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PROFESSOR: Ladies and gentlemen,
welcome to lecture
00:00:23.780 --> 00:00:25.420
number six.
00:00:25.420 --> 00:00:27.840
In this lecture I would like
to discuss with you the
00:00:27.840 --> 00:00:33.160
formulation and calculation of
isoparametric finite elements.
00:00:33.160 --> 00:00:36.860
We considered earlier already,
in lecture four, the
00:00:36.860 --> 00:00:39.300
calculation all finite
element matrices.
00:00:39.300 --> 00:00:42.050
But in that lecture we
considered the generalized
00:00:42.050 --> 00:00:45.930
coordinate finite
element models.
00:00:45.930 --> 00:00:49.020
The generalized coordinate
finite element models were the
00:00:49.020 --> 00:00:52.330
first finite elements derived.
00:00:52.330 --> 00:00:56.110
However, I now want to discuss
with you a more general
00:00:56.110 --> 00:01:00.210
approach of deriving the
required interpolation
00:01:00.210 --> 00:01:02.450
matrices and element matrices.
00:01:02.450 --> 00:01:07.560
And this more general approach
is the isoparametric finite
00:01:07.560 --> 00:01:09.462
element derivation.
00:01:09.462 --> 00:01:12.360
The isoparametric finite
elements that I will be
00:01:12.360 --> 00:01:15.750
discussing in this and the
next lecture are, in my
00:01:15.750 --> 00:01:19.920
opinion, the most effective
elements currently available
00:01:19.920 --> 00:01:24.320
for plane stress, plane strain,
axisymmetric analysis,
00:01:24.320 --> 00:01:28.750
three dimensional analysis,
thick and thin shell analysis.
00:01:28.750 --> 00:01:32.160
These elements are being used,
for example, in the computer
00:01:32.160 --> 00:01:33.810
program [? Aldena ?].
00:01:33.810 --> 00:01:37.050
They are also used in other
computer programs and
00:01:37.050 --> 00:01:39.790
represent a modern approach
to the solution
00:01:39.790 --> 00:01:42.920
off structure problems.
00:01:42.920 --> 00:01:46.270
In this lecture, I would like
to talk about the derivation
00:01:46.270 --> 00:01:49.470
of continuum elements.
00:01:49.470 --> 00:01:52.640
In the next lecture we will talk
about the derivation of
00:01:52.640 --> 00:01:55.840
structural elements, [INAUDIBLE]
and central
00:01:55.840 --> 00:01:58.190
elements, beam elements.
00:01:58.190 --> 00:02:01.510
The basic concept of
isoparametric finite element
00:02:01.510 --> 00:02:05.830
analysis, is that we interpolate
the geometry of an
00:02:05.830 --> 00:02:12.410
element, and the displacements
of an element in
00:02:12.410 --> 00:02:15.360
exactly the same way.
00:02:15.360 --> 00:02:18.730
Let us look at the geometry
interpolation.
00:02:18.730 --> 00:02:22.290
Here you see for three
dimensional analysis, the
00:02:22.290 --> 00:02:26.640
interpolation of the
x-coordinate within the
00:02:26.640 --> 00:02:32.560
element, where we use
interpolation functions h i.
00:02:32.560 --> 00:02:35.600
Of course these interpolation
functions are unknown and I
00:02:35.600 --> 00:02:37.840
will have to show you
how we derive
00:02:37.840 --> 00:02:40.390
them for certain elements.
00:02:40.390 --> 00:02:43.306
The x i as a nodal
point coordinate.
00:02:47.860 --> 00:02:52.870
The x i value for i equals 1,
for example, is nothing else
00:02:52.870 --> 00:02:58.160
than the x-coordinate
of nodal point one.
00:02:58.160 --> 00:02:58.780
And so on.
00:02:58.780 --> 00:03:03.330
So the x i, the nodal point
coordinates for all nodes,
00:03:03.330 --> 00:03:06.560
they are given, they are input
in the analysis the way we
00:03:06.560 --> 00:03:09.110
have been discussing it in
the previous lecture.
00:03:09.110 --> 00:03:16.190
And if we know the h i, we have
a direct relationship for
00:03:16.190 --> 00:03:21.580
the x-coordinates within the
element, as a function of the
00:03:21.580 --> 00:03:24.460
nodal point coordinates.
00:03:24.460 --> 00:03:29.210
Again, I have to show you how
we obtain the h i functions
00:03:29.210 --> 00:03:32.300
for the various elements
that we are using.
00:03:32.300 --> 00:03:36.310
Similarly, for the y
interpolation and similarly
00:03:36.310 --> 00:03:37.560
for the z interpolation.
00:03:40.550 --> 00:03:45.500
Having derived the h i functions
we are using in the
00:03:45.500 --> 00:03:51.760
isoparametric finite element,
the same functions also to
00:03:51.760 --> 00:03:54.300
interpolate the displacements.
00:03:54.300 --> 00:04:00.700
Notice that we're having here,
the nodal point displacements
00:04:00.700 --> 00:04:07.980
u i, v i, and w i, and here we
have the same interpolation
00:04:07.980 --> 00:04:11.560
functions that we already
used in the coordinate
00:04:11.560 --> 00:04:13.810
interpolations.
00:04:13.810 --> 00:04:18.680
And if the number of nodes
that are used in the
00:04:18.680 --> 00:04:20.420
description of the element--
00:04:20.420 --> 00:04:24.550
and, in fact, we will see that N
can be a variable, it can be
00:04:24.550 --> 00:04:30.870
equal to three, four, five, up
to a large number of nodes.
00:04:30.870 --> 00:04:36.870
And in practice we generally
don't go much further than 60.
00:04:36.870 --> 00:04:40.030
I mentioned that I want to
discuss in this lecture
00:04:40.030 --> 00:04:41.180
continuum elements.
00:04:41.180 --> 00:04:46.730
Well, the continuum elements
that we addressing ourselves
00:04:46.730 --> 00:04:51.190
to are the truss element, the
two d elements-- plane stress,
00:04:51.190 --> 00:04:54.320
plane strain, and axisymmetric
elements--
00:04:54.320 --> 00:04:57.550
and then, the 3 d elements for
three dimensional and thick
00:04:57.550 --> 00:04:59.640
shell analysis.
00:04:59.640 --> 00:05:01.700
These are continuum elements.
00:05:01.700 --> 00:05:06.720
We call them continuum elements,
because we only use
00:05:06.720 --> 00:05:11.940
u, here u and v, and
here, u v and w--
00:05:11.940 --> 00:05:12.740
the displacements--
00:05:12.740 --> 00:05:17.270
to describe the internal
element displacements.
00:05:17.270 --> 00:05:21.440
We're only using the nodal point
displacements u, u v,
00:05:21.440 --> 00:05:25.770
and u v w, for all of the nodal
points to describe the
00:05:25.770 --> 00:05:29.080
internal element
displacements.
00:05:29.080 --> 00:05:34.080
Structure elements, on the
other hand, also use the
00:05:34.080 --> 00:05:39.360
rotations at the nodal point
theta x, theta y, and theta z,
00:05:39.360 --> 00:05:40.940
in order to describe the
00:05:40.940 --> 00:05:43.050
deformations within the element.
00:05:43.050 --> 00:05:45.730
And I will be discussing
structural elements in the
00:05:45.730 --> 00:05:49.620
next lecture, so, for the
moment, we do not have any
00:05:49.620 --> 00:05:51.150
rotations at the nodes.
00:05:51.150 --> 00:05:57.730
Or what we will allow, u v and
w displacements at the nodes.
00:05:57.730 --> 00:06:03.130
Typical examples are given
on this Viewgraph.
00:06:03.130 --> 00:06:05.160
These are elements that are
available in the computer
00:06:05.160 --> 00:06:06.980
program [? Aldena ?].
00:06:06.980 --> 00:06:12.910
Here we have a truss element,
a two-noded truss element.
00:06:12.910 --> 00:06:17.480
The only displacement of
interest in the truss is the
00:06:17.480 --> 00:06:20.350
actual displacement here.
00:06:20.350 --> 00:06:24.390
Here we have a three-noded
truss, or cable element.
00:06:24.390 --> 00:06:27.470
Here we have a set of two
dimensional elements.
00:06:27.470 --> 00:06:31.120
Notice we have triangular
elements here, a triangular
00:06:31.120 --> 00:06:32.100
element here.
00:06:32.100 --> 00:06:35.640
We have here an eight node
element that is curved.
00:06:35.640 --> 00:06:37.920
We can construct curved
elements in the
00:06:37.920 --> 00:06:39.420
isoparametric approach.
00:06:39.420 --> 00:06:41.980
And here we have a rectangular,
eight node
00:06:41.980 --> 00:06:45.290
element, straight sides,
in other words.
00:06:45.290 --> 00:06:47.950
These elements are used in plane
stress, plane strain,
00:06:47.950 --> 00:06:49.295
and axisymmetric analysis.
00:06:52.100 --> 00:06:55.540
In three dimensional analysis
we might be talking about--
00:06:55.540 --> 00:06:57.050
we might be using--
00:06:57.050 --> 00:07:00.550
such elements such as the
eight node brick.
00:07:00.550 --> 00:07:06.940
Each node now has three
displacements, u, v, and w.
00:07:06.940 --> 00:07:10.190
This is here a higher order
element, where we have, in
00:07:10.190 --> 00:07:14.430
addition to the corner nodes,
we also have mid-side nodes.
00:07:14.430 --> 00:07:16.250
You don't need to
have a mid-side
00:07:16.250 --> 00:07:18.060
nodes along all sides.
00:07:18.060 --> 00:07:20.050
For example here,
I did not put a
00:07:20.050 --> 00:07:21.960
mid-side node as an example.
00:07:21.960 --> 00:07:25.610
In fact we could simply have
mid-side nodes here, and no
00:07:25.610 --> 00:07:28.030
mid-side nodes anywhere else.
00:07:28.030 --> 00:07:30.670
I will show you how we
deconstruct these
00:07:30.670 --> 00:07:35.500
interpolation matrices in
the next few minutes.
00:07:35.500 --> 00:07:41.220
Let us consider, as
a very simple
00:07:41.220 --> 00:07:44.500
example, a special case.
00:07:44.500 --> 00:07:48.510
And the special case that I
would like to look at is a
00:07:48.510 --> 00:07:52.940
truss element, which
is two units long.
00:07:52.940 --> 00:07:57.170
In other words, the length from
here to there, is equal
00:07:57.170 --> 00:08:01.720
to 1, and similarly the length
from here to there, is also
00:08:01.720 --> 00:08:02.970
equal to one.
00:08:05.460 --> 00:08:11.700
I'm describing the element with
a coordinate r that I set
00:08:11.700 --> 00:08:16.140
equal to zero at the midpoint
of the element, plus one at
00:08:16.140 --> 00:08:19.570
the right end and minus
one at the left end.
00:08:19.570 --> 00:08:24.350
This is special geometry and
we will later on have to
00:08:24.350 --> 00:08:29.240
generalize our approach to more
general geometries where
00:08:29.240 --> 00:08:31.750
the length is not equal
to two and the element
00:08:31.750 --> 00:08:33.720
might even be curved.
00:08:33.720 --> 00:08:35.880
But for instructive purposes,
let's look at
00:08:35.880 --> 00:08:38.250
this geometry first.
00:08:38.250 --> 00:08:43.330
Similarly, for two dimensional
analysis, I want to look at an
00:08:43.330 --> 00:08:47.940
element that has length two this
direction and length two
00:08:47.940 --> 00:08:50.050
that direction.
00:08:50.050 --> 00:08:53.390
In other words, a generalization
of the concept
00:08:53.390 --> 00:08:55.920
that we just looked
at for the truss.
00:08:55.920 --> 00:08:59.660
Now we have two dimensions and
in each dimension we have a
00:08:59.660 --> 00:09:01.060
length of two.
00:09:01.060 --> 00:09:06.690
I embed into this element
an rs-coordinate system.
00:09:06.690 --> 00:09:12.710
And this coordinate system now
has its origin here at r
00:09:12.710 --> 00:09:17.130
equals zero and s equals zero.
00:09:17.130 --> 00:09:22.310
The r-axis bisects this
side, and the s-axis
00:09:22.310 --> 00:09:25.560
bisects this side.
00:09:25.560 --> 00:09:29.280
Notice that this is a coordinate
system that is
00:09:29.280 --> 00:09:31.670
embedded into the element.
00:09:31.670 --> 00:09:37.370
We also call that the natural
coordinate system, or the
00:09:37.370 --> 00:09:40.300
Isoparametric coordinate
system.
00:09:40.300 --> 00:09:45.700
This element, two by two, would
lie in space in an x
00:09:45.700 --> 00:09:49.650
xy-coordinate system
as indicated here.
00:09:49.650 --> 00:09:52.880
Now for three dimensional
analysis we would proceed in
00:09:52.880 --> 00:09:56.750
the same way, then we would
consider an element that is
00:09:56.750 --> 00:09:59.700
two by two by two
units long into
00:09:59.700 --> 00:10:01.450
each coordinate direction.
00:10:01.450 --> 00:10:05.530
And then we would have an
rs-axis and a t-axis coming
00:10:05.530 --> 00:10:10.530
out of the transparency in
this particular case.
00:10:10.530 --> 00:10:15.260
Well, I want to look at the
truss element, the special
00:10:15.260 --> 00:10:19.920
truss element, and the special
two d element.
00:10:19.920 --> 00:10:23.010
And I want to show you how
we can construct the
00:10:23.010 --> 00:10:26.260
interpolation matrices, the
displacement interpolation
00:10:26.260 --> 00:10:29.210
matrices, h i.
00:10:29.210 --> 00:10:32.210
These are also the coordinate
interpolation matrices.
00:10:32.210 --> 00:10:34.550
I want to show you how we
construct them for the special
00:10:34.550 --> 00:10:39.220
elements, how we then can
calculate the strain
00:10:39.220 --> 00:10:41.490
displacement interpolation
matrices
00:10:41.490 --> 00:10:43.240
for the special elements.
00:10:43.240 --> 00:10:46.510
And then I want to go on and
show you how we generalize to
00:10:46.510 --> 00:10:50.750
concepts to curved elements.
00:10:50.750 --> 00:10:54.790
And once we have discussed the
two dimensional case, I think
00:10:54.790 --> 00:10:58.520
you can see yourself how the
concepts are generalized to
00:10:58.520 --> 00:11:00.985
the three dimensional case.
00:11:00.985 --> 00:11:05.440
Let's look then at our two-noded
truss first.
00:11:05.440 --> 00:11:12.000
Here we have once more, the
truss, and we have node one on
00:11:12.000 --> 00:11:14.900
the right hand side, node two
on the left hand side.
00:11:14.900 --> 00:11:18.820
Our r-coordinate system starts
in the middle of the truss.
00:11:18.820 --> 00:11:22.900
Notice what we want to obtain
is that we interpolate our u
00:11:22.900 --> 00:11:26.010
displacement via the
interpolations.
00:11:26.010 --> 00:11:31.470
U being equal to h i times u i,
the h i are the unknowns.
00:11:31.470 --> 00:11:34.430
I want to show you how
we obtain the h i.
00:11:34.430 --> 00:11:37.600
The u i are the nodal
point displacements.
00:11:37.600 --> 00:11:41.270
In this particular case it would
be here, u two and here
00:11:41.270 --> 00:11:42.595
we would have u one.
00:11:46.120 --> 00:11:48.780
I in other words, goes
from one to two for
00:11:48.780 --> 00:11:50.810
this particular case.
00:11:50.810 --> 00:11:55.140
The h i has to be
a function of r.
00:11:55.140 --> 00:12:02.230
For a given r, however, we can
evaluate the u displacement if
00:12:02.230 --> 00:12:04.530
we have u i is given.
00:12:04.530 --> 00:12:08.780
In other words, when u one and
u two are given, then for a
00:12:08.780 --> 00:12:12.140
given r, we can evaluate
directly from this relation
00:12:12.140 --> 00:12:15.810
the displacement at
that point r.
00:12:15.810 --> 00:12:20.650
Well, with two nodes, we
can only have a bi--
00:12:20.650 --> 00:12:24.784
only a linear representation
in the displacements.
00:12:24.784 --> 00:12:30.050
The h one, from this relation,
must be this function.
00:12:30.050 --> 00:12:35.310
Because if we look at this, and
we say let u two be equal
00:12:35.310 --> 00:12:38.390
to zero, we would simply
have that u is
00:12:38.390 --> 00:12:41.270
equal to h one u one.
00:12:41.270 --> 00:12:46.100
Well, therefore, our h one must
look this one, because u
00:12:46.100 --> 00:12:48.830
one has its full
strength here.
00:12:48.830 --> 00:12:53.690
If we put u one equal to one, we
would have u is equal to h
00:12:53.690 --> 00:12:57.700
one, and this would
be the variation.
00:12:57.700 --> 00:13:03.240
Similarly, for h two, in this
case, we would have u is equal
00:13:03.240 --> 00:13:07.290
to h two u two.
00:13:07.290 --> 00:13:10.160
H one is now equal to zero.
00:13:10.160 --> 00:13:13.990
If we put you two equal to one,
as I have done in this
00:13:13.990 --> 00:13:19.170
particular case, then our linear
variation is indicated
00:13:19.170 --> 00:13:22.680
as shown here , and
the function h two
00:13:22.680 --> 00:13:24.340
is given right here.
00:13:24.340 --> 00:13:30.310
Notice that h two is equal to
zero when r is equal to one.
00:13:30.310 --> 00:13:35.780
And h two is equal to one when
r is equal to minus 1.
00:13:35.780 --> 00:13:38.570
So, this gives us h two.
00:13:38.570 --> 00:13:41.740
The two-noded truss, therefore,
simply has this
00:13:41.740 --> 00:13:45.170
description here for the
displacement and the
00:13:45.170 --> 00:13:46.660
coordinates.
00:13:46.660 --> 00:13:49.850
Remember that we're using the
same h i for displacement and
00:13:49.850 --> 00:13:51.590
coordinates.
00:13:51.590 --> 00:13:54.720
It simply has this description
where h one and h two are
00:13:54.720 --> 00:13:57.080
defined as shown.
00:13:57.080 --> 00:14:02.320
Let us now say that we want to
add another node, that we want
00:14:02.320 --> 00:14:04.130
to put another node
right there.
00:14:04.130 --> 00:14:07.040
In other words, we want to go
from a two-noded description
00:14:07.040 --> 00:14:08.750
to a three-noded description.
00:14:08.750 --> 00:14:11.700
On one of the earlier
Viewgraphs, you could see a
00:14:11.700 --> 00:14:13.680
three-noded cable element.
00:14:13.680 --> 00:14:19.540
Well, the way we proceed in the
construction of the h i
00:14:19.540 --> 00:14:23.550
functions is as follows.
00:14:23.550 --> 00:14:29.556
The two-noded element simply
had this description here.
00:14:29.556 --> 00:14:32.470
H one is given as shown
here on the left
00:14:32.470 --> 00:14:35.100
side of the blue line.
00:14:35.100 --> 00:14:38.010
H two was simply this.
00:14:38.010 --> 00:14:42.680
And we knew that we could do
no better than a linear
00:14:42.680 --> 00:14:46.300
description in displacements
between two nodes.
00:14:46.300 --> 00:14:50.530
However, now we have a third
node right here.
00:14:50.530 --> 00:14:55.890
And a third node means that
we can use a parabolic
00:14:55.890 --> 00:14:57.960
description in displacements.
00:14:57.960 --> 00:15:04.230
Now h three must be equal to one
at the third node and zero
00:15:04.230 --> 00:15:05.840
at both sides.
00:15:05.840 --> 00:15:06.540
Why?
00:15:06.540 --> 00:15:12.270
Well, because, remember we have
u now equal to h i u I,
00:15:12.270 --> 00:15:15.220
where i equals one to three.
00:15:15.220 --> 00:15:21.180
And if we put u one equal to
zero and u to equal to zero,
00:15:21.180 --> 00:15:26.690
we simply have u equals h three
u three, and, therefore,
00:15:26.690 --> 00:15:29.060
h three is this function.
00:15:31.750 --> 00:15:35.640
I've written it down here when
r is equal to zero its one
00:15:35.640 --> 00:15:39.750
when r is equal to minus 1
or plus h three is zero.
00:15:42.890 --> 00:15:44.810
So this is our h three.
00:15:44.810 --> 00:15:51.320
However, if we now look back to
our earlier description of
00:15:51.320 --> 00:15:56.450
h one and h two for the
two-noded element, we remember
00:15:56.450 --> 00:15:58.850
that we had a linear
variation here.
00:15:58.850 --> 00:16:01.640
That we had a linear variation
here and here.
00:16:01.640 --> 00:16:04.560
This one was a linear
variation of h one--
00:16:04.560 --> 00:16:08.220
let me take here the green color
to show once more what
00:16:08.220 --> 00:16:09.370
we are talking about--
00:16:09.370 --> 00:16:11.560
that was our linear variation.
00:16:11.560 --> 00:16:16.140
And here, this one here was our
linear variation there.
00:16:16.140 --> 00:16:22.790
Well, now with the third node
there, we recognize that, at
00:16:22.790 --> 00:16:25.635
this third node our h two--
00:16:25.635 --> 00:16:28.650
our actual h two for the
three-noded truss--
00:16:28.650 --> 00:16:30.147
must be zero here.
00:16:32.730 --> 00:16:34.820
H one must be zero here.
00:16:34.820 --> 00:16:37.940
Well, how can we make
it zero here?
00:16:37.940 --> 00:16:42.550
We can make it zero by
subtracting from our two-noded
00:16:42.550 --> 00:16:47.690
truss h one, one half of this
parabolic description.
00:16:47.690 --> 00:16:49.910
And that's what I
have done here.
00:16:49.910 --> 00:16:55.525
Similarly, we have to subtract
it here, because then, what we
00:16:55.525 --> 00:16:59.350
are doing is we are taking one
half off this parabola, and we
00:16:59.350 --> 00:17:01.660
are putting it right on here.
00:17:01.660 --> 00:17:05.430
This is one half of the
bottom parabola.
00:17:05.430 --> 00:17:06.780
We are putting it on there.
00:17:06.780 --> 00:17:10.280
And that brings this point
back to that point.
00:17:10.280 --> 00:17:12.220
Remember this is equal to one.
00:17:12.220 --> 00:17:16.560
I want one half as a correction
here, and that's
00:17:16.560 --> 00:17:21.920
why I take one half of one minus
r squared, to bring this
00:17:21.920 --> 00:17:23.950
point back to there.
00:17:23.950 --> 00:17:30.100
This total description then,
this total part--
00:17:30.100 --> 00:17:33.060
this part here, all
of that together--
00:17:33.060 --> 00:17:38.500
is our h one for the three-noded
element, for the
00:17:38.500 --> 00:17:40.030
three-noded element.
00:17:40.030 --> 00:17:46.170
And similarly, we would
have four h two.
00:17:46.170 --> 00:17:51.510
This total part here is this
function here for our
00:17:51.510 --> 00:17:53.050
three-noded element.
00:17:53.050 --> 00:17:56.470
Now the important point that
I really would like you to
00:17:56.470 --> 00:18:02.940
understand is that, we have
started off with a two-noded
00:18:02.940 --> 00:18:05.230
element description--
00:18:05.230 --> 00:18:09.610
h one only the linear part, h
two only the linear part.
00:18:09.610 --> 00:18:13.180
We have constructed the
interpolation function for the
00:18:13.180 --> 00:18:18.470
third node, and then we have
corrected the earlier
00:18:18.470 --> 00:18:24.200
two-noded interpolation
functions via subtracting a
00:18:24.200 --> 00:18:27.990
certain part of the third
interpolation function, in
00:18:27.990 --> 00:18:31.480
order to obtain the new
h one and h two for
00:18:31.480 --> 00:18:33.950
the three-noded element.
00:18:33.950 --> 00:18:38.620
And this is indeed the actual
procedure that we can use very
00:18:38.620 --> 00:18:42.520
effectively in constructing
higher order elements.
00:18:42.520 --> 00:18:45.040
We are starting off with
the lower order element
00:18:45.040 --> 00:18:46.810
descriptions--
00:18:46.810 --> 00:18:51.070
the ones shown here dashed with
a dashed line, the linear
00:18:51.070 --> 00:18:52.690
part only--
00:18:52.690 --> 00:18:59.590
and we add in the higher order
description, and subtract the
00:18:59.590 --> 00:19:03.620
correction from the lower
order description.
00:19:03.620 --> 00:19:06.910
The correction has to be
subtracted to bring this point
00:19:06.910 --> 00:19:11.130
back to zero, because h one
total must now be equal to
00:19:11.130 --> 00:19:15.330
zero here, h two total must be
equal to zero here, and this
00:19:15.330 --> 00:19:19.000
way we have constructed our new
h one and h two for the
00:19:19.000 --> 00:19:22.500
three-noded truss element.
00:19:22.500 --> 00:19:28.950
Let's look at the four-noded
element in two dimensions now.
00:19:28.950 --> 00:19:32.120
Here we use very similar
concepts.
00:19:32.120 --> 00:19:35.500
These are the four nodes
for the element.
00:19:35.500 --> 00:19:39.980
The description along a side
is just like we have been
00:19:39.980 --> 00:19:44.450
discussing now for the truss,
linear here and linear here.
00:19:44.450 --> 00:19:47.320
Notice h one can directly
be written down.
00:19:47.320 --> 00:19:51.630
It has to be equal to one here
and zero everywhere else.
00:19:51.630 --> 00:19:57.400
This is a function which is
bilinear and which satisfies
00:19:57.400 --> 00:20:00.170
these conditions to be equal
to one here and.
00:20:00.170 --> 00:20:02.770
Zero at the other nodes.
00:20:02.770 --> 00:20:07.980
It is a linear variation along
this side, along this side,
00:20:07.980 --> 00:20:10.590
and also across the surface.
00:20:10.590 --> 00:20:18.040
In other words, for a given
value of r, we have a linear
00:20:18.040 --> 00:20:21.140
variation across here too.
00:20:21.140 --> 00:20:22.960
How do we obtain h two?
00:20:22.960 --> 00:20:27.980
Well, we simply have to change
the signs over here--
00:20:27.980 --> 00:20:32.810
the plus signs here-- in
an appropriate way.
00:20:32.810 --> 00:20:36.270
H two shall be equal to one
here zero at all the other
00:20:36.270 --> 00:20:40.270
nodes, while we see that this
function satisfies these
00:20:40.270 --> 00:20:44.030
conditions, let's put in
r equal to minus 1.
00:20:44.030 --> 00:20:47.920
That makes this two, divided
by four, gives us one half.
00:20:47.920 --> 00:20:50.790
S has to be equal to plus
1 at this point--
00:20:50.790 --> 00:20:52.070
you get another two in here--
00:20:52.070 --> 00:20:56.010
so h two is equal to one
right there, and its
00:20:56.010 --> 00:20:57.480
zero everywhere else.
00:20:57.480 --> 00:21:00.010
Similarly we construct
h three and h four.
00:21:00.010 --> 00:21:04.240
Indeed we can immediately
observe that these signs--
00:21:04.240 --> 00:21:09.720
plus in both cases here, a minus
here, plus there, minus,
00:21:09.720 --> 00:21:12.620
minus, plus, minus-- these signs
correspond to nothing
00:21:12.620 --> 00:21:17.570
else in the signs of the r- and
s-coordinates of the nodal
00:21:17.570 --> 00:21:19.590
point under consideration.
00:21:19.590 --> 00:21:23.320
R and s is plus here, you have
two plus signs here.
00:21:23.320 --> 00:21:26.460
R is negative here, you put
a negative sign here.
00:21:26.460 --> 00:21:29.940
But s is positive here, you
put a plus sign there.
00:21:29.940 --> 00:21:33.470
r and s both are negative here,
so we have two negative
00:21:33.470 --> 00:21:36.030
signs here, et cetera.
00:21:36.030 --> 00:21:40.590
Let us now see how we construct
from this basic four
00:21:40.590 --> 00:21:44.030
node element, which really
corresponds to our basic two
00:21:44.030 --> 00:21:46.520
node element in the case of
the truss, how we can
00:21:46.520 --> 00:21:48.620
construct higher
order elements.
00:21:48.620 --> 00:21:52.425
Well, we proceed in much the
same way as in the truss
00:21:52.425 --> 00:21:53.400
formulation.
00:21:53.400 --> 00:21:57.500
Here we have a four-noded basic
element and we have
00:21:57.500 --> 00:22:00.100
added a fifth node to it.
00:22:00.100 --> 00:22:02.800
Now let's look at
this in detail.
00:22:02.800 --> 00:22:06.950
Since we now have added a fifth
node here, we know that
00:22:06.950 --> 00:22:09.820
we can allow a parabolic
distribution.
00:22:09.820 --> 00:22:12.220
In fact, we should allow a
parabolic distribution of
00:22:12.220 --> 00:22:14.950
displacements along the side.
00:22:14.950 --> 00:22:19.780
Well, if its a parabola along
the side for s being plus 1,
00:22:19.780 --> 00:22:23.380
we immediately see that
this function here--
00:22:23.380 --> 00:22:27.500
this s equal to plus one, this
is equal to two, knocks out
00:22:27.500 --> 00:22:29.420
that one half, we have
a one minus r
00:22:29.420 --> 00:22:31.340
squared along the side--
00:22:31.340 --> 00:22:34.910
just the same function that we
already had in the truss.
00:22:34.910 --> 00:22:39.630
Well, along this direction we
can only vary things linearly
00:22:39.630 --> 00:22:43.550
because we have two nodes only
in along these directions.
00:22:43.550 --> 00:22:46.650
And this is the reason
why we have to put in
00:22:46.650 --> 00:22:48.720
a one plus s here.
00:22:48.720 --> 00:22:51.400
The linear variation is
given by one plus s.
00:22:51.400 --> 00:22:56.240
And we also notice that when s
is equal to minus 1, in other
00:22:56.240 --> 00:22:59.790
words, we are looking at this
side, this function is zero.
00:22:59.790 --> 00:23:02.960
When s is equal to plus one,
as we pointed out earlier,
00:23:02.960 --> 00:23:04.180
this function here--
00:23:04.180 --> 00:23:06.880
this part and that part--
00:23:06.880 --> 00:23:10.330
gives us a one together, and we
have simply a one minus r
00:23:10.330 --> 00:23:11.780
squared along here.
00:23:11.780 --> 00:23:15.990
So, here you can see these
triangles that show the linear
00:23:15.990 --> 00:23:18.790
variation along this side.
00:23:18.790 --> 00:23:23.150
These parabolas here run really
across here, but with a
00:23:23.150 --> 00:23:24.520
different intensity.
00:23:24.520 --> 00:23:29.250
The intensity off the parabola
goes down linearly from one at
00:23:29.250 --> 00:23:33.160
this end, to is zero
at that end.
00:23:33.160 --> 00:23:34.720
This is our h five.
00:23:34.720 --> 00:23:39.800
Well, if we have this as our h
five, we remember that our
00:23:39.800 --> 00:23:45.520
original h one of the of the
four-noded element had a
00:23:45.520 --> 00:23:49.310
linear variation along here,
and also a linear variation
00:23:49.310 --> 00:23:51.210
along here.
00:23:51.210 --> 00:23:57.590
We will now have to take this
h one and subtract some
00:23:57.590 --> 00:23:59.040
correction from it.
00:23:59.040 --> 00:24:05.610
In order to make this point here
have zero displacement,
00:24:05.610 --> 00:24:06.860
four h one.
00:24:06.860 --> 00:24:10.890
Well, what we will do is we take
this h five and subtract
00:24:10.890 --> 00:24:15.020
a multiple of h five
from h one.
00:24:15.020 --> 00:24:18.620
In fact, you can see since the
original h one is equal to one
00:24:18.620 --> 00:24:21.820
half here, we simply have to
subtract one half of this
00:24:21.820 --> 00:24:25.050
function to obtain
the new h one.
00:24:25.050 --> 00:24:28.740
And this is what I have done
on this Viewgraph.
00:24:28.740 --> 00:24:33.340
The h one now is the original
h one that we had.
00:24:33.340 --> 00:24:37.740
And we are subtracting one half
of h five, which is one
00:24:37.740 --> 00:24:43.240
half of the parabolic
distribution, to bring this
00:24:43.240 --> 00:24:46.490
point here down to zero
in displacements.
00:24:46.490 --> 00:24:48.010
And that's what we
have done here.
00:24:48.010 --> 00:24:52.760
The resulting function
then is shown here.
00:24:52.760 --> 00:24:57.290
And our h one for the five-noded
element is shown
00:24:57.290 --> 00:24:58.280
right here.
00:24:58.280 --> 00:25:01.070
Similarly, for h two--
00:25:01.070 --> 00:25:05.180
this is the h two function for
the five-noded element--
00:25:05.180 --> 00:25:10.500
interesting to note that h three
and h four are for the
00:25:10.500 --> 00:25:14.840
five-noded element, the same as
for the four-noded element.
00:25:14.840 --> 00:25:20.160
Because this fifth node lies
between nodes one and two.
00:25:20.160 --> 00:25:23.910
And there is no effect
along this side and
00:25:23.910 --> 00:25:25.440
along this side here--
00:25:25.440 --> 00:25:32.300
along that side four h three and
h four-- so, we have our
00:25:32.300 --> 00:25:38.060
original functions also for
the five-noded element.
00:25:38.060 --> 00:25:39.910
Let us look now at a
00:25:39.910 --> 00:25:42.210
generalization of this concept.
00:25:42.210 --> 00:25:46.320
Here we have a typical
nine-noded element.
00:25:46.320 --> 00:25:49.940
A very effective element for
many types of applications.
00:25:49.940 --> 00:25:53.970
I already show it here in its
curved form, but think of it,
00:25:53.970 --> 00:25:56.270
please, as follows.
00:25:56.270 --> 00:25:57.970
This is the x-axis.
00:25:57.970 --> 00:25:59.630
This is the r-axis.
00:25:59.630 --> 00:26:04.160
S is equal to zero
along this side.
00:26:04.160 --> 00:26:07.030
S is equal to plus one
along this side.
00:26:07.030 --> 00:26:10.670
S is equal to minus
1 along this side.
00:26:10.670 --> 00:26:13.440
R is equal to plus one
along this side.
00:26:13.440 --> 00:26:16.080
R is equal to minus one
along the side.
00:26:16.080 --> 00:26:18.830
And r is zero along this axis.
00:26:18.830 --> 00:26:24.020
So, in the r s description,
in the embedded coordinate
00:26:24.020 --> 00:26:30.480
system, this element is still a
two by two squared element.
00:26:30.480 --> 00:26:37.180
Well, if we look then at the
interpolation functions, for
00:26:37.180 --> 00:26:39.490
this element, they
look as follows.
00:26:39.490 --> 00:26:43.020
Now maybe you have difficulty
seeing all this information,
00:26:43.020 --> 00:26:47.590
so, please then refer to the
study guide where you'll find
00:26:47.590 --> 00:26:49.270
this Viewgraph.
00:26:49.270 --> 00:26:53.510
For the four-noded element,
we had these
00:26:53.510 --> 00:26:56.670
interpolation functions.
00:26:56.670 --> 00:27:02.300
If we want to deal with the
five-noded element, what we
00:27:02.300 --> 00:27:06.920
have to do is, we add this
interpolation function.
00:27:06.920 --> 00:27:07.620
And--
00:27:07.620 --> 00:27:09.060
as I pointed out earlier--
00:27:09.060 --> 00:27:14.880
we have to correct our
h one and h two.
00:27:14.880 --> 00:27:17.830
But that is all of the
correction that is required.
00:27:17.830 --> 00:27:20.900
H three and h four
not corrected--
00:27:20.900 --> 00:27:24.940
they are blank spots here, they
are blank spots here.
00:27:24.940 --> 00:27:29.460
So, our five-noded element would
have these interpolation
00:27:29.460 --> 00:27:32.800
functions now shown in red.
00:27:32.800 --> 00:27:35.410
For if we added a sixth node--
00:27:35.410 --> 00:27:39.050
and I now should go back to
our earlier picture--
00:27:39.050 --> 00:27:42.630
if I wanted to add, in addition
to the fifth node,
00:27:42.630 --> 00:27:44.250
also the sixth node.
00:27:44.250 --> 00:27:50.320
Well, then, I have to put
another interpolation function
00:27:50.320 --> 00:27:53.430
down here, which now
is parabolic in s
00:27:53.430 --> 00:27:54.900
and linear in r.
00:27:54.900 --> 00:27:58.900
And I have to correct h two and
h three again, these are
00:27:58.900 --> 00:28:00.740
the corrections.
00:28:00.740 --> 00:28:05.360
So, now I have in green here
shown to you the interpolation
00:28:05.360 --> 00:28:08.340
functions off a six-noded
element.
00:28:08.340 --> 00:28:12.850
And like that we can proceed
by adding interpolation
00:28:12.850 --> 00:28:17.230
functions and correcting
the earlier constructed
00:28:17.230 --> 00:28:20.150
interpolation functions
as shown.
00:28:20.150 --> 00:28:24.830
Like that we can proceed
to directly obtain the
00:28:24.830 --> 00:28:30.610
interpolation functions for
the five-noded, six-noded,
00:28:30.610 --> 00:28:33.370
seven-, eight-, and nine-noded
element.
00:28:33.370 --> 00:28:38.280
In fact, its also important to
notice that we could have this
00:28:38.280 --> 00:28:43.190
node, that node, this node, and
that node, and just h nine
00:28:43.190 --> 00:28:44.330
also added.
00:28:44.330 --> 00:28:47.690
We could have, in other words,
a five-noded element which
00:28:47.690 --> 00:28:48.520
looks like this.
00:28:48.520 --> 00:28:52.030
It has this node, that one, that
one, and that one, and
00:28:52.030 --> 00:28:53.390
that one in the middle.
00:28:53.390 --> 00:28:59.230
Another five-noded element would
be this one, this one,
00:28:59.230 --> 00:29:02.250
this one, this one with
that node in there.
00:29:02.250 --> 00:29:06.960
So there is no necessity in
having all the nodes below a
00:29:06.960 --> 00:29:07.910
certain number.
00:29:07.910 --> 00:29:12.220
But we can simply use four
nodes, and then add whichever
00:29:12.220 --> 00:29:17.480
nodes we want to have
into the element.
00:29:17.480 --> 00:29:21.690
This is, then, how we construct
00:29:21.690 --> 00:29:23.690
interpolation functions.
00:29:23.690 --> 00:29:29.020
And once we had the h i, we
directly can obtain the h
00:29:29.020 --> 00:29:33.840
matrix, the matrix that gives
the displacements in terms of
00:29:33.840 --> 00:29:37.070
the nodal point displacements.
00:29:37.070 --> 00:29:40.170
Notice that the h
matrix really is
00:29:40.170 --> 00:29:43.410
constructed from the hi's.
00:29:43.410 --> 00:29:45.910
And the elements of
the b matrix--
00:29:45.910 --> 00:29:48.490
the strain displacement
interpolation matrix--
00:29:48.490 --> 00:29:52.330
are the derivatives of
the h i or zero.
00:29:52.330 --> 00:29:54.810
And I will show you an
example right now.
00:29:54.810 --> 00:29:58.160
Because we are using four, we
are still looking at the
00:29:58.160 --> 00:30:02.150
special case of a two by two
by two element in a truss
00:30:02.150 --> 00:30:03.500
case, only this two.
00:30:03.500 --> 00:30:06.000
Plane stress, plane strain,
axisymmetry, two dimensional
00:30:06.000 --> 00:30:08.770
analysis, we have
these two two's.
00:30:08.770 --> 00:30:10.960
In other words, we're talking
about a two by two element.
00:30:10.960 --> 00:30:13.510
And in three dimensional
analysis we would talk about a
00:30:13.510 --> 00:30:15.120
two by two by two element.
00:30:15.120 --> 00:30:18.270
In these cases, we have x equal
to r, y equal to s, z
00:30:18.270 --> 00:30:19.270
equal to t.
00:30:19.270 --> 00:30:20.930
So the strains--
00:30:20.930 --> 00:30:24.440
which are derivatives with
respect to x the actual
00:30:24.440 --> 00:30:26.100
physical coordinates--
00:30:26.100 --> 00:30:28.790
can also directly be obtained
by simply taking the
00:30:28.790 --> 00:30:31.710
derivative with respect
to r, and then
00:30:31.710 --> 00:30:35.230
similarly for s and t.
00:30:35.230 --> 00:30:39.920
Let us look at a four node,
two dimensional element.
00:30:39.920 --> 00:30:42.760
This is really the element
that we have used in our
00:30:42.760 --> 00:30:46.860
earlier example of the
cantilever analysis.
00:30:46.860 --> 00:30:50.730
Here we would simply have
that u r s, v r s, are
00:30:50.730 --> 00:30:53.820
described as shown.
00:30:53.820 --> 00:30:58.100
Notice that in the first row we
are really saying nothing
00:30:58.100 --> 00:31:02.890
else than u is a summation
h i u i.
00:31:02.890 --> 00:31:06.150
Where i equals one to
four because we have
00:31:06.150 --> 00:31:07.725
a four-noded element.
00:31:11.510 --> 00:31:12.830
H i u i.
00:31:12.830 --> 00:31:15.930
In the second row we are really
saying nothing else
00:31:15.930 --> 00:31:22.450
than v being the summation of i
equals one to four h i v i.
00:31:22.450 --> 00:31:26.670
And all I have done is I have
taken these hi's and assembled
00:31:26.670 --> 00:31:30.545
them into a matrix form to
obtain our h matrix.
00:31:33.180 --> 00:31:35.290
This h matrix here--
00:31:35.290 --> 00:31:39.120
the entries in that h matrix--
are dependent on the ordering
00:31:39.120 --> 00:31:43.660
that you're using here for u
one, v one, u two, et cetera.
00:31:43.660 --> 00:31:46.100
With this ordering, these
are the entries.
00:31:46.100 --> 00:31:49.650
Notice there are zeroes here,
because the v degrees of
00:31:49.650 --> 00:31:54.860
freedom at the nodes have
no contribution to the u
00:31:54.860 --> 00:31:57.110
displacement in the element.
00:31:57.110 --> 00:32:01.470
Well, if we look at the plane
stress case and want to
00:32:01.470 --> 00:32:06.260
construct our b matrix, remember
the b matrix gives
00:32:06.260 --> 00:32:11.970
the strains in terms of the
nodal point displacements.
00:32:11.970 --> 00:32:18.190
And we remember also that our
epsilon xx is equal to a dy
00:32:18.190 --> 00:32:26.800
dx, our epsilon yy is equal to
dv dy and our gamma xy is
00:32:26.800 --> 00:32:32.370
equal to a du dy plus dv dx.
00:32:32.370 --> 00:32:36.260
Well, if we recognize
these facts.
00:32:36.260 --> 00:32:40.560
And we also note again that for
the two by two element, r
00:32:40.560 --> 00:32:44.330
is identical to x, s
is identical to y.
00:32:44.330 --> 00:32:53.630
Then, we can obtain the epsilon
rr or epsilon xx by
00:32:53.630 --> 00:32:58.750
simply taking the derivatives of
the hi's with respect to r.
00:32:58.750 --> 00:33:04.940
Notice here, since u is equal
to the summation of h i u i,
00:33:04.940 --> 00:33:10.060
dy dr, which is equal to du dx,
is nothing else then the
00:33:10.060 --> 00:33:16.730
summation of partial
h i dr u i.
00:33:16.730 --> 00:33:21.110
And this part, which runs
from one to four--
00:33:21.110 --> 00:33:22.800
i going from one to four--
00:33:22.800 --> 00:33:26.170
I simply put right in there.
00:33:26.170 --> 00:33:29.510
I proceed similarly
for epsilon yy.
00:33:29.510 --> 00:33:34.540
Now I'm talking about the v
displacements, which are
00:33:34.540 --> 00:33:37.170
stored after the u
displacements.
00:33:37.170 --> 00:33:38.620
That's why I'm looking
here at the second
00:33:38.620 --> 00:33:40.640
column, the last column.
00:33:40.640 --> 00:33:44.670
And here we have dh one ds,
because we are talking about
00:33:44.670 --> 00:33:47.870
the derivative with respect to
y or with respect to s, which
00:33:47.870 --> 00:33:49.050
is the same thing.
00:33:49.050 --> 00:33:54.540
So here we have the entries
for the epsilon ss part.
00:33:54.540 --> 00:34:02.610
Now for the strain part, we are
talking du dy or du ds, dv
00:34:02.610 --> 00:34:08.510
dx, dv dr. And all we have to
do in now is take this term
00:34:08.510 --> 00:34:12.530
and put it right in there, and
take this term and put it
00:34:12.530 --> 00:34:15.219
right in there.
00:34:15.219 --> 00:34:17.760
And then the last row
here gives us
00:34:17.760 --> 00:34:20.719
the shearing strength.
00:34:20.719 --> 00:34:23.080
So this is our b matrix for the
00:34:23.080 --> 00:34:26.639
special two by two element.
00:34:26.639 --> 00:34:30.040
It is constructed in
a very simple way.
00:34:30.040 --> 00:34:34.320
The h i are known, and if we
had five or six or seven
00:34:34.320 --> 00:34:36.889
nodes, we would proceed in
exactly the same way.
00:34:36.889 --> 00:34:40.550
All we would have to do is
include additional columns in
00:34:40.550 --> 00:34:45.300
the b matrix that would give us
the appropriate entries for
00:34:45.300 --> 00:34:50.000
strains generated by the
additional nodal point
00:34:50.000 --> 00:34:51.690
displacements.
00:34:51.690 --> 00:34:56.285
Let us now look how we can
generalize these concepts to
00:34:56.285 --> 00:35:01.710
the element that is not,
anymore, in the physical x and
00:35:01.710 --> 00:35:05.090
y space, two by two element.
00:35:05.090 --> 00:35:08.210
In the physical x and y
space, this element--
00:35:08.210 --> 00:35:10.330
four-noded element now--
00:35:10.330 --> 00:35:12.540
might look as shown here.
00:35:12.540 --> 00:35:15.870
However, what we do is we still
deal with the r- and
00:35:15.870 --> 00:35:18.930
s-coordinate system embedded
on the element.
00:35:18.930 --> 00:35:24.340
We are on s one here, minus one
s plus one here, r and s
00:35:24.340 --> 00:35:28.920
both minus one here, plus
one here s minus one.
00:35:28.920 --> 00:35:33.030
So in the natural coordinate
space, the rs space, we still
00:35:33.030 --> 00:35:34.715
have a two by two element.
00:35:38.409 --> 00:35:42.820
The interpolation of the
displacement therefore--
00:35:42.820 --> 00:35:45.610
the interpolation of
the displacements--
00:35:45.610 --> 00:35:49.360
even for the distorted element,
is exactly still as
00:35:49.360 --> 00:35:50.820
shown here.
00:35:50.820 --> 00:35:53.260
Provided we are entering
always with the
00:35:53.260 --> 00:35:55.176
appropriate r and s.
00:35:55.176 --> 00:35:57.460
H one is a function
of r and s.
00:35:57.460 --> 00:36:02.050
So if we look at a point in
the general element, if we
00:36:02.050 --> 00:36:05.420
look at a point in the general
element, here, for example, at
00:36:05.420 --> 00:36:06.380
such a point.
00:36:06.380 --> 00:36:09.960
Because that point has a
specific r and s value.
00:36:09.960 --> 00:36:12.760
And if we want to find the
displacement at that point,
00:36:12.760 --> 00:36:17.960
well, we would have to put the
r and s value of that point
00:36:17.960 --> 00:36:21.350
into h one, h two, h three, h
four, and that gives us then
00:36:21.350 --> 00:36:25.290
the displacement at that point,
in terms of the nodal
00:36:25.290 --> 00:36:26.440
point displacement.
00:36:26.440 --> 00:36:31.010
So, the displacement
interpolation for this element
00:36:31.010 --> 00:36:33.820
can still be done in
the r and s space.
00:36:33.820 --> 00:36:36.590
However, difficulties arise--
00:36:36.590 --> 00:36:39.480
or additional considerations
I should say rather--
00:36:39.480 --> 00:36:43.120
arise when we talk about
strains, because the physical
00:36:43.120 --> 00:36:47.150
strains that we have to deal
with are derivatives with
00:36:47.150 --> 00:36:51.850
respect to x and y, and
not r and s anymore.
00:36:51.850 --> 00:36:55.000
Well, so what we have to
do is, use a Jacobian
00:36:55.000 --> 00:36:56.960
Transformation.
00:36:56.960 --> 00:36:58.850
What we want are the
derivatives with
00:36:58.850 --> 00:37:01.710
respect to x and y.
00:37:01.710 --> 00:37:05.400
What we can find easily are
derivatives with respect to r
00:37:05.400 --> 00:37:08.570
and s on the displacements,
because the displacements are
00:37:08.570 --> 00:37:12.090
given in terms of
r and s values.
00:37:12.090 --> 00:37:22.480
So, remember u is equal to some
of h i u i and the h i is
00:37:22.480 --> 00:37:27.330
a function of r and s, so, we
can directly find derivatives
00:37:27.330 --> 00:37:29.950
with respect to r and s of u.
00:37:29.950 --> 00:37:35.240
What we cannot find easily are
derivatives with respect to x.
00:37:35.240 --> 00:37:37.960
This is the relationship
that we use.
00:37:37.960 --> 00:37:43.090
It gives us a transformation
from derivatives of x and y to
00:37:43.090 --> 00:37:45.560
derivatives r and s.
00:37:45.560 --> 00:37:49.240
A question must immediately be
in your mind, why do we not
00:37:49.240 --> 00:37:52.720
write down directly this
relationship, which is given
00:37:52.720 --> 00:37:53.860
by the [INAUDIBLE].
00:37:53.860 --> 00:37:58.440
If we want d dx of
displacements, why not just
00:37:58.440 --> 00:38:03.130
use d dx being equal to d
dr, dr dx, and so on.
00:38:03.130 --> 00:38:05.260
Well, the difficultly
is that we cannot
00:38:05.260 --> 00:38:08.510
find dr dx very easily.
00:38:08.510 --> 00:38:13.230
We have x being this function
of h i x i.
00:38:13.230 --> 00:38:17.960
This is the interpolation which
we have to use now.
00:38:17.960 --> 00:38:20.130
I mentioned earlier on the
first slide that we
00:38:20.130 --> 00:38:24.410
interpolate displacement and
coordinates in the same way.
00:38:24.410 --> 00:38:27.210
So, here we have a linear
interpolation of the
00:38:27.210 --> 00:38:30.960
coordinates, from this node
to that node, and in
00:38:30.960 --> 00:38:32.540
between here too.
00:38:32.540 --> 00:38:35.210
So we are using this
interpolation here on the
00:38:35.210 --> 00:38:36.580
coordinates.
00:38:36.580 --> 00:38:47.540
And we can easily dx dr, but we
cannot easily find dr dx.
00:38:47.540 --> 00:38:50.880
You would have to invert this
relationship somehow so that
00:38:50.880 --> 00:38:53.360
we have r in terms of x.
00:38:53.360 --> 00:38:56.820
Well, it is easier, therefore,
to write this relationship
00:38:56.820 --> 00:38:59.510
down, which is really
the chain rule.
00:38:59.510 --> 00:39:01.910
This is also the chain rule, its
a chain rule the other way
00:39:01.910 --> 00:39:10.180
around, which gives d dr being
equal to d dx, dx dr plus d dy
00:39:10.180 --> 00:39:13.030
dy dr, if we multiply
this out.
00:39:13.030 --> 00:39:16.950
And its this relationship that
we can use effectively.
00:39:16.950 --> 00:39:21.430
Well, this relationship in three
dimensional analysis
00:39:21.430 --> 00:39:24.780
would involve a third row
and third column.
00:39:24.780 --> 00:39:29.360
In one dimensional analysis we
only talk about one by one.
00:39:29.360 --> 00:39:31.620
In other words, in one dimension
analysis for a truss
00:39:31.620 --> 00:39:33.770
we just have that entry.
00:39:33.770 --> 00:39:36.700
In general we can write it in
this way, where j is the
00:39:36.700 --> 00:39:42.620
Jacobian transformation from the
xyz-coordinate system to
00:39:42.620 --> 00:39:46.030
the rst-coordinate system.
00:39:46.030 --> 00:39:48.890
And since we want these
derivative--
00:39:48.890 --> 00:39:53.180
because these derivatives give
us actual strains, we have to
00:39:53.180 --> 00:39:56.660
invert this relationship.
00:39:56.660 --> 00:40:02.320
Having constructed then, these
derivatives in terms of these
00:40:02.320 --> 00:40:06.050
derivatives, which we can find
just as we have done before,
00:40:06.050 --> 00:40:10.070
we can now establish the
b matrix in much
00:40:10.070 --> 00:40:11.320
the same way as earlier.
00:40:13.960 --> 00:40:19.020
And since we now have h and b
matrices for an element--
00:40:19.020 --> 00:40:22.890
these are a function
r, s, and t--
00:40:22.890 --> 00:40:25.970
we would perform the
integration.
00:40:25.970 --> 00:40:29.220
And now I'm referring to
the integration of
00:40:29.220 --> 00:40:30.290
the stiffness matrix.
00:40:30.290 --> 00:40:37.390
Remember k is equal to b,
transposed cb over the volume.
00:40:37.390 --> 00:40:41.880
Now notice that since the b
matrix that we are using in
00:40:41.880 --> 00:40:48.260
here is a function of r and s in
a two dimensional analysis.
00:40:48.260 --> 00:40:52.140
And r runs from minus one to
plus one, and s runs from
00:40:52.140 --> 00:40:54.940
minus one to plus one.
00:40:54.940 --> 00:41:02.090
We now have to use a
transformation also on dv to
00:41:02.090 --> 00:41:06.920
integrate over the r s volume.
00:41:06.920 --> 00:41:08.300
And that integration--
00:41:08.300 --> 00:41:11.900
and that dv element is expressed
as shown here--
00:41:11.900 --> 00:41:18.830
and that is given to us from
mathematical analysis.
00:41:18.830 --> 00:41:21.680
So basically what we are saying
here is that we are
00:41:21.680 --> 00:41:25.610
replacing this integral over
the physical volume by an
00:41:25.610 --> 00:41:31.160
integral minus one to plus one,
minus one to plus one.
00:41:31.160 --> 00:41:36.010
And that signifies from minus
one to plus one over r and
00:41:36.010 --> 00:41:38.270
over s, if we had a three
dimensional analysis we would
00:41:38.270 --> 00:41:41.780
have another integral
sign here.
00:41:41.780 --> 00:41:47.570
B transposed now r and s
function of r and s, cb,
00:41:47.570 --> 00:41:50.250
function of r and s.
00:41:50.250 --> 00:41:55.860
And then our dv, now in
terms of r and s.
00:41:55.860 --> 00:41:59.230
So this is how we really
do things.
00:41:59.230 --> 00:42:03.540
And remember this dv here, this
dv, is that one there.
00:42:06.750 --> 00:42:17.020
This integration is effectively
performed using
00:42:17.020 --> 00:42:22.540
numerical integration and
I will discuss later on.
00:42:22.540 --> 00:42:25.760
Let's look once at the Jacobian
transformation for
00:42:25.760 --> 00:42:29.250
some very simple examples--
00:42:29.250 --> 00:42:30.940
the Jacobian transformation for
00:42:30.940 --> 00:42:33.590
some very simple examples--
00:42:33.590 --> 00:42:36.780
just to make things
a little clearer.
00:42:36.780 --> 00:42:41.540
In this case we really have
taken our two by two element
00:42:41.540 --> 00:42:45.690
and we have stretched it
into the x- and y-axes.
00:42:45.690 --> 00:42:50.040
That stretching is giving us a
three here and a two there
00:42:50.040 --> 00:42:54.840
because our two by two element
has a length of two here, and
00:42:54.840 --> 00:42:58.490
six divided by two
gives us three.
00:42:58.490 --> 00:43:01.110
Similarly here we have stretched
the element by a
00:43:01.110 --> 00:43:03.220
factor of two.
00:43:03.220 --> 00:43:07.990
This relationship here is,
in general, calculated--
00:43:07.990 --> 00:43:11.040
the j is, in general,
calculated--
00:43:11.040 --> 00:43:12.260
as shown here.
00:43:12.260 --> 00:43:15.410
And the result of that, by
putting these interpolations
00:43:15.410 --> 00:43:20.130
into there are these
values here.
00:43:20.130 --> 00:43:23.760
Physically what these mean
is a stretching, in this
00:43:23.760 --> 00:43:25.990
particular case, into
the x- and y-axis.
00:43:29.050 --> 00:43:30.300
Somewhat--
00:43:32.360 --> 00:43:34.910
the case where we cannot
directly--
00:43:34.910 --> 00:43:39.910
not easily directly write down
to j matrix is this one.
00:43:39.910 --> 00:43:45.870
Here we would go through the
actual evaluation the way I
00:43:45.870 --> 00:43:47.250
have indicated it.
00:43:47.250 --> 00:43:49.770
In other words, we would
go through this actual
00:43:49.770 --> 00:43:54.850
evaluation, substituting from
here and of course for y also.
00:43:54.850 --> 00:43:57.640
And this would be the result
Notice that we now have a
00:43:57.640 --> 00:44:03.740
stretching here of three,
compression from a two length
00:44:03.740 --> 00:44:04.920
to a one length--
00:44:04.920 --> 00:44:07.300
therefore we have a one half
here-- and there's also an
00:44:07.300 --> 00:44:11.940
angle change that gives us the
off-diagonal element here.
00:44:11.940 --> 00:44:17.170
Another interesting case here,
as an example, we have the
00:44:17.170 --> 00:44:19.100
same lengths here--
00:44:19.100 --> 00:44:23.540
two same lengths here-- but a
distortion in the element
00:44:23.540 --> 00:44:27.920
because this node two
has come down from
00:44:27.920 --> 00:44:30.490
there to its midpoint.
00:44:30.490 --> 00:44:35.700
And the resulting Jacobian
is given here.
00:44:35.700 --> 00:44:40.060
Now notice that that Jacobian is
a function r and s, so the
00:44:40.060 --> 00:44:43.250
inverse, which is used in the
construction of the b matrix
00:44:43.250 --> 00:44:47.660
would also be a function
of r and s.
00:44:47.660 --> 00:44:53.650
A particularly interesting case
is the one where we shift
00:44:53.650 --> 00:44:56.300
nodes to advantage.
00:44:56.300 --> 00:44:59.430
See here we have
our original--
00:44:59.430 --> 00:45:02.070
our three-noded element that we
talked about already-- in
00:45:02.070 --> 00:45:03.800
the r space now.
00:45:03.800 --> 00:45:06.230
Its a truss element.
00:45:06.230 --> 00:45:11.020
And let's say that in our actual
physical space, we have
00:45:11.020 --> 00:45:14.490
this node there, the three
node there, and
00:45:14.490 --> 00:45:16.530
the two node there.
00:45:16.530 --> 00:45:19.580
The element in the actual
physical space
00:45:19.580 --> 00:45:22.100
has a length of l.
00:45:22.100 --> 00:45:25.650
We have taken this node and
shifted it to the quarter
00:45:25.650 --> 00:45:29.370
point of the element, to the
quarter point of the element.
00:45:29.370 --> 00:45:33.820
What I will show you right now
is that by having done so-- by
00:45:33.820 --> 00:45:37.620
having taken this node from its
midpoint and shift it over
00:45:37.620 --> 00:45:40.520
in the actual physical space
to the quarter point--
00:45:40.520 --> 00:45:46.850
we will find that the strain has
a singularity here of one
00:45:46.850 --> 00:45:49.540
over square root x.
00:45:49.540 --> 00:45:56.430
This is a very important point
which can be used in the
00:45:56.430 --> 00:46:00.360
analysis of fracture problems,
because we know that in the
00:46:00.360 --> 00:46:05.040
analysis of fracture problems,
we have a one over square root
00:46:05.040 --> 00:46:11.460
x singularity at a crack tip.
00:46:11.460 --> 00:46:15.400
And, if we want to predict the
actual stress there or the
00:46:15.400 --> 00:46:20.010
displacement around the crack
tip, it can be of advantage to
00:46:20.010 --> 00:46:25.460
use this fact, shifting nodes to
quarter points in order to
00:46:25.460 --> 00:46:28.920
capture the stress singularity
more accurately.
00:46:28.920 --> 00:46:32.450
Well let me show you then how
this strain or stress
00:46:32.450 --> 00:46:35.950
singularity comes about.
00:46:35.950 --> 00:46:41.310
If we look at this element
here and we use our
00:46:41.310 --> 00:46:45.450
interpolation on the
coordinates, this
00:46:45.450 --> 00:46:47.530
would be the result.
00:46:47.530 --> 00:46:51.910
Now, notice that we
have substituted
00:46:51.910 --> 00:46:54.050
the x i values here.
00:46:54.050 --> 00:47:00.205
X one is zero, x two is l,
x three is l over four.
00:47:00.205 --> 00:47:03.610
We have substituted those values
and directly come up
00:47:03.610 --> 00:47:05.770
with this result.
00:47:05.770 --> 00:47:09.300
Well, we can see that
this indeed is true.
00:47:09.300 --> 00:47:13.150
Let's put r equal
to plus one n.
00:47:13.150 --> 00:47:16.780
In other words, the right
hand side node--
00:47:16.780 --> 00:47:19.260
for the right hand side node--
and we would have a two here
00:47:19.260 --> 00:47:22.630
squared, gives us four, goes
out with that four.
00:47:22.630 --> 00:47:26.450
So at i equals plus 1 we have
x equal to l, which
00:47:26.450 --> 00:47:27.810
is correct of course.
00:47:27.810 --> 00:47:32.650
Let's put i equal to minus one
n, we find x is equal to zero.
00:47:32.650 --> 00:47:36.510
Let's put i equal to
zero n we find x is
00:47:36.510 --> 00:47:38.600
equal to l over four.
00:47:38.600 --> 00:47:43.030
In other words, this has been a
simple check in that we have
00:47:43.030 --> 00:47:44.350
the right interpolation--
00:47:44.350 --> 00:47:45.790
geometry interpolation--
00:47:45.790 --> 00:47:48.360
for this element.
00:47:48.360 --> 00:47:53.940
Our j now is simply, dx
dr, X is given here.
00:47:53.940 --> 00:47:55.960
If you take the differentiation
of that you
00:47:55.960 --> 00:47:59.000
get the two in front that gives
l over two times one
00:47:59.000 --> 00:48:04.390
plus r, this value here,
in other words.
00:48:04.390 --> 00:48:07.980
Then our b matrix is
constructed by the
00:48:07.980 --> 00:48:11.530
inverse of the j--
00:48:11.530 --> 00:48:13.160
that is this one--
00:48:13.160 --> 00:48:19.160
two times the r derivative of
the interpolation functions.
00:48:19.160 --> 00:48:22.100
Of course here we talk only
about one strain.
00:48:22.100 --> 00:48:26.420
Remember, in the truss, the only
displacement of concern
00:48:26.420 --> 00:48:32.020
is the u displacement and the
strain is simply epsilon xx, a
00:48:32.020 --> 00:48:34.120
strain into this
direction also.
00:48:36.790 --> 00:48:40.000
Well, this is our
b matrix then.
00:48:40.000 --> 00:48:43.560
And if we take our h one,
two, and three, and we
00:48:43.560 --> 00:48:45.680
differentiate these--
00:48:45.680 --> 00:48:49.500
as indicated here with respect
to r-- we directly obtain
00:48:49.500 --> 00:48:51.790
these functions here.
00:48:51.790 --> 00:49:00.260
If we now recognize that since
we have x related to r here,
00:49:00.260 --> 00:49:03.490
we can also invert this
relationship.
00:49:03.490 --> 00:49:06.830
We can write r in terms of x.
00:49:06.830 --> 00:49:10.810
And if we have done so we can
take that relationship and put
00:49:10.810 --> 00:49:12.320
it right in here.
00:49:12.320 --> 00:49:15.690
Then we would get
b in terms of x.
00:49:15.690 --> 00:49:17.770
And that's what I
have done here.
00:49:17.770 --> 00:49:23.260
The first line shows simply r in
terms of x now, and I have
00:49:23.260 --> 00:49:28.470
substituted that r value into
the b matrix and this is there
00:49:28.470 --> 00:49:29.090
is the result.
00:49:29.090 --> 00:49:34.940
Notice that we have a
strain singularity.
00:49:34.940 --> 00:49:38.730
This is the first
element here.
00:49:38.730 --> 00:49:40.830
The next element--
00:49:40.830 --> 00:49:41.600
this is here--
00:49:41.600 --> 00:49:46.470
the next element in b matrix,
and that is the third element
00:49:46.470 --> 00:49:47.660
in the b matrix.
00:49:47.660 --> 00:49:52.820
Notice that we have in this
element, that one, and that
00:49:52.820 --> 00:49:56.990
one the one over square root
x, which means that we have
00:49:56.990 --> 00:49:59.006
one over square root
x singularity
00:49:59.006 --> 00:50:01.780
at x equal to zero.
00:50:01.780 --> 00:50:07.450
Well, this fact is used very
effectively in fracture
00:50:07.450 --> 00:50:09.370
mechanics analysis.
00:50:09.370 --> 00:50:16.035
Assume that we have crack here
and that we want to analyze
00:50:16.035 --> 00:50:20.150
the stress conditions
around that crack.
00:50:20.150 --> 00:50:25.680
What we can do is we use now two
dimension elements, as its
00:50:25.680 --> 00:50:28.230
a plane stress situation.
00:50:28.230 --> 00:50:32.540
We would put a two dimensional
triangular element there and
00:50:32.540 --> 00:50:36.050
we shift the midpoint nodes.
00:50:36.050 --> 00:50:38.420
This is now very small
here, but I hope
00:50:38.420 --> 00:50:39.760
you can still follow.
00:50:39.760 --> 00:50:44.090
You shift these midpoint nodes
to the quarter point, just the
00:50:44.090 --> 00:50:48.810
way we have been
doing it here.
00:50:48.810 --> 00:50:51.815
You are putting the third node
the quarter point and the
00:50:51.815 --> 00:50:57.105
result is that at this crack tip
we have a one over square
00:50:57.105 --> 00:51:05.290
root x singularity, using
this element layout.
00:51:05.290 --> 00:51:10.020
And we know that in fact, there
is a one over square
00:51:10.020 --> 00:51:14.570
root x singularity in linear
fracture mechanics analysis.
00:51:14.570 --> 00:51:18.160
And so this is an effective
way of capturing this
00:51:18.160 --> 00:51:22.990
singularity, and has been used
or is currently being used
00:51:22.990 --> 00:51:26.060
very abundantly in practice.
00:51:26.060 --> 00:51:30.810
The important point that I
wanted to make really is that
00:51:30.810 --> 00:51:37.930
we can shift nodes in the
element to our advantage.
00:51:37.930 --> 00:51:42.470
But, we really do that in
specific applications such as
00:51:42.470 --> 00:51:43.720
fracture mechanics.
00:51:45.840 --> 00:51:50.170
In general we will see later
when I talk about modeling of
00:51:50.170 --> 00:51:53.730
finite element systems, in
general, it is most effective
00:51:53.730 --> 00:52:00.880
to leave the mid-side nodes at
their physical midpoints.
00:52:00.880 --> 00:52:03.510
In other words for an
eight-noded element, two
00:52:03.510 --> 00:52:07.380
dimensional analysis, we would
put this mid-node in the
00:52:07.380 --> 00:52:11.670
physical space also, actually
at the midpoint.
00:52:11.670 --> 00:52:13.150
We would not shift it.
00:52:13.150 --> 00:52:17.010
Then the element has good
convergence characteristics
00:52:17.010 --> 00:52:20.230
into all directions and this is
really how the element is
00:52:20.230 --> 00:52:23.860
used most effectively for
general applications.
00:52:23.860 --> 00:52:27.090
However, in specific
applications, such as fracture
00:52:27.090 --> 00:52:31.760
mechanics analysis, it can be
of advantage to shift these
00:52:31.760 --> 00:52:35.970
mid-side nodes to pick up
certain strain or stress
00:52:35.970 --> 00:52:40.590
singularities that
we know do exist.
00:52:40.590 --> 00:52:45.720
Now on the last transparency
that I wanted to show you, I
00:52:45.720 --> 00:52:49.850
wanted to indicate something to
you that I will be talking
00:52:49.850 --> 00:52:54.020
about in later lectures more
abundantly, namely the fact
00:52:54.020 --> 00:52:56.960
that we're using numerical
integration.
00:52:56.960 --> 00:53:01.860
B, for the k matrix as an
example, is once again now a
00:53:01.860 --> 00:53:05.220
function of r and s.
00:53:05.220 --> 00:53:09.210
This part here is also a
function of r and s.
00:53:09.210 --> 00:53:13.870
So we have here function
of r and s.
00:53:13.870 --> 00:53:17.130
and we have here also a
function of r and s.
00:53:17.130 --> 00:53:21.930
So this f matrix here is
a function of r and s.
00:53:21.930 --> 00:53:27.400
Notice that the b also includes
the inversion of the
00:53:27.400 --> 00:53:30.550
j, the Jacobian matrix.
00:53:30.550 --> 00:53:34.060
It includes the inversion of
the j, because we had to
00:53:34.060 --> 00:53:39.785
construct the x and y and z
derivatives from the r, s, and
00:53:39.785 --> 00:53:41.180
t derivatives.
00:53:41.180 --> 00:53:47.940
So, what we do in practical
analysis is that we use
00:53:47.940 --> 00:53:51.540
numerical integration to
evaluate the k matrix.
00:53:51.540 --> 00:53:54.690
I have indicated that here
schematically, if you look at
00:53:54.690 --> 00:53:56.670
this element here.
00:53:56.670 --> 00:54:02.900
What we do is, we evaluate the
f matrix-- this is a matrix.
00:54:02.900 --> 00:54:07.340
In two dimensional analysis we
would only run over i and j.
00:54:07.340 --> 00:54:08.630
I is this direction.
00:54:08.630 --> 00:54:09.610
J is that direction.
00:54:09.610 --> 00:54:12.640
In three dimensional analysis,
which is in general analysis,
00:54:12.640 --> 00:54:16.630
we run i, j, and k this way.
00:54:16.630 --> 00:54:21.130
We evaluate the f matrix
here at specific
00:54:21.130 --> 00:54:24.410
points, r, s, and t.
00:54:24.410 --> 00:54:26.850
T now being this axis.
00:54:26.850 --> 00:54:31.560
And then multiply that f matrix
by certain weight
00:54:31.560 --> 00:54:36.040
constants and sum these
contributions over all i, j,
00:54:36.040 --> 00:54:42.010
and k, in order to obtain an
accurate enough approximation
00:54:42.010 --> 00:54:43.815
to the actual stiffness
matrix.
00:54:46.370 --> 00:54:53.360
The order of approximation
with which we obtain the
00:54:53.360 --> 00:54:55.330
actual stiffness matrix--
00:54:55.330 --> 00:54:59.100
or rather how closely the
numerically calculated
00:54:59.100 --> 00:55:02.610
stiffness matrix approximates
the actual stiffness matrix--
00:55:02.610 --> 00:55:06.440
depends on, number one, how many
integration points we are
00:55:06.440 --> 00:55:09.260
using and what kind
of integration
00:55:09.260 --> 00:55:11.880
scheme we are using.
00:55:11.880 --> 00:55:15.460
These points here correspond
to the Gauss numerical
00:55:15.460 --> 00:55:16.370
integration.
00:55:16.370 --> 00:55:19.310
In this case for two dimensional
analysis, we would
00:55:19.310 --> 00:55:21.820
use a two by two integration.
00:55:21.820 --> 00:55:26.550
In other words, i and j would
both run from one to two.
00:55:26.550 --> 00:55:29.840
K is not applicable and we would
have altogether four
00:55:29.840 --> 00:55:34.190
evaluations off the Fs here.
00:55:34.190 --> 00:55:37.870
Multiply each of them by
weighting factors, which has
00:55:37.870 --> 00:55:42.870
been derived for us by Gauss
some long time ago.
00:55:42.870 --> 00:55:47.860
And summing up these
contributions gives us a close
00:55:47.860 --> 00:55:50.310
enough approximation
to the k matrix.
00:55:50.310 --> 00:55:53.650
Of course, the question of how
many points we have to use,
00:55:53.650 --> 00:55:56.960
what integration scheme we
should use, is a very
00:55:56.960 --> 00:55:58.360
important one.
00:55:58.360 --> 00:56:01.790
We must use enough integration
points to get a close enough
00:56:01.790 --> 00:56:05.240
approximation to the actual
stiffness matrix and I will be
00:56:05.240 --> 00:56:09.300
addressing those questions
in a later lecture.
00:56:09.300 --> 00:56:11.520
This is all I wanted to
say in this lecture.
00:56:11.520 --> 00:56:12.920
Thank you very much for
your attention.