d ynamics of m achinery u5mea19 prepared by mr.shaik shabbeer mr.vennishmuthu.v assistant professor,...

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

qq

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

THANK YOU