CHAPTER 5 TRANSVERSE VIBRATIONS-III: SIMPLE ROTOR SYSTEMS WITH GYROSCOPIC EFFECTS In previous chapter, we considered rotor-bearing systems for a single mass rotor with different level of complexity at supports. We analysed the rotor system for the translatory and rotary motions by considering the respective inertias. However, we neglected an important effect on dynamic behaviours of the rotor system so called the gyroscopic effect, which predominates especially for high speed rotor. In the present chapter also we shall still be dealing with a single-mass rotor system; and again with the assumption of rigid bearings. However, now we shall include the effect of gyroscopic effects and will explore the motion of the rotor for the synchronous as well as the asynchronous whirl. For the present case, we shall analyse the rotor system by two different approaches, firstly by the quasistatic analysis (which gives a better physical insight into the effect of gyroscopic effects, however, can be applied practically to simple systems only), and secondly by the dynamic analysis (which can be easily extended to multi-DOF systems). An important aspect, which we will observe from the present chapter, is that because of the gyroscopic effect the whirl natural frequency becomes dependent on the rotor spin speed. Another interesting phenomenon that can be observed is that the rotor can have the forward and backward whirling motion. Moreover, for the present case the distinction among the rotor spin speed, the whirl natural frequency, and the critical speed will be made clearer.

When a relatively large disc (or rotor) spins at a very high speed about its longitudinal (polar) axis, then it has a large angular momentum and it is the main characteristics of the rotors to carry a tremendous amount of rotary power. However, if it has precession (slow or fast) about its transverse (diametral) axes, which comes due to flexibility of bearings or of shafts itself, then it develops change in the angular momentum due to change in its direction. This leads to an inertia moment called the gyroscopic moment. Basically the gyroscopic moment develops due to the Coriolis acceleration component.

Figure 5.1(a) and (b) shows the motion of a disc in a simply supported shaft, when the disc is at the mid-span and away from it, respectively. For the latter case, the precession of the disc about its diametral axes takes place along with the spinning about the polar axis, which leads to the gyroscopic moment on the disc and it depends upon the spin speed among other parameters. Rotor systems with a point mass and with an appreciable mass moment of inertia are shown in Figures 5.1(c) and (d), respectively. The critical speeds of both the rotor systems will not the same. This is due to the fact

207

that centrifugal forces of particles of the disc do not lie in one plane during motion and thus from a moment tending to straighten the shaft and it will be discussed subsequently in more detail.

ω

ω

(a) A simply supported shaft with a disc at the mid-span

(b) A simply supported shaft with a disc near the bearing

(d) A cantilever shaft with a rigid disc at the free (c) A cantilever shaft with a point disc at the end free end Figure 5.1. The spinning and precession (whirling) motions of the disc

5.1 Angular Momentum The angular velocity is a vector quantity. A change in the magnitude and direction of angular velocity results in the angular acceleration. Let in Fig. 5.2, OA rotates about the z-axis in the x-y plane and OB is the position it takes after an infinitesimal time interval. Let ∆θ be the infinitesimal angular displacement of OA, it is the angular displacement vector along z-axis. Similarly, the angular velocity, angular acceleration and angular momentum are also vector quantity. The gyroscopic moment can be understood using the principle of angular momentum.

Figure 5.2. An angular displacement vector

208

A particle of mass, m, is moving with a velocity, v. From Figure 5.3, the linear momentum, L, is defined as L = mv

(5.1)

The direction and sense of the linear momentum are same as the linear velocity.

Figure 5.3. A particle in a linear motion

Now, referring to Figure 5.4(a) in which a point mass m is revolving at a radius r in a plane, the angular momentum is defined as the moment of linear momentum

(

)

Angular momentum = H = ( mv ) r = mr 2 ω = I pω

(5.2)

where Ip is the polar mass of inertia of particle of mass, m, about it’s axis of rotation; and ω is the angular velocity. The direction of the angular momentum will be same as angular velocity.

ω

(b) A flywheel in rotation

(a) A point mass in rotation

Figure 5.4 Angular momentum

Referring to Figure 5.4(b) in which a flywheel of the mass m and of the radius of gyration k is rotating with an angular speed of, ω , the angular momentum is given as H = ( mv ) k = ( mk 2 )ω = I pω

with

I p = mk 2

(5.3)

209

5.2 Gyroscopic Moments in Rotating Systems In the present section, the concept of the gyroscopic moment will be introduced with the help of simple rotating systems, e.g., discs and blades. These basic concepts you might have studied to some extent in the subject of dynamics of machinery (Wrigley, et al., 1969; Bevan, 1984; Mabie and Reinholts, 1987; Rao and Dukkipati, 1995).

5.2.1 Motion of a rotor mounted on two bearings Let us assume that a rotor with a flexible massless shaft carrying a disc is constraint to move in a vertical (single) plane. It is assumed that the constraint is not providing any friction forces during the motion of the shaft that plane. A rotor is spinning with a constant angular velocity, ω ; the angular momentum, H, will be given by I pω . Let x-y-z be the rectangular coordinate system (see Figure 5.5), where oz is the spin axis, ox is the precession axis, and oy is the gyroscopic moment axis.

Figure 5.5. A rotor mounted on two bearings with a single plane motion

Let the disc (or the spin axis) precession angle is ∆ϕ from z-axis as shown in Figure 5.5. The angular momentum will change from H (i.e., OC) to H ′ (i.e., OD), which can be written as (see the triangle ∆OCD which is in y-z plane)



  H ′ = H + ∆H

(5.4)

210

where ∆H is the change in angular momentum (i.e., CD). It is due to change in the direction of H. From ∆OCD, we have CD = ( OC ) ∆ϕ

or

∆H = H ( ∆ϕ )

or

∆H = I pω∆ϕ

with
   Now the rate of change of the angular momentum can be written as

M=

dH ∆ϕ = lim I pω = I pων ∆ t → 0 dt ∆t

(5.5)

where ν is the uniform angular velocity of precession, and M is the gyroscopic moment. The


 gyroscopic moment will have same sense and direction as ∆H , i.e. CD . In the vector form, equation can be written as  M = ν × H = I p (ν × ω )

(5.6)

From the right hand screw rule, we will get the direction of gyroscopic moment, i.e. along negative yaxis (the clockwise direction when seen from above, see Figure 5.5). Whenever an axis of rotation or spin axis changes its direction about another orthogonal axis then a gyroscopic moment acts about the third orthogonal axis. This is active moment acting on the disc, which means disc will experience this moment. In other word, if to a spinning disc a moment, M, is applied then precession take place in an attempt to align the angular momentum vector, Ipω, with the applied moment vector.

If we consider the free body diagram of the disc, the reaction from the shaft on to the disc-hub will be the active moment (i.e., in the negative y-axis direction). Hence, the shaft will experience a reactive moment from the disc hub in the opposite direction as the active moment on to the disc (i.e., in the positive y-axis direction). Let F be a force on the shaft from the bearing, then its direction due to gyroscopic moment will be as shown in Figure 5.5. A reactive moment will be experienced by bearings through the shaft in the opposite direction as the active gyroscopic moment on to the disc (i.e., a couple due to –F forces).

5.2.2 Gyroscopic moments though Coriolis component of accelerations: Figure 5.6 shows a disc that is spinning with a angular velocity, ω, and a precession angular velocity,

ν. Let z and y be the spin and precession axes, respectively. It is assumed here also that the disc has

211

precession about a single axis (i.e., y-axis). An infinitesimal mass, dm, at point P has coordinates

( r ,θ )

in polar coordinates or (x, y) in rectangular coordinates. From Figure 5.6(a), it can be seen that

the velocity of point P, i.e. ωr , will be perpendicular to OP. The velocity component along x-axis will be ω r sin θ = ω y and whereas along the y-axis is ω r cosθ = ω x . Particle P has the motion along the x-axis (i.e., parallel to the x-axis) and simultaneously it is rotating about y-axis as shown in Figure 5.6(c). Hence, a Coriolis component of acceleration, i.e. 2(ω y )ν , acts along positive z-axis direction (Figures 5.6(a) and (b)). Similarly for particle P ′, the Coriolis acceleration component will be 2(ω y )ν ; and it acts along negative z-axis direction as shown in Figures 5.6(a and b). The force due to acceleration of the particle P is given as F = dm(2ω yν )

(5.7)

Figure 5.6 The gyroscopic moment on a rotating disc (a) x-y plane (b) y-z plane and (c) z-x plane

Let us first consider the moment about x-axis, and due to the particle P it is dm(2ω y 2ν ) , hence the total moment about x-axis will be

212

M xx = ∫ 2ων y 2 dm = 2ων I xx = I pων

(5.8)

with

I xx = ∫ y 2 dm = 12 I p

(for thin disc)

(5.9)

where Ip (or Izz) is the polar moment of inertia. The component of the gyroscopic moment, Mxx, acts along the positive x-axis direction (Fig. 5.6(a)). Now consider the moment about y-axis and due to the particle P it is dm(2ω xyν ) , hence the total moment about y-axis will be

M yy = ∫ 2ων xydm = 2ων I xy

(5.10)

with

I xy = ∫ xydm = 0

(for symmetric disc)

(5.11)

It should be noted that if the disc were not symmetric then I xy ≠ 0 , so we would get M xx and M yy both non-zero. Hence, accelerating forces arising out of these Coriolis acceleration components due to motion of particles in x-direction, produce a net moment (or a couple) M, along positive x-axis direction only. There is no Coriolis component of acceleration when we analyze the motion of particle in the y-direction since it has no precession about x-axis, ω x = 0 (i.e., since the single plane precession is assumed) as shown in Figure 5.6(b).

5.2.3 Gyroscopic moments in a rotating thin blade In Figure 5.7, we have z-axis as the axis of spin and y-axis as the axis of precession. Let ξ and η are the two orthogonal principal axes of the thin rod, with ξ making an angle of θ with the x-axis. Due to the Coriolis component of acceleration the force at a point P, of the mass dm is given as dF = dm {2ν (ω r sin θ )}

(5.12)

which acts along the spin axis (positive z-axis direction in Figure 5.7b). The moment due this force about η-axis is given as (with the moment arm of r)

(

dMη = dFr = dm 2ων r 2 sin θ

)

(5.13)

The total moment of all particles above and below the η-axis is given as

Mη = ∫ ( 2ων sin θ ) r 2 dm = 2ων sin θ ∫ r 2 dm = 2ων sin θ Iη with

(5.14)

213

Iη = ∫ r 2 dm

(5.15)

Figure 5.7 A thin rod rotating about its centroid axis (a) x-y plane (b) y-z plane

From the parallel axis theorem, we have I p = Iξ + Iη ≈ Iη

(5.16)

where I p = I zz is the polar moment of inertia. Since the rod is thin, hence we have Iξ ≈ 0 . In view of equation (5.16), equation (5.14) reduces to Mη = 2 I pων sin θ

(5.17)

which is along the negative η-axis direction as shown in Figure 5.7. Taking component of moment M η along the x- and y- axes, as M xx = 2 I pων sin 2 θ = I pων (1 − cos 2θ )

(5.18)

M yy = 2 I pων sin θ cos θ = I pων sin 2θ

(5.19)

and

Figure 5.8 A two-bladed propeller

214

There are two gyroscopic moments, respectively, about the x- and y-axes. This comes because of the asymmetric body of revolution, i.e. Iη ≠ Iξ . From equations (5.18) and (5.19) it can be seen that M xx and M yy are varying with θ , i.e. M xx varies from 0 to 2 I pων ; and M yy varies from − I pων to I pων . The above analysis is applicable to the two-bladed propeller or the airscrew (Figure 5.8). The above analysis can be extended to a multi-bladed propeller (e.g., Fig. 5.9).

5.2.4 Gyroscopic moments in a multi-bladed propeller: Let n be the number of blades ( n ≥ 3 ) and

α = 2π / n is equally spaced angle between two blades. Let us consider one of the blades (designate as 1), which is inclined to an angle θ with x-axis as shown in Figure 5.10. Let the moment of inertia of each blade about η-axis (i.e. perpendicular to the blade) be equal to Iη which in turn is equal to the polar mass moment of inertia of blade 1 alone, i.e., I p1 ; since I p1 = Iξ + Iη ≈ Iη . Total polar moment of inertia of the airscrew about the axis of rotation (z-axis) is: I p = nI p1 .

Figure 5.9 A three-bladed propeller

Figure 5.10 One of the blade positions

Total moment about x-axis of blade 1 is given as

M x1 = 2ων ∫ y 2 dm = 2ων I p1 sin 2 θ = I p1 ων (1 − cos 2θ )

(5.20)

The location of other blades is given by the phase angle ( n − 1)α (the phase angle of various blades with respect to one of the reference blade would be (n − 1)α , where α is the phase difference between two blades). Noting equation (5.20), on summing up the moments due to all n blades, we have M x = I p1 ων [1 − cos 2θ ] + I p1 ων 1 − cos 2 (θ + α )  + I p1 ων 1 − cos 2 (θ + 2α )  + ...

+ I p1 ων 1 − cos 2 {θ + ( n − 1) α }

(5.21)

or

{

}

M x = I p1 ων n − cos 2θ + cos 2 (θ + α ) + cos 2 (θ + 2α ) + ... + cos 2{θ + ( n − 1) α }   

(5.22)

215

which can be simplified as

 cos 2{θ + 0.5 ( n − 1) α } sin nα  M x = I p1 ων  n −  sin α  

(5.23)

Since α = 2π / n for n > 2 , we have α = 2π / 3, 2π / 4,  and sin α ≠ 0 for all these values. Moreover, since nα = 2π for all value of n, we have sin nα = sin 2π = 0 . Hence for all values of n > 2 , from equation (5.23) we can write

M x = nI p1 ων

or

M x = I pων

(5.24)

For n = 2, we have sin α = sin π = 0 and sin nα = sin 2π = 0 , hence from equation (5.23), we get

M x = I p1 ων (1 − cos 2θ )

(5.25)

The gyroscopic moment about y-axis for a blade as shown in Figure 5.10, which makes angle θ with x-axis, is M y = I p1 ων sin 2θ

(5.26)

The total moment about y-axis for n blades with phase angles of (n − 1)α for n = 1, 2, , is

{

}

M yy = I p1 ων sin 2θ + sin 2 (θ + α ) + ... + sin 2 (θ + ( n − 1) α )   

(5.27)

The sine series in equation (5.27) will be zero for all values n > 2 , hence

M y = 0 for n > 2

(5.28)

So with equations (5.24) and (5.28), we can conclude that for the multi-bladed propeller with number of blades 3 and above is equivalent to a plane disc with the polar mass moment of inertia I p = nI p1 about the axis of rotation.

The objective of the present section was to have understanding of the gyroscopic moment in rotating components especially the direction of application. Now we shall deal with effects of gyroscopic moment, and the procedure of obtaining natural frequencies and critical speeds in simple single-mass rotor systems with the synchronous and asynchronous whirls (Den Hartog, 1984).

216

5.3 Synchronous Motion We are considering the case of perfectly balanced rotor, however, it is assumed to be whirling at its critical speed in slightly deflected position. The angular whirl frequency, v, of the centre of the shaft is assumed to be same as the angular velocity of rotation as of shaft ω (i.e., the spin speed). This implies that a particular point of the disc which is outside will always be outside; the inside point will always remains inside; it is called the synchronous motion. The shaft fibres in tension always remain in tension while whirling, and similarly the compression fibres always remain in compression. Thus any individual point of the disc moves in a circle in a plane perpendicular to the undistorted centre line of the shaft. We will consider two representative cantilever (overhung) rotor cases, firstly a thin disc at the free end and secondly a long stick at the free end.

5.3.1 A cantilever rotor with a thin disc In the present section, a thin disc attached to a flexible cantilever shaft at its free end is considered. In Fig. 5.11b, let points C and G are the shaft centre and the disc centre of gravity, respectively. Since no unbalance in the rotor, hence points C and G are coincident. Let δ and ϕx are the linear and angular displacements of the disc at the centre of shaft, C. The centrifugal force of a mass element dm at point P with coordinate (r, θ) is ω2r1 dm and is directed away from point B (a point on the bearing axis as shown in Figure 5.11b). It can be considered as two force vectors, the one when dm is assumed to be rotated about shaft center C (along CP) and the second force when C itself is rotating about B (along BC). Two similar triangles BCP and EFP can be constructed to represent these three forces (Fig. 5.11b). The component in the vertical direction is ω2δ dm and is directed vertical down. When these component forces are added together will give a force and no moment (Fig. 5.12 (a and b)). Whereas, the component in the radial direction is ω2r dm and is directed away from the disc centre C. When these component forces are added together will give zero force and the moment will be non-zero (Fig. 5.12 (c and d)). These will be illustrated now. The force ω2δ dm for various masses add together will give (Fig. 5.12 (a and b)) Fy = mω2δ

where m is the total mass of the disc. This force acts downward at C.

217

Figure 5.11 A cantilever rotor with centrifugal forces on the rigid disc

Figure 5.12 The centrifugal force on a particle of the disc (a and b) due to pure spinning motion, and (c and d) due to pure whirling motion

218

The force ω2r dm are all radiate from the center of the disc C. From Figure 5.12 (c and d), the ycomponent of the force, ω2r dm, is ω2rsinθ dm = ω2y dm (since y = rsinθ) and the moment arm of this elemental force is yϕ x . Thus the moment of the centrifugal force of a small particle dm is dMyz = (ω2y dm) yϕx = ω2y2ϕx dm

and it will act in the negative x-axis direction (Fig. 5.12c). The total moment Myz of centrifugal forces is

∫

∫

M yz = ω 2 y 2ϕx dm = ω 2ϕx y 2 dm = ω 2ϕx I d

(5.29)

where Id is the area moment of inertia of the disc about one of its diameter. The x-component forces

ω2rcosθ dm = ω2x dm (since x = rcosθ) will balance themselves since these forces are on the plane of the disc (Fig. 5.12(c and d)). Thus in totality the end of the shaft is subjected to a force, mω2δ, and to a moment, Idω2ϕx, under the influence of this it assumes a linear deflection δ and an angular deflection ϕx. This can happen only at a certain speed ω, i.e. at the critical speed. Thus the calculation of critical speed is reduced to a quasistatic problem (in which the dynamic forces are considered as time-independent), Now the objective is to find at which value of ω a shaft will deflect δ and ϕx under the influence of Fy = mω2δ and Myz =

ω2Idϕx.

Figure 5.13 A cantilever beam with loadings at the free end

The linear angular displacement of the free end of the cantilevered (fixed-free end conditions) beam as shown in Figure 5.13 will be (Timoshenko and Young, 1968)

δ=

Fy l 3 3EI

−

M yz l 2 2 EI

mω δ ) l I ω ϕ )l ( ( = − 2

3

2

d

3EI

2

x

2 EI

(5.30)

and

ϕx =

Fy l 2 2 EI

(= mω δ ) l − ( I ω ϕ ) l 2

−

M yz l EI

2

2

d

3EI

x

EI

(5.31)

219

It should be noted here that the above relations can also be developed for other boundary conditions such as simply supports, fixed-fixed, fixed-hinged etc. It requires calculation of influence coefficients by using the deflection theory of strength of materials. Now the above relations can be used to find critical speeds for the fixed-free boundary condition. Equations (5.30) and (5.31) can be rearranged as 3    l2  2 l − 1 δ +  − I d ω 2  mω ϕ x = 0 3 EI 2 EI    

(5.32)

2     2 l 2 l ω − m + 1 ϕ x = 0  δ +  I dω 2 EI EI    

(5.33)

and

This homogeneous set of equations can have a solution for δ and ϕx only when the determinant vanishes 3 2     2 l 2 l m I ω − 1 − ω     d 3EI 2 EI     =0 2   l l   2 2 + 1  −mω   Idω EI EI 2    

(5.34)

which gives the frequency equation as  12 EI  ml 2   12 E 2 I 2 I − =0 d  − 3  4   mI d l  mI d l  3

ω4 + ω2 

(5.35)

Defining the critical speed function, ω cr and the disc mass effect, µ , as

ω cr = ω

ml 3 EI

and

µ=

Id ml 2

(5.36)

Equation (5.35) can be written as

4

 12 − 12  − = 0 µ  µ

ω cr4 + ω cr2 

(5.37)

With the solution 2

  2 2  12 ω = 6−  ± 6 −  + µ  µ µ  2 cr

(5.38)

The positive sign will give a positive value for ω cr2 or a real value for ω cr and the negative sign will give a complex value of the critical speed, which has no physical significance. Plot of   versus µ is

given in Figure 5.14.

220

Figure 5.14 Variation of the critical speed function with the disc effect

It should be noted that the critical speed of the rotor increases with the disc mass effect, µ . That means the effective stiffness of the rotor system increases due to the thin disc rather than a point-mass disc. This can be seen from Fig. 5.11 that the effect of centrifugal forces is to resist the tilting of the disc thereby increasing the effective stiffness of the rotor system. Overall for the synchronous whirl condition due to the gyroscopic effect the (forward) critical speed of the system increases. It will be shown that for anti-synchronous whirl condition due to gyroscopic effect the (backward) critical speed of the system decreases. Two limiting cases of Fig. 5.14 are discussed as follows:

Case 1: A disc having point-mass For the disc effect µ = 0 (i.e., the concentrated mass of the disc) from equation (5.37), we have

4ωcr2 − 12 = 0 ⇒

ωcr2 = 3

(5.39)

Noting equation (5.36), above equation gives 2  ml 3   ωcr  =3   EI  

⇒ ωcr =

3EI ml 3

It gives the synchronous critical speed of disc having point-mass for the cantilever case.

(5.40)

221

Case 2: A disc having infinite size For µ → ∞ (i.e., a disc for which all the mass is concentrated at a relatively large radius, Id → ∞.) no finite angular displacement ϕx is possible, since it would require an infinite torque, which the shaft cannot furnish. The disc remains parallel to itself and the shaft is much stiffer than without the disc effect (i.e., µ = 0). From equation (5.37) for µ → ∞, we get

ωcr2 (ωcr2 − 12 ) = 0

since ωcr2 ≠ 0 hence ωcr2 = 12

⇒

ωcr =

12 EI

ml 3

(5.41)

It should be noted that for the present case one of the solution ωcr2 = 0 is considered as not feasible. However, when µ → - ∞ it is a feasible solution and the natural frequency of the system would be zero. This particular issue will be considered in the next section by replacing the thin disc with a long stick, which requires a separate analytical treatment; and it has been treated only for the pure rotational motion.

5.3.2 A cantilever rotor with a long stick For the present case, the disc at free end of a cantilever rotor has considerable amount of length as shown in Figure 5.15. The couple of centrifugal forces for this case are such; it tries to push away the rotor from the static equilibrium position angularly as shown in Figure 5.16. For the present case also the synchronous whirl condition is assumed. For the thin disc case, the couple of centrifugal forces try it to bring back to static equilibrium position angularly. In Figure 5.16 principal axes directions are (1) and (2). The coordinate of a point mass dm is (y, z).

Figure 5.15 A cantilevered rotor with a long stick at the free end

222

Figure 5.16 Centrifugal forces in a cantilevered rotor with a long stick at the free end For the present case, it is assumed that no unbalance is present in the rotor (i.e., the shaft geometrical centre C and the centre of gravity G are coincident) and the centre of gravity G of the body is positioned in the axis of rotation x (i.e., no linear displacement, δ = 0). Hence, there is no net centrifugal force, mω2δ, and only a moment is present (where m is the mass of the long stick). The force on a particle is ω2y dm and its moment arm about y-axis is z (Fig. 5.16), so that the moment is dMyz = ω2yz dm

(5.42)

For the thin disc we have z = y ϕ x so above equation reduces to dMyz = ω2y2 ϕ x dm (see equation (5.29) ). Now for the whole body, we have

  ω 

(5.43)

Let 1 and 2 be the principal axes along the longitudinal and transverse directions of the long stick (Fig. 5.16), respectively. Let the mass moment of inertia about these principal axes be I1 and I2, which are the polar and diametral mass moment of inertias, respectively. This set of axes is at an angle of ϕ x with respect to the y-z axes as shown in Figure 5.16. The product of inertia (it is identical to the shear stress in the subject of the strength of materials) about y-z axes is defined as

∫ yzdm =

I1 − I 2 sin 2ϕ x ≈ ( I1 − I 2 )ϕ x 2

(5.44)

Here the angular deformation, ϕ x , is assumed to be small. For the thin disc I1 = Ip = 2Id and I2 = Id so that equations (5.44) and (5.43) gives Myz = ω2 Id ϕ x , which is same as equation (5.29). However, for a disc of the diameter D and the thickness b (Fig. 5.16), we have for the present case

223

I1 =

mD 2 8

I2 =

and

mD 2 mb 2 + 16 12

(5.45)

On substituting equations (5.44) and (5.45) into equation (5.43), we get

        

  



    

(5.46)

For moment of the centrifugal forces is to be zero, from above equation, we have mD 2 mb 2 = 16 12

⇒b=

3 D 2

or b = 0.866 D

and it becomes negative for b > 0.866D (i.e., for the long stick) and it is positive for b < 0.866D ( i.e., for the thin disc). Table 5.1 gives the summary of gyroscopic moment with its sign for different ratio of b and D. Equation (5.46) can be written as

   ω  with

 mD 2 mb 2  Id =  −  12   16

(5.47)

It can be found that the net resultant force will also be same for the long stick and the thin disc, when we consider both the linear and angular motions simultaneously. Thus, the critical speed analysis of the previous section will be valid for the present case also, when we consider both the linear and angular motions simultaneously. Hence, Figure 5.14 of the previous case will still be applicable for the range of thin disc. However, plot from the above equation will represent both the long stick and thin disc cases as shown in Fig. 5.17, since the form of net moment equations (5.48) and (5.49) are identical. From equation (5.38), we have 2

  2 2  12 ω = 6−  ± 6 −  + µ  µ µ  2 cr

(5.50)

with

µ=

Id ml

2

and

 mD 2 mb 2  Id =  −  12   16

(5.51)

224

Figure 5.17 Variation of the critical speed function with the disc effect The condition for which the square root term in equation (5.50) remain always positive is

2

 2  12 6 −  + > 0 µ µ 

( 3µ − 1)

or

2

+ 3µ > 0

(5.52)

From above equation it can be observed that both terms are always positive for positive value of µ, since the first term is a square term and the second term is positive. For a negative value of µ, the first term in the above inequality is always positive; and it is always greater than the second term. Hence for all real value of µ, above inequality is true (for imaginary values of µ we may violate the inequality however, it is not a feasible case in real systems) It means we will get always a real root from equation (5.50) when we consider a positive sign in front of the square root. For µ → 0 again equation (5.40) would be valid. For µ → ±∞ from equation (5.50), critical speeds are ωcr2 → 12 and

ωcr2 → 0 , respectively, for the positive and negative signs. Table 5.1 Gyroscopic effects for different geometries of the disc/stick Relation between b and D

Sign of Id

Gyroscopic moment sign

b = 0.87 D

0

0

b > 0.87D (Long stick)

Negative

Negative

b < 0.87 D (Thin disc)

Positive

Positive

225

For the long stick case, it is assumed that the shaft extend to the centre of the cylinder without interference. If shaft is attached to the end of the cylinder, the elastic-influence coefficients are modified. The phenomenon described in the present section is generally referred to as a gyroscopic effect. Now with some numerical examples the calculation of critical speeds would be demonstrated.

Example 5.1 Obtain the transverse critical speed for the synchronous motion of a cantilever rotor as shown in Figure 5.18. Take mass of the thin disc, m, as 1 kg with the radius, r, as 3 cm. The shaft is assumed to be massless; and its length and diameter are 0.2 m and 0.01 m, respectively. Take shaft Young’s modulus of the shaft material as E = 2.1×1011 N/m2.

Figure 5.18 Solutions: Case I: For the disc effect µ = 0 (i.e., the concentrated mass of the disc) from equation (5.40), we have

3EI 3 × 2.1 × 1011 × 4.909 × 10−10 = = 196.61 rad/s ml 3 1 × 0.23

ωcr =

Answer

with

I=

π 64

(0.01)4 = 4.909 × 10−10 m 4 ;

d = 0.01 m;

m =1 kg;

l = 0.2 m

(a)

Case II: Considering the disc as rigid (i.e., µ ≠ 0 ), from equation (5.38), we have

2

2

  2 2  12 2  2  12   ωcr =  6 −  +  6 −  + = 6 − = 1.7431  + 6 −  + 0.0056  0.0056  0.0056 µ µ µ     (b)

with

µ=

Id ml 2

=

1 4

mr 2

ml 2

=

r2 0.032 = = 0.0056 ; 4l 2 4 × 0.22

Now, from equation (5.53), we have

(c)

226

ωcr = ωcr

ml 3 1 × 0.23 = ωcr = 0.00881ωcr EI 2.1 × 1011 × 4.909 × 10−10

(d)

On substituting value of   from equation (b) into equation (d), the critical speed is given by

ωcr =

ωcr 0.00881

=

1.7431 = 197.86 rad/s 0.00881

Answer

It should be noted that as compared to case I the critical speed is more, which is expected due to increase in the effective stiffness while considering the diametral mass moment of inertia of the disc For the present case I d = 14 mr 2 = 14 × 1 × 0.032 = 2.25 × 10 −4 kg-m2, which is very less that is why the increase in the critical speed is marginal. Reader can check the change in the critical speed, for example, with radius of the disc equals to 6 cm. The effect would be far more predominant in the limiting case, when disc is very large that is µ → ∞ , the critical speed can be calculated from equation (5.41), as

ωcr =

12 EI

ml 3

=

12 × 2.1 × 1011 × 4.909 × 10−10 = 393.22 rad/s 1× 0.23

Answer

Example 5.2 Obtain the transverse critical speed for the synchronous motion of a rotor as shown in Figure 5.19. The shaft is assumed to be fixed supported at one end. Take dimensions of the cylinder (stick) as (i) D = 0.2 m, b = 0.0041 m (ii) D = 0.0547 m, b = 0.0547 m (iii) D = 0.0361 m, b = 0.1649 m and (iv) D = 0.0547 m, b = 0.1093 m. Parameters D and b are the diameter and the length of the cylinder. The shaft is assumed to be massless and its length l and diameter d are 0.2 m and 0.01 m, respectively. Take the Young’s modulus of the shaft material as 2.1×1011 N/m2 and the density of the cylinder material as 7800 kg/m3.

Figure 5.19 A rotor with a long stick

227

Solution: For the long stick the critical speed is given as

ωcr = ω cr

EI ml 3

(a)

with 2

  2 2  12 ; ωcr =  6 −  +  6 −  + µ µ µ  

µ=

Id ml 2

;

 mD 2 mb 2  Id =  −  12   16

and

(b)

Above equation is valid for all value of µ except for µ = 0 (i.e., the thin disc). For µ = 0 we have the relation given by equation (5.40) and it is given as

ωcr =

3EI ml 3

ωcr =

so that

ωcr EI ml 3

=

3EI ml 3 = 3 = 1.732 EI ml 3

(c)

with

I=

π 64

d4 =

m = 1 kg,

π 64

0.014 = 7.854 × 10−9 m4;

EI = 1649.34 N-m2

l = 0.2 m

Hence, for µ = 0 (thin disc), we have

EI 1649.34 = = 454.06 3 ml 1× 0.23

and

ωcr =

3EI 3 ×1649.34 = = 786.45 3 ml 1× 0.23

(d)

Table 5.2 summarises the calculation of rotor parameters and critical speeds from the above equations. It should be noted the choice of the D and b are such that the mass of the disc remains the same for all cases. For the point mass parameters D and b are not defined. On comparing the cases of point mass and thin disc, the increasing trend in the critical speed is observed. For short stick (with D = b) the disc parameter, µ, becomes negative with a very low value. This leads to very high value of critical speed. However, for the long stick again the trend is towards decrease in critical speed, where the disc parameter, µ, remains negative with a relatively high value.

228

Table 5.2 Summary of rotor parameters and synchronous critical speeds Type of disc

D,

b,

(m)

(m)

m, (kg)

I d, 2

µ=

(kg-m )

Id

ωcr

ml 2

EI ml 3

ωcr , (rad/s)

(rad/s) Thin disc

0.2000

0.0041

1

0.0025

0.0625

1.86

454.06

845.00

Point mass

-

-

1

0.0000

0.0000

1.73*

454.06

786.43

Short stick

0.0547

0.0547

1

-6.23×10-5

-0.0016

50.89

454.06

23 107.11

Long stick

0.0316

0.1649

1

-2.2×10-3

-0.0550

9.06

454.06

4113.78

* From equation (5.41)

5.4 Asynchronous Rotational Motion In this section we will consider the asynchronous whirling motion of the spinning rotor. Consider a rotor, which is suspended practically at its centre of gravity by three very flexible torsional springs as shown in Figure 5.20. This will enable us to analyse the effect of rotational displacement of the rotor on its whirling frequencies, without complicating with the general motion in which both the linear and angular displacements take place simultaneously. Such general motion is quite complicated and will be considered in subsequent sections.

The aim is to calculate natural frequencies of modes of motion for which the centre of gravity O remain at rest and the shaft whirls about O in a cone of angle 2ϕx. Let the effective torsional stiffness of the support is kt. The disc on the motor shaft rotates very fast, and as the springs on which the motor is mounted are flexible, the whirling take place at a very slow rate than the shaft rotation. The angular momentum, H, is given as H = I pω

(5.54)

where Ip is the polar mass moment of inertia of the rotor (i.e., rotating parts of the motor and the disc). In the case when the whirl is in the same direction as the rotation (i.e., the forward whirl), the time rate of change of angular momentum will be directed from B to C (i.e., out of the plane of paper in Figure 5.20(c)). This is equal to the moment experienced by the motor frame from the disc. The reaction, i.e. the moment acting on the disc from the motor frame is pointing into the paper (in Figure 5.20(c)) and therefore tends to make ϕx smaller. This acts in an addition to the existing spring k. Hence, it is seen that the whirl in the direction of rotation (forward whirl) makes the natural frequency higher. In the same manner it can be reasoned that for the whirl opposite to the direction of rotation (i.e. the backward whirl), the natural frequency is made lower by the gyroscopic effect (Fig. 5.20d). To calculate the magnitude of the gyroscopic effect, we have

229

(

d I pω I pω

) = BC = BC AB = ν dtϕ OB

AB OB

(5.55)

x

(a) A motor on torsional springs

(b) A rotor system with motor suspensions

(c) A conical forward whirl of the rotor (d) A conical backward whirl of the rotor Figure 5.20 A motor and a rotor system supported on torsional springs

Equation (5.55) can be rearranged as

d I pω = νϕ x I pω dt

(

)

(5.56)

230

Equation (5.56) gives the gyroscopic moment. The elastic moment due to the springs k is equal to kϕx and the total moment is equal to (kt ± Ipων) ϕx

(5.57)

where the positive sign for a whirl in the same sense as the rotation, and the negative sign for a whirl in the opposite sense. In equation (5.57), the term in the parenthesis is the equivalent spring constant and hence the natural frequency will be (since we have ν = ωnf)

 ! 

"# $%& ''() %*

 ! +

or

%& ' %*



!

"#



%*

0

(5.58)

where Id is the diametral mass of whole motor assembly (including frame and disc). The solution of equation (5.58) can be given as



!

$

%& ' %*

%& '

$ -

%*

"

 . %#

(5.59)

*

From equation (5.59), it can be observed that the natural frequency of rotor system depends upon the spin speed of rotor, ω. The ± sign before the square root, only the positive sign need to be retained since the negative sign gives two values of ωnf, which are both negative, and equal and opposite to the two positive roots obtained with positive sign before the square root. This is due to the fact that natural frequencies remain same for the spining of rotor in either direction. Let us define non-

 dimensional terms as the frequency ratio as 

!

  ! /

!(0(120# ,

where 

!(0(120#

 345 / be

the non-rotating shaft natural frequency, i.e. without the gyroscopic effect; and the spin ratio  

60.5 /345  9. Equation (5.59) takes the following form  

!

 $  . √  . 1

(5.60)

Figure 6.9 shows the variation of the non-dimensional natural frequency, ωnf , with the nondimensional speed, ω , which govern by equation (5.60). It is seen that the natural frequency is split into two frequencies on account of the gyroscopic effect (i) a slow one whereby the whirl is opposed to the rotation (i.e. the backward whirl) and (ii) a fast one where the whirl and rotation directions are the same (i.e., the forward whirl). It can be seen that we will have two critical speeds one corresponding to the forward whirl and another backward whirl. These critical speeds can be obtained

231

by the condition that whenever the spin speed is equal to the natural frequency we will have critical speeds. It will be illustrated in the following numerical example.

Forward

υ

Backward

ϖ Figure 5.21 The natural frequency variation with the spin speed

Example 5.3 A long rigid symmetric rotor is supported at ends by two identical bearings. Let the shaft has the diameter of 0.2 m, the length of 1 m, and the material mass density equal to 7800 kg/m3. The bearing has dynamic parameters as follows: kxx = kyy = k = 1 kN/mm with other stiffness and damping terms equal to zero. By considering the pure tilting motion and the gyroscopic effect, obtain whirl natural frequencies of the system, if rotor is rotating at 10, 000 rpm. Obtain also the forward and backward critical speeds of the rotor-bearing system. Compare the critical speeds of rotor without considering the gyroscopic effect.

Solution: For the circular cylinder the polar moment of inertia, Ip, and the diameter moment of inertia, Id, are given as

I p = 12 mr 2 ,

and

I d = 121 m ( 3r 2 + l 2 )

where m is the mass of the cylinder, r is the radius of the cylinder, and l is the length of the cylinder. For the pure tilting motion of the rotor, we have the following rotor properties

I p = 12 mr 2 = 12 × 245.04 × (0.1) 2 = 1.2252 kgm 2

I d = 121 m(3r 2 + l 2 ) = 121 × 245.04{3 × (0.10) 2 + 12 } = 21.0326 kgm 2

232

 2π   = 1047.198 rad/s  60 

ω = 10,000 

The effective torsional stiffness due to bearings on the rotor can be obtained by considering Figure 5.22. A long rotor is supported at ends by two identical bearings; consider pure tilting of the rotor by an angle ϕ x , which gives 0.5lϕ x compression of the left bearing and the same amount of extension of the right bearing. This produces reaction forces at bearings with the magnitude of 0.5klϕ x and the direction as shown in Figure 5.22. The moment on to the rotor due these bearing forces would be

0.5kl 2ϕ x . Hence, the effective torsional stiffness would be 0.5kl 2 .

Figure 5.22 Free body diagram of the long rigid rotor supported on bearings

From equation (5.59), we have

ωnf

3,4

2

 I pω  keff =± +  +  2 I  2I d Id  d  I pω

(a)

with k eff = 12 kl 2 = 0.5 × 1 × 106 × 12 = 5 × 105 N/m

where the positive sign for the forward whirl and the negative sign for the backward whirl. On substituting values in equation (a), we get

ωnf

3,4

2  1.2252 5 × 105  1.2252    = × 1047.198 ± + × 1047.198  = [ ±30.50 + 157.17 ] 21.0326  2 × 21.0326  2 × 21.0326   

which gives

ωnf = 187.67 rad/s (forward whirl) and ωnf = 126.67 rad/s (backward whirl). 3

4

Answer.

233

It should be noted that these natural whirl frequencies change with the spin speed of the rotor. For obtaining the forward and backward critical speeds, we have, respectively, following conditions, i.e.,

ω = ωnf and ω = −ωnf . On substituting these conditions in equation (a), one at a time, we get the following expression for critical speeds

ωcrF , B =

kt ( Id ∓ I p )

(b)

where the negative sign is for the forward critical speed, ωcrF , and the positive sign for the backward critical speed, ωcrB . It should be noted from equation (b) that for I d < I p , the term in side the square bracket will be negative and there would not be any forward critical speed. For the present case, we have I d > I p and the corresponding critical speeds are

ωcrF =

5 × 105 = 158.88 rad/s ( 21.0326 − 1.2252 )

and

ωcrB =

5 × 105 = 149.88 rad/s ( 21.0326 + 1.2252 )

Answer.

Hence, the rotor is operating well above the critical speeds, since operating speed is 1047.2 rad/s. That means if we consider perfectly balanced rotor that is rotating at 1047.2 rad/s and if we perturb the rotor in forward whirl it will have whirl frequency equal to 187.67 rad/s. Moreover, if we perturb the rotor in backward whirl it will have whirl frequency equal to 126.67 rad/s. While rotor is coasting up from stationary position to operating speed then it will cross critical speeds, where large oscillations are expected. An alternative solution of the present problem will be presented in subsequent section based on dynamic analysis by considering the formulation of governing differential equations of the rotor.

When no gyroscopic effect is considered, ωnf1 and ωnf4 remain the same; and from equation (a) for I p = 0 , we get

ωnf = ± 3 ,4

keff Id

=

5×105 = 154.184 rad/s 21.0326

Answer.

234

which is between the forward and backward whirl frequencies (187.67, 126.67) rad/s obtained by considering the gyroscopic effect and these frequencies are now independent of the rotor spin speed.

If we consider the pure linear (translational) motion, then the critical speed will not experience any gyroscopic moments, hence corresponding critical speeds can be obtained as

ωcr = ωcr = 1

2

2k = m

2 × 1 × 106 = 90.34 rad/s 245.04

Answer.

with

m = ρ 4π d 2l = 7800 × 4π (0.2)2 × 1 = 245.04 kg where d is the diameter of the cylinder. In this case the natural frequency does not change with the spin speed of the rotor, and remains equal to the critical speeds (90.34 rad/s) as obtained above.

5.5 Asynchronous General Motion In the previous analysis of the cantilever rotor in section 5.3, the rotor was whirling (the whirling is defined as a circular motion of the deflected shaft centre line about its undeflected position with small amplitude) and spinning at the same angular speed and in the same direction. Cases have been observed where whirling and the spinning occur at different frequencies and sometimes in opposite directions as described in previous section. The aim of the present section is to calculate natural whirling frequencies, ν, of a shaft with a single disc on it at any speed of rotation, ω, in most general manners as shown in Figure 5.23, where ω is the shaft spin speed about deflected centerline and ν is the shaft whirling frequency about undeformed position OA. Since a general motion of the disc is very difficult to visualize hence the following three cases have been considered that is relatively easier to visualize, and the final motion will be superposition of some of these cases (i.e., cases I and II).

Figure 5.23 A cantilever rotor having a general motion of the spinning and the whirling

235

Case I: ν = ν0 with ω = 0 (Pure whirling motion) The shaft does not spin about its deflected centerline, but deflected centerline OC whirls with ν0 about bearing centre line OA.

Case II: ν = 0 with ω = ω0 (Pure spinning motion) The shaft is in the deflected position OC, and it spins with ω0 about the deflected centerline OC.

Case III: ν = ω (Synchronous whirling and spinning motions) The shaft fiber in tension remains in tension, and similarly shaft fiber in compression remains in compression, i.e. the synchronous whirl. It has already been discussed in detail in section 5.3 and it will not be discussed here again. With this combined ν and ω motions (Cases I and II), our first aim would be to obtain its angular momentum. For the case II when it does not whirl, but only spins, the angular momentum is equal to Ipω (along AC as shown in Figure 5.23) where Ip is the polar moment of inertia of the disc.

(a) The pure whirling motion

(b) The angular momentum due to the pure whirling motion Figure 5.24 A pure whirling motion of the rotor (case I)

236

For case I when no spinning ω = 0, but only a whirling, ν: The disc wobbles in the space (about its diameter) and it is difficult to visualize its (wobbling) angular speed. The visualization can be made easier by remarking that at the point C the shaft is always perpendicular to the disc, so that we can study the motion of a shaft segment near C instead of the disc. The line CA is tangent to the shaft at the point C. The piece ds of the shafting at the disc moves with the line AC, describing a cone with the point A as an apex as shown in Figure 5.24(a). The velocity of the point C for a whirling in the count clockwise direction, as seen from the right, is perpendicular and into the paper and its value is

νy, where y is the linear displacement of the disc centre. The line AC lies in the paper at time t = 0 (see Figure 5.24(a)), but at time dt later, point C is behind the paper by

CC′ = (νy)dt

(5.61)

The angle between two positions of line AC (i.e., AC and AC′ in Figure 5.24) is

CC′ ν ydt = = νϕ x dt AC AC

with

ϕx =

y AC

(5.62)

when ϕx is the angular displacement of the disc and it is considered to be small. From equation (5.63) the angle of rotation of AC in time dt is equal to νϕx dt. Hence the angular speed of AC (and of the disc) is equal to νϕx. The disc rotates about a diameter in the plane of the paper and perpendicular to AC at C, so that the appropriate moment of inertia is Id (= ½ Ip for the thin disc). The angular momentum vector of the disc due to whirl is Idνϕx and is shown in Figure 5.24(b). The direction of the angular momentum can be obtained by considering the tilting of the disc when point C is moving to C/. During this motion disc will try to tilt such that its left hand side face would be visible to the observer. It will be clearer when the disc centre will occupy position along the line OB and the disc is inside the plane of the paper (Fig. 5.25a). At this instant it tilts about its diameter, hence the observer will see motion of the disc tilting in the clockwise direction when looking the disc from the bottom along the diagonal. The disc centre will occupy same position on line OB when it is out side the plane of the paper in that case observer will be able to see the right hand side face of the disc (Fig. 5.25b). The total angular momentum is the vector sum of Ipω and Idνϕx, which have been obtained from cases I and II, respectively.

237

Fig. 5.25 Disc tilting during whirling disc is (a) into the plane of paper (b) outside the plane of paper

Figure 5.26 Angular momentums due to the whirling and the spinning of a rotor

Now our aim is to calculate the rate of change of the angular momentum vector, which has been obtained for Cases I and II, individually. For this purpose we resolve the vector into components parallel (the direction from O to A as positive) and perpendicular (the direction from B to C as positive) to line OA as shown in Figure 5.26; and are given, respectively, as

I pω cos ϕ x + I dνϕ x sin ϕ x ≈ I d ( 2ω + νϕ x2 )

(5.64)

I pω sin ϕ x − I dνϕ x cos ϕ x ≈ I d ϕ x ( 2ω − ν )

(5.65)

and

where I p ≈ 2 I d for thin disc, and for a small angular displacement, ϕx, of the disc such that cos ϕ x ≈ 1 and sin ϕ x ≈ ϕ x .

Components parallel to line OA rotates around line OA in a circle with a radius y and keeps the magnitude, and the direction of the angular momentum constant during the process so that its rate of change is zero as shown in Figure 5.27(a).

238

(a)

(b) Figure 5.27 Angular momentum components (a) parallel (b) perpendicular to the undeflected position of the shaft Angular momentum components perpendicular to line OA is a vector along the direction of line BC, and it rotates in a circle with the center as point B as shown in Figure 5.27(b). At time t = 0 this vector lies in the plane of paper, at time dt this vector moves behind the paper at an angle νdt (see Figure 5.27b). The increment in the vector is CC′ , which is directed perpendicular to the paper and into it, with the magnitude equal to the length of the vector itself, I d ϕ x ( 2ω − ν ) , multiplied by νdt. It is given as I d ϕ x ( 2ω − ν )ν dt

(5.66)

The rate of change of the angular momentum with time is then given as I d ϕ x ( 2ω − ν )ν

(5.67)

239

Hence, this is an active moment the disc would experience or in other words this is the moment exerted on the disc by the shaft (i.e., by action). The reaction moment exerted by the disc on the shaft is the equal and opposite, i.e. a vector directed out of the paper and perpendicular to it at C. Beside this moment there is a centrifugal force mω2y acting on the disc from case II as shown in Figure 5.28.

Figure 5.28 The inertia force and moment acting from the disc on the shaft caused by the shaft rotation, ω, and the shaft whirling, ν

Influence coefficients of the shaft can be defined as: α11 is the deflection y at the disc from 1 N force;

α12 is the angle ϕx at the disc from 1 N force or is the deflection y at the disc from1 N-m moment, i.e. α21= α12 (by the Maxwell’s theorem of strength of materials); and α22 is the angle ϕx at disc from 1N m moment. For the cantilever beam (Timoshenko and Young, 1968) with a concentrated load F and the moment M as shown in Figure 5.28, we have

α11 =

l3 l2 , α12 = α 21 = 3EI 2 EI

and α 22 =

l EI

(5.68)

It should be remembered that other boundary conditions can also be used; however, we have to obtain the relevant influence coefficients. The linear and angular deflections can be expressed as

y = Fα11 + M α12

and ϕ x = Fα12 + M α 22

(5.69)

It should be noted that the sign convention of M used for obtaining the influence coefficient is clockwise, however, the reactive moment from the disc to the shaft is counter clockwise, and hence the negative sign in the moment term. On substituting the force, F, and the moment, M, from Figure 5.28, we get

y = α11mν 2 y + α12 {− I d ϕ xν ( 2ω − ν )}

(5.70)

240

and

ϕ x = α12 mν 2 y + α 22 {− I d ϕ xν ( 2ω − ν )}

(5.71)

which can be rearranged in matrix form as

1 − α11mν 2 I d α12ν (2ω −ν )   y     = 0 2 1 + α 22 I dν (2ω −ν )  ϕ x   −α12 mν

(5.72)

Equations (5.72) are homogeneous equations in y and ϕx, on putting determinant of the matrix equal to zero, we get the frequency equation as

ν 4 ( − mα α I + mα I ) + ν 3 ( 2m α α I ω −2mα I ω ) + ν 2 (α 22 I d + mα11 ) + ν ( −α 22 I d 2ω ) − 1 = 0 11

22

d

2 12

d

11

22

d

2 12

d

(5.73)

Equation (5.73) contain seven system parameters: ω, ν, m, Id, α11, α12, and α22, which makes a good understanding of the solution very difficult. It is worthwhile to diminish the number of parameters as much as possible by the dimensional analysis. Introducing four new variables

The dimensionless frequency:

ν = ν α11m ;

The disc effect:

µ=

The elastic coupling:

α=

The dimensionless speed:

I dα 22 mα11

α 122 α11α 22

ω = ω mα11

(5.74)

With this new four non-dimensional variables, equation (5.73) becomes

ν 4 − 2ων 3 +

µ + 1 2 2ω 1 ν − ν− =0 µ (α − 1) α −1 µ (α − 1)

(5.75)

Equation (5.75) is the fourth degree polynomial in ν , so for a given α , carrying a given disc µ , and rotating at certain speed ω , there will be four natural frequencies of the whirl.

Case I: For a point mass of the disc, i.e., Id = 0 or µ = 0. Multiply equation (5.75) by µ and on substituting µ = 0, we get