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Introduction Mechanical Vibrations What is vibration? Vibrations are oscillations of a system about an equilbrium position. Vibration… It is also an everyday phenomenon we meet on everyday life Vibration … Useful Vibration Harmful vibration Compressor Noise Testing Destruction Wear Ultrasonic cleaning Fatigue Vibration parameters All mechanical systems can be modeled by containing three basic components: spring, damper, mass When these components are subjected to constant force, they react with a constant displacement, velocity and acceleration Free vibration When a system is initially disturbed by a displacement, velocity or acceleration, the system begins to vibrate with a constant amplitude and frequency depend on its stiffness and mass. This frequency is called as natural frequency, and the form of the vibration is called as mode shapes Equilibrium pos. Forced Vibration If an external force applied to a system, the system will follow the force with the same frequency. ’ However, when the force frequency is increased to the system’s natural frequency, amplitudes will dangerously increase in this region. This phenomenon called as “Resonance” Watch these … Bridge collapse: http://www.youtube.com/watch?v=j-zczJXSxnw Hellicopter resonance: http://www.youtube.com/watch?v=0FeXjhUEXlc Resonance vibration test: http://www.youtube.com/watch?v=LV_UuzEznHs Flutter (Aeordynamically induced vibration) : http://www.youtube.com/watch?v=OhwLojNerMU Modelling of vibrating systems Lumped (Rigid) Modelling Numerical Modelling Element-based methods (FEM, BEM) Statistical and Energybased methods (SEA, EFA, etc.) Because running in the International Space Station might cause unwanted vibrations, they have installed a Treadmill Vibration Isolation System. 2 - 10 • Mechanical vibration is the motion of a particle or body which oscillates about a position of equilibrium. Most vibrations in machines and structures are undesirable due to increased stresses and energy losses. • Time interval required for a system to complete a full cycle of the motion is the period of the vibration. • Number of cycles per unit time defines the frequency of the vibrations. • Maximum displacement of the system from the equilibrium position is the amplitude of the vibration. • When the motion is maintained by the restoring forces only, the vibration is described as free vibration. When a periodic force is applied to the system, the motion is described as forced vibration. • When the frictional dissipation of energy is neglected, the motion is said to be undamped. Actually, all vibrations are damped to some degree. 19 - 11 Free Vibrations of Particles. Simple • If a particle is displaced through a distance x from its Harmonic Motion equilibrium position and released with no velocity, the m particle will undergo simple harmonic motion, ma F W k st x kx mx kx 0 • General solution is the sum of two particular solutions, k k x C1 sin t C 2 cos t m m C1 sin n t C 2 cos n t • x is a periodic function and n is the natural circular frequency of the motion. • C1 and C2 are determined by the initial conditions: 19 - 12 x C1 sin nt C2 cos nt C2 x0 v x C1 n cos nt C2 n sin nt C1 v0 n Free Vibrations of Particles. Simple Harmonic Motion x xm sin nt xm v0 n 2 x02 amplitude tan 1 v0 x0 n phase angle n fn 19 - 13 2 n 1 n period n natural frequency 2 Free Vibrations of Particles. Simple • Velocity-time and acceleration-time curves can be Harmonic Motion represented by sine curves of the same period as the displacement-time curve but different phase angles. x xm sin nt v x xm n cos n t xm n sin n t 2 a x xm n2 sin n t xm n2 sin n t 19 - 14 Simple Pendulum (Approximate • Results obtained for the spring-mass system can be Solution) applied whenever the resultant force on a particle is proportional to the displacement and directed towards the equilibrium position. • Consider tangential components of acceleration and force for a simple pendulum, Ft mat : W sin ml g sin 0 l for small angles, g l 0 m sin n t n 19 - 15 2 n 2 l g Simple Pendulum (Exact Solution) g l An exact solution for sin 0 leads to l 2 d n 4 g 0 1 sin 2 2 sin 2 m which requires numerical solution. n 19 - 16 2K l 2 g Concept Question The amplitude of a vibrating system is shown to the right. Which of the following statements is true (choose one)? a) The amplitude of the acceleration equals the amplitude of the displacement b) The amplitude of the velocity is always opposite (negative to) the amplitude of the displacement c) The maximum displacement occurs when the acceleration amplitude is a minimum d) The phase angle of the vibration shown is zero 2 - 17 Sample Problem 19.1 SOLUTION: • For each spring arrangement, determine the spring constant for a single equivalent spring. • Apply the approximate relations for the harmonic motion of a spring-mass system. A 50-kg block moves between vertical guides as shown. The block is pulled 40mm down from its equilibrium position and released. For each spring arrangement, determine a) the period of the vibration, b) the maximum velocity of the block, and c) the maximum acceleration of the block. 19 - 18 Sample Problem 19.1 k1 4 kN m k2 6 kN m SOLUTION: • Springs in parallel: - determine the spring constant for equivalent spring - apply the approximate relations for the harmonic motion of a spring-mass system k 104 N/m n 14.14 rad s m 20 kg n P k1 k2 k P 2 n vm x m n 0.040 m 14.14 rad s k1 k2 10 kN m 10 N m 4 vm 0.566 m s am x m an2 0.040 m 14.14 rad s 2 19 - 19 n 0.444 s am 8.00 m s 2 Sample Problem 19.1 k1 4 kN m k2 6 kN m • Springs in series: - determine the spring constant for equivalent spring - apply the approximate relations for the harmonic motion of a spring-mass system n n k 2400N/m 6.93 rad s m 20 kg 2 n n 0.907 s vm x m n P k1 k2 k P k1 k2 10 kN m 104 N m 19 - 20 0.040 m 6.93 rad s vm 0.277 m s am x m an2 0.040 m 6.93 rad s 2 am 1.920 m s 2 Free Vibrations of Rigid Bodies • If an equation of motion takes the form x n2 x 0 or n2 0 the corresponding motion may be considered as simple harmonic motion. • Analysis objective is to determine n. • Consider the oscillations of a square plate W b sin mb I 1 m 2b 2 2b 2 2 mb 2 , W mg but I 12 3 3g 3g sin 0 5b 5b 3g 2 5b then n , n 2 5b n 3g • For an equivalent simple pendulum, 19 - 21 l 5b 3 Sample Problem 19.2 SOLUTION: k • From the kinematics of the system, relate the linear displacement and acceleration to the rotation of the cylinder. • Based on a free-body-diagram equation for the equivalence of the external and effective forces, write the equation of motion. A cylinder of weight W is suspended as • Substitute the kinematic relations to arrive shown. at an equation involving only the angular displacement and acceleration. Determine the period and natural frequency of vibrations of the cylinder. 19 - 22 Sample Problem 19.2 SOLUTION: • From the kinematics of the system, relate the linear displacement and acceleration to the rotation of the x r 2 x 2r cylinder. a r r a r • Based on a free-body-diagram equation for the equivalence of the external and effective forces, write the equation M Aof motion. M A eff : Wr T2 2r ma r I but T2 T0 k 12 W k 2r • Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration. Wr 1 W 2kr 2r mrr 1 mr 2 2 2 8k 0 3m 8k n 3m 19 - 23 n 2 n 2 3m 8k fn n 1 8k 2 2 3m Sample Problem 19.3 SOLUTION: • Using the free-body-diagram equation for the equivalence of the external and effective moments, write the equation of motion for the disk/gear and wire. W 20 lb n 1.13 s n 1.93 s • With the natural frequency and moment of inertia for the disk known, calculate the torsional spring constant. The disk and gear undergo torsional vibration with the periods shown. • With natural frequency and spring Assume that the moment exerted by the constant known, calculate the moment of wire is proportional to the twist angle. inertia for the gear. Determine a) the wire torsional spring • Apply the relations for simple harmonic constant, b) the centroidal moment of motion to calculate the maximum gear inertia of the gear, and c) the maximum velocity. angular velocity of the gear if rotated through 90o and released. 19 - 24 Sample Problem 19.3 W 20 lb n 1.13 s n 1.93 s SOLUTION: • Using the free-body-diagram equation for the equivalence of the external and effective moments, write the equation of motion for the disk/gear and wire. M O M O : K I eff K 0 I n K I n 2 n 2 I K • With the natural frequency and moment of inertia for the disk known, calculate the torsional spring constant. 2 1 20 8 2 I 12 mr 2 0.138 lb ft s 2 32.2 12 1.13 2 19 - 25 0.138 K K 4.27 lb ft rad Sample Problem 19.3 • With natural frequency and spring constant known, calculate the moment of inertia for the gear. I 1.93 2 I 0.403 lb ft s2 4.27 W 20 lb n 1.13 s • Apply the relations for simple harmonic motion to calculate the maximum gear velocity. n 1.93 s m sin nt mn sin nt m mn m 90 1.571 rad K I n n 2 n 2 K 4.27 lb ft rad 19 - 26 I K 2 2 1.571 rad 1 . 93 s n m m m 5.11rad s Group Problem Solving SOLUTION: • Using the free-body and kinetic diagrams, write the equation of motion for the pendulum. • Determine the natural frequency and moment of inertia for the disk (use the small angle approximation). • Calculate the period. A uniform disk of radius 250 mm is attached at A to a 650-mm rod AB of negligible mass which can rotate freely in a vertical plane about B. If the rod is displaced 2°from the position shown and released, determine the period of the resulting oscillation. 19 - 27 Group Problem Solving Draw the FBD and KD of the pendulum (mbar ~ 0). Bn Bt l r man mat I mg Determine the equation of motion. M B I B mgl sin I ml 2 2 - 28 *Note that you could also do this by using the “moment” from at, and that at = l mgl sin I lmat Group Problem Solving Find I, set up equation of motion using small angle approximation mgl sin I ml 2 I 1 2 mr , sin 2 1 2 2 mr ml 2 mgl 0 Determine the natural frequency n2 gl r2 2 l2 (9.81)(0.650) 2 2 1 (0.250) (0.650) 2 2 - 29 14.053 n 3.7487 rad/s Calculate the period n 2 n 1.676 s n 1.676 s Concept Question In the previous problem, what would be true if the bar was hinged at A instead of welded at A (choose one)? a) The natural frequency of the oscillation would be larger b) The natural frequency of the oscillation would be larger c) The natural frequencies of the two systems would be the same 19 - 30 Principle of Conservation of Energy • Resultant force on a mass in simple harmonic motion is conservative - total energy is conserved. T V constant 1 mx 2 2 2 12 kx 2 constant x n2 x 2 • Consider simple harmonic motion of the square plate, T1 0 V1 Wb 1 cos Wb 2 sin 2 m 2 12 Wb m2 2 T2 12 mvm2 12 I m 2 12 mbm 12 12 53 mb2 m2 23 mb2 m2 V2 0 T1 V1 T2 V2 0 12 Wb m2 12 19 - 31 53 mb2 m2 n2 0 n 3g 5b Sample Problem 19.4 SOLUTION: • Apply the principle of conservation of energy between the positions of maximum and minimum potential energy. • Solve the energy equation for the natural frequency of the oscillations. Determine the period of small oscillations of a cylinder which rolls without slipping inside a curved surface. 19 - 32 Sample Problem 19.4 SOLUTION: • Apply the principle of conservation of energy between the positions of maximum and minimum potential energy. T1 V1 T2 V2 T1 0 V1 Wh W R r 1 cos W R r m2 2 V2 0 2 T2 12 mvm2 12 I m 12 mR r m2 12 34 mR r 2m2 19 - 33 2 1 mr 2 R r 2 m 2 r Sample Problem 19.4 • Solve the energy equation for the natural frequency of the oscillations. T1 0 V1 W R r m2 2 T2 34 mR r 2m2 V2 0 T1 V1 T2 V2 0 W R r mg R r n2 19 - 34 m2 2 m2 2 2 g 3 Rr 34 m R r 2m2 0 34 m R r 2 m n 2m n 2 n 2 3 Rr 2 g Forced Vibrations Forced vibrations - Occur when a system is subjected to a periodic force or a periodic displacement of a support. f forced frequency F ma : Pm sin f t W k st x mx W k st x m sin f t mx mx kx Pm sin f t mx kx k m sin f t 19 - 35 Forced Vibrations x xcomplementary x particular C1 sin n t C2 cos n t xm sin f t Substituting particular solution into governing equation, m 2f xm sin f t kxm sin f t Pm sin f t xm Pm k m k m 2f 1 f n 2 1 f n 2 Pm mx kx Pm sin f t mx kx k m sin f t At f = n, forcing input is in resonance with the system. 19 - 36 Concept Question A small trailer and its load have a total mass m. The trailer can be modeled as a spring with constant k. It is pulled over a road, the surface of which can be approximated by a sine curve with an amplitude of 40 mm and a wavelength of 5 m. Maximum vibration amplitude occur at 35 km/hr. What happens if the driver speeds up to 50 km/hr? a) The vibration amplitude remains the same. b) The vibration amplitude would increase. c) The vibration amplitude would decrease. 2 - 37 Sample Problem 19.5 SOLUTION: • The resonant frequency is equal to the natural frequency of the system. • Evaluate the magnitude of the periodic force due to the motor unbalance. Determine the vibration amplitude from the frequency ratio at 1200 rpm. A motor weighing 350 lb is supported by four springs, each having a constant 750 lb/in. The unbalance of the motor is equivalent to a weight of 1 oz located 6 in. from the axis of rotation. Determine a) speed in rpm at which resonance will occur, and b) amplitude of the vibration at 1200 rpm. 19 - 38 Sample Problem 19.5 SOLUTION: • The resonant frequency is equal to the natural frequency of the system. m W = 350 lb k = 4(350 lb/in) 350 10.87 lb s2 ft 32.2 k 4750 3000 lb in 36,000 lb ft k 36,000 m 10.87 57.5 rad/s 549 rpm n Resonance speed = 549 rpm 19 - 39 Sample Problem 19.5 • Evaluate the magnitude of the periodic force due to the motor unbalance. Determine the vibration amplitude from the frequency ratio at 1200 rpm. f 1200 rpm 125.7 rad/s 1 1 lb 0.001941 lb s 2 ft m 1 oz 16 oz 32.2 ft s 2 W = 350 lb k = 4(350 lb/in) 57.5 rad/s Pm man mr 2 6 125.7 2 15.33 lb 0.001941 12 n xm Pm k 1 f n 2 15.33 3000 1 125.7 57.52 0.001352 in xm = 0.001352 in. (out of phase) 19 - 40 Damped Free Vibrations • All vibrations are damped to some degree by forces due to dry friction, fluid friction, or internal friction. • With viscous damping due to fluid friction, F ma : W k st x cx mx mx cx kx 0 • Substituting x = elt and dividing through by elt yields the characteristic equation, ml2 cl k 0 2 c k c l 2m m 2m • Define the critical damping coefficient such that 2 k cc 0 m 2m 19 - 41 cc 2m k 2m n m Damped Free Vibrations • Characteristic equation, 2 c k c l 2m m 2m ml2 cl k 0 cc 2m n critical damping coefficient • Heavy damping: c > cc x C1e l1t C2 e l2t - negative roots - nonvibratory motion • Critical damping: c = cc x C1 C 2t e nt - double roots - nonvibratory motion • Light damping: c < cc x e c 2m t C1 sin d t C2 cos d t 2 d n 19 - 42 c 1 damped frequency cc Concept Question 8 6 4 2 0 -2 -4 -6 -8 0 0.5 1 1.5 2 2.5 3 3. The graph above represents an oscillation that is… a) Heavily damped b) critically damped 2 - 43 c) lightly da Concept Question 8 6 4 2 0 -2 -4 -6 -8 0 0.5 1 1.5 2 2.5 3 The period for the oscillation above is approximately… a) 1.25 seconds b) 2.5 Hz c) 0.6 seconds Estimate the phase shift for the oscillationZero 2 - 44 3. Forced vibrations can be caused by a test machine, by rocks on a trail, by rotating machinery, and by earthquakes. Suspension systems, shock absorbers, and other energy-dissipating devices can help to dampen the resulting vibrations. 2 - 45 Damped Forced Vibrations mx cx kx Pm sin f t x xcomplementary x particular xm xm Pm k 1 tan 19 - 46 1 f 2c cc f n 1 f n 2 n 2c c 2 2 c f n 2 magnification factor phase difference between forcing and steady state response Group Problem Solving SOLUTION: • Determine the system natural frequency, damping constant, and the unbalanced force. • Determine the steady state response and the magnitude of the motion. A simplified model of a washing machine is shown. A bundle of wet clothes forms a mass mb of 10 kg in the machine and causes a rotating unbalance. The rotating mass is 20 kg (including mb) and the radius of the washer basket e is 25 cm. Knowing the washer has an equivalent spring constant k = 1000 N/m and damping ratio z = c/cc = 0.05 and during the spin cycle the drum rotates at 250 rpm, determine the amplitude of the motion. 19 - 47 Group Problem Solving Given: m= 20 kg, k= 1000 N/m, f= 250 rpm, e= 25 cm, mb= 10 kg Find: xm Calculate the forced circular frequency and the natural circular frequency f (2 )(250) 26.18 rad/s 60 n k 1000 7.0711 rad/s m 20 Calculate the critical damping constant cc and the damping constant c cc 2 km 2 (1000)(20) 282.84 N s/m 19 - 48 c c cc c (0.05)(141.42) 14.1421 N s/m Group Problem Solving Calculate the unbalanced force caused by the wet clothes Pm mb e 2f Pm (10 kg)(0.25 m)(26.18 rad/s)2 1713.48 N Use Eq 19.52 to determine xm mx cx kx Pm sin f t xm 19 - 49 Pm k m 2f 2 (c f )2 1713.48 [1000 (20)(26.18)2 ]2 [(141421)(26.18)]2 1713.48 (12,707.8)2 (370.24)2 1713.48 0.13478 m 12,713.2 xm 134.8 mm Concept Question The following parameters were found in the previous problem: f 26.18 rad/s n 7.0711 rad/s =0.05 What would happen to the amplitude xm if the forcing frequency f was cut in half? a) The vibration amplitude remains the same. b) The vibration amplitude would increase. c) The vibration amplitude would decrease. 2 - 50 Concept Question Case 1 f 26.18 rad/s n 7.0711 rad/s Case 2 f 13.09 rad/s n 7.0711 rad/s xm 2 - 51 Pm k m 2f 2 (c f )2 Electrical Analogues • Consider an electrical circuit consisting of an inductor, resistor and capacitor with a source of alternating voltage Em sin f t L di q Ri 0 dt C 1 Lq Rq q Em sin f t C • Oscillations of the electrical system are analogous to damped forced vibrations of a mechanical system. 19 - 52 Electrical Analogues • The analogy between electrical and mechanical systems also applies to transient as well as steadystate oscillations. • With a charge q = q0 on the capacitor, closing the switch is analogous to releasing the mass of the mechanical system with no initial velocity at x = x0. • If the circuit includes a battery with constant voltage E, closing the switch is analogous to suddenly applying a force of constant magnitude P to the mass of the mechanical system. 19 - 53 Electrical Analogues • The electrical system analogy provides a means of experimentally determining the characteristics of a given mechanical system. • For the mechanical system, m1x1 c1 x1 c2 x1 x 2 k1 x1 k 2 x1 x2 0 m2 x2 c2 x2 x1 k 2 x2 x1 Pm sin f t • For the electrical system, q q q L1q1 R1 q1 q 2 1 1 2 0 C1 C2 q2 q1 L2 q2 R2 q 2 q1 Em sin f t C2 • The governing equations are equivalent. The characteristics of the vibrations of the mechanical system may be inferred from the oscillations of the electrical system. 19 - 54 Degree of Freedom (DOF) • Mathematical modeling of a physical system requires the selection of a set of variables that describes the behavior of the system. • The number of degrees of freedom for a system is the number of kinematically independent variables necessary to completely describe the motion of every particle in the system DOF=1 DOF=2 Single degree of freedom (SDOF) Multi degree of freedom (MDOF) Equivalent model of systems Example 1: Example 2: SDOF MDOF DOF=1 DOF=2 Equivalent model of systems MDOF Example 3: DOF= 3 if body 1 has no rotation SDOF DOF=2 DOF= 4 if body 1 has rotation body 1 What are their DOFs? SDOF systems Helical springs Shear stress: Stiffness coefficient: F: Force, D: Diameter, G: Shear modulus of the rod, N: Number of turns, r : Radius Springs in combinations: Parallel combination Series combination Elastic elements as springs Moment of Inertia What are the equivalent stiffnesses? Example A 200-kg machine is attached to the end of a cantilever beam of length L= 2.5 m, elastic modulus E= 200x109 N/m2, and cross-sectional moment of inertia I = 1.8x10–6 m4. Assuming the mass of the beam is small compared to the mass of the machine, what is the stiffness of the beam? Damping Viscous Damping