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PROFESSOR: Ladies and gentlemen,
welcome to this
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lecture on nonlinear finite
element analysis of solids and
00:00:26.030 --> 00:00:27.190
structures.
00:00:27.190 --> 00:00:29.390
In this lecture, I would like
to continue with our
00:00:29.390 --> 00:00:33.280
discussion of the use of
constitutive relations used in
00:00:33.280 --> 00:00:35.020
nonlinear analysis.
00:00:35.020 --> 00:00:37.130
In the previous lecture, we
already considered the
00:00:37.130 --> 00:00:41.300
modeling of elastic and hyper
elastic materials, subjected
00:00:41.300 --> 00:00:45.470
to large displacements, large
rotations, and large strains.
00:00:45.470 --> 00:00:47.980
I'd like to not turn our
attention to the use of the
00:00:47.980 --> 00:00:49.680
updated Lagrangian
formulation.
00:00:49.680 --> 00:00:52.280
Of course, the kinematic
relations for the total and
00:00:52.280 --> 00:00:55.260
updated Lagrangian formulations,
we discussed in
00:00:55.260 --> 00:00:57.470
an earlier lecture already.
00:00:57.470 --> 00:01:00.920
In particular, I would like now,
in very general terms, to
00:01:00.920 --> 00:01:02.970
ask the following question--
00:01:02.970 --> 00:01:06.330
is it possible to obtain, using
the updated Lagrangian
00:01:06.330 --> 00:01:11.740
formulation, identically the
same numerical results for
00:01:11.740 --> 00:01:16.890
each load step and each
iteration, as are obtained
00:01:16.890 --> 00:01:19.720
when we use a total Lagrangian
formulation?
00:01:19.720 --> 00:01:23.020
Let's look at this question
here, now, in
00:01:23.020 --> 00:01:24.560
quite gentle terms.
00:01:24.560 --> 00:01:29.930
And I have rephrased it here
once more, so as to really
00:01:29.930 --> 00:01:32.010
make very clear what I mean.
00:01:32.010 --> 00:01:36.960
Let's say that we have a Program
1 that uses only the
00:01:36.960 --> 00:01:39.800
total Lagrangian formulation.
00:01:39.800 --> 00:01:43.130
And the constitutive relations
that we are employing in that
00:01:43.130 --> 00:01:46.620
total Lagrangian formulation
are defined as shown here.
00:01:46.620 --> 00:01:49.690
So total stress at time, t, of
course, second Piola-Kirchhoff
00:01:49.690 --> 00:01:53.890
stress, is given as a function
of displacements.
00:01:53.890 --> 00:01:56.970
And the tangent material
relationship, relating the
00:01:56.970 --> 00:01:59.470
increment in the second
Piola-Kirchhoff stress and the
00:01:59.470 --> 00:02:01.860
increment in the Green-Lagrange
strain is as
00:02:01.860 --> 00:02:04.280
shown here.
00:02:04.280 --> 00:02:07.720
The program results for a
particular physical problem
00:02:07.720 --> 00:02:11.610
might look systematically
as shown here.
00:02:11.610 --> 00:02:16.690
Of course, these two
relationships here have been
00:02:16.690 --> 00:02:19.900
obtained from a set of
physical laboratory
00:02:19.900 --> 00:02:22.480
experiments.
00:02:22.480 --> 00:02:24.900
This is how Program 1 works.
00:02:24.900 --> 00:02:29.900
Now we have also, say,
a Program 2.
00:02:29.900 --> 00:02:33.130
And this Program 2 only uses
the updated Lagrangian
00:02:33.130 --> 00:02:35.070
formulation.
00:02:35.070 --> 00:02:39.210
Now, remember our laboratory
physical experimental results
00:02:39.210 --> 00:02:42.590
have given us the considered
affiliations for the towards
00:02:42.590 --> 00:02:44.370
Lagrangian formulation.
00:02:44.370 --> 00:02:51.500
However, we now have a program
that only is operating on the
00:02:51.500 --> 00:02:52.890
updated Lagrangian
formulation.
00:02:52.890 --> 00:02:57.270
The constitutive relations are,
in this case, well, like
00:02:57.270 --> 00:03:00.860
that and like that, where I have
given here three dots,
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because we want to now study
what they should be looking
00:03:04.470 --> 00:03:11.010
like in order to obtain the
same results as we are
00:03:11.010 --> 00:03:14.200
attaining with Program 1.
00:03:14.200 --> 00:03:18.880
In other words, how can we
obtain, with this Program 2,
00:03:18.880 --> 00:03:22.110
identically the same results
as are obtained
00:03:22.110 --> 00:03:24.180
using Program 1.
00:03:24.180 --> 00:03:28.530
This is a question that
we really are asking.
00:03:28.530 --> 00:03:32.680
To answer this question, we
have to go back to some
00:03:32.680 --> 00:03:35.410
derivations that we discussed
in an earlier lecture.
00:03:35.410 --> 00:03:38.250
And there, we discussed that
this is here the governing
00:03:38.250 --> 00:03:41.190
continuum mechanics equation,
corresponding to the total
00:03:41.190 --> 00:03:42.930
Lagrangian formulation.
00:03:42.930 --> 00:03:47.310
Here we have the incremental
stress strain law going into
00:03:47.310 --> 00:03:48.500
the formulation.
00:03:48.500 --> 00:03:52.280
Notice this is a tangent
material tensor.
00:03:52.280 --> 00:03:56.170
Here we have the second
Piola-Kirchhoff stress, and
00:03:56.170 --> 00:03:57.610
here, as well.
00:03:57.610 --> 00:03:59.650
And of course, here on the right
hand side, we have to
00:03:59.650 --> 00:04:05.100
total external loads
entering the total
00:04:05.100 --> 00:04:07.350
external virtual work.
00:04:07.350 --> 00:04:11.100
In the updated Lagrangian
formulation, this is the
00:04:11.100 --> 00:04:13.300
equation that we derived
earlier.
00:04:13.300 --> 00:04:16.279
Here, again, is a total
external virtual work.
00:04:16.279 --> 00:04:19.959
And here, a constitutive
relation enters, corresponding
00:04:19.959 --> 00:04:21.959
to this formulation.
00:04:21.959 --> 00:04:27.290
And here we have the Cauchy
stress that is being used in
00:04:27.290 --> 00:04:30.150
the updated Lagrangian
formulation.
00:04:30.150 --> 00:04:33.720
The terms used in the
formulations are summarized on
00:04:33.720 --> 00:04:35.090
this view graph.
00:04:35.090 --> 00:04:37.750
In the TL formulation, of
course, we're integrating over
00:04:37.750 --> 00:04:39.870
the original volume.
00:04:39.870 --> 00:04:42.560
In the updated Lagrangian
formulation, we're integrating
00:04:42.560 --> 00:04:45.520
over the current
volume, time t.
00:04:45.520 --> 00:04:48.450
And there is this
transformation, where the mass
00:04:48.450 --> 00:04:53.890
density's at time t, and the
original mass density, enters.
00:04:53.890 --> 00:04:58.660
Notice here are listed the
linear and the nonlinear
00:04:58.660 --> 00:05:03.280
increment in the Green-Lagrange
strain.
00:05:03.280 --> 00:05:06.620
Here, there also is a linear and
a nonlinear increment in
00:05:06.620 --> 00:05:08.635
the Green-Lagrange, but for
the updated Lagrangian
00:05:08.635 --> 00:05:12.990
formulation, and here, for the
total Lagrangian formulation.
00:05:12.990 --> 00:05:15.210
There's a relationship
between these
00:05:15.210 --> 00:05:18.980
increments, as shown here.
00:05:18.980 --> 00:05:22.580
And this relationship I will
focus our attention upon
00:05:22.580 --> 00:05:25.760
further on the next view graph,
because I'd like you to
00:05:25.760 --> 00:05:29.530
clearly understand how we obtain
this relationship.
00:05:29.530 --> 00:05:34.530
But we notice is that the
deformation gradient enters in
00:05:34.530 --> 00:05:38.610
this equation, as well
as in this equation.
00:05:38.610 --> 00:05:41.460
The variations on these linear
strain increments and
00:05:41.460 --> 00:05:45.550
nonlinear strain increments
of the total Lagrangian
00:05:45.550 --> 00:05:48.660
formulation and the updated
Lagrangian formulation are
00:05:48.660 --> 00:05:52.550
also related via the deformation
gradient.
00:05:52.550 --> 00:05:55.110
In fact, twice the deformation
gradient enters here.
00:05:57.740 --> 00:06:01.870
Let's look at how we derive this
kinematic relationships.
00:06:01.870 --> 00:06:05.430
Well, we use that there is
a fundamental property of
00:06:05.430 --> 00:06:10.330
Green-Lagrange strain as
expressed in this equation.
00:06:10.330 --> 00:06:14.540
If you know a Green-Lagrange
strain at a point, then we can
00:06:14.540 --> 00:06:19.630
pick any material fiber
with these components.
00:06:19.630 --> 00:06:24.200
And we know, if these are the
origin of components of the
00:06:24.200 --> 00:06:31.160
material fiber, at that point,
that after motion, in other
00:06:31.160 --> 00:06:35.720
words, in configuration t,
corresponding to time t, that
00:06:35.720 --> 00:06:38.590
material fiber has this length,
where the original
00:06:38.590 --> 00:06:41.810
length is given here.
00:06:41.810 --> 00:06:45.960
This is a very basic equation
that we want to study a bit
00:06:45.960 --> 00:06:48.500
more on the next view graph.
00:06:48.500 --> 00:06:52.230
But let me already point out
that we can, directly, use
00:06:52.230 --> 00:06:55.370
this basic relation, and,
of course, apply it also
00:06:55.370 --> 00:06:57.590
corresponding to time
t, plus delta t.
00:06:57.590 --> 00:07:01.110
Notice here, t, here,
t plus delta t.
00:07:01.110 --> 00:07:03.320
Here t, here, t plus delta t.
00:07:03.320 --> 00:07:06.220
Otherwise, same relationship.
00:07:06.220 --> 00:07:09.720
And we also can apply
incrementally.
00:07:09.720 --> 00:07:13.750
Notice this a Green-Lagrange
strain increment, from time t,
00:07:13.750 --> 00:07:15.950
to time t plus delta t.
00:07:15.950 --> 00:07:18.230
But refer to time t.
00:07:18.230 --> 00:07:20.940
That's why we have the
little t down here.
00:07:20.940 --> 00:07:24.340
Up here, we always referred the
increments in strains, or
00:07:24.340 --> 00:07:27.220
the total strains, if it is
a total strain, to the
00:07:27.220 --> 00:07:29.040
configuration at time zero.
00:07:29.040 --> 00:07:32.450
Now we are referring the
quantity to time t.
00:07:32.450 --> 00:07:38.130
And, if we simply apply this
same basic relationship here,
00:07:38.130 --> 00:07:41.570
we directly obtain this
equation here for the
00:07:41.570 --> 00:07:44.630
increment in Green-Lagrange
strain from time, t, to time,
00:07:44.630 --> 00:07:46.240
t plus delta t.
00:07:46.240 --> 00:07:51.460
Refer to time t configuration,
this relationship holds.
00:07:51.460 --> 00:07:53.810
Let's look at this relationship
here.
00:07:53.810 --> 00:07:56.590
What does it really
mean physically?
00:07:56.590 --> 00:07:59.100
Well, it means the following--
00:07:59.100 --> 00:08:02.890
here we have our stationary
coordinate frame, x1, x2, x3.
00:08:02.890 --> 00:08:06.830
And at time zero, we have a
material fiber here, any
00:08:06.830 --> 00:08:07.760
material fiber.
00:08:07.760 --> 00:08:09.450
But let's just pick one.
00:08:09.450 --> 00:08:13.780
And that has a length of 0 ds.
00:08:13.780 --> 00:08:20.650
The components of this vector,
the vector that is shown here,
00:08:20.650 --> 00:08:25.570
of length 0 ds, the components
are shown here.
00:08:25.570 --> 00:08:29.515
And that vector we
call a d 0 x.
00:08:34.030 --> 00:08:41.030
This material fiber moves from
time 0 to time t, to this
00:08:41.030 --> 00:08:43.110
configuration.
00:08:43.110 --> 00:08:46.540
The new components
are listed here--
00:08:46.540 --> 00:08:50.500
dtx2, dtx1, dtx3.
00:08:50.500 --> 00:08:53.020
Of course, there's a
relationship between the new
00:08:53.020 --> 00:08:56.290
components and the original
components of
00:08:56.290 --> 00:08:57.390
the material fiber.
00:08:57.390 --> 00:09:02.660
And here, goes into the
deformation gradient.
00:09:02.660 --> 00:09:08.410
Now the equation that I just
referred to really relates
00:09:08.410 --> 00:09:11.930
this length and that length.
00:09:11.930 --> 00:09:16.220
And the way we obtain it is to
look at the definition of the
00:09:16.220 --> 00:09:16.890
Green-Lagrange strain.
00:09:16.890 --> 00:09:21.210
strain The Green-Lagrange strain
is defined as follows--
00:09:21.210 --> 00:09:23.200
I'm going to simply
write it in here.
00:09:27.150 --> 00:09:28.400
You see here, we have--
00:09:30.710 --> 00:09:31.040
oops.
00:09:31.040 --> 00:09:33.500
I should have not to use that
delta t, because I'm
00:09:33.500 --> 00:09:34.750
looking at time t.
00:09:40.840 --> 00:09:43.410
This is the equation for the
Green-Lagrange strain, from
00:09:43.410 --> 00:09:45.040
here to there.
00:09:45.040 --> 00:09:49.120
And what I'm now going to do
is to premultiply this
00:09:49.120 --> 00:09:52.450
relationship by d0x,
transposed, and
00:09:52.450 --> 00:09:55.400
postmultiplied by d0x.
00:09:55.400 --> 00:09:58.040
Of course, I have to do
the same thing here.
00:09:58.040 --> 00:10:02.220
d0x, transposed, goes in
front of the bracket.
00:10:02.220 --> 00:10:06.170
And d0x goes into the
back of the bracket.
00:10:06.170 --> 00:10:08.820
Let me make clear what
I'm doing here.
00:10:08.820 --> 00:10:10.550
I'm putting this in here.
00:10:10.550 --> 00:10:12.930
And I'm putting that
on the back.
00:10:12.930 --> 00:10:17.380
Notice I'm using a transposed
here and a transposed there.
00:10:17.380 --> 00:10:19.980
The basic relation of the
Green-Lagrange strain, once
00:10:19.980 --> 00:10:26.990
again, is this being equal to
that part with out the d0x
00:10:26.990 --> 00:10:31.330
transposed in front, and without
the d0x on the back.
00:10:31.330 --> 00:10:39.450
But now we notice, directly,
that this part here and this
00:10:39.450 --> 00:10:42.985
part here is nothing
else then dtx.
00:10:45.630 --> 00:10:51.750
If you recognize that this
is dtx, that is dtx, you
00:10:51.750 --> 00:10:56.200
immediately get on the right
hand side, one half,
00:10:56.200 --> 00:11:01.290
bracket open, d--
00:11:01.290 --> 00:11:07.150
well, I should say t ds,
squared, minus 0ds squared.
00:11:07.150 --> 00:11:09.310
That's what you have on
the right hand side.
00:11:09.310 --> 00:11:11.710
And on the left hand
side, you have what
00:11:11.710 --> 00:11:12.550
I showed you earlier.
00:11:12.550 --> 00:11:16.950
Let me go back one view graph,
just to make sure that we
00:11:16.950 --> 00:11:18.270
understand each other.
00:11:18.270 --> 00:11:21.040
On the right hand side, you
immediately obtain this term
00:11:21.040 --> 00:11:22.830
minus that term.
00:11:22.830 --> 00:11:27.070
And on the left hand side, you
end up with this term.
00:11:27.070 --> 00:11:31.570
So this is the relationship, a
very basic relationship, that
00:11:31.570 --> 00:11:34.960
we want to use in
our derivation.
00:11:34.960 --> 00:11:37.450
And it is obtained by simply
looking at the Green-Lagrange
00:11:37.450 --> 00:11:41.840
strain, the way it's defined,
and the way we have defined it
00:11:41.840 --> 00:11:45.200
in an earlier lecture, and
premultiplying by d0x
00:11:45.200 --> 00:11:48.580
transposed, and postmultiplying
by d0x.
00:11:48.580 --> 00:11:52.240
Of course, what you're seeing
there, in that relationship I
00:11:52.240 --> 00:11:57.250
showed you, you see the
components of this term here,
00:11:57.250 --> 00:12:01.660
written out in terms of
i and j components.
00:12:01.660 --> 00:12:05.470
Well, so what we are doing,
then, is to follow a material
00:12:05.470 --> 00:12:11.700
fiber, original length 0 ds,
through it's motion to t ds
00:12:11.700 --> 00:12:15.950
and to t plus delta t ds.
00:12:15.950 --> 00:12:19.640
And that's expressed by the
three formulas that are on the
00:12:19.640 --> 00:12:21.490
previous view graph.
00:12:21.490 --> 00:12:28.220
Hence, if we subtract from the
second formula, the first
00:12:28.220 --> 00:12:34.060
formula, we directly obtain
this relationship here.
00:12:34.060 --> 00:12:39.850
And we now use this term here,
or this equation, linking up
00:12:39.850 --> 00:12:46.980
dtx to d0x and expand
these two terms
00:12:46.980 --> 00:12:49.530
to obtain this equation.
00:12:49.530 --> 00:12:53.170
Now, notice that this equation
has to hold for
00:12:53.170 --> 00:12:55.160
any material fiber.
00:12:55.160 --> 00:12:59.620
And if it does hold for any
material fiber, it immediately
00:12:59.620 --> 00:13:03.660
follows that this equation here
has to hold it, because
00:13:03.660 --> 00:13:07.760
we can apply it to a material
fiber that is oriented along
00:13:07.760 --> 00:13:11.590
the x1 axis, then another one
that is oriented along the x2
00:13:11.590 --> 00:13:14.400
axis, then another one that
is oriented along the x3
00:13:14.400 --> 00:13:16.430
axis, and so on.
00:13:16.430 --> 00:13:22.380
And surely, this relationship
here must then hold.
00:13:22.380 --> 00:13:27.020
If we look at this relationship
more closely, we
00:13:27.020 --> 00:13:31.710
see that we can write it out as
a linear strain increment
00:13:31.710 --> 00:13:35.210
and as a nonlinear strain
increment on the left hand
00:13:35.210 --> 00:13:39.610
side and on the right hand
side of the equation.
00:13:39.610 --> 00:13:43.910
And this shows, then, directly
that the 0 eij must be equal
00:13:43.910 --> 00:13:49.240
to twice a deformation
gradient times ters.
00:13:49.240 --> 00:13:53.400
The reason being that this term
here is a function of the
00:13:53.400 --> 00:13:55.990
incremental displacement only.
00:13:55.990 --> 00:13:59.610
Linear functional incremental
displacement, so is this.
00:13:59.610 --> 00:14:03.710
And if you simply equate those
terms that are linear, we get
00:14:03.710 --> 00:14:06.670
that this part must be
equal to that part.
00:14:06.670 --> 00:14:10.970
And if you create all the
nonlinear parts, we find that
00:14:10.970 --> 00:14:15.260
0ei, e to ij, is equal to twice
the deformation gradient
00:14:15.260 --> 00:14:17.090
times te to rs.
00:14:17.090 --> 00:14:19.520
And those two equations
are given down here.
00:14:22.250 --> 00:14:25.750
If we now take variations on the
left hand side and on the
00:14:25.750 --> 00:14:29.830
right hand side, and if you
recognize that's a variation,
00:14:29.830 --> 00:14:33.590
is respect to the configuration
at time t plus
00:14:33.590 --> 00:14:39.220
delta t, it is clear that the
deformation gradient terms
00:14:39.220 --> 00:14:40.680
don't change.
00:14:40.680 --> 00:14:43.430
And, therefore taking the
variation on the left hand
00:14:43.430 --> 00:14:49.910
side, that variation is simply
applied to the ters, with
00:14:49.910 --> 00:14:51.570
these being constant terms.
00:14:51.570 --> 00:14:56.730
Similarly here, and this then,
completes the derivation of
00:14:56.730 --> 00:14:59.170
all the kinematic relationships
that I had in
00:14:59.170 --> 00:15:03.230
the earlier table and that
we want to use now.
00:15:03.230 --> 00:15:08.000
In addition, we have, also,
the important relationship
00:15:08.000 --> 00:15:10.705
between the second
Piola-Kirchhoff stress and the
00:15:10.705 --> 00:15:11.720
Cauchy stress.
00:15:11.720 --> 00:15:13.910
And that relationship
is shown here.
00:15:13.910 --> 00:15:18.720
Of course, we introduced that
one in an earlier lecture.
00:15:18.720 --> 00:15:21.910
Finally, we have a relationship
between the
00:15:21.910 --> 00:15:25.540
tangent material tensor,
corresponding to the total
00:15:25.540 --> 00:15:29.090
Lagrangian formulation and
corresponding to the updated
00:15:29.090 --> 00:15:30.770
Lagrangian formulation.
00:15:30.770 --> 00:15:33.340
Notice zero here, t there.
00:15:33.340 --> 00:15:36.460
And that relationship
is given here.
00:15:36.460 --> 00:15:40.070
It's given by a fourth order
tensor transformation, which
00:15:40.070 --> 00:15:42.630
I'd like to now discuss
with you further.
00:15:42.630 --> 00:15:45.610
This is an important
relationship that we want to
00:15:45.610 --> 00:15:47.319
spend a bit off time on.
00:15:52.370 --> 00:15:56.000
The relationship is derived
as follows--
00:15:56.000 --> 00:16:00.580
the increment in the second
Piola-Kirchhoff stress is
00:16:00.580 --> 00:16:03.700
clearly given by the tangent
material relation times
00:16:03.700 --> 00:16:06.140
increment in the Green-Lagrange
strain.
00:16:06.140 --> 00:16:09.620
And we're looking at
differential increments here.
00:16:09.620 --> 00:16:13.180
The increment in the second
Piola-Kirchhoff stress, refer
00:16:13.180 --> 00:16:18.020
to time t, is given by the right
hand side here, where
00:16:18.020 --> 00:16:20.320
this increment in Green-Lagrange
strain is
00:16:20.320 --> 00:16:23.050
refer to time t.
00:16:23.050 --> 00:16:27.800
We also have this relationship
between the stress increments.
00:16:27.800 --> 00:16:32.020
This is the same relationship
that holds between they total
00:16:32.020 --> 00:16:35.610
second Piola-Kirchhoff stress
at time t and the Cauchy
00:16:35.610 --> 00:16:37.620
stress at time t.
00:16:37.620 --> 00:16:41.440
And we have this relationship,
which we just derived.
00:16:41.440 --> 00:16:46.190
Now, using these four equations,
we directly obtain
00:16:46.190 --> 00:16:52.430
by substitution, that this left
hand side, substituted
00:16:52.430 --> 00:16:57.720
for d0 Sij, must be equal to
this right hand side, where we
00:16:57.720 --> 00:16:59.320
have substituted this
relationship
00:16:59.320 --> 00:17:02.600
for d0 epsilon rs.
00:17:02.600 --> 00:17:07.119
And this equation, then,
directly yields this equation,
00:17:07.119 --> 00:17:11.829
where the coefficient here must
be what we are looking
00:17:11.829 --> 00:17:14.920
for, namely tCabpq.
00:17:14.920 --> 00:17:19.660
This is the tangent material
tensor to be used in the
00:17:19.660 --> 00:17:22.060
updated Lagrangian formulation,
00:17:22.060 --> 00:17:23.369
referred to time t.
00:17:26.510 --> 00:17:32.050
The tangent material relation is
once more summarized here.
00:17:32.050 --> 00:17:35.805
And this is what we need to use
in the updated Lagrangian
00:17:35.805 --> 00:17:37.055
formulation.
00:17:39.360 --> 00:17:43.840
Notice it's a fourth order
tensor transformation on this
00:17:43.840 --> 00:17:47.050
value here, on this
tensor here.
00:17:47.050 --> 00:17:50.135
Now we can compare the updated
and total Lagrangian
00:17:50.135 --> 00:17:52.240
formulations.
00:17:52.240 --> 00:17:57.580
And the question that we ask
is under what condition is
00:17:57.580 --> 00:18:00.780
this left hand side, which we
use in the total Lagrangian
00:18:00.780 --> 00:18:06.060
formulation equal to this term
here, which we use in the
00:18:06.060 --> 00:18:07.990
updated Lagrangian
formulation.
00:18:07.990 --> 00:18:15.470
Well, we simply substitute here
for the stress, and for
00:18:15.470 --> 00:18:19.490
the strain, our earlier
relationships.
00:18:19.490 --> 00:18:22.820
We multiply out, and
immediately, we obtain the
00:18:22.820 --> 00:18:24.110
right hand side.
00:18:24.110 --> 00:18:28.300
Notice ij, of course, are dummy
indices, which can be
00:18:28.300 --> 00:18:31.520
substituted by m and n.
00:18:31.520 --> 00:18:37.410
It's interesting to upset here
that this transformation is a
00:18:37.410 --> 00:18:41.190
kinematic transformation, which
is actually buried in
00:18:41.190 --> 00:18:44.280
the proper use of the updated
Lagrangian formulation.
00:18:44.280 --> 00:18:49.150
This transformation is a stress
transformation that has
00:18:49.150 --> 00:18:53.560
to be enforced for any Cauchy
stress that we would be
00:18:53.560 --> 00:18:55.720
calculating in the
computer program.
00:18:59.320 --> 00:19:02.560
Let us look at the
second term.
00:19:02.560 --> 00:19:05.600
Here, we have, on the left hand
side, the term of the
00:19:05.600 --> 00:19:07.320
total Lagrangian formulation.
00:19:07.320 --> 00:19:12.280
And we ask whether this is
equal to the term that I
00:19:12.280 --> 00:19:14.320
showed you on the right hand
side, corresponding to the
00:19:14.320 --> 00:19:15.570
updated Lagrangian
formulation.
00:19:17.280 --> 00:19:20.890
Well, if we substitute our
definitions for the second
00:19:20.890 --> 00:19:24.940
Piola-Kirchhoff stress here, and
what we know to be a true
00:19:24.940 --> 00:19:30.940
kinematic relations, here, and
we multiply out, and also use,
00:19:30.940 --> 00:19:36.600
of course, this relationship
here, we directly obtain what
00:19:36.600 --> 00:19:39.180
we know we have to obtain for
the updated Lagrangian
00:19:39.180 --> 00:19:40.390
formulation.
00:19:40.390 --> 00:19:43.070
In other words, the left
hand side is equal to
00:19:43.070 --> 00:19:44.850
the right hand side.
00:19:44.850 --> 00:19:48.940
Notice, again, the kinematic
transformation here and the
00:19:48.940 --> 00:19:52.560
stress transformation here.
00:19:52.560 --> 00:19:58.380
Finally, we also have to look
at this term, the term in
00:19:58.380 --> 00:20:00.310
which the material
tensor enters.
00:20:00.310 --> 00:20:04.220
We ask again if this left hand
side equal to the right hand
00:20:04.220 --> 00:20:07.860
side using the proper
transformations.
00:20:07.860 --> 00:20:16.200
And if we substitute, as shown
here, we directly observe yes,
00:20:16.200 --> 00:20:20.500
if the term of the total
Lagrangian formulation is
00:20:20.500 --> 00:20:22.747
equal to the term of the
updated Lagrangian
00:20:22.747 --> 00:20:26.700
formulation, provided, of
course, we use the proper
00:20:26.700 --> 00:20:28.520
transformations.
00:20:28.520 --> 00:20:32.050
And here, in particular, the
constitutive transformation
00:20:32.050 --> 00:20:35.430
that we discussed just
a bit earlier.
00:20:35.430 --> 00:20:41.115
So we conclude then that the
updated Lagrangian terms are
00:20:41.115 --> 00:20:45.240
identically equal to the total
Lagrangian terms, provided we
00:20:45.240 --> 00:20:50.020
follow the transformation rules
that I have summarized.
00:20:50.020 --> 00:20:55.640
This means that in the finite
element analysis, if we use
00:20:55.640 --> 00:21:00.660
the same finite element
interpolation functions in the
00:21:00.660 --> 00:21:03.330
updated Lagrangian formulation,
as in the total
00:21:03.330 --> 00:21:04.820
Lagrangian formulation--
00:21:04.820 --> 00:21:08.920
of course we have to use the
same interpolation functions
00:21:08.920 --> 00:21:11.610
and the same finite element
assumptions--
00:21:11.610 --> 00:21:16.815
then, we can see directly that,
for the total Lagrangian
00:21:16.815 --> 00:21:21.520
formulation with this equivalent
equation and the
00:21:21.520 --> 00:21:25.320
updated Lagrangian formulation
with this equivalent equation,
00:21:25.320 --> 00:21:29.940
that these matrices are
identically equal.
00:21:29.940 --> 00:21:34.480
And that's a very important
observation, a very important
00:21:34.480 --> 00:21:35.660
observation.
00:21:35.660 --> 00:21:39.585
It means, then, going back to
the question that I asked
00:21:39.585 --> 00:21:43.550
earlier, regarding the use of
Program 1 and Program 2, it
00:21:43.550 --> 00:21:51.240
means that, to summarize, if we
use Program 2, then Program
00:21:51.240 --> 00:21:56.740
2 gives the same results as
Program 1, provided the Cauchy
00:21:56.740 --> 00:22:01.120
stresses are calculated from
this relationship.
00:22:01.120 --> 00:22:02.570
What does that mean?
00:22:02.570 --> 00:22:05.560
It means that the second
Piola-Kirchhoff stresses of
00:22:05.560 --> 00:22:08.690
course are given by the
relationships that were
00:22:08.690 --> 00:22:13.360
determined by physical
laboratory results.
00:22:13.360 --> 00:22:18.140
Remember, we know how old the
second Piola-Kirchhoff stress
00:22:18.140 --> 00:22:21.432
is defined as a function
of deformations, of the
00:22:21.432 --> 00:22:23.070
deformations in the material.
00:22:23.070 --> 00:22:26.360
That was our assumption because
that went into the use
00:22:26.360 --> 00:22:27.810
of Program 1.
00:22:27.810 --> 00:22:31.310
And we are asking now only how
can we use Program 2, which
00:22:31.310 --> 00:22:35.190
only contains the updated
Lagrangian formulation.
00:22:35.190 --> 00:22:41.230
So we know how St0Smn
is defined.
00:22:41.230 --> 00:22:45.340
And if we know this quantity,
we have to make this
00:22:45.340 --> 00:22:49.750
transformation to calculate
the Cauchy stress
00:22:49.750 --> 00:22:52.380
in the Program 2.
00:22:52.380 --> 00:22:57.230
Similarly, we also remember,
postulated that we know this
00:22:57.230 --> 00:23:01.330
tensor from the physical
laboratory test results.
00:23:01.330 --> 00:23:02.840
We know this tensor.
00:23:02.840 --> 00:23:07.790
And what I'm saying now is that
we have to transform this
00:23:07.790 --> 00:23:13.050
tensor, as shown here, to
obtain this tensor.
00:23:13.050 --> 00:23:17.000
And it is this one that would go
then into all our Program 2
00:23:17.000 --> 00:23:18.290
computations.
00:23:18.290 --> 00:23:24.170
If, in Program 2, we use this
Cauchy stress, defined by the
00:23:24.170 --> 00:23:25.970
second Piola-Kirchhoff stress,
as shown here.
00:23:25.970 --> 00:23:29.740
And this tangent material
relation, you find by this
00:23:29.740 --> 00:23:33.510
tensor, which we know, then
Program 2 will give
00:23:33.510 --> 00:23:37.670
identically the same results
as Program 1.
00:23:37.670 --> 00:23:40.960
The kinematic relations that
we talked earlier about are
00:23:40.960 --> 00:23:44.370
buried in the proper
implementation of the updated
00:23:44.370 --> 00:23:46.030
Lagrangian formulation
in Program 2.
00:23:49.030 --> 00:23:54.410
Conversely, if we say that the
material relationship for
00:23:54.410 --> 00:23:59.550
Program 2 is given, say, in
other words, the Cauchy stress
00:23:59.550 --> 00:24:03.110
is given and also the tangent
material relationship is given
00:24:03.110 --> 00:24:06.340
from laboratory test results.
00:24:06.340 --> 00:24:12.790
Then you can surely show that
Program 1 with the total
00:24:12.790 --> 00:24:15.540
Lagrangian formulation will
give identically the same
00:24:15.540 --> 00:24:20.550
results, provided these
transformations are performed
00:24:20.550 --> 00:24:21.640
in the program.
00:24:21.640 --> 00:24:24.530
Now, the Cauchy stress is given,
the tangent material
00:24:24.530 --> 00:24:27.970
relationship is given, and
what we have to do in the
00:24:27.970 --> 00:24:31.620
program, in every solution step,
in every iteration, is
00:24:31.620 --> 00:24:34.830
to transform this Cauchy
stress to the second
00:24:34.830 --> 00:24:38.160
Piola-Kirchhoff stress that's
going to be used in Program 1.
00:24:38.160 --> 00:24:42.760
And we have to transform this
tensor here, as shown here, to
00:24:42.760 --> 00:24:43.710
this tensor.
00:24:43.710 --> 00:24:46.810
And this tensor here, 0Cijs,
is going to be
00:24:46.810 --> 00:24:48.060
used in Program 1.
00:24:51.260 --> 00:24:55.370
Therefore we can really conclude
that's the choice of
00:24:55.370 --> 00:24:58.310
whether to use a total
Lagrangian formulation or the
00:24:58.310 --> 00:25:04.660
updated Lagrangian formulation
is based only merely on how
00:25:04.660 --> 00:25:10.700
effective computationally one
formulation is over the other.
00:25:10.700 --> 00:25:13.680
And then we note now that
the B matrix, the strain
00:25:13.680 --> 00:25:18.190
displacement matrix in the U.L.
formulation contains less
00:25:18.190 --> 00:25:20.790
entries than the strain
displacement matrix in the
00:25:20.790 --> 00:25:22.240
T.L. formulation.
00:25:22.240 --> 00:25:26.070
This means, of course, that the
product, B transpose CB,
00:25:26.070 --> 00:25:30.430
which is used in the stiffness
matrix, is cheaper to
00:25:30.430 --> 00:25:33.916
evaluate, less expensive to
evaluate, in the updated
00:25:33.916 --> 00:25:34.170
Lagrangian formulation.
00:25:34.170 --> 00:25:37.490
Across So there's a plus here
for the updated Lagrangian
00:25:37.490 --> 00:25:38.740
formulation.
00:25:40.270 --> 00:25:45.730
However, if the stress-strain
law is available in terms of
00:25:45.730 --> 00:25:48.750
the second Piola-Kirchhoff
stress, then it is most
00:25:48.750 --> 00:25:51.760
natural to use a total
Lagrangian formulations.
00:25:51.760 --> 00:25:56.630
And this is the case when we
analyze rubber type materials,
00:25:56.630 --> 00:25:59.270
for example we use the
Mooney-Rivlin material law,
00:25:59.270 --> 00:26:01.120
which we discussed very briefly
in the previous
00:26:01.120 --> 00:26:06.600
lecture, and, in particular,
when we want to analyze
00:26:06.600 --> 00:26:11.910
inelastic response and we
want to allow for large
00:26:11.910 --> 00:26:14.340
displacements and
large rotations.
00:26:14.340 --> 00:26:16.860
but we have small strain
conditions.
00:26:16.860 --> 00:26:20.760
I'd like to refer you now back
to the previous lecture, where
00:26:20.760 --> 00:26:24.150
we made a big point out of
the fact that the second
00:26:24.150 --> 00:26:27.520
Piola-Kirchhoff stress, the
Green-Lagrange strains, are
00:26:27.520 --> 00:26:31.120
the components of these two
tensors are invariant under
00:26:31.120 --> 00:26:36.160
rigid body rotation, and that
that means that we can
00:26:36.160 --> 00:26:39.280
directly use material
relationships that are
00:26:39.280 --> 00:26:43.780
applicable to infinitesimal
displacement and strains, in
00:26:43.780 --> 00:26:45.420
other words, the engineering
stress and
00:26:45.420 --> 00:26:47.680
engineering strain variables.
00:26:47.680 --> 00:26:50.720
You can use these material
relations directly in a large
00:26:50.720 --> 00:26:55.330
displacement, large rotation
analysis, provided we use the
00:26:55.330 --> 00:26:57.090
total Lagrangian formulation.
00:26:57.090 --> 00:26:59.870
And I explained that, I spent a
bit of time on that, in the
00:26:59.870 --> 00:27:00.870
previous lecture.
00:27:00.870 --> 00:27:04.600
Please refer back to
that information.
00:27:04.600 --> 00:27:08.340
Let us now look at a very
special case, namely, the case
00:27:08.340 --> 00:27:10.410
of elasticity.
00:27:10.410 --> 00:27:13.200
We discussed in the previous
lecture already that one way
00:27:13.200 --> 00:27:16.990
to proceed is to use this
relationship here in the total
00:27:16.990 --> 00:27:18.910
Lagrangian formulation, where
we have the second
00:27:18.910 --> 00:27:21.460
Piola-Kirchhoff stress on
the left hand side, the
00:27:21.460 --> 00:27:24.530
Green-Lagrange strain on the
right hand side, multiplying
00:27:24.530 --> 00:27:26.730
the constitutive relation,
corresponding to the total
00:27:26.730 --> 00:27:28.630
Lagrangian formulation.
00:27:28.630 --> 00:27:33.450
Now if we use this equation in
the relation that we just
00:27:33.450 --> 00:27:36.870
established, or that we talked
about, namely giving us of a
00:27:36.870 --> 00:27:43.380
Cauchy stress, in terms of the
second Piola-Kirchhoff stress,
00:27:43.380 --> 00:27:47.260
we obtain this equation here.
00:27:47.260 --> 00:27:49.482
And of course, we also have to
use in the updated Lagrangian
00:27:49.482 --> 00:27:51.300
formulation, this
00:27:51.300 --> 00:27:54.060
transformation, as we just discussed.
00:27:54.060 --> 00:27:56.770
Now notice that when we compare
these two right hand
00:27:56.770 --> 00:28:00.380
sides, we see the deformation
gradient entering twice here,
00:28:00.380 --> 00:28:03.330
whereas four times here.
00:28:03.330 --> 00:28:07.460
We would like to recast, now,
this equation on the Cauchy
00:28:07.460 --> 00:28:11.130
stress, so as to have the
deformation gradient here
00:28:11.130 --> 00:28:15.050
entering in the same form as
we have it entering here.
00:28:15.050 --> 00:28:18.560
And that is achieved by the
definition of the Almansi
00:28:18.560 --> 00:28:20.520
strain tensor.
00:28:20.520 --> 00:28:25.420
Here, now, we have the Cauchy
stress in terms of the Almansi
00:28:25.420 --> 00:28:29.990
strain tensor, multiplying a
new constitutive tensor.
00:28:29.990 --> 00:28:34.080
And that new constitutive tensor
is given down here.
00:28:34.080 --> 00:28:38.580
Notice the right hand side here
contains the constitutive
00:28:38.580 --> 00:28:43.750
tensor of the total Lagrangian
formulation, t0 here.
00:28:43.750 --> 00:28:47.430
And we are multiplying this
tensor four times by the
00:28:47.430 --> 00:28:49.720
deformation gradient
components.
00:28:49.720 --> 00:28:52.390
This is exactly what
we try to achieve.
00:28:52.390 --> 00:28:56.450
And we achieve it by using the
Almansi strain tensor, which
00:28:56.450 --> 00:28:59.860
is defined as follows.
00:28:59.860 --> 00:29:03.560
Here we have the definition of
the Almansi strain tensor, in
00:29:03.560 --> 00:29:05.870
terms of the Green-Lagrange
strain tensor.
00:29:05.870 --> 00:29:08.070
And here, we have the
inverse deformation
00:29:08.070 --> 00:29:10.750
gradient entering twice.
00:29:10.750 --> 00:29:14.480
The same definition is
also given here,
00:29:14.480 --> 00:29:17.340
but in matrix form.
00:29:17.340 --> 00:29:20.730
And if we substitute for the
inverse deformation gradient,
00:29:20.730 --> 00:29:25.140
in terms of displacements, and
multiply components out, we
00:29:25.140 --> 00:29:28.440
obtain directly this
relationship here for the
00:29:28.440 --> 00:29:34.870
Almansi strain tensor, where,
notice our notation is this t
00:29:34.870 --> 00:29:39.402
and that t are those
two t's here.
00:29:39.402 --> 00:29:44.970
We are taking the displacement
from time 0 to time t, and we
00:29:44.970 --> 00:29:48.880
differentiating those
displacements with respect to
00:29:48.880 --> 00:29:50.880
the current coordinates.
00:29:50.880 --> 00:29:53.310
And that's important.
00:29:53.310 --> 00:29:58.570
There's also a minus here, which
is a bit of a surprise,
00:29:58.570 --> 00:30:01.040
because in the Green-Lagrange
strain definition, of course,
00:30:01.040 --> 00:30:04.000
we have a plus here, but
different terms, of course,
00:30:04.000 --> 00:30:05.740
different terms.
00:30:05.740 --> 00:30:08.610
Anyway, this is the definition
of the Almansi strain tensor.
00:30:08.610 --> 00:30:14.300
And this tensor is very useful
in the way I just described.
00:30:14.300 --> 00:30:19.510
The Almansi strain tensor
is a symmetric tensor.
00:30:19.510 --> 00:30:23.980
The ij components are equal
to the ji components.
00:30:23.980 --> 00:30:27.770
The components are not invariant
under rigid body
00:30:27.770 --> 00:30:30.220
rotation of the material.
00:30:30.220 --> 00:30:32.270
This is a property, an important
property, of the
00:30:32.270 --> 00:30:35.190
Green-Lagrange strain tensor,
but it is not a property of
00:30:35.190 --> 00:30:36.440
the Almansi strain tensor.
00:30:38.610 --> 00:30:43.140
The Almansi strain tensor is
really, overall, not a very
00:30:43.140 --> 00:30:45.430
useful measure, strain
measure.
00:30:45.430 --> 00:30:48.340
But we wanted to introduce
it here briefly.
00:30:48.340 --> 00:30:53.650
And it is quite useful in the
one particular case, namely,
00:30:53.650 --> 00:30:58.430
in the analysis of isotropic
materials undergoing large
00:30:58.430 --> 00:31:01.050
displacements and large
rotations, but
00:31:01.050 --> 00:31:03.260
generally small strains.
00:31:03.260 --> 00:31:07.330
Let's look at one simple example
here that gives us a
00:31:07.330 --> 00:31:11.110
bit of insight, what's the
Almansi strain tensor looks,
00:31:11.110 --> 00:31:13.960
what the strain components
look like.
00:31:13.960 --> 00:31:17.960
Here we have a simple four node
element that is being
00:31:17.960 --> 00:31:21.430
pulled out into the
red configuration.
00:31:21.430 --> 00:31:24.690
Notice the original
length is 0L.
00:31:24.690 --> 00:31:26.640
The pullout is t delta.
00:31:26.640 --> 00:31:29.710
Of course, t delta could also be
negative, in which case we
00:31:29.710 --> 00:31:31.130
are pushing in here.
00:31:31.130 --> 00:31:34.790
And the current length is tL.
00:31:34.790 --> 00:31:40.930
Notice the Green-Lagrange strain
is given here, where we
00:31:40.930 --> 00:31:46.700
have a plus sign here and where
we have 0L down here.
00:31:46.700 --> 00:31:51.000
The Almansi strain 1, 1
component, of course, in this
00:31:51.000 --> 00:31:55.500
particular case, has this
relationship here.
00:31:55.500 --> 00:31:59.990
Notice t [? sigma ?] over
tL, versus 0L in the
00:31:59.990 --> 00:32:01.760
Green-Lagrange strain.
00:32:01.760 --> 00:32:05.830
And a minus sign here, whereas
we have a plus sign there.
00:32:05.830 --> 00:32:11.600
If we plot these strain tensor
components as a function of t
00:32:11.600 --> 00:32:16.740
delta over the original lengths
of the element, we
00:32:16.740 --> 00:32:20.610
find this blue curve for the
Green-Lagrange strain.
00:32:20.610 --> 00:32:24.900
That is very easy to see
that this is the
00:32:24.900 --> 00:32:25.770
Green-Lagrange strain.
00:32:25.770 --> 00:32:29.950
Notice here we have
1/2 minus 1/2.
00:32:29.950 --> 00:32:33.280
That minus 1/2 comes from
this relationship.
00:32:33.280 --> 00:32:36.520
This is minus one
at this point.
00:32:36.520 --> 00:32:38.440
And that is plus 1/2.
00:32:38.440 --> 00:32:40.380
So we get a minus 1/2 here.
00:32:40.380 --> 00:32:45.940
And at this point here, when
delta over L0 is one, we have
00:32:45.940 --> 00:32:49.730
3/2, verified by that formula.
00:32:49.730 --> 00:32:51.790
The engineering strain,
of course, is
00:32:51.790 --> 00:32:54.440
just a straight line.
00:32:54.440 --> 00:32:59.390
The Almansi strain looks as
shown here by the red curve.
00:32:59.390 --> 00:33:06.810
Notice that when delta is
compressive, going into this
00:33:06.810 --> 00:33:09.200
direction in the picture, in
other words, when we are
00:33:09.200 --> 00:33:12.810
compressing that element, of
course, tL becomes smaller and
00:33:12.810 --> 00:33:16.900
smaller, meaning this variable
down here becomes smaller and
00:33:16.900 --> 00:33:21.770
smaller, and our Almansi strain
component very rapidly
00:33:21.770 --> 00:33:23.195
becomes very large negative.
00:33:26.910 --> 00:33:32.470
It turns out that the use of
the Almansi strain tensor,
00:33:32.470 --> 00:33:36.832
corresponding with this material
relationship, t t
00:33:36.832 --> 00:33:42.750
Caijrs, which I defined on the
earlier view graph, is quite
00:33:42.750 --> 00:33:47.220
effective when we want to
analyze with the U.L.
00:33:47.220 --> 00:33:51.170
Formulation situations that
involve a linear isotropic
00:33:51.170 --> 00:33:54.300
material and large
displacements, large
00:33:54.300 --> 00:33:57.590
rotations, but small strains.
00:33:57.590 --> 00:34:05.040
In this case, we can use
directly that t t Caijrs is
00:34:05.040 --> 00:34:08.429
given by the right
hand side here.
00:34:08.429 --> 00:34:12.550
Notice here we have the Lame
constants, lambda and mu,
00:34:12.550 --> 00:34:15.139
which we already used in the
previous lecture, and the
00:34:15.139 --> 00:34:16.389
chronica deltas.
00:34:18.949 --> 00:34:24.080
The t 0 Cijrs tensor
is given as here on
00:34:24.080 --> 00:34:26.889
the right hand side.
00:34:26.889 --> 00:34:32.460
Again, Lame constants and
chronica delta entries.
00:34:32.460 --> 00:34:38.000
We would use here the same
Lame constants in both of
00:34:38.000 --> 00:34:40.600
these relationships.
00:34:40.600 --> 00:34:44.520
And we also use that the
incremental, the tangent
00:34:44.520 --> 00:34:48.250
material relationship, is
the same as the total
00:34:48.250 --> 00:34:52.199
stress-strain relationship
in both cases.
00:34:52.199 --> 00:35:00.860
If we use reformulating, if we
use this right hand side for
00:35:00.860 --> 00:35:06.220
that tensor and make this tensor
equal to that tensor,
00:35:06.220 --> 00:35:09.990
and if we use the same right
hand side as up there, down
00:35:09.990 --> 00:35:19.190
here, for the t 0 Cijrs and the
0Cijrs, same numbers, same
00:35:19.190 --> 00:35:22.740
Young's modulus, same Poisson
ratio, then for large
00:35:22.740 --> 00:35:26.620
displacement, large rotation,
but small strain analysis, we
00:35:26.620 --> 00:35:30.450
virtually obtain identically
the same results.
00:35:30.450 --> 00:35:35.030
I'd like to just demonstrate
to you with a very simple
00:35:35.030 --> 00:35:38.590
problem analysis what
I mean here.
00:35:38.590 --> 00:35:45.670
Let us turn to just one slide
in which we show a solution
00:35:45.670 --> 00:35:50.130
response of an arch that was
analyzed using the updated and
00:35:50.130 --> 00:35:51.380
total Lagrangian formulations.
00:35:53.730 --> 00:35:57.170
Here you now see the slide
that shows the arch.
00:35:57.170 --> 00:36:01.470
It is subjected to a point
load at it's apex.
00:36:01.470 --> 00:36:04.860
And we measure the displacement
at the apex.
00:36:04.860 --> 00:36:08.780
The displacement is w0, half of
the arch was modeled with
00:36:08.780 --> 00:36:09.930
12 eight-node elements.
00:36:09.930 --> 00:36:16.590
And as you can see, the TL and
UL solutions for this arch are
00:36:16.590 --> 00:36:18.670
practically, identically
the same.
00:36:18.670 --> 00:36:22.620
We can't see any difference to
the accuracy that we have
00:36:22.620 --> 00:36:24.530
plotted the response.
00:36:24.530 --> 00:36:28.610
Notice that in these TL and UL
solutions, we used exactly the
00:36:28.610 --> 00:36:30.850
same Young's modulus
and Poisson ratio.
00:36:33.450 --> 00:36:35.660
The reason that practically
the same response is
00:36:35.660 --> 00:36:39.190
calculated using the updated
Lagrangian formulation and the
00:36:39.190 --> 00:36:44.460
total Lagrangian formulation
lies in that the constitutive
00:36:44.460 --> 00:36:49.090
transformations that would have
to be applied to obtain
00:36:49.090 --> 00:36:52.870
the exact response, exactly the
same response with a two
00:36:52.870 --> 00:36:57.380
formations, really reduce
here to mere rotations.
00:36:57.380 --> 00:37:02.470
And that is the fact because,
for this type of problem,
00:37:02.470 --> 00:37:05.440
large displacement, large
rotation, but small strain
00:37:05.440 --> 00:37:10.090
problem, the mass density
remains constant and the
00:37:10.090 --> 00:37:13.230
deformation gradient, which
can always be written as a
00:37:13.230 --> 00:37:17.320
product all of a rotation
matrix, an orthogonal matrix,
00:37:17.320 --> 00:37:22.160
times another symmetric
matrix.
00:37:22.160 --> 00:37:24.930
This is, of course, here is a
polar decomposition of the
00:37:24.930 --> 00:37:28.690
deformation gradient, that for
this particular case, large
00:37:28.690 --> 00:37:32.080
displacement, large rotation,
but small strains, this
00:37:32.080 --> 00:37:36.090
deformation gradient is simply
almost equal to the rotation
00:37:36.090 --> 00:37:37.470
matrix only.
00:37:37.470 --> 00:37:41.210
In other words, the stretch
matrix is almost equal to the
00:37:41.210 --> 00:37:43.570
identity matrix.
00:37:43.570 --> 00:37:48.960
However, when we look at large
strain problems, large strain
00:37:48.960 --> 00:37:55.810
analysis, and we were to use
this relationship here for the
00:37:55.810 --> 00:37:59.660
Cauchy stress, in terms of the
Almansi strain, and this
00:37:59.660 --> 00:38:03.720
relationship here, giving us a
second Piola-Kirchhoff stress
00:38:03.720 --> 00:38:07.700
in terms of the Green-Lagrange
strain, with these two
00:38:07.700 --> 00:38:12.540
constitutive tensors, given
as shown here by
00:38:12.540 --> 00:38:15.190
the same Lame constants.
00:38:15.190 --> 00:38:18.640
In other words, Poisson ratio
and Young's modulus, being
00:38:18.640 --> 00:38:21.960
identically the same,
entering here.
00:38:21.960 --> 00:38:25.530
More specifically, you would put
in here a Young's modulus,
00:38:25.530 --> 00:38:29.490
say, of 30 million pounds per
square inch and the Poisson
00:38:29.490 --> 00:38:31.230
ratio of 0.3.
00:38:31.230 --> 00:38:34.260
These are the exact numbers that
would go into here, with
00:38:34.260 --> 00:38:37.300
these, of course, being
the chronica deltas.
00:38:37.300 --> 00:38:40.620
If we were to use this
relationship in the total
00:38:40.620 --> 00:38:43.540
Lagrangian formulation and
that relationship in the
00:38:43.540 --> 00:38:47.150
updated Lagrangian formulation
for large strains, then we
00:38:47.150 --> 00:38:51.070
would get very different
results.
00:38:51.070 --> 00:38:55.450
Let us look at the simple
problem that we already solved
00:38:55.450 --> 00:38:58.950
earlier ones, or that we
considered earlier ones in the
00:38:58.950 --> 00:39:02.230
previous lecture, using the
total Lagrangian formulation.
00:39:02.230 --> 00:39:07.513
I'd like to now consider the
same problem using the updated
00:39:07.513 --> 00:39:11.990
Lagrangian formulation with this
material relationship.
00:39:11.990 --> 00:39:16.595
Cauchy stress given in terms
of the Almansi strain, and
00:39:16.595 --> 00:39:21.870
this E curl, which we used
already before in the last
00:39:21.870 --> 00:39:24.890
lecture in this total Lagrangian
formulation.
00:39:24.890 --> 00:39:28.400
Notice this E curl here is given
on the right hand side
00:39:28.400 --> 00:39:33.670
here in terms of Young's modulus
and Poisson's ratio.
00:39:33.670 --> 00:39:37.890
The problem, very briefly
reviewed, is that we are
00:39:37.890 --> 00:39:45.000
looking at a bar with constant
cross sectional area A bar.
00:39:45.000 --> 00:39:49.790
And we will pull this bar out
and also compress it in and
00:39:49.790 --> 00:39:52.595
look at the force required
to pull the bar
00:39:52.595 --> 00:39:54.900
out or compress it.
00:39:54.900 --> 00:39:57.920
And we want to now look at
this problem using this
00:39:57.920 --> 00:40:00.890
material relationship here,
Cauchy stress in terms of
00:40:00.890 --> 00:40:05.450
Almansi strain with
E curl constant.
00:40:05.450 --> 00:40:10.040
Here we have, for this one
dimensional problem, the
00:40:10.040 --> 00:40:14.950
Almansi strain, given
as shown up here.
00:40:14.950 --> 00:40:20.880
Notice this differentiation here
of tU1 with respect to
00:40:20.880 --> 00:40:23.430
the coordinate 1 at time t.
00:40:23.430 --> 00:40:27.950
It's nothing else than
tL minus 0L over tL.
00:40:27.950 --> 00:40:30.570
Here we have that same
quantity squared.
00:40:30.570 --> 00:40:32.500
Of course, there's a
minus 1/2 in front.
00:40:32.500 --> 00:40:34.430
And the result is this.
00:40:34.430 --> 00:40:37.930
The Cauchy stress is directly
obtained in terms of the
00:40:37.930 --> 00:40:41.790
physical force applied to the
bar divided by the cross
00:40:41.790 --> 00:40:44.210
sectional area, which
is constant.
00:40:44.210 --> 00:40:47.340
If we use that current length
is given in terms of the
00:40:47.340 --> 00:40:51.200
original length plus the
extension and this material
00:40:51.200 --> 00:40:55.880
relationship here, we obtain
directly this force
00:40:55.880 --> 00:40:57.750
displacement response.
00:40:57.750 --> 00:41:01.370
Force plotted vertically up,
displacement plotted along
00:41:01.370 --> 00:41:05.570
here, t sigma is the
displacement.
00:41:05.570 --> 00:41:09.200
And notice that the force
displacement response is
00:41:09.200 --> 00:41:11.540
highly nonlinear--
00:41:11.540 --> 00:41:14.430
here is, by the way, the
actual expression--
00:41:14.430 --> 00:41:17.880
highly nonlinear, and looks
quite different from what we
00:41:17.880 --> 00:41:21.020
calculated earlier using the
total Lagrangian formulation
00:41:21.020 --> 00:41:25.063
with the same E curl, same E
curl used earlier in the total
00:41:25.063 --> 00:41:29.570
Lagrangian formulation, in which
case the response really
00:41:29.570 --> 00:41:32.120
looked something like this.
00:41:35.670 --> 00:41:39.270
So we get a totally different
description of the material
00:41:39.270 --> 00:41:43.590
response using these two
formulation when the material
00:41:43.590 --> 00:41:46.790
is subjected to large strains.
00:41:46.790 --> 00:41:52.250
Let us now also look at one
example that demonstrates this
00:41:52.250 --> 00:41:55.920
feature, the features that I
just discussed a bit more.
00:41:55.920 --> 00:42:01.270
Here we have a frame subjected
to a tip load
00:42:01.270 --> 00:42:03.090
over at this end.
00:42:03.090 --> 00:42:04.690
A is the tip load.
00:42:04.690 --> 00:42:09.500
The frame has a thickness h, L
length here, L length there,
00:42:09.500 --> 00:42:11.680
thickness h here, as well.
00:42:11.680 --> 00:42:15.440
The width of the frame
structure is b.
00:42:15.440 --> 00:42:19.570
And the geometric data
are given here.
00:42:19.570 --> 00:42:22.780
And the material data
are given here.
00:42:22.780 --> 00:42:25.870
Notice that h/L is 1/50.
00:42:28.960 --> 00:42:34.170
We modeled this structural using
51 two-dimensional eight
00:42:34.170 --> 00:42:38.700
node elements, plane
strain elements.
00:42:38.700 --> 00:42:41.990
Notice here is a typical
element shown,
00:42:41.990 --> 00:42:43.150
an eight node element.
00:42:43.150 --> 00:42:46.870
There are 25 elements
along here.
00:42:46.870 --> 00:42:50.090
There's one element there for
the corner and another 25
00:42:50.090 --> 00:42:53.680
elements in the column.
00:42:53.680 --> 00:43:00.680
Now we want to ask the question
what happens in the
00:43:00.680 --> 00:43:05.680
elastic analysis when we subject
this structure, in
00:43:05.680 --> 00:43:11.440
other words, to the load r and
we used once a TL and once a
00:43:11.440 --> 00:43:16.380
UL formulation, but with the
same material constants.
00:43:16.380 --> 00:43:19.530
In other words, the material
constants E and nu that I gave
00:43:19.530 --> 00:43:23.780
on the earlier view graph are
the same for the TL solution
00:43:23.780 --> 00:43:26.030
and the UL solution.
00:43:26.030 --> 00:43:30.040
And we do not make any
transformation, therefore, on
00:43:30.040 --> 00:43:31.240
the material relationships.
00:43:31.240 --> 00:43:34.380
We simply use the total
Lagrangian formulation with E
00:43:34.380 --> 00:43:36.630
and nu given, these two
constants plugged in.
00:43:36.630 --> 00:43:38.600
And these constants
are the same
00:43:38.600 --> 00:43:41.000
throughout the TL solution.
00:43:41.000 --> 00:43:43.380
We perform this way
the TL solution.
00:43:43.380 --> 00:43:47.500
And then, afterwards, we do the
UL solution, again, simply
00:43:47.500 --> 00:43:50.670
putting E and nu into the
analysis and keeping it
00:43:50.670 --> 00:43:54.580
constant throughout
that analysis.
00:43:54.580 --> 00:43:57.880
For large displacement, large
rotation, but small strain
00:43:57.880 --> 00:44:01.570
conditions, the TL and UL
formulation will give similar
00:44:01.570 --> 00:44:06.410
results, similar because
we don't make these
00:44:06.410 --> 00:44:08.810
transformations that I discussed
earlier in the
00:44:08.810 --> 00:44:13.310
lecture in order to obtain
exactly the same results.
00:44:13.310 --> 00:44:16.920
But for large displacement,
large rotation, small strain
00:44:16.920 --> 00:44:19.850
conditions, similar results
are obtained.
00:44:19.850 --> 00:44:24.470
Of course, we have shown that
already, by the solution that
00:44:24.470 --> 00:44:27.990
I showed you on the slide, the
analysis of the arch subjected
00:44:27.990 --> 00:44:29.000
to an apex load.
00:44:29.000 --> 00:44:32.530
But we will see that now again
here in this example.
00:44:32.530 --> 00:44:35.060
For large displacement, large
rotation, and large strain
00:44:35.060 --> 00:44:39.140
conditions, the TL and UL
formulations will gave quite
00:44:39.140 --> 00:44:41.600
different results.
00:44:41.600 --> 00:44:44.760
Well, let's look now at the
results that we obtained for
00:44:44.760 --> 00:44:45.920
this problem.
00:44:45.920 --> 00:44:50.650
Here we have plotted the force
vertically, meganewtons, and
00:44:50.650 --> 00:44:55.110
the vertical displacement of
the tip in terms of meter.
00:44:55.110 --> 00:45:01.770
The 2-D elements, TL and UL
formulations, give this result
00:45:01.770 --> 00:45:06.020
here, the black curve,
solid curve.
00:45:06.020 --> 00:45:09.020
And we also wanted to solve this
problem once using beam
00:45:09.020 --> 00:45:12.790
elements using the total
Lagrangian formulation.
00:45:12.790 --> 00:45:16.740
In fact, we used four node
isoparametric beam elements.
00:45:16.740 --> 00:45:18.270
We have not discussed
these yet.
00:45:18.270 --> 00:45:20.810
We will discuss the formulation
of these elements
00:45:20.810 --> 00:45:22.730
in a later lecture.
00:45:22.730 --> 00:45:25.800
And you can see that the beam
elements give a slightly
00:45:25.800 --> 00:45:28.830
different response solution.
00:45:28.830 --> 00:45:31.160
Of course, the assumptions, the
kinematic assumptions, in
00:45:31.160 --> 00:45:32.700
the beam elements
are different.
00:45:32.700 --> 00:45:35.500
And that explains the difference
in the response
00:45:35.500 --> 00:45:36.620
calculated.
00:45:36.620 --> 00:45:39.770
But the TL and UL formulations,
using the eight
00:45:39.770 --> 00:45:44.350
node isoparametric plane, the
strain elements, we obtain
00:45:44.350 --> 00:45:45.685
basically the same response.
00:45:48.670 --> 00:45:52.540
The formations are very
large for this frame.
00:45:52.540 --> 00:45:55.630
Notice here we have the
undeformed frame.
00:45:55.630 --> 00:45:57.190
There's the load.
00:45:57.190 --> 00:46:00.390
We are pressing down here.
00:46:00.390 --> 00:46:04.060
These are the deformations at
a load of one meganewton.
00:46:04.060 --> 00:46:07.690
And this is how the frame
looks at full load, five
00:46:07.690 --> 00:46:08.750
meganewton--
00:46:08.750 --> 00:46:11.100
very large deformations.
00:46:11.100 --> 00:46:15.100
In fact, the deformation are so
large for the frame, that
00:46:15.100 --> 00:46:19.270
in an actual practical problem,
probably the frame,
00:46:19.270 --> 00:46:21.180
of course, would have
undergone inelastic
00:46:21.180 --> 00:46:22.890
deformations.
00:46:22.890 --> 00:46:24.820
However, we look at this
problem as a numerical
00:46:24.820 --> 00:46:28.600
experiment just to demonstrate
what is happening and what we
00:46:28.600 --> 00:46:31.400
have been discussing
in the lecture.
00:46:31.400 --> 00:46:36.930
If you look at the maximum
deformations once more, a
00:46:36.930 --> 00:46:39.690
little bit closer, we find
that the vertical tip
00:46:39.690 --> 00:46:43.090
displacement in the TL
formulation for this problem
00:46:43.090 --> 00:46:48.330
is given as 15.289 meters, and
in the UL formulation, as
00:46:48.330 --> 00:46:50.200
15.282 meters.
00:46:50.200 --> 00:46:55.690
Notice that there is only a
change in the fifth digit.
00:46:55.690 --> 00:46:58.480
The displacements and rotations
have certainly been
00:46:58.480 --> 00:47:02.380
very large in the analysis of
this frame, but the strains
00:47:02.380 --> 00:47:04.400
are still quite small.
00:47:04.400 --> 00:47:08.920
If you look at the strain at the
base of the frame, maybe
00:47:08.920 --> 00:47:12.890
we have the maximum strain,
maximal moments there, we find
00:47:12.890 --> 00:47:17.390
that the moment is approximately
as shown here.
00:47:17.390 --> 00:47:20.800
And the strain, using strength
of material formulas, is
00:47:20.800 --> 00:47:23.390
approximately given by 3%.
00:47:23.390 --> 00:47:26.950
Now up to 2%, you certainly
would consider it a small
00:47:26.950 --> 00:47:27.970
strain problem.
00:47:27.970 --> 00:47:33.150
So here we are just at the limit
of going over into a lot
00:47:33.150 --> 00:47:37.080
strain region near the
base of the column.
00:47:37.080 --> 00:47:39.680
This is all I wanted to present
to you, discuss with
00:47:39.680 --> 00:47:40.790
you, in this lecture.
00:47:40.790 --> 00:47:42.340
Thank you very much for
your attention.