Namaste in today's tutorial we will discuss
how to calculate the eigenvectors and how
to check the model Orthogonality, so let us
go directly into the problem.
So here we have to check the model orthogonality
of eigenvector if so here k is given to us
the K matrix of three by three in nature is
given to us and the ?1 ?2 and ?3 value and
corresponding ? 1 and time period values are
given to us, so why making use of this we
will see how to calculate the Eigen vector
and then later in second half we will check
how to check the model orthogonality of a
Eigen vector.
So first of all calculation of Eigen vector
since we have to calculate the Eigen vector
it is nothing but the K -? M | K - ?| M into
Ø = 0 will give us the Eigen vector, so if
I put ? 1 as 911 .97 in this equation I will
get this 5 matrix now here Ø 1, Ø11 so 1
indicates the mode number and this suffix
indicate the floor level so Ø1, Ø 12 and
Ø 13 so when I multiply this matrix with
this one which is nothing but the k - ? M
into Ø 11 vector or the 3 by 1 matrix.
I will get this three equation first second
and third so how we get let us calculate one
equation so 18.6 7 x Ø 11 will give me first
1, -10 . 36 x Ø 1 2 + 0 x Ø 13 so since
it is zero there is no need to calculate so
this will give me the first equation, similarly
second equation will have - 10.36 x Ø11 +
18 .67 x Ø 12 and last term here we have
a positive value or some value other than
zero that is -10.36 x Ø 13.
We will get the second equation similarly
third equation now here we have three equations
and three and one we can easily find out.
Now since I want the vector in some normalized
form what I will do is I will assume the top
value which is nothing but the Ø 13 is a
one, now first of all I will substitute this
one to get one more unknown value from equation
3 so our equation 3 is nothing but the -106
x -10.36 a 12 which is at the second floor
plus 8.31 Ø 13 = 0 so this is third equation,
now if I substitute this Ø 1 3 as a 1 I will
get 10.36 Ø 12 + 8.31 = 0.
If I compute this one I will get the Ø 12
s 0.802, now let us see that what exactly
physical meaning of a12 now if I plot this
three-floor m1 m2 m3 so first of all for first
more this is the third position or the top
floor position Ø 13 assuming it as a 1 so
this I will get the first point now Ø 12
will be over here as a pointed 0.802, now
let us see how to calculate or how to get
the Ø11 now in order to calculate Ø 11 we
already have value of Ø 1 3 as a 1, Ø 12
as appointed 02.
Now we will substitute this in equation number
2, now what is equation number 2equation number
2 is nothing but the 106 - 10.36 Ø11 + 10.6
7 Ø 12 - 10.36 Ø 13 = 0, so this is our
equation number 2 now substituting values
10.36 Ø11 + 10.67 x 0.802 which is value
of Ø 1 2 – 10.36 = 0, so if I solve this
one I will get Ø11 as a 0.445 so once I plot
this 0.445 which is somewhere over here.
And I will connect all the dots so here I
will get the values this will give me the
mode shape, so here I we got the mode shape
of a mode shape of a building so this is for
? 1mode shape of a ?1.
Now similarly so this is the our final answer
0.445, 0.802 one we are writing from first
to third floor or the top floor top most floor
if I draw the diagram Ø 13 Ø 12 and Ø 11
here we have Ø 13 as a 1 and 0.802 and 0.445.
Now similarly we can calculate the board shape
that is a Ø vector for second mode by substituting
? 2 as actual value of 7150 9.72 so by substituting
? 2 I will get this to matrix and again multiplying
this 2 matrix with each other I will get equation
number one equation number 2 equation number
three again with the three unknown of Ø 21
Ø 22 and Ø 23.
The way we have consider assume the value
of Ø 13 here also we will assume value of
Ø 2 3 as a one now rewriting the equation
number three 10.36, Ø 22, -5. 75, Ø23 = 0
so this is my equation number three now10.36
Ø 22 which is unknown 5,771 x 1 = 0, so Ø
22 is equal to -0.555 so here you can observe
that the value is less than one but it is
having the negative sign similarly let us
write the equation number 2 which is nothing
but the 10.36 Ø 21 + 4.6 1Ø 22- 10.36 , Ø
23 this is a entire bracket is equal to zero.
So now I will substitute each and every value
one by one 4.61 x -0.5 Ø - 10.36 X 1 = 0
So I get Ø21s 1.247 so again negative value
but greater than one.
So when I plot this one you can see that I
will get one on the roof top as a Ø 23 and
then second one at the second floor which
is Ø 22 because of negative sign it is on
the left hand side 0.55 and then third value
at the first floor which is highest if we
just look at the absolute magnitude as a 1.2
47 but since it is negative we are drawing
it on the left hand side as Ø21 this was
the second mode.
Now the same way when we go for the third
calculation we get the two matrix this 2 matrix
multiplication of this two matrix will lead
us to a the equation where we have three unknowns
this time path.
When we assume a 33 as a 1 and rewriting the
third equation 106 – 10.36 532- 23.28 Ø33
= 0 this was my second equation sorry third
equation when I substitute again 10.36 Ø
32 is equal to 3.28 x 1 so here I will get
Ø32 s - 2.247 so this is a value at the second
floor, now when I substitute for 1st floor
in equation number 2, -10.36 Ø31-12. 92 Ø32
- 10.36 Ø33 = 0 this one is second one, - 10.36
Ø31 - 12.92 x Ø32value as - 2.2 47 – 10.36
= 0, Ø31 = 1. 802 so here we also got the
values of Ø for third one in first and the
second floor.
Now when I plot this one I will get on the
this kind of vector in this kind of diagram,
so this was the?1 ?2 ?3 value corresponding
? values and the corresponding time period
and this is the eigenvector which was for
? 1? 2 and ? 3, so this is a Ø matrix so
this can be called as a vector an entire matrix
we can say as a Ø matrix.
So this is a physical meaning of it so for
?1 Ø vector consists of Ø13, Ø12, Ø11
with corresponding values the magnitude will
give us from center of mass of that floor
how much far it is from center of center of
mass of that floor and the positive and negative
sign will give us about the in which direction
it is and when we when we draw it we will
get the mode shape so these are the mode shape
for third one.
So here we can see that the for the first
mode it will touch the vertical axis only
at the 1 point for second one it is touching
at the first point and at second point and
for third mode we are touching vertical axis
at the bottom at the middle one and the third
one, so this can be a cross check for the
calculation of Ø values now let us see how
to check the orthogonality of a Eigen vector.
Now when we see when we have to define the
model orthogonality of a matrix we have to
prove that Ø-1 K Ø should be diagonal matrix
what is a diagonal matrix, diagonal matrix
means we only have diagonal element as a sum
value nonzero values rest all values are 0
so if we can prove this we can say that the
Ø matrix is orthogonal.
So here now I am substituting the this is
the Ø matrix which we have just now calculated
and this one is a Ø transpose and this is
a K matrix, so when I multiply this Ø -1
K or and K into Ø ultimately I will get the
answer as this matrix., so this has a diagonal
element as a nonzero value and all non diagonal
element as a zero by this we can prove that
the modal orthogonality exist.
Till now we have checked the manual method
to find out or to check the model orthogonality
where we have computed the Ø -1 K Ø, so
product of Ø-1 K Ø gave us the diagonal
element or the diagonal matrix so that is
a 1 part or the one way to prove that the
model orthogonality now we will see how to
check the or the prove model orthogonality
for graphical method.
Now here we have Ø matrix which we have already
calculated every one of you know now when
I suppose these are the three axis this is
X, this is y, this is z now you can consider
this row as X this row as a y, this Row is
a Z now let me plot a single point so each
vector will give me one single point let us
say a1 now I will connect I will first of
all I will plot this a 11 in a graph or in
a space and then I will connect it from the
origin ,origin is nothing.
But the 000 now let us say x value for Ø
1 so Ø 1 in X Direction is around 0.45 say
this is 0.445 then in y direction it is 0.8
which is somewhere around here 0.8 and in,
in that direction it is one, I have to find
out that in space so this was X Direction
here it was y direction now if I go vertically
Z direction up to 1 so I got this point as
Ø 1.
Now if I connect it through origin till that
point I will get this vector from origin it
is going outward now when I compute the same
thing same point for Ø 2 and Ø 3 let us
look graphically how it will look.
Now here you can see these are the three vector
the way we have generated Ø 1 I am considered
plotting Ø 2 and Ø 3 so when I plot this
in any floating software available you can
see that from origin how they are going and
how they are orthogonal to each other they
are perpendicular to each other and they will
not cross each other in the space, so if we
can prove this or if you can draw this is
also one of the method to check the model
orthogonality.
