9.7 ROTATING UNBALANCE

Back

Many vibrating problems arise because of rotating unbalance in the piece of

machinery. The unbalance may be expressed in terms of the product of an

equivalent unbalanced mass m and the eccentricity of the mass ". The total

mass of the system is denoted by M.

Vibration Isolation for Noise Control 431

Copyright © 2003 Marcel Dekker, Inc.

Let us consider the system shown in Fig. 9-9. The distance yрtЮ is the

displacement of the center of mass of the system from the static equilibrium

position, and y1рtЮ is the displacement of the unbalanced mass with respect

to the same reference. The displacement of the unbalanced mass may be

written in the following form:

y1рtЮ ј yрtЮ ю " e j!t (9-110)

The quantity ! is the rotational frequency for the unbalanced mass.

If we apply Newton’s second law of motion to the system, we obtain

the following expression:

рM _ mЮ

d2y

dt2 ю m

d2y1

dt2 ю RM

dy

dt ю KSy ј 0 (9-111)

We may eliminate the term involving the displacement of the unbalanced

mass by using Eq. (9-110):

d2y1

dt2 ј

d2y

dt2 _ "!2 e j!t (9-112)

If we make the substitution from Eq. (9-112) into Eq. (9-111), the following

result is obtained:

M

d2y

dt2 ю RM

dy

dt ю KSy ј m"!2 e j!t ј Feq e j!t (9-113)

432 Chapter 9

FIGURE 9-9 Mechanical system with a rotating unbalance.

Copyright © 2003 Marcel Dekker, Inc.

We observe that Eq. (9-113) for the system driven by an unbalanced

mass is exactly the same as Eq. (9-69) for an external applied force, if we use

an equivalent force Feq, given by the following:

Feq ј m"!2 (9-114)

Based on this observation, we may write the following expression for the

magnitude of the maximum displacement of the system in terms of the

magnification factor MF:

ymax ј

FeqMF

KS ј

m"!2MF

KS

(9-115)

If we introduce the undamped natural frequency from KS ј M!2

n and the

damping ratio r ј !=!n into Eq. (9-115), we obtain the following result:

ymax ј рm"=MЮr2рMFЮ (9-116)

The magnitude of the force transmitted to the foundation may be

found from the definition of the transmissibility, using the equivalent

force as the driving force for the system:

FT ј FeqTr ј m"!2Tr ј рm"!2

nЮr2рTrЮ (9-117)

In the low-frequency region (the stiffness-controlled region), the magnification

factor and the transmissibility each approach unity. For small

frequency ratios (or for practical purposes, r < 0:2), the maximum amplitude

and the transmitted force are proportional to the square of the frequency:

ymax  рm"=MЮr2 рr _ 1Ю (9-118a)

FT  рm"!2

nЮr2 рr _ 1Ю (9-118b)

In the high-frequency region (the mass-controlled region), the magnification

factor is almost inversely proportional to the square of the frequency

ratio. For large frequency ratios (or for practical purposes, r > 6),

the maximum amplitude and transmitted force approach the following limiting

values:

ymax  m"=M рr _ 1Ю (9-119a)

FT  рm"!2

nЮр1 ю 4_2r2Ю1=2 рr _ 1Ю (9-119b)

If the speed of rotation for a system with rotational (or translational)

unbalance is constant, the magnitude of the equivalent driving force is constant.

On the other hand, for situations in which the speed of rotation is not

constant, such as during start-up or change in operating conditions for the

machine, the equivalent force is not constant, but varies directly propor-

Vibration Isolation for Noise Control 433

Copyright © 2003 Marcel Dekker, Inc.

tional to the square of the frequency. At low speeds, the equivalent force is

small, and consequently, the amplitude of vibration and the transmitted

force are both relatively small. However, at high speeds, the equivalent

force is large, and the amplitude of the transmitted force may be quite

large, depending on the value of the damping ratio. We note from Eq. (9-

119b) that the transmitted force is directly proportional to the damping

ratio at high frequencies, so it is desirable to have as little damping as

practical (_ less than 0.15) for vibration isolation systems used to isolate

vibration due to a rotary machine with variable speed.

Some subjective reactions to vibration of machines are presented in

Table 9-3. The velocity values are maximum or peak vibration velocity

values, vmax.

Example 9-6. A reciprocating air compressor has a total mass of 700 kg

(1543 lbm) and operates at a speed of 1800 rpm. The mass of the unbalanced

reciprocating parts is 12 kg (26.5 lbm), and the eccentricity is 100mm

(3.94 in). The damping ratio for the support system is 0.050. Determine

434 Chapter 9

TABLE 9-3 Subjective Response to Machine Vibration—

the Velocity Values are Peak or Maximum Vibrational

Velocitiesa

Subjective impression vmax range, mm/s Lv range, dB( p)

Very rough >16 >124

Rough 8–16 118–124

Slightly rough 4–8 112–118

Fair 2–4 106–112

Good 1–2 100–106

Very good 0.50–1 94–100

Smooth 0.25–0.50 88–94

Very smooth 0.125–0.25 82–88

Extremely smooth <0.125 <82

aThe velocity is related to the displacement and frequency by the

following expression for sinusoidal vibration:

vmax ј 2_fymax

The velocity level is defined by the following expression:

Lv ј 20 log10рv=vrefЮ

The reference velocity level is vref ј 10 nm/s. The designation dBр pЮ

denotes peak or maximum velocity levels.

Source: Fox (1971).

Copyright © 2003 Marcel Dekker, Inc.

the required spring constant for the support to achieve a transmissibility

level of _20 dB.

The value of the transmissibility is as follows:

Tr ј 10_20=20 ј 0:100

The value of the parameter _ may be calculated from its definition, Eq. (9-

109):

_ ј 1 ю р2Юр0:050Ю2р1 _ 0:1002Ю

р0:100Ю2 ј 1 ю 0:495 ј 1:495

The frequency ratio may be found from the value of the transmissibility and

Eq. (9-108):

рr2Ю2 _ р2Юр1:495Юr2 _

1 _ 0:1002

0:1002 ј 0

r2 ј 1:495 ю р1:4952 ю 99Ю1=2 ј 11:557

r ј р11:557Ю1=2 ј 3:40 ј f =fn

The required undamped natural frequency for the system may now be

found:

fn ј р1800=60Ю

р3:40Ю ј 8:825 Hz ј рKS=MЮ1=2

2_

The required spring constant for the supports may be calculated from the

value of the undamped natural frequency:

KS ј р4_2Юр8:825Ю2р700Ю ј 2:152 _ 106 N=m

ј 2:152MN=m р12,290 lbf=inЮ

If we use four springs (one at each corner) to support the compressor, the

spring constant for each spring would have the following value:

KS1 ј KS=4 ј р2152Ю=р4Ю ј 538 kN=m р3072 lbf=inЮ

Let us determine the other parameters for the vibrating system. The

equivalent force may be found from Eq. (9-114):

Feq ј m"!2

n ј р12Юр0:100Юр4_2Юр1800=60Ю2 ј 42:64 _ 103 N

Feq ј 42:64kN р9585 lbf Ю

The magnitude of the force transmitted into the foundation is found from

the transmissibility:

FT ј FeqTr ј р42:64Юр0:100Ю ј 4:264kN р958:5 lbf Ю

Vibration Isolation for Noise Control 435

Copyright © 2003 Marcel Dekker, Inc.

The magnification factor may be calculated from Eq. (9-81):

MF ј

1

fЅ1_3:402_2 юЅр2Юр0:050Юр3:40Ю_2g1=2 ј 0:0947 ј

KSymax

Feq

The maximum amplitude of vibration for the system is as follows:

ymax ј р0:0947Юр42:64Юр103Ю

р2:152Юр106Ю ј 0:00188m ј 1:88mm р0:074 inЮ

The damping coefficient for the system may be determined from Eq.

(9-25):

RM ј 2рKSMЮ1=2_ ј р2ЮЅр2:152Юр106Юр700Ю_1=2р0:050Ю

RM ј 3881 N-s/m ј 3:381 kN-s/m

The magnitude of the mechanical impedance may be determined from Eq.

(9-89):

jZMj ј р3881Юf1юЅ1=р2Юр0:050Ю_2Ѕ3:40_р1=3:40Ю_2g1=2

jZMj ј р3881Юр31:07Ю ј 120:6_103 N-s/m ј 120:6kN-s/m

The mechanical admittance or mobility is equal to the reciprocal of the

mechanical impedance:

jYMj ј

1

jZMj ј

1

р120:6Юр103Ю ј 8:292_10_6 m/s-N ј 8:292 mm/s-N

The maximum amplitude of the vibratory velocity of the system may

be found from the definition of the mechanical mobility:

vmax ј FeqjYMj ј р42:64Юр103Юр8:292Юр10_6Ю

ј 0:3536m=s р13:92in=secЮ

This level of vibration corresponds to ‘‘smooth’’ operation, as suggested by

the data in Table 9-3.

The maximum vibratory acceleration of the machine may be calculated

from the following:

amax ј 2_fvmax ј р2_Юр1800=60Юр0:3536Ю ј 66:65m=s2

amax=g ј р66:65Ю=р9:806Ю ј 6:80g

Навигация