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- 1. Ninth EditionCHAPTER VECTOR MECHANICS FOR ENGINEERS:19 DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Mechanical Vibrations Lecture Notes: J. Walt Oler Texas Tech University © 2010 The McGraw-Hill Companies, Inc. All rights reserved.
- 2. Vector Mechanics for Engineers: DynamicsEditionNinth Contents Introduction Sample Problem 19.4 Free Vibrations of Particles. Simple Harmonic Motion Forced Vibrations Simple Pendulum (Approximate Solution) Sample Problem 19.5 Simple Pendulum (Exact Solution) Damped Free Vibrations Sample Problem 19.1 Damped Forced Vibrations Free Vibrations of Rigid Bodies Electrical Analogues Sample Problem 19.2 Sample Problem 19.3 Principle of Conservation of Energy © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 2
- 3. Vector Mechanics for Engineers: DynamicsEditionNinth Introduction • 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 3
- 4. Vector Mechanics for Engineers: DynamicsEditionNinth Free Vibrations of Particles. Simple Harmonic Motion • If a particle is displaced through a distance xm from its equilibrium position and released with no velocity, the particle will undergo simple harmonic motion, ma = F = W − k ( δ st + x ) = − kx m + kx = 0 x • General solution is the sum of two particular solutions, k k x = C1 sin t + C 2 cos m m t = C1 sin ( ω n t ) + C 2 cos( ω n t ) • x is a periodic function and ωn is the natural circular frequency of the motion. • C1 and C2 are determined by the initial conditions: x = C1 sin ( ω n t ) + C 2 cos( ω n t ) C 2 = x0 v = x = C1ω n cos( ω n t ) − C 2ω n sin ( ω n t ) C1 = v0 ω n © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 4
- 5. Vector Mechanics for Engineers: DynamicsEditionNinth Free Vibrations of Particles. Simple Harmonic Motion v C1 = 0 ωn C 2 = x0 • Displacement is equivalent to the x component of the sum of two vectors C1 + C 2 which rotate with constant angular velocity ω n . x = xm sin ( ω n t + φ ) xm = ( v0 ω n ) 2 + x0 = amplitude 2 φ = tan −1 ( v0 x0ω n ) = phase angle 2π τn = = period ωn 1 ωn fn = = = natural frequency τ n 2π © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 5
- 6. Vector Mechanics for Engineers: DynamicsEditionNinth Free Vibrations of Particles. Simple Harmonic Motion • Velocity-time and acceleration-time curves can be represented by sine curves of the same period as the displacement-time curve but different phase angles. x = xm sin ( ω n t + φ ) v=x = xmω n cos( ω n t + φ ) = xmω n sin ( ω n t + φ + π 2 ) a = x = − xmω n sin ( ω n t + φ ) 2 = xmω n sin ( ω n t + φ + π ) 2 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 6
- 7. Vector Mechanics for Engineers: DynamicsEditionNinth Simple Pendulum (Approximate Solution) • Results obtained for the spring-mass system can be 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 θ = mlθ θ + g sin θ = 0 l for small angles, g θ + θ = 0 l θ = θ m sin ( ω n t + φ ) 2π l τn = = 2π ωn g © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 7
- 8. Vector Mechanics for Engineers: DynamicsEditionNinth Simple Pendulum (Exact Solution) g An exact solution for θ + sin θ = 0 l l π 2 dφ leads to τn = 4 ∫ g 0 1 − sin 2 (θ 2 ) sin 2 φ m which requires numerical solution. 2K l τn = 2π π g © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 8
- 9. Vector Mechanics for Engineers: DynamicsEditionNinth 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 9
- 10. Vector Mechanics for Engineers: DynamicsEditionNinth 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 ωn = = = 14.14 rad s m 20 kg 2π τn = τ n = 0.444 s ωn P = k1δ + k2δ vm = x m ω n P = ( 0.040 m )(14.14 rad s ) vm = 0.566 m s k= = k1 + k2 δ 2 = 10 kN m = 10 N m 4 am = x m an = ( 0.040 m )(14.14 rad s ) 2 am = 8.00 m s 2 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 10
- 11. Vector Mechanics for Engineers: DynamicsEditionNinth Sample Problem 19.1 • Springs in series: k1 = 4 kN m k2 = 6 kN m - determine the spring constant for equivalent spring - apply the approximate relations for the harmonic motion of a spring-mass system k 2400N/m ωn = = = 6.93 rad s m 20 kg 2π τn = τ n = 0.907 s ωn vm = x m ω n P = k1δ + k2δ = ( 0.040 m )( 6.93 rad s ) vm = 0.277 m s P k= = k1 + k2 2 am = x m an δ = 10 kN m = 104 N m = ( 0.040 m )( 6.93 rad s ) 2 am = 1.920 m s 2 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 11
- 12. Vector Mechanics for Engineers: DynamicsEditionNinth Free Vibrations of Rigid Bodies • If an equation of motion takes the form 2 2 + ω n x = 0 or θ + ω nθ = 0 x the corresponding motion may be considered as simple harmonic motion. • Analysis objective is to determine ωn. • Consider the oscillations of a square plate − W ( b sin θ ) = ( mbθ) + I θ [ ] but I = 12 m ( 2b ) 2 + ( 2b ) 2 = 2 mb 2 , W = mg 1 3 3 g sin θ ≅ θ + 3 g θ = 0 θ + 5b 5b 3g 2π 5b then ω n = , τn = = 2π 5b ωn 3g • For an equivalent simple pendulum, l = 5b 3 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 12
- 13. Vector Mechanics for Engineers: DynamicsEditionNinth Sample Problem 19.2 SOLUTION: • From the kinematics of the system, relate k 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 • Substitute the kinematic relations to arrive as shown. at an equation involving only the angular Determine the period and natural displacement and acceleration. frequency of vibrations of the cylinder. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 13
- 14. Vector Mechanics for Engineers: DynamicsEditionNinth Sample Problem 19.2 SOLUTION: • From the kinematics of the system, relate the linear displacement and acceleration to the rotation of the cylinder. x = rθ δ = 2 x = 2rθ α = θ a = rα = rθ a = rθ • Based on a free-body-diagram equation for the equivalence of the external and effective forces, write the equation of motion. ∑ M A = ∑ ( M A ) eff : Wr − T2 ( 2r ) = ma r + I α but T2 = T0 + kδ = 1 W + k ( 2rθ ) 2 • Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration. (2 ) Wr − 1 W + 2krθ ( 2r ) = m( rθ) r + 1 mr 2θ 2 8 kθ =0 θ + 3m 8k 2π 3m ωn 1 8k ωn = τn = = 2π fn = = 3m ωn 8k 2π 2π 3m © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 14
- 15. Vector Mechanics for Engineers: DynamicsEditionNinth 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 • With the natural frequency and moment τ n = 1.13 s τ n = 1.93 s of inertia for the disk known, calculate The disk and gear undergo torsional the torsional spring constant. 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 15
- 16. Vector Mechanics for Engineers: DynamicsEditionNinth 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. ∑ M O = ∑ ( M O ) eff : + Kθ = − I θ K θ + θ = 0 W = 20 lb I τ n = 1.13 s τ n = 1.93 s K 2π I ωn = τn = = 2π I ωn K • With the natural frequency and moment of inertia for the disk known, calculate the torsional spring constant. 2 1 20 8 I = 1 mr 2 = 2 = 0.138 lb ⋅ ft ⋅ s 2 2 32.2 12 0.138 1.13 = 2π K = 4.27 lb ⋅ ft rad K © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 16
- 17. Vector Mechanics for Engineers: DynamicsEditionNinth Sample Problem 19.3 • With natural frequency and spring constant known, calculate the moment of inertia for the gear. I 1.93 = 2π I = 0.403 lb ⋅ ft ⋅ s 2 4.27 • Apply the relations for simple harmonic motion to W = 20 lb calculate the maximum gear velocity. τ n = 1.13 s τ n = 1.93 s θ = θ m sin ω nt ω = θ mω n sin ω nt ω m = θ mω n θ m = 90° = 1.571 rad 2π 2π ω m = θ m = (1.571 rad ) τ K 2π I n 1.93 s ωn = τn = = 2π I ωn K ω m = 5.11rad s K = 4.27 lb ⋅ ft rad © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 17
- 18. Vector Mechanics for Engineers: DynamicsEditionNinth 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 + 1 kx 2 = constant 2 2 x2 + ωn x2 = 2 • Consider simple harmonic motion of the square plate, 0 T1 = [ V1 = Wb(1 − cosθ ) = Wb 2 sin 2 (θ m 2 ) ] 2 ≅ 1 Wbθ m 2 2 2 T2 = 1 mvm + 1 I ω m V2 = 0 2 2 2 2 2 3 ( = 1 m( bθm ) + 1 2 mb 2 ω m 2 ) ( = 1 5 mb 2 θm 2 3 2 ) T1 + V1 = T2 + V2 2 2 3 ( 2 2 ) 0 + 1 Wbθ m = 1 5 mb 2 θ mω n + 0 2 ω n = 3 g 5b © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 18
- 19. Vector Mechanics for Engineers: DynamicsEditionNinth 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 19
- 20. Vector Mechanics for Engineers: DynamicsEditionNinth 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 − cosθ ) ( ≅ W ( R − r) θm 2 2 ) 2 2 T2 = 1 mvm + 1 I ω m V2 = 0 2 2 ( ) 2 1 m( R − r )θ 2 R − r 2 2 = 2 m + 1 1 mr 2 2 θm r = 3 m( R − r ) 2 θm 4 2 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 20
- 21. Vector Mechanics for Engineers: DynamicsEditionNinth Sample Problem 19.4 • Solve the energy equation for the natural frequency of the oscillations. T1 = 0 ( V1 ≅ W ( R − r ) θ m 2 2 ) T2 = 3 m( R − r ) 2θm 4 2 V2 = 0 T1 + V1 = T2 + V2 2 θm 3 0 +W ( R − r) = 4 m( R − r ) 2 θm + 0 2 2 2 θm 3 ( mg )( R − r ) = 4 m( R − r ) 2 (θ mω n ) 2 m 2 2 2 g 2π 3 R−r ωn = τn = = 2π 3 R−r ωn 2 g © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 21
- 22. Vector Mechanics for Engineers: DynamicsEditionNinth Forced Vibrations Forced vibrations - Occur when a system is subjected to a periodic force or a periodic displacement of a support. ω f = forced frequency ∑ F = ma : Pm sin ω f t + W − k ( δ st + x ) = m x ( ) W − k δ st + x − δ m sin ω f t = m x m + kx = Pm sin ω f t x m + kx = kδ m sin ω f t x © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 22
- 23. Vector Mechanics for Engineers: DynamicsEditionNinth Forced Vibrations x = xcomplementary + x particular = [ C1 sin ω n t + C 2 cos ω n t ] + xm sin ω f t Substituting particular solution into governing equation, − mω 2 xm sin ω f t + kxm sin ω f t = Pm sin ω f t f Pm Pm k δm xm = = = ( ) k − mω 2 1 − ω f ω n 2 1 − ω f ω n 2 f ( ) m + kx = Pm sin ω f t x m + kx = kδ m sin ω f t x At ωf = ωn, forcing input is in resonance with the system. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 23
- 24. Vector Mechanics for Engineers: DynamicsEditionNinth 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 24
- 25. Vector Mechanics for Engineers: DynamicsEditionNinth Sample Problem 19.5 SOLUTION: • The resonant frequency is equal to the natural frequency of the system. 350 m= = 10.87 lb ⋅ s 2 ft 32.2 k = 4( 750 ) = 3000 lb in = 36,000 lb ft W = 350 lb k = 4(350 lb/in) k 36,000 ωn = = m 10.87 = 57.5 rad/s = 549 rpm Resonance speed = 549 rpm © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 25
- 26. Vector Mechanics for Engineers: DynamicsEditionNinth 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. ω f = ω = 1200 rpm = 125.7 rad/s 1 lb 1 m = (1 oz ) = 0.001941 lb ⋅ s 2 ft 16 oz 32.2 ft s 2 W = 350 lb Pm = man = mrω 2 k = 4(350 lb/in) = ( 0.001941) (12 )(125.7) 2 = 15.33 lb 6 ω n = 57.5 rad/s Pm k 15.33 3000 xm = = ( 1 − ω f ωn )2 1 − (125.7 57.5) 2 = −0.001352 in xm = 0.001352 in. (out of phase) © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 26
- 27. Vector Mechanics for Engineers: DynamicsEditionNinth 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 ( δ st + x ) − cx = m x m + cx + kx = 0 x • Substituting x = eλt and dividing through by eλt yields the characteristic equation, 2 c c k mλ2 + cλ + k = 0 λ=− ± − 2m 2m m • Define the critical damping coefficient such that 2 cc k k − =0 cc = 2 m = 2mω n 2m m m © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 27
- 28. Vector Mechanics for Engineers: DynamicsEditionNinth Damped Free Vibrations • Characteristic equation, 2 c c k mλ2 + cλ + k = 0 λ=− ± − 2m 2m m cc = 2mω n = critical damping coefficient • Heavy damping: c > cc x = C1e λ1t + C 2 e λ2t - negative roots - nonvibratory motion • Critical damping: c = cc x = ( C1 + C 2t ) e −ω nt - double roots - nonvibratory motion • Light damping: c < cc x = e −( c 2m ) t ( C1 sin ω d t + C 2 cos ω d t ) 2 c ω d = ω n 1 − = damped frequency c c © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 28
- 29. Vector Mechanics for Engineers: DynamicsEditionNinth Damped Forced Vibrations m + cx + kx = Pm sin ω f t x x = xcomplementary + x particular xm xm 1 = = = magnification Pm k δ [1 − (ω f ωn ) ] [ 2 2 ( )] + 2( c cc ) ω f ω n 2 factor ( 2( c cc ) ω f ω n ) tan φ = = phase difference between forcing and steady ( 1− ω f ωn ) 2 state response © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 29
- 30. Vector Mechanics for Engineers: DynamicsEditionNinth Electrical Analogues • Consider an electrical circuit consisting of an inductor, resistor and capacitor with a source of alternating voltage di q E m sin ω f t − L − Ri − = 0 dt C 1 Lq + Rq + q = Em sin ω f t C • Oscillations of the electrical system are analogous to damped forced vibrations of a mechanical system. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 30
- 31. Vector Mechanics for Engineers: DynamicsEditionNinth Electrical Analogues • The analogy between electrical and mechanical systems also applies to transient as well as steady- state 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 31
- 32. Vector Mechanics for Engineers: DynamicsEditionNinth Electrical Analogues • The electrical system analogy provides a means of experimentally determining the characteristics of a given mechanical system. • For the mechanical system, m11 + c1 x1 + c2 ( x1 − x2 ) + k1 x1 + k 2 ( x1 − x2 ) = 0 x m2 2 + c2 ( x2 − x1 ) + k 2 ( x2 − x1 ) = Pm sin ω f t x • For the electrical system, q q −q L1q1 + R1 ( q1 − q2 ) + 1 + 1 2 = 0 C1 C2 q −q L2 q2 + R2 ( q2 − q1 ) + 2 1 = Em sin ω 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 19 - 32

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