rtiwari rd book 01a

8/10/2019 Rtiwari Rd Book 01a

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Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])

2

When the rotor is not eccentric, however, a small imbalance mass, mi is attached at a relatively larger

radius of rias shown in Figure 1.2(c), the imbalance force can be written as

trmtF ii sin)(2= (1.2)

For the case when the rotor is eccentric and a small imbalance mass is attached as shown in Figure 2(d),

the imbalance force will be

)sin(sin)( 22 ++= trmtemtF ii (1.3)

No imbalance

Fig 1.2(a)Rotor geometrical centre and

centre 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 eccentricity

square of spin speed

Fig 1.2(c)Rotor geometrical centre, centre of gravity

and an additional imbalance mass

Fig 1.2(d)Rotor geometrical center, centre of gravity

and imbalance mass

Imbalance force is the vectoraddition of forces due to the rotor

and imbalance forces

Fig 1.1(a)

A rigid rotor mounted on flexible bearings

bkbk

2effk k=

Fig 1.1(b)

A flexible rotor mounted on rigid bearings

348 /effk EI L=

F(t)

Fig 1.1(c)An equivalent single degree of freedom

spring-mass system

effk

F(t)

Fig 1.1(d)Free body diagram of disc mass

effk y

y


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

the imbalance mass. Figure 1.3 shows the unbalance location on a rotor system. For a constant angular

velocity of the rotor, , the location of the unbalance is given as t = .

On application of the Newtons law on the free body of the rotor mass as shown in Figure 1.1(d), i.e.

equating sum of external forces to the mass of the rotor multiplied by the acceleration of the center of

gravity of the rotor mass, we have

ymtemykeff =+ sin2

(1.4)

where effk is the effective stiffness of the rotor system (see Figure 1.1). Equation (1.4) is a standard

equation of motion of a single DOF spring-mass system and can be written as

temykym eff sin2=+ (1.5)

From the free vibration of the when the external imbalance force is absent, the rotor mass will be having

oscillation and that will be given by

)sin()( tYty n= (1.6)

where n is the frequency of oscillation during the free vibration and that is called the natural frequency

of the system. On substituting equation (1.6) into the homogeneous part of equation of motion (1.5), it

gives

0)sin()( 2 =+ tYkm neffn (1.7)

For the non-trivial solution of equation (1.7), the natural frequency of the system can be written as

Figure 1.3 Unbalance location on a rotor

o

At time, t= 0

C G

o

At time, t

G

C


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mkeffnf /= (1.8)

The steady state forced response can be modeled as

)sin()( = tYty (1.9)

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

imbalance force. On substituting equation (1.9) into equation (1.5), the steady state forced response

amplitude can be written as

[ ]

2 2

2 2

( ) sin( ) sin( )

or

( ) sin cos cos sin sin( )

eff

eff

m k Y t me t

m k Y t t me t

+ =

+ =

On separating the sine and cosine terms, we have

2 2 2( ) cos and ( ) ( sin ) 0eff eff m k Y me m k Y + = + =

Since the term2( )effm k + in general may not be zero, from the second expression we have 0=

and the first expression can be simplified as

2 2 2

2 2 2 2

/1eff nf

mY Y e

k m

= = = =

with / nf = (1.10)

where Y is the non-dimensional unbalance response (ratio of the unbalance response to the eccentricity)

and is the frequency ratio (ratio of the spin speed of the rotor to the natural frequency of the rotor

system). The non-dimensional unbalance response (absolute value) is plotted with respect to the

frequency ratio as shown in Figure 1.4. The response changes its sign after critical speed, which means

that shaft deflects in the opposite direction to the unbalance eccentricity. Both linear and semi-log plot is

shown to have clarity of response variation. It can be seen that as the frequency ratio increases the non-

dimensional response asymptotically approaches to unity. It means unbalance response approaches to theeccentricity of the rotor. Physically it implies that the rotor rotates about its center of gravity at high

frequency ratio. From Figure 3 and equation (1.10) it should be noted that we have unbounded unbalance

response when the denominator2(1 ) becomes zero i.e when the spin speed is

1cr = or /cr eff nf k m = = (1.11)


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This is a resonance conditionand the spin speed corresponding to the resonance is defined as critical

speed. The subscript: crrepresent the critical. For the present case the critical speed is equal to the natural

frequency of the system as given in equation (8), the undamped steady state forced response amplitude

tends to infinity. The natural frequency concept has come from free vibrations and critical speed fromforced vibrations. It should be noted that in rotor dynamics in general the critical speed and the natural

frequency might not be same (e.g. when gyroscopic couple is considered in the analysis). The sign

represent that the rotor will have critical speed while rotating in either clockwise or counter clockwise.

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

speed). The analysis presented in this section can be applied to the transverse, torsional and axial

vibrations of rotors and accordingly critical speed can be termed by prefixing respective names of

vibrations. For torsional vibration care should be taken that mass will be replaced by the polar mass

moment of inertia of rotor and stiffness will be torsional stiffness. Similarly, for axial vibration mass willremain same as transverse vibration, however, the stiffness will be the axial stiffness.

(a) Linear plot (b) Semi-log plotFigure 1.4 Non-dimensional unbalance response versus frequency ratio

% Non-dimensional unbalance response wrt to freq ratio "Figure_1_4.m"

% Copywriters: Dr R Tiwari, Dept of Mechanical Engg., IIT Guwahati.% 13-01-2005clear all;deta_freq=0.005;

freq_ratio(1)=deta_freq;N_pt=1000;

for ii = 1:1:N_pty_resp(ii)=freq_ratio(ii)^2/(1-freq_ratio(ii)^2);if(ii


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freq_ratio(ii+1)=freq_ratio(ii)+deta_freq;end

endfigure(1)

plot(freq_ratio,abs(y_resp), 'k-', [0 5], [1 1], 'k--');xlabel('w/wn ---->');

ylabel('y/e ---->');figure(2)semilogy(freq_ratio,abs(y_resp), 'k-', [0 5], [1 1], 'k--');

xlabel('w/wn ---->');ylabel('y/e ---->');

Unbalance will always be present in a rotor. However, the unbalance response can be reduced by the

following methods.

(i) Correction at source i.e. balancing the rotor: 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 that the best

achievable state of balance tends to degrade during operation of a rotor (e.g. turbomachinery). There are

two type of unbalances (a) 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 1.5a. The rotor can be balanced by a

single plane balancing and (b) 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 1.5b. For balancing such rotors

minimum of two 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 1.5 Classification of unbalances for a short rigid rotor

G

F

G

GG


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(ii) Operate rotor away from the critical speed: i.e. during design itself or during operation by providing

temporary auxiliary support: Moving the machine operating speed farther away from the critical speed

can be achieved by changing the rotor operational speed or by changing the critical speed itself. The

critical speed can be changed either at the design stage or during operation. At design stage changingrotor mass or its distributions and dimensions of the rotor and its support lengths can alter the critical

speed. During operation auxiliary support can be provided to increase the effective stiffness of the rotor,

which in turn increases the critical speed. By this arrangement the actual rotor critical speed can be safely

traversed and then the auxiliary support can be withdrawn which brings the critical speed of the rotor

below the operation speed. In general, changing the critical speed is useful for constant operational speed

machines or for machines with a narrow range of operational speed.

(iii)Add damping to the system or 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 the synchronous whirl is to add damping. On the other hand the damping

(e.g. the shaft material or hysterisis 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

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 damping is retained but there is no capacity to provide

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

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 that their dynamic properties (i.e. damping and stiffness) can be controlled

by changing current or magnetic flux in 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 1.6.


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Figure 1.6 (a) Schematic diagram of squeeze film dampers

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

Figure 1.6 (c) Basic principle of active magnetic bearings

Bearing bush

Outer raceway of rolling

bearing (can displace

radially and constraint

not to rotate.Squeeze film

Rotor

Oil feed groove

Rotor

Rolling bearing

Electrodes

Teflon

Sensor

Rotor

Power Amplifier Electromagnet

Controller


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Example 1.1: A rotor has a mass of 10 kg. The operational speed of the rotor system is at (100 1) Hz.

What should be the bounds of rotor effective stiffness so that the critical speed should not fall within 5%

of operating speeds? Assume there is no damping in the rotor system.

Solution:Operation speed range: 99 to 101 Hz

5% of lower operational speed: 99-990.05=99(1-0.05)=990.95=99.05 Hz

5% of upper operational speed: 101+1010.05=101(1+0.05)=1011.05= 106.05 Hz

The effective stiffness corresponding to lower operating speed = 99.05210 = 98.1 kN/m

The effective stiffness corresponding to upper operating speed = 106. 05210 = 112.5 kN/m

So the effective stiffness should not fall in he range of 98.1 to 112.5 kN/m.

Example 1.2: A rotor has 100 Hz as critical speed and its operating speed is 120 Hz. If we want to avoid

altogether crossing of the critical speed then what should be the increase in the support stiffness by

auxiliary support system. We should have at least 5 Hz of gap between the operating speed and the

critical speed. The rotor has a mass of 10 kg.

Solution: Initial stiffness of the support = 100210 = 100 kN/m

First we will reach 95 Hz of rotor speed, which makes 5 Hz of gap with original critical speed of the

rotor. Now since we cannot increase the rotor speed further we need to increase the critical speed of the

rotor to al least 125 Hz.

Hence the corresponding effective of the support stiffness should be = 125210 = 156.25 kN/m.

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

1.2 Rankine Rotor Model

The single DOF rotor model has limitations that it cannot represent the orbital motion of the rotor in two

transverse directions. Rankine (1869) used a two DOF model to describe the motion of the rotor in two

transverse directions as shown in Figure 1.7(a). The shape of orbit produced depends upon the relative

amplitude and phase of the motions in two transverse directions and the orbit could of circular, elliptical

or straight line, inclined tox andyaxis, as shown in Figure 1.8. However, as shown in Figure 1.7(b) the

free body diagram of the rotor, for the constant spin speed the radius of the whirling of the rotor center


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will increase parabolically and will be given as kFr c /= whereFcis the centrifugal force and it can be

physically also visualized as there will not be any resonance condition, as 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 imbalance force.

1.3 Jeffcott Rotor Model

Figure 1.9 shows a typical Jeffcott rotor. 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. The shaft spin speed is, and the shaft whirls about the bearing axis with whirl frequency,

. For the present case synchronous whirl is assumed (i.e. = ). The synchronous motion occurs

between the earth and the moon. Because of the synchronous motion the moon orbital speed and its own

spin speed are equal as shown in Figure 1.10(a). The unbalance force in general leads to synchronous

whirl conditions. Other kinds of whirl condition, which can occur in real system, are: anti-synchronous

(i.e. = ) and asynchronous (i.e. ) as shown in Figure 1.10(b). The antisynchronous whirl

occurs when there are rubs between the rotor and stator, however, very rarely it occurs. Asynchronous

condition occurs when the speed conditions are high (e.g. when gyroscopic effects are predominates) or

Fig 1.7(b) Free body diagram of the

model

Fr

Fig 1.7(a) Rankine rotor model

(Two degree of freedom spring-

mass rotor model)

Fig 1.8(a)

Circular motion

efk

Fig 1.8(c)

Straight line motion

efk

Fig 1.8(b)

Elliptical motion

efk


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when rotor is asymmetric or when the bearing is anisotropic in dynamic properties. The stiffness of the

shaft is expressed as

3load/deflection 48 /k EI L= = (1.12)

Coordinates to define the position of the center of rotation of the rotor are xu and yu . The location of the

imbalance is given by . Thus, three dofs are needed to define the position of the Jeffcott rotor.

Figure 1.10(a) Synchronous whirl Figure 1.10(b) Anti-synchronous whirl

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

( )cos2

2

eudt

dmucku xxx += (1.13)

Fig 1.9(b)A Jeffcott rotor model iny-zplane

Fig 1.9(a)

A Jeffcott rotor model

v

CG e=

Fig 1.9(c)Free body diagram of the

disc inx-yplane

Shaft spin

direction

Shaft whirling

direction

Shaft

Shaft spin

direction

Shaft whirling

direction

Shaft


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( )sin2

2

eudt

dmmgucku yyy += (1.14)

and

dImge = cos (1.15)

Apart from restoring force contribution from the shaft, the damping force is also considered. The damping

force is idealized as viscous damperand it is mainly coming from the support and aerodynamic forces at

disc. The material damping of the shaft will not contribute viscous damping and it may leads to instability

in the rotor and that will be considered in detail latter.

For the case = ti.e. when the disc is rotating at constant spin speed, the Jeffcott rotor model is reduces

to two DOF rotor model. Neglecting the effect of gravity force, equations of motion in thexandycan be

written as

( )teudt

dmucku xxx cos2

2

+= (1.16)

and

( )teudt

dmucku yyy sin2

2

+= (1.17)

Equations of motion can be written in the standard from as

temkuucum xxx cos2=++ (1.18)

temkuucum yyy sin2=++ (1.19)

It should be noted that equations of motion are uncoupled and motion can be analysed independently in

two transverse planes. Noting equation (1.8), from the undamped free vibration analyses it can be seen

that since the rotor is symmetric rotor will be having two natural frequencies that are equal and given as

mknf /2,1 = (1.20)

The damping does not affect the natural frequency of the system appreciably. However, their effect is

more predominate for suppressing the resonance amplitude. By extending the previous method the steady

state forced response can be written as


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[ ]

cos( )

cos ( ( / 2 ) sin( )

x x

y y y

u U t

u U t U t

=

= + = (1.21)

where Uxand Uyare the steady state forced response amplitudes in the xandydirections, respectively.

is the phase lag of the displacement with respect to the imbalance force. The phase difference between thetwo direction responses will be of 900as two directions are orthogonal to each other. For the direction of

whirling shown in Figure 1.9 i.e. counter clockwise (ccw) for the present axis system the response in y

direction will lag behind xdirection response by /2 radians. Hence the lag of the ydirection response

with respect to the force will be (/2 + ). On taking the first and second derivatives of the response with

respect to time, t, we get

)cos(

)sin(

=

=

tUu

tUu

yy

xx

and

)sin(

)cos(

2

2

=

=

tUu

tUu

yy

xx

(1.22, 1.23)

On substituting equations (1.21) to (1.23) into equation (1.18) and separating the in-phase (i.e. cost) and

quadrature (i.e. sint) terms, we get

emkUcUUm xxx22

cossincos =++ (1.24)

0sincossin2 =+ xxx kUcUUm (1.25)

Equation (1.25) gives

2tan

mk

c

= (1.26)

which gives

( ) ( )222sin

cmk

c

+= and

( ) ( )222

2

cos

cmk

mk

+

= (1.27, 1.28)

Substituting equations (1.27) and (1.28) into equation (1.24), we get

( ) ( )222

2

cmk

emUx

+= (1.29)

Similarly, we can obtain response amplitude inydirection from equation (1.19) as


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2tan

mk

c

= (1.39)

On substitution of phase from equations (1.39) to (1.37) the whirl amplitude can be written as

( ) ( )222

2

cmk

emUr+

= (1.40)

Equations (1.39) and (1.40) are similar to previous results i.e. equations (1.26) to (1.30). The non-

dimensional form of equations (1.39) and (1.40) can be written as

21

2tan

= (1.41)

( ) ( )222

2

21/

+== eUU rr (1.42)

with

kmcccmk ccnn 2;/;/;/ ==== (1.43)

where is the frequency ratio, n is the natural frequency of non-rotating rotor, is the damping ratio

and ccis the critical damping of the system for which the damping ratio is equal to unity. Figure 1.11(a)

shows that the maximum amplitude occurs at slightly higher frequency than the n when damping is

present in the system, however maximum amplitude occurs at n for the undamped case. The increase in

the damping results in increase in the critical speed, however damping is the most important parameter for

reducing the whirl amplitude at critical speed. Since the measurement of amplitude of vibration at critical

speed is difficult, hence the determination of precise critical speed is difficult. To overcome this problem

the measurement of the phase is advantageous at least to determine the undamped natural frequency of the

system.


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% Computer code for generating Jeffcott Rotor Response (Figure 1.11)% Copywrites: Dr R Tiwari, Dept of Mechanical Engg., IIT Guwahati, 16-1-2005clear all;

jj=0;zeta=[0.0 0.01 0.05 0.1 0.4 1.0 2.0 10.0 100.0 10000.0];

for jj = 1:1:length(zeta);freq_ratio(1)=0;ii=1;

for ii = 2:1:300freq_ratio(ii)=freq_ratio(ii-1)+0.01;

phase(ii,jj)=(180/pi)*atan2(2*zeta(jj)*freq_ratio(ii),(1-freq_ratio(ii)^2));resp_ratio(ii,jj)=freq_ratio(ii)^2/sqrt((1-freq_ratio(ii)^2)^2+(2*zeta(jj)*freq_ratio(ii)));ii=ii+1;

endend

figure(1)plot(freq_ratio, phase);%title('Phase verus frequency ratio plot');xlabel('Frequnecy ratio');ylabel('Phase');

figure(2)%semilogy(freq_ratio, resp_ratio);

plot(freq_ratio, resp_ratio, [1 1], [0 10], 'b-');%title('Non-dimensional response verus frequency ratio plot');axis([0 freq_ratio(end) 0 10]);

xlabel('Frequency ratio');ylabel('Non-dimensional Response');

Fig 1.11(a)Variation of the non-dimensional

response versus frequency ratio for

different damping ratios

Fig 1.11(b)Variation of the phase versus frequency

ratio for different damping ratios


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As it can be seen from Figure 1.11(b) the phase angle is 900at frequency n even for the case of damped

system. For lightly underdamped system the phase angle changes from 0 0 to 900 as the spin speed is

increased and becomes 1800 as the spin speed is increased to higher frequency ratio. For very high-

overdamped system the phase angle always remain at 900before and after the resonance, which may be a

physically unrealistic case. As the spin speed crosses the critical speed the center of the mass of the disc

of Jeffcott rotor 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 spin speed approaches infinity the displacement of the shaft tends to the

equal to the disc eccentricity. The change in phase between the force and the response is also shown in

Figure 7 for three difference spin speeds i.e. below critical speed, at critical speed and above the critical

speed. Since for the present analysis the synchronous whirl condition is assumed, at a particular speed

shaft will not be having any flexural vibration and it (in a particular bend configuration) will be whirling

(orbiting) about its bearing axis as shown in Figure 1.10(a). It can be seen that the black point in the shaft

will be having compression during the whirling. However, it can be seen from Figure 1.10(b) for

antisynchronous whirl that the shaft (the black point in shaft) will be having reversal of the bending

stresses twice per whirling (orbiting) of the shaft. For asynchronous whirl the black point inside the shaft

will take all the time different positions during whirling of the shaft.

Fig 1.12(b)Phase angle between the force and

response vectors ( )n =

Fig 1.12(c)Phase angle between the force and

response vectors ( )n >

Fig 1.12(a)Phase angle between the force and

response vectors ( )n <

CG e=/ 2

imbF imbF

imbF


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Matrix Method

With the development in software, which can handle complex matrices, the following procedure may be

very helpful for numerical simulation of even the very complicated rotor systems also. Equations (1.18)

and (1.19) can be combined in the matrix form as

=

+

+

tem

tem

u

u

k

k

u

u

c

c

u

u

m

m

y

x

y

x

y

x

sin

cos

0

0

0

0

0

02

2

(1.44)

The force vector in equation (1.44) is expressed as

( )( )

( )

( )

222

22 2

cos sincosRe Re Re

sin cossin

x

y

j t

imb j t

j timb

me e Fm e t j t m e te

Fm e t j t m e t me je

+ = = =

(1.45)

xy imbimb jFF = (1.46)

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

F andyimb

F are the

imbalance force components in x and y directions respectively. On substituting equation (1.45) into

equation (1.44) and henceforth for brevity the symbol Re(.) will be removed and it can be written as

tj

imb

imb

y

x

y

x

y

x

eF

F

u

u

k

k

u

u

c

c

u

u

m

m

y

x

=

+

+

0

0

0

0

0

0

(1.47)

The relationship (1.46) is true for the present axis system and the direction of whirling of the imbalance

force vector chosen as shown in Figure 1.13(a). For this case theyimb

F lag behind theximb

F by 900. Let us

derive this relationship: Ifximb

F Fe = , then

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

j j j j j

imb imbF Fe Fe e Fe j jFe jF = = = + = =

For the direction of whirl opposite to the axis system, as shown in Figure 1.13(b), the following

relationship will hold

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

j j j j j

imb imbF Fe Fe e Fe j jFe jF += = = + = =

so that,


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xy imbimb jFF = (1.48)

in which case theyimb

F lead theximb

F by 900. It should be noted in equation (1.49) that the right hand

side force vector elements have significance of real parts only, which is quite clear from the expandedform of the force vector in equation (1.45). Equation (1.47) can be written in more compact form as

[ ]{ } [ ]{ } [ ]{ } { } tjimb eFuKuCuM =++ (1.49)

The solution can be chosen as

{ } { } tjeUu = (1.50)

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

The above equation gives

{ } { } { } { }2 andj t j tu j U e u U e = = (1.51)

On substituting equations (1.50) and (1.51) into equation (1.49), we get

[ ] [ ] [ ]( ){ } { }imbFUCjKM =++ 2 (1.52)

The above equation can be written as

[ ]{ } { }imbFUZ = (1.53)

with

[ ] [ ] [ ] [ ]( )CjKMZ ++= 2 (1.54)

Fig 1.13(a)The direction of whirl same as the

positive axis direction

Fig 1.13(b)

The direction of whirl opposite to the

positive axis direction

t

v

t

v

imbxFimbxF

imbyFimbyFimbFimbF

y xF jF= y xF jF=


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

{ } [ ] { }imbFZU1

= (1.55)

The above method is quite general in nature and it can be applied to multi-dof systems once equations of

motion in the standard form are available.

Example 1.3: Obtain the response of a rotor system with the following equations of motion.

temkuum xx cos2=+ and temkuum yy sin

2=+

Solution: The first equation can be written as

tj

xxx eFkuum =+ with 2meFx =

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

tj

xx eUu =

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

tj

xx eUu 2=

On substituting in equation of motion, we get

( ) 22 mekUUm xx =+

which gives

2

2

mk

meUx

=

Hence the solution becomes

2 2 2

2 2 2Re (cos sin ) cosj tx

me me meu e t j t t

k m k m k m

= = + =

Similarly for the second equation of motion can be written as

tj

yyy eFkuum =+ with 2jmeFy =

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


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tj

yy eUu =

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

tj

yy eUu 2=

On substituting in equation of motion, we get

( ) yyy FkUUm =+ 2 which gives 2mkF

U y

y

=

Hence the solution becomes

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 mej t t t

k m k m

= = = +

= + =

(Answer)

rtiwari rd book 01a

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