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 00:00:26,240 --> 00:00:27,940 PROFESSOR: Ladies and gentlemen, welcome to this 9 00:00:27,940 --> 00:00:30,617 course on nonlinear finite element analysis of solids and 10 00:00:30,617 --> 00:00:31,870 structures. 11 00:00:31,870 --> 00:00:34,600 A few years ago, we produced at MIT a course on the linear 12 00:00:34,600 --> 00:00:36,450 analysis of solids and structures. 13 00:00:36,450 --> 00:00:38,320 That course was quite well-received. 14 00:00:38,320 --> 00:00:40,590 And we obtained a number of requests to also produce a 15 00:00:40,590 --> 00:00:42,290 course on the nonlinear analysis of solids and 16 00:00:42,290 --> 00:00:43,330 structures. 17 00:00:43,330 --> 00:00:46,740 The present course is the answer to those requests. 18 00:00:46,740 --> 00:00:51,810 In this course, we will be using my book as a textbook. 19 00:00:51,810 --> 00:00:53,950 You might be already a bit familiar with this book. 20 00:00:53,950 --> 00:00:56,740 We will be referring to some of the sections 21 00:00:56,740 --> 00:00:58,830 in quite some detail. 22 00:00:58,830 --> 00:01:02,870 In addition, of course, you also have the study guide. 23 00:01:02,870 --> 00:01:06,630 Let me now share with you, briefly, some thoughts that I 24 00:01:06,630 --> 00:01:10,090 had while designing this course. 25 00:01:10,090 --> 00:01:13,070 The field of nonlinear, finite element analysis is a very 26 00:01:13,070 --> 00:01:14,780 large field. 27 00:01:14,780 --> 00:01:18,430 And, in fact, four large fields come to mind as feeding 28 00:01:18,430 --> 00:01:23,260 into nonlinear, finite element methods, continuum mechanics, 29 00:01:23,260 --> 00:01:26,540 finite element discretizations, numerical 30 00:01:26,540 --> 00:01:30,270 algorithms, software considerations. 31 00:01:30,270 --> 00:01:32,110 Because the field of nonlinear, finite element 32 00:01:32,110 --> 00:01:35,690 analysis is such a large field, I had to select certain 33 00:01:35,690 --> 00:01:40,140 topics as the topics of this finite series of lectures. 34 00:01:40,140 --> 00:01:42,880 I believe that the lectures provide a good introduction 35 00:01:42,880 --> 00:01:46,240 and foundation to a nonlinear, finite element analysis. 36 00:01:46,240 --> 00:01:48,890 Of course, the lectures cannot answer all the questions that 37 00:01:48,890 --> 00:01:50,250 you might have regarding nonlinear, 38 00:01:50,250 --> 00:01:51,910 finite element analysis. 39 00:01:51,910 --> 00:01:54,060 I believe, though, that they will be answering some of the 40 00:01:54,060 --> 00:01:57,510 questions and, hopefully, stimulate discussions that you 41 00:01:57,510 --> 00:02:00,400 might have while watching these lectures with your 42 00:02:00,400 --> 00:02:01,580 colleagues. 43 00:02:01,580 --> 00:02:04,220 We, at MIT, continue to work in nonlinear, 44 00:02:04,220 --> 00:02:05,610 finite element analysis. 45 00:02:05,610 --> 00:02:09,300 And we also offer, from time to time, short courses. 46 00:02:09,300 --> 00:02:12,050 It would be very nice to meet you at one of these short 47 00:02:12,050 --> 00:02:14,950 courses and to discuss any questions that you might have 48 00:02:14,950 --> 00:02:17,850 regarding these lectures, once you have seen them, and 49 00:02:17,850 --> 00:02:19,770 regarding your work on nonlinear, finite element 50 00:02:19,770 --> 00:02:22,420 analysis in general. 51 00:02:22,420 --> 00:02:24,570 Let me now return to what I like to talk 52 00:02:24,570 --> 00:02:26,040 about in this course. 53 00:02:26,040 --> 00:02:29,170 In this course, we want to concentrate on methods that 54 00:02:29,170 --> 00:02:34,710 are generally applicable, modern techniques and 55 00:02:34,710 --> 00:02:37,010 practical procedures. 56 00:02:37,010 --> 00:02:40,110 I believe it is important that we discuss in this course 57 00:02:40,110 --> 00:02:42,670 practical and effective procedures. 58 00:02:42,670 --> 00:02:46,070 In particular, methods that are or are now becoming an 59 00:02:46,070 --> 00:02:48,780 integral part of computer-aided design, 60 00:02:48,780 --> 00:02:51,180 computer-aided engineering software. 61 00:02:51,180 --> 00:02:54,090 In this course, I would like to discuss with you geometric 62 00:02:54,090 --> 00:02:56,980 and material nonlinear analysis, static and dynamic 63 00:02:56,980 --> 00:03:01,100 solutions, basic principles and their use, and share with 64 00:03:01,100 --> 00:03:02,950 you example solutions. 65 00:03:02,950 --> 00:03:05,280 I believe that the course will be of interest in many 66 00:03:05,280 --> 00:03:07,720 branches of engineering throughout the world. 67 00:03:07,720 --> 00:03:09,700 And in this spirit, we have designed a 68 00:03:09,700 --> 00:03:13,460 logo for this course. 69 00:03:13,460 --> 00:03:16,050 The logo is shown here. 70 00:03:16,050 --> 00:03:17,780 It shows the whole world as an 71 00:03:17,780 --> 00:03:20,480 assemblage off finite elements. 72 00:03:20,480 --> 00:03:23,380 If you're watching me from Chicago, you're watching me 73 00:03:23,380 --> 00:03:24,960 from about here. 74 00:03:24,960 --> 00:03:26,700 If you're watching from Munich, you're 75 00:03:26,700 --> 00:03:28,560 watching about there. 76 00:03:28,560 --> 00:03:30,920 And if you are in Tokyo, you're watching 77 00:03:30,920 --> 00:03:32,270 me from about there. 78 00:03:32,270 --> 00:03:34,940 Welcome to this course. 79 00:03:34,940 --> 00:03:38,750 I'd like to now start with the first lecture, which has the 80 00:03:38,750 --> 00:03:42,020 title Introduction to Nonlinear Analysis. 81 00:03:42,020 --> 00:03:46,100 And let me take off my jacket, because we have a lot of work 82 00:03:46,100 --> 00:03:49,200 ahead, and summarize to you what I would like to present 83 00:03:49,200 --> 00:03:51,490 to you in this first lecture. 84 00:03:51,490 --> 00:03:55,250 We discuss, first, some introductory view graphs and 85 00:03:55,250 --> 00:03:57,600 show some short movies. 86 00:03:57,600 --> 00:04:01,670 We then classify nonlinear analyses. 87 00:04:01,670 --> 00:04:04,295 We then discuss the basic approach of 88 00:04:04,295 --> 00:04:05,990 an incremental solution. 89 00:04:05,990 --> 00:04:09,620 And we share some example solutions. 90 00:04:09,620 --> 00:04:12,810 Let me walk over to my view graphs which I've prepared for 91 00:04:12,810 --> 00:04:16,339 this lecture so that we can start with a discussion of 92 00:04:16,339 --> 00:04:18,370 this material. 93 00:04:18,370 --> 00:04:21,279 Finite element nonlinear analysis in engineering 94 00:04:21,279 --> 00:04:26,730 mechanics can be an art, but it can also be a frustration. 95 00:04:26,730 --> 00:04:30,670 For those of you who have been doing some nonlinear analysis 96 00:04:30,670 --> 00:04:34,740 already, I think you will value that it can be an art 97 00:04:34,740 --> 00:04:38,280 and it can be a frustration because it can be a very 98 00:04:38,280 --> 00:04:39,700 difficult matter. 99 00:04:39,700 --> 00:04:43,070 But it's always provides a great challenge. 100 00:04:43,070 --> 00:04:46,170 And that, of course, is the exciting part of working in 101 00:04:46,170 --> 00:04:49,690 nonlinear, finite element analysis. 102 00:04:49,690 --> 00:04:53,930 Some important engineering phenomena can only be assessed 103 00:04:53,930 --> 00:04:56,920 using nonlinear analysis techniques. 104 00:04:56,920 --> 00:05:00,050 For example, the collapse or buckling of structures due to 105 00:05:00,050 --> 00:05:01,430 sudden overloads-- 106 00:05:01,430 --> 00:05:04,340 I'm thinking here, typically of shell structures, say-- 107 00:05:04,340 --> 00:05:07,010 the progressive damage behavior due to long-lasting 108 00:05:07,010 --> 00:05:10,400 severe loads such as, for example, high-temperature 109 00:05:10,400 --> 00:05:14,170 loads in nuclear reactor components and, for certain 110 00:05:14,170 --> 00:05:18,140 structures such as cables and transmission towers, nonlinear 111 00:05:18,140 --> 00:05:22,140 phenomena need be included in the analysis, even for service 112 00:05:22,140 --> 00:05:23,950 load calculations. 113 00:05:23,950 --> 00:05:25,970 The need for nonlinear analysis has certainly 114 00:05:25,970 --> 00:05:29,530 increased in recent years due to the need for the use of 115 00:05:29,530 --> 00:05:32,450 optimized structures, the use of new materials that are 116 00:05:32,450 --> 00:05:35,380 being introduced or have been introduced over the recent 117 00:05:35,380 --> 00:05:38,750 years, the addressing of safety-related issues of 118 00:05:38,750 --> 00:05:40,020 structures. 119 00:05:40,020 --> 00:05:44,270 There's more rigorous attention being given to such 120 00:05:44,270 --> 00:05:47,960 safety-related issues now and certain the corresponding 121 00:05:47,960 --> 00:05:50,660 benefits can be most important. 122 00:05:50,660 --> 00:05:54,380 Problems that are addressed by nonlinear, finite element 123 00:05:54,380 --> 00:05:58,000 analysis can be found in many branches of engineering. 124 00:05:58,000 --> 00:06:01,780 And I've listed here some such branches, nuclear engineering, 125 00:06:01,780 --> 00:06:05,290 earthquake engineering, the automobile industries, defense 126 00:06:05,290 --> 00:06:08,090 industries, in aeronautical engineering, mining 127 00:06:08,090 --> 00:06:11,560 industries, offshore engineering and so on. 128 00:06:11,560 --> 00:06:17,090 I'd like to now show you some movies that we have assembled 129 00:06:17,090 --> 00:06:21,810 regarding some finite element analysis in certain fields. 130 00:06:21,810 --> 00:06:24,580 And let us start with a movie related to the defense 131 00:06:24,580 --> 00:06:30,840 industries, then go on to a movie related to some work in 132 00:06:30,840 --> 00:06:35,530 the automobile industries, and then look at a movie related 133 00:06:35,530 --> 00:06:38,820 to some work in earthquake engineering. 134 00:06:38,820 --> 00:06:43,690 And finally, I'd like to also show you an interesting movie 135 00:06:43,690 --> 00:06:46,310 regarding a structural engineering problem. 136 00:06:46,310 --> 00:06:50,540 So let me now introduce you to the first movie, which is a 137 00:06:50,540 --> 00:06:55,220 movie in which a tank is modeled using a geometric 138 00:06:55,220 --> 00:06:58,480 model first and then setting up a finite element mesh. 139 00:07:01,290 --> 00:07:03,290 Here we see a tank in a maneuver. 140 00:07:03,290 --> 00:07:06,510 You surely have seen such structure before. 141 00:07:06,510 --> 00:07:08,900 And here you see, in the top part of the picture, the 142 00:07:08,900 --> 00:07:11,980 geometric model of a tank and, in the bottom part of the 143 00:07:11,980 --> 00:07:14,850 picture, a finite element discretization. 144 00:07:14,850 --> 00:07:18,380 The top model was generated using a geometric modeler. 145 00:07:18,380 --> 00:07:20,700 And the bottom model was obtained using a finite 146 00:07:20,700 --> 00:07:22,900 element mesh generator. 147 00:07:22,900 --> 00:07:25,660 These models were constructed by Structural Dynamics 148 00:07:25,660 --> 00:07:26,910 Research Corporation. 149 00:07:29,460 --> 00:07:32,230 The second movie relates to a problem that has obtained much 150 00:07:32,230 --> 00:07:35,970 attention in the automobile industry, namely the problem 151 00:07:35,970 --> 00:07:39,130 of what happens when a motor car crashes against 152 00:07:39,130 --> 00:07:40,130 another motor car. 153 00:07:40,130 --> 00:07:44,090 And tests have been performed of motor cars crashing against 154 00:07:44,090 --> 00:07:45,170 rigid walls. 155 00:07:45,170 --> 00:07:46,490 Test results have been assembled. 156 00:07:46,490 --> 00:07:49,740 And finite element methods may well be used in the analysis 157 00:07:49,740 --> 00:07:52,950 of such types of problems. 158 00:07:52,950 --> 00:07:56,520 Here we see a test performed by the Ford Motor Company of a 159 00:07:56,520 --> 00:07:59,250 car crashing against a rigid wall. 160 00:07:59,250 --> 00:08:02,260 And here, a close-up of what happens to the passengers not 161 00:08:02,260 --> 00:08:08,690 wearing seat belts, and another such test. 162 00:08:08,690 --> 00:08:10,660 Here now, in more detail, what happens to the 163 00:08:10,660 --> 00:08:14,570 front part of the car. 164 00:08:14,570 --> 00:08:17,140 There is since some time much interest in modeling these 165 00:08:17,140 --> 00:08:18,930 phenomena on the computer. 166 00:08:18,930 --> 00:08:21,480 And finite element methods can be applied and are very 167 00:08:21,480 --> 00:08:22,730 valuable here. 168 00:08:25,210 --> 00:08:27,250 This certain movie relates to the use of finite element 169 00:08:27,250 --> 00:08:28,950 methods in earthquake engineering. 170 00:08:28,950 --> 00:08:32,120 Here we look at the finite element model of a tank and 171 00:08:32,120 --> 00:08:35,870 the dynamic motions of that model. 172 00:08:35,870 --> 00:08:38,960 We are surely all aware that earthquake motions can cause 173 00:08:38,960 --> 00:08:41,940 severe structural vibrations and possibly collapse of 174 00:08:41,940 --> 00:08:43,570 structures. 175 00:08:43,570 --> 00:08:47,730 Here we see a transformer and the tank to be modeled on it. 176 00:08:47,730 --> 00:08:50,940 Here a close-up view of the tank. 177 00:08:50,940 --> 00:08:53,640 The finite element model of the tank was constructed by 178 00:08:53,640 --> 00:08:55,930 Asea, a Swedish company. 179 00:08:55,930 --> 00:08:58,620 And here we see some vibratory motions of the model. 180 00:09:02,650 --> 00:09:05,870 The fourth movie shows the dynamic response and collapse 181 00:09:05,870 --> 00:09:10,580 of the Tacoma Narrows bridge in 1940. 182 00:09:10,580 --> 00:09:14,690 The Tacoma Narrows bridge collapsed on November 7, 1940, 183 00:09:14,690 --> 00:09:18,530 about four months after its opening in winds of 40 to 45 184 00:09:18,530 --> 00:09:20,640 miles per hour. 185 00:09:20,640 --> 00:09:23,090 Here you see the dynamic motions of the bridge prior to 186 00:09:23,090 --> 00:09:26,680 its collapse, a side view, and now a view 187 00:09:26,680 --> 00:09:29,050 along its center line. 188 00:09:29,050 --> 00:09:32,610 Notice the high torsion of vibrations of the bridge. 189 00:09:32,610 --> 00:09:35,130 The bridge went through large dynamic motions for hours 190 00:09:35,130 --> 00:09:37,100 until its collapse. 191 00:09:37,100 --> 00:09:40,870 We only show a very small segment of that time. 192 00:09:40,870 --> 00:09:42,565 And here you see how the bridge collapsed. 193 00:09:53,600 --> 00:09:56,140 Of course, these movies only indicate, to some extent, 194 00:09:56,140 --> 00:09:58,260 where finite element methods might be applied in 195 00:09:58,260 --> 00:10:01,520 engineering practice, but I thought you might like, you 196 00:10:01,520 --> 00:10:03,700 might enjoy, seeing the movies. 197 00:10:03,700 --> 00:10:07,150 Let me now continue with the material that I have 198 00:10:07,150 --> 00:10:09,930 documented on the view graphs. 199 00:10:09,930 --> 00:10:13,760 For an effective nonlinear analysis, a good physical and 200 00:10:13,760 --> 00:10:16,840 theoretical understanding is most important. 201 00:10:16,840 --> 00:10:20,140 You want to have some good physical insight in the 202 00:10:20,140 --> 00:10:23,260 problem, setup, and mathematical formulation of 203 00:10:23,260 --> 00:10:24,360 finite element model. 204 00:10:24,360 --> 00:10:26,950 Solve that model, and that will enrich 205 00:10:26,950 --> 00:10:28,090 your physical insight. 206 00:10:28,090 --> 00:10:31,880 It is this interaction and mutual enrichment between the 207 00:10:31,880 --> 00:10:34,710 physical insight and mathematical formulation that 208 00:10:34,710 --> 00:10:37,010 can be most valuable. 209 00:10:37,010 --> 00:10:39,630 The best approach for a nonlinear finite element 210 00:10:39,630 --> 00:10:42,870 analysis is to use reliable and generally applicable 211 00:10:42,870 --> 00:10:45,150 finite elements. 212 00:10:45,150 --> 00:10:47,950 With such methods, we can establish models that we can 213 00:10:47,950 --> 00:10:51,570 understand, that we have confidence in. 214 00:10:51,570 --> 00:10:54,140 We start with simple models of nature. 215 00:10:54,140 --> 00:10:57,620 And we find these as need arises. 216 00:10:57,620 --> 00:11:00,880 In fact, I like to think of an engineer as developing a first 217 00:11:00,880 --> 00:11:03,490 model on the back of an envelope using, of course, 218 00:11:03,490 --> 00:11:04,840 simple equations. 219 00:11:04,840 --> 00:11:07,570 And these equations will give some insight into the 220 00:11:07,570 --> 00:11:09,230 structural response. 221 00:11:09,230 --> 00:11:11,890 If necessary, then the finite element model is set up. 222 00:11:11,890 --> 00:11:13,610 The first finite element model is set up. 223 00:11:13,610 --> 00:11:15,280 And this finite element model is then 224 00:11:15,280 --> 00:11:18,440 refined as need arises. 225 00:11:18,440 --> 00:11:21,630 To perform a nonlinear analysis, we want to, 226 00:11:21,630 --> 00:11:24,900 altogether, stay then with simple, relatively small and 227 00:11:24,900 --> 00:11:26,160 reliable models. 228 00:11:26,160 --> 00:11:29,070 We always want to perform a linear analysis first. 229 00:11:29,070 --> 00:11:33,440 I will show examples in this course particular related to 230 00:11:33,440 --> 00:11:34,620 this item here. 231 00:11:34,620 --> 00:11:38,282 It is very important, in my view, to first do a linear 232 00:11:38,282 --> 00:11:41,740 analysis and then only go on to a nonlinear analysis. 233 00:11:41,740 --> 00:11:44,030 As I mentioned already, we want to refine the model by 234 00:11:44,030 --> 00:11:48,900 introducing nonlinearities as desired and, once again, use 235 00:11:48,900 --> 00:11:51,950 reliable and well-understood models, obtain accurate 236 00:11:51,950 --> 00:11:54,200 solutions of the models. 237 00:11:54,200 --> 00:11:57,410 This is very important for possible proper interpretation 238 00:11:57,410 --> 00:11:59,150 of the results. 239 00:11:59,150 --> 00:12:01,440 Here we show schematically the finite 240 00:12:01,440 --> 00:12:03,060 element modeling process. 241 00:12:03,060 --> 00:12:04,670 We have a problem in nature. 242 00:12:04,670 --> 00:12:08,420 We model that problem, the kinematic conditions, the 243 00:12:08,420 --> 00:12:11,200 constitutive relations, the boundary conditions, the 244 00:12:11,200 --> 00:12:14,040 loads, and so on, using finite elements. 245 00:12:14,040 --> 00:12:17,290 We solve the model and interpret the results. 246 00:12:17,290 --> 00:12:20,170 Now this model surely can only approximate the 247 00:12:20,170 --> 00:12:21,840 actual problem in nature. 248 00:12:21,840 --> 00:12:25,490 And on interpretation of the results, we may find that we 249 00:12:25,490 --> 00:12:27,310 really should refine our model. 250 00:12:27,310 --> 00:12:31,340 And we do so, set up a new model, solve again. 251 00:12:31,340 --> 00:12:33,790 And like this, we may cycle a number of times 252 00:12:33,790 --> 00:12:35,930 through this process. 253 00:12:35,930 --> 00:12:39,250 Of course, traditionally, test results have been sought for 254 00:12:39,250 --> 00:12:42,200 very complex problems, laboratory test results have 255 00:12:42,200 --> 00:12:42,960 been sought. 256 00:12:42,960 --> 00:12:45,170 And this may still be necessary now. 257 00:12:45,170 --> 00:12:49,300 However, the finite element modeling process will 258 00:12:49,300 --> 00:12:53,060 certainly compliment these laboratory test results. 259 00:12:53,060 --> 00:12:55,570 Let's look at a typical nonlinear problem. 260 00:12:55,570 --> 00:12:57,410 Here we have a bracket of mild steel 261 00:12:57,410 --> 00:12:59,810 subjected to the load shown. 262 00:12:59,810 --> 00:13:02,670 And the possible questions that we might ask are, what is 263 00:13:02,670 --> 00:13:04,700 the yield load of this bracket? 264 00:13:04,700 --> 00:13:08,310 In other words, at what load do we see first plastic 265 00:13:08,310 --> 00:13:09,580 deformations? 266 00:13:09,580 --> 00:13:10,590 What is the limit load? 267 00:13:10,590 --> 00:13:13,270 What is the maximum load that this bracket can take? 268 00:13:13,270 --> 00:13:15,830 What are the plastic zones? 269 00:13:15,830 --> 00:13:19,310 What are the residual stresses when the load is removed? 270 00:13:19,310 --> 00:13:22,190 Is there adhering where the loads are applied 271 00:13:22,190 --> 00:13:23,890 right around here? 272 00:13:23,890 --> 00:13:26,440 What is the creep response of the bracket when the bracket 273 00:13:26,440 --> 00:13:29,270 is subjected to high temperature conditions and 274 00:13:29,270 --> 00:13:30,300 these loads? 275 00:13:30,300 --> 00:13:33,310 And what are the permanent deflections of 276 00:13:33,310 --> 00:13:34,670 the bracket, et cetera. 277 00:13:34,670 --> 00:13:37,370 There are many more questions that could be asked. 278 00:13:37,370 --> 00:13:42,280 And certainly, these questions here can only be answered by a 279 00:13:42,280 --> 00:13:44,840 nonlinear analysis. 280 00:13:44,840 --> 00:13:47,110 Possible analyses that we might 281 00:13:47,110 --> 00:13:48,740 consider are the following. 282 00:13:48,740 --> 00:13:51,890 First we, of course, would always perform, as I mentioned 283 00:13:51,890 --> 00:13:54,690 earlier, a linear elastic analysis. 284 00:13:54,690 --> 00:13:57,480 The linear elastic analysis would determine the total 285 00:13:57,480 --> 00:14:01,800 stiffness and the yield load of the bracket. 286 00:14:01,800 --> 00:14:04,420 Notice here we show the displacements in red. 287 00:14:04,420 --> 00:14:07,130 Actually, these displacements would be very small. 288 00:14:07,130 --> 00:14:10,890 And they are magnified in this view graph. 289 00:14:10,890 --> 00:14:14,710 Another analysis now would be to perform a plastic analysis, 290 00:14:14,710 --> 00:14:17,830 but assuming still small deformations. 291 00:14:17,830 --> 00:14:20,870 This analysis would determine sizes and shapes of the 292 00:14:20,870 --> 00:14:22,190 plastic zones. 293 00:14:22,190 --> 00:14:24,510 The displacements here are still small. 294 00:14:24,510 --> 00:14:27,140 We have not shown them at all actually here. 295 00:14:27,140 --> 00:14:29,170 They are very small, however. 296 00:14:29,170 --> 00:14:31,230 Then we might want to go on to a large 297 00:14:31,230 --> 00:14:33,380 deformation plastic analysis. 298 00:14:33,380 --> 00:14:36,985 This analysis would determine the ultimate load capacity of 299 00:14:36,985 --> 00:14:37,890 the bracket. 300 00:14:37,890 --> 00:14:41,060 And notice the displacements are now large, they are 301 00:14:41,060 --> 00:14:42,160 actually large. 302 00:14:42,160 --> 00:14:44,960 Here, we have had small displacement assumptions. 303 00:14:44,960 --> 00:14:49,530 Here, we have included the large deformation effects as 304 00:14:49,530 --> 00:14:52,850 also shown on the view graph. 305 00:14:52,850 --> 00:14:58,890 For analysis, it is very good to actually classify all the 306 00:14:58,890 --> 00:15:02,580 types of analysis that one might want to perform. 307 00:15:02,580 --> 00:15:06,450 And the first category of nonlinear analysis is the one 308 00:15:06,450 --> 00:15:08,570 that we call Materially-nonlinear-only 309 00:15:08,570 --> 00:15:11,440 analysis, MNO analysis. 310 00:15:11,440 --> 00:15:13,950 In this analysis, we assume that the displacements are 311 00:15:13,950 --> 00:15:17,800 infinitesimal, the strains are infinitesimal. 312 00:15:17,800 --> 00:15:19,530 In other words, both of these quantities 313 00:15:19,530 --> 00:15:21,120 are very, very small. 314 00:15:21,120 --> 00:15:23,730 And the stress-strain relationship is nonlinear. 315 00:15:23,730 --> 00:15:27,680 So all nonlinearities lie, really, in here, in the 316 00:15:27,680 --> 00:15:30,040 stress-strain relationship. 317 00:15:30,040 --> 00:15:31,690 Here, I'm showing a schematic example. 318 00:15:31,690 --> 00:15:34,470 You're looking at a four-node element that is subjected to 319 00:15:34,470 --> 00:15:37,840 the loads shown, and delta is the displacement. 320 00:15:37,840 --> 00:15:40,540 Notice that delta/l is very small. 321 00:15:40,540 --> 00:15:44,080 In fact, even this four is already relatively large. 322 00:15:44,080 --> 00:15:47,500 We should really have here, possibly, a two. 323 00:15:47,500 --> 00:15:50,060 So nonlinearity lies in the material description. 324 00:15:50,060 --> 00:15:52,400 The material is an elasto-plastic material, 325 00:15:52,400 --> 00:15:55,140 Young's modulus, E, yield stress, sigma y and 326 00:15:55,140 --> 00:15:58,060 strain-hardening modulus, Et. 327 00:15:58,060 --> 00:16:03,460 As long as the stresses do not exceed sigma y, we really have 328 00:16:03,460 --> 00:16:04,700 a linear analysis. 329 00:16:04,700 --> 00:16:07,650 So if you use a computer program with an MNO 330 00:16:07,650 --> 00:16:13,480 formulation and you subject your model to forces such that 331 00:16:13,480 --> 00:16:17,790 the stresses are below sigma yield everywhere in the model, 332 00:16:17,790 --> 00:16:22,990 then you should really obtain the linear analysis results. 333 00:16:22,990 --> 00:16:26,020 The next category of problems is the one of large 334 00:16:26,020 --> 00:16:30,670 displacements, large rotations, but small strains. 335 00:16:30,670 --> 00:16:33,610 Here, in other words, we still keep the assumption of small 336 00:16:33,610 --> 00:16:36,220 strain, but we allow large 337 00:16:36,220 --> 00:16:38,680 displacements and large rotations. 338 00:16:38,680 --> 00:16:42,760 The stress-strain relations can be linear or nonlinear. 339 00:16:42,760 --> 00:16:46,690 Schematically here, once again, our four-node element. 340 00:16:46,690 --> 00:16:49,830 This four-node element now would move, as shown, going 341 00:16:49,830 --> 00:16:52,690 through large displacements and large rotations. 342 00:16:52,690 --> 00:16:56,640 But the strains in the element, expressed by delta 343 00:16:56,640 --> 00:17:00,080 prime over l are still small. 344 00:17:00,080 --> 00:17:04,230 Once again, you may actually want to make this 0.02. 345 00:17:04,230 --> 00:17:08,400 As long as the displacements are very small, we have now, 346 00:17:08,400 --> 00:17:12,220 really, an MNO analysis. 347 00:17:12,220 --> 00:17:14,599 The third category of problems is the one of large 348 00:17:14,599 --> 00:17:17,450 displacements, large rotations and large strains. 349 00:17:17,450 --> 00:17:22,869 Here we have included all kinematic nonlinearities. 350 00:17:22,869 --> 00:17:26,030 And the stress-strain relation is probably also nonlinear 351 00:17:26,030 --> 00:17:30,050 because we are dealing with large strains. 352 00:17:30,050 --> 00:17:34,580 As an example, we have, again, our four-node element here 353 00:17:34,580 --> 00:17:35,950 moving as shown. 354 00:17:35,950 --> 00:17:40,890 Notice material fibers here, displaced by a large amount, 355 00:17:40,890 --> 00:17:45,010 rotate and are also stretched by a large amount. 356 00:17:45,010 --> 00:17:46,330 This, of course, is the most general 357 00:17:46,330 --> 00:17:48,580 formulation of a problem. 358 00:17:48,580 --> 00:17:52,020 However, still not considering any nonlinearities in the 359 00:17:52,020 --> 00:17:55,220 boundary conditions, nonlinearities in the boundary 360 00:17:55,220 --> 00:17:57,300 conditions provide contact problems. 361 00:17:57,300 --> 00:17:59,930 And here, we look schematically at a simple 362 00:17:59,930 --> 00:18:01,140 contact problem. 363 00:18:01,140 --> 00:18:04,370 The four-node element, once again, subjected to loads, 364 00:18:04,370 --> 00:18:07,500 there's a gap here between this element and the spring. 365 00:18:07,500 --> 00:18:11,720 And as soon as this gap is closed, the spring, of course, 366 00:18:11,720 --> 00:18:16,300 provides stiffness into the system, the system being now 367 00:18:16,300 --> 00:18:18,200 the four-node element. 368 00:18:18,200 --> 00:18:22,280 And, well, that difference has to be taken into account in 369 00:18:22,280 --> 00:18:23,460 the analysis. 370 00:18:23,460 --> 00:18:28,350 And there are procedures to solve contact problems. 371 00:18:28,350 --> 00:18:31,050 This is, of course, a very simple contact problem. 372 00:18:31,050 --> 00:18:33,970 But there are procedures to solve contact problems. 373 00:18:33,970 --> 00:18:37,530 Contact problems are very difficult problems to handle, 374 00:18:37,530 --> 00:18:40,230 particularly if you're dealing with frictional effects. 375 00:18:40,230 --> 00:18:43,350 We will, in this course, in this set of lectures, really 376 00:18:43,350 --> 00:18:47,520 not address contact problems, except that, in the last 377 00:18:47,520 --> 00:18:49,390 lecture, we actually look at one 378 00:18:49,390 --> 00:18:50,690 particular contact problem. 379 00:18:50,690 --> 00:18:54,480 We use a computer program for the analysis of a contact 380 00:18:54,480 --> 00:18:56,950 problem there. 381 00:18:56,950 --> 00:18:59,920 Let's turn to an example analysis. 382 00:18:59,920 --> 00:19:02,260 And I'd like to consider with you, briefly, the 383 00:19:02,260 --> 00:19:03,860 analysis of a bracket. 384 00:19:03,860 --> 00:19:06,800 Here is a bracket of the kind that we 385 00:19:06,800 --> 00:19:09,060 looked earlier at already. 386 00:19:09,060 --> 00:19:11,380 Notice the dimensions are as shown. 387 00:19:11,380 --> 00:19:14,170 The thickness of the bracket is one inch. 388 00:19:14,170 --> 00:19:20,900 And this bracket is going to be loaded to large loads. 389 00:19:20,900 --> 00:19:25,070 We make a material assumption, namely the one of an 390 00:19:25,070 --> 00:19:28,480 elasto-plastic material with isotopic 391 00:19:28,480 --> 00:19:30,430 hardening as shown here. 392 00:19:30,430 --> 00:19:33,610 The yield stress is 26,000 psi. 393 00:19:33,610 --> 00:19:35,660 Here you have the Young's modulus. 394 00:19:35,660 --> 00:19:38,960 And here you have the strain-hardening modulus. 395 00:19:38,960 --> 00:19:43,500 We use symmetry conditions to analyze the bracket the same 396 00:19:43,500 --> 00:19:46,520 way as we would do it, of course, in a linear analysis. 397 00:19:46,520 --> 00:19:49,680 And with the symmetry conditions, we consider on 398 00:19:49,680 --> 00:19:53,040 this view graph just the lower part of the bracket. 399 00:19:53,040 --> 00:19:56,400 You see a pin here, rollers there. 400 00:19:56,400 --> 00:19:59,410 And we're using eight-node, isopolymetric elements to 401 00:19:59,410 --> 00:20:02,770 model this part of the bracket. 402 00:20:02,770 --> 00:20:05,600 We apply the load as shown here. 403 00:20:05,600 --> 00:20:08,570 Notice no special considerations to the hole 404 00:20:08,570 --> 00:20:10,200 which the load is applied. 405 00:20:10,200 --> 00:20:13,230 Our interest really lies in predicting the stresses and 406 00:20:13,230 --> 00:20:17,100 strains in this region and to also predict the overall 407 00:20:17,100 --> 00:20:19,950 collapse of the bracket. 408 00:20:19,950 --> 00:20:23,990 We will use three kinematic formulations for the analysis. 409 00:20:23,990 --> 00:20:26,620 First of all, we use the Material-nonlinear-only 410 00:20:26,620 --> 00:20:28,780 analysis assumption. 411 00:20:28,780 --> 00:20:31,980 In other words, small displacements, small rotations 412 00:20:31,980 --> 00:20:34,340 and small strains are assumed in the 413 00:20:34,340 --> 00:20:36,300 analysis of this bracket. 414 00:20:36,300 --> 00:20:39,860 Then we will perform an analysis using the total 415 00:20:39,860 --> 00:20:41,290 Lagrangian formulation. 416 00:20:41,290 --> 00:20:44,120 We will discuss this formulation quite extensively 417 00:20:44,120 --> 00:20:45,850 in later lectures of the course. 418 00:20:45,850 --> 00:20:50,930 This formulation assumes large displacements, large rotations 419 00:20:50,930 --> 00:20:53,590 and large strains kinematically. 420 00:20:53,590 --> 00:20:55,690 Kinematically, these are the assumptions. 421 00:20:55,690 --> 00:21:01,510 However, we will point out that the material law that we 422 00:21:01,510 --> 00:21:04,930 are using, or the material law description that we are using, 423 00:21:04,930 --> 00:21:09,540 is only applicable with this formulation to small strains. 424 00:21:09,540 --> 00:21:12,750 So the overall analysis, using this total Lagrangian 425 00:21:12,750 --> 00:21:16,310 formulation, is then only applicable to model large 426 00:21:16,310 --> 00:21:18,190 displacements, large rotations, 427 00:21:18,190 --> 00:21:20,240 but only small strains. 428 00:21:20,240 --> 00:21:22,000 I get back to that just now. 429 00:21:22,000 --> 00:21:25,200 We will also use an updated Lagrangian formulation which 430 00:21:25,200 --> 00:21:28,600 kinetically includes large displacements, large rotations 431 00:21:28,600 --> 00:21:30,070 and large strains. 432 00:21:30,070 --> 00:21:33,700 And on the material model level, we also include there 433 00:21:33,700 --> 00:21:35,520 large strain effects. 434 00:21:35,520 --> 00:21:39,960 So once again here, as summarized on this view graph, 435 00:21:39,960 --> 00:21:42,600 the material used in conjunction with the total 436 00:21:42,600 --> 00:21:45,410 Lagrangian formulation is actually not applicable to 437 00:21:45,410 --> 00:21:49,200 large-strain situations, but only to large displacements, 438 00:21:49,200 --> 00:21:51,660 rotation in small-strain conditions. 439 00:21:51,660 --> 00:21:57,970 So once again, our total analysis is only applicable to 440 00:21:57,970 --> 00:22:00,780 this kind of situation. 441 00:22:00,780 --> 00:22:03,750 Whereas, the updated Lagrangian formulation does 442 00:22:03,750 --> 00:22:07,730 model, kinematically and on the stress-strain level, large 443 00:22:07,730 --> 00:22:11,910 displacements, large rotations and large strains. 444 00:22:11,910 --> 00:22:18,100 The analysis results obtained are shown here. 445 00:22:18,100 --> 00:22:22,770 Notice we are plotting here the force applied and the 446 00:22:22,770 --> 00:22:27,320 total deflection between points of load application. 447 00:22:27,320 --> 00:22:32,840 The three analyses give us three distinct curves at large 448 00:22:32,840 --> 00:22:34,170 displacements. 449 00:22:34,170 --> 00:22:36,730 But for small displacements, of course, these curves are 450 00:22:36,730 --> 00:22:38,780 indistinguishable. 451 00:22:38,780 --> 00:22:42,520 Notice here we have about 10% strain at point A. I will show 452 00:22:42,520 --> 00:22:44,630 you just now where point A is. 453 00:22:44,630 --> 00:22:47,810 The important point to notice is that these are three 454 00:22:47,810 --> 00:22:48,880 distinct curves. 455 00:22:48,880 --> 00:22:51,380 They are distinct because we have made different 456 00:22:51,380 --> 00:22:54,490 assumptions, kinematically and on the 457 00:22:54,490 --> 00:22:56,690 stress-strain load level. 458 00:22:56,690 --> 00:22:58,700 We will talk about such assumptions in the later 459 00:22:58,700 --> 00:23:00,760 lectures of this course. 460 00:23:00,760 --> 00:23:03,350 On this view graph, now you see this point A, which I 461 00:23:03,350 --> 00:23:04,660 mentioned earlier. 462 00:23:04,660 --> 00:23:08,410 And you see also the original mesh shown in dashed, black 463 00:23:08,410 --> 00:23:13,240 lines, and the deformed mesh, shown in red, corresponding to 464 00:23:13,240 --> 00:23:16,730 a level of load of 12,000 pounds. 465 00:23:16,730 --> 00:23:21,250 I'd like to now look with you at two animations regarding 466 00:23:21,250 --> 00:23:26,030 problems that show nonlinearities typical of the 467 00:23:26,030 --> 00:23:28,730 nonlinearities that we are talking about in this course. 468 00:23:28,730 --> 00:23:33,540 The first animation shows a plate with a hole that is 469 00:23:33,540 --> 00:23:36,290 subjected to high tensile forces. 470 00:23:36,290 --> 00:23:38,690 The plate undergoes plastic deformations. 471 00:23:38,690 --> 00:23:40,930 And, in fact, the load becomes so large that the plate 472 00:23:40,930 --> 00:23:42,800 basically ruptures. 473 00:23:42,800 --> 00:23:46,600 Let's look at this animation now. 474 00:23:46,600 --> 00:23:49,080 Here we see one quarter of the plate. 475 00:23:49,080 --> 00:23:51,800 The plate is subjected to a uniformly distributed tensile 476 00:23:51,800 --> 00:23:54,130 load along its upper edge. 477 00:23:54,130 --> 00:23:56,680 For the quarter model of the plate, symmetry boundary 478 00:23:56,680 --> 00:23:59,440 conditions are imposed along the left vertical and lower 479 00:23:59,440 --> 00:24:01,400 horizontal edges. 480 00:24:01,400 --> 00:24:03,500 The time code given above the plate gives the 481 00:24:03,500 --> 00:24:05,380 time of the load step. 482 00:24:05,380 --> 00:24:08,990 Also given is the load applied at that time. 483 00:24:08,990 --> 00:24:12,140 However, note that we perform a static analysis. 484 00:24:12,140 --> 00:24:14,600 Therefore, the time code merely denotes a load level, 485 00:24:14,600 --> 00:24:16,970 as we will discuss later in detail. 486 00:24:16,970 --> 00:24:19,970 Each time increment of one milli corresponds to a load 487 00:24:19,970 --> 00:24:23,390 increment of 12.5 megapascal. 488 00:24:23,390 --> 00:24:25,610 We will increase the load monotonically. 489 00:24:25,610 --> 00:24:27,870 The plate will become plastic at the hole. 490 00:24:27,870 --> 00:24:30,740 And this plasticity will then spread until, in essence, the 491 00:24:30,740 --> 00:24:32,590 plate ruptures. 492 00:24:32,590 --> 00:24:35,870 We used 288 eight-node plane stress finite elements for the 493 00:24:35,870 --> 00:24:37,440 quarter of the plate. 494 00:24:37,440 --> 00:24:39,810 And we show to spread of plasticity in the plate by 495 00:24:39,810 --> 00:24:43,510 darkening the areas that are plastic. 496 00:24:43,510 --> 00:24:46,740 Here now you see the time and load increasing. 497 00:24:46,740 --> 00:24:50,120 In the first load steps, the plate remains elastic and the 498 00:24:50,120 --> 00:24:52,740 deformations are small. 499 00:24:52,740 --> 00:24:54,400 Now you see the first plasticity 500 00:24:54,400 --> 00:24:56,660 developing near the hole. 501 00:24:56,660 --> 00:24:59,410 The plastic zone increases rapidly as the higher load 502 00:24:59,410 --> 00:25:03,220 levels are reached until a large portion of the plate is 503 00:25:03,220 --> 00:25:07,500 plastic and, in essence, the maximum load-carrying capacity 504 00:25:07,500 --> 00:25:10,170 has been reached. 505 00:25:10,170 --> 00:25:13,090 We will consider the analysis of this plate in more detail 506 00:25:13,090 --> 00:25:14,340 later in the course. 507 00:25:17,010 --> 00:25:19,810 The second animation shows a frame that is subjected to 508 00:25:19,810 --> 00:25:23,400 forces such that the frame undergoes very large 509 00:25:23,400 --> 00:25:24,650 displacements. 510 00:25:26,730 --> 00:25:29,560 Here we see the frame model, using beam elements that we 511 00:25:29,560 --> 00:25:32,020 discuss later in the course. 512 00:25:32,020 --> 00:25:35,120 The frame is loaded, as shown, by the force arrows. 513 00:25:35,120 --> 00:25:37,130 There are pin connections at the points of load 514 00:25:37,130 --> 00:25:39,350 application. 515 00:25:39,350 --> 00:25:42,370 The frame is assumed to be of an elastic material, hence, no 516 00:25:42,370 --> 00:25:45,190 plasticity is assumed to develop or that the frame will 517 00:25:45,190 --> 00:25:48,890 be subjected to very large displacements. 518 00:25:48,890 --> 00:25:51,860 The indicated loads will push the top of the frame down and 519 00:25:51,860 --> 00:25:54,820 the bottom up to such an extent that the points of load 520 00:25:54,820 --> 00:25:57,730 application cross over. 521 00:25:57,730 --> 00:25:59,630 You see above the frame a time and a 522 00:25:59,630 --> 00:26:01,530 corresponding load level. 523 00:26:01,530 --> 00:26:03,680 This is, again, a static analysis. 524 00:26:03,680 --> 00:26:06,740 And each time increment of one corresponds merely to a load 525 00:26:06,740 --> 00:26:10,630 increment of 250 pounds. 526 00:26:10,630 --> 00:26:12,300 Here now, you see the deformations of the frame 527 00:26:12,300 --> 00:26:16,060 develop as the load is increased. 528 00:26:16,060 --> 00:26:17,770 The displacements are very large at 529 00:26:17,770 --> 00:26:19,910 the higher load levels. 530 00:26:19,910 --> 00:26:22,580 Of course, this is only a numerical experiment at the 531 00:26:22,580 --> 00:26:25,820 high load levels, but an interesting one that indicates 532 00:26:25,820 --> 00:26:28,100 in a simple manner what can all be done in 533 00:26:28,100 --> 00:26:30,130 finite element analysis. 534 00:26:30,130 --> 00:26:32,730 We will consider this problem solution also in more detail 535 00:26:32,730 --> 00:26:33,980 later in the course. 536 00:26:36,820 --> 00:26:39,195 Let us now look at the basic approach of 537 00:26:39,195 --> 00:26:41,020 an incremental solution. 538 00:26:41,020 --> 00:26:44,440 We consider a body, a structure or solid subjected 539 00:26:44,440 --> 00:26:46,710 to force and displacement boundary conditions that are 540 00:26:46,710 --> 00:26:50,170 changing, and we describe the externally applied forces and 541 00:26:50,170 --> 00:26:51,840 the displacement boundary conditions as 542 00:26:51,840 --> 00:26:53,610 a function of time. 543 00:26:53,610 --> 00:26:59,240 Schematically, here we show a body, of course, supported as 544 00:26:59,240 --> 00:27:03,890 shown and subjected to a force varying as a function of time 545 00:27:03,890 --> 00:27:07,860 and a prescribed displacement varying as a function of time. 546 00:27:07,860 --> 00:27:11,360 The time variation of the force is shown down here. 547 00:27:11,360 --> 00:27:14,660 And the time variation of the prescribed displacement is 548 00:27:14,660 --> 00:27:16,250 shown here. 549 00:27:16,250 --> 00:27:20,370 Notice that here we have a particular time point, T. And 550 00:27:20,370 --> 00:27:23,880 here we have a time point T plus delta t. 551 00:27:23,880 --> 00:27:27,700 Here we have, similarly, the time point T and here, also, 552 00:27:27,700 --> 00:27:30,800 the time point T plus delta t. 553 00:27:30,800 --> 00:27:33,950 Since we anticipate nonlinearities, we use an 554 00:27:33,950 --> 00:27:37,380 incremental approach measured in load steps or time steps. 555 00:27:37,380 --> 00:27:40,530 And this means that the loads, the prescribed loads, the 556 00:27:40,530 --> 00:27:44,430 imposed loads and the prescribed displacements, are 557 00:27:44,430 --> 00:27:47,470 discretized as a function of time, as shown 558 00:27:47,470 --> 00:27:48,290 on this view graph. 559 00:27:48,290 --> 00:27:54,620 Notice here we have, at time T, the impulse load TRI. 560 00:27:54,620 --> 00:28:01,450 This upper, superscript T means at time T then delta t, 561 00:28:01,450 --> 00:28:04,830 in advance, time T plus delta t. 562 00:28:04,830 --> 00:28:08,580 We would have, in other words, T plus delta t ri. 563 00:28:08,580 --> 00:28:12,270 In other words this T would now be replaced by T plus 564 00:28:12,270 --> 00:28:13,920 delta t and so on. 565 00:28:13,920 --> 00:28:18,100 This is how we are inputting or prescribing the impulse 566 00:28:18,100 --> 00:28:22,190 loads and also the prescribed displacements. 567 00:28:22,190 --> 00:28:26,480 When they apply forces and displacements very slowly, 568 00:28:26,480 --> 00:28:29,170 meaning that the frequencies of the loads are much smaller 569 00:28:29,170 --> 00:28:31,740 than the natural frequencies of the structure, we have a 570 00:28:31,740 --> 00:28:32,980 static analysis. 571 00:28:32,980 --> 00:28:37,580 This, of course, means that the periods of the loads are 572 00:28:37,580 --> 00:28:41,320 very long when measured on the natural 573 00:28:41,320 --> 00:28:42,410 periods of the structure. 574 00:28:42,410 --> 00:28:45,250 In other words, you have a spring and you apply a load 575 00:28:45,250 --> 00:28:49,150 that varies very slowly. 576 00:28:49,150 --> 00:28:52,400 And this means we have a static analysis. 577 00:28:52,400 --> 00:28:55,840 If the load's very fast, and by that we mean that the 578 00:28:55,840 --> 00:28:58,100 frequencies of the loads are in the range of the natural 579 00:28:58,100 --> 00:29:00,140 frequency of the structure, then we 580 00:29:00,140 --> 00:29:03,350 have a dynamic analysis. 581 00:29:03,350 --> 00:29:05,040 Let us look a bit closer at the 582 00:29:05,040 --> 00:29:06,960 meaning of the time variable. 583 00:29:06,960 --> 00:29:10,630 Time is a pseudo-variable, only denoting the load level 584 00:29:10,630 --> 00:29:14,080 in nonlinear static analysis with time-independent material 585 00:29:14,080 --> 00:29:16,870 properties. 586 00:29:16,870 --> 00:29:18,610 As an example. 587 00:29:18,610 --> 00:29:23,590 here we have a cantilever subjected to a load, R, a tip 588 00:29:23,590 --> 00:29:27,880 load, R. And if we were to perform a run, 1, in which we 589 00:29:27,880 --> 00:29:30,450 prescribe the loads as shown here-- 590 00:29:30,450 --> 00:29:34,240 and notice, at time 1, a load of 100, at time 591 00:29:34,240 --> 00:29:36,380 2, a load of 200-- 592 00:29:36,380 --> 00:29:39,010 delta t is equal to 1. 593 00:29:39,010 --> 00:29:42,700 If we were to perform this analysis and, in addition, 594 00:29:42,700 --> 00:29:47,200 this analysis where delta t is equal to 2 and the load at 595 00:29:47,200 --> 00:29:51,030 time 2 is equal to 100, the load at time 4 is equal to 596 00:29:51,030 --> 00:29:55,850 200, then we would obtain identically the same results 597 00:29:55,850 --> 00:30:01,590 in run 1 and run 2 because it's a static analysis and the 598 00:30:01,590 --> 00:30:05,150 material properties are time-independent. 599 00:30:05,150 --> 00:30:07,840 So in this particular case, certainly, time is a 600 00:30:07,840 --> 00:30:08,710 pseudo-variable. 601 00:30:08,710 --> 00:30:13,810 And we can design a time-stepping. 602 00:30:13,810 --> 00:30:17,450 We can design many different time-stepping schemes using 603 00:30:17,450 --> 00:30:19,850 different time steps and always obtain the same 604 00:30:19,850 --> 00:30:24,020 results, provided, at the end of the first time step, we 605 00:30:24,020 --> 00:30:27,520 have the load 100 and at the end of the second time step we 606 00:30:27,520 --> 00:30:30,020 have the load 200. 607 00:30:30,020 --> 00:30:34,680 However, time is an actual variable in dynamic analysis 608 00:30:34,680 --> 00:30:37,980 and in nonlinear static analysis with time-dependent 609 00:30:37,980 --> 00:30:41,420 material properties, for example, when the material 610 00:30:41,420 --> 00:30:44,330 contains grid conditions. 611 00:30:44,330 --> 00:30:49,360 Now delta t must be chosen very carefully with respect to 612 00:30:49,360 --> 00:30:52,200 the physics of the problem, with respect to the numerical 613 00:30:52,200 --> 00:30:54,750 techniques used and the costs involved. 614 00:30:54,750 --> 00:30:58,700 If delta t is not chosen appropriately, you may have a 615 00:30:58,700 --> 00:31:01,790 very high cost for the analysis. 616 00:31:01,790 --> 00:31:05,580 And on the other hand, you may, with improper choice of 617 00:31:05,580 --> 00:31:08,870 delta t, also obtain very bad results. 618 00:31:08,870 --> 00:31:13,330 So it is very important to choose delta t judiciously for 619 00:31:13,330 --> 00:31:17,690 an accurate and cost-effective analysis. 620 00:31:17,690 --> 00:31:21,820 At the end of each load or time step, we need to satisfy 621 00:31:21,820 --> 00:31:25,820 the three basic requirements of mechanics, equilibrium, 622 00:31:25,820 --> 00:31:28,570 compatibility and the stress-strain law. 623 00:31:28,570 --> 00:31:31,920 These are the three fundamental requirements to be 624 00:31:31,920 --> 00:31:33,930 satisfied in mechanics. 625 00:31:33,930 --> 00:31:36,680 This is achieved in finite element analysis in an 626 00:31:36,680 --> 00:31:40,420 approximate manner using finite elements by the 627 00:31:40,420 --> 00:31:43,080 application of the principle of virtual work. 628 00:31:43,080 --> 00:31:45,420 Now there is a lot of information 629 00:31:45,420 --> 00:31:46,800 on this view graph. 630 00:31:46,800 --> 00:31:50,420 And we don't have time in this lecture to go into any depth. 631 00:31:50,420 --> 00:31:53,230 There is, of course, quite a bit of mathematics that has to 632 00:31:53,230 --> 00:31:57,190 be introduced for the discussion of all of what we 633 00:31:57,190 --> 00:31:58,440 see here on the view graph. 634 00:31:58,440 --> 00:32:00,700 And that's what we do in the later lectures. 635 00:32:00,700 --> 00:32:04,210 I do not have time now to go into these mathematics, but 636 00:32:04,210 --> 00:32:06,740 let us just very briefly look at the basic 637 00:32:06,740 --> 00:32:09,690 procedure that we using. 638 00:32:09,690 --> 00:32:15,990 We are saying that, at any time step, T plus delta t, the 639 00:32:15,990 --> 00:32:19,940 externally applied loads, the vector of externally applied 640 00:32:19,940 --> 00:32:25,250 loads and this vector, includes pressure loads. 641 00:32:25,250 --> 00:32:29,210 Concentrated loads and, in the dynamic analysis also inertial 642 00:32:29,210 --> 00:32:34,890 forces, this vector must at any time T plus delta t equal 643 00:32:34,890 --> 00:32:40,930 to the vector F at time T plus delta t where this vector F 644 00:32:40,930 --> 00:32:45,540 corresponds to the internal element stresses at time T 645 00:32:45,540 --> 00:32:46,930 plus delta t. 646 00:32:46,930 --> 00:32:49,200 We will talk in the later lectures about how we 647 00:32:49,200 --> 00:32:52,470 calculate this vector T plus delta tF. 648 00:32:52,470 --> 00:32:55,580 Of course, there are some very important considerations in 649 00:32:55,580 --> 00:32:59,490 the proper calculation of this vector F. 650 00:32:59,490 --> 00:33:03,080 Let us now assume that the solution at time T is known. 651 00:33:03,080 --> 00:33:07,440 Hence, the stresses, at time T, are known. 652 00:33:07,440 --> 00:33:11,480 The volume surface area and so on, all of the information 653 00:33:11,480 --> 00:33:14,910 corresponding to the body at time T is known. 654 00:33:14,910 --> 00:33:18,380 And we now want to obtain the solution corresponding to time 655 00:33:18,380 --> 00:33:22,300 T plus delta t, that is for the loads applied at time T 656 00:33:22,300 --> 00:33:23,360 plus delta t. 657 00:33:23,360 --> 00:33:25,910 Of course, this is a typical step of 658 00:33:25,910 --> 00:33:27,540 the incremental solution. 659 00:33:27,540 --> 00:33:31,600 Once we have the solution for time T plus delta t, we can 660 00:33:31,600 --> 00:33:34,770 use the same scheme to calculate the solution for 661 00:33:34,770 --> 00:33:38,540 time T plus 2 delta t and so on. 662 00:33:38,540 --> 00:33:40,850 For this purpose, we solve, in static 663 00:33:40,850 --> 00:33:43,420 analysis, this set of equations. 664 00:33:43,420 --> 00:33:47,130 TK is a tangent stiffness matrix of the system. 665 00:33:47,130 --> 00:33:50,810 Delta u is a vector of increments in the nodal point 666 00:33:50,810 --> 00:33:52,210 displacements. 667 00:33:52,210 --> 00:33:54,580 On the right-hand side, we have the externally applied 668 00:33:54,580 --> 00:33:59,590 loads corresponding to time T plus delta t in this R vector. 669 00:33:59,590 --> 00:34:05,580 And here, TF corresponds to or is the vector of nodal point 670 00:34:05,580 --> 00:34:09,250 forces that correspond to the internal element stresses at 671 00:34:09,250 --> 00:34:10,770 time T. 672 00:34:10,770 --> 00:34:14,880 Notice this is an out-of-balance load vector. 673 00:34:14,880 --> 00:34:17,980 And this out-of-balance load vector gives us an increment 674 00:34:17,980 --> 00:34:19,699 in displacements. 675 00:34:19,699 --> 00:34:22,679 This increment in displacements is added to the 676 00:34:22,679 --> 00:34:27,530 displacements at time T. And that gives us a displacement 677 00:34:27,530 --> 00:34:30,210 vector corresponding to time T plus delta t. 678 00:34:30,210 --> 00:34:33,350 Notice I have an approximation sign there. 679 00:34:33,350 --> 00:34:36,580 This approximation sign is a result of the fact that we 680 00:34:36,580 --> 00:34:41,110 obtained this set of equations by linearization. 681 00:34:41,110 --> 00:34:47,280 We will talk about how these equations are obtained in the 682 00:34:47,280 --> 00:34:48,230 later lectures. 683 00:34:48,230 --> 00:34:53,290 We will start from the basic principle of virtual work and 684 00:34:53,290 --> 00:34:57,210 use continuum mechanics considerations to derive, in a 685 00:34:57,210 --> 00:35:02,560 very consistent manner, this set of equations. 686 00:35:02,560 --> 00:35:05,050 Of course, there are many details involved. 687 00:35:05,050 --> 00:35:07,480 We will talk about the total Lagrangian formulations, the 688 00:35:07,480 --> 00:35:09,710 updated Lagrangian formulation, the 689 00:35:09,710 --> 00:35:13,720 Material-nonlinear-only analysis, the details that go 690 00:35:13,720 --> 00:35:16,160 into these formulations in terms of kinematic 691 00:35:16,160 --> 00:35:20,520 approximations, kinematic approximations related to 692 00:35:20,520 --> 00:35:24,800 large displacements, large rotations, large strains. 693 00:35:24,800 --> 00:35:27,590 We also will talk about the constitutive relations that 694 00:35:27,590 --> 00:35:30,340 enter in this tangent stiffness matrix and in this 695 00:35:30,340 --> 00:35:34,000 force vector and so on. 696 00:35:34,000 --> 00:35:38,420 Anyway, this set of equations gives us an approximation to 697 00:35:38,420 --> 00:35:41,870 the displacements at time T plus delta t because of the 698 00:35:41,870 --> 00:35:47,300 linearization process used to arrive at these equations. 699 00:35:47,300 --> 00:35:50,890 And, more generally, we want to solve this set of 700 00:35:50,890 --> 00:35:54,320 equations, tangent stiffness matrix, still on the left-hand 701 00:35:54,320 --> 00:36:00,200 side, times a displacement increment corresponding to the 702 00:36:00,200 --> 00:36:05,080 iteration, I. And on the right-hand side, we have the 703 00:36:05,080 --> 00:36:08,060 load vector corresponding to the loads at time T plus delta 704 00:36:08,060 --> 00:36:12,960 t and the nodal point force vector corresponding to the 705 00:36:12,960 --> 00:36:16,970 internal element stresses at time T plus delta t and at the 706 00:36:16,970 --> 00:36:19,590 end of iteration I minus 1. 707 00:36:19,590 --> 00:36:22,560 This is here the out-of-balance load vector 708 00:36:22,560 --> 00:36:28,050 corresponding to the start of iteration I. This 709 00:36:28,050 --> 00:36:30,950 out-of-balance load vector, and with that tangent 710 00:36:30,950 --> 00:36:34,920 stiffness matrix, gives us an increment in the nodal point 711 00:36:34,920 --> 00:36:39,160 displacement vector corresponding to iteration I. 712 00:36:39,160 --> 00:36:43,730 We take this one, add it to the previous displacements 713 00:36:43,730 --> 00:36:46,640 that we had already, namely those corresponding to time T 714 00:36:46,640 --> 00:36:51,070 plus delta t at the end of iteration I minus 1, and we 715 00:36:51,070 --> 00:36:55,640 obtain a better approximation to the displacements at time T 716 00:36:55,640 --> 00:36:57,130 plus delta t. 717 00:36:57,130 --> 00:37:01,010 Notice, in this iteration, we have initial conditions that 718 00:37:01,010 --> 00:37:06,060 I've given down here in iteration I equal to 1. 719 00:37:06,060 --> 00:37:09,830 On the right-hand side we need, of this equation here, 720 00:37:09,830 --> 00:37:11,650 we need this vector. 721 00:37:11,650 --> 00:37:16,570 And that vector is given by the vector TF, the nodal point 722 00:37:16,570 --> 00:37:18,900 force vector corresponding to the internal element stresses 723 00:37:18,900 --> 00:37:22,210 at time T. We also need initial conditions on the 724 00:37:22,210 --> 00:37:27,510 displacement vector, and those are given right here. 725 00:37:27,510 --> 00:37:30,610 Notice that, when I is equal to 1, this set of equations 726 00:37:30,610 --> 00:37:32,490 reduces to the equations that we had on the 727 00:37:32,490 --> 00:37:34,240 previous view graph. 728 00:37:34,240 --> 00:37:39,040 Once again, we will derive these equations from basic 729 00:37:39,040 --> 00:37:42,190 continuum mechanics principles in the later lectures. 730 00:37:42,190 --> 00:37:46,050 And we will also talk about iterative schemes to 731 00:37:46,050 --> 00:37:47,930 accelerate the convergence in the 732 00:37:47,930 --> 00:37:52,040 solution of these equations. 733 00:37:52,040 --> 00:37:54,650 This then brings me to the end of what I wanted to discuss 734 00:37:54,650 --> 00:37:58,050 this you on the view graphs. 735 00:37:58,050 --> 00:38:01,810 I'd like to now turn to an example solution, an 736 00:38:01,810 --> 00:38:03,130 interesting example solution. 737 00:38:03,130 --> 00:38:07,620 But, moreover, it displays the kinds of nonlinearalities that 738 00:38:07,620 --> 00:38:12,270 we will be talking about in this set of lectures. 739 00:38:12,270 --> 00:38:16,150 This example solution is documented on slides. 740 00:38:16,150 --> 00:38:21,070 So let me walk over here and put on the first slide. 741 00:38:21,070 --> 00:38:24,590 Here we show the structure that we want to consider. 742 00:38:24,590 --> 00:38:28,820 It's a spherical shell subjected to pressure loading. 743 00:38:28,820 --> 00:38:33,650 The material data for the shell are given here. 744 00:38:33,650 --> 00:38:37,260 Notice we also include the mass density because we will 745 00:38:37,260 --> 00:38:40,095 be talking about dynamic analysis of 746 00:38:40,095 --> 00:38:42,220 this shell as well. 747 00:38:42,220 --> 00:38:46,100 We will also introduce an imperfection of the shell. 748 00:38:46,100 --> 00:38:48,850 And that imperfection is given down here. 749 00:38:48,850 --> 00:38:53,340 Notice it is given as a function of the angle phi, 750 00:38:53,340 --> 00:38:58,580 which you see up here, as well as a thickness, H, of the 751 00:38:58,580 --> 00:39:06,340 shell, the Legendre polynomial given here and parameter data 752 00:39:06,340 --> 00:39:08,260 that we will be varying. 753 00:39:08,260 --> 00:39:10,920 First, we will consider a static analysis of the shell 754 00:39:10,920 --> 00:39:13,310 and then the dynamic analysis of the shell. 755 00:39:13,310 --> 00:39:17,720 On the next slide, now you see the model that was used for 756 00:39:17,720 --> 00:39:21,890 the analysis of the shell, twenty 8-node elements, 757 00:39:21,890 --> 00:39:25,030 axisymmetric elements subjected to the pressure 758 00:39:25,030 --> 00:39:30,040 loading, and, we are modeling, the pressure loading in 759 00:39:30,040 --> 00:39:33,260 deformation-dependent loading. 760 00:39:33,260 --> 00:39:37,020 That means that, as the shell deforms, the pressure will 761 00:39:37,020 --> 00:39:40,900 remain perpendicular to the shell surface 762 00:39:40,900 --> 00:39:42,730 to which it is applied. 763 00:39:42,730 --> 00:39:47,170 On the next slide now, we show the first analysis results. 764 00:39:47,170 --> 00:39:51,460 On this axis, we plot P/PCR where PCR is the buckling 765 00:39:51,460 --> 00:39:54,050 pressure of the shell, assuming elastic conditions, 766 00:39:54,050 --> 00:39:57,570 and calculated using analytical solution. 767 00:39:57,570 --> 00:40:00,410 On this axis, we plot the rate of displacement of the shell 768 00:40:00,410 --> 00:40:03,020 at phi equal to zero. 769 00:40:03,020 --> 00:40:08,030 The dotted line here shows the linear elastic solution to the 770 00:40:08,030 --> 00:40:09,130 shell problem. 771 00:40:09,130 --> 00:40:11,340 Notice that these dots, of course, could 772 00:40:11,340 --> 00:40:13,520 be continued here. 773 00:40:13,520 --> 00:40:16,610 The total Lagrangian formulation, which includes 774 00:40:16,610 --> 00:40:19,640 large displacement and large rotation effects, gives us a 775 00:40:19,640 --> 00:40:23,150 buckling pressure of 0.98 PRC. 776 00:40:23,150 --> 00:40:26,800 A Material-nonlinear-only analysis, including the 777 00:40:26,800 --> 00:40:29,970 elasto-plastic material conditions, but not including, 778 00:40:29,970 --> 00:40:32,220 in other words, large displacement, large rotation 779 00:40:32,220 --> 00:40:37,760 effects, gives us this solution here. 780 00:40:37,760 --> 00:40:41,650 Notice that, of course, this MNO analysis does not give us 781 00:40:41,650 --> 00:40:45,160 a proper bucking prediction for the shell. 782 00:40:45,160 --> 00:40:49,290 To obtain the elastic-plastic buckling load of the shell, we 783 00:40:49,290 --> 00:40:50,770 have to perform a total 784 00:40:50,770 --> 00:40:52,790 Lagrangian formulation solution. 785 00:40:52,790 --> 00:40:58,280 And we obtain this load level here as the bucking pressure. 786 00:40:58,280 --> 00:41:01,380 This analysis was performed with no 787 00:41:01,380 --> 00:41:03,260 imperfections on the shell. 788 00:41:03,260 --> 00:41:06,440 The next slide now shows the results of the 789 00:41:06,440 --> 00:41:08,100 same kind of analyses. 790 00:41:08,100 --> 00:41:13,280 But at an imperfection of delta 0.1, notice here the 791 00:41:13,280 --> 00:41:17,590 linear analysis, denoted as E, the Material-nonlinear-only 792 00:41:17,590 --> 00:41:22,770 analysis results, denoted as EP, the large displacement, 793 00:41:22,770 --> 00:41:25,300 large rotation analysis results. 794 00:41:25,300 --> 00:41:28,810 But assuming an elastic condition for the shell, 795 00:41:28,810 --> 00:41:32,970 denoted as E, TL, TL standing for total Lagrangian 796 00:41:32,970 --> 00:41:36,500 formulation and the elasto-plastic large 797 00:41:36,500 --> 00:41:41,500 displacement solution given here, the results given here 798 00:41:41,500 --> 00:41:43,730 and denoted as EP, TL. 799 00:41:46,330 --> 00:41:50,190 Notice that, because of the imperfection, the load, the 800 00:41:50,190 --> 00:41:54,310 maximum load-carrying capacity of the shell, is decreased. 801 00:41:54,310 --> 00:41:59,340 On the next slide now we show the maximum load-carrying 802 00:41:59,340 --> 00:42:03,410 capacities for delta equal to zero, delta equal to 0.1, 803 00:42:03,410 --> 00:42:06,550 delta equal to 0.2, delta equal to 0.4. 804 00:42:06,550 --> 00:42:10,090 In each case, we have used the total Lagrangian formulation 805 00:42:10,090 --> 00:42:12,150 including elasto-plastic conditions. 806 00:42:12,150 --> 00:42:15,800 In other words, of course, we had to include the large 807 00:42:15,800 --> 00:42:19,330 displacement effects in order to pick up the proper buckling 808 00:42:19,330 --> 00:42:25,070 load or load-carrying capacity of the shell, so these are the 809 00:42:25,070 --> 00:42:27,490 load-carrying capacities of the shell for different 810 00:42:27,490 --> 00:42:29,660 imperfection levels. 811 00:42:29,660 --> 00:42:33,590 The next slide now shows the results obtained using a 812 00:42:33,590 --> 00:42:35,630 dynamic analysis. 813 00:42:35,630 --> 00:42:40,400 At this time, we apply the pressure instantaneously as a 814 00:42:40,400 --> 00:42:43,410 step load, constant in time. 815 00:42:43,410 --> 00:42:46,450 Notice we are plotting here now mean displacement data of 816 00:42:46,450 --> 00:42:49,840 the shell and time along this axis. 817 00:42:49,840 --> 00:42:53,480 The Etl solution results are very close to 818 00:42:53,480 --> 00:42:54,790 the E solution results. 819 00:42:54,790 --> 00:42:57,910 I think you know now what I mean by E and Etl. 820 00:42:57,910 --> 00:43:01,130 The E Ptl solution results are quite close to 821 00:43:01,130 --> 00:43:03,170 the EP solution results. 822 00:43:03,170 --> 00:43:07,320 This means that the shell is stable under this load 823 00:43:07,320 --> 00:43:09,250 application. 824 00:43:09,250 --> 00:43:12,790 In other words, there is no buckling, no increase in 825 00:43:12,790 --> 00:43:17,000 deformations as time progresses. 826 00:43:17,000 --> 00:43:22,910 The next slide now shows, as we increase the imperfection 827 00:43:22,910 --> 00:43:29,990 level from delta equal to zero to delta equal to 0.1, the 828 00:43:29,990 --> 00:43:35,150 displacements of the shell using the E Ptl solution or 829 00:43:35,150 --> 00:43:37,870 formulation increase with time. 830 00:43:37,870 --> 00:43:42,790 Therefore, the shell is unstable under this pressure 831 00:43:42,790 --> 00:43:43,520 application. 832 00:43:43,520 --> 00:43:47,310 Of course, dynamic pressure application, no. 833 00:43:47,310 --> 00:43:51,430 Notice the other solution results are given here. 834 00:43:51,430 --> 00:43:55,140 Therefore, we now can play with the imperfection level 835 00:43:55,140 --> 00:43:58,960 and with the load level in order to estimate the 836 00:43:58,960 --> 00:44:02,400 load-carrying capacity of the shell for each of these cases. 837 00:44:02,400 --> 00:44:08,210 And on this slide, you're seeing for a constant 838 00:44:08,210 --> 00:44:11,840 imperfection level, we increase the load 839 00:44:11,840 --> 00:44:13,370 applied to the shell. 840 00:44:13,370 --> 00:44:16,930 And you can see that, at 0.4 PCR, the 841 00:44:16,930 --> 00:44:18,250 shell is still stable. 842 00:44:18,250 --> 00:44:21,950 At 0.5 PCR, the shell is unstable. 843 00:44:21,950 --> 00:44:24,720 On the final slide now, we show all the information that 844 00:44:24,720 --> 00:44:26,230 we have obtained in the analysis. 845 00:44:26,230 --> 00:44:29,310 On this axis, we are plotting the buckling load, on this 846 00:44:29,310 --> 00:44:31,460 axis, the amplitude of imperfection. 847 00:44:31,460 --> 00:44:33,470 This curve here corresponds to the static 848 00:44:33,470 --> 00:44:34,780 backing of the shell. 849 00:44:34,780 --> 00:44:36,840 And this curve corresponds to the dynamic 850 00:44:36,840 --> 00:44:39,100 buckling of the shell. 851 00:44:39,100 --> 00:44:41,230 Of course, we could not discuss all the details 852 00:44:41,230 --> 00:44:43,610 regarding these analyses in this short time. 853 00:44:43,610 --> 00:44:46,460 Please refer to the study guide in which you'll find a 854 00:44:46,460 --> 00:44:49,300 reference to the paper in which we have discussed the 855 00:44:49,300 --> 00:44:52,750 details of these analyses. 856 00:44:52,750 --> 00:44:55,940 This then brings me to the end of this lecture. 857 00:44:55,940 --> 00:44:57,800 And I hope you enjoy the course.