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PROFESSOR: Ladies and gentlemen,
welcome to
00:00:23.760 --> 00:00:25.600
lecture number 11.
00:00:25.600 --> 00:00:28.120
In this lecture, I would like
to continue discussing with
00:00:28.120 --> 00:00:32.240
you the solution of the dynamic
equilibrium equations.
00:00:32.240 --> 00:00:35.400
We discussed the solution of
dynamic equilibrium equations
00:00:35.400 --> 00:00:38.440
already in lecture number 10.
00:00:38.440 --> 00:00:41.950
In that lecture, we consider
direct integration methods.
00:00:41.950 --> 00:00:45.270
In this lecture, I would like to
consider with you, discuss
00:00:45.270 --> 00:00:49.980
with you the mode superposition
analysis.
00:00:49.980 --> 00:00:54.100
The basic idea in the mode
superposition analysis is the
00:00:54.100 --> 00:01:01.330
transformation from the U
displacements into a set of
00:01:01.330 --> 00:01:03.891
new displacements, X.
00:01:03.891 --> 00:01:07.380
P is an n by n matrix,
which is nonsingular.
00:01:10.260 --> 00:01:15.510
If P is nonsingular, we should
recognize that the vector U of
00:01:15.510 --> 00:01:20.075
lengths n is uniquely given
when we have the vector X,
00:01:20.075 --> 00:01:23.100
also of lengths n
and vice versa.
00:01:23.100 --> 00:01:26.600
In other words, if we have the
vector X, we can calculate U.
00:01:26.600 --> 00:01:30.420
Or if we have U, we can
calculate X. P, of course, has
00:01:30.420 --> 00:01:33.440
to be a nonsingular matrix.
00:01:33.440 --> 00:01:35.630
We call this a transformation
matrix.
00:01:35.630 --> 00:01:38.790
And we call the X displacement,
that I will be
00:01:38.790 --> 00:01:42.640
talking about, the generalized
displacement.
00:01:42.640 --> 00:01:49.040
Now, if we use this
transformation, we find that
00:01:49.040 --> 00:01:53.260
if we substitute this
transformation into this
00:01:53.260 --> 00:01:58.240
equation for U, and we postulate
that P shall not be
00:01:58.240 --> 00:01:59.690
a function of time.
00:01:59.690 --> 00:02:02.200
So that when we find the
derivative of U,
00:02:02.200 --> 00:02:03.610
with respect to time--
00:02:03.610 --> 00:02:06.020
let me denote it
by a dot here--
00:02:06.020 --> 00:02:11.120
we simply obtain that U dot
is equal to P times X dot.
00:02:11.120 --> 00:02:16.440
And similarly, U double dot is
equal to P times X double dot.
00:02:16.440 --> 00:02:20.450
If we then substitute for U,
U dot, and U double dot--
00:02:20.450 --> 00:02:22.240
in other words, for the
displacements, velocities, and
00:02:22.240 --> 00:02:23.450
accelerations--
00:02:23.450 --> 00:02:30.230
into this equation, we obtain
this equation, provided--
00:02:30.230 --> 00:02:32.540
and now, there's one
additional step--
00:02:32.540 --> 00:02:36.320
after the substitution for the
displacements, velocities, and
00:02:36.320 --> 00:02:41.170
accelerations, we also
premultiply the total equation
00:02:41.170 --> 00:02:42.760
by a P transposed.
00:02:42.760 --> 00:02:47.850
In other words, what we do is
we take a P transposed here,
00:02:47.850 --> 00:02:52.580
and we are substituting for
these vectors here from that
00:02:52.580 --> 00:02:54.580
relation here.
00:02:54.580 --> 00:02:57.586
The P transposed, of course,
gives us, also, a P transposed
00:02:57.586 --> 00:02:59.580
R on the right-hand side.
00:02:59.580 --> 00:03:03.890
So R curl here is defined
as shown down here.
00:03:03.890 --> 00:03:08.500
Notice that M curl, C
curl, and K curl are
00:03:08.500 --> 00:03:10.090
given as shown here.
00:03:10.090 --> 00:03:14.470
And they involve, of course,
the M, C, and K matrices.
00:03:14.470 --> 00:03:17.480
And this P here comes via this
00:03:17.480 --> 00:03:21.430
substitution into these vectors.
00:03:21.430 --> 00:03:26.120
This P transposed comes from
that P transposed here.
00:03:26.120 --> 00:03:28.900
Now, what have we been
doing so far?
00:03:28.900 --> 00:03:30.910
Basically, what we have
been doing is we
00:03:30.910 --> 00:03:32.360
have changed bases.
00:03:32.360 --> 00:03:36.680
We have come from the
displacement defined by the
00:03:36.680 --> 00:03:37.830
finite elements--
00:03:37.830 --> 00:03:39.240
by the finite element
nodal points--
00:03:39.240 --> 00:03:41.550
to generalized displacements.
00:03:41.550 --> 00:03:45.670
The generalized displacements
now govern the equilibrium of
00:03:45.670 --> 00:03:48.460
the system.
00:03:48.460 --> 00:03:55.650
Now, this transformation is
effective if we find that our
00:03:55.650 --> 00:04:00.890
M curl, C curl, and K curl have
a better structure than
00:04:00.890 --> 00:04:04.810
the matrices, M, C, and K.
And, of course, if it is
00:04:04.810 --> 00:04:08.775
simpler to find the P matrix
or a P matrix.
00:04:12.180 --> 00:04:14.500
Let's look at a very
simple case.
00:04:14.500 --> 00:04:19.550
If, for example, M curl, C curl,
and K curl would all be
00:04:19.550 --> 00:04:23.580
diagonal matrices, then the
solution of this set of
00:04:23.580 --> 00:04:26.440
equations would be
very simple.
00:04:26.440 --> 00:04:28.570
In fact, this is
our objective.
00:04:28.570 --> 00:04:32.960
It is our objective to come up
with a P matrix, which when
00:04:32.960 --> 00:04:36.590
used in this transformation,
gives us an M curl, C curl,
00:04:36.590 --> 00:04:39.070
and K curl matrix that
are all diagonal.
00:04:41.760 --> 00:04:44.790
And, of course, in addition we
want to find the P matrix with
00:04:44.790 --> 00:04:46.920
a minimum amount of effort.
00:04:46.920 --> 00:04:50.900
Once we have diagonal matrices
here on the left-hand side--
00:04:50.900 --> 00:04:53.370
in other words, M curl,
C curl, and K curl--
00:04:53.370 --> 00:04:57.600
then, we can use our direct time
integration schemes, the
00:04:57.600 --> 00:05:01.270
Newmark method, or the essential
difference method.
00:05:01.270 --> 00:05:03.760
Usually we use the
Newmark method.
00:05:03.760 --> 00:05:08.040
Or we might even be able to
solve the decoupled equations.
00:05:08.040 --> 00:05:11.980
They're decoupled because M
curl, CK curl, are diagonal.
00:05:11.980 --> 00:05:16.200
We might even solve these
equations in closed form.
00:05:16.200 --> 00:05:20.050
The question then is how do we
find P or [UNINTELLIGIBLE]
00:05:20.050 --> 00:05:21.940
P?
00:05:21.940 --> 00:05:23.960
We want to find it cheaply.
00:05:23.960 --> 00:05:29.460
It shall be economical to
calculate P. And we want to,
00:05:29.460 --> 00:05:34.000
of course, have that M curl,
C curl, K curl are ideally
00:05:34.000 --> 00:05:35.250
diagonal matrices.
00:05:37.530 --> 00:05:41.990
Well, the P matrix can
be constructed--
00:05:41.990 --> 00:05:45.630
or a P matrix can be
constructed, effectively, by
00:05:45.630 --> 00:05:48.820
looking at the free vibration
equilibrium
00:05:48.820 --> 00:05:51.980
equations of the system.
00:05:51.980 --> 00:05:56.340
If we look at these equilibrium
equations here, in
00:05:56.340 --> 00:05:59.960
which we neglect damping and we
neglect the right-hand side
00:05:59.960 --> 00:06:02.650
forces-- we're looking at the
equilibrium equations of the
00:06:02.650 --> 00:06:06.800
dynamic system without damping
and without forces applied.
00:06:06.800 --> 00:06:10.650
And if we look for a solution
to this system, then, we can
00:06:10.650 --> 00:06:13.420
assume a solution
of this form.
00:06:13.420 --> 00:06:16.150
U, of course, being our
displacement vector.
00:06:16.150 --> 00:06:22.110
Phi being all the vector that
is independent now of time.
00:06:22.110 --> 00:06:27.640
And here is the time dependency,
sin, omega, omega
00:06:27.640 --> 00:06:31.370
is a radiant per second
circular frequency.
00:06:31.370 --> 00:06:33.000
t, of course, being the time.
00:06:33.000 --> 00:06:35.440
And t 0 is a time shift.
00:06:35.440 --> 00:06:41.370
If we make this assumption,
substitute into here--
00:06:41.370 --> 00:06:43.830
when we do substitute, of
course, remember we have to
00:06:43.830 --> 00:06:46.520
take the second derivative of
this function here with
00:06:46.520 --> 00:06:48.910
respect to time because we're
taking the second
00:06:48.910 --> 00:06:51.340
derivative of U here.
00:06:51.340 --> 00:06:55.940
And that gives us a minus omega
squared sin, omega t
00:06:55.940 --> 00:07:01.680
minus t 0, substituting
from here into there.
00:07:01.680 --> 00:07:05.860
We immediately get this equation
because we can cancel
00:07:05.860 --> 00:07:09.410
out sin omega t minus
t 0 on both sides.
00:07:09.410 --> 00:07:11.530
This is the resulting
equation.
00:07:11.530 --> 00:07:15.050
So what we now want
is to solve for
00:07:15.050 --> 00:07:17.930
phi and omega squared.
00:07:17.930 --> 00:07:21.510
This is the generalized
eigenvalue problem.
00:07:21.510 --> 00:07:25.580
The generalized eigenvalue
problem because M is a matrix
00:07:25.580 --> 00:07:28.210
other than the identity
matrix in general.
00:07:28.210 --> 00:07:34.440
And we know that they are n
solutions to this problem if K
00:07:34.440 --> 00:07:38.920
and M are n by n matrices.
00:07:38.920 --> 00:07:40.990
There are n solutions
to this problem.
00:07:40.990 --> 00:07:45.180
We call these solutions
the eigensolutions.
00:07:45.180 --> 00:07:51.430
And we call omega 1 squared,
I should say, and phi 1 the
00:07:51.430 --> 00:07:54.100
first eigenpair.
00:07:54.100 --> 00:07:57.440
This is the second eigenpair,
this is the nth eigenpair.
00:07:57.440 --> 00:08:01.090
There are n such eigenpairs.
00:08:01.090 --> 00:08:07.370
Each of these eigenpairs
satisfies this equation here.
00:08:07.370 --> 00:08:10.870
Therefore, these are
a solution--
00:08:10.870 --> 00:08:13.070
we call it an eigensolution--
00:08:13.070 --> 00:08:15.380
to this equation.
00:08:15.380 --> 00:08:19.570
And in addition, we also can
prove-- we can show--
00:08:19.570 --> 00:08:25.470
that the phi I vectors--
these vectors here--
00:08:25.470 --> 00:08:28.430
are M orthogonal.
00:08:28.430 --> 00:08:33.840
M orthogonal because phi
I transposed M phi j--
00:08:33.840 --> 00:08:40.640
for example, phi 1 transposed
M phi 2 is equal to 0.
00:08:40.640 --> 00:08:44.159
And phi 1 transposed M phi j--
00:08:44.159 --> 00:08:50.010
sorry phi 1 transposed M
phi 1 is equal to 1.
00:08:50.010 --> 00:08:53.810
And because it's equal to 1,
they're not only M orthogonal,
00:08:53.810 --> 00:08:57.150
they're indeed even
M ortho normal.
00:08:57.150 --> 00:09:03.590
What we find is that if we look
at this eigenproblem, and
00:09:03.590 --> 00:09:08.470
we notice that a certain
vector satisfies this
00:09:08.470 --> 00:09:12.390
eigenproblem, then also a
multiple of that vector
00:09:12.390 --> 00:09:14.630
satisfies the eigenproblem.
00:09:14.630 --> 00:09:20.230
And we determine this constant,
this undetermined
00:09:20.230 --> 00:09:26.000
constant, in such a way as to
have phi I transposed M phi I
00:09:26.000 --> 00:09:28.190
equal to 1.
00:09:28.190 --> 00:09:33.430
So these are the eigenpairs and
the eigenvectors, phi 1 to
00:09:33.430 --> 00:09:37.460
phi n, satisfy this relation.
00:09:37.460 --> 00:09:42.020
The eigenvalues, omega 1 squared
to omega n squared can
00:09:42.020 --> 00:09:46.810
be ordered in this way where
we find that we might have
00:09:46.810 --> 00:09:48.240
multiple eigenvalues.
00:09:48.240 --> 00:09:51.590
Omega 1 might be equal
to omega 2.
00:09:51.590 --> 00:09:56.440
But if we do, we still have only
a total number of n such
00:09:56.440 --> 00:09:57.760
eigenvalues.
00:09:57.760 --> 00:10:01.700
I will discuss in the next
lecture with you the solution
00:10:01.700 --> 00:10:05.450
techniques that we're using to
calculate these eigenpairs.
00:10:05.450 --> 00:10:08.070
Let's for the moment assume that
we have calculated them.
00:10:08.070 --> 00:10:10.010
And we want to use them.
00:10:10.010 --> 00:10:12.830
Well, if we now define
the following
00:10:12.830 --> 00:10:15.640
matrix, phi capital phi.
00:10:15.640 --> 00:10:18.930
It's a capital phi because
we see crossbars
00:10:18.930 --> 00:10:21.220
top and bottom here.
00:10:21.220 --> 00:10:24.960
If we define this matrix
as phi 1 to phi n--
00:10:24.960 --> 00:10:27.550
in other words, we're taking
the first eigenvector, and
00:10:27.550 --> 00:10:31.340
list it as a first column
of this matrix here.
00:10:31.340 --> 00:10:34.500
The second eigenvector, we list
as the second column in
00:10:34.500 --> 00:10:35.990
this matrix and so on.
00:10:35.990 --> 00:10:38.640
So that we find, of course, that
this matrix here is an n
00:10:38.640 --> 00:10:40.670
by n matrix.
00:10:40.670 --> 00:10:43.860
And if we define a diagonal
matrix, omega squared.
00:10:43.860 --> 00:10:46.150
This is a capital omega.
00:10:46.150 --> 00:10:50.360
Omega squared, being
equal to 0s in the
00:10:50.360 --> 00:10:51.780
off diagonal elements.
00:10:51.780 --> 00:10:54.720
But the diagonal elements
being omega 1
00:10:54.720 --> 00:10:57.240
squared to omega n squared.
00:10:57.240 --> 00:11:02.980
Then, we can write the solutions
to the eigenproblem.
00:11:02.980 --> 00:11:04.670
The solutions to the
eigenproblem--
00:11:04.670 --> 00:11:06.350
and I'm still talking
about this
00:11:06.350 --> 00:11:08.230
eigenproblem problem here.
00:11:08.230 --> 00:11:12.900
We can write the solutions to
this e eigenproblem in the
00:11:12.900 --> 00:11:14.610
following form.
00:11:14.610 --> 00:11:18.960
Notice that this omega squared
here has to be on the back of
00:11:18.960 --> 00:11:20.330
the capital phi.
00:11:20.330 --> 00:11:24.010
In other words, this is the
solution to the problem, K phi
00:11:24.010 --> 00:11:30.110
equals omega squared M phi,
where the omega squared is a
00:11:30.110 --> 00:11:32.440
scalar and appears
here in front.
00:11:32.440 --> 00:11:33.930
We usually write it in front.
00:11:33.930 --> 00:11:35.690
Of course, we can put
it also at the back
00:11:35.690 --> 00:11:36.700
because it's a scalar.
00:11:36.700 --> 00:11:39.060
We can put in front
or the back.
00:11:39.060 --> 00:11:43.410
However, when we have listed the
solutions to this problem
00:11:43.410 --> 00:11:47.290
in this way, then, we have to
put the capital omega squared
00:11:47.290 --> 00:11:50.560
at the back of M times phi.
00:11:50.560 --> 00:11:53.410
You can verify that by simply
multiplying out
00:11:53.410 --> 00:11:54.930
the right-hand side.
00:11:54.930 --> 00:12:00.690
We also have this relationship
here, which is the M ortho
00:12:00.690 --> 00:12:06.670
normality, the phi ortho
normality with M. In other
00:12:06.670 --> 00:12:10.910
words, this here says nothing
else and phi I transposed M
00:12:10.910 --> 00:12:15.440
phi j is equal to delta I j.
00:12:15.440 --> 00:12:19.190
We call this the
chronica delta.
00:12:19.190 --> 00:12:22.450
And this chronica delta
is equal to 1 for I
00:12:22.450 --> 00:12:24.430
being equal to j.
00:12:24.430 --> 00:12:26.690
And it's 0 otherwise.
00:12:26.690 --> 00:12:30.560
So we are having this
relationship here.
00:12:30.560 --> 00:12:34.380
And if we use this relationship,
and we are
00:12:34.380 --> 00:12:39.000
premultiplying this equation by
phi transposed to capital
00:12:39.000 --> 00:12:41.960
phi transposed now,
both sides.
00:12:41.960 --> 00:12:43.890
We are finding that we
are getting here
00:12:43.890 --> 00:12:45.600
the identity matrix.
00:12:45.600 --> 00:12:49.160
And therefore, we're having that
this left-hand side is
00:12:49.160 --> 00:12:51.020
equal to omega squared.
00:12:51.020 --> 00:12:53.080
This matrix here.
00:12:53.080 --> 00:12:54.550
And this is an important fact.
00:12:54.550 --> 00:12:56.850
In other words, we find
that the vectors--
00:12:56.850 --> 00:12:58.400
the eigenvectors--
00:12:58.400 --> 00:13:01.990
satisfy these two
relationships.
00:13:01.990 --> 00:13:06.540
Now, if we look back at what
we started off with in our
00:13:06.540 --> 00:13:11.320
multiple position analysis
discussion, we wanted a matrix
00:13:11.320 --> 00:13:19.130
that we call P, which is such
that P transposed M, P ideally
00:13:19.130 --> 00:13:21.700
is a diagonal matrix.
00:13:21.700 --> 00:13:23.780
Well, we have no
such a matrix.
00:13:23.780 --> 00:13:29.850
If we take phi being equal to P.
Then, phi transposed M phi
00:13:29.850 --> 00:13:31.570
is in fact a diagonal matrix.
00:13:31.570 --> 00:13:34.710
In fact, a very nice diagonal
matrix, the identity matrix,
00:13:34.710 --> 00:13:36.390
which is a matrix that
is diagonal.
00:13:36.390 --> 00:13:39.330
It has just once on
the diagonal.
00:13:39.330 --> 00:13:43.200
Similarly, we wondered here
that P transposed KP is a
00:13:43.200 --> 00:13:44.320
diagonal matrix.
00:13:44.320 --> 00:13:48.720
Well, we have achieved that
by this phi matrix.
00:13:48.720 --> 00:13:53.150
So the phi matrix is, indeed,
a particular P
00:13:53.150 --> 00:13:56.160
matrix that we can use.
00:13:56.160 --> 00:13:59.850
So what we are then saying is
let us use a multiple position
00:13:59.850 --> 00:14:02.950
analysis this particular
P matrix.
00:14:02.950 --> 00:14:06.630
The P matrix, which is equal
to the phi matrix and phi
00:14:06.630 --> 00:14:11.340
storing the eigenvectors of the
generalized eigenproblem.
00:14:11.340 --> 00:14:15.610
Well, substituting from here, as
before, into the governing
00:14:15.610 --> 00:14:17.030
dynamic equilibrium
equations, we
00:14:17.030 --> 00:14:19.310
directly obtain this equation.
00:14:19.310 --> 00:14:23.990
However, we notice now that this
matrix here, in general,
00:14:23.990 --> 00:14:26.710
is not yet a diagonal matrix.
00:14:26.710 --> 00:14:29.420
We call that our C curl.
00:14:29.420 --> 00:14:31.960
And our C curl has not been
diagonalized yet.
00:14:31.960 --> 00:14:36.670
We will have to deal with that
matrix still a little later.
00:14:36.670 --> 00:14:39.040
How do we solve these
equations?
00:14:39.040 --> 00:14:42.220
Well, if we solve these
equations, we have to also
00:14:42.220 --> 00:14:45.080
recognize, of course, that we
have to include the initial
00:14:45.080 --> 00:14:46.610
conditions to the problem.
00:14:46.610 --> 00:14:49.610
And the initial conditions
mean that the initial
00:14:49.610 --> 00:14:52.670
displacements are defined
and the initial
00:14:52.670 --> 00:14:55.250
velocities are defined.
00:14:55.250 --> 00:14:57.390
These, of course, are the
initial conditions that are
00:14:57.390 --> 00:15:03.170
used in the direct integration
solution of the governing
00:15:03.170 --> 00:15:04.970
dynamic equilibrium equations.
00:15:04.970 --> 00:15:08.730
Well, if we are now performing
the solution of this set of
00:15:08.730 --> 00:15:11.510
equations, we will have to
transform these initial
00:15:11.510 --> 00:15:15.180
conditions to initial conditions
on the generalized
00:15:15.180 --> 00:15:16.380
displacements.
00:15:16.380 --> 00:15:19.780
And that is achieved via
these relations here.
00:15:19.780 --> 00:15:22.950
Let us see how we
derive these.
00:15:22.950 --> 00:15:26.140
Well, if we look at this
equation here, and if we
00:15:26.140 --> 00:15:28.310
notice that, of course,
this equation must be
00:15:28.310 --> 00:15:30.240
applicable 4 times 0.
00:15:30.240 --> 00:15:32.270
We can simply put a 0 there.
00:15:32.270 --> 00:15:35.270
And we put a 0 there.
00:15:35.270 --> 00:15:37.750
Then, of course, we also
recognize that we can
00:15:37.750 --> 00:15:44.100
premultiply this equation
with this relation here.
00:15:44.100 --> 00:15:46.880
And if we do that on the
right-hand side, we notice
00:15:46.880 --> 00:15:49.420
that phi transposed M
times that phi is
00:15:49.420 --> 00:15:52.000
the identity matrix.
00:15:52.000 --> 00:15:56.010
And so we have this relation
directly obtained.
00:15:56.010 --> 00:15:58.420
The left-hand side in this
equation is the right-hand
00:15:58.420 --> 00:15:59.740
side in this equation.
00:15:59.740 --> 00:16:03.920
And the right-hand side here
is simply this 0 x now.
00:16:03.920 --> 00:16:06.410
The same holds also for
the initial velocity.
00:16:06.410 --> 00:16:10.720
We can simply put dots on top
of the U here and of the X.
00:16:10.720 --> 00:16:12.750
And therefore, we have
this relation here.
00:16:12.750 --> 00:16:16.480
So the solution of this set of
equations and with these
00:16:16.480 --> 00:16:19.840
initial conditions subject to
these initial conditions,
00:16:19.840 --> 00:16:22.960
gives us a solution to of
the dynamic response.
00:16:22.960 --> 00:16:28.200
Now, I should mention here that
we still have to deal
00:16:28.200 --> 00:16:29.530
with the damping matrix.
00:16:29.530 --> 00:16:32.990
The damping matrix here needs
particular attention.
00:16:32.990 --> 00:16:36.060
We do not have, in general,
a diagonal matrix here.
00:16:36.060 --> 00:16:40.920
And we will want to construct
specific damping matrices with
00:16:40.920 --> 00:16:42.960
which we can deal,
effectively.
00:16:42.960 --> 00:16:45.050
And yet, of course, these
matrices must make,
00:16:45.050 --> 00:16:46.820
physically, sense.
00:16:46.820 --> 00:16:49.550
Let us look now, however, first
at the case where we
00:16:49.550 --> 00:16:51.480
neglect damping.
00:16:51.480 --> 00:16:55.020
When we neglect damping, we have
this set of equations and
00:16:55.020 --> 00:16:57.820
individual equations
of this form.
00:16:57.820 --> 00:17:01.590
Notice that these are individual
equations because
00:17:01.590 --> 00:17:03.950
this is a vector.
00:17:03.950 --> 00:17:06.609
This is a diagonal matrix.
00:17:06.609 --> 00:17:11.000
And on the right-hand side, we
simply have a vector off load,
00:17:11.000 --> 00:17:13.010
of course, time dependent.
00:17:13.010 --> 00:17:17.930
So we can look at the first row
in this matrix relation.
00:17:17.930 --> 00:17:20.859
And the first row in that matrix
relation will be this
00:17:20.859 --> 00:17:24.490
equation here with
I equal to 1.
00:17:24.490 --> 00:17:27.200
The I-th row in this relation
here is the
00:17:27.200 --> 00:17:29.730
I-th equation here.
00:17:29.730 --> 00:17:33.000
In other words, I subscripts
on the [UNINTELLIGIBLE].
00:17:33.000 --> 00:17:34.980
And, in general therefore,
we want to
00:17:34.980 --> 00:17:37.300
solve n such equations.
00:17:37.300 --> 00:17:42.490
I going from 1 to n, with this
being the load vector and, of
00:17:42.490 --> 00:17:44.760
course, with the initial
conditions.
00:17:44.760 --> 00:17:48.860
These initial conditions are
obtained by simply looking at
00:17:48.860 --> 00:17:54.300
this equation here and
extracting the I-th row from
00:17:54.300 --> 00:17:56.600
each of these two relations.
00:17:56.600 --> 00:17:58.510
So this is then the problem.
00:17:58.510 --> 00:18:05.230
Notice the following that if we
solve all n equations with
00:18:05.230 --> 00:18:09.740
these initial conditions, then,
if we use, for example,
00:18:09.740 --> 00:18:15.640
the Newmark direct integration
method on these equations, we
00:18:15.640 --> 00:18:21.330
would obtain exactly the same
solution as if we had solved
00:18:21.330 --> 00:18:23.270
the fully coupled equations.
00:18:23.270 --> 00:18:25.970
The fully coupled equations--
let me write them down once
00:18:25.970 --> 00:18:27.620
more here--
00:18:27.620 --> 00:18:29.900
plus KU equals R.
00:18:29.900 --> 00:18:34.030
In other words, whether we
apply the Newmark time
00:18:34.030 --> 00:18:37.710
integration scheme, a direct
integration scheme, to this
00:18:37.710 --> 00:18:42.100
set of equations with the
initial conditions on U and U
00:18:42.100 --> 00:18:47.460
dot or apply the Newmark direct
time integration scheme
00:18:47.460 --> 00:18:52.020
to this set of equations, or
rather this one here with
00:18:52.020 --> 00:18:55.900
these initial conditions, we
obtain exactly the same
00:18:55.900 --> 00:18:59.900
response provided we use the
same time step, delta t in
00:18:59.900 --> 00:19:01.700
both integrations.
00:19:01.700 --> 00:19:03.570
There has been no assumption
so far.
00:19:03.570 --> 00:19:06.770
All we have done is a
transformation from the U
00:19:06.770 --> 00:19:10.050
displacements to the
X displacements.
00:19:10.050 --> 00:19:14.430
And this transformation is
effective if we can find the
00:19:14.430 --> 00:19:17.550
phi matrix that we're using
in the transformation very
00:19:17.550 --> 00:19:20.090
cheaply, very economically.
00:19:20.090 --> 00:19:24.500
Well, the effectiveness, really,
of multiple position
00:19:24.500 --> 00:19:28.630
goes beyond what I have
just described.
00:19:28.630 --> 00:19:32.640
The important point in multiple
position analysis is
00:19:32.640 --> 00:19:36.120
that we do not need to solve
all of these equations.
00:19:36.120 --> 00:19:38.530
That we do not need to solve
and look at all of the
00:19:38.530 --> 00:19:41.370
decoupled equations.
00:19:41.370 --> 00:19:45.040
But let us look first once also
at how we can solve the
00:19:45.040 --> 00:19:47.680
decoupled equations
now exactly.
00:19:47.680 --> 00:19:50.480
Where we can solve them exactly
for example using a
00:19:50.480 --> 00:19:54.040
Duhamel integral formulation
as shown here.
00:19:54.040 --> 00:19:58.390
X i of t is given via
this relation here.
00:19:58.390 --> 00:20:01.670
And we would, of course,
substitute our ri of tau where
00:20:01.670 --> 00:20:07.020
we now assume that this part
here is given, directly.
00:20:07.020 --> 00:20:11.600
In a computer program, very
frequently r i would only be
00:20:11.600 --> 00:20:14.050
given as a segment of
straight lines.
00:20:14.050 --> 00:20:16.610
And if that is the case, then,
of course, we might have to
00:20:16.610 --> 00:20:20.190
still evaluate all of this
here numerically if the
00:20:20.190 --> 00:20:25.330
segment is very complicated,
complex, like
00:20:25.330 --> 00:20:26.710
an earthquake analysis.
00:20:26.710 --> 00:20:30.090
ri might be varying
very rapidly.
00:20:30.090 --> 00:20:32.930
And then, we might still want to
evaluate this, numerically.
00:20:32.930 --> 00:20:35.860
However in general, we can
evaluate here this
00:20:35.860 --> 00:20:37.590
relationship analytically.
00:20:37.590 --> 00:20:41.270
And we don't need to use
numerical integration.
00:20:41.270 --> 00:20:45.930
Having got xi, of course, we get
our U of t as shown here
00:20:45.930 --> 00:20:48.440
by our transforming back.
00:20:48.440 --> 00:20:52.980
Well, the important point that
I now would like to get back
00:20:52.980 --> 00:20:58.440
to you and discuss with you for
a little while is that we
00:20:58.440 --> 00:21:02.920
do not need to use or to solve
all of these equations in
00:21:02.920 --> 00:21:04.680
multiple position analysis.
00:21:04.680 --> 00:21:08.220
In fact, multiple position
analysis is only effective
00:21:08.220 --> 00:21:10.625
when we do not need
to look at all of
00:21:10.625 --> 00:21:12.510
these decoupled equations.
00:21:12.510 --> 00:21:16.360
If we have to solve all
decoupled equations, then, we
00:21:16.360 --> 00:21:19.510
would have to calculate, course,
first all eigenvalues
00:21:19.510 --> 00:21:23.140
and eigenvectors, which can
be an enormous expense.
00:21:23.140 --> 00:21:25.490
And then, the multiple
position analysis
00:21:25.490 --> 00:21:28.150
would not be effective.
00:21:28.150 --> 00:21:32.760
I would like to repeat, however,
if we do solve all
00:21:32.760 --> 00:21:36.560
decoupled equations, using a
specific time integration
00:21:36.560 --> 00:21:41.300
scheme with a specific
data t time step.
00:21:41.300 --> 00:21:44.810
Then, if we use the same time
integration scheme on
00:21:44.810 --> 00:21:48.730
decoupled equations with the
same time step, we must get
00:21:48.730 --> 00:21:51.560
exactly the same response.
00:21:51.560 --> 00:21:54.750
Then there has been no
assumption so far.
00:21:54.750 --> 00:21:58.465
Well, however, if we then
do neglect some of
00:21:58.465 --> 00:22:00.190
the equations here.
00:22:00.190 --> 00:22:02.230
We do not solve some
of the equations.
00:22:02.230 --> 00:22:05.270
Then, of course, our response
will be different that we are
00:22:05.270 --> 00:22:09.310
predicting via this approach
when we compare with the
00:22:09.310 --> 00:22:13.000
response predicted by solving
these equations.
00:22:13.000 --> 00:22:20.470
Well, let us look once at this
graph again, which we already
00:22:20.470 --> 00:22:22.740
looked at briefly in
the last lecture.
00:22:22.740 --> 00:22:26.770
And I would like to explain
to you now why it is not
00:22:26.770 --> 00:22:30.020
necessary to solve all of the
decoupled equations in
00:22:30.020 --> 00:22:31.460
multiple position analysis.
00:22:31.460 --> 00:22:34.450
Why it is not, in general,
not necessary?
00:22:34.450 --> 00:22:36.990
Well, if we look at a
single oscillator
00:22:36.990 --> 00:22:38.240
equation, as given here.
00:22:38.240 --> 00:22:40.130
I include a damp in here,
of course, side
00:22:40.130 --> 00:22:42.590
can be equal to 0.
00:22:42.590 --> 00:22:46.270
Where P is the driving
frequency, omega is the
00:22:46.270 --> 00:22:49.510
frequency of the oscillator
itself.
00:22:49.510 --> 00:22:55.490
And if we plot the maximum
dynamic response observed,
00:22:55.490 --> 00:22:59.460
which we define if the dynamic
load factor when we divide
00:22:59.460 --> 00:23:03.810
that maximum dynamic response by
the static response, we get
00:23:03.810 --> 00:23:05.550
this set of curves.
00:23:05.550 --> 00:23:10.370
Let us look at say the curved
side equal to 0
00:23:10.370 --> 00:23:12.010
a little bit closer.
00:23:12.010 --> 00:23:15.830
If we do that, and if we
solve this equation
00:23:15.830 --> 00:23:17.140
with side equals 0.
00:23:17.140 --> 00:23:19.390
This term now not being there.
00:23:19.390 --> 00:23:23.920
For certain values of P over
omega, we are getting this
00:23:23.920 --> 00:23:27.370
curve here and that
curve here.
00:23:30.100 --> 00:23:31.140
What do we notice?
00:23:31.140 --> 00:23:37.060
Well, we notice that when P over
omega is equal to 0, we
00:23:37.060 --> 00:23:38.720
are getting the static
response.
00:23:38.720 --> 00:23:40.780
Or P over omega very
small, we're
00:23:40.780 --> 00:23:43.260
getting the static response.
00:23:43.260 --> 00:23:46.990
This means, of course, that
omega, the system frequency,
00:23:46.990 --> 00:23:50.630
is much larger than the applied
frequency, which means
00:23:50.630 --> 00:23:54.150
really that this oscillator
is extremely stiff.
00:23:56.920 --> 00:24:03.380
Well, when compared to the
frequency of load application.
00:24:03.380 --> 00:24:04.850
So then, we are in this range.
00:24:04.850 --> 00:24:10.800
And we notice that say P over
omega in this range here,
00:24:10.800 --> 00:24:14.140
smaller than a certain value,
and this is here 0.5 say,
00:24:14.140 --> 00:24:15.970
smaller than 0.2.
00:24:15.970 --> 00:24:20.170
We would see that basically, the
maximum dynamic response
00:24:20.170 --> 00:24:24.270
is really just nothing else
than the static response.
00:24:24.270 --> 00:24:27.110
So we are talking in this
region here really about
00:24:27.110 --> 00:24:28.160
static response.
00:24:28.160 --> 00:24:31.060
And in this region here, we
are talking about a truly
00:24:31.060 --> 00:24:33.010
dynamic response.
00:24:33.010 --> 00:24:34.380
Now what does that mean?
00:24:34.380 --> 00:24:40.060
Well, if we look at our
decoupled equations here, it
00:24:40.060 --> 00:24:49.000
means that if i consist of
frequencies, basically, which
00:24:49.000 --> 00:24:54.630
when measured on the system of
the oscillator frequency are
00:24:54.630 --> 00:24:59.960
very small, we, basically, would
have a static solution.
00:24:59.960 --> 00:25:02.950
You would, basically, have a
static solution for this
00:25:02.950 --> 00:25:05.920
equilibrium equation.
00:25:05.920 --> 00:25:11.800
A truly dynamic solution will
only be obtained when P, in
00:25:11.800 --> 00:25:15.850
other words, the frequency of
the load divided by omega,
00:25:15.850 --> 00:25:20.030
falls into this range here.
00:25:20.030 --> 00:25:23.630
Well, this is what we
are using in most
00:25:23.630 --> 00:25:25.190
superposition analysis.
00:25:25.190 --> 00:25:32.590
We only need to consider those
frequencies of the system.
00:25:32.590 --> 00:25:35.760
And these are the frequencies
of the systems, the omegas,
00:25:35.760 --> 00:25:40.340
which are truly excited
by the dynamic loads.
00:25:40.340 --> 00:25:44.160
We only need to look at those
decoupled equilibrium
00:25:44.160 --> 00:25:49.250
equations for which we have that
the frequencies in the
00:25:49.250 --> 00:25:53.030
decoupled equations are truly
excited by the dynamic loads.
00:25:53.030 --> 00:25:55.900
Now, I only talked about
the P value here.
00:25:55.900 --> 00:25:58.340
Of course, in an actual
analysis, we would find that
00:25:58.340 --> 00:26:02.940
ri of P contains a number of
frequencies, a spectrum of
00:26:02.940 --> 00:26:03.920
frequencies.
00:26:03.920 --> 00:26:07.510
Well, what we have to do is look
at the highest frequency
00:26:07.510 --> 00:26:10.830
that is contained in ri.
00:26:10.830 --> 00:26:14.750
And we measure that highest
frequency on the frequencies
00:26:14.750 --> 00:26:16.600
of the system.
00:26:16.600 --> 00:26:21.120
If that highest frequency
divided by the frequency of
00:26:21.120 --> 00:26:24.510
the system is falling into
this range here, we are
00:26:24.510 --> 00:26:27.060
talking about a static
response, basically.
00:26:27.060 --> 00:26:35.160
And we do not need to consider
those equilibrium equations in
00:26:35.160 --> 00:26:38.340
are multiple position
analysis.
00:26:38.340 --> 00:26:41.870
Well, so what we are then saying
is let us all solve
00:26:41.870 --> 00:26:46.310
these equations not from i
equals 1 to n but from i
00:26:46.310 --> 00:26:55.260
equals 1 to P. And then, instead
of obtaining the true
00:26:55.260 --> 00:26:59.640
response, the "exact" response,
we are obtaining a
00:26:59.640 --> 00:27:06.970
response U P. i equals 1 to P.
Phi i, x i gives us U P, where
00:27:06.970 --> 00:27:10.685
U P is an approximation now
to the U displacements.
00:27:14.150 --> 00:27:18.250
And the U displacements, once
again, are the solution to
00:27:18.250 --> 00:27:19.830
this system of equations.
00:27:22.590 --> 00:27:24.370
How can we find out the error?
00:27:24.370 --> 00:27:29.710
Well, if U P is an approximate
solution to the system of
00:27:29.710 --> 00:27:33.670
equations that we want to solve,
then, let us substitute
00:27:33.670 --> 00:27:35.730
back into that system
of equations.
00:27:35.730 --> 00:27:37.270
Let me write it down
once more here.
00:27:37.270 --> 00:27:41.530
We have MU double dot
plus KU equals R
00:27:41.530 --> 00:27:42.830
that we want to solve.
00:27:42.830 --> 00:27:46.540
Well, we can also write that, of
course, MU double dot plus
00:27:46.540 --> 00:27:53.380
KU minus R. And this
shall be 0.
00:27:53.380 --> 00:27:56.620
So if we have an approximate
solution to this set of
00:27:56.620 --> 00:27:59.840
equations, why not just
substitute that approximate
00:27:59.840 --> 00:28:04.570
solution in here and see how
large the right-hand side is.
00:28:04.570 --> 00:28:09.100
If it's exactly 0, well, then
we have that our U P is,
00:28:09.100 --> 00:28:12.540
indeed, very close to U. And if
it's exactly 0 because U P
00:28:12.540 --> 00:28:13.910
would be equal to U.
00:28:13.910 --> 00:28:15.680
We will not find this to be 0.
00:28:15.680 --> 00:28:18.550
We will never find it to be 0
because in a computer, we
00:28:18.550 --> 00:28:21.620
would, of course, use finite
digit arithmetic to evaluate
00:28:21.620 --> 00:28:22.400
the left-hand side.
00:28:22.400 --> 00:28:24.020
And we would have at
least round off.
00:28:24.020 --> 00:28:27.840
But what we want is at the
right-hand side calculated,
00:28:27.840 --> 00:28:34.340
which we now might call some
value shall be close to 0.
00:28:34.340 --> 00:28:36.990
And that means really what
we want to measure
00:28:36.990 --> 00:28:39.230
is this part here.
00:28:39.230 --> 00:28:40.920
Now what have I written
down here?
00:28:40.920 --> 00:28:47.180
I have here epsilon P of t
being this vector here.
00:28:47.180 --> 00:28:48.020
This is a vector.
00:28:48.020 --> 00:28:51.680
And we can call this an out of
balance load vector divided by
00:28:51.680 --> 00:28:53.490
the applied load vector.
00:28:53.490 --> 00:28:59.630
Notice that what I'm doing here,
I take the norm of this
00:28:59.630 --> 00:29:02.580
vector because I want to
get a single number.
00:29:02.580 --> 00:29:07.570
I want to get a single number
here, which is representative
00:29:07.570 --> 00:29:09.680
of all the elements
in that vector.
00:29:09.680 --> 00:29:12.770
The same applies here.
00:29:12.770 --> 00:29:17.200
This norm is calculated by
taking each value in this
00:29:17.200 --> 00:29:21.790
vector, squaring that value,
adding the squares of the
00:29:21.790 --> 00:29:24.385
values up, and then, taking
a square root out
00:29:24.385 --> 00:29:26.160
of that final value.
00:29:26.160 --> 00:29:30.030
In other words, if we write it
down here, quickly, what we do
00:29:30.030 --> 00:29:35.180
is if we have a vector a, and
we want to find this norm.
00:29:35.180 --> 00:29:39.900
What we do is we take a 1
squared plus a 2 squared, et
00:29:39.900 --> 00:29:43.360
cetera, add all the individual
entries up
00:29:43.360 --> 00:29:44.620
of that vector here--
00:29:44.620 --> 00:29:46.300
squares of the entries up--
00:29:46.300 --> 00:29:48.010
and then, take the square
root out of it.
00:29:48.010 --> 00:29:50.170
And that gives us a single
number, which is
00:29:50.170 --> 00:29:53.080
representative of all the
elements in that vector.
00:29:53.080 --> 00:29:57.150
Notice that if one element is
large, well then, this value
00:29:57.150 --> 00:29:59.900
must be large, too, because
that one element will make
00:29:59.900 --> 00:30:01.870
this value large here.
00:30:01.870 --> 00:30:07.450
Well, if epsilon t is close to
0, assuming that r of t is not
00:30:07.450 --> 00:30:08.490
0, of course.
00:30:08.490 --> 00:30:09.650
Otherwise, you couldn't
use this.
00:30:09.650 --> 00:30:12.640
You would have to use some
equivalent approach.
00:30:12.640 --> 00:30:17.210
If epsilon P is close to 0,
then, we know that our
00:30:17.210 --> 00:30:19.880
equilibrium equation has been
solved quite accurately.
00:30:23.290 --> 00:30:26.820
In general, we might find that
it's not close to 0.
00:30:26.820 --> 00:30:30.430
And what we want to do is
a static correction.
00:30:30.430 --> 00:30:38.260
This static correction takes
into account the fact that we
00:30:38.260 --> 00:30:40.540
have neglected the
higher modes.
00:30:40.540 --> 00:30:42.590
And in the higher modes, of
course, we know we have
00:30:42.590 --> 00:30:45.290
basically static response.
00:30:45.290 --> 00:30:49.290
that's why we did not include
those equations in the
00:30:49.290 --> 00:30:51.140
multiple position analysis.
00:30:51.140 --> 00:30:54.050
So what do we do in that
static correction?
00:30:54.050 --> 00:30:57.540
We recognize that R could
be written in this form.
00:30:57.540 --> 00:31:01.780
This RI is exactly the value
that I defined earlier.
00:31:01.780 --> 00:31:04.130
M phi i is a vector.
00:31:04.130 --> 00:31:07.380
You see, I can take M and
multiply it by phi i, and I
00:31:07.380 --> 00:31:08.680
get a vector.
00:31:08.680 --> 00:31:11.380
Now they are in such
a vectors.
00:31:11.380 --> 00:31:13.360
We might not have calculated
all n.
00:31:13.360 --> 00:31:18.950
But you can say that indeed,
in theory there are n.
00:31:18.950 --> 00:31:21.700
So we can write this down.
00:31:21.700 --> 00:31:27.250
We also then can calculate the
delta R, which is that amount
00:31:27.250 --> 00:31:30.150
of the load vector, which has
not been included in the
00:31:30.150 --> 00:31:31.910
multiple position analysis.
00:31:31.910 --> 00:31:32.890
How do we obtain it?
00:31:32.890 --> 00:31:36.160
Well, we know R is the
total load vector.
00:31:36.160 --> 00:31:40.500
And if we now sum instead of i
equals 1 to n, only from i
00:31:40.500 --> 00:31:44.590
equals 1 to P, we are getting
this part here.
00:31:44.590 --> 00:31:48.200
But this part is exactly what
we have included in the
00:31:48.200 --> 00:31:50.650
multiple position analysis
already.
00:31:50.650 --> 00:31:53.820
So we are left with a vector on
the left-hand side, which
00:31:53.820 --> 00:31:56.410
we have not included in the
multiple position analysis.
00:31:56.410 --> 00:32:00.360
And we did not include it
because we knew at most, we
00:32:00.360 --> 00:32:03.230
can have static response,
corresponding to these loads.
00:32:03.230 --> 00:32:05.450
Let us now calculate that
static response.
00:32:05.450 --> 00:32:07.480
That's done right down here.
00:32:07.480 --> 00:32:10.830
Now, this static response has
to be added to the response
00:32:10.830 --> 00:32:15.630
which we predicted in the
multiple position analysis.
00:32:15.630 --> 00:32:20.010
Let us now go to the case
with damping included.
00:32:20.010 --> 00:32:22.050
So far we neglected that.
00:32:22.050 --> 00:32:26.180
Now, with damping included, we
have this part here, too.
00:32:26.180 --> 00:32:28.580
Notice phi transpose C
phi, in general, is
00:32:28.580 --> 00:32:29.710
not a diagonal matrix.
00:32:29.710 --> 00:32:31.180
And that's why I considered
first the
00:32:31.180 --> 00:32:32.690
case without damping.
00:32:32.690 --> 00:32:36.270
However, if we have
proportionate damping, and in
00:32:36.270 --> 00:32:39.950
structural analysis, frequently,
all that is
00:32:39.950 --> 00:32:44.090
necessary is to include
proportionate damping.
00:32:44.090 --> 00:32:46.360
Which we postulate to
be of this form.
00:32:46.360 --> 00:32:48.360
2 omega i xi i.
00:32:48.360 --> 00:32:52.710
Delta ij is again the chronica
delta, which is equal to 1
00:32:52.710 --> 00:32:56.950
when i is equal to j and is
equal to 0 otherwise.
00:32:56.950 --> 00:33:00.740
So if we are postulating
this, where omega i,
00:33:00.740 --> 00:33:02.080
of course, is given.
00:33:02.080 --> 00:33:05.860
That is the free vibration
frequency, which is stored in
00:33:05.860 --> 00:33:07.490
this matrix here.
00:33:07.490 --> 00:33:10.970
xi i is a value, which
we are not given yet.
00:33:10.970 --> 00:33:12.350
We have not discussed it yet.
00:33:12.350 --> 00:33:17.270
But if we postulate that this
matrix here, basically, or
00:33:17.270 --> 00:33:20.200
each entry in that matrix can
be written in this way.
00:33:20.200 --> 00:33:23.400
Then, indeed, we would have
decoupled equations.
00:33:23.400 --> 00:33:27.860
It turns out that in practice,
as I mentioned already, this
00:33:27.860 --> 00:33:32.810
is all that is necessary to
include damping appropriately.
00:33:32.810 --> 00:33:42.200
Well, if we know these xi i
values, then, we can also
00:33:42.200 --> 00:33:50.540
construct a C matrix with
the xi i values given.
00:33:50.540 --> 00:33:54.960
The xi i values are frequently
obtained from experiments.
00:33:54.960 --> 00:33:59.870
In other words, the structure
is taken, its vibrated, it's
00:33:59.870 --> 00:34:02.600
excited into certain
vibration modes.
00:34:02.600 --> 00:34:04.840
And from these experiments
then, we
00:34:04.840 --> 00:34:07.870
obtain the xi i value.
00:34:07.870 --> 00:34:11.070
So once we have the xi
i value given, we can
00:34:11.070 --> 00:34:12.280
calculate the C matrix.
00:34:12.280 --> 00:34:15.889
And one way of calculating is
to use the Caughey series.
00:34:15.889 --> 00:34:19.679
Here we have the Caughey series,
which was proposed for
00:34:19.679 --> 00:34:25.929
calculating the C matrix, which
satisfies this criteria.
00:34:25.929 --> 00:34:28.889
In other words, damping
is proportional.
00:34:28.889 --> 00:34:36.770
Well, if the psi i values are
given, we can calculate the
00:34:36.770 --> 00:34:40.020
omega i, of course
are also given.
00:34:40.020 --> 00:34:45.179
We can calculate using this
equation, the constants a 0 to
00:34:45.179 --> 00:34:46.860
ap minus 1.
00:34:46.860 --> 00:34:51.100
Once we have these constants,
we can substitute them into
00:34:51.100 --> 00:34:53.699
here and be coming up
with a C matrix.
00:34:53.699 --> 00:34:59.100
And this C matrix here,
indeed, satisfies the
00:34:59.100 --> 00:35:02.590
normality property shown here.
00:35:02.590 --> 00:35:07.510
So, in other words, given the
psi i values from experimental
00:35:07.510 --> 00:35:12.500
results, given the omega i
values, we can calculate a C
00:35:12.500 --> 00:35:16.820
matrix that, in fact, gives us
a diagonal matrix when we
00:35:16.820 --> 00:35:20.320
carry out phi i transposed
C phi j.
00:35:20.320 --> 00:35:24.000
A special case of this C matrix
is Rayleigh damping,
00:35:24.000 --> 00:35:25.980
used abundantly in practice.
00:35:25.980 --> 00:35:29.130
Where we just include the first
two terms, alpha and
00:35:29.130 --> 00:35:31.230
beta being constants now.
00:35:31.230 --> 00:35:33.700
Let us go through a particular
example.
00:35:33.700 --> 00:35:40.520
Let's say that we know the
critical damping psi 1 is 0.02
00:35:40.520 --> 00:35:43.190
and psi 2 is 0.10.
00:35:43.190 --> 00:35:46.340
In other words, 10% critical
damping in the second mode and
00:35:46.340 --> 00:35:48.590
2% critical damping
in the first mode.
00:35:48.590 --> 00:35:53.060
The corresponding frequencies
are 1 and 3.
00:35:53.060 --> 00:35:54.910
We want to calculate
our friend, beta.
00:35:54.910 --> 00:36:00.660
Well, what we do is use this C
matrix in the relation, phi i
00:36:00.660 --> 00:36:03.440
transposed, C times phi i.
00:36:03.440 --> 00:36:06.280
And that should give
us 2 omega psi i.
00:36:06.280 --> 00:36:10.490
Of course, we have two
i values, i equals 1
00:36:10.490 --> 00:36:11.740
and i equals 2.
00:36:11.740 --> 00:36:15.240
So we can apply this
equation twice.
00:36:15.240 --> 00:36:18.500
Well, when we calculate
it out, we get this.
00:36:18.500 --> 00:36:22.210
Applying it twice, we directly
obtain these
00:36:22.210 --> 00:36:24.110
two equations here.
00:36:24.110 --> 00:36:28.100
And we solve these two equations
for alpha and beta.
00:36:28.100 --> 00:36:33.440
Having alpha and beta, we can
substitute now back into this
00:36:33.440 --> 00:36:37.280
equation here and have
our C matrix.
00:36:37.280 --> 00:36:41.820
This would be, for example, a C
matrix that we could use in
00:36:41.820 --> 00:36:44.350
a direct integration analysis.
00:36:44.350 --> 00:36:48.040
You see, if we are performing a
multiple position analysis,
00:36:48.040 --> 00:36:51.230
we might know, from experience,
from experimental
00:36:51.230 --> 00:36:57.090
results, that psi 1 and psi
2 are of these two values.
00:36:57.090 --> 00:37:01.760
And then, if we know that we
have so much critical damping
00:37:01.760 --> 00:37:05.460
or so much damping in the first
and second vibration
00:37:05.460 --> 00:37:09.140
mode, and we want to perform
a step-by-step direct
00:37:09.140 --> 00:37:12.790
integration analysis, we would
have to construct a C matrix.
00:37:12.790 --> 00:37:17.880
And this is the procedure
to construct a C matrix.
00:37:17.880 --> 00:37:22.210
Then, of course, we would have
to ask ourselves well, knowing
00:37:22.210 --> 00:37:27.990
these psi values for omega 1 and
omega 2, what are now the
00:37:27.990 --> 00:37:32.780
psi values that we are
implicitly using for the
00:37:32.780 --> 00:37:37.230
higher modes when we employ
this specific C matrix?
00:37:37.230 --> 00:37:40.290
And we use now, this
equation here.
00:37:40.290 --> 00:37:44.010
This is the equation, which we
already derived on this view
00:37:44.010 --> 00:37:46.810
graph here by looking
at this relation.
00:37:46.810 --> 00:37:49.200
You are now using this equation
because this equation
00:37:49.200 --> 00:37:53.710
must be applicable not just for
psi 1 and psi 2 but for
00:37:53.710 --> 00:37:55.250
all psi values.
00:37:55.250 --> 00:38:00.120
Well, then using this equation,
we can solve for the
00:38:00.120 --> 00:38:01.360
psi values.
00:38:01.360 --> 00:38:05.350
And, of course, psi 1 and psi 2,
substituting alpha and beta
00:38:05.350 --> 00:38:10.340
in here and omega 1 and omega
2 would, indeed, be just 2%
00:38:10.340 --> 00:38:13.190
and 10%, 0.02 and 0.10.
00:38:13.190 --> 00:38:15.220
However, we can now use
this equation to
00:38:15.220 --> 00:38:16.860
get the high xi values.
00:38:16.860 --> 00:38:19.560
And there's something to be
observed, which is quite
00:38:19.560 --> 00:38:24.360
important, namely we notice that
the beta value that we
00:38:24.360 --> 00:38:26.840
calculated is associated
with the high
00:38:26.840 --> 00:38:28.420
frequencies, basically.
00:38:28.420 --> 00:38:31.460
And the alpha value goes away.
00:38:31.460 --> 00:38:34.620
Or this contribution goes away
when we're looking at the high
00:38:34.620 --> 00:38:35.280
frequencies.
00:38:35.280 --> 00:38:38.680
For high frequencies therefore,
the beta value is a
00:38:38.680 --> 00:38:40.480
dominant factor.
00:38:40.480 --> 00:38:44.730
And for low frequencies,
the alpha value is
00:38:44.730 --> 00:38:47.800
the dominant factor.
00:38:47.800 --> 00:38:52.960
In other words, once again,
looking at this equation here,
00:38:52.960 --> 00:38:56.840
in the high frequency response,
beta K, really, will
00:38:56.840 --> 00:38:57.990
determine the damping.
00:38:57.990 --> 00:39:01.430
And in the low frequency, the
response is really the alpha
00:39:01.430 --> 00:39:06.030
value that determines
the damping.
00:39:06.030 --> 00:39:08.690
Well, the response solution
now would be performed in
00:39:08.690 --> 00:39:11.120
exactly the same
way as before.
00:39:11.120 --> 00:39:13.195
As in the case of
no damping, we
00:39:13.195 --> 00:39:15.990
solve now again P equations.
00:39:15.990 --> 00:39:18.290
P equations only.
00:39:18.290 --> 00:39:23.880
And we are doing that by looking
at this equation and
00:39:23.880 --> 00:39:26.740
with, of course, ri being given
as shown here and the
00:39:26.740 --> 00:39:29.960
initial conditions
being given here.
00:39:29.960 --> 00:39:35.120
Solving this equation here via
numerical integration or exact
00:39:35.120 --> 00:39:38.110
integration, using, for
something, Duhamel integral,
00:39:38.110 --> 00:39:41.990
we obtain the xi values
as a function of time.
00:39:41.990 --> 00:39:45.670
We substitute into here to
get our U P, which is an
00:39:45.670 --> 00:39:49.070
approximation to the
solution of the
00:39:49.070 --> 00:39:51.120
dynamic equilibrium equations.
00:39:51.120 --> 00:39:53.680
The dynamic equilibrium
equations now being MU double
00:39:53.680 --> 00:39:59.410
dot plus CU dot plus KU,
being equal to r.
00:39:59.410 --> 00:40:02.090
Notice we have a C matrix
in here now.
00:40:02.090 --> 00:40:05.170
However, and this is an
important point, we have a
00:40:05.170 --> 00:40:13.620
specific C matrix constructed
such that the phi i transposed
00:40:13.620 --> 00:40:18.110
C phi j gives us a diagonal
matrix with the diagonal
00:40:18.110 --> 00:40:23.740
elements being equal to
2 omega i, psi i.
00:40:23.740 --> 00:40:26.090
And one of these matrices,
for example, would be
00:40:26.090 --> 00:40:28.180
the Rayleigh matrix.
00:40:28.180 --> 00:40:32.310
The matrix obtained using
Rayleigh damping, rather.
00:40:32.310 --> 00:40:35.640
The important point is, once
again, that all we have been
00:40:35.640 --> 00:40:44.000
doing so far is to make a
change of bases from the
00:40:44.000 --> 00:40:47.610
finite element coordinates
to the generalized
00:40:47.610 --> 00:40:50.230
displacements, xi.
00:40:50.230 --> 00:40:51.840
That was the first step.
00:40:51.840 --> 00:40:56.370
The second step was that we
are neglecting the modes P
00:40:56.370 --> 00:41:03.330
plus 1 to n by only solving the
first P equations here and
00:41:03.330 --> 00:41:08.260
using this as an approximation
to the actual solution.
00:41:08.260 --> 00:41:16.500
The important point, of course,
is now that we have to
00:41:16.500 --> 00:41:19.200
ask ourselves when is
multiple position
00:41:19.200 --> 00:41:20.775
analysis really effective?
00:41:23.360 --> 00:41:26.870
Some considerations I have here
summarized on the last
00:41:26.870 --> 00:41:27.860
view graph.
00:41:27.860 --> 00:41:31.410
The multiple position analysis
is effective when the response
00:41:31.410 --> 00:41:33.880
lies in only a few modes.
00:41:33.880 --> 00:41:35.240
What does this mean?
00:41:35.240 --> 00:41:41.970
It means that if we look at our
eigenvalue problem, K phi
00:41:41.970 --> 00:41:44.500
equals omega squared M phi.
00:41:44.500 --> 00:41:47.750
We only need to solve this
eigenvalue problem
00:41:47.750 --> 00:41:50.270
for a very few modes.
00:41:50.270 --> 00:41:53.680
In other words, P being much
smaller than M, I will only
00:41:53.680 --> 00:41:59.470
have to solve for phi 1 up to
phi P and omega 1 squared up
00:41:59.470 --> 00:42:02.890
to omega P squared.
00:42:02.890 --> 00:42:05.850
This is a very important
point.
00:42:05.850 --> 00:42:11.750
I assumed here that we are
taking the lowest frequencies,
00:42:11.750 --> 00:42:16.530
omega 1 to omega P. Of course,
in general analysis, it might
00:42:16.530 --> 00:42:20.960
just be necessary to calculate
certain frequencies, say omega
00:42:20.960 --> 00:42:23.480
10 to omega 20.
00:42:23.480 --> 00:42:26.660
We only want to find the
frequencies in a certain range
00:42:26.660 --> 00:42:30.570
because we know that the
excitation frequency only lies
00:42:30.570 --> 00:42:32.160
in a particular range.
00:42:32.160 --> 00:42:34.540
That would be, for example,
the case in vibration
00:42:34.540 --> 00:42:37.690
excitation analysis.
00:42:37.690 --> 00:42:41.040
This then is a requirement for
multiple position analysis,
00:42:41.040 --> 00:42:42.920
really, to be effective.
00:42:42.920 --> 00:42:48.250
When the response is obtained
over many time intervals, then
00:42:48.250 --> 00:42:51.640
multiple position analysis
also becomes effective.
00:42:51.640 --> 00:42:55.010
We should remember here that
if we perform a direct
00:42:55.010 --> 00:42:59.580
integration analysis, and if
that is an implicit direct
00:42:59.580 --> 00:43:01.600
integration analysis, using the
00:43:01.600 --> 00:43:03.390
Newmark method, for example.
00:43:03.390 --> 00:43:08.830
Then, the number of operations
that we talked about in the
00:43:08.830 --> 00:43:13.400
Newmark method was
1/2 nm squared.
00:43:13.400 --> 00:43:16.390
That is the initial triangular
factorization
00:43:16.390 --> 00:43:18.040
that we have to perform.
00:43:18.040 --> 00:43:20.270
m being the half bandwidth.
00:43:20.270 --> 00:43:23.450
Plus 2 nm.
00:43:23.450 --> 00:43:26.630
n is the order, of course,
of the system.
00:43:26.630 --> 00:43:28.550
m is the half bandwidth,
once again.
00:43:28.550 --> 00:43:31.910
Times t, the number
of time steps.
00:43:31.910 --> 00:43:35.460
This is an initial expense that
we have to perform, using
00:43:35.460 --> 00:43:39.220
the direct integration Newmark
method, for example.
00:43:39.220 --> 00:43:41.460
The same holds for the
Wilson Theta method.
00:43:41.460 --> 00:43:42.900
or the Houbolt method.
00:43:42.900 --> 00:43:44.130
This is an initial expense.
00:43:44.130 --> 00:43:47.170
And then, we perform this
number of operations
00:43:47.170 --> 00:43:48.700
for each time step.
00:43:48.700 --> 00:43:52.890
Now, if we have to run 5,000
time steps, then direct
00:43:52.890 --> 00:43:53.750
integration would be
00:43:53.750 --> 00:43:57.100
extremely, extremely expensive.
00:43:57.100 --> 00:44:03.600
And, in fact, you might not be
able to afford it, using
00:44:03.600 --> 00:44:06.550
implicit direct integration.
00:44:06.550 --> 00:44:10.210
Well, therefore, you can see
directed by looking at this
00:44:10.210 --> 00:44:11.640
operation column.
00:44:11.640 --> 00:44:15.090
What should make the analysis
more effective is to bring
00:44:15.090 --> 00:44:16.640
down the bandwidths.
00:44:16.640 --> 00:44:19.460
And, of course, this is exactly
what we're doing in
00:44:19.460 --> 00:44:20.840
multiple position analysis.
00:44:20.840 --> 00:44:25.240
We are going from a bandwidth
of m to a bandwidth of 0.
00:44:25.240 --> 00:44:28.770
You are having a diagonal
matrix.
00:44:28.770 --> 00:44:33.660
Remember, the bandwidths we
defined to be the number of
00:44:33.660 --> 00:44:36.540
off diagonal elements here
that we are having.
00:44:36.540 --> 00:44:40.320
We did not include
the diagonal.
00:44:40.320 --> 00:44:41.130
That is a half bandwidth.
00:44:41.130 --> 00:44:43.230
If you look at the
total band here--
00:44:43.230 --> 00:44:48.700
that total band here was
equal to 2m plus 1.
00:44:48.700 --> 00:44:53.140
So if we are reducing the
bandwidths, as we do in
00:44:53.140 --> 00:44:56.540
multiple position analysis, then
we can see immediately
00:44:56.540 --> 00:45:00.060
that this number of operations
here, which is directly
00:45:00.060 --> 00:45:02.790
proportional to the number
of times steps goes down.
00:45:02.790 --> 00:45:05.940
And that is exactly the
objective in multiple position
00:45:05.940 --> 00:45:08.060
analysis that I talked
about earlier.
00:45:08.060 --> 00:45:09.900
That is the first important
objective.
00:45:09.900 --> 00:45:13.130
We want to reduce the bandwidths
for the direct
00:45:13.130 --> 00:45:15.720
integration analysis.
00:45:15.720 --> 00:45:18.380
The second objective, of course,
is that we do not need
00:45:18.380 --> 00:45:20.860
to even consider all
of the equations.
00:45:20.860 --> 00:45:24.410
So looking at n.
00:45:24.410 --> 00:45:26.130
We are reducing n, also.
00:45:26.130 --> 00:45:28.300
We are not looking anymore
at n equations.
00:45:28.300 --> 00:45:30.020
We are looking at P equations.
00:45:30.020 --> 00:45:33.230
And we are reducing the
bandwidths as I said earlier.
00:45:33.230 --> 00:45:37.400
m goes to 0.
00:45:37.400 --> 00:45:40.670
And n goes to P. Of
course, this is an
00:45:40.670 --> 00:45:42.290
approximate count here.
00:45:42.290 --> 00:45:46.765
So basically, you can see that
in the solution using multiple
00:45:46.765 --> 00:45:53.120
position analysis, the number
of operations in the time
00:45:53.120 --> 00:45:56.540
integration of the decoupled
equations is
00:45:56.540 --> 00:45:58.850
relatively small, is little.
00:45:58.850 --> 00:46:02.920
And, in fact, this solution
here can be obtained very,
00:46:02.920 --> 00:46:04.060
very effectively.
00:46:04.060 --> 00:46:06.520
The time integration of the
decoupled equations is very
00:46:06.520 --> 00:46:11.590
effective in an actual
practical analysis.
00:46:11.590 --> 00:46:14.880
Where the expense lies in
multiple position is really in
00:46:14.880 --> 00:46:17.380
the calculation of the required
00:46:17.380 --> 00:46:20.150
frequencies and motions.
00:46:20.150 --> 00:46:23.100
And, of course, this is exactly
the subject of the
00:46:23.100 --> 00:46:26.390
next lecture where I want to
discuss with you the effective
00:46:26.390 --> 00:46:29.680
solution techniques for
calculating these vectors and
00:46:29.680 --> 00:46:30.430
frequencies.
00:46:30.430 --> 00:46:35.170
Multiple position analysis is
effective, altogether only,
00:46:35.170 --> 00:46:39.570
when we can calculate these
eigenvalues and eigenvectors
00:46:39.570 --> 00:46:41.310
in an effective way.
00:46:41.310 --> 00:46:44.600
I should also finally, once
again, emphasize that it may
00:46:44.600 --> 00:46:48.310
be important to calculate the
error, epsilon pt that I
00:46:48.310 --> 00:46:50.570
referred to.
00:46:50.570 --> 00:46:56.060
Or to add a static correction
to the predicted response in
00:46:56.060 --> 00:47:00.580
the multiple position analysis
when P is, of course, much
00:47:00.580 --> 00:47:01.470
smaller than n.
00:47:01.470 --> 00:47:04.760
We have neglected the high
frequency response here.
00:47:04.760 --> 00:47:07.650
That high frequency response
is a static
00:47:07.650 --> 00:47:09.420
response, in general.
00:47:09.420 --> 00:47:13.330
And should be taken into
account via say static
00:47:13.330 --> 00:47:16.780
correction, as I have been
discussing with you.
00:47:16.780 --> 00:47:19.030
This is all I wanted to
say in this lecture.
00:47:19.030 --> 00:47:20.380
Thank you very much for
your attention.