FBD of SDOF Base Excitation
2.4 Base Excitation
System Sketch
• Important class of vibration analysis
– Preventing excitations from passing
from a vibrating base through its mount
into a structure
• Vibration isolation
– Vibrations in your car
– Satellite operation
– Disk drives, etc.
x(t)
System FBD
m
m
k
y(t)
c
k ( x − y ) c(x − y)
base
∑ F =-k (x-y)-c(x-y)=mx
mx +cx + kx = cy + ky
(2.61)
1
SDOF Base Excitation (cont)
2
Particular Solution (sine term)
With a sine for the forcing function,
x +2ζωn x + ωn2x =f 0s sin ωt
x ps = As cos ωt + B s sin ωt = X s sin(ωt − φs )
Assume: y (t ) = Y sin(ωt ) and plug into Equation(2.61)
mx +cx + kx = cωY cos(ωt ) + kY sin(ωt ) (2.63)
harmonic forcing functions
For a car,
ω=
2π
τ
=
where
2πV
λ
The steady-state solution is just the superposition of the
two individual particular solutions (system is linear).
f0 c
f0 s
Bs =
x +2ζωn x + ω x =2ζωnωY cos(ωt ) + ω Y sin(ωt ) (2.64)
2
n
As =
2
n
3
−2ζωnωf 0s
(ωn2 − ω 2 )2 + ( 2ζωnω )
(ωn2 − ω2 )f 0s
2
(ωn2 − ω2 )2 + ( 2ζωnω )
Use rectangular form to
make it easier to add
the cos term
2
4
Magnitude X/Y
Particular Solution (cos term)
Now add the sin and cos terms to get the
magnitude of the full particular solution
With a cosine for the forcing function, we showed
x +2ζωn x + ωn2 x =f0 c cosωt
f0c2 + f0s2
= ω nY
x pc = Ac cosωt + Bc sinωt = X c cos(ωt −φc )
X =
where
where f0c = 2ζω nωY and f0s = ω n2Y
(ωn2 −ω 2 ) f0 c
Ac = 2 2 2
2
(ωn −ω ) + ( 2ζωnω )
if we define r =
Bc =
2ζωnω f0 c
(ωn2 −ω 2 )2 + ( 2ζωnω )
(ω n2 − ω 2 )2 + (2ζω nω )
2
ω
ωn
this becomes
(ω n2 − ω 2 )2 + (2ζω nω )
X =Y
1 + (2ζ r)2
X
=
2
Y
(1 − r 2 )2 + (2ζ r )
2
(2ζω )2 + ω n2
2
1+ (2ζ r)2
(1− r 2 )2 + (2ζ r )
2
(2.70)
(2.71)
5
The relative magnitude plot
of X/Y versus frequency ratio: Called the
Displacement Transmissibility
From the plot of relative Displacement
Transmissibility observe that:
40
ζ =0.01
ζ =0.1
ζ =0.3
ζ =0.7
30
X/Y (dB)
20
• X/Y is called Displacement
Transmissibility Ratio
• Potentially severe amplification at
resonance
• Attenuation for r > sqrt(2) Isolation Zone
• If r< sqrt(2) transmissibility decreases with
damping ratio Amplification Zone
• If r >> 1 then transmissibility increases
with damping ratio Xp~2Yζ/r
10
0
-10
-20
0
0.5
Figure 2.13
1
1.5
2
Frequency ratio r
2.5
6
3
7
8
Next examine the Force Transmitted to the
mass as a function of the frequency ratio
FT = − k ( x − y ) − c ( x − y ) = mx
Plot of Force Transmissibility (in dB)
versus frequency ratio
40
From FBD
At steady state, x ( t ) = X cos(ωt − φ ),
ζ =0.01
ζ =0.1
ζ =0.3
ζ =0.7
30
20
so x =-ω X cos(ωt − φ )
F/kY (dB)
2
x(t)
FT = mω 2 X = k r 2 X
m
10
0
FT
k
-10
c
-20
y(t)
base
9
0
0.5
Figure 2.14
1
1.5
2
Frequency ratio r
2.5
3
10
Example 2.4.1: Effect of speed
Figure 2.15 Comparison between force
and displacement transmissibility
on the amplitude of car vibration
Force
Transmissibility
Displacement
Transmissibility
11
12
Model the road as a sinusoidal input to
base motion of the car model
Approximation of road surface:
y(t) = (0.01 m)sin ω bt
1
⎞ ⎛ hour ⎞ ⎛ 2π rad ⎞
⎛
= 0.2909v rad/s
⎝ 0.006 km ⎟⎠ ⎜⎝ 3600 s ⎟⎠ ⎜⎝ cycle ⎟⎠
ω b = v(km/hr) ⎜
From the input frequency, input amplitude,
natural frequency and damping ratio use
equation (2.70) to compute the amplitude of
the response:
r=
ω b (20km/hr) = 5.818 rad/s
From the data give, determine the frequency and
damping ratio of the car suspension:
ωn =
ζ=
k
=
m
c
=
2 km 2
X =Y
2000 Ns/m
(4 × 10
4
)
N/m (1007 kg )
= 0.158
(
1 − (0.923)
)
2 2
+ (2 (0.158 )(0.923))
2
= 0.0319 m
14
2.5 Rotating
Unbalance
•
•
•
•
From (2.77) at r =1:
kY
⇒ FT =
1 + 4ζ 2
2ζ
900
c
From given m, c, and k: ζ = 2 km = 2 40, 000i3000 ≅ 0.04
Gyros
Cryo-coolers
Tires
Washing machines
m0
e
Machine of total mass m i.e. m0
included in m
e = eccentricity
mo = mass unbalance
ωr = rotation frequency
From measured excitation Y = 0.001 m:
FT =
2
13
transmitted to a machine through base motion
at resonance
1/2
1 + [2(0.158)(0.923)]
What happens as the car goes faster? See Table 2.1.
Example 2.4.2: Compute the force
FT ⎡ 1 + (2ζ )2 ⎤
=
kY ⎢⎣ (2ζ )2 ⎥⎦
1 + (2ζ r)2
(1 − r 2 )2 + (2ζ r)2
= (0.01 m )
4 × 10 4 N/m
= 6.303 rad/s ( ≈ 1 Hz)
1007 kg
ω b 5.818
=
ω 6.303
kY
(40, 000 N/m)(0.001 m)
1 + 4ζ 2 =
1 + 4(0.04)2 = 501.6 N
2ζ
2(0.04)
15
ωr t
k
c
16
Rotating Unbalance (cont)
Rx
ωr
Rotating Unbalance (cont)
What force is imparted on the
structure? Note it rotates
with x component:
m0
e
θ
xr = e sin ω r t
Ry
The problem is now just like any other SDOF
system with a harmonic excitation
m0eωr2sin(ωr t)
x(t)
⇒ a x = xr = −eω r2 sin ω r t
m
From sophomore dynamics,
Rx = m0 a x = −mo eωr2 sin θ = −mo eωr2 sin ωr t
or x + 2ζωn x + ωn2 x =
(2.82)
mo 2
eωr sin ωr t
m
k
c
Ry = m0 ay = −mo eω cosθ = −mo eω cos ωrt
2
r
mx + cx + kx = mo eωr2 sin ωr t
2
r
17
Note the influences on the
forcing function (we are assuming that
the mass m is held in place in the y direction as indicated
in Figure 2.18)
18
Figure 2.20: Displacement magnitude vs
Rotating Unbalance (cont)
frequency caused by rotating unbalance
• Just another SDOF oscillator with a
harmonic forcing function
• Expressed in terms of frequency ratio r
x p (t) = X sin(ω rt − φ )
(2.83)
r2
moe
X=
m (1− r 2 )2 + (2ζ r )2
(2.84)
⎛ 2ζ r ⎞
⎝ 1− r 2 ⎟⎠
φ = tan−1 ⎜
(2.85)
19
20
Example 2.5.1:Given the deflection at resonance (0.1m), ζ
= 0.05 and a 10% out of balance, compute e and the amount of added
mass needed to reduce the maximum amplitude to 0.01 m.
Given
Fig 2.21
k = 1 × 10 5 N/m
mtail = 60 kg
At resonance r = 1 and
1
1
0.1 m
1
mX
=
=
⇒ 10
=
= 10 ⇒ e = 0.1 m
e
2ζ
m0 e 2ζ 2(0.05)
mrot = 20 kg
m0 = 0.5 kg
Now to compute the added mass, again at resonance;
m⎛ X ⎞
⎜
⎟ = 10
m0 ⎝ 0.1 m ⎠
Example 2.5.2 Helicopter rotor unbalance
Fig 2.22
ζ = 0.01
Use this to find Δm so that X is 0.01:
Compute the deflection at
1500 rpm and find the rotor
speed at which the deflection is maximum
m + Δm
m + Δm ⎛ 0.01 m ⎞
= 100 ⇒ Δm = 9m
⎜⎝
⎟⎠ = 10 ⇒
(0.1)m
m0
0.1 m
Here m0 is 10%m or 0.1m
21
Example 2.5.2 Solution
The rotating mass is 20 + 0.5 or 20.5. The stiffness is provided by the
Tail section and the corresponding mass is that determined in Example
1.4.4. So the system natural frequency is
k
=
m
m + tail
3
The frequency of rotation is
ωn =
ω r = 1500 rpm = 1500
⇒ r=
22
Now compute the deflection at r =
3.16 and ζ =0.01 using eq (2.84)
X=
5
10 N/m
= 46.69 rad/s
60 kg
20.5 +
3
m0 e
r2
m (1 − r 2 )2 + (2ζ r)2
(0.5 kg )(0.15 m )
=
20.5 kg
rev min 2π rad
= 157 rad/s
min 60 s rev
(3.16 )2
2
(1 − (3.16)2 ) − (2(0.01)(3.16))2
= 0.004 m
At around r = 1, the max deflection occurs:
r = 1 ⇒ ω r = 49.69 rad/s = 49.69
157 rad/s
= 3.16
49.49 rad/s
At r = 1:
X=
23
(0.5 kg )(0.15 m )
20.5 kg
rad rev 60 s
= 474.5 rpm
s 2π rad min
1
= 0.183 m or 18.3 cm
2(0.01)
24
Base motion applied to
measurement devices
2.6 Measurement Devices
• A basic transducer
used in vibration
measurement is the
accelerometer.
• This device can be
modeled using the
base equations
developed in the
previous section
Let z (t ) = x (t ) − y (t ) (2.87):
⇒
mz + cz ( t ) + kz ( t ) = mωb2Y cosωb t (2.88)
Z
r2
⇒ =
Y
(1 − r 2 ) 2 + (2ζ r ) 2
and
∑ F =-k (x -y )-c(x -y )=mx
(2.90)
Accelerometer
2ζ r ⎞
⎟ (2.91)
⎝ 1− r2 ⎠
θ = tan −1 ⎛⎜
⇒ mx = -c ( x − y ) - k ( x − y )
(2.86) and (2.61)
Here, y(t) is the measured
response of the structure
These equations should be familiar
from base motion.
Here they describe measurement!
Strain Gauge
25
Magnitude and sensitivity
plots for accelerometers.
26
Home Work-Chapter 2
2.6, 2.7,2.9
2.11,2.15,2.19
2.24,2.28
2.30,2.35
2.40,2.44
2.51,2.52,2.53,2.57,2.58
Effect of damping on
proportionality constant
Fig 2.27
Fig 2.26
Magnitude plot showing
Regions of measurement
In the accel region, output voltage is
nearly proportional to displacement
27
28