CHAPTER 2 ANALYSIS OF SIMPLE ROTOR SYSTEMS

Transcription

1 CHAPTER ANALYSIS OF SIMPLE ROTOR SYSTEMS In previous chapter, a brief history and recent trends in the subject of rotor dynamics has been outlined. The main objective of the previous chapter was to have introduction of various phenomena in rotor dynamics so as to have an idea of them before detailed mathematical treatment is given to comprehensively understand these phenomena. It also introduced various mathematical methodologies, which are used to understand the dynamic behaviour of rotor systems, and briefed the recent and future requirements of the modern high-speed, high-power, and high-reliability rotating machineries. Rotating machines are extensively used in diverse engineering applications, such as power stations, marine propulsion systems, aircraft engines, machine tools, automobiles, household accessories and futuristic micro- and nano-machines. The design trend of such systems in modern engineering is towards lower weight and operating at super critical speeds. An accurate prediction of rotor system dynamic characteristics is vitally important in the design of any type of machinery. There have been many studies relating to the field of rotor dynamic systems during the past years (Biezeno and Grammel, 1959; Chen and Gunter, 005; Dimentberg, 1961; Tondl, 1965; Rieger, 1977; Dimargonas and Paipetis, 1983; Mahrenholtz, 1984; Vance, 1988; Goodwin, 1989; Childs, 1993; Krämer, 1993; Lee, 1993; Rao, 1996; Lalanne and Ferraris, 1998; Genta, 1999; Rao, 000; Admas, 001; Yamamoto and Ishida, 001; Robert, 003; Muszynska, 005; Genta, 005. Of the many published works, the most extensive portion of the literature on rotor dynamics analysis is concerned with determining critical speeds, natural whirl frequencies, the instability thresholds and bands, and the unbalance and transient responses. Apart from these analyses some works also cover balancing of rotors, the estimation of bearing dynamic parameters, the condition monitoring and the nonlinear analysis. For understanding basic phenomena of any dynamic system requires adequate modeling of the system. To start with only transverse vibrations of the rotor are considered. The torsional and axial vibrations will be considered in subsequent chapters. The dynamic system can be as simple as single degree of freedom (DOF. The rotor is considered as a single mass in the form of a point mass, a rigid disc or a long rigid shaft. In a three-dimensional space a particle and a rigid body can have utmost three and six DOFs, respectively. On neglecting the torsional and axial vibrations effects, the single mass rotor has at most four DOF. In the present chapter, mathematical treatment is performed of simple rotor models in use over

2 5 the years is detailed, e.g., the single-dof undamped and damped, two-dof Rankine s model, the two and three-dof Jeffcott rotor models, and four-dof rotor models derived from the Jeffcott rotor model. Various terminologies in use to explain the dynamic behaviour of the rotor system, e.g. the unbalance, the whirling and wobbling motions, the natural and excitation frequencies, the resonance, the critical speed, the synchronous, anti-synchronous and asynchronous motions, the phase, etc.; are introduced. The understanding of the present chapter would help in exploring more complex rotor models described in subsequent chapters..1 Single-DOF Undamped Rotor Model The simplest model of the rotor system can be a single DOF. Figure.1 shows two types of rotor model. In Figure.1(a the bearing (support is assumed to be flexible and the rotor (the shaft and the disc as rigid; whereas in Figure.1(b the bearing is assumed to be rigid (i.e., the simply supported and the shaft as flexible with the disc as rigid. The mass of the rotor is considered as that of the rigid disc that is mounted on the massless thin shaft. Both the cases can be idealized as a single DOF as shown in Figure.1(c. Fig.1(a A rigid rotor mounted on flexible bearings Fig.1(b A flexible shaft with a rigid disc mounted on rigid bearings Fig.1(c An equivalent single degree of freedom spring-mass system Fig.1(dThe free body diagram of the disc mass

3 6.1.1 Unbalance force model: If the rotor is perfectly balanced then theoretically speaking there will not be any unbalance force as shown in Figure.(a, where C and G are the center of rotation (or the geometrical center and the center of gravity of the rotor, respectively. However, in actual practice it is unattainable to have a perfectly balanced rotor as shown in Figures.(b-(d, where U is the location of an additional unbalance mass (for example a small screw attached to the disc. The unbalance may come due to manufacturing tolerances, operational wears and tears, thermal distortions, repair, etc. The rotor unbalance gives a sinusoidal force at the rotor spin speed. Thus, the unbalance force is modeled as a sinusoidal force f ( t = mω esinωt (.1 where m is the mass of the rotor, ω is the spin speed of the rotor, e is the eccentricity of the rotor (see Figure.(b and the product me is normally called the unbalance. The above unbalance force will come when the rotor is eccentric, i.e., the rotor center of rotation and the center of gravity are not coincident. This type of unbalance is called the residual or inherent unbalance. When the rotor is not eccentric (i.e., when C and G are coincident, and a small unbalance mass, m i is attached at a relatively larger radius of r i as shown in Figure.(c, the unbalance force can be written as f ( t = m ω r sinωt i i (. The above type of unbalance is called the trial or additional unbalance and it is often used for the residual unbalance calculation using dynamic balancing procedures. For the case when the rotor is eccentric and a small trial unbalance mass is attached as shown in Figure (d, the total unbalance force would be f ( t = mω esinωt + m ω r sin( ωt + φ i i (.3 where φ is the phase difference between the vectors of unbalance forces due to the rotor eccentricity and the trial unbalance mass.

4 7 C,G C G (a Rotor geometrical center and center of gravity are coincide. (b Rotor geometrical center and center of gravity are not coincide. U C,G U C G (c An additional unbalance mass location is not coincident with rotor geometrical center and center of gravity. (d Rotor geometrical center, center of gravity and unbalance mass location are not coincident. Figure. Different types of unbalances in a single plane Figure.3 shows positive conventions and variables to define the unbalance location on a rotor system at a particular instant of time, t. For a constant angular velocity of the rotor, ω, the location of the unbalance is given as θ = ωt (in general θ ω (the dot represents the derivative with respect to the time, i.e., when a rotor has some angular acceleration, e.g., for a constant angular acceleration θ = α and for zero initial conditions, we will have θ = αt and 1 θ = α. t Figure.3 Unbalance location on a rotor at a particular instant of time

5 8.1. Equations of Motion: On application of the Newton s second law of motion on the free body of the rotor mass as shown in Figure.1(d, i.e., on equating the sum of external forces to the mass of rotor multiplied by the acceleration of center of gravity of the rotor mass, we have F y = my or k eff y + mω esinωt = m y (.4 where k eff is the effective stiffness of the rotor system (see Figure.1 and keff y is the restoring force that acts opposite to the motion so negative sign. Equation (.4 is a standard form of equation of motion of a single DOF spring-mass system and could be rearranged as m y + keff y = mω esinωt ( Free Vibrations: From the free vibration, when the external unbalance force is absent (i.e., e = 0, it is generally assumed that the rotor mass will have a simple harmonic oscillation and the free response displacement is expressed as y( t = Y sinω nf t (.6 where ω nf is the frequency of oscillation during the free vibration and it is called the natural frequency of the system. On substituting equation (.6 (with y( t = Yω sinω t into the homogeneous part (i.e., with e = 0 of equation of motion (.5, it gives nf nf ( mω + k Y sinω t = 0 (.7 nf eff nf For a non-trivial solution (i.e., Y 0 or ω 0 of equation (.7, the natural frequency of the system can be written as nf ( mω + k = 0 or ω = m (.8 nf eff nf k eff /.1.4 Forced responses: Now to obtain the steady state forced response, it can be expressed as y( t = Y sin( ωt φ (.9 where Y is the amplitude of displacement and φ is the phase lag of displacement with respect to the unbalance force. On substituting equation (.9 with y( t = Yω sin( ωt φ steady state forced response amplitude can be written as into equation (.5, the

6 9 ( + eff sin( = sin mω k Y ωt φ meω ωt On expanding, we get [ ] ( + eff sin cos cos sin = sin mω k Y ωt φ ωt φ meω ωt On separating the sine and cosine terms of ω t, we have ( mω + keff Y cos φ = meω and ( mω + keff Y( sin φ = 0 Since terms ( ω m + k eff and Y in general may not be zero, hence from the second expression, we have φ = 0 and the first expression can be simplified as Y Y mω ω ω = = = = e k mω ω ω 1 ω eff nf with ω ω = and ω nf = k eff / m (.10 ω nf where Y is the non-dimensional unbalance response (ratio of the unbalance response to the eccentricity and ω is the frequency ratio (ratio of the spin speed of the rotor to the natural frequency of the rotor system. The absolute value of non-dimensional unbalance response, Y, is plotted with respect to the frequency ratio as shown in Figure.4. From Figure.4 and equation (.10 it should be noted that we have unbounded unbalance response when the denominator speed is ω becomes zero, i.e., when the spin (1 ω cr ωcr = = ± 1 or ω nf ω cr keff = ± = ± ωnf (.11 m This is a resonance condition and the spin speed corresponding to the resonance is defined as critical speed. The subscript: cr represent the critical. For the present case the critical speed is equal to the transverse natural frequency of the non-rotating rotor system as given by equation (.11. The undamped steady state forced response amplitude tends to be infinity at the critical speed. The natural frequency and the critical speed concepts have come from the free and forced vibrations, respectively. It should be noted that in rotor dynamics, in general, the natural frequency might not be a constant and might vary with the spin speed of the shaft (e.g., when gyroscopic couple is considered in the analysis when the spin speed of rotor is high, for the case of speed dependent bearing dynamic properties, etc.. The ± sign represent that the rotor will have critical speed while rotating in either the clockwise or counter clockwise sense. Since

7 30 the damping is not considered in the analysis the phase angle, φ, becomes zero (or after the critical speed. In Fig..4 the response changes its sign (i.e., the positive to the negative after the frequency ratio equal to unity, which corresponds to the critical speed (i.e., the spin speed of the shaft is equal to the natural frequency of the rotor system. It means that the phase difference between the force and the response becomes 180 0, which is 0 0 when the frequency ratio is less than unity, in the absence of damping. Both the linear and semi-log plots are shown to have clarity of the response variation, near and away from the critical speed. It can be seen that as the frequency ratio increases above unity the non-dimensional response asymptotically approaches to unity, which means unbalance response approaches to the eccentricity of the rotor. Physically it implies that the rotor rotates about its center of gravity at high frequency ratio. The analysis presented in this section can be applied to the transverse, torsional and axial vibrations of rotors and accordingly natural frequency can be termed by prefixing respective types of vibrations. For torsional vibrations care should be taken that the mass will be replaced by the polar mass moment of inertia of the rotor and the stiffness will be by the torsional stiffness. Similarly, for the axial vibrations the mass will remain the same as transverse vibrations, however, the stiffness will be the axial stiffness. More detailed treatment will be presented in subsequent chapters. Y Y ω (a Linear plot ω (b Semi-log plot Figure.4 Variation of the non-dimensional unbalance response versus the frequency ratio

8 31.1.5: Attenuation of Vibrations: The most common cause of the vibrations in rotors is the unbalance among the other faults and it will always be present in a rotor. However, the unbalance response can be reduced up to the desired level depending upon the applications by one or combination of following three methods. (i Correction at the source: Balancing the rotor is the most direct approach, since it attacks the problem at source. However, in practice a rotor cannot be balanced perfectly and the best achievable state of balance tends to degrade during operation of a rotor (e.g., the turbo-machinery. There are two type of unbalances (a the static unbalance: The principal axis of the polar mass moment of inertia of the rotor is parallel to the centerline of the shaft as shown in Figure.5a and b. The rotor can be balanced by a single plane balancing and (b the dynamic unbalance: The principal axis of the polar mass moment of inertia of the rotor is inclined to the centerline of the shaft as shown in Figure.5c and d. For balancing such (rigid rotors, minimum of two balancing planes are required. (a Perfectly balance (no force and moment (b Static unbalance (pure radial force (c Dynamic unbalance (pure moment (d Dynamic unbalance (both force and moment Figure.5 Classification of unbalances for a short rigid rotor (ii Operate the rotor away from the critical speed: This could be done during the design itself, or during operation by providing an auxiliary support. At the design stage the critical speed can be altered by changing the rotor mass and its distributions, and the effective stiffness (e.g., by changing dimensions of the shaft, i.e., the shaft diameter and length. During the operation an auxiliary support can be provided

9 3 to increase the effective stiffness of the rotor, which in turn increases the critical speed. For the case when the rated operational speed is above the critical speed, the actual rotor critical speed can be safely traversed by this arrangement (by temporarily increasing the critical speed and then the auxiliary support can be withdrawn which brings the critical speed of the rotor again below the rated operation speed (refer Example.. In general, changing the critical speed is useful for machines with a constant or with a narrow range of operational speed (e.g. the turbo-machinery in power plants. Example.1: A rotor has a mass of 10 kg and the operational speed of (100 ± 1 rad/s. What should be bounds of the effective stiffness of shaft so that the critical speed should not fall within 5% of operating speeds? Assume that there is no damping in the rotor system. Solution:The operational speed range is 99 to 101 rad/s. Now 5% of the lower operational speed would be = rad/s, and 5% of the upper operational speed would be = rad/s. Then the effective stiffness corresponding to the lower operating speed would be ω m = nf = kn/m and the effective stiffness corresponding to the upper operating speed would be ω m = nf = 11.5 kn/m. Hence, the effective stiffness of shaft should not fall in the range of 98.1 to 11.5 kn/m. It should be noted that the unit of ω nf is in rad/s when other quantities are in SI units (i.e. m in kg and k in N/m. Example.: A rotor system has 100 rad/s as the critical speed and its operating speed is 10 rad/s. If we want to avoid altogether crossing of the critical speed, then what should be the enhancement in the support stiffness by an auxiliary support system. To avoid excessive vibration, let us assume we should have at least 5 rad/s of gap between the operating speed and the critical speed. The rotor has a mass of 10 kg. Solution: The initial stiffness of the support is ω m = nf = 100 kn/m. First we can reach safely a rotor speed of 95 rad/s, which is 5 rad/s lower than the original critical speed of the rotor. Now since we cannot safely increase the rotor speed further, we need to increase the critical speed of the rotor to at least 15 rad/s. This will allow us to reach up to 10 rad/s that is 5 rad/s lower than 15 rad/s i.e. the new critical speed. The corresponding effective of the support stiffness should be = kn/m. Hence, the auxiliary support system should increase the effective stiffness by 56.5 kn/m.

10 33 Example.3: A kg mass of a overhung rotor (cantilever caused the deflection at the free end of 0.5 cm. What would be the stiffness and the natural frequency of the system? The stiffness of the spring = k = corresponding static deflection δ static force mg = = = N/m k mg / δ g 39.4 The natural frequency = ωn = = = = = rad/s =.3 Hz 3 m m δ 10 (iii Add damping to the system or the active control of the rotor: If a critical speed must be traversed slowly or repeatedly, or if machine operation near a critical speed can not be avoided, then the most effective way to reduce the amplitude of vibration is to add the damping. On the other hand the some form damping (e.g., the shaft material or hysteretic or internal damping may lead to rotor instability (self-excited vibration. The squeeze film and magnetic bearings are often used to control the dynamics of such rotor systems. Squeeze-film bearings (SFB are, in effect, fluid-film bearings in which both the journal and bearing are non-rotating. The ability to provide the damping is retained but there is no capacity to provide the stiffness as the latter is related to journal rotation. They are used extensively in applications where it is necessary to eliminate instabilities, and to limit rotor vibration and its effect on the supporting structures of rotor-bearing systems, especially in aeroengines. In recent years, advanced development of electromagnetic bearing technology has enabled the active control of rotor bearing systems through active magnetic bearings (AMB. In particular the electromagnetic suspension of a rotating shaft without the mechanical contact has allowed the development of supercritical shafts in conjunction with modern digital control strategies. With the development of smart fluids (for example electro and magneto-rheological fluids now new controllable bearings are in the primitive development stage. The basic premise of such smart fluids is that their dynamic properties (i.e. the damping and the stiffness can be controlled by changing the current or magnetic flux in a micro-second time. Schematics of typical passive and active (i.e., smart or controllable squeeze film dampers, and active magnetic bearings are shown in Figure.6.

11 34 Figure.6 (a A passive squeeze film damper Figure.6 (b Schematic diagram of a smart (active fluid-film damper Figure.6 (c Basic principles of active magnetic bearings

12 35. A Single-DOF Damped Rotor Model In the previous section the damping was ignored in the rotor model and now in the present section its effects would be considered. The simplest damping model is the viscous (or proportional damping and the damping force is expressed as fd ( t = cy (.1 where c is a constant of proportionality and is called the viscous damping coefficient. In the free body diagram (Fig..1d of the rotor with addition of damping force, it gives equations of motion as f ( t ky cy = my or my + cy + ky = f ( t (.13 For the free damped motion, equation (.13 takes the form my + cy + ky = 0 (.14 Let us assume a solution of the form y st = e (.15 where s is a constant (may be a complex number, so that (.15 in equation (.14, we get st and y = se st. On substituting equation y = s e st ( ms cs k e = (.16 Hence, from the condition that equation (.15 is a solution of equation (.14 for all values of t, equation (.16 gives the following characteristic equation s + c k s 0 m + m = (.17 which can be solved as s 1, c c k = ± m m m (.18 Hence, the following form of the general solution is obtained

13 36 s1t s t y( t = Ae + Be (.19 where A and B are constants to be determined from initial conditions of the problem (e.g., y(0 = y0 and y (0 = v. On substituting equation (.18 into equation (.19, we get 0 c k c k c t t t m m m m m y( t = e Ae + Be (.0 The term outside the bracket in right hand side is an exponentially decaying function for positive value of c. However, the term c k m m can have following three cases: (i c k > : Exponents of terms within the square bracket in equation (.0 are real numbers, which m m means there would not be any harmonic functions. Hence, no oscillation is possible and it is called the over-damped system (see Figure.7 Figure.7 Response of an over-damped system (ii c k < : Exponents of terms within the square bracket in equation (.0 are imaginary numbers m m k c ± j m m, which means, we can write Hence, equation (.0 takes the following form k c ± j t m m k c k c e = cos t ± jsin t. m m m m c t m k c k c y( t = e ( A + Bcos t j( A Bsin t m m + m m

14 37 Let a = ( A + B = X cosφ and b = i( A B = X sinφ, we get c t m k c x = Xe cos t φ m m ; with φ 1 = tan ( b / a; X = a + b (.1 For the present case oscillations are possible (with decaying type and it is called the under-damped system (Fig..8. The damped natural frequency is given as ω nf d k c = m m. Figure.8 Response of an under-damped system (iii c k = : Exponents of terms within the square bracket in equation (.0 are zero numbers. The m m damping corresponding to this case is called the critical damping, c c, which is defined as c = m k / m = mω = km (. c n Any other damping can be expressed in terms of the critical damping by a non-dimensional number ζ called the damping ratio, as ζ = c / cc (.3 The solution for the critical damped case, having two real repeated roots, can be expressed as ω [ ] nf t x( t = e A + Bt (.4

15 38 For this case either no oscillation (similar to Fig..7 or for specific initial conditions single crossing of the zero amplitude axis is possible. Figure.9 shows a response of the critically damped system with a single crossing. Figure.9 Response of an critically damped system with a single zero crossing To summarize, depending upon the value of damping ratio ζ we can have following cases (i ζ > 1: the over-damped condition (ii ζ < 1: the under-damped condition with the damped natural frequency as nf = 1 d nf (iii 1 ω ω ζ ζ = : the critical damping and (iv ζ = 0 : the undamped system. For all the three cases, the integration constants A and B are obtained from two initial conditions. Figure.10 shows a response of an unstable system with ζ < 0 in which exponential increase in the amplitude can be seen. More detailed treatment of the present section could be seen in the text on basics of vibrations (Thomson and Dahleh, Figure.10 Response of an unstable system

16 39.3 Rankine Rotor Model The single DOF rotor model has limitations that it cannot represent the orbital motion of the rotor in two transverse directions, which the case in practice. Rankine (1869 used a two DOF model to describe the motion of the rotor in two transverse directions as shown in Figure.11(a. The shape of orbit produced depends upon the relative amplitude and phase of the motions in two transverse directions (which in turn depend upon relative difference in stiffness in two transverse directions. The orbit could of a circular, elliptical or straight line, inclined to the x and y axes, as shown in Figure.1. The straight-line motion (Fig..1c could be considered as a single-dof system considered in Section.1, whereas, the elliptical orbital motion (Fig..1b may occur when the shaft has different stiffness in two orthogonal directions. (a Two- DOF spring-mass rotor model (b Free body diagram of the model Figure.11 Rankine rotor model (a Circular motion (b Elliptical motion (c Straight-line motion Figure.1 Orbital motion of the mass

17 40 For the circular orbital motion (Fig..1a may occur for a symmetrical shaft. It can be thought of a mass attached with a spring and is revolved about a point. From the free body diagram of the rotor as shown in Figure.11(b, for a constant spin speed, the radius of whirling of the rotor center will increase parabolically and is given as r = F / c k where F c is the centrifugal force. It can be physically also visualized as there will not be any resonance condition, as it is found in the single DOF model, when the spin speed is increased gradually. This is a serious limitation of the Rankine model. Moreover, this model does not represent the realistic rotating unbalance force..4 Jeffcott Rotor Model To overcome the limitations of the Rankine model, Jeffcott (1919 proposed a model and Figure.13 shows a typical Jeffcott (it is also called Föppl or Laval rotor model. It consists of a simply supported flexible massless shaft with a rigid disc mounted at the mid-span. The disc center of rotation, C, and its center of gravity, G, is offset by a distance, e, which is called the eccentricity. The shaft spin speed is, ω and the shaft whirls about the bearing axis with a whirl frequency, ν. For the present case, the synchronous whirl is assumed (i.e., ν = ω, which is prevalent in the case of unbalance responses. The synchronous motion also occurs between the earth and the moon, and due to this we see always the same face of the moon from the earth. In the synchronous motion the shaft, the orbital speed and its own spin speed are equal as shown in Figure.14(a. The sense of rotation of the shaft spin and the whirling are also same. The black spot on the shaft represent the unbalance location or any other mark on to the shaft. The unbalance force in general leads to synchronous whirl conditions hence this motion is basically a forced response. Other kinds of whirl motion, which may occur in real system, are: anti-synchronous (i.e., ν = ω ; as shown in Figure.14(b and asynchronous (i.e., ν ω. The anti-synchronous whirl may occur when there is rub between the rotor and the stator, however, it occurs very rarely. For this case, the sense of rotation of the shaft spin and the whirling are opposite. Asynchronous whirl motion may occur when speeds are high (e.g., when gyroscopic effects are predominate or when the rotor is asymmetric or when dynamic properties of the bearing are anisotropic. The asynchronous whirl motion may occur even in the perfectly balanced rotor, and due to this it will have whirl frequency as one of the natural frequency of the rotor system as long as the rotor linear model is considered. The black mark on to the shaft will not be so systematic as in Figure.14 and may occupy various positions depending upon the frequency of whirl.

18 41 The transverse stiffness, k, of a shaft simply supported shaft is expressed as Load P 48EI k = = = ( Deflection PL / 48EI L ( where E is the Young s modulus, I is the second moment of area of the shaft cross-section and L is the span of the shaft. Coordinates to define the position of the center of rotation of the rotor, C, are x and y (Fig..13c. The location of the unbalance is given by θ, which is measured from the x-axis in the counter clockwise direction. Thus, three geometrical coordinates (x, y, θ are needed to define the position of the Jeffcott rotor (i.e., it has three-dofs. Fig.13(a A Jeffcott rotor model Fig.13(b A Jeffcott rotor model in y-z plane Fig.13(c Free body diagram of the disc in x-y plane

19 4 Shaft whirling direction Shaft whirling direction Shaft Shaft Shaft spin direction Figure.14(a Synchronous whirl Shaft spin direction Figure.14(b Anti-synchronous whirl From Figure.13(c the force (or the moment balance in the x, y and θ directions can be written as d kx cx = m ( x + ecosθ (.6 dt and d ky cy mg = m ( y + esinθ (.7 dt mgecosθ = I θ (.8 p where ( x + ecosθ and ( y esinθ + are the position of the centre of gravity, G, of the rotor; m is the mass and I p is the polar mass moment of inertia of the disc. Apart from the restoring force contribution from the shaft, a damping force is also considered. The damping force is idealized as viscous damper and it is mainly coming from the support and aerodynamic forces at the disc. The material damping of the shaft is not considered, which may leads to the instability in the rotor and it will be considered in detail in subsequent chapters. For the case θ = ωt, i.e., when the disc is rotating at a constant spin speed, the Jeffcott rotor model reduces to two-dof rotor model. Physically it means that only the transverse vibration is considered and the torsional vibration is ignored. Neglecting also the effect of gravity force by considering the static equilibrium as the reference for axes system and this effect will be considered when we will discuss the

20 43 sub-critical speed phenomenon. Hence, equations of motion in the x and y directions, from equations (.6 and (.7, can be written as and mx + cx + kx = mω ecosωt (.9 my + cy + ky = mω esinωt (.30 It should be noted that above equations of motion are uncoupled, and motion can be analysed independently in two transverse planes. Noting equation (.8, from the undamped free vibration analyses it can be seen that since the rotor is symmetric, the rotor system will have two equal transverse natural frequencies in two orthogonal directions and are given as ω k m (.31 nf / = 1,.4.1: Steady-state response (Approach 1: The damping does not affect the natural frequency of the system appreciably. However, its effect is more predominate for suppressing the vibration amplitude at the resonance. Steady state forced responses of equations (.9 and (.30 can be assumed as and x = X cos( ωt φ (.3 [ ] y = Y cos ( ωt ( φ + π / = Y sin( ωt φ (.33 where X and Y are the steady state forced response amplitude in the x and y directions, respectively; ω is same as the excitation frequency due to the unbalance force and is equal to the shaft spin speed (synchronous condition is assumed, φ is the phase lag of the displacement, x(t, with respect to the unbalance force. The phase difference between the two orthogonal direction responses for the direction of spinning of the shaft chosen is π/ radian. For the direction of shaft whirling shown in Figure.13 (i.e., counter clockwise; ccw for the present axis system, the response in y-direction will lag the response in x- direction by π/ radian. Hence the lag of the y-direction response with respect to the force will be (π/ + φ. On taking the first and second time derivatives of the response, x(t, we get x = ω X sin( ωt φ and x = ω X cos( ωt φ (.34

21 44 On substituting equations (.33 and (.34 into equation (.9 and separating the in-phase (i.e., cosωt and quadrature (i.e., sinωt terms, we get and ω φ ω φ φ ω m X cos + cx sin + kx cos = m e (.35 ω φ ω φ φ m X sin cx cos + kx sin = 0 (.36 Equation (.36 can be solved for the phase, as cω tanφ = k mω (.37 which gives sinφ = cω ( k mω + ( cω and cosφ = k mω ( k mω + ( cω (.38 Substituting equation (.38 into equation (.35, it gives the displacement amplitude as X = mω e ( k mω + ( cω (.39 Similarly, we can obtain the response amplitude in the y-direction from equation (.30 as Y = mω e ( k mω + ( cω (.40 Above amplitudes could be plotted with respect to spin speed of the shaft for overall understanding of the response, and it will be seen in the next subsection. From equations (.3, (.33, (.39 and (.40, it can be seen that because of the symmetry of the rotor we have X = Y = R and the orbit is a circular in nature, i.e., x + y = R (.41.4.: Steady-state response (Approach : An alternative approach that is very popular in rotor dynamics analyses is to use the complex algebra to define the whirl radius as r = x + jy (.4

22 45 where j = 1. On multiplying equation (.30 by j and adding to equation (.9, we get jω + + = e t with mr cr kr meω = ω + ω (.43 jωt e cos t jsin t Now the steady state response can be assumed as r R ω φ j( t = e (.44 where R is the whirl amplitude (it is a real quantity, φ is the phase lag of response with respect to the unbalance force. On differentiating equation (.44 with respect to time, t, we get j( and r = jωre ωt φ j( r Re ωt = φ (.45 ω On substituting equations (.44 and (.45 into equation (.43, we get {( j } k m c R me e j φ ω + ω = ω (.46 Equation (.46 can be written as {( j }( cos j sin k m c R R me ω + ω φ φ = ω (.47 On separating the real and imaginary parts of equation (.47, we get and ( k mω Rcosφ ωcrsinφ meω + = (.48 ω φ ω φ ( k m Rsin + crcos = 0 (.49 From equation (.49, we get the phase as cω tanφ = k mω (.50 On substitution of the phase from equations (.50 into (.48, the whirl amplitude can be written as R = mω e ( k mω + ( cω (.51 Equations (.49 and (.51 are similar to previous results, i.e., equations (.37 and (.39. The nondimensional form of equations (.50 and (.51 can be written as ζω tanφ = 1 ω (.5

23 46 and with R R = = e ω ( 1 ω + ( ζω (.53 ω = ω / ω ; ω = k / m; ζ = c / c ; c = km (.54 nf nf c c where R is the whirl amplitude ratio, ω is the frequency ratio, ω nf is the natural frequency of nonrotating rotor system, ζ is the damping ratio, and c c is the critical damping for which the damping ratio is equal to unity. Fig..15 shows the whirl amplitude and the phase variation with the frequency ratio. Figure.15(a shows that the maximum amplitude (i.e., the location of the critical speed occurs at ω = 1 for the undamped case; however, at slightly higher frequency ratio than one (i.e., ω > 1, when damping is present in the system. It should be noted that we observed previously that damped natural frequency is lesser than the undamped case. It could be observed that the damping has the most important parameter for reducing the whirl amplitude at the critical speed. Figure.15 Plot of (a Non-dimensional response (b Phase versus frequency ratio ω

24 47 It can be seen from Figure.15(b, for a lightly under-damped system the phase angle changes from 0 0 to 90 0 as the spin speed is increased to ω nf (i.e., ω =1 and gradually becomes as the spin speed is increased to a higher frequency ratio. It should be noted that the phase angle is 90 0 at ω =1 even for the case of various level of damping in the rotor system. For highly over-damped system the phase angle always remain at 90 0 before and after ω =1, which might be a physically unrealistic case to attain. As the spin speed crosses the critical speed, the center of gravity of the disc comes inside of the whirl orbit and the rotor tries to rotate about the center of gravity. This can be seen from Figure.15(a as the spin speed approaches infinity the displacement of the shaft tends to the equal to the disc eccentricity ( R = 1. Since the measurement of amplitude of vibration at the critical speed is difficult, hence the determination of precise critical speed is difficult. To overcome this problem the measurement of the phase at critical speed is advantageous (since it remain constant at 90 0 irrespective of damping in the system. The change in phase between the force and the response is also shown in Figure.16 for three difference spin speeds, i.e., below the critical speed, at the critical speed, and above the critical speed. (a below critical speed (b at critical speed (c after critical speed Figure.16 Orientation of the unbalance force when damping is present in the rotor system Since for the present analysis the synchronous whirl condition is assumed, at a particular speed shaft will not have any flexural vibration and it (in a particular bend configuration will whirl (orbiting about its bearing axis as shown in Figure.14(a. It can be seen that the black point in the shaft will have compression during the whirling. However, it can be seen from Figure.14(b for anti-synchronous whirl that the shaft (the black point in shaft will have reversal of the bending stresses twice per whirling of the shaft. For asynchronous whirl the black point inside the shaft will take all the time different positions during whirling of the shaft.

25 48.4.3: Steady-state response (Approach 3: With the development in software, which can handle complex matrices, the following matrix procedure may be very helpful for the numerical simulation, which may be extended even for very complicated rotor systems. Equations (.9 and (.30 can be combined in the matrix form as m 0 x c 0 x k 0 x mω ecosωt m + + = y c y k y mω esinωt (.55 The force vector in equation (.55 could be expressed as ( jωt ( j ωt ω ( cosω + jsinω meω e F jωt e ( sin jcos mω e cosωt m e t t unbx Re Re Re = = = j mω esinωt mω e ωt ωt meω e Funb y (.56 F unby = jf (.57 unbx where Re(. represents the real part of the quantity inside the parenthesis. F unb x and F unb y are the unbalance force components in x and y directions, respectively. On substituting equation (.56 into equation (.55, it can be written as (henceforth for brevity the symbol Re(. will be omitted m 0 x c 0 x k 0 x Funb x e 0 m + y + = 0 c y 0 k y F unby jωt (.58 The relationship (.57 is true for the present axis system along with directions of the whirling (R and the unbalance force vectors chosen as shown in Figure.17(a. For this case F unb y lag behind F unb x by Let us derive this relationship: If Funb x j = Fe θ, then [ π π ] j( θ π / j θ ( π / j jθ jθ Funb = Fe = Fe e = Fe cos( / + jsin( / = jfe = jfunb y For the direction of whirl (R opposite to the axis system as shown in Figure.17(b, the following relationship holds so that, [ π π ] j( θ + π / j θ ( π / j jθ jθ Funb = Fe = Fe e = Fe cos( / + jsin( / = jfe = jfunb F y unby = jf (.59 unbx x x

26 49 in which case F unb y lead F unb x by It should be noted in equation (.58 that the right hand side force vector elements have significance of real parts only, which is quite clear from the expanded form of the force vector in equation (.56. Equation (.58 can be written in more a compact form as [ M ]{ x} + [ C]{ x} + [ K ]{ x} = { F } e jωt (.60 unb The solution can be chosen as { x} { X} e jωt = (.61 where elements of the vector {X} are, in general, complex. Fig.17(a The direction of whirl same as the positive axis direction Fig.17(b The direction of whirl opposite to the positive axis direction Equation (.61 can be differentiated to give ωt { } = j ω { } and { } = ω { } j jωt (.6 x X e x X e On substituting equations (.61 and (.6 into equation (.60, we get with [ Z ]{ X } = { F unb } (.63 ( [ Z ] ω [ M ] [ K ] jω [ C] = + + (.64 where [Z] is the dynamic stiffness matrix. The response can be obtained as

27 50 1 { } [ ] { } X = Z F unb (.65 The above method is quite general in nature and it can be applied to multi-dof systems also once equations of motion in the standard form are available. The following example illustrates the method discussed in the present section for a Jeffcott rotor. Example.4: Obtain the unbalance response of a rotor system with following equations of motion. mx + kx = mω ecosωt and my + ky = mω esinωt Solution: Since equations of motion are uncoupled hence both equations can be solved independently. The first equation can be written as j t + = x with mx kx F e ω F x = meω in which the real part of right hand side term has only meaning. The solution can be assumed as j t x = Xe ω where in general X is a complex quantity. The above equation gives j t x = ω Xe ω On substituting assumed solutions in the equation of motion, we get which gives ( ω m X + kx = meω X meω = k mω Hence the response in the x-direction becomes meω jωt meω meω x = e = Re (cos t jsin t cos t ω + ω = ω k mω k mω k mω Similarly, the second equation of motion can be written as + = with j t my ky Fye ω Fy = jmeω in which the real part of the right hand side term only has a meaning. The solution can be assumed as

28 51 j t y = Ye ω On substituting in the equation of motion, we get F y Y = k mω Hence the response in the y-direction becomes Fy jωt jmeω jωt meω meω y = e = e = Re ( jcos t sin t sin t ω + ω ω = k mω k mω k mω k mω (Answer Now the same problem is solved in the matrix form. We have so that m 0 k me ω = unb 0 m = 0 k = = 0 0 jmeω [ M ] ; [ K ] ; [ C] ; { F } [ ] ω [ ] [ ] ( k mω k mω 0 1/ 0 Z M K Z 0 k mω 0 1/ k m 1 = ( + = ; [ ] = 1 { X} [ Z ] { F } ( 1/ k mω 0 meω 1 meω = unb = = 0 1/ ( k mω jmeω ( k mω jmeω The response in the x- and y-directions can be written as ( ω { x} x 1 meω jωt 1 meω cosωt = = Re e = y ( k mω jmeω ( k mω meω sinωt (Answer.5 A Jeffcott Rotor Model with an Offset Disc Figure.18(a show a more general case of the Jeffcott rotor when the rigid disc is placed with some offset from the mid-span. With a and b locate the position of the disc in a shaft of length l. The spin speed of the shaft is considered as constant. For such rotors apart from two transverse displacements of the center of disc, i.e., x and y, the tilting of disc about the x and y-axis, i.e., ϕ x and ϕ y, also occurs; and it makes the rotor system as a four DOFs. For the present analysis, the effect of the gyroscopic moment has been neglected. In Fig..18(b points C and G represent the geometrical center and the center of gravity of disc, respectively. The angle, φ, represent the phase between the force and the response.

29 5 ϕ x Fig.18(a A Jeffcott rotor with a disc offset from the mid span in the y-z plane Fig.18(b Free body diagram of the shaft in the x-y plane Fig.18(c Free body diagram of the disc in the y-z plane Fig.18(d Free body diagram of the shaft in the y-x plane From Figure.18(b, we can have the following relations for the eccentricity e = CH = ecosωt and e = GH = esinωt (.66 x where e x and e y are components of the eccentricity, e, in the x and y -directions, respectively (in fact these components of eccentricity are in the plane of disc that is inclined. From Figure.18(c equations of motion of the disc in the y- and ϕ x directions can be written as y and d f y = m ( y + ey cosϕx (.67 dt f e ϕ M = I ϕ (.68 y y x yz d x

30 53 where m is the disc mass, I d is the diametral mass moment of inertia about the x-axis, f y is the reaction force and M yz is the reaction moment. It should be noted that the moment is taken about the point G. From above equations it can be observed that equations are non-linearly coupled with the angular (titling component of displacement, ϕ x. Figure.19(a shows the rotor in the z-x plane. From Figure.19(b, we can write equations of motion as and d fx = m ( x + ex cosϕ y (.69 dt f e ϕ M = I ϕ (.70 x x y zx d y where I d is the diametral mass moment of inertia about the y-axis, f x is the reaction force and M zx is the reaction moment. Equations (.69 and (.70 are also non-linearly coupled with the angular component of displacement, ϕ y. However, two transverse planes (i.e. y-z and z-x motions are not coupled and that will allow two-plane motion to analyze independent of each other, i.e., set of equations (.67 and (.68 and equations (.69 and (.70 can be solved independent of each other. Fig.19(a A Jeffcott rotor with a disc offset from the mid span in the z-x plane Fig.19(b Free body diagram of the disc in the z-x plane Fig.19(c Free body diagram of the shaft in the z-x plane

31 54 Unbalance forces can be simplified with the assumption of small angular displacement (i.e., cosϕ = cosϕ 1 and equations (.67 and (.69 can be simplified as x y + = sin (.71 my f y mω e ωt and (.7 mx + fx = mω ecosωt Now equations (.71, (.68, (.7 and (.70 are assembled as m y f 0 y mω esinωt 0 Id 0 0 ϕ x M f yz yeyϕ x = 0 0 m 0 x f 0 x mω ecosωt I ϕ d y M f zx xexϕ y 0 (.73 which can be written in matrix notation as [ M ]{ x} + { R } + { R } = { f } (.74 L NL unb with [ M ] m I 0 0 d = 0 0 m Id ; { x} y ϕ x = ; x ϕ y { R } L f y M yz = f x M zx ; { R } NL 0 f yeyϕ x = 0 f xexϕ y ; { f } unb mω esinωt 0 = mω ecosωt 0 where [M] represents the mass matrix, {f umb } is the unbalance force vector, {x} is the displacement vector, {R} is the reaction force/moment vector and subscripts: L and NL represent the linear and the nonlinear, respectively. It should be noted that the ordering of the displacement vector can be changed depending upon the convenience and accordingly elements of other matrices and vectors will change their positions. The reaction forces and moments onto the shaft can be expressed in terms of shaft displacements at the disc location with the help of influence coefficients as x = α f + α M 11 x 1 zx and y x zx ϕ = α1 f + α M (.75

32 55 where α ij represent the displacement at i th station due to a unit force at j th station. It should be noted that the displacement and force terms are used as general sense so that displacement can be a linear or an angular displacement whereas the force can be a force or a moment. The coupling of the force and the displacement in two orthogonal planes has not been considered because of the symmetry of the shaft. Equation (.75 can be written in a matrix form as x α11 α1 f x = ϕ α α M y 1 zx (.76 with 3 3a l a al a b α α EIl 3EIl = α α 1 3al 3a l ab( b a 3EIl 3EIl where EI is the beam flexure, parameters a and b are defined in Figure.19(a with l = a + b. From the simple beam deflection theory, we can get these influence coefficients (Timoshenko and Young, Equation (.76 can be written as 1 f x x x α11 α1 k11 k1 = = M α α ϕ k k ϕ zx 1 y 1 y (.77 where k ij is the stiffness coefficient and defined as force at i th station due to a unit displacement at j th station. Similarly, since the shaft is symmetric about its rotation axis, we can obtain f k k y M k k 1 ϕ x y = 11 1 yz (.78 Equations (.77 and (.78 can be combined in matrix form as with { R } [ K ]{ x} L = (.79

33 56 [ K ] k k k 11 1 k = 0 0 k11 k1 0 0 k k 1 ; { R } L f y M yz = f x M zx ; { x} y ϕ x = x ϕ y Noting equation (.79, the nonlinear reaction force vector takes the following form { R } NL 0 0 f yeyϕ x ( k11 y k1ϕ x eyϕ + x = = 0 0 f e ϕ ( k x + k ϕ e ϕ x x y 11 1 y x y (.80 Above equation contains product of the linear and angular displacements, which makes the system equations nonlinear in nature. The present analysis considers only linear systems, so contributions from these nonlinear terms can be ignored with the assumption of small displacements. On substituting reactions forces and moments from equation (.79 into equations of motion, i.e., equation (.74, we get [ ]{ } + [ ]{ } = { } with [ M ] M x K x f unb (.81 m I 0 0 d = 0 0 m Id ; [ K ] k k k 11 1 k = 0 0 k11 k1 0 0 k k 1 ; { x} y ϕ x = x ϕ y ; { f } unb mω esinωt 0 = mω ecosωt 0.5.1: Calculation of natural frequencies: For obtaining natural frequencies of the system the determinant of the dynamic stiffness matrix, [ Z ] ([ K] ω [ M ] =, should be equated to zero and solved for ω, which gives four natural frequencies of the rotor system. It will be illustrated through examples subsequently. More general method based on the eigen value problem will be discussed in subsequent sections..5.: Unbalance forced response: The unbalance forcing with frequency, ω, can be written as { f } { F } e jωt unb r i = with F = F + jf k = 1,,, N (.8 unb unb unb unb k k k

34 57 where { F unb} is the complex unbalance force vector and it contains the amplitude and the phase information, and N is the total DOFs of the system (N = 4 for the present case. The response of the system can be written as { x} = { X} e jωt so that { x} = ω { X} e jωt (.83 On substituting equations (.8 and (.83 into equation (.81, we get the unbalance response as 1 { } = [ ] { } X Z F umb ( with [ Z ] [ K ] ω [ M ] = (.84 where [Z] is the dynamic stiffness matrix. Similar to the force amplitude vector, the response vector will also have complex quantities and can be written as r i X = X + jx k = 1,,, N (.85 k k k which will give amplitude and phase information, as ( ( amp r i k = k + k X X X and phase 1 X tan ( i / k k = X k X k (.86 Equation (.84 is more a general form of the Jeffcott rotor response as that of the disc at mid-span. However, it is expected to provide four critical speeds corresponding to four-dofs of the rotor system. Most often it is beneficial to observer the amplitude and the phase of response rather than the time history and present method gives the response in frequency domain as well. When the damping term is also present, the above unbalance response procedure can easily handle additional damping term, and the dynamic stiffness will take the following form ( [ Z ] = [ K ] ω [ M ] + jω [ C] (.87 where [C] is the damping matrix. It should be noted [Z] is now a complex matrix and by simulation critical speed can be obtained by noticing peaks of responses while varying the spin speed of the shaft. The procedure for obtaining damped natural frequencies will be discussed subsequently...5.3: Bearing reaction forces: Forces transmitted through bearings are those, which are related to the deflection of the shaft as shown in Figure.0 on the y-z plane.

35 58 On taking moments about ends L (left and R (right of the shaft, we have and a l M L = 0 f ya M yz Ryl = 0 or Ry = f y M yz (.88 b l M R = 0 Ry1l f yb M yz = 0 or Ry1 = f y + M yz (.89 1 l 1 l Figure.0 Bearing reaction forces on the shaft in y-z plane From above equations, bearing reaction forces at the left and right sides are related to the loading on the shaft, f y and M yz. In matrix form equations (.88 and (.89 can be written as {f b } = [D] {f s } (.90 with jωt y1 = b = Ry { } { } f F e b R ; { } { } j ωt y f F e s f = s = M yz ; [ D] b l 1 l = a l 1 l where subcripts: b and s represent the bearing and the shaft, respectively. Complex vectors {F b } and {F s } are bearing forces at the shaft ends and shaft reaction forces at the disc, respectively. On using equations (.79 and (.84 into the form of equation (.90 for both plane motions (i.e., y-z and z-x, we get {F b } = [D] [K]{X}=[D] [K] [Z]{F unb }=[C]{F unb } (.91

36 59 with [C] = [D][K][Z] It should be noted that equation (.91 has been written for both plane motions (i.e., y-z and z-x, however they are uncoupled for the present case. Similar to the unbalance force amplitude vector, the bearing vector will also have complex quantities and can be written as r i F = F + jf k = 1,,, N (.9 b b b k k k which will give the amplitude and the phase information, as ( b ( b amp r i b = + k k k F F F phase 1 i r and Fb = tan ( Fb / Fb k k k (.93 It should be noted that for the case of no damping the phase remains zero between a force in one plane and a response in that plane. These procedures will be illustrated now with simple numerical examples. Example.5 Find the bending natural frequency of a rotor system shown in Figure.1. The disc is rigid and has mass of 10 kg with negligible diametral mass moment of inertia. Consider the shaft as massless and flexible with E = N/m. Take one plane motion only. Figure.1 Example.5 Solution: Figure. shows the deflected position of the shaft. For a simply supported beam, the influence coefficient is defined as (Timoshanko and Young, 1968 ( y( x bx l x b α = =, ( x a F 6EIl