Rtiwari Rd Book 10

8/10/2019 Rtiwari Rd Book 10

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

CHAPTER 10

EXPERIMENTAL ESTIMATION OF DYNAMIC PARAMETERS OF

BEARINGS

One of the important factors governing the vibration characteristics of rotating machinery is the

effective dynamic stiffness of the supports as seen by the rotor as shown in Figure 10.1. The dynamic

stiffness of the support is determined by the combined effects of flexibility of the bearing, the bearing

pedestal assembly (bearing housing) and the foundations on which the pedestal is mounted. For the

case of turbo generator rotors mounted on oil-film bearings might be three times more flexible as

compared to pedestals and foundations.

Figure 10.1(a) A simplified representation of a rotor-bearing-

foundation system

(b) Single-rotor-degree of

freedom idealisation

In the case of aeroengine compressor shafts are mounted on rolling element bearings, the foundation

of bearing is far more flexible. The theoretical models available for predicting the rotor support

stiffness are insufficiently accurate. It is for this reason that designers of high-speed rotating must rely

on empirically derived values (i.e. experimental) for support stiffness and damping in their design

calculations. Following methods are available which is classified in terms of type of forcing applied

(i) Static force method

(ii) Dynamic force method

o Use of an electromagnetic vibration or exciter

Complex receptance method

Direct complex impedance derivation

Multi-frequency testing


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o Use of centrifugal forces

Imbalance mass attached to the journal

Imbalance mass attached to an independent vibrator shaft

o Transient methods

Measurement on the running machine

o Forces inherent in the system (residual imbalance and random)

10.1 Static force method

It is possible to determine all four stiffness coefficients (i.e. xyyyxx KKK ,, and yxK ) of the bearing

oil film by application of static loads only. Unfortunately this method of loading does not enable the

oil-film damping coefficient to be determined.

The exact operating position of the shaft center on a particular bearing depends upon the Sommerfeld

number. Because the bearing oil-film coefficient are specific to a particular location of shaft center on

the static locus as shown in Figure 10.2. A static load must first be applied in order to establish

operation at the required point on the locus. The next step is to apply incremental loads in both the

horizontal and vertical directions, which will cause changes in the journal horizontal and vertical

displacement relative to the bearing bush (or more precisely with respect to its static equilibrium

position). By relating the measured changes in displacements to the changes in the static load it is

possible to determine four-stiffness coefficients on the bearing oil film. We have, increments in forces

as

Figure 9.2 (a) A bearing model (b) Steady state locus curve of the shaft center


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400

x xx xy

f k x k y= + ; and y yx yyf k x k y= + (10.1)

where xand yare the journal displacement in xandydirections, respectively (with respect to static

equilibrium position). If the displacement in theydirection is made to zero by application of suitable

loads x and yf then

/xx xk f x= ; /yx yk f x= (10.2)

Similarly if the displacement in thex-direction is made zero then

/xy x

k f y= ; /yy y

k f y= (10.3)

Determination of the oil-film coefficient in this way necessitates a test rig, which is capable of

applying loads to the journal in both the horizontal and vertical directions. The method is somewhat

tedious in the experimental stage since evaluation of the required loads to ensure zero change in

displacement in one or other direction is dependent on the application of trial loads.

Alternative method (i) Instead of applying loads in both x and y directions, to ensue zero

displacements in one of these directions, it is easier to simply apply a load in one direction only and

measure resulting displacements in both directions. Equations (10.1) can be written as

{ } [ ]{ }f K d= (10.4)

with

{ } [ ] { }; ;x xx xy

y yx yy

f k k xf K d

f k k y

= = =

If the [K]matrix is inverted then equation (10.4) can be written as

{ } [ ]{ }d f= (10.5)

with

[ ] [ ]1 xx xy

yx yy

K

= =


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where the quantities xx, xy, etc. are called the oil-film influence coefficients. If the force in they

direction is zero then

xx

x

x

f = ;

yx

x

y = (10.6)

Similarly, when the force in thexdirection is zero, we have

xyf

= ;yy

y

= (10.7)

The bearing stiffness coefficient may be obtained by inverting the influence coefficient matrix i.e.

[ ] [ ]1

K

= . This method still requires a test rig which is capable of providing loads on the bearing in

bothxandydirections (specially in the horizontal direction,x).

Alternative method(ii): If there is no facility on the test rig for applying loads transverse to the normal

steady-state load direction of the bearing, it is still possible to obtain approximate value of the

stiffness coefficients.

Figure 10.3 Shift in the journal center position due to a horizontal load

In Figure 10.3, eis the eccentricity, is the altitude angle,Ais the steady state position for a vertical

load w,Bis the additional imaginary force xF is applied to change its steady state running position to

B,Ris the resultant of wand xF , d is the angle ofRwith respect to vertical line i.e. w, d is the


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change in altitude angle due to additional xF , ede + is the new eccentricity after application of xF .

The influence coefficient can be obtained as

( ) ( )

( )( )

sin sin

sin cos cos sin sin

e

xx

x x x x

e

x

e d d ex PB PR PB SA

F F F F

e d d d e

F

+ + = = = =

+ + =

Since for small displacement, we have ( ) ede e + , dd =sin and 1cos =d . The influence

coefficient can be simplified to

( )sin cos sin ( )cosxx

x x

e d e e d

F F

+ (10.8)

A further simplification can be made if the resultant R is considered to be of vertically same

magnitude as the original load w, except that it has been turned through an angle d . We may write

w

Fddd x== tan

(10.9)

On substituting equation (10.9) into (10.8), it gives

cos cosxxx

x

dFe e

F w w w

= (10.10)

Similarly it may be shown that

sin xyx

de

w w

= = (10.11)

Since vertical load is easy to apply, one can getyy

yF

= andxy

y

y

F = . Stiffness coefficients can

be obtained as [ ] [ ]1

k

= .


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Example 10.1: Under particular operating conditions, the theoretical values of the stiffness

coefficients for a hydrodynamic bearing are found to be;Kxx=30 MN/m,Kxy=26.7 MN/m,Kyx=-0.926

MN/m, Kyy=11.7 MN/m. A testing is being designed so that these values can be confirmed

experimentally. What increment in horizontal (Fx) and vertical (Fy) loads must the rig is capable of

providing in order to provide (a) a displacement increment of 12 m in the horizontal direction whilst

that in the vertical direction is maintained zero and (b) a displacement increment of 12 m in the

vertical direction whilst that in horizontal direction is maintained zero.

Solution: From equation (10.1) static forces required in thexandydirections to a given displacement

can be obtained. For case (a) following forces are required

30 12 360 Nxf = = and 0.926 12 11.112 Nyf = =

For case (b) following forces are required

26.7 12 320.4 Nx

f = = and 11.7 12 140.4 Ny

f = =

The MATLAB code

INPUT FILE

% Input file name is input_qus_1_1.m% For the first condition

x1=12*10^-6; % displacement in horizontal direction when Fx is applied in m

x2=0; % displacement in horizontal direction when Fy is applied in m

% For the second conditiony1=0; % displacement in vertical direction when Fx is applied in my2=12*10^-6; % displacement in vertical direction when Fy is applied in m

%Fx= load in horizontal direction in N.%Fy= load in vertical direction in N.

Kxx=30*10^6; % dynamic stiffness coefficient for a bearing in N/m

Kxy=26.7*10^6; % dynamic stiffness coefficient for a bearing in N/mKyx=-.926*10^6; % dynamic stiffness coefficient for a bearing in N/m

Kyy=11.7*10^6; % dynamic stiffness coefficient for a bearing in N/m

MAIN FILE

clear all;

input_qus_1_1;

x=[x1x2; y1y2];

k=[kxxkxy;kyxkyy];

f=k*x;fprintf ('The loads to be applied in the first condition');fprintf ('\nfx1=');

fprintf (num2str (f (1,1))); fprintf (' N\n');

fprintf ('\nfy1=');


fprintf ('\nThe loads to be applied in the second condition');fprintf ('\nfx2=');


fprintf ('\nfy2=');



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OUTPUT

The loads to be applied in the first condition

fx1=360 N, fy1=-11.112 N

The loads to be applied in the second conditionfx2=320.4 N, fy2=140.4 N

Example 10.2: The test rig described in Example 10.1 is used to measure the hydrodynamic bearingstiffness coefficients by applying first of all a horizontal load of 360 N, which is then removed and

replaced by a vertical load of 320 N. The horizontal load produces displacement of 10.3 m and 3.3

m in the horizontal and vertical directions respectively, whilst the vertical load produces respective

displacements of 18.3 m and 19.7 m. Calculate the value of stiffness coefficients based on these

measurements.

Solution:For the horizontal load of 360 N alone from equation (10.4), we have

910.3 28.6 10 m/N360

xx = = ; 93.3 9.167 10 m/N

360yx

= =

For the vertical load of 320 N alone from equation (10.5), we have

18.357.188 m/N

320xy

= = ;

19.761.563 m/N

320yy = =

From equation (10.3), we can obtain stiffness coefficients as

[ ]1

9 928.6 57.188 0.0269 0.0250 26.9 25.0

10 10 MN/m9.167 61.563 -0.0040 0.0125 -4.0 12.5

xx xy

yx yy

k kK

k k

= = = =

The MATLAB code

INPUT FILE

% Input file name is input_qus_1_2.m

x1=10.3*10^-6; % displacement in horizontal direction when Fx is applied (in meter)

y1=3.3*10^-6; % displacement in vertical direction when Fx is applied (in meter)

Fx1=360; % load in horizontal direction (in N)Fy1=0; % load in vertical direction (in N)

x2=-18.3*10^-6; % displacement in horizontal direction when Fy is applied (in meter)

y2=19.7*10^-6; % displacement in vertical direction when Fy is applied (in meter)

Fx2=0; % load in horizontal direction (in N)

Fy2=320; % load in vertical direction (in N)

MAIN FILE

clear all;

input_qus_1_2;

f=[Fx1Fx2; Fy1Fy2];

X=[x1 x2; y1y2];


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a=X/f;

k=a^-1;

fprintf ('Dynamic stiffness coefficients are');

fprintf ('\nkxx=');

fprintf (num2str (k (1,1))); fprintf (' N/m\n');fprintf ('\nkxy=');

fprintf (num2str (k (1,2))); fprintf (' N/m\n');

fprintf ('\nkyx=');

fprintf (num2str (k (2,1))); fprintf (' N/m\n');fprintf ('\nkyy=');fprintf (num2str (k (2,2))); fprintf (' N/m\n');

OUTPUT

Dynamic stiffness coefficients are

kxx=26935055.0703 N/m, kxy=25020888.7201 N/m

kyx=-4010634.2575 N/m, kyy=12518040.2583 N/m

Exercise 10.1For the estimation of bearing stiffness coefficients by the static load method, the static

load of 400 N is applied in the vertical and horizontal directions, one at a time. When the load is

applied in the horizontal direction, it produces displacements of 22 m and 20 m in the vertical and

horizontal directions respectively, whilst the vertical load produces respective displacements of 4 mand 12 m. Obtained bearing stiffness coefficients from the above measurements.

Answer: The stiffness coefficients are kxx=-4.651 MN/m, kxy=13.953 MN/m, kyx=25.581 MN/m and

kyy=23.256 MN/m.

Exercise 10.2 A test rig is used to measure the hydrodynamic bearing stiffness coefficients by

applying first of all a horizontal load of 400 N. It produces displacements of 10 m and 4 m in the

horizontal and vertical directions, respectively. Then in second case only a vertical load of 300 N is

applied. It produces displacements of -20 m and 20 m in the horizontal and vertical directions,

respectively. Calculate the value of the stiffness coefficients based on these measurements.

10.2 Use of Electromagnetic Vibrator

In order to fully analyse the behavior of a bearing under dynamic loading it is necessary to cause the

journal to vibrate within the bearing bush under the action of a known exciting force as shown in

Figure 10.4(a). Alternatively, the bearing bush can be allowed to float freely on the journal as shown

in Figure 10.4(b), which is mounted on a slave bearings and the forcing is applied to the bush. By

measuring the resulting system vibrations and relating these to the force, it is possible to determine the

effective oil-film stiffness and damping coefficient. By varying the amplitude, frequency and shape of

the electrical signal input to the vibratior it is possible to exercise full control over the forcing applied

to the system.


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y(t)

x(t)

Journal

Floating bearing

bush

)0and0( bb yx

Figure 10.4(a). A fixed bearing and a rotating journal floating on the fluid

Figure 10.4(b). A fixed rotating shaft and a non-rotating bearing floating on the fluid

10.2.1 Complex Receptance Method

The method involves applying a sinusoidally varying force to the journal in the horizontal direction,

whilst the forcing in the vertical direction is zero, and measuring the resulting displacement

amplitudes in the horizontal and vertical directions together with their respective phase relative to the

exciting force. It is then necessary to repeat the procedure with the forcing applied only in the vertical

Fluid

y(t)

x(t)

Journal

Fluid

Fixed non-floating

bearing housing

)0and0( bb yx


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direction. The knowledge of force amplitude and measured displacement quantities, then enables the

eight oil-film coefficient to be derived. The force transmitted across the oil-film may be represented in

the form

x xx xy xx xyf k x k y c x c y= + + +

and y yx yy yx yyf k x k y c x c y= + + +

(10.12)

Assuming sinusoidal variations of x and y (i.e.tjXex = etc., where is frequency of forcing

function), equation (10.12) gives

( ) )ycjkxcjkfxyxyxxxxx

+++= and ) )ycjkxcjkfyyyyyxyxy

+++=

(10.13)

which can be written in matrix form as

=

y

x

ZZ

ZZ

f

f

yyyx

xyxx

y

x (10.14)

whereZis a complex stiffness coefficient given by Z k j c= +

=

y

x

yyyx

xyxx

f

f

RR

RR

y

x (10.15)

where [ ] [ ] 1= ZR is called the complex receptance matrix. For the case of forcing in horizontal

direction only equation (10.15) gives

x

xxf

xR = and

x

yxf

yR = (10.16)

where xandyare the measured displacement in the horizontal and vertical directions at a particular

time and xf is the force in the horizontal direction at that instant. For the case when forcing is in the

vertical direction the other reacceptance terms are derived as

y

xyf

xR = and

y

yyf

yR = (10.17)


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On inverting [ ]R elements of [ ]Z can be obtained. The elements of [ ]Z contain all eight bearing

stiffness and damping coefficients as defined in equation (10.14) i.e. Z k j c= + where is the

frequency of forcing function. The problems related to this method can be easily solved in the

MATLAB and is illustrated in the following example.

Example 10.3 A bearing is forced in the horizontal direction by a force Fx = 150 sin 200tN. The

resulting vibrations are x = 2010-6 sin(200t-0.2) m in the horizontal direction and y = 2010-6

sin(200t-0.32) m in the vertical direction. When the same forcing is applied in the vertical direction

the horizontal and vertical displacements take the respective formsx = 810-6

sin(200t+ 0.15) m and

y = 2610-6sin(200t-0.3) m. Determine dynamic coefficients of the bearing.

Solution:

We have two sets of measurements

(i) For 150sin 200 N and 0x y

F t F= =

67 10 sin(200 0.2) mx t= and 620 10 sin(200 0 .32) my t=

which can be written in complex plane as

Forj200150 t

xF e= alone, we have

6 j(200 -0.2) 6 j(200 -0.32)7 10 and 20 10t tx e y e = = (A)

and

(ii) 0xF = and 150sin 200 NyF t=

60.8 10 sin(200 0.15) mx t

= + and 626 10 sin(200 0.3) my t=

which can be written in complex plane as

Forj200150 tyF e= alone, we have

6 j(200 0.15) 6 j(200 -0.3)0.8 10 and 26 10t tx e y e + = = (B)


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Bearing dynamic coefficients are defined as

andx xx xy xx xy y yx yy yx yyF k x k y c x c y F k x k y c x c y= + + + = + + + (C)

On substituting the first set of measurement from equation (A) into equation (C), we have

j200 6 j( 200 -0.2 )

6 j( 200 -0.32 )

( j200 )( j200 )150 7 10

( j200 ) ( j200 )0 20 10

t txy xyxx xx

tyx yx yy yy

k ck ce e

k c k c e

++ =

+ +

(D)

Similarly on substituting the second set of measurement from equation (B) into equation (C), we have

6 j(200 0.15)

j200 6 j( 200 -0.3)

( j200 )( j200 )0 8 10

150 ( j200 ) ( j200 ) 26 10

txy xyxx xx

t tyx yx yy yy

k ck c e

e k c k c e

+

++

= + +

(E)

Let the dynamic stiffness is defined as

Z= k+ jc with = 200 rad/sec

First set of equations from equations (D) and (E), we have

6 j(200 -0.2) 6 j(200 -0.32) j2007 10 20 10 150t t txx xye Z e Z e

+ = (F)

and

6 j(200t-0.15) 6 j(200t-0.3)

xx xy8 10 e 26 10 e 0Z Z + = (G)

Equation (G) gives

6 j( 200 -0.3)0.45j

6 j( 200 -0.15 )

26 10 or -3.25

8 10

t

xx xy xx xyt

eZ Z Z e Z

e

= =

(H)

On substituting equation (H) into equation (F), we get


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j200

6 j(200 -0.32) 6 j(200 -0.2) 6 0..32j 6 0.2j

150 150

26 10 22.75 10 20 10 e 22.75 10 e

t

xy t t

eZ

e e = =

or

7 7

3.52 10 j 1.884 10xyZ = + (I)

On substituting equation (I) into equation (H), we get

7 87.64 10 j 1.049 10xxZ = (J)

Similarly from first set of equation (E), we have

( )6 j( 200 0..32)0.12 j

6 j(200 0.2)

20 10 2.8571

7 10

t

yx yy yx yyt

eZ Z Z e Z

e

= =

(K)

On substituting equation (K) into second equation of (E), we get

j200

6 j(200 0.3) 6 j(200 0.15) j0.3 j0.15

150 150 106

26 10 22.86 10 26 22.86

t

yy t t

eZ

e e e e +

= =

(L)

In simplification of equations (K) and (L), we get

6 7 7 72.6156 10 j 1.2987 10 and 1.186 10 j 3.594 10yy yx

Z Z= + = (M)

Stiffness and damping coefficients can be obtained by separating real and imaginary part of the

dynamic stiffness coefficients from equations (I), (J) and (M), as

7 7 7 67.64 10 N/m; 3.52 10 N/m; 1.186 10 N/m; 2.6156 10 N/m

524500 N/ m-sec ; 94200 N/m-sec ; 179700 N/m-sec; 64935 N/m-sec

xx xy yx yy

xx xy yx yy

k k k k

c c c c

= = = =

= = = =

MATLAB solution:

INPUT FILE

% The name of the input file is input_qus_5.m


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w=200; % Frequency of the applied force

% For the given 1st condition

Fx1=150; % Amplitude of the applied horizontal force (in N)

Fy1=0; % Amplitude of the applied vertical force (in N)

X1=20*10^-6; % amplitude of the horizontal vibration (in meter)1=-0.2; % Phase angle of horizontal vibration (in radian)

Y1=20*10^-6; %Amplitude of the vertical vibration (in meter)

2=-.32; % Phase angle of vertical vibration (in radian)

% For the given 2nd conditionFx2=0; % Amplitude of the applied horizontal force (in N)Fy2=150; % Amplitude of the applied vertical force (in N)

X2=8*10^-6; % amplitude of the horizontal vibration (in meter1=0.15; % Phase angle of horizontal vibration (in radian)


2=-0.3; % Phase angle of vertical vibration (in radian)

x1=X1*(cos(1)+i*sin(1));

y1=Y1*(cos(2)+i*sin(2));

x2=X2*(cos(1)+i*sin(1));y2=Y2*(cos(2)+i*sin(2));

MAIN FILE

clear all;

input_qus_1_5;

F=[Fx1Fx2; Fy1Fy2];

X=[x1x2; y1y2];K=X\F;

fprintf ('Dynamic coefficients of the bearing are');

fprintf ('\nkxx=');fprintf (num2str (real (K (1,1)))); fprintf (' N/m\n');

fprintf ('\nkxy=');fprintf (num2str (real (K (1,2)))); fprintf (' N/m\n');

fprintf ('\nkyx=');

fprintf (num2str (real (K (2,1)))); fprintf (' N/m\n');fprintf ('\nkyy=');

fprintf (num2str (real (K (2,2)))); fprintf (' N/m\n');

fprintf ('\ncxx=');

fprintf (num2str (imag (K (1,1)))); fprintf (' N/m-sec\n');

fprintf ('\ncxy=');fprintf (num2str (imag (K (1,2)))); fprintf (' N/m-sec\n');

fprintf ('\ncyx=');


fprintf ('\ncyy=');


OUTPUT

The dynamic coefficients of the bearing are

Kxx=9877721.7564 N/m

Kxy=-2269493.2286 N/m

Kyx=-7650433.7248 N/m

Kyy=7292188.2744 N/mCxx=3491110.953 N/m^2Cxy=-2289239.7727 N/m^2

Cyx=-2532978.0684 N/m^2Cyy=3430612.8087 N/m^2

The above problem we can be solved by an alternative method:

INPUT FILE% The input file name is input_altr_qus_1_5.m

w=200;t=pi/(4*w); % time of operation in second

% For the first condition

Fx1=150*sin(200*t); % applied force in horizontal direction (in N)

Fy1=0; % applied force in vertical direction (in N)

A1=-.2; % phase of the vibration in the horizontal direction for the first condition (in radian)B1=-.32; % phase of the vibration in the vertical direction for the first condition (in radian)


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Mx1=20*10^-6; % amplitude of the vibration in the horizontal direction for the first condition (in meter)

My1=20*10^-6; % amplitude of the vibration in the vertical direction for the first condition (in meter)

% for the second condition

Fx2=0; % applied force in horizontal direction (in N)

Fy2=150*sin(200*t); % applied force in vertical direction(in N)A2=.15; % phase of the vibration in the horizontal direction for the second condition (in radian)

B2=-.3; % phase of the vibration in the vertical direction for the second condition (in radian)

Mx2=8*10^-6; % amplitude of the vibration in the horizontal direction for the second condition (in meter)

My2=26*10^-6; % amplitude of the vibration in the vertical direction for the second condition (in meter)x1=Mx1*(sin(w*t)*cos(A1)+j*sin(A1)*cos(w*t)); % displacement in horizontal direction when Fxis applied (in meter)x2=Mx2*(sin(w*t)*cos(A2)+j*sin(A2)*cos(w*t)); % displacement in horizontal direction when Fyis applied (in meter)

y1=My1*(sin(w*t)*cos(B1)+j*sin(B1)*cos(w*t)); % displacement in vertical direction when Fxis applied (in meter)y2=My2*(sin(w*t)*cos(B2)+j*sin(B2)*cos(w*t)); % displacement in vertical direction when Fyis applied (in meter)

MAIN FILE

clear all;

input_altr_qus_1_5;

F=[Fx1Fx2; Fy1Fy2];X=[x1x2; y1y2];

K=X\F;fprintf ('Dynamic coefficients of the bearing are');

fprintf ('\nkxx=');


fprintf ('\nkxy=');

fprintf (num2str (real (K (1,2)))); fprintf (' N/m\n');fprintf ('\nkyx=');


fprintf ('\nkyy=');fprintf (num2str (real (K (2,2)))); fprintf (' N/m\n');

fprintf ('\ncxx=');fprintf (num2str (imag (K (1,1)))); fprintf (' N/m-sec\n');

fprintf ('\ncxy=');

fprintf (num2str (imag (K (1,2)))); fprintf (' N/m-sec\n');fprintf ('\ncyx=');


fprintf ('\ncyy=');


OUTPUT

The dynamic coefficients of the bearing are

kxx=9877721.7564 N/m

kxy=-2269493.2286 N/m

kyx=-7650433.7248 N/mkyy=7292188.2744 N/m

cxx=3491110.953 N/m-seccxy=-2289239.7727 N/m-sec

cyx=-2532978.0684 N/m-sec

cyy=3430612.8087 N/m-sec

Example 10.4 A bearing is forced in the horizontal direction by a force Fx = 150 sin 200tN. The

resulting vibrations are x = 710-6 sin(200t-0.2) meters in the horizontal direction and y = 2010-6

sin(200t-0.32) meters in the vertical direction. When the same forcing is applied in the vertical

direction the horizontal and vertical displacements take the respective formsx=810-6sin(200t +0.15)

meters andy=2610-6sin (200t-0.3) meters. Determine elements of complex receptance matrix for the

bearing.

Solution: The following measurement were done


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Case I: For 150sin 200 NxF t= and 0yF = , we have

6 67 10 sin(200 0.2) m and 20 10 sin(200 0.32) mx t y t = =

Case II: For 150sin 200 Ny

F t= and 0x

F = , we have

6 68 10 sin(200 0.15) m and 26 10 sin(200 0.3) mx t y t = + =

For a forceFxleading a displacementXby is shown in Figure 8.5.

Fxcos

Fx

Fysin

X

Figure 8.5 Phase between the displacement and force vectors

From Figure 8.5 the receptance can be expressed as

cos j sinxx

x x

XR

F F =

+

whereXandFxare displacement and force amplitudes. The displacement is lagging behind force by

angle.

Fx = 150 N

0.2 rad

X= 710-6

m

Figure 10.6 A typical force and displacement vectors

From Figure 10.6, we have

6-67.0 10 (0.04574 -j 0.00927) 10 m/N

cos j sin 150cos 0.2 j 150sin 0.2x

xx

x x

x XR

F F F

= = = =

+ +


17/33


414

( )6

626 10 0.1266 j 0.042 10 m/N150cos0.32 j150sin0.32x

yx

yR

f

= = =

+

( )

( )

66

66

26 100.046 j 0.007 10 m/N

150cos0.3 j 150sin0.3

8 100.127 j 0.0394 10 m/N

150cos0.15 j 150sin0.15

y

y

yy

xy

yR

f

xR

f

= = = +

+

= = =

+

Hence, the receptance matrix can be written as

(45.74 j 9.27) (52.73 j 7.97) m]

(126.56 j 41.94) (165.59 j 51.22) MN

xx xy

yx yy

R RR

R R

+ = = +

MATLAB Solution:

INPUT FILE

% The name of the input file is input_qus_1_6.m

w=200; % Frequency of the applied force% For the given 1st condition


Fy1=0; % Amplitude of the applied vertical force (in N)X1=7*10^-6; % amplitude of the horizontal vibration (in meter)

1=-0.2; % Phase angle of horizontal vibration (in radian)

Y1=20*10^-6; %Amplitude of the vertical vibration (in meter)2=-.32; % Phase angle of vertical vibration ( in radian)

% For the given 2nd condition


Fy2=150; % Amplitude of the applied vertical force (in N)X2=8*10^-6; % amplitude of the horizontal vibration (in meter1=0.15; % Phase angle of horizontal vibration (in radian)

Y2=26*10^-6; %Amplitude of the vertical vibration (in meter)2=-0.3; % Phase angle of vertical vibration (in radian)



MAIN FILE

clear all;

input_qus_1_6;F=[Fx1Fx2; Fy1Fy2];

X=[x1x2; y1y2];

K=X/F;fprintf ('The elements of complex receptance matrix for the bearing are');

fprintf ('\nRxx=');fprintf (num2str (K (1,1))); fprintf (' N/m\n');

fprintf ('\nRxy=');

fprintf (num2str (K (1,2))); fprintf (' N/m\n');fprintf ('\nRyx=');fprintf (num2str (K (2,1))); fprintf (' N/m\n');

fprintf ('\nRyy=');

fprintf (num2str (K (2,2))); fprintf (' N/m\n');


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415

OUTPUT

The elements of complex receptance matrix for the bearing are

Rxx=4.5736e-008-9.2712e-009i m/N

Rxy=5.2734e-008+7.97e-009i m/NRyx=1.2656e-007-4.1942e-008i m/N

Ryy=1.6559e-007-5.1224e-008i m/N

Exercise 10.3 A bearing is forced in the horizontal direction by a force tFx 150sin200= N. The

resulting journal vibrations are6

12 10 sin(150 0.35)x t= m (in the horizontal direction) and

620 10 sin(150 0.4)y t

= m (in the vertical direction). When the same force is applied in the

vertical direction the horizontal and vertical displacements take the respective forms

613 10 sin(150 0.3)x t= + and 625 10 sin(150 0.38)y t= . Determine elements of the complex

impedance matrix for the bearing.

MATLAB Solution:

INPUT FILE

% The name of the input file is input_qus_8_4.m

w=150; % Frequency of the applied force

% For the given 1st condition

Fx1=200; % Amplitude of the applied horizontal force (in N)Fy1=0; % Amplitude of the applied vertical force (in N)

X1=12*10^-6; % amplitude of the horizontal vibration (in meter)1=-0.35; % Phase angle of horizontal vibration (in radian)


2=-0.4; % Phase angle of vertical vibration (in radian)% For the given 2nd condition

Fx2=0; % Amplitude of the applied horizontal force (in N)Fy2=200; % amplitude of the applied vertical force (in N)

X2=13*10^-6; % Amplitude of the horizontal vibration (in meter)

1=0.3; % Phase angle of the horizontal vibration (in radian)Y2=25*10^-6; % amplitude of the vertical vibration (in meter)

2=-0.38; % Phase angle of the vertical vibration (in radian)

MAIN FILE

clear all;input_qus_1_4;


x2=X2*(cos(1)+i*sin(1));

y2=Y2*(cos(2)+i*sin(2));F=[Fx1Fx2; Fy1Fy2];

X=[x1x2; y1y2];

K=X\F;

fprintf ('The elements of complex impedance matrix for the bearing are');

fprintf ('\nKxx=');

fprintf (num2str (K (1,1))); fprintf (' N/m\n');fprintf ('\nKxy=');fprintf (num2str (K (1,2))); fprintf (' N/m\n');

fprintf ('\nKyx=');


fprintf ('\nKyy=');


OUTPUT

The elements of complex impedance matrix for the bearing areKxx=5061762.20178+27691680.0054i N/m


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416

Kxy=7007752.44721-12851849.7486i N/m

Kyx=-4491637.24935-22067930.6871i N/m

Kyy=2029852.21747+13358903.8911i N/m

Exercise 10.4. For the bearing dynamic parameter estimation, how many minimum numbers of

independent sets of force-response measurements are required? Justify your answer. (Assume there is

no residual imbalance in rotor.)

10.2.2 Direct Complex Impedance Derivation

It is possible to determine the complex stiffness coefficients ZZxx , etc. in equation (10.14) directly

without resorting to the use of receptances. This may be done provided that forcing in both the

horizontal and vertical directions be able provide simultaneously, and with independent control over

each input with respect to its amplitude and relative phase. The system force-displacement

relationship is given by equation (10.14). In the present approach it is to ensure that one of the

resulting system displacement vectors, say Y, is zero. This can be made to be the case by correctly

adjusting the amplitude of the force in they-direction and its phase relative to that in thex-direction.

Suitable values for these quantities can be found relatively easily by trial and error. The first line of

equation (10.14) thus gives

x

fZ xxx= (10.18)

which will allow the value ofxx

Z to be determined directly provided that the amplitude and phase of

the horizontal displacement relative to the horizontal force had been measured. If the amplitude and

phase of the force in they-direction were also noted then the value of yxZ could also be determined as

x

fZ

y

yx= (10.19)

In the above case, it is the phase of thex-direction displacement amplitude relative to the force in the

y-direction that is significant. Similarly by adjusting forcing amplitudes and relative phases so as to

ensure a zero horizontal displacement,x, then the values of xyZ and yyZ could also be determined.

This method requires more complicated experimental procedure. It may be more costly in terms of

equipment, since two vibrators and additional control units are required.

Some experimental measurement considerations are discussed now. Choice of forcing frequency (or

forcing frequency range) is an important parameter to choose. It depends upon the system resonance.


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417

If the system is excited close to its resonant frequency then a response of suitable magnitude may be

obtained for a lower force amplitude input. (since the bearing impedance changes with journal

vibration non-linearity effect will play a major roll). The advantage of exciting the system at a

frequency in the region of its resonant frequency, that is that phase lag will be generally greater than

zero. (between response and force). With this for small inaccuracies in their (phase) measurement are

less likely to substantially alter the magnitude of coefficients, which are derived. This is not the case

hen the lag angle is very small or when it is close to090 . In these cases (i.e. close to 00 and 090 ) ill

conditioning of the equation of motion results in significant changes in the magnitude of the derived

coefficient for even a change of only 3-4 degrees of phase (which may be about the accuracy to which

phase is measured). This coefficient derived using data generated well away from the critical speed

may well be considerably in accurate (of the order of 100% in some cases). Also since at critical

speeds it is observed that the orbit of the shaft center is elliptical in nature and that leads to well

conditioning of regression matrix.

10.2.3 Multi Frequency Testing

It has advantage that the certainty of exciting all system modes within the prescribe frequency range,

and inherent high noise rejection. The method involves forcing the system in bothxandydirections,

at all frequencies within the range of interest, simultaneously. The aim is to arrive at more accurate

values of the coefficients, which are assumed to be independent of frequency, by means of some

averaging procedure. When all (several) frequencies are excited simultaneously, the knowledge of

bearing behavior at many different frequencies should enable more accurate results to be obtained.

Also it saves the laboratory time. Fourier analysis can be used to convert measured input and output

signals from the time domain to the frequency domain. Recent advances in laboratory

instrumentation, for example, the emergence of spectrum analysers (FFT analysis) capable of carrying

out the Fourier transform, have helped the technique to evolve. In theory, any shape of input signal

with multi-frequency content can be used to force the system. For example an impulse signal (Figure

10.7) is actually composed of signals at all frequencies in coexistence. Because of the likely

concentration of the signal at the low-frequency end of the spectrum however, a impulse in practice

provides useful signals over only a relatively small frequency range. For higher frequency the signal/

noise ratio becomes too low. An alternative is a white noise signal, which contains all frequencies

within its spectrum Band-limited white noise, sometimes referred to as coloured noise, contains all

frequencies within a prescribed range. One way of producing such a signal is with pseudo random

binary sequences (PRBS) where the frequency range that is present is chosen to excite appropriate

modes in the system under test. Unfortunately, both with impulse and PRBS signals there is a danger

of saturating the system so that amplitudes at some frequencies are so large that non-linearity are

encountered and the test becomes invalid. These disadvantages can be overcome by using a signal


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418

mode up of equal-amplitude sinusoidal signals whose frequency are those which one wishes to excite

within a particular frequency range. One signal of this type is Schroeder phased harmonics.

Figure 10.7 Impulse and white noise signals in the time and frequency domains

If the system response to multi-frequency signal is recorded, bearing properties may be obtained as

follows. The displacement in thexandydirections occurring at a frequency are written in the form

j j and

t tx Xe y Ye

= = (10.20)

Thus

j j 2 j 2 jj ; j ; andt t t t Xe y Ye x Xe y Ye = = = = (10.21)

The forcing function may similarly be defined as

j t

x xf F e = and j ty yf F e

= (10.22)

Equations of motion of the journal in thexandydirections are

x xx xy xx xyf k x k y c x c y Mx = and y yx yy yx yyf k x k y c x c y My = (10.23)

where M is the mass of journal andxx

k ,xx

c etc. are oil film stiffness and damping coefficients.

Substituting equations (10.20) to (10.22) into equation (10.23), we get

( ) ( )

( ) ( )

2

2

xx xy x

yx yy y

Z Z F M XX

Z Z F M YY

+ = +

(10.24)


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419

where jZ k c= + . On separating the real and imaginary parts, we rearrange equation (10.24) as

2 2

2 2

1 0

0 1

xx yx

xx yx

r i r i r i r r

x y xy yyi r i r i r i i

x y xy yy

k k

c c

X X Y Y F F k k M X M Y X X Y Y F F c c X M Y

=

(10.25)

Equation (10.25) may be written for0 0 0 0, 2 , 3 , ,n = (Total of ntimes in all). Values of

and quantities in the first and last matrices of equation (23) are determined by the Fourier

transformation of time-domain signals. All of these equations (10.25) may be grouped as a single

matrix equation as

[ ] [ ] [ ] 222662 = nn AZD (10.26)

The contents of the [ ]Z matrix might be best obtained by means of a least squares estimator. This

involves recognizing that measurements obtained in the laboratory will be inaccurate and so there are

no values of the coefficient in the [ ]Z matrix which will satisfy all lines of equation (10.26). A

residual matrix is developed which defines the errors between the left hand and right hand sides of

equation (10.26) i.e.

[ ] [ ] [ ] [ ] 26622222 = ZDAE nnn (10.27)

The contents of the [ ]Z matrix are defined as being those values, which ensure that the sums of the

squares of the elements in the [ ]E matrix are minimized. On multiplying equation (10.27) by [ ]T nD 26 ,

we get

[ ] [ ][ ] [ ] [ ]ADZDD TT = or [ ] [ ] [ ][ ] [ ] [ ]ADDDZ TT 1= (10.28)

Since measured terms used to make up the [ ]D and [ ]A matrices are obtained via Fourier

transformation of the output and input signals. The noise occurring at a frequency greater than 0n is

automatically filtered out of the analysis.


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420

10.3 Use of Centrifugal Forces

One of the simplest ways of exciting a journal in a sinusoidal manner is by means of centrifugal

forcing, simply by attaching imbalance masses of known magnitude to a rotating shaft. Advantage

with this method is that there is no need for costly electromagnetic exciter. In present method the

processing of measured data in the time domain. Out of three methods in two methods an imbalance

of known magnitude attached to the journal (and are based on the assumption that inherent rotor

imbalance is insignificantly small). The third method involves use of a separate imbalance mass shaft,

which can rotate at frequencies independent of the journal rotational frequency.

(i)Imbalance mass attached to the journal

This means of exciting the journal can be used to determine the bearing oil-film damping coefficients

when the bearing stiffness coefficients are already known. (for example by static force method). The

experimental involves measurement of the horizontal and vertical displacement amplitudes of the

journal relative to the bearing, and of the bearing or pedestals itself relative to space (fixed

foundation). In addition, measurements of the corresponding phase lag angles of each of these

displacements behind the imbalance force vector are also made.

Figure 10.8 A rotor-bearing system with an unbalance

In addition to imbalance force as shown in Figure 10.8, the rotor also has oil-film forces acting on it,

these being transmitted to rotor by shaft. Governing equations of the rotor can be written as

( )2 2 2 2 2 2x xx xy xx yy pf k x k y c x c y M x x = +

and

( )2 2 2 2 2 2y yx yy yx yy pf k x k y c x c y M y y = + (10.29)


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421

where x and y are displacements of journal relative to bearing (or pedestal),p

x andp

y are the

displacement of pedestal (or bearing) relative to fixed foundation, 2M is the central rotor mass

(symmetric) andf is the known imbalance force on the rotor. We can write

j tx xf F e = ; j ty yf F e =

so that

j tx Xe = ; j ty Ye = ; j tp px X e = ; j tp pY e

= (10.30)

where , , , , andx y p p

F F X Y X Y are in general complex quantity and contain the amplitude and

phase information and is the rotational speed of the rotor. On substituting equation (10.30) into

equation (10.29), we get

( ) ( )

( ) ( )

( )

( )

2

2

x pxx xy

yx yy y p

F M X XZ Z X

Z Z Y F M Y Y

+ + = + +

(10.31)

On separating real and imaginary terms, we get

22

22

2 2

2 2

i

r

r

i

pixx xy xx xy

pryx yy yy yy

rxx yy xx xy p

iyx yy yx yyp

M XXM k k c c

F M YYk k M c c

Xc c k M k F M X

Yc c k k M M Y

+ = +

(10.32)

Quantities , , , , , , andx y p p

F F X Y X Y are either known or are measured during the course of

the experiment. , ,xx xxk c are unknown (eight for the present case). Equation (30) has four

equations so if four stiffness coefficient are known (by static force method) the remaining four

damping coefficient can be obtained from this. Alternatively, if bearing dynamic parameters are

speed-independent then measurement at least two speed will be sufficient to obtain all eight

coefficients or by changing F and rotating the rotor at same speed (for speed-dependent bearing

parameters) all eight coefficients can be obtained.

Imbalanced mass attached to an independent vibrator shaft


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422

In previous method equation (30) has eight unknowns ( , ,xx xxk c ) and it has four equations. If a test

rig capable of providing excitation by means of imbalance forcing, where the forcing frequency could

be varied without upsetting the journal rotational frequency. Thus a second equation (30) could be

obtained, resulting in eight simultaneous equations in all, by using a different value of without

upsetting the bearing Sommerfeld number. Two sets of measurements can be taken for two different

rotational frequency of secondary shaft, most convenient would be so that the steady state

position should not disturb (as shown in Figure 10.9b).

(a) Excitation unit arrangement

(b) Basic principle of the excitation unit

Figure 10.9 An anti-synchronous excitation by an auxiliary unbalance unit

Journal

Fluid

Fixed non-floating

bearing housing

)0and0( bb yx

Anti-synchronous

excitation


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423

Example 10.5 For estimation of bearing dynamic coefficients the following measurements were

made: (i)1

X and Y1for simultaneous application of1x

F and1y

F and (ii)2

X and Y2for simultaneous

application2x

F and2

F ; whereXand Yare displacements andFis force and in general they all are

complex in nature. If shapes of both the orbits of the shaft center are circular in shape, whether it

would be possible to estimate all bearing dynamic coefficients from these two measurements.

Solution: Consider a single bearing and use a complex stiffness,2 jZ k m c = + , at a single

frequency to describe the equation of motion in frequency domain, as

xx xy x

x yy y

Z Z f

Z Z f

=

(A)

Using two unbalance runs with corresponding responses 211 ,, xyx and 2y and right hand sides

1 1 2, ,x y xf f f and 2yf , equation (A) may be written as

1 21 2

1 21 2

xx xy x x

yx yy y y

Z Z f fx x

Z Z f fy y

=

(B)

The solution of equation (B) is obtained as

1 22 2

1 21 11 2 2 1

1

( )

xx xy x x

yx yy y y

Z Z f fy x

Z Z f fy xx y x y

=

(C)

For circular orbits 11 jxy = and 22 jxy = (or negative, depends on the definition of axes, and the

direction of rotation). Then the denominator of equation (C) becomes

0)()( 12211221 == jxxjxxyxyx (D)

and hence, equation (D) is ill-conditioned for circular orbits. Having a third unbalance run does not

help. For three unbalances equation (B) may be written as

1 2 31 2 3

1 2 31 2 3

xx xy x x x

yx yy y y y

Z Z f f fx x x

Z Z f f fy y y

=

(E)


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424

The least squares solution involves the following inversion

11 1 2 2 21 2 3 1 2 3 1 1 2 2 3 3

2 2 2 2 21 2 3 1 1 2 2 3 3 1 2 3

3 3

2 2 2

1 2 3 1 1 2 2 3 3

2 2 2 2 2 2 2 2 2

1 2 3 1 2 3 1 1 2 2 3 3 1 1 2 2 3 3 1 2 3

( )1

( )( ) ( )

x yx x x x x x x y x y x y

x yy y y x y x y x y y y y

x y

y y y x y x y x y

x x x y y y x y x y x y x y x y x y x x x

+ + + + =

+ + + +

+ + + += + + + + + + + + + +

(F)

Ifii

jxy = , then the denominator of the equation (F) becomes

( ) ( ) ( )

( ) ( ) ( )

22 2 2 2 2 2

1 2 3 1 2 3 1 1 2 2 3 3

22 2 2 2 2 2 2 2 2 2

1 2 3 1 2 3 1 2 3j j 0

x x x y y y x y x y x y

x x x x x x x x x

+ + + + + +

= + + + + + + =

(G)

and the circular orbits are ill-conditioned. There is another possibility when ill-conditioning may

occur, namely when 11 xy = and 22 xy = for any value of , where is a constant. Then the

denominator of equation (C) becomes zero, leading to ill-conditioning. This means that a change in

orbit from one unbalance to the next is required. The ill-conditioning due to a circular orbit may be

avoided by taking measurements in both the clockwise and anticlockwise directions of rotation of the

rotor. For this case 11 jxy = and 22 jxy = . Then the denominator of equation (C) becomes

0)()( 12211221 = jxxjxxyxyx (H)

and hence, equation (C) becomes well-conditioned.

Exercise 10.5 The eight bearing stiffness and damping coefficients are to be determined by using the

method described above. Experimental measurements of journal vibration amplitude and phase lag

angle are given in the Table 10.1; pedestal vibrations are found to be negligible. Determine the values

of oil-film coefficients implied by these measurements, and the maximum change in the direct cross-

coupling terms introduced by an error of +4 in the measurement of the phase recorded as 42.5.

Table 10.1 Some test data used to calculate bearing stiffness and damping coefficients


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425

Forward excitation Reverse excitation

Horizontal vibration amplitude 66.4 m 46.6m

Horizontal phase lag 42.5 20.9

Vertical vibration amplitude 55.5 m 38.4 m

Vertical phase lag 9.9 111

Force amplitude 1.0 KN 1.0 KN

Forcing frequency 12.6 Hz 12.6 HzJournal mass 150 kg 150 kg

MATLAB Solution:

INPUT FILE

% Name of this input file is input_qus_1_7.mX1=66.4*1.0e-6; % horizontal vibration amplitude (in meter)

A1=42.5; % horizontal phase lag (in degree)Y1=55.5*1.0e-6; % vertical vibration amplitude (in meter)

B1=9.9; % vertical phase lag (in degree)

F1=1*1.0e+3; % force amplitude (in N)

n1=12.6; % forcing frequency(in Hz)

M=150; % journal mass (in Kg)% For the reverse excitation condition.

X2=46.6*1.0e-6; % horizontal vibration amplitude (in meter)A2=-20.9; % horizontal phase lag (in degree)

Y2=38.4*1.0e-6; % vertical vibration amplitude (in meter)

B2=-111; % vertical phase lag (in degree)F2=1*1.0e+3; % force amplitude (in N)

n2=12.6; % forcing frequency (in Hz)

M=150; % journal mass (in Kg)

MAIN FILE

clear all;

input_qus_1_7;

w1=2*pi*n1;w2=2*pi*n2;

a1= A1*(pi/180);b1= B1*(pi/180);

a2= A2*(pi/180);

b2= B2*(pi/180);p=[-X1*sin(a1) Y1*cos(b1) 0 0 w1*X1*cos(a1) w1* Y1*sin(b1) 0 0;

0 0 - X1*sin(a1) Y1*cos(b1) 0 0 w1* X1*cos(a1) w1*Y1*sin(b1);

X1*cos(a1) Y1*sin(b1) 0 0 w1* X1*sin(a1) - w1*Y1*cos(b1) 0 0;0 0 X1*cos(a1) Y1*sin(b1) 0 0 w1* X1*sin(a1) - w1*Y1*cos(b1);

- X2*sin(a2) Y2*cos(b2) 0 0 w2* X2*cos(a2) w2*Y2*sin(b2) 0 0;

0 0 - X2*sin(a2) Y2*cos(b2) 0 0 w2* X2*cos(a2) w2*Y2*sin(b2);

X2*cos(a2) Y2*sin(b2) 0 0 w2* X2*sin(a2) - w2*Y2*cos(b2) 0 0;

0 0 X2*cos(a2) Y2*sin(b2) 0 0 w2* X2*sin(a2) - w2*Y2*cos(b2)];

f=[-M* w1^2* X1*sin(a1);

F1+M* w1^2*Y1*cos(b1);F1+M* w1^2* X1*cos(a1);

M* w1^2*Y1*sin(b1);

-M* w2^2*X2*sin(a2);

F2+M* w2^2*Y2*cos(b2);

F2+M* w2^2*X2*cos(a2);M* w2^2*Y2*sin(b2)];

k=p\f;

disp ('The bearing coefficients are');fprintf ('\nKxx='); fprintf (num2str (k (1,1))); fprintf (' N/m\n');

fprintf ('\nKxy='); fprintf (num2str (k (2,1))); fprintf (' N/m\n');

fprintf ('\nKyx='); fprintf (num2str (k (3,1))); fprintf (' N/m\n');fprintf ('\nKyy='); fprintf (num2str (k (4,1))); fprintf (' N/m\n');

fprintf ('\nCxx='); fprintf (num2str (k (5,1))); fprintf (' N/m\n');

fprintf ('\nCxy='); fprintf (num2str (k (6,1))); fprintf (' N/m\n');

fprintf ('\nCyx='); fprintf (num2str (k (7,1))); fprintf (' N/m\n');


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426

fprintf ('\nCyy='); fprintf (num2str (k (8,1))); fprintf (' N/m\n');

OUTPUT

The bearing stiffness and dynamic coefficients are

Kxx=32899302.6948 N/m

Kxy=14491075.2905 N/m

Kyx=-9134864.0493 N/m

Kyy=-19410190.4825 N/mCxx=115385.6165 N/mCxy=257052.0267 N/m

Cyx=403687.2122 N/mCyy=183041.7669 N/m

10.4 Transient Methods

In this method it is possible to take measurement on running machines. In this method the system

consists of a symmetrical rigid rotor mounted in two identical journal bearings. Transient vibration of

the rotor in the bearings is caused by applying a force impulse (as shown in Figure 10.10) to the rotor

center of gravity. In practice this is provided by striking the rotor with a calibrated hammer whose

head mass is known. This means of excitation results in an impulse, which lasts for a finite period of

time (typically a fraction of second). If an accelerometer is mounted in the hammerhead, it is possible

to determine the instantaneous force, which is applied to the rotor. The electrical output from the

hammer will then indicate the vibration of the applied force with time. An impulse can be considered

as made up of a number of sine waves of different frequencies, all occurring simultaneously. By

varying the hammer head mass, the stiffness of the hammer impact force (tip) and the initial hammer

head velocity, it is possible to vary the amplitude, frequency content and duration of the applied

impulse. EOM of the journal would be

xMyCxCykxkfxyxxxyxxx

= and yMyCxCykxkf yyyxyyyxy = (10.33)

Figure 10.10 An impulse in the time and frequency domains

Since forcing may be considered to be sinusoidal, albeit at several different frequencies, any one

component will be of the form in the horizontal and vertical directions


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427

j t

x xf F e = and j ty yf F e

= (10.34)

The corresponding horizontal and vertical displacementsxandywill be of the form

j tx Xe = and tYe (10.35)

so that

2 2j ; ; j ; andx x x x y y y y = = = = (10.36)

On substituting in EOM yields

( ) ( )

( ) ( )

2

2

j j

j j

xx xx xy xyx

yyx yx yy yy

k M C k C F X

F Yk C k M C

+ + = + +

(10.37)

which can written as

( ) ( )

( ) ( )

2

2

j j1

j j

yy yy xy xy x

yyx yx xx xx

k M C k C FX

FY D k C k M C

+ + = +

where

( ) ( ) ( ) ( )2 2j j j jxx xx yy yy yx yx xy xyD k M C k M C k C k C = + + + + (10.38)

Equation (36) is similar to the case of electromagnetic exciter method (first method) except in present

case the inertia force has now also been allowed for. If forcing is applied in one direction (for example

the hammer strikes the rotor in horizontal direction) then it is possible to define the reacceptance as :

(from equation 36)

( )2 jyy yyxx

x

k M C X

R D F

+

= = and

( )jyx yxyx

x

k C Y

R D F

+

= = (10.39)

Similarly if hammer strikes the rotor in they-direction then

( )jxy xyxy

y

k C XR

D F

+= = and

( )2 jxx xxyy

y

k M C YR

D F

+= = (10.40)


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428

The reacceptance terms defined in equations (10.39) and (10.40) are clearly functions of frequency

and so take a different value depending on the vibration frequency being considered. The

reacceptance terms are in general complex because displacement and force are not in-phase. The

method of determining the oil-film stiffness and damping coefficient makes use only of the modulus

of the reacceptance terms, however, doesnot use the data describing phase. In experiment the right

hand side of equations (10.39)-(10.40) exist at many different frequencies simultaneously, and the

corresponding receptance terms must be determined for each of these frequencies. The reacceptance

will be

( )

( )

==

dtetf

dtetx

f

xR

tj

x

tj

x

xx

oftransformFourier

oftransformFourier)( (10.41)

This may be obtained in experiment by spectrum analyzer and it will display reacceptance as shown

in Figure 10.11.

Figure 10.11 A typical variation of magnitude of receptance terms

The above reacceptances have been obtained from right hand side of equations (10.39)-(10.40) using

two independent forcing. Now our aim is to obtain xxk etc. so that when it is substituted back in left

hand side of equation (10.39)-(10.40) it should give the value of the right hand side of equation

(10.39)-(10.40). These processes can be repeated until appropriate values are found which results in

the difference between left hand side and right hand side of equations (10.39)-(10.40) being

minimized, for all frequencies under consideration. The least squares error criteria may be used so

that to minimize a scalar quantity

[ ] =i j

ijtheoryij RRs

2

exp)()( (10.42)


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429

Using filter synchronous the imbalance response must be subtracted as shown in Figure 10.12.

Figure 10.12 Effect of residual unbalance on the impulse response

Step function: This step function can be generated by giving gradual static load to the rotor and

suddenly releasing the load at well defined upper limit of the static load as shown in Figure 10.13.

Figure 10.13 A step function forcing and corresponding response

References[1] Goodwin, M.J., 1991, Experimental Techniques for Bearing Impedance Measurement,ASME

Journal of Engineering for Industry, Vol. 113, No. 3, 335-342.

[2] Mitchell, J.R., Holmes, R. and Ballegooyen, H.V., 1965-66, Experimental Determination of a

Bearing Oil Film Stiffness, in the 4thLubrication and Wear Convention,IMechE, Vol. 180,

No. 3K, 90-96.

[3] Parkins, D.W., 1979, Theoretical and Experimental Determination of the DynamicCharacteristics of a Hydrodynamic Journal Bearing, ASME Journal of Lubrication

Technology, Vol. 101, No. 2, 129-139.[4] Swanson, E.E. and Kirk, R.G., 1997, Survey of Experimental Data for Fixed Geometry

Hydrodynamic Journal Bearings,ASME Journal of Tribology, Vol. 119, No. 4, 704-710.[5] Tiwari, R., Lees, A.W. and Friswell, M.I., 2004, Identification Of Dynamic Bearing Parameters:

A Review, Shock & Vibration Digest, Vol. 36, No. 2, 99-124.

[6] Tripp, H. and Murphy, B. T., 1984, Eccentricity measurements on a tilting pad bearing, Trans

ASLE, Vol.28, No. 2, 217-224.


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[7] Woodcock, J.S. and Holmes, R., 1969-70, The Determination and Application of the DynamicProperties of a Turbo-Rotor Bearing Oil Film, Proceedings of IMechE, Vol. 184, No. 3L,

111-119.

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