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PROFESSOR: Ladies and gentlemen,
welcome to this

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00:00:23,500 --> 00:00:26,070
lecture on nonlinear finite
element analysis of solids and

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00:00:26,070 --> 00:00:27,610
structures.

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00:00:27,610 --> 00:00:29,800
In the previous lectures,
we discussed the general

12
00:00:29,800 --> 00:00:32,950
continuum mechanics formulations
that we use in

13
00:00:32,950 --> 00:00:35,200
nonlinear finite element
analysis.

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00:00:35,200 --> 00:00:37,910
And we also derived some
element matrices.

15
00:00:37,910 --> 00:00:39,770
In this lecture, I'd like
to discuss this

16
00:00:39,770 --> 00:00:41,840
you the truss element.

17
00:00:41,840 --> 00:00:45,650
The truss element is a very
interesting element and a very

18
00:00:45,650 --> 00:00:46,415
important element.

19
00:00:46,415 --> 00:00:49,890
It is important because it's
used in the analysis of many

20
00:00:49,890 --> 00:00:52,890
truss structures and also
in the analysis of cable

21
00:00:52,890 --> 00:00:53,940
structures.

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00:00:53,940 --> 00:00:57,080
It's an interesting element to
study from a theoretical point

23
00:00:57,080 --> 00:01:01,230
of view because we can use the
general continuum mechanics

24
00:01:01,230 --> 00:01:06,700
equations analytically and
derive directly the finite

25
00:01:06,700 --> 00:01:09,660
element equations and the finite
matrices corresponding

26
00:01:09,660 --> 00:01:11,650
to truss element.

27
00:01:11,650 --> 00:01:14,700
These are derived
in closed form.

28
00:01:14,700 --> 00:01:17,550
We can study these matrices
and get some insight, some

29
00:01:17,550 --> 00:01:20,850
physical insight, into what
really these individual terms

30
00:01:20,850 --> 00:01:24,010
in the continuum mechanics
equations mean.

31
00:01:24,010 --> 00:01:26,370
We want to study in this lecture
the updated Lagrangian

32
00:01:26,370 --> 00:01:27,860
formulation.

33
00:01:27,860 --> 00:01:31,010
And in the next lecture, the
total Lagrangian formulation.

34
00:01:31,010 --> 00:01:34,480
I mentioned earlier in a
lecture that these two

35
00:01:34,480 --> 00:01:38,870
formulations really reduce to
exactly the same matrices

36
00:01:38,870 --> 00:01:39,620
provided certain

37
00:01:39,620 --> 00:01:41,710
transformations routes are followed.

38
00:01:41,710 --> 00:01:45,580
And for the truss element, we
can actually very nicely

39
00:01:45,580 --> 00:01:48,890
analytically demonstrate
that all is true.

40
00:01:48,890 --> 00:01:51,280
So let us study now the updated
Lagrangian formulation

41
00:01:51,280 --> 00:01:52,730
and later on the total
Lagrangian

42
00:01:52,730 --> 00:01:55,310
formulation of the element.

43
00:01:55,310 --> 00:01:58,750
The assumptions that we are
using for the truss element

44
00:01:58,750 --> 00:02:01,440
are that the stresses are
transmitted only in the

45
00:02:01,440 --> 00:02:04,620
direction normal to
the cross section.

46
00:02:04,620 --> 00:02:07,520
The stress is constant over
the cross section.

47
00:02:07,520 --> 00:02:10,190
And the cross sectional area
remains constant doing

48
00:02:10,190 --> 00:02:12,450
deformations.

49
00:02:12,450 --> 00:02:15,820
Of course, these first two
assumptions here are those

50
00:02:15,820 --> 00:02:18,660
that we are also using
in linear analysis.

51
00:02:18,660 --> 00:02:21,740
This we add now as an
assumption, meaning that we

52
00:02:21,740 --> 00:02:24,840
really only consider small
strain problems.

53
00:02:24,840 --> 00:02:27,620
We will consider large rotation,
large displacement,

54
00:02:27,620 --> 00:02:30,740
but only small strain
problems.

55
00:02:30,740 --> 00:02:33,780
Here we have a typical
element.

56
00:02:33,780 --> 00:02:40,710
We align that to another truss
element with the x1 in its

57
00:02:40,710 --> 00:02:42,470
original configuration.

58
00:02:42,470 --> 00:02:45,420
And the elemental goes through
large deformations as you can

59
00:02:45,420 --> 00:02:48,680
see and the large
rotation, theta.

60
00:02:48,680 --> 00:02:52,950
This is node 1, originally
here, moves there.

61
00:02:52,950 --> 00:02:55,810
This node 2, originally
here, moves there.

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00:02:55,810 --> 00:02:59,050
Notice that the length of the
element is L in its original

63
00:02:59,050 --> 00:03:02,010
configuration and it
is also L in the

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00:03:02,010 --> 00:03:04,350
configuration at time t.

65
00:03:04,350 --> 00:03:09,840
We use Young's modulus, an
elastic material, in other

66
00:03:09,840 --> 00:03:12,620
words, we assume for
the element.

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00:03:12,620 --> 00:03:17,750
The cross sectional area is A.
And once again, we also use

68
00:03:17,750 --> 00:03:22,660
the fact that the element lies
in the x1, x2 space.

69
00:03:22,660 --> 00:03:26,815
The same kind of deviation that
I'm now following through

70
00:03:26,815 --> 00:03:28,940
could, of course, also be
generalized to the three

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00:03:28,940 --> 00:03:31,460
dimensional case, in other
words were you have three

72
00:03:31,460 --> 00:03:33,730
axes, x3 being included.

73
00:03:33,730 --> 00:03:36,840
But all of the relevant
information, particularly the

74
00:03:36,840 --> 00:03:40,190
physical insight that we want to
get from this deviation, is

75
00:03:40,190 --> 00:03:43,520
very well obtained by just
considering element lying in

76
00:03:43,520 --> 00:03:48,380
the x1, x2 plane and moving
in that plane.

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00:03:48,380 --> 00:03:51,960
If we now look at the
deformations of the element,

78
00:03:51,960 --> 00:03:55,720
we can see that at times 0
it was here once again.

79
00:03:55,720 --> 00:04:00,960
The displacement off node 2
is into the x1 direction

80
00:04:00,960 --> 00:04:06,490
signified by this symbol here,
time t, u, into the 1

81
00:04:06,490 --> 00:04:09,220
direction at node 2.

82
00:04:09,220 --> 00:04:13,320
Into the 2 direction,
we have u2.

83
00:04:13,320 --> 00:04:17,180
The lower 2 means 2 direction,
2 of course being upwards.

84
00:04:17,180 --> 00:04:19,610
The upper 2 means node 2.

85
00:04:19,610 --> 00:04:22,079
The t, once again,
means at time t.

86
00:04:24,850 --> 00:04:30,450
If you look at the left node
here, node 1, that one moves

87
00:04:30,450 --> 00:04:32,250
this way and that way.

88
00:04:32,250 --> 00:04:35,000
We use, of course, a
similar symbolism

89
00:04:35,000 --> 00:04:37,440
to denote this movement.

90
00:04:37,440 --> 00:04:41,990
So this is the deformation from
time 0 to times t given

91
00:04:41,990 --> 00:04:45,250
by these nodal point
displacements.

92
00:04:45,250 --> 00:04:51,530
Now from time t to time t plus
delta t, we obtain am addition

93
00:04:51,530 --> 00:04:52,410
deformation.

94
00:04:52,410 --> 00:04:55,990
And that additional deformation
is given by these

95
00:04:55,990 --> 00:05:02,870
displacements, at node 2 here
and at node 1 here.

96
00:05:02,870 --> 00:05:06,000
Notice that, in other words,
from time t to times t plus

97
00:05:06,000 --> 00:05:10,140
delta t, the truss has moved
from the red configuration to

98
00:05:10,140 --> 00:05:12,410
the blue configuration.

99
00:05:12,410 --> 00:05:15,740
Of course, note that these
displacements are measured in

100
00:05:15,740 --> 00:05:17,690
the stationary coordinate
frame.

101
00:05:17,690 --> 00:05:21,690
I made a big point in the
earlier lectures out of the

102
00:05:21,690 --> 00:05:24,980
fact that the coordinate frame
remains stationary and the

103
00:05:24,980 --> 00:05:28,760
elements move through the
coordinate frame.

104
00:05:28,760 --> 00:05:31,000
Of course, what we're interested
in doing is to

105
00:05:31,000 --> 00:05:33,720
calculate, in the finite
element solution, these

106
00:05:33,720 --> 00:05:36,720
incremental displacements.

107
00:05:36,720 --> 00:05:38,180
We assume that the
configuration

108
00:05:38,180 --> 00:05:39,780
at time t is known.

109
00:05:39,780 --> 00:05:42,760
We want to calculate the
new configuration.

110
00:05:42,760 --> 00:05:45,040
That means we want to calculate
these incremental

111
00:05:45,040 --> 00:05:46,810
displacements.

112
00:05:46,810 --> 00:05:50,680
And we achieve that by setting
up the appropriate matrices

113
00:05:50,680 --> 00:05:52,860
for the element.

114
00:05:52,860 --> 00:05:57,930
Well to develop the appropriate
matrices for the

115
00:05:57,930 --> 00:06:00,290
updated Lagrangian formulation
which I would like to discuss

116
00:06:00,290 --> 00:06:03,760
in this lecture, it is best
to introduce an auxiliary

117
00:06:03,760 --> 00:06:04,630
coordinate frame.

118
00:06:04,630 --> 00:06:08,540
And that auxiliary coordinate
frame is one that is aligned

119
00:06:08,540 --> 00:06:10,950
with one axis along
the element.

120
00:06:10,950 --> 00:06:15,780
We introduce x1 curl
and x2 curl.

121
00:06:15,780 --> 00:06:20,870
The curl always meaning body
aligned coordinate frame.

122
00:06:20,870 --> 00:06:23,990
Notice that this is now the
rotation here that the element

123
00:06:23,990 --> 00:06:25,250
has undergone.

124
00:06:25,250 --> 00:06:28,030
And of course, we really want to
get the stiffness matrix in

125
00:06:28,030 --> 00:06:31,640
the x1, x2 frame.

126
00:06:31,640 --> 00:06:34,700
x2, of course, being
perpendicular to x1, not shown

127
00:06:34,700 --> 00:06:37,500
on this viewgraph, but shown
on the earlier viewgraphs.

128
00:06:37,500 --> 00:06:41,310
We know that if we have
calculated the matrices in

129
00:06:41,310 --> 00:06:45,100
this curl coordinate frame, the
body attached coordinate

130
00:06:45,100 --> 00:06:48,270
frame, we very easily can
transform to the stationary

131
00:06:48,270 --> 00:06:49,810
coordinate frame.

132
00:06:49,810 --> 00:06:53,830
So let us now concentrate on
finding the K-matrix, the

133
00:06:53,830 --> 00:06:56,540
F-vector, corresponding
to this body

134
00:06:56,540 --> 00:06:57,790
attached coordinate frame.

135
00:07:00,560 --> 00:07:04,370
To do so, we look back to our
continuum mechanics equations

136
00:07:04,370 --> 00:07:06,910
that we developed in
earlier lectures.

137
00:07:06,910 --> 00:07:10,400
Here we have the basic continuum
mechanics equation,

138
00:07:10,400 --> 00:07:13,160
the principal of virtual work
written for the update

139
00:07:13,160 --> 00:07:16,910
Lagrangian formulation in the
curled coordinate frame, see

140
00:07:16,910 --> 00:07:19,370
that curl there.

141
00:07:19,370 --> 00:07:23,000
And this was the starting point
for the development off

142
00:07:23,000 --> 00:07:26,400
all of the equations that we're
using in the update

143
00:07:26,400 --> 00:07:27,490
Lagrangian formulation.

144
00:07:27,490 --> 00:07:32,150
The linearization resided
in this equation.

145
00:07:32,150 --> 00:07:35,930
And we talked about this
equation quite a bit.

146
00:07:35,930 --> 00:07:39,330
Except, of course, we did not
have to curls there because we

147
00:07:39,330 --> 00:07:45,110
were talking about the x1, x2,
x3 uncurled coordinate frame.

148
00:07:45,110 --> 00:07:47,500
Now we're talking about this
curled coordinate frame, so I

149
00:07:47,500 --> 00:07:49,620
simply put a curl on there.

150
00:07:49,620 --> 00:07:54,510
But otherwise, the terms
are identical.

151
00:07:54,510 --> 00:07:58,040
Because for the truss as I
state earlier the only

152
00:07:58,040 --> 00:08:02,690
non-zero stress is the stress
along the length of the truss

153
00:08:02,690 --> 00:08:05,670
which acts normally to the cross
section of the truss, we

154
00:08:05,670 --> 00:08:09,890
can simplify this general
equation to this equation.

155
00:08:09,890 --> 00:08:14,670
The only non-zero stress
is tau curl 1 1.

156
00:08:14,670 --> 00:08:19,050
The only strain that we call
stress, only small strain that

157
00:08:19,050 --> 00:08:22,710
we call stress, is e curl 1 1.

158
00:08:22,710 --> 00:08:25,070
Of course a t on the left-hand
side because it's referred to

159
00:08:25,070 --> 00:08:27,560
as the configuration
at time t.

160
00:08:27,560 --> 00:08:31,450
So this very general equation
reduces to a much simpler

161
00:08:31,450 --> 00:08:34,870
equation already
as shown here.

162
00:08:34,870 --> 00:08:39,429
What we are out now to do is to
evaluate these quantities

163
00:08:39,429 --> 00:08:43,590
here and there using the finite
element interpolation.

164
00:08:46,110 --> 00:08:50,630
First let us now look at what
are some of the terms that we

165
00:08:50,630 --> 00:08:52,470
easily can identify.

166
00:08:52,470 --> 00:08:55,470
We identify this tensile
term here to be

167
00:08:55,470 --> 00:08:57,470
simply Young's modulus.

168
00:08:57,470 --> 00:09:01,300
This stress term to be simply
the force in the truss divided

169
00:09:01,300 --> 00:09:03,010
by the cross sectional area.

170
00:09:03,010 --> 00:09:05,010
The volume of the truss
is given here.

171
00:09:05,010 --> 00:09:07,850
Notice, once again, the length
of the truss is assumed to be

172
00:09:07,850 --> 00:09:10,500
constant, therefore we only
consider small strain

173
00:09:10,500 --> 00:09:13,400
conditions, I mentioned
that earlier.

174
00:09:13,400 --> 00:09:17,060
If we now use this information
in the general equation, we

175
00:09:17,060 --> 00:09:19,220
directly arrive at this
equation here.

176
00:09:21,800 --> 00:09:24,450
You simply substitute and you
will see immediately that

177
00:09:24,450 --> 00:09:25,740
these are the terms.

178
00:09:25,740 --> 00:09:29,060
Of course, what we now have to
evaluate are these curled

179
00:09:29,060 --> 00:09:32,240
terms here.

180
00:09:32,240 --> 00:09:37,820
To proceed, we identify that
the e curl 1 1, the linear

181
00:09:37,820 --> 00:09:40,900
strain term, is simply given
by the linear strain

182
00:09:40,900 --> 00:09:44,230
displacement matrix with a curl
on it times the nodal

183
00:09:44,230 --> 00:09:46,330
point displacement vector.

184
00:09:46,330 --> 00:09:49,130
These are the nodal point
displacements.

185
00:09:49,130 --> 00:09:50,910
There are four such
displacements because we have

186
00:09:50,910 --> 00:09:54,540
two nodes and two displacements
per node.

187
00:09:54,540 --> 00:09:59,020
Notice that I have a curl here
signifying that we're talking

188
00:09:59,020 --> 00:10:02,150
about the curled coordinate
frame.

189
00:10:02,150 --> 00:10:05,930
And there's a hat on top off
that curl as you can see here.

190
00:10:05,930 --> 00:10:09,900
That hat means these are
discrete nodal point

191
00:10:09,900 --> 00:10:12,180
displacements.

192
00:10:12,180 --> 00:10:16,480
This term here is evaluated
via that term.

193
00:10:16,480 --> 00:10:21,900
Notice that we construct the
BNL in such a way that this

194
00:10:21,900 --> 00:10:25,570
right-hand side is equal
to that left-hand side.

195
00:10:25,570 --> 00:10:27,990
I mentioned that also in the
earlier lectures for the two

196
00:10:27,990 --> 00:10:29,610
dimensional and the
three dimensional

197
00:10:29,610 --> 00:10:31,560
finite element cases.

198
00:10:31,560 --> 00:10:35,950
Well the vector of nodal point
displacements, this vector

199
00:10:35,950 --> 00:10:38,910
here, is listed out here.

200
00:10:38,910 --> 00:10:41,350
Notice it contains the four
displacements that I referred

201
00:10:41,350 --> 00:10:42,260
to earlier.

202
00:10:42,260 --> 00:10:45,890
And these are the displacements,
u curl

203
00:10:45,890 --> 00:10:49,470
1 1, u curl 2 1.

204
00:10:49,470 --> 00:10:54,610
This upper 1 always denoting
nodal point, the lower 1 and 2

205
00:10:54,610 --> 00:10:56,630
meaning coordinate directions.

206
00:10:56,630 --> 00:10:58,175
Similarly for these two terms.

207
00:11:00,880 --> 00:11:05,540
To evaluate now these terms,
we recognize that the total

208
00:11:05,540 --> 00:11:08,520
strain, the total incremental
Green-Lagrange strain I should

209
00:11:08,520 --> 00:11:12,080
say, is given by this
relationship here from our

210
00:11:12,080 --> 00:11:14,990
general continuum mechanics
equation.

211
00:11:14,990 --> 00:11:17,970
The linear strain term
is given here.

212
00:11:17,970 --> 00:11:20,840
And the nonlinear strain
term is the rest.

213
00:11:20,840 --> 00:11:24,750
The rest meaning taking this
total, subtracting the linear

214
00:11:24,750 --> 00:11:27,460
one, you're left
with that one.

215
00:11:27,460 --> 00:11:29,640
And once again, these
expressions are directly

216
00:11:29,640 --> 00:11:33,490
obtained by just taking the
general continuum mechanics

217
00:11:33,490 --> 00:11:37,670
equations and varying the
indices the way you want to

218
00:11:37,670 --> 00:11:38,670
see them varied.

219
00:11:38,670 --> 00:11:43,310
1 1, for i and j, if you
would refer back.

220
00:11:43,310 --> 00:11:48,060
And k, the k that used earlier
goes over 1 and 2 in this

221
00:11:48,060 --> 00:11:49,260
particular case.

222
00:11:49,260 --> 00:11:52,050
In a three dimensional case,
of course, you would have k

223
00:11:52,050 --> 00:11:54,500
also going to 3.

224
00:11:54,500 --> 00:11:57,590
The variation on this term
resides directly in this

225
00:11:57,590 --> 00:11:59,170
expression.

226
00:11:59,170 --> 00:12:02,720
And we notice that this
expression can also be written

227
00:12:02,720 --> 00:12:05,020
in matrix form.

228
00:12:05,020 --> 00:12:08,840
One row vector times
one column vector.

229
00:12:08,840 --> 00:12:13,060
It is convenient to work now
with this matrix form because

230
00:12:13,060 --> 00:12:16,360
we want to bring it into a
form BNL transposed BNL.

231
00:12:19,010 --> 00:12:25,670
Before doing so, let us identify
one important point,

232
00:12:25,670 --> 00:12:28,510
namely that the displacement
derivatives are

233
00:12:28,510 --> 00:12:30,170
constant along the truss.

234
00:12:30,170 --> 00:12:34,310
The reason for it is that we
have only two nodes and these

235
00:12:34,310 --> 00:12:37,710
two notes can only specify
a linear variation and

236
00:12:37,710 --> 00:12:40,250
displacement between the two
nodes, meaning a linear

237
00:12:40,250 --> 00:12:44,380
variation along the truss and
the displacement derivatives

238
00:12:44,380 --> 00:12:46,050
are therefore constant.

239
00:12:46,050 --> 00:12:49,700
For example, this displacement
derivative is simply obtained

240
00:12:49,700 --> 00:12:52,730
by taking the difference in the
nodal point displacements,

241
00:12:52,730 --> 00:12:55,630
of course, in direction 1
because we're talking about 1

242
00:12:55,630 --> 00:12:59,220
here, over the length L.

243
00:12:59,220 --> 00:13:06,930
So this is directly obtained
as u curl 1 comma 1.

244
00:13:06,930 --> 00:13:10,970
Similarly, we also evaluate u
curl 2, 1 and you see the

245
00:13:10,970 --> 00:13:15,010
lower part here gives
us that expression.

246
00:13:15,010 --> 00:13:17,920
Of course, the upper row here
is nothing else than the

247
00:13:17,920 --> 00:13:20,610
rewriting of that equation.

248
00:13:20,610 --> 00:13:25,580
If you rewrite this equation in
matrix form, you directly

249
00:13:25,580 --> 00:13:28,650
obtain this relationship.

250
00:13:28,650 --> 00:13:32,120
Or another check would be simply
takes this vector,

251
00:13:32,120 --> 00:13:37,240
multiply this matrix by that
vector, and the upper part of

252
00:13:37,240 --> 00:13:40,730
that multiplication will
directly be this result.

253
00:13:43,580 --> 00:13:46,640
If we now use this information,
we directly

254
00:13:46,640 --> 00:13:52,130
extract the BL matrix, which of
course links up the linear

255
00:13:52,130 --> 00:13:56,240
strain increment with the nodal
point displacements.

256
00:13:56,240 --> 00:13:58,450
Here there is a small error.

257
00:13:58,450 --> 00:13:59,490
Let me just point it out.

258
00:13:59,490 --> 00:14:06,200
This bracket, of course, should
go to there because the

259
00:14:06,200 --> 00:14:09,530
B matrix does not contain
the nodal point

260
00:14:09,530 --> 00:14:10,680
displacement vector.

261
00:14:10,680 --> 00:14:13,470
It only goes up to there.

262
00:14:13,470 --> 00:14:17,860
Because the e curl 1 1 is equal
to B matrix times the

263
00:14:17,860 --> 00:14:20,250
nodal point displacement
vector.

264
00:14:20,250 --> 00:14:22,210
Similarly we can write
now the variation in

265
00:14:22,210 --> 00:14:24,760
eta 1 1 in this form.

266
00:14:24,760 --> 00:14:29,780
Notice that this part here
captures this amount and this

267
00:14:29,780 --> 00:14:32,910
part here captures
that amount.

268
00:14:32,910 --> 00:14:35,500
This one, of course, we had
already derived earlier.

269
00:14:35,500 --> 00:14:38,710
We just plug it in now.

270
00:14:38,710 --> 00:14:40,810
With this information
we can not directly

271
00:14:40,810 --> 00:14:42,890
develop the k matrices.

272
00:14:42,890 --> 00:14:46,850
We notice that the linear strain
stiffness matrix is

273
00:14:46,850 --> 00:14:49,140
obtained from this expression.

274
00:14:49,140 --> 00:14:50,530
And here you have it.

275
00:14:50,530 --> 00:14:54,320
You simply substitute for these
terms and obtain this

276
00:14:54,320 --> 00:14:56,890
matrix, much in the
same way as we are

277
00:14:56,890 --> 00:14:58,140
dealing in linear analysis.

278
00:15:04,040 --> 00:15:06,360
The nonlinear strain stiffness
matrix, that's the one we want

279
00:15:06,360 --> 00:15:10,870
to look at next, is obtained
from this relationship.

280
00:15:10,870 --> 00:15:14,540
And we simply now substitute
what we have derived for this

281
00:15:14,540 --> 00:15:19,180
term and directly obtain
this expression here.

282
00:15:19,180 --> 00:15:23,020
And what's under this blue
bracket, of course, is our

283
00:15:23,020 --> 00:15:24,570
nonlinear strain stiffness
matrix.

284
00:15:28,000 --> 00:15:32,770
Finally, the force vector is
obtained from this expression.

285
00:15:32,770 --> 00:15:36,960
And you, once again, simply
plug in and obtain this

286
00:15:36,960 --> 00:15:40,210
expression here where what's
under the blue

287
00:15:40,210 --> 00:15:42,480
is the force vector.

288
00:15:42,480 --> 00:15:46,430
The beauty is that we have
started with a fairly

289
00:15:46,430 --> 00:15:50,170
complicated continuum
mechanics equation.

290
00:15:50,170 --> 00:15:54,430
We have specialized it to the
truss element, recognizing

291
00:15:54,430 --> 00:15:57,460
that only certain terms really
need to be included or are

292
00:15:57,460 --> 00:15:59,730
included in this particular
case.

293
00:15:59,730 --> 00:16:03,280
We have evaluated them in a
fairly simple, straightforward

294
00:16:03,280 --> 00:16:07,980
manner, plugged in, and obtained
now matrices that can

295
00:16:07,980 --> 00:16:11,310
of course be analytically
evaluated the way that they

296
00:16:11,310 --> 00:16:12,430
have it done.

297
00:16:12,430 --> 00:16:16,050
And now we can in fact even
think about these matrices and

298
00:16:16,050 --> 00:16:18,790
try to physically interpret
their meaning.

299
00:16:18,790 --> 00:16:21,130
We will do that just now.

300
00:16:21,130 --> 00:16:24,990
However, before getting into
that, we want to get back to

301
00:16:24,990 --> 00:16:28,150
one point, namely that we have
sued so far a curled

302
00:16:28,150 --> 00:16:29,950
coordinate system.

303
00:16:29,950 --> 00:16:32,790
And as we mentioned earlier, we
will have to transform from

304
00:16:32,790 --> 00:16:35,780
that curled coordinate system
to the stationary global

305
00:16:35,780 --> 00:16:37,090
coordinate system.

306
00:16:37,090 --> 00:16:39,660
That is achieved by this
transformation.

307
00:16:39,660 --> 00:16:43,140
Notice that the curled
displacements are nothing else

308
00:16:43,140 --> 00:16:46,860
than transformation matrix
times the uncurled

309
00:16:46,860 --> 00:16:47,790
displacement.

310
00:16:47,790 --> 00:16:49,700
This transformation matrix
is very well

311
00:16:49,700 --> 00:16:52,550
known from linear analysis.

312
00:16:52,550 --> 00:16:54,650
It simply transforms
displacement from one

313
00:16:54,650 --> 00:16:57,750
coordinate system
into another.

314
00:16:57,750 --> 00:17:02,400
This is the one that holds
anywhere along the element for

315
00:17:02,400 --> 00:17:04,230
the continuous displacement.

316
00:17:04,230 --> 00:17:06,170
But it also holds
at the nodes.

317
00:17:06,170 --> 00:17:09,750
So if we apply that relationship
at the nodes, be

318
00:17:09,750 --> 00:17:13,000
directly obtain this
transformation here.

319
00:17:13,000 --> 00:17:16,619
The curled displacements at the
nodes are given in terms

320
00:17:16,619 --> 00:17:20,329
of the transformation matrix
times the uncurled

321
00:17:20,329 --> 00:17:22,010
displacement at the nodes.

322
00:17:22,010 --> 00:17:27,730
Of course, we want to get the
matrix expressions related to

323
00:17:27,730 --> 00:17:32,070
this uncurled displacement
vector.

324
00:17:32,070 --> 00:17:35,880
Well having obtained this
relationship, we go back to

325
00:17:35,880 --> 00:17:39,740
the basic equations, recognizing
that this part

326
00:17:39,740 --> 00:17:45,310
here is coming from the linear
strain stiffness matrix.

327
00:17:45,310 --> 00:17:50,500
We substitute for u curl,
the uncurled vectors

328
00:17:50,500 --> 00:17:51,780
with the T's in front.

329
00:17:51,780 --> 00:17:54,800
There is a transpose
appearing now here.

330
00:17:54,800 --> 00:17:57,320
This capital transpose has to
be there because there's a

331
00:17:57,320 --> 00:17:59,090
transpose there.

332
00:17:59,090 --> 00:18:04,700
And what's in here underlined
in red is the linear strain

333
00:18:04,700 --> 00:18:07,980
stiffness matrix of the element
corresponding to the

334
00:18:07,980 --> 00:18:11,330
stationery global coordinate
system.

335
00:18:11,330 --> 00:18:14,270
We proceed much the same way
for the nonlinear strain

336
00:18:14,270 --> 00:18:15,660
stiffness matrix.

337
00:18:15,660 --> 00:18:19,810
And what's in here now is what
we want to get, the nonlinear

338
00:18:19,810 --> 00:18:22,530
strain stiffness matrix
corresponding to the global

339
00:18:22,530 --> 00:18:24,520
stationary coordinate frame.

340
00:18:24,520 --> 00:18:27,680
Same thing holds of course
for the F-vector.

341
00:18:27,680 --> 00:18:31,750
And here we have the F-vector
now in the uncurled global

342
00:18:31,750 --> 00:18:33,530
stationary coordinate frame.

343
00:18:33,530 --> 00:18:40,050
So it's really this part, that
part, and this part that we're

344
00:18:40,050 --> 00:18:43,570
using in our finite element
analysis when we assemble

345
00:18:43,570 --> 00:18:47,330
truss elements into a global
assemblage of truss elements.

346
00:18:50,500 --> 00:18:53,590
Performing the indicated matrix
multiplications gives

347
00:18:53,590 --> 00:18:55,350
this expression here.

348
00:18:55,350 --> 00:18:59,140
In fact, this is the same
stiffness matrix that we use

349
00:18:59,140 --> 00:19:04,670
in linear analysis when we
have an element that is

350
00:19:04,670 --> 00:19:09,280
oriented at an angle theta
to the global x-axis, the

351
00:19:09,280 --> 00:19:11,840
horizontal x-axis.

352
00:19:11,840 --> 00:19:16,030
So you might have very well
seen this matrix before.

353
00:19:16,030 --> 00:19:21,060
The nonlinear strain stiffness
matrix looks this way.

354
00:19:21,060 --> 00:19:26,120
Notice there is a tP over
L and then these terms.

355
00:19:26,120 --> 00:19:28,050
Of course, it's also
symmetric.

356
00:19:28,050 --> 00:19:30,710
And this is the matrix
corresponding to the global

357
00:19:30,710 --> 00:19:32,550
coordinate frame now.

358
00:19:32,550 --> 00:19:36,040
We notice immediately that this
matrix is in fact the

359
00:19:36,040 --> 00:19:39,240
same matrix that we already had
evaluated in the curved

360
00:19:39,240 --> 00:19:39,960
coordinate frame.

361
00:19:39,960 --> 00:19:42,550
We will get back to
that just now.

362
00:19:42,550 --> 00:19:46,610
The F-vector comes
out to be this.

363
00:19:46,610 --> 00:19:50,310
Let's now look at these
terms physically.

364
00:19:50,310 --> 00:19:53,730
Here we have a single truss
element pinned at the

365
00:19:53,730 --> 00:19:55,490
left-hand side.

366
00:19:55,490 --> 00:20:00,270
And a load R is applied to that
truss at the other end.

367
00:20:00,270 --> 00:20:04,490
Of course, this direction of
the load must be along the

368
00:20:04,490 --> 00:20:06,030
truss element.

369
00:20:06,030 --> 00:20:09,750
Or rather the truss element
aligns its direction so that

370
00:20:09,750 --> 00:20:13,160
it balances this load
that is applied.

371
00:20:13,160 --> 00:20:15,760
The internal force is tP.

372
00:20:15,760 --> 00:20:18,000
tP balances tR.

373
00:20:18,000 --> 00:20:21,080
And tP, of course, acts along
the truss element.

374
00:20:21,080 --> 00:20:24,500
Now if you look at this node
here, node 2, we notice

375
00:20:24,500 --> 00:20:28,600
immediately that this tR vector
can be decomposed.

376
00:20:28,600 --> 00:20:33,420
Or this tR can be decomposed
into tR cosine theta along

377
00:20:33,420 --> 00:20:37,240
this axis and tR sin theta
along that axis.

378
00:20:37,240 --> 00:20:42,240
P is acting here and P can also
be decomposed as shown.

379
00:20:42,240 --> 00:20:45,920
Now if we look here, we find
therefore that the tR vector

380
00:20:45,920 --> 00:20:49,200
corresponding to the global
coordinate system contains

381
00:20:49,200 --> 00:20:53,590
these two entries shown here.

382
00:20:53,590 --> 00:20:58,450
And the tF from our earlier
viewgraph at node 2 was indeed

383
00:20:58,450 --> 00:21:00,720
simply this part.

384
00:21:00,720 --> 00:21:06,750
We notice therefore that tR is
equal to tF as show here or tR

385
00:21:06,750 --> 00:21:08,640
minus ttF equal to 0.

386
00:21:08,640 --> 00:21:12,000
And that is, of course, what
has to hold when we satisfy

387
00:21:12,000 --> 00:21:15,440
equilibrium at that node.

388
00:21:15,440 --> 00:21:16,730
So we have a nice

389
00:21:16,730 --> 00:21:19,340
interpretation of this F-vector.

390
00:21:19,340 --> 00:21:22,610
And in fact, we can see that our
arithmetic of getting that

391
00:21:22,610 --> 00:21:24,580
F-vector has been correct.

392
00:21:27,470 --> 00:21:31,190
Let's look now at the KNL
matrix, the nonlinear strain

393
00:21:31,190 --> 00:21:34,450
stiffness matrix.

394
00:21:34,450 --> 00:21:40,430
We already pointed out earlier
that the KNL in the uncurled

395
00:21:40,430 --> 00:21:45,420
frame was equal to the KNL
in the curled frame.

396
00:21:45,420 --> 00:21:48,560
And we can ask ourself, of
course, why is that so?

397
00:21:48,560 --> 00:21:50,790
Well the reason is
the following.

398
00:21:50,790 --> 00:21:56,060
Let's look again at this truss
element pinned here and we

399
00:21:56,060 --> 00:22:02,330
look at node 2, then we see that
this vector here, which

400
00:22:02,330 --> 00:22:08,150
is by Pythagoras the component
or the resultant vector of

401
00:22:08,150 --> 00:22:14,240
these two displacements,
has this length.

402
00:22:14,240 --> 00:22:17,720
Now this length can be evaluated
as shown here.

403
00:22:20,530 --> 00:22:23,730
Because the derivatives are
constants, we discussed that

404
00:22:23,730 --> 00:22:29,020
earlier, we can write this
derivative squared plus that

405
00:22:29,020 --> 00:22:32,790
derivative squared, square root
out of it, times L, being

406
00:22:32,790 --> 00:22:34,830
equal to that.

407
00:22:34,830 --> 00:22:37,890
But then we recognize that
what we're seeing here is

408
00:22:37,890 --> 00:22:42,010
nothing else than that term
there, in other words, our

409
00:22:42,010 --> 00:22:45,090
nonlinear strain increment.

410
00:22:45,090 --> 00:22:50,740
Which in this particular case
we can see, for the given

411
00:22:50,740 --> 00:22:54,820
displacement is constant and
independent of the coordinate

412
00:22:54,820 --> 00:22:56,130
frame used.

413
00:22:56,130 --> 00:23:00,500
And that is really the reason
why the KNL matrix is constant

414
00:23:00,500 --> 00:23:05,180
for any coordinate frame
that we are using.

415
00:23:05,180 --> 00:23:08,610
Let us now try to understand
what the elements in the KNL

416
00:23:08,610 --> 00:23:11,900
matrix physically mean.

417
00:23:11,900 --> 00:23:17,140
They, in fact, give the change
in force, the required change

418
00:23:17,140 --> 00:23:20,360
in the externally applied
force, when

419
00:23:20,360 --> 00:23:22,380
the element is rotated.

420
00:23:22,380 --> 00:23:26,940
Here we have an element at time
t, once again fixed at

421
00:23:26,940 --> 00:23:32,110
the left end, and subjected
to a force tR.

422
00:23:32,110 --> 00:23:34,590
Of course, this force tR
is balanced by the

423
00:23:34,590 --> 00:23:37,440
internal force tP.

424
00:23:37,440 --> 00:23:40,640
Let us now impose a displacement
such that the

425
00:23:40,640 --> 00:23:45,230
element rotates about
this point.

426
00:23:45,230 --> 00:23:49,710
We reach the configuration
at time t plus delta t by

427
00:23:49,710 --> 00:23:52,340
imposing this displacement
here, which we

428
00:23:52,340 --> 00:23:55,340
denote u curl 2 2.

429
00:23:55,340 --> 00:23:58,690
Now in this configuration,
a new force has

430
00:23:58,690 --> 00:24:02,290
to act on this element.

431
00:24:02,290 --> 00:24:05,810
And this new force, denoted
as t plus delta

432
00:24:05,810 --> 00:24:11,450
tR, is shown here.

433
00:24:11,450 --> 00:24:14,740
Here we have the force
t plus delta tR.

434
00:24:14,740 --> 00:24:20,600
Notice it is aligned with the
red element at time t plus

435
00:24:20,600 --> 00:24:23,330
delta t now.

436
00:24:23,330 --> 00:24:27,440
The change in the force from
configuration t to t plus

437
00:24:27,440 --> 00:24:31,830
delta t is given
by this vector.

438
00:24:31,830 --> 00:24:36,170
And the element in that vector
can be calculated by taking

439
00:24:36,170 --> 00:24:40,430
moment equilibrium
about this point.

440
00:24:40,430 --> 00:24:43,900
We do just that right here.

441
00:24:43,900 --> 00:24:46,980
This is the equation of moment
equilibrium about the fixed

442
00:24:46,980 --> 00:24:48,540
point of the truss.

443
00:24:48,540 --> 00:24:53,870
Notice that, of course, this
displacement is small.

444
00:24:53,870 --> 00:24:58,090
And from this, we directly
obtain that delta R is given

445
00:24:58,090 --> 00:25:00,340
as shown here on the
right-hand side.

446
00:25:00,340 --> 00:25:04,200
And this is actually the entry
4, 4 of the KNL matrix.

447
00:25:06,960 --> 00:25:10,090
Now the same information is
also expressed here on the

448
00:25:10,090 --> 00:25:15,010
right-hand side where we have
written the delta R as a

449
00:25:15,010 --> 00:25:20,990
matrix product of KNL times u
hat being equal to t plus

450
00:25:20,990 --> 00:25:22,240
delta tR minus tR.

451
00:25:26,490 --> 00:25:29,770
So this completes our discussion
of the finite

452
00:25:29,770 --> 00:25:33,400
element matrices of the truss
element corresponding to the

453
00:25:33,400 --> 00:25:35,790
updated Lagrangian
formulation.

454
00:25:35,790 --> 00:25:38,330
In the next lecture, we will
actually discuss the total

455
00:25:38,330 --> 00:25:40,420
Lagrangian formulation
of the truss element.

456
00:25:40,420 --> 00:25:43,850
And we will find that
identically the same matrices

457
00:25:43,850 --> 00:25:47,350
are obtained as in the updated
Lagrangian formulation.

458
00:25:47,350 --> 00:25:49,610
That will be a very interesting
theoretical point,

459
00:25:49,610 --> 00:25:51,980
as well as practical point.

460
00:25:51,980 --> 00:25:55,930
Let us now look at an example
of using the truss element.

461
00:25:55,930 --> 00:25:59,410
And the example is a
simple one, and yet

462
00:25:59,410 --> 00:26:01,220
quite practical one.

463
00:26:01,220 --> 00:26:03,470
We will analyze a prestressed
cable.

464
00:26:03,470 --> 00:26:07,110
The cable has a length 2L.

465
00:26:07,110 --> 00:26:11,460
At the midpoint, a load
of 2 tR is supplied.

466
00:26:11,460 --> 00:26:14,010
The Young's modulus of the
cable is E, the cross

467
00:26:14,010 --> 00:26:17,800
sectional area of
the cable is A.

468
00:26:17,800 --> 00:26:20,920
We can model one half of the
cable because of symmetry

469
00:26:20,920 --> 00:26:23,290
conditions as shown down here.

470
00:26:23,290 --> 00:26:25,930
Notice that, of course, we
assume the transverse

471
00:26:25,930 --> 00:26:29,470
displacement to be small, the
angle therefore to be small,

472
00:26:29,470 --> 00:26:32,690
because we assume that the
length of the cable element

473
00:26:32,690 --> 00:26:35,230
remains the same.

474
00:26:35,230 --> 00:26:38,450
In other words, it does not
change from time zero to time

475
00:26:38,450 --> 00:26:41,520
t to time t plus delta
t and so on.

476
00:26:41,520 --> 00:26:46,140
Using the U.L. Formation, we
obtain directly from the

477
00:26:46,140 --> 00:26:49,080
general matrices that we
discussed us now, this

478
00:26:49,080 --> 00:26:51,610
equation here.

479
00:26:51,610 --> 00:26:55,280
Which gives the tangent
stiffness matrix times the

480
00:26:55,280 --> 00:26:59,090
displacement increment equal to
the externally applied load

481
00:26:59,090 --> 00:27:03,650
minus the nodal point force
corresponding to the internal

482
00:27:03,650 --> 00:27:07,300
element stress, the internal
element force tP, of course,

483
00:27:07,300 --> 00:27:09,420
in this case.

484
00:27:09,420 --> 00:27:13,250
Notice that this is the
equilibrium equation that

485
00:27:13,250 --> 00:27:17,590
carries us from time t to
time t plus delta t.

486
00:27:17,590 --> 00:27:23,190
We would iterate on this, of
course, introducing an

487
00:27:23,190 --> 00:27:26,490
iteration counter here on
the right-hand side.

488
00:27:26,490 --> 00:27:29,670
We have discussed all that in
earlier lectures to obtain an

489
00:27:29,670 --> 00:27:34,100
accurate solution for time t
plus delta t, but these are

490
00:27:34,100 --> 00:27:36,490
some details that we don't
need to look at now.

491
00:27:36,490 --> 00:27:40,400
For the moment, let's just look
at this equation here, in

492
00:27:40,400 --> 00:27:46,370
which we imply basically a
simple step by step forward

493
00:27:46,370 --> 00:27:50,090
incrementation of load
without iteration.

494
00:27:50,090 --> 00:27:53,090
In practice once again, we would
actually iterate to make

495
00:27:53,090 --> 00:27:56,010
sure that we are satisfying
equilibrium at the end

496
00:27:56,010 --> 00:27:58,150
of each time step.

497
00:27:58,150 --> 00:28:01,490
Of particular interest is the
configuration at time 0.

498
00:28:01,490 --> 00:28:04,770
And there we recognized that
the linear strain stiffness

499
00:28:04,770 --> 00:28:07,060
matrix up here is 0.

500
00:28:07,060 --> 00:28:10,520
And that's all we have in terms
of stiffness is the

501
00:28:10,520 --> 00:28:12,860
nonlinear strain stiffness.

502
00:28:12,860 --> 00:28:17,430
And that is expressed in
this equation here.

503
00:28:17,430 --> 00:28:23,230
Notice that therefore if P0 is
equal to 0, in other words,

504
00:28:23,230 --> 00:28:26,060
there's no prestress in the
cable, then initially it has

505
00:28:26,060 --> 00:28:27,310
no stiffness.

506
00:28:29,710 --> 00:28:34,880
Of course, this element here,
let's look at it here at time

507
00:28:34,880 --> 00:28:39,840
t, this element here will
increase ad the deformations

508
00:28:39,840 --> 00:28:43,400
increase because tP
will increase.

509
00:28:43,400 --> 00:28:46,290
And this element here will
also increase as the

510
00:28:46,290 --> 00:28:47,540
deformations increase.

511
00:28:51,010 --> 00:28:54,770
And in fact, this is expressed
once more here.

512
00:28:54,770 --> 00:28:57,930
This total matrix will in
fact be increasing.

513
00:28:57,930 --> 00:29:00,900
We say the cable stiffens
and the load is applied.

514
00:29:04,440 --> 00:29:09,030
If theta becomes very large, in
fact if as theta goes to 90

515
00:29:09,030 --> 00:29:13,540
degrees, the stiffness becomes
quite large and the stiffness

516
00:29:13,540 --> 00:29:19,000
approaches as theta goes to 90
degrees EA over L. Now this,

517
00:29:19,000 --> 00:29:22,800
of course, is rather theoretical
because we assume

518
00:29:22,800 --> 00:29:25,880
in our formulation that the
length remains constant and

519
00:29:25,880 --> 00:29:27,950
therefore that the deformations
are small, as I

520
00:29:27,950 --> 00:29:30,830
pointed out earlier.

521
00:29:30,830 --> 00:29:35,130
Let us now look at the actual
load displacement curve that

522
00:29:35,130 --> 00:29:40,100
is calculated for this
particular truss structure,

523
00:29:40,100 --> 00:29:42,520
which is really a cable
structure, we could look also

524
00:29:42,520 --> 00:29:45,620
at it as a trust structure
because it is made up of two

525
00:29:45,620 --> 00:29:47,750
truss elements.

526
00:29:47,750 --> 00:29:51,070
Here we have the deflection
plotted.

527
00:29:51,070 --> 00:29:54,060
And here we have the applied
force plotted for

528
00:29:54,060 --> 00:29:55,310
a particular case.

529
00:29:57,810 --> 00:30:03,140
Notice that L is 120 inches,
A is 1 square inch.

530
00:30:03,140 --> 00:30:05,550
Notice that the maximum
displacement we're looking at

531
00:30:05,550 --> 00:30:08,460
here is about 2 inches.

532
00:30:08,460 --> 00:30:12,720
So 2 inches over the length, 120
inches, means really small

533
00:30:12,720 --> 00:30:13,970
deformations.

534
00:30:15,420 --> 00:30:20,290
But notice the stiffening effect
shown by the increase

535
00:30:20,290 --> 00:30:24,350
in the slope of this curve.

536
00:30:24,350 --> 00:30:28,440
We can actually show the
stiffness matrix components as

537
00:30:28,440 --> 00:30:29,790
a function of the
applied load.

538
00:30:29,790 --> 00:30:32,420
And this is a very interesting
viewgraph.

539
00:30:32,420 --> 00:30:39,510
Here we see that at 0 load,
the only stiffness that is

540
00:30:39,510 --> 00:30:42,110
there is given by KNL.

541
00:30:42,110 --> 00:30:45,270
We pointed that out
already earlier.

542
00:30:45,270 --> 00:30:48,050
KL is equal to 0.

543
00:30:48,050 --> 00:30:53,290
As the load increases,
KNL increases as

544
00:30:53,290 --> 00:30:56,220
shown by the blue curve.

545
00:30:56,220 --> 00:31:02,180
As a load increases, also KL
increases as shown here.

546
00:31:02,180 --> 00:31:05,210
And the sum of these two curves,
of course, gives us

547
00:31:05,210 --> 00:31:07,430
the red curve.

548
00:31:07,430 --> 00:31:13,330
And the red curve is the total
tangent stiffness matrix.

549
00:31:13,330 --> 00:31:18,840
At any particular applied force,
we can directly read

550
00:31:18,840 --> 00:31:21,560
off the stiffness
corresponding to

551
00:31:21,560 --> 00:31:22,640
that applied force.

552
00:31:22,640 --> 00:31:26,330
Or, more naturally, we would
of course look for the

553
00:31:26,330 --> 00:31:27,620
corresponding displacement.

554
00:31:27,620 --> 00:31:30,570
In other words, at a particular
applied force, we

555
00:31:30,570 --> 00:31:32,560
have a corresponding
displacement.

556
00:31:32,560 --> 00:31:35,890
And for that corresponding
displacement, we

557
00:31:35,890 --> 00:31:38,570
would have a stiffness.

558
00:31:38,570 --> 00:31:41,910
Of course, these curves are
directly obtained from the

559
00:31:41,910 --> 00:31:45,090
expression that I just
showed you earlier.

560
00:31:45,090 --> 00:31:47,710
Well, this brings us to the
end of this lecture.

561
00:31:47,710 --> 00:31:50,470
In the next lecture then, as I
pointed out already, I'd like

562
00:31:50,470 --> 00:31:53,670
to look with you at the total
Lagrangian formulation of the

563
00:31:53,670 --> 00:31:55,140
truss element.

564
00:31:55,140 --> 00:31:59,170
And as I mentioned earlier, we
will find something very

565
00:31:59,170 --> 00:32:03,080
interesting, namely that
reducing all of these complex

566
00:32:03,080 --> 00:32:06,650
continuum mechanics equations
down to what we really need

567
00:32:06,650 --> 00:32:10,570
for the truss element, we will
find that the same matrices in

568
00:32:10,570 --> 00:32:13,790
a total Lagrangian formulation
are obtained as in the updated

569
00:32:13,790 --> 00:32:16,140
Lagrangian formulation.

570
00:32:16,140 --> 00:32:17,470
Thank you very much for
your attention.