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Mod-05 Lec-06 Geared and Branched Systems with English subtitles   Complain

Today, we will study geared system and branched system, this is special kind of rotor system

on which we rotor with single shaft, but multiple shafts are there. They are connected by the

gears either reduce the speed or to increase the speed of the shaft. In this particular

case, because the shafts are rotating at different speed, we need to take a special care for

such system. Today, we will see in detail how we can able to analyze gear system and

expansion of branch system in which from one particular point multiple shaftings are connected

through gears.

So, let us see what are the things that we will be covering, so for gear system, basically

we will try to convert gear system to equivalent single shaft systems. So, that whatever the

previous analyses we did we can able to apply for this particular system for branched system,

because multiple shafts are connected at one particular junction.

So, for this particular case we will use transfer frequently and subsequently we will be using

finite element method in the substituent lecture. So, basic concept which we are introducing

here equivalent polar mass movement of inertia equivalent torsion stiffness junction point

for branch system gear ratio.

So, let us start with a branch system a gear system. So, in gear system we have one shaft

which is connected by thy wheel fly wheel and if we rise as a gear which is connected

to another gear and that is gear to another shaft and here another fly wheel is there

let us say polar movement of inertia is I P 2. So, basically import is form here and

we are getting output from this particular shaft system, let us say this stiffness of

this particular shaft which is between gear and this fly wheel is 2. Now, let us say we

are mentioning this as shaft 1 and this is shaft 2 and in this particular shaft let us

say is having nominal speed of omega 1. Apart from this because of the polar movement

of inertia, it is having some torsional oscillation. So, let us say that fe z 1 and because it

is transmuting power, so is having torque T 1, these are just before the this particular

gear. Basically, this we call it has premium and this gear and this is gear, now if we

to analyze this for caution vibration and we can convert this particular shaft system

to single shaft system. Let us say we have this particular movement

of inertia first fly wheel and another fly wheel is here another and they are connected

by uniform shaft single shaft. So, we will try to convert the previous actual system

to this equaling system in which single shaft is there. If we want to convert we already

such system in which this shaft system mounted on stiffness bearing we can able to obtain

the energy frequency for torsion for this particular case. So, basically you can able

to visualize this as if we enclose this particular gear system including this second shaft and

the second fly wheel into a black box kind of thing and same thing

is here. So, basically this particular polar

movement of inertia will not be same as previous one now we are talking about equivalent system.

So, let us say this is P equivalent I P equivalent and this shaft which was having stiffness

kg two let us say that we are mentioning k t e here we have up to this point gear. So,

before this whatever the stiffness is the as the same as previous one we are not converting

this particular shaft to equivalent system. So, basically we are trying to find out this

equivalent type of this particular shaft system and this polar movement of inertia with respect

to the output speed of the gear system. In this particular case, we are not considering

the polar movement of inertia of this gear we are neglecting them, but if we want to

consider then additional polar movement of inertia will be coming here.

So, that will be give us the system as three degree system from the two degree for simplicity

we at present ignore this, but that can be added subsequently. So, here our aim is to

obtain this equivalent system which we if we can find out what will be the new polar

movement of inertia of this fly wheel and what will be the new torsional stiffness in

this particular shaft. In this particular equivalent system in which single shaft is

there, so basically the principle will be will be equating the strain energy of this

two system the actual system and the equivalent system.

The kinetic energy of this actual system of the equivalent system, so this two energy

should be same then we can call these two system as equivalent system, so let us begin

with strain energy for calculation of strain energy. So, what we will be doing, we will

first we assume this particular disc we are giving we are attaching we are giving the

0 displacement means we are fixing this particular movement of inertia also here also in this

particular fly wheel. Now, we are giving input to the shaft 1 here and same input to the

shaft here also we are giving fe z 1, so you can able to see that because of that the effect

of the gear ratio.

So, the gear ratio is nothing but input velocity divided by the output shaft velocity, so in

our case if we are talking about angular displacement. So, this will be fe z 1 divided by fe z 2

or it will be equal to omega 1 nominally speed or the gear. So, let us take the angular displacement

as the gear ratio in which the input angular displacement divided by the output angular

displacement.

So, we can be able to see that when we are giving fe z 1 input to this particular gear

will be having opposite motion and the front angular displacement. Similarly, if we want

the nominal speed they will be different top will also be different, so these are all at

the location of the gear that is fe, that is the gear. So, you can able to see that

if we want to equate the two in this system for a particular input, so we giving a fe

z input, let us say we are giving a twist of fe z 1 s in the both the system and we

are trying to find out the strain system in the energy.

So, for first actual system we will be having strain energy that will be because the second

shaft is having the stiffness 52 and its angular displacement will be this square. So, this

strain energy system in the actual system fe z 1 by 1 full square, this is for the actual

system for equivalent system similar strain energy we can obtain that is because the tertian stiffness of the equivalent

system is k t e. Here, if we again see whatever the angular

displacement giving at this point the same displacement this particular shaft is getting.

So, here we will be having strain energy corresponding to the same displacement, now this two energy

because these two systems are same these two energies. We quit because these two energy

are same, so you can able to see if we equate this two we will get k t e after the cancelation

of angular displacements as k t 2 divided by n square. So, this is the equivalent stiffness

of the shaft new shaft which is given as k t 2 divide by gear ratio square.

So, you can be able to see that this particular torsional stiffness we can obtain by using

this. Now, in second case will release this fly wheel as we did earlier we fixed it, now

we are assuming they not fixed, but now we are considering the shaft as a recite for

second case, these two shaft are recite. That means whatever the displacement we are giving

here this particular fly wheel will be having same displacement because now we are considering

the shaft as recite.

Now, we will be obtaining the kinetic energy these two systems, so let us say for actual

system we have kinetic energy I P 2. The angular velocity, let us say this and for the equivalent

system the kinetic energy is I P equivalent, let us say the angular velocity of this. Now,

this angular velocity which we have it is having two component 1 is corresponding to

the nominal speed omega 2 of the second shaft plus angular displacement due to the tertian

angular velocity to the torsion. Similarly, the equivalent angular velocity or the angular

velocity of the equivalent system will be omega 1 because there is no gear in this.

So, whatever the input velocity will be transmitted to this particular shaft and then cautioner

displacement or cautioner velocity that will be also be same. Now, this particular angular

velocity we can able to we can able to substitute this in this equation and we substitute. So,

we will be having basically these two kinetic energy are the same, so we can equate them

also, half P 2 omega 2 plus fe z whole square is equal to half I P e omega 1 plus fe z 1

of whole square. Now, this displacement we can able to write

for pi z 1, we can able to write it as in terms of torque and the this for first system in which torque

this is the torque and the divided by the torsion stiffness. Then, angular displacement

then for second system and the actual system T 2 divided by k T 2 and using the gear ratio,

we can able to write this as in terms as 182, this displacement we can able to substitute

in this expression.

So, once we substitute this expression in kinetic energy of both the system which we

equated earlier, we will get omega 1 by n time derivate to of the angular displacement

which we are replacing by the torque divided by this. For the actual system, this is actual

system, this is the actual system I P 2 and this is I P e, this is for the equivalent

system whole square. So, basically we have done here we have substitute this and this

quantity here and we use the gear ratio to convert everything to convert omega 1 fe z

1. Now, you can able see that once we take one

by n common we will get I P 2, if we take 1 by n common, this will be because n by square.

We will get one by n square omega n plus n by 1, 2 here will be having T 1 by k 2 m square

is equal to I P e omega 1 plus d 1 by d 2 T 1 n square by k t e. Here, we have converted

this also into k T 2, k t e is nothing but k 2 divided by n square n 2, here now you

can able to see this quantity is getting cancelled. We are getting equivalent polar movement as

this one, so just we need to divide the gear ratio square to the actual movement polar

movement inertia to get this.

So, once we have done this, we can be able to see that, now we know this and this and

this particular shaft if we reserving the stiffness, this is k T 1, this is the two

shaft segment. Basically, they are connected in series, so we can able to find out the

total stiffness of this and then we can able to as we did earlier as for single shaft system.

We can be able to analyze for torsion vibration, so I am not repeating that particular things

here, only things will be molded sheet.

So, this the equivalent system two rotor system which we obtain, so if we are getting more

shape let us say one of the mode shape will be having one node here. So, let us say this

gear possession gear position is here and the node is here, so this is for the equivalent

system if we want to obtain the node shape for the system. So, what were the displacement

are there, angular displacement because this is nothing but angular displacement of z 1

fe z 2, so angular displacement should be divide by the gear ratio. So, we can see that

after the gear this is the gear location or to the displacement will be divided by n,

so again the respect will be change in the slope of the line and this particular thing

is nothing but fe z 2 by n. So, the mode shape will be solid line up to

the gear and this dotted line will be the modified node shape for the n shape all the

analyses will remain same as the previous one. Now, we will analyze the branch system

which is an extension of the gear system in this particular case we will find that at

one particular junction there will be more than two B s are attached. So, the power which

is getting transmitted from one particular shaft it is transmitted to are three many

shaft generally such application. We find when we are having one main motor and it is

driving several other shaft and this motions are given by the various gear.

So, let us see how this particular system can be analyze, let us see first what is the

branch system, so we have one shaft system in which we can have multiple disc. So, let

us say we have one 0, 1 several disc are that is fe number of term in a particular system,

so this is the single shaft system, now we are having another branch here. So, this gear

basically which is connecting another shaft system, so let us say this particular branch

we are giving the name as branch A this is and this is branch C and in top there is another

branch which is let us say branch B. So, basically this is pinion and is driving

this gear as well as this gear, so these are various fly wheels this or a disc in a rotor

system and we can have like for branch B, we can have q number of disc. Similarly, for

branch C we have the branch system of let us say R number of disc so there are so many

number of disc particular system, but main thing is this branching. We can able to see

that from one particular shaft the power is getting transmitted into two shaft we can

have more also, but for illustration we are showing two because this is a this is a multi

rotor system. So, will be using the transfer this method

and for this we have already obtain like state factor, let us say P th station of branch

a right side of the disc, so basically particularly this state factor is here. This is the branch

B and fe th station right side of the this particular gear is this particular state factor

we are relating this state factor with state factor at zero station, let us say overall

transform matrix this is A. So, this is a state factor at zero station at branch A,

so basically this equation representing the whole branch a similar expression we can able

to write for branch B. In this particular case, we have q number

of things, so this is equal to let us say overall translation of this is B and this

is zero station and branch B. Similarly, we can able to write for branch C, so write of

R th disc of branch C is equal to let us say that C is the overall transfer matrix of trans

C this is the state factor of zero station so here is the state factor this is right

of R th. So, you should write it correctly so this

state factor is right of R th disc, so these equations basically representing the bond

equation for the three branches, but they are not been connected as they are connected

by gear. Now, we will be taking one at the time these equations and we will try to relate

them basically we try to solve the bond three conditions which is there here or here or

here or at the junction point. These rotors basically we are assuming them as we did earlier,

they are mounted on frictionless support, so we can have the torsional motion like this.

So, let us start with the for branch A we can able to expand this as like this, let

us say overall transfer matrix for one I am representing as a 1 1 and a 1 2, a 2 1 and

a 2 2 fe z and torque at zero station of branch A. Now, the bonding condition at zero station

is that this particular torque is 0 because particular and of the shape of the pi torque

will be 0. Here, this placement will not be 0 and for simplicity because we are analyzing

the free vibration we are taking this particular station angular displacement as unity and

obtain this related what are the displacement of the others.

So, we will equate this equation one and this according to bonding condition bond zero,

so you can able to see that if we expand this we will get fe z and A as a 1 1 and second

equation a 1 1 as a 2 1. Now, we will take the branch B, sorry this is fe, these are

fe is equal to the fe number of discs are there in the branch A and in this we have

q number disc of b 1 1, b 1 2. These are the overall translation for branch B here torque

is there and here angular displacement is the illustration of branch B at junction point

between gear A and gear B.

Let us say that gear ratio is n A B, so that is nothing but fe z 1 that is P th station

of branch A divided by fe station. This is fe z of branch B, so this input this is output,

so you can able to see that.

You can able to replace this particular displacement is pi z divide by gear ratio that is A to

B and this we already obtain earlier that a 1 1. So, this can be written a 1 1 divide

by the n A B, if we see the other bonding condition of the branch B at this station

we have torque 0.

So, that means here torque is 0 and is we expand we can able to get pi z q B, this is

the state angular displacement of branch B at q th station is equal to B 1 1 first equation

A 1 1 divide by and A B plus B 1 2 into T naught B. So, this is the first equation second

equation will give us 0 is equal to a 2 1 a 1 1 by n A B plus a 2 2. So, we have two

equations, now we can able to see that we can able to this particular equation we can

solve for T naught B and then you can able to substitute this in the first equation.

So, we are solving for T naught B from previous equation, so we can able to see we will be

having minus 2 b 1 a 1 1 by n A B and then this will go in denominator b 2 2, this is

we got from second equation we substitute here. So, we can able to get fe z q B s B

as 1 1 a 1 1 divided by 1 A B, then b 1 2 and this T naught B will give us minus b 2

1 a 1 1 divided by n A b 2 2. We could able to get angular displacement at the branch

system at q th station that is the far end of B of this one. Now, we can able to take

up the third branch for third branch we can able to write the expanded form that is C branch.

Here, we have R C is equal to c 1 1 c 2 2 c 2 2 fe z and T and this is the zero station

of the branch C. So, here the bonding condition is because this end is free and this is 0

and gear ratio between branch A and C, we can able to write this as has we did earlier

for this here the gear ratio will change A 1 1 divided by A to C.

Earlier it was A to B, now we can able to see that is A to C can able to express again

the expressions fe z or C is equal to c 1 1 A 1 by n A C plus c 1 2 T naught c is the

first equation. Second equation will be zero c 1 1, a 1 2 n A C plus c 1 1 T not C, so

these are the two equations again we can able to see that we can able to solve for these

and you can element T naught C from here. Now, let us see apart from this at present

we could able to relate the branch A, branch B and branch B and C using the gear ratio.

We apply the bonding condition, now another condition we are not exploited that is the

junction point whatever the work will done we will be doing at the shaft a will be equal

to the work done at shaft a B and C. So, let us see this particular bonding condition or

this particular condition how we can able satisfy and once will satisfy this we will

get the frequency equation from the formulation.

So, this the last condition we need to satisfy at the junction, so this the work done at

branch A should be equal to the work done in branch B and C and you can able to solve

this for this particular torque. So, I have divide by the angular displacement of this

and this angular displacement ratio are nothing but the gear ratio so we can able to write

like this. Now, I am solving for T not C this expression T not C is multiplication of this

quantity and if we take in other side the quantity will be negative earlier we obtain

T naught B, so I am substituting this we obtain earlier T not B we obtain earlier.

So, substitute that expression well and T naught this particular we obtain earlier,

so that I substitute as a 2 1, so this A to C. This is the equation for branch C which

we earlier wrote this was the bonding condition this using gear ratio we converted and this

was T naught. So, basically this expression we are writing and now you can able see from

the second equation if we expand we will get this expression which is not containing any

other state factor like fe z or torque. They are containing the terms related to the

overall transformation of branch a B and the gear ratio so basically this transform ties

officiate they are the function of omega. So, this frequency equation basically and

we need to solve this using some kind of empirical technique which we have already explained

earlier frequency equation and this particular case we will be having many natural frequency.

So, in the branch we had P number of disc branch B we and q number of disc branch C

we had R number of disc. So, total pinup of degree x, but the angular displacement of

the branch B at this point is related by gear ratio. Similarly, for branch C is related

with this so that means two angular displacement are related, so the degree of film will be

that much less because we have two additional constant.

So, these many natural frequencies we expect from the previous frequency equation and once

we get the frequency equation we can use transformative to get the relative displacements. This place

mode shape for the corresponding to each of the natural frequencies, let us take one example

for branch system in which whatever the method we explain will be more clear.

So, here we are taking a branch.

The branch system is similar to the previous one only thing for the analysis only, we have

taken the only the one disc in branch A, similarly another disc branch this particular branch.

So, we are not using the multiple disc only single disc we are consider each of them and

in this particular case these are the polar movement of inertia various disc is this polar

movement of inertia gear. We are neglecting and the gear ratio between branch B and C

or between the B and C three that means B and C. Similarly, between B and D gear is

four various length are given of the previous of the figure diameter of the shafts are given

at the material quantity of the shaft is given.

Now, we can able to obtain the various the property of this particular problem, so this

are the polar movement of inertia three fly wheel which is given to us second movement

of area can be calculated or to the torsional stiffness of each of the shaft segment. We

can able to obtain this two shaft segment having same property so they are stiffness

is same.

Now, we can able to write the vector branch a in this particular case we can able to see

that overall transmuting two component what is the odd matrix and the field matrix and

we know the property of this. So, we can able to get the A matrix you can able to see that

this contain unknown as natural frequency, this is for branch A similarly for this particular

shaft third here this particular shaft segment. So, this state factor is the right of the

that particular shaft segment this is the overall transition for that branch this also

only point and filled matrix because we are not considering the inertia of the gear. They

can be obtained, so this is the for that particular branch overall transfer matrix for second

for third branch the lower one that means if we see the figure this particular branch

again we are writing overall transfer matrix.

So, this is also having one matrix and one field matrix and they can be multiple and

we can now obtain all the branches, overall transfer matrix.

Now, the frequency equation derived value was of this form of three branches and these

are various components of the matrix. So, this form a matrix first row first column

this is C matrix second one second column like this, so we can able to substitute this

various storms from three matrix which we obtain like this so we will get expression

like this. If we simplify, we get a polynomial of this form and you can able to see that

one of the natural frequency is 0 and other two natural frequencies will be getting quadratic

part.

So, one natural frequency is 0 another two are two natural frequencies corresponding

to the flexible mode which we are getting from this. Basically, we have three natural

frequencies, one of them is 0, natural frequency which is corresponding to recite body mode

and these are flexible mode we take up another example of a simple gear system in which two

gear are there. In this particular case we will obtain the frequency equation natural

frequency system of the torsion natural frequency of the system.

In this case also you can able to see the gear system in which this is the one of the

fly wheel and this connected by this particular gear to this gear this is the another gear.

The various property of this a gear and in this particular case these are the fly wheel

and in this particular space, we consider the inertia of the gear also and other property

of the shaft A and B are given here and the metallic company of the shafts are given.

Now, in this property we can be able to now I am writing the transfer matrix that is can

able to see is the from shaft A 0 th station of shaft a and is the right of one station.

This is the zero station of shaft A this is the 1, so we had related the state factor

which is here right of A 1 and this is the 0 station of branch A. So, we are related

by a matrix were a matrix is multiplication of three on matrix fill matrix odd matrix.

In this particular case, we can able to if we see that if we multiple this we will get

this expression condition for branch A for other shaft soon, we can able to overall transfer

matrix multiplication of this will give us this expression.

At gear location, we have this angular displacement these in the gear ratio and the torque are

related like this for and these two can be combined like this. So, basically this particular

matrix is nothing but gear ratio, so we can able to see that using m matrix we can transfer

the state vector from one shaft to another shaft in between gear is there having gear

ratio is n.

We can able to this m matrix to connect two equation like this, so now we can able to

see that we are related the state factor of branch a 2 branch B directly with the gear

ratio this is overall transfer matrix and this transfer multiplication is given here.

So, if we multiple them will get this transfer matrix and for the condition free condition

we already seen that if we apply abounded the condition equation, we get this as frequency

equation, so this particular form will be the frequency equation.

So, after multiplication of this we can get this expression, so this is the particular

frequency equation. So, if we expand this and substitute the value of this parameter,

we will get the polynomial of this form.

You can be able to see that we are getting frequencies which are 1 is 0 natural frequency

and another two are corresponding to the flexible mode just continue. In this particular case

consider the gear polar movement of inertia also, when we convert it when we transfer

matrix basically we got three degree of epidermis of this form because this is corresponding

to the gear system. So, that is why we are getting three natural frequency one of them

is 0 because free boundary condition, I will conclude now. So, today we have seen the analysis

of the gear system and branch system for tort analysis vibration case.

We obtain the frequency equation for both of cases for branch for gear system we did

some kind of equaling we converted the gear system to single shaft system. That we can

able to analyze with the method describe previously for shaft single shaft system which is much

easier to analyze. With transform method Trans system which we have multiple degree of freedom

the T m method is more advantages. We can able to get the recite body mode and flexible

mode depending upon of the bonding condition of the problem and in substituent class; we

will be studying the gear system and branch system analysis with finite element method

also.

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