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PROFESSOR: Ladies and gentlemen,
welcome to
00:00:24.000 --> 00:00:25.840
lecture number 3.
00:00:25.840 --> 00:00:29.000
In the previous two lectures,
we discussed some basic
00:00:29.000 --> 00:00:32.430
concepts related to finite
element analysis.
00:00:32.430 --> 00:00:35.760
In this lecture, I would like
to present to you a general
00:00:35.760 --> 00:00:38.640
formulation of the
displacement-based finite
00:00:38.640 --> 00:00:40.500
element method.
00:00:40.500 --> 00:00:42.800
This is a very general
formulation.
00:00:42.800 --> 00:00:47.510
We use it to analyze 1D, 2D,
three-dimensional problems,
00:00:47.510 --> 00:00:49.770
plate and shell structures.
00:00:49.770 --> 00:00:53.340
And it provides the basis of
almost all finite element
00:00:53.340 --> 00:00:57.470
analysis performed at
present in practice.
00:00:57.470 --> 00:01:00.010
We will see that the formation
is really a modern an
00:01:00.010 --> 00:01:03.410
application of the Ritz/Golerkin
procedures that
00:01:03.410 --> 00:01:06.010
we discussed in the
last lecture.
00:01:06.010 --> 00:01:09.400
We will consider in this lecture
static and dynamic
00:01:09.400 --> 00:01:13.030
conditions, but as I pointed out
earlier, we will only be
00:01:13.030 --> 00:01:16.750
concerned with linear
analysis.
00:01:16.750 --> 00:01:21.990
On this first view graph, I've
prepared schematically a
00:01:21.990 --> 00:01:25.220
sketch of a three-dimensional
body.
00:01:25.220 --> 00:01:28.900
This three-dimensional body
could represent, typically, a
00:01:28.900 --> 00:01:32.690
bridge, a shaft, a building--
00:01:32.690 --> 00:01:35.050
whatever structure we
want to analyze.
00:01:35.050 --> 00:01:38.710
And this three-dimensional body
here is subject to the
00:01:38.710 --> 00:01:41.920
following forces-- it is
subjected to concentrated
00:01:41.920 --> 00:01:51.190
forces, forces that have
components, Fxi, Fyi, Fzi, at
00:01:51.190 --> 00:01:55.210
one point i, and there are
many such points i.
00:01:55.210 --> 00:01:58.050
The body's also subjected
to body force
00:01:58.050 --> 00:02:03.910
components, Fbx, Fby, Fbz.
00:02:03.910 --> 00:02:07.830
These are forces per
unit volume.
00:02:07.830 --> 00:02:10.860
And we will see later on that we
include in these forces the
00:02:10.860 --> 00:02:14.970
d'Alembert forces, when we
consider dynamic analysis.
00:02:14.970 --> 00:02:20.200
The body is also subjected to
distributed surface forces,
00:02:20.200 --> 00:02:26.690
with components, Fsx,
Fsy, and Fsz.
00:02:26.690 --> 00:02:30.980
These surface forces would be,
for example, distributed water
00:02:30.980 --> 00:02:35.570
pressure in a dam, frictional
forces, et cetera.
00:02:35.570 --> 00:02:40.050
So we have, basically,
concentrated surface forces,
00:02:40.050 --> 00:02:45.740
we have distributed surface
forces, and volume forces--
00:02:45.740 --> 00:02:50.150
externally applied volume
forces, body forces.
00:02:50.150 --> 00:02:52.820
The body is, of course, also
properly supported.
00:02:52.820 --> 00:02:55.840
We have here, typically,
a support that prevents
00:02:55.840 --> 00:02:58.640
displacements in
any direction.
00:02:58.640 --> 00:03:01.410
And here, we have another
such support.
00:03:01.410 --> 00:03:03.910
Here, we have a roller support,
which prevents
00:03:03.910 --> 00:03:07.650
displacements only in
this direction.
00:03:07.650 --> 00:03:13.520
The body is defined in the
coordinate system, XYZ, and
00:03:13.520 --> 00:03:16.476
notice that I'm using
here capital XYZ's.
00:03:19.110 --> 00:03:24.240
And the displacements of the
body are measured as U, V, and
00:03:24.240 --> 00:03:29.100
W into the capital X,
Y, and Z directions.
00:03:29.100 --> 00:03:34.720
I'm using capital letters here
to denote global displacements
00:03:34.720 --> 00:03:36.780
and global coordinates.
00:03:36.780 --> 00:03:40.580
We will later on in the finite
element discretization also
00:03:40.580 --> 00:03:45.480
introduce small lowercase x, y
and z's, and u, v, and w's to
00:03:45.480 --> 00:03:49.270
measure the displacement in
the individual elements.
00:03:49.270 --> 00:03:53.300
So the problem is, other words,
that we have this body,
00:03:53.300 --> 00:03:57.950
this general structure,
subjected to certain forces,
00:03:57.950 --> 00:04:03.180
properly constrained, and
we want to calculate the
00:04:03.180 --> 00:04:06.600
displacements of the body, the
strains in the body, and the
00:04:06.600 --> 00:04:08.720
stresses, of course,
in the body.
00:04:08.720 --> 00:04:12.200
Well, on this view graph, here,
I listed the external
00:04:12.200 --> 00:04:18.010
forces once as vectors, here's
our force FB, the body force
00:04:18.010 --> 00:04:21.860
per unit volume, with components
into the x, y, and
00:04:21.860 --> 00:04:23.190
z directions.
00:04:23.190 --> 00:04:25.810
Here, we have the surface forces
with components in the
00:04:25.810 --> 00:04:27.690
x, y, and z directions.
00:04:27.690 --> 00:04:30.430
And here, we have a typical
concentrated force at that
00:04:30.430 --> 00:04:35.900
point i, with components
FX, FY, and FZ again.
00:04:35.900 --> 00:04:39.250
The displacements of the body
measured in the global
00:04:39.250 --> 00:04:43.130
coordinates are U, V and
W, as shown here.
00:04:43.130 --> 00:04:46.680
Of course, notice that these U,
V and W's are functions of
00:04:46.680 --> 00:04:50.570
the capital XYZ coordinates.
00:04:50.570 --> 00:04:54.180
The strains corresponding to
these displacements, which of
00:04:54.180 --> 00:04:56.405
course, are known,
are listed here.
00:04:56.405 --> 00:05:00.300
And the three-dimensional
analysis, we have six such
00:05:00.300 --> 00:05:04.830
known strains, from epsilon
XX to gamma ZX.
00:05:04.830 --> 00:05:07.100
Of course, the last three being
the shearing strains,
00:05:07.100 --> 00:05:10.050
and the first three being
the normal strains.
00:05:10.050 --> 00:05:13.720
The corresponding stresses
are listed here.
00:05:13.720 --> 00:05:18.770
Again, six components from
tau XX to tau ZX.
00:05:18.770 --> 00:05:21.290
Of course, if we were to
actually analyze a
00:05:21.290 --> 00:05:24.120
two-dimensional problem, such
as the plane stress problem,
00:05:24.120 --> 00:05:26.590
we would only use the
appropriate quantities from
00:05:26.590 --> 00:05:30.880
here and from there, as we'll
discuss later on.
00:05:30.880 --> 00:05:34.460
The starting point of our
analysis, in which we want to
00:05:34.460 --> 00:05:37.700
calculate the stresses, the
strains, and of course, the
00:05:37.700 --> 00:05:39.320
displacement also.
00:05:39.320 --> 00:05:42.560
The starting point is the
principle of virtual
00:05:42.560 --> 00:05:43.630
displacements.
00:05:43.630 --> 00:05:46.910
Now, this is the principle which
we already discussed
00:05:46.910 --> 00:05:49.050
very briefly in the
last lecture.
00:05:49.050 --> 00:05:52.830
Remember, please that we can
derive it by looking at the
00:05:52.830 --> 00:05:57.220
total potential of the system,
which is given as the strain
00:05:57.220 --> 00:06:02.170
energy, minus the potential of
the external loads, W, U being
00:06:02.170 --> 00:06:03.560
the strain energy.
00:06:03.560 --> 00:06:09.560
If we invoke the stationarity
of pi, and we use the
00:06:09.560 --> 00:06:11.830
essential boundary conditions,
which are the displacement
00:06:11.830 --> 00:06:16.340
boundary conditions, then we
can derive the governing
00:06:16.340 --> 00:06:18.870
differential equations of
equilibrium, and the force
00:06:18.870 --> 00:06:21.270
boundary conditions, the natural
boundary conditions,
00:06:21.270 --> 00:06:23.520
as I have shown in
the last lecture.
00:06:23.520 --> 00:06:27.570
Well, we will not derive these
boundary conditions and the
00:06:27.570 --> 00:06:30.260
governing differential equations
in this approach,
00:06:30.260 --> 00:06:33.770
but rather what we do is we
invoke this principle, we set
00:06:33.770 --> 00:06:37.860
del pi equal to 0, and that
gives us the principle of
00:06:37.860 --> 00:06:39.530
virtual displacements.
00:06:39.530 --> 00:06:43.040
And it is this principle which
is a starting point all our
00:06:43.040 --> 00:06:45.190
finite element analysis.
00:06:45.190 --> 00:06:48.150
Let's recall once what
does it mean.
00:06:48.150 --> 00:06:52.680
Well, here we have the body
forces that I applied to the
00:06:52.680 --> 00:06:56.360
body, the surface forces that
I applied to the body.
00:06:56.360 --> 00:07:00.180
These are externally applied
loads and concentrated forces
00:07:00.180 --> 00:07:06.120
that are also applied to the
body at the points, i.
00:07:06.120 --> 00:07:11.110
These forces are in
equilibrium with
00:07:11.110 --> 00:07:13.690
the stresses, tau.
00:07:13.690 --> 00:07:16.930
Let's assume that we know the
stresses, at this point.
00:07:16.930 --> 00:07:19.310
Then the principle states
the following--
00:07:19.310 --> 00:07:24.480
if we subject the body to
any arbitrary virtual
00:07:24.480 --> 00:07:27.680
displacements, listed
in here--
00:07:27.680 --> 00:07:29.965
and I'm saying any arbitrary
virtual displacements--
00:07:35.570 --> 00:07:39.190
excuse me, that however satisfy
the essential boundary
00:07:39.190 --> 00:07:41.710
conditions, and that means just
the displacement boundary
00:07:41.710 --> 00:07:43.910
conditions.
00:07:43.910 --> 00:07:49.540
Then the work done by the loads,
and that total work is
00:07:49.540 --> 00:07:50.260
given here.
00:07:50.260 --> 00:07:53.000
This is a virtual work because
we are taking virtual
00:07:53.000 --> 00:07:58.350
displacements and subject the
forces to these virtual
00:07:58.350 --> 00:07:59.560
displacements.
00:07:59.560 --> 00:08:04.840
Then the external virtual work
done is equal to the internal
00:08:04.840 --> 00:08:11.150
virtual work done, which is
obtained by multiplying the
00:08:11.150 --> 00:08:18.400
real stresses, which are in
equilibrium with these
00:08:18.400 --> 00:08:22.660
extraordinary applied forces.
00:08:22.660 --> 00:08:27.190
Multiplying the real stresses by
the virtual strains, which
00:08:27.190 --> 00:08:31.220
correspond to the virtual
displacements.
00:08:31.220 --> 00:08:33.500
So let me use here a
different color.
00:08:33.500 --> 00:08:39.159
These virtual strains correspond
to these virtual
00:08:39.159 --> 00:08:42.780
displacements, and of course,
these virtual displacements
00:08:42.780 --> 00:08:44.650
over the body--
00:08:44.650 --> 00:08:48.600
these are, of course, a function
of x, y and z.
00:08:48.600 --> 00:08:51.380
These virtual displacements
over the body give us also
00:08:51.380 --> 00:08:54.590
virtual displacements on the
surface of the body, which are
00:08:54.590 --> 00:08:55.910
listed in here.
00:08:55.910 --> 00:08:58.900
So let us put another
arrow in there.
00:08:58.900 --> 00:09:04.000
And these virtual displacement
also give us concentrated
00:09:04.000 --> 00:09:06.980
virtual displacements at those
points where we have
00:09:06.980 --> 00:09:09.760
concentrated load supply.
00:09:09.760 --> 00:09:14.580
So once again, if we take the
body and subject that body,
00:09:14.580 --> 00:09:20.820
who is in equilibrium under
Fb, Fs, and Fi, with tau--
00:09:20.820 --> 00:09:22.590
tau being the real stresses.
00:09:22.590 --> 00:09:26.440
If you take that body and
subject it to any arbitrary
00:09:26.440 --> 00:09:29.460
virtual displacement that
satisfy the displacement
00:09:29.460 --> 00:09:34.910
boundary conditions, then the
external virtual work is equal
00:09:34.910 --> 00:09:37.750
to the internal virtual work.
00:09:37.750 --> 00:09:40.140
The internal virtual work being
obtained by taking the
00:09:40.140 --> 00:09:43.830
real stresses, times the
virtual strains, which
00:09:43.830 --> 00:09:47.200
correspond to the virtual
displacements here, and
00:09:47.200 --> 00:09:51.120
integrating that product over
the volume of the body.
00:09:51.120 --> 00:09:54.800
And the external pressure work
is obtained by taking the real
00:09:54.800 --> 00:10:00.100
forces, multiplying these by the
virtual displacements, and
00:10:00.100 --> 00:10:02.380
integrating these contributions
over the
00:10:02.380 --> 00:10:04.770
complete body.
00:10:04.770 --> 00:10:06.720
Physically, what
does this mean?
00:10:06.720 --> 00:10:11.180
Here, we have again,
our general body.
00:10:11.180 --> 00:10:14.440
Let's see once, pictorially,
what we're doing.
00:10:14.440 --> 00:10:18.630
Well, let's take a certain
virtual displacement, which I
00:10:18.630 --> 00:10:20.210
depict here.
00:10:20.210 --> 00:10:23.650
Now, here, we have a boundary
condition, so this point, P
00:10:23.650 --> 00:10:27.010
can only move over to there.
00:10:27.010 --> 00:10:30.080
It could not move this way
because we have to satisfy, in
00:10:30.080 --> 00:10:34.150
the virtual displacements, the
actual displacement boundary
00:10:34.150 --> 00:10:35.480
conditions.
00:10:35.480 --> 00:10:39.160
Here, the point cannot move at
all, and here, this point can
00:10:39.160 --> 00:10:40.050
also not move.
00:10:40.050 --> 00:10:43.500
So a typical set of virtual
displacement
00:10:43.500 --> 00:10:44.940
might look like that.
00:10:49.030 --> 00:10:51.960
Just sketched in here.
00:10:51.960 --> 00:10:55.230
This roller out has
moved over there.
00:10:55.230 --> 00:11:00.040
So there's our new roller
right there.
00:11:00.040 --> 00:11:04.300
What I satisfy are the order
displacement conditions.
00:11:04.300 --> 00:11:06.600
Only horizontal movement
was possible here.
00:11:06.600 --> 00:11:09.260
No movement here, no
movement here.
00:11:09.260 --> 00:11:14.300
The virtual displacement vector
here is that one. u
00:11:14.300 --> 00:11:16.740
bar, for a particular point.
00:11:16.740 --> 00:11:19.170
And that is the point
that I'm looking at.
00:11:19.170 --> 00:11:22.230
Then what the principle says,
once again, is that if I take
00:11:22.230 --> 00:11:25.750
these virtual displacements,
multiply them by the real
00:11:25.750 --> 00:11:29.870
forces, integrate that product
over the total body--
00:11:29.870 --> 00:11:32.530
that is my external virtual
work, and that external
00:11:32.530 --> 00:11:35.460
virtual work shall be equal to
the internal virtual work,
00:11:35.460 --> 00:11:39.120
which is obtained by taking the
real stresses, which are
00:11:39.120 --> 00:11:42.330
in equilibrium with these
externally applied loads.
00:11:42.330 --> 00:11:45.230
And multiplying these real
stresses by the virtual
00:11:45.230 --> 00:11:48.860
strains that correspond to these
virtual displacements,
00:11:48.860 --> 00:11:52.640
and integrating that product,
as the internal virtual work
00:11:52.640 --> 00:11:54.260
over the whole body.
00:11:54.260 --> 00:11:57.680
This is an extremely powerful
principle and an extremely
00:11:57.680 --> 00:12:01.090
important principle, and
provides a basis of our finite
00:12:01.090 --> 00:12:02.610
element formulation.
00:12:02.610 --> 00:12:07.560
In our finite element analysis,
we are proceeding in
00:12:07.560 --> 00:12:11.320
the following way-- we say,
well, let us idealize this
00:12:11.320 --> 00:12:15.130
complete body as an assemblage
of elements, and what I've
00:12:15.130 --> 00:12:19.120
done here is to draw one
typical element.
00:12:19.120 --> 00:12:23.130
This is an 8-node element, a
brick element, a distorted
00:12:23.130 --> 00:12:27.590
brick element, to make it a
little bit more general.
00:12:27.590 --> 00:12:32.460
It is an 8-node element because
we have four nodes on
00:12:32.460 --> 00:12:35.700
the top surface, and four nodes
at the bottom surface.
00:12:35.700 --> 00:12:38.870
There's another node here.
00:12:38.870 --> 00:12:43.050
This element here undergoes
certain
00:12:43.050 --> 00:12:44.720
displacements, of course.
00:12:44.720 --> 00:12:48.100
And what I will be doing is I
will express the displacements
00:12:48.100 --> 00:12:51.480
in that element as a function
of the letter coordinates
00:12:51.480 --> 00:12:54.050
system, x, y, and z.
00:12:54.050 --> 00:12:55.760
The displacement in
the element being
00:12:55.760 --> 00:12:59.140
lower u, v, and w.
00:12:59.140 --> 00:13:01.250
If we idealize the
total body as an
00:13:01.250 --> 00:13:03.250
assemblage of such elements--
00:13:03.250 --> 00:13:05.820
in other words, there's another
element coming in from
00:13:05.820 --> 00:13:09.040
the top, and another element
coming in from the sides, from
00:13:09.040 --> 00:13:10.650
the four sides, and
another element
00:13:10.650 --> 00:13:12.930
coming in from the bottom.
00:13:12.930 --> 00:13:16.760
So if we idealize the total body
as an assemblage of such
00:13:16.760 --> 00:13:21.830
brick elements that lie next
to each other, et cetera.
00:13:21.830 --> 00:13:25.570
And we express the displacement
in each of these
00:13:25.570 --> 00:13:29.170
brick elements, as a function
of the nodal point
00:13:29.170 --> 00:13:32.120
displacements, of the
displacements of the corners
00:13:32.120 --> 00:13:36.620
of the bricks, then we can, of
course, express the total
00:13:36.620 --> 00:13:40.350
displacement in the body as a
function of the nodal point
00:13:40.350 --> 00:13:41.500
displacement.
00:13:41.500 --> 00:13:43.480
And that is the important
step in the
00:13:43.480 --> 00:13:44.820
finite element analysis.
00:13:44.820 --> 00:13:50.100
That the displacements in each
of these sub-domains or
00:13:50.100 --> 00:13:53.460
elements are expressed in
terms of nodal point
00:13:53.460 --> 00:13:54.930
displacements.
00:13:54.930 --> 00:13:59.690
These corner nodes, as shown
here, for the brick element.
00:13:59.690 --> 00:14:05.320
And then since the total body is
made up of an assemblage of
00:14:05.320 --> 00:14:09.480
such brick elements, we can
express the total displacement
00:14:09.480 --> 00:14:12.600
in the body as a functional
of these nodal point
00:14:12.600 --> 00:14:14.490
displacements.
00:14:14.490 --> 00:14:17.310
And invokes the principle of
virtual displacements.
00:14:17.310 --> 00:14:20.350
Now let's go into the
actual specifics.
00:14:20.350 --> 00:14:25.070
Well, for element m, this might
be element 10, m in that
00:14:25.070 --> 00:14:26.870
case would be equal to 10.
00:14:26.870 --> 00:14:29.210
Then we have the following
relationship--
00:14:29.210 --> 00:14:32.140
and this is the important
assumption of the finite
00:14:32.140 --> 00:14:33.350
element discretization.
00:14:33.350 --> 00:14:35.725
We say that the displacements--
00:14:35.725 --> 00:14:39.190
there are three displacements,
U, V, and W, of course, now.
00:14:39.190 --> 00:14:44.350
For element m, U, V and W are
listed in this vector u, are
00:14:44.350 --> 00:14:48.000
equal to a displacement
interpolation matrix, Hm,
00:14:48.000 --> 00:14:52.760
which is a function of x, y, and
z, times the nodal point
00:14:52.760 --> 00:14:53.910
displacements.
00:14:53.910 --> 00:14:57.980
And what I'm listing in this u
hat vector are all the nodal
00:14:57.980 --> 00:15:00.880
point displacements that I've
called in the finite element
00:15:00.880 --> 00:15:02.310
discretization.
00:15:02.310 --> 00:15:09.120
For this brick element here,
we have eight nodes, and 24
00:15:09.120 --> 00:15:10.440
nodal point displacements.
00:15:10.440 --> 00:15:16.260
At each node, we have U, V and
W. But notice, once again,
00:15:16.260 --> 00:15:20.260
there are other brick elements
on top of it, on the sides,
00:15:20.260 --> 00:15:21.890
and below of it.
00:15:21.890 --> 00:15:25.190
And each of these brick
elements, of course, has a set
00:15:25.190 --> 00:15:28.810
of such nodal point
displacements.
00:15:28.810 --> 00:15:31.870
We notice, however, that the
element below it here--
00:15:31.870 --> 00:15:35.630
if I take my pen here, and draw
in another element, we
00:15:35.630 --> 00:15:40.670
notice that that element
has the same
00:15:40.670 --> 00:15:42.350
node as the top element.
00:15:42.350 --> 00:15:45.310
In other words, this node here
is common to this top element
00:15:45.310 --> 00:15:46.580
and the bottom element.
00:15:46.580 --> 00:15:50.400
And that's where we have the
coupling between elements.
00:15:50.400 --> 00:15:53.540
We will see that more
distinctly later.
00:15:53.540 --> 00:15:57.070
So what I'm doing here is I
express the displacements of
00:15:57.070 --> 00:16:01.900
element m as a function of all
the nodal point displacements,
00:16:01.900 --> 00:16:06.420
and I'm listing here in u hat
these displacements for N,
00:16:06.420 --> 00:16:07.760
capital N nodal points.
00:16:07.760 --> 00:16:09.810
We have this vector.
00:16:09.810 --> 00:16:13.500
Now, in general, later on, we
would simply call all of these
00:16:13.500 --> 00:16:17.370
components, Ui's, and so our
u hat here, would be
00:16:17.370 --> 00:16:18.630
written in this way.
00:16:18.630 --> 00:16:22.122
Notice I use the transpose, the
capital T here, to denote
00:16:22.122 --> 00:16:23.510
the transpose of a vector.
00:16:23.510 --> 00:16:31.170
So this UN is equal to that W,
capital N. That's just for
00:16:31.170 --> 00:16:33.240
ease of notation.
00:16:33.240 --> 00:16:35.670
This is our major assumption.
00:16:35.670 --> 00:16:37.560
This is the major assumption
in the
00:16:37.560 --> 00:16:38.910
finite element analysis.
00:16:38.910 --> 00:16:43.370
We will have to define for each
element this displacement
00:16:43.370 --> 00:16:45.620
interpolation matrix.
00:16:45.620 --> 00:16:52.350
We notice that when we define
it, there will be many columns
00:16:52.350 --> 00:16:57.870
that are simply 0's because only
certain displacements in
00:16:57.870 --> 00:17:01.500
this vector listed here, in this
vector, really affect the
00:17:01.500 --> 00:17:02.930
displacement in an element.
00:17:02.930 --> 00:17:08.940
In other words, typically, for
this element here, if we look
00:17:08.940 --> 00:17:13.560
at this node, then the
displacement at this node do
00:17:13.560 --> 00:17:17.160
not affect the displacement in
this element because this node
00:17:17.160 --> 00:17:21.140
does not belong to
the element.
00:17:21.140 --> 00:17:25.089
That is recognized by the fact
that in this Hm matrix there
00:17:25.089 --> 00:17:28.720
will be many columns that
are simply 0's.
00:17:28.720 --> 00:17:32.710
In fact, the only non-zero
columns in this Hm matrix are
00:17:32.710 --> 00:17:37.850
those that correspond to nodal
points in this vector, or
00:17:37.850 --> 00:17:40.130
nodal point displacements
in this vector that
00:17:40.130 --> 00:17:43.170
belong to element m.
00:17:43.170 --> 00:17:46.170
Well, having laid down this
assumption, and we will
00:17:46.170 --> 00:17:48.960
define, once again, later
on the Hm matrix
00:17:48.960 --> 00:17:50.560
for specific elements--
00:17:50.560 --> 00:17:53.380
1D, 2D, 3D elements and so on.
00:17:53.380 --> 00:17:56.540
Having laid down this
assumption, we can derive the
00:17:56.540 --> 00:17:59.920
strains and the strains are
simply given by this
00:17:59.920 --> 00:18:04.610
relationship, with the Bm
matrix is the strain
00:18:04.610 --> 00:18:07.670
interpolation matrix,
of element m.
00:18:07.670 --> 00:18:13.150
Notice that the rows in this
matrix are obtained by using
00:18:13.150 --> 00:18:14.950
the rows in the Hm.
00:18:14.950 --> 00:18:19.810
And differentiating these rows
and combining these rows in
00:18:19.810 --> 00:18:22.380
the appropriate way.
00:18:22.380 --> 00:18:24.210
I show examples later on.
00:18:24.210 --> 00:18:27.930
There is no more assumption
in this step.
00:18:27.930 --> 00:18:33.320
This Bm matrix is simply
obtained from the Hm matrix.
00:18:33.320 --> 00:18:37.540
By recognizing what strains we
are talking about, and by
00:18:37.540 --> 00:18:40.840
recognizing that we can simply
use the rows here,
00:18:40.840 --> 00:18:44.770
differentiate them, linearally
combine them, if necessary, to
00:18:44.770 --> 00:18:46.860
obtain the Bm matrix.
00:18:46.860 --> 00:18:50.330
Of course, we also have to use
our stress strain law to
00:18:50.330 --> 00:18:53.430
obtain stresses from the
strains, and the stresses in
00:18:53.430 --> 00:18:55.730
element m are given
as shown here.
00:18:55.730 --> 00:18:58.120
These are the strains that I
talked about here already.
00:18:58.120 --> 00:19:00.660
This is our stress strain
law, which can vary
00:19:00.660 --> 00:19:03.370
from element to element.
00:19:03.370 --> 00:19:07.670
I also introduce here an initial
stress, which might
00:19:07.670 --> 00:19:09.260
already be in the body.
00:19:09.260 --> 00:19:12.590
This might be due to overburden
pressure in an
00:19:12.590 --> 00:19:16.440
underground structure, as
an example, and so on.
00:19:16.440 --> 00:19:20.030
So, this is our stress strain
law, which we have to satisfy
00:19:20.030 --> 00:19:21.520
for the body, of course.
00:19:21.520 --> 00:19:24.570
Our compatibility conditions
in the analysis
00:19:24.570 --> 00:19:26.090
will also be satisfied.
00:19:26.090 --> 00:19:28.680
The strain compatibility
conditions are satisfied
00:19:28.680 --> 00:19:34.210
because we are deriving the
strains from continuous
00:19:34.210 --> 00:19:38.190
displacements, within
the element.
00:19:41.880 --> 00:19:46.930
We'll impose on to the different
elements that they
00:19:46.930 --> 00:19:49.610
remain compatible under
deformations.
00:19:49.610 --> 00:19:53.610
By that, I mean if we have an
element coming in here and
00:19:53.610 --> 00:19:57.860
another element going out
there that the elements
00:19:57.860 --> 00:20:01.400
underloading, when the top
element here, and the bottom
00:20:01.400 --> 00:20:03.080
element, both of them
are loaded.
00:20:03.080 --> 00:20:07.290
No gap is opening up here
so that displacement
00:20:07.290 --> 00:20:10.490
compatibility between the
elements is satisfied.
00:20:10.490 --> 00:20:13.320
So if we look at the three
conditions that we have to
00:20:13.320 --> 00:20:16.580
satisfy in an analysis--
00:20:16.580 --> 00:20:18.910
the first one being the
stress strain law.
00:20:18.910 --> 00:20:21.410
That is satisfied because we
are using this equation.
00:20:21.410 --> 00:20:24.650
The second one being
compatibility.
00:20:24.650 --> 00:20:29.590
That is satisfied because we
using this relationship here
00:20:29.590 --> 00:20:37.010
to calculate our strains from
the Hm, via the Bm matrix.
00:20:37.010 --> 00:20:40.410
And we are satisfying, of
course, that the elements
00:20:40.410 --> 00:20:43.680
remain together, so no
gaps opening up.
00:20:43.680 --> 00:20:45.830
We will be talking about it
later on when we talk about
00:20:45.830 --> 00:20:47.770
the convergence requirements
also of
00:20:47.770 --> 00:20:49.360
finite element analysis.
00:20:49.360 --> 00:20:52.300
And finally, our equilibrium
condition has to be satisfied.
00:20:52.300 --> 00:20:54.990
That is the third condition
where that equilibrium
00:20:54.990 --> 00:20:59.240
condition is embodied in the
principle of virtual work.
00:20:59.240 --> 00:21:01.910
Here, I have it once
again written down.
00:21:01.910 --> 00:21:04.040
The equilibrium conditions
are in this
00:21:04.040 --> 00:21:06.500
principle of virtual work.
00:21:06.500 --> 00:21:11.280
And as I stated earlier, that if
this equation is satisfied
00:21:11.280 --> 00:21:15.440
for any and all the arbitrary
virtual displacements that
00:21:15.440 --> 00:21:18.560
satisfy the displacement
boundary conditions, the real
00:21:18.560 --> 00:21:23.050
displacement boundary
conditions, then tau is in
00:21:23.050 --> 00:21:25.910
equilibrium with
Fb, Fs, and Fi.
00:21:25.910 --> 00:21:30.200
Well, what we will be doing is
we will be applying this
00:21:30.200 --> 00:21:32.900
principle of virtual
displacements for our finite
00:21:32.900 --> 00:21:36.853
element discretization, which
means that in an integral
00:21:36.853 --> 00:21:40.710
sense, we satisfy equilibrium.
00:21:40.710 --> 00:21:45.250
However, if we look into an
element, then within the
00:21:45.250 --> 00:21:48.440
element, we will only satisfy
the differential equations of
00:21:48.440 --> 00:21:50.670
equilibrium in an
approximate way.
00:21:50.670 --> 00:21:52.160
We will not satisfy
them exactly.
00:21:52.160 --> 00:21:58.680
However, if we have a proper
finite element discretization,
00:21:58.680 --> 00:22:01.180
and by that, I mean if we
satisfy all the convergence
00:22:01.180 --> 00:22:07.150
requirements that have to be
satisfied in order to obtain a
00:22:07.150 --> 00:22:10.300
valid solution, or a reliable
solution in a finite element
00:22:10.300 --> 00:22:14.420
analysis, then we know that as
our elements become smaller
00:22:14.420 --> 00:22:18.040
and smaller and smaller,
we will be finally--
00:22:18.040 --> 00:22:19.780
and always, of course, applying
the principle of
00:22:19.780 --> 00:22:21.050
virtual displacements--
00:22:21.050 --> 00:22:23.980
we will be finally satisfying,
also, the differential
00:22:23.980 --> 00:22:27.830
equations of equilibrium locally
within each element.
00:22:27.830 --> 00:22:31.510
So stress strain law is
satisfied, compatibility is
00:22:31.510 --> 00:22:33.960
satisfied, both of
them exactly.
00:22:33.960 --> 00:22:36.735
The equilibrium requirements
are only satisfied in an
00:22:36.735 --> 00:22:40.230
integral sense, if we have a
coarse finite element mesh,
00:22:40.230 --> 00:22:43.750
but as the finite elements
become more and more, as we
00:22:43.750 --> 00:22:46.870
refine our finite element mesh,
we will be satisfying
00:22:46.870 --> 00:22:48.660
the equilibrium requirements.
00:22:48.660 --> 00:22:53.460
Also locally, within an element,
always closer and
00:22:53.460 --> 00:22:56.790
closer, and we will be
approximating, or we will be
00:22:56.790 --> 00:22:59.260
getting closer to the
satisfaction of the
00:22:59.260 --> 00:23:01.060
differential equation
of equilibrium.
00:23:03.870 --> 00:23:07.080
The first step now is to rewrite
this principle of
00:23:07.080 --> 00:23:11.640
virtual displacement, in this
form, namely as a sum of
00:23:11.640 --> 00:23:14.010
integrations over
the elements.
00:23:14.010 --> 00:23:15.030
There's no assumption yet.
00:23:15.030 --> 00:23:18.510
All I've done is since our total
body is idealized as a
00:23:18.510 --> 00:23:24.150
sum of volumes, namely the
volumes over the elements, I
00:23:24.150 --> 00:23:27.960
can rewrite the total integral
as a the sum
00:23:27.960 --> 00:23:30.030
over the element integrals.
00:23:30.030 --> 00:23:31.610
And that's what I
have done here.
00:23:31.610 --> 00:23:35.380
Notice we have now here, m,
denoting element m and we are
00:23:35.380 --> 00:23:37.110
summing over all of
the elements.
00:23:37.110 --> 00:23:39.260
There's no assumption
here yet.
00:23:39.260 --> 00:23:43.060
Now, however, I can substitute
our assumption.
00:23:43.060 --> 00:23:47.050
Namely, that Um is given in
this way, and epsilon m is
00:23:47.050 --> 00:23:47.850
given that way.
00:23:47.850 --> 00:23:49.750
Once again, this is actually
not an assumption.
00:23:49.750 --> 00:23:51.900
This is the major assumption.
00:23:51.900 --> 00:23:55.710
That epsilon m follows
from this assumption.
00:23:55.710 --> 00:23:59.640
This equation follows from
that equation entirely.
00:23:59.640 --> 00:24:03.550
Substituting now from here,
these equations into the
00:24:03.550 --> 00:24:07.710
principle of virtual
displacements, we directly
00:24:07.710 --> 00:24:10.420
obtain the following
equations.
00:24:10.420 --> 00:24:14.660
Here, I have on the
left hand side--
00:24:14.660 --> 00:24:17.660
let's go through this
equation in detail.
00:24:17.660 --> 00:24:22.560
On the left hand side, I have
the following part.
00:24:22.560 --> 00:24:25.980
I have an integral over all of
the elements, that is the
00:24:25.980 --> 00:24:30.090
integral here, summing over
element m and integrating over
00:24:30.090 --> 00:24:32.980
each of the element.
00:24:32.980 --> 00:24:39.310
This part here, Bm transposed,
times U hat T, is equal to the
00:24:39.310 --> 00:24:42.100
epsilon bar, mT.
00:24:42.100 --> 00:24:47.930
So notice that our epsilon bar
m, is a virtual strain, is
00:24:47.930 --> 00:24:49.220
given in this way.
00:24:52.200 --> 00:24:55.980
Before, I talked only about
the real strains.
00:24:55.980 --> 00:24:57.920
In other words, the
bar was not there.
00:24:57.920 --> 00:24:59.620
I put that now into
brackets here.
00:24:59.620 --> 00:25:01.840
This bar was not there.
00:25:01.840 --> 00:25:04.500
Well, what we're doing in the
finite element analysis is to
00:25:04.500 --> 00:25:08.220
use the same assumption for the
virtual strains as we use
00:25:08.220 --> 00:25:09.650
for real strains.
00:25:09.650 --> 00:25:12.790
In other words, we use this
equation without the bar and
00:25:12.790 --> 00:25:15.740
with the bar, on top of
the epsilon and on
00:25:15.740 --> 00:25:18.850
top of the u hat.
00:25:18.850 --> 00:25:22.210
So here, we have our epsilon
bar transposed.
00:25:22.210 --> 00:25:25.360
This part here is the stress.
00:25:25.360 --> 00:25:28.380
Tau m equals Cm epsilon m--
00:25:28.380 --> 00:25:29.330
that is our cm.
00:25:29.330 --> 00:25:33.330
The Bm here comes in
from the epsilon m.
00:25:33.330 --> 00:25:37.400
So Bm times U hat is equal to
epsilon m, once written down
00:25:37.400 --> 00:25:38.700
again here.
00:25:38.700 --> 00:25:43.190
So this total part here
is nothing else than--
00:25:43.190 --> 00:25:47.250
let's go back once more to
the previous view graph--
00:25:47.250 --> 00:25:52.530
nothing else than this part
here, than that part there.
00:25:52.530 --> 00:25:57.240
But with the initial stress,
tau i, m being on
00:25:57.240 --> 00:25:58.000
the right hand side.
00:25:58.000 --> 00:26:01.180
Since we do know the initial
stress, we put that one, of
00:26:01.180 --> 00:26:02.580
course, on the right
hand side.
00:26:02.580 --> 00:26:04.080
It is a load contribution.
00:26:04.080 --> 00:26:08.600
And here, you see it as a minus
sign because we put on
00:26:08.600 --> 00:26:12.360
the right hand side,
and this part here.
00:26:12.360 --> 00:26:15.150
This part times this u
hat, notice there's
00:26:15.150 --> 00:26:16.400
a big bracket here.
00:26:19.550 --> 00:26:26.120
That u hat bar here operating on
that Bm transposed, there's
00:26:26.120 --> 00:26:29.220
a transposed here, too, gives
us again the epsilon bar m
00:26:29.220 --> 00:26:31.310
transposed.
00:26:31.310 --> 00:26:34.700
So let us look now at what we
have on the right hand side.
00:26:34.700 --> 00:26:36.840
On the right hand side,
I want to have
00:26:36.840 --> 00:26:40.540
discretized this part here.
00:26:40.540 --> 00:26:43.910
Well, what we do is we
substitute here from our
00:26:43.910 --> 00:26:46.930
displacement interpolation,
here from our displacement
00:26:46.930 --> 00:26:51.230
interpolation, and each of these
integrals can directly
00:26:51.230 --> 00:26:54.380
be expressed, in terms of the
nodal point displacements.
00:26:54.380 --> 00:26:55.915
And that's what we
have done here.
00:26:55.915 --> 00:27:02.400
The Hm times the u hat bar is
the u bar m transposed.
00:27:02.400 --> 00:27:07.900
Here, we have the u bar s.
00:27:07.900 --> 00:27:14.110
That is the u hat bar
times the Hsm.
00:27:14.110 --> 00:27:16.790
Notice that this part, you
should not have been
00:27:16.790 --> 00:27:17.990
written that far.
00:27:17.990 --> 00:27:19.660
In other words, there
is the end.
00:27:19.660 --> 00:27:25.350
The u bar s m transposed only
goes up to there and it
00:27:25.350 --> 00:27:31.610
embodies this Hs m transposed
and the u hat bar transposed.
00:27:31.610 --> 00:27:33.830
This part we talked
about already.
00:27:33.830 --> 00:27:36.850
So what we have done then
is to rewrite--
00:27:36.850 --> 00:27:38.460
this is the important part--
00:27:38.460 --> 00:27:43.360
is to rewrite this principle of
virtual displacements, in
00:27:43.360 --> 00:27:46.670
which we had no assumption
yet.
00:27:46.670 --> 00:27:50.520
We rewrote this in terms of the
nodal point displacements
00:27:50.520 --> 00:27:55.260
and element interpolation
matrices that we use for our
00:27:55.260 --> 00:27:57.330
finite element discretization.
00:27:57.330 --> 00:27:59.800
Now, we of course, have the
assumption that the
00:27:59.800 --> 00:28:04.090
displacements within each
element are given by the Hm
00:28:04.090 --> 00:28:08.230
matrix, the strains are given
by the Bm matrix.
00:28:08.230 --> 00:28:12.120
So this is the result that
we have obtained.
00:28:12.120 --> 00:28:16.150
And at this point, we now
invoke the principle of
00:28:16.150 --> 00:28:17.560
virtual displacements.
00:28:17.560 --> 00:28:23.590
Since this principle here shall
hold for any arbitrary
00:28:23.590 --> 00:28:28.540
virtual displacements that
satisfy the displacement
00:28:28.540 --> 00:28:31.640
boundary conditions,
we can now invoke
00:28:31.640 --> 00:28:35.700
this principle n times.
00:28:35.700 --> 00:28:41.600
And by that, I mean, once by
imposing a unit displacement
00:28:41.600 --> 00:28:44.330
at the first displacement degree
of freedom, and leaving
00:28:44.330 --> 00:28:46.440
all the others 0.
00:28:46.440 --> 00:28:49.810
Second time around, imposing
a unit displacement at the
00:28:49.810 --> 00:28:53.160
second degree of freedom,
all the others being 0.
00:28:53.160 --> 00:28:56.830
Third time around, imposing a
unit displacement at the third
00:28:56.830 --> 00:28:59.670
degree of freedom, all the
other displacements
00:28:59.670 --> 00:29:02.070
being 0, and so on.
00:29:02.070 --> 00:29:06.040
And that really amounts to then
saying that this vector
00:29:06.040 --> 00:29:10.290
here becomes an identity
matrix.
00:29:10.290 --> 00:29:14.420
And similarly, this one here
becomes an identity matrix.
00:29:14.420 --> 00:29:20.430
And therefore, we can take
those two out, and our
00:29:20.430 --> 00:29:26.410
resulting equation then is
simply what we have left.
00:29:26.410 --> 00:29:29.670
Taking these two identity
matrices out, and that is our
00:29:29.670 --> 00:29:32.610
finite element equilibrium
equation, or I should rather
00:29:32.610 --> 00:29:36.370
say that these are n finite
element equilibrium equations,
00:29:36.370 --> 00:29:40.110
namely corresponding to the n
nodal point displacements that
00:29:40.110 --> 00:29:41.950
we are considering.
00:29:41.950 --> 00:29:46.810
In general, what one does most
effectively is to really
00:29:46.810 --> 00:29:51.110
derive these corresponding
to all displacements.
00:29:51.110 --> 00:29:57.290
Having removed the boundary
conditions, and then later on,
00:29:57.290 --> 00:30:02.190
one imposes the known bounded
displacements.
00:30:02.190 --> 00:30:06.440
And that's what I want to
discuss a little bit later in
00:30:06.440 --> 00:30:07.290
more detail.
00:30:07.290 --> 00:30:11.020
So this is the result then
that we obtained.
00:30:11.020 --> 00:30:15.030
And we obtained really in
shorthand, Ku equals r.
00:30:15.030 --> 00:30:18.770
Where K is this matrix.
00:30:18.770 --> 00:30:20.890
It's the structural
stiffness matrix.
00:30:20.890 --> 00:30:26.820
Notice that I'm summing here
over the elements.
00:30:26.820 --> 00:30:30.580
This being here, the element
stiffness matrix.
00:30:30.580 --> 00:30:35.350
Notice that this element
stiffness matrix here is an n
00:30:35.350 --> 00:30:43.450
by n matrix, has the same
order as this K matrix.
00:30:43.450 --> 00:30:47.530
However, we will recognize that
a large number of columns
00:30:47.530 --> 00:30:50.240
and rows in this matrix
are simply 0.
00:30:50.240 --> 00:30:57.030
In fact, all those columns and
rows are filled with 0's that
00:30:57.030 --> 00:31:00.880
do not correspond to a nodal
point displacement degree of
00:31:00.880 --> 00:31:04.640
freedom of element m.
00:31:04.640 --> 00:31:07.570
I show you later on
some examples.
00:31:07.570 --> 00:31:14.740
However, by using this Bm here,
and making the element
00:31:14.740 --> 00:31:18.260
stiffness matrix of the same
order as this total structural
00:31:18.260 --> 00:31:23.100
stiffness matrix, we can
directly sum over all of the
00:31:23.100 --> 00:31:24.800
elements stiffness matrices.
00:31:24.800 --> 00:31:28.130
And that, of course, is our
direct stiffness procedure,
00:31:28.130 --> 00:31:33.280
which already I pointed
out to you earlier.
00:31:33.280 --> 00:31:36.950
The direct stiffness procedure
means that we are adding the
00:31:36.950 --> 00:31:43.040
element stiffness matrices
into the total stiffness
00:31:43.040 --> 00:31:46.620
matrix via this summation
here.
00:31:46.620 --> 00:31:50.310
So the Km here, being what
I have here in the blue
00:31:50.310 --> 00:31:54.180
brackets, must be, of course,
of the same order as K in
00:31:54.180 --> 00:31:56.420
order to be able to
do that method
00:31:56.420 --> 00:31:57.730
theoretically, at least.
00:31:57.730 --> 00:32:01.190
Later on, we will see that we
indeed only work with the
00:32:01.190 --> 00:32:06.120
non-zero rows and columns in
the K matrix, and then use
00:32:06.120 --> 00:32:09.620
connectivity arrays to assemble
Km effectively into
00:32:09.620 --> 00:32:11.590
the actual K matrix.
00:32:11.590 --> 00:32:14.220
For the RB vector, we
have this part.
00:32:14.220 --> 00:32:17.490
Once again, we're using the
direct stiffness procedure to
00:32:17.490 --> 00:32:21.460
add the contributions of all
the elements in order to
00:32:21.460 --> 00:32:24.130
obtain the total RB vector.
00:32:24.130 --> 00:32:31.250
Once again, the rows now, or
rather, the elements because
00:32:31.250 --> 00:32:36.190
this of course, is a vector
here of n long now.
00:32:36.190 --> 00:32:39.580
Those elements that do not
correspond to nodal point
00:32:39.580 --> 00:32:41.645
degrees of freedom
will all be 0's.
00:32:44.660 --> 00:32:49.630
The RS vector, similarly, is
obtained as shown here.
00:32:49.630 --> 00:32:53.130
Now, we of course, sum the
element contributions as they
00:32:53.130 --> 00:32:58.390
arise from the surface forces
and the RI vector is obtained
00:32:58.390 --> 00:33:00.850
as shown here, and the
concentrated load vector is
00:33:00.850 --> 00:33:07.870
simply a vector listing all the
concentrated forces in F.
00:33:07.870 --> 00:33:14.580
Notice that this HSM matrix here
is directly obtained from
00:33:14.580 --> 00:33:19.670
this Hm matrix.
00:33:19.670 --> 00:33:22.930
Sometimes one has difficulties
visualizing what this
00:33:22.930 --> 00:33:24.340
matrix really is.
00:33:24.340 --> 00:33:27.980
Well, we will see later on if
this is an element here, and
00:33:27.980 --> 00:33:31.020
our coordinate system, say, lies
in that element this way,
00:33:31.020 --> 00:33:34.710
then the Hm matrix, of course,
gives us a displacement within
00:33:34.710 --> 00:33:35.910
the element.
00:33:35.910 --> 00:33:40.550
Whereas the HSM matrix gives us,
say, the displacement on
00:33:40.550 --> 00:33:43.420
this surface element, if
it is that surface
00:33:43.420 --> 00:33:45.430
that we want to consider.
00:33:45.430 --> 00:33:48.630
Now, to get the displacement on
the surface of the element
00:33:48.630 --> 00:33:52.170
when we know the displacement
within the total volume of the
00:33:52.170 --> 00:33:55.400
element, well, what we simply
have to do is we have to
00:33:55.400 --> 00:33:59.690
substitute the coordinates of
the surface in the Hm here to
00:33:59.690 --> 00:34:02.920
obtain the HSM.
00:34:02.920 --> 00:34:05.860
I will show you later
on some examples.
00:34:05.860 --> 00:34:10.210
Now, in dynamic analysis, of
course, the loads are time
00:34:10.210 --> 00:34:15.929
dependent and if we are
considering a truly dynamic
00:34:15.929 --> 00:34:19.520
analysis, then we have to
include inertia forces.
00:34:19.520 --> 00:34:23.659
And the inertia forces can
directly be taken care of, or
00:34:23.659 --> 00:34:27.060
can directly be included in
analysis if we use the
00:34:27.060 --> 00:34:28.800
d'Alembert principle.
00:34:28.800 --> 00:34:34.060
Here, we have the body loads,
which are the externally
00:34:34.060 --> 00:34:37.210
applied forces per
unit volume.
00:34:37.210 --> 00:34:40.690
And if we split these up into
those forces that are
00:34:40.690 --> 00:34:44.679
externally applied, and those
that are arising due to the
00:34:44.679 --> 00:34:47.070
d'Alembert forces,
as shown here.
00:34:47.070 --> 00:34:48.630
Then we directly have
the inertia
00:34:48.630 --> 00:34:50.219
effect in the analysis.
00:34:50.219 --> 00:34:53.739
We now can, of course, express
our accelerations in the
00:34:53.739 --> 00:34:58.120
element in terms of nodal point
accelerations again, and
00:34:58.120 --> 00:35:02.710
we are using here the same Hm
matrix that we use already for
00:35:02.710 --> 00:35:04.540
the displacement
interpolations.
00:35:04.540 --> 00:35:13.110
If we substitute from here and
here into the RB which I had
00:35:13.110 --> 00:35:14.820
written down here.
00:35:14.820 --> 00:35:20.790
If we substitute into this RB
here, this equation for fB,
00:35:20.790 --> 00:35:24.620
then we directly can write down
this equation here, M U
00:35:24.620 --> 00:35:28.780
double dot plus Ku equals R,
where the M matrix now is
00:35:28.780 --> 00:35:30.970
obtained as shown here.
00:35:30.970 --> 00:35:38.780
Notice that this R vector now
only contains this RB part.
00:35:38.780 --> 00:35:44.320
In other words, not anymore
the fB here, but
00:35:44.320 --> 00:35:46.280
rather an fB curl.
00:35:48.870 --> 00:35:52.670
Notice also that in this
analysis now, or in this view
00:35:52.670 --> 00:35:56.130
graph, I've dropped
the hat on the u.
00:35:56.130 --> 00:35:58.920
These are the nodal point
displacements, these are the
00:35:58.920 --> 00:36:00.470
nodal point accelerations.
00:36:00.470 --> 00:36:02.920
I've dropped the hat just
for convenience.
00:36:02.920 --> 00:36:06.430
I had already dropped it
actually here also.
00:36:06.430 --> 00:36:09.940
We earlier had the hat there.
00:36:09.940 --> 00:36:13.830
Here, we had the hat still
because I wanted to
00:36:13.830 --> 00:36:18.700
distinguish the actual nodal
point displacements from the
00:36:18.700 --> 00:36:23.240
continuous displacements in the
structure, or in the body.
00:36:23.240 --> 00:36:27.010
So here, the hat still
being there.
00:36:27.010 --> 00:36:30.630
And here, I dropped the hat
already, just for convenience
00:36:30.630 --> 00:36:35.540
of writing, and from now on,
when we have this vector, U
00:36:35.540 --> 00:36:39.290
here, then that means that
we're talking about the
00:36:39.290 --> 00:36:42.560
concentrated nodal point
displacements, or the actual
00:36:42.560 --> 00:36:45.130
note point displacements of
the finite element mesh.
00:36:48.010 --> 00:36:51.650
I mentioned earlier that it is
most convenient to include in
00:36:51.650 --> 00:36:56.420
the formulation all of the nodal
point displacements,
00:36:56.420 --> 00:37:00.750
including those that actually
might be 0.
00:37:00.750 --> 00:37:03.630
In other words, for our
three-dimensional body, to
00:37:03.630 --> 00:37:05.210
make a quick sketch here.
00:37:05.210 --> 00:37:11.430
If we have here our support in
an actual analysis, it is most
00:37:11.430 --> 00:37:15.140
effective to say well, let us
remove the support and assign
00:37:15.140 --> 00:37:19.260
a node there with three
unknown displacements.
00:37:19.260 --> 00:37:23.540
Once we have derived these
equations of equilibrium, of
00:37:23.540 --> 00:37:25.620
course, we now will have to
impose the fact that the
00:37:25.620 --> 00:37:28.380
displacements are 0's there.
00:37:28.380 --> 00:37:32.180
And that is then done
effectively, for example, as
00:37:32.180 --> 00:37:33.250
shown here.
00:37:33.250 --> 00:37:38.080
We have here the general
equations, M U double dot plus
00:37:38.080 --> 00:37:43.130
Ku equals R. And what we are
doing is we are listing the
00:37:43.130 --> 00:37:48.250
displacements and accelerations
into vectors, U
00:37:48.250 --> 00:37:53.870
double dot A, U A, and U double
dot b, and Ub, where
00:37:53.870 --> 00:37:59.430
the b components of the
displacements and
00:37:59.430 --> 00:38:01.680
accelerations are known.
00:38:01.680 --> 00:38:04.870
And now they might be 0, as I
showed here in this particular
00:38:04.870 --> 00:38:08.510
example, or they might
be actual values
00:38:08.510 --> 00:38:10.480
that we want to impose.
00:38:10.480 --> 00:38:13.580
If they are known, well, we
can look at the first
00:38:13.580 --> 00:38:17.980
equation, as shown here, and put
all the known quantities
00:38:17.980 --> 00:38:20.750
on the right hand side,
substitute for Ub and U double
00:38:20.750 --> 00:38:24.110
dot b, and we know then the
right hand side load vector.
00:38:24.110 --> 00:38:27.280
Thus, we can calculate Ua,
and U double dot a.
00:38:27.280 --> 00:38:30.530
Having now calculated the
velocities, the accelerations,
00:38:30.530 --> 00:38:33.390
and displacements, we can go
back and get the reactions.
00:38:33.390 --> 00:38:37.540
The reactions, of course,
being Rb.
00:38:37.540 --> 00:38:40.600
Here, I assumed that the
displacements which we are
00:38:40.600 --> 00:38:45.050
talking about in the vector
here are actually the ones
00:38:45.050 --> 00:38:47.740
that we also might
want to impose.
00:38:47.740 --> 00:38:50.590
Well, in some cases, of course,
we might have defined
00:38:50.590 --> 00:38:54.220
in our finite element
formulations the U and V
00:38:54.220 --> 00:38:56.830
displacement, as shown here.
00:38:56.830 --> 00:39:01.250
But a displacement that we want
to impose is actually
00:39:01.250 --> 00:39:04.380
this one here, namely, that
one might have to be
00:39:04.380 --> 00:39:07.470
restrained, and this one here
might have to be free.
00:39:07.470 --> 00:39:11.110
In that case, if our finite
element formulation has used
00:39:11.110 --> 00:39:15.230
the U and V displacement, we
have to make a transformation
00:39:15.230 --> 00:39:16.460
as shown here.
00:39:16.460 --> 00:39:21.810
A well known transformation
from the U to the U bar
00:39:21.810 --> 00:39:26.630
displacements, and this, in a
more general sense, is written
00:39:26.630 --> 00:39:28.340
down here once again.
00:39:28.340 --> 00:39:33.350
When we have many more degrees
of freedom, our T matrix would
00:39:33.350 --> 00:39:34.950
look as shown here.
00:39:34.950 --> 00:39:38.830
It's an identity matrix with
the little transformation
00:39:38.830 --> 00:39:40.520
matrix that I've shown here.
00:39:40.520 --> 00:39:44.140
The cosine minus sine
sine cosine matrix.
00:39:44.140 --> 00:39:47.920
Now, put into the appropriate
rows and column.
00:39:47.920 --> 00:39:52.720
The i'th column, j'th column,
i'th row, and j'th row would
00:39:52.720 --> 00:39:55.640
carry these 2x2 matrix.
00:39:55.640 --> 00:39:59.050
And otherwise, we just have
1's on the diagonal.
00:39:59.050 --> 00:40:02.430
So this is the more general
transformation that we are
00:40:02.430 --> 00:40:06.420
using when we have many more
degrees of freedom than just
00:40:06.420 --> 00:40:08.720
the two that we want
to modify.
00:40:08.720 --> 00:40:13.660
Substituting from here into our
equations of equilibrium M
00:40:13.660 --> 00:40:18.650
U double dot plus Ku equals R.
We directly obtained this
00:40:18.650 --> 00:40:23.510
equation, where M bar now is
shown here, K bar is shown
00:40:23.510 --> 00:40:25.710
here, and R bar is shown here.
00:40:25.710 --> 00:40:30.460
Let me mention here that this
looks like a former matrix
00:40:30.460 --> 00:40:31.760
multiplication.
00:40:31.760 --> 00:40:34.870
In fact, two former matrix
multiplications.
00:40:34.870 --> 00:40:38.870
M times T, And then the product
should be taken times
00:40:38.870 --> 00:40:41.880
T transposed, pre-multiplied
by T transpose.
00:40:41.880 --> 00:40:45.970
Well, in actuality, of course,
all we need to do is combine
00:40:45.970 --> 00:40:47.980
rows and columns.
00:40:47.980 --> 00:40:51.230
The i'th and j'th rows and
columns to obtain directly our
00:40:51.230 --> 00:40:53.830
M bar matrix, similarly
for the K bar
00:40:53.830 --> 00:40:57.240
and the R bar matrices.
00:40:57.240 --> 00:41:00.860
Another procedure that is
also used in practice--
00:41:00.860 --> 00:41:01.950
can be very effective--
00:41:01.950 --> 00:41:06.300
is an application of
the penalty method.
00:41:06.300 --> 00:41:12.800
In this procedure, we impose,
basically, physically a spring
00:41:12.800 --> 00:41:18.580
of very large stiffness,
where K is much larger
00:41:18.580 --> 00:41:20.660
than K bar i i.
00:41:20.660 --> 00:41:27.820
And then we supplement our basic
equations that are shown
00:41:27.820 --> 00:41:34.000
here by this equation here.
00:41:34.000 --> 00:41:41.730
So if this K is much larger
than K bar i i, and if we
00:41:41.730 --> 00:41:46.790
supplement this equation or add
this equation into this
00:41:46.790 --> 00:41:52.330
equation here, then we notice
that the spring stiffness will
00:41:52.330 --> 00:41:55.570
wipe out basically the other
stiffnesses that come into
00:41:55.570 --> 00:41:59.160
this degree of freedom, and our
solution will simply be
00:41:59.160 --> 00:42:03.000
that U i is equal to b, which
is the one that we want.
00:42:03.000 --> 00:42:07.850
So this penalty method can be
used to impose displacement
00:42:07.850 --> 00:42:11.390
degrees of freedom, and it
really physically amounts to
00:42:11.390 --> 00:42:14.440
adding a spring into the degree
of freedom where we
00:42:14.440 --> 00:42:17.230
want to impose a certain
displacement.
00:42:17.230 --> 00:42:20.860
And it's important, however,
that if we use that method
00:42:20.860 --> 00:42:25.130
that we are always dealing with
single degree of freedoms
00:42:25.130 --> 00:42:25.685
being imposed.
00:42:25.685 --> 00:42:27.730
And by that, I mean
the following--
00:42:27.730 --> 00:42:35.400
if we had a system, like this
one here, and our original
00:42:35.400 --> 00:42:39.060
displacement degrees of freedom
are these, then we
00:42:39.060 --> 00:42:41.990
first have to make the
transformation onto that
00:42:41.990 --> 00:42:44.010
finite element element
system to these
00:42:44.010 --> 00:42:45.840
degrees of freedom here.
00:42:45.840 --> 00:42:47.750
And now we add our spring in.
00:42:47.750 --> 00:42:52.690
In other words, we want to add
our spring into this system of
00:42:52.690 --> 00:42:58.880
equations because now there's
no coupling from this degree
00:42:58.880 --> 00:43:01.170
of freedom that we want to be
impose into other degrees of
00:43:01.170 --> 00:43:03.180
freedom, through that spring.
00:43:03.180 --> 00:43:06.510
This spring only enters on the
diagonal, and now it is a
00:43:06.510 --> 00:43:09.140
numerically stable process.
00:43:09.140 --> 00:43:14.060
However, if we were not to
perform this transformation--
00:43:14.060 --> 00:43:18.160
in other words, if we were
still to deal with these
00:43:18.160 --> 00:43:22.880
degrees of freedom, and then
add our spring in--
00:43:22.880 --> 00:43:25.370
of course, that spring now
would introduce coupling
00:43:25.370 --> 00:43:27.840
between these two degrees of
freedom, and numerical
00:43:27.840 --> 00:43:32.050
difficulties may arise
in the solution.
00:43:32.050 --> 00:43:36.220
So basically, then in summary,
if we do have degrees of
00:43:36.220 --> 00:43:39.900
freedom to be imposed, we
first go through this
00:43:39.900 --> 00:43:47.870
transformation to obtain the M
bar, U double dot bar, K bar,
00:43:47.870 --> 00:43:51.360
U bar equals R bar system
of equations where the
00:43:51.360 --> 00:43:55.530
displacement that we're talking
about are containing
00:43:55.530 --> 00:43:59.070
those displacements that we
actually want to impose.
00:43:59.070 --> 00:44:02.850
We then can impose these
displacements using the
00:44:02.850 --> 00:44:06.670
penalty method, which
is this one.
00:44:06.670 --> 00:44:13.790
Or we can impose these
displacements using the more
00:44:13.790 --> 00:44:18.050
conventional procedure, using,
in other words, simply this
00:44:18.050 --> 00:44:23.680
procedural of imposing Ub and
rewriting the equations into
00:44:23.680 --> 00:44:24.710
two equations.
00:44:24.710 --> 00:44:28.570
The first equation we solved for
the Ua now, of course, in
00:44:28.570 --> 00:44:32.290
this particular case, we would
now have all bars on there.
00:44:32.290 --> 00:44:35.630
And if we have done a
transformation, and in the
00:44:35.630 --> 00:44:37.830
second equation then, we
obtain the reactions.
00:44:40.380 --> 00:44:44.350
Let me now go through a simple
example to show you the
00:44:44.350 --> 00:44:48.550
application of what
I have discussed.
00:44:48.550 --> 00:44:51.330
This is a very simple
example, but a very
00:44:51.330 --> 00:44:53.310
illustrative example.
00:44:53.310 --> 00:44:56.880
In particular, it is also the
example that we talked about
00:44:56.880 --> 00:45:01.540
already earlier in lecture 2,
when we did a Ritz Analysis on
00:45:01.540 --> 00:45:03.390
this problem.
00:45:03.390 --> 00:45:06.620
In fact, our finite element
analysis that we are now
00:45:06.620 --> 00:45:10.310
pursuing, using the general
equation that I presented to
00:45:10.310 --> 00:45:13.530
you is really nothing else
than a Ritz Analysis.
00:45:13.530 --> 00:45:17.180
And in fact, if you look
at the earlier
00:45:17.180 --> 00:45:18.910
solutions that we obtained--
00:45:18.910 --> 00:45:25.390
solution 2, in the Ritz
analysis, corresponds to the
00:45:25.390 --> 00:45:26.790
finite element solution
that I will be
00:45:26.790 --> 00:45:28.520
discussing with you now.
00:45:28.520 --> 00:45:30.350
So here is a problem
once again.
00:45:30.350 --> 00:45:36.950
We have a bar of unit area, from
here to there, and then
00:45:36.950 --> 00:45:40.670
of changing area, from
here to there.
00:45:40.670 --> 00:45:44.790
The length here is 100,
the length here is 80.
00:45:44.790 --> 00:45:50.560
The bar in actuality, is
supported here, but as I
00:45:50.560 --> 00:45:54.360
mentioned earlier, we remove
that support in our finite
00:45:54.360 --> 00:45:57.880
element formulation, and
introduce, in fact, a
00:45:57.880 --> 00:46:00.030
displacement degree
of freedom there.
00:46:00.030 --> 00:46:03.930
So here, I want to put
down the first node.
00:46:03.930 --> 00:46:09.040
Now, the first step in any
finite element analysis must,
00:46:09.040 --> 00:46:14.350
of course, be the step of
idealizing the total structure
00:46:14.350 --> 00:46:16.060
as an assemblage of elements.
00:46:16.060 --> 00:46:17.710
And there are generally
choices--
00:46:17.710 --> 00:46:20.180
how many elements to take,
what type of element to
00:46:20.180 --> 00:46:21.650
take, and so on.
00:46:21.650 --> 00:46:26.130
In this particular case,
I know that there's a
00:46:26.130 --> 00:46:30.460
discontinuity in area here and
for that reason, intuitively,
00:46:30.460 --> 00:46:35.560
I will put one element
from here to there
00:46:35.560 --> 00:46:37.720
with a constant area.
00:46:37.720 --> 00:46:42.380
Also, let us consider for the
moment, this also as being one
00:46:42.380 --> 00:46:45.240
element, and this then will
correspond to the Ritz
00:46:45.240 --> 00:46:48.350
Analysis that we performed
earlier.
00:46:48.350 --> 00:46:54.400
Notice that we have here a bar
of unit area, a bar of
00:46:54.400 --> 00:46:56.070
changing area.
00:46:56.070 --> 00:47:01.950
This total bar assemblage is
subjected to a load of 100, a
00:47:01.950 --> 00:47:05.690
concentrated load of
100, as shown here.
00:47:05.690 --> 00:47:08.620
The only strains that
this bar can
00:47:08.620 --> 00:47:10.740
develop are normal strains.
00:47:10.740 --> 00:47:16.360
In other words, if a section
originally is here, that
00:47:16.360 --> 00:47:20.980
section we move over a certain
amount and by that amount.
00:47:20.980 --> 00:47:25.620
That is U. In the coordinate
system that we are using, the
00:47:25.620 --> 00:47:28.850
y-coordinate being in this
direction, this is the
00:47:28.850 --> 00:47:30.400
y-coordinate here.
00:47:30.400 --> 00:47:36.270
This would be our U of Y.
However, since we are dealing
00:47:36.270 --> 00:47:40.780
with two elements to analyze
this bar, what I will do is I
00:47:40.780 --> 00:47:44.170
will introduce a little
coordinate system here.
00:47:44.170 --> 00:47:48.470
And I used little y in
this particular case.
00:47:48.470 --> 00:47:50.850
So we put a little y here.
00:47:50.850 --> 00:47:54.120
There's also, for this
element, a little y.
00:47:54.120 --> 00:48:00.720
And the area, in this particular
element, is given
00:48:00.720 --> 00:48:05.150
as 1 plus y divided
by 40 squared.
00:48:05.150 --> 00:48:09.310
So this is the changing
area in this domain.
00:48:09.310 --> 00:48:14.050
However, also, remember please,
if in our analysis, if
00:48:14.050 --> 00:48:18.830
an original section in this area
was vertical like that,
00:48:18.830 --> 00:48:22.570
after deformations, due to the
load here, it will still be
00:48:22.570 --> 00:48:26.140
vertical, and now our
displacements in this element
00:48:26.140 --> 00:48:29.000
will also be given by a u.
00:48:29.000 --> 00:48:31.790
And that u is a function of--
00:48:31.790 --> 00:48:34.730
if we look at this little y,
if you use that little y of
00:48:34.730 --> 00:48:40.780
that y as this u here is
a function of this y.
00:48:40.780 --> 00:48:44.350
This y corresponds to the y
in this element, that y
00:48:44.350 --> 00:48:48.760
corresponds to the y in this
element because we use
00:48:48.760 --> 00:48:51.430
different coordinate systems
for each element.
00:48:51.430 --> 00:48:53.840
Now, this is actually an
important point that we can
00:48:53.840 --> 00:48:56.870
use for each element, a
different coordinate system.
00:48:56.870 --> 00:48:59.870
We could use Cartesian
coordinate systems for each
00:48:59.870 --> 00:49:02.540
element, different ones.
00:49:02.540 --> 00:49:04.840
In fact, that is most
generally done.
00:49:04.840 --> 00:49:10.030
If we have specific geometries,
we might use
00:49:10.030 --> 00:49:12.860
cylindrical coordinates systems
for certain elements,
00:49:12.860 --> 00:49:15.910
Cartesian coordinate systems for
other elements, and so on.
00:49:15.910 --> 00:49:18.450
This is an extremely important
point that we can have
00:49:18.450 --> 00:49:21.580
different coordinate systems for
different elements because
00:49:21.580 --> 00:49:24.350
that eases the calculation
of the
00:49:24.350 --> 00:49:26.890
element stiffness matrices.
00:49:26.890 --> 00:49:32.290
So here, our element 1,
here our element 2.
00:49:32.290 --> 00:49:36.970
And the displacements that we
are talking about are U 1 at
00:49:36.970 --> 00:49:40.840
this node, U 2 at this node,
U 3 at that node.
00:49:40.840 --> 00:49:45.980
These three displacements shall
give us the displacement
00:49:45.980 --> 00:49:46.960
distributions.
00:49:46.960 --> 00:49:50.220
Of course, only an approximate
displacement distribution in
00:49:50.220 --> 00:49:52.490
the complete element mesh.
00:49:52.490 --> 00:49:55.210
Two elements make
up our element--
00:49:55.210 --> 00:49:59.370
complete element idealization
or complete element mesh.
00:49:59.370 --> 00:50:03.410
Well, the equation that, of
course I will now be operating
00:50:03.410 --> 00:50:08.290
on is this one, KU equals R.
In this particular case, we
00:50:08.290 --> 00:50:11.000
recognize that we want to
calculate our stiffness
00:50:11.000 --> 00:50:15.320
matrix, K. Here, we have two
elements, so M, in this
00:50:15.320 --> 00:50:18.960
particular case, will
be equal to 1 and 2.
00:50:18.960 --> 00:50:22.590
We don't have a body force
vector, we don't have a
00:50:22.590 --> 00:50:26.570
surface force vector, we don't
have an initial stress vector.
00:50:26.570 --> 00:50:29.560
However, we have a concentrated
load vector.
00:50:29.560 --> 00:50:33.370
So what I will want to do then
is calculate our K matrix, and
00:50:33.370 --> 00:50:37.110
establish our concentrated
load vector.
00:50:37.110 --> 00:50:40.890
The K matrix embodies the
strain displacement
00:50:40.890 --> 00:50:43.730
interpolations, which are
obtained from the element
00:50:43.730 --> 00:50:45.310
displacement interpolations.
00:50:45.310 --> 00:50:48.300
Now, as I already pointed
out, we have
00:50:48.300 --> 00:50:49.980
now here, two elements.
00:50:49.980 --> 00:50:54.250
The first element shown here,
second element shown here.
00:50:54.250 --> 00:50:59.870
Notice that the U1, U2
correspond to these two
00:50:59.870 --> 00:51:02.550
displacements, U1, U2.
00:51:02.550 --> 00:51:10.030
The U2, U3 correspond to these
displacements, U2, U3.
00:51:10.030 --> 00:51:14.370
Notice that there's a coupling
between the elements because
00:51:14.370 --> 00:51:19.070
U2 is here a displacement of
that element, and is here the
00:51:19.070 --> 00:51:21.470
displacement of element 2.
00:51:21.470 --> 00:51:23.760
The length of the element
1 is 100, length of
00:51:23.760 --> 00:51:25.900
element 2 is 80.
00:51:25.900 --> 00:51:30.080
Well, if we have two
displacements to describe the
00:51:30.080 --> 00:51:34.020
displacement in an element, then
we recognize immediately
00:51:34.020 --> 00:51:39.160
that all that we can have
is a linear variation in
00:51:39.160 --> 00:51:41.920
displacement between
the two end points,
00:51:41.920 --> 00:51:43.410
between these two nodes.
00:51:43.410 --> 00:51:48.100
And so the element displacement
interpolations
00:51:48.100 --> 00:51:50.850
must involve these functions.
00:51:50.850 --> 00:51:55.510
For unit displacement at this
end of an element, Y over L is
00:51:55.510 --> 00:51:58.500
the interpolation of
the displacement.
00:51:58.500 --> 00:52:02.070
For unit displacement at this
end of the element, this is
00:52:02.070 --> 00:52:03.420
the interpolation.
00:52:03.420 --> 00:52:06.460
Notice that the actual
displacement, of course, goes
00:52:06.460 --> 00:52:09.860
into this direction, but I'm
plotting it upwards to show
00:52:09.860 --> 00:52:14.230
you the magnitude of
that displacement.
00:52:14.230 --> 00:52:19.180
If we have established these
interpolation functions,
00:52:19.180 --> 00:52:22.390
recognizing, of course, that for
element 1, L is 100, for
00:52:22.390 --> 00:52:26.070
element 2, L is equal to 80.
00:52:26.070 --> 00:52:30.270
Then we can directly write down
our H1 and H2 matrices.
00:52:30.270 --> 00:52:31.720
They are given as shown here.
00:52:31.720 --> 00:52:40.810
Notice that Um is equal to Hmu,
where U is equal to--
00:52:40.810 --> 00:52:42.530
I write down the transpose--
00:52:42.530 --> 00:52:46.310
U1, U2, U3.
00:52:46.310 --> 00:52:49.650
Now, we notice that
element 1--
00:52:49.650 --> 00:52:52.200
let's go back once more
for element 1.
00:52:52.200 --> 00:52:58.380
Only U1 and U2 influence the
displacements in that element.
00:52:58.380 --> 00:53:00.590
And that is shown here.
00:53:00.590 --> 00:53:06.670
U3 has 0 and does not influence
the displacement in
00:53:06.670 --> 00:53:08.950
the element.
00:53:08.950 --> 00:53:13.610
Similarly for H2, U1 does not
influence the displacement in
00:53:13.610 --> 00:53:15.190
that element.
00:53:15.190 --> 00:53:17.620
So we have these
displacements.
00:53:17.620 --> 00:53:22.170
Notice also that I've written
here, Um, of course, but that
00:53:22.170 --> 00:53:25.840
Um here, for our specific
case is simply this
00:53:25.840 --> 00:53:27.290
displacement, Vm.
00:53:27.290 --> 00:53:30.420
We have only one displacement
component.
00:53:30.420 --> 00:53:36.280
Taking the derivative of these
relations here, we get
00:53:36.280 --> 00:53:39.380
directly the strains.
00:53:39.380 --> 00:53:43.550
Notice that here, we should have
probably put an m there.
00:53:43.550 --> 00:53:49.420
This is the normal strain and
we obtain these matrices by
00:53:49.420 --> 00:53:51.400
simply taking the derivatives.
00:53:51.400 --> 00:53:55.260
Well, now we have the components
that we need to
00:53:55.260 --> 00:53:57.160
evaluate the K matrix.
00:53:57.160 --> 00:54:00.690
And as I mentioned earlier,
that is the total K matrix
00:54:00.690 --> 00:54:01.800
obtained by summing the
00:54:01.800 --> 00:54:03.810
contributions over the elements.
00:54:03.810 --> 00:54:08.420
This is coming from element 1,
this is coming from element 2.
00:54:08.420 --> 00:54:12.000
Notice that the area is 1,
Young's modulus of stress
00:54:12.000 --> 00:54:13.080
strains law.
00:54:13.080 --> 00:54:15.080
We are integrating
from 0 to 100.
00:54:15.080 --> 00:54:20.690
This is here, B1 transpose,
that is B1.
00:54:20.690 --> 00:54:23.820
This is here, B2 transpose,
that is B2.
00:54:23.820 --> 00:54:27.630
This is the area that I pointed
out to you earlier.
00:54:27.630 --> 00:54:31.330
Evaluating these two matrices,
we directly
00:54:31.330 --> 00:54:33.140
obtain these matrices.
00:54:33.140 --> 00:54:34.690
And notice the following--
00:54:34.690 --> 00:54:39.270
that there's no coupling from
the third degree of freedom
00:54:39.270 --> 00:54:40.660
into element 1.
00:54:40.660 --> 00:54:43.830
Similarly, there's no coupling
from the first degree of
00:54:43.830 --> 00:54:45.970
freedom into element 2.
00:54:45.970 --> 00:54:50.400
In fact, what we will do later
on is simply calculate the
00:54:50.400 --> 00:54:52.270
non-zero parts.
00:54:52.270 --> 00:54:57.160
We call these the compacted
element stiffness matrices.
00:54:57.160 --> 00:55:01.290
And knowing these non-zero parts
and knowing into which
00:55:01.290 --> 00:55:06.090
degrees of freedom they have to
put in the assemblage phase
00:55:06.090 --> 00:55:09.070
to obtain the total stiffness
matrix, we can directly
00:55:09.070 --> 00:55:10.700
assemble the stiffness matrix.
00:55:10.700 --> 00:55:14.300
In other words, if I know this
part here and I know that the
00:55:14.300 --> 00:55:17.630
first column corresponds to the
first column of the global
00:55:17.630 --> 00:55:21.050
stiffness matrix, the second
column corresponds to the
00:55:21.050 --> 00:55:24.060
second column in the global
stiffness matrix, then I can
00:55:24.060 --> 00:55:28.110
just add this contribution
into this part here.
00:55:28.110 --> 00:55:32.150
Similarly, I can simply add this
contribution here into
00:55:32.150 --> 00:55:34.800
that part there, without
carrying
00:55:34.800 --> 00:55:36.690
always these 0's along.
00:55:36.690 --> 00:55:38.770
And that is, of course,
a very important
00:55:38.770 --> 00:55:40.650
computational aspect.
00:55:40.650 --> 00:55:45.370
However, in theory, we are
really still performing these
00:55:45.370 --> 00:55:49.240
additions as shown here, we
really still perform the
00:55:49.240 --> 00:55:52.560
additions as we pointed
them out in the
00:55:52.560 --> 00:55:54.150
direct stiffness procedure.
00:55:54.150 --> 00:55:57.730
In other words, we still perform
this summation, as I
00:55:57.730 --> 00:55:59.470
pointed out to you earlier.
00:55:59.470 --> 00:56:04.280
The important point, however,
is that we now have
00:56:04.280 --> 00:56:07.810
established the K matrix,
corresponding to the system.
00:56:07.810 --> 00:56:10.720
Our R vector is simply, in this
00:56:10.720 --> 00:56:15.150
particular case, 0, 0, 100.
00:56:15.150 --> 00:56:17.440
Because we only have
100 applied at the
00:56:17.440 --> 00:56:18.670
third degree of freedom.
00:56:18.670 --> 00:56:22.870
We now have to impose that U1
is 0, we simply set U1 equal
00:56:22.870 --> 00:56:24.890
to 0 in the equilibrium
equations, as
00:56:24.890 --> 00:56:26.020
I pointed out earlier.
00:56:26.020 --> 00:56:31.310
We solve for U2 and U3, and
having obtained U2 and U3, we
00:56:31.310 --> 00:56:34.730
know the displacement in each
of these parts, and we know,
00:56:34.730 --> 00:56:36.730
therefore, the strains
and the stresses in
00:56:36.730 --> 00:56:38.410
each of these parts.
00:56:38.410 --> 00:56:42.460
The solution is plotted in the
example that I discussed with
00:56:42.460 --> 00:56:43.710
you in lecture 2.
00:56:46.820 --> 00:56:50.750
This example, really, showed
some off the basic points of
00:56:50.750 --> 00:56:52.050
finite element analysis.
00:56:52.050 --> 00:56:54.830
Of course, we have to discuss
much more how we actually
00:56:54.830 --> 00:56:59.950
obtain the Hm matrices for more
complex, more complicated
00:56:59.950 --> 00:57:03.160
elements that I use in actual
practical analysis.
00:57:03.160 --> 00:57:05.480
However, this is all I wanted
to mention in this lecture.
00:57:05.480 --> 00:57:06.730
Thank you for your attention.