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PROFESSOR: Ladies and gentlemen,
welcome to lecture
9
00:00:24,720 --> 00:00:26,040
number two.
10
00:00:26,040 --> 00:00:29,400
In this lecture, I would like to
discuss some basic concepts
11
00:00:29,400 --> 00:00:33,040
of finite element analysis with
regard to the analysis of
12
00:00:33,040 --> 00:00:35,590
continuous systems.
13
00:00:35,590 --> 00:00:38,710
We discussed in the first
lecture already some basic
14
00:00:38,710 --> 00:00:42,965
concepts of analysis of
discrete systems.
15
00:00:42,965 --> 00:00:47,080
However, in actuality, in the
analysis of a complex system,
16
00:00:47,080 --> 00:00:50,600
we are dealing with a continuous
system, and there
17
00:00:50,600 --> 00:00:54,660
are some additional basic
concepts that are used in
18
00:00:54,660 --> 00:00:57,310
analysis of continuous
systems, using
19
00:00:57,310 --> 00:00:58,870
finite element methods.
20
00:00:58,870 --> 00:01:02,150
And those additional concept
that are used, I want to
21
00:01:02,150 --> 00:01:04,069
discuss in this lecture.
22
00:01:04,069 --> 00:01:09,370
Well, when we talk about the
analysis of a continuous
23
00:01:09,370 --> 00:01:15,010
system, we can analyze that
system via a differential
24
00:01:15,010 --> 00:01:19,320
formulation or a variational
formulation.
25
00:01:19,320 --> 00:01:23,430
If we use a differential
formulation or variational
26
00:01:23,430 --> 00:01:27,920
formulation, of course we obtain
continuous variables,
27
00:01:27,920 --> 00:01:34,380
and we have an infinite number
of state variables, or rather,
28
00:01:34,380 --> 00:01:37,180
if we talk about displacements,
a U
29
00:01:37,180 --> 00:01:41,300
displacement, for example, of a
rod, as I will be discussing
30
00:01:41,300 --> 00:01:46,220
just now, we will have infinite
values of that
31
00:01:46,220 --> 00:01:48,520
displacement along the rod.
32
00:01:48,520 --> 00:01:51,100
In the differential formulation
or the variational
33
00:01:51,100 --> 00:01:54,330
formulation, we would have to
solve for that continuous
34
00:01:54,330 --> 00:01:55,970
variable along the rod.
35
00:01:55,970 --> 00:02:00,470
Well, we will also notice that
in the analysis of a complex
36
00:02:00,470 --> 00:02:04,940
system, we cannot solve the
differential equations that we
37
00:02:04,940 --> 00:02:09,539
are arriving at directly,
and we have to resort
38
00:02:09,539 --> 00:02:11,020
to numerical methods.
39
00:02:11,020 --> 00:02:14,840
Now, some concepts that have
been used for a long time are
40
00:02:14,840 --> 00:02:17,310
the weighted residual methods.
41
00:02:17,310 --> 00:02:19,980
These have been used by
Galerkin least squares
42
00:02:19,980 --> 00:02:24,410
approaches to solve the
differential equations that
43
00:02:24,410 --> 00:02:29,100
govern the equilibrium motion
of the system approximately.
44
00:02:29,100 --> 00:02:31,420
In the variational formulation,
the Ritz method
45
00:02:31,420 --> 00:02:35,580
has been used for quite
a long time.
46
00:02:35,580 --> 00:02:37,800
These are classical techniques,
therefore,
47
00:02:37,800 --> 00:02:40,280
weighted residual method
and the Ritz method.
48
00:02:40,280 --> 00:02:43,580
And what I want to show to you
in this lecture is how the
49
00:02:43,580 --> 00:02:47,780
finite element method is really
an extension off these
50
00:02:47,780 --> 00:02:51,110
methods, or how this method, the
finite element method, is
51
00:02:51,110 --> 00:02:55,050
related to these classical
techniques.
52
00:02:55,050 --> 00:03:00,180
Well, when we talk about the
differential formulation, we
53
00:03:00,180 --> 00:03:05,060
are looking at the differential
equilibrium, or
54
00:03:05,060 --> 00:03:07,020
the equilibrium of
a differential
55
00:03:07,020 --> 00:03:10,020
element of the system.
56
00:03:10,020 --> 00:03:15,350
Now, I want to show to you the
basic ideas of a differential
57
00:03:15,350 --> 00:03:20,240
formulation by looking at,
or analyzing, this rod.
58
00:03:20,240 --> 00:03:25,090
Here we have a rod that is fixed
at the left-hand side,
59
00:03:25,090 --> 00:03:27,120
on rollers.
60
00:03:27,120 --> 00:03:29,640
x is a variable along the rod.
61
00:03:29,640 --> 00:03:34,530
u is the displacement of that
rod into this direction.
62
00:03:34,530 --> 00:03:37,730
The rod is subjected, as shown
here, to a load , R0
63
00:03:37,730 --> 00:03:39,400
at its right end.
64
00:03:39,400 --> 00:03:44,420
Notice that in this analysis, we
assume that plane sections
65
00:03:44,420 --> 00:03:45,950
remain plane.
66
00:03:45,950 --> 00:03:52,300
In other words, a section that
was originally there at time t
67
00:03:52,300 --> 00:03:56,150
greater than 0 has
moved to here.
68
00:03:56,150 --> 00:03:59,830
And this movement here is
the u displacement.
69
00:03:59,830 --> 00:04:05,070
But notice that the vertical
section here remains vertical
70
00:04:05,070 --> 00:04:06,170
during the motion.
71
00:04:06,170 --> 00:04:12,280
So at every section, we have
only 1 degree of freedom.
72
00:04:12,280 --> 00:04:14,010
There's no rotation
of that section.
73
00:04:14,010 --> 00:04:17,810
However, this one degree
of freedom, u, varies
74
00:04:17,810 --> 00:04:20,320
continuously along the rod.
75
00:04:20,320 --> 00:04:23,900
Therefore, the rod itself has
really an infinite number of
76
00:04:23,900 --> 00:04:26,510
degrees of freedom.
77
00:04:26,510 --> 00:04:30,020
For this very simple
system, we could
78
00:04:30,020 --> 00:04:32,160
obtain an exact solution.
79
00:04:32,160 --> 00:04:38,080
However, I want to show you how
we proceed in analyzing
80
00:04:38,080 --> 00:04:41,580
this rod via differential
formulation, a variational
81
00:04:41,580 --> 00:04:45,700
formulation, and so on,
simply as an example.
82
00:04:45,700 --> 00:04:49,630
Therefore, the basic ideas,
really, that I will be putting
83
00:04:49,630 --> 00:04:52,090
forth to you, that I will be
discussing with you, are
84
00:04:52,090 --> 00:04:56,710
really the important things that
I want to expose to you.
85
00:04:56,710 --> 00:05:00,260
It's not the analysis of this
very specific problem.
86
00:05:00,260 --> 00:05:03,270
It's really the basic idea that
I want to clarify to you
87
00:05:03,270 --> 00:05:05,970
by looking at this
one problem.
88
00:05:05,970 --> 00:05:10,010
Well, for this problem here,
the governing differential
89
00:05:10,010 --> 00:05:13,210
equation of motion
is shown here.
90
00:05:13,210 --> 00:05:17,500
Notice u once again is the
displacement of a section.
91
00:05:17,500 --> 00:05:22,410
That is, this displacement here
at that coordinate x.
92
00:05:22,410 --> 00:05:26,540
c is given here as square root E
over rho, where E is Young's
93
00:05:26,540 --> 00:05:28,960
modulus of the material,
rho is the mass
94
00:05:28,960 --> 00:05:30,910
density of the material.
95
00:05:30,910 --> 00:05:32,320
t, of course, is the
time variable.
96
00:05:34,950 --> 00:05:37,850
Notice the cross-sectional
area a here, I have also
97
00:05:37,850 --> 00:05:39,150
written down here.
98
00:05:39,150 --> 00:05:43,320
This cross-sectional area
cancels out on both sides, as
99
00:05:43,320 --> 00:05:44,880
you will see just now.
100
00:05:44,880 --> 00:05:48,550
Now this question here is
obtained in the differential
101
00:05:48,550 --> 00:05:52,980
formulation by looking at the
equilibrium of an element.
102
00:05:52,980 --> 00:05:58,070
And we might consider this to be
the element that I will now
103
00:05:58,070 --> 00:06:00,590
focus our attention on.
104
00:06:00,590 --> 00:06:03,540
Here we have that element
drawn again.
105
00:06:03,540 --> 00:06:10,450
This element here of length dx
is subjected at its left side,
106
00:06:10,450 --> 00:06:17,420
because x comes from here as a
variable and brings us up to
107
00:06:17,420 --> 00:06:18,950
this station.
108
00:06:18,950 --> 00:06:22,360
At the left side here, the
element is subjected to sigma,
109
00:06:22,360 --> 00:06:24,110
the stress sigma.
110
00:06:24,110 --> 00:06:27,140
At the right side, we have
the stress sigma plus
111
00:06:27,140 --> 00:06:29,070
partial d sigma dx.
112
00:06:29,070 --> 00:06:33,370
This here really means nothing
else than a d sigma, an
113
00:06:33,370 --> 00:06:35,450
increment in the sigma.
114
00:06:35,450 --> 00:06:41,020
Well, the equilibrium
requirement for this element
115
00:06:41,020 --> 00:06:48,560
here is now that the force on
this side here, that is, sigma
116
00:06:48,560 --> 00:06:54,780
times a on this side, which is
actually this one here, I
117
00:06:54,780 --> 00:06:57,900
should have pointed
to this one--
118
00:06:57,900 --> 00:07:03,730
and this force here, which is
the force on the right side.
119
00:07:03,730 --> 00:07:07,430
If we subtract these two forces,
that must be equal to
120
00:07:07,430 --> 00:07:09,970
the force applied.
121
00:07:09,970 --> 00:07:14,030
Or rather, the d'Alembert
force.
122
00:07:14,030 --> 00:07:18,020
Notice that if we look at this
element, there's a force on
123
00:07:18,020 --> 00:07:20,540
the right, there's a
force on the left.
124
00:07:20,540 --> 00:07:24,890
And this being the force on the
right, let's call that,
125
00:07:24,890 --> 00:07:27,580
say, R1, and let's
call that R2.
126
00:07:27,580 --> 00:07:31,770
So I put here R1,
I put here R2.
127
00:07:31,770 --> 00:07:38,860
And R1 minus R2 must be equal to
the d'Alembert force, which
128
00:07:38,860 --> 00:07:41,930
is due to the inertia
of the material.
129
00:07:41,930 --> 00:07:45,750
This is the basic Newton's
law applied to this
130
00:07:45,750 --> 00:07:47,060
differential element.
131
00:07:47,060 --> 00:07:49,870
Now, if we referred back to
how we proceeded in the
132
00:07:49,870 --> 00:07:53,680
analysis of a discrete system,
we really proceeded in exactly
133
00:07:53,680 --> 00:07:54,650
the same way.
134
00:07:54,650 --> 00:07:58,470
But our element then was a
discrete element, a discrete
135
00:07:58,470 --> 00:07:59,380
spring element.
136
00:07:59,380 --> 00:08:03,200
We now use the same concept, but
apply those concepts to a
137
00:08:03,200 --> 00:08:05,340
differential element.
138
00:08:05,340 --> 00:08:10,120
So this is the equilibrium
equation of the element.
139
00:08:10,120 --> 00:08:12,660
The constitutive relation
is given here--
140
00:08:12,660 --> 00:08:15,670
that the stress is equal
to e, the Young's
141
00:08:15,670 --> 00:08:17,470
Modulus, times the strain.
142
00:08:17,470 --> 00:08:18,780
This is the strain.
143
00:08:18,780 --> 00:08:21,770
And notice once again, since we
are considering sections to
144
00:08:21,770 --> 00:08:26,720
be remaining plane, and simply
move horizontally, the only
145
00:08:26,720 --> 00:08:28,960
strain that we're talking
about is this one.
146
00:08:28,960 --> 00:08:32,370
If we combine these two
equations here, we directly
147
00:08:32,370 --> 00:08:34,110
obtain that equation here.
148
00:08:34,110 --> 00:08:37,270
Notice as I mentioned earlier,
a cancels out.
149
00:08:37,270 --> 00:08:41,100
If a is constant, that's why
a does not enter into this
150
00:08:41,100 --> 00:08:45,770
equation, and this substitution
for sigma into
151
00:08:45,770 --> 00:08:50,420
here gives us the second
derivative here.
152
00:08:50,420 --> 00:08:54,260
Of course, this part here
cancels out that part there,
153
00:08:54,260 --> 00:08:57,910
and that second derivative
is this one here.
154
00:08:57,910 --> 00:09:01,710
The E brought over to this
side gives us a 1 over c
155
00:09:01,710 --> 00:09:07,240
squared, where c, once again,
is defined as shown here.
156
00:09:07,240 --> 00:09:11,210
The important point, really, is
that we are looking here in
157
00:09:11,210 --> 00:09:15,035
the differential formulation at
a differential element of
158
00:09:15,035 --> 00:09:20,130
length dx at a particular
station x.
159
00:09:20,130 --> 00:09:22,670
That we're looking at this
element and we establish the
160
00:09:22,670 --> 00:09:26,620
equilibrium requirement
of that element.
161
00:09:26,620 --> 00:09:34,820
R1 minus R2 shall be equal to
the mass of the element times
162
00:09:34,820 --> 00:09:37,760
the acceleration.
163
00:09:37,760 --> 00:09:42,450
We also introduced a
constitutive relation.
164
00:09:42,450 --> 00:09:45,640
So far, clearly, we have used
two conditions for the
165
00:09:45,640 --> 00:09:46,830
solution of the problem.
166
00:09:46,830 --> 00:09:49,030
The first one is the equilibrium
condition.
167
00:09:49,030 --> 00:09:51,990
The second is the constitutive
condition, or constitutive
168
00:09:51,990 --> 00:09:52,810
requirement.
169
00:09:52,810 --> 00:09:55,960
We have to ask ourselves,
where do we satisfy the
170
00:09:55,960 --> 00:09:57,420
compatibility condition?
171
00:09:57,420 --> 00:09:59,750
Because there are always these
three conditions that we have
172
00:09:59,750 --> 00:10:00,840
to satisfy.
173
00:10:00,840 --> 00:10:03,940
Well, the compatibility
condition is satisfied by
174
00:10:03,940 --> 00:10:10,940
solving this differential
equation for this rod, and
175
00:10:10,940 --> 00:10:13,820
obtaining a u that
is continuous.
176
00:10:13,820 --> 00:10:19,080
In other words, a u that tells
that all the sections have
177
00:10:19,080 --> 00:10:19,930
remained together.
178
00:10:19,930 --> 00:10:22,730
We did not cut that bar apart.
179
00:10:22,730 --> 00:10:26,280
In the discrete system analysis
of the spring system
180
00:10:26,280 --> 00:10:30,330
of lecture 1, if you were to
think back to it, we had to
181
00:10:30,330 --> 00:10:34,510
satisfy the compatibility
condition explicitly, in
182
00:10:34,510 --> 00:10:36,950
establishing the equilibrium
equations, because we had to
183
00:10:36,950 --> 00:10:41,340
make sure that all the springs
remain attached to the carts.
184
00:10:41,340 --> 00:10:43,800
Here we satisfy the
compatibility condition by
185
00:10:43,800 --> 00:10:46,900
solving this equation
for a continuous u.
186
00:10:46,900 --> 00:10:49,070
Well, the boundary conditions,
of course,
187
00:10:49,070 --> 00:10:51,130
also have to be stated.
188
00:10:51,130 --> 00:10:54,610
And here we have a boundary
condition on the
189
00:10:54,610 --> 00:10:56,380
left end of the rod.
190
00:10:56,380 --> 00:11:00,930
Remember, please, that the rod
is fixed at its left end, so
191
00:11:00,930 --> 00:11:05,090
we have this condition here,
and clearly u must be 0 for
192
00:11:05,090 --> 00:11:08,240
all times t at x equals 0.
193
00:11:08,240 --> 00:11:11,930
At the right end, we
apply a load R0.
194
00:11:11,930 --> 00:11:17,160
And there we have, this being
here the area times the stress
195
00:11:17,160 --> 00:11:20,820
at x equals L, E
times a du dx.
196
00:11:20,820 --> 00:11:25,300
Notice that E du dx is, of
course, the stress, and so we
197
00:11:25,300 --> 00:11:29,160
have here this total force
at the right end
198
00:11:29,160 --> 00:11:31,600
being equal to R0.
199
00:11:31,600 --> 00:11:35,070
We also have initial conditions
for the solution of
200
00:11:35,070 --> 00:11:40,310
this equation that I showed to
you, this equation here.
201
00:11:40,310 --> 00:11:43,320
We have to have two spatial
conditions, two boundary
202
00:11:43,320 --> 00:11:46,640
conditions, one at the left
and one at the right end.
203
00:11:46,640 --> 00:11:48,150
The ones that I just
showed to you.
204
00:11:48,150 --> 00:11:50,870
We also have to have to have two
initial conditions, one on
205
00:11:50,870 --> 00:11:53,580
the displacement and one
on the velocity.
206
00:11:53,580 --> 00:11:56,570
Well, their initial conditions,
in this particular
207
00:11:56,570 --> 00:11:58,680
example, might be
as shown here.
208
00:11:58,680 --> 00:12:02,190
At time 0, all of the
displacement are 0.
209
00:12:02,190 --> 00:12:04,400
And at time 0, all
the velocities
210
00:12:04,400 --> 00:12:06,470
along the rod are 0.
211
00:12:06,470 --> 00:12:11,160
So the basic differential
equation given here, once
212
00:12:11,160 --> 00:12:16,740
again, plus these boundary
conditions, plus these initial
213
00:12:16,740 --> 00:12:20,700
conditions, define the
complete problem.
214
00:12:20,700 --> 00:12:24,540
I also like to point out here
that this boundary condition
215
00:12:24,540 --> 00:12:28,820
here, which does not involve any
derivative, is called an
216
00:12:28,820 --> 00:12:32,380
essential, or displacement
boundary condition.
217
00:12:32,380 --> 00:12:35,690
An essential boundary condition
because it does not
218
00:12:35,690 --> 00:12:40,090
involve any derivative when,
and this is important, the
219
00:12:40,090 --> 00:12:45,060
highest derivative in this
differential equation is 2.
220
00:12:45,060 --> 00:12:49,390
The right-hand side boundary
condition is called a natural
221
00:12:49,390 --> 00:12:51,110
force boundary condition.
222
00:12:51,110 --> 00:12:53,150
It's really involving forces.
223
00:12:53,150 --> 00:12:57,000
And it involves, as a highest
derivative, a derivative of
224
00:12:57,000 --> 00:13:02,280
order 1 when the differential
equation here involves as the
225
00:13:02,280 --> 00:13:06,740
highest derivative a derivative
of order 2.
226
00:13:06,740 --> 00:13:08,480
So in general--
227
00:13:08,480 --> 00:13:11,830
and this is a very
important point--
228
00:13:11,830 --> 00:13:14,360
we can say the following.
229
00:13:14,360 --> 00:13:17,970
If the highest order of the
spatial derivative in the
230
00:13:17,970 --> 00:13:24,650
problem governing differential
equation is 2m, in our case, m
231
00:13:24,650 --> 00:13:28,350
is equal to 1 for our problem.
232
00:13:28,350 --> 00:13:30,730
The highest order of the spacial
derivative in the
233
00:13:30,730 --> 00:13:33,840
essential boundary condition
is m minus 1.
234
00:13:33,840 --> 00:13:36,380
In other words, in this
case, of order 0.
235
00:13:36,380 --> 00:13:39,060
The highest order the spatial
derivative in the natural
236
00:13:39,060 --> 00:13:42,380
boundary conditions that I just
discussed is 2m minus 1,
237
00:13:42,380 --> 00:13:45,590
which is 1 in our
particular case.
238
00:13:45,590 --> 00:13:51,030
Then if we talk about this
problem, we talk about a C m
239
00:13:51,030 --> 00:13:54,190
minus 1 variational problem.
240
00:13:54,190 --> 00:13:56,440
It will become apparent
to you later on why we
241
00:13:56,440 --> 00:13:57,560
call it this way.
242
00:13:57,560 --> 00:14:02,980
C m minus 1 means continuity
of order m minus 1.
243
00:14:02,980 --> 00:14:05,790
In fact, in the Ritz analysis
that we will be performing
244
00:14:05,790 --> 00:14:08,570
later on, that I will discuss
with you later on, we find
245
00:14:08,570 --> 00:14:14,420
that we need, in the solution of
that kind of problem, only
246
00:14:14,420 --> 00:14:20,080
continuity in the Ritz functions
of order m minus 1.
247
00:14:20,080 --> 00:14:23,610
Well, let us now look at the
variational formulation.
248
00:14:23,610 --> 00:14:26,510
I mentioned earlier that we have
two different approaches.
249
00:14:26,510 --> 00:14:28,670
The first approach is a
differential formulation, the
250
00:14:28,670 --> 00:14:31,080
second approach is the
variational formulation.
251
00:14:31,080 --> 00:14:35,690
The variational formulation
operates in much the same way
252
00:14:35,690 --> 00:14:37,760
as I introduced it
to you for the
253
00:14:37,760 --> 00:14:39,320
analysis of discrete systems.
254
00:14:39,320 --> 00:14:43,620
We talk about pi a functional,
being equal to the strain
255
00:14:43,620 --> 00:14:47,150
energy minus the potential
of the loads.
256
00:14:47,150 --> 00:14:50,390
Now for the rod, the strain
energy is given here.
257
00:14:50,390 --> 00:14:55,260
Notice this is 1/2 times the
stress times the strain, and
258
00:14:55,260 --> 00:14:59,290
integrated over the total volume
of the element, or of
259
00:14:59,290 --> 00:15:01,320
the rod, I should rather say.
260
00:15:01,320 --> 00:15:06,240
The total potential of the
loads is given here.
261
00:15:06,240 --> 00:15:09,060
I could have written it this
way, with a minus out there
262
00:15:09,060 --> 00:15:10,550
and a plus in there--
263
00:15:10,550 --> 00:15:11,440
same thing.
264
00:15:11,440 --> 00:15:15,860
This, then, is really nothing
other than the loads
265
00:15:15,860 --> 00:15:18,620
multiplied by the total
displacement.
266
00:15:18,620 --> 00:15:20,590
And of course, there's an
integration involved here,
267
00:15:20,590 --> 00:15:23,480
because the body forces, the
body loads that I introduced
268
00:15:23,480 --> 00:15:28,560
here, fB, are varying along
the length of the rod.
269
00:15:28,560 --> 00:15:31,740
I introduce these fB body forces
because I want to use,
270
00:15:31,740 --> 00:15:35,560
later on, the d'Alembert
principle, put these, in other
271
00:15:35,560 --> 00:15:40,230
words, equal to minus the
acceleration forces, and can
272
00:15:40,230 --> 00:15:42,620
directly apply what I discussed
now, also to the
273
00:15:42,620 --> 00:15:45,860
dynamic analysis of this rod,
which we considered in the
274
00:15:45,860 --> 00:15:47,790
differential formulations.
275
00:15:47,790 --> 00:15:51,670
Together with stating pi as
shown here, we also have to
276
00:15:51,670 --> 00:15:54,620
state the left-hand boundary
condition, which is an
277
00:15:54,620 --> 00:15:56,550
essential boundary condition.
278
00:15:56,550 --> 00:15:59,570
We have to list all the
essential boundary conditions
279
00:15:59,570 --> 00:16:01,860
here, or the displacement
boundary conditions.
280
00:16:01,860 --> 00:16:04,020
Essential and displacement
mean the same
281
00:16:04,020 --> 00:16:05,220
thing, in that sense.
282
00:16:05,220 --> 00:16:10,410
Well, then we invoke the
stationality of pi.
283
00:16:10,410 --> 00:16:14,890
We are saying that del pi shall
be equal to 0 for any
284
00:16:14,890 --> 00:16:20,500
arbitrary variations of u that
satisfy, however, the
285
00:16:20,500 --> 00:16:22,590
essential boundary conditions.
286
00:16:22,590 --> 00:16:24,340
This boundary condition here.
287
00:16:24,340 --> 00:16:28,380
So this has to hold, this
statement shall hold, for any
288
00:16:28,380 --> 00:16:30,930
arbitrary variations in u.
289
00:16:30,930 --> 00:16:36,460
However, del u 0 shall be 0,
that satisfy, in other words,
290
00:16:36,460 --> 00:16:39,980
the essential boundary
condition.
291
00:16:39,980 --> 00:16:46,330
Well, if we apply this variation
on pi, we obtain
292
00:16:46,330 --> 00:16:49,920
directly this result here.
293
00:16:49,920 --> 00:16:56,600
Notice that this part here is
obtained by applying the
294
00:16:56,600 --> 00:17:00,750
variation on this part here.
295
00:17:00,750 --> 00:17:04,430
The variation operator operates,
del operates, much
296
00:17:04,430 --> 00:17:06,730
in the same way as a
differential operator.
297
00:17:06,730 --> 00:17:11,990
So this tool cancels this at
1/2, and we are left with EA
298
00:17:11,990 --> 00:17:15,849
del u del x times the variation
on del u del x.
299
00:17:15,849 --> 00:17:18,819
And that is given right there.
300
00:17:18,819 --> 00:17:26,119
The variation on this part here
gives us simply a del u
301
00:17:26,119 --> 00:17:32,370
times fB, and a del uL times R.
Of course, uL is equal to
302
00:17:32,370 --> 00:17:37,980
is the displacement at x equal
to L. And this is, therefore,
303
00:17:37,980 --> 00:17:39,290
the final result.
304
00:17:39,290 --> 00:17:45,360
Now if we look at this, we
recognize, really, that this
305
00:17:45,360 --> 00:17:48,360
is the principle of virtual
displacement.
306
00:17:48,360 --> 00:17:50,180
It's a principle of virtual
displacement
307
00:17:50,180 --> 00:17:51,420
governing the problem.
308
00:17:51,420 --> 00:17:55,760
In general, we can write this
principle as follows.
309
00:17:55,760 --> 00:17:59,400
Notice that here, we have
variations in strains, which
310
00:17:59,400 --> 00:18:02,270
is that part there.
311
00:18:02,270 --> 00:18:08,610
The real stresses are given
there, which are those.
312
00:18:08,610 --> 00:18:14,310
And here we have variations in
displacement, those operating
313
00:18:14,310 --> 00:18:17,120
on the body forces.
314
00:18:17,120 --> 00:18:19,030
There, there.
315
00:18:19,030 --> 00:18:22,640
Notice, I sum here over all
body forces in general.
316
00:18:22,640 --> 00:18:24,360
We have three components
in fB--
317
00:18:24,360 --> 00:18:27,030
the x, y, and z component.
318
00:18:27,030 --> 00:18:31,460
So I list these components in
a vector that I call fB.
319
00:18:31,460 --> 00:18:34,560
Similarly, of course, we have
three displacement components
320
00:18:34,560 --> 00:18:40,520
that appear here in this vector
U. Tau in general has 6
321
00:18:40,520 --> 00:18:41,870
components.
322
00:18:41,870 --> 00:18:44,750
Del epsilon also has
6 components.
323
00:18:44,750 --> 00:18:47,830
Putting their transpose on the
del epsilon vector means that
324
00:18:47,830 --> 00:18:49,380
we're summing the
product of the
325
00:18:49,380 --> 00:18:51,230
strains times these stresses.
326
00:18:51,230 --> 00:18:54,330
Similarly, we are summing
here the product of the
327
00:18:54,330 --> 00:18:57,620
displacement components times
the force components.
328
00:18:57,620 --> 00:19:00,445
We also have here
a contribution
329
00:19:00,445 --> 00:19:02,270
due to surface forces.
330
00:19:02,270 --> 00:19:04,420
fS are the surface forces.
331
00:19:04,420 --> 00:19:07,540
3 components, again, these are
the variations in the surface
332
00:19:07,540 --> 00:19:08,610
displacement.
333
00:19:08,610 --> 00:19:13,510
So this part here, del uLR,
really corresponds to this
334
00:19:13,510 --> 00:19:16,200
part here, involving
surface forces.
335
00:19:16,200 --> 00:19:19,450
So the surface forces read
again, and the variations in
336
00:19:19,450 --> 00:19:22,430
the surface displacements,
are those here.
337
00:19:22,430 --> 00:19:26,240
Now notice that this principle
here, or this equation, I
338
00:19:26,240 --> 00:19:30,540
should rather say, once again
has to be satisfied for any
339
00:19:30,540 --> 00:19:35,810
arbitrary variations in
displacements that satisfy the
340
00:19:35,810 --> 00:19:37,610
essential boundary conditions.
341
00:19:37,610 --> 00:19:42,100
For the problem of the rod,
these del u's have to satisfy
342
00:19:42,100 --> 00:19:45,840
the condition that the variation
at the left-hand
343
00:19:45,840 --> 00:19:48,670
boundary on u is 0, because
that is the
344
00:19:48,670 --> 00:19:50,390
essential boundary condition.
345
00:19:50,390 --> 00:19:54,880
Also, of course, notice that
this variation in the u's
346
00:19:54,880 --> 00:19:59,750
corresponds to these variations
in the strains.
347
00:19:59,750 --> 00:20:05,230
In other words, these strains
here are obtained from the
348
00:20:05,230 --> 00:20:08,650
variations in the
displacement.
349
00:20:08,650 --> 00:20:09,460
That's important.
350
00:20:09,460 --> 00:20:10,630
They are linked together.
351
00:20:10,630 --> 00:20:13,980
So if we impose certain
variations in the
352
00:20:13,980 --> 00:20:17,520
displacement, we have to impose
here the corresponding
353
00:20:17,520 --> 00:20:19,981
variations in strains.
354
00:20:19,981 --> 00:20:22,770
Of course, these variations in
the displacements give us also
355
00:20:22,770 --> 00:20:24,590
variations in surface
displacement.
356
00:20:24,590 --> 00:20:28,580
So these here are again
linked up with that.
357
00:20:28,580 --> 00:20:32,090
Later on in our formulation of
the finite element method, we
358
00:20:32,090 --> 00:20:35,990
will write the variations in
the strains here as virtual
359
00:20:35,990 --> 00:20:40,210
strains, arbitrary
virtual strains.
360
00:20:40,210 --> 00:20:42,490
And we are talking about the
strain vector with a
361
00:20:42,490 --> 00:20:43,830
bar on top of it.
362
00:20:43,830 --> 00:20:46,650
Similarly, a bar here, a
bar here, instead of
363
00:20:46,650 --> 00:20:48,050
the variation sign.
364
00:20:48,050 --> 00:20:50,900
However, the meaning is
quite identical--
365
00:20:50,900 --> 00:20:53,030
what we are saying here, and
this is the principle of
366
00:20:53,030 --> 00:20:57,030
virtual displacement that we
will be talking about later
367
00:20:57,030 --> 00:21:01,050
on, when we formulate the finite
element equations, what
368
00:21:01,050 --> 00:21:04,190
we are saying here really is
that this equation has to be
369
00:21:04,190 --> 00:21:09,860
satisfied for any arbitrary
virtual displacements and
370
00:21:09,860 --> 00:21:12,550
corresponding virtual strains.
371
00:21:12,550 --> 00:21:15,260
However, the virtual
displacements have to satisfy
372
00:21:15,260 --> 00:21:17,590
the displacement boundary
conditions, the essential
373
00:21:17,590 --> 00:21:19,280
boundary conditions.
374
00:21:19,280 --> 00:21:23,880
Well, from this, or rather that,
which of course these
375
00:21:23,880 --> 00:21:25,810
equations are completely
equivalent, we
376
00:21:25,810 --> 00:21:29,120
can go one step further.
377
00:21:29,120 --> 00:21:34,600
And if we go one step further,
by applying integration by
378
00:21:34,600 --> 00:21:42,050
parts and recognizing that this
part here is completely
379
00:21:42,050 --> 00:21:46,780
equal to that, by that I mean
taking the variation on the
380
00:21:46,780 --> 00:21:52,518
derivative of x is the same as
taking the sorry... by taking
381
00:21:52,518 --> 00:21:58,290
variation on the derivative of
u with respect to x, that is
382
00:21:58,290 --> 00:22:03,020
completely identical to first
taking the variation on u, and
383
00:22:03,020 --> 00:22:05,760
then taking the differentiation
of that
384
00:22:05,760 --> 00:22:08,530
variation on u with
respect to x.
385
00:22:08,530 --> 00:22:10,770
Well, if we recognize that
these two things are
386
00:22:10,770 --> 00:22:19,200
identical, then using
integration by parts on this
387
00:22:19,200 --> 00:22:24,210
equation here, which means,
really, on that equation here,
388
00:22:24,210 --> 00:22:26,560
for the special case of
the rod that we're now
389
00:22:26,560 --> 00:22:32,680
considering, we directly obtain
this equation here.
390
00:22:32,680 --> 00:22:38,060
The integration by parts is
performed by integrating this
391
00:22:38,060 --> 00:22:41,830
relation here first, and notice
that if we do integrate
392
00:22:41,830 --> 00:22:45,910
this, we want to lower the
differentiation of the virtual
393
00:22:45,910 --> 00:22:50,230
part here and increase the order
of differentiation on
394
00:22:50,230 --> 00:22:52,210
this part here.
395
00:22:52,210 --> 00:22:58,620
And that then directly gives
us these two terms
396
00:22:58,620 --> 00:23:01,700
and that term here.
397
00:23:01,700 --> 00:23:08,830
If we then also list this part
here together with what we
398
00:23:08,830 --> 00:23:13,810
obtain from that part, we
directly obtain this equation
399
00:23:13,810 --> 00:23:18,680
here, and we proceed similarly
for the coefficients of del
400
00:23:18,680 --> 00:23:21,500
uL, and we obtain
this part here.
401
00:23:21,500 --> 00:23:26,230
So this equation here is
obtained by simply using
402
00:23:26,230 --> 00:23:30,970
integration by parts on, I
repeat, this equation here.
403
00:23:30,970 --> 00:23:34,420
We have not used
any assumption.
404
00:23:34,420 --> 00:23:37,980
All we did is a mathematical
manipulation of this equation
405
00:23:37,980 --> 00:23:39,980
in a different form.
406
00:23:39,980 --> 00:23:42,500
And this is the new form
that we obtained.
407
00:23:42,500 --> 00:23:49,260
Del u0, of course, is the
variation of u at x equals 0.
408
00:23:49,260 --> 00:23:53,500
Now when we look at this
relation, we can extract now
409
00:23:53,500 --> 00:23:57,310
the differential equation of
equilibrium and the natural or
410
00:23:57,310 --> 00:23:59,230
force boundary conditions.
411
00:23:59,230 --> 00:24:00,350
How do we extract them?
412
00:24:00,350 --> 00:24:03,970
Well, the first point is that
I mentioned already earlier,
413
00:24:03,970 --> 00:24:09,690
this part here is 0, because
del u0 is imposed to be 0.
414
00:24:09,690 --> 00:24:11,310
So that part is 0.
415
00:24:11,310 --> 00:24:13,850
We can strike it out directly.
416
00:24:13,850 --> 00:24:17,770
Now when we look at this part
here and that part here and
417
00:24:17,770 --> 00:24:23,570
recognize that del u is now
arbitrary, we can directly
418
00:24:23,570 --> 00:24:28,420
extract the relation that this
must be 0 and that must be 0.
419
00:24:28,420 --> 00:24:29,430
How this is done?
420
00:24:29,430 --> 00:24:31,890
Well, first of all, we
recognize that this
421
00:24:31,890 --> 00:24:37,370
integration here really goes
from 0 plus to L minus, if you
422
00:24:37,370 --> 00:24:39,320
want to be really exact.
423
00:24:39,320 --> 00:24:43,020
Because at the boundary, we have
the boundary conditions.
424
00:24:43,020 --> 00:24:48,140
Of course, this 0 plus means it
is infinitesimally close to
425
00:24:48,140 --> 00:24:53,350
0, and L minus, we are
integrating up to a distance
426
00:24:53,350 --> 00:24:59,680
infinitesimally small to
actually L. So putting the 0
427
00:24:59,680 --> 00:25:03,010
plus and L minus here is just
for conceptual understanding
428
00:25:03,010 --> 00:25:04,260
really necessary.
429
00:25:04,260 --> 00:25:07,950
Well, if we then impose the
following variations, say that
430
00:25:07,950 --> 00:25:14,680
del uL is 0, is exactly 0,
then this part is out.
431
00:25:14,680 --> 00:25:18,190
And then we only have to
look at this part.
432
00:25:18,190 --> 00:25:25,810
Now we can apply any arbitrary
variation on u from 0 to L
433
00:25:25,810 --> 00:25:30,590
minus, and this, then, this
total integration, must be
434
00:25:30,590 --> 00:25:31,490
equal to 0.
435
00:25:31,490 --> 00:25:33,920
It must satisfy the
0 condition.
436
00:25:33,920 --> 00:25:40,060
And that can only be true when
this part here, what is in the
437
00:25:40,060 --> 00:25:41,340
brackets, is 0.
438
00:25:41,340 --> 00:25:45,470
Because if this is not 0, I can
always select a certain
439
00:25:45,470 --> 00:25:50,310
delta u which, when multiplied
by that and integrated as
440
00:25:50,310 --> 00:25:51,920
shown here--
441
00:25:51,920 --> 00:25:54,080
remember, this part
is not there--
442
00:25:54,080 --> 00:25:55,840
will not give us 0.
443
00:25:55,840 --> 00:25:58,810
So therefore, this
part must be 0.
444
00:25:58,810 --> 00:26:01,950
And that is our first condition,
which is the
445
00:26:01,950 --> 00:26:05,960
equilibrium condition of
a differential element.
446
00:26:05,960 --> 00:26:09,340
It's the equilibrium condition
of a differential element.
447
00:26:09,340 --> 00:26:13,710
Now, if we say, let us look at
the right-hand side boundary,
448
00:26:13,710 --> 00:26:17,520
and put del u0 everywhere along
the length of the rod,
449
00:26:17,520 --> 00:26:28,310
except at x equals exactly L.
Then this part would be 0, and
450
00:26:28,310 --> 00:26:33,780
this part here is non-zero,
provided--
451
00:26:36,350 --> 00:26:38,460
or rather, this part
would be present.
452
00:26:38,460 --> 00:26:42,850
And the only way that this part
can be 0, as it must be,
453
00:26:42,850 --> 00:26:46,850
is that this part here, the
coefficient on del uL, is
454
00:26:46,850 --> 00:26:48,020
actually 0.
455
00:26:48,020 --> 00:26:52,810
So this way, we have extracted
two points, two conditions.
456
00:26:52,810 --> 00:26:55,920
This part must be 0 and
that part must be 0.
457
00:26:55,920 --> 00:26:58,830
We extract these conditions
by looking at specific
458
00:26:58,830 --> 00:27:00,340
variations on u.
459
00:27:00,340 --> 00:27:05,610
First we look at the variation
on u where del uL is 0, and
460
00:27:05,610 --> 00:27:11,000
otherwise arbitrary from
x equals 0 to L minus.
461
00:27:11,000 --> 00:27:14,200
And then we can directly
conclude this must be 0, and
462
00:27:14,200 --> 00:27:19,730
second time around, we say,
let del u equal 0 from x
463
00:27:19,730 --> 00:27:23,380
equals 0 to L minus, which makes
all of this integral 0,
464
00:27:23,380 --> 00:27:27,910
and we can focus all our
attention on this part here,
465
00:27:27,910 --> 00:27:30,630
and we directly can conclude
that this part now
466
00:27:30,630 --> 00:27:32,070
must also be 0.
467
00:27:32,070 --> 00:27:36,000
So this way, then, we extracted
the differential
468
00:27:36,000 --> 00:27:39,490
equation of equilibrium and
the natural boundary
469
00:27:39,490 --> 00:27:40,370
conditions.
470
00:27:40,370 --> 00:27:41,580
Very important.
471
00:27:41,580 --> 00:27:44,700
Differential equation of
equilibrium and the natural
472
00:27:44,700 --> 00:27:47,540
boundary condition.
473
00:27:47,540 --> 00:27:49,780
And now we really recognize.
474
00:27:49,780 --> 00:27:54,280
of course, that if we put fB
equal to the d'Alembert force,
475
00:27:54,280 --> 00:27:59,920
or minus the d'Alembert force,
equal to this right-hand side,
476
00:27:59,920 --> 00:28:03,910
substitute from here back into
there, cancel out A, we
477
00:28:03,910 --> 00:28:06,840
directly obtain this
differential equation.
478
00:28:06,840 --> 00:28:09,565
Notice that c is E divided
by rho square root
479
00:28:09,565 --> 00:28:10,860
E divided by rho.
480
00:28:10,860 --> 00:28:15,070
Now, the important point really
is that by having
481
00:28:15,070 --> 00:28:25,400
started off with this pi
functional and that condition,
482
00:28:25,400 --> 00:28:30,150
this condition also, and that
the variations on u at x
483
00:28:30,150 --> 00:28:35,620
equals 0 shall be identically
0, we directly can extract
484
00:28:35,620 --> 00:28:38,820
from this pi functional the
differential equation of
485
00:28:38,820 --> 00:28:41,850
equilibrium and the natural
boundary conditions.
486
00:28:41,850 --> 00:28:44,070
The natural boundary conditions,
in fact, are
487
00:28:44,070 --> 00:28:46,200
contained right in there.
488
00:28:46,200 --> 00:28:48,450
That's where the natural
boundary conditions appear.
489
00:28:48,450 --> 00:28:51,250
The essential boundary
conditions are there, and have
490
00:28:51,250 --> 00:28:53,940
to be satisfied by
the variations.
491
00:28:53,940 --> 00:29:02,350
So in general, then, we find
the following points.
492
00:29:02,350 --> 00:29:06,860
The important points are that
by invoking del pi equals 0
493
00:29:06,860 --> 00:29:13,730
and using the essential boundary
conditions only, we
494
00:29:13,730 --> 00:29:16,790
generate the principle of
virtual displacement, an
495
00:29:16,790 --> 00:29:18,880
extremely important fact.
496
00:29:18,880 --> 00:29:23,680
And this will be the starting
equation that we will be using
497
00:29:23,680 --> 00:29:27,210
to generate our finite element
equations later on.
498
00:29:27,210 --> 00:29:29,570
We also can extract the
problem governing
499
00:29:29,570 --> 00:29:32,030
differential equation.
500
00:29:32,030 --> 00:29:34,200
Therefore, the problem governing
differential
501
00:29:34,200 --> 00:29:36,700
equation is contained in the
principle of virtual
502
00:29:36,700 --> 00:29:40,990
displacement, and this one is
contained in del pi equals 0,
503
00:29:40,990 --> 00:29:42,880
in the del pi equals
0 condition.
504
00:29:42,880 --> 00:29:47,610
Of course also satisfying the
essential boundary condition.
505
00:29:47,610 --> 00:29:50,460
We also can extract the natural
boundary conditions.
506
00:29:50,460 --> 00:29:54,770
So these are also contained,
in essence, in pi, and as I
507
00:29:54,770 --> 00:29:58,270
showed to you, they're contained
really in w.
508
00:29:58,270 --> 00:29:59,640
That's where they appear.
509
00:29:59,640 --> 00:30:03,310
Now in the derivation of the
problem governing differential
510
00:30:03,310 --> 00:30:06,630
equation, we used integration
by parts.
511
00:30:06,630 --> 00:30:09,050
And the highest spatial
derivative in
512
00:30:09,050 --> 00:30:11,700
pi is of order m.
513
00:30:11,700 --> 00:30:17,430
We used integration by parts m
times, and what we're finding
514
00:30:17,430 --> 00:30:20,970
is that the highest spatial
derivative in the problem
515
00:30:20,970 --> 00:30:24,350
governing differential
equation is then 2m.
516
00:30:24,350 --> 00:30:27,840
And this then puts together the
complete structure of the
517
00:30:27,840 --> 00:30:29,970
equations that we're
talking about.
518
00:30:29,970 --> 00:30:32,730
Here I have another view graph
that summarizes the
519
00:30:32,730 --> 00:30:34,880
process once more.
520
00:30:34,880 --> 00:30:38,060
We are starting with the total
potential pi of the system
521
00:30:38,060 --> 00:30:41,350
being equal to the strain
energy minus the total
522
00:30:41,350 --> 00:30:42,910
potential of the loads.
523
00:30:42,910 --> 00:30:46,070
We're using this condition and
the essential boundary
524
00:30:46,070 --> 00:30:47,250
condition--
525
00:30:47,250 --> 00:30:48,610
very important--
526
00:30:48,610 --> 00:30:51,030
to generate the principle
of virtual displacement.
527
00:30:51,030 --> 00:30:54,920
At this stage, we can solve the
problem, and we will do so
528
00:30:54,920 --> 00:30:57,870
by using finite element
methods.
529
00:30:57,870 --> 00:31:02,190
We can, however, also go on from
the principle of virtual
530
00:31:02,190 --> 00:31:06,920
displacement using integration
by parts, and then we would
531
00:31:06,920 --> 00:31:09,665
derive the differential equation
of equilibrium and
532
00:31:09,665 --> 00:31:12,050
the natural boundary
conditions.
533
00:31:12,050 --> 00:31:13,820
We can solve the problem.
534
00:31:13,820 --> 00:31:17,250
Well, we can solve the problem
at this level really only when
535
00:31:17,250 --> 00:31:21,010
we can solve the differential
equation of equilibrium,
536
00:31:21,010 --> 00:31:23,150
subject to the natural
boundary conditions.
537
00:31:23,150 --> 00:31:27,290
And that we can really only do
for very simple systems.
538
00:31:27,290 --> 00:31:32,550
Therefore, this process here,
from here onward, can really
539
00:31:32,550 --> 00:31:36,000
be followed only for relatively
simple systems.
540
00:31:36,000 --> 00:31:40,760
For complex systems, shell
structures, complicated shell
541
00:31:40,760 --> 00:31:45,640
beam structures, plane stress
systems, real engineering
542
00:31:45,640 --> 00:31:49,820
structural analysis systems,
this is not possible, and we
543
00:31:49,820 --> 00:31:53,760
stop right there, and we solve
our problem this way, using
544
00:31:53,760 --> 00:31:55,200
finite element methods.
545
00:31:55,200 --> 00:32:00,320
Now, the one important point,
however, I want to make once
546
00:32:00,320 --> 00:32:01,600
more clear--
547
00:32:01,600 --> 00:32:05,910
when we proceed this way, we are
deriving the differential
548
00:32:05,910 --> 00:32:10,470
equation of equilibrium for
each differential element.
549
00:32:10,470 --> 00:32:14,410
And this means that when we go
this route for the simple
550
00:32:14,410 --> 00:32:19,045
systems that we can go this
route, we satisfy the
551
00:32:19,045 --> 00:32:23,200
equilibrium condition on each
differential element.
552
00:32:23,200 --> 00:32:26,910
However, when we go this route,
we will find that we
553
00:32:26,910 --> 00:32:31,430
only satisfy the equilibrium
conditions in a global sense,
554
00:32:31,430 --> 00:32:36,150
in an integrated sense, using
finite element methods, and
555
00:32:36,150 --> 00:32:42,300
that to actually satisfy the
equilibrium conditions in a
556
00:32:42,300 --> 00:32:47,130
local sense, meaning for each
differential element, we have
557
00:32:47,130 --> 00:32:52,670
to use many elements, and only
then, of course, our finite
558
00:32:52,670 --> 00:32:56,040
element solution will converge
to the solution that we would
559
00:32:56,040 --> 00:32:59,660
have obtained solving the
differential equation of
560
00:32:59,660 --> 00:33:00,950
equilibrium.
561
00:33:00,950 --> 00:33:05,270
Therefore, when we start from
the principle of virtual
562
00:33:05,270 --> 00:33:09,880
displacement, if we, as I will
show you, use many elements,
563
00:33:09,880 --> 00:33:13,440
we really obtain the same
solution as here.
564
00:33:13,440 --> 00:33:16,040
However, if we do not use
many elements, if we
565
00:33:16,040 --> 00:33:18,140
use a coarse mesh--
566
00:33:18,140 --> 00:33:20,470
we will talk about a coarse
mesh later on--
567
00:33:20,470 --> 00:33:23,790
then we will see that we only
satisfy the equilibrium
568
00:33:23,790 --> 00:33:26,605
conditions in a global sense for
the complete structure, in
569
00:33:26,605 --> 00:33:29,410
an integrated sense for the
complete structure.
570
00:33:29,410 --> 00:33:32,390
For each finite element, we will
satisfy the equilibrium
571
00:33:32,390 --> 00:33:35,180
conditions, but we will not
satisfy the equilibrium
572
00:33:35,180 --> 00:33:40,610
conditions accurately for each
differential element--
573
00:33:40,610 --> 00:33:46,760
dx, dy, dz being arbitrarily
small--
574
00:33:46,760 --> 00:33:50,230
in the continuous body.
575
00:33:50,230 --> 00:33:52,650
That will become clearer, then,
when we actually go
576
00:33:52,650 --> 00:33:54,120
through an example.
577
00:33:54,120 --> 00:34:00,120
Now I mentioned earlier that
a whole class of classical
578
00:34:00,120 --> 00:34:02,850
methods are the weighted
residual methods.
579
00:34:02,850 --> 00:34:05,730
In the weighted residual
methods, we proceed in the
580
00:34:05,730 --> 00:34:06,910
following way.
581
00:34:06,910 --> 00:34:10,139
We consider the steady state
problem, which is given here,
582
00:34:10,139 --> 00:34:11,860
with these boundary
conditions.
583
00:34:11,860 --> 00:34:15,840
Now this operator L2m
phi equals R--
584
00:34:15,840 --> 00:34:20,830
this L2m is the operator that
governs the problem.
585
00:34:20,830 --> 00:34:24,920
Like in our particular rod
problem, L2m would be equal to
586
00:34:24,920 --> 00:34:28,460
delta u delta x squared.
587
00:34:28,460 --> 00:34:32,500
So the highest derivative in
this spacial operator being
588
00:34:32,500 --> 00:34:36,830
2m, in this case, 2,
for our example.
589
00:34:36,830 --> 00:34:39,560
The boundary conditions can
be written this way.
590
00:34:39,560 --> 00:34:43,380
And the basic step, then, in the
weighted residual and the
591
00:34:43,380 --> 00:34:46,730
Ritz analysis is to assume a
solution all of this form,
592
00:34:46,730 --> 00:34:51,550
where phi bar gives us the
assumed solution ai are
593
00:34:51,550 --> 00:34:56,560
parameters that are unknown,
and fi are bases functions.
594
00:34:56,560 --> 00:34:59,800
These are functions that
have to be assumed.
595
00:34:59,800 --> 00:35:05,000
Well, there is, of course,
considerable concern on what
596
00:35:05,000 --> 00:35:06,750
kind of functions to choose.
597
00:35:06,750 --> 00:35:10,220
In the weighted residual method,
these functions here,
598
00:35:10,220 --> 00:35:14,070
if you directly operate on this
equation here, should
599
00:35:14,070 --> 00:35:16,730
satisfy all boundary
conditions.
600
00:35:16,730 --> 00:35:19,830
Notice that in the weighted
residual method, here I'm
601
00:35:19,830 --> 00:35:24,940
really talking about going on
this view graph through the
602
00:35:24,940 --> 00:35:27,480
differential equations of
equilibrium and natural
603
00:35:27,480 --> 00:35:30,870
boundary conditions, deriving
them, if you want, via this
604
00:35:30,870 --> 00:35:34,090
route, and then trying to
solve the problem here
605
00:35:34,090 --> 00:35:36,370
numerically.
606
00:35:36,370 --> 00:35:38,340
We will actually, as I said
earlier, in the finite element
607
00:35:38,340 --> 00:35:43,050
process, go this route, but we
can also go that route with
608
00:35:43,050 --> 00:35:44,710
the weighted residual methods.
609
00:35:44,710 --> 00:35:49,930
In fact, there is a close
relationship between using
610
00:35:49,930 --> 00:35:52,860
weighted residual methods at
this level and the Ritz method
611
00:35:52,860 --> 00:35:56,230
at that level, the way I will
be describing it later on.
612
00:35:56,230 --> 00:35:59,440
But in the weighted residual
method, this is the
613
00:35:59,440 --> 00:36:01,340
assumption.
614
00:36:01,340 --> 00:36:06,000
And then if we look at this
equation here, we can
615
00:36:06,000 --> 00:36:08,990
construct an error
R, substituting
616
00:36:08,990 --> 00:36:10,780
from here into there.
617
00:36:10,780 --> 00:36:16,270
And that capital R error is
given by this equation.
618
00:36:16,270 --> 00:36:18,540
Notice that this is, of course,
our trial function
619
00:36:18,540 --> 00:36:19,550
that we have.
620
00:36:19,550 --> 00:36:24,290
And if the right-hand side is 0,
everywhere over the domain,
621
00:36:24,290 --> 00:36:28,130
then of course our error would
be 0, and we would have solved
622
00:36:28,130 --> 00:36:32,570
our equation that we're
looking at here.
623
00:36:32,570 --> 00:36:34,470
That is the equation
we want to solve.
624
00:36:34,470 --> 00:36:38,400
Since if all of these functions
fi satisfy all of
625
00:36:38,400 --> 00:36:42,490
the boundary conditions, then
we are satisfying these
626
00:36:42,490 --> 00:36:45,170
equations, and all we have to
worry about further is to
627
00:36:45,170 --> 00:36:47,630
satisfy this equation.
628
00:36:47,630 --> 00:36:50,390
If R is 0, we would also satisfy
the differential
629
00:36:50,390 --> 00:36:51,880
equation of equilibrium,
and we would have,
630
00:36:51,880 --> 00:36:53,510
in fact, the solution.
631
00:36:53,510 --> 00:36:57,560
However, that, of course, would
be a very lucky choice
632
00:36:57,560 --> 00:36:59,150
on the fi functions.
633
00:36:59,150 --> 00:37:02,350
In general, R will not
be identically 0
634
00:37:02,350 --> 00:37:03,890
all over the domain.
635
00:37:03,890 --> 00:37:09,910
In fact, how much R, or how
close R will be to 0 will of
636
00:37:09,910 --> 00:37:11,890
course depend on the ai.
637
00:37:11,890 --> 00:37:17,270
And this is then basically
our objective, namely, to
638
00:37:17,270 --> 00:37:21,310
calculate ai values that are
making this R, this left-hand
639
00:37:21,310 --> 00:37:24,350
side capital R, as close
as possible to 0.
640
00:37:24,350 --> 00:37:25,660
And that can be achieved
via the
641
00:37:25,660 --> 00:37:28,170
Galerkin method, for example.
642
00:37:28,170 --> 00:37:29,380
This is the basic process.
643
00:37:29,380 --> 00:37:33,150
Here we're substituting the R.
These are the trial functions,
644
00:37:33,150 --> 00:37:36,290
and we're integrating the
product of these over the
645
00:37:36,290 --> 00:37:37,830
total domain.
646
00:37:37,830 --> 00:37:41,340
The domain here, in the case
of our rod, would simply be
647
00:37:41,340 --> 00:37:43,880
the volume of the rod.
648
00:37:43,880 --> 00:37:48,780
This is the mechanism that
generates to us n equations in
649
00:37:48,780 --> 00:37:52,550
the trial parameters ai.
650
00:37:52,550 --> 00:37:55,770
In another approach, the least
squares method, we would
651
00:37:55,770 --> 00:37:59,900
operate on the square of the
error, and minimize the square
652
00:37:59,900 --> 00:38:02,750
of the error when integrated
over the total domain with
653
00:38:02,750 --> 00:38:06,910
respect to the trial
parameters ai.
654
00:38:06,910 --> 00:38:10,720
That again gives us n equations,
and we set up these
655
00:38:10,720 --> 00:38:13,410
n equations just as we're
doing here, to
656
00:38:13,410 --> 00:38:14,730
solve for the ai.
657
00:38:14,730 --> 00:38:19,820
Knowing, then, the ai, we can
back substitute into this
658
00:38:19,820 --> 00:38:22,510
assumption here, and now
we have our approximate
659
00:38:22,510 --> 00:38:24,320
solution, phi bar.
660
00:38:24,320 --> 00:38:27,590
If the trial functions have been
selected to satisfy the
661
00:38:27,590 --> 00:38:30,770
boundary conditions, then of
course phi bar will satisfy
662
00:38:30,770 --> 00:38:33,500
the boundary conditions.
663
00:38:33,500 --> 00:38:36,650
However, what phi bar will not
satisfy exactly is this
664
00:38:36,650 --> 00:38:37,836
equation here.
665
00:38:37,836 --> 00:38:41,110
However, we have minimized the
error in the satisfaction of
666
00:38:41,110 --> 00:38:45,790
this equation in some sense
using the Galerkin method or
667
00:38:45,790 --> 00:38:47,280
the least squares method.
668
00:38:47,280 --> 00:38:52,950
These methods can also be
extended when the fi trial
669
00:38:52,950 --> 00:38:57,460
functions do not satisfy all
of the boundary conditions,
670
00:38:57,460 --> 00:38:59,620
namely, not the natural
boundary conditions.
671
00:38:59,620 --> 00:39:02,410
They can be extended to their
places, but classically, they
672
00:39:02,410 --> 00:39:06,640
have been used with trial
functions that satisfy all
673
00:39:06,640 --> 00:39:08,140
boundary conditions.
674
00:39:08,140 --> 00:39:14,360
When we were to extend the
Galerkin method for the case
675
00:39:14,360 --> 00:39:18,510
where the trial functions do
not satisfy the natural
676
00:39:18,510 --> 00:39:22,430
boundary conditions, then we
really talk basically about
677
00:39:22,430 --> 00:39:27,050
already a Ritz analysis, and
that is the next procedure
678
00:39:27,050 --> 00:39:28,860
that I want to introduce
to you.
679
00:39:28,860 --> 00:39:35,670
Now of course, if we wanted to
really start deriving the Ritz
680
00:39:35,670 --> 00:39:39,310
method from the Galerkin
approach--
681
00:39:39,310 --> 00:39:42,020
in other words, if we wanted to
derive this Ritz analysis
682
00:39:42,020 --> 00:39:44,580
method from the Galerkin
approach, we would have to
683
00:39:44,580 --> 00:39:47,460
extend, first of all, this
Galerkin approach to include
684
00:39:47,460 --> 00:39:49,440
the natural boundary conditions,
and then we would
685
00:39:49,440 --> 00:39:54,340
have to perform integrations on
this equation, and we would
686
00:39:54,340 --> 00:39:58,510
obtain the Ritz analysis
method.
687
00:39:58,510 --> 00:40:03,240
The actual way of starting, of
introducing the Ritz analysis
688
00:40:03,240 --> 00:40:07,790
method, is to introduce
it as a separate tool.
689
00:40:07,790 --> 00:40:10,750
And it is introduced in
the following way.
690
00:40:10,750 --> 00:40:15,370
Let pi be the functional of the
C m minus 1 variational
691
00:40:15,370 --> 00:40:18,690
problem that is equivalent to
the differential formulation
692
00:40:18,690 --> 00:40:20,360
that we talked about earlier.
693
00:40:20,360 --> 00:40:23,960
Now, the differential
formulation that I talked
694
00:40:23,960 --> 00:40:26,120
about here is this one.
695
00:40:26,120 --> 00:40:27,380
That's a differential
formulation.
696
00:40:30,440 --> 00:40:34,380
And as an example once again,
the operator L2m is del 2 u
697
00:40:34,380 --> 00:40:37,400
del x squared.
698
00:40:37,400 --> 00:40:38,430
There's a constant here.
699
00:40:38,430 --> 00:40:40,610
We could also put an EA
in front here, but
700
00:40:40,610 --> 00:40:42,080
that only is a constant.
701
00:40:42,080 --> 00:40:45,280
It doesn't change the character
of the operator.
702
00:40:45,280 --> 00:40:47,690
But this is basically the
operator for the problem that
703
00:40:47,690 --> 00:40:50,660
we considered, and of course,
our boundary conditions are
704
00:40:50,660 --> 00:40:51,380
also there.
705
00:40:51,380 --> 00:41:00,270
So if we have a pi functional
that is equivalent to the
706
00:41:00,270 --> 00:41:03,120
differential formulation given
in those two equations that I
707
00:41:03,120 --> 00:41:06,350
just pointed out to you again,
then in the Ritz method, we
708
00:41:06,350 --> 00:41:10,320
substitute the trial functions,
phi bar--
709
00:41:10,320 --> 00:41:12,780
let us look at them again,
these are the trial
710
00:41:12,780 --> 00:41:14,230
functions--
711
00:41:14,230 --> 00:41:21,870
into pi, and we generate n
simultaneous equations for the
712
00:41:21,870 --> 00:41:27,660
parameters that appear in this
assumption here by invoking
713
00:41:27,660 --> 00:41:29,770
the stationality of pi.
714
00:41:29,770 --> 00:41:36,890
Notice that by invoking that
del pi is 0, we really say
715
00:41:36,890 --> 00:41:44,130
that del pi ai is 0 for all i.
716
00:41:44,130 --> 00:41:47,820
And that gives us the condition
which we used to set
717
00:41:47,820 --> 00:41:51,740
up the individual equations
that we need to set up to
718
00:41:51,740 --> 00:41:55,900
solve for the trial
parameters ai.
719
00:41:55,900 --> 00:41:58,980
Well, let us look at some
of the properties.
720
00:41:58,980 --> 00:42:02,080
The trial functions used in the
Ritz analysis need only
721
00:42:02,080 --> 00:42:05,550
satisfy the essential boundary
conditions, an extremely
722
00:42:05,550 --> 00:42:06,840
important fact.
723
00:42:06,840 --> 00:42:10,260
In the classical weighted
residual method, as I pointed
724
00:42:10,260 --> 00:42:13,020
out, the trial function should
satisfy all boundary
725
00:42:13,020 --> 00:42:13,730
conditions.
726
00:42:13,730 --> 00:42:16,540
Therefore, they can be very
difficult to choose.
727
00:42:16,540 --> 00:42:19,750
In the Ritz analysis method,
we only need to satisfy the
728
00:42:19,750 --> 00:42:22,080
essential boundary conditions.
729
00:42:22,080 --> 00:42:25,780
The application of del pi
equals 0 generates the
730
00:42:25,780 --> 00:42:28,010
principle of virtual
displacement.
731
00:42:28,010 --> 00:42:30,640
I mentioned that to you
earlier already.
732
00:42:30,640 --> 00:42:34,460
And therefore, in effect, we use
in the Ritz analysis this
733
00:42:34,460 --> 00:42:37,270
principle of virtual
displacement.
734
00:42:37,270 --> 00:42:41,380
By invoking del pi equals 0,
we minimize basically the
735
00:42:41,380 --> 00:42:45,670
violation of the internal
equilibrium requirements and
736
00:42:45,670 --> 00:42:49,070
the violation of the natural
boundary conditions.
737
00:42:49,070 --> 00:42:55,050
Well, remember that invoking
del pi equal to 0 and then
738
00:42:55,050 --> 00:42:58,400
using integration by parts, we
actually generate, we could
739
00:42:58,400 --> 00:43:02,020
generate the differential
equations of equilibrium and
740
00:43:02,020 --> 00:43:04,160
the natural boundary conditions
the way I've shown
741
00:43:04,160 --> 00:43:07,270
it to you, for the simple
bar structure.
742
00:43:07,270 --> 00:43:09,910
Now, what I'm saying here is
that we do not want to
743
00:43:09,910 --> 00:43:12,230
generate these differential
equations of equilibrium and
744
00:43:12,230 --> 00:43:13,180
natural boundary conditions.
745
00:43:13,180 --> 00:43:17,140
However, please recognize that
they are contained in the
746
00:43:17,140 --> 00:43:19,980
equation del pi equals 0.
747
00:43:19,980 --> 00:43:23,500
Therefore, by substituting our
trial functions, we violate
748
00:43:23,500 --> 00:43:26,150
the internal equilibrium
requirements and the natural
749
00:43:26,150 --> 00:43:29,410
boundary conditions, but
we will see that we are
750
00:43:29,410 --> 00:43:34,630
minimizing that violation in
these conditions here.
751
00:43:34,630 --> 00:43:36,930
Also, we will see that we
generate a symmetric
752
00:43:36,930 --> 00:43:40,960
coefficient matrix, and K and
the governing equations then,
753
00:43:40,960 --> 00:43:44,370
our KU equals R, that
we want to solve.
754
00:43:44,370 --> 00:43:49,670
And that is really the basis of
the finite element method
755
00:43:49,670 --> 00:43:52,370
for the analysis of continuous
systems.
756
00:43:52,370 --> 00:43:56,180
Let me now go in detail
through an example.
757
00:43:56,180 --> 00:44:01,330
Here we have is simple bar
structure which has an area 1
758
00:44:01,330 --> 00:44:06,860
square centimeter from A to B,
and from B to C, B being this
759
00:44:06,860 --> 00:44:08,995
point here where the
area changes and C
760
00:44:08,995 --> 00:44:10,500
being that point there.
761
00:44:10,500 --> 00:44:13,850
From B to C, we have
a varying area.
762
00:44:13,850 --> 00:44:16,420
This variation in the
area is shown here.
763
00:44:16,420 --> 00:44:21,750
It's 1 plus y/40 squared is the
area at any station y, y
764
00:44:21,750 --> 00:44:25,290
being measured from point
B, as you can see here.
765
00:44:25,290 --> 00:44:27,200
The length here is
100 centimeters.
766
00:44:27,200 --> 00:44:29,340
This length is 80 centimeters.
767
00:44:29,340 --> 00:44:33,340
The structure is subjected to
a load of 100 newtons here.
768
00:44:33,340 --> 00:44:36,930
Notice that this arrow really
lies on top of that dashed
769
00:44:36,930 --> 00:44:40,390
line we just separated out for
you to understand that there
770
00:44:40,390 --> 00:44:41,290
is this arrow.
771
00:44:41,290 --> 00:44:43,810
So this is the load
applied at the
772
00:44:43,810 --> 00:44:46,820
mid-line of this structure.
773
00:44:46,820 --> 00:44:51,690
We assume once again for the
structure also that there is
774
00:44:51,690 --> 00:44:54,840
only the following displacement
mechanism.
775
00:44:54,840 --> 00:45:00,670
If a section was originally
there, and it is a vertical
776
00:45:00,670 --> 00:45:04,260
section to the midline, then
it has moved over, and I
777
00:45:04,260 --> 00:45:08,730
grossly exaggerate now, to this
position, where this is
778
00:45:08,730 --> 00:45:11,530
the displacement that we're
talking about u.
779
00:45:11,530 --> 00:45:15,050
Grossly exaggerated,
of course.
780
00:45:15,050 --> 00:45:18,400
So we're having a bar structure
subjected to a
781
00:45:18,400 --> 00:45:21,550
concentrated load fixed
at the left end.
782
00:45:21,550 --> 00:45:27,360
And our objective now is to
solve this structure, to solve
783
00:45:27,360 --> 00:45:32,300
for the unknown displacement u,
being 0 here, of course as
784
00:45:32,300 --> 00:45:35,180
a function of x, when
this structure is
785
00:45:35,180 --> 00:45:37,130
subjected to that load.
786
00:45:37,130 --> 00:45:41,690
Well, in the calculation of
this example, I want to
787
00:45:41,690 --> 00:45:44,590
display to you as many
of the concepts
788
00:45:44,590 --> 00:45:47,090
that we just discussed.
789
00:45:47,090 --> 00:45:50,580
Here we have pi being equal
to this value here.
790
00:45:50,580 --> 00:45:53,050
The strain energy
is given here.
791
00:45:53,050 --> 00:45:57,090
Notice this is 1/2 times the
stress times the strain
792
00:45:57,090 --> 00:46:00,190
integrated over the volume
of the structure.
793
00:46:00,190 --> 00:46:04,800
The integration goes from 0 to
180, because that is the
794
00:46:04,800 --> 00:46:06,750
length of that structure.
795
00:46:06,750 --> 00:46:11,860
The total potential of the
external load is 100, which is
796
00:46:11,860 --> 00:46:15,270
the intensity of the load times
the displacement at x
797
00:46:15,270 --> 00:46:18,910
equal to 180... at
x equal to 180.
798
00:46:18,910 --> 00:46:22,120
Well, the essential boundary
condition is that u is
799
00:46:22,120 --> 00:46:25,180
0 at x equals 0.
800
00:46:25,180 --> 00:46:28,260
I'd like to now consider
two different cases
801
00:46:28,260 --> 00:46:29,880
for the Ritz analysis.
802
00:46:29,880 --> 00:46:34,580
In the case one I want to use a
function that spans u, which
803
00:46:34,580 --> 00:46:36,550
spans continuously--
804
00:46:36,550 --> 00:46:38,430
and let me draw it out here--
805
00:46:38,430 --> 00:46:44,760
from u, from x equals
0, to x equals 180.
806
00:46:44,760 --> 00:46:46,530
That is the endpoint.
807
00:46:46,530 --> 00:46:51,510
So here we have this function,
this part here and that part
808
00:46:51,510 --> 00:46:53,800
there, these are the two trial
parameters that we
809
00:46:53,800 --> 00:46:55,010
want to solve for.
810
00:46:55,010 --> 00:46:58,200
And we will select them, we will
calculate them, rather,
811
00:46:58,200 --> 00:47:00,820
using the Ritz analysis.
812
00:47:00,820 --> 00:47:09,860
Case two, I also use trial
functions, but notice now that
813
00:47:09,860 --> 00:47:11,740
I'm performing the following.
814
00:47:11,740 --> 00:47:14,190
We have a domain AB--
815
00:47:14,190 --> 00:47:15,680
let me go back once more--
816
00:47:15,680 --> 00:47:19,510
a domain AB, and a domain BC.
817
00:47:19,510 --> 00:47:24,280
And I want to now use one
function for AB and one
818
00:47:24,280 --> 00:47:26,820
function for BC.
819
00:47:26,820 --> 00:47:35,020
The AB function is simply this
one here, a linear variation
820
00:47:35,020 --> 00:47:36,280
up to this point.
821
00:47:36,280 --> 00:47:40,220
Now notice that UB is our
trial parameter--
822
00:47:40,220 --> 00:47:43,700
that's the one we don't know,
our trial function parameter.
823
00:47:43,700 --> 00:47:46,150
x/100 is simply is the function
824
00:47:46,150 --> 00:47:47,210
that I'm talking about.
825
00:47:47,210 --> 00:47:50,100
And notice that this function
only is applicable for this
826
00:47:50,100 --> 00:47:54,580
domain where, let me put down
here the length that is x
827
00:47:54,580 --> 00:47:58,270
equal to 100, and this is
here x equal to 180.
828
00:47:58,270 --> 00:48:04,650
Now for this part here, I
use this function here.
829
00:48:04,650 --> 00:48:07,680
Now notice what this
function does.
830
00:48:07,680 --> 00:48:13,130
Well, if we look at this part
here in front, it involves uB,
831
00:48:13,130 --> 00:48:15,980
which is also there,
and it involves uC.
832
00:48:15,980 --> 00:48:19,770
uB, by the way, is the physical
displacement right
833
00:48:19,770 --> 00:48:21,460
here into this direction.
834
00:48:21,460 --> 00:48:24,390
Of course, I'm plotting u
upwards here to be able to
835
00:48:24,390 --> 00:48:25,420
show it to you.
836
00:48:25,420 --> 00:48:31,610
But the displacement, uB, is the
displacement of this point
837
00:48:31,610 --> 00:48:33,340
B to the right.
838
00:48:33,340 --> 00:48:38,460
uC is the displacement of this
point C to the right.
839
00:48:38,460 --> 00:48:42,730
Then we recognize that this part
here corresponds really
840
00:48:42,730 --> 00:48:45,490
to a variation such as that.
841
00:48:45,490 --> 00:48:51,590
Notice when x is equal to 100,
which is that point, this
842
00:48:51,590 --> 00:48:53,560
function here is 1.
843
00:48:53,560 --> 00:49:00,380
When x is equal to 180, which is
that point there, this part
844
00:49:00,380 --> 00:49:04,870
is equal to 0, because 180 minus
100 is 80, divided by 80
845
00:49:04,870 --> 00:49:07,480
is 1, and 1 minus 1 is 0.
846
00:49:07,480 --> 00:49:10,580
So this dashed line
corresponds to
847
00:49:10,580 --> 00:49:12,400
this function here.
848
00:49:12,400 --> 00:49:16,080
Let me put a dashed line
underneath there.
849
00:49:16,080 --> 00:49:21,470
Well, if we now look at this
part here, we notice that this
850
00:49:21,470 --> 00:49:26,890
part is 0, or this trial
function here is 0 at this
851
00:49:26,890 --> 00:49:31,540
point B. And it varies linearly
like that across,
852
00:49:31,540 --> 00:49:36,660
where this part here, of
course, denotes uC.
853
00:49:36,660 --> 00:49:41,270
That is this one here, solid
black line, and here also,
854
00:49:41,270 --> 00:49:42,530
solid black line.
855
00:49:42,530 --> 00:49:49,750
Now the superposition of both
these functions, the dashed
856
00:49:49,750 --> 00:49:55,760
blue and the solid black line,
give us this function here.
857
00:49:55,760 --> 00:50:01,420
So the actual function that I'm
talking about is a linear
858
00:50:01,420 --> 00:50:04,950
variation along here, and a
linear variation along here,
859
00:50:04,950 --> 00:50:07,610
where I plot it vertically
up here--
860
00:50:07,610 --> 00:50:12,300
uB and uC here.
861
00:50:12,300 --> 00:50:16,070
Now, this is a specific case
that I want to draw your
862
00:50:16,070 --> 00:50:19,330
attention on, because this
really corresponds, as we
863
00:50:19,330 --> 00:50:22,720
shall see, to a true finite
element analysis.
864
00:50:22,720 --> 00:50:25,730
And the reason for it is that
we're talking about one domain
865
00:50:25,730 --> 00:50:27,960
here and another domain there.
866
00:50:27,960 --> 00:50:32,890
And both of these domains are
identified as finite elements.
867
00:50:32,890 --> 00:50:39,200
Well, the first step now is
to use pi, invoke the
868
00:50:39,200 --> 00:50:43,250
stationality condition as we did
earlier, and this gives us
869
00:50:43,250 --> 00:50:45,910
the principle of virtual
displacement.
870
00:50:45,910 --> 00:50:47,620
I mentioned it earlier
already.
871
00:50:47,620 --> 00:50:49,710
Our virtual strains are here.
872
00:50:49,710 --> 00:50:51,750
The stresses are here.
873
00:50:51,750 --> 00:50:55,290
The virtual work is
on this side.
874
00:50:55,290 --> 00:50:57,380
I discussed it earlier
already.
875
00:50:57,380 --> 00:51:01,650
We do not want to go
now via this route.
876
00:51:01,650 --> 00:51:05,300
We first of all want to now
obtain the exact solution.
877
00:51:05,300 --> 00:51:09,860
The exact solution is obtained
by using integration by parts
878
00:51:09,860 --> 00:51:16,150
on delta pi, being, of course,
equal to 0, and extracting the
879
00:51:16,150 --> 00:51:18,850
differential equation of
equilibrium for each
880
00:51:18,850 --> 00:51:22,450
differential element
in this structure.
881
00:51:22,450 --> 00:51:24,560
This means that if we are
talking here about the
882
00:51:24,560 --> 00:51:27,050
differential element
equilibrium of each
883
00:51:27,050 --> 00:51:30,880
differential element dx long
anywhere along the structure,
884
00:51:30,880 --> 00:51:33,170
in other words, the equilibrium
of typically an
885
00:51:33,170 --> 00:51:36,860
element like that.
886
00:51:36,860 --> 00:51:38,980
That is a differential equation
of equilibrium.
887
00:51:38,980 --> 00:51:41,980
And we also, of course, have
the natural boundary
888
00:51:41,980 --> 00:51:42,520
conditions.
889
00:51:42,520 --> 00:51:44,970
We can also derive the natural
boundary conditions.
890
00:51:44,970 --> 00:51:47,910
The solution to this is obtained
by integration, and
891
00:51:47,910 --> 00:51:50,380
this is the solution given.
892
00:51:50,380 --> 00:51:55,110
Well, the stresses, then, of
course are obtained by
893
00:51:55,110 --> 00:51:58,430
differentiation of the u's to
get strains, and multiplying
894
00:51:58,430 --> 00:52:01,100
those by E, and these are
the stresses in the bar.
895
00:52:01,100 --> 00:52:04,490
These are the exact stresses
in the bar that satisfy the
896
00:52:04,490 --> 00:52:07,190
differential equations of
equilibrium and the natural
897
00:52:07,190 --> 00:52:08,260
boundary conditions.
898
00:52:08,260 --> 00:52:12,290
This is the exact solution of
this bar problem, the way I
899
00:52:12,290 --> 00:52:13,750
have formulated it.
900
00:52:13,750 --> 00:52:15,970
Now will perform our
Ritz analysis.
901
00:52:15,970 --> 00:52:19,840
In case one, we use
pi equal to this.
902
00:52:19,840 --> 00:52:22,730
Notice that I have substituted
now our trial functions
903
00:52:22,730 --> 00:52:27,350
corresponding to case one
into the functional pi.
904
00:52:27,350 --> 00:52:32,910
That gives us this term, that
term here, and that term here.
905
00:52:32,910 --> 00:52:36,930
Notice that I have broken up the
integration from 0 to 100,
906
00:52:36,930 --> 00:52:44,010
and 100 to 180, because the area
changes from over this
907
00:52:44,010 --> 00:52:47,200
length here, and only for that
reason, really, I have broken
908
00:52:47,200 --> 00:52:49,520
up the integrations.
909
00:52:49,520 --> 00:52:50,740
Now this is pi.
910
00:52:50,740 --> 00:52:56,260
And if we now invoke, we can
integrate this out, and then
911
00:52:56,260 --> 00:52:59,600
invoke that del pi
shall be 0, we
912
00:52:59,600 --> 00:53:01,870
obtain this set of equations.
913
00:53:01,870 --> 00:53:05,850
We solve for a1 and a2,
substitute back into our
914
00:53:05,850 --> 00:53:09,700
assumption that we had earlier,
and we got this u.
915
00:53:09,700 --> 00:53:16,330
Notice that of course this u
displacement does satisfy the
916
00:53:16,330 --> 00:53:17,800
essential boundary conditions.
917
00:53:17,800 --> 00:53:19,940
It does satisfy the
essential boundary
918
00:53:19,940 --> 00:53:22,040
conditions at x equals 0.
919
00:53:22,040 --> 00:53:24,660
You can just substitute x equals
0, and you would see
920
00:53:24,660 --> 00:53:25,890
that u is 0.
921
00:53:25,890 --> 00:53:30,560
It does not satisfy, however,
the natural boundary condition
922
00:53:30,560 --> 00:53:33,190
at x equal to 180.
923
00:53:33,190 --> 00:53:37,300
Sigma is given here, obtained
by calculating the strains
924
00:53:37,300 --> 00:53:39,260
from here and multiplying
by E--
925
00:53:39,260 --> 00:53:43,240
this is our approximate solution
to the problem.
926
00:53:43,240 --> 00:53:46,880
We are satisfying the
compatibility conditions,
927
00:53:46,880 --> 00:53:50,280
because the bar has
remained together.
928
00:53:50,280 --> 00:53:54,340
No material has been
cut away from it.
929
00:53:54,340 --> 00:53:59,380
Also, we are satisfying the
constitutive relations, but we
930
00:53:59,380 --> 00:54:03,920
do not satisfy the internal
equilibrium on a differential
931
00:54:03,920 --> 00:54:06,190
local elements sense.
932
00:54:06,190 --> 00:54:09,700
We do not satisfy the
differential equilibrium, and
933
00:54:09,700 --> 00:54:11,880
we do not satisfy the natural
boundary conditions.
934
00:54:11,880 --> 00:54:15,330
But we satisfy them in
an approximate sense.
935
00:54:15,330 --> 00:54:16,370
Case two.
936
00:54:16,370 --> 00:54:21,450
Here now we're talking about
our two linear functions.
937
00:54:21,450 --> 00:54:24,620
And here we naturally integrate
from 0 to 100 for
938
00:54:24,620 --> 00:54:29,040
the first linear function, and
from 100 to 180 for the second
939
00:54:29,040 --> 00:54:30,760
linear function.
940
00:54:30,760 --> 00:54:36,040
Notice this is, again, the area,
and notice that this is
941
00:54:36,040 --> 00:54:37,435
here the strain squared.
942
00:54:40,190 --> 00:54:42,980
It's strain squared here because
our E is out there,
943
00:54:42,980 --> 00:54:45,510
which would give
us the stress.
944
00:54:45,510 --> 00:54:51,310
And the area here, of course,
is equal to 1, which we did
945
00:54:51,310 --> 00:54:52,280
not write down.
946
00:54:52,280 --> 00:54:55,040
The important point is that this
is now our pi for these
947
00:54:55,040 --> 00:54:55,650
two functions.
948
00:54:55,650 --> 00:54:59,690
We again invoke del
pi equal to 0.
949
00:54:59,690 --> 00:55:01,820
We obtain now this
set of equations.
950
00:55:01,820 --> 00:55:04,010
We are solving from this
set of equations uB
951
00:55:04,010 --> 00:55:07,000
and uC, given here.
952
00:55:07,000 --> 00:55:10,320
Having got uB and uC, of
course we now have the
953
00:55:10,320 --> 00:55:14,240
complete displacements along the
bar, because we only need
954
00:55:14,240 --> 00:55:19,500
to substitute back into our
original approximations that
955
00:55:19,500 --> 00:55:22,280
we looked at earlier.
956
00:55:22,280 --> 00:55:25,320
Let me just get them
once more here.
957
00:55:25,320 --> 00:55:27,370
We had them here.
958
00:55:27,370 --> 00:55:31,840
If we now substitute from
uB and uC into these two
959
00:55:31,840 --> 00:55:33,360
equations, we have the complete
960
00:55:33,360 --> 00:55:35,170
displacement solution.
961
00:55:35,170 --> 00:55:36,510
Of course, this is
an approximate
962
00:55:36,510 --> 00:55:37,720
displacement solution.
963
00:55:37,720 --> 00:55:40,650
And similarly, our stresses
are approximate.
964
00:55:40,650 --> 00:55:45,800
Now on these last few graphs,
I have plotted the solution.
965
00:55:45,800 --> 00:55:50,650
And notice that this is
here the direction x.
966
00:55:50,650 --> 00:55:55,200
Here we have the point B, here
we have the point C, here we
967
00:55:55,200 --> 00:56:00,400
have the point A. Our exact
solution, which satisfies the
968
00:56:00,400 --> 00:56:04,300
constitutive relations,
compatibility relations, and
969
00:56:04,300 --> 00:56:07,270
the differential equations of
equilibrium, and all bounded
970
00:56:07,270 --> 00:56:09,780
conditions, is the
solid line here.
971
00:56:09,780 --> 00:56:13,150
Our solution one, case one, Ritz
analysis, is the dashed
972
00:56:13,150 --> 00:56:17,790
line here, and the solution
two is this dashed dotted
973
00:56:17,790 --> 00:56:19,350
line, down there.
974
00:56:19,350 --> 00:56:22,610
Notice that we are quite close
in our Ritz analysis to the
975
00:56:22,610 --> 00:56:25,370
exact solution in the
displacement.
976
00:56:25,370 --> 00:56:29,100
However, the strains and
stresses are obtained by the
977
00:56:29,100 --> 00:56:32,530
differentiation of these
displacement solutions, and
978
00:56:32,530 --> 00:56:36,240
here I show to you the
calculated stresses.
979
00:56:36,240 --> 00:56:40,620
Again, point A here, point
B here, point C there.
980
00:56:40,620 --> 00:56:43,430
The important point
is the following.
981
00:56:43,430 --> 00:56:49,430
In the exact solution, we have
the stress of 100 in domain
982
00:56:49,430 --> 00:56:52,810
AB, and then we have
this curve here, a
983
00:56:52,810 --> 00:56:54,670
very high slope there.
984
00:56:54,670 --> 00:56:59,460
And in our solution one, we had
this variation in stress.
985
00:56:59,460 --> 00:57:02,340
Notice that it goes continuously
over the complete
986
00:57:02,340 --> 00:57:08,620
domain, because our assumed
displacement function was
987
00:57:08,620 --> 00:57:11,640
continuous also over this
domain, and its first
988
00:57:11,640 --> 00:57:15,100
derivative was continuous over
this complete domain.
989
00:57:15,100 --> 00:57:18,860
So that's why our solution one
is continuous there, and in
990
00:57:18,860 --> 00:57:21,220
fact, we're seeing just the
straight line there, because
991
00:57:21,220 --> 00:57:23,660
our displacement approximation
was parabolic.
992
00:57:23,660 --> 00:57:29,300
Our solution two is exact
here, 100, and very
993
00:57:29,300 --> 00:57:32,350
approximate here for
the displacement.
994
00:57:32,350 --> 00:57:35,830
But notice that at the midpoint
between B and C, we
995
00:57:35,830 --> 00:57:39,300
get very good results.
996
00:57:39,300 --> 00:57:43,180
Now the important point really
is shown here on
997
00:57:43,180 --> 00:57:44,820
the last view graph.
998
00:57:44,820 --> 00:57:48,520
We note that in this last
analysis, we use trial
999
00:57:48,520 --> 00:57:50,840
functions that do not satisfy
the natural boundary
1000
00:57:50,840 --> 00:57:54,680
condition, and I'm talking now
about the piecewise linear
1001
00:57:54,680 --> 00:57:57,660
functions, in other words, from
A to B and B to C each,
1002
00:57:57,660 --> 00:57:59,030
just a straight line.
1003
00:57:59,030 --> 00:58:01,440
We use trial functions that
do not satisfy the natural
1004
00:58:01,440 --> 00:58:02,490
boundary conditions.
1005
00:58:02,490 --> 00:58:05,560
The trial functions themselves
are continuous, but the
1006
00:58:05,560 --> 00:58:10,720
derivatives are discontinuous
at point B. Notice our
1007
00:58:10,720 --> 00:58:15,080
stresses here are discontinuous
at point B.
1008
00:58:15,080 --> 00:58:18,700
For a cm minus 1 variational
problem, the way I've defined
1009
00:58:18,700 --> 00:58:22,150
it, we only need continuity in
the m minus first derivatives
1010
00:58:22,150 --> 00:58:23,830
of the functions.
1011
00:58:23,830 --> 00:58:27,610
In this problem, m is 1, and
therefore we only need
1012
00:58:27,610 --> 00:58:35,300
continuity in the functions
themselves, and not in any
1013
00:58:35,300 --> 00:58:39,330
derivatives, because we only
need continuity in the m minus
1014
00:58:39,330 --> 00:58:41,050
first derivative.
1015
00:58:41,050 --> 00:58:46,390
The domains A and B and B and C
are finite elements, and in
1016
00:58:46,390 --> 00:58:49,850
actuality, we've performed a
finite element analysis.
1017
00:58:49,850 --> 00:58:51,990
This is all I wanted to
say in this lecture.
1018
00:58:51,990 --> 00:58:53,240
Thank you for your attention.