MECHANICAL VIBRATIONS ME65

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MECHANICAL VIBRATIONS (ME65)

• Session 10
• By
• Dr. P. Dinesh
• Sambhram Institute of Technology
• Bangalore

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In this session 10

• Vibration Isolation
• Vibration transmissibility
• Energy Dissipated by Damping
• Sharpness of Resonance
• Problems

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• Vibration Isolation
• Vibratory forces are unavoidable in machines
• Their effect on machine or supporting structure
  can be minimised by proper isolation
• The isolation is of force or motion

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• The effectiveness of isolation is measured in
  terms of minimum force or motion transmitted to
  support or surroundings
• Materials used for isolation have elastic and
  damping properties
• Materials Cork, rubber, felt, metal springs

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• Transmissibility

F sin?t
F
c?x
FT
machine
a
kx
cx
kx
x
F
Foundation
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• Only spring force and damping force act on mass
  as well as foundation
• From vector diagram
• Ft v(kX)2 (c?X)2 Xv(k2 c?2 )
• As X (F/K)/v((1-(?/?n)2)2 (2??/?n)2)
• Ft
• (F/K)/v((1-(?/?n)2)2 (2??/?n)2)(v(k2c?2)

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• Defining Transmissibility TR as
• TR Ft / F , where F m?2e
• TR v(1(2??/?n)2)/(1-(?/?n)2)2 (2??/?n)2
• (F - a) , the angle by which Ft lags the
  impressed force is
• tan-1((2??/?n)/(1-(?/?n)2))tan-1(2??/?n)

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• Characteristic curves

? 0
TR
1
? 0.6
?/?n
v2
1
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• Characteristic curve

?0.125
(F-a)
60º
?1
?/?n
1
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• Highlights of Curves
• Vibration isolation begins when TR is less
  than 1 and freq. ratio is greater than v2. For
  low values of ? the freq. ratio is made large by
  adding mass to system so that ?n is small.

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• Damping is important at resonance to avoid large
  values of TR
• For ideal operating conditions TR should be zero
  or freq. ratio should be as great as possible or
  ?n should be small

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• Energy dissipated during damping
• Damping is present in vibrating system which
  removes energy from system
• This is due to internal friction, fluid
  resistance or resistance offered by air also.
• The energy dissipated per cycle for a spring mass
  damper system excited by a harmonic force is

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• Energy Dissipated per cycle pc?X2
• From the above equation the power required to
  vibrate a system can be computed.

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• Sharpness of Resonance
• It is used as a measure for the sharpness of
  resonance for various vibrating systems

a,b are called half power points
xp
b
a
0.707xp
x
?p
?1
?2
?
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• ?p ?n is the condition at resonance
• ?1 and ?2 are referred to as side bands
• Q , the quality factor is a measure of
  sharpness of resonance is defined as
• Q (?p /?2 ?1)

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• Problem on Forced vibration
• 1) An engine weighing 1000 N including
  reciprocating parts is mounted on springs. The
  weights of the reciprocating parts is 22 N and
  the stroke is 90 mm. The engine speed is 720 rpm,
  i) neglecting damping , find the stiffness of
  springs so that the force transmitted to the
  foundation is 5 of the amplitude force, ii)if
  under the actual working condition the damping
  reduces the amplitude of successive vibrations by
  25, determine the force transmitted at 720 rpm.

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• GivenW 1000 N, w 22N, e90 x 10-3 / 2 45 x
  10-3 m, N 720 rpm
• i) Spring stiffness
• TR v(1(2??/?n)2)/(1-(?/?n)2)2 (2??/?n)2
• From above for no damping and as ?/?n has to be
  more than v2,hence
• TR 1 / ((?/?n)2 -1)
• Spring stiffness from ?n vk/m

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• TR Ftr / F 0.005F/F 0.005
• TR 1 / ((?/?n)2 -1)
• ? 2p720 / 60 75.398 rad/sec.
• From above eqn, ?n 16.453 rad/sec.
• As ?n vk/m , 16.453 vk/(1000/9.81)
• k 27595.2 N/m is the spring stiffness

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• ii) Force transmitted at 720 rpm
• Knowing TR, using the relation TR Ftr /F the
  force transmitted is obtained.
• TR v(1(2??/?n)2)/(1-(?/?n)2)2 (2??/?n)2
• As obtained earlier, ? 75.398 rad/sec and ?n
  16.453 rad/sec.
• With log. decrement known ? can be determined as
  d 2p/v1- ?2

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• As d ln(x1/x2) and (x1/x2) (x1/0.75x1) 1.33
  ie, d ln1.33 0.288
• Therefore, from, 0.288 2p/v1- ?2,
• damping factor ? 0.0458 and from
• TR v(1(2??/?n)2)/(1-(?/?n)2)2 (2??/?n)2
• TR 0.0542 and also Ftr /F TR, Disturbing
  force, Fm?2e (w/g) ?2e
• F(22/9.81)(75.398)2(45 x 10-3) 573.7 N

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• As TR 0.0542 Ftr /F and F 573.7 N
• Ftr TR x F 31.1 N is the transmitted force.

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• 2) A machine of mass 1000 kg, is acted upon by an
  external force of 2450 N at a frequency of 1500
  rpm. To reduce the effects of vibration,
  isolators of rubber having a static deflection of
  2 mm under the machine load and an estimated
  damping factor of 0.2 are used. Determine i)
  Force transmitted to the foundation ii) Amplitude
  of vibration of the machine iii) Phase lag of the
  transmitted force w.r.t. the external force.

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• m 1000 kg, F 2450 N, N 1500 rpm,
• d 2 mm 2 x 10-3 m, ? 0.2
• We use,
• TR v(1(2??/?n)2)/(1-(?/?n)2)2 (2??/?n)2
• Also TR Ftr /F
• ? 2pN/60 157.08 rad/sec.
• ?n v(k/m) vg/ d v9.81/2x10 -3
• 
  70 rad/sec.

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• Subs for ?, ?, ?n into expr. for TR
• TR 0.325, as TR Ftr /F , Ftr 797 N
• The amplitude X is from the relation
• X (F/k) /v(1-(?/?n)2)2 (2??/?n)2
• As k mg/d 1000 x 9.81/2 x 10 -3
• 4.905 x
  106 N/m.
• That gives X 1.21x 10-4 m

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• The phase lag of transmitted force wrt F is
• (F - a) tan-1((2??/?n)/(1-(?/?n)2))
• tan-1(2??/?n)
• 125.55º

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• End of Session 10