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Dr. R. Tiwari ([email protected]) 1
ByDr. Rajiv Tiwari
Department of Mechanical EngineeringIndian Institute of Technology Guwahati 781039
Under AICTE Sponsored QIP Short Term Course on
Theory & Practice of Rotor Dynamics(15-19 Dec 2008)
IIT Guwahati
DYNAMIC BALANCING OF ROTORS
Dr. R. Tiwari ([email protected]) 2
IntroductionThe 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 wherethe downtime is very expensive [6,7].
Dr. R. Tiwari ([email protected]) 3
Static balancing: Single plane balancingDynamic balancing :
(i) Two plane balancing: For rigid rotors only. (ii) Flexible rotor balancing : If the shaft deflects, and the
deflection changes with speed, as it does in the vicinity of critical speeds .
)( crωω <
)( crωω >
Dr. R. Tiwari ([email protected]) 4
Dr. R. Tiwari ([email protected]) 5
Dr. R. Tiwari ([email protected]) 6
Dr. R. Tiwari ([email protected]) 11
Balancing of Rigid RotorCradle 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 ([email protected]) 12
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 ([email protected]) 13
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 consuming.
Dr. R. Tiwari ([email protected]) 14
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 � = ±90°, where � 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 ).
(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 ([email protected]) 15
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 ([email protected]) 16
Construction Procedure :
Figure 3 Construction procedure
D
A
B
O
E
C
E’
~900
φ
α
Dr. R. Tiwari ([email protected]) 17
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 theparallelogram 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).
The magnitude of trial mass Wt is proportional to AB.
Dr. R. Tiwari ([email protected]) 18
The unbalance OA can be obtained accordingly in the mass term. The location of unbalance is 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.
OAB∠
Dr. R. Tiwari ([email protected]) 19Figure 4
00 900EB
OLocation of fourth measurement for trial mass at 900
(ccw dir. +ve)
900
00
2700
1800
+ve
Unbalance position at shaft location, (ccw direction +ve)
A
Location of fourth measurement for trial mass at 900
(cw dir. +ve)
900
00
2700
1800
+ve
Unbalance position at shaft location (ccw dir. +ve)
00
2700
E’
B
OLocation of fourth measurement for trial mass at 2700
(ccw dir. +ve)
900
00
2700
1800
+ve
Unbalance position at shaft location, (ccw direction +ve)
Location of fourth measurement for trial mass at 2700
(cw dir. +ve)
900
00
2700
1800
+ve
Unbalance position at shaft location (ccw dir. +ve)
E900
00B
O
2700
E’B
O00
Dr. R. Tiwari ([email protected]) 20
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
(ii) amplitude with unit trial mass at an angle of
(iii) amplitude with unit trial mass at an angle of
(iv) x = amplitude with 1 at an angle of and without residual imbalance(i.e. ), and
RU ϕ∠
00
0180
000RU = 0OA a= 1AB a= 2AC a=
Dr. R. Tiwari ([email protected]) 21
Figure 3.5 shows variation parameters involved in the present problem. From OAB∆
, we have
2 2 20 1
0
cos2
a OB aa OB
ϕ + −=
(A)
and
( )2 2 20 2
0
cos2
a OC aa OC
π ϕ + −− = (B)
Since OB OC= , we have
2 2 20 2cos
20
a OB a
a OBϕ
+ −− =
(C)
Dr. R. Tiwari ([email protected]) 22
0aB
A
C
x
x
D
Figure .5 Geometrical constructions for determination of unbalance
Reference line 2a
1a
0ϕ
On equating equations (A) and (C), we get
2 2 2 2 2 20 0 2 0 12 cos ( ) ( )a OB a OB a a OB aϕ = − + − = + −
which gives
2 2 2 20 1 22 2 ( )a OB a a+ − +
(D)
Dr. R. Tiwari ([email protected]) 23
OB x= , since OB (or OC ) are effect of trial mass of unit magnitude.
Hence equation (D) gives
2 2 2 21 2 0( ) / 2x a a a= + − or 1 2 2 2
1 2 02( )x a a a= ± + −
Equations (A) and (C) gives (noting that OC OB x= = ),
2 2 20 0 22 cosx a x a aϕ= − − +
2 2 20 0 12 cosx a x a aϕ= − +
and
On equating equations (F) and (G), we get
2 21 2 0cos ( ) / 2a a a xϕ = − +
(E)
(F)
(G)
(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.
Dr. R. Tiwari ([email protected]) 24
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” amplitude 6.0 µmRun 2; with 5gm. at 0 deg. amplitude 5.0 µmRun 3; with 5 gm. at 180 deg. amplitude 10.0 µmRun 4; with 5gm. at 90 deg. amplitude 10.5 µm
Find the weight and location of the correction. Take the trial and balancingmasses at the same radius.
Dr. R. Tiwari ([email protected]) 25
Answer:
O
A
D
B
C
OA = AD = 6 cmDB =10 cmOB = 5 cmOE = 10.5 cmAB = 6.3 cmImbalance position = angle BAO = 71 deg. CCW
���
E
Reference line
AB = 6.3 cm ≡ 5 gmHence, the residual imbalance is given asOA = AD = 6 cm ≡ 4.762 gm
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
Figure 3.6 Geometrical constructions
Dr. R. Tiwari ([email protected]) 26
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
1 2 2 21 2 02
( ) 4.09 gmx a a a= ± + − =
and
2 21 2 0cos ( ) / 2 0.6065a a a xϕ = − + =
52.8degϕ =
Dr. R. Tiwari ([email protected]) 27
(7)
Definition of Influence coefficients (i) Only force F1 (ii) Only force F2
y11 = displacement at station 1 due to y12 = force F1 at station1 = y22 =
y21 = displacement at station 2 due to force F1 at station1 =
1F1
y11 y21
2 1F2
y12
y22
2
111Fα
121Fα
212Fα
212Fα
The Influence Coefficient Method
Dr. R. Tiwari ([email protected]) 28
(iii) When both F1 and F2 are present
Figure 7
Influence coefficients can be obtained by experimentation or by strength of formulae i.e.
F2
y1 y2
F1
1 11 12 11 1 12 2
2 21 22 21 1 22 2
1 11 12 1
2 21 22 2
y y y F F
y y y F F
y For
y F
α αα α
α αα α
= + = += + = +
� � � � � �=� � � �� �
� �
1 1 2 11 1 2 1
1 1
, .y y
e t cF F
α α= =
Dr. R. Tiwari ([email protected]) 29
The Influence Coefficient Method
R
a bL1
�1
L1L2
�2
R1
�1
R1
R2
�2
�aR
�bR
Trial mass TL
L1
�aL
L3
�3
�bL
R1R3
�3
Figure 5 Bearing measurements and influence coefficients for a rigid rotor
L
No trial mass
Dr. R. Tiwari ([email protected]) 30
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 ([email protected]) 31Figure 6
Signal from station a shaft
Spike due to notch in the shaft surface reference signal
Phase = 0 with respect to notch
T t1
t2
Phase lead = radians
Phase lag = radians
Tt12π
Tt22π
Dr. R. Tiwari ([email protected]) 32
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
(1)
where ‘�’ represent vector since displacement has magnitude and phase information. Similarly
(2)
We remove the trial mass from plane R and place in plane L and repeat the test to obtain the measured values
bRα
RbR TRR����
/)( 12 −=α
RaR TLL����
/)( 12 −=α
LT�
Dr. R. Tiwari ([email protected]) 33
(3,4)With the help of equations (1) to (4), we can obtain influence coefficient experimentally.
Let the correct balance masses be and . Since the original unbalance response is R1 and L1 as measured in right and left planes, we can write
(5)
Correction masses will produce vibration equal and opposite to the vibration due to unbalance masses. Hence,
(6)
These can be calculated either by a graphical method or analytical method of vectors (complex algebra) i.e.
3 1 3 1( ) / ( ) /bL L aL LR R T and L L Tα α= − = −� � � � � �� �
RW�
LW�
1 1 and R bR L bL R aR L aLR W W L W Wα α α α− = + − = +� � � � � �� � � �
��
����
���
�−=
��
��
L
R
aLaR
bLbR
W
W
L
Rαααα��
��
�
�
1
1
Dr. R. Tiwari ([email protected]) 34
which gives
11
w h ere ( )a b d b
a d cbc d c a
− −� � � �= ∆ = −� � � �−∆� � � �
1 1 1 1. . . .. . . .
bL aL aR bRR L
bR aL aR bL bR aL aR bL
L R R LW and W
α α α αα α α α α α α α
− −= =− −
� � � �� � � �
� � � � � � � �
Dr. R. Tiwari ([email protected]) 35
Experimental set-up for influence coefficient method of balancing
Figure 8 Experimental set-up for influence coefficient method of balancing
ChargeAmplifier
Vibration meter
Phasemeter
Accelerometer
MeasurementPlane a
MeasurementPlane b
Phase mark on shaft
L R
Photo electric Probe
Accelerometer (or proximity probes on the shaft near to the bearing)
�
Oscilloscope
Hardware or virtual instrumentationPhoto sensitive mark
Photo electric probe
Maximum displacement location
Dr. R. Tiwari ([email protected]) 36
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 atend 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: 1
00.3 kg at 30 phaseRT = , which can be written as
( )1
0.3(cos30 j sin 30) 0.2598 j 0.15 KgRT = + = +
Displacement in plane 1: 02 50 �m at 61 phaseR = , which can be written as
( )2 24.2404 j 43.73R = + µm
Displacement in plane 2: 02 42 �m at 130 phaseL = , which can be written as
( )2 26.997 j 32.173L = − + µm
Dr. R. Tiwari ([email protected]) 37
Measured responses due to residual imbalances are
In plane 1: 01 30�m at 230 phaseR = − ( )1 19.2836 j 22.98R≡ = + µm
In plane 2: 01 70�m at 330 phaseL = − ( )1 60.621 j 35L≡ = − + µm
We have, influence coefficients as
1
62 111
-(48.8919 j 51.63766) 10bR
R
R RT
α α −= = = + × µm/kg
and
1
62 112
-� (92.3559 j 69.1995) 10aR
R
L LT
α −= = = − × µm/kg
It is given that machine is symmetric about centreline.
21 12 22 11 and α α α α= =
Dr. R. Tiwari ([email protected]) 38
Measurements, influence coefficients and correction mass are related as
1 11 12
1 21 22
R
L
R w
L w
α αα α
−� � � � � �=� � � �� �− � �
which can be simplified as
{ } { }1 12 1 22 1 21 1 11
1 1 and R Lw L R w R Lα α α α= − = −
∆ ∆
with ( )2 2 611 12 12 22 11 12� � � � � � 468.066 j 16907.73 10−∆ = − = − = − + × (µm/kg)2
which gives the balancing mass and its angular position as
3 3 -3 0R 3.3519 10 j 7.123 10 7.893 10 kg at 295w − −= − × + × ≡ ×
and3 3 -3 0
L 3.90356 10 j 2.7136 10 4.7541 10 kg at -35w − −= × − × ≡ ×
Dr. R. Tiwari ([email protected]) 39
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 methodA 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 ([email protected]) 40
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 speed.
Dr. R. Tiwari ([email protected]) 41
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.
1 2 Rigid body modes
Dr. R. Tiwari ([email protected]) 42
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)
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).
x
z
y
bearing axis
Dr. R. Tiwari ([email protected]) 43
The deflection y(x) can be written in terms of summation of mode shapes as(8)
where 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 areI Mode II Mode
i th Mode
= )()( xYxy iiφ
)(xYi
lx
xyπ
sin)(1 =lx
xyπ2
sin)(2 =
lxi
xyiπ
sin)( =
Dr. R. Tiwari ([email protected]) 44
For other end conditions mode shapes can be obtained by free vibration analysis. Modal series for eccentricity can also be written as
(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 and integrate from 0 to 1.
(10)
noting the orthogonality condition of mode shapes
(11)
=i
ii xYxa )()( α
( )jY x
� � ��
���
�=l l
ji
iij dxxYxYxmxYxyxm0 0
)()()()()()( φ
0
( ) ( ) ( ) 0l
i jm x Y x Y x dx for all i j= ≠�
Dr. R. Tiwari ([email protected]) 45
Equation (10) gives
(12)
which can be written as (13)
with the generalized mass in jth mode is given as
The generalized mass can be obtained by knowing can be obtained by free vibration analysis and theoretically speaking can be found by experiment.
��� ==l
jj
l
jj
l
j dxxYxmdxxYxmdxxYxyxm0
2
0
2
0
)()()()()()()( φφ
( )�=l
jj
j dxxYxyxmM 0
)()(1φ
( )jM=
�=l
ji dxxYxmM0
2 )()(
jM ( )and ( ); ( )j jm x Y x Y x
Dr. R. Tiwari ([email protected]) 46
Governing equation for shaft motion is
(14)
where � is the rotor speed. On substituting for y(x) and a(x) from equation (8) and (9), we get
(15)
Noting the orthogonality condition and multiply both sides by and integrate over the length of shaft, the left hand side of equation (15) gives
On performing integration by parts, we get
[ ] [ ])()()()()( 22
2
xaxyxmxyxEIdx
d +=′′ ω
{ }���
�
���
�+
��
��
=���
�
���
�
��
��
′′ )()()()()( 22
2
xYxYxmxYxEIdx
dii
iii
iii αφωφ
)(xYi
� ��
�
�
��
�
�
�
���
�
��� ″
l
j
i
ii dxxYxYxEIdx
d
0
2
2
)()()( φ
Dr. R. Tiwari ([email protected]) 47
First term vanishes (since it is zero for all boundary conditions), on taking again integration by parts of the second term, we get
First term again vanishes for all boundary conditions. On noting the orthogonality condition equation (11), we get
(16)
where the generalized stiffness in jth mode is defined as
{ }[ ] dxxYxYxEIdxd
xYxEIdxd
xY j
l
i
ii
l
iij )()()()()()(0
0′′
��
�
�
��
�
�
�
���
�
���
′′−′′ � φφ
{ }[ ] dxxYxYxEIxYxEIxYl
j
i
ii
l
iij � ��
�
�
��
�
�′′
�
���
�
���
′′+′′′−0
0)()()()()()( φφ
[ ] jj
l
jj KdxxYxEI φφ =′′= �0
2)()(
[ ] dxxYxEIKl
jj � ′′=0
2)()(
Dr. R. Tiwari ([email protected]) 48
From right hand side of equation (15), noting equation (8) and (9), we have
(17)Therefore form (15), noting equations (16) and (17), we have
which can be rearranged as
(18)
Once is obtained, then distribution of eccentricity a(x) can be found from equation (9). Equation (18) requires m(x), y(x), .
The m(x) can be accurately found out, y(x) is difficult to obtained, is obtained by eigen analysis and is natural frequency in ith mode = which can be obtained by eigen value analysis.
( ) jjj Mαφω +2
( ) jjjjj MK αφωφ += 2
)(/
2
2
jjj
j
MKφ
ωω
α−
=
jα( ) /j j j jY x and p K M=
( )jY xjp K j
M j
� �� �� �� �
Dr. R. Tiwari ([email protected]) 49
p
x
q
z
Influence coefficient method
Figure 11
Figure 12
2
1212
F1
F1
21α 11α
1p<<ω21α ′ 11α ′
1p≈ω
11α ′′
21α ′′pω ≈ 2
Rigid shaft
Flexible shaft
Measuring planes
Balancing planes
Dr. R. Tiwari ([email protected]) 50
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
where is the vibration measurement at the measuring plane. Measurements are taken at number of speeds.
On writing equation (19) for each of the speeds
pUUU�
���
,,, 21
��
��
�
�
��
��
�
�
�����
�
�
�����
�
�
=
��
��
�
�
��
��
�
�
pqpq
p
p
q U
U
U
v
v
v
��
�
�
��
��
��
��
��
��
��
�2
1
1
22221
11211
2
1
αα
αααααα
υ
(19)
Dr. R. Tiwari ([email protected]) 51
(20)
Once the influence coefficients are known for all speeds equation (20) can be used to obtained unbalances :
(21)
1 1 1 11 1 1 1 2 11 1 1 12 2 1 2 2 2
1 1 1 11 2
2 2 2 21 1 1 1 2 1
2 2 2 21 2
1 1 1 1 2 1
1 2 1 2 2 2
p
p
q q q q p
p
q q q q p
n n n np
n n n np
nq
v
v
v
v
v
v
v
v
α α αα α α
α α αα α α
α α α
α α αα α α
� �� �� �� �� �� �� �� �� � =� �� �� �� �� �� �� �� �� �
� � ���
� � ���
� � � �� � ��
�� � ��
�
� � � � �� � ��
�
� � � � �� � ��
�� � ��
�
� � ��
{ }
1
2
3
1 2
[ ] { }
p
n n nq q q p
U
U
o r UU
U
ν α
α α α
� �� �� �� �� �� � � �� � � �� � � �� � � � =� �� �
� �� �� �� �� �� � � �
� �� �� �� �� �
�
�
�
��
� �� � �
�
][α
( ) }{][][}{1
vU T −= αα
Dr. R. Tiwari ([email protected]) 52
Influence coefficient matrix can be obtained by attaching trial masses and measuring displacements,
from equation (19), we get for a particular speed
(22)
On subtracting equation (22) from first q equation in equation (19), we get
(23)
Equation (23) gives(24)
��
��
�
�
��
��
�
� +
������
�
�
������
�
�
=
��
��
�
�
��
��
�
�
p
R
qpqq
p
p
q U
U
TU
v
v
v
��
�
��
��
������
��
��
��
��
��
�2
1
112
1
1
1
2
1
22
1
21
1
1
1
12
1
11
11
121
111
ααα
ααα
ααα
��
��
�
�
��
��
�
�
������
�
�
������
�
�
=
��
��
�
�
��
��
�
�
−
−−
0
0
11
2
1
1
1
2
1
22
1
21
1
1
1
12
1
11
111
12
121
11
111
�
�
��
������
��
��
��
��
���
��
�R
qpqq
p
p
T
vv
vv
vv
ααα
ααα
ααα
,11
1111
11RT
vv�
�� −
=α ,,12
1211
21 ��
��
RT
vv −=α
R
q T
vv�
��
1111
1
−=α
Dr. R. Tiwari ([email protected]) 53
Similarly by attaching a trial mass on plane 2 we get second column of the influence coefficient matrix in equation (24), theabove analysis should be done at a constant speed.
Similarly we can find the influence coefficient-matrix for different speeds.
Dr. R. Tiwari ([email protected]) 54
Dr. R. Tiwari ([email protected]) 55
Dr. R. Tiwari ([email protected]) 56
Dr. R. Tiwari ([email protected]) 57
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 ([email protected]) 58
[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 ([email protected]) 59
[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