1
00:00:00,040 --> 00:00:02,470
The following content is
provided under a Creative

2
00:00:02,470 --> 00:00:03,880
Commons license.

3
00:00:03,880 --> 00:00:06,920
Your support will help MIT
OpenCourseWare continue to

4
00:00:06,920 --> 00:00:10,570
offer high quality educational
resources for free.

5
00:00:10,570 --> 00:00:13,470
To make a donation or view
additional materials from

6
00:00:13,470 --> 00:00:17,400
hundreds of MIT courses, visit
MIT OpenCourseWare at

7
00:00:17,400 --> 00:00:18,650
ocw.mit.edu.

8
00:00:21,580 --> 00:00:23,760
PROFESSOR: Ladies and gentlemen,
welcome to

9
00:00:23,760 --> 00:00:25,600
lecture number 11.

10
00:00:25,600 --> 00:00:28,120
In this lecture, I would like
to continue discussing with

11
00:00:28,120 --> 00:00:32,240
you the solution of the dynamic
equilibrium equations.

12
00:00:32,240 --> 00:00:35,400
We discussed the solution of
dynamic equilibrium equations

13
00:00:35,400 --> 00:00:38,440
already in lecture number 10.

14
00:00:38,440 --> 00:00:41,950
In that lecture, we consider
direct integration methods.

15
00:00:41,950 --> 00:00:45,270
In this lecture, I would like to
consider with you, discuss

16
00:00:45,270 --> 00:00:49,980
with you the mode superposition
analysis.

17
00:00:49,980 --> 00:00:54,100
The basic idea in the mode
superposition analysis is the

18
00:00:54,100 --> 00:01:01,330
transformation from the U
displacements into a set of

19
00:01:01,330 --> 00:01:03,891
new displacements, X.

20
00:01:03,891 --> 00:01:07,380
P is an n by n matrix,
which is nonsingular.

21
00:01:10,260 --> 00:01:15,510
If P is nonsingular, we should
recognize that the vector U of

22
00:01:15,510 --> 00:01:20,075
lengths n is uniquely given
when we have the vector X,

23
00:01:20,075 --> 00:01:23,100
also of lengths n
and vice versa.

24
00:01:23,100 --> 00:01:26,600
In other words, if we have the
vector X, we can calculate U.

25
00:01:26,600 --> 00:01:30,420
Or if we have U, we can
calculate X. P, of course, has

26
00:01:30,420 --> 00:01:33,440
to be a nonsingular matrix.

27
00:01:33,440 --> 00:01:35,630
We call this a transformation
matrix.

28
00:01:35,630 --> 00:01:38,790
And we call the X displacement,
that I will be

29
00:01:38,790 --> 00:01:42,640
talking about, the generalized
displacement.

30
00:01:42,640 --> 00:01:49,040
Now, if we use this
transformation, we find that

31
00:01:49,040 --> 00:01:53,260
if we substitute this
transformation into this

32
00:01:53,260 --> 00:01:58,240
equation for U, and we postulate
that P shall not be

33
00:01:58,240 --> 00:01:59,690
a function of time.

34
00:01:59,690 --> 00:02:02,200
So that when we find the
derivative of U,

35
00:02:02,200 --> 00:02:03,610
with respect to time--

36
00:02:03,610 --> 00:02:06,020
let me denote it
by a dot here--

37
00:02:06,020 --> 00:02:11,120
we simply obtain that U dot
is equal to P times X dot.

38
00:02:11,120 --> 00:02:16,440
And similarly, U double dot is
equal to P times X double dot.

39
00:02:16,440 --> 00:02:20,450
If we then substitute for U,
U dot, and U double dot--

40
00:02:20,450 --> 00:02:22,240
in other words, for the
displacements, velocities, and

41
00:02:22,240 --> 00:02:23,450
accelerations--

42
00:02:23,450 --> 00:02:30,230
into this equation, we obtain
this equation, provided--

43
00:02:30,230 --> 00:02:32,540
and now, there's one
additional step--

44
00:02:32,540 --> 00:02:36,320
after the substitution for the
displacements, velocities, and

45
00:02:36,320 --> 00:02:41,170
accelerations, we also
premultiply the total equation

46
00:02:41,170 --> 00:02:42,760
by a P transposed.

47
00:02:42,760 --> 00:02:47,850
In other words, what we do is
we take a P transposed here,

48
00:02:47,850 --> 00:02:52,580
and we are substituting for
these vectors here from that

49
00:02:52,580 --> 00:02:54,580
relation here.

50
00:02:54,580 --> 00:02:57,586
The P transposed, of course,
gives us, also, a P transposed

51
00:02:57,586 --> 00:02:59,580
R on the right-hand side.

52
00:02:59,580 --> 00:03:03,890
So R curl here is defined
as shown down here.

53
00:03:03,890 --> 00:03:08,500
Notice that M curl, C
curl, and K curl are

54
00:03:08,500 --> 00:03:10,090
given as shown here.

55
00:03:10,090 --> 00:03:14,470
And they involve, of course,
the M, C, and K matrices.

56
00:03:14,470 --> 00:03:17,480
And this P here comes via this

57
00:03:17,480 --> 00:03:21,430
substitution into these vectors.

58
00:03:21,430 --> 00:03:26,120
This P transposed comes from
that P transposed here.

59
00:03:26,120 --> 00:03:28,900
Now, what have we been
doing so far?

60
00:03:28,900 --> 00:03:30,910
Basically, what we have
been doing is we

61
00:03:30,910 --> 00:03:32,360
have changed bases.

62
00:03:32,360 --> 00:03:36,680
We have come from the
displacement defined by the

63
00:03:36,680 --> 00:03:37,830
finite elements--

64
00:03:37,830 --> 00:03:39,240
by the finite element
nodal points--

65
00:03:39,240 --> 00:03:41,550
to generalized displacements.

66
00:03:41,550 --> 00:03:45,670
The generalized displacements
now govern the equilibrium of

67
00:03:45,670 --> 00:03:48,460
the system.

68
00:03:48,460 --> 00:03:55,650
Now, this transformation is
effective if we find that our

69
00:03:55,650 --> 00:04:00,890
M curl, C curl, and K curl have
a better structure than

70
00:04:00,890 --> 00:04:04,810
the matrices, M, C, and K.
And, of course, if it is

71
00:04:04,810 --> 00:04:08,775
simpler to find the P matrix
or a P matrix.

72
00:04:12,180 --> 00:04:14,500
Let's look at a very
simple case.

73
00:04:14,500 --> 00:04:19,550
If, for example, M curl, C curl,
and K curl would all be

74
00:04:19,550 --> 00:04:23,580
diagonal matrices, then the
solution of this set of

75
00:04:23,580 --> 00:04:26,440
equations would be
very simple.

76
00:04:26,440 --> 00:04:28,570
In fact, this is
our objective.

77
00:04:28,570 --> 00:04:32,960
It is our objective to come up
with a P matrix, which when

78
00:04:32,960 --> 00:04:36,590
used in this transformation,
gives us an M curl, C curl,

79
00:04:36,590 --> 00:04:39,070
and K curl matrix that
are all diagonal.

80
00:04:41,760 --> 00:04:44,790
And, of course, in addition we
want to find the P matrix with

81
00:04:44,790 --> 00:04:46,920
a minimum amount of effort.

82
00:04:46,920 --> 00:04:50,900
Once we have diagonal matrices
here on the left-hand side--

83
00:04:50,900 --> 00:04:53,370
in other words, M curl,
C curl, and K curl--

84
00:04:53,370 --> 00:04:57,600
then, we can use our direct time
integration schemes, the

85
00:04:57,600 --> 00:05:01,270
Newmark method, or the essential
difference method.

86
00:05:01,270 --> 00:05:03,760
Usually we use the
Newmark method.

87
00:05:03,760 --> 00:05:08,040
Or we might even be able to
solve the decoupled equations.

88
00:05:08,040 --> 00:05:11,980
They're decoupled because M
curl, CK curl, are diagonal.

89
00:05:11,980 --> 00:05:16,200
We might even solve these
equations in closed form.

90
00:05:16,200 --> 00:05:20,050
The question then is how do we
find P or [UNINTELLIGIBLE]

91
00:05:20,050 --> 00:05:21,940
P?

92
00:05:21,940 --> 00:05:23,960
We want to find it cheaply.

93
00:05:23,960 --> 00:05:29,460
It shall be economical to
calculate P. And we want to,

94
00:05:29,460 --> 00:05:34,000
of course, have that M curl,
C curl, K curl are ideally

95
00:05:34,000 --> 00:05:35,250
diagonal matrices.

96
00:05:37,530 --> 00:05:41,990
Well, the P matrix can
be constructed--

97
00:05:41,990 --> 00:05:45,630
or a P matrix can be
constructed, effectively, by

98
00:05:45,630 --> 00:05:48,820
looking at the free vibration
equilibrium

99
00:05:48,820 --> 00:05:51,980
equations of the system.

100
00:05:51,980 --> 00:05:56,340
If we look at these equilibrium
equations here, in

101
00:05:56,340 --> 00:05:59,960
which we neglect damping and we
neglect the right-hand side

102
00:05:59,960 --> 00:06:02,650
forces-- we're looking at the
equilibrium equations of the

103
00:06:02,650 --> 00:06:06,800
dynamic system without damping
and without forces applied.

104
00:06:06,800 --> 00:06:10,650
And if we look for a solution
to this system, then, we can

105
00:06:10,650 --> 00:06:13,420
assume a solution
of this form.

106
00:06:13,420 --> 00:06:16,150
U, of course, being our
displacement vector.

107
00:06:16,150 --> 00:06:22,110
Phi being all the vector that
is independent now of time.

108
00:06:22,110 --> 00:06:27,640
And here is the time dependency,
sin, omega, omega

109
00:06:27,640 --> 00:06:31,370
is a radiant per second
circular frequency.

110
00:06:31,370 --> 00:06:33,000
t, of course, being the time.

111
00:06:33,000 --> 00:06:35,440
And t 0 is a time shift.

112
00:06:35,440 --> 00:06:41,370
If we make this assumption,
substitute into here--

113
00:06:41,370 --> 00:06:43,830
when we do substitute, of
course, remember we have to

114
00:06:43,830 --> 00:06:46,520
take the second derivative of
this function here with

115
00:06:46,520 --> 00:06:48,910
respect to time because we're
taking the second

116
00:06:48,910 --> 00:06:51,340
derivative of U here.

117
00:06:51,340 --> 00:06:55,940
And that gives us a minus omega
squared sin, omega t

118
00:06:55,940 --> 00:07:01,680
minus t 0, substituting
from here into there.

119
00:07:01,680 --> 00:07:05,860
We immediately get this equation
because we can cancel

120
00:07:05,860 --> 00:07:09,410
out sin omega t minus
t 0 on both sides.

121
00:07:09,410 --> 00:07:11,530
This is the resulting
equation.

122
00:07:11,530 --> 00:07:15,050
So what we now want
is to solve for

123
00:07:15,050 --> 00:07:17,930
phi and omega squared.

124
00:07:17,930 --> 00:07:21,510
This is the generalized
eigenvalue problem.

125
00:07:21,510 --> 00:07:25,580
The generalized eigenvalue
problem because M is a matrix

126
00:07:25,580 --> 00:07:28,210
other than the identity
matrix in general.

127
00:07:28,210 --> 00:07:34,440
And we know that they are n
solutions to this problem if K

128
00:07:34,440 --> 00:07:38,920
and M are n by n matrices.

129
00:07:38,920 --> 00:07:40,990
There are n solutions
to this problem.

130
00:07:40,990 --> 00:07:45,180
We call these solutions
the eigensolutions.

131
00:07:45,180 --> 00:07:51,430
And we call omega 1 squared,
I should say, and phi 1 the

132
00:07:51,430 --> 00:07:54,100
first eigenpair.

133
00:07:54,100 --> 00:07:57,440
This is the second eigenpair,
this is the nth eigenpair.

134
00:07:57,440 --> 00:08:01,090
There are n such eigenpairs.

135
00:08:01,090 --> 00:08:07,370
Each of these eigenpairs
satisfies this equation here.

136
00:08:07,370 --> 00:08:10,870
Therefore, these are
a solution--

137
00:08:10,870 --> 00:08:13,070
we call it an eigensolution--

138
00:08:13,070 --> 00:08:15,380
to this equation.

139
00:08:15,380 --> 00:08:19,570
And in addition, we also can
prove-- we can show--

140
00:08:19,570 --> 00:08:25,470
that the phi I vectors--
these vectors here--

141
00:08:25,470 --> 00:08:28,430
are M orthogonal.

142
00:08:28,430 --> 00:08:33,840
M orthogonal because phi
I transposed M phi j--

143
00:08:33,840 --> 00:08:40,640
for example, phi 1 transposed
M phi 2 is equal to 0.

144
00:08:40,640 --> 00:08:44,159
And phi 1 transposed M phi j--

145
00:08:44,159 --> 00:08:50,010
sorry phi 1 transposed M
phi 1 is equal to 1.

146
00:08:50,010 --> 00:08:53,810
And because it's equal to 1,
they're not only M orthogonal,

147
00:08:53,810 --> 00:08:57,150
they're indeed even
M ortho normal.

148
00:08:57,150 --> 00:09:03,590
What we find is that if we look
at this eigenproblem, and

149
00:09:03,590 --> 00:09:08,470
we notice that a certain
vector satisfies this

150
00:09:08,470 --> 00:09:12,390
eigenproblem, then also a
multiple of that vector

151
00:09:12,390 --> 00:09:14,630
satisfies the eigenproblem.

152
00:09:14,630 --> 00:09:20,230
And we determine this constant,
this undetermined

153
00:09:20,230 --> 00:09:26,000
constant, in such a way as to
have phi I transposed M phi I

154
00:09:26,000 --> 00:09:28,190
equal to 1.

155
00:09:28,190 --> 00:09:33,430
So these are the eigenpairs and
the eigenvectors, phi 1 to

156
00:09:33,430 --> 00:09:37,460
phi n, satisfy this relation.

157
00:09:37,460 --> 00:09:42,020
The eigenvalues, omega 1 squared
to omega n squared can

158
00:09:42,020 --> 00:09:46,810
be ordered in this way where
we find that we might have

159
00:09:46,810 --> 00:09:48,240
multiple eigenvalues.

160
00:09:48,240 --> 00:09:51,590
Omega 1 might be equal
to omega 2.

161
00:09:51,590 --> 00:09:56,440
But if we do, we still have only
a total number of n such

162
00:09:56,440 --> 00:09:57,760
eigenvalues.

163
00:09:57,760 --> 00:10:01,700
I will discuss in the next
lecture with you the solution

164
00:10:01,700 --> 00:10:05,450
techniques that we're using to
calculate these eigenpairs.

165
00:10:05,450 --> 00:10:08,070
Let's for the moment assume that
we have calculated them.

166
00:10:08,070 --> 00:10:10,010
And we want to use them.

167
00:10:10,010 --> 00:10:12,830
Well, if we now define
the following

168
00:10:12,830 --> 00:10:15,640
matrix, phi capital phi.

169
00:10:15,640 --> 00:10:18,930
It's a capital phi because
we see crossbars

170
00:10:18,930 --> 00:10:21,220
top and bottom here.

171
00:10:21,220 --> 00:10:24,960
If we define this matrix
as phi 1 to phi n--

172
00:10:24,960 --> 00:10:27,550
in other words, we're taking
the first eigenvector, and

173
00:10:27,550 --> 00:10:31,340
list it as a first column
of this matrix here.

174
00:10:31,340 --> 00:10:34,500
The second eigenvector, we list
as the second column in

175
00:10:34,500 --> 00:10:35,990
this matrix and so on.

176
00:10:35,990 --> 00:10:38,640
So that we find, of course, that
this matrix here is an n

177
00:10:38,640 --> 00:10:40,670
by n matrix.

178
00:10:40,670 --> 00:10:43,860
And if we define a diagonal
matrix, omega squared.

179
00:10:43,860 --> 00:10:46,150
This is a capital omega.

180
00:10:46,150 --> 00:10:50,360
Omega squared, being
equal to 0s in the

181
00:10:50,360 --> 00:10:51,780
off diagonal elements.

182
00:10:51,780 --> 00:10:54,720
But the diagonal elements
being omega 1

183
00:10:54,720 --> 00:10:57,240
squared to omega n squared.

184
00:10:57,240 --> 00:11:02,980
Then, we can write the solutions
to the eigenproblem.

185
00:11:02,980 --> 00:11:04,670
The solutions to the
eigenproblem--

186
00:11:04,670 --> 00:11:06,350
and I'm still talking
about this

187
00:11:06,350 --> 00:11:08,230
eigenproblem problem here.

188
00:11:08,230 --> 00:11:12,900
We can write the solutions to
this e eigenproblem in the

189
00:11:12,900 --> 00:11:14,610
following form.

190
00:11:14,610 --> 00:11:18,960
Notice that this omega squared
here has to be on the back of

191
00:11:18,960 --> 00:11:20,330
the capital phi.

192
00:11:20,330 --> 00:11:24,010
In other words, this is the
solution to the problem, K phi

193
00:11:24,010 --> 00:11:30,110
equals omega squared M phi,
where the omega squared is a

194
00:11:30,110 --> 00:11:32,440
scalar and appears
here in front.

195
00:11:32,440 --> 00:11:33,930
We usually write it in front.

196
00:11:33,930 --> 00:11:35,690
Of course, we can put
it also at the back

197
00:11:35,690 --> 00:11:36,700
because it's a scalar.

198
00:11:36,700 --> 00:11:39,060
We can put in front
or the back.

199
00:11:39,060 --> 00:11:43,410
However, when we have listed the
solutions to this problem

200
00:11:43,410 --> 00:11:47,290
in this way, then, we have to
put the capital omega squared

201
00:11:47,290 --> 00:11:50,560
at the back of M times phi.

202
00:11:50,560 --> 00:11:53,410
You can verify that by simply
multiplying out

203
00:11:53,410 --> 00:11:54,930
the right-hand side.

204
00:11:54,930 --> 00:12:00,690
We also have this relationship
here, which is the M ortho

205
00:12:00,690 --> 00:12:06,670
normality, the phi ortho
normality with M. In other

206
00:12:06,670 --> 00:12:10,910
words, this here says nothing
else and phi I transposed M

207
00:12:10,910 --> 00:12:15,440
phi j is equal to delta I j.

208
00:12:15,440 --> 00:12:19,190
We call this the
chronica delta.

209
00:12:19,190 --> 00:12:22,450
And this chronica delta
is equal to 1 for I

210
00:12:22,450 --> 00:12:24,430
being equal to j.

211
00:12:24,430 --> 00:12:26,690
And it's 0 otherwise.

212
00:12:26,690 --> 00:12:30,560
So we are having this
relationship here.

213
00:12:30,560 --> 00:12:34,380
And if we use this relationship,
and we are

214
00:12:34,380 --> 00:12:39,000
premultiplying this equation by
phi transposed to capital

215
00:12:39,000 --> 00:12:41,960
phi transposed now,
both sides.

216
00:12:41,960 --> 00:12:43,890
We are finding that we
are getting here

217
00:12:43,890 --> 00:12:45,600
the identity matrix.

218
00:12:45,600 --> 00:12:49,160
And therefore, we're having that
this left-hand side is

219
00:12:49,160 --> 00:12:51,020
equal to omega squared.

220
00:12:51,020 --> 00:12:53,080
This matrix here.

221
00:12:53,080 --> 00:12:54,550
And this is an important fact.

222
00:12:54,550 --> 00:12:56,850
In other words, we find
that the vectors--

223
00:12:56,850 --> 00:12:58,400
the eigenvectors--

224
00:12:58,400 --> 00:13:01,990
satisfy these two
relationships.

225
00:13:01,990 --> 00:13:06,540
Now, if we look back at what
we started off with in our

226
00:13:06,540 --> 00:13:11,320
multiple position analysis
discussion, we wanted a matrix

227
00:13:11,320 --> 00:13:19,130
that we call P, which is such
that P transposed M, P ideally

228
00:13:19,130 --> 00:13:21,700
is a diagonal matrix.

229
00:13:21,700 --> 00:13:23,780
Well, we have no
such a matrix.

230
00:13:23,780 --> 00:13:29,850
If we take phi being equal to P.
Then, phi transposed M phi

231
00:13:29,850 --> 00:13:31,570
is in fact a diagonal matrix.

232
00:13:31,570 --> 00:13:34,710
In fact, a very nice diagonal
matrix, the identity matrix,

233
00:13:34,710 --> 00:13:36,390
which is a matrix that
is diagonal.

234
00:13:36,390 --> 00:13:39,330
It has just once on
the diagonal.

235
00:13:39,330 --> 00:13:43,200
Similarly, we wondered here
that P transposed KP is a

236
00:13:43,200 --> 00:13:44,320
diagonal matrix.

237
00:13:44,320 --> 00:13:48,720
Well, we have achieved that
by this phi matrix.

238
00:13:48,720 --> 00:13:53,150
So the phi matrix is, indeed,
a particular P

239
00:13:53,150 --> 00:13:56,160
matrix that we can use.

240
00:13:56,160 --> 00:13:59,850
So what we are then saying is
let us use a multiple position

241
00:13:59,850 --> 00:14:02,950
analysis this particular
P matrix.

242
00:14:02,950 --> 00:14:06,630
The P matrix, which is equal
to the phi matrix and phi

243
00:14:06,630 --> 00:14:11,340
storing the eigenvectors of the
generalized eigenproblem.

244
00:14:11,340 --> 00:14:15,610
Well, substituting from here, as
before, into the governing

245
00:14:15,610 --> 00:14:17,030
dynamic equilibrium
equations, we

246
00:14:17,030 --> 00:14:19,310
directly obtain this equation.

247
00:14:19,310 --> 00:14:23,990
However, we notice now that this
matrix here, in general,

248
00:14:23,990 --> 00:14:26,710
is not yet a diagonal matrix.

249
00:14:26,710 --> 00:14:29,420
We call that our C curl.

250
00:14:29,420 --> 00:14:31,960
And our C curl has not been
diagonalized yet.

251
00:14:31,960 --> 00:14:36,670
We will have to deal with that
matrix still a little later.

252
00:14:36,670 --> 00:14:39,040
How do we solve these
equations?

253
00:14:39,040 --> 00:14:42,220
Well, if we solve these
equations, we have to also

254
00:14:42,220 --> 00:14:45,080
recognize, of course, that we
have to include the initial

255
00:14:45,080 --> 00:14:46,610
conditions to the problem.

256
00:14:46,610 --> 00:14:49,610
And the initial conditions
mean that the initial

257
00:14:49,610 --> 00:14:52,670
displacements are defined
and the initial

258
00:14:52,670 --> 00:14:55,250
velocities are defined.

259
00:14:55,250 --> 00:14:57,390
These, of course, are the
initial conditions that are

260
00:14:57,390 --> 00:15:03,170
used in the direct integration
solution of the governing

261
00:15:03,170 --> 00:15:04,970
dynamic equilibrium equations.

262
00:15:04,970 --> 00:15:08,730
Well, if we are now performing
the solution of this set of

263
00:15:08,730 --> 00:15:11,510
equations, we will have to
transform these initial

264
00:15:11,510 --> 00:15:15,180
conditions to initial conditions
on the generalized

265
00:15:15,180 --> 00:15:16,380
displacements.

266
00:15:16,380 --> 00:15:19,780
And that is achieved via
these relations here.

267
00:15:19,780 --> 00:15:22,950
Let us see how we
derive these.

268
00:15:22,950 --> 00:15:26,140
Well, if we look at this
equation here, and if we

269
00:15:26,140 --> 00:15:28,310
notice that, of course,
this equation must be

270
00:15:28,310 --> 00:15:30,240
applicable 4 times 0.

271
00:15:30,240 --> 00:15:32,270
We can simply put a 0 there.

272
00:15:32,270 --> 00:15:35,270
And we put a 0 there.

273
00:15:35,270 --> 00:15:37,750
Then, of course, we also
recognize that we can

274
00:15:37,750 --> 00:15:44,100
premultiply this equation
with this relation here.

275
00:15:44,100 --> 00:15:46,880
And if we do that on the
right-hand side, we notice

276
00:15:46,880 --> 00:15:49,420
that phi transposed M
times that phi is

277
00:15:49,420 --> 00:15:52,000
the identity matrix.

278
00:15:52,000 --> 00:15:56,010
And so we have this relation
directly obtained.

279
00:15:56,010 --> 00:15:58,420
The left-hand side in this
equation is the right-hand

280
00:15:58,420 --> 00:15:59,740
side in this equation.

281
00:15:59,740 --> 00:16:03,920
And the right-hand side here
is simply this 0 x now.

282
00:16:03,920 --> 00:16:06,410
The same holds also for
the initial velocity.

283
00:16:06,410 --> 00:16:10,720
We can simply put dots on top
of the U here and of the X.

284
00:16:10,720 --> 00:16:12,750
And therefore, we have
this relation here.

285
00:16:12,750 --> 00:16:16,480
So the solution of this set of
equations and with these

286
00:16:16,480 --> 00:16:19,840
initial conditions subject to
these initial conditions,

287
00:16:19,840 --> 00:16:22,960
gives us a solution to of
the dynamic response.

288
00:16:22,960 --> 00:16:28,200
Now, I should mention here that
we still have to deal

289
00:16:28,200 --> 00:16:29,530
with the damping matrix.

290
00:16:29,530 --> 00:16:32,990
The damping matrix here needs
particular attention.

291
00:16:32,990 --> 00:16:36,060
We do not have, in general,
a diagonal matrix here.

292
00:16:36,060 --> 00:16:40,920
And we will want to construct
specific damping matrices with

293
00:16:40,920 --> 00:16:42,960
which we can deal,
effectively.

294
00:16:42,960 --> 00:16:45,050
And yet, of course, these
matrices must make,

295
00:16:45,050 --> 00:16:46,820
physically, sense.

296
00:16:46,820 --> 00:16:49,550
Let us look now, however, first
at the case where we

297
00:16:49,550 --> 00:16:51,480
neglect damping.

298
00:16:51,480 --> 00:16:55,020
When we neglect damping, we have
this set of equations and

299
00:16:55,020 --> 00:16:57,820
individual equations
of this form.

300
00:16:57,820 --> 00:17:01,590
Notice that these are individual
equations because

301
00:17:01,590 --> 00:17:03,950
this is a vector.

302
00:17:03,950 --> 00:17:06,609
This is a diagonal matrix.

303
00:17:06,609 --> 00:17:11,000
And on the right-hand side, we
simply have a vector off load,

304
00:17:11,000 --> 00:17:13,010
of course, time dependent.

305
00:17:13,010 --> 00:17:17,930
So we can look at the first row
in this matrix relation.

306
00:17:17,930 --> 00:17:20,859
And the first row in that matrix
relation will be this

307
00:17:20,859 --> 00:17:24,490
equation here with
I equal to 1.

308
00:17:24,490 --> 00:17:27,200
The I-th row in this relation
here is the

309
00:17:27,200 --> 00:17:29,730
I-th equation here.

310
00:17:29,730 --> 00:17:33,000
In other words, I subscripts
on the [UNINTELLIGIBLE].

311
00:17:33,000 --> 00:17:34,980
And, in general therefore,
we want to

312
00:17:34,980 --> 00:17:37,300
solve n such equations.

313
00:17:37,300 --> 00:17:42,490
I going from 1 to n, with this
being the load vector and, of

314
00:17:42,490 --> 00:17:44,760
course, with the initial
conditions.

315
00:17:44,760 --> 00:17:48,860
These initial conditions are
obtained by simply looking at

316
00:17:48,860 --> 00:17:54,300
this equation here and
extracting the I-th row from

317
00:17:54,300 --> 00:17:56,600
each of these two relations.

318
00:17:56,600 --> 00:17:58,510
So this is then the problem.

319
00:17:58,510 --> 00:18:05,230
Notice the following that if we
solve all n equations with

320
00:18:05,230 --> 00:18:09,740
these initial conditions, then,
if we use, for example,

321
00:18:09,740 --> 00:18:15,640
the Newmark direct integration
method on these equations, we

322
00:18:15,640 --> 00:18:21,330
would obtain exactly the same
solution as if we had solved

323
00:18:21,330 --> 00:18:23,270
the fully coupled equations.

324
00:18:23,270 --> 00:18:25,970
The fully coupled equations--
let me write them down once

325
00:18:25,970 --> 00:18:27,620
more here--

326
00:18:27,620 --> 00:18:29,900
plus KU equals R.

327
00:18:29,900 --> 00:18:34,030
In other words, whether we
apply the Newmark time

328
00:18:34,030 --> 00:18:37,710
integration scheme, a direct
integration scheme, to this

329
00:18:37,710 --> 00:18:42,100
set of equations with the
initial conditions on U and U

330
00:18:42,100 --> 00:18:47,460
dot or apply the Newmark direct
time integration scheme

331
00:18:47,460 --> 00:18:52,020
to this set of equations, or
rather this one here with

332
00:18:52,020 --> 00:18:55,900
these initial conditions, we
obtain exactly the same

333
00:18:55,900 --> 00:18:59,900
response provided we use the
same time step, delta t in

334
00:18:59,900 --> 00:19:01,700
both integrations.

335
00:19:01,700 --> 00:19:03,570
There has been no assumption
so far.

336
00:19:03,570 --> 00:19:06,770
All we have done is a
transformation from the U

337
00:19:06,770 --> 00:19:10,050
displacements to the
X displacements.

338
00:19:10,050 --> 00:19:14,430
And this transformation is
effective if we can find the

339
00:19:14,430 --> 00:19:17,550
phi matrix that we're using
in the transformation very

340
00:19:17,550 --> 00:19:20,090
cheaply, very economically.

341
00:19:20,090 --> 00:19:24,500
Well, the effectiveness, really,
of multiple position

342
00:19:24,500 --> 00:19:28,630
goes beyond what I have
just described.

343
00:19:28,630 --> 00:19:32,640
The important point in multiple
position analysis is

344
00:19:32,640 --> 00:19:36,120
that we do not need to solve
all of these equations.

345
00:19:36,120 --> 00:19:38,530
That we do not need to solve
and look at all of the

346
00:19:38,530 --> 00:19:41,370
decoupled equations.

347
00:19:41,370 --> 00:19:45,040
But let us look first once also
at how we can solve the

348
00:19:45,040 --> 00:19:47,680
decoupled equations
now exactly.

349
00:19:47,680 --> 00:19:50,480
Where we can solve them exactly
for example using a

350
00:19:50,480 --> 00:19:54,040
Duhamel integral formulation
as shown here.

351
00:19:54,040 --> 00:19:58,390
X i of t is given via
this relation here.

352
00:19:58,390 --> 00:20:01,670
And we would, of course,
substitute our ri of tau where

353
00:20:01,670 --> 00:20:07,020
we now assume that this part
here is given, directly.

354
00:20:07,020 --> 00:20:11,600
In a computer program, very
frequently r i would only be

355
00:20:11,600 --> 00:20:14,050
given as a segment of
straight lines.

356
00:20:14,050 --> 00:20:16,610
And if that is the case, then,
of course, we might have to

357
00:20:16,610 --> 00:20:20,190
still evaluate all of this
here numerically if the

358
00:20:20,190 --> 00:20:25,330
segment is very complicated,
complex, like

359
00:20:25,330 --> 00:20:26,710
an earthquake analysis.

360
00:20:26,710 --> 00:20:30,090
ri might be varying
very rapidly.

361
00:20:30,090 --> 00:20:32,930
And then, we might still want to
evaluate this, numerically.

362
00:20:32,930 --> 00:20:35,860
However in general, we can
evaluate here this

363
00:20:35,860 --> 00:20:37,590
relationship analytically.

364
00:20:37,590 --> 00:20:41,270
And we don't need to use
numerical integration.

365
00:20:41,270 --> 00:20:45,930
Having got xi, of course, we get
our U of t as shown here

366
00:20:45,930 --> 00:20:48,440
by our transforming back.

367
00:20:48,440 --> 00:20:52,980
Well, the important point that
I now would like to get back

368
00:20:52,980 --> 00:20:58,440
to you and discuss with you for
a little while is that we

369
00:20:58,440 --> 00:21:02,920
do not need to use or to solve
all of these equations in

370
00:21:02,920 --> 00:21:04,680
multiple position analysis.

371
00:21:04,680 --> 00:21:08,220
In fact, multiple position
analysis is only effective

372
00:21:08,220 --> 00:21:10,625
when we do not need
to look at all of

373
00:21:10,625 --> 00:21:12,510
these decoupled equations.

374
00:21:12,510 --> 00:21:16,360
If we have to solve all
decoupled equations, then, we

375
00:21:16,360 --> 00:21:19,510
would have to calculate, course,
first all eigenvalues

376
00:21:19,510 --> 00:21:23,140
and eigenvectors, which can
be an enormous expense.

377
00:21:23,140 --> 00:21:25,490
And then, the multiple
position analysis

378
00:21:25,490 --> 00:21:28,150
would not be effective.

379
00:21:28,150 --> 00:21:32,760
I would like to repeat, however,
if we do solve all

380
00:21:32,760 --> 00:21:36,560
decoupled equations, using a
specific time integration

381
00:21:36,560 --> 00:21:41,300
scheme with a specific
data t time step.

382
00:21:41,300 --> 00:21:44,810
Then, if we use the same time
integration scheme on

383
00:21:44,810 --> 00:21:48,730
decoupled equations with the
same time step, we must get

384
00:21:48,730 --> 00:21:51,560
exactly the same response.

385
00:21:51,560 --> 00:21:54,750
Then there has been no
assumption so far.

386
00:21:54,750 --> 00:21:58,465
Well, however, if we then
do neglect some of

387
00:21:58,465 --> 00:22:00,190
the equations here.

388
00:22:00,190 --> 00:22:02,230
We do not solve some
of the equations.

389
00:22:02,230 --> 00:22:05,270
Then, of course, our response
will be different that we are

390
00:22:05,270 --> 00:22:09,310
predicting via this approach
when we compare with the

391
00:22:09,310 --> 00:22:13,000
response predicted by solving
these equations.

392
00:22:13,000 --> 00:22:20,470
Well, let us look once at this
graph again, which we already

393
00:22:20,470 --> 00:22:22,740
looked at briefly in
the last lecture.

394
00:22:22,740 --> 00:22:26,770
And I would like to explain
to you now why it is not

395
00:22:26,770 --> 00:22:30,020
necessary to solve all of the
decoupled equations in

396
00:22:30,020 --> 00:22:31,460
multiple position analysis.

397
00:22:31,460 --> 00:22:34,450
Why it is not, in general,
not necessary?

398
00:22:34,450 --> 00:22:36,990
Well, if we look at a
single oscillator

399
00:22:36,990 --> 00:22:38,240
equation, as given here.

400
00:22:38,240 --> 00:22:40,130
I include a damp in here,
of course, side

401
00:22:40,130 --> 00:22:42,590
can be equal to 0.

402
00:22:42,590 --> 00:22:46,270
Where P is the driving
frequency, omega is the

403
00:22:46,270 --> 00:22:49,510
frequency of the oscillator
itself.

404
00:22:49,510 --> 00:22:55,490
And if we plot the maximum
dynamic response observed,

405
00:22:55,490 --> 00:22:59,460
which we define if the dynamic
load factor when we divide

406
00:22:59,460 --> 00:23:03,810
that maximum dynamic response by
the static response, we get

407
00:23:03,810 --> 00:23:05,550
this set of curves.

408
00:23:05,550 --> 00:23:10,370
Let us look at say the curved
side equal to 0

409
00:23:10,370 --> 00:23:12,010
a little bit closer.

410
00:23:12,010 --> 00:23:15,830
If we do that, and if we
solve this equation

411
00:23:15,830 --> 00:23:17,140
with side equals 0.

412
00:23:17,140 --> 00:23:19,390
This term now not being there.

413
00:23:19,390 --> 00:23:23,920
For certain values of P over
omega, we are getting this

414
00:23:23,920 --> 00:23:27,370
curve here and that
curve here.

415
00:23:30,100 --> 00:23:31,140
What do we notice?

416
00:23:31,140 --> 00:23:37,060
Well, we notice that when P over
omega is equal to 0, we

417
00:23:37,060 --> 00:23:38,720
are getting the static
response.

418
00:23:38,720 --> 00:23:40,780
Or P over omega very
small, we're

419
00:23:40,780 --> 00:23:43,260
getting the static response.

420
00:23:43,260 --> 00:23:46,990
This means, of course, that
omega, the system frequency,

421
00:23:46,990 --> 00:23:50,630
is much larger than the applied
frequency, which means

422
00:23:50,630 --> 00:23:54,150
really that this oscillator
is extremely stiff.

423
00:23:56,920 --> 00:24:03,380
Well, when compared to the
frequency of load application.

424
00:24:03,380 --> 00:24:04,850
So then, we are in this range.

425
00:24:04,850 --> 00:24:10,800
And we notice that say P over
omega in this range here,

426
00:24:10,800 --> 00:24:14,140
smaller than a certain value,
and this is here 0.5 say,

427
00:24:14,140 --> 00:24:15,970
smaller than 0.2.

428
00:24:15,970 --> 00:24:20,170
We would see that basically, the
maximum dynamic response

429
00:24:20,170 --> 00:24:24,270
is really just nothing else
than the static response.

430
00:24:24,270 --> 00:24:27,110
So we are talking in this
region here really about

431
00:24:27,110 --> 00:24:28,160
static response.

432
00:24:28,160 --> 00:24:31,060
And in this region here, we
are talking about a truly

433
00:24:31,060 --> 00:24:33,010
dynamic response.

434
00:24:33,010 --> 00:24:34,380
Now what does that mean?

435
00:24:34,380 --> 00:24:40,060
Well, if we look at our
decoupled equations here, it

436
00:24:40,060 --> 00:24:49,000
means that if i consist of
frequencies, basically, which

437
00:24:49,000 --> 00:24:54,630
when measured on the system of
the oscillator frequency are

438
00:24:54,630 --> 00:24:59,960
very small, we, basically, would
have a static solution.

439
00:24:59,960 --> 00:25:02,950
You would, basically, have a
static solution for this

440
00:25:02,950 --> 00:25:05,920
equilibrium equation.

441
00:25:05,920 --> 00:25:11,800
A truly dynamic solution will
only be obtained when P, in

442
00:25:11,800 --> 00:25:15,850
other words, the frequency of
the load divided by omega,

443
00:25:15,850 --> 00:25:20,030
falls into this range here.

444
00:25:20,030 --> 00:25:23,630
Well, this is what we
are using in most

445
00:25:23,630 --> 00:25:25,190
superposition analysis.

446
00:25:25,190 --> 00:25:32,590
We only need to consider those
frequencies of the system.

447
00:25:32,590 --> 00:25:35,760
And these are the frequencies
of the systems, the omegas,

448
00:25:35,760 --> 00:25:40,340
which are truly excited
by the dynamic loads.

449
00:25:40,340 --> 00:25:44,160
We only need to look at those
decoupled equilibrium

450
00:25:44,160 --> 00:25:49,250
equations for which we have that
the frequencies in the

451
00:25:49,250 --> 00:25:53,030
decoupled equations are truly
excited by the dynamic loads.

452
00:25:53,030 --> 00:25:55,900
Now, I only talked about
the P value here.

453
00:25:55,900 --> 00:25:58,340
Of course, in an actual
analysis, we would find that

454
00:25:58,340 --> 00:26:02,940
ri of P contains a number of
frequencies, a spectrum of

455
00:26:02,940 --> 00:26:03,920
frequencies.

456
00:26:03,920 --> 00:26:07,510
Well, what we have to do is look
at the highest frequency

457
00:26:07,510 --> 00:26:10,830
that is contained in ri.

458
00:26:10,830 --> 00:26:14,750
And we measure that highest
frequency on the frequencies

459
00:26:14,750 --> 00:26:16,600
of the system.

460
00:26:16,600 --> 00:26:21,120
If that highest frequency
divided by the frequency of

461
00:26:21,120 --> 00:26:24,510
the system is falling into
this range here, we are

462
00:26:24,510 --> 00:26:27,060
talking about a static
response, basically.

463
00:26:27,060 --> 00:26:35,160
And we do not need to consider
those equilibrium equations in

464
00:26:35,160 --> 00:26:38,340
are multiple position
analysis.

465
00:26:38,340 --> 00:26:41,870
Well, so what we are then saying
is let us all solve

466
00:26:41,870 --> 00:26:46,310
these equations not from i
equals 1 to n but from i

467
00:26:46,310 --> 00:26:55,260
equals 1 to P. And then, instead
of obtaining the true

468
00:26:55,260 --> 00:26:59,640
response, the "exact" response,
we are obtaining a

469
00:26:59,640 --> 00:27:06,970
response U P. i equals 1 to P.
Phi i, x i gives us U P, where

470
00:27:06,970 --> 00:27:10,685
U P is an approximation now
to the U displacements.

471
00:27:14,150 --> 00:27:18,250
And the U displacements, once
again, are the solution to

472
00:27:18,250 --> 00:27:19,830
this system of equations.

473
00:27:22,590 --> 00:27:24,370
How can we find out the error?

474
00:27:24,370 --> 00:27:29,710
Well, if U P is an approximate
solution to the system of

475
00:27:29,710 --> 00:27:33,670
equations that we want to solve,
then, let us substitute

476
00:27:33,670 --> 00:27:35,730
back into that system
of equations.

477
00:27:35,730 --> 00:27:37,270
Let me write it down
once more here.

478
00:27:37,270 --> 00:27:41,530
We have MU double dot
plus KU equals R

479
00:27:41,530 --> 00:27:42,830
that we want to solve.

480
00:27:42,830 --> 00:27:46,540
Well, we can also write that, of
course, MU double dot plus

481
00:27:46,540 --> 00:27:53,380
KU minus R. And this
shall be 0.

482
00:27:53,380 --> 00:27:56,620
So if we have an approximate
solution to this set of

483
00:27:56,620 --> 00:27:59,840
equations, why not just
substitute that approximate

484
00:27:59,840 --> 00:28:04,570
solution in here and see how
large the right-hand side is.

485
00:28:04,570 --> 00:28:09,100
If it's exactly 0, well, then
we have that our U P is,

486
00:28:09,100 --> 00:28:12,540
indeed, very close to U. And if
it's exactly 0 because U P

487
00:28:12,540 --> 00:28:13,910
would be equal to U.

488
00:28:13,910 --> 00:28:15,680
We will not find this to be 0.

489
00:28:15,680 --> 00:28:18,550
We will never find it to be 0
because in a computer, we

490
00:28:18,550 --> 00:28:21,620
would, of course, use finite
digit arithmetic to evaluate

491
00:28:21,620 --> 00:28:22,400
the left-hand side.

492
00:28:22,400 --> 00:28:24,020
And we would have at
least round off.

493
00:28:24,020 --> 00:28:27,840
But what we want is at the
right-hand side calculated,

494
00:28:27,840 --> 00:28:34,340
which we now might call some
value shall be close to 0.

495
00:28:34,340 --> 00:28:36,990
And that means really what
we want to measure

496
00:28:36,990 --> 00:28:39,230
is this part here.

497
00:28:39,230 --> 00:28:40,920
Now what have I written
down here?

498
00:28:40,920 --> 00:28:47,180
I have here epsilon P of t
being this vector here.

499
00:28:47,180 --> 00:28:48,020
This is a vector.

500
00:28:48,020 --> 00:28:51,680
And we can call this an out of
balance load vector divided by

501
00:28:51,680 --> 00:28:53,490
the applied load vector.

502
00:28:53,490 --> 00:28:59,630
Notice that what I'm doing here,
I take the norm of this

503
00:28:59,630 --> 00:29:02,580
vector because I want to
get a single number.

504
00:29:02,580 --> 00:29:07,570
I want to get a single number
here, which is representative

505
00:29:07,570 --> 00:29:09,680
of all the elements
in that vector.

506
00:29:09,680 --> 00:29:12,770
The same applies here.

507
00:29:12,770 --> 00:29:17,200
This norm is calculated by
taking each value in this

508
00:29:17,200 --> 00:29:21,790
vector, squaring that value,
adding the squares of the

509
00:29:21,790 --> 00:29:24,385
values up, and then, taking
a square root out

510
00:29:24,385 --> 00:29:26,160
of that final value.

511
00:29:26,160 --> 00:29:30,030
In other words, if we write it
down here, quickly, what we do

512
00:29:30,030 --> 00:29:35,180
is if we have a vector a, and
we want to find this norm.

513
00:29:35,180 --> 00:29:39,900
What we do is we take a 1
squared plus a 2 squared, et

514
00:29:39,900 --> 00:29:43,360
cetera, add all the individual
entries up

515
00:29:43,360 --> 00:29:44,620
of that vector here--

516
00:29:44,620 --> 00:29:46,300
squares of the entries up--

517
00:29:46,300 --> 00:29:48,010
and then, take the square
root out of it.

518
00:29:48,010 --> 00:29:50,170
And that gives us a single
number, which is

519
00:29:50,170 --> 00:29:53,080
representative of all the
elements in that vector.

520
00:29:53,080 --> 00:29:57,150
Notice that if one element is
large, well then, this value

521
00:29:57,150 --> 00:29:59,900
must be large, too, because
that one element will make

522
00:29:59,900 --> 00:30:01,870
this value large here.

523
00:30:01,870 --> 00:30:07,450
Well, if epsilon t is close to
0, assuming that r of t is not

524
00:30:07,450 --> 00:30:08,490
0, of course.

525
00:30:08,490 --> 00:30:09,650
Otherwise, you couldn't
use this.

526
00:30:09,650 --> 00:30:12,640
You would have to use some
equivalent approach.

527
00:30:12,640 --> 00:30:17,210
If epsilon P is close to 0,
then, we know that our

528
00:30:17,210 --> 00:30:19,880
equilibrium equation has been
solved quite accurately.

529
00:30:23,290 --> 00:30:26,820
In general, we might find that
it's not close to 0.

530
00:30:26,820 --> 00:30:30,430
And what we want to do is
a static correction.

531
00:30:30,430 --> 00:30:38,260
This static correction takes
into account the fact that we

532
00:30:38,260 --> 00:30:40,540
have neglected the
higher modes.

533
00:30:40,540 --> 00:30:42,590
And in the higher modes, of
course, we know we have

534
00:30:42,590 --> 00:30:45,290
basically static response.

535
00:30:45,290 --> 00:30:49,290
that's why we did not include
those equations in the

536
00:30:49,290 --> 00:30:51,140
multiple position analysis.

537
00:30:51,140 --> 00:30:54,050
So what do we do in that
static correction?

538
00:30:54,050 --> 00:30:57,540
We recognize that R could
be written in this form.

539
00:30:57,540 --> 00:31:01,780
This RI is exactly the value
that I defined earlier.

540
00:31:01,780 --> 00:31:04,130
M phi i is a vector.

541
00:31:04,130 --> 00:31:07,380
You see, I can take M and
multiply it by phi i, and I

542
00:31:07,380 --> 00:31:08,680
get a vector.

543
00:31:08,680 --> 00:31:11,380
Now they are in such
a vectors.

544
00:31:11,380 --> 00:31:13,360
We might not have calculated
all n.

545
00:31:13,360 --> 00:31:18,950
But you can say that indeed,
in theory there are n.

546
00:31:18,950 --> 00:31:21,700
So we can write this down.

547
00:31:21,700 --> 00:31:27,250
We also then can calculate the
delta R, which is that amount

548
00:31:27,250 --> 00:31:30,150
of the load vector, which has
not been included in the

549
00:31:30,150 --> 00:31:31,910
multiple position analysis.

550
00:31:31,910 --> 00:31:32,890
How do we obtain it?

551
00:31:32,890 --> 00:31:36,160
Well, we know R is the
total load vector.

552
00:31:36,160 --> 00:31:40,500
And if we now sum instead of i
equals 1 to n, only from i

553
00:31:40,500 --> 00:31:44,590
equals 1 to P, we are getting
this part here.

554
00:31:44,590 --> 00:31:48,200
But this part is exactly what
we have included in the

555
00:31:48,200 --> 00:31:50,650
multiple position analysis
already.

556
00:31:50,650 --> 00:31:53,820
So we are left with a vector on
the left-hand side, which

557
00:31:53,820 --> 00:31:56,410
we have not included in the
multiple position analysis.

558
00:31:56,410 --> 00:32:00,360
And we did not include it
because we knew at most, we

559
00:32:00,360 --> 00:32:03,230
can have static response,
corresponding to these loads.

560
00:32:03,230 --> 00:32:05,450
Let us now calculate that
static response.

561
00:32:05,450 --> 00:32:07,480
That's done right down here.

562
00:32:07,480 --> 00:32:10,830
Now, this static response has
to be added to the response

563
00:32:10,830 --> 00:32:15,630
which we predicted in the
multiple position analysis.

564
00:32:15,630 --> 00:32:20,010
Let us now go to the case
with damping included.

565
00:32:20,010 --> 00:32:22,050
So far we neglected that.

566
00:32:22,050 --> 00:32:26,180
Now, with damping included, we
have this part here, too.

567
00:32:26,180 --> 00:32:28,580
Notice phi transpose C
phi, in general, is

568
00:32:28,580 --> 00:32:29,710
not a diagonal matrix.

569
00:32:29,710 --> 00:32:31,180
And that's why I considered
first the

570
00:32:31,180 --> 00:32:32,690
case without damping.

571
00:32:32,690 --> 00:32:36,270
However, if we have
proportionate damping, and in

572
00:32:36,270 --> 00:32:39,950
structural analysis, frequently,
all that is

573
00:32:39,950 --> 00:32:44,090
necessary is to include
proportionate damping.

574
00:32:44,090 --> 00:32:46,360
Which we postulate to
be of this form.

575
00:32:46,360 --> 00:32:48,360
2 omega i xi i.

576
00:32:48,360 --> 00:32:52,710
Delta ij is again the chronica
delta, which is equal to 1

577
00:32:52,710 --> 00:32:56,950
when i is equal to j and is
equal to 0 otherwise.

578
00:32:56,950 --> 00:33:00,740
So if we are postulating
this, where omega i,

579
00:33:00,740 --> 00:33:02,080
of course, is given.

580
00:33:02,080 --> 00:33:05,860
That is the free vibration
frequency, which is stored in

581
00:33:05,860 --> 00:33:07,490
this matrix here.

582
00:33:07,490 --> 00:33:10,970
xi i is a value, which
we are not given yet.

583
00:33:10,970 --> 00:33:12,350
We have not discussed it yet.

584
00:33:12,350 --> 00:33:17,270
But if we postulate that this
matrix here, basically, or

585
00:33:17,270 --> 00:33:20,200
each entry in that matrix can
be written in this way.

586
00:33:20,200 --> 00:33:23,400
Then, indeed, we would have
decoupled equations.

587
00:33:23,400 --> 00:33:27,860
It turns out that in practice,
as I mentioned already, this

588
00:33:27,860 --> 00:33:32,810
is all that is necessary to
include damping appropriately.

589
00:33:32,810 --> 00:33:42,200
Well, if we know these xi i
values, then, we can also

590
00:33:42,200 --> 00:33:50,540
construct a C matrix with
the xi i values given.

591
00:33:50,540 --> 00:33:54,960
The xi i values are frequently
obtained from experiments.

592
00:33:54,960 --> 00:33:59,870
In other words, the structure
is taken, its vibrated, it's

593
00:33:59,870 --> 00:34:02,600
excited into certain
vibration modes.

594
00:34:02,600 --> 00:34:04,840
And from these experiments
then, we

595
00:34:04,840 --> 00:34:07,870
obtain the xi i value.

596
00:34:07,870 --> 00:34:11,070
So once we have the xi
i value given, we can

597
00:34:11,070 --> 00:34:12,280
calculate the C matrix.

598
00:34:12,280 --> 00:34:15,889
And one way of calculating is
to use the Caughey series.

599
00:34:15,889 --> 00:34:19,679
Here we have the Caughey series,
which was proposed for

600
00:34:19,679 --> 00:34:25,929
calculating the C matrix, which
satisfies this criteria.

601
00:34:25,929 --> 00:34:28,889
In other words, damping
is proportional.

602
00:34:28,889 --> 00:34:36,770
Well, if the psi i values are
given, we can calculate the

603
00:34:36,770 --> 00:34:40,020
omega i, of course
are also given.

604
00:34:40,020 --> 00:34:45,179
We can calculate using this
equation, the constants a 0 to

605
00:34:45,179 --> 00:34:46,860
ap minus 1.

606
00:34:46,860 --> 00:34:51,100
Once we have these constants,
we can substitute them into

607
00:34:51,100 --> 00:34:53,699
here and be coming up
with a C matrix.

608
00:34:53,699 --> 00:34:59,100
And this C matrix here,
indeed, satisfies the

609
00:34:59,100 --> 00:35:02,590
normality property shown here.

610
00:35:02,590 --> 00:35:07,510
So, in other words, given the
psi i values from experimental

611
00:35:07,510 --> 00:35:12,500
results, given the omega i
values, we can calculate a C

612
00:35:12,500 --> 00:35:16,820
matrix that, in fact, gives us
a diagonal matrix when we

613
00:35:16,820 --> 00:35:20,320
carry out phi i transposed
C phi j.

614
00:35:20,320 --> 00:35:24,000
A special case of this C matrix
is Rayleigh damping,

615
00:35:24,000 --> 00:35:25,980
used abundantly in practice.

616
00:35:25,980 --> 00:35:29,130
Where we just include the first
two terms, alpha and

617
00:35:29,130 --> 00:35:31,230
beta being constants now.

618
00:35:31,230 --> 00:35:33,700
Let us go through a particular
example.

619
00:35:33,700 --> 00:35:40,520
Let's say that we know the
critical damping psi 1 is 0.02

620
00:35:40,520 --> 00:35:43,190
and psi 2 is 0.10.

621
00:35:43,190 --> 00:35:46,340
In other words, 10% critical
damping in the second mode and

622
00:35:46,340 --> 00:35:48,590
2% critical damping
in the first mode.

623
00:35:48,590 --> 00:35:53,060
The corresponding frequencies
are 1 and 3.

624
00:35:53,060 --> 00:35:54,910
We want to calculate
our friend, beta.

625
00:35:54,910 --> 00:36:00,660
Well, what we do is use this C
matrix in the relation, phi i

626
00:36:00,660 --> 00:36:03,440
transposed, C times phi i.

627
00:36:03,440 --> 00:36:06,280
And that should give
us 2 omega psi i.

628
00:36:06,280 --> 00:36:10,490
Of course, we have two
i values, i equals 1

629
00:36:10,490 --> 00:36:11,740
and i equals 2.

630
00:36:11,740 --> 00:36:15,240
So we can apply this
equation twice.

631
00:36:15,240 --> 00:36:18,500
Well, when we calculate
it out, we get this.

632
00:36:18,500 --> 00:36:22,210
Applying it twice, we directly
obtain these

633
00:36:22,210 --> 00:36:24,110
two equations here.

634
00:36:24,110 --> 00:36:28,100
And we solve these two equations
for alpha and beta.

635
00:36:28,100 --> 00:36:33,440
Having alpha and beta, we can
substitute now back into this

636
00:36:33,440 --> 00:36:37,280
equation here and have
our C matrix.

637
00:36:37,280 --> 00:36:41,820
This would be, for example, a C
matrix that we could use in

638
00:36:41,820 --> 00:36:44,350
a direct integration analysis.

639
00:36:44,350 --> 00:36:48,040
You see, if we are performing a
multiple position analysis,

640
00:36:48,040 --> 00:36:51,230
we might know, from experience,
from experimental

641
00:36:51,230 --> 00:36:57,090
results, that psi 1 and psi
2 are of these two values.

642
00:36:57,090 --> 00:37:01,760
And then, if we know that we
have so much critical damping

643
00:37:01,760 --> 00:37:05,460
or so much damping in the first
and second vibration

644
00:37:05,460 --> 00:37:09,140
mode, and we want to perform
a step-by-step direct

645
00:37:09,140 --> 00:37:12,790
integration analysis, we would
have to construct a C matrix.

646
00:37:12,790 --> 00:37:17,880
And this is the procedure
to construct a C matrix.

647
00:37:17,880 --> 00:37:22,210
Then, of course, we would have
to ask ourselves well, knowing

648
00:37:22,210 --> 00:37:27,990
these psi values for omega 1 and
omega 2, what are now the

649
00:37:27,990 --> 00:37:32,780
psi values that we are
implicitly using for the

650
00:37:32,780 --> 00:37:37,230
higher modes when we employ
this specific C matrix?

651
00:37:37,230 --> 00:37:40,290
And we use now, this
equation here.

652
00:37:40,290 --> 00:37:44,010
This is the equation, which we
already derived on this view

653
00:37:44,010 --> 00:37:46,810
graph here by looking
at this relation.

654
00:37:46,810 --> 00:37:49,200
You are now using this equation
because this equation

655
00:37:49,200 --> 00:37:53,710
must be applicable not just for
psi 1 and psi 2 but for

656
00:37:53,710 --> 00:37:55,250
all psi values.

657
00:37:55,250 --> 00:38:00,120
Well, then using this equation,
we can solve for the

658
00:38:00,120 --> 00:38:01,360
psi values.

659
00:38:01,360 --> 00:38:05,350
And, of course, psi 1 and psi 2,
substituting alpha and beta

660
00:38:05,350 --> 00:38:10,340
in here and omega 1 and omega
2 would, indeed, be just 2%

661
00:38:10,340 --> 00:38:13,190
and 10%, 0.02 and 0.10.

662
00:38:13,190 --> 00:38:15,220
However, we can now use
this equation to

663
00:38:15,220 --> 00:38:16,860
get the high xi values.

664
00:38:16,860 --> 00:38:19,560
And there's something to be
observed, which is quite

665
00:38:19,560 --> 00:38:24,360
important, namely we notice that
the beta value that we

666
00:38:24,360 --> 00:38:26,840
calculated is associated
with the high

667
00:38:26,840 --> 00:38:28,420
frequencies, basically.

668
00:38:28,420 --> 00:38:31,460
And the alpha value goes away.

669
00:38:31,460 --> 00:38:34,620
Or this contribution goes away
when we're looking at the high

670
00:38:34,620 --> 00:38:35,280
frequencies.

671
00:38:35,280 --> 00:38:38,680
For high frequencies therefore,
the beta value is a

672
00:38:38,680 --> 00:38:40,480
dominant factor.

673
00:38:40,480 --> 00:38:44,730
And for low frequencies,
the alpha value is

674
00:38:44,730 --> 00:38:47,800
the dominant factor.

675
00:38:47,800 --> 00:38:52,960
In other words, once again,
looking at this equation here,

676
00:38:52,960 --> 00:38:56,840
in the high frequency response,
beta K, really, will

677
00:38:56,840 --> 00:38:57,990
determine the damping.

678
00:38:57,990 --> 00:39:01,430
And in the low frequency, the
response is really the alpha

679
00:39:01,430 --> 00:39:06,030
value that determines
the damping.

680
00:39:06,030 --> 00:39:08,690
Well, the response solution
now would be performed in

681
00:39:08,690 --> 00:39:11,120
exactly the same
way as before.

682
00:39:11,120 --> 00:39:13,195
As in the case of
no damping, we

683
00:39:13,195 --> 00:39:15,990
solve now again P equations.

684
00:39:15,990 --> 00:39:18,290
P equations only.

685
00:39:18,290 --> 00:39:23,880
And we are doing that by looking
at this equation and

686
00:39:23,880 --> 00:39:26,740
with, of course, ri being given
as shown here and the

687
00:39:26,740 --> 00:39:29,960
initial conditions
being given here.

688
00:39:29,960 --> 00:39:35,120
Solving this equation here via
numerical integration or exact

689
00:39:35,120 --> 00:39:38,110
integration, using, for
something, Duhamel integral,

690
00:39:38,110 --> 00:39:41,990
we obtain the xi values
as a function of time.

691
00:39:41,990 --> 00:39:45,670
We substitute into here to
get our U P, which is an

692
00:39:45,670 --> 00:39:49,070
approximation to the
solution of the

693
00:39:49,070 --> 00:39:51,120
dynamic equilibrium equations.

694
00:39:51,120 --> 00:39:53,680
The dynamic equilibrium
equations now being MU double

695
00:39:53,680 --> 00:39:59,410
dot plus CU dot plus KU,
being equal to r.

696
00:39:59,410 --> 00:40:02,090
Notice we have a C matrix
in here now.

697
00:40:02,090 --> 00:40:05,170
However, and this is an
important point, we have a

698
00:40:05,170 --> 00:40:13,620
specific C matrix constructed
such that the phi i transposed

699
00:40:13,620 --> 00:40:18,110
C phi j gives us a diagonal
matrix with the diagonal

700
00:40:18,110 --> 00:40:23,740
elements being equal to
2 omega i, psi i.

701
00:40:23,740 --> 00:40:26,090
And one of these matrices,
for example, would be

702
00:40:26,090 --> 00:40:28,180
the Rayleigh matrix.

703
00:40:28,180 --> 00:40:32,310
The matrix obtained using
Rayleigh damping, rather.

704
00:40:32,310 --> 00:40:35,640
The important point is, once
again, that all we have been

705
00:40:35,640 --> 00:40:44,000
doing so far is to make a
change of bases from the

706
00:40:44,000 --> 00:40:47,610
finite element coordinates
to the generalized

707
00:40:47,610 --> 00:40:50,230
displacements, xi.

708
00:40:50,230 --> 00:40:51,840
That was the first step.

709
00:40:51,840 --> 00:40:56,370
The second step was that we
are neglecting the modes P

710
00:40:56,370 --> 00:41:03,330
plus 1 to n by only solving the
first P equations here and

711
00:41:03,330 --> 00:41:08,260
using this as an approximation
to the actual solution.

712
00:41:08,260 --> 00:41:16,500
The important point, of course,
is now that we have to

713
00:41:16,500 --> 00:41:19,200
ask ourselves when is
multiple position

714
00:41:19,200 --> 00:41:20,775
analysis really effective?

715
00:41:23,360 --> 00:41:26,870
Some considerations I have here
summarized on the last

716
00:41:26,870 --> 00:41:27,860
view graph.

717
00:41:27,860 --> 00:41:31,410
The multiple position analysis
is effective when the response

718
00:41:31,410 --> 00:41:33,880
lies in only a few modes.

719
00:41:33,880 --> 00:41:35,240
What does this mean?

720
00:41:35,240 --> 00:41:41,970
It means that if we look at our
eigenvalue problem, K phi

721
00:41:41,970 --> 00:41:44,500
equals omega squared M phi.

722
00:41:44,500 --> 00:41:47,750
We only need to solve this
eigenvalue problem

723
00:41:47,750 --> 00:41:50,270
for a very few modes.

724
00:41:50,270 --> 00:41:53,680
In other words, P being much
smaller than M, I will only

725
00:41:53,680 --> 00:41:59,470
have to solve for phi 1 up to
phi P and omega 1 squared up

726
00:41:59,470 --> 00:42:02,890
to omega P squared.

727
00:42:02,890 --> 00:42:05,850
This is a very important
point.

728
00:42:05,850 --> 00:42:11,750
I assumed here that we are
taking the lowest frequencies,

729
00:42:11,750 --> 00:42:16,530
omega 1 to omega P. Of course,
in general analysis, it might

730
00:42:16,530 --> 00:42:20,960
just be necessary to calculate
certain frequencies, say omega

731
00:42:20,960 --> 00:42:23,480
10 to omega 20.

732
00:42:23,480 --> 00:42:26,660
We only want to find the
frequencies in a certain range

733
00:42:26,660 --> 00:42:30,570
because we know that the
excitation frequency only lies

734
00:42:30,570 --> 00:42:32,160
in a particular range.

735
00:42:32,160 --> 00:42:34,540
That would be, for example,
the case in vibration

736
00:42:34,540 --> 00:42:37,690
excitation analysis.

737
00:42:37,690 --> 00:42:41,040
This then is a requirement for
multiple position analysis,

738
00:42:41,040 --> 00:42:42,920
really, to be effective.

739
00:42:42,920 --> 00:42:48,250
When the response is obtained
over many time intervals, then

740
00:42:48,250 --> 00:42:51,640
multiple position analysis
also becomes effective.

741
00:42:51,640 --> 00:42:55,010
We should remember here that
if we perform a direct

742
00:42:55,010 --> 00:42:59,580
integration analysis, and if
that is an implicit direct

743
00:42:59,580 --> 00:43:01,600
integration analysis, using the

744
00:43:01,600 --> 00:43:03,390
Newmark method, for example.

745
00:43:03,390 --> 00:43:08,830
Then, the number of operations
that we talked about in the

746
00:43:08,830 --> 00:43:13,400
Newmark method was
1/2 nm squared.

747
00:43:13,400 --> 00:43:16,390
That is the initial triangular
factorization

748
00:43:16,390 --> 00:43:18,040
that we have to perform.

749
00:43:18,040 --> 00:43:20,270
m being the half bandwidth.

750
00:43:20,270 --> 00:43:23,450
Plus 2 nm.

751
00:43:23,450 --> 00:43:26,630
n is the order, of course,
of the system.

752
00:43:26,630 --> 00:43:28,550
m is the half bandwidth,
once again.

753
00:43:28,550 --> 00:43:31,910
Times t, the number
of time steps.

754
00:43:31,910 --> 00:43:35,460
This is an initial expense that
we have to perform, using

755
00:43:35,460 --> 00:43:39,220
the direct integration Newmark
method, for example.

756
00:43:39,220 --> 00:43:41,460
The same holds for the
Wilson Theta method.

757
00:43:41,460 --> 00:43:42,900
or the Houbolt method.

758
00:43:42,900 --> 00:43:44,130
This is an initial expense.

759
00:43:44,130 --> 00:43:47,170
And then, we perform this
number of operations

760
00:43:47,170 --> 00:43:48,700
for each time step.

761
00:43:48,700 --> 00:43:52,890
Now, if we have to run 5,000
time steps, then direct

762
00:43:52,890 --> 00:43:53,750
integration would be

763
00:43:53,750 --> 00:43:57,100
extremely, extremely expensive.

764
00:43:57,100 --> 00:44:03,600
And, in fact, you might not be
able to afford it, using

765
00:44:03,600 --> 00:44:06,550
implicit direct integration.

766
00:44:06,550 --> 00:44:10,210
Well, therefore, you can see
directed by looking at this

767
00:44:10,210 --> 00:44:11,640
operation column.

768
00:44:11,640 --> 00:44:15,090
What should make the analysis
more effective is to bring

769
00:44:15,090 --> 00:44:16,640
down the bandwidths.

770
00:44:16,640 --> 00:44:19,460
And, of course, this is exactly
what we're doing in

771
00:44:19,460 --> 00:44:20,840
multiple position analysis.

772
00:44:20,840 --> 00:44:25,240
We are going from a bandwidth
of m to a bandwidth of 0.

773
00:44:25,240 --> 00:44:28,770
You are having a diagonal
matrix.

774
00:44:28,770 --> 00:44:33,660
Remember, the bandwidths we
defined to be the number of

775
00:44:33,660 --> 00:44:36,540
off diagonal elements here
that we are having.

776
00:44:36,540 --> 00:44:40,320
We did not include
the diagonal.

777
00:44:40,320 --> 00:44:41,130
That is a half bandwidth.

778
00:44:41,130 --> 00:44:43,230
If you look at the
total band here--

779
00:44:43,230 --> 00:44:48,700
that total band here was
equal to 2m plus 1.

780
00:44:48,700 --> 00:44:53,140
So if we are reducing the
bandwidths, as we do in

781
00:44:53,140 --> 00:44:56,540
multiple position analysis, then
we can see immediately

782
00:44:56,540 --> 00:45:00,060
that this number of operations
here, which is directly

783
00:45:00,060 --> 00:45:02,790
proportional to the number
of times steps goes down.

784
00:45:02,790 --> 00:45:05,940
And that is exactly the
objective in multiple position

785
00:45:05,940 --> 00:45:08,060
analysis that I talked
about earlier.

786
00:45:08,060 --> 00:45:09,900
That is the first important
objective.

787
00:45:09,900 --> 00:45:13,130
We want to reduce the bandwidths
for the direct

788
00:45:13,130 --> 00:45:15,720
integration analysis.

789
00:45:15,720 --> 00:45:18,380
The second objective, of course,
is that we do not need

790
00:45:18,380 --> 00:45:20,860
to even consider all
of the equations.

791
00:45:20,860 --> 00:45:24,410
So looking at n.

792
00:45:24,410 --> 00:45:26,130
We are reducing n, also.

793
00:45:26,130 --> 00:45:28,300
We are not looking anymore
at n equations.

794
00:45:28,300 --> 00:45:30,020
We are looking at P equations.

795
00:45:30,020 --> 00:45:33,230
And we are reducing the
bandwidths as I said earlier.

796
00:45:33,230 --> 00:45:37,400
m goes to 0.

797
00:45:37,400 --> 00:45:40,670
And n goes to P. Of
course, this is an

798
00:45:40,670 --> 00:45:42,290
approximate count here.

799
00:45:42,290 --> 00:45:46,765
So basically, you can see that
in the solution using multiple

800
00:45:46,765 --> 00:45:53,120
position analysis, the number
of operations in the time

801
00:45:53,120 --> 00:45:56,540
integration of the decoupled
equations is

802
00:45:56,540 --> 00:45:58,850
relatively small, is little.

803
00:45:58,850 --> 00:46:02,920
And, in fact, this solution
here can be obtained very,

804
00:46:02,920 --> 00:46:04,060
very effectively.

805
00:46:04,060 --> 00:46:06,520
The time integration of the
decoupled equations is very

806
00:46:06,520 --> 00:46:11,590
effective in an actual
practical analysis.

807
00:46:11,590 --> 00:46:14,880
Where the expense lies in
multiple position is really in

808
00:46:14,880 --> 00:46:17,380
the calculation of the required

809
00:46:17,380 --> 00:46:20,150
frequencies and motions.

810
00:46:20,150 --> 00:46:23,100
And, of course, this is exactly
the subject of the

811
00:46:23,100 --> 00:46:26,390
next lecture where I want to
discuss with you the effective

812
00:46:26,390 --> 00:46:29,680
solution techniques for
calculating these vectors and

813
00:46:29,680 --> 00:46:30,430
frequencies.

814
00:46:30,430 --> 00:46:35,170
Multiple position analysis is
effective, altogether only,

815
00:46:35,170 --> 00:46:39,570
when we can calculate these
eigenvalues and eigenvectors

816
00:46:39,570 --> 00:46:41,310
in an effective way.

817
00:46:41,310 --> 00:46:44,600
I should also finally, once
again, emphasize that it may

818
00:46:44,600 --> 00:46:48,310
be important to calculate the
error, epsilon pt that I

819
00:46:48,310 --> 00:46:50,570
referred to.

820
00:46:50,570 --> 00:46:56,060
Or to add a static correction
to the predicted response in

821
00:46:56,060 --> 00:47:00,580
the multiple position analysis
when P is, of course, much

822
00:47:00,580 --> 00:47:01,470
smaller than n.

823
00:47:01,470 --> 00:47:04,760
We have neglected the high
frequency response here.

824
00:47:04,760 --> 00:47:07,650
That high frequency response
is a static

825
00:47:07,650 --> 00:47:09,420
response, in general.

826
00:47:09,420 --> 00:47:13,330
And should be taken into
account via say static

827
00:47:13,330 --> 00:47:16,780
correction, as I have been
discussing with you.

828
00:47:16,780 --> 00:47:19,030
This is all I wanted to
say in this lecture.

829
00:47:19,030 --> 00:47:20,380
Thank you very much for
your attention.