Ppt Mech 5sem DOM

8/11/2019 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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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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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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Critical damping

No real oscillation

Time taken for the displacement to become effective

zero is a minimum


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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

Ppt Mech 5sem DOM

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