MECHANICAL VIBRATIONS ME 65

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MECHANICAL VIBRATIONS (ME 65)

• Session 9
• By
• Dr. P.Dinesh
• Sambhram Institute of Technology
• Bangalore

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In this session 9

• Characteristic Curves
• Highlights of curves
• Solution by complex algebra
• Rotating and Reciprocating Unbalance
• Problem

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• Characteristic Curves.
• The plot between (?/?n) the frequency ratio and
  (X/ Xst), the magnification factor for various
  values of damping is Frequency response curve and
  plot between frequency ratio and phase angle, is
  
• Phase Frequency response curve.

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• We know,
• (X / Xst) Magnification Factor/Amplitude
  Ratio MF
• 1/v ((1-(?/?n )2 )2 (2? ?/?n )2)
• X Max. steady state amplitude
• Xst zero freq. deflection F/k
• tan F (2? ?/?n) / (1-(?/?n )2)

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Charateristic Curve MF Vs Freq. Ratio
MF
Low damping
1
High damping
1
Freq. Ratio
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• Highlights
• At zero freq. ratio MF is 1 and all curves start
  from it.
• At resonance freq. MF is infinity and damping is
  zero
• With increase of damping the MF reduces for all
  values of damping
• Max. amplitude occurs to the left of the
  resonance freq.

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• Characteristic Curves F Vs Freq. Ratio

Very low ?
180º
F
High ?
90º
1
Freq. ratio
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• For zero damping the phase angle is 0 or 180
  degrees.
• At resonant frequency the phase angle is 90 deg.

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• Effect of Freq. Ratio on Forces

m?2X
Freq. Ratio less than 1
F
c?X
kx
Freq. Ratio equal to 1
m?2X
c?X
F
kX
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• Freq. ratio greater than 1

m?2X
c?X
kx
F
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• For values of freq. ratios. very less than unity
  the inertia and damping forces are small , hence
  F kX
• For freq. ratio of unity or Resonance,
• F c?X, and phase angle is 90 deg.

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• For values of Freq. ratio very large compared to
  unity , inertia force is very large and damping
  and spring forces are small. Phase angle is close
  to 180 deg.

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Solution by complex algebra

• Representing harmonic force in complex form as F
  Fei?t and with mxcxkx Fei?t
• The frequency response being defined as ratio of
  X to Xst ,
• X/Xst 1/v(k-m?2)2(c?)2
• and F tan-1((c?)/(k-m?2))

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Rotating and reciprocating unbalance

• All rotating and reciprocating machines have some
  unbalance in their rotating and reciprocating
  elements
• Such unbalance is due to an eccentric mass mo
  rotating at a distance e from the CG of machine

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• Rotating Equipment

mo?2e
e
mo
m Total mass mo unbalanced mass
m
x
x Displ. Of non rotating mass
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• When machine rotates the displ. of mo is sum of
  displ. x and e sin?t
• The eqn. of motion is
• (m-mo)x mod2(xesin?t)/dt2 -kx - cx
• On simplification
• mx cxkx mo?2esin?t

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• This is same as mx cxkx Fsin?t
• i.e., F mo?2e
• As per earlier discussion,
• X steady state amplitude is
• (mo?2e /k) / v((1-(?/?n )2 )2 (2? ?/?n )2)
• As ?n2 k/m, k ?n2m
• X(mo?2e/?n2m)/v((1-(?/?n )2 )2 (2? ?/?n )2)

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• (X/(mo e / m))
• (?/?n)2 /v((1-(?/?n )2 )2 (2? ?/?n
  )2)
• and tan F (2? ?/?n) / (1-(?/?n )2)
• where, F, is the phase angle
• Reciprocating Masses
• The same analysis can be carried out for
  reciprocating masses for which the unbalanced
  force again is mo?2esin?t.

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• Characteristic Curves for Rotating Masses

X/(moe/m)
1
Freq. Ratio
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• Discussion on characteristic curves
• Damping factor plays an important role in
  controlling the amplitudes during resonance.
• For low values of frequency ratio (?/?n), X tends
  to zero.
• At high speeds of operation, damping effects are
  negligible.

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• The peak amplitudes occur to right of resonance
  unlike for balanced systems.
• At resonance, ? ?n ie (X /moe/m) 1/2?
• Also, (X ) resonance moe / 2m?
• From the plot of (X / moe/m) v/s ?/?n , it is
  observed that at low speeds, because the inertia
  force is small, all the curves start from zero.

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• At resonance (X / moe/m ) 1/2? and the
  amplitude of such vibrations can be controlled by
  the damping provided in the system. For very
  large freq.ratio,(X/moe/m) tends to one.
• The phase angle Vs Freq. ratio plot is same as
  that for forced vibration

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• Problem 1) A single cylinder engine has a total
  mass of 100kg and is mounted on a steel frame.The
  deflection of frame is 3 mm. The reciprocating
  masses of the engine amounts to 10 kg and the
  stroke of the engine is 80 mm. A dashpot with c
  2 N sec/mm is used. Determine a) amplitude of
  vibration if driving shaft is rotated at 1000 rpm
  and b) speed of shaft at resonance.

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• Given m 100 kg, mo 10 kg, e stroke/2 80/2
  40 mm 40 x 10-3 m,
• d 3x10-3 m, c2 N sec/mm 2000 N sec/m
• N 1000 rpm
• Amplitude at 1000 rpm
• N 1000 rpm as N speed/no.of cylinders
• ?, freq. of forced vib. 2pN/60
• 104.72
  rad/sec,

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• ?n vk/mv(mg/d)/m
• vg/static def. v9.81/3x10-3 57.2 rad/sec
• The amplitude is obtained from
• (X/(mo e / m))
• (?/?n)2 /v((1-(?/?n )2 )2 (2? ?/?n
  )2)
• In the above eqn. ? is to be determined from
  the relation c 2m ?n ?
• 2000 2x100x57.2x ?
• ie., ? 0.1748

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• Substituting we have ,
• X 3 x 10-3 m
• b) Speed of shaft at resonance
• At resonance ? ?n
• ? 2pN/60
• N (57.2 x 60) /2p
• 546.22 rpm

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• End of session 9