1 00:00:00,040 --> 00:00:02,470 The following content is provided under a Creative 2 00:00:02,470 --> 00:00:03,880 Commons license. 3 00:00:03,880 --> 00:00:06,920 Your support will help MIT OpenCourseWare continue to 4 00:00:06,920 --> 00:00:10,570 offer high-quality educational resources for free. 5 00:00:10,570 --> 00:00:13,470 To make a donation or view additional materials from 6 00:00:13,470 --> 00:00:17,400 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,400 --> 00:00:18,650 ocw.mit.edu. 8 00:00:20,520 --> 00:00:22,440 PROFESSOR: Ladies and gentlemen, welcome to this 9 00:00:22,440 --> 00:00:25,320 lecture on non-linear finite element analysis of solids and 10 00:00:25,320 --> 00:00:26,420 structures. 11 00:00:26,420 --> 00:00:28,810 In this lecture, I'd like to continue with our discussion 12 00:00:28,810 --> 00:00:30,970 of the use of constitutive relations 13 00:00:30,970 --> 00:00:32,930 in non-linear analysis. 14 00:00:32,930 --> 00:00:36,270 In the previous two lectures, we talked about modeling 15 00:00:36,270 --> 00:00:38,250 elastic materials in non-linear 16 00:00:38,250 --> 00:00:40,100 finite element analysis. 17 00:00:40,100 --> 00:00:43,720 We talked about the use of linear stress-strain loss and 18 00:00:43,720 --> 00:00:46,720 the use of non-linear stress-strain laws as regard 19 00:00:46,720 --> 00:00:50,110 to the T.L. and U.L. Formulations. 20 00:00:50,110 --> 00:00:53,790 I'd like to now discuss with you the modeling of inelastic 21 00:00:53,790 --> 00:00:57,790 materials, in particular elastoplasticity and creep in 22 00:00:57,790 --> 00:01:00,350 finite element analysis. 23 00:01:00,350 --> 00:01:02,920 Let us proceed as follows. 24 00:01:02,920 --> 00:01:07,200 Let us first briefly discuss inelastic material behavior as 25 00:01:07,200 --> 00:01:10,910 we observe them in the laboratory test results. 26 00:01:10,910 --> 00:01:15,760 Let us then briefly discuss how we model these laboratory 27 00:01:15,760 --> 00:01:19,810 test results in one-dimension analysis. 28 00:01:19,810 --> 00:01:23,210 And then I'd like to generalize these modeling 29 00:01:23,210 --> 00:01:28,830 considerations to 2D and 3D stress situations. 30 00:01:28,830 --> 00:01:32,460 So let me now walk over here to the view graph projector to 31 00:01:32,460 --> 00:01:34,880 share the view grass with you that I've 32 00:01:34,880 --> 00:01:37,490 prepared for this lecture. 33 00:01:37,490 --> 00:01:41,000 The most distinguishing feature of an inelastic 34 00:01:41,000 --> 00:01:47,560 material is that the total stress is not uniquely related 35 00:01:47,560 --> 00:01:50,430 to the current total strain. 36 00:01:50,430 --> 00:01:54,270 of course, for an elastic material, the total stress is 37 00:01:54,270 --> 00:01:57,430 given uniquely by the total current strain. 38 00:01:57,430 --> 00:02:00,340 For the inelastic material, this is not the case. 39 00:02:00,340 --> 00:02:05,650 And to calculate the response of the material, we need the 40 00:02:05,650 --> 00:02:07,050 response history. 41 00:02:07,050 --> 00:02:11,070 We need to work with stress increments and strain 42 00:02:11,070 --> 00:02:15,870 increments throughout the total solution. 43 00:02:15,870 --> 00:02:20,630 If we look at the equation that we will be operating upon 44 00:02:20,630 --> 00:02:26,570 in the analysis, we have on the left-hand side here a 45 00:02:26,570 --> 00:02:30,640 differential stress increment given via the elastic 46 00:02:30,640 --> 00:02:35,680 constitutive tensor times the total strain increment, 47 00:02:35,680 --> 00:02:39,240 differential total strain increment, I should say, minus 48 00:02:39,240 --> 00:02:43,590 the differential inelastic strain increment. 49 00:02:43,590 --> 00:02:46,960 Now, notice that this here is really is the differential 50 00:02:46,960 --> 00:02:48,930 elastic strain increment. 51 00:02:48,930 --> 00:02:50,950 Total strain minus inelastic strain 52 00:02:50,950 --> 00:02:52,670 gives us elastic strain. 53 00:02:52,670 --> 00:02:55,410 Of course, always the differential of course. 54 00:02:55,410 --> 00:03:00,290 And this, therefore, is the basic relation that we will be 55 00:03:00,290 --> 00:03:03,070 using, that we are using, in inelastic response 56 00:03:03,070 --> 00:03:04,480 calculations. 57 00:03:04,480 --> 00:03:07,380 Notice that I did not include temperature effects here, 58 00:03:07,380 --> 00:03:11,360 which would mean I would have to subtract, in addition, the 59 00:03:11,360 --> 00:03:14,940 thermal differential strain increment. 60 00:03:14,940 --> 00:03:20,250 Later on, we will also consider such relation. 61 00:03:20,250 --> 00:03:25,550 The inelastic response may occur rapidly all slowly in 62 00:03:25,550 --> 00:03:31,170 time, depending on the type of problem you're dealing with, 63 00:03:31,170 --> 00:03:33,020 depending on the type of problem you 64 00:03:33,020 --> 00:03:35,780 want to model in nature. 65 00:03:35,780 --> 00:03:42,360 If the inelastic response occurs very rapidly, then we 66 00:03:42,360 --> 00:03:45,050 model it using plasticity rules. 67 00:03:45,050 --> 00:03:48,820 In other words, in plasticity, the model assumes that the 68 00:03:48,820 --> 00:03:53,420 inelastic strain increment occurs instantaneously. 69 00:03:53,420 --> 00:04:00,450 In creep, however, the model assumes that the inelastic 70 00:04:00,450 --> 00:04:05,480 strain occurs as a function of time, and in fact, this time 71 00:04:05,480 --> 00:04:10,750 scale here might be hours, days, or years. 72 00:04:10,750 --> 00:04:13,440 Of course, the actual response is a combination of these two 73 00:04:13,440 --> 00:04:18,540 phenomena in nature, and we therefore use plasticity and 74 00:04:18,540 --> 00:04:23,800 creep, or alternatively, we may also use a viscoplastic 75 00:04:23,800 --> 00:04:29,620 material model to model what happens in actual nature. 76 00:04:29,620 --> 00:04:36,820 In our discussion now, we want to consider materially 77 00:04:36,820 --> 00:04:39,000 non-linear only conditions. 78 00:04:39,000 --> 00:04:42,440 In other words, we only would focus on material 79 00:04:42,440 --> 00:04:43,690 non-linearities. 80 00:04:43,690 --> 00:04:46,790 We know that, of course, all the information that we are 81 00:04:46,790 --> 00:04:51,250 discussing can directly be applied in an MNO analysis, 82 00:04:51,250 --> 00:04:54,190 Materially Non-Linear Only analysis, and it can also 83 00:04:54,190 --> 00:04:58,520 directly be applied in a large displacement, large rotation, 84 00:04:58,520 --> 00:05:01,580 by small-strain analysis, by using the total Lagrangian 85 00:05:01,580 --> 00:05:02,550 formulation. 86 00:05:02,550 --> 00:05:05,400 We talked in an earlier lecture about the fact that 87 00:05:05,400 --> 00:05:08,180 using second Piola-Kirchhoff stresses and Green-Lagrange 88 00:05:08,180 --> 00:05:13,140 Strains, we can directly use the total Lagrangian 89 00:05:13,140 --> 00:05:18,000 formulation with these stress and strain measures to model a 90 00:05:18,000 --> 00:05:23,700 large displacement, large rotation response of materials 91 00:05:23,700 --> 00:05:27,980 by using the same material relationship that we use to 92 00:05:27,980 --> 00:05:32,390 employ in material non-linear only analysis, but with these 93 00:05:32,390 --> 00:05:36,870 two new stress and strain measures. 94 00:05:36,870 --> 00:05:42,760 Well, this is once more summarized here, that in other 95 00:05:42,760 --> 00:05:46,300 words, in the total Lagrangian information, we use the 96 00:05:46,300 --> 00:05:49,450 Green-Lagrange strain component instead of the 97 00:05:49,450 --> 00:05:52,530 engineering strain component, and the second Piola-Kirchhoff 98 00:05:52,530 --> 00:05:55,310 stress component instead of the engineering stress 99 00:05:55,310 --> 00:06:00,420 component, and then we directly can, with the 100 00:06:00,420 --> 00:06:03,510 material relationship that we have established for an M&O 101 00:06:03,510 --> 00:06:07,490 analysis, directly with these material relationships, we can 102 00:06:07,490 --> 00:06:10,340 model large displacement, large rotation, but small 103 00:06:10,340 --> 00:06:12,000 strain problems. 104 00:06:14,800 --> 00:06:19,760 Let us now start by considering some observations 105 00:06:19,760 --> 00:06:23,880 off the material response in laboratory tests. 106 00:06:23,880 --> 00:06:29,130 We only want to look at these observations schematically, 107 00:06:29,130 --> 00:06:33,450 and therefore, we do not give any details. 108 00:06:33,450 --> 00:06:37,470 We also note that regarding the notation that I want to 109 00:06:37,470 --> 00:06:42,000 use now, we will not anymore have the t superscripts on the 110 00:06:42,000 --> 00:06:45,520 stress and t superscripts on the strain. 111 00:06:45,520 --> 00:06:49,130 We will not have these t superscripts when we look at 112 00:06:49,130 --> 00:06:51,420 laboratory test results. 113 00:06:51,420 --> 00:06:56,410 So whenever we have a material law, a laboratory material law 114 00:06:56,410 --> 00:06:59,830 that we're looking on, or a material law that is 115 00:06:59,830 --> 00:07:03,630 identified by laboratory test results, or, in fact, a 116 00:07:03,630 --> 00:07:07,580 material law that we will enter into a finite element 117 00:07:07,580 --> 00:07:11,750 analysis, I will not have the t superscript on stress and t 118 00:07:11,750 --> 00:07:13,540 superscript on strain anymore. 119 00:07:13,540 --> 00:07:17,480 We will, however, use those superscripts again when we 120 00:07:17,480 --> 00:07:20,480 take that material law and actually apply it in finite 121 00:07:20,480 --> 00:07:22,800 element analysis in an incremental solution. 122 00:07:26,090 --> 00:07:31,450 The material behavior is obtained by performing a 123 00:07:31,450 --> 00:07:36,020 tensile test in the laboratory, schematically 124 00:07:36,020 --> 00:07:36,930 shown as here. 125 00:07:36,930 --> 00:07:40,500 We have a force p up there, a force p here. 126 00:07:40,500 --> 00:07:43,020 There is a gauge length. 127 00:07:43,020 --> 00:07:45,930 Given here, the initial length is l0. 128 00:07:45,930 --> 00:07:49,240 There's cross-sectional area A0, and of course, the 129 00:07:49,240 --> 00:07:53,690 engineering strain, notice I now leave out the t here, the 130 00:07:53,690 --> 00:07:55,830 upper superscript t. 131 00:07:55,830 --> 00:07:58,130 I leave that out, as I just explained. 132 00:07:58,130 --> 00:08:03,040 The engineering strain at any time is given as shown here. 133 00:08:03,040 --> 00:08:05,980 The stress is given, shown here. 134 00:08:05,980 --> 00:08:10,470 Where l now is the length of this gauge 135 00:08:10,470 --> 00:08:12,460 length at that time. 136 00:08:12,460 --> 00:08:17,480 And here we have, shown in red, how the specimen actually 137 00:08:17,480 --> 00:08:19,160 were [? deformed. ?] 138 00:08:19,160 --> 00:08:24,460 Notice A0 here and l right there. 139 00:08:24,460 --> 00:08:29,820 So l minus l0 is really the extension of the gauge length. 140 00:08:29,820 --> 00:08:32,610 We divide that by the original length, and we get the 141 00:08:32,610 --> 00:08:37,559 engineering strain, the small strain, measure that we are so 142 00:08:37,559 --> 00:08:41,280 used to seeing in linear analysis. 143 00:08:41,280 --> 00:08:44,630 And here the engineering stress that we are so used to 144 00:08:44,630 --> 00:08:46,830 seeing in linear analysis, as well. 145 00:08:46,830 --> 00:08:51,730 So this is the test from which we obtain our information 146 00:08:51,730 --> 00:08:53,590 regarding the material behavior. 147 00:08:53,590 --> 00:08:59,150 And let's look at, systematically, test results. 148 00:08:59,150 --> 00:09:02,790 Here we have seen engineering stress plotted vertically, and 149 00:09:02,790 --> 00:09:05,950 the engineering strain plotted horizontally. 150 00:09:05,950 --> 00:09:09,150 And the test result that we would have obtained are shown 151 00:09:09,150 --> 00:09:11,520 systematically in red. 152 00:09:11,520 --> 00:09:14,390 Notice fracture occurs here corresponding 153 00:09:14,390 --> 00:09:17,100 to an ultimate strain. 154 00:09:17,100 --> 00:09:19,750 Notice that the test, of course, is performed in the 155 00:09:19,750 --> 00:09:22,530 tensile region in an actual analysis. 156 00:09:22,530 --> 00:09:26,952 We also want to apply a constitutive relation in the 157 00:09:26,952 --> 00:09:28,090 compressive region. 158 00:09:28,090 --> 00:09:32,270 And it is assumed that the same behavior that we have in 159 00:09:32,270 --> 00:09:34,820 the tensile region is also applicable in 160 00:09:34,820 --> 00:09:36,150 the compressive region. 161 00:09:36,150 --> 00:09:39,020 So we simply reflect this curve into the compressive 162 00:09:39,020 --> 00:09:42,360 region as shown by the blue dashed line. 163 00:09:42,360 --> 00:09:45,810 These are the test resides obtained at a constant 164 00:09:45,810 --> 00:09:48,130 temperature. 165 00:09:48,130 --> 00:09:51,830 Now let us look at the effect of the strain rate. 166 00:09:51,830 --> 00:09:57,190 If the strain rate increases, in other words, de dt 167 00:09:57,190 --> 00:10:00,890 increases, we can see that the test results would change 168 00:10:00,890 --> 00:10:02,390 schematically, as shown here. 169 00:10:05,450 --> 00:10:07,550 Also, there is an effect of temperature. 170 00:10:07,550 --> 00:10:11,950 If the temperature is increasing, but of course at a 171 00:10:11,950 --> 00:10:16,700 constant temperature, we perform the test, then as the 172 00:10:16,700 --> 00:10:20,170 temperature increases for the different tests, we would see 173 00:10:20,170 --> 00:10:24,260 that the behavior of the stress-strain law is shown 174 00:10:24,260 --> 00:10:27,955 here by these red curves, schematically. 175 00:10:30,680 --> 00:10:34,000 This is the instantaneous response of the material. 176 00:10:34,000 --> 00:10:36,510 Now let's look at the material behavior in 177 00:10:36,510 --> 00:10:38,350 time dependent response. 178 00:10:38,350 --> 00:10:42,560 In other words, over a very large time. 179 00:10:42,560 --> 00:10:46,760 Now at a constant stress, inelastic strains develop. 180 00:10:46,760 --> 00:10:49,730 And this is an important effect for materials when the 181 00:10:49,730 --> 00:10:52,320 temperatures are high. 182 00:10:52,320 --> 00:10:59,050 A typical response curve, engineering strain, plotted 183 00:10:59,050 --> 00:11:03,800 vertically here now, and time plotted horizontally, is shown 184 00:11:03,800 --> 00:11:05,020 on this view graph. 185 00:11:05,020 --> 00:11:08,540 Notice there is, instantaneously, of course, 186 00:11:08,540 --> 00:11:12,130 inelastic or elastoplastic strain that occurs. 187 00:11:12,130 --> 00:11:13,940 That's the one we just talked about. 188 00:11:13,940 --> 00:11:18,230 And then with time, there is an increase in strain, as 189 00:11:18,230 --> 00:11:21,800 shown by the red curve here. 190 00:11:21,800 --> 00:11:26,700 We talk about the primary range of the creep curve, a 191 00:11:26,700 --> 00:11:29,390 secondary range, and a tertiary range 192 00:11:29,390 --> 00:11:31,800 until fracture occurs. 193 00:11:31,800 --> 00:11:35,720 This is a curve at which sigma is constant and the 194 00:11:35,720 --> 00:11:39,770 temperature is constant, as well. 195 00:11:39,770 --> 00:11:47,860 If we increase the stress, we, of course, get an increase in 196 00:11:47,860 --> 00:11:51,820 the instantaneous strain, as shown here, but also the 197 00:11:51,820 --> 00:11:55,670 actual creep strain will change. 198 00:11:55,670 --> 00:12:01,390 And that is indicated by these red curves. 199 00:12:01,390 --> 00:12:03,540 Once again, fracture up here. 200 00:12:03,540 --> 00:12:07,630 And these are the results now at a constant temperature, but 201 00:12:07,630 --> 00:12:12,570 at different levels of constants stresses. 202 00:12:12,570 --> 00:12:18,230 The levels of constant stress means here, one stress, there, 203 00:12:18,230 --> 00:12:21,690 another stress, there, another stress, and another stress. 204 00:12:21,690 --> 00:12:26,170 But this stress is constant with time. 205 00:12:26,170 --> 00:12:29,570 There's also a very significant effect off the 206 00:12:29,570 --> 00:12:32,665 temperature on the creep strength, and that is shown on 207 00:12:32,665 --> 00:12:33,830 this view graph. 208 00:12:33,830 --> 00:12:38,670 Once again, strain plotted here times there. 209 00:12:38,670 --> 00:12:41,440 There's an initial strain, of course, 210 00:12:41,440 --> 00:12:45,050 instantaneously reached. 211 00:12:45,050 --> 00:12:47,140 That's the one we talked about earlier. 212 00:12:47,140 --> 00:12:51,420 And then at a very low temperature, almost no creep 213 00:12:51,420 --> 00:12:53,250 effect, as shown here. 214 00:12:53,250 --> 00:12:55,530 Basically a straight line. 215 00:12:55,530 --> 00:12:58,300 Then as the temperature increases, the material 216 00:12:58,300 --> 00:13:05,940 creeps, and of course, doing this for this curve, the 217 00:13:05,940 --> 00:13:07,400 temperature is constant. 218 00:13:07,400 --> 00:13:08,680 For this curve, the temperature 219 00:13:08,680 --> 00:13:11,130 is constant in time. 220 00:13:11,130 --> 00:13:14,270 For this curve, the temperature is also constant, 221 00:13:14,270 --> 00:13:17,070 but at a higher level. 222 00:13:17,070 --> 00:13:19,980 And for this curve, the temperature is also constant 223 00:13:19,980 --> 00:13:22,150 in time, but still at a higher level. 224 00:13:22,150 --> 00:13:24,910 And once again, fracture occurring here, and of course, 225 00:13:24,910 --> 00:13:27,250 sigma is constant, as well. 226 00:13:27,250 --> 00:13:34,180 For each of these curves, we have the same stress. 227 00:13:34,180 --> 00:13:37,740 But a different temperature here, then there, then there 228 00:13:37,740 --> 00:13:41,270 and there, the temperature being constant in time for 229 00:13:41,270 --> 00:13:44,650 each of these curves. 230 00:13:44,650 --> 00:13:49,820 these are some typical laboratory results 231 00:13:49,820 --> 00:13:51,110 that we would get. 232 00:13:51,110 --> 00:13:54,760 And now we want to look at, how would we model these 233 00:13:54,760 --> 00:13:58,570 laboratory test results in a finite element solution? 234 00:13:58,570 --> 00:14:03,250 And let's look first at the models we might use for a 235 00:14:03,250 --> 00:14:05,950 one-dimensional stress situation. 236 00:14:05,950 --> 00:14:08,630 So we consider a one-dimensional situation as 237 00:14:08,630 --> 00:14:10,790 shown schematically here-- 238 00:14:10,790 --> 00:14:16,060 a bar under a load p giving a displacement u. 239 00:14:16,060 --> 00:14:20,160 Notice now we go back to introduce our left superscript 240 00:14:20,160 --> 00:14:23,580 time t, because we want to look now at this [? grade ?] 241 00:14:23,580 --> 00:14:25,210 times t. 242 00:14:25,210 --> 00:14:28,700 The stress at any time is given by this formula. 243 00:14:28,700 --> 00:14:31,343 The area being constant, of course, the cross-section of 244 00:14:31,343 --> 00:14:32,490 the area being constant. 245 00:14:32,490 --> 00:14:35,930 The strain at any time is given by the displacement 246 00:14:35,930 --> 00:14:38,870 measured here, at that time, divided by 247 00:14:38,870 --> 00:14:40,000 the original length. 248 00:14:40,000 --> 00:14:44,890 Remember, we're talking about infinitesimally small strain 249 00:14:44,890 --> 00:14:46,950 conditions. 250 00:14:46,950 --> 00:14:49,510 Let us assume also that the load has now increased 251 00:14:49,510 --> 00:14:53,980 monotonically to a final value, p star. 252 00:14:53,980 --> 00:14:58,010 And let us assume that the time is long, so that inertia 253 00:14:58,010 --> 00:15:00,760 effects are negligible, so we can talk 254 00:15:00,760 --> 00:15:03,735 about a static analysis. 255 00:15:03,735 --> 00:15:09,390 If you look at the response that we would 256 00:15:09,390 --> 00:15:12,530 see, we see the following. 257 00:15:12,530 --> 00:15:18,230 The load increases as shown here, and here we 258 00:15:18,230 --> 00:15:20,670 have the time axis. 259 00:15:20,670 --> 00:15:22,930 Notice T star is small. 260 00:15:22,930 --> 00:15:26,090 It's a very small time interval. 261 00:15:26,090 --> 00:15:31,590 And in this time interval, creep effects would not take 262 00:15:31,590 --> 00:15:34,010 place because T star is so small. 263 00:15:34,010 --> 00:15:38,360 If they do take place, their effect is negligible. 264 00:15:38,360 --> 00:15:39,590 That's the assumption. 265 00:15:39,590 --> 00:15:44,330 So plasticity effects predominate here, and those we 266 00:15:44,330 --> 00:15:46,140 take into account. 267 00:15:46,140 --> 00:15:51,450 And then from here onwards, there will be no more plastic 268 00:15:51,450 --> 00:15:53,510 deformations in the material. 269 00:15:53,510 --> 00:15:59,760 Now the material will creep with time. 270 00:15:59,760 --> 00:16:03,570 And so all the inelastic strains are modeled from here 271 00:16:03,570 --> 00:16:06,780 onwards as creep strains, and from here to 272 00:16:06,780 --> 00:16:09,440 there as plastic strains. 273 00:16:09,440 --> 00:16:14,780 Now let's see how we would model these strains. 274 00:16:14,780 --> 00:16:21,420 Here we see how we would model the plastic strains in this 275 00:16:21,420 --> 00:16:23,850 uniaxial situation. 276 00:16:23,850 --> 00:16:28,520 Notice we assume a bilinear material description, bilinear 277 00:16:28,520 --> 00:16:31,730 meaning a Young's modulus until the yield stress is 278 00:16:31,730 --> 00:16:39,420 reached, and then another constant value of a new 279 00:16:39,420 --> 00:16:41,330 Young's modulus, if you want to. 280 00:16:41,330 --> 00:16:44,060 It's not really Young's modulus. 281 00:16:44,060 --> 00:16:45,510 We call it ET. 282 00:16:45,510 --> 00:16:48,470 It's a strain-hardening model. 283 00:16:48,470 --> 00:16:53,450 Notice that ET is generally considerably smaller than E. 284 00:16:53,450 --> 00:16:57,190 Typically it might be 1% of E. 285 00:16:57,190 --> 00:16:59,480 So we're coming up here. 286 00:16:59,480 --> 00:17:02,030 After the yield stress, we have a constant Young's 287 00:17:02,030 --> 00:17:05,630 modulus, and beyond the yield stress, we have a 288 00:17:05,630 --> 00:17:10,480 strain-hardening modulus, ET, to describe the change in 289 00:17:10,480 --> 00:17:14,180 stress as a function of the strain. 290 00:17:14,180 --> 00:17:18,440 Notice now that if we had moved up to here and we were 291 00:17:18,440 --> 00:17:24,244 to unload from this level of stress, we would follow the 292 00:17:24,244 --> 00:17:30,340 original Young's modulus in the unloading that we had in 293 00:17:30,340 --> 00:17:31,630 the loading. 294 00:17:31,630 --> 00:17:35,850 In other words, here we had the same E in unloading, 295 00:17:35,850 --> 00:17:39,660 indicated by the dashed line, as we have for loading here. 296 00:17:42,250 --> 00:17:45,990 In our particular modeling situation, we are coming up 297 00:17:45,990 --> 00:17:52,710 here to the yield stress, then we proceed, going along ET, up 298 00:17:52,710 --> 00:17:56,650 to that point, and that point corresponds to our time t 299 00:17:56,650 --> 00:17:59,150 star, where P star has been reached. 300 00:17:59,150 --> 00:18:02,310 The maximum load has been reached. 301 00:18:02,310 --> 00:18:08,830 Notice that this part here is a plastic strain corresponding 302 00:18:08,830 --> 00:18:14,830 to t star, and this part here is the elastic strain, 303 00:18:14,830 --> 00:18:18,200 corresponding to t star. 304 00:18:18,200 --> 00:18:23,270 And that is obtained by simply dropping a vertical line down 305 00:18:23,270 --> 00:18:28,220 from that point t star onto the horizontal axes. 306 00:18:28,220 --> 00:18:33,200 The stress is then given by the Young's modulus E, that's 307 00:18:33,200 --> 00:18:39,590 E, or the slope is E here, times te at 308 00:18:39,590 --> 00:18:41,620 any particular time. 309 00:18:41,620 --> 00:18:43,855 In other words, we could have here, instead of 310 00:18:43,855 --> 00:18:45,445 the t, the t star. 311 00:18:45,445 --> 00:18:49,300 At any particular time this holds, and the total inelastic 312 00:18:49,300 --> 00:18:53,030 strain is given by simply the plastic strain, because 313 00:18:53,030 --> 00:18:57,290 remember in our model, we assume that t star is so small 314 00:18:57,290 --> 00:19:01,430 that all the inelastic strains can be modeled using the 315 00:19:01,430 --> 00:19:02,840 plasticity assumption. 316 00:19:02,840 --> 00:19:05,850 This assumption. 317 00:19:05,850 --> 00:19:09,810 Of course, once again, these relationships hold for any 318 00:19:09,810 --> 00:19:15,670 particular point on this curve here, in particular, the point 319 00:19:15,670 --> 00:19:18,610 time t star, where these are the values 320 00:19:18,610 --> 00:19:21,790 that we're using here. 321 00:19:21,790 --> 00:19:26,290 Now from time t star onwards, we have creep effects, and 322 00:19:26,290 --> 00:19:28,460 those would be typically modeled as shown 323 00:19:28,460 --> 00:19:29,710 on this view graph. 324 00:19:29,710 --> 00:19:34,800 Here is t star, which is very small, almost equal to 0. 325 00:19:34,800 --> 00:19:40,370 And here we have the time axis, the creep strain axis, 326 00:19:40,370 --> 00:19:44,100 shown here, and here we have a typical creep law that is 327 00:19:44,100 --> 00:19:46,210 widely used in engineering practice, the 328 00:19:46,210 --> 00:19:48,290 power law, shown here. 329 00:19:48,290 --> 00:19:51,770 A0, A1, A2 are constant. 330 00:19:51,770 --> 00:19:54,980 That has to be determined from the laboratory tests, and they 331 00:19:54,980 --> 00:19:58,100 will change for different materials. 332 00:19:58,100 --> 00:20:03,100 Notice that the stress is still generated by the Young's 333 00:20:03,100 --> 00:20:06,390 modulus times the elastic strain, and the total 334 00:20:06,390 --> 00:20:09,690 inelastic strain now is made up of the creep strain 335 00:20:09,690 --> 00:20:13,640 obtained by this formula plus the plastic strain, the 336 00:20:13,640 --> 00:20:16,260 plastic strain already identified on the 337 00:20:16,260 --> 00:20:17,510 earlier view graph. 338 00:20:20,770 --> 00:20:25,310 Another way of modeling the instantaneous inelastic 339 00:20:25,310 --> 00:20:28,590 strains, which we call plastic strains, and the 340 00:20:28,590 --> 00:20:31,850 time-dependent inelastic strains, which we call it 341 00:20:31,850 --> 00:20:36,590 creep strains, is to use a viscoplastic material model. 342 00:20:36,590 --> 00:20:41,560 And in this case, we have this basic relationship here where 343 00:20:41,560 --> 00:20:46,710 we say that the total strain times a time derivative of the 344 00:20:46,710 --> 00:20:51,060 total strain, is given by the time derivative of the stress 345 00:20:51,060 --> 00:20:55,010 divided by the Young's modulus plus a fluidity parameter 346 00:20:55,010 --> 00:20:58,060 times this bracket here. 347 00:20:58,060 --> 00:21:04,900 And this bracket here needs to be read as shown down here. 348 00:21:04,900 --> 00:21:08,660 Notice it is equal to 0. 349 00:21:08,660 --> 00:21:12,320 If sigma is smaller than sigma yield, the yield stress, and 350 00:21:12,320 --> 00:21:15,690 is equal to sigma minus sigma yield, sigma is greater than 351 00:21:15,690 --> 00:21:17,630 sigma yield. 352 00:21:17,630 --> 00:21:21,140 This part here is the time derivative of the 353 00:21:21,140 --> 00:21:23,760 viscoplastic strain. 354 00:21:23,760 --> 00:21:27,400 Of course, gamma depends on the material-- 355 00:21:27,400 --> 00:21:29,540 on the particular material that you're looking at. 356 00:21:29,540 --> 00:21:38,880 And also, sigma yield will change possibly with time, 357 00:21:38,880 --> 00:21:41,810 because there might be strain hardening in the material, the 358 00:21:41,810 --> 00:21:46,570 same kind of strain hardening that we have observed in our 359 00:21:46,570 --> 00:21:49,920 plasticity description. 360 00:21:49,920 --> 00:21:54,410 Typical solutions for the 1-D specimen would be, if you have 361 00:21:54,410 --> 00:21:58,700 a non-hardening material, in other words, the sigma yield, 362 00:21:58,700 --> 00:22:02,870 the value does not increase, then we would have here in 363 00:22:02,870 --> 00:22:10,800 elastic strain, and with gamma, determined by gamma, a 364 00:22:10,800 --> 00:22:15,410 viscoplastic strain that, of course, varies with time. 365 00:22:15,410 --> 00:22:18,540 And you can see the total strain here is made up of this 366 00:22:18,540 --> 00:22:22,690 elastic strain plus a viscoplastic strain, which 367 00:22:22,690 --> 00:22:25,610 also depends on gamma. 368 00:22:25,610 --> 00:22:30,820 If we have a hardening material, then the situation 369 00:22:30,820 --> 00:22:34,770 for the total strain looks as shown on this 370 00:22:34,770 --> 00:22:36,240 side of the view graph. 371 00:22:36,240 --> 00:22:39,310 Total strain plotted, again, vertically, time here. 372 00:22:39,310 --> 00:22:42,070 We have an elastic strain. 373 00:22:42,070 --> 00:22:48,660 And with time, a viscoplastic strain that, however, will 374 00:22:48,660 --> 00:22:53,180 reach a steady state value, an ultimate value. 375 00:22:53,180 --> 00:22:58,570 This amount here depends on the increase of the yield 376 00:22:58,570 --> 00:23:01,630 stress as a function of the viscoplastic strain. 377 00:23:01,630 --> 00:23:06,550 Notice that also, this curve, this red curve, depends on the 378 00:23:06,550 --> 00:23:08,130 value of gamma. 379 00:23:08,130 --> 00:23:11,970 As gamma increases, you are getting a curve that goes more 380 00:23:11,970 --> 00:23:17,020 sharply up to this green line, this steady-state solution. 381 00:23:20,230 --> 00:23:23,520 Let us now look at plasticity in more detail. 382 00:23:26,220 --> 00:23:30,110 In other words, how we are more dealing plasticity 383 00:23:30,110 --> 00:23:31,250 relationships. 384 00:23:31,250 --> 00:23:36,310 plastic materials, phenomena of plasticity in finite 385 00:23:36,310 --> 00:23:38,140 element analysis. 386 00:23:38,140 --> 00:23:41,880 So far, we've only considered loading conditions. 387 00:23:41,880 --> 00:23:46,140 However, we of course, in an actual analysis, might have 388 00:23:46,140 --> 00:23:50,200 that the material is loading, unloading, again 389 00:23:50,200 --> 00:23:51,750 reloading and so on. 390 00:23:51,750 --> 00:23:56,290 So we have to look briefly at, how do we take such situations 391 00:23:56,290 --> 00:23:57,730 into account? 392 00:23:57,730 --> 00:24:01,710 Well, on this view graph, systematically, we see what 393 00:24:01,710 --> 00:24:04,210 kind of load history a material might 394 00:24:04,210 --> 00:24:05,480 be subjected to. 395 00:24:05,480 --> 00:24:08,020 Here we have a load increasing. 396 00:24:08,020 --> 00:24:11,960 Say, up to here, we have elastic conditions, then we go 397 00:24:11,960 --> 00:24:13,210 into plasticity. 398 00:24:15,380 --> 00:24:21,070 At this point, we unload elastically, and right here, 399 00:24:21,070 --> 00:24:25,300 we go back into plasticity. 400 00:24:25,300 --> 00:24:29,190 From here to there, plasticity, and then elastic 401 00:24:29,190 --> 00:24:30,250 behavior again. 402 00:24:30,250 --> 00:24:35,450 So how would we model such a load situation on 403 00:24:35,450 --> 00:24:36,990 the material level? 404 00:24:36,990 --> 00:24:40,770 That is what we like to discuss now. 405 00:24:40,770 --> 00:24:44,990 Here we show a material assumption, or a material 406 00:24:44,990 --> 00:24:49,670 relationship, that would be typically used. 407 00:24:49,670 --> 00:24:54,070 Stress plotted up, strain plotted horizontally. 408 00:24:54,070 --> 00:25:00,150 Here we load the material, going to plasticity in the 409 00:25:00,150 --> 00:25:03,910 loading cycle, in the loading regime, and now we unload 410 00:25:03,910 --> 00:25:05,160 elastically. 411 00:25:06,840 --> 00:25:10,712 Up to this point, we are elastic. 412 00:25:10,712 --> 00:25:14,980 We now are plastic, and we unload again elastically, from 413 00:25:14,980 --> 00:25:16,550 the compression region. 414 00:25:16,550 --> 00:25:20,110 Now notice in this particular material description, we 415 00:25:20,110 --> 00:25:22,340 assume isotropic hardening. 416 00:25:24,900 --> 00:25:29,870 And this assumption means that once we have loaded up to this 417 00:25:29,870 --> 00:25:36,230 point, the new yield stress is given by this value here. 418 00:25:36,230 --> 00:25:40,760 And before we would yield again, in compression, 419 00:25:40,760 --> 00:25:46,860 typically, or if we unload here and come back, of course, 420 00:25:46,860 --> 00:25:52,870 intention, we would have to have a gain as stress that is 421 00:25:52,870 --> 00:25:55,070 equal to this yield stress. 422 00:25:55,070 --> 00:26:00,290 So notice in compression, we have this value being here, 423 00:26:00,290 --> 00:26:02,310 the same as that one. 424 00:26:02,310 --> 00:26:06,320 And that means we have to go all the way down here until we 425 00:26:06,320 --> 00:26:10,720 will get back into plasticity, and at this point, we go into 426 00:26:10,720 --> 00:26:13,150 plasticity, as shown here. 427 00:26:13,150 --> 00:26:16,930 The total plastic strains is a sum of this 428 00:26:16,930 --> 00:26:19,070 part plus that part. 429 00:26:19,070 --> 00:26:21,420 All of these are plastic strains that have been 430 00:26:21,420 --> 00:26:24,090 accumulated in the loading cycle. 431 00:26:24,090 --> 00:26:30,550 Notice, once we have loaded in compression further, we have a 432 00:26:30,550 --> 00:26:34,800 new yield stress, sigma yield 2, which is larger 433 00:26:34,800 --> 00:26:37,200 than sigma yield 1. 434 00:26:37,200 --> 00:26:41,720 So whenever we go into the plasticity regime and load 435 00:26:41,720 --> 00:26:46,250 further, our yield stress increases, and that means we 436 00:26:46,250 --> 00:26:48,280 assume isotropic hardening. 437 00:26:48,280 --> 00:26:51,530 I should also point out at this point that we make an 438 00:26:51,530 --> 00:26:55,660 assumption of the bilinear material behavior, namely, 439 00:26:55,660 --> 00:27:00,100 bilinear in having one Young's modulus here and one 440 00:27:00,100 --> 00:27:02,150 strain-hardening modulus here. 441 00:27:02,150 --> 00:27:05,490 If you have a complex stress-strain relationship 442 00:27:05,490 --> 00:27:10,610 measured in laboratory, you may want to use one et up to a 443 00:27:10,610 --> 00:27:16,320 certain stress level, and then another et from that stress 444 00:27:16,320 --> 00:27:20,120 level on to the next stress level, and maybe another et 445 00:27:20,120 --> 00:27:21,110 beyond that. 446 00:27:21,110 --> 00:27:23,520 So you have a multilinear description of 447 00:27:23,520 --> 00:27:25,180 the material behavior. 448 00:27:25,180 --> 00:27:29,630 We don't have time to go into such details. 449 00:27:29,630 --> 00:27:32,460 It's an extension of this material description. 450 00:27:35,370 --> 00:27:39,700 Instead of using isotropic hardening assumption, we may 451 00:27:39,700 --> 00:27:42,580 want to use a kinematic hardening assumption. 452 00:27:42,580 --> 00:27:45,880 And that assumption is depicted on this view graph. 453 00:27:45,880 --> 00:27:50,650 Notice stress, again, upwards, strain, horizontally, total 454 00:27:50,650 --> 00:27:54,300 strain horizontally noted, and here we're coming up in our 455 00:27:54,300 --> 00:28:00,740 loading cycle to this point, and the unload now, same way 456 00:28:00,740 --> 00:28:03,130 as before in isotropic hardening-- 457 00:28:03,130 --> 00:28:07,000 the difference is right now occurring in this regime, 458 00:28:07,000 --> 00:28:13,070 namely our yield stress that will make us yield in 459 00:28:13,070 --> 00:28:18,720 compression is not the same yield stress that we talked 460 00:28:18,720 --> 00:28:20,630 about in isotropic hardening. 461 00:28:20,630 --> 00:28:25,480 Instead, this difference in stress is simply twice the 462 00:28:25,480 --> 00:28:27,820 initial yield stress. 463 00:28:27,820 --> 00:28:32,100 So we reach yield in compression much earlier than 464 00:28:32,100 --> 00:28:34,430 in isotopically hardening. 465 00:28:34,430 --> 00:28:40,730 Namely, when the stress has decreased by twice this amount 466 00:28:40,730 --> 00:28:45,070 here, we come into the regime of yielding again. 467 00:28:45,070 --> 00:28:48,770 And now we would be yielding all the way along here, up to 468 00:28:48,770 --> 00:28:52,420 this point, and now we unload elastically again. 469 00:28:52,420 --> 00:28:55,900 Notice, this is now the plastic strain that we 470 00:28:55,900 --> 00:28:58,920 accumulate in the compression region. 471 00:28:58,920 --> 00:29:03,990 This is the plastic strain that we accumulate in the 472 00:29:03,990 --> 00:29:06,960 tensile region. 473 00:29:06,960 --> 00:29:09,880 These were a few words regarding how we model 474 00:29:09,880 --> 00:29:14,260 one-dimensional elastoplastic behavior, and let us now look 475 00:29:14,260 --> 00:29:19,030 at the multi-axial plasticity behavior. 476 00:29:19,030 --> 00:29:21,610 How we model, in other words, plastic behavior in 477 00:29:21,610 --> 00:29:24,040 multi-axial stress conditions. 478 00:29:24,040 --> 00:29:28,550 In that case, we need a yield condition, a flow rule, and a 479 00:29:28,550 --> 00:29:33,240 hardening rule, and I'd like to consider now isothermal, in 480 00:29:33,240 --> 00:29:36,810 other words, constant temperature conditions. 481 00:29:36,810 --> 00:29:41,400 The yield condition flow rule and hardening rule are 482 00:29:41,400 --> 00:29:45,590 expressed using a stress function. 483 00:29:45,590 --> 00:29:50,680 And very widely used stress functions are the Von Metus 484 00:29:50,680 --> 00:29:53,830 function and the Drucker-Prager function. 485 00:29:53,830 --> 00:29:58,930 The Von Mises function is given on this view graph here. 486 00:29:58,930 --> 00:30:05,160 tF is equal to 1/2 tsij tsij minus t kappa. 487 00:30:05,160 --> 00:30:10,340 Of course, we're summing here over all stress components sij 488 00:30:10,340 --> 00:30:14,510 where the stress component sij is called the deviatoric 489 00:30:14,510 --> 00:30:18,190 stress, and is given as t sigma ij minus t 490 00:30:18,190 --> 00:30:21,110 sigma mm over 3. 491 00:30:21,110 --> 00:30:25,330 This is summing all the normal stress components, in other 492 00:30:25,330 --> 00:30:29,690 words, t sigma 11 plus t sigma 2 2 plus t sigma 3 3, and 493 00:30:29,690 --> 00:30:31,160 dividing by 3. 494 00:30:31,160 --> 00:30:32,830 This is the Kronecker delta that we 495 00:30:32,830 --> 00:30:34,890 encountered already earlier. 496 00:30:34,890 --> 00:30:39,270 The t kappa value is given here as 1/3 497 00:30:39,270 --> 00:30:41,330 t sigma yield squared. 498 00:30:41,330 --> 00:30:45,880 Notice the yield stress here changes as a function of 499 00:30:45,880 --> 00:30:48,150 loading, as a function of time. 500 00:30:48,150 --> 00:30:51,730 And in fact, we are here using isotropic hardening 501 00:30:51,730 --> 00:30:53,360 conditions. 502 00:30:53,360 --> 00:30:58,230 In the Drucker-Prager stress function, assumption we have 503 00:30:58,230 --> 00:30:59,760 this equation here. 504 00:30:59,760 --> 00:31:05,210 tF is equal to 3 alpha t sigma m plus t sigma bar minus k. 505 00:31:05,210 --> 00:31:08,450 Notice t sigma m is defined as here. 506 00:31:08,450 --> 00:31:13,620 Notice the sum here over all the stress is ii, in other 507 00:31:13,620 --> 00:31:17,740 words, t sigma 1 1 plus t sigma 2 2 plus t sigma 3 3. 508 00:31:17,740 --> 00:31:20,750 That is t sigma i i. 509 00:31:20,750 --> 00:31:24,860 And here we have t sigma bar defined, again in terms of the 510 00:31:24,860 --> 00:31:26,360 deviatoric stresses. 511 00:31:26,360 --> 00:31:31,320 We will talk in this lecture more about the Von Mises yield 512 00:31:31,320 --> 00:31:34,790 condition, which is very widely used for 513 00:31:34,790 --> 00:31:36,590 modeling metal behavior. 514 00:31:36,590 --> 00:31:40,340 The Drucker-Prager yield condition is widely used to 515 00:31:40,340 --> 00:31:47,090 model rock and solid mechanics types of problems. 516 00:31:47,090 --> 00:31:49,020 We'll not talk in this lecture about this 517 00:31:49,020 --> 00:31:50,460 condition any further. 518 00:31:50,460 --> 00:31:53,640 Please refer to the textbook, where there is a bit more 519 00:31:53,640 --> 00:31:55,470 information given about the 520 00:31:55,470 --> 00:32:00,000 Drucker-Prager stress function. 521 00:32:00,000 --> 00:32:03,360 We will also use, in our discussion, both matrix 522 00:32:03,360 --> 00:32:06,220 notation and index notation. 523 00:32:06,220 --> 00:32:10,290 In matrix notation, for example, the vector DEP of 524 00:32:10,290 --> 00:32:13,570 plastic strains is defined as shown here. 525 00:32:13,570 --> 00:32:18,050 Notice we are summing these two components here for the 526 00:32:18,050 --> 00:32:20,630 shear strains. 527 00:32:20,630 --> 00:32:28,500 The stress vector is shown here, and if we use index 528 00:32:28,500 --> 00:32:33,710 notation, we imply these components. 529 00:32:33,710 --> 00:32:39,750 dep 1 2 and dep 2 1. 530 00:32:39,750 --> 00:32:43,080 Notice, once again, in these vector or matrix notations, 531 00:32:43,080 --> 00:32:46,920 these two components are added to get the total plastic shear 532 00:32:46,920 --> 00:32:48,190 strain, 1, 2. 533 00:32:51,270 --> 00:32:56,120 The basic equations are, then, for the Von Mises yield 534 00:32:56,120 --> 00:32:59,130 condition, as shown here. 535 00:32:59,130 --> 00:33:00,980 This is a yield condition. 536 00:33:00,980 --> 00:33:04,300 tF, t sigma ij, t kappa. 537 00:33:04,300 --> 00:33:07,800 t kappa is a function of the plastic strains. 538 00:33:07,800 --> 00:33:09,670 These are the current stresses. 539 00:33:09,670 --> 00:33:10,580 And [UNINTELLIGIBLE] 540 00:33:10,580 --> 00:33:13,290 yielding tF is equal to 0. 541 00:33:13,290 --> 00:33:16,450 That's what we're saying here. 542 00:33:16,450 --> 00:33:19,590 The equivalent for the one-dimensional response, in 543 00:33:19,590 --> 00:33:21,460 other words, would be this equation. 544 00:33:24,350 --> 00:33:29,490 Notice once again that tS is equal to 0 throughout the 545 00:33:29,490 --> 00:33:33,290 total plastic response, throughout all of the yielding 546 00:33:33,290 --> 00:33:35,800 that we observe. 547 00:33:35,800 --> 00:33:39,760 The flow rule, and here we use the associative flow rule, 548 00:33:39,760 --> 00:33:43,700 because tF is here the same tF that we also use 549 00:33:43,700 --> 00:33:45,160 in the yield condition. 550 00:33:45,160 --> 00:33:47,630 The flow rule is given by this equation. 551 00:33:47,630 --> 00:33:50,050 Notice we have on the left-hand side the plastic 552 00:33:50,050 --> 00:33:51,910 strain increments. 553 00:33:51,910 --> 00:33:54,600 Here we have a scalar, t lambda. 554 00:33:54,600 --> 00:33:58,490 And the partial of tF with respect 555 00:33:58,490 --> 00:34:00,990 to the current stresses. 556 00:34:00,990 --> 00:34:05,300 t lambda is a positive scalar. 557 00:34:05,300 --> 00:34:08,909 In one-dimension analysis, we would have that the plastic 558 00:34:08,909 --> 00:34:14,280 strains 1 1, 2 2, and 3 3 are as shown here. 559 00:34:14,280 --> 00:34:17,600 And notice, of course, they only depend on the 560 00:34:17,600 --> 00:34:19,469 one-dimensional stress, or the one stress 561 00:34:19,469 --> 00:34:20,630 that is only active. 562 00:34:20,630 --> 00:34:26,600 And notice that the sum of these plastic strains is 0. 563 00:34:26,600 --> 00:34:31,060 This stress-strain relationship is given here on 564 00:34:31,060 --> 00:34:32,280 this view graph-- 565 00:34:32,280 --> 00:34:36,120 stress increment on the left-hand side, elastic 566 00:34:36,120 --> 00:34:38,330 stress-strain, material [UNINTELLIGIBLE] 567 00:34:38,330 --> 00:34:39,920 go into the CE. 568 00:34:39,920 --> 00:34:43,850 Here's the total strain increment minus the plastic 569 00:34:43,850 --> 00:34:45,360 strain increment. 570 00:34:45,360 --> 00:34:48,260 The 1-D equivalent, in other words, for the one-dimensional 571 00:34:48,260 --> 00:34:53,310 stress situation, we would have this equation here. 572 00:34:53,310 --> 00:35:01,430 Now our goal is to determine a CEP matrix such that we have 573 00:35:01,430 --> 00:35:04,640 on the left-hand side the stress increment, and on the 574 00:35:04,640 --> 00:35:07,760 right-hand side, that CEP matrix times the 575 00:35:07,760 --> 00:35:11,460 total strain increment. 576 00:35:11,460 --> 00:35:14,880 Of course, the kind of stress conditions, yielding, et 577 00:35:14,880 --> 00:35:17,510 cetera, conditions must all go into the 578 00:35:17,510 --> 00:35:21,570 determination off CEP. 579 00:35:21,570 --> 00:35:26,780 Let us now look at the general derivation of this CEP matrix. 580 00:35:26,780 --> 00:35:29,720 This is quite a general derivation. 581 00:35:29,720 --> 00:35:31,340 It's also given in the textbook. 582 00:35:31,340 --> 00:35:35,790 You may want to refer to the textbook as well while we 583 00:35:35,790 --> 00:35:38,970 discuss this deviation or afterwards. 584 00:35:38,970 --> 00:35:44,350 We define these quantities here, tqij being the partial 585 00:35:44,350 --> 00:35:47,650 of F with respect to the current stresses with the 586 00:35:47,650 --> 00:35:51,030 plastic strains fixed. 587 00:35:51,030 --> 00:35:56,090 We also define tpij equal to minus partial F with respect 588 00:35:56,090 --> 00:36:00,380 to the plastic strains with the stress fixed. 589 00:36:00,380 --> 00:36:02,670 These are two convenient definitions for the 590 00:36:02,670 --> 00:36:03,910 derivation. 591 00:36:03,910 --> 00:36:08,960 If we use matrix notation, we assemble in 592 00:36:08,960 --> 00:36:10,630 this vector at tq-- 593 00:36:10,630 --> 00:36:13,960 we are writing actually, the transposed here, just for ease 594 00:36:13,960 --> 00:36:16,450 of writing-- 595 00:36:16,450 --> 00:36:19,930 these components are assembled in this vector. 596 00:36:19,930 --> 00:36:25,660 Notice we have 2 times tq12, 2 times tq 2 0 3, and 597 00:36:25,660 --> 00:36:27,810 2 times tq 3 1. 598 00:36:27,810 --> 00:36:32,070 And these tools here result from our definition of the 599 00:36:32,070 --> 00:36:34,480 plastic strain and stress increment vectors. 600 00:36:34,480 --> 00:36:41,180 Remember in the plastic strain vector, we had the sum of Dep 601 00:36:41,180 --> 00:36:48,380 1 2 plus Dep 2 1 components, we got the total plastic share 602 00:36:48,380 --> 00:36:54,460 strain assembled in that plastic strain vector. 603 00:36:54,460 --> 00:36:57,620 And because we have defined our plastic strain vector and 604 00:36:57,620 --> 00:37:02,370 our stress increment vectors as shown earlier on the view 605 00:37:02,370 --> 00:37:06,340 graph, we now carry these two along here. 606 00:37:06,340 --> 00:37:10,780 tp, transposed again, is written out here. 607 00:37:13,320 --> 00:37:18,640 We know determine t lambda in terms of de. 608 00:37:18,640 --> 00:37:20,350 That is our first step. 609 00:37:20,350 --> 00:37:24,090 Because if we have this relationship, then we can say, 610 00:37:24,090 --> 00:37:28,210 we have the plastic strain increment in terms of a total 611 00:37:28,210 --> 00:37:29,610 strain increment. 612 00:37:29,610 --> 00:37:34,400 Using tf equal to 0, throughout the plastic 613 00:37:34,400 --> 00:37:37,310 deformations, because we know that our yield condition has 614 00:37:37,310 --> 00:37:41,090 to be satisfied throughout all of the plastic deformations, 615 00:37:41,090 --> 00:37:44,540 we can directly write down this equation here. 616 00:37:44,540 --> 00:37:47,520 And with our definition just given, we rewrite this 617 00:37:47,520 --> 00:37:50,160 right-hand side as shown here. 618 00:37:50,160 --> 00:37:56,080 And we also substitute for dep, t lambda times tq. 619 00:37:56,080 --> 00:38:04,370 Notice, once again, we have now in the fourth fifth, and 620 00:38:04,370 --> 00:38:09,390 sixth component of tq this factor two, because of the 621 00:38:09,390 --> 00:38:12,930 definition of dep shown earlier. 622 00:38:12,930 --> 00:38:18,180 Well, there's nothing else than a matrix rewriting of 623 00:38:18,180 --> 00:38:22,070 this relationship here, of this right-hand side here. 624 00:38:22,070 --> 00:38:28,950 And we now use this relationship in our further 625 00:38:28,950 --> 00:38:30,710 developments. 626 00:38:30,710 --> 00:38:34,050 But before doing so, let us see how we can rewrite this 627 00:38:34,050 --> 00:38:36,080 left-hand side here. 628 00:38:36,080 --> 00:38:38,550 tq times [UNINTELLIGIBLE] t sigma. 629 00:38:38,550 --> 00:38:41,370 We can write this left-hand side also as shown on the 630 00:38:41,370 --> 00:38:45,240 right-hand side here, because we can substitute for t sigma 631 00:38:45,240 --> 00:38:47,830 simply the stress-strain law. 632 00:38:47,830 --> 00:38:51,060 The flow rule, of course, gives us this relationship 633 00:38:51,060 --> 00:38:54,350 here for the plastic strain increment. 634 00:38:54,350 --> 00:39:01,630 Hence, if we use what we have up here and substitute for the 635 00:39:01,630 --> 00:39:05,600 plastic strain increments, we directly obtain what I'm 636 00:39:05,600 --> 00:39:06,850 circling here. 637 00:39:09,140 --> 00:39:14,560 Next we use what we derived on the previous view graph, 638 00:39:14,560 --> 00:39:19,340 namely, tq times post t sigma is equal to what's shown here 639 00:39:19,340 --> 00:39:21,860 on the right-hand side, what I'm circling now. 640 00:39:21,860 --> 00:39:25,870 And this together then gives us an equation that we can 641 00:39:25,870 --> 00:39:27,810 solve for t lambda. 642 00:39:27,810 --> 00:39:30,980 Having solved for t lambda, and I will show you the result 643 00:39:30,980 --> 00:39:34,550 on the next few graph charts now, we can substitute 40 644 00:39:34,550 --> 00:39:39,300 lambda back here, and we get our plastic strains directly. 645 00:39:39,300 --> 00:39:41,250 Well, let's look at the result. 646 00:39:41,250 --> 00:39:46,620 Here we have t lambda, notice, given in terms of the 647 00:39:46,620 --> 00:39:49,670 differential total strain increment. 648 00:39:49,670 --> 00:39:54,650 And now substituting, as I said just now, 40 lambda, we 649 00:39:54,650 --> 00:39:57,410 get the plastic strains in terms of the 650 00:39:57,410 --> 00:39:58,660 total strain increment. 651 00:40:02,020 --> 00:40:06,150 This relationship is now used in the 652 00:40:06,150 --> 00:40:08,570 stress-strain law again. 653 00:40:08,570 --> 00:40:12,490 Notice the stress-strain law is shown here. 654 00:40:12,490 --> 00:40:19,820 And we directly obtain, by expressing dep in terms of de 655 00:40:19,820 --> 00:40:23,640 by the relation that we just have obtained on the previous 656 00:40:23,640 --> 00:40:28,625 view graph, we directly obtain this final expression for CEP. 657 00:40:31,200 --> 00:40:36,940 Of course the current stresses go in there, and the current 658 00:40:36,940 --> 00:40:39,690 material conditions go in there. 659 00:40:39,690 --> 00:40:45,260 And we will apply this general expression to an example just 660 00:40:45,260 --> 00:40:49,290 now, namely the example of the Von Mises flow 661 00:40:49,290 --> 00:40:51,600 rule and yield condition. 662 00:40:51,600 --> 00:40:56,330 However, before doing so, we know need to break, because we 663 00:40:56,330 --> 00:40:59,201 have to continue our discussion on a second reel. 664 00:41:04,390 --> 00:41:07,460 So let us now apply this general relationship here 665 00:41:07,460 --> 00:41:10,670 which we just derived to the case of the Von Mises yield 666 00:41:10,670 --> 00:41:14,070 condition with isotropic hardening. 667 00:41:14,070 --> 00:41:18,600 In that case, yielding is reached when the principal 668 00:41:18,600 --> 00:41:22,460 stress is here on the right-hand side, substituted 669 00:41:22,460 --> 00:41:26,375 into this right-hand side, give us the yield stress, t 670 00:41:26,375 --> 00:41:27,110 sigma yield. 671 00:41:27,110 --> 00:41:32,240 Or another way of saying it, our tf function is given right 672 00:41:32,240 --> 00:41:36,290 down here, where we have here the deviatoric stresses 673 00:41:36,290 --> 00:41:40,160 squared, t kappa is shown here. 674 00:41:40,160 --> 00:41:43,470 We had in fact this relationship already on an 675 00:41:43,470 --> 00:41:44,570 earlier view graph. 676 00:41:44,570 --> 00:41:49,260 Notice that this relationship on TF is really the same as 677 00:41:49,260 --> 00:41:51,530 what you see up here. 678 00:41:51,530 --> 00:41:54,390 But of course, here we are talking about the use of 679 00:41:54,390 --> 00:41:57,660 principal stresses, whereas here we are talking about the 680 00:41:57,660 --> 00:41:58,910 use of deviatoric stresses. 681 00:42:02,610 --> 00:42:06,520 Pictorally, we can represent the yield 682 00:42:06,520 --> 00:42:08,850 surface as shown here. 683 00:42:08,850 --> 00:42:12,760 In plain stress analysis, this would be the yield surface. 684 00:42:12,760 --> 00:42:18,200 In other words, t sigma 2 and t sigma 1 are the principle of 685 00:42:18,200 --> 00:42:24,120 stresses that are generally non-zero, but t sigma 3 is 0. 686 00:42:24,120 --> 00:42:29,070 Here we have the yield stress, and on the right-hand side, we 687 00:42:29,070 --> 00:42:31,450 see the [? general ?] 688 00:42:31,450 --> 00:42:32,520 yield condition in 689 00:42:32,520 --> 00:42:34,660 three-dimensional stress space. 690 00:42:34,660 --> 00:42:39,200 Notice initially we have sigma yield, [UNINTELLIGIBLE] 691 00:42:39,200 --> 00:42:46,240 radius of this cylinder, and that radius expands with time, 692 00:42:46,240 --> 00:42:50,280 and it expands uniformly, as shown, with the green circle 693 00:42:50,280 --> 00:42:54,520 here to t sigma yield. 694 00:42:54,520 --> 00:42:59,440 In other words, the cylinder grows uniformly, the radius 695 00:42:59,440 --> 00:43:02,370 grows with isotopically hardening. 696 00:43:02,370 --> 00:43:05,810 Notice that for the plane stress case, the hardening 697 00:43:05,810 --> 00:43:11,050 would be stress by having this ellipse grow, as briefly 698 00:43:11,050 --> 00:43:15,370 indicated, by this green line, which of course goes all the 699 00:43:15,370 --> 00:43:16,620 way around it. 700 00:43:18,850 --> 00:43:21,350 Let us now compute the derivative 701 00:43:21,350 --> 00:43:23,050 that we need to have. 702 00:43:23,050 --> 00:43:25,540 First we consider tpij. 703 00:43:25,540 --> 00:43:29,100 tpij was defined as shown right here. 704 00:43:29,100 --> 00:43:33,190 And simply substituting for tf in terms of the deviatoric 705 00:43:33,190 --> 00:43:36,180 stress components and the current yield stress, and 706 00:43:36,180 --> 00:43:40,590 differentiating, we obtain this part here. 707 00:43:40,590 --> 00:43:43,530 Notice t sigma ij is fixed. 708 00:43:43,530 --> 00:43:48,710 Therefore, these are constant here, and of course drop out. 709 00:43:48,710 --> 00:43:53,120 We get this result, and all we have to evaluate the partial 710 00:43:53,120 --> 00:43:57,790 of the yield stress with respect to the plastic strain. 711 00:43:57,790 --> 00:44:03,600 That is achieved by the concept of the plastic work. 712 00:44:03,600 --> 00:44:07,980 The plastic work per unit volume is the amount of energy 713 00:44:07,980 --> 00:44:12,790 that is unrecoverable when the material is unloaded. 714 00:44:12,790 --> 00:44:17,320 This energy, in other words, has been used up to create a 715 00:44:17,320 --> 00:44:19,030 plastic deformation. 716 00:44:19,030 --> 00:44:23,080 What we will do is use this concept for the 717 00:44:23,080 --> 00:44:26,110 one-dimensional case, and then generalize it to the 718 00:44:26,110 --> 00:44:29,150 multidimensional situation. 719 00:44:29,150 --> 00:44:34,480 Pictorally, in the 1D example, in other words, only one 720 00:44:34,480 --> 00:44:38,650 stress component being non-zero, one normal stress 721 00:44:38,650 --> 00:44:42,972 component to be non-zero, we have this stress-strain law. 722 00:44:42,972 --> 00:44:46,630 You're looking at the bilinear case, which 723 00:44:46,630 --> 00:44:48,080 we discussed earlier. 724 00:44:48,080 --> 00:44:51,340 And the plastic work is simply obtained by 725 00:44:51,340 --> 00:44:52,960 this integral here. 726 00:44:52,960 --> 00:44:55,980 Stress times plastic strain. 727 00:44:55,980 --> 00:44:58,300 Integrate it over the whole path. 728 00:44:58,300 --> 00:45:02,680 So this shaded area is equal to the plastic work. 729 00:45:02,680 --> 00:45:07,090 In general, of course, we have the current stress times the 730 00:45:07,090 --> 00:45:09,920 plastic strain and in component forms. 731 00:45:09,920 --> 00:45:11,630 And we are summing over all the components. 732 00:45:14,140 --> 00:45:17,750 If we look at the one-dimensional case, we can 733 00:45:17,750 --> 00:45:22,830 directly express twp as shown here, in terms of the Young's 734 00:45:22,830 --> 00:45:25,880 modulus and the strain-hardening modulus, and 735 00:45:25,880 --> 00:45:32,410 the yield stress at time t, this one right there, and the 736 00:45:32,410 --> 00:45:36,550 initial yield stress, that one right there. 737 00:45:36,550 --> 00:45:39,430 This expression here gives the shaded area. 738 00:45:39,430 --> 00:45:43,290 We can now differentiate and obtain the yield stress 739 00:45:43,290 --> 00:45:46,170 differentiated with respect to the plastic work, this 740 00:45:46,170 --> 00:45:49,460 expression down here. 741 00:45:49,460 --> 00:45:53,990 Using this expression in all our general multidimensional 742 00:45:53,990 --> 00:45:59,420 dimensional case, we directly obtain tpij. 743 00:45:59,420 --> 00:46:03,500 In terms of what's shown here on the right-hand side, notice 744 00:46:03,500 --> 00:46:07,640 we take sigma yield stress at time t with respect to the 745 00:46:07,640 --> 00:46:10,910 plastic work corresponding to time t, and then partial 746 00:46:10,910 --> 00:46:14,740 plastic work with respect to the plastic strains. 747 00:46:14,740 --> 00:46:20,740 Substituting here, we get this bracket for the sigma yield 748 00:46:20,740 --> 00:46:24,970 with respect to the plastic work, and plastic work with 749 00:46:24,970 --> 00:46:27,320 respect to the plastic strains is equal 750 00:46:27,320 --> 00:46:29,800 to the current stress. 751 00:46:29,800 --> 00:46:33,040 And immediately we arrive at this final 752 00:46:33,040 --> 00:46:35,395 relationship here for tpij. 753 00:46:38,510 --> 00:46:42,290 Alternatively, we could have also used the effective stress 754 00:46:42,290 --> 00:46:44,290 in effective plastic strain. 755 00:46:44,290 --> 00:46:47,390 The effective stress is defined as shown here, using 756 00:46:47,390 --> 00:46:51,140 the deviatoric stresses, and the effect of plastic strain 757 00:46:51,140 --> 00:46:52,860 is defined here. 758 00:46:52,860 --> 00:46:55,940 Notice this is a differential increment in the effect of 759 00:46:55,940 --> 00:46:58,200 plastic strain. 760 00:46:58,200 --> 00:47:01,650 And if we use these two quantities, we can also obtain 761 00:47:01,650 --> 00:47:07,740 tpij via this relationship, and the result is of course 762 00:47:07,740 --> 00:47:10,040 the same that I just showed you on the 763 00:47:10,040 --> 00:47:13,050 previous view graph. 764 00:47:13,050 --> 00:47:16,150 Let's now consider tqij, because that is the other 765 00:47:16,150 --> 00:47:17,660 quantity we need. 766 00:47:17,660 --> 00:47:20,560 And here we recognize we have to take the partial of f with 767 00:47:20,560 --> 00:47:23,520 respect to the current stresses with the plastic 768 00:47:23,520 --> 00:47:25,200 strains fixed. 769 00:47:25,200 --> 00:47:31,510 We substitute for f, and we differentiate, recognizing now 770 00:47:31,510 --> 00:47:33,740 with the plastic strains fixed, the 771 00:47:33,740 --> 00:47:36,530 yield stress is a constant. 772 00:47:36,530 --> 00:47:39,640 Therefore, all we need to do is differentiate the product 773 00:47:39,640 --> 00:47:43,860 of the deviatoric stresses, and that is achieved here. 774 00:47:43,860 --> 00:47:48,160 The substitute for the deviatoric stress is shown on 775 00:47:48,160 --> 00:47:55,600 the right-hand side here, and recognize that the t sigma kl 776 00:47:55,600 --> 00:48:00,160 differentiates with respect to t sigma ij is nothing else 777 00:48:00,160 --> 00:48:04,510 than the product of these two Kronecker deltas. 778 00:48:04,510 --> 00:48:07,970 Delta ik, of course being equal to 1 when i is equal to 779 00:48:07,970 --> 00:48:13,820 k and equal to 0 when i is not equal to k. 780 00:48:13,820 --> 00:48:17,800 And we also recognize that this relationship here reduces 781 00:48:17,800 --> 00:48:23,320 directly to that relationship, and the whole right-hand side 782 00:48:23,320 --> 00:48:26,700 then becomes simply the deviatoric stress at time t. 783 00:48:29,300 --> 00:48:35,300 With these relationships for tqij and tpij substituting 784 00:48:35,300 --> 00:48:41,480 into our general derivation of CEP, we directly obtain this 785 00:48:41,480 --> 00:48:45,270 CEP matrix for the three-dimensional case of the 786 00:48:45,270 --> 00:48:49,060 Von Mises flow condition which isotropic strain-hardening. 787 00:48:49,060 --> 00:48:50,610 Let's look at some of these terms. 788 00:48:50,610 --> 00:48:52,950 In front, we have the Young's modulus and the Poisson's 789 00:48:52,950 --> 00:48:54,570 ratio down here. 790 00:48:54,570 --> 00:48:56,880 Young's modulus here, Poisson's ratio here. 791 00:48:56,880 --> 00:49:00,390 Notice here, again, Poisson's ratio, the same way as we have 792 00:49:00,390 --> 00:49:07,450 it in elastic analysis, here we have now the deviatoric 793 00:49:07,450 --> 00:49:09,750 stress component entering. 794 00:49:09,750 --> 00:49:13,890 Similarly here, the deviatoric stress component entering. 795 00:49:13,890 --> 00:49:16,580 And so on. 796 00:49:16,580 --> 00:49:20,250 There's also a factor beta that goes in there, and beta 797 00:49:20,250 --> 00:49:24,830 is defined down here in terms of the yield stress, Young's 798 00:49:24,830 --> 00:49:28,040 modulus and strain-hardening modulus, and Poisson's ratio. 799 00:49:31,790 --> 00:49:34,840 So this is a general expression, and let's now see 800 00:49:34,840 --> 00:49:37,270 how good we use that general expression. 801 00:49:37,270 --> 00:49:42,520 Well, it would enter in the evaluation of the stresses 802 00:49:42,520 --> 00:49:45,410 from time t to time t plus delta t. 803 00:49:45,410 --> 00:49:49,440 We have a stress increment which is added to the current 804 00:49:49,440 --> 00:49:53,690 stress at time t to obtain the stress at time t plus delta t. 805 00:49:53,690 --> 00:49:58,780 This relationship here is expressed as shown here with 806 00:49:58,780 --> 00:49:59,910 the CEP matrix. 807 00:49:59,910 --> 00:50:04,070 In other words, d sigma is given as CEP times de. 808 00:50:04,070 --> 00:50:08,430 Now, this integration has to be performed at every 809 00:50:08,430 --> 00:50:12,700 integration point used in the finite element. 810 00:50:12,700 --> 00:50:16,390 Here we're showing 3 by 3 integration used in an 8-node, 811 00:50:16,390 --> 00:50:18,310 two-dimensional element. 812 00:50:18,310 --> 00:50:21,330 And notice once again that this integration has to be 813 00:50:21,330 --> 00:50:23,620 performed at every one of these integration points, 814 00:50:23,620 --> 00:50:26,840 which of course can be quite expensive. 815 00:50:26,840 --> 00:50:30,480 So we need to perform the integration as effective as 816 00:50:30,480 --> 00:50:33,100 just possible. 817 00:50:33,100 --> 00:50:37,600 One simple scheme to use is Euler forward integration. 818 00:50:37,600 --> 00:50:40,810 And without subimplementation-- 819 00:50:40,810 --> 00:50:43,150 I will explain subimplementation just now-- 820 00:50:43,150 --> 00:50:48,570 without subincrementation, we evaluate the integral on the 821 00:50:48,570 --> 00:50:53,480 left-hand side here by simply saying, let us evaluate CP 822 00:50:53,480 --> 00:50:55,620 corresponding to time t. 823 00:50:55,620 --> 00:50:59,050 Since we know the stress at time t, we know the deviatoric 824 00:50:59,050 --> 00:51:03,830 stress at time t, we can directly evaluate CEP, 825 00:51:03,830 --> 00:51:07,390 entering into that matrix equation that I gave you on 826 00:51:07,390 --> 00:51:09,380 one of the previous view graphs. 827 00:51:09,380 --> 00:51:12,530 You can get CEP at time t, since we know the stress at 828 00:51:12,530 --> 00:51:15,440 time t, and simply multiply it by the total 829 00:51:15,440 --> 00:51:18,920 increment in strain. 830 00:51:18,920 --> 00:51:23,700 Now, this would be OK if the strain increment is very 831 00:51:23,700 --> 00:51:28,350 small, and consequently also, say it leads to a very small 832 00:51:28,350 --> 00:51:29,580 stress increment. 833 00:51:29,580 --> 00:51:36,130 But in general, this linearization involved in 834 00:51:36,130 --> 00:51:40,135 using the Euler forward method can introduce a too-large 835 00:51:40,135 --> 00:51:44,250 error that we don't want to see in the analysis, and 836 00:51:44,250 --> 00:51:48,740 therefore, we use what we call the subincrementation. 837 00:51:48,740 --> 00:51:53,550 Within subincrements, the subincrementation proceeds as 838 00:51:53,550 --> 00:51:57,160 shown this view graph for the evaluation of this integral. 839 00:51:57,160 --> 00:52:04,590 Notice CEP, times t, times delta e the over n, where n is 840 00:52:04,590 --> 00:52:07,310 the number of subincrements, once again. 841 00:52:07,310 --> 00:52:12,910 This means that we now can evaluate the stress at the end 842 00:52:12,910 --> 00:52:15,180 of the first subincrement. 843 00:52:15,180 --> 00:52:18,870 We use that stress, which corresponds to 844 00:52:18,870 --> 00:52:21,240 t plus delta tau-- 845 00:52:21,240 --> 00:52:22,710 tau is delta t over n-- 846 00:52:22,710 --> 00:52:27,080 to evaluate this CEP matrix and multiply the result again 847 00:52:27,080 --> 00:52:30,390 by delta e/n. 848 00:52:30,390 --> 00:52:35,200 Now this then, together, gives us this stress at the end of 849 00:52:35,200 --> 00:52:37,730 the second subincrement. 850 00:52:37,730 --> 00:52:42,780 And like that, we proceed to forward march the stress over 851 00:52:42,780 --> 00:52:49,730 each of the subincrements up to time t plus delta t. 852 00:52:49,730 --> 00:52:53,850 This is nothing else really but the application of the 853 00:52:53,850 --> 00:52:58,370 previous formula, the one on the previous view graph, over 854 00:52:58,370 --> 00:53:01,260 each subincrement, always updating, of course, the 855 00:53:01,260 --> 00:53:06,360 stress that goes into the evaluation of the CEP matrix. 856 00:53:06,360 --> 00:53:10,790 Pictorally, we have the following situation. 857 00:53:10,790 --> 00:53:16,800 Along here, we are plotting time t we have here, time t 858 00:53:16,800 --> 00:53:20,340 here, time t plus delta t there at the end. 859 00:53:20,340 --> 00:53:26,165 Here we have the subincrements 1, 2, 3, et cetera. 860 00:53:26,165 --> 00:53:28,000 You know that by the dot. 861 00:53:28,000 --> 00:53:31,820 And notice at the end of the first subincrement, we have 862 00:53:31,820 --> 00:53:33,840 time t plus delta tau. 863 00:53:33,840 --> 00:53:36,610 At the end of the second subincrement, we have time t 864 00:53:36,610 --> 00:53:39,200 plus 2 delta tau, and so on. 865 00:53:39,200 --> 00:53:42,800 The subimplementation schematically proceeds by 866 00:53:42,800 --> 00:53:46,790 calculating this stress increment so that we have t 867 00:53:46,790 --> 00:53:54,190 plus delta tau sigma, then evaluating a new stress-strain 868 00:53:54,190 --> 00:54:01,000 law, based on this stress, multiplying by the 869 00:54:01,000 --> 00:54:04,430 subincrement in strain, total strain, to 870 00:54:04,430 --> 00:54:06,750 get a stress increment. 871 00:54:06,750 --> 00:54:13,630 Thus you get this stress, and like that, you proceed, 872 00:54:13,630 --> 00:54:17,260 forward marching, again and again, until you have the 873 00:54:17,260 --> 00:54:21,020 final stress at time t plus delta t. 874 00:54:21,020 --> 00:54:24,150 Let us now look at the summary of the procedure the way that 875 00:54:24,150 --> 00:54:27,350 you would use it in a computer program, and that summary is 876 00:54:27,350 --> 00:54:29,950 given on the following view graph. 877 00:54:29,950 --> 00:54:32,755 We assume that the strain corresponding to time t plus 878 00:54:32,755 --> 00:54:34,190 delta t is known. 879 00:54:34,190 --> 00:54:35,930 We have calculated that. 880 00:54:35,930 --> 00:54:39,180 The stress corresponding to time t is also known, and of 881 00:54:39,180 --> 00:54:41,060 course the strain corresponding to time t is 882 00:54:41,060 --> 00:54:42,680 also known. 883 00:54:42,680 --> 00:54:44,720 We then, as a first, 884 00:54:44,720 --> 00:54:47,040 meta-calculate a strain increment. 885 00:54:47,040 --> 00:54:49,830 DELEPS is equal to STRAIN minus EPSILON. 886 00:54:49,830 --> 00:54:52,287 This is a strain increment from time t to time 887 00:54:52,287 --> 00:54:54,790 t plus delta t. 888 00:54:54,790 --> 00:54:57,740 We use that strain increment to calculate a stress 889 00:54:57,740 --> 00:55:00,980 increment, assuming elastic behavior. 890 00:55:00,980 --> 00:55:08,710 CE denotes a matrix of elastic material behavior. 891 00:55:08,710 --> 00:55:13,200 In other words, in this matrix go the Young's modulus and 892 00:55:13,200 --> 00:55:17,600 Poisson's ratio, and of course this one varies depending on 893 00:55:17,600 --> 00:55:20,320 what kind of conditions, plane-stress, plane-strain, 894 00:55:20,320 --> 00:55:22,810 axis-symmetric, or three-dimensional analysis 895 00:55:22,810 --> 00:55:23,570 you're looking at. 896 00:55:23,570 --> 00:55:27,390 But I think you know what I mean by that. 897 00:55:27,390 --> 00:55:33,060 Having delsigma, we calculate tau by updating sigma for that 898 00:55:33,060 --> 00:55:34,930 stress increment. 899 00:55:34,930 --> 00:55:40,170 With this stress now, we check whether the strain was 900 00:55:40,170 --> 00:55:42,280 actually taken elastically. 901 00:55:42,280 --> 00:55:45,070 In other words, whether our assumption that we have made 902 00:55:45,070 --> 00:55:46,690 up here was correct. 903 00:55:46,690 --> 00:55:49,440 And that is done by entering into the yield 904 00:55:49,440 --> 00:55:51,710 condition with that tau. 905 00:55:51,710 --> 00:55:55,640 If f of tau is smaller or equals 0, then the strain 906 00:55:55,640 --> 00:56:00,500 increment is elastic, and in this case, tau was correct, 907 00:56:00,500 --> 00:56:05,810 and we return If that was not the case, then we have to 908 00:56:05,810 --> 00:56:10,300 continue in our solution procedure, 909 00:56:10,300 --> 00:56:13,040 and we do so as follows. 910 00:56:13,040 --> 00:56:17,080 If the previous state of stress was plastic, we set 911 00:56:17,080 --> 00:56:22,060 this ratio to 0 and go to g, because the whole strain, 912 00:56:22,060 --> 00:56:26,070 then, was taken up plastically, 913 00:56:26,070 --> 00:56:27,570 also in this step. 914 00:56:27,570 --> 00:56:34,890 Otherwise, if this condition is not true, we have to find 915 00:56:34,890 --> 00:56:39,510 that stress value at which yielding started. 916 00:56:39,510 --> 00:56:45,030 And we do also by entering into the yield function with 917 00:56:45,030 --> 00:56:48,550 sigma plus ratio times delsigma, where 918 00:56:48,550 --> 00:56:50,610 ratio is now unknown. 919 00:56:50,610 --> 00:56:54,430 Delsigma is known, sigma is known, so we can calculate 920 00:56:54,430 --> 00:56:57,570 from this equation here the value of ratios. 921 00:56:57,570 --> 00:57:04,080 And ratio gives us that stress increment, which brings us to 922 00:57:04,080 --> 00:57:07,770 initiation of yielding. 923 00:57:07,770 --> 00:57:14,160 Having ratio, we update the stress from sigma too tau, a 924 00:57:14,160 --> 00:57:19,710 new value of tau, where ratio goes in here, and we calculate 925 00:57:19,710 --> 00:57:24,250 the elastic-plastic strain increment, using also ratio. 926 00:57:24,250 --> 00:57:28,100 And that is a total elastic-plastic strain 927 00:57:28,100 --> 00:57:29,350 increment here. 928 00:57:31,510 --> 00:57:37,590 This value depth now is used integrate out the stress 929 00:57:37,590 --> 00:57:41,160 occurring during the plastic deformations. 930 00:57:41,160 --> 00:57:45,340 And that is done by the subincrementation process that 931 00:57:45,340 --> 00:57:46,640 we discussed earlier. 932 00:57:46,640 --> 00:57:52,100 We divide depth into subincrements DDEPS, and we 933 00:57:52,100 --> 00:57:57,640 calculate this tau value by cycling through this 934 00:57:57,640 --> 00:58:01,670 subincrementation here, which we described on an earlier 935 00:58:01,670 --> 00:58:02,620 view graph. 936 00:58:02,620 --> 00:58:06,300 And this then completes the calculation of the stress 937 00:58:06,300 --> 00:58:11,760 value corresponding to time t plus t. 938 00:58:11,760 --> 00:58:16,680 So this is a very compact summary of how the stress 939 00:58:16,680 --> 00:58:19,500 integration from time t to time t 940 00:58:19,500 --> 00:58:21,080 plus delta t is performed. 941 00:58:21,080 --> 00:58:24,360 Of course, there are many important details here that we 942 00:58:24,360 --> 00:58:27,900 haven't really looked at in depth, but I think it gives 943 00:58:27,900 --> 00:58:31,630 you a summary that is quite valuable, because it tells, 944 00:58:31,630 --> 00:58:35,660 overall, how the integration is being performed. 945 00:58:35,660 --> 00:58:42,340 Let us now look at two example solutions. 946 00:58:42,340 --> 00:58:45,630 First, an example solution for which we have prepared some 947 00:58:45,630 --> 00:58:49,380 slides, and after that, an example solution for which we 948 00:58:49,380 --> 00:58:52,180 have prepared an animation. 949 00:58:52,180 --> 00:58:56,520 Let me walk over here to share with you the information on 950 00:58:56,520 --> 00:58:57,680 the slides. 951 00:58:57,680 --> 00:59:01,190 And here we consider the following problem. 952 00:59:01,190 --> 00:59:06,780 You have a rigid punch here subjected to a load, lying on 953 00:59:06,780 --> 00:59:08,030 an infinite half-space. 954 00:59:10,460 --> 00:59:16,650 As the load increases, plasticity develops down here, 955 00:59:16,650 --> 00:59:20,310 and there is an ultimate load, a limit load, 956 00:59:20,310 --> 00:59:22,410 that we want to calculate. 957 00:59:22,410 --> 00:59:26,730 In other words, there's a load here at which there will be 958 00:59:26,730 --> 00:59:30,330 very large deformations down here, and the material can 959 00:59:30,330 --> 00:59:32,520 take no higher load. 960 00:59:32,520 --> 00:59:36,320 This plane strain punch problem was considered by many 961 00:59:36,320 --> 00:59:41,330 investigators, but Hill actually predicted the plastic 962 00:59:41,330 --> 00:59:45,200 zones, and also the ultimate limit load. 963 00:59:45,200 --> 00:59:48,830 And we want to compare our finite element solution with 964 00:59:48,830 --> 00:59:51,040 the solution given by Hill. 965 00:59:51,040 --> 00:59:56,240 On the next slide now, we see the finite element mesh used. 966 00:59:56,240 --> 01:00:03,090 Notice we are only idealizing 1/2 of the domain, as shown by 967 01:00:03,090 --> 01:00:05,220 this boundary here. 968 01:00:05,220 --> 01:00:10,420 Notice also that we have moved this boundary here and that 969 01:00:10,420 --> 01:00:15,610 boundary here far enough away from the load application. 970 01:00:15,610 --> 01:00:18,650 Notice that we have a fine mesh here, 971 01:00:18,650 --> 01:00:21,230 a coarse mesh there. 972 01:00:21,230 --> 01:00:24,450 In the fine mesh, we use 8-node elements, in the coarse 973 01:00:24,450 --> 01:00:26,450 mesh, we use 4-node elements. 974 01:00:26,450 --> 01:00:32,150 And there is a transition region here, there, from the 975 01:00:32,150 --> 01:00:39,390 8-node over to 7-node, into 6-node, and then finally into 976 01:00:39,390 --> 01:00:41,980 4-node elements. 977 01:00:41,980 --> 01:00:44,680 It's an interesting small point here. 978 01:00:44,680 --> 01:00:49,670 There is an incompatibility here between this element and 979 01:00:49,670 --> 01:00:51,370 these two elements. 980 01:00:51,370 --> 01:00:55,030 An incompatibility because this element here, being a 981 01:00:55,030 --> 01:00:58,140 6-node element, has a parabolic displacement 982 01:00:58,140 --> 01:01:02,810 distribution along this line, whereas this element here, 983 01:01:02,810 --> 01:01:08,190 being a 7-node element and a linear assumption along this 984 01:01:08,190 --> 01:01:11,690 line here for displacement, only admits a linear 985 01:01:11,690 --> 01:01:13,070 displacement distribution. 986 01:01:13,070 --> 01:01:16,390 So there is small incompatibility here, and 987 01:01:16,390 --> 01:01:17,510 similarly here. 988 01:01:17,510 --> 01:01:22,580 However, that incompatibility we deem to be small, that it 989 01:01:22,580 --> 01:01:26,280 would not affect the results very considerably. 990 01:01:26,280 --> 01:01:30,540 Now, in an elastoplastic plastic, analysis like in any 991 01:01:30,540 --> 01:01:33,500 non-linear analysis that you would like to perform, I 992 01:01:33,500 --> 01:01:35,820 pointed out in an earlier lecture, it is always 993 01:01:35,820 --> 01:01:40,150 important to start with the linear solutions, just to see 994 01:01:40,150 --> 01:01:45,970 how well your mesh performs and whether you have any data 995 01:01:45,970 --> 01:01:47,470 input errors, possibly, even. 996 01:01:47,470 --> 01:01:51,710 It is good to perform first a linear solution where you can 997 01:01:51,710 --> 01:01:56,240 compare with analytical or other results. 998 01:01:56,240 --> 01:01:58,840 This is what we have done here, and let me show you on 999 01:01:58,840 --> 01:02:01,850 the next slide what we have considered. 1000 01:02:01,850 --> 01:02:07,710 We have applied a point load here to the half space, used 1001 01:02:07,710 --> 01:02:12,880 the same finite element mesh that I just showed you, and 1002 01:02:12,880 --> 01:02:14,350 looked at the stresses. 1003 01:02:14,350 --> 01:02:21,910 At this depth A, below the surface, it's 0.697 inches. 1004 01:02:21,910 --> 01:02:25,290 At that depth, we're looking at the stresses along this 1005 01:02:25,290 --> 01:02:30,780 line here on a horizontal line. 1006 01:02:30,780 --> 01:02:36,030 These stresses here are plotted on this graph for 1007 01:02:36,030 --> 01:02:37,930 different components. 1008 01:02:37,930 --> 01:02:43,120 Sigma zz, Sigma yy, tau yz. 1009 01:02:43,120 --> 01:02:46,020 And they are compared with the theoretical solution, the 1010 01:02:46,020 --> 01:02:47,100 [UNINTELLIGIBLE] 1011 01:02:47,100 --> 01:02:49,200 theoretical solution. 1012 01:02:49,200 --> 01:02:53,610 Notice excellent comparison, obtained really for the stress 1013 01:02:53,610 --> 01:02:58,510 components, and this means that our mesh is quite 1014 01:02:58,510 --> 01:03:02,700 reasonable, and we can proceed with the 1015 01:03:02,700 --> 01:03:04,600 elastoplastic analysis. 1016 01:03:04,600 --> 01:03:06,790 Notice please that we have used here two-point 1017 01:03:06,790 --> 01:03:09,870 integration, 2-by-2 integration, for all the 1018 01:03:09,870 --> 01:03:12,940 elements, including the 8-node element. 1019 01:03:12,940 --> 01:03:16,600 We then performed the same analysis, also one, using 1020 01:03:16,600 --> 01:03:19,820 three-point integration, and these results are shown on 1021 01:03:19,820 --> 01:03:21,370 this slide. 1022 01:03:21,370 --> 01:03:25,160 Same picture here. 1023 01:03:25,160 --> 01:03:29,530 Here stress is as a function of the distance from the 1024 01:03:29,530 --> 01:03:31,460 center line in inches. 1025 01:03:31,460 --> 01:03:34,910 Notice with three-point integration, also excellent 1026 01:03:34,910 --> 01:03:38,370 comparison for the stress components. 1027 01:03:38,370 --> 01:03:42,800 On the next slide now, you'll see the elastoplastic results. 1028 01:03:42,800 --> 01:03:47,420 And here we see the applied load, normalized by a factor 1029 01:03:47,420 --> 01:03:51,380 plotted versus the displacement of the punch, 1030 01:03:51,380 --> 01:03:53,620 also normalized by factor. 1031 01:03:53,620 --> 01:03:56,530 And we see two curves here, one corresponding to 1032 01:03:56,530 --> 01:03:59,910 three-point integration and one corresponding to two-point 1033 01:03:59,910 --> 01:04:01,500 integration. 1034 01:04:01,500 --> 01:04:03,530 Now notice that the three-point integration 1035 01:04:03,530 --> 01:04:07,940 results are higher than those of the 1036 01:04:07,940 --> 01:04:09,760 two-point integration results. 1037 01:04:09,760 --> 01:04:12,090 And that's a very interesting phenomenon. 1038 01:04:12,090 --> 01:04:17,350 It is due to the fact that the three-point integration puts a 1039 01:04:17,350 --> 01:04:20,980 higher incompressibility constraint into the mesh. 1040 01:04:20,980 --> 01:04:24,220 In fact, a too-high incompressibility constraint.. 1041 01:04:24,220 --> 01:04:26,910 This is a very interesting phenomenon, but it is best 1042 01:04:26,910 --> 01:04:29,190 discussed in another lecture. 1043 01:04:29,190 --> 01:04:34,640 Notice that however our solution results all give us a 1044 01:04:34,640 --> 01:04:37,760 horizontal line here for the two-point integration and 1045 01:04:37,760 --> 01:04:41,730 three-point integration, and certainly, the two-point 1046 01:04:41,730 --> 01:04:43,700 integration results are quite close to 1047 01:04:43,700 --> 01:04:45,190 the theoretical value. 1048 01:04:45,190 --> 01:04:48,630 If we were to use the finer mesh and still use three-point 1049 01:04:48,630 --> 01:04:51,070 integration, we would get also closer to 1050 01:04:51,070 --> 01:04:53,730 the theoretical value. 1051 01:04:53,730 --> 01:04:56,950 Of course, we have here, I should have pointed out, an 1052 01:04:56,950 --> 01:05:01,140 elastic, perfectly plastic material assumption for the 1053 01:05:01,140 --> 01:05:03,320 material under the footing. 1054 01:05:03,320 --> 01:05:07,430 Please refer to the details of this example solution given in 1055 01:05:07,430 --> 01:05:12,370 the paper in which we have reported this sample solution. 1056 01:05:12,370 --> 01:05:14,980 And the reference to that paper is given 1057 01:05:14,980 --> 01:05:17,250 in the study guide. 1058 01:05:17,250 --> 01:05:22,780 The second problem that I'd like to discuss with you 1059 01:05:22,780 --> 01:05:26,590 concerns the analysis of a plate with a hole that we are 1060 01:05:26,590 --> 01:05:31,760 subjecting to very large tensile forces. 1061 01:05:31,760 --> 01:05:35,380 And there is one interesting feature that we observe in 1062 01:05:35,380 --> 01:05:40,340 that analysis which we have not seen here on the slides, 1063 01:05:40,340 --> 01:05:43,820 namely, the spread of plasticity through the domain. 1064 01:05:43,820 --> 01:05:50,010 So let me introduce you to this problem using just two 1065 01:05:50,010 --> 01:05:55,070 view graphs, and then show you the animation which gives the 1066 01:05:55,070 --> 01:05:56,270 results to the problem. 1067 01:05:56,270 --> 01:06:01,210 Here we have the plate with a hole. 1068 01:06:01,210 --> 01:06:04,900 It's subjected to tensile forces, so you can see. 1069 01:06:04,900 --> 01:06:07,440 And we want to calculate, really, the ultimate limit 1070 01:06:07,440 --> 01:06:11,470 load of the plate, and we want to see now the spread of 1071 01:06:11,470 --> 01:06:12,780 plasticity. 1072 01:06:12,780 --> 01:06:17,600 Of course, that spread of plasticity also occurs in the 1073 01:06:17,600 --> 01:06:23,500 analysis of the rigid punch on an elastoplastic footing, but 1074 01:06:23,500 --> 01:06:25,810 I did not show you that spread of placiticity there. 1075 01:06:25,810 --> 01:06:27,260 I'd like to show you the spread the 1076 01:06:27,260 --> 01:06:29,090 plasticity now here. 1077 01:06:29,090 --> 01:06:32,160 And you will see that the plasticity actually spread 1078 01:06:32,160 --> 01:06:35,310 through the elements, integration point by 1079 01:06:35,310 --> 01:06:38,370 integration point. 1080 01:06:38,370 --> 01:06:42,530 And of course initially, as a load increase is not so rapid, 1081 01:06:42,530 --> 01:06:45,830 but then all of a sudden you see a very rapid spread of the 1082 01:06:45,830 --> 01:06:50,700 plasticity, right through this domain here, that domain 1083 01:06:50,700 --> 01:06:53,280 there, and so on, by symmetry conditions. 1084 01:06:53,280 --> 01:06:56,330 Now, because of symmetry conditions, we also will look 1085 01:06:56,330 --> 01:06:58,690 at only one quarter of the plate-- 1086 01:06:58,690 --> 01:07:01,310 this quarter right here. 1087 01:07:01,310 --> 01:07:09,270 The material assumption that is used in the analysis is 1088 01:07:09,270 --> 01:07:10,850 shown on this view graph. 1089 01:07:10,850 --> 01:07:14,180 Stress here, strain there. 1090 01:07:14,180 --> 01:07:17,900 Notice Young's modulus, as shown here, Poisson ratio 1091 01:07:17,900 --> 01:07:20,520 there, and the strain-hardening modulus is 1092 01:07:20,520 --> 01:07:21,450 shown here. 1093 01:07:21,450 --> 01:07:23,070 We assume isotropic hardening. 1094 01:07:25,660 --> 01:07:29,660 The same problem would also be considered further in a later 1095 01:07:29,660 --> 01:07:34,300 lecture, when we actually use the ADINA program to show an 1096 01:07:34,300 --> 01:07:35,460 application. 1097 01:07:35,460 --> 01:07:38,580 So you will encounter this problem there again. 1098 01:07:38,580 --> 01:07:41,860 Notice that in this particular analysis that you will be 1099 01:07:41,860 --> 01:07:46,140 seeing now, in the animation, we assumed a materially 1100 01:07:46,140 --> 01:07:48,310 non-linear only condition. 1101 01:07:48,310 --> 01:07:51,500 In other words, we assumed infinitesimally small 1102 01:07:51,500 --> 01:07:53,840 displacement and strains. 1103 01:07:53,840 --> 01:07:58,440 However, in the very last load steps, just before the plate 1104 01:07:58,440 --> 01:08:04,030 ruptures, we encounter, actually, larger strains, but 1105 01:08:04,030 --> 01:08:06,950 only those last load steps. 1106 01:08:06,950 --> 01:08:12,080 And in order to capture this very last behavior of the 1107 01:08:12,080 --> 01:08:16,229 plate very accurately, one would have to perform a large 1108 01:08:16,229 --> 01:08:21,250 strain analysis of the plate, which we did not do here, and 1109 01:08:21,250 --> 01:08:24,340 which really would be the object off 1110 01:08:24,340 --> 01:08:26,410 another lecture, as well. 1111 01:08:26,410 --> 01:08:30,270 So let's look now at the response of the plate. 1112 01:08:30,270 --> 01:08:35,180 Yes, it is subjected to the tensile forces and how the 1113 01:08:35,180 --> 01:08:39,779 plasticity spread through the plate until final rupture. 1114 01:08:39,779 --> 01:08:41,630 Let's look at the animation. 1115 01:08:41,630 --> 01:08:45,720 We will look first at a fast animation, where the response 1116 01:08:45,720 --> 01:08:48,899 occurs very fast, and then we'll look at the same 1117 01:08:48,899 --> 01:08:52,870 animation again but slower, so that I can better talk about 1118 01:08:52,870 --> 01:08:55,600 it while you're watching it, and you can better understand 1119 01:08:55,600 --> 01:08:58,660 what I'm saying. 1120 01:08:58,660 --> 01:09:02,420 Here we see one quarter of the plate discretized using 288 1121 01:09:02,420 --> 01:09:05,670 eight-node isoparametric plane stress elements. 1122 01:09:05,670 --> 01:09:07,890 The time course gives the load step. 1123 01:09:07,890 --> 01:09:10,920 We perform the static analysis and the load corresponding to 1124 01:09:10,920 --> 01:09:13,660 each solution step is also shown. 1125 01:09:13,660 --> 01:09:16,479 Here you see now the load increasing, and in the first 1126 01:09:16,479 --> 01:09:18,254 load step, the plate remains elastic. 1127 01:09:24,859 --> 01:09:28,069 Plasticity then developed as a whole and spreads rapidly 1128 01:09:28,069 --> 01:09:29,799 through the plate as the load increases. 1129 01:09:38,109 --> 01:09:40,010 The deformations of the plate are given here 1130 01:09:40,010 --> 01:09:42,729 to a magnified scale. 1131 01:09:42,729 --> 01:09:45,910 This was a rather fast look at the plate behavior, so let's 1132 01:09:45,910 --> 01:09:49,750 look at it again and over a somewhat longer time scale. 1133 01:09:49,750 --> 01:09:52,120 Here we have again the plate at time 0-- 1134 01:09:52,120 --> 01:09:55,300 that is at zero-load application. 1135 01:09:55,300 --> 01:09:57,790 Now the load application starts, and initially, the 1136 01:09:57,790 --> 01:09:59,040 plate remains elastic. 1137 01:10:23,470 --> 01:10:25,760 Be sure that an element integration point has gone 1138 01:10:25,760 --> 01:10:28,200 plastic by darkening the regions around it. 1139 01:10:37,900 --> 01:10:41,560 Since we use 3-by-3 Gauss integration, considering an 1140 01:10:41,560 --> 01:10:44,780 element on an average 1/9 of the element [? area ?] 1141 01:10:44,780 --> 01:10:47,070 corresponds to an integration point. 1142 01:10:50,060 --> 01:10:52,130 Here you can see how the plasticity spreads from 1143 01:10:52,130 --> 01:10:54,670 element to element, and indeed, from integration point 1144 01:10:54,670 --> 01:10:57,610 to integration point. 1145 01:10:57,610 --> 01:10:59,400 We will discuss the analysis of the plate 1146 01:10:59,400 --> 01:11:01,510 again in lecture 22. 1147 01:11:01,510 --> 01:11:04,700 We will also perform analyses and study the results, 1148 01:11:04,700 --> 01:11:08,510 including large displacement and large strain effects. 1149 01:11:08,510 --> 01:11:11,890 Once again, the deformations of the plate are shown here in 1150 01:11:11,890 --> 01:11:14,410 the animation to a magnified scale. 1151 01:11:14,410 --> 01:11:16,865 The actual displacements are discussed in lecture 22. 1152 01:11:20,690 --> 01:11:23,420 So this brings me then to the end of this lecture. 1153 01:11:23,420 --> 01:11:25,580 I discussed everything I wanted to discuss with you in 1154 01:11:25,580 --> 01:11:26,300 this lecture. 1155 01:11:26,300 --> 01:11:27,920 Thank you very much for your attention.