Transcript ert 452 - vibration

ERT 452 - VIBRATION
Munira Bt Mohamed Nazari
School of Bioprocess, UniMAP 2012
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LECTURE 1:
FUNDAMENTALS OF VIBRATION
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Topic Outline
 Introduction
 Basic Concepts of Vibration
 Classification of Vibration
 Vibration Analysis Procedure
 Spring Elements
 Mass or Inertia Elements
 Damping Elements
 Harmonic Motion
 Harmonic Analysis
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INTRODUCTION
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Brief History of Vibration
Phenomenon
of Vibration
Musical instrument
(string)
Use monochord
Observed that if 2 string of
different length are subject to
the same tension, the shorter
one emits a higher note.
Pythagoras
(582 - 507 BC)
Frequency of vibration
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Brief History of Vibration
(1564 – 1642)
Galileo Galilei
- Founder of modern experimental science.
- Started experimenting on simple pendulum.
- Study the behavior of a simple pendulum (observe
pendulum movement of a lamp).
- Describing resonance, frequency, length, tension
and density of a vibrating stretched string.
(1642 – 1727)
Sir Isaac Newton
- Derive the equation of motion of a vibrating body.
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Brief History of Vibration
(1902 – 1909)
Frahm
- Investigate the importance of torsional vibration
study in the design of the propeller shafts of
steamships.
- Propose the dynamic vibration absorber, which
involves the addition of a secondary spring-mass
system to eliminate the vibration of main system.
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Importance of the Study of Vibration
WHY???
 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.
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Importance of the Study of Vibration
EXAMPLE OF PROBLEMS
 Vibrational problems of prime movers due to inherent
unbalance in the engine.
 Wheel of some locomotive rise more than centimeter
off the track – high speeds – due to imbalance.
 Turbines – vibration cause spectacular mechanical
failure.
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Importance of the Study of Vibration
DISADVANTAGES
 Cause rapid wear.
 Create excessive noise.
 Leads
to poor surface finish (eg: in metal cutting
process, vibration cause chatter).
 Resonance – natural frequency of vibration of a
machine/structure coincide with the frequency of the
external excitation (eg: Tacoma Narrow Bridge –
1948)
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Applications
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BASIC CONCEPTS OF VIBRATION
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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.
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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:
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Basic Concepts of Vibration
 Examples of single degree-of-freedom systems:
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Basic Concepts of Vibration
 Examples of Two degree-of-freedom systems:
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Basic Concepts of Vibration
Examples of Three degree of freedom systems:
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Basic Concepts of Vibration
Example of Infinite number of degrees of freedom
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.
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CLASSIFICATION OF VIBRATION
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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
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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
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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
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Classification of Vibration
Examples of deterministic and random excitation:
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VIBRATION ANALYSIS PROCEDURE
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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
Derive
system/component
Free body diagram
(FBD)
Find the response
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(solve problem
Response (result):
method)
Displacement, velocities
& acceleration
Vibration Analysis Procedure
Example of the modeling of a forging hammer:
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Example 1.1
Mathematical Model of a Motorcycle
Figure below 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.
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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.(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.
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Example 1.1
Solution
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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.(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.(d) can be
obtained.
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Example 1.1
Solution
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Example 1.1
Solution
Note that the models shown in Figs.(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.(e) can
be obtained instead of Fig.(c).
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SPRING ELEMENTS
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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)
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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
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(x) 2  ...
x*
1.3
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Spring Elements
Static deflection of a beam at the free end is
given by:
Wl 3
 st 
3EI
1.6
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
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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
1.11
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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
1.17 
n
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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.
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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:
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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
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GJ
G ( D  d ) (80  10 ) (0.25  0.15 )
k 


l
32l
32(3)
23
4
4
23
23
t 23
23
23
 8.9012  10 N - m/rad
6
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9
4
4
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
t 23
teq
6
t12
t 23
 6.5997  10 N - m/rad
6
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6
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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.
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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) cos135  151.426
 l1  12.3055 m
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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
E.1
U  k1( x cos 45)  k2[ x cos( 90   )]2
2
2
A1E1 (100  106 )(207  109 )
k1 

 1.6822  106 N/m
l1
12.0355
A2 E2 (2500  106 )(207  109 )
7
k2 

 5.1750  10 N/m
l2
10
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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
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MASS OR INERTIA ELEMENTS
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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.
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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.
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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
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l3
x3  x1
l1
1.18
Mass or Inertia Elements
and,
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
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2
 l2 
 l3 
 m1    m2    m3
 l1 
 l1 
1.20
1.21
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
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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
T  mx  J 0
2
2
1.22
Kinetic energy of the equivalent mass is given by:
Teq
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1
2
 meq xeq
2
1.23
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
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J eq  J 0  mR 2
1.25
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 cam-follower system
by assuming the location of meq as (i) pt A and (ii) pt C.
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Example 1.7
Cam-Follower Mechanism
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Example 1.7
Solution
The kinetic energy of the system (T) is:
1
2 1
2 1
2 1
2

T  m p x p  mv xv  J r r  mr xr
2
2
2
2
E.1
If meq denotes equivalent mass placed at pt A, with
xeq  x the kinetic energy equivalent mass system
Teq is:
Teq
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1
2
 meq xeq
2
E.2
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
2
l1
 mv
l22
2
l1
 mr
l32
2
l1
E.3
Similarly, if equivalent mass is located at point C,
xeq  xv , hence,
Teq
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1
1
2
 meq xeq  meq xv2
2
2
E.4
Example 1.7
Solution
Equating (E.4) and (E.1) gives
meq
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 l1 
 mv  2  m p  
l2
 l2 
Jr
2
2
2
 l3 
 mr  2 
 l1 
E.5
DAMPING ELEMENTS
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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.
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Damping Elements
Hysteresis loop for elastic materials:
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Damping Elements
Construction of Viscous Dampers
µ
Velocity of intermediate fluid layers
are assumed to vary linearly
Fixed plane
Plate be moved with a velocity v in its own plane
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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 and c  hA
is the damping constant.
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Damping Elements
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*
1.29
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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.
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Example 1.10
Equivalent Spring and Damping Constants of a Machine
Tool Support
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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
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Example
1.10 Solution
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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
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d1
d2
d3
d4
(E.2)
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
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1
2
3
4
Parallel
(E.4)
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 i
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
i
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HARMONIC MOTION
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Harmonic Motion
Periodic Motion: motion repeated after equal
intervals of time
Harmonic Motion: simplest type of periodic
motion
Displacement (x):
x  Asin  Asin t
1.30
(On horizontal axis)
Velocity:
dx
  A cos t
dt
Acceleration: d x
2
dt
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2
1.31
  2 A sin t   2 x
1.32
Harmonic Motion
• Scotch yoke
mechanism:
The similarity
between cyclic
(harmonic) and
sinusoidal
motion.
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Harmonic Motion
Complex number representation of harmonic
motion:

X  a  ib
1.35
where i = √(–1) and a and b denote the real and
imaginary x and y components of X, respectively.
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Harmonic Motion
Also, Eqn. (1.36) can be expressed as

X  A cos   iA sin 

X  Acos   i sin    Aei
1.36
1.43
A j  (a 2j  b 2j ); j  1, 2
1.47 
Thus,
1 b j

 j  tan  ; j  1, 2
aj 
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1.48
Harmonic Motion
Operations on Harmonic Functions:
Rotating Vector,

X  Aeit
it
Displacement  Re[ Ae
]  A cos t
Velocity  Re[iAeit ]  A sin t
 A cos t  90
1.51
1.54
1.55
Accelerati on  Re[  2 Aeit ]
  2 A cos t
  2 A cos t  180
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Where Re denotes the real part.
1.56
Harmonic Motion
• Displacement, velocity, and accelerations as
rotating vectors
• Vectorial addition of
harmonic functions
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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
81
2
(E.1)
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
2
2
(E.4)
82
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)
83
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
84
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 cycles per unit
time
1 
1.60
f  
T 2
85
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
x1  A1 sin t
x2  A2 sin t   
1.61
1.62
86
Harmonic Motion
Definitions of Terminology:
Beats are formed when two harmonic motions, with
frequencies close to one another, are added
87
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.
88
HARMONIC ANALYSIS
ERT 452
89
Harmonic Analysis
• A periodic function:
90
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
n
n
(1.70)
91
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   .
92
Harmonic Analysis
•Complex Fourier Series:
The Fourier series can also be represented in terms of
complex numbers.
e  cos t  i sin t
e  cos t  i sin t
it
and
Also,
it
e e
cos t 
2
e e
sin t 
2i
it
it
93
(1.78)
(1.79)
 it
(1.80)
it
(1.81)
Harmonic Analysis
•Frequency Spectrum:
Harmonics plotted as vertical lines on a diagram of
amplitude (an and bn or dn and Φn) versus frequency
(nω).
94
Harmonic Analysis
• Representation of a function in time and
frequency domain:
95
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 1
n
Odd function & its Fourier
series expansion
x(t )   x(t )
(1.89)
x(t )   b sin nt
(1.90)

n 1
n
96
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.
97
Harmonic Analysis
• Numerical Computation of
Coefficients.
If x(t) is not in a simple
form, experimental
determination of the
amplitude of vibration and
numerical integration
procedure like the
trapezoidal or Simpson’s
rule is used to determine
the coefficients an and bn.
2
a  x
N
2nt
2
a   x cos
N

2nt
2
b   x sin
N

N
0
i 1
(1.97 )
i
N
n
i 1
i
N
i
n
i 1
i
i
(1.98)
(1.99)
98
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.
99
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  
2

.
100
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  

2 /
0
 At 
A dt 
 

  2
t
2
2 /
A
(E.4)
0
101
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
102
2
2
0
(E.6)
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

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.
103
(E.7)
ERT 452
104