Last class we were discussing the torsion
equation which I wrote; here I am just giving
you a phi double dot minus l omega square
I m… sine theta naught plus phi cosine theta
naught… plus phi minus l omega square I
m eta… xi cosine square minus plus M A;
Suppose in the, if we say phi is small; phi
is small such that theta naught is not small,
that is the pitch angle then, you can approximate
sine theta naught plus phi cosine theta naught
and then cosine as cosine theta naught minus
phi and this you have to substitute. You can
collect the terms as phi because otherwise,
in this form the equation is non-linear even
isolated torsional equation is a non-linear
equation, but your pitch angle which you set
that is theta 0 that can be large angle that
is the finite input.
But, the phi is the elastic twist that you
say that is the small angle. So, that is the
assumption you are making here and then you
substitute back then collect the terms of
phi separately and then leave out all the
other terms. If you write that that equation
will be that is you substitute and then rearrange
all the terms.
Then you will have, the equation will be l
I m xi xi I m eta eta phi double dot plus
l omega I m xi xi minus I m eta eta cosine
square theta naught minus sine square theta
naught plus k phi over l omega square minus
I m eta xi 4 sin theta naught cosine theta
naught phi equals aerodynamic moments minus
l omega square I m xi xi minus I m eta eta
sine theta naught cosine theta naught plus
l omega square I m eta zeta cosine square
theta naught minus sine square theta naught
(Refer the above slide at the reference time
for the equation). This is my equation; it
is the linear differential equation because
there is a phi double dot and there is a phi.
So, this like m s double dot plus k x equal
some, but you look at the external loading
and a is the aerodynamic loading there is
one constant term. Please understand this
constant term is there, which is a function
of pitch angle. Now, you when you we wrote
the phi square term, this one, yeah that see,
always phi is small. Note that you make that
assumption so that, the torsional deformation
is small. So, pitch angle is large; large
means it can be any 10 degrees, 15 degrees,
5 degrees, but phi is a small angle. So, the
idea is this kind of tells you even if you
do not have any dynamics; you keep the blade
at the pitch angle theta naught and rotating
theta, no aerodynamics in vacuum. This is
a constant term which is the function of theta
naught if theta naught is 0 this term is what
this is 0. Only this will be there this is
a cross product if it is a symmetric cross
section then that will also go off, but if
it is a non-symmetric cross section you will
have some term.
But, usually looking its symmetric cross section
of the aerofoil then you can always find what
is the twist - elastic twist phi due to a
pitch angle and that will be because this
is usually a small number. This is, you can
neglect this and even this quantity is larger
than this quantity. You will find without
dynamics in the sense without any epsilon,
aerodynamic loads there is no phi double dot
equilibrium torsional deformation of the blade.
Your phi equal phi equilibrium equals this
quantity divided by this quantity and that
defines what is the torsion.
Suppose, if it is a hinge point there is no
spring then you will find the answer will
be phi, will be exactly equal to minus theta
naught; that means, the blade will always
come flap. In the sense, if you keep the blade
like this and rotate it the blade will always
come down it will be flat. This is because
that is an actual equilibrium position; you
see there is always a twisting moment; twisting
moment because of inertia load.
Now, you see the assignment which I gave will
be rotating at the higher r p m; you can have
in elastic twist of the blade due to control
system, stiffness plus due to the blade elastic
stiffness. That is the one can actually reverse
calculate that should be the k phi assuming
that this must be my pitch angle. So, usually
what you set it, pilot makes you a command
this is my pitch angle but, what the blade
experiences is not the same just dynamics
no aerodynamics comes that will also change
angle of attack.
But, here because of the elastic twist you
will find a loss in the angle of attack of
the pitch angle that is due to the k phi k
phi plays a major role. That is why the industries
they calculate what is the control system
stiffness. You will have swap plates and then
there is a hydraulic jack at the below the
swash plate above the control rods. You calculate
the stiffness of that entire unit; do a fringe
model some approximate calculation anything,
but come up with the stiffness usually of
course, control rods are very small.
So, they are by e a over l, but the swap plate
will have its owns stiffness that nothing
reached them so, that will give this. There
will be a loss in pitch angle and now I will
not write any expression for M A because,
this is basically aerodynamics load. You can
use quasi steady aerodynamic load or very
simple instantaneous angle of attack. Take
it, put it quasi static; these are all terminologies
in the aero-elastic analysis. This is purely
simple torsional dynamic set up from here.
You can see if I m eta zeta is 0. What is
my frequency - natural frequency of the blade
in torsion like you have the flap frequency,
you have the lag frequency now you have the
torsion frequency. If you take the torsion
frequency, omega r that is basically you have
the stiffness term you have the massive term.
You will have omega bar that will be essentially
this term divided by this term. So, you will
have I m I m eta zeta cosine square theta
minus plus k phi l omega square I am because
l omega square this you make it non-dimensional.
So, you will get another omega square will
come all the l omega square will go off I
have I m xi xi plus I m this is nothing, but
the mass moment of inertia of the blade lower
section per unit section that is all this
is sum of this is the difference.
Now, you see when theta is 0 that is theta
naught when you set the pitch angle of 0 this
term will… and usually, this is much larger
than because you need thin aerofoil - thin
aerofoil this is what you can much smaller
than I m xi in the inertia calculation. Therefore,
you simplify this as theta naught is 0; you
said then you will get a highly simplified
expression.
This is this and this 1 plus I will put an
approximation k phi over I m xi omega square
this is my very simple form; that means, this
is you can take it the blade torsional inertia
because, I took this is a constant l into
this will be torsional inertia of the blade
about the and you are not with respect to
omega square. This is like a simple torsional
spring and this is due to inertia 1 plus something
like you had flap motion where you always
get 1 plus something here also this one will
come.
But, this you have to be very careful; how
it comes? it comes because of this term and
then and sometimes people make mistakes because
these are all very small quantities in a coupled
torsion equation. Of course, but usually you
know control system has a sufficiently large
disturbance because you do not want the control
platform is soft because, whenever you give
an input it should fake fully go immediately
and there should not be any phase neutrons
between this and that. So, usually try to
design that this is stiff control rod stiffness,
but we take an actual blade. You have a control
rod stiffness you have the elastic g j of
the blade both will contribute to that those
equations there you have write the elastic
blade here.
We assume rigid blade, why do you place the
frequency of torsion? If I flap is around
1.09 it is a 4 0 3 like that lead lag is 0.25
or 0.7 and the non-dimensional torsion you
keep it far away far away means usually in
the around, but 3.5 is low torsional stiffness.
So, this is around 3.5 to 4.8 something like
that around this range it will be 4.54 3.5
is very low torsionally it soft blade, but
you see this frequency is far away from flap
and lag where you know I want to bring this
and I have problem; this is mainly dominated
by control system.
So, if we want to analyze the blade dynamics
you need to know the control system stiffness.
So, that has to be calculated separately by
a stress analysis that you have to get some
estimate of the torsion stiffness and I will
not go into the details of this particular
term. I wrote the tennis racquet quick that
is basically this; see even if k phi is 0
it is like a centrally hinged flap.
So, torsion also we will have,, if you disturb
it will alternate like this as it goes around
once per revolution that is all it is like
a fight. See, that is why when you analyze
torsion it is not a line it is actually a
2-D about the axis to blade tangent oscillate
like this. that is why that is with this we
kind of pretty much analyzed; this is of course,
you know this is the non-rotating torsion
frequency because the spring inertia that
is all.
This is non-rotating non-rotating plus 1,
that is what I wrote the, but omega non-rotating
torsion plus 1 omega t. now, this all for
isolated system but, if you ask what about
the damping in torsion lead lag, we discuss
the damping is very small and duplicate external
damper is provided. If torsion the main damping
comes because of aerodynamics usually it is
about 10 percent. That is why there is no
damper provided in torsion. Flap is heavily
damped lead lag you put and external damper
torsion it also damped out isolate torsion,
but further you need to know unsteady aerodynamics
not the steady then only you will get the
phi dot like term. You need to know the unsteady
aerodynamics how do you get pitching moment
that a several theory.
That theory is in the aero-elasticity fixed
wing theory. Fixed wing theory, is it applicable
to rotary-wing? We use it; applicability is
something different, but, that is the theory
available and it is used but, with the correction
for the inflow that is all.
Sir fixed wing and rotary wing other than
the actual forces is acting to what would
be the
Torsion is pretty much there is no rotation
that is all this is free externally only this
there is no rotation this will not be there.
a rotating blade and a fixed blade fixed blade
there is that bending torsion problem what
you want in the structure one course that
is it.
Even we will be adding axial load. Yeah axial
load, but I has made it as a rigid blade.
Please understand in all my assumptions, in
all these things I said my blade is rigid
but, this question is, what we have a centrifugal
force what you will do? That is an axial deformation
with that it is there, but, then you have
to include the axial frequency axial mode.
Now, treatment of axial degree of freedom
itself has several approaches.
How would if done is one is you make it as
a rigid blade approximation that is it why
is it valid the reason is axial stiffness
is always much higher it is e a than bending
and torsions therefore, solve it then in the
bending problem you have an axial because
it is a combined if you want to treat axial
motion centrifugal force you have to bring
it to a effect if you make it as a rigid blade
it is taken care because you are not bother
about that only thing is what is the effect
of the centrifugal load in flap motion you
consider lead lag motion you consider torsional
motion you consider.
But if you say, no I want to include elastic
deformation in axial direction, then, you
write the axial equation. Then there are certain
approximations made which is usually what
normally make in traditional bending problems.
What you do you draw a line reference line
and then it deforms you see the length of
that line does not change any material above
that this is what you will not you take cantilever
beam then you take a line. Then, you say what
this deforms, what is the length of this line,
is it the same or different or what. You make
an assumption that the reference we call neutral
axis. Neutral axis does not extend, that is
basically called extensionality condition.
If you impose that condition then you can
sum; so, that is also done. Suppose, you say,
no I do not want to do also then, you know
that problem. You include axial mode in your
formulation that it is really missing it will
complicate. The intense of aerodynamics, what
is the difference in between a static wing
and rotary wing aerodynamic?
Aerodynamic static wing and rotary-wing plenty
of first of all through velocity is varying
across the span you will have radial flow.
Yeah and you have a inflow everything is that
flow is in order. Inflow can be accounted
how to what.
That is your doubt how do you take phi dot?
Yes. Is that, yes, phi dot…
If you give because, that is not taught to
you. Phi dot is suppose you have to take the
theory, there is in the unsteady aerodynamic
theory. You take an aerofoil; this is a quarter
dot. You take the half point and you take
some a elastic axis this is the most simplest
model and then mass center. So, you take this
point can come down and it can rotate this
is the elastic axis. So, this is the 2-D therefore,
now the theory associated with a 2-D is the
theory in the subsonic in cumbersome and that
theory takes care of both the h and alpha
h is the skewing motion alpha is the pitching
motion. So, you will get the lift and moment
expressions are given about sometimes what
and these expressions will be a function of
alpha - alpha dot, alpha double dot, h - h
dot, h double dot, h triple dot.
So, you find everything and this theory is
the vortex model, this is the unsteady aerodynamic
theory, this is not torque; suppose, this
is what I am telling now if you want to get
into aero-elastic problems some rotary-wing
you start teaching. This you teach, make your
all approximations and go and do it. That
is, you start with fixed wing aero-elasticity
is that, that is actually unsteady. Aerodynamic
theory of a 2-D aerofoil alright, then you
learned then from there, how do you incorporate
that in your rotor lead? You can solve the
problem this was the 1937 or something like
that theta theory but, subsequently it was
modified by and there is a Louis theory, but
still theta theory is because it is experimentally
who want to be valid for this.
Now, you can have start c f d, but c f d how
to because this gives a form of the equation
which you can handle the stability of the
problem c f d is a numerical calculation.
So, it is only time here you were analyzing
in the frequency domain and get the stability
directly. So, there is a lot of difference.
How do you see the c f d for a stability problem?
Then, you will have time domain integration
and then the damping still it may be a question
I have not seen any publication in that they
try to use for loads, but not for stability.
These are differences because this is going
more deeper into the modeling and then predicting
what is happening as we have shown here into
first level, I will call it, this is a helicopter
dynamics is the first level, but you learn
isolated effect. You see what each blade does
in flap motion, lead lag motion, torsion motion
and then you know the frequency you keep;
what range they are. These are all isolated
motion to understand what influences the frequency.
Now, you have learnt up to this next what
because, that is a where you built the subject.
I will see after having analyzed the individual
dynamics of the rotor blade, what does the
various topics how people can go because isolated
blade here you have flap lag torsion? Now,
when you did this, what you have learnt is
rigid blade, rigid blade, rigid blade. This
is the assumption you have made. Suppose,
you say, no I do not make this blade; I want
to take it is as an elastic blade then, you
have to go. That means, model in itself will
lead to different direction that is elastic
blade. These suppose I think one of you were
using that how do you model elastic blade?
Whatever he learnt he will give this by itself
is separate study. Modeling an elastic blade
elastic rotor blade with all complexities
please understand you will have it is a twisted
blade it is flexible in bending flap bending
lag bending torsion axial. So, you will have
axial-flap-lag-torsion problem complete all
of that. One is structural model then you
have to have that aerodynamic, sorry, the
inertia model then it comes aerodynamic. Aerodynamics
I leave it as it is purely blade modeling
is only a large structural and inertia it
is a structural inertia only this. Actually,
people spend their entire some of the people
research career only on modeling structural.
When you say blade modeling you have to say
what material. I said material composite material,
how do you go, how do you model it, how do
you get it, etcetera, this people have spent
their whole career on. This is one direction;
another one aerodynamic modeling is required
if you want to solve aero-elastic problem.
Traditionally I will put it, traditionally
people use a relatively simplified aero-elastic
aerodynamic model of course, simplified in
the sense quasi steady with the dynamic stall
with a some dynamic way etcetera all because
that what we have used. Now, today c f d is
entering into that it will have worked on
that c f d problems some of them and they
trying to relate, but it is have some approximation
still it is not fully advanced to the level
that you can do everything.
Now, that is it is a aerodynamic model. So,
you can get into aerodynamic modeling purely
when I say aerodynamic modeling, what is that
I do? In the rotary wing specifically I have
to take care of the wing which is a structural
means the blade gives you can have lifting
line theory, but here when you go steady unsteady
good unsteady aerodynamic model and then the
wake that is the inflow which is generated
at the rotor disk. We always use constant
inflow in some uniform, but in hover only
we had the very good forward flight is only
constant inflow.
Now, here the inflow constant it can with
varying time then above how do you model that.
So, inflow modeling itself has gone through
very interesting, I would say development
that is an interesting developing and today
is that we have certain models which were
developing. That is a dynamic inflow, dynamic
wake etcetera these are all global theories
still I am telling it is not a c f d per say
but still, it is able to give you the variation
of inflow.
When there is a variation in the rotor blade
loading, the whole concept started like that
if there is a change in load inflow will change
when the inflow changes, load changes. how
do you relate these two? This is one line
of problems we will have developed the time
on modeling that another one is unsteady.
Aerodynamic model for getting lift drag moment
etcetera, one is please note one is getting
only the inflow another one is getting lift
drag moment on sectional.
So, if you want to include stall include stall
models develop stall models 2-D and then use
it in 3-D. So, one can get into this also,
but when you talk about aero-elastic you need
to have all. So, will you make? If you do
steady aero-elasticity then may get rid of
this and then we have good model in everything
and then use it for aero-elastic problems.
You can get into rotary-wing, aero-elasticity
this is one line when I talk about rotary
wing aero-elasticity. I have flap lag torsion;
initially, development if you see that was
not done with elastic blade rigid blade, but
put flap lag combine then put flap torsion.
So, flap lag problem, flap-lag flap-torsion
problem, flap-lag then flap-torsion these
two problems are treated initially that is
one line because that you wanted to understand
when the blade will go unstable etcetera.
Subsequently, because this particular problem
flaps torsion do slight bending torsional
problem of a fiction. Only thing is this is
the rotary wing, the aero-elastic problem
in bending torsion for fixed wing. Here it
is flap-torsion subsequently they said that
nothing we should have everything. So, flap-lag
torsion problems but, these results were generated
with various coupling various things.
I will show one diagram not experimental set
up because it was a essentially see this was
a model - model rotor. So, they put because
that model was built to represent the theoretical
assumption please understand that is why you
see the other diagram not that they made a
model and the you start modeling is actually
whatever theoretical assumption you make we
corporate in the experimental thing.
So, what they did was they put a… it is
actually nice kind of combination of flap
and lag flexure at the same location because,
you already say spring model and the spring
model was putting and there is no torsion
blade of course, there it a little bit stiff.
In the sense, I am not like a rigid blade,
but then they have pitch-lag coupling they
also introduced pitch-flap coupling everything
in this model.
I am just studying only the stability of that
and of course, ground resonance problem also
we will talk later that again coupling. So,
this is analyzing aero-elastic sometimes you
may require stall please understand because
stall may be required in this stall modeling
then stall modeling you have to bring in even
here because, even aerodynamics may not be
good. So, it is like a valid term structural
model was up to the minimum check all this
couplings effect of aerodynamic modeling that
was this, but this is isolated blade you are
only solving blades problem, but in a helicopter
it is a rotor. That means, I have to go for
coupled rotor fuselage dynamics. That means,
fuselage is shaking blade is moving you have
to study this problem this is the actual problem,
but in this you have like you have aero-elastic
stability and response here I will call it
here aero-elastic stability and response only
these two yes. How do the flap motion?
Experimental measurement of flap motion means
what exactly you would have to think in. They
usually have some string hinge in the root
of the blade. So, that will give you what
is the bending moment flap bending moment,
lead lag bending moment and may be torsional
moment. Torsional moment normally goes to
the pitch link load.
So, pitch link load they measure what is the
load on pitch link. So, flap link they put
a string hinge where they will measure the
root bending moment. Please understand, it
is not that entire motion every section they
will do, but recently means may be about 10
years 1 u h 60 helicopter that U S army a
experimental program was conducted on which
they had some 220 ports some. So, many meters
I do not know I have the figure on the blade
both span wise pressure codes meters and they
measured the in-flight load; that is only
one test and that is what people have been
using it for predicting their models.
Even yesterday, there was a seminar; of course,
it was a seminar and because that data they
are not supplying to everybody it is a flight
test data and they did it for certain monomers
and then they are trying to predict what theory,
how could it be experimental even now that
research papers skip because that has a brought
certain problems. Before that the response
was not properly predicted, those things were
identified now. See, with other models we
explain certain phenomenon, with other models
even though we do not have that data.
If you change your aerodynamic model from
simple to a more sophisticated, so we have
five different models. So, we said if you
use the most simplest model what will be the
response; as you complicate the model how
the response really changes because, there
was some phase shift phase shift was where
the minimum some load happens on advancing
side something like that. They found out one
of the measurement and theoretical was actually
shown somewhere else.
So, they want to know what is really happening;
that means, I still have not understood, but
those problems have been not solved this is
only the way otherwise there is something
called a tracking and balancing tracking means
you want to track the tip path because, all
the blades must do same motion. Because, if
all of them are identical right no blade is,
no 2 blades are identical. There is always
some reflection and they have a criteria by
which they will bring the mass and the first
moment between some limit after that they
put the blade they rotate it they have color
code.
Earlier they put some chalk and then they
bring a cloth; they will know where it hits
each color chalk. Then, they will know this
blade is high that blade is low then they
change it. Another way is, they also have
a vibration machine inside because, if blades
are not tracking same position you will have
a one per revolution vibration. They will
try to minimize that and that is the way.
Otherwise, you can put some other wireless
system and then measure. These are all measurement
systems; what industry does? They will take
and then earlier only what is that the slip
ring data - will come to the slip ring.
Now, you can have wireless also, but that
you can have; but then, technology has to
be implemented on real life things and then
take the data. That is what I think if you
want add anything you can add, that is the
vibration machine. If you tell which blade
to adjust that is all. So, blade stiffness
is what blade stiffness is e I; e i is flap
but, e i is the variable it is not the uniform
see the aerodynamic section to almost near
the tip it is a constant value when you come
near the root the value will change.
So, it is essentially a variable property
beam; it is not the same value it is not uniform
beam it is an nonuniform beam with property
varying near the root section drastically
down varies they change very large small range.
Sir this spring which we are modeling for
flap the spring constant would be lower than
the
No, spring constant these you trying to estimate
spring constant estimate please understand
estimate the spring constant from frequency
that is all and the hinge offset that is how.
We measure that spring constant with respect
to?
What is the question of measurement of spring
constant? There is no measurement.
sir.
You measure only frequency
Yeah.
From the frequency because you know that suppose
you take flap flap is 1 plus what M m x c
z e over I b right plus k beta over right
this is omega rotating flap square suppose
if you say non-rotating blade flap frequency
you can call it omega N R flap that is only
this quantity this you can estimate because
in a hingeless blade it is very difficult
to know where exactly is that there is no
point.
So, it is like from the frequency you try
to estimate what is this and what is e if
you assume this you can get this equation
is only one suppose if you have non-rotating
frequency then you can say this is my k and
then once I know that then I can go and get
there is another way or else they will say
if my k beta is 0 what is my hinge offset;
that means, that will be an equivalent articulated
blade this is the what is the hinge offset
estimating of orders less than.
No estimate dimensionwise you have to find
dimensionwise this is Newton meter per radians
that is what e i is what Newton meter square
where what is the connection no connection
is you can you see one is the different unit
that is the different unit you can directly
put it there of course, that affects this
you are looking a numerical value.
My question was, is the thing we have to take
the most of the bending which is so that,
different and blade hinge? The blade see please
understand not that your question is blade
also will bend.
Yes.
But the second mode third mode first mode
majority of the deformation if you say this
is the hub majority of the deformation will
happen like this is the first mode usually
you will find the this is the if it is the
hingeless blade, but you can also get something
here
, but it is a hingeless blade if it is a articulated
blade then you have to calculate because the
root hinge moment is 0 then the blade may
have some other point bend moment will be
higher, but that is usually not going you
also make a measurement you put a spring in
there and make measurement only for because
that test I told you they measured and 50
percent 55 percent something like that bending
moment that 50 percent and then try to correlate
how your theoretical value matches with the
flight test. But, number wise I do not see
your question really is it a numerical value
you are saying not a numerical value then
It is a b blade stiffness is say E i blade
stiffness is e i.
Yes.
See these are all equivalent model equivalent
model that is all if your actual analysis
if you want to do aero-elastic load then directly
put the e i you do not go to this model see
this model gives an idea of what is my flap
frequency what is my hinge offset that is
all not this k beta you adjust it to match
that.
If you want to represent your blade as a rigid
blade with a root string then you go and then
put this; that means what you are doing elastic
blade you are representing by a rigid blade
with root spring. So, the modeling if you
want you may use this and then you estimate
the value beta and e that estimation. The
industriously root is going to where we.
Always, everywhere the root only it takes
only the root see if you really no no blade
middle point root only have all the loads
see one is the centrifugal load centrifugal
load is of the order of 100 1000 Newton phenomenally
large. But the flap load will not be large;
see, if you take flap shear load, if it is
a 4000 kg helicopter it is 40,000 newton;
for 4 blades that means, per blade it is 10,000.
You support the weight of the helicopter by
4 blades, 4 shear load 10,000. But, centrifugal
load in order of 100, 1000, but that and then
in flap load is useless because, that will
give you the fatigue because it will oscillate
all the vibratory load comes here. Non inverse
the flap because it is a bending problem load
wise may be magnitude may not be very large,
but this cyclic load vibratory part this is
the one which is the critical mean value and
then vibration mean is usually not tremendously
large because you of you take what 4000 kg
helicopter means and 10000 Newton for four
blades that is all that is the mean value
in the shear is it clear. Now, I will go back
to this because, I am giving only the type
of problems this requires now fuselage motion.
So, you will have fuselage motion in addition
to flap-lag torsion; that means, hub is going
to move this problem is more complicated.
But all flight dynamics problems come under
this because you want to you want to do something
then you need to have a model for the in addition
there are certain problems where you face
instability that is I will just list them
out one is lag and fuselage I will put it
only lag fuselage you will have air ground
and air resonance ground air resonance and
then flap this is a gross approximation flap
and fuselage this is flight dynamic usually
that is what in that.
But if you want vibration then, you have to
have flap lag torsion; flap lag torsion axial
everything you can put axial also and then
fuselage this is vibration now you see there
are different types of problems you want to
you can say the most sophisticated analysis
is coupled flap lag torsion axial fuselage
motion and fuselage also elastic modes of
the fuselage flag is the motion, but this
not perfectly that it becomes it is really
today it is not done its high complex and
then if you want to take transmission tail
rotor fuselage aerodynamic interaction, that
is a problem is really very complicated, but
how can the industry is very demand go ahead
and make a helicopters you have to make simplifications
the simplification ground resonance when they
analyze they take only lead lag motion of
the blade and the fuselage as a rigid body.
You understand and then solve this problem
ground resonance problem; so, that means,
the modeling that now changed only I take
lag and fuselage motion suppose you say I
want flight dynamics flap motion and fuselage
motion because the flap controls your loads
hub loads your monomer everything.
So, you included plus there is also one more
what is the frequency range you are interested
usually the rigid body dynamics the frequency
is low in the sense it is less than one two
or may be at the most three thirds not all
hertz whereas, this will be about 3 5 hertz
fibers this problem you will have from 20
hertz 40 hertz everything all the frequency
will be there now you see depending on the
frequency range of interest your modeling
also changes, but technically if you have
a complete model you can solve all problems,
but then when you solve the problem what approach
you will use.
So, that mathematical technique becomes very
important if you want analyze stability then
you have to know how do I analyze stability
of the problem of the system if I want response
control response how do I solve the control
response problem.
Now, these are now going much deeper into
the helicopter problems. So, depending on
the problem you address that issue. So, you
find the modeling will change which of these
I am going to do. These are to a large extent,
these are all research problems even though
the flap is, but if you want to add all of
them and then vibration and then flight dynamics
and control characteristics everything today
models are getting developed everywhere people
are improving their fundamental understanding
based on some flight test or experimental
data improve keep improving, but to have a
simple understanding let us take just flap-lag
aero-elastic stability.
That means, you are having two motions of
the blade it is an isolated blade if you want
to go to rotor now please understand this
is one blade analysis this you have to take
all the blades and all the blades they are
all kept at different different azimuth locations
you have to take everything into account and
there are certain techniques associated with
treating this problem how do it. We have developed
equation for one blade, but I am putting four
blades or three blades two anything as the
number of blades you have to have. So, many
equations then you have fuselage motion then
there is something like let we not look at
individual blade I will start looking at this
as a rotar disk I do not worry about individual
blade I see how the disk is moving
Now, again in the flap motion how the disk
will move in the lead lag motion similarly
that is where from a rotating frame to a non-rotating
frame some transformations only if you make
if you want to analyze certain stability problems
otherwise it becomes complicated of course,
they do some approximations. We all said that
the flap motion is like this if it is cone
in forward tilt it is like suppose if this
itself is oscillating; that means, they there
is an equilibrium position of the flap about
that it is oscillating.
So, now you see my flap motion itself is not
a fixed value it is going to be if I say beta
k which is a function of time I am going to
write it actually not approximation I put
it equal to I will say there is a beta k equilibrium
which can be a function of time plus this
can also we please note that this can be also
be a function of time this can also be a function
of time this is a perturbation. If I want,
if you just the... So, this is the first thing
you looks that a learning perturbation analysis
and your analyzing either rotor fuselage stability
or rotary wing aero-elasticity stability you
have to know perturbation analysis. So, what
is the perturbation analysis you have developed
your flap-lag equations as you know that inertia
load aerodynamic load and root spring model.
We have developed a coupled flap-lag equations,
but those equations should be non-linear then
how you solve stability equilibrium what you
do is you take those equations those equations
may have time dependent coefficients everything
itself non-linear complete my c equation.
If it is a rigid blade approximation those
equations are ordinary differential equations
a rigid blade ODE. ODE of course, non-linear
time varying everything you can add of the
I will tell you one by one. If it is a elastic
blade this is p d e there will be partial
differential equations elastic blade. Now,
what are the various other additional characteristics
if you say, I am just giving you only the
description?
This ODE will be non-linear time varying coefficients
this also non-linear time varying. So, you
have everything you can analyze you look at
the equation you will not know what is what
I am talking about in the sense the coefficients
of you all know that there is very simple
in the ODE may know that these are the x plus
k x f of t right, but you know these are time
varying; that means, they are also functions
of time right in whole only the time varying
is gone, but in forward flight time varying
will appear.
If a, if you solve non time varying simple
case its always hover problem in the earlier
then people are just over case, but then if
a time varying system comes how we analyze
stability even a simple case how do I solve
stability problem in a time varying situations
systems. So, that part I will cover in the
last week that is not really helicopter, but
that is not usually talked in any other course
a time varying systems stability problem you
will develop from systematic things that will
take above 2-Days you will add that there
is something which is totally different from
what you learnt from any of the course.
Now, PDE ODE; how do you… saw which you
were if you are taking elastic blade; that
means, you have to have a very systematic
approach for addressing response stability
problems. The PDE problems are converted into
ODE unless you have studied vibration. Vibration
means not spring mass, it is the theory of
vibration or continuous system vibration.
You will not know how do I convert this to
this; that is you have to learn that something
separately.
On the other hand, you will say I am starting
directly rigid blade that the rigid blade
I assume then I will get the directly ODE
equation straight away I go to ODE now in
the ODE how do I solve. So, the first thing
is all the degrees of freedom that degree
of freedom is flap degree of freedom is lag
if you are solving flap-lag problem each degree
of freedom that is an equilibrium there is
a perturbation. So, you always a flap equilibrium
perturbation and in the lag equilibrium and
perturbation.
This is for isolated blade I am writing. So,
isolated blade means that k will only doubly
index it need not be even there because there
is no 1 or 2 3 you are giving at only one
blade. So, the k does not come into picture
on the other hand if you go to this you will
have k index has to be there. Then how you
treat the class of problems how do you treat
this class of problems. So, isolated blade
the index scale has no meaning only one side
that what you do is you have your equation
of motion that is the non-linear ODE it is
time varying you substitute this is the equation
and then the product of perturbation neglect
it that is if you have a term like this beta
zeta is normally with the… you write it
as beta k plus and then plus theta k t delta
beta k.
The product of these perturbation; you throw
it out you say it is a higher order term I
do not cancel that; then, you collect all
the terms which do not have the perturbation
quantities. Write them, you collect all the
terms which do not have the perturbation quantities.
Write that separately and all the terms which
contain the perturbation quantities except
both of them equal to independently to 0;
then you will get 2 sets of equations. So,
what you started with one set of equation
which is the flap-lag non-linear, everything
you now have 2 sets of equations. One set
of equation is the equilibrium that is the
term which contains this; that is called equilibrium
equation and these are perturbation, you will
have like this. If you first solve the equilibrium
equation to get data and zeta, first of this
can be a differential equation. But, you solve
once you solve that, then you take that than
varying here; then you solve the perturbation
equation first.
This is how the helicopter problems are solved.
Stability is analyzed and the perturbation
equilibrium is analyzed in the actual non-linear
set up and I think this because you cannot
solve any one problem. That is because you
go into the flap-lag stability I have to give
a fully equation; you did the full equations,
then you do it.
The nature of equations, please understand,
the nature of the equation can be in this
non-linear time varying ODE flaps stability
linear time varying coefficients. Now, if
you take the case of hover in condition then
what will happen is the ODE will become non-linear
algebraic equations. That will be this will
not come; so, this will go up.
So, you will have a non-linear algebraic equation;
so, flap-lag in hover to solve that. Once
you know what is the flap equation what is
the lag equation you will substitute here,
you will have the linear stability equation
constant coefficients and you solve for eigen
values. I do not know that the matrix solve
for the eigen values and you will get instability
results. This is the general procedure and
this is consistently whether you do numerically
all this calculation by hand, this is the
procedure.
If it becomes time varying coefficients, how
do you solve the stability? That part will
one approximation where you convert this time
varying into another frame that is called
a multi-blade coordinate. That is not a exact
transformation, but it is the approximation.
Convert the time varying coefficient to constant
coefficient; then you another one is you applied
theory. That part I will teach in the last
week I think. I leave you, if you have any
questions you can ask. With this pretty much
your course, I am stopping.
