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PROFESSOR: Ladies and gentlemen,
welcome to this

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00:00:23,610 --> 00:00:26,730
lecture on non-linear finite
element analysis.

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00:00:26,730 --> 00:00:30,050
The title of this lecture is
"Basic Considerations in

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00:00:30,050 --> 00:00:35,690
Non-Linear Analysis." In this
lecture I would like to

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00:00:35,690 --> 00:00:39,790
discuss with you first the
application of the principle

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00:00:39,790 --> 00:00:41,190
of virtual work.

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00:00:41,190 --> 00:00:45,955
We will spend quite a bit of
time on the basic principle of

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00:00:45,955 --> 00:00:49,770
virtual work, because it's such
an important ingredient

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00:00:49,770 --> 00:00:52,510
for non-linear finite
element analysis.

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00:00:52,510 --> 00:00:54,670
In fact, the displacement-based
finite

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00:00:54,670 --> 00:00:57,900
element method, which is widely
used for non-linear

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00:00:57,900 --> 00:01:02,890
analysis, is based on an
application of the principle

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00:01:02,890 --> 00:01:05,060
of virtual work.

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00:01:05,060 --> 00:01:09,500
I also would like to discuss
with you and particularly

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00:01:09,500 --> 00:01:13,760
emphasize some basic
requirements of mechanics.

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00:01:13,760 --> 00:01:18,540
There are a few very important
requirements of mechanics that

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00:01:18,540 --> 00:01:23,470
it is important to always keep
in mind, and I'd like to share

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00:01:23,470 --> 00:01:27,080
some experiences with you
regarding these requirements.

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00:01:27,080 --> 00:01:32,240
And finally, to emphasize all
the points that I have tried

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00:01:32,240 --> 00:01:35,780
to make in the lecture, I like
to present and share with you

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00:01:35,780 --> 00:01:40,340
some experiences regarding
some actual analyses.

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00:01:40,340 --> 00:01:42,930
And I've chosen two analyses.

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00:01:42,930 --> 00:01:46,590
First we will look at the
analysis of a plate with a

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00:01:46,590 --> 00:01:49,860
hole, and then we will look
at the analysis of a

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00:01:49,860 --> 00:01:51,952
plate with a crack.

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00:01:51,952 --> 00:01:56,470
Let me now go over to my view
graph and start with a

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00:01:56,470 --> 00:02:00,270
discussion of the principle
of virtual work.

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00:02:00,270 --> 00:02:04,040
The principle of virtual work,
in essence, says that this

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00:02:04,040 --> 00:02:08,759
integral here on the left-hand
side is equal to what is on

37
00:02:08,759 --> 00:02:10,419
the right-hand side.

38
00:02:10,419 --> 00:02:13,420
And we should, of course,
discuss now in detail what we

39
00:02:13,420 --> 00:02:17,110
have on the left- and on
the right-hand side.

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00:02:17,110 --> 00:02:20,260
On the right-hand side,
we have this tR--

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00:02:20,260 --> 00:02:25,540
this is a script R, an unusual
symbol, but some time ago,

42
00:02:25,540 --> 00:02:28,360
some long time ago, I decided
to select this

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00:02:28,360 --> 00:02:30,970
symbol for this quantity.

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00:02:30,970 --> 00:02:37,340
This script t, superscript t
R, contains two integrals.

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00:02:37,340 --> 00:02:42,250
The first integral is an
integral over the volume tv.

46
00:02:42,250 --> 00:02:49,970
Notice it's the current volume
of the body of tfiB, the body

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00:02:49,970 --> 00:02:54,880
forces at time t multiplied
by the virtual

48
00:02:54,880 --> 00:02:59,100
displacement, del ui.

49
00:02:59,100 --> 00:03:02,210
This product here is
integrated over the

50
00:03:02,210 --> 00:03:04,750
volume at time t.

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00:03:04,750 --> 00:03:08,630
We add to this integral another
integral, which is an

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00:03:08,630 --> 00:03:12,690
integral over the surface
area of the body.

53
00:03:12,690 --> 00:03:20,420
And here we have surface forces
tfiS per unit area, and

54
00:03:20,420 --> 00:03:24,710
these are virtual displacements,
delta uiS.

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00:03:24,710 --> 00:03:31,320
We integrate this product over
the total area of the body.

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00:03:31,320 --> 00:03:33,970
So this defines the
right-hand side.

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00:03:33,970 --> 00:03:39,510
On the left-hand side, we have
one quantity here, which is t

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00:03:39,510 --> 00:03:45,390
tau ij, t once again referring,
of course, to time.

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00:03:45,390 --> 00:03:49,590
tau is a stress, and ij being
the components of the stress.

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00:03:49,590 --> 00:03:53,780
This stress is called
the Cauchy stress.

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00:03:53,780 --> 00:03:58,030
It's a force per unit
area at time t.

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00:03:58,030 --> 00:04:02,740
And on the left-hand side, we
have also a quantity which is

63
00:04:02,740 --> 00:04:06,760
a virtual strain.

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00:04:06,760 --> 00:04:12,610
Delta here meaning virtual,
teij meaning the strain.

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00:04:12,610 --> 00:04:16,459
And it's defined
as shown here.

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00:04:16,459 --> 00:04:20,880
Now let us look briefly at what
these quantities are.

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00:04:20,880 --> 00:04:25,710
We have a partial of del ui with
respect to the current

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00:04:25,710 --> 00:04:31,700
coordinate txj, and we have a
partial of del uj with respect

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00:04:31,700 --> 00:04:35,330
to the current coordinate txi.

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00:04:35,330 --> 00:04:40,830
Here I have prepared a view
graph which separates out,

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00:04:40,830 --> 00:04:43,310
once more, these important
quantities.

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00:04:43,310 --> 00:04:48,690
And on the top here in red, we
see the virtual displacement

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00:04:48,690 --> 00:04:50,460
and the virtual strain.

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00:04:50,460 --> 00:04:55,380
Notice that the virtual strain
here carries a subscript t, a

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00:04:55,380 --> 00:04:57,210
left subscript t.

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00:04:57,210 --> 00:05:00,900
This left subscript t means that
it's a strain referred to

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00:05:00,900 --> 00:05:03,720
the current configuration.

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00:05:03,720 --> 00:05:06,580
del always meaning virtual,
of course.

79
00:05:06,580 --> 00:05:09,450
And it's important to recognize
the following.

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00:05:09,450 --> 00:05:12,830
It's important to recognize
that these are the virtual

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00:05:12,830 --> 00:05:17,350
strains corresponding to these
virtual displacements.

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00:05:17,350 --> 00:05:20,180
That's the most important
point to recognize.

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00:05:20,180 --> 00:05:23,650
In green here, we have some
other quantities--

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00:05:23,650 --> 00:05:26,070
the current volume, the current

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00:05:26,070 --> 00:05:28,500
surface area of the body.

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00:05:28,500 --> 00:05:32,720
Current meaning always
at time t.

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00:05:32,720 --> 00:05:39,780
And here in blue, we see the
externally applied forces.

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00:05:39,780 --> 00:05:45,730
tfiB being the body forces, the
volume forces, and tfiS

89
00:05:45,730 --> 00:05:50,420
being the surface forces per
unit volume, per unit surface.

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00:05:50,420 --> 00:05:53,060
These are externally
applied forces.

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00:05:53,060 --> 00:05:56,360
So we have here quite different
quantities--

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00:05:56,360 --> 00:05:59,860
namely, virtual quantities
and real quantities.

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00:05:59,860 --> 00:06:04,510
So let me go back to this view
graph to discuss with you in

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00:06:04,510 --> 00:06:08,320
more detail what the principle
really tells us.

95
00:06:08,320 --> 00:06:12,890
It tells that the internal
virtual work consisting of the

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00:06:12,890 --> 00:06:16,320
product of the actual
stresses--

97
00:06:16,320 --> 00:06:19,670
t tau ij, which in
general analysis

98
00:06:19,670 --> 00:06:21,950
are, of course, unknown--

99
00:06:21,950 --> 00:06:28,920
times the virtual strains, which
we impose, this internal

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00:06:28,920 --> 00:06:31,300
virtual work--

101
00:06:31,300 --> 00:06:33,580
there's an integration over
the whole volume--

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00:06:33,580 --> 00:06:36,290
must be equal to the external
virtual work.

103
00:06:36,290 --> 00:06:40,200
And the external virtual work,
once again, is given here.

104
00:06:40,200 --> 00:06:43,020
This equation must hold--

105
00:06:43,020 --> 00:06:45,750
and this is an important
point--

106
00:06:45,750 --> 00:06:50,580
for arbitrary variations in
displacement, or arbitrary

107
00:06:50,580 --> 00:06:55,550
virtual displacements that
satisfy the geometric boundary

108
00:06:55,550 --> 00:06:58,890
conditions, that satisfy the
essential boundary conditions,

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00:06:58,890 --> 00:07:00,690
or you might say, the
displacement boundary

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00:07:00,690 --> 00:07:01,870
conditions.

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00:07:01,870 --> 00:07:05,760
In other words, this is an
equation that must hold for

112
00:07:05,760 --> 00:07:11,120
any virtual displacements and
corresponding virtual strains.

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00:07:11,120 --> 00:07:15,020
Remember, I just said that for
virtual displacement, we

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00:07:15,020 --> 00:07:16,770
always have corresponding
virtual strains.

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00:07:19,500 --> 00:07:22,220
This equation must always
hold, provided the

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00:07:22,220 --> 00:07:25,250
displacements satisfy the
displacement boundary

117
00:07:25,250 --> 00:07:28,000
conditions.

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00:07:28,000 --> 00:07:31,790
Here we have, once again, the
unknown forces, or stresses, I

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00:07:31,790 --> 00:07:33,940
should say, forces
per unit area.

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00:07:33,940 --> 00:07:36,650
Notice these are Cauchy
stresses.

121
00:07:36,650 --> 00:07:39,190
Here we have the word Cauchy.

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00:07:39,190 --> 00:07:42,460
In an infinitesimal displacement
analysis, we

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00:07:42,460 --> 00:07:45,610
really only talk about
one kind of stress--

124
00:07:45,610 --> 00:07:48,520
the engineering stress.

125
00:07:48,520 --> 00:07:51,230
In large displacement analysis,
large strain

126
00:07:51,230 --> 00:07:54,440
analysis, we have different
kind of stress measures.

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00:07:54,440 --> 00:07:57,880
And the stress that I now talk
about here is the Cauchy

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00:07:57,880 --> 00:08:03,070
stress, which is the force per
unit area at time t, which is,

129
00:08:03,070 --> 00:08:05,570
in other words, an actual
physical stress.

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00:08:05,570 --> 00:08:09,870
Later on, we will introduce
another kind of stress, a very

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00:08:09,870 --> 00:08:11,620
important stress measure,
which, however,

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00:08:11,620 --> 00:08:13,570
is not quite physical.

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00:08:13,570 --> 00:08:15,510
This one here is a physical
stress, the

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00:08:15,510 --> 00:08:17,240
force per unit area.

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00:08:17,240 --> 00:08:19,450
And that is the stress that
you, for example, in a

136
00:08:19,450 --> 00:08:22,860
computer program, would like
to get printed out.

137
00:08:22,860 --> 00:08:25,690
This is the force per unit area
that you would design

138
00:08:25,690 --> 00:08:28,260
your structure with.

139
00:08:28,260 --> 00:08:32,400
This here is, once again,
the virtual strain.

140
00:08:32,400 --> 00:08:38,400
And notice that this strain
is referred to the current

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00:08:38,400 --> 00:08:42,789
geometry, the coordinate
at time t.

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00:08:42,789 --> 00:08:46,580
Therefore, stress at time t
times virtual strain at time

143
00:08:46,580 --> 00:08:51,200
t-- that's what we're looking at
here, and that gives us the

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00:08:51,200 --> 00:08:53,710
total internal virtual
work when we

145
00:08:53,710 --> 00:08:56,340
integrate over the original--

146
00:08:56,340 --> 00:08:59,730
over the current volume,
current, I should say, very

147
00:08:59,730 --> 00:09:00,570
important--

148
00:09:00,570 --> 00:09:02,720
the current volume
of the body.

149
00:09:02,720 --> 00:09:06,990
So if we have a body that
moves through space and

150
00:09:06,990 --> 00:09:10,110
undergoes large displacements,
large rotations, large

151
00:09:10,110 --> 00:09:14,630
strains, then this here is the
quantity of the internal

152
00:09:14,630 --> 00:09:19,880
virtual work with the current
stresses at time t in the

153
00:09:19,880 --> 00:09:22,600
configuration at time t.

154
00:09:22,600 --> 00:09:26,180
Let us look at a schematic
example, because there are so

155
00:09:26,180 --> 00:09:27,956
many important considerations.

156
00:09:30,460 --> 00:09:33,380
Here we have, in a
three-dimensional Cartesian

157
00:09:33,380 --> 00:09:40,190
coordinate frame, denoted with
x1, x2, x3 as coordinate axes,

158
00:09:40,190 --> 00:09:44,520
a body in which we are showing
two material particles.

159
00:09:44,520 --> 00:09:48,950
Notice one of them in the body
itself, in the volume of the

160
00:09:48,950 --> 00:09:51,690
body, rather, and one
on the surface.

161
00:09:51,690 --> 00:09:55,670
The body is here supported,
and this is the

162
00:09:55,670 --> 00:09:58,590
configuration at time 0.

163
00:09:58,590 --> 00:10:04,680
Now this body moves through
a certain amount.

164
00:10:04,680 --> 00:10:07,680
It undergoes large
displacements, large

165
00:10:07,680 --> 00:10:11,470
rotations, large strains,
and takes on the red

166
00:10:11,470 --> 00:10:13,830
configuration here.

167
00:10:13,830 --> 00:10:17,640
Notice that in that red
configuration, we have certain

168
00:10:17,640 --> 00:10:21,570
body forces, which we already
talked about earlier, and

169
00:10:21,570 --> 00:10:24,910
surface forces, that we also
talked about earlier.

170
00:10:24,910 --> 00:10:28,750
Notice that the particles
have now moved to new

171
00:10:28,750 --> 00:10:31,550
configurations, of course,
to new positions.

172
00:10:31,550 --> 00:10:36,320
The surface particle has moved
there, and the particle within

173
00:10:36,320 --> 00:10:40,360
the volume has moved there.

174
00:10:40,360 --> 00:10:46,970
We denote this as tui, the
displacement at time t, or the

175
00:10:46,970 --> 00:10:49,930
displacement from configuration
02,

176
00:10:49,930 --> 00:10:53,540
configuration t, for this
particular particle.

177
00:10:53,540 --> 00:10:59,950
Similarly, here we denote this
displacement as tuiS.

178
00:10:59,950 --> 00:11:04,590
Now the principle of virtual
work says that we will be

179
00:11:04,590 --> 00:11:06,940
imposing on to this red
configuration the

180
00:11:06,940 --> 00:11:10,270
configuration at time t, a
variation in displacement,

181
00:11:10,270 --> 00:11:12,520
which is the virtual
displacement.

182
00:11:12,520 --> 00:11:17,740
And here we see now one such
virtual displacement.

183
00:11:17,740 --> 00:11:23,110
Notice that it is shown by the
blue line for the whole body,

184
00:11:23,110 --> 00:11:25,870
and it's a virtual displacement
from the

185
00:11:25,870 --> 00:11:28,950
configuration at time t.

186
00:11:28,950 --> 00:11:33,010
Notice that the particle, which
was originally here, or

187
00:11:33,010 --> 00:11:37,020
which was at time t here, I
should more precisely say, has

188
00:11:37,020 --> 00:11:42,010
moved to this point in the
virtual displacement.

189
00:11:42,010 --> 00:11:47,110
And similarly, the particle that
was here at time t has

190
00:11:47,110 --> 00:11:50,880
moved to this configuration,
to this point here.

191
00:11:50,880 --> 00:11:52,260
Now, this is one variation.

192
00:11:52,260 --> 00:11:54,310
However, the principle of
virtual work, of course,

193
00:11:54,310 --> 00:11:57,920
states that the left-hand side
must be equal to the

194
00:11:57,920 --> 00:11:59,230
right-hand side--

195
00:11:59,230 --> 00:12:00,890
I think you know now what
I mean by left-

196
00:12:00,890 --> 00:12:02,800
and right-hand sides--

197
00:12:02,800 --> 00:12:04,120
for any variation.

198
00:12:04,120 --> 00:12:09,150
So we have here another
variation.

199
00:12:09,150 --> 00:12:13,180
All such variations, however,
must satisfy the displacement

200
00:12:13,180 --> 00:12:15,860
boundary conditions, and those
displacement boundary

201
00:12:15,860 --> 00:12:19,360
conditions are given
right down here.

202
00:12:19,360 --> 00:12:23,070
Now notice that the particle,
which I looked at earlier

203
00:12:23,070 --> 00:12:27,090
already, has now moved to this
configuration, or in this

204
00:12:27,090 --> 00:12:28,900
variation of displacement.

205
00:12:28,900 --> 00:12:32,640
Similarly, this particle
has moved here.

206
00:12:32,640 --> 00:12:36,710
Now the principle of virtual
work states once again that

207
00:12:36,710 --> 00:12:41,130
for any variation that is
satisfying the displacement

208
00:12:41,130 --> 00:12:45,130
boundary conditions, the
left-hand side, the internal

209
00:12:45,130 --> 00:12:48,230
virtual work, must be equal to
the right-hand side, the

210
00:12:48,230 --> 00:12:50,250
external virtual work.

211
00:12:50,250 --> 00:12:54,090
And these are two variations
that we looked at.

212
00:12:54,090 --> 00:12:58,080
Of course, there are
many, many more.

213
00:12:58,080 --> 00:13:01,250
In fact, we could think of
millions, billions such

214
00:13:01,250 --> 00:13:02,410
variations.

215
00:13:02,410 --> 00:13:07,670
And we will, in finite element
analysis, however, discretize

216
00:13:07,670 --> 00:13:11,160
the whole structure using finite
elements, and then only

217
00:13:11,160 --> 00:13:13,430
allow certain sets
of variations.

218
00:13:13,430 --> 00:13:17,010
Namely, those contained in the
interpolation functions of the

219
00:13:17,010 --> 00:13:19,390
finite elements.

220
00:13:19,390 --> 00:13:23,680
Well, let us now look at
one important point.

221
00:13:23,680 --> 00:13:27,340
Namely that if we were to
integrate the principle of

222
00:13:27,340 --> 00:13:31,570
virtual work by parts, we would
obtain the governing

223
00:13:31,570 --> 00:13:35,920
differential equations of
motion plus the force or

224
00:13:35,920 --> 00:13:40,030
natural boundary conditions,
just like an infinitesimal

225
00:13:40,030 --> 00:13:41,250
displacement analysis.

226
00:13:41,250 --> 00:13:44,780
However, the difference, of
course, is that we have

227
00:13:44,780 --> 00:13:49,250
referred all our static and
kinematic variables now to

228
00:13:49,250 --> 00:13:53,150
time t, to the current
configuration of the body.

229
00:13:53,150 --> 00:13:55,400
Let's look now at an example.

230
00:13:55,400 --> 00:13:58,350
It's a simple example, in
some sense, but it still

231
00:13:58,350 --> 00:14:03,380
illustrates many of the basic
points that I just mentioned.

232
00:14:03,380 --> 00:14:05,940
The example is a truss
stretching

233
00:14:05,940 --> 00:14:08,160
under its own weight.

234
00:14:08,160 --> 00:14:10,590
Here we have the truss
in its original

235
00:14:10,590 --> 00:14:14,880
configuration shown in black.

236
00:14:14,880 --> 00:14:19,010
It has an original length L0.

237
00:14:19,010 --> 00:14:22,640
It's subjected, of course,
to gravity, g.

238
00:14:22,640 --> 00:14:27,950
A typical cross-sectional
area is given here, 0A.

239
00:14:27,950 --> 00:14:33,070
Notice that this bar, under its
own weight, moves down to

240
00:14:33,070 --> 00:14:36,090
the configuration
at time t here.

241
00:14:36,090 --> 00:14:40,180
tA is this cross-sectional
area now.

242
00:14:40,180 --> 00:14:43,350
In other words, this area has
now moved to that area.

243
00:14:43,350 --> 00:14:47,140
The length of the truss, of
course, has changed, and in

244
00:14:47,140 --> 00:14:50,500
fact has changed to tL.

245
00:14:50,500 --> 00:14:53,000
In order to analyze this
problem, we have to make

246
00:14:53,000 --> 00:14:54,800
certain assumptions.

247
00:14:54,800 --> 00:14:58,640
The first assumption is, plane
cross-sections remain plane.

248
00:14:58,640 --> 00:14:59,600
Second assumption--

249
00:14:59,600 --> 00:15:03,550
constant uni-axial stress
on each cross-section.

250
00:15:03,550 --> 00:15:06,710
And then if we make these two
assumptions, then we have a

251
00:15:06,710 --> 00:15:10,000
one-dimensional analysis.

252
00:15:10,000 --> 00:15:13,800
Using these assumptions, we then
obtain directly, from the

253
00:15:13,800 --> 00:15:16,280
general principle of virtual
work, the following.

254
00:15:16,280 --> 00:15:19,400
Here we have, on the left-hand
side, the general left-hand

255
00:15:19,400 --> 00:15:26,060
side of the principle of virtual
work, Cauchy stress,

256
00:15:26,060 --> 00:15:30,000
the virtual strains, integrated
over the current

257
00:15:30,000 --> 00:15:34,660
volume at time t, which then
becomes, on the right-hand

258
00:15:34,660 --> 00:15:38,430
side here, the one stress only,
because we have only one

259
00:15:38,430 --> 00:15:43,670
stress in this bar, the vertical
stress, times the

260
00:15:43,670 --> 00:15:47,120
virtual strain corresponding
to that motion of the bar,

261
00:15:47,120 --> 00:15:49,380
namely, vertically downwards.

262
00:15:49,380 --> 00:15:53,620
And this product is integrated
over the current area at time

263
00:15:53,620 --> 00:15:59,430
t, and of course, the
longitudinal coordinate tdx.

264
00:15:59,430 --> 00:16:03,090
Notice that here we are
integrating over the current

265
00:16:03,090 --> 00:16:08,310
volume, because we have
a tA and a tL here.

266
00:16:08,310 --> 00:16:12,130
The right-hand side becomes,
in this particular case, an

267
00:16:12,130 --> 00:16:16,310
integration over the current
length of the gravity force,

268
00:16:16,310 --> 00:16:20,590
which involves the mass density
at time t, in other

269
00:16:20,590 --> 00:16:22,580
words, the current
mass density.

270
00:16:22,580 --> 00:16:24,100
Notice the mass density,
of course,

271
00:16:24,100 --> 00:16:27,890
changes as the bar extends.

272
00:16:27,890 --> 00:16:29,550
g is the gravity constant.

273
00:16:29,550 --> 00:16:35,190
And here we have the virtual
displacement, area at time t,

274
00:16:35,190 --> 00:16:39,990
and of course the differential
increment tdx.

275
00:16:39,990 --> 00:16:43,220
Once again, we here also
integrate over the

276
00:16:43,220 --> 00:16:47,000
length Lt at time t.

277
00:16:47,000 --> 00:16:51,640
Hence, if we now equate these
two quantities, we directly

278
00:16:51,640 --> 00:16:54,730
obtain what's shown here.

279
00:16:54,730 --> 00:16:58,390
Now it's important to recognize
that in this

280
00:16:58,390 --> 00:17:03,940
equation, these here are the
virtual strains corresponding

281
00:17:03,940 --> 00:17:07,890
to these virtual
displacements.

282
00:17:07,890 --> 00:17:10,336
And once again, we apply
the principle of

283
00:17:10,336 --> 00:17:13,520
course, at time t.

284
00:17:13,520 --> 00:17:18,890
So stresses at time t, meaning
forces per unit area at time

285
00:17:18,890 --> 00:17:22,520
t, area at time t, and so on.

286
00:17:22,520 --> 00:17:28,790
This virtual strain, also used
to describe this quantity, is

287
00:17:28,790 --> 00:17:30,720
given right here.

288
00:17:30,720 --> 00:17:34,560
Notice that here we have
the partial of del u

289
00:17:34,560 --> 00:17:37,125
with respect to tx.

290
00:17:37,125 --> 00:17:40,050
tx, the coordinate at time t.

291
00:17:40,050 --> 00:17:46,410
And this del u is exactly the
del u that we see right there.

292
00:17:46,410 --> 00:17:50,780
Now this principle must hold for
any arbitrary variation in

293
00:17:50,780 --> 00:17:55,990
displacement, any arbitrary
virtual displacement, that

294
00:17:55,990 --> 00:17:58,900
satisfy the displacement
boundary conditions.

295
00:17:58,900 --> 00:18:00,980
And the displacement boundary
condition here, of course, is

296
00:18:00,980 --> 00:18:04,930
simply that the displacement
is 0 at the top of the bar.

297
00:18:04,930 --> 00:18:09,700
If we use, now, integration by
parts on this equation that I

298
00:18:09,700 --> 00:18:14,380
just discussed with you, we
obtain this equation here.

299
00:18:14,380 --> 00:18:17,860
And that equation contains
really two terms,

300
00:18:17,860 --> 00:18:19,440
two parts to it.

301
00:18:19,440 --> 00:18:24,460
First, an integral that is taken
over the length L of the

302
00:18:24,460 --> 00:18:31,910
bar, and then a quantity that
corresponds to the length of

303
00:18:31,910 --> 00:18:37,720
the position 0 and L, tL.

304
00:18:37,720 --> 00:18:41,890
And all of the sum of these
two must be equal to 0.

305
00:18:41,890 --> 00:18:45,660
Since the variation del u is
arbitrary, except, as I

306
00:18:45,660 --> 00:18:49,440
mentioned, right at the top of
the bar, we directly obtain

307
00:18:49,440 --> 00:18:53,110
that this equation must hold,
which is really extracting

308
00:18:53,110 --> 00:18:56,240
this term here from
the integral.

309
00:18:56,240 --> 00:18:57,720
And this, of course,
is the governing

310
00:18:57,720 --> 00:19:00,010
differential equation.

311
00:19:00,010 --> 00:19:05,630
And also, we must have that at
the lower part of the bar,

312
00:19:05,630 --> 00:19:10,120
this force or natural boundary
condition must be satisfied.

313
00:19:10,120 --> 00:19:13,630
Therefore, this is an example
which shows, really, how the

314
00:19:13,630 --> 00:19:16,930
principle of virtual work
contains the basic

315
00:19:16,930 --> 00:19:18,740
differential equations
of equilibrium.

316
00:19:18,740 --> 00:19:22,240
Here it is just one, because we
have only one direction for

317
00:19:22,240 --> 00:19:23,260
the motion.

318
00:19:23,260 --> 00:19:27,110
And also the stress boundary
conditions, the natural or

319
00:19:27,110 --> 00:19:30,880
force stress boundary
conditions.

320
00:19:30,880 --> 00:19:33,890
The finite element application
of the principle of virtual

321
00:19:33,890 --> 00:19:36,580
work proceeds as follows.

322
00:19:36,580 --> 00:19:43,780
Here, once again, virtual
strains, virtual displacements

323
00:19:43,780 --> 00:19:47,950
in the volume of the body and
on the surface of the body.

324
00:19:47,950 --> 00:19:50,670
These are the virtual
quantities.

325
00:19:50,670 --> 00:19:55,170
The actual applied forces,
externally applied forces, are

326
00:19:55,170 --> 00:19:58,300
these blue quantities.

327
00:19:58,300 --> 00:20:01,390
We deliberately chose a
different color to emphasize

328
00:20:01,390 --> 00:20:05,290
that these are real forces,
blue, applied to the body, and

329
00:20:05,290 --> 00:20:08,660
these are virtual

330
00:20:08,660 --> 00:20:11,430
displacements and virtual strains.

331
00:20:11,430 --> 00:20:14,700
The stresses here at time t,
of course, are unknown, and

332
00:20:14,700 --> 00:20:17,790
those we want to calculate.

333
00:20:17,790 --> 00:20:22,710
In the finite element method,
what's shown in this big box,

334
00:20:22,710 --> 00:20:27,680
we interpolate displacements,
virtual displacements.

335
00:20:27,680 --> 00:20:32,950
We calculate from those
interpolations, strains, real

336
00:20:32,950 --> 00:20:35,830
strains, virtual strains.

337
00:20:35,830 --> 00:20:39,950
Via the stress-strain law, we
got stresses, and we will see

338
00:20:39,950 --> 00:20:44,130
in the later lectures that in
fact, what we are arriving at

339
00:20:44,130 --> 00:20:50,460
is a force vector tF and an
external load vector.

340
00:20:50,460 --> 00:20:57,130
Now, this tF here, shown in
green, is a result of this

341
00:20:57,130 --> 00:21:00,200
stress t tau ij.

342
00:21:00,200 --> 00:21:06,500
And it's really the nodal
point force vector that

343
00:21:06,500 --> 00:21:13,070
corresponds to these internal
element stresses, t tau ij.

344
00:21:13,070 --> 00:21:18,100
This R here is a nodal point
force vector that corresponds

345
00:21:18,100 --> 00:21:21,140
to these externally
applied forces.

346
00:21:23,760 --> 00:21:30,210
This del UT, on left-hand side
and on the right-hand side, is

347
00:21:30,210 --> 00:21:34,980
a vector of virtual
displacements, nodal point

348
00:21:34,980 --> 00:21:39,190
displacements, and which, in the
finite element analysis,

349
00:21:39,190 --> 00:21:43,730
we then set equal to the
identity matrix, invoking the

350
00:21:43,730 --> 00:21:46,490
principle of virtual work in
turn for each degree of

351
00:21:46,490 --> 00:21:50,290
freedom, just like we do in
linear elastic analysis.

352
00:21:50,290 --> 00:21:52,770
This becomes the identity
matrix, that becomes the

353
00:21:52,770 --> 00:21:56,160
identity matrix, and the result
is, of course, simply

354
00:21:56,160 --> 00:22:01,310
that tF must be equal to tR.

355
00:22:01,310 --> 00:22:05,390
The procedure that we are
following in this box, of

356
00:22:05,390 --> 00:22:07,800
course, is of much interest
in non-linear analysis.

357
00:22:07,800 --> 00:22:11,080
And what's happening in this box
here, we will be talking

358
00:22:11,080 --> 00:22:17,690
about in some considerable
detail in the later lectures.

359
00:22:17,690 --> 00:22:20,900
But for the moment, let's now
assume that the solution at

360
00:22:20,900 --> 00:22:22,750
time t is known.

361
00:22:22,750 --> 00:22:26,880
Hence the stress at time t,
the volume at time t, et

362
00:22:26,880 --> 00:22:31,130
cetera, are known, and we want
to obtain the solution at time

363
00:22:31,130 --> 00:22:33,180
t plus delta t.

364
00:22:33,180 --> 00:22:36,850
In other words, we want to move
ahead 1 increment in time

365
00:22:36,850 --> 00:22:40,720
delta t and establish the
solution at that time.

366
00:22:40,720 --> 00:22:44,770
The principle of virtual work
applied now at time t plus

367
00:22:44,770 --> 00:22:48,170
delta t in exactly the same way
as we just applied it at

368
00:22:48,170 --> 00:22:53,210
time t results in this
set of equations.

369
00:22:53,210 --> 00:22:56,520
A vector of nodal point forces
that is equivalent to the

370
00:22:56,520 --> 00:22:59,200
current internal element
stresses at time t

371
00:22:59,200 --> 00:23:00,970
plus delta t now.

372
00:23:00,970 --> 00:23:04,580
And these are the nodal point
externally applied forces at

373
00:23:04,580 --> 00:23:07,280
time t plus delta t.

374
00:23:07,280 --> 00:23:11,580
Well, of course, we don't know
t plus delta tF, and for the

375
00:23:11,580 --> 00:23:16,740
solution, we now say that t plus
delta tF is equal to tF,

376
00:23:16,740 --> 00:23:20,980
which we do know, plus
a matrix tK--

377
00:23:20,980 --> 00:23:25,040
we call this as tangent
stiffness matrix--

378
00:23:25,040 --> 00:23:32,620
times del U. If we set this
left-hand side equal to t plus

379
00:23:32,620 --> 00:23:36,890
delta tR, the externally applied
loads, we directly

380
00:23:36,890 --> 00:23:38,950
obtain this equation.

381
00:23:38,950 --> 00:23:42,060
We are putting the unknowns on
the left-hand side, and the

382
00:23:42,060 --> 00:23:45,550
unknowns, of course, are the
incremental displacements.

383
00:23:45,550 --> 00:23:47,845
The tangent stiffness
matrix is known.

384
00:23:47,845 --> 00:23:49,930
It's the matrix corresponding
to the

385
00:23:49,930 --> 00:23:52,892
configuration at time t.

386
00:23:52,892 --> 00:23:57,980
The load vector is known, and
tF is assumed to be known.

387
00:23:57,980 --> 00:24:01,110
Now, this is here an
out-of-balance load vector,

388
00:24:01,110 --> 00:24:05,430
and this set of equations then
result into the solution of an

389
00:24:05,430 --> 00:24:09,380
increment in nodal point
displacements delta U. This

390
00:24:09,380 --> 00:24:12,290
increment in nodal point
displacement is added to the

391
00:24:12,290 --> 00:24:15,640
displacement at time t to obtain
the displacement at

392
00:24:15,640 --> 00:24:17,100
time t plus delta t.

393
00:24:17,100 --> 00:24:18,050
But--

394
00:24:18,050 --> 00:24:20,390
and here is the big but--

395
00:24:20,390 --> 00:24:23,030
by writing this equation
up here, we had to

396
00:24:23,030 --> 00:24:24,940
linearize the problem.

397
00:24:24,940 --> 00:24:28,480
You see, we are dealing with a
large amount of non-linearity

398
00:24:28,480 --> 00:24:33,890
that can come from the
kinematics, from the material

399
00:24:33,890 --> 00:24:35,620
relationships and so on.

400
00:24:35,620 --> 00:24:37,020
And we had to linearize.

401
00:24:37,020 --> 00:24:41,410
We linearized here about the
configuration at time t to

402
00:24:41,410 --> 00:24:45,850
obtain this term, and this means
that down here in this

403
00:24:45,850 --> 00:24:49,930
equation, we only can have
an approximation sign.

404
00:24:49,930 --> 00:24:52,680
In other words, what we have
calculated are not the exact

405
00:24:52,680 --> 00:24:55,100
increments in nodal point
displacement, but an

406
00:24:55,100 --> 00:24:56,690
approximation to it.

407
00:24:56,690 --> 00:24:59,930
More generally, then, we
repeat that process.

408
00:24:59,930 --> 00:25:05,260
We apply the same process in an
iterative way to obtain a

409
00:25:05,260 --> 00:25:06,630
very accurate solution.

410
00:25:06,630 --> 00:25:09,030
And this is done in
the following way.

411
00:25:09,030 --> 00:25:12,730
On the right-hand side now, we
have a load vector, the one

412
00:25:12,730 --> 00:25:16,810
that we discussed already
before, minus the nodal point

413
00:25:16,810 --> 00:25:21,170
forces corresponding to
time t plus delta t in

414
00:25:21,170 --> 00:25:23,300
iteration i minus 1.

415
00:25:23,300 --> 00:25:26,810
I should say at the beginning
of iteration i, or end of

416
00:25:26,810 --> 00:25:29,050
iteration i minus 1.

417
00:25:29,050 --> 00:25:31,335
This gives us an out-of-balance
load vector.

418
00:25:31,335 --> 00:25:35,790
We're calculating an increment
in displacements, delta Ui.

419
00:25:35,790 --> 00:25:39,140
And that increment in
displacement is added to the

420
00:25:39,140 --> 00:25:41,450
displacements which were
known already.

421
00:25:41,450 --> 00:25:44,580
At the end of iteration i minus
1 corresponding to time

422
00:25:44,580 --> 00:25:47,370
t plus delta t, we knew
these displacements.

423
00:25:47,370 --> 00:25:50,590
We add the increment that we
calculated, and we obtain a

424
00:25:50,590 --> 00:25:55,420
better guess, so to say, for
the current displacement at

425
00:25:55,420 --> 00:25:57,610
time t plus delta t.

426
00:25:57,610 --> 00:26:00,400
Now of course in this iteration
here, we need some

427
00:26:00,400 --> 00:26:02,670
initial conditions, and
the initial conditions

428
00:26:02,670 --> 00:26:04,340
are given down here.

429
00:26:04,340 --> 00:26:08,540
The t plus delta t F 0, in other
words, when i is equal

430
00:26:08,540 --> 00:26:14,730
to 1, then we need t plus delta
t F 0 is given by tF.

431
00:26:14,730 --> 00:26:20,490
The displacement t plus delta t
U0, when i is equal to 1, is

432
00:26:20,490 --> 00:26:22,370
given by tU.

433
00:26:22,370 --> 00:26:26,020
These initial conditions are
used in the iteration here,

434
00:26:26,020 --> 00:26:29,390
and we notice that this set
of equations in the first

435
00:26:29,390 --> 00:26:33,200
iteration, when i is equal to
1, reduces to the equations

436
00:26:33,200 --> 00:26:36,300
that I already discussed
with you with the

437
00:26:36,300 --> 00:26:38,270
previous view graph.

438
00:26:38,270 --> 00:26:42,610
Well, if we look at this
set of equations, we

439
00:26:42,610 --> 00:26:44,230
recognize the following.

440
00:26:44,230 --> 00:26:47,970
Nodal point equilibrium is
satisfied when the equation t

441
00:26:47,970 --> 00:26:51,170
plus delta tR minus t plus
delta tF, i minus

442
00:26:51,170 --> 00:26:53,080
1, is equal to 0.

443
00:26:53,080 --> 00:26:57,910
When the externally applied
loads are equal to the nodal

444
00:26:57,910 --> 00:27:02,160
point forces corresponding to
the stresses at time t plus

445
00:27:02,160 --> 00:27:08,860
delta t, when these two vectors
are equal, or rather,

446
00:27:08,860 --> 00:27:13,520
when R minus F is equal to 0,
then of course we have nodal

447
00:27:13,520 --> 00:27:15,250
point equilibrium.

448
00:27:15,250 --> 00:27:18,400
Compatibility is satisfied
provided a compatible element

449
00:27:18,400 --> 00:27:20,360
mesh is used.

450
00:27:20,360 --> 00:27:23,700
The stress-strain law enters
in the calculation of the

451
00:27:23,700 --> 00:27:27,450
tangent stiffness matrix and the
nodal point force vector

452
00:27:27,450 --> 00:27:31,790
F, which corresponds to the
current stresses at time t

453
00:27:31,790 --> 00:27:35,300
plus delta t, iteration
i minus 1.

454
00:27:35,300 --> 00:27:40,890
Well, we will get back to these
considerations just now.

455
00:27:40,890 --> 00:27:46,630
But most important is to
recognize that we have to

456
00:27:46,630 --> 00:27:52,940
calculate F t plus delta ti
minus 1 very accurately from

457
00:27:52,940 --> 00:27:56,190
the displacements t plus
delta t Ui minus 1.

458
00:27:56,190 --> 00:27:59,340
The general procedure is
depicted on this view graph.

459
00:27:59,340 --> 00:28:03,670
Here we have the displacement
at time t plus delta t ui

460
00:28:03,670 --> 00:28:07,180
minus 1, which give
us strains.

461
00:28:07,180 --> 00:28:10,120
We can directly, via
differentiations of the

462
00:28:10,120 --> 00:28:15,350
displacements, get strains,
which gives us stresses if we

463
00:28:15,350 --> 00:28:16,970
enter here with the appropriate

464
00:28:16,970 --> 00:28:18,560
constitutive relation.

465
00:28:18,560 --> 00:28:21,760
Of course, we have to use the
constitutive relation

466
00:28:21,760 --> 00:28:24,910
corresponding to the problem
that we want to solve.

467
00:28:24,910 --> 00:28:28,670
We get stresses, and the
stresses, then, give us this

468
00:28:28,670 --> 00:28:31,600
nodal point force vector
that I talked about.

469
00:28:31,600 --> 00:28:37,200
Notice that in this relationship
here, we have to

470
00:28:37,200 --> 00:28:40,050
perform an integration, and that
integration is expressed

471
00:28:40,050 --> 00:28:43,140
here, where we obtain the
stresses at time t plus delta

472
00:28:43,140 --> 00:28:49,600
t in iteration i minus 1 by
taking the stresses at time t

473
00:28:49,600 --> 00:28:53,670
and integrating the
stress-strain law times the

474
00:28:53,670 --> 00:28:59,330
incremental strains from time
t to time t plus delta t,

475
00:28:59,330 --> 00:29:03,570
where we use, of course, the
strains at time t, and the

476
00:29:03,570 --> 00:29:06,010
strains at time t plus delta t

477
00:29:06,010 --> 00:29:08,680
corresponding to that iteration.

478
00:29:08,680 --> 00:29:13,410
This step is most important,
because you have to calculate

479
00:29:13,410 --> 00:29:21,580
the F vector correctly so that
when you say that this

480
00:29:21,580 --> 00:29:28,220
equation is satisfied, that
indeed you have obtained the

481
00:29:28,220 --> 00:29:31,590
right configuration for the
finite element mesh that

482
00:29:31,590 --> 00:29:32,510
you're considering.

483
00:29:32,510 --> 00:29:35,870
So it is most important to
calculate the F correctly.

484
00:29:35,870 --> 00:29:38,060
Of course, the tangent stiffness
matrix should also

485
00:29:38,060 --> 00:29:42,000
be calculated appropriately, and
we will talk about that as

486
00:29:42,000 --> 00:29:43,250
well later on.

487
00:29:45,770 --> 00:29:49,420
In all this discussion, I should
point out we assumed

488
00:29:49,420 --> 00:29:52,470
that the nodal point loads are
independent of the structural

489
00:29:52,470 --> 00:29:53,680
deformations.

490
00:29:53,680 --> 00:29:57,150
In other words, they vary only
as a function of time.

491
00:29:57,150 --> 00:29:59,600
Here's a small problem
that schematically

492
00:29:59,600 --> 00:30:01,510
shows what we mean.

493
00:30:01,510 --> 00:30:07,500
The load R is always acting
vertically, never mind how

494
00:30:07,500 --> 00:30:10,070
much this beam has displaced.

495
00:30:10,070 --> 00:30:13,790
And that load variation is shown
here on the graph as a

496
00:30:13,790 --> 00:30:15,040
function of time.

497
00:30:17,730 --> 00:30:22,860
We satisfy the basic
requirements of mechanics if

498
00:30:22,860 --> 00:30:26,730
we perform this finite element
analysis appropriately.

499
00:30:26,730 --> 00:30:30,570
We satisfy the stress-strain law
because we are evaluating

500
00:30:30,570 --> 00:30:34,350
the stresses correctly
corresponding

501
00:30:34,350 --> 00:30:36,080
from the given strains.

502
00:30:36,080 --> 00:30:39,490
We also satisfy the
compatibility requirements

503
00:30:39,490 --> 00:30:43,110
provided we use compatible
elements, and of course we

504
00:30:43,110 --> 00:30:47,130
satisfy the displacement
boundary conditions.

505
00:30:47,130 --> 00:30:50,840
Equilibrium we also satisfy
corresponding to the finite

506
00:30:50,840 --> 00:30:54,190
element nodal point degrees
of freedom.

507
00:30:54,190 --> 00:30:57,040
But there is one important
point.

508
00:30:57,040 --> 00:30:59,920
This means that we only
satisfy global

509
00:30:59,920 --> 00:31:01,970
equilibrium in general.

510
00:31:01,970 --> 00:31:07,870
We satisfy equilibrium locally
only if we use a fine enough

511
00:31:07,870 --> 00:31:09,840
finite element mesh.

512
00:31:09,840 --> 00:31:12,870
In other words, I'd like to
distinguish now between

513
00:31:12,870 --> 00:31:16,060
satisfying global equilibrium
and local equilibrium.

514
00:31:18,670 --> 00:31:21,930
Local equilibrium means that
we want to satisfy the

515
00:31:21,930 --> 00:31:25,740
differential equations of
equilibrium at every

516
00:31:25,740 --> 00:31:27,390
point in the mesh.

517
00:31:27,390 --> 00:31:31,360
And that is only achieved if we
have a large enough finite

518
00:31:31,360 --> 00:31:33,540
elements for the
given problem--

519
00:31:33,540 --> 00:31:37,120
in other words, if our
discretization is fine enough.

520
00:31:37,120 --> 00:31:41,540
A check on whether we are
satisfying local equilibrium

521
00:31:41,540 --> 00:31:47,280
well enough is to look at the
stresses along the boundary

522
00:31:47,280 --> 00:31:49,350
and see whether the
stress boundary

523
00:31:49,350 --> 00:31:51,900
conditions are satisfied.

524
00:31:51,900 --> 00:31:52,670
That is one check.

525
00:31:52,670 --> 00:31:58,850
Another check is to test whether
along boundaries of

526
00:31:58,850 --> 00:32:03,910
the elements, there are small
stress jumps only.

527
00:32:03,910 --> 00:32:08,310
In other words, whether the
stresses from one element to

528
00:32:08,310 --> 00:32:11,110
another element do not
jump too much.

529
00:32:11,110 --> 00:32:16,510
We call that the stress jumping
because as we will

530
00:32:16,510 --> 00:32:20,300
just talk further, when we
calculate the stress at a node

531
00:32:20,300 --> 00:32:23,100
for an element, we get,
of course, one value.

532
00:32:23,100 --> 00:32:26,510
And if at that same node, the
stresses are calculated using

533
00:32:26,510 --> 00:32:28,130
another element, you
would get, in

534
00:32:28,130 --> 00:32:29,520
general, another value.

535
00:32:29,520 --> 00:32:35,020
This stress difference should
not be very large, and of

536
00:32:35,020 --> 00:32:37,780
course, if we have a homogeneous
material, if we

537
00:32:37,780 --> 00:32:40,320
have other conditions satisfied
as well, in other

538
00:32:40,320 --> 00:32:44,370
words, if typically the exact
solution would have a

539
00:32:44,370 --> 00:32:48,070
continuous stress variation,
then the stress jump should be

540
00:32:48,070 --> 00:32:51,400
very small for an accurate
solution.

541
00:32:51,400 --> 00:32:54,220
And this is a good check, an
indication of whether we

542
00:32:54,220 --> 00:32:58,050
indeed have obtained
a valid solution.

543
00:32:58,050 --> 00:33:02,810
Well, I like to now demonstrate
to you some of

544
00:33:02,810 --> 00:33:06,800
these considerations by showing
you the results of two

545
00:33:06,800 --> 00:33:08,370
example analyses.

546
00:33:08,370 --> 00:33:12,740
But it is best if we show these
examples on a different

547
00:33:12,740 --> 00:33:15,390
reel, on a second reel
of this lecture.

548
00:33:15,390 --> 00:33:16,640
Thank you very much.

549
00:33:19,900 --> 00:33:23,590
We just discussed some issues
regarding how equilibrium is

550
00:33:23,590 --> 00:33:25,560
satisfied in a
displacement-based finite

551
00:33:25,560 --> 00:33:27,270
element analysis.

552
00:33:27,270 --> 00:33:30,890
We discussed briefly that global
equilibrium is always

553
00:33:30,890 --> 00:33:32,650
satisfied for any finite
element mesh

554
00:33:32,650 --> 00:33:33,980
that you have selected.

555
00:33:33,980 --> 00:33:38,340
By global equilibrium, we mean
equilibrium at the nodal

556
00:33:38,340 --> 00:33:41,000
points of the finite
element mesh.

557
00:33:41,000 --> 00:33:46,972
However, locally, equilibrium is
only satisfied if the mesh

558
00:33:46,972 --> 00:33:48,330
is fine enough.

559
00:33:48,330 --> 00:33:52,860
And an indication to see whether
a mesh is fine enough

560
00:33:52,860 --> 00:33:55,750
is given, for example, by
whether these stressed

561
00:33:55,750 --> 00:33:59,520
boundary conditions are well
satisfied, and whether there

562
00:33:59,520 --> 00:34:02,760
are no stress jumps
between elements.

563
00:34:02,760 --> 00:34:06,520
I would like now to share with
you experiences relating to

564
00:34:06,520 --> 00:34:07,550
two problems.

565
00:34:07,550 --> 00:34:11,080
These are actually linear
elastic analysis problems that

566
00:34:11,080 --> 00:34:16,360
exemplify how we are looking
at stress jumps in order to

567
00:34:16,360 --> 00:34:21,100
choose an appropriate
finite element mesh.

568
00:34:21,100 --> 00:34:26,520
The first problem is
the analysis of a

569
00:34:26,520 --> 00:34:29,550
plate with a whole.

570
00:34:29,550 --> 00:34:33,130
The plate has certain geometric
dimensions, of

571
00:34:33,130 --> 00:34:35,980
course, and it has also
material data.

572
00:34:35,980 --> 00:34:38,710
The material data are
given right there.

573
00:34:38,710 --> 00:34:41,790
Notice that we also give
material data for an

574
00:34:41,790 --> 00:34:43,750
elastoplastic analysis.

575
00:34:43,750 --> 00:34:46,710
We will not perform now the
elastoplastic analysis.

576
00:34:46,710 --> 00:34:48,719
We will consider it later.

577
00:34:48,719 --> 00:34:52,560
For the moment, we just need
these two material data.

578
00:34:52,560 --> 00:34:56,150
The| geometric data for the
plate are given here.

579
00:34:56,150 --> 00:34:59,540
Now, this plate with
a hole has a

580
00:34:59,540 --> 00:35:01,880
doubly symmetric structure.

581
00:35:01,880 --> 00:35:04,680
In other words, there are two
symmetry lines, as shown here.

582
00:35:04,680 --> 00:35:08,590
So we only need to look at 1/4
of the plate in the analysis.

583
00:35:08,590 --> 00:35:11,780
Notice that the plate is
subjected to a pulling force

584
00:35:11,780 --> 00:35:16,360
up here and similarly
down here.

585
00:35:16,360 --> 00:35:19,480
The purpose of the analysis is
to accurately determine the

586
00:35:19,480 --> 00:35:24,390
stresses in the plate, and of
course, once again, assuming

587
00:35:24,390 --> 00:35:27,180
that the loads are small enough
to warrant a linear

588
00:35:27,180 --> 00:35:29,350
elastic analysis.

589
00:35:29,350 --> 00:35:33,140
The considerations that are
valid in linear elastic

590
00:35:33,140 --> 00:35:36,510
analysis regarding the choice
of a mesh are of course also

591
00:35:36,510 --> 00:35:40,490
valid in a non-linear analysis,
and we more easily

592
00:35:40,490 --> 00:35:43,400
can discuss them now, just
using a linear elastic

593
00:35:43,400 --> 00:35:47,100
analysis as an example.

594
00:35:47,100 --> 00:35:50,760
Using symmetry, as I pointed out
already, we can just focus

595
00:35:50,760 --> 00:35:53,640
our attention on 1/4
of the plate.

596
00:35:53,640 --> 00:35:55,590
Here is that quarter
of the plate.

597
00:35:55,590 --> 00:35:57,190
Here is the hole.

598
00:35:57,190 --> 00:36:00,040
Notice we have roller boundary
conditions here, roller

599
00:36:00,040 --> 00:36:03,233
boundary conditions there, and
the load is applied up there.

600
00:36:06,480 --> 00:36:11,410
Accuracy considerations in
displacement-based finite

601
00:36:11,410 --> 00:36:14,900
element analysis are that
first of all, of course,

602
00:36:14,900 --> 00:36:16,840
compatibility must
be satisfied.

603
00:36:16,840 --> 00:36:19,520
And that is achieved by
using a compatible

604
00:36:19,520 --> 00:36:21,220
finite element mesh.

605
00:36:21,220 --> 00:36:24,640
The material law must also
be satisfied, and that is

606
00:36:24,640 --> 00:36:28,910
achieved by just using the
appropriate material data.

607
00:36:28,910 --> 00:36:32,760
Of course, in non-linear
analysis, this item needs a

608
00:36:32,760 --> 00:36:36,030
lot more attention, but we are
performing now a linear

609
00:36:36,030 --> 00:36:40,510
elastic analysis, so this is a
very simple condition, and

610
00:36:40,510 --> 00:36:44,100
it's satisfied by simply
selecting the proper Young's

611
00:36:44,100 --> 00:36:47,370
modulus, Poisson's ratio
for the model.

612
00:36:47,370 --> 00:36:52,280
Equilibrium, once again, is
locally only satisfied if we

613
00:36:52,280 --> 00:36:54,060
choose a fine enough mesh.

614
00:36:54,060 --> 00:36:57,850
Otherwise we only approximate
equilibrium locally.

615
00:36:57,850 --> 00:37:01,110
And as I pointed out earlier, we
can observe the equilibrium

616
00:37:01,110 --> 00:37:04,840
error by plotting stress
discontinuities.

617
00:37:04,840 --> 00:37:12,200
Here we have the results
of a two-element

618
00:37:12,200 --> 00:37:13,960
mesh for this problem.

619
00:37:13,960 --> 00:37:18,330
Notice this quarter of the
plate, idealized using just

620
00:37:18,330 --> 00:37:20,330
two 8-node elements.

621
00:37:20,330 --> 00:37:24,570
This is the line z equal to 0,
and we will later on look at

622
00:37:24,570 --> 00:37:26,480
stresses along this line.

623
00:37:26,480 --> 00:37:29,260
This is a line y equal to z.

624
00:37:29,260 --> 00:37:31,980
We will also look at stresses
along this line.

625
00:37:31,980 --> 00:37:35,750
But we will plot the stresses
along this line here as a

626
00:37:35,750 --> 00:37:38,490
function of y.

627
00:37:38,490 --> 00:37:40,290
You should keep that in mind.

628
00:37:40,290 --> 00:37:42,470
In other words, the first stress
quality that we will

629
00:37:42,470 --> 00:37:46,690
see will actually start
somewhere here, because we are

630
00:37:46,690 --> 00:37:51,130
plotting this stress
as a function of y.

631
00:37:51,130 --> 00:37:55,810
The displacements, somewhat
amplified after load

632
00:37:55,810 --> 00:37:58,910
application, are shown here.

633
00:37:58,910 --> 00:38:02,140
Notice the maximum displacement
here, uz, is

634
00:38:02,140 --> 00:38:05,980
0.0285 millimeters.

635
00:38:05,980 --> 00:38:11,010
The maximum stress is at this
point here, 281 MPa,

636
00:38:11,010 --> 00:38:12,290
megapascal.

637
00:38:12,290 --> 00:38:18,090
Well, if we plot the stresses
along the line z equal to 0,

638
00:38:18,090 --> 00:38:20,640
and we're evaluating the
stresses, I should also say,

639
00:38:20,640 --> 00:38:24,830
at the nodal points of the
elements, then we obtain these

640
00:38:24,830 --> 00:38:26,310
values here.

641
00:38:26,310 --> 00:38:29,740
One value there, one value
there, one value there.

642
00:38:29,740 --> 00:38:36,220
And once again, we are plotting
the stresses along

643
00:38:36,220 --> 00:38:41,920
the y-axis, measured along here,
distance, and the stress

644
00:38:41,920 --> 00:38:45,210
magnitude is measured
along here.

645
00:38:45,210 --> 00:38:46,720
Tau zz.

646
00:38:46,720 --> 00:38:48,180
These are nodal point
stresses.

647
00:38:48,180 --> 00:38:51,060
And we simply have drawn a
smooth curve through these

648
00:38:51,060 --> 00:38:52,390
nodal point stresses.

649
00:38:52,390 --> 00:38:56,310
The applied stress, far away
from the hole, is 100 MPa.

650
00:38:59,590 --> 00:39:04,660
If we plot the stresses along
the line y equals z, along

651
00:39:04,660 --> 00:39:11,910
this line here, we notice that
we have a choice of choosing

652
00:39:11,910 --> 00:39:13,850
either the nodal point stresses
from the element

653
00:39:13,850 --> 00:39:17,460
below it or from the
element above it.

654
00:39:17,460 --> 00:39:20,940
Well, the nodal point stresses
from the element below it are

655
00:39:20,940 --> 00:39:24,600
given by the crosses here.

656
00:39:24,600 --> 00:39:28,770
The nodal point stresses from
the element above it--

657
00:39:28,770 --> 00:39:31,820
and notice, these are the nodal
point stresses at these

658
00:39:31,820 --> 00:39:35,700
nodes here are given
by this point.

659
00:39:35,700 --> 00:39:38,420
I should say, actually, that
we are plotting here the

660
00:39:38,420 --> 00:39:41,040
principal stress, the maximum
principal stress.

661
00:39:41,040 --> 00:39:42,550
We only look at one stress.

662
00:39:42,550 --> 00:39:45,840
Of course, you could plot such
curves for all three stress

663
00:39:45,840 --> 00:39:49,550
components, but here we're
looking just at sigma 1 being

664
00:39:49,550 --> 00:39:52,560
the maximum principal stress.

665
00:39:52,560 --> 00:39:54,820
Now there we can see the
stress discontinuity.

666
00:39:54,820 --> 00:39:59,110
We see a stress discontinuity
right at that point.

667
00:39:59,110 --> 00:40:03,570
Let's see once, what does this
stress continuity result from?

668
00:40:03,570 --> 00:40:06,480
Notice that there
is a node here--

669
00:40:06,480 --> 00:40:09,920
and I just like to put
it in one color--

670
00:40:09,920 --> 00:40:12,900
there's a node here coming
from this element.

671
00:40:12,900 --> 00:40:17,890
And there's another node coming
in from that element.

672
00:40:17,890 --> 00:40:22,680
Of course, this red point
corresponds to this mark here.

673
00:40:22,680 --> 00:40:28,190
So this node here gives us a
stress in the stress graph

674
00:40:28,190 --> 00:40:30,180
given by that symbol.

675
00:40:30,180 --> 00:40:37,880
This blue point there is a
node corresponding to the

676
00:40:37,880 --> 00:40:40,600
element with the cross, and
there we get also a stress

677
00:40:40,600 --> 00:40:44,420
value, and that stress value
corresponds to the cross mark

678
00:40:44,420 --> 00:40:45,020
on the graph.

679
00:40:45,020 --> 00:40:47,340
So let's go back to the
graph once more.

680
00:40:47,340 --> 00:40:52,120
Once again, same location,
because this blue node lies

681
00:40:52,120 --> 00:40:55,760
really on top of the red node.

682
00:40:55,760 --> 00:40:58,930
They are put together.

683
00:40:58,930 --> 00:41:02,120
And yet although we're looking
at the same material particle,

684
00:41:02,120 --> 00:41:04,700
we have this stress
discontinuity.

685
00:41:04,700 --> 00:41:07,720
And this stress discontinuity
tells us that the mesh is

686
00:41:07,720 --> 00:41:09,420
really not fine enough.

687
00:41:09,420 --> 00:41:14,010
So we go to a finer mesh.

688
00:41:14,010 --> 00:41:18,330
And here now we have selected
a 64-element mesh.

689
00:41:18,330 --> 00:41:22,520
Once again, 8-node elements,
as shown here, for that

690
00:41:22,520 --> 00:41:25,180
quarter of the plate
with the hole.

691
00:41:25,180 --> 00:41:27,250
Here is the typical
8-node element.

692
00:41:27,250 --> 00:41:31,280
This is the undeformed mesh, and
when we apply the loads,

693
00:41:31,280 --> 00:41:34,660
we obtain this deformed shape.

694
00:41:34,660 --> 00:41:41,850
The maximum stress is now 345
MPa, and the displacement

695
00:41:41,850 --> 00:41:48,770
right at that point here
is 0.0296 millimeter.

696
00:41:48,770 --> 00:41:51,170
Well, we can now once
again plot stresses.

697
00:41:51,170 --> 00:41:54,240
We can plot stresses
along this line--

698
00:41:54,240 --> 00:42:00,240
or I should rather show it here,
along this line, and

699
00:42:00,240 --> 00:42:04,610
along that line, same way
as we have done before.

700
00:42:04,610 --> 00:42:08,790
And let's look at the stresses
along the line z equal to 0.

701
00:42:11,370 --> 00:42:15,830
Of course, y is the coordinate
along which we are plotting

702
00:42:15,830 --> 00:42:20,340
the stresses, and you can see
now here a stress jump, a

703
00:42:20,340 --> 00:42:21,950
stress discontinuity.

704
00:42:21,950 --> 00:42:27,760
This is where two elements
adjoin each other at one node.

705
00:42:27,760 --> 00:42:30,090
Otherwise, we see very little
stress discontinuity.

706
00:42:30,090 --> 00:42:34,300
In fact, too small to
be visible by eye.

707
00:42:34,300 --> 00:42:38,760
If we plot stresses along
the line y equal to z--

708
00:42:38,760 --> 00:42:40,930
and here I should be
more specific.

709
00:42:40,930 --> 00:42:43,570
In fact, we only plot,
once again, the

710
00:42:43,570 --> 00:42:46,390
maximum principal stress.

711
00:42:46,390 --> 00:42:50,750
We find that we get
these curves here.

712
00:42:50,750 --> 00:42:54,880
Now notice that at that
location, we get four distinct

713
00:42:54,880 --> 00:42:57,260
stress values.

714
00:42:57,260 --> 00:43:01,940
The reason being that at this
node, four elements couple

715
00:43:01,940 --> 00:43:05,980
into that node, and that
gives us four stress

716
00:43:05,980 --> 00:43:09,140
values at that node.

717
00:43:09,140 --> 00:43:12,730
If you go further away, we find
that there are very few

718
00:43:12,730 --> 00:43:14,535
or very small stress
discontinuities.

719
00:43:17,110 --> 00:43:21,170
So however, since we have still
one stress discontinuity

720
00:43:21,170 --> 00:43:27,890
there at that one location that
I pointed out, we went

721
00:43:27,890 --> 00:43:33,400
ahead and analyzed an even finer
mesh, 288-element mesh.

722
00:43:33,400 --> 00:43:38,620
This mesh is shown now here,
8-node elements, once again,

723
00:43:38,620 --> 00:43:43,030
in the undeformed shape, and
here in the deformed shape.

724
00:43:43,030 --> 00:43:47,700
Notice that the maximum stress
is now 337 megapascals and the

725
00:43:47,700 --> 00:43:53,530
maximum displacement here
is 0.0296 millimeters.

726
00:43:53,530 --> 00:43:59,040
We can again plot stresses along
this line and along that

727
00:43:59,040 --> 00:44:02,470
diagonal line as a
function of y.

728
00:44:02,470 --> 00:44:08,900
And if we do so, we find that
along the line z equal to 0,

729
00:44:08,900 --> 00:44:13,600
we find for tau zz virtually no
more stress discontinuity.

730
00:44:13,600 --> 00:44:17,080
A nice, smooth stress curve.

731
00:44:17,080 --> 00:44:21,660
And along the line y equal to
z, for the maximum principal

732
00:44:21,660 --> 00:44:25,240
stress, also almost
no discontinuity.

733
00:44:29,140 --> 00:44:33,560
You can see here a slight
discontinuity, but that is

734
00:44:33,560 --> 00:44:37,320
really a very small one that
in engineering analysis, we

735
00:44:37,320 --> 00:44:40,410
generally can live with.

736
00:44:40,410 --> 00:44:44,900
Of course, these stress jump
plots are very useful and very

737
00:44:44,900 --> 00:44:47,410
helpful in identifying
whether a mesh is

738
00:44:47,410 --> 00:44:49,480
actually a good mesh.

739
00:44:49,480 --> 00:44:53,180
However, we should realize
that in order to identify

740
00:44:53,180 --> 00:44:57,580
whether a mesh is really good,
we would have to plot stress

741
00:44:57,580 --> 00:45:02,820
jumps along many lines in the
mesh, and that can, of course,

742
00:45:02,820 --> 00:45:05,490
provide quite some cost.

743
00:45:05,490 --> 00:45:09,680
It also means that we have to
interpret a lot of data.

744
00:45:09,680 --> 00:45:14,570
And one perhaps more convenient
way to look at

745
00:45:14,570 --> 00:45:19,400
whether a mesh is a good mesh
is to plot pressure bands.

746
00:45:19,400 --> 00:45:24,700
We have been lately using this
procedure to identify whether

747
00:45:24,700 --> 00:45:29,610
meshes are good, and I'd like
to just share that latest

748
00:45:29,610 --> 00:45:31,140
experience with you.

749
00:45:31,140 --> 00:45:34,650
You plot bands of constant
pressure, where the pressure

750
00:45:34,650 --> 00:45:39,960
is defined by taking the mean
of the stresses tau xx, tau

751
00:45:39,960 --> 00:45:41,500
yy, and tau zz.

752
00:45:41,500 --> 00:45:45,920
Of course, pressure meaning a
negative sign in front here.

753
00:45:45,920 --> 00:45:51,170
The two-element mesh gives us
this pressure band plot.

754
00:45:51,170 --> 00:45:55,470
Notice that these are bands
of 5 MPa difference.

755
00:45:55,470 --> 00:46:00,130
In other words, here you have a
jump from this point to that

756
00:46:00,130 --> 00:46:03,870
point of 5 MPa, and
once again, from

757
00:46:03,870 --> 00:46:05,390
here to there again.

758
00:46:05,390 --> 00:46:07,610
5 MPa.

759
00:46:07,610 --> 00:46:10,860
These are the pressure
band plots for

760
00:46:10,860 --> 00:46:12,750
the two-element mesh.

761
00:46:12,750 --> 00:46:16,060
Here we have it as the same
kind of plot for the

762
00:46:16,060 --> 00:46:17,910
64-element mesh.

763
00:46:17,910 --> 00:46:20,710
Let's look at this pressure
band plot a

764
00:46:20,710 --> 00:46:22,970
little bit more carefully.

765
00:46:22,970 --> 00:46:27,645
Here we have, as you can see,
again, differences of 5 MPa.

766
00:46:30,480 --> 00:46:36,330
This is a non-dark area,
unshaded area,

767
00:46:36,330 --> 00:46:38,260
provides a band of 5 MPa.

768
00:46:38,260 --> 00:46:41,820
That shaded area provides
a band of 5 MPa.

769
00:46:41,820 --> 00:46:45,370
Now, if we have a good mesh,
then the pressure band should

770
00:46:45,370 --> 00:46:47,500
be continuous.

771
00:46:47,500 --> 00:46:51,470
In other words, this band here
should be continuous, and

772
00:46:51,470 --> 00:46:57,390
there should be no pressure
band break.

773
00:46:57,390 --> 00:46:59,480
Let's look at this one here.

774
00:46:59,480 --> 00:47:02,500
It's nice and continuous,
however, there's a break.

775
00:47:02,500 --> 00:47:06,980
And so there is no continuity
in stresses here, which is

776
00:47:06,980 --> 00:47:11,220
quite simply displayed
by the pressure band

777
00:47:11,220 --> 00:47:12,930
picture shown here.

778
00:47:12,930 --> 00:47:17,850
For a continuous stress field,
we would have no breaks in the

779
00:47:17,850 --> 00:47:19,230
pressure bands.

780
00:47:19,230 --> 00:47:21,310
We have selected here 5 MPa.

781
00:47:21,310 --> 00:47:25,230
Of course, you could also select
1 MPa, and that would

782
00:47:25,230 --> 00:47:27,280
be then a much tighter
criteria.

783
00:47:27,280 --> 00:47:30,410
You could also select 10 MPa.

784
00:47:30,410 --> 00:47:34,810
Of course, then that would be a
very loose criterion, and 5

785
00:47:34,810 --> 00:47:37,310
MPa for this particular problem,
since you can see,

786
00:47:37,310 --> 00:47:40,030
we're getting quite a number
of bands along here, is a

787
00:47:40,030 --> 00:47:42,730
reasonable way to proceed.

788
00:47:42,730 --> 00:47:46,560
If you go to the 288-element
mesh, the pressure band plots

789
00:47:46,560 --> 00:47:47,670
look like this.

790
00:47:47,670 --> 00:47:49,345
Beautifully smooth.

791
00:47:49,345 --> 00:47:52,150
You can see here the smoothness
in the pressure

792
00:47:52,150 --> 00:47:56,910
bands, no discontinuity
at all.

793
00:47:56,910 --> 00:48:02,670
And this shows really that this
mesh certainly satisfies

794
00:48:02,670 --> 00:48:06,920
the criterion of smooth
pressure bands, smooth

795
00:48:06,920 --> 00:48:09,510
pressure variations, over
the whole mesh.

796
00:48:09,510 --> 00:48:12,660
And that is a strong indication
that the stresses

797
00:48:12,660 --> 00:48:16,360
are smooth, and therefore that
the mesh is really an

798
00:48:16,360 --> 00:48:17,610
acceptable mesh.

799
00:48:19,980 --> 00:48:22,620
We see, therefore, that stress
discontinuities are

800
00:48:22,620 --> 00:48:25,400
represented by breaks in
the pressure bands.

801
00:48:25,400 --> 00:48:28,380
As the mesh is refined, of
course, these breaks in the

802
00:48:28,380 --> 00:48:32,220
pressure bands disappear, or
certainly become smoother

803
00:48:32,220 --> 00:48:34,450
until finally they disappear.

804
00:48:34,450 --> 00:48:38,040
The stress state everywhere in
the mesh is represented by one

805
00:48:38,040 --> 00:48:40,220
picture, and that is, of course,
the beauty of this

806
00:48:40,220 --> 00:48:41,530
whole procedure.

807
00:48:41,530 --> 00:48:47,100
If you plot stress jumps, stress
component jumps, then

808
00:48:47,100 --> 00:48:54,490
you have to plot for each
component along all the lines

809
00:48:54,490 --> 00:48:57,010
where elements join
each other, these

810
00:48:57,010 --> 00:48:58,120
components of stresses.

811
00:48:58,120 --> 00:49:00,980
And there is a voluminous amount
of information that is

812
00:49:00,980 --> 00:49:02,000
being generated.

813
00:49:02,000 --> 00:49:07,720
Here we have just one picture
that basically gives us a lot

814
00:49:07,720 --> 00:49:10,220
of information about
the smoothness

815
00:49:10,220 --> 00:49:12,400
of the stress situation.

816
00:49:12,400 --> 00:49:15,170
Of course, the pressure band is
plotted, maybe plotted by a

817
00:49:15,170 --> 00:49:17,610
computer program.

818
00:49:17,610 --> 00:49:21,470
Actual magnitudes of pressure
are not given, but we are not

819
00:49:21,470 --> 00:49:25,040
really interested in seeing
those from these pictures.

820
00:49:25,040 --> 00:49:28,090
What we would like to see are
simply whether the pressure

821
00:49:28,090 --> 00:49:30,070
bands are smooth, and whether,
therefore, the mesh is an

822
00:49:30,070 --> 00:49:32,040
acceptable mesh.

823
00:49:32,040 --> 00:49:34,570
Let's summarize the results.

824
00:49:34,570 --> 00:49:41,190
Here we now performed three
analyses of this plate.

825
00:49:41,190 --> 00:49:45,410
One two-element mesh was used,
a 64-element mesh was used,

826
00:49:45,410 --> 00:49:48,200
and a 288-element
mesh was used.

827
00:49:48,200 --> 00:49:51,330
The number of degrees of
freedom are given here.

828
00:49:51,330 --> 00:49:54,400
Notice that the number of
degrees of freedom goes up

829
00:49:54,400 --> 00:49:57,930
very rapidly as the number
of elements goes up.

830
00:49:57,930 --> 00:50:01,380
The relative cost is given in
this column, and you can see

831
00:50:01,380 --> 00:50:05,350
here that we have normalized
the cost to the cost of the

832
00:50:05,350 --> 00:50:10,220
analysis using the
64-element mesh.

833
00:50:10,220 --> 00:50:14,330
So the two-element mesh is very
cheap, 8% of the cost,

834
00:50:14,330 --> 00:50:18,910
and the 288-element mesh
solution is quite expensive.

835
00:50:18,910 --> 00:50:24,060
The displacement at the top of
the plate is given in this

836
00:50:24,060 --> 00:50:28,530
column, and we notice that with
a 64-element mesh, we

837
00:50:28,530 --> 00:50:31,590
really obtain a very good
displacement prediction.

838
00:50:31,590 --> 00:50:33,810
In fact, there is no
change going to a

839
00:50:33,810 --> 00:50:36,210
larger number of elements.

840
00:50:36,210 --> 00:50:40,970
And the stress concentration
factor, interestingly enough,

841
00:50:40,970 --> 00:50:45,270
increases first and then
slightly decreases.

842
00:50:45,270 --> 00:50:47,620
Notice this is not an
infinite plate--

843
00:50:47,620 --> 00:50:52,150
it's a finite plate, and
therefore you do not get the

844
00:50:52,150 --> 00:50:54,860
value of three, for example,
that you would see in an

845
00:50:54,860 --> 00:51:01,500
infinite plate, for a hole
in an infinite plate.

846
00:51:01,500 --> 00:51:09,390
So the two-element mesh cannot
be used for stress predictions

847
00:51:09,390 --> 00:51:11,020
is certainly a conclusion.

848
00:51:11,020 --> 00:51:12,520
It's not accurate enough.

849
00:51:12,520 --> 00:51:18,020
The 64-element mesh gives
reasonable results for

850
00:51:18,020 --> 00:51:21,210
stresses, and certainly very
accurate results for the

851
00:51:21,210 --> 00:51:22,620
displacements.

852
00:51:22,620 --> 00:51:27,270
The 288-element mesh is probably
overrefined for

853
00:51:27,270 --> 00:51:29,000
linear elastic stress
analysis.

854
00:51:29,000 --> 00:51:34,640
However, for other types of
analysis, for example, an

855
00:51:34,640 --> 00:51:37,560
elastoplastic analysis, if you
want to have very accurate

856
00:51:37,560 --> 00:51:40,120
results, we might want
to use, actually,

857
00:51:40,120 --> 00:51:42,570
this fineness of mesh.

858
00:51:42,570 --> 00:51:45,430
One question that one, of course
might have is, why did

859
00:51:45,430 --> 00:51:47,860
he not use 9-node elements?

860
00:51:47,860 --> 00:51:49,880
We used 8-node elements.

861
00:51:49,880 --> 00:51:52,660
Here we show now a mesh
of 9-node elements.

862
00:51:52,660 --> 00:51:56,520
Notice each of these elements
now is a 9-node element.

863
00:51:56,520 --> 00:51:59,690
Would the solution significantly
improve if we

864
00:51:59,690 --> 00:52:02,310
had used 9-node elements?

865
00:52:02,310 --> 00:52:06,070
In other words, one 9-node
element for each 8-node

866
00:52:06,070 --> 00:52:08,420
element that was used before?

867
00:52:08,420 --> 00:52:10,560
Well, the answer is no.

868
00:52:10,560 --> 00:52:13,310
The solution does not
improve very much.

869
00:52:13,310 --> 00:52:17,270
If you look at this column here,
let's look at the 64

870
00:52:17,270 --> 00:52:23,300
8-node elements solution, in
which case we had 416 degrees

871
00:52:23,300 --> 00:52:24,670
of freedom.

872
00:52:24,670 --> 00:52:29,500
With a 9-node element, we will
have 544 degrees of freedom.

873
00:52:29,500 --> 00:52:33,420
The displacements at the top
are unchanged between the

874
00:52:33,420 --> 00:52:37,110
8-node and 9-node
element results.

875
00:52:37,110 --> 00:52:40,000
The stress concentration
factor is very

876
00:52:40,000 --> 00:52:43,290
little effected here.

877
00:52:43,290 --> 00:52:46,750
Only in the fourth digit,
as a matter of fact.

878
00:52:46,750 --> 00:52:49,320
And the stress jump and pressure
band plot do not

879
00:52:49,320 --> 00:52:50,872
change significantly, either.

880
00:52:53,590 --> 00:52:56,710
Well, the second example that
I'd then like to discuss with

881
00:52:56,710 --> 00:53:01,740
you is the analysis of
a plate with a crack.

882
00:53:01,740 --> 00:53:06,030
We will once again look at
different meshes for this

883
00:53:06,030 --> 00:53:10,890
problem, and we will
also discuss

884
00:53:10,890 --> 00:53:12,230
one additional aspect--

885
00:53:12,230 --> 00:53:16,830
namely that for certain types of
answers that you're looking

886
00:53:16,830 --> 00:53:21,520
for, you actually can use
sometimes very coarse meshes.

887
00:53:21,520 --> 00:53:25,480
Whereas if you're looking for
the answer of a much more

888
00:53:25,480 --> 00:53:29,440
difficult question, of course
you will have to use a much

889
00:53:29,440 --> 00:53:30,580
finer mesh.

890
00:53:30,580 --> 00:53:33,370
Here we have a plate
with a crack.

891
00:53:33,370 --> 00:53:38,330
Notice the plate is subjected
to a pull of 100 MPa right,

892
00:53:38,330 --> 00:53:41,130
and at the left-hand
side, same pull.

893
00:53:41,130 --> 00:53:44,270
The crack is lying
right there.

894
00:53:44,270 --> 00:53:47,370
Notice it's not a a
symmetric crack.

895
00:53:47,370 --> 00:53:54,960
This is a crack tip A, that's
the crack tip B. To model this

896
00:53:54,960 --> 00:53:57,980
plate, we have to really
model half of it.

897
00:53:57,980 --> 00:54:00,890
We have to model half of it,
because the crack is not a

898
00:54:00,890 --> 00:54:02,620
symmetric crack.

899
00:54:02,620 --> 00:54:05,370
The material data for this
problem are given down here.

900
00:54:05,370 --> 00:54:11,060
E, Young's modulus, nu,
Poisson's ratio, Kc is the

901
00:54:11,060 --> 00:54:13,410
fracture toughness.

902
00:54:13,410 --> 00:54:18,880
And here we have the fact that
the thickness of the plate

903
00:54:18,880 --> 00:54:20,920
shall be just 1 centimeter.

904
00:54:20,920 --> 00:54:23,280
It's a plane stress condition
that we're looking at.

905
00:54:23,280 --> 00:54:26,440
And the question that we want
to address here is, will the

906
00:54:26,440 --> 00:54:27,690
crack propagate?

907
00:54:29,870 --> 00:54:34,040
Well, let us look at some
background information.

908
00:54:34,040 --> 00:54:38,240
Since we want to perform a
linear elastic analysis and we

909
00:54:38,240 --> 00:54:44,520
want to apply linear elastic
fracture mechanic theory when

910
00:54:44,520 --> 00:54:50,720
considering whether the crack
will propagate in the plate,

911
00:54:50,720 --> 00:54:54,610
we calculate a stress intensity
factor, and we have

912
00:54:54,610 --> 00:55:00,130
a mode 1 stress condition,
so we calculate K1.

913
00:55:00,130 --> 00:55:03,760
K1 determines the strength of
the 1 over square root r

914
00:55:03,760 --> 00:55:06,720
stress singularity at the crack
tip, r of course being

915
00:55:06,720 --> 00:55:08,690
the radius from the crack tip.

916
00:55:08,690 --> 00:55:11,540
And as very well established
in linear elastic fracture

917
00:55:11,540 --> 00:55:18,070
mechanics, when K1 is
greater than Kc,

918
00:55:18,070 --> 00:55:19,870
the crack will propagate.

919
00:55:19,870 --> 00:55:22,910
Kc is a property of
the material.

920
00:55:22,910 --> 00:55:29,800
The computation of K1 is
performed by this here.

921
00:55:29,800 --> 00:55:36,560
It's equal to the square root
out of E times G. G is

922
00:55:36,560 --> 00:55:40,510
obtained as the partial
differentiation of pi with

923
00:55:40,510 --> 00:55:45,640
respect to a, where pi is a
total potential energy of the

924
00:55:45,640 --> 00:55:49,290
structure, and a is the area
of the crack surface.

925
00:55:49,290 --> 00:55:53,150
G is known as the energy release
rate for the crack.

926
00:55:53,150 --> 00:55:56,210
This is quite well established,
and what we want

927
00:55:56,210 --> 00:55:59,900
to do, of course, in the finite
element analysis, is to

928
00:55:59,900 --> 00:56:02,780
calculate this G value.

929
00:56:02,780 --> 00:56:05,180
The question is, how
do we do that?

930
00:56:05,180 --> 00:56:09,460
Well, in this finite element
analysis, each crack is

931
00:56:09,460 --> 00:56:12,320
represented by a node.

932
00:56:12,320 --> 00:56:17,140
And if you look at a cut through
the plate where this

933
00:56:17,140 --> 00:56:21,600
is here the crack, we have
this here as the node

934
00:56:21,600 --> 00:56:23,890
at the crack tip.

935
00:56:23,890 --> 00:56:27,410
This node in the finite element
analysis is, so to

936
00:56:27,410 --> 00:56:32,500
say, moved forward a
differential amount, so that

937
00:56:32,500 --> 00:56:36,750
the change in crack
area is given via

938
00:56:36,750 --> 00:56:39,060
this little bit here.

939
00:56:39,060 --> 00:56:42,310
Now that the original crack tip
location is here, the new

940
00:56:42,310 --> 00:56:45,260
crack tip location
is right there.

941
00:56:45,260 --> 00:56:51,480
Then we can, in this process,
evaluate del pi over del a as

942
00:56:51,480 --> 00:56:55,930
del pi del l times
1/t, where t is a

943
00:56:55,930 --> 00:56:59,020
thickness of that plate.

944
00:56:59,020 --> 00:57:02,230
Of course, this here is only
conceptually being done, that

945
00:57:02,230 --> 00:57:04,500
we're actually moving
the crack.

946
00:57:04,500 --> 00:57:07,880
What we're really doing is the
differentiation of del pi with

947
00:57:07,880 --> 00:57:12,910
respect to l, where l is the
length of the crack and del l

948
00:57:12,910 --> 00:57:22,050
is the differentiation of l
into this direction here.

949
00:57:22,050 --> 00:57:26,170
The quantities del pi del l may
be efficiently calculated

950
00:57:26,170 --> 00:57:32,310
by an algorithm that we have
just recently been publishing.

951
00:57:32,310 --> 00:57:36,320
Of course, there are different
algorithms available as well,

952
00:57:36,320 --> 00:57:40,000
but we would like to use this
procedure to calculate

953
00:57:40,000 --> 00:57:41,550
data pi over l.

954
00:57:41,550 --> 00:57:43,950
And you can look up how
it's being done in

955
00:57:43,950 --> 00:57:46,640
this particular paper.

956
00:57:46,640 --> 00:57:48,040
Let's look at the results.

957
00:57:48,040 --> 00:57:53,140
Finite element analyses have
been performed using a

958
00:57:53,140 --> 00:57:56,830
17-element mesh, and this
is the mesh here.

959
00:57:56,830 --> 00:57:59,410
We call this a coarse mesh.

960
00:57:59,410 --> 00:58:01,490
Notice the crack
is right there.

961
00:58:04,540 --> 00:58:09,840
Notice that we have only two
elements above the crack, to

962
00:58:09,840 --> 00:58:14,050
the boundary of the plate, and
we have here once again also

963
00:58:14,050 --> 00:58:16,850
only two elements below it.

964
00:58:16,850 --> 00:58:21,710
Notice that these nodes around
the crack tip B have been

965
00:58:21,710 --> 00:58:25,160
moved to the quarter point.

966
00:58:25,160 --> 00:58:28,020
And similarly, they've been
moved here to the quarter

967
00:58:28,020 --> 00:58:35,040
point, meaning that this node is
not at the midpoint of this

968
00:58:35,040 --> 00:58:37,445
element side, but at the
quarter point of

969
00:58:37,445 --> 00:58:38,500
that element side.

970
00:58:38,500 --> 00:58:43,220
And similarly for the other
nodes around the crack tips.

971
00:58:46,300 --> 00:58:50,510
If we look at the results, the
stress results, maybe plus the

972
00:58:50,510 --> 00:58:57,400
stresses on the line of
symmetry, we find that tau yy

973
00:58:57,400 --> 00:59:04,550
varies as shown here, then of
course increases very rapidly

974
00:59:04,550 --> 00:59:05,440
near the crack.

975
00:59:05,440 --> 00:59:06,690
This is where the crack is.

976
00:59:08,990 --> 00:59:15,680
Increases very rapidly, then
drops down to very close to 0.

977
00:59:15,680 --> 00:59:19,030
Of course, it should
be 0 right in here.

978
00:59:19,030 --> 00:59:24,320
And here we have, coming from
the other side to crack tip A,

979
00:59:24,320 --> 00:59:26,380
this very rapid variation
in stresses.

980
00:59:29,290 --> 00:59:31,490
These are the stresses tau yy.

981
00:59:31,490 --> 00:59:34,740
If you look at the pressure
band plot--

982
00:59:34,740 --> 00:59:37,400
I introduced you to those
kinds of plots

983
00:59:37,400 --> 00:59:39,200
in the earlier example--

984
00:59:39,200 --> 00:59:41,370
we find that the pressure
jumps are really

985
00:59:41,370 --> 00:59:43,640
larger than five MPa.

986
00:59:43,640 --> 00:59:46,300
For example, you can see it
right here, you can see it

987
00:59:46,300 --> 00:59:49,550
right there, you can
see it right there.

988
00:59:49,550 --> 00:59:52,830
So it's a coarse mesh, and
certainly the stresses are not

989
00:59:52,830 --> 00:59:55,420
very well represented,
not very well picked

990
00:59:55,420 --> 00:59:57,360
up with this mesh.

991
00:59:57,360 --> 01:00:01,910
However, if we look at the
stress intensity factors that

992
01:00:01,910 --> 01:00:04,430
we wanted to estimate in this
analysis, because they will

993
01:00:04,430 --> 01:00:09,020
tell us whether the crack will
propagate or not, we find that

994
01:00:09,020 --> 01:00:13,500
for the stress intensity
factors, for example, KA, we

995
01:00:13,500 --> 01:00:16,810
get an excellent solution.

996
01:00:16,810 --> 01:00:21,080
For KB, the solution is not
as good, but perhaps still

997
01:00:21,080 --> 01:00:22,330
acceptable.

998
01:00:25,810 --> 01:00:29,490
We now go to a finer mesh,
a much finer mesh.

999
01:00:29,490 --> 01:00:33,360
128 elements are used
in this mesh.

1000
01:00:33,360 --> 01:00:37,050
And we can see once again
here, 1/2 the plate.

1001
01:00:37,050 --> 01:00:42,080
Here is point A, crack tip A.
Here is point B, crack tip B.

1002
01:00:42,080 --> 01:00:46,190
Notice the crack lies
in between.

1003
01:00:46,190 --> 01:00:48,920
And we will be plotting stresses
along this line.

1004
01:00:51,440 --> 01:00:55,580
This is a line of symmetry here,
and we will plot tau yy,

1005
01:00:55,580 --> 01:01:02,710
which is the stress across
here, as a function of z.

1006
01:01:02,710 --> 01:01:05,570
But let's first look a little
bit closer at the mesh.

1007
01:01:05,570 --> 01:01:10,190
If we look at one detail closer
to the crack tip A, we

1008
01:01:10,190 --> 01:01:14,910
see this is the mesh that we
use, and we can go still

1009
01:01:14,910 --> 01:01:22,710
further and show the elements
just around the crack tip,

1010
01:01:22,710 --> 01:01:26,880
right A. Here is A. Notice
that we use triangular

1011
01:01:26,880 --> 01:01:30,160
elements, these are
isoparametric degenerate

1012
01:01:30,160 --> 01:01:34,250
elements, and that we have
moved, once again, the nodal

1013
01:01:34,250 --> 01:01:37,960
points from the midpoint
to the quarter point.

1014
01:01:37,960 --> 01:01:42,620
Because we know that if we do
so, we are generating in the

1015
01:01:42,620 --> 01:01:46,740
finite element the 1 over square
root r singularity that

1016
01:01:46,740 --> 01:01:50,610
we would like to have in the
analysis, because we know that

1017
01:01:50,610 --> 01:01:53,420
the stresses vary as
a function of 1

1018
01:01:53,420 --> 01:01:54,450
over square root r.

1019
01:01:54,450 --> 01:01:57,480
So it's of benefit to shift
the nodes to the quarter

1020
01:01:57,480 --> 01:02:01,870
points, as was done
in this example.

1021
01:02:01,870 --> 01:02:04,120
Now we look here at
the stresses.

1022
01:02:04,120 --> 01:02:08,370
Stresses tau yy, as I pointed
out, is a function of z.

1023
01:02:08,370 --> 01:02:16,440
And you can see here a stress
that very rapidly grows, then

1024
01:02:16,440 --> 01:02:20,980
on the other side comes down
to practically 0 in the

1025
01:02:20,980 --> 01:02:27,190
cracked area, and on the other
side, coming again now from

1026
01:02:27,190 --> 01:02:31,120
the right here, a stress that
varies very rapidly up close

1027
01:02:31,120 --> 01:02:32,750
to tip A.

1028
01:02:32,750 --> 01:02:35,070
We also can see here that
the stress variations

1029
01:02:35,070 --> 01:02:37,140
are nice and smooth.

1030
01:02:37,140 --> 01:02:39,350
Of course, that indicates to us
that the mesh is really a

1031
01:02:39,350 --> 01:02:41,650
good mesh, but we can
also look at our

1032
01:02:41,650 --> 01:02:43,440
pressure bands again.

1033
01:02:43,440 --> 01:02:51,610
We find that the pressure jumps
are quite small if we go

1034
01:02:51,610 --> 01:02:53,750
not too close to
the crack tip.

1035
01:02:53,750 --> 01:02:58,110
In other words, here we have
some pressure band variation,

1036
01:02:58,110 --> 01:03:04,200
some pressure band breaks, but
we have still quite smooth

1037
01:03:04,200 --> 01:03:08,840
bands that tell us that the
mesh is quite fine.

1038
01:03:08,840 --> 01:03:16,580
If we go very close to the crack
tip A, then we find that

1039
01:03:16,580 --> 01:03:21,330
the pressure jumps are, of
course, large right here.

1040
01:03:21,330 --> 01:03:24,310
You can see here breaks
in the pressure bend.

1041
01:03:24,310 --> 01:03:27,390
Right here, for example, a
typical example, if you look

1042
01:03:27,390 --> 01:03:30,890
very closely here, there's
a black band coming

1043
01:03:30,890 --> 01:03:32,150
into a white band.

1044
01:03:32,150 --> 01:03:36,540
So there's clearly a break
in the pressure band.

1045
01:03:36,540 --> 01:03:41,510
So here, then, in this area,
the stresses are not

1046
01:03:41,510 --> 01:03:45,730
continuous, and would also not
be very accurately predicted.

1047
01:03:45,730 --> 01:03:49,530
However, we know that our scheme
of calculating the

1048
01:03:49,530 --> 01:03:54,690
stress intensity factor works
for even a coarser mesh, and

1049
01:03:54,690 --> 01:03:58,640
for this fine mesh, we've got
indeed excellent results, as

1050
01:03:58,640 --> 01:04:04,930
you can see for KA and for KB.

1051
01:04:04,930 --> 01:04:08,020
Both are very well predicted.

1052
01:04:08,020 --> 01:04:11,780
The point of this analysis is
really that the degree off

1053
01:04:11,780 --> 01:04:16,290
refinement needed in a mesh
depends on what kind of

1054
01:04:16,290 --> 01:04:20,080
question you have, what kind of
problem you're solving, and

1055
01:04:20,080 --> 01:04:23,980
what kind of question
you're asking.

1056
01:04:23,980 --> 01:04:27,930
If you're only interested in
predicting displacements

1057
01:04:27,930 --> 01:04:30,510
accurately, a coarse
mesh may do.

1058
01:04:30,510 --> 01:04:34,085
If you want to calculate stress
intensity factors, also

1059
01:04:34,085 --> 01:04:36,500
a coarse mesh may do.

1060
01:04:36,500 --> 01:04:39,540
If you want to calculate lowest
natural frequencies,

1061
01:04:39,540 --> 01:04:43,220
associate mode shapes in a
dynamic analysis, once again,

1062
01:04:43,220 --> 01:04:45,620
a coarse mesh might
very well do.

1063
01:04:45,620 --> 01:04:47,860
However, if you want to
calculate stresses very

1064
01:04:47,860 --> 01:04:51,000
accurately, then you may
need a fine mesh.

1065
01:04:51,000 --> 01:04:54,390
And in general non-linear
analysis, we need accurate

1066
01:04:54,390 --> 01:04:58,975
stresses, because the stresses
influence the material

1067
01:04:58,975 --> 01:05:02,250
relationship that has to be
used as we go through the

1068
01:05:02,250 --> 01:05:03,650
incremental solution.

1069
01:05:03,650 --> 01:05:07,740
And therefore, frequently we
need, in non-linear analysis,

1070
01:05:07,740 --> 01:05:12,540
a finer mesh than what would do
well in a linear analysis.

1071
01:05:12,540 --> 01:05:19,040
So we should keep that in mind,
that in fact what might

1072
01:05:19,040 --> 01:05:22,800
be a sufficiently fine mesh in
linear elastic analysis may

1073
01:05:22,800 --> 01:05:28,160
not be sufficiently fine in
a non-linear elastoplastic

1074
01:05:28,160 --> 01:05:29,740
geometrically non-linear
analysis.

1075
01:05:29,740 --> 01:05:33,650
Of course, how fine the mesh
has to be in a general

1076
01:05:33,650 --> 01:05:37,770
non-linear analysis depends very
much on many criteria,

1077
01:05:37,770 --> 01:05:41,760
criteria that we will still
discuss in the next lectures

1078
01:05:41,760 --> 01:05:43,460
of this course.

1079
01:05:43,460 --> 01:05:47,530
But I hope this gave you a bit
of an overview of what are

1080
01:05:47,530 --> 01:05:51,220
some important considerations
in linear as well as

1081
01:05:51,220 --> 01:05:52,890
non-linear analysis.

1082
01:05:52,890 --> 01:05:54,140
Thank you for your attention.