Applications of force vibration

• Rotating unbalance • Base excitation • Vibration measurement devices

Rotating unbalance (1)

Rotating unbalance: Vibration caused by irregularities in the distribution of the mass in the rotating component.

Rotating unbalance (2)

e

Fr

kx

m0 Fr

trω

)()( txtx r+

xc

0mm −

)(tx

FBD 1 FBD 2

FBD 1 rr Fxxm −=+ )(0

FBD 2 kxxcFxmm r −−=− )( 0

00 =+++ kxxcxmxm r

tex rr ω= sin

tex rrr ωω−= sin2

temkxxcxm rr ωω=++ sin20

m0

m

Rotating unbalance (3)

temkxxcxm rr ωω=++ sin20

)](Im[)( tztx =tj rZetz ω=)( ,

tm

emxxx rrnn ωω=ω+ζω+ sin2 202

tjrnn

rem

emzzz ωω=ω+ζω+ 2022

ζ+−

=

ωζω+ω−ω

ω=

rjrr

mem

jmemZ

rnrn

r

212 2

20

22

20

)sin()()( 0 θ+ωω= tHm

emtx r

)(ωH

21

12tan

rr

−ζ

−=θ −;

Rotating unbalance (4)

)sin()sin()()( 0 θ+ω=θ+ωω= tXtHm

emtx rr

emmXH

0

)( =ω

Example: washing machine

A model of a washing machine is illustrated in the figure. A bundle of wet clothes form a mass of 10 kg (m0) and causes a rotating unbalance. The rotating mass id 20 kg (including m0) and the diameter of the washer basket (2e) is 50 cm. Assume that the spin cycle rotates at 300 rpm. Let k be 1000 N/m and ζ =0.01. (a) Calculate the force transmitted to the sides of the washing machine. (b) The quantities m, m0 e and ω are all fixed by the previous design. Design the isolation system so that the force transmitted to the side of the washing machine is less than 100 N.

Example: Unbalance motor

An electric motor has an eccentric mass of 1 kg (10% of the total mass) and is set on two identical springs (k = 3.2 N/mm). The motor runs at 1750 rpm, and the mass eccentricity is 100 mm from the center. The springs are mounted 250 mm apart with the motor shaft in the center. Neglect damping and determine the amplitude of vertical vibration.

Base excitation (intro)

Examples

• Sensitive equipment placed on vibrating foundation

• An automotive suspension system • Buildings subjected to earthquakes

Base excitation (1)

FBD

0)()( =−+−+ yxkyxcxm EOM

Assumption: the base moves harmonically

tYty bω= cos)(

tkYtcYkxxcxm bbb ω+ωω−=++ cossin

1st harmonic input 2nd harmonic input

1st method: superposition

(xp)1 (xp)2 xp = +

Base excitation (2)

]Re[cos)( tjb

bYetYty ω=ω=

yyxxx nnnn22 22 ω+ζω=ω+ζω+

2nd method: Frequency response method

kyyckxxcxm +=++

)](Re[)( tztx =tj bZetz ω=)( ,

tjn

tjbnnn

bb YeYejzzz ωω ω+ωζω=ω+ζω+ 22 22

Sub. into EOM to change to complex form

Complex EOM

( ) ( ) tjnbn

tjnbnb

bb YejZej ωω ω+ωζω=ω+ωζω+ω− 222 22

Yjrr

jrYj

jZnbnb

nbn

ζ+−

ζ+=

ωζω+ω+ω−ωζω+ω

=)2(1

)2(1)2(

)2(222

2

nbr ωω=;

Base excitation (3)

)](Re[)( tztx =

tj bZetz ω=)(

YHYjrr

jrZ ⋅ω=⋅

ζ+−

ζ+= )(

)2(1)2(1

2

YeHZ j ⋅⋅ω= θ)(

)()()()( θ+ωωθ ⋅ω=⋅⋅⋅ω= tjtjj bb YeHeYeHtz

ζ+−

ζ+=ω

jrrjrH)2(1

)2(1)( 2

)cos()()( θ+ω⋅ω= tYHtx b

)cos()( θω += tXtx b YHX ⋅= )(ω；

Displacement transmissibility (1)

Disp. transmissibility = Input disp.

Output disp.

222

2

)2()1()2(1)(

rrrH

YX

ζζω+−

+==

Disp. Transmissibity specifies how motion is transmitted from the base to the mass at various driving frequency ω.

Displacement transmissibility (2)

222

2

)2()1()2(1)(

rrrH

YX

ζζω+−

+==

1. r ≈ 0 ⇒ D.T. ≈ 1 2. r ≈ 1 ⇒ D.T. → large 3. r < ⇒ D.T. > 1 Large ζ, small D.T. r = ⇒ D.T. = 1 r > ⇒ D.T. < 1 Large ζ, large D.T.

2

22

Increase ζ

Increase ζ

r

Force transmissibility (1)

The force transmitted to the mass is done through the spring and damper.

0)()( =−+−+ yxkyxcxm

)()()( yxcyxktF −+−=

From EOM

)()()()( txmyxkyxctF −=−+−=

)cos()()( θ+ω⋅ω= tYHtx bFrom

)cos()()( 2 θωωω +⋅⋅−= tYHtx bb

)cos()cos()()( 2 θωθωωω +=+⋅⋅= tFtYHmtF bTbb∴

Force transmissibility (2)

)cos()cos()()( 2 θωθωωω +=+⋅⋅= tFtYHmtF bTbb

YHmF bT ⋅⋅= )(2 ωω

222

22

)2()1()2(1

rrrr

kYFT

ζζ+−

+⋅=

Yrr

rmF bT ⋅+−

+⋅= 222

22

)2()1()2(1ζ

ζω

222

22

)2()1()2(1

rrrkYrFT ζ

ζ+−

+⋅=

F.T. tells how much force is transmitted from the base to the mass at various driving frequencies

Force transmissibility (3)

222

22

)2()1()2(1

rrrr

kYFT

ζζ+−

+⋅=

Example (Base excitation)

Determine the effect of speed on the amplitude of displacement of the automobile as well as the effect of the value of the car’s mass. Assume that the suspension system provides an equivalent stiffness of 4×105 N/m and damping of 20×103 N.s/m.

tty bωsin)01.0()( =

vb 2909.0=ω

m

rad/s

v (km/h)

Example 2 (Base excitation)

Determine the amplitude of vertical vibration of the spring-mounted trailer as it travels at a velocity of 25 km/h over the road whose contour may be expressed by a sinusoid. The mass of the trailer is 500 kg and that of the wheels alone may be neglected. During loading, each 75 kg added to the load caused the trailer to sag 3 mm on its springs. Assume that the wheels are in contact with the road at all times and neglect damping. At what critical speed vc is the vibration of the trailer greatest? [J. L. Meriam & L. G. Kraige 8/71]

Measurement devices (1)

FBD

0)()( =−+−+ yxkyxcxm EOM

Assumption: the base moves harmonically

tYty ωcos)( =

y(t) : Actual vibration of the interested object x(t) : Vibration of the mass m inside a measurement device Measured value : w(t) = x(t) – y(t)

ymkwwcwm −=++

Measurement devices (2)

ymkwwcwm −=++ ywww nn −=++ 22 ωζω

]Re[cos)( tjYetYty ωω ==

)](Re[)( tztw =tjZetz ω=)( ,

Sub. into EOM to change to complex form

tjnn Yezzz ωωωζω 222 =++ Complex EOM

tjtjnn YeZej ωω ωωωζωω 222 )2( =++−

Yrjr

rYj

Znn

+−

=

++−

=ζωωζωω

ω212 2

2

22

2

nr ωω=

)(ωH )(ωTor Frequency response

Measurement devices (3)

)()()()( θωωθω ωω +=== tjtjjtj YeTYeeTZetz

)cos()cos()()](Re[)( θωθωω +=+== tWtYTtztw

222

2

)2()1()(

rrrT

YW

ζω

+−==

YeTYTYrjr

rZ jθωωζ

)()(21 2

2

=⋅=

+−

=

Actual vibration Measured vibration

Displacement Transmissibility

Differ from “Displacement transmissibility” of base excitation

Seismometer

• Is used to measure displacement of vibration • The value of |T| approaches unity when r is large → Seismometer

region (r > 3) • Low natural frequency (large mass or small k) • Can measure in a range from 10 to 500 Hz with its natural

frequency of 1 to 5 Hz

YWT =)(ω

r

Increasing ζ

Range for seismometer

Accelerometer (1)

• Is used to measure acceleration of vibration • Use piezoelectric effect: Force (acceleration) ∝ Voltage

Quartz or ceramic crystals

222

2

)2()1()(

rrrT

YW

ζω

+−==

222

2

222

2

)2()1()2()1(

)(

rr

A

rrYW nyn

ζ

ω

ζ

ωω

+−=

+−=

222

2

)2()1( rr

AW y

nζ

ω+−

=⋅

Accelerometer (2)

Increasing ζ

r

222

2

)2()1( rr

AW y

nζ

ω+−

=⋅

• |T| approaches unity when r is small → accelerometer region • Accelerometer range is widest when damping ratio = 0.65-7

(0 < r < 0.6) • High natural frequency (small mass and large k) • Can measure in a range from 0 to over 10 kHz with its natural

frequency of 30 to 50 kHz and weight less than 20 gm.

Example : (Accelerometer)

An accelerometer has a suspended mass of 0.01 kg with a damped natural frequency of vibration of 150 Hz. When mounted on an engine undergoing an acceleration of 1 g at an operating speed of 6000 rpm, the acceleration is recorded as 9.5 m/s2 by the instrument. Find the damping constant and the spring stiffness of the accelerometer. [Singiresu S. Rao, Ex 10.3]