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PROFESSOR: Ladies and gentlemen,
welcome to this

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00:00:23,670 --> 00:00:26,330
lecture on non-linear finite
element analysis of solids and

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00:00:26,330 --> 00:00:27,420
structures.

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00:00:27,420 --> 00:00:29,740
In this lecture, I'd like to
continue with our discussion

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00:00:29,740 --> 00:00:32,570
of structural elements, and
in particular, I'd like to

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00:00:32,570 --> 00:00:36,670
concentrate our attention first
on the formulation of

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00:00:36,670 --> 00:00:37,980
beam elements.

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00:00:37,980 --> 00:00:41,080
After that, I'd like to talk
more about the formulation of

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00:00:41,080 --> 00:00:44,730
shell elements, and then
show you applications.

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00:00:44,730 --> 00:00:48,690
When we talk about beam
elements, we might think about

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00:00:48,690 --> 00:00:52,640
the usual Hermitian beam
elements, in which the

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00:00:52,640 --> 00:00:55,850
transverse placements cubically
interpolated, and

19
00:00:55,850 --> 00:00:59,390
the longitudinal displacements
are linearly interpolated.

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00:00:59,390 --> 00:01:02,250
This is an element that is
usually used in linear

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00:01:02,250 --> 00:01:04,310
analysis, and very
widely used--

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00:01:04,310 --> 00:01:07,260
you might have used it quite
a bit yourself--

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00:01:07,260 --> 00:01:11,490
and it is also very effective
in the analysis of linear

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00:01:11,490 --> 00:01:13,760
response of structures.

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00:01:13,760 --> 00:01:19,030
However, in my discussion now,
I will talk about the

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00:01:19,030 --> 00:01:22,970
isoparametric beam element,
and how we formulate the

27
00:01:22,970 --> 00:01:25,630
isoparametric beam element, in
other words, and how we apply

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00:01:25,630 --> 00:01:27,830
it in the analysis
of structures.

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00:01:27,830 --> 00:01:31,190
So when I talk about the beam
element, I really mean this

30
00:01:31,190 --> 00:01:35,460
isoparametric beam element that
we are formulating in

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00:01:35,460 --> 00:01:37,300
this lecture.

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00:01:37,300 --> 00:01:41,080
The isoparametric beam
formulation can be very

33
00:01:41,080 --> 00:01:45,270
effective for the analysis of
curve themes, for the analysis

34
00:01:45,270 --> 00:01:48,640
of geometrically nonlinear
problems, and for the analysis

35
00:01:48,640 --> 00:01:51,120
of stiffened shell structures,
where we couple the

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00:01:51,120 --> 00:01:54,660
isoparametric beam with the
isoparametrically formulated

37
00:01:54,660 --> 00:01:56,450
shell elements.

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00:01:56,450 --> 00:01:59,490
The formulation is quite
analogous to the formulation

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00:01:59,490 --> 00:02:04,150
of the isoparametric degenerate
shell elements that

40
00:02:04,150 --> 00:02:07,590
we discussed in the previous
lecture, and so we can go a

41
00:02:07,590 --> 00:02:11,900
bit fast over the formulation
of the beam element here.

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00:02:11,900 --> 00:02:14,640
I'm showing a typical
beam element,

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00:02:14,640 --> 00:02:16,800
rectangular cross section.

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00:02:16,800 --> 00:02:18,810
Here we show three nodes.

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00:02:18,810 --> 00:02:21,350
Actually, the beam element can
be used with two nodes, three

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00:02:21,350 --> 00:02:24,950
nodes, or four nodes, as we will
discuss further later on.

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00:02:24,950 --> 00:02:30,990
We notice that the depth off
the element is ak, the

48
00:02:30,990 --> 00:02:33,770
thickness of the
element is bk.

49
00:02:33,770 --> 00:02:40,451
And notice that at these nodes,
we measure certain

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00:02:40,451 --> 00:02:42,460
nodal point variables.

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00:02:42,460 --> 00:02:45,710
and here I'm showing the nodal
point variables that we do

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00:02:45,710 --> 00:02:47,260
measure at the nodes.

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00:02:47,260 --> 00:02:51,300
Three displacements, incremental
displacements, and

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00:02:51,300 --> 00:02:55,260
three incremental rotations.

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00:02:55,260 --> 00:03:00,880
Notice that the description of
the beam element follows much

56
00:03:00,880 --> 00:03:02,860
the same as the geometric
description of the beam

57
00:03:02,860 --> 00:03:06,510
element follows, much the same
way as we are dealing with the

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00:03:06,510 --> 00:03:07,560
shell element.

59
00:03:07,560 --> 00:03:10,160
We introduce director vectors.

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00:03:10,160 --> 00:03:11,710
Now we have two director
vectors.

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00:03:11,710 --> 00:03:13,400
In the shell, of course,
we had only one

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00:03:13,400 --> 00:03:14,890
director vector per node.

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00:03:14,890 --> 00:03:16,220
Now we have two.

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00:03:16,220 --> 00:03:23,940
Here we show tvsk, the director
vector at time t into

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00:03:23,940 --> 00:03:26,940
the S-direction, corresponding
to node k.

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00:03:26,940 --> 00:03:29,910
Notice here tvtk.

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00:03:29,910 --> 00:03:33,870
The director vector at time t
corresponding to node k into

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00:03:33,870 --> 00:03:35,140
the t direction.

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00:03:35,140 --> 00:03:41,820
this t means direction t that
t opting out means time t

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00:03:41,820 --> 00:03:47,600
Notice that corresponding to
this direction t here, we have

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00:03:47,600 --> 00:03:51,050
the t isoparametric
coordinate.

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00:03:51,050 --> 00:03:56,250
And of course, this t
isoparametric coordinate also

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00:03:56,250 --> 00:03:59,580
runs into the thickness ak.

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00:03:59,580 --> 00:04:02,990
Notice that corresponding to
bk, we have the s-direction

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00:04:02,990 --> 00:04:06,490
coordinate, and that s-direction
coordinate,

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00:04:06,490 --> 00:04:10,860
s-isoparametric coordinate, runs
into the direction of the

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00:04:10,860 --> 00:04:12,393
director vector, tvsk.

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00:04:15,480 --> 00:04:20,620
So these are the variables used
to describe the geometry

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00:04:20,620 --> 00:04:25,040
and the displacement of
the beam element.

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00:04:25,040 --> 00:04:29,980
We start with the general
equation for the

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00:04:29,980 --> 00:04:32,100
geometry at time t.

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00:04:32,100 --> 00:04:33,780
Here we have txi.

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00:04:33,780 --> 00:04:37,450
The coordinates of a material
particle within the

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00:04:37,450 --> 00:04:40,580
element at time t--

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00:04:40,580 --> 00:04:44,370
and remember, these are the
coordinates measured in the

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00:04:44,370 --> 00:04:47,880
stationary coordinate system.

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00:04:47,880 --> 00:04:50,610
And these coordinates are
obtained as shown on the

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00:04:50,610 --> 00:04:51,260
right-hand side.

89
00:04:51,260 --> 00:04:52,890
Let's have a close look
at what we see on

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00:04:52,890 --> 00:04:53,950
the right-hand side.

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00:04:53,950 --> 00:04:56,940
We have the hk interpolation
function.

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00:04:56,940 --> 00:04:59,660
This is the interpolation
function corresponding to the

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00:04:59,660 --> 00:05:02,710
r-direction running through
the nodes of the element,

94
00:05:02,710 --> 00:05:05,470
along the mid-surface
of the element.

95
00:05:05,470 --> 00:05:08,290
And it's a one-dimensional
interpolation function, the

96
00:05:08,290 --> 00:05:11,290
same one that we used to see
for the truss element.

97
00:05:11,290 --> 00:05:15,160
These are the nodal point
coordinates at time t.

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00:05:15,160 --> 00:05:18,920
We are summing over all the
nodal points, k from 1 to n,

99
00:05:18,920 --> 00:05:24,020
and n can be equal to 2, 3, or
4, as I mentioned earlier.

100
00:05:24,020 --> 00:05:28,280
Then we add to this term a term
that comes in because of

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00:05:28,280 --> 00:05:34,130
the sickness of the beam into
the t-direction, ak, thickness

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00:05:34,130 --> 00:05:35,610
off the beam into the
t-direction--

103
00:05:35,610 --> 00:05:38,095
here is a t-coordinate.

104
00:05:38,095 --> 00:05:42,730
The isoparametric coordinates
going from minus 1 to plus 1.

105
00:05:42,730 --> 00:05:45,800
Here we have hk, same
interpolation function as

106
00:05:45,800 --> 00:05:50,760
there, and here we have the
direction, cosines of the

107
00:05:50,760 --> 00:05:55,970
director vector in the
t-direction at time t.

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00:05:55,970 --> 00:05:59,950
A similar term, similar to this
one, is added for the

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00:05:59,950 --> 00:06:02,190
f-direction.

110
00:06:02,190 --> 00:06:05,080
s, the isoparametric coordinate,
running from minus

111
00:06:05,080 --> 00:06:10,060
1 to plus 1, bk being the
thickness into the s-direction

112
00:06:10,060 --> 00:06:13,240
at node k, hk, same
interpolation function that we

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00:06:13,240 --> 00:06:16,410
have seen up here already, and
these are the direction

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00:06:16,410 --> 00:06:21,720
cosines of the director vector
in the s-direction,

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00:06:21,720 --> 00:06:26,410
corresponding to node
k at time t.

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00:06:26,410 --> 00:06:29,510
This equation is very similar
to the equation that we used

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00:06:29,510 --> 00:06:30,960
in the shell element
formation.

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00:06:30,960 --> 00:06:33,830
Of course, in the shell element
formulation, we

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00:06:33,830 --> 00:06:37,630
interpolated here in a
two-dimensional domain over

120
00:06:37,630 --> 00:06:39,010
the shell mid-surface.

121
00:06:39,010 --> 00:06:42,980
Now we only run along the
neutral axis of the beam.

122
00:06:42,980 --> 00:06:46,330
And we had only one term in the
shell corresponding to the

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00:06:46,330 --> 00:06:47,970
director vector.

124
00:06:47,970 --> 00:06:51,160
There was only one direction,
namely, the one normal to the

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00:06:51,160 --> 00:06:52,720
mid-surface of the shell.

126
00:06:52,720 --> 00:06:57,040
Here we have two directions
to deal with.

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00:06:57,040 --> 00:07:00,640
Since the displacements can
directly be obtained from the

128
00:07:00,640 --> 00:07:03,840
geometry interpolation at time
t minus the geometry

129
00:07:03,840 --> 00:07:07,730
interpolation at time 0,
of course this geometry

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00:07:07,730 --> 00:07:10,630
interpolation is obtained by
simply substituting in the

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00:07:10,630 --> 00:07:15,750
equation that we just looked
at for time t, the time 0.

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00:07:15,750 --> 00:07:20,990
Since we obtain tui, as shown
here, this is the result.

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00:07:20,990 --> 00:07:26,080
We simply apply our geometry
interpolation at time t, and

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00:07:26,080 --> 00:07:28,140
at time 0, subtract.

135
00:07:28,140 --> 00:07:31,950
This is what you get on
the right-hand side.

136
00:07:31,950 --> 00:07:37,220
Nodal point displacement from
time 0 to time t, measured in

137
00:07:37,220 --> 00:07:41,370
the stationary global Cartesian
coordinate system.

138
00:07:41,370 --> 00:07:48,780
Here change in the director
vector cosines, or I should

139
00:07:48,780 --> 00:08:00,400
say, more precisely maybe, the
change in the cosine of the

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00:08:00,400 --> 00:08:05,520
angle director vector time 0,
director vector time t.

141
00:08:05,520 --> 00:08:09,320
Of course, remember here we
have direction cosines

142
00:08:09,320 --> 00:08:10,710
corresponding to time t.

143
00:08:10,710 --> 00:08:14,280
Here we have direction cosines
corresponding to time 0.

144
00:08:14,280 --> 00:08:18,960
We subtract, and that
goes in here.

145
00:08:18,960 --> 00:08:23,030
Here, same kind of term,
but for direction s.

146
00:08:26,320 --> 00:08:29,460
Also the incremental
displacements are obtained

147
00:08:29,460 --> 00:08:34,549
from this relationship here, and
if you apply the geometry

148
00:08:34,549 --> 00:08:37,539
interpolation at time t plus
delta t and at time t, you

149
00:08:37,539 --> 00:08:40,549
directly obtain this
right-hand side.

150
00:08:40,549 --> 00:08:46,340
These are the increment in the
direction cosine from time t

151
00:08:46,340 --> 00:08:50,790
to time t plus delta t of the
director vector in the

152
00:08:50,790 --> 00:08:51,850
t-direction.

153
00:08:51,850 --> 00:08:55,040
Similar here for the
s-direction.

154
00:08:55,040 --> 00:08:58,420
Of course, we are used very well
to deal with increment

155
00:08:58,420 --> 00:09:01,750
and nodal point displacement,
but we cannot very well deal

156
00:09:01,750 --> 00:09:06,850
with these quantities
here, vti and vsi.

157
00:09:06,850 --> 00:09:09,540
We want to deal with nodal
point rotation.

158
00:09:09,540 --> 00:09:13,600
And so what we do is, we express
this quantity and that

159
00:09:13,600 --> 00:09:17,660
quantity in terms of nodal
point rotations, and that

160
00:09:17,660 --> 00:09:21,820
those expressions are
given down here.

161
00:09:21,820 --> 00:09:23,000
These are the two expressions.

162
00:09:23,000 --> 00:09:25,840
You can simply multiply out, and
you would immediately see

163
00:09:25,840 --> 00:09:28,560
that they indeed hold.

164
00:09:28,560 --> 00:09:30,800
With these geometry and
displacement interpolations

165
00:09:30,800 --> 00:09:34,060
established, we can now
directly calculate or

166
00:09:34,060 --> 00:09:36,720
establish the strain
displacement matrices

167
00:09:36,720 --> 00:09:40,460
corresponding to the Cartesian
strain components, and then a

168
00:09:40,460 --> 00:09:43,040
simple transformation, standard
transformation gives

169
00:09:43,040 --> 00:09:47,460
us the strain displacement
matrices corresponding to the

170
00:09:47,460 --> 00:09:50,550
eta zeta psi directions.

171
00:09:50,550 --> 00:09:59,520
Of course, these are the
directions with the components

172
00:09:59,520 --> 00:10:02,330
of strains and stresses that we
need to deal with for the

173
00:10:02,330 --> 00:10:03,830
beam formulation.

174
00:10:03,830 --> 00:10:09,040
Notice that zeta here is a
physical coordinate running

175
00:10:09,040 --> 00:10:13,220
into the direction, so to say,
of the a thickness of the

176
00:10:13,220 --> 00:10:19,460
beam, and zeta corresponds to
the t-isoparametric coordinate

177
00:10:19,460 --> 00:10:21,350
that we just talked about.

178
00:10:21,350 --> 00:10:25,350
Psi here corresponds to the
s-isoparametric coordinate,

179
00:10:25,350 --> 00:10:29,190
and runs into the b width,
so to say, of the beam.

180
00:10:29,190 --> 00:10:34,640
Eta here corresponds to the
r-isoparametric coordinate.

181
00:10:34,640 --> 00:10:40,050
Of course, these zeta, eta, and
psi are actual physical

182
00:10:40,050 --> 00:10:42,680
coordinates that we
are dealing with.

183
00:10:42,680 --> 00:10:49,570
We have to, of course, calculate
the tau psi eta, the

184
00:10:49,570 --> 00:10:54,970
tau eta eta, and the tau vita
psi stress components and

185
00:10:54,970 --> 00:10:57,770
corresponding strain components
for the beam

186
00:10:57,770 --> 00:10:58,810
formulation.

187
00:10:58,810 --> 00:11:02,860
And the stress-strain law,
corresponding to these

188
00:11:02,860 --> 00:11:05,260
components, is given here.

189
00:11:05,260 --> 00:11:10,140
Notice tau eta eta is obtained
by taking the Young's modulus

190
00:11:10,140 --> 00:11:13,850
and multiplying the Young's
modulus by epsilon

191
00:11:13,850 --> 00:11:16,050
eta eta, and so on.

192
00:11:16,050 --> 00:11:19,780
Notice here we have one share
component and another share

193
00:11:19,780 --> 00:11:23,930
component corresponding to this
column and corresponding

194
00:11:23,930 --> 00:11:27,620
to that column, and here we
have k factor, the share

195
00:11:27,620 --> 00:11:32,980
correction factor, that we
already used from the shell

196
00:11:32,980 --> 00:11:37,930
analysis formulation that
we discussed earlier.

197
00:11:37,930 --> 00:11:45,480
Notice that via the same
approach, we can also directly

198
00:11:45,480 --> 00:11:47,430
obtain a beam element for

199
00:11:47,430 --> 00:11:49,860
elastoplastic and creep analysis.

200
00:11:49,860 --> 00:11:52,180
In this particular case, of
course, the only stress

201
00:11:52,180 --> 00:11:55,250
component that would be non-zero
again would be tau

202
00:11:55,250 --> 00:12:00,710
eta eta, tau eta zeta,
and tau eta psi.

203
00:12:00,710 --> 00:12:06,710
The approach here is very
similar to what we have

204
00:12:06,710 --> 00:12:11,690
discussed earlier regarding the
shell element formation.

205
00:12:11,690 --> 00:12:14,700
There is one important point
in the beam element

206
00:12:14,700 --> 00:12:15,330
formulation.

207
00:12:15,330 --> 00:12:19,910
Namely that with the kinematic
assumptions used so far for

208
00:12:19,910 --> 00:12:25,270
the beam element, we do not
allow cross-sectional

209
00:12:25,270 --> 00:12:26,650
out-of-plane displacement.

210
00:12:26,650 --> 00:12:28,930
We do not allow warping.

211
00:12:28,930 --> 00:12:33,220
Our interpolation does not
contain the effect of warping.

212
00:12:33,220 --> 00:12:35,450
In torsion or loading, however,
we know that warping

213
00:12:35,450 --> 00:12:39,580
is very important, and for that
reason, we need to amend

214
00:12:39,580 --> 00:12:43,970
our displacement assumptions
by two or more displacement

215
00:12:43,970 --> 00:12:45,900
patterns, so to say.

216
00:12:45,900 --> 00:12:50,960
We allow these displacements to
take place in the element.

217
00:12:50,960 --> 00:12:54,950
u eta is equal to alpha
times psi beta.

218
00:12:54,950 --> 00:12:57,710
This is the exact warping
displacement for infinitely

219
00:12:57,710 --> 00:12:59,900
narrow sections.

220
00:12:59,900 --> 00:13:05,460
And this term here, with beta
a constant, is the exact

221
00:13:05,460 --> 00:13:09,440
warping displacement for
a square cross section.

222
00:13:09,440 --> 00:13:13,660
Now the actual section might
not be infinitely narrow,

223
00:13:13,660 --> 00:13:19,670
might not be square, but we can
easily test, and I will

224
00:13:19,670 --> 00:13:22,060
show you just now an example
of that in fact, a

225
00:13:22,060 --> 00:13:26,750
superposition of these functions
and the evaluation

226
00:13:26,750 --> 00:13:32,310
of alpha and beta by static
condensation actually yields

227
00:13:32,310 --> 00:13:37,080
good results, even for
cross-sections that are not

228
00:13:37,080 --> 00:13:41,100
infinitely narrow, and that
are not exactly square.

229
00:13:41,100 --> 00:13:44,900
Let's look at an example,
linear elasticity.

230
00:13:44,900 --> 00:13:49,130
If we take a section and simply
put it into torsion

231
00:13:49,130 --> 00:13:54,760
with b/a equal to 1, the
analytical value given by

232
00:13:54,760 --> 00:14:03,190
Timoshenko for this k constant
here in that formula is 0.141.

233
00:14:03,190 --> 00:14:06,080
And we obtain with
this formulation

234
00:14:06,080 --> 00:14:07,850
exactly that same value.

235
00:14:07,850 --> 00:14:11,680
We need to get it, because we
have embedded the proper

236
00:14:11,680 --> 00:14:15,710
interpolation for u eta into
our formulation, and then u

237
00:14:15,710 --> 00:14:18,320
static condensation, of course,
to condense out these

238
00:14:18,320 --> 00:14:22,100
two constants, alpha and beta,
that I just talked about.

239
00:14:22,100 --> 00:14:28,090
For b/a, a very large, very
narrow section, we also get

240
00:14:28,090 --> 00:14:31,895
the exact result in our finite
element solution, the exact

241
00:14:31,895 --> 00:14:33,400
analytical result.

242
00:14:33,400 --> 00:14:38,550
In between, we have a small
arrow, an arrow that, however,

243
00:14:38,550 --> 00:14:39,965
we believe is quite
acceptable.

244
00:14:43,060 --> 00:14:48,400
Let's look at one example that
I'd like to show you the

245
00:14:48,400 --> 00:14:49,540
results of.

246
00:14:49,540 --> 00:14:53,270
You might be interested
in this example.

247
00:14:53,270 --> 00:14:57,010
Here we have a narrow ring.

248
00:14:57,010 --> 00:14:59,680
Here you have a plan view of
that ring with the dimensions

249
00:14:59,680 --> 00:15:03,360
given, and Young's modulus and
Poisson's ratio given for that

250
00:15:03,360 --> 00:15:07,370
ring, subject to these
bending moments.

251
00:15:07,370 --> 00:15:12,160
Of course, the twisting
capability or capacity of the

252
00:15:12,160 --> 00:15:15,780
beam is of utmost importance
now, has to be properly

253
00:15:15,780 --> 00:15:20,810
modelled under this
loading condition.

254
00:15:20,810 --> 00:15:23,120
The model that we use
to analyze the

255
00:15:23,120 --> 00:15:27,700
problem is shown here.

256
00:15:27,700 --> 00:15:32,550
And you can see, at one side of
the ring, we have fixed it,

257
00:15:32,550 --> 00:15:35,210
and at the other side,
we prescribed a

258
00:15:35,210 --> 00:15:37,400
rotation, see eta x.

259
00:15:37,400 --> 00:15:42,090
We use here four node
elements as shown.

260
00:15:44,660 --> 00:15:48,190
This analysis was actually
inspired by us looking around

261
00:15:48,190 --> 00:15:52,240
in our office and seeing a
magnetic tape lying there.

262
00:15:52,240 --> 00:15:55,780
And if you look around, you
might have yourself a magnetic

263
00:15:55,780 --> 00:15:56,660
tape there.

264
00:15:56,660 --> 00:15:59,610
And you pick up this little ring
from the magnetic tape.

265
00:15:59,610 --> 00:16:02,470
And here I'd like to just
demonstrate to you briefly

266
00:16:02,470 --> 00:16:03,650
what we are doing.

267
00:16:03,650 --> 00:16:07,480
You see here, we have this ring,
much alike of what I've

268
00:16:07,480 --> 00:16:09,260
shown you on the view graph.

269
00:16:09,260 --> 00:16:14,320
I'm fixing it on the one side,
and on my right hand now,

270
00:16:14,320 --> 00:16:17,910
these two fingers, I'm going
to put a rotation onto that

271
00:16:17,910 --> 00:16:22,120
ring, and I'm going to turn
my fingers 180 degrees.

272
00:16:22,120 --> 00:16:23,760
So please watch closely.

273
00:16:23,760 --> 00:16:25,460
This is what I'm doing.

274
00:16:25,460 --> 00:16:29,060
I'm just turning my fingers
180 degrees.

275
00:16:29,060 --> 00:16:32,840
And by turning my fingers 180
degrees, if you watch very

276
00:16:32,840 --> 00:16:39,010
closely, you can see that the
top of the ring does not touch

277
00:16:39,010 --> 00:16:39,960
the bottom of the ring.

278
00:16:39,960 --> 00:16:42,720
There is actually a little gap
in between, of course, because

279
00:16:42,720 --> 00:16:44,920
of the elasticity in the ring.

280
00:16:44,920 --> 00:16:50,490
So this is what I've tried to
do now on the computer.

281
00:16:50,490 --> 00:16:54,400
And what I'm particularly
interested to see is the

282
00:16:54,400 --> 00:16:59,160
amount of moment required in my
right hand, so to say, to

283
00:16:59,160 --> 00:17:03,180
be transmitted to the ring
in order to twist it.

284
00:17:03,180 --> 00:17:10,849
Well, here we have the results
for the computer solution.

285
00:17:10,849 --> 00:17:14,839
Notice here, we have a force
deflection, curve, rotation

286
00:17:14,839 --> 00:17:19,109
here, up to 180 degrees.

287
00:17:19,109 --> 00:17:22,540
Moment plotted here
in pounds, inch.

288
00:17:22,540 --> 00:17:27,000
And you can see that there's
fairly linear behavior first,

289
00:17:27,000 --> 00:17:30,770
but then a highly non-linear
behavior as the rotation

290
00:17:30,770 --> 00:17:33,270
becomes very large.

291
00:17:33,270 --> 00:17:36,690
Here we use the TL formulation
for the beam element.

292
00:17:36,690 --> 00:17:40,760
And once again, we rotated it
180 degrees, as I have shown

293
00:17:40,760 --> 00:17:43,720
you with my little
experiment here.

294
00:17:43,720 --> 00:17:51,550
If you look at the beam, you see
a top view like that, and

295
00:17:51,550 --> 00:17:53,540
the side view looks this way.

296
00:17:53,540 --> 00:17:56,610
I've pointed out that
top and bottom do

297
00:17:56,610 --> 00:17:57,910
not touch each other.

298
00:17:57,910 --> 00:17:59,290
There's a little air
gap in between.

299
00:18:02,690 --> 00:18:07,000
I'd like to show you now another
solution regarding the

300
00:18:07,000 --> 00:18:09,320
torsion in an analysis.

301
00:18:09,320 --> 00:18:14,080
And let me walk over to the
slides here, where I actually

302
00:18:14,080 --> 00:18:16,610
have two slides regarding
that solution.

303
00:18:16,610 --> 00:18:20,170
The interesting part of this
problem is that we now, in

304
00:18:20,170 --> 00:18:23,940
this particular analysis, are
using the same displacement

305
00:18:23,940 --> 00:18:28,420
interpolation functions for
which I've shown you already--

306
00:18:28,420 --> 00:18:32,530
two analyses in which the
material remained elastic.

307
00:18:32,530 --> 00:18:35,690
The same interpolation functions
are now also used to

308
00:18:35,690 --> 00:18:38,570
analyze an elastoplastic
problem.

309
00:18:38,570 --> 00:18:42,160
And the problem is described
here, where we have now a

310
00:18:42,160 --> 00:18:46,520
square section subjected
to a torque.

311
00:18:46,520 --> 00:18:52,950
The material data that Greenberg
used are given here.

312
00:18:52,950 --> 00:18:58,370
And in our analysis, we used
this material data, where the

313
00:18:58,370 --> 00:19:02,550
yield stress and the
strain-hardening modulus are

314
00:19:02,550 --> 00:19:06,780
matched to this stress-strain
law here.

315
00:19:06,780 --> 00:19:10,600
Now notice what is going to
happen is, we are increasing

316
00:19:10,600 --> 00:19:14,870
this torque incrementally, and
we want to bring, of course,

317
00:19:14,870 --> 00:19:19,800
this section into the
elastoplastic regime to see

318
00:19:19,800 --> 00:19:23,070
how the section will
take the torque.

319
00:19:23,070 --> 00:19:26,890
On this slide now, we have
plotted the torque vertically

320
00:19:26,890 --> 00:19:31,730
and the rotation horizontally,
both quantities with some

321
00:19:31,730 --> 00:19:34,690
constants attached,
as you can see.

322
00:19:34,690 --> 00:19:39,610
And the Greenberg result and
the Adina result are indeed

323
00:19:39,610 --> 00:19:40,750
very close.

324
00:19:40,750 --> 00:19:43,850
Please look at the reference
that is given in the study

325
00:19:43,850 --> 00:19:47,540
guide if you're interested in
seeing and reading more about

326
00:19:47,540 --> 00:19:48,810
this problem.

327
00:19:48,810 --> 00:19:52,120
Let me now walk back to the view
graph and continue with

328
00:19:52,120 --> 00:19:57,040
the discussion of the material
relating to isoparametric

329
00:19:57,040 --> 00:19:59,510
beams and shell elements.

330
00:19:59,510 --> 00:20:02,900
The next topic that I like to
talk to you about is the

331
00:20:02,900 --> 00:20:05,240
actual use of these elements.

332
00:20:05,240 --> 00:20:11,180
Of course, the shell element
that I'm here talking about,

333
00:20:11,180 --> 00:20:16,030
or that I will share some more
experiences with you upon, is

334
00:20:16,030 --> 00:20:19,320
the shell element that I have
been already discussing in the

335
00:20:19,320 --> 00:20:20,080
earlier lecture.

336
00:20:20,080 --> 00:20:25,810
So please refer back to that
lecture with respect to this

337
00:20:25,810 --> 00:20:28,300
discussion.

338
00:20:28,300 --> 00:20:31,310
One interesting point is that
these elements can all be

339
00:20:31,310 --> 00:20:34,340
programmed for use with
different numbers of nodes.

340
00:20:34,340 --> 00:20:37,320
For the beam, we can have two,
three, or four nodes, as

341
00:20:37,320 --> 00:20:40,530
already mentioned earlier, and
for the shell, we could have

342
00:20:40,530 --> 00:20:45,670
4, 8, 9, up to 16 nodes,
as I mentioned also

343
00:20:45,670 --> 00:20:47,410
in the earlier lecture.

344
00:20:47,410 --> 00:20:50,770
The elements with these nodal
point configurations can be

345
00:20:50,770 --> 00:20:54,670
employed for analysis of
moderately thick structures.

346
00:20:54,670 --> 00:20:56,240
Remember, shear deformations
are

347
00:20:56,240 --> 00:20:58,920
approximately taken into account.

348
00:20:58,920 --> 00:21:03,830
However, if the sink of the
analysis of thin structures,

349
00:21:03,830 --> 00:21:07,700
then we have to be very careful,
and only certain

350
00:21:07,700 --> 00:21:10,610
elements of the ones that
I just referred

351
00:21:10,610 --> 00:21:12,240
to should be employed.

352
00:21:12,240 --> 00:21:16,190
For shells, we can only
recommend to use a 16-node

353
00:21:16,190 --> 00:21:19,620
element with 4-by-4 Gauss
integration over the

354
00:21:19,620 --> 00:21:20,730
mid-surface.

355
00:21:20,730 --> 00:21:22,570
This element is depicted here.

356
00:21:22,570 --> 00:21:28,260
1, 2, 3, 4, 4 times gives
us 16 nodes, and 4-by-4

357
00:21:28,260 --> 00:21:32,540
integration indicated
by the blue crosses.

358
00:21:32,540 --> 00:21:37,870
For beams, we'd best use the
2-node beam element with

359
00:21:37,870 --> 00:21:42,650
1-point integration along the
r-direction, or the 3-node

360
00:21:42,650 --> 00:21:45,200
beam element with 2-point
integration along the

361
00:21:45,200 --> 00:21:48,940
r-direction, all or with Gauss
integration, or the 4-node

362
00:21:48,940 --> 00:21:51,890
beam element with 3-point Gauss
integration along the

363
00:21:51,890 --> 00:21:53,680
r-direction.

364
00:21:53,680 --> 00:21:56,500
We have to ask, of course,
now, why is that so?

365
00:21:56,500 --> 00:22:02,190
Well, the reason is that the
other elements become overly

366
00:22:02,190 --> 00:22:08,570
and artificially stiff when we
use them to model thin and

367
00:22:08,570 --> 00:22:10,400
curved structures.

368
00:22:10,400 --> 00:22:11,580
Two phenomena occur--

369
00:22:11,580 --> 00:22:13,600
shear locking and membrane
locking.

370
00:22:13,600 --> 00:22:15,550
Most interesting phenomena.

371
00:22:15,550 --> 00:22:18,490
And I'd like to share with you
now some information regarding

372
00:22:18,490 --> 00:22:20,950
these phenomena.

373
00:22:20,950 --> 00:22:24,570
The 2-, 3-, and 4-node beam
elements with 1-, 2-, and

374
00:22:24,570 --> 00:22:28,960
3-point Gauss integration along
the beam axes do not

375
00:22:28,960 --> 00:22:31,280
display these phenomena,
these shear- and

376
00:22:31,280 --> 00:22:33,880
membrane-locking phenomena.

377
00:22:33,880 --> 00:22:37,280
But please notice, I'm using
1-, 2-, and 3-point Gauss

378
00:22:37,280 --> 00:22:39,170
integration corresponding
to the 2-, 3-,

379
00:22:39,170 --> 00:22:41,160
and 4-node beam element.

380
00:22:41,160 --> 00:22:45,250
The 16-node element with 4-by-4
integration, Gauss

381
00:22:45,250 --> 00:22:50,400
integration, is also relatively
immune to shear and

382
00:22:50,400 --> 00:22:51,990
membrane locking.

383
00:22:51,990 --> 00:22:53,920
However, we have to
be careful here.

384
00:22:53,920 --> 00:22:56,510
The element should not be
distorted for best predictive

385
00:22:56,510 --> 00:22:58,250
capability.

386
00:22:58,250 --> 00:23:01,720
Let us study what this shear
locking phenomenon is.

387
00:23:01,720 --> 00:23:06,600
And to do so, let us look at a
2-noded beam, a very simple

388
00:23:06,600 --> 00:23:09,190
beam, in just 2-dimensional
action.

389
00:23:09,190 --> 00:23:13,680
So we have one node here and
one node here, and the

390
00:23:13,680 --> 00:23:18,710
defomations of the beam are
described by w1 and theta 1 at

391
00:23:18,710 --> 00:23:24,170
this node, and w2 and theta
2 at this node.

392
00:23:24,170 --> 00:23:27,040
Notice the deformations of the
beam, of course, are given by

393
00:23:27,040 --> 00:23:31,160
w, the transverse displacement,
and beta, the

394
00:23:31,160 --> 00:23:33,860
section rotation.

395
00:23:33,860 --> 00:23:38,370
The transverse displacement and
section rotation, w and

396
00:23:38,370 --> 00:23:42,880
beta, are interpolated linearly,
as shown here, via

397
00:23:42,880 --> 00:23:45,790
the nodal point displacements
and rotations.

398
00:23:45,790 --> 00:23:47,870
Notice the linear interpolation,
of course, is

399
00:23:47,870 --> 00:23:51,320
given here via this expression
and that expression, and

400
00:23:51,320 --> 00:23:53,730
similarly here via this
expression and that

401
00:23:53,730 --> 00:23:54,760
expression.

402
00:23:54,760 --> 00:23:59,430
Now there's one important point,
namely that w and beta

403
00:23:59,430 --> 00:24:05,460
are independent quantities.

404
00:24:05,460 --> 00:24:10,140
If you are familiar, as
you probably are, with

405
00:24:10,140 --> 00:24:15,655
Euler-Bernoulli beam theory,
you recognize that in

406
00:24:15,655 --> 00:24:18,620
Euler-Bernoulli beam theory,
they are not independent

407
00:24:18,620 --> 00:24:19,190
quantities.

408
00:24:19,190 --> 00:24:25,430
In fact, beta is dw dr, or dw
dx, x being the physical

409
00:24:25,430 --> 00:24:27,310
coordinate.

410
00:24:27,310 --> 00:24:31,650
Here in our formation,
isoparametric degenerate

411
00:24:31,650 --> 00:24:37,210
element, we have that beta and
w, these two quantities, once

412
00:24:37,210 --> 00:24:44,430
again, are independent and
linearly interpolated in this

413
00:24:44,430 --> 00:24:47,330
particular beam element.

414
00:24:47,330 --> 00:24:53,190
Of course, these two quantities
are tied together

415
00:24:53,190 --> 00:24:57,000
by the physics, and that is
by the shear deformation.

416
00:24:57,000 --> 00:24:59,230
So let's look at that next.

417
00:24:59,230 --> 00:25:05,010
Here we have that gamma shear
strain is equal to a del w del

418
00:25:05,010 --> 00:25:07,150
x minus beta.

419
00:25:07,150 --> 00:25:10,750
And this is shown here
on the picture.

420
00:25:10,750 --> 00:25:15,500
Notice this is a neutral axis of
the beam, which originally

421
00:25:15,500 --> 00:25:18,790
of course was horizontal.

422
00:25:18,790 --> 00:25:24,580
Here we have a vertical line
shown in green, and the

423
00:25:24,580 --> 00:25:29,000
section rotation is shown
at the angle beta here.

424
00:25:29,000 --> 00:25:32,930
These are the material particles
shown in red that

425
00:25:32,930 --> 00:25:37,090
were originally, of course, at
right angles to the neutral

426
00:25:37,090 --> 00:25:42,300
axes, and that were originally
lying vertically up.

427
00:25:42,300 --> 00:25:45,280
In other words, this line here,
this red line was a

428
00:25:45,280 --> 00:25:49,850
vertical line originally, and
was originally at right angles

429
00:25:49,850 --> 00:25:51,430
to the neutral axis.

430
00:25:51,430 --> 00:25:55,090
That right angle is not
preserved because gamma is

431
00:25:55,090 --> 00:25:59,040
there, and gamma is, in
general, a non-zero.

432
00:25:59,040 --> 00:26:03,610
This is here of course dw dx.

433
00:26:03,610 --> 00:26:07,320
So this picture here
shows what we are

434
00:26:07,320 --> 00:26:10,390
calculating up here.

435
00:26:10,390 --> 00:26:14,900
And you can see that gamma
ties together w and beta,

436
00:26:14,900 --> 00:26:18,960
which are, however, by
themselves, interpolated

437
00:26:18,960 --> 00:26:21,270
independently.

438
00:26:21,270 --> 00:26:25,390
Consider now the simple case of
a cantilever subjected to a

439
00:26:25,390 --> 00:26:27,210
tip-bending moment.

440
00:26:27,210 --> 00:26:30,380
And we use one 2-node beam
element to model the

441
00:26:30,380 --> 00:26:31,920
cantilever.

442
00:26:31,920 --> 00:26:36,780
Clearly at this end, the section
rotation is 0, and w1,

443
00:26:36,780 --> 00:26:39,600
the transverse displacement
is also 0.

444
00:26:39,600 --> 00:26:41,750
So these are the bounded
conditions at this end of the

445
00:26:41,750 --> 00:26:43,100
cantilever.

446
00:26:43,100 --> 00:26:47,720
And if we use these bounded
conditions and put them into

447
00:26:47,720 --> 00:26:50,550
our general equation, we get
directly that beta is given as

448
00:26:50,550 --> 00:26:56,380
shown here, and gamma is
given as shown here.

449
00:26:56,380 --> 00:26:58,520
Now look at this equation
carefully.

450
00:26:58,520 --> 00:27:03,880
We have a constant term and
a linear varying term.

451
00:27:03,880 --> 00:27:08,850
The exact solution to this
problem, analytical solution

452
00:27:08,850 --> 00:27:16,550
to this problem, if we use
elementary beam theory, we

453
00:27:16,550 --> 00:27:20,760
would immediately recognize that
the shear strain must be

454
00:27:20,760 --> 00:27:25,940
0 for this element, or for
this beam situation.

455
00:27:25,940 --> 00:27:27,150
The shear strain should be 0.

456
00:27:27,150 --> 00:27:31,830
Now, if we look at this
carefully, we can see that

457
00:27:31,830 --> 00:27:38,790
gamma can only be 0 if w2
is 0, and theta 2 is 0.

458
00:27:38,790 --> 00:27:43,610
But if theta 2 and w2 are 0,
rotation and transverse

459
00:27:43,610 --> 00:27:46,705
displacement right here at this
end, then the beam would

460
00:27:46,705 --> 00:27:48,080
have not deflected.

461
00:27:48,080 --> 00:27:52,020
Of course, that is not
borne out in physics.

462
00:27:52,020 --> 00:27:54,520
We know that if we take a beam
and put a bending moment on

463
00:27:54,520 --> 00:27:57,760
it, then we get a deflection
here and a

464
00:27:57,760 --> 00:27:59,400
rotation at this end.

465
00:27:59,400 --> 00:28:03,440
So you can see already from this
equation here that there

466
00:28:03,440 --> 00:28:04,800
will be some difficulty.

467
00:28:04,800 --> 00:28:08,250
There will be some difficulty
because gamma should be 0

468
00:28:08,250 --> 00:28:12,880
physically, and this equation
tells us that gamma can't

469
00:28:12,880 --> 00:28:15,010
really be exactly 0.

470
00:28:15,010 --> 00:28:18,630
Well, that is once more
summarized in

471
00:28:18,630 --> 00:28:20,460
this sentence here--

472
00:28:20,460 --> 00:28:23,590
clearly, gamma cannot be 0 at
all the points along the beam

473
00:28:23,590 --> 00:28:27,840
unless theta 2 and
w2 are both 0.

474
00:28:27,840 --> 00:28:31,000
But then also beta would be 0,
and there would be no bending

475
00:28:31,000 --> 00:28:34,050
of the beam.

476
00:28:34,050 --> 00:28:35,590
Let's look now at this point.

477
00:28:35,590 --> 00:28:38,060
Since for the beam, the bending
strain energy is

478
00:28:38,060 --> 00:28:41,610
proportional to h cubed,
elementary beam theory again,

479
00:28:41,610 --> 00:28:45,350
the shear strain energy
is proportional to h--

480
00:28:45,350 --> 00:28:46,860
elementary beam theory--

481
00:28:46,860 --> 00:28:50,720
we see immediately that any
arrow in the shear strains due

482
00:28:50,720 --> 00:28:53,800
to the finite element
interpolation functions

483
00:28:53,800 --> 00:28:56,270
becomes increasingly
more detrimental

484
00:28:56,270 --> 00:28:57,800
is h becomes small.

485
00:28:57,800 --> 00:28:59,240
Why is that the case?

486
00:28:59,240 --> 00:29:04,680
If h becomes small, then
this number becomes

487
00:29:04,680 --> 00:29:06,510
very rapidly small.

488
00:29:06,510 --> 00:29:09,990
This number here become small,
but this one goes much faster

489
00:29:09,990 --> 00:29:12,140
to us, very small, so to say.

490
00:29:12,140 --> 00:29:16,250
And if there's an error in the
shear strain energy, that

491
00:29:16,250 --> 00:29:19,530
error, because of this
proportionality sign, is going

492
00:29:19,530 --> 00:29:23,300
to be magnified relative to any
error that we would see

493
00:29:23,300 --> 00:29:27,340
here, and it's going to be
detrimental in our solution.

494
00:29:27,340 --> 00:29:30,720
Let's look at an example.

495
00:29:30,720 --> 00:29:33,900
For the cantilever example, this
very simple example that

496
00:29:33,900 --> 00:29:37,660
we just studied already a bit,
the shear strain energy should

497
00:29:37,660 --> 00:29:42,510
be 0, and as h decreases, the
relative error in the shear

498
00:29:42,510 --> 00:29:46,630
strain increases rapidly, and
in effect introduces an

499
00:29:46,630 --> 00:29:52,270
artificial stiffness which
we identify as

500
00:29:52,270 --> 00:29:54,100
locking of the model.

501
00:29:54,100 --> 00:29:57,230
This is the terminology that
is quite widely used now.

502
00:29:57,230 --> 00:29:59,050
We say that the model locks.

503
00:29:59,050 --> 00:30:00,280
Let's look at some results.

504
00:30:00,280 --> 00:30:04,420
In this table, we show the
results for h/L with L equal

505
00:30:04,420 --> 00:30:08,400
to 100 for these three values.

506
00:30:08,400 --> 00:30:12,330
The analytical value for the
rotation at the end where the

507
00:30:12,330 --> 00:30:16,500
bending moment is applied
is given here.

508
00:30:16,500 --> 00:30:19,380
Bernoulli beam theory.

509
00:30:19,380 --> 00:30:22,390
The finite element solution,
using, of course, exact

510
00:30:22,390 --> 00:30:23,070
integration--

511
00:30:23,070 --> 00:30:25,630
I always assume I should
maybe point that very

512
00:30:25,630 --> 00:30:27,020
strongly still out.

513
00:30:27,020 --> 00:30:31,400
We assume that we use exact
integration of the K matrix.

514
00:30:31,400 --> 00:30:37,700
Meaning in this particular
case, we're using 2-point

515
00:30:37,700 --> 00:30:40,370
Gauss integration along the
length of the beam.

516
00:30:40,370 --> 00:30:47,480
With that scheme then, we obtain
with a h/L being 0.5.

517
00:30:47,480 --> 00:30:51,310
Well, an error, but something
still reasonable.

518
00:30:51,310 --> 00:30:54,390
After all, we're only
using one element.

519
00:30:54,390 --> 00:30:57,130
Of course, we are pretty far
off, but still something

520
00:30:57,130 --> 00:30:58,050
reasonable.

521
00:30:58,050 --> 00:31:06,070
But now notice, as you decrease
h, h/L, of course,

522
00:31:06,070 --> 00:31:08,170
becomes much smaller.

523
00:31:08,170 --> 00:31:10,200
Theta analytical goes up.

524
00:31:10,200 --> 00:31:13,290
The beam basically
bends, is very

525
00:31:13,290 --> 00:31:15,840
flexible as it gets thinner.

526
00:31:15,840 --> 00:31:19,580
And the finite element solution
also becomes more

527
00:31:19,580 --> 00:31:22,930
flexible, gives us a more
flexible beam, but is much too

528
00:31:22,930 --> 00:31:26,660
stiff when compared to the
analytical solution.

529
00:31:26,660 --> 00:31:29,810
Notice for orders of magnitudes
difference.

530
00:31:29,810 --> 00:31:33,100
If you plot it, this is
basically 0 compared to that

531
00:31:33,100 --> 00:31:36,130
value, and that means
the beam locks.

532
00:31:36,130 --> 00:31:41,740
It doesn't really at all
bend anymore as h

533
00:31:41,740 --> 00:31:43,720
becomes very thin.

534
00:31:43,720 --> 00:31:46,750
This behavior is, of course,
also observed when you take

535
00:31:46,750 --> 00:31:49,970
more than 1 element
to model the beam.

536
00:31:49,970 --> 00:31:54,380
Here we show more elements than
just one, and notice that

537
00:31:54,380 --> 00:31:57,590
each of these elements
should still carry a

538
00:31:57,590 --> 00:31:59,150
constant bending moment.

539
00:31:59,150 --> 00:32:02,130
If it cannot do so, it develops,
in other words,

540
00:32:02,130 --> 00:32:03,750
these furious shear strains.

541
00:32:03,750 --> 00:32:07,780
We are going to have a very
stiff model, even when we use

542
00:32:07,780 --> 00:32:09,200
that many elements.

543
00:32:09,200 --> 00:32:12,600
Let's look at this particular
example close off.

544
00:32:12,600 --> 00:32:18,830
And here we have an example
beam, in other words,

545
00:32:18,830 --> 00:32:22,650
cantilever beam, subjected
to the bending moment.

546
00:32:22,650 --> 00:32:24,190
l equals 10 meter.

547
00:32:24,190 --> 00:32:27,590
Square cross section,
height 0.1 meter.

548
00:32:27,590 --> 00:32:30,520
We use the 2-noded beam
elements to model this

549
00:32:30,520 --> 00:32:36,070
cantilever, and we plot the tip
deflection as a function

550
00:32:36,070 --> 00:32:37,390
of the number of elements.

551
00:32:37,390 --> 00:32:40,270
A most interesting graph.

552
00:32:40,270 --> 00:32:44,570
Look, here is the beam
theory solution.

553
00:32:44,570 --> 00:32:49,640
And here we plot the number
of elements--

554
00:32:49,640 --> 00:32:53,850
here we talk about 400 elements,
because there's

555
00:32:53,850 --> 00:32:55,900
times 10 to the 2 down here.

556
00:32:55,900 --> 00:32:58,950
400 elements at this point.

557
00:32:58,950 --> 00:33:03,540
600 elements, 800 elements,
200 elements.

558
00:33:03,540 --> 00:33:06,060
At this point we have
100 elements.

559
00:33:06,060 --> 00:33:08,160
The height of an element
is equal to the

560
00:33:08,160 --> 00:33:10,750
length of the element.

561
00:33:10,750 --> 00:33:17,150
And at this point, we still
don't even get a good

562
00:33:17,150 --> 00:33:19,900
approximation to the beam
theory solution.

563
00:33:19,900 --> 00:33:24,620
We need basically something like
400 to 500, 600 elements

564
00:33:24,620 --> 00:33:26,870
to get close to the beam
theory solution.

565
00:33:26,870 --> 00:33:30,530
However, if we take a very large
number of elements, we

566
00:33:30,530 --> 00:33:34,150
actually do converge, but we
need a lot of elements, as you

567
00:33:34,150 --> 00:33:35,520
can see here.

568
00:33:35,520 --> 00:33:39,390
Of course, it is quite
impractical to use this kind

569
00:33:39,390 --> 00:33:43,970
of 2-noded beam element to model
thin structures, then

570
00:33:43,970 --> 00:33:48,080
beams. Once again, I must point
out that we use 2-point

571
00:33:48,080 --> 00:33:50,430
Gauss integration into the
r-direction, and therefore,

572
00:33:50,430 --> 00:33:56,410
we're dealing with the exact K
matrix corresponding to the

573
00:33:56,410 --> 00:33:59,720
interpolation that we have
embedded into the element.

574
00:33:59,720 --> 00:34:04,690
In fact, if we look in this
regime down here, we find that

575
00:34:04,690 --> 00:34:09,270
as shown on this view graph,
the solution really becomes

576
00:34:09,270 --> 00:34:13,280
beta very slowly as the number
of elements is increased.

577
00:34:13,280 --> 00:34:16,840
Notice that the solution really
doesn't increase in

578
00:34:16,840 --> 00:34:23,230
accuracy much at all as we go
from 10, 20, to 30 elements.

579
00:34:23,230 --> 00:34:27,790
A remedy for the 2-node beam
element is to only use 1-point

580
00:34:27,790 --> 00:34:30,690
integration along
the beam axis.

581
00:34:30,690 --> 00:34:34,020
This then corresponds to
assuming a constant transverse

582
00:34:34,020 --> 00:34:35,270
shear strain.

583
00:34:35,270 --> 00:34:38,030
However, by using 1-point
integration, we still

584
00:34:38,030 --> 00:34:41,159
integrate the bending
strain exactly.

585
00:34:41,159 --> 00:34:44,929
Because remember, beta is
linear, del beta del x is

586
00:34:44,929 --> 00:34:47,449
constant, and 1-point
integration will pick up that

587
00:34:47,449 --> 00:34:49,010
term exactly.

588
00:34:49,010 --> 00:34:53,909
If we look at the results, now
once again, of this problem,

589
00:34:53,909 --> 00:34:57,610
cantilever problem, the
cantilever beam is subjected

590
00:34:57,610 --> 00:34:59,760
to an end moment,
tip end moment.

591
00:34:59,760 --> 00:35:05,850
We find now which h/l, as shown
here, theta analytical,

592
00:35:05,850 --> 00:35:09,480
shown here, and our finite
elements solution with just

593
00:35:09,480 --> 00:35:12,680
one element, using one element,
of course now with

594
00:35:12,680 --> 00:35:16,230
1-point integration in
r-direction, gives us

595
00:35:16,230 --> 00:35:17,220
excellent results.

596
00:35:17,220 --> 00:35:22,010
In fact, the exact results,
as you can see here.

597
00:35:22,010 --> 00:35:26,440
The 3- and 4-note beam elements,
evaluated using 2-

598
00:35:26,440 --> 00:35:29,840
and 3-point integration,
are similarly affected.

599
00:35:29,840 --> 00:35:33,540
In fact, they give us also the
exact results for this problem

600
00:35:33,540 --> 00:35:34,820
that we've considered.

601
00:35:34,820 --> 00:35:39,100
We should note now that these
theme elements, based on what

602
00:35:39,100 --> 00:35:44,250
one might reduced integration,
are actually reliable

603
00:35:44,250 --> 00:35:48,510
elements, because they do not
possess any spurious zero

604
00:35:48,510 --> 00:35:49,880
energy modes.

605
00:35:49,880 --> 00:35:57,440
They have, in other words, only
six zero energy modes

606
00:35:57,440 --> 00:36:01,780
corresponding to the actual
physical rigid body mode that

607
00:36:01,780 --> 00:36:05,470
the elements should contain,
should be able to represent,

608
00:36:05,470 --> 00:36:07,550
in its three-dimensional
space.

609
00:36:07,550 --> 00:36:12,590
So they only have those six zero
eigenvalues in this 3D

610
00:36:12,590 --> 00:36:15,450
analysis that they should have,
and no spurious zero

611
00:36:15,450 --> 00:36:17,000
energy modes.

612
00:36:17,000 --> 00:36:19,610
The formulation can actually
be interpreted as a mixed

613
00:36:19,610 --> 00:36:23,500
interpolation of displacements
and transferred shear strains.

614
00:36:23,500 --> 00:36:30,790
And this is really the key for
developing all the low-order

615
00:36:30,790 --> 00:36:32,550
effective shell elements.

616
00:36:32,550 --> 00:36:37,170
I will get back to that in
just a little while.

617
00:36:37,170 --> 00:36:40,170
So far, we have talked
about shear locking.

618
00:36:40,170 --> 00:36:43,490
I mentioned earlier also that
there is a phenomenon of

619
00:36:43,490 --> 00:36:45,710
membrane-locking.

620
00:36:45,710 --> 00:36:49,010
Membrane-locking means that in
addition to not exhibiting

621
00:36:49,010 --> 00:36:54,130
erroneous shear strains, the
beam model must also not

622
00:36:54,130 --> 00:36:59,350
contain erroneous mid-surface
membrane strains, particularly

623
00:36:59,350 --> 00:37:02,460
when we analyze curved
structures.

624
00:37:02,460 --> 00:37:06,660
The beam elements that I just
mentioned, 2-noded beam with

625
00:37:06,660 --> 00:37:10,040
1-point r-integration, 3-noded
beam with 2-point

626
00:37:10,040 --> 00:37:14,190
r-integration, 4-noded beam with
3-point r-integration, do

627
00:37:14,190 --> 00:37:16,690
also not membrane-lock.

628
00:37:16,690 --> 00:37:20,630
Let's consider here, once again,
a simple example--

629
00:37:20,630 --> 00:37:24,730
a simple curve beam subjected
to a bending moment,

630
00:37:24,730 --> 00:37:28,190
tip-bending moment, fixed
at the other end.

631
00:37:28,190 --> 00:37:33,380
Here is an angle alpha that
the beam spans out.

632
00:37:33,380 --> 00:37:39,390
The exactly integrated 3-noded
beam element, when curved,

633
00:37:39,390 --> 00:37:42,690
does contain erroneous shear
strains and erroneous

634
00:37:42,690 --> 00:37:44,430
mid-surface membrane strains.

635
00:37:44,430 --> 00:37:47,080
In other words, shear locks
and membrane locks.

636
00:37:47,080 --> 00:37:50,780
And that is shown in
this table here, at

637
00:37:50,780 --> 00:37:51,940
least to some extent.

638
00:37:51,940 --> 00:37:54,860
What I can't show is how much
there is shear-locking and how

639
00:37:54,860 --> 00:37:56,560
much there is membrane-locking.

640
00:37:56,560 --> 00:37:59,530
You would see that both effects
are very significant.

641
00:37:59,530 --> 00:38:05,300
If you were to run, for example,
this beam element

642
00:38:05,300 --> 00:38:10,440
with a computer program, using,
for this beam element,

643
00:38:10,440 --> 00:38:15,740
3-point integration into the r
direction and printing out the

644
00:38:15,740 --> 00:38:20,040
stresses, print out all the
stress components and look at

645
00:38:20,040 --> 00:38:23,850
the shear component and the
membrane components, in other

646
00:38:23,850 --> 00:38:25,250
words, the normal stress
components,

647
00:38:25,250 --> 00:38:26,730
along the neutral axes.

648
00:38:26,730 --> 00:38:29,390
If you do so, you would see that
they are very significant

649
00:38:29,390 --> 00:38:32,760
shear stresses and very
significant membrane stresses,

650
00:38:32,760 --> 00:38:35,050
which of course should be
0 for this problem.

651
00:38:35,050 --> 00:38:39,710
And these are, of course,
causing the locking effect.

652
00:38:39,710 --> 00:38:46,690
Here we have the results in
terms of just rotations, h/R

653
00:38:46,690 --> 00:38:50,720
now, with R equal to 100, given
here, analytical value

654
00:38:50,720 --> 00:38:54,070
given here, finite element
solutions, 3-node element with

655
00:38:54,070 --> 00:38:57,760
3-point integration, you can
see we have a locking

656
00:38:57,760 --> 00:39:01,110
phenomenon right here.

657
00:39:01,110 --> 00:39:03,900
Once again, how much there is
membrane, and how much shear

658
00:39:03,900 --> 00:39:06,170
locking you would see by
actually running as a problem

659
00:39:06,170 --> 00:39:07,890
and looking at the
shear stress and

660
00:39:07,890 --> 00:39:10,050
the membrane stress.

661
00:39:10,050 --> 00:39:13,040
And the finite element
solution with 2-point

662
00:39:13,040 --> 00:39:16,315
integration gives excellent
results.

663
00:39:19,610 --> 00:39:22,620
So this is, of course, the
element that we would be using

664
00:39:22,620 --> 00:39:24,660
in engineering practice.

665
00:39:24,660 --> 00:39:26,660
Similarly, we can study
the use of the

666
00:39:26,660 --> 00:39:28,470
4-node cubic beam element.

667
00:39:28,470 --> 00:39:34,800
And here we find that for h/R,
these values, analytical value

668
00:39:34,800 --> 00:39:38,280
once more listed here, we find
that the 4-node element with

669
00:39:38,280 --> 00:39:43,850
4-point integration gives also
very good results, and the

670
00:39:43,850 --> 00:39:48,320
3-point integration gives
also excellent results.

671
00:39:48,320 --> 00:39:52,580
So we note that the cubic beam
element performs well, even

672
00:39:52,580 --> 00:39:55,750
when using full integration.

673
00:39:55,750 --> 00:39:59,220
It is not susceptible to
membrane and shear locking.

674
00:39:59,220 --> 00:40:04,610
But notice, we have our
third-point node exactly at

675
00:40:04,610 --> 00:40:06,000
the third point of
the elements.

676
00:40:06,000 --> 00:40:09,290
Once we start shifting these
nodes, then we would also not

677
00:40:09,290 --> 00:40:11,410
get as good results
as shown here.

678
00:40:13,920 --> 00:40:16,590
We would not get these good
results once you shift the

679
00:40:16,590 --> 00:40:20,500
node at the third point away
from their physical third

680
00:40:20,500 --> 00:40:22,960
point locations.

681
00:40:22,960 --> 00:40:26,800
Now, considering the analysis
of shells, the phenomenon of

682
00:40:26,800 --> 00:40:29,700
shear and membrane locking
are also present, but the

683
00:40:29,700 --> 00:40:33,380
difficulty lies in that this
simple reduced integration

684
00:40:33,380 --> 00:40:39,250
approach that we're using for
the beam elements cannot be

685
00:40:39,250 --> 00:40:43,940
directly applied, because the
resulting elements contain

686
00:40:43,940 --> 00:40:46,040
spurious zero energy modes.

687
00:40:46,040 --> 00:40:52,590
For example, if you integrate
the 4-node shell element just

688
00:40:52,590 --> 00:40:56,320
with 1-point integration,
contains 6 spurious zero

689
00:40:56,320 --> 00:40:57,000
energy modes.

690
00:40:57,000 --> 00:41:00,780
In other words, it has
altogether 12 zero eigenvalues

691
00:41:00,780 --> 00:41:03,600
where it should only have 6.

692
00:41:03,600 --> 00:41:06,320
The 6 of them are
spurious modes.

693
00:41:06,320 --> 00:41:11,070
And such spurious energy modes
can lead to very large errors

694
00:41:11,070 --> 00:41:15,200
in the solution that, unless
we have a comparison with

695
00:41:15,200 --> 00:41:19,560
accurate results, are not known,
and therefore, it can

696
00:41:19,560 --> 00:41:23,580
be very dangerous to
use such elements.

697
00:41:23,580 --> 00:41:27,270
We want to have elements that
are reliable in general

698
00:41:27,270 --> 00:41:31,300
applications, and these elements
should only contain

699
00:41:31,300 --> 00:41:34,380
the actual physical rigid
body modes, no

700
00:41:34,380 --> 00:41:36,430
spurious zero energy modes.

701
00:41:36,430 --> 00:41:38,870
And of course, should also
have high predictive

702
00:41:38,870 --> 00:41:43,520
capability, not have any shear
or membrane locking.

703
00:41:43,520 --> 00:41:47,920
For this reason, we can only
recommend the 16-node shell

704
00:41:47,920 --> 00:41:50,510
element with 4-by-4
Gauss integration.

705
00:41:50,510 --> 00:41:55,640
The cubic beam element result
that I just showed you

706
00:41:55,640 --> 00:41:59,720
regarding the analysis of the
curve beam already indicated

707
00:41:59,720 --> 00:42:02,410
that the cubic element
does quite well.

708
00:42:02,410 --> 00:42:04,550
In fact, that is also
borne out by the

709
00:42:04,550 --> 00:42:06,260
16-node shell element.

710
00:42:06,260 --> 00:42:10,710
Using 4-by-4 Gauss integration
on the shell mid-surface, we

711
00:42:10,710 --> 00:42:14,470
have an element that is
relatively immune to membrane

712
00:42:14,470 --> 00:42:17,190
and shear locking, but I
mentioned already that the

713
00:42:17,190 --> 00:42:19,320
element should not
be distorted.

714
00:42:19,320 --> 00:42:23,010
And if it's not distorted,
it performs best.

715
00:42:23,010 --> 00:42:25,130
I'd just like to share now
with you some thoughts

716
00:42:25,130 --> 00:42:29,260
regarding development that
we have recently pursued.

717
00:42:29,260 --> 00:42:32,550
Namely, recently we have
developed elements based on

718
00:42:32,550 --> 00:42:35,800
the mixed interpolation of
tensorial components.

719
00:42:35,800 --> 00:42:38,030
This sounds very complicated.

720
00:42:38,030 --> 00:42:41,240
Actually, if you look closely at
what we're doing, it is not

721
00:42:41,240 --> 00:42:42,790
that complicated.

722
00:42:42,790 --> 00:42:45,920
The elements, however, do not
lock in shear or membrane

723
00:42:45,920 --> 00:42:49,100
action, and also do not contain
any spurious zero

724
00:42:49,100 --> 00:42:49,810
energy mode.

725
00:42:49,810 --> 00:42:54,040
And of course, that is the key
of success of an element.

726
00:42:54,040 --> 00:42:59,760
We will use later on in the
example solutions a 4-node

727
00:42:59,760 --> 00:43:03,650
element that is based on this
mixed interpolation of

728
00:43:03,650 --> 00:43:05,080
tensorial components.

729
00:43:05,080 --> 00:43:10,030
We call that element the MITC4,
Mixed Interpolation

730
00:43:10,030 --> 00:43:15,150
Tensorial Component with 4 nodes
element, and I want to

731
00:43:15,150 --> 00:43:17,470
discuss this element just
briefly with you now.

732
00:43:17,470 --> 00:43:20,480
We also have developed
an element that

733
00:43:20,480 --> 00:43:23,890
we call MITC8 element.

734
00:43:23,890 --> 00:43:27,475
It's an 8-node element based
on mixed interpolation of

735
00:43:27,475 --> 00:43:28,770
tensorial components.

736
00:43:28,770 --> 00:43:31,530
A most interesting
development.

737
00:43:31,530 --> 00:43:34,380
If you're interested in that
development, please refer to

738
00:43:34,380 --> 00:43:37,710
the paper that is given, or that
is being referred to, in

739
00:43:37,710 --> 00:43:39,440
the study guide.

740
00:43:39,440 --> 00:43:43,370
I'd like to discuss with you
now the MITC4 element, and

741
00:43:43,370 --> 00:43:45,110
then show you example
solutions.

742
00:43:45,110 --> 00:43:49,200
However, we have, in order to be
able to do that, to change

743
00:43:49,200 --> 00:43:53,550
now the reel, so let us just do
so, and then continue with

744
00:43:53,550 --> 00:43:54,800
this discussion.

745
00:43:58,390 --> 00:44:03,240
So let us look then at the MITC4
element briefly, the

746
00:44:03,240 --> 00:44:06,240
element that I mentioned before
we had the break.

747
00:44:06,240 --> 00:44:11,320
The element is described
using four nodes.

748
00:44:11,320 --> 00:44:15,200
We have the same isoparametric
coordinate system that we are

749
00:44:15,200 --> 00:44:16,450
usually having.

750
00:44:16,450 --> 00:44:20,310
Notice here director vector,
same description as for the

751
00:44:20,310 --> 00:44:22,160
shell element that we discussed

752
00:44:22,160 --> 00:44:24,860
in the earlier lecture.

753
00:44:24,860 --> 00:44:29,510
Each node has 5 degrees of
freedom, 2 rotational degrees

754
00:44:29,510 --> 00:44:30,530
of freedom, and 3

755
00:44:30,530 --> 00:44:32,330
translational degrees of freedom.

756
00:44:32,330 --> 00:44:34,755
Once again, the way we have
been discussing it in the

757
00:44:34,755 --> 00:44:35,710
earlier lecture.

758
00:44:35,710 --> 00:44:38,660
We use the element for analysis
of plates and for the

759
00:44:38,660 --> 00:44:42,440
analysis of moderately thick
shells, and also thin shells.

760
00:44:42,440 --> 00:44:43,950
And that is the important
point.

761
00:44:43,950 --> 00:44:47,410
This element is directly
applicable in a very effective

762
00:44:47,410 --> 00:44:50,340
manner to thin shell analysis.

763
00:44:50,340 --> 00:44:53,310
The key step in the formulation
of the element is

764
00:44:53,310 --> 00:44:57,560
to interpolate the geometry
and the displacement as

765
00:44:57,560 --> 00:44:59,340
earlier described,
as we discussed

766
00:44:59,340 --> 00:45:01,110
in the earlier lecture.

767
00:45:01,110 --> 00:45:04,640
But to interpret the transverse
shear strain tensor

768
00:45:04,640 --> 00:45:09,040
components separately, and
these interpolations are

769
00:45:09,040 --> 00:45:13,710
selected judiciously, to tie,
then, the intensities of these

770
00:45:13,710 --> 00:45:16,600
components to the values
evaluated using the

771
00:45:16,600 --> 00:45:19,030
displacement interpolations.

772
00:45:19,030 --> 00:45:23,370
Let's look at what I'm saying
here now pictorially.

773
00:45:23,370 --> 00:45:27,860
The strain tensor interpolation
that we are

774
00:45:27,860 --> 00:45:36,100
using is shown here in blue
on the black element.

775
00:45:36,100 --> 00:45:38,960
These are the rt transfer
shear strain tensor

776
00:45:38,960 --> 00:45:41,230
components.

777
00:45:41,230 --> 00:45:44,390
This is one interpolation
function, and this is the

778
00:45:44,390 --> 00:45:46,720
other interpolation function.

779
00:45:46,720 --> 00:45:53,240
Notice that if we look at the
r- and s-axes, the rt strain

780
00:45:53,240 --> 00:45:57,100
tensor component is constant
in r, but varies

781
00:45:57,100 --> 00:45:59,280
linearly with s.

782
00:45:59,280 --> 00:46:03,270
Here, too, constant in 4,
varying linearly with s.

783
00:46:03,270 --> 00:46:08,670
Now it's these intensities, one
here and one there, that

784
00:46:08,670 --> 00:46:14,820
we tie to the nodal point
displacements and rotations.

785
00:46:14,820 --> 00:46:19,720
Here for this intensity, we
use these nodal point

786
00:46:19,720 --> 00:46:23,870
descriptions, and for this
intensity, we use these nodal

787
00:46:23,870 --> 00:46:27,880
point descriptions, or nodal
point degrees of freedom.

788
00:46:27,880 --> 00:46:32,710
So if you recognize that these
are, it seems, unknown

789
00:46:32,710 --> 00:46:36,570
quantities that enter into the
formulation since we're using

790
00:46:36,570 --> 00:46:38,190
these two interpolations--

791
00:46:38,190 --> 00:46:41,370
this one is an unknown quantity
that enters into the

792
00:46:41,370 --> 00:46:44,100
formulation, that one is an
unknown quantity that enters

793
00:46:44,100 --> 00:46:45,380
into the formulation--

794
00:46:45,380 --> 00:46:48,940
if you recognize that, then you
have to ask yourself, what

795
00:46:48,940 --> 00:46:50,810
do we do with these unknown
quantities?

796
00:46:50,810 --> 00:46:54,580
Well, we eliminate these
unknown quantities by

797
00:46:54,580 --> 00:46:57,860
expressing them in terms of
the nodal point degrees of

798
00:46:57,860 --> 00:47:01,650
freedom, these two nodes here
and these two nodes.

799
00:47:01,650 --> 00:47:08,020
And thus we eliminate these
tensor component intensities.

800
00:47:08,020 --> 00:47:12,520
And in the K matrix and the F
vector, we end up having then

801
00:47:12,520 --> 00:47:14,160
only the nodal point

802
00:47:14,160 --> 00:47:15,770
displacement degrees of freedom.

803
00:47:15,770 --> 00:47:18,380
And that's the key part.

804
00:47:18,380 --> 00:47:21,620
For the ft transfer shear strain
tensor components, we

805
00:47:21,620 --> 00:47:24,640
use these two interpolations.

806
00:47:24,640 --> 00:47:29,240
Now we notice that we have a
linear description in r and

807
00:47:29,240 --> 00:47:33,690
constant in s for both of
these two components.

808
00:47:33,690 --> 00:47:37,460
We proceed, of course, to
eliminate this intensity and

809
00:47:37,460 --> 00:47:40,350
that intensity in terms
of the nodal

810
00:47:40,350 --> 00:47:42,630
point degrees of freedom.

811
00:47:42,630 --> 00:47:45,510
Same way as we are
proceeding here.

812
00:47:45,510 --> 00:47:50,130
This element, the MITC4 element,
has only 6 zero

813
00:47:50,130 --> 00:47:52,690
eigenvalues, no spurious
zero energy modes.

814
00:47:52,690 --> 00:47:57,760
It passes a patch test. And
this is a very important

815
00:47:57,760 --> 00:48:01,290
point-- that the element process
the patch test. What

816
00:48:01,290 --> 00:48:03,400
do we mean by the patch test?

817
00:48:03,400 --> 00:48:08,190
Well, the idea in the patch
test is that any arbitrary

818
00:48:08,190 --> 00:48:11,280
patch of elements should be
able to represent constant

819
00:48:11,280 --> 00:48:13,860
stress conditions.

820
00:48:13,860 --> 00:48:19,100
Let's see how we perform that
patch test. Here we take an

821
00:48:19,100 --> 00:48:21,460
arbitrary patch of elements.

822
00:48:21,460 --> 00:48:26,630
Some of the elements in that
patch of elements would be

823
00:48:26,630 --> 00:48:28,580
geometrically distorted.

824
00:48:28,580 --> 00:48:32,150
And we subject this patch to
the minimum displacement

825
00:48:32,150 --> 00:48:34,450
rotation boundary conditions
to eliminate the physical

826
00:48:34,450 --> 00:48:38,590
rigid body molds, and then
constant boundary tractions

827
00:48:38,590 --> 00:48:40,380
corresponding to the
constant stress

828
00:48:40,380 --> 00:48:43,600
condition that is tested.

829
00:48:43,600 --> 00:48:51,440
These two items we apply to
the patch of elements, and

830
00:48:51,440 --> 00:48:54,100
then we calculate
all nodal point

831
00:48:54,100 --> 00:48:56,840
displacement and element stresses.

832
00:48:56,840 --> 00:48:59,510
The patch test is passed if
the calculated element

833
00:48:59,510 --> 00:49:01,000
internal stresses
and nodal point

834
00:49:01,000 --> 00:49:03,000
displacements are correct.

835
00:49:03,000 --> 00:49:05,460
In other words, let's
go through this

836
00:49:05,460 --> 00:49:06,830
thought once again.

837
00:49:06,830 --> 00:49:08,660
What we're doing,
really, is we're

838
00:49:08,660 --> 00:49:11,210
looking at this structure.

839
00:49:11,210 --> 00:49:15,670
In essence, we are taking out
of this big mesh of elements

840
00:49:15,670 --> 00:49:18,960
in the structure a certain
set of elements.

841
00:49:18,960 --> 00:49:22,040
We call that set of elements
a patch of elements.

842
00:49:22,040 --> 00:49:26,320
We apply to this patch of
elements the minimum boundary

843
00:49:26,320 --> 00:49:31,910
conditions to eliminate the
rigid body molds and traction

844
00:49:31,910 --> 00:49:36,550
along the boundary that should
result in constant stress

845
00:49:36,550 --> 00:49:37,800
conditions within
the elements.

846
00:49:40,310 --> 00:49:44,800
If in fact, our solution for
that patch of elements gives

847
00:49:44,800 --> 00:49:49,330
us constant stress conditions
and nodal point displacement

848
00:49:49,330 --> 00:49:52,320
that correspond to these
constant stress conditions,

849
00:49:52,320 --> 00:49:53,870
then the patch test is passed.

850
00:49:53,870 --> 00:49:57,190
and this means that we would
have an element that

851
00:49:57,190 --> 00:50:00,930
ultimately, as the mesh is made
finer and finer with that

852
00:50:00,930 --> 00:50:04,430
element, the solution
will converge

853
00:50:04,430 --> 00:50:06,450
to the correct solution.

854
00:50:06,450 --> 00:50:10,200
That's the patch test, and
the most important test,

855
00:50:10,200 --> 00:50:13,670
particularly for shell elements,
when we have these

856
00:50:13,670 --> 00:50:18,470
complications of various kinds
of interpolations in the

857
00:50:18,470 --> 00:50:20,530
element formulation,
and so on.

858
00:50:20,530 --> 00:50:26,210
Well, here, schematically, I
show such patch of elements.

859
00:50:26,210 --> 00:50:31,070
x1-axis, x2-axes, flat
patch of elements.

860
00:50:31,070 --> 00:50:33,870
Notice the elements
not distorted,

861
00:50:33,870 --> 00:50:35,570
not rectangular elements.

862
00:50:35,570 --> 00:50:37,750
We take certain material
properties,

863
00:50:37,750 --> 00:50:39,380
assign a certain thickness.

864
00:50:39,380 --> 00:50:43,250
We make it a thin patch
of elements.

865
00:50:43,250 --> 00:50:47,940
The plate that we're looking at
here is, in essence, thin.

866
00:50:47,940 --> 00:50:54,340
Notice thickness 0.01, width
10 in both directions.

867
00:50:54,340 --> 00:50:58,460
And so if the elements would
lock do to shear or membrane

868
00:50:58,460 --> 00:51:01,350
locking, the way we discussed
earlier, we would certainly

869
00:51:01,350 --> 00:51:05,000
see it by subjecting this patch
of elements to bending

870
00:51:05,000 --> 00:51:07,640
moment boundary conditions.

871
00:51:07,640 --> 00:51:11,580
The membrane test would
correspond to these stress

872
00:51:11,580 --> 00:51:13,070
conditions.

873
00:51:13,070 --> 00:51:16,760
The bending twisting test would
correspond to these

874
00:51:16,760 --> 00:51:18,520
stress conditions.

875
00:51:18,520 --> 00:51:25,050
These are the externally applied
forces, tractions that

876
00:51:25,050 --> 00:51:29,340
the patch of element is
subjected for bending twisting

877
00:51:29,340 --> 00:51:33,880
tests, here for membrane
tests.

878
00:51:33,880 --> 00:51:41,700
And once again, we adjust the
rigid body modes from the

879
00:51:41,700 --> 00:51:45,240
patch of elements by setting
the appropriate number of

880
00:51:45,240 --> 00:51:47,520
degrees of freedom, nodal point
degrees of freedom,

881
00:51:47,520 --> 00:51:51,720
equal to zero, and then subject
that patch of elements

882
00:51:51,720 --> 00:51:59,670
to the tractions corresponding
to these stress conditions,

883
00:51:59,670 --> 00:52:01,660
the six ones that I'm
showing here.

884
00:52:01,660 --> 00:52:06,740
And we measure the internal
stresses, and find whether

885
00:52:06,740 --> 00:52:10,030
these internal stresses indeed
correspond to the constant

886
00:52:10,030 --> 00:52:11,450
stress situation.

887
00:52:11,450 --> 00:52:15,770
Also, since only those nodal
point degrees of freedom have

888
00:52:15,770 --> 00:52:19,810
been removed on the boundary
that take out the rigid body

889
00:52:19,810 --> 00:52:23,910
modes, we should also see that
the other nodal point degrees

890
00:52:23,910 --> 00:52:27,800
of freedom, of course, have
taken on the exact analytical

891
00:52:27,800 --> 00:52:30,340
value corresponding to
the constant stress

892
00:52:30,340 --> 00:52:31,850
condition that is tested.

893
00:52:31,850 --> 00:52:36,150
This one, that one, that one,
or the other three here.

894
00:52:36,150 --> 00:52:37,930
That's the patch test.

895
00:52:37,930 --> 00:52:43,390
Well, let us now to look at some
examples regarding the

896
00:52:43,390 --> 00:52:45,830
elements that we have discussed
in the previous

897
00:52:45,830 --> 00:52:47,590
lecture and this lecture.

898
00:52:47,590 --> 00:52:51,200
And the first example that I
like to look at with you is

899
00:52:51,200 --> 00:52:54,130
the analysis of a spherical
shell with a hole.

900
00:52:54,130 --> 00:52:57,920
There's a hole here, and you
see the sphere here.

901
00:52:57,920 --> 00:52:59,720
Here's the backside
of the sphere,

902
00:52:59,720 --> 00:53:02,160
shown by a dashed line.

903
00:53:02,160 --> 00:53:07,200
And this sphere is subjected to
concentrated forces, shown

904
00:53:07,200 --> 00:53:08,730
by these arrows.

905
00:53:08,730 --> 00:53:11,780
Radius given here, thickness
given here, and the material

906
00:53:11,780 --> 00:53:16,750
property of the sphere
given right there.

907
00:53:16,750 --> 00:53:21,490
The first step is to select
director vectors.

908
00:53:21,490 --> 00:53:24,890
We talked in the previous
lecture very heavily about the

909
00:53:24,890 --> 00:53:27,580
fact that we use director
vectors, and we have to define

910
00:53:27,580 --> 00:53:29,110
these director vectors.

911
00:53:29,110 --> 00:53:31,430
The initial director vectors,
or the director vectors for

912
00:53:31,430 --> 00:53:33,180
the initial configuration.

913
00:53:33,180 --> 00:53:36,290
And we in fact, in this
particular case, can quite

914
00:53:36,290 --> 00:53:39,700
easily generate the director
vectors for each node.

915
00:53:39,700 --> 00:53:42,970
The director vector for each
node, in this particular case,

916
00:53:42,970 --> 00:53:46,550
is chosen to be parallel
to the radial vector.

917
00:53:46,550 --> 00:53:52,010
So if we look, this is the skin
of the shell from the

918
00:53:52,010 --> 00:53:56,110
midpoint out via the radial
vector, we would see the

919
00:53:56,110 --> 00:53:59,840
director vector right here
in two dimensions, of

920
00:53:59,840 --> 00:54:03,140
course, x and y.

921
00:54:03,140 --> 00:54:06,020
So we generate these
director vectors.

922
00:54:06,020 --> 00:54:08,690
In ADINA, that can be
done automatically.

923
00:54:08,690 --> 00:54:12,300
And then we would select the
boundary, displacement

924
00:54:12,300 --> 00:54:14,520
boundary conditions.

925
00:54:14,520 --> 00:54:21,730
If we look at what happens, for
example, in the zx phase,

926
00:54:21,730 --> 00:54:25,120
using symmetry conditions, of
course we would only look at a

927
00:54:25,120 --> 00:54:27,230
particular part of the shell.

928
00:54:27,230 --> 00:54:29,990
The part of the shell, in fact,
maybe I should show you

929
00:54:29,990 --> 00:54:35,640
that one first, that we want
to look at is shown here.

930
00:54:35,640 --> 00:54:39,450
Notice here we have the part of
the shell that we analyze

931
00:54:39,450 --> 00:54:41,100
because of symmetry
conditions.

932
00:54:41,100 --> 00:54:43,470
We have here a symmetry
boundary and

933
00:54:43,470 --> 00:54:45,500
there a symmetry boundary.

934
00:54:45,500 --> 00:54:51,120
Notice that we have x, y, and z,
the coordinate system used

935
00:54:51,120 --> 00:54:58,230
here, and that this part of the
shell really lies in the

936
00:54:58,230 --> 00:55:01,690
part of the coordinate system
that we're looking at.

937
00:55:01,690 --> 00:55:08,920
In other words, zy, y here,
z up there, xz, x

938
00:55:08,920 --> 00:55:11,530
here, z comes out there.

939
00:55:11,530 --> 00:55:15,930
So here we have y equal to 0 on
this face, and on this face

940
00:55:15,930 --> 00:55:18,890
here, we have x equal to 0.

941
00:55:18,890 --> 00:55:21,760
And what I'd like to now talk
briefly about is about the

942
00:55:21,760 --> 00:55:23,510
selection of the boundary
conditions

943
00:55:23,510 --> 00:55:25,285
along these two faces.

944
00:55:27,810 --> 00:55:30,550
Well, these boundary conditions,
if we look at the

945
00:55:30,550 --> 00:55:37,710
zx face, in other words, this
face here, we see on this view

946
00:55:37,710 --> 00:55:41,520
graph here the x face--
here is the shell.

947
00:55:41,520 --> 00:55:45,260
Here we would have our director
vector at time 0, and

948
00:55:45,260 --> 00:55:48,640
at time t, that director vector
would have moved as

949
00:55:48,640 --> 00:55:52,530
indicated by the red here.

950
00:55:52,530 --> 00:55:58,910
Notice that with this movement,
the only rotation

951
00:55:58,910 --> 00:56:03,400
that we can have is the
theta y rotation.

952
00:56:03,400 --> 00:56:08,000
And that is now shown on
the next view graph.

953
00:56:08,000 --> 00:56:13,110
Notice theta y here is free,
a free rotation.

954
00:56:13,110 --> 00:56:16,180
Theta x must be 0.

955
00:56:16,180 --> 00:56:18,100
Theta z must be 0.

956
00:56:18,100 --> 00:56:23,050
And of course uy must be 0,
because these are material

957
00:56:23,050 --> 00:56:26,270
particles, and in particular
that one there is a material

958
00:56:26,270 --> 00:56:28,390
particle on the zx phase.

959
00:56:28,390 --> 00:56:30,390
So uy is 0.

960
00:56:30,390 --> 00:56:32,970
uz and ux are have free.

961
00:56:32,970 --> 00:56:34,810
The free degrees of freedom,
once again, uz,

962
00:56:34,810 --> 00:56:38,940
ux, and theta y.

963
00:56:38,940 --> 00:56:47,060
Well, to impose, then, the
boundary condition, we would

964
00:56:47,060 --> 00:56:51,270
proceed as shown for this
material particle along all

965
00:56:51,270 --> 00:56:55,500
the nodes on this face, and
similarly, of course, on the

966
00:56:55,500 --> 00:56:56,900
other face--

967
00:56:56,900 --> 00:56:58,570
the face on the zy.

968
00:57:02,640 --> 00:57:06,780
To prevent the rigid body
translations, we would also

969
00:57:06,780 --> 00:57:11,960
have to take 1 degree of freedom
out, namely one z

970
00:57:11,960 --> 00:57:15,450
degree of freedom out, one
displacement degree of freedom

971
00:57:15,450 --> 00:57:17,780
corresponding to the
z-direction out.

972
00:57:17,780 --> 00:57:21,960
And that then would basically
give us all the boundary

973
00:57:21,960 --> 00:57:25,140
conditions that we have to
apply for this problem.

974
00:57:25,140 --> 00:57:29,640
So let's look at this view graph
here to show the mesh

975
00:57:29,640 --> 00:57:30,890
that we are Using.

976
00:57:33,260 --> 00:57:38,550
And we show here that at this
node, we set uz equal to 0 to

977
00:57:38,550 --> 00:57:41,295
prevent the rigid body motion
into this direction.

978
00:57:44,530 --> 00:57:51,690
The linear elastic analysis
results for this problem are

979
00:57:51,690 --> 00:57:53,110
given here.

980
00:57:53,110 --> 00:57:55,580
Notice displacement at
the point of load

981
00:57:55,580 --> 00:57:58,340
application is 0.0936.

982
00:57:58,340 --> 00:58:00,790
The analytical value
is shown here.

983
00:58:00,790 --> 00:58:04,310
Pictorially, the original mesh
is shown in black, and the

984
00:58:04,310 --> 00:58:08,720
deformed mesh is shown in red.

985
00:58:08,720 --> 00:58:11,890
The important point of this
problem, and I'd like to go

986
00:58:11,890 --> 00:58:16,720
back to it once more, is really
that when we look at

987
00:58:16,720 --> 00:58:22,710
this mesh, that we define the
director vectors at each of

988
00:58:22,710 --> 00:58:24,120
these nodes--

989
00:58:24,120 --> 00:58:26,950
and that can be done
automatically, they can be

990
00:58:26,950 --> 00:58:28,760
generated--

991
00:58:28,760 --> 00:58:33,590
that at each all of these
internal nodes, we use 5

992
00:58:33,590 --> 00:58:39,625
degrees of freedom, we have two
locally aligned alpha and

993
00:58:39,625 --> 00:58:42,945
beta degrees of freedom, the one
that I talked about in the

994
00:58:42,945 --> 00:58:46,360
earlier lecture, at each of
these internal nodes.

995
00:58:46,360 --> 00:58:50,500
I like to distinguish the
internal nodes here from the

996
00:58:50,500 --> 00:58:52,380
boundary nodes.

997
00:58:52,380 --> 00:58:57,460
On the boundary nodes, we of
course have also only 5

998
00:58:57,460 --> 00:59:01,770
natural degrees of freedom, but
we assign now 6 degrees of

999
00:59:01,770 --> 00:59:02,290
freedom there.

1000
00:59:02,290 --> 00:59:05,880
Meaning that the alpha and beta
degrees of freedom are

1001
00:59:05,880 --> 00:59:10,600
rotated into the three Cartesian
coordinate axes so

1002
00:59:10,600 --> 00:59:15,400
as to have three rotations at
each of these nodes on the

1003
00:59:15,400 --> 00:59:16,850
symmetry faces.

1004
00:59:16,850 --> 00:59:20,460
And then we can very easily
impose the symmetry boundary

1005
00:59:20,460 --> 00:59:23,230
condition that I just
talked about.

1006
00:59:23,230 --> 00:59:26,540
That's the important point
of this problem, really--

1007
00:59:26,540 --> 00:59:30,110
how we model the boundary
conditions, and what degrees

1008
00:59:30,110 --> 00:59:35,020
of freedom are assigned
to each of the nodes.

1009
00:59:35,020 --> 00:59:39,480
Well as a second example, I'd
like to consider with you the

1010
00:59:39,480 --> 00:59:44,720
analysis of an open box,
a 5-sided open box.

1011
00:59:44,720 --> 00:59:51,370
Here we show the box upside
down, or open site down.

1012
00:59:51,370 --> 00:59:54,640
The open side, in other
words, is down here.

1013
00:59:54,640 --> 01:00:00,420
And the box lies on a rigid,
frictionless surface.

1014
01:00:00,420 --> 01:00:04,400
On top here, we have pressure
applied to the face.

1015
01:00:04,400 --> 01:00:08,200
The box is modelled using
shell elements.

1016
01:00:08,200 --> 01:00:13,690
The point of this problem is to
discuss with you how we are

1017
01:00:13,690 --> 01:00:18,180
modelling this box using the
concept of director vectors, 5

1018
01:00:18,180 --> 01:00:21,460
degrees of freedom, 6 degrees of
freedom, that we discussed

1019
01:00:21,460 --> 01:00:22,710
in the earlier lecture.

1020
01:00:24,830 --> 01:00:28,570
So what we need to do here is
choose initial director

1021
01:00:28,570 --> 01:00:32,970
vectors, choose 5 or 6 degrees
of freedom for each node, and

1022
01:00:32,970 --> 01:00:34,800
choose the appropriate
boundary conditions.

1023
01:00:37,650 --> 01:00:41,760
The way we proceed there
is as follows.

1024
01:00:41,760 --> 01:00:46,490
We recognize that we could deal
with director vectors at

1025
01:00:46,490 --> 01:00:52,580
every one of the nodes, but if
we do so, at a corner, you

1026
01:00:52,580 --> 01:00:56,550
would have no unique
director vectors.

1027
01:00:56,550 --> 01:01:00,060
You could, of course, assign
there a mean normal.

1028
01:01:00,060 --> 01:01:04,720
In other words, if you look
closely here, my hands show

1029
01:01:04,720 --> 01:01:08,740
the corner, and the pointer
shows the mean

1030
01:01:08,740 --> 01:01:12,960
normal at that corner.

1031
01:01:12,960 --> 01:01:16,170
And that mean normal, mean in
quotes, would be the direction

1032
01:01:16,170 --> 01:01:17,740
of the director vector.

1033
01:01:17,740 --> 01:01:20,250
Let me show it from
up like that.

1034
01:01:20,250 --> 01:01:23,990
Here you have the right angle
between my hands, and you have

1035
01:01:23,990 --> 01:01:26,220
the pointer giving you
the director vector.

1036
01:01:26,220 --> 01:01:28,880
That would be one possibility.

1037
01:01:28,880 --> 01:01:32,030
However, if you use ADINA, then,
in this particular case,

1038
01:01:32,030 --> 01:01:35,350
it is more effective to not
input director vectors.

1039
01:01:35,350 --> 01:01:39,550
Because if you don't do so
with ADINA, then for each

1040
01:01:39,550 --> 01:01:43,790
node, ADINA will generate
automatically a mid-surface

1041
01:01:43,790 --> 01:01:45,570
normal vector.

1042
01:01:45,570 --> 01:01:49,190
In other words, if no director
vector has input for a node,

1043
01:01:49,190 --> 01:01:52,620
then ADINA generates for each
element connected to the node

1044
01:01:52,620 --> 01:01:55,960
a nodal point mid-surface this
normal vector at that node

1045
01:01:55,960 --> 01:01:58,260
from the element geometry.

1046
01:01:58,260 --> 01:01:59,290
Now, what does this mean?

1047
01:01:59,290 --> 01:02:05,090
It means that if you use this
option, there will then be as

1048
01:02:05,090 --> 01:02:09,020
many different nodal point
mid-surface normal vectors at

1049
01:02:09,020 --> 01:02:12,550
the node as there are elements
connected to the node.

1050
01:02:12,550 --> 01:02:15,040
Now we will have to look at
this statement much more

1051
01:02:15,040 --> 01:02:18,490
closely for this particular
example.

1052
01:02:18,490 --> 01:02:22,170
In this particular case, using
the option of automatic

1053
01:02:22,170 --> 01:02:24,760
generation of element nodal
point mid-surface normal

1054
01:02:24,760 --> 01:02:30,650
vectors, we find that at a node
away from an edge, we

1055
01:02:30,650 --> 01:02:34,620
will end up with just one
mid-surface normal vector.

1056
01:02:34,620 --> 01:02:39,860
Because see, this is a flat
surface, and the program will

1057
01:02:39,860 --> 01:02:43,470
calculate for this element and
that element and that element

1058
01:02:43,470 --> 01:02:46,440
and this element, corresponding
to that node,

1059
01:02:46,440 --> 01:02:48,830
the same normal vector.

1060
01:02:48,830 --> 01:02:50,870
And that is the normal
vector that the

1061
01:02:50,870 --> 01:02:52,630
program will then use.

1062
01:02:52,630 --> 01:02:57,160
However, if we look at an edge,
along an edge, and pick

1063
01:02:57,160 --> 01:03:01,540
a node there, then at this node,
there will be 1 normal

1064
01:03:01,540 --> 01:03:05,750
vector calculated for this
element and that element at

1065
01:03:05,750 --> 01:03:12,060
that node, shown by this red
arrow, and one normal vector

1066
01:03:12,060 --> 01:03:15,690
at this node calculated for this
element and that element,

1067
01:03:15,690 --> 01:03:18,690
shown by this red arrow.

1068
01:03:18,690 --> 01:03:22,210
Of course here, you would have
three normal vectors.

1069
01:03:22,210 --> 01:03:27,410
One on that phase, that
phase, and this phase.

1070
01:03:27,410 --> 01:03:32,340
Well, with these normal vectors,
then, given, you

1071
01:03:32,340 --> 01:03:37,090
recognize that there will be no
stiffness corresponding to

1072
01:03:37,090 --> 01:03:41,720
the rotation about this vector,
because these elements

1073
01:03:41,720 --> 01:03:45,620
don't carry stiffness in this
rotational degree of freedom.

1074
01:03:45,620 --> 01:03:49,150
We have to be careful with that,
and we have got to make

1075
01:03:49,150 --> 01:03:54,430
sure that we solve the problem
with this rotation, the

1076
01:03:54,430 --> 01:03:57,510
rotation corresponding to this
vector here or about this

1077
01:03:57,510 --> 01:04:00,700
vector, being deleted
in the model.

1078
01:04:00,700 --> 01:04:05,140
Whereas at this node here, we
assign 6 degrees of freedom to

1079
01:04:05,140 --> 01:04:10,850
the model, because the rotation
corresponding around

1080
01:04:10,850 --> 01:04:14,840
this vector will have stiffness,
around this vector

1081
01:04:14,840 --> 01:04:19,580
will have stiffness, and if you
think of a rotation in the

1082
01:04:19,580 --> 01:04:24,550
direction of that vector but
at this node, about this

1083
01:04:24,550 --> 01:04:26,710
direction, you will have
also stiffness.

1084
01:04:26,710 --> 01:04:30,120
So we need 6 degrees of freedom
being assigned at a

1085
01:04:30,120 --> 01:04:34,130
node that corresponds to
an edge or corner.

1086
01:04:34,130 --> 01:04:37,300
We have 5 degrees of freedom
to be assigned

1087
01:04:37,300 --> 01:04:39,780
to all other nodes.

1088
01:04:39,780 --> 01:04:42,880
This is once more
summarized here.

1089
01:04:42,880 --> 01:04:49,530
5 degrees of freedom at all
the nodes inside a face.

1090
01:04:49,530 --> 01:04:53,230
6 degrees of freedom along
an edge, and of

1091
01:04:53,230 --> 01:04:56,620
course also at a corner.

1092
01:04:56,620 --> 01:04:59,950
Let us look now overall at the
boundary conditions that we

1093
01:04:59,950 --> 01:05:02,620
have to assign to the box
in order to be able

1094
01:05:02,620 --> 01:05:04,540
to solve the problem.

1095
01:05:04,540 --> 01:05:07,540
Here we show now the
box open side up.

1096
01:05:07,540 --> 01:05:11,500
So this is really the side on
which the box is placed, but

1097
01:05:11,500 --> 01:05:15,460
we just switch the box around
so that we can more easily

1098
01:05:15,460 --> 01:05:18,290
look at what happens on
this boundary here.

1099
01:05:18,290 --> 01:05:24,960
Well, we have nodes that we
might call internal nodes

1100
01:05:24,960 --> 01:05:28,620
along here, nodes that
are not corner nodes.

1101
01:05:28,620 --> 01:05:32,860
And these nodes would have the
following boundary condition.

1102
01:05:32,860 --> 01:05:38,160
First of all, all the rotational
degrees of freedom

1103
01:05:38,160 --> 01:05:40,820
shown by this arrow,
this double arrow,

1104
01:05:40,820 --> 01:05:42,390
must have been deleted.

1105
01:05:42,390 --> 01:05:44,060
This I just discussed.

1106
01:05:44,060 --> 01:05:46,660
It is due to the fact that
there is no stiffness

1107
01:05:46,660 --> 01:05:52,110
corresponding to this rotation
at that node, because all the

1108
01:05:52,110 --> 01:05:54,550
elements coming into that node
don't have stiffness

1109
01:05:54,550 --> 01:05:58,880
corresponding to this rotational
degree of freedom.

1110
01:05:58,880 --> 01:06:00,320
This degree of freedom
is free.

1111
01:06:00,320 --> 01:06:01,540
That one is free.

1112
01:06:01,540 --> 01:06:03,230
This rotation is free.

1113
01:06:03,230 --> 01:06:04,870
That rotation is free.

1114
01:06:04,870 --> 01:06:07,120
This translation is
also deleted.

1115
01:06:07,120 --> 01:06:11,120
That is due to the fact that the
box can't move vertically.

1116
01:06:11,120 --> 01:06:13,280
It's placed on the
rigid surface.

1117
01:06:15,910 --> 01:06:20,930
If we look at a node that is a
corner node-- and let's pick a

1118
01:06:20,930 --> 01:06:21,920
more typical one.

1119
01:06:21,920 --> 01:06:23,840
Here we have one.

1120
01:06:23,840 --> 01:06:28,520
Then at this node, we only take
out, as you can see, the

1121
01:06:28,520 --> 01:06:30,790
z displacement degree
of freedom.

1122
01:06:30,790 --> 01:06:34,370
Because, once again,
the box lies on

1123
01:06:34,370 --> 01:06:35,860
the rigid flat surface.

1124
01:06:35,860 --> 01:06:39,730
But notice that these
rotations, all three

1125
01:06:39,730 --> 01:06:41,400
rotations, are left free.

1126
01:06:41,400 --> 01:06:45,750
They all carry stiffness the
way I already discussed it.

1127
01:06:45,750 --> 01:06:49,790
Also these translations
are free.

1128
01:06:49,790 --> 01:06:54,890
Typical corner node, another
typical corner node, the same

1129
01:06:54,890 --> 01:06:57,250
action as this corner node.

1130
01:06:57,250 --> 01:06:59,820
Now here we see something
in addition.

1131
01:06:59,820 --> 01:07:02,960
We see that this degree of
freedom is taken out, the same

1132
01:07:02,960 --> 01:07:05,290
way as over there.

1133
01:07:05,290 --> 01:07:06,870
But we also see that
this degree of

1134
01:07:06,870 --> 01:07:08,200
freedom is taken out.

1135
01:07:08,200 --> 01:07:11,550
This one is taken out in order
to prevent a rigid body

1136
01:07:11,550 --> 01:07:15,830
rotation about my
pointer here.

1137
01:07:20,510 --> 01:07:22,710
That's why we take
this one out.

1138
01:07:22,710 --> 01:07:27,260
And we take this one out and
that one out to also prevent

1139
01:07:27,260 --> 01:07:31,040
the rigid body rotation that I
just mentioned, and to prevent

1140
01:07:31,040 --> 01:07:32,840
the rigid body translation.

1141
01:07:32,840 --> 01:07:36,550
See, this translation has to be
taken out as well, and that

1142
01:07:36,550 --> 01:07:39,650
translation has to be taken out,
and this rotation has to

1143
01:07:39,650 --> 01:07:40,880
be taken out.

1144
01:07:40,880 --> 01:07:44,710
I repeat, this translation,
that translation, and that

1145
01:07:44,710 --> 01:07:47,090
rotation, all of those
have to be taken out.

1146
01:07:47,090 --> 01:07:50,490
And that is achieved by knocking
out this degree of

1147
01:07:50,490 --> 01:07:54,220
freedom, that one,
and that one.

1148
01:07:54,220 --> 01:07:57,500
So these are the boundary
conditions that we apply to

1149
01:07:57,500 --> 01:08:02,340
the nodes on the edges
that are lying

1150
01:08:02,340 --> 01:08:05,340
on the rigid surface.

1151
01:08:05,340 --> 01:08:08,950
If we perform now a linear
elastic static analysis of the

1152
01:08:08,950 --> 01:08:12,010
box with a uniform pressure
applied to the top off the

1153
01:08:12,010 --> 01:08:19,529
box, using our 4-node MITC4
element, here is a typical

1154
01:08:19,529 --> 01:08:22,029
element shown, this is
[? the measure ?]

1155
01:08:22,029 --> 01:08:29,590
that we use, we find that the
deformations are magnified,

1156
01:08:29,590 --> 01:08:32,240
quite highly magnified,
looking like that.

1157
01:08:32,240 --> 01:08:35,170
Very reasonable deformation.

1158
01:08:35,170 --> 01:08:38,189
We could not compare here
with any other solution.

1159
01:08:38,189 --> 01:08:40,390
But it's still a very
interesting demonstrative

1160
01:08:40,390 --> 01:08:46,700
example, because of the way we
are dealing with the director

1161
01:08:46,700 --> 01:08:49,600
vectors, which actually
we don't define.

1162
01:08:49,600 --> 01:08:51,180
We let the program generate

1163
01:08:51,180 --> 01:08:53,090
mid-surface nodal point vectors.

1164
01:08:53,090 --> 01:08:56,819
And because we could discuss the
use of the shell element

1165
01:08:56,819 --> 01:09:00,450
degrees of freedom, freedom is
the imposition of boundary

1166
01:09:00,450 --> 01:09:02,510
conditions, and other aspects.

1167
01:09:02,510 --> 01:09:04,450
This, then, brings me to the
end of what I wanted to

1168
01:09:04,450 --> 01:09:07,979
discuss this you related to the
view graph, and I'd like

1169
01:09:07,979 --> 01:09:11,240
to now share some further
experiences with you that I

1170
01:09:11,240 --> 01:09:12,819
documented on the slides.

1171
01:09:12,819 --> 01:09:17,850
So let me walk over here and let
us discuss the information

1172
01:09:17,850 --> 01:09:19,899
that is even on the slides.

1173
01:09:19,899 --> 01:09:25,380
The first slide here relates
to the analysis of a simply

1174
01:09:25,380 --> 01:09:29,540
supported plate under
uniform pressure.

1175
01:09:29,540 --> 01:09:33,040
This plate is modelled
as shown here.

1176
01:09:33,040 --> 01:09:36,710
We only model 1/4 of the plate
because of symmetry

1177
01:09:36,710 --> 01:09:37,740
conditions.

1178
01:09:37,740 --> 01:09:44,310
And as you can see here, n is
equal to 2 for 4 elements.

1179
01:09:44,310 --> 01:09:48,270
We analyze this plate and
we calculate the center

1180
01:09:48,270 --> 01:09:52,670
displacement, which means the
displacement at that point,

1181
01:09:52,670 --> 01:09:55,230
and compare that center
displacement to

1182
01:09:55,230 --> 01:09:57,640
the analytical solution.

1183
01:09:57,640 --> 01:10:02,400
Notice that in this particular
case, l/H is equal to 1000, so

1184
01:10:02,400 --> 01:10:05,370
it's a very thin plate.

1185
01:10:05,370 --> 01:10:09,690
And certainly we would get very
bad results if we had a

1186
01:10:09,690 --> 01:10:12,640
shear locking phenomenon in the
element that we're using.

1187
01:10:12,640 --> 01:10:15,670
As I pointed out earlier, the
MITC4 element does not have

1188
01:10:15,670 --> 01:10:16,730
any shear locking.

1189
01:10:16,730 --> 01:10:22,940
And as you can see now here, as
we increase N, as shown in

1190
01:10:22,940 --> 01:10:27,320
this axes, we get excellent
convergence behavior to the

1191
01:10:27,320 --> 01:10:29,720
analytical solution.

1192
01:10:29,720 --> 01:10:32,610
Now the next slide shows
the analysis of a

1193
01:10:32,610 --> 01:10:34,240
very similar plate.

1194
01:10:34,240 --> 01:10:39,270
Again, a square plate, l/H
still being 1,000.

1195
01:10:39,270 --> 01:10:41,310
Simply supported plate.

1196
01:10:41,310 --> 01:10:46,360
But now the plate is subjected
to a concentrated load.

1197
01:10:46,360 --> 01:10:51,720
And in this particular case,
we see the solution results

1198
01:10:51,720 --> 01:10:55,520
obtained with the MITC4 element
as a function of N--

1199
01:10:55,520 --> 01:11:01,510
same number N that I referred
to in the earlier slide.

1200
01:11:01,510 --> 01:11:05,060
We see that we get also very
good convergence behavior in

1201
01:11:05,060 --> 01:11:06,010
this particular case.

1202
01:11:06,010 --> 01:11:09,700
Not quite as good as for the
uniformly distributed pressure

1203
01:11:09,700 --> 01:11:14,580
case, but still very
good as well here.

1204
01:11:14,580 --> 01:11:21,690
The next slide shows the plate
now clamped at its edges.

1205
01:11:21,690 --> 01:11:23,990
Not any more simply supported.

1206
01:11:23,990 --> 01:11:28,810
l/H still 1,000, so still
a very thin plate.

1207
01:11:28,810 --> 01:11:31,340
And here we once again
calculate the center

1208
01:11:31,340 --> 01:11:34,190
displacement as a function
of N, the

1209
01:11:34,190 --> 01:11:36,130
number of elements used.

1210
01:11:36,130 --> 01:11:39,430
Remember, N equal 1
means 4 elements.

1211
01:11:39,430 --> 01:11:43,440
Remember N equals 4
means 16 elements.

1212
01:11:43,440 --> 01:11:48,180
Once again, we see excellent
convergence behavior to the

1213
01:11:48,180 --> 01:11:51,250
exact analytical solution.

1214
01:11:51,250 --> 01:11:54,420
And finally, this was a case
for uniform pressure.

1215
01:11:54,420 --> 01:11:57,890
Now we look at the case for a
concentrated load applied at

1216
01:11:57,890 --> 01:11:58,840
the center.

1217
01:11:58,840 --> 01:12:01,250
And here we see this convergence
behavior.

1218
01:12:01,250 --> 01:12:04,650
Of course, this is quite a
complicated, difficult problem

1219
01:12:04,650 --> 01:12:07,440
to analyze, because of the
concentrated load and the

1220
01:12:07,440 --> 01:12:10,680
clamp boundary conditions.

1221
01:12:10,680 --> 01:12:16,200
Here in this analysis, we used
nice square elements.

1222
01:12:16,200 --> 01:12:20,100
A very important consideration
in a practical analysis is to

1223
01:12:20,100 --> 01:12:23,610
identify how well an
element performs

1224
01:12:23,610 --> 01:12:25,410
when it becomes distorted.

1225
01:12:25,410 --> 01:12:32,110
And the next slides now relates
to that question.

1226
01:12:32,110 --> 01:12:37,100
Here we now model 1/4 of the
plate using mesh 1, shown

1227
01:12:37,100 --> 01:12:40,400
here, and using mesh
2, shown here.

1228
01:12:40,400 --> 01:12:45,670
Notice the elements that we're
now using in these analyses is

1229
01:12:45,670 --> 01:12:47,490
still the MITC4 element.

1230
01:12:47,490 --> 01:12:50,490
Each of these elements is an
MITC4 element, but it's a

1231
01:12:50,490 --> 01:12:51,610
distorted element.

1232
01:12:51,610 --> 01:12:56,190
And our objective is really to
identify what displacement

1233
01:12:56,190 --> 01:12:59,660
predictions we obtain at the
center, at point c of the

1234
01:12:59,660 --> 01:13:05,250
plate, when the plate is subject
to uniform pressure,

1235
01:13:05,250 --> 01:13:08,320
and the plate is also
simply supported.

1236
01:13:08,320 --> 01:13:12,650
And what displacement
predictions do we obtain for

1237
01:13:12,650 --> 01:13:14,870
these two measures at
these two points?

1238
01:13:14,870 --> 01:13:19,930
How much do they deteriorate
when compared to the usage of

1239
01:13:19,930 --> 01:13:21,480
nice square elements?

1240
01:13:21,480 --> 01:13:24,310
Well, the displacement
predictions are given here.

1241
01:13:24,310 --> 01:13:27,910
Finite element over Kirchhoff
theory results.

1242
01:13:27,910 --> 01:13:33,830
Mesh 1, 0.93, mesh 2, 1.01.

1243
01:13:33,830 --> 01:13:36,240
Notice mesh 2 has
just 1% error.

1244
01:13:36,240 --> 01:13:37,870
Mesh 1 has 7% error.

1245
01:13:37,870 --> 01:13:42,040
But with mesh 1, we have a very
coarse mesh, and highly

1246
01:13:42,040 --> 01:13:43,712
distorted elements.

1247
01:13:43,712 --> 01:13:47,280
It's of course also of interest
to see what moment

1248
01:13:47,280 --> 01:13:50,770
predictions one obtains, and
here we compare the finite

1249
01:13:50,770 --> 01:13:55,160
element moment versus the
Kirchhoff moment at point c.

1250
01:13:55,160 --> 01:14:01,800
And you can see 15% error for
mesh 1, 2% error for mesh 2.

1251
01:14:01,800 --> 01:14:05,030
Actually also very good
results for moment

1252
01:14:05,030 --> 01:14:06,460
prediction, as well.

1253
01:14:06,460 --> 01:14:09,950
I should also emphasize, please,
that l/H is equal to

1254
01:14:09,950 --> 01:14:14,810
1,000, so we are still talking
about this very thin plate.

1255
01:14:14,810 --> 01:14:18,440
The next slide now shows how
the element behaves in the

1256
01:14:18,440 --> 01:14:20,550
analysis of a skew plate.

1257
01:14:20,550 --> 01:14:24,270
It's also now a distorted
element mesh, or rather, the

1258
01:14:24,270 --> 01:14:28,590
elements used in the meshing
of this plate are distorted

1259
01:14:28,590 --> 01:14:31,590
elements, but they're a little
bit more regular distorted.

1260
01:14:31,590 --> 01:14:34,170
In fact, they are parallelogram
elements.

1261
01:14:34,170 --> 01:14:38,400
Notice, please, that this is a
top view, so we are seeing a

1262
01:14:38,400 --> 01:14:40,810
plan view of the plate.

1263
01:14:40,810 --> 01:14:42,060
And--

1264
01:14:58,140 --> 01:15:01,450
SPEAKER: Pick up lecture
20, tape B, slide

1265
01:15:01,450 --> 01:15:05,950
20-7, time code 20:33:18.

1266
01:15:05,950 --> 01:15:07,200
Take one.

1267
01:15:17,760 --> 01:15:20,300
PROFESSOR: The next analysis
that I like to consider now is

1268
01:15:20,300 --> 01:15:22,050
the analysis of the
skew plate.

1269
01:15:22,050 --> 01:15:25,590
Here we show a top view of the
skew plate with an angle of

1270
01:15:25,590 --> 01:15:27,030
skew of 30 degrees.

1271
01:15:27,030 --> 01:15:29,120
So it's a highly skewed plate.

1272
01:15:29,120 --> 01:15:31,180
And this is the kind of
mesh we want to use.

1273
01:15:31,180 --> 01:15:36,340
Here we show the 4-by-4 mesh
of MITC4 elements.

1274
01:15:36,340 --> 01:15:39,150
Notice the elements are
distorted, but sort of

1275
01:15:39,150 --> 01:15:41,370
regularly distorted, because
they are having a

1276
01:15:41,370 --> 01:15:43,720
parallelogram shape.

1277
01:15:43,720 --> 01:15:46,780
The boundary conditions on the
plate are simply supported

1278
01:15:46,780 --> 01:15:51,030
boundary conditions, which I
should point out are modelled

1279
01:15:51,030 --> 01:15:55,580
using the isoparametric
degenerate plate elements,

1280
01:15:55,580 --> 01:15:58,550
shell elements, that
we talked about.

1281
01:15:58,550 --> 01:16:01,790
These boundary conditions are
modelled by simply setting

1282
01:16:01,790 --> 01:16:05,550
only the transverse displacement
at the nodes, on

1283
01:16:05,550 --> 01:16:07,480
the boundary equal to 0.

1284
01:16:07,480 --> 01:16:09,840
In other words, the
displacements corresponding to

1285
01:16:09,840 --> 01:16:15,250
the z-axis, pointing out of the
slide, are set equal to 0,

1286
01:16:15,250 --> 01:16:20,010
but the rotations on the
boundary nodes are left free.

1287
01:16:20,010 --> 01:16:23,010
This is a very interesting
point, most interesting point.

1288
01:16:23,010 --> 01:16:27,600
And if you like to read up on
it, please refer to the

1289
01:16:27,600 --> 01:16:30,010
reference that is given
in the study guide

1290
01:16:30,010 --> 01:16:32,400
related to this slide.

1291
01:16:32,400 --> 01:16:36,760
Well, we use these kinds of
meshes to analyze the plate.

1292
01:16:36,760 --> 01:16:39,500
Notice the material data
are given here.

1293
01:16:39,500 --> 01:16:42,640
The thickness of the plate
is given here.

1294
01:16:42,640 --> 01:16:48,770
B is equal to 1, so we have a
thickness with ratio of 1/100.

1295
01:16:48,770 --> 01:16:50,510
It's a fairly thin plate.

1296
01:16:50,510 --> 01:16:55,780
And we are considering uniform
pressure applied to the plate.

1297
01:16:55,780 --> 01:16:59,680
The next slide now shows
analysis results.

1298
01:16:59,680 --> 01:17:05,470
4-by-4 mesh, 8-by-8, 16-by-16,
32-by-32 meshes were used.

1299
01:17:05,470 --> 01:17:10,040
We compare our finite element
solution at the point C, the

1300
01:17:10,040 --> 01:17:14,150
midpoint of the plate,
versus the Morley

1301
01:17:14,150 --> 01:17:15,780
solution at that point.

1302
01:17:15,780 --> 01:17:20,580
And you can see that we really
have obtained very nice

1303
01:17:20,580 --> 01:17:24,060
conversions to the
Morley solution.

1304
01:17:24,060 --> 01:17:28,060
Notice here, we're looking
at the principal moments,

1305
01:17:28,060 --> 01:17:31,310
actually the maximum and the
minimum moments, at that

1306
01:17:31,310 --> 01:17:35,410
midpoint of the plate again,
point C, and we're comparing

1307
01:17:35,410 --> 01:17:37,380
finite element solutions
with Morley solutions.

1308
01:17:37,380 --> 01:17:40,610
And here once again, quite
good convergence for the

1309
01:17:40,610 --> 01:17:42,910
moment, as well.

1310
01:17:42,910 --> 01:17:48,720
Now in this analysis, we use the
kind of mesh, this 4-by-4

1311
01:17:48,720 --> 01:17:51,920
mesh, that we had on
the previous slide.

1312
01:17:51,920 --> 01:17:54,920
And one can ask, of course, is
that a good mesh to use?

1313
01:17:54,920 --> 01:17:57,330
Well, we deliberately wanted
to use this mesh in this

1314
01:17:57,330 --> 01:18:00,020
analysis to test the capability
of the element.

1315
01:18:00,020 --> 01:18:01,660
Of course, one could use
a different mesh.

1316
01:18:01,660 --> 01:18:05,110
And on the next slide now, I'm
showing the results obtained

1317
01:18:05,110 --> 01:18:09,010
by using just four elements,
a 2-by-2 mesh, to

1318
01:18:09,010 --> 01:18:11,220
model 1/4 of the plate.

1319
01:18:11,220 --> 01:18:14,630
And you can see, with this
2-by-2 mesh, we get already

1320
01:18:14,630 --> 01:18:18,550
excellent predictions of the
displacement at the point C.

1321
01:18:18,550 --> 01:18:20,980
This is where point
C is, once again.

1322
01:18:20,980 --> 01:18:24,400
Excellent prediction of the
displacement at point C when

1323
01:18:24,400 --> 01:18:26,460
compared with Morley solution.

1324
01:18:26,460 --> 01:18:30,700
The moment is quite poorly
predicted, still.

1325
01:18:30,700 --> 01:18:33,830
The two moments that
I defined earlier.

1326
01:18:33,830 --> 01:18:37,780
If we go to a finer mesh, 4-by-4
mesh, the moments, of

1327
01:18:37,780 --> 01:18:41,750
course, are much better, and
so is the displacement

1328
01:18:41,750 --> 01:18:42,160
prediction.

1329
01:18:42,160 --> 01:18:44,240
But the displacement prediction
was already goods,

1330
01:18:44,240 --> 01:18:47,000
so there is not that much
of an improvement.

1331
01:18:47,000 --> 01:18:49,880
But in general, one can say,
this here, certainly for

1332
01:18:49,880 --> 01:18:54,010
displacement predictions, is a
much better mesh to use than

1333
01:18:54,010 --> 01:18:57,970
the earlier meshes that
we used for the

1334
01:18:57,970 --> 01:18:59,710
earlier table results.

1335
01:18:59,710 --> 01:19:03,050
Now I'd like to look at another
problem, the analysis

1336
01:19:03,050 --> 01:19:08,020
of a cantilever subjected to
an end-bending moment, and

1337
01:19:08,020 --> 01:19:10,520
undergoing very large
displacement.

1338
01:19:10,520 --> 01:19:12,600
Here are the data given
corresponding to the

1339
01:19:12,600 --> 01:19:16,200
cantilever, an elastic
structure that

1340
01:19:16,200 --> 01:19:17,230
we're looking at.

1341
01:19:17,230 --> 01:19:23,390
And we will be measuring v, phi,
and u, and u, v, and phi

1342
01:19:23,390 --> 01:19:25,730
would take on large values.

1343
01:19:25,730 --> 01:19:28,920
In fact, we will actually
normalize u and v to the

1344
01:19:28,920 --> 01:19:31,740
length of the cantilever.

1345
01:19:31,740 --> 01:19:35,830
The next slide now shows
the results obtained.

1346
01:19:35,830 --> 01:19:41,440
v/l, u/l, and phi over
2 pi as a function

1347
01:19:41,440 --> 01:19:43,020
of the moment applied.

1348
01:19:43,020 --> 01:19:46,440
We're using here moment
parameter, but of course, the

1349
01:19:46,440 --> 01:19:50,340
moment increases along this
axis, and here we are plotting

1350
01:19:50,340 --> 01:19:54,110
u/l, v/l, and phi over 2 pi.

1351
01:19:54,110 --> 01:19:58,730
Notice that even for very large
displacement, v/l being

1352
01:19:58,730 --> 01:20:04,510
0.7, with 1 cubic element, cubic
in the direction of the

1353
01:20:04,510 --> 01:20:08,790
cantilever, excellent results.

1354
01:20:08,790 --> 01:20:11,740
This is, in other words, the use
of the cubic element, in

1355
01:20:11,740 --> 01:20:14,920
general the 16-node
shell element, for

1356
01:20:14,920 --> 01:20:17,520
these types of analyses.

1357
01:20:17,520 --> 01:20:20,060
And it shows that the 16-node
shell element is

1358
01:20:20,060 --> 01:20:22,750
really quite effective.

1359
01:20:22,750 --> 01:20:25,100
The next slide now shows
how we would

1360
01:20:25,100 --> 01:20:27,050
apply the MITC4 element.

1361
01:20:27,050 --> 01:20:30,440
Here we use just two elements
to model the cantilever,

1362
01:20:30,440 --> 01:20:33,680
bending moment applied
here, clamped

1363
01:20:33,680 --> 01:20:35,380
boundary conditions here.

1364
01:20:35,380 --> 01:20:39,810
And you can see that here we
are plotting u/l, w/l, phi

1365
01:20:39,810 --> 01:20:43,350
over 2 pi, as before, moment
parameter along here.

1366
01:20:43,350 --> 01:20:45,550
And you can see that we're
getting really excellent

1367
01:20:45,550 --> 01:20:50,360
results just with 2 elements for
w/l 0.6, or certainly up

1368
01:20:50,360 --> 01:20:53,570
0.5, very close to the
analytical solution.

1369
01:20:53,570 --> 01:20:58,560
The analytical solution is
given by the solid line.

1370
01:20:58,560 --> 01:21:02,430
The next slide now shows the
result obtained using three

1371
01:21:02,430 --> 01:21:06,330
MITC4 elements to model
the same cantilever.

1372
01:21:06,330 --> 01:21:10,470
Notice these are now the nodes
that we also have used when

1373
01:21:10,470 --> 01:21:13,860
using the cubic element
in our first result.

1374
01:21:13,860 --> 01:21:17,120
And note this excellent
comparison with the analytical

1375
01:21:17,120 --> 01:21:21,750
solution for u/l, w/l,
and phi over 2 pi.

1376
01:21:21,750 --> 01:21:24,770
Very large displacement under
very large displacement

1377
01:21:24,770 --> 01:21:26,910
conditions.

1378
01:21:26,910 --> 01:21:32,880
The next slide now shows the
analysis of an I-section

1379
01:21:32,880 --> 01:21:35,590
clamped at this end
and subjected to a

1380
01:21:35,590 --> 01:21:37,330
torque at this end.

1381
01:21:37,330 --> 01:21:40,590
This is actually quite a
complicated problem if you

1382
01:21:40,590 --> 01:21:44,450
think deeper about the problem
and recognize that you have

1383
01:21:44,450 --> 01:21:49,010
torsional conditions, restrained
warping here,

1384
01:21:49,010 --> 01:21:49,710
[UNINTELLIGIBLE]

1385
01:21:49,710 --> 01:21:55,490
torsion here, then of course you
have bending in the beam,

1386
01:21:55,490 --> 01:21:56,700
et cetera, et cetera.

1387
01:21:56,700 --> 01:22:00,190
It's not a very easy
problem if you look

1388
01:22:00,190 --> 01:22:01,780
deeply into the problem.

1389
01:22:01,780 --> 01:22:06,020
And we wanted to obtain some
relatively good prediction for

1390
01:22:06,020 --> 01:22:10,520
this structural response when
the structure goes into the

1391
01:22:10,520 --> 01:22:12,200
elastoplastic regime.

1392
01:22:12,200 --> 01:22:15,640
Here we have the material data
used in the analysis.

1393
01:22:15,640 --> 01:22:21,040
You can see perfectly plastic
conditions by using tangent

1394
01:22:21,040 --> 01:22:24,620
material modulus equal to 0.

1395
01:22:24,620 --> 01:22:30,210
The slide here now shows one
model that we use, namely the

1396
01:22:30,210 --> 01:22:37,770
use of altogether 9
isoparametric cubic elements.

1397
01:22:37,770 --> 01:22:40,900
These are rectangular
elements.

1398
01:22:40,900 --> 01:22:43,950
Notice this is 1 isoparametric
element.

1399
01:22:43,950 --> 01:22:48,310
This is the mid-surface or the
neutral axis of that one

1400
01:22:48,310 --> 01:22:51,990
isoparametric cubic rectangular
section element.

1401
01:22:51,990 --> 01:22:55,280
We are modeling the whole
I-beam by assembling the

1402
01:22:55,280 --> 01:22:58,150
I-beam using 9 such elements.

1403
01:22:58,150 --> 01:23:02,590
Of course we discussed earlier
that these elements contain

1404
01:23:02,590 --> 01:23:04,950
the proper warping conditions.

1405
01:23:04,950 --> 01:23:09,600
We have supplemented our
isoparametric interpolation by

1406
01:23:09,600 --> 01:23:12,660
warping functions, and those
now of course will be

1407
01:23:12,660 --> 01:23:15,840
activated in picking up the
torsional response, or

1408
01:23:15,840 --> 01:23:18,150
modelling the torsional
response.

1409
01:23:18,150 --> 01:23:20,610
This is a free end, that's
the fixed end.

1410
01:23:20,610 --> 01:23:23,570
This was one model that we used,
and in addition, we used

1411
01:23:23,570 --> 01:23:26,320
another model, which
is shown here--

1412
01:23:26,320 --> 01:23:28,770
a model of shell elements,
9-node shell

1413
01:23:28,770 --> 01:23:32,200
elements, 9 such elements.

1414
01:23:32,200 --> 01:23:35,460
Of course, a better response
would have been obtained by

1415
01:23:35,460 --> 01:23:38,880
using the 16-node element, but
that of course would have been

1416
01:23:38,880 --> 01:23:40,820
a more expensive analysis.

1417
01:23:40,820 --> 01:23:45,600
So in this particular case, we
wanted to test this model

1418
01:23:45,600 --> 01:23:48,500
versus the isoparametric beam
model that I just showed you.

1419
01:23:48,500 --> 01:23:53,360
Notice one interesting point,
how we are modelling this

1420
01:23:53,360 --> 01:23:55,170
connection here.

1421
01:23:55,170 --> 01:23:59,030
You see a blown-up
figure of it.

1422
01:23:59,030 --> 01:24:04,610
At this point node c, we have
just 5 degrees of freedom.

1423
01:24:04,610 --> 01:24:09,260
No rotational stiffness about
this vector here.

1424
01:24:09,260 --> 01:24:12,310
So that is not a degree
of freedom.

1425
01:24:12,310 --> 01:24:15,060
Here we have 6 degrees of
freedom, 3 translational

1426
01:24:15,060 --> 01:24:18,610
degrees of freedom and 3
rotational degrees of freedom.

1427
01:24:18,610 --> 01:24:20,670
Here we have 5 degrees
of freedom.

1428
01:24:20,670 --> 01:24:24,200
No degree of freedom, no
stiffness corresponding to the

1429
01:24:24,200 --> 01:24:27,440
rotation about this
vector here.

1430
01:24:27,440 --> 01:24:30,170
That's one of the important
points that we discussed

1431
01:24:30,170 --> 01:24:31,580
earlier already.

1432
01:24:31,580 --> 01:24:34,860
Well, to proceed with the
analysis, we actually use the

1433
01:24:34,860 --> 01:24:37,940
longitudinal element dimensions
shown here.

1434
01:24:37,940 --> 01:24:41,890
Notice that we use a shorter
element at the fixed end than

1435
01:24:41,890 --> 01:24:46,160
here at the free end, because we
anticipate heavy plasticity

1436
01:24:46,160 --> 01:24:50,070
here that we wanted to model
quite accurately.

1437
01:24:50,070 --> 01:24:53,480
Once again, the torque is
applied here, and theta x is

1438
01:24:53,480 --> 01:24:54,410
measured here.

1439
01:24:54,410 --> 01:24:59,170
And on the next slide now, we
show the variation of the

1440
01:24:59,170 --> 01:25:03,470
torque as a function of
the theta x [? rate ?]

1441
01:25:03,470 --> 01:25:04,760
rotation.

1442
01:25:04,760 --> 01:25:07,580
Notice here you have the isobeam
model solution, and

1443
01:25:07,580 --> 01:25:10,190
here you have the shell
model solution.

1444
01:25:10,190 --> 01:25:13,840
We compare our results with
a sand heap solution and a

1445
01:25:13,840 --> 01:25:15,730
Merchant's upper
bound solution.

1446
01:25:15,730 --> 01:25:17,500
These are simplified analytical

1447
01:25:17,500 --> 01:25:20,530
solutions to this problem.

1448
01:25:20,530 --> 01:25:25,590
Our results here are somewhat
close, but not very close.

1449
01:25:25,590 --> 01:25:28,440
Of course, one could make a
much deeper study of this

1450
01:25:28,440 --> 01:25:30,600
problem, a very interesting
problem.

1451
01:25:30,600 --> 01:25:34,790
But the way we now already have
concluded our study and

1452
01:25:34,790 --> 01:25:36,660
the way I show you the results,
see, I think it's

1453
01:25:36,660 --> 01:25:39,900
already quite valuable, because
we have learned a bit

1454
01:25:39,900 --> 01:25:44,640
regarding the modeling of
this type of problem.

1455
01:25:44,640 --> 01:25:49,380
Well, the next slide now shows
a final problem for which I

1456
01:25:49,380 --> 01:25:51,810
wanted to show you some
solution results.

1457
01:25:51,810 --> 01:25:56,310
Here we have a cylindrical
shell supported on rigid

1458
01:25:56,310 --> 01:25:59,910
diaphragms at the end,
and the shell is

1459
01:25:59,910 --> 01:26:01,930
subjected to uniform pressure.

1460
01:26:01,930 --> 01:26:06,000
We measure here wB, the
displacement at that

1461
01:26:06,000 --> 01:26:10,840
particular point B. The data
corresponding to the shell are

1462
01:26:10,840 --> 01:26:12,500
given here.

1463
01:26:12,500 --> 01:26:16,310
Notice that we model the shell
as an elastic, perfectly

1464
01:26:16,310 --> 01:26:22,070
plastic material, and notice
that in the shell, we modeled

1465
01:26:22,070 --> 01:26:25,790
only this part of the shell,
this quarter part of the

1466
01:26:25,790 --> 01:26:29,540
shell, using a 9-by-9
uniform mesh of the

1467
01:26:29,540 --> 01:26:32,950
4-node MITC4 element.

1468
01:26:32,950 --> 01:26:38,500
The results are now given
on the slide shown here.

1469
01:26:38,500 --> 01:26:43,130
Pressure applied plotted along
this axis, wB plotted along

1470
01:26:43,130 --> 01:26:46,950
this axis, and this is a
non-linear solution response

1471
01:26:46,950 --> 01:26:51,445
we have obtained using the MITC4
element, and we compare

1472
01:26:51,445 --> 01:26:54,940
our solution with the solution
given by Krakeland.

1473
01:26:54,940 --> 01:26:57,360
I might point out that in this
analysis, we used our

1474
01:26:57,360 --> 01:27:01,750
automatic load-stepping scheme
that we discussed when we

1475
01:27:01,750 --> 01:27:04,970
talked about the solution of the
non-linear finite element

1476
01:27:04,970 --> 01:27:07,480
equations in an earlier
lecture.

1477
01:27:07,480 --> 01:27:10,050
Very good comparison
up to this point

1478
01:27:10,050 --> 01:27:11,590
with Krakeland's solution.

1479
01:27:11,590 --> 01:27:14,830
We did not go much further
than that solution given,

1480
01:27:14,830 --> 01:27:16,910
because we could not
compare with any

1481
01:27:16,910 --> 01:27:19,250
other solution anyway.

1482
01:27:19,250 --> 01:27:22,510
Well, this brings me now to the
end of what I wanted to

1483
01:27:22,510 --> 01:27:25,720
discuss with you regarding
shell elements.

1484
01:27:25,720 --> 01:27:28,570
Of course, please remember
there is a lot more

1485
01:27:28,570 --> 01:27:32,340
information that we could have
discussed and that we could

1486
01:27:32,340 --> 01:27:36,920
have studied together, but we
only have two lectures, and

1487
01:27:36,920 --> 01:27:39,820
this is the information that I
wanted to share with you in

1488
01:27:39,820 --> 01:27:40,820
these two lectures.

1489
01:27:40,820 --> 01:27:43,410
I hope you enjoyed it.

1490
01:27:43,410 --> 01:27:47,190
Please also refer to the papers
that are given as

1491
01:27:47,190 --> 01:27:49,830
references for these two
lectures, particularly for the

1492
01:27:49,830 --> 01:27:53,250
slides, where you will find much
more information, many

1493
01:27:53,250 --> 01:27:56,530
more details regarding this
material that we discussed.

1494
01:27:56,530 --> 01:27:57,840
Thank you very much for
your attention.