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Chapter 9 Vibration Control

9. Chapter 9 Vibration Control. Chapter Outline. 9.1 Introduction 9.2 Vibration Nomograph and Vibration Criteria 9.3 Reduction of Vibration at the source 9.4 Balancing of Rotating Machines 9.5 Whirling of Rotating Shafts 9.6 Balancing of Reciprocating Engines

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Chapter 9 Vibration Control

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  1. 9 Chapter 9Vibration Control

  2. Chapter Outline • 9.1 Introduction • 9.2 Vibration Nomograph and Vibration Criteria • 9.3 Reduction of Vibration at the source • 9.4 Balancing of Rotating Machines • 9.5 Whirling of Rotating Shafts • 9.6 Balancing of Reciprocating Engines • 9.7 Control of Vibration • 9.8 Control of Natural Frequencies • 9.9 Introduction of Damping • 9.10 Vibration Isolation • 9.11 Vibration Absorbers

  3. 9.1 9.1Introduction

  4. 9.1 Introduction • Vibration leads to wear of machinery and discomfort of humans, thus we want to eliminate vibration • Designer must compromise between acceptable amount of vibration and manufacturing cost • We shall consider various techniques of vibration control in this chapter.

  5. 9.2 9.2Vibration Nomograph and Vibration Criteria

  6. 9.2 Vibration Nomograph and Vibration Criteria • Vibration nomograph displays the variations of displacement, velocity and acceleration amplitudes wrt frequency of vibration • Harmonic motion: • Velocity: • Acceleration: • Amplitude of velocity: • Amplitude of acceleration:

  7. 9.2 Vibration Nomograph and Vibration Criteria Taking log of Eq. 9.3 and Eq. 9.4: When X is constant, ln vmax varies linearly with ln(2πf) When amax is constant, ln vmax varies linearly with ln(2πf) This is shown as a nomograph in the next slide. Every pt on the nomograph denotes a specific sinusoidal vibration.

  8. 9.2 Vibration Nomograph and Vibration Criteria

  9. 9.2 Vibration Nomograph and Vibration Criteria Vibration severity of machinery is defined in terms of the root mean square (rms) value of vibration velocity. (ISO 2372) Vibration severity of whole building vibration (ISO DP 4866) Vibration limits for human (ISO 2631)

  10. 9.2 Vibration Nomograph and Vibration Criteria • Example 9.1Helicopter Seat Vibration ReductionThe seat of a helicopter, with the pilot, weights 1000N and is found to have a static deflection of 10 mm under self-weight. The vibration of the rotor is transmitted to the base of the seat as harmonic motion with frequency 4 Hz and amplitude 0.2 mm. • a) What is the level of vibration felt by the pilot? • b) How can the seat be redesigned to reduce the effect of vibration?

  11. 9.2 Vibration Nomograph and Vibration Criteria • Example 9.1Helicopter Seat Vibration ReductionSolutionMass = m = 1000/9.81 = 101.9368 kg • Stiffness = k = W/δst = 1000/0.01 = 105N/m • Natural frequency = ωn = • Frequency ratio = r = • Amplitude of vibration felt by pilot: • where Y is the amplitude of base displacement

  12. 9.2 Vibration Nomograph and Vibration Criteria Example 9.1Helicopter Seat Vibration ReductionSolution At 4 Hz, the amplitude of 0.3616 mm may not cause much discomfort. However the velocity and acceleration at 4 Hz are not acceptable for a comfortable ride. Try to bring amax down to 0.01 m/s2

  13. 9.2 Vibration Nomograph and Vibration Criteria Example 9.1Helicopter Seat Vibration ReductionSolution Either use softer material for seat or increase mass of seat.

  14. 9.3 9.3Reduction of Vibration at the Source

  15. 9.3 Reduction of Vibration at the Source • Try to alter the source so that it produces less vibration • E.g. balance rotating or reciprocating machines, use close tolerances or better surface finish • Some sources cannot be eliminated e.g. turbulence, engine combustion instability, road roughness

  16. 9.4 9.4Balancing of Rotating Machines

  17. 9.4 Balancing of Rotating Machines • Unbalanced mass in rotating disc will cause vibration. • Can be eliminated by removing the unbalanced mass or adding equal mass to cancel out vibration • Need to determine the amount and location of the eccentric mass experimentally • We shall consider 2 types of balancing: single-plane balancing and 2-plane balancing

  18. 9.4 Balancing of Rotating Machines Single-Plane Balancing When center of mass is displaced from the axis of rotation, the element is statically unbalanced. To determine whether a disc is balanced, mount it as shown below. Rotate the disc and let it come to rest. Mark the lowest point. Repeat a few times.

  19. 9.4 Balancing of Rotating Machines Single-Plane Balancing If the disc is unbalanced, the markings will coincide (static unbalance). Static unbalance can be corrected by removing material at the mark or adding material 180° from the mark. Amount of unbalance can be found by rotating the disc at a known speed ω and measuring the reactions at the 2 bearings.

  20. 9.4 Balancing of Rotating Machines Single-Plane Balancing If the unbalanced mass m is located at radius r, the centrifugal force will be mrω2. Measured bearing reactions:

  21. 9.4 Balancing of Rotating Machines Single-Plane Balancing Another method for single-plane balancing uses a vibration analyzer as shown:

  22. 9.4 Balancing of Rotating Machines Single-Plane Balancing Turn the rotor and fire a stroboscopic light at the same frequency ω. A marking on the rotor will appear stationary but positioned at an angle θ from the mark on the stator. The amplitude Au caused by the unbalance is also noted by the vibration analyzer.

  23. 9.4 Balancing of Rotating Machines Single-Plane Balancing Add a known trial weight W to the rotor and repeat the procedure. The new angle of the marking, φ and the new amplitude Au+w are noted. Construct vector diagram:

  24. 9.4 Balancing of Rotating Machines Single-Plane Balancing The difference vector is the unbalance vector due to trial weight W. Original unbalance is at angle α from position of trial weight. Magnitude of original unbalance WO=(AU/AW)•W

  25. 9.4 Balancing of Rotating Machines Two-Plane Balancing If rotor is as shown, unbalance can be anywhere along the length Can be balanced by adding weights in any 2 planes, most common planes being the end planes of the rotor

  26. 9.4 Balancing of Rotating Machines Two-Plane Balancing Consider a rotor with unbalanced mass as shown: Force due to unbalance, F=mω2RReplace unbalanced mass m by m1 and m2 located at the ends of the rotor as shown:

  27. 9.4 Balancing of Rotating Machines Two-Plane Balancing Forces exerted due to m1 and m2 are F1=m1ω2R and F2=m2ω2R For equivalence of forces: mω2R=m1ω2R+m2ω2R or m = m1+ m2 Taking moments at the right end: Thus m1= m/3, m2= 2m/3 Thus any unbalanced mass can be replaced by 2 unbalanced mass at the end planes.

  28. 9.4 Balancing of Rotating Machines Two-Plane Balancing Vibration analyzer Replace unbalance weight by UL and UR as shown: Measure vibration amplitude and phase of original unbalance at A and B

  29. 9.4 Balancing of Rotating Machines Two-Plane Balancing Vibration analyzer Add known trial weight WL in left plane at known position. Subtract Eq. 9.13 and 9.14 from Eq. 9.15 and 9.16:

  30. 9.4 Balancing of Rotating Machines Two-Plane Balancing Vibration analyzer Remove WL and add known trial weight WR in right plane at known position. Subtract Eq. 9.13 and 9.14 from Eq. 9.19 and 9.20:

  31. 9.4 Balancing of Rotating Machines Two-Plane Balancing Vibration analyzer Once are known, Eq 9.13 and Eq 9.14 can be solved to find the unbalance vectors. Rotor can now be balanced by adding equal and opposite weights in each plane.

  32. 9.4 Balancing of Rotating Machines • Example 9.2 • Two-Plane Balancing of Turbine Rotor • In the 2-plane balancing of a turbine rotor, the data obtained from measurement of the original unbalance, the right-plane trial weight, and the left-plane trial weight are shown below. Determine the size and location of the balance weights required.

  33. 9.4 Balancing of Rotating Machines • Example 9.2 • Two-Plane Balancing of Turbine Rotor • Solution

  34. 9.4 Balancing of Rotating Machines Example 9.2 Two-Plane Balancing of Turbine Rotor Solution

  35. 9.4 Balancing of Rotating Machines Example 9.2 Two-Plane Balancing of Turbine Rotor Solution

  36. 9.4 Balancing of Rotating Machines Example 9.2 Two-Plane Balancing of Turbine Rotor Solution Thus the required balance weights are

  37. 9.5 9.5Whirling of Rotating Shafts

  38. 9.5 Whirling of Rotating Shafts • In many applications, a heavy rotor is mounted on a light, flexible shaft supported in bearings. • Unbalances and other effects will cause a shaft to bend at certain speeds known as whirling speeds • Whirling is the rotation of the plane made by the line of centers of the bearings and the bent shaft.

  39. 9.5 Whirling of Rotating Shafts Consider a shaft as shown below. Forces acting on rotor: inertia force, spring force, damping forces

  40. 9.5 Whirling of Rotating Shafts Equations of Motion Let O be the equilibrium position of the shaft when balanced perfectly. Shaft (line CG) rotates at velocity ω. During rotation the rotor deflects by a distance A. CG of rotor (G) is at distance a from geometric centre C.

  41. 9.5 Whirling of Rotating Shafts Equations of Motion Angular velocity of line OC, is known as the whirling speed and is in general not equal to ω.

  42. 9.5 Whirling of Rotating Shafts Equations of Motion Substitute into Eq. 9.26: These equations are coupled. Define w as w=x+iy Add Eq 9.32 to Eq 9.33 and multiply by i:

  43. 9.5 Whirling of Rotating Shafts Critical Speeds When frequency of rotation of shaft = one of the natural frequencies of the shaft, critical speed of undamped system: When ω = ωn, rotor undergoes large deflections. Slow transition of rotating shaft through the critical speed aids development of large amplitudes.

  44. 9.5 Whirling of Rotating Shafts Response of the System Assuming ci=0, Solution: Substituting the steady-state part into equation of motion, we can find amplitude of whirl: Phase angle

  45. 9.5 Whirling of Rotating Shafts Response of the System Φ=0° for small ω Phase lag is 90° at resonance

  46. 9.5 Whirling of Rotating Shafts Stability Analysis Assume w(t)=est, characteristic equation: ms2 + (ci + c)s + k – iωci=0 With s=iλ, -mλ2 + (ci + c)s + k –iωci=0 This is a particular case of (p2+iq2)λ2+ (p1+iq1)λ+ (p0+iq0)=0 For this system to be stable, p2 = -m, p1 = q2 = 0, q1 = ci + c, p0 = k, q0 = -ωci Therefore m(ci + c)>0 and km(ci + c)2–m2ω2ci2

  47. 9.5 Whirling of Rotating Shafts Stability Analysis Internal and external friction can cause instability at rotating speeds above the 1st critical speed.

  48. 9.5 Whirling of Rotating Shafts • Example 9.3 • Whirl Amplitude of a Shaft Carrying an Unbalanced Rotor • A shaft, carrying a rotor of mass 50kg and eccentricity 2mm, rotates at 12000 rpm. Determine (a) the steady-state whirl amplitude and (b) the maximum whirl amplitude during start-up conditions of the system. Assume the stiffness of the shaft as 40MN/m and the external damping ratio as 0.1.

  49. 9.5 Whirling of Rotating Shafts Example 9.3 Whirl Amplitude of a Shaft Carrying an Unbalanced Rotor Solution Forcing frequency of rotor: Natural frequency: Frequency ratio:

  50. 9.5 Whirling of Rotating Shafts Example 9.3 Whirl Amplitude of a Shaft Carrying an Unbalanced Rotor Solution a) Steady-state amplitude: b) During start-up conditions, ω passes through ωn. Using r=1, we obtain whirl amplitude:

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