This document discusses base excitation in vibration analysis and provides examples. It begins by introducing base excitation as an important class of vibration analysis that involves preventing vibrations from passing through a vibrating base into a structure. Examples of base excitation include vibrations in cars, satellites, and buildings during earthquakes. The document then provides mathematical models and equations to analyze single degree of freedom base excitation systems. Graphs of transmissibility ratios are presented and examples are worked through, such as calculating car vibration amplitude at different speeds. Rotating unbalance is also covered as another source of vibration excitation.

1. 2.4 Base Excitation
• 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
– Building under earthquake
– Disk drives, etc.
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1/27
@ProfAdhikari, #EG260

2. FBD of SDOF Base Excitation
System Sketch
x(t)
m
k
y(t)
System FBD
m
c
 
k ( x − y ) c(x − y)
base
 
x
∑ F =-k(x-y)-c(x-y)=m

m x + kx = cy + ky
x+c 
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2/27
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(2.61)

3. SDOF Base Excitation (cont)
Assume: y(t) = Y sin(ω t) and plug into Equation(2.61)
m x + kx = cω + kY sin(ω t) (2.63)
x+c 
Y cos(ω t) 



harmonic forcing functions
For a car,
ω=
2π 2πV
=
τ
λ
The steady-state solution is just the superposition of the
two individual particular solutions (system is linear).
f0
s
 
 
2
2
 ζω n x + ω n x = 2ζω nω Y cos(ω t) + ω n Y sin(ω t)

x+2
 

 
f0 c
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3/27
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(2.64)

4. Particular Solution (sine term)
With a sine for the forcing function,
2
 ζω n x + ω n x =f0 s sin ω t

x+2
x ps = As cos ω t + Bs sin ω t = X s sin(ω t − φ s )
where
As =
Bs =
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4/27
−2ζω nω f0 s
(ω − ω ) + ( 2ζω nω )
2
n
2 2
2
(ω − ω ) f0 s
2
n
2 2
2
(ω − ω ) + ( 2ζω nω )
2
n
2
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Use rectangular form to
make it easier to add
the cos term

5. Particular Solution (cos term)
With a cosine for the forcing function, we showed
2
 ζω n x + ω n x =f0c cos ω t

x+2
x pc = Ac cosω t + Bc sin ω t = Xc cos(ω t − φ c )
where
Ac =
Bc =
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(ω − ω ) f0c
2
n
2 2
2
(ω − ω ) + ( 2ζω nω )
2
n
2
2ζω nω f0c
(ω − ω ) + ( 2ζω nω )
2
n
2 2
2
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6. Magnitude X/Y
Now add the sin and cos terms to get the
magnitude of the full particular solution
X =
f 02c + f 02s
(ω − ω ) + ( 2ζω nω )
2
n
2 2
2
= ω nY
(2ζω ) 2 + ω n2
(ω − ω ) + ( 2ζω nω )
2
n
2 2
2
where f 0 c = 2 ζω nωY and f 0 s = ω n2Y
if we define r = ω ωn this becomes
X =Y
(1 − r ) + ( 2ζ r )
X
1 + (2ζ r ) 2
=
2
2 2
Y
(1 − r ) + ( 2ζ r )
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1 + (2ζ r ) 2
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2 2
(2.71)
2
(2.70)

7. The relative magnitude plot
of X/Y versus frequency ratio: Called the
Displacement Transmissibility
40
ζ =0.01
ζ =0.1
ζ =0.3
ζ =0.7
30
X/Y (dB)
20
10
0
-10
-20
0
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7/27
0.5
Figure 2.13
1
1.5
2
Frequency ratio r
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2.5
3

8. From the plot of relative Displacement
Transmissibility observe that:
• 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~2Yξ /r
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9. Next examine the Force Transmitted to the
mass as a function of the frequency ratio
 
FT = −k(x − y) − c( x − y ) = m
x
From FBD
At steady state, x(t) = X cos(ω t − φ ),
 ω 2 X cos(ω t − φ )
so x=x(t)
FT = mω X = kr X
2
m
2
FT
k
c
y(t)
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9/27
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base

10. Plot of Force Transmissibility (in dB)
versus frequency ratio
40
ζ =0.01
ζ =0.1
F/kY (dB)
30
ζ =0.3
ζ =0.7
20
10
0
-10
-20
0
0.5
Figure 2.14
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10/27
1
1.5
2
Frequency ratio r
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2.5
3

11. Figure 2.15 Comparison between force
and displacement transmissibility
Force
Transmissibility
Displacement
Transmissibility
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12. Example 2.4.1: Effect of speed
on the amplitude of car vibration
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12/27
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13. 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 
ω b = v(km/hr) 


 = 0.2909v rad/s
 0.006 km   3600 s   cycle 
ω b (20km/hr) = 5.818 rad/s
From the data given, determine the frequency and
damping ratio of the car suspension:
ωn =
k
=
m
c
ζ=
=
2 km 2
© Eng. Vib, 3rd Ed.
13/27
4 × 10 4 N/m
= 6.303 rad/s ( ≈ 1 Hz)
1007 kg
2000 Ns/m
(4 × 10
4
)
N/m (1007 kg )
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= 0.158

14. 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 5.818
=
ω 6.303
1 + (2ζ r)2
X =Y
(1 − r 2 )2 + (2ζ r)2
= (0.01 m )
1 + [2(0.158)(0.923)]
2
(1 − (0.923) ) + (2 (0.158)(0.923))
2 2
2
= 0.0319 m
What happens as the car goes faster? See Table 2.1.
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14/27
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15. Example 2.4.2: Compute the force
transmitted to a machine through base
motion at resonance
From (2.77) at r =1:
1/2
FT  1 + (2ζ ) 
=
kY  (2ζ )2 

2
⇒ FT =
kY
1 + 4ζ 2
2ζ
From given m, c, and k: ζ =
c
900
=
≅ 0.04
2 km 2 40, 000g
3000
From measured excitation Y = 0.001 m:
kY
(40, 000 N/m)(0.001 m)
2
FT =
1 + 4ζ =
1 + 4(0.04)2 = 501.6 N
2ζ
2(0.04)
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15/27
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16. 2.5 Rotating
Unbalance
•
•
•
•
Gyros
Cryo-coolers
Tires
Washing machines
m0
e
Machine of total mass m i.e. m0
ω rt
included in m
e = eccentricity
mo = mass unbalance
ω r t =rotation frequency
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16/27
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k
c

17. Rotating Unbalance (cont)
Rx
ω rt
What force is imparted on the
structure? Note it rotates
with x component:
m0
e
θ
Ry
xr = e sin ω r t
⇒ a x = r = −eω r2 sin ω r t
x
From the dynamics,
R 0 = es = es r
= x m i −ω t
a
ω m iω
n
θ
n
x m−
2
o r
2
o r
2
o r
2
o r
R 0 = ec = ec r
= y m o −ωω
a
ωθm ot
s
s
y m−
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18. Rotating Unbalance (cont)
The problem is now just like any other SDOF
system with a harmonic excitation
m0 eω r2 sin(ω r t)
 + cx + kx = mo eω r2 sin ω r t
mx 
x(t)
m
k
c
© Eng. Vib, 3rd Ed.
18/27
(2.82)
mo 2

or  + 2ζω n x + ω x =
x
eω r sin ω r t
m
2
n
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)
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19. Rotating Unbalance (cont)
• Just another SDOF oscillator with a
harmonic forcing function
• Expressed in terms of frequency ratio r
x p (t ) = X sin(ω r t − φ )
(2.83)
moe
r2
X=
m (1 − r 2 )2 + (2ζ r )2
(2.84)
 2ζ r 
φ = tan 
2
1− r 
(2.85)
−1
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20. Figure 2.20: Displacement magnitude vs
frequency caused by rotating unbalance
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21. 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.
At resonance r = 1 and
mX
1
1
0.1 m
1
=
=
⇒ 10
=
= 10 ⇒ e = 0.1 m
m0 e 2ζ 2(0.05)
e
2ζ
Now to compute the added mass, again at resonance;
m X 

 = 10
m0  0.1 m 
Use this to find Δm so that X is 0.01:
m + ∆m  0.01 m 
m + ∆m
= 100 ⇒ ∆m = 9m

 = 10 ⇒
m0  0.1 m 
(0.1)m
Here m0 is 10%m or 0.1m
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22. Example 2.5.2 Helicopter rotor unbalance
Given
Fig 2.21
k = 1 × 10 5 N/m
mtail = 60 kg
mrot = 20 kg
m0 = 0.5 kg
ζ = 0.01
Fig 2.22
Compute the deflection at
1500 rpm and find the rotor
speed at which the deflection is maximum
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23. 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 the example
of a heavy beam. So the system natural frequency is
k
ωn =
=
33
m+
mtail
140
The frequency of rotation is
10 5 N/m
= 53.72 rad/s
33
20.5 +
60 kg
140
rev min 2π rad
ω r = 1500 rpm = 1500
= 157 rad/s
min 60 s rev
157 rad/s
⇒ r=
= 2.92
53.96 rad/s
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23/27
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24. Now compute the deflection at r =
2.91 and ζ =0.01 using eq (2.84)
m0 e
r2
X=
m (1 − r 2 )2 + (2ζ r)2
(0.5 kg )(0.15 m )
=
34.64 kg
(2.92 )2
2 2
(1 − (2.92) ) − (2(0.01)(2.92))2
= 0.002 m
At around r = 1, the max deflection occurs:
rad rev 60 s
r = 1 ⇒ ω r = 53.72 rad/s = 53.72
= 515.1 rpm
s 2π rad min
At r = 1:
m0 e 1 (0.5 kg )(0.15 m ) 1
X=
=
= 0.108 m or 10.8 cm
meq 2ζ
34.34 kg
2(0.01)
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25. 2.6 Measurement Devices
• A basic transducer
used in vibration
measurement is the
accelerometer.
 
x
• This device can be ∑ F = - k(x-y) - c(x-y) = m
modeled using the ⇒ m = -c( x − y) - k(x − y)
 
x
base equations
(2.86) and (2.61)
developed in the
Here, y(t) is the measured
previous section
response of the structure
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26. Base motion applied to
measurement devices
Let z(t) = x(t) − y(t) (2.87) :
⇒
2
m + cz(t) + kz(t) = mω b Y cosω bt (2.88)
z 
Z
r2
⇒ =
Y
(1− r 2 )2 + (2ζ r)2
(2.90)
Accelerometer
and
 2ζ r 
θ = tan −1 
2÷
 1− r 
(2.91)
These equations should be familiar
from base motion.
Here they describe measurement!
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Strain Gauge
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27. Magnitude and sensitivity
plots for accelerometers.
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
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