Wang X. Vehicle Noise and Vibration Refinement

Vehicle vibration measurement and analysis 53

the integration at the point of recording the signal. These often become

unreliable for double integration of acceleration, limiting signal-to-noise

ratio and suffering from DC drift.

For simple harmonic motion analysed in the frequency domain, the inte-

gral may be obtained by dividing the amplitude of a spectral component by

iω where i is the square root of −1 and ω = 2πf (f is the frequency in Hz).

3.8.4 Assessing vibration levels – human response

to vibration

Two ISO standards deal with the human response to vibration. They are:

• ISO 2631 Part 1 (1985) – Mechanical Vibration and Shock: Evaluation

of Human Exposure to Whole Body Vibration – Part 1: General

Requirements.

• ISO 5349 (1986) – Mechanical Vibration Guidelines for the Measurement

of Human Exposure to Hand-transmitted Vibration.

The ISO 2631 Part 1 (General Requirements) offered user-friendly guid-

ance on the effects of vibration acceleration amplitude (1–80 Hz). Three

boundaries are given for various exposure periods between 1 minute and

24 hours. These are:

• Reduced comfort boundary

• Fatigue-decreased proﬁ ciency boundary

• Exposure limits (for health and safety).

The boundaries often still form part of contemporary performance speciﬁ -

cations for vehicles. Part 1 of ISO 2631 was revised in 1997. The guidance

on safety, performance and comfort boundaries was removed. In its place,

there are the following:

• The main body of the text describes a means of measuring a weighted

rms acceleration according to:

a

T

at t

T

ww

d=

(

)

⎡

⎣

⎢

⎤

⎦

⎥

∫

1

2

0

12

(3.10)

where a

w

(t) is the weighted acceleration time history in m/s

2

, and T is

the duration in seconds.

• Different weightings are given for health, comfort perception, the dif-

ferent axes and the position of the human subject (standing, seated and

recumbent).

A weighting W

k

is used for assessing the effects of vibration on the

comfort and perception of a standing person with vertical acceleration in

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

54 Vehicle noise and vibration reﬁ nement

the head-to-toe axis. Weighting W

k

suggests that a standing person is most

susceptible to vertical vibration in the frequency range 4–8 Hz. Annexes to

the main text give limited guidance on the health, comfort and perception

and motion sickness effects of vibration. This guidance is in terms of

weighted rms acceleration. The vibration dose value (VDV) is introduced

but it is not used in any guidance on the likely effects of vibration. VDV is

deﬁ ned as:

VDV d

w

4

=

(

)

[]

⎡

⎣

⎢

⎤

⎦

⎥

∫

at t

T

0

14

(3.11)

where the units of VDV are m s

−1.75

.

If the VDV is repeated n times during a period, the total VDV period is

then:

V

DV VDV

total

=

⎛

⎝

⎞

⎠

×

=

∑

t

i

n

i

14

1

(3.12)

Approximate indications of the likely reaction to various magnitudes of

frequency-weighted rms acceleration are listed according to ISO 2631 Part

1 as:

• <0.315 m s

−2

not comfortable

• 0.315–0.63 m s

−2

a little uncomfortable

• 0.5–1.0 m s

−2

fairly uncomfortable

• 0.8–1.6 m s

−2

uncomfortable

• 1.25–2.5 m s

−2

very uncomfortable

• >2.0 m s

−2

extremely uncomfortable

Responses of human behaviour to vibration in different frequency ranges

are shown in Fig. 3.14 which shows that motion sickness occurs between

0.1 and 0.63 Hz. Major body resonances occur between 1 and 80 Hz.

Vibration disturbs speech at 2–20 Hz, and vibration causes potential annoy-

ance between 0.1 and about 400 Hz. Vibration may cause spinal damage at

about 5 to 20 Hz. Figure 3.15 shows that the acceleration of the human

body can be written as a weighted acceleration sum:

W

AS =

(

)

+

()

+14 14

22

2

..aaa

xyz

(3.13)

If a human is exposed to three different vibration environments, the dose

is deﬁ ned as:

D

ose =×

=

∑

t

i

n

τ

1

100%

(3.14)

where t

i

is elapsed time and τ

i

is allowed time.

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

Vehicle vibration measurement and analysis 55

EFFECTS ON HUMANS

FREQUENCY - Hz

INFRASONICS

*

+

^

KEY

STRUCTURE-BORNE VIBRATION

AIRBORNE NOISE

AUDIBLE RANGE ULTRASONICS

0.1 1 10 100 10

3

10

4

10

5

Tonal auditory sensation

Pulsatile auditory sensation

Non-auditory sensation

Vibratory sensation

Disequilibrium

Motion sickness

Major body resonances

Disturbance of speech

Blurring of vision

Interference of tasks

Potential annoyance

Subjective after-failure

Spinal damage

+

Frequency range

Uncertain range of occurrence

Pressure Injury (barotrauma) at extreme intensities

Chronic disorders associated clinically with repeated occupational exposure to whole body vibration

occupational exposure to vibration from hand-held power tools.

Chronic diseases affecting the blood vessels, bones and joints of the hand and forearm associated with cumulative

Gastrointestinal disorders

+

Reynaud’s phenomenon etc.

^

Potential annoyance

Subjective after-failure

Speech Interference

Loss of work efficiency

Hearing loss

Middle ear damage*

3.14 Response of human hearing behaviour to vibration.

a

z

a

y

a

x

L

y

L

x

L

z

3.15 Acceleration of the human body.

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

56 Vehicle noise and vibration reﬁ nement

τ

i

a

t

i

=

⎛

⎝

⎜

⎞

⎠

⎟

×

0

2

0

eq

(3.15)

where

a

0

2

28 3 485=×=..ms

(a

0x

= a

0y

= 2 m/s

2

, a

0z

= 2.8 m/s

2

) and energy

equivalent acceleration is deﬁ ned as:

aT

T

at

T

eq rms

d

()

=

∫

1

2

0

(3.16)

where T is sampling time and a

rms

is root mean square acceleration.

If t

0

= 10 minutes, the four-hour energy equivalent acceleration is:

aat

T

eq eq

() ( )4

4

0

=

(3.17)

where

Tt

i

n

=

∑

1

is total exposed time in one event.

The exposure from several events can be calculated from:

aT

T

ii

i

n

i

n

eq

(

)

=

∑

2

1

(3.18)

where T

i

is the total elapsed time in the ith event and a

i

is the equivalent

acceleration in the ith event.

Vibration exposure is deﬁ ned by:

• Vibration + time = vibration exposure

• Vibration exposure + time = tissue damage.

Vibration exposure is measured according to national and international

standards. In order to avoid breaking occupational health legislation, at

lowest cost, you will need to conduct the following monitoring and risk

assessment:

• Is there a problem?

• How big is the problem?

• What causes the problem?

• How do we reduce the problem?

• How do we prevent the problem?

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

Vehicle vibration measurement and analysis 57

3.9 Vibration isolation

Consider the situation when the support to a single-degree-of-freedom

system has a motion applied to it as shown in Fig. 3.16. In Fig. 3.16 both

the support and the mass are moving rather than just the mass, and so it is

the relative displacement between the mass and the support that is impor-

tant. The equation of motion for free vibration becomes:

m



x

= −k(x − y) − c(



x

−



y

) (3.19)

Declaring that:

z = x − y,



z

=



x

−



y

,



z

=



x

− ÿ (3.20)

then:

m(



z

+ ÿ) = −kz − c



z

(3.21)

m



z

+ kz + c



z

= −mÿ (3.22)

and, if it is a harmonic excitation:

y = Ye

iωt

(3.23)

Then:



y

= iωYe

iωt

(3.24)

ÿ = −ω

2

Ye

iωt

(3.25)

and so:

mω

2

Ye

iωt

= m



z

+ kz + c



z

(3.26)

Now we can assume that:

z = Ze

(iωt−iφ)

(3.27)

Then:



z

= iωZe

(iωt−iφ)

(3.28)



z

= −ω

2

Ze

(iωt−iφ)

(3.29)

mω

2

Ye

iωt

= −mω

2

Ze

(iωt−iφ)

+ kZe

(iωt−iφ)

+ ciωZe

(iωt−iφ)

m

k

x

y

c

3.16 A single-degree-of-freedom system being excited at the base

support.

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

58 Vehicle noise and vibration reﬁ nement

Z

mY

km c

t

e

i

ii

i

ωφ

ω

ωω

−

(

)

=

−+

2

Z

mY

km c

e

i

i−

=

−+

φ

ω

ωω

2

(3.30)

Now:

xyz YZ

Y

mY

km c

kc

km c

t

=+= +

()

=+

−+

⎛

⎝

⎜

⎞

⎠

⎟

=

+

−+

−

ee

i

e

i

ii

i

φω

ω

ωω

ω

2

ii

e

i

ω

Y

t

(3.31)

Now:

x = Xe

iωt−φi

(3.32)

X

kc

km c

Y

tt

e

i

e

ii i

ωφ ω

ω

ωω

−

=

+

−+

2

(3.33)

X

Y

kc

km c

kk m c mc

km c

ei

i

i−

=

+

−+

=

−

()

+−

−

()

+

φ

ω

ωω

ωω ω

ωω

2

222 3

2

22

(3.34)

X

Y

kc

km c

=

+

()

−+

()

⎡

⎣

⎤

⎦

2

Z

ZZ

()

2

(3.35)

Dividing the numerator and denominator by k

2

:

X

Y

c

k

m

k

c

k

=

+

()

−

⎛

⎝

⎜

⎞

⎠

⎟

+

()

⎡

⎣

⎢

⎤

⎦

⎥

1

2

ω

ωω

(3.36)

where

m

k

n

=

⎛

⎝

⎞

⎠

1

2

ω

(3.37)

and

c

k

n

= 2

1

ξ

ω

(3.38)

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

Vehicle vibration measurement and analysis 59

Then Equation 3.36 becomes:

X

Y

n

nn

=

+

⎛

⎝

⎞

⎠

−

⎛

⎝

⎜

⎞

⎠

⎟

+

⎛

⎝

⎞

⎠

⎡

⎣

⎢

⎤

⎦

⎥

1

2

1

2

ξω

ω

ξω

ω

(3.39)

From Equation 3.34:

tan −

(

)

=

−

()

+

φ

ω

ωω

mc

kk m c

3

222

(3.40)

φ

ξω

ω

ξω

ω

=

+

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

+

⎛

⎝

⎞

⎠

−

tan

1

3

2

1

2

n

nn

(3.41)

Equations 3.40 and 3.41 have been used to generate Fig. 3.17.

4

3

2

1

0

mod X/Y

01 2 3450.5

(a)

(b)

Damping ratio = 0.15

Damping ratio = 0.25

Damping ratio = 0.5

1.5 2.5

Frequency ratio w

/w

n

2

4.53.5

√

140

120

100

80

60

40

20

0

Phase angle

01 2 3450.5 1.5 2.5

Frequency ratio w /w

n

4.53.5

3.17 The response of a single-degree-of-freedom system when forced

at its support.

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

60 Vehicle noise and vibration reﬁ nement

In many systems of the type shown in Fig. 3.17 we are interested in

transmitting as little vibration as possible to the base. This problem can

become critical when the excitation is harmonic. Clearly the force transmit-

ted to the base is through springs and dampers. The transmitted force can

be written as:

F

tr

= m



x

= −mω

2

Xe

−iφ

e

iωt

(3.42)

F

kY

kc

km c

t

n

tr

i

e

i

=

+

−+

−

⎛

⎝

⎞

⎠

⎡

⎣

⎢

⎤

⎦

⎥

ω

ωω

ω

2

(3.43)

Assume F

0

= kYe

iωt

is the amplitude of the actual excitation force.

FF

n

nn

tr

i

=−

⎛

⎝

⎞

⎠

+

⎛

⎝

⎞

⎠

−

⎛

⎝

⎞

⎠

+

⎛

⎝

⎞

⎠

0

2

12

ω

ξ

ω

ξ

ω

(3.44)

TR

i

tr

==

−−

⎛

⎝

⎞

⎠

−

⎛

⎝

⎞

⎠

+

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

F

n

nn

n0

2

12

ξ

ω

ξ

ω

(3.45)

TR =

+

⎛

⎝

⎞

⎠

−

⎛

⎝

⎞

⎠

⎛

⎝

⎜

⎞

⎠

⎟

+

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

12

2

ξ

ω

ξ

ω

n

nn

n

(3.46)

When damping is negligible:

TR =

−

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

=

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

1

2

0

2

0

ω

n

X

g

X

g

Δ

(3.47)

From Equation 3.36:

X

Y

X

g

n

=

−

⎛

⎝

⎞

⎠

=

−

1

2

0

ω

Δ

(3.48)

ΔX

0

is the static deﬂ ection of the mass on the spring stiffness k. To reduce

transmissibility at a given frequency the stiffness may be increased. This

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

Vehicle vibration measurement and analysis 61

will reduce the static deﬂ ection of the mass, and therefore reduce the

transmissibility.

3.10 The vibration absorber

By tuning the two-degrees-of-freedom system in Fig. 3.18 to the frequency

of the exciting force, the system acts as a vibration absorber, and in the

ideal case reduces the motion of the main mass m

1

to zero.

The equation of motion for the system shown in Fig. 3.18 can be written

in matrix form as:

m

2



x

2

+ k

2

(x

2

− x

1

) = 0 (3.49)

m

1



x

1

+ k

1

x

1

+ k

2

(x

1

− x

2

) = F

0

sin ωt (3.50)

m

x

kk k

kk

x

Ft

1

2

1

2

12 2

22

1

2

0

00

⎡

⎣

⎢

⎤

⎦

⎥

{}

+

+−

−

⎡

⎣

⎢

⎤

⎦

⎥

{}

=

{



sin

ω

}}

(3.51)

Consider excitation force in the complex domain:

F = F

0

e

iωt

(3.52)

x

1

= X

1

e

iωt

, x

2

= X

2

e

iωt

kkm k

kkm

X

F

12 1

2

222

2

1

2

0

+− −

−−

⎡

⎣

⎢

⎤

⎦

⎥

{}

=

{}

ω

(3.53)

X

kkm k

kkm

F

1

2

12 1

2

222

2

1

0

{}

=

+− −

−−

⎡

⎣

⎢

⎤

⎦

⎥

{}

−

ω

(3.54)

F

0

sin wt

k

1

k

2

m

1

m

2

3.18 The vibration absorber.

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

62 Vehicle noise and vibration reﬁ nement

The determinant of the left-hand matrix of Equation 3.53 can be written

as:

Det[K] = (k

1

+ k

2

− mω

2

)(k

2

− m

2

ω

2

) − k

2

(3.55)

Two natural frequencies of the system can be calculated from:

Det[K] = (k

1

+ k

2

− mω

2

)(k

2

− m

2

ω

2

) − k

2

= 0 (3.56)

Z

ZZZZZ

12

2

11

2

1

22

2

11

2

1

22

2

1

114

=

+

⎛

⎝

⎜

⎞

⎠

⎟

+± +

⎛

⎝

⎜

⎞

⎠

⎟

+

⎡

⎣

⎢

⎤

⎦

⎥

−

k

11

2

22

2

Z

(3.57)

ω

2

1

< ω

2

22

< ω

2

(3.58)

where ω

2

11

= k

1

/m

1

and ω

2

22

= k

2

/m

2

.

X

km F

kkm km k

1

22

2

0

12 1

2

22

2

=

−

()

+−

()

−

()

−

ω

ωω

(3.59)

X

kF

kkm km k

2

20

12 1

2

22

2

=

+−

()

−

()

−

ωω

(3.60)

Dividing the numerator and the denominator of Equation 3.59 by k

1

k

2

gives:

X

k

m

kk

F

kk

m

kk k

m

kk

1

2

12

0

21

1

2

12 2

2

12

1

11 1

=

−

⎛

⎝

⎜

⎞

⎠

⎟

+−

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

ω

ωω

⎝⎝

⎜

⎞

⎠

⎟

−

k

2

1

(3.61)

X

k

F

k

1

2

22

2

0

2

1

2

11

2

22

2

1

11

=

−

⎛

⎝

⎜

⎞

⎠

⎟

+−

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

−

Z

kk

1

(3.62)

X

F

k

2

0

1

2

1

2

11

2

22

2

1

11

=

+−

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

−

ω

(3.63)

In Equation 3.55, if

ωω

==

22 2 2

km

, the amplitude X

1

= 0 but the absorber

mass has a displacement equal to

X

F

k

2

0

2

=

−

(3.64)

k

2

and m

2

depend on the maximum allowable X

2

. Note that although the

vibration absorber is effective at ω = ω

22

there are two natural frequencies

Copyrighted Material downloaded from Woodhead Publishing Online

Delivered by http://woodhead.metapress.com

ETH Zuerich (307-97-768)

Sunday, August 28, 2011 12:01:01 AM

IP Address: 129.132.208.2

Wang X. Vehicle Noise and Vibration Refinement

Подождите немного. Документ загружается.