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 Wirling 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

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.

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:

9.2 Vibration Nomograph and Vibration Criteria Taking log of Eq. 9.1and Eq. 9.2: 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.

9.2 Vibration Nomograph and Vibration Criteria

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)

Example 9.1 The 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?

Solution Mass = 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

Solution 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

Solution Either use softer material for seat or increase mass of seat.

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

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

9.4.1 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.

9.4.1 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.

9.4.1 Single-Plane Balancing If the unbalanced mass m is located at radius r, the centrifugal force will be mrω2. Measured bearing reactions: Another method for single-plane balancing uses a vibration analyzer as shown:

9.4.1 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.

9.4.1 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:

9.4.1 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

9.4.2 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

9.4.2 Two-Plane Balancing Consider a rotor with unbalanced mass as shown: Force due to unbalance, F=mω2R Replace unbalanced mass m by m1 and m2 located at the ends of the rotor as shown:

9.4.2 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.

9.4.2 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

9.4.2 Two-Plane Balancing Add known trial weight WL in left plane at known position. Subtract Eq. 9.3 and 9.4 from Eq. 9.5 and 9.6:

9.4.2 Two-Plane Balancing Remove WL and add known trial weight WR in right plane at known position. Subtract Eq. 9.3 and 9.4 from Eq. 9.7 and 9.8:

9.4.2 Two-Plane Balancing Once are known, Eq 9.3 and Eq 9.4 can be solved to find the unbalance vectors. Rotor can now be balanced by adding equal and opposite weights in each plane.

Example 9.2 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.

Solution

Solution

Solution Thus the required balance weights are

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.

9.5.1 Equations of motion Consider a shaft as shown below: Forces acting on rotor: inertia force, spring force, damping forces

9.5.1 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.

9.5.1 Equations of motion Angular velocity of line OC, is known as the whirling speed and is in general not equal to ω.

9.5.1 Equations of motion Therefore

9.5.1 Equations of motion Substitute into Eq. 9.9: These equations are coupled. Define w as w=x+iy Add Eq 9.10 to Eq 9.11 and multiply by i:

9.5.2 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.

9.5.3 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

9.5.3 Response of the system Differentiate A wrt ω and set result equal to 0, maximum whirl amplitude occurs when ω=ωn when damping c =0 A at low speeds is determined by k.

9.5.3 Response of the system Φ=0° for small ω Phase lag is 90° at resonance

9.5.4 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

9.5.4 Stability Analysis Internal and external friction can cause instability at rotating speeds above the 1st critical speed.

Example 9.3 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.

Solution Forcing frequency of rotor: Natural frequency Frequency ratio a) Steady-state amplitude

Solution b) During start-up conditions, ω passes through ωn. Using r=1, we obtain whirl amplitude:

9.6 Balancing of Reciprocating Engines Moving elements: Piston, crank, connecting rod Vibrations due to Periodic variations of gas pressure in cylinder Inertia forces associated with moving parts

9.6.1 Unbalanced forces due to fluctuations in gas pressure Expanding gas in cylinder exerts force F on piston F can be resolved into F/cos Φ and F tan Φ. F/cos Φ induces torque Mt which rotates the crank shaft

9.6.1 Unbalanced forces due to fluctuations in gas pressure Figure shows equilibrium forces on stationary parts of the engine Total resultant force =0 Resultant torque MQ=FhtanΦ where Torque induced at crank shaft is felt at engine support.

9.6.2 Unbalanced forces due to inertia of moving parts Acceleration of Piston

9.6.2 Unbalanced forces due to inertia of moving parts

9.6.2 Unbalanced forces due to inertia of moving parts Acceleration of the Crankpin Vertical and horizontal displacements of crankpin C:

9.6.2 Unbalanced Forces due to Inertia of the moving parts Vertical component of inertia force for one cylinder:

9.6.2 Unbalanced Forces due to Inertia of the moving parts Vertical component of inertia force for one cylinder:

9.6.3 Balancing of Reciprocating Engines mc can be made zero by counterbalancing the crank. Hence Fy can be reduced to zero. mp always positive hence vertical unbalanced force Fx always exists. Thus single cylinder engine is inherently unbalanced. In multi-cylinder engine, can balance inertia forces by proper arrangement of cranks Following figure shows arrange of 6-cylinder engine.

9.6.3 Balancing of Reciprocating Engines

9.6.3 Balancing of Reciprocating Engines For force balance,

9.6.3 Balancing of Reciprocating Engines Assume (mp)i = mp and (mc)i = mc for i=1,2,…,N When t=0, conditions for total force balance: Moments about the z and x-axes:

9.6.3 Balancing of Reciprocating Engines Necessary conditions for balancing of moments: Thus we can arrange the cylinders of a multicylinder reciprocating engine to satisfy Eq 9.12 and 9.13 The engine will be completely balanced against the inertia forces and moments.

9.7 Control of Vibration Some import methods to control vibrations: Control ωn and avoid resonance under external excitations. Introduce damping mechanism to prevent excessive response of system Use vibration isolators to reduce transmission of excitation forces from one part of the machine to another Add an auxiliary mass neutralizer or vibration absorber to reduce response of system

9.8 Control of Natural Frequencies Resonance Large displacements large strains and stresses failure of system Often the excitation frequency cannot be controlled. Hence must control natural frequency by varying mass m or stiffness k to avoid resonance. Practically mass cannot be changed easily. Hence we change stiffness k by altering the material or number and location of bearings.

9.9 Introduction of Damping System may be required to operate over a range of speed, hence cannot avoid resonance Can use material with high internal damping to control the response. Can also use bolted or riveted joints to increase damping. Bolted or riveted joints permit slip between surfaces and dissipate more energy compared to welded joints. However they also reduce stiffness of structure, produce debris and cause fretting corrosion.

9.9 Introduction of Damping Equation of motion of 1-DOF system with internal damping under excitation: Amplitude of response at resonance:

9.9 Introduction of Damping Viscoelastic materials have larger values of η and are used to provide internal damping. Disadvantage is their properties change with temperature, frequency and strain. Sandwich viscoelastic material between elastic layers – Constrained layer damping Material with largest η will be subjected to the smallest stresses.

9.10 Vibration Isolation Insert isolator between vibrating mass and vibration source to reduce response Passive isolators: springs, cork, felt etc. E.g. Mounting of high-speed punch press

9.10 Vibration Isolation Active isolator comprised of servomechanism with sensor, signal processor and actuator. Effectiveness given in terms of transmissibility Tr which is the ratio of amplitude of the transmitted force to that of the exciting force 2 types of isolation situations: Protect base of vibrating machine against large unbalanced or impulsive forces Protect system against motion of its foundation

9.10 Vibration Isolation Protect base of vibrating machine against large unbalanced or impulsive forces Protect system against motion of its foundation

9.10.1 Vibration isolation system with rigid foundation Resilient member placed between vibrating machine and rigid foundation Member is modeled as a spring k and a dashpot c as shown:

9.10.1 Vibration isolation system with rigid foundation Reduction of Force transmitted to foundation: Equation of motion: Steady state solution: where Force Ft transmitted to the foundation: Magnitude of total transmitted force FT:

9.10.1 Vibration isolation system with rigid foundation Transmissibility Following graphs shows the variation of Tr with r.

9.10.1 Vibration isolation system with rigid foundation

9.10.1 Vibration isolation system with rigid foundation Reduction of Force transmitted to Mass: Displacement transmissibility Td is also the ratio of the maximum steady-state accelerations of the mass and the base.

9.10.2 Isolation of source of vibration from surroundings By defining

9.10.2 Isolation of source of vibration from surroundings

9.10.2 Isolation of source of vibration from surroundings Reduction of force transmitted to foundation due to rotating unbalance Excitation force Force transmissibility

9.10.3 Vibration Isolation System with Flexible Foundation If the foundation moves, the system has 2 DOF Equations of motion: Assuming solution of the form xj=Xjcosωt, j=1,2

9.10.3 Vibration Isolation System with Flexible Foundation Natural frequencies given by roots of Amplitude of m1 at steady-state:

9.10.3 Vibration Isolation System with Flexible Foundation Amplitude of m2 at steady-state: Force transmitted to structure:

9.10.3 Vibration Isolation System with Flexible Foundation Transmissibility of isolator where ω2 is the natural frequency of the system Ft decrease as ω2 decrease

Example 9.4 An exhaust fan, rotating at 1000rpm, is to be supported by 4 springs, each having a stiffness of K. If only 10% of the unbalanced force of the fan is to be transmitted to the base, what should the value of K? Assume the mass of the exhaust fan to be 40kg.

Solution Transmissibility = 0.1 Forcing frequency

Solution Natural frequency Assuming ζ=0, To avoid imaginary values,

Example 9.5 A vibrating system is to be isolated from its supporting base. Find the required damping ratio that must be achieve by the isolator to limit the transmissibility at resonance to Tr=4. Assume the system to have a single degree of freedom. Solution: Setting ω=ωn,

Example 9.6 A stereo turntable, of mass 1kg, generates an excitation force at a frequency of 3Hz. If it is supported on a base through a rubber mount, determine the stiffness of the rubber mount to reduce the vibration transmitted to the baes by 80%.

Solution Using N=3x60=180 cpm and R=0.8,

9.10.4 Vibration isolation system with partially flexible foundation Base of isolator is partially flexible as shown: Mechanical impedance of base:

9.10.4 Vibration isolation system with partially flexible foundation Equations of motion Harmonic solution

9.10.4 Vibration isolation system with partially flexible foundation Amplitude of transmitted force: Transmissibility of isolator Z(ω) can be found experimentally by measuring the displacement produced by a vibrator.

9.10.5 Shock Isolation Shock load is a force load applied for less than one natural time period of the system Impulse Velocity imparted to the mass, i.e. application of shock load is equivalent to giving an initial velocity to the system Initial conditions:

9.10.5 Shock Isolation Free vibration solution: Force transmitted to the foundation due to spring and damper:

Example 9.7 Isolation under Shock An electronic instrument of mass 20kg is subjected to a shock in the form of a step velocity of 2m/s. If the maximum allowable values of deflection (due to clearance limit) and acceleration are specified as 20mm and 25g respectively, determine the spring constant of an undamped shock isolator.

Solution Magnitude of velocity of mass: Magnitude of acceleration of mass: where X is the displacement amplitude Selecting the value of ωn as 105.3681,

Example 9.8 Isolation under step load A sensitive electronic instrument of mass 100 kg is supported on springs and packaged for shipment. During shipping, the package is dropped from a height that effectively applied a shock load of intensity F0 to the instrument, as shown below.

Example 9.8 Isolation under step load Determine the stiffness of the springs used in the package if the maximum deflection of the instrument is required to be less than 2 mm. the response spectrum of the shock load is shown below with F0 =1000 N and t0 = 0.1s.

Solution Response spectrum Making use of the known data, The root can be found by MATLAB.

9.10.6 Active Vibration Control An active vibration isolation system is shown below.

9.10.6 Active Vibration Control System maintains a constant distant between vibrating mass and referee Depending on the types of sensor, signal processor and actuator used, the system can be electromechanical, electrofluidic, electromagnetic, piezoelectric or fluidic.

9.11 Vibration Absorbers When the excitation freq coincides with the ωn, the system may experience excessive vibration. Dynamic vibration absorber is another spring mass system designed to shift ωn of the resulting system away from the excitation freq.

9.11.1 Undamped Dynamic Vibration Absorber

9.11.1 Undamped Dynamic Vibration Absorber Assuming Amplitude of masses: We want to reduce X1. Thus set numerator of X1 to zero.

9.11.1 Undamped Dynamic Vibration Absorber X1 and X2 can be rewritten as:

9.11.1 Undamped Dynamic Vibration Absorber 2 peaks correspond to 2 ωn of composite sys.

9.11.1 Undamped Dynamic Vibration Absorber At X1=0, ω= ω1, Size of absorber can be found from: Absorber introduces 2 resonant frequencies Ω1 and Ω2, at which the amplitudes are infinite. Values of Ω1 and Ω2 can be found by noting

9.11.1 Undamped Dynamic Vibration Absorber Setting denominator of 2 roots of the equation:

Example 9.9 A diesel engine, weighing 3000 N, is supported on a pedestal mount. It has been observed that the engine induces vibration into the surrounding area through its pedestal at an operating speed of 6000 rpm. Determine the parameters of the vibration absorber that will reduce the vibration when mounted on the pedestal. The magnitude of the exciting force is 250 N, and the amplitude of motion of the auxiliary mass is to be limited to 2 mm.

Solution Amplitude of motion of auxiliary mass is equal and opposite to that of the exciting force.

Example 9.10 A motor-generator set shown below is designed to operate in the speed range of 2000 to 4000 rpm. However, the set is found to vibrate violently at a speed of 3000 rpm due to a slight unbalance in the rotor. It is proposed to attached a cantilever mounted lumped mass absorber system to eliminate the problem. When a cantilever carrying a trial mass of 2 kg tuned to 3000 rpm is attached to the set, the resulting natural frequencies of the system are found to be 2500 rpm and 3500 rpm. Design the absorber to be attached (by specifying its

Example 9.10 mass and stiffness) so that the natural frequencies of the total system fall outside the operating speed range of the motor-generator set.

Solution

Solution

Solution

9.11.2 Damped Dynamic Vibration Absorber Amplitude of machine can be reduced by adding a damped vibration absorber as shown.

9.11.2 Damped Dynamic Vibration Absorber Equations of motion Assume solution: Steady-state solutions:

9.11.2 Damped Dynamic Vibration Absorber

9.11.2 Damped Dynamic Vibration Absorber

9.11.2 Damped Dynamic Vibration Absorber If c2=ζ=0, resonance occurs at 2 undamped resonant frequencies If ζ=∞, m2 and m1 are clamped together and system behaves as 1-DOF system. Resonance occurs at

9.11.2 Damped Dynamic Vibration Absorber All curves intersect at pt A and B which can be located by The most efficient absorber (tuned vibration absorber) is one where pts A and B coincides, i.e.

9.11.2 Damped Dynamic Vibration Absorber Make curve horizontal at either A or B as shown below.

9.11.2 Damped Dynamic Vibration Absorber Set slope =0 at A and B:

9.11.2 Damped Dynamic Vibration Absorber Average value of ζ2 used in design: