Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
CHAPTER 9
DYNAMIC BALANCING OF ROTORS
The unbalance in rotors will not only cause rotor vibrations, but also transmit rotating forces to the
bearings and to the foundation structure. The force thus transmitted may cause damage to the machine
parts and its foundation. If the transmitted force is large enough, it might affect even the neighbouring
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 [9.1-9.5]. Now the trend in the unbalance estimation is to reduce the
number of test runs required, especially for the application of large turbogenerators where the
downtime is very expensive [9.6, 9.7].
•
Static balancing: Singe plane balancing (refer undergraduate dynamics of machinery course)
•
Dynamic balancing:
o
Two plane balancing: For rigid rotors only (ω < ω cr ) .
o
Flexible rotor balancing: If the shaft deflects, and the deflection changes with speed,
as it does in the vicinity of critical speeds (ω > ω cr ) .
9.1 Balancing of Rigid Rotor
9.1.1 Cradle balancing machine: The rotor is placed in the bearings of a cradle as shown in Fig. 9.1.
F1
I
II
+
+
F
F
F2
Figure 9.1 Craddle balancing machine
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
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
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 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. Use a trial mass at a
Amplitude of
vibration
chosen location and determine the amplitude of vibration.
Minimum
amplitude
Trial mass location θ
Figure 9.2 Plot of vibration amplitudes versus trial mass locations
Make a plot of this amplitude for different location of the same trial mass (see Fig. 9.2). 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 achived. Similar
procedure can be repeated bey Fixing F2 and releasing F1. This procedure is tedious and sometimes
may be time 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 θ = 00 (iii) with a trial mass at 1800 and (iv) with same trial mass at θ = ±900, where θ is
measured from a conveniently chosen location. This procedure has to be repeated for two cases (e.g.
when fulcruming at F1 and then for F2). Let OA is the amplitude measured with trial run (1), OB is the
amplitude measured in trial run (2) by addition of a trial mass Wt at 00 (arbitrary
chosen location on
rotor). 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). OC is the vibration measured in trial run (3), with the trial
mass at 1800. So we will have AB = AC with 1800 phase difference between them (Hence AC vector
is also the effect of trial mass Wt so the magnitudes AB = AC and phase will be 1800). However we
know only OA, OB & OC from test run (1), (2) & (3) respectively & conditions AB = AC with 1800
phase. From these information we have to construct or locate points O, A, B & C on a plane.
378
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
Construction procedure
D
E’
C
A
~900 α
B
φ
E
O
Figure 9.3 A construction procedure for finding the unbalance vector
Erect a line OAD equal to 2OA. With O a center and OB & OC as radii and D as center and OC &
OB as radii draw arcs to intersect at B & C. Draw a circle with BC as diameter and A as center.
Construct the parallelogram OBDC (point B and C we will be obtained by above construction). Now
AB represent 00 position (i.e. reference line) and AC 1800 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’ (+900 or –900). 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). The magnitude of trial mass Wt is proportional to AB. The unbalance OA can be obtained
accordingly in 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 pivoted at FII and measurements 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. The
machines are either soft support or hard support machines.
379
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
900
0
E
O
Location of fourth
90
00
0
0
270
Unbalance position at shaft
location, (ccw direction +ve)
900
0
0B
A
1800
A
0
measurement for trial mass
at 900 (ccw dir. +ve)
0B
00
φ
2700
O
Location of fourth
measurement for trial
mass at 900 (cw dir. +ve)
Unbalance position
at shaft location
(ccw dir. +ve)
900
+ve
φ
1800
A
B
00
0
measurement for trial mass
at 2700 (ccw dir. +ve)
+ve
φ
00
φ
O
Location of fourth
+ve
φ
1800
φ
E’
2700
900
900E
φ
A
0B
+ve
270
Unbalance position at shaft
location, (ccw direction +ve)
φ
O
Location of fourth
measurement for trial
mass at 2700 (cw dir.
+ve)
1800
E’
2700
00
2700
Unbalance position
at shaft location
(ccw dir. +ve)
Figure 9.4 Procedure of obtaining the sense of the unbalance mass angular position
Example 9.1 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 a0 , a1 and a2 show that in
this answer there is an ambiguity sign. Thus four runs are necessary to solve the problem completely.
Answer: Measurements are
(i) a 0 = amplitude of vibration with residual unbalance U R ∠ϕ
(ii) a1 = amplitude with unit trial mass at an angle of 0 0
(iii) a 2 = amplitude with unit trial mass at an angle of 180 0
(iv) x = amplitude with 1 at an angle of 0 0 and without residual imbalance (i.e. U R = 0 )
OA = a 0 , AB = a1 and AC = a 2
380
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
Figure 9.5 shows variation parameters involved in the present problem. From ∆OAB , we have
cos ϕ =
a02 + OB 2 − a12
2a0 OB
(A)
and
cos (π − ϕ ) =
a02 + OC 2 − a22
2a0 OC
(B)
Since O B = O C , we have
− cos ϕ =
a02 + OB 2 − a22
2a0OB
(C)
D
C
x
x
Reference
line
0
ϕ a0
a2
B
a1
A
Figure 7.5 Geometrical constructions for
determination of unbalance
On equating equations (A) and (C), we get
2a0 OB cos ϕ = −(a02 + OB 2 − a22 ) = (a02 + OB 2 − a12 )
which gives
2a 02 + 2O B 2 − (a12 + a 22 )
(D)
O B = x , since O B (or O C ) are effect of trial mass of unit magnitude. Hence equation (D) gives
x 2 = (a12 + a 22 ) / 2 − a 02
or
x=±
381
1
2
(a12 + a22 ) − a02
(E)
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
Equations (A) and (C) gives (noting that O C = O B = x ),
x 2 = 2a 0 x cos ϕ − a 02 + a12
(F)
x 2 = −2a 0 x cos ϕ − a 02 + a 22
(G)
and
On equating equations (F) and (G), we get
cos ϕ = (− a12 + a 22 ) / 2a 0 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 obtained from only above measurements.
Example 9.2. 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:
amplitude 6.0 µm
amplitude 5.0 µm
amplitude 10.0 µm
amplitude 10.5 µm
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.
Find the weight and location of the correction. Take the trial and balancing masses at the same radius.
Answer:
E
D
C
A
line B
nce
e
r
e
Ref
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
71°
Figure 9.6 Geometrical constructions
382
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
Figure 9.6 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
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 = ? kg
and
cos ϕ = ( − a12 + a 22 ) / 2a 0 x = ? degree
Exercise 9.1 A rotor is being balanced in the cradle-balancing machine by pivoting rotor about a
fulcrum (e.g. F1). The following amplitudes of vibrations are observed at the critical speed: (i) 14 µm
for the rotor without additional weights, (ii) 10 µm with 5 gm placed in location 0 deg, (iii) 22 µm
with 5 gm placed in location 90 deg and (iv) 22 µm with 5 gm placed in location 180 deg. Find the
amount and angular location of the necessary correction mass.
Exercise 9.2 A rotor is being balanced in the cradle-balancing machine by pivoting rotor about a
fulcrum (e.g. F1). The following amplitudes of vibrations are observed at the critical speed:
1. 14 µm for the rotor without additional weights
2. 10 µm with 5 gm placed in location 0 deg.
3. 22 µm with 5 gm placed in location 90 deg
4. 22 µm with 5 gm placed in location 180 deg
Find the amount and angular location of the necessary correction mass.
383
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
9.1.2 The Influence Coefficient Method
Definition of Influence coefficients: Figure 9.7(a) shows that when a force F1 is applied at station 1
and due to this force the beam deflections at stations 1 and 2 are given as
y11 = displacement at station 1 due to force F1 at station 1 = α 11 F1
and
y21 = displacement at station 2 due to force F1 at station1 = α 21 F1
where α is the influence coefficient and its first subscript represents the displacement station and
second represents the force station. Similarly for Figure 9.7(b), we have
y12 = α 12 F2
1
y22 = α 22 F2
and
F1
y11
F2
1
2
y22
y12
(i) Only force F1
2
y22
(ii) Only force F2
F1
F2
y1
y2
(iii) When both F1 and F2 are present
Figure 9.7 Definition of influence coefficients
In Figure 9.7(c), we have
y1 = y12 + y12 = α11 F1 + α12 F2
⇒
y2 = y21 + y22 = α 21 F1 + α 22 F2
 y1  α11 α12   F1 
 =
 
 y2  α 21 α 22   F2 
Influence coefficients can be obtained by experimentation or by strength of formulae i.e.
α11 =
y11
y
, α 21 = 21 etc.
F1
F1
384
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
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 converetd 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) as
shown in Figure 9.7. Let L1 and R1 be the initial readings of vibration levels (displacement, velocity
or acceleration) measured with phase angle γ 1 and δ 1 respectively.
L
L1
a
γ1
L1
R
b
No trial mass
αaR
R1
L2
γ2 Trial mass TR
δ2
L1
R3
Trial mass TL
αbR
R2
αbL
γ3
R1
δ1
δ3
R1
αaL
L3
Figure 7.7 Bearing measurements and influence
coefficients for a rigid rotor
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 γ 2 and δ 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 .
α bR = ( R2 − R1 ) / TR
(9.1)
where " →" represent vector since displacement has magnitude and phase information. Similarly
385
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
α aR = ( L2 − L1 ) / TR
(9.2)
We remove the trial mass from plane R and place TL in plane L and repeat the test to obtain the
measured values
α bL = ( R3 − R1 ) / TL
α aL = ( L3 − L1 ) / TL
and
(9.3, 9.4)
With the help of equations (9.1) to (9.4), 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
− R1 = WRα bR + WLα bL
and − L1 = WRα aR + WLα aL
(9.5)
Correction masses will produce vibration equal and opposite to the vibration due to unbalance masses.
Hence,
 R1 
α bR
  = − α aR
 L1 
α bL  WR 
 
α aL  WL 
(9.6)
These can be calculated either by a graphical method or analytical method of vectors (complex
algebra) i.e.
a b 
c d 


which gives
−1
=
1  d − b
∆ − c a 
where ∆ = (ad − cb)
L1 .α bL − R1 .α aL
WR = α bR .α aL − α aR .α bL
and
386
R1 .α aR − L1 .α bR
WL = α bR .α aL − α aR .α bL
(9.7)
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
Phase mark
on shaft
Accelerometer
Measurement
Plane ‘a’
Accelerometer
(or proximity
probes on the
shaft near to the
bearing)
R
L
Photo electric
Probe
Measurement
Plane ‘b’
θ
Charge
Amplifier
Phase
meter
Oscilloscope
θ
Maximum
displacement
location
Vibration
meter
Photo sensitive
mark
Hardware or virtual instrumentation
Photo electric probe
Figure 9.8 Experimental set-up for the influence coefficient method of balancing
Signal from station a shaft
t2
T
Spike due to
notch in the
shaft surface
reference signal
Phase lag = 0 with respect to the
notch
t1
2πt1
radians and
T
2πt 2
Phase lag =
radians
T
Phase lead =
Figure 9.9 A procedure of experimental phase measurement with the help of oscilloscope
387
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
Example 9.3 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: TR = 0.3 kg at 300 phase , which can be written as
1
TR1 = 0.3(cos30 + j sin 30) = ( 0.2598 + j 0.15 ) kg
Displacement in plane 1: R2 = 50 µm at 610 phase , which can be written as
R2 = ( 24.2404 + j 43.73) µm
Displacement in plane 2: L2 = 42 µm at 1300 phase , which can be written as
L2 = ( −26.997 + j 32.173) µm
Measured responses due to residual imbalances are
In plane 1: R1 = −30µm at 2300 phase ≡ R1 = (19.2836 + j 22.98 ) µm
In plane 2: L1 = −70µm at 3300 phase ≡ L1 = ( −60.621 + j 35 ) µm
We have, influence coefficients as
α11 = α bR =
R2 - R1
= (48.8919 + j 51.63766) × 10−6 µm/kg
TR1
α12 = α aR =
L2 - L1
= (92.3559 − j 69.1995) × 10−6 µm/kg
TR1
and
388
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
It is given that machine is symmetric about centreline.
α 21 = α12 and α 22 = α11
Measurements, influence coefficients and correction mass are related as
 − R1  α11 α12   wR 

=
 
 − L1  α 21 α 22   wL 
which can be simplified as
wR =
1
1
{L1α12 − R1α 22 } and wL = {R1α 21 − L1α11}
∆
∆
with
2
2
∆ = α11α12 − α12 α 22 = α11
− α12
= ( −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 × 10-3 kg at ???0
Exercise 9.3 Obtain the correction masses in the right and left planes for a rigid rotor by using the
influence coefficients balancing method for the following observations: (i) The original unbalance
vectors as measured for a turbine rotor are: L1 = 9.144×10-3 mm peak to peak at 900 and R1 = 10.16
×10-3 mm peak to peak at 450 (ii) The trial mass in right plane TR = 6.8 gm at 22.50: L2 = 5.08 ×10-3
mm peak to peak at 270 and R2 = 6.35 ×10-3 mm peak to peak at 990 (iii) The trial mass in left plane TL
= 6.8 gm at 360: L3 = 9.40×10-3 mm peak to peak at 00 and R3 = 30.5 ×10-3 mm peak to peak at 990.
Exercise 9.4 For finding unbalance in the rotor the following measurements were made in the right
and left balancing plane (i). Without trial mass: L1 = 0.0015 mm ∠ 1500 ; R1 = 0.0007 mm ∠ 1620 (ii)
Trial mass at left plane TL = 29 gm ∠00 : L2 = 0.0015 mm ∠1100 ; R2 = 0.0015 mm < 1700 (iii) Trial
389
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
mass at right plane TR = 29 gm ∠00 : L3 = 0.0025 ∠ 2000; R3 = 0.0015 ∠ 1700. Obtain the influence
coefficient and correction masses in the right and left balancing planes.
Exercise 9.5 For the dynamic balancing of a rotor by using influence coefficients method, if the
uncertainty in the measurement of the vibration amplitude and phase are 2 and 5 percent, respectively.
Calculate the uncertainty of the magnitude and the phase of influence coefficients and correction
masses. Give all the intermediate formulations
9.2 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.
9.2.1 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. 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.
(a) Rigid body modes
1
(b) First flexible mode
2
(c) Second flexible mode
Figure 9.10 Typical rigid and flexible modes of rotor-bearing systems
390
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
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 speed.
Note the readings. Add a pair of trial masses 1800 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. Similarly higher modes can be balanced i.e. upto 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.
z
x
bearing axis
y
Figure 7.11 Rotor-brearing axis system
Assume that all unbalance is distributed only in the x-y plane (Figure 9.11). Let the rotor speed be ω
and the deflection of the rotor be y(x). The deflection y(x) can be written in terms of summation of
mode shapes as
y ( x) = ∑ φiYi ( x)
(9.8)
where Yi ( x) is the mode shape in the ith mode and φ i is an unknown constant. The deflection y(x) can
be measured experimentally. For example, for simply supported end conditions, mode shapes are
πx
I - mode
y1 ( x) = sin
II - mode
y2 ( x) = sin
2π x
l
ith - mode
yi ( x) = sin
iπ x
l
l
391
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
For other end conditions mode shapes can be obtained by the free vibration analysis. Modal series for
eccentricity can also be written as
a( x) = ∑ α iYi ( x )
(9.9)
i
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 α i for the eccentricity distribution to be known. Multiplying (9.8) by the mass per unit
length m(x) and the mode shape Y j ( x) would shape and integrate from 0 to l.


∫0 m( x) y( x)Y j ( x) = ∫0 m( x) ∑i φiYi ( x)Y j ( x)dx
l
l
(9.10)
Noting the orthogonality condition of mode shapes, we have
l
∫ m( x)Y ( x)Y ( x)dx = 0 for all i ≠
i
j
j
(9.11)
0
Equation (9.10) gives
l
l
l
∫ m( x) y( x)Y j ( x)dx = ∫ m( x)φ jY j ( x)dx =φ j ∫ m( x)Y j ( x)dx
2
0
2
0
(9.12)
0
which can be written as
l
φj =
1
m( x) y ( x )Y j ( x)dx
M j ∫0
(9.13)
with
l
M i = ∫ m( x)Y j2 ( x)dx
0
where M j is the generalized mass in jth mode and it can be obtained by knowing m( x) and Y j ( x) .
The m( x) can be obtained by free vibration analysis and y ( x) theoretically speaking can be found by
experiment. Governing equation for the shaft motion is given as
392
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
d2
{EI ( x) y′′( x)} = ω 2 m( x){ y ( x) + a( x)}
dx 2
(9.14)
where ω is the rotor speed. On substituting for y(x) and a(x) from equations (9.8) and (9.9)
respectively, we get


d2 



EI ( x ) ∑ φiYi ′′( x)  = ω 2 m( x )  ∑ φiYi ( x )  + {∑ α iYi ( x )}
2 
dx 
 i


 i

(9.15)
Noting the orthoganality condition and multiply both sides by Y j ( x) and integrate over the length of
shaft, the left hand side of equation (9.15) gives
l
d2 


EI ( x ) ∑ φiYi″ ( x )   Y j ( x)dx
2 
 i


∫ dx
0
On performing integration by parts, we get
Y j ( x)
l
l
d
d 


 EI ( x ) {∑ φiYi ′′( x)} − ∫  EI ( x) ∑ φiYi ′′( x) Y j′′( x)dx


0
dx
 i

0 dx 
First term vanishes (since it is zero for all boundary conditions), on taking integration by parts, we get
[
− Y j′ ( x) EI ( x)
{∑φ Y ′′( x)}] + ∫ EI ( x)∑φ Y ′′( x)Y ′′( x)dx




l
i i
l
i i
0
0
j
i
First term again vanishes for all boundary conditions. On noting the orthogonality condition equation
(9.11), we get
l
[
∫
]
= φ j EI ( x) Y j′′( x) 2 dx = φ j K j
(9.16)
0
where the generalized stiffness in jth mode is defined as
l
Kj =
∫ EI ( x)[Y ′′( x)] dx
2
j
0
393
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
From right hand side of equation (9.15), noting equation (9.8) and (9.9), we have
ω 2 (φ j + α j )M j
(9.17)
Therefore from (9.15), noting equations (9.16) and (9.17), we have
φ j K j = ω 2 (φ j + α j )M j
which can be rearranged as
αj =
K j / M j −ω2
ω2
φj
(9.18)
where α j is an known, then distribution of eccentricity a (x) can be found from equation (9.9). The
equation (9.18) requires m(x), y(x), Y j ( x) and p j = K j / M j . The m(x) can be accurately found out,
y(x) is difficult to obtained, Y j ( x) is obtained by the eigen analysis and p j ( x) is natural frequency in
ith mode = K j / M j which can be obtained by the eigen analysis.
9.7.2 Influence Coefficient Methods
Figure 9.12 shows the effect of deflection of the shaft on influence coefficients. It can be seen that
they depend upon shaft speeds (especially near critical speeds).
F1
(a)
2
1
α 21
α11
ω << p1
2
1
′
α 21
′
α11
ω ≈ p1
(b)
(c)
2
′′
α 21
ω = p2
Figure 7.12 Effect of mode shapes on influence coefficients
394
F1
′′
α11
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
p
x
q
z
Figure 9.13 Measurement locations (q) and balancing planes (p)
Choose p number of balancing planes (where mass can be added or chip-off) where p>2, q number is
measuring planes; generally it is two i.e. at bearing planes. Let the unbalance ( me ) in each of the
balancing planes be U 1 , U 2 , … , U p
 v1  α11 α12
v  α
 2   21 α 22
 =
   vq  α q1 α q 2
… α1 p   U 1 
 
… α 2 p  U 2 
 
 
 α qp  U p 
(9.19)
where v is the vibration measurement at the measuring plane. Measurements are taken at number of
speeds. On writing equation (9.19) for each of the speeds, we get
{ν } = [α ]{U }
(9.20)
with
395
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
1
α11
 1
α 21
  1
α q1
α 2
 11
 [α ] = α 2
 q1
  n
α11
α 21n

 α n
 q1
 v11 
 1 
 v2 

 1 
 vq 
v 2 
1

{v} =  2  ;
vq 

 n 
v1 
v n 
1

v n 
 q
α121 … α11p 
1

α 22
… α 21 p 

1
1 
α q 2 … α qp 
α122 … α12p 


;
α q22 … α qp2 



α12n … α1np 
α 22n α 2n p 


α qn2 … α qpn 
 U1 
 
U2 
[U ] = U 3 
 
 
U p 
Once the influence coefficients [α ] are known for all speeds equation (9.20) can be used to obtained
unbalances :
(
)
−1
{U } = [α ]T [α ] {v}
(9.21)
Influence coefficient matrix can be obtained by attaching trial mass and measuring displacement, from
equation (9.19), we get for a particular speed
1
1
 v111  α 11 α 12
1
 1   1
v21  α 21 α 22
 =
   1
vq11   1
  α q1 α q 2
… α 1  U + T 
1p
1
R
1   
… α   U2 
2p


  
1   U p 

… α qp
 
(9.22)
On subtracting equation (9.22) from first q equation in equation (9.19), we get
α 1 α 1
 v111 − v11   11
12
 1 1  α1 α1
v21 − v2 
21
22

=

  vq11 − vq1  α1 α1

  q1
q2

… α1  1 p T 
1   R 
… α  0 
2p
 
 

… α 1   0 
qp 
Equation (9.23) gives
396
(9.23)
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, (rtiwari@iitg.ernet.in)
vq11 − vq1
1
v111 − v11 1 v211 − v21
α 11 =
, α 21 =
,… , α =
q1
TR
TR
TR
1
(9.24)
Similarly by attaching a trial mass on plane 2 we get second column of the influence coefficient
matrix in equation (9.24), the above analysis should be done at a constant speed. Similarly we can
find the influence coefficient-matrix for different speeds.
Exercise 9.6 Justify your answer for the following cases: a) For the dynamic balancing of a flexible
rotor at one particular speed how many minimum numbers of balancing-planes are required? b) For
the dynamic balancing of a flexible rotor, in general, how many minimum balancing-planes are
required? c) For flexible shaft whether imbalance changes with the shaft speed? d) Apart from
balancing the rotor what are the other methods by which the amplitude of the synchronous whirl can
be reduced?
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, Balancing of High-Speed Machinery, Springer-Verlag.
[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.
[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
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