chapter 2 dynamic motion

7/30/2019 Chapter 2 Dynamic Motion

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9

C H A P T E R 2

Dynamic Motion 2

Many mechanical problems are init iallyrecognized by a change in machinery vibration amplitudes. In order to under-

sta nd, an d correctly dia gnose the vibratory chara cterist ics of rota ting ma chinery,

it is essential for the machinery diagnostician to understand the physics ofdynamic motion. This includes the influence of stiffness and damping on the fre-

quency of an oscillat ing mass as well as the interrelationship between fre-

quency, displacement, velocity, a nd a ccelerat ion of a body in motion.

MALFUNCTION CONSIDERATIONSAND CLASSIFICATIONS

Before examining the intricacies of dynamic motion, it must be recognized

tha t ma ny fa cets of a mecha nical problem must be considered t o achieve a suc-

cessful and acceptable diagnosis in a timely manner. For instance, the following

list identifi es some of the r elated considerations for addr essing and realist ically

solving a machinery vibration problem:

r Economic Im pact

r Machinery Type and Construction

r Ma chinery H istory Trends Failures

r Frequency Distribution

r Vibratory Motion Dist ribution a nd Direction

r Forced or Free Vibration

The economic impact is directly associated with the criticality of the

machinery. A problem on a main process compressor would receive immediate

at tention, whereas a seal problem on a fully spared reflux pump would receive a

lower priority. Clearly, the types of machinery, the historical trends, and failure

histories are all important pieces of information. In addition, the frequency of

the vibration, plus the location and direction of the motion are indicators of theproblem type and severity. Traditionally, classifications of forced and free vibra-

tion are used to identify the origin of the excitation. This provides considerable

insight into potential corrective actions. For purposes of explanation, the follow-

ing lists identify some common forced and free vibration mechanisms.


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10 Chapter-2

Forced Vibration Mechanisms Free Vibration Mechanisms

r Mass Unbalance u Oil Whir l

r Misalignment u Oil or S tea m Whipr

Sha f t B ow

u

Int ernal Friction

r

Gyroscopic

u

Rotor Resonance

r

Gear Contact

u

St ructural Resonances

r

Rotor Rubs

u

Acoustic Resonances

r

Electrical Excitat ions

u

Aerodynamic Excitations

r

External Excitat ions

u

Hydrodynamic Excitat ions

Forced vibra tion problems a re generally s olved by removing or reducing the

exciting or driving force. These problems a re ty pica lly eas ier to identify a nd s olve

than free vibration problems. Free vibration mechanisms are self-excited phe-

nomena tha t a re dependent upon the geometry, mass, st iffness, and da mping of

the mechanical system. Corrections to free vibration problems may require phys-

ical modifi cat ion of th e ma chinery. As such, these t ypes of problems ar e often dif-ficult to correct. Success in treating self-excited problems are directly related to

the diagnosticians ability to understand, and apply the appropriate physical

principles. To add ress t hese funda ment a l concepts of dyna mic motion, including

free and forced vibrat ion, the follow ing chapt er is presented for considerat ion.

It should be mentioned tha t m uch of the equat ion str ucture in this chapter

wa s sum ma rized from t he classica l textbook by William T. Thomson

1

, entitled

M echani cal Vi brati ons

. For m ore information, and detailed equa tion derivation,

the reader is encouraged to reference this source directly. The same basic equa-

tion structure is also described in his newer text entit led Th eory of Vibr ati on

wi th Appl icat ions

2

. Regardless of the vintage, at least one copy of Thomson

should be part of the reference library for every diagnostician.

F

UNDAMENTAL

C

ONCEPTS

Initially, consider a simple system consisting of a one mass pendulum as

shown in Fig. 2-1. Assume that the pendulum mass M is a concrete block sus-pended by a weightless a nd rigid cable of length L

. Fur ther a ssume that t he sys-

tem operates without frictional forces to dissipate system energy. Intuitively, if

the pendulum is displaced from the vertical equilibrium position, it will oscillate

back and forth under the influence of gravity. The mass will move in the same

path , and w ill require the sa me am ount of t ime to return t o any specified refer-

ence point. Du e to the frictionless environment , the a mplitude of the motion will

remain constant. The time required for one complete oscillation, or cycle, is

called th e Period

of the m otion. The t otal num ber of cycles completed per un it of

1 William Tyrell Thomson, Mechanical Vibrations

, 2nd Edition, 9th Printing, (EnglewoodCliffs, New J ersey: P rent ice Ha ll, Inc., 1962), pp.1-75

2 Willi a m T. Thom son , Theory of Vibrat ion wi th Appli cations

, 4th Edition, (Englewood Cliffs,New J ersey: P rentice H all, 1993), pp. 1-91.


3/58

Fundamental Concepts 11

t ime is the Frequency

of the oscillat ion. Hence, frequency is simply t he reciprocal

of the period a s show n in t he following expression:

(2-1)

The box around this equation identifies this expression as a significant or

important concept. This same identification scheme will be used throughout this

text. Within equation (2-1), period is a time measurement with units of hours,

minutes or seconds. Frequency carries corresponding units such as Cycles per

Hour, Cycles per Minute (CPM), or Cycles per Second (CPS or Hz). Understand-

a bly, the oscillatory motion of the pendulum is repetitive, a nd periodic. As sh own

in M arks Hand book

3

, Fourier proved that periodic functions can be expressed

w ith circula r funct ions (i.e., a series of sines an d cosines) where th e frequency

for each term in the equation is a multiple of the fundamental. I t is common to

refer to periodic motion as harmonic motion. Although many types of vibratory

motions are harmonic, it should be recognized that harmonic motion must beperiodic, but periodic motion does not necessarily ha ve to be ha rmonic.

3 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), pp. 2-36.

Fig. 21 Oscillating Pendu-

lum Displaying Simple Har-monic Motion

Frequency 1Pe r i od-------------------=

A C

B

Mass

Negative Positive

Max. Neg. Displ.Zero VelocityMax. Pos. Accel.

Max. Pos. Displ.Zero VelocityMax. Neg. Accel.

Zero DisplacementMaximum VelocityZero Acceleration

W=MG

Wsin

Wcos

Equilibrium

Stationary I-Beam

CableLength

-L


4/58

12 Chapter-2

In a rotating system, such as a centrifugal machine, frequency is normally

expressed a s a circular rotat iona l frequency

. Since one complete cycle consists

of one revolution, and one revolution is equ a l to 2

radians, the following conver-

sion a pplies:

(2-2)

Combining (2-1) and (2-2), the rotational frequency

may be expressed in

terms of the Period

as follows:

(2-3)

The frequency unit s for

in equa tion (2-3) ar e Ra dia ns per Second, or Ra di-

ans per Minute. Again, this is dependent upon the time units selected for the

period. Although these are simple concepts, they are continually used through-

out this text . Hence, a clear and definitive understanding of period and fre-

quency are ma nda tory for a ddressing virtually a ny vibrat ion problem.Returning to the pendulum of Fig. 2-1, a gravitational force is constantly

a cting on th e ma ss. This vert ical force is th e weight of the block. From physics it

is known that weight W

is equal t o the product of mass M

, and th e accelerat ion of

gravity G

. As the pendulum oscillates through an angular displacement

, this

force is resolved into two perpendicular components. The cosine term is equal

a nd opposite to th e tension in the string, an d the sine component is the RestoringForce

acting to bring the mass back to the vertical equilibrium position. For

sma ll values of angular displacement, sin

is closely approximat ed by the ang le

expressed in radians. Hence, this restoring force may be represented as:

(2-4)

Similarly, the maximum distance traveled by the mass may also be deter-mined from pla ne geometry. As sh own in Fig. 2-1, the cable length is known , an d

the a ngular displacement is specified by

. The actual change in lateral position

for th e mass is the dista nce from A

t o B

, or from B

t o C

. In either case, this dis-

ta nce is equal to L

sin

.

Once more, for sm a ll an gles, si n

in radians, and the

total defl ection from th e equilibrium posit ion ma y be sta ted a s:

(2-5)

This repetitive restoring force acting over the same distance has a spring

like quality . In actuality , this characterist ic may be defined as the horizontal

st iffness K

of this simple mechanical system as follows:

(2-6)

If equa tions (2-4) a nd (2-5) a re subs tit uted into (2-6), a nd if t he w eight W

is

replaced by the equivalent mass M

t imes th e acceleration of gravity G

, the fol-

lowing expression is produced:

2 Frequency 2 F= =

2Pe r i od-------------------=

Res to r i ngF or ce W

D ef l ect i on L

St i f f n ess K Fo rce

Def lec t ion------------------------------= =


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(2-7)

La ter in this chapter it will be shown t ha t th e nat ural frequency of oscilla-tion for an undamped single degree of freedom system is determined by equation

(2-44) as a function of mass M

and st iffness K

. If equation (2-7) is used for the

stiffness t erm w ithin equa tion (2-44), the following rela tionship result s:

(2-8)

Equation (2-8) is often presented within the literature for describing the

na tur a l frequency of a simple pendulum. A direct example of this concept ma y be

illustrated by considering the motion of the pendulum in a grandfathers clock.

Typically, the pendu lum r equires 1.0 second to tr av el one half of a str oke, or 2.0

seconds to transverse a complete stroke (i.e., one complete cycle). The length L

of

th e pendulum ma y be det ermined by combining equa tions (2-3) a nd (2-8):

If the period is represented in terms of the pendulum length L

, the above

expression ma y be stat ed as:

(2-9)

Equation (2-9) is a common expression for characterizing a simple pendu-

lum. The validity of this equation may be verified in technical references such as

M arks Handbook

4

. For the specific problem at hand, equation (2-9) may be

solved for th e pendulum length

. Performing this manipulation, an d inserting thegravita t ional constant G

, plus th e period of 2.0 seconds, the followin g is obta ined:

Thus, th e pendulum lengt h in a gra ndfa thers clock should be 39.12 inches.

This va lue is accura te for a concentra ted ma ss, and a weightless support ar m. In

an actual clock, the pendulum is often ornate, and weight is distributed along

the length of the support a rm. This ma kes it diffi cult to accurately determine the

loca tion of the cent er of gra vity of the pendulum m a ss. Neverth eless, even rough

measurements reveal tha t t he pendulum length is in th e vicinity of 40 inches. In

ad dit ion, clock ma kers normally provide a calibration screw a t t he bottom of the

pendulum t o allow the owner to a djust the clock accura cy. By turning t his a djust-

ment s crew, the effective length of the pendulum m a y be alt ered. From th e previ-

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. 3-15.

KFo rce

Def lec t ion------------------------------

W L --------------- W

L-----

M GL

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

KM------

M GL

-----------------1

M------ G

L----= = =

2Pe r i od-------------------

GL----= =

Per iod 2 LG----=

LG Per i od

2

42---------------------------------

386.1Inches/Second2( ) 2.0Seconds( )2

42-------------------------------------------------------------------------------------------- 39.12 Inches= = =


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14 Chapter-2

ous equa tions, it is clear tha t cha nging the pendulum length w ill alter th e period

of the pendulum. By moving the weight upwa rd, and decreas ing the arm length,

the clock w ill run fa ster (i.e., higher frequency w ith a shorter period). Conversely,

by lowering the main pendulum mass, the length of the arm will be increased,a nd t he clock will run s lower (i.e., a lower frequency w ith a longer period).

Alth ough the gra ndfa ther clock is a simple applicat ion of periodic motion, it

does provide a realistic example of the fundamental concepts. Additional com-

plexity will be incorporated later in this text when the behavior of a compound

pendulum is discussed. It should be noted that a compound pendulum is a

mechanical system that normally contains two degrees of freedom. This addi-

tional fl exibility m ight be obtained by a dding fl exible members such a s springs,

or a ddit iona l masses to a simple system. In a tw o mass system, each mass might

be capable of moving independently of the other mass. For this type of arrange-

ment, each mass must be tracked with an independent coordinate system, and

this would be considered as a two degree of freedom system.

The number of independent coordinates required to accurately define the

motion of a system is termed the Degr ee of Fr eedomof that system. Processmachinery displays many degrees of freedom, and accurate mathematical

description of these systems increases proportionally to the number of required

coordina tes. How ever, in th e case of th e simple pendulum, only one coordina te is

required to describe the motion and the pendulum is a single degree of free-

dom system exhibiting harmonic motion. More specifically, this is an example of

basic dynamic motion where the restoring force is proportional to the displace-

ment. This is commonly referred to as Simpl e H arm onic M ot ion(SHM). Otherdevices such as the undamped spring mass (Fig. 2-7), the torsional pendulum

(Fig. 2-25), the part icle rota ting in a circula r pat h, an d a fl oatin g cork bobbing up

and down in the wa ter a t a constant ra te a re a l l examples of SHM.

B efore expa nding t he discussion to more complex syst ems, it is desira ble to

conclude the discussion of the simple pendulum. Once again, the reader isreferred ba ck to the exa mple of the oscilla ting pendulum d epicted in Fig. 2-1. On

this diagram, it is meaningful to mentally trace the posit ion of the mass during

one complete cycle. Starting at the vertical equilibrium position B, the displace-

ment is zero at t ime equal to zero. One quarter of a cycle later, the mass has

moved to the ma ximum positive position C. This is followed by a zero crossing a t

point Bas t he mass a pproaches the ma ximum negative value at posit ion A . The

last quarter cycle is completed as the mass returns from the A location back to

the origina l equilibrium, or center rest point B.

Intuit ively, the mass achieves zero velocity as it swings back and forth to

the ma ximum displacement points A a nd C(i.e., the mass comes to a complete

stop). In addition, the maximum positive velocity occurs as the mass moves

through point Bfrom left t o right , combined wit h a ma ximum negat ive velocity

as t he mass moves through Bgoing from right to left. Finally, the mass must de-accelerate going from Bt o C, and accelerate from Cback to point A . Then the

mass will de-accelerate as it moves from A back to the origina l equilibrium point

Btha t displays zero lat eral a ccelera tion.

Another w ay to compa re an d correla te th e displacement, velocity, and a ccel-


7/58


eration characterist ics of this pendulum would be a t ime domain examination.

Although a meaningful visualization of the changes in displacement, velocity,

an d a ccelera tion with respect t o t ime may be difficult a ma thema tical descrip-

tion simplifi es this t ask. For instan ce, assume t ha t the periodic displacement ofthe mass may be described by the following fundamental equation relating dis-

placement a nd t ime:

(2-10)

where: Displacement = Instantaneous DisplacementD = Maximum Displacement (equal to pendulum position A or C)F = Frequency of Oscillationt = Time

In a rotat ing system, such a s a centrifuga l ma chine, this expression can be

simplified somewhat by substituting the rotational frequency tha t wa s previ-ously defi ned in equa tion (2-2) to yield:

(2-11)

The instantaneous velocity of this periodic motion is the time derivative of

displacement . Velocity ma y now be determ ined a s follows :

B y converting th e cosine t o a sine fun ction, expression (2-12) is derived:

(2-12)

Note that velocity leads displacement by /2 or 90 . Another w ay to st a tethe same concept is that displacement lags behind velocity by 90 in the t ime

domain. The same procedure can now be repeated to examine the relationshipbetween velocity and acceleration. Since acceleration is the time rate of change

of velocity, the first time derivative of velocity will yield acceleration. The same

result may be obtained by taking the second derivative of displacement with

respect to time to obtain acceleration:

By a dding to the sine term, the negative sign is removed, and the follow-ing expression is obta ined:

(2-13)

Accelerat ion lea ds displa cement by or 180 , and it leads velocity by 90 . Itma y a lso be sta ted th at displacement lags acceleration by 180 in t ime. The rela-

tionship between displacement, velocity, and acceleration may be viewed graphi-

cally in the polar coordinate format of Fig. 2-2. This diagram reveals that

D isp l a cem en t D 2 F t( )sin=

D isp l acem en t D t( )sin=

Ve loc i t ytddD isp l acem en t D t( )cos= =

V el oci t y D t 2+( )sin=

Acce le ra t iont

2

2

d

dD i sp l a cem en t D 2 t( )sin= =

A ccel er at i on D 2 t +( )sin=


8/58

16 Chapter-2

mecha nical systems in m otion do display a consistent a nd defi na ble relat ionship

betw een frequency, an d t he respective displacement, velocity, a nd a ccelerat ion of

the body in motion.

Understanding the t iming between vectors is mandatory for diagnosing

machinery behavior. It is very easy to become confused between terms such as

leadinga nd lagging, and the diagnostician might inadvertently make a 90 or a180 m ista ke. In s ome insta nces, th is type of error might go unnoticed. However,

during rotor bala ncing, a 180 error in w eight placement m ight result in exces-

sive vibration or even physical damage to the machine. This type of error is

totally unnecessary, and it may be prevented by establishing and maintaining a

consistent timing or phase convention.

From Fig. 2-2, it is noted that time is shown to increase in a counterclock-wise direction. If this diagram represented a rotating shaft , t ime and rotation

would m ove togeth er in a count erclockwise direction. As discussed in succeeding

chapters, phase is measured from the peak of a vibration signal backwards in

time to the reference trigger point. This concept is illustrated in Fig. 2-3 that

depicts a rotating disk with a series of angles marked off at 45 increments.

Assume that the disk is turning counterclockwise on the axial centerline. If this

rotating disk is observed from a sta t ionary viewing posit ion, the a ngles will move

past th e viewing point in consecutive order.

That is, as the disk turns, the angles progress in a 0-45-90-135-180-225-

270-315 consecutive num eric order pa st th e fi xed viewing position. How ever, if

the angles increased with rotation, the observed viewing order would be back-

wa rds. Since this does not ma ke good physical sense, the direction of numerically

increasing a ngles are alwaysset a gainst sha ft rota tion as in Fig. 2-3. This a ngu-lar convention will be used th roughout t his text , and vector a ngles will alwa ys be

considered as degrees of phase lag. This convention applies to shaft and casing

vibration vectors, balance weight vectors, balance sensitivity vectors, plus all

Fig. 22 Timing Relationship Between Displacement Velocity, and Acceleration

Displacement VectorDispl . = Dsin( t)Velocity VectorVel . = D sin(t+ /2)

Acceleration VectorAccel.= D2 sin(t+ )

t+

t+ /2Time

Phase

t


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analytically calculated vectors. In short , angles and the associated phase aremeasured a gainst rotat ion ba sed upon th is physical relat ionship.

For proper identifi cat ion, pha se an gles should be specifi ed as a pha se lag, or

provided with a negative sign. In most cases, it is convenient to ignore the nega-

tive sign, and recognize that these angles are phase lag values. Using this con-

vention, phas e between t he 3 vibrat ion vectors in F ig. 2-2 ma y be converted by :

(2-14)

(2-15)

(2-16)

If a velocity phase angle occurs at 225, it is determined from (2-14) thatthe displacement phase a ngle is computed by: 225+ 90 = 315. Simila rly, the

velocity phase may be converted to an acceleration phase from equation (2-16)

a s: 225 -90 = 135 . If the phase la g negat ive sign is used, the a ngle conversions in

equa tions (2-14) to (2-16) mu st a lso be nega tiv e (i.e., -90 a nd -180 ). In eith er

case, consistency is necessa ry for a ccura te a nd r epeata ble results.

In addit ion t o phase, the vibration ma gnitude of an object m ay be converted

from displacement to velocity or acceleration at a constant frequency. This

requires a conversion of unit s w ithin th e motion equa tions (2-12) and (2-13). For

example, consider the following definition of English units for these parameters:

D= Displacement Mils,peak to peak = Mils ,p-p

V= Velocity In ches/Second, zero to peak = I P S ,o-p

A = Accelera tion G s,zero to peak = Gs,o-p

F= Fr equen cy Cy cles/S econd (Hz )

Reinstalling 2Ffor the frequency , and considering the peak values of the

Fig. 23 Traditional Angle Designation On A Rotating Disk

StationaryViewingPosition

Angle orPhase

Direction

Timeand

Rotation

180

0

90

270

315

45

135

225

Axis

of

Rotatio

n

Phasedisplacement Phasevelocity 90+=

Phasedisplacement Phaseacceleration 180+=

Phasevelocity Phaseacceleration 90+=


10/58

18 Chapter-2

terms (i.e., sin= 1), equa tion (2-12) may be resta ted a s follow s:

Since velocity is generally defined as zero to peak (o-p), and displacement is

typically considered as peak t o peak (p-p), the displacement value must be halved

to be consistent with the velocity wave. Applying the appropriate physical unit

conversions, the following expression evolves:

Which simplifi es to the follow ing common equa tion:

(2-17)

Next, consider t he relationship between a ccelera tion a nd displacement a sdescribed by equa tion (2-13), an d expand ed wit h proper engineering u nits to th e

following expression:

Accelera tion un its for th e above conver sion a re In ches/S econd 2. Measure-

ment unit s of Gs can be obtained by dividing this la st expression by the a cceler-

at ion of gravity a s follows:

This conversion expression ma y be simplifi ed to the followin g forma t:

(2-18)

The relationship betw een a ccelera tion an d velocity ma y be st at ed a s:

Expanding this expression, and including dimensional units, the following

equa tion for converting velocity a t a specifi c frequency to accelerat ion evolves:

V D D 2 F= =

VD2----Mils

1Inch

1 000Mils,-------------------------

2RadiansCycle

------------------- FCycles

Second-----------------

=

VD F318.31----------------=

A D 2 D2----Mils

1Inch

1 000Mils,-------------------------

2RadiansCycle

------------------- FCycles

Second-----------------

2

= =

AD F

250.661------------------

InchesSecond

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

AD F

250.661------------------

InchesSecond

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

1G

386.1Inches/Second2

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

AD F

219 560,------------------

D F139.9-------------

2= =

A V V 2 F= =

A VInches

Second---------------

2RadiansCycle

------------------- FCycles

Second-----------------

1G386.1Inches/Second

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

=


11/58


Sim plifying this expression, the follow ing common equa tion is derived:

(2-19)

The last three equations allow conversion between displacement, velocity,

and acceleration at a fixed frequency measured in Cycles per Second (Hz). A set

of expressions for frequency measured in Cycles per Minute (CPM) may also be

developed. Since machine speeds are measured in Revolutions per Minute

(RP M), this a ddit ional conversion is quite useful in ma ny inst an ces. Performing

this frequency conversion on equations (2-17), (2-18), and (2-19) produces the

next t hree common conversion equa tions:

(2-20)

(2-21)

(2-22)

The vibration units for equations (2-20), (2-21), and (2-22) are identical to

the English engineering units previously defined. However, the frequency for

th ese la st t hree equat ions ca rry t he units of Revolutions per Minute (i.e., RP M or

Cy cles per Min ute).

The simultaneous existence of three parameters (i.e., displacement, veloc-

ity, and acceleration) to describe vibratory motion can be confusing. This is fur-

ther complicat ed by the fa ct tha t inst rumenta tion vendors a re often specialized

in the manufacture of a single type of transducer. Hence, one company may pro-

mote the use of displacement probes, whereas another vendor may strongly

endorse velocity coils, a nd a th ird supplier may cultiva te th e application of accel-

erometers. The specific virtues and limitations of each of these types of trans-

ducer syst ems a re discussed in great er detail in cha pter 6 of this t ext . H owever,

for the purposes of this current discussion, it is necessary to recognize that dis-

placement, velocity, and acceleration of a m oving body a re alw ays related by the

frequency of the motion.

This relationship between variables may be expressed in various ways. For

example, consider an element vibrating at a frequency of 100 Hz (6,000 CPM)

an d a velocity of 0.3 IP S,o-p. From eq ua tion (2-17) the rela tionship bet ween veloc-

ity a nd displacement ma y be used to solve for th e displacement a s follows:

Similarly, the equivalent acceleration of this mechanical element may be

determin ed from equa tion (2-19) in t he followin g ma nner:

A

V F61.45---------------=

VD RPM

19 099,--------------------------=

A D RPM8 391,---------------

2=

AV RPM

3 687,--------------------------=

D318.31 V

F---------------------------

318.31 0.3IPSo-p100Hz

-------------------------------------------- 0.955Milsp-p= = =


12/58

20 Chapter-2

Thus, the displacement and acceleration amplitudes for this velocity maybe computed for any given frequency. Another way to view this interrelationship

between parameters is to extend this calculation procedure to a large range of

frequencies, and plot the results as shown in Fig. 2-4. Within this diagram, the

velocity is maintained at a constant magnitude of 0.3 IPS, o-p and the displace-

ment a nd a ccelera tion am plitudes calculated a nd plotted for several frequencies

betw een 1 a nd 20,000 Hz (60 and 1,200,000 CP M).

Fig. 2-4 shows that displacement is large at low frequencies, and accelera-

tion is larger at high frequencies. From a measurement standpoint , displace-

ment would be used for lower frequencies, and acceleration would be desirable

for high frequency da ta . Again, specific t ran sducer char acterist ics must also be

considered, and the rea der is referred t o cha pter 6 for a ddit ional deta ils on t he

actua l operat ing ran ges of tra nsducers.

For purposes of completeness, it sh ould be recognized tha t t he circular fun c-

tions previously discussed can be replaced by an exponential form. For instance,

equa tion (2-23) is a norma l forma t for t hese expressions:

(2-23)

In t his equat ion, i is equal to the squa re root of minus 1 and e is the nat -

ural log base that has a value of 2.71828. This expression will satisfy the same

equations, and produce identical results to the circular formats. However, it is

Fig. 24 Equivalent Displacement, Velocity, and Acceleration Amplitudes V. Frequency

AV F61.45---------------

0.3IPSo-p 100Hz61.45

------------------------------------------- 0.488 Gso-p= = =

J

J

J

J

J

J

J

J

J

J

J

J

J

J

HHHHHHHHHHHHHH

B

B

B

B

B

B

B

B

B

B

B

B

B

B

0.001

0.01

0.1

1

10

100

110

100

1,000

10,000

20,000

1 10

100

1000

10000

20000

VibrationAmplitude(Mils,IPS,G's

)

Frequency (Hertz)

Velocity

Acceleration

60

600

6,0

00

60,000

600,000

1,200,000Frequency (Cycles/Minute)

Displacement

D isp l a cem en t D ei

t=


13/58

Vector Manipulation 21

sometimes easier to manipulate equations using an exponential form rather

than a circular function. For reference, the relationship between the exponential

an d t he circular function is shown as follows:

(2-24)

The cos(t) term is often referred to as the Real, or the In-Phasecompo-nent. The i sin(t) term is the projection of the vector on the imaginary axis.This is normally called the Imag inary, or the Quadra tu recomponent. Theseterms a re used intercha ngeably. It sh ould be understood th at the form, and not

the intent of the equations has been altered. It should also be mentioned that

both the Reala nd Imag inary(In-Phasea nd Quadra tu re) components must sat-isfy the equa tion of motion for t he mechanical syst em.

VECTOR MANIPULATION

Many physical characterist ics of machines are described with vectors. A

magnitude is joined with a directional component to provide a parameter with

real physical significance. These vector quantities are routinely subjected to var-

ious types of mathematical operations. More specifically, the addition, subtrac-

tion, multiplication, and division of vectors must be performed as an integral

part of vibration and modal analysis, rotor balancing, analytical modeling, plus

instrumentation calibration.

For reference purposes, it is necessar y t o defin e the m ethods used for vector

manipulation. The different vector operations may be performed with a hand

held calculator, they may be executed with the math tools incorporated in

spreadsheets, or they may be included as subroutines into computer programs.

In addit ion, some Dynamic Signal Analyzers (DSA) use vector math as part of

the signal processing and computational capabilit ies. In all cases, these funda-

menta l ma th operations must be performed in a consistent ma nner.

From an explanatory standpoint , the specific vector equations will be

shown, and a numeric example will be presented for each type of operation. The

examples will be performed with circular coordinates, however an exponential

form will provide an identical solution. For consistency, the following pair of

polar coordinate vectors will be used throughout this series of explanations:

(2-25)

(2-26)

The fi rst vector (2-25) has of a ma gnit ude A , occurring a t a n a ngle . Simi-larly, the second vector (2-26) has an amplitude of B, and an angle . As previ-ously discussed, these vectors may be represented in a Cartesian coordinate (X-

Y) system by the following pair of equations:

ei t t( )cos i t( )sin+=

Va A =

Vb B =


14/58

22 Chapter-2

Multiplying the amplitude by the cosine and sine of the associated angle

will a llow conversion from polar t o recta ngula r coordina tes. The cosine term rep-

resents the ma gnitude on the X-Axis, and the sine term identifi es the a mplitude

on the Y-Axis. From the last pair of equations, the individual Cartesian ampli-

tudes for each vector component m ay be summa rized as:

(2-27)

(2-28)

(2-29)

(2-30)

This conversion of the initial vectors now provides the format to allow the

addit ion a nd subtra ction of two vector quan tit ies. Vector addition is performedby summing t he individual X an d Y components, and converting from Ca rtesian

back to polar coordinates. The summation of X-Axis components is achieved by

a dding equa tions (2-27) and (2-29) in th e following ma nner:

(2-31)

Similarly, the Y-Axis summation component is obtained by addition of the

previously described equations (2-28) and (2-30) as follows:

(2-32)

The X and Y summation components are now be converted back into polar

coordina tes of a mplitude a nd a ngle a s show n in equa tions (2-33) and (2-34):

(2-33)

(2-34)

Vector addition is used with many different types of calculations. Forinsta nce, consider the insta llat ion of two weights into the ba lance ring on a tur-

bine rotor. If the weights are both installed in the same hole, the effective weight

correction would be the simple sum of the two weights. However, if the weights

are screwed into two different holes, the effective balance correction must be

Va VaxVay

+ A cos A sin+= =

Vb Vbx Vby+ B cos B sin+= =

VaxA cos=

VayA sin=

VbxB cos=

VbyB sin=

Vad dxVax

Vbx+ A cos B cos+= =

Vad dyVay

Vby+ A sin B sin+= =

Vad d Va d dx( )2 Vad dy( )

2+=

ad dVad dyVad dx----------------

a t a n=


15/58


determined by vector addition of the two individual weight vectors. For demon-

stra tion purposes, assum e that 50 G ram s wa s inserted into a hole at 60 , and 40G ram s w as insta lled a t the 80 hole as described in Fig. 2-5. The init ial w eight

vectors a re represented wit h equa tions (2-25) and (2-26) as:

The summation of horizontal vector components in the X-Axis is deter-

mined wit h eq ua tion (2-31):

Similarly, the summation of vertical vector components in the Y-Axis may

be comput ed w ith equa tion (2-32) as follows:

The calculat ed X and Y ba lan ce weights ident ify th e combined effect of both

weights in the horizontal and vertical directions. These weights are actually X

an d Y coordina tes tha t m ay be converted to a polar coordinat e magnit ude usingequa tion (2-33) in th e following m a nner:

Fig. 25 Vector AdditionOf Two Balance Weights

50Grams @60

40Grams @80

88.6Grams @69

r

Va =

r

Vb=

r

Vadd

=

0

90

270

180

Effective Weight Vector

First Weight Vector

Second Weight Vector

Va A 50 Grams 60= =

Vb B 40 Grams 80= =

Vad dxA cos B cos+=

Va d dx50 60cos 40 80cos+=

Vad dx50 0.500 40 0.174+ 25.00 6.95+ 31.95 Grams= = =

Va d dyA sin B sin+=

Vad dx50 60sin 40 80sin +=

Va d dx50 0.866 40 0.985+ 43.30 39.39+ 82.69 Grams= = =


16/58

24 Chapter-2

Finally, the angle of the resultant vector may now be determined from

equa tion (2-34) as shown :

Thus, the 50 Gram weight installed at 60 plus the 40 Gram weight at 80

are vectorially equivalent to 88.6 Grams at 69 (as shown in Fig. 2-5). The mag-

nitude of this vector sum is the net effective weight t ha t s hould be used for a ddi-tional balancing calculations such as centrifugal force. The effective angle of this

weight pa ir is necessar y information for intermediate ba lancing response calcu-

lations, as well as the documentation of final results. For more information on

this t ype of calculat ion, please refer to cha pter 11 of this text.

The sa me ba sic approach is used for vector subtraction, with one signifi -cant difference. Instead of adding Cartesian coordinates, the X and Y compo-

nents a re subtra cted. Tha t is, by subtracting th e Bvector from the A vector, theX-Axis change is obtained by subtracting equation (2-29) from (2-27):

(2-35)

In a simila r ma nner, th e Y-Axis component is obta ined by subt ra ction of the

previously ident ifi ed equa tion (2-30) from (2-28) a s follows:

(2-36)

Calculation of the differential vector is achieved with equations (2-37) and

(2-38) tha t a re identica l in form t o the vector addit ion conversions:

(2-37)

(2-38)

This type of vector computation is extremely useful for performing routinetasks such as runout subtraction on proximity probe displacement signals. For

inst a nce, Fig. 2-6 displays a synchronous vibra tion vector at full operat ing speed

of 2.38 Mils,p-p at an an gle of 134. Assume tha t t he slow speed 1X runout wa s

Vad d Va d dx( )2 Vad dy( )

2+=

Va d d 31.95( )2

82.69( )2

+ 7 858, 88.64 Grams= = =

ad dVad dyVad dx

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

a t a n=

ad d82.69

31.95-------------

a t a n 2.588( )a t a n 68.9= = =

Vsu bxVax

Vbx A cos B cos= =

Vsu byVay

Vby A sin B sin= =

Vsu b Vsu bx( )2 Vsu by( )

2+=

su bVsu byVsu bx---------------

a t a n=


17/58


measured to be 0.94 Mils,p-p, at 78. Subtraction of the slow roll from the full

speed vector yields a compensa ted, or a runout corrected vector.

Mat hemat ically, the init ial vibra tion at run ning speed may be identifi ed as

t he A vector, a nd th e slow roll runout ma y be represented by t he Bvector. Subst i-tut ion of the d efi ned vibr at ion vectors int o equa tions (2-25) and (2-26) provides

the following vectors for subtraction:

The difference betw een horizonta l X-Axis vector components is d etermined

w ith eq ua tion (2-35) in th e follow ing ma nner:

Sim ilar ly, the d ifference of vector components in t he vert ical Y-Axis ma y be

computed with equation (2-36):

The negative value for the horizontal component is perfectly normal, and

acceptable. This negative sign, combined with the positive sign on the vertical

component, identifies that the final vector will reside in the upper left polar

quadrant (i.e., angle between 90 and 180). The computed X and Y coordinates

may now be converted to polar coordinates using equation (2-37) to determine

the magnitude of the runout corrected vector:

Fig. 26 Vector Subtrac-tion Of Shaft Runout FromRunning Speed Vector

r

Va =

r

Vb =

0

90

270

180

0.94Mils,p-p

@782.38Mils,p-p

@134

2.01Mils,p-p@157Runout Corrected Vectorr

Vsub=

Initial Vector Runout Vector

Va A 2.38 Milsp-p 134= =

Vb B 0.94 Milsp-p 78= =

Vsu bxA cos B cos=

Vsu bx 2.38 134cos 0.94 78cos=Vsu bx

2.38 0.695( ) 0.94 0.208 1.654 0.196 1.850 Milsp-p

= = =

Vsu byA sin B sin=

Vsu by2.38 134sin 0.94 78sin =

Vsu by2.38 0.719 0.94 0.978 1.711 0.919 0.792 Mils

p-p= = =


18/58

26 Chapter-2

The angle of the runout compensated vector may now be calculated from

equation (2-38) as follows:

The 180 addition to the angle is a quadrant correction. Thus, subtracting a

runout of 0.94 Mils,p-p at 78 from the full speed vector of 2.38 Mils, p-p at 134

yields a runout compensated vector quantity of 2.01 Mils,p-p a t 157. This ca lcu-

lated result is in full agreement with the vector diagram shown in Fig. 2-6. This

compensated vector represents the actual dynamic motion (i.e., vibration) of the

shaft. For more information on runout compensation, please refer to chapters 6,

7, 8, an d 11.

The major complexity associated with vector addition and subtraction is

due to the necessity for converting from polar to Cartesian coordinates, perform-

ing a simple operation, and then converting from Cartesian back to polar coordi-

nates. Fortunately, this multiple conversion is not required for vector

multiplication and division.

Vector multiplication of tw o vector qua nt ities ma y be executed by sim ply

multiplying amplitudes, an d a dding th e respective phase a ngles as follows:

(2-39)

This ma nipulation is easy to perform, a nd t he only cautionary note resides

with the va lue of the a ngle. In ma ny cases, this ma y exceed 360, due to the size

of angles a nd . When a full circle has been exceeded (i.e., final angle greaterthan 360), the size of the angle may be reduced by 360 to yield a physically

meaningful angle between 0 an d 360.

Vector multiplication is necessary in the machinery diagnosis business. For

example, consider the situation of determining the required balance weight to

correct the 1X vibration response of a machine. Presuming that the unit has a

properly defined balance sensit ivity vector, the required balance weight and

a ngle can be determined fr om equa tion (2-39). This req uires a vector m ultiplica-tion between the m easured vibra tion, and the sensit ivity vector. For demonstra -

tion purposes, assume that the measured vibration vector is 2.0 Mils, p-p a t a n

an gle of 40 . Further a ssume tha t t he rotor bala nce sensit ivity vector is equal to

Vsu b Vsu bx( )2 Vsu by( )

2+=

Vsu b 1.850( )2

0.792( )2

+ 4.050 2.01 Milsp-p= = =

su bVsu byVsu bx---------------

a t a n=

su b0.792

1.850----------------

a t a n 0.428( )a t a n 23.2= = =

su b 23.2 180+ 156.7= =

Vm u l Va Vb A B( ) +( )= =


19/58


150.0 G ra ms/Mil,p-p at an an gle of 190. B ased on this da ta , the opera ble vectors

for this vector ma nipulat ion a re identifi ed as:

Multiplicat ion of th ese tw o vectors is performed w ith eq ua tion (2-39) a s:

This vector product indicat es tha t t he insta llat ion of a 300 Gra m w eight at

an angle of 230 will balance the measured synchronous response of 2.0 Mils, p-pat 40. Naturally, the accuracy of this value is dependent upon the correctness of

the balance sensitivity vector.

As described in further detail in chapter 11, a vector summation between

the calculated vibration from the weight, plus the current vibration vector will

result in a predicted vibration vector with the weight at tached. An addit ional

vector summa tion with the sha ft runout w ill produce an uncompensated 1X vec-

tor. For a perfectly linear mechanical system, this would be the vibration ampli-

tude and phase displayed by a synchronous tracking filter . Although this

discussion is somewhat premature within the sequence of this text , the main

point is t ha t vector calculations ma y involve a string of ma nipulat ions t o achieve

the necessar y result .

Vector division represents the final category of vector math. Referringback to the initial vectors, equations (2-25) and (2-26), vector division is per-

formed by dividing the am plitudes, and subtra cting the a ngles as follows:

(2-40)

This kind of manipulation is also easy to perform, and again a cautionary

note resides with the fi na l value of the angle. In ma ny cases, this a ngle may drop

below 0 , due to the rela tive size of an gles a nd . When t he zero point is crossed(i.e., negative angle), the size of the angle may be increased by 360 to yield a

physically mea ningful angle betw een 0 a nd 360.

Vector division is widely used for various types of machinery calculations.

For instance, the computation of a balance sensitivity vector requires the divi-

sion of a ca libra tion w eight vector by a d ifferentia l vibra tion response vector. The

technical details associated with this calculation are in chapter 11. However,

from a pure computational standpoint, consider the following initial pair of vec-

tors for division.

Va A 2.0 Milsp-p 40= =

Vb B 150 Grams/Milp-p 190= =

Vm u l A B( ) +( )=

Vm u l 2.0 Milsp-p 150 Grams/Milp-p( ) 40 190+( ) 300 Grams 230= =

Vd i vVa

Vb

-------AB----

( )= =


20/58

28 Chapter-2

Division of th ese tw o vectors is performed w ith equa tion (2-40) as follows :

This calculation identifies a single balance sensitivity vector based upon a

mea sured d ifferentia l response vector of 5.00 Mils,p-p a t a n a ngle of 60. This vec-

tor cha nge in shaft vibrat ion response wa s due to the insta llat ion of a 400 G ra m

weight a t an an gle of 230. Vector division of the w eight by th e differential vibra-

tion vector yields th e bala nce sensitivit y vector of 80.0 Gr a ms/Mil,p-p a t a n ang leof 170. This unbalance sensitivity vector may now be used to compute balance

corrections in a manner similar to the earlier example of vector multiplication.

These simplified rules for vector mult iplica tion an d division may be verifi ed

by performing the same operations using exponential functions instead of the

presented polar coordinates. The results will be identical, and this will reinforce

the concept tha t t he vector ma th m ay be successfully executed using either expo-

nential or circular functions. In all cases, these vector manipulations are contin-

ually used throughout the field of machinery analysis, and these procedures

must be ma stered to a llow progression t o the real ma chinery t opics.

UNDAMPED FREE VIBRATIONExpanding upon the concepts of the previous section, again consider the

single mass pendulum of Fig. 2-1. Within this earlier mechanical system, the

mass of the concrete block was identified as the only significant element in the

system. If this concrete block remains constant , and if the weightless cable is

replaced by a coil spring, the simple spring mass system of Fig. 2-7 is produced.

Assume that the spring is suspended from a totally rigid I-Beam, and consider

the mass to be confined to movement only in the vertical direction. Since damp-

ing is not involved, this is considered as an undamped mechanical system. In

addition, there are no external forces applied to this system, so it must be classi-

fi ed as a system tha t exhibits free vibrat ion wh en it is displaced, an d allowed to

oscillat e in th e vertical plane. The resultan t motion is defin ed as unda mped free

vibra tion of this one degree of freedom mecha nical sys tem.If th is physical example is converted into a t rad it iona l physics diagra m, the

sketches shown in Fig. 2-8 evolve. The left diagram shows the main mechanical

elements, and the right sketch displays the Free Body Diagram. Normally, this

mechanical system would remain at rest (i.e., no motion). For this system to

Va A 400 Grams 230= =

Vb B 5.00 Milp-p 60= =

Vd i vVa

Vb

-------AB----

( )= =

Vd i v400 Grams

5.00 Milp-p

-------------------------- 230 60( ) 80.0 Grams/Mil

p-p170= =


21/58

Undamped Free Vibration 29

move, some type of initial disturbance is required. Furthermore, when this

mechanical system is in motion, the free body diagram (Fig. 2-8) reveals two

active forces; a spring force, and the gravitational term. The general equation of

motion for th is body is simply the equa lity of active forces a s follows:

B y rea rra nging terms, the following summa tion of forces is obtained:

Substituting a simpler alpha identification for each of the four variables,the equation of motion for this simple spring mass system may be stated in the

ma nn er t ha t W. T. Thomson used:

(2-41)

If equation (2-41) is divided by the mass, the resultant expression contains

a system m echa nical consta nt (i.e. , K / M), plus the interrelated acceleration anddisplacement of th e body:

(2-42)

Eq ua tion (2-42) can be sa tisfi ed by either of th e previously discussed circu-

lar or exponential functions. For simplicity, assume that an exponential functiona s d efin ed in equa tion (2-23) is subst itut ed int o (2-42) to yield th e followin g ver-

sion of the equa tion of motion:

Fig. 27 Spring MassMechanical System

Fig. 28 Equivalent Spring Mass Mechanical System AndAssociated Free Body Diagram

Mass

Coil Spring

StationaryI-BeamSpring with

Stiffness =K

EquilibriumD = 0

Mass = M

Spring Force =- StiffnessxDisplacement

Gravity Force =MassxAcceleration

Displacement = +D

M a ss A ccel er a t i on ( ) St i f f ness D isp l acemen t( )=

M a ss A ccel er a t i on ( ) S t i f f n ess D i s pl a cemen t ( )+ 0=

M A( ) K D( )+ 0=

AKM------

D+ 0=


22/58

30 Chapter-2

Extr acting t he common terms from t his equa tion, the following is obta ined:

(2-43)

Equation (2-43) is satisfied for all values of time twhen the terms withinthe bra ckets a re equat ed to zero:

This may now be solved for the natural or critical frequency c as follows:

(2-44)

Another common form of this expression is obta ined by converting t he rota -

tional frequency c units of Radians per Second to Cycles per Second in accor-da nce wit h equ at ion (2-2) to yield th e following:

(2-45)

Clearly, the frequency of oscillation is a function of the spring constant, and

the ma ss. This is the und am ped na tura l frequency of the mecha nical system. It

is also commonly called the undamped critical frequency, and the subscript c

has been added to identify frequencies c a nd Fc. In a ll cases, following a n init ia ldisturba nce, the ma ss w ill oscillate (or vibrat e) at this na tura l frequency, and the

amplitude of the motion will gradually decay as a function of time. This reduc-

tion in amplitude is due to energy dissipation within a real mechanical system.

Although this result is simple in format , it does represent an extra ordina r-

ily importa nt concept in t he fi eld of vibrat ion a na lysis. Tha t is, the na tura l fre-

quency of a mechanical resonance will respond to an alteration of the st iffness

and the mass. Often, the diagnostician has limited information on the effective

stiffness, or equivalent mass of the mechanical system. However, changes in

stiffness or mass will behave in the manner described by equation (2-44). In

ma ny inst an ces, this knowledge of th e proper relat ionship between para meters

will allow a respectable solution to a mechanical problem.

Init ially , the existence of a unique natural frequency that is a function of

the mechanical system mass and stiffness may appear to be only of academicinterest . In reality , there a re fi eld applications of this physical r elat ionship that

may be used to provide solutions for mechanical problems. For instance, if a

mecha nical system is excited by a periodic force at a frequency t ha t approaches a

natural resonant frequency of the mechanical system the resultant vibratory

D 2 ei t K

M------

D ei t+ 0=

D ei t 2 K

M------+

0=

2 KM------+

0=

cKM------=

Fc1

2------

KM------=


23/58


motion may be excessive, or even destructive. Three potential solutions to this

type of problem were identified by J. P. Den Hartog 5, in his text MechanicalVibrat ions. Quoting from page 87 of this book:

I n ord er t o im pr ove such a situ ati on, we mi ght first att empt to eli mi nat eth e force. Qui te oft en t hi s is not pr acti cal or even possibl e. Th en w e may changeth e mass or th e spri ng constant of the system in an at tempt to get aw ay fr om th er esonance cond it ion, but i n some cases thi s is also im pr actical. A thi rd possibil i t yli es in th e appli cati on of th e dynam ic vibr ati on absor ber, in vent ed by Frahm i n1909The vibra ti on absor ber consists of a compar ati vely smal l vibr atory system

k, m att ached to th e ma in mass M . Th e nat ur al fr equency of th e att ached

absorber i s chosen t o be equal to th e fr equency of th e distur bing force. It w i l l beshown th at th en th e mai n mass M does not vibr ate at al l, and t hat t he smal l sys-tem k , m vibr ates in such a w ay tha t i ts spr in g for ce is at al l i nstances equal andopposit e to Posin t. Thu s there is no net force actin g on M and th er efore th e

mass does not vibr ate

In h is text book, Den Ha rtog proceeds to derive a detailed equat ion set t ha tsupports t he above stat ements. He also exam ines torsional systems, and da mped

vibra tion a bsorbers. Thomson6 also discussed the utilizat ion of both la teral a nd

torsional vibration absorbers. However, for this discussion, the application of a

simple lat eral unda mped spring ma ss vibra tion absorber will be reviewed. The

fundamental engineering principles behind an absorber installat ion are illus-

tra ted w ith t he following case history.

Case History 1: Piping System Dynamic Absorber

The mechanical system under consideration consists of a pair of product

tra nsfer pumps tha t w ere subjected to a modificat ion of the discha rge piping to

span across a new r oadwa y. These essentia l pumps were motor driven at a con-stant speed of 1,780 RPM. The pumps had a successful eight year operating his-

tory, with only minor seal problems, and one coupling failure. During a plant

revision, the pump discharge piping was rerouted to a new pipe rack. Due to the

design of the new rack, the discharge line was poorly supported, and problems

began t o appear on both pumps shortly aft er the piping modificat ion.

Multiple seal failures were combined with repetitive bearing, and coupling

failures. These two pumps that previously received maintenance attention only

once or twice a year were now subjected to overhauls on a monthly basis. This

increased maintenance passed unnoticed for a long t ime. Unfortunately, one

night the ma in pump failed wh en the spa re pump wa s out for repairs. This coin-

cidence of mechanical failures forced a plant outage, and this event focused man-

agement attention upon the reduced reliability of these pumps.

Vibrat ion a na lysis of the pumps a nd t he associat ed piping revealed a domi-

5 J.P. Den Hart og, Mechani cal Vibrati ons, 4th edition, (New York: McGra w-Hill B ook Company,1956), p. 87.

6 Willia m T. Thom son, Theory of Vibrat ion wi th Appl i cat ions, 4th Edition, (Englewood Cliffs,New J ersey : Pr ent ice Ha ll, 1993), pp. 150-159.

k m


24/58

32 Chapter-2

na nt m otion a t t he pump running speed of 1,780 RP M. Compar ison w ith hist ori-

cal data revealed 1X vibration amplitudes on the pump and motor were ten to

tw enty t imes higher tha n previously measured. This ma chinery abnorma lity wa s

coincident with vertical vibration levels in excess of 25 Mils,p-p at the middle ofthe un support ed dischar ge line (i.e., midspa n of the roa d crossing).

A temporary brace was fabricated, and placed below the discharge line.

This support reduced the piping vibration, and also resulted in a drop in the

pump synchronous motion. Considering the positive results of this test, and

some preliminary calculations on the natural frequency of the piping span, it

was concluded that the pump running speed was very close to a lateral natural

frequency of the new discha rge pipe.

Since a brace in the middle of the road was unacceptable as a long-term

solution, other possibilities were examined and discarded. Finally, the applica-

tion of a t uned spring mass vibra tion absorber wa s considered a s a potentia l and

practical solution. For this problem a simple horizontal cantilevered vibration

absorber was designed to resemble the diagram in Fig. 2-9.

This device consists of a fa bricat ed pipe sad dle tha t is securely bolted to the

outer diameter of the discharge pipe. It is physically located at the point of high-

est vibration (i.e., center of the piping span). Since the pipe vibrates vertically,

the a bsorber is posit ioned horizonta lly so that the cant ilevered w eight ma y a lso

vibrate vertically. In this case, the spring consists of flat bar stock that has the

most flexible axis placed in the direction of the desired motion. The overhung

ma ss is bolted to the fla t ba r stock spring, and it ma y be moved back and forth to

allow a djustment of the nat ural frequency.

B y inspection of this damper a ssembly, it is apparent t ha t t he st iffness and

mass of the spring, plus the overhung mass are equivalent to a simple spring

mass system. The problem in designing an appropriate vibration absorber is now

reduced to a reasonable selection of physical dimensions t o obta in a na tura l fre-

quency of 1,780 CPM for this installed assembly.

Several a pproaches ma y be used to determine an acceptable set of a bsorber

Fig. 29 Typical TunedSpring Mass VibrationAbsorber AssemblyFor Piping System

FabricatedPipe Saddle

Sliding Overhung Mass

Spring

Pipe I.D.


25/58


dimensions. For example, a Finite Element Analysis (FEA) could be performed.

However, an FEA approach may become unnecessarily complicated and t ime

consuming. Use of published beam na tura l frequency equa tions ma y a lso be con-

sidered. However, one must be careful of published canned equations where theassumptions and boundary conditions may not be clearly explained or under-

stood. Fortunately, a practical approach for performing these calculations was

presented by J ohn D . Raynesford7, in his H ydr ocar bon Pr ocessin gar ticle on t hissubject . In this a rt icle, he considered th e system a s a simple spring ma ss a ssem-

bly. The dimensions of the spring were combined wit h t he overhun g ma ss t o pro-

vide the basic elements for the absorber design. Specifically, Raynesford

considered the total static deflection Ytot a l of the vibration absorber to be associ-

a ted with the weight W, mass M, gravitat ional constant G, and the spring con-s t an t Kof the assembly in t he following ma nner:

(2-46)

This is t he sam e general st iffness relat ionship that wa s previously a pplied

to the simple pendulum in equation (2-7). If equation (2-46) for stiffness is placed

into the previously developed natural frequency equation (2-44), the following

substitut ion a nd changes ma y be performed:

Solving for the total deflection Ytot a l, the following equation is obtained:

(2-47)

The total end point deflection of the vibration absorber was presumed to be

due to a combination of the uniformly distributed weight of the spring, plus the

cantilevered mass on a weightless beam. Tradit ional deflection equations for

these two elements may be extracted from various references. For example,

deflection of a beam with a uniformly distributed load ma y be obta ined from ref-

erences such as Shigley 8, or Roa rk9 as follows:

(2-48)

7 J ohn D. Rayn esford, Use D yna mic Absorbers to Reduce Vibration, H ydr ocar bon Processin g,Vol. 54, N o. 4, (April 1975), pp. 167-171.

8 J oseph E. Shigley an d Cha rles R. Mischke, Standar d H andbook of Machin e Design, (NewYork: McGraw-Hill Book Company, 1986), pp 11.5-11.6.

9 Warren C. Young, Roarks Form ulas for Stress & Strai n, 6th edition, (New York: McGra w-HillB ook C o., 1989), pp. 100-102.

KW

Yt o t a l

----------------M G

Yt o t a l

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

cKM------

M GYt o t a l -----------------

1

M------ G

Yt o t a l ----------------= = =

Yt o t a l G

c

2------=

Ys p r i n g Ws p r i n g L

38 E I

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


26/58

34 Chapter-2

where: Yspring = End Deflection of Spring (Inches)

Wspring = Weight of Spring (Pounds)

L = Length of Spring (Inches)

E = Modulus of Elasticity (= 30 x 106 Pounds / Inch2 for steel)

I = Spring Area Moment of Inertia (Inches4)

Similarly, the deflection of a cantilevered mass on a weightless beam may

be extr a cted from either S higley, or Roa rk, as follows:

(2-49)

where: Ymass = End Deflection at Mass (Inches)

Wmass = Weight of Mass (Pounds)

The total deflection due to the weight of the spring plus the cantilevered

ma ss is obta ined by superposition (a ddit ion) of these well proven beam defl ection

equat ions a s follows:

(2-50)

Substituting equations (2-47), (2-48), and (2-49) into the total deflection

equation (2-50) yields the following combined result:

(2-51)

At this point , the Raynesford art icle begins a tr ial and error solution to

arr ive at t he vibration absorber dimensions. Another wa y to obtain a set of real-

istic dimensions is to pursue a further simplification of the equation. For

inst a nce, equa tion (2-51) ma y be solved for the weight of the overhung m a ss a s:

(2-52)

In equation (2-52), the area moment of inertia Ifor the flat bar stock usedfor t he spring is determined by the next equa tion for a recta ngular cross section:

(2-53)

where: b = Width of Rectangular Spring (Inches)h = Height of Rectangular Spring (Inches)

As a lway s, the spring weight is calculat ed simply by multiplying volume by

the ma terial density as follows:

Ymas s Wmass L

33 E I

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

Yt o t a l Ys p r i n g Ymass+=

G

c2

------Ws p r i n g L

38 E I

-----------------------------------Wmass L

33 E I

-------------------------------+=

Wmass3 G E I

L3 c

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

3

8--- Ws p r i n g

=

Ib h

312

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


27/58


(2-54)

where: = Material Density (= 0.283 Pounds/Inches3 for steel)

Equations (2-53) and (2-54) will now be substituted back into (2-52), and

simplifi ed to yield the following expression for the overhung ma ss:

This expression cont a ins th e known q ua nt ities of th e accelerat ion of gravit y

G, the modulus of elasticity E, the density of the spring ma terial . I f a spring isconst ructed from fl a t st ock tha t is 1 inch wide by 1/2 inch th ick, then d imensions

ba nd hare also defined. The undamped natural frequency of the system cshould be equal to the measured excitation frequency of 1,780 CPM. Performing

these numerical substitut ions int o the la st expression y ields:

Performing these calculations, the following simplified result is obtained:

(2-55)

Eq ua tion (2-55) correla tes t he weight of the overhung m a ss to th e overhung

length for the defi ned condit ions. The gra ph shown in Fig. 2-10 is a plot of equa -

tion (2-55). It describes this s pecifi c relat ionship betw een the lengt h of th e springand the magnitude of the overhung mass. From this plot it is obvious that the

longer the spring, the less mass required. Conversely, as the spring is shortened,

the overhung weight must be increased. For this particular piping problem, a

spring length of 12 inches wa s selected with an overhung w eight calculat ed from

equation (2-55) of 5.4 pounds. This same weight value could also be extracted

from t he curve plotted in Fig. 2-10 for a spring length of 12 inches.

To allow fi ne tunin g of th e absorber resonan t frequency, the spring wa s fab-

ricated to be 15 inches long. This additional length does slightly violate the

developed equation array, but the error is small. In addition, it must be recog-

nized tha t th e developed equat ions do not const itut e a rigorous solution, but they

do provide an acceptable solution. Thus, the extra spring length allows the abil-

ity to perform a final adjustment of the natural frequency to correct for varia-

tions in the calculations, the fabrication process, or t he fi eld att achment.Normally, it is desirable to bench test the vibration absorber in the shop

before insta llat ion, and perform most a djustments before insta lling the device in

the field. In most instances, a simple hammer test with an accelerometer and

Ws p r i n g b h L =

WmassG E b h3

4 L 3 c2

------------------------------------- 3 b h L

8------------------------------------------

=

Wmas s

386.1In

Sec2

--------- 30610 Lb

In2

------ 1In 0.5In( )3

4 L 3 1 780CycleMin

------------, 2 RadCycle-------------- Min

60Sec--------------

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

3 1In 0.5In 0.283 Lb

In3

------ L

8-------------------------------------------------------------------------

=

Wmas s 21.84

L-------------

3Pound-Inch3 0.053 L( )PoundsInch

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


28/58

36 Chapter-2

spectrum a na lyzer will identify th e nat ura l frequency of the a bsorber. If the na t-

ural frequency is low, then t he overhung w eight should be moved in t owa rds t he

support. The opposite is also true. That is, if the measured natural frequency of

the absorber in the shop test is on the high side, then the overhung weight

should be moved aw ay from t he support .

In this case, the 5.4 pound weight was finally positioned at 12.5 inches from

the base during the shop frequency response test. Another minor adjustment

was made after the absorber was bolted into place on the discharge line. This

device proved to be successful, and piping vibration was reduced from levels in

excess of 25 Mils,p-p, to a fi na l condit ion at th e pipe midspa n of 1.5 to 2.0 Mils,p-p.

More significan tly, the vibrat ion a mplitudes on t he tw o tran sfer pumps returned

to previous hist orica l levels, and t he failur es cea sed.

The article by Raynesford also offers the following two important rules

regarding the at ta chment and fabricat ion of absorbers:

1. Tr y to att ach th e absorber at th e point of maxi mum vibr ati on and i n sucha way as to vibrat e in th e same plan e. Th at i s, if t he bear in g housing vi brat es inth e hori zont al pl ane, moun t t he absorber verti cally so it can also vibr ate in th ehorizontal pl ane. Adj ust the weight in an d out un ti l m ini mum vi brati on on theun it (maximum on the absorber) i s achi eved.

2. A r i gid att achment i s essent i al -th e wa nd must flex, not t he att achment . Becareful when usin g welds. Th ey ar e pr one to fai l ur e in th e heat effected zone. Makegenerous use of lar ge ra di i at t he junctu re of t he wand and th e base or att ach-ment

To th is pair of recommenda tions, it would also be advisa ble to suggest t ha t

the a bsorber be shop tuned to the desired na tura l frequency. This is a lwa ys eas -

ier to perform in the machine shop versus the field. In addit ion, the vibration

absorber should be installed with a permanent safety chain loosely connecting

the assembly with some adjacent rigid structure. If the support saddle or theattachment welds fail , this safety chain would restrain the spring mass assem-

bly, and significantly minimize the potential for any personnel injury.

Overall, it must be recognized that a vibration absorber provides a cost-

Fig. 210 Spring LengthVersus Overhung WeightFor Piping System LateralVibration Absorber

0

2

4

6

8

10

12

14

16

18

20

8 10 12 14 16 18 20

OverhungWeight(Pounds)

Length of Spring (Inches)

Vibration Absorber for

1,780 CPM1" x 1/2" Flat Bar Spring

Selected Configuration of12" Long with 5.4 Pound Weight


29/58

Free Vibration with Damping 37

effective solution to some difficult problems, and it demonstrates a practical

applicat ion of an unda mped mecha nical system. On the other hand, a vibrat ion

absorber is certainly not a universal solution for all machinery vibration prob-

lems. It is often ineffective w hen used t o correct a rotor resonance, or a n a coust icresonance problem. In most instances, this type of vibration absorber is useful

for addressing certain types of structural resonance problems and it should

always be applied with good engineering judgment and common sense.

FREE VIBRATIONWITH DAMPING

Now consider an expansion upon the concepts of the undamped system by

including a nother t ype of vibra tion a bsorber. At t his point, consider a single con-

crete block suspended from a rigid I-Beam by a spring, plus an automotive type

shock absorber as shown in Fig. 2-11. Again, the weight is allowed to move in a

vertical direction, and the equivalent spring ma ss da mper system is depicted in

Fig. 2-12. Since damping is now involved, this is a damped mechanical system.

As before, there is no external force applied, and the behavior of this system

must be classifi ed as free vibration wh en it is a llowed to oscillate.

The shock absorber is a viscous damper that displays the fundamental

property of a damping force that is proportional to velocity times the damping

coefficient. This rela tionship is quit e clear w hen th e ana logy of an old versus n ew

shock absorber is considered. Specifica lly, an old worn out sh ock a bsorber will be

quit e loose, and t he inner rod will move easily in a nd out of the ma in body. How-ever, a new shock absorber will be tight, and extension or compression of the

inner rod requires the application of a slow steady force. If an individual

attempts to rapidly move the inner rod, they will find that this higher speed

motion is resisted by a significan tly la rger force.

Fig. 211 Spring MassDamper System

Fig. 212 Equivalent Spring Mass Damper Mechanical Sys-tem And Associated Free Body Diagram

Coil Spring

Damper

Mass

M

O

N

R

O

E

StationaryI-Beam

Spring withStiffness=K

Di splacement = +D

EquilibriumD=0

Mass = M

DampingForce=

-C

oe

ff.xV

eloc

ity

Gravity Force =M assxAccel.

Damper withCoeff.=C

SpringForce=

-S

tiffnessxD

isp

l.


30/58

38 Chapter-2

Functionally, the shock absorber or damper removes energy from the sys-

tem. To stat e it a nother wa y, the da mper provides the funda menta l means of

energy dissipat ion for th e mecha nical syst em. If this physical representa tion is

converted into a traditional physics free body diagram, Fig. 2-12 evolves. Oncemore, the vertical motion must be init iated by an init ial disturbance, and the

system now reveals a spring force, a da mping force, plus th e necessary gravita -

tional term. From this free body diagram, the force balance yields the following

equat ion of motion for this da mped mecha nical system:

Moving all t erms t o the left side of the equa tion, the expression becomes:

Substituting a simpler alpha identification for the six physical variables,

the equat ion of motion may be sta ted as:

(2-56)

Again , in th e ma nn er used b y W. T. Thomson, t he periodic displa cement of

this damped spring mass system may be defined with an exponential function

similar to equation (2-23). If the displacement is identified by D, and th e t ime isspecified by t, and Sis a constant that has to be determined, an appropriateexponential equation would have the following form:

(2-57)

As demonst ra ted ea rlier in this cha pter, accelerat ion, velocity, and d isplace-

ment a re integrally relat ed, and equa tion (2-56) may be rewrit ten in t erms of dis-

placement by the substitution of equation (2-57). Certainly the displacement

term may be inserted directly. The velocity and acceleration terms are obtainedby taking the first and second time derivatives of equation (2-57) to yield the fol-

lowing equa tion of motion for this da mped single degree of freedom sys tem:

This expression ma y be simplifi ed by factoring out t he common exponent ial

term, and dividing by the mass Mto yield the next form of the motion equat ion:

(2-58)

As discovered in the undamped case previously discussed, equation (2-58)

may be satisfi ed for a ll values of t ime t, when the follow ing occurs:

M a ss A ccel ( ) St i f f ness D i s p l ( ) D am pi ng Coef f Ve loc i t y( )+=

M ass A ccel ( ) Damp i n g Coef f V el oci t y ( ) St i f f ness D i sp l ( )+ + 0=

M A( ) C V( ) K D( )+ + 0=

D eS t

=

M S2

eS t( ) C S e S t( ) K eS t( )+ + 0=

eS t

S2 C

M------

S KM------

+ +

0=

S2 C

M------

S KM------

+ +

0=


31/58


This last expression takes the dist inctive form of a quadratic equation.

From bas ic algebra, it is known tha t t his expression ma y be solved for the con-

s t an t Sin the following t rad it iona l ma nner:

(2-59)

Two solutions a re produced ( rad ical), an d t he general equat ion mus t be

expanded to correspond with this dual root. Hence, the periodic displacement

described by equa tion (2-57) is redefi ned in t he followin g ma nner:

(2-60)

Constants A a nd Bdepend on how t he oscillat ion wa s sta rted. The behaviorof a damped system is dependent on whether the radical from equation (2-59) is

rea l (+ ), ima gina ry (-), or zero (0). The simplest case is th e zero value for t he ra d-

ical, and t his term is defined as crit ical damping Cc, as follow s:

(2-61)

B y rear ran ging terms, the following int ermediat e result is obtained:

By taking the square root of both sides of the equation, and substituting

equa tion (2-44), th e followin g is obta ined:

(2-62)

For convenience, a da mping r a tio of will be defined as t he actual da mpingCdivided by t he crit ical dam ping Ccas follows:

(2-63)

Combining th e da mping r a tio from equa tion (2-63), an d equa tion (2-62), the

term C/2Mma y be reconfi gured as:

(2-64)

B ased on these derived expressions, the solution t o the qua dra tic equat ion

(2-59) ma y now be rewrit ten a s:

S1 2,C

2M---------

C2M---------

2 KM------

=

D A eS1t

B eS2t

+=

Cc2M---------

2 K

M------

0=

Cc2M---------

2 K

M------

=

Cc

2M--------- KM------ c= =

CCc------=

C2M---------

C2M---------

CcCc------

Cc2M---------

CCc------

c = = =


32/58

40 Chapter-2

If the undam ped na tura l frequencyc

is fa ctored out of th is expression, the

following result is obtained:

(2-65)

Interestingly enough, the solution for constants S1 a nd S2 reveals a rela-

tionship between the undamped natural frequency c, and the damping ra t io .The transition between oscillatory and non-oscillatory motion is referred to as

critical damping. For this case, C= Cc, = 1, a nd equ a tion (2-65) simplifi es to:

(2-66)

Subst ituting t his crit ical da mping solution for S1 a nd S2 back into the gen-eral equa tion of motion, equa tion (2-60) produces th e following result:

(2-67)

This function contains only one constant (A+ B), and the solution lacks therequired number of independent constants to properly represent the general

solution. In this case, an expression in the form of will sat isfy the equa-

tion. Upon substitution of this new form, the general solution of equation (2-67)

can be correctly w rit ten in the following ma nner:

(2-68)

The significance of a critically damped system is depicted in Fig. 2-13. In

this dia gra m, the displacement of the mass is plotted a gainst t ime. For numeri-

cal simplicity, the constants A a nd Bwere a ssigned va lues of + 10 a nd -5 respec-

Fig. 213 Time DomainAmplitude Response OfCritically Damped, AndOverdamped MechanicalSystems

S1 2, c c ( )2 c( )

2=

S1 2, c 2

1

=

S1 2, c=

D A B+( ) e ct=

t ect

D A B t +( ) e ct=

0

2

4

6

8

10

0.0 0.5 1.0 1.5 2.0 2.5

Displacement

Time

Critically Damped, =1.0

Overdamped, =2.0

Constants:

A = 10, B = -5, c= 5.0


33/58


t ively, and the natural frequency c was set equal to 5.0. The solid linerepresents a critically damped system. The resultant motion is aperiodic, and

this crit ically damped system retur ns to rest in t he shortest t ime wit hout oscilla-

tion (vibration) of the mass. Stated in another way, a crit ically damped system

conta ins t he minimum am ount of dam ping necessary for a periodic motion.

If the system damping is greater than crit ical damping, the system is con-

sidered to be overdamped. Conversely, if the mechanical system has less than

crit ical da mping, the syst em is underdam ped, an d it w ill oscillate or vibrat e with

time. It should be noted tha t m ost process ma chines are underda mped, an d sus-

tained motion of the rotating or reciprocating elements is normal behavior.

For a better understanding of damping, consider an overdamped system

(equivalent to a new shock absorber). In this case, the damping ratio would be

greater tha n one (>1), and the Sterms in the quadratic solution equation (2-65)ma y be specifi ed as:

Combining these expressions with the general equation produces the fol-

lowing equat ion for th e motion of a n overdamped mechanical syst em:

(2-69)

This function describing an overdamped mechanical system is plotted as

th e dotted line in Fig. 2-13. The displacement chan ge wit h t ime is th e sum of tw o

decaying exponential functions, and system vibra tion is not ma inta ined. Motion

is aperiodic, and the body returns to rest without oscillation. It is also clear from

this composite diagram that the overdamped system of equation (2-69) does not

return t o rest a s ra pidly a s th e previously discussed crit ically da mped case.

Fina lly, consider the situa tion of a mecha nical system w ith sma ll damping

(equivalent to a worn out shock absorber). This is generally referred to as an

underdam ped system where


34/58

42 Chapter-2

This expression ma y be simplified to a more understanda ble forma t a s:

(2-71)

B y ins pection, equa tion (2-71) consists of the superposition of an oscillat ing

sine wave plus an exponential term. In most cases, the amplitude of the sine

wave is decreased by the exponential function with increasing time. The variable

Y in this equation represents the peak intersection between the exponentialfunction and zero time. The term is the t iming lag between the oscillatory

curve and a zero time starting point. For demonstration purposes, a responsecurve for an under damped system is plotted in Fig. 2-14. A displacement value

of 10.0 was assigned to Y, the timing offset was set equal to zero, and a con-sta nt va lue of 5.0 was used for the undamped na tura l frequency c. This syst emexhibits a n oscillatory motion with respect to t ime, and this is referred to a s free

vibration of the under damped mecha nical system.

Another interesting point from examining (2-71) is that the term

is multiplied by the t ime tto determine the number of radians. This

suggests that the undamped natural frequency c is altered by the dampingra t io to produce a new frequency. In fact, this is commonly identified as thedam ped natura l frequency or damped cri t icalfor th e mecha nical system, and it isdefined in t he following m an ner:

(2-72)

Fig. 214 Time DomainAmplitude Response OfAn Under DampedMechanical System

D e ct

A e

ct i 1 2

B ect i 1

2

+=

D Y e ct ct 1

2 +( )sin

=

-8

-6

-4

-20

2

4

6

8

10

0 1 2 3 4 5

Displacement

Time

Underdamped, =+0.1

Constant: c= 5.0

Y1=8.5

81

Y2=4.5

60

Y3=2.4

25

Y4=1.2

89

c 1 2

dampedc r i t i c a l c 1 2

=


35/58


This is a very importa nt equat ion because it directly influences the impact

of damping upon a resonance. As shown in the forthcoming Fig. 2-18 for forced

vibration variat ions in damping produce major changes in amplitude and

phase th rough a resona nce. However, there is also a subtle shift in t he resona ntfrequency a s th e dam ping is varied. Thus, the ma chinery dia gnostician must be

fully aware of the fact that changes in system damping will alter the behavior

through a resona nce in th e following t hree ways:

r Signifi cant chan ges in the peak amplitude at t he resona nce.

r Signifi cant va riat ions in the phase angle cha nge across the resona nce.

r Subtle change in the damped natural frequency (i.e., damped critical speed).

Fig. 2-14 for an under damped system shows that the oscillatory motion

decays with t ime. Examination of a longer t ime record would reveal that the

amplitude decrease is actually an exponential decay. The rate of this exponential

decay ma y be qua ntifi ed by the log decrement w hich is defi ned as follows:

(2-73)

In equation (2-73), Y1 a nd Y2 represent any two successive amplitudes in

the decaying dynamic signal. The natural logarithm of this rat io defines the

da mping as the log decrement . In some cases, particularly w ith lightly da mpedor short dura tion experimenta l da ta , it is necessary to examine multiple cycles of

the decaying signal to determine the log decrement. For these situations, the

right hand side of equation (2-73) may be used. Within this part of the expres-

sion, the initial peak amplitude is still specified by Y1, and the a mplitude follow-

in g Nnumber of cycles is identifi ed as YN+ 1. The va lidity of this r elat ionship may

be checked by calculating the log decrement for different combinations. Forinstance, in Fig. 2-14, the first amplitude peak Y1 ha s a ma gnitude of 8.581, and

the second peak Y2 is equal to 4.560. Using the first part of equation (2-73), the

log decrement ma y be computed from these values in t he following m an ner:

The same calculation may be performed using the first three cycles in Fig.

2-14. For t his ca se the t hird pea k Y4 has an amplitude of 1.289, and the log dec-

rement ma y be computed w ith the r ight s ide of equa tion (2-73) a s follows :

The sa me result of = 0.632 has been reached using a single cycle an d mult i-

L ogDecrement Y1Y2-------

ln 1N-----

Y1YN 1+----------------

ln= = =

L ogDecrement Y1Y2-------

ln 8.5814.560-----------

chapter 2 dynamic motion

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