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PROFESSOR: Ladies
and gentlemen,
00:00:23.210 --> 00:00:25.740
welcome to lecture four.
00:00:25.740 --> 00:00:28.240
In the last lecture, I presented
to you a general
00:00:28.240 --> 00:00:32.000
formulation for finite
element analysis.
00:00:32.000 --> 00:00:35.570
In this lecture, I would like
to talk to you about the
00:00:35.570 --> 00:00:38.400
derivation of specific finite
element matrices.
00:00:41.260 --> 00:00:44.870
The formulation that we will be
using, or the derivation I
00:00:44.870 --> 00:00:48.240
should say, that we will be
using, leads us to generalized
00:00:48.240 --> 00:00:50.770
finite element models.
00:00:50.770 --> 00:00:53.220
I will be talking later on
about what we mean by
00:00:53.220 --> 00:00:54.870
generalized.
00:00:54.870 --> 00:00:58.450
Let's remember, let's recall
what we arrived
00:00:58.450 --> 00:01:00.340
at in the last lecture.
00:01:00.340 --> 00:01:05.840
We had that the element
stiffness matrix for element m
00:01:05.840 --> 00:01:11.040
is obtained by this integration
here where we
00:01:11.040 --> 00:01:15.450
integrate the product of a
strain-displacement matrix
00:01:15.450 --> 00:01:20.010
times a stress-strain matrix
times the strain-displacement
00:01:20.010 --> 00:01:24.390
matrix over the volume
of the element.
00:01:24.390 --> 00:01:28.270
We also had derived the
expression for a body force
00:01:28.270 --> 00:01:32.200
vector, B. m, again,
denoting element m.
00:01:32.200 --> 00:01:36.600
And here, we integrated over the
volume of the element the
00:01:36.600 --> 00:01:41.560
product of the displacement
interpolation matrix times the
00:01:41.560 --> 00:01:46.740
body forces in the element.
00:01:46.740 --> 00:01:52.050
We also had derived a surface
force vector of the element in
00:01:52.050 --> 00:01:56.960
which we integrate the product
of the surface interpolation
00:01:56.960 --> 00:02:02.100
matrix times the surface forces
applied to the element.
00:02:02.100 --> 00:02:07.110
So in essence, we see then if we
look at these expressions,
00:02:07.110 --> 00:02:11.750
we see that we need really a B
matrix, a strain-displacement
00:02:11.750 --> 00:02:14.290
matrix for the element.
00:02:14.290 --> 00:02:15.230
Of course, we need the
00:02:15.230 --> 00:02:17.970
stress-strain law for the element.
00:02:17.970 --> 00:02:21.250
And that will really be given to
us from the serial strength
00:02:21.250 --> 00:02:23.980
of materials.
00:02:23.980 --> 00:02:27.880
We also want the displacement
interpolation
00:02:27.880 --> 00:02:31.010
matrix for the element.
00:02:31.010 --> 00:02:34.410
Once we have this displacement
interpolation matrix, we can
00:02:34.410 --> 00:02:37.550
directly obtain the
interpolation matrix for the
00:02:37.550 --> 00:02:42.590
surface displacement of the
element by simply substituting
00:02:42.590 --> 00:02:47.740
into this matrix here, the
surface coordinates.
00:02:47.740 --> 00:02:53.590
In summary then, we need the H
matrix, the B matrix, and the
00:02:53.590 --> 00:02:57.620
stress-strain law for every
element that we have in the
00:02:57.620 --> 00:02:59.790
assemblage.
00:02:59.790 --> 00:03:03.690
Notice that the stress-strain
law, of course, must depend on
00:03:03.690 --> 00:03:06.330
what kind of element
we are looking at.
00:03:06.330 --> 00:03:10.090
How we have mechanically
idealized our system.
00:03:10.090 --> 00:03:13.490
In a truss structure, we will
have simply the Young's
00:03:13.490 --> 00:03:15.120
modulus in here.
00:03:15.120 --> 00:03:17.740
In a plane stress idealization,
we will have a
00:03:17.740 --> 00:03:19.310
plane stress law.
00:03:19.310 --> 00:03:21.420
In a plate bending idealization,
we will have the
00:03:21.420 --> 00:03:28.960
plate bending stress-strain
law in there, and so on.
00:03:28.960 --> 00:03:32.940
Well, I want to talk today
about these matrices.
00:03:32.940 --> 00:03:36.240
How do we obtain
these matrices?
00:03:36.240 --> 00:03:41.030
Once we have talked about that
subject, I finally want to
00:03:41.030 --> 00:03:43.990
also spend a little bit of
time talking about the
00:03:43.990 --> 00:03:46.170
convergence of the
analysis results.
00:03:46.170 --> 00:03:49.540
Of course, in finite element
analysis, we must be concerned
00:03:49.540 --> 00:03:52.960
about the accuracy of the
analysis results that we are
00:03:52.960 --> 00:03:57.620
obtaining, and about the
convergence of the results to
00:03:57.620 --> 00:04:03.610
the actual analytically,
theoretically accurate result
00:04:03.610 --> 00:04:06.000
that we want to obtain.
00:04:06.000 --> 00:04:10.245
Well, let us then first talk
about the derivation of the H,
00:04:10.245 --> 00:04:12.480
B, and C matrices.
00:04:12.480 --> 00:04:17.990
And the H, B, and C matrix,
of course, depend on the
00:04:17.990 --> 00:04:21.711
particular problem that
you're looking at.
00:04:21.711 --> 00:04:24.730
Let us therefore categorize
the different types of
00:04:24.730 --> 00:04:28.320
problems that we encounter
in structural analysis.
00:04:28.320 --> 00:04:33.060
Here we have a truss structure,
and a truss
00:04:33.060 --> 00:04:38.430
structure is really an
assemblage of truss elements.
00:04:38.430 --> 00:04:43.710
And in each truss element that
lies between the nodes, we
00:04:43.710 --> 00:04:47.090
have one-dimensional
stress conditions.
00:04:47.090 --> 00:04:51.470
In other words, the only stress
that we see on the
00:04:51.470 --> 00:04:55.780
section AA would be
the tau xx stress.
00:04:55.780 --> 00:04:59.230
And that is the only stress, no
other stress is involved.
00:04:59.230 --> 00:05:01.640
Of course, just going a little
bit ahead, we immediately
00:05:01.640 --> 00:05:05.480
would also say that the only
strain we would be interested
00:05:05.480 --> 00:05:08.770
in is epsilon xx and the
stress-strain law is simply
00:05:08.770 --> 00:05:11.650
Young's modulus.
00:05:11.650 --> 00:05:15.400
The second kind of analysis that
we might be performing is
00:05:15.400 --> 00:05:18.290
a plane stress analysis.
00:05:18.290 --> 00:05:21.430
In plane stress conditions,
the only stresses that are
00:05:21.430 --> 00:05:26.420
non-zero are tau xx,
tau yy, and tau xy.
00:05:26.420 --> 00:05:29.600
Remember in lectures three, I
had listed all six stress and
00:05:29.600 --> 00:05:30.680
strain components.
00:05:30.680 --> 00:05:34.100
And what I'm doing now is I take
out just those stress and
00:05:34.100 --> 00:05:37.290
strain components that are
really applicable for the kind
00:05:37.290 --> 00:05:40.260
of problem that I
want to look at.
00:05:40.260 --> 00:05:43.460
Plane stress situations we will
encounter in the analysis
00:05:43.460 --> 00:05:44.860
of a beam, for example.
00:05:44.860 --> 00:05:47.670
Here's a simple cantilever
beam subjected to
00:05:47.670 --> 00:05:50.090
surface loading p.
00:05:50.090 --> 00:05:57.370
And if we look at an element
here, it is redrawn here.
00:05:57.370 --> 00:05:59.790
We would find that
tau xx would be a
00:05:59.790 --> 00:06:01.280
stress into this direction.
00:06:01.280 --> 00:06:03.610
Tau xy is the shear
stress, and tau yy
00:06:03.610 --> 00:06:06.180
is that normal stress.
00:06:06.180 --> 00:06:08.760
The cantilever, of course, can
be subjected to a surface
00:06:08.760 --> 00:06:12.360
loading, a bending moment, tip
loading, whatever loading.
00:06:12.360 --> 00:06:15.310
The stress components, the only
stress components that we
00:06:15.310 --> 00:06:20.720
assume of course, act in the
structure would be these three
00:06:20.720 --> 00:06:22.480
stress components.
00:06:22.480 --> 00:06:26.180
In fact, of course, in the
actual physical situation, we
00:06:26.180 --> 00:06:30.340
would find that around here,
depending on the specific
00:06:30.340 --> 00:06:33.660
support conditions that we are
using, we would have a much
00:06:33.660 --> 00:06:36.310
more complicated stress
situation than the plane
00:06:36.310 --> 00:06:38.170
stress situation assumed.
00:06:38.170 --> 00:06:40.670
However, if we are only
interested in the tip
00:06:40.670 --> 00:06:45.050
displacement here, or the
displacements away from the
00:06:45.050 --> 00:06:48.250
support condition from
the support.
00:06:48.250 --> 00:06:50.550
Similarly, if we are only
interested in the stresses and
00:06:50.550 --> 00:06:54.580
strains away from the support,
certainly it is accurate
00:06:54.580 --> 00:06:55.810
enough to--
00:06:55.810 --> 00:06:58.790
it yields an accurate enough
model to assume a
00:06:58.790 --> 00:07:01.070
plane stress situation.
00:07:01.070 --> 00:07:04.190
Another plane stress situation
occurs here, for example,
00:07:04.190 --> 00:07:06.470
where we have a sheet
with a hole.
00:07:06.470 --> 00:07:10.730
If the sheet is thin, the only
stresses that we would see on
00:07:10.730 --> 00:07:14.440
an element, such as shown here,
would be z stresses, the
00:07:14.440 --> 00:07:17.385
plane stress stresses.
00:07:17.385 --> 00:07:20.950
A next category of analysis
that we would encounter in
00:07:20.950 --> 00:07:25.190
practice is a plane
strain condition.
00:07:25.190 --> 00:07:31.970
Plane strain condition means
that in the space xyz shown
00:07:31.970 --> 00:07:37.500
here with the displacements uvw,
the displacement w is 0.
00:07:37.500 --> 00:07:42.450
It's identically 0 if our x,
y-coordinate system lies in
00:07:42.450 --> 00:07:47.020
the plane of a typical section
of the total problem.
00:07:47.020 --> 00:07:50.560
And let me show you
one such typical--
00:07:50.560 --> 00:07:53.000
or describe to you one such
difficult problem.
00:07:53.000 --> 00:07:56.740
Here we have a dam that
is very long.
00:07:56.740 --> 00:07:59.550
In fact, we might assume
it infinitely long.
00:07:59.550 --> 00:08:05.020
And it is supported on this
face here, so that w, the
00:08:05.020 --> 00:08:08.300
displacement, normal
to that face is 0.
00:08:08.300 --> 00:08:10.320
And the same support
conditions are
00:08:10.320 --> 00:08:12.480
at the other face.
00:08:12.480 --> 00:08:16.400
If the dam is subject to uniform
water loading, water
00:08:16.400 --> 00:08:20.800
pressure loading all along its
side, then by symmetry it
00:08:20.800 --> 00:08:24.980
follows that w, the
displacement, at any section
00:08:24.980 --> 00:08:27.430
of the dam, the displacement
normal to that
00:08:27.430 --> 00:08:29.490
section must be 0.
00:08:29.490 --> 00:08:35.030
Therefore, we have w 0 meaning
directly epsilon zz 0.
00:08:35.030 --> 00:08:39.270
The strain into the z-direction
is also 0.
00:08:39.270 --> 00:08:46.870
Similarly, the shear stresses,
tau zx and tau zy are 0.
00:08:46.870 --> 00:08:49.140
And the only stresses that we
are left with which are
00:08:49.140 --> 00:08:51.990
non-zero are those shown here.
00:08:51.990 --> 00:08:59.840
These are the stresses acting on
a typical slice of the dam.
00:08:59.840 --> 00:09:05.040
A slice say, of unit length
would be subjected to just
00:09:05.040 --> 00:09:07.070
these stress conditions.
00:09:07.070 --> 00:09:11.190
Therefore, in that array of six
stresses and strains that
00:09:11.190 --> 00:09:14.410
I talked to you about in lecture
three, we really only
00:09:14.410 --> 00:09:17.590
are concerned--
00:09:17.590 --> 00:09:21.410
you only need to calculate these
four components of those
00:09:21.410 --> 00:09:24.600
six that I talked about there.
00:09:24.600 --> 00:09:27.100
Another stress-strain condition
that we are
00:09:27.100 --> 00:09:31.370
interested in frequently in
practice is an axisymmetric
00:09:31.370 --> 00:09:32.830
analysis condition.
00:09:32.830 --> 00:09:37.100
This case we have an axis of
revolution for the structure,
00:09:37.100 --> 00:09:40.490
and here we look at a
cylinder, which is
00:09:40.490 --> 00:09:45.370
axisymmetric in geometry about
that center line here, the
00:09:45.370 --> 00:09:46.750
axis of revolution.
00:09:46.750 --> 00:09:50.590
And the loading is also assumed
in this particular
00:09:50.590 --> 00:09:52.470
case to be also axisymmetric.
00:09:52.470 --> 00:09:56.040
So typically, we are talking
about a cylinder, circular, of
00:09:56.040 --> 00:09:59.010
course, when looking from a
plane view subjected to
00:09:59.010 --> 00:10:03.220
internal pressure, which
is the same all around.
00:10:03.220 --> 00:10:07.410
And of course, the geometry is
also axisymmetric when we look
00:10:07.410 --> 00:10:11.810
at the axis of revolution, the
center line of the cylinder.
00:10:11.810 --> 00:10:16.930
In this particular case, the
only stress conditions that we
00:10:16.930 --> 00:10:19.710
need to calculate are
shown here for an
00:10:19.710 --> 00:10:21.890
element in that cylinder.
00:10:21.890 --> 00:10:25.890
We have tau xx, tau
yy, tau zz.
00:10:25.890 --> 00:10:29.510
This, of course, being
the hoop stress.
00:10:29.510 --> 00:10:32.290
And the shear stress, tau xy.
00:10:32.290 --> 00:10:34.150
These are the stress conditions
that we need to
00:10:34.150 --> 00:10:36.090
calculate for this
particular case.
00:10:36.090 --> 00:10:42.080
Similarity, because we are
talking about an axisymmetric
00:10:42.080 --> 00:10:46.260
condition, an axisymmetric
condition where really, every
00:10:46.260 --> 00:10:51.080
one of these elements behaves
in the same way, we can look
00:10:51.080 --> 00:10:57.205
at one radian of the total
circumference in the analysis.
00:10:59.740 --> 00:11:04.420
So this is an axisymmetric
analysis, which we also are
00:11:04.420 --> 00:11:08.450
concerned with frequently
in practice.
00:11:08.450 --> 00:11:13.330
Finally, the very large
number of analyses--
00:11:13.330 --> 00:11:17.040
in a very large number of
analyses, we consider plates
00:11:17.040 --> 00:11:19.080
and shells.
00:11:19.080 --> 00:11:23.930
If we look at a plate, an
element in that plate, we
00:11:23.930 --> 00:11:30.240
would find that in general we
would have the tau xx, tau yy.
00:11:30.240 --> 00:11:32.400
These are the normal stresses.
00:11:32.400 --> 00:11:37.530
And tau xy shearing stresses
in the plane of the plate.
00:11:37.530 --> 00:11:40.060
We would also have transfer
shearing stresses,
00:11:40.060 --> 00:11:43.230
tau xz and tau yz.
00:11:43.230 --> 00:11:47.290
Sorry, tau yz, shown
here as the
00:11:47.290 --> 00:11:49.450
transfer shearing stresses.
00:11:49.450 --> 00:11:52.490
Of course, these transfers
shearing stresses give us
00:11:52.490 --> 00:11:56.020
shear forces.
00:11:56.020 --> 00:12:03.890
And these stresses here give
us membrane forces plus
00:12:03.890 --> 00:12:07.810
bending moments and
twisting moments.
00:12:07.810 --> 00:12:11.330
Elementary theory of plates
shows us, tells us, that these
00:12:11.330 --> 00:12:14.370
are the significant stresses
in a plate.
00:12:14.370 --> 00:12:18.080
The important point is that in
a plate, as well as in a
00:12:18.080 --> 00:12:23.790
shell, the stress normal
to the plate is 0.
00:12:23.790 --> 00:12:25.320
The same holds for the shell.
00:12:25.320 --> 00:12:29.600
The stress normal to the shell
is 0, and we have the stress
00:12:29.600 --> 00:12:30.330
components.
00:12:30.330 --> 00:12:35.426
However, we have to now remember
that the xx and yy
00:12:35.426 --> 00:12:39.210
axes are aligned with the
curvature of the shelf.
00:12:39.210 --> 00:12:41.420
Are aligned with the curvature
of the shell.
00:12:41.420 --> 00:12:45.330
And then these are the stress
components that are of
00:12:45.330 --> 00:12:48.010
interest in the analysis
of shells.
00:12:48.010 --> 00:12:51.330
In many cases, when we look at
the analysis of plates, the
00:12:51.330 --> 00:12:53.300
membrane forces--
00:12:53.300 --> 00:12:56.690
the total membrane
forces are 0.
00:12:56.690 --> 00:13:01.400
In which case, of course, we
assume that tau yy has a
00:13:01.400 --> 00:13:04.480
linear variation through the
thickness of the plate.
00:13:04.480 --> 00:13:10.230
As stipulated, for example, in
the Kirchhoff plate theory.
00:13:10.230 --> 00:13:15.270
This much then for the static
components, or rather the
00:13:15.270 --> 00:13:18.300
stress components of plates and
shells when we look at the
00:13:18.300 --> 00:13:21.350
kinematics of plates
and shells.
00:13:21.350 --> 00:13:25.910
Just to remind you, we assume
that there's a normal to the
00:13:25.910 --> 00:13:27.800
mid-surface.
00:13:27.800 --> 00:13:31.520
And that normal in Kirchhoff
plate theory remains during
00:13:31.520 --> 00:13:35.510
deformation normal to
the plate surface.
00:13:35.510 --> 00:13:40.550
And all material particles on
that normal AB remain on a
00:13:40.550 --> 00:13:41.480
straight line.
00:13:41.480 --> 00:13:42.460
These are the two things.
00:13:42.460 --> 00:13:46.720
I repeat, all material particles
on the line AB
00:13:46.720 --> 00:13:51.450
remain during deformation on the
line AB, and the line AB
00:13:51.450 --> 00:13:53.955
remains normal to the
mid-surface in Kirchhoff plate
00:13:53.955 --> 00:13:57.750
theory, which does not exclude
shear deformations.
00:13:57.750 --> 00:14:01.200
However, if we use a theory
that does include shear
00:14:01.200 --> 00:14:04.970
deformations, we would still
have that the material
00:14:04.970 --> 00:14:09.450
particles on line AB will still
lie after deformations
00:14:09.450 --> 00:14:14.230
on the line A prime
B prime say.
00:14:14.230 --> 00:14:16.850
But the line A prime B prime--
00:14:16.850 --> 00:14:20.240
B is still a straight line-- is
not anymore normal to the
00:14:20.240 --> 00:14:21.090
mid-surface.
00:14:21.090 --> 00:14:23.680
In this case, we have included
shear deformations.
00:14:23.680 --> 00:14:26.830
We will see in the next lecture
that, in fact, it is
00:14:26.830 --> 00:14:30.030
simpler to use a theory
including shear deformations
00:14:30.030 --> 00:14:33.980
to derive effective
finite elements.
00:14:33.980 --> 00:14:39.380
Well, this then is a review
for the kinds of stress
00:14:39.380 --> 00:14:43.010
components and strain components
that we are
00:14:43.010 --> 00:14:43.910
interested in.
00:14:43.910 --> 00:14:47.920
Of course, that information
which I discussed so far is
00:14:47.920 --> 00:14:53.050
available from the strengths
of material theory and
00:14:53.050 --> 00:14:54.760
continue mechanic theory.
00:14:54.760 --> 00:14:58.140
And as an engineer, however, we
have to be able to decide
00:14:58.140 --> 00:15:00.470
what kind of stress and strain
00:15:00.470 --> 00:15:02.230
situations you want to analyze.
00:15:02.230 --> 00:15:04.520
And there, of course, is already
one very important
00:15:04.520 --> 00:15:05.010
assumption.
00:15:05.010 --> 00:15:09.430
A very important assumption in
laying out or starting with
00:15:09.430 --> 00:15:13.210
the right model for the actual
physical structure.
00:15:13.210 --> 00:15:16.340
Whether we want to model the
actual physical structure as a
00:15:16.340 --> 00:15:21.240
truss assemblage, as an
assemblage of plates, beams,
00:15:21.240 --> 00:15:26.810
et cetera, all of those
situations can be covered in
00:15:26.810 --> 00:15:28.090
the finite element theory.
00:15:28.090 --> 00:15:31.630
But we have to make the right
assumptions right from the
00:15:31.630 --> 00:15:34.380
start in starting with
the right model.
00:15:34.380 --> 00:15:36.460
So here I listed ones.
00:15:36.460 --> 00:15:38.850
In one table, the particular
problems
00:15:38.850 --> 00:15:40.100
that we might encounter.
00:15:40.100 --> 00:15:42.640
We might have a bar.
00:15:42.640 --> 00:15:45.770
The displacement components
then being simply u.
00:15:45.770 --> 00:15:51.020
In a beam, which I have not
given earlier, but here we
00:15:51.020 --> 00:15:53.200
have a transverse
displacement, w.
00:15:53.200 --> 00:15:57.350
In plane stress situations, we
have two displacements, u and
00:15:57.350 --> 00:15:58.100
v.
00:15:58.100 --> 00:16:01.090
In a plane strain situation,
u and v.
00:16:01.090 --> 00:16:03.230
In an axisymmetric situation,
u and v.
00:16:03.230 --> 00:16:06.360
Three-dimensional situation,
we have u, v, and w.
00:16:06.360 --> 00:16:08.760
In plate bending, simply the
transverse displacements.
00:16:11.590 --> 00:16:15.600
In analysis of shells, in the
analysis of shells we would,
00:16:15.600 --> 00:16:20.260
of course, have the transverse
placements and also the
00:16:20.260 --> 00:16:21.770
membrane displacements
included.
00:16:21.770 --> 00:16:25.710
We will talk about that
in the next lecture.
00:16:25.710 --> 00:16:29.120
These are the displacement
components for the particular
00:16:29.120 --> 00:16:31.270
problem that we are
looking at.
00:16:31.270 --> 00:16:35.810
And correspondingly, we have
particular strains.
00:16:35.810 --> 00:16:39.170
The strains that we are talking
about are in the bar,
00:16:39.170 --> 00:16:40.480
simply the normals strains.
00:16:40.480 --> 00:16:44.640
In a beam, the second derivative
of the transverse
00:16:44.640 --> 00:16:45.680
displacements.
00:16:45.680 --> 00:16:49.310
Here we have kappa xx defined.
00:16:49.310 --> 00:16:50.660
This is kappa xx defined.
00:16:50.660 --> 00:16:53.120
In plane stress, we have
these three components.
00:16:53.120 --> 00:16:54.860
Plane strain, these
three components.
00:16:54.860 --> 00:16:57.550
Axisymmetric these three
components, and so on.
00:16:57.550 --> 00:17:00.500
And it is here, of course,
where the engineer has to
00:17:00.500 --> 00:17:04.450
decide what kind of problem he
has in hand when he actually--
00:17:04.450 --> 00:17:07.880
when he looks at an actual
physical situation.
00:17:07.880 --> 00:17:10.849
Corresponding to these strains,
00:17:10.849 --> 00:17:12.690
we have also stresses.
00:17:12.690 --> 00:17:15.010
And here we are listing
the stresses.
00:17:15.010 --> 00:17:19.970
Notice that for the beam, we
are talking about a bending
00:17:19.970 --> 00:17:21.760
moment, which we might
consider to be
00:17:21.760 --> 00:17:23.150
a generalized stress.
00:17:23.150 --> 00:17:25.810
In plate bending, we talk
similarly about bending
00:17:25.810 --> 00:17:28.380
moments, which you also
might consider as
00:17:28.380 --> 00:17:30.440
being generalized stresses.
00:17:30.440 --> 00:17:34.680
Otherwise, we have the actual
stresses that correspond to
00:17:34.680 --> 00:17:36.490
the strains.
00:17:36.490 --> 00:17:39.420
Having decided what stresses
and strains we
00:17:39.420 --> 00:17:40.920
want to look at--
00:17:40.920 --> 00:17:43.390
in other words, what kind of
problem we have at hand, we
00:17:43.390 --> 00:17:45.630
also have to select the
appropriate material matrix.
00:17:45.630 --> 00:17:48.840
Again, this one is given form
strengths of material.
00:17:48.840 --> 00:17:52.140
Here, for a bar, a truss
structure, we simply have
00:17:52.140 --> 00:17:52.820
Young's modulus.
00:17:52.820 --> 00:17:55.360
For a beam, we have the
flexural rigidity.
00:17:55.360 --> 00:18:00.300
In plane stress analysis, this
would be the material law that
00:18:00.300 --> 00:18:03.380
you would find in many
textbooks on
00:18:03.380 --> 00:18:05.470
strengths of materials.
00:18:05.470 --> 00:18:09.640
Well, having therefore resolved
the problem of how to
00:18:09.640 --> 00:18:12.420
establish the stress-strain
law, or the stress-strain
00:18:12.420 --> 00:18:17.530
matrix c, we now go on to
discuss the construction of
00:18:17.530 --> 00:18:20.860
the displacement interpolation
matrix.
00:18:20.860 --> 00:18:22.770
For a one-dimensional
bar element,
00:18:22.770 --> 00:18:25.570
we would have simply--
00:18:25.570 --> 00:18:29.430
we can assume simply a
polynomial with unknown
00:18:29.430 --> 00:18:34.980
coefficients, alpha 1,
alpha 2, alpha 3.
00:18:34.980 --> 00:18:37.805
And this is simply
a polynomial nx.
00:18:37.805 --> 00:18:41.620
x being, of course, the
coordinate that is being used
00:18:41.620 --> 00:18:43.610
to describe the displacement
for the element.
00:18:43.610 --> 00:18:46.230
For two-dimensional elements,
here we're talking about plane
00:18:46.230 --> 00:18:47.890
stress, plane strain.
00:18:47.890 --> 00:18:51.360
Axisymmetric analysis, we would
have, similarly, these
00:18:51.360 --> 00:18:52.390
assumptions here.
00:18:52.390 --> 00:18:58.060
alphas for the u displacements
and betas for the v
00:18:58.060 --> 00:19:00.260
displacements.
00:19:00.260 --> 00:19:05.880
For the plate bending element
in which we only discretize
00:19:05.880 --> 00:19:09.080
the w displacement, the
transverse displacement, we
00:19:09.080 --> 00:19:12.150
would have this assumption
here.
00:19:12.150 --> 00:19:15.130
For three-dimensional solid
elements, similarly, we have
00:19:15.130 --> 00:19:16.540
these assumptions.
00:19:16.540 --> 00:19:22.660
Notice that these gammas,
alphas, betas are unknowns for
00:19:22.660 --> 00:19:23.250
the element.
00:19:23.250 --> 00:19:26.990
And we will actually relate
these coefficients to the
00:19:26.990 --> 00:19:29.040
nodal point displacements
of an element.
00:19:31.930 --> 00:19:35.905
If we look, therefore, in
general, at the description of
00:19:35.905 --> 00:19:40.330
an element, we really have on
the left-hand side here, the
00:19:40.330 --> 00:19:44.350
actual displacements, the
continuous displacements in
00:19:44.350 --> 00:19:53.260
the element being given by
matrix phi, which contains the
00:19:53.260 --> 00:19:58.020
constant and x, x squared,
x cubed terms, and so on.
00:19:58.020 --> 00:20:02.510
And this alpha vector here
contains the generalized
00:20:02.510 --> 00:20:06.100
coordinates, the alphas,
betas, gammas.
00:20:06.100 --> 00:20:08.910
Typically, therefore, just to
focus our attention on a
00:20:08.910 --> 00:20:13.840
particular case, we might have
if we say that u of x for a
00:20:13.840 --> 00:20:19.785
truss structure is equal to
alpha 0 plus alpha 1x plus
00:20:19.785 --> 00:20:21.960
alpha 2x squared.
00:20:21.960 --> 00:20:27.110
In this particular case, we
would have that we can write
00:20:27.110 --> 00:20:32.730
this, of course, as 1 x x
squared in matrix form times
00:20:32.730 --> 00:20:36.610
alpha 0 alpha 1, alpha 2.
00:20:36.610 --> 00:20:42.150
And here, I'm talking about this
being equal to the phi,
00:20:42.150 --> 00:20:45.090
and this being equal to
the vector alpha.
00:20:45.090 --> 00:20:50.360
So just to give you an example
for this particular case.
00:20:50.360 --> 00:20:55.970
If we now want to evaluate
the alpha values.
00:20:55.970 --> 00:20:58.650
Of course, in two-dimension
analysis as I shown earlier to
00:20:58.650 --> 00:21:00.780
you, we have also betas here.
00:21:00.780 --> 00:21:03.220
In three-dimensional analysis,
we also would have gammas in
00:21:03.220 --> 00:21:04.640
this vector.
00:21:04.640 --> 00:21:08.580
If you want to evaluate these
alpha values, then we can
00:21:08.580 --> 00:21:12.880
simply use this relation here,
or that relation here, and
00:21:12.880 --> 00:21:15.880
apply it to the element
nodal points.
00:21:15.880 --> 00:21:19.620
To the displacements at the
nodal points of the elements.
00:21:19.620 --> 00:21:23.800
If we do that, in other words,
we substitute into here, into
00:21:23.800 --> 00:21:27.770
the phi matrix, the actual
coordinate of the nodal
00:21:27.770 --> 00:21:30.820
points, and on the left-hand
side, of course, the
00:21:30.820 --> 00:21:34.160
displacement at that nodal
point, we directly generate
00:21:34.160 --> 00:21:36.030
this relation.
00:21:36.030 --> 00:21:38.870
We can then invert this relation
to obtain alpha.
00:21:38.870 --> 00:21:41.420
And here, we have now the
generalized coordinates in
00:21:41.420 --> 00:21:46.030
terms of the nodal point
displacements.
00:21:46.030 --> 00:21:49.650
This then would really complete
the evaluation of the
00:21:49.650 --> 00:21:53.620
continuous displacements u in
terms of the nodal points
00:21:53.620 --> 00:21:56.850
displacement u hat.
00:21:56.850 --> 00:22:00.620
If we have laid down this
assumption, we can directly
00:22:00.620 --> 00:22:04.420
also obtain the relation for the
strains of the elements.
00:22:04.420 --> 00:22:08.165
We have decided what kinds of
strains we want to look at
00:22:08.165 --> 00:22:09.870
that we have to include.
00:22:09.870 --> 00:22:13.210
And by proper differentiation of
the rows here, we directly
00:22:13.210 --> 00:22:14.620
generate this relation.
00:22:14.620 --> 00:22:16.820
Of course, this is a
[UNINTELLIGIBLE]
00:22:16.820 --> 00:22:21.310
relation then for the material
of the element.
00:22:21.310 --> 00:22:25.500
And we then directly have
tau in terms of alpha.
00:22:25.500 --> 00:22:29.060
But alpha was already given
in terms of u hat here.
00:22:29.060 --> 00:22:33.980
In summary, therefore, we have
our displacement interpolation
00:22:33.980 --> 00:22:40.470
matrix here, given as
phi times a inverse.
00:22:40.470 --> 00:22:44.130
This is obtained by simply
substituting from here,
00:22:44.130 --> 00:22:47.790
substitute from here,
directly into there.
00:22:47.790 --> 00:22:52.920
And you get that u is
equal to H u hat.
00:22:52.920 --> 00:22:56.380
And our H being phi a inverse.
00:22:56.380 --> 00:23:00.280
Once again, the u is the
continuous displacement in the
00:23:00.280 --> 00:23:02.240
elements. u hat has nodal point
00:23:02.240 --> 00:23:03.900
displacements in the elements.
00:23:03.900 --> 00:23:08.650
Similarly, if we substitute
from here into there and
00:23:08.650 --> 00:23:13.990
recognize that our epsilon
strain is equal to B times u
00:23:13.990 --> 00:23:18.570
hat, we directly obtain
that B must be equal
00:23:18.570 --> 00:23:21.970
to E times A inverse.
00:23:21.970 --> 00:23:25.500
Well, let us now look at a
particular case, a particular
00:23:25.500 --> 00:23:30.350
example to demonstrate how we
proceed for a particular case.
00:23:30.350 --> 00:23:34.380
Here I'm looking at the analysis
of a cantilever
00:23:34.380 --> 00:23:36.100
subjected to a load.
00:23:36.100 --> 00:23:38.510
And I've idealized that
cantilever as an assemblage of
00:23:38.510 --> 00:23:40.560
four elements.
00:23:40.560 --> 00:23:44.040
Each element is a four
nodal element.
00:23:44.040 --> 00:23:49.370
So for element 1, we have nodes
1, 2, 4, and 5 of the
00:23:49.370 --> 00:23:52.260
global assemblage.
00:23:52.260 --> 00:23:55.840
Notice that all together
we have 9 nodes.
00:23:55.840 --> 00:23:58.700
And once again, 4 elements.
00:23:58.700 --> 00:24:05.160
The cantilever structure is SiN
and we believe that it is
00:24:05.160 --> 00:24:09.600
appropriately modeled using
a plane stress analysis.
00:24:09.600 --> 00:24:14.160
In which case, the only stresses
that are of concern
00:24:14.160 --> 00:24:17.345
are the stresses in this
plane, tau yy,
00:24:17.345 --> 00:24:19.280
tau xx, and tau xy.
00:24:19.280 --> 00:24:24.320
So these other stress components
here are 0.
00:24:24.320 --> 00:24:28.660
If we look now at a particular
element to focus our attention
00:24:28.660 --> 00:24:31.420
on one element, because
all these elements
00:24:31.420 --> 00:24:32.530
are really the same.
00:24:32.530 --> 00:24:34.220
Basically, the same.
00:24:34.220 --> 00:24:36.190
Let's look at element 2.
00:24:36.190 --> 00:24:44.030
And we notice that element 2
being a four nodal element is
00:24:44.030 --> 00:24:49.980
shown once more here with the
displacements U1, V1 at the
00:24:49.980 --> 00:24:54.110
local nodal point of
the element 1.
00:24:54.110 --> 00:24:59.440
V2, U2 being the displacements
of the local nodal point 2.
00:24:59.440 --> 00:25:02.920
And here, V3, U3 similarly
for nodal point 3.
00:25:02.920 --> 00:25:06.200
And V4, U4 for nodal point 4.
00:25:06.200 --> 00:25:09.350
We have taken this element of
the assemblage and what we
00:25:09.350 --> 00:25:15.330
want to really do is derive the
displacement and strain
00:25:15.330 --> 00:25:19.790
interpolation matrices for this
element corresponding to
00:25:19.790 --> 00:25:25.360
U1, V1; U2, V2; U3,
V3; and U4, V4.
00:25:25.360 --> 00:25:30.080
If we have done that, we
recognize, of course, that U1
00:25:30.080 --> 00:25:32.130
corresponds to a global
displacement.
00:25:32.130 --> 00:25:35.410
Similarly, V1 corresponds
to a global placement.
00:25:35.410 --> 00:25:40.015
Then we can directly construct
the displacement and strain
00:25:40.015 --> 00:25:45.520
interpolations corresponding to
the global displacements.
00:25:45.520 --> 00:25:48.800
But the finite element procedure
is really to look
00:25:48.800 --> 00:25:50.480
locally at an element.
00:25:50.480 --> 00:25:51.900
Locally at an element.
00:25:51.900 --> 00:25:54.810
And work in terms of local
displacements.
00:25:54.810 --> 00:25:57.400
Later on then, we relate the
local displacements to the
00:25:57.400 --> 00:25:58.720
global displacements.
00:25:58.720 --> 00:26:02.270
The advantage of proceeding this
way is that we really--
00:26:02.270 --> 00:26:04.920
in looking at one element,
we look at all four
00:26:04.920 --> 00:26:06.170
simultaneously.
00:26:09.510 --> 00:26:16.950
The procedure then is, once more
in summary, we want this
00:26:16.950 --> 00:26:21.820
matrix here, the H2 matrix
corresponding to element 2.
00:26:21.820 --> 00:26:25.970
The vector U lists all of the
displacements, all of the 18
00:26:25.970 --> 00:26:27.170
displacements.
00:26:27.170 --> 00:26:37.365
However, we will look at the
element 2 with local nodal
00:26:37.365 --> 00:26:39.980
point displacements, forgetting
about the global
00:26:39.980 --> 00:26:41.340
nodal point displacement.
00:26:41.340 --> 00:26:47.600
And our objective is to derive
a relationship of u equals H
00:26:47.600 --> 00:26:54.860
times u hat, where u hat
lists U1, V1; U2, V2;
00:26:54.860 --> 00:26:58.800
U3, V3; U4 and V4.
00:26:58.800 --> 00:27:05.100
Once we have this H, which is,
of course, a 2 by 8 matrix,
00:27:05.100 --> 00:27:08.856
because there are 2
displacements, u and v. The u
00:27:08.856 --> 00:27:12.960
and v displacement
for the element.
00:27:12.960 --> 00:27:16.060
And there are 8 entries
in u hat.
00:27:16.060 --> 00:27:18.970
Therefore, this H matrix
must be a 2 by 8.
00:27:18.970 --> 00:27:23.930
Once we have obtained this H
matrix, we can construct
00:27:23.930 --> 00:27:27.730
directly our H2 matrix that
we talked about earlier.
00:27:30.470 --> 00:27:36.390
Well, let's proceed then in the
way that I discussed in
00:27:36.390 --> 00:27:38.640
general form earlier.
00:27:38.640 --> 00:27:43.240
Since we have a four nodal
element, let me just sketch it
00:27:43.240 --> 00:27:44.210
once more here.
00:27:44.210 --> 00:27:49.170
Since we have a four nodal
element, and we are talking
00:27:49.170 --> 00:27:55.490
about u and v displacements, we
really can only have four
00:27:55.490 --> 00:27:59.140
constants, four generalized
coordinates.
00:27:59.140 --> 00:28:02.870
Generalized coordinates
corresponding to the u and
00:28:02.870 --> 00:28:05.570
corresponding to the
v displacements.
00:28:05.570 --> 00:28:11.450
Remember that the only unknown
u displacements are these 4.
00:28:11.450 --> 00:28:18.850
And these 4 u displacements will
give us alpha 1, alpha 2,
00:28:18.850 --> 00:28:20.620
alpha 3, and alpha 4.
00:28:20.620 --> 00:28:23.560
The same holds for the
v displacements.
00:28:23.560 --> 00:28:28.290
This is a bilinear approximation
of the u
00:28:28.290 --> 00:28:31.520
displacements in the element.
00:28:31.520 --> 00:28:35.510
We write this in matrix
form as shown here.
00:28:35.510 --> 00:28:40.300
Noting that this capital Phi,
notice this is a capital Phi
00:28:40.300 --> 00:28:42.190
because there are these
cross bars.
00:28:42.190 --> 00:28:45.660
It's easy to confuse the capital
Phi with the lowercase
00:28:45.660 --> 00:28:48.150
phi, which doesn't have
the cross bars.
00:28:48.150 --> 00:28:50.730
So this capital Phi was
the cross bars here
00:28:50.730 --> 00:28:52.580
is defined in here.
00:28:52.580 --> 00:28:56.290
Where the lowercase phi
is simply an array
00:28:56.290 --> 00:29:00.020
of 1, x, y, x, y.
00:29:00.020 --> 00:29:03.190
This relationship here is
nothing else than that
00:29:03.190 --> 00:29:06.340
relationship, but written
in matrix form.
00:29:06.340 --> 00:29:11.020
The alpha lists all the
generalized coordinates.
00:29:11.020 --> 00:29:12.050
The generalize coordinates.
00:29:12.050 --> 00:29:14.850
And this is the reason why we
call this the generalized
00:29:14.850 --> 00:29:19.300
coordinate finite element
formulation for this
00:29:19.300 --> 00:29:20.940
particular element.
00:29:20.940 --> 00:29:24.520
Let us now define the
u hat vector,
00:29:24.520 --> 00:29:26.280
which I refer to earlier.
00:29:26.280 --> 00:29:31.140
This case, the u hat vector
involves 4 u nodal point
00:29:31.140 --> 00:29:35.420
displacements and 4 v nodal
point displacements.
00:29:35.420 --> 00:29:37.910
Now if we apply--
00:29:37.910 --> 00:29:40.700
and this is really one
important point.
00:29:40.700 --> 00:29:47.410
If we apply this relationship
here, which of course must
00:29:47.410 --> 00:29:49.580
hold all over the element.
00:29:49.580 --> 00:29:52.470
In particular, it must also hold
for the nodal points of
00:29:52.470 --> 00:29:53.770
the element.
00:29:53.770 --> 00:29:58.220
If we apply this relationship
for the nodal points of the
00:29:58.220 --> 00:30:02.730
elements, we can apply it eight
times because we have 8
00:30:02.730 --> 00:30:05.430
nodal point displacements.
00:30:05.430 --> 00:30:10.320
We directly generate this
relationship here.
00:30:10.320 --> 00:30:14.470
u hat equals A times alpha,
where A involve the
00:30:14.470 --> 00:30:17.600
coordinates of the
nodal points.
00:30:17.600 --> 00:30:21.670
Hence, we have our H Phi times
A inverse the way I have
00:30:21.670 --> 00:30:24.240
derived it earlier.
00:30:24.240 --> 00:30:28.920
So this is a very systematic
approach of deriving the
00:30:28.920 --> 00:30:32.190
displacement interpolation
matrix.
00:30:32.190 --> 00:30:35.530
It consists of evaluating
the A matrix.
00:30:35.530 --> 00:30:39.250
And of course, we assume that
that A matrix can be inverted.
00:30:39.250 --> 00:30:42.670
This is, for this element,
certainly the case, as long as
00:30:42.670 --> 00:30:47.070
we are talking about the
rectangular element or not
00:30:47.070 --> 00:30:48.890
very highly distorted
elements.
00:30:48.890 --> 00:30:52.000
We can for the four nodal
element, invert this matrix
00:30:52.000 --> 00:30:55.140
and directly get this
relationship.
00:30:55.140 --> 00:31:00.130
Let's look now, once more back
at the analysis that we wanted
00:31:00.130 --> 00:31:01.260
to perform.
00:31:01.260 --> 00:31:05.450
Remember that we looked at this
element here, and we have
00:31:05.450 --> 00:31:10.910
now obtained the H matrix, the
displacement interpolation
00:31:10.910 --> 00:31:13.200
matrix for this element.
00:31:13.200 --> 00:31:15.810
Once more, here is
the H matrix.
00:31:15.810 --> 00:31:19.270
In fact, this H matrix here,
since we have no superscript
00:31:19.270 --> 00:31:23.680
attached to it yet, really holds
for every one of the 4
00:31:23.680 --> 00:31:25.290
elements that we are
talking about.
00:31:25.290 --> 00:31:32.130
Provided we put into the u hat
here the appropriate nodal
00:31:32.130 --> 00:31:34.350
point displacements
of the elements.
00:31:34.350 --> 00:31:39.170
We can directly say that this
matrix here holds for any one
00:31:39.170 --> 00:31:40.420
of the elements.
00:31:42.400 --> 00:31:45.230
If the elements, if the 4
elements are not completely
00:31:45.230 --> 00:31:48.630
identical, of course, the A
inverse will change from
00:31:48.630 --> 00:31:49.880
element to element.
00:31:52.290 --> 00:31:59.620
Having gone through that step
and having looked at element 2
00:31:59.620 --> 00:32:05.330
once more, we directly recognize
how to establish now
00:32:05.330 --> 00:32:08.600
the H2 matrix.
00:32:08.600 --> 00:32:14.750
Remember our u1 will be related
to the global U11.
00:32:14.750 --> 00:32:17.626
Our v1 will be related
to the--
00:32:17.626 --> 00:32:23.050
or will correspond I should
rather say, to the global U12.
00:32:23.050 --> 00:32:25.760
Let's look at why this
is the case.
00:32:25.760 --> 00:32:29.570
Why did I write down U11
here and U12 there?
00:32:29.570 --> 00:32:33.360
Well, let's look at
this point there.
00:32:33.360 --> 00:32:36.710
Which is of course, the
local nodal point 1.
00:32:36.710 --> 00:32:41.000
The local nodal point 1, right
upperhand corner of element 1,
00:32:41.000 --> 00:32:45.100
corresponds here to
nodal point 6.
00:32:45.100 --> 00:32:49.590
Well, the global nodal point
displacements of
00:32:49.590 --> 00:32:53.870
nodal point 6 are U11.
00:32:53.870 --> 00:32:55.180
Let me put them in.
00:32:55.180 --> 00:32:56.760
And U12.
00:32:56.760 --> 00:32:58.220
Why is that the case?
00:32:58.220 --> 00:33:00.970
Well, if we start numbering
here-- and I'm going to
00:33:00.970 --> 00:33:05.770
include now, I'm going to
include now the displacement
00:33:05.770 --> 00:33:08.590
of these nodal points, although
we know that they are
00:33:08.590 --> 00:33:11.200
actually 0, that boundary
condition we can
00:33:11.200 --> 00:33:13.910
impose later on.
00:33:13.910 --> 00:33:20.035
So if I start numbering here, I
would have U1, U2, 3, 4, 5,
00:33:20.035 --> 00:33:24.940
6, 7, 8, 9, 10, 11, 12.
00:33:24.940 --> 00:33:29.990
So the global nodal point
displacements are U11, U12
00:33:29.990 --> 00:33:32.270
corresponding to
nodal point 9.
00:33:32.270 --> 00:33:36.460
And those are the ones that
appear right here
00:33:36.460 --> 00:33:41.400
corresponding to the local
U1 and the local V1.
00:33:41.400 --> 00:33:46.290
Well, if we proceed in the same
way for all of the other
00:33:46.290 --> 00:33:50.700
nodal points for element 2,
and these are the results.
00:33:50.700 --> 00:33:56.810
You can look these up in the
textbook figure 4.6, or in the
00:33:56.810 --> 00:33:58.970
accompanying notes.
00:33:58.970 --> 00:34:05.100
We can directly construct the
H2 matrix for the element.
00:34:05.100 --> 00:34:13.239
And now the H2 matrix, of
course, involves as the vector
00:34:13.239 --> 00:34:19.520
the global nodal point
displacements, U1 to U18.
00:34:19.520 --> 00:34:22.690
The H matrix for element 2.
00:34:22.690 --> 00:34:26.239
And if the 4 elements are
identical, we will have the
00:34:26.239 --> 00:34:30.300
same H matrix for all
the elements.
00:34:30.300 --> 00:34:34.400
The H matrix here looks
as shown here.
00:34:34.400 --> 00:34:36.800
Notice that we have a
one quarter there.
00:34:36.800 --> 00:34:39.139
It's a little arrow here.
00:34:39.139 --> 00:34:41.750
The bracket should've
been there.
00:34:41.750 --> 00:34:42.679
Bracket should have
been there.
00:34:42.679 --> 00:34:45.780
So we have a one quarter
outside of the bracket.
00:34:45.780 --> 00:34:49.679
And we have as the first
entry, this one here.
00:34:49.679 --> 00:34:52.570
This corresponds to U1.
00:34:52.570 --> 00:34:54.600
This is the entry corresponding
to U1.
00:34:54.600 --> 00:34:56.330
I left the entries
corresponding
00:34:56.330 --> 00:34:59.350
to U2, U3, U4 out.
00:34:59.350 --> 00:35:02.030
This is the entry corresponding
to V1.
00:35:02.030 --> 00:35:06.800
And I left out the entries
corresponding to V2, V3, V4.
00:35:06.800 --> 00:35:09.450
This is our H matrix here.
00:35:09.450 --> 00:35:13.200
And now we basically want to
blow it up to obtain the H2
00:35:13.200 --> 00:35:15.920
matrix corresponding
to element 2.
00:35:15.920 --> 00:35:22.020
Well, if we recognize as we just
did that U1, the local
00:35:22.020 --> 00:35:29.910
U1, the local V1, correspond
to capital U11 and capital
00:35:29.910 --> 00:35:34.350
U12, and I just showed you that
that is indeed the case.
00:35:34.350 --> 00:35:38.480
We would simply take the first
element here, this element
00:35:38.480 --> 00:35:42.270
here, and put it right there.
00:35:42.270 --> 00:35:45.990
This element here would
go right there.
00:35:45.990 --> 00:35:50.410
This element here would
go right there.
00:35:50.410 --> 00:35:53.630
And this element here would
go right there.
00:35:53.630 --> 00:35:57.410
In fact, H1 5 is 0 as
you can see here.
00:35:57.410 --> 00:35:58.910
And H2 1 is 0.
00:35:58.910 --> 00:36:01.970
So H1 1 is this term here.
00:36:01.970 --> 00:36:04.960
And H2 5 is that term here.
00:36:04.960 --> 00:36:08.090
Those are two non-zero entries
in the complete H matrix.
00:36:10.590 --> 00:36:13.840
Well, this is then the way how
we actually can proceed to
00:36:13.840 --> 00:36:16.330
construct the H2 matrix.
00:36:16.330 --> 00:36:21.510
I pointed out earlier in lecture
three already that we
00:36:21.510 --> 00:36:24.760
do not perform this step
computationally.
00:36:24.760 --> 00:36:28.040
In fact, what we do
computationally is we stay
00:36:28.040 --> 00:36:29.740
with this H matrix.
00:36:29.740 --> 00:36:33.220
We calculate a compacted
stiffness matrix, a compacted
00:36:33.220 --> 00:36:36.950
load vector, which the stiffness
matrix will be of
00:36:36.950 --> 00:36:38.680
order 8 by 8.
00:36:38.680 --> 00:36:44.390
And then we assemble these
element matrices with
00:36:44.390 --> 00:36:48.040
identification arrays and
connectivity arrays into the
00:36:48.040 --> 00:36:50.540
global structural stiffness
matrix, into
00:36:50.540 --> 00:36:53.690
the global load vectors.
00:36:53.690 --> 00:36:57.650
So we really don't perform
this step of
00:36:57.650 --> 00:37:00.370
blowing up the H matrix.
00:37:00.370 --> 00:37:03.660
And I will show you in the next
lecture how we actually
00:37:03.660 --> 00:37:08.130
perform the computations of
assembling element matrices.
00:37:08.130 --> 00:37:12.420
However, for theoretical
purposes, it is neat if you
00:37:12.420 --> 00:37:17.030
understand how we can construct
this H2 matrix from
00:37:17.030 --> 00:37:18.570
this H matrix.
00:37:18.570 --> 00:37:23.460
And of course, similarly, we
could construct H1, H3, and H4
00:37:23.460 --> 00:37:26.480
from this basic H matrix.
00:37:26.480 --> 00:37:33.180
Once we have done that, we can
directly write that k being
00:37:33.180 --> 00:37:36.060
equal to the sum of the km.
00:37:36.060 --> 00:37:39.670
In other words, the global
structure stiffness matrix is
00:37:39.670 --> 00:37:43.400
equal to the sum of the element
matrices, stiffness
00:37:43.400 --> 00:37:51.490
matrices, where each of these
km's has the same order as the
00:37:51.490 --> 00:37:52.960
structure stiffness matrix k.
00:37:56.690 --> 00:38:02.810
Let us now look briefly at the
development of the strain
00:38:02.810 --> 00:38:04.390
displacement matrix.
00:38:04.390 --> 00:38:08.430
In plane stress conditions,
we have three entries.
00:38:08.430 --> 00:38:10.590
The strains are given
as shown here.
00:38:10.590 --> 00:38:13.460
Elementary strengths materials
tells us that
00:38:13.460 --> 00:38:15.210
these are the strains.
00:38:15.210 --> 00:38:21.480
And I showed earlier that the
B matrix is constructed by
00:38:21.480 --> 00:38:23.950
multiplying the E times
the A inverse.
00:38:23.950 --> 00:38:26.950
The A inverse was already used
in the calculation of the
00:38:26.950 --> 00:38:30.060
displacement interpolation
matrix.
00:38:30.060 --> 00:38:36.630
The E matrix is obtained by
appropriately differentiating
00:38:36.630 --> 00:38:43.110
the entries in the displacement
interpolations.
00:38:43.110 --> 00:38:50.970
And this differentiation is
performed on the phi matrix
00:38:50.970 --> 00:38:52.240
right here.
00:38:52.240 --> 00:38:55.750
Or rather, on these
entries here.
00:38:55.750 --> 00:38:59.650
By simply differentiating these
entries here, we would
00:38:59.650 --> 00:39:06.390
obtain the strains and writing
those in matrix form, we
00:39:06.390 --> 00:39:09.940
obtain the E matrix here.
00:39:09.940 --> 00:39:16.000
Once we have the B matrix, which
in this particular case
00:39:16.000 --> 00:39:22.390
is 3 by 8 matrix for a typical
element, we can, of course,
00:39:22.390 --> 00:39:29.080
construct the B2 matrix for
element 2 just in the same way
00:39:29.080 --> 00:39:31.000
as the H2 matrix.
00:39:31.000 --> 00:39:33.190
Of course, we would again,
blow up the B
00:39:33.190 --> 00:39:35.205
matrix from a 3 to 8.
00:39:35.205 --> 00:39:40.400
3 by 8 I should rather say, to
a 3 by 18 now because we have
00:39:40.400 --> 00:39:42.360
actually 18 degrees
of freedom in the
00:39:42.360 --> 00:39:44.910
complete structural model.
00:39:44.910 --> 00:39:47.190
This then completes what I
wanted to say about the
00:39:47.190 --> 00:39:49.830
construction of the
displacement and
00:39:49.830 --> 00:39:51.960
strain-displacement
interpolation matrix for the
00:39:51.960 --> 00:39:55.470
element, the H2 and
B2 matrices.
00:39:55.470 --> 00:39:59.380
However, let me just spend
a few moments on the
00:39:59.380 --> 00:40:02.680
construction of the surface
displacement
00:40:02.680 --> 00:40:03.880
interpolation matrix.
00:40:03.880 --> 00:40:06.980
In other words, the
interpolation of the
00:40:06.980 --> 00:40:10.270
displacements along the
surface of an element.
00:40:10.270 --> 00:40:14.860
Here, assume for example, that
this is a surface of the
00:40:14.860 --> 00:40:16.890
elements that we are
concerned with.
00:40:16.890 --> 00:40:19.890
And that there would be some
loading applied to that
00:40:19.890 --> 00:40:21.510
surface of the element.
00:40:21.510 --> 00:40:26.080
So that we need, in other words,
the HSM matrix the way
00:40:26.080 --> 00:40:28.650
I have been pointing
it out earlier.
00:40:28.650 --> 00:40:33.300
What we would do to obtain this
matrix here, is really
00:40:33.300 --> 00:40:38.330
simply use our earlier
results for the HS--
00:40:38.330 --> 00:40:41.890
sorry, for the HM matrix.
00:40:41.890 --> 00:40:46.830
And here we have the
H matrix and the H2
00:40:46.830 --> 00:40:49.380
matrix for element 2.
00:40:49.380 --> 00:40:52.570
What we would do is we would use
this matrix, or this one
00:40:52.570 --> 00:40:55.820
of course, this matrix is
contained in this one.
00:40:55.820 --> 00:41:00.830
And simply, substitute the
particular coordinates along
00:41:00.830 --> 00:41:01.790
the surface.
00:41:01.790 --> 00:41:04.440
Now, going back once more
to the element under
00:41:04.440 --> 00:41:05.950
consideration.
00:41:05.950 --> 00:41:10.380
If we are looking at this
surface, then y is constant
00:41:10.380 --> 00:41:13.700
and equal to this
distance here.
00:41:13.700 --> 00:41:19.060
So knowing y, we would simply
substitute that value of y
00:41:19.060 --> 00:41:27.480
into this function here, and
thus obtain the HS matrix.
00:41:27.480 --> 00:41:31.480
So, in general, therefore the
HS matrix is simply obtained
00:41:31.480 --> 00:41:36.700
from the H matrix, or the HS2
matrix here would simply be
00:41:36.700 --> 00:41:42.050
obtained from the H2 matrix by
substituting the coordinates
00:41:42.050 --> 00:41:46.470
along the line, or along
the surface that we are
00:41:46.470 --> 00:41:47.720
considering.
00:41:49.410 --> 00:41:53.860
Let us now to look at the
overall process of the finite
00:41:53.860 --> 00:41:55.820
element solution.
00:41:55.820 --> 00:41:59.990
Here I have summarized what we
all are concerned about.
00:41:59.990 --> 00:42:02.800
Of course we have the actual
physical problem that we want
00:42:02.800 --> 00:42:04.350
to analyze.
00:42:04.350 --> 00:42:08.450
We have a geometric domain that
we have to discretize.
00:42:08.450 --> 00:42:11.270
A material that has to be
represented, a loading and the
00:42:11.270 --> 00:42:12.750
boundary conditions.
00:42:12.750 --> 00:42:16.000
All of those have to
be represented.
00:42:16.000 --> 00:42:20.150
In the mechanic idealization,
the idealization step that is
00:42:20.150 --> 00:42:23.870
a very important step that I
referred to earlier, we would
00:42:23.870 --> 00:42:26.890
idealize the kinematics as being
a truss, plane stress,
00:42:26.890 --> 00:42:29.930
three-dimensional, Kirchhoff
plate, kinematics.
00:42:29.930 --> 00:42:34.000
The material isotopic elastic,
the loading, and the boundary
00:42:34.000 --> 00:42:34.910
conditions.
00:42:34.910 --> 00:42:37.560
And this mechanical idealization
really gives us a
00:42:37.560 --> 00:42:41.180
differential equation
or equilibrium for
00:42:41.180 --> 00:42:45.160
one-dimensional stress
situation, a bar.
00:42:45.160 --> 00:42:49.190
We will have an equation
such as this one.
00:42:49.190 --> 00:42:52.420
In the finite element solution
then, we operate on this
00:42:52.420 --> 00:42:53.810
equation basically.
00:42:53.810 --> 00:42:57.840
We want to solve the governing
differential equation of the
00:42:57.840 --> 00:43:00.920
mechanical idealization.
00:43:00.920 --> 00:43:05.340
The errors that we are
therefore, concerned with are
00:43:05.340 --> 00:43:08.580
discretization errors by the
use of the finite element
00:43:08.580 --> 00:43:09.840
interpolations.
00:43:09.840 --> 00:43:13.070
And then, a number
of other errors.
00:43:13.070 --> 00:43:14.550
Numerical integration
in space.
00:43:14.550 --> 00:43:16.990
If we use numerical integration,
there would be
00:43:16.990 --> 00:43:18.530
other errors introduced.
00:43:18.530 --> 00:43:21.270
In the evaluation of the
constitutive relations, again,
00:43:21.270 --> 00:43:22.800
errors would be introduced.
00:43:22.800 --> 00:43:25.380
The solution of the dynamic
equilibrium equations, by
00:43:25.380 --> 00:43:26.810
[UNINTELLIGIBLE] time
integration, mode
00:43:26.810 --> 00:43:29.890
superposition further errors
would be introduced.
00:43:29.890 --> 00:43:32.780
If you solved the governing
equilibrium equations, ku
00:43:32.780 --> 00:43:36.180
equals r, by iteration there
would be further errors
00:43:36.180 --> 00:43:37.120
introduced.
00:43:37.120 --> 00:43:40.130
And of course, finally there
are round-off errors on the
00:43:40.130 --> 00:43:41.270
digital computer.
00:43:41.270 --> 00:43:46.720
What I want to do now in
discussing convergence, I want
00:43:46.720 --> 00:43:50.260
to say that all of these
errors are negligible.
00:43:50.260 --> 00:43:53.770
In other words, we are basically
evaluating the
00:43:53.770 --> 00:43:56.380
integrations in space exactly.
00:43:56.380 --> 00:43:59.600
There is no round-off because we
have an infinite precision
00:43:59.600 --> 00:44:01.200
machine, and so on.
00:44:01.200 --> 00:44:04.760
So all the errors that we are
looking at now are the
00:44:04.760 --> 00:44:06.500
discretization errors,
00:44:06.500 --> 00:44:08.860
And if we look only at
the discretization
00:44:08.860 --> 00:44:12.210
errors, we can say some--
00:44:12.210 --> 00:44:15.540
make some very fundamental
remarks about the convergence
00:44:15.540 --> 00:44:18.310
of the finite elements scheme.
00:44:18.310 --> 00:44:22.330
I've summarized here on the
blackboard some of these very
00:44:22.330 --> 00:44:24.200
important concepts.
00:44:24.200 --> 00:44:26.300
When we talk about convergence,
let us assume
00:44:26.300 --> 00:44:30.560
first now that we are having a
compatible element layout.
00:44:30.560 --> 00:44:33.270
In other words, we're using a
compatible element layout.
00:44:33.270 --> 00:44:37.480
And then, we know that we have
monotonic convergence to the
00:44:37.480 --> 00:44:40.410
solution of the
problem-governing differential
00:44:40.410 --> 00:44:45.460
equations provided the elements
that we are using
00:44:45.460 --> 00:44:46.420
contain:
00:44:46.420 --> 00:44:49.800
Firstly, all required
rigid body modes.
00:44:49.800 --> 00:44:52.950
This means, for example, that
for in a plane stress
00:44:52.950 --> 00:44:56.850
analysis, the element must be
able to undergo three rigid
00:44:56.850 --> 00:45:00.230
body modes, two translations,
horizontally and vertically,
00:45:00.230 --> 00:45:03.040
and a rigid body rotation.
00:45:03.040 --> 00:45:07.100
And the element must be able
to represent the required
00:45:07.100 --> 00:45:08.790
constant strain states.
00:45:08.790 --> 00:45:12.220
In a plane stress analysis,
three constant strain states.
00:45:12.220 --> 00:45:17.170
Pulling this way, pulling that
way, and the constant shear.
00:45:17.170 --> 00:45:20.600
If these two conditions are
satisfied, we have monotonic
00:45:20.600 --> 00:45:23.790
convergence in a compatible
element layout.
00:45:23.790 --> 00:45:26.380
What do we mean by a compatible
element layout?
00:45:26.380 --> 00:45:30.930
Well, here I have show a small
picture showing two elements,
00:45:30.930 --> 00:45:32.470
a white and an orange element.
00:45:32.470 --> 00:45:37.190
And these elements are
compatible because we have
00:45:37.190 --> 00:45:41.050
three nodes along this line
for each of the elements.
00:45:41.050 --> 00:45:44.440
So the white element
displacements basically vary
00:45:44.440 --> 00:45:48.040
parabolically along this line,
and so do the orange element
00:45:48.040 --> 00:45:49.820
displacements.
00:45:49.820 --> 00:45:53.460
No gap can develop between
these two lines.
00:45:53.460 --> 00:45:55.950
Here we have an incompatible
element layout.
00:45:55.950 --> 00:45:59.940
Two nodes that describe only
a linear variation
00:45:59.940 --> 00:46:02.370
for the white element.
00:46:02.370 --> 00:46:06.760
But three nodes for the orange
element describing a parabolic
00:46:06.760 --> 00:46:08.010
variation in displacement.
00:46:08.010 --> 00:46:11.660
Therefore a gap can open up and
this is an incompatible
00:46:11.660 --> 00:46:13.740
element layout.
00:46:13.740 --> 00:46:17.090
Well, let us assume then that
we have a compatible element
00:46:17.090 --> 00:46:18.940
layout fist of all.
00:46:18.940 --> 00:46:22.260
Then, typically for an analysis
such as this one, a
00:46:22.260 --> 00:46:26.230
simple cantilever subjected to
a tip load, the convergence
00:46:26.230 --> 00:46:29.140
that we would observe would
be the following.
00:46:29.140 --> 00:46:32.240
As we increase the number of
elements that idealize the
00:46:32.240 --> 00:46:36.490
cantilever, the displacement
measured here would
00:46:36.490 --> 00:46:41.700
monotonically approach from
below the exact displacement.
00:46:41.700 --> 00:46:44.130
What do we mean by exact
displacement?
00:46:44.130 --> 00:46:46.220
Notice I put it into quotes.
00:46:46.220 --> 00:46:48.940
Well, the exact displacement
that I'm talking about here is
00:46:48.940 --> 00:46:52.190
the tip displacements that we
would calculate from the
00:46:52.190 --> 00:46:54.580
problem-governing differential
equation.
00:46:57.670 --> 00:46:59.870
Of course, that
problem-governing differential
00:46:59.870 --> 00:47:03.940
equation also corresponds to a
mechanical idealization of the
00:47:03.940 --> 00:47:06.840
actual physical situation.
00:47:06.840 --> 00:47:09.050
And depending on which
problem-governing differential
00:47:09.050 --> 00:47:13.230
equation we choose, we would
have a different dashed line.
00:47:13.230 --> 00:47:16.850
But assuming now that we have
made up our mind which one to
00:47:16.850 --> 00:47:19.945
choose, which mechanical
idealization to choose, and we
00:47:19.945 --> 00:47:23.760
are choosing a corresponding
finite element idealization,
00:47:23.760 --> 00:47:26.890
this would be the convergence
that we would observe.
00:47:26.890 --> 00:47:29.630
And it would be monotonically
from below.
00:47:29.630 --> 00:47:31.560
In other words, our finite
element model,
00:47:31.560 --> 00:47:33.210
actually is too stiff.
00:47:33.210 --> 00:47:36.860
It is stiffer than the actual
physical structure.
00:47:36.860 --> 00:47:41.300
Well, why I think we can
intuitively understand that
00:47:41.300 --> 00:47:44.190
quite well if we just recognize
that we are
00:47:44.190 --> 00:47:50.040
constraining the displacements
within each element to only be
00:47:50.040 --> 00:47:57.130
able to basically, take patterns
on that are contained
00:47:57.130 --> 00:47:57.840
in the element.
00:47:57.840 --> 00:48:01.155
In other words, for the four
nodal element here, we really
00:48:01.155 --> 00:48:05.110
have for the U displacements and
for the V displacements,
00:48:05.110 --> 00:48:08.560
only 4 displacement
patterns each that
00:48:08.560 --> 00:48:09.840
the element can undergo.
00:48:09.840 --> 00:48:13.660
So we are constraining the
exact displacements to be
00:48:13.660 --> 00:48:18.820
really, approximated by these
displacement patterns.
00:48:18.820 --> 00:48:21.940
And that makes the actual finite
element model stiffer
00:48:21.940 --> 00:48:25.350
than the actual physical
structure.
00:48:25.350 --> 00:48:28.370
If we have an incompatible
layout, and I showed earlier
00:48:28.370 --> 00:48:32.190
what we mean by that,
then, in addition,
00:48:32.190 --> 00:48:34.830
every patch of elements--
00:48:34.830 --> 00:48:38.490
by patch of elements we mean
small assemblage of elements--
00:48:38.490 --> 00:48:42.010
must be able to represent the
constant strain states.
00:48:42.010 --> 00:48:44.150
It's not just that each element
must be able to
00:48:44.150 --> 00:48:46.090
represent the constant
strain state, but
00:48:46.090 --> 00:48:47.990
each patch of elements.
00:48:47.990 --> 00:48:53.250
Also, then if this holds true,
then we have convergence.
00:48:53.250 --> 00:48:55.640
But non-monotonic convergence.
00:48:55.640 --> 00:48:58.940
By non-monotonic convergence,
I mean the following.
00:48:58.940 --> 00:49:02.140
We might get this result for
a certain set of elements.
00:49:02.140 --> 00:49:04.760
And as we increase the number
of elements, we are
00:49:04.760 --> 00:49:09.450
oscillating about the correct
result, the exact result in
00:49:09.450 --> 00:49:12.570
quotes again, and finally
we converge.
00:49:12.570 --> 00:49:16.270
This is non-monotonic
convergence because we can
00:49:16.270 --> 00:49:20.700
have solutions lying on either
side of the exact result.
00:49:20.700 --> 00:49:31.490
Well, let me now show to you
some of the facts that I just
00:49:31.490 --> 00:49:33.900
talked about on this
view graph.
00:49:33.900 --> 00:49:35.480
What do these mean?
00:49:35.480 --> 00:49:39.770
Well, I said already that each
element in every case must be
00:49:39.770 --> 00:49:43.480
able to represent the
rigid body modes.
00:49:43.480 --> 00:49:46.910
Here for a plane stress analysis
of a cantilever beam
00:49:46.910 --> 00:49:49.790
shown here, we have 3
rigid body modes.
00:49:49.790 --> 00:49:53.430
And as I mentioned earlier
briefly already, the element
00:49:53.430 --> 00:49:57.260
must be able to go over, go up
as a rigid body and rotate as
00:49:57.260 --> 00:49:59.370
a rigid body.
00:49:59.370 --> 00:50:00.610
Why is this the case?
00:50:00.610 --> 00:50:03.060
Well, if we look at an analysis
of the cantilever
00:50:03.060 --> 00:50:06.280
where we have not a tip load,
but a load say, at this end
00:50:06.280 --> 00:50:09.650
here, we know that the bending
moment looks this way.
00:50:09.650 --> 00:50:16.700
Hence, these elements over here
must be able to freely
00:50:16.700 --> 00:50:18.790
rotate and displace.
00:50:18.790 --> 00:50:22.940
In other words, we know that the
exact condition, the exact
00:50:22.940 --> 00:50:25.630
physical condition, there should
be no stress here.
00:50:25.630 --> 00:50:29.450
The stress should be
0 in this element.
00:50:29.450 --> 00:50:33.600
And this means that the tip
element here must be able to
00:50:33.600 --> 00:50:35.780
translate freely as
a rigid body and
00:50:35.780 --> 00:50:37.340
rotate as a rigid body.
00:50:37.340 --> 00:50:39.040
If this were not the
case, then we would
00:50:39.040 --> 00:50:40.600
get stresses here.
00:50:40.600 --> 00:50:44.640
And of course, our analysis or
finite element model can never
00:50:44.640 --> 00:50:48.250
predict the actual physical
conditions accurately or
00:50:48.250 --> 00:50:52.450
closely, even if we take an
infinite number of elements.
00:50:52.450 --> 00:50:56.860
So this somehow explains to you
intuitively why the rigid
00:50:56.860 --> 00:50:59.330
body more criteria must
be satisfied.
00:50:59.330 --> 00:51:02.260
How about the constant strain
state criteria?
00:51:02.260 --> 00:51:06.940
Well, if we assume that we take
more and more elements--
00:51:06.940 --> 00:51:10.820
and let me draw in here
such elements.
00:51:10.820 --> 00:51:14.020
For example, now we have gone
from a certain number of
00:51:14.020 --> 00:51:17.940
elements to a larger
number of elements.
00:51:17.940 --> 00:51:21.400
These elements, as we keep on
refining the mesh, would
00:51:21.400 --> 00:51:23.290
become very small, and
would go down.
00:51:23.290 --> 00:51:25.810
In fact, each element would
go down to a point.
00:51:25.810 --> 00:51:29.860
But at a point we only have one
constant stress basically.
00:51:29.860 --> 00:51:33.310
So the element must be able in
the limit, to represent the
00:51:33.310 --> 00:51:36.240
constant stress at a point.
00:51:36.240 --> 00:51:40.270
And therefore, we have the
constant stress criteria also.
00:51:40.270 --> 00:51:44.330
One way of finding out whether
an element satisfies these
00:51:44.330 --> 00:51:48.180
criteria, the constant stress
criteria and the rigid body
00:51:48.180 --> 00:51:50.580
mode criteria is to calculate
its eigenvalues.
00:51:50.580 --> 00:51:52.260
Here, we have taken ones.
00:51:52.260 --> 00:51:54.770
A four node plane stress element
and calculate the
00:51:54.770 --> 00:51:55.560
eigenvalues.
00:51:55.560 --> 00:52:00.790
Notice the lowest three
eigenvalues are 0,
00:52:00.790 --> 00:52:03.710
representing the rigid
body modes.
00:52:03.710 --> 00:52:07.600
And then, we have a flexural
mode here, the fourth
00:52:07.600 --> 00:52:09.310
eigenvalue.
00:52:09.310 --> 00:52:12.170
The fifth eigenvalue is again
the flexural mode.
00:52:12.170 --> 00:52:14.610
And here is the constant
sharing mode.
00:52:14.610 --> 00:52:17.500
And here are the two
constant, one
00:52:17.500 --> 00:52:19.350
compression and tension modes.
00:52:19.350 --> 00:52:23.310
So these are the constant strain
states that the element
00:52:23.310 --> 00:52:24.800
can represent therefore.
00:52:24.800 --> 00:52:29.800
And earlier I showed you the
three rigid body modes.
00:52:29.800 --> 00:52:34.660
So surely, this element
therefore satisfies both of
00:52:34.660 --> 00:52:36.890
the criteria.
00:52:36.890 --> 00:52:43.770
Let's look now at what happens
when we don't have a
00:52:43.770 --> 00:52:46.220
compatible mesh.
00:52:46.220 --> 00:52:49.600
We have an additional condition
that every patch of
00:52:49.600 --> 00:52:52.810
elements must also be able
to represent the
00:52:52.810 --> 00:52:55.030
constant strain states.
00:52:55.030 --> 00:52:58.000
Only then we will have
convergence.
00:52:58.000 --> 00:53:00.820
But we will not necessarily have
monotonic convergence.
00:53:00.820 --> 00:53:03.870
We can only talk about
non-monotonic convergence.
00:53:03.870 --> 00:53:08.150
Well, here I've taken
a patch of elements.
00:53:08.150 --> 00:53:12.640
Each element by itself satisfies
the constant strain
00:53:12.640 --> 00:53:15.890
criterion and the rigid
body more criterion.
00:53:15.890 --> 00:53:21.620
In this case here, we use a
compatible element layout.
00:53:21.620 --> 00:53:26.440
Compatible because along each
side, the adjacent elements
00:53:26.440 --> 00:53:29.210
shares the nodes.
00:53:29.210 --> 00:53:32.410
And we are subjecting this patch
of elements to a total
00:53:32.410 --> 00:53:36.180
stress here of 100 Newton
per square centimeters.
00:53:36.180 --> 00:53:38.880
Of course, the stress in each
of these elements would be
00:53:38.880 --> 00:53:42.880
simply a stress this
direction of 1,000.
00:53:42.880 --> 00:53:45.690
Excuse me, 1,000 not 100.
00:53:45.690 --> 00:53:49.570
The other stress, the vertical
stress and the shear stress is
00:53:49.570 --> 00:53:52.240
0 for this patch of elements.
00:53:52.240 --> 00:53:54.840
Let us now perform the
following experiment.
00:53:54.840 --> 00:54:00.530
Let us say that we are assigning
two nodes, two
00:54:00.530 --> 00:54:05.570
different nodes to element
4 and 5 at this location.
00:54:05.570 --> 00:54:08.570
Here they share the same node.
00:54:08.570 --> 00:54:10.890
Let us do the same
at this end.
00:54:10.890 --> 00:54:14.090
And let us say that node
17 only belongs to
00:54:14.090 --> 00:54:16.160
this element 4.
00:54:16.160 --> 00:54:18.730
Node 18 belongs only
to element 6.
00:54:18.730 --> 00:54:22.170
Node 20 and 19 belong
to element 5.
00:54:22.170 --> 00:54:28.130
In other words, node 17 can
basically gives this
00:54:28.130 --> 00:54:29.080
displacement pattern.
00:54:29.080 --> 00:54:30.950
And 19 gives this displacement
pattern.
00:54:30.950 --> 00:54:33.880
An incompatibility
can reside here.
00:54:33.880 --> 00:54:35.940
And similarly, along
this line.
00:54:35.940 --> 00:54:41.910
If we subject this patch of
elements to the stress state
00:54:41.910 --> 00:54:45.920
of the external forces that we
used here too, we would find
00:54:45.920 --> 00:54:50.230
that the stresses at the points
A, B, C, D, E shown
00:54:50.230 --> 00:54:53.580
along here, which here were,
of course, exactly equal to
00:54:53.580 --> 00:55:01.530
1,000 for this stress tau yy
and 0 in this direction.
00:55:01.530 --> 00:55:02.820
And the sheer stress was 0.
00:55:02.820 --> 00:55:07.480
If you look at the tau yy
at A, B, C, D, E now the
00:55:07.480 --> 00:55:11.540
incompatible element mesh
layout, we would find the
00:55:11.540 --> 00:55:13.450
following stress result.
00:55:13.450 --> 00:55:16.380
Notice that the stresses
now are not 1,000.
00:55:16.380 --> 00:55:18.930
In fact, they vary drastically
from 1,000.
00:55:18.930 --> 00:55:22.300
359 instead of 1,000.
00:55:22.300 --> 00:55:27.480
This means that this patch
of elements with the
00:55:27.480 --> 00:55:32.820
incompatibility here, does not
represent exactly the constant
00:55:32.820 --> 00:55:36.560
stress state that it should
represent because we are
00:55:36.560 --> 00:55:40.080
subjecting it to a constant
stress state.
00:55:40.080 --> 00:55:43.980
Therefore, this patch of element
does not satisfy the
00:55:43.980 --> 00:55:47.500
requirement for convergence.
00:55:47.500 --> 00:55:51.450
Not monotonic convergence and
not non-monotonic convergence.
00:55:51.450 --> 00:55:57.350
It just should not be used in an
actual practical analysis.
00:55:57.350 --> 00:56:01.610
This completes what I wanted to
say then in this lecture.
00:56:01.610 --> 00:56:03.310
Thank you very much for
your attention.