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Dr. R. Tiwari ([email protected])

ByDr. Rajiv Tiwari

Department of Mechanical Engineering

Indian Institute of Technology Guwahati 781039

Under AICTE Sponsored QIP Short Term Course on

Theory & Practice of Rotor Dynamics

(15-19 Dec 2008)

IIT Guwahati

ANALYSIS OF

SIMPLE ROTOR SYSTEMS

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INTRODUCTION

Rotating machines are extensively used in diverseengineering applications, such as

power stations

marine propulsion systems

aircraft engines

machine tools

automobiles and

household accessories

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Electrical motorRotor of an electrical motor

Rolling bearing

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CNC machine equipped with 12 rotating tools and two spindles.

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Different stages of a turbomachinery

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Different stages of a turbomachinery

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Balancing of a big rotating machinery

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Dynamic balancing center

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A turbo-machinery

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Pumps, motors and rotating machines can be monitored for signs of

poor lubrication, shaft misalignment or bearing failure.

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The design trend of such systems in modernengineering is towards

lower weight

operating at super critical speeds

Of the many published works, the most extensive

portion of the literature on rotor dynamics is concernedwith determining

critical speeds

natural whirl frequenciesinstability thresholds and

imbalance response

INTRODUCTION

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SINGLE MASS ROTOR MODELS

For understanding basic phenomena of any dynamicsystem requires adequate modeling of the system.

The rotor is considered as single mass in the form of apoint mass, a rigid disc or a long rigid shaft.

In this section we present simple rotor models andanalyze them to illustrate their behavior.

Single DOF Rotor Model

Rankine Rotor Model

Jeffcott Rotor Model

Rigid Rotor Supported on Flexible Bearings

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The simplest model of the rotor systemcan be single DOF.

Two types of rotor model are shown here

In Figure 1.1(a) the bearing (support) isassumed to be rigid (simply supported)and the shaft as flexible.

The mass of the rotor is considered asthat of rigid disc that is mounted on themassless flexible shaft.

In Figure 1.1(b) the bearing is

assumed to be flexible and the rotor asrigid.

Both the cases can be idealized as asingle DOF as shown in Figure 1.1(c).

1.1 Single DOF Rotor Model

Fig 1.1(b)

A rigid rotor

mounted on flexible

bearings

Fig 1.1(a)

A flexible rotor

mounted on rigidbearings

Fig 1.1(c)

An equivalent

single degree of

freedom spring-

mass system

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If the rotor is perfectly balanced then theoreticallyspeaking there will not be any imbalance force asshown in Figure 1.2(a). In actual practice it is

impossible to have a perfectly balanced rotor.

The rotor imbalance gives a sinusoidal force at therotor rotational frequency. Thus, the imbalanceforce is modeled as sinusoidal force

where m is the mass of the rotor, is the spinspeed of the rotor and e is the eccentricity of therotor

When the rotor is not eccentric, however, a smallimbalance mass, is attached at a relativelylarger radius of (see Figure 1.2(c)), theimbalance force can be written as

For the case when the rotor is eccentric and asmall imbalance mass is attached as shown inFigure 1.2(d), the imbalance force will be

where is the phase difference between thevectors of imbalance forces due to the rotoreccentricity and the imbalance mass

ω

ir im

t emt F ω ω sin)(2

= (1)

t r mt F ii ω ω sin)(2

=(2)

)sin(sin)( 22 α ω ω ω ω ++= t r mt emt F ii (3)

No imbalanceFig 1.2(a)

Rotor geometrical centre andcentre of gravity coincident

Fig 1.2(b)Rotor geometrical centre

and centre of gravity not

coincident

Imbalance force =

mass of rotor × eccentricity

× square of spin speed

Imbalance force =

mass of rotor × radius ×square of spin speed

Fig 1.2(c)

Rotor geometrical centre,

centre of gravity and an

additional imbalance mass

Imbalance force is the

vector addition of forces

due to the rotor and

imbalance forces

Fig 1.2(d)

Rotor geometrical center, centre ofgravity and imbalance mass

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On application of the Newton’s law on the freebody of the rotor mass as shown in Figure 1.1(d),

i.e. equating sum of external forces to the mass ofthe rotor multiplied by the acceleration of thecenter of gravity of the rotor mass, we have

where is the effective stiffness of the rotor

systemEquation (4) is a standard equation of motion of asingle DOF spring-mass system and can bewritten as

For the free vibration, when the externalimbalance force is absent, the rotor mass will behaving oscillation and that will be given by

where is the frequency of oscillation duringthe free vibration and that is called the naturalfrequency of the system. On substituting equation(6) into the homogeneous part of equation ofmotion (5), it gives

Fig 1.1(d) Free body diagram

of the disc mass

ymt em yk eff =+− ω ω sin2

(4)

eff k

t em yk ym eff ω ω sin2

=+ (5)

)sin()( t Y t y n= (6)

nω

0)sin()(2

=+− t Y k m neff n ω ω (7)

y

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For the non-trivial solution of equation (7), the natural frequency of the system canbe written as

The steady state forced response can be modeled as

where Y is the amplitude of displacement and is the phase lag of the

displacement with respect to the imbalance force.

On substituting equation (9) into equation (5), the steady state forced responseamplitude can be written as

with

From equation (10) it should be noted that when the spin speed is equal to thenatural frequency of the system as given in equation (8), the undamped steadystate forced response amplitude tends to infinity. This is a resonance condition andthe spin speed corresponding to the resonance is defined as critical speed . Sincedamping is not considered in the analysis phase angle, , becomes zero.

/ n eff k mω = (8)

)sin()( ϕ −= t Y t y (9)

ϕ

2

2

ω

ω

mk

meY

eff −= 0ϕ = (10)

ϕ

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The analysis presented in this section can be applied to thetransverse, torsional and axial vibrations of rotors and accordinglycritical speed can be termed by prefixing respective names of

vibrations.

For torsional vibrations care should be taken that mass will bereplaced by the polar mass moment of inertia of rotor and stiffness willbe the torsional stiffness.

Similarly, for axial vibrations mass will remain same as transverse

vibration, however, the stiffness will be the axial stiffness.

The critical speed is given by

1cr ω = ± / cr eff nf k mω ω = ± = ±or (11)

The ± sign represent that the rotor will have critical speed while rotating in

either clockwise or counter clockwise.

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Dr. R. Tiwari ([email protected])Fig 1.4 Non-dimensional unbalanceresponse versus frequency ratio

(a) Linear plot

(b) Semi-log plot

% Non-dimensional unbalance response wrt to

freq ratio "Figure_1_4.m"% Copywriters: Dr R Tiwari,

Dept of Mechanical Engg., IIT Guwahati.% 13-01-2005

clear all;deta_freq=0.005;

freq_ratio(1)=deta_freq;N_pt=1000;for ii = 1:1:N_pt

y_resp(ii)=freq_ratio(ii)^2/(1-freq_ratio(ii)^2);if(ii

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The unbalance response can be reduced by the following methods.

Correction at source i.e. balancing the rotor :Balancing the rotor is the most direct approach, since it attacks theproblem at source.

However, in practice a rotor cannot be balanced perfectly and that

the best achievable state of balance tends to degrade duringoperation of a rotor (e.g. turbomachinery).

There are two type of unbalances

Static unbalance : : The principal axis of the polar mass moment ofinertia of the rotor is parallel to the centerline of the shaft as shown in

Figure 1.5b. The rotor can be balanced by a single plane balancing

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 inFigure 1.5c & d. For balancing such rotors minimum of two planes arerequired.

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Dr. R. Tiwari ([email protected])Fig 1.5 Classification of unbalances for a short rigid rotor

F

• G

• GM

• GM

• G

(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)

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Operate rotor away from the critical speed :

(i.e. during design itself or during operation by providingtemporary auxiliary support)

Moving the machine operating speed farther away from the criticalspeed can be achieved by changing the rotor operational speed orby changing the critical speed itself.

The critical speed can be changed either at the design stage orduring operation.

At design stage changing rotor mass or its distributions anddimensions of the rotor and its support lengths can alter the criticalspeed.

During operation auxiliary support can be provided to increase theeffective stiffness of the rotor, which in turn increases the criticalspeed

By this arrangement the actual rotor critical speed can be safelytraversed and then the auxiliary support can be withdrawn whichbrings the critical speed of the rotor below the operation speed.

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Add damping to the system or active control of the rotor :

If a critical speed must be traversed slowly or repeatedly, orif machine operation near a critical speed can not beavoided, then the most effective way to reduce theamplitude of the synchronous whirl is to add damping.

The squeeze film and magnetic bearings are often used to

control the dynamics of such systems.

o Squeeze-film bearings (SFB) are, in effect, fluid-film bearings in

which both the journal and bearing are non-rotating.

o In recent years, advanced development of electromagnetic bearingtechnology has enabled the active control of rotor bearing systemsthrough active magnetic bearings (AMB).

With the development of smart fluids (for example electroand magneto-rheological fluids) now new controllablebearings are in the primitive development stage.

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Schematics of typical passive and active (i.e. smart or controllable) squeeze

film dampers and active magnetic bearings are shown in Figure 1.6.

Bearing bush

Outer raceway ofrolling bearing (can

displace radially andconstraint not torotate.

Squeeze film

Rotor

Oil feed groove

Rotor

Rolling bearing

Electrodes

Teflon

Sensor

Rotor

Power Amplifier Electromagnet

Controller

Fig 1.6 (a) Schematic diagram of squeeze film dampers

Fig 1.6 (b) Smart (active) fluid-film dampers

Fig 1.6 (c) Basic principle of active magnetic bearings

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The single DOF rotor model haslimitations that it cannot represent theorbital motion of the rotor in twotransverse directions.

Rankine (1869) used a two DOF modelto describe the motion of the rotor intwo transverse directions as shown in

Figure 1.7(a).

The shape of orbit produced dependsupon the relative amplitude and phaseof the motions in two transverse

directions and the orbit could be ofcircular, elliptical or straight line,inclined to x and y axis, as shown inFigure 1.8.

Fig 1.7(a)

Two degree of freedom springmass rotor model

Fig 1.8 (a)Circular motion

Fig 1.8 (b)Elliptical motion

Fig 1.8 (c)

Straight line motion

1.2 RANKINE ROTOR MODEL

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However, as shown in Figure 1.7(b) the freebody diagram the radius of whirling of the

rotor center will increase parabolically withspin speeds and will be given as

where is the centrifugal force

It can be physically also visualized as therewill not be any resonance condition, as found

in the single DOF model, when the spinspeed is increased gradually. This is aserious limitation of the Rankine model.

Moreover, this model does not represent therealistic rotating imbalance force.

Fig 1.7(b)Free body diagram of the model

k F r c / =(11)

cF

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1.3 JEFFCOTT ROTOR MODEL

Figure 1.9 shows a typical Jeffcott rotor.It consists of a simply supported flexiblemassless shaft with a rigid disc mountedat the mid-span.

The disc center of rotation, C, and itscenter of gravity, G, is offset by adistance, e.

The shaft spin speed is and the shaftwhirls about the bearing axis with whirlfrequency is, . For present case

synchronous condition has beenassumed I.e. ν = ω (see Figure 1.10a).

The stiffness of the shaft is expressed as

Fig 1.9(a)A Jeffcott rotor model

Fig 1.9(b)A Jeffcott rotor model in y-z

plane

Fig 1.9(c)

Free body diagram of thedisc in x-y plane

ω

υ

3load/deflection 48 / k EI L= = (12)

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Coordinates to define the position of thecenter of rotation of the rotor are and .

The location of the imbalance is given by .

Thus, three dofs are needed to define theposition of the Jeffcott rotor.

From Figure 1.9(c) the force balance in ,and directions can be written as

and

xu

θ

yu

θ

xu

( )2

2cos

x x x

d ku cu m u e

dt θ − − = + (13)

( )2

2sin

y y y

d ku cu mg m u e

dt θ − − − = + (14)

cosd mge I θ θ − =

(15)

yu

Shaft spindirection

Shaft whirlingdirection

Shaft

Shaft spindirection

Shaft whirlingdirection

Shaft

Fig 1.10(a) Synchronous whirl

Fig 1.10(b) Anti-synchronous whirl

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Apart from restoring force contribution from the shaft, the dampingforce is also considered. The damping force is idealized as viscousdamper and it is mainly coming from the support and aerodynamicforces at disc.

The material damping of the shaft will not contribute viscous dampingand it may leads to instability in the rotor and it is not considered here.

For the case i.e. when the disc is rotating at constant spinspeed, the Jeffcott rotor model is reduces to two DOF rotor model.Neglecting the effect of gravity force, equations of motion in the x andy can be written as

and

( )2

2cos x x x

d ku cu m u e t

dt

ω − − = + (16)

t θ ω =

( )2

2sin y y y

d ku cu m u e t

dt ω − − = + (17)

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• Equations of motion can be written in the standard from as

• It should be noted that equations of motion are uncoupled and motioncan be analyzed independently in two transverse planes.

• Noting equation (8), from the undamped free vibration analyses it canbe seen that since the rotor is symmetric rotor hence it will be havingtwo natural frequencies that are equal and given as

• The damping does not affect the natural frequency of the systemappreciably. However, their effect is more predominate forsuppressing the resonance amplitude.

2 cos x x x

mu cu ku m e t ω ω + + = (18)

2 sin y y ymu cu ku m e t ω ω + + = (19)

1,2 / nf k mω =

(20)

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• The steady state forced response can be written as

where and are the steady state forced response amplitudes in thex and y directions, respectively. is the phase lag of the x -direction

displacement with respect to the imbalance force.

• The phase difference between the two direction responses will be of900 as two directions are orthogonal to each other. For the direction ofwhirling shown in Figure 1.5 i.e. counter clockwise (ccw ) for the

present axis system the response in the y direction will lead the xdirection response by radians. Hence the lead of the y directionresponse with respect to the force will be .

• On taking the first and second derivatives of the response with

respect to time, t , we get

and

xu

sin( )

cos( )

x x

y y

u U t

u U t

ω ω ϕ

ω ω ϕ

= − −

= −

(22,23)

yu

ϕ

/ 2π / 2π ϕ −

2

2

cos( )

sin( )

x x

y y

u U t

u U t

ω ω ϕ

ω ω ϕ

= − −

= − −

[ ]cos( )cos ( ( / 2 ) sin( )

x x

y y y

u U t u U t U t

ω ϕ ω π ϕ ω ϕ

= −= + − = −

(21)

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• On substituting equations (21) to (23) into equation (18) and

separating the in-phase (i.e. ) and quadrature (i.e. ) terms,we get

• Equation (25) gives

which gives

and

cos t ω

2 2cos sin cos x x xm U cU kU m eω ϕ ω ϕ ϕ ω − + + = (24)

( ) ( )

2

2 22

cos k m

k m c

ω ϕ

ω ω

−=

− +

( ) ( )2 22

sin c

k m c

ω ϕ

ω ω

=

− +

2

sin cos sin 0 x x xm U cU kU ω ϕ ω ϕ ϕ − − + =

2tan

c

k m

ω ϕ

ω =

−

sin t ω

(25)

(26)

(27)

(28)

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• Substituting equations (27) and (28) into equation (24), we get

• Similarly, we can obtain response amplitude in the y -direction fromequation (19) as

• From equations (29) and (30) it can be seen that because of thesymmetry of the rotor the orbit is circular in nature. An alternativeapproach that is very popular in rotor dynamics analyses is to use thecomplex algebra to define the whirl radius as

where

( ) ( )

2

2 22 x

m eU

k m c

ω

ω ω

=

− +

(29)

( ) ( )

2

2 22 y

m eU

k m c

ω

ω ω

=

− +(30)

r x yu u ju= + (31)

1 j = −

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• On multiplying equation (19) by j and adding to equation (18), we get

• The steady state response can be assumed as

where is the whirl amplitude (it is a real quantity), is the phase lagof response with respect to the imbalance force.

• On differentiating equation (33) with respect to time, t , we get

• On substituting equations (33) and (34) into equation (32), we get

2e j t r r r mu cu ku me ω ω + + = (32)

( )e j t r r u U ω ϕ −= (33)

ϕ r U

( ) 2 ( )e ; e j t j t r r r r u j U u U ω ϕ ω ϕ ω ω − −= = − (34)

( )2 2e jr k m j c U meϕ ω ω ω − − + = (35)

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Equation (35) can be written as

• On separating the real and imaginary parts of equation (36), we get

• From equation (38), we get the phase

• On substitution of phase from equations (39) to (37) the whirlamplitude can be written as

( ) ( )2 2cos sinr r k m j c U jU meω ω ϕ ϕ ω − + − = (36)

2 2( ) cos sinr r k m U cU meω ϕ ω ϕ ω − + = (37)

2tan

c

k m

ω ϕ

ω =

−

(39)

( ) ( )

2

2 22r

m eU

k m c

ω

ω ω

=

− +

(40)

2( ) sin cos 0r r k m U cU ω ϕ ω ϕ − − + = (38)

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;

with

whereis the frequency ratio,

is the natural frequency of non-rotating rotor,

is the damping ratio and

is the critical damping of the system for which the damping ratio isequal to unity.

( ) ( )

2

2 22

/

1 2

r r U U e ω

ω ζω

= =

− +

(41,42)

/ ; / ; / ; 2n n c ck m c c c kmϖ ω ω ω ζ = = = = (43)

nω

cc

ϖ

ζ

• Equations (39) and (40) are similar to previous results i.e. equations(26) to (30). The non-dimensional form of equations (39) and (40) can bewritten as

2

2tan

1

ζϖ ϕ

ϖ

=

−

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Fig 1.11(a)

Variation of the non-

dimensional response versus

frequency ratio for different

damping ratios

Figure 1.11(a) shows that the maximumamplitude occurs at slightly higher frequencythan when the damping is present in thesystem, however maximum amplitude occursat ω n for the undamped case.

The increase in the damping results inincrease in the critical speed, howeverdamping is the most important parameter forreducing the whirl amplitude at critical speed.

Since the measurement of the amplitude ofvibration at critical speed is difficult, hencedetermination of the precise critical speed isdifficult.

To overcome this problem the measurementof the phase is advantageous at least todetermine the undamped natural frequencyof the system.

A it b f Fi 1 11(b) th

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Fig 1.11(b)Variation of the phase versus

frequency ratio for different

damping ratios

As it can be seen from Figure 1.11(b) thephase angle is 900 at frequency even for thecase of damped system.

For lightly underdamped system the phaseangle changes from 00 to 900 as the spinspeed is increased and becomes 1800 as thespin speed is increased to higher frequencyratio.

For very high-overdamped system the phaseangle always remain at 900 before and afterthe resonance, which may be a physically

unrealistic case.

As the spin speed crosses the critical speedthe center of the mass of the disc of Jeffcottrotor comes inside of the whirl orbit and rotor

tries to rotate about the center of gravity.

As can be seen from the graph at the spinspeed approaches infinity the displacement ofthe shaft tends to the equal to the disc

eccentricity.

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The change in phase between the force and the response is also shown

in Figure 1.7 for three difference spin speeds i.e. below the criticalspeed, at the critical speed and above the critical speed.

Fig 1.12(a)

Phase angles between the

force and response vectorsbelow critical speed

Fig 1.12(b)


force and response vectors

at critical speed

Fig 1.12(c)


force and response vectorsabove critical speed

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With the development in the software, which can handle complex matrices, thefollowing procedure may be very helpful for numerical simulation of even verycomplicated rotor systems also.

Equations (18) and (19) can be combined in the matrix form as

The force vector in equation (44) is expressed as

where the represents the real part of the quantity inside the parenthesisand are the imbalance force components in x and y directions,

respectively.

2

2

0 0 0 cos

0 0 0 sin

x x x

y y y

u u um c k m e t

u u um c k m e t

ω ω

ω ω

+ + =

(44)

( )

( )

( )

( )

222

22 2

cos sincosRe Re Re

sin cossin

j t

x j t

j t y

me e F m e t j t m e t e

F m e t j t m e t me je

ω

ω

ω

ω ω ω ω ω ω

ω ω ω ω ω ω

+ = = =

− −

(45)

y ximb imbF jF = − (46)

ximbF

yimbF

Re(.)

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On substituting equation (45) into equation (44) and henceforth for

brevity the symbol Re(.) will be removed and it can be written as

The relationship (46) is true for thepresent axis system and the directionof whirling of the imbalance force vectorchosen (see Figure 1.8(a)). For this case

leads by 900

.

For the direction of whirl opposite tothe axis system as shown in Figure (8(b))

the following relationship will hold

in which case the lags by 900.

0 0 0

0 0 0

x

y

imb x x x j t

y y y imb

F u u um c k e

u u u F m c k

ω

+ + =

(47)

y ximb imbF jF = (48)

Fig 1.8(a)

The direction of whirl same as the

positive axis direction

yimbF

ximbF

yimbF

ximbF Fig 1.8(b)

The direction of whirl opposite

to the positive axis direction

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Equation (47) can be written in more compact form as

The solution can be chosen as

where the vector elements are, in general, complex quantity.

The above equation gives

On substituting equations (50) and (51) into equation (49), we get

The above equation can be written as

[ ]{ } [ ]{ } [ ]{ } { j t imb M u C u K u F e ω

+ + = (49)

{ } { { {2 and j t j t u j U e u U eω ω ω ω = = − (51)

{ } { }

j t

u U e

ω =

{U

[ ] [ ] [ ]( ){ } {2 imb M K j C U F ω ω − + + =

(50)

(52)

[ ]{ } { }imb Z U F = (53)

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with

where is the dynamic stiffness matrix.

The response can be obtained as

where the vector elements are, in general, complex.

The above method is quite general in nature and it can be applied tomulti-dof systems once equations of motion in the standard form areavailable.

[ ] [ ] [ ] [ ]( )2 Z M K j C ω ω = − + + (54)

{ [ ] { }1

imbU Z F −

=

}{ xU

[ ] Z

(55)

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Obtain the response for the following form of equations of motion

and

The first equation can be written as

with

in which the real part of the right hand side term has meaning. The

solution can be assumed as

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

On substituting in equation of motion, we get

j t

x xu U e ω

=

2

xF meω = j t

x x xmu ku F e ω + =

2 cos x xmu ku m e t ω ω + =2 sin y ymu ku m e t ω ω + =

xU

2 j t

x xu U e ω

ω = −

( )2 2 x xm U kU meω ω − + =

Example

which gives

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which gives

Hence the solution becomes

Similarly for the second equation of motion can be written as

with

in which the real part of the right hand side term only has meaning.

The solution can be assumed as

where in general is a complex quantity. The above equationgives

j t

y yu U e

ω =

2

yF jmeω = − j t

y y ymu ku F e ω

+ =

2

2 x

meU

k m

ω

ω

=

−

2 2 2

2 2 2Re (cos sin ) cos j t x

me me meu e t j t t

k m k m k m

ω ω ω ω ω ω ω

ω ω ω

= = + =

− − −

yU

2 j t

y yu U e ω ω = −

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On substituting in equation of motion, we get

which gives

Hence the solution becomes

(Answer)

( )2 y y ym U kU F ω − + = 2 y

y

F U

k mω =

−

2 2

2 2 2

2 2

2 2

(cos sin )

Re ( cos sin ) sin

y j t j t

y

F jme jmeu e e t j t

k m k m k m

me me j t t t

k m k m

ω ω ω ω ω ω ω ω ω

ω ω ω ω ω

ω ω

− −= = = +

− − −

= − + =

− −

More generalized model of Jeffcott rotor

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Fig 1.14(b)

Free body diagram of the disc

in the x-y plane

Disc offset from the midspan in the y -z plane

Fig 1.14(a)

A Jeffcott rotor with a disc

offset from the midspan in

the y-z plane

Fig 1.14(c)

Free body diagram of the disc

in the y-z plane

Fig 1.14(d)

Free body diagram of the shaft

in the y-z plane

More generalized model of Jeffcott rotor

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Disc offset from the midspan in the z -x plane

Fig 1.15(a)

A Jeffcott rotor with a disc offset

from the midspan in the z-x plane

g

Fig 1.15(b)

Free body diagram of the

shaft in the z-x plane

Fig 1.15(c)

Free body diagram of the

disc in the z-x plane

M li d J ff tt R t M d l

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More generalized Jeffcott Rotor Model

Figures 14 and 15 show a more general case of Jeffcott rotor whenthe rigid disc is placed with some offset from the mid-span,respectively in y -z and x -z planes.

For such rotors apart from two transverse displacements of center ofthe disc i.e. and , the tilting of the disc about the x and y axis i.e.

and , also occurs and makes the rotor system as four dofs.

In figure the point C is the geometrical center and G is the center ofgravity of the disc.

From geometry the component, we can have the following relations

From Figure 9(b) equations of motion of the disc can be written as

and

xu

(56)cos ; sin x y

e e t e e t ω ω = =

yu

yϕ xϕ

(57,58)( )

2

2cos

y d y y x

d F m u e

dt φ − = + yz x x I φ − =

From above equations it can be observed that equations are coupled

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q q pwith titling component of the displacement,

Similarly, from Figure 10(b) we can write equations of motion as

and

Equations (59) and (60) are coupled with titling component of thedisplacement,

However, two transverse plane motions are not coupled and that willallow two-plane motion to analyze independent of each other i.e. set ofequations (57 and 58) and equations (59 and 60) can be solvedindependent of each other.

The analyses can be further simplified with the assumption of smalltilting angle i.e. and equations (57 and 59) can be

simplified as

xϕ

(59,60) zx y y I φ − =

( )2

2cos

x d x x y

d F m u e

dt φ − = +

yϕ

cos cos 1 x y

φ φ = ≈

2 sind y y d m u F m e t ω ω + = (61)

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and

Equations (61, 58, 62, 60) can be assembled as

which can be written in matrix notation as

with

where is the reaction force/moment vector.

2

cosd x x d m u F m e t ω ω + = (62)

2

2

0 0 0 sin

0 0 0 0

0 0 0 cos

0 0 0 0

d y y d

x x yz

d x x d

y y zx

m u F m e t

I M

m u F m e t

I M

ω ω

φ

ω ω

φ

+ =

(63)

[ ]{ { { imb M u R f + = (64,65)[ ]

0 0 0

0 0 0

0 0 0

0 0 0

d

x

d

y

m

I M

m

I

=

{ R

The reaction forces and moments onto the shaft can be expressed in

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terms of the shaft displacements at disc location with the help ofinfluence coefficients as

where represent the displacement at station due to a unit force

at station.

Equation (66) can be written in the matrix form as

which gives

where is the stiffness coefficients and defined as force at stationdue to a unit displacement at station.

11 12

21 22

x x zx

y x zx

u F M

F M

α α

φ α α

= +

= +

(66)

ijα th

ith

j

11 12

21 22

x x

y zx

u F

M

α α

φ α α

=

(67)

1

11 12 11 12

21 22 21 22

x x x

y y zx

u uF k k

M k k

α α

φ φ α α

−

= =

ijα thith

j

Similarly, since the shaft is symmetric about its rotational axis, we canbt i

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obtain

Equations (67) and (68) can be combined as

which can be written in matrix notation as

with

11 12

21 22

y y

yz x

F uk k M k k φ

=

(68)

11 12

21 22

11 12

21 22

0 00 0

0 0

0 0

y y

x yz

x x

y zx

uF k k M k k

uF k k

M k k

φ

φ

=

(69)

{ } [ ]{ R K u= (70,71)[ ]

11 12

21 22

11 12

21 22

0 0

0 0

0 00 0

k k

k k K

k k k k

=

On substituting reaction forces and moments from equations (70) intoti f ti i ti (64) t

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equations of motion i.e. equation (64), we get

In general, for the simple harmonic vibration, we can write

On substituting in equation (73), we get the response as

with

where is the dynamic stiffness matrix, in general, elements of thismatrix are complex quantity, however, since the damping is notconsidered here they are real quantities.

[ ]{ } [ ]{ } { }imb M u K u f + = (72)

{ { }2u uω = − (73)

{ [ ] { }1

imb

u Z f −

=

(74,75)[ ] [ ] [ ]( )2

Z K M ω = −

[ ] Z

Example 1:

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Finding the bearing critical speed of a rotor system shown in Figure 1.16Take E = 2.1×1011 N/m2.

Fig 1.16Solution :

Fig 1.17

The influence coefficient is given as

( )2 2 2( ), ( )

6

bx l x b y x x a

F EIL

α − −

= = ≤

8

1 11

1 12316.83 rad/s

10 1.863 10n

mω

α −= = =

× ×

2 2 2

80.611 11

0.4 11 4

0.4 0.6 1 0.6 0.4 1.863 10 m/N

6 2.1 10 (0.1) 164

x ab

α α π

−= =

=

× × − − = = = ×× × × × ×

Then calculate the natural frequency as

For obtaining 11α

we have x =0.6 m, l = 1.0 m and b = 0.4 m.

Example 1:

Example 2:Obtain the transverse critical speeds of a Jeffcot rotor system as shown in Figure 1

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Obtain the transverse critical speeds of a Jeffcot rotor system as shown in Figure 1.

Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, I d = 0.02

kg-m2 and the disc is placed at 0.25 m from the right support. The shaft is havingdiameter of 10 mm and total length of the span of 1 m. The shaft is assumed to bemassless. Use one of these methods (i) mechanical Impedance or (ii) dynamic

stiffness. Take shaft Young’s modulus E = 2.1 × 1011 N/m2. Neglect the gyroscopiceffects. Take one plane motion only.

Figure 1.18 A Jeffcott rotor system

2 24

11 1.137 103

a b

EIlα −= = ×

( )2 3 2 412 3 2 3 3.03 10a l a al EIlα −

= − − − = − ×

4

21

( ) 3 3.03 10ab b a EIlα −= − = − ×

( )2 2 322 3 3 3 1.41 10al a l EIlα −

= − − − = ×

For the present problem only single planemotion is considered. For free vibration, from

equation (70), we get

1

11 12

21 22

0 0

0 0d

m x x

I

α α

α α φ φ

− +

Since it will execute the SHM, we have1

11 122

21 22

0 0

0 0

nf

d

m x

I

α α ω

α α φ

− − + =

Solution:

Example 2 contd…

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Previous equation is an eigen value problem. For non-trial solution, we have

1

11 122

21 22

00

0nf

d

m

I

α α ω

α α

− − + =

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

( ) ( )4 2 211 22 12 11 22 1 0d nf nf d mI m I ω α α α ω α α − − + + =

On substituting the present problem parameters values, it gives

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

4.291

=n

2902 =nω

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

rad/sec and rad/sec

Example 2 contd…

Example 3:The rotor of a turbine 13 6 kg in mass is supported at the mid span of a shaft

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The rotor of a turbine 13.6 kg in mass is supported at the mid span of a shaft

with bearings 0.4064 m. apart. The rotor is known to have an unbalance of0.2879 kg-cm. Determine the forces exerted on the bearings at speed of 6000rpm if the diameter of the steel shaft is 2.54 cm. Assume the shaft to besimply supported at the bearings. Take E = 200 GNm-2.

Solution:

The following data are available

U = me = 0.2879 kg-cm,

M = 13.6 kg;

e = 0.0211 cm,ω= 6000 rpm,

D = 2.54 cm,

E = 200×109 N/m2

( )2

2 3 2 60000.2879 10 113.66N60

me− ×= = × × =

56.83cos 200 = 56.83cos200 133.4 N

56.83sin 200 N

F t mg t x

F t y

= + +

=

(i). For rigid rotor & rigid bearings .Unbalance force

Force at each bearing (amplitude)

= 113.66/2 = 56.83 NThe component of forces in vertical &

horizontal directions are given

(ii) For flexible rotor and rigid bearings.

Example 3 contd..

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/ 2 / 2 A B y R R F ky= = =

( ) ( ) ( )4 33 11 6

48 / 48 2.0 10 0.0254 / 0.4064 2.92 10 N/m64

k EI l π

= = × × × = ×

2me ky myω − = 2

y yω = −

( ) ( )

( )

2224

22 6

0.2879 10 2004.64 10 m

2.92 10 13.6 200

me y

k m

ω

ω

−

−× ×

= = = − ×− × − ×

( ) g gThe bearing reaction forces can be written as

since F y = ky (A)

EOM of the disc,

from the free body diagram of the disc is given as

For simple harmonic motion

The above equation can be written as

The stiffness is given as

Figure 1.20

( )6 4 / 2 2.92 10 4.64 10 / 2 677.6 N A R ky −

= = × × − × = −

677.6 cos200 133.4 N and 677.7 sin200 N A A x y

R t R t = × + = ×

From the equation (A), we have

The component of the forces in the vertical

& horizontal direction can be obtained as

Example 3 contd..

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(iii) From EOM of the disc,

we have

( )2 yF m y e= +

2 2 413.6 (200) [0.0211 10 ( 4.64 10 ] 1358.4 N

− −×= × × + − × = −

/ 2 679.2 N A B y R R F = = = −

2211 12 11

221 22 210

A

B

R C C C meme

R C C C me

ω ω

ω

= =

[ ][ ][ ]{ } [ ]{ }P K Z F C F = =

Hence bearing forces are

(iv) Bearing forces are given as

1 1 2 1 0.4064[ ]

1 1 2 1 0.4064

b l lP

a l l

− − = =

( ) ( )2 2 2 3 2311 22 12 21

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

; , 0; 0

48 3 (12 ) 3 3

al a l a l a all l ab b a

EI EIl EI EIl EIl

α α α α = = = = = = =

where

Example 3 contd..

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1 3 6

1111 12

512 22

22

1 048 / 0 2.92 10 0[ ]

0 1 0 12 / 0 1.21 10 EI lK

EI l

α α α α α α

− × = = = =

×

[ ]

1 1 22 21111 12 11

2 2

21 22 222

22

7

6

10

0

10

0

4.08 10 0

0 1.369 10

d d

d

k mk m k k m Z

k k I k I

k I

ω ω ω

ω ω

ω

− −

=

−

−

− − − = = − − −

− ×=

×

[ ]

( ) ( )

( ) ( )

2

1111

222

22

11 22112 22

1111 22

22 11 222

2 222

11 22

10

0

1[ ] [ ] [ ]11 0

0

10

11 10

d

d

d d

k

k mb l lC P K Z a l l k

k I

k b l k lk

k m k I k mb l l

k a l l k a l k l

k I k m k I

ω

ω

ω ω ω

ω ω ω

−−

= = −

− − −−−

= = − − −

Example 3 contd..

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On substituting from equation (10), we have

( ) ( )112 2 2 211;11 21 2

211

11

A B

k a l k a l R C me me R C me me

k mk mω ω 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 )677.6 N

A

B

R

R

π

π

−× ×= × × × = −

× − ×

= −

Example 4.Find the transverse natural frequency of a rotor system as shown in Figure1 21 C id h f l d i d f l i h 2 1 (10)11 N/ 2 f

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1.21. Consider shaft as massless and is made of steel with 2.1 (10)11 N/m2 of

Young’s modulus, E , and 7800 kg/m3 of mass density, ρ . The disc has 10 kgof mass. The shaft is simply supported at ends (In the diagram all dimensions arein cm).

Solution:

Fig 1.21

Fig 1.22

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

influence coefficients ) for the present problem using energy method.11α

On taking force and moment balance, we have

0 0v A BF F F F + = + − =

0 1 0.6 0 A B M F F + = × − × =which gives reaction forces as 0.6 and 0.4 B AF F F F = =

Example 4 contd...

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Fig 1.23 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 above of the shaft segment for 0.0 ≤ x ≤ 0.6, we get

1 10 0.4 0; or 0.4 , 0 0.6

A x x M M Fx M Fx x= − − = = − ≤ ≤

Fig 1.24 Free body diagram of shaft segment

for 0.6 ≤≤≤≤ x ≤≤≤≤ 1.0

On taking the moment balance in the free body diagram as shown above 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 1.0 x x x M M F x Fx M F x x= − + − − = = − − ≤ ≤

The strain energy is expressed as

Example 4 contd...

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1 1

0.6 1.02 2

1 10 0.62 2

x x M dx M dx

U EI EI

= +

( ) ( )211 2

0.6 0.6

0 0

1

2

/ / x x x x M M F dx M M F dxU

F EI EI

δ ∂ ∂∂ ∂∂

= = +∂

{ }{ }0.6 1

1 2 1 20 0.6

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

EI EI EI EI δ

− − − − − −= + = +

The strain energy is expressed as

The linear displacement is expressed as

On substituting bending moment expression obtained earlier, we get

The stiffness is given as1

7

0.01152 0.007688.45 10 N/m

1 2

F k

EI EI δ

−

= = + = ×

Example 4 contd...

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11 2 4 6 4 3 4 4 ; ; 2

12 10 N/m 0.1 4.907 10 m 0.3 3.976 10 m

64 64 E I I

π π − −= × = × = × = × = ×

78.45 102906.81 rad/s

10

k p

m

×= = =

where

which gives natural frequency as

It should be noted that the tilting motion of the disc has not considered

More general form of the Jeffcott rotor response is equation(75) , theforcing is assumed to be from the imbalance only and then it can bee pressed as

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expressed as

with

where superscripts and represent the terms real and imaginary,

respectively. The vector contains amplitude and phaseinformation of the imbalance forcing with respect to some convenientshaft location and is the total dofs of the system.

The solution can be written as

The response can be obtained as

with

{ } { j t imb imb f F e ω

= (76,77)

{ { } j t u U e ω =

N

k k k

r i

imb imb imbF F jF = + 1, 2. ,k N =

[ ] [ ] [ ]( )2 Z K M ω = −

(78)

{ imbF

ir

{ } 1[ ] { } j t imbu Z F e ω −

=(79,80)

Similar to the force amplitude vector, the response vector will alsohave complex quantities and can be written as

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which will give amplitude and phase information, as

and

If displacement is defined, as

and

Equations of motion (72) can be written as

with

r i

k k k U U jU = + (81)1, 2, ,k N =

( ) ( )2 2

amp r i

k k k U U U = + (82,83)( )1

tan / phase i k

k k k U U U −

=

r x yu u ju= +

(84,85)r y x jφ φ φ = +

2

11 12

j t

d r r r d m u k u k m e e ω φ ω + + = (86)

21 220d r r r I k u k φ φ + + =

(87)

d x y I I I = =

Let the solution be

( )j tω ϕ− (88)( )j tω ϕ−

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and

where and are whirl amplitude and phase respectively, so that

and

On substituting above solutions into equations of motion (86-87), weget

Equation (90) can be expressed as

On substituting the above equation into equation (89), we get

( )ur j t

r r u U e

ω ϕ = (88)

( )r

j t

r r e φ ω ϕ

φ

−

= Φ

( )2 ur j t

r r u U e

ω ϕ ω

−= −

( )2 r j t

r r e φ ω ϕ φ ω

−= − Φ

r U r Φ

( )( )2 2

11 12

ur r j j t

d r r d k m U e k e m eφ ϕ ω ϕ ω ω

− −

− + Φ = (89)

( )( )2

21 22 0ur r

j j t

r r r k U e k I e φ ϕ ω ϕ ω

− −

+ − Φ =

(90)

( )

( ) 21

222

ur r j t j

r r r

k

e U ek I

φ ω ϕ ϕ

ω

− −−Φ =

−

(91)

( )( )

( )

2 2

11 22 12 21 2

2

22

ur d r j

r d

r

k m k I k k U e m e

k I

ϕ ω ω

ω ω

− − − − =

−

(92)

On equating the real and imaginary parts of both sides of equation(92), we get

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and

From the second equation, we get

which means there will not be any phase difference.On substituting phase information in equation (93), we get

which is the whirl amplitude.

( ) ( )( )

2 211 22 12 21 2

2

22

cosr

d r

r u d

r

k m k I k k U m e

k I

ω ω ϕ ω

ω

− − − =−

(93)

( )( )( )

2 2

11 22 12 21

2

22

sin 0r

d r r u

r

k m k I k k U k I

ω ω ϕ ω

− − − =

−

(94)

sin 0; i.e. 0r r u uϕ ϕ = =

( )( ) ( )

2 222

2 2

11 22 12 21

d r

r

d r

m e k I U

k m k I k k

ω ω

ω ω

−=

− − −(95)

The condition of resonance can be obtained by equating denominatorof equation (95) to zero

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By defining

and

Equation (96) can be written as

The solution of the above polynomial can be expressed as

which represents the critical speeds of the rotor system.

It can be seen that term inside the square root is always positive i.e.

which can be written as

( ) ( )2 211 22 12 21 0d cr r cr k m k I k k ω ω − − − = (96)

2 11 ,ud

k

m

ω = (97)

( ) ( )4 2 2 2 2 2 2 2 0cr u cr u u uφ φ φ φ ω ω ω ω ω ω ω ω − + + − =

( ) ( )2

2 2 2 2 2 24 0u u u uφ φ φ φ ω ω ω ω ω ω + − − >

(100)

2 22

r

k

I φ ω =

2 21u

r

k

I

φ ω =2 12u

d

k

m

φ ω =

(98)

( ) ( ) ( )1,2

21 12 2 2 2 2 2 2 2 2

2 24cr u u u u uφ φ φ φ φ ω ω ω ω ω ω ω ω ω = + ± + − − (99)

( )2

2 2 2 24 0u u uφ φ φ ω ω ω ω − + >

It can be seen that above condition be always true since all individualterms , , , and are real quantity. However, the following terminside the square root can be

uω φ ω uφ ω uφ ω

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inside the square root can be

which gives two critical speeds. The above term can be

which gives only one critical speed since one pair root will be complexconjugate. Figure 1.11 gives these two cases.

( )2 2 2 2 0u u uφ φ φ ω ω ω ω − > (101)

( )2 2 2 2 0u u uφ φ φ ω ω ω ω − < (102)

( ) 02222 >− φ φ φ ω ω ω ω uuu

Figure 1.25(a)

( ) 02222

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12 21 0k k = =

( )

( )( ) ( )

2 2 222

2 2 2

11 22 11

d r d r

d r d

m e k I m eU

k m k I k m

ω ω ω

ω ω ω

−= =

− − −

which gives

which is same as discussed in the previous section for Jeffcott rotor.The response is shown in Figure 1.25(c).

Fig 1.25(c) Amplitude versus spin speed for 12 21 0k k = =

(103)

2 2

On substituting equation (95) into equation (91), we get

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( )( )

( )( )

2 2

2221

2 2 222 11 22 12 21

r d r j

r

r d r

m e k I k ek I k m k I k k

φ ϕ ω ω ω ω ω

− −

−Φ = − − − −

sin 0; i.e. 0r r

φ φ ϕ ϕ = =

( )( )

2

21

2 211 22 12 21

d

r d r

m ek

k m k I k k

ω

ω ω Φ = −

− − −

From the above equation, we get

which means there will not be any phase difference. On substituting phaseinformation in equation (104), we get

(104)

(105)

(106)

which is the whirl amplitude of angular displacement and the conditionof resonance can be obtained by equating the denominator of equation(106) to zero, which is same as the previous case. For disc at the center

of the shaft span, we have 12 21 0k k = = , which gives0

r Φ = (107)

which is very obvious since when the disc is at the center of the shaftspan, it will not produce any moments and hence there will not be tilting of

the disc take place.

Bearing reaction forces :

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The forces transmitted through bearings are those are related to thedeflection of the shaft.

Fig 1.26 Bearing reaction forces on the shaft

10 0 or A ya yz B B y yz

a M F M R l R F M l l

= − − = = −

10 0 or B A y yz A y yz

b M R l F b M R F M

l l= − − = = +

On taking moment about ends A and B of the shaft,we have

(108)

(109)

From above equations, the bearing reaction forces at A and B are relatedto the loading on the shaft Fy and Myz , in matrix form as follows

1

1

y A

yz B

F R b l l

M R a l l

=

−

(110)

or {R } = [D ] {p } with [ ]1

1

b l l D

a l l

=

− (111)

Using equations (70) and (80), equation (111) can be written as

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{R } = [D ] [K ] {d }=[D ] [K ] [Z ] {f }=[C ] {f } (112)

( )( ) ( )

( )( ) ( )

222 2

11 12 22 21 12 2212 12 11

2 221 22 21 11 21 22 21 12 22

0

d A

B d

bk k k I k bk k R C C me C me me

R C C lme C ak k k I k ak k

ω ω ω ω

ω ω

+ − − + = = =

∆ − − − −

with ( ) ( )2 211 22 12 21d cr r cr k m k I k k ω ω ∆ = − − −

which can be expanded as

(113)

Bearing reaction forces will be having similar variation as of theresponse, since it has the same denominator,

It can be shown from equation (113) that the forces transmitted throughthe bearings are also a maximum at the system critical speed. These

forces are dynamic forces and are superimposed on any steady loads,which may be present, due to gravity loading for example.

∆ as that of the response.

1.4 Symmetrical rigid shaft in flexible anisotropic bearings

In systems where the bearings are far more flexible than the shaft it is the

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bearings, which will have the greatest influence on the motion of the rotor.Such rotors may be idealized as rigid rotor.

Fig 1.28(b) Positive conversions of angular displacements.

Assumptions:

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It is assumed that the shaft has no flexibility The bearings are assumed to behave as linear springs having a stiffness

kx in the horizontal direction and ky in the vertical direction. m is the rotor

mass.The center of gravity is offset from geometrical center by distances e and d

as shown in Figure 1.28(a). x and y are the linear displacements of the rotor

(geometrical center) in the horizontal and vertical directions respectively.

φ and θ are the angular displacement of the rotor (geometrical center line)

in the z-x and y -z planes, respectively.

Since for the present case there is no coupling between variousdisplacements i.e. x , y , φ and θ . Hence free body diagrams and equations ofmotion have been obtained by giving such displacements independent ofeach other.

2 cos -me t k x mxω ω =

EOM in the x and y directions are

(114)

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x

2 sin yme t k y myω ω − =

2 2cos x d me d t k l I ω ω φ φ − − =

2 2sin y d

me d t k l I ω ω θ θ − − =

and

EOM in the φ and θ directions are

and

(117)

(116)

(115)

( )

Fig 1.29(a) Free body diagram of the rotor in y -z plane

Fig 1.29(b) Free body diagram of therotor in x-y plane

Fig 1.29(c) Free body diagram of therotor in z-x plane

For sinusoidal vibrations, we can write

2 2 2 2, , and x x y yω ω φ ω φ θ ω θ = − = − = − = − (118)

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2 2

2 2

cos cos ; sin sin x y

me me x t X t y t Y t

k m k m

ω ω ω ω ω ω

ω ω = = = =

− −

2 2

2 2 2 2cos cos and sin sin

x d y d

me d me d t t t t

k l I k l I

ω ω φ ω ω θ ω ω

ω ω = = Φ = = Θ

− −

Substituting equation (118) into equations (114) to (117),

the unbalance response can be expressed as

(119)

Critical speeds can be written as

1 2 3 4

22

; ; ; and y y x x

cr cr cr cr

d d

k k lk k l

m m I I ω ω ω ω = = = = (120)

From equation (119) on squaring x and y and adding, it gives

2 2

2 21

x y

X Y

+ =(121)

(It is an equation of ellipse. )

Similarly from equation (119), we get

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2 2

2 21

φ θ

θ + =

Φ(122)

Equation (122), relating to the angular motion of the rotor, is also the equation of

an ellipse.

o This means that there is an elliptical orbital trajectory of the rotor ends dueto angular motion of the rotor.

o This rotor motion is caused by the imbalance couple meω 2d acting on therotor, and it is superimposed on the lateral motion described previously.

o A reversal of the direction of the orbit associated with this motion alsooccurs, between two critical speeds associated with angular motion of the

rotor (i.e.

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Figure 1.30 Whirl directions with respect to the shaft spin frequency

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Figure 1.31 Mode Shapes for a rigid rotor mounted on flexible bearings

Th lit d f th f t itt d t th b i i diff t i

Important Notes:

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The amplitude of the force transmitted to the bearings is different inhorizontal and vertical directions, as well as at each end of the rotor.

The force transmitted is that which causes the bearings to deformand is given by the product of spring stiffness and rotor deflection atthe bearing.

The bearing force amplitudes are

in horizontal and vertical direction respectively.

The + sign refers to the angular motion of the rotor causes the rotorend to deflect in the same direction to the lateral deflections of the

rotor and the - sign refers to the angular motion of the rotor causes

the rotor end to deflection in the opposite direction to the lateral

deflections of the rotor.These bearing forces must take on maximum values when thesystem is operated at the critical speeds, where x , y , φ and θ aremaximum.

( ) ( ); and2 2

y x x y

k k F x l F y lφ θ = ± = ± (125)

Example 8.A long rigid symmetric rotor is supported at ends by two identical

b i L t th h ft h th di t f 0 2 th l th f h ft

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bearings. Let the shaft has the diameter of 0.2 m, the length of shaftis 1 m and the mass density of the shaft material equal to 7800kg/m3. The bearing dynamic characteristics are as follows: kxx = ky y= 1 kN/mm with rest of the stiffness and damping terms equal to zero.By considering the gyroscopic effect negligible also, obtain the

natural frequencies of the system.Since cross-coupled stiffness coefficients is x and y directions are zeroand no gyroscopic effect is considered, hence single plane motion can beconsidered one at a time. For the present analysis there is no coupling isconsidered between the linear and rotational displacements and since

stiffness in x and y direction is same hence natural frequencies in thesedirections can be written as

Similarly natural frequencies corresponding to the tilting motion can bewritten as

1,2

62 2 1 1090.34 rad/s

245.04

k

mω

× ×= ± = ± = ±

2 2 2

3,4

1 10 1154.184 rad/s

2 2 21.0326d

kl

I ω

× ×= ± = ± = ±

×

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Solution (b):

Example 9 contd..

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With different stiffness properties in horizontal & vertical directions, thenatural frequencies are given as

1 2

3 4

2 2

2 216.73 rad/s; 11.84 rad/s

2 2189.32 rad/s; and 133.87 rad/s

ver hozn n

ver hozn n

d d

k k m m

k l k l

I I

ω ω

ω ω

= = = =

= = = =

Answer

Solution (b):

Example 10:

Find critical speeds of a rotor system as shown in Figure 1 33

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Find critical speeds of a rotor system as shown in Figure 1.33

1.1 kN/mm A x

k = 1.8 kN/mm A y

k = 3.1 kN/mm B

k x = 3.8 kN/mm B y

k =

Take the bearing stiffness properties as:

; ; and

.

2 2cos and sin A B A B x x y y

me t k x k x mx me t k y k y myω ω − − = − − =

2 2 2

sin and A B y y d me d t k l k l I ω θ θ θ − − =

Solution :Equations of motion in x & y directions are

Equations of motion in θ and φ directions are

2 2 2 sin A B y y d

me d t k l k l I ω φ φ θ − − =

The steady state force vibration responses

can be obtain as2

2

2cos ; sin

2 A B A B x x y y

me me x t y t

k k m k k m

ω ω ω ω

ω ω = =

+ − + −

( ) ( )

2 2

2 2 2 2

cos ; sin

A B A B x x d y y d

me d me d t t

k k l I k k l I φ θ

−

= =

+ + −

Example 10 contd..:

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On equating determinates of responses, critical speeds can be obtained as

( )

( )

6

1

6

2

26 2

3

26 2

4

(1.1 3.1) 10648 rad/s;

10

(1.8 3.8) 10748.3 rad/s

10

(1.1 3.1) 10 1 6480.7 rad/s;0.1

(1.8 3.8) 10 1748.3 rad/s

0.1

A B

A B

A B

A B

x x

y y

x x

d

y y

d

pm

pm

l p I

l p

I

k k

k k

k k

k k

+ ×= = =

+ ×= = =

+ × ×= = =

+ × ×= = =

+

+

+

+Answer

1.5 Symmetrical rigid shaft in flexible anisotropic

bearings with damping and cross coupling

For the case of oil-film lubricated bearings the bearing have associatedd i i ll i iff i

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For the case of oil-film lubricated bearings the bearing have associateddamping properties as well as spring stiffness properties.

Furthermore in case of hydrodynamic bearings the shaft motion in thehorizontal direction is coupled with that in the vertical direction.

However, coupling between the translational and tilting motion has notbeen considered since the rotor is symmetric.

In most applications the properties of such bearings are described interms of the eight linearised bearing stiffness and damping coefficients.

Symmetrical rigid shaft in flexible anisotropic bearings will be identical to Fig 1.28

with cross-coupled terms. The EOM for rotor are given by

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2 2 2

2 2 2 2

-

-

x xx xy xx xy

y yx yy yx yy

xz xx xy xx xy d

yz yx yy yx yy d

F k x k y c x c y mx

F k x k y c x c y my

M k l k l c l c l I

M k l k l c l c l I

φ θ φ θ φ

φ θ φ θ θ

− − − − =

− − − − =

− − − =

− − − =

(126)

It is assumed that there is no coupling between the linear (i.e. x and y ) and

angular displacements (i.e. φ and θ ) due to symmetry of the rotor.

The imbalance force meωωωω2 is located same distance from the rotor

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The imbalance force me ω ωω ω is located same distance from the rotorgeometrical center.

Out of balance forces in the horizontal and vertical directions may then bewritten as

( ) ( )2 2 j jcos Re Ret t x xF me t me e F eω ω ω ω ω = = =

2 xF meω =

( ) ( )2 2 j jsin Re j Ret t y yF me t me e F eω ω ω ω ω = = − =

2 j yF meω = −

with

with(127)

xF yF where and are complex forces (which contains amplitude andphase informations) in the x and y directions. These forces are acting atthe center of gravity.

Moments about the rotor geometrical center caused by these forces are

( ) ( )2 2 j jcos Re Ret t xz xz M me d t me de M eω ω ω ω ω = = = 2 xz M me d ω =with

( ) ( )2 2 j jsin Re j Ret t yz yz M me d t me de M eω ω ω ω ω = = − =

2 -j yz M me d ω =with

(128)

The response can be assumed as

j j j j

; ; ; =t t t t

x Xe y Ye e eω ω ω ω

φ θ= = = Φ Θ (129)

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; ; ; = x Xe y Ye e eφ θ = = = Φ Θ

where X , Y , Φ and Θ are complex displacements.

Equations of motion (126) can be written as

[ ]{ } [ ]{ } [ ]{ } { }( ) x C x K x f t + + =

[ ] [ ] [ ]

{ } { }

2 2 2 2

2 2 2 2

0 0 0 00 0 0

0 0 0 00 0 0; C ; ;

0 0 0 00 0 0

0 0 0 00 0 0

( ) ; ( )

xx xy xx xy

yx yy yx yy

xx xy xx xyd

yx yy yx yyd

x

y

xz

c c k k m

c c k k m M K

l c l c l k l k I

l c l c l k l k I

F x

F y x t f t

M φ

θ

= = =

= =

yz M

The response takes the following form

{ } { } { { } { } { j j 2 j; so that j andt t t x X e x X e x X eω ω ω ω ω = = = −

(129)

(130)

(131)

On substituting equations (127), (128) and (131) into equations of motion (130)

we get

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we get

[ ] [ ] [ ]( ){ { }2 j C K X F ω ω − + + =with

{ } { };

x

y

xz

yz

F X

F Y X F

M

M

= = Φ

Θ

[ ]{ { D X F = [ ] [ ] [ ] [ ]( )2 j D K M C ω ω = − +which can be written as with

The response can be obtained as {X } = [D ]-1 {F }

The displacement amplitude of the rotor will be given by

2 2 2 2 2 2 2 2, , ,r i r i r i r i X X X Y Y Y = + = + Φ = Φ + Φ Θ = Θ + Θ

(132)

(133)

(134)

(135)

Phase lag will be given by

1 1 1 1i i i iX Y

β δ Φ Θ

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-1 -1 -1 -1tan , tan , tan , tani i i i

r r r r

X Y

X Y α β γ δ

Φ Θ= = = =

Φ Θ

The resulting shaft whirl orbit can be plotted using equation (129) and (134)

j j andt t x Xe y Yeω ω = =and in general will be found to take the form as shown in Figure 1.34.

Fig 1.34 Rotor whirl orbit

Example 11.

Obtain the bending critical speeds and mode shapes of a rigid rotor,

consist of massless rigid shaft of 2 m of span, 5 kg mass and diametral

mass moment of inertia of 0.1 kg-m2, supported by flexible bearings as

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g , pp y gshown in Figure 128. The bearing properties are: kxx = 2.0××××104 N/m, kyy = 8.8××××104 N/m, kxy = 1.0××××103 N/m, kyx = 1.5××××103 N/m, cxx = 1.0 N-s/m, cyy = 1.0 N-s/m, cxy = 1.0××××10-1 N-s/m and kxx = 1.0××××10-1 N-s/m. Obtain theunbalance response (amplitude and phase) at bearing locations when the

radial eccentricity of 0.1 mm and axial eccentricity of 1 mm is present inthe rotor and locate critical speeds.

Solution :

Fig 1.35 shows the unbalance responses both for the linear and

angular displacements. Both the amplitude and phase has beenplotted. It can be observed that in the plot of linear and angulardisplacement two peaks appears and they correspond to the criticalspeeds of the system. Since the linear and angular displacements areuncoupled for the present case and hence corresponding criticalspeeds appears in respective plots. There are four critical speeds: 70rad/s, 120 rad/s, 480 rad/s and 920 rad/s

Example 11 contd...

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Dr. R. Tiwari ([email protected])Fig 1.35 Amplitude and phase variation with respect to spin speeds

Exercise Problem E1.6.Obtain the critical speeds for transverse vibrations of rotor-bearingsystem as shown in Figure E1.6. Consider shaft as rigid. The shaft is of 1

m of span and the diameter is 0.05 m with the mass density of 7800kg/m3 The shaft is supported at ends by flexible bearings Consider the

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p ykg/m3. The shaft is supported at ends by flexible bearings. Consider themotion in both the vertical and horizontal planes. Take the followingbearing properties:

For bearing A: kxx = 20 MN/m, kyy = 15 MN/m, kxy = -1.5 MN/m, kyx = 25

MN/m, cxx = 200 kN-s/m, cxy = 150 kN-s/m, cyx = 140 kN-s/m, cyy = 400kN-s/m,For bearing B: kxx = 24 MN/m, kyy = 17 MN/m, kxy = -2.5 MN/m, kyx =

30 MN/m, cxx = 210 kN-s/m, cxy = 160 kN-s/m, cyx = 135 kN-s/m, cyy =380 kN-s/m.

Bearing forces:

The forces, which are transmitted to the bearings, are those, whichdeform the bearing lubricant film, and do not include rotor inertia terms.

In general bearing forces will lag behind the imbalance force such thatthe bearing horizontal and vertical force components, at one end A of themachine, can be represented as

A bx A xx A xy A xx A xy A xx A xy A xx A xy

A by A yx A yy A yx A yy A yx A yy A yx A yy

f k x k y c x c y k l k l c l c l

f k x k y c x c y k l k l c l c lf k x k y c x c y k l k l c l c l

φ θ φ θ

φ θ φ θ φ θ φ θ

= + + + + + + +

= + + + + + + +

+ + + + + + +

(137)

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B bx B xx B xy B xx B xy B xx B xy B xx B xy

B by B yx B yy B y

f k x k y c x c y k l k l c l c l

f k x k y c

φ θ φ θ = + + + + + + +

= + +

x B yy B yx B yy B yx B yy x c y k l k l c l c lφ θ φ θ + + + + +

where 2, 2 A xx xx B xx xxk k k k = = , etc.

Equation (137) is for more general case, which can be written in matrix form as

{ } [ ]{ } [ ]{ }b b b f c x k x= +

{ } { } { }

[ ] [ ]

; ; ;

;

A bx

A by

b

B bx

B by

A xx A xy A xx A xy A xx A xy A xx

A yx A yy A yx A yy

b b

B xx B xy B xx B xy

B yx B yy B yx B yy

f x x

f y y f x x

f

f

c c c l c l k k k

c c c l c lc k

c c c l c l

c c c l c l

φ φ

θ θ

= = =

= =

A xy

A yx A yy A yx A yy

B xx B xy B xx B xy

B yx B yy B yx B yy

l k l

k k k l k l

k k k l k l

k k k l k l

with

( )

(138)

(139){ } { { { } { {

j j j

; j ;

t t t

b b x X e x X e f F e

ω ω ω

ω = = =

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On substituting equation (139) into equation (138), we get

{ [ ] [ ]( ){ jb b bF k c X ω = +

This can be used to evaluate bearing forces.

The amplitude of forces transmitted to bearings are then given by

2 2 2 2 2 2 2 2; ; ;r i r i r i r i

A bx A bx A bx A by A by A by B bx B bx B bx B by B by B byF F F F F F F F F F F F + + + += = = =

with corresponding phase angles are given by

1 1 1 1tan ; tan ; tan ; tani i i i

r r r r

A bx A by B bx B by

A bx A by B bx B by

F F F F F F F F

ε ς η λ − − − − = = = =

(140)

(141)

{ } { { { } { {

1.6 Asymmetrical flexible shaft in flexible anistropic bearings

with damping and cross coupling

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Fig 1.36

A flexible shaft in flexible bearings

For the present case both the shaft and bearings are flexible.

The analysis allows for different instantaneous displacements of the shaft

at the disc and at bearings.

The system will behave in a similar manner to that described in previoussection, except that the flexibility of shaft will increase the overall flexibility

of the support system as seen by the disc.

An equivalent set of system stiffness and damping coefficients is firstevaluated, which allows for the flexibility of the shaft in addition to that of

bearings, and is used in place of the bearing coefficients in previoussection analysis.

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section analysis.

The total deflection of the disc is the vector sum of the deflection of thedisc relative to the shaft ends plus that of the shaft ends to thefoundation. For disc we observe the displacement of geometrical center

of the disc.

The deflection of shaft ends in bearings is related to the forcetransmitted through bearings by the bearing stiffness and dampingcoefficients as

bx xx xy xx xy

by yx yy yx yy

f k m k n c m c n

f k m k n c m c n

= + + +

= + + +

where m and n are instantaneous displacements of shaft ends relative tobearings in the horizontal and vertical directions respectively, and take theform

j j;t t m Me n Neω ω = = (143)

(142)

j j

;

t t

bx bx by by f F e f F e

ω ω = =

The bearing forces have the following form

On substituting in equation of motion (59) we get

(145)

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j j

j j

bx xx xy xx xy

by yx yy yx yy

F k M k N c M c N

F k M k N c M c N

ω ω

ω ω

= + + +

= + + +

{ } [ ]{ }bF K V =

On substituting in equation of motion (59), we get

which can be written in matrix form as for both bearings A and B as

{ } [ ]

( ) ( )

( ) ( )

( ) ( )

( ) ( )

{ }

j j 0 0

j j 0 0;

0 0 j j

0 0 j j

xx xx xy xy A A A bx

yx yx yy yy A by A A

b

B bx xx xx xy xy B B

B by

yx yx yy yy B B

T

A A B B

k c k cF

k c k cF F K

F k c k c

F k c k c

V M N M N

ω ω

ω ω

ω ω

ω ω

+ + + + = =

+ +

+ +

=

with

(146)

(147)

The magnitude of reaction forces transmitted by bearings can also beevaluated in terms of the forces applied to the shaft by the rotor. (Hence shaft

is not assumed to be rigid and moment balance is considered).

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Fig 1.37 Free body diagram of the shaft

0 ( ) B A by y yz M l f F l d M = = − −

( ) ( )1 1 A by y yz f d l F l M = − −

0 A B by y yz M l f dF M = = +

( ) ( )1 B by y yz f d l F l M = +

or (148)

or

(149)

( ) ( )1- 1 A bx x xz f d l F l M = +

( ) ( )1 B bx x xz f d l F l M = +

Similarly forces in the horizontal direction may be written as

(150)

(151)

Equations (148-151) can be combined in the matrix form as

{ [ ]{b s f A f =(152)

{ } { } { { j j andt t b b s s f F e f F eω ω

= =

For an unbalance excitation, we have

On substituting equation (153) in equation (147), we get

(153)

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{ } [ ]{ }b sF A F =

where subscript b refers to the bearing and s refers to the shaft.

In above equation bearing forces are related to reaction forces at the shaft

Equating equation (147) and (154), we get

[K ] {V } = [A]{F s } or {V } = [K ]-1 [A] {F s } (155)

(154)

The deflection at the location of the disc due to movement of theshaft end can be obtained as follows.

Consider the shaft to be rigid for some instant and assuming shaft enddeflections in horizontal direction be Am and B m at ends A and B,ti l h i Fi 1 38

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respectively as shown in Figure 1.38.

Fig 1.38 Rigid body movement of the shaft in z -x plane

Slope in x -z plane of the shaft will be

( )1-

B A

A A B

m m d d m d m m

l l l

− = + = +

( )- B Am m lφ =

For motion in y - direction and y -z plane,we have

( ) ( )1- A B y d l m d l m= +

( )- A Bn n lθ =

Equations (156-159) can be combined in a matrix form as

{ } [ ]{ }1s

u B v=

(156)

(157)

(158)

(159)

(160)

{ } { } { { }1 1 j j

andt t

s su U e v V eω ω

= =

For unbalance excitation (or for free vibration analysis), shaftdisplacements at bearing locations and at disc center vary sinusoidally suchthat

(161)

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{ } { }

{ [ ]{ }1s

BU V =

{ } [ ][ ] [ ]{ } [ ]{ }1

1

s sB A sU K F C F −

= =

On substituting in equation (

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