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PROFESSOR: Ladies and gentlemen,
welcome to this

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00:00:23,850 --> 00:00:26,370
lecture on Nonlinear Finite
Element Analysis of Solids and

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00:00:26,370 --> 00:00:27,700
Structures.

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00:00:27,700 --> 00:00:29,770
In the previous lectures, we
discussed quite general

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00:00:29,770 --> 00:00:33,220
kinematic formulations and
numerical algorithms that we

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00:00:33,220 --> 00:00:36,210
use for nonlinear finite
element analysis.

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00:00:36,210 --> 00:00:40,610
We also referred to and used a
stress strain matrix C, but we

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00:00:40,610 --> 00:00:43,580
did not discuss this matrix
in any depths.

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00:00:43,580 --> 00:00:45,990
This is the topic of
this lecture and

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00:00:45,990 --> 00:00:48,420
the following lectures.

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00:00:48,420 --> 00:00:52,340
The stress strain matrix C is
used in the calculation of the

19
00:00:52,340 --> 00:00:56,180
K matrix, and is schematically
shown on this graph.

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00:00:56,180 --> 00:00:58,730
And of course, the stress strain
matrix is also used in

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00:00:58,730 --> 00:01:04,069
the evaluation of the force
vector F, as shown here.

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00:01:04,069 --> 00:01:06,510
The stress strain matrix, of
course, will be different for

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00:01:06,510 --> 00:01:08,180
different material behaviors.

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00:01:08,180 --> 00:01:09,980
It will be different for
different kinematic

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00:01:09,980 --> 00:01:11,500
formulations that we're using.

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00:01:11,500 --> 00:01:15,260
And all these aspects we now
have to discuss in this and

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00:01:15,260 --> 00:01:17,650
the following lectures.

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00:01:17,650 --> 00:01:23,680
We notice that when we do finite
element analysis, we

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00:01:23,680 --> 00:01:26,540
have the choice of different
kinds of formulations.

30
00:01:26,540 --> 00:01:30,160
And we talked about the
formulation that takes only

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00:01:30,160 --> 00:01:33,030
into account infinitesimally
small displacements.

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00:01:33,030 --> 00:01:35,920
But of course, material
nonlinear behavior, we

33
00:01:35,920 --> 00:01:38,870
actually call that formation
a materially nonlinear only

34
00:01:38,870 --> 00:01:41,770
formation, an MNO formulation.

35
00:01:41,770 --> 00:01:44,720
We may have or use a formulation
that takes into

36
00:01:44,720 --> 00:01:48,580
account large displacements,
large rotations, but only

37
00:01:48,580 --> 00:01:51,350
small strains.

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00:01:51,350 --> 00:01:55,320
This, for example, would be a
total Lagrangian formulation.

39
00:01:55,320 --> 00:02:00,290
And we have also formulations
that take into account large

40
00:02:00,290 --> 00:02:03,200
displacements, large locations,
and large strains.

41
00:02:03,200 --> 00:02:06,170
And here we use, once again,
the total Lagrangian

42
00:02:06,170 --> 00:02:09,750
formulation or an updated
Lagrangian formulation.

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00:02:09,750 --> 00:02:12,020
We discussed these formulations,
of course,

44
00:02:12,020 --> 00:02:14,610
earlier in the earlier
lectures.

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00:02:14,610 --> 00:02:18,350
The applicability of material
descriptions is also very well

46
00:02:18,350 --> 00:02:21,830
categorized in these
three categories.

47
00:02:21,830 --> 00:02:24,400
In other words, there are
certain material description

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00:02:24,400 --> 00:02:28,000
constitutive relations that hold
for infinitesimally small

49
00:02:28,000 --> 00:02:31,560
displacements and small strains,
and that are not

50
00:02:31,560 --> 00:02:35,020
directly applicable
to large strains.

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00:02:35,020 --> 00:02:37,430
If, however, we have a material
description that is

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00:02:37,430 --> 00:02:41,240
applicable to infinitesimally
small displacements and small

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00:02:41,240 --> 00:02:45,050
strains, then we will see that
same material description can

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00:02:45,050 --> 00:02:48,250
directly be used also in the
large displacement, large

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00:02:48,250 --> 00:02:52,520
rotation, but small strain
analysis, provided we use the

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00:02:52,520 --> 00:02:53,920
proper formulation.

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00:02:53,920 --> 00:02:56,970
We will discuss this
aspect just now.

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00:02:56,970 --> 00:03:00,740
In fact, the formulation that we
are using in this category

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00:03:00,740 --> 00:03:05,330
of problems is a total
Lagrangian formulation.

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00:03:05,330 --> 00:03:10,010
Let us very briefly once more
recall that we have the

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00:03:10,010 --> 00:03:11,790
materially-nonlinear-only
formulation, as I just

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00:03:11,790 --> 00:03:14,580
mentioned, the total Lagrangian
formulation and the

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00:03:14,580 --> 00:03:16,570
updated Lagrangian formulation,
which we

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00:03:16,570 --> 00:03:20,120
discussed in earlier lectures.

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00:03:20,120 --> 00:03:25,960
And that really kinematically,
these formulations here

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00:03:25,960 --> 00:03:29,120
kinematically, for the analysis
of two dimensional,

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00:03:29,120 --> 00:03:32,150
three dimensional solids, using
isoparametric finite

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00:03:32,150 --> 00:03:36,690
element formations, we include
all large displacement, large

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00:03:36,690 --> 00:03:39,380
rotation, and large
strain effects.

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00:03:39,380 --> 00:03:42,730
Of course, whether the
formulation will actually be

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00:03:42,730 --> 00:03:47,550
applicable to model a particular
problem, will also

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00:03:47,550 --> 00:03:51,680
depends on whether we use and
have the right constitutive

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00:03:51,680 --> 00:03:53,450
relation for that problem.

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00:03:53,450 --> 00:03:56,460
And that is an aspect that we
will concentrate on now in

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00:03:56,460 --> 00:03:58,480
this and the following
lectures.

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00:03:58,480 --> 00:04:02,530
But notice that kinematically
these formations include all

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00:04:02,530 --> 00:04:03,780
nonlinearities.

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00:04:05,920 --> 00:04:10,780
If we look very briefly at
various material descriptions,

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00:04:10,780 --> 00:04:15,340
we have here typical models that
I used and examples of

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00:04:15,340 --> 00:04:20,529
applications, materials,
practically all materials

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00:04:20,529 --> 00:04:24,370
behave elastically as long as
the stress and the strains are

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00:04:24,370 --> 00:04:26,348
small enough.

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00:04:26,348 --> 00:04:28,620
A hyperelastic material
would be, for

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00:04:28,620 --> 00:04:30,696
example, a rubber material.

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00:04:30,696 --> 00:04:33,540
A hypoelastic material
would be the

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00:04:33,540 --> 00:04:36,780
concrete material model.

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00:04:36,780 --> 00:04:40,870
And elasto-plastic material
would be applicable to model

88
00:04:40,870 --> 00:04:44,900
metals, soils, rocks under
high stresses.

89
00:04:44,900 --> 00:04:48,800
A creep material model would be
applicable to model metals

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00:04:48,800 --> 00:04:50,430
at high temperatures.

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00:04:50,430 --> 00:04:54,820
When the temperatures are
high, then metals are

92
00:04:54,820 --> 00:04:57,110
subjected very easily
to creep.

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00:04:57,110 --> 00:05:01,070
A viscoplastic material model
really takes into account the

94
00:05:01,070 --> 00:05:03,670
elasto-plasticity and the
creeping effect of the

95
00:05:03,670 --> 00:05:08,720
material and is used frequently
to idealize or to

96
00:05:08,720 --> 00:05:12,280
represent polymers and metals.

97
00:05:12,280 --> 00:05:16,140
We will talk about a number of
these material models in this

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00:05:16,140 --> 00:05:17,660
and the following lecture.

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00:05:17,660 --> 00:05:23,670
We will not go into depths of
concrete material models and

100
00:05:23,670 --> 00:05:27,130
rock material models, but still
whatever we are talking

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00:05:27,130 --> 00:05:32,770
about is applicable to a wide
range of material models.

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00:05:32,770 --> 00:05:36,130
Let us first look now, in this
lecture, at the elastic

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00:05:36,130 --> 00:05:39,120
material behavior.

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00:05:39,120 --> 00:05:44,250
Elasticity means that for a
particular strain, you are

105
00:05:44,250 --> 00:05:46,710
given a particular stress.

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00:05:46,710 --> 00:05:50,010
The history does not enter
in the solution.

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00:05:50,010 --> 00:05:51,770
And a linear elastic
stress strain

108
00:05:51,770 --> 00:05:54,380
relationship is shown here.

109
00:05:54,380 --> 00:06:00,450
Notice for the strain, tE, we
are given a stress t sigma.

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00:06:00,450 --> 00:06:02,320
And it's a unique stress.

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00:06:02,320 --> 00:06:03,840
It is a unique stress.

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00:06:03,840 --> 00:06:06,520
And of course, the history
of the formation does

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00:06:06,520 --> 00:06:08,920
not enter at all.

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00:06:08,920 --> 00:06:13,010
Notice that in this particular
case, we would have that

115
00:06:13,010 --> 00:06:20,126
typically t sigma here is
equal to E times te.

116
00:06:20,126 --> 00:06:24,340
te being of course this value
here, t sigma there.

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00:06:24,340 --> 00:06:27,430
Notice that an increment in
stress is given by d sigma

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00:06:27,430 --> 00:06:29,370
equal to E times de.

119
00:06:29,370 --> 00:06:33,260
Linear elasticity, a linear
elastic material description,

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00:06:33,260 --> 00:06:34,695
means E is constant.

121
00:06:38,200 --> 00:06:42,880
For nonlinear, as a nonlinear
elastic material description,

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00:06:42,880 --> 00:06:46,600
we might have schematically in
a one-dimensional analysis,

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00:06:46,600 --> 00:06:49,440
this relationship here,
the red curve.

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00:06:49,440 --> 00:06:54,560
Notice that, once again, for a
particular strain, we pick up

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00:06:54,560 --> 00:06:57,660
a unique value of stress.

126
00:06:57,660 --> 00:07:03,420
And this relationship here
shows that t sigma is

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00:07:03,420 --> 00:07:05,804
given tC times te.

128
00:07:05,804 --> 00:07:10,320
tC is the slope of
this line here.

129
00:07:10,320 --> 00:07:15,020
Now, an increment in stress
is obtained by this

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00:07:15,020 --> 00:07:16,510
relationship here.

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00:07:16,510 --> 00:07:18,680
d sigma equals C times de.

132
00:07:18,680 --> 00:07:23,220
Notice that C is not
the same as tC.

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00:07:23,220 --> 00:07:26,690
This is here, relating
the total stress

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00:07:26,690 --> 00:07:27,930
to the total strain.

135
00:07:27,930 --> 00:07:30,790
Here we talk about an increment
in stress related to

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00:07:30,790 --> 00:07:32,430
an increment in strain.

137
00:07:32,430 --> 00:07:36,820
It's important to note that
C is not the same as tC.

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00:07:36,820 --> 00:07:40,750
Unless, of course, we have a
linear elastic relationship.

139
00:07:43,370 --> 00:07:48,420
In a computer program, we would
represent such a curve

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00:07:48,420 --> 00:07:53,510
typically by a series of
straight lines, as shown here.

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00:07:53,510 --> 00:07:59,330
And the input, of course, would
be these points here,

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00:07:59,330 --> 00:08:03,460
and the corresponding points
on the stress curve.

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00:08:03,460 --> 00:08:07,670
Notice that if you take a large
number of such straight

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00:08:07,670 --> 00:08:11,010
segments, you can really
approximate a continuous curve

145
00:08:11,010 --> 00:08:14,620
such as this one here
quite well.

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00:08:14,620 --> 00:08:18,930
We can generalize this elastic
material behavior using the

147
00:08:18,930 --> 00:08:23,350
total Lagrangian formulation,
as shown here.

148
00:08:23,350 --> 00:08:27,320
Notice that the total stress,
second Piola-Kirchhoff

149
00:08:27,320 --> 00:08:29,380
stress-- we talked about
this stress measure

150
00:08:29,380 --> 00:08:31,580
in an earlier lecture--

151
00:08:31,580 --> 00:08:35,679
is given by the stress strain
law times the total strain.

152
00:08:35,679 --> 00:08:38,400
Green-Lagrange strain-- once
again we talked about this

153
00:08:38,400 --> 00:08:41,610
strain measure in an
earlier lecture.

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00:08:41,610 --> 00:08:44,130
Incrementally, we have
this relationship.

155
00:08:44,130 --> 00:08:46,510
The increment in the second
Piola-Kirchhoff stress is

156
00:08:46,510 --> 00:08:51,130
related to the increment in the
Green-Lagrange strain via

157
00:08:51,130 --> 00:08:52,890
this constitutive law.

158
00:08:52,890 --> 00:08:57,690
Notice that this tensor or the
components of this tensor are

159
00:08:57,690 --> 00:09:02,400
not the same as those components
in general.

160
00:09:02,400 --> 00:09:06,790
This material description is
frequently employed in two

161
00:09:06,790 --> 00:09:12,460
types of analysis, the usual
constant material moduli for

162
00:09:12,460 --> 00:09:15,570
infinitesimal displacement
analysis.

163
00:09:15,570 --> 00:09:20,120
Or for rubber-type analysis,
rubber-type materials, I

164
00:09:20,120 --> 00:09:23,890
should say, where of course this
matrix is not constant.

165
00:09:23,890 --> 00:09:27,090
And that matrix is not
constant either.

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00:09:27,090 --> 00:09:29,550
I like to now spend
time on discussing

167
00:09:29,550 --> 00:09:33,410
these type of analyses.

168
00:09:33,410 --> 00:09:36,980
The use of the constant material
moduli means, for

169
00:09:36,980 --> 00:09:41,500
example, for an isotropic
material, that we use these

170
00:09:41,500 --> 00:09:46,610
matrices given via this
relationship here.

171
00:09:46,610 --> 00:09:50,750
Notice lambda and mu are the
Lame constants, that you're

172
00:09:50,750 --> 00:09:54,780
probably familiar with, and data
ij is the Kronecker data,

173
00:09:54,780 --> 00:09:57,570
defined as shown down here.

174
00:09:57,570 --> 00:10:00,430
These are the same constants
that are used

175
00:10:00,430 --> 00:10:03,410
also in linear analysis.

176
00:10:03,410 --> 00:10:09,530
So no new parameters are
introduced here.

177
00:10:09,530 --> 00:10:15,100
As examples, here we have the
stress strain law for the 2-D

178
00:10:15,100 --> 00:10:18,190
plane stress analysis.

179
00:10:18,190 --> 00:10:22,880
I've written it in matrix form,
and this zero here means

180
00:10:22,880 --> 00:10:25,640
total Lagrangian formulation.

181
00:10:25,640 --> 00:10:29,550
And notice that there could
also be a t up here.

182
00:10:29,550 --> 00:10:35,150
In other words, we use the same
matrix for the increment

183
00:10:35,150 --> 00:10:39,440
in the stresses and for
the total stresses.

184
00:10:39,440 --> 00:10:45,590
So I could very easily put a t
up there, and we notice then

185
00:10:45,590 --> 00:10:51,640
that the last row corresponds
to this relationship here.

186
00:10:51,640 --> 00:10:54,770
It's important to recognize that
on the right hand side

187
00:10:54,770 --> 00:10:57,970
here, we have the Green-Lagrange
strain, t0 zero

188
00:10:57,970 --> 00:11:01,240
epsilon 12 plus t0 21.

189
00:11:01,240 --> 00:11:05,200
In other words, we take the
total sharing strain into

190
00:11:05,200 --> 00:11:06,510
account here.

191
00:11:06,510 --> 00:11:09,340
And that gives us then
this row and

192
00:11:09,340 --> 00:11:10,590
that column, of course.

193
00:11:13,070 --> 00:11:17,090
For 2-D, access symmetric
analysis, we would use this

194
00:11:17,090 --> 00:11:20,180
stress-strain law, once again,
for the increments in the

195
00:11:20,180 --> 00:11:22,670
stresses and for the
total stresses.

196
00:11:25,480 --> 00:11:29,850
For an orthotropic material, we
would typically have this

197
00:11:29,850 --> 00:11:34,316
stress strain law in plane
stress, plane strain analysis.

198
00:11:34,316 --> 00:11:37,630
And there you see
again, the zero.

199
00:11:37,630 --> 00:11:42,880
Notice Ea, Eb are the material
and nu ab are the material

200
00:11:42,880 --> 00:11:46,790
moduli corresponding to the
a and b, directions and

201
00:11:46,790 --> 00:11:51,020
similarly the sheer
modulis Gab.

202
00:11:51,020 --> 00:11:56,880
The material being orthotropic
means Ea is not equal to Eb.

203
00:11:56,880 --> 00:12:00,680
This once again is the material
law that we are quite

204
00:12:00,680 --> 00:12:04,610
familiar with from
linear analysis.

205
00:12:04,610 --> 00:12:08,240
Now let's look at the sample
problem, the sample analysis.

206
00:12:08,240 --> 00:12:11,950
And I would like to choose a
very simple problem to just

207
00:12:11,950 --> 00:12:16,050
demonstrate some of the aspects
of using a total

208
00:12:16,050 --> 00:12:19,980
Lagrangian formulation with
a constant material law.

209
00:12:19,980 --> 00:12:27,230
Here we have a bar, with cross
sectional area A bar.

210
00:12:27,230 --> 00:12:31,380
We use the material constants E
and nu, the ones that we are

211
00:12:31,380 --> 00:12:35,530
used to you in liner analysis.

212
00:12:35,530 --> 00:12:40,550
And we want to use this stress
strain relationship.

213
00:12:40,550 --> 00:12:45,650
We assume that the area A bar
remains constant when we pull

214
00:12:45,650 --> 00:12:49,740
the bar out, or when
we push it in.

215
00:12:49,740 --> 00:12:54,370
This E curl here is given via
this relationship here, E

216
00:12:54,370 --> 00:12:58,750
being the Young's modulis nu
being the Poisson ratio.

217
00:12:58,750 --> 00:13:04,090
In tension, the bar would be
pulled out as shown here.

218
00:13:04,090 --> 00:13:06,370
Notice the area, the
cross sectional

219
00:13:06,370 --> 00:13:08,740
area, remains constant.

220
00:13:08,740 --> 00:13:14,860
And the displacement is
measured as t delta.

221
00:13:14,860 --> 00:13:22,050
In compression, the bar would be
pushed into, as shown here.

222
00:13:22,050 --> 00:13:27,930
Notice here now t delta and the
bar has been pushed in as

223
00:13:27,930 --> 00:13:29,530
shown by the blue outline.

224
00:13:32,470 --> 00:13:39,810
If we now apply this stress
strain law, to the problem and

225
00:13:39,810 --> 00:13:45,700
develop a relationship between
the force required here to

226
00:13:45,700 --> 00:13:51,190
pull the bar out or push
it in, by delta,

227
00:13:51,190 --> 00:13:54,400
we obtain the following.

228
00:13:54,400 --> 00:13:57,330
Here we have the Green-Lagrange
strain.

229
00:13:57,330 --> 00:14:00,830
And of course, the right hand
side involves derivatives that

230
00:14:00,830 --> 00:14:04,410
we have been discussing in
the previous lecture.

231
00:14:04,410 --> 00:14:09,950
Notice t0u1,1 is nothing else
than this relationship here.

232
00:14:09,950 --> 00:14:14,070
It's given by the current length
and the original length

233
00:14:14,070 --> 00:14:15,630
of the bar.

234
00:14:15,630 --> 00:14:20,470
Substituting into here, we get
directly this right hand side.

235
00:14:20,470 --> 00:14:25,340
We also remember that we can
calculate or we have a very

236
00:14:25,340 --> 00:14:29,300
clear relationship between the
Cauchy stress, which is the

237
00:14:29,300 --> 00:14:31,910
force per unit area, which
we are interested

238
00:14:31,910 --> 00:14:33,250
in calculating here.

239
00:14:33,250 --> 00:14:37,560
And the second, Piola-Kirchhoff
stress, this

240
00:14:37,560 --> 00:14:40,750
is the inverse of the
deformation gradient.

241
00:14:40,750 --> 00:14:46,100
We substitute for here, the
Cauchy stress of course being

242
00:14:46,100 --> 00:14:51,490
simply the force divided by
A bar, the inverse of the

243
00:14:51,490 --> 00:14:53,440
deformation gradient
being given by this

244
00:14:53,440 --> 00:14:55,400
relationship here.

245
00:14:55,400 --> 00:14:58,740
And therefore we can evaluate
the second Piola-Kirchhoff

246
00:14:58,740 --> 00:15:02,730
stress in of the externally
applied force, the physical

247
00:15:02,730 --> 00:15:07,650
force, divided by the constant
area, cross-sectional area, A

248
00:15:07,650 --> 00:15:12,520
bar, and the original length
and the current length.

249
00:15:12,520 --> 00:15:15,370
We therefore have now the
Green-Lagrange strain and the

250
00:15:15,370 --> 00:15:20,020
second Piola-Kirchhoff stress,
in terms of current lengths

251
00:15:20,020 --> 00:15:25,160
and original lengths
and force applied--

252
00:15:25,160 --> 00:15:27,900
and of course the cross section
area goes in there.

253
00:15:27,900 --> 00:15:30,950
Substituting these relationship
and that

254
00:15:30,950 --> 00:15:34,680
relationship into our stress
strain relationship of the

255
00:15:34,680 --> 00:15:43,530
previous view graph and using
this fact here, we directly

256
00:15:43,530 --> 00:15:49,600
obtain this red curve in the--

257
00:15:49,600 --> 00:15:54,750
which gives us the force
applied to the bar as a

258
00:15:54,750 --> 00:15:59,240
function of the displacements
at the end of the bar.

259
00:15:59,240 --> 00:16:03,340
Now notice that this is
a nonlinear curve--

260
00:16:03,340 --> 00:16:07,190
the expression is actually
given here--

261
00:16:07,190 --> 00:16:11,270
and that it is quite
unrealistic, certainly in this

262
00:16:11,270 --> 00:16:12,300
region here.

263
00:16:12,300 --> 00:16:17,340
It's not a realistic material
description for large strains.

264
00:16:17,340 --> 00:16:24,070
You see here at this point, we
have a 0 force to be applied

265
00:16:24,070 --> 00:16:29,060
when the displacement, the end
of the displacement, is minus

266
00:16:29,060 --> 00:16:31,010
the original lengths
off the bar.

267
00:16:31,010 --> 00:16:32,920
Quite unrealistic, of course.

268
00:16:32,920 --> 00:16:38,410
And therefore, we recognize
that really this use of a

269
00:16:38,410 --> 00:16:43,730
constant E and nu, going in
here, constant Young's modulus

270
00:16:43,730 --> 00:16:46,710
and Poisson's ratio,
going in here.

271
00:16:46,710 --> 00:16:53,060
It's really not a proper or
realistic way to proceed for

272
00:16:53,060 --> 00:16:54,310
large strains.

273
00:16:57,010 --> 00:17:00,260
The usual isotopical and also
orthotopic material

274
00:17:00,260 --> 00:17:05,020
relationships, meeting constant
E, nu, Ea, et cetera,

275
00:17:05,020 --> 00:17:09,020
the way I just showed it on a
few previous view graphs are

276
00:17:09,020 --> 00:17:12,520
really almost exclusively
employed in large

277
00:17:12,520 --> 00:17:16,234
displacement, large rotation,
but small strain analysis.

278
00:17:18,800 --> 00:17:23,300
We should now recall one very
important namely that the

279
00:17:23,300 --> 00:17:26,849
components of the second
Piola-Kirchhoff stress tensor,

280
00:17:26,849 --> 00:17:30,470
and of the Green-Lagrange strain
tensor are invariant

281
00:17:30,470 --> 00:17:35,610
very under a rigid body motion
of the material.

282
00:17:35,610 --> 00:17:41,560
And this is most important
because we use this fact to

283
00:17:41,560 --> 00:17:45,770
apply material descriptions that
are applicable to small

284
00:17:45,770 --> 00:17:51,090
strain analysis directly also
to large displacement, large

285
00:17:51,090 --> 00:17:53,210
rotation analysis.

286
00:17:53,210 --> 00:17:58,350
You see, E nu, Ea are material
constant that applicable to

287
00:17:58,350 --> 00:17:59,940
small strain analysis.

288
00:17:59,940 --> 00:18:04,230
And we can directly them some
also in large displacement,

289
00:18:04,230 --> 00:18:06,710
large quotation analysis,
but small strain

290
00:18:06,710 --> 00:18:09,410
analysis using this fact.

291
00:18:13,930 --> 00:18:16,950
And this means--

292
00:18:16,950 --> 00:18:21,590
another way of saying it is
that the actual straining

293
00:18:21,590 --> 00:18:24,990
increases only the components
of the Green-Lagrange strain

294
00:18:24,990 --> 00:18:29,060
tensor through the material
relationships.

295
00:18:29,060 --> 00:18:31,610
Let's look at this
equation here.

296
00:18:31,610 --> 00:18:35,810
We have here the nodal point
force vector being equal to

297
00:18:35,810 --> 00:18:36,980
the stress strain--

298
00:18:36,980 --> 00:18:42,280
the strain displacement matrix
times a stress vector, the

299
00:18:42,280 --> 00:18:45,030
second Piola-Kirchhoff
stress vector.

300
00:18:45,030 --> 00:18:49,310
Now this vector is invariant
under rigid body rotation.

301
00:18:49,310 --> 00:18:51,670
And of course, this vector
is calculated using the

302
00:18:51,670 --> 00:18:53,750
constitutive relation.

303
00:18:53,750 --> 00:18:57,030
This strain displacement
matrix takes

304
00:18:57,030 --> 00:18:59,570
into account the rotation.

305
00:18:59,570 --> 00:19:05,690
Let's look at this fact
a little bit closer.

306
00:19:05,690 --> 00:19:09,050
Here we have pictorially an
element, a four-node element

307
00:19:09,050 --> 00:19:14,050
say, that originally is in this
configuration, the black

308
00:19:14,050 --> 00:19:24,110
configuration, and we are
pulling it out and sharing it.

309
00:19:24,110 --> 00:19:27,850
If this length here is 1
originally, and that length

310
00:19:27,850 --> 00:19:32,550
here is 1 originally, then this
distance here corresponds

311
00:19:32,550 --> 00:19:36,870
to the Green-Lagrange strain in
the 1:1 direction, in the

312
00:19:36,870 --> 00:19:38,220
one direction.

313
00:19:38,220 --> 00:19:41,890
In other words, it's equal
to 1 0 epsilon 1 1.

314
00:19:41,890 --> 00:19:46,560
Similar, this distance here is
equal to 1 0 epsilon 2 2, and

315
00:19:46,560 --> 00:19:50,800
the total sharing is equal
to the sum of these two

316
00:19:50,800 --> 00:19:53,790
components here.

317
00:19:53,790 --> 00:19:59,170
This is the deformation that
brings the element, say, from

318
00:19:59,170 --> 00:20:03,730
the original configuration
to state 1.

319
00:20:03,730 --> 00:20:10,470
Now, going from state 1 to state
2, we rotate the element

320
00:20:10,470 --> 00:20:14,290
rigidly, as shown here.

321
00:20:14,290 --> 00:20:19,740
Then we know that this distance
here is still 1 0

322
00:20:19,740 --> 00:20:21,690
epsilon 1 1.

323
00:20:21,690 --> 00:20:24,230
Same as what we have had here.

324
00:20:24,230 --> 00:20:30,010
This distance here is the same
as what we have seen here.

325
00:20:30,010 --> 00:20:35,810
And the total sharing is still
the same as what the total

326
00:20:35,810 --> 00:20:39,060
sharing was here.

327
00:20:39,060 --> 00:20:44,140
Notice the Green-Lagrange
strain components

328
00:20:44,140 --> 00:20:48,480
corresponding to the stationary
coordinate axes.

329
00:20:48,480 --> 00:20:52,300
1 and 2 have not changed.

330
00:20:52,300 --> 00:20:59,170
Because here this 1 and that 2
refer still too the stationary

331
00:20:59,170 --> 00:21:00,420
coordinate frame.

332
00:21:04,210 --> 00:21:06,425
On this view graph, we summarize
this information

333
00:21:06,425 --> 00:21:10,330
once more for small strains,
we have of course this

334
00:21:10,330 --> 00:21:16,570
relationship here, which means
that these strain components

335
00:21:16,570 --> 00:21:19,280
are much smaller than 1.

336
00:21:19,280 --> 00:21:23,340
And we can directly say that
the second Piola-Kirchhoff

337
00:21:23,340 --> 00:21:26,710
stress in configuration 1 is
given by the Green-Lagrange

338
00:21:26,710 --> 00:21:29,030
strains in configuration 1.

339
00:21:29,030 --> 00:21:32,290
Notice the second
Piola-Kirchhoff stress is

340
00:21:32,290 --> 00:21:36,860
approximately equal to the
Cauchy stress in state 1.

341
00:21:36,860 --> 00:21:41,680
Now since state 2 is reached by
just a rigid body motion,

342
00:21:41,680 --> 00:21:47,620
we find that this holds, as I
already pointed out on the

343
00:21:47,620 --> 00:21:48,850
earlier view graph.

344
00:21:48,850 --> 00:21:52,290
We also find that this holds.

345
00:21:52,290 --> 00:21:55,575
And of course, the Cauchy
stresses in state 2 are

346
00:21:55,575 --> 00:22:01,040
obtained by a rotation of the
Cauchy stresses in state 1.

347
00:22:01,040 --> 00:22:04,320
But the second Piola-Kirchhoff
stresses and Green-Lagrange

348
00:22:04,320 --> 00:22:07,620
strains, the components of
these two tensors are

349
00:22:07,620 --> 00:22:10,790
invariant under the
rigid body motion.

350
00:22:10,790 --> 00:22:14,270
And that means that we can
directly apply material

351
00:22:14,270 --> 00:22:18,360
relationships that are
applicable for small strains

352
00:22:18,360 --> 00:22:20,710
and infinitesimal
displacements.

353
00:22:20,710 --> 00:22:24,750
Those that we are, say, already
familiar with from an

354
00:22:24,750 --> 00:22:26,660
MNO analysis.

355
00:22:26,660 --> 00:22:30,070
We can directly apply these to
large displacement, large

356
00:22:30,070 --> 00:22:34,160
rotation, but small strain
analysis, by simply using

357
00:22:34,160 --> 00:22:38,000
Green-Lagrange strains instead
of engineering strain

358
00:22:38,000 --> 00:22:40,590
components, and second
Piola-Kirchhoff stresses

359
00:22:40,590 --> 00:22:43,950
instead of the engineering
stress components.

360
00:22:46,780 --> 00:22:51,270
Applications are really all
large displacement, large

361
00:22:51,270 --> 00:22:53,290
rotation, small strain
analyses of

362
00:22:53,290 --> 00:22:54,700
beams, plates, and shells.

363
00:22:57,450 --> 00:23:02,565
We can use for this analysis
frequently simply the 2D and

364
00:23:02,565 --> 00:23:06,040
3D elements that we
discussed earlier.

365
00:23:06,040 --> 00:23:10,070
But we may want to use beam and
shell elements, and those

366
00:23:10,070 --> 00:23:14,070
we will discuss in
later lectures.

367
00:23:14,070 --> 00:23:18,860
We also have a very large
application area here, namely,

368
00:23:18,860 --> 00:23:22,700
the linearized buckling analysis
of structures.

369
00:23:22,700 --> 00:23:28,520
We have already performed the
analysis of shells, or shown

370
00:23:28,520 --> 00:23:31,250
how we analyze shelves
and beams

371
00:23:31,250 --> 00:23:34,670
using continuum elements.

372
00:23:34,670 --> 00:23:36,800
You might recall it from
the earlier lectures.

373
00:23:36,800 --> 00:23:41,500
Here you have just schematically
how we would

374
00:23:41,500 --> 00:23:45,530
model say a frame using
plain stress elements.

375
00:23:45,530 --> 00:23:47,960
And here you should see
schematically how we are

376
00:23:47,960 --> 00:23:51,170
modeling using axisymmetric
continuum elements and

377
00:23:51,170 --> 00:23:52,420
axisymmetric shell.

378
00:23:55,060 --> 00:24:01,100
Here we use 3D elements to
model a general shell.

379
00:24:01,100 --> 00:24:05,590
But as I pointed out, we will
talk in later lectures about

380
00:24:05,590 --> 00:24:09,290
actual beam and shell elements
that are frequently much more

381
00:24:09,290 --> 00:24:12,135
effective in the analysis of
beam and shell structures.

382
00:24:15,050 --> 00:24:18,970
This completes the first large
item that I wanted to talk to

383
00:24:18,970 --> 00:24:24,220
you about, namely, the use of
constant material moduli, E,

384
00:24:24,220 --> 00:24:31,590
nu, Ea, Eb, et cetera in large
displacement, large rotation

385
00:24:31,590 --> 00:24:34,900
analysis, but small
strain analysis.

386
00:24:34,900 --> 00:24:39,210
We found that the applicability
of constant

387
00:24:39,210 --> 00:24:43,610
material moduli to large strain
analysis is really very

388
00:24:43,610 --> 00:24:46,260
questionable.

389
00:24:46,260 --> 00:24:52,340
For large strain analysis, we
use frequently a hyperelastic

390
00:24:52,340 --> 00:24:54,210
material model.

391
00:24:54,210 --> 00:24:58,180
And that is very widely used to
analyze rubber materials.

392
00:24:58,180 --> 00:25:02,670
Here we have the relationship
that is basic to this

393
00:25:02,670 --> 00:25:03,500
description.

394
00:25:03,500 --> 00:25:08,160
We have the Piola-Kirchhoff
stress, given in terms of the

395
00:25:08,160 --> 00:25:12,710
partial of the strain energy
density function, W, this

396
00:25:12,710 --> 00:25:15,510
respect to the Green-Lagrange
strains.

397
00:25:15,510 --> 00:25:20,030
So we get one component here,
by taking this t0W and

398
00:25:20,030 --> 00:25:23,460
differentiating it with respect
to the corresponding

399
00:25:23,460 --> 00:25:25,930
component of the Green-Lagrange
strain.

400
00:25:25,930 --> 00:25:27,810
And of course this right
hand side gives

401
00:25:27,810 --> 00:25:29,460
us this tensor t0Cijrs.

402
00:25:32,460 --> 00:25:35,860
Incrementally, then, we have
an increment in the second

403
00:25:35,860 --> 00:25:40,010
Piola-Kirchhoff stress is given
by this stress strain

404
00:25:40,010 --> 00:25:46,400
law here, involving a tangent
material description and an

405
00:25:46,400 --> 00:25:48,140
increment, a differential
increment, in the

406
00:25:48,140 --> 00:25:49,960
Green-Lagrange strain.

407
00:25:49,960 --> 00:25:54,720
This 0Cijrs is nothing else
but the second partial

408
00:25:54,720 --> 00:25:59,210
derivative of W with respect to
these Green-Lagrange strain

409
00:25:59,210 --> 00:26:00,670
components.

410
00:26:00,670 --> 00:26:03,390
W is a strain energy density
function per

411
00:26:03,390 --> 00:26:05,170
unit original volume.

412
00:26:05,170 --> 00:26:08,700
And of course that has to be
defined for the particular

413
00:26:08,700 --> 00:26:11,200
material that you're looking
at and will involve certain

414
00:26:11,200 --> 00:26:13,180
material constants.

415
00:26:13,180 --> 00:26:17,290
For rubber, we generally use
this kind of description here,

416
00:26:17,290 --> 00:26:23,140
where W is a function of I1,
I2, I3, and the Is are the

417
00:26:23,140 --> 00:26:26,340
invariants of the Cauchy-Green
deformation tensor.

418
00:26:26,340 --> 00:26:29,975
We discuss the Cauchy-Green
deformation tensor in an

419
00:26:29,975 --> 00:26:31,720
earlier lecture.

420
00:26:31,720 --> 00:26:35,600
Of course, it's given in terms
of the deformation gradient.

421
00:26:35,600 --> 00:26:38,280
So I1 is equal to this.

422
00:26:38,280 --> 00:26:40,590
I2 is equal to this relationship
here on

423
00:26:40,590 --> 00:26:41,700
the right hand side.

424
00:26:41,700 --> 00:26:44,530
And I3 is simply the
determinant of the

425
00:26:44,530 --> 00:26:48,900
Cauchy-Green deformation
tensor.

426
00:26:48,900 --> 00:26:52,665
For example, one material law
that is quite widely used is

427
00:26:52,665 --> 00:26:56,050
the Mooney-Rivlin material
law, where two

428
00:26:56,050 --> 00:26:59,260
constants enter here.

429
00:26:59,260 --> 00:27:02,140
And of course, we have
here I1 and 12.

430
00:27:02,140 --> 00:27:05,270
These are material constants
that have to be determined

431
00:27:05,270 --> 00:27:07,960
depending on what rubber
you're dealing with.

432
00:27:07,960 --> 00:27:12,880
And the incompressibility
constraint is imposed onto the

433
00:27:12,880 --> 00:27:16,720
formulation by having
I3 equal to 1.

434
00:27:16,720 --> 00:27:20,930
Of course, rubber is an
incompressible material, at

435
00:27:20,930 --> 00:27:24,060
least almost an incompressible
material, and is modeled as an

436
00:27:24,060 --> 00:27:25,240
incompressible material.

437
00:27:25,240 --> 00:27:28,930
And so we want to have
I3 equal to 1.

438
00:27:28,930 --> 00:27:33,100
In general, if you deal with
incompressible materials, it

439
00:27:33,100 --> 00:27:38,260
is effective to not use the
displacement-based finite

440
00:27:38,260 --> 00:27:40,700
element formulations the way
I've been discussing it in the

441
00:27:40,700 --> 00:27:45,210
previous lectures, but to amend
that formulation, you

442
00:27:45,210 --> 00:27:48,340
should really extend that
formulation to have an

443
00:27:48,340 --> 00:27:50,630
effective formulation
to deal with

444
00:27:50,630 --> 00:27:54,120
incompressibility in the analysis.

445
00:27:54,120 --> 00:27:58,430
Except for the plane stress
case, we can directly use what

446
00:27:58,430 --> 00:28:00,680
we discussed earlier.

447
00:28:00,680 --> 00:28:05,570
The formation for
incompressibility, or for

448
00:28:05,570 --> 00:28:10,840
analysis of incompressible
materials, such formulations

449
00:28:10,840 --> 00:28:15,700
of course are very valuable
but the discussion of such

450
00:28:15,700 --> 00:28:20,010
formulations really deserves
another lecture.

451
00:28:20,010 --> 00:28:26,680
Well, we like to now show how
this material law is used in

452
00:28:26,680 --> 00:28:28,400
the plane stress case.

453
00:28:28,400 --> 00:28:32,130
There we can directly employ
the formulation that we

454
00:28:32,130 --> 00:28:35,010
discussed in the earlier
lectures.

455
00:28:35,010 --> 00:28:39,360
Time 0, for example, we would
have a piece of material as

456
00:28:39,360 --> 00:28:40,330
shown here.

457
00:28:40,330 --> 00:28:44,210
At time t, this piece of
material has deformed, as

458
00:28:44,210 --> 00:28:46,230
shown here.

459
00:28:46,230 --> 00:28:49,070
Notice plane stress of course
means stress is

460
00:28:49,070 --> 00:28:52,300
0 through the thickness.

461
00:28:52,300 --> 00:28:56,630
For this two-dimensional
problem, the Cauchy-Green

462
00:28:56,630 --> 00:29:01,100
deformation tensor
is as shown here.

463
00:29:01,100 --> 00:29:04,840
0 components here.

464
00:29:04,840 --> 00:29:07,890
And we imposed the
incompressibility of the

465
00:29:07,890 --> 00:29:14,980
material by simply setting the
determinant of this matrix

466
00:29:14,980 --> 00:29:16,170
equal to 1.

467
00:29:16,170 --> 00:29:20,770
And this gives us then a
constraint on T0C33.

468
00:29:20,770 --> 00:29:24,330
Basically, we are saying
physically that we constrain

469
00:29:24,330 --> 00:29:30,110
the material to shrink into this
33 direction such that

470
00:29:30,110 --> 00:29:34,000
for the applied stresses in the
plane of the material, the

471
00:29:34,000 --> 00:29:37,270
material acts as
incompressible.

472
00:29:37,270 --> 00:29:40,770
And that's what's being done
here mathematically.

473
00:29:40,770 --> 00:29:45,010
We can now evaluate with this
constraint I1 and I2.

474
00:29:45,010 --> 00:29:47,280
Here you see the expressions.

475
00:29:49,950 --> 00:29:53,965
And if we have I1 and I2, we
can go back to the material

476
00:29:53,965 --> 00:29:59,070
law, substitute and directly
octane obtain our second

477
00:29:59,070 --> 00:30:05,940
Piola-Kirchhoff stresses, in
terms of the Cauchy-Green

478
00:30:05,940 --> 00:30:09,160
deformation tensor components.

479
00:30:09,160 --> 00:30:12,290
And of course, we also have
a relationship between the

480
00:30:12,290 --> 00:30:15,100
Cauchy-Green deformation tensor
components and the

481
00:30:15,100 --> 00:30:17,690
Green-Lagrange strain
components.

482
00:30:17,690 --> 00:30:19,380
Those we use.

483
00:30:19,380 --> 00:30:24,370
That relationship we use, and
that actually brings into it

484
00:30:24,370 --> 00:30:25,800
this relationship here.

485
00:30:28,500 --> 00:30:31,640
And if you performed then the
differentiation, the final

486
00:30:31,640 --> 00:30:36,220
result is shown on
this view graph.

487
00:30:36,220 --> 00:30:39,730
This is the stress strain
relationship that gives the

488
00:30:39,730 --> 00:30:45,280
total stresses in terms
of the Cij components.

489
00:30:50,670 --> 00:30:56,010
You similarly proceed to also
calculate the tangent

490
00:30:56,010 --> 00:30:58,850
relationship, the tangent
stress strain law.

491
00:30:58,850 --> 00:31:01,780
As I pointed out earlier, you
need to now take second

492
00:31:01,780 --> 00:31:05,290
derivatives of W with respect
to the Green-Lagrange strain

493
00:31:05,290 --> 00:31:06,740
components.

494
00:31:06,740 --> 00:31:09,310
And if you also use the
relationships between the

495
00:31:09,310 --> 00:31:11,530
Green-Lagrange strain components
and the components

496
00:31:11,530 --> 00:31:13,590
of the Cauchy-Green deformation
tensor, you

497
00:31:13,590 --> 00:31:15,590
directly arrive at
this expression.

498
00:31:15,590 --> 00:31:19,090
And of course now, you would
have to substitute as we did

499
00:31:19,090 --> 00:31:23,600
earlier for calculating the
total stress strain law, in

500
00:31:23,600 --> 00:31:27,580
other words, the stress strain
giving us a total stress as a

501
00:31:27,580 --> 00:31:29,930
function of the total
deformations.

502
00:31:29,930 --> 00:31:34,110
You proceed in the same way, and
obtain the tangent stress

503
00:31:34,110 --> 00:31:35,790
strain law.

504
00:31:35,790 --> 00:31:39,080
Let us now look at an example.

505
00:31:39,080 --> 00:31:43,280
In this example, we are
considering the analysis of a

506
00:31:43,280 --> 00:31:45,080
tensile test specimen.

507
00:31:45,080 --> 00:31:46,830
The specimen is shown here.

508
00:31:46,830 --> 00:31:49,490
All dimensions are given
in millimeters.

509
00:31:49,490 --> 00:31:53,270
The thickness of the specimen
is one millimeter, and the

510
00:31:53,270 --> 00:31:57,540
Mooney-Rivlin constants
are given here.

511
00:31:57,540 --> 00:32:00,720
We need to model only one
quarter of the specimen

512
00:32:00,720 --> 00:32:02,850
because of symmetry
conditions.

513
00:32:02,850 --> 00:32:07,930
And in the analysis, we used
14 eight-node elements to

514
00:32:07,930 --> 00:32:12,370
represent this quarter
of the specimen.

515
00:32:12,370 --> 00:32:15,550
Notice that on the right hand
side, we constrained all

516
00:32:15,550 --> 00:32:20,370
material particles to move
together and only move to the

517
00:32:20,370 --> 00:32:24,840
right, as shown by delta here.

518
00:32:24,840 --> 00:32:27,890
This represents 1/2 the gauge
length of the specimen, and

519
00:32:27,890 --> 00:32:31,040
one is particularly interested
in seeing the deformations in

520
00:32:31,040 --> 00:32:33,990
that gauge length.

521
00:32:33,990 --> 00:32:37,880
This view graph now shows
the calculated response.

522
00:32:37,880 --> 00:32:40,510
The applied node
is given here.

523
00:32:40,510 --> 00:32:42,930
The extension is given here.

524
00:32:42,930 --> 00:32:45,240
This is the gauge response.

525
00:32:45,240 --> 00:32:47,330
And that gives us--

526
00:32:47,330 --> 00:32:50,910
this curve gives us the
total response.

527
00:32:50,910 --> 00:32:54,620
Of course, quite nonlinear,
because the material is a

528
00:32:54,620 --> 00:32:57,840
nonlinear elastic material.

529
00:32:57,840 --> 00:33:02,900
The deformed mesh is shown on
the next view graph here.

530
00:33:02,900 --> 00:33:05,975
At the force 4 Newton, we
have this deformed mesh

531
00:33:05,975 --> 00:33:08,110
here shown in red.

532
00:33:08,110 --> 00:33:12,640
And notice here, we have the
undeformed mesh in dashed

533
00:33:12,640 --> 00:33:15,090
black line.

534
00:33:15,090 --> 00:33:17,860
Notice that the displacements
are large, and the strains

535
00:33:17,860 --> 00:33:23,010
certainly are very large
in this problem.

536
00:33:23,010 --> 00:33:26,270
Finally I would like to share
with you some analysis results

537
00:33:26,270 --> 00:33:29,400
that we obtained in the analysis
of a rubber sheet

538
00:33:29,400 --> 00:33:30,640
with a hole.

539
00:33:30,640 --> 00:33:34,590
And these analysis results are
shown on the slides, so let me

540
00:33:34,590 --> 00:33:40,240
walk over here so that we can
share that information.

541
00:33:40,240 --> 00:33:43,330
Here, we show the
rubber sheet.

542
00:33:43,330 --> 00:33:47,970
It's a square sheet, 20 inch
by 20 inch, that contains a

543
00:33:47,970 --> 00:33:51,970
hole of diameter 6-inch.

544
00:33:51,970 --> 00:33:56,460
Notice the sheet is subjected
to a force on the left hand

545
00:33:56,460 --> 00:33:59,460
side, the distributed
pressure, uniformly

546
00:33:59,460 --> 00:34:01,960
distributed pressure on the
left hand side and on the

547
00:34:01,960 --> 00:34:04,330
right hand side.

548
00:34:04,330 --> 00:34:09,590
The mass density is given over
here also, and the thickness

549
00:34:09,590 --> 00:34:11,940
off the sheet is given
here as well.

550
00:34:11,940 --> 00:34:15,545
We will use the mass density
later on when we perform also

551
00:34:15,545 --> 00:34:19,960
or look at the results of
a dynamic analysis.

552
00:34:19,960 --> 00:34:22,980
The next slide now shows the
finite element mesh that we

553
00:34:22,980 --> 00:34:27,120
used to model one quarter
of the sheet.

554
00:34:27,120 --> 00:34:30,590
We only needed to model one
quarter because of symmetry

555
00:34:30,590 --> 00:34:31,969
conditions.

556
00:34:31,969 --> 00:34:36,860
Notice we used four-node
elements in this mesh, and

557
00:34:36,860 --> 00:34:41,040
that the distributive pressure
on this right hand side is

558
00:34:41,040 --> 00:34:47,590
modeled by concentrated
loads at these nodes.

559
00:34:47,590 --> 00:34:51,350
The next slide shows
now the solution

560
00:34:51,350 --> 00:34:54,050
results that we obtained.

561
00:34:54,050 --> 00:34:58,250
Notice down here, we have a
little key for the points that

562
00:34:58,250 --> 00:35:00,070
we want to look at.

563
00:35:00,070 --> 00:35:07,330
Here is point B. Here is point
A. And here is point C. The

564
00:35:07,330 --> 00:35:11,500
graphs here shows the
displacements at point B, at

565
00:35:11,500 --> 00:35:17,110
point A, and point C. In each
case, the W displacement,

566
00:35:17,110 --> 00:35:23,180
which at point B is into this
direction, at point A is into

567
00:35:23,180 --> 00:35:28,700
this direction, but at
point C is downwards.

568
00:35:28,700 --> 00:35:33,210
Notice the very large
displacements, W here, going

569
00:35:33,210 --> 00:35:39,050
up to 10, 11 inches for
the load applied.

570
00:35:39,050 --> 00:35:44,500
Remember the sheet is a square
sheet of 20 inches, so we have

571
00:35:44,500 --> 00:35:46,215
very large deformations here.

572
00:35:46,215 --> 00:35:50,990
We are computing our solution
results obtained in the total

573
00:35:50,990 --> 00:35:55,960
Lagrangian formulation, with the
hyperelastic Mooney-Rivlin

574
00:35:55,960 --> 00:36:00,320
material law, with the results
obtained by Iding.

575
00:36:00,320 --> 00:36:04,220
In fact, there is almost no
difference between our

576
00:36:04,220 --> 00:36:07,020
results, and the results of
Iding, so that we could not

577
00:36:07,020 --> 00:36:07,830
plot the difference.

578
00:36:07,830 --> 00:36:11,040
And these curves here
refer to our results

579
00:36:11,040 --> 00:36:12,290
and the Iding results.

580
00:36:14,810 --> 00:36:18,150
The next slide now shows the
deformations that the sheet

581
00:36:18,150 --> 00:36:21,490
has undergone at maximum load.

582
00:36:21,490 --> 00:36:25,870
Here is the original
configuration of the sheet,

583
00:36:25,870 --> 00:36:27,560
the initial configuration.

584
00:36:27,560 --> 00:36:31,670
And here, we show the deformed
configuration of the sheet.

585
00:36:34,520 --> 00:36:36,720
Notice that the sheet has
undergone very large

586
00:36:36,720 --> 00:36:40,750
displacements and very large
strains, particularly of

587
00:36:40,750 --> 00:36:42,620
course here in the region
of the hole.

588
00:36:45,900 --> 00:36:50,900
The final slide then shows the
dynamic response that the

589
00:36:50,900 --> 00:36:55,740
sheet undergoes when we subject
the sheet to a step

590
00:36:55,740 --> 00:37:00,980
loading, where the static
loading was applied before.

591
00:37:00,980 --> 00:37:06,340
Notice the step loading is
applied over 10 delta-t where

592
00:37:06,340 --> 00:37:10,560
delta-a is 0.0015 seconds.

593
00:37:10,560 --> 00:37:16,850
It goes up linearly to 75 pounds
per square inch, and

594
00:37:16,850 --> 00:37:19,440
then it stays constant.

595
00:37:19,440 --> 00:37:23,730
Once again, we're looking at the
displacements at point B,

596
00:37:23,730 --> 00:37:29,450
point A, W displacements here,
and at point C downward

597
00:37:29,450 --> 00:37:33,860
displacement, again denoted
as W. Here you see

598
00:37:33,860 --> 00:37:35,470
the solution response.

599
00:37:35,470 --> 00:37:42,700
Point B, point A, and point C.
In this particular analysis,

600
00:37:42,700 --> 00:37:47,640
we used first the Newmark method
with data equal to 1/2

601
00:37:47,640 --> 00:37:49,870
and alpha equal to 1/4.

602
00:37:49,870 --> 00:37:52,450
In fact, that is nothing else
than the trapezoidal rule,

603
00:37:52,450 --> 00:37:55,090
which we discussed in
an earlier lecture.

604
00:37:55,090 --> 00:37:58,430
We also performed the analysis
once with the Wilson theta

605
00:37:58,430 --> 00:38:00,680
method, theta being 1.4.

606
00:38:00,680 --> 00:38:03,720
This method we did not discuss
in the earlier lecture, but it

607
00:38:03,720 --> 00:38:06,020
is described in the textbook.

608
00:38:06,020 --> 00:38:09,080
Notice that the results obtained
using the Newmark

609
00:38:09,080 --> 00:38:12,600
method and the Wilson theta
method are really very close

610
00:38:12,600 --> 00:38:15,280
to each other.

611
00:38:15,280 --> 00:38:18,820
It's also interesting to see the
static response that the

612
00:38:18,820 --> 00:38:23,130
sheet undergoes for this load
level at the points A, B, and

613
00:38:23,130 --> 00:38:28,940
C, and that is given by these
horizontal lines.

614
00:38:28,940 --> 00:38:32,790
Once again, you have here a
large displacement and very

615
00:38:32,790 --> 00:38:34,890
large strain problem.

616
00:38:34,890 --> 00:38:37,550
This brings me to the end of
what I wanted to discuss with

617
00:38:37,550 --> 00:38:38,580
you in this lecture.

618
00:38:38,580 --> 00:38:39,900
Thank you very much for
your attention.