Mechanical Vibrations
Chapter 3
Peter Avitabile
Mechanical Engineering Department
University of Massachusetts Lowell
22.457 Mechanical Vibrations - Chapter 3
1
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
SDOF Definitions
Assumptions
• lumped mass
x(t)
• stiffness proportional
to displacement
m
• damping proportional to
velocity
k
• linear time invariant
c
• 2nd order differential
equations
22.457 Mechanical Vibrations - Chapter 3
2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Harmonic Vibration
Consider the SDOF system with
a sinusoidally varying forcing
function applied to the mass as
shown
F=F0sinωt
From the Newton’s Second Law,
∑ f = ma
⇒ m&x& + cx& + kx = F0 sin ωt
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.1.1)
Forced Harmonic Vibration
The solution consists of the complementary
solution (homogeneous solution) and the particular
solution. The complementary part of the solution
has already been discussed in Chapter 2.
The particular solution in the one of interest here.
Since the oscillation of the response is at the
same frequency as the excitation, the particular
solution will be of the form
x = X sin (ωt − φ )
22.457 Mechanical Vibrations - Chapter 3
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(3.1.2)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Harmonic Vibration
Substituting this into the differential equation,
the solution is of the form
F0
cω  (3.1.4)
−1 
(3.1.3)
X=
φ
=
tan

2
2
2 2
 k − mω 
k − mω + (cω)
(
)
Note that this is also seen graphically as
(recall that the velocity and acceleration are 90 and 180 degrees ahead of the displacement)
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Harmonic Vibration
This is expressed in nondimensional form as
F0

 cω
1
−

k 
k
(3.1.5) φ = tan
X=
 1 − mω2 k 
2
2
2
1 − mω  + cω




k
k

( )
(3.1.6)
and can be further reduced recalling the following
expressions for a SDOF
ωn = k
m
22.457 Mechanical Vibrations - Chapter 3
c c = 2mωn
6
c
ζ=
cc
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Harmonic Vibration
The nondimensional expression is
Xk
1
=
2
F0
2
  ω     ω  2
 1 −    +  2ζ   
  ωn     ωn  


φ = tan
22.457 Mechanical Vibrations - Chapter 3
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−1
 ω
2ζ  
 ωn 
 ω
1−  
 ωn 
2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.1.7)
(3.1.8)
Forced Harmonic Vibration
This yields the popular plot of forced response
22.457 Mechanical Vibrations - Chapter 3
8
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Harmonic Vibration
The complex force vector also yields useful
information for interpretation of the results
22.457 Mechanical Vibrations - Chapter 3
9
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Harmonic Vibration
The differential equation describing the system
&x& + 2ζωn x& + ω2n x =
F0
sin ωt
m
(3.1.10)
and the complete solution of this problem is given
as
F
sin(ωt − φ)
x(t) = 0
k 
2
2 2
(3.1.11)
1 −  ω   +  2ζ ω  
  ωn     ωn  


(
+ X1e −ζω t sin 1 − ζ 2 ωn t + φ1
n
22.457 Mechanical Vibrations - Chapter 3
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)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Complex Frequency Response Function
The Complex FRF - real and imaginary parts
h ( jω) =
−j
 ω
1−  
 ωn 
2
(3.1.17)
2 2
2
  ω
1 −    +  2ζ ω  
  ωn     ωn  


 ω
2ζ  
 ωn 
2 2
2
  ω
1 −    +  2ζ ω  
  ωn     ωn  


22.457 Mechanical Vibrations - Chapter 3
11
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Rotating Unbalance
The effects of unbalance is a common problem in
vibrating systems.
Consider a one
dimensional system
with an unbalance
represented by an
eccentric mass, m,
with offset, e,
rotating at some
speed, ω, as shown
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Rotating Unbalance
Let x be the displacement of the non-rotating
mass (M-m) about the equilibrium point, then the
displacement of the eccentric mass is
x + e sin ωt
and the equation of motion becomes
d2
(M − m)&x& + m 2 (x + e sin ωt ) = − kx − cx&
dt
22.457 Mechanical Vibrations - Chapter 3
13
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Rotating Unbalance
This can easily be cast as
(
)
M&x& + cx& + kx = meω2 sin ωt
(3.2.1)
which is essentially identical to (3.1.1) with the
substitution of F0=meω2
The steady-state solution just developed is
applicable for this solution
X=
meω2
(k − Mω )
2 2
(3.2.2)
+ (cω)2
22.457 Mechanical Vibrations - Chapter 3
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 cω 
φ = tan −1 
2
 k − Mω 
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.2.3)
Forced Response - Rotating Unbalance
The differential equation describing the system
&x& + 2ζωn x& + ω2n x =
F0
sin ωt
m
(3.1.10)
and the complete solution of this problem is given
as
F
sin(ωt − φ)
x(t) = 0
k 
2
2 2
(3.1.11)
1 −  ω   +  2ζ ω  
  ωn     ωn  


(
+ X1e −ζω t sin 1 − ζ 2 ωn t + φ1
n
22.457 Mechanical Vibrations - Chapter 3
15
)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Rotating Unbalance
Manipulating into nondimensional form
 ω
 
 ωn 
2
MX
=
2
m e
  ω 2    ω  2
 1 −    +  2ζ   
  ωn     ωn  


φ = tan −1
22.457 Mechanical Vibrations - Chapter 3
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 ω
2ζ  
 ωn 
 ω
1−  
 ωn 
2
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.2.4)
(3.2.5)
Forced Response - Rotating Unbalance
This yields the popular plot of forced response
22.457 Mechanical Vibrations - Chapter 3
17
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Rotating Unbalance
The differential equation describing the system
&x& + 2ζωn x& + ω2n x =
F0
sin ωt
m
(3.1.10)
and the complete solution of this problem is given
as
F
sin(ωt − φ)
x(t) = 0
k 
2
2 2
(3.1.11)
1 −  ω   +  2ζ ω  
  ωn     ωn  


(
+ X1e −ζω t sin 1 − ζ 2 ωn t + φ1
n
22.457 Mechanical Vibrations - Chapter 3
18
)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Rotating Unbalance
The complete solution of this problem is given as
sin(ωt − φ)
x ( t ) = meω2
(k − Mω )
2 2
(
+ (cω)2
+ X1e −ζω t sin 1 − ζ 2 ωn t + φ1
n
22.457 Mechanical Vibrations - Chapter 3
19
(3.2.6)
)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Support Motion
Many times a system is excited at the location of
support commonly called ‘base excitation’
22.457 Mechanical Vibrations - Chapter 3
20
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Support Motion
With the motion of the base denoted as ‘y’ and
the motion of the mass relative to the intertial
reference frame as ‘x’, the differential equation
of motion becomes
m&x& = − k ( x − y) − c( x& − y& )
(3.5.1)
z=x−y
(3.5.2)
Substitute
into the equations to give
m&z& + cz& + kz = − m&y& = mω2 Y sin ωt
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.5.3)
Forced Response - Support Motion
This is identical in form to equation 3.2.1 where z
replaces x and mω2Y replaces meω2
Thus the solution can be written by inspection as
Z=
mω2 Y
(k − mω )
2 2
(3.5.4)
+ (cω)2
cω 
φ = tan 
2
 k − mω 
−1 
22.457 Mechanical Vibrations - Chapter 3
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(3.5.5)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Forced Response - Support Motion
The steady state amplitude and phase from this
equation can be written as
X
=
Y
k 2 + (cω)2
(k − mω )
2 2
(3.5.8)
+ (cω)2
3


mc
ω

tan φ = 
2
2
 k k − mω − (cω) 
(
22.457 Mechanical Vibrations - Chapter 3
)
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.5.9)
Forced Response - Support Motion
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Vibration Isolation
Dynamical response can be minimized through the
use of a proper isolation design.
An isolation system attempts either to protect
delicate equipment from vibration transmitted to it
from its supporting structure or to prevent
vibratory forces generated by machines from
being transmitted to its surroundings.
The basic problem is the same for these two
objectives - reducing transmitted force.
22.457 Mechanical Vibrations - Chapter 3
25
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Vibration Isolation - Force Transmitted
Notice that motion transmitted from the
supporting structure to the mass m is less than
one when the frequency ratio is greater that
square root 2.
This implies that the
natural frequency of the
supported system must be
very small compared to the
disturbing frequency.
A soft spring can be used
to satisfy this requirement.
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Vibration Isolation - Force Transmitted
Another problem is to reduce the force transmitted
by the machine to the supporting structure which
essentially has the same requirement.
The force to be isolated is transitted through the
spring and damper as shown
22.457 Mechanical Vibrations - Chapter 3
27
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Vibration Isolation - Force Transmitted
The force to be isolated is transitted through the
spring and damper is
FT =
(kX ) + (cωX )
2
2
 2ζω 
= kX 1 + 

 ωn 
2
(3.6.1)
With the disturbing force equal to F0sinwt this
equation becomes
X=
F0
k
2 2
2
  ω
1 −    +  2ζ ω  
  ωn     ωn  


22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.6.1a)
Vibration Isolation - Force Transmitted
The transmissibility TR, defined as the ratio of
the transmitted force to the disturbing force, is
  ω 
1 +  2ζ  
  ωn  
2
FT
=
TR =
2
2
F0
  ω     ω  2
 1 −    +  2ζ   
  ωn     ωn  


and when damping is small becomes
FT
1
TR =
=
F0  ω  2
  −1
 ωn 
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(3.6.2)
(3.6.3)
Sharpness of Resonance
The peak amplitude of response occurs at
resonance. In order to find the sharpness of
resonance, the two side bands at the half power
points are required.
At the half power points,
2
1 1 
1
=
 
2
2  2ζ 
2
  ω     ω  2
 1 −    +  2ζ   
  ωn     ωn  


22.457 Mechanical Vibrations - Chapter 3
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(3.10.1)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Sharpness of Resonance
Solving yields
2
(
)
 ω
  = 1 − 2ζ 2 ± 2ζ 1 − ζ 2
 ωn 
(3.10.2)
and if the damping is assumed to be small
2
 ω
  = 1 ± 2ζ
 ωn 
(3.10.3)
Letting the two frequencies corresponding to the
roots of 3.10.3 gives
ω2 − ω1
ω22 − ω12
≅2
4ζ =
2
ωn
ωn
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Sharpness of Resonance
The Q factor is defined as
ωn
1
=
Q=
ω2 − ω1 2ζ
22.457 Mechanical Vibrations - Chapter 3
32
(3.10.4)
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB2_3
VIBRATION TOOLBOX EXAMPLE 2_3
function VTB2_3(z,rmin,rmax,opt)
% VTB2_3 Steady state magnitude and phase of a
% single degree of freedom damped system.
>> vtb2_3([0.02:.02:.1],0.5,1.5,1)
>> vtb2_3([0.02:.02:.1],0.5,1.5,3)
Normalized Amplitude
10
10
10
10
Normalized Amplitude vers us Frequency Ratio
2
ζ=
ζ=
ζ=
ζ=
ζ=
1
0.02
0.04
0.06
0.08
0.1
0
-1
0.5
0.6
0.7
0.8
0.9
1
1.1
Frequency Ratio
1.2
1.3
1.4
1.5
P has e vers us Frequency Ratio
180
P has e lag (°)
157.5
135
112.5
90
ζ=
ζ=
ζ=
ζ=
ζ=
67.5
45
22.5
0
0.5
22.457 Mechanical Vibrations - Chapter 3
0.6
0.7
0.8
33
0.9
1
1.1
Frequency Ratio
1.2
1.3
1.4
0.02
0.04
0.06
0.08
0.1
1.5
Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB1_4
VIBRATION TOOLBOX EXAMPLE 1_4
>> clear
>> x0=0; v0=0; m=1; d=.1; k=2; dt=.01;
>> t=0:dt:n*dt; u=[sin(t)];
>> [x,xd]=VTB1_4(n,dt,x0,v0,m,d,k,u);
>>
>> plot(t,u); % Plots force versus time.
>> plot(t,x); % Plots displacement versus time.
n=10000;
2
1.5
1
0.5
0
-0.5
-1
-1.5
0
22.457 Mechanical Vibrations - Chapter 3
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory