CHAPTER 2

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, 2005; 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, 2000; Admas, 2001; Yamamoto and Ishida, 2001; Robert,

2003; Muszynska, 2005; Genta, 2005). 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

25

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.

2.1 Single-DOF Undamped Rotor Model

The simplest model of the rotor system can be a single DOF. Figure 2.1 shows two types of rotor model.

In Figure 2.1(a) the bearing (support) is assumed to be flexible and the rotor (the shaft and the disc) as

rigid; whereas in Figure 2.1(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

2.1(c).

Fig 2.1(a) A rigid rotor mounted on flexible

bearings

Fig 2.1(b) A flexible shaft with a rigid disc

mounted on rigid bearings

Fig 2.1(c) An equivalent single degree of

freedom spring-mass system Fig 2.1(d)The free body diagram of the

disc mass

26

2.1.1 Unbalance force model: If the rotor is perfectly balanced then theoretically speaking there will not

be any unbalance force as shown in Figure 2.2(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 2.2(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

( ) sinf t m e tω ω= 2 (2.1)

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 2.2(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, mi is

attached at a relatively larger radius of ri as shown in Figure 2.2(c), the unbalance force can be written as

( ) sini if t m r tω ω= 2 (2.2)

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 2(d), the total unbalance force

would be

( ) sin sin( )i if t m e t m r tω ω ω ω φ= + +2 2 (2.3)

where φ is the phase difference between the vectors of unbalance forces due to the rotor eccentricity and

the trial unbalance mass.

27

(a) Rotor geometrical center and center of gravity

are coincide.

(b) Rotor geometrical center and center of gravity

are not coincide.

(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 2.2 Different types of unbalances in a single plane

Figure 2.3 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 tθ ω= (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 tθ α=� and 1 2

2tθ α= ).

Figure 2.3 Unbalance location on a rotor at a particular instant of time

C,G

G

C

C,G

U U

C G

28

2.1.2 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 2.1(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

yF my=∑ �� or ymtemykeff��=+− ωω sin

2 (2.4)

where effk is the effective stiffness of the rotor system (see Figure 2.1) and

effk y is the restoring force

that acts opposite to the motion so negative sign. Equation (2.4) is a standard form of equation of motion

of a single DOF spring-mass system and could be rearranged as

temykym eff ωω sin2=+�� (2.5)

2.1.3 Free Vibrations: From the free vibration, when the external unbalance force is absent (i.e., 0e = ), it

is generally assumed that the rotor mass will have a simple harmonic oscillation and the free response

displacement is expressed as

( ) sin nfy t Y tω= (2.6)

where nfω is the frequency of oscillation during the free vibration and it is called the natural frequency of

the system. On substituting equation (2.6) (with ( ) sinnf nfy t Y tω ω= −��2

) into the homogeneous part (i.e.,

with 0e = ) of equation of motion (2.5), it gives

( ) sinnf eff nfm k Y tω ω− + =2 0 (2.7)

For a non-trivial solution (i.e., 0Y ≠ or 0nfω ≠ ) of equation (2.7), the natural frequency of the system

can be written as

( )nf effm kω− + =2 0 or mkeffnf /=ω (2.8)

2.1.4 Forced responses: Now to obtain the steady state forced response, it can be expressed as

( ) sin( )y t Y tω φ= − (2.9)

where Y is the amplitude of displacement and φ is the phase lag of displacement with respect to the

unbalance force. On substituting equation (2.9) with ( ) sin( )y t Y tω ω φ= − −��2

into equation (2.5), the

steady state forced response amplitude can be written as

29

2 2( ) sin( ) sineffm k Y t me tω ω φ ω ω− + − =

On expanding, we get

[ ]2 2( ) sin cos cos sin sineffm k Y t t me tω ω φ ω φ ω ω− + − =

On separating the sine and cosine terms of tω , we have

2 2 2( ) cos and ( ) ( sin ) 0eff effm k Y me m k Yω φ ω ω φ− + = − + − =

Since terms 2( )effm kω− + 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

2 2 2

2 2 2 21eff nf

Y mY

e k m

ω ω ωω ω ω ω

= = = =− − −

with nf

ωω

ω= and mkeffnf /=ω (2.10)

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 2.4. From Figure 2.4 and equation (2.10) it should be noted that we

have unbounded unbalance response when the denominator 2(1 )ω− becomes zero, i.e., when the spin

speed is

1crcr

nf

ωω

ω= = ± or

eff

cr nf

k

mω ω= ± = ± (2.11)

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 (2.11). 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

30

the damping is not considered in the analysis the phase angle, φ, becomes zero (or 1800 after the critical

speed).

In Fig. 2.4 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 1800, 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.

(a) Linear plot

(b) Semi-log plot

Figure 2.4 Variation of the non-dimensional unbalance response versus the frequency ratio

ω ω

Y

Y

31

2.1.5: 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 2.5a 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 2.5c 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 2.5 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

32

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 2.2). 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 2.1: 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

99-99×0.05 = 94.05 rad/s, and 5% of the upper operational speed would be 101+101×0.05 = 106.05 rad/s.

Then the effective stiffness corresponding to the lower operating speed would be 2

nfmω = 94.05

2×10 =

88.45 kN/m and the effective stiffness corresponding to the upper operating speed would be 2

nfmω =

106.052×10 = 112.5 kN/m. Hence, the effective stiffness of shaft should not fall in the range of 98.1 to

112.5 kN/m. It should be noted that the unit of nfω is in rad/s when other quantities are in SI units (i.e. m

in kg and k in N/m).

Example 2.2: A rotor system has 100 rad/s as the critical speed and its operating speed is 120 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 2

nfmω = 100

2×10 = 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 125 rad/s. This will allow us to reach up to 120 rad/s that is 5 rad/s lower

than 125 rad/s i.e. the new critical speed.

The corresponding effective of the support stiffness should be 1252×10 = 156.25 kN/m. Hence, the

auxiliary support system should increase the effective stiffness by 56.25 kN/m.

33

Example 2.3: A 2 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 = 3

3

static force 2 10 9.8139.24N/m

corresponding static deflection 0.5 10

mg

δ

−

−

× ×= = =

×

The natural frequency = 3

/ 39.24140.1 rad/s = 22.3 Hz

2 10n

k mg g

m m

δω

δ −= = = = =

×

(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 2.6.

34

Figure 2.6 (a) A passive squeeze film damper

Figure 2.6 (b) Schematic diagram of a smart (active) fluid-film damper

Figure 2.6 (c) Basic principles of active magnetic bearings

35

2.2 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

( )df t cy= � (2.12)

where c is a constant of proportionality and is called the viscous damping coefficient. In the free body

diagram (Fig. 2.1d) 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+ + =�� � (2.13)

For the free damped motion, equation (2.13) takes the form

0my cy ky+ + =�� � (2.14)

Let us assume a solution of the form

st

y e= (2.15)

where s is a constant (may be a complex number), so that st

y se=� and 2 st

y s e=�� . On substituting equation

(2.15) in equation (2.14), we get

2( ) 0stms cs k e+ + = (2.16)

Hence, from the condition that equation (2.15) is a solution of equation (2.14) for all values of t, equation

(2.16) gives the following characteristic equation

2 0

c ks s

m m+ + = (2.17)

which can be solved as

2

1,22 2

c c ks

m m m

= − ± −

(2.18)

Hence, the following form of the general solution is obtained

36

( )s t s t

y t Ae Be= +1 2 (2.19)

where A and B are constants to be determined from initial conditions of the problem (e.g., 0(0)y y= and

0(0)y v=� ). On substituting equation (2.18) into equation (2.19), we get

2 2

2 22( )

c k c kt tc

t m m m mmy t e Ae Be

− − − −

= +

(2.20)

The term outside the bracket in right hand side is an exponentially decaying function for positive value of

c. However, the term

2

2

c k

m m

−

can have following three cases:

(i)

2

2

c k

m m

>

: Exponents of terms within the square bracket in equation (2.20) are real numbers, which

means there would not be any harmonic functions. Hence, no oscillation is possible and it is called the

over-damped system (see Figure 2.7)

Figure 2.7 Response of an over-damped system

(ii)

2

2

c k

m m

<

: Exponents of terms within the square bracket in equation (2.20) are imaginary numbers

2

j2

k c

m m

± −

, which means, we can write

22 2

j2

cos jsin2 2

k ct

m m k c k ce t t

m m m m

± −

= − ± −

.

Hence, equation (2.20) takes the following form

2 2

2( ) ( )cos j( )sin2 2

ct

m k c k cy t e A B t A B t

m m m m

= + − + − −

37

Let ( ) cosa A B X φ= + = and ( ) sinb i A B X φ= − = , we get

2

2 cos2

ct

m k cx Xe t

m mφ

= − −

; with 1 2 2tan ( / ); b a X a bφ −= = + (2.21)

For the present case oscillations are possible (with decaying type) and it is called the under-damped

system (Fig. 2.8). The damped natural frequency is given as

2

2dnf

k c

m mω = −

.

Figure 2.8 Response of an under-damped system

(iii)

2

2

c k

m m

=

: Exponents of terms within the square bracket in equation (2.20) are zero numbers. The

damping corresponding to this case is called the critical damping, cc , which is defined as

kmmmkmc nc 22/2 === ω (2.22)

Any other damping can be expressed in terms of the critical damping by a non-dimensional number ζ

called the damping ratio, as

/ cc cζ = (2.23)

The solution for the critical damped case, having two real repeated roots, can be expressed as

[ ]( ) nf tx t e A Bt

ω−= + (2.24)

38

For this case either no oscillation (similar to Fig. 2.7) or for specific initial conditions single crossing of

the zero amplitude axis is possible. Figure 2.9 shows a response of the critically damped system with a

single crossing.

Figure 2.9 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) 1ζ > : the

over-damped condition (ii) 1ζ < : the under-damped condition with the damped natural frequency as

21dnf nf

ω ω ζ= − (iii) 1ζ = : the critical damping and (iv) 0ζ = : the undamped system. For all the three

cases, the integration constants A and B are obtained from two initial conditions. Figure 2.10 shows a

response of an unstable system with 0ζ < 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, 1998).

Figure 2.10 Response of an unstable system

39

2.3 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 2.11(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 2.12. The straight-line motion

(Fig. 2.12c) could be considered as a single-DOF system considered in Section 2.1, whereas, the

elliptical orbital motion (Fig. 2.12b) may occur when the shaft has different stiffness in two orthogonal

directions.

Figure 2.11 Rankine rotor model

(a) Circular motion (b) Elliptical motion (c) Straight-line motion

Figure 2.12 Orbital motion of the mass

(a) Two- DOF spring-mass rotor model (b) Free body diagram of the model

40

For the circular orbital motion (Fig. 2.12a) 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 2.11(b), for a constant spin speed, the radius of whirling of the rotor center will increase

parabolically and is given as kFr c /= where Fc 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.

2.4 Jeffcott Rotor Model

To overcome the limitations of the Rankine model, Jeffcott (1919) proposed a model and Figure 2.13

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 2.14(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 2.14(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 2.14 and may occupy various positions depending upon the frequency of whirl.

41

The transverse stiffness, k, of a shaft simply supported shaft is expressed as

( )3 3

Load 48

Deflection / 48

P EIk

PL EI L= = = (2.25)

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. 2.13c). 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 2.13(a) A Jeffcott rotor model

Fig 2.13(b) A Jeffcott rotor model in y-z plane Fig 2.13(c) Free body diagram of the disc

in x-y plane

42

Figure 2.14(a) Synchronous whirl

Figure 2.14(b) Anti-synchronous whirl

From Figure 2.13(c) the force (or the moment) balance in the x , y and θ directions can be written as

( )2

2cos

dkx cx m x e

dtθ− − = +� (2.26)

( )2

2sin

dky cy mg m y e

dtθ− − − = +� (2.27)

and

cos pmge Iθ θ− = �� (2.28)

where ( )cosx e θ+ and ( )siny e θ+ are the position of the centre of gravity, G, of the rotor; m is the

mass and Ip 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

Shaft whirling

direction

Shaft spin

direction

Shaft

Shaft whirling

direction

Shaft spin

direction

Shaft

43

sub-critical speed phenomenon. Hence, equations of motion in the x and y directions, from equations

(2.26) and (2.27), can be written as

2 cosmx cx kx m e tω ω+ + =�� � (2.29)

and 2 sinmy cy ky m e tω ω+ + =�� � (2.30)

It should be noted that above equations of motion are uncoupled, and motion can be analysed

independently in two transverse planes. Noting equation (2.8), 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

mknf /2,1

=ω (2.31)

2.4.1: 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 (2.29) and (2.30) can be assumed as

cos( )x X tω φ= − (2.32)

and

[ ]cos ( ( / 2) sin( )y Y t Y tω φ π ω φ= − + = − (2.33)

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/2 radian. For the direction of shaft whirling shown in Figure 2.13 (i.e.,

counter clockwise; ccw) for the present axis system, the response in y-direction will lag the response in x-

direction by π/2 radian. Hence the lag of the y-direction response with respect to the force will be (π/2 +

φ). On taking the first and second time derivatives of the response, x(t), we get

sin( )x X tω ω φ= − −� and 2 cos( )x X tω ω φ= − −�� (2.34)

44

On substituting equations (2.33) and (2.34) into equation (2.29) and separating the in-phase (i.e., cosωt)

and quadrature (i.e., sinωt) terms, we get

2 2cos sin cosm X cX kX m eω φ ω φ φ ω− + + = (2.35)

and 2 sin cos sin 0m X cX kXω φ ω φ φ− − + = (2.36)

Equation (2.36) can be solved for the phase, as

2tan

c

k m

ωφ

ω=

− (2.37)

which gives

( ) ( )2 22

sinc

k m c

ωφ

ω ω=

− + and

( ) ( )

2

2 22

cosk m

k m c

ωφ

ω ω

−=

− + (2.38)

Substituting equation (2.38) into equation (2.35), it gives the displacement amplitude as

( ) ( )

2

2 22

m eX

k m c

ω

ω ω=

− + (2.39)

Similarly, we can obtain the response amplitude in the y-direction from equation (2.30) as

( ) ( )

2

2 22

m eY

k m c

ω

ω ω=

− + (2.40)

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 (2.32), (2.33), (2.39) and (2.40), 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.,

2 2 2x y R+ = (2.41)

2.4.2: Steady-state response (Approach 2): An alternative approach that is very popular in rotor

dynamics analyses is to use the complex algebra to define the whirl radius as

jr x y= + (2.42)

45

where j 1= − . On multiplying equation (2.30) by j and adding to equation (2.29), we get

2 je t

mr cr kr meωω+ + =�� � with

j cos jsinte t tω ω ω= + (2.43)

Now the steady state response can be assumed as

j( )e t

r Rω φ−= (2.44)

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 (2.44) with respect to time, t, we get

j( )j e t

r Rω φω −=� and 2 j( )e t

r Rω φω −= −�� (2.45)

On substituting equations (2.44) and (2.45) into equation (2.43), we get

( ){ }2 j 2j ek m c R meφω ω ω−− + = (2.46)

Equation (2.46) can be written as

( ){ }( )2 2j cos j sink m c R R meω ω φ φ ω− + − = (2.47)

On separating the real and imaginary parts of equation (2.47), we get

2 2( ) cos sink m R cR meω φ ω φ ω− + = (2.48)

and 2( ) sin cos 0k m R cRω φ ω φ− − + = (2.49)

From equation (2.49), we get the phase as

2tan

c

k m

ωφ

ω=

− (2.50)

On substitution of the phase from equations (2.50) into (2.48), the whirl amplitude can be written as

( ) ( )

2

2 22

m eR

k m c

ω

ω ω=

− + (2.51)

Equations (2.49) and (2.51) are similar to previous results, i.e., equations (2.37) and (2.39). The non-

dimensional form of equations (2.50) and (2.51) can be written as

2

2tan

1

ζωφ

ω=

− (2.52)

46

and

( ) ( )

2

2 221 2

RR

e

ω

ω ζω= =

− + (2.53)

with

/ ; / ; / ; 2nf nf c ck m c c c kmω ω ω ω ζ= = = = (2.54)

where R is the whirl amplitude ratio, ω is the frequency ratio, nfω is the natural frequency of non-

rotating rotor system, ζ is the damping ratio, and cc is the critical damping for which the damping ratio is

equal to unity. Fig. 2.15 shows the whirl amplitude and the phase variation with the frequency ratio.

Figure 2.15(a) shows that the maximum amplitude (i.e., the location of the critical speed) occurs at 1ω =

for the undamped case; however, at slightly higher frequency ratio than one (i.e., 1ω > ), 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 2.15 Plot of (a) Non-dimensional response (b) Phase versus frequency ratio ω

47

It can be seen from Figure 2.15(b), for a lightly under-damped system the phase angle changes from 00 to

900 as the spin speed is increased to

nfω (i.e., ω =1) and gradually becomes 1800 as the spin speed is

increased to a higher frequency ratio. It should be noted that the phase angle is 900 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 900 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 2.15(a) as the

spin speed approaches infinity the displacement of the shaft tends to the equal to the disc eccentricity (

1R = ). 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 900 irrespective of damping in the

system). The change in phase between the force and the response is also shown in Figure 2.16 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 2.16 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 2.14(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 2.14(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.

48

2.4.3: 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 (2.29) and (2.30) can be combined in the

matrix form as

cos

sin

m x c x k x m e t

m y c y k y m e t

ω ωω ω

+ + =

2

2

0 0 0

0 0 0

�� �

�� � (2.55)

The force vector in equation (2.55) could be expressed as

( )( )

( )( )

2 j22

j

22 2 j

cos jsincosRe Re Re

sin jcossin j

x

y

t

unb t

tunb

me e Fm e t tm e te

Fm e t tm e t me e

ω

ω

ω

ωω ω ωω ωω ω ωω ω ω

+ = = = − −

(2.56)

jy xunb unb

F F= − (2.57)

where Re(.) represents the real part of the quantity inside the parenthesis. xunbF and

yunbF are the

unbalance force components in x and y directions, respectively. On substituting equation (2.56) into

equation (2.55), it can be written as (henceforth for brevity the symbol Re(.) will be omitted)

j0 0 0

0 0 0

x

y

unb t

unb

Fm x c x k xe

Fm y c y k y

ω

+ + =

�� �

�� � (2.58)

The relationship (2.57) is true for the present axis system along with directions of the whirling (R) and the

unbalance force vectors chosen as shown in Figure 2.17(a). For this case yunbF lag behind

xunbF by 900.

Let us derive this relationship: If j

xunbF Feθ= , then

[ ]j( / 2) j ( / 2) j j cos( / 2) jsin( / 2) j jy x

j

unb unbF Fe Fe e Fe Fe Fθ π θ π θ θπ π− −= = = − + − = − = −

For the direction of whirl (R) opposite to the axis system as shown in Figure 2.17(b), the following

relationship holds

[ ]j( / 2) j ( / 2) j j cos( / 2) jsin( / 2) j jy x

j

unb unbF Fe Fe e Fe Fe Fθ π θ π θ θπ π+= = = + = =

so that,

jy xunb unbF F= (2.59)

49

in which case yunbF lead

xunbF by 900. It should be noted in equation (2.58) 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 (2.56). Equation (2.58) can be written in more a compact form as

[ ]{ } [ ]{ } [ ]{ } { } j t

unbM x C x K x F eω+ + =�� � (2.60)

The solution can be chosen as

{ } { } j tx X e

ω= (2.61)

where elements of the vector {X} are, in general, complex.

Equation (2.61) can be differentiated to give

{ } { } { } { }j 2 jj and t tx X e x X e

ω ωω ω= = −� �� (2.62)

On substituting equations (2.61) and (2.62) into equation (2.60), we get

[ ]{ } { }unbZ X F= (2.63)

with

[ ] [ ] [ ] [ ]( )2 jZ M K Cω ω= − + + (2.64)

where [Z] is the dynamic stiffness matrix. The response can be obtained as

Fig 2.17(a) The direction of whirl

same as the positive axis direction

Fig 2.17(b) The direction of whirl

opposite to the positive axis direction

50

{ } [ ] { }1

unbX Z F−

= (2.65)

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 2.4: Obtain the unbalance response of a rotor system with following equations of motion.

2 cosmx kx m e tω ω+ =�� and

2 sinmy ky m e tω ω+ =��

Solution: Since equations of motion are uncoupled hence both equations can be solved independently.

The first equation can be written as

j t

xmx kx F eω+ =�� with

2ωmeFx =

in which the real part of right hand side term has only meaning. The solution can be assumed as

j t

x Xeω=

where in general X is a complex quantity. The above equation gives

2 j t

x Xeωω= −��

On substituting assumed solutions in the equation of motion, we get

( )2 2m X kX meω ω− + =

which gives

2

2

meX

k m

ωω

=−

Hence the response in the x-direction becomes

2 2 2

j

2 2 2Re (cos jsin ) costme me me

x e t t tk m k m k m

ωω ω ωω ω ω

ω ω ω

= = + = − − −

Similarly, the second equation of motion can be written as

j t

ymy ky F eω+ =�� with 2

jyF meω= −

in which the real part of the right hand side term only has a meaning. The solution can be assumed as

51

j ty Ye

ω=

On substituting in the equation of motion, we get

2

yFY

k mω=

−

Hence the response in the y-direction becomes

2 2 2

j j

2 2 2 2

jRe ( jcos sin ) sin

y t tF me me me

y e e t t tk m k m k m k m

ω ωω ω ωω ω ω

ω ω ω ω −

= = = − + = − − − −

(Answer)

Now the same problem is solved in the matrix form. We have

[ ] [ ] [ ] { }2

2

0 0 0 0; ; ;

0 0 0 0 junb

m k meM K C F

m k me

ωω

= = = = −

[ ] [ ] [ ]( ) [ ]( )

( )

22

12

2 2

1/ 00;

0 0 1/

k mk mZ M K Z

k m k m

ωωω

ω ω

− − − = − + = = − −

so that

{ } [ ] { }( )

( ) ( )

22 2

1

2 222

1/ 0 1

j j0 1/unb

k m me meX Z F

me mek mk m

ω ω ωω ωωω

− − = = = − −−−

The response in the x- and y-directions can be written as

{ }( ) ( )

2 2

j

2 22 2

cos1 1Re

j sin

tx me me t

x ey me me tk m k m

ωω ω ωω ω ωω ω

= = = −− −

(Answer)

2.5 A Jeffcott Rotor Model with an Offset Disc

Figure 2.18(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., ϕx and ϕy, 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. 2.18(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.

52

ϕx

Fig 2.18(a) A Jeffcott rotor with a disc offset Fig 2.18(b) Free body diagram of the shaft

from the mid span in the y-z plane in the x-y plane

Fig 2.18(c) Free body diagram of the disc Fig 2.18(d) Free body diagram of the shaft

in the y-z plane in the y-x plane

From Figure 2.18(b), we can have the following relations for the eccentricity

cosx

e CH e tω= = and sinye GH e tω= = (2.66)

where x

e and ye 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 2.18(c) equations of motion of the disc in the y- and x

ϕ directions can be written as

( )2

2cosy y x

df m y e

dtϕ− = + (2.67)

and

y y x yz d xf e M Iϕ ϕ− − = �� (2.68)

53

where m is the disc mass, d

I is the diametral mass moment of inertia about the x-axis, fy is the reaction

force and Myz 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, ϕx. Figure 2.19(a) shows the rotor in the z-x plane. From Figure 2.19(b), we

can write equations of motion as

( )2

2cosx x y

df m x e

dtϕ− = + (2.69)

and

x x y zx d yf e M Iϕ ϕ− − = �� (2.70)

where d

I is the diametral mass moment of inertia about the y-axis, fx is the reaction force and Mzx is the

reaction moment. Equations (2.69) and (2.70) are also non-linearly coupled with the angular component

of displacement, ϕy. 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 (2.67) and (2.68)

and equations (2.69) and (2.70) can be solved independent of each other.

Fig 2.19(c) Free body diagram of the shaft in the z-x plane

Fig 2.19(a) A Jeffcott rotor with a disc offset

from the mid span in the z-x plane

Fig 2.19(b) Free body diagram of the disc in

the z-x plane

54

Unbalance forces can be simplified with the assumption of small angular displacement (i.e.,

cos cos 1x yϕ ϕ= ≈ ) and equations (2.67) and (2.69) can be simplified as

2

sinymy f m e tω ω+ =�� (2.71)

and 2 cosxmx f m e tω ω+ =�� (2.72)

Now equations (2.71), (2.68), (2.72) and (2.70) are assembled as

2

2

00 0 0 sin

0 0 0 0

00 0 0 cos

0 0 0 0

y

x y y xd yz

x

y x x yd zx

ym f m e t

f eI M

xm f m e t

f eI M

ω ωϕ ϕ

ω ωϕ ϕ

+ + =

��

��

��

��

(2.73)

which can be written in matrix notation as

[ ]{ } { } { } { }L NL unbM x R R f+ + =�� (2.74)

with

[ ]

0 0 0

0 0 0

0 0 0

0 0 0

d

d

m

IM

m

I

=

; { } x

y

y

xx

ϕ

ϕ

=

��

��

����

��

;

{ }

y

yz

L

x

zx

f

MR

f

M

=

; { }

0

0

y y x

NL

x x y

f eR

f e

ϕ

ϕ

=

; { }

2

2

sin

0

cos

0

unb

m e t

fm e t

ω ω

ω ω

=

where [M] represents the mass matrix, {fumb} 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 zxx f Mα α= +11 12 and

y x zxf Mϕ α α= +21 22 (2.75)

55

where αij represent the displacement at ith 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 (2.75) can be written in a matrix form as

11 12

21 22

x

y zx

x f

M

α αϕ α α

=

(2.76)

with

2 3 22 2

11 12

2 221 22

3 2

3 3

3 3( )

3 3

a l a ala b

EIl EIl

al a lab b a

EIl EIl

α α

α α

− − −

=− − −−

where EI is the beam flexure, parameters a and b are defined in Figure 2.19(a) with l a b= + . From the

simple beam deflection theory, we can get these influence coefficients (Timoshenko and Young, 1968).

Equation (2.76) can be written as

x

y yzx

x xf k k

M k k

α αϕ ϕα α

−

= =

1

11 12 11 12

21 22 21 22

(2.77)

where kij is the stiffness coefficient and defined as force at ith station due to a unit displacement at j

th

station. Similarly, since the shaft is symmetric about its rotation axis, we can obtain

y

yz x

f yk k

M k k ϕ

=

11 12

21 22

(2.78)

Equations (2.77) and (2.78) can be combined in matrix form as

{ } [ ]{ }LR K x= (2.79)

with

56

[ ]

11 12

21 22

11 12

21 22

0 0

0 0

0 0

0 0

k k

k kK

k k

k k

=

; { }

y

yz

L

x

zx

f

MR

f

M

=

; { } x

y

y

xx

ϕ

ϕ

=

Noting equation (2.79), the nonlinear reaction force vector takes the following form

{ } 11 12

11 12

0 0

( )

0 0

( )

y y x x y x

NL

x x y y x y

f e k y k eR

f e k x k e

ϕ ϕ ϕ

ϕ ϕ ϕ

+

= = +

(2.80)

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 (2.79) into equations of motion, i.e., equation (2.74), we get

[ ]{ } [ ]{ } { }unbM x K x f+ =�� (2.81)

with

[ ]

0 0 0

0 0 0

0 0 0

0 0 0

d

d

m

IM

m

I

=

; [ ]

11 12

21 22

11 12

21 22

0 0

0 0

0 0

0 0

k k

k kK

k k

k k

=

; { } x

y

y

xx

ϕ

ϕ

=

; { }

2

2

sin

0

cos

0

unb

m e t

fm e t

ω ω

ω ω

=

2.5.1: Calculation of natural frequencies: For obtaining natural frequencies of the system the determinant

of the dynamic stiffness matrix, ( )2[ ] [ ] [ ]Z K Mω= − , 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.

2.5.2: Unbalance forced response: The unbalance forcing with frequency, ω, can be written as

{ } { } j t

unb unbf F e

ω= with jk k k

r i

unb unb unbF F F= + 1, 2, ,k N= � (2.82)

57

where { }unbF 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

{ } { } j tx X e

ω= so that { } { } j tx X e

ωω= − 2�� (2.83)

On substituting equations (2.82) and (2.83) into equation (2.81), we get the unbalance response as

{ } [ ] { }umbX Z F

−=

1 with [ ] [ ] [ ]( )MKZ 2ω−= (2.84)

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

jr i

k k kX X X= + 1, 2, ,k N= � (2.85)

which will give amplitude and phase information, as

( ) ( )amp r ik k kX X X= +

2 2

and ( )tan /phase i kk k kX X X−= 1 (2.86)

Equation (2.84) 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

[ ] [ ] [ ] [ ]( )jZ K M Cω ω= − +2 (2.87)

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.

2..5.3: Bearing reaction forces: Forces transmitted through bearings are those, which are related to the

deflection of the shaft as shown in Figure 2.20 on the y-z plane.

58

On taking moments about ends L (left) and R (right) of the shaft, we have

2 2

10 0 or

L y yz y y y yz

aM f a M R l R f M

l l= ⇒ − − = = −∑ (2.88)

and

1 1

10 0 or

R y y yz y y yz

bM R l f b M R f M

l l= ⇒ − − = = +∑ (2.89)

Figure 2.20 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, fy and Myz. In matrix form equations (2.88) and (2.89) can be written as

{fb} = [D] {fs} (2.90)

with

{ } { } 1j

2

yt

b b

y

Rf F e

R

ω = =

; { } { } j yt

s s

yz

ff F e

M

ω = =

; [ ]

1

1

b l lD

a l l

= −

where subcripts: b and s represent the bearing and the shaft, respectively. Complex vectors {Fb} and {Fs}

are bearing forces at the shaft ends and shaft reaction forces at the disc, respectively. On using equations

(2.79) and (2.84) into the form of equation (2.90) for both plane motions (i.e., y-z and z-x), we get

{Fb} = [D] [K]{X}=[D] [K] [Z]{Funb}=[C]{Funb} (2.91)

59

with

[C] = [D][K][Z]

It should be noted that equation (2.91) 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

jk k k

r i

b b bF F F= + 1, 2, ,k N= � (2.92)

which will give the amplitude and the phase information, as

( ) ( )k kk

amp r ib bb

F F F= +2 2

and ( )tan /k kk

phase i rb bb

F F F−= 1 (2.93)

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 2.5 Find the bending natural frequency of a rotor system shown in Figure 2.21. 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 = 2.1×1011

N/m2. Take one plane motion only.

Figure 2.21 Example 2.5

Solution: Figure 2.22 shows the deflected position of the shaft. For a simply supported beam, the

influence coefficient is defined as (Timoshanko and Young, 1968)

( )2 2 2

( ), ( )

6

bx l x by xx a

F EIlα

− −= = ≤

60

Figure 2.22 A shaft in the deflected position

For obtaining 11α (which is defined as the deflection at station 1 for the unit force at station 1), hence, we

have x = 0.6 m with l = 1.0 m and b = 0.4 m. Hence, it can be obtained as

( )2 2 2

80.611 11

0.4 11 4

0.4 0.6 1 0.6 0.41.863 10 m/N

6 2.1 10 (0.1) 164

x ab

α απ

−= ==

× × − −= = = ×

× × × × ×

Considering a single plane motion and neglecting the angular displacement components, the natural

frequency can be obtained as (refer equation (2.83) for the dynamic stiffness matrix)

nfZ mω

α= − =2

11

10

which gives

8

1 11

1 12316.83 rad/s

10 1.863 10n

mω

α −= = =× ×

or 368.92 Hz (Answer)

Example 2.6: Obtain transverse natural frequencies of an offset Jeffcott rotor system as shown in Figure

2.23. Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, Id = 0.02 kg-m2 and the

disc is placed at 0.25 m from the right support. The shaft has the diameter of 10 mm and total length of

the span is 1 m. The shaft is assumed to be massless. Use the influence coefficient method. Take shaft

Young’s modulus E = 2.1 × 1011

N/m2. Neglect the gyroscopic effect and take one plane motion only.

61

Solution: Influence coefficients for a linear and angular diaplacements (y, ϕ) correspoding to a force (f)

and a moment (M) acting at the disc are defined as

2 241.137 10

3yf

a b

EIlα −= = × m/N;

( )2 3 2

43 2

3.03 103

yM

a l a al

EIlα −

− −= − = − × m/N

4( )3.03 10

3f

ab b a

EIlϕα −−

= = − × m/N; ( )2 2

33 3

1.41 103

M

al a l

EIlϕα −

− −= − = × m/N

Figure 2.23 Example 2.6

For the present problem only single plane motion is considered. For free vibration, from equation (2.81),

we get

1

0 0

0 0

yf yM

f Md

m x x

I ϕ ϕ

α αα αϕ ϕ

−

+ =

��

��

Since it will execute the SHM for the free vibration, we have

1

20 0

0 0

yf yM

nf

f Md

m x

I ϕ ϕ

α αω

α α ϕ

− − + =

where nfω is the natural frequency of the system. Above equation is an eigen value problem. For non-

trivial solution, we have

1

20

00

yf yM

nf

f Md

m

I ϕ ϕ

α αω

α α

− − + =

62

which gives a frequency equation in the form of a polynomial, as

( ) ( )4 2 2 1 0d nf yf M yM nf yf M d

mI m Iϕ ϕω α α α ω α α− − + + =

On substituting values of the present problem parameters, it gives

4 4 2 78.505 10 7.3 10 0nf nfω ω− × + × =

It can be solved to give two natural frequency of the system as

129.4nfω = rad/sec and

2290nfω = rad/sec (Coupled translation and tilting motions)

For the present problem the linear and angular displacements in a single plane are coupled. Since natural

frequencies obtained are system natural frequencies and hence are not as such related to the pure

translational or pure rotational motions. If we consider these two motions are uncoupled, then

corresponding natural frequencies can be obtained as

. ..

nf

yfmω

α −= = = =

× ×1 4

1 1879 5 29 65

10 1 137 10 rad/s (Pure translation motion on the disc)

and

2 3

1 1188

0.02 1.4146 10nf

d MI ϕ

ωα −= = =

× × rad/s (Pure tilting motion of the disc)

It can be seen that there is a small difference in the fundamental natural frequency due to pure translation

motion (29.65 rad/s) with that of the fundamental natural frequency of the coupled system (29.4 rad/s),

and a large difference in the natural frequency for the pure tilting motion (188 rad/s) with the second

natural frequency of the coupled system (290 rad/s).

Example 2.7: A disc of mass 13.6 kg and the polar mass moment of inertia 0.02 kg-m2, is mounted at the

mid-span of a shaft with a span length of 0.4064 m. Assume the shaft to be simply supported at bearings.

The rotor is known to have an unbalance of 0.2879 kg-cm. Determine forces exerted on bearings at the

spin speed of 6000 rpm. The diameter of the steel shaft is 2.54 cm with E = 200 GNm-2

.

Solution: The following data are available:

U = mass of rotor × eccentricity = 0.2879 kg-cm, m = 13.6 kg, e = U/m = 0.0211 cm,

ω = 6000 rpm, d = 2.54 cm, E = 200×109 N/m

2

63

Figure 2.24 Example 2.7

Bearing forces are obtained by considering firstly, shaft as rigid and then by considering shaft as flexible.

In both cases, bearings are considered as rigid in transverse directions.

(i). For the rigid shaft & rigid bearings (Fig. 2.24):

The unbalance force ( )22 3 2π 60000.2879 10 113.66 N

60meω − × ×= = × =

The force at each bearing (amplitude) = 113.66/2 = 56.83 N

The component of forces in the vertical & horizontal directions are given as

( )

56.83cos200π = 56.83cos200π 133.4 N

56.83sin 200π 56.83sin 200π 133.4 N

f

f

t tx

t mg ty

= +

= + = +

(ii) For the flexible shaft and rigid bearings (Method 1) (Fig. 2.25):

Since fy = ky, bearing reaction forces can be written as (Figure 2.25c)

/ 2 / 2A B yR R f ky= = =

The stiffness of the rotor system as shown in Figure 2.24 is given as

( )( )

11 4

33

6

48 2.0 10 0.025448 64 2.92 10 N/m

0.4064

EIk

l

π× × ×

= = = ×

64

(a) A flexible rotor system

(b) Free body diagram of disc (c) Free body diagram of shaft

Figure 2.25 Example 2.7

EOM of the disc, from the free body diagram of the disc (Figure 2.25b), is given as

2me ky myω − = ��

For the simple harmonic motion, we have 2y yω= −�� , hence the above equation can be written as

( ) ( )( )

2224

22 6

0.2879 10 200π4.64 10 m

2.92 10 13.6 200π

mey

k m

ωω

−−

××= = = − ×

− × − ×

Hence the bearing reaction force can be obtained as

( )6 4 / 2/ 2 2.92 10 4.64 10 677.6 NAR ky −= = × × − × = −

65

Components of bearing force in the vertical & horizontal directions can be obtained as

677.6cos200AxR πt= and ( )677.7sin200 133.4 NAy

R πt= +

(iii) For the flexible shaft and rigid bearings (Method 2): Now the influence coefficient method is used.

Bearing forces are given as

1/ 2

1/ 2

1 1 0.4064

1 1 0.4064

A y y

zx zxB

f f

M M

R b l l

R a l l

=

− −=

where the reaction forces from the disc can be expressed as

11 12

21 22

y

zx x

f y

M

k k

k k ϕ

=

with

1 1

11

1

22

3 611 12

512 22

0 48 / 0 2.92 10 0[ ]

0 0 12 / 0 1.21 10

EI lK

EI l

α

α

α α

α α

− −

−

×= = = =

×

where

2 2 2 3 23

11 22 12 21;

- 3 -3 - - 3 -2 -( - )

, 0; 048 3 (12 ) 3 3

al a l a l a all l ab b a

EI EIl EI EIl EIlα α α α

= = = = = = =

Displacement vectors are related with the unbalance force as

1 211 12

21 22 0x

Zy

Z Z

Z me

ϕω

−

=

with

[ ]1 22 7

1 1111

2 622

222

10

0 4.08 10 0

10 0 8.84 100d

d

I

I

k mk mZ

k

k

ωω

ωω

− −−

=−

= =

−− − ×

− ×−

In view of above equations, bearing reaction forces can be written as

66

( ) ( )

( ) ( )

( )

( )

11

222 11

11 2

2

11

11 22 22 2 2

11

11 22

2 211 22

1

01

dA

B

d

k a l

k m

k a lme

k m

k b l k lme

k m k IR me

R k a l k l

k m k I

ω

ωω

ωω ω ω

ω ω

− = −

−− −

=

− −

From above equations, we have

62 2

6 2

2.92 10 (1/ 2)(0.2879 10 ) (200 ) 677.6 N

2.92 10 13.6 (200 )AR π

π−×

= −×

= × × ×× − ×

and

677.6 NB

R = −

which is same as by previous method. It would be interesting to vary the spin speed and plot the bearing

forces with it. It should be noted since the disc is at the mid-span, hence there is no contribution of the

diametral mass moment of inertia on to bearing reactions. If there had been couple unbalance then the

diametral mass moment of inertia would have affected bearing reactions. As an exercise take the disc

location from the left support a = 0.3 l and obtain bearing bearings for the same.

Example 2.8. Find the transverse natural frequency of a rotor system as shown in Figure 2.27. Consider

the shaft as massless and is made of steel with 2.1(10)11

N/m2 of the Young’s modulus, and 7800 kg/m

3 of

the mass density. The disc has 10 kg of the mass. The shaft is simply supported at ends.

Figure 2.27 Example 2.8 (all dimensions are in cm)

Solution: Considering only the linear displacement, first we will obtain the stiffness (or the influence

coefficient, 11α ) for Figure 2.28 using the energy method. On taking the force and moment balances, we

have

67

0 0A BF F F F+ ↑ = ⇒ + − =∑ and 0 1 0.6 0A BM F F+ = ⇒ × − × =∑

which gives reaction forces as

0.4A

F F= and 0.6B

F F=

Figure 2.28 (dimensions are in m)

Figure 2.29 Free body diagram of shaft segment for 0 ≤ x ≤ 0.6

Bending moments are obtained at various segments of the shaft to get the strain energy of the system. On

taking the moment balance in the free body diagram as shown in Figure 2.29 of a shaft segment for 0.0 ≤

x ≤ 0.6, we get

1 1 0 0.4 0; or 0.4 , 0 0.6A

M M Fx M Fx x+ = ⇒ − = = ≤ ≤∑ (a)

Figure 2.30 Free body diagram of shaft segment for 0.6 ≤ x ≤ 1.0

68

On taking the moment balance in the free body diagram as shown in Figure 2.30 of the shaft segment for

0.6 ≤ x ≤ 1.0, we get

2 2 0 ( 0.6) 0.4 0 or 0.6 (1 ); 0.6 1.0A

M M F x Fx M F x x+ = ⇒ + − − = = − ≤ ≤∑ (b)

The strain energy is expressed as

2 2

1 2

1 2

0.6 1.0

0 0.62 2

M dx M dx

EI EIU += ∫ ∫

The linear displacement is expressed as

( ) ( )1.0

211

1 2

0.6

2

0 0.6

// M FM F M dxM dxU

FEI EI

δ∂ ∂∂ ∂∂

= = +∂

∫ ∫

On substituting bending moment expression from equations (a) and (b), we get

{ }{ }1 2

0.6 1

1 20 0.6

0.6 (1 ) ( 0.6(1 ) 0.01152 0.00768( 0.4 )( 0.4 ) F x x dxFx x dxF

EI EI EI EIδ

=

− − − −− −= + +∫ ∫

The stiffness is given as

1 2

1

70.01152 0.007688.45 10 N/m

Fk

EI EIδ

−

= = + = ×

where 11 2 4 6 4 3 4 4

; ; 21

2 10 N/m 0.1 4.907 10 m 0.3 3.976 10 m64 64

E I Iπ π− −= × = × = × = × = ×

which gives the natural frequency as

78.45 102906.81 rad/s

10nf

k

mω ×

= = = (answer)

It should be noted that the tilting motion of the disc has not considered. For the coupled linear and angular

motions, natural frequencies of the system can be obtained as an exercise by obtaining corresponding

influence coefficients.

69

Example 2.9 Obtain the bending natural frequency for the synchronous motion of a rotor as shown in

Figure 2.31. The rotor is assumed to be fixed supported at one end. Take mass of the disc m = 1 kg. 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 E = 2.1×1011

N/m2.

Figure 2.31

Solution: Let us assume for simplicity that there is no coupling between the linear and angular motions.

Considering only the linear displacement, the transverse stiffness for this case would be

11 104

3 3

3 3 2.1 10 4.909 103.866 10

0.2yf

f EIk

y l

−× × × ×= = = = × N/m (a)

with

4 4 100.01 4.909 10

64 64I d

π π −= = = × m4 (b)

Hence, the natural frequency would be

43.866 10

196.621

nf

k

mω

×= = = rad/s (answer)

2.6 Suppression of Critical Speeds

In the present section, an interesting phenomenon will be dealt in which a critical speed will be shown to

be eliminated by suitably choosing system parameters. For this purpose, the Jeffcott rotor model with a

disc offset has been chosen. Now, for a detailed in depth analysis a closed form expression for the

response is obtained by the defining the following complex displacements

jr x y= + and jr y xϕ ϕ ϕ= + (2.94)

70

Equations of motion (2.74) can be written as

j t

d r dm r k r k m e eωϕ ω+ + = 2

11 12�� (2.95)

and

d r rI k r kϕ ϕ+ + =21 22 0�� (2.96)

with x yd d dI I I= = . Let the solution be

j( )rtr Re ω φ−= and

j( )r

t

r reϕω φϕ −

= Φ (2.97)

where R and Φr are the linear and angular whirl amplitudes, respectively; φr and ϕφ are the phase of the

linear and angular whirl amplitudes, respectively (these are all real quantities); so that

j( )rtr Re ω φω −= − 2�� and

j( )t

r reϕω φϕ ω −= − Φ2

�� (2.98)

On substituting equations (2.97) and (2.98) into equations of motion (2.95)-(2.96), we get

( ) jj2 2

11 12r

d r dk m Re k e m eϕφφω ω−−− + Φ = (2.99)

( ) jj 2

21 220r

d rk Re k I e ϕφφ ω −− + − Φ = (2.100)

Equation (2.100) can be expressed as

( )j j21

2

22

r

r

d

ke Re

k I

ϕφ φ

ω− −−

Φ =−

(2.101)

On substituting equation (2.101) into equation (2.99), we get

( )( )( )

j rd d

d

d

k m k I k kRe m e

k I

φω ω

ωω

− − − − = −

2 211 22 12 21 2

222

(2.102)

On equating the real and imaginary parts of both sides of equation (2.102), we get

71

( )( )( )

cosd d

r d

d

k m k I k kR m e

k I

ω ωφ ω

ω

− − − = −

2 211 22 12 21 2

222

(2.103)

and

( )( )( )

sind d

r

d

k m k I k kR

k I

ω ωφ

ω

− − − = −

2 211 22 12 21

222

0 (2.104)

From equation (2.104), we get

sin ; i.e. r rφ φ= =0 0 (2.105)

which means there will not be any phase difference. On substituting phase information in equation

(2.104), we get

( )( )( )

d d

d d

m e k IR

k m k I k k

ω ω

ω ω

−=

− − −

2 222

2 211 22 12 21

(2.106)

which is the whirl amplitude and the condition of resonance can be obtained by equating the denominator

of equation (2.106) to zero

( )( )d cr d crk m k I k kω ω− − − =2 211 22 12 21 0 (2.107)

where crω represents the critical speed. By defining

2 11r

d

k

mω = ,

2 22

d

k

Iϕω = ,

2 12r

d

k

mϕω = , and

2 21r

d

k

Iϕω = (2.108)

Equation (2.107) can be written as

( ) ( )4 2 2 2 2 2 2 2 0cr r cr r r rϕ ϕ ϕ ϕω ω ω ω ω ω ω ω− + + − = (2.109)

The solution of the above polynomial can be expressed as

( ) ( ) ( )1,2

21 12 2 2 2 2 2 2 2 2

2 24cr r r r r rϕ ϕ ϕ ϕ ϕω ω ω ω ω ω ω ω ω= + ± + − −

72

or

( ) ( ) ( )1,2

21 12 2 2 2 2 2 2 2

2 24cr r r r r rϕ ϕ ϕ ϕ ϕω ω ω ω ω ω ω ω ω= ± + ± + − − (2.110)

which gives critical speeds of the rotor system (the outer most negative sign has no meaning since

frequency can not be negative). Hence, for the case when the rotor is not mounted at the mid-span, there

are two critical speeds due to coupling of the linear and angular displacements. The above solution

(i.e., equation (2.110)) can be more critically analysed as follows. It can be seen that terms inside the first

square root is always positive, i.e., ( ) ( )22 2 2 2 2 2

4 0r r r rϕ ϕ ϕ ϕω ω ω ω ω ω+ − − > , since it can be rearranged as

( )22 2 2 2

4 0r r rϕ ϕ ϕω ω ω ω− + > (2.111)

It can be seen that the above condition be always true since all individual terms ωr, ωϕ, ωrϕ, and ωφr are

the real quantity. However, if the following condition is valid for terms inside the first square root

( )2 2 2 2 0r r rϕ ϕ ϕω ω ω ω− > (2.112)

then, it gives two real critical speeds (1,2cr

ω ), since equation (2.104) gives two real roots. However, if the

following condition prevails

( )2 2 2 2 0r r rϕ ϕ ϕω ω ω ω− < (2.113)

then, it gives only one real critical speed since the other root will be complex. Figures 2.32 (a) and (b)

gives these two cases, respectively. It can be seen that for the first case two distinct peaks corresponds to

two critical speeds. For the second case only one critical speed is observed. However, there is anti-

resonance with very low amplitude of vibrations. The following data is taken for the simulation: the disc

mass = 1 kg, the unbalance mass eccentricity = 0.0001 m, the diametral mass moment of inertia = 0.03

kgm2, k11 = 1000 N/m, k22 = 6 N/m, k12 = 100 N/m and k21 = 0.5 N/m. For the disc at the center of the

shaft span, we have 12 21 0k k= = , which gives

( )( )( ) ( )

2 2 222

2 2 2

11 22 11

d d d

d d d

m e k I m eR

k m k I k m

ω ω ω

ω ω ω

−= =

− − − (2.114)

73

which is same as discussed in the previous section for the Jeffcott rotor. The response is shown in Figure

2.32(c). It can be observed that it has only one critical speed, which may not coincide with the critical

speeds obtained by equation (2.110) in Figures 2.32(a) and (b). However, there will be another critical

speed corresponding to angular displacement and it is illustrated subsequently.

Figure 2.32(c) Variation of the amplitude (R) versus the spin speed (ω) for 12 21 0k k= =

Figure 2.32(a) Variation of the

amplitude (R) versus the spin speed (ω) for

( )2 2 2 2 0r r rϕ ϕ ϕω ω ω ω− >

Figure 2.32(b) Variation of the

amplitude (R) versus the spin speed

(ω) for ( )2 2 2 2 0r r rϕ ϕ ϕω ω ω ω− <

74

On substituting equation (2.106) into equation (2.101), we get

( )( )

( )( )

2 2

22j 21

2 2 222 11 22 12 21

d d

r

d d d

m e k Ike

k I k m k I k k

ϕφ ω ω

ω ω ω

− −−Φ =

− − − −

(2.115)

On equating imaginary parts of equation (2.115), we get

sin 0; i.e. 0ϕ ϕφ φ= = (2.116)

which means there will not be any phase difference in the angular displacement also. On substituting

phase information in equation (2.115), we get

( )( )2

21

2 2

11 22 12 21

d

r

d d

m ek

k m k I k k

ω

ω ωΦ = −

− − −

(2.117)

which is the whirl amplitude of angular displacement and the condition of resonance can be obtained by

equating the denominator of equation (2.117) to zero, which is same as in equations (2.106) and (2.110)

for the linear displacement. For the disc at the center of the shaft span, we have 12 21 0k k= = , equation

(2.117) becomes

( )( )2 2

11 220

d dk m k Iω ω− − = (2.118)

which gives critical speeds as

1

11cr

d

k

mω = and

2

22cr

d

k

Iω = (2.119)

which is the case when the disc is at the center of the shaft span, and the linear and angular displacements

are uncoupled.

For the single plane motion from equation (2.91), we have

75

( )( ) ( )

( )( ) ( )

222 2

11 12 22 21 12 2212 12 11

2 221 22 21 11 21 22 21 12 22

0

A

B

db

b d

bk k k I k bk kF c c me cme me

c cF lme c ak k k I k ak k

ωωω ωω ω

+ − − + = = = ∆ − − − −

with ( )( )2 2

11 22 12 21d cr d crk m k I k kω ω∆ = − − − (2.120)

The bearing force amplitude and phase can be obtained from equation (2.120). Bearing reaction forces

will have similar trend in the variation with spin speed as that of the response, since it has the same

denominator, ∆ , as that of the response. It can be shown from equation (2.120) that forces transmitted

through bearings are also a maximum at system critical speeds. These forces are dynamic forces and are

superimposed on any steady loads, which may be present, for example due to gravity loading. In real

systems which are designed to operate above their critical speeds, the machine would normally be run

through the critical speed very quickly so that very large vibrations and forces associated with the

resonance do not have sufficient time to build up. Same is true during the run-down where some form of

braking may be employed. If the system is to run at the critical speed and vibrations are allowed to build

up then either the shaft will fracture and a catastrophic failure will result, or there may be sufficient

damping in the system to simply limit the vibration and force amplitudes to some very large (however,

tolerable) value.

Concluding Remarks: The present chapter explains various simple rotor models in use to describe some of

the important rotor behaviour, especially natural frequencies and critical speeds (i.e., the shaft spin at

which the amplitude of rotor is maximum). Basic terminologies generally used to describe the rotor

dynamic characteristics are introduced. For a single-DOF system the natural frequency and hence the

critical speed decrease by small amount due to damping. However, in the Jeffcott rotor model it is shown

that critical speed increases slightly due the increase in damping in the system. Apart from the amplitude

of the rotor vibrations, it is shown that the phase between the force and the response is also important

parameters to understand the rotor behaviour, especially at the critical speeds, where it changes of the

order of 1800. The damping is shown to be an important factor in suppressing the rotor vibrations at the

resonance. It is shown that the Jeffcott rotor is very a basic model to understand several important

phenomena of the rotor system. However, several other phenomena also emanate from supports, and for

this the basic understanding support dynamics is very important. The motivation of the next chapter

would be to find out dynamic parameters of the rolling element and hydrodynamic bearings, and seals in

isolation to the shaft. This will help in understanding some of the instabilities, which occurs due to

support dynamics.

76

Exercise Problems

Exercise 2.1: For a single degree of freedom damped rotor system, obtain an expression for the frequency

ratio ( / nfω ω ω= ) for which damped response amplitude becomes maximum (i.e. location of the critical

speed). Show that it is always more than the undamped natural frequency of the system. What is the

maximum feasible value of damping ratio for under-damped system is possible.

[Hint: Differential the denominator of the unbalance response (Y/e) expression with respect to the

frequency ratio ω and equate is to zero. Answer: 21/ 1 2ω ζ= − , 1/ 2ζ < ]

Exercise 2.2: Let us define a new frequency ratio in terms of the damped natural frequency, i.e.,

/dd nfω ω ω= with 21

dnf nfω ω ζ= − . Obtain an expression for the amplitude ratio (Y/e) and the phase,

φ, in terms of the new frequency ratio defined. Plot the amplitude ratio and the phase versus the new

frequency ratio and discuss the results. Obtain an expression for the frequency ratio ( /dd nfω ω ω= ) for

which damped response amplitude becomes maximum. What is the maximum feasible value of damping

ratio for under-damped system is possible.

[Answer: ( )

( ){ } ( )

2 2

22 2 2 2 2

1

1 1 4 1

d

d d

yy

e

ω ζ

ω ζ ζ ω ζ

−= =

− − + −

, ( )

( )

0.52

1

2 2

2 1tan

1 1

d

d

ζω ζφ

ω ζ− −

= − −

, 2

2

1 2

1d

ζω

ζ−

=−

; for

1/ 2 1ζ< < : dω is a complex quantity. The maximum feasible value of damping ratio for under-

damped system will remain the same 1/ 2ζ < ].

Exercise 2.3: Obtain transverse critical speeds of an overhung rotor system as shown in Figure E2.3.

Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, Id = 0.02 kg-m2. The shaft

diameter is 10 mm and total length of the span is 0.2 m. The shaft is assumed to be massless and its

Young’s modulus E = 2.1 × 1011

N/m2. Neglect the gyroscopic effect and take one plane motion only.

Influence coefficients are given as 3 2/ 3 ; / 2 ; /yf yM f Ml EI l EI l EIϕ ϕα α α α= = = =

77

Figure E2.3 A cantilever shaft with a disc at free end

[Answer: With the diametral mass moment of inertia effect two natural frequencies will exist: 1nfω = 5.55

rad/s and 2nfω = 144.12 rad/s. If the linear and angular motion is uncoupled then 0yM fϕα α= = :

1nfω =

5.56 rad/s and 2nfω = 71.80 rad/s. In case diametral mass moment of inertia is zero and no coupling

between the linear and angular motion 1nfω = 9.91 rad/s].

Exercise 2.4: Obtain the transverse critical speed of a rotor system as shown in Figure E2.4. Take the

mass of the disc, m = 5 kg and the diametral mass moment of inertia, Id = 0.02 kg-m2. Take shaft length a

= 0.3 m and b = 0.7 m. The diameter of the shaft is 10 mm. Neglect the gyroscopic effect.

Figure E2.4 An overhung rotor system

For the present case, influence coefficients are given as 2 ( )

3yf

a a b

EIα

+= ,

(3 )

3M

a b

EIϕα

+= , and

(3 2 )

6f yM

a a b

EIϕα α

+= = .

[Answer: 1nfω = 12.14 rad/s and

2nfω = 110.24 rad/s. With negligible diametral mass moment of inertia

1nfω = 5.98 rad/s]

78

Exercise 2.5: Obtain the bearing reaction forces and moments of an overhung rotor at rotor speeds of (i)

0.51nfω , (ii) 0.5(

2nfω +1nfω ) and (iii) 1.5

2nfω ; where 1nfω and

2nfω are the first and second bending natural

frequencies, respectively. Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, Id =

0.02 kg-m2. The disc has a residual unbalance of 25 g-cm. The shaft diameter is 10 mm and the total

length of the span is 0.5 m. The shaft is assumed to be massless and its Young’s modulus E = 2.1 × 1011

N/m2. Take one plane motion only.

Influence coefficients are given as 3 2/ 3 ; / 2 ; /yf yM f Ml EI l EI l EIϕ ϕα α α α= = = = .

[Answer: 1nfω = 15.60 rad/s and

2nfω = 203.76 rad/s.; (i) AR = 1.4568×10

9 N,

AM = -1.1008 Nm (ii) AR =

-3.2363×1012

N, AM = 2.4240×10

12 Nm (iii)

AR = -2.1125×104 N,

AM = 1.5831×104 Nm].

Exercise 2.6: Find transverse natural frequencies of an overhung rotor system as shown in Figure E2.6.

Consider the shaft as massless and is made of steel with the Young’s modulus of 2.1(10)11

N/m2

. A disc

is mounted at the free end of the shaft with the mass of 10 kg and the diametral mass moment of inertia of

0.04 kg-m2. In the diagram all dimensions are in cm.

Figure E2.6 A stepped shaft with cantilever end conditions

[Answer: For the pure translator motion: 1200.7 rad/s; For pure rotary motion: 6561.9 rad/s; For the

coupled translator and rotary motion: 1186.0 rad/s and 17278.3 rad/s]

Exercise 2.7: (a) While the Jeffcott rotor is whirling, with the help of the center of gravity, the center

spinning of the disc and the bearing axis, draw their relative positions in an axial plane when the rotor is

(i) below the critical speed (ii) at critical speed and (iii) above the critical speed. (b) Define following

terms: natural frequency and critical speed of a rotor; synchronous and asynchronous whirls.

79

Exercise 2.8: In a design stage of a rotor-bearing system it has been found that its one of the critical speed

is very close to the fixed operating speed of the rotor. List what are the design modifications a designer

can do to overcome this problem.

Exercise 2.9: A cantilever shaft of 1 m length (l) and 30 mm diameter (d) has a 5 kg mass (m) attached at

its free end, with negligibly small diametral mass moment of inertia. The shaft has a through hole parallel

to the shaft axis of diameter 3 mm (di) , which is vertically below the shaft center, with the distance

between the centers of the shaft and the hole as 6 mm (e). Consider no cross coupling in two orthogonal

directions as well as between the linear and angular displacements; and obtain the transverse natural

frequencies of the shaft system in two principal planes. Consider the shaft as massless and Young’s

modulus E = 2.1× 1011

N/m2.

[Hint: Find the equivalent stiffness of the shaft in two principal directions and then obtain natural

frequencies:1 1

/nf eq

k mω = and 2 2

/nf eq

k mω = . 1

1

3

3eq

EIk

l= and

2

2

3

3eq

EIk

l= ;

( )4 4 2 2 8

1 16 3.95 1064

i iI d d d e

π −= − − = × m4; ( )4 4 8

2 3.98 1064

iI d d

π −= − = × , 1

70.55nfω = rad/s,

270.78nfω = rad/s].

Exercise 2.10: For the Jeffcott rotor consider pure rotational displacement (tilting) of the disc (without

linear displacement) and obtain transverse natural frequency for the tilting motion.

[Hint: The stiffness due to titling motion would be ( )2 2

6

6 6 2t

EIlk

al a l=

− − and the diametral mass moment

of inertia would be 1 2

4dI mr= . Hence, the natural frequency would be /nf t d

k Iω = .]

Exercise 2.11 What should be the length and the diameter of a cantilever shaft if the bending critical

speed has to fixed at 100 Hz and has 2 kg of mass at its free end? Because of the space limitation the

length of the shaft must be less than 30 cm. [Answer: l = 0.3m and d = 0.0288 m].

Exercise 2.12 The transverse critical speed of a rotor system as shown in Figure 1 is to be fixed at 5.98

rad/s. Take disc as a point mass with m = 5 kg. What should be the overhung shaft length, a? Take shaft

length b = 0.7 m. The diameter of the shaft is 10 mm. Neglect the gyroscopic effect. [Answer: 0.291 m]

80

Figure E2.12 An overhung rotor system

[Solution:

Exercise 2.13 The transverse critical speed of a rotor system as shown in Figure 1 is to be fixed at 5.98

rad/s. Take disc as a point mass with m = 5 kg. What should be the diameter of the uniform shaft, d? Take

shaft length a = 0.3, b = 0.7 m. Neglect the gyroscopic effect. [Answer: 0.01m]

Figure E2.12 An overhung rotor system

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Overhang shaft length (m)

Fle

xib

ility

coeff

icie

nt

(m/N

)

(0.291, 0.0056)

81

Exercise 2.14 The transverse critical speed of a rotor system as shown in Figure 1 is to be fixed at 5.98

rad/s. Take disc as a point mass with m = 5 kg. What should be the diameter of the uniform shaft, d? Take

shaft length 2a = b = 0.7 m. Neglect the gyroscopic effect. [Hint: 3 /(8 )b EIα = ]

Figure E2.12 An overhung rotor system

Exercise 2.15 For a Jeffcott rotor with a disc at the mid-span, influence coefficients are given as:

3 / (48 ); 0; / (12 )yf yM f Ml EI l EIϕ ϕα α α α= = = = , where l is the span length and EI is the modulus

of rigidity of the shaft. Let m and Id be the mass and the diametral mass moment of inertia, respectively,

of the disc. Obtain the natural frequencies of the rotor system.

[Answer: 1

31/ ( ) 48 / ( )nf yf m EI mlω α= = and

21 / ( ) 12 / ( )

nf M dm EI I lϕω α= = ] .

Exercise 2.16 Choose a single correct answer from the multiple choice questions:

(i) The critical speed phenomenon of a rotor is a

(A) free vibration (B) forced vibration (C) transient vibration (D) unstable vibration

(ii) A rigid body is defined as

(A) a body with no deformation (B) a body with their particles have fixed distances

(C) both (A) and (B) (D) either (A) or (B)

(iii) A particle has how many degrees of freedom

(A) 1 (B) 2 (C) 3 (D) more than 3

(iv) A rigid body has how many degrees of freedom

(A) 1 (B) 3 (C) 6 (D) more than 6

82

(v) A flexible body has how many degrees of freedom

(A) 1 (B) 3 (C) 6 (D) infinite

(vi) If three points have fixed relative distances between them, then it represents a system of

(A) single particle (B) a rigid body (C) a flexible body (D) none of the above

(vii) A system consists of three particles with their relative distances as constant, then it has how many

degrees of freedom

(A) 1 (B) 3 (C) 6 (D) infinite

[Hint: A single particle has three DOF, one additional particle in the system will add only two more DOF,

since it has one constraint to maintain the fixed distance. The third particle will add only one more DOF,

since it has two constraints to maintain fixed distances from the first as well as the second particle. Hence,

a system consists of three particles with their relative distances as constant have total six DOF, i.e., same

as a rigid body.]

(viii) A perfectly balanced Jeffcott rotor (i.e., a flexible shaft with a disc at mid span) when it is rotating at

a particular speed, if it is perturbed in transverse plane from its equilibrium then the frequency of whirl

would be equal to

(A) the shaft spin speed (B) the transverse natural frequency

(C) the more transverse natural frequency (D) the less transverse natural frequency

(ix) In a Jeffcott rotor with an off-set disc (i.e., not at the mid-span) and if the disc has a tilt in the

transverse plane. The shaft would experience

(A) A gyroscopic couple (B) an external moment (C) either (A) or (B)

(D) both (A) and (B)

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