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Vibrations and stability of constrained rotating strings and disks at supercritical speeds Yang, Longxiang 1995

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VIBRATIONS AND STABILITY OF CONSTRAINED ROTATING STRINGS ANDDISKS AT SUPERCRITICAL SPEEDSByLongxiang YangB.A.Sc, Beijing Institute of Chemical TechnologyM.A.Sc. University of British ColumbiaA THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE FACULTY OF GRADUATE STUDIESDEPARTMENT OF MECHANICAL ENGINEERINGWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIASeptember 1995© Longxiang Yang, 1995In presenting this thesis in partial fulfilment of the requirements for an advanced degree atthe University of British Columbia, I agree that the Library shall make it freely available forreference and study. I further agree that permission for extensive copying of this thesis forscholarly purposes may be granted by the head of my department orby his or her representatives.It is understood that copying or publication of this thesis for financial gain shall not be allowedwithout my written permission.Department of Mechanical EngineeringThe University of British Columbia2324 Main MallVancouver, CanadaV6T 1Z4Date:Ct-. i/ IffAbstractThis thesis presents an analysis of the vibration and stability characteristics of rotatingstrings and disks at supercritical speeds.The dynamic interactions between an idealized rotating circular string and a stationaryconstraint consisting of a spring, a damper, a mass or a frictional restraint are studied. Themethod of traveling waves is applied to develop the characteristic equation. The physics ofthe interactions between the string and the restraints are discussed in depth. The nonlinearvibrations of an elastically-constrained rotating string are investigated. The nonlinearities ofthe string deformation and the spring stiffness are considered. Butenin’s method is adoptedto develop a closed-form analytical solution for single-mode oscillations of the system. Theanalysis shows that the geometric nonlinearity restrains the flutter instability of the string atsupercritical speeds.The effects of rigid-body motions on free oscillations of an elastically-constrained rotatingdisk are studied. The coupling between the translational rigid-body motion and the flexiblebody deformation is shown to reduce the divergence instability of the disk, but the tilting rigid-body motion does not change the stability characteristics. An analysis of nonlinear vibrationsof an elastically-constrained rotating flexible disk is developed. The equations of motion arepresented by using von Karman thin plate theory. The stress function is analytically solvedby assuming a multi-mode transverse displacement field. The study shows that the geometricnonlinearity generates hardening effects on the dynamics of a rotating disk, and the unboundedmotions at divergence and flutter speeds predicted by the existing linear analyses do not takeplace because of the large-amplitude vibrations.11Table of ContentsAbstract iiTable of Contents iiiList of Tables viList of Figures viiNomenclature xiiAcknowledgements Xvii1 Introduction 11.1 Background 11.2 Objectives and Scope 42 Interactions between a Rotating String and Stationary Constraints 62.1 Introduction 62.2 Equations of Motions 72.3 Solution of Equations of Motion 92.3.1 Traveling Wave Method 92.3.2 Solution at Critical Speed 122.3.3 Approximate Solution 152.4 Results and Discussion 162.4.1 Effects of a Spring Restraint 161112.4.2 Effects of an Inertial Restraint 202.4.3 Effects of a Damping Restraint 222.4.4 Effects of a Friction Restraint 242.5 Summary 273 Nonlinear Vibrations of an Elastically-Constrained Rotating String 293.1 Introduction 293.2 Equations of Motion 303.3 Solution 353.3.1 Single-Mode Solution 353.3.2 Discussion of Single-Mode Approximation 393.3.3 Multi-Mode Simulation 473.4 Summary 504 Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 524.1 Introduction 524.2 Analysis of a Rigid Rotating Disk 534.2.1 Equations of Motion 534.2.2 CASE I: K =0 554.2.3 CASE H: Effects of Spring Constraints 564.3 Effects of Rigid-Body Motions on a Flexible Disk 584.3.1 Modeling of Rigid-Body Motions in Conventional Analyses 584.3.2 Effects of Rigid-Body Motions 624.3.3 Experiment 664.4 Summary 67iv6 ConclusionsBibliographyAppendicesA Equation of Motion of a String Including Speed Dependent TensionB Effects of Constraints on EigenvaluesC Functions in Butenin SolutionD Formulation of Rigid-Body MotionsE Parameters of the DisksF Selection of Rmn(r)6969707887901001021051101101111131161191205 Nonlinear Vibrations of an Elastically-Constrained Rotating Disk5.1 Introduction5.2 Equations of Motion5.3 Stress Function Solution5.4 Time Governing Equations5.5 Numerical Results5.6 SummaryVList of Tables2.1 Eigenvalues calculated from analytical and numerical solutions 175.2 Comparison of linear frequencies, spring atr = 1,0 = 0, M = 0, N = 3. . . 90B.3 Eigenvalues calculated from equation (2.16) (analytical) and equations (B. 147)—(B.150) (approximate), = 0.1, n = 1, negative wave (w = n(1 — Q*)) . . 112viList of Figures1.1 Schematic of a splined guided saw 11.2 Eigenvalue-speed diagram of an elastically-constrained rotating disk 22.3 A rotating circular string subject to constraints of a spring, a damper, a massand a friction 82.4 Effects of a spring K = 0.5 on the eigenvalue spectrum, approximate solutionwith N = 18; (a) frequency, (b) real part, —: constrained, - -: free 182.5 Backward and forward traveling wave contribution to modes 2N (- -) and iF(—) around crossing point speed SV = 1/3 192.6 Effects of a mass M* = 0.2 on the eigenvalue spectrum, approximate solutionwith N = 18; (a) frequency, (b) real part,—: constrained, - -: free 212.7 Effects of a damper C* = 0.1 on the eigenvalue spectrum, approximate solutionwith N = 18; (a) frequency, (b) real part, —: constrained, - -: free, a: single-mode 222.8 Effects of a friction = 0.1 on the eigenvalue spectrum, approximate solutionwith N = 18; (a) frequency, (b) real part, —: constrained, - -: free 252.9 Effects of a friction on the stability of traveling waves; (a) forward wave, (b)backward wave, (c) reflected wave 273.10 Displacements of an element rd6 of a string in a planar deformation 313.11 Time history of C, e = 0.3, V = 0.2, K* = 0.0, e2 = 0.0; (a) numerical, (b)Butenin 40vii3.12 Time history of C1, e = 0.3, V = 0.9, K = 0.0, 62 = 0.0; (a) numerical, (b)Butenin 423.13 Internal resonance (time history of C1) by Butenin method, 61 = 0.2, fz* = 0.5,K* =0.0,62 = 0.0 433.14 Effects of &i on frequency-speed diagram of mode 2, K* = 0.0,62 = 0.0. . . 443.15 Time history of C1, 62 = 0.1, = 0.2, K* = 1.0,61 = 0.0; (a) numerical, (b)Butenin 453.16 Internal resonance (time history of C1) by Butenin method, 62 = 0.5, $.V = 0.5,K=0.2,e= 0 463.17 Effects of 62 on frequency-speed diagram of mode 2, K* = 5.0,61 = 0.0. . 473.18 FFT spectrum of mode 1, = 0.2,e = 0.1, K* = 0.0,62 = 0.0 483.19 Linear response at flutter speed V = 1.5, K* = 1.0,61 = 0.0, 62 = 0.0 493.20 Nonlinear response at flutter speed V = 1.5, K* = 1.0,61 = 0.05,62 = 0.0. . . 494.21 Schematic of a rigid rotating disk subject to elastic constraints 534.22 Frequency-speed diagram of a rigid disk (Disk 1), K1 = K2 = K3 = 5000 N/rn,r1=r30.2rn,6 0°,62 120°, 63 = 240° 564.23 Frequency-speed diagram of a rigid disk (Disk 1), K = 5000 N/rn, K2 =3000 N/rn, K3 = 1000 N/rn, r1 = T2 = = 0.2 rn, 6 = 0°, 62 = 120°,63 =240° 574.24 Comparison of frequency-speed diagram between a rigid disk (Disk 1) and asplined saw (Disk 3) without constraints; x: rigid disk, —: splined saw. . . . 59viii4.25 Comparison of frequency-speed diagram between a rigid disk (Disk 1) and asplinedsaw (Disk 3); (a) one spring, K = 10000 N/rn, r = 0.2 m, 6 = 00; (b)two springs, K1 = K2 = 10000 N/rn, r1 = T2 = 0.2 m, 6 = 00, 62 = 120°; (c)three springs, K1 = K2 = K3 = 10000 N/rn, ri = = = 0.2 m, 8 = 00,62 = 120°, 63 = 240°; : rigid disk, —: splined saw 614.26 Eigenvalue-speed diagram of a flexible disk (Disk 2) subject to one spring,K = 210142 N/rn, r = 0.305 rn, 8 = 270° 624.27 Eigenvalue-speed diagram of a flexible disk (Disk 2) subject to two springs,K1 = K2 = 210142 N/rn, r1 = ‘2 = 0.305 rn, 6 = 270°, 62 = 30° 634.28 Effects of removal of mode (0,0) on the eigenvalue-speed diagram of a flexibledisk (Disk 2) subject to one spring, K = 210142 N/rn, r = 0.305 rn, 6 = 270° 644.29 Effects of removal of mode (0,1) on the eigenvalue-speed diagram of a flexibledisk (Disk 2) subject to one spring, K = 210142 N/rn, r = 0.305 m, 8 = 270° 644.30 Real parts of the eigenvalues of a splined saw (Disk 3) with finite guides, guidedimension: 0.152 rn x 0.14 rn, stiffness: 4.23 x 108 N/rn/rn2,—: withfree-free edges, - -: with grooved restraint 654.31 Comparison of frequency-speed diagram between experiment and analysis ofa splined saw (Disk 3) without constraints; 0: experiment, —: analysis 675.32 Comparison of stress function of mode (0,2) 915.33 Nonlinear inpiane radial stresses at different cross sections, six-mode approximation, Coo = Coi = C02 = So1 =... = So5 = 1.0, C03 = C = Co5 = 1.0. . . 925.34 Nonlinear inpiane shear stresses at different cross sections, six-mode approximation, Coo = Co1 = C02 = So = S05 = 1.0, Co3 = Co = Co5 = —1.0. . . 925.35 Nonlinear inplane tangential stresses at different cross sections, six-mode approximation, Coo = Coi = C02 = So =... = So5 = 1.0, C03 = C = Co5 = —1.0. 93lx5.36 Effects of nonlinearity on frequencies at Q4’ = 0, no spring, single-mode approximation, C01., = S0,d0 = Son = 0 945.37 Effects of nonlinearity on frequencies of forward wave of mode (0,3), no spring,single-mode approximation, C03 = So3,d03 = 503 = 0 945.38 Effects of nonlinearity on frequencies ofbackward and reflected waves of mode(0,3), no spring, single-mode approximation, Co3 = 503, d03 = 503 = 0 955.39 Ratio of nonlinear and linear frequencies of mode (0,3) observed in the rotatingdisk, no spring, single-mode approximation, C03 = So3,d03 = 503 = 0 965.40 Frequencies of mode (0,3) calculated from single-mode and multi-mode (sixmodes) approximation, = 6, no spring, for single-mode: Co3 = So3 = 5.0,for multi-mode: Coo ... = So5 = 5.0, all derivatives are zero 975.41 Linear response of Coo and Co2 at flutter speed = 6.5803, one spring,K = 1.0, r = 1.0, 6 = 0°, four-mode approximation, C00 = = So = 1.0,all derivatives are zero 985.42 Nonlinear response of Coo and Co2 at flutter speed V = 6.5803, one spring,K = 1.0, r = 1.0, 9 = 0°, four-mode approximation, C00 = ... = So3 = 1.0,all derivatives are zero 985.43 Linear response of Co2 at divergence speed = 5.34, four springs, K = K =K = K = 2.0, rt = = 1.0, r = = 0.6, 6 = 62 = 0°, 93 = 94 = 30°,single-mode approximation, C02 = S02 = 1.0, C02 = So2 = 1.0 995.44 Nonlinear response of Co2 at divergence speed l* = 5.34, four springs, K =K;=K;=K:=2.o,r=r=1.0,r=r;=0.6,6i=80°,9930°single-mode approximation, C02 = S02 = 1.0, C02 = S02 = 1.0 100D.45 Sequence of rotations 117D.46 Coordinate rotations in each plane; (a) about X axis, (b) about Y’ axis 118xF.47 Second derivative of Rmn, R/R0 = 0.4, mode (0,3) 121F.48 Axisymmetric parts of radial and tangential stresses, R/R0 = 0.4, single-modeapproximation, mode (0,3); (a) radial stress, (b) tangential stress 121xiNomenclaturea, a2 = amplitude coefficients in Butenin solution (3.62)amnpqtj = coefficients defined by (5.136)A = cross sectional area of the stringA1, A2 = coefficients in equations (2.25), (2.28) and (2.30)Bnpq = coefficients defined by (5.123)b, f = backward and forward traveling waves (2.10)B1, B2 = constants in equation (3.68)c01,c02,. .. , = constants in homogeneous stress function (5.124)C = damping coefficient of the stringC, C2 = amplitude coefficients in equation (2.17)= complex conjugates of C1 and C2C,., Sn = time dependent generalized coordinates of the stringCmn, Smn = time dependent generalized coordinates of the diskD = Eh3/[12(1—j.)]D1—D6 = constants in equations (4.85)—(4.87)E = Young’s modulus= coefficients in radial modal functions (4.91) and (5.12 1)fo(6), go(G) = initial conditions defined in (2.8) and (2.9)= nonlinear functions in Butenin equation (C. 151)= average values of F and (C.156)C = E/[2(1+v)Jh = disk thicknessxliI = moment of inertia of the rigid diskI,ngl, Lmngi, Lmgi = coefficients defined by (5.143)K = spring stiffness of the stringK = stiffness of ith spring of the diskK1, IC2 = linear frequencies of the string in (3.62)£, £2 = differential operators defined by (5.143)M = inertial coefficient of the stringMd = mass of the rigid diskM77,M79,M98 = transverse bending moments defined by (5.109)n2 = parameters defined by (3.61)N77,N78,N88 = inpiane forces of the disk defined by (5.110)p1’ P2, qi, q2 = coefficients defined by (3.66)P = tension in the undeformed stringP0 . = tension in the string due to rotation defined by (A.145)P = total tension in the string defined by (A. 144)Q = transverse external load in the diskQrr = transverse force defined by (5.109)(r,O,z) = space-fixed polar coordinatesRb = amplitude ratio of backward and forward wavesR1 = amplitude ratio of forward and backward wavesR.j, R0 = inner and outer radii of the diskRmn(r) = radial modal function defined by (4.91) or (5.121)=R/R0S = linear wave speed in the string defined by (2.4)t = timexiiiT = kinetic energyTmnpq,(t) = time functions defined by (5.126)u, v, w displacements in r, S and z directionsV = potential energyW = work done by constraintsX = coefficient defined by (2.14)c1, a2 = coefficients defined by (C. 154)a, a5 = coefficients defined by (2.34)amnpq,/3mnpq, 7mnpq, 8mnpq = coefficients defined by (5.123),62 = phase angles in Butenin solution (3.62)7 = real parts of the eigenvaluesr = coefficient defined by (2.14)So = Dirac delta function= parameter defined by (3.61)= longitudinal strain of the string defined by (3.39)err, f99, i-e, = total inpiane strains of the disk defined by (5.95)= linear membrane strains of the disk defined by (5.96)= nonlinear membrane strains of the disk defined by (5.97)err, 6, , = bending strains of the disk defined by (5.98)£1 = normalized nonlinearity of the string defined by (3.54)= normalized nonlinearity of the spring defined by (3.54)= parameters defined by (C. 157)Cc, C., ec, C. = amplitude coefficients in equation (2.19)= coefficient defined by (2.14)xiv= parameter defined by (3.61)‘ ,= parameters in Butenin equations (C. 151)= eigenvalues of the system =-y+ iw)“1, 2 = eigenvalues in equations (2.25) and (2.28)A = parameter defined by (3.61)= frictional coefficient of the stringp = Poisson’s ratio= nonlinearity of the spring in equation (3.42)Cnpq, ?7flpq = coefficients defined by (5.126)p = mass per unit length for the string,or mass per unit volume for the diskJ8, = inpiane stresses of the disk defined by (5.99)r = normalized timeso = linear stress function defined by (5.138)= nonlinear stress function defined by (5.139)= stress function= homogeneous and particular stress functions4mnpqij(7’) = particular stress functions defined by (5.126)4’mnpqi,(l’) = homogeneous stress functions defined by (5.135)x = m+n+p+q+i+jbx, iJ = rigid-body rotations in the diskw = imaginary parts of the eigenvalues (frequencies)= nonlinear frequencies of the string= rotation speedxvSubscripts= partial differentiationg, 1, m, n, p. q = mode numberSuperscripts= whole differentiation* = nondimensional parametersi,j = indexinE(E)xviAcknowledgementsI greatly appreciate the support, the trust and the advice which I have received from mysupervisor, Professor Stan Hutton, throughout my graduate studies. Professors Avrum Soudackand Gary Schajer deserve special thanks for their encouragement and assistance which I haveenjoyed in the past. I thank Professors Hilton Ramsey and Mervyn Olson for many usefuldiscussions during the development of this work. Dr. John Kishimoto is acknowledged formany technical consultations.I would like to express my gratitude for the financial support of the Natural Sciences andEngineering Research Council of Canada and the Science Council of British Columbia throughgraduate fellowship program. I am grateful to Mr. Jan Aune of MacMillan Bloedel Researchfor being my industrial supervisor and for teaching me the practical knowledge of splined saws.My final but foremost thanks go to my wife, Cindy, my son, Ian, and my parents whoseunderstanding and support made this work possible. To all of them, I dedicate this thesis.xviiChapter 11.1 BackgroundIntroductionA constrained rotating circular disk has many industrial applications, including guided saws forcutting wood and computer recording disks. The present work is particularly concerned withsawblade applications. Figure 1.1 shows schematically a guided splined saw of a type widelyused to perform the tasks of secondary breakdown and resawing in North American sawmills.Figure 1.1: Schematic of a splined guided saw.The saw is driven by a splined arbor, and can freely slide at the arbor. A pair of guides is placedat the blade surface to position the blade. The clearance between the blade and the guides isapproximately 0.05 mm and is lubricated with water or an air/oil mixture. The guides increase‘:2guide padsawbiade splined arbor1Chapter 1. Introduction 2the transverse stiffness in the cutting region, and the thickness of the sawbiade used can thusbe reduced. As a result, the cutting kerf is decreased, and significant savings can be generated.However, the critical speed of a rotating disk is decreased when its thickness is reduced, andthe disk may run near or even above the critical speed if the production (feed rate) is maintainedat the same level as before. Existing linear analyses show that a constrained rotating disk isunstable in the supercritical speed region [1, 2, 3]. For example, an elastically-constrained diskhas divergence and flutter instability regions at supercritical speeds, as shown in Figure 1.2.The divergence instability involves one mode which has zero frequency (imaginary part of the4030>%C)Cci)2Oa)U-100Cua)01.0Rotation SpeedFigure 1.2: Eigenvalue-speed diagram of an elastically-constrained rotating disk.eigenvalue) and a positive real part of the eigenvalue. Flutter instability involves two modeswhich have an identical nonzero frequency and positive real parts of the eigenvalues. BecauseChapter 1. Introduction 3of these instabilities, the operation of a constrained rotating disk at supercritical speeds hasnot been recommended in practice. However, it was found from recent experimental work[4, 5] that some thin splined saws do not exhibit the predicted instabilities and run smoothlyin the supercritical speed region. It was also reported [6] that splined saws have been usedat supercritical speeds in some sawmills, and produced good cutting performance. Thus thedynamic behavior of a constrained rotating disk is much more complicated in the supercriticalspeed region than is explained by the existing linear analyses. Therefore the existing linearanalyses may have overlooked some factors that are important in determining the stabilitycharacteristics of a constrained rotating disk.Experimental work showed that the existing analyses can accurately evaluate the vibrationsof an unconstrained rotating disk, such as a conventional collared saw, at both subcritical andsupercritical speeds [7, 10]. Obviously, the introduction of constraints complicates the dynamic responses of a rotating disk. Therefore it is important to study the dynamic interactionsbetween a rotating disk and constraints in order to better understand the stability characteristicsat supercritical speeds. Particularly, the following fundamental questions need to be addressed:(1) what are the traveling wave components in each vibration mode; (2) what are the effects ofdifferent types of constraints, such as spring, mass, damper and friction, on the stability characteristics. The vibration analysis of a rotating circular disk subject to a stationary transverseload system is complex, and detailed understanding of the interactions of the various travelingwaves is difficult. For this reason, it is desirable to seek a simpler model that contains thefundamental physics of interest. Such physics involves the interaction of a rotating flexiblemedium with a stationary constraint in which the inertial, gyroscopic, and centripetal acceleration effects together with the stiffness effects of the medium are in dynamic equilibriumwith the forces generated by the constraint. Moreover, the system should be non-dispersivein order to simplify the wave effects. A rotating circular string has all these features while itsmathematics is much simpler than that of a disk [8]. Hence the string model will be used toChapter 1. Introduction 4study the dynamic interactions.A rotating disk with free-free edges, such as a splined saw, has one translational and tworotational rigid-body motions. It was reported that the inclusion of the rigid-body motions cansignificantly improve the stability characteristics [41]. However, the report did not specifywhich rigid-body motion reduces the instabilities, and therefore the result has limited use inpractice. In this thesis, the effects of each rigid-body motion on the instabilities will be studied.The linear analysis cannot correctly provide the dynamic responses of a constrained rotatingdisk at supercritical speeds. In the linear analysis, the transverse displacement is assumed tobe small, and does not change the inpiane stresses of the disk. It is suspected, however, thatwhen the instability occurs, the transverse displacement becomes larger and may subsequentlyaffect the inplane stresses. As a result, the stability characteristics of the disk are changed.Large-amplitude vibrations of highly flexible rotating disks have been observed in experimentalwork [4, 24]. It was also reported that in panel flutter problem nonlinear effects (midplanestretching) generally restrain the unstable motion to a bounded limit cycle, especially forsupersonic airfiows [32]. Therefore, the effects of the geometric nonlinearity on the instabilitiesof a constrained rotating disk should be studied.1.2 Objectives and ScopeThe objectives of this research work are as follows• To understand the dynamic interactions between a rotating circular string and a stationaryconstraint consisting of a spring, a damper, a mass or a frictional restraint;• To study how the rigid-body motions affect the divergence instability of an elasticallyconstrained rotating disk;Chapter 1. Introduction 5• To investigate the effects of geometric nonlinearity on both divergence and flutter instabilities of an elastically-constrained rotating disk.In Chapter 2, the dynamic interactions between an idealized rotating circular string and astationary constraint consisting of a spring, a damper, a mass or a frictional restraint are studied.The method of traveling waves is applied to develop the characteristic equation. An approximatesolution is generated using Galerkin’s method to solve the eigenvalues effectively. The physicsof the stability characteristics and the interactions between the string and the restraints arediscussed based on both analytical and approximate solutions.In Chapter 3, the nonlinear vibrations of an elastically-constrained rotating string are studied.The nonlinearities of the string deformation and the spring stiffness are considered in thegoverning equations. Butenin’s method is adopted to develop a closed-form analytical solutionfor single-mode oscillations of the system. Multi-mode responses of the string are investigatedby means of numerical integration. The effects of the geometric nonlinearity on the flutterinstability at supercritical speeds are discussed.In Chapter 4, a conventional modal analysis is used to examine the effects of rigid-bodymotions on the divergence instability of an elastically-constrained rotating disk at supercriticalspeeds. The analysis shows how the translational rigid-body motion affects the divergenceinstability.In Chapter 5, an analysis of nonlinear vibrations of an elastically-constrained rotating diskis developed. The equations of motion, which are two coupled nonlinear partial differentialequations corresponding to transverse force equilibrium and to deformation compatibility, are• first developed by using von Karman thin plate theory. Then the stress function is analyticallysolved from the compatibility equation by assuming a multi-mode transverse displacement field.Finally, numerical integration is used to solve the time governing equations, and the effects ofnonlinearity on both divergence and flutter instabilities of a rotating disk are investigated.Chapter 2Interactions between a Rotating String and Stationary Constraints2.1 IntroductionMany complex problems arise in the study of the dynamic interactions between rotating flexibledisks and stationary constraints. At speeds above the lowest critical speed the dynamics of thedisks becomes very complicated. The vibration analysis of a rotating circular disk subject toa stationary transverse load system is complex, and detailed understanding of the interactionsof the various traveling waves is difficult. For this reason, it is desirable to seek a simplermodel that contains the fundamental physics of interest. Such physics involves the interactionof a rotating flexible medium with a stationary constraint in which the inertial, gyroscopic,and centripetal acceleration effects together with the stiffness effects of the medium are indynamic equilibrium with the forces generated by the constraint. Moreover, the system shouldbe non-dispersive in order to simplify the wave effects. With these considerations in mind ananalysis of a somewhat artificial rotating circular string with a general stationary restraint isconsidered in this chapter.Schajer [8] first introduced the model of a rotating circular string subject to an elasticrestraint, and studied the vibrations of the string in the subcritical speed region. The exactsolutions for K = 0 and K = oo were given, and the approximate solution for finite K wasderived by Rayleigh-Ritz method. Curve veering in the subcritical speed region was observed.Perkins and Mote [91 and Yang [10] presented the exact solution for the same model with finiteK. Xiong and Hutton [11] investigated the effects of dispersion due to distributed springs on6Chapter 2. Interactions between a Rotating String and Stationary Constraints 7the vibrations of a rotating string, and the force boundary condition used erroneously in theprevious analyses [8]—[1O] was correctly presented by including the momentum flux of thestring.The objective of this chapter is to study the dynamic interactions between a rotating circularstring and different point restraints. Considered are the effects due to a spring, viscous damper,mass and friction, as have been studied in the case of a rotating disk [1, 2, 3, 12]. The exactsolution for the string is first developed by the traveling wave method. Because the exactsolution is singular at the critical speed, the vibration characteristics at this speed are examined.The characteristic equation of the exact solution comprises complex exponential functions,and an effective computer program for solving this equation has not been found. Therefore,Galerkin’s method is adopted to develop a closed-form approximation. Finally, the vibrationcharacteristics of the string are discussed based on both analytical and approximate solutions,and comparisons are made between this analysis and the analytical results of Chen and Bogy[13, 14].2.2 Equations of MotionsFigure 2.3 depicts a rotating circular string of displacement w(8, t) constrained by a stationaryload system with stiffness, inertial, damping and frictional characteristics. The parameters rand p are the radius and mass per unit length of the string. The rotation speed of the string is 2.The following assumptions are adopted in developing the governing equation of the problem:1. The circumferential tension P in the string is constant [8];2. The string is restrained at one point by an eyelet connected to a spring of stiffness K anda viscous damper of damping coefficient C [1, 3, 8];Chapter 2. Interactions between a Rotating String and Stationary Constraints 8The following nondimensional parameters are then introduced= St, = w/r, fl = S2/S,K*= Kr/F, = P/F, C = C//P,3. The mass of the eyelet is M, and the eyelet generates a constant friction force p on thestring [1, 3];4. The velocity of the string in the9 direction has constant value rf1.wK CFigure 2.3: A rotating circular string subject to constraints of a spring, a damper, a mass and afriction.It is noted that assuhiption 1 is artificial. However, as shown in Appendix A, this assumptionis made in order to induce a critical speed for the string.In the following, Hamilton’s principlet2SI (T—V)dt=OJtIis used to develop the governing equation and boundary conditions. Applying the precedingassumptions, equation (2.1) is written asjt2{S j[F(!f)2 — p(f22r2 + (W,t +f2w,)2JrdS + S[Kw2(O, t) — (0, t)]-i-(Cw, +p’2)Sw(0, t)}dt = 0. (2.2)M=M/(pr),(2.1)(2.3)Chapter 2. Interactions between a Rotating String and Stationary Constraints 9where the wave speed S is given byS2=. (2.4)prPerforming the variation operation, and using the nondimensional parameters, leads to thefollowing nondimensional governing equation and boundary conditionsgoverning equationW*,tt. ÷2cl*w*,.8—(1 — r Z)w*,Ø9 = 0, (2.5)boundary conditionsw*(0, t) — w*(27r, t) = 0, (2.6)(1 *2)[w*,9 (27r, t) — w (0, t*)]+K*w*(0, t*) + C*w*,.. (0, t*) + M*w*,t.t. (0, t) + iu* w* ,o (0, t*) = 0. (2.7)The initial conditions are assumed as followsw*(6, 0) = fo(9), (2.8)(2.9)2.3 Solution of Equations of Motion2.3.1 Traveling Wave MethodThe exact solution of equations (2.5)—(2.7) can be derived from the method of separation ofvariables as has been done for the case of a spring restraint [9, 10, 11]. In this section, themethod of traveling waves is used to obtain the characteristic equation for the given problem,and it will be seen that this method is more general than the method of separation of variables.The traveling wave solution of equation (2.5) is given byw*(O,t*)= f(6 — (1 + V)t*) + b(9 + (1 — fZ*)t*), (2.10)Chapter 2. Interactions between a Rotating String and Stationary Constraints 10where f and bare the forward and backward traveling waves respectively. Substituting equation(2.10) into boundary conditions (2.6) and (2.7) yields,f(—(1 + f2*)t*) + b((1 — — f(27r — (1 + cl*)t*) — b(27r + (1 — = 0, (2.11)(1 — V2)[f’(—(l + *)t*) + b’((l — *)j*) — f’(2ir — (1 + *)t*) — b’(2ir + (1 —__K*[f(_(1 + *)j*) + b((1 —_C*[_(1 + 2)f’(—(1 + 2*)t*) + (1 — fZ*)1J((1 —_JI4*[(1 + fZ*)2hI(_(1 + l*)t*) + (1 — fl*)2bhI((1 —[f’((1 + l*)t*) + b’((l — *)t*)J = 0. (2.12)In order to derive the characteristic equation, the boundary conditions (2.11) and (2.12) arecombined into one equation by eliminating either function f or b. The following equation isderived through eliminating the forward wave f2F[b’(2ir + 27ri, + LX) — b’(2irt + tX) — b’(27r + tX) + b’(tX)] + [K*b(27rt + tX)+FC*bl(21rt + tX) +F2M*bhl(21rz, + tX)] — [K*b(21r + tX) + FC*bI(27r + iX)+F2M*bll(2uc + tX)] + u* [b’(2rt + LX) — z.b’(27r + tX) — (1 — L)b’(LX)] = 0, (2.13)whereF = 1 —= F/(—(1 + 1*)), (2.14)Equation (2.13) is a differential-difference equation, which has second-order differential andfourth-order difference, for the backward traveling wave function b in terms of X. The generalsolution is assumed in the form [15]b(y) = C1e’, (2.15)Chapter 2. Interactions between a Rotating String and Stationary Constraints 11where X is a characteristic root and C1 is a complex constant. Unlike a second-order differentialequation which only has two characteristic roots, equation (2.13) may have an infinite number,hence, solution (2.15) can be composed of an infinite number of exponential functions. Assuming A = )/F in keeping with the results in [9] and [11], and substituting equation (2.15)into (2.13), leads to the following characteristic equation2A(1 — e’X1 — eiWr) — (K + CA + M*A2)(eT — e)_i*2_(1+f*)e_W_(1_f2*)eT*1=o. (2.16)Equation (2.16) is identical to that given in [11] if C* = M* = = 0. The same result canbe obtained by applying similar operations to the forward traveling wave function f. Thecharacteristic roots of the transcendental equation (2.16) are generally complex numbers, andare solved numerically by a computer code based on Muller’s method [16].It should be noted that the assumption that the variables are separable is not used in thepreceding development. Hence, this method is more general than the method of separationof variables, and may be applied to problems which cannot be solved by the latter method.Moreover, if we assume y = 8 + (1 — *)t* in equation (2.15), and y = 8 — (1 + Q*)t* for theforward wave function f(y) =C2e”’°’ where C2 is a complex constant, then, the requireddisplacement w (2.10) becomesw*(8,t*) = et*(CIeT* +C2eT). (2.17)Equation (2.17) is a separable form of the solution which may be expressed in terms oftrigonometric functions as:w*(8,t*) = eY(t*+T[ccosw(t*+18)+ssinw(t*+1+ et*i&[ccosw(t*— ‘1 ++ c sin w(t*—+ (2.18)Chapter 2. Interactions between a Rotating String and Stationary Constraints 12where= ‘y+iw,cc == i(Ci — C1), (2.19)cc ==and , are complex conjugates of C1, C2 respectively. Coefficients , E, C and aredetermined from boundary conditions (2.6) and (2.7) and initial conditions (2.8) and (2.9).Closed-form solutions to the non-homogeneous form of equation (2.5), expressed in terms ofa state space modal expansion, and in terms of a Green’s function formulation, have beenpresented by Wickert and Mote [17].2.3.2 Solution at Critical SpeedEquations (2.16) and (2.17) are not valid at the critical speed because of the singularity. Hence,a solution for the case of the critical speed must be developed. Since the terms associated with(1 — * 2) are zero, governing equation (2.5) and the force boundary condition (2.7) becomeW*,pt. +2w*,t*9 = 0, (2.20)K*w*(0, t) + (0, t*) + Ifwt. (0, t*) + (0, j*) = 0. (2.21)The traveling wave solution (2.10) becomesw(6, ) = f(6 — 2t*) + b(6). (2.22)Substituting equation (2.22) into equations (2.8) and (2.9), integrating, and regrouping, gives1 e—2w*(8, t*)=f(6 — 2t*) + b(6) = fo(8) — j go(i,b)d’i/.’. (2.23)Chapter 2. Interactions between a Rotating String and Stationary Constraints [3Equation (2.23) presents a general solution at the critical speed provided the boundary conditions(2.6) and (2.21) are satisfied. This means that selection of the initial conditions (2.8) and (2.9) isdependent on the boundary conditions. The following discussion will be focused on the effectsof constraints on the vibration characteristics of the string subject to the boundary conditions(2.6) and (2.21).CaseA:K*O, C’O, M*O, t’OSubstitution of equation (2.23) into (2.21), letting T = _2t*, and differentiating with respectto r, leads toM’g(r) + (C* — *)g(T) — K*go(T) = 0. (2.24)Therefore,go(T) = Aje” +A2e2T, (2.25)where A1 and A2 are constants, and= 2M*— 1*)+ /(C — L)2 + 2K*M*],= 2M*— 1)—iJ(C* — *)2 + 2K*M*].Obviously, and )2 are real numbers. Then, the displacement (2.23) becomesw*(6, t*) = !_[e°(1 — e_2A1t*)] + _[e’26(1 —e2)12t)] + fo(8). (2.26)Substitution of equation (2.26) into continuity condition (2.6) shows that constants A, and A2must be zero. Accordingly, the displacement w’ is independent of time t*, and a standing wavecorresponding to the initial displacement is developed at the critical speed.Case B: K* 510, C ‘ 0, IL* 1 0, M* = 0If W = 0, equation (2.24) becomes(C* — L*)g(T) — K*go(T) = 0. (2.27)Chapter 2. Interactions between a Rotating String and Stationary Constraints 14Then,go(T) AieT, (2.28)where A1 is a constant, and A1 = K*/[2(C* — p*/2)]. Similarly, application of continuitycondition (2.6) indicates that A1 = 0. For the case of C* = p*/2 equation (2.27) shows thatgo(r) = 0. Therefore, the characteristic is the same as in Case A.Case C: M 1 0, K = C = =0Substitution of equation (2.23) into (2.21) yieldsw*,t*t* (0,t*) = _2g(_2t*) = 0. (2.29)Thus, go(T) = A1 where A1 is a constant, andw*(6, t) = Ait* +f0(e). (2.30)The term Ait* represents the rigid-body motion which is caused by the absence of a physicalsupport in the system. If A1 = 0, a standing wave is recovered from solution (2.30).Further examination of different constraint combination demonstrates that either go(r) = 0,or go(T) = constant. Consequently, at the critical speed the string subject to constraints doesnot vibrate, and the displacement is bounded if the rigid-body motion is not considered. Thisconclusion is different from the one presented by Mote [18] that a linear solution of an axiallymoving string does not exist at the critical speed. It is however recognized that the restrictionthat the initial velocity of the string must be zero (go(T) = 0) for a linear, nonzero, unforcedresponse at the critical speed can be removed if geometric nonlinearities are considered. In thepresent analysis the tension has been assumed constant. If the nonlinear terms associated withthe speed-dependent tension are considered the response characteristics will be significantlydifferent from the present work [1 8]—[2 1].The method of separation of variables cannot be used to solve the given problem becauseChapter 2. Interactions between a Rotating String and Stationary Constraints 15it only provides the trivial solution. Therefore, the traveling wave method is more general thanthe method of separation of variables in solving this specific problem.2.3.3 Approximate SolutionThe eigenvalues, which are a function of speed and restraint parameters, can be solved fromthe characteristic equation (2.16). However, this equation consists of exponential functions,and the roots can be complex numbers. An effective root solver, which can calculate all thespecified roots for a given speed in one run, has not been found. Moreover, numerical instabilitycan occur near the critical speed because of the singularity of the term 1/(1 — LV). In viewof these difficulties, an approximate solution is developed using Galerkin’s method, whicheliminates the above problems.To this end, the boundary condition (2.7) and equation (2.5) are combined such that* ‘n* * ,i n*2 *W ** W ,t6 k’ — )W 99+(K*w* + +111* W* +I.L*w ,e )6(8) = 0, (2.31)where 8(8) is the Dirac delta function. The solution of equation (2.31) is assumed to be of theformNw*(8, t*) = [Cn(t*)cosn6 + S(t) sinn6], (2.32)n=Owhere n is the mode number, and C(t*) and S(t*) are unknown functions to be determined.Following Galerkin’s method yields:NcxjCrn + 2m12*Sm + m2(1 cr2)Cm] + (K*C + C*Cn + M*C + ,*flS) =0,n=O— 2mQ*Cm +m2(l — 1r2)Sm] = 0,(2.33)Chapter 2. Interactions between a Rotating String and Stationary Constraints 16where1 2w (0) ifm=0,a(a3)= (2.34)7r (ir) ifm/0.Assuming that C(t*) = and S(t*) = sneAt* , where c and s are constants and A is theeigenvalue, transforms equation (2.33) from 2N +1 ordinary differential equations into 2N +1algebraic equations of the form(A2[m] + A[c] + [kJ){x} = 0. (2.35)The eigenvalues for given parameters can be computed from equation (2.35). The calculatedeigenvalues have been compared with the analytical results from equation (2.16) for differentrestraint conditions in Table 2.1. The comparison shows that for the lowest 5 modes theapproximations are almost identical to the analytical results for the case when N = 15.2.4 Results and DiscussionIn this section the characteristics of the string are discussed with a view to developing betterunderstanding of the physical interactions taking place. Chen and Bogy [13] analyticallystudied the rate of change of the eigenvalues for gyroscopic systems subject to a stationarypoint restraint, and the results of this analysis for the current problem are presented in AppendixB. Their analysis is not valid in regions where repeated eigenvalues occur (i.e. at “crossingpoints”). The exact solution (2.16) is then used to discuss the behavior of the string at crossingpoints.2.4.1 Effects of a Spring RestraintBefore discussing the results, the terminology used in the discussion is defined. As shown inequation (2.18), a vibration mode of a constrained string is composed of both backward andforward traveling waves. A mode which is dominated by a forward wave is called a “positiveChapter 2. Interactions between a Rotating String and Stationary Constraints 17Table 2.1: Eigenvalues calculated from analytical and numerical solutions.Q Restraint Solution Mode 1 Mode 2 Mode 3 Mode 4 Mode 5Analytical 0.238i 0.540i 1.02i 1.50i 1.55iK =0.5Numerical 0.239i 0.541i 1.02i 1.50i l.55iAnalytical 0.496i 0.992i 1.45i 1.50i 1.99iM5 =0.2Numerical 0.496i 0.992i 1.45i 1.50i 1.99i0.5 —0.00398 —0.00398 —0.0159 —0.00398Analytical 0.500i 1.OGi 1.50i 1.50i 2.OOiC = 0.1 —0.00398 —0.00398 —0.0159 —0.00398Numerical 0.500i 1.OOi 1.50i 1.50i 2.OOi—0.00853 —0.00853 —0.00853Analytical 0.500i 1.OOi 1.50i 1.50i 2.OOi= 0.1 —0.00796 —0.00796 —0.00796Numerical 0.500i 1.OOi 1.50i 1.50i 2.OOi——0.0775 0.0775Analytical 0.334i 0.334i 0.778i 1. 19i 1.59iKS= 0.5 —0.0763 0.0763Numerical 0.333i 0.333i 0.778i 1.19i 1.59iAnalytical 0.403i 0.805i 1.2li 1.61i 2.02iM5=0.2Numerical 0.403i 0.805i 1.21i 1.61i 2.02i1.4 0.00318 0.00318 0.00318 0.00318 0.00318Analytical 0.400i 0.800i 1.20i 1.60i 2.OOiC5 = 0.1 0.00318 0.00318 0.00318 0.00318 0.00318Numerical 0.400i 0.800i 1.20i 1.60i 2.OOi—0.00757 —0.00757 —0.00757 —0.00757 —0.00757Analytical 0.400i 0.800i 1.20i 1.60i 2.OOi= 0.1 —0.00796 —0.00797 —0.00797 —0.00796 —0.00796Numerical 0.400i 0.800i 1.20i 1.60i 2.OOiChapter 2. Interactions between a Rotating String and Stationary Constraints 18Figure 2.4: Effects of a spring K* = 0.5 on the eigenvalue spectrum, approximate solutionwith N = 18; (a) frequency, (b) real part, constrained, - -: free.model chosen correspond to those of a string with the same tension rotating at different speeds.Figure 2.4(a) compares the frequencies of the constrained string with those of a free string. Thenatural frequency of all negative and positive waves is increased, while that of reflected waveswave mode”, and a “negative wave mode” consists predominantly of a backward wave. Amode composed mainly of a reflected forward traveling wave at supercritical speeds is calleda “reflected wave mode”. A point at which a backward wave intersects a forward wave for thecase of the free string is called a “crossing point”, and the corresponding speed as the “crossingpoint speed”.Figure 2.4 plots the eigenvalues as a function of speed for the string constrained by aspring of stiffness K* = 0.5. It should be born in mind that the results of the mathematical3.02.O(a)LL0.00.00.1Cl)0.0 (b)0a)-0.10.00.5 1.0 1.5 2.00.5 1.0 1.5 2.0Speed (2/S)Chapter 2. Interactions between a Rotating String and Stationary Constraints 19is decreased due to the presence of the spring as shown in Figure 2.4(a). This observation issimilar to the result for an elastically-constrained disk presented by Chen and Bogy [13, 14],and is consistent with the result given in Appendix B. Figure 2.4(a) shows that the frequencycurves avoid crossing in the subcritical speed region. An important feature of curve veeringthat has been observed in previous studies is that the mode shapes are interchanged in theprocess of veering [9, 22]. The change is continuous, but could be abrupt. The physics of thecurve veering phenomenon can be illustrated with the help of solution (2.18). The amplitudesof a backward and forward traveling wave in any vibration mode are + and +respectively. Figure 2.5 shows the amplitude changes of the backward and forward waves inmode 2N and 1P shown in Figure 2.4(a) by plotting the ratios Rb = +f32/i./2 + andFigure 2.5: Backward and forward traveling wave contribution to modes 2N (- -) and iF (—)around crossing point speed V = 1/3.R1 = 1/Rb as a function of speed about the crossing point speed = 1/3. It is shown that for2N, (Rb)2N>> 1 and (Rf)2N 0 when speed V is far below 0.32. Therefore, the backward10.00C)5.00.ESpeed (IS)Chapter 2. Interactions between a Rotating String and Stationary Constraints 20traveling wave is dominant in the mode. When fl4 is increased, (Rb)2N is reduced and (Rf)2N isincreased, which indicates that the influence of the backward wave on the vibration is decreased,and that of the forward wave is increased. The ratio (&)2N (Rf)2N 1 at 11 = 0.32, hence,the backward and the forward waves have comparable contributions to the vibration of mode2N. As the speed is further increased, R1 becomes larger than Rb. Consequently, the forwardwave exerts more influence on the vibration than the backward wave, and mode 2N becomesmode iF. Similar observation applies to mode iF.In the supercritical speed region, the frequency curves for reflected and positive modescouple as they intersect. Further increase of the rotation speed separates the coupled frequencycurves. Inspection of the real parts of eigenvalues shown in Figure 2.4(b) indicates that one ofthe real parts is positive during such coupling, hence, flutter instability takes place.It is known from previous analysis [8] that the frequency curves of the positive modes passthrough the crossing points at subcritical speeds. Figure 2.4(a) indicates that the frequencycurves of reflected waves also pass through the crossing points at supercritical speeds. Withthe exact solution (2.18), one can easily show that a nodal point is developed at the constraintdue to the modulation of the forward and backward waves. The procedure will be discussedwhen the effect of a frictional constraint is considered.2.4.2 Effects of an Inertial RestraintThe effects of a point mass restraint are examined in this section. Figure 2.6 presents eigenvaluespeed plots for an eyelet with a nondimensional mass of 0.2. From Figure 2.6(a), it is observedthat the frequencies of the negative and positive modes are decreased, while the frequency ofthe reflected mode is increased. The cause of the changes is similar to that of a rotating diskas discussed in [13] (see equation (B.148) in Appendix B). The frequency curve of a negativemode does not intersect that of a positive mode at subcritical speeds, and curve veering takesChapter 2. Interactions between a Rotating String and Stationary Constraints 213.0Cl)>%(a)gi.oLL0.00.0 2.00.1Cl)0.0 (b)00)0.100 0.5 1.0 1:5 2.0Speed (21S)Figure 2.6: Effects of a mass M = 0.2 on the eigenvalue spectrum, approximate solution withN = 18; (a) frequency, (b) real part, —: constrained, - -: free.place as in the case of an elastically-constrained string. The frequency curves of the negativeand reflected modes pass through the crossing points because a nodal point is developed at thepoint mass at these speeds. In the supercritical speed region, the curves of the positive and thereflected modes become coupled when they intersect. Further increase of the speed separatesthe coupled frequency curves. Some real parts of the eigenvalues become positive when thefrequency curves are coupled as shown in Figure 2.6(b), and flutter instability occurs.A general characteristic of curve veering in the frequency-speed loci is that veering occursonly when all frequencies are either increased, as in the case of an elastically-constrained stringin subcritical speed region, or decreased, as in the case of a string with a point mass. However,the frequency curves cross when the change of the positive modes is different from that of0.5 1.0 1.5ooQ p)Chapter 2. Interactions between a Rotating String and Stationary Constraints 22the reflected modes. In Figure 2.4(a), for example, frequencies of the positive modes of theelastically-constrained string are increased in the supercritical speed region, while those of thereflected waves are decreased, and crossing then occurs.2.4.3 Effects of a Damping RestraintFigure 2.7 shows the effects of damping on the eigenvalues of a rotating string. The nondimensional damping coefficient is 0.1. From Figure 2.7(a), note that the frequencies are negligibly changed with this amount of damping, which is consistent with the prediction of3.0Cl)2.001)D1.0Ii-0.5 1.0 1.5(a)(b)2.00.00.00.01 —-Cl)0.000.01-0.02oAA4J-0.030.0 0.5 1.0 1.5 2.0Speed (a/S)Figure 2.7: Effects of a damper C* = 0.1 on the eigenvalue spectrum, approximate solutionwith N = 18; (a) frequency, (b) real part,—: constrained, - -: free, 0: single-mode.equation (B. 149) in Appendix B. All frequency curves pass the crossing points, unlike theChapter 2. Interactions between a Rotating String and Stationary Constraints 23cases of the elastic and inertia constraints.The variation of the real parts of the eigenvalues shown in Figure 2.7(b) is complicated. Thereal part of the eigenvalue of the negative (reflected) wave mode iN (1R) increases linearlywith the speed except near zero speed. It is negative in the subcritical speed region and becomespositive at supercritical speeds. Accordingly, the reflected wave is destabilized at supercriticalspeeds. It is noted that this mode does not interact with other modes, and the instability doesnot involve mode interaction. If the eigenvalues of mode 4N (4R) are considered, the real partcannot be distinguished from that of mode iN (11?) apart from the crossing points. However,this value becomes zero when the mode encounters other modes at crossing points, for instance,when interacting with mode 1P at f2 = 3/5 and V = 5/3. For the positive wave iF, thereal part decreases linearly with speed apart from the crossing points and remains negative atall speeds. The value is reduced to minima when interacting with negative wave modes atsubcritical speed, e.g., mode 2N at V = 1/3, and increased to maxima when encounteringreflected modes at supercritical speed, e.g., mode 4R at = 5/3. If lines are used to connectthese minimum and maximum points of mode 1 P at crossing points, they turn out to be astraight line with constant value as shown in Figure 2.7(b). The variation of other positivemodes is similar to mode iF.The preceding results at crossing points can be verified by the exact solution (2.16). Atcrossing points, assuming that A = -y + in(1 — 1) = y + im(1 + i*), and substituting A intoequation (2.16), yields2w., 2w2(1 — eTz?Ew)(1 — eTh+n ) — C*(eI—n — eTh+n) = 0. (2.36)Equation (2.36) leads to the following observations. First, the natural frequencies of the stringwith a damping restraint at the crossing points are the same as those of the free string. Second,the real part 7 has two roots. One root is zero which is associated with a negative wave modeof zero displacement at the constrained point. The other is negative which corresponds to aChapter 2. Interactions between a Rotating String and Stationary Constraints 24minimum (maximum) of a positive wave mode, and is about —C/27r for small C*.In Figure 2.7(b), the real parts of the eigenvalues as a function of speed for a single mode(N = 1 and m = 1 in equation (2.33)) are also plotted. The real parts of the mode are linearfunctions of the speed when Q’ > 0.05, and coincide with those from multi-mode solution apartfrom the crossing points. The slopes of the lines seem to be of the same magnitude, but withopposite signs. Computation on other single modes gives the same results. The results indicatethat the real parts of the eigenvalues of a mode are linearly proportional to speed if it doesnot interact with other modes. Near crossing points, a positive (negative) mode is no longerdominated by the forward (backward) wave, but mixed with a backward (forward) wave. Asa consequence, the real part is influenced by both waves, and deviates from the straight lines.At crossing points, a positive mode consists of both forward and backward traveling waveswith comparable amplitudes, and its real part is the sum of the two waves. Moreover, the realparts at crossing points should barely change because the slopes of the lines of the forward andbackward waves have the same amplitudes, but with opposite signs, which has been observedfrom Figure 2.7(b). The real part of a negative mode becomes zero at crossing points becausea nodal point is developed at 6 = 0.2.4.4 Effects of a Friction RestraintThe effects of a frictional force are studied in this section. Figure 2.8 presents the eigenvaluesas a function of speed for a nondimensional friction coefficient 0.1. The frictional force hasnegligible effect on the frequencies of the string as seen in Figure 2.8(a), and all frequencycurves pass through the crossing points, in a similar manner to the string constrained by adamper. However, inspection of Figure 2.8(b) shows that the real parts of the eigenvaluesbecome zero at crossing points. These observations can be analytically verified by studying thecharacteristic equation (2.16) as shown in the development of equation (2.36).Chapter 2. Interactions between a Rotating String and Stationary Constraints 253.0Cl)2.0(a)U00.00.0 2.00.0150.000 (b)ci)-0.0150.0 0.5 1.0 1.5 2.0Speed (21S)Figure 2.8: Effects of a friction = 0.1 on the eigenvalue spectrum, approximate solutionwith N = 18; (a) frequency, (b) real part,—: constrained, - -: free.The real parts of the eigenvalues shown in Figure 2.8(b) exhibit different behavior from theresults discussed in the previous cases. The real part of a negative (reflected) mode, e.g., mode4N (4R), remains as a negative constant apart from the crossing points, and becomes zero atcrossing point speeds, e.g., at l’ = 1/3 (V = 5/3). The value of a positive wave, e.g., modeiF, remains as a positive constant apart from the crossing points, and becomes zero at crossingpoint speeds, e.g., at i* = 1/3. The magnitudes of the constants appear to be the same, andare about 0.008 in this case, which are consistent with the result (t*/47r) of equation (B.150)in Appendix B. Inspections of other modes demonstrate similar characteristics. Therefore,the motion of a negative (reflected) mode is stable at all speeds, but that of a positive mode is0.5 1.0 1.5JTTITA1cTTChapter 2. Interactions between a Rotating String and Stationary Constraints 26unstable apart from crossing point speeds.The eigenvalues at a crossing point are ) = in(1—= im(l + LV). Substituting= in(1 — Q*) and ). = irn(1 + l*) into the terms containing eIO_fl*) and e/*) in tv’s(2.17) respectively, then boundary condition (2.6) is automatically satisfied. Consequently, thecoefficients, ,and in w (2.18) can be arbitrarily defined. Taking = 0, = A1,= 0 and = —nAi/m, and substituting them together with )e = in(1 — = im(1 + 1*)into the derivative of w (2.18) with respect to 8, leads ton+m . n—rnw (8, t) = —2nA1 sin( 2 sin(wt + 2 0). (2.37)Obviously, w (0, t) = 0. As a consequence, the friction has no effect on the dynamics ofthe string, and the eigenvalue is the same as that of the free string. As to the second mode,it is known that both the forward and backward waves make equivalent contributions at thecrossing points, hence, the real part is equally affected by the two waves. The amplitudes ofthe real parts of these waves are equal but of opposite signs as discussed before, as a result,the sum of them becomes zero, which suggests that the real part of the second mode is zero atcrossing points. The above characteristics of the real parts of the string at crossing points werealso observed in the disk case [3, 14].Next consider the mechanism which is responsible for the stability of the system. Figure 2.9schematically illustrates the effects of the friction on the stability of traveling waves. It can beseen from Figure 2.9(a) that the vertical component of the frictional force is always in phasewith the forward wave motion, therefore, the friction does positive work to the wave, anddestabilizes the motion. The frictional force and the displacement of the backward wave areout of phase at subcritical speed as shown in Figure 2.9(b), consequently, the friction forcedoes negative work to the system and stabilizes the motion. In the supercritical speed region,the backward wave becomes a forward wave, while the direction of the friction force alsochanges sign. As a result, the friction force and the motion are as well out of phase as shownChapter 2. Interactions between a Rotating String and Stationary Constraints 27(a)F FFigure 2.9: Effects of a friction on the stability of traveling waves; (a) forward wave, (b)backward wave, (c) reflected wave.in Figure 2.9(c), and the friction will stabilize the motion. The preceding schematic illustrationalso applies to the damping case as shown for a stationary disk subject to a rotating damper byShen and Mote [23].2.5 SummaryThe characteristic equation of an idealized rotating circular string subject to a point constraint isdeveloped from the method of traveling waves. This method is general because the assumptionthat the variables are separable is not used in the development. The solution shows that eachvibration mode of the constrained string consists ofboth forward and backward traveling waves.Away from crossing points, the mode is dominated either by a backward or by a forward wave,and at crossing points, the mode is composed of equal contributions of backward and forwardtraveling waves. The response of the string at the critical speed is analyzed. It is found that atChapter 2. Interactions between a Rotating String and Stationary Constraints 28critical speed the string does not vibrate, and the displacement is bounded if rigid-body motionis excluded.For the string with a spring (mass) constraint, frequency curves avoid crossing if the changesin frequency of all modes are in the same direction, such as the increase in frequency due to thespring restraint and the decrease in frequency due to the mass restraint in the subcritical speedregion. However frequency curves will cross if the directions of the frequency changes aredifferent, such as at supercritical speeds, then the corresponding real parts of the eigenvaluesbecome positive and flutter instability occurs.For the string with a damping constraint, the real parts of the eigenvalues vary linearly withspeed away from crossing points. The damper does negative work on both backward andforward traveling waves and thus stabilizes the motion of these waves. However the damperdoes positive work on a reflected wave and thus destabilizes the motion. At crossing points,the real part of one eigenvalue becomes zero because a node is developed at the constrainedpoint, while the real part of the other eigenvalue adopts a constant negative value as a result ofthe equal contributions from the forward and backward (reflected) waves.For the string with a frictional constraint, the amplitude of the real parts of the eigenvaluesis constant away from crossing points. The friction does negative work on the backward andreflected traveling waves and stabilizes the motion of these waves. However it destabilizes themotion of a forward wave by doing positive work. At crossing points, the real parts of theeigenvalues all become zero due to. the modulation of both forward and backward (reflected)waves.The vibration and stability characteristics of the idealized model considered are very similarto those of a rotating disk [1, 3, 13, 14]. However physical insight into these characteristicsmay be better understood by studying the simpler string problem.Chapter 3Nonlinear Vibrations of an Elastically-Constrained Rotating String3.1 IntroductionIt was found from recent experimental work [4, 24] that a highly flexible rotating disk exhibitslarge-amplitude vibrations in the supercritical speed region. As mentioned in Chapter 1, aguided circular saw does not become unstable at supercritical speeds as predicted by the existinglinear analyses. The nonlinearity may affect the stability characteristics of a constrained rotatingdisk. A rotating circular string bears an analogy to a rotating disk while its mathematics ismuch simpler than that of a disk. Accordingly, it is useful to study the nonlinear behavior of arotating circular string in order to understand the nonlinear phenomena of a rotating disk.In the past, some studies have been conducted on the nonlinear oscillations of a straightstring which is relevant to the current research. Fundamental studies of the nonlinear dynamicsof a stationary string has been conducted by many researchers [25]—[34]. In these studies, bothfree and forced vibrations of undamped or damped strings have been extensively investigatedand the dynamics of such a string are well understood. Mote [181 discussed the nonlinearoscillations of an axially moving sthng and indicated that the relationship between smallnessof displacement and linearity for the stationary string cannot be extrapolated to the axiallymoving string. Ames [191 analyzed the nonlinear vibrations of a traveling string and conductedexperiments, which showed that the plane motion of the string is truly nonlinear and doeshave a hard spring jump. Shih [35, 361 further studied the motion of elliptic ballooning for atraveling string based on experimental work [19]. Kim [37] discussed the nonlinear vibrations29Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 30of traveling strings using the method of characteristics. Bhat [38] numerically studied thedynamic behavior of a belt moving on an elastic foundation and supported on two pulleys atthe ends, and indicated that in the absence of damping the system is unstable for any belt speedlarger than the wave speed in the belt.An analysis of the nonlinear, transverse, free response of a rotating circular string subjectedto a stationary elastic constraint is presented in this chapter. The nonlinearities considered arethose arising from the effect of displacement on string tension and the effect of a nonlinearspring restraint. First, equations of motion governing the nonlinear behavior of the stringare generated using Hamilton’s principle. Then Galerkin’s method is adopted to transformthe governing equations from a nonlinear partial differential equation into a set of couplednonlinear ordinary differential equations. A closed-form approximation for a pair of single-mode equations is derived using Butenin’s method. Finally, the multi-mode vibrations arestudied numerically, and characteristics of the nonlinear response are discussed.3.2 Equations of MotionIn addition to the assumptions presented in Chapter 2, the motion of the string is further assumedto be two dimensional, i.e., only transverse motion w and tangential motion v are considered.Consider an element rd8 of a string as shown in Figure 3.10. After deformation, the lengthof the element becomesds = [(1 + v, /r)2 + w, /r2J rd6 (3.38)[1 + +v,9+w,8)——(2rv,9+v,9 -i-tv,9) + j_g(2rve-i-v,9 -i-tv,9) I rd6.Hence, the strain of the element is= ds — rd8 (3.39)rd8= j!-(2rv,e +v, -l-w,) — —(2rv,e +v, )2 + j (2rv,9+v +w )3Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 31wFigure 3.10: Displacements of an element rd6 of a string in a planar deformationIn the following, Hamilton’s principle (2.1) is applied to develop the governing equation andthe boundary conditions for the system. The kinetic energy of the string is1 p2wT=—p I [(W,t +iw,9 )2 + (V,t +Qv,0+rIl)2]rd8. (3.40)2 JoThe potential energy ist2w 1 twoV= J (Pc + —EAe2)rd8 + J Fdwo. (3.41)o 2 owhere e is given by equation (3.39), P is the initial tension in the string, and F is the forcegenerated by a stationary constraint attached to the string at 8 = 0. wO is the displacement ofthe string at 0 = 0. The restraint in this model is an analogy to the guide of a circular saw. Themodeling of guides is difficult due to the complex interaction between the guides and the disk.The guides are usually modeled as discrete linear springs. As a result, only the elastic force isincluded in this study. A “hardening” cubic nonlinearity v1 is added to the linear elastic force inorder to obtain some perspective on the nonlinear effects of the spring. Hence, the restrainingforce is of the form0F(wo) = Kw(0, t) + v1Kw3(0, t). (3.42)Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 32Substituting equations (3.40) and (3.41) into equation (2.1), integrating, and neglecting theterms containing v,, v,, v,, and leads to the following governingequations and boundary conditions:transverse motiongoverning equationw,+2f2w,t+fl,e ,2W,O82 3 2 3 4—,2 [—w,e(v,e——v,o+—-w,91jv 9w,3,)],e= 0, (3.43)boundary conditionsw(0,t) = w(2ir,t), (3.44)(- — f2)[w, (0, t) — w,S (27r, t)J + EA P {[v,e (0, t)w,9 (0, t)—:, (2, t)w,6 (2, t)J — (0, t)w,9(0 t) 2 (2, t)w,9(2, t)](0, t) — w, (27r, t)} — _-[v,9(0, t)w, (0, t) — v, (27r, t)w, (27r, t)]—-_[w, (0, t) — w, (27r, t)J} = [w(0, t) + viw3(0, t)]; (3.45)tangential motiongoverning equation2 EA EA—P 1 2 2 2 3 4v,tt+21v et+fZ V,9 ——--j-v,se— 2 [(w,8——v,w,9——-jw,9)],o = 0, (3.46)pr p’r 2r r 4rboundary conditionsv(0, t) = v(27r, t), (3.47)—pr22)[v,e(0,t)—v,(2ir,t)]+ EA2 1’{!_[w,(O,t) — w,(2ir,t)] (3.48)—-[v, (0, t)w, (0, t)—v,e (27r, t)w, (2r, t)J — j[w, (0, t) — w, (2w, t)J} = 0.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 33Equations (3.43)—(3.48) formulate the nonlinear dynamics of the string. It is noted that equations(3.43) and (3.46) are the same as the governing equations of a straight string given by Anand[311 and Nayfeh [32] except for some higher order terms (v,e w,, w, and w,).The transverse motion w is of most interest. Therefore, this analysis will be focused on thetransverse vibrations. To this end, the following assumptions are adopted [31, 32]:1. The longitudinal inertia terms are negligibly small;2. The string is made of metallic material, hence, EA/P = 0(400 — 1000);3. The rotation speed fl is of the same order as the transverse wave speed (/7pr2)), ie,4. v,=0(w,).Applying the preceding assumptions, neglecting the higher order terms (w, v,s, w,) in equations (3 .46)—(3 .48), integrating equation (3.46) with respect to6, and using boundary condition(3.47), leads to the relationship between w and v in the form ofv(8, t) = 0 / w, dO — --- / w, dO. (3.49)4’ffro 2roSubstituting equation (3.49) into (3.43)—(3.45), and neglecting the higher order terms (w,8 v,,v,, w,) in equations (3.43)—(3.45), leads toP EA ,2W,tt +2I1w, +12 W,9 2W,9—4(j w, dS)w,08 = 0, (3.50)pr 47rpr ow(0,t) = w(2ir,t), (3.51)— 122) + EA j2w, dSJ[w,e (0, t) — (2ir, t)J = [w(O, t) +v1w3(0, t)]. (3.52)pr 47rpr 0 PTEquations (3.50)—(3.52) represent the analytical model of the nonlinear transverse oscillationof a guided rotating circular string.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 34Governing equation (3.50) is a second order nonlinear partial differential equation, whichis difficult to handle directly. Galerkin’s method is therefore applied to transform the partialdifferential equation into ordinary differential equations. Adopting a similar procedure to thatpresented in Chapter 2, the governing equation (3.50) is rewritten asw,t +2Qw,9—(S2 —f2)w,68— EA (J w, dO)w,99 +S(6)(w +v1w3) = 0, (3.53)47rpr 0 prwhere S is the wave speed given by (2.4), and S(8) is the Dirac delta function. Before proceedingto apply Galerkin’s method, the following nondimensional parameters are introducedt= 4J Wr ‘=—i-, 6 = K = -i--. (3.54)Applying these parameters, equation (3.53) is nondimensionalized asW*,t*t* +2*w*,e—(1 — V2)w, _!_L(J w*, d8)w,6947r 0+K*6(6)(w* ÷ 62W* 3) = 0. (3.55)The approximate solution of equation (3.55) is given by equation (2.32). ApplyingGalerkin’s method leads to the following ordinary differential equations for time functionsCm(t) and Sm(t)Ncc[Cm+ 2mIrSm: m2(1 — 2)Cm] + irm2eiCmn2(C+ S)+K*[ C, + 62( C)3] = 0,n=0 n=O— 2m11*Orn + m2(1 — 2)Sm] + rm26jSm n2(C + S) = 0,m=0,l,2,•..,N, (3.56)where a and a,, are given by equation (2.34). The initial conditions are as followsC0(0) = C,Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 35Cm(O) = Cm0,Sm(0) Sm0,Cm(0)= C’m0, (3.57)Sm(0) = Sm0,m=l,2,,N.Equations (3.56) together with conditions (3.57) determine the dynamic responses of the string.3.3 SolutionEquations (3.56) are composed of 2N + 1 nonlinear ordinary differential equations. Development of an analytical solution for N 2 can be lengthy, and even impractical. In this analysis,a single-mode solution is first develop by the Butenin method. Then Runge-Kutta numericalintegration [40] is used to solve equations (3.56), and the features which cannot be observedfrom the single-mode solution are demonstrated based on the numerical results.3.3.1 Single-Mode SolutionThe governing equation for the single-mode vibration is(i). Form=0,C0 + —(Co +e2C) = 0. (3.58)27r(ii). Form’0,Cm+2m12*Sm +m2(1 — fZ*2)Cm+ mei(C+SCm)+ Cm+e2)= 0,Sm — 2mVOm +m2(1 — r2)S + me1(S+ C,Sm) = 0. (3.59)The solution of equation (3.58) is a Jacobian Elliptic Cosine and is of little interest here. Thefollowing discussion will be focused on solving equation (3.59). An important characteristicChapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 36of equation (3.59) is that it includes the gyroscopic forces 2mf*Sm and _2mQ*Cm. Butenin’smethod [39] can be applied to solve these types of nonlinear differential equations. To this end,equation (3.59) is rewritten in the form ofCmiSm+?Cm = A(C6iSCm),= AFSi(S+CjSm)1, (3.60)where= _2m*,K*= (1_f2*)+_,= m2(1iV), V14 K*A = —me+—2,4— m4ei/4Si — . (3.61)m4ej/4 + K*e2/7rThe solution of equation (3.60) is assumed in the form of (see Appendix C for detailed Buteninsolution)Cm = a1 sin(K1t + 3) + a2 sin(Kt* + /32),Sm = a1 cos(Kit*+/31)+acos(K:t*+132), (3.62)where 1C and IC2 are natural frequencies of the linear system (3.60) (A = 0) which are obtainedfrom equation (C.153), a1 and a2 are constants determined by equation (C.154), a1 and a2 arevibration amplitude, and and /32 are phase angles. Parameters a1,a2, /3 and /32 are functionsof time t*. Therefore,p == —(a1 sin + a2 sin)3 —S1(aia cos +a2cosi)2(a sin + a2 sini),Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 37C = —S1(S+C )= —81[(aacos +c2acosi)3 + (a1 sin + a2 sin)(cIajcosC+c2a2cos?7)], (3.63)where = KIt +/3 and, = K2t* +132. Thus, the average values ofF and are= 0,F1 = 0,C2 = 0,F2 = 0,C3 = _Si[L(1 +3a)a+ !(1 +3a)aa], (3.64)F3 =C4 = —S[(1 + 3a)a + (1 + 3a)aa2],F4 = —(3 +S1a)a — (3 +S1c)aa2,which lead toda1— U,aTda2——= p1a+p2, (3.65)= qia2+qa1,arwhere33 3P1 =,c 1 3 2 aS1P2 = —-(—aiSi + —a1a25+ — + —),2 2aj 2a1Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 3833 3qi = ——c2S1+ +ic 1 3 2 3 aS1q2 = —---.(—a2S1+ —a2cz15+ — + —). (3.66)n12 2 2a2 2a2Thus, the first two equations of (3.65) yielda1 = constant,a2 = constant. (3.67)This result shows that the vibration amplitudes are independent of time which is apparentlycorrect because the system (3.59) is conservative. Integrating the last two equations of (3.65)leads toi3 = (yia+p2a)r+Bi,/32 = (qia+qa)T B. (3.68)It is clear that i3 and /32 are linear functions of time. Constants a1, a2, B1 and B2 can bedetermined by initial conditions (3.57). Substituting (3.68) into (3.62) and (3.60), condition(3.57) leads toCm =a1sinB+a2,Sm0 = czosB+aCmO =a1frCcosB+a2l,SmO = —cz1aKsinB—a2KsinB. (3.69)So thatfl?(Sm0+a2KCmo)tanBi = •KiA2(aCmo— K2Sm0)fl(Sm0+ ailCiCmo)tanB2= iKaiOmo—KiSmo’Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 39—— 2Cm0 — K2SmO-a1cosB1(aKa)’aiCmo—KiSmoa2 = . (3.70)cosB2(1AC —Accordingly, the solution of system (3.59) is given byCm(t) = a1 sin(wmlt* + B1)+a2sin(wmt*+ B2),Sm(t*) = ajalcos(wmlt* +Bi)+acos(wm +B2), (3.71)whereAn3WmI = K1+ 2(K? K2)(Plai+P2a2)1An3Wm2 = K2+ 2(K?2)(Ia2 + q2a1), (3.72)are the frequencies of free vibration of the system.3.3.2 Discussion of Single-Mode ApproximationIn this section, the closed-form solution (3.71) is compared with the numerical simulation ofequation (3.56) over a range of parameters. The mode number m = 1. The initial conditionsare fixed in the computation, and their values are, Cm = Sm = 1, Cm = Sm0 = 0. Basedon the results of the Butenin method, the characteristics of the nonlinear response of a rotatingcircular string guided by a spring are discussed.3.3.2.1 Effects of Nonlinearity 61The nonlinearity e is caused by the stretching of the string due to the transverse displacement.In the following discussion, the stiffness K* and the nonlinearity 62 are set equal to zero.In Figure 3.11, time histories of the closed-form solution are compared with those ofnumerical simulation for e = 0.3 when the normalized speed * = 0.2. Figure 3.11 indicatesChapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 403J : (a)100 200 300 500 6003...W2(b)100 200 300 400 600Time (t/(pr2/P))Figure 3.11: Time history of C1, & = 0.3, 2* = 0.2, K* = 0.0, e2 = 0.0; (a) numerical, (b)Butenin.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 41that the approximate results are almost the same as the simulation results. Further computationshows that the accuracy of the closed-form solution degrades as e1 increases. The solutioncannot provide accurate results when r1 is greater than 0.6. It is a typical weakness of theButenin method that for higher values of El, the phase errors degrade the solution as timeincreases. It should be pointed out that the accuracy of the solution also depends on the modenumber m. For instance, for m = 2 the closed-form approximation only gives reasonableresults when i < 0.1. Phase error in time history becomes more significant for the highermodes because these modes possess higher natural frequencies. Therefore, satisfactory resultsare only obtained for smaller e1 for the higher modes. However, El is the ratio of the crosssectional area and the square of the radius of the string in this specific problem and should bea very small number. Hence, the closed-form solution derived from the Butenin method isadequate for this work.Figure 3.12 shows the comparison of time history between the analysis and the simulationfor E = 0.3 when the speed = 0.9. The results indicate that the analysis can adequatelypredict the natural response fore1 = 0.3. It is found however that the deviation of the analyticalresults from the numerical computation becomes more visible than that of = 0.2 as shownin Figure 3.11. Further study indicates that the analytical solution presents poorer results athigher speeds for the given time range. The reason for this is that the frequency of the forwardtraveling wave becomes larger as the rotation speed increases. Subsequently, the phase errordecreases the accuracy of the solution for a given Ei and time period.It is known from the linear analysis that the natural frequency of the forward traveling waveis three times of that of the backward traveling wave for all modes, i.e., Wm2 = 3Wml, whenthe rotation speed = 0.5. Since the system possesses a cubic nonlinearity, the phase andthe amplitude may modulate the oscillation, and internal resonance may occur at this speed.Figure 3.13 shows the time history for the system with e1 = 0.2 at = 0.5. A beating-likebehavior develops in the time history, which demonstrates the characteristics of the internalChapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 423Lii 2I : A (a)io 200 400 500 6003W2I : IN\I100 200 400 500 600Time (t/’J(pr2/P))Figure 3.12: Time history of C1, ei = 0.3, 1 = 0.9, K* = 0.0, e2 = 0.0; (a) numerical, (b)Butenin.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 433100 200 •300 600Time (tbl(pr2/P))Figure 3.13: Internal resonance (time history of C1) by Butenin method, e = 0.2, 2 = 0.5,K*= 0.0, 2 = 0.0.resonance. More numerical experiments show that the internal resonance seems to be bounded.It should be mentioned that the Butenin method cannot provide explicit analysis on the internalresonance as was done by Nayfeh [32] through the method of multiple time scale. However, itcan be used to demonstrate the internal resonance phenomenon.The effects of e on the frequencies of the string are shown in Figure 3.14. Note that thefrequencies of both forward and backward traveling waves are increased as the nonlinearity £1increases. An interesting point is that the critical speed of the string is increased when e istaken into account. It can be observed from Figure 3.14 that the critical speed is 1 for e = 0,however, the speed is greater than 1 in the case of 6 > 0. The critical speed is essentiallythe wave speed in the string. It is noted from equation (3.55) that for the nonlinear string thewave speed becomes Ji + 1/47rfw, dO instead of 1 in the corresponding linear system.Therefore, the critical speed should be greater than 1 for e > 0.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 44C,)Ca)Figure 3.14: Effects of e on frequency-speed diagram of mode 2, K* = 0.0, & = 0.0.3.3.2.2 Effects of Nonlinearity 62The parameter 62 comes from the nonlinearity of the spring. The nonlinearity e is set to bezero in the following discussion.The comparison between the closed-form solution and the numerical integration is plottedin Figure 3.15 for K* = 1.0 and 62 = 0.1 at speed f* = 0.2. It shows that the analysis providesexactly the same results as the simulation does. Similar to the effects of e, the accuracy of theapproximation in the given time period decreases as the nonlinearity 62 increases. Computationshows that the analysis cannot present reasonable results for the above time range when 62> 1.The accuracy is also dependent on the mode number. The solution gives poorer response for thehigher modes when keeping the same 62. The stiffness K* should also be taken into accountfor the accuracy discussion. The stiffer spring increases the frequencies, consequently, theaccuracy degrades as stiffness K* increases.The linear natural frequencies are nearly commensurable at the ratio of 1 : 3 at = 0.5Speed (2/S)Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 453£i12(a)100 200 400 500 600(b)0 100 200 300 400 500 600Time (t/-J(pr2/P))Figure 3.15: Time history of C1, E2 = 0.1, V = 0.2, K* = 1.0, e = 0.0; (a) numerical, (b)Butenin.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 463• 100 ‘ 200 400 500 ‘ ‘600Time (t/’(pr2/P))Figure 3.16: Internal resonance (time history of C1) by Butenin method, 62 = 0.5, V = 0.5,K*= 0.2, e = 0.0.when stiffness K* is small. Accordingly, an internal resonance is observed at this speed asshown in Figure 3.16. The internal resonance caused by 62 is essentially the same as that inFigure 3.13 because both of them are generated by the cubic nonlinearity. The two frequenciesbecome less commensurable for larger values of K*, hence, the internal resonance appearsto be less distinct. Further numerical study shows that the internal resonance caused by 62 isstable.Figure 3.17 presents the effects of 62 on the frequency-speed plot. 62 does not change thefrequency of the backward traveling wave when stationary, i.e. V = 0 because the spring isat a nodal point of the mode so that it cannot increase the potential energy of the system. Thecritical speed remains unchanged since 62 does not change the the wave speed in the string.The rest of the frequency of the backward traveling wave is slightly increased by 62. Thefrequencies of the forward traveling waves are increased as 62 increases.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 47U-1.0Figure 3.17: Effects of 62 on frequency-speed diagram of mode 2, K* = 5.0, e1 = 0.0.3.3.3 Multi-Mode SimulationIn this section, numerical integration is used to solve equations (3.56), and the characteristicsof the multi-mode vibrations are subsequently discussed.The effects of e on the dynamics of the string are first examined. To this end, K* and 62are set to be zero. Equation (3.56) shows that Cm and Sm are multiplied by EIJn2(C, + S) inthe nonlinear terms. Therefore, if only one mode is given nonzero initial conditions, equations(3.56) just generate the motion of this mode. As a consequence, the single-mode analysiscan provide satisfactory results if there is only one mode with nonzero initial conditions. Inreality, however, the initial conditions are much more complicated by way of the involvement ofmultiple modes. Thus the single-mode solution cannot appropriately solve the dynamics of sucha system. For the sake of illustration, let all Cm = Sm = 1 and Cm = Sm =0 (m = 0, 1,.. . , N),and compute the cases of N = 1,3 respectively. Figure 3.18 shows the FF1’ spectrum from thetime history of C1 for each case. It is seen that the natural frequencies depend on the initial0.2 0.4 0.6Speed (2/S)0.8Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 480.50.4Eb. 0.3C)U)0.20.1OlioFigure 3.18: FF1’ spectrum of mode 1, = 0.2, e = 0.1, K* = 0.0, 62 = 0.0.conditions used, and the frequencies increase as more initial conditions are applied. Therefore,the multi-mode solution should be applied for the system with complex initial conditions.It is shown in Chapter 2 that an elastically-constrained rotating string may possess complexeigenvalues with positive real parts in the supercritical speed region resulting in a flutterinstability. Figure 3.19 shows the liner response of Co and C1 with K = 1. It is clear thatthe motion grows exponentially. However numerical study shows that the motion of the abovesystem is bounded when the nonlinearity e is taken into account. Figure 3.20 shows thetime history of the same parameters as in Figure 3.19 except that e = 0.05. The response isbounded with a much larger time scale and much smaller amplitude scale compared to Figure3.19. The nonlinearity c changes the mode coupling and restrains the growth of the motion.This observation is very important. It may partiy explain the contradiction in the instability atsupercritical speed predicted by the linear theory to the experimental observation for a guidedcircular saw, i.e., the nonlinearity generated by the mid-plane deformation might limit theincrease of the oscillations. The effects of geometric nonlinearity on flutter instability of anFrequency (oIS)Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 49500250U)00.U)a,cc -250500w 10 20 30 40Time (t/(pr2/P))Figure 3.19: Linear response at flutter speed V = 1.5, K* = 1.0,5025U)0a.Cl)a,CC-25-50Time (IJ’J(pr2/P))Figure 3.20: Nonlinear response at flutter speed = 1.5, K = 1.0, e = 0.05, 2 = 0.0.ClCo=0.0,62 = 0.0.Chapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 50elastically-constrained rotating disk will be studied in Chapter 5. Numerical study shows thatthe spring nonlinearity 62 does not suppress the instability. It should be mentioned that thesingle-mode analysis cannot demonstrate these characteristics because two different modes areassociated with the flutter instability.3.4 SummaryThe nonlinear response of a rotating circular string is modeled to include the deformationnonlinearity e and the guide stiffness nonlinearity 62. Applying the Galerkin method, thegoverning equation is transformed from a nonlinear partial differential equation into a set ofcoupled nonlinear ordinary differential equations. The Butenin method is adopted to developa closed-form solution for the single-mode approximation of the system. Then numericalintegration is used to study the multi-mode vibrations.The study shows that the Butenin method can present a satisfactory analysis for small ei.The accuracy of the solutions is affected by the rotation speed and mode number. Increaseof the speed or use of higher modes degrades the accuracy for a given i• Comparison withthe multi-mode simulation indicates that the single-mode approximation can provide correctresults if only this mode is given nonzero initial conditions.The closed-form solution can give reasonable accuracy for small 62. The accuracy decreaseswhen analyzing the higher modes.String nonlinearity e increases the wave speed, and consequently, the critical speed isincreased. Frequencies are also increased by the presence of ei. Spring nonlinearity 62increases the natural frequencies of the system, but it does not change the critical speed aswould be expected, since a point spring does not change the wave speed.Internal resonance is observed for both e and £2 at speed 1 = 0.5, and appears to be stable.String nonlinearity e the mechanism ofmode coupling for an elastically-constrainedChapter 3. Nonlinear Vibrations of an Elastically-Constrained Rotating String 51string at supercritical speeds, and as a consequence, the corresponding instability in the unear analysis is restrained. Spring nonlinearity e2 cannot restrain the growth of the motion atsupercritical speeds.Chapter 4Effects of Rigid-Body Motions on Vibrations of a Rotating Disk4.1 IntroductionRecently, Yang [41] formulated the equations of motion for a rotating disk constrained by onestationary spring with coupled translational and tilting rigid-body motions, and reported thatthe inclusion of the rigid-body motions can significantly improve the stability characteristics.However, he did not specify which rigid-body motion can reduce the instabilities, and hisobservation is somewhat limited as will be discussed in this study. Chen and Bogy [42]studied the effects of rigid-body tilting on vibrations of a rotating disk with a fixed center point,and indicated that the tilting does not eliminate any instability induced by the stationary loadsystem. In these analyses, explicit expressions for the rigid-body motions were derived, andmodal functions were used to approximate the flexible body displacement.The objectives of this chapter are to verify that the conventional eigenfunction analyses,such as [1,2, 12, 13,44], can properly account for the rigid-body motions of a rotating disk, andto study how the rigid-body motions affect the divergence instability of elastically-constrainedrotating disks. The governing equations of a rigid rotating disk are first formulated, and thedynamics of the rigid disk is discussed. Then the eigenvalues of a flexible rotating disk arecomputed based on the eigenfunction expansion method, and the rigid-body motions of thedisk are compared with those of a rigid disk. Finally, the effects of translational and tiltingrigid-body motions on the divergence instability are analyzed.52Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 53Figure 4.21: Schematic of a rigid rotating disk subject to elastic constraints.4.2 Analysis of a Rigid Rotating DiskIn this section, the dynamics of a rigid rotating disk is studied. Figure 4.21 shows a rigid rotatingdisk subject to stationary elastic constraints. Axes XYZ are fixed in space. The disk can freelyslide along Z axis with displacement WO, and rotate about the axes X and Y with angles ‘çband ? separately. The elastic restraints are modeled by I1 at point (r1 cos 6 r sin 6 0).4.2.1 Equations of MotionThe displacement field at the point {X Y Z}T(= {r cos 6 r sin 6 O}T) is given by’u = r(cosb cos 8 + sin’b sin’b, sin 8),v = rsin8cosb,w = r(sin &,, cos 6 — sin ib cos Ø,, sin 8) + w0. (4.73)‘Formulation of the rigid-body motions is presented in Appendix D.‘TI“KYxChapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 54The angular velocities of the disk about X, Y and Z axes arew =‘bcos’b,= (4.74)= Q+&Siflb.Hence the total kinetic energy of the disk is given byT = + sin b)2 + (b2 + 2 cos2b)] + Mdtho2, (4.75)where I is the moment of inertia with respect to diametrical axis, Md is the mass of the disk.The potential energy isV_ZK:w. (4.76)If the oscillations are small, one hassin’ib, cosb, 1, sinb cos 1. (4.77)Application of Lagrange’s equation, use of relationship (4.77), and neglecting second andhigher order terms, yields,N3Mdz10 + K1w =0, (4.78)N3Ib + 2f2It?,b — Kr1 sin62w = 0, (4.79)N3Itcb—2f2I’çb + K1rcos61w = 0, (4.80)where= r(cOSO:2,by — sin 6i/’) +wo. (4.81)Equations (4.78)—(4.80) are linearized governing equations of a rigid rotating disk subject toelastic constraints. Next, the vibration characteristics of the disk are discussed.Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 554.2.2 CASE I: K, =0For the disk without any constraint, equations (4.78)—(4.80) are simplified aswo=O, (4.82)(4.83)— 2}b =0. (4.84)The solutions of these equations are given bywo(t) = D1t+ D2, (4.85)b(t) = D3 sin(2t) + 134 cos(2ft) +135, (4.86)= —D3cos(22t) +134 sin(2ft) + D6, (4.87)where D1—D6are constants dependent on the initial conditions.This solution shows that the translational mode is uncoupled from the tilting modes, anddoes not oscillate, therefore, the natural frequency of this mode is zero. The displacement ofthe tilting modes is composed of two motions: oscillation with a natural frequency of 2 andthe constant tilting with a natural frequency of zero.Since the system equations (4.82)—(4.84) do not have external elastic constraints, one mayask why the system oscillates and where the restoring force comes from. In order to clarify thispoint, transform the ground-fixed coordinates and b1, into the disk-fixed coordinates ‘P andwhich rotate with the disk at as follows= cos(t) — sin(t) I’X (4.88)I. J sin(1t) +cos(t) ( ‘IJ, JSubstitution of equation (4.88) into (4.83) and (4.84) leads to+ =0,+= 0. (4.89)Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 56302520z150•‘- 10502000Figure 4.22: Frequency-speed diagram of a rigid disk (Disk 1), K1 = K2 = K3 = 5000 N/rn,r1 = = = 0.2 rn, 6 = 0°, 62 = 120°, 83 = 240°.It is clear that in the rotating coordinates the tilting modes are uncoupled and the frequencyof each mode is 2. Although there are no physical elastic constraints to support the disk, thegyration generates an inertial force on the disk, and allows the system to oscillate at frequencyin the rotating coordinates. It should be noted that the transformation does not change thefrequency of the translational mode.4.2.3 CASE II: Effects of Spring ConstraintsFor a disk with elastic constraints, the governing equations (4.78)—(4.80) are all coupled, andit is difficult to get analytical solutions. Hence numerical techniques are applied to solve thesystem eigenvalues. The parameters of the disk are given in Appendix E.Figure 4.22 presents the frequency-speed relationship with a stationary observer for Disk1 supported by three springs which are equally spaced and of the same stiffness. The constantcurve is the frequency of the translational mode, and the two other curves correspond to the0 500 1000 1500Rotation Speed (rpm)Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 57302520I15LL 10500 2000Figure 4.23: Frequency-speed diagram of a rigid disk (Disk 1), K1 = 5000 N/rn,K2 = 3000 N/rn, K3 = 1000 N/rn, r1 = r2 = r3 = 0.2 rn, 6 = 00, 62 = 120°, 83 = 240°.tilting modes. The difference between two frequency curves of the tilting modes is 2l. Itis noted that as the rotation speed increases, the frequency of the backward traveling waveasymptotically approaches to zero, consequently, no critical speed exists for a rigid rotatingdisk, which was also observed by Chen and Bogy [42]. Because of the symmetry of the springsthe frequency curves of different modes intersect each other. Computation shows that all realparts of the eigenvalues are zero, therefore, the system is stable at any speed.The frequencies of the disk with unsymmetric springs are plotted in Figure 4.23. It is notedthat each frequency curve may correspond to one specific mode only when the rotation speed isgreater than 250 rpm. For example, the constant curve corresponds to the translational mode.However, each frequency curve represents a mode combination when the speed is below 250rpm. As the rotation speed increases, the elastic force generated by the gyration overcomes theunsymmetry of the springs. Then the external springs mainly contribute to the stiffness of thetranslational mode, and the gyroscopic force controls the vibration of the tilting modes. Curve500 1000 1500Rotation Speed (rpm)Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 58veering phenomenon is observed in Figure 4.23. It is noted that veering takes place around thespeed of 110 rpm at which the frequency curves cross for the disk with three symmetric springsas shown in Figure 4.22.4.3 Effects of Rigid-Body Motions on a Flexible Disk4.3.1 Modeling of Rigid-Body Motions in Conventional AnalysesIn the existing analyses, such as [1, 2, 12, 13, 44], the displacement field of a constrainedrotating disk is approximated by the following modal expansionNMw(r, 8, t) = {Rmn(T)[Cmn(t) cos(n8) + Smn(t) sin(nO)]}, (4.90)n-.O mOwhere w(r, 8,t) is the displacement in Z direction (transverse) as shown in Figure 4.21,Rmn(r) cos(nS) and Rmn() sin(n8) are the mode shapes of an unconstrained disk, and Cmn(t)and Smn(t) are time functions to be determined. m is the number of nodal circles, and n is thenumber of nodal diameters. Following the conventional notation, (m,n) hereinafter is used torepresent the vibration mode of m nodal circles and ii nodal diameters of the disk. Rmn(r) isusually approximated by Bessel functions, or by some polynomial functions. The analysis [12]assumed that Rmn(r) is a polynomial in the form of4Rmn(r) = Erm, (4 91)where are constants which are determined by the boundary conditionsWI(R,9,t) = 0,W,,. I(R,6,t) = 0,.M,.,. 1n,e,t = 0,1 1(N77w,,.+Nr6W,8+Qrr + M78,8)I(R,8,t) = 0,r rR=Rj or R0; (4.92)Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 59NI>.0Ca,LI1000Figure 4.24: Comparison of frequency-speed diagram between a rigid disk (Disk 1) and asplined saw (Disk 3) without constraints; L: rigid disk, splined saw.and the normalizing conditionRmn(Ro) = 1, (4.93)where 14, and Ri are the outer and inner radius of the disk. Rmn given by (4.91) is used inthis chapter. It should be noted that all the above-mentioned analyses with modal expansionmethod can present similar linear results although different Rnrn(r) is used.The frequency-speed diagram of an unconstrained flexible rotating disk with free-freeedges, such as an unguided splined saw, is compared with that of a rigid disk in Figure 4.24.The frequencies of mode (0,0) and mode (0,1) in the flexible disk coincide with those of thetranslational mode and the tilting mode in the rigid disk. Hence, the rigid-body motions of anunconstrained flexible disk with free-free edges are modeled by mode (0,0) and mode (0,1) inequation (4.90). Obviously, mode (0,0) models the translation w0 and mode (0,1) models therotations andThe existence of the rigid-body motions in a free-free disk is dependent on the number of0 200 400 600 800Rotation Speed (rpm)Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 60external constraints. The disk has three rigid-body motions if it is subjected to no constraints, asdiscussed previously. Adding a constraint to the system will eliminate one rigid-body motion.The system has no rigid-body motion if the disk is guided by more than two constraints.However, a rigid disk always has three rigid-body motions no matter how many constraintsit is subjected to. Figure 4.25 illustrates the frequency-speed relationship for a free-free diskwith a different number of springs. It is noted that the natural frequencies of mode (0,0) andmode (0,1) of the flexible disk have similar tendency as those of the rigid disk, especially athigh speeds. This observation indicates that the rigid-body motions in a flexible disk subject toconstraints are modeled by mode (0,0) and mode (0,1). However, the vibration characteristicsof mode (0,0) and mode (0,1) in the flexible disk are quite different from those of the motions inthe rigid disk. First, the frequencies are quite different at low speeds. Second, the eigenvaluesof mode (0,0) and mode (0,1) may become complex with positive real parts at supercriticalspeeds, but those of the rigid disk are always pure imaginary. Therefore the motions in arigid disk are not exactly the same as mode (0,0) and mode (0,1) in a flexible disk subject toconstraints.The eigenvalue-speed diagram for a free-free disk with one point spring, which is the sameas that in Figure 3 of Yang [41], is presented in Figure 4.26. Comparison between Figure 4.26and Figure 3 of Yang [41] shows that the conventional modal analysis can present the sameresults as in [41]. Consequently, the analysis [41] is essentially the same as the conventionalanalyses, and it appears that it is not necessary to model rigid-body motion and flexible bodydeformation separately as done by Yang [41], and Chen and Bogy [42]. It is noted that the firstflutter instability speed 6.2 given in [41] is erroneous. It is shown in Figure 4.26 that the flutterinstability occurs about the speed of 5.4. It is believed that analysis [41] can present the sameresult if smaller speed incremental is used in the computation.The conclusion that the coupling between the rigid-body motions and flexible body deformation in a constrained free-free disk eliminates the divergence instability [411 needs to beChapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 61NI>%C.)(a)00)LLNI>%C)0)0LLNI>0C0,0)IU-2000Rotation Speed (rpm)Figure 4.25: Comparison of frequency-speed diagram between a rigid disk (Disk 1) anda splined saw (Disk 3); (a) one spring, K = 10000 N/rn, r = 0.2 rn, 8 = 00; (b) twosprings, K1 = K2 = 10000 N/rn, r1 = r2 = 0.2 m, 81 = 0, 9 = 120°; (c) three springs,K1 = K2 = K3 = 10000 N/rn, r = = 1’3 = 0.2 rn, 81 = 00, 82 = 120°, 83 = 240°; LS: rigiddisk, —: splined saw.0 500 1000 1500Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 6220Q15I105Li02Q0.0 2.5 5.0 7.5 10.0Speed (3tI/3O/’J(D/phR))Figure 4.26: Eigenvalue-speed diagram of a flexible disk (Disk 2) subject to one spring,K = 210142 N/rn, r = 0.305 rn, 8 = 2700.clarified. For the disk guided by one spring, the divergence instability indeed disappears at thefirst critical speed. However, one cannot draw the conclusion from this result that the rigid-bodymotions can eliminate the divergence instability of a constrained disk in the supercritical speedregion. Figure 4.27 plots the eigenvalues as a function of speed for the same disk subject totwo springs. It is seen that divergence instability occurs at first critical speed. One rigid-bodymotion (backward traveling wave of mode (0,1)) exists in this case, but it does not improvethe stability characteristics. Moreover, one spring guide used in the discussion is not practical,such as in the case of splined guided saws, hence the result is of limited use.4.3.2 Effects of Rigid-Body MotionsAn interesting aspect of the work [41] is that it indicated that rigid-body motions can affectthe divergence instability of the disk, although it did not specify which rigid-body motion ismore influential. From the practical point of view, it is of interest to find out which rigid-bodyChapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 63201510.0 2.5 5.0 7.5 10.0Speed (7tcv30/-.I(D/phR))Figure 4.27: Eigenvalue-speed diagram of a flexible disk (Disk 2) subject to two springs,K1 = K2 = 210142 N/rn, r1 = r2 = 0.305 rn, 0 = 270°, 02 = 30°.motion changes the stability characteristics of a constrained rotating disk.It is known from the preceding discussion that modes (0,0) and (0,1) correspond to thetranslational and tilting rigid-body motions respectively. Hence, removal of mode (0,0) ormode (0,1) from the modal analysis is equivalent to the elimination of the corresponding rigid-body motion. In Figure 4.28, the eigenvalue-speed diagram of a constrained disk with free-freeedges, which is the same as the one used in Figure 4.26, is calculated without mode (0,0).Because of the absence of mode (0,0), the frequencies below 2 in Figure 4.28 are differentfrom those in Figure 4.26. But the critical speed (4.2) of the disk is the same as that in Figure4.26. The real parts of the eigenvalues above the first critical speed in Figure 4.28 are positive,hence the divergence instability occurs. The numerical observation indicates that restrictionof the translational rigid-body motion can cause divergence instability. The eigenvalue-speeddiagram without mode (0,1) is shown in Figure 4.29. Comparison with Figure 4.26 showsthat the frequencies below 2 in Figure 4.29 are different, and the numerical results also showChapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk20Q.15ICaa)a)105210—1-20.0 2.5 5.0Speed (,t2130/’J(D/phR))7.5 10.01564Figure 4.28: Effects of removal of mode (0,0) on the eigenvalue-speed diagram of a flexibledisk (Disk 2) subject to one spring, K = 210142 N/rn, r = 0.305 m, 8 = 2700.20Q110I5210—1-20.0 2.5 5.0Speed (,rl3O/’J(D/phR))7.5 10.0Figure 4.29: Effects of removal of mode (0,1) on the eigenvalue-speed diagram of a flexibledisk (Disk 2) subject to one spring, K = 210142 N/rn, r = 0.305 m, 8 = 270°.oQpQChapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 650.200.10 ---V0.00---0.10-0.200.995 1.000 1.005 1.010 1.015 1.020crFigure 4.30: Real parts of the eigenvalues of a splined saw (Disk 3) with finite guides, guidedimension: 0.152 m x 0.14 m, stiffness: 4.23 x i0 N/rn/rn2,—: with free-free edges, - -:with grooved restraint.that there is no zero frequency mode which corresponds to the backward wave of mode (0,1).Inspection of the real parts of the eigenvalues indicates that there is no divergence instabilityabove the first critical speed, therefore restriction of the tilting rigid-body motion does not causethe divergence instability. It can be concluded that only the coupling between the translationalrigid-body motion and the flexible body deformation reduces the divergence instability, whilethe tilting rigid-body motion does not change the divergence characteristics. This result isconsistent with the observation of Chen and Bogy [42].The application of this result to splined guided saws is that adequate translational rigid-bodymotion should be allowed in order to reduce the divergence instability at supercritical speeds.The splined saws used in sawmills are constrained by a pair of guides which have relativelylarge contact area with the blades. The guides reduce the translational rigid-body motion tosome extent, as a result, the divergence instability can occur at supercritical speeds. Figure 4.30presents the real parts of the eigenvalues as a function of speed for a splined saw with free-freeChapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 66edges restrained by finite guides which are modeled as distributed springs. The plot shows thatthe divergence instability takes place at the critical speed although it is not significant. It isoften the case in the operation of splined saws that a groove is developed on the spline arbor dueto wear, and restricts the sliding of the inner edge of the saw blade. The boundary conditionsof the saw can then be modeled as simply-supported and free edges. The real parts of theeigenvalues of a splined guided saw with a grooved restraint are also presented in Figure 4.30.Comparison shows that the grooved restraint increases the unstable region, consequently thegrooved spline arbor should be avoided in the operation of splined guided saws at supercriticalspeeds.4.3.3 ExperimentA splined saw (Disk 3) without any constraint was tested in the laboratory in order to observe theexistence of the rigid-body motions. An electromagnet is applied to excite the sawbiade withpseudo random signals. A proximity transducer is used to measure the displacement. B&K2034 spectrum analyzer is then used to obtain the transfer function of the vibration response ofthe sawbiade. Figure 4.31 presents both experimental and numerical frequencies as a functionof rotation speed. The tilting rigid-body motion (forward traveling wave of mode (0,1)) of thedisk is clearly observed in the experiment. Therefore, a splined saw does possess rigid-bodymotions if it is unconstrained. The experiment also shows that the conventional analyses canpresent accurate results in predicting the dynamics of unguided circular saws. It should bementioned that it is very hard to get measurements below 200 rpm because the sawbiade doesnot have enough stiffness to resist the attraction of the electromagnet at low speeds and tendsto contact the electromagnet. The measurement also becomes difficult above 800 rpm due tothe increasing vibrations of the test rig and aerodynamic effects, thus, some test data are notconsistent with the analytical results.Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 675040N30>..C.)Ca,0•U-1000 200 400 600 800 1000Rotation Speed (rpm)Figure 4.31: Comparison of frequency-speed diagram between experiment and analysis of asplined saw (Disk 3) without constraints; o: experiment, —: analysis.4.4 SummaryWhen a rigid rotating disk is unconstrained, it has two zero natural frequencies correspondingto the translational and the tilting rigid-body motions, and one frequency of 212 correspondingto the tilting rigid-body motion. Observation in the coordinate system rotating at 12 shows thatthe tilting mode oscillates at frequency of 12. The gyroscopic force acts as an elastic constraintto the tilting mode and cause it to vibrate at frequency of 212.When a rigid rotating disk is subjected to unsymmetric elastic springs, “curve veering”phenomenon is observed in the frequency-speed diagram. Each frequency curve in the veeringregion represents a mode combination. At high speeds, the vibration of the tilting modes iscontrolled by the gyroscopic force, and the oscillation of the translation mode is dominated bythe external springs.The rigid-body motions of an unconstrained flexible rotating disk with free-free edges areobserved in the frequency-speed diagram computed from the conventional modal analyses.Chapter 4. Effects of Rigid-Body Motions on Vibrations of a Rotating Disk 68The rigid-body motions are correctly modeled by the displacements of mode (0,0) and mode(0,1). When a flexible disk is subjected to elastic constraints, mode (0,0) and mode (0,1) arenot equivalent to the motions of a rigid disk.The coupling between the translational rigid-body motion and the flexible body deformationcan reduce the divergence instability of an elastically-constrained rotating disk at supercriticalspeeds, but the tilting rigid-body motion does not change the stability characteristics.A spline arbor with worn groove should be avoided in order to achieve stable operation ofsplined guided saws at supercritical speeds.The experiment on an unconstrained splined saw verifies the existence of the rigid-bodymotions, and proves the validity of the conventional analyses for unguided rotating disks.Chapter 5Nonlinear Vibrations of an Elastically-Constrained Rotating Disk5.1 IntroductionLarge-amplitude vibrations of highly flexible rotating disks have been observed in experimentalwork [4, 24]. The analysis on the nonlinear vibrations of an elastically-constrained rotatingstring presented in Chapter 3 shows that the geometric nonlinearity can restrain the flutterinstability at supercritical speeds. It was reported that in panel flutter problem nonlinear effects(midplane stretching) generally restrain the unstable motion to a bounded limit cycle, especiallyfor supersonic airfiows [32]. Therefore, the effects of the geometric nonlinearity of a rotatingdisk may be important in determining the stability characteristics in the supercritical speedregion.The nonlinear vibrations of an unconstrained rotating disk have been studied in the past.The first formulation of the nonlinear vibrations of a rotating disk using von Karman fieldequations was presented by Nowinski [45], and the vibration of a rotating membrane associatedwith mode (0,2) was studied in detail. Nowinski [46] also studied the nonlinear vibrations of anelastic disk rotating in a viscous fluid, and analyzed the hydrodynamic aspects of the problem.Young [47] studied the nonlinear transverse vibrations and stability of rotating disks withnonconstant spinning rate. The analysis of nonlinear oscillations of imperfect circular diskscan be found in references [48]—[5 1]. All of these studies dealt with rotating disks withoutconstraints: Moreover, only a single-mode vibration was discussed in these analyses due to themathematical complexity of the problem. Flutter instability involves two different vibration69Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 70modes, accordingly, multiple modes have to be included in the nonlinear analysis in order tostudy the stability characteristics.In this chapter, the effects of geometric nonlinearity on the vibrations of an elastically-constrained rotating disk are studied. The equations ofmotion, which are two coupled nonlinearpartial differential equations corresponding to transverse force equilibrium and to deformationcompatibility, are first formulated by using von Karman thin plate theory. Then the stress function is analytically solved from the compatibility equation by assuming a multi-mode transversedisplacement field. Galerkin’s method is applied to transform the force equilibrium equationinto a set of coupled nonlinear ordinary differential equations in terms of time functions, andthe Runge-Kutta algorithm of Chapter 3 is used to solve these equations numerically. Finally,numerical examples are presented to verify the accuracy of the developed analysis and toillustrate the nonlinear response characteristics of a rotating disk.5.2 Equations of MotionThe space-fixed coordinate system (r, 8, z) is used in the following development. The disk isin the plane defined by the polar coordinate (r, 8) The assumptions, which are used to definethe problem and the range of validity of the analysis, are as follows:1. Tractions on surfaces parallel to the middle plane of the disk are negligibly small compared with the inplane stresses, and the inplane displacements are linear functions ofcoordinate z [52]—[54];2. The linear strains are much smaller than the angles of rotation’ which are small comparedto unity [52]—[54];3. The angle of rotation about z axis is much smaller than those about r and 8 axes [52]—[54];‘Definitions of linear strains and angles of rotation are given on page 6 in reference [54].Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 714. The normal stress in z direction is much smaller than those in r and 6 directions (planestress) [52]—[54];5. The inpiane inertia terms are negligibly small [32].Application of assumption 1 leads to the following displacement field in the disk€(r, 6, z, t) = u(r, 6, t) — zw,,.£i(r,6,z,t)=v(r,6,t)— W,8, (5.94)where u, v and w are respectively the radial, tangential and transverse displacements in themiddle plane of the disk. And assumptions 2 and 3 lead to the strain-displacement relationshipin the form of— 4 -n -E. — 6,.,. + 6,.,. + 6,.,.,= + + E, (5.95)4 -n -= 6e + E,. + E,.,where the linear membrane strains-1 —6,.,.—U,,.= (u+v,), (5.96)=the nonlinear membrane strains== (5.97)-n 1= W,,. W,Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 72and the bending strains€77. = —ZW,,.,.o=_(r’w,,. +w,), (5.98)= —-—(rw,,. —‘w,o).7.The stress-strain relationship is given by (assumption 4)Ea,.,.— 2,E80),1—v= E+ ye,.,.), (5.99)1—v=where E and ii are Young’s modulus and Poisson’s ratio, and G = E/[2(l + v)].The potential energy of the disk is2,r R, h/2 1U= 10 JR {1_h,2 [2 + + a,.9€,.8)— prfl2u] dz — Qw} rdrd8N3— / (—K1wjdw, (5.100)where p is the density of the disk, is the rotation speed, K. is the stiffness of the ith elasticconstraint, Q is the normal force on the surface of the disk, h is the thickness, and R0 (Ri) isouter (inner) radius.The kinetic energy is2ir R0 h/2 1= 1 L, 1 h/2 [(W,t +f2w, )2 + (U,t )2+(‘v,t +flv, -i-rf)2]rdzdrd6. (5.101)Substitution of equations (5.100) and (5.101) into Hamilton equation (2.1), integrating, andregrouping, yields the following governing equations and boundary conditions:Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 73transverse displacementgoverning equationN,DV4w + ph(w,tt +2fw,e+12w,) + KwS(6 — 61)S(r — r2) — Q“rNrrW,r +Nr9W,6),r —--(Ne9w,e+rNrsw,r ),e = 0, (5.102)boundary conditionsW(R,6,t) = 0,W,,. I(R,9,t) = 0,M,.r (R,9,t) = 0,1 1(NrrW,r +Nr9W, +Qrr + )I(R,9,t) = 0,r rR=Rj or 1?; (5.103)radial displacementgoverning equation -Nrr,r +N68+(Nrr — N96) + phrfl2 = 0, (5.104)boundary conditionsN,.,. I(R,8,t) = 0,UI(R,9,t) = 0,R=Rj or R0; (5.105)tangential displacementgoverning equationN,.9,,.+-N96+N,. 0, (5.106)Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 74boundary conditionsNr8I(R,e,e) = 0,VI(R,9,t) = 0,R=R or (5.107)where the inpiane forcesN71. = J 77dZ= 1_v2V12= 1 _2{(U,r+w,r)+_(U+v,o+_w,e)j,N99= j cj99dz == —2 [—(U + V, +w,9) + v(u,7+w,7)] , (5.108)h/2N79= J-h/2= Gh(9+)= GhTthe bending moments and transverse force= —D +-(w, -i-rw,,. )j= —D +—.(w,99 +T’W,7 )]M79=D(1 z’)(rw,7—w,), (5.109)Q7... — — ( w),,.and V2 is harmonic operator, and D = Eh3/[12(1 —Equations (5.102)—(5. 107) describe the nonlinear vibrations of an elastically-constrainedrotating disk in terms of three displacements u, v and w. It is noted that these equations aresimilar to those of a stationary disk presented in [54] except the terms associated with rotationand spring K. The motion of the disk is determined once the three displacements areChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 75solved from equations (5.1 02)—(5. 107). However, solving these three coupled nonlinear partialdifferential equations is very difficult. It is more convenient to utilize an inplane stress functionto reduce the number of governing equations from three to two. By doing this, the analytical andcomputational work is substantially reduced as reported in [56]. The stress function approachwill be adopted in this study. To this end, the inpiane forces are written in terms of a stressfunction as followsN,.,. = 4),,. +-,99 _phcl2r2,N98 = h,,.,. _phfZ2r2, (5.110)N,.8 =Then equations (5.104) and (5.106) are automatically satisfied by N,.,., N99 and N,.9 defined in(5.110). The condition of compatibility,—(r48),,.s+(r29,,. ),,. —r.,.,,. = 0 (5.111)is then used to derive the second equation. Substitution of equations (5.95)—(5.98) into equation(5.111) leads toe,.,., —(rc,.9),,.+ r29),,. —r,.,.,,.= (w,,.9 _w,)2 — (5.112)By using relationships (5.99), (5.108) and (5.110), equation (5.112) is rewritten in terms ofstress function as followsV4 — 2pf2(1—v) = E4- [(W,i.ø -_—W, )2—w,,.,. (rw,,. +w,99)] . (5.113)Equations (5.102) and (5.113) are the governing equations for transverse displacement w andstress function . Comparison shows that equations (5.102) and (5.113) are the same as thoseof an unconstrained rotating disk presented in [45] if the elastic constraint is not included in theChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 76current formulation. Once w and 4> are solved, the displacements u and v can be determinedas follows= f {, [-‘r +-4>,oe —v4>,- _pfl2r2(1 — v)] — } di’, (5.114)v =j{ [—v(,7+—4,6s ) + 4>,,.,. _pf22r2(l — v)] — u — } dS. (5.115)Although the stress function method has some advantages over the displacement equationmethod, such as less analytical and computational work involved, the drawbacks of this methodshould also be addressed. First, the boundary conditions associated the inplane displacementsu and v become more complicated because the integration of 4> is required to calculate uand v as shown in equations (5.114) and (5.115). Second, the inplane displacements cannot be accurately evaluated by the stress function method because the inplane inertia termsin displacement equations (5.104) and (5.106) have to be neglected in order to make stressfunction 4> satisfy these two equations.Next the equations of motion (5.102) and (5.113) will be nondimensionalized, so that theycan be solved in a general fashion, by introduction of the following nondimensional parameters:= r/Ro, w = w/(h/), t =ci. = 4> = 4>/ç,), K = K/(), Q = Q/(/),N,.*,. =N7/(/), N;8= Nso/(&), N,.9 =M:r = Mg6 =M88/(/), M,9 = Me/(iJ),= Qrr/(/) s = 6/(), E = 12(1 — ,2), 6=jJ.Applying these parameters, governing equations (5.102) and (5.113) and boundary conditions(5.103), (5.105) and (5.107) are rewritten as followsChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 77transverse displacementgoverning equationN,-V4w* + (W*,t.t* +21Z*w* +cr2W,99 ) + K:w* 5* (6 — 6:)S*(r* — r)—Q + r 2ç. 2(!v2W* + W*,r* ) — ....._ [*,r*r* (T*4)*,r +4’,% ) (5.116)+*,r*r* (r*w*,r. +w*,ss) — 2r* 2(!W*,9),r (__4*, ),r* j 0,boundary conditionsl3Lr*r* (1,6,t) U,(Q*r* + —lI1.s)I(I,e,t.) =•Mr*r (.,8,t) = 0, (5.117)(Q7 + .M*e)I(et*) = 0;stress functiongoverning equation—-_2 2(1 — z.’) = E*e__j {(w,r.8 __w*,9 )2 — W*,7*r* (rw,r* +w*,eo )] , (5.118)boundary conditionsN*r (1,8,t) = +_- 4,*66 2r 2)I(1,9,t) = 0,Wr*91(l,B,t ) = _(.....cI*,9),r (1,8,t) = 0,N:.r. IGR,8,V) =4,* r 2)I(R,o,t*) = 0, (5.119)lV:.9I(A,o,t*) = _(__4*,e ),r I(,8,t) = 0;where t = R/R0. This study will be focused on the stability characteristics of a splined guidedsaw, and boundary conditions (5.117) and (5.119) are associated with annular disks which haveChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 78a free inner edge. It should be mentioned that the nonlinear term N,*r*W*,r* +N.8w’,/r inthe transverse shear force in equation (5.103) is neglected because N.7. and N, vanish alongthe inner and outer edges (boundary condition (5.119)).Equations (5.116) and (5.118) are coupled nonlinear partial differential equations, and theexact solutions have not been reported. As a result, an approximate solution will be developedin this study. The procedure adopted to solve equations (5.116) and (5.118) is first to assumethe transverse displacement w in the form ofNMw*(r*, 6, t) = > {R,nn(r*)[Cmn(t*) cos(nS) + Smn(t) sin(ri6)}}, (5.120)n=O rn—UwhereRmn(r) = Er mt (5.121)and Crnn(t) and Smn(t) are time functions to be determined later. The stress function 4’ isdetermined in terms of Cmn(t*) and Smn(t) from equation (5.118) by applying the assumed w.Finally, the governing equations for Cmn(t*) and Srnn(t) are developed from equation (5.116)by applying Galerkin’s method. It should be mentioned that the radial modal function Rmn(r*)(5.121) is different from equation (4.91). The reason for choosing this form is discussed inAppendix F.5.3 Stress Function SolutionThe transverse displacement w is approximated by expression (5.120). Similar to the linearproblem, the coefficients in R,, are determined by the boundary conditions (5.117)together with the normalizing condition Rmn(1) = 1.Substitution of w (5.120) into equation (5.118), and regrouping the terms according to thetrigonometric functions, yieldsM M N N 4 4= 2i2 2(1—v)+ E*e E1D [i3rnnpqAnpq cos(n + q)6rn=O p=O n=O q=O iO 3=0Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 79cos(n — q)8 +7mnA sin(n + q)8 (5.122)+8 B”mnpq mnpq5hll(fl — q)O] rwherea,nnpq = (CmnCpq+SmnSpq),I3mnpq = ( CmnCpq + SmnSpq),7mnpq = (CmnSpq+SmnCpq), (5.123)8mnpq = (CmnSpq+SmnCpq),= (q+p+j— 1) [nq(nIm.Ii_1)+(n+m+i_n2)(q.ip+j\l E’ E’mn pq’= (q+p+j_1)[nq(n+m+i_1)_(n+m+i_n2)(q+p+j)]EE.mnpqThe homogeneous solution of 4) is of the form ([52, 55])= cOIT*2+cOlnr*4)Ic= (ch1cr*3+c12c_)cosO,4)18 = (ch18r*3+c128_)sin8, (5.124)ckc = (ccr* k+2 + ck2r* k + ck3r* —k+2 +c1r_k) cos kS,ks = (cklsr* k+2 + ck2sr* k + ck3ar* —k+2 + ck4sr* _k) sin kO,k 2,where c01,c02,. . . , are constants which will be determined from the inpiane force boundarycondition (5.119). The particular solutions are given by1 MMNN44= —(1 — v)Q*2r*4+ Ee_____________m=O p=O n=O q=O i=O =OIn±qI=OM M N N 4 4= Ee > •(r)T’ .(t*) cos k8, (5.125)mnpq:j ‘ mnpqm=O p=O n=O qfJ i=O j=Ok=n±qI> IChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 80M M N N 4 4= E*e 4k8 (r*Tc8.(t’) sin k6,“ / mnpqçjm0 pO nO q=O i=O j=O k=n±q> 1whereif x=1,if x=3,otherwise,if x=0,if x=2,if x=4,otherwise,I 1n2r if x=OO (re) = r*21n2* if x=2mnpqtj [ r X otherwise,I r1nrmnpqij(r*)= I r31nr1c (r*) = 4,18mnpq:3 [I 1nr,2c (r*) = 2smnpq:3mnpqij(7*) =I r1nr[= J k In r if x = k,4,ks * r* k+2 in r* if x = k + 2,mnpqij [ r X otherwise,,kc (* —mnpq:r710 (4* —.L mnpqt3V’plc (* —mnpq23(5.126)+ OmnpqBnpq)(/3mnpqAnpq +a .Ahj 1.fr’mnpq mnpq+‘—mnpq mnpqA 23 t13 13çmapq + I_lmnpq çmnpq + ?7mnpq— (/9mnpqApq +/3mnpqAnpq + a13,)mnpqA,npqA 3 13 23çmnpq + mnpq çmnpq + ?mnpqif x=O,if x=2otherwise,if x=l,if x=3,otherwise,Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 81I - [7mnpq4npq + sign(n — q)S 13i 1 if x = 1,mnpq mnpqjjg [ymnpqAnpq + sign(n — q)5mB )1mnpqjI 7mnpqAnpq + SmnpqB_____________mapz otherwise,sign(n— q)..I npq + Lnpq Emnpq ÷I (8mnpqAnpq + 0mnnpq) X =0,if X2,18mnAnpq+amnpqBnpq) 1’ x=4,I 3mnpq4nn + otherwise,.ij iiI. tmnpq + LSmnpq mnpq + ?lmnpqI [7mnpqAnpq + sign(n — q)Smn,j3” 1 =0,mnpqj_j {mnpqAnpq + sign(n — q)S B” 1 if x = 2,mnpq mnpqJ1 [7mnpqAnpq + sign(n — q)SB” 1 if x =4,mnpq]I 7m Anpq+. ‘5mnpql3”_____mnpqsgn(n— q),otherwise,I npq + ZSnpq nnpq +( I3mnpqApq + amnpqB if x =I — 8k(k—1)I3mnpqAnpq + mnpqBnpq8k(k+1) if =k+2,_ _a,flflp(l3mn qotherwise,‘rnpq + L4npq + C +( ‘y,nnpqA{npq + .sign(n —I 8k(k—1)I 7mnpqAnpq + sign(n — q)SmnBI 8k(k÷1)I 7mnpqAnpq + . SmnpqB’mnpq.szgn(n— q) ,I nnpq + Znpq Emnpq +x{(x— l)[(x— i)(x—2)— 1]÷1},—(n— q)2 {2X(X — 2) + 4 — (n — q)2j,—(n + q)2 [2X(X — 2) + 4 — (n +pIs (4* —£ mnpq:jT2C (4* —mnpq:3\T2 (t —mnpq:3”pkc (4* —£ mnpqljrpks (4*£ mnpqijV’mapqTlmnpqmnpqif Xk,if =k+2,otherwise,Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 821+i if x>O,s’Lgn(x) =[—1 if x<O,x = m+n+p+q+i+j,k>l.It is seen from equation (5.126) that particular solution cImnj has different expressionsdepending on the value of x. The fundamental particular solution is r* X which can be used formost values of. However, r X is either the same as the homogeneous term, or does not satisfyequation (5.122) when x is of some specific values. Take as an example. r*O is nota solution of equation (5.122), and r*2 has already been contained in cOlr*2 of homogeneoussolution (5.124). Accordingly, the solutions which are different fromr* X need to be developed.The complete stress function is given by*(r*8t*) = (r*,t*) + O(r*,t*) +> [kc(r*,8,t*) + kc(r*, 8,t*)+8(r*, 8, t*) + i1c8(r*, 8 j*)j (5.127)The coefficients in the homogeneous part of the stress function (5.124) will next be determined from the boundary condition (5.119). Substituting (5.127) into (5.119), and isolating theterms with the same trigonometric functions, leads tok=o+ O),r* __!r* ç2 = 0,2eat r = R and r* = 1. (5.128)k = 1 and cos 0+ 4)IC),r* + (41c + 4)Ic),99 =at r* = R and r* = 1; (5.129)Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 83or(__c1,8+_I,8 = 0,at r = F and r* = 1. (5.130)Inspection shows that equations (5.129) and (5.130) are identical, hence, either of them can beused to solve the unknown coefficients.k = i and sin 8+ 4I3), +(4” + 418), = 0,at r = 1? and r* = 1. (5.131)Ic 2and cos 84Icc+ +(4’ + i’80 = 0,(±IC,e 0at r* = 1? and r = 1. (5.132)Ic 2andsin8+ ‘),r* + ka),69 = 0,(..L41c,0 ),r 0,at r* = . and r* = 1. (5.133)Solving equations (5.128)—(5.133) yields the homogeneous part of the stress function in theform of=16& 8eM M N N 4 4_______________m=O p=O n=O q=O i=O j0 In±qOChapter 5 Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 84MMNN 44kc= E kc cos kO, (5.134)mapq:j ‘ mnpqm=Op=On=Oq=0 i=0 j=0’k=In±qI> IM M N N 4 4ks= Ee ks .(r)T .(t”) sin kS,mnpq:j TflflP23m0p=O n=0 q0 :0 j=0 S_______________________k=In±q> Iwhere { 011 *2amnjr + in r if x =0,4,0 (i’) = 021 *2mnpqij amnpqtjT + in r if x = 2,03! *2amnjr + in r otherwise,I a if x 1,1c (re) = 1s ..(r*=J a121 if x=3mnpqj mnpq:j ‘/mnpqija131 •,‘ + otherwise,mnpqtja201 •r + a202 r 2mnpqt3 mnpqt+ a204 •r -2 if x = 0,mnpqj mnpqta21 •r4+a212 ••r2mnpqt mnpqlj214 *—22c (re) = ,2a= + amnjr if x = 2,mnpq:3\ / (5.135)- mnpqtja221 •?‘ +a222 •r 2mnpq:j mnpq:+a223 + a2 •r 2 if x = 4,mnpqt mnpq:231 4+a232 •r2amnjr mnpq+a233 + a234 •r 2 otherwise,mnpqij mnpqtI ku *k+2+akl .r*ki mnpqj+akl3 •r —k+2 + aId4 •r —k if x = k,mnpqj mnpqijk21 *k+2+ak2 rkkc (re) = (re)= amnjr mnpqtjmnpq:2mnpqtj+ak23 •r k+2 +a1c24 •r —k if x = k + 2,mnpq:3 mnpqijk31 * k+2 + a1’32 •r kamnjjr+alC33•*—k+2+ ak34 r —k otherwise;mnpqtj mnpqijChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 85andmRamnj=—1’2lnR012________amn,=— 1’021 R21nR(lnR+ 1)=—_______2RInR(lnR+ 1)022____ _ ______ _ _amnj= 2_1031 Xamnpqtj= 2(R — 1)(1? — 1),032 —amnj= (R2— 1)(R — 1),2inapqjj=R21n112___amnj=1—4(21n+ 1)121_am,pqj,= 2(R4 — 1)122 R1nRamnj =—— 1’131 X1= 2(. — (R —132 X1amnpquj= 2(.— 1)(R —1),201amn, = 2 [(p2 —R2)lnR — (R — R_1)2] /2,202=— [4(R4 —R2)lnR —3i4+3R2— 3_2 +3] /2,203amnpquj = [(8R4 — 6.2 2k)in R — — 2R4 + 6R2 — -2 — 2] /2,204amnj = [4(_R4+.2)1nkuamnzj = [2k(. — 1)(R + 1)(k — 1) in j — —2k÷2 + 2 — 2k + i} /1x,,k12amnjj = [(2k2 + 2R — 2kR — 2R2)ln(R) — 2R + kR4 — 2kR+k + —2k+2 + 2k+2] /‘k, (5.136)k13amnj = [2k(W’ — 1) in R + (k + 1)(R2k+ 2 — 2k+2 — 1)] /Ak,Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 86k 14amj = [2(_.R2k+2 + — 1)in R — + — + 2k)] /tXk,k21am,= [(2k+2+ 2R + 2kk — 2kR)in — — -2k+2+2R + k— 2k + kk9 /Ikk,k22amni,q = [_2k.2(.R — 1)(R + 1)(k + 1)1n — + — + -2k÷2] /k,k23amny,qt,= [_2(_2k+2+ + 1) In 1? + k(—R2’+ 2k+2 + — 1)] /L1k,k24amnjj = [2k(_.2k+2 + 4) in]? — (R22 + — — ]?2k+4)(k — 1)] /L,k3 1amnj = 1 + k)(k + x) — ]?_2k+(k + x) — ]?x(_ 1 + k)(k + x)+ x — 2) ]?k+x(_k + x) + k(k + x — 2)] /‘k,k32amnpqjj = [kkk + x) — R2(k + 1)(k + x — 2) — — x +2)+]?‘xkk + x) — ]?_k+x-I-2( + 1)(k + x — 2) — ]?k+x+2(— x + 2)] /k,k33amnpqj, = {2(k + 1)(—k + x) — ]?—k+x(k + x) + Rk÷x(k + 1)(—k + x)_]?k+2+xk(_k +y — 2) — ]?2k+(k+x) — k(—k + x —2)] /k,k34amnj = []?4k(k— x) — ]?2(_ 1 + k)(k— x +2) + ]?_k+x+2(k + x — 2)— x) — ]?k+X+2(_ 1 + k)(k— x + 2) + ]?22(k + x — 2)] ,i,= 2 [_k2(1— R2)+ ]?2(]?k —k = 2,3,...,2N.The stress function (5.127) is then completely determined by (5.125) and (5.134). For convenience, the stress function (5.127) is further written in terms of linear and nonlinear componentsas follows4*(r*, 6, t*) = t*) + Ee4’(r, 6, t), (5.137)where SQ(r*, t*) and #(r*, 6, t*) are the linear and nonlinear stress components having the form= (1*2]?2(3j,)1fl*, (5.138)Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 87g(r*, 8,*)={cc(r*, 6, t*) + 8(r*, 6,t*)], (5.139)andMMNN 44Oc(r*, 6, t*) = [ npqjjV) + 4O(r*)]m=O p=O n=O q=O :0 =O= 0, (5.140)M M N N 4 4çt,kc(r*9t*) = > [nj(r) + 4nj(r*)] T(t*)cosk6,m=Op=O n=O q=O i=O =OM M N N 4 44,ks(r* 6,t*) = + T,(t*)sink6.m=O pO n=O q=O :0 j=OThe stress function (5.137) is similar to those given by Nowinski [45] and Efstathiades [49] if asingle-mode expansion with R = r + Er*2+ Er* 4) is substituted into (5.137).5.4 Time Governing EquationsThe time functions Cm(t*) and Smn(t) will be computed from (5.116). By using stressfunction (5.137), equation (5.116) is rewritten in the formN.V4w + (W*,t*t* +2l*w*,et* +*2w*,) + K:w*S*(6 — 6jS*(r* — :) —2Q* 2(...v2w* + W,r* ) — cp,,.. +ço,,..,.. (r*w*,r. _I_w*,89)] (5.141)—[* (T*&r* +,ee ) + (r*ws * +W*,88) — 2r* 2(±w* 8 (&e )r.] = 0.The term containing e2 in equation (5.141) makes nonlinear contributions to the given problem.If this term is removed, the governing equation for linear vibrations of a constrained rotatingdisk, such as that presented in [12], will be recovered.This study will be focused on the natural responses of the disk. Therefore, the externalload Q* is neglected in the following discussion. Applying Galerkin’s method as in Chapter2 and 3, and using the orthogonality of trigonometric functions, yields the following coupledChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 88nonlinear ordinary differential equations[ac(i91dz+ IgjSrnl + JjCmi) + E(L,’nngiCmn + LngiSmn)_E*e2(z91+ = 0, (5.142)[a(i91Sa — IjjCi +I9iSmi) + (LngiCmn + LngiSmi)+ =0,g=O,1,...,M; l=O,1,...,N;where and a5 are defined by equation (2.34), and= Jr*RmiRgzdr*,I,2,igi = 2112* J r*RmjRgldr*,Igi= J rR1 [ica, — i2cr2p1 + r* 212*C(PLml)),r + 2RmiPr*r*] dr,L91 = KrR91(r)R(r)cos l8 cosn8,= K’rRgi(r)Rmn(r) C05 l8 sin n81,Lngi = KrR91(r)Rmn(r’) sin l9 COS n8,= KrR91(r)Rmn(r) sin l8 sin n8, (5.143)N 2N 2= J cos nO cos kO cos lOdOnOk—.O 0I; {CmnRmnr*r* _..kC) + Cmn9C,r*r* (Rmn,r* “1?mn)_2nkSmn(Rmn),r* j r*Rgjdr*,N 2N 2wInsg1 = J sin nO sin kO cos LOdOn=0k0 0Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 89J [SmnRmnr*r* (±lk8,. _ij,ks) + SmnijC8,r*r (Rmn,r* 14Rmn)_2nkCmn(Rmn),r* (±4kc)7]r*Rgzdr*,N2N 21gi = J COS fl8 Sfl k6 Sfl 16d8n=Ok=O 0f [CmnRmnrT* _ks) + Cmn,r*r* (Rmn,r “Rmn)+2flkSmn(!R,nn),r* r*Rgjdr*,N 2N 2it= f sin nO cos kO sin lOdOn=Ok=0 0j’ [SmnRmn.*r (__kC,r* _i.kc) + Smnb,r*r* (-Rmn,r —--R,)+2nkCmn(—-Rmn),r* (±4/’8) r jr*Rgzdr*,d4 2 d3 1+212 & 1÷212 d 14_412L = —+————— —+ —+dr*4 r*dr*3 i*2 dr2 r3 dr& 3 d£2 = —+————.dr*2 r*dr* r*2Equations (5.142) are the governing equations for functions Cma and Smn. The nonlinearterms are composed of cubic functions of Cmn and Smn. These equations are multiple, coupled,nonlinear, ordinary differential equations, and the nonlinear terms are very complicated. Itis impractical to solve these equations analytically in this thesis due to the time restraint. Anumerical simulation will be used to study the nonlinear vibrations of a constrained rotatingdisk, in particular, the effects of geometric nonlinearity on the stability characteristics atsupercritical speeds. The Runge-Kutta numerical scheme as in Chapter 3 is employed to solvethese equations.Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 905.5 Numerical ResultsThe nonlinear vibrations of an elastically-constrained rotating disk will be discussed in thissection. Disk 4 is used for the computation, and its parameters are presented in Appendix B.First, the linear natural frequencies of the disk are calculated by neglecting the nonlinear termsin equation (5.142), and are compared with the existing analysis [12] in Table 5.2. Comparisonshows that the current results are very similar to those obtained by Hutton et at. [121. Therefore,modal function (5.120) represents a good approximation for the linear dynamic responses of arotating disk.Table 5.2: Comparison of linear frequencies, spring at r = 1, 8 = 0°, M = 0, N = 3=0K*=0 K=1 K*=0 K*=lMode Author 0 0.19688 0 0.08837(0,0) Hutton [5] 0 0.19752 0 0.08920Backward Author 0 0 0 0Mode Hutton 0 0 0 0(0,1) Forward Author 0 0 0.95493 0.97308Hutton 0 0 0.95493 0.97280Backward Author 0.74102 0.74102 0.16785 0.17965Mode Hutton 0.74556 0.74556 0.17609 0.18706(0,2) Forward Author 0.74102 0.77061 2.0777 2.0872Hutton 0.74556 0.77373 2.0859 2.0950Backward Author 2.0019 2.0019 0.87065 0.87576Mode Hutton 1.9552 1.9552 0.86003 0.86425(0,3) Forward Author 2.0019 2.0166 3.7354 3.7418Hutton 1.9552 1.9672 3.7248 3.7299The nonlinear stress function of mode (0,2) is compared with previous results [45, 49] inFigure 5.32. Only the axisymmetric terms are compared because the asymmetric terms are zeroin Nowinski’s formulation [45]. The comparison shows that the current result is very similarto the existing analyses.Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 91oe0tzLLa,U)1.0Radius (r/R0)Figure 5.32: Comparison of stress function of mode (0,2).Figures 5.33—5.35 present the nonlinear inpiane stresses of the disk at different crosssections. These stresses are calculated from equation (5.110) by using the nonlinear stressfunction (5.139) and removing phfl2r/2term. Six modes are used in the computation. Theradial and shear stresses are zero at inner and outer edges, and thus satisfy the inpiane stressboundary conditions (5.119). Numerical examination shows that the inplane forces and inpianemoments at any cross section are in equilibrium. Take &—f plane as an example, the resultantsof the shear and tangential stresses are zero, and the moment generated by the tangential stressis also zero. Therefore, the modeling of the nonlinear stress function appears to be correct.The inplane stress distribution is complex because it is affected by displacements w as well asu and v. The radial stress is mainly compressive along the radial direction. The shear stresscan be compressive or tensile depending on the angle of the cross sections, and may changesign along the same cross section, such as section a and b. The tangential stress is mainlycompressive near the inner edge, and becomes tensile at the outer edge. The tangential stress ismuch larger than the radial and shear stresses, and may have dominant effects on the nonlinear0.2 0.4 0.6 0.8Chapter .5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 921--__0— .....---- __-e.. ----::--._--_-) 4’---- ‘0S S ————— ,tIS — — - — — — — — — — — - - -—i \ d———5 ----- --- “ ,,- - - - -9------ - .. - - - -- _--‘C ——0 ‘S S_ ——-2 — — — — — ——,j’SSSC 4’d b —:0405,O6=::,0.9 1.0Radius (r/R0)Figure 5.33: Nonlinear inpiane radial stresses at different cross sections, six-mode approximation, Coo = C01 = C02 = S01 =•- = S05 = 1.0, C03 = C04 = C05 = —1.0..4Cd _—+---... be— 3—)-a9—9 --1 -- :-------% .-;..-—-: ..a -0s_Z:. e _:c. -2U,.1. .1.0.4 0.5 0.6 0.7 0.8 0.9 1.0Radius (r/R0)Figure 5.34: Nonlinear inplane shear stresses at different cross sections, six-mode approximation, Coo = C01 = Co2 = S01 = = So5 = 1.0, C03 = C04 = C05 = —1.0.Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 9330:: ea15 ,‘//1010Radius (r/R0)Figure 5.35: Nonlinear inpiane tangential stresses at different cross sections, six-mode approximation, C = C01 = C02 = S01 = = So5 = 1.0, Co3 Co = Co5 = —1.0.behavior of the disk. The stress results are very similar to the single-mode results presentedby Nowinski [45]. It should be mentioned that the results depicted in Figures 5.33—5.35 arecalculated from a specific displacement field. If the displacements are changed, the stresseswill be different. However, further computation shows that the general characteristics of theinpiane stresses described here are the same for different displacement fields.Figures 5.36—5.38 illustrate the effects of geometric nonlinearity on the frequencies of arotating disk at different speeds. In these plots, time history is first calculated from the single-mode solution for a given initial displacement field. An FF1’ is then used to compute thespectrum of the time response. The frequencies of modes (0,2), (0,3) and (0,4) at zero speedas a function of relative displacement w/h are presented in Figure 5.36. It is seen that thefrequencies increase as the displacement becomes larger. Accordingly, the large displacementmakes the disk stiffer, and the nonlinearity is of the hardening type. The frequency-amplituderelationship of the forward traveling wave of mode (0,3) is plotted in Figure 5.37. The linearChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 941 ..•o. -.- •. .a.C I I I0.00 0.25 0.50 0.75 1.00 1.25Displacement (w/h)Figure 5.36: Effects of nonlinearity on frequencies at 2 = 0, no spring, single-mode approximation, Con = Son, C0 = S0 =0.120.00 025 0.50 0.75 1.00 125Displacement (wlh)Figure 5.37: Effects of nonlinearity on frequencies of forward wave of mode (0,3), no spring,single-mode approximation, Co3 = So3, Co3 = So3 =0.Chapter .5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 95ia)0a)U-Figure 5.38: Effects of nonlinearity on frequencies of backward and reflected waves of mode(0,3), no spring, single-mode approximation, C03 = S03, C03 = S03 = 0.critical speed of mode (0,3) is 6.8. The plot shows that the frequencies of the forward travelingwaves increase with amplitude at both subcritical (fl* = 6) and supercritical (f* = 9) speeds.Figure 5.38 presents the frequency-amplitude plot of the backward and reflected waves of mode(0,3). The increase of displacement increases the frequency of a backward wave at subcriticalspeeds. However, at supercritical speeds, the frequency of a reflected wave decreases as thedisplacement increases. When the frequency becomes zero, the rotation speed is the newcritical speed corresponding to the large-amplitude vibration. For example, the critical speedfor mode (0,3) changes from 6.8 to 9 when the relative amplitude becomes 0.95. This result isconsistent with the observation that the geomethc nonlinearity increases the critical speed ofa rotating string as discussed in Chapter 3. Further increase of the displacement changes thereflected wave to a backward wave, and hence the frequency increases with displacement.It is noted from equation (5.137) that the stress function is composed of linear and nonlinearterms. The linear term is a function of 2, but the nonlinear term is independent of rotationDisplacement (wlh)1.25Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 96i=oo 0=6p =gJaJ0.00 0.25 0.50 0.75 1.00 1.25Displacement (w/h)Figure 5.39: Ratio of nonlinear and linear frequencies of mode (0,3) observed in the rotatingdisk, no spring, single-mode approximation, C03 = S03, Co3 = S03 = 0.speed. As a result, the effects of the nonlinear inpiane stresses on the dynamics of a rotatingdisk become insignificant at very high speeds compared to the centrifugal inpiane stresses.Figure 5.39 presents the ratio of nonlinear frequency w,. and linear frequency w of mode (0,3)as a function of the relative amplitude. The frequencies are observed in a coordinate systemrotating with the disk. The plot shows that for a given amplitude the increase of the frequencybecomes less notable as the speed increases. For example, for w/h = 1.25, the increase infrequency at = 0 is 38%, while the increase at V = 9 is 6%. This result is very similar tothat obtained by Nowinski [45].Figure 5.40 presents the spectrum of time response do3 at V = 6 calculated from bothsingle and multi-mode solutions. The dominant peaks are the frequencies of the backward andforward traveling waves of mode (0,3). It is noted that the frequencies of the multi-mode arehigher than those of the single-mode. When large amplitude vibration occurs, all modes arecoupled, and make contribution to the motions of other modes. The modal coupling is alsoChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 9712single-mode10 multi-mode8E. 6I:002 4 12Frequency (wI’I(D/(phR)))Figure 5.40: Frequencies of mode (0,3) calculated from single-mode and multi-mode (sixmodes) approximation, = 6, no spring, for single-mode: C03 = S03 = 5.0, for multi-mode:C =•. = S05 = 5.0, all derivatives are zero.shown by the small peaks in the multi-mode plot. These peaks are the frequencies of othermodes, and cannot been seen in the single-mode plot. As discussed in Chapter 3, multiplemodes should be used in order to obtain more accurate results if the system has complex initialconditions.In linear analysis, flutter instability occurs in an elastically-constrained rotating disk atsupercritical speeds. The time responses of flutter modes become unbounded as shown in Figure5.41. However, the responses remain finite at flutter speeds when nonlinearity is considered.Figure 5.42 shows the nonlinear time history of the flutter modes. The initial displacements usedin the computation are about 1/15 of the disk thickness, and are within the linear range. Becauseof this, the nonlinearity has very small effect on the flutter instability at the beginning of theresponse, and the displacements grow as shown in Figure 5.42. However, as the displacementsincrease, the effect of nonlinearity becomes important. The large displacements then stiffenChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 98C’)=00.0-U)=00ci)a,-10020010020Time (ti’J(phR/D))-200Figure 5.41: Linear response of C00 and Co2 at flutter speed = 6.5803, one spring, K = 1.0,= 1.0, 0 = 0°, four-mode approximation, C00 = -. = S03 = 1.0, all derivatives are zero.200cooCo2100-II ,‘ II —0-100-200 I I0 25 50 75 100 125 150Time (tJ’J(phRID))Figure 5.42: Nonlinear response of C00 and C02 at flutter speed 2 = 6.5803, one spring,K = 1.0, r = LO, 8 = 00, four-mode approximation, C00 = . - - = S03 = 1.0, all derivativesare zero.Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 99100—100 I I0 10 20 30 40 50Time (t/(phR/D))Figure 5.43: Linear response of Co2 at divergence speed V = 5.34, four springs,K = K = K = K = 2.0, r = = 1.0, r =rZ = 0.6, 6 = 82 = 00, 83 = 64 = 300,single-mode approximation, C02 = So2 = 1.0, C02 = S02 = 1.0.the disk, and change the frequencies of the flutter modes. The condition of flutter instability isthat the two modes must have the same frequency. Therefore, the instability will be stopped ifthe frequencies of the flutter modes become different. Computation upto nondimensional time250 shows that the nonlinear response does not exceed the magnitudes shown in Figure 5.42.The nonlinearity also changes the linear response at divergence speeds. In Figure 5.43,the linear response of mode (0,2) becomes infinite with time. Figure 5.44 shows the nonlinearresponse of the divergence mode. It is seen that the response does not become infinite when thenonlinearity is considered. The nonlinearity increases the frequency of the mode, and henceincreases the critical speed. Consequently, the speed is not in the divergence region, and themotion becomes restricted.Chapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 1002010A-10—20I,,,I,,,,I,,.,I I,,,,0 25 50 75 100 125 150 175 200Time (tI(phR:/D))Figure 5.44: Nonlinear response of C02 at divergence speed 12* = 5.34, four springs,K = K’ = K = K = 2.0, r = = 1.0, r =r = 0.6, 6i = 62 = 00, 63 = 64 = 30°,single-mode approximation, C02 = So2 = 1.0, Co2 = S02 = 1.0.5.6 SummaryA general analysis of the nonlinear vibrations of an elastically-constrained rotating disk isdeveloped by using von Karman thin plate theory. The stress function of multiple modes isanalytically solved from the condition of compatibility. The importance of the formulationis that it can be used to study the nonlinear dynamic characteristics associated with multiplemodes, such as flutter instability.The geometric nonlinearity generates hardening effects on the dynamics of a rotating disk.Frequencies increase with the transverse displacements. The effects of nonlinearity becomeless significant when the rotation speed is very high. A multi-mode solution should be used inorder to obtain more accurate frequency results.The unbounded motions at divergence and flutter speeds predicted by the existing linearanalyses will not take place because of the geometric nonlinearity. The large displacementsChapter 5. Nonlinear Vibrations of an Elastically-Constrained Rotating Disk 101change the frequencies of the unstable modes. As a result, the divergence and flutter motionsbecome restricted.Chapter 6ConclusionsThe linear analysis of a constrained rotating circular string shows that each vibration modeconsists of both forward and backward traveling waves. Away from the crossing points in thefrequency-speed diagram, a mode is dominated either by a backward or by a forward wave.At a crossing point, the mode is composed of equal contributions of backward and forwardwaves. For a string with a spring (mass) constraint, frequency curves avoid crossing if thechanges in frequency of all modes are in the same direction, such as the increase (decrease) infrequency due to the spring (mass) restraint in the subcritical speed region. However frequencycurves will cross if the directions of the frequency changes are different for each mode, suchas at supercritical speeds, then the corresponding real parts of the eigenvalues become positiveand flutter instability occurs. For a string with a damping constraint, the real parts of theeigenvalues vary linearly with rotation speed away from crossing points. At crossing points,the real part of one eigenvalue becomes zero because a node is developed at the constrainedpoint, while the real part of the other eigenvalue adopts a constant negative value as a result ofthe equal contributions from the forward and backward (reflected) waves. For a string with africtional constraint, the amplitude of the real parts of the eigenvalues is constant away fromcrossing points. At crossing points, the real parts of the eigenvalues all become zero due to themodulation of both forward and backward (reflected) waves.The nonlinear analysis of an elastically-constrained string shows that geometric nonlinearityincreases the wave speed, and consequently, the critical speed is increased. Frequencies are•also increased by geometric nonlinearity. Spring nonlinearity increases the natural frequencies102Chapter 6. Conclusions 103of the system, but it does not change the critical speed, since a point spring does not change thewave speed over a finite length of the string. An internal resonance is observed in the stringwith both geometric nonlinearity and spring nonlinearity when rotation speed is 0.5 of the wavespeed, and appears to be stable. Geometric nonlinearity changes the frequency of the fluttermodes to different values, and as a consequence, the corresponding instability predicted by thelinear analysis is restrained. Spring nonlinearity cannot restrain the growth of the motion atsupercritical speeds.The rigid-body motions of an unconstrained flexible rotating disk with free-free edges areobserved in the frequency-speed diagram computed from the conventional modal analyses,and also confirmed by the experimental results. The rigid-body motions are correctly modeledby the displacements of mode (0,0) and mode (0,1). The coupling between the translationalrigid-body motion and the flexible body deformation can reduce the divergence instabilityof an elastically-constrained rotating disk at supercritical speeds, but the tilting rigid-bodymotion does not change the stability characteristics. This result implies that a spline arbor withworn groove should be avoided in order to achieve stable operation of splined guided saws atsupercritical speeds.A general analysis of the nonlinear vibrations of an elastically-constrained rotating disk isdeveloped by using von Karman thin plate theory. The stress function of multiple modes isanalytically solved from the condition of compatibility. The importance of the formulation isthat it can be used to study the nonlinear dynamic characteristics associated with multiple modes,such as flutter instability. The analysis shows that the frequencies of a forward and backwardwave increase with the transverse displacement, while the frequencies of a reflected wavedecrease when the displacement becomes larger. The frequencies measured in a coordinatesystem rotating with the disk always increase with the displacement, for example, the frequencyof mode (0,3) is increased by 30% at nondimensional speed of 6 when the displacement is 1.25times of the plate thickness. Thus the geometric nonlinearity generates stiffening effects onChapter 6. Conclusions 104the vibrations of a rotating disk. The effects of nonlinearity become less significant whenrotation speed is very high. Take mode (0,3) as an example. When the displacement is 1.25times of the plate thickness, the increase in frequency at 0 speed is 38%, while the increaseat speed of 9 is 6%. This study also clearly demonstrates that the unbounded motions atdivergence and flutter speeds do not take place as predicted by the existing linear analyseswhen the effects of geometric nonlinearity are considered. The large displacements changethe frequencies of the unstable modes. As a consequence, the divergence and flutter motionsbecome restricted. This result partially explains why some thinner splined saws can successfullyoperate at supercritical speeds. Thus the divergence and flutter instabilities should not be themajor concerns for supercritical speed operation of splined saws.Bibliography[1] Iwan, W.D., and Moeller, T.L., 1976, “The Stability of a Spinning Elastic Disk with aTransverse Load System,” ASME Journal ofApplied Mechanics, Vol. 43, pp. 485—490.[2] Mote, C.D. 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B., 1993, “Natural Frequencies and Stability of a FlexibleSpinning Disk-Stationary Load System With Rigid-Body Tilting,” ASME Journal ofApplied Mechanics, Vol. 60, pp. 470—477.[43] Shames, I.H., 1966, Engineering Mechanics: Dynamics, Prentice-Hall Inc., EnglewoodCliffs, New Jersey.[44] Schajer, G.S., 1985, “CSAW — Guided Circular Saw Vibration and Stability Program,”Research Report 045-1516, Weyerhaeuser Company, Tacoma, WA, USA.[45] Nowinski, J.L., 1964, “Nonlinear Transverse Vibrations of a Spinning Disk,” ASMEJournal ofApplied Mechanics, Vol. 31, pp. 72—78.[46] Nowinski, J.L., 1984, “Nonlinear Vibrations of an Elastic Disk Rotating in a ViscousFluid,” Ingenieur-Archiv, Vol. 54, pp. 291—300.[47] Young, T.H., 1992, “Nonlinear Transverse Vibrations and Stability of Spinning Diskswith Nonconstant Spinning Rate,” ASME Journal of Vibration and Acoustics, Vol. 114,pp. 506—513.[48] Tobias, S.A., and Arnold, R.N., 1957, “Influence of Dynamical Imperfection on theVibration of Rotating Disks,” Proceedings of the institution of Mechanical Engineers,Vol. l’7l,pp. 669—690.[491 Efstathiades, G.J., 1971, “A New Approach to the Large-Deflection Vibrations of Imperfect Circular Disks Using Galerkin’s Procedure,” Journal of Sound and Vibration, Vol.16, pp. 23 1—253.[50] Dugdale, D.S., 1979, “Non-linear Vibration of a Centrally Clamped Rotating Disk,”International Journal ofEngineering Science, Vol. 17, pp. 745—756.[51] Yokochi, H., Tsuchikawa, S., and Kimura, 5., 1993, “Vibration Characteristics of a Rotating Circular Saw III — Non-linear Vibration and Coupled Vibration,” Mokuzai Gakkaishi,Vol. 39, pp. 776—782.[52] Timoshenko, S.P., and Goodier, J.N., 1970, Theoiy ofElasticity, McGraw-Hill Inc., NewYork.[53] Yu, Y.Y., 1964, “Generalized Hamilton’s Principle and Variational Equation of Motionin Nonlinear Elasticity Theory, with Application to Plate Theory,” Journal ofAcousticalSociety ofAmerica, Vol. 36, pp. 111—120.[54] Chia, C.Y., 1980, Nonlinear Analysis ofPlates, McGraw-Hill Inc., New York.[55] Filonenko-Borodich, M., 1965, Theoiy ofElasticity, Dover Publications Inc., New York.Bibliography 109[56] Sathyamoorthy, M., 1987, “Nonlinear Vibration Analysis of Plates: A Review and Surveyof Current Developments,” Applied Mechanics Review, Vol. 40, pp. 1553—1561.[57] Sathyamoorthy, M., 1983, “Nonlinear Vibrations of Plates—A Review,” Shock VibrationDigest, Vol. 15, pp. 3—16.[58] Leissa, A.W., 1978, “Recent Research in Plate Vibrations: 1973—1976. ComplicatingEffcets,” Shock Vibration Digest, Vol. 10, pp. 2 1—35.[59] Leissa, A.W., 1981, “Plate Vibrations Research: 1976—1980. Complicating Effcets,”Shock Vibration Digest, Vol. 13, pp. 19—36.[60] Leissa, A.W., 1987, “Recent Studies in Plate Vibrations: 198 1—1985. ComplicatingEffcets,” Shock Vibration Digest, Vol. 19, pp. 10—24.Appendix AEquation of Motion of a String Including Speed Dependent TensionThe total tension of a rotating circular string can be written asP=P÷P0, (A.144)where P is the initial tension of constant value, and .P0 is the centrifugal tension of the formPci=pr212. (A.145)Therefore, the governing equation becomesW,tt+22w,—S2w,99 = 0. (A. 146)The string does not have a critical speed since the centripetal term disappears from the governingequation.110Appendix BEffects of Constraints on EigenvaluesChen and Bogy [13] studied the effects of the restraint parameters on the eigenvalues apartfrom the crossing points of a rotating disk using modal analysis. In the following this approachis adopted to examine the changes of the eigenvalues of a string subject to various constraints.(1).K*O, M*C*1u*=OThe effects of an elastic restraint on the eigenvalues are given bynegative and positive waves,= 4nir (B. 147)—— reflected waves,4n7rwhere g is the constraint load, and g0 represents the unconstrained system.(2). M*O, K*=C*=*=OThe changes of the eigenvalues due to an inertial restraint are given bymCi —— negative waves,47rÔA in(1 +8MQ=QO =— 4positive waves, (13.148)in(f — 1)2+ reflected waves.47r(3). C ‘O, K* = M = =0For the case of a damping restraint—negative waves,=positive waves, (B. 149)+— 1reflected waves.47riiiAppendix B. Effects of Constraints on Eigenvalues 112(4). tt*, K’ =M*C*=OFor the case of a frictional restraintI — —-- negative and reflected waves,= . (B.150)( +— positive waves.The effects of the restraints on the eigenvalues demonstrated in equations (B.147)—(B.150)are consistent with the observations discussed in this paper. Moreover, the eigenvalue changesdue to the constraints can be accurately evaluated from equations (B. 147)—(B. 150) if the restraintparameters are small as shown in Table B.3.Table B.3: Eigenvalues calculated from equation (2.16) (analytical) and equations (B.1 47)—(B.150) (approximate), 1 = 0.1, n = 1, negative wave (w = n( 1 —K=0.5 M*=0.2 CI=0.1 p=O.O2Analytical 0.9305i 0.8865i —0.007165 —0.001608Approximate 0.9398i 0.887 ii —0.007162 —0.001592Appendix CFunctions in Butenin SolutionThe general form of equation (3.59) is as follows== (C.151)where ic1, ic2, n, n are constants; A is a small parameter which characterizes the closeness ofthe system to a linear conservative one; .F and are nonlinear functions in terms of x, th, y, i.When A = 0, the system (C.151) has the solutionz = a1 sin(Kit* + 3) + a2 sin(Kt* + /32),y = cx1a cos(K1t + 3) +a2cos(K2t* +132), (C.152)where K and 1(2 are natural frequencies which are determined byK4-i- [(n) + (+n) — ,c1 + nn = 0. (C. 153)The coefficients a1 and a2 are determined by the formulas— (n)+1(? —_______a1—— (Fr4)+1(?’(nf) +____a2= = (F7) +1(•(C.154)When A 1 0, the solution of the system (C. 151) is assumed to be of the same form as solution(C. 152) but where a1, a2, and /32 are functions of time. These functions are determined as113Appendix C. Functions in Butenin Solution 114followsda1 1—= —--(iG — —F1),d-rda2 1—---= — —F2),ar n1 a2d31 1a1— = ——(iC3+ F),dr n1 a1a2— = -(icG4+F), (C.155)dr n1 a2where1 p2w p2w —C1 =— I I QsinCdCd,27r JO Jo1 p2w p2w —C2 =— I I csini,dcdri,2irJo Jot2w p2w —C3 =— I I ccosdd,1,27r j0 j01 p2w p2w —C4 =— I I cosddii,27r Jo Jo1 p2w p2w —F1= —j J Fcosdd,27r 0 01 p2w p2w —F2= —J J FcosiidCd,27r 0 01 p2w p2w —F3 =— I I Fsinddi,2irJo Jo1 p2w p2w —F4 =— I I Fsini’dd,2ir j0 j0T= 2(K?—K) (C.156)and= F(ai sinC+a2sinr, a1,C cosC+a2Ks?7,a,a1COsC+a2a2CoS,—asin—ai),= c(a1 sinC +a2sin?7, atKicosC+acosii,aiaj cos+a2acos, —aia1K sinC —a2Ksini),Appendix C. Functions in Butenin Solution 115C = K1t*+/3,1 = K2t+/3. (C.157)Solving equations (C.155) gives the solution (C.152) for the system (C.151).Appendix DFormulation of Rigid-Body MotionsThe rigid disk has three degrees of freedom, the rotations and b1, about X and Yaxes’, and the translation wo along Z axis. Following the approach of classical rigid-bodydynamics [43], the transformation between ground-fixed coordinate system (X, Y, Z) and disk-fixed coordinate system (x, y, z) is first developed. To this end, the following transformationsequences are adopted to generate the new coordinate system:(1) rotation about X axis, (X, Y, Z) = (X’, Y’, Z’);(2) rotation ‘b,i about Y’ axis, (X’, Y’, Z’) (X”, Y”, Z”);(3) translation wo 1ong Z” axis, (X”, Y”, Z”) =l (x, y, z).The corresponding rotation sequences are shown in Figure D.45. The transformation matricesfor the rotations are given byRotation about X axis (Figure D.46(a)):{r’} = [li]{r}, (D.158)where {r’} = {X’ Y’ Z?}T, {r} = {X Y Z}T and1 0 0[li] = 0 cosb sinb0 —sinb cosbRotation about Y’ axis (Figure D.46(b)):{r”} = [12]{r’}, (D.159)‘It should be mentioned that these angles are not Euler’s angles.116Appendix D. Formulation of Rigid-Body Motions 117‘lixx, x’where {r”} = {X” Y” ZII}T andcos b1 0 — sin1 00 cosb,’{r”} = [l]{r}, (D.160)where[1] = [12][l1]COS byI Sfl Sfl ?,byi — COS ‘4hz sin 11i’y?= 0 cos’4b sin’4bSfl 4hyi — Sfl ‘4bx COS ‘4bi COS COSThe final transformation is derived by superposing the translation wo in the form ofz,, ‘Vx,’Figure D.45: Sequence of rotations.[121= 0VSuperposition of the two rotations leads to{r”} = [l]{r} + {0 0 wo}T, (D.161)Appendix D. Formulation of Rigid-Body Motions 118O(Y’,Y”)(a) (b)Figure D.46: Coordinate rotations in each plane; (a) about X axis, (b) about Y’ axis.where {r”} = {x y z}T.The displacements at the point {X Y Z}T(= {r cos 6 r sinS O}T) are then derived asfollowsu = r(cos ‘b,’ cos 6 + sinS sin ib sin bi),V = rsinScoSb,w = r(sin sb,’ cos 6 — sin cos ?/ sinS) + w0. (D. 162)The angular velocities of the disk about its geometric axes (x, y, z) are derived by projectingl, and b1, on these axes as follows= bcosb’,WY == +sinb. (D.163)If the rigid rotation angles are small, angle in equations (D.162) and (D.163) can bereplaced by which is defined in the ground-fixed coordinate system.z, z x,, x,‘1’ z,,OQ(X’) V z,Appendix EParameters of the DisksDisk 1 Disk 2 Disk 3 Disk 4type circular disk circular disk splined saw splined sawmaterial steel steel steel steelthickness (m) 0.00183 0.003 0.00183 0.001outer dia. (m) 0.768 0.6 10inner dia. (m) 0.122 0.150tip-to-tip dia. (m) 0.813 0.508spline No. 2 No. 4119Appendix FSelection of Rmn(r)The numerical formulation developed in Chapter 5 is based on the radial modal function(5.121). Function (4.91) was originally used to construct the solutions. However, the numericalresults calculated from equation (4.91) show that with large amplitude, the frequency of mode(0,2) is increased, but those of modes (0,3) and above are decreased. The results are notconsistent with the existing analyses [56]—[60] which showed that the geometric nonlinearityis of hardening type and frequencies increase with vibration amplitude. A test on a stationarydisk with free-free edges conducted in Wood Machining Laboratory also indicated that allfrequencies increase when the vibration amplitude becomes larger. Therefore, Rmn(r) givenby (4.91) can not be used to present sensible results for the nonlinear analysis.The difference between radial modal functions (4.91) and (5.121) is that their secondand higher order derivatives are completely different for mode (0,3) and above. Figure F.47compares the second derivatives of old R,,rn (4.91) and new Rmn (5.121). It is seen thatthey have different profiles. The second derivative of R,, determines Accordingly,w,7*,.. will be different when different R,, is used. Because of the difference ofthe stress function calculated from equation (5.118) will be different. Figure F.48 presents theaxisymmetric parts of the radial and tangential stresses determined from both old and newThe plot shows that the inplane stresses are completely different. Further examination indicatesthat the work done by the inplane stresses calculated from old R is negative for mode (0,3)and above, as a result, the corresponding frequencies are decreased. The work calculated from•new R is positive for all modes, and the frequencies are increased.120Appendix F. Selection of Rmn(T)0IFigure F.47: Second derivative of R,,, Ri/Ro = 0.4, mode (0,3).zU)-0.5-0.4 0.5 0.6 0.7Radius (r/R0)121Figure F.48: Axisymmetric parts of radial and tangential stresses, Ri/R0 = 0.4, single-modeapproximation, mode (0,3); (a) radial stress, (b) tangential stress.Radius (r/R0)1.0 -050.0zCl)Snew0.8 0.9 1, D(a)(b)Radius (rIB0)

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