Rtiwari Rd Book 09

8/10/2019 Rtiwari Rd Book 09

1/21

Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])

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 )( cr< .

o Flexible rotor balancing: If the shaft deflects, and the deflection changes with speed,

as it does in the vicinity of critical speeds )( 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.

Figure 9.1 Craddle balancing machine

The cradle is placed on two springs and can be fulcrum aboutF1orF2to 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

F F

I II

F1 F2+ +


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378

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 planeI, by fixingF1and releasingF2. Run the rotor to resonance,

observing the maximum amplitude to the right of fulcrum F2. This vibration is due to all the

unbalance in planeII, since the unbalance in planeIhas no moment aboutF1. Use a trial mass at a

chosen location and determine the amplitude of vibration.

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 FixingF2and releasingF1. This procedure is tedious and sometimes

may be time consuming. A procedure to determine the correction mass and location can be laid downas follows, based on four observations of amplitude : (i) without any addition to the rotor (ii) with a

trial mass at = 00(iii) with a trial mass at 1800and (iv) with same trial mass at = 900, where is

measured from a conveniently chosen location. This procedure has to be repeated for two cases (e.g.

when fulcruming atF1and then forF2). 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 Wtat 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 180

0phase difference between them (Hence AC vector

is also the effect of trial mass Wtso 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.

Minimumam litude

Amplitudeof

vibration

Trial mass location


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379

Construction procedure

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 Wtis 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 cradlepivoted at FIIand 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.

D

B

O

E

C

E

~900


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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 a1but 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 . Solvexand in terms of 0 1,a a and 2a showthat in

this answer there is an ambiguity sign. Thus four runs are necessary to solve the problem completely.

Answer:Measurements are

(i) =0a amplitude of vibration with residual unbalance RU

(ii) =1a amplitude with unit trial mass at an angle of00

(iii) =2a amplitude with unit trial mass at an angle of0180

(iv)x= amplitude with 1 at an angle of00 and without residual imbalance (i.e. 0RU = )

0aOA= , 1aAB= and 2aAC=

00 900E

BO

Location of fourthmeasurement for trial mass

at 900(ccw dir. +ve)

900

00

2700

1800

+ve

Unbalance position at shaft

location, (ccw direction +ve)

A

Location of fourthmeasurement for trial

mass at 900(cw dir. +ve)

900

00

2700

1800

+ve

Unbalance positionat shaft location

(ccw dir. +ve)

0

2700

E

B

OLocation of fourthmeasurement for trial mass

at 2700(ccw dir. +ve)

90

00

270

1800

+ve

Unbalance position at shaft

location, (ccw direction +ve)

A

Location of fourthmeasurement for trial

mass at 2700(cw dir.

+ve)

90

00

270

1800

+ve

Unbalance positionat shaft location

(ccw dir. +ve)

E900

00B O

270

EB

O0


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Figure 9.5 shows variation parameters involved in the present problem. From OAB , we have

2 2 2

0 1

0

cos2

a OB a

a OB

+ = (A)

and

( )2 2 2

0 2

0

cos2

a OC a

a OC

+ = (B)

Since COBO = , we have

2 2 2

0 2

0

cos2

a OB a

a OB

+ = (C)

On equating equations (A) and (C), we get

2 2 2 2 2 2

0 0 2 0 12 cos ( ) ( )a OB a OB a a OB a= + = +

which gives

)(22 222

1

22

0 aaBOa ++ (D)

xBO = , since BO (or CO ) are effect of trial mass of unit magnitude. Hence equation (D) gives

2

0

2

2

2

1

2 2/)( aaax += or 1 2 2 21 2 02 ( )x a a a= + (E)

B

C

0a 2a

1a

0

D

Figure 7.5 Geometrical constructions fordetermination of unbalance

Reference

line


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382

Equations (A) and (C) gives (noting that xBOCO == ),

2

1

2

00

2 cos2 aaxax += (F)

and

2

2

2

00

2 cos2 aaxax += (G)

On equating equations (F) and (G), we get

xaaa 02

2

2

1 2/)(cos += (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:

Run 1; rotor as is amplitude 6.0 mRun 2; with 5gm. at 0 deg. amplitude 5.0 mRun 3; with 5 gm. at 180 deg. amplitude 10.0 mRun 4; with 5gm. at 90 deg. amplitude 10.5 m

Find the weight and location of the correction. Take the trial and balancing masses at the same radius.

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

71

E

Refere

nceline

Figure 9.6 Geometrical constructions


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383

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 = 710CCW direction.

The unbalance magnitude and phase can be also obtained from equations (E) and (H), we have

1 2 2 2

1 2 02( )x a a a= + = ? kg

and

xaaa 02

2

2

1 2/)(cos += = ? 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 m

for the rotor without additional weights, (ii) 10 m with 5 gm placed in location 0 deg, (iii) 22 m

with 5 gm placed in location 90 deg and (iv) 22 m with 5 gm placed in location 180 deg. Find the

amount and angular location of the necessary correction mass.

Exercise 9.2A 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 m for the rotor without additional weights

2. 10 m with 5 gm placed in location 0 deg.

3. 22 m with 5 gm placed in location 90 deg

4. 22 m with 5 gm placed in location 180 deg

Find the amount and angular location of the necessary correction mass.


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384

9.1.2The Influence Coefficient Method

Definition of Influence coefficients: Figure 9.7(a) shows that when a forceF1is 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 forceF1at station 1 = 111F

and

y21= displacement at station 2 due to forceF1at station1 = 121F

where 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= 212F and y22= 222F

(i) Only forceF1 (ii) Only forceF2

(iii) When bothF1andF2are present

Figure 9.7 Definition of influence coefficients

In Figure 9.7(c), we have

1 12 12 11 1 12 2

2 21 22 21 1 22 2

y y y F F

y y y F F

= + = +

= + = +

1 11 12 1

2 21 22 2

y F

F

=

Influence coefficients can be obtained by experimentation or by strength of formulae i.e.

11 2111 21

1 1

,y

F F = = etc.

1F1

11 22

2 1F2

1222

2

F2

1 2

F1


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385

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 1 and 1 respectively.

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 L2and R2 with phase 2 and 2 respectively in the a

& b planes. The difference between R2and 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 bR .

RbR TRR

/)( 12= (9.1)

where "" represent vector since displacement has magnitude and phase information. Similarly

R

a bL1

1

L1L2

2

R1

1

R1

R2

2

aR

bRTrial mass TR

Trial mass TL

L1

aL

L3

3

bL

R1R3

3

Figure 7.7 Bearing measurements and influence

coefficients for a rigid rotor

L

No trial mass


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386

RaR TLL

/)( 12= (9.2)

We remove the trial mass from plane R and place LT

in plane L and repeat the test to obtain the

measured values

LbL TRR

/)( 13= and LaL TLL

/)( 13= (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 RW

and LW

. Since the original unbalance response is R1 and L1 as

measured in right and left planes, we can write

1 1 andR bR L bL R aR L aLR W W L W W = + = +

(9.5)

Correction masses will produce vibration equal and opposite to the vibration due to unbalance masses.

Hence,

=

L

R

aLaR

bLbR

W

W

L

R

1

1 (9.6)

These can be calculated either by a graphical method or analytical method of vectors (complex

algebra) i.e.

=

ac

bd

dc

ba 11

)(where cbad=

which gives

bLaRaLbR

aLbLR

RLW

..

.. 11

= and

bLaRaLbR

bRaRL

LRW

..

.. 11

= (9.7)


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387

Signal from station a shaft

Spike due tonotch in the

shaft surface

reference signal

Phase lag = 0 with respect to the

notch

T t1

t2

Phase lead =T

t12 radians and

Phase lag =Tt22 radians

Figure 9.8 Experimental set-up for the influence coefficient method of balancing

Figure 9.9 A procedure of experimental phase measurement with the help of oscilloscope

Charge

AmplifierVibration

meterPhase

meter

Accelerometer

Measurement

Plane a

Measurement

Plane b

Phase mark

on shaftL R

Photo electricProbe

Accelerometer(or proximity

probes on theshaft near to the

bearing)

Oscilloscope

Hardware or virtual instrumentationPhoto sensitive

mark

Photo electric probe

Maximum

displacement

location


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388

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 300relative to some reference position; this causes changes in vibration vectors of 50 m at 610at

end 1 and 42 m 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 measuredimbalance vibrations are 30 m at 2300at end 1 and 70 m at 3300at 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(cos 30 j sin 30) 0.2598 j 0.15RT = + = + kg

Displacement in plane 1:0

2 50 m at 61 phaseR = , which can be written as

( )2 24.2404 j 43.73R = + m

Displacement in plane 2: 02 42 m at 130 phaseL = , which can be written as

( )2 26.997 j 32.173L = + m

Measured responses due to residual imbalances are

In plane 1: 01 30m at 230 phaseR = ( )1 19.2836 j 22.98R = + m

In plane 2: 01 70m at 330 phaseL = ( )1 60.621 j 35L = + m

We have, influence coefficients as

1

62 111

-(48.8919 j 51.63766) 10bR

R

R R

T

= = = + m/kg

and

1

62 112

- (92.3559 j 69.1995) 10aR

R

L L

T = = = m/kg


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389

It is given that machine is symmetric about centreline.

21 12 22 11and = =

Measurements, influence coefficients and correction mass are related as

1 11 12

1 21 22

R

L

R w

L w

=

which can be simplified as

{ } { }1 12 1 22 1 21 1 111 1

andR Lw L R w R L = =

with

( )2 2 611 12 12 22 11 12 468.066 j 16907.73 10 = = = + (m/kg)2

which gives the balancing mass and its angular position as

3 3 -3 0

R 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 ???w =

Exercise 9.3Obtain 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= 9.14410-3mm peak to peak at 900andR1= 10.16

10-3mm 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 270andR2= 6.35 10-3mm peak to peak at 990(iii) The trial mass in left plane TL

= 6.8 gm at 360

:L3= 9.4010-3

mm peak to peak at 00

andR3= 30.5 10-3

mm peak to peak at 990

.

Exercise 9.4For 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 162

0 (ii)

Trial mass at left plane TL= 29 gm 00 :L2= 0.0015 mm 110

0 ;R2= 0.0015 mm


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390

mass at right plane TR= 29 gm 00:L3= 0.0025 200

0; R3= 0.0015 170

0. 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.

(b) First flexible mode (c) Second flexible mode

Figure 9.10 Typical rigid and flexible modes of rotor-bearing systems

1 2

(a) Rigid body modes


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391

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 Nthmode

can be balanced byNbalancing planes. Instead ofNplane correction, Kellenberger suggested that the

rotor should be corrected inN+2 planes, so as not to disturb the rigid body balancing.

Assume that all unbalance is distributed only in thex-yplane (Figure 9.11). Let the rotor speed be

and the deflection of the rotor be y(x). The deflectiony(x) can be written in terms of summation of

mode shapes as

( ) ( )i iy x Y x= (9.8)

where )(xYi is the mode shape in the ithmode and i is an unknown constant. The deflectiony(x) can

be measured experimentally. For example, for simply supported end conditions, mode shapes are

I - mode 1( ) sinx

y xl

=

II - mode2

2( ) sin

xy x

l

=

ithmode ( ) sinii x

y xl

=

z

bearing axis

Figure 7.11 Rotor-brearing axis system


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393

{ } { }2

2

2( ) ( ) ( ) ( ) ( )

dEI x y x m x y x a x

dx = + (9.14)

where is the rotor speed. On substituting for y(x) and a(x) from equations (9.8) and (9.9)

respectively, we get

{ }2

2

2( ) ( ) ( ) ( ) ( )i i i i i i

i i

dEI x Y x m x Y x Y x

dx

= + (9.15)

Noting the orthoganality condition and multiply both sides by ( )jY x and integrate over the length of

shaft, the left hand side of equation (9.15) gives

2

2

0( ) ( ) ( )

l

i i j

i

dEI x Y x Y x dxdx

On performing integration by parts, we get

{ }0

0

( ) ( ) ( ) ( ) ( ) ( )l

l

j i i i i j

i

d dY x EI x Y x EI x Y x Y x dx

dx dx

First term vanishes (since it is zero for all boundary conditions), on taking integration by parts, we get

{ }[ ] dxxYxYxEIxYxEIxYl

j

i

ii

l

iij

+

0

0)()()()()()(

First term again vanishes for all boundary conditions. On noting the orthogonality condition equation

(9.11), we get

[ ] jj

l

jj

KdxxYxEI ==

02

)()( (9.16)

where the generalized stiffness injthmode is defined as

[ ] dxxYxEIKl

jj =0

2)()(


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394

From right hand side of equation (9.15), noting equation (9.8) and (9.9), we have

( ) jjj M +2 (9.17)

Therefore from (9.15), noting equations (9.16) and (9.17), we have

( ) jjjjj MK += 2

which can be rearranged as

2

2

/j jj j

K M

= (9.18)

where j is an known, then distribution of eccentricity )(xa can be found from equation (9.9). The

equation (9.18) requires m(x),y(x), ( )jY x and /j jK M= . The m(x) can be accurately found out,

y(x) is difficult to obtained, ( )jY x is obtained by the eigen analysis and ( )j x is natural frequency in

ithmode = /

jK M 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).

(a) (b)

(c)2

21

1

11

2

21

1


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395

Figure 9.13 Measurement locations (q) and balancing planes (p)

Choosepnumber of balancing planes (where mass can be added or chip-off) wherep>2, qnumber is

measuring planes; generally it is two i.e. at bearing planes. Let the unbalance )( em in each of the

balancing planes be pUUU

,,, 21

1 11 12 1 1

2 21 22 2 2

1 2

p

p

q q q qp p

v U

v U

v U

=

(9.19)

where vis 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

{ } }]{[ U = (9.20)

with

q

z


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396

{ } [ ]

1 1 1 1

1 11 12 1

1 1 1 1

2 21 22 2

1 1 1 1

1 2

2 2 2 2

1 11 12 1

2 2 2 2

1 2

1 11 12

1

;

p

p

q q q qp

p

q q q qp

n n n

n

n

q

v

v

v

v

vv

v

v

v

= =

[ ]

1

2

3

1

21 22 2

1 2

;

pn

p

n n n

p

n n n

q q qp

U

UU U

U

=

Once the influence coefficients ][ are known for all speeds equation (9.20) can be used to obtained

unbalances :

( ) }{][][}{ 1 vU T = (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 11

11 12 111 1

1 1 1 121 221 22 2

11 1 1

11 2

p R

p

q pq q qp

v U T

v U

v U

+ =

(9.22)

On subtracting equation (9.22) from first qequation in equation (9.19), we get

1 1 11 1

11 12 111 1

1 1 11 1

21 2 21 22 2

1 11 1 1

11 2

0

0

pR

p

q qq q qp

v v T

v v

v v

=

(9.23)

Equation (9.23) gives


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1 11 11 1

11,

R

v v

T

=

1 11 21 221

, ,R

v v

T

=

1 1

11

1

q q

qR

v v

T

=

(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 canfind the influence coefficient-matrix for different speeds.

Exercise 9.6Justify 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+2planes?

[2] J. Drechsler 1980 Institution of Mechanical Engineers Conference on Vibrations in RotatingMachinery, Cambridge, UK, 65-70. Processing surplus information in computer aidedbalancing of large flexible rotors.

[3] P. Gnilka 1983Journal of Vibration 90, 157-172. Modal balancing of flexible rotors without testruns: an experimental investigation.

[4] J.M. Krodkiewski, J. Ding and N. Zhang 1994Journal of Vibration169, 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 Matricesfor Simultaneous Estimation of the Residual Unbalance and Bearing Dynamic Parameters (inpress).

[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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