rtiwari rd book 05a

8/10/2019 Rtiwari Rd Book 05a

1/22

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

CHAPTER 5

GYROSCOPIC EFFECTS IN ROTORS

5.1 Synchronous motionFigure 1 shows the motion of a disc when the disc is at the mid-span and away from it. For the latter

case the precession of the disc takes place along with the spinning, which leads to the gyroscopic

couple on the disc. The critical speed of a rotor system with disc having point mass and with

appreciable mass moment of inertia are not the same. This is due to the fact that centrifugal forces of

the particles of the disc do not lie in one plane and thus from a couple tending to straighten the shaft

as shown in Figure 5.2. We are considering the case of completely balanced rotor and whirling at its

critical speed in slightly deflected position. The angular whirl velocity vof the centre of the shaft is

assumed to be same as the angular velocity of rotation as of shaft (i.e. spin speed). This implies that

a particular point of the disc which is outside in Figure 5.2(b) will always be outside; the inside point

will always remains inside; it is called the synchronous motion. The shaft fibers in tension always

remain in tension while whirling, and similarly the compression fibers always remain in compression.

Thus any individual point of disc moves in a circle in a plane perpendicular to the undistorted centre

line of the shaft.

(a) When disc is at middle of the span, (b) When disc is at near the bearing,it will whirl in its own plane it will not whirl in its own plane

Figure 5.1 Disc motion in rotor systems

(a) (b)Figure 5.2 Critical speeds of case (a) and case (b) are not equal, even if the shafts

are identical and the masses at the end are equal


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B

r1 S, G

r 2r dm

2dm

2r1dm

Figure 5.3 Frequency of whirl vis equal to frequency of spin (i.e. synchronous whirl)

Centrifugal force of a mass element dm is 2r1dmand is directed away from point B. Component in

vertical direction is 2dm and is directed vertical down (added together will give a force and no

moment.) and component in radial dir is 2

rdmand is directed away from the disc from shaft center

or shaft centre S (components added together will give no force and only moment) (i.e. two force

vectors one when dmis assumed to be rotated about S and second force when S itself is rotating about

B). The force 2dmfor various masses add together will give m

2(where mis the total mass of the

rotor) acting downward at S (since S and G are same).

2rdm

2ydm y y

2ydm

r Shaft y

x S

y= rcos

Figure 5.4 Centrifugal force on a particle of the disc

The force 2rdmall radiate from the center of the disc S. They-component of

2rdmis 2ydmand the

moment arm of this elemental force isyand x-component will balance themself. Thus the moment of

the centrifugal force a small particle dm is (2ydmy) and total momentMof centrifugal forces is


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2 2 2 2 2

dy ydm dm I = = = (5.1)

where Idis the area moment of inertia of the disc about one of its diameter. Thus end of the shaft is

subjected to a force m2 and to a moment Id2 , under the influence of this it assumes a linear

deflection and an angular deflection . This can happen only at a certain speed i.e. critical speeds.

Thus the calculation of critical speed is reduced to a static problem, namely that of finding at which

value of a shaft will deflect and under the influence of P = m2andM =

2Id.

P

EI

M

l

Figure 5.4 A cantilevered beam with loadings at free end

The linear angular displacement of the free end of the cantilevered beam as shown in Figure 5.4 will

be

( ) ( )2 3 2 2233 2 3 2

dm l I l

MlPl

EI EI EI EI

= = (5.2)

( ) ( )2 2 222 3

dm l I l

MlPl

EI EI EI EI

= = (5.3)

Equations (5.2) & (5.3) can be rearranged as

232 2

22 2

13 2

1 12

0

0

d

d

l lm I

EI EI

l lm I

EI EI

+

+ =

+ =

(5.4)

This homogeneous set of equations can have a solution for and only when the determinantvanishes

232 2

22 2

13 2

12

0

d

d

l lm I

EI EI

l lm I

EI EI

+

=


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which gives

24 2

3 4

2 212

3

120d

dd

EI ml ImI l

E I

mI l

+ = (5.5)

Defining

3

kl

EI= as the critical speed function.

and2

dI

Dml

= as the disc effect (5.6)

Equation (5.5) can be written as

4 2 4 1212 0k k

D D

+ = (5.7)

With the solution

2

2 2 2 126 6kD D D

= + (5.8)

For which only + sign will give a positive value for k2or a real value for k. Plot of k2versusDis given

in Figure 5.5.

12

32 2mlk

EI

=

EI

ml

32 12=

9

3

EI

ml

32 3=

0 1 2 3 4

2dID

ml=

Figure 5.5 Variation of the critical speed function with the disc effect


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Case I: For disc effectD= 0 (i.e. concentrated mass of the rotor) from equation (5.7), we have

4 2 24 12 0Dk k Dk+ =

For D = 0 above equation gives

2 2 124 12 0 34

k k = = =

Noting equation (5.6), we get

2 32

3

33

EI ml

EIml

= = (5.9)

Equation (5.9) gives critical speed of a point-mass disc for the overhang case.

Case II: For D(i.e. a disc for which all mass is concentrated at a large radius) no finite angle is

possible, since it would require an infinite torque, which shaft cannot furnish. The disc remains

parallel to itself and the shaft is much stiffer than without the disc effect (i.e. D =0). From equation

(5.7) for D

( )3

2 2 2 2 2

3

1212 0 0 hencesince 12 12 or k k k

EI

EI mlk

ml = = = = (5. 10)

which is the critical speed forId.

Case III:When the end disc of an overhang rotor is having considerable amount of length, as shown

in Figure 5.6, the couple of centrifugal forces for this case is such, it tries to push away from home

angularly as shown in Figure 5.7. For the thin disc case the couple of centrifugal forces tries it to urge

back home angularly.

Rigid mass attached at endof cantilever beam.

Figure 5.6 A cantilevered rotor with a long stick at free end


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y 2ydm

(2) x (1)

D

x

Gb

(x, y)

Figure 5.7 Centrifugal force in a cantilevered rotor with a long stick at free end

Here it is assumed that the centre of gravity G of the body is placed already in the axis of rotation x

(i.e. =0). So there is no total centrifugal force (m2) and only a moment. The force on a particle is

2ydm and its moment arm about G is x, so that the moment is

dM= 2xydm (5.11)

and for the whole body

2M xydm= (5.12)

For thin discx=y (see equation (5.1) i.e. dM= 2ydm). 1 and 2 are principal axes. Let moment of

inertia about those axes be I1 and I2.This set of axes is at angle with respect to the x-yaxes. The

product of inertia aboutx-yaxes is

( )1 2 1 2sin22

I Iydm I I

= (5.13)

For thin discI1 = 2IdandI2 =Idso that equations (5.13) and (5.12) givesM = 2Id, which is same as

equation (5.1). For a disc of diameterDand thickness b, we have

2 2 2

1 216 12and8

mD mD mbI I= = + (5.14)

Substituting equations (13) and (14) into equation (12), we get

( )2 2

2 2

1 216 12

mD mbM I I

= = (5.15)


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For thin stick case it is assumed that the shaft extend to the centre of the cylinder without interference.

If shaft is attached to the end of the cylinder, the elastic-influence coefficients are modified. The

phenomenon described is generally referred to as a gyroscopic effect.

5.2 Asynchronous Motion (Rotational Motion only)

Now consider small rotor is suspended practically at its centre of gravity by three very flexible

torsional springs as shown in Figure 5.8.

Torsional

spring

Motor Disc

Angular

momentum

vectorIpO .

Motor Disc

(a) Motor on torsional springs (b) A rotor system with motor suspension

Time rate change ofangular momentum B

Ip

C dt

A

(c) A conical whirl of the rotor

Figure 5.8 A motor-rotor system supported on torsional springs

The following system have been used: is the Spin speed of disc, is the angular velocity of whirl of

the shaft centre line, Id is the Moment of inertia of stationary & rotating parts about an axis

perpendicular to paper,Ip is the Moment of inertia of rotating parts about shaft axis,Kis the torsional

spring of shaft system. We want to calculate natural frequencies of modes of motion for which the


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centre of gravity O remain at rest and the shaft whirls about O in a cone of angle 2. The disc on the

motor shaft rotates very fast, and as the springs on which the motor is mounted are flexible, the

whirling take place at a very slow rate than the shaft rotation. The angular momentum is given as

Angular momentum =Ip

(5.17)

In case whirl is in the same direction as rotation: The time rate of change of angular momentum will

be directed from B to C (i.e. out of the paper). This is equal to the moment exerted by the motor frame

on the disc. The reaction i.e. the moment acting on the motor is pointing into the paper and therefore

tends to make smaller. This acts in an addition to the existing springK, so that it is seen that the

whirl in the direction of rotation makes the natural frequency higher. In the same manner it can be

reasoned that for whirl opposite to the direction of rotation the frequency is made lower by gyroscopic

effect. To calculate the magnitude of the gyroscopic effect, we have

( ) BC BC ABOB AB OB

p

p

d Idt

I

= = =

which can be written as

( ) ppd

I Idt

= (5.18)

which gives the gyroscopic moment. The elastic moment due to the springsKis equal toKand the

total moment is equal to

(KIp) (5.19)

where the positive sign for a whirl in the same sense as the rotation and negative sign for a whirl in

the opposite sense as the rotation. In equation (5.19), the term in the parenthesis is the equivalent

spring constant, the natural frequency will be (of the whirl i.e. = n)

2 n p

n

d

K II = (5.20)

which can be written as

2 0p

n n

dd

I K

II

=


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A B

O , y

C

IpFigure 5.10 An overhang rotor

Case I:= 0 with = 0

Shaft is not rotating about its deflected center line but deflected center line OC is rotating with 0

Case II: = 0 with = 0

Shaft will be in the deflected position and rotating with 0 about the deflected centerline OC.

Case III: =

The shaft fiber in tension will remain in tension and similarly shaft fiber in compression will remain

in compression i.e. synchronous whirl.

With this combined and motion, our aim is to obtain its angular momentum. If it does not whirl,

but only rotates, the angular momentum is equal toIp (as shown in Figure 5.10) whereIp is discs

polar momentof inertia about the (deflected) shaft centerline.

For no rotation = 0, but only a whirl : The disc wobbles in space (about its diameter) and it is

difficult to visualize its angular speed. The visualization can be made easier by remarking that at C

shaft is always perpendicular to the disc always, so that we can study the motion of the shaft instead

of the disc. The line CA is tangent to the shaft at C. The piece dsof the shafting at disc moves with

line AC, describing a cone with A as an apex as shown in Figure 5.11. Speed of point C for a whirling

count clockwise seen from the right, is perpendicular to the paper into the paper and its value is y.


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O A B

C/ at time dt

C

(a) The pure whirling motion dt

O A

Id

C

(b) The angular momentum

Figure 5.11 Pure whirling motion of the rotor

The line AC lies in the paper now, but at time dtlater, point C is behind the paper by

CC/= (y)dt (5.22)

The angle between two positions of line AC then is

CC ydt

AC AC

= with

AC= (5.23)

when is small, the angle of rotation of AC in time dtis equal to dt. Hence the angular speed of

AC (and of the disc) is equal to . The disc rotates about a diameter in the plane of the paper and

perpendicular to AC at C, so that the appropriate moment of inertia isId(= Ipfor thin disc.). The

angular momentum vector of the disc due to whirl is Idand is shown in Figure 5.11(b). The total

angular momentum is the vector sum ofIpandId.


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O A B

y Id OA

Parallel to OA

C

Ip

Figure 5.12 Angular momentum due to whirling and spinning of the rotor

Now we want to calculate the rate of change of this angular momentum vector, for that purpose we

resolve the vector into components and to OA (as shown in Figure 5.12) The component toOA rotates to OA around the OA in a circle with radiusyand keeps its length during the process so

that its rate of change is zero as shown in Figure 5.13.

O A

(Ip+Id2

)

Figure 5.13 Angular momentum components parallel to the shaft undeflected position

The component to OA is a vector along the direction BC (and it is the rotating radius of a circle

with center B) as shown in Figure 5.14.

B

O A

dt

CC

Id(2-)

Figure 5.14 Angular momentum components perpendicular to the shaft undeflected position

From Figure 5.14 angular momentum components will be:


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( )cos 2p d dI I I = (5.24)

and in the direction from B to C. At time t = 0 this vector lies in the plane of the paper; at time dt this

vector goes behind the paper at angle dt (see figure). The increment in the vector (directed to

paper and into it) is the length of the vector itself multiplied by dt, and is given as

Id(2-)dt (5.25)

The rate change of angular momentum is

Id(2-) (5.26)

By the main theorem of mechanics this is the moment exerted on the disc by the shaft (i.e. by action).

The reaction, the moment exerted by the disc on the shaft is the equal and opposite i.e. a vector

directed out of the paper and perpendicular to it at C. Beside this couple there is a centrifugal force

m2yacting on the disc as shown in Figure 5.15.

y

M= -Id(2-)

F= m2y

Figure 5.15 The inertia force and couple acting from the disc on the shaft caused by a shaft rotation,

, and a shaft whirling,

Influence coefficient of the shaft can be defined as: 11is the deflectionyat the disc from 1 N force;

12is the angle at the disc from 1 N force or is the deflectionyat the disc from1 N-m moment i.e.

21= 12 (by Maxwells theorem) and 22 is the angle at disc from 1N m moment. For cantilever

beam, we have

3 2

11 12 21 22 and,3 2

l l l

EI EI EI = = = = (5.27)


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The linear and angular deflections can be expressed as

11 12 12 22andy F M F M = + = +

On substituting the force,F, and moment,Mfrom Figure 5.15, we get

( )

( )

2

11 12

2

12 22

2

and 2

d

d

y m y I

m y I

= +

= +

(5.28)

which can be rearranged as

2

11 121 ] (2 )][ [ 0dm Iy + =

and

2

12 22] 1 (2 )][ [ 0dm Iy + + = (5.29)

Equation (5.29) is a homogeneous equation inyand , on putting determinant equal to zero, we get

4 2 3 2

11 22 11 2212 12

[ ] [2 2 ]d d d d m I m I m I m I + + 2

22 11 22[ [ 0] 2 ]d dI m I + =+ +

(5.30)

Equation (5.30) contain seven system parameters: , , m, Id, 11, 12, and 22,which makes a good

understanding of the solution very difficult. It is worthwhile to diminish the number of parameters as

much as possible by dimensional analysis. Introducing four new variables:

11F m = the dimensionless frequency ;22

11

dDI

m

= the disc effect

2

12

11 22

E

= the elastic coupling and 11S m = the dimensionless speed. (5.31)

With this new four variables equation (5.30) becomes

4 3 21 25 125 0( 1) 1 ( 1)

DF F F F

D E E D E

+ + =

(5.32)


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Equation (5.32) is the fourth order polynomial inF, so for a givenE, carrying a given disc d, and

rotating at certain speed S, there will be four natural frequencies of whirl.

Case I: For a point mass of the disc. i.e.Id = 0 orD= 0. Multiply equation (5.32) byDand substitute

D= 0, we get

22

11

1 10 or 1 or 1 or 1

1

DF F F m

E E I

+ = = = =

(5.33)

11 1/ in the usual natural notation, we havesince k=

1m k

k m= =

the result is very familiar.

Case II:E= 0 (i.e. no elastic coupling) or a force causes a deflection only without rotation , while a

couple causes on rotation only without deflectiony. This is the case of shaft with simple supports

with a disc in the mid-span point as shown in Figure 5.16.

(a) (b)

Figure 5.16 A Jeffcott rotor (a) pure translation (b) pure rotational motions

Equation (5.32) reduces to

4 3 21 125 25 0D

F F F FD D

+ + + = (5.34)

Equation (5.34) can be written as

2 1( 1)( 1) 25 0F F F FD

+ = (5.35)

Equation (5.35) gives


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238

frequency only becauseF= +1 corresponding to counter clockwise whirl andF= -1 corresponding to

clockwise whirl.

Case III:Plot of the frequency equation (5.32) for the general case.

Taking a case of a disc on a cantilever shaft, for which

22

2

12

3

11 22

2

3

3

4

lEI

l lEI EI

E

= = = and taking againD= 1.

ForE= andD= 1, equation (32) becomes

4 3 22 8 8 4 0F SF F SF + + = (5.37)

Equation (5.37) is the fourth degree polynomial with four rootsF=f (S). It can be solved in two ways

(i) Take a value of Sand solve fourth degree equation inFor (ii) Take a value ofF and solve a linear

equation in Si.e.

4 2

3

8 4

2 8

F FS

F F

+=

(5.38)

ForF= 1 we get S = 0.5 etc. and it will result in Figure 5.18. Plot can be obtain for other value ofD&

E(i.e. other discs and other shafts). It is seen that for S= 0 there are only two natural frequenciesF=

0.74 andF= 2.73 corresponding to 2-Dofsy& of a non-rotating disc. When the disc rotates all

four natural frequencies are different. The curves are symmetrical about the verticalF-axis, which

means that for +Sand Sthe sameF-values occur, in other word four natural frequencies remain same

for discs clockwise rotation or counter clockwise rotation. For = line: It intersects the curve at A.

At the point A there is a forward whirl at the same speed as the rotation. Previously a plot of these

intersection points A for various values of the disc effect D is plotted (with the assumption of

synchronous motion). This kind of disturbance is obviously excited by unbalance, for synchronous

whirl. It is a resonance phenomenon and the vibration amplitude at this critical is proportional to the

amount of unbalance.


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5 =3

F 4

=3

E2

1 DA

-5 -4 -3 -2 -1 1 2 3 4 5

O S

-1 B

-2C

-3

-4 =-

-5

Figure 5.18 Whirl frequency versus spin speed variation

In summary the single mass system with gyroscopic effects have been considered for the following

cases:

Case I:

Figure 5.19 Overhang rotors with (a) point mass (b) rigid disc and (c) long stick

For = i.e. synchronous whirl one critical speed has been obtained.

Case II:

Figure 5.20 A flexibly mounted motor carrying a rotor


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It has two kinds of whirls: fast whirl (forward) in same sense as and slow whirl (backward) in

opposite sense as .

Case III:

Figure 5.21 An overhang rotor

Four whirl natural frequencies with two forward and two backward. One of the forward whirl is same

as for the case I. In Figure 5.18 B and C represent the backward whirl at the same speed as the spin

with opposite sense of rotation.

Example 5.1Obtain transverse forward and backward synchronous critical speeds of a rotor system

as shown in Figure 5.22. Take the mass of the disc, m= 10 kg, the diametral mass moment of inertia,

Id= 0.02 kg-m2, the polar mass moment of inertia,Ip= 0.04 kg-m

2. The disc is placed at 0.25 m from

the right support. The shaft is having diameter of 10 mm and total span length of 1 m. The shaft is

assumed to be massless. Take shaft Youngs modulus E= 2.1 1011N/m2. Consider the gyroscopic

effects and take two plane motions.Use the influence coefficient method.

a b

l = a + b

Figure 5.22 A rotor system

Influence coefficients are defined as:

11 12

21 22

y F

=

with

( )

( )

2 2 2 3 2

11 12

2 2

21 22

/ 3 ; 3 2 / 3

( ) / 3 ; 3 3 / 3

a b EIl a l a al EIl

ab b a EIl al a l EIl

= =

= =

Solution:

From equation (5.32), the frequency equation is given by

4 3 21 2 12 0( 1) 1 ( 1)

D SF SF F FD E E D E

+ + = (A)

we have

3

22

4

11

0.02 1.4146 100.0249

10 1.137 10

dIDm

= = =


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241

and

2 4 2

12

4 3

11 22

(3.031 10 )0.5712

1.137 10 1.4146 10E

= = =

On substituting values ofDandEin equation (A), we get

4 3 22 95.99 4.664 93.658 0F SF F SF + + = (B)

For the forward whirlF= +S, hence from equation (B)

4 4 2 2 4 22 95.99 4.664 93.66 0 91.33 93.66 0F F F F F F + + = + =

which can be solved as

22 291.33 91.33 4 93.66 91.33 93.36 92.35

2 2F F

+ = = = and 1.015

Negative value is not a feasible solution, taking positive value only, we get

4111.0075 1.137 10 10 29.88F F Fcr cr cr F m = = = = rad/sec

For the backward whirl, we haveF= -S, hence from equation (B), we get

4 4 2 2 4 22 95.99 4.664 93.66 0 33.55 31.22 0F F F F F F+ + = + =

which can be solved as

22 33.55 33.55 4 31.22 33.55 31.63

2 2F = =

96.02 =F and 32.59 F= mBcr 11 = 0.98 and 5.71

Hence two possible backward whirls are possible


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0.98 29.66 28.47B

cr = = rad/s (B1) and 169.34 rad/s (B2).

Figure 5.23 shows the corresponding Campbell diagram (not to the scale).

Exercise 5.1 Obtain the forward and backward synchronous bending critical speed of a rotor as

shown in Figure 1. The rotor is assumed to be fixed supported at one end. Take mass of the disc m= 2

kg, polar mass moment of inertia Ip= 0.01 kg-m2and diametral moment of inertiaId= 0.005 kg-m

2.

The shaft is assumed to be massless and its length and diameter are 0.2 m and 0.1 m, respectively.

Take shaft Youngs modulus E = 2.1 1011N/m2. Using the finite element method and considering

the mass of the shaft with material density = 7800 kg/m3 obtain first two forward and backward

synchronous bending critical speeds by drawing the Campbell diagram.

Figure 1

Disc

ShaftShaft

Bearings Figure E5.1

B1=28.47 rad/s

F1=29.88 rad/s

B2=169.34 rad/s

S

F

Figure 5.23 Campbell diagram

rtiwari rd book 05a

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