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DRK INSTITUTE OF SCIENCE AND TECHNOLOGY(Approved by AICTE, Permitted by Govt of A.P and Affiliated to JNTU)

Bowrampet (v), via Air force Academy, Hyderabad -500043

CERTIFICATE

This is to certify that the project work entitledDYNAMIC ANALYSIS FOR DESIGN

OPTIMIZATION OF A GAS TURBINE ROTOR BLADE is submitted in partialfulfillment of the requirements for the award of degree ofBACHELOR OFTECHNOLOGY in MECHANICAL ENGINEERING is a record of bonafied work

carried out by

M.V.RAMA KRISHNA

YG.PRAVEEN

This project is submitted in partial fulfillment of the requirement for the award of degree

Bachelor of Technology in Mechanical Engineering from Jawaharlal Nehru

Technological University.

Internal Guide He

Mr.CH.V.K.S.N.S.MOORTHY Dr.L.BHASKARA RAO

Associate Professor, Professor & Head of Dept.

Department of Mechanical Engineering, Dept. of Mechanical Engg,

DRK Institute of Science & Technology. DRK Institute of Science & Tech.

External Examiner:

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ACKNOWLEDGEMENT

I would like to take this opportunity and express my heartfelt thanks to all those who

helped me in the course of this project work.

I am very much grateful to BHARAT HEAVY ELECTRICALS LIMITED for

providing with real data regarding the functioning of organization and I take this opportunity

to express my heartfelt gratitude to, Mr.UDAY KUMAR, AGM (TC & GT), B.H.E.L, for

having permitted me to undertake the project in the organization and encouragement in

completing this project successfully.

I am most thankful to,Mr.PANKAJ KUMAR MISHRA, Design Engineer.

(TC & GT), B.H.E.L, for their constant support and providing necessary techn

information regarding the project work.

I express my deep sense of gratitude and heartiest thanks to my guide

Mr.CH.V.K.S.N.MOORTHY, associate professor of Mechanical Engineering , DRK

Institute of Science and Technology , Hyderabad for his esteem guidance , meticulous

attention and constructive suggestions in completing this project work in a systematic way.

Its a great pleasure to express my gratitude toMr.L.BHASKAR RAO, HOD

Mechanical Engineering , DRK Institute of Science and Technology , Hyderabad for his

timely guidance to carry out the project work.

I offer my special thanks to Mr.K.SUBBARAO, Principal ,DRK Institute of Science

and Technology ,Hyderabad for his valuable support and rendered me during the crucial

stage of my project.

I also extend my sincere thanks to all the staff members of Mechanical Engineering

Department ,DRK Institute of Science and Technology,Hyderabad.

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CONTENTS

TITLE PAGE PAGE

CERTIFICATE ii

ACKNOWLEDGEMENT iv

ABSTRACT v

CONTENTS vi

LIST OF FIGURES viii

LIST OF TABLES x

NOMENCLATURE xi

1. INTRODUCTION 1

2. LITERATURE REVIEW 4

2.1 Vibration analysis of free-standing turbine blade

2.2 Forced vibration of turbine blade 7

2.3 Vibration Analysis of Blade-Disc Assembly 7

2.4 Life assessment of turbine blade 8

2.5 Remarks 9

MODELING & DYNAMICS OF UNIFORM CROSS-

3. 10

SECTION BLADE

3.1 Modeling of turbine blade 10

3.2 Modal analysis 12

3 .3 Aerodynamic excitation forces 15

3.4 Steady stress due to rotation 16

3.5 Critical speeds 19

3.6 Dynamic stress analysis

3.7 Life assessment 23

3.8 Remarks 26

MODELING & DYNAMICS OF BLADE WITH PRE

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4. 27 TWIST AND TAPER

4.1 Modeling of turbine blade 27

4.2 Modal analysis 28

4.3 Critical speeds 31

4.4 Steady stress due to rotation 33

4.5 Dynamic stress analysis 34

4.6 Life assessment 37

4.7 Remarks 37

5. CONCLUSION AND SCOPE FOR FUTURE WORK

REFERENCES

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LIST OF FIGURES

FIGURE DESCRIPTION

1.1 Steps in Blade and Blade-Disc Assembly Design

3.1 Uniform blade

3.2 Blade with taper and pre-twist

3.3 Blade-disc assembly

3.4 Asymmetric cross-section of the blade

3 .5 Meshed geometry of uniform blade

3.6 Flap wise bending mode at 512.54Hz

3.7 Chord wise bending mode at 1773 Hz

3.8 Third mode at3O45.2Hz

3.9 Torsional modeat3l4O.4Hz

3.10 Forcing functions

3.11 Steady stress experienced in the uniform blade due to

Rotation

3.12 Variation of Steady Stress with speed of rotation

3.13 Campbell diagram ofuniform blade

3.14 Dynamic stress experienced in uniform blade

3.15 Stress variation at the root ofthe blade at various speeds

3.16 Stress variation at the root ofthe blade at various speeds

on log- scale

3.17 Fatiguefailuresurface

4.1 Bladewithpre-twistandtaper

4.2 Meshing

4.5 Firstmodeat583.66Hz

4.6 Secondmodeatl632.4Hz

4.7 Thirdmodeat2884.lHz

4.8 Fourthmodeat477l.3Hz

4.9 Campbell diagram of tapered and twisted blade

4.10 Steady stresses experienced in the blade due to rotation

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4.11 Variation of Steady Stress with speed of rotation

4.12 Dynamic stresses experienced in the blade with taper and pre- twist 35

4.13 Stress variation at the root of the blade 35

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LIST OF TABLES

TABLE DESCRIPTION PAGE

3 . 1 Natural frequencies and type of modes shapes of turbine blade

3 .2 Validation of Natural frequencies of the uniform blade

3 .3 Excitation force amplitudes

3 .4 Variation of Steady Stress with speed of rotation 18

3.5 Critical speeds obtained from the Campbell diagram 20

3.6 Stresses developed at root of the uniform blade due to 1st 22

harmonic excitation at different speeds

3.7 Stresses developed at root of the uniform blade due to 2nd 23

harmonic excitation at different speeds

4.1 Natural frequencies of tapered and twisted blade 30

4.2 Comparison of natural frequencies of uniform cross-section and 30

tapered - twisted blade

4.3 Critical speeds obtained from the Campbell diagram 32

4.4 Variation of Steady Stress with speed of rotation 33

4.5 Stresses developed at root of the tapered blade due to 1st 36

harmonic excitation

4.6 Stresses developed at root of the tapered blade due to 2nd 36

harmonic excitation

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NOMENCLATURE

[C] Damping matrixD Nodal diameter

E Modulus of elasticity

Fx, Fr Forcing functions

{F} Centrifugal force vector

f(t) Nozzle Excitation

f Excitation force frequency

k Multiple of running speed

ka Surface factor

kb Size factor

kc Reliability factor

kd Temperature factor

ke Stress concentration factor

kf Miscellaneous effects factor

[K] Stiffness matrix

[M] Mass matrix

m Harmonic number

N Number of stress cycles to failure

nz Number of nozzles

Rf Endurance limit modifying factor

S Turbine speed

Se Endurance limit of the standard rotating beam fatigue

Se specimen made of blade material for io3 cycles

Se Endurance limit of the standard rotating beam fatigue

specimen made of blade material for N cycles

S1 Blade endurance limit for 103 stress cycles

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S2 Blade endurance limit for 106 stress cycles

Su Ultimate tensile strength

[U] Displacement vector

[] Velocity vector

[] Acceleration vector

{i} Eigen vector

Modal damping ratio

Mass density yp Yield strength af Failure value of alternating stress a Failure value of alternating stress e Blade endurance limit corresponding to N stress cycle m Basic mean stress mf Failure value of mean stress Running speed excitation frequency i Eigen value

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ABSTRACT

The Core design effort in engineering section resolves around dynamic analysis of the

system. The continuous tendency to reduce the weight and cost of the structure and toincrease the power to the weight ratio of machines, engine and vehicles has brought to the

force. The need to predict the vibration level in many fields of engineering.

The large amplitude of vibration generated by fast moving dynamic part givesexcessive deformation damage accumulations, fracture loss of stability, fatigue and loss of

life. The performance and reliability of a dynamical structural systems are governed by its

possible failures mode during design life .So, its highly desirable to predict the EIGENvalue, Eigen factor a stress level, the response property for effectively design and stabilized

the mechanical system.

A structural members used in Jet and Rocket propulsions, high speed aircrafts,

submarines,messiles,Rotor system and large scale engineering and civil structure undergoesvarious kinds of dynamic loads during its life period. In many systems, structure not only

experience harmonic excitations, but large amplitude of Random pressure fields also, acts on

the structure which many times produce non-linearity in the system, which produce

complex, dynamical phenomena in the system.In this project work, my objective was focused on various dynamical phenomena and

study of a Rotor & Blade, design of Rotor dynamical system of all rotating machines.Assume significance from the point of Reliability and performance. While

Conventional design rotors have provided good results in this areas, this need is to be

optimized further to provide least cost solutions.

The core design effort in rotor dynamic analysis resolves around dynamic analysis of

rotor like calculation of Imbalance responses, Eigen value, Eigen vector, model analysis,

and stress calculations fatigue life estimations, various dynamical phenomena like rubbing,and bifurcations in the rotor systems discs or a Rotor will be not coincide with its Geometric

Centre. In such situations, a high speed high unbalanced force will be generated which will

be lead to high vibrations, in the whole rotor system. Resulting into ultimate failure of theRotor systems

.

Compared to rotor design, Blade design is very complex. Emerging blade technologyis finding it increasingly essential to correlate blade vibration to blade Fatigue in order to

access the residual life of existing blade and for development of newer designs. This project

was converted Model Analysis, for Vibration calculation of Eigen Value, Eigen Factor,

Stress analysis of turbine blade. Free Vibration analysis is carried out to compute the naturalpreparation and mode cheaper. The resonance condition is predicted using CAMPBELL

DIAGRAM. Dynamic stess analyses are performed using ANSYS SOFTWARE.

In many situations due to tool design there may be rubbing appears between rotatingand stationary part will get excessive heat at contact point. Rubbing also damages machine,

stator, and blade. During rubbing for a moment at contact point high stiffness formed due to

momentary joint between rotator and stationery due to natural frequency of system andhence the resonance behavior is identified.

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3. THEORY

3.1 Gas TurbineA gas turbine is an engine where fuel is continuously burnt with compressed air to produce a

stream of hot, fast moving gas. This gas stream is used to power the compressor that

supplies the air to the engine as well as providing excess energy that may be used to do other

work. The engine consists of three main parts viz., compressor, combustor and turbine.

The compressor usually sits at the front of the engine. There are two main types of

compressor, the centrifugal compressor and the axial compressor. The compressor will draw

in air and compress it before it is fed into the combustion chamber. In both types thecompressor rotates and it is driven by a shaft that passes through the middle of the engine

and is attached to the turbine.

The combustor is where fuel is added to the compressed air and burnt to produce high

velocity exhaust gas. Down the middle of the combustor runs the flame tube. The flame tubehas a series of holes in it to allow in the compressed air. It is inside the flame tube that fuel

is injected and burnt. There will be one or more igniters that project into the flame tube to

start the mixture burning. Air and fuel are continually being added into the combustor once

the engine is running. Combustion will continue without the use of the igniters once theengine has been started. The combustor and flame tube must be very carefully designed to

allow combustion to take place efficiently and reliably. This is especially difficult given the

large amount of fast moving air being supplied by the compressor. The holes in the flame

tube must be carefully sized and positioned. Smaller holes around where the fuel is addedprovide the correct mixture to burn. This is called the primary zone. Holes further down the

flame tube allow in extra air to complete the combustion. This is the secondary zone. A finalset of hole just before the entry to the turbine allow the remainder of the air to mix with the

hot gases to cool them before they hit the turbine. This final zone is known as the dilution

zone. The exhaust gas is fed from the end of the flame tube into the turbine.

The turbine extracts energy from the exhaust gas. The turbine can, like the compressor, becentrifugal or axial. In each type the fast moving exhaust gas is used to spin the turbine.

Since the turbine is attached to the same

shaft as the compressor at the front of the

engine the turbine and compressor willturn together. The turbine may extract just

enough energy to turn the compressor. The

rest of the exhaust gas is left to exit therear of the engine to provide thrust as in a

pure jet engine. Or extra turbine stages

may be used to turn other shafts to powerother machinery such as the rotors of a

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Fi . 3.1: Sim lified as turbine


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helicopter, the propellers of a ship or electrical generators in power stations the end of the

flame tube into the turbine.

Cold air is drawn in from the left into the compressor (blue). The compressed air (light blue)

then goes into the combustor. From the outside of the combustor the air goes through holes

(purple) into the flame tube (yellow). Fuel is injected (green) into the flame tube and ignited.

The igniters are not show here. The hot exhaust gas flows from the end of the flame tube

past the turbine (red) rotating it as it passes. From there the exhaust exits the engine. The

turbine is connected via a shaft (black) to the compressor. Hence as the turbine rotates the

compressor rotates with it drawing in more air to continue the cycle.

Fig.3.2: Gas turbine

Energy is released when airis mixed with fuel and ignited in the combustor. The resulting

gases are directed over the turbine's blades, spinning the turbine and, cyclically, powering

the compressor. Finally, the gases are passed through a nozzle, generating additional thrust

by accelerating the hot exhaust gases by expansion back to atmospheric pressure.

Energy is extracted in the form of shaft power, compressed air and thrust, in any

combination, and used to poweraircraft,trains, ships, electrical generators, and even tanks.

3.1.1 Theory of operation

Gas turbines are described thermodynamically by the Brayton cycle, in which air iscompressed isentropically, combustion occurs at constant pressure, and expansion over the

turbine occurs isentropically back to the starting pressure.

In practice, friction and turbulence cause:

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a) non-isentropic compression for a given overall pressure ratio, the compressor delivery

temperature is higher than ideal.

b) non-isentropic expansion although the turbine temperature drop necessary to drive thecompressor is unaffected, the associated pressure ratio is greater, which decreases the

expansion available to provide useful work.

c) pressure losses in the air intake, combustor and exhaust reduces the expansion availableto provide useful work.

Fig. 3.3 Idealized Brayton Cycle

As with all cyclic heat engines, higher combustion temperature means greaterefficiency.The limiting factor is the ability of the steel, nickel, ceramic, or other materials that make up

the engine to withstand heat and pressure. Considerable engineering goes into keeping the

turbine parts cool. Most turbines also try to recover exhaust heat, which otherwise is wastedenergy. Recuperators are heat exchangers that pass exhaust heat to the compressed air, prior

to combustion. Combined cycle designs pass waste heat to steam turbine systems and

combined heat and power(co-generation) uses waste heat for hot water production.

Since neither the compression nor the expansion can be truly isentropic, losses through the

compressor and the expander represent sources of inescapable working inefficiencies. In

general, increasing the compression ratio is the most direct way to increase the overallpower output of a Brayton system.

3.1.2 Rotor

The compressor portion of the gas turbine rotor is an assembly of wheels, a speed ring, ties

bolts, the compressor rotor blades, and a forward stub shaft .Each wheel has slots broached

around its periphery. The rotor blades and spacers are inserted into these slots and held inaxial position by staking at each end of the slot. The wheels are assembled to each other

with mating rabbets for concentricity control and are held together with tie bolts. Selectivepositioning of the wheels is made during assembly to reduce balance correction.

After assembly, the rotor is dynamically balanced. The forward stub shaft is machined to

provide the thrust collar which carries the forward and aft thrust loads. The stub shaft alsoprovides the journal for the No. 1 bearing, the sealing surface for the No. 1 bearing oil seals

and the compressor low-pressure air seal

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Of the many factors affecting the efficient working of a simple gas turbine, includingunbalanced forces, vibrations are the prominent that lead to development of cyclic stresses

which inturn results in fatigue failure.

Machines in the best of operating condition will have some vibration

small, minor defects. Therefore, each machine will have a level of vibration that may be

regarded as normal or inherent. However, when machinery vibration increases or becomesexcessive, some mechanical trouble is usually the reason. Vibration does not increase or

become excessive for no reason at all. Something causes it - unbalance, misalignment, worn

gears or bearings, looseness, etc.

3.2 VIBRATION

Vibration is considered with the oscillating motions of the bodies and the forces associatedwith them. All bodies possessing mass and elasticity are capable of vibration. The mass is

inherent of the body and the elasticity is due to relative motion of the parts of the body. The

objective of the designer is to control the vibration when it is objectionable. Objectionablevibrations in a machine may cause the loosening of the parts, its malfunctioning or its

eventual failure. The ultimate goal in the study of the vibration is to determine its effect on

the performance and safety of the system under consideration. The performance of manyinstruments depends on the proper control of the vibrational characteristics of the devices.

FIG.3.4 Elements of Vibrating System

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The elements that constitute a vibratory system are shown in figure. They are idealized and

called mass, spring, damper and the excitation force. The first three elements describe the

physical system.

Energy may be stored in the mass and spring, and dissipated in the damper in the form of

heat. Energy enters the system through theapplication of an excitation. The mass may gain

or lose kinetic energy in accordance with the velocity change of the body. The spring

possesses elasticity and is capable of storing the potential energy under deformation. The

damper has neither mass nor elasticity and is capable of dissipating the energy. Viscous

damping in which the damping force is proportional to the velocity is generally assumed in

engineering. There are two general classes of vibrations viz., free vibrations and Forced

vibrations.

Free vibrations takes place when a system oscillates under the action of forces inherent in

the system itself, and when external impressed forces are absent. The system under free

vibration will vibrate at one or more of its natural frequencies, which are properties of the

dynamic system established by its mass and stiffness distribution.

Vibration that takes place under the excitation of the external forces is called forced

vibration. When the excitation is oscillatory, the system is forced to vibrate at the excitation

frequency. If the frequency of excitation coincides with one of the natural frequencies of the

system, a condition of resonance is encountered, and dangerously large oscillations may

result. The failure of the major structures such as bridges, buildings or airplane wings is an

awesome possibility under resonance. Thus, the calculation of the natural frequencies is of

major importance in the study of the vibration.

Vibrating systems are all subject to damping to some degree because friction and other

resistances dissipate energy. If the damping is small, it has very little influence on the

natural frequencies of the system, and hence the calculations for the natural frequencies are

generally made on the basis of no damping. On the other hand, damping is of great

importance in limiting the amplitude of oscillation at resonance.

3.2.1 SINGLE DEGREE FREEDOM (SDOF) SYSTEMS

The number of independent coordinates required to describe the motion of a

system is called degrees of freedom of the system. The system shown in figure has one

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degree of freedom and hence called SDOF system. The SDOF system is the keystone for

more advanced studies in vibrations of multi-degree (MDOF) systems or

structures. The idealized elements in fig-1 form a model of a vibrating system, which in

reality can be quite complex. The spring shown in fig-1 may possess mass and elasticity. In

order to consider it as an idealized spring, either its mass is assumed negligible or an

appropriate portion of its mass is lumped together with the other masses of a system. For

example, a beam has its mass and elasticity inseparably distributed along its length. The

vibrational characteristics of a beam or more generally of an elastic body or a continuous

system are approximated by a finite number of lumped parameters. This method is a

practical approach to the study of some very complicated structures, such as an aircraft or

missile.

3.2.2 EQUATION OF MOTION FOR SDOF

The equation of motion for the SDOF system shown in fig-1 can be formulated using

either energy method or Newtons laws of motion. The equation of motion for the system is

given by-

)(...

tFkxxcxm =++

The equation of motion for undamped free vibrations is-

0

0

2..

..

=+

=+

xx

kxxm

n

Where,

m

kn =2

, is called circular frequency.

The solution of the equation is-

tAtAx nn sincos 21 +=

Where 1A and 2A are arbitrary constants to be evaluated by initial conditions )0(x and

)0(.

x .

Since the components of the solution are harmonic of the same frequency, their sum is also

harmonic and can be written as-

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)sin( += tAx n

Where 22

2

1 AAA += is the amplitude of the motion

And )(tan2

11

AA= is the phase angle.

The equation indicates that once the system is set into motion it will vibrate with simple

harmonic motion and the amplitude A of the motion will not diminish with time. The system

oscillates because it possesses two types of the energy storage elements, namely, the mass

and the spring. The rate of energy interchange between these elements is the natural

frequency nfof the system which is given by

m

kf nn

2

1

2==

The above equation shows that the natural frequency is a property of the system. It is a

function of the values of m and k and is independent of the amplitude of oscillation or the

manner by which the system is set into motion. It can be observed that the amplitude A and

the phase angle are dependent on the initial conditions.

Since the natural frequency of any body subjected to vibration is directly proportional to the

stiffness of the body k, and inversely proportional to the mass possessed by the body, m,

the effect of increasing the stiffness of this body by some means will increase both the

natural frequency and the mass of the body.

With the damping, the equation of the motion is modified as follows-

)sin( += tAex nt

Where is the damping factor and the 21 = nd

The above equation shows that the amplitude decreases exponentially with time.

3.2.3 MULTI-DEGREE FREEDOM (MDOF) SYSTEM:

As n degree of freedom system is described a set of n simultaneous equations

ordinary differential equations of second order. The system has as many natural frequencies

as the degrees of freedom. A mode of vibration is associated with each natural frequency.

Since the equations of motions are coupled, the motions of the masses are the combination

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of the motions of the individual modes. If the equations are uncoupled by the proper choice

of the coordinates, each mode can be examined as an independent one degree of freedom.

The basic steps in solving the equation of motion of MDOF systems are given below.

1. Finding the characteristic equation from the equation of motion

2. Solving the characteristic equation for the natural frequencies

3. Obtaining the modal matrix in order to uncouple the equations of motion

4. Solving the corresponding one-degree of freedom system.

For the free vibration of the un-damped system, the equations of motions are expressed in

matrix form as-

[ ] [ ]{ } { }0..

=+

xKxM

Where, [ ]M= Mass matrix

[ ]K= Stiffness matrix

{ }x = Displacement matrix

By multiplying the above equation by [ ] 1M and denoting [ ] [ ] [ ]= MM1 (unit matrix),

[ ] [ ] [ ]AKM =1 (A system matrix), we obtain-

[ ] [ ][ ] 0..

=+

xAx

The matrix [ ]A is referred as the system matrix, or the dynamic matrix, since this matrixdefines the dynamic properties of the system.

Assuming harmonic motion [ ]xx =

.., where 2= , the above equation becomes-

[ ] [ ]{ }{ } 0= xA

The characteristic equation of a system are called Eigen values, and the natural frequencies

of the system are determined from them by the relationship,2

ii =

By substituting iinto the matrix equation, we can obtain the corresponding mode shape

iX which is called eigenvector. Thus for an n degree of freedom system, there will be n

Eigen values and n eigenvectors.

Damping in moderate amounts has little influence on the natural frequency and may be

neglected in its calculation. The system has then been considered to be conservative & the

principle of conversation of energy offers another approach to the calculation of natural

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frequency. The effect of damping is mainly evident in the diminishing of the vibration

amplitude with time.

3.2.4 Vibration Model

The basic vibration model of a simple oscillatory system consists of a mass, a mass less

spring & a damper. The mass is considered to be lumped & measured in SI system as

kilograms

g

Wm = lb s 2 / m

The spring supporting the mass is assumed to be a negligible mass. Its free deflection

relationship is considered to be linear, following Hooks law

kxF=

Where, k = stiffness in N/m or lb/in

The viscous damping generally represented by a dashpot is described by a force proportional

to the velocity or

.

xCF=

The damping coefficient C is measured in Newton/ m/s or pound/inch/ second

3.2.5 EQUATIONS OF MOTION:

A simple un-damped spring mass system which is assumed to move only along the vertical

direction has one degree of freedom (DOF) because its motion is described by a single

coordinate x .When placed into motion, oscillation will take place at the natural frequency

nf, which is property of the system.

Newtons second law is the first basis for examining the motion of the system. As shown in

figure the deformation of the spring in the static equilibrium position is and the spring

fore x is equal to the gravitational force W acting on mass m

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mgWk == -------------- (3.2.5.1)

By measuring the displacement x from the static equilibrium position, the forces acting on m

are ( ))(( xk + andW. With x chosen to be positive in the down ward direction, all

quantities force, velocity & acceleration are also positive in the downward direction.

We now apply Newtons second law of motion to the mass m,

+== )(..

xkWFxm

And because k =w, we obtain

kxxm =..

------ (3.2.5.2)

It is evident that the choice of the static equilibrium position as reference for x has

eliminated w, the force due to gravity, and the static spring force k fro the equation of

motion, and the resistant force on m is simply the spring force due to the displacement x.

By defining the circular frequency nby the equation

m

kn =2 ------- (3.2.5.3)

Equation (2) can be written as

02..

=+ xx n ------ (3.2.5.4)

And we conclude by comparison with equation ( xx 2..

= ) that the motion is harmonic

Equation (4), a homogeneous second-order linear differential equation, has the following

general solution.

X = A sin tn + B cos tn-------- (3.2.5.5)

Where A and b are the two necessary constants these constants are evaluated from initial

conditions x(0) and X. and equation (5) can be shown to reduce to

n

xx

.

)0(= txt nn cos)0(sin + -------- (3.2.5.6)

The natural period of the oscillation is established from 2=n or

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k

m 2= ---------------- (3.2.5.7)

And the natural frequency is

m

kf

n 2

11

==------------ (3.2.5.8)

These quantities can be expressed in terms of the static expressed in terms of the static

deflection by observing equation (3.2.5.1), mgk= . Thus equation (3.2.5.8) can be

expressed in terms of the static deflection as

gfn

2

1= -------------- (3.2.5.9)

Note that, nfand ndepend only on the mass and stiffness of the system, which are

properties of the system

Although our discussion was in terms of the spring mass system of figure (1) the results are

applicable to all single DOF system, including rotation. The spring can be a beam or

torsional member and the mass can be replaced by a mass moment of inertia.

3.2.6 VISCOUSLY DAMPED FREE VIBRATION:

Viscous damping force is expressed by the equation

.

xCFd= -------- (3.2.6.1)

Where c is a constant of proportionality symbolically, it is designated by a dashpot as shown

in figure.

)(...

tFkxxCxm =++ -------------------- (3.2.6.2)

The solution of this equation has two parts. If F(t) = 0, we have the homogeneous

differential equation where solution corresponds physically to that of freedamped vibration

with F(t) 0, we obtain the particular solution that is due to the excitation irrespective of the

homogeneous solution. We will first examine the homogeneous equation that will give us

some information about the role of damping.

0...

=++ kxxCxm ---------- (3.2.6.3)

21


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The traditional approach is to assume a solution of the form

steX= ---------- (3.2.6.4)

Where s is a constant upon substitution upon substitution into the differential equation we

obtain

0)( 2 =++ stekCsms

Which is satisfied for all values of t when

02 =++m

ks

m

Cs --------(3.2.6.5)

Equation (5) which is known as the characteristic equation has two roots

m

k

m

C

m

Cs

=2

2,1

22

----- (3.2.6.6)

Hence, the general solution is given by the equation

tstsBeAex

21 += ------- (3.2.6.7)

Where A and B are constants to be evaluated from the initial conditions x (0) and x (0)

Equation (3.2.6.6) substituted into equation (3.2.6.7) gives

tm

k

m

Ct

m

k

m

C

tm

c

BeAeex

+=

22

222 (

--------- (3.2.6.8)

The first term e tmC

2 is simply an exponentially decaying function of time. The behavior of

the terms in the parentheses, however, depends on whether the numerical value with in the

radical is positive, zero or negative.

When the damping term2

2

m

Cis less than k/m, the exponent becomes an imaginary

number ( tm

C

m

ki

2

2

Because

22


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tm

C

m

kit

m

C

m

ke

tm

C

m

ki 22

2

2sin

2cos

2

=

The term of equation (3.2.6.8) with in the parentheses is oscillatory. We refer to this case asunder damped.

In the limiting case b/w the oscillatory and non oscillatory motion,m

k

m

C =

2

2and the

radical is zero. The damping corresponding to this case is called critical damping, cC

kmmm

kmC nc 222 === ----------------- (3.2.6.9)

Any damping can then be expressed in terms of the critical damping by a non dimensional

numbers , called the damping ratio

cC

C= --------------------- (3.2.6.10)

And we can also express S1,2 in terms of as follows

n

c

m

C

m

C =

=22

Equation (3.2.6.6) then becomes( ns 122,1 = ----------- (3.2.6.11)

The three cases of damping discussed here now depend on whetheris grater than, less

than, or equal to unity further more the differential equation of motion can now be expressed

in terms ofand nas

( )tFm

xxxn

n

12

2...

=++ --------------- (3.2.6.12)

This form of the equation for single DOF systems will be found to be helpful in identifying

the natural frequency and the damping of the system. We will frequently encounter this

equation in the modal summation for multi DOF systems

23


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Figure shows equation (3.2.6.11) plotted in a complex plane with along the horizontal axis.

If 0= equation (3.2.6.11) reduces to is

n

=

2,1so that the roots on the imaginary axis

correspond to the un-damped case. For 0 1 equation (3.2.6.11) can be written as

22,1 1

= is

n

The roots s1 and s2 are then conjugate complex pts on a circular arc converging at the point

12,1 =n

s

. As increases beyond unity, the roots separate along the horizontal axis and

remains real numbers with this diagram in mind, we are now ready to examine the solution

given by equation (3.2.6.8)

3.3 INTRODUCTION TO FINITE ELEMENT METHOD

The finite element method is a numerical analysis technique for obtaining approximatesolutions to a wide variety of engineering problems. Because of its diversity and flexibility

as an analysis tool, it is receiving much attention in engineering schools and in industries. In

more and more engineering situations today, we find that it is necessary to obtain

approximate numerical solutions to problems rather than exact closed form solution.

It is not possible to obtain analytical mathematical solutions for many engineering problems.An analytical solution is a mathematical expression that gives the values of the desired

unknown quantity at any location in a body, and as a consequence it valid for an infinite

number of locations in the body. For problems involving complex material properties and

boundary conditions, the engineer resorts to numerical methods that provide approximate,but acceptable, solutions.

The finite element method has become a powerful tool for the numerical solution of a wide

range of engineering problems. It has developed simultaneously with the increasing use ofhigh-speed electronic digital computers and with the growing emphasis on numericalmethods for engineering analysis. This method started as a generalization of the structural

idea to some problems of elastic continuum, is well-established numeric

applicable to any continuum problem, stated in terms of differential equations or as anextrinnum problem.

24


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The fundamental areas that have to be learned for working capability of finite element

method includes:

1. Matrix algebra.2. Solid mechanics.

3. Variational methods.

4. Computer skills.

Matrix techniques are definitely the most efficient and systematic way to handle algebra of

finite element method. Basically, matrix algebra provides a scheme by which a large numberof equations can be stored and manipulated. Since vast majority of the literature on the finite

element method treats problems in structural and continuum mechanics, including soil and

rock mechanics, the knowledge of these fields became necessary.

It is useful to consider the finite element procedure basically as variational approach. This

conception has contributed significantly to the convenience in formulating the method and

to its generality.

3.3.1 PROCESS OF FINITE ELEMENT ANALYSIS

25

Physical

ProblemChange of Physical Problem

Mathematical Model

Governed by DifferentEquations, Assumption in

GeometryKinematicsMaterial LawLoading

Boundary Conditionsetc.,

Finite Element

SolutionChoice of

Finite Element

Mesh Density

Solution Parametric

Representation of

Loadin

Assessment of Accuracy

of Finite Element

Solution ofMathematical ModelInterpolation of ResultsDesign improvement Improve MathematicalModel

Define MeshDefine Analysis


26/60

3.3.2 GENERAL DESCRIPTION OF THE METHOD

In brief the basis of the finite element method is the representation of a body as a structure

by an assemblage of sub-divisions called finite elements. The elements are consideredinterconnected at joints, which are called nodes or nodal points. Simple functions are chosen

to approximate the distribution or variation as the actual displacement functions or

displacement models. The unknown functions are the displacements at nodal points.

26


27/60

Hence final solution will yield the approximate displacements at discrete locations in the

body, the nodal points. A displacement model can be expressed in various simple forms,

such as polynomials and trigonometric functions. Since polynomials offer mathematical manipulation, they have been employed commonly in finite

applications.

A variational principle of mechanics, such as the principle of minimum potential energy, is

usually employed to obtain the set of equilibrium equations for each element. The potential

energy of a loaded elastic body or structure is represented by a sum of internal energy storedas a result of deformations and the potential energy of the external loads.

The equilibrium equations for the entire body are then obtained by combining equations for

the individual elements in such away that the continuity of displacements is preserved at

interconnecting nodes.

The theory of the finite element method may be divided into two phases:

1. The first phase consists of the study of the individual element.

2. The second phase is the study of the assemblage of elements representing the entire body.

3.3.4 Interpolation Polynomials

The basic idea of the finite element method is piecewise approximation, that is the solution

is obtained by dividing the region of interest into small regions (finite elements) andapproximating the solution over each sub region by a simple function. Thus a necessary and

important step is that of choosing simple functions for the behavior of the solution with in an

element are called interpolation function or approximating functions or interpolation models.Polynomial type of interpolation function have been most widely used due to the following

reasons:

1. It is easier too formulate and computerize the finite element equations with polynomial

type of interpolation functions. Specifically, it is easier to perform differentiation or

integration with polynomials.2. It is possible to improve the accuracy of the results by increasing the order of the

polynomial, as shown in figure.

Theoretically a polynomial of infinite order corresponds to the exact solution. But inpractice we take polynomials of finite order only as approximation. Infigure

Approximation polynomials of the general forms.

F (x) = a1 + a2x + a3 (x)2 + +an (x)

(n-1) +1(x) n

The greater the number of terms included in the approximation the more closely the exact

solution is represented. In equation numberThe coefficient of polynomial as are known as generalized co-ordinates and n is a degree

of polynomial.

The above equation is for one-dimensional model

for two and three-dimensional finite element the polynomial forms is given below.

27


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Two-dimensional:

F(x,y) = a1+a2x+a3y + a4x2 + a5y

2 + a6xy + . + an+1xn

From the above three equations, for each order of polynomial we can have three equationsfor each one. In most of the practical application the order of the polynomial in the

interpolation function is taken as one, two, or three.

3.3.5 SUMMARY OF ANALYSIS PROCEDURE

The solution of a general continuum by the finite element method always follows an orderly

step by step procedure. The step by step procedure for static structural problem can be stated

as follows:

1. Discretizations of structure (Domain).

The first step in the finite element method is to divide the structure or solution region into

sub-divisions or elements.

2. Selection of a proper interpolation model.

Since the displacement (field variable) solution of a complex structure under any specified

load conditions cannot be predicted exactly. We assume some suitable solution with in anelement to approximate the unknown solution. The assumed solution must be simple from

computation point of view, and it should satisfy certain convergence requirements.

3. Derivation of element stiffness matrices (characteristic matrices) and load vectors.

From assumed displacement model the stiffness matrix [K (e)] and the load vector P (e) ofelement e are to be derived either by using equilibrium conditions or a suitable variation

principle.

4. Assemblage of element equations to obtain the overall equilibrium equations.

Since the structure is composed of several finite element stiffness matrices and load vectors,

they are to be assembled in suitable manner and the overall equilibrium equation have to beformulated as

[K]= P

28


29/60

Where, [K] is called assembled stiffness matrix,

is called the vector of nodal displacement and

P is the vector of nodal forces for the complete structure.

4. IMBALANCE RESPONSE OF ROTOR

29


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Rotor dynamics is the study of rotating machines and has a very important part to play

throughout the modern industrial world. Rotating machinery is used in many applications

such as:

- Turbo-machines.

- Power stations.

- Machine tools.

- Automobiles.

- Household machines.

- Aerospace applications.

- Marine propulsion.

- Medical equipment.

The interactions these machines have in their surroundings is of great importance as if these

machines are not operating at the correct speed ranges, vibration can occur which may

ultimately cause failure. Failure of machines in applications such as aero-engines, turbo-

machines, space vehicles, etc creates enormous repair costs and more importantly may put

human life in danger. This means governments and industries put a great deal of resources

into the study of rotor dynamics to calculate the safe operating ranges before the machine

goes into service and also methods of detecting imminent failure.

Rotor dynamics is a collective term for rotating machines and can be split into the sub-

groups that make it up. These are rotating shafts, bearings, seals, out of balance systems,

instability and condition monitoring.

The material presented in this chapter was selected to introduce and explain the nature of

rotor dynamic phenomena from comparatively simple analytic models as

expected. The phenomena demonstrated by flexible rotors and the techniques employed fortheir analysis is basically similar to other areas of vibration and structural dynamics.

Specifically, one is generally concerned with linear response phenomena (natural frequency

and frequency response calculation) linear instabilities, parametric instabilities and forced

steady state and transient non linear response.

Considering these topics separately, the following sections consider. The

characteristics of flexible rotors due to imbalance and product of inertia disturbances,gravity loading rotor distribution of how rotor forward and backward critical speed are

defined and these relation ships to rotor natural frequencies are examined. The phenomenon

of rotor sub synchronous whirl (linear instability) is introduced using internal rotor dampingas a representative destabilizing mechanism.

30


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The potential for improving stability by adding damping at bearing or by introducing

orthotropic stiffness characteristics at bearing supports is examined. Rotor shaft- stiffnessorthotropic is used as an example of mechanism which has. The potential for parametric

excitation of motor instabilities non symmetric clearance effects at bearing or rubbing due to

a rammed or off centered seal are also shown to have the potential for rotor parametricexcitation. Further these mechanisms are also shown to supports fractional frequency sub

harmonic rotor resonance additional non linear mechanisms examined here are

a) symmetric synchronic now response of rotors in bearing clearances,

b) Inter action of a rotor and stator across a clearance

c) A transient solution of a critical speed transition

A thorough under standing of material presented in this chapter in necessary (and generally

sufficient) to understand the desired design behavior of rotating machinery and to recognize

the distinctive characteristics of undesirable dynamic behavior

------------- (4.1.16)

Where

YX jRRr += ------------------- (4.1.17)

The solution is

JeCar )(= --------------- (4.1.18)

Which shows that synchronous shaft motion is on oscillatory to an observer moving with therotor fixed x, y, z coordinate system. In physical terms, synchronous rotor motion does not

cause alternating stress in the shaft.

4.2 GYROSCOPIC EFFECTS

Machines where rotor is mounted so that any shaft deflection changes the

shaft slope (rotor is not mounted in the center of a symmetrical shaft or is overhung). When

shaft starts rotating it has a conical path until speed reaches the limit at which the centripetal

forces straighten the shaft, figure 4.2.1

31


32/60

an instance of time t, the rotor exhibits an angular velocity, about the OY axis. The

angular momentum vectors are shown in figure 4.2.1(b), for a moment of time and at a

period time a small interval later. This shows the change in angular velocity is , where

is the polar moment of inertia of the rotor.

In figure 2.5(a) the gyroscopic couple applied by the rotor that reacts with the support

structure, g , must act about the axis xO when viewed from O. the net moment about

this axis is:-

..

dgx IMM =

Where xM is the moment applied to the rotor by the shaft

dIis the moment of inertia of the rotor about its diameter

As.

pg IM= , equation 2.14 becomes )1.2.4(...

+= dpx IIM

32


33/60

Similarly in the xz plane )2.2.4(...

+= dpy IIM

As the angular displacements in general of the rotor are periodic:-

)3.2.4(sincos .1.2 += tt

)4.2.4(sincos .2.

1 += tt

If however the system is isentropic (rotor support characteristics are the same in both

directions) then the amplitude of and will be equal and the equations will be

)5.2.4(cos . = t

)6.2.4(sin . = t

Where the motion of one lags half a phase behind the other one in a perpendicular direction.

Differentiating equations 2.19 and 2.20 then substituting them into equations 2.15 and 2.16

gives the magnitude of the net moment applied to the shaft as,

)7.2.4()( 2 = dp IIM

For the isentropic system the deflection, z, and slope, , of the shaft are given by:

)9.2.4(

)8.2.4(

43

21

+=+=

MCFC

MCFCz

Where ,, 21 CC etc are coefficients applied to the shaft by the rotor relating forces and

moments to the shaft deformations. For an overhung rotor mounted in rigid bearings (or

rotor with small length to diameter ratio, dp II 2= ) then,

)10.2.4(23

23

+=

EI

Fl

EI

Flz

)11.2.4(2

2

+=EI

lM

EI

Fl

Substituting equation 2.21, 2.24, and 2.25 and 2mzF= into equations 2.22 and 2.23

gives:-

)12.2.4(023

12232

=

+

EI

lIz

EI

lm d

)13.2.4(012

2222

=

++

EI

lIz

EI

lm d

33


34/60

These equations are satisfied when:-

)14.2.4(022

13

12222232

=

+

+

EI

lI

EI

lm

EI

lI

EI

lm dd

For a point mass rotor, the natural frequency is: ,3 2

1

3

=

ml

EIp if 2

3

ml

Id=

the equation becomes:

)15.2.4(04

11

4

4

224 =

+

p

p

Showing the relationship graphically in figure 2.6, it can be seen that the natural frequency

is almost twice as great for rotors with disc-masses than those for point-mass rotor systems.

Gyroscopic effects are significant when there is a heavy overhung rotor or the rotor is

running at high speed. The analysis from X can be extended to allow for gyroscopic effects

by modifying the right hand side of the equation to allow the gyroscopic couple terms.

4.6 ROTATING RIM or RING

The pulleys, flywheels, rotating wheels etc are quite commonly used in the engineering

field. For designing the same it is necessary to find out the stresses developed therein on

account of rotation. In the analysis given the effect of spokes is negative.

Consider a wheel rim of mean radius r, thickness t and of unit width, rotation about its centre

with a uniform angular velocity . Let c be the circumferential i.e., the hoop stress

produced in the rim on account of this rotation. Consider an element of the wheel rim

making an angle at the centre. The forces acting on this element are shown in figure.

34


35/60

Fig 4.6.1. forces acting on the element.

Ifis the density of the rim material,

Centrifugal force =rtr 2. )1(

= tr22 ---------( i )

For equilibrium of the element shown, equating forces in the direction of the centrifugal

force,

Centrifugal force =2sin)]1([2

tc

=2

2

tc ( is small)= tc -------(ii)

from ( i ) and ( ii ), 222 == rc --------( iii )

Where is the peripheral velocity = r

It is very important to understand the units of various quantities in equation (4.6a). since

Newton is the force which produces in mass of 1 kilogram, an acceleration of 2sec1meter ,

for consistency the following units be used in equation (4.6b):

Hoop stress, )(, 2 pam

Nc

35


36/60

Material density, 3, mkg

Mean radius of rim, m

and Peripheral velocity,

sec,mv

4.6.1 THIN ROTATING DISC

It is well known that components rotating at high speed such as stream or gas turbine rotors,

grinding wheels, circular saw, etc are subjected to stresses. As such the highest speed at

which such component may be allowed to rotate would be limited by the safe allowable

stress for its material. The problem of the rotational stresses in discs, therefore, becomes

quite important. In such a case the deformations produced are symmetrical about the rotation

axis. For simplicity, we assume that the disc under consideration is a thin disc in which case

it becomes a plane stress problem which means the radial and the circumferential stresses

are constant through the disc thickness and stress in the axial direction is zero.

Fig:4.6.1

Let the thin disc of density and unit thickness be rotating about its centre with angular

velocity (fig. 4.6.2). Then,

=r radial stress produced at radius r

=+ rr radial stress produced at radius )( rr+

=c hoop stress at radius r, assumed constant overr

=u radial shift ( displacement ) at radius r

=+ uu radial shift or displacement at radius )( rr +

36


37/60

Various stresses/ forces acting on the small element of the disc considered are shown in fig.

4.6.3, whereas the straining of the fibers at radiirand )( rr + are shown in fig.4.6.4.

Fig:4.6.2

( i ) The circumferential strain

=r

u

r

rur=

+

2

2)(2

Which is also equal to )(1 rc vE

rc vr

Eu=

or rvrEu rc = --------(4.6.1)

( ii ) from fig.4.6.2, radial shift at radius ris u , whereas at radius )( rr + , it

is )( uu +

Radial stress =r

u

Which is also equal to )(1

cr vE

cr vr

uE

=

37


38/60

In the limit, cr vdr

duE = -------(4.6.2)

( iii ) Differentiating Eq.(4.6.1), w.r.t. r,

d r

dv rv

d r

dr

d r

d uE

r

r

c

c

+=

= cr v [from Eq.(4.6.3)]

0=+++

dr

dvr

dr

drvv rc

rcrc

Or 0)1)(( =++dr

dvr

dr

drv rcrc

-------(4.6.3)

Fig:4.6.3

( iv ) Now, the centrifugal force on the element (fig.4.6.3) is,

= rrr 2)1(

= rr22

Equating the forces on the element in the direction of the centrifugal force, we have,

]))[((22 rrrr rr +++

2

2

rr cr +=

=

22sin

38


39/60

or rrrrrr crrr +=+++ ))((22

orrrrrrrrr crrrrr +=++++

22

or crr

rrr

=++22

( rr being small compared to other terms)

orr

rr rrc

+= 22

or in the limit, ,0r

dr

drr rrc

+= 22 --------(4.6.4)

( v ) Substituting )( rc from Eq. (4.6.3) into Eq. (4.6.4),

0)1(22 =++

+dr

dvr

dr

drv

dr

drr rcr

or 0)1(22 =++++

dr

dvr

dr

dr

dr

drv

dr

drvr rcrr

or )1(2

vrdr

d

dr

d rc +=+

Integrating w.r.t. r,

Ar

vrc 22

)1(2

2 ++=+ --------(4.6.5)

Where 2A is constant of integration.

Now Eq. (4.6.5) Eq. (4.6.4) gives,

dr

drrA

rv rr

++= 22

22

22

)1(2

orAr

v

dr

dr r

r2

2

)3(2

22

+

+

=+

or Arv

rdr

d

rr 2

2

3)(

1 222 ++

=

or Arrv

rdr

dr 2

2

3)(

322 ++

=

Integrating w.r.t. r,

39


40/60

BArrv

rr ++

= 24228

3

Where -B is a constant of integration

or22

2

8

3r

v

r

BAr

+= -----------(4.6.6)

from Eq. (4.6.5)

rc Ar

v +

+= 2

2

1 22

++

+= 22

2

22

8

32

2

1r

v

r

BAAr

v

22

2

8

31r

v

r

BAc

++= ------------(4.6.7)

4.6.2 SOLID DISC WITH UNLOADED BOUNDRY

For the solid disc, constant B in Eq. (4.6.6) and (4.6.7) is zero, since otherwise stresses

would become infinite at 0=r . Constant A can be found from the condition that 0=r

when 0rr= ( outside radius)

from Eq. (4.6.6), 2028

30 r

vA +=

20

2

8

3r

vA

+=

substituting this value of A in Eq. (15.7) and (15.8) and B=0,

222

0

2

8

3

8

3r

vr

vr

+

+=

i.e., )(8

3 220

2rr

vr

+= --------(4.6.8)

and222

0

2

8

31

8

3r

vr

vc

++=

i.e., =])31()3[(8

22

0

2

rvrv ++

------(4.6.9)

The distribution of rand c is shown in fig.4.6.4

40


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Maximum stresses occur at the centre and are,

maxmax cr

= ( at 0=r ),

=2

0

2

8

3r

v+ --------(4.6.10)

Also at the outer radius,

])31()3[(8

2

0

2

0

2

0

rvrvrrc ++=

=

2

0

2

4

1r

v

= ----------(4.6.11)

4.6.3 DISC WITH A CENTRAL HOLE AND UNLOADED BOUNDARIES:

Constants A and B can be found out from the boundary conditions that,

0=r when irr=

and 0= when 0rr=

from Eq. (4.6.6),

22

2 8

30 i

i

rv

r

BA

+=

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42/60

and2

0

2

2

0 8

30 r

v

r

BA

+=

which give,

2

0

22

8

3rr

vB

i

+

=

and )(8

3 20

22 rrv

A i +

=

+

+= 2

2

2

0

2

2

0

22

8

3r

r

rrrr

v iir ------(4.6.12)

and

+

+++

+= 2

2

2

0

22

0

22

3

31

8

3r

v

v

r

rrrr

v iic ------(4.6.13)

for rto be maximum, 0=dr

dr

let value ofrat which ris maximum be R

then differentiating Eq.(4.6.11) and equating the same to zero,

022

8

33

2

0

2

2 =

+= R

R

rrv

dr

d ir

i.e.,2

0

24 rrR i=

or 0rrR i=

i.e., ris maximum at that value of radius which is the geometric mean of the inner and

the outer radii of the disc.

Substituting 0rrr i= in Eq. (4.6.2)

2

0

2

0

0

2

0

2

2

0

22

)(8

3

8

3max

i

i

i

i

ir

rrv

rrrr

rrrr

v

+

=

+

+=

----------(4.6.14)

Also, it may be seen from Eq. (4.6.12) that c is maximum when ris minimum, i.e.,

when irr=

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+

++++

+= 2

2

2

0

2

2

0

22

3

31

8

3max i

i

i

ic rv

v

r

rrrr

v

+

++

= 2202

3

1

4

3ir

v

vr

v -------(4.6.15)

The distribution of rand cas given by Eq. (4.6.11) and (4.6.12) is shown in fig.4.6.5

5. RESULTS & DISCUSSIONS

x=0:.001:10;

y=x.*exp(-x);

plot(x,y)

axis([0 10 0 0.5])gtext('E=1');

xlabel('WnT............');

ylabel('XWn/Vo..............');

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clear all

x=0:0.01:0.99;

y=1.01:0.01:4;

for i=1:length(x)

f1(i) = abs(x(i)^2/(1-x(i)^2));

end

for i=1:length(y)

f2(i) = abs(y(i)^2/(1-y(i)^2));

end

plot(x,f1);

hold on

plot(y,f2);

xlabel('p');

ylabel(abs(p^2/(1-p^2))');

44


45/60

p=0:.1:4;

e=[0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8];

for i=1:length(e)

for j= 1:length(p)

f(i,j)=1/sqrt(((1-p(j)^2)^2)+(2*e(i)*p(j))^2);

end

end

plot(p,f);

hold on

grid on

gtext('e=0.1');

gtext('e=0.2');

gtext('e=0.3');

gtext('e=0.4');

gtext('e=0.5');

gtext('e=0.6');gtext('e=0.7');

gtext('e=0.8');

xlabel('p');

ylabel('Magnification Factor');

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A=[0.02 0.1 0.5];

for i=1:1:3;

b=(1-A(1,i).^2).^0.5;

x=0:0.001:100;

y=exp(-A(1,i).*x).*sin(x.*b)./b;

plot(x,y)

hold on

end

axis([0 50 -1 1])

grid on

gtext('E=0.02');

gtext('E=0.1');

gtext('E=0.5');

xlabel('Wnt.........');

ylabel('XWn/Vo...........');

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A=[1.5 2];

for i=1:1:2;

x=0:0.001:10;

b=(A(1,i).^2-1).^0.5;

y=(exp((-A(1,i)+b).*x)-exp((-A(1,i)-b).*x))./(2.*b);

plot(x,y)

hold on

end

axis([0 10 0 .3])

grid on

gtext('gita=1.5');

gtext('gita=2');

xlabel('WnT.......');

ylabel('XWn/Vo.........');

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p=0:.1:4;

e=[0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8];

for i=1:length(e)for j= 1:length(p)

f(i,j)=p(j)^2/sqrt(((1-p(j)^2)^2)+(2*e(i)*p(j))^2);

end

end

plot(p,f);

hold on

grid on

gtext('e=0.1');

gtext('e=0.2');gtext('e=0.3');

gtext('e=0.4');

gtext('e=0.5');

gtext('e=0.6');

gtext('e=0.7');

gtext('e=0.8');

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xlabel('p');

ylabel('Magnification Factor');

clear all

x=0:0.01:0.99;y=1.01:0.01:4;

for i=1:length(x)

f1(i) = abs(1/(1-x(i)^2));

end

For i=1:length(y)

f2 (i) = abs(1/(1-y(i)^2));

End

Plot(x,f1);

Hold on

Plot(y,f2);

Xlabel('p');

Ylabel('Magnification Factor');

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Fig No: Normal Stress (SX) in X - Direction

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Fig No 3: Vonmises Stress in Rotor

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