Chapter 3 Mode Shapes

7/30/2019 Chapter 3 Mode Shapes

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67

C H A P T E R 3

Rotor Mode Shapes 3

The analysis of machinery vibration char-a cteristics must be based upon a solid w orking knowledge of the dy na mics asso-

ciated with a rotating mechanical system. Items that influence rotor motion

include the sha ft constr uction, support locat ions, and the dist ribution of massessuch as impellers, spacers, and couplings. Certainly the stiffness and damping

characteristics of the bearings and machine support structure will play an

important role in influencing rotor behavior. In addition, the relationship

between the operat ing speed range, and t he system critica ls are very importa nt.

These factors are parameters for establishing the response characteristics

of the rotor system, and the associated dynamic rotor mode shapes. I t is mean-

ingful to recognize that many machine characteristics can be adequately

described by a pplying t he ba sic concepts of mass a nd support distribution, sys-

tem viscous da mping, plus stiffness of th e ma jor elements a nd support structur e.

It must also be recognized that mode shapes may be forced deflections based

upon t he a ctive system forces. They ma y a lso be na tur a l modes of vibrat ion (res-

onances), plus a combination of the two to yield a forced resonant response.

MASSAND SUPPORT DISTRIBUTION

The distribution of weight and supports along the length of a rotor estab-

lishes the static deflections, plus the static bearing loads. For example, consider a

consta nt dia meter sha ft tha t is simply support ed betw een tw o points a s shown in

Fig. 3-1. This shaft will have a maximum deflection at the midspan, and each

support locat ion w ill ca rry one ha lf of the sha ft w eight.

If one support is moved to the rotor mids pan , the condit ion described in Fig.

3-2 will occur. In this case, the maximum deflection occurs at the unsupported

end of the shaft. The load applied to each support will now be dependent upon

the support characteristics. Specifically, if the left support is a free support, it

will not ha ve a st a tic loa d. Under t his condition, the center support w ill ca rry t heentire weight, and the shaft will be balanced on this center pivot. On the other

hand, if the left support is connected to the rotor (e.g., a bearing), it will produce

a vertical rea ction if t he middle support is not perfectly centered. For t his rea son,


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68 Chapter-3

overhung ma chines such as power tur bine rotors for dual sha ft ga s tur bines, or

overhung blowers must be carefully examined to determine the static anddyna mic bea ring loads, and directions.

Next consider the a ddition of a concentr a ted load (e.g. , a n impeller) a t the

middle of the rotor. The diagram presented in Fig. 3-3 represents the deflection

associated with the additional force applied at the midspan of the simply sup-

ported shaft. Clearly the center deflection must increase when the additional

loa d is applied. In a ddition, wit h th e supports locat ed at the sha ft ends, it is rea -

sonable to conclude that the total weight (shaft plus midspan load) will be

equally sha red betw een the tw o support s.

Fig. 3-4 illustr a tes t he condition of an overhung rotor wit h t he a ddition of a

center weight. In this configuration, one support is located directly below the

concentrated midspan load. In this case, the mode shape, and the maximum

defl ection a re identica l to Fig. 3-2 wit h zero externa l loa d. However, th e force bal-

ance has been altered, and the center support must now carry the shaft weight

plus the center load.

Fig. 31 Simply Supported Shaft WithStatic Deflection Due To Beam Weight

Fig. 32 Overhung Shaft With StaticDeflection Due To Beam Weight

Fig. 33 Simply Supported Shaft Deflected ByBeam Weight And Center Load

Fig. 34 Overhung Shaft Deflected ByBeam Weight And Center Load

1/2 Weight1/2 Weight

Uniform Shaft

Maximum Deflection

20

MaximumDeflection

Uniform Shaft

20

Maximum Deflection

ConcentratedLoad

35

ConcentratedLoad

MaximumDeflection

20


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Mass and Support Distribution 69

Next, consider the static mode shapes displayed in Figs. 3-5 and 3-6 that

describe the infl uence of moving the concentr a ted load from t he midspan to the

end of the rotor. For the shaft simply supported between bearings (Fig. 3-5), the

mode sha pe return s ba ck to the init ia l condit ion (Fig. 3-1). The a ddit ional load isdirectly transmitted to the right hand support. Under this configuration, the

deflected mode shape is dependent only on the shaft weight, but the support

loa ds a re clearly d ifferent.

Fina lly, for t he overhu ng ca se of Fig. 3-6, th e cant ilevered loa d a t t he end of

the sha ft results in a n increase in th e maximum deflection. This ty pe of behavior

certainly makes intuitive sense, and it is representative of real overhung

ma chines. I t should a lso be recognized tha t the a pplica tion of this loa d t o the free

end of the rotor will result in a downward vertical restraining force at the left

end support. Aga in, this is consistent w ith the forces a nd m oments encountered

in ma chines such as power t urbines and overhung blowers.

Overall , it is r ecognized tha t the st a tic support forces (i.e. , at the bea rings),

plus the location a nd ma gnitude of the maximum deflection a re dependent upon

the support characteristics. I t is apparent that the addition of elements such as

impellers, couplings, balance pistons, and spacers will directly influence the

resulta nt support forces, a nd the a ssociat ed maximum st a tic deflection.

Simple mechanical systems can often be modeled as a uniform weight dis-

tribution for the shaft, combined with concentrated loads for the impellers. The

static deflections can be calculated with beam theory, and the static bearing

loads determined by summation of moments. Reference books such as Roarks

Form ul as for Stress and Str ain

1

provide characteristic equations for many typi-

cal mecha nical systems w ithout resorting t o deta iled beam calculat ions.For more complex rotors, it is necessa ry t o divide th e rotor into discrete a nd

Fig. 35 Simply Supported ShaftDeflected By Beam Weight And End Load

Fig. 36 Overhung Shaft Deflected ByBeam Weight And End Load

1 Warren C. Young, Roarks Form ulas for Str ess & Strai n

, Sixth Edition, (New York: McGraw-Hill B ook Compan y, 1989).

ConcentratedLoad

Maximum Deflection

20

MaximumDeflection

ConcentratedLoad

50


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70 Chapter-3

defina ble segments. Ba sed upon th e dimensions a nd ma teria l density, it is possi-

ble to calculate the weight for each station. From this weight distribution, the

sta tic loa ds a t both bea rings can be computed. This a pproach identifies t he ma ss

distribution a long t he rotor, plus t he resulta nt bearing forces in a sta tic position.

The w eight of each section or port ion of a rotor is dependent upon t he phys-ica l dimensions, plus t he density of th e sha ft m a teria l. For example, the hollow

circular cylinder d epicted in F ig. 3-7 is dimensionally specified by a n outer dia m-

eter D

o

, an inner diameter D

i

, and an overall length L

. The shaft radius to the

outer diameter is R

o

, and the internal bore radius is identified as R

i

. Based on

these dimensions, a variety of necessary calculations may be performed. For

instance, from plane geometry, the cross sectional area of this annulus is com-

puted by subtr a cting circula r a reas in th e following ma nner:

(3-1)

The volume of the a nnulus ma y be determined by multiplying the cylinder

length L

times t he cross sectiona l a rea A

annu lus

a s sta ted in the next expression:

(3-2)

The weight of the annulus may be calculated by multiplying the cylinder

mat erial densi ty

times the tota l volume V

annu lus

a s show n in eq ua tion (3-3):

(3-3)

For reference purposes, the densities of many common metals have been

summa rized in Appendix B of this t ext. This t a bulat ion a lso includes the modu-

lus of elasticity, the shear modulus, and the coefficient of thermal expansion for

each meta l. These fundam enta l properties a re referred to t hroughout this book,and it is convenient to have typical values readily available. I t should be men-

tioned th a t some ma teria ls do not h ave t otally uniq ue properties. For critical cal-

cula tions the precise physica l properties should be obta ined from a meta llurgica l

reference source, or specifi c tests of the met a l.

Fig. 37 Hollow Circular CylinderWith Physical Dimensions

Length(L)

Outer

Diameter(D

o)

Inner

Diam.(

Di)

Do=2 R

o

Di=2 R

i

A a n n u l u s Ro2 Ri

2 4--- Do

2Di

2( )= =

Va n n u l u s L A a n n u l u s L

4------------- Do

2Di

2( )= =

Wa n n u l u s Va n n u l u s L

4----------------------- Do

2Di

2( )= =


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For some calculations, it is necessary to know the mass of the mechanical

part. For the annulus under discussion, the mass is easily determined by divid-

ing the weight from equation (3-4) by the acceleration of gravity G

a s follows:

(3-4)

In ma ny insta nces, ma chinery sha fts a re not hollow, a nd they a re fabricat ed

of solid metal. I f the inner diameter D

i

is set equal to zero in equations (3-1)

through (3-4), the following expressions for solid machine shafts with an outer

diam eter of D

are easily developed:

(3-5)

(3-6)

(3-7)

(3-8)

In order to be perfectly clear on the dimensional aspects of equations (3-1)

to (3-8), the defi ned va ria bles a nd t heir respective English engineering unit s a re

summa rized as follows:

L

= Cylinder Length (Inches)

D

= Solid Cylinder Outer Diameter (Inches)

D

o

= Hollow Cylinder Outer Diameter (Inches)

D

i

= Hollow Cylinder Inner Diameter (Inches)

R

o

= Hollow Cylinder Outer Radius (Inches)

R

i

= Hollow Cylinder Inner Radius (Inches)

= Cylinder Material Density (Pounds / Inch

3

)

A

= Cylinder Cross Sectional Area (Inches

2

)

V

= Cylinder Volume (Inches

3

)

W

= Cylinder Weight (Pounds)

M

= Cylinder Mass (Pounds-Second

2/ Inch)

G

= Acceleration of Gravity (= 386.1 Inches / Second

2

)

These variables and associated units will be used in the next sections on

inertia , plus thr oughout the rema inder of this t ext.

Ma n n u l u s

Wa n n u l u s

G--------------------------

L 4 G

----------------------- Do2

Di2

( )= =

A so l i d D2

4-----------------=

Vso l i d L D2

4---------------------------=

Wso l i d L D2

4-------------------------------------=

Mso l i d L D2

4 G-------------------------------------=


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72 Chapter-3

Case History 3: Two Stage Compressor Rotor Weight Distribution

Consider th e compressor r otor depicted in Fig. 3-8. This h igh speed rota tin g

a ssembly consists of a sha ft w ith va rious dia meters, plus two midspan impellers,a nd a thr ust collar. Since the impellers are mounted ba ck to back, this part icular

design does not include a bala nce piston. As indicat ed on the dia gra m, the r otor

weights 282 pounds, and there is a moderate overhang on the thrust end of the

rotor. Norma l opera tin g speed va ries betw een 15,000 a nd 17,500 RP M. This com-

pressor is a drive through unit with a stream turbine coupled to one end, and

a nother compressor coupled to the opposite (thr ust ) end. Hist orically, the bea ring

on t he drive end of the rotor seldom exhibited any da ma ge, but t he journal bea r-

ing located at the thrust end was often found to be in a distressed or damaged

condition. On other occasions, the bea rings w ould fa il during opera tion, and the

rotor would be severely damaged. Depending on the severity of the failure, the

sha ft w a s often chrome pla ted. After severe fa ilures, the entire rota ting a ssembly

wa s t ota l ly replaced.

This machine operates in a very dirty and corrosive environment. The rotorwas constantly subjected to large unbalance forces due to the accumulation of

Fig. 38 Two Stage Cen-trifugal Compressor RotorWith Weight And MomentDiagram

7.49"

17.96"

24.03"

27.24"

33.43"

40.68"43.66"

45.92"

49.25"

53.06"

2.62"

1.11#

4.83#

10.58#

36.41#

19.21#

37.18#

32.40#

37.18#

17.37#

32.67#

10.40#

9.13#

17.51#

13.84#

2.09#

Seal

Thrust EndJournal

Wa Wb

Impeller

ThrustCollar

Seal

Impeller

Drive EndJournal

BearingJournal3.62"

Seal Seal

ThrustCollar

Impellers14.5"

Total Rotor Weight282 Pounds

40.68"

14.51"4.40"

21.17"


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material in the impellers, plus random corrosion of the impeller vanes. During a

typical run, t he compressor w ould st a rtup sm oothly a fter a n overha ul. How ever,

synchronous running speed vibration response would always deteriorate with

the pa ssa ge of time.An initia l step in the a na lysis of this problem required th e determinat ion of

sta tic bearing loa ds. To achieve this goal, the rotor w a s divided into fi fteen sec-

tions, a nd th e weight of each section wa s calcula ted. Since this wa s a solid shaft,

equation (3-7) was used to compute the weight of each shaft section. The weight

of the thrust collar was determined with (3-3), and the two impellers were

weighed separately. Next, the thrust collar and wheel weights were combined

with their respective shaft section weights, and the weight distribution summa-

riz ed in F ig. 3-8.

This information was then combined with the distance from the center of

the drive end bearing to the centroid of each rotor segment. The complete array

of weights a nd dist a nces a re shown in Fig. 3-8. From t his sketch, it is possible to

perform a summation of moments around the center or transverse axis of the

drive end journa l in t he following ma nner:

(3-9)

Since the total rotor weight is 282 pounds, the next force balance applies:

Thus, the drive end bear ing ha s a 116 pound sta tic loa d, and t he thrust end

journ a l car ries 166 pounds. Alth ough t he different ia l force is only 50 pounds, it is

an appreciable percentage difference. Ultimately, it was determined that the

bearings w ere only ma rgina lly sized t o accommoda te t he rotor weight. However,

they were considerably undersized when the additional unbalance forces due to

foreign objects were included. It was also determined that the available load

capacity for the drive end wa s bar ely accepta ble, whereas t he loa d carry ing ca pa-

bility for the thrust end journal bearing was unacceptable. Based upon these

conclusions, both bearings were increased in size and load capacity.

The lar ger bearings r educed the num ber of ma chine fa ilures per yea r, a nd

overall reliability was substantially improved. Further improvements in

ma chine longevity w ould r equire changes in t he chemical plan t process. Unfortu-

nately, the required alterations to the processing scheme could not be economi-

cally justified. Hence, the compressor rotor was occasionally sacrificed to meet

Momen t sccw Momen t scw=17.54 Inch-Pounds Wb 40.68 Inches+ 6 768.74 Inch-Pounds,=

Wb 40.68 Inches 6 751.20 Inch-Pounds,=

Wb 166 Pounds=

Wa Wb+ 282 Pounds=

Wa 166 Pounds+ 282 Pounds=

Wa 116 Pounds=


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74 Chapter-3

production quota s.

From this example, it is clear that even a simple analysis of rotor weight

distribution and bearing static loads may be beneficial. In some cases this may

solve a problem, or it may provide insight into prospective solutions. It must alsobe recognized tha t the dy na mics of the r otat ing syst em must be considered. This

includes the effects of mass unbalance that serves to deform the mode shape,

plus the effects of inertia, stiffness, and damping of machine elements.

INERTIA CONSIDERATIONSAND CALCULATIONS

As shown in the previous example, the calculation of rotor weights is easily

achieved for rotors that can be segmented into a series of cylinders. For a homo-

geneous material, the volume and weight of each section may be computed. The

summa tion of individual cylindrical section w eights yields the tota l rotor w eight.

In most inst a nces, this is an a ccepta ble and a chieva ble computa tional procedure.

However, problems occur wit h ma chines t ha t h a ve non-symmetr ic shaft s, or a recomposed of multiple materials. For instance, the rotor weight of electric

machines, such as motors and generators, may be indeterminate due to an

unknown combina tion of iron, insulat ion, an d copper w ithin t he rotor windings.

In t hese situa tions, the rotor a ssembly ca n be weighed on a scale, a nd t he distri-

bution of weight between bearing journals estimated. Any additional elements

such as a cooling fan , flyw heel, or overhung exciter ma y t hen be added t o deter-

mine the total rota ting syst em weight. Hence, wh en accura te calcula tions ca nnot

be performed, the machinery diagnostician can revert to the traditional experi-

menta l technique of weighing the rotor.

In addition to weight or mass, the inertia properties of a rotating system

must a lso be considered during a ny dy na mic ana lysis. Due to th e complexity of

inertia calculations, and the strong potential for confusion between area and

ma ss m oment of inertia s, significant errors a re possible. The physics definit ion of

inert ia is t he property of a body to resist a ccelerat ion. This includes the t endency

of a body a t rest to remain a t r est, or th e tendency of a body in motion to stay in

motion. This reference to a physical body implies mass, which suggests the pres-

ence of a three-dimensional object. Hence, an area moment of inertia is some-

wh a t of a misnomer since a plane a rea ha s no depth, and t herefore no mass, nor

the potentia l for inertia . Nevertheless, area moments of inertia abound in th e lit-

erature, and for the sake of completeness, they will be discussed.

It should a lso be mentioned tha t confusion regarding inertia is further com-

pounded by some of the technica l litera ture th a t t ra nsposes pola r w ith t he tra ns-

verse inertias. In other cases, the authors fail to identify the specific type of

inertia calcula tions, much less the reference inertia a xis. In a n effort to circum-

vent these potential sources of confusion, the following discussion of inertia isoffered for considera tion.

It is desira ble to begin wit h a review of the a rea m oment of inertia. In Fig.

3-9, Sketch A shows an irregular plane surface with axes x-x a nd y-ypassing

through the a rea designat ed as A . These recta ngular a xes may be in any location


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Inertia Considerations and Calculations 75

w ith respect t o the ar ea. If t he ar ea w a s symm etric (e.g. , circle or recta ngle), and

if the axes were axes of symmetry, they would be called the principal inertiaaxes. However, the general solution for inertia about each axis is determined

from str ength of mat erials in tegrals a s :

(3-10)

(3-11)

The Xa nd Yterm s in (3-10) and (3-11) are dist a nces from ea ch respective

axis to the incrementa l area identifi ed as dA . This definition of area moment of

inertia is consistent with references such as Roark2, or Shigley3. In both equa-

tions, inertia is calculated by multiplying each area by t he squa re of the dista nce

to th e respective reference a xis. The a rea is measur ed in inches squa red, and thisis multiplied by the square of the distance to the axis in inches. This product

yields an a rea m oment of inert ia w ith E nglish engineering units of Inches4.

In Sketch Bof Fig. 3-9, a t hird a xis z-zha s been a dded. This new a xis is per-

pendicular to the plane of the area A , and it passes through the intersection of

axes x-xa nd y-y. The dist a nce from the int ersection of a xis z-zwith the a rea A, t oany incrementa l area dA is identifi ed by the ra dius R.The inert ia a bout t his a xisis termed the a rea polar m oment of inertia Jtha t is g iven by:

(3-12)

From plane geometry i t is known tha t the ra dius R, may be expressed in

terms of Xa nd Ycoordinat es as:

Fig. 39 Area Moment Of Inertia For An Irregular Plane Surface

2 Warren C. Young, Roarks Form ulas for Str ess & Strai n, Sixth Edition, (New York: McGraw-Hi ll B ook Compa ny, 1989), p. 59.

3 Joseph E. Shigley and Charles R. Mischke, Standar d H andbook of Machin e Design, (NewYork: McGraw-Hill Book Company, 1986), p. 9.13.

x

y

y

x x

y

y

x

y0

y0

XOffset

z

x

y

y

x

z

Sketch A Sketch C Sketch B

Area (A)IncrementalArea (dA)

At Coordinates(Xand Y)

IncrementalArea (dA)

At Radius (R)from Axis z-z

Ixx Y2

Ad=

Iyy X2

Ad=

Jzz R2

Ad=


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76 Chapter-3

(3-13)

E qua tion (3-13) may be insert ed int o (3-12), an d expa nded a s follow s:

Substituting (3-10) and (3-11) into the above expression, the polar inertia

Jzzis equa ted to the summat ion of the transverse area inert ia s Ixxa nd Iyya s:

(3-14)

For a symmetrical a rea such as a circle, the tra nsverse area inert ia s Ixxa nd

Iyya re equa l. Thus, setting inertia s Ixx=Iyy=I, equa tion (3-14) ma y be rest a ted in

the following simplified format :

(3-15)

Since most rotating shafts are circular, equation (3-15) is common within

the machinery business. The applicability of this geometric simplification will be

better a pprecia ted during the forthcoming discussion of ma ss moment of inertia.

However, before a ddressing tha t t opic, the tr a nslat ion of the inertia a xis should

be reviewed. Specifically, Sketch Cin Fig. 3-9 identifies a new axis yo-yotha t i s

para llel to the previously defin ed vertica l a xis y-y. The consta nt dista nce betw een

the tw o axes is identifi ed as Xoffset. I t ca n be shown th a t t he area m oment of iner-

tia a bout t his new a xis is given by the expression:

(3-16)

The same argument may be made in the perpendicular plane for the area

polar moment of inertia. Consider a new polar axis zo-zo that is paral le l to the

previously defined z-zaxis. If the distance between the two axes is identified as

Roffset, the area moment of inert ia a bout t his new polar a xis is given by:

(3-17)

The moment of inertia of a n a rea a bout a part icular a xis may be converted

to a mass moment of inertia by including the thickness, and the density of the

body. This convert s a tw o-dimensiona l into a t hree-dimensiona l problem, a nd t he

complexity of the associated equation structure increases proportionally. For

explan a tion purposes, a ssume tha t the plan e ar ea sh own in F ig. 3-9 is expandedby ad ding a fi nite depth. If a m a teria l density is a pplied to the resulting volume,

th e ma ss s hown in Fig. 3-10 evolves. Sketch A in th is diagra m reveals tha t a tw o-

dimensional x-ycoordinate system is inadequate for locating a point within the

R2

X2

Y2

+=

Jzz X2

Y2

+( ) Ad=Jzz X

2Ad Y2 Ad+=

Jzz Ixx Iyy+=

Jzz I I+ 2 I= =

Iyoyo Iyy A Xoffset( )

2

+=

JzozoJzz A Roffset( )

2+=


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m ass M. Hence, the three-dimensional x-y-z coordinate system in Sketch Bof

Fig. 3-10 is necessary to locate an incremental mass dMwit hin the boundaries of

the object. The equat ions for ma ss moment of inert ia a round the t hree axes a re

defined by sources such a s M arks Hand book4 a s follows:

(3-18)

(3-19)

(3-20)

The translation of one axis to another parallel axis also applies to mass

inertia a s w ell a s to the previously discussed ar ea moment of inertia . For exam -

ple, the polar mass moment of inertia around axis z-zmay be translated to the

para l lel a xis zo-zowith the following common expression:

(3-21)

Within t his equat ion, the dista nce Xoffsetrepresents t he para llel offset a long

t he x-xaxis as shown in Sketch Cof Fig. 3-10. If the offset of the new axis was

further displaced from both the x-x a nd y-y axes, the polar moment of inertia

along the new axis identifi ed as z1-z1w ould be given by t he following:

(3-22)

The dista nce betw een t he z-zaxis and the z1-z1 axis is defi ned by the ra dial

offset Roffset. From (3-13) it is know n t ha t Roffset2= Xoffset

2+ Yoffset2. The va lidity

Fig. 310 Mass Moment Of Inertia For An Irregular Solid Object

4 Eugene A. Avallone and Theodore Baumeister III, Marks Standard H andbook for M echani-cal E ngineers, Tent h E dit ion, (New York: McG ra w -Hi ll, 1996), p. 3-9.

x

y

y

x x

y

y

x

z0

z0

XOffset

z

x

y

y

x

z

Sketch A Sketch C Sketch B

z

z

Mass (M)

IncrementalMass (dM)

At Coordinates(X, Yand Z)

z1

z1

YOffset

Ixx Y2

Z+2

( ) Md=

Iyy X2

Z+2

( ) Md=

Izz X2

Y+2

( ) Md=

JzozoJzz M Xoffset( )

2+=

Jz1z1Jzz M Roffset( )

2+=


12/82

78 Chapter-3

of this is demonstrated by allowing Yoffsetto equal zero, and noting that (3-22)

th en r everts ba ck to equ a tion (3-21). In a ll ca ses, (3-21) an d (3-22) ar e a na logous

to the parallel axis equation (3-17) for a plane area. In fact, this similarity in

ma thema tical forma ts is t he source for some of the confusion a ssociat ed wit h th etopic of inert ia. In order to mainta in a proper distinction betw een a rea a nd ma ss

moment of inertia, the diagnostician should always pay close attention to the

engineering units. Recall that the area moment of inertia carries English engi-

neering units of Inches4, and the ma ss inertia ha s units of Pound-Inch-Second2.

The integrals shown on the last few pages are interesting calculus topics,

but t hey do not help the fi eld diagnosticia n unt il they a re solved for specifi c geo-

metric sha pes. For inst a nce, it is meaningful t o develop the equa tions for a circu-

lar shaft cross section. In order to maintain continuity, the hollow circular

cylinder from Fig. 3-7 will be reused. For the calculation of inertia, the principal

a xes pa ssing through the a xial centerline o-o, a nd the midspan diamet ra l center-

line b-bhave been shown in Fig. 3-11. For this geometric figure, the cross sec-tional area moment of inertia Iareaalong a diameter of the hollow circle is given

by Spotts5, or Ha rris6in the following common format:

(3-23)

The area polar moment of inertia Jareais computed about an axis that is

perpendicula r t o the circular cross section. On Fig. 3-11 th is w ould be the center-

line axis o-oof the cylinder. The polar inertia of this circular area is equal to

twice the inertia along a diameter (i .e. , J =2xI) as evident from (3-15). Multiply-

Fig. 311 Hollow CircularCylinder With PrincipalAxes Identified

5 M.F. Spotts, Design of M achin e Elements, 6th E dit ion, (Englewood Cliffs , New J ersey: Pren-tice-Ha ll, In c., 1985), p. 18.

6 Cyril M. Harris , Shock and Vibr ation H andbook, Fourth edition, (New York: McGraw-Hill,1996), p. 1.12.

Length(L)

Outer

Diameter(D

o)

Inner

Diam.(

Di)

b

b

oo

DiameterCenterline

AxialCenterline

Do=2 R

o

Di=2 Ri

Ia reaa n n u l u s

Do4

Di4

( )64

-----------------------------------=


13/82


ing equa tion (2-23) by t w o, the next expression for the a rea pola r inert ia evolves:

(3-24)

The va lidity of equa tion (3-24) is supported by Spott s, page 150, an d other s.

Note that the last two expressions are plane area moments of inertia (i .e. , no

depth, no density, a nd no w eight). However, in ma ny m a chinery ca lculat ions it is

ma nda tory t o consider th e ma ss of the element. When ma ss is included, the cal-

cula t ions become more complex. Specifica lly, if th e hollow cy lind er in Fi g. 3-11 is

considered to be a rotating machinery shaft, the mass polar moment of inertia

a long t he axis of rota tion (a xis o-o) is defi ned by G ieck7 a nd others. If these equa -

tions are placed into the nomenclature used in Fig. 3-11, the general expression

for th e mass pola r moment of inertia of the cylinder ma y be sta ted a s:

(3-25)

Equation (3-25) is the mass inertia term of a rotating shaft that is often

described as t he WR2 of the rotor. In a ctuality, this q ua ntit y should be identifi ed

as the mass polar moment of inertia, but common nomenclature sometimes

supersedes technica l a ccura cy. Neverth eless, if the dia meters a re used inst ead of

the radii, and if (3-4) for the annulus mass is substituted into (3-25); the follow-

ing ma nipulation ma y be performed t o reach a common equa tion used for calcu-

lat ion of the ma ss pola r m oment of inertia for a hollow cylinder.

Simplifica tion of the last expression yields th e following equa tion:

(3-26)

This mass polar inert ia is a s importa nt t o a torsional a nalysis as t he mass

is necessa ry for a lat eral a na lysis. Specifi cally, equat ion (2-102) identifi es the ut i-

lizat ion of the polar inert ia t o compute a t orsional na tura l frequency in the same

way that the mass is used in (2-44) to compute a lateral natural frequency. As

show n in Ta ble 2-1, th e inertia l term in t he genera l equa tion of motion is gov-

erned by ma ss in a latera l system, and polar inert ia in a torsional syst em. Dur-ing the analysis or modeling of real machinery both the polar inertia and the

7 Kurt Gieck and Reiner Gieck, Engineer in g Form ulas, 6th edition, (New York: McGraw-HillInc., 1990), p. M3.

Ja reaa n n u l u s

Do4

Di4

( )32-----------------------------------=

Jmassa n n u l u s

Ma n n u l u s

2-------------------------- Ro2

Ri2

+( )=

Jmassa n n u l u s

Ma n n u l u s 2

-------------------------- Ro2

Ri2

+( )Ma n n u l u s

2--------------------------

Do2

4-------

Di2

4-------+

==

Jmassa n n u l u s

Ma n n u l u s

8-------------------------- Do

2Di

2+( ) L

4 G 8----------------------- Do

2Di

2( ) Do

2Di

2+( )==

Jmassa n n u l u s L

32 G----------------------- Do

4Di

4( )=


14/82

80 Chapter-3

transverse inertia are utilized. Since (3-26) defines the mass polar moment of

inertia Jmassthrough the axial centerline o-o, i t is now reasonable to define the

transverse mass Inert ia Imassof the hollow cylinder. Although any defined axis

may be used for the calculations, the customary midspan diameter axis b-b

depicted in Fig. 3-11 will be used for the following exercise. Again, extracting a

sta ndard mass inert ia equation from G ieck, a nd modifying the t erms t o be con-sistent with Fig. 3-11, the general expression for the mass moment of inertia on

the cylinder diam eter axis b-bpassing t hrough the center of gravit y is a s follow s:

(3-27)

If diameters are used instead of radii, and if the mass equation (3-4) is

included, the tra nsverse inertia equa tion (3-27), ma y be m odifi ed a s shown :

If t he ma ss pola r m oment of inert ia from (3-26) is substit uted int o the last

expression, the tra nsverse inertia equa tion may be simplified a s follow s:

(3-28)

If the length L of the annulus is sm a ll compa red to the outer diam eter, the

influence of the far right hand term in equation (3-28) is significantly dimin-

ished, and the following a pproximat ion is often used:

(3-29)

This result is consistent wit h t he previous relationship described by equa -

tion (3-15) between the ratio of inertias for a plane circular cross section. In

actual practice during the analytical modeling of a rotor system, the shaft sta-

tion lengths are normally kept fairly short, and equation (3-29) may be a good

Imassa n n u l u s

Ma n n u l u s

12-------------------------- 3Ro

23Ri

2L

2+ +( )=

Imassa n n u l u s

Ma n n u l u s

12-------------------------- 3Ro2

3Ri2

L2

+ +( )

Ma n n u l u s

12--------------------------

3Do2

4----------

3Di2

4---------- L2

+ +

= =

Imassa n n u l u s

Ma n n u l u s

16-------------------------- Do

2Di

2 4L2

3----------+ +

=

Imassa n n u l u s

Ma n n u l u s

16-------------------------- Do

2Di

2+( )

Ma n n u l u s

12-------------------------- L

2+=

Imassa n n u l u s L 16 4 G-------------------------- Do

2Di

2( ) Do

2Di

2+( ) L

12 4 G-------------------------- Do

2Di

2( ) L 2+=

Imassa n n u l u s L

64 G----------------------- Do

4Di

4( )

L 3

48 G-------------------------- Do

2Di

2( )

+=

Imassa n n u l u s

Jmassa n n u l u s 2

-------------------------------- L 3

48 G-------------------------- Do

2Di

2( )

+=

Imassa n n u l u s

Jmassa n n u l u s 2

--------------------------------


15/82


a pproxima tion. For longer sh a ft s ections, th e more complex equa tion (3-28) must

be used. Equa tion (3-29) is part icular ly useful for estima ting the t ra nsverse iner-

tia of wheels that have a large diameter, and a comparatively short length. For

example, a centrifugal compressor impeller may be 24 inches in diameter, with adisk and a cover thickness of only 0.25 inches each. In this situation, one half of

the mass polar moment of inertia Jmasswill be very close to the deta iled tr a ns-

verse mass moment of inertia Imass.

As mentioned earlier in this chapter, the diagnostician must always be

aw a re of the potentia l dilemma in the a pplica tion of inertia w ithin t echnical doc-

uments, and computer programs. For instance, one set of software uses weight

inertia instead of mass inertia. The output inertia values from these programs

carry inertia units of Pounds-Inches 2 instead of Pound-Inch-Second2. Although

the difference betw een t he tw o inertia values is only th e accelerat ion of gra vity,

the results ca n be confusing to the unprepared. In a ll ca ses, it is ma nda tory t o be

completely knowledgeable of a ll a spects of a ny inertia calcula tions. Within this

text, inertias an d their respective English engineering units a re as follows:

Iarea = Area Moment of Inertia on Diameter (Inches4)

Jarea = Area Polar Moment of Inertia (Inches4)

Imass = Mass Transverse Moment of Inertia (Pound-Inch-Second2)

Jmass = Mass Polar Moment of Inertia (Pound-Inch-Second2)

In ma ny insta nces, ma chinery sha fts a re not hollow, a nd they a re fabricat ed

of solid meta l. I f the inner dia meter Diis set equal to zero in equations (3-23), (3-

24), (3-26), and (3-28), the following inertia expressions for solid shafts with a

diam eter of Dare easily developed:

(3-30)

(3-31)

(3-32)

(3-33)

Once aga in th ese expressions may be verifi ed from va rious sources such a s

Spotts, Marks, or S higley. Plea se reca ll tha t the developed equa tions a re bas ed

upon a circular cross section. If t he cross sectiona l a rea is not circula r, then the

equa tions must be modified based upon the original int egrals used to define iner-

tia. Often this type of calculation is not practical due to the complexity of the

mechanical part. In these cases, the diagnostician must resort to other tech-

Ia reaso l i d D4

64

-----------------=

Ja reaso l i d D4

32-----------------=

Jmassso l i d L D4

32 G-------------------------------------=

Imassso l i d

Jmassso l i d 2

------------------------- L 3 D2

48 G---------------------------------------

+=


16/82

82 Chapter-3

niques to determine th e inertia properties of the ma chine element.

In the same way that a scale may be used to determine the weight o f a

rotor, there a re experimental techniques tha t m ay be applied to determine iner-

tia properties. For example, consider the machine part shown in Fig. 3-12. Thiscould be a bla nk for a bull gear or a fl yw heel, or a ny other ma chine element for

wh ich a ma ss polar moment of inertia ma y be required. Due to the complexity of

the par t it ma y not be fea sible to calculat e the inertia directly, but it is possible to

experimenta lly determine the inert ia.

One technique consists of suspending th e part from a h orizontal support a s

shown in Fig. 3-12. Idea lly, this s upport member sh ould be a knife edge, but more

realistically, it w ill probably be a solid circular rod a s depicted. If t he ma ss pola r

moment of inertia thr ough t he axia l centerline of the element is desired, then t he

distance between that centerline and the supporting pivot point Xoffset must be

accurately measured. In addition, the part should be weighed to determine thetotal weight Win pounds. The machine element may now be rocked back and

forth a s a compound pendulum. By measur ing t he period of the m otion, the iner-

tia about the pivot point may be determined. This offset inertia may now be

translated back to the centerline of the machine part by applying the parallel

axis equations previously developed.

In actual practice, the Inertia of this compound pendulum may be deter-

mined from M arks Hand book8. Ext ra cting the appropriat e equa tion, a nd pla cing

it in terms used with in this t ext, the following equa tion for overa ll inertia a bout

the pivot point m ay be stat ed:

(3-34)

Fig. 312 MechanicalArrangement For RockingTest To Determine MassPolar Moment of Inertia

8 Eugene A. Avallone and Theodore Baumeister III, M arks Standard H andbook for M echani-cal Engi neers, Tent h E dit ion, (New York: McG ra w -Hil l, 1996), p. 3-16.

ElementWeight (W)

RockingOscillation40 to 60

Fixed andRigid

SupportX

0ffset

Jzo

zo

W Xoffset Per iod2

4 2-----------------------------------------------------------=


17/82


If equation (3-34) is inserted into the axis translation equation (3-21), the

following is obtained:

Factoring out th e common terms, t he a bove may be simplified t o:

(3-35)

This is a useful expression since all of the variables are easily determined.

Specifically, the weight Wof the machine part can be measured on a scale, and

the dist a nce betw een t he pivoting point a nd t he geometric center of the element

Xoffset is easily measured. The acceleration of gravity Gis constant , and the

Periodof the swinging motion is determined with a stopwatch in seconds. Nor-

ma lly, a series of runs a re ma de to determine the a verage period of oscillat ion.

Experimental techniques are always more credible if the validity of the

equations, and the experimental procedures can be verified with a controlledtest. With an inertia experiment, the object to be tested should be of simple

geometry such as the solid cylinder shown in Fig. 3-13. As indicated on the dia-

gram, the average cylinder diameter Dis equa l to 6.312 Inches, and t he avera ge

length L is 3.735 Inches. A shop scale indicated that the weight Wof the speci-

Fig. 313 Rocking Test ToDetermine Mass PolarMoment Of Inertia Of ASolid Brass Cylinder

Jzozo Jzz M Xoffset( ) 2+=

Jzz JzozoM Xoffset( )

2=

Jzz

W Xoffset Per iod2

4 2----------------------------------------------------------- M Xoffset( )

2=

Jzz

W Xoffset Per iod2

4 2-----------------------------------------------------------

W

G----- Xoffset( )

2=

Jzz W XoffsetPer iod

2 -------------------

2 XoffsetG

------------------ =

BrassCylinderWeight

(W=36 Lbs.)

RockingOscillation40 to 60

Fixed andRigid

Support

X0ffset

=5.

07"

Length

(L =3.735")Diameter

(D=6.312")


18/82


19/82


bra ss cylinder ha d a tendency to wa lk down the r od. This problem wa s corrected

by re-leveling t he support rod a nd t he cylinder.

Follow ing th ese test setup m odifi cations, the a ctua l test wa s conducted. The

bra ss cylinder w a s displaced a bout 30 from th e vert ica l centerline and released.The peak of the motion at one extremity was visually sighted, and a stopwatch

was used to measure the time required for multiple back and forth cycles. The

fi nal measured test da ta is summarized a s fol lows:

Rocking Run #1 7.83 Seconds for 10 cycles








Rocking Run #9 7.93 Seconds for 10 cyclesRocking Run #10 7.89 Seconds for 10 cycles

Total Time = 79.19 Seconds for 100 cycles

Average Time for 1 Cycle = 0.7919 Seconds = Period

It is easy t o lose tra ck of the cycle count, or miss a timing point, a nd nega te

the a ccura cy of a da ta set. These types of errors are evident during th e dat a col-

lection work, and erroneous times are identified and discarded. For instance,

approximately tw enty runs w ere made to col lect t he dat a in the a bove tabular

summa ry. Ten of the timing runs w ere not used due to obvious errors in the da ta

accumulation. The ten acceptable test runs reveal an average period of 0.7919

Seconds. This is considered to be a consistent value, and the experimental mass

pola r moment of inert ia ma y now be computed from equa tion (3-35).

The experimental value for the polar inertia is 0.502 versus the calculated

value for this brass cylinder of 0.464 Pound-Inch-Seconds2. The difference of

0.038 represents an 8%error of the experimental versus the computed value. In

some instances this level of deviation is perfectly acceptable. For example, if thepart under test wa s a coupling hub tha t w ill be mounted on a pow er turbine with

an inertia of 40.0 Pound-Inch-Seconds2, the small differential of 0.038 Pound-

Inch-Seconds 2 would be insignifi cant . However, if the pa rt under test wa s one of

Jzz W XoffsetPer iod

2 -------------------

2 XoffsetG

------------------

=

Jzz 36.0 Pounds 5.07 Inches0.7919 Seconds

2 ------------------------------------

2 5.07 Inches386.1 Inches/Second

2------------------------------------------------

=

Jzz 182.52 Pound-Inches 0.01588 0.01313( )Second2 0.502 Pound-Inch-Second 2= =


20/82

86 Chapter-3

eight impellers to be mounted on a slender shaft, the cumulative error may be

unacceptable. In order to explain this error, it is necessary to re-examine the

component equations used for the inertia test calculation. Specifically, (3-34) for

the tota l inertia a bout t he pivot point ma y be solved as follows:

Substituting the test inertia from the above calculation back into equation

(3-21), and performing the axis translation, the following result is obtained:

As expected, this result is identical to the previous answer obtained by

using the composite equation (3-35). The interesting point of the above calcula-

tions is tha t inert ia due t o the axis tr a nslat ion is equa l to 2.397, versus the over-

all test inertia of 2.899 Pound-Inch-Seconds2. Hence, the cylinder inertia is only

a bout 20%of the inertia due to the a xis tra nslat ion. This is not a desirable condi-

tion since the a xis tra nslat ion is th e domina nt term. The geometrical confi gura -

tion displayed in Fig. 3-12 would not be as error prone since the XOffsetdistance

resides within the body of the element, and the axis translation term would not

dominate the test. Hence, the diagnostician should always be concerned about

trust ing this type of experimenta l inertia test w ith a long XOffsetdista nce between

the pivot a xis a nd t he desired principal polar moment of inert ia a xis. The other

lesson to be lear ned is th a t sim plified expressions such as (3-35) ma y not provide

full visibility concerning t he potentia l a ccura cy of the fi na l results. In some cases

it is necessary to revert back to the basic equations, and reexamine the entire

calcula tion a nd/or experiment a l test procedure.

With respect to the brass cylinder, it is concluded that improvement of the

test a ccura cy will require a r eduction or elimina tion of the axis tr a nslat ion term.

This could be accomplished with a test that consisted of suspending the mass

from cables, and then measuring the period as the mass oscillated in a twistingmanner (Fig. 3-14). Since the axis of oscillation is the axial geometric centerline

of the element, th ere is no axis tra nsla tion involved, and test a ccura cy should be

improved. This type of inertia test is ideal for machine parts such as compressor

impellers or turbine disks that contain complex geometrical cross sections. For

Jzozo

W Xoffset Per iod2

4 2-----------------------------------------------------------=

Jzozo

36.0 Pounds 5.07 Inches 0.7919 Seconds( )2

4 2-------------------------------------------------------------------------------------------------------------- 2.899 Pound-Inch-Second

2= =

JzozoJzz M Xoffset( )

2+=

2.899 Pound-Inch-Second2

JzzWG----- Xoffset( )

2+=

Jzz 2.899 Pound-Inch-Second2 36.0 Pounds

386.1 Inches/Second2

------------------------------------------------ 5.07 Inches( )2=

Jzz 2.899 2.397( ) Pound-Inch-Second2

0.504 Pound-Inch-Second2

= =


21/82


insta nce, the interna l sta r pa tt ern shown in Fig. 3-14 might be diffi cult t o model

with an equivalent inner diameter for the ma chine par t .

For this procedure, the machine element is suspended from three thin

cables (3 points determine a plane), spaced at 120 apart. The test piece must be

leveled as precisely as possible. If it is not level, then any induced twisting oscil-

lations will cause the machine element to wobble during the test. This wobble

not only negates the test accuracy, it can prove to be dangerous for parts with

a ny a pprecia ble physica l size a nd w eight. After leveling, the avera ge suspension

cable length L s-cand the cable ra dius Rs-ca re accura tely measured a nd recorded.

For best results, each of the suspension cable lengths should be equal, and the

radius for all three cables should be identical. As before, the machine element to

be tested is weighed on a n a ccura te scale in English units of Pounds.

During execution of this t est, the ma chine element is m a nua lly displa ced in

a twisting manner, and released. The machine part will torsionally twist back

and forth, and the period of the twisting oscillations will be measured with a

stopwatch. Since friction should not be major problem, the part will oscillate

back a nd forth for ma ny cycles. I t is not unusua l to observe thirty or more cycles

resulting from one initial displacement. B a sed upon th ese measur ed para meters

the ma ss polar moment of inertia ma y be computed a s follows:

(3-36)

The genera l form of (3-36) wa s extr a cted from t he Shock & Vibrat ion H and-

book9,a nd it wa s converted t o the nomenclatur e used in this t ext. As previously

Fig. 314 MechanicalArrangement For TwistingTest To Determine Mass

Polar Moment of Inertia

9 Cyril M. Harris , Shock and Vibr ation H andbook, Fourth edition, (New York: McGraw-Hill,1996), p. 38.5.

ElementWeight (W)

SuspensionCable

Length(L

s-c

)

Element AxialCenterline

Suspension

Cable

TwistingOscillation

30 to 40

Radius(R

s-c)

Jzz

W Rs c2 Per iod2

42 L s c-------------------------------------------------------=


22/82

88 Chapter-3

discussed on the rocking inertia test, it is mandatory to validate the test proce-

dure wit h a n a ctual t est on a known geometric sha pe. For compa ra tive purposes,

the solid brass cylinder used for the rocking test will be used for this twisting

inertia t est as shown in Fig. 3-15. From this dia gra m it is noted tha t t he avera gesuspension cable ra dius Rs-cwas 3.00 Inches, and the average cable length L s-cwa s 33.73 Inches. As before, the tota l cylinder w eight w a s 36.0 Pound s.

The fi rst test confi gura tion used 48 Inch long suspension ca bles in a n effort

to increase the period of the oscillation, and improve the time measurement

accuracy. Conceptually this was a good idea, but it turned out to be impractical

since the long cables ha d a tendency to w ra p a round each other. This proved t o be

a n unma na geable situa tion, a nd t he support cable lengths w ere reduced to 33.73Inches. During the acquisition of test data, the cylinder was twisted about 20

circumsta ntia lly from rest, an d released. The pea k of the motion a t one extrem-

ity w as visual ly sighted, and a stopwa tch wa s used to measure the t ime required

for complete back an d forth cycles. The test da ta is summa rized as follow s:

Twisting Run #1 13.83 Seconds for 10 cycles










Total Time = 140.18 Seconds for 100 cycles

Average Time for 1 Cycle = 1.4018 Seconds = Period

Fig. 315 Twisting Test ToDetermine Mass PolarMoment Of Inertia Of ASolid Brass Cylinder

BrassCylinderWeight

(W=36.0 Lbs.)

SuspensionCableLength

(Ls-c

=33.73")

AxialCenterline

Suspension

CableRadius

TwistingOscillation30 to 40

(Rs-c=3.00")


23/82


This is a much smoother test tha n t he rocking inertia previously discussed.

The number of miscounts and aborted runs were substantially reduced, and

a pproximat ely fifteen runs were ma de to collect the dat a sh own in the above ta b-

ular summary. Five of the timing runs were not used due to obvious errors indata accumulation. The ten consistent test runs reveal an average period of

1.4018 Seconds. This w a s considered to be a consist ent va lue for t his experimen-

tal procedure, and the mass polar moment of inertia may be computed from

equation (3-36) as follows:

The experimental polar inertia from this twisting procedure of 0.478 isquit e close to t he previously calcula ted v a lue of 0.464 Pound-Inch-Seconds2. The

3% deviation is q uite a ccepta ble for m ost rotor dyna mics ca lculat ions. This is

particularity true for smaller components that are stacked on a shaft to achieve

a final rotor assembly. I t should be recognized that both the rocking and the

tw isting inertia t ests ha ve their own d omain of a pplica tion tha t is dependent on

the size an d geometry of the ma chine element.

J ust a s the w eights of individua l components a re summed up to determine

a tota l rotor weight, the inertia of the component pieces may be added t o deter-

mine the overall rotor polar inertia . The origin of the inertia values ma y be from

calcula tions of defined geometries, or from experientia lly determined inertia va l-

ues. In any case, as long as the engineering units and the inertia axis are com-

mon, the numeric inertia values ma y be summed up to determine the ma ss polar

moment of inertia for th e entire rota ting a ssembly.In some insta nces, there is minimal opport unity to determine the inertia of

rotor components since the unit cannot be disa ssembled or unst a cked. In th ese

situations, the general inertia characteristics may be estimated based upon

ava ilable dimensions and probable ma teria ls of constr uction. In other cases, the

complexity of the rotor may not a llow a reasona ble segmenta tion and estima tion

of inertia properties. This is particularly true for rotors that are constructed of

multiple materials, plus rotors that contain complicated geometric configura-

tions. In these instances, another experimental technique may be employed to

determine the overall ma ss pola r m oment of inertia of the rotor.

This technique is based upon the familiar college physics experiment

depicted in Fig. 3-16. In this diagram, a cylinder or drum is mounted in rigid

bearings tha t a llow rota tion of the cylinder, but restrict a ny lat eral or tra nslat ion

movement of the cylinder. A cord is wrapped around the cylinder at a shaft

ra dius of Rs. I t is assumed that this cord is of insignificant weight and diameter,

and that it will not stretch with the application of axial tension. Next, a known

weight (mass M) is attached to the end of the cord, and allowed to free fall. The

Jzz

W Rs c2 Per iod2

42 L s c-------------------------------------------------------=

Jzz36.0 Pounds 3.00 Inches( )2 1.4018 Seconds( )2

42 33.73 Inches---------------------------------------------------------------------------------------------------------------------- 0.478 Pound-Inch-Second

2= =


24/82

90 Chapter-3

t ime Trequired to fall a distance Dis measur ed wit h a stop wa tch. The experi-

ment norma lly consists of determining th e cylinder ma ss polar moment of iner-

t ia Jmassbased on the four known quantities of radius Rs, mass M, fall distance

D, and the a verage fal l t ime T.

This basic physics problem may be solved by constructing free body dia-

gra ms of the cylinder an d t he falling ma ss, developing equat ions of motion, an d

then solving for t he polar inertia term. Another wa y t o achieve the sa me result is

by performing a n energy ba lan ce on t he syst em shown in Fig. 3-16. Conservat ion

of energy requires that the change in potential energy is equal to the change in

kinetic energy. In this case, the change in potential energy is simply the eleva-

tion cha nge in the ma ss (Mx Gx D). The overa ll cha nge in kin etic energy is com-

posed of a cha nge in tra nsla tional energy of the fa lling mass (Mx V2/2), plus t he

change in the rota tional kinetic energy of the rotor (Jmass2/2). These t ra dit iona lphysica l concepts ma y be represented mat hemat ica lly in the following m a nner:

(3-37)

The velocity Vin equa tion (3-37) represents t he a vera ge velocity of the fa ll-

ing mass M . In all cases, the falling weight Wis equa l to the mass Mt imes theacceleration of gravity G. Substi tuting the quanti ty W/ Gfor the mass Mpro-

duces the follow ing:

Fig. 316 TraditionalMechanical ArrangementFor Polar Moment Of Iner-tia Experiment Based OnFalling Mass Attached ToA Rotating Cylinder

Radius

(Rs

)

Fall Distance (D)

during Fall Time (T)

Mass(M)

Cylinder with Mass PolarMoment of Inertia (J

mass)

AngularVelocity ()

M G D M V2

2-------------------

Jmass 2

2-----------------------------+=


25/82


This may now be solved for the mass polar moment of inertia Jmassa s:

(3-38)

The average velocity Vof the ma ss Mduring free fall may be determined

from the fundamental equations of motion for rectilinear motion with constant

acceleration. More specifically, all physics books agree that the average velocity

multiplied t imes the drop time Twill determine the fall distan ce Da s follow s:

(3-39)

If the initial starting velocity Vois equal to zero, then equation (3-39) may

be simplified t o represent t he velocity Vin terms of the drop distan ce D, and thetota l ela psed drop time Tas shown in the next equation:

(3-40)

The last conversion to be performed consists of an expression for the angu-lar velocity of the cylinder in terms of the known experimental parameters.Since ta ngentia l velocity Vdivided by the ra dius Rsis equal t o the angula r veloc-

it y , equa tion (3-40) may be used t o determine in the following m a nner:

(3-41)

Equation (3-38) for the polar inertia may now be clarified into the known

experimental data by substituting equations (3-40) and (3-41) back into equation

(3-38), an d performing t he following s implifi cat ion of ter ms:

W D W V2

2 G-------------------

Jmass 2

2-----------------------------+=

2W D W V2

G------------------- Jmass

2+=

Jmass 2 2W D W V

2G

-------------------=

JmassW

2------ 2D

V2

G-------

=

DVo V+

2------------------

T=

V2D

T--------=

VRs-------

2D

Rs T-----------------= =


26/82

92 Chapter-3

The ca ncellat ion of common t erms result s in equa tion (3-42).

(3-42)

where: Jmass = Mass Polar Moment of Inertia (Pounds-Inches-Seconds2)

W = Falling Weight (Pounds)Rs = Shaft Radius (Inches)

D = Drop Distance (Inches)T = Drop Time (Seconds)G = Acceleration of Gravity (= 386.1 Inches / Second2)

With a few minor modifi cat ions, this tra ditional physics experiment ma y be

used to determine the polar inertia of a complete rotor assembly. One implemen-

tation of this concept is the technique by Michael Calistrat 10 that is i l lustrated

in Fig. 3-17. This diagram depicts a rotor resting in the rollers of a two bearing

shop bala ncing ma chine. To minimize errors, the rotor a ssembly s hould be prop-

erly ba lanced, and the r ollers in t he bala ncing m achine should be in good condi-

tion. As shown in Fig. 3-17, an overhead pulley is supported by an external

structur e, or the overhead cra ne tha t is used to move rotors in and out of the ba l-

a ncing ma chine. The dia meter of th e overhea d pulley does not ma tt er, but t his

pulley should be in good condition, and it s hould tu rn ea sily. The pulley redirects

the gra vita tional force to pull upwa rd on th e rotor, but it does not diminish t he

effect of th e falling ma ss. A plast ic covered braided st eel cable is a tt a ched t o the

shaf t wi th Duct ta pe a t a rad ius of Rshaft, and w ra pped around the shaf t for ten

or fi fteen turns. I t s probably a good idea t o keep the cable aw ay from any of the

rotor bearing journals, or the rotor sections observed by proximity probes. The

thickness of this cable is generally sma ll compar ed to the diam eter of the sha ft.

Aga in, as s hown in Fig. 3-17, the st eel ca ble passes overhea d, thr ough the pulley,and then is secured with a pair of vertical weights. The first weight is a tare

10 Michael M. Calistrat , Fl exibl e Coupl in gs, their design selection an d use, (Houston: CarolinePublishing, 1994), p 464.

JmassW

2------ 2D

V2

G-------

=

JmassW

2 D Rs T( )2

------------------------------------------- 2D2 D T( )2

G------------------------------

=

Jmass

W Rs2

T2

4 D 2-------------------------------- 2D

4 D 2

G T2

-----------------

=

Jmas s W Rs2 2D T

2

4 D2

---------------- 4 D2

G T2

-----------------T

2

4 D2

----------------

=

Jmass W Rs2 T

2

2D--------

1

G----

=


27/82


weight that is used to counterbalance the rotor, and keep the cable taunt. This

tare weight should just barely allow the rotor to rotate, and overcome system

friction.

The ta re weight is determined by tria l and err or, and sma ll cha nges in this

w eight w ill have a big influence on t he rotor. Since this w ork is performed in a

balancing machine, it makes sense to use balancing clay to get a good tare

weight established. The second weight Wis the experimenta l weight tha t a ctu-

a lly tur ns t he rotor. I t ma y be compromised of bala ncing clay, a sta ck of wa shers,

or any reasona ble combina tion thereof. In a ll ca ses, the dia gnostician must h ave

the a bi li ty to a ccura te ly weigh t his mass Wat the end of the test .

As w ith t he tw o previous inertia measur ements, the ma jority of the time is

spent in sett ing up the t est. The test execution requires minima l time, and this

rotor inertia procedure is no exception. Following the setup per Fig. 3-17, the

cable is wound up on the shaft, and the weight is naturally raised in elevation.

At t his time the shaft is released, a nd the time Tfor t he weight t o fa ll through a

predetermined dista nce Dis measured w ith a stopwat ch. The inertia calcula tion

described by equation (3-42) may be used directly. However, improved accuracy

w ill result by including the cable radius Rcablewith the shaf t ra dius Rshafta s rec-

ommended by C a listra t. This a ddition is not required for very thin cables, but it

does improve accura cy for t hicker cables, since the moment a rm is rea lly the sum

of the shaft plus the cable radius. Hence, (3-42) may be rewritten as follows:

(3-43)

The nomenclat ure in eq ua tion (3-43) has been expanded to be more under-

sta ndable, and yet ma intain consistency with t he general terms used within th is

text. Again, t he experimenta l w eight Wis in pounds, the shaft radius Rshaft, the

Fig. 317 Mechanical TestArrangement For PolarMoment Of Inertia Measure-ment Of Full Rotor Assem-bly Mounted In A ShopBalancing Machine, Tech-nique by M. M. Calistrat

MachineR

otorMounte

din

TwoBearing

BalancingM

achine

Overhead Pulley

Shaft WithRadius (Rshaft)

Weight (W)

Tare Weight

Axial ShaftCenterline

Cable WithRadius (Rcable)

Jmass Wei gh t R sha f t Rcab l e +( )2 T i m e

2

2 D i s ce tan-----------------------------------

1

G----

=


28/82

94 Chapter-3

cable ra dius Rcable, and the fal l Distancea re a ll in inches. The mea sured T imefor

the free fall through the predetermined distance is defined as the time in sec-

onds, an d Gis the gravi tat ional consta nt .

As wit h a ny experiment, the validity sh ould be checked with a test m a ndrelof known characteristics. For this test, a section of 4140 steel was selected with a

dens it y of 0.283 Poun d/In ch3. The len gt h L of the specimen wa s 59.81 inches, an d

the ends were squa red up in a la the. In addition, the outer diameter Dof th e test

piece wa s clea ned up to nomina lly 5.980 inches across t he length of th e element .

Once more equa tion (3-7) is used to comput e the ma ndr el weight :

This calcula ted w eight of 475 pounds w a s lower tha n t he uncalibrat ed shop

scale reading of 490 pounds. Although a better a greement in w eights w ould ha vebeen comforting, the 15 pound deviation was considered to be within the mea-

surement a ccura cy of th e shop scale. Next, the ma ss polar moment of inertia of

this st eel sha ft m ay be computed w ith equat ion (3-32) in t he following ma nner:

Prior to testing, the runout (eccentricity) along the length of the mandrel

was confirmed to be less than 1.0 Mil. In addition, a check balance was per-

formed, and the residual unbalance was minor. For the actual test runs, a falldistance of 60 Inches was established on a vertical reference stand. The shaft

radius Rshaftwa s 2.99 Inches, an d th e ca ble diam eter w a s 1/8, for a Rcableof 0.06

inches. The t a re w eight wa s found to be 0.696 Pounds (315.7 gra ms). The experi-

mental w eight Wwa s a djusted until the free fall t ime wa s a bout 5 seconds. This

required a tota l of 3.43 Pounds (1,556 gram s). The timed t est da ta wa s obta ined

by releasing the shaft and measuring the time required for the weight to fall 60

inches. A stopwa tch wa s used to measure the t ime, a nd th e test da ta is follow s:

Falling Weight Run #1 4.52 Seconds




Falling Weight Run #5 4.56 SecondsTotal Time = 22.81 Seconds for 5 Drops

Average Time for 1 Drop of 60 Inches = 4.562 Seconds

Wso l i d L D2

4-------------------------------------=

Wso l i d 59.81 Inches 0.283 Pounds/Inch3 5.98 Inches( )2

4--------------------------------------------------------------------------------------------------------------------------------- 475 Pounds= =

Jmassso l i d L D4

32 G-------------------------------------=

Jmassso l i d 59.81 In. 0.283 Pounds/In.3 5.98 In.( )4

32 386.1 Inches/Second2

------------------------------------------------------------------------------------------------------------- 5.504 Pound-Inch-Second2

= =


29/82


Eight different t est runs w ere made, and t he obviously incorrect times w ere

discarded. The a bove arr ay wa s quit e consistent, a nd t here wa s good confi dence

in both t he validity a nd th e accura cy of this experimenta l dat a . Hence, equa tion

(3-43) was now applied to compute the measured mass polar moment of inertia:

Note tha t t here is excellent a greement betw een th e experimenta l inertia of

5.450 and the calculated polar inertia of this steel shaft of 5.504 Pound-Inch-

Second2

. The a ctua l deviat ion of 1%is considered to be quite a ccepta ble accura cyfor t his ty pe of measurement. I t is often desirable to perform this t est w ith differ-

ent weights, and different fall distances to verify the consistency of the proce-

dure . In al l cases, the test weight should be substantial ly larger than the tare

weight , and please recal l that the in i t ial s tart ing veloci ty is zero at the start

time. For bigger and heavier rotors that have a tendency to sag within the bal-

ancing machine, consideration should be given to rolling the rotors for several

hours before at tempting t his type of inert ia t est. In add ition, dial indicat or mea-

surements should be made at the rotor midspan before beginning the inertia

tests, an d a fter th e conclusion of the last run. Any a pprecia ble sag to the sha ft of

the rotor under test might negate the test results, and require a repeat of the

slow roll and the t est procedure.

Fina lly, it should be recognized tha t other experiment a l procedures exist for

determination of mass moment of inertia. In addition, many computation pro-grams provide the capability for three-dimensional calculations of the inertia

properties of complex bodies. As always, the machinery diagnostician should be

aware of the engineering units, and test cases of known geometries should

a lwa ys be run to verify the technique.

Jmass Wei gh t R sha f t Rcab l e +( )2 T i m e

2

2 D i s ce tan-----------------------------------

1

G----

=

Jmass 3.43 Pounds 2.99 0.06 Inches+( )2 4.562 Seconds( )

2

2 60 Inches----------------------------------------

1

386.1 Inches/Sec2

-----------------------------------------

=

Jmass 3.43 9.302 0.1734 0.0026( ) 5.450 Pound-Inch-Second2

= =


30/82

96 Chapter-3

DAMPING INFLUENCE

Three basic types of damping occur in a machinery system. These damping

types a re commonly referred to as viscousdamping, coulombdamping, and solidor structuraldamping.

Viscous damping is encount ered by solid bodies moving th rough a viscousfluid. In this type of damping, the resistance force is proportional to the velocity

of the moving object. As a n exa mple of viscous da mping, consider th e situa tion of

a cook stirring a pot of soup versus a pot of molasses. I t is self-evident tha t stir-

ring t he mola sses is considerably m ore diffi cult due to the t hickness a nd higher

viscosity of the molasses as compared to the thin soup. The required force is

directly proportional to the velocity of the stirring spoon. In most cases the cook

would stir the molasses at a much slower rate than he would the soup, simply

beca use it w ould t a ke too much strengt h or energy to stir it ra pidly.

The sa me t ype of physical propert y, i.e., viscous d a mping, is encount ered in

the bearings a nd oil seals of la rge rotat ing ma chines. In t his case, the da mping is

provided by the lubricating oil, and the rotating shaft is the rigid body movingth rough th e viscous fluid. The process flu ids ha ndled by th e machine a lso provide

damping to the rotor system. For liquid handling machines such as pumps and

hydraulic turbines, this is significant. However, for gas handling machines such

a s t urbines or centrifuga l compressors t his is a minor considerat ion.

Fig. 318 CalculatedBode Plot Of UnbalanceResponse For A SimpleMechanical System WithVariations In Damping

180

150

120

90

60

30

0

Phase

Lag(Degrees)

=0.2

Low Damping, =0.1

=0.5

=2.0

=1.0

High Damping,=2.0

=0.1

0

1

2

3

4

5

0.0 0.5 1.0 1.5 2.0 2.5 3.0

AmplitudeRatio

Critical Speed Frequency Ratio

Low Damping, =0.1

=0.2

=0.5

=2.0=1.0

()


31/82

Damping Influence 97

The next type of damping is coulomb damping, which arises from thesliding of one dry surface upon another (rub condition). The coulomb friction

force is nea rly consta nt, a nd it depends on the na ture of the sliding surfaces, a nd

the perpendicular pressure between the surfaces. This type of force is generallydominant in dam ped systems during t he fina l sta ges of motion when other types

of da mping become negligible.

The third category of damping is often referred to as solid or structuraldamping. This is due to interna l friction wit hin th e ma teria l, and it differs fromviscous damping in tha t it is independent of frequency, and proport iona l to t he

maximum stress of the vibratory cycle. Since stress and strain are proportional

in the elastic range, it can be stated that solid damping force is proportional to

deflection. Structural damping in rotating machinery is small when compared

w ith viscous da mping, but it does exist.

The ma jor contribution tha t positive dam ping makes to a rota ting ma chin-

ery syst em is t he dissipation of energy. This infl uence is most dr a ma tically illus-

trated when a mechanical system passes through a resonance as in Fig. 3-18.

This calculated Bode diagram was duplicated from chapter 2. The unbalance

response plot of frequency ratio versus amplitude ratio and phase of a damped

system provides a good perspective of the actual influence of damping. The fam-

ily of curves in this diagra m a re plotted wit h a da mping rat io (or damping factor)

extending from =0.1 to 2.0. Recall tha t t his ra tio is the actua l dam ping dividedby the critical damping. Note, that with a lightly damped system of = 0.1, theresponse at the resonance is quite high. This translates to the fact that there is

little energy dissipat ion und er th is condition. The syst em is under da mped, a nd

it is susceptible to insta bility due to a lack of an energy dissipa tion. Conversely,

when the damping factor is large, = 2.0, th e system is over damped, responsethr ough the resonance is restra ined, a nd overa ll sta bility of the system is high.

In ma ny insta nces of free vibrat ion, rotor insta bility ca n be rela ted to a la ck

of damping. The system damping may be assessed by examining the criticalspeed response on Bode plots (synchronous 1X vectors versus speed). Typically,

the amplification factor Qthr ough the resona nce is used to qua ntify t he severity

of the resonance, plus the damping ratio. A large amplification factor is associ-

a ted w ith a poorly damped, high amplitude resona nce. Conversely, a low a mplifi-

cat ion factor is genera lly associat ed with a well damped resona nce, tha t displays

sma ll amplitudes at the peak of the resona nce.

Ext ra ction of the am plifi cation factor from the B ode plots m a y be performed

in several different ways. Unfortunately, there is disagreement within the tech-

nical community a s to the best ma nner to obta in this informa tion from the mea-

sured vibration data. For the sake of completeness, three separate approaches

for determina tion of this dimensionless am plifi cation fa ctor w ill be presented.

The first technique consists of visually comparing the measured vibration

response data with a set o f calculat ed curves, and estimat ing a damping rat io .The rotor am plifica tion fa ctor for th e specifi c resona nce is then comput ed directly

from the damping ratio. As a practical example of an industrial machine, con-

sider the compressor Bode plot shown in Fig. 3-19 (same as Fig. 2-21). This vari-

able speed vibration data is runout compensated, and it exhibits a clean


32/82

98 Chapter-3

transition through a single critical speed. If this data is viewed in conjunction

wit h t he computed fa mily of curves shown in F ig. 3-18, a suit a ble damping ra tio

may be selected. Specifically, the shape of the amplitude versus speed, and the

phase versus speed are compared between the measured (Fig. 3-19) and the cal-

culat ed plots (Fig. 3-18). It is rea sona ble to conclude tha t t he calcula ted plot wit h

a damping ra t io of 0.2 is the closest match to the measured machine responsedata. From the previously developed equation (2-91), the amplification factor

(i.e., amplitude ratio) Qma y be determined from the damping ra tio as follow s:

(3-44)

Subst ituting t he previously identifi ed value for t he dam ping rat io of = 0.2,the following result is obtained:

This visual comparison between curve shapes provides any easy way to

estimate Qfor a clea nly d efi ned resona nce. H owever, for m ore complex response

characteristics, other techniques are available. For instance, the second

approach for computation of the amplification factor is derived from electrical

engineering terminology. This technique has also been adopted by various

mechanical standards organizations such as the American Petroleum Institute.

In this procedure, the center frequency of the resonance is divided by the reso-

nance bandwid th a t the H alf Power Poin tin a ccorda nce with the following:

(3-45)

Fig. 319 Measured BodePlot Of A Centrifugal Com-pressor Startup

QCu r v e F i t Amp l i f i c a t i o n F act or Am p l i t u d e Ra t i o 1

2 ------------= = =

QCu r v e F i t1

2 ------------

1

2 0.2----------------

1

0.4------- 2.5= = = =

QH a l f Power CenterFrequency

FrequencyB a n dw i d t h @-3dB---------------------------------------------------------------------------------------=


33/82


34/82

100 Chapter-3

ha ndle with the fi rst a pproach. Whereas, the second scheme would allow a better

qua ntifi cat ion of the Qfor each response peak.

One objection to the second approach is that Qvaries with changes to the

center frequency of the resonance. For instance, if the shape of the responsecurve is maintained, and the center frequency is reduced to 3,000 RPM, the Q

dr ops to 3.3 (= 3,000/900). By t he sa me t oken, if the crit ica l speed occurs a t

12,000 RPM, the Qnow increa ses t o 13.3 (= 12,000/900). This cha nge in a mplifi -

cation factors with a constant shape to the response peak may be quite confus-

ing, a s w ell a s contr a dictory t o the desired definit ion.

The third approach for calculation of the amplification factor consists of

dividing the amplitude at the resonance by the amplitude at a speed far above

the r esona nce. Aga in referring to t he sa me B ode plot example, this a pproach is

illustrated in Fig. 3-21. As before, the magnitude of 1.95 Mils,p-p a t the t rans la-

tional critical speed of 6,100 RPM is identified. Above this resonance, the phase

and amplitude enter a plateau region where the 1X vector remains reasonably

constant as speed increases. This is normal behavior, and the synchronous

response above a critical is typically flat until some external force, or a higher

order resonance influences the motion. Within the context of this example, the

region a t 8,000 RP M is suffi ciently removed from the critical speed, and the rota -

tional speed vibration amplitude in this region is 0.85 Mils,p-p. The a mplifi cat ion

factor determined wit h t his th ird technique is computed by simply dividing the

maximum vibration amplitude at the critical speed by the amplitude measured

well above the resonance in accordance with the following expression:

(3-46)

Fig. 321 AmplificationFactor from a Bode plotUsing A Simple Amplitude

Ratio Between the Reso-nance and a Speed FarAbove the Resonance

QResonance/ Above

Am p l i t u d e A tResonance

Am p l i t u d e AboveResonance---------------------------------------------------------------------------=


35/82

Damping Influence 101

Using the vibration amplitudes identified on Fig. 3-21, the amplification

factor determined by t his method is easily computed a s:

This value of 2.3 is indicative of a n a dequa tely da mped machine, a nd a sta -

ble mecha nical syst em. Due to the differences betw een ma chines, it is diffi cult t o

precisely cat egorize the va ria tions between good a nd ba d Qfactors. However, it

is reasona ble to identify severa l categories of amplifica tion factors versus da mp-

ing a nd st a bility chara cteristics. Summa ry Ta ble 3-1 at tempts t o provide some

realistic guidelines for these interrela ted par a meters:

Machines that fall into the last category may be potentially dangerous, and

may not survive a maiden startup. Machines within this group often require

extensive modifica tions to bear ings a nd/or seals t o increase sy stem da mping. In

some instances, modifications such as the installation of squeeze film damper

bearings ma y be required to provide adequa te da mping for the syst em.

It should a lso be noted th a t t he amplifi cation factors referred to herein a reassociated with shaft vibration measurements where the oil film viscous damp-

ing is dominant (i .e. , between the journal and bearing). Casing vibration mea-

surements would typically be more receptive to structural damping, and not so

sensitive to viscous da mping in t he bear ings. Thus, casing m easurements gener-

ally display Qs that are much higher than shaft measurements, simply due to

the lack of dam ping within t he ca sing and t he support st ructure.

Another consideration that must be applied to any evaluation of variable

speed da ta is the a ccelera tion ra te, or more specifi cally th e ra te of speed change

of the rota ting syst em. During ma chinery tra in coastdown s, there is usually min-

imal, if any, control of the deceleration. However, during startup, the rate of rotor

acceleration is often controllable on variable speed drivers such as steam tur-

bines. On ma chines wit h older contr ol systems, the st a rtup r a te is often depen-

dent upon the skill and knowledge of the operator handling the trip & throttleva lve. On newer speed control systems, the sta rtup r a te is usua lly contr olled by

a n electr onic governor w ith predetermined sta rt up speed ra mps. U nfortuna tely,

some electr onic governors suppliers a re not w ell versed in the a ccepta ble sta rt up

ra tes for va rious types of machinery tr a ins. These vendors often set a bnormally

Table 31 Shaft Amplification Factor Versus Damping And Anticipated Rotor Stability

Amplification Factor Damping Stability

Q = < 2 Well D a mped E xt remely S t a ble

Q = 2 t o 8 Adeq ua t ely D a mped Norma l S t a bilit y

Q = 8 t o 15 P oor ly D a mped Ma rgina l S t a bilit y

Q = > 15 Insuffi cient D a mping Inherent ly U nst a ble

QResonance/ AboveAm p l i t u d e A tResonance

Am p l i t u d e AboveResonance--------------------------------------------------------------------------- 1.95 Milsp p

0.85 Milsp p------------------------------ 2.3= = =


36/82

102 Chapter-3

fast s ta rtup rat es in an ef fort t o snapthe r otor(s) thr ough a critica l speed ra nge.

This tendency has resulted in machinery damage, and more than one rotor fail-

ure. Hence, the end user should alwa ys verify t ha t t he proposed sta rtup a cceler-

a tion rat es are reasonable for the ma chinery in question.Intuitively, the passage of a rotor system through a critical speed region

should be performed in a direct a nd knowledgea ble manner. If th e sta rtu p rat e is

inordina tely slow, the ma chine ma y hang upin a resona nce, and ca use mecha ni-

cal damage due to the high vibration levels. An example of this type of occur-

rence is briefly discussed in the t urbine genera tor case hist ory 39 in chapter 11.

On the other hand, if the speed acceleration rate through a resonance is exces-

sive, the machine may self-destruct after it reaches operating speed and

a tt empts to reba lance itself about t he ma ss center. Although this t ype of occur-

rence is ra re, it is certa inly a voida ble, and tota lly unnecessary.

The speed tra nsition rat e through a r esona nce w ill a lter the char a cteristics

of the vibrat ion response dat a . A slow sta rt up will show a higher pea k at the res-

onance, combined with a broader bandwidth. Conversely, a rapid startup will

produce a lower peak amplitude at the critical speed, plus a smaller resonant

bandwidth. This attenuated response characteristic has erroneously led many

individuals into a false sense of security by ra mping through critical speeds at a

high ra te. Hence, the dia gnostician must be aw a re of the tra nsition rat e through

the system critical speed(s), and any evaluation of the resonant characteristics

(e.g., Q) should be weighed by t his speed chan ge ra te.

It is impossible to fully quantify proper startup acceleration rates for all

cla sses of ma chinery opera ting

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