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Mod-04 Lec-01 Gyroscopic Effects : Synchronous whirl of a Rotor Systems with a thin Disc

Previous lecture we have seen some simple rotor
model in which we considered the flexibility of the bearing. Specially, when we considered
the hydro dynamic, specially when we considered the hydro dynamic bearing model there is eight
linearized coefficient of the stiffness and damping. We showed that those coefficients,
we can able to calculate at equilibrium positions and that equilibrium position changes with
speed. So, these coefficients in actual practice they will vary with speed, so whatever the
analysis we did earlier that can considered the speed dependency of such stiffness and
damping coefficients of the bearings. Today, we will study effect of gyroscopic
couple on the natural frequency or the critical speed of the system. This gyroscopic effect
generally occurs when the disc is spinning at high speed and it is wobbling. That means
when simultaneously tilting of the disc is taking place during the whirling motion and
because of this particular gyroscopic effect we will see that even the critical the natural
frequency the whirl natural frequency will now no longer will be equal to the spin speed.
So, if we are considering a perfectly balance rotor and if we are operating at certain speed
the whirl frequency may not be equal to the spin speed.
So, for this particular case we will see that as we are changing the speed the gyroscopic
moment changes and that changes the whirl frequency. So, the whirl frequency basically,
now it will depend up on the speed of the rotor itself. Two things we will observe specially
regarding the natural frequency that in this particular case when gyroscopic effect is
there, the splitting of the natural frequency will observed. These particular phenomena,
we will see in the subsequent lecture. Today, we will concentrate on a synchronous whirl
condition in which we will try to calculate the critical speed rather than obtaining the
whirl frequency. So, with this particular module of the gyroscopic
couple now the definition of the whirl frequency, critical speed they will be quite distinct.
Then we will observe that we will be having phenomena of forward whirl and backward whirl
also because of the splitting of the natural frequency. Corresponding we will be having
critical speeds also which will be having forward whirl and backward whirl in which
case the spin direction or the spin sense of rotation and the whirl sense of rotation
may be same or different. So, for this particular case, this is the
overview of the lecture in which we will be doing the analysis of a simple cantilever
rotor within a thin disc at the end. We will not consider the mass of the shaft. Already
the elasticity of the shaft will be considered and disc will be considering as a rigid, it
is having no working of the disc in this particular analysis. In this we will see the gyroscopic
couple effect onto the critical speed. Basically, we will be calculating what is the critical
speed, because of this gyroscopic couple is… Some concepts which we will be covering in
this lecture is the gyroscopic effect disc effect or the diametral mass moment of inertia,
how it changes in the critical speed of the rotor system? Now, with some simple example
let us see when this gyroscopic couple will be predominant. Let us take a simply supported case in which
this is the bearing axis and we have a rigid bearing that is simply supported condition
and this is the shaft elastic line. Let us say disc is at the center of the mid span
of the shaft. So, during whirling we will observe the disc would not tilt. It is spinning
about its own axis with omega, but disc tilting will not take place because it is there at
their mid span and at mid span the shaft slope is 0.
So, during whirling of this, there will not be any tilting of the disc would take place.
So, once this disc is not tilting about it diameter there will not be any gyroscopic
couple effect on to this particular system. But if the same system is there with the disc
at offset position, so let us say here in this particular case because the disc at any
particular position should always be perpendicular to the shaft elastic line.
So, we will find that the disc will tilting about its one of the diameter in this particular
plane, so not only we are having spinning of the shaft, but the disc itself is having
some kind of precision frequency. That is the whirling frequency of the disc. So, this
wobbling motion of the disc along with the spinning of the shaft would give the gyroscopic
couple, so we expect this gyroscopic couple effect will be present in this particular
model. Another case if we consider, let us say a
cantilever shaft and this shaft is mass less, only elasticity is there. Let us say disc
is in the form of a point mass m. If this is spinning at certain frequency during whirling
we expect this will occupy this position, so this is one of the cases. If this particular
cantilever beam is having a large thin disc of the same mass, but having diametral mass
moment of inertia of the disc in this particular because this is point mass the material mass
moment of inertia will be nearly 0. When this is spinning we will find that during whirling
this disc would be tilting about its diameter. So, in this particular case we expect a gyroscopic,
so you can able to see there will be precision frequency of disc would be there.
So, if we consider this particular model of the rotor and in this in which the mass at
the free end are same, but only difference is here its point mass here it is disc. So,
we expect the critical speed of this system and this system should not be same. Especially
when this particular speed is high we will find that the gyroscopic couple will make
the critical speed of the system and this system is different.
So, with this example it is very clear that when the gyroscopic couple will be present
in the system and when it will not be present. So, when point mass is there in a cantilever
beam we cannot expect a gyroscopic couple in that because its diametral mass momentum
of inertia is 0 now. Now, we will consider a simple cantilever beam we will not be considering
any unbalance of that in that particular disc. We are assuming that there is no unbalance
in the disc and we are operating near the critical speed of the system, so that the
natural frequency or the whirl frequency is equal to the spin speed as I mentioned. When
gyroscopic couple is there the whirl frequency and the spin speed may not be same. In this
particular case we are considering a very special case in which we are trying to find
out at what whirl frequency the spin speed at what speed it is the whirl frequency is
equal to the spin speed. So, that means we are trying to find out directly the critical
speed of the system and in this particular case let us try to analyze a cantilever beam
with a large disc at the free end. So, if this is the disc this shaft is spinning
and disc is also spinning with the same speed and because of this we will see that for this
particular configuration. Let us say when the shaft is at the bottom most position during
whirling, if you try to see for this particular instant how the centrifugal force of various
masses of the discs are acting. For this particular instant we will see that the mass element
which is away from the spins axis will be having large centrifugal force because this
be lesser and lesser like this. Now, if you see the effect of this centrifugal force what
they are trying to do to the disc? They are trying to apply a couple which is acting in
counter clockwise direction. So, it is trying to orient the disc to its
original vertical position because the tilt of the disc during whirling is this angle
and this particular moment is trying to oppose that tilting which is coming from the centrifugal
force. So, we will see that because of this moment, effective stiffness of the shaft against
the rotation is becoming higher because this particular moment is trying to prevent the,
it is making this particular angular displacement lesser and lesser because of the this moment.
You can expect that when the centrifugal force is high at if we are increasing the speed
as we are increasing the speed this will be more and more and we expect this moment will
be more and because of that effective stiffness of the system increases.
As we know from fundamental that their natural frequency is defined as root square k by m,
so if stiffness is increasing mass is same. So, we expect that the natural frequency of
the system should be increasing. So, we will find that in this particular case when we
are rotating the rotor at certain speed. We are trying to find out the condition for at
what speed the whirl frequency is equal to spin speed for a balanced rotor. The effective
stiffness is increasing and because of that the natural frequency is also increasing.
So, this particular phenomena which I have illustrated through the centrifugal force
concept is we will see in more detail through the concept of the quasi static analysis.
We will be finding out what is the effective the force and moment this particular centrifugal
force are applying at the free end of the shaft. From there we will try to get the frequency
equation and calculate the critical speed of the system. Now, we will analyze how the
whirl frequency of the system increases with the spring speed of the shaft because of the
centrifugal force. Let us for that particular case, we will take
the same example and now we will see this particular disc from side how this particular
system is giving us the. So, let us say this is bearing axis B to and this particular axis
bearing axis we are representing as z this angle is or let us say this direction is y.
So, this angle we will represent as phi x this is the static deflection let us say delta
of the centre of the shaft on the disc. So, here let us say the intersection of this axis
to the disc is I am writing as B. This is the shaft centre which is coming from here
and this B to C is distance that is translating displacement of the shaft from its equilibrium
position of the bearing axis. Now, if we consider a small mass let us say
this is at reduce r 1 from the bearing axis. So, if we see the motion of this particular
disc and as I told we are looking into the condition of synchronous whirl. So, let us
see this particular motion what it represent, so in the synchronous whirl condition the
spin speed that is the spinning of the disc and the whirling frequency. Let us say this
is the cantilever beam, so this whirling frequency and the spin should be same now when we are
considering this synchronous whirl condition. So, what will happen for this particular case
the shaft will bend? Once it is bent, it will be rotating about bearing axis in the rigid
body mount and without bending of the shaft. So, bending of the shaft if you see, the fibers
in this particular shaft which is in tension will be always be in tension during the whirling
and which is compression will always be in compression during whirling for the synchronous
whirl condition. So, it will be as a rigid body, it will be rotating about the bearing
axis. So, that is the synchronous whirl condition
and in this particular case we see that each and every particle on to the shaft will be
making a, if shaft is symmetry again circular path. It will be making a circular path about
the bearing axis. Also on the disc, if you take any element or the mass that also will
be making a circular path may be the radius of the circular path will be different for
each of them. So, in this particular figure the r 1 which I have shown here, so this particular
disc during the synchronous whirl will be making a circle with r 1 as a radius and bearing
axis as the centre. So, it will be making a circular motion, so we can expect that because
of this the centrifugal force which will be acting on this mass.
Let us say this mass is small d m, so this will be along the direction of the r 1 only.
So, let us say this is centrifugal force and d m omega square r 1; this is acting in along
the direction of r 1, now this particular motion of the whirling again. Let us see this
particular whirling motion, so this whirling motion, which I told for synchronous whirl
is taking as a rigid body about the bearing axis, so and it is spinning also. Both are
taking place because of that synchronous whirl is taking place.
If we consider this motion in two parts, one is pure spinning at omega RPM or radians per
second and then without rotation only whirling, so basically in actual case both spinning
and whirling take place simultaneously. But now, I am considering spinning separately
and whirling separately, so in this particular case the disc is not spinning only it is whirling.
So, in this particular case we will see that the two motions which was spinning and whirling
is simultaneously which was taking place beyond split in two parts, so when we are considering
the spinning. So, we can able to see that this particular
mass for that particular rotation when we are considering pure spinning, it is spinning
about the centre of the shaft, so that means we can able to take the centrifugal force
because of this. Let us say radius of the disc is r, this will be d m omega square r
and another case is this one, the pure whirling. In that particular case we expect this particular
mass and this particular configuration will be again.
If I can able to see here this when pure whirling is taking place, so when disc is coming from
let us say top to bottom. So, when it has reached at the bottom the direction of the
centrifugal force of that mass due to this whirling will be toward the downward direction.
It is coming, so at this position the centrifugal force should be downward direction that means
vertically downward. So, this component is that particular direction and this will be
d m omega square delta. Now, when the disc has reached at the bottom,
each and every particle of the shaft is having delta displacement. So, we will be having
this particular force centrifugal force, so you can able to see that the total displacement
which is total centrifugal force which is d m omega square r 1. Summation of these two
components due to pure spinning and pure whirling this particular forces which now we have got
now we will take the forces component of this in two planes.
We will try to find out because of these forces what are the resultant force and moment which
is actually coming on to the rotor system for which I am drawing, let us say in one
of the plane. Let us say this is z axis, bearing axis and y vertical axis and disc is this
one. So, this is the disc already I am showing it and let us say there is a particle here
having mass d m. Let us say that point is P and this is the shaft which is attached
to the disc. This is the shaft now, this is the centre of the disc, C here tilt is about
x axis of phi x. Now, this is the case for a pure whirling case, pure whirling motion,
there is no spinning. So, if you see from side of the spin of
the disc let us say this is y axis and this
is x axis. Particle P is here is at a radius r
and this is the centre of the disc, now if
you see this particular force which is acting downward direction. So, for this particular
configuration not only this mass, each and every mass will be having same inertia force,
because each and every particle of the disc is having delta displacement.
So, this particular mass d m which is there at p will be having a force m omega square
or d m omega square delta. So, each and every particle will be having same force. This particular
force on this plane is in this direction, so this is d m omega square delta. If we considered
another point vertically opposite to this here P prime, this will also be having same
direction force same magnitude, because that particle is d n here P prime. If we consider
it will be having downward direction d n omega square delta.
Now, you can able to see that if we consider all the masses in the system they will add
up and they are not cancelling each other. Basically, all forces are downward, so if
we add all such forces, this is for the elemental mass. If we add for all such masses of the
system will get in net force and the direction of that force will be in the y direction.
So, this is the net force which will be acting due to the pure modeling case pure modeling
motion regarding the in this particular case there are adding up to give the force.
In this, what about the moment you can able to see this particular particle and this particular
will produce about the center moment, which will cancel each other because they are opposite
to each other, So there will not be any net moment due to this or here also, if we try
to take moment about this point this two will cancel each other. So, the net moment due
to the whirl motion will be 0, so moment will not be there due to the whirling motion. Only
this particular total mass of the disk in to omega square delta force will be acting
on this particular disk. Now, let us see the pure rolling motion, pure spinning motion,
now will consider the pure spinning motion of the disk. So, in this particular case I am again drawing
the free body diagram of the disk in, let us say z y plane. z is the bearing axis, disk
is here, shaft is this one. Now, if we take the diagram in other plane will see the disk
like, this is center of the disk C, these are the y axis and x axis. P particle is let
us say here and radius of that is r. Let us say this inclination with respect to axis
theta here. Because of the pure spinning we have centrifugal force omega square or d m,
which is acting readily out ward. This particular force we can able to take
component in two directions, this particular form here we can able to relate let us say
this distance is y and this distance position from the position of the mass is x. So, we
know that x is equal to r cos theta and y is equal to r sin theta, so the component
which is there in the horizontal direction here will be having value omega square r d
m cos theta, this will be acting in the x direction. If we combine the r in cos theta
this can be written as omega x d m. Similarly, in the y direction force component
will give us omega square r d m sin theta that will give us omega square y d m because
half n sin theta is y. Now, this particular force in this plane if we want to see the
y component is acting in the direction of y x. So, if the component is acting in the
direction of x axis, so if a particle is here the force one is there in this direction.
The magnitude is omega square y d m other force the x direction, x direction is that
is perpendicular to the screen because according to the right hand rule I will be inside the
screen, x will be inside the screen. So, this particular force component which
is in x direction will be in the plane of the disk, so we cannot able to see that particular
force here that will be on the plane of the disk, but the y component is along the y direction.
So, we can able to see that this particular component is not on the plane of the disk,
now this particular distance we can able to see that this is y distance this angle is
phi x. So, this distance will be y into phi x because
this is y and the angle is phi x, so this distance will be phi x, so the component of
this particular force will be making y phi x distance from the x y axis. Now, if we see
this particular particle x component which is this direction if you take a particle here
opposite here, so this particular force and this particular mass force will balance each
because they are in the same plane. If we consider another point, let us say below
this particular force in the y direction and this particular force if we consider here
P prime let us say which is below this will act because spinning, so it will act in this
direction. The magnitude of this will be same as the upper one, so you can able to see that
they will make a couple, they will balance each other, but they will make a couple and
the magnitude of the couple if you can able to find out will be. So, let us say for single
mass d m the couple about the bearings the shaft centre will be that is in y z plane,
omega square y d m this is the force and momentum is y phi x this is the momentum this one.
So, this is a couple due to mass d m, if we sum such moments we can able to get the total
moment due to the various forces and that will be giving us this and because omega is
constant it will come out this is also independent of this integration. So, we will be having
y square d m. We know y square d m is nothing but diametral mass moment of inertia of the
shaft, sorry of the disc about its diameter, because in this particular case when we are considering
the pure spinning I am let us speak on here. So, for the case of pure spinning we found
that the x component forces are cancelling each other whereas, the forces in the y direction
they are giving a pure moment and that moment is the gyroscopic couple moment. In the previous
case, the pure whirling case we saw that the net force centrifugal force, we were getting
so effectively, because of the pure spinning and pure whirling. We got one force and one
moment and now we will be using this to get the frequency equation. So, from the previous analysis of the pure
whirling and pure spinning if we remove the disc on to the shaft, we will be having inertia
force
and in moment
and direction of moment. In the previous slide, we can able to see is this particular force
is producing a counter clockwise in y z plane. So, that particular direction we have retained
here now effectively. Now, our problem is to, you have converted the dynamic forces
into a static force, so that means basically we have converted a dynamic system to a quasi
static system. Now, we will be obtaining what is the deflection and angle because of these
particular forces. From the strength of material deflection theory
we have such relations like for displacement, the linear displacement is given as because
of the force
and because of moment. So, this is the linear displacement and force and moment if we are
applying, we will get this relation. Similarly, for angular displacement we have relation
F y l square to E I then due to moment because we are dealing with the linear system. So,
because of this moment and force displacements can be added up to get these relations. So,
this is coming from the deflection theory of the strength of material, so any strength
of material book can be referred for such relations.
Now, we need to substitute the centrifugal force and gyroscopic moment in this expression,
so we can able to substitute like this gyroscopic momentum. Hence, similarly in the deflection
rotational displacement. Now, we will see that these two equation this equation one
and two basically each and every term. They contain either linear displacement or translatory
displacement or angular displacement. Here also delta is there, here also phi x
is there, so all terms containing so basically, they are homogenous equation. We can able
to rearrange these equations like this m omega square l cube by E I minus one delta. So,
I have taken this in one side and I have taken the delta common plus minus I d omega square
l square by 2 a phi x is equal to 0, so this have to obtain from equation one. Similarly,
from equation two we can able to get delta plus I d omega square l by E I minus plus
one phi x is equal to 0. Now, express this series equation three and four, so equation
three and four now you can able to put any matrix form. So, we are able to write this as y phi x and
because this homogeneous equation right hand side is 0, so here we have m omega square
l cube by 3 E I minus 1. First term then minus I d omega square l square by 2 E I and here
minus omega square l square by 2 E I and here I d omega square l by E I plus 1. Now, this
homogeneous equation one solution is when both displacements are 0. So, that is the
case when we are not considering the motion at all, so for non trivial solution for non
zero displacement, the determinant of this matrix should be 0. If we put the determinant
of this matrix 0 will get a polynomial in terms of the omega which is the critical speed
of the system because we started with whirl frequency is equal to omega, so we were looking
for the critical speed of the system directly. So, we will get a polynomial in terms of omega
four basically it is a quadratic n omega square, so this equation will be after rearrangement
of this equation this is square term then a constant term for equal to 0. So, this is
a polynomial and we can close the bracket we can able to solve this for omega square
the polynomial which we have got is basically, a frequency equation. It could able to solve
in terms of omega square, but in this particular expression several variables are there.
Now, we will try to reduce the variables, so that we can able to do better interpretation
of this particular equation, so I am defining two terms non dimensional terms. So, first
is the critical speed effect
critical speed function, so that is non dimensional term defined as spin speed m l cube by E I.
Another is a mass effect disk, mass effect this is mu that I am defining is diametral
mass moment of inertia divided by m l square. So, this also this two are the non dimensional
terms we can able to use this to rearrange this polynomial and if we do it will get a
much simpler expression
in terms of the non dimensional parameter. We defined this, so this is much simpler and
in this you can able to see there is only single disk parameter is there, this are included
in the non dimensional terms. Now, in the solution of this, we can able to get the disk
can be solved in close form, so the solution will be of this form. We are getting two roots
at this, but if we see carefully this expression within the bracket term, with in the square
root term this particular disk effect is positive, so this will be positive. This is a positive,
so we can able to see that this term, which is within square root is always greater than
this one. Now, if we take the negative sign, we will
get this particular critical speed as imaginary quantity, so that is not feasible one. So,
we can neglect the negative one and will consider only the positive sign here and with this
a phi plot the omega critical verse mu at mu is equal to 0. Let us say, so for mu is
equal to 0 we cannot able to solve from here we need to go here, we need to multiply each
quantity by mu, so you can able to see that this term will vanish, this term will give
us for mu is equal to 0. From the frequency equation will get 4 omega
bar critical square and we get minus 12 is equal to 0. So, omega critical square is equal
to 3, so here for 0 mu we are getting value of this is 3. Another case when mu is infinity
tends to infinity, we can able to see that this terms will vanish and for infinity. If
you simplify will get this as equal to 12, so that is here let us say this is twelve,
so mu is equal to infinity means, very large disk or the disk with large mass moment of
inertia, because the material mass moment of inertia because mu defined as I d by m
l square. So, if this is large then this should be very
large, so for that particular case if we plot this particular equation, we will get a curve
like this. Asymptotically, it will disk critical speed will be equal to infinity even will
be equal to 12 when this particular frequency ratio this disk effect very large. So, you
can able to see if we see this one when mu is 0 what is this case mu is 0 means if you
see here I d is 0. So, there is no mass moment of inertia of the disk that means that particular
disk is a point mass, so this is representing is a point mass disk is when is having infinite
diametral mass whirl of inertia. So, inertia will be so high the disk will
not tilt in that particular case and this the general case of critical speed in which
it is varying with the disk effect. So, if disk effect is more and more if diametral
mass moment of inertia is more and more will be seeing the critical speed is increasing
continuously, but it is having some maximum value is equal to 12. The critical speed is
square is 12, this non dimensional parameter. Today’s lecture we have seen that how a
critical speed for synchronize whirl condition of a cantilever been, we obtained with this
particular case we have consider the a very special case, in which the not only the motion
is the synchronize fault that is the spin speed and the wall frequencies are same they
are direction is also same. We have seen that as we are increasing the disk effect we are
increasing the gyroscopic moment and because of that we are having increase in the critical
speed. So, as we have seen at the beginning of the lecture the centrifugal force tries
to kill the disk to its original position. So, it effectively increases the stiffness
of the system and because of that the critical speed should increase. So, finally we have
seen that that particular effect as we are increasing the disk effect the gyroscopic
moment will be more and more and will be having increase in the critical speed of the system
due to gyroscopic effect.

4 comments on “Mod-04 Lec-01 Gyroscopic Effects : Synchronous whirl of a Rotor Systems with a thin Disc”

1. Sami Meshal says:

Thank you so much NPTEL .. This lecture was very helpful ..

2. GERALDO CARVALHO BRITO JUNIOR says:

Very nice lecture…

3. Saurabh Pandey says:

i thought teacher is reading a script with no emotion..like machine