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We have seen that it is possible to utilize principle of 2 degree of freedom system in
neutralizing the vibration of an absorber. To demonstrate how, this kind of tuned vibration
absorbers are designed, we will check up with specific examples.
This is a vacuum pump; we need not go into the details of the story behind it. Main thing
is that the machine which is being driven by a motor is suspended or supported on two
strings. It is capable of making vertical movements whose frequency will be same as
the unbalanced force which is generated in the machine, may be a vacuum pump of cylinder
type. So, you have to design a vibration absorber. Vibration absorber means that another spring
mass system is to be connected here or here . So, what is given in this case is that the
machines run with nominal speed at fifteen hundred cycles per minute. We can easily find
out the corresponding omega and obviously, the frequency of the force generated will
be same; that means this related to omega. The magnitude of the force at this speed is
50 Newton. The rattle space is nothing but twice the amplitude of the absorber mass wherever
it is, if it is here or if it is here ; I have kept here, as larger space is here. So,
the maximum amount of displacement ma can have on both sides of its equilibrium position
that is called rattle space; that is given as 15 millimeter.
Now, it is also very important to see that after adding the absorber mass or absorber
system, the natural frequencies of the resulting 2 degree of freedom system should be wide
apart. In this case, it has been told that omega n1 should be below 1350 cpm and omega
n2, the larger one should be above 1650 cpm. We can easily find out radians per second
and we can convert. So, that means the two natural frequencies should correspond to values.
They should lie outside this region where the nominal operational speed of the system
is this. Obviously, we know that this will be also the resonating frequency of the primary
system. So, we have to find out ka and ma. So, the three conditions will be: one is that
square root of ka by ma is the natural frequency of the absorber system, should be equal to
the natural frequency of the original system and which must also match with the operating
frequency which is 1500 cpm; that is equivalent to 157 radians per second - that is one condition.
Second condition; xa must be less than equal to half into rattle space that is 15 millimeter.
Third thing is that omegan1 and omegan2 must lie outside this.
How do you proceed? First thing will be we have found out and we know that omegaa equal
to omegan is equal to 157 radians per second which is same as square root of Ra by ma,
which is actually omegana. Next, we also know that omegan1 by omegan equal to chi1, omegan2
by omegan we have found out chi2 and we also know Next, therefore what is done generally
in such cases that we first take a trial auxiliary system.
So, we first attach a trial absorber and when we attach the trial absorber, say of mass
ma trial is 0.1 kg. Whenever we say that absorber mass, it has to be tuned. Though it is not
mentioned, it is obvious that ka t should be such that it should be omega n square into
ma t, where omegan is the operational speed that is 157 radians per second and trial mass
is here; so this can be found. We also know that once we attach this trial absorber, the
resulting omegan1 t and omegan2 t are 1400 cpm and 1607 cpm respectively. Since the two
resulting natural frequencies are not outside the region, obviously this trial absorber
is not satisfactory. But with this experiment we can check the result; that means, omegan1
t is this much; therefore, trial value of chi1 t will be 1400 by 1500; so that is equal
to 0.933 and chi2 t we can find out either as 1607 by 1500 which will be 1.071. We also
know that this is nothing but 1 by chi1, because chi1 into chi2 must be always equal to 1;
that we have derived. We should also mention that chi12 can be written as half square root of
4 plus v, plus or minus root v. So, now we know
the trial values of this and from this we can easily find out now trial mass, because
we know that the trial value of m is nothing but trial value of the absorber mass by m,
will be nothing but from this relation we get chi2 trial minus chi1 trial square, which
is 0.019. So, we can definitely find out from this,
what is the value of your original mass. This can give us the value, because, the m of original
system is nu trial into ma trial. If necessary, you may want to find out, we are not required
to do it, but we can easily find out how much it is going to be. Anyhow, we will not do
that; rather, let us find out the design absorber, final one if you want to do.
Now, we know there are many constraints; let us start satisfying one constraint exactly.
What we may say, let us take this constraint to be satisfied, that means, with design absorber
chi1 d will be 1350 which is just the limit by 1500, which is equal to 0.9. Obviously,
with this we have to find out that chi2 d will be then simply 1 by chi1 d, that relation
and that will lead to omegan2 in the design stage will be equal to here actually this
was design here, so this will be 1 by chi1 d into omegana, which is nothing but 157 radians
per second or which is equivalent to 1500 cpm. So, this will give us 1667. So, when
we put the trial mass here or a design absorber here, in such a way that when the two frequencies
are splitting up, the lower frequency is just equal to this.
Then the upper frequency which we will get will be above this; this means the frequency
condition is being satisfied. But there is another condition which you have to now define.
So, two conditions we are satisfying; that is, natural frequency is same as the exciting
frequency and also which is equal to the natural frequency of the original system. We have
also put a design absorber in such a way that if you put a design absorber to make the lower
natural frequency exactly equal to the limiting value of the natural frequency, we avoid any
large oscillation when the exciting frequency varies, till the upper frequency is above
the other thing; that means we are satisfying that condition.
So, the other condition, the rattle space, we have to find out. To find out the rattler
space we have to find Xa and Xa means the amplitude of the absorber mass is simply F0
by ka; that we have derived in the previous lecture. That is xa is nothing but . So, what
is F0 is given. So, we have to find out ka. If we can find out ka, we can find out xa.
Now, how to do that? You know design value of this nu is simply design value of chi2
minus design value of chi1 square and that is equal to 0.0446. Since chi1 is this; chi2
is 1 by 0.9; so, we can easily find out. Design value of nu by trial value of nu becomes
trial value of nu which is equal to 2.347. Now, since nu is nothing but ma by m; so,
nu d by nu t will be ma d by ma t . This will give us the design value of the absorber mass
will be equal to 2.347 into trial value of the absorber mass which was 0.1 kg; so the
whole thing is 0.2347. So, we have found out the mass of the design absorber, not the trial
mass. So, once we find out that, we know the natural
frequency. Since we know the natural frequency, we can easily find out ka d will be equal
to simply 157 square and that will be equal to 5785 Newton per meter; that is the stiffness
of the absorber. Now, we know xa is going to be with the design absorber 50 Newton divided
by this , which is equivalent to 8.64 millimeter; we know that our limiting value is 7.5. So,
this is more than 7.5 millimeter; so, this condition is not satisfied; so, we cannot
use this; we have to now do the design again. This is a straining design, a mechanical design
is not a straightforward problem applying formally and finding out values. It is always
trying to fit in the constraints along with the desired conditions to be satisfied.
So here, we have found that starting from this it will not be possible. Now, one thing,
should we start satisfying this condition . Now without going through the calculation
I can tell you and one can verify, that if we satisfy this then the lower value of omegan
which will be resulting here, the way we have found out will be more than 1350; that is,
the frequency condition will not be satisfied. Therefore, let us not start from that side;
rather now, let us try to see if we satisfy the rattle space what happens. So, this is
the second design mass or second design attempt. So, our next attempt is this. Now, let us
start designing the other work starting from the rattle space condition; that means, let
xa d be exactly equal to 7.5 millimeter. Now, with a 15 Newton force, so this is nothing
but equivalent to 50 by ka into 1000, where ka is in Newton per millimeter. So, this will
give us the value of the stiffness of the absorber springs the final design which we
are trying and this will become something like 6666.7 Newton per millimeter. So, once
we know the stiffness, we know the frequency has to always match, because, it is 157 radians;
we get absorber mass also will be equal to 0.27 kilograms. So, ka by ma is equal to square
of this. We can easily calculate and find out. Now, once we get this, we can find out
the value of nu. What will be the mass of the system? So, with 0.1 kg nu as this much
, so with so much kg what will be the value of nu? Obviously, 0.27 by 0.2347 into 0.046
and this becomes 0.051. We have been able to find out nu. Once we
know nu, we should be able to find out chi1, how? We just use this relation putting the
value of nu here we get chi; when it is minus sign it is chi1; when it is plus sign it is
chi2. So, using this calculation gives us chi2 will be 1.119. Since chi1 is nothing
but omegan1 by omegan and chi2 is nothing but omegan2 by this. So, directly they will
give us the value of omegan1 which is equal to or equivalent to 1340 cpm and which will
be equivalent to 1679 cpm. Thus, we find that we start by satisfying the rattle space condition, like this, find out the stiffness
of the absorber satisfying this and the resulting ma because it has to have the same natural
frequency as this. Once you know the mass of the absorber and we know for a given mass
of the absorber what the nu was, we can find out what should be the nu in this case.
For example, if we had nu here in this, the mass of the system, I think we can easily
find out from this without any problem. Now, that means, for example if 0.1 kg mass of
the absorber nu found out trial mass any how we do not require that. So, it was this much
0.019. So, if nu is this much what will be the value of the trial mass with this? I want
to know nu directly. This relation gives us the chi values and the chi values are nothing
but the natural frequencies divided by the natural frequency of the primary frequency
or the exciting frequency or the absorber frequency, because, they are all kept same.
So, this now we find, the lower one is below 1350 and the upper one above. So, it is satisfying
this frequency condition also. So, this is just a simple example of indicating that how
an absorber is actually designed in the real life situation. It is always a trial and error
back and forth satisfying the equation. Now, if a system consists more than 2 degrees
of freedom, we do not give a separate 3 degree of freedom or 4 degree of freedom, what we
now consider in general cases of multiple degrees of freedom. 2 degrees of freedom was
a special case of multiple degrees of freedom system; nevertheless, all the important basic
concepts like natural mode, natural frequencies that we have already discussed in 2 degrees
of freedom system and they will be all applicable in case of multiple degrees of freedom system.
So, therefore in N degree of freedom system, we will have N number of natural frequencies
and N number of modes. So, now, it will be necessary to know new concepts. But to present
or to handle the whole analysis or discussion in a very compact manner, it will be necessary
to utilize metric representation. So, to indicate that how matrix equations are written, I will
take a simple example of a 2 degree freedom system.
This is a 2 degree of freedom system. We have already seen this; it is familiar to us. What
will be the equations of motion? Equations of motion will be m1 x1 2 dot will be equal
to k2 into x2 minus x1 minus k1 x1. We can reorganize and
we can finally write, or same thing can be written as m1 x1 2 dot plus k1 plus k2; we
have done this before. For the second mass, we can write in this manner . This is simple
reorganization and obviously, we are discussing the problem of free vibration. Now, how we
can represent this set of equations in matrix form? The same equation you will find can
be written as this. So, we use second bracket for column matrixes and third bracket for
square or rectangular matrixes. So, you know matrix representation. Therefore,
there is no need to explain in detail, but you will find this is the matrix representing
the mass of the whole system. This square matrix represents the stiffness of the whole
system; this column matrix represents the mode shape. You can say when it is oscillating
in a natural mode, this will represent the column of the natural mode oscillation and
this is the acceleration direction.
Therefore, in compact form we write that mass matrix into acceleration matrix plus stiffness
matrix into displacement matrix is equal to column normal, where m is the mass matrix.
If we take the coordinates from equilibrium position, it will be always a diagonal matrix
and k is the stiffness matrix. This will always be a symmetric matrix; diagonal matrix also,
of course is a symmetric matrix, because, all diagonal terms are equal to 0. Here the
diagonal term with this and this; that means, one half of the matrix that is above the diagonal
term will be same as the terms in the locations below the diagonal. So, this is a general
characteristic of all multiple degree freedom system.
Now, if this system oscillates in a natural mode with frequency omega and amplitude by
a column matrix X like capital X1, capital X2 and capital X3. These are amplitudes of
the mass of the coordinates 1, 2, 3 and so on; frequency is same for all.
So, therefore then displacement of the system can be written as and obviously, acceleration
will be this. If you substitute in this equation, what we get? We get this term into minus omega
square, cosine omega t has been cancelled because it was common to both terms. Now,
pre-multiplying both sides, all terms, by M inverse, what happens? When you do it, M
inverse becomes unit matrix, so it is 1. So, unit matrix into X symmetric plus M inverse
into k stiffness matrix into column matrix X is equal to 0. Now, this matrix we can give
a name say A matrix which you call it as dynamic matrix.
So, we can write the equation in the form A X is equal to omega square X for a particular
natural mode. Remember, if it is the I th mode, this frequency will be omegaI and mode
shape will be XI. Now, this is a typical equation which perhaps familiar to you. Now, this matrix
equation is satisfied only for specific sets of Xs, what I mean to say that when there
is a column matrix, you multiply the column matrix by square matrix, you get back the
column matrix only with a constant equation. This is possible only for some specific column
matrices and for some resulting values of some specific coefficient of this one. This
is called the Eigen values problem and the value which satisfies this equation here with
omega are called Eigen values and these columns are also called vectors.
So, these vectors are called Eigen vectors which are nothing but the normal mode and
the Eigen values are nothing but the squares of the natural mode. Of course, these can
be written in equation form also.
We get a state of homogeneous equation just like previous stages and the condition for
non-trivial solution will be the determinant of this matrix . I is the unit matrix, which
is again nothing but the characteristic equation solution which will be nothing but the values
of omegas of the natural frequency. That we have already seen. So, for each natural mode,
that means, if natural mode is identified by this I, corresponds to a natural frequency
omegani or we can write it in omegai. We do not want to write all the time omegan, because
now we will be writing this too often, so we may write this simply as omegai. This is
nothing but the Ith matrix. So, that n subscript we can remove from all the part. So, if there
are N number of degrees of freedom then i will vary from 1, 2 up to N; N sets of values
that means x1 x2 up to xN. So, there are N coordinates and for the ith matrix, this is
the natural mode of the ith matrix and this corresponds to omegai.
Now, we can see these natural modes are really nothing but Eigen vectors of this equation.
Now, another very important thing; we will find that natural mode cannot be just arbitrarily
anything. We will now discover some extremely important properties of this natural mode
which will be extremely useful and let us see this.
So, let us consider the ith mode. For ith mode, what will be the equation of motion?
You can see, it will be minus omega I square, let us call this as equation 1. For the jth
mode equation will be this and
we call this as equation 2. Now pre-multiply 1 by x of j transpose; I am quite sure you
understand transpose; automatically just exchanging the columns and rows. So, if we pre-multiply
this equation by this, what we will get? We get this. Let us call it as equation 3. Now,
pre-multiply 2 by x of i transpose. So, that equation will be this.
Now, it has been already mentioned that m and k are both symmetric matrix; of course,
m is a diagonal matrix. So, it is obviously , but this is a symmetric matrix. Therefore,
since m and k are symmetric matrices, they satisfy this condition, where x of j transpose
m x i will be same as x of i transpose m into j and this . j transpose into k is a
property of symmetric matrix. When a symmetric matrix is pre-multiplied by the transpose
of a column matrix and post-multiplied by another column matrix that is equal to the
transpose of the column matrix post-multiplied by m and here it is like this, both for m
and k will be like this. So, now subtract equation 4 from equation 3; if you subtract
this from this, what we will get? Omegaj square minus omegai square because this is equal
to this. So, we can write any one of them. So, we will write this j transpose m Xi transpose
and this equal to this. So, subtracting it will be 0.
When i is not equal to j, in general, omegai is not equal to omegaj. So, the only way this
condition can be satisfied is like this . Now if this is 0, this term is 0, then this has
to be also 0; So, it follows. This is a very special and important property; that means
that two different modes when post-multiplied and pre-multiplied, then both the mass matrix
and pre matrix, it becomes 0. They are not 0 only when i and j are equal and they are
called generalized mass and generalized stiffness; that means, if I put Xi transpose m Xi this
will be a quantity which we may call it as mi and we can give the name generalized mass.
Similarly, xi transpose k Xi will be again 0 quantity and we call it generalized stiffness.
This property is called Orthogonality of Normal Modes. These are two very important properties
and at later stage, you will find they will be extremely useful to us. Now, I think our
task is determination of the natural mode and natural frequency. If we directly solve
these and try to get the values, of course, it is fine and we will get the values as we
have already done in one or two cases. However, when a system is large, quite often what happens?
A regular normal actual system is modeled in the form of a parameter system, but to
make the model very realistic resulting in acceptable values of the answer, so the number
of degrees of freedoms is very large. So, for example, a machine tool or car, vehicle
all such kinds of actual systems, the presentation will be meaningful only when the numbers of
degrees of freedom are large. So, when capital N, that is degrees of freedom is very large
then solving this equation directly, it becomes serious one and that is why many methods are
being revised. It can be applied to simple engineering design problems and this method
which now we will take up rather than solving the direct equations and one type of method
called as iteration method; we will iterate the matrix and we will get the value. That
is of course, a repetitive process, but it converges to the answer after few steps. Only
thing that computationally it is far less intensive than solving, say for example, a
10 degree of freedom system directly solving 10 degree polynomials.
We will next take up the methods of solving this multiple degree system. First, we will
discuss matrix iteration method, and subsequently, we will just discuss some approximate methods
for resulting in very quick values giving an idea about the order of magnitude of the
fundamental natural frequency that is the lowest natural frequency.