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
rt ω
) ( )
(t x t
x + r
x c m0
m−
(t) x
FBD 1 FBD 2
FBD 1
r
r F
x x
m0(+ ) = −
FBD 2 (m − m0)x= Fr − cx − kx mx+ m0xr + cx + kx = 0 t
e
xr = sin ωr
t e
xr = − ωr2 sin ωr
t e
m kx
x c x
m+ + = 0 ωr2 sin ωr
m0
m
Rotating unbalance (3)
t e
m kx
x c x
m+ + = 0 ωr2 sin ωr
)]
( Im[
)
(t z t
x =
t j r
Ze t
z( ) = ω ,
m t e x m
x
x+ 2ζωn + ωn2 = 0 ωr2 sin ωr
t r j
n
n e r
m e z m
z
z+ 2ζω + ω2 = 0 ω2 ω
ζ +
= −
ω ζω +
ω
− ω
= ω
r j
r r m
e m j
m e Z m
r n r
n
r 2 1 2 2
0 2 2
2 0 2
) sin(
) ( )
( =
0H ω ω t + θ
m e t m
x
r) (ω H
2 1
1 tan 2
r r
−
− ζ
=
θ
−;
Rotating unbalance (4)
) sin(
) sin(
) ( )
( = 0 H ω ω t + θ = X ω t + θ
m e t m
x r r
e m H mX
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 )
( )
( − + − =
+ c x y k x y x
m
EOM
Assumption: the base moves harmonically
t Y
t
y( ) = cosωb
t kY
t cY
kx x
c x
m+ + = − ωb sin ωb + cosωb
1st harmonic input 2nd harmonic input
1st method: superposition
(x
p)
1(x
p)
2x
p= +
Base excitation (2)
] Re[
cos )
(t Y bt Yej bt
y = ω = ω
y y
x x
x+ 2ζωn + ωn2 = 2ζωn + ωn2
2nd method: Frequency response method ky
y c kx
x c x
m+ + = +
)]
( Re[
)
(t z t
x =
t j b
Ze t
z( ) = ω ,
t n j
t b j
n n
nz z j Ye b Ye b
z+ 2ζω + ω2 = 2ζω ω ω + ω2 ω
Sub. into EOM to change to complex form
Complex EOM
(
−ωb2 + j2ζωnωb + ωn2)
Ze jωbt =(
j2ζωnωb + ωn2)
Yejωbt j Yr r
j Y r
j Z j
n b n
b
n b
n
ζ +
−
ζ
= + ω
ζω +
ω + ω
−
ω ζω +
= ω
) 2 ( 1
) 2 ( 1 )
2 (
) 2
(
2 2
2 2
n
r = ωb ω
;
Base excitation (3)
)]
( Re[
)
(t z t
x =
t j b
Ze t
z( ) = ω
Y H
j Y r r
j
Z r ⋅ = ω ⋅
ζ +
−
ζ
= + ( )
) 2 ( 1
) 2 ( 1
2
Y e
H
Z = (ω) ⋅ jθ ⋅
)
) (
( )
( )
(t = H ω ⋅e jθ ⋅Y ⋅ejωbt = H ω ⋅Yej ωbt+θ z
ζ +
−
ζ
= +
ω r r j
j H r
) 2 ( 1
) 2 ( ) 1
( 2
) cos(
) ( )
(t = H ω ⋅Y ω t + θ
x b
) cos(
)
(t = X ω t +θ
x b ； X = H(ω) ⋅Y
Displacement transmissibility (1)
Disp. transmissibility =
Input disp.
Output disp.
2 2
2
2
) 2 ( )
1 (
) 2 ( ) 1
( r r
H r Y
X
ζ ω ζ
+
−
= +
=
Disp. Transmissibity specifies how motion is transmitted from the base to the mass at various driving frequency ω.
Displacement transmissibility (2)
2 2
2
2
) 2 ( ) 1
(
) 2 ( ) 1
( r r
H r Y
X
ζ ω ζ
+
−
= +
=
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 )
( )
( − + − =
+ c x y k x y x
m
) (
) (
)
(t k x y c x y F = − + −
From EOM
) ( )
( )
( )
(t c x y k x y mx t
F = − + − = − )
cos(
) ( )
(t = H ω ⋅Y ω t + θ
x b
From
) cos(
) ( )
(t = −ω2 ⋅ H ω ⋅Y ω t +θ
x b b
) cos(
) cos(
) ( )
(t = mω2 ⋅ H ω ⋅Y ω t +θ = F ω t +θ
F b b T b
∴
Force transmissibility (2)
) cos(
) cos(
) ( )
(t = mω2 ⋅ H ω ⋅Y ω t +θ = F ω t +θ
F b b T b
Y H
m
FT = ωb2 ⋅ (ω) ⋅
2 2
2 2 2
) 2 ( )
1 (
) 2 ( 1
r r
r r kY
FT
ζ ζ
+
−
⋅ +
= r Y
r m r
FT b ⋅
+
−
⋅ +
= 2 2 2 2 2
) 2 ( )
1 (
) 2 ( 1
ζ ω ζ
2 2
2 2 2
) 2 ( )
1 (
) 2 ( 1
r r
kYr r FT
ζ ζ
+
−
⋅ +
=
F.T. tells how much force is transmitted from the base to the mass at various driving frequencies
Force transmissibility (3)
2 2
2 2 2
) 2 ( ) 1
(
) 2 ( 1
r r
r r kY
FT
ζ ζ +
−
⋅ +
=
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.
t t
y( ) = (0.01)sinωb
b = 0.2909v 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 EOM mx+ c(x − y) + k(x − y) = 0
Assumption: the base moves harmonically t
Y t
y( ) = cosω
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)
y m kw
w c w
m + + = −
Measurement devices (2)
y m kw
w c w
m + + = − w + 2ζωnw +ωn2w = −y ]
Re[
cos )
(t Y t Yej t
y = ω = ω
)]
( Re[
)
(t z t
w =
t
Ze j
t
z( ) = ω ,
Sub. into EOM to change to complex form
t n j
nz z Ye
z+ 2ζω +ω2 = ω2 ω Complex EOM
t j t
n j
n Ze Ye
j ζω ω ω ω ω ω
ω2 2 2) 2
(− + + =
r Y j
r Y r
Z j
n
n
+
= −
+ +
= −
ζ ω
ω ζω ω
ω
2 1
2 2
2 2
2
2
r = ω ωn
) (ω
H or T(ω) Frequency response
Measurement devices (3)
)
) (
( )
( )
(t = Ze jωt = T ω e jθYe jωt = T ω Ye j ωt+θ z
) cos(
) cos(
) ( )]
( Re[
)
(t = z t = T ω Y ωt +θ =W ωt +θ w
2 2
2 2
) 2 ( )
1 ) (
( r r
T r Y
W
ω ζ
+
= −
=
Y e T
Y T
r Y j
r
Z r ω ω jθ
ζ ( ) ( )
2
1 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
Y T(ω) = W
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
2 2
2 2
) 2 ( )
1 ) (
( r r
T r Y
W
ω ζ
+
= −
=
2 2
2
2 2
2 2
2
) 2 ( )
1 ( )
2 ( )
1 (
) (
r r
A r
r
W n Y y n
ζ ω ζ
ω ω
+
= − +
= −
2 2
2 2
) 2 ( )
1
( r r
W n Ay
ω ζ
+
= −
⋅
Accelerometer (2)
Increasing ζ
r
2 2
2 2
) 2 ( )
1
( r r
W n Ay
ω ζ
+
= −
⋅
• |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]
