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PROFESSOR: Ladies and gentlemen,
welcome to
00:00:23.260 --> 00:00:25.220
Lecture Number 10.
00:00:25.220 --> 00:00:27.600
In this lecture, I would like
to discuss a solution of
00:00:27.600 --> 00:00:29.520
dynamic equilibrium equations.
00:00:29.520 --> 00:00:32.240
And in particular, I would like
to talk about the direct
00:00:32.240 --> 00:00:35.860
integration solution of the
equilibrium equations in
00:00:35.860 --> 00:00:37.790
dynamic analysis.
00:00:37.790 --> 00:00:39.980
These, of course, are the
equilibrium equations that we
00:00:39.980 --> 00:00:42.100
derived already earlier.
00:00:42.100 --> 00:00:45.270
M as a mass matrix, C as a
damping matrix, K as a
00:00:45.270 --> 00:00:48.490
stiffness matrix, U as a
displacement vector, and here
00:00:48.490 --> 00:00:50.490
we have the velocities and the
00:00:50.490 --> 00:00:52.535
accelerations at the null points.
00:00:52.535 --> 00:00:56.440
R, of course, being the load
vector and this load vector is
00:00:56.440 --> 00:00:59.320
now time dependent.
00:00:59.320 --> 00:01:02.200
I also should draw your
attention on the fact that I
00:01:02.200 --> 00:01:04.170
would like to talk about direct
00:01:04.170 --> 00:01:06.500
integration of these equations.
00:01:06.500 --> 00:01:10.140
We will talk about the modes
of opposition procedures in
00:01:10.140 --> 00:01:12.100
the next lecture.
00:01:12.100 --> 00:01:16.010
By direct integration, we mean
that we integrate these
00:01:16.010 --> 00:01:18.970
equations directly without--
00:01:18.970 --> 00:01:20.120
that is, without--
00:01:20.120 --> 00:01:23.280
a transformation of these
equations into a different
00:01:23.280 --> 00:01:25.740
form prior to the integration.
00:01:25.740 --> 00:01:28.090
In modes of opposition analysis,
now you will see in
00:01:28.090 --> 00:01:30.620
the next lecture, we're actually
transforming these
00:01:30.620 --> 00:01:35.040
equations first, and then we
integrate the response.
00:01:35.040 --> 00:01:38.150
So the procedures that I'd
like to talk about are
00:01:38.150 --> 00:01:42.410
explicit, implicit integration
procedures that directly
00:01:42.410 --> 00:01:45.980
operate on these equations
without a transformation and
00:01:45.980 --> 00:01:48.670
integrate, solve these
equations.
00:01:48.670 --> 00:01:51.670
I also would like to talk
about some computational
00:01:51.670 --> 00:01:55.530
considerations with regard to
these explicit and implicit
00:01:55.530 --> 00:01:58.750
integration procedures, the
number of operations involved,
00:01:58.750 --> 00:02:00.810
the costs involved in
performing these
00:02:00.810 --> 00:02:02.820
integrations and so on.
00:02:02.820 --> 00:02:06.450
And then there is an important
consideration, namely the
00:02:06.450 --> 00:02:09.900
selection of the solution
time step, delta-t.
00:02:09.900 --> 00:02:12.320
We will see that when we
integrate the dynamic
00:02:12.320 --> 00:02:15.050
equilibrium equations, we have
to select the time step
00:02:15.050 --> 00:02:19.000
delta-t with which we perform
the integration.
00:02:19.000 --> 00:02:23.560
And this time step has to be
selected in order to preserve
00:02:23.560 --> 00:02:26.990
stability of the integration and
to preserve of course or
00:02:26.990 --> 00:02:29.960
to obtain accuracy in
the integration.
00:02:29.960 --> 00:02:32.870
Finally, I'd like to talk
about some modeling
00:02:32.870 --> 00:02:36.640
considerations pertaining to
the solution of the dynamic
00:02:36.640 --> 00:02:38.160
equilibrium equations.
00:02:38.160 --> 00:02:44.600
Well, here on this view graph,
I'm showing once again the
00:02:44.600 --> 00:02:47.570
dynamically equilibrium
equations here, noticing once
00:02:47.570 --> 00:02:51.820
more that R, the load vector,
is now a function of time.
00:02:51.820 --> 00:02:55.080
The same equations here have
been rewritten once here,
00:02:55.080 --> 00:02:59.750
where FI denotes inertia
forces, now time
00:02:59.750 --> 00:03:00.930
dependent, of course.
00:03:00.930 --> 00:03:03.740
The mass matrix is constant,
but the accelerations of
00:03:03.740 --> 00:03:05.750
course depend on time.
00:03:05.750 --> 00:03:10.770
Then this part here, which are
the damping forces, has been
00:03:10.770 --> 00:03:13.820
expressed here as FD a function
of time again.
00:03:13.820 --> 00:03:17.480
The C matrix is constant with
time, but the velocities, of
00:03:17.480 --> 00:03:18.580
course, vary with time.
00:03:18.580 --> 00:03:20.340
So here we have damping
forces.
00:03:20.340 --> 00:03:22.700
It's a nodes of the finite
element system.
00:03:22.700 --> 00:03:27.020
And these are the elastic
forces, KU, which are really
00:03:27.020 --> 00:03:30.110
due to the internal
element stresses.
00:03:30.110 --> 00:03:34.370
And on the right hand side, we
have the external forces.
00:03:34.370 --> 00:03:39.070
So what we're saying is that
we want to solve this force
00:03:39.070 --> 00:03:42.810
equilibrium over all times t.
00:03:42.810 --> 00:03:45.910
The inertia forces plus the
damping forces plus the
00:03:45.910 --> 00:03:51.180
elastic forces, it's the nodes
of the finite element system,
00:03:51.180 --> 00:03:55.340
of course, calculated in the
virtual work sense, the way
00:03:55.340 --> 00:03:58.040
we've been talking about in
the earlier lectures.
00:03:58.040 --> 00:04:00.750
The sum of these forces here
must be equal to the
00:04:00.750 --> 00:04:03.890
externally applied
nodal point logs.
00:04:03.890 --> 00:04:07.940
Well, the procedure
that we will be
00:04:07.940 --> 00:04:10.550
following is the following.
00:04:10.550 --> 00:04:16.019
We will consider time steps
delta-t a part at which we
00:04:16.019 --> 00:04:19.180
want to calculate the response
of the system.
00:04:19.180 --> 00:04:20.810
In order to do so, of
course, we have to
00:04:20.810 --> 00:04:22.710
define our loads first.
00:04:22.710 --> 00:04:26.970
And here, I have schematic shown
once the description of
00:04:26.970 --> 00:04:31.770
the load Ri, that is, the load
at degree of freedom i as a
00:04:31.770 --> 00:04:33.510
function of time t.
00:04:33.510 --> 00:04:39.490
Notice that we are having here
this distribution of that load
00:04:39.490 --> 00:04:43.990
component as a function
of time.
00:04:43.990 --> 00:04:46.120
The distribution is
quite simple.
00:04:46.120 --> 00:04:48.690
In a computer program, of
course, all we would have to
00:04:48.690 --> 00:04:52.670
put in is this point, this
intensity here, that load
00:04:52.670 --> 00:04:54.930
intensity, at a specified
time.
00:04:54.930 --> 00:04:58.620
That load intensity, at a
specified time, and that load
00:04:58.620 --> 00:05:00.780
intensity at a specified time.
00:05:00.780 --> 00:05:04.570
In between, these intensities,
the computer program would
00:05:04.570 --> 00:05:08.880
simply interpolate linearly.
00:05:08.880 --> 00:05:12.370
This is the load at the
component i, displacement
00:05:12.370 --> 00:05:13.370
component i.
00:05:13.370 --> 00:05:16.160
This is the load at displacement
component j.
00:05:16.160 --> 00:05:18.860
Notice here, of course, we
would have in general a
00:05:18.860 --> 00:05:22.470
different curve at degree of
freedom j, then the curve is
00:05:22.470 --> 00:05:25.570
used at degree of freedom i.
00:05:25.570 --> 00:05:31.140
I show here time steps, discrete
time steps, delta-t1,
00:05:31.140 --> 00:05:33.880
delta-t2, delta-t3.
00:05:33.880 --> 00:05:37.260
These of course apply for the
complete system, for the i
00:05:37.260 --> 00:05:41.250
component and the j component of
displacements or loads, and
00:05:41.250 --> 00:05:45.160
for all other displacements
and loads also.
00:05:45.160 --> 00:05:47.970
Notice that they can
vary with time.
00:05:47.970 --> 00:05:51.760
In many analysis, however, we
simply pick one time step and
00:05:51.760 --> 00:05:54.500
keep it constant all along.
00:05:54.500 --> 00:05:56.740
The objective is
the following.
00:05:56.740 --> 00:06:01.200
In the direct integration of the
equations of equilibrium,
00:06:01.200 --> 00:06:04.740
the objective is knowing
the initial conditions.
00:06:04.740 --> 00:06:07.980
In other words, knowing the
displacement at time 0, the
00:06:07.980 --> 00:06:11.190
velocities at time 0.
00:06:11.190 --> 00:06:15.540
We want to calculate the
displacements, velocities, and
00:06:15.540 --> 00:06:21.120
accelerations at these
discrete time points.
00:06:21.120 --> 00:06:24.920
We are matching here with the
time interval delta-t1.
00:06:24.920 --> 00:06:29.780
And here we are changing the
time interval to delta-t2.
00:06:29.780 --> 00:06:32.970
Knowing all the velocities,
accelerations, and
00:06:32.970 --> 00:06:36.120
displacement at this time, we
want to calculate the same
00:06:36.120 --> 00:06:40.200
quantities at the discrete
time increments of
00:06:40.200 --> 00:06:42.180
delta-t2, and so on.
00:06:42.180 --> 00:06:45.100
So basically what we are saying
is that knowing the
00:06:45.100 --> 00:06:51.010
initial conditions, we want to
calculate in a step-by-step
00:06:51.010 --> 00:06:54.860
procedure, the displacements,
velocities, and accelerations
00:06:54.860 --> 00:06:57.850
at these discrete time points.
00:06:57.850 --> 00:07:02.790
Of course, the real problem then
reduces to the following.
00:07:02.790 --> 00:07:08.980
Given the initial conditions at
the particular time t, we
00:07:08.980 --> 00:07:13.220
might call this time t, and by
initial conditions, I mean we
00:07:13.220 --> 00:07:15.640
know the displacements,
velocities, accelerations, we
00:07:15.640 --> 00:07:19.580
know all the quantities
at time t.
00:07:19.580 --> 00:07:24.740
Our objective is to calculate
the unknown quantities at time
00:07:24.740 --> 00:07:29.080
t plus delta-t1 in this
particular case.
00:07:29.080 --> 00:07:35.230
So this point is t
plus delta-t1.
00:07:35.230 --> 00:07:41.230
In general, or in many cases, we
keep a constant time step,
00:07:41.230 --> 00:07:44.070
and so we will simply
talk about t plus
00:07:44.070 --> 00:07:46.810
delta-t from now onwards.
00:07:46.810 --> 00:07:50.610
We should remember, however, the
delta-t might change doing
00:07:50.610 --> 00:07:51.910
the response solution.
00:07:51.910 --> 00:07:54.980
I repeat once more, this is an
important point, given the
00:07:54.980 --> 00:07:58.290
initial conditions or given the
conditions at time t, our
00:07:58.290 --> 00:08:01.900
objective is to calculate the
conditions, displacements,
00:08:01.900 --> 00:08:07.100
velocities, accelerations,
at time t plus delta-t.
00:08:07.100 --> 00:08:13.380
In the explicit integration,
we proceed in
00:08:13.380 --> 00:08:14.980
the solution as follows.
00:08:14.980 --> 00:08:22.590
We use the equations of dynamic
equilibrium at time t
00:08:22.590 --> 00:08:26.500
to obtain the solution at
time t plus delta-t.
00:08:26.500 --> 00:08:27.820
This is important.
00:08:27.820 --> 00:08:31.620
In explicit integration, we
use the condition, the
00:08:31.620 --> 00:08:34.590
equilibrium equations, the
equilibrium equation, the
00:08:34.590 --> 00:08:39.200
dynamic equilibrium equation,
and by that I mean, just to
00:08:39.200 --> 00:08:41.830
refresh your memory, this
equation-- we use that
00:08:41.830 --> 00:08:46.760
equation at time t in explicit
integration to obtain the
00:08:46.760 --> 00:08:49.980
solution at time
t plus delta-t.
00:08:49.980 --> 00:08:53.210
Once we have the solution at
time t plus delta-t, of course
00:08:53.210 --> 00:08:55.550
we can march ahead in
the same way again.
00:08:55.550 --> 00:08:58.980
In implicit integration,
however, we are looking at the
00:08:58.980 --> 00:09:03.430
equilibrium equations at time t
plus delta-t t to obtain the
00:09:03.430 --> 00:09:05.610
solution at time
t plus delta-t.
00:09:05.610 --> 00:09:09.345
In other words, if you look at
once more at the equations
00:09:09.345 --> 00:09:12.410
here, at the equilibrium
equations, these equations
00:09:12.410 --> 00:09:15.280
would be applied at time t plus
delta-t to obtain the
00:09:15.280 --> 00:09:18.190
solution at time
t plus delta-t.
00:09:18.190 --> 00:09:21.500
This is indeed the fundamental
difference between explicit
00:09:21.500 --> 00:09:22.980
and implicit integration.
00:09:22.980 --> 00:09:26.790
In explicit integration, we
are looking at equilibrium
00:09:26.790 --> 00:09:29.270
equations at time t to
obtain the solution
00:09:29.270 --> 00:09:31.150
at time t plus delta-t.
00:09:31.150 --> 00:09:33.640
And in implicit integration,
we are looking at the
00:09:33.640 --> 00:09:37.090
equilibrium equations at time
t plus delta-t to attain the
00:09:37.090 --> 00:09:39.970
solution at time
t plus delta-t.
00:09:39.970 --> 00:09:42.830
In the following now, I would
like to show to you an
00:09:42.830 --> 00:09:45.740
explicit and an implicit
integration scheme.
00:09:45.740 --> 00:09:48.720
The explicit integration scheme
that is abundantly used
00:09:48.720 --> 00:09:51.640
in practice is the central
difference method.
00:09:51.640 --> 00:09:53.950
I will later on then discuss
how this method
00:09:53.950 --> 00:09:55.550
is effectively used.
00:09:55.550 --> 00:09:58.710
In the central difference
method, we are applying, as I
00:09:58.710 --> 00:10:02.980
said earlier, the equilibrium
equations at time t, because
00:10:02.980 --> 00:10:05.780
it's an explicit integration
scheme.
00:10:05.780 --> 00:10:09.460
The two additional questions
that we're using to solve for
00:10:09.460 --> 00:10:13.580
the unknown displacements at
time t plus delta-t, is
00:10:13.580 --> 00:10:17.180
firstly an equation for the
accelerations at time t.
00:10:17.180 --> 00:10:22.030
And notice that this equation
involves as the unknown, as
00:10:22.030 --> 00:10:24.340
the only unknown,
the displacement
00:10:24.340 --> 00:10:27.110
at time t plus delta-t.
00:10:27.110 --> 00:10:29.960
In addition, of course, we are
using an equation for the
00:10:29.960 --> 00:10:34.690
velocities at time t and again
this equation only involves as
00:10:34.690 --> 00:10:36.500
the only unknown the
displacement
00:10:36.500 --> 00:10:38.300
at time t plus delta-t.
00:10:38.300 --> 00:10:42.000
Substituting these two equations
into the equilibrium
00:10:42.000 --> 00:10:48.060
equation at time t gives us one
equation in the unknown t,
00:10:48.060 --> 00:10:50.620
displacement at time
t plus delta-t.
00:10:50.620 --> 00:10:54.280
We can simply solve for
it, as shown, now on
00:10:54.280 --> 00:10:56.240
the next view graph.
00:10:56.240 --> 00:10:59.530
You can see here the equation
that is obtained by
00:10:59.530 --> 00:11:03.420
substituting the relations for
the acceleration and for the
00:11:03.420 --> 00:11:07.350
velocities in the equilibrium
equation at time t.
00:11:07.350 --> 00:11:08.990
The unknown appears here.
00:11:08.990 --> 00:11:11.430
This is the displacement
at time t plus
00:11:11.430 --> 00:11:14.270
delta-t, which is unknown.
00:11:14.270 --> 00:11:17.310
And there is a coefficient
matrix here, that involves the
00:11:17.310 --> 00:11:19.830
mass matrix and the
damping matrix.
00:11:19.830 --> 00:11:22.730
And on the right hand side, of
course, we are calculating a
00:11:22.730 --> 00:11:23.840
load vector.
00:11:23.840 --> 00:11:27.270
And that load vector, notice
once again, corresponds to the
00:11:27.270 --> 00:11:29.650
conditions at time t.
00:11:29.650 --> 00:11:32.420
That's why it's an explicit
integration scheme.
00:11:32.420 --> 00:11:35.780
The difference operator, the
essential difference operator
00:11:35.780 --> 00:11:40.045
that I've been used here, simple
operators which you
00:11:40.045 --> 00:11:44.060
have seen on the previous slide,
that involve only the
00:11:44.060 --> 00:11:48.000
displacement at time t
minus delta-t and the
00:11:48.000 --> 00:11:49.500
displacement at time t.
00:11:49.500 --> 00:11:52.660
And notice that here
are these two
00:11:52.660 --> 00:11:55.990
displacement vectors appearing.
00:11:55.990 --> 00:12:00.803
Notice that this equation
is cheaply--
00:12:00.803 --> 00:12:04.930
or can be used in a very cheap
way to calculate the
00:12:04.930 --> 00:12:07.450
displacement, the unknown
displacement at time t plus
00:12:07.450 --> 00:12:12.370
delta-t, if the mass matrix
is a diagonal matrix.
00:12:12.370 --> 00:12:14.920
And, say, the damping
is neglected.
00:12:14.920 --> 00:12:17.520
In other words, this term not
being there, this being a
00:12:17.520 --> 00:12:22.390
diagonal mass matrix, then the
solution is very trivial.
00:12:22.390 --> 00:12:25.620
In fact, all we need to do is
calculate the right hand side
00:12:25.620 --> 00:12:30.470
load vector and divide each
component of that load vector
00:12:30.470 --> 00:12:35.390
by this coefficient here which,
if m once again is a
00:12:35.390 --> 00:12:38.080
diagonal mass matrix,
is simply Mii
00:12:38.080 --> 00:12:40.540
divided by delta-t squared.
00:12:40.540 --> 00:12:44.310
And in fact, this is really
how was the procedure is
00:12:44.310 --> 00:12:47.580
usually applied, namely,
with a diagonal mass
00:12:47.580 --> 00:12:49.370
matrix and no damping.
00:12:49.370 --> 00:12:54.050
The damping usually is not
taken into account, or at
00:12:54.050 --> 00:12:57.450
most, we have diagonal dampers
also, and the mass matrix
00:12:57.450 --> 00:13:01.450
being a diagonal matrix,
then the solution is
00:13:01.450 --> 00:13:02.760
very cheaply performed.
00:13:02.760 --> 00:13:05.530
There's one further
important point.
00:13:05.530 --> 00:13:11.120
This load vector of course is
constructed in the usual way.
00:13:11.120 --> 00:13:16.630
This multiplication here is
quite cheap if the mass matrix
00:13:16.630 --> 00:13:18.410
is a diagonal mass matrix.
00:13:18.410 --> 00:13:20.010
K times tU.
00:13:20.010 --> 00:13:23.470
Well, K here is of course
a banded matrix.
00:13:23.470 --> 00:13:27.770
It's the stiffness matrix
of the system, constant.
00:13:27.770 --> 00:13:31.270
Well, there are some operations
involved here.
00:13:31.270 --> 00:13:35.280
And if we look at this operation
here in detail, we
00:13:35.280 --> 00:13:38.590
see here how it can be
effectively performed.
00:13:38.590 --> 00:13:40.270
We have a stiffness
matrix times the
00:13:40.270 --> 00:13:41.780
displacement vector here.
00:13:41.780 --> 00:13:44.920
And of course, the stiffness
matrix remember is obtained by
00:13:44.920 --> 00:13:47.350
summing the element
contributions.
00:13:47.350 --> 00:13:51.500
So you can write this K
matrix in this way.
00:13:51.500 --> 00:13:55.540
Since tU is independent of the
elements, we can also take it
00:13:55.540 --> 00:13:57.230
into the summation sign.
00:13:57.230 --> 00:14:02.270
And if we look at this product
closely, we find that is
00:14:02.270 --> 00:14:04.970
nothing else than tFm.
00:14:04.970 --> 00:14:13.900
It's a nodal point vector of
loads or elements forces that
00:14:13.900 --> 00:14:15.260
correspond--
00:14:15.260 --> 00:14:18.940
element elastic forces-- that
correspond to the current
00:14:18.940 --> 00:14:20.420
displacements.
00:14:20.420 --> 00:14:25.280
And this tF vector is much more
cheaply calculated than
00:14:25.280 --> 00:14:27.390
the Km times tU.
00:14:27.390 --> 00:14:30.740
And in fact, this is the way
how we actually perform the
00:14:30.740 --> 00:14:33.370
calculation of this
product here.
00:14:33.370 --> 00:14:37.130
Now notice the following
important further point.
00:14:37.130 --> 00:14:41.980
If we were to leave this K times
tU the way it stands
00:14:41.980 --> 00:14:44.800
here, we would have to assemble
a global stiffness
00:14:44.800 --> 00:14:46.710
matrix of the system.
00:14:46.710 --> 00:14:49.700
However, if we write that
product this way, we never
00:14:49.700 --> 00:14:52.060
need to assemble a global
stiffness matrix
00:14:52.060 --> 00:14:53.380
of the total system.
00:14:53.380 --> 00:14:58.010
In fact, all we need to do is
calculate element forces and
00:14:58.010 --> 00:15:01.860
add these elements forces via
the direct stiffness procedure
00:15:01.860 --> 00:15:03.940
the way I've been discussing
earlier.
00:15:03.940 --> 00:15:09.410
So this is a very effective way
of calculating K times tU.
00:15:09.410 --> 00:15:12.560
We never need to assemble a
global stiffness matrix.
00:15:12.560 --> 00:15:17.060
In fact, this is one of the main
reasons why the central
00:15:17.060 --> 00:15:20.040
difference method
is effective.
00:15:20.040 --> 00:15:25.250
It's this reason that we can
calculate the KtU part, this
00:15:25.250 --> 00:15:29.530
times that, very effectively
via this procedure.
00:15:29.530 --> 00:15:30.500
That is the first reason.
00:15:30.500 --> 00:15:34.980
And the second reason is that
if M is diagonal, we really
00:15:34.980 --> 00:15:37.160
never solve equations.
00:15:37.160 --> 00:15:41.550
The equation solving here is so
trivial, that we might just
00:15:41.550 --> 00:15:43.270
say that we do not
solve equations.
00:15:43.270 --> 00:15:45.530
There is no factorization
involved.
00:15:45.530 --> 00:15:49.910
There's no factorization of a
stiffness matrix or another
00:15:49.910 --> 00:15:52.900
coefficient matrix involved,
because if M is diagonal, and
00:15:52.900 --> 00:15:57.300
as I pointed out, we simply
divide the right hand side by
00:15:57.300 --> 00:16:01.310
Mii divided by delta-t
squared.
00:16:01.310 --> 00:16:05.550
Notice that if we start the
integration process, we know
00:16:05.550 --> 00:16:07.040
of course the initial
conditions.
00:16:07.040 --> 00:16:08.540
We know tU.
00:16:08.540 --> 00:16:11.930
And we do not know--
00:16:11.930 --> 00:16:14.830
usually directly the
t minus delta-tU.
00:16:14.830 --> 00:16:17.570
That is the displacement
prior to the solution.
00:16:17.570 --> 00:16:22.160
And to obtain those, we can
use this formula, which is
00:16:22.160 --> 00:16:27.420
really the central difference
method applied at time t,
00:16:27.420 --> 00:16:29.790
where we however now involve
the accelerations
00:16:29.790 --> 00:16:31.940
corresponding to time 0.
00:16:31.940 --> 00:16:35.270
We involve the accelerations
corresponding to time zero
00:16:35.270 --> 00:16:38.290
right here, and the velocities
corresponding to times 0, and
00:16:38.290 --> 00:16:40.580
the displacements corresponding
to time zero.
00:16:40.580 --> 00:16:42.670
And of course, these quantities
must all be given
00:16:42.670 --> 00:16:46.070
as the initial conditions
to the solution.
00:16:46.070 --> 00:16:48.840
I pointed out already that we're
using this technique
00:16:48.840 --> 00:16:51.130
mostly with the lump
mass matrix.
00:16:51.130 --> 00:16:53.880
In addition I should also point
out that we're using it
00:16:53.880 --> 00:16:56.250
primarily with lower
order elements,
00:16:56.250 --> 00:16:57.890
with lower order elements.
00:16:57.890 --> 00:17:04.430
The reason for that is the
following, namely, one of
00:17:04.430 --> 00:17:08.190
stability and accuracy of the
central difference method.
00:17:08.190 --> 00:17:15.460
The method can only be applied
if delta-t, the time step that
00:17:15.460 --> 00:17:18.200
I'm talking about here,
is smaller than a
00:17:18.200 --> 00:17:20.119
critical time step.
00:17:20.119 --> 00:17:25.240
And that critical time step is
given by the smallest period
00:17:25.240 --> 00:17:30.180
in the system divided by
pi, 3.14 et cetera.
00:17:30.180 --> 00:17:32.730
The smallest period in
the system can be
00:17:32.730 --> 00:17:34.930
of course very small.
00:17:34.930 --> 00:17:39.140
And therefore the critical time
step that we're talking
00:17:39.140 --> 00:17:42.410
about here can be very small.
00:17:42.410 --> 00:17:46.060
The smallest period in the
system, just to give you an
00:17:46.060 --> 00:17:50.210
example, would be extremely
small if you have a bar
00:17:50.210 --> 00:17:52.990
element, and you have
some very short
00:17:52.990 --> 00:17:54.470
truss element in there.
00:17:54.470 --> 00:17:58.000
The shorter the truss element,
the stiffer of course it is.
00:17:58.000 --> 00:18:02.380
AU over L is the stiffness
of a truss element, and
00:18:02.380 --> 00:18:06.860
therefore, a very small period
would be in that
00:18:06.860 --> 00:18:10.470
assemblage of elements.
00:18:10.470 --> 00:18:14.240
Therefore, what is required
really is to have a mesh that
00:18:14.240 --> 00:18:19.760
is more or less uniform, so that
there is no artificially
00:18:19.760 --> 00:18:23.320
small time step required in
the integration of the
00:18:23.320 --> 00:18:24.770
equilibrium equations.
00:18:24.770 --> 00:18:27.060
Because there is this
condition that
00:18:27.060 --> 00:18:28.650
delta-t must be smaller--
00:18:28.650 --> 00:18:30.980
we should really say must
be smaller or equal--
00:18:30.980 --> 00:18:36.380
to delta-t critical, but to be
on the conservative side, we
00:18:36.380 --> 00:18:40.200
are selecting delta-t smaller
than delta-t critical in
00:18:40.200 --> 00:18:43.310
practical analysis of
complex meshes.
00:18:43.310 --> 00:18:46.130
Because this condition has to
be satisfied, we're talking
00:18:46.130 --> 00:18:50.320
about a conditionally
stable scheme.
00:18:50.320 --> 00:18:55.380
It is only stable when delta-t
is smaller than that value.
00:18:55.380 --> 00:18:59.920
If we were to select delta-t
larger than delta-t critical,
00:18:59.920 --> 00:19:05.140
we would see that after just a
few time steps, the solution--
00:19:05.140 --> 00:19:06.770
it gets out of bounds.
00:19:06.770 --> 00:19:10.860
And after 10 or 20 or 30 time
step, in a computer program,
00:19:10.860 --> 00:19:12.150
you would all of
a sudden notice
00:19:12.150 --> 00:19:13.560
overflow of the numbers.
00:19:13.560 --> 00:19:16.380
So the solution wouldn't make
any sense whatsoever anymore.
00:19:16.380 --> 00:19:18.490
And this is an absolute
condition that
00:19:18.490 --> 00:19:20.380
does have to be satisfied.
00:19:20.380 --> 00:19:23.370
There's no way you can use a
time step larger than delta-t
00:19:23.370 --> 00:19:26.570
critical, and still get an
acceptable, reasonable
00:19:26.570 --> 00:19:31.380
solution to your equations
of motion.
00:19:31.380 --> 00:19:35.010
In practice, of course, we
have to estimate, in a
00:19:35.010 --> 00:19:37.210
conservative way, a time step.
00:19:37.210 --> 00:19:39.380
And that means we have
to have an estimate
00:19:39.380 --> 00:19:41.680
also of delta-t critical.
00:19:41.680 --> 00:19:43.440
For continuum elements--
00:19:43.440 --> 00:19:46.350
by continuum elements, I mean
plain stress, plain strain,
00:19:46.350 --> 00:19:48.560
axis a matrix, three-dimensional
element, the
00:19:48.560 --> 00:19:50.670
way we have been discussing
it earlier.
00:19:50.670 --> 00:19:57.430
A good way of obtaining time
step delta-t is to use this
00:19:57.430 --> 00:20:03.650
formula here, delta-t being
delta-L over C. C is the wave
00:20:03.650 --> 00:20:05.800
velocity through the system.
00:20:05.800 --> 00:20:06.390
[UNINTELLIGIBLE]
00:20:06.390 --> 00:20:10.890
is divided by rho, square root
out of it gives us C. And
00:20:10.890 --> 00:20:19.210
delta-L is an element length
that has to be defined.
00:20:19.210 --> 00:20:24.290
In fact, it's the smallest
distance between nodes.
00:20:24.290 --> 00:20:27.990
The smallest distance between
nodes if we talk about low
00:20:27.990 --> 00:20:29.040
order elements.
00:20:29.040 --> 00:20:30.590
What I mean by that
is the following.
00:20:30.590 --> 00:20:34.250
If we have a mesh here
that looks like that.
00:20:34.250 --> 00:20:38.125
And I deliberately show elements
that are not equal in
00:20:38.125 --> 00:20:42.960
size, then delta-E in this
particular case would be this
00:20:42.960 --> 00:20:43.860
distance here.
00:20:43.860 --> 00:20:48.050
That is our delta-L, sorry,
our delta-L is
00:20:48.050 --> 00:20:51.610
that distance there.
00:20:51.610 --> 00:20:55.320
This, of course, as I have here
in the heading, assumes
00:20:55.320 --> 00:20:57.040
that we have low-order
elements.
00:20:57.040 --> 00:21:00.460
That means only corner nodes
in the elements.
00:21:00.460 --> 00:21:02.240
For higher-order elements--
00:21:02.240 --> 00:21:05.650
let me put down here one
higher-order element, where we
00:21:05.650 --> 00:21:11.420
have these nodes now, say
an 8 node element.
00:21:11.420 --> 00:21:14.310
In this particular case, we
take the smallest distance
00:21:14.310 --> 00:21:19.710
between nodes, again, just
as in this case.
00:21:19.710 --> 00:21:21.940
And in this particular case,
it would be, say, this
00:21:21.940 --> 00:21:23.140
distance here.
00:21:23.140 --> 00:21:26.200
That is our smallest distance
between nodes.
00:21:26.200 --> 00:21:29.460
However, we have to also divide
this distance by a
00:21:29.460 --> 00:21:31.800
relative stiffness factor.
00:21:31.800 --> 00:21:34.430
And why does that relative
stiffness factor
00:21:34.430 --> 00:21:35.860
come into the picture?
00:21:35.860 --> 00:21:39.320
Well, it does come into the
picture, because remember that
00:21:39.320 --> 00:21:44.190
the stiffness at an interior
node is larger than the
00:21:44.190 --> 00:21:47.570
stiffness at a corner
for the element.
00:21:47.570 --> 00:21:50.870
The amount it is larger depends
on the particular
00:21:50.870 --> 00:21:53.600
element used.
00:21:53.600 --> 00:21:57.490
But since there's a lot of
stiffness here, we have to put
00:21:57.490 --> 00:22:01.430
here a relative stiffness factor
into the calculation of
00:22:01.430 --> 00:22:06.970
delta-L, the effective lengths
that we're using in this
00:22:06.970 --> 00:22:10.230
particular formula.
00:22:10.230 --> 00:22:14.450
Now, for this element here,
for example, the relative
00:22:14.450 --> 00:22:16.910
stiffness factor would
be conservatively
00:22:16.910 --> 00:22:20.410
just equal to 4.
00:22:20.410 --> 00:22:23.690
But for other elements, and
particular also when there's
00:22:23.690 --> 00:22:27.260
node shifting involved and so
on, the relative stiffness
00:22:27.260 --> 00:22:30.330
factor has to be calculated,
has to be estimated.
00:22:30.330 --> 00:22:33.950
And we should always remember
that it should be estimated
00:22:33.950 --> 00:22:41.320
here in a conservative way,
because a smaller delta-L gave
00:22:41.320 --> 00:22:44.570
us a smaller delta-t.
00:22:44.570 --> 00:22:47.680
And then we are then
conservatively applying the
00:22:47.680 --> 00:22:50.900
fact that delta-t shall be
smaller than the delta-t
00:22:50.900 --> 00:22:55.220
critical, value that I have
been referring hereto.
00:22:55.220 --> 00:23:00.120
The method is used mainly for
wave propagation analysis.
00:23:00.120 --> 00:23:03.650
The reason being that we have
to use a fairly large number
00:23:03.650 --> 00:23:07.260
of time steps to predict
the solution.
00:23:07.260 --> 00:23:11.220
Large, because a time step
delta-t is usually very small.
00:23:11.220 --> 00:23:13.940
And in wave propagation
analysis, many modes of the
00:23:13.940 --> 00:23:16.590
system are being excited.
00:23:16.590 --> 00:23:20.340
So we really want to pick
up a large number of
00:23:20.340 --> 00:23:22.170
modes in the mesh.
00:23:22.170 --> 00:23:27.990
And the small time step
integrating accurately many
00:23:27.990 --> 00:23:31.540
modes in the mesh goes together
with the fact that we
00:23:31.540 --> 00:23:35.180
want to predict many modes in
the mesh or a large number of
00:23:35.180 --> 00:23:38.960
modes in the mesh, and therefore
the method is
00:23:38.960 --> 00:23:42.060
effectively used for wave
propagation analysis.
00:23:42.060 --> 00:23:46.500
The number of operations are
proportional to the number of
00:23:46.500 --> 00:23:52.042
elements that we are using and
the number of time steps.
00:23:52.042 --> 00:23:55.830
Of course, this is an
approximate proportionality
00:23:55.830 --> 00:23:58.550
here that I'm talking about.
00:23:58.550 --> 00:24:06.340
This fact derives directly from
the consideration that we
00:24:06.340 --> 00:24:07.440
talked about here.
00:24:07.440 --> 00:24:10.640
Notice that as we include more
and more elements in the
00:24:10.640 --> 00:24:13.880
solution process, of course,
we have to sum over more
00:24:13.880 --> 00:24:15.560
elements here.
00:24:15.560 --> 00:24:18.800
And as we go more and more time
steps, we have to apply
00:24:18.800 --> 00:24:23.720
this equation as many times
also more and more.
00:24:23.720 --> 00:24:27.440
The important point is that we
don't assemble a K matrix, so
00:24:27.440 --> 00:24:30.850
the bandwidth off the system,
the bandwidth of the system
00:24:30.850 --> 00:24:33.300
does not appear in an
operation count.
00:24:33.300 --> 00:24:36.130
It's the number of elements
that do appear in the
00:24:36.130 --> 00:24:37.380
operation count.
00:24:40.010 --> 00:24:48.330
The next procedure that
I like to talk about
00:24:48.330 --> 00:24:50.100
is the Newmark method.
00:24:50.100 --> 00:24:55.600
And in the Newmark method,
this being an implicit
00:24:55.600 --> 00:24:59.710
integration scheme now,
we proceed as follows.
00:24:59.710 --> 00:25:00.610
We apply--
00:25:00.610 --> 00:25:02.730
because it's an implicit
integration scheme--
00:25:02.730 --> 00:25:05.450
the equilibrium equation, the
dynamic equilibrium equation
00:25:05.450 --> 00:25:08.150
at time t plus delta-t.
00:25:08.150 --> 00:25:12.450
Here it is given the additional
formula that we're
00:25:12.450 --> 00:25:16.560
using is this one here for the
velocity at time t plus
00:25:16.560 --> 00:25:19.140
delta-t, and this one here for
the displacement of time t
00:25:19.140 --> 00:25:20.530
plus delta-t.
00:25:20.530 --> 00:25:25.380
Now notice that with this set of
equations, with these three
00:25:25.380 --> 00:25:31.070
equations, knowing the
displacements and velocities
00:25:31.070 --> 00:25:37.620
at time t and accelerations at
time t, we have with these
00:25:37.620 --> 00:25:40.410
three equations,
three unknowns.
00:25:40.410 --> 00:25:44.440
The velocity at time t plus
delta-t, the acceleration at
00:25:44.440 --> 00:25:47.415
time t plus delta-t, and
the displacement
00:25:47.415 --> 00:25:49.720
at time t plus delta-t.
00:25:49.720 --> 00:25:52.165
These are the only three
unknowns that appear.
00:25:52.165 --> 00:25:55.540
There are three linearly
independent equations.
00:25:55.540 --> 00:25:57.550
And we can simply use them to
00:25:57.550 --> 00:25:59.515
calculate these three unknowns.
00:25:59.515 --> 00:26:02.030
Of course, we want to do so
in a very effective way.
00:26:02.030 --> 00:26:04.320
And I will show you
how we proceed.
00:26:04.320 --> 00:26:11.640
The important point is that with
this scheme, we are using
00:26:11.640 --> 00:26:16.170
an implicit approach, as
I defined it earlier.
00:26:16.170 --> 00:26:21.070
And when we do substitute from
here into there, we will find
00:26:21.070 --> 00:26:25.540
that on the left hand side, we
have a coefficient matrix that
00:26:25.540 --> 00:26:30.500
involves a bandwidth even if the
mass matrix is diagonal.
00:26:30.500 --> 00:26:33.415
And this coefficient matrix
here is written
00:26:33.415 --> 00:26:35.730
here down as K hat.
00:26:35.730 --> 00:26:40.640
This coefficient matrix
invoice this K matrix.
00:26:40.640 --> 00:26:43.350
I should say maybe a few words
to these formulae.
00:26:43.350 --> 00:26:46.640
These formulae can be obtained
by Taylor Series expansion,
00:26:46.640 --> 00:26:49.970
around the conditions at
times t plus delta-t.
00:26:49.970 --> 00:26:53.070
They have been proposed
originally by Newmark.
00:26:53.070 --> 00:26:56.330
That's why it's called now
the Newmark method.
00:26:56.330 --> 00:27:00.420
Newmark in particular used the
values of delta equals 1/2 and
00:27:00.420 --> 00:27:02.430
alpha equal to 1/4.
00:27:02.430 --> 00:27:06.800
And then it's a constant average
acceleration scheme.
00:27:06.800 --> 00:27:14.960
The constant average
acceleration schemes use these
00:27:14.960 --> 00:27:17.700
values as I pointed out
already earlier.
00:27:17.700 --> 00:27:20.310
The method is unconditionally
stable.
00:27:20.310 --> 00:27:25.550
This means that the time step
delta-t does not to have to be
00:27:25.550 --> 00:27:28.480
smaller than a certain value,
as it was the case in the
00:27:28.480 --> 00:27:30.490
central difference method.
00:27:30.490 --> 00:27:33.970
The method is primarily was
for analysis of structural
00:27:33.970 --> 00:27:36.410
dynamic problems.
00:27:36.410 --> 00:27:40.030
I will talk about why
this is so just now.
00:27:40.030 --> 00:27:42.560
And the number of operations
that we're talking about here
00:27:42.560 --> 00:27:48.110
are an initial factorization
of the K hat matrix.
00:27:48.110 --> 00:27:51.610
This is the K hat matrix, And
that initial factorization
00:27:51.610 --> 00:27:55.900
involves this K matrix, so the
bandwidth of the K hat matrix
00:27:55.900 --> 00:27:59.130
is the same as the bandwidth
of the K matrix.
00:27:59.130 --> 00:28:04.170
And therefore the operations
here are one half NM squared.
00:28:04.170 --> 00:28:11.160
And for each solution, in time,
because this equation
00:28:11.160 --> 00:28:16.380
here has to be solved for each
time step, you're starting
00:28:16.380 --> 00:28:19.870
with delta-t, and then we go
to 2 delta-t, 3 delta-t, 4
00:28:19.870 --> 00:28:22.090
delta-t, and so on.
00:28:22.090 --> 00:28:25.820
And since this solution is to be
formed for each time step,
00:28:25.820 --> 00:28:28.490
however, remember the
factorization is only done
00:28:28.490 --> 00:28:36.520
once because this K hat matrix
is independent of the time.
00:28:36.520 --> 00:28:39.410
Therefore, we have a forward
reduction and back
00:28:39.410 --> 00:28:41.710
substitution in each
time step.
00:28:41.710 --> 00:28:45.500
And the number of operations in
each time step are 2nm. t
00:28:45.500 --> 00:28:47.600
is the number of time
steps here.
00:28:47.600 --> 00:28:51.650
t is the number of
time steps here.
00:28:51.650 --> 00:28:54.320
Well, let us look then
at the accuracy
00:28:54.320 --> 00:28:56.600
considerations that we have.
00:28:56.600 --> 00:28:59.770
I mentioned already the method
is unconditionally stable,
00:28:59.770 --> 00:29:04.470
which means that the delta-t
time step that we're using can
00:29:04.470 --> 00:29:08.360
be any amount, and we will never
have a blow up of the
00:29:08.360 --> 00:29:11.460
solution, a blow up of the
solution due to round-off
00:29:11.460 --> 00:29:13.440
errors that I involved.
00:29:13.440 --> 00:29:15.350
In the central difference
method, of course, the
00:29:15.350 --> 00:29:17.650
explicit time integration
scheme, I mentioned that
00:29:17.650 --> 00:29:20.730
delta-t has to be smaller than
a critical time step.
00:29:20.730 --> 00:29:24.760
So in the implicit Newmark
integration scheme, what we
00:29:24.760 --> 00:29:29.600
can do is we select our of
time step based solely on
00:29:29.600 --> 00:29:32.290
accuracy considerations.
00:29:32.290 --> 00:29:35.270
Let us consider then these
equations that we want to
00:29:35.270 --> 00:29:39.220
solve, and I'm now leaving out
the damping term in order to
00:29:39.220 --> 00:29:42.410
obtain some understanding of how
we would select the time
00:29:42.410 --> 00:29:43.700
step, delta-t.
00:29:43.700 --> 00:29:47.200
Of course, remember the larger
the time step, the less normal
00:29:47.200 --> 00:29:50.440
operations I involved
in the solution.
00:29:50.440 --> 00:29:52.410
So we want to select the
large time step.
00:29:52.410 --> 00:29:55.500
However, if we make the time
step extremely large, then
00:29:55.500 --> 00:29:58.610
surely we cannot expect any
accuracy in the solution.
00:29:58.610 --> 00:30:02.070
In fact, if delta-t is very
large, we simply predict the
00:30:02.070 --> 00:30:03.670
static solution.
00:30:03.670 --> 00:30:05.430
This is one property of
the Newmark method.
00:30:05.430 --> 00:30:07.950
If you were to use a computer
program with the Newmark
00:30:07.950 --> 00:30:10.160
method, if you're interested in
it, why don't you try, say,
00:30:10.160 --> 00:30:12.860
for example, using ADINA, in
which we have the Newmark
00:30:12.860 --> 00:30:15.520
method, and just put delta-t
very large, and all you would
00:30:15.520 --> 00:30:17.660
be getting out is a
static solution.
00:30:17.660 --> 00:30:20.360
Well, since we're interested
in the dynamic response, we
00:30:20.360 --> 00:30:22.840
have to therefore select
delta-t small enough to
00:30:22.840 --> 00:30:28.550
integrate accurately the
solution response.
00:30:28.550 --> 00:30:31.740
And these are the equations that
I want to now consider,
00:30:31.740 --> 00:30:36.500
and I want to show you how can
we select delta-t rationally
00:30:36.500 --> 00:30:39.340
and to obtain the required
accuracy.
00:30:39.340 --> 00:30:44.040
The displacement vector here can
be expressed in terms of
00:30:44.040 --> 00:30:48.630
the mode shape vectors
times xi values.
00:30:48.630 --> 00:30:53.360
I will be talking about how
these phi i vectors are
00:30:53.360 --> 00:30:57.780
calculated in the
next lectures.
00:30:57.780 --> 00:31:00.350
This involves the solution
of the eigenvalue problem
00:31:00.350 --> 00:31:01.890
written down here.
00:31:01.890 --> 00:31:04.460
The free vibration eigenvalue
a problem corresponding to
00:31:04.460 --> 00:31:06.380
this equation.
00:31:06.380 --> 00:31:11.090
I like to defer the discussion
at this point, because we will
00:31:11.090 --> 00:31:13.390
be talking about it later
on in detail.
00:31:13.390 --> 00:31:17.520
All I like to say is that this
transformation here, phi i
00:31:17.520 --> 00:31:21.550
being a vector time independent
and xi being a
00:31:21.550 --> 00:31:23.830
scalar, dependent on time.
00:31:23.830 --> 00:31:29.560
This transformation can be used
in this solution here, or
00:31:29.560 --> 00:31:32.100
can be used rather to transform
this system of
00:31:32.100 --> 00:31:36.330
equations into a
different form.
00:31:36.330 --> 00:31:42.390
And in that transformation, we
use the fact that when we
00:31:42.390 --> 00:31:45.480
calculate phi transpose
K times 5--
00:31:45.480 --> 00:31:47.510
notice these are capital phis.
00:31:47.510 --> 00:31:50.280
I've crossed bars here,
capital phis.
00:31:50.280 --> 00:31:54.090
This is a lower phi, no crossed
bars top and bottom.
00:31:54.090 --> 00:31:58.770
This capital phi here stores
all of the lowercase phi
00:31:58.770 --> 00:32:02.000
vectors, and their n
of such vectors.
00:32:02.000 --> 00:32:07.520
If I use this capital phi and
calculate this product here, I
00:32:07.520 --> 00:32:08.900
get a diagonal matrix.
00:32:08.900 --> 00:32:12.620
In fact, the diagonal matrix
here stores on its diagonal
00:32:12.620 --> 00:32:15.030
the frequency squared, the free
00:32:15.030 --> 00:32:16.920
vibrational frequency squared.
00:32:16.920 --> 00:32:21.870
Also phi transposed M phi gives
us an identity matrix.
00:32:21.870 --> 00:32:27.000
These, once again, the vectors
in this matrix, capital phi,
00:32:27.000 --> 00:32:31.170
are these phi i vectors
which are the
00:32:31.170 --> 00:32:33.590
eigenvectors of this problem.
00:32:33.590 --> 00:32:37.450
And I will talk later on about
how we calculate them.
00:32:37.450 --> 00:32:41.280
Well, if we apply this
transformation--
00:32:41.280 --> 00:32:44.000
whoops, let me put it right once
more-- if we apply this
00:32:44.000 --> 00:32:52.090
transformation to this system of
equations, and we use this
00:32:52.090 --> 00:32:58.100
property right here, we
obtain n decoupled
00:32:58.100 --> 00:33:01.670
equations of this form.
00:33:01.670 --> 00:33:05.600
X double ought i plus omega
i squared, xi equals phi i
00:33:05.600 --> 00:33:09.150
transposed times R. And they
are in such equations, in
00:33:09.150 --> 00:33:13.340
other words I run
c from 1 to n.
00:33:13.340 --> 00:33:16.900
In fact, if we use multiple
position analysis, we actually
00:33:16.900 --> 00:33:19.780
perform this transformation.
00:33:19.780 --> 00:33:22.570
What I now want to do is
simply refer to it
00:33:22.570 --> 00:33:23.270
theoretically.
00:33:23.270 --> 00:33:25.150
I do not want to transform it.
00:33:25.150 --> 00:33:27.600
You're talking about direct
integration of the solution of
00:33:27.600 --> 00:33:30.470
equations, in which there is
no transformation involved.
00:33:30.470 --> 00:33:33.000
So I do not want to go through
this transformation.
00:33:33.000 --> 00:33:38.300
However, I want to refer
to it theoretically.
00:33:38.300 --> 00:33:43.750
And it gives me a tremendous
insight into what happens.
00:33:43.750 --> 00:33:48.250
So basically, what I can say
theoretically is that the
00:33:48.250 --> 00:33:51.890
direct step-by-step solution of
this system of equations,
00:33:51.890 --> 00:33:54.590
using the explicit integration
scheme, the central difference
00:33:54.590 --> 00:33:58.150
method, or the Newmark implicit
integration scheme,
00:33:58.150 --> 00:34:01.980
corresponds completely to the
direct step-by-step solution
00:34:01.980 --> 00:34:07.630
of n equations, of this form,
where the displacements here,
00:34:07.630 --> 00:34:11.940
U, are given in this way.
00:34:11.940 --> 00:34:15.980
x being now a function of time,
so if I differentiate
00:34:15.980 --> 00:34:21.179
here, your double dot would be
obtained by differentiating xi
00:34:21.179 --> 00:34:24.710
also twice, with respect
to time.
00:34:24.710 --> 00:34:26.620
Once again, this is
an important,
00:34:26.620 --> 00:34:28.350
very important, fact.
00:34:28.350 --> 00:34:32.969
I look at these equations, and
I say that's a solution of
00:34:32.969 --> 00:34:37.280
these set of coupled
differential equations.
00:34:37.280 --> 00:34:42.310
It's completely equivalent to
the solution of n equations of
00:34:42.310 --> 00:34:48.478
this form with this
transformation used.
00:34:48.478 --> 00:34:50.950
No dots here or dots there.
00:34:50.950 --> 00:34:53.820
In other words, this equation
holds for displacements
00:34:53.820 --> 00:34:55.820
without the red dots and
for the accelerations
00:34:55.820 --> 00:34:58.280
with the red dots.
00:34:58.280 --> 00:34:59.820
It is equivalent.
00:34:59.820 --> 00:35:03.270
This solution is equivalent
to that solution.
00:35:03.270 --> 00:35:06.990
However, in direct integration,
I never perform
00:35:06.990 --> 00:35:08.420
actually the transformation.
00:35:08.420 --> 00:35:11.820
All I know is that when I
operate on this system of
00:35:11.820 --> 00:35:14.570
equations, using the Newmark
method or the central
00:35:14.570 --> 00:35:18.180
difference method, I, in
fact, operate on n
00:35:18.180 --> 00:35:20.130
such decoupled equations.
00:35:20.130 --> 00:35:21.590
That's important.
00:35:21.590 --> 00:35:25.270
And this means that for an
analysis on our direct
00:35:25.270 --> 00:35:29.420
integration scheme, we can in
fact analyze our direct
00:35:29.420 --> 00:35:33.310
integration scheme by looking at
these equations, instead of
00:35:33.310 --> 00:35:34.700
these equations.
00:35:34.700 --> 00:35:37.190
You see if I were to analyze my
direct integration scheme
00:35:37.190 --> 00:35:40.450
using these equations, I would
have to deal with this huge
00:35:40.450 --> 00:35:45.720
matrix K, the huge matrix M. All
I now need to deal with,
00:35:45.720 --> 00:35:50.150
in the analysis of the direct
integration scheme, is with n
00:35:50.150 --> 00:35:53.900
such simple one degree
of freedom equations.
00:35:53.900 --> 00:35:57.810
And in fact, I don't even need
to deal with n such equations,
00:35:57.810 --> 00:36:02.020
for the analysis purpose, I
only need to deal with one
00:36:02.020 --> 00:36:04.210
equation, where I
make now omega a
00:36:04.210 --> 00:36:06.710
variable and r a variable.
00:36:06.710 --> 00:36:13.070
So my analysis on this
equation, using this
00:36:13.070 --> 00:36:17.000
particular time integration
scheme that I want to analyze,
00:36:17.000 --> 00:36:22.140
is completely giving me all the
information that I need to
00:36:22.140 --> 00:36:25.780
know when I use my direct
integration scheme in the
00:36:25.780 --> 00:36:28.200
solution of this equation.
00:36:28.200 --> 00:36:32.020
For analysis purposes of my
direct integration scheme, I
00:36:32.020 --> 00:36:35.330
can simply analyze my direct
integration scheme using this
00:36:35.330 --> 00:36:38.030
equation and get all
the relevant
00:36:38.030 --> 00:36:40.840
information from that analysis.
00:36:40.840 --> 00:36:44.140
This is an extremely
important fact.
00:36:44.140 --> 00:36:48.060
Once again, I do not actually do
perform the transformation.
00:36:48.060 --> 00:36:50.190
We don't do the transformation.
00:36:50.190 --> 00:36:53.780
Only in theory for analysis
purposes.
00:36:53.780 --> 00:36:57.610
I'm looking at the solution
of this equation using the
00:36:57.610 --> 00:36:59.100
Newmark integration scheme.
00:36:59.100 --> 00:37:02.290
And I get all the relevant
information on the accuracy,
00:37:02.290 --> 00:37:07.650
stability, et cetera of the
integration scheme by looking
00:37:07.650 --> 00:37:10.520
at the solution of this equation
with that direct
00:37:10.520 --> 00:37:12.030
integration scheme.
00:37:12.030 --> 00:37:15.030
Now, even in the analysis
of this
00:37:15.030 --> 00:37:17.000
equation, of course, remember--
00:37:17.000 --> 00:37:20.710
or in the use of this equation,
remember that r is
00:37:20.710 --> 00:37:22.170
available now.
00:37:22.170 --> 00:37:25.660
And I have some choice what
to put in there and so on.
00:37:25.660 --> 00:37:30.220
But the important point is that
what I want to identify
00:37:30.220 --> 00:37:34.650
is the characteristics of
the integration scheme.
00:37:34.650 --> 00:37:39.620
And the characteristics of the
integration scheme are really
00:37:39.620 --> 00:37:44.150
quite well seen already if we
look at a very simple case,
00:37:44.150 --> 00:37:45.640
namely this case.
00:37:45.640 --> 00:37:48.260
I set r deliberately
equal to zero.
00:37:48.260 --> 00:37:52.090
And I look at the case where I
have initial condition x, the
00:37:52.090 --> 00:37:54.950
displacements being one, the
velocities being zero.
00:37:54.950 --> 00:37:57.420
this means, of course, that the
accelerations are minus
00:37:57.420 --> 00:37:59.050
omega squared.
00:37:59.050 --> 00:38:04.680
If I use the Newmark method, the
one that I just talked to
00:38:04.680 --> 00:38:10.870
you about, on that system of
equations, then I can plot
00:38:10.870 --> 00:38:13.090
errors that reside in
that integration.
00:38:13.090 --> 00:38:17.400
And what I've plotted here the
percentage period elongation,
00:38:17.400 --> 00:38:23.810
pe divided by t times 100, which
is this amount here that
00:38:23.810 --> 00:38:27.530
we see when we integrate the
differential equation that I
00:38:27.530 --> 00:38:32.260
showed you via as the Newmark
method with different delta-t
00:38:32.260 --> 00:38:35.300
over t values.
00:38:35.300 --> 00:38:41.820
On the bottom here, we are
plotting delta-t divided by t,
00:38:41.820 --> 00:38:49.510
the period t, of course, being
given for the particular
00:38:49.510 --> 00:38:51.250
oscillator that we're
looking at.
00:38:51.250 --> 00:38:52.870
We are starting off with one.
00:38:52.870 --> 00:38:56.570
The exact solution, I should
take better now another color,
00:38:56.570 --> 00:39:01.760
the exact solution would look
something like that,
00:39:01.760 --> 00:39:03.410
finishing off here.
00:39:03.410 --> 00:39:04.710
That would be the
exact solution.
00:39:04.710 --> 00:39:10.030
And what we're seeing now
is an amplitude decay.
00:39:10.030 --> 00:39:10.865
It's already there.
00:39:10.865 --> 00:39:13.260
It would come in right there.
00:39:13.260 --> 00:39:15.090
That is the exact solution.
00:39:15.090 --> 00:39:16.140
We'll stop here.
00:39:16.140 --> 00:39:19.190
The period is of course
this t here, as shown.
00:39:19.190 --> 00:39:23.280
This would be the exact solution
for one period.
00:39:23.280 --> 00:39:27.130
What we are seeing as an
amplitude decay, a drop in the
00:39:27.130 --> 00:39:30.650
numerical solution, and
period elongation in
00:39:30.650 --> 00:39:32.150
the numerical solution.
00:39:32.150 --> 00:39:35.470
And the percentage period
elongation is plotted here.
00:39:35.470 --> 00:39:39.470
We notice that the Newmark
method corresponds to this
00:39:39.470 --> 00:39:40.480
curve here.
00:39:40.480 --> 00:39:45.650
So with delta-t over
t being 0.10.
00:39:45.650 --> 00:39:47.020
We have about--
00:39:47.020 --> 00:39:49.830
let us just look it up here--
00:39:49.830 --> 00:39:54.550
we have about 3%, 3%
period elongation
00:39:54.550 --> 00:39:56.050
for the Newmark method.
00:39:56.050 --> 00:39:58.110
I'm also showing other
techniques here, which are
00:39:58.110 --> 00:40:01.320
also implicit techniques, the
Wilson Theta method with Theta
00:40:01.320 --> 00:40:07.970
1.4 has about 5% period
elongation for delta-t over t
00:40:07.970 --> 00:40:11.050
being 0.10.
00:40:11.050 --> 00:40:14.380
Another point of interest
is of course also is the
00:40:14.380 --> 00:40:15.990
amplitude decay.
00:40:15.990 --> 00:40:20.520
And that amplitude decay is
plotted on this graph here,
00:40:20.520 --> 00:40:23.190
percentage amplitude decay
as a function of
00:40:23.190 --> 00:40:25.370
delta-t over t again.
00:40:25.370 --> 00:40:26.540
Delta-t over t.
00:40:26.540 --> 00:40:30.020
And notice that the Newmark
method does not appear.
00:40:30.020 --> 00:40:34.080
In fact, the Newmark method
has no amplitude decay.
00:40:34.080 --> 00:40:36.860
The solution is right
down here.
00:40:36.860 --> 00:40:41.230
No amplitude decay for any
delta-t over t, whereas the
00:40:41.230 --> 00:40:45.690
Wilson methods and the Houbolt
method is also shown here,
00:40:45.690 --> 00:40:49.170
have the amplitude decay
shown by these curves.
00:40:49.170 --> 00:40:52.660
Notice that the Wilson method
has less amplitude decay then
00:40:52.660 --> 00:40:54.040
the Houbolt method.
00:40:54.040 --> 00:40:58.990
The period elongation for the
Houbolt method was also shown
00:40:58.990 --> 00:41:01.370
on this view graph here.
00:41:01.370 --> 00:41:06.840
So the Newmark method does have
period elongation, but it
00:41:06.840 --> 00:41:14.410
does not have any amplitude
decay in this solution.
00:41:14.410 --> 00:41:17.150
I should also mention this, of
course, as one important
00:41:17.150 --> 00:41:20.000
point, I forgot to mention that
I'm looking here at the
00:41:20.000 --> 00:41:24.530
Newmark method with delta
1/2 alpha equal to 1/4.
00:41:24.530 --> 00:41:29.900
This is the constant average
acceleration method.
00:41:29.900 --> 00:41:32.540
Well, let us look then
at how can we
00:41:32.540 --> 00:41:35.460
actually perform the solution?
00:41:35.460 --> 00:41:40.020
How can we select an appropriate
time step delta-t?
00:41:40.020 --> 00:41:47.770
Here we notice that the dynamic
load factor that we
00:41:47.770 --> 00:41:53.890
are knowing that we can derive
for the solution of this
00:41:53.890 --> 00:41:56.660
differential equation here--
00:41:56.660 --> 00:41:58.510
here I included damping--
00:41:58.510 --> 00:42:01.620
the dynamic load factor plotted
along here as a
00:42:01.620 --> 00:42:03.590
functional of p over omega.
00:42:03.590 --> 00:42:07.760
p being the exciting free
frequency, omega being the
00:42:07.760 --> 00:42:10.290
frequency of the system.
00:42:10.290 --> 00:42:13.280
The dynamic load factor
looks as shown here.
00:42:13.280 --> 00:42:17.800
If we have no damping, you're
talking about this curve.
00:42:17.800 --> 00:42:19.800
Now notice the important
point.
00:42:19.800 --> 00:42:23.140
The dynamic load factor, of
course, is 1 for static
00:42:23.140 --> 00:42:26.900
response, and it really gives
us a magnification of the
00:42:26.900 --> 00:42:28.680
static response.
00:42:28.680 --> 00:42:32.290
In a static response, we would
have this solution.
00:42:32.290 --> 00:42:40.670
And if the p over omega value
is say 2, we would have that
00:42:40.670 --> 00:42:44.990
maximum solution in the dynamic
response analysis.
00:42:44.990 --> 00:42:48.500
If p over omega is
equal to 0.5.
00:42:48.500 --> 00:42:52.920
we would have, for zero damping,
this maximum solution
00:42:52.920 --> 00:42:55.430
in the dynamic response
history.
00:42:55.430 --> 00:42:58.130
Or rather this would be the
maximum dynamic response that
00:42:58.130 --> 00:43:00.620
you would see in the
dynamic solution.
00:43:00.620 --> 00:43:04.370
This is how the dynamic load
factor is defined.
00:43:04.370 --> 00:43:08.140
Of course, at p over omega equal
to 1, we have resonance.
00:43:08.140 --> 00:43:11.090
In practice, we have always some
damping, and if you look
00:43:11.090 --> 00:43:15.430
at this amount of damping, this
would be the maximum with
00:43:15.430 --> 00:43:20.470
p over omega equal to 1, we
would have this as the maximum
00:43:20.470 --> 00:43:23.520
response that is ever
measured in the
00:43:23.520 --> 00:43:26.900
dynamic response solution.
00:43:26.900 --> 00:43:33.610
Well, if we look at this graph,
we recognize of course
00:43:33.610 --> 00:43:38.920
that as p over omega goes
against 0, in other words as
00:43:38.920 --> 00:43:45.020
the exciting frequency, as
over the frequency of the
00:43:45.020 --> 00:43:49.060
system goes against 0, we
really talk about static
00:43:49.060 --> 00:43:49.980
conditions.
00:43:49.980 --> 00:43:54.580
As p over omega is very large,
we are talking about no
00:43:54.580 --> 00:43:56.050
response whatsoever.
00:43:56.050 --> 00:44:00.390
And the important dynamic
response really lies somewhat
00:44:00.390 --> 00:44:04.640
in this region here with that
being the tail end of it.
00:44:04.640 --> 00:44:08.350
But the important dynamic
response really lies in here,
00:44:08.350 --> 00:44:11.250
that we have to include
in the solution.
00:44:11.250 --> 00:44:14.030
Out here we have basically
a static response.
00:44:14.030 --> 00:44:18.150
Let us look at two
simple cases.
00:44:18.150 --> 00:44:25.420
For p over omega 0.05, we can
see that the static response
00:44:25.420 --> 00:44:28.640
here shown in a dashed line, the
dynamic response showing
00:44:28.640 --> 00:44:29.890
as a solid line.
00:44:29.890 --> 00:44:31.100
They're basically the same.
00:44:31.100 --> 00:44:33.710
In other words, a dynamic
response being equal to the
00:44:33.710 --> 00:44:37.360
static response, practically
equal to the static response.
00:44:37.360 --> 00:44:40.960
We are down at this
range here.
00:44:40.960 --> 00:44:44.815
As another example, if we have
p over omega equal to 3, we
00:44:44.815 --> 00:44:47.570
are down over here.
00:44:47.570 --> 00:44:53.070
Then the response, the static
response is shown here in the
00:44:53.070 --> 00:44:57.030
dashed line, and the dynamic
response is shown
00:44:57.030 --> 00:44:58.380
as the solid line.
00:44:58.380 --> 00:45:01.110
Very little response
here at all.
00:45:01.110 --> 00:45:05.280
Because as p over omega goes
to infinity, there is no
00:45:05.280 --> 00:45:07.870
response for the system
at all anymore.
00:45:07.870 --> 00:45:10.780
Basically, what we're saying
then is that the frequency of
00:45:10.780 --> 00:45:13.230
the applied loading is
so fast, that the
00:45:13.230 --> 00:45:14.950
structure can;t just move.
00:45:14.950 --> 00:45:16.330
It just cannot move at all.
00:45:16.330 --> 00:45:18.960
And we don't have any response
in the structure.
00:45:18.960 --> 00:45:27.240
Well, the important point is
that if we know the frequency
00:45:27.240 --> 00:45:33.720
content of the loading, then
what we have to do in the use
00:45:33.720 --> 00:45:39.220
of the Newmark method is to
calculate a delta-t small
00:45:39.220 --> 00:45:42.700
enough to pick up the loading,
of course, because that is
00:45:42.700 --> 00:45:48.960
given in terms of this great
steps or this great input
00:45:48.960 --> 00:45:52.240
points to the computer program,
and the time step has
00:45:52.240 --> 00:45:56.360
to be small enough to integrate
the dynamic response
00:45:56.360 --> 00:45:58.260
of importance accurately.
00:45:58.260 --> 00:46:02.090
The static response, the static
response is contained
00:46:02.090 --> 00:46:09.120
in the solution anyway, because
the K matrix is on the
00:46:09.120 --> 00:46:11.270
left hand side of
the solution.
00:46:11.270 --> 00:46:15.390
So the complete solution process
therefore can be
00:46:15.390 --> 00:46:17.060
summarized as follows.
00:46:17.060 --> 00:46:21.850
Identify the frequencies
contained in the loading, and
00:46:21.850 --> 00:46:24.910
using a Fourier analysis,
if necessary.
00:46:24.910 --> 00:46:28.220
Choose a finite element mesh
that accurately represents all
00:46:28.220 --> 00:46:32.350
frequencies up to about four
times the highest frequency
00:46:32.350 --> 00:46:35.210
omega u contained
in the loading.
00:46:35.210 --> 00:46:37.670
The highest frequency contained
in the loading is
00:46:37.670 --> 00:46:42.920
really the cutoff frequency that
will give us a time step.
00:46:42.920 --> 00:46:46.630
And that then, the time step is
in fact obtained via these
00:46:46.630 --> 00:46:47.340
consideration.
00:46:47.340 --> 00:46:48.690
Perform a direct integration
analysis.
00:46:48.690 --> 00:46:52.640
The time step delta-t for this
solutions should equal about 1
00:46:52.640 --> 00:46:56.930
over 20 Tu, where Tu is 2 pi
over omega u, which is the
00:46:56.930 --> 00:46:59.320
highest frequency contained
in the loading.
00:47:02.340 --> 00:47:05.260
What we are saying basically
is that we want to select a
00:47:05.260 --> 00:47:09.160
finite element mesh to
represent the highest
00:47:09.160 --> 00:47:13.600
frequency in the loading
correctly, accurately, and
00:47:13.600 --> 00:47:17.650
then we want to select a time
step in the data integration
00:47:17.650 --> 00:47:22.250
solution using the Newmark
method that integrates the
00:47:22.250 --> 00:47:25.900
highest frequency contained in
the loading also accurately,
00:47:25.900 --> 00:47:28.360
or the response in that highest
frequency contained in
00:47:28.360 --> 00:47:29.610
the loading also accurately.
00:47:32.000 --> 00:47:36.760
And this then gives us basically
a step-by-step
00:47:36.760 --> 00:47:41.090
procedure for the selection of
the finite element mesh and
00:47:41.090 --> 00:47:42.070
the times step.
00:47:42.070 --> 00:47:46.310
Notice that the mesh selection
and the time step selection
00:47:46.310 --> 00:47:51.360
both of course hinge on the
structure and the particular
00:47:51.360 --> 00:47:53.290
frequency content
of the loading.
00:47:53.290 --> 00:47:55.070
That's important.
00:47:55.070 --> 00:47:58.906
This is the modeling procedure
that we would use for a
00:47:58.906 --> 00:48:01.340
structural vibration problem.
00:48:01.340 --> 00:48:04.890
We would use this time step
here in the implicit
00:48:04.890 --> 00:48:08.850
integration, because we get
all the accuracy we want.
00:48:08.850 --> 00:48:11.990
In an explicit integration,
we would have to use a
00:48:11.990 --> 00:48:13.320
smaller time step.
00:48:13.320 --> 00:48:16.390
The sentence down here "or be
smaller for stability reasons"
00:48:16.390 --> 00:48:19.290
really refers to the explicit
time integration.
00:48:19.290 --> 00:48:22.780
In the Newmark method, we would
not use that condition.
00:48:22.780 --> 00:48:28.010
We would simply use Tu, being 2
pi omega u and delta-t being
00:48:28.010 --> 00:48:30.280
1 over 20 times Tu.
00:48:30.280 --> 00:48:34.400
So in the Newmark method, I
blank this out, and this is
00:48:34.400 --> 00:48:37.760
here the delta-t that
we would be using.
00:48:37.760 --> 00:48:43.610
And just to summarize once the
modeling of wave propagation
00:48:43.610 --> 00:48:46.520
problem here, here what we would
do is we assume that the
00:48:46.520 --> 00:48:48.870
wavelength is Lw.
00:48:48.870 --> 00:48:53.180
The total time for the wave to
travel past a point is then tw
00:48:53.180 --> 00:48:57.500
equal Lw over c, where
c is the wave speed.
00:48:57.500 --> 00:49:00.500
Assuming that n time steps are
necessary to represent the
00:49:00.500 --> 00:49:05.750
wave, we get our delta-t and the
effective length, which I
00:49:05.750 --> 00:49:08.800
talked about earlier, of the
finite element would be then
00:49:08.800 --> 00:49:09.990
given by this formula.
00:49:09.990 --> 00:49:14.260
So the procedure here is
somewhat different in that if
00:49:14.260 --> 00:49:18.870
we have a wave that looks like
this say, we now select a
00:49:18.870 --> 00:49:23.370
certain number of times steps,
delta-t, to pick up the wave
00:49:23.370 --> 00:49:24.730
accurately.
00:49:24.730 --> 00:49:31.280
And this being the length Lw,
the time for the wave to
00:49:31.280 --> 00:49:33.500
travel past the point is tw.
00:49:33.500 --> 00:49:38.400
And delta-t then is the number
of time points, or time steps,
00:49:38.400 --> 00:49:40.880
that are used to represent
the wave accurately.
00:49:43.870 --> 00:49:47.420
And once we have the time step,
we select the effective
00:49:47.420 --> 00:49:48.530
length of an element.
00:49:48.530 --> 00:49:52.030
And then, of course it depends
on whether we have a low-order
00:49:52.030 --> 00:49:56.120
element or a high-order element,
that will give us
00:49:56.120 --> 00:50:00.990
then the procedure of
how to apply the
00:50:00.990 --> 00:50:03.620
effective lengths criteria.
00:50:03.620 --> 00:50:06.440
I like to finally now summarize
the complete
00:50:06.440 --> 00:50:07.690
solution process.
00:50:09.680 --> 00:50:15.560
This is a table that I put
together for you to show how
00:50:15.560 --> 00:50:19.740
the complete solution process
for explicit and implicit time
00:50:19.740 --> 00:50:22.110
integration can be very
effectively implemented in a
00:50:22.110 --> 00:50:23.860
computer program.
00:50:23.860 --> 00:50:27.340
We form a linear stiffness
matrix K, the mass matrix M,
00:50:27.340 --> 00:50:30.230
the damping matrix C, whichever
applicable.
00:50:30.230 --> 00:50:31.870
We calculate the following
constants.
00:50:31.870 --> 00:50:34.510
If we select the Newmark
method, these are the
00:50:34.510 --> 00:50:37.040
constants that you would use.
00:50:37.040 --> 00:50:38.610
Let's not go into details.
00:50:38.610 --> 00:50:43.270
These constants are given
in the textbook.
00:50:43.270 --> 00:50:46.270
They are given in the ADINA
manual, if you are using the
00:50:46.270 --> 00:50:47.490
ADINA computer program.
00:50:47.490 --> 00:50:52.000
They are widely published.
00:50:52.000 --> 00:50:55.000
We simply cast them
in this form.
00:50:55.000 --> 00:50:56.520
In the central difference
method, we
00:50:56.520 --> 00:50:59.810
would use these constants.
00:50:59.810 --> 00:51:02.420
So there is some initialization
involved.
00:51:02.420 --> 00:51:05.710
Then in the central difference
method, we also want to
00:51:05.710 --> 00:51:13.150
calculate our first
displacement, delta-t U, as
00:51:13.150 --> 00:51:15.330
shown here.
00:51:15.330 --> 00:51:18.910
And in the Newmark method, we
don't need to do that, because
00:51:18.910 --> 00:51:22.340
we simply march ahead from time
t to time t plus delta-t.
00:51:22.340 --> 00:51:25.820
You only need the condition at
the conditions at time t to
00:51:25.820 --> 00:51:29.090
obtain the solution at
times t plus delta-t.
00:51:29.090 --> 00:51:32.070
In the central difference
method, remember we need two
00:51:32.070 --> 00:51:33.420
previous values.
00:51:33.420 --> 00:51:37.150
We then form an effective linear
coefficient matrix.
00:51:37.150 --> 00:51:39.570
In implicit time integration,
this is
00:51:39.570 --> 00:51:40.780
the coefficient matrix.
00:51:40.780 --> 00:51:42.590
The stiffness matrix
being there, as I
00:51:42.590 --> 00:51:43.840
showed to you earlier.
00:51:43.840 --> 00:51:47.890
The mass matrix and damping
matrix being there.
00:51:47.890 --> 00:51:56.340
If we look at this a0, we
notice that a0 is 1 over
00:51:56.340 --> 00:51:59.700
delta-t squared and there's
a constant alpha here.
00:51:59.700 --> 00:52:02.660
The Newmark constant.
00:52:02.660 --> 00:52:05.080
But it's 1 over delta-t
squared.
00:52:05.080 --> 00:52:06.640
That is the important point.
00:52:06.640 --> 00:52:12.280
So as delta-t gets larger and
larger, larger and larger,
00:52:12.280 --> 00:52:15.680
this effect goes against 0.
00:52:15.680 --> 00:52:19.030
The same holds for
the a1 constant.
00:52:19.030 --> 00:52:21.060
As delta-t gets larger and
larger, this effect
00:52:21.060 --> 00:52:22.150
goes against 0.
00:52:22.150 --> 00:52:25.560
And this is the reason why the
Newmark method really goes
00:52:25.560 --> 00:52:31.550
down or collapses to the static
analysis, if we have
00:52:31.550 --> 00:52:32.970
very large time steps.
00:52:32.970 --> 00:52:38.320
In explicit integration, we use
this coefficient matrix.
00:52:38.320 --> 00:52:43.740
And now for each integration
step, or for each time step,
00:52:43.740 --> 00:52:46.320
we perform the following
calculations.
00:52:46.320 --> 00:52:51.780
We calculate the t plus
delta-t R hat.
00:52:51.780 --> 00:52:55.780
And implicit time integration,
the Newmark method, in
00:52:55.780 --> 00:52:59.050
explicit time integration,
we calculate this vector.
00:52:59.050 --> 00:53:02.450
Notice at tF here,
no k times tU.
00:53:02.450 --> 00:53:06.170
k times 2u is equal to tF.
00:53:06.170 --> 00:53:08.890
In implicit time integration,
having calculated for each
00:53:08.890 --> 00:53:15.530
time step this load step, we
now use the equations K hat
00:53:15.530 --> 00:53:20.320
times t plus delta-t u, equals
t plus delta-t R hat to
00:53:20.320 --> 00:53:22.850
calculate t plus delta-t u.
00:53:22.850 --> 00:53:27.010
Of course, we would triangulize
the K hat matrix
00:53:27.010 --> 00:53:29.310
prior to going into this loop.
00:53:29.310 --> 00:53:31.750
We do it once and forward.
00:53:31.750 --> 00:53:36.790
In explicit time integration,
we use this matrix or this
00:53:36.790 --> 00:53:39.850
vector here, I should say, this
vector now, and we can
00:53:39.850 --> 00:53:44.850
calculate directly for the
t plus delta-t U also.
00:53:44.850 --> 00:53:48.300
The two equations that I used
are summarized here.
00:53:48.300 --> 00:53:50.120
In implicit time integration,
this is the
00:53:50.120 --> 00:53:53.410
equation that we use.
00:53:53.410 --> 00:53:55.750
Notice we only need
to factorize this
00:53:55.750 --> 00:53:57.080
K hat matrix once.
00:53:57.080 --> 00:53:59.910
We do it outside the
step-by-step integration, as I
00:53:59.910 --> 00:54:01.470
pointed out earlier.
00:54:01.470 --> 00:54:05.330
And then this gives us the
increment in displacements.
00:54:05.330 --> 00:54:09.230
In explicit time integration,
this is the equation we use.
00:54:09.230 --> 00:54:13.580
If M and C are diagonal, it's
a trivial operation as I
00:54:13.580 --> 00:54:15.410
pointed out earlier.
00:54:15.410 --> 00:54:18.070
With this then, we have
the new displacements.
00:54:18.070 --> 00:54:21.990
And on the last view graph, I
show you the formulae that we
00:54:21.990 --> 00:54:25.490
are then using to calculate
the accelerations in the
00:54:25.490 --> 00:54:29.050
Newmark method from the
incremental displacements and
00:54:29.050 --> 00:54:30.880
variables that we
knew already.
00:54:30.880 --> 00:54:33.780
Having now calculated the
accelerations, we can get the
00:54:33.780 --> 00:54:36.370
velocities at time
t plus delta-t.
00:54:36.370 --> 00:54:38.990
And of course, we have already
our displacement at time t
00:54:38.990 --> 00:54:39.980
plus delta-t.
00:54:39.980 --> 00:54:42.520
In the central difference
method, we use our basic
00:54:42.520 --> 00:54:46.010
equations that we already
applied earlier in the
00:54:46.010 --> 00:54:48.780
development of the method to
calculate the velocities and
00:54:48.780 --> 00:54:51.510
the accelerations at time t.
00:54:51.510 --> 00:54:56.750
This completes the procedure
that is effectively used in a
00:54:56.750 --> 00:55:00.400
computer program for the
integration of the dynamic
00:55:00.400 --> 00:55:01.680
equilibrium equations.
00:55:01.680 --> 00:55:06.460
I like to just now mention
that I've placed heavy
00:55:06.460 --> 00:55:09.910
emphasis on the Newmark method
in an implicit integration.
00:55:09.910 --> 00:55:11.900
Of course, there are other
implicit integration formulae
00:55:11.900 --> 00:55:16.570
available, the Wilson method,
the Houbolt method, and so on.
00:55:16.570 --> 00:55:18.760
The Newmark method
has shown, has
00:55:18.760 --> 00:55:20.750
proven to be quite effective.
00:55:20.750 --> 00:55:24.030
And it's certainly a good
example to demonstrate to you
00:55:24.030 --> 00:55:28.150
how the actual procedure is
implemented and how an
00:55:28.150 --> 00:55:30.370
implicit integration scheme
is being used.
00:55:30.370 --> 00:55:33.560
Other implicit integration
methods would be used in much
00:55:33.560 --> 00:55:34.790
the same way.
00:55:34.790 --> 00:55:37.990
This is all I wanted to mention
to you, discuss with
00:55:37.990 --> 00:55:39.180
you in this lecture.
00:55:39.180 --> 00:55:40.530
Thank you very much for
your attention.