dynamic field balancing

Karen Grace

Techncial Associates Field Dynamic Balancing Copyright 2001 Techncial Associates of Charlotte, P.C.

Section Subject Page

1. INTRODUCTION TO DYNAMIC BALANCING ...................................................... 1-1

2. TYPES OF UNBALANCE ........................................................................................ 2-1A. Static Unbalance ............................................................................................. 2-2B. Couple Unbalance .......................................................................................... 2-3C. Quasi-Static Unbalance .................................................................................. 2-4D. Dynamic Unbalance ....................................................................................... 2-5

3. TYPES OF BALANCE PROBLEMS ........................................................................ 3-1A. Rigid Vs Flexible Rotors .......... 3-1B. Critical Speeds.......... 3-4

4. HOW TO ENSURE THE DOMINANT PROBLEM IS UNBALANCE. ... 4-1A. Review of Typical Spectra and Phase Behaviors for Common Machinery Problems ................................................................................... 4-1

1. Mass Unbalance ....................................................................................... 4-12. Eccentric Rotor .......................................................................................... 4-13. Bent Shaft ................................................................................................. 4-34. Misalignment ............................................................................................ 4-35. Resonance................................................................................................ 4-36. Mechanical Looseness/Weakness ........................................................... 4-4

B. Summary of Phase Relationships for Various Machinery ........................ 4-41. Force (or Static) Unbalance....................................................................... 4-42. Couple Unbalance .................................................................................... 4-43. Dynamic Unbalance ................................................................................. 4-44. Angular Misalignment............................................................................... 4-65. Parallel Misalignment ............................................................................... 4-66. Bent Shaft ................................................................................................. 4-67. Resonance................................................................................................ 4-68. Rotor Rub.................................................................................................. 4-69. Mechanical Looseness/Weakness Due to Base/Frame Problems or Loose Hold Down Bolts ....................................................... 4-610. Mechanical Looseness Due to a Cracked Frame.................................... 4-6

C. Summary of Normal Unbalance Symptoms .............................................. 4-61. Special Characteristics ............................................................................. 4-62. Centrifugal Force Due to Unbalance ......................................................... 4-63. Unbalance Force Directivity ...................................................................... 4-74. Radial/Axial Vibration Comparison .......................................................... 4-75. Overhung Rotor Unbalance Directivity ...................................................... 4-76. Steadiness & Repeatability of Phase Due To Unbalance ......................... 4-87. Resonant Amplitude Magnification ........................................................... 4-88. Phase Behavior For Dominant Static,Couple & Dynamic Unbalance ....... 4-8

TABLE OF CONTENTS AND SEMINAR AGENDAField Dynamic Balancing

Copyright 2001 Techncial Associates of Charlotte, P.C.Techncial Associates Field Dynamic Balancing


5. CAUSES OF UNBALANCE .................................................................................... 5-1A. Assembly Errors ............................................................................................. 5-1B. Casting Blow Holes ......................................................................................... 5-1C. Fabrication Tolerance Problems ...................................................................... 5-1D. Key Length Problems ...................................................................................... 5-1E. Rotational Distortion ........................................................................................ 5-3F. Deposit Buildup or Erosion .............................................................................. 5-3G. Unsymmetrical Design .................................................................................... 5-3

6. WHY DYNAMIC BALANCING IS IMPORTANT .................................................... 6-1

7. UNITS OF EXPRESSING UNBALANCE ................................................................ 7-1

8. VECTORS ............................................................................................................... 8-1

9. DYNAMIC FIELD BALANCING TECHNIQUES..................................................... 9-A. Recommended Trial Weight Size .................................................................... 9-B. How a Strobe-Lit Mark On a Rotor Moves When a Trial Weight is Added ........ 9-C. Single-Plane Balancing Using a Strobe Light And a Swept-Filter Analyzer .... 9-D. Single-Plane Method of Balancing .................................................................. 9-E. Balancing in One Run ..................................................................................... 9-F. Two-Plane Balancing Techniques .................................................................... 9-G. Cross-Effects ................................................................................................... 9-H. Single-Plane Method For Two-Plane Balancing .............................................. 9-I. Vector Calculations For Two-Plane Balancing .................................................. 9-J. Rotor Balancing By Static Couple Derivation .................................................. 9-K. Single-Plane Balancing With Remote Phase And A Data Collector ................. 9-L. Taking Phase Readings With A Data Collector................................................. 9-M. Single-Plane Balancing Using A Data Collector ............................................. 9-N. Two-Plane Balancing Using A Data Collector .................................................. 9-O. Overhung Rotors ............................................................................................. 9-P. Multi-Plane Balancing .................................................................................... 9-Q. Splitting Balance Correction Weights .............................................................. 9-R. Combining Balance Correction Weights Using Vectors .................................. 9-S. Effect of Angular Measurement Errors of Potential Unbalance Reduction ....... 9-

1. Effect of Phase Angle Measurement Errors By Instruments ...................... 9-2. Effect of Angular Measurement Errors When Attaching Balance Correction Weights .................................................................... 9-




10. Balancing Machines - Soft-Bearing Vs Hard-Bearing Machines ................... 10-A. Soft-Bearing Machine ................................................................................... 10-B. Hard-Bearing Machine ................................................................................ 10-

11. Recommended Vibration And Balance Tolerances .......................................... 11-A. Vibration Tolerances ..................................................................................... 11-1. Recommended Overall Vibration Specifications .......................................... 11-2. Synopsis Of Spectral Alarm Band Specifications ......................................... 11-

B. Balance Tolerances On Allowable Residual Unbalance ................................. 11-1. ISO 1940 Balance Quality Grades ................................................................ 11-

a. Application of Tolerances to Single-Plane Problems .......................... 11-b. Application of Tolerances to Two-Plane Problems.............................. 11-c. Application of Tolerances to Special Rotor Geometries ...................... 11-

C. How to Determine Residual Unbalance Remaining in a Rotor After Balancing .................................................................................................... 11-D. Comparison of ISO 1940 With API and Navy Balance Specifications .......... 11-

APPENDIX A Balancing Terminology

APPENDIX B Weight Removal Charts

APPENDIX C Conversion Chart for Converting Inches of Flat Stock # 1020 Steel toOunces of Weight

APPENDIX D Three-Point Method of Balancing



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Techncial Associates Field Dynamic Balancing Copyright 2001 Techncial Associates of Charlotte, P.C. 1-1

CHAPTER1

INTRODUCTION TO DYNAMIC BALANCINGINTRODUCTION TO DYNAMIC BALANCING

Unbalance has been found to be one of the most common causes of machineryvibration, present to some degree on nearly all rotating machines. Vibration due to anunbalance while the rotor is rotating is the result of a heavy spot located at a radius fromthe mass centerline producing a centrifugal force. The amount of centrifugal force willbe the result of the weight of the heavy spot, the radius of such heavy spot and thespeed the rotor is rotating. Thus, unbalance can be described as centrifugal forces thatdisplace the rotors mass center from the rotors rotating center. Another way to state thisin general is unbalance is a condition, which exists when vibratory forces or motion isapplied to the bearings of a rotor as a result of centrifugal forces, particularly withrespect to a rigid rotor. How far and in what manner this displacement takes place willbe discussed later in this text.

In order to reduce the amount of forces generated by this imbalance there are severalfactors that we will have to understand. Before a part can be balanced certainconditions must be met.

1. The vibration must be due to unbalance. A complete vibration analysisneeds to be performed to make sure that unbalance is the primary causeof the vibration forces.

2. We must be able to start and stop the rotor.

3. We must be able to add or remove weight.

In most instances, weight corrections can be made with the rotor mounted in its normalinstallation, operating as it normally does. This process of balancing a part withouttaking it out of the machine is called IN-PLACE BALANCING. Balancing in-placeeliminates costly disassembly and eliminates the possibility of damage during thetransportation of the rotor. Rotors that are totally enclosed such as some motors, pumpsand compressors, can be removed and transported to a balancing machine. Theprinciples of balancing are similar either way. Before we discuss balancing, we shouldfirst understand unbalance, where it comes from and what must be done to correct it.


TYPES OF UNBALANCE

Figure 1 helps illustrate unbalance. Here, assuming a perfectly balanced rotor, a 5 ounce(141.75 gram) weight is placed on the rotor at a 10-inch (254 mm) radius. This produces anunbalance of 50 oz-in (36000 gram-mm). Note that the same 5 ounce (141.75 gram) weightplaced at 5 inches (127 mm) from the center would produce only a 25 oz-in (18000 gram-mm)unbalance. Figure 2 illustrates the same rotor as that in Figure 1 but with a correction madesuch that the rotor is balanced by countering the 50 oz-in (36000 gram-mm) originalunbalance by placing a correction of 50 oz-in (36000 gram-mm) (by placing an identicalweight at a 10 inch (254 mm) radius directly opposite the original weight).

FIGURE 1. ILLUSTRATING UNBALACING

FIGURE 2. BALANCE CORRECTION

CHAPTER2

Techncial Associates Field Dynamic Balancing Copyright 2001 Techncial Associates of Charlotte, P.C.2-2

Static unbalance is sometimes known as force unbalance or kinetic unbalance. Staticunbalance is a condition where the mass centerline is displaced from and parallel to the shaftcenterline as shown in Figure 4. This is the simplest type of unbalance, which has classicallybeen corrected for many years by placing a fan rotor on knife-edges and allowing it to roll to thebottom. That is, when the fan wheel is released, if the heavy spot is angularly displaced from thebottom (or 6:00 position), it will tend to roll to the bottom hopefully ending up in the 6:00 position ifthe rotor was sufficiently free to rotate. So-called correction of this unbalance was thenaccomplished by placing a weight opposite this location (or at about 12:00).

Referring to Figure 3, in a perfectly balanced rotor, both the shaft and mass centerlines wouldcoincide with one another with equal mass distribution throughout the rotor.

As mentioned before, unbalance occurs when the mass centerline does not coincide with theshaft centerline as shown in Figure 3. The mass centerline can be thought of as an axis aboutwhich the weight of the rotor is equally distributed. The mass centerline is also the axis aboutwhich the part would like to rotate if free to do so. However, if the rotor itself is restricted in itsbearing, vibration will occur if the shaft and mass centerlines do not coincide. Following belowwill be a discussion on each of the four major types of unbalance, which include - STATIC,COUPLE, QUASI-STATIC and DYNAMIC UNBALANCE. Each of these types of unbalancewill be defined by the relationship between the shaft and mass centerlines of the rotor.

A. Static Unbalance

FIGURE 3. STATIC UNBALANCE


FIGURE 4. STATIC UNBALANCE

Actually, there are two types of static unbalance as shown in Figure 4A and Figure 4B. InFigure 4A, the unbalance is centered directly above the rotor center of gravity (CG). Figure 4Blikewise shows static unbalance, but with equal masses placed at identical distances from themass centerline and rotor CG on each end. Whether the static unbalance occurs as in eitherFigures 4A or 4B, each can be corrected by placement of a correction weight in only one planeat the CG, or by attaching two weights with one half the total weight at either end assuming theCG is equidistant from each bearing.

NOTE: This text is being written with the assumption that we would be able to attach a weightof suitable size at the appropriate radius. If this cannot be accomplished, then the appropriateamount of weight can be removed from the heavy spot.

B. Couple Unbalance

Couple unbalance is a condition where the mass centerline intersects the shaft centerline atthe rotor center of gravity as shown in Figure 5. Here, a couple is created by placement ofequal weights 180 opposite each other and equidistant from the CG in opposite directions. Acouple is simply two equal and parallel forces acting opposite one another, but not in thesame plane. Instead, they are offset from one another, which would tend to rotate the rotor.Significant couple unbalance can introduce severe instability to the rotor causing it to wobbleback and forth (like a seesaw) with the fulcrum at the rotor CG.

Unlike static unbalance, couple unbalance only becomes apparent when the shaft rotates. Inother words, if the rotor is placed on knife-edges, it would not tend to rotate no matter whatposition it is placed since it would be statically balanced. Like static unbalance, coupleunbalance likewise causes high vibration at 1X RPM. Unlike static unbalance, coupleunbalance will bring about a very different phase behavior, which will be discussed in SectionIV. Unlike static unbalance, couple unbalance must be corrected in two planes with corrections180 opposite each other.

FIGURE 5. COUPLE UNBALANCE


The axial location of the correction couple will not matter as long as it is equal in magnitude,but opposite in direction to the unbalance couple. For example, looking at Figure 6, placementof the two 5 oz (141.75 gram) weights at an axial distance with 10 inches (254 mm) to the leftand right of the CG as shown will create a clockwise couple unbalance. This can becounteracted either by placing identical 5 oz (141.75 gram) weights at a 10-inch (254 mm)distance directly opposite the original weights or by placing 10 oz (283.5 gram) weights at anaxial distance only 5 inches (127 mm) from the CG.

Only a very few cases will a rotor have true static or true couple unbalance. Normally, anunbalanced rotor will have some of each type. Combination of static and couple unbalance isfurther classified as quasi-static and dynamic unbalance.

FIGURE 6. CORRECTION OF COUPLE UNBALANCE

C. Quasi-Static Unbalance

Quasi-static unbalance represents a specific combination of static and couple unbalancewhere the static unbalance is directly in line with one of the couple moments as shown inFigure 7. Quasi-static unbalance is that condition where the mass centerline intersects with theshaft, but at a point other than the rotor center of gravity (CG). In Figure 7, the Figures 7A and7B illustrate quasi-static unbalance. In Figure 7A, the unbalance mass is placed at a locationother than the CG which introduces both static and couple unbalance. In reality, Figure 7Brepresents the same unbalance as that in Figure 7A. The two unbalance masses actingopposite one another close to the CG counteract one another statically, but do not compensatefor the unbalance introduced by the unbalance mass on the top left-hand side of the rotor.

FIGURE 7A FIGURE 7B

FIGURE 7. QUASI-STATIC BALANCE


Figure 8 illustrates another type of quasi-static unbalance often not even considered byanalysts. In this case, assuming you had a perfectly balanced rotor, when this is connected toan unbalanced coupling, a quasi-static unbalance is created. This is a very common type ofunbalance since most couplings are not balanced unless they are of great size or speed.Similarly, quasi-static unbalance can be introduced by inserting the wrong size key into theshaft or pump impeller, which again will create both a static and a couple unbalance. In eachcase, the required correction in addition to a static correction at the same location as thecouple component nearest the coupling, key, etc.

FIGURE 8. UNBALANCED COUPLING CAUSING QUASI-STATIC UNBALANCE

D. Dynamic Unbalance

Dynamic unbalance is the most common type of unbalance and can only be corrected bymass correction in at least two planes. Figure 9 illustrates dynamic unbalance which again is acombination of both static and couple unbalance, but with unbalance masses at differentangular positions from one another as shown in Figure 9. Because the unbalance masses areat different angular positions, dynamic unbalance is that condition where the shaft centerlineand mass centerline do not intersect with one another, nor are they parallel with one another.As will be pointed out in Section III, dynamic unbalance causes phase differences betweenthe horizontal on one bearing versus the horizontal on the other bearing to be far different fromeither 0 or 180. That is, the horizontal phase difference may be 60 or 180, or most anything.However, if the horizontal phase difference is 60, the vertical phase difference should be thesame as the horizontal within one clock position (+/- 30).

FIGURE 9. DYNAMIC UNBALANCE


TYPES OF BALANCE PROBLEMS

CHAPTER3

In order to achieve a satisfactory balance with the minimum number of start-stop operations itnot only is important that we recognize the type of balance problem we have (static, couple,quasi-static or dynamic), it should now be obvious that not all balancing can be achieved bybalancing in a single correction plane. A guide to determining whether single plane, twoplane or multi-plane will be required will be determined by the ratio of the length to diameterof the rotor along with the speed of the rotor. It is also very important to recognize whether therotor is flexible (one that bows)or rigid (one that maintains it geometric shape.) The L/D ratio iscalculated using the dimensions of the rotor exclusive of the supporting shaft. See Table 1.

The selection of single plane versus two-plane balancing based on the L/D ratio and rotorspeed is offered only as a guide and may not hold true in all cases. Experience reveals thatsingle plane balancing is normally acceptable for rotors such as grinding wheels, single-sheave pulleys, and similar parts even through their operating speed may be greater than1000 RPM.

TABLE 1

A. Rigid Vs Flexible Rotors

Only a few rotors are made of a single disc, but instead they are made of several discson a common shaft, often times in complex shapes and sizes. This makes it practicallyimpossible to know which disc(s) the heavy spot is located. The unbalance could be in anyplane or planes located along the length of a rotor, and it would be most difficult and timeconsuming to determine where. In addition, it is not always possible to make weightcorrections in just any plane. Therefore, the usual practice is to compromise by making weightcorrections in the two most convenient planes available. This is possible because anycondition of unbalance can be compensated for by weight corrections in any two balancingplanes. This is true only if the rotor and shaft are rigid and do not bend or deflect due to theforces caused by unbalance.


Whether or not a rotor is classified as rigid or flexible depends on the relationship between therotating speed (RPM) of the rotor and its natural frequency. You will recall that every objectincluding the rotor and shaft of a machine has a natural frequency, or a frequency at which itlikes to vibrate. When the natural frequency of some part of a machine is also equal to therotating or some other exciting frequency of vibration, there is a condition of resonance.

A flexible rotor balanced at one operating speed may not be balanced when operating atanother speed. If a rotor were first balanced below 70% of the its first critical speed with thecorrection weights added in the two end planes, the two correction weights added wouldcompensate for all sources of unbalance distributed throughout the rotor. If the rotor wereincreased to above 70% of the critical speed, the rotor would deflect due to the centrifugalforce of the unbalance located at the center of the rotor as shown in Figure 10. As the rotorbends or deflects, the weight of the rotor is moved out away from the rotating centerlinecreating a new unbalance condition. It would then be necessary to rebalance the two endplanes at this new operating speed, and then the rotor would be out of balance at the sloweroperating speed. The only solution to insure smooth operation at all speeds is to make thebalance correction in the actual planes of unbalance, thus a multi-plane balance.

This subject will be discussed in greater detail later in the course.

FIGURE 10. ROTOR DEFLECTION DUE TO UNBALANCE ABOVE CRITICAL SPEED

FIGURE 11. ROTOR FLEXURAL MODES

Fig. 10ARotor with dynamic unbalance,balanced in two planes below

critical speed

Fig. 10BOperating above critical

speed the rotor deflects dueto unbalance in the center


The rotor in Figure 10 represents the more simple type of flexible rotor. A Rotor can deflect inseveral ways depending on its operating speed and the distribution of unbalance through outthe rotor. Figure 11 illustrates the first, second and third flexural modes a rotor could take whilegoing through the first, second and third criticals. These rotors may require that balancecorrections be made in several planes to insure smooth operation through all speed ranges.

Whether a flexible rotor requires multi-plane balancing depends on the normal operatingspeeds of the rotor and the significance of rotor deflection on the functional requirements of themachine. Flexible rotors generally fall into one of the following categories:

1. If the rotor operates at only one speed and a slight amount of deflection willnot accelerate wear or hamper the productivity of the machine, thenbalancing in any two correction planes to minimize bearing vibration maybe all that is required.

2. If a flexible rotor operates at only one speed, but it is essential that rotordeflection be minimized, then multi-plane balancing may be required.

3. If it is essential that a rotor operate smoothly over a broad range of speedswhere the rotor is rigid at lower speeds and flexible at higher speed, thenmulti-plane balancing is required.

B. Critical Speeds

The rotating speed at which the rotor itself goes into bending resonance is called a criticalspeed. Depending on how many bending modes the rotor goes through is dependant uponthe number of operating speeds coincide with the rotors natural frequency. In general, rotorsoperating below 70% of their natural frequency are considered to be rigid rotors and above70% of their natural frequency are considered to be flexible rotors.

When a rotor bends or deflects due to operating through its critical speed, the weight of therotor is moved out away from the rotating centerline creating a new unbalanced condition. Thisrotor could be corrected by rebalancing in the two end planes; however, the rotor would thenbe out of balance at slower speeds where there is no deflection. The only solution to insuresmooth operation at each speed is to make the corrections in the planes of unbalance, thusmulti-plane balancing. Remember, any unbalance can be corrected by making corrections inany two balance planes, but only if the rotor is a non-flexible rotor.


HOW TO ENSURE THE DOMINANT PROBLEM ISUNBALANCE

CHAPTER4

Before analysts begin balancing a machine, they should always ensure that the dominant problemis in fact unbalance before they begin. Vibration consultants commonly report that on over one-halfthe jobs on which they are requested to balance machinery, they do not in fact perform anybalancing, but find other problems requiring different corrective measures instead. An analystshould always employ both spectral and phase behaviors for some of the more commonmachinery problems, each of which can cause high vibration at 1X RPM, including eccentric rotor,bent shaft, misalignment, resonance and even certain types of mechanical looseness/weakness.While there are still other problems that generate 1X RPM vibration, a review of Table II will helpthe analyst distinguish which problem is at hand.

It should be pointed out that the column entitled TYPICAL SPECTRUM in Table II means just that- that is, these spectra are not intended to be all-inclusive. For example, it is quite possible formisalignment to generate only high 1X RPM vibration in certain cases, however, they most oftengenerate a noticeable 2X RPM peak. Therefore, such a spectrum is shown under the TYPICALSPECTRUM column. Following below will be a quick review of Table II pointing out the morecommon spectral and phase behaviors of the problems shown. Later, a more detailed look will betaken specifically on unbalance symptoms.

A. REVIEW OF TYPICAL SPECTRA AND PHASE BEHAVIORS FOR COMMONMACHINERY PROBLEMS

1. Mass Unbalance: Table II shows that mass unbalance always generates highvibration at 1X RPM. The centrifugal forces caused by unbalance always act in the radialdirection, but can sometimes generate high axial vibration in the case of overhung rotorslike in Unbalance Case C of Table II. Pure force, or static unbalance, is evidenced by identicalphase in the radial direction on both the outboard and inboard bearings supporting the rotor.On the other hand, pure couple unbalance is evidenced by a 180 phase difference in theradial direction between the outboard and inboard bearings (the horizontals will be 180 out ofphase with one another as well as the outboard and inboard verticals with one another in purecouple unbalance). Overhung rotors represent a special case of unbalance on which highaxial vibration can be generated which is in phase between the inboard and outboardbearings supporting the overhung rotor as shown in Table II.

2. Eccentric Rotor: Like unbalance, an eccentric rotor will generate high vibration at1X RPM of the eccentric rotor itself with the highest vibration normally being in a directionthrough the centers of the two rotors as shown in Table II under ECCENTRIC ROTOR.However, the main difference between an eccentric rotor and an unbalanced one is withrespect to phase behavior-pure unbalance will normally cause the phase difference in the


horizontal and vertical directions to be about 90 while in the case of an eccentric rotor, thehorizontal and vertical phase difference will normally be either approximately 0 or 180 (eachof which indicate straight-line motion). One of the problems with an eccentric rotor occurs ifone attempts to balance the eccentric rotor. What will often result is that the balance exercisemay in fact reduce vibration in one radial direction, but increase it in the other, depending onthe amount of eccentricity.

3. Bent Shaft: A bent shaft will most always generate high axial vibration with thegreatest component being 1X RPM if bent near the shaft center, but can create a high 2X RPMcomponent if bent near the coupling. One of the things that sets apart bent shaft symptomsfrom those of unbalance is with respect to phase behavior - a bent shaft will cause axialvibration on the outboard bearing of a rotor to approximately 180 out of phase with respect tothat of the inboard rotor bearing, while unbalance will normally cause axial outboard andinboard phase to be about the same.

4. Misalignment: Although misalignment normally generates a 2X RPM componentgreater than or equal to 30% of the amplitude at 1X RPM, it can sometimes cause only a high1X RPM component, particularly in the axial direction. However, one of the things that againdifferentiate it from unbalance is its phase behavior - misaligned shafts will cause phaseacross the coupling to be approximately 180 different, whereas unbalance will normallycause almost equal phase on either side of the coupling. As Table II shows, angularmisalignment is evidenced by a 180 phase change across the coupling in the axial directionwhereas parallel, (or offset misalignment), causes a 180 difference in the radial directionacross the coupling. Finally, a misaligned bearing cocked on the shaft generates spectra verysimilar to that of shaft misalignment. However, it can be detected by measuring at each of 4points in the axial direction on each bearing as shown in Table II. This measurement shouldshow that the phase is almost the same at each of the 4 points around the clock if the bearingis properly oriented. If there is a 180 phase difference across either points 1 and 3, orbetween 2 and 4 as shown in Table II, a cocked bearing is indicated.

5. Resonance: Resonance occurs when a forcing frequency coincides with asystem natural frequency and can cause excessive vibration amplitudes. Even a small amountof unbalance, for example, can be greatly amplified if the rotor is operating at or near a naturalfrequency. Such a resonant problem is evidenced if the phase changes dramatically for only asmall change in speed (Figure 12 shows that a rotor will experience almost a full 180 phasechange when its speed passes completely through a natural frequency). At the same time, theamplitude first increases dramatically and then decreases as the rotor passes through thenatural frequency (as shown in Table II).


6. Mechanical Looseness/Weakness: Table II shows three different types ofmechanical looseness, one of which is lesser known, but causes high radial vibrationpredominantly at 1X RPM which again causes a spectrum almost identical to an unbalancevibration spectrum as shown under mechanical looseness Type A. Type A looseness iscaused by a looseness or weakness of machine feet, base plate, foundation, loose hold-downbolts at the base, distortion at the frame or base, etc. Again, the thing which sets it apart fromunbalance is its phase behavior. Referring to MECHANICAL LOOSNESS Type A in Table Inote that a problem is evidenced between the base plate and its base by a 180 phasechange between these two sections. In other words, when a phase measurement is taken, ifeverything is moving together as it should, the phase should be almost identical as onemoves his probe in the vertical direction from the foot to the baseplate, and then down to thebase. One of the most important points about this type of looseness/weakness problem is thateven if one is able to temporarily correct the problem by balancing and alignment procedures,the vibration will likely reoccur when even the least bit of unbalance or misalignmentsymptoms return. They must first correct the looseness/weakness problem, then balance oralign if any further correction measures are still required.

B. SUMMARY OF PHASE RELATIONSHIPS FOR VARIOUS MACHINERY

Section A above summarized the typical spectral and phase relationships for some of themore common machinery problems. One of the most important points that this sectionhopefully made was that the key parameter that helped differentiate one problem from anotherwas phase. Therefore, because of the importance of phase, following below will be asummary showing how phase generally behave for each particular problem scenario (seeTable II):

1. Force (or static) unbalance is evidenced by nearly identical phase in theradial direction on each bearing of a machine.

2. Couple unbalance shows approximately a 180 out-of-phase relationship whencomparing the outboard and inboard horizontal, or the outboard and inboardvertical direction on the same machine.

3. Dynamic Unbalance is indicated when the phase difference is well removedfrom either 0 or 180 but importantly is nearly the same in the horizontal andvertical directions. That is, the horizontal phase difference could be almostanything ranging from 0 to 180 between the outboard and inboard bearings;but the key point is that the vertical phase difference should be almost identicalto the horizontal phase difference (+/- 30). For example, if the horizontal phasedifference between the outboard and inboard bearings is 60, and dominantproblem is dynamic unbalance, then the vertical phase difference between thesetwo bearings should also be about 60 (+/- 30). If the horizontal phasedifference varies greatly from the vertical phase difference when high 1X RPMvibration is present, this strongly suggests the dominant problem is notunbalance.

4. Angular misalignment is indicated by approximately a 180 phase differenceacross the coupling, with measurements in the axial direction.


FIG

UR

E

12C

HA

NG

E

OF

V

IB

RA

TI

ON

D

IS

PL

AC

EM

NT

A

ND

P

HA

SE

L

AG

W

IT

H

RP

M

AB

OV

E,

BE

LO

WA

ND

A

T

RO

TO

R

RE

SO

NA

NC

E


5. Parallel misalignment causes radial phase differences across the coupling tobe approximately 180 out of phase with respect to one another.

6. Bent shaft causes axial phase on the same shaft of a machine to approach a180 difference when comparing axial measurements on the outboard withthose on the inboard bearing of the same rotor.

7. Resonance is shown by exactly a 90 phase change at the point when theforcing frequency coincides with a natural frequency, and approaches a full 180phase change when the machine passes through the natural frequency(depending on the amount of damping present).

8. Rotor rub causes significant, instantaneous changes in phase.

9. Mechanical looseness/weakness due to base/frame problems or loosehold-down bolts is indicated by nearly a 180 phase change when one movesthe probe from the machine foot down to its baseplate and support base.

10. Mechanical looseness due to a cracked frame, loose bearing or looserotor causes phase to be unsteady with widely differing phase measurementsfrom one measurement to the next. The phase measurement may noticeablydiffer every time you speed up the machine.

C. Summary of Normal Unbalance Symptoms

Sections A and B above summarized how the analyst can ensure that the dominant problem isunbalance. Following below will be a more detailed look at the symptoms normally presentwhen some type of unbalance is the major problem:

1. Special Characteristics - unbalance is always indicated by high vibration at 1X RPMof the unbalanced part. Normally, this 1X RPM vibration will dominate the spectrum. Infact, the amplitude at 1X RPM will normally be greater than or equal to 80% of the overallamplitude when the problem is limited to unbalance (may be only 50% to 80% of theoverall if other problems exist in addition to unbalance).

2. Centrifugal Force Due to Unbalance - Mass unbalance produces centrifugalforces proportional to the following equation:


For example, assuming a sample rotor with a 1 oz (28.35 grams) unbalance at an 18 inch(457.2 mm) radius (U= 18 oz-in) (12,962 gram-mm) turning 6000 RPM.

FC = (.000001775)(18 oz-in)(6000 RPM)2

FC

= 1150 lbs (from centrifugal force due to unbalance alone)

That is, only a 1 oz (28.35 gram) unbalance on a 3 foot (914.4 mm) diameter wheel turning6000 RPM would introduce a centrifugal force of 1150 lbs (521.6 kg) that would have to besupported by the bearings in addition to the static rotor weight they must support. Importantly,note that the centrifugal force varies with the square of RPM (that is, tripling the speed willresult in an increase in unbalance vibration by a factor of 9 times).

3. Unbalance Force Directivity - Mass unbalance generates a uniform rotating forcethat is continually changing direction, but is evenly applied in all radial directions. As aresult, the shaft and supporting bearings tend to move in somewhat a circular orbit.However, due to the fact that vertical bearing stiffness is normally higher than that in thehorizontal direction, the normal response is a slightly elliptical orbit. Subsequently,horizontal vibration is normally somewhat higher than that in the vertical, commonlyranging between 1.5 and 2 times higher. When the ratio of horizontal to vertical is higherthan about 5 to 1, it normally indicates problems other than unbalance, particularlyresonance.

4. Radial/Axial Vibration Comparison - When unbalance is dominant, radial vibration(horizontal and vertical) will normally be quite higher than that in the axial direction(except for overhung rotors).

5. Overhung Rotor Unbalance Directivity - Generally causes high 1X RPM vibration inboth the axial and radial directions. Overhung rotors most often have both static and coupleimbalance, which will normally require correction in at least two planes.

FC = mrw2

gC

FC = .000001775 Un

2 = .00002841 Wrn

2

where:

Fc = Centrifugal Force (lb)u = Unbalance of Rotating Part (oz-in)w = Weight of Rotating Part (lb)r = eccentricity of the rotor (in)n = Rotating Speed (RPM)

(EQUATION 1)= Wr(386)(16)

2pn

60

2


6. Steadiness & Repeatability of Phase Due to Unbalance - Unbalanced rotorsnormally exhibit steady and repeatable phase in radial directions. When the rotor is trimbalanced, phase can begin to dwell back and forth under a strobe light as youachieve a better and better balance, particularly if problems other than unbalance arepresent.

7. Resonant Amplitude Magnification - The effects of unbalance may sometimes beamplified by resonance. Only a slight unbalance vibration can increase by a factor of 10up to as much as 50 times if the rotor is operated at or near resonance with a systemnatural frequency.

8. Phase Behavior for Dominant Static, Couple and Dynamic Unbalance - Figure13 illustrates typical phase measurements for a machine which has either static (TableA), couple (Table B) or dynamic (Table C) unbalance.

Static Unbalance Phase - Table A shows a machine having dominant static unbalance. Notethat the horizontal phase difference between the #1 and #2 bearings is about 5 (30 minus25) compared to a vertical phase difference of about 10 (120 - 110). Similarly, over on thepump, the horizontal phase difference at positions 3 and 4 is about 10 and the vertical phasedifference is about 15.

Couple Unbalance Phase - Table B illustrates typical couple unbalance phase readings. Notethe 180 phase difference between positions 1 and 2 horizontal (210 - 30), and the 175phase difference between positions 1 and 2 vertical (295 - 120).

Dynamic Unbalance - Table C illustrates typical behavior for dynamic unbalance. Note that thehorizontal phase difference between outboard and inboard bearings can be anything from 0to 180. However, whatever the phase difference in horizontal, the phase difference in thevertical should then be almost identical (within one clock position or +/- 30). In the Figure 13example in Table C, note the 60 phase difference between positions 1 and 2 in both thehorizontal and vertical directions; while over on the pump at positions 3 and 4, the 10difference in the pump horizontal readings compared to the 5 difference in the vertical (170 -165).

Key Point About Unbalance Phase Behavior - Whatever the phase difference between theoutboard and inboard horizontal phase measurements on a rotor, the vertical phase differencebetween outboard and inboard bearings must be about the same (within +/- 30), or thedominant problem is not unbalance. If, for example, the horizontal phase difference on amotor between its outboard and inboard bearings were 30, while the outboard and inboardvertical phase difference was approximately 150, an analyst would likely waste much timeand effort if he attempted to balance the rotor.


FIGURE 13TYPICAL PHASE MEASUREMENTS WHICH WOULD INDICATE EITHER

STATIC, COUPLE OR DYNAMIC UNBALANCE

TABLE A

TABLE B

TABLE C


CAUSES OF UNBALANCE

CHAPTER5

There are a variety of causes of unbalance. These can be summarized as follows:

A. Assembly Errors - Sometimes occur after assembly when the mass center of rotation ofone part does not line up with the mass center of rotation of the part to which it wasassembled. For example, even if both a pump impeller and the pump shaft were separatelyprecision balanced and then assembled, this can happen if the pump impeller had beenbalanced on a balancing shaft that fit its bore within 1 mil, but then was mounted on the shaftwhich itself allows a clearance of over 3 mils. This would shift the mass of the impeller/shaftrotor away from the shaft center which would throw the assembly out of balance, or at leastcause it to have noticeably more unbalance than that when each part was separatelybalanced.

B. Casting Blow Holes - Cast parts occasionally will be left with blow holes within them thatmight not be detectable by visual inspection means. Depending on the diameter of the rotoras well as its speed, this can throw it considerably out of balance.

C. Fabrication Tolerance Problems - A common problem with parts such as a sheave dealswith stack up of clearance tolerance. In this case, since the bore of the sheave is necessarilylarger than that for the shaft diameter, when a key or setscrews is employed, the take-up inclearance shifts the rotating centerline of the sheave away from that of the shaft on which it ismounted.

D. Key Length Problems - Use of no key or the wrong size of key can cause noticeableunbalance problems. Mr. Ralph Buscarello of Update International points out the greatimportance of employing a half-key (full key length, but half-key depth) when balancingcouplings, impellers, sheaves, etc. Figure 14 helps explain why this is important. Mr.Buscarello recommends that a tag like the one shown in the figure should be attached to thefinish balanced rotor any time a machine part is to be balanced and then mounted on a shaft.For example, if the coupling shown in Figure 14 had a B dimension of about 4 inches (101.6mm) and an A dimension of 8 inches (203.2 mm), Mr. Buscarello recommends a final keylength of about 6 inches [1/2 X (8 + 4) inches] or 152.4 mm [1/2 X (203.2 + 101.6) mm].

To further illustrate, assume a machine is to be outfitted with a 1/4" X 1/4" X 6"(6.4 mm X 6.4 mm X 152.4 mm) key, then both the coupling and the shaft should be outfittedwith1/4" X 1/8" X 6" (6.4 mm X 6.4 mm X 152.4 mm) keys when balancing. Also assume thiscoupling is perfectly balanced, weighs 5 lbs (2.3 kg), will operate at 1800 RPM and will bemounted on a 4" (101.6 mm) shaft diameter. The following will illustrate the effect of not usingthe proper half-key:


(b) Weight of 4" long key = (1.698)(4) = 1.132 oz (4" key) 6

(c) Unused key weight if used only 4" half-key = 1.698 - 1.132 = .283 oz (unused half-key weight)

2

(d) Distance of Key CG from shaft center = 2" radius + (1/2 x 1/8") = 2.0625"

(e) So, if 6" half-key rather than a 4" half-key used, unbalance introduced when youinsert 6" full key will be: 2.0625" x .283 lb. = .584 oz-in(unbalance introduced by wrong half-key length used to balance the coupling).

Then, referring to the ISO balance tolerance table shown in Figure 54 (on page 11-16), let ussee how this would affect an otherwise perfectly balanced coupling installed on a rotor turning1800 RPM. Assuming the coupling weight of 5 lbs and the unbalance of .584 oz-in introducedby the key, this corresponds to a residual unbalance of .1168 oz-in/lb which equals .0073 lb-in/lb. Referring to Figure 54 at 1800 RPM, this would degrade the perfectly balanced couplingdown to an ISO Balance Quality G 40, or one with a poor balance quality grade.

Of course, if no half-key were used at all when balancing the coupling, this would introduceeven more unbalance to the system. And, one of the real problems with this being a couplingis that the weight would be overhung from the motor bearing meaning that it could introduceconsiderable couple unbalance. This fact is often overlooked, particularly when dealing withcouplings, most of which are not even factory balanced unless specifically requested by theend user.

Figure 14 SUGGESTED TAG THAT SHOULD ACCOMPANY FINISHBALANCED KEYED ROTOR

(Ref. Practical Solutions to Machinery and Maintenance Vibration Problems, Update International).

Coupling outfitted with a 1/4" x 1/8" x 4" key

(a) Final key weight = (1/4" x 1/4" x 6" ) x .238 lb/in3 x 16 oz/lb = 1.698 oz (6 key)


E. Rotational Distortion - Sometimes, a part might be well balanced, but might distort whenrotating due to stress relieving or thermal distortion. Parts fabricated by welding process orshaped by pressing, drawing, bending and so forth will sometimes have high internalresidual stresses. If not relieved during the fabrication, they may begin doing so over a periodof time when operating, distorting slightly and taking on a new shape. This can throw the rotorout of balance. In addition, some machines have problems with thermal distortion caused bysuch problems as uneven thermal expansion of parts when brought up to operatingtemperatures. This sometimes mandates that the rotor be balanced at its normal elevatedoperating temperature.

F. Deposit Buildup or Erosion - Fan or impeller wheels are often thrown well out of balancedue to buildup of deposits of dirt or other foreign matter brought into them by the pumpingfluid or air. When small pieces of these deposits break away, it can sometimes introduceserious unbalance. On the other hand, some high-speed centrifugal compressor rotors aresusceptible to erosion from small droplets of water traveling at very high speeds which impactthe impeller rotors. This can cause uneven erosion of impeller surfaces and eventually canintroduce considerableunbalance.

G. Unsymmetrical Design - Unbalance can be introduced if good symmetry is notmaintained in all parts. For example, rotor windings on electric motors are sometimes difficultto keep symmetrical; the thickness in sheaves sometimes vary from on side to the other; thedensity of coating finishes sometimes varies around the rotor periphery. Other problems canaffect rotor symmetry, each of which can detrimentally affect rotor balance.

In summary, all of the above causes of unbalance can exist to some degree in a rotor.However, the vector summation of all unbalance can be considered as a concentration at apoint termed the heavy spot. Balancing, then, is the technique for determining the amountand location of this heavy spot so that an equal amount of weight can be removed at thislocation or an equal amount of weight added directly opposite.


CHAPTER6

WHY DYNAMIC BALANCING IS IMPORTANT

The forces created by unbalance can be among the most destructive forces in rotatingmachinery if left uncorrected. Not only will these forces damage the bearing but they havebeen know to crack foundations, break welds, etc. In addition, the vibration displacement dueto unbalance can be detrimental to product quality in many applications. The amount of forcecreated by unbalance depends on the speed of rotation and the weight of the heavy spot.Figure 15 represents a rotor with a heavy spot (W) located at some radius (R) from the rotatingcenterline.

If the unbalance weight, radius and machine RPM are known, the force (F) generated can befound using the following formula:

F = 1.77 x (RPM/1000)2 x ounce-inches (EQUATION 2)

In this formula the unbalance is expressed in oz-inches and (F) is the force in pounds. Theconstant 1.77 is required to make the formula dimensionally correct. When the unbalance isexpressed in terms of gram-inches, the force (F) in pounds can be found by using thefollowing formula:

F = 1/16 x (RPM/1000)2 x gram-inches (EQUATION 3)

For unbalance expressed in gram-mm, the force (F) in kg can be calculated using thefollowing formula:

F = 0.001 x (RPM/1000)2 x gram-mm (EQUATION 4)

From these formulas it can be seen that the centrifugal force due to unbalance actuallyincreases by the square of the rotor RPM. For example, from Figure 16 we see that the forcecreated by a 3 ounce weight attached at a radius of 30 inches (90 oz-in unbalance) androtating at 3600 RPM is over 2000 lbs (907 kg). By doubling the speed to 7200 RPM, theunbalance force is increased to over 8000 pounds (3630 kg). From this we can see, especiallyon high-speed machines, a small amount of unbalance can create a tremendous amount offorce.


CHAPTER7

UNITS OF EXPRESSING UNBALANCE

Units of unbalance in a rotating work piece is normally expressed as the product of theunbalance weight (lbs., oz., grams, etc.) and its distance from the rotating centerline (inches,mm, etc.), see Figure 15. The units for expressing unbalance are generally oz-inches, gram-inches, gram-mm, etc. For example, a 1 oz (28.35 gram) weight located at 10" (254 mm) fromthe rotating centerline would be 10 oz-inches, (7200 gram-mm) and a 2 oz (56.7 gram) weightlocated 6" (152.4 mm) from the rotating centerline would be 12 oz-in (8641 gram-mm). Figure16 represents other examples of unbalance expressed as the product of weight and distance.

FIGURE 15. THE FORCE DUE TO UNBALANCE CAN BE FOUND IF THEUNBALANCE WEIGHT (W), RADIUS (R) AND ROTATING SPEED ARE KNOWN

FIGURE 16. CENTRIFUGAL FORCE EXERTED BYUNBALANCE (OZ-IN) AT VARIOUS SPEEDS


FIGURE 17. UNITS OF UNBALANCE ARE EXPRESSED AS THE PRODUCT OF THEUNBALANCE WEIGHT AND ITS DISTANCE FROM THE ROTATIONAL CENTER

UNBALANCE UNITS


CHAPTER8

Scalar quantities such as mass, time, volume, or force may be represented by a length of asingle line in any arbitrarily chosen direction. A quantity, which has both magnitude anddirection, is called a vector quantity. Describing a vector is giving it magnitude (length) anddirection.

Unbalance forces generate a magnitude equivalent to a certain number of ounces of weight orounce-inches and an angular direction with respect to a reference point on the rotor, can berepresented by a vector. It should be apparent that unbalance forces that tend to move therotor away from its axis of rotation cause a certain magnitude. These forces and their exactlocation on the rotor cannot be measured directly. However, their effects on the rotor and/orbearing supports can be measured.

An unbalance vector, then, can be described as a straight line whose length is proportional tothe amount of unbalance and the angular direction measured from a reference point.

The combined effect of several unbalances or balance weights can be determined by vectorcalculations. Examples of several vectors are shown in Figures 18. In Figure 18A, vectors aredrawn to represent the radial location of weights. The length of the vector represents the radiusin inches. Figure 18B vectors are shown to represent the weight in ounces. In Figure 18C, thevectors represent the amount of unbalance in ounce-inches.

Balancing vectors are used to represent the amount and angular location of unbalance, aswell as to measure the effect of trial weight when solving balancing problems.

VECTORS


FIGURE 18B. VECTOR WEIGHTFIGURE 18A. VECTOR RADIUS (LENGTH)

FIGURE 18C. VECTOR UNBALANCE (RADIUS X WEIGHT)

FIGURE 18


CHAPTER9

DYNAMIC FIELD BALANCING TECHNIQUES

Generally, it is best to balance a good majority of rotating machines in place since it can bedone under the actual operating conditions and speed which exist during operation, in its ownbearings and on its own foundation. In addition, balancing in-place eliminates the possibledamage to the rotor during disassembly and transportation to a balancing machine. Followingwill be information on what techniques should be mastered in order to best accomplish fieldbalancing using portable balancing equipment.

Information on recommended techniques on performing field balancing including single-plane, two-plane, multi-plane and over-hung rotors will be discussed. In addition, instructionswill be provided on directly related topics such as how to properly size trial weights, how tosplit balance correction weights when is not possible to place a single weight at the angularlocation specified by the solution, and how to vectorially combine the effects of severalweights into one correction weight of just the right size and at just the right location.

To begin with we will be discussing balancing using the vector method of balancing.Although there are many instruments with balancing programs in them on the market today,the mastering of the vector solution will give us a very good understanding of the effects thatwe should get and how to read the vector to determine if we made an error in our weightselection and location. We will later discuss the use of instruments with built in balancingprograms.

Although it is possible to balance any object with amplitude alone, we will begin ourdiscussion of balancing using conventional vectors, both amplitude and phase. At the end ofthis chapter you will find the instructions for balancing using just the amplitude, called theFour Point Method of Balancing.

A. Recommended Trial Weight Size

It is important that the size of the trial weight be carefully chosen as well as the location atwhich the trial weight will be placed. If the trial weight is too large, damage may be done tothe machine if the trial weight happens to be installed at or close to the rotor heavy spotproducing even more vibration, particularly if the rotor is operating above critical speed. Onthe other hand, if the trial weight is too small, it may bring about no significant change inamplitude or phase that can cause significant error when calculations for the proper correctionweight and location are made.


As a general rule, a trial weight should produce either/or both a 30% change in amplitude or a30 phase change. In order to provide a sufficiently large trial weight effect, but without riskingdamage to the rotor, it is recommended that a trial weight which will produce an equivalentunbalance force at each bearing of about 10% of the rotor weight supported by each bearingshould be installed. Therefore, referring to Equation (1), a similar equation can be derived tohelp the analyst choose a proper trial weight:

FC =.000001775 Un2 = .00002841 Wrn2 (EQUATION 1 Repeated)

Solving for U:

U = 563,380 FC (EQUATION 5)

Now, assuming the trial weight should cause a 10% effect (.10 X U),

TW = .10 X U = .10 (563,380) W = 56,338 W (with W=Bearing Load at this point) n2 n2

In order to make the equation easier for the analyst to use, double the constant (56,338) so thatW can be considered the full rotor weight.

Therefore,

TW= 112,676W (EQUATION 6)

Where:TW =Recommended Trial Weight Effect (oz-in) W =Weight of Rotating Part (lb) n =Rotating Speed (RPM)

n2

n2


Now, if the radius at which the trial weight will be placed is known, the trial weight size thatshould be employed can be calculated as per the following:

TW = U = mr

Therefore:

m = TW (EQUATION 7)

Where: m = Trial Weight Size (oz or grams)r = Radius at which Trial Weight will be placed (in)TW = Trial Weight Effect (oz-in or gram-in)

r

An example will serve to illustrate the use of these equations:

Example - The rotor shown in Figure 19 is to be balanced. It has a weight of 453.6 kg.,operates at 1800 RPM and has a 24" (609.6 mm) diameter wheel. To determinethe recommended trial weight size (oz),

TW = 112,676 W = (112,676)(1000) = 34.78 oz-in n2 (1800)2 (at 12" radius)

Then,m = TW = 34.78 = 2.90 oz (Record trial weight size) r 12

The centrifugal force that would be developed by this 2.90 oz (82.2 gram) trial weight is:

Centrifugal Force = (.000001775)(34.78 oz-in)(1800)2 = 200.0 lb.


is operated. In addition, the machine casings and inspection doors should be closed beforeoperation in case the trial weights do accidentally come off the rotor. If it is not possible toclose the casing or inspection door, a shield should be placed between the machine andthe analyst for protection. The analyst and all others should place themselves to the side ofthe machine away from direction of rotation when the machine is operated. When attachingtemporary clips or set-screwed trial weights, attempt to fasten these so that the centrifugalforce is working for you to hold the weights (for example, if an analyst desires to attach abalance clip to fan blades, fasten them on the inside of the blades so that the throat restsagainst the blades inside surface).

Finally, it is a good idea to identify the location of the trial weights by marking them in casethey do happen to come off.

B. How A Strobe-Lit Mark on a Rotor Moves When a Trial Weight Is Moved

Figure 20 shows an important concept about how a phase reference mark moves relative tothe movement of a trial weight. This often confuses analysts, but really is a simple concept ifone takes a close look. In Figure 20A, a rotor is shown with the key weight at the top (or 0).Most analysts will put their phase reference mark in line with the key weight or some otherconvenient reference point, but it really does not matter exactly where the reference mark isapplied. If the pickup is located at point A, the 90 position, and has a zero response time (noelectronic lag), the strobe light will flash when the heavy spot is at the 90 position, and thephase mark will be seen at the top or 0 position. Now please refer to Figure 20B where theweight has been moved 90 clockwise to point B at the 180 position. Again, note that thephase mark is still at the 0 location, or 180 away from point B where the trial weight is nowlocated. If the strobe light now flashes when B is at the pickup (90 position), this means thatpoint A written on the rotor is at the 0 position while the phase mark is over at the 270position (180 away from the heavy spot). Note what happened. The weight was moved 90clockwise, but the phase mark moved 90 counterclockwise. The direction of rotation does notmatter. You get the same results. The point is this: If you want to move a phasereference mark clockwise, move the trail weight counterclockwise and vise versa.The reference mark always will shift in a direction opposite a shift of the heavy spot;and the angle that the reference mark shifts is equal to the angle that the heavy spothas shifted.

FIGURE 19. EXAMPLE ROTOR TO BE BALANCED


Instruments used to measure phase may or may not have an electronic lag, however, the effectswill still be the same as discussed above. The fact that the phase shift is predictable can be usedand the lag figured once the trial weight effect has been calculated. This will be discussed in moredetail under Balancing in One Run later in this text.

FIGURE 20.HOW A STROBE-LIT REFERENCE MARK MOVES WHEN A TRIAL WEIGHT IS MOVED

FIGURE 20A.TRIAL WEIGHT AT LOCATION A (90 CLOCKWISE FROM PHASE

REFERENCE MARK)

FIGURE 20A.TRIAL WEIGHT MOVED TO LOCATION B

(180 CLOCKWISE FROM PHASEREFERENCE MARK)

C. Single-Plane Balancing Using A Strobe Light And A Swept-Filter Analyzer

At the start of a balancing problem we have no idea how large the heavy spot is, nor dowe know where on the part it is located. The unbalance in the part at the start of our problem iscalled the ORIGINAL UNBALANCE, and the vibration amplitude and phase readings thatrepresent the unbalance are called our ORIGINAL READINGS.

In the beginning we must tune our analyzer to a frequency of 1X RPM at which time ourstrobe light will flash at a rate equal to 1X RPM. When in the filtered mode on the analyzer, thisflashing strobe will appear to freeze the rotor and our reference mark will appear to bestopped.

For example, the part in Figure 21 has an original unbalance of 5.0 mils (127 microns)at 120. Once the original unbalance has been noted and recorded, the next step is to changethe original unbalance by adding a TRIAL WEIGHT to the part. The resultant unbalance in thepart will be represented by a new amplitude and phase of vibration. The change caused bythe trial weight can be used to learn the size and location of the original unbalance, or wherethe trial weight must be placed to be opposite the original unbalance heavy spot, and howlarge the trial weight must be to be equal to the original heavy spot.


FIGURE 21.THIS ROTOR HAS AN ORIGINAL UNBALANCE OF 5.0 MILS (127 microns) AND 120 PHASE

By adding a trial weight to the unbalanced part, one of three things might happen:

1) First, if we are lucky, we might add the trial weight right on the heavy spot. If we do, thevibration will increase, but the reference mark will appear in the same position it did onthe original run. To balance the part, all we have to do is move the trail weight directlyopposite its first position, and adjust the amount of the weight until we achieve asatisfactory balance.

2) The second thing that could happen is that we could add the trial weight in exactly the rightlocation opposite the heavy spot. If the trial weight were smaller than the unbalance, wewould see a decrease in vibration, and the reference mark would appear in the sameposition as seen on the original run. To balance the part, all we would have to do isincrease the weight until we reached a satisfactory vibration level.

If the trial weight were larger than the unbalance, then its position would now be the heavyspot, and the reference mark would shift 180, or directly opposite where it was originally.In this case, all we would have to do to balance the part is reduce the amount of the trialweight until we achieved a satisfactory level.

3) The third thing that can happen by adding a trial weight is the usual one where the trialweight is added neither at the heavy spot, nor opposite it. When this happens, thereference mark shifts to a new position, and the vibration amplitude may change to a newamount. In this case, the angle and direction the trial weight must be moved, and howmuch the weight must be increased or decreased to be equal and opposite the originalunbalance heavy spot, is determined by making a VECTOR DIAGRAM.


D. Single-Plane Vector Method Of Balancing

A vector is simply a line whose length represents the amount of unbalance and whosedirection represents the angle of the unbalance. For example, if the vibration amplitude is5.0 mils (127 microns) and the phase reference mark position is 120, the unbalance can berepresented by a line with an arrowhead (a vector) 5.0 divisions long pointed at 120 asillustrated in Figure 22. To simplify drawing vectors, polar coordinate graph paper like thatshown is normally used. The radial lines, which radiate from the center, or origin, represent theangular position of the vector and are scaled in degrees increasing in the clockwise direction.The concentric circles with a common center at the origin are spaced equally for plotting thelength of vectors.

When a trail weight is added to a part, we actually add to the original unbalance. The resultantunbalance will be at some new position between the trail weight and original unbalance. Wesee this resultant unbalance as a new vibration amplitude and phase reading. In Figure 22,our ORIGINAL unbalance was represented by 5.0 mils (127 microns) and a phase of 120.After adding a trial weight, Figure 23A, the unbalance due to both the ORIGINAL PLUS THETRIAL WEIGHT is represented by 8 mils (203 microns) and a phase of 30. These tworeadings can be represented by vectors. Using polar graph paper, the ORIGINAL unbalancevector is plotted by drawing a line from the origin at the same angle as the reference mark, or120, as shown in Figure 22. A convenient scale is selected for the length of the vector. In thisexample, each major division equals 1.0 mil (25.4 microns). Thus, the ORIGINAL unbalancevector is drawn 5 major divisions in length to represent 5 mils (127 microns). The vector for theORIGINAL unbalance is labeled O.

Next, the vector representing the ORIGINAL PLUS THE TRIAL WEIGHT unbalance is drawnto the same scale at the new phase angle observed. For our example, this vector will bedrawn 8 major divisions in length to represent 8.0 mils (203 microns) at an angular position of30 that was the new phase angle. The ORIGINIAL PLUS THE TRIAL WEIGHT vector islabeled O + T in Figure 23A. These two vectors, together with the known amount of trialweight, are all thats needed to determine the required balance correction - both weightamount and location.

FIGURE 22AN UNBALANCE OF 5 MILS (127 Microns) AT 120 CAN BE REPRESENTED BY A

VECTOR DRAWN 5 DIVISIONS LONG AND POINTING AT 120


To solve the balancing problem, the next step is to draw a vector connecting the end of theO vector to the end of the O + T vector as illustrated in Figure 23 B. This connectingvector is labeled T and represents the difference between vectors O and O + T(O + T) - (O) = T. Thus, vector T represents the effect of the trial weight alone. By measuringthe length of the T vector using the same scale used for O and O + T, the effect of thetrial weight in terms of vibration amplitude is determined. For example, vector T in Figure 3Bis 9.4 mils (239 microns) in length. This means that the trial weight added to the rotorproduced an effect equal to 9.4 mils (239 microns) of vibration. This relationship can now beused to determine how much weight is required to be equivalent to the original unbalance,O. The correct balance weight is found following the formula:

Correction weight = Trial weight x O (EQUATION 8)

For our example, assume that the amount of trial weight added to the rotor in Figure 21 is 10grams. From the vector diagram, Figure 23B, we know that O = 5.0 mils (127 microns) andT = 9.4 mils (239 microns). Therefore:

Correction weight = 10 grams x 5 mils = 5.3 grams 9.4 mils

orCorrection weight = 10 grams x 127 microns = 5.3 grams

239 microns

To balance a part, our objective is to adjust vector T to make it equal in length and pointingdirectly opposite the original unbalance vector O. In this way, the effect of the correctionweight will serve to cancel out the original unbalance, resulting in a balanced rotor. Adjustingthe amount of weight according to the correct formula will make vector T equal in length tothe O vector. The next step is to determine the correct angular position of the weight.

The direction in which the trial weight acts with respect to the original unbalance isrepresented by the direction of vector T. See Figure 23B. Vector T can always be thoughtof as pointing away from the end of the O vector. Therefore, vector T must be shifted by theincluded angle (O) between vector O and vector T in order to be opposite vector O. Ofcourse, in order to shift vector T the required angle, it will be necessary to move the trialweight by the same angle. From the vector diagram, Figure 23B, the measured angle (O)between O and T is 58. Therefore, it will be necessary to move the weight 58.

A BFIGURE 23. THE SINGLE-PLANE VECTOR SOLUTION

T


Remember, the trial weight is moved from its position on the part through the angledetermined by the vector diagram. This is not an angle from the reference mark, but is theangle from the initial position of the trial weight to the required position.

To determine which direction we must move the weight, i.e., clockwise or counterclockwise,you will recall from our experiment in Figure 20 that the reference mark shifts in a directionopposite a shift of the heavy spot. Therefore, the following rule should be used to determinewhich direction the weight must be shifted:

Always shift the trial weight in the direction opposite the observed shift of the referencemark from O to O + T. Thus if the reference mark shifts counterclockwise from O toO + T, the trial weight must be moved in a clockwise direction. Or, if the observed phaseshift is clockwise, then the weight must be moved counterclockwise. This rule appliesregardless of the direction of rotation of the rotor.

By following these instructions carefully, the part should now be balanced. However, verysmall errors in measuring the phase angle, in shifting the weight, and adjusting the weight tothe proper amount can result in some remaining vibration still due to unbalance.

If further correction is required, simply observe and record the new amplitude and phase ofvibration. For example, assume that the balance correction applied according to the vectordiagram in Figure 24 resulted in a new amplitude reading of 1.0 mil (25.4 microns) and a newphase reading of 270. Plot this new reading as a new O+T vector on the polar graph paperalong with the original unbalance vector O as shown in Figure 24. Next, draw a lineconnecting the end of the original O vector to the end of the new O+T vector to find thevector T. Measure the length of the new T vector. In the example, Figure 24, T = 5.9 mils(150 microns). Using the new value for vector T proceed to find the new balance correctionweight using the familiar formula:

CORRECT WEIGHT = TRIAL WEIGHT X O T

FIGURE 24UNBALANCE CAN BE FURTHER REDUCED BY MAKING A VECTOR DIAGRAM USING

THE NEW O+T VECTOR ALONG WITH THE ORIGINAL O VECTOR


Remember that the value for the trial weight applied to this formula is the amount of weightpresently on the rotor and not the value of the trial weight applied on the first trial run. In thisexample, the original trial weight was 10 grams; however, this was adjusted to 5.3 grams as aresult of our first vector solution, Figure 23. Therefore, to solve for the new correct weight theformula becomes:

CORRECT WEIGHT = 5.3 grams X 5.0 mils = 4.5 grams 5.9 mils

orCORRECT WEIGHT = 5.3 grams X 127 microns = 4.5 grams

150 microns

To determine the new location for the correction weight, measure the included anglebetween the original vector O and the new T vector. In the example, Figure 24, thismeasured angle is approximately 5, and since the phase shift from O to the new O+Tis clockwise, the weight must be shifted 5 counterclockwise.

Applying this new balance correction should further reduce the unbalance vibration. Thisprocedure may be repeated as many times as necessary using the new O+T and trialweight value, but always using the original O vector.

E. Balancing in One Run

At the start of a balancing problem, we have no way of knowing exactly how much weight isrequired or where the weight must be added to balance the part. However, once a part hasbeen balanced, it is possible to determine how much and where weight must be added(or removed) to balance the unit or similar units in the future - in only one run.

We have learned that the amplitude of vibration is directly proportional to the unbalanceweight. Further, we also know how much vibration will result from a given amount ofunbalance. We have also learned that the phase of the reference mark moves in a directionopposite the shift of the heavy spot.

Should it be necessary to rebalance this rotor or a like rotor in the future, it will be a simplematter to determine the amount of correction weight needed. We may find that we have manylike rotors in our plant and this One Run Balancing will save extensive hours attempting tobalance rotors.

An unbalance constant can be worked out for any rotor. After you have successfullybalanced the part the first time using the 4-step or vector method, simply divide the finalbalance weight by the original amplitude of vibration. For example, if the original amplitude ofvibration was say, 12 mils (305 microns), and after balancing you note that a correction weightof 18 grams has been added, then this rotor has an unbalance constant or rotor sensitivity of:

18 grams = 1.5 grams/mil12 mils

or18 grams = 0.059 grams/micron305 microns


If this rotor requires rebalancing in the future, the amount of balance weight needed can easilybe determined by simply multiplying the new original amplitude times the constant of1.5 grams/mil. In addition to the UNBALANCE WEIGHT/VIBRATION AMPLITUDE constant,there is another constant relationship, which can be determined for finding the location of theunbalance. The position of the heavy spot in a rotor relative to the vibration pickup is definedas the FLASH ANGLE of the system. The flash angle of a rotor is the angle, measured in thedirection of shaft rotation, between the point where the vibration pickup is applied and theposition of the heavy spot when the strobe light flashes. See Figure 25.

FIGURE 25. FLASH ANGLE

The reference mark has nothing to do with this relationship since it can be placedanywhere on the rotor. The reference mark simply allows us to see the position of the rotorwhen the strobe light flashes.

To find the flash angle for a part, proceed as follows:

1. Note the original unbalance readings and proceed to balance the partusing the vector or 4-step method.

2. After the rotor has been balanced successfully, stop the work piece and turn ituntil the reference mark is in the same position observed under the strobelight on the original run.

3. With the rotor in this position, note the location of your applied balanceweight. This represents the location of the original light spot of the rotor. Ofcourse, 180 away or directly opposite the original light spot is the originalheavy spot.

4. Following the direction of shaft rotation, note the angle between the pointwhere the vibration pickup is applied and the position of the heavy spot.This measured angle is the flash angle.


After the weight constant and flash angle for a part have been learned, it is a simple matter torebalance the part in the future. The information learned by balancing one rotor can be usedfor other rotors. The RPM, pickup location and machine configuration (i.e. mass, stiffness, etc.)must be the same each time. To balance a part in one run proceed as follows:

1. Operate the machine and record the unbalance data - amplitude and phase.

2. Stop the machine and turn the rotor until the reference mark is in the sameposition observed under the strobe light.

3. With the rotor in this position, measure off the flash angle from the pickup inthe direction of shaft rotation to find the heavy spot of the rotor.

4. Next, multiply the unbalance constant times the amplitude of unbalancevibration to find the amount of weight, which must be either removed from theheavy spot or added on the light spot directly opposite.

F. Two-Plane Balancing Techniques

Two-plane balancing is done in much the same manner as single-plane balancing. Thereare, however, a number of balancing techniques commonly in use, which will yield goodresults depending upon the type of unbalance problem encountered. The choice ofbalancing techniques will depend on several factors, such as unbalance configuration,length-to-diameter ratio, balance speed compared to operating speed of the rotor, rotorflexibility and amount of cross-effect.

Two-plane balancing techniques are:

1. Separate single-plane approach - used when the rotor length-to-diameterratio is large.

2. Simultaneous single-plane approach - used when the rotor length-to-diameter ratio is larg

dynamic field balancing

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