vibration fundamental -basic
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Mechanical
VibrationsSingiresu S. Rao
SI Edition
Chapter 1
Fundamentals of
Vibration
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8. Continuous Systems
9. Vibration Control
10. Vibration Measurement and Applications
11. Numerical Integration Methods in VibrationAnalysis
12. Finite Element Method
13. Nonlinear Vibration
14. Random Vibration
Course Outline
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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
Chapter Outline
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1.8 Mass or Inertia Elements
1.9 Damping Elements
1.10 Harmonic Motion
1.11 Harmonic Analysis
Chapter Outline
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1.1 Preliminary Remarks
Brief History of vibration
Examination of vibrations important role
Vibration analysis of an engineering system
Definitions and concepts of vibration
Concept of harmonic analysis for generalperiodic motions
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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 ascientific basis. He conducted
experiments on a vibrating string by
using a simple apparatus called a
monochord. He further developed
the concept of pitch.
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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. 132Zhang Heng invented the worldsfirst seismograph to measure
earthquakes
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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 ConcerningTwo New Sciences, in 1638, describingresonance, frequency, length, tension anddensity of a vibrating stretched string
Robert Hooke (1635 1703)- found relation between pitch and frequency ofvibration of a string
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1.2 Brief History of Vibration
Joseph Sauveur (1653 1716)
- coined the word acoustics for the science ofsound
- 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
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1.2 Brief History of Vibration
Joseph Lagrange (1736 1813)- found the analytical solution of the vibratingstring and the wave equation
Simeon Poisson (1781 1840)
- solved the problem of vibration of arectangular flexible membrane
R.F.A. Clebsch (1833 1872)- studied the vibration of a circular membrane
Lord Baron Rayleigh
- founded Rayleigh-Ritz method, used to findfrequency of vibration of a conservativesystem and multiple natural frequencies
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1.2 Brief History of Vibration
Recent contributions:1902 Frahm investigated the importance of
torsional vibration study in the design ofpropeller 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 bladesC.G.P. De Laval (1845 1913)
- presented a practical solution to the problemof vibration of an unbalanced rotating disk
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1.2 Brief History of Vibration
1892 Lyapunov laid the foundations of modernstability theory which is applicable to all
types of dynamical systems
1920 Duffling and Van der Pol brought the first
definite solutions into the theory ofnonlinear vibrations and drew attention to
its importance in engineering
Introduction of the correlation function byTaylor
1950 advent of high-speed digital computers
generate approximate solutions
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1.2 Brief History of Vibration
1950s developed finite element method enabledengineers to conduct numerically detailed
vibration analysis of complex mechanical,
vehicular, and structural systems
displaying thousands of degrees offreedom with the aid of computers
Turner, Clough, Martin and Topp presented thefinite element method as known today
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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 mountingsTo utilize profitably in several consumer and
industrial applications
To improve the efficiency of certain machining,
casting, forging & welding processesTo stimulate earthquakes for geological
research and conduct studies in design ofnuclear reactors
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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 transferof potential energy to kinetic
energy and vice versa
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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 ofsingle degree-of-freedomsystems:
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1.4 Basic Concepts of Vibration
Examples ofsingle degree-of-freedom
systems:
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1.4 Basic Concepts of Vibration
Examples ofTwo degree-of-freedom systems:
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1.4 Basic Concepts of Vibration
Examples ofThree degree of freedom systems:
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1.4 Basic Concepts of Vibration
Example ofInfinite-number-of-degrees-of-
freedom system:
Infinite number of degrees of freedom system
are termed continuous ordistributedsystems
Finite number of degrees of freedom are
termed discrete orlumpedparameter systems
More accurate results obtained by increasing
number of degrees of freedom
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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
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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 infriction or other resistance during oscillations
LinearVibration:
When all basic components of a vibratorysystem, i.e. the spring, the mass and the
damper behave linearly
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1.5 Classification of Vibration
NonlinearVibration:
Ifany of the components behave nonlinearly
Deterministic Vibration:
If the value or magnitude of the excitation (force
or motion) acting on a vibratory system isknown at any given time
Nondeterministic orrandom Vibration:
When the value of the excitation at a giventime cannot be predicted
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1.5 Classification of Vibration
Examples of deterministic and random
excitation:
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1.6 Vibration Analysis Procedure
Example of the modeling of a forging hammer:
E l 1 1
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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 thevertical 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.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. Theequivalent 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. 1.18(c). In this model, the mass of the vehicle
body (mv) and the mass of the rider (mr) areshown 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 beobtained.
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Example 1.1 Solution
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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 shownin Fig. 1.18(e) can be obtained instead of Fig.
1.18(c).
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1.1kxF
F= spring force,
k= spring stiffness or spring constant, and
x = deformation (displacement of one end
with respect to the other)
1.7 Spring Elements
Linearspring is a type of mechanical link that is
generally assumed to have negligible mass and
damping
Spring force is given by:
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Work done(U) in deforming a spring or the
strain (potential) energy is given by:
When an incremental force Fis added to F:
2.12
1 2kxU
3.1...)(!2
1
)()(
)(
22
2
*
*
*
*
x
dx
Fd
xdx
dFxF
xxFFF
x
x
1.7 Spring Elements
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1.7 Spring Elements
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1.7 Spring Elements
Static deflection of a beam at the free end is
given by:
Spring Constantis given by:
6.13
3
EI
Wlst
W= mg is the weight of the mass m,
E = Youngs Modulus, andI = moment of inertia of cross-section of beam
7.13
3l
EIWk
st
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1.7 Spring Elements
Combination of Springs:
1) Springs inparallel if we have n springconstants k1, k2, , kn inparallel, then theequivalent spring constant keq is:
11.121 ... neq kkkk
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1.7 Spring Elements
Combination of Springs:
2) Springs in series if wehave n spring constants k1,
k2, , kn in series, then the
equivalent spring constantkeq is:
17.11
...
111
21 neqkkkk
Example 1 3
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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 asAA orBB) 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) aregiven by
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Example 1.3 Solution
m/rad-N109012.8
)3(32
)15.025.0()1080(
32
)(
6
449
23
4
23
4
23
23
23
23
l
dDG
l
GJk
t
m/rad-N105255.25
)2(32
)2.03.0()1080(
32
)(
6
449
12
4
12
4
12
12
12
12
l
dDG
l
GJk
t
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Example 1.3 Solution
Since the springs are in series, Eq. (1.16) gives
m/rad-N105997.6
)109012.8105255.25(
)109012.8)(105255.25(
6
66
66
2312
2312
tt
tt
t
kk
kkk
eq
Example 1 5
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Example 1.5
Equivalent kof a Crane
The boomAB of crane is a uniform steel bar of
length 10 m and x-section area of 2,500 mm2.A weight Wis suspended while the crane is
stationary. Steel cable CDEBFhas 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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m3055.12
426.151135cos)10)(3(2103
1
2221
l
l
Example 1.5 Solution
A vertical displacementx of pt B will cause the
spring k2 (boom) to deform byx2 =x cos 45 andthe spring k1 (cable) to deform by an amount
x1 =x cos (90). Length of cable FB, l1 is asshown.
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Example 1.5 Solution
The angle satisfies the relation:
The total potential energy (U):
0736.35,8184.0cos
10cos)3)((23 2122
1
ll
1.E)]90cos([2
1)45cos(
2
1 22
21 xkxkU
N/m106822.10355.12
)10207)(10100( 696
1
111
l
EAk
N/m101750.510
)10207)(102500( 796
2
222
l
EAk
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Example 1.5 Solution
Potential Energy of the equivalent spring is:
By setting U = Ueq, hence:
2.E2
1 2xkU eqeq
N/m104304.26 6eq k
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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-masssystem as shown.
1.8 Mass or Inertia Elements
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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 verticalmembers denote the
spring elements.
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Case 1: Translational Masses Connected by a
Rigid Bar
Velocities of masses can be expressed as:
18.111
331
1
22 x
l
lxx
l
lx
1.8 Mass or Inertia Elements
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Case 1: Translational Masses Connected by a
Rigid Bar
By equating the kinetic energy of the system:
19.11xxeq
20.12
1
2
1
2
1
2
1 2eqeq
233
222
211 xmxmxmxm
21.13
2
1
32
2
1
21eq m
l
lm
l
lmm
1.8 Mass or Inertia Elements
1 8
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meq = single equivalent translational mass
= translational velocity= rotational velocity
J0 = mass moment of inertia
Jeq = single equivalent rotational mass
x
1.8 Mass or Inertia Elements
Case 2: Translational and Rotational Masses
Coupled Together
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Case 2: Translational and Rotational Masses
Coupled Together
1. Equivalent translational mass:
Kinetic energy of the two masses is given
by:
Kinetic energy of the equivalent mass is
given by:
22.12
1
2
1 20
2 JxmT
23.12
1 2eqeqeq xmT
1.8 Mass or Inertia Elements
1 8 M I ti El t
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Case 2: Translational and Rotational Masses
Coupled Together
2. Equivalent rotational mass:
Here, and , equating Teq
and Tgives
eq Rx
25.1
2
1
2
1
2
1
20eq
20
22eq
mRJJor
JRmJ
24.120
eqR
Jmm
Since and , equating Teq & T
gives R
x xx eq
1.8 Mass or Inertia Elements
Example 1.6
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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 thepulley and rotates with it.
E l 1 6 S l i
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Example 1.6 Solution
Assuming small displacements, the equivalent
mass (meq) can be determined using theequivalence of the kinetic energies of the two
systems. When the mass m is displaced by a
distance , the pulley and the rigid link 1 rotate by
an angle . This causes the rigid link 2
and the cylinder to be displaced by a distance
. Since the cylinder rolls without
slippage, it rotates by an angle .The kinetic energy of the system (T) can be
expressed (for small displacements) as:
x
pprx /
1
pprxllx /
112
cpcc rrxlrx // 12
E l 1 6 S l ti
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Example 1.6 Solution
)E.1(2
1
2
1
2
1
2
1
2
1
2
12
2
22
22
2
11
22
xmJxmJJxmTcccpp
where Jp, J1, and Jc denote the mass moments of
inertia of the pulley, link 1 (about O), and cylinder,
respectively, indicate the angularvelocities of the pulley, link 1 (about O), and
cylinder, respectively, and represent the
linear velocities of the mass m and link 2,
respectively. Noting that ,Equation (E.1) can be rewritten as
cp and, 1
2and xx
2/2cccrmJ 3/and
2
111lmJ
E l 1 6 S l ti
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Example 1.6 Solution
)E.2(2
1
22
1
21
321
21
21
2
1
2
1
2
2
1
2
2
2
11
2
2
p
c
cp
cc
ppp
p
r
lxm
rr
lxrm
rlxm
rxlm
rxJxmT
By equating Equation (E.2) to the kinetic energy of
the equivalent system
E.3)(21 2xmT
eq
E l 1 6 S l ti
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Example 1.6 Solution
)E.4(2
1
3
12
2
1
2
2
1
2
2
12
2
2
11
2
p
c
p
c
ppp
p
eq
r
lm
r
lm
r
lm
r
lm
r
Jmm
we obtain the equivalent mass of the system as
Example 1.7
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p
Cam-Follower Mechanism
A cam-follower mechanism is
used to convert the rotary motionof a shaft into the oscillating orreciprocating motion of a valve.
The follower system consists of a
pushrod of mass mp, a rocker armof mass mr, and mass moment ofinertia Jrabout its C.G., a valve ofmass mv, and a valve spring of
negligible mass.Find the equivalent mass (meq) of this cam-follower system by assuming the location ofmeqas (i) ptA and (ii) pt C.
E l 1 7 S l ti
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The kinetic energy of the system (T) is:
Ifmeq denotes equivalent mass placed at ptA,
with , the kinetic energy equivalent masssystem Teq is:
1.E2
1
2
1
2
1
2
1 2222rrrrvvpp xmJxmxmT
xx eq
2.E21 2
eqeqeq xmT
Example 1.7 Solution
Example 1 7 Solution
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By equating Tand Teq, and note that
Similarly, if equivalent mass is located at point C,
hence,
11
3
1
2 and,,,l
x
l
lxx
l
lxxxx rrvp
3.E21
2
321
2
221
eql
lm
l
lm
l
Jmm rvrp
,eq v
xx
4.E2
1
2
1 2eq
2eqeqeq vxmxmT
Example 1.7 Solution
Example 1 7 Solution
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Equating (E.4) and (E.1) gives
5.E
2
21
23
2
2
122
eq
l
lm
l
lm
l
Jmm rp
rv
Example 1.7 Solution
1 9 Damping Elements
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1.9 Damping Elements
Viscous Damping:
Damping force is proportional to the velocity ofthe vibrating body in a fluid medium such as air,water, gas, and oil.
Coulomb orDry Friction Damping:
Damping force is constant in magnitude butopposite in direction to that of the motion of thevibrating body between dry surfaces
Material orSolid orHysteretic Damping:Energy is absorbed or dissipated by materialduring deformation due to friction betweeninternal planes
1 9 Damping Elements
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Hysteresis loop for elastic materials:
1.9 Damping Elements
1 9 Damping Elements
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Shear Stress () developed in the fluid layer at
a distance y from the fixed plate is:
where du/dy = v/h is the velocity gradient. ShearorResisting Force (F) developed at the
bottom surface of the moving plate is:
where A is the surface area of the moving plate.
26.1dy
du
27.1cvhAv
AF
Where A is the surface area of the moving
plate and is the damping constanth
A
c
1.9 Damping Elements
1 9 Damping Elements
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1.9 Damping Elements
and
is called the damping constant.
If a damper is nonlinear, a linearization process
is used about the operating velocity (v*) and theequivalent damping constant is:
28.1
h
Ac
29.1*vdv
dFc
Example 1.9
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Piston-Cylinder Dashpot
Develop an expression for the damping constant
of the dashpot as shown in Fig. 1.36(a).
Example 1 9 Solution
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Example 1.9 Solution
The damping constant of the dashpot can be
determined using the shear stress equation forviscous fluid flow and the rate of fluid flow
equation. As shown in Fig. 1.36(a), the dashpotconsists of a piston diameterD and length l,
moving with velocity v0 in a cylinder filled with a
liquid of viscosity . Let the clearance between the
piston and the cylinder wall be d. At a distance y
from the moving surface, let the velocity and shearstress be vand , and at a distance (y + dy) let the
velocity and shear stress be (v dv) and (+ d),
respectively (see Fig. 1.36b).
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Example 1.9 Solution
The negative sign fordvshows that the velocity
decreases as we move toward the cylinder wall.The viscous force on this annular ring is equal to
E.1)(dy
dy
dDlDldF
But the shear stress is given by
(E.2)dy
dv
where the negative sign is consistent with a
decreasing velocity gradient. Using Eq. (E.2) in
Eq. (E.1), we obtain
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Example 1.9 Solution
E.3)(2
2
dy
vd
DldyF
E.4)(4
4
22 D
P
D
Pp
The force on the piston will cause a pressure
difference on the ends of the element, given by
Thus the pressure force on the end of the elementis
E.5)(4
dyD
PDdyp
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where denotes the annular area between y
and (y+ dy). If we assume uniform mean velocityin the direction of motion of the fluid, the forces
given in Eqs. (E.3) and (E.5) must be equal. Thus
we get
Example 1.9 Solution
2
2
4dyvdDldydy
DP
Ddy
or
E.6)(4
22
2
lD
P
dy
vd
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Example 1 9 Solution
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Example 1.9 Solution
The volume of the liquid flowing through the
clearance space per second must be equal to thevolume 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
E.9)(
4
2
0
D
Qv
Equations (E.8) and (E.9) lead to
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Example 1.9 Solution
E.10)(4
213
03
3
vd
D
dlD
P
By writing the force as P = cv0, the damping
constant c can be found as
E.11)(2143
3
3
Dd
dlDc
Example 1.10 Equivalent Spring and Damping
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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 springconstant, 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.
Example 1.10 Equivalent Spring and Damping
C f S
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Constants of a Machine Tool Support
Example 1 10 Solution
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The free-body diagrams of the four springs and
four dampers are shown in Fig. 1.37(c). Assumingthat 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, , and all thedampers will be subject to the same relative
velocity , where and 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
Example 1.10 Solution
x
x x x
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Example 1.10 Solution
Example 1 10 Solution
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Let the total forces acting on all the springs and all
the dampers be Fsand Fd, respectively (see Fig.
1.37d). The force equilibrium equations can thus
be expressed as
Example 1.10 Solution
E.1)(4,3,2,1;4,3,2,1;
ixcFixkF
idi
isi
E.2)(4321
4321
ddddd
sssss
FFFFF
FFFFF
Example 1 10 Solution
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Example 1.10 Solution
E.3)(xcF
xkF
eqd
eqs
where Fs + Fd= W, with Wdenoting the total
vertical force (including the inertia force) acting onthe milling machine. From Fig. 1.37(d), we have
Equation (E.2) along with Eqs. (E.1) and (E.3),
yield
E.4)(4
4
4321
4321
cccccc
kkkkkk
eq
eq
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where ki= kand ci= cfori= 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 and the ith damper experiences
a velocity of where and can be related to
the displacement and velocity 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.
Example 1.10 Solution
x x
ix
ix
ix
ix
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30.1sinsin tAAx
31.1cos tAdt
dx
32.1sin 222
2
xtAdt
xd
1.10 Harmonic Motion
Periodic Motion: motion repeated after equal
intervals of time
Harmonic Motion: simplest type of periodic
motion
Displacement (x): (on horizontal axis)Velocity:
Acceleration:
1 10 Harmonic Motion
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1.10 Harmonic Motion
Scotch yokemechanism:
The similarity
between cyclic
(harmonic) andsinusoidal
motion.
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Complex numberrepresentation of harmonic
motion:
where i = (1) and a and bdenote the real and
imaginaryx andy components ofX,respectively.
35.1ibaX
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Also, Eqn. (1.36) can be expressed as
Thus,
43.1sincos
36.1sincos
iAeiAX
iAAX
47.12,1;)( 22 jbaA jjj
48.12,1;tan 1
ja
b
j
jj
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Operations on Harmonic Functions:
Rotating Vector,
where Re denotes the real part
51.1tiAeX
54.1cos]Re[ntDisplaceme tAAe ti
55.190cossin]Re[Velocity
tAtAAei
ti
56.1180cos
cos
]Re[onAccelerati
2
2
2
tA
tA
Ae ti
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Displacement, velocity, and accelerations as
rotating vectors
Vectorial addition ofharmonic functions
Example 1.11
Additi f H i M ti
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Addition of Harmonic Motions
E.1)()()()cos()(21txtxtAtx
Find the sum of the two harmonic motions
).2cos(15)(andcos10)(21 ttxttx
Solution:
Method 1: By using trigonometric relations: Since
the circular frequency is the same for bothx1(t)
andx2(t), we express the sum as
Example 1.11 Solution
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p
E.2)()2sinsin2cos(cos15cos10
)2cos(15cos10sinsincoscosttt
ttttA
E.3)()2sin15(sin
)2cos1510(cos)sin(sin)cos(cos
t
tAtAt
That is,
That is,
By equating the corresponding coefficients of
costand sinton both sides, we obtain
E.4)(1477.14
)2sin15(2cos1510
2sin15sin2cos1510cos
22
A
AA
Example 1.11 Solution
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p
E.5)(5963.74
2cos1510
2sin15tan 1
and
Method 2: By using vectors: For an arbitrary value
oft, the harmonic motionsx1(t) and x2(t) can be
denoted graphically as shown in Fig. 1.43. By
adding them vectorially, the resultant vectorx(t)can be found to be
E.6)()5963.74cos(1477.14)( ttx
Example 1.11 Solution
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p
E.7)(15ReRe)(
10ReRe)()2()2(
22
11
titi
titi
eeAtx
eeAtx
Method 3: By using complex number
representation: the two harmonic motions can bedenoted in terms of complex numbers:
The sum ofx1(t) andx2(t) can be expressed as
E.8)(Re)(
)(
ti
Aetx
whereA and can be determined using Eqs. (1.47)
and (1.48) asA = 14.1477 and = 74.5963
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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
Frequency of oscillation (f) is the no. of
cycles per unit time
59.1
2
T
60.12
1
Tf
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Definitions of Terminology:
Natural frequency is the frequency which a
system oscillates without external forces
Phase angle () is the angular differencebetween two synchronous harmonic motions
62.1sin
61.1sin
22
11
tAx
tAx
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Definitions of Terminology:
Beats are formed when two harmonic
motions, with frequencies close to one
another, are added
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Definitions of Terminology:
Decibel is originally defined as a ratio ofelectric powers. It is now often used as a
notation of various quantities such as
displacement, velocity, acceleration,
pressure, and power
1.69)(log20dB
1.68)(log10dB
0
0
XX
P
P
where P0 is some reference value of power
andX0 is specified reference voltage.
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y
Fourier Series Expansion:Ifx(t) is a periodic function with period , itsFourier Series representation is given by
1
0
21
21
0
)70.1()sincos(2
...2sinsin
...2coscos
2
)(
nnn
tnbtnaa
tbtb
tataa
tx
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y
Gibbs Phenomenon:An anomalous behavior observed from aperiodic 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 errorin amplitude remains at
approximately 9 percent,
even when .k
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Complex Fourier Series:
The Fourier series can also be represented interms of complex numbers.
)79.1(sincos
)78.1(sincos
tite
tite
ti
ti
Also,
)81.1(2
sin
)80.1(2
cos
i
eet
eet
titi
titi
and
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Frequency Spectrum:Harmonics plotted as vertical lines on a diagram
of amplitude (an and bn ordn and n) versus
frequency (n)
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A periodic function:
Representation of a function in time andfrequency domain:
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Even and odd functions:
1
0
)88.1(cos2
)(
)87.1()()(
nn tnaatx
txtx
Even function & its Fourierseries expansion
Odd function & its Fourier
series expansion
1
)90.1(sin)(
)89.1()()(
nn
tnbtx
txtx
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Half-Range Expansions:
The function is extended to include
the interval as shown in the
figure. The Fourier series
expansions ofx1(t) andx2(t) areknown as half-range expansions.
0to
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Numerical Computation
of CoefficientsIfx(t) is not in a simple
form, experimental
determination of the
amplitude of vibrationand numerical
integration procedure
like the trapezoidal or
Simpsons rule is usedto determine the
coefficients an and bn. )99.1(2
sin2
)98.1(2cos2
)97.1(2
1
1
10
iN
iin
iN
iin
N
ii
tnx
Nb
tnxN
a
xN
a
Example 1.12
Fourier Series Expansion
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Fourier Series Expansion
Determine the Fourier series expansion of the
motion of the valve in the cam-follower systemshown in the Figure.
Example 1.12 Solution
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E.1)()()(
)()(tan
1
2
21
tyl
ltx
l
tx
l
ty
E.2)(0;)(
tt
Yty
Ify(t) denotes the vertical motion of the pushrod,
the motion of the valve,x(t), can be determinedfrom the relation:
or
where
and the period is given by .
2
Example 1.12 Solution
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1
2
l
Yl
A
E.3)(0;)( t
t
Atx
By defining
x(t) can be expressed as
Equation (E.3) is shown in the Figure.
To compute the Fourier coefficients an and bn, we
use Eqs. (1.71) to (1.73):
E.4)(2
)(
/2
0
2
/2
0
/2
00A
tAdt
tAdttxa
Example 1.12 Solution
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E.5)(..2,,1,0
sincos
2cos
coscos)(/2
0
22
/2
0
/2
0
/2
0
n
n
tnt
n
tnAdttnt
A
dttn
t
Adttntxan
E.6)(..2,,1,
cossin
2sin
sinsin)(
/2
0
22
/2
0
/2
0
/2
0
nn
A
n
tnt
n
tnAdttnt
A
dttnt
Adttntxbn
Example 1.12 Solution
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Therefore the Fourier series expansion ofx(t) is
E.7)(...3sin
3
12sin
2
1sin
2
...2sin2
2sin2
)(
tttA
tA
tAA
tx
The first three terms of the
series are shown plotted in the
figure. It can be seen that theapproximation reaches the
sawtooth shape even with a
top related