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PROFESSOR: Ladies and gentlemen,
welcome to this

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lecture on nonlinear finite
element analysis of solids and

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structures.

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In this lecture I'd like
to discuss with you

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some example solutions.

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In the last lectures, we
discussed quite a bit of

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theory related to finite element
formulations and

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numerical algorithms, and I
thought that you might like

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now to share with me
some experiences

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00:00:42,900 --> 00:00:45,090
regarding example solutions.

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The examples that I'd like to
discuss with you are first,

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00:00:49,550 --> 00:00:53,720
the analysis of a truss arch,
the snap-through analysis of a

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truss arch.

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Then the collapse analysis of
an elasto-plastic cylinder.

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Then the large displacement
solution of a spherical shell.

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00:01:05,470 --> 00:01:08,770
Then we want to look at the
analysis of a cantilever, the

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00:01:08,770 --> 00:01:11,850
large displacement analysis
of a cantilever when that

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structure is subjected
to pressure loading.

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00:01:14,370 --> 00:01:17,780
Once deformation independent
pressure loading, and once

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00:01:17,780 --> 00:01:20,280
deformation dependent loading.

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00:01:20,280 --> 00:01:24,740
And then we would like to look
at the analysis of a

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00:01:24,740 --> 00:01:27,440
diamond-shaped frame.

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00:01:27,440 --> 00:01:30,330
And finally, I'd like to discuss
with you the failure

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00:01:30,330 --> 00:01:34,670
and repair of a beam/cable
structure.

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00:01:34,670 --> 00:01:38,690
So let me know walk over here to
some view graphs which give

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00:01:38,690 --> 00:01:42,840
us the information regarding
all the analyses.

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00:01:42,840 --> 00:01:46,540
The first example you're looking
at is a simple truss

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00:01:46,540 --> 00:01:50,090
structure which is shown
here that is

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00:01:50,090 --> 00:01:52,480
subjected to a load r here.

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00:01:52,480 --> 00:01:56,350
Notice this is a two degree
of freedom system only.

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00:01:56,350 --> 00:02:00,780
Actually, if you use symmetry
conditions you would have a

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00:02:00,780 --> 00:02:02,380
one degree of freedom system.

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00:02:02,380 --> 00:02:05,332
Notice the dimension
l is equal to 10.

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00:02:05,332 --> 00:02:09,570
k, the stiffness, ea over
l, is given here.

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00:02:09,570 --> 00:02:12,880
You want to perform a
post-buckling analysis of the

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00:02:12,880 --> 00:02:15,500
structure using the automatic
load step incrementation

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00:02:15,500 --> 00:02:19,420
algorithm, and then perform a
linearized buckling analysis.

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00:02:19,420 --> 00:02:23,970
Notice that the two solution
algorithms that we are using

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00:02:23,970 --> 00:02:26,560
to perform these two analyses
were discussed in

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00:02:26,560 --> 00:02:27,790
the previous lecture.

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00:02:27,790 --> 00:02:32,410
I assume in this lecture that
you're quite familiar with

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00:02:32,410 --> 00:02:35,640
what I have been discussing,
particularly in the previous

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00:02:35,640 --> 00:02:38,290
two lectures.

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00:02:38,290 --> 00:02:41,080
Let us start then with the
post-buckling analysis.

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Analytical solution to this
problem is given here, and

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00:02:45,200 --> 00:02:48,020
actually, this solution is also
given in the textbook.

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So please refer to the textbook
for the derivation of

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this equation.

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The response can be plotted, and
here you see is a plot of

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the load r as a function of
the displacement data.

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Notice this, of course, is a
critical point, a limit point.

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And the difficulty particularly
is to march up

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here the solution algorithms,
march across that limit point,

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00:03:12,610 --> 00:03:15,620
and then down here, and
further up here.

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The automatic load step
incrementation procedure that

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we discussed was employed with
this starting displacement.

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One data is, of course, equal to
1u, and we set it equal to

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plus 0.1, and the solution
obtained is given here.

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Notice this is a finite element
solution, which of

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course is very close to the
analytical solution.

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The solution details for load
step seven are given in this

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table here.

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Let's look at this table
now a bit closer.

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At the beginning of that load
step we have for the

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displacement, this value, and
for the load, that value.

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As we iterate, i, of
course, increases.

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We obtain this solution for
the displacement, this

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00:04:12,360 --> 00:04:18,839
solution here for the load, and
these are here, the actual

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load applied, corresponding to
t plus delta t lambda i, in

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other words.

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These are the incremental
displacements, and these are

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the incremental load.

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Now, as we discussed in the
previous lecture, in the

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algorithm using the constant
arc length, we see that the

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displacement here increases.

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The total displacement
increases, of course.

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The total load here actually
decreases.

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The incremental displacement
for this step, as

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you can see, increases.

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The incremental load
decreases.

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00:05:00,840 --> 00:05:04,650
And that gives us, of course,
this behavior that we talked

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about, namely that the
displacement and load

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00:05:07,440 --> 00:05:10,140
increments lie on this arc.

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00:05:10,140 --> 00:05:14,300
That's why the constant
arc length algorithm.

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00:05:14,300 --> 00:05:18,060
Pictorially, for this load step
we can see the following.

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00:05:18,060 --> 00:05:22,260
Notice here is our equilibrium
configuration at which we are

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00:05:22,260 --> 00:05:24,480
starting with iteration.

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00:05:24,480 --> 00:05:28,300
Notice the final load increment
that we want to

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00:05:28,300 --> 00:05:31,620
obtain is given by lambda r.

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00:05:31,620 --> 00:05:35,240
Notice the final displacement
increment that we want to

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00:05:35,240 --> 00:05:36,970
obtain is u.

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00:05:36,970 --> 00:05:39,820
And we're iterating towards
this u, and towards that

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00:05:39,820 --> 00:05:44,830
lambda r, as shown
further here.

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00:05:44,830 --> 00:05:48,460
In the first iteration
we get lambda 1 of r,

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and lambda and u1.

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00:05:51,920 --> 00:05:54,210
That gives us this point.

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In the second iteration
we get that point.

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00:05:57,490 --> 00:05:59,870
In this third iteration
we get that point.

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00:05:59,870 --> 00:06:04,680
And after the completion of the
iteration, at convergence

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00:06:04,680 --> 00:06:06,310
we end up on that point.

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00:06:06,310 --> 00:06:12,040
Notice that if you take this
point here and connect it with

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00:06:12,040 --> 00:06:16,040
the iteration i point, you would
get an arc around here.

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00:06:16,040 --> 00:06:20,280
And that is, of course, a
constant arc length algorithm.

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00:06:20,280 --> 00:06:23,490
Also please, I should point out
here that we have broken

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the scale as shown down here.

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00:06:26,860 --> 00:06:28,930
So we are starting here
already with 0.7

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00:06:28,930 --> 00:06:33,070
and here with 13.000.

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00:06:33,070 --> 00:06:37,030
The solution details for load
step here, load step eight,

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00:06:37,030 --> 00:06:38,860
are as follows.

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00:06:38,860 --> 00:06:42,610
We used, for this particular
load step, the increment of

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00:06:42,610 --> 00:06:45,070
external work algorithm.

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00:06:45,070 --> 00:06:47,480
The modified Newton iteration,
as well.

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00:06:47,480 --> 00:06:50,710
And these were the starting
values now.

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00:06:50,710 --> 00:06:53,930
And notice that we have
immediate convergence.

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00:06:53,930 --> 00:06:56,660
I pointed out in the previous
lecture that for a single

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00:06:56,660 --> 00:07:00,650
degree of freedom with the
constant increment of external

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work algorithm, we will
have convergence

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00:07:04,660 --> 00:07:06,520
after the second iteration.

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00:07:06,520 --> 00:07:12,100
In other words, the effective
delta u2 is going to be 0.

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00:07:12,100 --> 00:07:18,240
And that is shown right here in
the table, and also in this

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picture here.

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00:07:19,570 --> 00:07:23,630
Notice here is load
step seven.

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00:07:23,630 --> 00:07:29,700
The equilibrium point at which
we are starting off now, we

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want to of course walk
along this red curve.

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00:07:34,090 --> 00:07:38,890
Notice in the first iteration we
come to this point c here.

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00:07:38,890 --> 00:07:40,410
i is equal to 1.

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00:07:40,410 --> 00:07:44,560
And then immediately we shoot
down in the second iteration

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to the actual red curve.

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So we immediately converge, as
pointed out in the earlier

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00:07:53,750 --> 00:07:56,300
lecture, and on the
earlier u graph.

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The u displacement is equal to
u of 1 because delta u2 is

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equal to zero.

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00:08:06,710 --> 00:08:09,970
Let us now go and perform
linearized buckling analysis

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00:08:09,970 --> 00:08:13,420
for the same problem to estimate
the collapse load for

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00:08:13,420 --> 00:08:14,610
the truss arch.

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00:08:14,610 --> 00:08:18,060
We talked about the algorithm
that we're using for this

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00:08:18,060 --> 00:08:21,320
linearized buckling analysis
in the previous lecture.

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00:08:21,320 --> 00:08:24,830
Here we have the actual load
displacement curve.

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00:08:24,830 --> 00:08:29,160
This is the collapse load that
we would like to estimate in

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00:08:29,160 --> 00:08:31,720
our linearized buckling
analysis.

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00:08:31,720 --> 00:08:38,150
Now, if we use delta t of r
1,000, our predictive collapse

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load is 25,600.

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00:08:42,120 --> 00:08:46,130
Of course, our starting value,
our t minus data

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tk is equal to 0k.

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00:08:49,580 --> 00:08:51,920
I'm now referring, in other
words, to the view graph that

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00:08:51,920 --> 00:08:55,030
we saw in the earlier lecture,
and the t minus delta t of k

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00:08:55,030 --> 00:08:58,120
there is equal to 0k.

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00:08:58,120 --> 00:09:03,890
The delta, the t of k is equal
to data t of k now.

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00:09:03,890 --> 00:09:08,590
And delta t of r corresponding
to delta t of k

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00:09:08,590 --> 00:09:10,290
is given right here.

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Notice that this number is much
larger than the number

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that we want to estimate.

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00:09:16,170 --> 00:09:19,690
The reason being that the
pre-buckling displacements are

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not small in this
particular case.

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00:09:21,710 --> 00:09:24,640
And I pointed out in the earlier
lecture that that is

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00:09:24,640 --> 00:09:28,180
really necessary, the necessary
condition to obtain

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good results, acceptable
results, in a linearized

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buckling analysis.

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00:09:32,960 --> 00:09:34,910
It is not the case here,
and that's why we are

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00:09:34,910 --> 00:09:38,540
overestimating the buckling
load, the collapse load, by

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00:09:38,540 --> 00:09:40,190
quite a bit.

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00:09:40,190 --> 00:09:45,196
If we, however, increase our
delta t value such delta t of

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00:09:45,196 --> 00:09:52,600
r is 10,000, then you would
see that our predictive

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00:09:52,600 --> 00:09:58,940
collapse load is decreasing, and
gets closer to the value

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00:09:58,940 --> 00:10:00,850
that we want to obtain.

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00:10:00,850 --> 00:10:05,180
As we increase our delta t of
r value, and that means of

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00:10:05,180 --> 00:10:07,930
course that we are getting
closer and closer to the

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00:10:07,930 --> 00:10:12,480
collapse load by this value
here, we find that we are

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00:10:12,480 --> 00:10:16,010
getting, in our prediction for
the linearized buckling load,

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00:10:16,010 --> 00:10:20,530
closer and closer to the value
that we want to obtain.

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00:10:20,530 --> 00:10:22,210
These values here, by
the way, have been

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00:10:22,210 --> 00:10:24,360
given to three digits.

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00:10:24,360 --> 00:10:27,670
You notice that 15,000 certainly
is very close to

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00:10:27,670 --> 00:10:32,550
this value, but notice that the
delta t of r value is very

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00:10:32,550 --> 00:10:35,420
close to the buckling
load itself, to the

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00:10:35,420 --> 00:10:37,560
collapse load itself.

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00:10:37,560 --> 00:10:40,670
This is something
to keep in mind.

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00:10:40,670 --> 00:10:44,320
The basic conclusion here
is that the pre-bucking

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00:10:44,320 --> 00:10:49,090
displacements should be small in
the analysis in order to be

187
00:10:49,090 --> 00:10:52,785
able to obtain good results.

188
00:10:52,785 --> 00:10:55,890
By good I mean that the
linearized buckling load will

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00:10:55,890 --> 00:10:57,400
be close to the actual

190
00:10:57,400 --> 00:11:00,120
collapse load of the structure.

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00:11:00,120 --> 00:11:03,860
Let's look at an extension
off this example.

192
00:11:03,860 --> 00:11:06,510
And that extension is given
on this view graph.

193
00:11:06,510 --> 00:11:10,340
Here we have still our truss
structure, but now we

194
00:11:10,340 --> 00:11:12,780
introduce here this spring.

195
00:11:12,780 --> 00:11:14,720
Let's called that
spring k star.

196
00:11:14,720 --> 00:11:17,670
And this is the value that
we give to that spring.

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00:11:17,670 --> 00:11:20,920
If we trace out the load
displacement response, that

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00:11:20,920 --> 00:11:23,950
load displacement response
is shown here.

199
00:11:23,950 --> 00:11:24,890
Load vertically.

200
00:11:24,890 --> 00:11:26,770
Displacement horizontally.

201
00:11:26,770 --> 00:11:30,380
And we notice that our stiffness
matrix is always

202
00:11:30,380 --> 00:11:33,680
positive definite in
all of this region.

203
00:11:33,680 --> 00:11:38,730
So we don't really have a
singular matrix, and we don't

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00:11:38,730 --> 00:11:43,700
have what we might call
a true collapse load.

205
00:11:43,700 --> 00:11:46,660
The structure is stable
in some sense.

206
00:11:46,660 --> 00:11:50,140
Of course when we talk about a
stable structure, we really

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00:11:50,140 --> 00:11:53,750
have to also define what
we mean by stability.

208
00:11:53,750 --> 00:11:57,210
In a practical analysis you
might say, well, this

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00:11:57,210 --> 00:12:02,030
displacement is just too much,
and therefore I already

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00:12:02,030 --> 00:12:04,890
consider this structure
to be unstable at

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00:12:04,890 --> 00:12:06,920
this particular location.

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00:12:06,920 --> 00:12:08,730
You might want to say
that, depending

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00:12:08,730 --> 00:12:10,800
on your design criteria.

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00:12:10,800 --> 00:12:14,820
However, one thing to keep
in mind is that, in this

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00:12:14,820 --> 00:12:17,490
particular example,
we do not have a

216
00:12:17,490 --> 00:12:19,030
single stiffness matrix.

217
00:12:19,030 --> 00:12:24,770
The matrix stays positive all
the way as we march up on that

218
00:12:24,770 --> 00:12:27,060
load displacement response.

219
00:12:27,060 --> 00:12:29,950
Let's perform a linearized
buckling analysis of this

220
00:12:29,950 --> 00:12:31,190
structure now.

221
00:12:31,190 --> 00:12:35,280
And when we take delta t
of r equal to 10,000--

222
00:12:35,280 --> 00:12:39,130
in other words, down here,
our delta t of r--

223
00:12:39,130 --> 00:12:44,790
we get tau r as a collapse
load, 60,700.

224
00:12:44,790 --> 00:12:52,350
And if you take delta t of r
40,000, we get tau r 44,100.

225
00:12:52,350 --> 00:12:57,670
Notice that this is predicted
by the computer program.

226
00:12:57,670 --> 00:13:03,530
However, we also know that there
is no true collapse load

227
00:13:03,530 --> 00:13:07,630
in the sense of obtaining a
single stiffness matrix.

228
00:13:07,630 --> 00:13:11,930
So, here we learn something
more from this example.

229
00:13:11,930 --> 00:13:15,940
Namely that, in the linearized
buckling analysis we may

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00:13:15,940 --> 00:13:21,850
predict the collapse load, but
there may not actually be a

231
00:13:21,850 --> 00:13:26,130
collapse load in the structure
response corresponding to a

232
00:13:26,130 --> 00:13:28,580
single stiffness matrix.

233
00:13:28,580 --> 00:13:31,640
Whether you really want to
accept this here as a collapse

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00:13:31,640 --> 00:13:36,850
load depends on your
design criteria.

235
00:13:36,850 --> 00:13:40,430
Let's now look at another
example, the second example.

236
00:13:40,430 --> 00:13:49,340
And in this example we want to
consider the analysis of a

237
00:13:49,340 --> 00:13:53,630
cylinder that can undergo
enough elasto-plastic

238
00:13:53,630 --> 00:13:55,260
conditions.

239
00:13:55,260 --> 00:13:58,490
The material data of the
cylinder are given here.

240
00:13:58,490 --> 00:14:02,190
The geometric data of the
cylinder are given here.

241
00:14:02,190 --> 00:14:05,420
Our goal is to determine the
limit load of the cylinder.

242
00:14:08,210 --> 00:14:12,490
The finite element method we're
using here is shown on

243
00:14:12,490 --> 00:14:14,160
this view graph.

244
00:14:14,160 --> 00:14:16,550
Notice we have 8-node
elements here.

245
00:14:16,550 --> 00:14:19,350
Four 8-node elements.

246
00:14:19,350 --> 00:14:24,370
And in fact, if we look closely
at the response of the

247
00:14:24,370 --> 00:14:29,980
cylinder, we can identify that
these middle nodes really are

248
00:14:29,980 --> 00:14:34,300
not necessary, because the
response in the cylinder--

249
00:14:34,300 --> 00:14:36,880
We assume the cylinder to
be infinitely long.

250
00:14:36,880 --> 00:14:39,500
For that reason we have
introduced plane strain

251
00:14:39,500 --> 00:14:41,980
conditions.

252
00:14:41,980 --> 00:14:46,070
The response of the cylinder is
such that the same stress

253
00:14:46,070 --> 00:14:48,710
condition is obtained
throughout the

254
00:14:48,710 --> 00:14:50,075
lengths of the cylinder.

255
00:14:50,075 --> 00:14:52,690
There is no variation in
stresses along the length of

256
00:14:52,690 --> 00:14:53,440
the cylinder.

257
00:14:53,440 --> 00:14:56,360
And for that reason we could
actually have analyzed the

258
00:14:56,360 --> 00:14:58,390
structure with 6-noded
elements.

259
00:14:58,390 --> 00:15:00,140
In other words, we could
have dispensed

260
00:15:00,140 --> 00:15:01,570
with these nodes here.

261
00:15:04,370 --> 00:15:08,030
The internal pressure is
shown here to which

262
00:15:08,030 --> 00:15:09,490
the cylinder is subjected.

263
00:15:09,490 --> 00:15:13,500
And once again we want to
predict the ultimate load of

264
00:15:13,500 --> 00:15:16,290
this cylinder.

265
00:15:16,290 --> 00:15:19,260
Since the displacements are
small, we use here an MNO

266
00:15:19,260 --> 00:15:22,030
formulation.

267
00:15:22,030 --> 00:15:26,150
In this load displacement, the
response calculation, if we

268
00:15:26,150 --> 00:15:30,340
don't use the automatic load
step incrementation- the one

269
00:15:30,340 --> 00:15:32,880
that we discussed in the
previous lecture--

270
00:15:32,880 --> 00:15:40,600
the analyst has to prescribe a
load level, or a load curve,

271
00:15:40,600 --> 00:15:42,210
prior to the analysis.

272
00:15:42,210 --> 00:15:45,290
And here we have chosen
this load curve.

273
00:15:45,290 --> 00:15:48,920
Notice time is plotted
along this axis.

274
00:15:48,920 --> 00:15:51,380
Pressure is plotted
along this axis.

275
00:15:51,380 --> 00:15:55,710
And notice also that the elastic
limit load is 7.42.

276
00:15:55,710 --> 00:15:58,530
Therefore this kink here
is just above the

277
00:15:58,530 --> 00:15:59,910
elastic limit load.

278
00:15:59,910 --> 00:16:02,960
There's very little point in
using a lot of steps here,

279
00:16:02,960 --> 00:16:07,560
because we know that an MNO
formulation will reduce to a

280
00:16:07,560 --> 00:16:13,185
linear analysis until the
heave point is reached.

281
00:16:13,185 --> 00:16:15,860
Or the elastic limit
load is reached.

282
00:16:15,860 --> 00:16:19,510
So, we can go basically in one
step right up to the elastic

283
00:16:19,510 --> 00:16:23,450
limit load, and then we take a
series of steps to trace out

284
00:16:23,450 --> 00:16:26,345
the nonlinear response beyond
the elastic limit load.

285
00:16:29,330 --> 00:16:34,720
The purpose of the analysis was
really to compare various

286
00:16:34,720 --> 00:16:36,300
solution algorithms--

287
00:16:36,300 --> 00:16:39,260
the full Newton method with line
searches, the full Newton

288
00:16:39,260 --> 00:16:43,530
method without line searches,
the BFGS method, the modified

289
00:16:43,530 --> 00:16:46,420
Newton method with line
searches, the modified Newton

290
00:16:46,420 --> 00:16:49,840
method without line searches,
and the initial stress method.

291
00:16:49,840 --> 00:16:53,080
Now please recall that all
of these techniques were

292
00:16:53,080 --> 00:16:56,530
discussed in the previous
lecture, and I'd like to refer

293
00:16:56,530 --> 00:17:00,460
you back to that lecture to read
up on this material if

294
00:17:00,460 --> 00:17:04,280
you're not very familiar
with what I mean here.

295
00:17:04,280 --> 00:17:08,894
The convergence criteria that we
are using for the solution

296
00:17:08,894 --> 00:17:10,819
are given on this view graph.

297
00:17:10,819 --> 00:17:15,200
Notice here our ETOL,
and here our RTOL.

298
00:17:15,200 --> 00:17:21,650
We saw these formulas in the
previous lectures, and I don't

299
00:17:21,650 --> 00:17:24,230
want to go now through
these formulas again.

300
00:17:24,230 --> 00:17:28,470
The only point that I should
mention here is that this is a

301
00:17:28,470 --> 00:17:29,940
value you have to select.

302
00:17:29,940 --> 00:17:32,210
That is a value that
you have to select.

303
00:17:32,210 --> 00:17:35,040
In the computer program that
we are using, the ADINA

304
00:17:35,040 --> 00:17:38,530
program, of course, at default
values, which are really

305
00:17:38,530 --> 00:17:41,330
working quite well over a
large range of nonlinear

306
00:17:41,330 --> 00:17:46,200
problems, and those, of
course, are used very

307
00:17:46,200 --> 00:17:47,450
frequently.

308
00:17:47,450 --> 00:17:49,560
The RNORM here you
have to select.

309
00:17:49,560 --> 00:17:53,320
And this is a reasonable number
when you compare it

310
00:17:53,320 --> 00:17:57,960
with the total load applied
to the cylinder.

311
00:17:57,960 --> 00:18:01,330
Notice that we're looking here
at one radian of the cylinder.

312
00:18:01,330 --> 00:18:02,670
It's an axisymmetric analysis.

313
00:18:05,310 --> 00:18:09,510
When these procedures are
used to calculate the

314
00:18:09,510 --> 00:18:14,650
force-deflection curve, we find
that for p equal to 14,

315
00:18:14,650 --> 00:18:18,020
none of these procedures
converges.

316
00:18:18,020 --> 00:18:23,870
The response is shown here that
is calculated up to p

317
00:18:23,870 --> 00:18:27,120
equals 13.5.

318
00:18:27,120 --> 00:18:30,240
Of course, we have here drawn
a smooth curve through the

319
00:18:30,240 --> 00:18:34,620
discrete solution points that
have been obtained.

320
00:18:34,620 --> 00:18:38,610
If we compare now the different
solution techniques

321
00:18:38,610 --> 00:18:42,150
in terms of their cost
effectiveness, we find that

322
00:18:42,150 --> 00:18:47,150
normalizing to the full Newton
method solution time, we find

323
00:18:47,150 --> 00:18:51,080
that the full Newton method with
line searches takes 1.2

324
00:18:51,080 --> 00:18:55,620
times the time that is used
in the full Newton method.

325
00:18:55,620 --> 00:18:58,460
The BFGS method has here 0.9.

326
00:18:58,460 --> 00:19:02,420
The modified Newton method
with line searches 1.1.

327
00:19:02,420 --> 00:19:04,820
The modified Newton
method 1.1.

328
00:19:04,820 --> 00:19:07,450
And the initial stress
method 2.2.

329
00:19:07,450 --> 00:19:12,810
Showing that the BFGS is just a
tiny bit more effective than

330
00:19:12,810 --> 00:19:15,690
the full Newton method, hardly
worthwhile mentioning.

331
00:19:15,690 --> 00:19:19,490
But we also see that the initial
stress method is twice

332
00:19:19,490 --> 00:19:25,160
as expensive as virtually all
of the other methods.

333
00:19:25,160 --> 00:19:28,770
Once again please refer back
to the previous lecture in

334
00:19:28,770 --> 00:19:31,120
which I have been discussing
these various solution

335
00:19:31,120 --> 00:19:33,050
techniques.

336
00:19:33,050 --> 00:19:37,650
If we use the automatic load
step instrumentation we do not

337
00:19:37,650 --> 00:19:41,420
need to specify a load
function anymore.

338
00:19:41,420 --> 00:19:43,130
We discussed that earlier.

339
00:19:43,130 --> 00:19:46,050
And the softening in the
force-deflection curve is

340
00:19:46,050 --> 00:19:49,531
automatically taken
into account.

341
00:19:49,531 --> 00:19:55,400
With this technique, these are
the tolerances that we used.

342
00:19:55,400 --> 00:19:58,070
Default tolerances
in ADINA here.

343
00:19:58,070 --> 00:20:00,760
And this is the same RNORM
value that we used in the

344
00:20:00,760 --> 00:20:02,790
earlier analysis.

345
00:20:02,790 --> 00:20:08,240
Now the displacement response
is calculated as shown here.

346
00:20:08,240 --> 00:20:14,770
The p versus displacement
curve follows this path.

347
00:20:14,770 --> 00:20:17,210
And notice that the computed
limit load is

348
00:20:17,210 --> 00:20:20,560
really p equals 13.8.

349
00:20:20,560 --> 00:20:26,370
In the earlier analysis we
could not go beyond 13.5.

350
00:20:26,370 --> 00:20:29,650
In other words, for a value of
p equals 14 we did not reach

351
00:20:29,650 --> 00:20:30,630
convergence.

352
00:20:30,630 --> 00:20:32,510
We did not reach convergence
because the

353
00:20:32,510 --> 00:20:35,360
limit load is 13.8.

354
00:20:35,360 --> 00:20:37,580
And this is the difficulty
that I referred to at the

355
00:20:37,580 --> 00:20:39,710
start of last lecture.

356
00:20:39,710 --> 00:20:42,120
You see, if you use the earlier
solution algorithms,

357
00:20:42,120 --> 00:20:46,430
you have to very slowly this
restarting, or a very

358
00:20:46,430 --> 00:20:50,330
judicious choice of a load
function, get yourself closer

359
00:20:50,330 --> 00:20:52,170
and closer to the limit load.

360
00:20:52,170 --> 00:20:54,730
And that can be time
consuming.

361
00:20:54,730 --> 00:20:59,810
That means it also consumes
quite a bit of cost.

362
00:20:59,810 --> 00:21:02,700
Computer cost as well as
cost of your time.

363
00:21:02,700 --> 00:21:04,980
And therefore, it is so much
more convenient to use an

364
00:21:04,980 --> 00:21:08,460
automatic solution algorithm,
that means you don't need to

365
00:21:08,460 --> 00:21:12,160
specify a load function, that's
simply marches up here

366
00:21:12,160 --> 00:21:19,650
and gives you is the actual
collapse load quite easily, as

367
00:21:19,650 --> 00:21:21,400
shown here.

368
00:21:21,400 --> 00:21:24,510
Notice an interesting point is
that we actually march along

369
00:21:24,510 --> 00:21:31,390
this plateau when we use this
automatic solution algorithm.

370
00:21:31,390 --> 00:21:34,350
Let's look now at the
next example.

371
00:21:34,350 --> 00:21:40,600
And this pertains to the
solution of a spherical shell.

372
00:21:40,600 --> 00:21:44,760
Here we have a plan
view of the shell.

373
00:21:44,760 --> 00:21:46,800
The edges are clamped.

374
00:21:46,800 --> 00:21:48,810
And here we have a side
view of the shell.

375
00:21:48,810 --> 00:21:53,510
The shell is subjected to a
concentrated load at its apex.

376
00:21:53,510 --> 00:21:58,580
And we will increase that load,
as shown later on, to a

377
00:21:58,580 --> 00:22:04,850
very large value, so that this
shell, schematically shown

378
00:22:04,850 --> 00:22:08,720
here once again, will undergo
large displacements here.

379
00:22:08,720 --> 00:22:12,780
In fact, this point here will
move, as my pointer shows

380
00:22:12,780 --> 00:22:17,460
here, all the way down
to this level.

381
00:22:17,460 --> 00:22:20,040
Notice that this
distance here--

382
00:22:22,820 --> 00:22:24,660
and let me show it a
little bit clearer

383
00:22:24,660 --> 00:22:27,580
here on the view graph--

384
00:22:27,580 --> 00:22:32,430
is equal to .0859 inches.

385
00:22:32,430 --> 00:22:35,860
Here you see the other geometric
data of the shell,

386
00:22:35,860 --> 00:22:38,150
and here you see the
material data.

387
00:22:38,150 --> 00:22:43,480
We perform an axisymmetric
analysis of this shell.

388
00:22:43,480 --> 00:22:47,440
The finite element mesh is
shown in this view graph.

389
00:22:47,440 --> 00:22:54,740
Here we have the original mesh
subjected to the load.

390
00:22:54,740 --> 00:22:57,320
Notice that since we're looking
at an axisymmetric

391
00:22:57,320 --> 00:23:02,360
analysis, and since we are using
ADINA, we apply here a

392
00:23:02,360 --> 00:23:06,580
load of p, the total load
divided by 2 pi, because we

393
00:23:06,580 --> 00:23:09,980
only look at one radian of
the total structure.

394
00:23:09,980 --> 00:23:13,110
This 2 pi comes from the fact
that we are only analyzing one

395
00:23:13,110 --> 00:23:15,150
radian of the structure.

396
00:23:15,150 --> 00:23:19,010
The deformed configuration for
p equal 200 pounds is shown

397
00:23:19,010 --> 00:23:20,260
here in red.

398
00:23:23,180 --> 00:23:27,650
Notice we undergo large
displacements here, and that

399
00:23:27,650 --> 00:23:32,340
is also shown on the next view
graph, which gives the force

400
00:23:32,340 --> 00:23:37,640
deflection curve obtained with
this finite element mesh.

401
00:23:37,640 --> 00:23:39,850
And that force deflection
curve is shown here.

402
00:23:43,510 --> 00:23:46,550
The linear analysis would
predict this response.

403
00:23:46,550 --> 00:23:49,220
Of course, way off the actual
response when the

404
00:23:49,220 --> 00:23:53,100
displacements become large.

405
00:23:53,100 --> 00:24:00,520
For this problem, the maximum
load is 100 pounds, and we can

406
00:24:00,520 --> 00:24:04,180
apply that maximum load in one
load step, or in a sequence of

407
00:24:04,180 --> 00:24:05,180
load steps.

408
00:24:05,180 --> 00:24:10,300
If we apply the full load in 10
equal steps, we find that

409
00:24:10,300 --> 00:24:11,980
the full Newton method
with line

410
00:24:11,980 --> 00:24:15,670
searchers works very nicely.

411
00:24:15,670 --> 00:24:17,950
The full Newton without line
searches, in fact, is even

412
00:24:17,950 --> 00:24:20,490
more effective.

413
00:24:20,490 --> 00:24:24,340
Notice the BFGS method, the
modified Newton method with

414
00:24:24,340 --> 00:24:26,490
line searches and without
line searches.

415
00:24:26,490 --> 00:24:30,350
These three methods
did not work.

416
00:24:30,350 --> 00:24:34,700
Of course, 10 steps to full
load with these very large

417
00:24:34,700 --> 00:24:39,570
displacements means really
using a very coarse load

418
00:24:39,570 --> 00:24:43,090
incrementation, or using, in
other words, a very small

419
00:24:43,090 --> 00:24:44,640
number of load steps.

420
00:24:44,640 --> 00:24:47,210
These were 10 equal load steps,
I should also say.

421
00:24:51,050 --> 00:24:55,600
If we apply the full load in 50
equal steps, we find that

422
00:24:55,600 --> 00:24:57,490
the full Newton method
with line searches

423
00:24:57,490 --> 00:24:59,990
is still most effective.

424
00:24:59,990 --> 00:25:05,420
And these four methods
here converge.

425
00:25:05,420 --> 00:25:10,240
They are not quite as effective
as the full Newton

426
00:25:10,240 --> 00:25:13,190
method, but the modified Newton
method without line

427
00:25:13,190 --> 00:25:15,520
searches does not
even converge.

428
00:25:15,520 --> 00:25:20,260
You would have to use many
more steps to make the

429
00:25:20,260 --> 00:25:25,110
modified Newton method without
line searches also converge.

430
00:25:25,110 --> 00:25:29,360
Here, the BFGS method compares
quite favorably with these

431
00:25:29,360 --> 00:25:31,340
methods here.

432
00:25:31,340 --> 00:25:35,350
The convergence criteria
employed for this problem are

433
00:25:35,350 --> 00:25:38,000
shown here.

434
00:25:38,000 --> 00:25:42,860
Notice ETOL is given here
as 0.001 default

435
00:25:42,860 --> 00:25:44,260
in the ADINA program.

436
00:25:44,260 --> 00:25:46,610
We used a maximum number of
iterations that we're

437
00:25:46,610 --> 00:25:48,940
permitted, which
is very large.

438
00:25:48,940 --> 00:25:52,180
Very large in this
particular case.

439
00:25:52,180 --> 00:25:55,400
We, of course, may also perform
the analysis using the

440
00:25:55,400 --> 00:25:58,790
automatic load stepping
algorithm.

441
00:25:58,790 --> 00:26:05,460
Here we use as this tolerance
ETOL, and RTOL this tolerance.

442
00:26:05,460 --> 00:26:07,070
Our norm is shown here.

443
00:26:07,070 --> 00:26:09,740
These are once again
default tolerances

444
00:26:09,740 --> 00:26:12,510
in the ADINA program.

445
00:26:12,510 --> 00:26:18,080
The response that we calculate
is shown on this view graph.

446
00:26:18,080 --> 00:26:19,850
Here, the applied
load once again.

447
00:26:19,850 --> 00:26:21,830
Displacement at the apex.

448
00:26:21,830 --> 00:26:24,800
And this is the response
calculated.

449
00:26:24,800 --> 00:26:33,140
Using as the initial
displacement 0.001 inches.

450
00:26:33,140 --> 00:26:35,090
You get the solid curve.

451
00:26:35,090 --> 00:26:38,390
And using as initial
displacement 0.01 inches, we

452
00:26:38,390 --> 00:26:41,560
get these points here.

453
00:26:41,560 --> 00:26:46,020
Notice, or recall please, that
when we use the automatic load

454
00:26:46,020 --> 00:26:51,130
stepping algorithm, we start
off that algorithm by

455
00:26:51,130 --> 00:26:54,270
specifying the displacement
for the first

456
00:26:54,270 --> 00:26:55,930
load step at one node.

457
00:26:55,930 --> 00:26:58,700
And these are the values that
I'm talking about here.

458
00:27:01,870 --> 00:27:05,490
As a next example, I'd like to
look with you at the analysis

459
00:27:05,490 --> 00:27:08,080
of the cantilever under
pressure loading.

460
00:27:08,080 --> 00:27:10,920
Here we have the cantilever.

461
00:27:10,920 --> 00:27:14,070
This is the pressure
loading applied.

462
00:27:14,070 --> 00:27:17,100
Notice it's quite a long
cantilever, when

463
00:27:17,100 --> 00:27:20,390
measured on the gaps.

464
00:27:20,390 --> 00:27:22,400
Here are the material data.

465
00:27:22,400 --> 00:27:28,830
We analyze this problem using
plane strain situation, and we

466
00:27:28,830 --> 00:27:32,860
want to perform the analysis for
p equal to 1 megapascal.

467
00:27:36,280 --> 00:27:41,390
The interesting point here is
to perform the analysis once

468
00:27:41,390 --> 00:27:45,880
with a deformation-independent
loading, p the vertical and

469
00:27:45,880 --> 00:27:50,820
remaining vertical, throughout
the whole analysis, and once

470
00:27:50,820 --> 00:27:53,790
with a deformation-dependent
loading assumption, in which

471
00:27:53,790 --> 00:27:58,390
the pressure loading remains
perpendicular to the

472
00:27:58,390 --> 00:27:59,955
mid-surface of the cantilever.

473
00:28:02,660 --> 00:28:05,390
I think you can appreciate
that if you have this

474
00:28:05,390 --> 00:28:10,110
situation, the deformations
will be larger.

475
00:28:10,110 --> 00:28:15,370
Well, we performed both of these
analyses, and once again

476
00:28:15,370 --> 00:28:20,540
the purpose here was really to
contrast the assumption of

477
00:28:20,540 --> 00:28:23,865
deformation-independent loading
with the assumption of

478
00:28:23,865 --> 00:28:26,760
deformation-dependent loading.

479
00:28:26,760 --> 00:28:30,280
The finite element model that
we used consisted of 25

480
00:28:30,280 --> 00:28:33,730
two-dimensional 8-node elements,
one layer through

481
00:28:33,730 --> 00:28:36,730
the thickness, and they
were evenly spaced.

482
00:28:36,730 --> 00:28:38,620
And here we list the
solution details.

483
00:28:38,620 --> 00:28:43,200
We used the full Newton method
without line searches.

484
00:28:43,200 --> 00:28:46,550
And the convergence tolerance
is shown here.

485
00:28:46,550 --> 00:28:51,150
I should also add that we used
10 steps for the total load

486
00:28:51,150 --> 00:28:52,430
that we wanted to reach.

487
00:28:52,430 --> 00:28:54,960
10 equal steps.

488
00:28:54,960 --> 00:28:58,950
The force deflection curve is
shown on this view graph.

489
00:28:58,950 --> 00:29:03,540
For small deflections, down
here, there's basically no

490
00:29:03,540 --> 00:29:07,155
difference with respect
to the assumption of

491
00:29:07,155 --> 00:29:08,760
deformation-dependent
loading or

492
00:29:08,760 --> 00:29:10,500
deformation-independent loading.

493
00:29:10,500 --> 00:29:15,360
However, as the displacements
become larger, the two curves

494
00:29:15,360 --> 00:29:18,995
part from each other and,
as expected, the

495
00:29:18,995 --> 00:29:23,200
deformation-dependent loading
gives larger displacements.

496
00:29:25,790 --> 00:29:29,400
I should add here, if you look
back at what we discussed in

497
00:29:29,400 --> 00:29:34,890
our earlier lectures, I should
add that regarding the

498
00:29:34,890 --> 00:29:39,030
solution for the
deformation-dependent loading

499
00:29:39,030 --> 00:29:44,790
case, we used a stiffness matrix
that is symmetric.

500
00:29:44,790 --> 00:29:47,130
The same stiffness matrix, as a
matter of fact, that we are

501
00:29:47,130 --> 00:29:50,860
usually using when
we are analyzing

502
00:29:50,860 --> 00:29:52,770
deformation-independent
loading.

503
00:29:52,770 --> 00:29:57,160
The deformation dependency of
the loading was only taken

504
00:29:57,160 --> 00:30:00,925
into account by updating the
right hand side load vector t

505
00:30:00,925 --> 00:30:02,840
plus delta t of r.

506
00:30:02,840 --> 00:30:07,145
In our earlier discussions, we
very largely assumed that t

507
00:30:07,145 --> 00:30:10,760
plus delta t of r is constant,
and prescribed by the analyst.

508
00:30:10,760 --> 00:30:14,410
But I also pointed out in the
earlier lectures that, if we

509
00:30:14,410 --> 00:30:17,485
have deformation-dependent
loading, you would have a t

510
00:30:17,485 --> 00:30:21,720
plus delta t of r, superscript
i minus one, where that i

511
00:30:21,720 --> 00:30:27,560
minus 1 then refers to the
configuration corresponding to

512
00:30:27,560 --> 00:30:31,620
the i at the end of i minus
first iteration.

513
00:30:31,620 --> 00:30:34,790
So, in this particular solution,
when we dealt with

514
00:30:34,790 --> 00:30:38,880
the deformation-dependent
loading, we kept this as the

515
00:30:38,880 --> 00:30:39,890
matrix difference
[UNINTELLIGIBLE]

516
00:30:39,890 --> 00:30:43,580
on the left hand side, but we
updated the load vector t plus

517
00:30:43,580 --> 00:30:46,850
delta t of r in the notation
that we used in our earlier

518
00:30:46,850 --> 00:30:51,250
lectures for the configuration,
and for the

519
00:30:51,250 --> 00:30:54,760
iteration, that we actually
are dealing with.

520
00:30:54,760 --> 00:31:01,060
And when we did so, we found
that to obtain this total

521
00:31:01,060 --> 00:31:05,640
curve here, to obtain this total
curve, we needed about

522
00:31:05,640 --> 00:31:12,880
20% more solution time than for
attaining this curve here.

523
00:31:12,880 --> 00:31:16,830
So, taking into account the
deformation-dependency of the

524
00:31:16,830 --> 00:31:20,840
loading in this particular
example just meant 20% more

525
00:31:20,840 --> 00:31:24,510
solution time with the
computer program.

526
00:31:24,510 --> 00:31:29,240
Here on the next view graph now,
we show pictorially how

527
00:31:29,240 --> 00:31:31,090
the cantilever has deformed.

528
00:31:31,090 --> 00:31:33,350
Here is the original mesh.

529
00:31:33,350 --> 00:31:36,870
And here is the deformed
mesh when we use the

530
00:31:36,870 --> 00:31:39,630
deformation-independent loading
assumption, and here

531
00:31:39,630 --> 00:31:42,880
is the deformed mesh when we
use a deformation-dependent

532
00:31:42,880 --> 00:31:44,040
loading assumption.

533
00:31:44,040 --> 00:31:47,650
Notice, of course, that the
deformations are larger when

534
00:31:47,650 --> 00:31:50,460
we have assumed that
the loading

535
00:31:50,460 --> 00:31:53,180
is deformation dependent.

536
00:31:53,180 --> 00:31:55,800
Let us now look at
the next example.

537
00:31:55,800 --> 00:31:58,965
And in this example
we consider a

538
00:31:58,965 --> 00:32:00,810
diamond-shaped frame.

539
00:32:00,810 --> 00:32:03,260
That frame is shown here.

540
00:32:03,260 --> 00:32:08,810
Notice we have a frictionless
hinge up there and down here.

541
00:32:08,810 --> 00:32:12,690
We have a rigid weld
here and over here.

542
00:32:12,690 --> 00:32:16,100
Notice that the frame is pushed
down with a load p

543
00:32:16,100 --> 00:32:19,510
here, and is pushed up here.

544
00:32:19,510 --> 00:32:23,040
Of course, as deformations of
the frame would be symmetric

545
00:32:23,040 --> 00:32:27,610
about this line here, notice
that the beam cross section is

546
00:32:27,610 --> 00:32:30,570
one inch by one inch, and
here are the material

547
00:32:30,570 --> 00:32:32,310
data for the beam.

548
00:32:32,310 --> 00:32:35,240
An elastic large displacement
analysis is

549
00:32:35,240 --> 00:32:37,440
what we want to perform.

550
00:32:37,440 --> 00:32:40,530
We use twenty 3-node isobeam
elements to model

551
00:32:40,530 --> 00:32:42,540
the complete frame.

552
00:32:42,540 --> 00:32:45,880
We will talk about the
isoparametic, or isobeam

553
00:32:45,880 --> 00:32:49,470
elements later on in
a later lecture.

554
00:32:49,470 --> 00:32:53,400
Notice this is here 15 inches,
and so is that.

555
00:32:53,400 --> 00:32:55,100
The deformations will
be very large.

556
00:32:55,100 --> 00:32:59,360
In fact, this node here will
go across all the way up to

557
00:32:59,360 --> 00:33:02,000
here, and that node
we go across all

558
00:33:02,000 --> 00:33:05,070
the way down to here.

559
00:33:05,070 --> 00:33:09,820
The displacement of the top
hinge of the frame is shown on

560
00:33:09,820 --> 00:33:11,610
this view graph.

561
00:33:11,610 --> 00:33:15,190
Notice here we are plotting the
load p vertically up, and

562
00:33:15,190 --> 00:33:18,080
the displacement of the top
hinge horizontally.

563
00:33:18,080 --> 00:33:22,100
Notice as the displacements are
very large, you're going

564
00:33:22,100 --> 00:33:26,870
up to almost 25 inches here.

565
00:33:26,870 --> 00:33:30,970
Notice that for this solution,
we used a load increment of

566
00:33:30,970 --> 00:33:33,400
250 pounds.

567
00:33:33,400 --> 00:33:40,160
So we used 80 steps to come from
zero to 20,000 pounds.

568
00:33:40,160 --> 00:33:43,100
Notice this, of course, is a
very large number of steps.

569
00:33:43,100 --> 00:33:47,240
You would have used another
80 steps here and so on.

570
00:33:47,240 --> 00:33:50,700
We wanted to use a small load
incrementation because we

571
00:33:50,700 --> 00:33:54,980
wanted to obtain many solution
points along this curve to be

572
00:33:54,980 --> 00:33:59,970
able to produce an animation
of the frame response.

573
00:33:59,970 --> 00:34:03,660
And I'd like to now share
with you this animation.

574
00:34:03,660 --> 00:34:05,860
We'd like to look at
the animation.

575
00:34:05,860 --> 00:34:10,500
And I'd like to really look
at the animation twice.

576
00:34:10,500 --> 00:34:13,210
First we will look at the
animation of the frame

577
00:34:13,210 --> 00:34:18,770
response in very large
deformations, very fast.

578
00:34:18,770 --> 00:34:23,120
And then we want to look at the
same animation, but in a

579
00:34:23,120 --> 00:34:25,750
slower movement, so that
I can also talk a

580
00:34:25,750 --> 00:34:27,070
little bit about it.

581
00:34:27,070 --> 00:34:30,780
Because the fast version is
going so fast that I can't

582
00:34:30,780 --> 00:34:32,250
really say very much.

583
00:34:32,250 --> 00:34:33,690
So let's now look at
this animation.

584
00:34:36,389 --> 00:34:38,280
Here we see the frame
is discretized

585
00:34:38,280 --> 00:34:40,460
by the isobeam elements.

586
00:34:40,460 --> 00:34:43,606
The two thick arrows show
the load application.

587
00:34:43,606 --> 00:34:48,139
Here you see how the frame go
through the last deformations.

588
00:34:48,139 --> 00:34:50,520
Let's now look at the same
animation again with a

589
00:34:50,520 --> 00:34:53,429
somewhat longer time scale.

590
00:34:53,429 --> 00:34:56,120
We perform, of course,
a static analysis.

591
00:34:56,120 --> 00:34:58,930
The time code above the
picture of the frame

592
00:34:58,930 --> 00:35:00,510
gives the load step.

593
00:35:00,510 --> 00:35:03,296
You also see the load applied
for each step.

594
00:35:03,296 --> 00:35:06,840
Here you see once again how
the frame deforms when

595
00:35:06,840 --> 00:35:09,480
subjected to the loads.

596
00:35:09,480 --> 00:35:12,770
Certainly the frame undergoes
very large displacements, and

597
00:35:12,770 --> 00:35:15,770
for the higher load levels the
analysis represents only a

598
00:35:15,770 --> 00:35:17,690
numerical experiment.

599
00:35:17,690 --> 00:35:19,120
But surely an interesting one.

600
00:35:27,940 --> 00:35:32,550
This completes our discussion of
the frame analysis example,

601
00:35:32,550 --> 00:35:36,800
and I would like now to look
with you at the last example

602
00:35:36,800 --> 00:35:40,200
in this lecture, namely the
analysis of the failure and

603
00:35:40,200 --> 00:35:43,190
repair of a beam cable
structure.

604
00:35:43,190 --> 00:35:47,930
For this analysis, we use one
option in the ADINA program

605
00:35:47,930 --> 00:35:52,240
that is quite valuable, namely
is the option of element birth

606
00:35:52,240 --> 00:35:54,590
and element death.

607
00:35:54,590 --> 00:36:00,210
What this means is that we can
assign certain elements that

608
00:36:00,210 --> 00:36:05,080
are born in the analysis as
this step by step solution

609
00:36:05,080 --> 00:36:09,280
proceeds, and we can assign to
certain element, so to say,

610
00:36:09,280 --> 00:36:10,350
their deaths.

611
00:36:10,350 --> 00:36:12,980
They are disappearing at
a particular time.

612
00:36:12,980 --> 00:36:16,610
This kind of analysis option
is very useful to simulate,

613
00:36:16,610 --> 00:36:20,310
for example, construction
processes and repairs of

614
00:36:20,310 --> 00:36:21,250
structures.

615
00:36:21,250 --> 00:36:25,130
As well known, structures can,
of course, collapse during the

616
00:36:25,130 --> 00:36:28,860
repair, or even during their
construction sequence.

617
00:36:28,860 --> 00:36:32,070
And therefore it can be
important to actually simulate

618
00:36:32,070 --> 00:36:34,410
this sequence of construction,
or the repair of the office

619
00:36:34,410 --> 00:36:39,380
structure, also on the computer,
to find out the

620
00:36:39,380 --> 00:36:42,090
particular construction sequence
or repair of a

621
00:36:42,090 --> 00:36:47,230
structure that one wants
to proceed with.

622
00:36:47,230 --> 00:36:52,950
Here we have a simple cable
beam structure.

623
00:36:52,950 --> 00:36:57,380
This is a cantilever beam
subjected to gravity loading,

624
00:36:57,380 --> 00:37:03,035
and here the cable is hinged at
this end of the beam and up

625
00:37:03,035 --> 00:37:07,030
there so as to support
the beam.

626
00:37:07,030 --> 00:37:10,770
Notice the material data for the
cable are given here, and

627
00:37:10,770 --> 00:37:13,050
there's no pretension
in the cable.

628
00:37:13,050 --> 00:37:16,210
The beam itself has these
material data.

629
00:37:16,210 --> 00:37:19,840
So we model the elasto-plastic
response of the beam.

630
00:37:19,840 --> 00:37:22,650
Here is the mass density
of to beam.

631
00:37:22,650 --> 00:37:26,360
The cross section is
shown over here.

632
00:37:26,360 --> 00:37:30,280
In this analysis, we simulate
the failure and repair of the

633
00:37:30,280 --> 00:37:33,100
cable in the following way.

634
00:37:33,100 --> 00:37:38,730
In load step one, the beams sags
under its weight, but is

635
00:37:38,730 --> 00:37:39,980
supported by the cable.

636
00:37:42,600 --> 00:37:46,960
From load step one to
two the cable snaps.

637
00:37:46,960 --> 00:37:49,830
In other words, for example,
if delta t is equal to 1.

638
00:37:49,830 --> 00:37:53,330
The time step delta t that we
talked about in the earlier

639
00:37:53,330 --> 00:37:58,430
lectures is equal to 1, then
at 1 this would be the load

640
00:37:58,430 --> 00:38:07,300
condition, and at time 1.5, the
cable would snap, so that

641
00:38:07,300 --> 00:38:11,920
at time 2 at the end of the
second load step, the cable is

642
00:38:11,920 --> 00:38:15,640
not anymore in the finite
element system, and

643
00:38:15,640 --> 00:38:18,060
the beam now sags.

644
00:38:18,060 --> 00:38:22,880
It undergoes plastic flow
at the built-in end.

645
00:38:22,880 --> 00:38:26,630
Now, in the next steps, two
to four, the new cable is

646
00:38:26,630 --> 00:38:30,830
installed and is tensioned
until the tip of the beam

647
00:38:30,830 --> 00:38:35,540
returns to its original
location, namely the location

648
00:38:35,540 --> 00:38:39,500
that it had taken at the
end of load step one.

649
00:38:39,500 --> 00:38:43,190
This is quite easily simulated
with this element birth and

650
00:38:43,190 --> 00:38:48,050
death option available in the
computer program you're using.

651
00:38:48,050 --> 00:38:52,430
Notice one element dies, namely
the cable element dies

652
00:38:52,430 --> 00:38:59,950
at time 1.5, say, and then at
2.5 the cable is again there,

653
00:38:59,950 --> 00:39:04,070
and at time 3 and at time
4 we see that the

654
00:39:04,070 --> 00:39:07,630
cable has been tensioned.

655
00:39:07,630 --> 00:39:10,450
Here's the finite
element model.

656
00:39:10,450 --> 00:39:15,900
We used two truss elements
here to model the cable.

657
00:39:15,900 --> 00:39:20,020
And we used five 2-node
Hermitian beam elements to

658
00:39:20,020 --> 00:39:22,470
model the beam.

659
00:39:22,470 --> 00:39:25,500
Notice for the beam element,
we use five Newton-Cotes

660
00:39:25,500 --> 00:39:29,210
integration points in the r
direction, three Newton-Cotes

661
00:39:29,210 --> 00:39:32,250
integration points in
the s direction.

662
00:39:32,250 --> 00:39:36,330
In load step one, the active
truss is number one.

663
00:39:36,330 --> 00:39:38,390
It's there.

664
00:39:38,390 --> 00:39:45,710
At the end of load step two,
no truss element is there,

665
00:39:45,710 --> 00:39:51,350
because remember, between load
step one and load step two, at

666
00:39:51,350 --> 00:39:58,990
time 1.5, if delta t is equal to
1, this one cable, this one

667
00:39:58,990 --> 00:40:01,285
truss element disappears.

668
00:40:01,285 --> 00:40:03,760
It dies, so to say.

669
00:40:03,760 --> 00:40:07,780
And then there is no more truss
connecting this point to

670
00:40:07,780 --> 00:40:11,190
that point at the end
of load step two.

671
00:40:11,190 --> 00:40:12,940
At time two.

672
00:40:12,940 --> 00:40:19,650
Now, in the next load step a
new truss element is born.

673
00:40:19,650 --> 00:40:21,680
Truss number two.

674
00:40:21,680 --> 00:40:25,770
It might be born, for example,
at a specific time 2.5, if

675
00:40:25,770 --> 00:40:28,600
delta t is equal to 1.

676
00:40:28,600 --> 00:40:30,450
And now the truss
is in position.

677
00:40:30,450 --> 00:40:31,890
It has been born.

678
00:40:31,890 --> 00:40:35,290
And we are tensioning it
from load step three

679
00:40:35,290 --> 00:40:36,540
to load step four.

680
00:40:39,660 --> 00:40:43,210
In the solution we use the UL
formulation for the truss

681
00:40:43,210 --> 00:40:45,790
elements and the
beam elements.

682
00:40:45,790 --> 00:40:48,680
The convergence tolerances
are listed here.

683
00:40:48,680 --> 00:40:51,850
Notice that because we have
rotational degrees of freedom

684
00:40:51,850 --> 00:40:56,710
for the beam, we also use a
tolerance RMNORM the way I

685
00:40:56,710 --> 00:41:00,090
describe it in the earlier
lecture when we talked about

686
00:41:00,090 --> 00:41:02,490
convergence tolerances.

687
00:41:02,490 --> 00:41:05,890
Let us now look at the
solution algorithm

688
00:41:05,890 --> 00:41:07,910
performances.

689
00:41:07,910 --> 00:41:12,330
If we use a full Newton method
with line searches all load

690
00:41:12,330 --> 00:41:15,070
steps are successfully
calculated.

691
00:41:15,070 --> 00:41:18,396
The normalized solution
time, say, is one.

692
00:41:18,396 --> 00:41:24,600
The full Newton method without
line searches did not work.

693
00:41:24,600 --> 00:41:26,080
We could not use
the full Newton

694
00:41:26,080 --> 00:41:28,660
method for this problem.

695
00:41:28,660 --> 00:41:33,300
The BFGS method in this
particular case worked quite

696
00:41:33,300 --> 00:41:35,630
nicely, but it needed
more time.

697
00:41:35,630 --> 00:41:38,570
Two and a half times the amount
of time that the full

698
00:41:38,570 --> 00:41:41,640
Newton method required
with line searches.

699
00:41:41,640 --> 00:41:46,320
Of course, the BFGS method
always has line searches.

700
00:41:46,320 --> 00:41:49,280
And if you use the modified
Newton method, with or without

701
00:41:49,280 --> 00:41:52,260
line searches, we did not
converge in load step two.

702
00:41:55,600 --> 00:41:58,540
So, this shows that really the
full Newton method with line

703
00:41:58,540 --> 00:42:02,630
searches is very powerful, and
in this particular case it is

704
00:42:02,630 --> 00:42:04,460
even the cheapest.

705
00:42:04,460 --> 00:42:08,730
The BFGS method did also
quite well, but is

706
00:42:08,730 --> 00:42:11,530
quite a bit more expensive.

707
00:42:11,530 --> 00:42:13,530
The results in all of
this analysis are

708
00:42:13,530 --> 00:42:15,390
listed in this table.

709
00:42:15,390 --> 00:42:18,650
Load step one, the displacement
at the tip of the

710
00:42:18,650 --> 00:42:20,540
cantilever is shown here.

711
00:42:20,540 --> 00:42:22,100
The stress in the cable.

712
00:42:22,100 --> 00:42:24,560
And the moment at the
built-in end.

713
00:42:24,560 --> 00:42:28,340
Load step two, this is
the tip displacement.

714
00:42:28,340 --> 00:42:31,020
Please see the very large
increment in the tip

715
00:42:31,020 --> 00:42:34,100
displacement, because
the cable is gone.

716
00:42:34,100 --> 00:42:36,130
There is no more cable.

717
00:42:36,130 --> 00:42:40,800
And the moment at the built-in
end has also increased a lot.

718
00:42:40,800 --> 00:42:45,350
Now we have installed the repair
cable, so to say, and

719
00:42:45,350 --> 00:42:49,540
in load step three we have
reduced the tip displacement

720
00:42:49,540 --> 00:42:54,540
from 0.63 0.31 meters.

721
00:42:54,540 --> 00:42:56,380
The stress in the cable
is given here.

722
00:42:56,380 --> 00:42:58,920
And the moment at the built-in
end is given here.

723
00:42:58,920 --> 00:43:01,890
And in the final load step,
we've tightened the cable

724
00:43:01,890 --> 00:43:05,160
enough to get back to the
displacement that the

725
00:43:05,160 --> 00:43:09,490
cantilever had at its
tip originally.

726
00:43:09,490 --> 00:43:10,990
Same numbers here.

727
00:43:10,990 --> 00:43:13,030
And the stress in the
cable is given here.

728
00:43:13,030 --> 00:43:14,860
And the moment at the
built-in end of the

729
00:43:14,860 --> 00:43:17,130
cantilever is given here.

730
00:43:17,130 --> 00:43:20,340
The last view graph here shows
the deformations of the

731
00:43:20,340 --> 00:43:22,570
structure pictorially.

732
00:43:22,570 --> 00:43:27,020
Here you see the deformations
at load step one--

733
00:43:27,020 --> 00:43:29,870
here the beam, here the cable.

734
00:43:29,870 --> 00:43:32,250
Notice that these displacements
are magnified by

735
00:43:32,250 --> 00:43:34,710
a factor of 10.

736
00:43:34,710 --> 00:43:39,740
These are the deformations
of the beam at the end

737
00:43:39,740 --> 00:43:40,790
of load step two.

738
00:43:40,790 --> 00:43:44,180
Notice there is no cable here,
and also these deformations

739
00:43:44,180 --> 00:43:45,810
have not been magnified.

740
00:43:45,810 --> 00:43:48,130
Very large deformations.

741
00:43:48,130 --> 00:43:52,950
Load step three, at the end of
load step three, we see the

742
00:43:52,950 --> 00:43:56,180
cable here, the new cable
that has been installed.

743
00:43:56,180 --> 00:43:59,630
And we see also that the tip
displacement of the cantilever

744
00:43:59,630 --> 00:44:04,510
has been reduced by tightening
that new cable already up.

745
00:44:04,510 --> 00:44:08,370
Now, at the end of load step
four, the displacement at the

746
00:44:08,370 --> 00:44:12,330
tip of the cantilever are the
same displacement as we have

747
00:44:12,330 --> 00:44:15,710
measured at the end
of load step one.

748
00:44:15,710 --> 00:44:18,550
Here again we have magnified
the displacements

749
00:44:18,550 --> 00:44:22,180
by a factor of 10.

750
00:44:22,180 --> 00:44:25,190
This then brings us to the
end of this example.

751
00:44:25,190 --> 00:44:27,220
I wanted to show you this
example really as a

752
00:44:27,220 --> 00:44:31,490
demonstrative example to
indicate to you how this

753
00:44:31,490 --> 00:44:35,800
element birth and death option
can be used quite effectively

754
00:44:35,800 --> 00:44:39,290
to simulate repairs
of structures.

755
00:44:39,290 --> 00:44:43,720
In actual analysis you, of
course, might want to use a

756
00:44:43,720 --> 00:44:47,180
finer finite element
discretization and also a

757
00:44:47,180 --> 00:44:51,110
finer integration scheme to
pick up the spread of

758
00:44:51,110 --> 00:44:53,490
plasticity more accurately.

759
00:44:53,490 --> 00:44:57,460
However, this example as a
demonstrative example really

760
00:44:57,460 --> 00:45:01,370
shows to us how we can use this
element birth and death

761
00:45:01,370 --> 00:45:04,620
option to simulate such
processes as repairs of

762
00:45:04,620 --> 00:45:05,260
structures.

763
00:45:05,260 --> 00:45:09,100
And I think you can easily
visualize how this element

764
00:45:09,100 --> 00:45:11,820
birth and death option can
also be used to simulate

765
00:45:11,820 --> 00:45:14,760
construction sequences
of structures.

766
00:45:14,760 --> 00:45:16,700
This then brings me to the
end of this lecture.

767
00:45:16,700 --> 00:45:18,280
Thank you very much for
your attention.