DYNAMIC BALANCING OF ROTORS
By
Dr. Rajiv Tiwari
Department of Mechanical Engineering
Indian Institute of Technology Guwahati 781039
Under AICTE Sponsored QIP Short Term Course on
Theory & Practice of Rotor Dynamics
(15-19 Dec 2008)
IIT Guwahati
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
1
Introduction
The unbalance in rotors will not only cause rotor vibrations, but also transmit
rotating forces to bearings and to the foundation structure.
The force thus transmitted may cause damage to machine parts and its
foundation. If the transmitted force is large enough, it might affect even the
neighboring machines and structures.
Thus, it is necessary to remove the unbalance of a rotor, to as large an
extend as possible, for its smooth running. The residual unbalance
estimation in rotor-bearing system is an age-old problem.
From the state of the art of the unbalance estimation, the unbalance can be
obtained with fairly good accuracy [1-5].
Now the trend in the unbalance estimation is to reduce the number of test
runs required, especially for the application of large turbo generators where
Dr. R. Tiwari
(rtiwari@iitg.ernet.in)
2
the downtime is very expensive
[6,7].
Static balancing:
Single plane balancing
Dynamic balancing :
(i) Two plane balancing: For rigid rotors only. ( ω < ω cr )
(ii) Flexible rotor balancing : If the shaft deflects, and the
deflection changes with speed, as it does in the vicinity of
(ω > ω cr )
critical speeds .
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Dr. R. Tiwari (rtiwari@iitg.ernet.in)
5
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
6
Balancing of a flexible rotor
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Basic Principles of Static Rigid Rotor Balancing
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Basic Principles of Couple Rigid Rotor Balancing
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Basic Principles of Dynamic Rigid Rotor Balancing
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Balancing of Rigid Rotor
Cradle balancing machine :
The rotor is placed in the bearings of a cradle as shown in Fig.
1.
+
Figure 1 Craddle balancing machine
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
11
The cradle is placed on two springs and can be fulcrum about F1 or F2 to
form a simple vibrating system.
Two fulcrum can be located at two chosen balance planes (i.e. I and II),
where the correction mass to be added.
The rotor can be driven by a motor through a belt pulley arrangement.
If the spring system is such that the natural frequency of the system is in the
range of motor speed, the phase angle or the location of the unbalance
mass in either plane can be determined as follows.
Fulcrum the cradle in plane I, by fixing F1 and releasing F2. Run the rotor to
resonance, observing the maximum amplitude to the right of fulcrum F2.
This vibration is due to all the unbalance in plane II, since the unbalance in
plane I has no moment about F1.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
12
Use a trial mass at a chosen location and determine the amplitude of
vibration.
Figure 2 Plot of vibration amplitudes versus trial mass locations
Make a plot of this amplitude for different location of the same trial mass
(see Fig. 2). The trial mass for correction is added at the location where the
amplitude of vibration is minimum.
Increase or decrease the trial mass at the same locations, until the desired
level of balance is achieved. Similar procedure can be repeated by Fixing F2
and releasing F1. This procedure is tedious and sometimes may be time
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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consuming.
A procedure to determine the correction mass and location can be laid
down as follows, based on four observations of amplitude :
(i) without any addition to the rotor
(ii) with a trial mass at
= 0°
(iii) with a trial mass at 180°and
(iv) with same trial mass at
conveniently chosen location.
= ±90°, where
is measured from a
This procedure has to be repeated for two cases (e.g. when fulcruming at F1
and then for F2 ).
(1) Let OA is the amplitude measured with trial run
(2) OB is the amplitude measured in trial run by addition of a trial mass Wt
at 0°(arbitrary chosen location on rotor).
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Hence the vector AB will represent the effect of trial mass Wt.
(At this stage we do not know the location of vector OA on the rotor).
(3) OC is the vibration measured in the trial run with the trial mass at 180°.
So we will have AB = AC with 180°phase difference between them.
(Hence AC vector is also the effect of trial mass Wt so the magnitudes AB =
AC and phase will be 180°).
However we know only OA, OB & OC from test run (1), (2) & (3)
respectively & conditions AB = AC with 180°phase.
From these information we have to construct or locate points O, A, B & C on
a plane.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
15
Construction Procedure :
D
E’
C
A
α
~900
B
φ
E
O
Figure 3 Construction procedure
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Erect a line OAD equal to 2×OA. With O as the center and OB & OC as radii
and D as center and OC & OB as radii draw arcs to intersect at B & C (point
B and C we will be obtained by above construction). .
Draw a circle with BC as diameter and A as center. Construct the
parallelogram OBDC.
Now AB represent 0° position (i.e. reference line) and AC 180° position on
the rotor (AO is actual unbalance). The angular measurement may be
clockwise or CCW and is determined from the fourth observation.
The observation could be either OE or OE’ (+90° or –90° ). If the value
observed is in the vicinity of OE, then the angle to be measured CCW.
However it will be CW if OE’ is the reading observed in test (the fourth run
also checks the validity of the linearity used in the balancing procedure).
Dr. R.W
Tiwari
(rtiwari@iitg.ernet.in)
The magnitude of trial mass
t is proportional to AB.
17
The unbalance OA can be obtained accordingly in the mass term. The
location of unbalance is ∠OAB and the direction from figure (i.e. CW or
CCW).
The test is repeated by making the cradle fulcrum bed at FII and
measurements are made in plane I.
This procedure is very time consuming and also restricts the mass and size
of the rotor.
Modern balancing machines use amplitude and phase measurement in two
planes for balancing a rotor.
Machines are either soft support or hard support machines.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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900
1800
00
B
E
O
+ve
2700
00
Unbalance position
at shaft location,
(ccw direction +ve)
1800
B
O
Location of fourth
measurement for
trial mass at 2700
(ccw dir. +ve)
1800
B
O
2700
Location of
fourth
measurement for
trial mass at 900
(cw dir. +ve)
Unbalance
position at
shaft location
(ccw dir. +ve)
900
00
2700
Unbalance position
at shaft location,
(ccw direction +ve)
+ve
00
+ve
E’
00
E
A
900
2700
900
00
900
Location of fourth
measurement for
trial mass at 900
(ccw dir. +ve)
900
1800
00
+ve
00
E’
B
O
2700
Location of
fourth
measurement for
trial mass at 2700
(cw dir. +ve)
Dr. R. TiwariFigure
(rtiwari@iitg.ernet.in)
4
2700
Unbalance
position at
shaft location
(ccw dir. +ve)
19
Example 3.1 In the balancing process we make the following observations:
(i) ao = amplitude of vibration of the unbalanced rotor “as is”
(ii) a1 = amplitude with an additional one-unit correction at the location 0 deg and
(iii) a2 = same as a1 but now at 180 deg.
The ideal rotor, unbalanced only with a unit unbalance (and thus not containing the
residual unbalance), will have certain amplitude, which we cannot measure.
Call that amplitude x. Let the unknown location of the original unbalance be ϕ.
Solve x and ϕ in terms of and show that in this answer there is an ambiguity sign.
Thus four runs are necessary to solve the problem completely.
Answer: Measurements are
(i) amplitude of vibration with residual unbalance U R ∠ϕ
(ii) amplitude with unit trial mass at an angle of 0 0
0
(iii) amplitude with unit trial mass at an angle of 180
0
(iv) x = amplitude with 1 at an angle of 0 and without residual imbalance
(i.e. U R = 0 ), OA = a0 AB = a1 and AC = a2
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Figure 3.5 shows variation parameters involved in the present problem. From ∆OAB
, we have
a02 + OB 2 − a12
cos ϕ =
2a0OB
(A)
and
a02 + OC 2 − a22
cos (π − ϕ ) =
2a0OC
Since
OB =OC
(B)
, we have
a 2 + OB 2 − a 2
2
− cosϕ = 0
2a OB
0
(C)
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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D
C
x
0
x
ϕ
Reference
line
B
a1
a2
a0
A
Figure .5 Geometrical constructions for
determination of unbalance
On equating equations (A) and (C), we get
2a0OB cos ϕ = −(a02 + OB 2 − a22 ) = (a02 + OB 2 − a12 )
(D)
which gives
2a02 + 2OB 2 − ( a12 + a22 )
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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OB = x, since OB
(or OC ) are effect of trial mass of unit magnitude.
Hence equation (D) gives
x 2 = ( a12 + a22 ) / 2 − a02
or
x=±
1
2
( a12 + a22 ) − a02
(E)
Equations (A) and (C) gives (noting that OC = OB = x ),
x 2 = 2a0 x cos ϕ − a02 + a12
and
x 2 = −2a0 x cos ϕ − a02 + a22
(F)
(G)
On equating equations (F) and (G), we get
cos ϕ = ( − a12 + a22 ) / 2a0 x
(H)
Equation (E) gives the magnitude of the unbalance and equation (H) gives the
magnitude of the phase angle, the direction or sense of the phase cannot be
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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obtained from only above measurements.
Example 3.2. A short rotor or flywheel has to be balanced. Observations of
the vibration at one of the bearings are made in four runs as follows:
Run 1; rotor “as is”
Run 2; with 5gm. at 0 deg.
Run 3; with 5 gm. at 180 deg.
Run 4; with 5gm. at 90 deg.
amplitude 6.0 µm
amplitude 5.0 µm
amplitude 10.0 µm
amplitude 10.5 µm
Find the weight and location of the correction. Take the trial and balancing
masses at the same radius.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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E
D
Answer:
C
A
e
Ref
line B
e
c
ren
O
OA = AD = 6 cm
DB =10 cm
OB = 5 cm
OE = 10.5 cm
AB = 6.3 cm
Imbalance position
= angle BAO
= 71 deg. CCW
Figure 3.6 Geometrical constructions
Figure 3.6 shows the geometrical construction of the present problem with lengths
of various arcs. From this the net effect of the imbalance is given as
AB = 6.3 cm ≡ 5 gm
Hence, the residual imbalance is given as
OA = AD = 6 cm ≡ 4.762 gm
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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AB is the reference line. The fourth observation is intersecting at E, hence angle to be
measured in the CW direction (i.e.∠BAE ). Hence, the unbalance position is given as
∠BA0 = 710 CCW direction.
The unbalance magnitude and phase can be also obtained from equations (E) and
(H), we have
x=±
1
2
( a12 + a22 ) − a02 = 4.09 gm
and
cos ϕ = (− a12 + a22 ) / 2a0 x = 0.6065
ϕ = 52.8deg
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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The Influence Coefficient Method
Definition of Influence coefficients
(i) Only force F1
1
F1
y11
2
y21
y11 = displacement at station 1 due to
force F1 at station1 = α 11 F1
y21 = displacement at station 2 due to
force F1 at station1 = α 21 F1
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
(7)
(ii) Only force F2
F2
1
y12
2
y22
y12 = α12 F2
y22 = α12 F2
27
(iii) When both F1 and F2 are present
F1
F2
y1
y2
y1 = y11 + y12 = α 11 F1 + α 12 F2
y 2 = y21 + y 22 = α 21 F1 + α 22 F2
y1
α 11 α 12
=
y2
α 21 α 22
or
F1
F2
Figure 7
Influence coefficients can be obtained by experimentation or by strength of
formulae i.e.
α 11
y11
=
,
F1
α
21
y 21
=
F1
e tc.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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The Influence Coefficient Method
L
L1
R
a
b
No trial mass
1
L1
aR
R1
1
R1
L2
2
2
bR
R2
bL
L1
R3
3
3
R1
Trial mass TL
aL
L3
Figure 5 Bearing measurements and influence coefficients for a rigid rotor
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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In soft support machines, the resonant frequency of the rotor support
system is low and the rotor runs at a speed above the resonance of the
support system. Vibratory amplitudes are measured, which are then
converted to forces.
In hard support system, the support natural frequency is very high and they
measure the rotor unbalance forces directly, independent of rotor mass and
configuration.
The balancing procedure is based on influence coefficient measurement.
We choose two convenient planes L and R for trial mass and two
measurement planes a and b (can be chosen as bearing locations).
Let L1 and R1 be the initial readings of vibration levels (displacement,
velocity or acceleration) measured with phase angle 1 and 1 respectively.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Signal from station a shaft
t2
T
t1
Spike due to
notch in the
shaft surface
reference signal
Phase = 0 with respect to notch
Phase lead =
2πt1
radians
T
2πt 2
Phase lag =
radians
T
Figure 6
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The phase angles are measured with the same reference during the test
and their relative locations with respect to rotor is initially known.
In the second run, place a trial mass TR at a convenient location in plane R
and let the observations be L2 and R2 with phase 2 and 2 respectively in
the a & b planes.
The difference between R2 and R1 will be the effect of trial mass in right
plane R on the measurement made in plane b. We can denote this as an
influence coefficient α bR
(1)
α bR = ( R 2 − R 1 ) / T R
where ‘ ’ represent vector since displacement has magnitude and phase
information. Similarly
(2)
α
aR
= ( L 2 − L1 ) / T
R
We remove the trial mass from plane R and place TL in plane L and repeat
Dr. R. Tiwari
(rtiwari@iitg.ernet.in)
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the test to obtain the measured
values
α bL = ( R3 − R1 ) / TL
and α aL = ( L3 − L1 ) / TL
(3,4)
With the help of equations (1) to (4), we can obtain influence coefficient
experimentally.
Let the correct balance masses be W R and W L . Since the original
unbalance response is R1 and L1 as measured in right and left planes,
we can write
(5)
−R = W α +W α
and − L = W α +W α
1
R bR
L bL
1
R aR
L aL
Correction masses will produce vibration equal and opposite to the
vibration due to unbalance masses. Hence,
R1
L1
α bR
=−
α aR
α bL
α aL
WR
(6)
WL
These can be calculated either by a graphical method or analytical
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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method of vectors (complex
algebra) i.e.
a
b
c
d
−1
1
=
∆
d
−b
−c
a
w h e re
∆ = (ad − cb )
which gives
WR =
L1 .α b L − R1 .α a L
α b R .α a L − α aR .α b L
and
WL =
R1 .α a R − L1 .α b R
α b R .α a L − α aR .α bL
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Experimental set-up for influence coefficient method of balancing
Phase mark
on shaft
Accelerometer
Measurement
Plane a
R
L
Accelerometer (or
proximity probes
on the shaft near
to the bearing)
Photo electric
Probe
Charge
Amplifier
Measurement
Plane b
Vibration
meter
Phase
meter
Oscilloscope
Maximum
displacement
location
Photo sensitive mark
Hardware or virtual instrumentation
Photo electric probe
Figure 8 Experimental set-up for influence coefficient method of balancing
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Example 3.3 A rigid rotor machine is exhibiting vibration problems caused by imbalance.
The machine is symmetric about its center-line. A trial balance mass of 0.3 kg is sited at
end 1 at an angle of 300 relative to some reference position; this causes changes in
vibration vectors of 50 µm at 610 at end 1 and 42 µm at 1300 at end 2.
Determine the influence coefficients for use in balancing the machine, and calculate the
balance mass required at each end of the machine if the measured imbalance vibrations
are –30 µm at 2300 at end 1 and –70 µm at 3300 at end 2.
Solution: Given data are
Trial mass in plane 1: TR1 = 0.3 kg
at 300 phase , which can be written as
TR1 = 0.3(cos30 + j sin 30) = ( 0.2598 + j 0.15 ) Kg
0
Displacement in plane 1: R2 = 50 m at 61 phase , which can be written as
R2 = ( 24.2404 + j 43.73)
Displacement in plane 2:
µm
L2 = 42 m at 1300 phase
, which can be written as
L2 = ( −26.997 + j 32.173)
µm
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Measured responses due to residual imbalances are
In plane 1: R1 = −30 m at 2300 phase ≡ R1 = (19.2836 + j 22.98 ) µm
In plane 2:
L1 = −70 m at 3300 phase ≡ L = ( −60.621 + j 35 ) µm
1
We have, influence coefficients as
α11 = α bR =
R2 - R1
= (48.8919 + j 51.63766) × 10−6
TR1
µm/kg
and
12
= α aR =
L2 - L1
= (92.3559 − j 69.1995) × 10−6 µm/kg
TR1
It is given that machine is symmetric about centreline.
α 21 = α12 and α 22 = α11
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Measurements, influence coefficients and correction mass are related as
− R1
α11 α12
=
α 21 α 22
− L1
wR
wL
which can be simplified as
wR =
1
1
{L1α12 − R1α 22 } and wL = {R1α 21 − L1α11}
∆
∆
with ∆ =
11 12
−
12
22
=
2
11
−
2
12
= ( −468.066 + j 16907.73 × 10−6 )
(µm/kg)2
which gives the balancing mass and its angular position as
wR = −3.3519 × 10−3 + j 7.123 × 10−3 ≡ 7.893 × 10-3 kg at 2950
and
wL = 3.90356 ×10−3 − j 2.7136 ×10−3 ≡ 4.7541×10-3kg at -350
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Balancing of Flexible Rotors :
As long as the rotor experiences no deformations i.e. it remains as a rigid
rotor, the balancing procedure discussed earlier is effective.
Once the rotor bends while approaching a critical speed, the bend center
line whirls around and additional centrifugal forces are set-up and the rigid
rotor balancing becomes ineffective. (sometimes rigid rotor balancing
worsens bending mode whirl amplitude).
Two different techniques are generally employed
(i) Modal balancing technique. Bishop, Gladwell & Parkinson and
(ii) Influence coefficient method. Tessarrik, Badgley and Rieger.
Modal balancing method
A practical procedure to balance the rotor by model correction, masses
equal in number to the flexible mode shapes, N, known as N-plane method.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
39
Run the rotor in a suitable hard bearing balancing machine, to a safe speed
approaching the first critical speed and record the bearing vibrations or
forces.
Choose an appropriate location for the trial mass.
For first critical speed, this should be roughly in the middle for a
symmetrical rotor in its axial distribution of mass. Record the readings at
the same speed as before.
Using the above two readings, the correct mass and location can be
determined. (single plane balancing). With this correction mass, it should be
possible to run the rotor through the first critical speed without appreciable
vibration.
Next, run the rotor to a safe speed approaching the second critical speed, if
the operating speed is near the second critical or above the second critical
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
40
speed.
1
2
Rigid body modes
Figure 9
Note the readings. Add a pair of trial masses 180° apart in two planes
without a affecting the first mode. (in fact if we try to balance one particular
mode it will not affect balancing of other modes).
Note the readings at the same speed near the second critical speed.
Two readings can be used to determined the correction mass required.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
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Similarly higher modes can be balanced i.e. up to Nth mode can be
balanced by N balancing planes. Instead of N plane correction,
Kellenberger suggested that the rotor should be corrected in N+2 planes,
so as not to disturb the rigid body balancing.
Modal Balancing (formulations)
z
x
bearing axis
y
Figure 10
Assume that all unbalance is distributed only in the x-y plane.
Let the rotor speed be
and the deflection of the rotor be y(x).
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
42
The deflection y(x) can be written in terms of summation of mode shapes as
(8)
y(x) =
φ Y (x)
i
i
where Yi (x) is the mode shape in the ith mode and øi is unknown constant.
The deflection y(x) can be measured experimentally.
For example for simply supported end conditions the mode shapes are
I Mode
II Mode
y1 ( x) = sin
πx
y 2 ( x) = sin
l
2πx
l
i th Mode
iπx
y i ((rtiwari@iitg.ernet.in)
x) = sin
Dr. R. Tiwari
l
43
For other end conditions mode shapes can be obtained by free vibration
analysis. Modal series for eccentricity can also be written as
a( x) =
i
α i Yi ( x)
(9)
It can be written in terms of mode summation since the y(x) is the result of
a(x). The main objective is to find out for the eccentricity distribution to be
known.
Multiplying (8) by the mass per unit length m(x) and the mode shape
Y j ( x) and integrate from 0 to 1.
l
0
l
φiYi ( x) Y j ( x)dx
m( x) y( x)Y j ( x) = m( x)
(10)
i
0
noting the orthogonality condition of mode shapes
l
0
m ( x )Yi ( x )Y j ( x ) dx = 0
for all i ≠ j
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
(11)
44
Equation (10) gives
l
l
l
m( x) y( x)Y j ( x)dx = m( x)φ j Y j ( x)dx =φ j m( x)Y j ( x)dx
0
2
0
2
(12)
0
which can be written as
φ
j
1
=
M
l
j
m ( x ) y ( x )Y j ( x ) dx
(13)
0
with the generalized mass in jth mode (= M j ) is given as
M
i
=
l
m ( x )Y
2
j
( x ) dx
0
The generalized mass M jcan be obtained by knowing m(x)andYj (x); Yj (x) can
be obtained by free vibration analysis and theoretically speaking can be
found by experiment.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
45
Governing equation for shaft motion is
d2
dx
2
′
′
=
[
EI
(
x
)
y
(
x
)
]
ω
m ( x )[ y ( x ) + a ( x ) ]
2
(14)
where is the rotor speed. On substituting for y(x) and a(x) from equation
(8) and (9), we get
d2
EI ( x )
2
dx
i
φ i Yi′′( x )
= ω 2 m ( x)
i
φ i Yi ( x ) + { α i Yi ( x )}
(15)
Noting the orthogonality condition and multiply both sides by Yi (x) and
integrate over the length of shaft, the left hand side of equation (15) gives
l
0
d2
dx 2
φ iYi ″ ( x )
EI ( x )
Y j ( x ) dx
i
On performing integration by parts, we get
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
46
[
d
EI ( x )
dx
Y j (x)
{
φ i Y i ′′( x )
}]
l
0
l
d
dx
−
0
φ i Y i ′′( x )
EI ( x )
Y j′′( x ) dx
i
First term vanishes (since it is zero for all boundary conditions), on taking
again integration by parts of the second term, we get
[
− Y j′ ( x ) EI ( x )
{
φ i Y i ′′( x )
}]
l
0
l
+
φ i Y i ′′( x ) Y j′′( x ) dx
EI ( x )
0
i
First term again vanishes for all boundary conditions. On noting the
orthogonality condition equation (11), we get
l
= φ
[
EI ( x ) Y j′′( x )
j
]2 dx
= φ jK
j
(16)
0
where the generalized stiffness in jth mode is defined as
l
K
j
[
EI ( x ) Y j′′ ( x )
=
0
]
2
dx
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
47
From right hand side of equation (15), noting equation (8) and (9), we have
ω 2 (φ j + α j )M j
(17)
Therefore form (15), noting equations (16) and (17), we have
φ j K j = ω 2 (φ j + α j )M j
which can be rearranged as
α
j
=
K
j
/M
ω
j
2
−ω
2
(φ j )
(18)
Once α j is obtained, then distribution of eccentricity a(x) can be found from
equation (9). Equation (18) requires m(x), y(x), Yj (x) and pj = Kj / Mj .
The m(x) can be accurately found out, y(x) is difficult to obtained, Y j ( x) is
obtained by eigen analysis and p j is natural frequency in ith mode = K j M
j
which can be obtained by eigen value analysis.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
48
Influence coefficient method
F1
2
1
α 21
α11
2
′
α 21
ω << p1
1
ω ≈ p1
′
α11
F1
Flexible shaft
Rigid shaft
′′
α11
2
Figure 11 α ′′
21
p
ω ≈ p2
Balancing planes
x
q
z
Measuring planes
Dr. R. Tiwari
(rtiwari@iitg.ernet.in)
Figure
12
49
Choose p number of balancing planes (where mass can be added or chip
off) where p > 2, q number is the measuring planes, generally it is two i.e. at
bearing planes.
Let the unbalance in each of the balancing planes be U 1 , U 2 ,
v1
v2
vq
=
α 11
α 21
α q1
α 12
α 22
α1p
α2p
U1
U2
α qp
Up
where υ is the vibration measurement at
Measurements are taken at number of speeds.
,U
p
(19)
the
measuring
plane.
On writing equation (19) for each of the speeds
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
50
v 21
α
α
1
11
1
21
α
α
1
12
1
22
α
α
1
1 p
1
2 p
v q1
v 12
α
α
1
q1
2
11
α
α
1
q 2
2
12
α
α
1
qp
2
1 p
α
2
q1
α
2
q 2
α
2
qp
v 1n
v 1n
α
α
n
11
n
21
α
α
n
12
n
22
α
α
n
1 p
n
2 p
v qn
α
n
q1
α
n
q 2
α
n
qp
v 11
v
2
q
=
U
1
U
2
U
3
U
p
or
{ν } = [ α
] {U }
(20)
Once the influence coefficients [α ] are known for all speeds equation (20)
can be used to obtained unbalances :
(
{U } = [α ] [α ]
T
)
−1
{v}
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
(21)
51
Influence coefficient matrix can be obtained by attaching trial masses and
measuring displacements,
from equation (19), we get for a particular speed
1
v11
1
v 21
=
vq11
α1
α1
α1
α
α
α
11
1
21
α1
12
1
22
α q1 2
q1
U 1 + TR
1p
1
U2
2p
α1
(22)
Up
qp
On subtracting equation (22) from first q equation in equation (19), we get
1
v11
− v11
1
v 21
− v 21
v q11 − v q1
=
α1
α1
α1
α
α
α
11
1
21
α1
q1
12
1
22
α1
q2
Equation (23) gives
1p
1
2p
α1
qp
TR
0
(23)
0
1
1
v
−
v
1
α21 = 21 2 ,
α1 =
,
11 Dr. R. T
TR
Tiwari
(rtiwari@iitg.ernet.in)
R
1
v11
− v11
,
α
1
q1
=
v q11 − v q1
TR
(24)
52
Similarly by attaching a trial mass on plane 2 we get second
column of the influence coefficient matrix in equation (24), the
above analysis should be done at a constant speed.
Similarly we can find the influence coefficient-matrix for
different speeds.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
53
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
54
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
55
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
56
References
[1] W. Kellenburger 1972 Transactions of the American Society of Mechanical
Engineers, Journal of Engineering for Industry 94, 584-560. Should
a
flexible rotor be balanced in N or N+2 planes?
[2] J. Drechsler 1980 Institution of Mechanical Engineers Conference on
Vibrations in Rotating Machinery, Cambridge, UK, 65-70. Processing
surplus information in computer aided balancing of large flexible rotors.
[3] P. Gnilka 1983 Journal of Vibration 90, 157-172. Modal balancing of
flexible rotors without test runs: an experimental investigation.
[4] J.M. Krodkiewski, J. Ding and N. Zhang 1994 Journal of Vibration 169,
685-698. Identification of unbalance change using a non-linear
mathematical model for rotor bearing systems.
[5] M.S. Darlow 1989, Springer – Verlag, Balancing of High-Speed Machinery,
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
57
[6] S. Edwards, A.W. Lees and M.I. Friswell 2000 Journal of Sound and
Vibration, 232(5), 963-992. Experimental Identification of Excitation and
Support Parameters of a Flexible Rotor-Bearings-Foundation System from
a Single Run-Down.
[7] R. Tiwari, 2005, Mechanical System and Signal Processing, Conditioning
of Regression Matrices for Simultaneous Estimation of the Residual
Unbalance and Bearing Dynamic Parameters (in press).
[8] IS 5172 : 1969 Specification for Balancing Bench,
[9] IS 13274 : 1992/ISO 1925 : 1990 Mechanical vibration – Balancing
– Vocabulary.
[10] IS 13275 : 1992/ISO 2371 : 1974 Description and evaluation of field
balancing equipment.
[11] IS 13277 : 1992/ISO 2953 : 1985 Balancing machine - Description
and evaluation.
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
58
[12] IS 13278 : 1999 /ISO 3719 : 1994 Mechanical Vibration - Symbols
for Balancing Machines and Associated Instrumentation.
[13] IS 13280 : 1992/ISO 5406 : 1980 Mechanical balancing of flexible
rotors.
[14] IS 14280 : 1995/ISO 8821 : 1989 Mechanical vibration - Balancing
- Shaft and fitment key convention.
[15] IS 14734 : 1999 /ISO 7475 : 1984 Balancing Machines Enclosures and Other Safety Measures.
[16] IS 14918 : 2001 Mechanical Vibration - Methods and Criteria for
the Mechanical Balancing of Flexible Rotors
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
59
Thank you
Dr. R. Tiwari (rtiwari@iitg.ernet.in)
60