533581676
96 學年研一機械振動
授課教師：
李炯達
教師研究室：
2202C
電話分機號碼：
3211
學分：
3
適修年級：
大四或研一
預修課程：
動力學、材料力學、工程數學
選別：
選修
評分方式：
期中考二次各佔 30%、期末考 40%
上課教室：
51212-2
上課時間：
星期二第 2,3,4 堂
授課目標：
振動學主要探討系統因初始位移或受到外在作用力所產生的振動現象。本
課程提供學生對振動學的基本認識，經由振動概念、單自由度、雙自由度、至
多自由度之離散系統，介紹振動學的基本學理，而後延伸至連續系統的振動分
析以及振動量測與應用。
授課大綱：
1 Fundamentals of Vibration
2 Free Vibration of single Degree of Freedom Systems
3 Harmonically Excited Vibration
4 Vibration Under General Forcing Conditions
5 Two Degree of Freedom Systems
6 Multi-degree of Freedom Systems
7 Continuous Systems
8 Vibration measurements and Applications
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教科書 (Text)
Rao, Singiresu S., 1998, "Mechanical Vibrations," 4th edition, Addison Wesley.
(台北)
This book is used for an introduction course in vibration engineering.
Path through the text:
Chapters 1-5, portions of Chapters 6, 7, 8, and 10 will be used.
參考書 (References)
1. Meirovitch, Leonard, 1986, "Elements of Vibration Analysis," 2nd edition,
McGraw-Hill. (民全書局)
2. Inman, Daniel J., 1994, "Engineering Vibration," Prentice Hall, Englewood
Cliffs, New Jersey. (民全書局)
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課程大綱與進度 Outline of Course
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
C.D. Lee
Topics
1.Fundamentals of Vibration
2. Free Vibration of single Degree of Freedom Systems
1st Midterm
3. Harmonically Excited Vibration
4. Vibration Under General Forcing Conditions
5. Two Degree of Freedom Systems
2nd Midterm Exam
6. Multi-degree of Freedom Systems
7. Continuous Systems
8. Vibration measurements and Applications
Final Exam
Course Review and Solution of Final Exam
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CH1. Fundamentals of Vibration
1.1 Preliminary Remarks
The chapter presents the brief introduction of the mechanical vibrations. Any
motion that repeats itself for an interval of time is called vibration or oscillation. All
bodies possessing mass and elasticity are capable of vibration. Thus, vibratory
behavior of engineering machines needs to be taken into account when engineering
machines are designed.
1.2 Brief History of Vibration
1.2.1 Origins of Vibration
Since ancient times, people were interested in vibration, sought out the rules and
laws of sound production, and used them in improving musical instruments. Music
was highly developed and was much appreciated by the Chinese, Hindus, Japanese,
Egyptians and Greeks.
Musical instruments:
 whistles or drums (the first musical instruments)
 stringed instruments (harps, Egyptian) from about 3000 B.C. (hunter’s bow)
 monochord in 550 B.C. (on a scientific basis as shown in Fig 1.2, Greek)
 treatises on music and sound around 350 B.C (Greeks).
 seismograph in A.D. 132 (Zhang Heng, Chinese)
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1.2.2 From Galilei to Rayleigh
Galilei (1564-1642, Italian):
 Being considered to be the founder of modern experimental science.
 Finding that the time period of a pendulum is independent of the amplitude
of swings but the length of the pendulum (  n 
1 2
l

 2
, n 
f n n
g
g
)
l
 Indicating the relationship between the frequency, length, tension, and
density of a vibrating stretched string.
Marin Mersenne (1588-1648, French) is the first one who published the laws of
vibrating strings
and is considered as the father of acoustics. Galilei discovered the same laws but
they were not made public until several years later.
Sauveur (1653-1716, French):
 Coining the word “acoustic” for the science of sound
 Observing the phenomenon of mode shapes, nodes, fundamental frequency
and the higher frequencies harmonics.
Newton (1642-1727, English): describing the law of universal gravitation as
well as the three laws of motion.
Taylor (1685-1731, English):
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
Finding the theoretical solution of the problem of the vibrating string

Presenting the Taylor’s theorem on infinite series.
In fact, the seventeen century is often considered the “century of genius”, since
the foundations of modern philosophy and science were laid during that period.
Bernoulli (1700-1782) D’alembert (1717-1783) Euler (1707-1783): Partial
derivatives in the equations of motion (1750). Euler- Bernoulli beam theory
Fouries (1768-1830) proved series of expansion.
Lagrange (1736-1813) presented analytical solution of the vibrating string.
Kirchhoff (1824-1887) gave the correct boundary conditions for the vibration of
plates.
Rayleigh (1877) found the fundamental frequency of vibration of a conservative
system (principle of conservation of energy ；Rayleigh’s method)
Ritz provided an extension of the Rayleigh’s method, which can be used to find
multiple natural frequencies (Rayleigh-Ritz Method).
1.2.3 Recent Contributions
Timoshenko (1878-1972) presented an improved theory of vibration of beams,
which has become known as the Timoshenko or Thick beam theory, by considering
the effect of rotary inertia and shear deformation.
Nayfeh surveyed the modern methods of perturbation theory for nonlinear
vibrations.
Taylor, Wiener and Khinchin introduced the correlation function and the spectral
density which opened the theory of random vibrations.
1.3 Importantce of the Study of Vibration

Early scholars concentrated their efforts on understanding the natural
phenomena and developing mathematical theories to describe the vibration of
physical systems.

In recent times, most investigators involved on the engineering applications
of vibration, such as the design of machines, foundations, structures, engines,
turbines, and control systems.
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Most machines have vibrational problems due to the inherent unbalance in the
power sources or structures.

The structure or machine components fail because of material fatigue
resulting from the cyclic variation of the induced stress.

The vibration causes more rapid wear of machine parts (bearings and gears).

The Vibration causes fasteners to become loose.

Vibration can cause chatter which leads to a poor surface finish.

Resonance leads to excessive deflections and failure.
Purposes of vibration study

Reduce vibration through proper design of machines and their mountings to
prevent failure.
Applications

Vibration is applied in vibratory conveyors, sieves, compactors, washing
machines, dentist’s drill, and massaging units, pile driving, testing of
materials, and finishing process etc.
1.4 Basic Concept of Vibration
1.4.1 Vibration
Any motion which repeats itself after an interval of time is called vibration or
oscillation. All bodies possessing mass and elasticity are capable of vibration.
Examples: The swinging of a pendulum
The motion of a plucked string
The Drum etc.
The theory of vibration deals with the study of oscillatory motions of bodies and
the forces associated with them.
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1.4.2 Elementary Parts of Vibrating Systems
A vibratory system, in general, includes a means for storing potential energy
(spring or elasticity), a means for storing kinetic energy (mass or inertia), and a
means by which energy is gradually lost (damper).
The vibration of a system involves the transfer of its potential energy to kinetic
energy and kinetic energy to potential energy, alternately. If the system is damped,
some energy is dissipated in each cycle of vibration and must be replaced by an
external source if a state of steady vibration is to be maintained.
Hook’s Law
F  Kx
Newton’s Law F  mx
Damper F  cx
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1.4.3 Degree of Freedom
The minimum number of independent coordinates required to determine
completely the positions of all parts of a system at any instant of time defines the
degree of freedom of the system.
1.4.4 Discrete and Continuous Syatems

Discrete or lumped parameter systems: Systems with a finite number of
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degrees of freedom.

Continuous or distributed systems: Systems with an infinite number of
degrees of freedom.
1.5 Classification of Vibration
1.5.1 Free and Forced Vibration

Free Vibration: No external force acts on the system.

Forced Vibration: The system is subjected to an external force.
1.5.2 Undamped and Damped Vibration

Undamped vibration: no energy is lost or dissipated.

Damped vibration: energy is lost.
However, consideration of damping becomes extremely important in analyzing
vibratory systems near resonance.
1.5.3 Linear and Nonlinear Vibration

Linear vibration: The principle of superposition holds and the mathematical
techniques of analysis are well developed.

Nonlinear vibration: Since all vibratory systems tend to behave nonlinear
with increasing amplitude of oscillation, some knowledge of nonlinear
vibration is desirable.
Ex. use a simple pendulum as an example
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ml 2  mgl sin   0
   2  0
  g / l 
1
2

3 
  0
ml 2  mgl 
6


1 

   2    3   0
6 

1.5.4 Deterministic and Random Vibration

Deterministic vibration: the value or magnitude of the excitation (force or
motion) is known.

Random or nondeterministic vibration: the value of the excitation at a given
time cannot be predicted.
1.6 Vibration Analysis Procedure

Step 1: Mathematical Modeling

Step 2:Derivation of Governing Equations

Step 3: Solution of the Governing Equations

Step 4: Interpretation of the Results
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 EX 1.1 Mathematical Model of a Motorcycle
1.7 Spring Elements
Hook’s Law
F  kx (the linear relation is assumed)
(1.1)
k is the stiffness or spring constant
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The work (U ) done by the spring is considered as strain or potential energy.
U
1 2
kx
2
(1.2)
When the stress exceeds the yield point of the material and the forcedeformation relation becomes nonlinear.
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Linearization process
Using Taylor’s series expansion about the static equilibrium position x * as
F  F  F ( x *  x)
dF
 F (x ) 
dx
*
1 d 2F
(x) 
2! dx 2
x*
1 d 3F
(x) 
3! dx 3
x*
(x 3 )
2
(1.3)
x*
For small values of x , then x  0
F  F  F ( x * ) 
dF
dx
(x)
(1.4)
x*
F  kx
k
dF
dx
(1.5)
x*
Elastic elements like beams also behave as springs. As shown in Fig. 1.21. The
mass of the cantilever beam is neglected for simplicity. The static deflection of the
beam at the free end is given by
 st
Wl 3

3EI
(1.6)
Where W  mg is the mass m. Hence the stiffness is
k
W
 st

3EI
l3
(1.7)
See Appendix B to find the spring constants of beams and plates.
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1.7.1 Combination of Springs
Case 1: Springs in Parallel. Parallel springs undergo a same static deflection
 st as shown in Fig. 1.22(b). The free body diagram, shown in FIG 1.22(c), gives the
equilibrium equation
(1.8)
W  k1 st  k 2 st
W  k eq  st
We have
(1.9)
k eq  k1  k 2
It gives
(1.10)
In general, the equivalent spring constant of n parallel springs can be obtained as:
k eq  k1  k 2  ................  k n
(1.11)
Case 2: Springs in Series. Under the action of a load W, springs 1 and 2 undergo
elongations 1 and  2 , respectively, as shown in Fig. 1.23(b). The total elongation of
 st   1   2
the system  st is given by
(1.12)
We have
(1.13)
W  k11  k 2 2
W  k eq  st
and
(1.14)
It gives
1
1 1
 
k eq k1 k 2
(1.16)
In general, the equivalent spring constant of n springs in series can be obtained as:
1
1 1
1
   .................. 
k eq k1 k 2
kn
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(1.17)
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 Example 1.2 Equivalent k of a Suspension System
Find the equivalent spring constant of the suspension if the each spring has a
shear modulous G=80  10 9 N/ m 2 and five effective turns, mean coil diameter
D=20cm, and wure diameter d=2cm.


d 4 G 0.02  80  10 9

 40,000 N / m and  eq  3k  120,000 N / m
8D 3 n
8(0.2) 3  5
4
Solution: k 
 Example 1.3 Torsional Spring Constant of a Propeller Shaft
Determine the torsional spring constant of the steel propeller shaft as spring in
Fig. 1.25.
Solution: We can consider the segments 12 and 23 of the shaft as springs in
combination and under the equal torque T. Hence the two springs are
considered as series springs. Therefore
k t12 

 


GJ12 G D124  d124
80  10 9  0.34  0.2 4


 25.5255  10 6 N  m / rad
l12
32l12
322
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k t 23 
k teq 

 


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4
4
GJ 23 G D23
 d 23
80  10 9  0.25 4  0.15 4


 8.9012  10 6 N  m / rad
l 23
32l 23
323
k t12 k t23
k t12  k t23

25.2525  8.9012
 10 6  6.5997  10 6 N-m/rad
25.5255  8.9012
 Example 1.4 Equivalent k of Hoisting Drum
A hoisting drum, carrying a steel wire rope, is mounted at the end of the end of
a cantilever beam as shown in the Fig. 1.26(a). Determined the equivalent spring
constant of the system.
Solution: The spring constant of the cantilever beam is given
kb 
3EI 3E  1 3  Eat 3
 3  at  
3
b3
b  12
 4b
(E.1)
The stiffness of the wire rope is
kr 
AE d 2 E

l
4l
(E.2)
Since both the wire rope and the cantulever beam experiece the same load W,
they can be molded as springs in series. Then
1
1
1
4b 3
4l
 

 2
3
k eq k1 k 2 Eat
d E
C.D. Lee
and
k eq
E  at 3 d 2 

  2 3
4  d b  lat 3 
(E.3)
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 Example 1.5 The boom AB of the crane shown in Fig. 1.27(a) is a uniform
steel bar of length 10 m and area of cross section 2,500 mm 2 mm The cable is
made of steel and has a cross-section area of 100mm mm 2 . Negecting the
effect of the cable CDEB, find the equivalent spring constant of the system in
the vertical direction.
:
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Solution: The equivalent spring constant can be found using the equivalent of
the potential energies of the two systems. A vertial displacement x of point B wil
cause the sping k 2 to deform by an amount x2  x cos 45o and the spring k1 to
deform by an amount x1  x cos90 o   . A11nd
l12  32  10 2  2310cos135o  151.426, l1  12.3055 m and the angle  satisfies the
relation l12  32  2l1 3cos   10 2 , cos  0.8184 ,   35.0736 o .
The potential energy in the spring can be expressed as



U
1
k1 x cos 45 o
2
k2 
A2 E2
 5.1750  10 7 N/m
l2
2


1
k 2 x cos 90 o  
2

2
where k1 
A1 E1
 1.6822  10 6 N / m ,
l1
1
2
Since U  k eq xeq2 , we obtain the equivalent spring
constant of the system as k eq  26.4304 10 6 N/m.
1.8 Mass or Inertia Elements
The mass or inertia element is assumed to be a rigid body; it can gain or lose
kinetic energy whenever the velocity of the body changes. In most cases, we must
use a mathematical model to represent the actual vibrating system, and there are
often several possible models.
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1.8.1 Combination of Masses
Case 1: Translational Masses Connected by a Rigid Bar. As shown in Fig.
1.29(a), if the small angular displacement is assumed and let the location of the
equivalent mass to be that of mass m1 . As
x 2 
l
l2
x1 , x 3  3 x1
l1
l1
then we obtain
x eq  x1
(1.18)
(1.19)
1
1
1
1
meq x eq2  m1 x12  m2 x 22  m3 x 32
2
2
2
2
l
This equation gives meq  m1   2
 l1
2
(1.20)
2

l 
 m2   3  m3

 l1 
(1.21)
Case 2: Translational and Rotational Masses Coupled Together. As shown in Fig.
1.30, the mass m having a translational velocity x is coupled to another mass (of
mass moment of inertia ) having a rotational velocity  . We can find either (1) a
single equivalent translational mass m eq or (2) a single equivalent rotational mass J eq .
1. Equivalent translatonal mass.
1
1
mx 2  J o 2
(1.22)
2
2
1
1
1
gives meq x 2  mx 2  J o x R
2
2
2
1
and T  meq x eq2 Since x eq  x and   x R . It
2
T
 
2
that is meq  m 
Jo
R2
(1.24)
2. Equivalent rotational mass.
eq   and x  R and T 
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 
2
1
1
1
J eq 2  m R  J o 2 that is J eq  J o  mR 2 .(1.25)
2
2
2
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 Example 1.6 Equivalent Mass of a System. Find the equivalent mass of the
system shown in Fig. 1.31 where the rigid link 1 is attached to the pulley and
rotates with it.
Solution:
When the m is displaced by a distance x, the pulley and rigid link 1 rotate by an
angle  p  1  x / rp . Rigid link 2 and the cylinder displaced a distance
x 2   p l1  xl1 / rp . The cylinder rotates by an angle  c  x 2 / rc  xl1 / rp rc . The kinetic
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1
2
1
2
1
2
1
2
1
2
1
2
J 1  m1l12 / 3
energy T  mx 2  J p p2  J 112  m2 x 22  J cc2  mc x 22
J c  mc rc2 / 2
1
1  x
T  mx 2  J p 
2
2  rp
T
and
2
2 



  1  m1l1  x

2  3  rp

1
meq x 2 . We obtain
2
2
2
(E.1) Noting that
.
2

 

 
  1 m2  xl1   1  mc rc  xl1

2  rp 
2  2  rp rc

J p m l2 m l2 m l2
l2
meq  m  2  1 21  22 1  c 21  mc 12
rp
3rp
rp
2rc
rp
2
Therefore

 
  1 mc  xl1 

2  rp 

2
(E.2)
(E.4)
 Example 1.7 Cam-Follower Mechanism. Find the equivalent mass( m eq ) of the
cam follower system by assuming the location of the m eq as (i) point A (ii)
point C.
Solution: Due to a vertical displacement x of the pushrod, the rocker arm
rotates by an angle  r  x / l1 about the pivot point, the valve move downward by
xv   r l 2  xl2 / l1 and the C.G. of the roker arm moves downward by
xr   r l3  xl3 / l1 . So T 
C.D. Lee
1
1
1
1
m p x 2p  J rr2  mr x r2  mv x v2
2
2
2
2
(E.1)
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1
2
(i) Point A: x eq  x , and T  meq x eq2
(E.2) we obtain meq  m p 
(ii) Point C: x eq  x v and it gives meq  mv 
l32
Jr
l 22

m

m
p 2
r 2
l 22
l2
l2
l32
Jr
l 22

m

m
v 2
r 2
l12
l1
l1
(E.5)
1.9 Damping Elements
The mechanism by which the vibrational energy is gradually converted into heat
or sound is known as damping. Damping is molded as one of the following type.
Viscous Damping. When mechanical systems vibrate in a fluid medium such as
air, gas, water, and oil, we call it viscous damping. The damping force is assumed to
be proportional to the velocity of the vibrating body.
Coulomb or Dry Friction Damping. The damping force is constant in magnitude.
Material or Solid or Hysteretic Damping. When material are deformed, energy is
dissipated and absorbed by the material.
1.9.1 Construction of Viscous Dampers
A viscous damping can be molded as two parallel plates separated by a distance
h, with a fluid of viscosity between the plates.
 : shear stress
v : velocity of the moving plate
 : viscousity of the fluid
F: shear force
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 
du
dy
(1.26), F  A 
Av
h
 cv (1.27),
c
A
h
(1.28)
1.9.2 Combination of Dampers.
Example 1.8 Clearance in a bearing. Determine the clearance between the plates
(Fig. 1.35) if the F  400N , v  10m / s ,   0.3445Pa  s .
Solution: c 
C.D. Lee
F 400
A

 40 N .s / m , and c 
, thus h=0.86125mm.
v
10
h
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Example 1.9 Piston-Cylinder Dashpot. Develop an expression for the damping
constant of the dashpot shown in Fig. 1.36(a).
Solution: At a distance y from the moving surface, let the velocity and the shear
stress be v and  , and at a distance (y+dy) let the velocity and the shear stress be
(v-dv) and   d  . The viscous forceof the annular ring is equal to
F  Dld   Dl
d
dy (E.1)
dy
and the shear stress
  
dv
dy
(E.2) we obtain
d 2v
(E.3) The pressure force on the ends of the element, given by
dy 2
P
4P
(E.4) Thus the pressure force on the end of the element is
p

2
 D  D 2


4


4P
p Ddy  
dy (E.5)
If the uniform mean velocity is assumed, we get
D
d 2v
4P

(E.6) Integrating this equaton twice and using the boundary
2
dy
D 2 l
F  Dl dy
conditions v  v0 at y=0 and v=0 at y=d. we obtain
v


2P
y

yd  y 2   0 1   (E.7) The rate of flow through the clearance space
2
D l
 d
can be obtained by integrating the rate of the rate of flow through an element
d
 2 Pd 3
v d
 0 
2
2 
 6d l
between the limits y=0 and y=d: Q  0 vDdy  D 
C.D. Lee
(E.8)
thus
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v0 
Q
 D 2

 4



(E.9)

2d  
3 
 3D l 1  D  

  v
Eq. (E.8) and (E.9) lead to P  
0
3
4d






(E.10)
By writing the force as P=c  0 the damping constant c can be found as
 3D 3l  2d 
c  
1

3 
D 
 4d 
(E.11)
1.10 Harmonic Motion
If the motion is repeated after equal intervals of time, it is called periodic motion.
The simplest type of periodic motion is harmonic motion. As shown in the Fig. 1.38,
the mass m has a harmonic motion of
x  Asin   Asin t
dx
 x  A cos t
dt
(1.30). The velocity of the mass m at time t is given by
(1.31). And the acceleration is given by
d 2x
 x   2 A sin t   2 x (1.32).
2
dt
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1.10.1 Vectorial Representation of Harmonic Motion

Harmonic motion can be represented as a vector X  OP , it’s projection on the
vertical axis is given by y  A sin t and on the horizontal axis x  A cos t .
1.10.2 Complex Number Represention of Harmonic Motion
It is more convenient to represent harmonic function using a complex number

representation. X  a  ib (1.35) where i   1 and a and b denote the x and y



components of X , respectively. X also can be expressed as X  A cos  iA sin  (1.36)
with A  a 2  b 2  2
1
C.D. Lee
(1.37) and   tan 1
b
a
(1.38)
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e i  cos   i sin 
(1.41)
e  i  cos   i sin 

X  Acos   i sin    Ae i
Thus
(1.42)
(1.43)
1.10.5 Definitions and Terminology
Cycle:
Amplitude: The maximum displacement of a vibrating body from its equilibrium
position is called the amplitude of vibration.
Period of oscillation: The time taken to finish one cycle of motion is known as
the period of oscillation or time period.  
2

,  is called the circular frequency.
Frequency of oscillation: The number of cycles per unit time is called the
1
frequency. f  


(Hz), and  (rad/s).
2
Phase angle: Two vibratory motions x1  A1 sin t and x2  A2 sin t    are called
synchronous because they have same frequency. In fig. 1.44, the second vector OP2
leads the first one OP1 by an angle  , known as phase angle.
Natural frequency: When placed into motion, oscillation will take place at the
nature frequency f n or  n 
k
, which is a property of the system.
m
Beats: When two harmonic motion, with frequencies close to one another, are
added, the resulting motion exhibits a phenomenon known as beats. For example, if
(1.64) where  is a small quantity, the
x1  X cos t (1.63) x2  X cos   t
addition of these motions yields x(t )  2 X cos
C.D. Lee
t


cos   t .
2
2

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Octave: When the maximum value of a range of frequency is twice its minimum
value, it is known as an octave band. For example 75-150Hz, 150-300Hz, and 300600Hz can be called octave band.
Decibel: It is a unit of measurement that is frequently used in vibration
x
p 
measurement. It is defined in terms of a power ratio. dB= 10 log 10  1  = 10 log 10  1
 x2
 p2 



2
The second equation results from the fact that power is proportional to the square of
the amplitude or voltage. The decibel is often expressed in terms of the first power of
 x1 
 .
 x2 
amplitude or voltage as dB= 20 log 10 
1.11 Harmonic Analysis
Harmonic motion is the simplest motion of vibratory systems. However, any
periodic function of time can be represented by Fourier series as an infinite of sine
and cosine terms.
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1.11.1 Fourier Series Expansion
If x is a period function with a period of , its Fourier series representation is
a0
 a1 cos t  a 2 cos 2t  ...........
2
 b1 sin t  b2 sin 2t  ............
x(t ) 
given by

(1.70)

a0
  a n cos nt  bn sin nt 
2
n 1
Where   2 /  we obtain
a0 
 2 
2 
x
(
t
)
dt

x(t )dt
 0
 0
(1.71)
an 
 2 
2 
x(t ) cos ntdt   x(t ) cos ntdt

0

 0
(1.72)
bn 
 2 
2 
x(t ) sin ntdt   x(t ) sin ntdt

 0
 0
(1.73)
It can also be represented by the series of sine or cosine terms only and given as
x(t )  d 0  d1 cost  1   d 2 cos2t  2   ........
(1.74)
Where d 0  a0 / 2
(1.75)

d n  an2  bn2

 bn
 an
 n  tan 1 
1
2



(1.76)
(1.77)
1.11.2 Complex Fourier Series
The Fourier series can also be represented in terms of complex numbers. From
Eqs. (1.41) and (1.42), that
e it  cos t  i sin t
(1.78)
e it  cos t  i sin t
(1.79)
and
e int  e int
2
(1.80)
e int  e int
sin  n t 
2i
(1.81)
cos  n t 
thus
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a0   1
1

   a n  ibn e int  a n  bn e int 
2 n 1  2
2


a
 0   c n e int  c n* e int
2 n 1
xt  



(1.82)
n 
c e 
i
n  
nt
n
where
c0 
a0
2
cn 
1
a n  ibn 
2
c n* 
1
a n  bn 
2
1.11.3 Frequency Spectrum
When the coefficients of the Fourier series are plotted against frequency the
result is a series of discrete lines called the Fourier spectrum. Generally plotted are
 bn
 an
the absolute values d n  an2  bn2  2 and  n  tan 1 
1

 . Figure 1.48 shows a typical

frequency spectrum.
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1.11.4 Time and Frequency Domain Representation
 Example 1.12 Fourier Series Expansion. Determine the Fourier series
expansion of the motion of the valve in the cam-follower system shown in Fig.
1.53.
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l2
t
where y (t )  Y , 0  t  
y(t ) (E.1)

l1
Yl
2
t
and the period   . By defining A  2 that x(t )  A , 0  t  
(E.3)


l1
Solution: tan  
y (t ) x(t )

l1
l2
or x(t ) 
Eq. (E.3) is shown in Fig. 1.46(a). We use Eqs. (1.71) and (1.73) to compute:
2
 2
 2
t
 At2 
 
a0    x(t )dt    A dt 
 0
 0

   2  0

A
(E.4)
an 
 2 
 2  t
x
(
t
)
cos
n

tdt

A cos ntdt  0, n  1,2,3......
 0
 0

bn 
 2 
 2  t
A
x
(
t
)
sin
n

tdt

A sin ntdt  
, n  1,2,3,...... (E.6)


0
0



n
(E.5)
Therefore the Fourier series expansion of x(t ) is
A A
A
 sin t 
sin 2t 
2 
2
A  
1
1

   sin t  sin 2t  sin 3t 
 2 
2
3

x(t ) 
(E.7)
It can be seen from Fig. 1.46(b) that the approximation reaches the sawtooth
shape even only three terms are computed.
Vibration Literaturees
 Journal of Vibration, Acoustics, Stress, and Reliability in Design
 Journal of Applied Mechanics te
 Journal of Sound and Vibration
 AIAA Journal
 ASCE Journal of Engineering Mechanics
 Earthquake Engineering and Structural Dynamics
 Bulletin of the Japan Society of Mechanical Engineers
 International Journal of Solids and Structures
 International Journal of Numerical Methods in Engineering
 Journal of the Acoustical Society of America
 Sound and Vibration
C.D. Lee
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