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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)
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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
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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
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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.
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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
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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.
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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.
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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
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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
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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.
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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
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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
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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
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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.
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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
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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 +
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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
=
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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
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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,
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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
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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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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))');
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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)=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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