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
meω 2 cos ωt - 2k x x = mx
and
meω 2 sin ωt − 2k y y = my
(4.1)
From Figure 4.2(d and e) equations of motion in the ϕx and ϕy directions are
−meω 2ez sin ωt − k y l 2ϕ x / 2 = I d ϕx
and
meω 2 ez cos ωt − k xl 2ϕ y / 2 = I d ϕy
(4.2)
For sinusoidal vibrations, we can write
x = −ω 2 x , y = − ω 2 y, ϕx = −ω 2ϕ x and ϕy = −ω 2ϕ y
(4.3)
166 On substituting equation (4.3) into equations (4.1) to (4.2), the unbalnce response can be expressed as
x=
meω 2 cos ωt = X cos ωt ; 2k x − mω 2
meω 2 sin ωt = Y sin ωt 2k y − mω 2
(4.4)
meω 2 ez cos ω t = Φ y cos ωt 0.5k x l 2 − ω 2 I d
(4.5)
y=
and
ϕx =
meω 2 ez sin ωt = Φ x sin ωt 0.5k y l 2 − ω 2 I d
and ϕ y =
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
k ωcr1 = x ; m
ωcr = 2
2k y m
;
ωcr
3
0.5k xl 2 = ; Id
and
ωcr = 4
0.5k y l 2 Id
(4.6)
From equation (4.4) on squaring expressions for x and y, and on adding, it gives
x2 X
+
2
y2 Y
=1
(4.7)
2
It is an equation of the ellipse. Similarly from equation (4.5), we get
ϕ x2 Φ 2x
+
ϕ y2 Φ 2y
=1
(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., ωcr1 < ωcr2 ).
Case I: When the rotor operates below the first critical speed, i.e., ω < ωcr1 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 ωcr1 < ω < ωcr2 , 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 ω > ωcr2 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
(ii) Anti-synchronous whirl,
(i) Synchronous whirl,
(iii) Synchronous whirl,
ωcr < ω < ωcr
ω < ωcr
1
1
ω > ωcr
2
2
(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
ω < ωcr and ω > ωcr the forward synchornous whirl persists. A reversal of the direction of the orbit 3
4
occurs for the rotor spin speed between two critical speeds associated with angular motion of the rotor (i.e., ωcr3 < ω < ωcr4 ) 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
f x = k x ( x ± 0.5lϕ y ) ;
and
f y = k y ( y ± 0.5lϕ x )
(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: k = 1 × 106 N/m,
m = ρπ r 2l = 7800π × 0.12 × 1 = 245.04 kg
and I d = 161 md 2 + 121 ml 2 = 161 × 245.04 × 0.22 + 121 × 245.04 × 12 = 0.6126 + 20.42 = 21.0326 kg-m2
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
ωcr = 1,2
2k = m
2 × 1× 106 = 90.34 rad/s 245.04
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
ωcr = 3,4
kl 2 = 2Id
1×106 ×12 = 154.184 rad/s 2 × 21.0326
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
mx + 2k x x = 0
and I d ϕy + 0.5k xl 2ϕ y = 0
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 m = 2.51 kg and I d = 0.00504 kg-m 2 . We have kx = k = 175 N/m and l = 0.508m. Hence, natural frequencies can be written as
ωnf = 1
2k = 11.81 rad/s and ωnf2 = m
kl 2 = 66.93 rad/s 2Id
Answer
(b) With different stiffness properties in the horizontal and vertical directions, natural frequencies are given as
ωnf =
2kver 2khoz = 16.70 rad/s; ωnf2 = = 11.81 rad/s m m
ωnf =
kver l 2 = 94.66 rad/s; and ωnf4 = 2Id
1
3
khoz l 2 = 66.93 rad/s 2Id
Answer
Example 4.3: Find critical speeds of a rotor system as shown in Figure 4.6. Take bearing stiffness properties as: k xA = 1.1 kN/mm ; k y A = 1.8 kN/mm ; k xB = 3.1 kN/mm and k yB = 3.8 kN/mm . 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
meω 2 cos ωt − k xA x − k xB x = mx and
meω 2 sin ωt − k y A y − k yB y = my
(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, meez , are
meω 2 ez cos ωt − 0.5k xA l 2ϕ y − 0.5k xB l 2ϕ y = I d ϕy and
meω 2ez sin ωt − 0.5k yA l 2ϕ x − 0.5k yB l 2ϕ x = I d ϕx (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
meω 2
x=
cos ωt ;
y=
k xA + k xB − mω 2
meω2 ez
ϕy =
(
)
0.5 k xA + k xB l
cos ωt ; 2
−
2
meω 2 sin ωt k yA + k yB − mω 2
ϕx =
ω Id
(c)
meω2ez
(
)
sinωt (d) 2
2
0.5 k y A + k yB l − ω I d
On equating determinates of responses in equations (c) and (d) to zero, critical speeds can be obtained as
ωcr = 1
ωcr = 2
ωcr =
kx + kx A
B
m ky +ky A
=
(1.1+ 3.1)×106 = 648.1 rad/s ; 10
=
(1.8+ 3.8)×106 = 748.3 rad/s ; 10
B
m
(
)
0.5 k x + k x l 2 A
B
Id
3
=
0.5(1.1+ 3.1)×106 ×12 = 4582.58 rad/s ; 0.1
=
0.5(1.8+ 3.8)×106 ×12 = 5291.50 rad/s 0.1
and
ωcr = 4
(
)
0.5 k y + k y l 2 A
Id
B
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
f x (t ) − 2k xx x − 2k xy y − 2cxx x − 2cxy y = mx
(4.10)
f y (t ) − 2 k yx x − 2 k yy y − 2c yx x − 2c yy y = my
(4.11)
(
)
(
)
(
)
(
)
M xz (t ) − k xx 0.5l 2ϕ y − k xy 0.5l 2ϕ x - cxx 0.5l 2ϕ y − cxy 0.5l 2ϕ x = I d ϕy and
(4.12)
175
(
)
(
)
(
)
(
)
M yz (t ) − k yx 0.5l 2ϕ y − k yy 0.5l 2ϕ x - c yx 0.5l 2ϕ y − c yy 0.5l 2ϕ x = I d ϕx
(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
(
)
(
f x = meω 2 cos ω t = Re meω 2 e jωt = Re Fx e jωt
)
with
Fx = meω 2
(4.14)
with
Fy = − jmeω 2
(4.15)
and
(
)
(
f y = meω 2 sin ω t = Re − jmeω 2 e jωt = Re Fy e jωt
)
where Fx and Fy 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
(
)
(
M xz = meω 2 ez cos ωt = Re meω 2 ez e jωt = Re M xz e jωt
(
)
(
)
M yz = meω 2 ez sin ωt = Re − jmeω 2 ez e jωt = Re M yz e jωt
)
with
M xz = meω 2 ez
(4.16)
with
M yz = -jmeω 2 ez
(4.17)
where M xz and M yz are complex moments (which contain the amplitude and phase information) about the y and x axes, respectively. Then the response can be assumed as
x = Xe jωt ;
y = Ye jωt ;
ϕ y = Φ y e jωt ;
ϕ x =Φ x e jωt
(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 )} with
(4.19)
176
0 0 ; 0 Id
2cxx 2c [C] = 0yx 0
2k xy
0
2k yy
0
0
0.5l 2 k xx
0
0.5l 2 k yx
0 ; 0.5l 2 k xy 0.5l 2 k yy
m 0 0 0 m 0 [ M ] = 0 0 I d 0 0 0 2k xx 2k [ K ] = 0yx 0
0
2cxy
0
2c yy
0
0
0.5l 2cxx
0
0.5l 2 c yx
0 ; 0.5l 2 cxy 0.5l 2 c yy 0
x y { x(t )} = ; ϕ y ϕ x
fx fy { f (t )} = M xz M yz
The response takes the following form
{ x} = { X } e jωt ;
so that
{ x} = jω { X } e jωt
and
{x} = −ω 2 { X } e jωt
(4.20)
On substituting equations (4.14)-(4.17) and (4.20) into equations of motion (4.19), we get
( −ω [ M ] + jω [C ] + [ K ]){ X } = {F } 2
(4.21)
with
X Y { X } = ; Φ y Φ x
Fx Fy {F } = M xz M yz
which can be written as
[ D ]{ X } = { F }
with
[ D ] = ([ K ] − ω 2 [ M ] + jω [C ])
(4.22)
The response can be obtained as {X} = [D]-1 {F}
(4.23)
Displacement amplitudes of the rotor will be given by
X=
X r2 + X i2 , Y = Yr2 + Yi 2 ,
and corresponding phase lag will be given by
Φ y = Φ 2yr + Φ 2yi , Φ x = Φ 2xr + Φ 2xi
(4.24)
177
Φ yi Φy r
Xi -1 Yi , β = tan , Xr Yr
γ = tan -1
α = tan -1
Φx -1 , δ = tan i Φ xr
(4.25)
The resulting shaft whirl orbit can be plotted using equations (4.18) and (4.23) (i.e., x = Xe jωt and
y = Ye jωt ), 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:
2 k xx = k11 , 2k xy = k12 , 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
m = π r 2l ρ = π × 0.0252 × 1× 7800 = 15.32 kg
(
)
(
)
I d = 121 m 3r 2 + l 2 = 121 × 15.32 × 3 × 0.0252 + 12 = 1.279 kg-m2
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 − ( 0.5k xx lϕ y + 0.5k xy lϕ x ) l = I d ϕy
(a)
− ( 0.5k yy lϕ x + 0.5k yx lϕ y ) l = I d ϕx
(b)
and
In matrix form above equations can be written as
2I d 0
0 ϕy k xx l + 2 I d ϕx k yx l
k xy l ϕ y 0 = k yy l ϕ x 0
(c)
k xy l ϕ y 0 = k yy l ϕ x 0
(d)
For free vibration, it takes the form
2 2I d ω nf 0
0 k xx l + 2 I d k yx l
which gives frequency equation as 4 I d2ωnf4 − 2 I d l ( k xx + k yy ) ωnf2 + ( k xx k yy − k xy k yx ) l 2 = 0
(e)
180
For the present problem the frequency equation becomes
2
4 × (1.279 ) ωnf4 − 2 × 1.279 × 1 × ( 200 + 150 ) × 106 ωnf2 + {200 × 150 − 15 × 10} × 106 × 12 = 0 or 6.543ωnf4 − 8.953 × 108 ωnf2 + 2.985 × 1016 = 0
(f)
which gives ωnf1 = 7584.28 rad/s and ωnf2 = 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 = 8.8×104 N/m, kxy = 1.0×103 N/m, kyx = 1.5×103 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
f bx = A k xx x + A k xy y + A c xx x + A cxy y + A k xx lϕ y + A k xy lϕ x + A cxx lϕ y + A cxy lϕ x
(4.26)
A
f by = A k yx x + A k yy y + A c yx x + A c yy y + A k yx lϕ y + A k yy lϕ x + A c yx lϕ y + A c yy lϕ x
(4.27)
B
f bx = B k xx x + B k xy y + B c xx x + B cxy y + B k xx lϕ y + B k xy lϕ x + B cxx lϕ y + B cxy lϕ x
(4.28)
B
f by = B k yx x + B k yy y + B c yx x + B c yy y + B k yx lϕ y + B k yy lϕ x + B c yx lϕ y + B c yy lϕ x
(4.29)
182
where A k xx = k xx ,
B
k xx = k xx , etc. Hence equations (4.26)-(4.29) can be written in the matrix form as
{ fb } = [cb ]{ x} + [ kb ]{ x}
(4.30)
with
A fbx f f = { b } A fby ; B bx B f by A cxx c [cb ] = A cyx B xx B c yx
x y x = { } ; ϕ y ϕ x
c A c yy
c l A c yx l
c B c yy
c l B c yx l
A xy
B xy
A xx
B xx
x y x = { } ; ϕ y ϕ x
c l A c yy l ; B cxy l B c yy l A xy
A k xx k [ kb ] = A k yx B xx B k yx
k xy A k yy
k xx l A k yx l
k xy B k yy
k xx l B k yx l
A
A
B
B
k xy l A k yy l B k xy l B k yy l A
For the unbalance excitation with a frequency ω, the bearing responses and forces can be expressed as
{ x} = { X } e jωt ;
{ x} = jω { X } e jωt ;
{ fb } = {Fb } e jωt
(4.31)
On substituting equation (4.31) into equation (4.33), we get
{Fb } = ([ kb ] + jω [ cb ]) { X }
(4.32)
This can be used to evaluate bearing forces. Amplitudes of forces transmitted through bearings are then given by
A
Fbx =
A
Fbx2r
+ A
Fbx2i ;
A
Fby =
A
Fby2 r
+ A
Fby2i ;
B
Fbx =
B
Fbx2r
+ B
Fbx2i ;
B
Fby =
B
Fby2 r
+ B
Fby2i (4.33)
with corresponding phase angles are given by
A Fbxi A Fbyi ; ς = tan −1 ; A Fbx r A Fbyr
ε = tan −1
B Fbxi B Fbyi ; λ = tan −1 B Fbxr B Fbyr
η = tan −1
(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
f bx = k xx xb + k xy yb + cxx xb + cxy y b
and
f by = k yx xb + k yy yb + c yx xb + c yy y b
(4.35)
where xb and yb are instantaneous displacements of shaft ends relative to bearings in the horizontal and vertical directions, respectively; and they take the following form
xb = X b e jωt ;
yb = Yb e jωt
(4.36)
where X b and Yb are complex displacements in x and y directions, respectively. Equation (4.36) can be differentiated once with respect to time, to give
184
xb = jω X b e jωt ;
y b = jωYb e jωt
(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
f bx = Fbx e jωt
fby = Fby e jωt
and
(4.38)
where Fbx and Fby are complex displacements in x and y directions, respectively. On substituting in equation of motion (4.35), we get
Fbx = k xx X b + k xyYb + jω cxx X b + jω cxyYb
(4.39)
Fby = k yx X b + k yyYb + jω c yx X b + jω c yyYb
(4.40)
and
which can be written in the matrix form for both bearings A and B, as
{Fb } = [ K ]{ X b }
(4.41)
with
( k xx + jω cxx ) A ( k yx + jω c yx ) [ K ] = A 0 0
A Fbx F {Fb } = A Fby ; B bx B Fby
A
A
(k (k
xy
+ jω cxy )
0
yy
+ jω c yy )
0
0 0
B
( k xx + jωcxx )
B
(k
yx
+ jω c yx )
0 ( k xy + jωcxy ) B ( k yy + jωc yy ) B 0
and
{ X b } = A X b
Y
A b
B
Xb
Y T
B b
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
∑M
B
=0 ⇒
A
fby l = f y (l − a) − M zx
or
∑M
A
=0 ⇒
B
fby l = f y a + M zx
or
f = f y (1 − a l ) − M zx (1 l )
(4.42)
fby = f y ( a l ) + M zx (1 l )
(4.43)
A by
and
B
Similarly, forces in the horizontal direction may be written as
f = f x (1- a l ) − M yz (1 l )
(4.44)
fbx = f x ( a l ) + M yz (1 l )
(4.45)
A bx
and B
Equations (4.42)-(4.45) can be combined in a matrix form as
{ fb } = [ A]{ f s }
(4.46)
with
A f bx f { fb } = A fby ; B bx B f by
fx fy { fs } = ; M yz M zx
(1 − a / l ) 0 [ A] = a/l 0
0 −1/ l 0 (1 − a / l ) 0 −1/ l 0 1/ l 0 a/l 0 1/ l
For the unbalance excitation, we have
{ fb } = {Fb } e jωt
and
{ f s } = {Fs } e jωt
(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
{Fb } = [ A]{Fs }
(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}
{ Xb} = [K]-1 [A] {Fs}
or
(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
x = A xb +
( B xb − A xb ) a = 1- a l
a A xb + B xb l l
(4.50)
The angular displacement of the shaft in x-z plane will be
ϕy =
( B xb - A xb ) l
(4.51)
187 Similarly, for the linear and angular displacements in the y- direction and in the y-z plane, respectively; we have
a a y = 1- A yb + B yb l l
(4.52)
and
ϕx =
( A yb - B yb )
(4.53)
l
Equations (4.50)-(4.53) can be combined in a matrix form as
{u } = [ B ]{x } s1
(4.54)
b
with
{u } s1
x y = ; ϕ y ϕ x s1
(1- a / l ) 0 [ B ] = 1/ l 0
A xb y { xb } = A b ; B xb B yb
0 (1- a / l )
a/l 0
0 1/ l
−1/ l 0
0 a/l 0 −1/ l
For the unbalance excitation (or for the free vibration analysis), shaft displacements at bearing locations and at the disc centre vary sinusoidally such that
{u } = {U } e s1
jωt
s1
and
{ xb } = { X b } e jωt
(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
{U } = [ B ]{ X } s1
(4.56)
b
On substituting equation (4.49) into equation (4.56), we get
{U } = [ B][ K ] [ A ]{F } = [C ]{F } −1
s1
s
s
(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
{u } = [α ]{ f } s2
(4.58)
s
with
{u } s2
x y = ; ϕ y ϕ x s2
fx f { f s } = y ; M yz M zx
α11 0 α12 0 α 0 α 0 [α ] = 021 α 022 α 11 12 0 α 21 0 α 22
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
{u } = {U } e s2
jω t
s2
and
{ f s } = {Fs } e jωt
(4.59)
On substituting equation (4.59) into equation (4.58), we get
{U } = [α ]{F } s2
(4.60)
s
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
{U s } = {U s } + {U s } = ([C ] + [α ]) {Fs } = [ D ]{Fs } 1
2
(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
{Fs } = [ D ] {U s } = [ E ]{U s }
(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
meω 2 cos ωt − f x = mx
and
- M zx = I d ϕy
(4.63)
Similarly, equations of motion, in the y-direction and on the y-z plane (see Figure 4.15b), can be written as
meω 2 sin ωt − f y = my
and
- M yz = I d ϕx
(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
[ M ]{u} + { f s } = { funb } (4.65) with
m 0 0 0 m 0 [M ] = 0 0 I d 0 0 0
0 0 ; 0 Id
x y {u} = ; ϕ y ϕ x
fx fy { fs } = M ; xz M yz
meω 2 2 − jmeω jωt jω t { funb } = e = { Funb } e 0 0
Equations of motion take the following form
−ω 2 [ M ]{U s } + {Fs } = { Funb }
(4.66)
190 Noting equation (4.62), equation (4.66) becomes
−ω 2 [ M ]{U s } + [ E ]{U s } = {Funb }
(4.67)
which gives
{U s } = [ H ]{Funb }
(4.68)
with
[ H ] = ( −ω 2 [ M ] + [ E ])
−1
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 { Fb } . 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
{ fb } = [ K ]{ xb }
or
{Fb } = [ K ]{ X b }
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
f bx − k fx x f − c fx x f = mb xf
(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
f by − k fy y f − c fy y f = mb yf
(4.70)
where yf is the vertical displacement of bearing. The displacement of the bearing will take the form
x f = X f e jωt
and y f = Y f e jωt
(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 D ]{ A X f } = { A Fb }
(4.72)
with
k f x
[ A D ] = 0
{
0 m 2 b −ω k f y 0
X f X = } A f Y A f
and
c f x 0 + j ω mb 0
{ A Fb } =
0 ; c f y
Fbx F by A
For both bearings A and B, equations of form (4.72) can be combined as
{Fb } = [ D ]{ X f }
(4.73)
with
Fb ; B Fb
{Fb } = A
D
[ D ] = A0
0 ; B D
AXf B X f
{X } = f
which gives relative displacements between bearings and the foundation, as
193
{ X } = [ D] {F } −1
f
(4.74)
b
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.
{W } = { X b } + { X f } = [ K ]
−1
+ [ D]−1 {Fb } = [α ′]{ Fb }
(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 −1
stiffness matrix [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
f fx = k fx x f + c fx x f
and
f fy = k fy y f + c fy y f
(4.77)
f fy = Ffy e jωt
(4.78)
For the unbalance excitation, we have
f fx = Ffx e jωt
and
On substituting equation (4.78) into equation (4.77), we get
Ff x k f x = 0 F f y
0 c fx + jω k fy 0
0 X f c f y Y f
(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 F fx1 , F fx2 , F fy1 and F fy2 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 rotorbearing-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.
B1
B2
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.
B1
B2
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 midspan. 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: k f x = k f y =100 MN/m and c f x = c f y = 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, L1 and L2 are distances of bearings 1 and 2 from the center of gravity of the rotor with L = L1 + L2 , R1 and R2 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} = { f }Runb where ω is the shaft rotational speed,
{ f }Runb
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 ml22 + it 0 [M ] = ml1l2 − it 0
k 1xx 1 k [ K ] = yx 0 0
0
ml1l2 − it
ml22 + it 0
0 ml + it
ml1l2 − it
0
k 1xy k 1yy
0 0
0
k
0
k
2 xx 2 yx
2 1
c1xx 1 c [C ] = yx 0 0
ml1l2 − it ; 0 2 ml1 + it 0
0 0 ; k xy2 k yy2
{ f }Runb
=
{q} = x1
y1
c1xy c1yy
0 0
0
cxx2
0
2 c yx
x2
y2
0 0 cxy2 2 c yy
T
u1Ω2s (l2 + r1 )sin(Ω s t + φ1 ) + u2 Ω2s (l2 − r2 )sin(Ω s t + φ2 ) u Ω 2 (l + r ) cos(Ω t + φ ) + u Ω2 (l − r ) cos(Ω t + φ ) s 1 2 s 2 2 s 2 1 s 2 1 2 2 u1Ω s (l1 − r1 )sin(Ω s t + φ1 ) + u2 Ω s (l1 + r2 )sin(Ω s t + φ2 ) 2 2 u1Ω s (l1 − r1 ) cos(Ω s t + φ1 ) + u2 Ω s (l1 + r2 ) cos(Ω s t + φ2 ) with li = Li / L ,
ri = Ri / R ,
it = I t / L2 ,
i p = I p / L2 ;
i = 1,2 ]
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 Rotational speed, ω
10 mm 100 Hz
200 Mass, m Length of rotor, L Distance of bearings from centre of rotor Distance of discs from centre of rotor Transverse mass moment of inertia, Id
4 kg 0.425 m 0.2125 m 0.130 m 0.0786 kg-m2
Rigid discs Inner diameter Outer diameter Thickness
10 mm 74 mm 25 mm
Bearings Diameter Length to diameter ratio Radial clearance, cr of Bearing 2 Kinetic viscosity Temperature of lubricant Specific gravity of lubricant
25.4 mm 1 0.075 mm 20.11 centi-Stokes 40oC 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 ( xG , ϕ z ) , where the linear displacement of the centre of gravity is xG, and tilting of the rotor from the horizontal (i.e., z-axis) is ϕ z . (ii)
If we choose generalized coordinates as ( xE , ϕ z ) , 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., k A l AE = k B lBE ) is xE, and tilting of the rotor remains same as for the first case.
(iii) If we choose generalized coordinates as ( x A , ϕ z ) , 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 ( x A , xB ) , 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 ( xC , ϕ z ) , where the linear displacement of the mid-span is xC, and tilting of the rotor from the horizontal (i.e., z-axis) is ϕ z . (vi) If we choose generalized coordinates as ( xE , xG ) , where displacements have similar meanings as defined previously. [Answer: xG k A + k B m 0 (i) 0 I ϕ + k l − k l d z A AG B BG
k B lBG − k Al AG xG 0 = ; where the subscript in l represents 2 2 k Al AG + k B lBG ϕ z 0
corresponding length. The mass matrix is uncoupled and the stiffness matrix is coupled, i.e., the static coupling exists. (ii)
m ml EG
mlEG xE k A + k B + I d ϕz 0
xE 0 = ; The stiffness matrix is uncoupled and + k l ϕ z 0 0
2 A AE
k l
2 B BE
the mass matrix is coupled, i.e., the dynamic coupling exists. (iii)
m ml AG
ml AG xA k A + kB + I d ϕz k B l
k B l x A 0 = ; The mass and stiffness matrices are k B l 2 ϕ z 0
uncoupled, i.e., both the static and dynamic coupling exists. (iv)
mlBG I d
ml AG x A k Al + − I d xB k All AG
k B l x A 0 = ; The mass and stiffness matrices are non−k B llBG xB 0
symmetric and coupled. (v)
xC ( k A − k B ) m mlCG + 0 I d ϕz ( −k Al AG − k B lBG )
matrices are non-symmetric and coupled.
xC 0 = ; The mass and stiffness + k B lBC lBG ) ϕ z 0
( −k AlAC + k BlBC ) ( k Al AC lAG
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 ϕy) or/and (y and ϕx) (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-bearingfoundation 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 10261029. 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.
----------------------------------************************------------------------------
