Friends welcome to the 15th lecture in module
2. In the previous lectures we discussed about different numerical methods to compute the
natural frequency and mode shape of a multi degree freedom system. We have also shown
you the comparison of the results and we realize that the values are tallying and they are
in good agreement. In this lecture we will explain you how a
computer code is being written to estimate the natural frequency and mode shape we will
again compare the rules of a specific problem what we did in the last lecture behind, compare
the results with computer codes then we will see that they are again in better agreements. So, we will now discuss about the frequency
and mode shape of multi degree freedom system model. We will take an example problem. The problem demands to obtain all mode shapes
and a corresponding frequency of a system shown the system has got 3 stiffness K K K
and m which are equal. And now marking the degrees of freedom this is my x 1 this is
my x 2, this is my x 3, I can say even x 1 of t x 2 of t and x 3 of t. So, one can very
clearly see here m 1 equals m 2 equals m 3 is simply m maybe some numeric value similarly
k 1, k 2 and k 3 are taken as k some numeric value. This problem we solved in the last
lecture behind compared the results by different methods we found them that they are in good
agreement. So, easily when the degrees of freedom are
marked at the point where mass is lumped the mass matrix will be the diagonally dominant
matrix whereas, off diagonal elements will all be 0 which we already know and there is
no confusion in writing the mass matrix readily. To write the stiffness matrix we need to formulate
the equation of motion which we already said by Newton’s method let us try to do that
quickly for this specific problem, let us compare this and see how we are able to get
the stiffness matrix. So, the system what we have is m m and m, k k and k, this is x 1 x 2 and
x 3. So, we got a free body diagram let us say unit displacement, this is going to be
k times of x 1 and when this spring will now oppose k times of x 1 minus x 2 and we are
giving unit displacement in the degree of freedom x 1. Similarly for the second degree
we give unit displacement along x 2 and this spring will try to oppose the movement which
will be k of x 2 minus x 1 and this spring will now again oppose the moment of k of x
2 minus x 3. And we have the third degree this is also m m and m. So, this is again
x 3, so this spring will try to oppose the moment which is x 3 minus x 2.
Let us write the force balance equation for the first case. So, force is equal to ma.
So, m x 1 double dot, x 1 double dot should be equal
to minus of kx 1 minus k x 1 minus x 2 which is simplify as minus kx 1 minus kx 1 plus
kx 2 which is minus 2 kx 1 plus kx 2. So, m x 1 double dot plus 2 kx 1 minus k x 2 is
0 that is my first equation of motion. Now, for the second degree similarly m x 2
double dot will be equal to minus k of x 2 minus x 1 minus k of x 2 minus x 3 which will
be minus kx 2 plus k x 1 minus k x 2 plus k x 3 which is say kx 1 minus 2 kx 2 plus
kx 3 that is m x 2 double dot. So, I can say m x 2 double dot minus kx 1 plus 2 kx 2 minus
kx 3 is 0, that is my second equation of motion. Let us say the third equation which I write
here m x 3 double dot is minus k x 3 minus x 2. So, m x 3 double dot is minus kx 3 plus
k x 2. So, m x 3 double dot minus k x 2 plus kx 3 is 0 that is my third equation. Now,
writing all these equations in a simultaneous form let us say m x 1 double dot plus 2 k
x 1 minus k x 2 is 0, m x 2 double dot minus kx 1 plus 2 k x 2 minus k x 3 is 0, m x 3
double dot minus kx 2 plus k x 3 is 0. I can now convert these equations into a matrix
form, I can now say this can be expressed as m times of 1 0 0, 0 1 0, 0 0 1 of x 1 double
dot x 2 double dot x 3 double dot plus k times of 2 minus 1 0, minus 1 2 minus 1, 0 minus
1 1 of x 1 x 2 x 3 is 0. So, this is my mass matrix and this is my
stiffness matrix. So, if you look at the file what we just now showed to you the mass matrix
and the stiffness matrix are exactly in the same style, so k times of or m times of 1
0 0 0 1 0 0 0 1 and k times of 2 minus 1 0, minus 1 2 minus 1, 0 minus 1 1. So, getting
a k matrix as an input to the program is not difficult we can write equations of motion,
getting mass matrix and stiffness matrix as an input for a given program is not a problem. Having said this let us try to write the program
by 4 methods Eigen Solver, Dunkerley, Influence coefficients, Stodola and Rayleigh Ritz. Let us take the Eigen solver method the coding
is available on the screen. Now the hatch box shows you the input file actually inputs
are given in multiple formats the first input refers to the degree of freedom, second line
of the input refers to the mass matrix value in kg you are going to give the value with
a space bar and separate each row by a semicolon. The third line of the program input is for
stiffness the units are in Newton per meter similarly you have to enter row wise the stiffness
matrix what we just now saw is 2 minus 1 0 is a first row. So, you can see here 2 minus
1 0 with a space bar in between and a semicolon separating the lines open with the bracket
and closed with a bracket. So, you can enter the input file as degree
of freedom of your choice mass matrix of your choice and k matrix of your choice. Once you
do this the program is now going to evaluate the eigen value and eigen vector for this
problem. So, it calculates the stiffness matrix it prints the stiffness matrix. Then it computes
the eigen values and eigen vectors, it also estimates the modal matrix. This method gives all frequencies and mode
shapes the mode shapes are combined in a modal matrix is written here. So, in a model matrix
of 3 by 3 because the 3 degree freedom system model, each column will refer to the each
vector so this is phi 1, this column will be phi 2 and this column will be phi 3 together
this matrix is called phi matrix or modal matrix. Then orthogonality condition is also checked
for this given problem and if the modes are not orthogonal the modes are made orthogonal
and one can find out that. So, now normalization is being done and we
get ultimately the normalized modal matrix with respect to mass. You can normalize the
matrix modal matrix either with respect to mass or with respect to stiffness. So, in
this case it is normalized with respect to mass. So, the last section of the code actually
gives you the mode shape plot also you can see here, the mode shape plot is being given
in this section, it even plots the mode shape. And there is a sample input mass matrix is
been given stiffness matrix has been printed and it calculates all the 3 frequencies it
calculates the corresponding omega 1 this is phi 1, omega 2 this is phi 2 and omega
3 this is phi 3. We can see here phi 1 does not have any zero
crossing phi 2 has 1 zero crossing and phi 3 has 2 zero crossings. So, these are first
mode shape, second mode shape and third mode shape. These modes are naturally not orthogonal
they are now normalized and you get the normalized mode shape like this. They are also plotted
the blue one, the blue one shows mode 1 the green one shows mode 2 and the red one of
course, shows the mode 3. So, this has been obtained with the sample program which has
been given to you on the screen I will also give you the complete coding at the end of
this presentation. Let us move further let us take the Dunkerleys
method. Dunkerley method is a shortcut method which again requests you to give the input
as degree of freedom, mass and stiffness. Then it calculates the influence coefficient
matrix by inverting by inverting the stiffness matrix, inverse the standard subroutine in
MATLAB. So, once k matrix is designated can invert the matrix to get influence coefficient.
So, friends please remember influence coefficient matrix is nothing, but the flexibility matrix.
So, inverse, influence coefficient matrix is obtained once you have this coding with
you then it estimates only the fundamental frequency with of course, a high degree of
approximation let us see the sample input. So, the sample input for the mass and stiffness
is same this becomes my flexibility matrix now which otherwise a call as influence coefficient
matrix and based on the Dunkerleys algorithm which says 1 by omega square is actually sum
of mi alpha ii it calculates omega 1 it gives you only the fundamental frequency with a
serious level of approximation. The value what you obtained is 0.408, in the previous
case the value what we obtained is 0.445 for the first mode shape and frequency. So, there
is an approximation.

mod02lec15 (part-1): Computer methods of dynamic analysis