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We have seen that determination of the natural frequencies and the natural modes of multi-degree
freedom systems are quite computationally intense, particularly when the number of degrees
of freedom increases. In many situations, the designers will be happy to have some approximate
idea about the fundamental frequency, that is, the lowest natural frequency.
In this lecture, we would like to present techniques of quickly and approximately finding
out the lowest natural frequency. There are quite a few methods and we would like to discuss
two methods: one method, which will give us the lower bound, that means, the actual natural
frequency of the first mode will be higher than the value we find; another method by
which we will get the upper bound, that means the value of the actual natural frequency
will be lower than the approximate value. Therefore, if we apply both the methods it
will be possible to get quickly a band and the actual natural frequency will be in that
band or that range and that can sometimes be very useful in the designing stage. The
first method which we will discuss is called Dunkerley�s Approach.
We will take up a particular case as an example, but whatever we discuss and derive will be
quite general as you will realize. Let us take this 3 degree freedom system. For this
particular system, for example, we can find out the influence coefficients. Now one thing
is perhaps clear to you - that there will be two kinds of equations while finding the
flexibility matrix and using the influence coefficients upwards. The cases which we solve
like this, masses are there, but the other end is free. Those cases are relatively simpler
for finding out the flexibility matrix because they are statically determinate. We can just
apply the load anywhere, immediately we know the load under which each individual is being
subjected. But a case like this when this end is also connected it becomes statically
indeterminate. Still, it is possible to find out the deflection by applying unit load,
only thing that it is slightly more involved, and therefore, I think one has to be careful.
The influence coefficients are a11 or aij. Now, we know that under normal mode oscillation,
each coordinate or displacement can be written as xi and say the mode is jth mode is equal
to Xi jth mode cosine omegaj t, which means, all i varying from 1, 2, 3 for the three stations
and j is a particular natural mode. So, in this case, we will be mostly concerned with
the fundamental frequency in the approximate method. We may consider this to be 1 and now
onwards for the sake of simplicity, we will not put this as first, but we will understand
that what we are writing is represented as first natural frequency or the first mode
of oscillation. So, we can write that xi is equal to capital Xi cosine omega t. Omega,
obviously, is then the first natural frequency because that is our concern; x two doti will
be minus omega square Xi omega t and the inertia force at each station is Fi is equal to, this
i we should not substitute, so we call it F; F will be minus mi xi two dot at the station
I, which will be nothing but omega square mi Xi cosine omega t. Now, we also know that
all the stations, all the masses reached their extreme position either way at the same instant
when their velocities become 0 and of course acceleration becomes maximum.
So, maximum inertia force equal to omega square mi Xi. At this instant, the deflection at
the ith location will be due to all these inertia forces, because, there is no other
external force acting. Therefore, it will be omega square mj Xj aij, this is the maximum
inertia force at this instant at the jth station and the deflection at the ith station due
to this force at jth station is multiplied by aij. This we sum up from j equal to 1 to
N; this N is the number of degrees of freedom. In this particular case, N is equal to 3 or
this we have done . The equation will be this ..
So, for three cases N is equal to 3, the equation in expanded form m1 a11 minus 1 by omega square.
It is the same set of equations which we have derived earlier and you all understand now
that a non-trivial solution will be possible with determinant being 0. So, for non-trivial
solution the determinant will be 0, this we have done.
So now, if you write the determinant the characteristic equation becomes like this . It becomes 1
by omega square power 3 plus a11 ; you need not go to the higher modes. So, for the general
case, N degree of freedom system, this will become 1 by omega square power N ; we have
not derived it, we have just written the calculation, but one can easily do the whole calculation
for N degree of freedom system case and the characteristic equation for the first two
terms will be like this. Solutions of this equation are the natural frequency of the
system. So, if the solutions, which represent the natural frequencies be omegan1, that is
the first natural frequency; omegan2, that is, the second natural frequency and so on;
omegan3 is the third natural frequency in this particular case; in general, it will
be then these are the routes of � that means, the 1 by omegan square, 1 by omegan2 square,
1 by omegann square these will be the routes of this characteristic equations.
We already know that sum of the routes; we can write them like this: 1 by omegan1 square
plus 1 by omegan2 square plus for this one it will be omegan3 square will be equal to
this . This is the known result of algebra, but one can easily do this by this technique.
That means this equation for 3 degrees of freedom then it becomes 1 by omega. Now, this
is an equation of the same order and if we expand it, the routes of this equation are
very obviously this, this and this. Therefore, this equation is same as the characteristic
equation for 3 degrees of freedom system; that means that equation will be same as this
. When you compare these two equations, you will find that first term will be the same
and this second term, whose power is 1 by omega square to the power 2, its coefficient
will be this; that you can see easily. From that the coefficient will be this and that
equation coefficient will be this . Therefore, the� will be�.
Now, comes the main logic of Dunkerley�s equation. Generally, omegan1 much less compared
to omegan2 which is again much less compared to omegan3; generally, they are gradually
increasing. So, 1 by omegan1 square is much more than 1 by omegan2 square. So, this is
the largest term when compared to these. These are approximately equal to this ; that means,
these two terms can be ignored since they are much smaller than this. This, of course,
can be written like this .
For a general case, Dunkerley's equation says that the fundamental natural frequency, that
means, 1 by omegan1 square is approximately equal to sigma1 aii mi I equal to 1. From
this, we can calculate the approximate value of fundamental natural frequency. Now this
value is going to be what, upper bound or lower bound? We can see here that, of course,
this 1 by omegan1 square is less than this, because, two terms we have ignored; we have
assumed this whole thing to be 1 by omegan1 square. So the real omega1 by omegan1 square,
actual value will be less than this. When you take the inverse, the actual omegan1 square
will be more than the value of omegan squared what we get from here; that means it is nothing
but a lower bound . We can say that the real natural frequency for the first mode is never
going to be less than this. It will be always more than that and approximately equal to
this. We can apply this to our one example we have
taken. Let us see that how we get that. Let us take the same problem which we have already solved;
then you can compare the results. This is the problem which you have solved and its
first natural frequency is this. If you remember it is equal to this; we found out by matrix
equation method and which if it is done accurately, it will give an exact answer. Let us apply
this: this is x1, this is x2 and this is x3. So what will be a11 we have already found
out earlier, but let us find again. a11 is the deflection of session one when unit load
is applied to session one itself which is nothing but the stretch of this. This will
also be the deflection at session two due to a unit force at one, also deflection at
� because this shift just like a rigid body. Again, we know that deflection at here due
to unit load here and here are going to be same. That is why, actually the flexibility
matrix here. a22 is equal to stretch of this string plus stretch of this string that will
be the deflection here if a unit load is applied at station two. This is 1 by k plus 1 by 2k
and 3 by 2k equals a33 because we will not require others; we will require only a11,
a22, and a33. This will be 1 by k plus 1 by 2k plus stretch of this string 1 by k equal
to 5 by 2k. So by Dunkerley's approach which is this - 1
by omegan1 square is approximately equal to a11 m1 plus a22 m2 plus a33 m3. We also know
that m1 equal to m, m2 is equal to 2m and m3 equal to 3m. So this will be equal to a11
is 1 by k into m which is simply 1 by k; a22 is 3 by 2k and this is 2m, it will be simply
3 m by k plus a33 is 5 by 2k and m 3 is m, it will be 5m by 2k and this is equal to . If
we find out omegan1 from this we will get square root of 1 by 6.5. When you compare
this result, the actual value is here. We find that this is slightly lower than the
actual value, but we find it is quite close; it is 0.4 here and it is 0.425 here; that
means the difference is in the second and that too not much. You can calculate the percentage
error and also that point is revealed here that it is a lower bound. That means the real
value is above. You can see that this is a very quick method of estimating a very approximate
value of fundamental frequencies of a system, because, calculating the influence coefficient,
it is generally very simple if the case is statically indeterminate like this. Then also
it can be done; only it will involve little bit more analysis, not very complicated. Mass
matrix is a diagonal matrix and finding the mass is very simple; just by observation and
calculation of this term is also a straightforward calculation. Thus, we can solve or get the
approximate value very quickly. Here, the advantage may not be very clear, but if you
have a 10 degree freedom system, you will find that this is a very quick method compared
to the earlier one.
Now let us take up the other method. This is an extremely important approach and we
will refer to it at a later stage also, but understand the case here. Now, under normal
mode oscillation condition, we have seen that all masses are oscillating with the same frequency
and almost they are either same phase or exactly opposite phase. It means that each mass attains
its maximum speed at the same instant of time which is the maximum velocity of it. If the
normal mode can be represented like this , then xi dot is velocity omega Xi sine omega t and
this maximum value of this t dot of each one is x dotimax is omega Xi is this, because
maximum value of sine omega theta will be only 1.
Similarly, all these masses attain their extreme position which represents the extreme stretching
of all the elastic members. Or if there is gravity then the extreme position in the gravitational
potential energy as well. Therefore, we know that extreme positions are all equal and they
happen at the same instant and the values are nothing but the capital Xi. So, that is
also quite straightforward, because, the maximum value of the kinetic energy is 1.What will
be maximum kinetic energy of the whole system? The maximum kinetic energy of ith mass will
be half mi into velocity is going to be half omega square mi Xi square. So, for all the
total kinetic energy of the system will be just sum of all this , it has N degrees of
freedom or N masses. This can be in matrix form. It can be written like this . This is
a straightforward demonstration of this and I am avoiding it to find out that this is
nothing but this . Similarly, maximum potential energy of the
system, each one will be stiffness matrix. If stiffness matrix be k , then this can be
also easily seen through a little bit analysis that total potential energy maximum for this
stiffness. We know that for a conservative system where there is no dispersion and now
when the kinetic energy is maximum; the potential energy is 0. Similarly, when the potential
energy is maximum; the kinetic energy is 0. Therefore, both of these two must be equal
to the total mechanical energy of the system and so they must be same; that we have used
even in the solution of single degree freedom. Remember, all these things that we are doing
are not mentioned during the case of normal mode oscillation, then only this is valid;
then only they all will be attaining 0 velocity and maximum velocity at the same instant.
Though we are not showing any i here as in particular mode, this is only valid for natural
mode oscillation. So if it is natural mode we can show like this. Therefore, for the
first mode omegan1 square can be expressed as this because then it will be natural mode
technique. Now, Rayleigh's principle says that if we assume column matrix X as the first
natural mode and evaluate this quantity we will get a quantity which we call as Rayleigh's
quotient. Rayleigh's quotient R is for an assumed mode
say, X transpose k X which need not be exactly same as the first mode and divide this by
this quantity . Then R tends to omegan1 as X approaches first mode. The most important
thing is it will approach from the higher side and when the assumed x becomes identical with
the actual first mode then R becomes equal to omegan1. So for an assumed shape X, whatever
value of the Rayleigh's quotient we get, it will be always slightly higher than the real
natural frequency. Of course, I must say R is not . So, determination of fundamental
frequency using Rayleigh's principle will always give us an upper bound.
There is another very important point to be noted here is that the value of Rayleigh's
quotient as it is nearer to omegan1 is somewhat insensitive to the choice of X, that means,
even if the choice of X is somewhat different or quite different from the first mode, the
value of this quotient will be somewhat nearer to this. That means a large error in X will
not be reflected as a large error in this natural frequency. Therefore R is also . Now,
let us solve the same problem using Rayleigh's.
So Rayleigh's technique says . So, let us find out what we get for solving the same
case. As you can see, we will need the k matrix and the m matrix; m matrix is quite straightforward.
Let us solve the same example. To find out the k matrix, best way for us will be to derive
the equations of motion. So, for mass one force acting are 2k into x2 minus x1, for
mass one it is m; for mass2 it is 2m; this force will be same as this and for mass3 it is again
the same. The equations of motion can be written. For the three masses it will be m x1 two dot
equal to 2k into x2 minus x1 minus k into x1; 2m x two dot. The second mass is k into
x3 minus x1 minus 2k into x2 minus x1 and for
the third mass it is this . So, now we rewrite them in a manner and for normal mode oscillation.
So that we can�.. omega square m X1 plus 3k X1 minus 2k X2 plus 0 X3 is 0. This one
will be 2 omega square m , here, this X1 term will be plus k X1 plus X2 term will be 2k
X2 and k X3 and third one will be minus omega square m X3, . There will be no X1 term; so
0 X1, for X2 it will be minus k X2 and for X3 plus k X3 and so we are not interested.
This becomes minus omega square m X plus k X equal to 0. This is the equation in matrix
form. So, obviously, this is nothing but m 0 0; 0 2m 0; 0 0 m into X1 X2 X3 plus k matrix:
3, minus 2, 0; minus 2, 3, minus 1; 0, minus 1, 1; m matrix is this and k matrix is this
. Now, from the application of this, the choice
of x is true that we will assume some value of x. If we make the choice somewhat logical
then obviously the results are going to be better; otherwise, if we take some extremely
wild choice deliberately, we will get something. Here, when it moves in the first mode we will
find the inertial load here is proportional to m; here it is proportional to 2m, but not
exactly because the amplitudes are different and amplitude here is more. Again here it
will be. Therefore, everywhere you will find the force acting is somewhat related to the
mass and the deflection is also related to the stiffness. So, one quick way of having
a reasonable is�. Let us see what is the phase of each position or displacement at
each position due to a gravitational pull; that means if we hang it what will be the
deflection of these in the static equilibrium position? That is one very common technique
used . Gravitational deflection that means if it
hangs what will be the X1? X1 will be something like mg, 2mg and mg; I will not do the detailed
calculations. So here it will be 4mg by k; X2 will be 11mg by k; X3 will be 13mg by k.
So we can take.. it will be 1 1.38 1.3�; it is in that ratio 4 11 13. If divided by
4, first one will be 1, second one will be 1.38 in this ratio. Let us use this as first
mode which is just due to gravitational problem; using this we get� I will not calculate
the whole thing. R is going to be k by m and 1.35 in the top and 7.47 in the numerator.
This will be equal to .181 and R is supposed to be square of the natural frequency. So
first natural frequency we get approximately equal to square root of this becomes 0.42
. Now, this is extremely revealing, because, even using an approximate method to use the
first mode by just using the gravitational pull and the subsequent information, the natural
frequency we get is very close, rather, almost same, but there may be difference in the higher
order third, fourth and fifth. You can see� Now again to demonstrate, as I mentioned,
this value accuracy is not very sensitively dependant on the correctness of the choice.
If I now say let us have quick method of finding out this. Even for finding the gravitational
pull and doing it you have to do some calculation.
Let us say that X1 X2 X3 be 1. Then this one being the same force we apply everywhere for
example, then this will stretch by another half amount of this. So, the next one will
be 1 plus half of it, so it will be 1.5. The other one, again, we approximately say another
half will go; so therefore this. This is a very crude way of as you can see. If we use
this, we will get Rayleigh's quotient as k by m into 1.75 by 9.5 equal to 0.1842 k by
m and this will result in approximate value of this . So surprise - that even if we make
such a crude way, we assume the mode shape you can see it is only in the third place.
This is the actual merit of Rayleigh's principle; otherwise, this expression is not the real
thing, but it is actually insensitivity to the errors in the choice of X in the result.
That is the main merit of this Rayleigh's principle and as you said that you will always
get it. Here it was very close. You did not see the difference, but here you can find
we have made some errors in the choice, more errors compared to the previous one. This
is also a static deflection case; this is not the real mode.
We have found out the real mode. In the previous lecture, you can compare and find out what
was the real mode. Even if you make this, you will get some value which is always more
than the real mode; that is why, we get always an upper bound. The proof that why it is in
upper bound shows that truth can be found in . Thus, for a designer, the quick way of
estimating the first natural frequency even for a complicated system can be done without
going to computer and running a whole program. Just by this which gives some ideas. So therefore,
if we say that this is the Dunkerley's method and this is the Rayleigh's method, then we
know that it must be between this and this . We narrow the gap and the designer can satisfy
whether this is in dangerous zone or not, that means whether there is some excitation
from outside is possible .