CHAPTER 4

TRANSVERSE VIBRATIONS-II:

SIMPLE ROTOR-BEARING-FOUNDATION SYSTEMS

In chapter 2, we studied dynamic behaviours rotors with a rigid disc with the flexible massless shaft.

These simple rotor models have advantage in that the mathematical modeling is simple and it predicts

some of the vital phenomena with relatively ease. However, in actual case as the previous chapter has

demonstrated that supports of rotors, i.e., bearings as well as the foundation are flexible; and have

considerable amount of damping. Consequently, they play vital role in predicting dynamic behaviour of

rotor systems. In the present chapter, we will incorporate bearing dynamic parameters in the mathematical

model of the rotor system. In previous chapter, it is shown that a bearing has normally eight dynamic

parameters (four for the stiffness and four for the damping, with direct and cross coupled terms). To start

with, first we will consider a long rigid rotor mounted on flexible anisotropic bearings (without damping

and without cross-coupled stiffness terms). Next, in the long rigid rotor system we will incorporate a

more general bearing model with eight bearing dynamic coefficients. Subsequently, along with the

flexibility of bearing, the shaft flexibility with rigid discs is also considered. Finally, the flexibility of the

shaft, bearings, and foundations has been included for prediction of the dynamic behaviour of the rotor

system and the forces transmitted through the supports. The approximate method is not used in the

present chapter; however, wherever equations are large in number the matrix and vector forms are

preferred.

4.1 Symmetrical Long Rigid Shaft in Flexible Anisotropic Bearings

In rotor systems where 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 the rigid rotor. It is assumed

that the shaft has no flexibility, and bearings are assumed to behave as linear springs having the stiffness,

kx and ky, the horizontal and vertical directions, respectively. The center of gravity of the rotor mass, m, is

offset from the geometrical center by distances e and ez as shown in Figure 4.1(a), respectively in the

radial and axial directions. x and y are the linear displacement of the rotor (the geometrical center) in the

horizontal and vertical directions, respectively; whereas, ϕy and ϕx are the angular displacement of the

rotor (the geometrical center line) in the z-x and y-z planes, respectively. Figure 4.1(b) shows the positive

convension for the angular displacement. For the present case, there is no coupling between various

displacements, i.e., x, y, ϕy and ϕx. Hence, free body diagrams (Figure 4.2) and equations of motion have

been obtained by giving such displacements independent of each other. It is assumed that the linear and

163

angaulr displacemnets are small, and the direction of the unbalnce force and moment do not change with

the tilting of the shaft (i.e., due to ϕy and ϕx). In fact without this assumptions, it will give equations of

motion with force containing unknown angular displacements (i.e., ϕy and ϕx) and equations would be

parametrically excited systems (i.e., differntial equations with time dependent coefficients).

Figure 4.1(a) A rigid rotor mounted on a flexible bearing

Figure 4.1(b) Positive conventions of angular displacements

164

Figure 4.2(a) A free body diagram of the rotor in y-z plane for a pure translator motion

Figure 4.2(b) A free body diagram of the rotor in y-z plane for a pure translator motion

Figure 4.2(c) A free body diagram of the rotor in x-y plane

165

Figure 4.2(d) A free body diagram of the rotor in y-z plane for the pure rotational motion

Figure 4.2(e) A free body diagram of the rotor in z-x plane for the pure rotational motion

From Figure 4.2(a, b and c) equations of motion in the x and y directions are

2 cos - 2 xme t k x mxω ω = and 2sin 2 yme t k y myω ω − = (4.1)

From Figure 4.2(d and e) equations of motion in the ϕx and ϕy directions are

2 2sin / 2z y x d xme e t k l Iω ω ϕ ϕ− − = and

2 2cos / 2z x y d yme e t k l Iω ω ϕ ϕ− = (4.2)

For sinusoidal vibrations, we can write

2 2 2 2, , and x x y yx x y yω ω ϕ ω ϕ ϕ ω ϕ= − = − = − = − (4.3)

166

On substituting equation (4.3) into equations (4.1) to (4.2), the unbalnce response can be expressed as

2 2

2 2cos cos ; sin sin

2 2x y

me mex t X t y t Y t

k m k m

ω ωω ω ω ω

ω ω= = = =

− − (4.4)

and 2 2

2 2 2 2sin sin and cos cos

0.5 0.5

z zx x y y

y d x d

me e me et t t t

k l I k l I

ω ωϕ ω ω ϕ ω ω

ω ω= = Φ = = Φ

− − (4.5)

where X and Y are linear displacement amplitudes, and Φx and Φy are angular displacement amplitudes.

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

linear displacements in the x- and y- directions, and two for angular displacements in the x-z and y-z

planes. Critical speeds can be written as

1 2 3 4

222 0.50.5; ; ; and

y yx xcr cr cr cr

d d

k k lk k l

m m I Iω ω ω ω= = = = (4.6)

From equation (4.4) on squaring expressions for x and y, and on adding, it gives

2 2

2 21

x y

X Y+ = (4.7)

It is an equation of the ellipse. Similarly from equation (4.5), we get

22

2 21

yx

x y

ϕϕ+ =

Φ Φ (4.8)

It would be interesting to observe the whirl direction (i.e., the clochwise or the counter clockwise) with

respect to the spin speed direction. Let us first consider the linear displacement only, i.e., equation (4.4).

and assume that kx < ky (i.e., 1 2cr crω ω< ).

Case I: When the rotor operates below the first critical speed, i.e., 1crω ω< from equation (4.4) both X

and Y are positive. Hence, the rotor whirls in the same direction as the spin of shaft a shown in Figures

4.3(a) and 4.4(a-c). This type of whirl is called the forward synchornous whirl.

167

Figure 4.3 Mode shapes for a rigid rotor mounted on flexible bearings

Case II: For 1 2cr cr

ω ω ω< < , from equation (4.4) the displacement amplitude X becomes negative

indicating that the horizontal displacement of the rotor is always in the opposite direction to that in Case I.

It follows from this that the rotor must whirl in the opposite direction to the spining of shaft as shown in

Figures 4.3(c) and 4.4(d-f). This type of whirl is called the backward synchronous whirl.

Case III: At higher shaft speeds 2cr

ω ω> both X and Y are negative which leads to both displacements in

opposite direction to that in Case I. Figures 4.3(a) and 4.4(g-i) reveals that the rotor whirl once more in

the same direction as the spin of shaft. (the phase may be different). In all three cases if we observe

motion of a point P on the orbit at time instants t = t1 and t2 the point will move from P1 to P2 as shown in

Figures 4.4(c, f and i), which clearly show the direction of whirl.

168

(i) Synchronous whirl,

1crω ω<

(ii) Anti-synchronous whirl,

1 2cr cr

ω ω ω< <

(iii) Synchronous whirl,

2cr

ω ω>

(a) Plot of x versus ωt

(d) Plot of x versus ωt

(g) Plot of x versus ωt

(b) Plot of y versus ωt

(e) Plot of y versus ωt

(h) Plot of y versus ωt

(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 4.4 Whirl directions with respect to the shaft spin frequency

169

Equation (4.8), relating to the angular motion of the rotor, is also equation an ellipse. This means that

there is an elliptical orbital trajectory of the rotor ends due to the angular motion of the rotor as shown in

Figures 4.4(b and d). This rotor motion is caused by the unbalance couple meω2d acting on the rotor. For

3crω ω< and 4crω ω> the forward synchornous whirl persists. A reversal of the direction of the orbit

occurs for the rotor spin speed between two critical speeds associated with angular motion of the rotor

(i.e., 3 4cr crω ω ω< < ) as shown in Figure 4.4d. In general, the motion of the rotor will be combination of

both the linear and angular dispalcemnets and at or near critical speeds only such whirling will be

distinguisable.

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

directions, as well as at each end of the rotor. The force transmitted is given by the product of spring

stiffness and rotor deflection at the bearing. Bearing force amplitudes are

( ) ( )0.5 ; and 0.5x x y y y x

f k x l f k y lϕ ϕ= ± = ± (4.9)

in the horizontal and vertical directions, respectively. In equation (4.9) the + sign refers to the angular

displacement of the rotor which causes its end to deflect in the same direction to the linear displacement

and the - sign refers to the angular displacement of the rotor which causes its end to deflect in the opposite

direction to the linear displacement. These bearing forces must take on maximum values when the system

is operated at any of the critical speeds, whenever either of x, y, ϕx and ϕy are maximum.

Example 4.1 A long rigid symmetric rotor is supported at ends by two identical bearings. The shaft has

0.2 m of diameter, 1 m of length, and 7800 kg/m3 of mass density. Bearing dynamic parameters are as

follows: kxx = kyy = 1 kN/mm with rest of the stiffness and damping terms equal to zero. By considering

the gyroscopic effect negligible, obtain transverse critical speeds of the system.

Solution: We have the following data:

61 10k = × N/m, 2 27800 0.1 1 245.04m r lρπ π= = × × = kg

and

1 1 1 12 2 2 2

16 12 16 12245.04 0.2 245.04 1 0.6126 20.42 21.0326dI md ml= + = × × + × × = + = kg-m

2

170

Since cross-coupled stiffness coefficients is the x and y directions are zero and no gyroscopic effect is

considered, hence a single plane motion can be considered. For the present analysis the coupling is not

considered between the linear and rotational displacements. The stiffness in the x and y directions are

same, hence critical speeds corresponding to the linear motion can be written as

1,2

62 2 1 10

90.34 rad/s245.04

cr

k

mω

× ×= = =

In which only the positive frequency has the physical meaning, the negative sign has been ignored.

Similarly critical speeds corresponding to the titling motion can be written as

3,4

2 6 21 10 1

154.184 rad/s2 2 21.0326

cr

d

kl

Iω

× ×= = =

× Answer

Example 4.2 (a) Find transverse natural frequencies of a system shown in Figure 4.5. The mass and the

diametral mass moment of inertia of the rotor are 2.51 kg and 0.00504 kg-m2, respectively. The total span

of the shaft between bearings is 508 mm. Treat the shaft as rigid. Bearings have equal flexibility in all

directions, the stiffness constant for either one of them being kx = 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 kver = 350 N/m for each of the bearings.

Figure 4.5 A rigid rotor on flexible supports

Solution: (a) Considering a single plane (e.g., the horizontal) motion with the assumption of uncoupled

linear and angular motions, EOM for free vibrations can be written as

2

2 0 and 0.5 0x d y x ymx k x I k lϕ ϕ+ = + =

171

where x is the linear displacement of the centre of gravity of the shaft and φy is the angular displacement

of shaft centre line with the z-axis in z-x plane. The mass and mass moment of inertia of the disc are given

as 2.51 kgm = and 20.00504 kg-mdI = . We have kx = k = 175 N/m and l = 0.508m. Hence, natural

frequencies can be written as

1 2

2211.81 rad/s and 66.93 rad/s

2nf nf

d

k kl

m Iω ω= = = = Answer

(b) With different stiffness properties in the horizontal and vertical directions, natural frequencies are

given as

1 2

2 216.70 rad/s; 11.81 rad/sver hoz

nf nf

k k

m mω ω= = = =

3 4

2 2

94.66 rad/s; and 66.93 rad/s2 2

ver hoz

nf nf

d d

k l k l

I Iω ω= = = = Answer

Example 4.3: Find critical speeds of a rotor system as shown in Figure 4.6. Take bearing stiffness

properties as: 1.1 kN/mmAxk = ; 1.8 kN/mm

Ayk = ; 3.1 kN/mmBxk = and 3.8 kN/mm

Byk = . The

disc has m =10 kg and Id = 0.1 kg-m2.

Figure 4.6 A long rigid rotor on flexible bearings

Solution: Considering two plane (e.g., the horizontal and the vertical) motions with the assumption of

uncoupled linear and angular motions, equations of motion in the x and y directions with the radial

unbalance, me, are given as

172

2 2

cos and sinA B A Bx x y yme t k x k x mx me t k y k y myω ω ω ω− − = − − = (a)

where x and y are the linear displacements of the centre of gravity of the shaft. Bearings at ends A and B

are not same; however, it is assumed that the difference in bearing dynamic parameters is small and the

shaft remains horizontal in the static equilibrium position. Equations of motion in the ϕy and ϕx directions

with the moment-unbalance, zmee , are

2 2 2 2 2 2cos 0.5 0.5 and sin 0.5 0.5

A B A Bz x y x y d y z y x y x d xme e t k l k l I me e t k l k l Iω ω ϕ ϕ ϕ ω ω ϕ ϕ ϕ− − = − − =

(b)

where φy and φx are angular displacements of shaft centre . The steady state unbalanced (forced) vibration

responses from equations (a) and (b) can be obtain as

2 2

22

cos ; sin

A BA B

y yx x

me mex t y t

k k mk k m

ω ωω ω

ωω= =

+ −+ − (c)

( ) ( )

2 2

2 2 2 2

ω ωcosω ; sinω

0.5 ω 0.5 ωA B A B

z zy x

x x d y y d

me e me et t

k k l I k k l Iϕ ϕ

−

= =+ + −

(d)

On equating determinates of responses in equations (c) and (d) to zero, critical speeds can be obtained as

1

6(1.1 3.1) 10

648.1 rad/s10

A Bx x

cr

k k

mω

+ + ×== = ;

2

6(1.8 3.8) 10

748.3 rad/s10

A By y

cr

k k

mω

+ + ×== = ;

( )

3

2 6 20.5(1.1 3.1) 10 1

4582.58 rad/s0.1

0.5A Bx x

cr

d

k k l

Iω

+ + × ×== = ;

and

( )

4

2 6 20.5(1.8 3.8) 10 1

5291.50 rad/s0.1

0.5A By y

cr

d

k k l

Iω

+ + ×=

×= = Answer

173

4.2 A Symmetrical Long Rigid Shaft on Anisotropic Bearings

Fluid-film bearings impart the damping as well as stiffness forces to the rotor system. The shaft motion in

the horizontal direction is coupled with that in the vertical direction in the presence of fluid-film bearings.

That means a vertical force may produce both the vertical and horizontal linear displacements; similarly, a

horizontal force may produce both the vertical and horizontal linear displacements. The same is valid for

angular displacements due to moments. However, the coupling between the translational (i.e., x and y)

and tilting (i.e., ϕy and ϕx) motion has not been considered. That means a force on a rigid rotor may not

produce angular displacement and similarly, a moment may not produce a linear displacement. As

discussed in previous chapter, in most applications the properties of such bearings are described in terms

of the eight linearised bearing stiffness and damping coefficients. Both bearings are assumed to be

identical. The symmetrical long rigid shaft on anisotropic bearings will be similar to Figure 4.1 with

additional damping and cross-coupled force terms at shaft ends in free body digrams (Fig. 4.7).

Fig. 4.7 (a) A free body diagram of the shaft for the pure translatory motion in y-z plane

Fig. 4.7 (b) A free body diagram of the shaft for the pure translatory motion in z-x plane

174

Fig. 4.7 (c) A free body diagram of the shaft for the pure tilting motion in y-z plane

Fig. 4.7 (d) A free body diagram of the shaft for the pure tilting motion in z-x plane

Equations of motion for the rotor can be written as from free body diagram (Fig. 4.7), as

( ) 2 2 2 2x xx xy xx xyf t k x k y c x c y mx− − − − = (4.10)

( ) 2 2 2 2y yx yy yx yyf t k x k y c x c y my− − − − = (4.11)

( ) ( ) ( ) ( )2 2 2 2( ) 0.5 0.5 - 0.5 0.5xz xx y xy x xx y xy x d y

M t k l k l c l c l Iϕ ϕ ϕ ϕ ϕ− − − = (4.12)

and

175

( ) ( ) ( ) ( )2 2 2 2( ) 0.5 0.5 - 0.5 0.5yz yx y yy x yx y yy x d x

M t k l k l c l c l Iϕ ϕ ϕ ϕ ϕ− − − = (4.13)

in the x, y, ϕy, and ϕx directions, respectively. Here f and M represent the external force and moment

(e.g., due to an unbalance). kij, cij (i, j = x, y) are the eight linearised bearing stiffness and damping

coefficients.

4.2.1 Unbalance response: The unbalance, me, is located at an axial distance, ez, from the rotor

geometrical center (Fig. 4.2c). Unbalance forces in the horizontal and vertical directions can then be

written as

( ) ( )2 2 j jcos Re Ret t

x xf me t me e F e

ω ωω ω ω= = = with 2 xF meω= (4.14)

and

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

y yf me t me e F e

ω ωω ω ω= = − = with 2

jyF meω= − (4.15)

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

directions, respectively. These forces are acting at the center of gravity. Moments about the rotor

geometrical center caused by these forces are

( ) ( )2 2 j jcos Re Ret t

xz z z xzM me e t me e e M e

ω ωω ω ω= = = with 2 xz zM me eω= (4.16)

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

yz z z yzM me e t me e e M e

ω ωω ω ω= = − = with 2

-jyz zM me eω= (4.17)

where xzM and yzM are complex moments (which contain the amplitude and phase information) about

the y and x axes, respectively. Then the response can be assumed as

j j j j

; ; ; =t t t t

y y x xx Xe y Ye e eω ω ω ωϕ ϕ= = = Φ Φ (4.18)

where X, Y, Φy and Φx are complex displacements. Equations of motion (4.10)-(4.13) can be written as

[ ] [ ] [ ] ( )M x C x K x f t+ + = (4.19)

with

176

[ ] [ ]

[ ]

2 2

2 2

2 2

2 2

2 2 0 00 0 0

2 2 0 00 0 0; C ;

0 0 0.5 0.50 0 0

0 0 0.5 0.50 0 0

2 2 0 0

2 2 0 0; ( )

0 0 0.5 0.5

0 0 0.5 0.5

xx xy

yx yy

xx xyd

yx yyd

xx xy

yx yy

xx xy

yx yy

c cm

c cmM

l c l cI

l c l cI

k k

k kK x t

l k l k

l k l k

= =

= =

; ( )

x

y

xzy

yzx

fx

fyf t

M

M

ϕ

ϕ

=

The response takes the following form

j j 2 j; so that j and

t t tx X e x X e x X e

ω ω ωω ω= = = − (4.20)

On substituting equations (4.14)-(4.17) and (4.20) into equations of motion (4.19), we get

[ ] [ ] [ ]( ) 2 jM C K X Fω ω− + + = (4.21)

with

;

x

y

xzy

yzx

FX

FYX F

M

M

= =

Φ Φ

which can be written as

[ ] D X F= with [ ] [ ] [ ] [ ]( )2 jD K M Cω ω= − + (4.22)

The response can be obtained as

X = [D]-1

F (4.23)

Displacement amplitudes of the rotor will be given by

2 2 2 2 2 2 2 2, , ,r i r ir i r i y y y x x xX X X Y Y Y= + = + Φ = Φ + Φ Φ = Φ + Φ (4.24)

and corresponding phase lag will be given by

177

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

r r

y xi i

r r y x

X Y

X Yα β γ δ

Φ Φ = = = = Φ Φ

(4.25)

The resulting shaft whirl orbit can be plotted using equations (4.18) and (4.23) (i.e., j t

x Xeω= and

j ty Ye ω= ), and in general for a stable rotor-bearing system the orbit will take the form as shown in

Figure 4.8.

Figure 4.8 A rotor whirl orbit

The form of the orbit is still elliptical as, it is in previous section, however, the major and minor axes no

longer along the x and y directions, respectively. A typical force vector is also shown on the diagram, and

which is shown to precede the displacement vector in the presence of damping. However, during crossing

of critical speeds it may change the phase.

In the present case the coupling is considered between the vertical and horizontal planes, and no coupling

is considered between the linear and tilting motions. It can be observed that equations (4.10) and (4.11)

(without damping) is similar in form as in chapter 2, i.e., the case when we considered no coupling

between the vertical and horizontal planes and the coupling is considered the between linear and tilting

motions (i.e., a Jeffcott rotor with an offset disc). Mathematically we can write:

11 122 , 2 , xx xyk k k k= = and rest of the analysis and interpretations will be similar as discussed

previous chapter. Similar analyses can be performed by considering equations (4.12) and (4.13). Now

through some examples the present section analysis will be illustrated.

178

Example 4.4 Obtain transverse natural frequencies of a rotor-bearing system as shown in Figure 4.9 for

pure tilting motion of the shaft. Consider the shaft as a rigid and the 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 both bearing A &B: kxx =

200 MN/m, kyy = 150 MN/m, kxy = 15 MN/m, kyx = 10 MN/m.

Figure 4.9 A rigid rotor mounted on two dissimilar bearings

Solution: The following data are given for the present problem

2 20.025 1 7800 15.32m r lπ ρ π= = × × × = kg

( ) ( )1 12 2 2 2

12 123 15.32 3 0.025 1 1.279

dI m r l= + = × × × + = kg-m

2

Equations of motion would be obtained by considering free body diagram of the rotor as shown in Fig.

4.10. Pure tilting of the shaft is considered with ϕx and ϕy be the angular displacements in y-z and z-x

plane, respectively. Let l be the total length of the rotor.

Fig. 4.10 (a) A free body diagram of a rigid rotor in the y-z plane

179

Fig. 4.10 (b) A free body diagram of a rigid rotor in the z-x plane

Equations of motion in ϕx and ϕy directions (Fig. 4.10), respectively, can be written as

( )0.5 0.5xx y xy x d y

k l k l l Iϕ ϕ ϕ− + = (a)

and

( )0.5 0.5yy x yx y d x

k l k l l Iϕ ϕ ϕ− + = (b)

In matrix form above equations can be written as

2 0 0

0 2 0

xx xyd y y

yx yyd x x

k l k lI

k l k lI

ϕ ϕ

ϕ ϕ

+ =

(c)

For free vibration, it takes the form

2

2 0 0

0 2 0

xx xyd y

nf

yx yyd x

k l k lI

k l k lI

ϕω

ϕ

+ =

(d)

which gives frequency equation as

( ) ( )2 4 2 24 2 0d nf d xx yy nf xx yy xy yx

I I l k k k k k k lω ω− + + − = (e)

180

For the present problem the frequency equation becomes

( ) ( ) 2 4 6 2 6 24 1.279 2 1.279 1 200 150 10 200 150 15 10 10 1 0nf nfω ω× − × × × + × + × − × × × =

or

4 8 2 166.543 8.953 10 2.985 10 0nf nfω ω− × + × = (f)

which gives 1nfω = 7584.28 rad/s and

2nfω = 8905.72 rad/s.

Example 4.5 Obtain transverse critical speeds of a long rigid rotor supported on two identical fluid-film

bearings at ends, which has 2 m of the span, 5 kg of mass, and 0.1 kg-m2 of the diametral mass moment of

inertia. Equivalent dynamic properties of both bearings 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 cyx = 1.0×10

-1

N-s/m. Obtain the unbalance response (the amplitude and the phase) with the spin speed of shaft 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 4.11 shows unbalance responses both for the linear (left side) and angular (right side)

displacements. Both the amplitude and phase is plotted. It can be observed that in the plot of linear and

angular displacements two peaks appears and they correspond to critical speeds of the rotor-bearing

system. Since the linear and angular displacements are uncoupled for the present case and hence

corresponding critical speeds appear in respective plots only. There are four critical speeds at: 63 rad/s,

132.5 rad/s, 450 rad/s and 932.5 rad/s. The change in phase at critical speeds can be seen in each of the

plots.

181

Figure 4.11 Amplitude and phase variation with respect to spin speeds (left) linear (right) angular

displacements

4.2.2 Bearing forces

Forces transmitted through bearings are due to deformation of the bearing lubricant film. However, these

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

the bearing horizontal and vertical force components, at both ends A and B of the machine, can be

represented as

A bx A xx A xy A xx A xy A xx y A xy x A xx y A xy xf k x k y c x c y k l k l c l c lϕ ϕ ϕ ϕ= + + + + + + + (4.26)

A by A yx A yy A yx A yy A yx y A yy x A yx y A yy xf k x k y c x c y k l k l c l c lϕ ϕ ϕ ϕ= + + + + + + + (4.27)

B bx B xx B xy B xx B xy B xx y B xy x B xx y B xy xf k x k y c x c y k l k l c l c lϕ ϕ ϕ ϕ= + + + + + + + (4.28)

B by B yx B yy B yx B yy B yx y B yy x B yx y B yy xf k x k y c x c y k l k l c l c lϕ ϕ ϕ ϕ= + + + + + + + (4.29)

182

where , A xx xx B xx xxk k k k= = , etc. Hence equations (4.26)-(4.29) can be written in the matrix form as

[ ] [ ] b b bf c x k x= + (4.30)

with

; ; ;

A bx

A by

b

B bx y y

B by x x

f x x

f y yf x x

f

f

ϕ ϕ

ϕ ϕ

= = =

[ ] [ ];

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

A yx A yy A yx A yy A yx A yy A yx A yy

b b

B xx B xy B xx B xy B xx B xy B xx B xy

B yx B yy B yx B yy B yx B yy B yx B yy

c c c l c l k k k l k l

c c c l c l k k k l k lc k

c c c l c l k k k l k l

c c c l c l k k k l k l

= =

For the unbalance excitation with a 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= = = (4.31)

On substituting equation (4.31) into equation (4.33), we get

[ ] [ ]( ) jb b b

F k c Xω= + (4.32)

This can be used to evaluate bearing forces. Amplitudes of forces transmitted through 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+ + + += = = =

(4.33)

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− − − −

= = = =

(4.34)

It should be noted that bearing forces will be maximum whenever rotor-bearing system is rotating at or

near critical speeds.

183

4.3 A Symmetrical Flexible Shaft on Anisotropic Bearings

For the present case, both the shaft and bearings are flexible as shown in Figure 4.12. The motion in two

orthogonal planes will be considered simultaneously. The analysis allows finding 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 experienced by the rigid 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 bearing coefficients of the previous section analysis. The total deflection of the

disc is the vector sum of the deflection of the disc relative to the shaft ends, plus that of shaft ends in

bearings. For the disc, we observe the displacement of its geometrical centre.

Figure 4.12 A flexible shaft on flexible bearings

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

bearing stiffness and damping coefficients as

bx xx b xy b xx b xy bf k x k y c x c y= + + + and by yx b yy b yx b yy bf k x k y c x c y= + + + (4.35)

where bx and

by are instantaneous displacements of shaft ends relative to bearings in the horizontal and

vertical directions, respectively; and they take the following form

j j; t t

b b b bx X e y Y eω ω= = (4.36)

where bX and

bY are complex displacements in x and y directions, respectively. Equation (4.36) can be

differentiated once with respect to time, to give

184

j jj ; jt t

b b b bx X e y Y eω ωω ω= = (4.37)

It should be noted that bearings are modelled as a point connection with the shaft and only linear

displacements is considered since they support mainly radial loads. Bearing forces have the following

form

j t

bx bxf F eω= and

j t

by byf F eω= (4.38)

where bxF and byF are complex displacements in x and y directions, respectively. On substituting in

equation of motion (4.35), we get

j jbx xx b xy b xx b xy bF k X k Y c X c Yω ω= + + + (4.39)

and

j jby yx b yy b yx b yy bF k X k Y c X c Yω ω= + + + (4.40)

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

[ ] b bF K X= (4.41)

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

k c k cF

k c k cFF K

F k c k c

Fk c k c

ω ω

ω ω

ω ω

ω ω

+ + + + = =

+ + + +

and

T

b A b A b B b B bX X Y X Y=

The magnitude of reaction forces transmitted by bearings can also be evaluated in terms of forces applied

to the shaft by the disc.

185

Figure 4.13 A free body diagram of the shaft

From Figure 4.13, the moment balance will be

A0 ( )B by y zxM f l f l a M= ⇒ = − −∑ or ( ) ( )1 1A by y zxf f a l M l= − − (4.42)

and

B0 A by y zxM f l f a M= ⇒ = +∑ or ( ) ( )1B by y zxf f a l M l= + (4.43)

Similarly, forces in the horizontal direction may be written as

( ) ( )1- 1A bx x yzf f a l M l= − (4.44)

and

( ) ( )1B bx x yzf f a l M l= + (4.45)

Equations (4.42)-(4.45) can be combined in a matrix form as

[ ] b sf A f= (4.46)

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 yz

B by zx

f f a l l

f f a l lf f A

f M a l l

f M a l l

− −

− − = = =

For the unbalance excitation, we have

j j and

t t

b b s sf F e f F eω ω= = (4.47)

186

where subscript b refers to the bearing and s refers to the shaft. On substituting equation (4.47) into

equation (4.41), we get

[ ] b sF A F= (4.48)

In equation (4.48) bearing forces are related to reaction forces at the shaft by the disc. On equating

equation (4.41) and (4.48), we get

[K] Xb = [A]Fs or Xb = [K]-1

[A] Fs (4.49)

Equation (4.49) relates the shaft end deflections to the reaction forces and moments on the shaft by the

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

Consider the shaft to be rigid for some instant and let us denote shaft end deflections in the horizontal

direction to be Axb and Byb at ends A and B, respectively, as shown in Figure 4.14. These displacements

are assumed to be small.

Figure 4.14 Rigid body movement of the shaft in z-x plane

The linear displacement in the x-direction can be written as

( )1-

B b A b

A b A b B b

x x a ax x a x x

l l l

− = + = +

(4.50)

The angular displacement of the shaft in x-z plane will be

( )-B b A b

y

x x

lϕ = (4.51)

187

Similarly, for the linear and angular displacements in the y- direction and in the y-z plane, respectively;

we have

1-A b B b

a ay y y

l l

= +

(4.52)

and

( )-A b B b

x

y y

lϕ = (4.53)

Equations (4.50)-(4.53) can be combined in a matrix form as

[ ] 1s bu B x= (4.54)

with

[ ]

( )( )

1

1

1- / 0 / 0

0 1- / 0 /; ;

1/ 0 1/ 0

0 1/ 0 1/

A b

A b

s b

y B b

x B bs

x x a l a l

y y a l a lu x B

x l l

y l l

ϕ

ϕ

= = = − −

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

and at the disc centre vary sinusoidally such that

1 1

j j and t t

s s b bu U e x X e

ω ω= = (4.55)

where ω is the spin speed (or natural frequency in case of free vibrations). On substituting equation (4.55)

into equation (4.54), we have

[ ] 1s b

U B X= (4.56)

On substituting equation (4.49) into equation (4.56), we get

[ ][ ] [ ] [ ] 1

1

s sB As

U K F C F−

= = (4.57)

which gives the deflection of the disc due to the unbalance, when the shaft is rigid. Equation (4.57) will

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

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

188

have to add the deflection due to the deformation of the shaft also in equation (4.57). The deflection

associated with flexure of the shaft alone has already been calculated in Chapter 2, which can be

combined in a matrix form as

[ ] 2s s

u fα= (4.58)

with

[ ]2

2

11 12

21 22

11 12

21 22

0 0

0 0; ;

0 0

0 0

x

y

s s

y yz

x zxs

x f

y fu f

M

M

α α

α αα

ϕ α α

ϕ α α

= = =

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

and disc displacements vary sinusoidally, and can be expressed as

2 2

j j and t t

s s s su U e f F e

ω ω= = (4.59)

On substituting equation (4.59) into equation (4.58), we get

[ ] 2s s

U Fα= (4.60)

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

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

flexure of the shaft, is then given by

1 2s s s

U U U= + [ ] [ ]( ) [ ] s sC F D Fα= + = (4.61)

Equation (4.61) describes the displacement of the shaft at the disc under the action of sinusoidal forces

and moments applied at the disc (it is similar to the influence coefficient matrix). Equation (4.61) can be

written as

[ ] [ ] 1

s s sF D U E U

−= = (4.62)

Equations of motion of the disc can be written in the x-direction and on the z-x plane (see Figure 4.15a),

as

189

2 cos xme t f mxω ω − = and - zx d yM I ϕ= (4.63)

Similarly, equations of motion, in the y-direction and on the y-z plane (see Figure 4.15b), can be written

as

2sin yme t f myω ω − = and - yz d xM I ϕ= (4.64)

Figure 4.15 Free body diagram of the disc (left) in z-x plane (right) in y-z plane

Equations of motion (4.63) and (4.64) of the disc can be written in matrix form as

[ ] s unbM u f f+ =

(4.65)

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 unb unb

xzd y

yzd x

fm x me

fm y meM u f f e F e

MI

MI

ω ω

ω

ω

ϕ

ϕ

− = = = = =

Equations of motion take the following form

[ ] 2

s s unbM U F Fω− + = (4.66)

190

Noting equation (4.62), equation (4.66) becomes

[ ] [ ] 2

s s unbM U E U Fω− + = (4.67)

which gives

[ ] s unbU H F= (4.68)

with

[ ] [ ] [ ]( )1

2H M Eω−

= − +

where [ ]H is the equivalent dynamic stiffness matrix, as experienced by the disc, of the shaft and the

bearing system. Once the response of the disc has been obtained the loading applied to the shaft by the

disc can be obtained by equation (4.62). Then from equation (4.49) we can get shaft ends deflections Xb

at each bearings, which is substituted in equation (4.41) to get bearing forces bF . Alternately, bearing

forces can be used directly from equation (4.48). Displacements and forces have the complex form; the

amplitude and the 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 displacements can be evaluated by

calculating arctangent of the ratio of the imaginary to real components as given by equations (4.33) and

(4.34).

4.4 A Rotor on Flexible Bearings and Foundations

In some rotating machines, e.g. turbines, bearings themselves may be mounted on flexible foundations

(Figure 4.16), which may in turn influence the motion of disc masses. In the present section, a very simple

model of the foundation is considered by ignoring cross-coupled terms of the stiffness and the damping.

For more detailed treatment on foundation effects, it can be referred to Krämer (1993).

191

Figure 4.16 A flexible rotor-bearing-foundation system

The net displacement of the disc is given by the vector sum of (i) the disc displacement relative to shaft

ends, plus (ii) that of shaft ends relative to the bearing, plus (iii) that of the bearing relative to the space.

The theoretical analysis of the disc, the shaft and the bearing responses, 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. Additional governing equations related to the foundation are derived, and how to relate them with

governing equations of the disc and bearings are detailed here.

The relationship between forces transmitted through bearings and displacements of shaft ends is governed

by the bearing stiffness and damping coefficients. The form of governing equation is given by equation

(4.41), which is

[ ] b bf K x= or [ ] b bF K X=

where xb is the shaft end displacement relative to the bearing. Displacements of bearings with respect to

foundations and forces transmitted through bearings are shown in Figure 4.17.

Figure 4.17 A bearing block mounted on a foundation

192

The bearing will respond in the horizontal direction for an external force fbx, which is governed by the

following equation

bx fx f fx f b ff k x c x m x− − = (4.69)

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

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

vertical direction to a force fby is given as

by fy f fy f b ff k y c y m y− − = (4.70)

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

j j

and t t

f f f fx X e y Y eω ω= = (4.71)

On substituting equation (4.71) in equations of motion (4.69) and (4.70), and on combining in the matrix

form (for bearing A), it gives

[ ] A A f A bD X F= (4.72)

with

[ ] 20 00

j ;0 00

x x

y y

f fb

A

f fb

k cmD

k cmω ω

= − +

and x

y

bf

A f A b

f bA A

FXX F

Y F

= =

For both bearings A and B, equations of form (4.72) can be combined as

[ ] b fF D X= (4.73)

with

A b

b

B b

FF

F

=

; [ ]0

0

A

B

DD

D

=

; A f

f

B f

XX

X

=

which gives relative displacements between bearings and the foundation, as

193

[ ] 1

f bX D F−

= (4.74)

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

individual displacements Xb (by equation (4.75)) and Xf (by equation (4.74)), i.e.

[ ] [ ] 1 1 [ ]b f b bW X X K D F Fα

− − ′= + = + =

(4.76)

where [ ]α ′ is a system equivalent dynamic receptance matrix describing the overall shaft support

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

may now proceed in the same manner as described in the previous section except the equivalent dynamic

stiffness matrix [ ]1

α−

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

is possible to substitute back and obtain Fs, Xb and Fb. Forces transmitted to foundations are given

as

fx fx f fx ff k x c x= + and fy fy f fy ff k y c y= + (4.77)

For the unbalance excitation, we have

j t

fx fxf F eω= and

j

t

fy fyf F eω= (4.78)

On substituting equation (4.78) into equation (4.77), we get

0 0j

0 0

x x x

y y y

f f f f

ff f f

F k c X

YF k cω

= +

(4.79)

Forces transmitted through foundations will not be the same as forces transmitted through bearings. Since

bearing masses (i.e., inertia forces) will absorb some forces towards its acceleration. If bearing masses are

negligible then bearings and foundations will transmit same amount of forces, however, may be with

some phase lag due to damping. The amplitude and the phase of forces transmitted through foundations

can be obtained from 1 2 1, , fx fx fyF F F and

2fyF as usual procedure described in previous sections.

More detailed study on the foundation effects is beyond the scope of the present book; however, various

studies have incorporated foundation effects in a rotor-bearing system analysis and some of them are

194

summarized here. Smith (1933) investigated the Jeffcott rotor with internal damping to include a

massless, damped and flexible support system. Lund (1965) and Gunter (1967) showed that damped and

flexible supports may improve the stability of high-speed rotors. Also, Lund and Sternlicht (1962),

Dworski (1964), and Gunter (1970) demonstrated that a significant reduction in the transmitted force

could be achieved by the proper design of a bearing support system. Kirk and Gunter (1972) analyzed the

steady state and transient responses of the Jeffcott rotor for elastic bearings mounted on damped and

flexible supports. Gasch (1976) dealt with the flexible rotating shaft of a large turbo-rotor by the finite

element analysis. He introduced foundation dynamics into the rotor equations via receptance matrices,

which were obtained from modal testing and modal analysis. Vance et al. (1987) provided comparison

results for computer predictions and experimental measurements on a rotor-bearing test apparatus. They

have modeled the rotor-bearing system to include foundation impedance effects by using the transfer

matrix method. Stephenson and Rouch (1992) have utilized the finite element method to analyze rotor-

bearing-foundation systems. They provided a procedure using modal analysis techniques, which could be

applied in measuring frequency response functions to include the dynamic effects of the foundation

structure. Kang et al. (2000) studied of foundation effects on the dynamic characteristics of rotor-bearing

systems. The modeling and analysis of rotor-bearing-foundation systems based on the finite element

method were discussed. A substructure procedure which included the foundation effects in the motion

equations and the application of the dynamic solver of a commercial package was addressed.

A good model of rotor and reasonably accurate model of fluid journal bearings may be constructed using

the FE method or any other reliable method. Indeed, a number of FE based software codes are available

for such modeling. However, a reliable FE model for the foundation is extremely difficult to construct

due to number of practical difficulties (Lees and Simpson, 1983). Experimental modal analysis (Ewins,

2000) is a possible solution, but this requires that the rotor be removed from the foundation, which is not

practical for an existing power station. With these difficulties it is unlikely that the techniques of FE

model updating (Friswell and Mottershead, 1995) could be used, and the direct estimation of the

foundation model from measured responses at the bearing pedestals from machine run-down data has

been accepted as a viable alternative technique (Lees, 1988 and Smart et al., 2000). The estimation

technique assumes that the state of unbalance is known from balancing runs, either by the difference in

the response from two run-downs, or by the estimated unbalance from a single run-down (Edwards, 2000

and Sinha et al., 2002).

195

Concluding Remarks

In the present chapter, we dealt mainly with dynamic responses (critical speeds and unbalance responses)

of a single mass rotor with flexible supports. Dynamic parameters of supports not only provide the

stiffness and damping forces to the rotor, but it also provides asymmetry in these dynamic parameters in

two orthogonal directions. The translatory and conical whirl motions are resulted in for the long rigid

rotor. It is found that the orbit of the shaft center not only becomes elliptical but its major axis becomes

inclined to both orthogonal axes. The forward and backward whirls are observed of the rigid rotor

mounted on anisotropic bearings. The flexibility of the foundation resulted in increase in the effective

flexibility experience by the rotor system, which is expected to decrease the critical speeds. Overall

complexity of the dynamic analysis procedure becomes cumbersome while considering bearings and

foundations even with a single mass rotor. It demands more systematic methods for the dynamic analysis

of multi-mass rotors. In subsequent chapters, while considering the torsional and transverse vibrations of

multi-DOF rotor systems two representative methods called the transfer matrix method (TMM) and the

finite element method (FEM) will be dealt in detail. In the next chapter we will still consider a single

mass rotor only, however, now the effect of gyroscopic moments would be included.

196

Exercise Problems

Exercise 4.1 Obtain bending critical speeds of a rotor as shown in Figure E4.1. It consists of a massless

rigid shaft (1 m of span with 0.7 m from the disc to the left bearing), a rigid disc (5 kg of the mass and 0.1

kg-m2 of the diametral mass moment of inertia) and supported on two identical flexible bearings (1 kN/m

of stiffness for each bearing). Consider the motion in vertical plane only. Is there is any difference in

critical speeds when the disc is placed at the centre of the rotor? If NO then justify the same and if YES

then obtain the same. [Hint: When the disc at the centre of the shaft-span then the uncoupled linear and

angular motions would take place.]

Figure E4.1

Exercise 4.2: Consider a long rigid rotor, R, supported on two identical bearings, B1 and B2, as shown in

Figure E4.2. The direct stiffness coefficients of both bearings in the horizontal and vertical directions are

equal, i.e. K. Take the direct damping, and the cross-coupled stiffness and damping coefficients of both

bearings negligible. The mass of the rotor is m, the span of the rotor is l, and the diametral mass moment

of inertia is Id. Derive equations of motion, and obtain natural frequencies of whirl. Neglect the

gyroscopic effect.

1B2

B

Figure E4.2

Exercise 4.3 Find critical speeds of the rotor bearing system shown in Figure E4.3. The shaft is rigid and

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

197

B1 and B2 are identical bearings and have following properties: kxx = 1.1 kN/m, kyy = 1.8 kN/m, kxy = 0.2

kN/m, and kyx = 0.1 kN/m. Take: B1D = 75 mm, and DB2 = 50 mm.

1B 2B

Figure E4.3

Exercise 4.4 For exercise 4.3 take 25 g-mm of the unbalance in the disc at 380 from a shaft reference

point. Plot the disc response amplitude and phase to show all critical speeds. Plot the variation of bearing

forces with the spin speed of rotor.

Exercise 4.5 Obtain transverse critical speeds of a rotor-bearing system as shown in Figure E4.5.

Consider the shaft as a rigid and the 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 = 200 MN/m, kyy = 150 MN/m, kxy = 15

MN/m, kyx = 10 MN/m, cxx = 200 kN-s/m, cyy = 150 kN-s/m, cxy = 14 kN-s/m, cyx = 21 kN-s/m, and for

bearing B: kxx = 240 MN/m, kyy = 170 MN/m, kxy = 12 MN/m, kyx = 16 MN/m, cxx = 210 kN-s/m, cyy =

160 kN-s/m, cyx = 13 kN-s/m, cyy = 18 kN-s/m. Use a numerical simulation to get the unbalance response

to cross check the critical speeds for an assumed unbalance.

Figure E4.5

Exercise 4.6 For exercise 4.5 consider the shaft as flexible and attach a rigid disc of 10 kg on the shaft at

a distance of 0.6 m from the end A. Obtain the transverse critical speeds of the system by attaching an

unbalance on the disc. Take 40 g-mm of the unbalance in the disc at 1300 from a shaft reference point.

198

Exercise 4.7 For exercise 4.5 obtain critical speeds of the rotor-bearing-foundation system when the

foundation has the following dynamic characteristics: xf

k =yfk =100 MN/m and

x yf fc c= = 50 kN-s/m.

Take the mass of each bearing as 2 kg. Plot the unbalance response amplitude and phase of the shaft end

and the bearing at A with respect to the spin speed of shaft to show all critical speeds of the system. Take

25 g-mm of the unbalance in the disc at 380 from a shaft reference point. Plot also the variation of the

bearing and foundation forces at A with the spin speed.

Exercise 4.8 Consider a simple rigid rotor-bearing system as shown in Figure E4.7. The rotor is

supported on two different flexible bearings. In Figure E4.7, 1L and

2L are distances of bearings 1 and 2

from the center of gravity of the rotor with 1 2L L L= + ,

1R and 2R are distances of balancing planes (i.e.,

rigid discs) from the center of gravity of the rotor, and u is the unbalance. Obtain bearing dynamic

parameters based on the short bearing approximations.

Let m be the mass of the rotor, It is the transverse mass moment of inertia of the rotor about an axis

passing through the center of gravity, Ip is the polar mass moment of inertia of the rotor, k and c are

respectively the stiffness and damping parameters, fx(t) and fy(t) are respectively the impulse in the

horizontal and vertical directions, u is the unbalance, φ is the phase, x and y are linear displacements in

the horizontal and vertical directions respectively, t is the time, and subscripts 1 and 2 represent the right

and left sides from the mid-span of the rotor, respectively. Obtain equations of motion of the rotor-bearing

system in terms of linear displacements (four in numbers, i.e., x1, y1, x2, y2) at two bearings. The

motivation behind obtaining the equations of motion in terms of bearing response is that in practical

situation often these responses can only be accessible to the practicing engineers.

Figure E4.7 A rigid rotor on flexible bearings

[Hint: The linearised equation of motion of the rotor-bearing system is given as

199

[ ]M q + [ ]C q + [ ]K q = Runb

f

where ω is the shaft rotational speed, Runb

f is the residual unbalance force vector, q is the

displacement response vector, and matrices [ ]M , [ ]C and [ ]K are the mass, damping, and stiffness

matrices and are given as

2

2 1 2

2

2 1 2

2

1 2 1

2

1 2 1

0 0

0 0[ ]

0 0

0 0

t t

t t

t t

t t

ml i ml l i

ml i ml l iM

ml l i ml i

ml l i ml i

+ −

+ − = − +

− +

;

1 1

1 1

2 2

2 2

0 0

0 0[ ]

0 0

0 0

xx xy

yx yy

xx xy

yx yy

c c

c cC

c c

c c

=

1 1

1 1

2 2

2 2

0 0

0 0[ ]

0 0

0 0

xx xy

yx yy

xx xy

yx yy

k k

k kK

k k

k k

=

; Runb

f =

2 2

1 2 1 1 2 2 2 2

2 2

1 2 1 1 2 2 2 2

2 2

1 1 1 1 2 1 2 2

2 2

1 1 1 1 2 1 2 2

( )sin( ) ( )sin( )

( )cos( ) ( )cos( )

( )sin( ) ( )sin( )

( )cos( ) ( )cos( )

s s s s

s s s s

s s s s

s s s s

u l r t u l r t

u l r t u l r t

u l r t u l r t

u l r t u l r t

φ φ

φ φ

φ φ

φ φ

Ω + Ω + + Ω − Ω + Ω + Ω + + Ω − Ω +

Ω − Ω + + Ω + Ω +

Ω − Ω + + Ω + Ω +

with

/i il L L= , /i ir R R= , 2/t ti I L= , 2

/p pi I L= ; 1,2i = ]

Exercise 4.9 Consider equations of motion of exercise 4.7 and numerical data given in Table E4.8. Obtain

the response (i.e., the amplitude and the phase) of the bearings with respect to the rotor speed and list

down critical speeds of the rotor-bearing system.

Table E4.8 Details of the rotor model for the numerical example

Property Numerical value

Rotor

Rotor shaft diameter 10 mm

Rotational speed, ω 100 Hz

1 1 2 2 T

q x y x y=

200

Mass, m 4 kg

Length of rotor, L 0.425 m

Distance of bearings from centre of rotor 0.2125 m

Distance of discs from centre of rotor 0.130 m

Transverse mass moment of inertia, d

I 0.0786 kg-m2

Rigid discs

Inner diameter 10 mm

Outer diameter 74 mm

Thickness 25 mm

Bearings

Diameter 25.4 mm

Length to diameter ratio 1

Radial clearance, cr of Bearing 2 0.075 mm

Kinetic viscosity 20.11 centi-Stokes

Temperature of lubricant 40oC

Specific gravity of lubricant 0.87

Exercise 4.10: For the case when a rigid rotor, mounted on two bearings at ends, as shown in Fig. E410,

has varying cross section along the longitudinal axis (e.g., a tapered rotor). For this case the centre of

gravity, G, of the rotor will be offset from the mid-span of the rotor, C. It is assumed that the rotor is

perfectly balanced (i.e., it has no external radial force and corresponding external moment). Let m be the

mass, Id be the diametral mass moment of inertia of the rotor about centre of gravity, kA and kB are

stiffness of bearings A and B respectively, and l is the length of the rotor. Obtain governing equations of

motion for the following three sets of chosen generalized coordinates for a single plane motion of the

rotor.

Fig. E4.10 An axially asymmetric shaft mounted on flexible dissimilar bearings

201

(i) If we choose generalized coordinates as ( , )G zx ϕ , where the linear displacement of the centre of

gravity is xG, and tilting of the rotor from the horizontal (i.e., z-axis) is zϕ .

(ii) If we choose generalized coordinates as ( , )E zx ϕ , where the linear displacement of a point on the

rotor where if a transverse force is applied then it produces pure translation of the rotor (i.e.,

A AE B BEk l k l= ) is xE, and tilting of the rotor remains same as for the first case.

(iii) If we choose generalized coordinates as ( , )A zx ϕ , where the linear displacement of the extreme left

end of the rotor is xA, and tilting of the rotor remains same as for the first case.

(iv) If we choose generalized coordinates as ( , )A Bx x , where the linear displacement of the extreme left

and right ends of the rotor are xA, and xB, respectively.

(v) If we choose generalized coordinates as ( , )C zx ϕ , where the linear displacement of the mid-span is xC,

and tilting of the rotor from the horizontal (i.e., z-axis) is zϕ .

(vi) If we choose generalized coordinates as ( , )E Gx x , where displacements have similar meanings as

defined previously.

[Answer:

(i) 2 2

0 0

0 0

A B B BG A AGG G

d B BG A AG A AG B BGz z

m k k k l k lx x

I k l k l k l k lϕ ϕ

+ − + = − +

; where the subscript in l represents

corresponding length. The mass matrix is uncoupled and the stiffness matrix is coupled, i.e., the static

coupling exists.

(ii) 2 2

0 0

0 0

EG E A B E

EG d z A AE B BE z

m ml x k k x

ml I k l k lϕ ϕ

+ + = +

; The stiffness matrix is uncoupled and

the mass matrix is coupled, i.e., the dynamic coupling exists.

(iii) 2

0

0

AG A A B B A

AG d z B B z

m ml x k k k l x

ml I k l k lϕ ϕ

+ + =

; The mass and stiffness matrices are

uncoupled, i.e., both the static and dynamic coupling exists.

(iv) 0

0

BG AG A BA A

d d A AG B BGB B

ml ml k l k lx x

I I k ll k llx x

+ = − −

; The mass and stiffness matrices are non-

symmetric and coupled.

(v) ( ) ( )

( ) ( )0

0 0

CG A B A AC B BCC C

d A AG B BG A AC AG B BC BGz z

m ml k k k l k lx x

I k l k l k l l k l lϕ ϕ

− − + + = − − +

; The mass and stiffness

matrices are non-symmetric and coupled.

202

Exercise 4.11 For a perfectly balanced rigid rotor mounted on flexible bearings as shown in Fig. E4.11

the following data are given: m = 10 kg, Id = 0.015 kg-m2, l = 1m, lAG = 0.6m, kA = 120 kN/m, kB = 140

kN/m. Consider one plane motion with two-DOFs and coupling in the generalized coordinates. Obtain the

transverse natural frequencies and mode shapes of the rotor-bearing system.

Fig. E4.11 An axially asymmetric shaft mounted on flexible dissimilar bearings

Exercise 4.12 Obtain transverse natural frequencies of a rotor-bearing system as shown in Figure 4.10 for

translatory motion of the shaft. Consider the shaft as a rigid and the 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 both bearing A &B: kxx =

200 MN/m, kyy = 150 MN/m, kxy = 15 MN/m, kyx = 10 MN/m.

Figure E4.12 A rigid rotor mounted on two dissimilar bearings

Exercise 4.13 Choose a single correct answer from the multiple choice questions:

(i) A rigid long rotor supported on flexible anisotropic bearings can have transverse natural frequencies

(a) 1 (b) 2 (c) 3 (d) 4 (e) more than 4

203

(ii) A rigid long rotor supported on flexible anisotropic bearings can have reversal of the orbit direction as

the spin speed of the rotor is increased.

(a) True (b) False

(iii) For a rigid rotor mounted on fluid-film bearings would have coupling of motions in

(a) linear displacements (x, y) only (b) angular displacements (ϕx, ϕy) only

(c) between the linear and angular displacements (x and ϕy) or/and (y and ϕx)

(d) both (a) and (b)

(iv) For a flexible rotor (e.g., a Jeffcott rotor with an offset disc) mounted on rigid bearings would have

coupling of motions in

(a) linear displacements (x, y) only (b) angular displacements (ϕx, ϕy) only

(c) between the linear and angular displacements (x and ϕϕϕϕy) or/and (y and ϕϕϕϕx)

(d) both (a) and (b)

(v) For a flexible rotor (e.g., a Jeffcott rotor with disc at mid span) mounted on rigid bearings would have

coupling of motions in

(a) linear displacements (x, y) only (b) angular displacements (ϕx, ϕy) only

(c) between the linear and angular displacements (x and ϕy) or/and (y and ϕx)

(d) none of the displacement would be coupled

(vi) For a flexible rotor (e.g., a Jeffcott rotor) mounted on flexible bearings would have coupling of

motions in

(a) linear displacements (x, y) only (b) angular displacements (ϕx, ϕy) only

(c) between the linear and angular displacements (x and ϕy) or/and (y and ϕx)

(d) all linear and angular displacement would be coupled

204

References

Dworski, J., 1964, Journal of Engineering for Power. Transactions ASME, Series A 86, 149-160. High

speed rotor suspension formed by fully floating hydrodynamic radial and thrust bearings.

Edwards, S., Lees, A.W., and Friswell, M.I., 2000, Journal of Sound and Vibration 232(5), 963–992.

Experimental identification of excitation and support parameters of a flexible rotor-bearing-

foundation system from a single run-down.

Ewins, D.J., 2000, Modal Testing: Theory, Practice and Application (2nd Edition ed.), Research Studies

Press Ltd, Hertfordshire, UK.

Friswell, M.I., and Mottershead, J.E., 1995, Finite Element Model Updating in Structural Dynamics,

Kluwer Academic Publishers, Dodrecht.

Gunter, E.J., 1967, Journal of Engineering for Industry. Transactions ASME, Series B 89, 683-688. The

influence of internal friction on the stability of high speed rotors.

Gunter, E. J., 1970, Journal of Lubrication Technology, Transactions ASME, Series F 92, 59-75.

Influence of flexibly mounted rolling element bearing on rotor response, Part I- linear analysis.

Gasch, R., 1976, Journal of Sound and Vibration 47, 53-73. Vibration of large turborotors in fluid-film

bearing on an elastic foundation.

Kang, Y., Chang, Y.-P., Tsai, J.-W., Mu, L.-H., and Chang, Y.-F., 2000, Journal of Sound and Vibration,

231(2), 343-374. An investigation in stiffness effects on dynamics of rotor-bearing-foundation

systems.

Kirk, R.G., and Gunter, E.J., 1972, ASME Journal of Engineering for Industries 94, 221-232. The effect

of support flexibility and damping on the synchronous response of a single-mass flexible rotor.

Krämer E., 1993, Dynamics of Rotors and Foundations, Springer-Verlag, New York.

Lees, A.W., Simpson, I.C., 1983, IMechE Conference on Vibrations in Rotating Machinery, Paper C6/83,

London, UK, pp. 37–44. The dynamics of turbo-alternator foundations.

Lees, A.W., 1988, IMechE Conference on Vibrations in Rotating Machinery, Paper C306/88, UK, pp.

209–216. The least squares method applied to identified rotor/foundation parameters.

Lund, J.W., and Sternlicht, B., 1962, Journal of Basic Engineering, Transactions ASME, Series D 84,

491-502. Rotor-bearing dynamics with emphasis on attenuation.

Lund, J.W., 1965, Journal of Applied Mechanics, Vol. 32, Transactions ASME, Series E 87, 911-920. The

stability of an elastic rotor in journal bearings with flexible, damped supports.

Pilkey, W.D., Wang, B.P., and Vannoy, D., 1976, ASME Journal of Engineering for Industries 1026-

1029. Efficient optimal design of suspension systems for rotating shafts.

205

Sinha, J.K., Lees, A.W., and Friswell, M.I., 2002, Mechanical Systems and Signal Processing 16(2–3),

255–271. The identification of the unbalance and the foundation model of a flexible rotating

machine from a single run down.

Smart, M.G., Friswell, M.I., and Lees, A.W., 2000, Proceedings of the Royal Society of London, Series A:

Mathematical, Physical and Engineering Sciences 456, 1583–1607. Estimating turbogenerator

foundation parameters—model selection and regularisation.

Smith, D.M., 1933, Proceedings of the Royal Society, Series A 142, 92. The motion of a rotor carried by a

flexible shaft in flexible bearing.

Stephenson, R.W., and Rouch, K.E., 1992, Journal of Sound and Vibration 154, 467-484. Generating

matrices of the foundation structure of a rotor system from test data.

Vance, J.M., Murphy, B.T., and Tripp, H.A., 1987, ASME Journal of Vibration Acoustics, Stress,

Reliability in Design 109, 8-14. Critical speeds of turbomachinery: computer predictions vs.

experimental measurements-Part II: effect of tilt-pad bearing and foundation dynamics.

----------------------------------************************------------------------------