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The design of anti-vibration systems

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1 The design of anti-vibration systems
VIBRATION ISOLATION The design of anti-vibration systems

2 Introduction Most mechanical devices such as fans, pumps, etc. produce vibration. Vibration is easily transmitted through solid materials and may ‘appear’ elsewhere producing troublesome vibration and/or noise. To prevent this, vibrating machines must be de-coupled from their supports and other connections

3 DEFINITIONS Free Vibration Forced Vibration Resonance Damping
A mass on a spring, a tuning fork or cymbal all vibrate at their natural frequency, f0 when displaced or struck. Forced Vibration A system excited by the application of an external periodic force of frequency, f will be forced into vibration. Resonance When a system is forced into vibration at its natural frequency, i.e. when f = f0 (The amplitude of vibration increases with each cycle) Damping Any influence which extracts energy is known as damping.

4 Damping Ratio Critical damping is when the mass returns to zero in the shortest time without overshoot. Damping Ratio = Damping Present Critical Damping

5 TRANSMISSIBILITY The performance of a spring isolation system is given by its ability to transmit the forces due to vibration of a machine through the anti-vibration springs. This is the FORCE TRANSMISSIBILITY, T T = Force Transmitted Force Applied The VIBRATION ISOLATION EFFICIENCY,  is given by:  = (1-T) x 100 % Decibel reduction (N dB) by an isolator is given by:- N (dB) = 20 lg (1/T) spring isolator Force Applied Force Transmitted

6 Example Find the transmissibility and the reduction in dB of a spring isolator that gives a vibration isolation efficiency of 99%  = (1-T) x 100 % so, T = 1 - /100 T = 1 – 99/100 T = 0.01 Reduction in level = 20lg(1/0.01) = 40 dB

7 Static Deflection The static deflection, s is the deflection causes when a mass is placed upon a spring isolator Gravity force = m.g Spring force = k.s Equating forces we get k.s = m.g or k = m.g/ s spring isolator free height static deflection (s )

8 Natural Frequency, f0 The natural frequency of a spring mass system (from SHM) is: Substituting the static deflection for k (k =m.g/s) we get, Cancelling and combining constants, f0 is found from:

9 Force Transmissibility Graph
Transmissibility, T Amplification, T 1.0 1.0 Isolation, T  1.0 f/f0 (f = f0) 1.414 Isolation only when f is greater than f02

10 Example Find the force transmissibility for a machine running at 3000 rpm if it is on a spring isolator with a natural frequency of 10 Hz. The force transmissibility equation f = 3000/60 = 50 Hz, f0 = 10 Hz T = 0.042, Isolation Efficiency,  = (1-T) x 100  96 %

11 PROBLEMS The amplitude of vibration at resonance can cause mechanical damage More of a problem with machines which run-up slowly, or that operate at different frequencies which may be close to resonance. Solution – damping or amplitude limiters Amplitude limiter with rubber buffers Friction damping Visco-elastic damping (dash-pot)

12 EFFECT OF DAMPING If damping is increased the resonance amplitudes are reduced, but unfortunately so is the isolation at higher frequencies f / f0

13 ISOLATOR DESIGN Start with required isolation efficiency, 
Calculate the required transmissibility, T - From =(1-T)x100 therefore, T = 1- [/100] Use the transmissibility equation and the forcing frequency, f to find the natural frequency, f0 if the isolator From T= 1/{(f/f0)2 -1) therefore, Calculate the static deflection required From therefore

14 /Continued A suitable vibration isolation mount can then be selected from a manufacturer’s catalogue which give the static deflections for different machine masses. Alternatively, the spring stiffness can be found from: k=m.g/s Remember, that this will be split between the number of mounts used (usually 4, one at each corner)

15 Example Find the static deflection of the spring isolators required to give 90% isolation efficiency for a pump with mass of 10kg running at 3000 rpm & give the reduction in dB Isolation efficiency of 90% means transmission of 10%, so required T = 0.1 Reduction = 20lg(1/0.1) = 20 dB f = 3000 rpm = 3000/60 = 50 Hz mass = 10kg (2.5kg per spring)

16 From before natural frequency needed
Substituting for f & T static deflection required = m or 1mm

17 The isolator can then be selected from a manufacturers catalogue for the mass of the machine.
Need a spring that will deflect by 1mm under the weight of 2.5kg (assuming 4 springs supporting the pump) Petrol driven water pump with rubber isolation mounts

18 INERTIA BASES Machines are often isolated using an inertia base
Spring Isolators Machines are often isolated using an inertia base (concrete) These add mass which reduces the vibration amplitude Increased mass allows stiffer springs which will help prevent rocking The extra mass reduces the centre of gravity also preventing rocking Flexible coupling

19 CENTRING THE MACHINE MASS OVER THE ISOLATORS
Inertia Base Centre of Gravity Centre of Gravity INCREASE IN MASS REDUCES THE VIBRATION AMPLITUDES EXTENDING THE FRAME TO CENTRE THE MASS TO PREVENT ROCKING

20 Dynamic Vibration Damping
Electronically controlled (active) with force applied to oppose that creating the vibration and thus prevent metal fatigue or Stockbridge dampers – use an additional spring-mass system tuned to the natural frequency of the vibrating system – parasitic spring-mass vibrates instead of actual system. Masses on springy arms used for power lines and the suspension cables on the old Severn Bridge

21 Wobbly Millennium Bridge

22 THE END


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