# Dynamic Analysis of Flexible Shaft and Elastic Disk Rotor System Based on the Effect of Alford Force

Authors: Wenjun Yang; Huiqun Yuan; Li Hui; Kaifeng Zhang
Journal: Shock and Vibration
 (2019)

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## Abstract

Taken the flexible shaft and elastic disk rotor system as the research object, vibration differential equations are established by the modal synthesis method. Based on the Runge–Kutta method of direct integration, dynamic equations are solved successfully. Then, the effects of Alford force caused by the airflow excitation are researched on the stability and dynamic response. The results show modal vibration of the disk takes a great effect on the coupled natural frequency of shaft transverse vibration in the high rotational speed region. The effect of shaft-disk coupling makes rotor frequency to decrease, and different rotational speeds have little effect on rotor frequency. Eccentric distance is sensitive to the amplitude of rotor response, but it has little effect on rotor stability, motion law, and relative relationship of two response frequencies. The proportion of rotor responses caused by aerodynamic excitation force becomes larger with the increase of rotational speed and disk radius. However, the proportion decreases with the second support moving to the right, and it is helpful for the stable operation of the rotor system.


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## Body

## 1. Introduction

As a kind of large rotating machinery, a gas turbine is widely applied in industrial fields such as aviation, aerospace, transportation, and electric power. It takes an important role in national economy.A gas turbine works under the high-speed environment, and various nonlinear factors are shown on strong coupling, high dimension, and unsteady state. There will be hidden serious problems in operation, and it is often huge and even disastrous if the accidents occur.For a discontinuous system of the mass continuous distribution axis, it is difficult to express the dynamic characteristics of all parts in an equation. Gao and Meng [1] studied the vibration and stability characteristics of elastically supported cantilever beam by the transfer matrix method. But, this research was only on a continuous beam and shaft system, the effect of the disk was neglected on dynamic characteristics. By combining the modal synthesis method with the transfer matrix method, Han et al. [2] solved the dynamic characteristics of a complex rotor system.With the appearance of optimization design techniques, the rotor structure is continuously improved. The wheel disk is made much thinner so that coupling vibration is more obvious between blade and disk. So elastic disk modeling is very important for the research on the gas turbine disk-shaft system. Zhang [3] established the perturbation motion equation of the flexible shaft-elastic disk rotor. Based on the finite element method, Meek et al. [4, 5] analyzed the dynamic characteristics of a rigid disc-flexible shaft rotor system. Hili et al. [6] developed an elastic disk-rigid shaft model with an elastic support, and the effects of elastic deformation and elastic support are discussed on rotor responses when the disk is under axial load.In addition, fluid-solid coupling problems affect rotor stable operation seriously. It easily leads to the instability phenomenon by airflow excitation in gas turbine working [7, 8]. Because there is always the relative eccentricity between rotor blade and casing, the clearance of the blade tip is unevenly distributed along the circumferential direction. This makes the distributions of blade efficiency, pressure, and aerodynamic force different.So except for a torque, blade circumferential aerodynamic force is also combined with Alford force, which is acting on the axis and perpendicular to eccentric shifting [9]. Alford force is a cyclic force. Once it is larger than the external damping to maintain the system stable, positive work will be performed on the rotor system. That is inputting energy to promote the development of rotor whirl motion. The self-excited vibration will occur when its amplitude increases continuously.A steam turbine was made by GE corporation, and a strong fluid gap vibration was produced when the load was increased [10]. This is the first time that the phenomenon of aerodynamic excitation was discovered in a rotating machine. Soon after, Thomas also reported a similar vibration phenomenon [11]. He first proposed the concept of gap excitation and suggested the calculation formula for turbine gap excitation force. However, the problem did not attract enough attention at that time.In 1957, the first ultra-supercritical pressure units were operated at American Ohio Ferrer Power Plant. After that, the former Soviet Union, West Germany, and Japan had also developed and taken into operation the ultra-supercritical pressure units. In working operation, there are quite a few aerodynamic excitation problems on several supercritical units. Because of the higher steam parameters, aerodynamic excitation of a high-pressure rotor is increased on high-power ultra-supercritical pressure units.In the 1960s, the problem of aerodynamic excitation occurred again. At this period, the gap excitation mainly took place in the turbine of high-power ultra-supercritical pressure units which were manufactured by the Federal Republic of Germany. In the study of aeroengines stability, American scholar J. S. Alford achieved the important progress in theory. In his research, the theory of gap-flow excitation was proposed for the first time, and the calculation formula of turbine excitation force was given. Then, the problem of gap excitation was given enough attention.For experimental verification of Alford’s force, Urlichs and Wohkab [12, 13], respectively, developed the experimental research on turbine engines. Vance and Laudadio [14] performed the test of lateral force acting on the blower impeller and confirmed the presence of Alford force. In the sealing force models proposed by Thomas and Alford, circumferential flow velocity is ignored at the inlet, which is the primary factor of crosscut forces.Soviet scholars studied the spiral effect of aerodynamic flow in a high-pressure turbine rotor and found that the inertial effect of aerodynamic flow is far greater than the friction effect in the sealing gap [15]. At the end of the twentieth century, more researchers began to study aerodynamic excitation forces and the interaction forces between impeller and flow section [16, 17]. Shanghai Power Equipment Research Institute conducted an experiment on the aerodynamic oscillation of a gas turbine [18]. Considering the excitation of aerodynamic force, a small-scale modeling test rig was established by the single-span flexible rotor. Then, a simplified calculation method was proposed for the airflow excitation stiffness. The stability and unstable rotation speed were analyzed on the 200 MW turbine power generators.Chai et al. [19] developed a further research on the airflow exciting force. Taken from the fluid dynamics, the Alford force formula of the uniform flow field was derived on a straight blade and a twisted blade. Chen and Sun [20] proposed an entirely new method of oscillatory fluid dynamics, and it was used for solving the airflow stability and dynamic coefficient of air seal.Lu et al. [21] analyzed aerodynamic excitation in the cylinder of 200 MW high-pressure turbine and discussed the effect of aerodynamic excitation on the stability of the rotor system. For modifying the formula of Alford force, Yang et al. [22] decomposed rotor displacement into static displacement and dynamic displacement. This provided a method for adjusting rotor static displacement to reduce the exciting force.In this paper, vibration differential equations of an elastic disk-flexible shaft rotor are established by the modal synthesis method. Based on the Runge–Kutta method of direct integration, dynamic equations are solved successfully. Then, the effects of Alford force caused by the airflow excitation are researched on the stability and dynamic response.


## 2. Vibration Characteristics of Disk-Shaft Coupling System

Figure1 is the model of the elastic disk with inner diameter a and outer diameter b. For keeping the disk orthogonal to the shaft in all modes, it is assumed that the contact between disk and shaft is rigid. The differential equation of transverse elastic vibration can be derived by Hamilton’s principle. The free vibration equation of the disk is expressed as(1)ρDhD∂2uD∂t2+D∇4uD−1r∂∂rrNr∂uD∂r−1r2Nθ∂2uD∂θ2=0.Figure 1
Schematic diagram of the disk.In order to simulate dynamic characteristics of a flexible shaft and an elastic disk rotor system, the elastic circular plate with an inner clamp edge and outer freedom edge is adopted in the research.Bending displacement of the elastic disk is expressed as(2)uDr,θ,t=∑i=1∞Uir,θqit,where Uir,θ is the orthogonal mode of the disk and qit is the time-related part of generalized coordinates. Here, it is the simple harmonic vibration, so q¨it=−ωi2qit, where ωi is the i-order natural frequency of the elastic disk. Uir,θ can be expressed as(3)Uir,θ=Umnr,θ=Rmnramncosmθ+bmnsinmθ,m,n=0,1,…,∞,where Umn is indicated as the orthogonal mode with m nodal diameters and n circles. Uir,θ can be expressed as Umn by the reordering of natural frequencies increasing. Rmnr is the radial distribution of mode shapes; amn and bmn are constants which determine the initial angle of the disk combined mode.For a full symmetric disk, theoretical analysis and experiments have shown there is no fixed initial angle for the disk mode whether it is uniform or not. To simplify the following formula, it is generally assumed that the initial angle is zero. When boundary conditions are certain,Rmnr has the inherent form.Inertia force of the vibrating disk is expressed as(4)Fp=−ρDhD∑i=1∞ωi2qit∫02π∫abRmnrcosmθrdrdθ.In the formula,(5)∫02πcosmθdθ=2π,m=0,0,m≠0.The direction of inertia moment atp point is perpendicular to the axis. It is expressed as(6)Mp=−ρDhD∑i=1∞ωi2qit∫02π∫abRmnrcosmθcosθr2drdθ.In the formula,(7)∫02πcosmθcosθdθ=π,m=1,0,m≠1.From equations (4) and (6), it can be found that the zero-diameter mode of the disk is coupled with longitudinal vibration of the shaft by the effect of inertia force Fp. In addition, one pitch diameter mode of the disk is coupled with bending vibration of the shaft by the effect of inertia moment. However, other pitch diameter modes of the disk are not coupled with bending vibration of the shaft. So the bending coupled vibration is only discussed, and one pitch diameter mode of the disk is selected.


## 3. Dynamic Modeling of Flexible Shaft and Elastic Disk Rotor System

### 3.1. Rotor Dynamics Model of Flexible Shaft and Elastic Disk

The model of the flexible disk-flexible rotor system is shown in Figure2. The disk is located at the leftmost end of the shaft. The distance from the first spring support to the disc is l1, the distance between the two spring supports is l2, and the distance from shaft end to the second spring support is l3.Figure 2
Rotor model of the flexible shaft and the elastic disk.For describing motion state of the shaft, four variablesvs1, ws1 and vs2, ws2 are taken in two planes which are relatively vertical. Among them, vs1 and ws1 indicate small lateral deformation by the effect of bending, and vs2 and ws2 indicate small lateral deformation by the effect of shearing. Therefore, total deformation of the shaft is expressed as(8)vs=vs1+vs2,ws=ws1+ws2.In the rotating coordinate system, potential energy of the shaft due to bending and shearing can be expressed as(9)Vs=12∫0SkGsAs∂vs2∂x2+∂ws2∂x2+EsIs∂2vs1∂x22+∂2ws1∂x22dx,where k is the shape factor which is a constant related to the shaft cross-sectional shape. It can characterize the distribution of uneven shear stress in the cross-sectional area. Es, Gs, and As are axis Young’s modulus, shear mode, and cross-sectional area, respectively. Is is the axis inertia moment related to diameter.Due to translation and rotation, kinetic energy of the axis can be expressed as(10)Ts=12∫0SρsAsv˙s2+w˙s2+2Ωw˙svs−wsv˙s+Ω2vs2+ws2dx+12∫0SρsIsθ˙sξ2+θ˙sη2+2Ωθ˙sξθsη+θsξθ˙sη−Ω2θsξ2+θsη2dx+∫0SρsIsΩ2dx,where ρs is the density of the shaft. θsξ and θsη denote the small-angle rotation of the shaft cross section due to bending, and they can be expressed as(11)θsξ=−∂ws1∂x,θsη=∂vs1∂x.Kinetic energy of the disk is expressed as(12)TD=12mDv˙2+w˙2+2Ωw˙v−v˙w+Ω2v2+w2+12JDyθ˙ξ−Ωθη2+θ˙η+Ωθξ2+12JDx2Ωθηθ˙ξ−Ω2θξ2+θη2+Ω2+12ρDhD∫02π∫abu˙D2+2rsinθθ˙ξ−Ωθη−cosθθ˙η+Ωθξu˙D−2Ωrsinθθ˙η+Ωθξ+cosθθ˙ξ−Ωθηu˙Drdrdθ,where mD is the mass of the disk, JDx and JDy are the disc inertia moments of axis and radial direction, v and w are the shaft displacements at the disc location, and θξ and θη denote the small rotation angle caused by shaft bending. The tiny rotation angle is ignored which is caused by the rigid body displacements because it will be eliminated in the Lagrangian equation.Strain energy of the disk can be expressed as(13)VD=D2∫02π∫ab∇2uD2−21−vD∂2uD∂r21r∂uD∂r+1r2∂2uD∂θ2+21−vD∂∂r1r∂uD∂θ2rdrdθ+12∫02π∫abNr∂uD∂r2+Nθ1r∂uD∂θ2rdrdθ.For elastic support in the rotor system, it can be calculated by the principle of virtual work. The formula is shown as follows:(14)δWt=−∑i=1Nk1vsδvs+wsδws+k2wsδvs−vsδwsx=bi.The elastic disk-shaft rotor is approximately analyzed by the modal synthesis method. Ritz base of each substructure is obtained with the subsystem hypothesis morphological method. Eigenfunctions of transverse vibrations from flexible beam and shear beam with free ends are selected as the hypothetical function of continuous shaft transverse deflection induced by bending and shearing. Free vibration eigenfunction of the balanced disk is applied as the hypothesis function of the elastic disk in the system.Transverse deformation of the elastic disk is expressed as(15)R1nr=AnJ1βnrb+BnY1βnrb+CnI1βnrb+DnK1βnrb,n=0,1,2,…,where corner marker “1” indicates the first section diameter mode, corner marker “n” indicates the number of pitch circles (not including pitch circle subjected to the inner circle clamp). J1 and Y1 are the Bessel functions of the first and second types, respectively, and I1 and K1 are the improved Bessel functions of the first and second types, respectively. An, Bn, Cn, and Dn are the constants which are determined by boundary conditions, the number of pitch circles, and the condition of orthogonality. βn is the dimensionless parameter related to natural frequency ωn, and the relation is expressed as(16)βn4=ωn2ρDhDb4D.Coefficient matrix determinant of linear equations is zero by the boundary conditions, soβn can be obtained.According to the Lagrange equation:(17)ddt∂L∂q˙i−∂L∂qi=0,where L=Ts+TD−Vs−VD, and the following is obtained:(18)Mq¨+Cq˙+Gq˙+Kq=0.As can be seen from Figure2, each boundary or connection condition is shown as follows.The boundary condition of O point is expressed as(19)μ1G1A1Θ1−f1′x1=0=mDω2f1x1=0,−E1I1Θ1′x1=0=Jd1ω2Θ1−Jp1ΩωΘ1x1=0.Connection condition of A point is expressed as(20)μ1G1A1Θ1−f1′x1=l1−μ2G2A2Θ2−f2′x2=0=k1f2x2=0,∂Θ1∂x1x1=l1=∂Θ2∂x2x2=0,f1x1=l1=f2x2=0,Θ1x1=l1=Θ2x2=0.The boundary condition of B point is expressed as(21)μ2G2A2Θ2−f2′x2=l2−μ3G3A3Θ3−f3′x3=0=k2f3x3=0,∂Θ2∂x2x2=l2=∂Θ3∂x3x3=0,f2x2=l2=f3x3=0,Θ2x2=l2=Θ3x3=0.The boundary condition of free end is expressed as(22)∂Θ3∂x3x3=l3=0,Θ3−∂f3∂x3x3=l3=0.By solving the equation set formed by the above equations, natural frequency and mode shape of the system can be obtained.


### 3.2. Differential Equations of Rotor System under Airflow Excitation Force

In order to analyze the effect of Alford force on rotor stability, the equivalent cross stiffness is applied to indicate the action of Alford force. Figure3 illustrates the process of instability caused by equivalent cross stiffness. O point represents the axis before the deformation, and O′ point represents the axis after deformation. R is the the radius of casing, and δ is the clearance between rotor and casing. Assuming that the rotor is precessing by the effect of unbalance, the direction of precession is shown by the arrow in Figure 3. Fc is the inertia force, Fe is the elastic recovery force, Fd is the external damping force, and Fa is induced by the Alford force. If Fa is greater than the force of outer damping Fd on the turbine, it will accelerate the positive precession. So the inertial force increases, the deflection increases, the unstable force increases, and the precession is faster. With this development, vibration amplitude of self-energized motion increases until it is limited by the structure or reaches the limit circle (or has caused damage before).Figure 3
The model of rotor damage-instability caused by aerodynamic excitation force.By the action of rotational dragging, the fluid will generate a circular flow in the sealing gap, which plays a significant role in rotor vibration. In a certain conditions, it will cause the rotor instable and generate self-excited lateral vibration. In 1987, Muszynska and Bently [23] proposed the Muszynska model based on a large number of experiments. This model introduces the average circumferential speed ratio τ to characterize the overall motion of fluid in the gap between the bearing and sealing. Rotational effect of seal excitation force is the main factor of rotor instability. The reaction force caused by sealing excitation force rotates around the rotor at a fixed angular velocity. In fixed coordinates, this reaction force can be described as the following mathematical equation:(23)−FxFy=K−mfτ2ω2τωD−τωDK−mfτ2ω2xy+D2τmfω−2τmfωDx˙y˙+mfmfx¨y¨.where K, D, and mf indicate the stiffness, damping, and inertia effects of fluid’s disturbing motion on the rotor. It is generally taken that cyclic force is rotor tangential driving force constituted by the cross stiffness term τωD. According to the equation, larger fluid damping D is easy to make rotor instable. Reducing the average circumferential velocity ratio τ can improve the stability of the rotor-seal system. A large number of experiments [24] have proved the assertion that the application of antiswirling structure can improve the stability of the sealing system. In addition, the main stiffness term in the model, that is, rotor radial stiffness K−mfτ2ω2, can become a negative value at the effect of fluid centrifugal inertia force.Perturbation analysis and experimental data show that [25] the effect of fluid centrifugal inertia force becomes very obvious, when the ratio of seal gap and rotor radius is small. This has a great influence on the stiffness matrix. So the radial stiffness term can easily reach a negative value, and it is also a factor that causes rotor instable. The Muszynska model correctly reflects the basic dynamic characteristics of sealing force. Equation (23) not only is applied in the condition of small displacement at the rotor center position but also can indicate the nonlinear sealing force at larger displacement. Experimental and numerical results confirm that [26] τ, K, and D are the nonlinear functions of disturbed displacements x and y. It can be expressed as follows:(24)K=K01−e2−n,D=D01−e2−n,n=12∼3,τ=τ01−e2b,0<b<1.where e=x2+y2/Cr is rotor relative partial displacement and Cr is the sealing clearance. Specific seal is approximately described by selecting the appropriate exponents n and b. Due to the effects of axial flow, surface roughness, secondary flow, and other factors, in general τ0<1/2. K0, D0, and mf can be calculated by the combination of Childs seal dynamic coefficient equation and equation (24):(25)K0=μ3μ0,D0=μ1μ3T,mf=μ2μ3T2.The Muszynska sealing force model is mainly based on a large number of experiments and abstracted from the model. It has a strong generality and grasps the main mechanical characteristics of sealing force. The correctness of the Muszynska model has been widely recognized in the application. The Muszynska sealing force model is adopted as the model of airflow excitation force, and the dynamic differential equation of the elastic disc-flexible shaft rotor is established by the modal synthesis method.Applying the principle of virtual work, virtual work caused by the Alford force can be obtained as(26)δWt=−k1vsδvs+wsδws+k2wsδvs−vsδws+c1v˙sδvs+w˙sδws+c2w˙sδvs−v˙sδws−c1Ωwsδvs−vsδws+c2Ωvsδvs−wsδwsmfv¨sδvs+w¨sδws+mfΩw˙sδvs−v˙sδws+mfΩ2vsδvs+wsδwsd,where(27)k1=K−mfτ2ω2,k2=τωD,c1=D,c2=2τmfω.According to the modal synthesis method, the dynamic equation of the rotor can be expressed as(28)Mq¨+Cq˙+Gq˙+Kq=Fq+eΩ2Fe.where Fe is the imbalance force caused by eccentric mass and Fq is the force of bearing oil film. Fq is expressed as(29)Fq=−Kqq+Cqq˙+Gqq˙+Mqq¨,where(30)Φ=Φs1TdΦs1dΦs1TdΦs2dΦs2TdΦs1dΦs2TdΦs2d0.SoKq, Cq, Gq, and Mq in equation (29) are expressed as(31)Kq=Kq1Kq2−Kq2TKq1,Kq1=k1+c2Ω+mfΩ2Φ,Kq2=k2+c1ΩΦCq=Cq1Cq2−Cq2TCq1,Cq1=c1Φ,Cq2=c2Φ,Gq=Gq1−Gq1T,Gq1=mfΩΦ.Assuming thatq¯=q/Cr, q¯˙=dq¯/dτdτ/dt=Ωq¯′, and e¯=e/Cr, equation (28) is expressed as(32)M+Mqq″¯+C+CqΩq¯′+G+GqΩq¯′+K+KqΩ2q¯=e¯Fe.


## 3.1. Rotor Dynamics Model of Flexible Shaft and Elastic Disk

The model of the flexible disk-flexible rotor system is shown in Figure2. The disk is located at the leftmost end of the shaft. The distance from the first spring support to the disc is l1, the distance between the two spring supports is l2, and the distance from shaft end to the second spring support is l3.Figure 2
Rotor model of the flexible shaft and the elastic disk.For describing motion state of the shaft, four variablesvs1, ws1 and vs2, ws2 are taken in two planes which are relatively vertical. Among them, vs1 and ws1 indicate small lateral deformation by the effect of bending, and vs2 and ws2 indicate small lateral deformation by the effect of shearing. Therefore, total deformation of the shaft is expressed as(8)vs=vs1+vs2,ws=ws1+ws2.In the rotating coordinate system, potential energy of the shaft due to bending and shearing can be expressed as(9)Vs=12∫0SkGsAs∂vs2∂x2+∂ws2∂x2+EsIs∂2vs1∂x22+∂2ws1∂x22dx,where k is the shape factor which is a constant related to the shaft cross-sectional shape. It can characterize the distribution of uneven shear stress in the cross-sectional area. Es, Gs, and As are axis Young’s modulus, shear mode, and cross-sectional area, respectively. Is is the axis inertia moment related to diameter.Due to translation and rotation, kinetic energy of the axis can be expressed as(10)Ts=12∫0SρsAsv˙s2+w˙s2+2Ωw˙svs−wsv˙s+Ω2vs2+ws2dx+12∫0SρsIsθ˙sξ2+θ˙sη2+2Ωθ˙sξθsη+θsξθ˙sη−Ω2θsξ2+θsη2dx+∫0SρsIsΩ2dx,where ρs is the density of the shaft. θsξ and θsη denote the small-angle rotation of the shaft cross section due to bending, and they can be expressed as(11)θsξ=−∂ws1∂x,θsη=∂vs1∂x.Kinetic energy of the disk is expressed as(12)TD=12mDv˙2+w˙2+2Ωw˙v−v˙w+Ω2v2+w2+12JDyθ˙ξ−Ωθη2+θ˙η+Ωθξ2+12JDx2Ωθηθ˙ξ−Ω2θξ2+θη2+Ω2+12ρDhD∫02π∫abu˙D2+2rsinθθ˙ξ−Ωθη−cosθθ˙η+Ωθξu˙D−2Ωrsinθθ˙η+Ωθξ+cosθθ˙ξ−Ωθηu˙Drdrdθ,where mD is the mass of the disk, JDx and JDy are the disc inertia moments of axis and radial direction, v and w are the shaft displacements at the disc location, and θξ and θη denote the small rotation angle caused by shaft bending. The tiny rotation angle is ignored which is caused by the rigid body displacements because it will be eliminated in the Lagrangian equation.Strain energy of the disk can be expressed as(13)VD=D2∫02π∫ab∇2uD2−21−vD∂2uD∂r21r∂uD∂r+1r2∂2uD∂θ2+21−vD∂∂r1r∂uD∂θ2rdrdθ+12∫02π∫abNr∂uD∂r2+Nθ1r∂uD∂θ2rdrdθ.For elastic support in the rotor system, it can be calculated by the principle of virtual work. The formula is shown as follows:(14)δWt=−∑i=1Nk1vsδvs+wsδws+k2wsδvs−vsδwsx=bi.The elastic disk-shaft rotor is approximately analyzed by the modal synthesis method. Ritz base of each substructure is obtained with the subsystem hypothesis morphological method. Eigenfunctions of transverse vibrations from flexible beam and shear beam with free ends are selected as the hypothetical function of continuous shaft transverse deflection induced by bending and shearing. Free vibration eigenfunction of the balanced disk is applied as the hypothesis function of the elastic disk in the system.Transverse deformation of the elastic disk is expressed as(15)R1nr=AnJ1βnrb+BnY1βnrb+CnI1βnrb+DnK1βnrb,n=0,1,2,…,where corner marker “1” indicates the first section diameter mode, corner marker “n” indicates the number of pitch circles (not including pitch circle subjected to the inner circle clamp). J1 and Y1 are the Bessel functions of the first and second types, respectively, and I1 and K1 are the improved Bessel functions of the first and second types, respectively. An, Bn, Cn, and Dn are the constants which are determined by boundary conditions, the number of pitch circles, and the condition of orthogonality. βn is the dimensionless parameter related to natural frequency ωn, and the relation is expressed as(16)βn4=ωn2ρDhDb4D.Coefficient matrix determinant of linear equations is zero by the boundary conditions, soβn can be obtained.According to the Lagrange equation:(17)ddt∂L∂q˙i−∂L∂qi=0,where L=Ts+TD−Vs−VD, and the following is obtained:(18)Mq¨+Cq˙+Gq˙+Kq=0.As can be seen from Figure2, each boundary or connection condition is shown as follows.The boundary condition of O point is expressed as(19)μ1G1A1Θ1−f1′x1=0=mDω2f1x1=0,−E1I1Θ1′x1=0=Jd1ω2Θ1−Jp1ΩωΘ1x1=0.Connection condition of A point is expressed as(20)μ1G1A1Θ1−f1′x1=l1−μ2G2A2Θ2−f2′x2=0=k1f2x2=0,∂Θ1∂x1x1=l1=∂Θ2∂x2x2=0,f1x1=l1=f2x2=0,Θ1x1=l1=Θ2x2=0.The boundary condition of B point is expressed as(21)μ2G2A2Θ2−f2′x2=l2−μ3G3A3Θ3−f3′x3=0=k2f3x3=0,∂Θ2∂x2x2=l2=∂Θ3∂x3x3=0,f2x2=l2=f3x3=0,Θ2x2=l2=Θ3x3=0.The boundary condition of free end is expressed as(22)∂Θ3∂x3x3=l3=0,Θ3−∂f3∂x3x3=l3=0.By solving the equation set formed by the above equations, natural frequency and mode shape of the system can be obtained.


## 3.2. Differential Equations of Rotor System under Airflow Excitation Force

In order to analyze the effect of Alford force on rotor stability, the equivalent cross stiffness is applied to indicate the action of Alford force. Figure3 illustrates the process of instability caused by equivalent cross stiffness. O point represents the axis before the deformation, and O′ point represents the axis after deformation. R is the the radius of casing, and δ is the clearance between rotor and casing. Assuming that the rotor is precessing by the effect of unbalance, the direction of precession is shown by the arrow in Figure 3. Fc is the inertia force, Fe is the elastic recovery force, Fd is the external damping force, and Fa is induced by the Alford force. If Fa is greater than the force of outer damping Fd on the turbine, it will accelerate the positive precession. So the inertial force increases, the deflection increases, the unstable force increases, and the precession is faster. With this development, vibration amplitude of self-energized motion increases until it is limited by the structure or reaches the limit circle (or has caused damage before).Figure 3
The model of rotor damage-instability caused by aerodynamic excitation force.By the action of rotational dragging, the fluid will generate a circular flow in the sealing gap, which plays a significant role in rotor vibration. In a certain conditions, it will cause the rotor instable and generate self-excited lateral vibration. In 1987, Muszynska and Bently [23] proposed the Muszynska model based on a large number of experiments. This model introduces the average circumferential speed ratio τ to characterize the overall motion of fluid in the gap between the bearing and sealing. Rotational effect of seal excitation force is the main factor of rotor instability. The reaction force caused by sealing excitation force rotates around the rotor at a fixed angular velocity. In fixed coordinates, this reaction force can be described as the following mathematical equation:(23)−FxFy=K−mfτ2ω2τωD−τωDK−mfτ2ω2xy+D2τmfω−2τmfωDx˙y˙+mfmfx¨y¨.where K, D, and mf indicate the stiffness, damping, and inertia effects of fluid’s disturbing motion on the rotor. It is generally taken that cyclic force is rotor tangential driving force constituted by the cross stiffness term τωD. According to the equation, larger fluid damping D is easy to make rotor instable. Reducing the average circumferential velocity ratio τ can improve the stability of the rotor-seal system. A large number of experiments [24] have proved the assertion that the application of antiswirling structure can improve the stability of the sealing system. In addition, the main stiffness term in the model, that is, rotor radial stiffness K−mfτ2ω2, can become a negative value at the effect of fluid centrifugal inertia force.Perturbation analysis and experimental data show that [25] the effect of fluid centrifugal inertia force becomes very obvious, when the ratio of seal gap and rotor radius is small. This has a great influence on the stiffness matrix. So the radial stiffness term can easily reach a negative value, and it is also a factor that causes rotor instable. The Muszynska model correctly reflects the basic dynamic characteristics of sealing force. Equation (23) not only is applied in the condition of small displacement at the rotor center position but also can indicate the nonlinear sealing force at larger displacement. Experimental and numerical results confirm that [26] τ, K, and D are the nonlinear functions of disturbed displacements x and y. It can be expressed as follows:(24)K=K01−e2−n,D=D01−e2−n,n=12∼3,τ=τ01−e2b,0<b<1.where e=x2+y2/Cr is rotor relative partial displacement and Cr is the sealing clearance. Specific seal is approximately described by selecting the appropriate exponents n and b. Due to the effects of axial flow, surface roughness, secondary flow, and other factors, in general τ0<1/2. K0, D0, and mf can be calculated by the combination of Childs seal dynamic coefficient equation and equation (24):(25)K0=μ3μ0,D0=μ1μ3T,mf=μ2μ3T2.The Muszynska sealing force model is mainly based on a large number of experiments and abstracted from the model. It has a strong generality and grasps the main mechanical characteristics of sealing force. The correctness of the Muszynska model has been widely recognized in the application. The Muszynska sealing force model is adopted as the model of airflow excitation force, and the dynamic differential equation of the elastic disc-flexible shaft rotor is established by the modal synthesis method.Applying the principle of virtual work, virtual work caused by the Alford force can be obtained as(26)δWt=−k1vsδvs+wsδws+k2wsδvs−vsδws+c1v˙sδvs+w˙sδws+c2w˙sδvs−v˙sδws−c1Ωwsδvs−vsδws+c2Ωvsδvs−wsδwsmfv¨sδvs+w¨sδws+mfΩw˙sδvs−v˙sδws+mfΩ2vsδvs+wsδwsd,where(27)k1=K−mfτ2ω2,k2=τωD,c1=D,c2=2τmfω.According to the modal synthesis method, the dynamic equation of the rotor can be expressed as(28)Mq¨+Cq˙+Gq˙+Kq=Fq+eΩ2Fe.where Fe is the imbalance force caused by eccentric mass and Fq is the force of bearing oil film. Fq is expressed as(29)Fq=−Kqq+Cqq˙+Gqq˙+Mqq¨,where(30)Φ=Φs1TdΦs1dΦs1TdΦs2dΦs2TdΦs1dΦs2TdΦs2d0.SoKq, Cq, Gq, and Mq in equation (29) are expressed as(31)Kq=Kq1Kq2−Kq2TKq1,Kq1=k1+c2Ω+mfΩ2Φ,Kq2=k2+c1ΩΦCq=Cq1Cq2−Cq2TCq1,Cq1=c1Φ,Cq2=c2Φ,Gq=Gq1−Gq1T,Gq1=mfΩΦ.Assuming thatq¯=q/Cr, q¯˙=dq¯/dτdτ/dt=Ωq¯′, and e¯=e/Cr, equation (28) is expressed as(32)M+Mqq″¯+C+CqΩq¯′+G+GqΩq¯′+K+KqΩ2q¯=e¯Fe.


## 4. Analysis and Discussion of the Results

In order to explain the effects of different parameters on rotor vibration responses, rotor parameters are taken as follows if no special description is given. The geometric parameters of rotor structure are indicated asl1=0.35, l2=0.95m, and l3=0.65m. The inner and outer radii, thickness, density, elastic modulus, and Poisson’s ratio of the elastic disk are, respectively, indicated as a=0.11m, b=0.445m, hD=0.088m, ρD=4640kg/m3, ED=1.13×1011N/m2, and vD=0.28. The radius, density, Poisson’s ratio, and elastic modulus of the flexible shaft are indicated as rs=0.11m, ρs=7900kg/m3, vs=0.28, and Es=2.06×1011N/m2. The stiffness coefficient of elastic support is indicated as k11=k21=6.0×108N/m. The damping coefficient of the support is indicated as c11=c21=3000N⋅s/m.In the model of aerodynamic excitation force, the circumferential loss coefficient of sealed inlet airflow is indicated asξ=0.1. The seal length is indicated as l=0.088. The radial seal clearance is indicated as δ=2mm. Fluid average axial velocity in the sealing cavity is indicated as v=5m/s. The sealing radius is indicated as R=0.445m. The coefficient of fluid viscosity is indicated as μ=1.8×10−5pa⋅s. The empirical coefficients are indicated as m0=−0.25, n0=0.079, τ0=0.25, and n=2. Axial seal pressure drop is indicated as ΔP=ρv21+ξ+2δ/2, where ρ is the density of gas.


### 4.1. Natural Vibration Analysis of Flexible Shaft and Elastic Disk Rotor

#### 4.1.1. Comparison of Results at Different Modes

Take the shaft bending mode number, shaft shear mode number, and the disk modal number are, respectively, as 25, 25, and 4. The first eight natural frequencies of the rigid disk-flexible shaft rotor system can be obtained based on the boundary and connection conditions. Similarly, the first eight natural frequencies of the elastic disk-flexible shaft rotor system can be obtained, as shown in Table1.Table 1
First eight natural frequencies of the rigid disk and elastic disk rotor models.
Frequency (rad·s−1)
ω
1
ω
2
ω
3
ω
4
ω
5
ω
6
ω
7
ω
8
Rigid disk rotor
701.26
931.53
1491.22
2879.45
4918.80
8051.80
11967.90
16246.80
Elastic disk rotor
700.51
931.04
1474.40
2585.82
4553.73
7546.20
11250.92
15203.63The corresponding mode shapes are shown in Figure4.Figure 4
Mode shape comparison of the different rotor models: first eight mode shapes of (a) flexible shaft and rigid disk rotor system and (b) flexible shaft and elastic disk rotor system.
(a)
(b)From modal vibration shapes in Figures4(a) and 4(b), it can be seen that the vibration of the disk is not obvious at the low-order frequency. As the frequency increases, the vibration of the disk gradually becomes prominent.The above derivation is performed in the rotating coordinate of shaft and the rotating coordinate of the disc, so the obtained natural frequencies are also relative to the rotating coordinate system. In order to compare with the experimental data, it is necessary to convert the obtained natural frequency into an inertial coordinate system. For the transverse vibration of the shaft and disk, the conversion relationship is different between rotating coordinate system and inertial coordinate system. Hence, there are two different natural frequencies in the inertial coordinate system.


#### 4.1.2. Analysis of Coupled Natural Frequency in Transverse Vibration

For transverse vibration of the shaft, conversion relation is expressed as(33)ω0=ω+Ω,where ω is the natural frequency in the rotating coordinate system, Ω is the angular velocity of the shaft, and ω0 is the natural frequency in the inertial coordinate system.For transverse vibration of the disk, the transformation is difficult from the rotating coordinate system to the inertial coordinate system. Compared with the rotation angleΩt, declination angles θξ and θη of transverse vibration can be ignored in the transformation of the coordinate system. So, the relationship between two angular coordinate systems can be expressed as(34)θ0=θ+Ωt.Excitation force of the disc isF0cosω0t, and vibration response at r0=ρ and θ0=0 is expressed as(35)u=Kρ,mF0frcosmθcosω0+mΩtω2−ω0+mΩ2+cosω0−mΩtω2−ω0−mΩ2,where ω is the natural frequency in the rotating coordinate system and m is the number of mode nodal diameters in the disk. It can be seen from the above equation that there are two resonant frequencies in the inertial coordinate system:(36)ω0=ω+mΩ,ω0=ω−mΩ,ω≥mΩ,−ω−mΩ,ω<mΩ.Plotting the chart of frequency-rotational angular velocity,ω−mΩ and −ω−mΩ are the continuous curves. Therefore, natural frequencies are generally divided into two parts when rotating. That is, at a certain rotational angular velocity, two different excitation frequencies will excite the same response mode in the inertial space.For the specific problem, natural frequency of disk-shaft coupling is indicated asω. The antiprecession mode of the rotating system may be negative, and only the first nodal diameter mode of the disk will couple with the vibration of shaft, so m=1. Equation (36) can be expressed as(37)ω0=ω+Ω,ω0=ω−Ω,ω≥mΩ,−ω−Ω,ω<mΩ.According to equation (28), the coupled natural frequency of shaft transverse vibration is shown as Figure 5. Here, 5(a) is the diagram of coupled natural frequency in the flexible shaft and rigid disk, and 5(b) is the diagram of coupled natural frequency in the flexible shaft and elastic disk.Figure 5
Coupled natural frequency of the shaft transverse vibration: (a) flexible shaft and rigid disk system and (b) flexible shaft and elastic disk system.
(a)
(b)In Figure5, the difference between Figures 5(a) and 5(b) is obvious when the rotational speed is large, and the difference between two models is smaller in the region with lower rotational speed. It is illustrated that modal vibration of the shaft occupies a dominant position in the low rotational speed region. Modal vibration of the disk has little effect on the coupled natural frequency of shaft transverse vibration, while it takes a great effect in the high rotational speed region. Hence, the difference between two models is quite obvious.According to equation (37), the coupled natural frequency of elastic disk transverse vibration is shown as Figure 6. Here, 6(a) is the diagram of forward natural frequency, and 6(b) is the diagram of backward natural frequency.Figure 6
Coupled natural frequency of transverse vibration of the disk: (a) forward processional modes and (b) backward processional modes.
(a)
(b)Mass matrix and stiffness matrix of the rotor system will be affected by the coupling effect of shaft and disk, but the gyroscopic matrix and damping matrix are not affected. Without considering the physical background, the disk-shaft coupling term can be directly removed in the mass matrix and stiffness matrix. So, it is to analyze the effect of the disk-shaft coupling on rotor natural frequency.From Figure7, it can be seen that the effect of shaft-disk coupling makes rotor frequency decrease, and different rotational speeds have little effect on rotor frequency.Figure 7
The effect of disk-shaft coupling term on rotor natural frequency: (a)Ω=0rad/s, (b) Ω=500rad/s, and (c) Ω=1000rad/s.
(a)
(b)
(c)The effects of different structural parameters are discussed on rotor frequency. Figure8(a) shows the frequency of the rotor system decreases with the increase of outer diameter. From Figure 8(b), the first-order frequency of the rotor system decreases when the first support moves to the right, but the third-order frequency increases. In Figure 8(c), the second-order frequency of the rotor system decreases obviously when the second support moves to the right, and it has little effect on the frequencies of other orders.Figure 8
The effect of structure parameters on rotor natural frequency: (a) radius of the disk, (b) position of the first support, and (c) position of the second support.
(a)
(b)
(c)


### 4.2. Stability Analysis of Elastic Disk-Flexible Shaft at Aerodynamic Force

The effects of different parameters are discussed on rotor time-domain chart, axis orbit, and spectrum graph. Based on the variation of rotor responses, the stability of rotor system is analyzed. Here, the dimensionless displacement in they direction is denoted as Y, the dimensionless displacement in the z direction is denoted as Z, the ratio of average circumferential speed is denoted as τ, the frequency of rotor response is denoted as f, and the amplitude of rotor response is denoted as A.


#### 4.2.1. The Effect of Eccentric Distance and Rotational Speed

Figures9 and 10 show the responses of the rotor system at different imbalance and rotational speed. According to the spectrum graph, the responses of the rotor system include the frequencies caused by aerodynamic excitation force and rotor imbalance. Summarizing the above figures, rotor frequency caused by aerodynamic excitation force can be obtained, and it is about 50 Hz.Figure 9
Rotor responses of the time domain, axis orbit, and spectrum at eccentric distancese¯=0.001 (a) and e¯=0.01 (b) and rotational speed Ω=800rad/s.
(a)
(b)Figure 10
Rotor responses of time domain, axis orbit, and spectrum at eccentric distancese¯=0.001 (a) and e¯=0.01 (b) and rotational speed Ω=1000rad/s.
(a)
(b)Comparing with rotor responses at different eccentric distances, there is no clear change in rotor motion law and relative relationship between two response frequencies. It illustrates eccentric distance has little effect on rotor stability, and only the amplitude of rotor response is changed.With the increase of rotational speed, the value of vibration frequency caused by the imbalance increases, and the proportion of vibration response caused by aerodynamic excitation force becomes larger. Ultimately, the rotor system is caused unstable when rotational speed increases to be a certain value.


#### 4.2.2. The Effect of Disk Radius

Figure11 shows the responses of the rotor system at different disk radii. According to spectrum graphs, the proportion of rotor responses caused by aerodynamic excitation force becomes larger with the increase in the disk radius and gradually covers the response caused by the imbalance. Comparing with the charts of the time domain, vibration amplitude of the rotor system increases rapidly with the increase in the disk radius. The rotor system becomes unstable when the disk radius increases to be a certain value.Figure 11
Rotor responses of time domain, axis orbit, and spectrum at different disk radii. (a)b=0.4m, (b) b=0.5m, and (c) b=0.6m.
(a)
(b)
(c)


#### 4.2.3. The Effect of Support Position

Figure12 shows the responses of the rotor system at different positions of the second support. According to spectrum graphs, the proportion of rotor responses caused by aerodynamic excitation force decreases with the second support moving to the right. Comparing with the charts of the time domain, vibration amplitude of the rotor system becomes smaller rapidly with the second support moving to the right. It is because that support center of the whole rotor system is also moving to the right; the mass of disk and Alford force can be balanced. The rotor system would be stable when the second support moves to the right at a certain position.Figure 12
Rotor responses of time domainat, axis orbit, and spectrum at different second support positions. (a)l2=1.0 m, (b) l2=1.1m, and (c) l2=1.2m.
(a)
(b)
(c)


## 4.1. Natural Vibration Analysis of Flexible Shaft and Elastic Disk Rotor

### 4.1.1. Comparison of Results at Different Modes

Take the shaft bending mode number, shaft shear mode number, and the disk modal number are, respectively, as 25, 25, and 4. The first eight natural frequencies of the rigid disk-flexible shaft rotor system can be obtained based on the boundary and connection conditions. Similarly, the first eight natural frequencies of the elastic disk-flexible shaft rotor system can be obtained, as shown in Table1.Table 1
First eight natural frequencies of the rigid disk and elastic disk rotor models.
Frequency (rad·s−1)
ω
1
ω
2
ω
3
ω
4
ω
5
ω
6
ω
7
ω
8
Rigid disk rotor
701.26
931.53
1491.22
2879.45
4918.80
8051.80
11967.90
16246.80
Elastic disk rotor
700.51
931.04
1474.40
2585.82
4553.73
7546.20
11250.92
15203.63The corresponding mode shapes are shown in Figure4.Figure 4
Mode shape comparison of the different rotor models: first eight mode shapes of (a) flexible shaft and rigid disk rotor system and (b) flexible shaft and elastic disk rotor system.
(a)
(b)From modal vibration shapes in Figures4(a) and 4(b), it can be seen that the vibration of the disk is not obvious at the low-order frequency. As the frequency increases, the vibration of the disk gradually becomes prominent.The above derivation is performed in the rotating coordinate of shaft and the rotating coordinate of the disc, so the obtained natural frequencies are also relative to the rotating coordinate system. In order to compare with the experimental data, it is necessary to convert the obtained natural frequency into an inertial coordinate system. For the transverse vibration of the shaft and disk, the conversion relationship is different between rotating coordinate system and inertial coordinate system. Hence, there are two different natural frequencies in the inertial coordinate system.


### 4.1.2. Analysis of Coupled Natural Frequency in Transverse Vibration

For transverse vibration of the shaft, conversion relation is expressed as(33)ω0=ω+Ω,where ω is the natural frequency in the rotating coordinate system, Ω is the angular velocity of the shaft, and ω0 is the natural frequency in the inertial coordinate system.For transverse vibration of the disk, the transformation is difficult from the rotating coordinate system to the inertial coordinate system. Compared with the rotation angleΩt, declination angles θξ and θη of transverse vibration can be ignored in the transformation of the coordinate system. So, the relationship between two angular coordinate systems can be expressed as(34)θ0=θ+Ωt.Excitation force of the disc isF0cosω0t, and vibration response at r0=ρ and θ0=0 is expressed as(35)u=Kρ,mF0frcosmθcosω0+mΩtω2−ω0+mΩ2+cosω0−mΩtω2−ω0−mΩ2,where ω is the natural frequency in the rotating coordinate system and m is the number of mode nodal diameters in the disk. It can be seen from the above equation that there are two resonant frequencies in the inertial coordinate system:(36)ω0=ω+mΩ,ω0=ω−mΩ,ω≥mΩ,−ω−mΩ,ω<mΩ.Plotting the chart of frequency-rotational angular velocity,ω−mΩ and −ω−mΩ are the continuous curves. Therefore, natural frequencies are generally divided into two parts when rotating. That is, at a certain rotational angular velocity, two different excitation frequencies will excite the same response mode in the inertial space.For the specific problem, natural frequency of disk-shaft coupling is indicated asω. The antiprecession mode of the rotating system may be negative, and only the first nodal diameter mode of the disk will couple with the vibration of shaft, so m=1. Equation (36) can be expressed as(37)ω0=ω+Ω,ω0=ω−Ω,ω≥mΩ,−ω−Ω,ω<mΩ.According to equation (28), the coupled natural frequency of shaft transverse vibration is shown as Figure 5. Here, 5(a) is the diagram of coupled natural frequency in the flexible shaft and rigid disk, and 5(b) is the diagram of coupled natural frequency in the flexible shaft and elastic disk.Figure 5
Coupled natural frequency of the shaft transverse vibration: (a) flexible shaft and rigid disk system and (b) flexible shaft and elastic disk system.
(a)
(b)In Figure5, the difference between Figures 5(a) and 5(b) is obvious when the rotational speed is large, and the difference between two models is smaller in the region with lower rotational speed. It is illustrated that modal vibration of the shaft occupies a dominant position in the low rotational speed region. Modal vibration of the disk has little effect on the coupled natural frequency of shaft transverse vibration, while it takes a great effect in the high rotational speed region. Hence, the difference between two models is quite obvious.According to equation (37), the coupled natural frequency of elastic disk transverse vibration is shown as Figure 6. Here, 6(a) is the diagram of forward natural frequency, and 6(b) is the diagram of backward natural frequency.Figure 6
Coupled natural frequency of transverse vibration of the disk: (a) forward processional modes and (b) backward processional modes.
(a)
(b)Mass matrix and stiffness matrix of the rotor system will be affected by the coupling effect of shaft and disk, but the gyroscopic matrix and damping matrix are not affected. Without considering the physical background, the disk-shaft coupling term can be directly removed in the mass matrix and stiffness matrix. So, it is to analyze the effect of the disk-shaft coupling on rotor natural frequency.From Figure7, it can be seen that the effect of shaft-disk coupling makes rotor frequency decrease, and different rotational speeds have little effect on rotor frequency.Figure 7
The effect of disk-shaft coupling term on rotor natural frequency: (a)Ω=0rad/s, (b) Ω=500rad/s, and (c) Ω=1000rad/s.
(a)
(b)
(c)The effects of different structural parameters are discussed on rotor frequency. Figure8(a) shows the frequency of the rotor system decreases with the increase of outer diameter. From Figure 8(b), the first-order frequency of the rotor system decreases when the first support moves to the right, but the third-order frequency increases. In Figure 8(c), the second-order frequency of the rotor system decreases obviously when the second support moves to the right, and it has little effect on the frequencies of other orders.Figure 8
The effect of structure parameters on rotor natural frequency: (a) radius of the disk, (b) position of the first support, and (c) position of the second support.
(a)
(b)
(c)


## 4.1.1. Comparison of Results at Different Modes

Take the shaft bending mode number, shaft shear mode number, and the disk modal number are, respectively, as 25, 25, and 4. The first eight natural frequencies of the rigid disk-flexible shaft rotor system can be obtained based on the boundary and connection conditions. Similarly, the first eight natural frequencies of the elastic disk-flexible shaft rotor system can be obtained, as shown in Table1.Table 1
First eight natural frequencies of the rigid disk and elastic disk rotor models.
Frequency (rad·s−1)
ω
1
ω
2
ω
3
ω
4
ω
5
ω
6
ω
7
ω
8
Rigid disk rotor
701.26
931.53
1491.22
2879.45
4918.80
8051.80
11967.90
16246.80
Elastic disk rotor
700.51
931.04
1474.40
2585.82
4553.73
7546.20
11250.92
15203.63The corresponding mode shapes are shown in Figure4.Figure 4
Mode shape comparison of the different rotor models: first eight mode shapes of (a) flexible shaft and rigid disk rotor system and (b) flexible shaft and elastic disk rotor system.
(a)
(b)From modal vibration shapes in Figures4(a) and 4(b), it can be seen that the vibration of the disk is not obvious at the low-order frequency. As the frequency increases, the vibration of the disk gradually becomes prominent.The above derivation is performed in the rotating coordinate of shaft and the rotating coordinate of the disc, so the obtained natural frequencies are also relative to the rotating coordinate system. In order to compare with the experimental data, it is necessary to convert the obtained natural frequency into an inertial coordinate system. For the transverse vibration of the shaft and disk, the conversion relationship is different between rotating coordinate system and inertial coordinate system. Hence, there are two different natural frequencies in the inertial coordinate system.


## 4.1.2. Analysis of Coupled Natural Frequency in Transverse Vibration

For transverse vibration of the shaft, conversion relation is expressed as(33)ω0=ω+Ω,where ω is the natural frequency in the rotating coordinate system, Ω is the angular velocity of the shaft, and ω0 is the natural frequency in the inertial coordinate system.For transverse vibration of the disk, the transformation is difficult from the rotating coordinate system to the inertial coordinate system. Compared with the rotation angleΩt, declination angles θξ and θη of transverse vibration can be ignored in the transformation of the coordinate system. So, the relationship between two angular coordinate systems can be expressed as(34)θ0=θ+Ωt.Excitation force of the disc isF0cosω0t, and vibration response at r0=ρ and θ0=0 is expressed as(35)u=Kρ,mF0frcosmθcosω0+mΩtω2−ω0+mΩ2+cosω0−mΩtω2−ω0−mΩ2,where ω is the natural frequency in the rotating coordinate system and m is the number of mode nodal diameters in the disk. It can be seen from the above equation that there are two resonant frequencies in the inertial coordinate system:(36)ω0=ω+mΩ,ω0=ω−mΩ,ω≥mΩ,−ω−mΩ,ω<mΩ.Plotting the chart of frequency-rotational angular velocity,ω−mΩ and −ω−mΩ are the continuous curves. Therefore, natural frequencies are generally divided into two parts when rotating. That is, at a certain rotational angular velocity, two different excitation frequencies will excite the same response mode in the inertial space.For the specific problem, natural frequency of disk-shaft coupling is indicated asω. The antiprecession mode of the rotating system may be negative, and only the first nodal diameter mode of the disk will couple with the vibration of shaft, so m=1. Equation (36) can be expressed as(37)ω0=ω+Ω,ω0=ω−Ω,ω≥mΩ,−ω−Ω,ω<mΩ.According to equation (28), the coupled natural frequency of shaft transverse vibration is shown as Figure 5. Here, 5(a) is the diagram of coupled natural frequency in the flexible shaft and rigid disk, and 5(b) is the diagram of coupled natural frequency in the flexible shaft and elastic disk.Figure 5
Coupled natural frequency of the shaft transverse vibration: (a) flexible shaft and rigid disk system and (b) flexible shaft and elastic disk system.
(a)
(b)In Figure5, the difference between Figures 5(a) and 5(b) is obvious when the rotational speed is large, and the difference between two models is smaller in the region with lower rotational speed. It is illustrated that modal vibration of the shaft occupies a dominant position in the low rotational speed region. Modal vibration of the disk has little effect on the coupled natural frequency of shaft transverse vibration, while it takes a great effect in the high rotational speed region. Hence, the difference between two models is quite obvious.According to equation (37), the coupled natural frequency of elastic disk transverse vibration is shown as Figure 6. Here, 6(a) is the diagram of forward natural frequency, and 6(b) is the diagram of backward natural frequency.Figure 6
Coupled natural frequency of transverse vibration of the disk: (a) forward processional modes and (b) backward processional modes.
(a)
(b)Mass matrix and stiffness matrix of the rotor system will be affected by the coupling effect of shaft and disk, but the gyroscopic matrix and damping matrix are not affected. Without considering the physical background, the disk-shaft coupling term can be directly removed in the mass matrix and stiffness matrix. So, it is to analyze the effect of the disk-shaft coupling on rotor natural frequency.From Figure7, it can be seen that the effect of shaft-disk coupling makes rotor frequency decrease, and different rotational speeds have little effect on rotor frequency.Figure 7
The effect of disk-shaft coupling term on rotor natural frequency: (a)Ω=0rad/s, (b) Ω=500rad/s, and (c) Ω=1000rad/s.
(a)
(b)
(c)The effects of different structural parameters are discussed on rotor frequency. Figure8(a) shows the frequency of the rotor system decreases with the increase of outer diameter. From Figure 8(b), the first-order frequency of the rotor system decreases when the first support moves to the right, but the third-order frequency increases. In Figure 8(c), the second-order frequency of the rotor system decreases obviously when the second support moves to the right, and it has little effect on the frequencies of other orders.Figure 8
The effect of structure parameters on rotor natural frequency: (a) radius of the disk, (b) position of the first support, and (c) position of the second support.
(a)
(b)
(c)


## 4.2. Stability Analysis of Elastic Disk-Flexible Shaft at Aerodynamic Force

The effects of different parameters are discussed on rotor time-domain chart, axis orbit, and spectrum graph. Based on the variation of rotor responses, the stability of rotor system is analyzed. Here, the dimensionless displacement in they direction is denoted as Y, the dimensionless displacement in the z direction is denoted as Z, the ratio of average circumferential speed is denoted as τ, the frequency of rotor response is denoted as f, and the amplitude of rotor response is denoted as A.


### 4.2.1. The Effect of Eccentric Distance and Rotational Speed

Figures9 and 10 show the responses of the rotor system at different imbalance and rotational speed. According to the spectrum graph, the responses of the rotor system include the frequencies caused by aerodynamic excitation force and rotor imbalance. Summarizing the above figures, rotor frequency caused by aerodynamic excitation force can be obtained, and it is about 50 Hz.Figure 9
Rotor responses of the time domain, axis orbit, and spectrum at eccentric distancese¯=0.001 (a) and e¯=0.01 (b) and rotational speed Ω=800rad/s.
(a)
(b)Figure 10
Rotor responses of time domain, axis orbit, and spectrum at eccentric distancese¯=0.001 (a) and e¯=0.01 (b) and rotational speed Ω=1000rad/s.
(a)
(b)Comparing with rotor responses at different eccentric distances, there is no clear change in rotor motion law and relative relationship between two response frequencies. It illustrates eccentric distance has little effect on rotor stability, and only the amplitude of rotor response is changed.With the increase of rotational speed, the value of vibration frequency caused by the imbalance increases, and the proportion of vibration response caused by aerodynamic excitation force becomes larger. Ultimately, the rotor system is caused unstable when rotational speed increases to be a certain value.


### 4.2.2. The Effect of Disk Radius

Figure11 shows the responses of the rotor system at different disk radii. According to spectrum graphs, the proportion of rotor responses caused by aerodynamic excitation force becomes larger with the increase in the disk radius and gradually covers the response caused by the imbalance. Comparing with the charts of the time domain, vibration amplitude of the rotor system increases rapidly with the increase in the disk radius. The rotor system becomes unstable when the disk radius increases to be a certain value.Figure 11
Rotor responses of time domain, axis orbit, and spectrum at different disk radii. (a)b=0.4m, (b) b=0.5m, and (c) b=0.6m.
(a)
(b)
(c)


### 4.2.3. The Effect of Support Position

Figure12 shows the responses of the rotor system at different positions of the second support. According to spectrum graphs, the proportion of rotor responses caused by aerodynamic excitation force decreases with the second support moving to the right. Comparing with the charts of the time domain, vibration amplitude of the rotor system becomes smaller rapidly with the second support moving to the right. It is because that support center of the whole rotor system is also moving to the right; the mass of disk and Alford force can be balanced. The rotor system would be stable when the second support moves to the right at a certain position.Figure 12
Rotor responses of time domainat, axis orbit, and spectrum at different second support positions. (a)l2=1.0 m, (b) l2=1.1m, and (c) l2=1.2m.
(a)
(b)
(c)


## 4.2.1. The Effect of Eccentric Distance and Rotational Speed

Figures9 and 10 show the responses of the rotor system at different imbalance and rotational speed. According to the spectrum graph, the responses of the rotor system include the frequencies caused by aerodynamic excitation force and rotor imbalance. Summarizing the above figures, rotor frequency caused by aerodynamic excitation force can be obtained, and it is about 50 Hz.Figure 9
Rotor responses of the time domain, axis orbit, and spectrum at eccentric distancese¯=0.001 (a) and e¯=0.01 (b) and rotational speed Ω=800rad/s.
(a)
(b)Figure 10
Rotor responses of time domain, axis orbit, and spectrum at eccentric distancese¯=0.001 (a) and e¯=0.01 (b) and rotational speed Ω=1000rad/s.
(a)
(b)Comparing with rotor responses at different eccentric distances, there is no clear change in rotor motion law and relative relationship between two response frequencies. It illustrates eccentric distance has little effect on rotor stability, and only the amplitude of rotor response is changed.With the increase of rotational speed, the value of vibration frequency caused by the imbalance increases, and the proportion of vibration response caused by aerodynamic excitation force becomes larger. Ultimately, the rotor system is caused unstable when rotational speed increases to be a certain value.


## 4.2.2. The Effect of Disk Radius

Figure11 shows the responses of the rotor system at different disk radii. According to spectrum graphs, the proportion of rotor responses caused by aerodynamic excitation force becomes larger with the increase in the disk radius and gradually covers the response caused by the imbalance. Comparing with the charts of the time domain, vibration amplitude of the rotor system increases rapidly with the increase in the disk radius. The rotor system becomes unstable when the disk radius increases to be a certain value.Figure 11
Rotor responses of time domain, axis orbit, and spectrum at different disk radii. (a)b=0.4m, (b) b=0.5m, and (c) b=0.6m.
(a)
(b)
(c)


## 4.2.3. The Effect of Support Position

Figure12 shows the responses of the rotor system at different positions of the second support. According to spectrum graphs, the proportion of rotor responses caused by aerodynamic excitation force decreases with the second support moving to the right. Comparing with the charts of the time domain, vibration amplitude of the rotor system becomes smaller rapidly with the second support moving to the right. It is because that support center of the whole rotor system is also moving to the right; the mass of disk and Alford force can be balanced. The rotor system would be stable when the second support moves to the right at a certain position.Figure 12
Rotor responses of time domainat, axis orbit, and spectrum at different second support positions. (a)l2=1.0 m, (b) l2=1.1m, and (c) l2=1.2m.
(a)
(b)
(c)


## 5. Conclusions

Vibration differential equations of the elastic disk-flexible shaft rotor are established by the modal synthesis method. Based on the Runge–Kutta method of direct integration, vibration characteristics of the elastic disk and the flexible shaft rotor were analyzed successfully. Then, the effects of Alford force caused by the airflow excitation are further researched on the stability and dynamic response. The results show the following:(1)
Comparing with vibration characteristics of different rotor models, modal vibration of the disk takes a great effect on the coupled natural frequency of shaft transverse vibration in the high rotational speed region. The effect of shaft-disk coupling makes rotor frequency decrease, and different rotational speeds have little effect on rotor frequency. The frequency of the rotor system decreases with the increase in outer diameter. The first-order frequency decreases with the first support moving to the right, but the third-order frequency increases. When the second support moves to the right, the second-order frequency of the rotor system decreases obviously.(2)
Analysis of rotor responses at different structure parameters was performed, and eccentric distance is sensitive to the amplitude of rotor response, but it has little effect on rotor stability, motion law, and relative relationship of two response frequencies. The proportion of rotor responses caused by aerodynamic excitation force becomes larger with the increase of rotational speed and disk radius. The rotor system becomes unstable when rotational speed or disk radius increases to a certain value. However, the proportion decreases with the second support moving to the right, and it is helpful for the stable operation of the rotor system.



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*Source: 3545939-2019-02-14.xml*