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.