Ppt Mech 5sem DOM
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Transcript of Ppt Mech 5sem DOM
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DYNAMICSOFMACHINERY
U5MEA19
Preparedby
Mr.Shaik Shabbeer Mr.Vennishmuthu.VAssistant Professor, Mechanical Department
VelTech Dr.RR & Dr.SR Technical University
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UNIT I : FORCE ANALYSIS
Rigid Body dynamics in general plane motionEquations ofmotion - Dynamic force analysis - Inertia force and Inertia torque
DAlembertsprinciple - The principle of superposition -
Dynamic Analysis in Reciprocating EnginesGas Forces -
Equivalent masses - Bearing loads - Crank shaft Torque -
Turning moment diagrams - Fly wheelsEngine shaking Forces
- Cam dynamics - Unbalance, Spring, Surge and Windup.
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Static force analysis.
If components of a machine accelerate, inertia is
produced due to their masses. However, the magnitudes
of these forces are small compares to the externally
applied loads. Hence inertia effect due to masses are
neglected. Such an analysis is known as static force
analysis
What is inertia?
The property of matter offering resistance to any change
of its state of rest or of uniform motion in a straight line is
known as inertia.
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conditions for a body to be in static and dynamic
equilibrium?
Necessary and sufficient conditions for static and dynamic
equilibrium are
Vector sum of all forces acting on a body is zero
The vector sum of the moments of all forces acting about any
arbitrary point or axis is zero.
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Static force analysis and dynamic force analysis.
If components of a machine accelerate, inertia forces are
produced due to their masses. If the magnitude of these forces
are small compared to the externally applied loads, they can be
neglected while analysing the mechanism. Such an analysis is
known as static force analysis.
If the inertia effect due to the mass of the component is also
considered, it is called dynamic force analysis.
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DAlemberts principle.
DAlemberts principle states that the inertia forces and torques,
and the external forces and torques acting on a body together
result in statical equilibrium.
In other words, the vector sum of all external forces and inertia
forces acting upon a system of rigid bodies is zero. The vector
sum of all external moments and inertia torques acting upon a
system of rigid bodies is also separately zero.
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The principle of super position states that for linear systems
the individual responses to several disturbances or driving
functions can be superposed on each other to obtain the total
response of the system.
The velocity and acceleration of various parts of reciprocating
mechanism can be determined , both analytically and
graphically.
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Dynamic Analysis in Reciprocating Engines-Gas Forces
Piston efforts (Fp): Net force applied on the piston , along the
line of stroke In horizontal reciprocating engines.It is also known
as effective driving force (or) net load on the gudgeon pin.
crank-pin effort.
The component of FQperpendicular to the crank is known as
crank-pin effort.
crank effort or turning movement on the crank shaft? It is the product of the crank-pin effort (FT)and crank pin
radius(r).
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Forces acting on the connecting rod
Inertia force of the reciprocating parts (F1) acting along the line
of stroke.
The side thrust between the cross head and the guide barsacting at right angles to line of stroke.
Weight of the connecting rod.
Inertia force of the connecting rod (FC)
The radial force (FR) parallel to crank and The tangential force (FT) acting perpendicular to crank
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Determination of Equivalent Dynamical System of Two
Masses by Graphical Method
Consider a body of mass m, acting at G as
shown in fig 15.15. This mass m, may be replaced by two masses m1 and m2 so that the system becomes
dynamical equivalent. The position of mass m1 may be fixed
arbitrarily atA. Now draw perpendicular CG at G, equal in
length of the radius of gyration of the body, kG .Then join AC
and draw CB perpendicular toAC intersecting AG produced in
B. The point B now fixes the position of the second
mass m2. The trianglesACG and BCG are similar. Therefore,
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Turning movement diagram or crank effort diagram?
It is the graphical representation of the turning movement or
crank effort for various position of the crank.
In turning moment diagram, the turning movement is taken asthe ordinate (Y-axis) and crank angle as abscissa (X axis).
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UNIT II : BALANCINGStatic and dynamic balancing - Balancing of rotating
massesBalancing reciprocating masses-
Balancing a single cylinder Engine - Balancing
Multi-cylinder Engines, Balancing V-engines, -Partial balancing in locomotive Engines-Balancing
machines.
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STATIC AND DYNAMIC BALANCING
When man invented the wheel, he very quickly learnt that ifit wasnt completely round and if it didnt rotate evenly
about its central axis, then he had a problem!
What the problem he had?
The wheel would vibrate causing damage to itself and itssupport mechanism and in severe cases, is unusable.
A method had to be found to minimize the problem. The
mass had to be evenly distributed about the rotating
centerline so that the resultant vibration was at a minimum.
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UNBALANCE:
The condition which exists in a rotor when vibratory
force or motion is imparted to its bearings as a result
of centrifugal forces is called unbalance or the
uneven distribution of mass about a rotors rotating
centreline.
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BALANCING:
Balancing is the technique of correcting or eliminating
unwanted inertia forces or moments in rotating orreciprocating masses and is achieved by changing the
location of the mass centres.
The objectives of balancing an engine are to ensure:
1. That the centre of gravity of the system remains stationeryduring a complete revolution of the crank shaft and
2. That the couples involved in acceleration of the different
moving parts balance each other.
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Types of balancing:
a) Static Balancing:
i) Static balancing is a balance of forces due to action of gravity.
ii) A body is said to be in static balance when its centre of gravity
is in the axis of rotation.
b) Dynamic balancing:
i) Dynamic balance is a balance due to the action of inertia forces.
ii) A body is said to be in dynamic balance when the resultant
moments or couples, which involved in the acceleration ofdifferent moving parts is equal to zero.
iii) The conditions of dynamic balance are met, the conditions of
static balance are also met.
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BALANCING OF ROTATING MASSES
When a mass moves along a circular path, it
experiences a centripetal acceleration and a force is
required to produce it. An equal and opposite force
called centrifugal force acts radially outwards and is
a disturbing force on the axis of rotation. The
magnitude of this remains constant but the directionchanges with the rotation of the mass.
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In a revolving rotor, the centrifugal force remains balanced as long as
the centre of the mass of rotor lies on the axis of rotation of the shaft.
When this does not happen, there is an eccentricity and an unbalance
force is produced. This type of unbalance is common in steam turbine
rotors, engine crankshafts, rotors of compressors, centrifugal pumpsetc.
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The unbalance forces exerted on machine members are time varying, impart
vibratory motion and noise, there are human discomfort, performance of the
machine deteriorate and detrimental effect on the structural integrity of the
machine foundation.
Balancing involves redistributing the mass which may be carried out by
addition or removal of mass from various machine members. Balancing of
rotating masses can be of
1. Balancing of a single rotating mass by a single mass rotating in the same
plane.
2. Balancing of a single rotating mass by two masses rotating in different
planes.
3. Balancing of several masses rotating in the same plane
4. Balancing of several masses rotating in different planes
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BALANCING OF A SINGLE ROTATING MASS BY A SINGLE
MASS ROTATING IN THE SAME PLANE
Consider a disturbing mass m1 which is attached to a shaft rotating at rad/s.
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r = radius of rotation of the mass m
The centrifugal force exerted by mass m1 on the shaft is given by,
F = m r c 1 1
This force acts radially outwards and produces bending moment on the shaft. In
order to counteract the effect of this force Fc1 , a balancing mass m2 may be
attached in the same plane of rotation of the disturbing mass m1 such that thecentrifugal forces due to the two masses are equal and opposite.
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BALANCING OF A SINGLE ROTATING MASS BY TWO MASSES ROTATING
There are two possibilities while attaching two balancing masses:
1. The plane of the disturbing mass may be in between the planes ofthe two balancing masses.
2. The plane of the disturbing mass may be on the left or right side of
two planes containing the balancing masses.
In order to balance a single rotating mass by two masses rotating in different
planes which are parallel to the plane of rotation of the disturbing mass i) the
net dynamic force acting on the shaft must be equal to zero, i.e. the centreof the masses of the system must lie on the axis of rotation and this is the
condition for static balancing ii) the net couple due to the dynamic forces
acting on the shaft must be equal to zero, i.e. the algebraic sum of the
moments about any point in the plane must be zero. The conditions i) and ii)
together give dynamic balancing.
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Balancing Multi-cylinder Engines, Balancing V-engines
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Problem 1.
Four masses A, B, C and D are attached to a shaft and revolve in the
same plane. The masses are 12 kg, 10 kg, 18 kg and 15 kg
respectively and their radii of rotations are 40 mm, 50 mm, 60 mm
and 30 mm. The angular position of the masses B, C and D are 60,
135 and 270from mass A. Find the magnitude and position of thebalancing mass at a radius of 100 mm.
Problem 2:
The four masses A, B, C and D are 100 kg, 150 kg, 120 kg and 130 kg
attached to a shaft and revolve in the same plane. The correspondingradii of rotations are 22.5 cm, 17.5 cm, 25 cm and 30 cm and the angles
measured from A are 45, 120 and 255. Find the position andmagnitude of the balancing mass, if the radius of rotation is 60 cm.
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UNIT III : FREE VIBRATION
Basic features of vibratory systems - idealizedmodels - Basic elements and lumping of
parameters - Degrees of freedom - Single degree
of freedom - Free vibration - Equations of motion -
natural frequency - Types of Damping - Dampedvibration critical speeds of simple shaft - Torsional
systems; Natural frequency of two and three rotor
systems
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INTRODUCTION
19 - 36
Mechanical vibrationis the motion of a particle or body which
oscillates about a position of equilibrium. Most vibrations in
machines and structures are undesirable due to increased stressesand energy losses.
Time interval required for a system to complete a full cycle of the
motion is theperiodof the vibration.
Number of cycles per unit time defines the frequencyof the vibrations.
Maximum displacement of the system from the equilibrium position is the
amplitudeof the vibration.
When the motion is maintained by the restoring forces only, the vibration
is described as free vibration. When a periodic force is applied to thesystem, the motion is described as forced vibration.
When the frictional dissipation of energy is neglected, the motion
is said to be undamped. Actually, all vibrations are dampedto
some degree.
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FREEVIBRATIONSOFPARTICLES. SIMPLEHARMONICMOTION
19 - 37
If a particle is displaced through a distancexmfrom its
equilibrium position and released with no velocity, the
particle will undergo simple harmonic motion,
0
kxxm
kxxkWFma st
General solution is the sum of twoparticular solutions,
tCtC
tm
kCt
m
kCx
nn cossin
cossin
21
21
xis aperiodic functionand nis the natural circular
frequencyof the motion.
C1and C2are determined by the initial conditions:
tCtCx nn cossin 21 02 xC
nvC 01 tCtCxv nnnn sincos 21
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FREEVIBRATIONSOFPARTICLES. SIMPLEHARMONICMOTION
19 - 38
txx nm sin
n
n
2period
2
1 n
nnf natural frequency
202
0 xvx nm amplitude
nxv 001
tan phase angle
Displacement is equivalent to thexcomponent of the sum of two vectors
which rotate with constant angular velocity21 CC
.n
02
01
xC
vCn
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FREEVIBRATIONSOFPARTICLES. SIMPLEHARMONICMOTION
19 - 39
txx nmsin
Velocity-time and acceleration-time curves can be
represented by sine curves of the same period as thedisplacement-time curve but different phase angles.
2sin
cos
tx
tx
xv
nnm
nnm
tx
tx
xa
nnm
nnm
sin
sin
2
2
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SIMPLEPENDULUM(APPROXIMATESOLUTION)
19 - 40
Results obtained for the spring-mass system can be
applied whenever the resultant force on a particle is
proportional to the displacement and directed towardsthe equilibrium position.
for small angles,
g
l
t
l
g
nn
nm
22
sin
0
:tt
maF
Consider tangential components of acceleration and
force for a simple pendulum,
0sin
sin
l
g
mlW
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SIMPLEPENDULUM(EXACTSOLUTION)
19 - 41
0sin
l
gAn exact solution for
leads to
2
0 22
sin2sin14
m
nd
g
l
which requires numerical solution.
g
lKn
2
2
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SAMPLEPROBLEM
19 - 42
A 50-kg block moves between vertical
guides as shown. The block is pulled
40mm down from its equilibrium position
and released.
For each spring arrangement,
determine a) the period of the vibration,
b) the maximum velocity of the block,
and c) the maximum acceleration of the
block.
For each spring arrangement, determinethe spring constant for a single
equivalent spring.
Apply the approximate relations for the
harmonic motion of a spring-mass
system.
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SAMPLEPROBLEM
19 - 43
mkN6mkN4 21 kk
Springs in parallel:
- determine the spring constant for equivalent
spring
mN10mkN10 4
21
21
kkPk
kkP
- apply the approximate relations for the
harmonic motion of a spring-mass system
nn
n m
k
2
srad14.14kg20
N/m104
s444.0n
srad4.141m040.0
nmm xv
sm566.0mv
2sm00.8ma
2
2
srad4.141m040.0
nmm axa
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SAMPLEPROBLEM
19 - 44
mkN6mkN4 21 kk Springs in series:
- determine the spring constant for equivalent
spring- apply the approximate relations for the harmonic
motion of a spring-mass system
nn
nm
k
2
srad93.6kg20
400N/m2
s907.0n
srad.936m040.0
nmm xv
sm277.0mv
2sm920.1ma
2
2
srad.936m040.0
nmm axa
mN10mkN10 4
21
21
kkPk
kkP
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FREEVIBRATIONSOFRIGIDBODIES
19 - 45
If an equation of motion takes the form
0or0 22
nn
xx
the corresponding motion may be considered
as simple harmonic motion.
Analysis objective is to determine n.
mgWmbbbmI ,22but 23222
121
0
5
3sin
5
3
b
g
b
g
g
b
b
g
nnn
3
52
2,
5
3then
For an equivalent simple pendulum,
35bl
Consider the oscillations of a square plate
ImbbW sin
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SAMPLEPROBLEM
19 - 46
k
A cylinder of weight Wis suspended as
shown.
Determine the period and natural
frequency of vibrations of the cylinder.
From the kinematics of the system, relate
the linear displacement and acceleration
to the rotation of the cylinder.
Based on a free-body-diagram equation for
the equivalence of the external and
effective forces, write the equation ofmotion.
Substitute the kinematic relations to arrive
at an equation involving only the angular
displacement and acceleration.
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SAMPLEPROBLEM
19 - 47
From the kinematics of the system, relate the linear
displacement and acceleration to the rotation of the cylinder.
rx rx 22
rra
ra
Based on a free-body-diagram equation for the equivalence of
the external and effective forces, write the equation of motion.
: effAA MM IramrTWr 22
rkWkTT 2but21
02
Substitute the kinematic relations to arrive at an equation
involving only the angular displacement and acceleration.
0
3
8
22 2
2121
m
k
mrrrmrkrWWr
m
kn
3
8
k
m
nn
8
32
2
m
kf nn
3
8
2
1
2
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 48
s13.1
lb20
n
W
s93.1n
The disk and gear undergo torsional
vibration with the periods shown.
Assume that the moment exerted by the
wire is proportional to the twist angle.
Determine a) the wire torsional springconstant, b) the centroidal moment of
inertia of the gear, and c) the maximum
angular velocity of the gear if rotated
through 90oand released.
Using the free-body-diagram equation for
the equivalence of the external and
effective moments, write the equation of
motion for the disk/gear and wire.
With the natural frequency and moment of
inertia for the disk known, calculate thetorsional spring constant.
With natural frequency and spring
constant known, calculate the moment of
inertia for the gear.
Apply the relations for simple harmonic
motion to calculate the maximum gear
velocity.
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 49
s13.1
lb20
n
W
s93.1n
Using the free-body-diagram equation for the equivalence of
the external and effective moments, write the equation ofmotion for the disk/gear and wire.
: effOO MM
0
I
K
IK
K
I
I
K
nnn
2
2
With the natural frequency and moment of inertia for
the disk known, calculate the torsional spring
constant.222
21 sftlb138.0
12
8
2.32
20
2
1
mrI
K
138.0213.1 radftlb27.4 K
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 50
s13.1
lb20
n
W
s93.1n
radftlb27.4 K
K
I
I
K
nnn
22
With natural frequency and spring constant
known, calculate the moment of inertia for the
gear.
27.4293.1 I 2sftlb403.0 I
Apply the relations for simple harmonic motion
to calculate the maximum gear velocity.
nmmnnmnm tt sinsin
rad571.190 m
s93.1
2rad571.1
2
n
mm
srad11.5m
PRINCIPLE OF CONSERVATION OF ENERGY
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PRINCIPLEOFCONSERVATIONOFENERGY
19 - 51
Resultant force on a mass in simple harmonic motion
is conservative - total energy is conserved.
constantVT
222
2212
21 constant
xx
kxxm
n
Consider simple harmonic motion of the square plate,
01T
2
21
21 2sin2cos1
m
m
Wb
WbWbV
22352122
32
212
21
2
212
21
2
m
mm
mm
mb
mbbm
IvmT
02 V
00 22235
212
21
2211
nmm mbWb
VTVT
bgn 53
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 52
Determine the period of small
oscillations of a cylinder which rolls
without slipping inside a curved
surface.
Apply the principle of conservation of
energy between the positions of maximum
and minimum potential energy.
Solve the energy equation for the natural
frequency of the oscillations.
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 53
01T
2
cos1
2
1
mrRW
rRWWhV
22
43
22
2
2
1
2
12
2
1
2
212
21
2
m
mm
mm
rRm
r
rRmrrRm
IvmT
02 V
Apply the principle of conservation of energy
between the positions of maximum and minimum
potential energy.
2211 VTVT
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SAMPLEPROBLEM
19 - 54
2211
VTVT
02
0 22
43
2
mm rRmrRW
22
4
32
2 mnm
m
rRmrRmg
rR
gn
3
22g
rR
nn
2
32
2
01T 221 mrRWV
2243
2 mrRmT 02 V
Solve the energy equation for the natural frequency
of the oscillations.
FORCED VIBRATIONS
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FORCEDVIBRATIONS
19 - 55
:maF
xmxkWtP stfm sin
tPkxxm fm sin
xmtxkW fmst sin
tkkxxm fm sin
Forced vibrations- Occurwhen a system is subjected
to a periodic force or a
periodic displacement of a
support.
f forced frequency
FORCEDVIBRATIONS
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19 - 56
txtCtC
xxx
fmnn
particulararycomplement
sincossin21
222
11 nf
m
nf
m
f
mm
kP
mk
Px
tkkxxm fm sin
tPkxxmfm
sin
At f= n, forcing input is in
resonancewith the system.
tPtkxtxm fmfmfmf sinsinsin2
Substituting particular solution into governing equation,
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 57
A motor weighing 350 lb is supported by
four springs, each having a constant 750
lb/in. The unbalance of the motor is
equivalent to a weight of 1 oz located 6
in. from the axis of rotation.
Determine a) speed in rpm at which
resonance will occur, and b) amplitude of
the vibration at 1200 rpm.
The resonant frequency is equal to the
natural frequency of the system.
Evaluate the magnitude of the periodic
force due to the motor unbalance.
Determine the vibration amplitude from
the frequency ratio at 1200 rpm.
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 58
The resonant frequency is equal to the natural
frequency of the system.
ftslb87.102.32
350 2m
ftlb000,36
inlb30007504
k
W= 350 lb
k = 4(350
lb/in)
rpm549rad/s5.57
87.10
000,36
m
kn
Resonance speed = 549
rpm
SAMPLE PROBLEM
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SAMPLEPROBLEM
19 - 59
W= 350 lb
k = 4(350
lb/in) rad/s5.57n
Evaluate the magnitude of the periodic force due to
the motor unbalance. Determine the vibration
amplitude from the frequency ratio at 1200 rpm.
ftslb001941.0sft2.32
1
oz16
lb1oz1
rad/s125.7rpm1200
2
2
m
f
lb33.157.125001941.0 2126
2
mrmaP nm
in001352.0
5.577.1251
300033.15
1 22
nf
mm
kPx
xm= 0.001352 in. (out of
phase)
DAMPEDFREEVIBRATIONS
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19 - 60
With viscous dampingdue to fluid friction,
:maF
0
kxxcxm
xmxcxkW st
Substitutingx = eltand dividing through by eltyields the characteristic equation,
m
k
m
c
m
ckcm
22
22
0 lll
Define the critical damping coefficient such that
ncc m
m
kmc
m
k
m
c220
2
2
All vibrations are damped to some degree by
forces due to dry friction, fluid friction, or internal
friction.
DAMPEDFREEVIBRATIONS
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19 - 61
Characteristic equation,
m
k
m
c
m
c
kcm
22
220 lll
nc mc 2 critical damping coefficient
Heavy damping: c > cc
tteCeCx 21 21ll - negative roots
- nonvibratory motion
Critical damping: c = cc
tnetCCx 21 - double roots- nonvibratory motion
Light damping: c < cc tCtCex dd
tmc cossin 212
2
1c
ndc
c damped frequency
DAMPEDFORCEDVIBRATIONS
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19 - 62
2
222
1
2tan
21
1
nf
nfc
nfcnf
m
m
m
cc
cc
xkP
x
magnification
factor
phase difference between forcing and steady
state response
tPkxxcxm fm sin particulararycomplement xxx
ELECTRICALANALOGUES
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19 - 63
Consider an electrical circuit consisting of an inductor,
resistor and capacitor with a source of alternating
voltage0sin
C
qRi
dt
diLtE fm
Oscillations of the electrical system are analogous to
damped forced vibrations of a mechanical system.
tEqC
qRqL fm sin1
ELECTRICAL ANALOGUES
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ELECTRICALANALOGUES
19 - 64
The analogy between electrical and mechanical
systems also applies to transient as well as steady-
state oscillations.
With a charge q = q0on the capacitor, closing the
switch is analogous to releasing the mass of the
mechanical system with no initial velocity atx = x0.
If the circuit includes a battery with constant voltage
E, closing the switch is analogous to suddenly
applying a force of constant magnitude Pto the
mass of the mechanical system.
ELECTRICAL ANALOGUES
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ELECTRICALANALOGUES
19 - 65
The electrical system analogy provides a means of
experimentally determining the characteristics of a given
mechanical system.
For the mechanical system,
0212112121111 xxkxkxxcxcxm
tPxxkxxcxm
fm sin
12212222
For the electrical system,
02
21
1
121111
C
qq
C
qqqRqL
tEC
qqqqRqL fm sin
21212222
The governing equations are equivalent. The characteristics
of the vibrations of the mechanical system may be inferred
from the oscillations of the electrical system.
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UNIT IV : FORCED VIBRATION
Response to periodic forcing - Harmonic Forcing -
Forcing caused by unbalance - Support motion
Force transmissibility and amplitude transmissibility
- Vibration isolation.
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DAMPING
a process whereby energy is taken from thevibrating system and is being absorbed by thesurroundings.
Examples of damping forces:
internal forces of a spring, viscous force in a fluid,
electromagnetic damping in galvanometers,
shock absorber in a car.
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DAMPEDVIBRATION(1)
The oscillating system is opposed by dissipativeforces.
The system does positive work on the surroundings.
Examples:
a mass oscillates under water
oscillation of a metal plate in the magnetic field
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DAMPEDVIBRATION(2)
Total energy of the oscillator decreases with time
The rate of loss of energy depends on theinstantaneous velocity
Resistive force instantaneous velocity i.e. F = -bv where b = damping coefficient Frequency of damped vibration < Frequency of
undamped vibration
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TYPESOFDAMPEDOSCILLATIONS(1)
Slight damping (underdamping)
Characteristics:
- oscillations with reducing amplitudes - amplitude decays exponentially with time
- period is slightly longer
- Figure
-
constanta.......4
3
3
2
2
1 a
a
a
a
a
a
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TYPESOFDAMPEDOSCILLATIONS(2)
Critical damping
No real oscillation
Time taken for the displacement to become effective
zero is a minimum
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TYPESOFDAMPEDOSCILLATIONS(3)
Heavy damping (Overdamping)
Resistive forces exceed those of
critical damping
The system returns very slowly tothe equilibrium position
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EXAMPLE: MOVINGCOILGALVANOMETER
the deflection of the pointer is critically damped
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EXAMPLE: MOVINGCOILGALVANOMETER
Damping is due to inducedcurrents flowing in themetal frame
The opposing couple
setting up causes the coilto come to rest quickly
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FORCEDOSCILLATION
The system is made to oscillate by periodic impulses
from an external driving agent
Experimental setup:
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CHARACTERISTICSOFFORCEDOSCILLATION
Same frequency as the driver system
Constant amplitude
Transient oscillations at the beginning which eventually
settle down to vibrate with a constant amplitude (steadystate)
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CHARACTERISTICSOFFORCEDOSCILLATION
In steady state, the system vibrates at the frequency of
the driving force
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ENERGY
Amplitude of vibration is fixed for a specific drivingfrequency
Driving force does work on the system at the same
rate as the system loses energy by doing work
against dissipative forces Power of the driver is controlled by damping
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AMPLITUDE
Amplitude of vibration depends on
the relative values of the natural
frequency of free oscillation
the frequency of the driving force
the extent to which the system is
damped
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EFFECTSOFDAMPING
Driving frequency for maximum amplitude becomesslightly less than the natural frequency
Reduces the response of the forced system
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PHASE(1)
The forced vibration takes on the frequency of thedriving force with its phase lagging behind
If F = F0cos t, then
x = A cos (t - )
where is the phase lag of x behind F
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PHASE(2)
Figure 1. As f 0, 0
2. As f ,
3. As f f0, /2
Explanation When x = 0, it has no tendency to move. maximum
force should be applied to the oscillator
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PHASE(3)
When oscillator moves away from the centre, the drivingforce should be reduced gradually so that the oscillator
can decelerate under its own restoring force
At the maximum displacement, the driving force
becomes zero so that the oscillator is not pushed any
further
Thereafter, F reverses in direction so that the oscillator
is pushed back to the centre
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PHASE(4)
On reaching the centre, F is amaximum in the opposite direction
Hence, if F is applied 1/4 cycle
earlier than x, energy is supplied tothe oscillator at the correct moment.
The oscillator then responds with
maximum amplitude.
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FORCEDVIBRATION
Adjust the position of the load on the drivingpendulum so that it oscillates exactly at a frequency
of 1 Hz
Couple the oscillator to the driving pendulum by the
given elastic cord Set the driving pendulum going and note the
response of the blade
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FORCEDVIBRATION
In steady state, measure the amplitude of forcedvibration
Measure the time taken for the blade to perform 10
free oscillations
Adjust the position of the tuning mass to change thenatural frequency of free vibration and repeat the
experiment
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FORCEDVIBRATION
Plot a graph of the amplitude of vibration at differentnatural frequencies of the oscillator
Change the magnitude of damping by rotating the
card through different angles
Plot a series of resonance curves
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RESONANCE(1)
Resonance occurs when an oscillator is acted upon by asecond driving oscillator whose frequency equalsthenatural frequency of the system
The amplitude of reaches a maximum
The energy of the system becomes a maximum The phase of the displacement of the driver leadsthat of
the oscillator by 90
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RESONANCE(2)
Examples Mechanics:
Oscillations of a childs swing
Destruction of the Tacoma Bridge
Sound: An opera singer shatters a wine glass
Resonance tube
Kundts tube
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RESONANCE
Electricity Radio tuning
Light
Maximum absorption of infrared waves by a NaCl crystal
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RESONANTSYSTEM
There is only one value of the driving frequency forresonance, e.g. spring-mass system
There are several driving frequencies which give
resonance, e.g. resonance tube
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RESONANCE: UNDESIRABLE
The body of an aircraft should not resonate with thepropeller
The springs supporting the body of a car should not
resonate with the engine
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DEMONSTRATIONOFRESONANCE
Resonance tube Place a vibrating tuning fork above the mouth of the
measuring cylinder
Vary the length of the air column by pouring water into
the cylinder until a loud sound is heard The resonant frequency of the air column is then equal
to the frequency of the tuning fork
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DEMONSTRATIONOFRESONANCE
Sonometer Press the stem of a vibrating tuning fork against the
bridge of a sonometer wire
Adjust the length of the wire until a strong vibration is
set up in it The vibration is great enough to throw off paper riders
mounted along its length
Osc i l lat ion o f a metal plate in the
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Osc i l lat ion o f a metal plate in the
magnetic f ield
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SLIGHTDAMPING
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CRITICALDAMPING
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HEAVYDAMPING
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AMPLITUDE
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PHASE
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BARTONSPENDULUM
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DAMPEDVIBRATION
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RESONANCECURVES
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RESONANCETUBE
A glass tube has a
variable water level
and a speaker at itsupper end
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UNIT V : GOVERNORS AND GYROSCOPES
Governors - Types - Centrifugal governors - Gravity
controlled and spring controlled centrifugal
governorsCharacteristics - Effect of friction -
Controlling Force .Gyroscopes - Gyroscopic forces and Torques -
Gyroscopic stabilization - Gyroscopic effects in
Automobiles, ships and airplanes
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GOVERNORS
Engine Speed controlThis presentation is from Virginia Tech and has not been edited by
Georgia Curriculum Office.
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GOVERNORS
Governors serve three basic purposes: Maintain a speed selected by the operator which is
within the range of the governor.
Prevent over-speed which may cause engine
damage. Limit both high and low speeds.
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GOVERNORS
Generally governors are used to maintain a fixedspeed not readily adjustable by the operator or to
maintain a speed selected by means of a throttle
control lever.
In either case, the governor protects against
overspeeding.
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HOWDOESITWORK?
If the load is removed on an operating engine, thegovernor immediately closes the throttle.
If the engine load is increased, the throttle will be
opened to prevent engine speed form being
reduced.
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EXAMPLE
The governor on your
lawnmower maintains
the selected engine
speed even when you
mow through a clump
of high grass or when
you mow over no grass
at all.
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PNEUMATICGOVERNORS
Sometimes called air-
vane governors, they
are operated by the
stream of air flow
created by the cooling
fins of the flywheel.
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AIR-VANEGOVERNOR
When the engine experiences sudden increases inload, the flywheel slows causing the governor to
open the throttle to maintain the desired speed.
The same is true when the engine experiences a
decrease in load. The governor compensates andcloses the throttle to prevent overspeeding.
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CENTRIFUGALGOVERNOR
Sometimes referred to
as a mechanical
governor, it uses
pivoted flyweights that
are attached to a
revolving shaft or gear
driven by the engine.
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MECHANICALGOVERNOR
With this system, governor rpm is always directly
proportional to engine rpm.
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MECHANICALGOVERNOR
If the engine is subjected to a sudden load thatreduces rpm, the reduction in speed lessens
centrifugal force on the flyweights.
The weights move inward and lower the spool and
governor lever, thus opening the throttle to maintainengine speed.
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VACUUMGOVERNORS
Located between the carburetor and the intake
manifold.
It senses changes in intake manifold pressure
(vacuum).
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VACUUMGOVERNORS
As engine speed increases or decreases the
governor closes or opens the throttle respectively
to control engine speed.
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HUNTING
Hunting is a condition whereby the engine speedfluctuate or is erratic usually when first started.
The engine speeds up and slows down over and
over as the governor tries to regulate the engine
speed. This is usually caused by an improperly adjusted
carburetor.
S
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STABILITY
Stability is the ability to maintain a desired enginespeed without fluctuating.
Instability results in hunting or oscillating due to
over correction.
Excessive stability results in a dead-beat governoror one that does not correct sufficiently for load
changes.
S
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SENSITIVITY
Sensitivity is the percent of speed change requiredto produce a corrective movement of the fuel
control mechanism.
High governor sensitivity will help keep the engineoperating at a constant speed.
S
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SUMMARY
Small engine governors are used to:
Maintain selected engine speed.
Prevent over-speeding.
Limit high and low speeds.
S
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SUMMARY
Governors are usually of the following types: Air-vane (pneumatic)
Mechanical (centrifugal)
Vacuum
S
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SUMMARY
The governor must have stability and sensitivity inorder to regulate speeds properly. This will prevent
hunting or erratic engine speed changes depending
upon load changes.
Gyroscope
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A gyroscope consists of a rotor mounted in the inner gimbal. The inner
gimbal is mounted in the outer gimbal which itself is mounted on afixed frame as shown in Fig. When the rotor spins about X-axis with
angular velocity rad/s and the inner gimbal precesses (rotates)
about Y-axis, the spatial mechanism is forced to turn about Z-axis
other than its own axis of rotation, and the gyroscopic effect is thus
setup. The resistance to this motion is called gyroscopic effect.
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GYROSCOPIC COUPLE
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Consider a rotary body of mass m having radius of gyration k mounted on the
shaft supported at two bearings. Let the rotor spins (rotates) about X-axis with
constant angular velocity rad/s. The X-axis is, therefore, called spin axis, Y-
axis, precession axis and Z-axis, the couple or torque axis .
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GYROSCOPIC EFFECT ON SHIP
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THANK YOU