations of Motion

51 y measured from unstrained spring position My + cy + ky = Mg x measured from static equilibrium position static displacement equilibrium δ = Mg / k y = x δ st st Mx + cx + kx = 0 Weight is balanced at all times by spring force kδst Imbalance in the washer + + = = Mx cx kx F me sin t ω ω 1.3 Equations of Motion

52 Ex. 10 Pitch & Vertical motion simple model of a car 1.3 Equations of Motion

53 sf ( b θ ) ( f sf b θ f ) ( θ ) ( θ ) + sf ( b θ f ) + sf ( b θ f ) ( ) ( ) ( θ ) ( θ ) ( + θ ) + ( + θ ) x F k x a x c x a x k x + b x c x + b x = m x sr b r sr b r b b Fc k x a x c x a x a ksr xb + bθ xr + csr xb + bθ xr b= I Cθ k x a x + c x a x k x = m x sf b f sf b f f f f f k x b x c x b x k sr b r sr b r = m x r r r r mb xb csf + csr csf a+ csrb csf csr x b 0 IC 0 0 θ csf a+ csrb csf a + csrb csf a csrb θ mf 0 xf csf csf a csf 0 x f mr xr csr csrb 0 csr x r ksf + ksr ksf a+ ksrb ksf ksr xb F ksf a+ ksrb ksf a + ksrb ksf a ksrb θ Fc + = ksf ksf a ksf 0 x f 0 ksr ksrb 0 ksr xr Equations of Motion

54 Ex. 11 EOM w/ springs in series A mass m is suspended on a massless beam of uniform flexural rigidity EI. Derive the differential EOM. 1.3 Equations of Motion

55 19EI equivalent stiffness of the beam at the midspan is 3 L From the arrangement, it is connected in series with the spring 19kEI keq = 3 19EI + kl 19kEI mx + x = EI + kl k 1.3 Equations of Motion

56 Ex. 1 EOM w/ springs in parallel Devise a lumped model for the n-storey building subjected to a horizontal earthquake excitation. 1.3 Equations of Motion

57 x > x > x i+ 1 i i 1 ( ) ki xi xi 1 ( ) k x x i+ 1 i+ 1 i x > x and x > x i i+ 1 i i 1 ( ) ki xi xi 1 ( ) k x x i+ 1 i i Equations of Motion

58 Each column acts as clamped-clamped beam 1EI with an end in horizontal deflection k =. 3 H Two columns for each storey behave like spring in parallel. k = 4EI i eqi 3 Hi Schematic model is a string of spring/mass. At each mass or ( ) ( ) F = ma mx = k x x k x x, x > x > x i i i+ 1 i+ 1 i i i i 1 i+ 1 i i 1 ( ) ( ) F = ma mx = k x x k x x, x > x and x > x i i i+ 1 i i+ 1 i i i 1 i i+ 1 i i 1 () = ( ) + () with the constraint x t x 0 a t dt o o t t Equations of Motion

59 1.4 Equilibrium and Linearization Linear Time Invariant (LTI) system is convenient to understand and analyzed. It can be described as the system of ordinary differential equations with constant coefficients of the form: m m 1 d x d x d x dx 0i m 1i m 1 1 mi a + a + + a + a + a x= f t dt dt dt dt ( m ) i ( m ) i () Very often, the EOM is nonlinear which is difficult to manipulate. Linearization makes the system become LTI using the Taylor s series expansion around an interested point. The equilibrium point is commonly chosen. 1.4 Equilibrium

60 How to find the equilibrium point? The solution does not change if the system is at the equilibrium. Let that point be x = x e and at that point x = x= = 0. Substitute into EOM and solve for x e. How to linearize the model? Apply the Taylor s series expansion to any nonlinear expressions around x e. Assume small motion, which allows one to ignore the nonlinear terms in the series. In other words, only the constant and linear terms are remained. d 1 d f x = f x + f x x x + f x x x + ( ) ( ) ( ) ( ) ( ) ( ) e e e dx x= x! dx e x= x e 1.4 Equilibrium

61 Ex. 13 Inverted pendulum Determine the EOM and linearize it. 1.4 Equilibrium

62 l l MO = IOα k sθ cθ + mglsθ = ml θ Find the equilibrium position θ = θe kl mg sθecθe + mglsθe = 0, sθe = 0 or cθe = kl Linearize about θe = 0 for which θ = θe + φ mg l l kl k sθ cθ + mglsθ = 0+ + mgl φ kl ml φ + mgl φ = 0, φ measured from θ = 0 mg 4m g Linearize about θe such that cθe =, sθe = 1 for which θ = θ e + φ kl k l l l kl 6m g k sθ cθ + mglsθ = 0+ + φ k kl 6m g 1 ml mg φ + φ = 0, φ measured from θ = cos k kl k(l/)sθ (l/)cθ 1.4 Equilibrium

63 l l M O = I k + mgl = ml kl Oα θ θ θ θ + θ = ml mgl 0 mg k(l/)θ (l/) small angle approx. 1.4 Equilibrium

64 Ex. 14 Determine the EOM and linearize it. 1.4 Equilibrium

65 kl 1 θ mg small angle approx. M O = I mgl kl l = ml Oα θ 1θ 1 θ 1.4 Equilibrium

66 Ex. 15 Determine the EOM and linearize it. equilibrium position 1.4 Equilibrium

67 Let the equilibrium position the spring deforms by δ. Fy = 0 mg kδeqcα = 0 1 Let y be the displacement measured from the equilibrium, corresponding to the spring being deflected by angle θ. Fy = my mg cy k( Δ+ δeq) c( α θ) = my ( ) From the geometry, Δ is the difference of the spring length between two postures. h h h Δ= ( htanα ) + ( h+ y) = + hy+ y cα c α cα From the geometry, h h+ y h tanα sα = s( α θ) tan ( α θ) = cα c α θ h+ y ( ) 1 h+ y c β = c ( α θ) 1 tan β = + h + hy + y c α eq () 1.4 Equilibrium

68 ( ) ( ) Subs. the findings into and recognize 1 : h h h + mg cy k + hy + y + δ y eq = my c α cα h + hy + y c α kh h + y mg h + y EOM: my + cy + k ( h + y) = mg cα c h α h + hy + y + hy + y c α c α Linearization about equilibrium y = 0 I ( h+ y)( h+ y) h + hy + y c α h hy y kh + + kh I kh k c α = + y 0 O y h + c h cα α + hy + y c α c α I kc α y II y= 0 ( ) ( ) 1.4 Equilibrium

69 ( h+ y)( h+ y) h + hy + y c α h hy y mgh mg + + II mg c α = ( y 0) O ( y ) h + c h cα α + hy + y c α c α y = 0 mg II s α y h Subs. I and II into EOM, mg my + cy + kc α s α y = 0 h If α θ α, i.e. θ is very small, which requires small enough α, then becomes ( δeq ) mg cy k Δ+ cα = my () Make use of 1, it becomes my + cy + kδ cα = 0 Subs. Δ and linearize about y = 0 to get my + cy + kc α y = 0 Note that the linearized Δ= ycα which makes sense from the figure by projecting y onto the line of spring. ( ) 1.4 Equilibrium

70 1.4 Equilibrium