1.4 Symmetrical rigid shaft in flexible anisotropic … Symmetrical rigid shaft in flexible...

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

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1.4 Symmetrical rigid shaft in flexible anisotropic bearings

In systems where the bearings are far more flexible than the shaft it is the bearings, which will have the

greatest influence on the motion of the rotor. Such rotors may be idealized as rigid rotor.

Rotor mass center - z

Imbalance force meω2 x

Rotor geometric center line B

kx/2

2l e

d ky/2

l

Rotor geometric center

A

kx/2

ky/2

Figure 1.28(a) A rigid rotor mounted on a flexible bearing

x

y

φ z z θ

Figure 1.28(b) Positive convertions of angular displacements.

It is assumed that the shaft has no flexibility, and 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


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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 various displacements i.e. x,

y, φ and θ. Hence free body diagrams and equations of motion have been obtained by giving such

displacements independent of each other.

EOM in the x and y directions are

2 cos - xme t k x mxω ω = �� and

2 sin yme t k y myω ω − = �� (114, 115)

EOM in the φ and θ directions are

2 2cos x dme d t k l Iω ω φ φ− − = �� and 2 2sin y dme d t k l Iω ω θ θ− − = �� (116, 117)

y meω2sinωt

lθ /2 z

θ lθ /2 ky lθ/2 d y

ky lθ/2

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

y meω2

G

kxx e ωt

C

y kyy

x

x

Figure 1.29(b) Free body diagram of the rotor in x-y plane


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x

meω2cosωt

d

φ 2xk lφ

y z

2xk lφ

l l

Figure 1.29(c) Free body diagram of the rotor in z-x plane

For sinusoidal vibrations, we can write

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

Substituting equation (118) into equations (114) to (117), the unbalnce response can be expressed as

2 2

2 2cos cos ; sin sin

x y

me mex t X t y t Y t

k m k m

ω ωω ω ω ω

ω ω= = = =

− −

2 2

2 2 2 2cos cos and sin sin

x d y d

me d me dt t t t

k l I k l I

ω ωφ ω ω θ ω ω

ω ω= = Φ = = Θ

− − (119)

where X and Y are linear displacement amplitudes and Φ and Θ are angular displacement amplitudes.

From the denominators of amplitudes it can be seen that the system has four critical speeds, two for

lateral vibration in the horizontal and vertical directions and two for angular vibration in the x-z and y-z

planes. Critical speeds can be written as

1 2 3 4

22

; ; ; and y yx x

cr cr cr cr

d d

k k lk k l

m m I Iω 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)

If kx < ky (i.e. 1 2cr crω ω< ) and

1crω ω< i.e when rotor is having low speed and operating below first

critical speed. From equation (119) both X and Y are positive. So rotor whirls in the same direction as the

rotation of the shaft. For 1 2cr crω ω ω< < i.e. the speed of rotation is in between the horizontal and vertical

direction critical speeds, then from equation (119) displacement X becomes negative indicating that the

rotor horizontal displacement is always in the opposite direction to that it adopted at low speeds. It

follows from this that the rotor must whirl in the opposite direction to that in which the shaft is rotating

and it called backward whirl. At higher shaft speeds 2crω ω> both X and Y are negative and with both

displacements in opposite direction to that shown in Figure 1.16 reveals that the rotor whirl once more in

the same direction as its rotation. ( phase may be different). It can have forward or backward whirl. The

orbital trajectory of the rotor must therefore take the form as shown in Figure 1.30.

Equation (122), relating to the angular motion of the rotor, is also the equation of an ellipse.This means

that there is an elliptical orbital trajectory of the rotor ends due to angular motion of the rotor. This rotor

motion is caused by the imbalance couple meω2d acting on the rotor, and it is superimposed on the lateral

motion described previously. A reversal of the direction of the orbit associated with this motion also

occurs, between two critical speeds associated with angular motion of the rotor (i.e. 3 4cr crω ω ω< < ).


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(i) Synchronous whirl,

1crω ω<

(ii) Anti-synchronous whirl,

1 2cr crω ω ω< <

(iii) Synchronous whirl,

2crω ω>

(a) Plot of x versus ωt

fplot('cos(x)', [0 2*pi], 'k');

(d) Plot of x versus ωt

fplot('-cos(x)', [0 2*pi], 'k');

(g) Plot of x versus ωt

fplot('-cos(x)', [0 2*pi], 'k');

(b) Plot of y versus ωt

fplot('sin(x)', [0 2*pi], 'k')

(e) Plot of y versus ωt

fplot('sin(x)', [0 2*pi], 'k')

(h) Plot of y versus ωt

fplot('-sin(x)', [0 2*pi], 'k');

(c) Plot of x-y

(the shaft center orbit) (f) Plot of x-y

(the shaft center orbit)

(i) Plot of x-y

(the shaft center orbit)

Figure 1.30 Whirl directions with respect to the shaft spin frequency

ω ω ω


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(a) Synchronous translation whirl (b) Synchronous conical whirl

(c) Anti-synchronous translation whirl (d) Anti-synchronous conical whirl

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

The amplitude of the force transmitted to the bearings is now different in horizontal and vertical

directions, as well as at each end of the rotor.The force transmitted is that which causes the bearings to

deform and is given by the product of spring stiffness and rotor deflection at the bearing. The bearing

force amplitudes are

( ) ( ); and 2 2

yxx y

kkF x l F y lφ θ= ± = ± (125)

in horizontal and vertical direction respectively. The + sign refers to the angular motion of the rotor

causes the rotor end 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 the system is

operated at the critical speeds, where x, y, φ and θ are maximum.

Example 1.8. A long rigid symmetric rotor is supported at ends by two identical bearings. Let the shaft

has the diameter of 0.2 m, the length of shaft is 1 m and the mass density of the shaft material equal to

7800 kg/m3. The bearing dynamic characteristics are as follows: kxx = kyy = 1 kN/mm with rest of the

ω

ω ω

ω

ω

ω ω

ω


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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 zero and no gyroscopic effect is

considered, hence single plane motion can be considered one at a time. For the present analysis there is no

coupling is considered between the linear and rotational displacements and since stiffnesses in x and y

direction is same hence natural frequencies in these directions can be written as

1,2

62 2 1 1090.34 rad/s

245.04

k

mω × ×

= ± = ± = ±

Similarly natural frequencies corresponding to the titling motion can be written as

2 2 2

3,4

1 10 1154.184 rad/s

2 2 21.0326d

kl

Iω

× ×= ± = ± = ±

×

Example 1.9. (a) Find the bending critical speed of the system shown in Figure 1.32 in which the disc is

made of solid steel with a diameter of 127 mm, and a thickness of 25.4 mm. The mass density of the steel

is 7800 kg/m3. The disc D is placed in the middle of a shaft and the total length of the steel shaft between

bearings is 508 mm, and its diameter is 12.7 mm. The bearings have equal flexibility in all directions, the

constant for either one of them being k = 175 N/m. (b) Solve the same problem as part (a) except that the

bearings have different vertical and horizontal flexibilities: khoz = 175 N/m and khoz = 350 N/m for each of

the bearings. Neglect the cross-coupled stiffness coefficients and the mass of the shaft and treat it as rigid.

Figure 1.32 A rigid rotor on flexible support

Solution (2) (a) EOM can be written as

20 and 0x d xmx k x I k lφ+ = + =��

Since 2xk k= , we have

Z

Y

X

D B2 B1


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22 0 and 0dmx kx I klφ φ+ = + =��

The mass and mass moment of inertia of the disc are given as

2π7800 (0.127) (0.0254) 2.51 kg

4m = × × = ; ( )21 12 2

2 22.51 0.127 / 2 0.00504 kg-mdI mr= = × × =

The natural frequencies are given as

1 2

22 211.836 rad/s and 133.87 rad/sn n

k kl

m mω ω= = = =

2(b) With different stiffness properties in horizontal & vertical directions, the natural 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

Example 1.10: Find critical speeds of a rotor system as shown in Figure 1.33.

m =10 kg, Id = 0.1 kg-m2

2 m kx

A

ky

B

Rigid rotor

kx

A

ky

A

Figure 1.33 A rigid rotor on flexible bearings

Take the bearing stiffness properties as: 1.1 kN/mmAx

k = ; 1.8 kN/mmAy

k = ; 3.1 kN/mmB

kx = and

3.8 kN/mmBy

k = .


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Solution: Equations of motion in x & y directions are

2 2cos and sinA B A Bx x y yme t k x k x mx me t k y k y myω ω− − = − − =��

Equations of motion in θ and φ directions are

2 2 2 2 2 2sin and sinA B A By y d y y dme d t k l k l I me d t k l k l Iω θ θ θ ω φ φ θ− − = − − =��

The steady state force vibration responses can be obtain as

2

2

2cos ; sin

2A B A Bx x y y

me mex t y t

k k m k k m

ω ωω ω

ω ω= =

+ − + −

( ) ( )2 2

2 2 2 2

ω ωcosω ; sinω

ω ωA B A Bx x d y y d

me d me dt t

k k l I k k l Iφ θ

−

= =+ + −

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 16480.7 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

lp

I

lp

I

k k

k k

k k

k k

+ ×= = =

+ ×= = =

+ × ×= = =

+ × ×= = =

+

+

+

+

Answer

Exercise 1.3. Obtain the bending critical speeds and mode shapes of a rotor, consist of massless rigid

shaft (1 m of span with 0.7m from the disc to the left bearing) and a rigid disc (5 kg), supported by

flexible bearings as shown in Figure E1.3 Consider the motion in vertical plane only. Is there is any

difference in critical speeds when the disc is placed at the center of the rotor? If NO then justify the same

and if YES then obtain the same. Take Id = 0.1 kg-m2.


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Disc

massless rigid shaft

20 kN/m

bearings

Figure E1.3 Exercise 1.4: Consider a long rigid rotor, R, supported on two identical bearings, B1 and B2, as shown in

Figure 1. Take the direct damping and the cross-coupled stiffness and damping coefficients of both

bearings negligible. The direct stiffness coefficients of both bearings in the horizontal and vertical

directions are equal i.e. K. The mass of the rotor is m, the span of the rotor is l, the dimetral and polar

mass moment of inertia are, respectively, Id and Ip and ω is the shaft rotational frequency. Derive

equations of motion and obtain the natural frequencies of whirl. Give a brief discussion of results.

Z

Y

X

R B2 B1

Figure E1.4


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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 associated damping properties as well as

spring stiffness properties. Furthermore in case of hydrodynamic bearings the shaft motion in the

horizontal direction is coupled with that in the vertical direction. However, coupling between the

translational and tilting motion has not been considered since the rotor is symmetric. In most applications

the properties of such bearings are described in terms of the eight linearised bearing stiffness and damping

coefficients. Symmetrical rigid shaft in flexible anisotropic bearings will be identical to Figure 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)

in horizontal and vertical directions and about the y and x axes passing through the geometrical center,

respectively. kxx, kxy, etc are eight linearised bearing stiffness and damping coefficients of both bearings

(twice those for one bearing, assuming a symmetrical system). 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 meω2 is located same distance from the rotor geometrical center. Out of balance

forces in the horizontal and vertical directions may then be written as

( ) ( )2 2 j jcos Re Ret t

x xF me t me e F eω ωω ω ω= = = with 2 xF meω=

( ) ( )2 2 j jsin Re j Ret t

y yF me t me e F eω ωω ω ω= = − = with 2 jyF meω= − (127)

where xF and yF are complex forces (which contains amplitude and phase informations) in the x and y

directions. These forces are acting at the center of gravity. The moments about the rotor geometrical

center caused by these forces are

( ) ( )2 2 j jcos Re Ret t

xz xzM me d t me de M eω ωω ω ω= = = with 2 xzM me dω=


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( ) ( )2 2 j jsin Re j Ret t

yz yzM me d t me de M eω ωω ω ω= = − = with 2 -jyzM me dω= (128)

where xzM and yzM are complex moments (which contains amplitude and phase informations) about the

x and y directions. The response can be assumed as

j j j j; ; ; =t t t tx Xe y Ye e eω ω ω ωφ θ= = = Φ Θ (129)

where X, Y, Φ and Θ are complex displacements. Equations of motion (126) can be written as

[ ]{ } [ ]{ } [ ]{ } { }( )M x C x K x f t+ + =�� (130)

with

[ ] [ ] [ ]

{ } { }

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 km

c c k kmM K

l c l c l k l kI

l c l c l k l kI

Fx

Fyx t f t

Mφθ

= = =

= = yzM

The response takes the following form

{ } { } { } { } { } { }j j 2 j; so that j and t t tx X e x X e x X eω ω ωω ω= = = −� �� (131)

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

[ ] [ ] [ ]( ){ } { }2 jM C K X Fω ω− + + = (132)

with

{ } { };

x

y

xz

yz

FX

FYX F

M

M

= =

Φ Θ

which can be written as


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[ ]{ } { }D X F= with [ ] [ ] [ ] [ ]( )2 jD K M Cω ω= − + (133)

The response can be obtained as

{X} = [D]-1 {F} (134)

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 iX X X Y Y Y= + = + Φ = Φ +Φ Θ = Θ +Θ (135)

and corresponding phase lag will be given by

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

r r r r

X Y

X Yα β γ δ

Φ Θ= = = = Φ Θ

(136)

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

j j and t tx Xe y Yeω ω= =

and in general will be found to take the form as shown in Figure 1.34.

Instantaneous force on the rotor

y

φ - Phase lag angle

x

Instantaneous rotor

displacement

Orbital trajectory

Figure 1.34 Rotor whirl orbit

The form of the orbit is still elliptical but the major and minor axes no longer line up with the bearing x

and y axes. The force vectors of equation (127) are also shown on the diagram and can be seen to precede

the displacement vectors.


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Since in the present case the coupling is considered between the vertical and horizontal planes and no

coupling is considered between the translational and tilting motions, it can be observed that first two set

of equations from equation (126) (without damped) is similar in form as the case when we considered no

coupling between the vertical and horizontal planes and coupling is considered between the translational

and tilting motion (i.e. flexible shaft with a disc away from mid-span). Mathematically we can write:

11 12, , xx xyk k k k= = � and rest of the analysis and interpretations are similar to as discussed previously.

The similar analyses can be obtained by considering last two set of equations (126).

Example 1.11. 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 shown in Figure 128. The bearing properties are: kxx = 2.0×104 N/m, kyy = 8.8×10

4

N/m, kxy = 1.0×103 N/m, kyx = 1.5×10

3 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 the unbalance response (amplitude and phase) at bearing locations when the

radial eccentricity of 0.1 mm and axial eccentricity of 1 mm is present in the rotor and locate critical

speeds.

Solution: Figure 1.35 shows the unbalance responses both for the linear and angular displacements. Both

the amplitude and phase has been plotted. It can be observed that in the plot of linear and angular

displacement two peaks appears and they correspond to the critical speeds of the system. Since the linear

and angular displacements are uncoupled for the present case and hence corresponding critical speeds

appears in respective plots. There are four critical speeds: 70 rad/s, 120 rad/s, 480 rad/s and 920 rad/s ?.


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Figure 1.35 Amplitude and phase variation with respect to spin speeds

%----------------------------------------------------------------------------

% Response plot for “Symmetrical rigid shaft in flexible anisotropic bearings “

% with damping and cross coupling. 22-1-2005

% Copywrites: Dr R Tiwari, Dept of Mechanical Engg., IIT Guwahati.

clear all;

%Rotor mass and diametral mass moment of inertia

rotor_mass=5; % kg

rotor_id=0.1; %kg-m2

radial_ecc=1.0e-4; %m

axial_ecc=1e-3; %m

%Bearing Stiffness and damping properties

kxx=2e4; kyy=8.8e4; kxy=1e3; kyx=1.5e3;%N/m

cxx=1; cyy=1; cxy=1e-1; cyx=1e-1;%N-s/m

l_span=1; %m Half the length of span.

ii=0;

for omega=1:0.5:1200; %rad/s

ii=ii+1;


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Fx=rotor_mass*radial_ecc*omega^2; %N

Fy=-j*Fx; %N

Mxz=rotor_mass*radial_ecc*(omega^2)*axial_ecc; %Nm

Myz=-j*Mxz; %Nm

F=[Fx Fy Mxz Myz].';

M=[rotor_mass 0 0 0;

0 rotor_mass 0 0;

0 0 rotor_id 0;

0 0 0 rotor_id];

C=[ cxx cxy 0 0;

cyx cyy 0 0;

0 0 cxx*l_span^2 cxy*l_span^2;

0 0 cyx*l_span^2 cyy*l_span^2];

K=[ kxx kxy 0 0;

kyx kyy 0 0;

0 0 kxx*l_span^2 kxy*l_span^2;

0 0 kyx*l_span^2 kyy*l_span^2];

D=K-omega^2*M+j*omega*C;

X(:,ii)=D\F;

omega_vector(ii)=omega;

end

figure(1)

semilogy(omega_vector,abs(X(1,:)), 'k-');

xlabel('w ---->');

ylabel('y ---->');

figure(2)

semilogy(omega_vector,abs(X(3,:)), 'k-');

xlabel('w ---->');

ylabel('phi ---->');

figure(3)

plot(omega_vector,phase(X(1,:)), 'k-');

xlabel('w ---->');

ylabel('phase (y) ---->');

figure(4)

plot(omega_vector,phase(X(3,:)), 'k-');

xlabel('w ---->');

ylabel('phase (phi) ---->');

-----------------------------------------------------------

Exercise Problem 1.5. Find the critical speeds of the rotor bearing system shown in Figure E1.5. The

shaft is made of steel with Young’s modulus E = 2.1 (10)11 N/m

2 and uniform diameter d = 10 mm. Treat

the shaft as rigid and massless. The mass of the disc is: md = 1 kg with negligible diamentral mass


57

moment of inertia. Bearings B1 and B2 are identical bearings and having the following properties: kyy =

1.1 kN/m, kzz = 1.8 kN/m, kyz = 0.2 kN/m and kzy = 0.1 kN/m. The various shaft lengths are as follows:

B1D = 75 mm, and DB2 = 50 mm.

Figure E1.5

Exercise Problem 1.6. Obtain the critical speeds for transverse vibrations of rotor-bearing system 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 7800 kg/m3. The shaft is supported at ends by flexible bearings. Consider the motion

in both the vertical and horizontal planes. Take the following bearing 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 = 400 kN-s/m,

and 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, which deform the bearing lubricant film, and

do not include rotor inertia terms. In general bearing forces will lag behind the imbalance force such that

the bearing horizontal and vertical force components, at one end A of the machine, 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

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

φ θ φ θ

φ θ φ θ

= + + + + + + +

= + + + + + + +

= + + + + + + +

= + +

� ��

� ��

� ��

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

(137)

where 2, 2A xx xx B xx xxk k k k= = , etc. So equation (137) is for more general case, which can be written

in matrix form as

{ } [ ]{ } [ ]{ }b b bf c x k x= +� (138)

Z

Y

X

D B2 B1


58

with

{ } { } { }

[ ] [ ]

; ; ;

;

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 yf 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 φθ 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

For unbalance excitation with frequency ω, the bearing responses and forces can be expressed as

{ } { } { } { } { } { }j j j; j ; t t t

b bx X e x X e f F eω ω ωω= = =� (139)

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

{ } [ ] [ ]( ){ }jb b bF k c Xω= +

This can be used to evaluate the bearing forces. The amplitude of forces transmitted to the bearings are

then given by

2 2 2 2 2 2 2 2; ; ;r i r i r i r iA 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+ + + += = = =

(140)

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 Fε ς η λ− − − −

= = = =

(141)


59

1.6 Asymmetrical flexible shaft in flexible anistropic bearings with damping and cross coupling.

Figure 1.36 A flexible shaft in flexible bearings

For the present case both the shaft and bearings are flexible as shown in Figure 1.36. The analysis must

allow 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 previous section, 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 first evaluated, which allows for the flexibility of the shaft in

addition to that of bearings, and is used in place of the bearing coefficients in previous section analysis.

The total deflection of the is the vector sum of the deflection of the disc relative to the shaft ends plus that

of the shaft ends in the bearings. For disc we observe the displacement of geometrical center of the disc.

The deflection of the shaft ends in the bearings is related to the force transmitted through bearings by the

bearing stiffness and damping coefficients 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

= + + +

= + + +

� �

� � (142)

where m and n are instantaneous displacements of shaft ends relative to bearings in the horizontal and

vertical directions respectively, and take the form

j j; t tm Me n Neω ω= = (143)

which can be differentiated once with respect to time to give

j jj ; jt tm Me n Neω ωω ω= =� � (144)

Rigid foundation

Bearings

Shaft Disc


60

The bearing forces have the following form

j j; t t

bx bx by byf F e f F eω ω= = (145)

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

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

= + + +

= + + + (146)

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

{ } [ ]{ }bF K V= (147)

with

{ } [ ]

( ) ( )( ) ( )

( ) ( )( ) ( )

{ }

j j 0 0

j j 0 0;

0 0 j j

0 0 j j

xx xx xy xyA AA bx

yx yx yy yyA by A A

b

B bxxx xx xy xyB B

B by

yx yx yy yyB B

T

A A B B

k c k cF

k c k cFF K

F k c k c

Fk c k c

V M N M N

ω ω

ω ω

ω ω

ω ω

+ + + + = = + + + +

=

The magnitude of the reaction forces transmitted by the bearings can also be evaluated 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).

Fy

d Myz

l

Afby Bfby


61

Figure 1.37 Free body diagram of the shaft

0 ( )B A by y yzM l f F l d M= ⇒ = − −∑ or ( ) ( )1 1A by y yzf d l F l M= − − (148)

0 A B by y yzM l f dF M= ⇒ = +∑ or ( ) ( )1B by y yzf d l F l M= + (149)

Similarly forces in the horizontal direction may be written as

( ) ( )1- 1A bx x xzf d l F l M= + (150)

( ) ( )1B bx x xzf d l F l M= + (151)

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

{ } [ ]{ }b sf A f= (152)

with

{ } { } [ ]

( )( )

1 / 0 1/ 0

0 1 / 0 1/; ;

/ 0 1/ 0

0 / 0 1/

A bx x

A by y

b s

B bx xz

B bxy yz

f F d l l

f F d l lf f A

f M d l l

f M d l l

− − − = = =

For an unbalance excitation, we have

{ } { } { } { }j j and t t

b b s sf F e f F eω ω= = (153)

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

{ } [ ]{ }b sF A F= (154)

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 by the disc. Equating equation (147) and (154), we get

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

Above equation relates the shaft end deflections in terms of reaction forces and moments by the disc on

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


62

follows. Consider the shaft to be rigid for some instant and assuming shaft end deflection in horizontal

direction be Am and Bm at ends A and B, respectively as shown in Figure 1.38.

l B

d

A x

z

Am

Bm

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

( )1-

B A

A A B

m m d dx m d m m

l l l

− = + = +

(156)

Slope in x-z plane of the shaft will be

( )-B Am m lφ = (157)

Similarly for motion in y- direction and y-z plane, we have

( ) ( )1- A By d l m d l m= + (158)

( )-A Bn n lθ = (159)

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

{ } [ ]{ }1s

u B v= (160)

with

{ } { } [ ]

( )( )

1

1- / 0 / 0

0 1- / 0 /; ;

1/ 0 1/ 0

0 1/ 0 1/

A

A

s

B

B

mx d l d l

ny d l d lu v B

m l l

n l l

φθ

= = = − −

For unbalance excitation (or for free vibration analysis), shaft displacements at bearing locations and at

disc center vary sinusoidally such that


63

{ } { } { } { }1 1

j j and t t

s su U e v V eω ω= = (161)

On substituting in equation (160), we have

{ } [ ]{ }1s

BU V= (162)

Substituting equation (155) in above equation, we get

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

1

s sB A sU K F C F−

= = (163)

which gives the deflection of the disc due to unbalance when shaft is rigid. Above equation will give

deflection of the disc (geometrical center) that is caused by only by movement of the shaft ends (rigid

body movement ) in flexible bearings. In order to obtain the net rotor deflection under a given load, we

have to add the deflection due to deformation of the shaft to that which have been calculated in equation

(163). The deflection associated with flexure of the shaft alone has already been calculated i.e. in x-

direction & x-z direction plane and in y-direction & y-z plane, respectively

11 12

21 22

x

xy

Fx

M

α αα αφ

=

and

11 12

21 22

y

yz

Fy

M

α αα αθ

=

(164)

Equation (164) can be combined in a matrix form as

{ } [ ]{ }2s su fα= (165)

with

{ } { } [ ]2

11 12

21 22

11 12

21 22

0 0

0 0; ;

0 0

0 0

X

y

s s

xz

yz

Fx

Fyu f

M

M

α αα α

αα αφα αθ

= = =

For unbalance excitation (or for free vibration analysis), shaft reaction forces at disc location and

displacement at disc center vary sinusoidally such that

{ } { } { } { }2 2

j j and t t

s s s su U e f F eω ω= = (166)


64

On substituting above equations in equation (165), we get

{ } [ ]{ }2s sU Fα= (167)

which is the deflection of disc due to shaft flexure of the shaft alone without considering bearing

flexibility. The net deflection of the rotor that caused by deflection of the bearings plus that due to flexure

of the shaft is then given by

{ } { } { }1 2s sU U U= + [ ] [ ]( ){ } [ ]{ }s sC F D Fα= + = (168)

The above equation describes the displacement of the shaft at the disc under the action of sinusoidal

forces and moments applied at the disc (identical to the influence coefficient matrix). Equation (168) can

be written as

{ } [ ] { } [ ]{ }1

sF D U E U−

= = (169)

EOM of the disc can be written in the x-direction and on the x-z plane, as

2 cos xme t F mxω ω − = �� and - xz dM I φ= �� (170)

Similarly the EOM in the y-direction and the y-z plane

2 sin xme t F myω ω − = �� and - yz dM I θ= �� (171)

x meω2cosωt y meω2

sinωt

Mxz Myz

φ

θ

x Fx y Fy

z z

(a) (b)


65

Figure 1.39 Free body diagram of the disc

Equations of motion of the disc can be written in matrix form as

[ ]{ } { } { }sM u f f+ =�� (172)

with

[ ] { } { } { } { }

2

2

j j

0 0 0

0 0 0 j; ; ;

0 0 0 0

0 0 0 0

x

y t t

s

xzd

yzd

Fm x me

Fm y meM u f f e F e

MI

MI

ω ω

ωω

φθ

− = = = = =

Equations of motion take the form

[ ]{ } { } { }2

sM U F Fω− + = (173)

From equation (169) above equation takes the form

[ ]{ } [ ]{ } { }2 M U E U Fω− + = (174)

which gives

{ } [ ] [ ]( ) { }1

2U M E Fω−

= − + (175)

where [ ] [ ]( ) 12 M Eω

−− + is the equivalent bearing stiffness & damping coefficients. Once response of

the rotor has been obtained the loading applied to the shaft by rotor can be obtained by equation (168)

{ } [ ]{ }( )sF E U= , which is substituted in equation (154) { }-1. . { } [ ] [ ]{ }si e V K A F= to get shaft ends

deflections {V} at each bearings, which is substituted in equation (147) { }( )[ ]{ }bF K V= to get bearing

forces. (or {Fb] = [A]{Fs} can be used directly to obtain bearing forces) equation (154). The

displacements and forces have the complex form; the amplitude and phase information can be extracted

from the real and imaginary parts. Amplitudes will be the modulus of complex numbers and phase angles

of all these parameters can be evaluated by calculating arctangent of the ratio of the imaginary to real

components as given by equations (140) and (141).


66

Exercise 1.6. Find the critical speeds of the rotor bearing system shown in Figure E1.6. The shaft is made

of steel with Young’s modulus E = 2.1 (10)11 N/m

2 and uniform diameter d = 10 mm. Treat the shaft as

flexible and massless. The mass of the disc is: md = 1 kg with negligible diamentral mass moment of

inertia. Bearings B1 and B2 are identical bearings and having the following properties: kyy = 1.1 kN/m, kzz

= 1.8 kN/m, kyz = 0.2 kN/m and kzy = 0.1 kN/m. The various shaft lengths are as follows: B1D = 75 mm,

and DB2 = 50 mm.

Figure E1.6 A flexible rotor mounted on a flexible bearings

Exercise Problem 1.7. Obtain the critical speeds for transverse vibrations of rotor-bearing system as

shown in Figure E1.7. Consider shaft as flexible and whole mass of the shaft is assumed to be

concentrated at its mid-span. The shaft is of 1 m of span and the diameter is 0.05 m with the mass density

of 7800 kg/m3. The shaft is supported at ends by flexible bearings. Consider the motion in both the

vertical and horizontal planes. Take the following bearing 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 = 400 kN-s/m,

and 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.

shaft

bearings

A B

Figure E1.7

X

Z

Y

D B2 B1


67

1.7 Effects of flexible foundations

In some machines bearings themselves may be mounted on flexible foundations (Figure 1.40), which may

in turn influence the motion of disc masses. The net displacement of the disc is given by the vector sum of

(i) disc displacement relative to the shaft ends, plus (ii) that of the shaft ends relative to the bearing, plus

(iii) that of bearing relative to space. The theoretical analysis of the disc, the shaft and the bearing

response and that of the force transmissibility of such a system, can be carried out in a similar manner to

that described in the previous section.

Figure 1.40 A flexible rotor-bearing-foundation system

The relationship between the force transmitted through bearings and displacements of the shaft ends is

governed by the bearing stiffness & damping coefficients and is given by (from equation 147 of previous

section)

{ } [ ]{ }bf K v= or { } [ ]{ }bF K V= (176)

where {v} is the shaft end displacement relative to the bearing. For obtaining a relationship between

displacements of bearings on their foundations and the force transmitted through bearings.

Bearings

Foundations

Shaft Disc


68

Force being transmitted Bearing

to bearing through shaft

kfx

fbx

cfx

mb

cfy kfy Bearing mounting

system

Figure 1.41 A bearing mounted on foundation

The bearing will respond in the horizontal direction for an external force fbx by the following equation

bx fx fx bf k a c a m a− − =� �� (177)

where a is the horizontal displacement of the bearing, mb is the bearing mass of one bearing and kfx, cfx,

kfy, cfy are foundation stiffness & damping coefficients. Similarly the response of the bearing in the

vertical direction to a force fby is

by fy fy bf k b c b m b− − =� �� (178)

where b is the bearing vertical displacement. The displacement of the bearing will take the form

j j and t ta Ae b Beω ω= = (179)

Substituting in equations (179), equations of motion (177) and (178) can be combined in the matrix form

as (for bearing A)

[ ]{ } { }A A A bD e F= (180)

with

[ ] { } { }20 00

j ; and 0 00

x x x

y y y

f f bb

A A A b

f f bb A A

k c Fm AD e F

k c Fm Bω ω

= − + = =


69

For both bearings equations of the form as equation (180), can be combined as

{ }{ }

[ ][ ]

{ }{ }

0

0

A b A A

B b B B

F D e

F D e

=

or { } [ ]{ }bF D E= (181)

which gives relative displacements between the bearing and foundation, as

{ } [ ] { }1

bE D F−

= (182)

The total displacement of the shaft ends under the action of an applied force {Fb} is given by summation

of individual displacements {V} and {E} i.e.

{ } { } [ ] { } [ ]{ }1 1{ } [ ] b bW V E K D F K F− − ′= + = + = (183)

where [ ]K ′ is a system equivalent stiffness matrix describing the overall shaft support characteristics and

allows for the flexibilities of both the bearings and their foundations. The study of the disc motion may

now proceed in the same manner as indicated in previous section except the equivalent stiffness matrix

[ ]K ′ should be substituted for [ ]K . Once the disc displacement vector {U} is known, it is possible to

substitute back and obtain {Fs}, {V} and {Fb} as in the previous section. The force transmitted to the

foundation are given as

fx fx fxf k a c a= + � and fy fy fyf k b c b= + � (184)

For unbalance excitation or free vibration analysis, we have

j j and t t

fx fx fy fyf F e f F eω ω= = (185)

On substituting above equations into equation (184), we get

0 0

0 0

x x x

y x x

f f f

f f f

F c c Aj

F c c Bω

= +


70

The force transmitted through the foundation will not be same as force transmitted through the

bearings, because bearing mass (i.e. the inertia force) will absorb some force towards its

acceleration. If bearing mass is not there then bearing and foundation will transmit same amount

of force , however, may be with some phase lag. Amplitude and phase of the force transmitted

through the foundation can be obtained from 1 2 1, , fx fx fyF F F and

2fyF as usual.

1.4 Symmetrical rigid shaft in flexible anisotropic … Symmetrical rigid shaft in flexible...

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