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PROFESSOR: Ladies and gentlemen,
welcome to this

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00:00:23,630 --> 00:00:26,410
lecture on nonlinear finite
element analysis of solids and

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structures.

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00:00:27,640 --> 00:00:30,460
In this lecture, I'd like to
continue with our discussion

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00:00:30,460 --> 00:00:33,450
of inelastic material
descriptions we use in finite

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element analysis.

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00:00:35,050 --> 00:00:38,850
Particularly, I'd like to now
discuss with you creep, creep

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00:00:38,850 --> 00:00:39,990
of materials.

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00:00:39,990 --> 00:00:43,890
We considered already in the
earlier lecture that a typical

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00:00:43,890 --> 00:00:47,430
creep law used in engineering
practice very widely is the

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00:00:47,430 --> 00:00:49,990
power creep law, shown here.

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00:00:49,990 --> 00:00:54,880
Here we plot creep strains as a
function of time, and curves

20
00:00:54,880 --> 00:00:58,840
that we would typically obtain
using this creep law for

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00:00:58,840 --> 00:01:04,010
constant stress values with
time are shown here.

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00:01:04,010 --> 00:01:08,210
Notice that as, of course, the
stress value increases, which

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00:01:08,210 --> 00:01:12,050
however is constant in
time, we obtain a

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00:01:12,050 --> 00:01:14,830
larger creep value.

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00:01:14,830 --> 00:01:17,730
Creep strains increase, the
creep strains increase, in

26
00:01:17,730 --> 00:01:21,670
other words, as the stress
increases clearly shown also

27
00:01:21,670 --> 00:01:22,800
by this law.

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00:01:22,800 --> 00:01:27,720
Notice a1 and a2 of course, and
a0 are constants that are

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00:01:27,720 --> 00:01:32,110
determined from laboratory
test results.

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00:01:32,110 --> 00:01:35,340
There are a number of
other creep laws.

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00:01:35,340 --> 00:01:38,090
Two are listed here.

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00:01:38,090 --> 00:01:44,140
One here, shown, and another
one shown here.

33
00:01:44,140 --> 00:01:48,550
Notice in this one here, the
temperature, which affects

34
00:01:48,550 --> 00:01:55,630
creep strains quite heavily,
enters explicitly right here.

35
00:01:55,630 --> 00:01:58,770
In the other creep laws, the
first two ones, the one that

36
00:01:58,770 --> 00:02:02,400
you saw on the earlier view
graph, and this one, these

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00:02:02,400 --> 00:02:07,680
constants, of course, a0, a1,
a2, and so on, would depend on

38
00:02:07,680 --> 00:02:08,669
the temperature.

39
00:02:08,669 --> 00:02:11,950
And if the temperature is
constant, you would simply

40
00:02:11,950 --> 00:02:15,620
select them depending on that
temperature condition, and

41
00:02:15,620 --> 00:02:17,550
model creep strains this way.

42
00:02:17,550 --> 00:02:20,770
However, as I pointed out,
here temperature enters

43
00:02:20,770 --> 00:02:22,730
explicitly.

44
00:02:22,730 --> 00:02:25,590
We will not discuss these
creep laws further.

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00:02:25,590 --> 00:02:29,470
The numerical computations with
these creep laws are very

46
00:02:29,470 --> 00:02:33,210
similar to the numerical
computations that we perform

47
00:02:33,210 --> 00:02:34,650
when using the power
creep law.

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00:02:34,650 --> 00:02:36,950
So I will now use the
power creep law as

49
00:02:36,950 --> 00:02:40,470
an example to discuss.

50
00:02:40,470 --> 00:02:45,440
The creep strain formula, given
here once again, cannot

51
00:02:45,440 --> 00:02:49,770
directly be applied to varying
stress situations because the

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00:02:49,770 --> 00:02:55,080
stress history does not enter
directly into the formula.

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00:02:55,080 --> 00:02:58,210
Let us look at that
much closer.

54
00:02:58,210 --> 00:03:03,190
Here we have a very simplified
example where we start off

55
00:03:03,190 --> 00:03:06,590
with a certain stress level,
and then that stress level

56
00:03:06,590 --> 00:03:10,010
drops down to this
stress level.

57
00:03:10,010 --> 00:03:13,910
Notice that if we were to apply
the formula sort of

58
00:03:13,910 --> 00:03:18,300
"blindly," blindly in quotes, we
would obtain creep strains

59
00:03:18,300 --> 00:03:25,080
that vary, depending on this
stress level, as shown here.

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00:03:25,080 --> 00:03:28,900
And then we might say suddenly
the creep strain would drop

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00:03:28,900 --> 00:03:34,270
down, and we would carry on
going on this curve here.

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00:03:34,270 --> 00:03:38,120
To creep curve, creep strain
curve, that corresponds to

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00:03:38,120 --> 00:03:40,590
this stress level.

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00:03:40,590 --> 00:03:44,450
Now, this is really quite
unrealistic that suddenly when

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00:03:44,450 --> 00:03:47,970
our stresses decrease, the
creep strains decrease.

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00:03:47,970 --> 00:03:53,970
Creep strains are accumulated
along this curve here and they

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00:03:53,970 --> 00:03:57,530
should certainly be staying
right here.

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00:03:57,530 --> 00:04:01,690
And the ensuing creep strain
should, of course, now depend

69
00:04:01,690 --> 00:04:06,070
on the curve in which sigma 1
enter, but we should not see

70
00:04:06,070 --> 00:04:08,270
the sudden decrease
in creep strain.

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00:04:08,270 --> 00:04:12,960
And we need to introduce the
concept of strain hardening,

72
00:04:12,960 --> 00:04:15,500
the assumption of strain
hardening.

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00:04:15,500 --> 00:04:17,300
And this says the following.

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00:04:17,300 --> 00:04:20,380
The material creep behavior
depends only on the current

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00:04:20,380 --> 00:04:21,640
stress level and the

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00:04:21,640 --> 00:04:24,830
accumulated total creep strain.

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00:04:24,830 --> 00:04:30,590
And we establish the ensuing
creep strain by solving for an

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00:04:30,590 --> 00:04:32,300
effective time.

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00:04:32,300 --> 00:04:34,590
Let's look at that closely.

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00:04:34,590 --> 00:04:38,050
Here we have that at a
particular time, the creep

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00:04:38,050 --> 00:04:41,000
strain is given via
this formula.

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00:04:41,000 --> 00:04:44,510
Now, if we know the creep strain
corresponding to a

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00:04:44,510 --> 00:04:49,830
particular time, then we of
course know the stress as

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00:04:49,830 --> 00:04:52,760
well, and we know
these constants.

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00:04:52,760 --> 00:04:57,690
Then we can solve for what is
called here an effective time.

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00:04:57,690 --> 00:05:02,350
This effective time is not the
physical time, it's a time

87
00:05:02,350 --> 00:05:05,210
that is simply used for the
numerical solution.

88
00:05:05,210 --> 00:05:09,840
So we solve for this tbar, and
then knowing this tbar we can

89
00:05:09,840 --> 00:05:14,720
proceed with the creep
calculations at a different

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00:05:14,720 --> 00:05:15,840
stress level.

91
00:05:15,840 --> 00:05:18,840
Let's look how we would
use that tbar.

92
00:05:18,840 --> 00:05:23,400
First, we would enter here.

93
00:05:23,400 --> 00:05:29,620
Now in this equation, knowing
tbar, simply substitute into

94
00:05:29,620 --> 00:05:35,030
here, in terms of the creep
strain that we are given from

95
00:05:35,030 --> 00:05:39,270
the previous history,
and we now get a new

96
00:05:39,270 --> 00:05:41,170
creep strain rate.

97
00:05:41,170 --> 00:05:43,810
This is how we do the
actual computation.

98
00:05:43,810 --> 00:05:47,050
Let's look at what this
means pictorially.

99
00:05:47,050 --> 00:05:51,470
Pictorially, we now look again
at the situation where we have

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00:05:51,470 --> 00:05:55,220
a certain stress value, and that
stress value drops down,

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00:05:55,220 --> 00:05:59,300
say, to here with time.

102
00:05:59,300 --> 00:06:04,970
As long as we are having this
stress, our creep strain

103
00:06:04,970 --> 00:06:07,780
accumulates as shown here.

104
00:06:07,780 --> 00:06:12,390
As soon as this stress drops
down, we are going over to

105
00:06:12,390 --> 00:06:14,740
this curve.

106
00:06:14,740 --> 00:06:18,520
So there's no sudden decrease
in creep strain, say down

107
00:06:18,520 --> 00:06:23,030
here, down to here and then
increase in creep strain, the

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00:06:23,030 --> 00:06:24,840
way we looked at it earlier.

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00:06:24,840 --> 00:06:30,570
Instead we continuously increase
the creep strain.

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00:06:30,570 --> 00:06:34,560
Let's look at how do we
get this curve here.

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00:06:34,560 --> 00:06:38,340
Well we talked about the
evaluation of tbar.

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00:06:38,340 --> 00:06:41,860
tbar is an effective time.

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00:06:41,860 --> 00:06:49,710
With this creep strain given at
the time that this stress

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00:06:49,710 --> 00:06:54,870
value drops down to here, with
this creep strain value given,

115
00:06:54,870 --> 00:07:01,220
we enter into the formula
corresponding to this creep

116
00:07:01,220 --> 00:07:06,910
strain accumulation and
calculate tbar.

117
00:07:06,910 --> 00:07:09,600
tbar, the effective
time, would be

118
00:07:09,600 --> 00:07:12,220
this value right here.

119
00:07:12,220 --> 00:07:19,610
We use now this formula, shown
by this black line here, to

120
00:07:19,610 --> 00:07:25,350
calculate the corresponding
increase in creep strain,

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00:07:25,350 --> 00:07:29,670
corresponding to this
level of stress.

122
00:07:29,670 --> 00:07:36,800
This means that if I draw a
little triangle in here, that

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00:07:36,800 --> 00:07:40,990
this line here, for example,
is equal to that line.

124
00:07:40,990 --> 00:07:44,420
And this slope, or this
distance here is

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00:07:44,420 --> 00:07:47,210
equal to that distance.

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00:07:47,210 --> 00:07:50,910
In other words, I can mark this
as, say, a and b, and I

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00:07:50,910 --> 00:07:53,100
have here a and b as well.

128
00:07:53,100 --> 00:07:59,360
So basically, we are translating
this curve here to

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00:07:59,360 --> 00:08:00,660
right there.

130
00:08:00,660 --> 00:08:05,190
And that is shown by
these blue arrows.

131
00:08:05,190 --> 00:08:09,060
Notice all of this is achieved
by the evaluation of the

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00:08:09,060 --> 00:08:10,740
effective time.

133
00:08:10,740 --> 00:08:14,060
The effective time corresponding
to this creep

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00:08:14,060 --> 00:08:17,480
strain, and this effective time
is applied to the creep

135
00:08:17,480 --> 00:08:23,200
strain accumulation formula
corresponding to this sigma 1

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00:08:23,200 --> 00:08:25,350
stress level.

137
00:08:25,350 --> 00:08:30,050
Well the increase in stress
is modeled similarly.

138
00:08:30,050 --> 00:08:36,159
Here we have initially stress
sigma 1, and suddenly an

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00:08:36,159 --> 00:08:39,090
increase to sigma 2 level.

140
00:08:39,090 --> 00:08:42,470
We would follow this
curve here,

141
00:08:42,470 --> 00:08:44,790
corresponding to sigma 1.

142
00:08:44,790 --> 00:08:52,600
And now the increase in stress
is captured by solving for the

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00:08:52,600 --> 00:08:54,470
effective time.

144
00:08:54,470 --> 00:08:59,570
Using this level of creep
strain, we enter into the

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00:08:59,570 --> 00:09:06,390
curve, shown here in black, to
obtain the effective time

146
00:09:06,390 --> 00:09:09,900
corresponding to this
curve here.

147
00:09:09,900 --> 00:09:12,470
This is the curve, of course,
corresponding to

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00:09:12,470 --> 00:09:14,780
this sigma 2 level.

149
00:09:14,780 --> 00:09:21,350
And we translate these points
over, as shown by the blue

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00:09:21,350 --> 00:09:26,830
arrows, to obtain this material
response curve.

151
00:09:26,830 --> 00:09:31,000
So we are now using the
effective time again to

152
00:09:31,000 --> 00:09:37,010
evaluate the ensuing creep
strain when we are going from

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00:09:37,010 --> 00:09:42,040
sigma 1 stress level to
sigma 2 stress level.

154
00:09:42,040 --> 00:09:45,530
If we have cyclic conditions,
we proceed similarly.

155
00:09:45,530 --> 00:09:50,670
Here, we have first sigma
1, and then as stress

156
00:09:50,670 --> 00:09:53,610
reverses to sigma 2.

157
00:09:53,610 --> 00:09:59,980
Notice sigma 1 brings us along
this curve, and at this

158
00:09:59,980 --> 00:10:04,310
particular time, we now
reverse the stress.

159
00:10:04,310 --> 00:10:10,640
And here we have to be careful
to recognize that this curve

160
00:10:10,640 --> 00:10:16,420
here, which I'm showing now in
green, corresponds to that

161
00:10:16,420 --> 00:10:19,360
curve there.

162
00:10:19,360 --> 00:10:28,980
In other words, as I proceed
along this curve here, I would

163
00:10:28,980 --> 00:10:31,680
proceed along that
curve there.

164
00:10:31,680 --> 00:10:37,130
The reversal in stress is picked
up by saying that the

165
00:10:37,130 --> 00:10:41,380
accumulated creep strain in one
direction corresponding to

166
00:10:41,380 --> 00:10:46,000
sigma 1 tensile stress is sort
of, say, forgotten, and you

167
00:10:46,000 --> 00:10:50,360
simply take this curve
here to be the

168
00:10:50,360 --> 00:10:54,150
reflection of that curve.

169
00:10:54,150 --> 00:10:59,050
So this is the condition or the
modeling of cyclic loading

170
00:10:59,050 --> 00:11:03,680
conditions as regards
to creep strain.

171
00:11:03,680 --> 00:11:08,350
Of course, these were now the
one dimensional situation, and

172
00:11:08,350 --> 00:11:13,200
in a multiaxial stress state,
we have to generalize these

173
00:11:13,200 --> 00:11:18,000
considerations to the multiaxial
conditions.

174
00:11:18,000 --> 00:11:23,330
Let's look at the multiaxial
creep, how we proceed there.

175
00:11:23,330 --> 00:11:26,760
Here we have the stress at time
t plus theta t is equal

176
00:11:26,760 --> 00:11:30,590
to the stress at time t plus an
integration of the stress

177
00:11:30,590 --> 00:11:35,580
strain law times the elastic
strain here, differential

178
00:11:35,580 --> 00:11:37,240
elastic strain increment--

179
00:11:37,240 --> 00:11:40,660
this is the elastic stress
strain law--

180
00:11:40,660 --> 00:11:44,320
from time t to time
t plus theta t.

181
00:11:44,320 --> 00:11:49,020
If this matrix is constant,
we can, of course, pull it

182
00:11:49,020 --> 00:11:53,080
outside the integration sign.

183
00:11:53,080 --> 00:12:01,720
The stress strain integration
is performed for the

184
00:12:01,720 --> 00:12:07,200
multiaxial stress state using
the concepts that we discussed

185
00:12:07,200 --> 00:12:11,140
for the uniaxial stress state
and generalizing them to the

186
00:12:11,140 --> 00:12:13,360
multiaxial stress state.

187
00:12:13,360 --> 00:12:18,410
We defined for this purpose an
effective stress shown here.

188
00:12:18,410 --> 00:12:23,310
These are deviatoric stresses
that we already encountered in

189
00:12:23,310 --> 00:12:25,900
the discussion of plasticity.

190
00:12:25,900 --> 00:12:32,520
An effective strain, shown here,
and we use now these two

191
00:12:32,520 --> 00:12:38,500
quantities in this material law,
the power material law.

192
00:12:38,500 --> 00:12:44,710
We only consider now the
situation of stresses

193
00:12:44,710 --> 00:12:48,320
monotonically increasing
or being constant.

194
00:12:48,320 --> 00:12:51,220
We do not go into depth
regarding cyclic loading

195
00:12:51,220 --> 00:12:57,980
conditions because we don't
have really time to do so.

196
00:12:57,980 --> 00:13:01,070
The assumption that the creep
strain rate are proportional

197
00:13:01,070 --> 00:13:04,780
to the current deviatoric
stresses is giving this

198
00:13:04,780 --> 00:13:07,890
equation, and that equation is
very similar to what we have

199
00:13:07,890 --> 00:13:11,520
seen in von Mises plasticity
where gamma

200
00:13:11,520 --> 00:13:14,890
is given down here.

201
00:13:14,890 --> 00:13:19,860
Notice that the effective creep
strain rate is given via

202
00:13:19,860 --> 00:13:24,700
formula where the effective
time goes in here.

203
00:13:24,700 --> 00:13:29,380
This is the effective time
calculated much the same way

204
00:13:29,380 --> 00:13:29,990
[UNINTELLIGIBLE PHRASE]

205
00:13:29,990 --> 00:13:34,820
in the way that I discussed
earlier for the uniaxial

206
00:13:34,820 --> 00:13:36,280
conditions.

207
00:13:36,280 --> 00:13:39,520
Except that we have to now deal
with the effective stress

208
00:13:39,520 --> 00:13:42,520
here in the effective
quantities on the

209
00:13:42,520 --> 00:13:44,030
creep strain as well.

210
00:13:44,030 --> 00:13:46,720
In other words, we calculate
this effective time just the

211
00:13:46,720 --> 00:13:49,130
same way as we did it in
uniaxial conditions, but

212
00:13:49,130 --> 00:13:52,210
always using effective stress
and effective strain

213
00:13:52,210 --> 00:13:53,980
quantities.

214
00:13:53,980 --> 00:13:56,980
Using matrix notation we can
then write that the creep

215
00:13:56,980 --> 00:13:59,670
strain increment, the
differential creep strain

216
00:13:59,670 --> 00:14:02,880
increment, is given via
this right-hand side.

217
00:14:02,880 --> 00:14:07,540
D is an operator shown here for
3D analysis that gives us

218
00:14:07,540 --> 00:14:10,480
the deviatoric stresses from
the actual stresses.

219
00:14:10,480 --> 00:14:13,140
In the analysis of creep
problems we perform a time

220
00:14:13,140 --> 00:14:16,790
integration, and this time
integration can be difficult

221
00:14:16,790 --> 00:14:20,680
due to the high exponent
on the stress.

222
00:14:20,680 --> 00:14:24,140
In fact, solution stability
arise if we, for example, use

223
00:14:24,140 --> 00:14:27,810
the Euler forward integration
with too large a time step.

224
00:14:27,810 --> 00:14:33,390
A rule of thumb is given here
where we want to allow only

225
00:14:33,390 --> 00:14:36,440
that the effective creep strain
increment is smaller,

226
00:14:36,440 --> 00:14:41,700
equal to 1/10 of the
current at time t,

227
00:14:41,700 --> 00:14:44,360
effective elastic strain.

228
00:14:44,360 --> 00:14:47,990
This is a very good rule of
thumb, however, it can be

229
00:14:47,990 --> 00:14:49,660
conservative.

230
00:14:49,660 --> 00:14:52,850
In some cases, you can really
increase this coefficient

231
00:14:52,850 --> 00:14:54,090
quite a bit.

232
00:14:54,090 --> 00:14:58,400
Alternatively, we can use
implicit integration in which

233
00:14:58,400 --> 00:15:02,390
when we use the alpha method,
we use this formula here.

234
00:15:02,390 --> 00:15:05,510
And we would use an implicit
integration typically alpha

235
00:15:05,510 --> 00:15:08,430
greater equal to 1/2.

236
00:15:08,430 --> 00:15:12,620
Alpha can, in this formula,
vary between 0 and 1.

237
00:15:12,620 --> 00:15:15,880
Of course, when alpha is equal
to 0, we simply have the Euler

238
00:15:15,880 --> 00:15:17,410
forward method.

239
00:15:17,410 --> 00:15:23,710
Let's look at how we would use
this formula when alpha is

240
00:15:23,710 --> 00:15:26,576
greater, equal to 1/2.

241
00:15:26,576 --> 00:15:31,130
We would calculate the stresses
corresponding to the

242
00:15:31,130 --> 00:15:34,070
i minus first iteration.

243
00:15:34,070 --> 00:15:38,550
And here we have to pause a
minute to remember to recall

244
00:15:38,550 --> 00:15:43,190
that this i minus first
iteration is the iteration

245
00:15:43,190 --> 00:15:46,830
that we have been talking about
in earlier lectures when

246
00:15:46,830 --> 00:15:53,230
we iterate for the equilibrium
of R equal F. Or we are

247
00:15:53,230 --> 00:15:59,690
evaluating in each iteration
k times theta U is equal to

248
00:15:59,690 --> 00:16:04,330
theta R, and the theta U carries
the i superscript, and

249
00:16:04,330 --> 00:16:07,870
the theta R carries the
i minus 1 superscript.

250
00:16:07,870 --> 00:16:11,580
Remember, we talked about the
iteration for nodal point

251
00:16:11,580 --> 00:16:16,460
equilibrium of external forces
with nodal point forces that

252
00:16:16,460 --> 00:16:19,500
are corresponding to the
internal element stresses.

253
00:16:19,500 --> 00:16:24,600
That i minus 1 up here refers
to that iteration.

254
00:16:24,600 --> 00:16:28,750
This k down here refers to the
iteration that is an addition

255
00:16:28,750 --> 00:16:32,760
necessary at each integration
point level due to the fact

256
00:16:32,760 --> 00:16:35,120
that we use an implicit
scheme.

257
00:16:35,120 --> 00:16:38,310
Well this left-hand side is
calculated by taking the

258
00:16:38,310 --> 00:16:41,830
stress at time t, which we know
already, and adding to it

259
00:16:41,830 --> 00:16:45,220
an increment in stress obtained
via the elastic

260
00:16:45,220 --> 00:16:46,790
material law here.

261
00:16:46,790 --> 00:16:50,200
The total strain increment from
time t to time t plus

262
00:16:50,200 --> 00:16:52,770
theta t, iteration i minus 1.

263
00:16:52,770 --> 00:16:56,880
This i minus 1 corresponds
to that i minus 1.

264
00:16:56,880 --> 00:16:59,090
We know this value, of course.

265
00:16:59,090 --> 00:17:04,970
Minus the creep strain
calculated based on the stress

266
00:17:04,970 --> 00:17:07,910
at the end of iteration
k minus 1.

267
00:17:07,910 --> 00:17:09,770
That one goes right in here.

268
00:17:09,770 --> 00:17:11,700
It goes in there as well.

269
00:17:11,700 --> 00:17:14,950
And therefore, these are the
creep strains corresponding to

270
00:17:14,950 --> 00:17:17,640
the end of iteration
k minus 1.

271
00:17:17,640 --> 00:17:21,589
We calculate it right inside and
obtain an updated value of

272
00:17:21,589 --> 00:17:25,240
stress corresponding
to iteration k.

273
00:17:25,240 --> 00:17:29,420
This is the iteration that has
to be performed at every one

274
00:17:29,420 --> 00:17:34,120
of the integration points in
the finite element mesh.

275
00:17:34,120 --> 00:17:38,380
Here we show nine-node element
with 3 by 3 integration.

276
00:17:38,380 --> 00:17:41,300
So is the kind of iteration that
has to be performed at

277
00:17:41,300 --> 00:17:44,510
every one of these integration
point stations.

278
00:17:44,510 --> 00:17:49,190
When alpha is greater or equal
to 1/2, we obtain a stable

279
00:17:49,190 --> 00:17:50,950
integration algorithm.

280
00:17:50,950 --> 00:17:54,710
This is a statement arrived at
from a linearized stability

281
00:17:54,710 --> 00:17:59,800
analysis, so in practice
we use frequently

282
00:17:59,800 --> 00:18:02,190
alpha equal to 1.

283
00:18:02,190 --> 00:18:05,380
Also, to accelerate this
iteration at the integration

284
00:18:05,380 --> 00:18:08,940
point level, it is frequently
effective to use some form of

285
00:18:08,940 --> 00:18:11,850
Newton-Raphson iteration.

286
00:18:11,850 --> 00:18:15,670
The choice of the time step
using the implicit integration

287
00:18:15,670 --> 00:18:21,720
scheme is now governed by
two considerations.

288
00:18:21,720 --> 00:18:25,110
The first one, of course, you
want to converge in the

289
00:18:25,110 --> 00:18:31,560
iteration, which had the
iteration counter k minus 1 to

290
00:18:31,560 --> 00:18:34,310
k you want to converge
in there.

291
00:18:34,310 --> 00:18:38,690
And of course also, even if we
do converge, we want to have

292
00:18:38,690 --> 00:18:41,260
an accurate integration.

293
00:18:41,260 --> 00:18:47,370
Our errors in the integration
should not be too large.

294
00:18:47,370 --> 00:18:49,330
Of course, we can also
use some form of

295
00:18:49,330 --> 00:18:52,775
subincrementation, the way we
have been discussing it when

296
00:18:52,775 --> 00:18:56,600
we talked in the last lecture
about plasticity.

297
00:18:56,600 --> 00:19:01,860
If we compare time steps that
are usable, that are

298
00:19:01,860 --> 00:19:06,250
realistically usable with the
implicit scheme with alpha

299
00:19:06,250 --> 00:19:11,000
greater/equal to 1/2 with the
time step that we are bound to

300
00:19:11,000 --> 00:19:13,960
use when we use alpha
equal to 0.

301
00:19:13,960 --> 00:19:18,170
Because we need to have a time
step small enough to have

302
00:19:18,170 --> 00:19:19,630
stability of the integration.

303
00:19:19,630 --> 00:19:22,720
We find that with alpha
greater/equal to 1/2, we can

304
00:19:22,720 --> 00:19:26,070
use usually, frequently,
much larger time steps.

305
00:19:26,070 --> 00:19:29,500
But not always.

306
00:19:29,500 --> 00:19:32,160
We will see later on in example
where alpha equal 0

307
00:19:32,160 --> 00:19:36,530
performs just as well as the
scheme of alpha equal to 1/2.

308
00:19:39,050 --> 00:19:43,360
If we deal with thermoplasticity
and creep, we

309
00:19:43,360 --> 00:19:48,660
have to model the plastic strain
components, and here we

310
00:19:48,660 --> 00:19:53,350
have schematically shown how
we can model those strain

311
00:19:53,350 --> 00:19:54,220
components.

312
00:19:54,220 --> 00:19:58,070
Notice for different
temperatures, for different

313
00:19:58,070 --> 00:20:02,480
temperature levels, we have
different stress strain laws.

314
00:20:02,480 --> 00:20:05,250
In each case, we have assumed
the bilinear assumption.

315
00:20:05,250 --> 00:20:08,400
Notice that the yield stress
drops down as the temperature

316
00:20:08,400 --> 00:20:10,360
increases, of course.

317
00:20:10,360 --> 00:20:15,870
And with increase in
temperature, our creep curves

318
00:20:15,870 --> 00:20:17,030
looks this way.

319
00:20:17,030 --> 00:20:20,870
Notice that our creep strain,
of course, increases at a

320
00:20:20,870 --> 00:20:25,930
given stress when the
temperature increases.

321
00:20:25,930 --> 00:20:30,650
To evaluate the stresses in
thermoplastic and creep

322
00:20:30,650 --> 00:20:35,010
analysis we use this
equation here.

323
00:20:35,010 --> 00:20:38,030
Notice elastic stress
strain law.

324
00:20:38,030 --> 00:20:41,320
Of course, now changing
with time, with

325
00:20:41,320 --> 00:20:43,380
temperature that is.

326
00:20:43,380 --> 00:20:47,750
And here we have the elastic
differential strain increment.

327
00:20:47,750 --> 00:20:50,590
Thermal strain also enter
now here as well.

328
00:20:50,590 --> 00:20:53,930
Creep strain, plastic
strain, total

329
00:20:53,930 --> 00:20:56,910
differential strain increment.

330
00:20:56,910 --> 00:21:00,270
If we use the alpha method,
this integration without

331
00:21:00,270 --> 00:21:07,660
subincrementation, or, say, a
subincrement of 1, going from

332
00:21:07,660 --> 00:21:13,040
t plus theta t in one step, we
would have that this equation

333
00:21:13,040 --> 00:21:15,570
reduces directly to
that equation.

334
00:21:15,570 --> 00:21:17,940
Notice this is here now
the stress strain law.

335
00:21:17,940 --> 00:21:22,580
At time t plus theta t,
temperature corresponding to

336
00:21:22,580 --> 00:21:25,280
that time would go in here.

337
00:21:25,280 --> 00:21:31,000
And here we have the total
strain increment from time t

338
00:21:31,000 --> 00:21:32,250
to t plus theta t.

339
00:21:35,410 --> 00:21:38,920
We subtract the plastic strain
increment over the time

340
00:21:38,920 --> 00:21:42,190
increment, the creep strain
increment over the time

341
00:21:42,190 --> 00:21:45,440
increment and thermal
strain increment.

342
00:21:45,440 --> 00:21:48,870
And these are here, the
corresponding quantities,

343
00:21:48,870 --> 00:21:52,130
total strain plastic strain,
creep strain, thermal strain

344
00:21:52,130 --> 00:21:55,790
corresponding to time t.

345
00:21:55,790 --> 00:22:03,100
Notice that ep and ec and e
thermal have to be evaluated

346
00:22:03,100 --> 00:22:08,500
in this integration, and we
evaluate them as shown here.

347
00:22:08,500 --> 00:22:11,940
Notice here, t plus alpha
theta t sigma goes in.

348
00:22:11,940 --> 00:22:16,700
And if alpha is equal to
greater/equal 1/2, typically

349
00:22:16,700 --> 00:22:20,410
we would use alpha 1/2
or alpha equals 1.

350
00:22:20,410 --> 00:22:23,430
For an implicit integration, we
need, again, to integrate

351
00:22:23,430 --> 00:22:28,000
at each integration point
to satisfy the

352
00:22:28,000 --> 00:22:30,950
stress strain equation.

353
00:22:30,950 --> 00:22:33,070
Of course, t alpha is a
coefficient of thermal

354
00:22:33,070 --> 00:22:34,800
expansion at time t.

355
00:22:34,800 --> 00:22:37,980
Please don't confuse this
value with the alpha

356
00:22:37,980 --> 00:22:42,910
integration parameter that
we talked about earlier.

357
00:22:42,910 --> 00:22:47,880
The final iterative equation, if
we use alpha greater/equal

358
00:22:47,880 --> 00:22:51,970
to 1/2 we would find
looks this way.

359
00:22:51,970 --> 00:22:54,650
Stress strain law at time
t plus theta t--

360
00:22:54,650 --> 00:22:57,210
elastic stress strain
law, I should say--

361
00:22:57,210 --> 00:23:00,940
goes in here, and on the
right-hand side, we have the

362
00:23:00,940 --> 00:23:04,990
total strain corresponding
to time t plus theta t,

363
00:23:04,990 --> 00:23:06,650
iteration i minus 1.

364
00:23:06,650 --> 00:23:10,560
End of iteration i minus 1,
plastic strain, creep strain,

365
00:23:10,560 --> 00:23:15,110
thermal strain, which are given
corresponding to time t.

366
00:23:15,110 --> 00:23:21,540
And we subtract here the plastic
strain increment, the

367
00:23:21,540 --> 00:23:26,030
creep straining increment, both
based on the stress at

368
00:23:26,030 --> 00:23:30,200
the end of iteration
k minus 1.

369
00:23:30,200 --> 00:23:33,640
And of course, the thermal
strain increment.

370
00:23:33,640 --> 00:23:37,640
So this is the equation that
we would solve at each

371
00:23:37,640 --> 00:23:41,030
integration point, at each
integration point in the

372
00:23:41,030 --> 00:23:43,690
finite element mesh
by iterating until

373
00:23:43,690 --> 00:23:44,920
convergence is reached.

374
00:23:44,920 --> 00:23:48,280
Until, in other words, this
stress here is basically equal

375
00:23:48,280 --> 00:23:49,290
to that stress.

376
00:23:49,290 --> 00:23:52,550
And as I pointed out earlier,
we really need to use some

377
00:23:52,550 --> 00:23:55,430
form of Newton-Raphson
iteration in order to

378
00:23:55,430 --> 00:24:00,620
accelerate this iteration
when theta t is not--

379
00:24:00,620 --> 00:24:03,130
unless theta t is very,
very small.

380
00:24:03,130 --> 00:24:06,540
Of course, we can also use
subincrementation.

381
00:24:06,540 --> 00:24:11,970
And then the convergence will be
increased because our time

382
00:24:11,970 --> 00:24:15,190
step over the subincrement
is smaller.

383
00:24:15,190 --> 00:24:20,080
But I should also mention, as
I did already earlier, it is

384
00:24:20,080 --> 00:24:23,290
very effective frequently to
use this new algorithm, the

385
00:24:23,290 --> 00:24:26,510
effective stress function
algorithm, to solve basically

386
00:24:26,510 --> 00:24:27,590
this equation.

387
00:24:27,590 --> 00:24:30,540
And please look at the reference
given in the study

388
00:24:30,540 --> 00:24:34,900
guide if you're interested in
reading up about that method.

389
00:24:34,900 --> 00:24:37,210
Let's look now at some
example solutions.

390
00:24:37,210 --> 00:24:39,970
And the first problem that I'd
like to discuss with you is a

391
00:24:39,970 --> 00:24:45,940
very simple problem of a bar
under a uniaxial stress sigma.

392
00:24:45,940 --> 00:24:49,715
The creep law that we want
to use is given here.

393
00:24:49,715 --> 00:24:55,980
The stress is method in
megapascal, time in hours, and

394
00:24:55,980 --> 00:25:01,670
E and nu, the elasticity
constants are given down here.

395
00:25:01,670 --> 00:25:06,190
What we want to do here is to
look at various solutions that

396
00:25:06,190 --> 00:25:09,430
have been obtained with
alpha equals 0.

397
00:25:09,430 --> 00:25:12,020
We don't use subimplementation.

398
00:25:12,020 --> 00:25:13,140
Alpha equals 1.

399
00:25:13,140 --> 00:25:15,760
The effective stress function
procedure was used here.

400
00:25:15,760 --> 00:25:20,710
In all cases, we assume an MNO
formulation, or we use an MNO

401
00:25:20,710 --> 00:25:21,910
formulation.

402
00:25:21,910 --> 00:25:25,350
And we use Full-Newton iteration
without line

403
00:25:25,350 --> 00:25:27,900
searches, with these tolerances,
and we discussed

404
00:25:27,900 --> 00:25:29,210
the meaning of these tolerances

405
00:25:29,210 --> 00:25:31,400
in an earlier lecture.

406
00:25:31,400 --> 00:25:35,300
In the response predictions for
which we used the ADINA

407
00:25:35,300 --> 00:25:38,730
program, of course, you will be
seeing the elastic strain

408
00:25:38,730 --> 00:25:41,460
plus the creep strain.

409
00:25:41,460 --> 00:25:45,180
Let's first look at the
response when we put a

410
00:25:45,180 --> 00:25:52,180
constant load of 100 MPa onto
the specimen when the creep

411
00:25:52,180 --> 00:25:57,050
material law is shown here, or
with this material creep law.

412
00:25:57,050 --> 00:26:00,550
Notice that displacements are
measured upwards, time

413
00:26:00,550 --> 00:26:02,600
measured horizontally here.

414
00:26:02,600 --> 00:26:06,300
And the time step that we used
in this particular case was 10

415
00:26:06,300 --> 00:26:09,910
hours with alpha equal to 1.

416
00:26:09,910 --> 00:26:15,880
So this is the total strain
accumulation, but measured in

417
00:26:15,880 --> 00:26:17,470
terms of displacement.

418
00:26:17,470 --> 00:26:21,250
Of course, the bar is 5 meters
long, so you could directly

419
00:26:21,250 --> 00:26:23,670
obtain the strain by simply
taking this displacement

420
00:26:23,670 --> 00:26:26,690
divided by 5.

421
00:26:26,690 --> 00:26:33,830
If we increase the stress from
100 MPa to 200 MPa, you obtain

422
00:26:33,830 --> 00:26:37,510
this response curve for the
displacement at the right end

423
00:26:37,510 --> 00:26:40,110
of the bar.

424
00:26:40,110 --> 00:26:44,080
We use, again, the alpha equals
1 parameter with the

425
00:26:44,080 --> 00:26:46,510
effective stress function
algorithm, and a

426
00:26:46,510 --> 00:26:49,520
time step of 10 hours.

427
00:26:49,520 --> 00:26:54,290
In this particular case, we had
a change from 100 MPa to

428
00:26:54,290 --> 00:27:00,490
200 MPa in the applied stress,
and that is modeled as shown

429
00:27:00,490 --> 00:27:06,690
on this view graph, constant
stress up to time 500 hours.

430
00:27:06,690 --> 00:27:12,860
And then the stress increment
taken over 10 hours.

431
00:27:12,860 --> 00:27:15,600
And then the stress is
constant at 200 MPa.

432
00:27:18,280 --> 00:27:21,980
Stress reversal from 100
MPa to minus 100

433
00:27:21,980 --> 00:27:24,695
MPa gives this response.

434
00:27:28,830 --> 00:27:32,630
Notice we are not crossing
exactly at that point because

435
00:27:32,630 --> 00:27:33,880
of elastic strains.

436
00:27:37,850 --> 00:27:42,450
A constant load of 100 MPa,
but now with a different

437
00:27:42,450 --> 00:27:50,390
exponent on the time for the
creep law gives this response.

438
00:27:50,390 --> 00:27:54,290
Theta t 10 hours, alpha
equals 1 still.

439
00:27:54,290 --> 00:28:00,250
If we increase the stress from
100 to 200 MPa with this new

440
00:28:00,250 --> 00:28:03,960
creep law, or new exponent I
should say, you get this

441
00:28:03,960 --> 00:28:09,330
response for the displacement
at the right end of the bar.

442
00:28:09,330 --> 00:28:15,920
And finally, the stress reversal
from 100 to minus 100

443
00:28:15,920 --> 00:28:18,280
MPa gives this response.

444
00:28:18,280 --> 00:28:24,210
Notice we have here, of course,
a or a displacement at

445
00:28:24,210 --> 00:28:26,910
the end of the bar due to
the elastic effects.

446
00:28:30,480 --> 00:28:34,060
Let's consider now, the use of
alpha equals 0 for the stress

447
00:28:34,060 --> 00:28:41,210
increase from 100 Mpa to 200 MPa
problem, and see how our

448
00:28:41,210 --> 00:28:45,600
alpha equals theta solution
performs when compared to

449
00:28:45,600 --> 00:28:46,500
alpha equals 1.

450
00:28:46,500 --> 00:28:49,610
You can see that with theta t
equals 10 hours, we get a

451
00:28:49,610 --> 00:28:53,890
response that is very much the
same of the two responses

452
00:28:53,890 --> 00:28:56,870
calculated with alpha equals
0, and alpha equals 1/2--

453
00:28:56,870 --> 00:28:59,710
very close to each other.

454
00:28:59,710 --> 00:29:02,780
If you use a lot of time steps,
theta t equals 50

455
00:29:02,780 --> 00:29:07,710
hours, both algorithms converge,
but the solution

456
00:29:07,710 --> 00:29:12,680
becomes less accurate with
alpha equals 0 in this

457
00:29:12,680 --> 00:29:14,890
particular problem, in this
particular problem.

458
00:29:17,530 --> 00:29:20,090
Notice our baseline solution
that we are comparing with

459
00:29:20,090 --> 00:29:24,060
here corresponds to alpha equals
1 and theta t equals

460
00:29:24,060 --> 00:29:28,690
10, which is a solid line.

461
00:29:28,690 --> 00:29:32,190
When we take and even larger
time step, theta t equals 100

462
00:29:32,190 --> 00:29:35,940
hours, we find that alpha equals
0 does not converge

463
00:29:35,940 --> 00:29:39,280
anymore at t equals 600 hours.

464
00:29:39,280 --> 00:29:42,410
Whereas the use of
alpha equals 1

465
00:29:42,410 --> 00:29:44,295
gives very good results.

466
00:29:44,295 --> 00:29:48,870
Alpha equals 1 gives these
triangles here, and they lie

467
00:29:48,870 --> 00:29:52,870
exactly on the response obtained
with alpha equals 1

468
00:29:52,870 --> 00:29:55,490
theta t equals 10 hours.

469
00:29:55,490 --> 00:29:59,330
So excellent results with alpha
equals 1, even when the

470
00:29:59,330 --> 00:30:01,710
time step is 10 times larger.

471
00:30:01,710 --> 00:30:05,655
With alpha equals 0, we could
only obtain the response up to

472
00:30:05,655 --> 00:30:08,310
this point, and then
we did not converge

473
00:30:08,310 --> 00:30:09,750
anymore in the solution.

474
00:30:13,310 --> 00:30:15,000
This was a very simple
problem, but a

475
00:30:15,000 --> 00:30:17,350
demonstrative problem.

476
00:30:17,350 --> 00:30:21,950
Let us look at another problem
now, another example, that has

477
00:30:21,950 --> 00:30:26,460
a bit more complexity, and I
think that you also will enjoy

478
00:30:26,460 --> 00:30:27,360
looking at.

479
00:30:27,360 --> 00:30:31,580
Here we have a column subjected
to a compressive

480
00:30:31,580 --> 00:30:36,090
load, R, which is not applied
exactly at the center of the

481
00:30:36,090 --> 00:30:37,755
column, but is applied
with an offset.

482
00:30:40,450 --> 00:30:44,490
We analyzed this column as
a plane stress problem.

483
00:30:44,490 --> 00:30:46,800
The material constant,
elastic material

484
00:30:46,800 --> 00:30:48,550
constants are given here.

485
00:30:48,550 --> 00:30:52,200
In elastic buckling
load, calculate it

486
00:30:52,200 --> 00:30:55,100
using analytical formulas.

487
00:30:55,100 --> 00:30:58,870
The Euler buckling load
is given here as 4,100

488
00:30:58,870 --> 00:31:01,520
kilonewton.

489
00:31:01,520 --> 00:31:04,800
The goal of the analysis is
to determine the collapse

490
00:31:04,800 --> 00:31:08,640
response using different
material assumptions.

491
00:31:08,640 --> 00:31:12,330
First of all, is using an
elastic material assumption.

492
00:31:12,330 --> 00:31:14,960
Then, assuming
elasto-plasticity, and

493
00:31:14,960 --> 00:31:18,700
finally, assuming creep
in the material.

494
00:31:18,700 --> 00:31:22,330
We use in each case the total
Lagrangian formulation.

495
00:31:22,330 --> 00:31:24,810
If you were to try to use a
material [UNINTELLIGIBLE]

496
00:31:24,810 --> 00:31:27,530
your only formulation for this
problem, of course, you would

497
00:31:27,530 --> 00:31:30,390
never predict the actual
collapse of the column.

498
00:31:30,390 --> 00:31:32,660
You have to include large
displacement,

499
00:31:32,660 --> 00:31:34,110
large rotation effect.

500
00:31:34,110 --> 00:31:37,710
All of the assumption of small
strain is quite reasonable

501
00:31:37,710 --> 00:31:40,300
even for this problem.

502
00:31:40,300 --> 00:31:44,510
The solution procedure that we
use will be the full Newton

503
00:31:44,510 --> 00:31:46,920
method without line searches.

504
00:31:46,920 --> 00:31:52,150
And these are the tolerance
values that we're using, which

505
00:31:52,150 --> 00:31:56,420
we discussed already in
an earlier lecture.

506
00:31:56,420 --> 00:32:00,920
The mesh used is a 10 eight-node
quadrilateral

507
00:32:00,920 --> 00:32:02,620
element mesh.

508
00:32:02,620 --> 00:32:05,250
Here you see the eight-node
elements, 10 of them.

509
00:32:05,250 --> 00:32:06,980
We don't show the nodes.

510
00:32:06,980 --> 00:32:11,880
But for each of these elements
we use also 3 by 3 Gauss

511
00:32:11,880 --> 00:32:12,610
integrations.

512
00:32:12,610 --> 00:32:15,870
So notice through the thickness
of the total column,

513
00:32:15,870 --> 00:32:20,600
we would have six integration
point stations.

514
00:32:20,600 --> 00:32:23,350
First we predict the
elastic response.

515
00:32:23,350 --> 00:32:26,200
And in the elastic response
using the total Lagrangian

516
00:32:26,200 --> 00:32:29,810
formulation, we enter with
this stress strain law.

517
00:32:29,810 --> 00:32:33,130
Second Piola-Kirchhoff stresses
are related to the

518
00:32:33,130 --> 00:32:37,310
greener ground strains via the
elasticity constants, and we

519
00:32:37,310 --> 00:32:39,450
discussed this stress strain
law quite a bit

520
00:32:39,450 --> 00:32:41,640
in an earlier lecture.

521
00:32:41,640 --> 00:32:44,840
The response predicted
is shown here.

522
00:32:44,840 --> 00:32:50,550
Applied force in kilonewton,
and here the lateral

523
00:32:50,550 --> 00:32:54,090
displacement at the
top of the column.

524
00:32:54,090 --> 00:32:56,430
Notice the Euler buckling
load, calculated from

525
00:32:56,430 --> 00:33:01,570
classical analysis, is shown by
the blue line here, and our

526
00:33:01,570 --> 00:33:03,370
analysis yield this response.

527
00:33:06,980 --> 00:33:12,060
If we perform the elasto-plastic
analysis with a

528
00:33:12,060 --> 00:33:14,030
strain hardening model
is equal to 0.

529
00:33:14,030 --> 00:33:17,390
In other words, we assume
perfect plasticity, a yield

530
00:33:17,390 --> 00:33:20,390
stress of 3,000 kilopascal.

531
00:33:20,390 --> 00:33:23,970
And we use it for von Mises
yield criterion.

532
00:33:23,970 --> 00:33:30,600
In this formula, we would
predict with this formula

533
00:33:30,600 --> 00:33:34,640
here, quite adequately,
the large displacement

534
00:33:34,640 --> 00:33:39,970
elastic-plastic response, and
of course, this will be for

535
00:33:39,970 --> 00:33:41,230
this column, the large

536
00:33:41,230 --> 00:33:44,180
displacement collapse response.

537
00:33:44,180 --> 00:33:48,590
Notice that CEP 0 is the
incremental elasto-plastic

538
00:33:48,590 --> 00:33:53,360
constitutive matrix calculated
the way we discussed it in the

539
00:33:53,360 --> 00:33:54,660
previous lecture.

540
00:33:54,660 --> 00:33:59,160
But instead of using the
engineering stresses, we enter

541
00:33:59,160 --> 00:34:04,280
now with the components of the
second Piola-Kirchhoff stress,

542
00:34:04,280 --> 00:34:07,130
right in there.

543
00:34:07,130 --> 00:34:11,510
The plastic buckling that is
calculated using this approach

544
00:34:11,510 --> 00:34:13,440
is shown here.

545
00:34:13,440 --> 00:34:17,130
Notice this is the
elasto-plastic response curve,

546
00:34:17,130 --> 00:34:19,830
applied force, lateral
displacement at

547
00:34:19,830 --> 00:34:21,465
the top of the column.

548
00:34:21,465 --> 00:34:24,230
The elastic response curve
is shown here.

549
00:34:27,659 --> 00:34:30,630
Finally, we want to calculate
the creep response.

550
00:34:30,630 --> 00:34:34,540
And for this analysis we have
to select a creep law.

551
00:34:34,540 --> 00:34:38,020
This is the creep law
that we selected.

552
00:34:38,020 --> 00:34:42,080
A power creep law with the
exponent up here equal to 1.

553
00:34:42,080 --> 00:34:45,139
On the time, in other words,
exponent from 1.

554
00:34:45,139 --> 00:34:48,150
Notice that we are entering
here, of course, with

555
00:34:48,150 --> 00:34:50,590
effective quantities, effective
creep strain,

556
00:34:50,590 --> 00:34:53,969
effective stresses, the way
I discussed it earlier.

557
00:34:53,969 --> 00:34:58,100
We do not include, in this
analysis, plasticity effects.

558
00:34:58,100 --> 00:35:02,430
We apply in this particular
analysis, a load, a constant

559
00:35:02,430 --> 00:35:06,870
load of 2,000 kilonewton to the
column, and simply watch

560
00:35:06,870 --> 00:35:08,560
what happens to the column.

561
00:35:08,560 --> 00:35:14,870
Due to the creep strains, the
displacement of the column

562
00:35:14,870 --> 00:35:20,810
will increase, and we want
to measure, calculate the

563
00:35:20,810 --> 00:35:23,410
increase of the column
displacement.

564
00:35:23,410 --> 00:35:26,680
And we say that the column
has collapsed.

565
00:35:26,680 --> 00:35:29,440
This, of course, is somewhat
arbitrary, but we say that it

566
00:35:29,440 --> 00:35:33,070
has collapsed when the lateral
displacement at the top of the

567
00:35:33,070 --> 00:35:36,880
column is 2 meters.

568
00:35:36,880 --> 00:35:41,690
This corresponds, by the way, to
a strain of about 2% at the

569
00:35:41,690 --> 00:35:45,230
base of the column, and we know
that our total Lagrangian

570
00:35:45,230 --> 00:35:48,360
formulation is quite applicable
up to that

571
00:35:48,360 --> 00:35:50,730
percentage range.

572
00:35:50,730 --> 00:35:59,400
We want to investigate varying
time steps, and varying alpha.

573
00:35:59,400 --> 00:36:03,690
Alpha equals 0 is the Euler
forward method, and alpha

574
00:36:03,690 --> 00:36:07,330
equals 0.5, and 1 corresponds,
of course, to an implicit

575
00:36:07,330 --> 00:36:09,900
integration scheme.

576
00:36:09,900 --> 00:36:13,880
If we do so, we obtain
the collapse times

577
00:36:13,880 --> 00:36:16,080
given in this table.

578
00:36:16,080 --> 00:36:19,380
using theta t equals 0.5.

579
00:36:19,380 --> 00:36:22,740
With alpha equals
0 we get 100.

580
00:36:22,740 --> 00:36:26,410
With alpha equals 0.5,
we get also 100.

581
00:36:26,410 --> 00:36:29,250
And with alpha equals
1, we get 98.5.

582
00:36:29,250 --> 00:36:34,730
So the collapse time predicted
with theta t equals 0.5 is

583
00:36:34,730 --> 00:36:41,220
very close using any one of
these integration schemes.

584
00:36:41,220 --> 00:36:45,940
When the time step gets larger,
the collapse time

585
00:36:45,940 --> 00:36:50,410
changes for alpha equals
0, for alpha 0.5,

586
00:36:50,410 --> 00:36:51,840
and for alpha 1.

587
00:36:51,840 --> 00:36:56,160
In fact, the difference between
these collapse times

588
00:36:56,160 --> 00:37:01,390
for theta t equals 0.5 is quite
a bit here, showing that

589
00:37:01,390 --> 00:37:08,810
this time step is a bit large
for this analysis.

590
00:37:08,810 --> 00:37:14,110
The column looks this
way pictorially.

591
00:37:14,110 --> 00:37:19,370
At time 1 hour, the creep
effects are still negligible.

592
00:37:19,370 --> 00:37:22,380
At time 50 hours, well
there we have some

593
00:37:22,380 --> 00:37:23,680
creep effects already.

594
00:37:23,680 --> 00:37:27,690
And at 100 hours, collapse
occurs of the column.

595
00:37:27,690 --> 00:37:31,330
In other words, the column
displacement being two meters

596
00:37:31,330 --> 00:37:34,200
or larger at the top.

597
00:37:34,200 --> 00:37:37,060
In this particular analysis,
we used theta t equals 0.5

598
00:37:37,060 --> 00:37:41,400
hours, and alpha equals 1/2.

599
00:37:41,400 --> 00:37:46,530
If we compare once more the
different responses predicted,

600
00:37:46,530 --> 00:37:50,460
we find that with theta t equals
0.5 hours, the lateral

601
00:37:50,460 --> 00:37:57,120
displacement as a function of
time predicted using alpha 1,

602
00:37:57,120 --> 00:38:04,130
alpha 0, alpha equals 0.5 as
shown, as given here by these

603
00:38:04,130 --> 00:38:06,000
curves, and notice these
curves are very

604
00:38:06,000 --> 00:38:08,440
close to each other.

605
00:38:08,440 --> 00:38:14,310
If we use the large time step,
theta t equals 5 hours, we can

606
00:38:14,310 --> 00:38:18,490
see here that alpha equals 0,
the alpha equals 0 curve is

607
00:38:18,490 --> 00:38:22,620
very close to the alpha
equals 0.5 curve.

608
00:38:22,620 --> 00:38:26,700
Where the alpha equals 1 curve
is quite a bit away.

609
00:38:26,700 --> 00:38:33,000
So in this particular case, we
really have a much larger area

610
00:38:33,000 --> 00:38:37,940
accumulation when we use alpha
equals 1, than we have with

611
00:38:37,940 --> 00:38:41,830
alpha equals 0.5 or
alpha equals 0.

612
00:38:41,830 --> 00:38:45,170
We conclude then for this
problem that as the time step

613
00:38:45,170 --> 00:38:50,480
is reduced, the collapse times
predicted using alpha equals

614
00:38:50,480 --> 00:38:53,940
0, alpha 1/2, and alpha
1 are very, very

615
00:38:53,940 --> 00:38:55,450
close to each other.

616
00:38:55,450 --> 00:38:59,470
In fact, for theta t equals
2.5, the difference in the

617
00:38:59,470 --> 00:39:03,110
collapse times is less than
two hours using the three

618
00:39:03,110 --> 00:39:04,780
integration schemes.

619
00:39:04,780 --> 00:39:09,690
And we also notice that for a
reasonable time step, in this

620
00:39:09,690 --> 00:39:12,890
particular problem, we certainly
did not encounter

621
00:39:12,890 --> 00:39:16,510
any solution instability,
particularly with respect to

622
00:39:16,510 --> 00:39:21,900
the alpha equals 0 integration
method where one might have

623
00:39:21,900 --> 00:39:23,910
instabilities, as I pointed
out earlier.

624
00:39:23,910 --> 00:39:26,440
For this problem, we did
not encounter any such

625
00:39:26,440 --> 00:39:28,060
instability.

626
00:39:28,060 --> 00:39:30,230
This concludes then what I
wanted to share with you in

627
00:39:30,230 --> 00:39:33,750
terms of experiences as
documented on the view graph.

628
00:39:33,750 --> 00:39:36,850
I'd like to now share some
further experiences with you

629
00:39:36,850 --> 00:39:41,350
regarding an interesting
analysis that we performed

630
00:39:41,350 --> 00:39:45,860
some time ago regarding a
heat treatment process.

631
00:39:45,860 --> 00:39:49,410
And that analysis is documented
on the slides, so

632
00:39:49,410 --> 00:39:55,390
let me walk over here and
discuss these sides with you.

633
00:39:55,390 --> 00:39:59,300
Here we have a cylinder, and
it is this cylinder that is

634
00:39:59,300 --> 00:40:04,160
subjected to heat treatment
process, namely initially the

635
00:40:04,160 --> 00:40:08,520
cylinder is at 900 degrees
Celsius, and it suddenly

636
00:40:08,520 --> 00:40:11,330
cooled down to 20
degrees Celsius.

637
00:40:11,330 --> 00:40:14,315
The objective of the analysis
was first of all, to predict

638
00:40:14,315 --> 00:40:17,950
the temperature distributions in
the cylinder as a function

639
00:40:17,950 --> 00:40:20,610
of time, as the cylinder
cools down.

640
00:40:20,610 --> 00:40:23,820
And then to predict the
corresponding stresses in the

641
00:40:23,820 --> 00:40:27,470
cylinder, and most importantly,
to predict the

642
00:40:27,470 --> 00:40:30,720
finer, the residual stresses.

643
00:40:30,720 --> 00:40:33,140
Here, once again,
the cylinder.

644
00:40:33,140 --> 00:40:35,400
Here, the finite
element model.

645
00:40:35,400 --> 00:40:39,210
Notice that we modeled one
typical section through the

646
00:40:39,210 --> 00:40:42,240
cylinder, and that section
is shown here.

647
00:40:44,870 --> 00:40:47,380
The center line of the cylinder
is here, and we

648
00:40:47,380 --> 00:40:50,490
perform an axisymmetric
analysis.

649
00:40:50,490 --> 00:40:54,020
Notice that we have a higher
density of elements at the

650
00:40:54,020 --> 00:40:58,260
outer face or near the outer
face of the cylinder than at

651
00:40:58,260 --> 00:41:00,030
the inside of the cylinder.

652
00:41:00,030 --> 00:41:02,920
So then now because we are
predicting higher temperature

653
00:41:02,920 --> 00:41:08,180
gradients, stress gradients at
this end, notice that in this

654
00:41:08,180 --> 00:41:13,160
particular analysis, we
constrain the top face to move

655
00:41:13,160 --> 00:41:16,500
just vertically up to basically

656
00:41:16,500 --> 00:41:19,200
remain a straight line.

657
00:41:19,200 --> 00:41:21,680
That was achieved using
constraint equations.

658
00:41:21,680 --> 00:41:24,760
Of course, all these
nodes along the top

659
00:41:24,760 --> 00:41:28,900
face can move over.

660
00:41:28,900 --> 00:41:33,060
The assumption of this top face
to move only vertically

661
00:41:33,060 --> 00:41:40,460
up is due to the fact that the
cylinder will reflect the fact

662
00:41:40,460 --> 00:41:45,910
that the cylinder is assumed
to be infinitely long.

663
00:41:45,910 --> 00:41:52,460
The next slide now shows the
material properties used for

664
00:41:52,460 --> 00:41:53,990
the temperature analysis.

665
00:41:53,990 --> 00:41:58,780
Here we have the heat capacity
as a function of temperature.

666
00:41:58,780 --> 00:42:02,130
And here we have the
conductivity as a function of

667
00:42:02,130 --> 00:42:03,460
temperature.

668
00:42:03,460 --> 00:42:07,370
This data was entered in the
heat transfer analysis in

669
00:42:07,370 --> 00:42:10,660
which we wanted to predict
the temperature.

670
00:42:10,660 --> 00:42:17,420
These material properties
here, Young's modulus, a

671
00:42:17,420 --> 00:42:21,530
strain hardening modulus,
Poisson's ratio are entered in

672
00:42:21,530 --> 00:42:24,400
the stress analysis.

673
00:42:24,400 --> 00:42:31,900
Notice the large variation in
Et, Poisson ratio, and E as a

674
00:42:31,900 --> 00:42:37,360
function of time over the range
from 0 degree Celsius to

675
00:42:37,360 --> 00:42:40,090
900 degrees Celsius.

676
00:42:40,090 --> 00:42:45,070
Next we see the variation
of the yield stress as

677
00:42:45,070 --> 00:42:46,580
a function of time.

678
00:42:46,580 --> 00:42:49,440
Notice very marked variation.

679
00:42:49,440 --> 00:42:54,900
And notice this here is a region
of a face change where

680
00:42:54,900 --> 00:42:57,425
we have a tremendous drop
in the use stress.

681
00:43:00,200 --> 00:43:04,080
The next slide now shows the
coefficient of thermal

682
00:43:04,080 --> 00:43:07,610
expansion as a function
of temperature.

683
00:43:07,610 --> 00:43:11,820
Notice that the coefficient is
positive in this regime and

684
00:43:11,820 --> 00:43:16,120
that regime, and is negative
here to take into account the

685
00:43:16,120 --> 00:43:19,200
face change effects.

686
00:43:19,200 --> 00:43:22,860
On the next slide now we see
the prediction of the

687
00:43:22,860 --> 00:43:25,610
temperature as calculated.

688
00:43:25,610 --> 00:43:29,160
Notice initially, the
whole cylinder is

689
00:43:29,160 --> 00:43:32,000
at 900 degrees Celsius.

690
00:43:32,000 --> 00:43:37,310
We measure here the axes along
the axis of the cylinder.

691
00:43:37,310 --> 00:43:39,460
And here, temperature.

692
00:43:39,460 --> 00:43:48,260
And as time progresses, the
temperature changes down to 20

693
00:43:48,260 --> 00:43:52,870
degrees Celsius at
300 seconds.

694
00:43:52,870 --> 00:43:55,360
Let's look a little bit
more closely here.

695
00:43:55,360 --> 00:43:59,800
Notice at point 0.05 seconds,
this here is the temperature

696
00:43:59,800 --> 00:44:01,906
distribution in the cylinder.

697
00:44:01,906 --> 00:44:05,210
At 0.5 seconds, this is the
temperature distribution in

698
00:44:05,210 --> 00:44:06,550
the cylinder.

699
00:44:06,550 --> 00:44:10,940
And at 3.5 seconds, we see this
temperature distribution.

700
00:44:10,940 --> 00:44:16,510
At 18.5 seconds, we see this
temperature distribution.

701
00:44:16,510 --> 00:44:20,640
Notice the last temperature
gradient right here in this

702
00:44:20,640 --> 00:44:24,360
area, and this is the area where
we had a denser finite

703
00:44:24,360 --> 00:44:26,820
element mesh.

704
00:44:26,820 --> 00:44:30,020
Then close to the center
line of the cylinder.

705
00:44:30,020 --> 00:44:34,700
The next slide now shows the
temperature as a function of

706
00:44:34,700 --> 00:44:41,110
time, here time, here
temperature

707
00:44:41,110 --> 00:44:43,290
at a number of locations.

708
00:44:43,290 --> 00:44:48,510
First of all, at the axes of the
cylinder, and notice there

709
00:44:48,510 --> 00:44:52,480
we have a calculated curve,
the solid curve, and a

710
00:44:52,480 --> 00:44:55,390
measured curve, the
dash curve.

711
00:44:55,390 --> 00:44:59,200
They are remarkably close
to each other.

712
00:44:59,200 --> 00:45:04,690
This is here the calculated
temperature at element 14

713
00:45:04,690 --> 00:45:06,610
within the cylinder.

714
00:45:06,610 --> 00:45:11,410
And this is here the surface
temperature , calculated

715
00:45:11,410 --> 00:45:13,520
because there we did not
have any measurement.

716
00:45:13,520 --> 00:45:17,780
The only measurement that we
had to compare with was the

717
00:45:17,780 --> 00:45:20,730
temperature at the axes
of the cylinder.

718
00:45:20,730 --> 00:45:28,720
Notice that at 300 seconds,
the axes, as well as the

719
00:45:28,720 --> 00:45:33,600
surface, basically all points
in the cylinder have reached

720
00:45:33,600 --> 00:45:36,390
20 degrees Celsius.

721
00:45:36,390 --> 00:45:40,000
The next slide now shows the
results obtained in the

722
00:45:40,000 --> 00:45:43,380
laboratory regarding the
residual stress field.

723
00:45:43,380 --> 00:45:46,050
Here we see as a functional
of the radius through the

724
00:45:46,050 --> 00:45:50,310
cylinder, the residual stresses,
circumferential

725
00:45:50,310 --> 00:45:55,430
stress, sigma phi phi,
longitudinal stress, sigma zz,

726
00:45:55,430 --> 00:45:59,890
and radius stress, sigma rr,
once again, as obtained in the

727
00:45:59,890 --> 00:46:00,720
laboratory.

728
00:46:00,720 --> 00:46:04,620
We wanted to compare our
calculated results obtained

729
00:46:04,620 --> 00:46:06,480
with the finite element
solution with

730
00:46:06,480 --> 00:46:08,290
these laboratory results.

731
00:46:08,290 --> 00:46:11,680
And the next slide shows the
results obtained in the finite

732
00:46:11,680 --> 00:46:12,640
element analysis.

733
00:46:12,640 --> 00:46:17,360
Sigma phi phi, sigma
zz, and sigma rr.

734
00:46:17,360 --> 00:46:21,270
And if you compare these results
with the results

735
00:46:21,270 --> 00:46:24,170
obtained in the laboratory,
you see a very close

736
00:46:24,170 --> 00:46:25,250
correspondence.

737
00:46:25,250 --> 00:46:28,050
In fact, an excellent
correspondence for this very

738
00:46:28,050 --> 00:46:30,670
complex problem that
is considered here.

739
00:46:30,670 --> 00:46:34,980
Notice that we are making
various assumptions on the

740
00:46:34,980 --> 00:46:36,930
material model level.

741
00:46:36,930 --> 00:46:40,240
In other words, a bilinear
material assumption, et

742
00:46:40,240 --> 00:46:44,050
cetera, et cetera, for
this analysis.

743
00:46:44,050 --> 00:46:47,660
If you are interested in the
details, please refer to the

744
00:46:47,660 --> 00:46:50,710
paper in which we are describing
this analysis in

745
00:46:50,710 --> 00:46:52,760
much more depth, and the
references given

746
00:46:52,760 --> 00:46:54,260
in the study guide.

747
00:46:54,260 --> 00:46:57,590
This then brings me to the end
of what I wanted to discuss

748
00:46:57,590 --> 00:46:59,900
with you in this lecture.

749
00:47:03,260 --> 00:47:07,650
At closing I like to really
mention to you that we have,

750
00:47:07,650 --> 00:47:11,660
in these two lectures in which
we discussed elasto-plasticity

751
00:47:11,660 --> 00:47:15,180
and creep response as modeled in
finite element analysis, we

752
00:47:15,180 --> 00:47:18,100
have in these two lectures
really covered only quite a

753
00:47:18,100 --> 00:47:21,740
bit of the material that is
worthwhile looking at, and

754
00:47:21,740 --> 00:47:23,800
studying, and getting
familiar with.

755
00:47:23,800 --> 00:47:26,850
The elasto-plasticity and creep
response of structures

756
00:47:26,850 --> 00:47:30,320
is, of course, a very large
field, and we have just taken

757
00:47:30,320 --> 00:47:35,270
two lectures to cover some of
the aspects as we use them in

758
00:47:35,270 --> 00:47:36,650
finite element analysis.

759
00:47:36,650 --> 00:47:38,985
There's a lot more we
could talk about.

760
00:47:38,985 --> 00:47:40,235
But thank you for
your attention.