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PROFESSOR: Ladies and gentlemen,
welcome to this
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00:00:23,920 --> 00:00:26,870
lecture on nonlinear finite
element analysis of solids and
10
00:00:26,870 --> 00:00:27,870
structures.
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00:00:27,870 --> 00:00:30,050
In this lecture, I would like
to continue with our
12
00:00:30,050 --> 00:00:34,090
discussion of solution methods
that we use to solve the
13
00:00:34,090 --> 00:00:37,580
finite element equations in
nonlinear static analysis.
14
00:00:37,580 --> 00:00:40,020
We considered already in the
previous lecture a number of
15
00:00:40,020 --> 00:00:43,720
solution techniques to solve
this set of equations.
16
00:00:43,720 --> 00:00:47,350
t plus delta tr is equal to t
plus delta tf, where t plus
17
00:00:47,350 --> 00:00:51,720
delta tr is the load vector of
externally applied loads at
18
00:00:51,720 --> 00:00:56,410
time t plus delta t, and t plus
tf is a nodal point force
19
00:00:56,410 --> 00:00:59,210
vector corresponding to the
internal element stresses at
20
00:00:59,210 --> 00:01:01,080
time t plus delta t.
21
00:01:01,080 --> 00:01:04,010
We talked about the full
Newton-Raphson method, the
22
00:01:04,010 --> 00:01:07,270
modified Newton-Raphson method,
the BFGS method, the
23
00:01:07,270 --> 00:01:08,920
initial stress method,
and we discussed
24
00:01:08,920 --> 00:01:11,110
also convergence criteria.
25
00:01:11,110 --> 00:01:13,890
I've summarized here the
equations corresponding to the
26
00:01:13,890 --> 00:01:16,260
modified Newton iteration.
27
00:01:16,260 --> 00:01:19,700
Now, a distinguishing feature
of these solution techniques
28
00:01:19,700 --> 00:01:25,610
is that the analyst has to
prescribe the externally
29
00:01:25,610 --> 00:01:26,680
applied load vector
30
00:01:26,680 --> 00:01:30,520
corresponding to all load steps.
31
00:01:30,520 --> 00:01:36,240
This means schematically that
if you have this kind of
32
00:01:36,240 --> 00:01:40,380
displacement load curve, or load
displacement curve, shown
33
00:01:40,380 --> 00:01:44,350
here in red, the analyst has
to prescribe prior to the
34
00:01:44,350 --> 00:01:50,760
analysis the load levels at
which the response is sought.
35
00:01:50,760 --> 00:01:54,800
I've here indicated 1r as a load
level corresponding to
36
00:01:54,800 --> 00:01:58,560
load step one, 2r as a load
level corresponding to load
37
00:01:58,560 --> 00:02:01,150
step two, et cetera.
38
00:02:01,150 --> 00:02:04,130
Of course, the corresponding
solutions that the analyst is
39
00:02:04,130 --> 00:02:08,509
looking for are the
displacements.
40
00:02:08,509 --> 00:02:11,810
Now, what might very well happen
in practical analysis
41
00:02:11,810 --> 00:02:16,560
is that the analyst chooses
certain load levels, and at
42
00:02:16,560 --> 00:02:19,570
the particular load level
convergence difficulties are
43
00:02:19,570 --> 00:02:20,470
encountered.
44
00:02:20,470 --> 00:02:23,550
Too many iterations are required
to converge, for
45
00:02:23,550 --> 00:02:28,650
example, in a reasonable cost.
46
00:02:28,650 --> 00:02:33,210
And typically, say, if this
happens at this load step
47
00:02:33,210 --> 00:02:36,800
here, the analyst decides
to restock.
48
00:02:36,800 --> 00:02:41,700
And this means that a solution
corresponding to r4 here was
49
00:02:41,700 --> 00:02:43,730
not obtained.
50
00:02:43,730 --> 00:02:45,160
I'm scratching that out.
51
00:02:45,160 --> 00:02:50,400
And the analyst restarts now
with a smaller load step,
52
00:02:50,400 --> 00:02:54,270
namely indicated by these
green lines here.
53
00:02:54,270 --> 00:02:58,030
The solution is obtained
corresponding to that
54
00:02:58,030 --> 00:03:01,850
configuration, corresponding
to that configuration, that
55
00:03:01,850 --> 00:03:04,570
configuration, which corresponds
in this particular
56
00:03:04,570 --> 00:03:08,290
case now, to the level of r4,
because we have three load
57
00:03:08,290 --> 00:03:10,400
steps to reach now r4.
58
00:03:10,400 --> 00:03:15,470
And say, at this point now,
again, for example, we may
59
00:03:15,470 --> 00:03:17,970
encounter solution
difficulties.
60
00:03:17,970 --> 00:03:21,400
The analyst is not able to
obtain this solution here.
61
00:03:21,400 --> 00:03:25,540
Once again, the load step
has to be smaller.
62
00:03:25,540 --> 00:03:29,370
To continue the analysis, a
restart is necessary at this
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00:03:29,370 --> 00:03:31,390
particular configuration.
64
00:03:31,390 --> 00:03:35,020
And like this the analyst tries
to get closer and closer
65
00:03:35,020 --> 00:03:37,600
to the collapse load.
66
00:03:37,600 --> 00:03:42,190
The conclusion is that we have
difficulties in calculating
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00:03:42,190 --> 00:03:46,890
the collapse loads when we
use the techniques that I
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00:03:46,890 --> 00:03:50,690
described in the last lecture.
69
00:03:50,690 --> 00:03:55,860
Of course, if you restart
enough, if you use small
70
00:03:55,860 --> 00:03:59,000
enough to load steps, you will
ultimately get very close to
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00:03:59,000 --> 00:04:02,030
this collapse load, but this
is tedious in a practical
72
00:04:02,030 --> 00:04:04,730
analysis, and we would like to
have really an automatic
73
00:04:04,730 --> 00:04:09,900
scheme that directly can trace
out the collapse load and you
74
00:04:09,900 --> 00:04:14,460
can also go into the
post-collapse response, which
75
00:04:14,460 --> 00:04:19,740
is the response beyond this
ultimate limit load here.
76
00:04:19,740 --> 00:04:24,830
Well, I've prepared some view
graphs to show you, discuss
77
00:04:24,830 --> 00:04:29,730
with you, a solution technique
that actually can be employed
78
00:04:29,730 --> 00:04:34,680
to trace automatically load
displacement response out, and
79
00:04:34,680 --> 00:04:37,980
go beyond the collapse
load, as well.
80
00:04:37,980 --> 00:04:41,620
The idea is that we want to
obtain more rapid convergence
81
00:04:41,620 --> 00:04:44,170
in each load step.
82
00:04:44,170 --> 00:04:48,230
We want to have the program
automatically select load
83
00:04:48,230 --> 00:04:50,200
increments.
84
00:04:50,200 --> 00:04:53,440
And we want to also be
able to solve for the
85
00:04:53,440 --> 00:04:55,120
post-buckling response.
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00:04:55,120 --> 00:04:59,980
Now, this here means, of course,
that a priori, prior
87
00:04:59,980 --> 00:05:03,400
to the analysis, you may not
know, in fact you do not know,
88
00:05:03,400 --> 00:05:06,880
for which loads you will
obtain the solution.
89
00:05:06,880 --> 00:05:10,190
We get the total load
displacement response curve,
90
00:05:10,190 --> 00:05:13,190
but the discrete load levels
at which the response was
91
00:05:13,190 --> 00:05:18,230
calculated is being decided
by the program.
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00:05:18,230 --> 00:05:21,380
An effective solution would
proceed as follows,
93
00:05:21,380 --> 00:05:23,020
schematically, of course.
94
00:05:23,020 --> 00:05:25,280
Here we have the load axis.
95
00:05:25,280 --> 00:05:27,850
Here we have the displacement
axis.
96
00:05:27,850 --> 00:05:30,430
The solution scheme would
start with large load
97
00:05:30,430 --> 00:05:35,190
increments in the region where
the response is almost linear,
98
00:05:35,190 --> 00:05:40,720
then cut the load increments
to a smaller size, and
99
00:05:40,720 --> 00:05:44,830
capture, of course, the ultimate
limit load here, and
100
00:05:44,830 --> 00:05:48,650
then decrease the load, as shown
here, to go into the
101
00:05:48,650 --> 00:05:51,080
post-buckling, post-collapse
response.
102
00:05:51,080 --> 00:05:55,000
And the solution scheme, of
course, could continue up this
103
00:05:55,000 --> 00:05:57,240
branch as well.
104
00:05:57,240 --> 00:06:01,280
And the solution scheme should
automatically select the load
105
00:06:01,280 --> 00:06:05,190
step sizes, depending on the
convergence that has been
106
00:06:05,190 --> 00:06:10,620
encountered in the previous
load steps.
107
00:06:10,620 --> 00:06:16,040
We compute now, t plus delta tr
using this equation here.
108
00:06:16,040 --> 00:06:19,810
Notice that in this equation,
t plus delta t lambda is an
109
00:06:19,810 --> 00:06:22,600
unknown scaler that is going
to be determined by the
110
00:06:22,600 --> 00:06:26,920
program, and r is a vector that
gives a particular load
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00:06:26,920 --> 00:06:28,095
distribution.
112
00:06:28,095 --> 00:06:30,280
r is constant.
113
00:06:30,280 --> 00:06:32,500
It may contain the contributions
of soft surface
114
00:06:32,500 --> 00:06:37,050
pressures, of concentrated
loads, et cetera.
115
00:06:37,050 --> 00:06:40,030
The point is that r is constant
and the program will
116
00:06:40,030 --> 00:06:42,390
automatically adjust lambda.
117
00:06:42,390 --> 00:06:46,340
Therefore, we assume in our
collapse analysis that the
118
00:06:46,340 --> 00:06:52,120
loads are increasing and
decreasing all in the same way
119
00:06:52,120 --> 00:06:54,190
as decided by one scale.
120
00:06:54,190 --> 00:06:58,170
Of course, one could extend
such an algorithm by
121
00:06:58,170 --> 00:07:02,460
introducing another vector,
calling this r1, introducing a
122
00:07:02,460 --> 00:07:05,020
vector r2, with another
scaler.
123
00:07:05,020 --> 00:07:08,000
And then have two scalers
that have to be adjusted
124
00:07:08,000 --> 00:07:09,830
automatically by the program.
125
00:07:09,830 --> 00:07:12,300
And perhaps one can even think
of three scalers, and so on,
126
00:07:12,300 --> 00:07:15,350
but then the algorithm would
become quite complicated.
127
00:07:15,350 --> 00:07:19,040
So we look at a scheme that only
has one scaler here, and
128
00:07:19,040 --> 00:07:22,680
one load vector that
remains constant
129
00:07:22,680 --> 00:07:24,400
throughout the solution.
130
00:07:24,400 --> 00:07:28,830
As an example here, you
see in blue the load
131
00:07:28,830 --> 00:07:32,020
intensity, t lambda r.
132
00:07:32,020 --> 00:07:35,740
Notice a concentrated load here,
a pressure load here,
133
00:07:35,740 --> 00:07:40,610
and at the time t plus delta t
we have the red intensity, t
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00:07:40,610 --> 00:07:45,750
plus delta t lambda r, and of
course this concentrated load
135
00:07:45,750 --> 00:07:50,530
has increased in the same way as
the pressure has increased,
136
00:07:50,530 --> 00:07:53,080
as shown right here.
137
00:07:53,080 --> 00:07:57,260
The basic approach of the
solution scheme is shown on
138
00:07:57,260 --> 00:07:58,700
this view graph.
139
00:07:58,700 --> 00:08:01,650
Here we have the low
displacement curve that we are
140
00:08:01,650 --> 00:08:05,150
looking for, shown
in red again.
141
00:08:05,150 --> 00:08:07,880
Of course, loads plotted
vertically, displacements
142
00:08:07,880 --> 00:08:09,640
plotted horizontally.
143
00:08:09,640 --> 00:08:15,250
Now, notice that if this is an
equilibrium point, the red
144
00:08:15,250 --> 00:08:19,950
dot, and if you as the analyst
were to chose an increment in
145
00:08:19,950 --> 00:08:24,480
load shown by this
black line--
146
00:08:24,480 --> 00:08:27,440
I'm putting my pointer
now onto it--
147
00:08:27,440 --> 00:08:32,140
then you can see that this black
line defined by this
148
00:08:32,140 --> 00:08:39,860
load increment would mean that
we have a large number of
149
00:08:39,860 --> 00:08:41,750
iterations necessary.
150
00:08:41,750 --> 00:08:45,390
If you need a large number of
iterations, to converge to the
151
00:08:45,390 --> 00:08:47,480
new equilibrium configuration.
152
00:08:47,480 --> 00:08:52,230
In fact, you can see that this
red curve is almost parallel
153
00:08:52,230 --> 00:08:56,470
to the black line, so we can
anticipate large convergence
154
00:08:56,470 --> 00:09:00,530
difficulties trying to get to
a solution with the load
155
00:09:00,530 --> 00:09:06,660
increment being equal from
here to that black line.
156
00:09:06,660 --> 00:09:10,330
The important feature of the
algorithm that I want to
157
00:09:10,330 --> 00:09:16,000
present to you now, is that this
load level is the first
158
00:09:16,000 --> 00:09:19,190
load level which we start in
the iteration, but then the
159
00:09:19,190 --> 00:09:24,130
algorithm automatically cuts
down this load level, as shown
160
00:09:24,130 --> 00:09:30,920
by this arc here, by the blue
arc, until this equilibrium
161
00:09:30,920 --> 00:09:33,230
point is solved for.
162
00:09:33,230 --> 00:09:38,650
And this means that the
convergence is much increased.
163
00:09:38,650 --> 00:09:41,280
In other words, the algorithm
does not have great
164
00:09:41,280 --> 00:09:45,370
difficulties calculating this
particular equilibrium point.
165
00:09:45,370 --> 00:09:48,130
And from here, of course,
the same is repeated.
166
00:09:48,130 --> 00:09:52,640
And like this the algorithm
goes along this red
167
00:09:52,640 --> 00:09:57,160
low-displacement curve, and
traces this whole curve out.
168
00:09:57,160 --> 00:10:00,680
Let us look a bit at the
notation that I will be using,
169
00:10:00,680 --> 00:10:03,020
because it's important
that we get familiar
170
00:10:03,020 --> 00:10:05,600
with it at this stage.
171
00:10:05,600 --> 00:10:11,010
Notice t lambda r, of course,
is the load at time t.
172
00:10:11,010 --> 00:10:13,120
The corresponding displacement
is tu.
173
00:10:13,120 --> 00:10:15,650
We used that notation
earlier already.
174
00:10:15,650 --> 00:10:19,250
Notice that the increment in
displacement from time t to
175
00:10:19,250 --> 00:10:25,010
time t plus delta t is u,
shown here in green.
176
00:10:25,010 --> 00:10:29,110
Notice that the increment in the
load from time t to time t
177
00:10:29,110 --> 00:10:32,660
plus delta t is, of course,
given by the change in this
178
00:10:32,660 --> 00:10:34,920
load factor here.
179
00:10:34,920 --> 00:10:37,200
And that change in the load
factor is lambda.
180
00:10:40,560 --> 00:10:46,310
Now, this lambda value is also
given in this equation here.
181
00:10:46,310 --> 00:10:47,580
It's buried in there.
182
00:10:47,580 --> 00:10:51,380
Notice this lambda of i here.
183
00:10:51,380 --> 00:10:54,700
We are writing it as a sum all
of delta lambda k's We will
184
00:10:54,700 --> 00:11:01,300
solve for all these delta lambda
k's And this lambda i,
185
00:11:01,300 --> 00:11:05,940
when i becomes very large,
converges to the lambda that I
186
00:11:05,940 --> 00:11:07,670
have up here.
187
00:11:07,670 --> 00:11:11,660
Similarly, this ui which is
written as the sum of the
188
00:11:11,660 --> 00:11:17,930
delta uk's this ui converges
to the u
189
00:11:17,930 --> 00:11:19,890
that I'm showing here.
190
00:11:19,890 --> 00:11:25,730
So, u and lambda, shown in
green here, are the exact
191
00:11:25,730 --> 00:11:29,270
values, the values that you
want to converge to with
192
00:11:29,270 --> 00:11:32,560
lambda i and ui.
193
00:11:32,560 --> 00:11:36,150
Notice once again, delta lambda
k, summing all these
194
00:11:36,150 --> 00:11:39,640
delta lambda k's up we get
lambda i, and summing all the
195
00:11:39,640 --> 00:11:42,050
delta uk's up, we get ui.
196
00:11:42,050 --> 00:11:45,520
We should keep that in mind for
the discussion that you
197
00:11:45,520 --> 00:11:47,870
want to go through just now.
198
00:11:47,870 --> 00:11:52,090
The governing equations
are now as follows.
199
00:11:52,090 --> 00:11:55,420
On the left hand side we have
a tension stiffness matrix.
200
00:11:55,420 --> 00:11:57,890
In the modified Newton iteration
we would have tau
201
00:11:57,890 --> 00:11:59,670
equal to t.
202
00:11:59,670 --> 00:12:02,500
Delta ui is our displacement
increment vector.
203
00:12:02,500 --> 00:12:05,680
This is here, this
scaler, t plus
204
00:12:05,680 --> 00:12:10,270
delta t lambda i, unknown.
205
00:12:10,270 --> 00:12:12,460
This one we know.
206
00:12:12,460 --> 00:12:15,960
And that one we want to
calculate in the iteration i.
207
00:12:15,960 --> 00:12:18,740
In other words, this is an
increment that is unknown at
208
00:12:18,740 --> 00:12:20,940
the beginning of
the iteration.
209
00:12:20,940 --> 00:12:24,060
Here we have the nodal point
force vector corresponding to
210
00:12:24,060 --> 00:12:27,510
the elements stresses at time t
plus delta t, and at the end
211
00:12:27,510 --> 00:12:29,970
of iteration i minus 1.
212
00:12:29,970 --> 00:12:34,240
Notice that this is a set of
linear equations in the
213
00:12:34,240 --> 00:12:37,680
unknowns that are used.
214
00:12:37,680 --> 00:12:42,030
There are n such unknowns, and
there's one more unknown here.
215
00:12:42,030 --> 00:12:46,410
So, we have n equations
in n, plus 1 unknowns.
216
00:12:46,410 --> 00:12:48,080
We need, therefore,
one more equation.
217
00:12:48,080 --> 00:12:52,655
And that equation is given by
this constraint equation.
218
00:12:52,655 --> 00:12:59,300
The f here on delta lambda and
delta u, some constraint
219
00:12:59,300 --> 00:13:02,650
equation here, gives us the
additional equation that we
220
00:13:02,650 --> 00:13:07,380
need for the solution these n
plus 1 coupled equations.
221
00:13:07,380 --> 00:13:10,780
The unknowns, of course,
are these values here--
222
00:13:10,780 --> 00:13:13,630
n plus 1 unknowns.
223
00:13:13,630 --> 00:13:18,320
To solve these equilibrium
equations, we can rewrite them
224
00:13:18,320 --> 00:13:19,570
as shown here.
225
00:13:21,850 --> 00:13:25,660
In other words, the equations
on the finite element
226
00:13:25,660 --> 00:13:29,550
displacements can be split up,
if you look at the previous
227
00:13:29,550 --> 00:13:33,150
view graph, directly
into two sets of
228
00:13:33,150 --> 00:13:35,240
equations as shown here.
229
00:13:35,240 --> 00:13:40,280
The interesting point is that,
this equation here does not
230
00:13:40,280 --> 00:13:42,780
involve a delta lambda.
231
00:13:42,780 --> 00:13:47,380
This equation here does not
involve a delta lambda either.
232
00:13:47,380 --> 00:13:51,260
Now, having calculated from this
equation this vector, and
233
00:13:51,260 --> 00:13:55,240
from that equation that vector,
we can directly write
234
00:13:55,240 --> 00:13:59,240
delta u, as shown here, of
course, now involving the
235
00:13:59,240 --> 00:14:01,550
delta lambda.
236
00:14:01,550 --> 00:14:06,060
And this is a form of equation
for delta u that we will be
237
00:14:06,060 --> 00:14:10,150
using a little later.
238
00:14:10,150 --> 00:14:14,710
And the first constraint
equation I'd like to introduce
239
00:14:14,710 --> 00:14:18,250
you to, the one that I briefly
mentioned earlier already,
240
00:14:18,250 --> 00:14:22,430
where we talked about an arc
which is used to swing the
241
00:14:22,430 --> 00:14:27,300
load level around, so as to get
to the low displacement
242
00:14:27,300 --> 00:14:31,500
curve very quickly, and that
constraint equation is this
243
00:14:31,500 --> 00:14:35,290
spherical constant arc-length
criterion,
244
00:14:35,290 --> 00:14:37,030
which is written here.
245
00:14:37,030 --> 00:14:42,060
Notice here lambda i, here ui.
246
00:14:42,060 --> 00:14:45,600
Of course both of these
quantities create, in essence,
247
00:14:45,600 --> 00:14:48,640
beta factor and delta
l squared.
248
00:14:48,640 --> 00:14:56,760
Delta l is set at the beginning
of the load step.
249
00:14:56,760 --> 00:14:58,890
Delta l, in other words,
is constant
250
00:14:58,890 --> 00:15:00,640
throughout the load step.
251
00:15:00,640 --> 00:15:03,670
The value of delta l is chosen
based on what has happened in
252
00:15:03,670 --> 00:15:05,670
the previous load steps.
253
00:15:05,670 --> 00:15:10,410
Having chosen delta l, you can
calculate the right hand side,
254
00:15:10,410 --> 00:15:14,450
and on the left hand side, we
have a constraint between the
255
00:15:14,450 --> 00:15:19,340
increment in lambda, and the
increment in delta u.
256
00:15:19,340 --> 00:15:23,990
Remember, lambda I is the sum of
all of the delta lambda k's
257
00:15:23,990 --> 00:15:28,180
and ui is the sum of all of the
delta uk's I just showed
258
00:15:28,180 --> 00:15:30,170
that on the previous
view graph.
259
00:15:30,170 --> 00:15:33,820
Therefore, if we substitute
here, we get a constraint
260
00:15:33,820 --> 00:15:39,560
equation between the delta uk,
and the delta lambda k.
261
00:15:39,560 --> 00:15:43,490
Well, here we have the
definitions once more.
262
00:15:43,490 --> 00:15:48,780
Notice beta is a normalizing
factor which is applied in
263
00:15:48,780 --> 00:15:50,840
order to make these terms
here dimensionless.
264
00:15:54,540 --> 00:15:57,490
The equation may be
solved as follows.
265
00:15:57,490 --> 00:16:02,380
Using that lambda is equal
to this equation here.
266
00:16:02,380 --> 00:16:05,940
ui is given as that.
267
00:16:05,940 --> 00:16:09,370
Of course, this here can be
expanded using the information
268
00:16:09,370 --> 00:16:12,000
that we discussed earlier.
269
00:16:12,000 --> 00:16:15,640
And substituting for this value
and that value into the
270
00:16:15,640 --> 00:16:18,710
constraint equation, we directly
obtain a quadratic
271
00:16:18,710 --> 00:16:23,400
equation in delta lambda i
Remember that these two
272
00:16:23,400 --> 00:16:28,540
vectors here are, of
course, known.
273
00:16:28,540 --> 00:16:31,110
These two vectors are known,
and that one is known also,
274
00:16:31,110 --> 00:16:33,980
because that was established
in the previous iteration.
275
00:16:37,150 --> 00:16:44,460
If we geometrically interpret
this solution scheme, we find
276
00:16:44,460 --> 00:16:45,390
the following.
277
00:16:45,390 --> 00:16:47,390
Here we have our equilibrium
point.
278
00:16:47,390 --> 00:16:50,670
Here is delta lambda, which
is constant throughout the
279
00:16:50,670 --> 00:16:52,430
solution step.
280
00:16:52,430 --> 00:16:54,865
And let's assume the
use of modified
281
00:16:54,865 --> 00:16:57,330
Newton-Raphson iteration.
282
00:16:57,330 --> 00:17:02,260
We start off with this tangent
here at this solution point,
283
00:17:02,260 --> 00:17:08,420
the equilibrium point, and we
iterate around this arc using
284
00:17:08,420 --> 00:17:14,630
the constant stiffness matrix
at each of these points.
285
00:17:14,630 --> 00:17:19,740
In the first iteration,
we get to this point.
286
00:17:19,740 --> 00:17:21,359
Second iteration, that point.
287
00:17:21,359 --> 00:17:22,920
Third iteration, that point.
288
00:17:22,920 --> 00:17:27,109
And that signified also
looking here to the
289
00:17:27,109 --> 00:17:28,060
displacements.
290
00:17:28,060 --> 00:17:32,010
Notice t plus delta t u1 is
the displacement value
291
00:17:32,010 --> 00:17:35,210
calculated in the
first iteration.
292
00:17:35,210 --> 00:17:38,030
This is the one calculated
in the second iteration.
293
00:17:38,030 --> 00:17:41,820
That is the one calculated
in the third iteration.
294
00:17:41,820 --> 00:17:44,220
We use a constant stiffness
matrix here.
295
00:17:44,220 --> 00:17:47,660
Of course, we could also change
the stiffness matrix,
296
00:17:47,660 --> 00:17:51,290
go in other words, to a
stiffness matrix that has been
297
00:17:51,290 --> 00:17:55,230
updated in the iteration
corresponding to the current
298
00:17:55,230 --> 00:17:57,970
stress and displacement
conditions, and then
299
00:17:57,970 --> 00:18:00,110
convergence would
be more rapid.
300
00:18:00,110 --> 00:18:03,090
But of course there would also
be a higher cost involved,
301
00:18:03,090 --> 00:18:05,590
because whenever we calculate
with a new stiffness matrix,
302
00:18:05,590 --> 00:18:07,780
and we triangularize that
stiffness matrix, there's a
303
00:18:07,780 --> 00:18:10,880
considerable cost involved
in doing so.
304
00:18:10,880 --> 00:18:14,080
This is one scheme, and we
find this scheme quite
305
00:18:14,080 --> 00:18:19,300
effective, except that when we
want to calculate limit loads,
306
00:18:19,300 --> 00:18:23,330
we find that near the limit load
the scheme can require
307
00:18:23,330 --> 00:18:24,860
quite a number of iterations.
308
00:18:24,860 --> 00:18:27,430
And so, we have been looking for
other schemes that might
309
00:18:27,430 --> 00:18:31,140
do better in the range
of the limit load.
310
00:18:31,140 --> 00:18:35,960
Well, here we have one other
scheme, namely the "constant"
311
00:18:35,960 --> 00:18:39,320
increment of external
work criterion.
312
00:18:39,320 --> 00:18:42,700
We have here quotes on the
constant, because we are
313
00:18:42,700 --> 00:18:47,860
setting w on the right hand side
in the first iteration,
314
00:18:47,860 --> 00:18:51,410
as shown here, and in the
next iteration then,
315
00:18:51,410 --> 00:18:53,100
we have a zero here.
316
00:18:53,100 --> 00:19:00,040
Notice w is a value that is
calculated in each load step,
317
00:19:00,040 --> 00:19:05,890
and as soon as we find that
the first scheme-- and
318
00:19:05,890 --> 00:19:09,740
obviously, constant arc lengths
criterion scheme--
319
00:19:09,740 --> 00:19:13,800
has difficulties converging, or
marches too slow, then we
320
00:19:13,800 --> 00:19:16,370
switch to this "constant"
increment of external work
321
00:19:16,370 --> 00:19:21,160
criterion, with a w from the
previous step, the increment
322
00:19:21,160 --> 00:19:24,200
of external work from
the previous step.
323
00:19:24,200 --> 00:19:28,620
We use that for the current
step, and then this equation
324
00:19:28,620 --> 00:19:32,290
here gives us a constraint
on delta lambda 1.
325
00:19:32,290 --> 00:19:37,430
Notice that this equation here
means really that we are
326
00:19:37,430 --> 00:19:44,390
looking at is this area here,
the shaded area, being w, and
327
00:19:44,390 --> 00:19:49,130
we want to have the delta lambda
1 in such a way that
328
00:19:49,130 --> 00:19:55,360
this shaded area w here is equal
to what we have had as
329
00:19:55,360 --> 00:20:00,400
the increment of external work
in the previous load step.
330
00:20:00,400 --> 00:20:05,010
Well, this gives us delta
lambda (1), and having
331
00:20:05,010 --> 00:20:09,930
obtained delta lambda (1),
we can go into the next
332
00:20:09,930 --> 00:20:15,440
iterations, 2 to 3, and so on
for i, and as I mentioned
333
00:20:15,440 --> 00:20:18,940
already earlier, we then have a
zero on the right hand side.
334
00:20:18,940 --> 00:20:21,670
Otherwise the same
left hand side.
335
00:20:21,670 --> 00:20:25,230
And this left hand side
has now two solutions.
336
00:20:25,230 --> 00:20:30,330
This solution here is
disregarded, because this
337
00:20:30,330 --> 00:20:34,380
solution would reverse the
direction of the load, and we
338
00:20:34,380 --> 00:20:35,790
don't want to admit that.
339
00:20:35,790 --> 00:20:39,930
We want to basically have the
load increase more and more
340
00:20:39,930 --> 00:20:42,830
until the total collapse of the
structure is reached, and
341
00:20:42,830 --> 00:20:45,980
then of course it may decrease
to go into the post-buckling
342
00:20:45,980 --> 00:20:47,960
response, but we don't want
to have the load totally
343
00:20:47,960 --> 00:20:52,780
reversed, and so this is a
solution that we now use for
344
00:20:52,780 --> 00:20:55,370
delta lambda (i).
345
00:20:55,370 --> 00:21:00,120
Notice that for a single degree
of freedom system, we
346
00:21:00,120 --> 00:21:04,390
would immediately find that
delta u (2) must be zero.
347
00:21:04,390 --> 00:21:04,730
Why?
348
00:21:04,730 --> 00:21:08,270
Because we notice that the load
vector is orthogonal to
349
00:21:08,270 --> 00:21:09,990
the displacement vector.
350
00:21:09,990 --> 00:21:12,450
Now, for a single degree of
freedom system, of course,
351
00:21:12,450 --> 00:21:19,230
each of these are just numbers,
and since the r is
352
00:21:19,230 --> 00:21:22,330
not zero, because that is of
course a prescribed value
353
00:21:22,330 --> 00:21:25,890
prior to the analysis, this
is not zero, delta
354
00:21:25,890 --> 00:21:28,340
u (2) must be zero.
355
00:21:28,340 --> 00:21:31,740
And that shows already the
effectiveness, just show it a
356
00:21:31,740 --> 00:21:33,845
bit, of this algorithm.
357
00:21:36,690 --> 00:21:40,310
All the algorithm altogether
is now as follows.
358
00:21:40,310 --> 00:21:43,060
We specify r.
359
00:21:43,060 --> 00:21:47,520
In other words, the analyst
has to specify the load
360
00:21:47,520 --> 00:21:52,210
distribution, concentrated
loads, distributed loads, that
361
00:21:52,210 --> 00:21:56,620
shall be dealt with in this
collapse analysis, and also
362
00:21:56,620 --> 00:22:00,850
the displacement at one degree
of freedom corresponding to
363
00:22:00,850 --> 00:22:02,730
delta t lambda.
364
00:22:02,730 --> 00:22:07,120
This is done to just start
the algorithm.
365
00:22:07,120 --> 00:22:12,160
Once the analyst has specified
the displacement at one degree
366
00:22:12,160 --> 00:22:17,900
of freedom, the program can
solve for delta tu--
367
00:22:17,900 --> 00:22:20,680
in other words, all the
other displacements--
368
00:22:20,680 --> 00:22:25,470
and this delta tu corresponds
to delta t lambda.
369
00:22:25,470 --> 00:22:29,750
We feel that to start the
algorithm it is easier for the
370
00:22:29,750 --> 00:22:34,500
analyst to prescribe the
displacement at one node
371
00:22:34,500 --> 00:22:39,290
corresponding to delta
t lambda--
372
00:22:39,290 --> 00:22:43,440
Rather, to prescribe this
displacement than to prescribe
373
00:22:43,440 --> 00:22:47,170
the first load level, because
if you don't have much of an
374
00:22:47,170 --> 00:22:50,600
idea how the structure really
behaves, it can be very
375
00:22:50,600 --> 00:22:54,500
difficult to give a load level
that is reasonable, and is not
376
00:22:54,500 --> 00:22:57,040
too close to the ultimate limit
load, or is already
377
00:22:57,040 --> 00:22:59,040
perhaps even beyond the
ultimate limit load.
378
00:22:59,040 --> 00:23:03,980
So we feel that this is
a good way to proceed.
379
00:23:03,980 --> 00:23:07,600
For example, if you analyze the
collapse of a shell, you
380
00:23:07,600 --> 00:23:12,130
may want to pick a node that
you know will have non-zero
381
00:23:12,130 --> 00:23:17,510
displacement and assign at that
node a displacement, say,
382
00:23:17,510 --> 00:23:20,250
one third of the thickness
of the shell.
383
00:23:20,250 --> 00:23:23,070
That displacement then would
give you delta t lambda, and
384
00:23:23,070 --> 00:23:26,320
surely would be less
than the ultimate
385
00:23:26,320 --> 00:23:28,860
limit load of the shell.
386
00:23:28,860 --> 00:23:32,020
This means, once you have solved
for delta t lambda and
387
00:23:32,020 --> 00:23:35,620
delta tu, we can set delta l.
388
00:23:35,620 --> 00:23:38,260
The arc lengths for the
next load steps.
389
00:23:38,260 --> 00:23:40,475
And now the program does
everything automatically.
390
00:23:40,475 --> 00:23:46,730
It selects the increment in
displacements and loads using
391
00:23:46,730 --> 00:23:50,340
this arc-lengths criterion
that we just discussed.
392
00:23:50,340 --> 00:23:53,880
We use 1, which is the
arc-lengths criterion, for the
393
00:23:53,880 --> 00:23:55,280
next load steps.
394
00:23:55,280 --> 00:24:00,280
We calculate w for
each load step.
395
00:24:00,280 --> 00:24:03,480
When w does not change
appreciably, or when there are
396
00:24:03,480 --> 00:24:06,830
difficulties with the
arc-lengths criterion, with 1,
397
00:24:06,830 --> 00:24:10,310
in other words, then we switch
to our constant increment and
398
00:24:10,310 --> 00:24:11,640
external work criterion.
399
00:24:11,640 --> 00:24:13,680
We call that the 2 criterion.
400
00:24:16,490 --> 00:24:21,820
Once again, notice that delta
l is calculated in each load
401
00:24:21,820 --> 00:24:28,380
step, and it is adjusted based
on the number of iterations
402
00:24:28,380 --> 00:24:32,870
that we have been using in
the previous load steps.
403
00:24:32,870 --> 00:24:38,390
In other words, if you go, say,
up to load step three,
404
00:24:38,390 --> 00:24:41,860
and you have, say, obtained
a solution up to load step
405
00:24:41,860 --> 00:24:48,830
three, the program has used a
certain delta in load step
406
00:24:48,830 --> 00:24:52,310
three, and now the program
looks at the number of
407
00:24:52,310 --> 00:24:55,980
iterations that were performed
to get to the equilibrium
408
00:24:55,980 --> 00:24:59,020
position at the end of
load steps three.
409
00:24:59,020 --> 00:25:02,420
Based on these number of
iterations, the program then
410
00:25:02,420 --> 00:25:08,200
cuts delta l down, or makes
delta l larger, depending on
411
00:25:08,200 --> 00:25:11,580
whether the iterations were very
many that you used, or
412
00:25:11,580 --> 00:25:15,110
were very few to get to the
equilibrium configuration at
413
00:25:15,110 --> 00:25:16,900
load step three.
414
00:25:16,900 --> 00:25:20,430
So, the a program automatically
adjusts delta l.
415
00:25:20,430 --> 00:25:24,660
Also notice that the stiffness
matrix is recalculated when
416
00:25:24,660 --> 00:25:26,860
convergence is slow.
417
00:25:26,860 --> 00:25:30,320
In fact, a full Newton-Raphson
iteration is performed
418
00:25:30,320 --> 00:25:34,050
automatically, the program
switched automatically from
419
00:25:34,050 --> 00:25:35,810
the modified Newton-Raphson
iteration to the full
420
00:25:35,810 --> 00:25:38,220
Newton-Raphson iteration
when it is
421
00:25:38,220 --> 00:25:41,090
deemed to be more effective.
422
00:25:41,090 --> 00:25:45,250
This is the automatic load
incrementation scheme that I
423
00:25:45,250 --> 00:25:47,220
wanted to discuss with you.
424
00:25:47,220 --> 00:25:51,210
And I'd now like to go over
to one other topic, a very
425
00:25:51,210 --> 00:25:54,680
important topic as well in the
analysis of the nonlinear
426
00:25:54,680 --> 00:25:58,900
response of structures, and
this is the linearized
427
00:25:58,900 --> 00:26:01,390
buckling analysis.
428
00:26:01,390 --> 00:26:07,170
In the linearized buckling
analysis, we want to predict a
429
00:26:07,170 --> 00:26:11,930
collapse load, the buckling load
of the structure via this
430
00:26:11,930 --> 00:26:14,740
criterion here.
431
00:26:14,740 --> 00:26:19,590
The determinant we know at
the collapse point of the
432
00:26:19,590 --> 00:26:23,600
stiffness matrix is zero.
433
00:26:23,600 --> 00:26:25,400
The matrix is, another
words, singular.
434
00:26:28,050 --> 00:26:34,770
The criterion that we use here
determined of k is equal to
435
00:26:34,770 --> 00:26:38,390
zero, of course, means
once again
436
00:26:38,390 --> 00:26:39,940
the k matrix is singular.
437
00:26:39,940 --> 00:26:44,440
And that means that we really
have a solution to this set of
438
00:26:44,440 --> 00:26:46,880
equations that is nontrivial.
439
00:26:46,880 --> 00:26:49,120
The trivial solution, of course,
would always be u star
440
00:26:49,120 --> 00:26:53,010
is equal to zero, but that is
not a real solution that we're
441
00:26:53,010 --> 00:26:54,620
interested in.
442
00:26:54,620 --> 00:27:00,870
A solution, a nontrivial
solution exists to this set of
443
00:27:00,870 --> 00:27:04,450
equations only in the case
when tau k is singular.
444
00:27:04,450 --> 00:27:07,330
And, of course, that is the
same as saying that the
445
00:27:07,330 --> 00:27:12,810
determined of k, tau
k is equal to zero.
446
00:27:12,810 --> 00:27:19,470
If tau k is singular the
structure is unstable, and the
447
00:27:19,470 --> 00:27:26,660
collapse load situation
has been reached.
448
00:27:26,660 --> 00:27:29,970
Let's look at what this
means physically.
449
00:27:29,970 --> 00:27:31,360
If we have here a beam.
450
00:27:31,360 --> 00:27:34,890
If we look at a beam, pinned
at this end, pinned at that
451
00:27:34,890 --> 00:27:39,710
end, subjected to a certain
load, then if the load takes
452
00:27:39,710 --> 00:27:44,930
on a certain value, the smallest
imbalance of the
453
00:27:44,930 --> 00:27:49,830
load, a very small load this
way, would immediately result
454
00:27:49,830 --> 00:27:51,870
in very large displacements.
455
00:27:51,870 --> 00:27:55,790
That's what the buckling
criterion tells us.
456
00:27:55,790 --> 00:28:01,640
In other words, at tau r, the
buckling load, the structure
457
00:28:01,640 --> 00:28:06,650
is unstable to any smallest
load imbalance.
458
00:28:06,650 --> 00:28:09,890
Of course, the material data for
the structure are given.
459
00:28:09,890 --> 00:28:12,410
And in this particular
case, we use the data
460
00:28:12,410 --> 00:28:15,490
of an elastic beam.
461
00:28:15,490 --> 00:28:18,460
In the linearized buckling
analysis we want to predict
462
00:28:18,460 --> 00:28:23,110
the load level, and the
buckling load shape
463
00:28:23,110 --> 00:28:30,420
corresponding to the buckling
situation, by linearizing
464
00:28:30,420 --> 00:28:33,860
about a particular configuration
that is not
465
00:28:33,860 --> 00:28:38,010
necessarily very close to
the buckling load level.
466
00:28:38,010 --> 00:28:42,340
In fact, we assume that tau k is
given via this relationship
467
00:28:42,340 --> 00:28:46,520
here, and tau r is given
via this relationship.
468
00:28:46,520 --> 00:28:53,040
Notice here t minus delta tk and
tk are known values, known
469
00:28:53,040 --> 00:28:56,960
stiffness matrices, and
similarly, this load vector,
470
00:28:56,960 --> 00:29:00,920
and that load vector, and that
one, are known values.
471
00:29:00,920 --> 00:29:02,750
Lambda is a scaler.
472
00:29:02,750 --> 00:29:05,220
So basically, what we are trying
to do in the linearized
473
00:29:05,220 --> 00:29:09,510
buckling analysis is to
linearize about the
474
00:29:09,510 --> 00:29:14,190
configuration, not necessarily
very close to the collapse
475
00:29:14,190 --> 00:29:19,990
load configuration, and
establish with this linearized
476
00:29:19,990 --> 00:29:27,270
relationship an equation from
which we can calculate an
477
00:29:27,270 --> 00:29:35,410
increment in load that gives
us an approximation to the
478
00:29:35,410 --> 00:29:38,160
actual collapse load.
479
00:29:38,160 --> 00:29:41,750
Pictorially, we are
doing this here.
480
00:29:41,750 --> 00:29:46,520
We have here the load
displacement curve in red.
481
00:29:46,520 --> 00:29:51,390
We have at displacement
t minus delta
482
00:29:51,390 --> 00:29:54,440
tu, this load level.
483
00:29:54,440 --> 00:29:59,090
At displacement tu we have
that load level.
484
00:29:59,090 --> 00:30:04,360
And if you linearize about these
configurations here,
485
00:30:04,360 --> 00:30:07,130
using the stiffness matrices
corresponding to these two
486
00:30:07,130 --> 00:30:13,510
load levels, we can predict a
collapse load, which because
487
00:30:13,510 --> 00:30:17,500
of this linearization is, of
course, not exactly the
488
00:30:17,500 --> 00:30:20,990
collapse load, but we hope
that we are close to that
489
00:30:20,990 --> 00:30:22,240
actual collapse load.
490
00:30:24,970 --> 00:30:28,430
Pictorially here, we
do the following.
491
00:30:28,430 --> 00:30:33,500
Notice here we have the
k plotted and the
492
00:30:33,500 --> 00:30:36,070
lambda value plotted.
493
00:30:36,070 --> 00:30:37,890
Along this axis.
494
00:30:37,890 --> 00:30:43,010
At a particular value,
lambda equal to 1, we
495
00:30:43,010 --> 00:30:44,870
would have this point.
496
00:30:44,870 --> 00:30:48,250
Now let's look first along
this axis here.
497
00:30:48,250 --> 00:30:50,170
Here we have tk.
498
00:30:50,170 --> 00:30:52,710
Here we have t minus delta tk.
499
00:30:52,710 --> 00:30:56,260
Notice tk is smaller than
t minus delta tk.
500
00:30:56,260 --> 00:30:58,640
We are looking here, of course,
at a single degree of
501
00:30:58,640 --> 00:31:00,220
freedom case again.
502
00:31:00,220 --> 00:31:02,980
This means that the structure
becomes softer, we are getting
503
00:31:02,980 --> 00:31:06,730
closer to the collapse load.
504
00:31:06,730 --> 00:31:09,910
In our linearized buckling
analysis we put really through
505
00:31:09,910 --> 00:31:15,070
this point and that point
a straight line.
506
00:31:15,070 --> 00:31:20,560
That projects us to a point,
lambda 1, at which tau k is
507
00:31:20,560 --> 00:31:23,740
equal to zero.
508
00:31:23,740 --> 00:31:26,800
Corresponding to what we're
doing here with the stiffness
509
00:31:26,800 --> 00:31:30,570
matrix, we have here also
picture for the loads.
510
00:31:30,570 --> 00:31:37,500
Notice t minus delta tr here, tr
there, and a straight line
511
00:31:37,500 --> 00:31:40,670
through it brings
us up to tau r.
512
00:31:40,670 --> 00:31:43,750
Tau r is the collapse load.
513
00:31:43,750 --> 00:31:47,490
Tau r is the collapse load,
linearized collapse load,
514
00:31:47,490 --> 00:31:52,290
corresponding to the value
that we have here
515
00:31:52,290 --> 00:31:53,910
obtained for lambda 1.
516
00:31:56,690 --> 00:32:00,930
The problem of solving for
lambda such that the
517
00:32:00,930 --> 00:32:06,610
determinant of tau k is zero is
really nothing else than an
518
00:32:06,610 --> 00:32:07,690
eigenproblem.
519
00:32:07,690 --> 00:32:12,060
And the eigenproblem that we
have to consider there is
520
00:32:12,060 --> 00:32:13,560
written down here.
521
00:32:13,560 --> 00:32:17,970
Notice phi is an eigenvector,
lambda is the eigenvalue--
522
00:32:17,970 --> 00:32:20,400
unknown at this point--
523
00:32:20,400 --> 00:32:28,080
and here we have a difference in
matrices that we know, and
524
00:32:28,080 --> 00:32:29,950
once again the eigenvector.
525
00:32:29,950 --> 00:32:31,740
Notice one interesting point.
526
00:32:31,740 --> 00:32:35,590
That to establish this
difference in matrix, all you
527
00:32:35,590 --> 00:32:38,520
need to do in the computer
program is to store the
528
00:32:38,520 --> 00:32:42,630
previous difference matrix, and
subtract from the previous
529
00:32:42,630 --> 00:32:45,750
difference matrix corresponding
to load step t
530
00:32:45,750 --> 00:32:49,220
minus delta t, the current
difference matrix.
531
00:32:49,220 --> 00:32:55,550
This is interesting to note that
because we don't need to
532
00:32:55,550 --> 00:32:57,740
separate out element difference
matrices.
533
00:32:57,740 --> 00:33:00,710
We don't need to talk about
nonlinear strain stiffness
534
00:33:00,710 --> 00:33:03,670
matrices coming from the
elements, and linear strain
535
00:33:03,670 --> 00:33:06,410
stiffness matrices coming
from the elements.
536
00:33:06,410 --> 00:33:10,300
We talked about, of course,
these particular element
537
00:33:10,300 --> 00:33:12,870
stiffness matrices in
previous lectures.
538
00:33:12,870 --> 00:33:16,160
We don't separate
any out here.
539
00:33:16,160 --> 00:33:19,390
In fact, what we're doing,
we simply take all of the
540
00:33:19,390 --> 00:33:23,110
contributions from the elements
into account here,
541
00:33:23,110 --> 00:33:26,180
and all of the contribution
towards element into account
542
00:33:26,180 --> 00:33:30,070
here, of course corresponding to
time t minus delta t here,
543
00:33:30,070 --> 00:33:33,280
and corresponding
to time t here.
544
00:33:33,280 --> 00:33:35,440
This therefore, is a very
simple operation in the
545
00:33:35,440 --> 00:33:36,690
computer program.
546
00:33:38,870 --> 00:33:44,380
And having established this
eigenvalue problem, we can
547
00:33:44,380 --> 00:33:48,990
schematically look at the
eigenvalues that would become
548
00:33:48,990 --> 00:33:50,610
calculated.
549
00:33:50,610 --> 00:33:54,320
Here we have some positive
eigenvalues indicated, and
550
00:33:54,320 --> 00:33:57,710
here we have some negative
eigenvalues indicated.
551
00:33:57,710 --> 00:34:00,930
Notice that there's a
possibility of negative
552
00:34:00,930 --> 00:34:03,450
eigenvalues as well.
553
00:34:03,450 --> 00:34:07,360
A physical problem that shows
how you can get a negative
554
00:34:07,360 --> 00:34:10,500
eigenvalue is shown in
this view graph.
555
00:34:10,500 --> 00:34:13,800
Here you have a structure, a
frame structure, so to say,
556
00:34:13,800 --> 00:34:15,960
that is subjected to
a compressive load
557
00:34:15,960 --> 00:34:17,739
and a tensile load.
558
00:34:17,739 --> 00:34:21,330
Now, notice that this
compressive load would of
559
00:34:21,330 --> 00:34:24,230
course initiate buckling here,
and there would be a
560
00:34:24,230 --> 00:34:26,690
corresponding to a positive
eigenvalue.
561
00:34:26,690 --> 00:34:30,239
However, this tensile load has
to reverse its sign in order
562
00:34:30,239 --> 00:34:34,060
for this member to buckle, and
that would correspond then to
563
00:34:34,060 --> 00:34:36,360
a negative eigenvalue.
564
00:34:36,360 --> 00:34:40,030
Now when we try to solve
eigenvalue problems that have
565
00:34:40,030 --> 00:34:43,360
negative and positive
eigenvalues, there can be
566
00:34:43,360 --> 00:34:46,280
difficulties with the eigenvalue
solution method
567
00:34:46,280 --> 00:34:47,320
that you're using.
568
00:34:47,320 --> 00:34:51,230
It is easier for a solution
method to simply deal with
569
00:34:51,230 --> 00:34:53,330
eigenvalues that are
only positive.
570
00:34:53,330 --> 00:34:58,360
And for that reason we are
reformulating the basic
571
00:34:58,360 --> 00:35:02,770
eigenvalue equation
into this form.
572
00:35:02,770 --> 00:35:06,560
Here you have now the eigenvalue
problem simply
573
00:35:06,560 --> 00:35:13,240
rewritten, really, with gamma
being the eigenvalue, and phi
574
00:35:13,240 --> 00:35:16,350
still being the eigenvector,
gamma of course, being a
575
00:35:16,350 --> 00:35:19,580
function of lambda.
576
00:35:19,580 --> 00:35:25,260
With this eigenvalue problem
then, we have only positive
577
00:35:25,260 --> 00:35:30,090
eigenvalues on gamma, of course,
and we are interested
578
00:35:30,090 --> 00:35:33,960
in finding the smallest
gamma value.
579
00:35:33,960 --> 00:35:36,450
Because the smallest gamma
value corresponds to the
580
00:35:36,450 --> 00:35:40,330
smallest lambda 1
positive value.
581
00:35:40,330 --> 00:35:43,900
And of course, sometimes you
also want to calculate gamma 2
582
00:35:43,900 --> 00:35:49,100
corresponding to the second
smallest positive lambda
583
00:35:49,100 --> 00:35:50,900
value, and so on.
584
00:35:50,900 --> 00:35:53,540
The negative lambda values
lie over here.
585
00:35:56,960 --> 00:36:00,650
The value of the linearized
buckling analysis can be
586
00:36:00,650 --> 00:36:02,400
summarized as follows.
587
00:36:02,400 --> 00:36:05,680
The buckling analysis is
not very expensive.
588
00:36:05,680 --> 00:36:10,340
It gives insight into possible
modes of failure.
589
00:36:10,340 --> 00:36:13,730
For applicability, however, the
pre-buckling displacements
590
00:36:13,730 --> 00:36:16,830
should be small.
591
00:36:16,830 --> 00:36:21,060
And very important it is that
the buckling analysis yields
592
00:36:21,060 --> 00:36:25,300
modes, buckling mode shapes,
that can be very effectively
593
00:36:25,300 --> 00:36:30,460
used to impose imperfections
onto a structure and to study
594
00:36:30,460 --> 00:36:33,970
the sensitivity of the structure
to imperfections.
595
00:36:33,970 --> 00:36:37,440
Many structures are very
sensitive, particularly shell
596
00:36:37,440 --> 00:36:40,350
structures, are very sensitive
to initial imperfections in
597
00:36:40,350 --> 00:36:44,620
the geometry, and this
linearized buckling analysis
598
00:36:44,620 --> 00:36:49,220
gives us the mechanisms and
means to calculate mode shapes
599
00:36:49,220 --> 00:36:52,580
that we can impose onto the
perfect structure as a
600
00:36:52,580 --> 00:36:56,010
geometric imperfection, onto
the geometry of the perfect
601
00:36:56,010 --> 00:37:00,420
structure, to study then the
behavior of the structure
602
00:37:00,420 --> 00:37:04,480
subject to these
imperfections.
603
00:37:04,480 --> 00:37:10,550
But it is very important that
the procedure be only employed
604
00:37:10,550 --> 00:37:14,100
with great care, because the
results can be quite
605
00:37:14,100 --> 00:37:14,840
misleading.
606
00:37:14,840 --> 00:37:17,280
The buckling loads that you
might calculate in the
607
00:37:17,280 --> 00:37:20,440
linearized buckling analysis,
these buckling loads may be
608
00:37:20,440 --> 00:37:24,490
way higher than the actual
buckling loads that you should
609
00:37:24,490 --> 00:37:26,330
be using in your design.
610
00:37:26,330 --> 00:37:30,560
And we will actually show some
examples in the next lecture
611
00:37:30,560 --> 00:37:35,020
pertaining to this particular
problem here.
612
00:37:35,020 --> 00:37:38,840
And one should always keep in
mind that the procedure really
613
00:37:38,840 --> 00:37:43,120
predicts only physically
realistic buckling loads when
614
00:37:43,120 --> 00:37:46,910
we have structures that
behave close to the
615
00:37:46,910 --> 00:37:48,460
Euler column type.
616
00:37:48,460 --> 00:37:53,150
In other words, that buckle
with very small initial
617
00:37:53,150 --> 00:37:57,640
displacements, or pre-buckling
displacements, I should say.
618
00:37:57,640 --> 00:37:59,930
Let's look at an example.
619
00:37:59,930 --> 00:38:03,930
Here we have the example
of an arch
620
00:38:03,930 --> 00:38:07,540
subjected to uniform pressure.
621
00:38:07,540 --> 00:38:11,050
The geometric data of the
arch are given here.
622
00:38:11,050 --> 00:38:12,890
The material data are
given as well.
623
00:38:12,890 --> 00:38:14,890
Notice it's an elastic arch.
624
00:38:14,890 --> 00:38:21,250
And it's an arch with cross
section b and h here, being
625
00:38:21,250 --> 00:38:23,680
both equal to 1.
626
00:38:23,680 --> 00:38:25,970
We also want to consider only
the two-dimensional
627
00:38:25,970 --> 00:38:27,200
motion of the arch.
628
00:38:27,200 --> 00:38:28,980
In other words, two-dimensional
action.
629
00:38:28,980 --> 00:38:33,700
We do not allow out-of-plane
bucking for this arch.
630
00:38:33,700 --> 00:38:40,240
Now the finite element model
that we select for the arch is
631
00:38:40,240 --> 00:38:43,800
a model of 10 2-node
isoparametric beam elements.
632
00:38:43,800 --> 00:38:46,780
We will talk about these
isoparametric beam elements--
633
00:38:46,780 --> 00:38:49,190
we call them also isobeam
elements--
634
00:38:49,190 --> 00:38:50,840
in a later lecture.
635
00:38:50,840 --> 00:38:55,780
And for the two-dimensional
motion of the arch, we model
636
00:38:55,780 --> 00:38:58,200
the complete arch.
637
00:38:58,200 --> 00:39:01,810
The purpose of the analysis is
to determine the collapse
638
00:39:01,810 --> 00:39:05,580
mechanism, the collapse load
level of the structure.
639
00:39:05,580 --> 00:39:11,250
And not only this particular
part, but we also want to
640
00:39:11,250 --> 00:39:16,150
calculate the post-collapse
response of the structure.
641
00:39:16,150 --> 00:39:22,750
Let us now go through the
solution that was performed
642
00:39:22,750 --> 00:39:24,140
for this arch.
643
00:39:24,140 --> 00:39:29,180
As a first step we calculated
the linearized buckling loads
644
00:39:29,180 --> 00:39:30,390
and corresponding mode shapes.
645
00:39:30,390 --> 00:39:32,480
And we calculated two.
646
00:39:32,480 --> 00:39:35,620
Interesting to note that the
first mode corresponds to this
647
00:39:35,620 --> 00:39:42,000
pressure, and it's an
anti-symmetric mode.
648
00:39:42,000 --> 00:39:48,010
The second buckling mode
shape looks like this.
649
00:39:48,010 --> 00:39:50,370
It's a symmetric buckling
mode shape.
650
00:39:50,370 --> 00:39:54,430
Notice that, in other words,
the anti-symmetric buckling
651
00:39:54,430 --> 00:39:59,660
mode corresponds to a lower
load than the symmetry
652
00:39:59,660 --> 00:40:02,320
buckling mode shape.
653
00:40:02,320 --> 00:40:05,350
Having performed the linearized
buckling analysis,
654
00:40:05,350 --> 00:40:08,580
we next use our automatic load
stepping algorithm to
655
00:40:08,580 --> 00:40:11,630
calculate the displacement
response of the arch as the
656
00:40:11,630 --> 00:40:13,000
load increases.
657
00:40:13,000 --> 00:40:16,630
Here on this view graph, we
have plotted the pressure
658
00:40:16,630 --> 00:40:19,680
vertically up here, and the
displacement of the center of
659
00:40:19,680 --> 00:40:21,410
the arch horizontally.
660
00:40:21,410 --> 00:40:26,790
Notice this black curve here
shows a computed response
661
00:40:26,790 --> 00:40:30,210
using about 60 steps
in the analysis.
662
00:40:30,210 --> 00:40:34,060
Notice it is a collapse load, of
course, here, and this is a
663
00:40:34,060 --> 00:40:36,950
post-collapse response.
664
00:40:36,950 --> 00:40:40,480
It's interesting to note that
the collapse load predicted
665
00:40:40,480 --> 00:40:42,950
using the buckling analysis
and linearized buckling
666
00:40:42,950 --> 00:40:48,240
analysis given by this blue line
lies below this actual
667
00:40:48,240 --> 00:40:52,600
ultimate load predicted for
using the automatic load
668
00:40:52,600 --> 00:40:54,870
stepping algorithm.
669
00:40:54,870 --> 00:40:58,470
And we have to ask ourselves
why is that the case?
670
00:40:58,470 --> 00:41:05,740
Well, the computed response of
the the perfect symmetry arch
671
00:41:05,740 --> 00:41:09,900
does not allow the
anti-symmetric displacements
672
00:41:09,900 --> 00:41:15,130
to take place, and it is the
anti-symmetric displacements
673
00:41:15,130 --> 00:41:20,550
that, of course, initialize,
so to say, anti-symmetric
674
00:41:20,550 --> 00:41:22,880
buckling response.
675
00:41:22,880 --> 00:41:25,910
We have computed the response
of a perfect symmetric arch
676
00:41:25,910 --> 00:41:31,580
that is subjected to a perfectly
symmetric load, and
677
00:41:31,580 --> 00:41:36,810
this does not allow the
anti-symmetric buckling mode
678
00:41:36,810 --> 00:41:39,800
of the arch to come
into effect.
679
00:41:39,800 --> 00:41:45,020
However, a real structure will
contain imperfections, and
680
00:41:45,020 --> 00:41:51,990
hence it will go into the
anti-symmetric behavior, and
681
00:41:51,990 --> 00:41:56,450
the actual collapse load will be
below the one that we have
682
00:41:56,450 --> 00:41:59,630
predicted in our first analysis
using the automatic
683
00:41:59,630 --> 00:42:02,180
load stepping algorithm, and
the results of which
684
00:42:02,180 --> 00:42:05,470
I've just shown you.
685
00:42:05,470 --> 00:42:09,640
Therefore, to really obtain a
realistic collapse load, we
686
00:42:09,640 --> 00:42:16,410
have to now impose onto the
perfect symmetry arch a
687
00:42:16,410 --> 00:42:22,550
geometric imperfection which
allows the anti-symmetric
688
00:42:22,550 --> 00:42:25,030
behavior to take place.
689
00:42:25,030 --> 00:42:31,470
And we do so by adding to the
geometry of the arch to the
690
00:42:31,470 --> 00:42:36,040
nodal point coordinates of the
arch a multiple of the
691
00:42:36,040 --> 00:42:39,580
anti-symmetric mode shape.
692
00:42:39,580 --> 00:42:43,400
This collapse mode is scaled
so that the magnitude of
693
00:42:43,400 --> 00:42:47,600
imperfection is less than 100.
694
00:42:47,600 --> 00:42:51,240
This resulting imperfect arch
is, of course, no longer
695
00:42:51,240 --> 00:42:55,830
symmetric, and we now solve for
the response of that arch
696
00:42:55,830 --> 00:43:00,210
using our automatic load
stepping algorithm again.
697
00:43:00,210 --> 00:43:03,700
This is the response
that you'll see.
698
00:43:03,700 --> 00:43:07,280
Pressure and displacement
of the arch.
699
00:43:07,280 --> 00:43:12,850
Notice this is well below the
linearized buckling analysis
700
00:43:12,850 --> 00:43:16,420
solution, the blue solution,
that I showed you on the
701
00:43:16,420 --> 00:43:19,380
earlier response graph.
702
00:43:19,380 --> 00:43:24,520
And this is here a much more
realistic estimate for the
703
00:43:24,520 --> 00:43:26,060
collapse load of the arch.
704
00:43:26,060 --> 00:43:29,270
This is the realistic estimate
that you would be using, for
705
00:43:29,270 --> 00:43:32,860
example, in the design of
the actual structure.
706
00:43:32,860 --> 00:43:35,540
It's also interesting to look at
the deflective shape of the
707
00:43:35,540 --> 00:43:40,390
structure, and here we have the
deflective shape of the
708
00:43:40,390 --> 00:43:45,400
perfect arch, of being
symmetric at this
709
00:43:45,400 --> 00:43:47,350
displacement value.
710
00:43:47,350 --> 00:43:52,430
And here we have a deflective
shape of the imperfect arch at
711
00:43:52,430 --> 00:43:54,500
this particular displacement
value.
712
00:43:54,500 --> 00:43:56,750
Of course, this one
is not a totally
713
00:43:56,750 --> 00:43:59,230
symmetric deflective shape.
714
00:43:59,230 --> 00:44:02,640
Well, we have discussed now
quite a number of solution
715
00:44:02,640 --> 00:44:07,290
schemes that we use to solve the
finite element equations
716
00:44:07,290 --> 00:44:09,160
in static non-linear analysis.
717
00:44:09,160 --> 00:44:13,100
We have looked in the previous
lecture at some schemes-- the
718
00:44:13,100 --> 00:44:15,880
full Newton-Raphson method, the
modified Newton-Raphson
719
00:44:15,880 --> 00:44:19,320
method, the BFGS method, the
initial stress method--
720
00:44:19,320 --> 00:44:21,535
in which we have to prescribe
the load levels.
721
00:44:21,535 --> 00:44:26,780
For each load step we have to
prescribe the load level prior
722
00:44:26,780 --> 00:44:30,570
to the analysis, by the input
of the computer program, and
723
00:44:30,570 --> 00:44:33,390
in this lecture I have shared
with you some experiences
724
00:44:33,390 --> 00:44:35,940
regarding an automatic load
stepping scheme regarding
725
00:44:35,940 --> 00:44:37,960
linearized buckling analysis.
726
00:44:37,960 --> 00:44:40,560
What we have not done yet,
really, in my opinion is to
727
00:44:40,560 --> 00:44:43,510
look at enough examples, and
that's what I would like to do
728
00:44:43,510 --> 00:44:44,630
in the next lecture.
729
00:44:44,630 --> 00:44:46,300
Thank you very much for
your attention.