Transcript x 1 - UniMAP Portal

Mechanical
Vibrations
Singiresu S. Rao
SI Edition
Chapter 1
Fundamentals of
Vibration
Course Outline
1.
2.
3.
4.
5.
6.
7.
Fundamentals of Vibration
Free Vibration of Single DOF Systems
Harmonically Excited Vibration
Vibration under General Forcing
Conditions
Two DOF Systems
Multidegree of Freedom Systems
Determination of Natural Frequencies
and Mode Shapes
© 2005 Pearson Education South Asia Pte Ltd.
2
Course Outline
8.
Continuous Systems
9.
Vibration Control
10.
Vibration Measurement and Applications
11.
Numerical Integration Methods in Vibration
Analysis
12.
Finite Element Method
13.
Nonlinear Vibration
14.
Random Vibration
© 2005 Pearson Education South Asia Pte Ltd.
3
Chapter Outline
1.1 Preliminary Remarks
1.2 Brief History of Vibration
1.3 Importance of the Study of Vibration
1.4 Basic Concepts of Vibration
1.5 Classification of Vibration
1.6 Vibration Analysis Procedure
1.7 Spring Elements
© 2005 Pearson Education South Asia Pte Ltd.
4
Chapter Outline
1.8 Mass or Inertia Elements
1.9 Damping Elements
1.10 Harmonic Motion
1.11 Harmonic Analysis
© 2005 Pearson Education South Asia Pte Ltd.
5
1.1 Preliminary Remarks
•
•
•
•
•
Brief History of vibration
Examination of vibration’s important role
Vibration analysis of an engineering system
Definitions and concepts of vibration
Concept of harmonic analysis for general
periodic motions
© 2005 Pearson Education South Asia Pte Ltd.
6
1.2 Brief History of Vibration
• Origins of vibration:
582-507 B.C. –
Pythagoras, the Greek philosopher
and mathematician, is the first to
investigate musical sounds on a
scientific basis. He conducted
experiments on a vibrating string by
using a simple apparatus called a
monochord. He further developed
the concept of pitch.
© 2005 Pearson Education South Asia Pte Ltd.
7
1.2 Brief History of Vibration
Around 350 B.C. –
Aristotle wrote treatises on music and sound
In 320 B.C. –
Aristoxenus wrote a three-volume work entitled
Elements of Harmony
In 300 B.C. –
Euclid wrote a treatises Introduction to Harmonics
A.D. 132 –
Zhang Heng invented the world’s
first seismograph to measure
earthquakes
© 2005 Pearson Education South Asia Pte Ltd.
8
1.2 Brief History of Vibration
• Galileo to Rayleigh:
Galileo Galilei (1564 – 1642)
- founder of modern experimental science
- started experimenting on simple pendulum
- published a book, Discourses Concerning
Two New Sciences, in 1638, describing
resonance, frequency, length, tension and
density of a vibrating stretched string
Robert Hooke (1635 – 1703)
- found relation between pitch and frequency of
vibration of a string
© 2005 Pearson Education South Asia Pte Ltd.
9
1.2 Brief History of Vibration
Joseph Sauveur (1653 – 1716)
- coined the word “acoustics” for the science of
sound
- founded nodes, loops, harmonics and the
fundamental frequency
- calculated the frequency of a stretched string
from the measured sag of its middle point
Sir Isaac Newton (1642-1727)
- published his monumental work, Philosophiae
Naturalis Principia Mathematica, in 1686,
discovering three laws of motion
© 2005 Pearson Education South Asia Pte Ltd.
10
1.2 Brief History of Vibration
Joseph Lagrange (1736 – 1813)
- found the analytical solution of the vibrating
string and the wave equation
Simeon Poisson (1781 – 1840)
- solved the problem of vibration of a
rectangular flexible membrane
R.F.A. Clebsch (1833 – 1872)
- studied the vibration of a circular membrane
Lord Baron Rayleigh
- founded Rayleigh-Ritz method, used to find
frequency of vibration of a conservative
system and multiple natural frequencies
© 2005 Pearson Education South Asia Pte Ltd.
11
1.2 Brief History of Vibration
• Recent contributions:
1902 – Frahm investigated the importance of
torsional vibration study in the design of
propeller shafts of steamships
Aurel Stodola (1859 – 1943)
- contributed to the study of vibration of beams,
plates, and membranes.
- developed a method for analyzing vibrating
beams which is applicable to turbine blades
C.G.P. De Laval (1845 – 1913)
- presented a practical solution to the problem
of vibration of an unbalanced rotating disk
© 2005 Pearson Education South Asia Pte Ltd.
12
1.2 Brief History of Vibration
1892 – Lyapunov laid the foundations of modern
stability theory which is applicable to all
types of dynamical systems
1920 – Duffling and Van der Pol brought the first
definite solutions into the theory of
nonlinear vibrations and drew attention to
its importance in engineering
– Introduction of the correlation function by
Taylor
1950 – advent of high-speed digital computers
– generate approximate solutions
© 2005 Pearson Education South Asia Pte Ltd.
13
1.2 Brief History of Vibration
1950s – developed finite element method enabled
engineers to conduct numerically detailed
vibration analysis of complex mechanical,
vehicular, and structural systems
displaying thousands of degrees of
freedom with the aid of computers
– Turner, Clough, Martin and Topp presented the
finite element method as known today
© 2005 Pearson Education South Asia Pte Ltd.
14
1.3 Importance of the Study of Vibration
• Why study vibration?
 Vibrations can lead to excessive deflections
and failure on the machines and structures
 To reduce vibration through proper design of
machines and their mountings
 To utilize profitably in several consumer and
industrial applications
 To improve the efficiency of certain machining,
casting, forging & welding processes
 To stimulate earthquakes for geological
research and conduct studies in design of
nuclear reactors
© 2005 Pearson Education South Asia Pte Ltd.
15
1.4 Basic Concepts of Vibration
 Vibration = any motion that repeats itself after
an interval of time
 Vibratory System consists of:
1) spring or elasticity
2) mass or inertia
3) damper
 Involves transfer of potential energy to kinetic
energy and vice versa
© 2005 Pearson Education South Asia Pte Ltd.
16
1.4 Basic Concepts of Vibration
 Degree of Freedom (d.o.f.) =
min. no. of independent coordinates required
to determine completely the positions of all
parts of a system at any instant of time
 Examples of single degree-of-freedom
systems:
© 2005 Pearson Education South Asia Pte Ltd.
17
1.4 Basic Concepts of Vibration
 Examples of single degree-of-freedom
systems:
© 2005 Pearson Education South Asia Pte Ltd.
18
1.4 Basic Concepts of Vibration
 Examples of Two degree-of-freedom systems:
© 2005 Pearson Education South Asia Pte Ltd.
19
1.4 Basic Concepts of Vibration
Examples of Three degree of freedom systems:
© 2005 Pearson Education South Asia Pte Ltd.
20
1.4 Basic Concepts of Vibration
Example of Infinite-number-of-degrees-offreedom system:
Infinite number of degrees of freedom system
are termed continuous or distributed systems
Finite number of degrees of freedom are
termed discrete or lumped parameter systems
More accurate results obtained by increasing
number of degrees of freedom
© 2005 Pearson Education South Asia Pte Ltd.
21
1.5 Classification of Vibration
Free Vibration:
A system is left to vibrate on its own after an
initial disturbance and no external force acts on
the system. E.g. simple pendulum
Forced Vibration:
A system that is subjected to a repeating
external force. E.g. oscillation arises from diesel
engines
Resonance occurs when the frequency of the
external force coincides with one of the
natural frequencies of the system
© 2005 Pearson Education South Asia Pte Ltd.
22
1.5 Classification of Vibration
Undamped Vibration:
When no energy is lost or dissipated in friction
or other resistance during oscillations
Damped Vibration:
When any energy is lost or dissipated in
friction or other resistance during oscillations
Linear Vibration:
When all basic components of a vibratory
system, i.e. the spring, the mass and the
damper behave linearly
© 2005 Pearson Education South Asia Pte Ltd.
23
1.5 Classification of Vibration
Nonlinear Vibration:
If any of the components behave nonlinearly
Deterministic Vibration:
If the value or magnitude of the excitation (force
or motion) acting on a vibratory system is
known at any given time
Nondeterministic or random Vibration:
When the value of the excitation at a given
time cannot be predicted
© 2005 Pearson Education South Asia Pte Ltd.
24
1.5 Classification of Vibration
Examples of deterministic and random
excitation:
© 2005 Pearson Education South Asia Pte Ltd.
25
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
© 2005 Pearson Education South Asia Pte Ltd.
26
1.6 Vibration Analysis Procedure
Example of the modeling of a forging hammer:
© 2005 Pearson Education South Asia Pte Ltd.
27
Example 1.1
Mathematical Model of a Motorcycle
Figure 1.18(a) shows a motorcycle with a rider.
Develop a sequence of three mathematical
models of the system for investigating vibration in
the vertical direction. Consider the elasticity of the
tires, elasticity and damping of the struts (in the
vertical direction), masses of the wheels, and
elasticity, damping, and mass of the rider.
© 2005 Pearson Education South Asia Pte Ltd.
28
Example 1.1
Solution
We start with the simplest model and refine it
gradually. When the equivalent values of the
mass, stiffness, and damping of the system are
used, we obtain a single-degree of freedom model
of the motorcycle with a rider as indicated in Fig.
1.18(b). In this model, the equivalent stiffness (keq)
includes the stiffness of the tires, struts, and rider.
The equivalent damping constant (ceq) includes
the damping of the struts and the rider. The
equivalent mass includes the mass of the wheels,
vehicle body and the rider.
© 2005 Pearson Education South Asia Pte Ltd.
29
Example 1.1
© 2005 Pearson Education South Asia Pte Ltd.
Solution
30
Example 1.1
Solution
This model can be refined by representing the
masses of wheels, elasticity of tires, and elasticity
and damping of the struts separately, as shown in
Fig. 1.18(c). In this model, the mass of the vehicle
body (mv) and the mass of the rider (mr) are
shown as a single mass, mv + mr. When the
elasticity (as spring constant kr) and damping (as
damping constant cr) of the rider are considered,
the refined model shown in Fig. 1.18(d) can be
obtained.
© 2005 Pearson Education South Asia Pte Ltd.
31
Example 1.1
© 2005 Pearson Education South Asia Pte Ltd.
Solution
32
Example 1.1
Solution
Note that the models shown in Figs. 1.18(b) to (d)
are not unique. For example, by combining the
spring constants of both tires, the masses of both
wheels, and the spring and damping constants of
both struts as single quantities, the model shown
in Fig. 1.18(e) can be obtained instead of Fig.
1.18(c).
© 2005 Pearson Education South Asia Pte Ltd.
33
1.7 Spring Elements
Linear spring is a type of mechanical link that is
generally assumed to have negligible mass and
damping
Spring force is given by:
F  kx
1.1
F = spring force,
k = spring stiffness or spring constant, and
x = deformation (displacement of one end
with respect to the other)
© 2005 Pearson Education South Asia Pte Ltd.
34
1.7 Spring Elements
Work done (U) in deforming a spring or the
strain (potential) energy is given by:
1 2
U  kx
2
1.2
When an incremental force ΔF is added to F:
F  F  F ( x*  x)
dF
 F (x ) 
(x)
dx x*
*
1 d 2F

2! dx2
© 2005 Pearson Education South Asia Pte Ltd.
(x) 2  ...
x*
1.3
35
1.7 Spring Elements
© 2005 Pearson Education South Asia Pte Ltd.
36
1.7 Spring Elements
Static deflection of a beam at the free end is
given by:
Wl 3
1.6
 st 
3EI
W = mg is the weight of the mass m,
E = Young’s Modulus, and
I = moment of inertia of cross-section of beam
Spring Constant is given by:
W
3EI
k 

l
3
1.7 
st
© 2005 Pearson Education South Asia Pte Ltd.
37
1.7 Spring Elements
Combination of Springs:
1) Springs in parallel – if we have n spring
constants k1, k2, …, kn in parallel, then the
equivalent spring constant keq is:
keq  k1  k2  ...  kn
© 2005 Pearson Education South Asia Pte Ltd.
1.11
38
1.7 Spring Elements
Combination of Springs:
2) Springs in series – if we
have n spring constants k1,
k2, …, kn in series, then the
equivalent spring constant
keq is:
1 1 1
1
   ... 
k
k k
k
eq
1
2
© 2005 Pearson Education South Asia Pte Ltd.
1.17 
n
39
Example 1.3
Torsional Spring Constant of a Propeller Shaft
Determine the torsional spring constant of the
speed propeller shaft shown in Fig. 1.25.
© 2005 Pearson Education South Asia Pte Ltd.
40
Example 1.3
Solution
We need to consider the segments 12 and 23 of
the shaft as springs in combination. From Fig.
1.25, the torque induced at any cross section of
the shaft (such as AA or BB) can be seen to be
equal to the torque applied at the propeller, T.
Hence, the elasticities (springs) corresponding to
the two segments 12 and 23 are to be considered
as series springs. The spring constants of
segments 12 and 23 of the shaft (kt12 and kt23) are
given by
© 2005 Pearson Education South Asia Pte Ltd.
41
Example 1.3
Solution
GJ
G ( D  d ) (80  10 ) (0.3  0.2 )
k 


l
32l
32(2)
12
4
4
12
12
9
4
4
t12
12
12
 25.5255  10 N - m/rad
6
GJ
G ( D  d ) (80  10 ) (0.25  0.15 )
k 


l
32l
32(3)
23
4
4
23
23
9
4
4
t 23
23
23
 8.9012  10 N - m/rad
6
© 2005 Pearson Education South Asia Pte Ltd.
42
Example 1.3
Solution
Since the springs are in series, Eq. (1.16) gives
k k
(25.5255  10 )(8.9012  10 )
k 

k k
(25.5255  10  8.9012  10 )
6
t12
6
t 23
teq
6
t12
6
t 23
 6.5997  10 N - m/rad
6
© 2005 Pearson Education South Asia Pte Ltd.
43
Example 1.5
Equivalent k of a Crane
The boom AB of crane is a uniform steel bar of
length 10 m and x-section area of 2,500 mm2.
A weight W is suspended while the crane is
stationary. Steel cable CDEBF has x-sectional
area of 100 mm2. Neglect effect of cable CDEB,
find equivalent spring constant of system in the
vertical direction.
© 2005 Pearson Education South Asia Pte Ltd.
44
Example 1.5
Solution
A vertical displacement x of pt B will cause the
spring k2 (boom) to deform by x2 = x cos 45º and
the spring k1 (cable) to deform by an amount
x1 = x cos (90º – θ). Length of cable FB, l1 is as
shown.
l12  32  102  2(3)(10) cos 135  151.426
 l1  12.3055 m
© 2005 Pearson Education South Asia Pte Ltd.
45
Example 1.5
Solution
 The angle θ satisfies the relation:
l12  32  2(l1)(3) cos   102
cos   0.8184,
  35.0736
 The total potential energy (U):
1
2 1
U  k1( x cos 45)  k2[ x cos( 90   )]2
2
2
E.1
A1E1 (100  106 )(207  109 )
k1 

 1.6822  106 N/m
l1
12.0355
A2 E2 (2500  106 )(207  109 )
k2 

 5.1750  107 N/m
l2
10
© 2005 Pearson Education South Asia Pte Ltd.
46
Example 1.5
Solution
 Potential Energy of the equivalent spring is:
U eq
1
 keq x 2
2
E.2
 By setting U = Ueq, hence:
keq  26.4304  106 N/m
© 2005 Pearson Education South Asia Pte Ltd.
47
1.8
Mass or Inertia Elements
Using mathematical model to represent the
actual vibrating system
E.g. In figure below, the mass and damping
of the beam can be disregarded; the system
can thus be modeled as a spring-mass
system as shown.
© 2005 Pearson Education South Asia Pte Ltd.
48
1.8
Mass or Inertia Elements
Combination of Masses
E.g. Assume that the
mass of the frame is
negligible compared to
the masses of the floors.
The masses of various
floor levels represent the
mass elements, and the
elasticities of the vertical
members denote the
spring elements.
© 2005 Pearson Education South Asia Pte Ltd.
49
1.8
Mass or Inertia Elements
Case 1: Translational Masses Connected by a
Rigid Bar
Velocities of masses can be expressed as:
l2
x2  x1
l1
© 2005 Pearson Education South Asia Pte Ltd.
l3
x3  x1
l1
1.18
50
1.8
Mass or Inertia Elements
Case 1: Translational Masses Connected by a
Rigid Bar
1.19
xeq  x1
By equating the kinetic energy of the system:
1
2 1
2 1
2 1
2
m1x1  m2 x2  m3 x3  meq xeq
2
2
2
2
2
meq
2
 l2 
 l3 
 m1    m2    m3
 l1 
 l1 
© 2005 Pearson Education South Asia Pte Ltd.
1.20
1.21
51
1.8
Mass or Inertia Elements
Case 2: Translational and Rotational Masses
Coupled Together
meq = single equivalent translational mass
x = translational velocity
 = rotational velocity
J0 = mass moment of inertia
Jeq = single equivalent rotational mass
© 2005 Pearson Education South Asia Pte Ltd.
52
1.8
Mass or Inertia Elements
Case 2: Translational and Rotational Masses
Coupled Together
1. Equivalent translational mass:
Kinetic energy of the two masses is given
by:
1 2 1 2
1.22
T  mx  J 0
2
2
Kinetic energy of the equivalent mass is
given by:
1
2
1.23
Teq  meq xeq
2
© 2005 Pearson Education South Asia Pte Ltd.
53
1.8
Mass or Inertia Elements
Case 2: Translational and Rotational Masses
Coupled Together
x

Since   and xeq  x , equating Teq & T
R
gives
J0
1.24
meq  m  2
R
2. Equivalent rotational mass:
Here, eq   and x  R , equating Teq
and T gives 1
2 1
2 1


J eq  mR   J 0 2
2
2
2
or
© 2005 Pearson Education South Asia Pte Ltd.
J eq  J 0  mR 2
1.25
54
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.
© 2005 Pearson Education South Asia Pte Ltd.
55
Example 1.6
Solution
Assuming small displacements, the equivalent
mass (meq) can be determined using the
equivalence of the kinetic energies of the two
systems. When the mass m is displaced by a
distance x , the pulley and the rigid link 1 rotate by
an angle     x / r . This causes the rigid link 2
and the cylinder to be displaced by a distance
x   l  xl / r . Since the cylinder rolls without
slippage, it rotates by an angle   x / r  xl / r r .
The kinetic energy of the system (T) can be
expressed (for small displacements) as:
p
2
p 1
1
1
p
p
c
© 2005 Pearson Education South Asia Pte Ltd.
2
c
1
p
c
56
Example 1.6
Solution
1
1  1  1
1  1
T  mx  J   J   m x  J   m x
2
2
2
2
2
2
2
2
p
2
p
1
2
1
2
2
2
c
c
c
(E.1)
2
2
where Jp, J1, and Jc denote the mass moments of
inertia of the pulley, link 1 (about O), and cylinder,
respectively,  ,  and  indicate the angular
velocities of the pulley, link 1 (about O), and
cylinder, respectively, and x and x represent the
linear velocities of the mass m and link 2,
respectively. Noting that J  m r / 2 and J  m l / 3 ,
Equation (E.1) can be rewritten as
p
1
c
2
2
2
c
© 2005 Pearson Education South Asia Pte Ltd.
c
c
1
1 1
57
Example 1.6
Solution
 x  1  m l
   
 r  2 3
2
1
1
T  mx  J
2
2
2
1 1
p
p
1m r
 
2 2
c
c
2
 x  1  xl
   m 
 r  2  r



 xl

 r r



2
2
1
2
p
p
1
p
c
 1  xl
  m 
 2 r
2
1
c
p
2
2
(E.2)
By equating Equation (E.2) to the kinetic energy of
the equivalent system
1
T  m x
2
eq
© 2005 Pearson Education South Asia Pte Ltd.
2
(E.3)
58
Example 1.6
Solution
we obtain the equivalent mass of the system as
J 1ml m l 1 m l
l
m m 


m
r 3 r
r
2 r
r
2
eq
2
2
p
1 1
2 1
c 1
2
2
2
2
p
© 2005 Pearson Education South Asia Pte Ltd.
p
p
p
2
1
c
2
(E.4)
p
59
Example 1.7
Cam-Follower Mechanism
A cam-follower mechanism is
used to convert the rotary motion
of a shaft into the oscillating or
reciprocating motion of a valve.
The follower system consists of a
pushrod of mass mp, a rocker arm
of mass mr, and mass moment of
inertia Jr about its C.G., a valve of
mass mv, and a valve spring of
negligible mass.
Find the equivalent mass (meq) of this camfollower system by assuming the location of meq
as (i) pt A and (ii) pt C.
© 2005 Pearson Education South Asia Pte Ltd.
60
Example 1.7
Solution
The kinetic energy of the system (T) is:
1
2 1
2 1
2 1
2

E.1
T  m p x p  mv xv  J r r  mr xr
2
2
2
2
If meq denotes equivalent mass placed at pt A,
with xeq  x , the kinetic energy equivalent mass
system Teq is:
Teq
© 2005 Pearson Education South Asia Pte Ltd.
1
2
 meq xeq
2
E.2
61
Example 1.7
Solution
By equating T and Teq, and note that
xl2
xl3
x

x p  x, xv 
, xr 
, and  r 
l1
l1
l1
meq  m p 
Jr
l12
 mv
l22
l12
 mr
l32
l12
E.3
Similarly, if equivalent mass is located at point C,
xeq  xv , hence,
Teq
1
1
2
 meq xeq  meq xv2
2
2
© 2005 Pearson Education South Asia Pte Ltd.
E.4
62
Example 1.7
Solution
Equating (E.4) and (E.1) gives
meq
 l1 
 mv  2  m p  
l2
 l2 
Jr
© 2005 Pearson Education South Asia Pte Ltd.
2
2
2
 l3 
 mr  2 
 l1 
E.5
63
1.9
Damping Elements
Viscous Damping:
Damping force is proportional to the velocity of
the vibrating body in a fluid medium such as air,
water, gas, and oil.
Coulomb or Dry Friction Damping:
Damping force is constant in magnitude but
opposite in direction to that of the motion of the
vibrating body between dry surfaces
Material or Solid or Hysteretic Damping:
Energy is absorbed or dissipated by material
during deformation due to friction between
internal planes
© 2005 Pearson Education South Asia Pte Ltd.
64
1.9
Damping Elements
Hysteresis loop for elastic materials:
© 2005 Pearson Education South Asia Pte Ltd.
65
1.9
Damping Elements
Shear Stress ( ) developed in the fluid layer at
a distance y from the fixed plate is:
du
1.26
 
dy
where du/dy = v/h is the velocity gradient.
• Shear or Resisting Force (F) developed at the
bottom surface of the moving plate is:
Av
1.27
F  A  
 cv
h
where A is the surface area of the moving plate.
© 2005 Pearson Education South Asia Pte Ltd.
Where A is the surface area of the moving
A
plate and c 
is the damping constant
h
66
1.9
Damping Elements
and
c
A
h
1.28
is called the damping constant.
If a damper is nonlinear, a linearization process
is used about the operating velocity (v*) and the
equivalent damping constant is:
dF
c
dv v*
© 2005 Pearson Education South Asia Pte Ltd.
1.29
67
Example 1.9
Piston-Cylinder Dashpot
Develop an expression for the damping constant
of the dashpot as shown in Fig. 1.36(a).
© 2005 Pearson Education South Asia Pte Ltd.
68
Example 1.9 Solution
The damping constant of the dashpot can be
determined using the shear stress equation for
viscous fluid flow and the rate of fluid flow
equation. As shown in Fig. 1.36(a), the dashpot
consists of a piston diameter D and length l,
moving with velocity v0 in a cylinder filled with a
liquid of viscosity µ. Let the clearance between the
piston and the cylinder wall be d. At a distance y
from the moving surface, let the velocity and shear
stress be v and τ, and at a distance (y + dy) let the
velocity and shear stress be (v – dv) and (τ + dτ),
respectively (see Fig. 1.36b).
© 2005 Pearson Education South Asia Pte Ltd.
69
Example 1.9 Solution
The negative sign for dv shows that the velocity
decreases as we move toward the cylinder wall.
The viscous force on this annular ring is equal to
d
F  Dld  Dl dy
(E.1)
dy
But the shear stress is given by
dv
  
dy
(E.2)
where the negative sign is consistent with a
decreasing velocity gradient. Using Eq. (E.2) in
Eq. (E.1), we obtain
© 2005 Pearson Education South Asia Pte Ltd.
70
Example 1.9 Solution
dv
F  Dldy
dy
2
2
(E.3)
The force on the piston will cause a pressure
difference on the ends of the element, given by
P
4P
p

 D  D


 4 
2
2
(E.4)
Thus the pressure force on the end of the element
is
4P
pDdy  
© 2005 Pearson Education South Asia Pte Ltd.
D
dy
(E.5)
71
Example 1.9 Solution
where Ddy denotes the annular area between y
and (y + dy). If we assume uniform mean velocity
in the direction of motion of the fluid, the forces
given in Eqs. (E.3) and (E.5) must be equal. Thus
we get
4P
dv
D
or
dy  Dldy
dv
4P

dy
D l
dy
2
2
2
© 2005 Pearson Education South Asia Pte Ltd.
2
2
(E.6)
72
Example 1.9 Solution
Integrating this equation twice and using the
boundary conditions v  v at y = 0 and v = 0 at
y = d, we obtain
0
2P
y

 yd  y   v 1  
v
D l
 d
(E.7)
2
2
0
The rate of flow through the clearance space can
be obtained by integrating the rate of flow through
an element between the limits y = 0 and y = d:
1 
 2 Pd
Q   vDdy  D 
 v d
 6D l 2 
3
d
0
© 2005 Pearson Education South Asia Pte Ltd.
2
0
(E.8)
73
Example 1.9 Solution
The volume of the liquid flowing through the
clearance space per second must be equal to the
volume per second displaced by the piston. Hence
the velocity of the piston will be equal to this rate of
flow divided by the piston area. This gives
v 
0
Q


D


4 
(E.9)
2
Equations (E.8) and (E.9) lead to
© 2005 Pearson Education South Asia Pte Ltd.
74
Example 1.9 Solution

 2d  
 3D l 1  D  
P
 v
4d




3
3
0
(E.10)
By writing the force as P = cv0, the damping
constant c can be found as
 3D l  2d 
c  
1  
D 
 4d 
3
3
© 2005 Pearson Education South Asia Pte Ltd.
(E.11)
75
Example 1.10
Equivalent Spring and Damping
Constants of a Machine Tool Support
A precision milling machine is supported on four
shock mounts, as shown in Fig. 1.37(a). The
elasticity and damping of each shock mount can
be modeled as a spring and a viscous damper, as
shown in Fig. 1.37(b). Find the equivalent spring
constant, keq, and the equivalent damping
constant, ceq, of the machine tool support in terms
of the spring constants (ki) and damping constants
(ci) of the mounts.
© 2005 Pearson Education South Asia Pte Ltd.
76
Example 1.10
Equivalent Spring and Damping
Constants of a Machine Tool Support
© 2005 Pearson Education South Asia Pte Ltd.
77
Example 1.10 Solution
The free-body diagrams of the four springs and
four dampers are shown in Fig. 1.37(c). Assuming
that the center of mass, G, is located
symmetrically with respect to the four springs and
dampers, we notice that all the springs will be
subjected to the same displacement, x , and all the
dampers will be subject to the same relative
velocity x , where x and x denote the
displacement and velocity, respectively, of the
center of mass, G. Hence the forces acting on the
springs (Fsi) and the dampers (Fdi) can be
expressed as
© 2005 Pearson Education South Asia Pte Ltd.
78
Example 1.10 Solution
© 2005 Pearson Education South Asia Pte Ltd.
79
Example 1.10 Solution
F  k x;
i  1,2,3,4
F  c x;
i  1,2,3,4
si
di
i
i
(E.1)
Let the total forces acting on all the springs and all
the dampers be Fs and Fd, respectively (see Fig.
1.37d). The force equilibrium equations can thus
be expressed as
F F F F F
s
s1
s2
s3
s4
F F F F F
d
© 2005 Pearson Education South Asia Pte Ltd.
d1
d2
d3
d4
(E.2)
80
Example 1.10 Solution
where Fs + Fd = W, with W denoting the total
vertical force (including the inertia force) acting on
the milling machine. From Fig. 1.37(d), we have
F k x
s
eq
F  c x
d
(E.3)
eq
Equation (E.2) along with Eqs. (E.1) and (E.3),
yield
k  k  k  k  k  4k
eq
1
2
3
4
c  c  c  c  c  4c
eq
© 2005 Pearson Education South Asia Pte Ltd.
1
2
3
4
(E.4)
81
Example 1.10 Solution
where ki = k and ci = c for i = 1, 2, 3, 4.
Note: If the center of mass, G, is not located
symmetrically with respect to the four springs and
dampers, the ith spring experiences a
displacement of x and the ith damper experiences
a velocity of x where x and x can be related to
the displacement x and velocity x of the center of
mass of the milling machine, G. In such a case,
Eqs. (E.1) and (E.4) need to be modified suitably.
i
i
© 2005 Pearson Education South Asia Pte Ltd.
i
i
82
1.10 Harmonic Motion
Periodic Motion: motion repeated after equal
intervals of time
Harmonic Motion: simplest type of periodic
motion
1.30
x  Asin  Asin t
Displacement (x): (on horizontal axis)
Velocity:
dx
  A cos t
dt
1.31
Acceleration:
d 2x
dt
2
  2 A sin t   2 x
© 2005 Pearson Education South Asia Pte Ltd.
1.32
83
1.10 Harmonic Motion
• Scotch yoke
mechanism:
The similarity
between cyclic
(harmonic) and
sinusoidal
motion.
© 2005 Pearson Education South Asia Pte Ltd.
84
1.10 Harmonic Motion
Complex number representation of harmonic
motion: 
1.35
X  a  ib
where i = √(–1) and a and b denote the real and
imaginary x and y components of X,
respectively.
© 2005 Pearson Education South Asia Pte Ltd.
85
1.10 Harmonic Motion
Also, Eqn. (1.36) can be expressed as

1.36
X  A cos   iA sin 

1.43
X  Acos   i sin    Aei
Thus,
A j  (a 2j  b 2j ); j  1, 2
1 b j

 j  tan  ; j  1, 2
aj 
© 2005 Pearson Education South Asia Pte Ltd.
1.47 
1.48
86
1.10 Harmonic Motion
Operations on Harmonic Functions:

Rotating Vector, X  Aeit
1.51
it
Displacement  Re[ Ae
]  A cos t
Velocity  Re[iAeit ]  A sin t
 A cos t  90
1.54
1.55
Accelerati on  Re[  2 Aeit ]
  A cos t
2
  2 A cos t  180
where Re denotes the real part
© 2005 Pearson Education South Asia Pte Ltd.
1.56
87
1.10 Harmonic Motion
• Displacement, velocity, and accelerations as
rotating vectors
• Vectorial addition of
harmonic functions
© 2005 Pearson Education South Asia Pte Ltd.
88
Example 1.11
Addition of Harmonic Motions
Find the sum of the two harmonic motions
x (t )  10 cos t and x (t )  15 cos(t  2).
1
2
Solution:
Method 1: By using trigonometric relations: Since
the circular frequency is the same for both x1(t)
and x2(t), we express the sum as
x(t )  A cos(t   )  x (t )  x (t )
1
© 2005 Pearson Education South Asia Pte Ltd.
2
(E.1)
89
Example 1.11 Solution
That is,
Acos t cos   sin t sin    10 cos t  15 cos(t  2)
 10 cos t  15(cos t cos 2  sin t sin 2)
(E.2)
That is,
cos t ( A cos  )  sin t ( A sin  )  cos t (10  15 cos 2)
 sin t (15 sin 2)
(E.3)
By equating the corresponding coefficients of
cosωt and sinωt on both sides, we obtain
A cos   10  15 cos 2
A sin   15 sin 2
A
10  15 cos 2  (15 sin 2)
 14.1477
© 2005 Pearson Education South Asia Pte Ltd.
2
2
(E.4)
90
Example 1.11 Solution
and
 15 sin 2 
  tan 

10

15
cos
2


 74.5963 
1
(E.5)
Method 2: By using vectors: For an arbitrary value
of ωt, the harmonic motions x1(t) and x2(t) can be
denoted graphically as shown in Fig. 1.43. By
adding them vectorially, the resultant vector x(t)
can be found to be
x(t )  14.1477 cos( t  74.5963 )
(E.6)
© 2005 Pearson Education South Asia Pte Ltd.
91
Example 1.11 Solution
Method 3: By using complex number
representation: the two harmonic motions can be
denoted in terms of complex numbers:
x (t )  Re A e
1
1
i t
x (t )  Re A e
2
2
  Re10e 
  Re15e 
it
i ( t  2 )
i ( t  2 )
(E.7)
The sum of x1(t) and x2(t) can be expressed as
x(t )  ReAe
i ( t  )

(E.8)
where A and α can be determined using Eqs. (1.47)
and (1.48) as A = 14.1477 and α = 74.5963º
© 2005 Pearson Education South Asia Pte Ltd.
92
1.10 Harmonic Motion
Definitions of Terminology:
Amplitude (A) is the maximum displacement
of a vibrating body from its equilibrium
position
Period of oscillation (T) is time taken to
complete one cycle of motion
2
1.59
T

Frequency of oscillation (f) is the no. of
1 
cycles per unit time
1.60
f  
T 2
© 2005 Pearson Education South Asia Pte Ltd.
93
1.10 Harmonic Motion
Definitions of Terminology:
Natural frequency is the frequency which a
system oscillates without external forces
Phase angle () is the angular difference
between two synchronous harmonic motions
1.61
x1  A1 sin t
x2  A2 sin t   
© 2005 Pearson Education South Asia Pte Ltd.
1.62
94
1.10 Harmonic Motion
Definitions of Terminology:
Beats are formed when two harmonic
motions, with frequencies close to one
another, are added
© 2005 Pearson Education South Asia Pte Ltd.
95
1.10 Harmonic Motion
Definitions of Terminology:
Decibel is originally defined as a ratio of
electric powers. It is now often used as a
notation of various quantities such as
displacement, velocity, acceleration,
pressure, and power
P
dB  10 log 
P 
X 
dB  20 log 
X 
(1.68)
0
(1.69)
0
where P0 is some reference value of power
and X0 is specified reference voltage.
96
© 2005 Pearson Education South Asia Pte Ltd.
1.11 Harmonic Analysis
• Fourier Series Expansion:
If x(t) is a periodic function with period  , its
Fourier Series representation is given by
a
x(t )   a cos t  a cos 2t  ...
2
b sin t  b sin 2t  ...
0
1
2
1
2
a
   (a cos nt  b sin nt )
2

0
n 1
© 2005 Pearson Education South Asia Pte Ltd.
n
n
(1.70)
97
1.11 Harmonic Analysis
•Gibbs Phenomenon:
An anomalous behavior observed from a
periodic function that is being represented by
Fourier series.
As n increases, the
approximation can be seen
to improve everywhere
except in the vicinity of the
discontinuity, P. The error
in amplitude remains at
approximately 9 percent,
even when k   .
© 2005 Pearson Education South Asia Pte Ltd.
98
1.11 Harmonic Analysis
•Complex Fourier Series:
The Fourier series can also be represented in
terms of complex numbers.
e  cos t  i sin t
(1.78)
e  cos t  i sin t
(1.79)
it
and
Also,
it
e e
cos t 
2
e e
sin t 
2i
i t
i t
© 2005 Pearson Education South Asia Pte Ltd.
 i t
(1.80)
 i t
(1.81)
99
1.11 Harmonic Analysis
•Frequency Spectrum:
Harmonics plotted as vertical lines on a diagram
of amplitude (an and bn or dn and Φn) versus
frequency (nω)
© 2005 Pearson Education South Asia Pte Ltd.
100
1.11 Harmonic Analysis
• A periodic function:
• Representation of a function in time and
frequency domain:
© 2005 Pearson Education South Asia Pte Ltd.
101
1.11 Harmonic Analysis
• Even and odd functions:
Even function & its Fourier
series expansion
x(t )  x(t )
(1.87 )
a
x(t )    a cos nt (1.88)
2

0
n
n 1
Odd function & its Fourier
series expansion
x(t )   x(t )
(1.89)
x(t )   b sin nt
(1.90)

n 1
n
© 2005 Pearson Education South Asia Pte Ltd.
102
1.11 Harmonic Analysis
• Half-Range Expansions:
The function is extended to include
the interval   to 0 as shown in the
figure. The Fourier series
expansions of x1(t) and x2(t) are
known as half-range expansions.
© 2005 Pearson Education South Asia Pte Ltd.
103
1.11 Harmonic Analysis
• Numerical Computation
of Coefficients
If x(t) is not in a simple
form, experimental
determination of the
amplitude of vibration
and numerical
2
integration procedure a  N  x
like the trapezoidal or
2nt
2
Simpson’s rule is used a  N  x cos 
to determine the
2nt
2
coefficients an and bn. b  N  x sin 
N
0
i 1
(1.97 )
i
N
n
i 1
i
N
i
n
© 2005 Pearson Education South Asia Pte Ltd.
i 1
i
i
(1.98)
(1.99)
104
Example 1.12
Fourier Series Expansion
Determine the Fourier series expansion of the
motion of the valve in the cam-follower system
shown in the Figure.
© 2005 Pearson Education South Asia Pte Ltd.
105
Example 1.12 Solution
If y(t) denotes the vertical motion of the pushrod,
the motion of the valve, x(t), can be determined
from the relation:
y (t ) x(t )
tan  

l
l
or
l
x(t )  y (t )
(E.1)
l
where
t
y (t )  Y ; 0  t  
(E.2)
1
2
2
1

and the period is given by  
© 2005 Pearson Education South Asia Pte Ltd.
2

.
106
Example 1.12 Solution
By defining
Yl
A
l
2
1
x(t) can be expressed as
t
x(t )  A ; 0  t  
(E.3)

Equation (E.3) is shown in the Figure.
To compute the Fourier coefficients an and bn, we
use Eqs. (1.71) to (1.73):

a  

0
2 /
0

x(t )dt  

© 2005 Pearson Education South Asia Pte Ltd.
2 /
0
 At 
A dt 
 

  2
t
2
2 /
A
(E.4)
0
107
Example 1.12 Solution


t
a   x(t ) cos nt  dt   A cos nt  dt



A
A  cos nt t sin nt 


 t cos nt  dt 



2  n
n
n
2 /
2 /
0
0
2 /
2 /
0
 0,
2
2
n  1, 2, ..
(E.5)


t
b   x(t ) sin nt  dt   A sin nt  dt



A
A  sin nt t cos nt 


 t sin nt  dt 



2  n
n
n
0
2 /
2 /
0
0
2 /
2 /
0
A

, n  1, 2, ..
n
© 2005 Pearson Education South Asia Pte Ltd.
2
2
0
(E.6)
108
Example 1.12 Solution
Therefore the Fourier series expansion of x(t) is
A A
A
x(t )   sin 2t 
sin 2t  ...
2 
2
A  
1
1

   sin t  sin 2t  sin 3t  ...
 2 
2
3

(E.7)
The first three terms of the
series are shown plotted in the
figure. It can be seen that the
approximation reaches the
sawtooth shape even with a
small number of terms.
© 2005 Pearson Education South Asia Pte Ltd.
109