Dynamics of Machines

Prof. Amitabha Ghosh

Indian Institute of Technology, Kanpur

Module - 12 Lecture - 6
Forced Vibration of Multiple Degrees of Freedom Systems

Once the free vibration problem of a multi degree freedom system is solved, it is possible
to solve forced vibration problem of the same system. Of course, earlier we have solved a
2 degree freedom systems forced vibration, but that was with a very specific objective to
its application vibration absorber. Now in this presentation, we would like to discuss the
matter of forced vibration of a multi degree freedom system. There are a couple of
approaches, one approach of course is a transfer matrix method which is applied to - very
often we apply to - torsional oscillation problem with a number of discs and typically the
method is known as Holzer’s method and its standard text books can be consulted to find
that. What I would like to present here is a very generalized approach through the post-
vibration problem of multi degree freedom system.

(Refer Slide Time: 01:40)

So, once we solve the free vibration of a multi degree freedom system, we know the
natural mode where i is equal to 1, 2, 3, up to N, which is the number of degrees of
freedom. We also normalize the natural mode in a manner (Refer Slide Time: 02:55)
otherwise, for free vibration amplitudes we do not have any definite value, so to get some
kind of normalized approach, we use this generalized mass equal to 1 and obviously in
that case (Refer Slide Time: 03:24) it will be equal to omegai to the power 2. Of course
we know that (Refer Slide Time: 03:38). Therefore, after you normalize you get some
unique values of the normal modes; of course, you also know the natural frequency. With
the approach I present it is possible to convert the forced vibration of an n degree
freedom system into n single degree freedom systems forced vibration problem? Since
single degree freedom systems have been solved and you know this solution, we can use
that and then retransform the problem to get the actual desired value.

(Refer slide Time: 05:02)

Let us first define the modal matrix, where the columns of this (Refer Slide Time: 05:18)
are nothing but the columns of this square matrix; they represent the natural mode. This is
the first natural mode, this is the second natural mode and for an n degree freedom
system, there will be n natural modes. So, there will be n columns and obviously x1, x2
and x3, it will go up to xn there will be n rows. This represents a square matrix of order n.
This we call modal matrix P. Next transform (Refer Slide Time: 6:36), how we transform
x P.
So, this is the way we transform and get another set of coordinates in case of x this
transformation rule and this y is called generally the principle coordinate.

(Refer Slide time: 07:50)

Now, we know our equation of motion, it is (Refer Slide time: 07:50). This is the forcing
function and of course, there are n such equations and they are all coupled, we know m
and k matrix. Therefore the second derivative in P matrices is independent of time; P
being a constant, P constituting constant elements; so a time derivative that means the
second derivative will be this. In terms of principle coordinates, we get this equation
(Refer Slide Time: 09:10). Replace x 2 dot by Py dot x by Py then pre-multiply both sides
by P transpose. (Refer Slide time: 09:57) Now, what will be this? We will find that P
matrix is this. Therefore, when we transpose it, this first column becomes the first row
and so on and here you will find because of the orthogonality condition, all the cross
terms will go; only the terms representing this and this will remain. So, the results of this
will be (Refer Slide time: 11:31).
(Refer Slide Time: 14:00)

All the terms will be gone except the diagonal terms; this will be a square matrix, only
the diagonal terms will be there and because of normalization all the diagonal terms will
be one. It will be a unit matrix (11:58) and here again, it will be a diagonal matrix where
the diagonal terms will be this and the rest will be 0. Here, it will be a column matrix.
These are the transformed (12:55). Therefore, what we are getting is nothing but some
decoupled equations, the first equation will be or what we will get yi two dot plus omegai
squared yi equal to Qit and i varying from 1, 2 and so on up to N. So, you get a simple
equation which is for the single degree freedom system; we are getting n number of such
equations.

This constitutes only y1 two dot and omega1 squared and y1 and Q1, next one is only in
terms of y2, next one is y3. So, they all will be decoupled equations and therefore each
one representing the solution for a single degree freedom system. Therefore, you know
the solution to this equation and we can find it out and once we find out and Q also we
have to find out by this multiplication. Once we find out y, (Refer Slide Time: 14:17) we
pre-multiply that by P matrix and then we get the actual solution wanted in terms of x.
So, this is the general procedure and I will explain now the procedure in detail with the
help of an example.
(Refer Slide Time: 19:36)

This is the problem; let us solve it. So, there are three masses connected by 3 springs: this
is 1 kilogram; this is 1 kilogram; this is 2 kilogram; this is 1000 Newton per meter; this is
1000 Newton per meter; this is 1000 Newton per meter and mass 3 using acted upon by
an external force which is 50 into cosine 20 t. The forcing circular frequency is twenty
radians per second and the magnitude of the exciting force is 50 Newton per meter. Now,
the equations of motion of the three masses can be written as this (Refer Slide Time:
18:35). m is 1; k is equal to 1000. Now, in matrx form equations can be represented
(Refer Slide Time: 19:40). So it is now clear that we are writing the equation of motion
for a normal mode of vibration.
(Refer Slide Time: (22:40)

Once it is normal mode vibration, we know that each one will oscillate with the same
frequency and x can be written as x1 cosine omega t and so on. (Refer Slide Time: 21:00)
Since it is normal mode oscillation, it is free oscillation that you can see; 0 means this is
the vibration for free oscillation. Our objective is as I mentioned that before we solve the-
forced vibration problem, we have to find out the natural mode and the natural
frequencies. So, first we have to solve the free vibration problem; that is what precisely
we are doing. The characteristic equation now, as this can be written as minus omega
squared, this is the m matrix and this is the k matrix, then m inverse will also be - you
can easily see even by inspection - inverse matrix will also be a diagonal matrix.
Multiplying by m inverse, the equation becomes (Refer Slide Time: 23:08) minus omega
squared, m inverse into k will be A matrix and A matrix in this particular case will be
1000, we can take common outside, this unit will be second square as we know, because
we are now solving a numerical problem. We have to be careful about the units.
(Refer Slide Time: 24:36)

So, the characteristic equation will be - because these are set of three homogenous
equations so the determinant must be 0, that means (Refer Slide Time: 24:54), when we
write in a standard form, becomes 2000 minus omega squared minus 1000 0 minus 1000,
(Refer Slide Time: 25:35) (poor audio) omega squared. Solving this, we will get three
roots or three values and we can tell you that the three roots of natural frequency will be
(Refer Slide Time: 25:59), second natural frequency will be and third natural frequency
will be.
(Refer Slide Time: 29:55)

Three natural frequencies after solving this equation will be (Refer Slide Time: 26:41).
Once we know the natural frequencies, now to find out the modes. First let us find out the
first mode. This procedure I will show and the same procedure will lead to other modes.
We have seen the equation, this equation A is this 1000 into 2 minus 1 0, minus 1 3
minus 2 0 minus 2 1. For first mode now, this is equal to, these are all… is equal to, when
you take the other side, minus becomes plus omegan1 squared for the first mode is 139.2
(Refer Slide Time: 28:39). This is the equation for the first mode. We have to find out x1,
x2 and x3 for this first mode. We have already normalized x1, x2, x3 - otherwise has no
meaning - now you have to normalize it using that condition you remember x transpose
m x equal to 1.
(Refer Slide Time: 35:27)

What we have done is X1 one X2 one X3 one, that is the transpose of X one, m matrix was
1 0 0, 0 1 0, 0 0 2, that was the m matrix you must remember and this was X1 one X2 one
X3 one, the first mode and that is equal to 1. That is normalization. Now, there are three
equations here and one equation here, in total four - we do not need four - so what we can
do is use any two equations from here and this equation from here of the state A and
equation B. Let us solve the first two; 2000X1 minus 1000X2 equal to 139.2X1. That was
the first equation, the second equation will be minus 1000X1 plus 3000X2 minus 2000X3
equal to 139.2 X2, that is the second equation, we do not need the third, we will have to
do this equation X1 squared plus X2 squared plus 2X3 squared equal to 1, so these three
equations we can solve to find out X1 one X2 one.

Since, it represents a particular mode, it should be, it is not a general equation, so, from
this one it has an equation of only X1 X2 that will give us X2 in terms of X1 this will give
us X2 one equal to 1.861 X1 one. If we use this here: X2 will be in terms of X1, and X3
will be there. So here this will give X3 in terms of X1 straight forward. Then using both
X2 and X3 in terms of this, we can get an equation in terms of X1 only and this will give
13.811 X1 one squared equal to 1 and the result will be X1 one equal to 0.269, X2 one is
just 1.861 (33:25) this is 0.501, X3 one is equal to 0.586.
(Refer Slide Time: 38:45)

So, we have found out X1 the first column of the modal matrix. (Refer Slide Time: 34:06)
All the time we have to remember, we are getting these unique numerical values; because
you have normalized the natural mode, otherwise there is no meaning of talking about
amplitude in a free vibration problem. Similarly, X2 will be following the same
procedure; only thing we have to do is we have to use for the second mode and this is
going to be the square of second mode that is 1745.8. So, the equations will be different
and solving that set of equations we will get (Refer Slide Time: 35:08). Again after
normalization only we will get numerical values like this. The third mode again we will
find out by using the third natural frequency 4115 in place of 139.2 that and this equation
will remain the same. Then we will get the third mode the numerical value and our modal
column matrix, this is our important matrix (Refer Slide Time: 36:39) which will
transform x into the principle coordinate y. Therefore Q(t) will be also found out by pre-
multiplying the forcing matrix which is nothing but this (Refer Slide Time: 37:11).

This will be 0.269 0.501 0.586 will be first row in place of first column, second column
will be the second row and third column will be the third row. This is P transpose and f t
matrix if you remember force acting on station 1 is 0 force acting on station 2 is 0 force
acting on station 3 is 50 cosine 20t so this is f t matrix.
(Refer Slide Time: 39:04)

Now after, we do this Q (t) becomes (39:33). This will be neutral. So, the equations of
motion in principle coordinate will become free decoupled equations like this.(Refer
Slide Time: 40:10) We know the solution of this steady state(41:10) - of course you have
to consider - steady state solution for each one (41:10) and so we get from here y1 equal
to (Refer Slide Time: 41:22). This we bring in from our memory steady state solution
(41:58). We know that, this will be 0 by you know that 1 by r.

(Refer Slide Time: 42:39)

Once we get this y, x will become f2 pre-multiply y by t; so, it will be .269, .501 (Refer
Slide Time: 42:53) and y we have found out and the question is now what is the
dimensions we are following the si unit, so this dimension will be in meters .This finally
will become (Refer Slide Time: 43:52). So, this is the solution to find out x1 x2 x3 as
functions of time of course in steady state solution for a post vibration problem, we
always know that the (44:34) will go and steady state oscillation will have to be in the
same frequency which is the frequency of… This is the general procedure for which I
think, now the question is that if there are two forces then what we can do, we can
separately solve and super impose because the system is linear. That way we will get the
solution by super-imposing the solutions separately obtained in this form.

Thus, we come to the end of our discussion on systems with multi degree freedom
system. Multi degree freedom systems are obviously - as we have been mentioning -
some modeling of the real system; that means, a beam with various cross section or a
machine or a structure. We can modulate in the form of a lump parameter system, which
will be approximately representing the system, the number of lump parameters or the
blocks we use, more number of mass to be utilized, obviously, we can have a better
approximation. Therefore, better accuracy or better results will demand larger number of
masses for a better approximation I heard there are situations where it is possible not to
break into lumps but to split the actual system as it is. Therefore, now we will look into
the phases where, we do not lump the parameters like this we treat the system as it is.

Say, for example, a cantilever beam, a bar or a shaft in torsional mode - if we hit it, it will
vibrate, you know that. So, they are all systems which are very often, rather the actual
systems are composed of such elements and therefore, our next attempt should be how to
handle cases where the system is not lumped and we have been doing so far what the
system is treated as it is. We will take up obviously the cases like the simple elements
like a uniform bar uniform beam (47:12) and how to solve the problem. So, from next
session we will take up the vibration problem of a continuous system and at least a few
cases (47:28). There will be some new concepts as you will understand and those new
concepts also, we will explain in the class.