Azzam Daood

8/13/2019 Azzam Daood

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BEARING DYNAMICS ANALYSIS USINGFINITE ELEMENT METHOD

A THESISSUBMITTED TO THE

COLLEGE OF ENGINEERING-UNIVERSITY O BASRAHIN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR

THE DEGREE OFMASTER OF SCIENCE

IN

ZZ M D OOD HASSANB. Sc.)

September 2000


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IIf

ijj

,II':II,jlII

:L ~


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CertificationI certify that this thesis is prepared under my supervision at the

university of Basrah, as a partial requirement for the degree ofMaster ofScience in Mechanical Engineering.

7Signature: ~ r d } Ytt cJ ? ____Supervisor: Dr. Ameen A NassarDate: 6 t e c. :> C

In view of the available recommendations, I forward this thesis fordebate by the examining committee.

Signature: W 1 /4 Z lName: Dr. R R Al-Rab)l

Head ofMech. Eng. Dept.)Date: /q /? -...s---


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Committee Report

We certify that we have read this thesis entitled Bearing DynamicsAnalysis Using Finite Element Method as examining committee andhave examined the student in its contents. n our opinion it is adequate forthe partial fulfillment of the Degree ofMaster of Science in MechanicalEngineering.

Signature:1\btoiTF '.Hrr.- A. S. Shihab

MemberDate: '1--1-/ If 12

Signature: i,i } iVC7Name: Dr. S. H. Muhoder

MemberDate:2f-t/ I 12

On behalf ofthe University:

Signature: - ~ - r-----:.,-Name: Dr. A. S. ResenDean

Date: /2000

S i g n a t u r e J J ~Name: Dr. A. A. Al-Edani

SupervisorDate1J-f /2000

Signature: " _ . ;= -Name: Dr. A. S. Resen

ChairmanDate: /2000


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cliHOJvledtJement

I wish to acknowledge my supervisorDr. Ameen A. Nassar for his useful suggestionswhich helped me during the preparation of thisthesis.

Also I would like to thank the Dean ofEngineering College Dr. A S. Resen and thecomputer center for their cooperation.

Thanks for all staff of mechanical engineeringdepartment.

Finally I wish to express my gratitude to myfamily and my friends for their encouragement.


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BSTR CT

Fluid film bearings commonly used in heavy rotating machines playa significant role in the dynamic behaviour o the rotor because the thinfilm that separates the moving surfaces supports the rotor load.

The thin film was represented as a linearized values o the stiffnessand damping coefficients. It has been shown that the stiffness and dampingproperties o the oil film significantly alter the critical speeds and out obalance response o a rotor.

n addition rotor instability occurs which is a self excited vibrationarising out o the bearing fluid film effects nd this is n important factorto be considered in the rotor design.

This work presented a dynamic modeling o rotor bearing systemswith rigid disks distributed parameter finite rotor elements and t1exiblediscrete multibearings. The mathematical model takes into account therotary inertia gyroscopic moment axial load internal viscous andhysteretic damping and transverse shear defonnations linear s well snonlinear stiffness and damping for the finite bearing usingTIMOSHENKO element.A computer program is developed in this work to calculate the whirlfrequency and critical damping speed and to evaluate the dynamicunbalance response and rotor stability.


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ONTENTSitle I Page

Abstract I IContents INomenclature IVChapter One: Introduction

1 1 General Introduction 11.2 Methods to Analyze the Rotor-Bearing Systems 21 3 Instability of Rotors Mounted on Fluid Film Bearing 31.4 Eigenvalues Problems i 61.5 Scope of the Thesis I 6

Chapter Two: Literature Review2 1 Introduction I 72.2 Literature Review I 7 I2 3 Conclusions I 10

Chapter Three: Theory3 1 Introduction I 11 I32 Beam Theories 11 I3.2.1 Beam-Bending Theory Rayleigh Beam Theory) I 123 2 .2 Timoshenko Beam Theory I 133.3 Shape Functions I 43.4 Fluid Film Bearing Element I 16 I3.5 General Dynamic Equations ; 173.5.a Strain Energy i 183.5.b Kinetic Energy 193.5.c Dissipation Energy I 203.5.d Potential Energy I 203.6 Element Equations for Finite Rotor Systems 203.6.a Kinetic Energy of Rotor Element I 203.6.b Potential Energy with Internal Damping I 223.6.c Dissipation Function 253.7 Total Energy of Rotor I 253.8 Effect of Initial Applied Force and Torque I 263.9 Rigid Disk Element I 273.10 Unbalance Response I 273.11 Solution of System Equations I 283 1l a Whirl Speeds I 283.1l.b Unbalance Response I 283.1l c Ruge-Kutta Scheme I 29


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Title Pag_eChaQ_ter Four: Programmin

4.1 Introduction 314.2 Master Program 314.3 Free Vibration Program 32 i4.4 Damping Program 324.5 Time Domain Program 334.6 Unbalance Response Program 33

Chapter Five: Results and Discussion5 1 Introduction 355.2 Case Study 35 I5.2.1 Critical and Whirl Speeds 355.2.2 Unbalance Response 385.2.3 Forced Vibration Response 395.2.4 Transient Response 405.3 Summary 41

Chapter Six: Conclusions and Recommendation i6 1 General Introduction 59 l6.2 Conclusions 59 I6.3 Recommendation 60References 62Appendices IAppendix A:Al System Matrices ComponentsA2 Unbalance Nodal Force VectorAppendixB:Cases Study


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omenclatureSymbol Description I Units I'[D) IStress-Strain of dynamic system. I

[B x,y,z)] ICoordinate matrix of dynamic system. I II[M],[C],[K] Mass, damping and stiffness matrix.[N x,y,z)] Shape function matrix of dynamic system[I] Identity matrix[F(t)] External force vector of dynamic system N IUx,Uy,Uz Displacement components at point (x,y,z) inside I m Ieam.K Shear function of the cross section IVy,Vz Shear forces in y,z direction I N iNtiNri Translation and rotation displacement shape function, I1=1,2,3,4v(x, t),oo(x, t) Centerline displacement in y,z directionG Shear modulus N/m2EI Bending stiffness per unit curvature N.m2 I

Cross section area , IA m- IL Element length m III0 ,IP Diametral and polar mass moment of inenia per unit Ilength Kg .m IIp Polar mass moment of inenia for element I Kg .m IMz,My Internal moment about z, y direction I N.m IInitially axially force I Ip N IT Initially axially torque N.m ICyy,CyzCzy,Czz Elements of bearing damping matrix N.s/mKyy,Kyz Elements ofbearing stiffness matrix N/m IK;cy,Kzz Ify.fz Interaction force of bearing in y,z direction Nmd Mass of disk Kg I

IV


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Symbol Description UnitsDiametral inertia and polar inertia of disk. where IId Id =112 md 3R 2 +H 2 )) Kg.m

1

Io, PI ~ 1/2 mdR2 )

H,R Disk thickness and radius mOu n Unbalance nodal force vector NQs,Qc Cosine and sine components of distributed unbalance Nforce vector.

Element mass center location in rotor element, whereZ(x),Y(x) (Zc, Yc)= (Z(O), Y(O)) m

(ZR, YR)= (Z(L), Y(L)) Iyd,zd Disk mass eccentricity in y,z direction m

Greek LettersSymbol Description Units

[cr],[] Stress and strain vectors I[o Displacement vector of dynamic system :ITransverse shear effect i

EJ Internal viscous damping Sec IIEH Internal hysteretic dampingYH Loss angle rad

~ Small angle rotation about (y,z) axis radn Natural frequency of rotating system rad/sec1 1. Mass per unit length Kg/m

Total energy of system I(J) Whirl speed rad/seca Complex eigenvalue

v


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Subscriptsn x n Order o matrix.

Derivative with respect to time.Second derivative with respect to time.Derivative with displacement.Second derivative with displacement.

r Transpose matrixd Refer to disk element.b Refer to bearing element.

AbbreviationK Kinetic energy.DE Dissipation energy.u Strain energy.v Potential energy.

V


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Chapter neINTRODUCTION

1 1 General IntroductionThe phenomenon o natural frequency and critical speed o rotating

shafts is the most common problem that is discussed by a vibrationengineer. Some o the rotors weight as much as 1 tons as in the case o

large steam turbine and obviously they deserve at most attention in thisregard. The rotor have always some amount o residual unbalance, howeverwell they are balanced, and will get into resonance when they rotate atspeed equal to their critical speeds. These speeds are called as critical speedby R NKLINE in 1869.

In 1895 GUSTAVE DELAVAL became the first person todemonstrate experimentally thatasteam turbine was capable o sustainedoperation above the rotor s lowest resonance speed (first critical speed).With this previously held restriction on maximum rotor speed removed,designers were free to increase pertain speeds, thereby utilized from all thepotential benefits o higher speeds, with this new found freedom, it did nottake designers too long to discover a true upper limit for safe operatingspeeds. Namely, the THRESHOLD SPEED o rotor in film journal bearing.

In 1925 it has become evident that an important class o rotorbearing dynamic phenomena cannot be studied without accounting for thehighly non-linear forces produced by fluid film journal bearings underlarge amplitude vibrations (amplitudes approaching the bearing clearance).

Rotor unbalance, rotor stability and torsional dynamics o drivetrains are the major aspects o the rotor bearing dynamics problem.


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Reference sources at which information on rotor bearing dynamics hasbeen presented are:-

(1) IUT M Denmark Symposium Proceeding, September, 1974.2) ASME Vibration Conference, Washington, D.C., September, 1975.3) Rochester Symposium Proceedings, July, 1975.4) ASME Flexible Rotor-Bearing Dynamics, Monographs 1973.

(5) SVIC Monographs on Computer Programs, and on Balancing 1972

The major cause of excessive vibrations in rotating shafts is theresidual unbalance. The unbalance in the rotor come from materialinhomogeneites, manufacturing processes, key ways, slots, . . . etc. Inaddition the rotor deteriorates its balance condition during the operatingdue to wear, thermal bendii)g, process dirt collection etc. Naturally, it isimportant to determine the response of a rotor due to specified unbalance,to study its dynamic behaviour so as to determine, whether is, a rebalanceis necessary during the rotor life.

Rotor formulations vary widely from simple discrete mass/beamrotors through complex distributed mass-elasticity rotors. someapplications of the finite element method for unbalance response analysishave been developed by Lund [1] and by Zorzi Nelson [2].1.2 Methods to Analyze the Rotor Bearing Systems

The two main methods employed currently are the transfer matrixmethod and the finite element technique. Both these procedures adopt acontinuous representation of element parameters which is the main reasonfor their relatively high accuracy compared to the conventional lumpedparameter modeling, as was shown by Ruhland Booker [3].

The transfer matrix method as modified by Lund and Orcutt [4].Allows for a continuous representation at the shaft section and was found

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Fi = LRro(c+e)the flow out of the wedge is

F0 = LRro(c-e)

Fig. 1 1 Journal bearing

... (1.1)

... (1.2)

If pressure is developed in the film, when the bearing is operatingunder steady conditions, the flow-in is reduced and flow-out is increased bythe pressure induced flow, which balances i and F0 for maintaining theflow continuity. However, ifthe load is small, in the absence of pressure asmall whirl velocity is induced to maintain the flow balance. If theinstantaneous angular velocity is v for the journal center j then theinduced velocity is ev as shown in Fig. 1.1. By lifting off the journal fromits steady state position, the film volume increases by,

F=2LRevwhere 2LR is the projected area of the bearing, therefore,

LRro(c +e)= LRro(c -e)+ 2L Rev

.. .( 1.3)

... (1.4)


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and1.5)

Hence the rotor tries to whirl at a frequency of half the speed ofrotation to maintain the flow balance. We notice that if v >fro theoutward flow is more and therefore pressure is developed in the film and

.the bearing becomes stable. If, however v


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1 4 Eigenvalue ProblemsThe theory for the eigenvalue problem for non-spinning systems with

elastic restoring and dissipation forces is well developed, while theeigenvalue problem for spinning systems containing elastic and dissipationparts has received very little attention. The main difference between nonspinning and spinning systems is the damping matrix which have a skewsymmetry properties due to the vector gyroscopic effect.

The equation ofmotion for typical rotor systems free vibration) canbe written in the following compact and matrix forms:

whereMX+Cx+Kx=O

M: Mass matrixC Damping matrixK: Stiflhess matrix

1.6)

Also in same cases the damping matrix and stiffness matrix takenonsymmetrical form and its mathematical insights are discussed byAdams [7].1 5 Scope o the Thesis

The scope of this work can be summarized as follows:Chapter Two summarized the important previous work in the rotor

dynamic field. The basic theories, energy equation of rotor elementformulation, forced load formulation and numerical solution method arereviewed in Chapter Three.

The properties and construction of programming are reviewed inChapter Four.

Chapter Five contain the discussion of selected cases.Finally, the major conclusions of this work and recommendation for

future work are summarized in Chapter Six.

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Chapter TwoLITER TURE REVIEW

2 1 IntroductionWith the ever increase in demand for large size and velocity in

modem machines rotor dynamics became more and more an importantsubject in the mechanical engineering design.

The rotor-bearing system is contain many difficult problems such asstability natural frequency unbalance response .. . etc consequently thereare many researches tried to treatment these problems. When mathematicalmodels are presented the researchers succeeded to treat such type oanalysis.

At first transfer matrix was indicated a continuous success in thatfield at last finite element method was produced as an optimum solution tosolve that type o complex system.

Recent advances in computer technology have allowed morerealistic i.e. more complex system problems to be examined as opposed to~ idealized solutions obtained by earlier authors. Technology proceduresin these problem areas are now becoming firmly established as designroutines.2 2 Literature Review

In order to study the effect o bearing supports on the dynamicbehaviour o the rotor the Jeffcot model can be used e.g. Morton 8] andKirk Gunter [9]. Rao 10] used Jeffcot model analysis to study theconditions for backward synchronous whirl o a flexible rotor inhydrodynamic bearings. In this study he has shown that there is no distinct

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critical speed as it conventionally known when one of the cross coupledstiffuess coefficients of he bearing is negative and that a specific relation issatisfied when the expression under a square roote in the frequencyequation becomes negative. It mentioned then that the rotor becomesunstable but no details were presents. Rao Bhat and Snakar [11] extendedthis study to take into account the effect of bearing damping on thesynchronous whirl of a rotor in hydrodynamic bearings.

The instability of a rotor in fluid film bearings is generally attributedto the selfexcited vibration arising out of fluid film bearings and has beenobserved first y NewKirk and Taylor [ ] who called it as oil whip of arotor. In this instability region the whirling frequency of the rotor isslightly less than half the rotational speed and therefore is known as halffrequency whirl instability. The unbalance response of a general rotor withproperties defmed at several stations. Transfer matrices and finite elementmethods are popular e.g. Rao [12]. A detailed analytical treatment forinstability speeds of rigid and flexible rotors on different types of bearing isgiven by Rao [13].

Riegen and Cundiff [14] studies unbalance response of a compressorrotor which did not show the two typical peaks corresponding to the majorand minor speeds of a rotor in fluid film bearings in the region of rigidbearing critical speeds. Bhat [15] used modal analysis to determine theunbalance response of a single mass rotor on fluid film bearings.

Rao [16] studied the instability of a rotor mounted in fluid filmbearings it is demonstrated that instability of a rotor mounted onhydrodynamic bearings can occur under super synchronous whirlconditions. t is shown that such an instability condition arises only whenone of the cross-coupled stiffuess coefficients is negative and the problemis formulated accordingly. In this study Rao used analytical method toformulate the problem therefore many factors were neglected in his study.

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Sharan and Rao [17] presented the dynamic response of a multi-disk rotorsystem supported by fluid film bearing using the method of influencecoefficients. The stiffness and the damping coefficients have beenconsidered as speed dependent. The two bearings, which support the rotorshaft, are dissimilar. The dynamic response is calculated by varyingparameters such as spacing between the disks, the bearing clearance ratioand the relative mass of the rotor disks. Wong and Shih [18] used anoptimization technique to fmd the optimum diameters of shaft elements sothat the optimized rotor can sustain maximum fluid leakage excitation.A rotor-bearing system is modeled as an assemblage of concentratedrigid disks, discrete bearings, and rotor segments with distributed mass, andit would also contain the model of leakage excitation in this study. Theweak in this study came from this application of finite element techniqueshad well made by Nelson in 1976.

The works of uhl [19] and Ruhl Booker [3] are the first examplesof the studies using finite elements in rotor dynamics. In these investigationthe effect of rotary inertia, gyroscopic moments, shear deformations, axialload, and internal damping have been neglected. These studies deals withthe instability of rotating system and unbalance response and their resultcompared with classical lumped mass. Thorkildsen [2 ] has included rotaryinertia and gyroscopic moment. Polk [21] has used a Rayleigh beam finiteelement in his work. In 1979 Nelson [22] utilized Timoshenko beam theoryfor establishing the shape functions. He derived the system matricesincluding the effects of rotary inertia, gyroscopic moments, axial load, andshear deformations. A flexible rotor bearing system is represented inHashish and Sankar [23] studies. The mathematical model takes intoaccount many factors using Timoshenko element. As an application thedifferent effects of the bearing lining flexibility and the bearing supportflexibility on the rotor stability behaviour is studied and discussed.

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Nevzat nd Levent [24] are present dynamic modeling of rotorbearing system with rigid disks, distributed parameter finite rotor elementsand flexible, discrete multibearings. A computer program is developed inthis work to calculate the forward and backward whirl speed, thecorresponding mode shapes, the dynamic unbalance response ofmultibearing rotor system.2 3 Conclusions

From the previous researchers we can take some notes on their workas follows:-

1. All the researches neglected a number of important factors likeaxial force and torque, viscous and hysteretic damping in theformulation therefore, we consider these factors in the formulation .

2. The numerical methods that used in the previous researches havem ny weekness points such as linearize any rotor-bearing system tosingle mass and massless shaft Jeffcot model). n this work, it hasbeen consider the finite elements technique to prevent thisweekness.

3. The study of bearing and its effect on the rotor stability is notconsidered in a good attention.

Also, from the above discussion we found that all researchers not takeinto account the effect of cross-coupled bearing coefficients on the systemwhile this factor have very important effect on the system as shown later.

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3 1 Introduction

Chapter ThreeTH ORY

In this investigation a flexible rotor bearing system is represented bya finite element model taking into account the gyroscopic moments rotaryinertia shear deformation stiffness and damping for the finite bearing. Asimple Timoshenko beam is utilized and for rotor element two nodes andeight degree o freedom was considered because o many reasons list asfollows: 1 Capacity o addition o many significant factors to the element

. model.2. Reducing the order o system matrices.3. Two nodes element have high accuracy and that proved in study o

Ruhl [5].The present formulation allows many nonlinear effects to be studied

and their influence on the dynamic response o flexible rotors. Theunbalance response o rotor supported in multilobe or tilting pad bearing isreadily obtained. n addition the effects o fluid film dampers on rotorperformance may be investigated.3.2 Beam Theories

In the structural studies the assumptions primarily aboutdisplacement fields. Also we usually make assumptions as to certainaspects o the constitutive law to be employed.

The next paragraph will be explained in details the Rayliegh beamtheory and Timoshenko beam theory.

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3.2.1 Beam Bending TheoryConsider a long beam, which defonns in the x-y plane to shear forces

(in the y-direction) and bending moments (in the z-direction), with the x-axis being its neutral axis, as shown in fig. 3.1. The Raleigh beam theory isbased upon the following assumptions and approximations:1. The beam material is homogenous, isotropic, and linearly elastic, that

is, the generalized Hook s equations are valid for such a case.2. The strain components at any point inside the beam are infinitesimal.

3. The variation of the literal deflection across the beam thickness isnegligible.

4. Transverse normal stress is negligible. . Transverse shear strain is negligible compared with rotations due to

bending. It can be deduced from this assumption that plane crosssection of the beam remain plane after loading.

Consider the continuous beam shown in fig. 3 With forces andmoments bending the beam in the x-y plane, it is clear that thedisplacement cqmponent in the z direction Uz) is negligible, and it can alsodeduced from the previous assumption that the displacement component inthe y-direction (the literal deflection Uy) can be approximated at any point(x,y,z) inside the beam as a function of x) only defonned on the x-y planedue to shear forces and bending moments, can be approximated as follows[25];

dvUx(x,y,z)::::: - xUy(x,y,z)::::: -v x)Uz(X, y,z)::::: 0Also, it can be shown:that the relationship between moment and shear

force withy direction deflection is as [25]:

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.. .(3.1)

... (3.2)

XzM (Bending moment)

Fig 3.1 Beam under bending in the x-y plane

3 2 2 Timoshenko Beam TheoryThe beam bending theory was not considered the effect of shearstress and strains, which may be lead to inaccurate results for short beams.

The beam bending equation which take account of transverse sheardeformation and rotary inertia was given by Timoshenko.

Also the first four assumptions of bending theory are also employedin Timoshenko s theory. The deflection and stress equation are summarizedas follows:

dv- = jl x) + x )dx ... (3.3)where \ji X) is rotating angle due to bending only and ~ x ) is rotatingangle due to shear deflection effect only.

Ux(x,y,z) = y\ji X) = y [ : ~ x ) ]Uy(x,y,z) = v x)Uz(x,y,z) =0

3

.. .(3.4)


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shear force and bending moment can define as:-d jlM E Idx

V x) = G A ~ x )... 3.5)

where K is shear factor of the cross-sectional, which consider recognizingthe n o n ~ u n i o r m shear stress distribution at a section and for alternationdeviation for shear factors.3.3 Shape unctions

A typical rotor-bearing system consists of a rotor composed ofdiscrete disks and rotor segments, and discrete elastic bearings. Figure 3.2shows a typical rotor element of length I with the coordinates used todescribe the end point displacements q). Each rotor element is modeled asan eight degree of freedom element with two rotations and two translationsat each end.

zFig. 3.2 Finite rotor element.

The displacements of a differential disk of thickness dx placed at adistance x from the element end are denoted by v{x, t ), ro x, t), J3 x, t andy x, t , as shown in fig. 3.2. Translations { v x, t)} and rotations {co x , t)}of such a differential disk internal to the element can be related to the end

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Also we can divided these shape functions into two parts, one forbending contribution and the other due to shear deformation contribution.

[Nt],; [Nbt]+ [Nst]Nlbt = - 1 - ( 1 - 3 ~ 2 + 2 ~ 3 ) , Nl t - - { 1 - ~ )1+ s 1+

N3bt = - 1 - ( 3 ~ 2 - 2 ~ 3 , N3st = - ' - ~1 ~ 1 ~N4bt = ~ ~ ( - ~ + ~ N4st 2 ~ ) ( - ~ + ~

the rotational shape function can be written as [23]N rl L(1: 2Nr2 = -L - 1 - 4 ~ - 4 > ~ + 3 ~(l+ 4>N rJ L l ~ - 2Nr4 = - L - ( - 2 ~ + ~ + 3 ~ 2 )1 )

3 4 Fluid Film Bearing Element

.. . (3.10a)

(3.10b)

For the modeling of bearings, the classical linearized model witheight spring and damping coefficients [26], [22], [27] is used, fig . 3.3. Inthis model, the force at each bearings is obtained from,

... (3.11)where C ij and Kij are the elements of damping and stiffness matrices forthe bearings and lFb is the vector of baring forces.

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Fig. 3.3. Modeling o bearings

The non-linearity problems in this m o e l ~ the coefficients Kij, Cijwere a function o ournal speed [27], [28], whereas, the values o Kij, Cijvaried with rotating speed and then the stiffness damping matrices becomea function to speed and that make a big problems for many investigations.

Many researchers tries to solve the previous problems and some othem like ADAMS [ ] and DIMAROGONAS [29]; their solutions wereclosed for their case studies only and t cannot apply for the universalstudies.

The present work depend upon charts which gives by Rao [27]: thischart give the direct and cross-coupled stiffness and damping coefficientwith different speed.3 5 General Dynamic Equations

Consider engineering dynamic system, geometrically described bymeans o a number o finite elements in the x-y-z Cartesian space, and it issubjected to dynamic loading.

At any instance o time t), the nodal displacement vector o thestructure is a function o time represented by o t), in order to formulate the

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dynamic equations for the system, the energy balance principle may beemployed.

This principle states that: At any instant of time the summation of thesystem energies is stationary, that is the summation of the kinetic energyKE, the dissipation energy DE, the strain energy U, and the potentialenergy Vis stationary, or:

KE +ED+ U + V =Stationary ... 3-12)If these energies are defined in terms of the nodal displacement vector o t)then:

~ K E + D E + U + V ) = O .. . 3-13)Each term in the previous equation will be discussed briefly, for a generaltype of dynamic system.3.5.a Strain Energy

Defining cr e as the vectors of stress and strain components at anypoint, inside a finite element, then for a linear elastic material, stress, strainmatrix D can be formulated, such that:

cr t)= De t)Hence, the strain energy can be expressed as follows :

m mU = t L JJJ 1crdxydz=t:L JJJe1Dedxdydz

e=lelement e=le)ementwhere m is the total number of elements.

f here exist a matrix of coordinates B x,y,z) such that:t x,y,z,t)= B x,y,z)o t)

then the strain energy can be written in the following form:u = to 1f JJJB1DBdxdydz]o

e=lelement

18

.. . 3.14)

... 3.15)

... 3.16)

... 3.17)


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which can be reduced as follows:

where K represents the stiffness matrix of the system.Hence, it can be deduced that: or its defined as:K= JfJBtDBdxdydz

e=l element .u- = K o t )o

3 5 b Kinetic Energy

... 3.18)

. .. 3-18)

Defining the velocity vector of an infmitesimal mass dm inside thesystem as q, then the total kinetic energy of the system can be explained asfollows:

KE=t q tqdm=t r fffMqtqdxdydzsystem e=lelement

If there exists a matrix of coordinate N, such that:q x,y,z, t =N x, , z ~ t )

where 8 t)= o t )dt

... 3.19)

. . . 3.20)

Then the kinetic energy of the system can be written in the followingmatrix form:

.. . 3.21)

Hence, it can be shown that:... 3.22)

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3.5.c Dissipation EnergyThe dissipation energy depends upon the nature of the damping, and

for the case of viscous damping, a matrix C can be defmed, such that:DE= 1 t)c8 t)

Hence, it can be deduced that:aoE = c8 t)o

3.S.d Potential Energy

.. . 3 .23)

... 3 .24)

In the absence of external fields, the potential energy of the system atan instant of time t) may be expressed in terms of the work done by theapplied forces at that instant, which are represented by an equivalent nodalforce vector, i.e.

V =-WORK= -o 1 t)F t) ... 3 .25)from the above eq. We can write:

avo = -F t) .. . 3.26)Substituting Eqs. 3.18), 3.22), 3.24) and 3.26) in 3.13) gives;

M8 t)+CB t)+Ko t)=F t) ... 3.27)which is the dynamic finite element matrix equation of the system, and itrepresents a system of simultaneous second order differential equation withrespect to time t).3.6 Element Equations for Finite Rotor Systems3.6.a Kinetic Energy of Rotor Element

Figure 3 4 explained the differential disk located at x) and thecross-section spin about rotating references.

We can write the kinetic energy equation as:

~ ] [ ] d x ... 3.28)I p l)c

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where 1 and Ip are the diametrical and polar inertia per unit length, J..l:mass/length.

Fig. 3.4. Cross section spins aboutrotating references.

z

To transfer from (abc) reference to (xyz) reference using Euler'sAngles formula:

d k = [ : n ~ ~ I : J d x + I p r o d x +[ ~ n ~ ~ ~ : m d x - r o Y ~ l p d x

by substituting Eqs. (3.6) and (3.28) in (3.29) we get:-dk = J..L4T[Ntf[Nt]qdx+ Ipro2dx + Io4T[Nrf[Nr]qdx

- rolpq T[N rY [NrP ]qdxwhere:

[Nry]= [Nrl 0 0 Nr2 Nr3 0 0 Nr4][NrP]=[O -Nr1 Nr2 0 0 -Nr3 Nr4 0]

and by take the integration over the length of element to obtain:K = 4T[Mt]+ [Mr }q + Ipro2 roqT[R]q

21

... (3.29)

.. .(3.30)

... (3.31)

... (3.32)


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where- 2Ip = 112)mR

I[ t) = J1 [N t JT [N t ]dx0I[ rJ Jio [N rJT [N r]dx

I[R ] = J p [ N r JT [ N r P dx where dx = ld

0

... 3.33) .

[Mt] and [Mr] are symmetric matrices, which are listed in Appendix A.3.6.b Potential Energy with Internal Damping

In addition to external sources of damping there is very large numberof mechanism where by vibration n r ~ y can be dissipation, Ahid et. al.[30] discussed large type of internal damping model in his books.

In present thesis we just take the final model of material damping fortwo important type of dampings:-1. Viscous damping: this type due to internal viscosity of material which

discussed in Ahid [30] and it represented as a simple dashpot modelwith a damping coefficient.

2. Hysteretic dampings the stress leads to corresponding strain by anangle YH) which is a material property. The hysteretic loop in thestress- strain plane is an ellipse whose area, proportional to sin Y .

Gives a measure of the dissipated energy such that: EHsmyH -vi+ ~

where EH is hysteretic loss factor, and for more information or detailsabout this type of damping, Ahid [30], Pao [27] gives a good investigationsabout this subject in their researches.

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und [31] described the relation of internal moments with internalviscous and hysteric damping.

[ ~ : ] = B e n d i n g term dissipation termwhere Bending term= EI[e1 [::]

Dissipation term= Ele1[N11[::]So the potential energy with damping effect:

d, =tEl{[ ~ r [.,1[:ndx ~ [J[,l }}where,

[e2l= { ~ ~ ] e b [ ~ ]2]= 8 [I] eb[N2]

[N2J=[ ~ ~ ]

23

. .. 3.34) .

. . . 3.35)

.. . 3.36)

. .. 3.37)


36/90

Substitute Eq. (3.37) into Eq. (3.36) then,

dp = ~ ~ { r [1][:ndx ~ r [N2l[ d x~ HJ 11[::Jdx E.[ r N2l[::JdxJ} . J.J?)

By integrating Eq. (3.37),

where

then,

Pe =tUqT Ea[kb]+ Eb[kcbllk + q T Ea[ks]+ Eb[kcsllh}IKcb]= El J Nbtf N2] Nbt]dx0

I[kcsl= ~ f [ N ~ t f [ N ] [ N ~ t ] d x0

[kcb]= N3] kb][kcsl= N3][ks]

where [N3] is:

~ G i \ 0 0 -1 0 0 0 0

[N ]= . 0 0 , 0 _ _ ~3 0 0 . 0 -1 0 00 0 1 0 1 1 0 0 0

0 0 0 0 ) 0 0 - 10, 0 0 0 1 0.

Substitute Eqs. (3.40), (3.41) in Eq. (3.38) then:

... (3.38)

. .. (3.39)

.. . (3.40)

... (3.41)

pe =t {lqT Ea Ikxs+Eb N 3lllkb]+ lqTea 1]+ Eb N3llj kJI} ... 3.42)24


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3 6 c Dissipation FunctionWe can write the Dissipation Function as a summation of dissipation

due to bending and dissipation due to shear,

and by integration ofEq. 3.43) gives:D = elqT kb + ks)q

3 7 Total energy o Rotor ElementTotal Energy= Potential Energy+ Kinetic Energy

+ Dissipation Energysubstitute Eqs. 3.32), 3.42) and 3.44) in Eq. 3.45):Total Energy [x] = fq T [Ea [I]+ dN 3 ]]x [kb + ks]q

+t&Icl[kb +ksR +tqT(MT +MR]q+ lpro2 +roqT[R)q

Using Lagrangian equation to give the equation of motion [32]

So the differential equations of rotor element can be written as:M T + M R + ( [k b + k r)- ro [G ]q +

[ a [ I 1 x s [ e l r o ] [ N 3 ] [ k b + k s J I = 0}+ H }+ Hwhere G=[R]-[Rf.

25

.. . 3.43)

.. . 3.44)

... 3.45)

. .. 3.46)

... 3.47)

... 3.48)


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3 8 Effect o Initial Apolied Force and TorgueThe potential energy due to initial force (P) is:

_ 1 Rv ]T[P O][v ]p dx2 ro 0 P ro0and by using Eq. (3.6) in Eq. (3.49) we get:

kp =-tqT A)qwhere [A] is described as follows:

IA]=P [ N ~ f N ~ ] d x0

The potential energy due to initial torque T) is:_ 1 [w]T[T o][w]T dx2 o y 0 T y

Also using Eq. (3.7) in Eq. (3.52) we get:kT = - t q T(T]q

I[T]=T [ N ~ f [ N ~ ] d x

0So the final differential equation is:

[MT ]+ [M R ll i + [EI[kb + k 5 ] - ro[G llci +

... (3.49)

... (3.50)

... (3.51)

...(3.52)

... (3 .53)

... (3.54)

[ a k b +k5) - A ) - T ) + g ] N J k b +k5 ]lq =0 . .. (3.55)}+f:H l+EHAnd the all matrices that were denoted in the above equation are listed inAppendix A.

26


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3.9 Ridgid Disk ElementThe kinetic energy and element matrix for ridgid disk element is

presented as follows:lMd jqd - rolod Jl d J= [o] ... 3.56)

where: [qdF [vropy],and lMd j and lad jmatrices are listed in Appendix A.3.10 Unbalance Force Vector

A- For unbalance on rotor element, assuming a linear distribution ofthe mass center location in rotor element, the mass eccentricities in ydirection is y x) and in z direction is z x), we can drive force nodal vectorfrom virtual work principle:

[Qun]= J ~ r o [ N f [ y x ) ] c o s r o t d x + [ N f [ - z x ) ] s i n r o t d x ... 3.57)0 z x) 0 y x)

Eq. 3.57) can be written as the summation of two components:[Qc]= JrNtf[y x)]dx

0 z x)5 ]= J [ N ~ f [ - z x ) J d xo y x)

... 3.58)

Nevzat and Levent [24] take mass unbalance distribution over the element:y x) Y L l - ~ ) + Y ~ )z x) zL l- ~ ) + z R ~ ) ... 3.59)

where YL YR zL and zR are the mass eccentricities of the left and rightends in Y and Z direction respectively.

B- For unbalance on disk element we can write nodal force vector:

... 3.60)

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So, Eq. 3.60) can b summarized as follows:lo: n J= o ~ Jcos rot+ o ~ Jsin rot

where yd and zd are the disk mass eccentricities in Y and Z directionrespectively.

3.11 Solution of System Equations3 1l a Whirl Speeds

Whirl speeds can be determined from the solution of the eigenvalueproblem resulting from the free vibration equation,

[M][q] [c][q] [K][q]= [o] . . . 3 .61)We can write Eq. 3.61) as

[ [o] -[Mn[[qn [[M] [ol][[qn[M] [c] J [q]J+ [o] [K] [q] = [o] 3.62)

Adams and Padovan [9] formed the quotient to solve the aboveequation. It yields both forward and backward whirl speeds from the sameEigenvector. n order to cut down the computation time and to guaranteethe right convergence, the eigenvalues and eigenvectors of thecorresponding undamped system are taken as the starting values for thecomplex eigensolution iterations.

The eigenvalues are found in the form,a A.+iro . . . 3.63)

where ro is the whirl speed.3.11.b Unbalance Response

The unbalance response is come from solution of the equation 3.61)after adding the forcing vector [F] which is the vector of unbalance forcesand can be written as the summation of two components,

F] = Qc]cosrot Q 5 ]sinrot . . . 3 .64)28


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where

... 3.65)

3.ll .c Runge Kutta SchemeThe Runge-Kutta procedure applied to second order differential

equations. The following procedure can be applied to the finite elementdynamic matrix equation given as,

MX+Cx+Kx=F t)where F t) is applied force vector.

. . . 3.66)

At initial time the following parameters are assumed to be knowninitially:

x1 t0 )=x 0 , x1 t0 )=x 0 , F1 t0)=F0the equation 3.66) can be written as,

MX =F-Cx -Kx A vector g is defined, such that:

g = .:ltX or g = .:ltxthen we can calculate:

29

. .. 3.67)

. .. 3 .68)


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where F2 =F t tfrom the above we find:

also,X4 =X0 { X0 g;x4 =x0 +g3 , 4 = F-Cx 3 -Kx3g4 = ~ where F4 = F t0

At the last we can find the displacement as follows,x =x0 + t [ x +i g, + gz + g3)]

30

.. . 3 .69)

... 3.70)

... 3.71)


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4 1 Introduction

Chapter FourPROGR MMING

The cost o experimental testing makes it practical to consider computersimulation to verify rotor bearing system designs without the necessity obuilding each design variation considered.

During the past few years, a number o program packages have beenpublished, for example the thesis which is presented by Mushtaq [33] inmechanical department, University o Basrah and NASTRAN package. In thiswork a computer package program is developed in FORTRAN language which iscalculate the mass, stiffness and damping matrices then, using the anotherprograms, we can calculate the forward and whirl speeds forced and unbalanceresponse .

On the other hand, the package is presented n a way, which we can makecomparison between the results to make sure from the accuracy o the program.4 2 Master Program

It is the main program, which calculate the mass, damping and stiffnessmatrices o the rotor bearing system.Master program divide into many subroutines which give the matrices oall parameters that are considered and then assembly these matrices in form ogeneral matrices which represent as mass, damping and stiffness matrix.

These subroutines are listed as follows :-SMT and StviR are give the mass matrix o the rotor element. SKBKSgives the stiffness matrix o the rotor element. STand SAM are give the matrixwhich is add as shows in eq. 3 .55 . These two subroutines are take into accountthe effect o the axial torque and force respectively. SGB is subroutine to enter

31


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effect o gyroscopic effect. SMTD and S.MRD are give the mass matrix forelement. SGB gives the damping matrix for disk element. SCB and SKB arethe damping and stiffness matrix for bearing element. Then after generated

these matrices, assembler is held and gives a general form o matrices.

These programs are carried out to calculate the whirl frequency for rotatingsystems.

Gupta [34] and Chandrosekaran [35] and others are published manyresearches about the stability and development o eigenvalues solver for structurewhile in rotor dynamics solvers, it is found that 0 Sami et. al. [36] were therecognized researches studied in that field where 0 Sami used undamped systemand isotropic bearing in his investigation.

n that program, it is prefer to select the QZ method [23] in order to find theeigenvalues and this selection was success by giving high accuracy and ability toseparation between closed eigenvalues [33].4 4 Damping Vibration Program

This program is not different from the free vibration program in method osolution, but before solving we made change on the differential equation which isdiscussed in section 3.11.a).

This change gives a complex eigenvalues which are deals with the dampingcritical speed for rotating systems. The complex eigenvalues refer to difficult insolution after the change on the differential equation and non-symmetric systemmatrices. The select o QZ solver in order to find the eigenvalues is success tosolve this problem and give high accuracy complex eigenvalues .4 5 Time Domain Response Program

These programs are developed to calculate the time response o rotatin gsystem. These solver which used is Runge-Kutta scheme. It is discussed insection 3 . .c).

These solvers have very critical stability, which require small timeincrement which take long calculation time. Response program di vided into two

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subroutines. MA subroutine is format the matrices by product the initialdisplacement with stiffness matrix and product the initial velocity with dampingmatrix.

In the subroutine FI we can use it for forced vibration and also forunbalance response in rotor or disk element.4 6 Unbalance Response Program

Normally, these programs are developed to find the response of unbalancedrotating systems. Its consists of two parts : Generation and assembler ofmatricessystem and solve this matrices. Mushtaq [33] used a complex static solver to findthe response. He is converted the complex equations as below:-

where

let

[M][X]=[V][M] : Complex matrix[X]: Complex variable[V] : complex vector[M] = [MR] + i [MI][X]= [XR] + i [XI][V] = [VR] + i [VI]

sub. In eq. 94.1)

... 4.1)

.. . 4 .2)

{ [ R ] + i[MI]}CXR + iXI) = VR + iVI) ... 4.3){[MR]XR- [MI]}+i{[MR]XI + [MI]XR}= VR + iVI) ... 4.4)so,

[MR]XR- [MI]XI = VR[MR ]XI + [MI]XR =VI

then solving equations 4.5) to find [XR] and [XI] ... . 4 .5)

In this work another method is take which can summarized by solving thegeneral matrix

Mx + Cx + Kx = F t .. 4 .6)by take F t) as unbalance forcewhich gives in section J. ll .b).

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( StartI E Generation [m], [c] and [k]I Matrices for rotor elementIICe Generation [m] and [c]Matrices for disk elementa

Generation [c] and [k]Q ) Matrices for bearing element......I Ull ro Assembly the all matricesI:? In general formI - .5- l iQ Non-damping systemaQDeferential equation Make the matrices QZ) solversolver (Runge-kutta)scheme as in eq .3 .62) M=k

ITirhe response due to Calculation ofTime response due unbalance in rotor or eigenvaluesto transient vibration disk element and eigenvectors

End

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hapterFiveRESULTS DISCUSSION

5.1 IntroductionThe widely application of rotors in industry caused that the vectors

have many types, shapes, size, complexity, supporting methods, etc. Someof rotors have a large length and high weight such s turbine's shafts andpower shafts, the other are have an intermediate length with small size likein the small generators.

There is another kind of difference depends upon the type of loadingand operation condition. Some of rotors are exposed to high axial torqueand force such as power shafts but the other types does not effect by thisproblem such as supported shafts.

In order to prove the rightful performance for the mathematicalmodel a modified comparisons were carried out, also for illustrate theaccuracy of the program developed and to investigate the combined effectof cross-coupled s ~ i f f u s s and damping coefficients and whirl speed on theinstability of rotor.5.2 Cases Study 5.2.1 Critical and Whirl Speeds

In order to illustrate the accuracy of the program developed and toi_ ? vestigate the effect of bearing properties on whirl speeds and instabilitythreshold, case (1) and case (2) is used.

As can be seen from table (5.1), a close agreement between thevalues listed in the table and the corresponding values [24] demonstratesthe accuracy of the program in calculating the complex eigenvalues.Furthermore, using 8-element give a good results as shown in the table.

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Also, the theoretical wliirl speed for case 1) is 568.1 rad/sec, whichis calculated from equations gives in [2 7] and this another .comparison toshow the accuracy o the program.

When the cross stiffness coefficients take into account we canconclude from table 5.2) these coefficients are increase the whirl speeds.Also the direct damping increased the whirl speeds, as a result thecombmed effect o direct damping and cross-coupled stiffness coefficientare increase the whirl speeds.

Fig. 5.1 and Fig. 5.2 explain the first and second mode whirl speedsand the effect o cross-coupled coefficients on it. Case 2) is used and thefollowing two cases are taken for analysis:

A-I) Internal viscous damping = 0.0002 sec. with no bearing damping.A-II) Internal viscous damping =0.0002 sec. with bearing damping.

Fig. 5.1 illustrates the stability for case A-I) . It is show theinstability o the first mode occurred at speed o 520.6 rad/sec and thesecond occurred at speed o 1039.1 rad/sec. When the cross stiffnesscoefficients taken into account the first and second mode occurred atspeeds of531.686 and 1504.9 rad/sec, respectively.

Fig. 5.2 illustrates the stability for case A-II). t shows that the directdamping increase the whirl speed. From Fig. 5.1 and Fig. 5.2 we canconclude that the cross-coupled coefficients improve the critical andthreshold speed for rotor bearing system and the table (5 .3) show thethreshold speed compared with references [5] and [33].

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Table 5.1) Whirl frequencies of the simply supported beam flexiblebearings) for case 1)

Whirl frequencies rad/s)c =0.0002, c =0.0

Mode 2-elementFirst F) 535.0First B) 535.0Second F) 1607.5Second B) 1607.5Third F) 2250.3

F: Forward modeB: Backward mode

5-element 8-element500.0 512.9500.0 512.91033.5 1050.01033.5 1050.02772.0 2291.7

Ref. [24]520.1521.81096.01095.32222.7

Table 5.2) Whirl frequencies and effect of cross-coupled coefficientsMode Cross stiffuess Direct dampingFirst F) 628.2 281.3First B) 628.2 281.3Third F) 1219.4 1601.2Third B) 1219.4 1601.2

Table 5.3) Threshold speed comparisonThreshold speed rad/sec) The source

942.47 Rouch and Kao [ ]1045.00 Mushtaq 33]1279.8 Present Work case A-II)

37

\\


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The critical speed is calculated by meet the synchronous line infigures 5.1 and 5.2. From Fig. 5.2 the critical speed without crosscoefficient is 531.2 rad/sec while it is 564.9 rad/sec with cross coefficient.The last value is true, but there is another critical speed which is . found531.686 rad/sec appear because the cross stiffness is negative , which giveinstability in super synchronous condition. This result is similar to Rao sstudy [16].5 2 2 Unbalance Response

In reality, we cannot define critical speeds, as we do for a rigidbearing rotors, because the bearing coefficients are function o the speed orotors. Consequently, it is always better to study the out o balanceresponse to locate the critical speed and to study the effect o bearingproperties on unbalance response.

Case (3) was considered for this purpose. Fig. 5.3 shows this effects.From this figure, it is clear that the critical speed in damped case shiftstowards the right. In addition, the cross-coupled peak response is higherthan direct damping peak response. Similar results were found in 17]and [27].

This case have a negative cross-coupled stiffness coefficient and fora positive cross-coupled, Case (4) was considered, Fig. 5.4 shows the directbearing damping reduces the unbalance response in the region o thecritical speeds and the peaks are shifted to the right from these criticalspeed. From the two damped response curves it is observed that the crosscoupled damping increase the response. This is because the predominantcross-coupled damping is negative. Fig. 5.5, 5.6, 5.7 and 5.8 shows aperiodic response with time and effect o cross-coupled on it whichincrease as shown Fig. 5.5 and Fig. 5.6 show the response in criticalspeed where Fig. 5.7 and Fig. 5.8 show the response at speeds (3000 rpmand 2000 rpm).

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5 2 3 Forced Vibration ResponseFig. 5.9 shows the response of forced rotor n bearing position with

appsense of cross-coupled bearing coefficients. From the above figure weconclude that the rotor is stable [33]. Fig. 5.10 and 5.11 illustrates theresponse of the same rotor n disk and right bearing position. In fig. 5.12illustrates the response of rotor when effect of cross-coupled is take intoaccount. Its show that the rotor is unstable and this result also clear in Fig.5.13 and 14. This result is true after we notice that the force frequency955 rpm) which is closed to the rotating speed 1000 rpm) also after the

calculation of the cross-coupled coefficients [27] we notice that the crosscoupled stiffness is negative, this gives another instability in this case asshown in reference [ 6] which demonstrate that instability of a rotor canoccur under super synchronous whirl conditions.

Another forced case we take case 6). Fig. 5.15, 5.16 and 5 1 7illustrated the response of the rotor in bearing position without crosscoupled coefficients. Fig. 5.18, 5.19, 5.20 illustrated the response withcross-coupled coefficients for the same case. From these figures it is clearthat the rotor is stable in the two states. The difference between them is theresponse in first state without cross-coupled) is less than the second statewith cross-coupled). Also there are peaks in the second state which

di.sappear in first state. This gives another conclusion, that, when the crosscoupled damping is positive as in this case, the response is reduced asshown in the above figures.

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5 2 4 Transient ResponseCase 1) is taken to study the transient response of the rotor bearing

system. Fig. 5.27 gives a response of the rotor with no cross-coupled whilefig. 5.22 gives a response with cross-coupled. Fig. 5.27 is similar to theresult of Ref. [32]. By observation of the two figures it can concluded thatthe cross-coupled coefficients is made the rotor instable or on the otherhand it made a self excited which arising the response to instability. For theseek of comparison, the result of fig. 5.23 have been compared withRef. [32] similar result which shows that the rotor also become instable,fig. 5.24. fig. 5.25 and fig. 5.26 show the response of the rotor when isloaded with axial torque.

Fig. 5.29 shows the response without shear deformation and withcross-coupled coefficients. It can be observed that the cross-coupled isreduce the time to reach the instability condition. In addition bycomparison of fig. 5.22 and fig. 5.29 we notice that the shear deformationhas no effect on the response, this is similar to the result which given byreferences [27] and [33]. Figures 5.30, 5.32 and 5.3 , 5.33 were comparedto show the effect of tysteretic damping on the response of the rotor.Which has no effect on the response in the two cases without and withcross coefficients).After we calculate the whirl speed for this case, which is 4973 .3 rpm ,while the rotating speed is 4000 rpm). Thus the speed 4973 .3 rpm iscritical speed and the rotor is unstable because it runs near to the criticalspeed as we show in the transient response.

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5.3 ummaryIt is clear that the analyzed case studies in this work have a proved

the accuracy of the results.Section 5.2.1) shows the accuracy ofth program by a test which

uses case 1) .for this purpose. The effect of the c ~ o s s c o u p l e d coefficientswhich increase the whirl speed as found in table 5.2). From Fig. 5.1 andFig. 5.2 it can be concluded that the cross coefficients also increase thecritical speed, which increase directly the threshold speed.

In the unbalance response, Fig. 5.3 and 5.4, we can notice that thepositive cross-coupled damping reduce the response and the negative crosscoupled damping increase the response and this occur at the region of thecritical speed. Also, the positive cross stiffness increase the response whilethe negative cross stiffness reduces the response. Figures 5.5 and 5.6 areshow the response of the rotor with time at the critical speed of 900 rpm)and they shows the effect of the cross-coupled wh ich reduce the response.while figures 5.7 and 5.8 are show the response of the rotor at differentspeed.

Section 5 .2 .3) shows the forced response of the rotor using two casestudies. We can concluded that the bearing coefficients made the rotorunstable as shown in Figs. 5.12 , 5.13 and 5.14 . This for the rst case, butfor the second case we concluded that the stability of a rotor can beimproved by the use of anistropy of bearings. This is clear in Figs. 5 18 ,5.19 and 5.20 .

Section 5 .2.4) shows the transient response and the effect of thecross-coupled coefficients, which made the rotor unstable as shown inFigs. 5.22, 5.23 and 5.24 .

41


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55/90

.: 1.5E-003....J~~ 1.2E-003

~r q~ 9 . 0 E 0 0 4'I~ 6.0E-004;

II III IIt I

/1 Ifo

I \II ' 'iiI *' ~ - - '=-.._.

3.0E-004 / I/ - - - _..390 490 590 690 790 890 990 1090 1190 1290SPEED(RPM)

FIC.5.3 UNBALANCE RESPONSE OROTOR IN BEARING POSITION

4 3 ~ .K+Kc

7

K4-0irect stiffnesKc-Coupled stiffnesC4-0irect dampin9C,-Coupled dampmg

I '

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ r r r r r r r r ~1000 2000 3000 . 4000 5000 6000 7000 8000SPEED RPM)FIC.5.4 UNBALANCE RESPONSE O ROTORIN BEARING POSITION

43


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1

2 . 2 E 0 0 7 . ~

1.5E-007

3E-007

NO CROSS-COUPLED3I3BEIEl WITH CROSS-COUPLED

2 . OE- 007 , . . , . .T"T" l ' T"TT, . , . . , r rr"T"TT"TT"m""T"Tm-rr-TTT"T"T""rTTT"T" l -rrm. . , . . j0.00 0.06 0 12 0 18 0.24 0.30 0.36 0.42 0.48TIME Sec.)

FIG.5 5 UNBALANCE RESPONSE O ROTORIN X-DIRECTION W = 9 ~ 0 RPM)

4 BE 008

3 4E-008

Z.OE-008

60E-009~~ -B.OE-009

2.2 E-008

3 6E-008

0 E - 0 0 8 ~ r n . . , . . , m ~ . . , . . , m ~ r r r r n . . , . . , , . . , . . . , . . . , . . , m ~ - . - . , . . , . . . , . . . , . . , n ;000 0.06 0 12 0 18 0.24 0.30 0.36 0 42 048TIME Sec.)

FIC5.6 UNBALANCE RESPONSE O ROTORIN Y-DIRECTJON {W=900 RPM)


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s.oB-oosr ------------------.- - WITHOUT CROSS COUPLEDI - WITH CROSS COUPI.0 rz.oB oo5

t.OB 005

4.2E-02f

.f,OE-005

.JlOE-005

I II I I II I I I II I

III I

1

IJ I I

I II I\J

I II I

\I-3.0E-005 -h- 'T T ..,..,,...,. 'T T T 'f'rrr'T'T rr-rT T 1r-rr-rrr-rr.,..,...,n-rTTl-r-r..-n-r-rr-r-l

0.00 0.06 0.12 0.18 0.24 0.30 0.36 0.42 0.48TIME Sec.)

FIC.5. 7 EFFECT OF CROSS-COUPLED O UNBALANCE.RESPONSE OF ROTOR W=3000 RPM)

3.0E-005 . :

1.8E-005

- - WrTHOUT CROSS COUPLED- WrTH CROSS COUPLED\

I II

rI II II I

r 6.0E-006li 6.0E-006 \ II I

- t .BE-005\ I\

II II II Iv

I II II I

-3.0E-005 33 o39 o 440.17 0.22 0.28 0.0.00 0.06 0. f f TIME{Sec.)OUPLED O UNBALANCEFIC.5.8 EFFECT OF CROSS-C W-ZOOO RPM)RESPONSE OF ROTOR -

45


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4 .0E -00 4 - - - -- - - -- - - -- - - -- - - -

M M2 .4E-004

)1 B.OE-005~ -B .OE-005

I.4E-004

-4. 0E 004 +-r.,....,....,..,.,...,.,...,rr-r.,-.,......-,.....-,..,...,..,....,..r-TO.,...,...,..-.,-,-,rro...,...,...,--,J0.00 0.05 0. 10 0.15 0.20 0 .25 0. 30 0.35 0.40TIME Sec.)

FIG .5 . 9 FORCED RESPONSE OF ROTOR INBEARING 1) WITHOUT CROSS-COUPLED

5.5E-004-.----------------------

3.3E-004 N vI t . tE 004

~~ t . t E 0 0 4

3 .3E 004 rv

5 .5E- 004 .:t-r.,....,..,.....,...,-,-T T 1.,...,..,..rr-r...,-,-.,....,..rrr...,-,-.,...,--rr-r-.-r T T rT l.,..,.,0 .00 0 .05 0.10 0 . 15 0. 2 0 0. 2 5 0.30 0. 3 5 0. 4 0TIME Se c )

FIC.5.10 FORCED RESPONSE OF ROTOR OF.DISK WITHOUT CROSS-COUPLED46


59/90

1-e ......

'..:

1

9.0E-005

4.5E-005 ~ 1\ ~N

N.7E-020

N4.5E-005

9, 0 0 0 5 -h-...-,..-,-,-,-,...,--,T T '1rr-T..,--,T T '1n-r T T rr-r..,--,T T '1-rT T' T' m---.---rl0.00 0.05 0.10 0. 15 0.20 0.25 0.30 0.35 0.40TIME(Sec.)

FIG.S. 11 FORCED RESPONSE O ROTOR INBEARING 2) WITHOUT CROSS-COUPLED

6 .3E+OOO . . . r - - - - - - - - - - - - - - - - - - - ,

4.3E+000

2.3E+000e ......a:::".'..:

.7E+OOOv

37E 00 oo ~ o T ' T T ' T T T T T o ,..,o T o6rrn TTTrro ,..,o,..t2rrrr TTT rrono T'TB TTTT'TT'1o T ro2r4 TTT rrno rrlo3oTIME{Sec.)

FIC.5 . 12 FORCED RESPONSE O BEARING )WITH CROSS-COUPLED7


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5.60E OOO ~ ~

8.60E OOO

1 1.60E OOO... -4.00E-001

-2.40E OOO

-4.40E OOO

-6. 40E 000 f n r - r n f T T T T T f T T T T T f T T ' T T T T T T 1 T f ' r T T T T T T T T T T T T n - n - r T T T T T T T T ~0.00 0.00 0.01 0.01 0.01 0.02 0.02 0.02 0.02 0.03 0.03TIME Sec.)

FIC.5.13 FORCED RESPONSE O DISKWITH CROSS-COPULED

3.5E 001 . . . . . . - - - - - - - - - - - - - - - - - - - - .

2.6E 00f

1 1.7E 00f;:: 8.3E OOO

~-5.0E-00f ] ....

-9.3E OOO

vf. BE 00 f +nrr rrTTTorr r rrn -nTTTTTT rTTlTTTTTTrTTl- r rTTTTTTTrrnTTTTTTrr i0.00 0.00 O.Ot 0.01 0.01 0.02 0.02 0.02 0.02 0.03 0.03TIME{Sec.)

FIG.5.14 FORCED RESPONSE O BEARJNG 2)WITH CROSS-COUPLED8


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2.4E-005 r----------=--:--:-----___

t .OE-005

1 4 .0 E 0 0 6_ YDIRECTION::: t .BE-005Q:

-3 .2E-005

-4.6E-005 Z .:.DIRECTION

- 6 0 E - 005 -t-r....,.,....,,..-OT '.,'T T''T T-,-,.-,-,....,,..-r-r-T T T T r-r-r-r-T T ',...,IOoOOO 0.025 Oo050 Oo075 0. 100 Oo125 0.150 0.175TIME Seco

FIC.5. 15 RESPONSE OF ROTOR OF GRINDINGMACHINE WITHOUT CROSS-COUPLED W=400 RPM)

7.0E-006

-5 .0-006 Y-DIRECTION

t .?E-005_::: -2o9E-005i:i:::s 4o1E-005

-5 .3-005Z IREC ON

- 6 5E - 005 .j..,......,..,....T'T',...,.-I'T'l,...,..... T T-rr'T T',...,.-rr-r...,:,..,....,..,-rT-rr..,.-, o Oo02 Oo04 Oo06 OoO 0010 Oof2 0. 14 0.16TIME Seco)

FIC.5o16 RESPONSE OF ROTOR OF GRINDIN GMACHINE WITHOUT CROSS-COUPLED W=600 RPM9


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1.4E-005 y ~

t .BE-020Y-DIRECTION

? - t . 4E-005

t; -2 .8E-005' 'l~--.:

-4 .2E-005

-5 .6E-005Z-DIRECTION

7.0E - 0 0 5 - j , . . , , . , . , . , - r r - r . , . , . - n - r T T T r T T T T T T n T T T T r r T 1 C T T T T T T T T T ' r T T T T T T T T T r r n ~0.000 0.020 0.040 0.060 0.080 0.100 0.120TIME Sec.)

FIG.5. 17 RESPONSE OF ROTOR OF GRINDINGMACHINE WITHOUT CROSS-COUPLED W= 1600 RPM)

3.0E-006 ...----------------------

2.1E-021

? -3.0E-006Pc:t; -6 .0E-006

.:-9.0E-006

- t .ZE-005

1.SE- 005 r m - r n r - r r r r r T T J - r n r - r , - m T T l . . , r r r m T T T T T l r r r ~ , - r ~0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10TIME Sec .

FIG 5 18 RESPONSE OF ROTOR O GRINDINGMACHINE WITH CROSS-COUPLED W=400 RPM)50


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2.0E-006 ] ( - - - - - - - - - - - - - - - - , - - - - - - ~

1.2 E 0 0 6

-- 4.0E-007.;...~t: 4 .0E-007.:

1 .2E 006

-.2 .0E-006

2 .BE- 0 0 6 t - r T T T T T . . . , . . , . . , . r - r - r r - T T ' . . . , . . , . . r r o . . . , . . , . . , ~ - - , J0.00 0. 03 0.06 0.09 0. 12 0. 15 0. 18 0.21 0. 2 4 0 27 0.30 0.33TIME Sec .

FIG.5.21 DYNAMIC RESPONSE OF ROTOR INBEARING WITHOUT CROSS-COUPLED COEFFICIENT

1.2E O O O .., . .. - - - - - - -- - - - - - - - - - - - ,W=4 DO RPW = 4 40 0 RP_......W=4 600 RPB.OE-001

1 4 .0E-001_~ Z-DIRECTIONt: 1 1 016 .:a __ .. ... .t f : ~ b - - - + - - t ~ : : : : : : : : : - - ~'le:.:

Y-DIRECTION

-4 .0E-001

-B.OE-001

- 1 Z 0 0 0 .:t-r-rrTT rr'l-rT T TTT'T'T' ,..,.,n-r-r'r-rrTT rT'lrr::r T ':':r7 : :J0.000 0.008 0 .005 0 .007 0 .010 0 .01 2 0. 015 0.01 7TIME Sec .

FIG.5.22 DYNAMIC RESPONSE OF ROTORWITH CROSS-COUPLED52


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3 . 2 E O O O r - - - - - - - - - - - - - - - - - - ~ - - - .

2.4E OOO

...... .W=4200 RPM- W=4400 RPMG eEl W=4600 RPM

1~ 1.6E+OOO~~:s~

B.OE-001

Z-DIRECTIONY-DIRECTION

-B.OE-001

1 6E 0 00 -h...,. T T...,....,....,....--r-,-,-..,--r-.--r--,-,--,-,,.,..-,....,....,....,....,....,....r-r-r-r-..-l0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18

TIME Sec.)

FIG.5.23 DYNAMIC RESPONSE OF ROTOR INSIDE BEARLNWITHOUT CROSS-:-COUPLED AND WITH INITIAL FORCE

1.2E OOO

. 9.0E-001~ W 4 2 0 0 RPM. . .. . W=4400 RPMGBB8El W=4600 RPM

?,.:. 6.0E-0015 1 S.OE-001t3 Z DIRECTION~ 2 BE- 0 1 7 ~ l l l l l t i E ~ : : : :Y DIRECTION

~-3.0E-001

-6.0E-001

- 9 . 0 E - 0 0 1 ~ - - r - T - - r - T T T T T T T T T T T T T T T T ' - r - l ~ ' T T , ; ' T T I ' ~0.00 0.00 0.00 0.01 0.01 0.01 0.01 0.01 0.02 0.02Tii.tE Sec.)FIG 5 24 DTNAMIC RESPONSE OF ROTOR INSIDE BEARIN GWI FH. CROSS COUPLED COEFFICIENT AND IN ITIAL FORCE

53


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t .tE OOO . , - - ~ 1 ;::: ::: :::: W:;:=:-: ;4 2 '00:;- ; : : ;Rr;;PM..--------:-------- ,

7.3E-00t

3 .8E-00t

.2E-001

6 .7E-001

I BE 9El W= 400 RPMW- 4-600 RPM

Z-DIRECTION

Y-DfRECTION

- 1 E -j-r-,-,--r-rr-T T .,..,.-.-r-rT..,.--r...,....,-,-,,-,-T T'.-......,..,..,.....,....,.,0. 00 0. 00 0 .01 0.01 0 .01 0.01 0.02 0.02TIME Sec.)

FIG.5.Z5 DYNAMIC RESPONSE OF ROTOR INSIDE BEARINGWITH CROSS COUPLED AND INITIAL TORQUE

5 .8E-009

4 .2E - 009

2 .6E-009

t .OE-009i6 .0E-010

-2.2E-009

-3.8E-009

-5.4E-009

- 7 . 0E 0 0 9 . + , . , ~ ~ T T T T T T T T T T I T I T T T I T T T T T T J T 1 m T T f T T T T r T T T T T T I l T T ' m T T f T T T T T T T I T T T 1 o r /0.00 0.02 0.04 0.06 0.08 0. 11 0. 18 0. 15 0. 17 0.19 0 .2 1 0. 2 3TIME Sec .)

FIG.5.Z6 RESPONS OF ROTOR WITH CROSS- COUPLEDAND INITIAL APPLIED TORQUE 20Kn.rn)54


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6 . 5 E o o e r_ _ _ _ _ _ _ _ _ _ _ _ : __ _ _ __

4.5E-008

2.5E- 008

[ 5.0E-009l ~ V V V V V vv v vvv v v v4

1 .5E-008l

-3 .5E-008

-5.5E-008

- 7 5E- 0 0 8 -f-r ... . .. . .. .. . ... . .. - .-.r . .. .. .TTT.. ...r-r-rT-rl.... r-rr-rr-.-.-.J0.00 0.05 0.11 0. 17 0.22 0.28 0.33 0.39 0.44 0.50TIME (Sec.)

FIC.5.27 DYNAMIC RESPONSE OF ROTGR INSIDEBEARING WITH NO BEARING DAMPING .1.4E -001 . . - - - -- - - - - - - - - :- - - - - - - - -

9.0E-002

4.0E-002 - - - - ~ ~ ~ \-1 .0E-002 V v6 .0E--002 v

-1 . 1E-001

-1 .6E-00f

-2. f E 0 0 f +-,.. T T''T T ....-r'T T .,...,-,.....-,...,...,..,.....-r-r...,......-rr,.....-.-,-,..,......,.,....,....,0.00 0.03 0 .06 0.08 O. ff 0.14 0.17 0.20 0.22 0.25TIME {Sec.)

FIC.5.28 RESPONSE OF ROTOR INSIDE BEARINGWITH CROSS-COUPLED COEFFICIENT55


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W = 4 20 0 RPM~ W = 4 4 RPMGI3I3ElEl W=4600 RPMB.OE-001

Z DIRECTION

Y DIRECTION

-B.OE-001

- .2E+0 0 0 -t-r"'r'T...,.,...'T"'T".,.-r-T"T"'T"T"'1r"'T"' -,--,....,.,....,-,-,.,..T"T"'1-,-,.-,--,.'T"'T".,-,-T"T"'1r-r10.000 0.004 0.008 0.012 0.016 0.020TIME Sec.FIG .5.29 DYNAMIC RESPONSE O ROTOR WITHNO SHEAR DEFORMATION AND CROSS-COUPLED

B E 0 0 6 . .

5E-006

f ZE - 006'-'r qt: f E 0 0.,.;j:

4 E 0 0

7 E 0 0

f E 0 0 ~ ~ ~ T r n T r n o n ~ n o r n o n ~ n T r n T n T r n T ~ ~0.00 0. tO 0.20 0.30 0.40 0.50TIME{Sec .)FIG.5.30 DYNAMIC RESPONSE O ROTOR WITHNO CROSS-COUPLED EI=O.O)

6


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B E - 0 0 6 : ~ - - - - - - - - ~ - - ~ - - - - - - - - - - - - - - - - - - ~

5E-006

'? 2E-006~t: t E 0 0-4

.:-4E-OO

-7E-OO

0.10 0.20 0.30 0.40 0.50TIME Sec.)

FIG.5 .3 t DYNAMIC RESPONSE OF ROTOR WITHCROSS-COUPLED EI=O.O,EH=O .O

B E - 0 0 6 : - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - ~

5E-006

'? 2E-006~t: t E 0 0 -1

.: -4E-00

-7E-00

1 0 0 I . ~ ~ ~ ~ T n T M ~ ~ ~ ~ ~ ~ ~ ~ ~ T M ~ ~~ ~ 1 ~ 2 ~ 3 ~ 4 ~ 5TIME Sec.)FIG.5.32 DYNAMIC RESPONSE OF ROTOR WITHNO CROSS-COUPLED EI=O .O,EH=O.O)

57


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1

t E - 0 0 5 ~ - - - - - - - - - - - - - - - - - -

6E-006

- 2E-00

-6E-00

- t E - 0 0 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~0.00 0.10 0.20 0.30 0.40 0.50TIME Sec.)FIG.5 .33 DYNAMIC RESPONSE O ROTOR WITHCROSS-COUPLED EI=O.O)

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Chapter SixCONCLUSION ND RECOMMEND TION

6 1 IntroductionThe present work gave an investiagtion about the practical problems

and mathematical insights for Bearing rotor which provide a developedmodel for this system listed in details in chapter three. Also different typeof cases were presented in chapter five to give a distinct view aboutprogramming package ability to solve Bearing rotor problems. This workproduced important results in system vibrative properties and the stabilityof the rotor in fluid film bearings. Also it shown that the mathematicalmodel programming procedure and numerical methods that appliedsuccessfully with this type of system.6 2 Conclusions

We can summarized the conclusions as follows:- . When the cross-coupled stiffness is negative the response is decrease

in the region of critical speed and it gives more instability to the rotorwhich may occur under super synchronous condition. Also the directdamping is shift the critical speed to the right in the region of criticalspeed on the other hand it is increase the critical speed and whencross-coupled damping is negative the response is increase.

2 When the cross-coupled stiffness is positive the response is increase inthe region of critical speed and the direct damping shifted the criticalspeed to the right and reduced the response of the rotor. The negativecross-coupled damping increase the response.

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3. The cross-coupled coefficients are very important factors becausethese effects on stability, which related on whirl speed and criticalspeed, and the response of the rotor.

4. This work proved that the stability of a tmbomachine can be improvedby the use of the anisotropy of bearings or bearing supports as incase 5).

5 e can neglected the shear deformation and the hysteretic damping inthe time response studies.

6.3 ecommendationFor further work we can summarised the following points:

I . Develop a method to calculate the fluid film bearings coefficients withdifferent speeds.

2. Develop another model for bearing element.3. e can use this work to study the fluid leakage which another

excitation.4. Enter another factor to the mathematical model like misalignment and

its effect on rotor stability.

6


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REFEREN E

[1] Lund, J W., Computer programs for unbalance response andstability , Part V AFAPL Technical report, TR65-45, Wrightpatterson Air Force base, Dayton ohio ( 1965).

[2] Zorzi, E. S and Nelson, H., Finite element simulation of rotorbearing system with internal damping . ASME gas turbine conf.Neworleans (1976).

[3] Ruhl, R and Booker, J F., A Finite element model for distributedparameter turborotor systems , ASME Journal of engineering forIndustry, vol. 94, 1972, p. 126.

[4] Lund, J. W., and Orcutt, F. k., Calculation and experiments on theunbalance of a flexible rotor , ASME J of engineering forindustry, Vol. 89, No. 4, 1967, P.785.

[5] Rouch, K. E. and Kao, J S., Dynamic reduction in rotor dynamics bythe finite element methods , ASME, J ofMechanical design,vol. 102, 1980 p.360.

[6] Newkirk, B. L., and Taylor, H. D., Oil film whirl-An investigation ofdisturbances on oil films in journal bearings , General electricreview, vol. 28, 1925, p. 559.

[7] Adams, M. L., and j. Padvan, Insights into linearized rotordynamics , J. of sound and vibration, vol. 76, No.1, 1981, p. l29-142.

[8] Morton, P G., Influence of coupled asymmetric bearings on themotion of a massive flexible rotor , Proc. Inst. Mech. Engrs . 182(13), 255 (1967-68).

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[9] Kirk R G. and Gunter, E. J. The effect of flexibility and dampingon the Synchronous response of a single mass flexible rotor , J.Eng. For industry, AS:ME 94, 221 (1972).

(10] Rao, J. S., Conditions for backward synchronous whirl of a flexiblerotor in hydrodynamic bearings , mechanism and machine theory17 (2), 143 (1982).

[11] Rao, J. S; Bhat, R. B. and Sankar, Effect of damping on thesynchronous whirl of a rotor in hydrodynamic bearings. Trans.CS:ME 6(3), 155 (1981).

(12] Rao, J. S., Out of balance response of turbo alternator rotors ,computer programs, Bharat heavy, Electricals limited,Hyderabad, India (1980).

(13] Rao, J. S. Rao, Instability of rotors in fluid film bearings , ASME, J.of vibration, Acoustics, Stress, and Reliability in design, vol.105, July, 1983.

(14] Rieger, N. F. and Cundiff, R. A., Discussion ofpaper by Morton, P.G., Influence of coupled asymmetric bearings on the motion ofa massive flexible rotor , Proc. Inst. Mech. Engineers. 182 (13),217 (1967-1968).

[15] Bhat, R. B., Unbalance response of a single mass rotor on fluid filmbearings using modal analysis , conf. Orlando, Florida ( 1982).

[16] Rao, J. S., Instability of rotors mounted in fluid film bearings with anegative cross-coupled coefficient , Mechanism and Machinetheory. Vol. 20, No.1, p. 181-187, 1985.

[17] Sharan, A. M and Rao, J. S., Unbalance response of rotor diskssupported by fluid film bearings with a negative cross-coupledstiffness using influence coefficient method , Mechanism andMachine theory. Vol. 20, No.5, p. 415-426, 1985 .

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[18] Wang, J. H., and, Shih, F. M., Improve the stability o rotor subjectedto fluid leakage by optimum diameters design , ASME, J. ovibration and Acoustics, Vol. 112, January 1990.

[19] Ruhl, R. L., Dynamics o distributed parameter rotor system:Transfer matrix and fmite Element Techniques , Ph. D.dissertation cornell university, 1970.

[20] Thorkildsen, T., Solution o a distributed mass and unbalance rotorsystem using a consistent mass matrix approach , MSEEngineering Report, Arizona state university, June 1972.

[21] Polk, S. R., Finite element formulation and solution o flexible rotorrigid disk systems for natural frequencies and critical speeds ,MSE Engineering Report, Arizona state university, May 1974.

[22] Nelson, H. D., Finite rotating shaft element using Timoshenko beamtheory , ASME, J. ofMechanical design, Vol. 102, 1980.

[23} Hashish, E., and Snakar, T. S. Finite element and modal analysis orotor-bearing systems under stochastic loading conditions ,ASME, J. o vibration, Acoustics, stress, and Reliability inDesign, Vol. 106 January, 1984.

[24] Nevzat, Ozguven, H., and Levent Ozkan, Z., Whirl speeds andunbalance response ofmultibearing rotors using finite elements.ASME, J. o vibration, Acoustics, stress, and Reliability inDesign, Vol. 106 January 1984.

[25] Clive L. Dym, and Irving H. shames, Solid mechanics: a variationalapproach , McGraw Hill, New York, P. 370-377, 1973 .

[26] Soni, A H., and Srinivasan, V., Seismic analysis o a gyroscopicmechanical system , ASME, J. o vibration, Acoustics, stress,and Reliability in Design, Vol. 105, October 1983.

[27] Rao, J. S., Rotor dynamics , Wiley Eastern limited, India, 1985 .

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[28] Bhat, R. B., and Sankar, T. S., "Dynamic behaviour of a simple rotorwith dissimilar hydro-dynamic bearing by modal analysis",ASME, J. of vibration, Acoustics, stress, and Reliability inDesign, Vol. 107, April1985.

[29] Dimarognas, A. D., "Interval analysis of vibrating systems", J. ofsound and vibration, vol. 183, No.4, P. 739-749, 1995.

[30] Ahid, D. N., David, I. G., John, P. H., "Vibration damping", JohnWiley and sons, New York, 1985.

[31] Lund, J. W., "Stability and damped critical speeds of a flexible rotorin fluid film bearings", J. ofEng. For industry, Vol., 96, No.2,1974.'

[32] Rao, S. S., The finite element method in engineering", Pergamonpress., U.S. A., 1982.

[33] Mushtaq, K. A., A programming package for vibration analysis inrotating ports system", M. Sc. Thesis, university of Basrah,October 1999.

[34] Gupta, K. K., "Development of a unified numerical procedure for freevibration analysis of structural", International journal fornumerical method in engineering, vol. 17, No. 2, 1981.

[35] Chandrosekaran, A. R., Pall, D. K., and Agarwal, B L. "Complexenginproblem for flutter analysis of structures", J of structures,vol. 43, No.4, 1992.

[36] Osami Matsushita, et a ., "Solution method for eigenvalue problem ofrotor bearing system", Bulletin ofthe JSME, Vol. 23, No. 185,1980.

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S stem Matrices Com onentsThe general form of differential equation of rotating system in matrices form is;

~ M T +(M.J)q + (E,(k, + k,)-ro(GJ)q +[ R- k, + k,]-(AJ-(T)+( , ro+ &1 1 J(N,](k, +k,]lq =(OJl+ n l+&n

ich thesematrices came form system elements;1- Rotor element.2- Disk element.3- Bearing element.1-Rotor element matricesi- Mass matrix componentshe matrices [MT],[M,] is symmetricomponents of [MT]MT(1,1)= KMt(156+294+ 140< >2)MT(2,1)= 0.0MT(3,1)= 0.0 .

MT(4,1)= KMt(22L+38.5L


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Mr 5,5)= KM1156+294+140< >2)Mr 6,5)= 0.0Mr 7,5)= 0.0 .Mr 8,5)= -KM122L+38.5L+ 17 .5L2)Mr 6, 6)= KMt 156+294+ 1402)Mr 7,6)= KMt 22L+38.5L+ 17.5L2)Mr 8,6)= 0.0Mr 7,7)= KM14L2+7L2+3.5L2j 2)Mr 8, 7)= 0.0Mr 8,8)= KMt 4L2+7L2 3 5L2j 2)Components o [M,]Mr 1,1)= 36KMrM, 2,1)= 0.0M, 3,1)= 0.0M, 4,1)= KMr 3L-15L)M, 5,1)= -36KMrM, 6,1)= 0.0M, 7,1)= 0.0M, 8, 1 = KMr 3L-15L)M, 2,2)= 36KMrM, 3,2)= -KM. 3L-15L)M, 4,2)= 0.0M, 5,2)= 0.0M, 6,2)= -36KMrM, 7,2)= -KM, 3L-15L)M, 8,2)= 0.0M. 3,3)= KMr 4L 2+5L2+10L2j 2)M, 4,3)= 0.0M, 5,3)= 0.0M, 6,3 )= KMr 3 L-15L)M, 7,3)= KMr -L2-5L2+5L2 >2)M, 8,3)= 0.0M, 4,4)= M r 4 L 5 L c ~ l O LM, 5,4)= -KMr 3L-15L)M, 6,4)= 0.0M, 7,4)= 0.0M, 8,4)= KMr -L 2-5L2+5L2j 2)M, 5,5)= 36KMrM, 6,5)= 0.0M, 7,5)= SKMrM, 8,5)= -KMr 3L-15L)M, 6,6)= 36KMrM, 7,6)= KM, 3L-15L)M, 8,6)= 0.0 .

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Mr(7,7)= ~ r 4 L + 5 L < j > + I O LM,(8,7)= 0.0 - _M,(8,8)= r 4 L + 5 L 2+IOL2j 2)

IlL J.lr2Where M 1 , and Mr :..:..:420(1 < > 2 120L(l < > 2ii- stiffuess matrix components

The matrices [Kb],[K1],[A]and [T] is symmetricComponents o [Kb]Kb 1,1 )= 2 K ~ c . bKb(2,1)=0.0Kb(3,1)=0.0Kb 4,1 = 6 L K ~ c . bKb 5, 1 = - 1 2 K ~ c . bKb(6,1)=0.0Kb(7,1)=0.0Kb 8, 1 = 6 L K ~ c . bKb(2,2)= 2 K ~ c . bKb 3 ,2)=6LKI


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Kb 8,8)=4L2K cbEI EI qWhere Kkb , ~ c s = - - - )L3 1+2)A 8,7)=0.0A 8,8)=KA 4L2+5L2+2.5L2 >2)

Where KA p30L l +


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Components of [T]T l,l)=l2KTT(2,1)=0.0T(3,1)=0.0T 4,1)=6LKrT(5,1)=-12KTT(6,1)=0.0T(7,1)=0.0T _8, 1)=6LKTT(2,2)=12KTT(3,2)=-6LKT T(4,2)=0.0T(5,2)=0.0T(6,2)=-12KTT(7,2)=-6LKTT(8,2)=0.0T(3,3)=KT(4L2+2L2P+L2q,2)T(4,3)=0.0T(5,3)=0.0T(6,3)=6LKTT(7,3)=KT(2L2-2eq,-L2 2) .T(8,3)=0.0tc4,4)=KTC4L +2eq,+L 2>T(5,4)=-6LKTT(6,4)=0.0T(7,4)=0.0T(8,4)=KT(2L2-2L2P-L2 2)T(5,5)=12KTT(6,5)=0.0T(7,5)=0.0T(8,5)=-6LKTT(6,6)=12KTT(7,6)=6LKTT(8,6)=0.0T(7,7)=KT(4L2+2L2P+L2 2)T(8,7)=0.0T(8,S)=KT(4L 2+2L2P+L2 2)T .WhereKr 3L 1+)

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2 Disk element matricesi Mass matrixmd 0.0 0.0 0.0] [0.0d 0.0 md 0.0 0.0 [ M ~ ] = 0 .0[MT]= 0.0 0.0 0.0 0.0 0.00.0 0.0 0.0 0.0 0.0

ii Damping matrix0.0 0.0

d 0.0 0.0 0.0 0.0[G ]= 0.0 0.0 IPd.0[0.0

0.0 0.0 - IPd0.01

0.03 Bearing element matricesi Stiffuess matrix

[

Kyy KyzKzy Kzz

Kb] = 0.0 0.00.0 0.0

ii Damping matrix

[c Cyz[Cb]= Czy Czz0.0 0.00.0 0.0

0.0 0 010.0 0.00.0 0.00.0 0.0

0.00 010.0 0.00.0 0.0

0.0 0.0

A

0.0 0.00.0]0.0 0.0 0.0

md.0 0.00.0 0.0 md


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A.2 Unbalance Nodal Force VectorFor linear mass unbalance distribution on rotor element nodal force vector was;

2[Qc] = U:O

Z L ZRL- 7 20 ZLL)- 3/20 ZRL) + _L + --3 6Y L YRL7 /20 YLL) + 3/20 YRL) + -}-+ - 6- )L2 L2-1/20 YLL2)-1/30 YRL2 )+ - )

Z L2 Z L2-1/20 ZLL2)-1/30 ZRL2)+ ~ - )[Qsl= J 10l 2 Z L ZRL- 3/20 ZLL) -7/20 ZRL) + --{-+ - 3- )y L YRL3/20 YLL) + 7/20 YRL) + < I > + + - 3- )L2 L21/30 YLL2)+1/20 YRL2)+ - ~ + ~ ) .

Z L2 ZRL21/30 ZLL2)+1/20 ZRL2)+ - + ~

AS


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PPENDIXB

The cases, which are used in chapter five, are listed with their detailsnd figures in the present appendix.

Case 1) supported on fluid film bearing:

E = 2.068 10 11 N/m2 10.16 em j _ ~ o ~ oo op = 7833 kg/m3G = 7.9538 10 10 N/m2ro = 4000 rpm

Isotropic bearing with coefficient:Kyy= Kzz = 1.7513 107 N/mKyz = Kzy = 1.6 107 N/mC = zz = 1.7513 103 N.s/m

The internal damping of rotor materialInternal viscous damping = 0.0002 s).

This case used by Nevzat and Levant [24] .Case 2) three Supported on fluid film bearing.

27 em ll

E = 2.068 10 11 N/m2

p = 7680 kg/m3 10 o111m _ j _ e O ~ Do o 127 em lllsotropic bearing with coefficients: oefficients:

Kyy=Kzz = 1.75 107 N/mKyz = -1.7 107 N/mKzy= 1.7 107 N/mCyy = Czz = 1.75 103 N.s/m

Internal viscous d a m p i n g ~ ; , ) = 0.0002 s)Bl


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Case 4) supported on fluid film bearing:

E= 3.534 * 1012 N/m2G = 13.598 * 1012 N/m 2 54cm To

o = 4500 rpm Isotropic bearing with coefficient

. 7Kyy = 1.01 * 10 N/m Cyy = 1930 N.s/mKzz = 4.16 * 101 N/mKyz = 4.16 * 105 N/mKzy = 3.12* 107 N/m

This case used by Rao [27].Case 5)

Czz = 70000 N.s/mCyz = 4100 N.s/mzy = -4000 N.s/m

1000 m

- supported on fluid film bearingE = 20 * 1010 N/m2 F = 10 sin 100 t kNG = 80 * 1 10 N/m2=7800kg/m 3c o = 1000 rpmDisk element:MD= 14.5 kgIp = 7.31 kg.m210 = 5.25 kg.m2- Istropic bearing with coefficients:Kyy = Kzz = 4.378 * 1 7 N/mKyz = -2189 * 104 N/mKzy = 81 * 106 N/mCyy = Czz = 1.752 * 1 3 N.s/mCyz = Czy = 1.752 * 1 3 N.s/mThis case used by Mushtaq [33] .

0.6

83

00I


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Case (6)- Supported on fluid film bearings

E = 20 * 1010 N/m2 .G = 80 * I 09 N/m2

= 7800kg/m3Disk Element:MD=2kgp = 0.01388 kg.m2In= 0.02122 kg.m

2

0.1

- Orthotropic bearing with coefficients:

ro = 2400 rpmBearing (1)

Kyy 15374 * 103 N/mK 2 21256 * IOTN/mKyz 33300 * I TN/mKzy -14431 * IOJ N/mCyy 238 * 103 N.s/mc72 I97 * 103 N.s/mCyz 124 * 103 N.s/mCzy 94 * toj N.s/m

B4

I) 2)[ = = J c JWheelc:::::;::::] c J ElementI0.215 0.075 i : 0.09T I =,

0.02

Bearing (2)7028 * I Oj N/mI5573 * IOj N/m

23272 * I Oj N/m-I7092 * IO' N/m

55 I03 N.s/mI95 IO N.s/m76 * IO:r N.s/m57* I03 N.s/m


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ro = 6000 rpmBearing (1) Bearing 2)

K 10396 10j N/m 6043 * 1o N/mK Z 19555 * to N/m t2901 * I0 N/mKyz 47085 10 3 N/m 4t557 to N/mKzy -39765 10j N/m -4t262 * 1Q N/mCyy 144 10j N.s/m t24 103 N.s/m

143 tOj N.s/m 144 JO N.s/mCyz 42 * 10j N.s/m 25 * JO N.s /mCzy 32 * 10j N.s/m t8 JO N.s/m

ro =9600 rpmBearing (1) Bearing 2 )

Kyy tOOOO * tOj N/m 6185 JO N/mK Z 19776 tOj N/m 12592 * 10 N/mKyz 677t5 10j N/m 63582 * 1o N/mKzy -63172 * 1Oj N/m -63172 I0 N/mCyy 132 * 10j N.s/m 122 10 N. s/m

133 10 j N.s/m 135 JO N.s/mCyz 26 t 3 N.s/m 15 10 N.s/mCzy 20 * tO j N.s/m 11 JO N.s/m

This case used by Hashish and Sankar [23].

B5


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Azzam Daood

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