Rotating unbalance Base excitation Vibration measurement ...
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Applications of force vibration • Rotating unbalance • Base excitation • Vibration measurement devices
Transcript of Rotating unbalance Base excitation Vibration measurement ...
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]
