DYNAMICS OF MACHINERY U5MEA19 Prepared by Mr.Shaik Shabbeer Mr.Vennishmuthu.V Assistant Professor, Mechanical Department VelTech Dr.RR & Dr.SR Technical University
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DYNAMICS OF MACHINERYU5MEA19
Prepared by Mr.Shaik Shabbeer Mr.Vennishmuthu.VAssistant Professor, Mechanical DepartmentVelTech Dr.RR & Dr.SR Technical University
UNIT I : FORCE ANALYSIS Rigid Body dynamics in general plane motion –
Equations of motion - Dynamic force analysis - Inertia force and Inertia torque – D’Alemberts principle - The principle of superposition - Dynamic Analysis in Reciprocating Engines – Gas Forces - Equivalent masses - Bearing loads - Crank shaft Torque - Turning moment diagrams - Fly wheels –Engine shaking Forces - Cam dynamics - Unbalance, Spring, Surge and Windup.
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.
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.
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.
D’Alembert’s principle. D’Alembert’s 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.
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.
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 FQ perpendicular 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).
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
bars acting 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
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 at A. 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 to AC intersecting AG produced in
B. The point B now fixes the position of the second mass m2. The triangles ACG and BCG are similar.
Therefore,
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 as the ordinate (Y-axis) and crank angle as abscissa (X axis).
UNIT II : BALANCING Static and dynamic balancing - Balancing of rotating masses –Balancing reciprocating masses- Balancing a single cylinder Engine - Balancing Multi-cylinder Engines, Balancing V-engines, - Partial balancing in locomotive Engines-Balancing machines.
STATIC AND DYNAMIC BALANCING
When man invented the wheel, he very quickly learnt that if it wasn’t completely round and if it didn’t rotate evenly about it’s central axis, then he had a problem!What the problem he had?The wheel would vibrate causing damage to itself and it’s support 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.
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 rotor’s rotating centreline.
BALANCING:
Balancing is the technique of correcting or eliminating unwanted inertia forces or moments in rotating or reciprocating 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 stationery during a complete revolution of the crank shaft and2. That the couples involved in acceleration of the different moving parts balance each other.
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 of different moving parts is equal to zero.iii) The conditions of dynamic balance are met, the conditions of static balance are also met.
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 direction changes with the rotation of the mass.
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 pumps etc.
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 of1. 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 plane4. Balancing of several masses rotating in different planes
BALANCING OF A SINGLE ROTATING MASS BY A SINGLEMASS ROTATING IN THE SAME PLANE
Consider a disturbing mass m1 which is attached to a shaft rotating at rad/s.
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 the centrifugal forces due to the two masses are equal and opposite.
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 of the 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 centre of 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.
Balancing Multi-cylinder Engines, Balancing V-engines
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 270˚ from mass A. Find the magnitude and position of the balancing 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 corresponding radii 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 and magnitude of the balancing mass, if the radius of rotation is 60 cm.
UNIT III : FREE VIBRATIONBasic features of vibratory systems - idealized models - Basic elements and lumping of parameters - Degrees of freedom - Single degree of freedom - Free vibration - Equations of motion - natural frequency - Types of Damping - Damped vibration critical speeds of simple shaft - Torsional systems; Natural frequency of two and three rotor systems
INTRODUCTION
19 - 36
• Mechanical vibration is 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 stresses and energy losses.
• Time interval required for a system to complete a full cycle of the motion is the period of the vibration.
• Number of cycles per unit time defines the frequency of the vibrations.
• Maximum displacement of the system from the equilibrium position is the amplitude of 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 the system, 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 damped to some degree.
FREE VIBRATIONS OF PARTICLES. SIMPLE HARMONIC MOTION
19 - 37
• If a particle is displaced through a distance xm from 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 two particular solutions,
tCtC
tm
kCt
m
kCx
nn cossin
cossin
21
21
• x is a periodic function and n is the natural circular frequency of the motion.
• C1 and C2 are determined by the initial conditions:
tCtCx nn cossin 21 02 xC
nvC 01 tCtCxv nnnn sincos 21
FREE VIBRATIONS OF PARTICLES. SIMPLE HARMONIC MOTION
19 - 38
txx nm sin
n
n 2
period
2
1 n
nnf natural frequency
20
20 xvx nm amplitude
nxv 00
1tan phase angle
• Displacement is equivalent to the x component of the sum of two vectorswhich rotate with constant angular velocity
21 CC
.n
02
01
xC
vC
n
FREE VIBRATIONS OF PARTICLES. SIMPLE HARMONIC MOTION
19 - 39
txx nm sin
• Velocity-time and acceleration-time curves can be represented by sine curves of the same period as the displacement-time curve but different phase angles.
2sin
cos
tx
tx
xv
nnm
nnm
tx
tx
xa
nnm
nnm
sin
sin2
2
SIMPLE PENDULUM (APPROXIMATE SOLUTION)
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 towards the equilibrium position.
for small angles,
g
l
tl
g
nn
nm
22
sin
0
:tt maF
• Consider tangential components of acceleration and force for a simple pendulum,
0sin
sin
l
g
mlW
SIMPLE PENDULUM (EXACT SOLUTION)
19 - 41
0sin l
gAn exact solution for
leads to
2
022 sin2sin1
4
m
nd
g
l
which requires numerical solution.
g
lKn
2
2
SAMPLE PROBLEM
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, determine the spring constant for a single equivalent spring.
• Apply the approximate relations for the harmonic motion of a spring-mass system.
SAMPLE PROBLEM
19 - 43
mkN6mkN4 21 kk• Springs in parallel:
- determine the spring constant for equivalent spring
mN10mkN10 4
21
21
kkP
k
kkP
- apply the approximate relations for the harmonic motion of a spring-mass system
nn
n m
k
2
srad14.14kg20
N/m104
s 444.0n
srad 4.141m 040.0 nmm xv
sm566.0mv
2sm00.8ma 22
srad 4.141m 040.0
nmm axa
SAMPLE PROBLEM
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
n m
k
2
srad93.6kg20
400N/m2
s 907.0n
srad .936m 040.0 nmm xv
sm277.0mv
2sm920.1ma 22
srad .936m 040.0
nmm axa
mN10mkN10 4
21
21
kkP
k
kkP
FREE VIBRATIONS OF RIGID BODIES
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
05
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
SAMPLE PROBLEM
19 - 46
k
A cylinder of weight W is 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 of motion.
• Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration.
SAMPLE PROBLEM
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.
03
8
22 221
21
m
k
mrrrmrkrWWr
m
kn 3
8 k
m
nn 8
32
2
m
kf nn 3
8
2
1
2
SAMPLE PROBLEM
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 spring constant, b) the centroidal moment of inertia of the gear, and c) the maximum angular velocity of the gear if rotated through 90o and 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 the torsional 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
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 of motion 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.
22
221 sftlb 138.0
12
8
2.32
20
2
1
mrI
K
138.0213.1 radftlb27.4 K
SAMPLE PROBLEM
19 - 50
s13.1
lb20
n
W
s93.1n
radftlb27.4 K
K
I
I
K
nnn
2
2
• With natural frequency and spring constant known, calculate the moment of inertia for the gear.
27.4293.1
I 2sftlb 403.0 I
• Apply the relations for simple harmonic motion to calculate the maximum gear velocity.
nmmnnmnm tt sinsin
rad 571.190 m
s 93.1
2rad 571.1
2 n
mm
srad11.5m
PRINCIPLE OF CONSERVATION OF ENERGY
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,
01 T 2
21
21 2sin2cos1
m
m
Wb
WbWbV
2235
21
2232
212
21
2212
21
2
m
mm
mm
mb
mbbm
IvmT
02 V
00 22235
212
21
2211
nmm mbWb
VTVT
bgn 53
SAMPLE PROBLEM
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
19 - 53
01 T 2
cos12
1
mrRW
rRWWhV
22
43
22
221
212
21
2212
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
SAMPLE PROBLEM
19 - 54
2211 VTVT
02
0 2243
2 m
m rRmrRW
2243
2
2 mnmm rRmrRmg
rR
gn
3
22g
rR
nn
2
32
2
01 T 221 mrRWV
2243
2 mrRmT 02 V
• Solve the energy equation for the natural frequency of the oscillations.
FORCED VIBRATIONS
19 - 55
:maF
xmxkWtP stfm sin
tPkxxm fm sin
xmtxkW fmst sin
tkkxxm fm sin
Forced vibrations - Occur when a system is subjected to a periodic force or a periodic displacement of a support.
f forced frequency
FORCED VIBRATIONS
19 - 56
txtCtC
xxx
fmnn
particulararycomplement
sincossin 21
222 11 nf
m
nf
m
f
mm
kP
mk
Px
tkkxxm fm sin
tPkxxm fm sin
At f = n, forcing input is in resonance with the system.
tPtkxtxm fmfmfmf sinsinsin2
Substituting particular solution into governing equation,
SAMPLE PROBLEM
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
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)
rpm 549rad/s 5.5787.10
000,36
m
kn
Resonance speed = 549 rpm
SAMPLE PROBLEM
19 - 59
W = 350 lb
k = 4(350 lb/in) rad/s 5.57n
• 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
oz 16
lb 1oz 1
rad/s 125.7 rpm 1200
22
m
f
lb 33.157.125001941.0 2126
2
mrmaP nm
in 001352.0
5.577.1251
300033.15
1 22
nf
mm
kPx
xm = 0.001352 in. (out of phase)
DAMPED FREE VIBRATIONS
19 - 60
• With viscous damping due to fluid friction,
:maF 0
kxxcxm
xmxcxkW st
• Substituting x = et and dividing through by et yields the characteristic equation,
m
k
m
c
m
ckcm
22
220
• Define the critical damping coefficient such that
ncc m
m
kmc
m
k
m
c 2202
2
• All vibrations are damped to some degree by forces due to dry friction, fluid friction, or internal friction.
DAMPED FREE VIBRATIONS
19 - 61
• Characteristic equation,
m
k
m
c
m
ckcm
22
220
nc mc 2 critical damping coefficient
• Heavy damping: c > cc
tt eCeCx 2121 - negative roots
- nonvibratory motion
• Critical damping: c = cc
tnetCCx 21 - double roots - nonvibratory motion
• Light damping: c < cc
tCtCex ddtmc cossin 21
2
2
1c
nd c
c damped frequency
DAMPED FORCED VIBRATIONS
19 - 62
2
222
1
2tan
21
1
nf
nfc
nfcnf
m
m
m
cc
cc
x
kP
x
magnification
factor
phase difference between forcing and steady state response
tPkxxcxm fm sin particulararycomplement xxx
ELECTRICAL ANALOGUES
19 - 63
• Consider an electrical circuit consisting of an inductor, resistor and capacitor with a source of alternating voltage
0sin 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
19 - 64
• The analogy between electrical and mechanical systems also applies to transient as well as steady-state oscillations.
• With a charge q = q0 on the capacitor, closing the switch is analogous to releasing the mass of the mechanical system with no initial velocity at x = x0.
• If the circuit includes a battery with constant voltage E, closing the switch is analogous to suddenly applying a force of constant magnitude P to the mass of the mechanical system.
ELECTRICAL ANALOGUES
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 sin12212222
• For the electrical system,
02
21
1
121111
C
C
qqqRqL
tEC
qqqqRqL fm sin
2
1212222
• The governing equations are equivalent. The characteristics of the vibrations of the mechanical system may be inferred from the oscillations of the electrical system.
UNIT IV : FORCED VIBRATION
Response to periodic forcing - Harmonic Forcing - Forcing caused by unbalance - Support motion – Force transmissibility and amplitude transmissibility - Vibration isolation.
DAMPING a process whereby energy is taken from
the vibrating system and is being absorbed by the surroundings.
Examples of damping forces: internal forces of a spring, viscous force in a fluid, electromagnetic damping in galvanometers, shock absorber in a car.
DAMPED VIBRATION (1)
The oscillating system is opposed by dissipative forces.
The system does positive work on the surroundings.
Examples: a mass oscillates under water oscillation of a metal plate in the magnetic field
DAMPED VIBRATION (2)
Total energy of the oscillator decreases with time
The rate of loss of energy depends on the instantaneous velocity
Resistive force instantaneous velocity i.e. F = -bv where b = damping
coefficient Frequency of damped vibration < Frequency of
undamped vibration
TYPES OF DAMPED OSCILLATIONS (1)
Slight damping (underdamping)
Characteristics: - oscillations with reducing amplitudes - amplitude decays exponentially with
time - period is slightly longer - Figure
- constant a.......
4
3
3
2
2
1 a
a
a
a
a
a
TYPES OF DAMPED OSCILLATIONS (2)
Critical damping No real oscillation Time taken for the displacement to become
effective zero is a minimum
TYPES OF DAMPED OSCILLATIONS (3)
Heavy damping (Overdamping)
Resistive forces exceed those of critical damping
The system returns very slowly to the equilibrium position
EXAMPLE: MOVING COIL GALVANOMETER
the deflection of the pointer is critically damped
EXAMPLE: MOVING COIL GALVANOMETER
Damping is due to induced currents flowing in the metal frame
The opposing couple setting up causes the coil to come to rest quickly
FORCED OSCILLATION
The system is made to oscillate by periodic impulses from an external driving agent
Experimental setup:
CHARACTERISTICS OF FORCED OSCILLATION
Same frequency as the driver system Constant amplitude Transient oscillations at the beginning which
eventually settle down to vibrate with a constant amplitude (steady state)
CHARACTERISTICS OF FORCED OSCILLATION
In steady state, the system vibrates at the frequency of the driving force
ENERGY
Amplitude of vibration is fixed for a specific driving frequency
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
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
EFFECTS OF DAMPING
Driving frequency for maximum amplitude becomes slightly less than the natural frequency
Reduces the response of the forced system
PHASE (1)
The forced vibration takes on the frequency of the driving force with its phase lagging behind
If F = F0 cos t, then x = A cos (t - ) where is the phase lag of x behind F
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
PHASE (3)
When oscillator moves away from the centre, the driving force 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
PHASE (4)
On reaching the centre, F is a maximum in the opposite direction
Hence, if F is applied 1/4 cycle earlier than x, energy is supplied to the oscillator at the ‘correct’ moment. The oscillator then responds with maximum amplitude.
FORCED VIBRATION
Adjust the position of the load on the driving pendulum 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
FORCED VIBRATION
In steady state, measure the amplitude of forced vibration
Measure the time taken for the blade to perform 10 free oscillations
Adjust the position of the tuning mass to change the natural frequency of free vibration and repeat the experiment
FORCED VIBRATION
Plot a graph of the amplitude of vibration at different natural frequencies of the oscillator
Change the magnitude of damping by rotating the card through different angles
Plot a series of resonance curves
RESONANCE (1)
Resonance occurs when an oscillator is acted upon by a second driving oscillator whose frequency equals the natural 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 leads
that of the oscillator by 90
RESONANCE (2)
Examples Mechanics:
Oscillations of a child’s swing Destruction of the Tacoma Bridge
Sound: An opera singer shatters a wine glass Resonance tube Kundt’s tube
RESONANCE
Electricity Radio tuning
Light Maximum absorption of infrared waves by a NaCl
crystal
RESONANT SYSTEM
There is only one value of the driving frequency for resonance, e.g. spring-mass system
There are several driving frequencies which give resonance, e.g. resonance tube
RESONANCE: UNDESIRABLE
The body of an aircraft should not resonate with the propeller
The springs supporting the body of a car should not resonate with the engine
DEMONSTRATION OF RESONANCE
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
DEMONSTRATION OF RESONANCE
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
Oscillation of a metal plate in the magnetic field
SLIGHT DAMPING
CRITICAL DAMPING
HEAVY DAMPING
AMPLITUDE
PHASE
BARTON’S PENDULUM
DAMPED VIBRATION
RESONANCE CURVES
RESONANCE TUBE
A glass tube has a variable water level and a speaker at its upper end
UNIT V : GOVERNORS AND GYROSCOPES
Governors - Types - Centrifugal governors - Gravity controlled and spring controlled centrifugal governors –Characteristics - Effect of friction - Controlling Force .Gyroscopes - Gyroscopic forces and Torques - Gyroscopic stabilization - Gyroscopic effects in Automobiles, ships and airplanes
GOVERNORS
Engine Speed control This presentation is from Virginia Tech and has not been edited
by Georgia Curriculum Office.
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.
GOVERNORS
Generally governors are used to maintain a fixed speed 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.
HOW DOES IT WORK?
If the load is removed on an operating engine, the governor immediately closes the throttle.
If the engine load is increased, the throttle will be opened to prevent engine speed form being reduced.
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.
PNEUMATIC GOVERNORS
Sometimes called air-vane governors, they are operated by the stream of air flow created by the cooling fins of the flywheel.
AIR-VANE GOVERNOR
When the engine experiences sudden increases in load, 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 and closes the throttle to prevent overspeeding.
CENTRIFUGAL GOVERNOR
Sometimes referred to as a mechanical governor, it uses pivoted flyweights that are attached to a revolving shaft or gear driven by the engine.
MECHANICAL GOVERNOR
With this system, governor rpm is always directly proportional to engine rpm.
MECHANICAL GOVERNOR
If the engine is subjected to a sudden load that reduces 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 maintain engine speed.
VACUUM GOVERNORS
Located between the carburetor and the intake manifold.
It senses changes in intake manifold pressure (vacuum).
VACUUM GOVERNORS
As engine speed increases or decreases the governor closes or opens the throttle respectively to control engine speed.
HUNTING
Hunting is a condition whereby the engine speed fluctuate 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.
STABILITY
Stability is the ability to maintain a desired engine speed without fluctuating.
Instability results in hunting or oscillating due to over correction.
Excessive stability results in a dead-beat governor or one that does not correct sufficiently for load changes.
SENSITIVITY
Sensitivity is the percent of speed change required to produce a corrective movement of the fuel control mechanism.
High governor sensitivity will help keep the engine operating at a constant speed.
SUMMARY
Small engine governors are used to:
Maintain selected engine speed. Prevent over-speeding. Limit high and low speeds.
SUMMARY
Governors are usually of the following types: Air-vane (pneumatic) Mechanical (centrifugal) Vacuum
SUMMARY
The governor must have stability and sensitivity in order to regulate speeds properly. This will prevent hunting or erratic engine speed changes depending upon load changes.
Gyroscope
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 a fixed frame as shown in Fig. When the rotor spins about X-axis with angular velocity ω 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.
GYROSCOPIC COUPLE
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 .
GYROSCOPIC EFFECT ON SHIP
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