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 --> 00:00:18,825 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:18,825 --> 00:00:20,075 ocw.mit.edu. 8 00:00:21,808 --> 00:00:23,920 PROFESSOR: Ladies and gentlemen, welcome to this 9 00:00:23,920 --> 00:00:26,870 lecture on nonlinear finite element analysis of solids and 10 00:00:26,870 --> 00:00:27,870 structures. 11 00:00:27,870 --> 00:00:30,050 In this lecture, I would like to continue with our 12 00:00:30,050 --> 00:00:34,090 discussion of solution methods that we use to solve the 13 00:00:34,090 --> 00:00:37,580 finite element equations in nonlinear static analysis. 14 00:00:37,580 --> 00:00:40,020 We considered already in the previous lecture a number of 15 00:00:40,020 --> 00:00:43,720 solution techniques to solve this set of equations. 16 00:00:43,720 --> 00:00:47,350 t plus delta tr is equal to t plus delta tf, where t plus 17 00:00:47,350 --> 00:00:51,720 delta tr is the load vector of externally applied loads at 18 00:00:51,720 --> 00:00:56,410 time t plus delta t, and t plus tf is a nodal point force 19 00:00:56,410 --> 00:00:59,210 vector corresponding to the internal element stresses at 20 00:00:59,210 --> 00:01:01,080 time t plus delta t. 21 00:01:01,080 --> 00:01:04,010 We talked about the full Newton-Raphson method, the 22 00:01:04,010 --> 00:01:07,270 modified Newton-Raphson method, the BFGS method, the 23 00:01:07,270 --> 00:01:08,920 initial stress method, and we discussed 24 00:01:08,920 --> 00:01:11,110 also convergence criteria. 25 00:01:11,110 --> 00:01:13,890 I've summarized here the equations corresponding to the 26 00:01:13,890 --> 00:01:16,260 modified Newton iteration. 27 00:01:16,260 --> 00:01:19,700 Now, a distinguishing feature of these solution techniques 28 00:01:19,700 --> 00:01:25,610 is that the analyst has to prescribe the externally 29 00:01:25,610 --> 00:01:26,680 applied load vector 30 00:01:26,680 --> 00:01:30,520 corresponding to all load steps. 31 00:01:30,520 --> 00:01:36,240 This means schematically that if you have this kind of 32 00:01:36,240 --> 00:01:40,380 displacement load curve, or load displacement curve, shown 33 00:01:40,380 --> 00:01:44,350 here in red, the analyst has to prescribe prior to the 34 00:01:44,350 --> 00:01:50,760 analysis the load levels at which the response is sought. 35 00:01:50,760 --> 00:01:54,800 I've here indicated 1r as a load level corresponding to 36 00:01:54,800 --> 00:01:58,560 load step one, 2r as a load level corresponding to load 37 00:01:58,560 --> 00:02:01,150 step two, et cetera. 38 00:02:01,150 --> 00:02:04,130 Of course, the corresponding solutions that the analyst is 39 00:02:04,130 --> 00:02:08,509 looking for are the displacements. 40 00:02:08,509 --> 00:02:11,810 Now, what might very well happen in practical analysis 41 00:02:11,810 --> 00:02:16,560 is that the analyst chooses certain load levels, and at 42 00:02:16,560 --> 00:02:19,570 the particular load level convergence difficulties are 43 00:02:19,570 --> 00:02:20,470 encountered. 44 00:02:20,470 --> 00:02:23,550 Too many iterations are required to converge, for 45 00:02:23,550 --> 00:02:28,650 example, in a reasonable cost. 46 00:02:28,650 --> 00:02:33,210 And typically, say, if this happens at this load step 47 00:02:33,210 --> 00:02:36,800 here, the analyst decides to restock. 48 00:02:36,800 --> 00:02:41,700 And this means that a solution corresponding to r4 here was 49 00:02:41,700 --> 00:02:43,730 not obtained. 50 00:02:43,730 --> 00:02:45,160 I'm scratching that out. 51 00:02:45,160 --> 00:02:50,400 And the analyst restarts now with a smaller load step, 52 00:02:50,400 --> 00:02:54,270 namely indicated by these green lines here. 53 00:02:54,270 --> 00:02:58,030 The solution is obtained corresponding to that 54 00:02:58,030 --> 00:03:01,850 configuration, corresponding to that configuration, that 55 00:03:01,850 --> 00:03:04,570 configuration, which corresponds in this particular 56 00:03:04,570 --> 00:03:08,290 case now, to the level of r4, because we have three load 57 00:03:08,290 --> 00:03:10,400 steps to reach now r4. 58 00:03:10,400 --> 00:03:15,470 And say, at this point now, again, for example, we may 59 00:03:15,470 --> 00:03:17,970 encounter solution difficulties. 60 00:03:17,970 --> 00:03:21,400 The analyst is not able to obtain this solution here. 61 00:03:21,400 --> 00:03:25,540 Once again, the load step has to be smaller. 62 00:03:25,540 --> 00:03:29,370 To continue the analysis, a restart is necessary at this 63 00:03:29,370 --> 00:03:31,390 particular configuration. 64 00:03:31,390 --> 00:03:35,020 And like this the analyst tries to get closer and closer 65 00:03:35,020 --> 00:03:37,600 to the collapse load. 66 00:03:37,600 --> 00:03:42,190 The conclusion is that we have difficulties in calculating 67 00:03:42,190 --> 00:03:46,890 the collapse loads when we use the techniques that I 68 00:03:46,890 --> 00:03:50,690 described in the last lecture. 69 00:03:50,690 --> 00:03:55,860 Of course, if you restart enough, if you use small 70 00:03:55,860 --> 00:03:59,000 enough to load steps, you will ultimately get very close to 71 00:03:59,000 --> 00:04:02,030 this collapse load, but this is tedious in a practical 72 00:04:02,030 --> 00:04:04,730 analysis, and we would like to have really an automatic 73 00:04:04,730 --> 00:04:09,900 scheme that directly can trace out the collapse load and you 74 00:04:09,900 --> 00:04:14,460 can also go into the post-collapse response, which 75 00:04:14,460 --> 00:04:19,740 is the response beyond this ultimate limit load here. 76 00:04:19,740 --> 00:04:24,830 Well, I've prepared some view graphs to show you, discuss 77 00:04:24,830 --> 00:04:29,730 with you, a solution technique that actually can be employed 78 00:04:29,730 --> 00:04:34,680 to trace automatically load displacement response out, and 79 00:04:34,680 --> 00:04:37,980 go beyond the collapse load, as well. 80 00:04:37,980 --> 00:04:41,620 The idea is that we want to obtain more rapid convergence 81 00:04:41,620 --> 00:04:44,170 in each load step. 82 00:04:44,170 --> 00:04:48,230 We want to have the program automatically select load 83 00:04:48,230 --> 00:04:50,200 increments. 84 00:04:50,200 --> 00:04:53,440 And we want to also be able to solve for the 85 00:04:53,440 --> 00:04:55,120 post-buckling response. 86 00:04:55,120 --> 00:04:59,980 Now, this here means, of course, that a priori, prior 87 00:04:59,980 --> 00:05:03,400 to the analysis, you may not know, in fact you do not know, 88 00:05:03,400 --> 00:05:06,880 for which loads you will obtain the solution. 89 00:05:06,880 --> 00:05:10,190 We get the total load displacement response curve, 90 00:05:10,190 --> 00:05:13,190 but the discrete load levels at which the response was 91 00:05:13,190 --> 00:05:18,230 calculated is being decided by the program. 92 00:05:18,230 --> 00:05:21,380 An effective solution would proceed as follows, 93 00:05:21,380 --> 00:05:23,020 schematically, of course. 94 00:05:23,020 --> 00:05:25,280 Here we have the load axis. 95 00:05:25,280 --> 00:05:27,850 Here we have the displacement axis. 96 00:05:27,850 --> 00:05:30,430 The solution scheme would start with large load 97 00:05:30,430 --> 00:05:35,190 increments in the region where the response is almost linear, 98 00:05:35,190 --> 00:05:40,720 then cut the load increments to a smaller size, and 99 00:05:40,720 --> 00:05:44,830 capture, of course, the ultimate limit load here, and 100 00:05:44,830 --> 00:05:48,650 then decrease the load, as shown here, to go into the 101 00:05:48,650 --> 00:05:51,080 post-buckling, post-collapse response. 102 00:05:51,080 --> 00:05:55,000 And the solution scheme, of course, could continue up this 103 00:05:55,000 --> 00:05:57,240 branch as well. 104 00:05:57,240 --> 00:06:01,280 And the solution scheme should automatically select the load 105 00:06:01,280 --> 00:06:05,190 step sizes, depending on the convergence that has been 106 00:06:05,190 --> 00:06:10,620 encountered in the previous load steps. 107 00:06:10,620 --> 00:06:16,040 We compute now, t plus delta tr using this equation here. 108 00:06:16,040 --> 00:06:19,810 Notice that in this equation, t plus delta t lambda is an 109 00:06:19,810 --> 00:06:22,600 unknown scaler that is going to be determined by the 110 00:06:22,600 --> 00:06:26,920 program, and r is a vector that gives a particular load 111 00:06:26,920 --> 00:06:28,095 distribution. 112 00:06:28,095 --> 00:06:30,280 r is constant. 113 00:06:30,280 --> 00:06:32,500 It may contain the contributions of soft surface 114 00:06:32,500 --> 00:06:37,050 pressures, of concentrated loads, et cetera. 115 00:06:37,050 --> 00:06:40,030 The point is that r is constant and the program will 116 00:06:40,030 --> 00:06:42,390 automatically adjust lambda. 117 00:06:42,390 --> 00:06:46,340 Therefore, we assume in our collapse analysis that the 118 00:06:46,340 --> 00:06:52,120 loads are increasing and decreasing all in the same way 119 00:06:52,120 --> 00:06:54,190 as decided by one scale. 120 00:06:54,190 --> 00:06:58,170 Of course, one could extend such an algorithm by 121 00:06:58,170 --> 00:07:02,460 introducing another vector, calling this r1, introducing a 122 00:07:02,460 --> 00:07:05,020 vector r2, with another scaler. 123 00:07:05,020 --> 00:07:08,000 And then have two scalers that have to be adjusted 124 00:07:08,000 --> 00:07:09,830 automatically by the program. 125 00:07:09,830 --> 00:07:12,300 And perhaps one can even think of three scalers, and so on, 126 00:07:12,300 --> 00:07:15,350 but then the algorithm would become quite complicated. 127 00:07:15,350 --> 00:07:19,040 So we look at a scheme that only has one scaler here, and 128 00:07:19,040 --> 00:07:22,680 one load vector that remains constant 129 00:07:22,680 --> 00:07:24,400 throughout the solution. 130 00:07:24,400 --> 00:07:28,830 As an example here, you see in blue the load 131 00:07:28,830 --> 00:07:32,020 intensity, t lambda r. 132 00:07:32,020 --> 00:07:35,740 Notice a concentrated load here, a pressure load here, 133 00:07:35,740 --> 00:07:40,610 and at the time t plus delta t we have the red intensity, t 134 00:07:40,610 --> 00:07:45,750 plus delta t lambda r, and of course this concentrated load 135 00:07:45,750 --> 00:07:50,530 has increased in the same way as the pressure has increased, 136 00:07:50,530 --> 00:07:53,080 as shown right here. 137 00:07:53,080 --> 00:07:57,260 The basic approach of the solution scheme is shown on 138 00:07:57,260 --> 00:07:58,700 this view graph. 139 00:07:58,700 --> 00:08:01,650 Here we have the low displacement curve that we are 140 00:08:01,650 --> 00:08:05,150 looking for, shown in red again. 141 00:08:05,150 --> 00:08:07,880 Of course, loads plotted vertically, displacements 142 00:08:07,880 --> 00:08:09,640 plotted horizontally. 143 00:08:09,640 --> 00:08:15,250 Now, notice that if this is an equilibrium point, the red 144 00:08:15,250 --> 00:08:19,950 dot, and if you as the analyst were to chose an increment in 145 00:08:19,950 --> 00:08:24,480 load shown by this black line-- 146 00:08:24,480 --> 00:08:27,440 I'm putting my pointer now onto it-- 147 00:08:27,440 --> 00:08:32,140 then you can see that this black line defined by this 148 00:08:32,140 --> 00:08:39,860 load increment would mean that we have a large number of 149 00:08:39,860 --> 00:08:41,750 iterations necessary. 150 00:08:41,750 --> 00:08:45,390 If you need a large number of iterations, to converge to the 151 00:08:45,390 --> 00:08:47,480 new equilibrium configuration. 152 00:08:47,480 --> 00:08:52,230 In fact, you can see that this red curve is almost parallel 153 00:08:52,230 --> 00:08:56,470 to the black line, so we can anticipate large convergence 154 00:08:56,470 --> 00:09:00,530 difficulties trying to get to a solution with the load 155 00:09:00,530 --> 00:09:06,660 increment being equal from here to that black line. 156 00:09:06,660 --> 00:09:10,330 The important feature of the algorithm that I want to 157 00:09:10,330 --> 00:09:16,000 present to you now, is that this load level is the first 158 00:09:16,000 --> 00:09:19,190 load level which we start in the iteration, but then the 159 00:09:19,190 --> 00:09:24,130 algorithm automatically cuts down this load level, as shown 160 00:09:24,130 --> 00:09:30,920 by this arc here, by the blue arc, until this equilibrium 161 00:09:30,920 --> 00:09:33,230 point is solved for. 162 00:09:33,230 --> 00:09:38,650 And this means that the convergence is much increased. 163 00:09:38,650 --> 00:09:41,280 In other words, the algorithm does not have great 164 00:09:41,280 --> 00:09:45,370 difficulties calculating this particular equilibrium point. 165 00:09:45,370 --> 00:09:48,130 And from here, of course, the same is repeated. 166 00:09:48,130 --> 00:09:52,640 And like this the algorithm goes along this red 167 00:09:52,640 --> 00:09:57,160 low-displacement curve, and traces this whole curve out. 168 00:09:57,160 --> 00:10:00,680 Let us look a bit at the notation that I will be using, 169 00:10:00,680 --> 00:10:03,020 because it's important that we get familiar 170 00:10:03,020 --> 00:10:05,600 with it at this stage. 171 00:10:05,600 --> 00:10:11,010 Notice t lambda r, of course, is the load at time t. 172 00:10:11,010 --> 00:10:13,120 The corresponding displacement is tu. 173 00:10:13,120 --> 00:10:15,650 We used that notation earlier already. 174 00:10:15,650 --> 00:10:19,250 Notice that the increment in displacement from time t to 175 00:10:19,250 --> 00:10:25,010 time t plus delta t is u, shown here in green. 176 00:10:25,010 --> 00:10:29,110 Notice that the increment in the load from time t to time t 177 00:10:29,110 --> 00:10:32,660 plus delta t is, of course, given by the change in this 178 00:10:32,660 --> 00:10:34,920 load factor here. 179 00:10:34,920 --> 00:10:37,200 And that change in the load factor is lambda. 180 00:10:40,560 --> 00:10:46,310 Now, this lambda value is also given in this equation here. 181 00:10:46,310 --> 00:10:47,580 It's buried in there. 182 00:10:47,580 --> 00:10:51,380 Notice this lambda of i here. 183 00:10:51,380 --> 00:10:54,700 We are writing it as a sum all of delta lambda k's We will 184 00:10:54,700 --> 00:11:01,300 solve for all these delta lambda k's And this lambda i, 185 00:11:01,300 --> 00:11:05,940 when i becomes very large, converges to the lambda that I 186 00:11:05,940 --> 00:11:07,670 have up here. 187 00:11:07,670 --> 00:11:11,660 Similarly, this ui which is written as the sum of the 188 00:11:11,660 --> 00:11:17,930 delta uk's this ui converges to the u 189 00:11:17,930 --> 00:11:19,890 that I'm showing here. 190 00:11:19,890 --> 00:11:25,730 So, u and lambda, shown in green here, are the exact 191 00:11:25,730 --> 00:11:29,270 values, the values that you want to converge to with 192 00:11:29,270 --> 00:11:32,560 lambda i and ui. 193 00:11:32,560 --> 00:11:36,150 Notice once again, delta lambda k, summing all these 194 00:11:36,150 --> 00:11:39,640 delta lambda k's up we get lambda i, and summing all the 195 00:11:39,640 --> 00:11:42,050 delta uk's up, we get ui. 196 00:11:42,050 --> 00:11:45,520 We should keep that in mind for the discussion that you 197 00:11:45,520 --> 00:11:47,870 want to go through just now. 198 00:11:47,870 --> 00:11:52,090 The governing equations are now as follows. 199 00:11:52,090 --> 00:11:55,420 On the left hand side we have a tension stiffness matrix. 200 00:11:55,420 --> 00:11:57,890 In the modified Newton iteration we would have tau 201 00:11:57,890 --> 00:11:59,670 equal to t. 202 00:11:59,670 --> 00:12:02,500 Delta ui is our displacement increment vector. 203 00:12:02,500 --> 00:12:05,680 This is here, this scaler, t plus 204 00:12:05,680 --> 00:12:10,270 delta t lambda i, unknown. 205 00:12:10,270 --> 00:12:12,460 This one we know. 206 00:12:12,460 --> 00:12:15,960 And that one we want to calculate in the iteration i. 207 00:12:15,960 --> 00:12:18,740 In other words, this is an increment that is unknown at 208 00:12:18,740 --> 00:12:20,940 the beginning of the iteration. 209 00:12:20,940 --> 00:12:24,060 Here we have the nodal point force vector corresponding to 210 00:12:24,060 --> 00:12:27,510 the elements stresses at time t plus delta t, and at the end 211 00:12:27,510 --> 00:12:29,970 of iteration i minus 1. 212 00:12:29,970 --> 00:12:34,240 Notice that this is a set of linear equations in the 213 00:12:34,240 --> 00:12:37,680 unknowns that are used. 214 00:12:37,680 --> 00:12:42,030 There are n such unknowns, and there's one more unknown here. 215 00:12:42,030 --> 00:12:46,410 So, we have n equations in n, plus 1 unknowns. 216 00:12:46,410 --> 00:12:48,080 We need, therefore, one more equation. 217 00:12:48,080 --> 00:12:52,655 And that equation is given by this constraint equation. 218 00:12:52,655 --> 00:12:59,300 The f here on delta lambda and delta u, some constraint 219 00:12:59,300 --> 00:13:02,650 equation here, gives us the additional equation that we 220 00:13:02,650 --> 00:13:07,380 need for the solution these n plus 1 coupled equations. 221 00:13:07,380 --> 00:13:10,780 The unknowns, of course, are these values here-- 222 00:13:10,780 --> 00:13:13,630 n plus 1 unknowns. 223 00:13:13,630 --> 00:13:18,320 To solve these equilibrium equations, we can rewrite them 224 00:13:18,320 --> 00:13:19,570 as shown here. 225 00:13:21,850 --> 00:13:25,660 In other words, the equations on the finite element 226 00:13:25,660 --> 00:13:29,550 displacements can be split up, if you look at the previous 227 00:13:29,550 --> 00:13:33,150 view graph, directly into two sets of 228 00:13:33,150 --> 00:13:35,240 equations as shown here. 229 00:13:35,240 --> 00:13:40,280 The interesting point is that, this equation here does not 230 00:13:40,280 --> 00:13:42,780 involve a delta lambda. 231 00:13:42,780 --> 00:13:47,380 This equation here does not involve a delta lambda either. 232 00:13:47,380 --> 00:13:51,260 Now, having calculated from this equation this vector, and 233 00:13:51,260 --> 00:13:55,240 from that equation that vector, we can directly write 234 00:13:55,240 --> 00:13:59,240 delta u, as shown here, of course, now involving the 235 00:13:59,240 --> 00:14:01,550 delta lambda. 236 00:14:01,550 --> 00:14:06,060 And this is a form of equation for delta u that we will be 237 00:14:06,060 --> 00:14:10,150 using a little later. 238 00:14:10,150 --> 00:14:14,710 And the first constraint equation I'd like to introduce 239 00:14:14,710 --> 00:14:18,250 you to, the one that I briefly mentioned earlier already, 240 00:14:18,250 --> 00:14:22,430 where we talked about an arc which is used to swing the 241 00:14:22,430 --> 00:14:27,300 load level around, so as to get to the low displacement 242 00:14:27,300 --> 00:14:31,500 curve very quickly, and that constraint equation is this 243 00:14:31,500 --> 00:14:35,290 spherical constant arc-length criterion, 244 00:14:35,290 --> 00:14:37,030 which is written here. 245 00:14:37,030 --> 00:14:42,060 Notice here lambda i, here ui. 246 00:14:42,060 --> 00:14:45,600 Of course both of these quantities create, in essence, 247 00:14:45,600 --> 00:14:48,640 beta factor and delta l squared. 248 00:14:48,640 --> 00:14:56,760 Delta l is set at the beginning of the load step. 249 00:14:56,760 --> 00:14:58,890 Delta l, in other words, is constant 250 00:14:58,890 --> 00:15:00,640 throughout the load step. 251 00:15:00,640 --> 00:15:03,670 The value of delta l is chosen based on what has happened in 252 00:15:03,670 --> 00:15:05,670 the previous load steps. 253 00:15:05,670 --> 00:15:10,410 Having chosen delta l, you can calculate the right hand side, 254 00:15:10,410 --> 00:15:14,450 and on the left hand side, we have a constraint between the 255 00:15:14,450 --> 00:15:19,340 increment in lambda, and the increment in delta u. 256 00:15:19,340 --> 00:15:23,990 Remember, lambda I is the sum of all of the delta lambda k's 257 00:15:23,990 --> 00:15:28,180 and ui is the sum of all of the delta uk's I just showed 258 00:15:28,180 --> 00:15:30,170 that on the previous view graph. 259 00:15:30,170 --> 00:15:33,820 Therefore, if we substitute here, we get a constraint 260 00:15:33,820 --> 00:15:39,560 equation between the delta uk, and the delta lambda k. 261 00:15:39,560 --> 00:15:43,490 Well, here we have the definitions once more. 262 00:15:43,490 --> 00:15:48,780 Notice beta is a normalizing factor which is applied in 263 00:15:48,780 --> 00:15:50,840 order to make these terms here dimensionless. 264 00:15:54,540 --> 00:15:57,490 The equation may be solved as follows. 265 00:15:57,490 --> 00:16:02,380 Using that lambda is equal to this equation here. 266 00:16:02,380 --> 00:16:05,940 ui is given as that. 267 00:16:05,940 --> 00:16:09,370 Of course, this here can be expanded using the information 268 00:16:09,370 --> 00:16:12,000 that we discussed earlier. 269 00:16:12,000 --> 00:16:15,640 And substituting for this value and that value into the 270 00:16:15,640 --> 00:16:18,710 constraint equation, we directly obtain a quadratic 271 00:16:18,710 --> 00:16:23,400 equation in delta lambda i Remember that these two 272 00:16:23,400 --> 00:16:28,540 vectors here are, of course, known. 273 00:16:28,540 --> 00:16:31,110 These two vectors are known, and that one is known also, 274 00:16:31,110 --> 00:16:33,980 because that was established in the previous iteration. 275 00:16:37,150 --> 00:16:44,460 If we geometrically interpret this solution scheme, we find 276 00:16:44,460 --> 00:16:45,390 the following. 277 00:16:45,390 --> 00:16:47,390 Here we have our equilibrium point. 278 00:16:47,390 --> 00:16:50,670 Here is delta lambda, which is constant throughout the 279 00:16:50,670 --> 00:16:52,430 solution step. 280 00:16:52,430 --> 00:16:54,865 And let's assume the use of modified 281 00:16:54,865 --> 00:16:57,330 Newton-Raphson iteration. 282 00:16:57,330 --> 00:17:02,260 We start off with this tangent here at this solution point, 283 00:17:02,260 --> 00:17:08,420 the equilibrium point, and we iterate around this arc using 284 00:17:08,420 --> 00:17:14,630 the constant stiffness matrix at each of these points. 285 00:17:14,630 --> 00:17:19,740 In the first iteration, we get to this point. 286 00:17:19,740 --> 00:17:21,359 Second iteration, that point. 287 00:17:21,359 --> 00:17:22,920 Third iteration, that point. 288 00:17:22,920 --> 00:17:27,109 And that signified also looking here to the 289 00:17:27,109 --> 00:17:28,060 displacements. 290 00:17:28,060 --> 00:17:32,010 Notice t plus delta t u1 is the displacement value 291 00:17:32,010 --> 00:17:35,210 calculated in the first iteration. 292 00:17:35,210 --> 00:17:38,030 This is the one calculated in the second iteration. 293 00:17:38,030 --> 00:17:41,820 That is the one calculated in the third iteration. 294 00:17:41,820 --> 00:17:44,220 We use a constant stiffness matrix here. 295 00:17:44,220 --> 00:17:47,660 Of course, we could also change the stiffness matrix, 296 00:17:47,660 --> 00:17:51,290 go in other words, to a stiffness matrix that has been 297 00:17:51,290 --> 00:17:55,230 updated in the iteration corresponding to the current 298 00:17:55,230 --> 00:17:57,970 stress and displacement conditions, and then 299 00:17:57,970 --> 00:18:00,110 convergence would be more rapid. 300 00:18:00,110 --> 00:18:03,090 But of course there would also be a higher cost involved, 301 00:18:03,090 --> 00:18:05,590 because whenever we calculate with a new stiffness matrix, 302 00:18:05,590 --> 00:18:07,780 and we triangularize that stiffness matrix, there's a 303 00:18:07,780 --> 00:18:10,880 considerable cost involved in doing so. 304 00:18:10,880 --> 00:18:14,080 This is one scheme, and we find this scheme quite 305 00:18:14,080 --> 00:18:19,300 effective, except that when we want to calculate limit loads, 306 00:18:19,300 --> 00:18:23,330 we find that near the limit load the scheme can require 307 00:18:23,330 --> 00:18:24,860 quite a number of iterations. 308 00:18:24,860 --> 00:18:27,430 And so, we have been looking for other schemes that might 309 00:18:27,430 --> 00:18:31,140 do better in the range of the limit load. 310 00:18:31,140 --> 00:18:35,960 Well, here we have one other scheme, namely the "constant" 311 00:18:35,960 --> 00:18:39,320 increment of external work criterion. 312 00:18:39,320 --> 00:18:42,700 We have here quotes on the constant, because we are 313 00:18:42,700 --> 00:18:47,860 setting w on the right hand side in the first iteration, 314 00:18:47,860 --> 00:18:51,410 as shown here, and in the next iteration then, 315 00:18:51,410 --> 00:18:53,100 we have a zero here. 316 00:18:53,100 --> 00:19:00,040 Notice w is a value that is calculated in each load step, 317 00:19:00,040 --> 00:19:05,890 and as soon as we find that the first scheme-- and 318 00:19:05,890 --> 00:19:09,740 obviously, constant arc lengths criterion scheme-- 319 00:19:09,740 --> 00:19:13,800 has difficulties converging, or marches too slow, then we 320 00:19:13,800 --> 00:19:16,370 switch to this "constant" increment of external work 321 00:19:16,370 --> 00:19:21,160 criterion, with a w from the previous step, the increment 322 00:19:21,160 --> 00:19:24,200 of external work from the previous step. 323 00:19:24,200 --> 00:19:28,620 We use that for the current step, and then this equation 324 00:19:28,620 --> 00:19:32,290 here gives us a constraint on delta lambda 1. 325 00:19:32,290 --> 00:19:37,430 Notice that this equation here means really that we are 326 00:19:37,430 --> 00:19:44,390 looking at is this area here, the shaded area, being w, and 327 00:19:44,390 --> 00:19:49,130 we want to have the delta lambda 1 in such a way that 328 00:19:49,130 --> 00:19:55,360 this shaded area w here is equal to what we have had as 329 00:19:55,360 --> 00:20:00,400 the increment of external work in the previous load step. 330 00:20:00,400 --> 00:20:05,010 Well, this gives us delta lambda (1), and having 331 00:20:05,010 --> 00:20:09,930 obtained delta lambda (1), we can go into the next 332 00:20:09,930 --> 00:20:15,440 iterations, 2 to 3, and so on for i, and as I mentioned 333 00:20:15,440 --> 00:20:18,940 already earlier, we then have a zero on the right hand side. 334 00:20:18,940 --> 00:20:21,670 Otherwise the same left hand side. 335 00:20:21,670 --> 00:20:25,230 And this left hand side has now two solutions. 336 00:20:25,230 --> 00:20:30,330 This solution here is disregarded, because this 337 00:20:30,330 --> 00:20:34,380 solution would reverse the direction of the load, and we 338 00:20:34,380 --> 00:20:35,790 don't want to admit that. 339 00:20:35,790 --> 00:20:39,930 We want to basically have the load increase more and more 340 00:20:39,930 --> 00:20:42,830 until the total collapse of the structure is reached, and 341 00:20:42,830 --> 00:20:45,980 then of course it may decrease to go into the post-buckling 342 00:20:45,980 --> 00:20:47,960 response, but we don't want to have the load totally 343 00:20:47,960 --> 00:20:52,780 reversed, and so this is a solution that we now use for 344 00:20:52,780 --> 00:20:55,370 delta lambda (i). 345 00:20:55,370 --> 00:21:00,120 Notice that for a single degree of freedom system, we 346 00:21:00,120 --> 00:21:04,390 would immediately find that delta u (2) must be zero. 347 00:21:04,390 --> 00:21:04,730 Why? 348 00:21:04,730 --> 00:21:08,270 Because we notice that the load vector is orthogonal to 349 00:21:08,270 --> 00:21:09,990 the displacement vector. 350 00:21:09,990 --> 00:21:12,450 Now, for a single degree of freedom system, of course, 351 00:21:12,450 --> 00:21:19,230 each of these are just numbers, and since the r is 352 00:21:19,230 --> 00:21:22,330 not zero, because that is of course a prescribed value 353 00:21:22,330 --> 00:21:25,890 prior to the analysis, this is not zero, delta 354 00:21:25,890 --> 00:21:28,340 u (2) must be zero. 355 00:21:28,340 --> 00:21:31,740 And that shows already the effectiveness, just show it a 356 00:21:31,740 --> 00:21:33,845 bit, of this algorithm. 357 00:21:36,690 --> 00:21:40,310 All the algorithm altogether is now as follows. 358 00:21:40,310 --> 00:21:43,060 We specify r. 359 00:21:43,060 --> 00:21:47,520 In other words, the analyst has to specify the load 360 00:21:47,520 --> 00:21:52,210 distribution, concentrated loads, distributed loads, that 361 00:21:52,210 --> 00:21:56,620 shall be dealt with in this collapse analysis, and also 362 00:21:56,620 --> 00:22:00,850 the displacement at one degree of freedom corresponding to 363 00:22:00,850 --> 00:22:02,730 delta t lambda. 364 00:22:02,730 --> 00:22:07,120 This is done to just start the algorithm. 365 00:22:07,120 --> 00:22:12,160 Once the analyst has specified the displacement at one degree 366 00:22:12,160 --> 00:22:17,900 of freedom, the program can solve for delta tu-- 367 00:22:17,900 --> 00:22:20,680 in other words, all the other displacements-- 368 00:22:20,680 --> 00:22:25,470 and this delta tu corresponds to delta t lambda. 369 00:22:25,470 --> 00:22:29,750 We feel that to start the algorithm it is easier for the 370 00:22:29,750 --> 00:22:34,500 analyst to prescribe the displacement at one node 371 00:22:34,500 --> 00:22:39,290 corresponding to delta t lambda-- 372 00:22:39,290 --> 00:22:43,440 Rather, to prescribe this displacement than to prescribe 373 00:22:43,440 --> 00:22:47,170 the first load level, because if you don't have much of an 374 00:22:47,170 --> 00:22:50,600 idea how the structure really behaves, it can be very 375 00:22:50,600 --> 00:22:54,500 difficult to give a load level that is reasonable, and is not 376 00:22:54,500 --> 00:22:57,040 too close to the ultimate limit load, or is already 377 00:22:57,040 --> 00:22:59,040 perhaps even beyond the ultimate limit load. 378 00:22:59,040 --> 00:23:03,980 So we feel that this is a good way to proceed. 379 00:23:03,980 --> 00:23:07,600 For example, if you analyze the collapse of a shell, you 380 00:23:07,600 --> 00:23:12,130 may want to pick a node that you know will have non-zero 381 00:23:12,130 --> 00:23:17,510 displacement and assign at that node a displacement, say, 382 00:23:17,510 --> 00:23:20,250 one third of the thickness of the shell. 383 00:23:20,250 --> 00:23:23,070 That displacement then would give you delta t lambda, and 384 00:23:23,070 --> 00:23:26,320 surely would be less than the ultimate 385 00:23:26,320 --> 00:23:28,860 limit load of the shell. 386 00:23:28,860 --> 00:23:32,020 This means, once you have solved for delta t lambda and 387 00:23:32,020 --> 00:23:35,620 delta tu, we can set delta l. 388 00:23:35,620 --> 00:23:38,260 The arc lengths for the next load steps. 389 00:23:38,260 --> 00:23:40,475 And now the program does everything automatically. 390 00:23:40,475 --> 00:23:46,730 It selects the increment in displacements and loads using 391 00:23:46,730 --> 00:23:50,340 this arc-lengths criterion that we just discussed. 392 00:23:50,340 --> 00:23:53,880 We use 1, which is the arc-lengths criterion, for the 393 00:23:53,880 --> 00:23:55,280 next load steps. 394 00:23:55,280 --> 00:24:00,280 We calculate w for each load step. 395 00:24:00,280 --> 00:24:03,480 When w does not change appreciably, or when there are 396 00:24:03,480 --> 00:24:06,830 difficulties with the arc-lengths criterion, with 1, 397 00:24:06,830 --> 00:24:10,310 in other words, then we switch to our constant increment and 398 00:24:10,310 --> 00:24:11,640 external work criterion. 399 00:24:11,640 --> 00:24:13,680 We call that the 2 criterion. 400 00:24:16,490 --> 00:24:21,820 Once again, notice that delta l is calculated in each load 401 00:24:21,820 --> 00:24:28,380 step, and it is adjusted based on the number of iterations 402 00:24:28,380 --> 00:24:32,870 that we have been using in the previous load steps. 403 00:24:32,870 --> 00:24:38,390 In other words, if you go, say, up to load step three, 404 00:24:38,390 --> 00:24:41,860 and you have, say, obtained a solution up to load step 405 00:24:41,860 --> 00:24:48,830 three, the program has used a certain delta in load step 406 00:24:48,830 --> 00:24:52,310 three, and now the program looks at the number of 407 00:24:52,310 --> 00:24:55,980 iterations that were performed to get to the equilibrium 408 00:24:55,980 --> 00:24:59,020 position at the end of load steps three. 409 00:24:59,020 --> 00:25:02,420 Based on these number of iterations, the program then 410 00:25:02,420 --> 00:25:08,200 cuts delta l down, or makes delta l larger, depending on 411 00:25:08,200 --> 00:25:11,580 whether the iterations were very many that you used, or 412 00:25:11,580 --> 00:25:15,110 were very few to get to the equilibrium configuration at 413 00:25:15,110 --> 00:25:16,900 load step three. 414 00:25:16,900 --> 00:25:20,430 So, the a program automatically adjusts delta l. 415 00:25:20,430 --> 00:25:24,660 Also notice that the stiffness matrix is recalculated when 416 00:25:24,660 --> 00:25:26,860 convergence is slow. 417 00:25:26,860 --> 00:25:30,320 In fact, a full Newton-Raphson iteration is performed 418 00:25:30,320 --> 00:25:34,050 automatically, the program switched automatically from 419 00:25:34,050 --> 00:25:35,810 the modified Newton-Raphson iteration to the full 420 00:25:35,810 --> 00:25:38,220 Newton-Raphson iteration when it is 421 00:25:38,220 --> 00:25:41,090 deemed to be more effective. 422 00:25:41,090 --> 00:25:45,250 This is the automatic load incrementation scheme that I 423 00:25:45,250 --> 00:25:47,220 wanted to discuss with you. 424 00:25:47,220 --> 00:25:51,210 And I'd now like to go over to one other topic, a very 425 00:25:51,210 --> 00:25:54,680 important topic as well in the analysis of the nonlinear 426 00:25:54,680 --> 00:25:58,900 response of structures, and this is the linearized 427 00:25:58,900 --> 00:26:01,390 buckling analysis. 428 00:26:01,390 --> 00:26:07,170 In the linearized buckling analysis, we want to predict a 429 00:26:07,170 --> 00:26:11,930 collapse load, the buckling load of the structure via this 430 00:26:11,930 --> 00:26:14,740 criterion here. 431 00:26:14,740 --> 00:26:19,590 The determinant we know at the collapse point of the 432 00:26:19,590 --> 00:26:23,600 stiffness matrix is zero. 433 00:26:23,600 --> 00:26:25,400 The matrix is, another words, singular. 434 00:26:28,050 --> 00:26:34,770 The criterion that we use here determined of k is equal to 435 00:26:34,770 --> 00:26:38,390 zero, of course, means once again 436 00:26:38,390 --> 00:26:39,940 the k matrix is singular. 437 00:26:39,940 --> 00:26:44,440 And that means that we really have a solution to this set of 438 00:26:44,440 --> 00:26:46,880 equations that is nontrivial. 439 00:26:46,880 --> 00:26:49,120 The trivial solution, of course, would always be u star 440 00:26:49,120 --> 00:26:53,010 is equal to zero, but that is not a real solution that we're 441 00:26:53,010 --> 00:26:54,620 interested in. 442 00:26:54,620 --> 00:27:00,870 A solution, a nontrivial solution exists to this set of 443 00:27:00,870 --> 00:27:04,450 equations only in the case when tau k is singular. 444 00:27:04,450 --> 00:27:07,330 And, of course, that is the same as saying that the 445 00:27:07,330 --> 00:27:12,810 determined of k, tau k is equal to zero. 446 00:27:12,810 --> 00:27:19,470 If tau k is singular the structure is unstable, and the 447 00:27:19,470 --> 00:27:26,660 collapse load situation has been reached. 448 00:27:26,660 --> 00:27:29,970 Let's look at what this means physically. 449 00:27:29,970 --> 00:27:31,360 If we have here a beam. 450 00:27:31,360 --> 00:27:34,890 If we look at a beam, pinned at this end, pinned at that 451 00:27:34,890 --> 00:27:39,710 end, subjected to a certain load, then if the load takes 452 00:27:39,710 --> 00:27:44,930 on a certain value, the smallest imbalance of the 453 00:27:44,930 --> 00:27:49,830 load, a very small load this way, would immediately result 454 00:27:49,830 --> 00:27:51,870 in very large displacements. 455 00:27:51,870 --> 00:27:55,790 That's what the buckling criterion tells us. 456 00:27:55,790 --> 00:28:01,640 In other words, at tau r, the buckling load, the structure 457 00:28:01,640 --> 00:28:06,650 is unstable to any smallest load imbalance. 458 00:28:06,650 --> 00:28:09,890 Of course, the material data for the structure are given. 459 00:28:09,890 --> 00:28:12,410 And in this particular case, we use the data 460 00:28:12,410 --> 00:28:15,490 of an elastic beam. 461 00:28:15,490 --> 00:28:18,460 In the linearized buckling analysis we want to predict 462 00:28:18,460 --> 00:28:23,110 the load level, and the buckling load shape 463 00:28:23,110 --> 00:28:30,420 corresponding to the buckling situation, by linearizing 464 00:28:30,420 --> 00:28:33,860 about a particular configuration that is not 465 00:28:33,860 --> 00:28:38,010 necessarily very close to the buckling load level. 466 00:28:38,010 --> 00:28:42,340 In fact, we assume that tau k is given via this relationship 467 00:28:42,340 --> 00:28:46,520 here, and tau r is given via this relationship. 468 00:28:46,520 --> 00:28:53,040 Notice here t minus delta tk and tk are known values, known 469 00:28:53,040 --> 00:28:56,960 stiffness matrices, and similarly, this load vector, 470 00:28:56,960 --> 00:29:00,920 and that load vector, and that one, are known values. 471 00:29:00,920 --> 00:29:02,750 Lambda is a scaler. 472 00:29:02,750 --> 00:29:05,220 So basically, what we are trying to do in the linearized 473 00:29:05,220 --> 00:29:09,510 buckling analysis is to linearize about the 474 00:29:09,510 --> 00:29:14,190 configuration, not necessarily very close to the collapse 475 00:29:14,190 --> 00:29:19,990 load configuration, and establish with this linearized 476 00:29:19,990 --> 00:29:27,270 relationship an equation from which we can calculate an 477 00:29:27,270 --> 00:29:35,410 increment in load that gives us an approximation to the 478 00:29:35,410 --> 00:29:38,160 actual collapse load. 479 00:29:38,160 --> 00:29:41,750 Pictorially, we are doing this here. 480 00:29:41,750 --> 00:29:46,520 We have here the load displacement curve in red. 481 00:29:46,520 --> 00:29:51,390 We have at displacement t minus delta 482 00:29:51,390 --> 00:29:54,440 tu, this load level. 483 00:29:54,440 --> 00:29:59,090 At displacement tu we have that load level. 484 00:29:59,090 --> 00:30:04,360 And if you linearize about these configurations here, 485 00:30:04,360 --> 00:30:07,130 using the stiffness matrices corresponding to these two 486 00:30:07,130 --> 00:30:13,510 load levels, we can predict a collapse load, which because 487 00:30:13,510 --> 00:30:17,500 of this linearization is, of course, not exactly the 488 00:30:17,500 --> 00:30:20,990 collapse load, but we hope that we are close to that 489 00:30:20,990 --> 00:30:22,240 actual collapse load. 490 00:30:24,970 --> 00:30:28,430 Pictorially here, we do the following. 491 00:30:28,430 --> 00:30:33,500 Notice here we have the k plotted and the 492 00:30:33,500 --> 00:30:36,070 lambda value plotted. 493 00:30:36,070 --> 00:30:37,890 Along this axis. 494 00:30:37,890 --> 00:30:43,010 At a particular value, lambda equal to 1, we 495 00:30:43,010 --> 00:30:44,870 would have this point. 496 00:30:44,870 --> 00:30:48,250 Now let's look first along this axis here. 497 00:30:48,250 --> 00:30:50,170 Here we have tk. 498 00:30:50,170 --> 00:30:52,710 Here we have t minus delta tk. 499 00:30:52,710 --> 00:30:56,260 Notice tk is smaller than t minus delta tk. 500 00:30:56,260 --> 00:30:58,640 We are looking here, of course, at a single degree of 501 00:30:58,640 --> 00:31:00,220 freedom case again. 502 00:31:00,220 --> 00:31:02,980 This means that the structure becomes softer, we are getting 503 00:31:02,980 --> 00:31:06,730 closer to the collapse load. 504 00:31:06,730 --> 00:31:09,910 In our linearized buckling analysis we put really through 505 00:31:09,910 --> 00:31:15,070 this point and that point a straight line. 506 00:31:15,070 --> 00:31:20,560 That projects us to a point, lambda 1, at which tau k is 507 00:31:20,560 --> 00:31:23,740 equal to zero. 508 00:31:23,740 --> 00:31:26,800 Corresponding to what we're doing here with the stiffness 509 00:31:26,800 --> 00:31:30,570 matrix, we have here also picture for the loads. 510 00:31:30,570 --> 00:31:37,500 Notice t minus delta tr here, tr there, and a straight line 511 00:31:37,500 --> 00:31:40,670 through it brings us up to tau r. 512 00:31:40,670 --> 00:31:43,750 Tau r is the collapse load. 513 00:31:43,750 --> 00:31:47,490 Tau r is the collapse load, linearized collapse load, 514 00:31:47,490 --> 00:31:52,290 corresponding to the value that we have here 515 00:31:52,290 --> 00:31:53,910 obtained for lambda 1. 516 00:31:56,690 --> 00:32:00,930 The problem of solving for lambda such that the 517 00:32:00,930 --> 00:32:06,610 determinant of tau k is zero is really nothing else than an 518 00:32:06,610 --> 00:32:07,690 eigenproblem. 519 00:32:07,690 --> 00:32:12,060 And the eigenproblem that we have to consider there is 520 00:32:12,060 --> 00:32:13,560 written down here. 521 00:32:13,560 --> 00:32:17,970 Notice phi is an eigenvector, lambda is the eigenvalue-- 522 00:32:17,970 --> 00:32:20,400 unknown at this point-- 523 00:32:20,400 --> 00:32:28,080 and here we have a difference in matrices that we know, and 524 00:32:28,080 --> 00:32:29,950 once again the eigenvector. 525 00:32:29,950 --> 00:32:31,740 Notice one interesting point. 526 00:32:31,740 --> 00:32:35,590 That to establish this difference in matrix, all you 527 00:32:35,590 --> 00:32:38,520 need to do in the computer program is to store the 528 00:32:38,520 --> 00:32:42,630 previous difference matrix, and subtract from the previous 529 00:32:42,630 --> 00:32:45,750 difference matrix corresponding to load step t 530 00:32:45,750 --> 00:32:49,220 minus delta t, the current difference matrix. 531 00:32:49,220 --> 00:32:55,550 This is interesting to note that because we don't need to 532 00:32:55,550 --> 00:32:57,740 separate out element difference matrices. 533 00:32:57,740 --> 00:33:00,710 We don't need to talk about nonlinear strain stiffness 534 00:33:00,710 --> 00:33:03,670 matrices coming from the elements, and linear strain 535 00:33:03,670 --> 00:33:06,410 stiffness matrices coming from the elements. 536 00:33:06,410 --> 00:33:10,300 We talked about, of course, these particular element 537 00:33:10,300 --> 00:33:12,870 stiffness matrices in previous lectures. 538 00:33:12,870 --> 00:33:16,160 We don't separate any out here. 539 00:33:16,160 --> 00:33:19,390 In fact, what we're doing, we simply take all of the 540 00:33:19,390 --> 00:33:23,110 contributions from the elements into account here, 541 00:33:23,110 --> 00:33:26,180 and all of the contribution towards element into account 542 00:33:26,180 --> 00:33:30,070 here, of course corresponding to time t minus delta t here, 543 00:33:30,070 --> 00:33:33,280 and corresponding to time t here. 544 00:33:33,280 --> 00:33:35,440 This therefore, is a very simple operation in the 545 00:33:35,440 --> 00:33:36,690 computer program. 546 00:33:38,870 --> 00:33:44,380 And having established this eigenvalue problem, we can 547 00:33:44,380 --> 00:33:48,990 schematically look at the eigenvalues that would become 548 00:33:48,990 --> 00:33:50,610 calculated. 549 00:33:50,610 --> 00:33:54,320 Here we have some positive eigenvalues indicated, and 550 00:33:54,320 --> 00:33:57,710 here we have some negative eigenvalues indicated. 551 00:33:57,710 --> 00:34:00,930 Notice that there's a possibility of negative 552 00:34:00,930 --> 00:34:03,450 eigenvalues as well. 553 00:34:03,450 --> 00:34:07,360 A physical problem that shows how you can get a negative 554 00:34:07,360 --> 00:34:10,500 eigenvalue is shown in this view graph. 555 00:34:10,500 --> 00:34:13,800 Here you have a structure, a frame structure, so to say, 556 00:34:13,800 --> 00:34:15,960 that is subjected to a compressive load 557 00:34:15,960 --> 00:34:17,739 and a tensile load. 558 00:34:17,739 --> 00:34:21,330 Now, notice that this compressive load would of 559 00:34:21,330 --> 00:34:24,230 course initiate buckling here, and there would be a 560 00:34:24,230 --> 00:34:26,690 corresponding to a positive eigenvalue. 561 00:34:26,690 --> 00:34:30,239 However, this tensile load has to reverse its sign in order 562 00:34:30,239 --> 00:34:34,060 for this member to buckle, and that would correspond then to 563 00:34:34,060 --> 00:34:36,360 a negative eigenvalue. 564 00:34:36,360 --> 00:34:40,030 Now when we try to solve eigenvalue problems that have 565 00:34:40,030 --> 00:34:43,360 negative and positive eigenvalues, there can be 566 00:34:43,360 --> 00:34:46,280 difficulties with the eigenvalue solution method 567 00:34:46,280 --> 00:34:47,320 that you're using. 568 00:34:47,320 --> 00:34:51,230 It is easier for a solution method to simply deal with 569 00:34:51,230 --> 00:34:53,330 eigenvalues that are only positive. 570 00:34:53,330 --> 00:34:58,360 And for that reason we are reformulating the basic 571 00:34:58,360 --> 00:35:02,770 eigenvalue equation into this form. 572 00:35:02,770 --> 00:35:06,560 Here you have now the eigenvalue problem simply 573 00:35:06,560 --> 00:35:13,240 rewritten, really, with gamma being the eigenvalue, and phi 574 00:35:13,240 --> 00:35:16,350 still being the eigenvector, gamma of course, being a 575 00:35:16,350 --> 00:35:19,580 function of lambda. 576 00:35:19,580 --> 00:35:25,260 With this eigenvalue problem then, we have only positive 577 00:35:25,260 --> 00:35:30,090 eigenvalues on gamma, of course, and we are interested 578 00:35:30,090 --> 00:35:33,960 in finding the smallest gamma value. 579 00:35:33,960 --> 00:35:36,450 Because the smallest gamma value corresponds to the 580 00:35:36,450 --> 00:35:40,330 smallest lambda 1 positive value. 581 00:35:40,330 --> 00:35:43,900 And of course, sometimes you also want to calculate gamma 2 582 00:35:43,900 --> 00:35:49,100 corresponding to the second smallest positive lambda 583 00:35:49,100 --> 00:35:50,900 value, and so on. 584 00:35:50,900 --> 00:35:53,540 The negative lambda values lie over here. 585 00:35:56,960 --> 00:36:00,650 The value of the linearized buckling analysis can be 586 00:36:00,650 --> 00:36:02,400 summarized as follows. 587 00:36:02,400 --> 00:36:05,680 The buckling analysis is not very expensive. 588 00:36:05,680 --> 00:36:10,340 It gives insight into possible modes of failure. 589 00:36:10,340 --> 00:36:13,730 For applicability, however, the pre-buckling displacements 590 00:36:13,730 --> 00:36:16,830 should be small. 591 00:36:16,830 --> 00:36:21,060 And very important it is that the buckling analysis yields 592 00:36:21,060 --> 00:36:25,300 modes, buckling mode shapes, that can be very effectively 593 00:36:25,300 --> 00:36:30,460 used to impose imperfections onto a structure and to study 594 00:36:30,460 --> 00:36:33,970 the sensitivity of the structure to imperfections. 595 00:36:33,970 --> 00:36:37,440 Many structures are very sensitive, particularly shell 596 00:36:37,440 --> 00:36:40,350 structures, are very sensitive to initial imperfections in 597 00:36:40,350 --> 00:36:44,620 the geometry, and this linearized buckling analysis 598 00:36:44,620 --> 00:36:49,220 gives us the mechanisms and means to calculate mode shapes 599 00:36:49,220 --> 00:36:52,580 that we can impose onto the perfect structure as a 600 00:36:52,580 --> 00:36:56,010 geometric imperfection, onto the geometry of the perfect 601 00:36:56,010 --> 00:37:00,420 structure, to study then the behavior of the structure 602 00:37:00,420 --> 00:37:04,480 subject to these imperfections. 603 00:37:04,480 --> 00:37:10,550 But it is very important that the procedure be only employed 604 00:37:10,550 --> 00:37:14,100 with great care, because the results can be quite 605 00:37:14,100 --> 00:37:14,840 misleading. 606 00:37:14,840 --> 00:37:17,280 The buckling loads that you might calculate in the 607 00:37:17,280 --> 00:37:20,440 linearized buckling analysis, these buckling loads may be 608 00:37:20,440 --> 00:37:24,490 way higher than the actual buckling loads that you should 609 00:37:24,490 --> 00:37:26,330 be using in your design. 610 00:37:26,330 --> 00:37:30,560 And we will actually show some examples in the next lecture 611 00:37:30,560 --> 00:37:35,020 pertaining to this particular problem here. 612 00:37:35,020 --> 00:37:38,840 And one should always keep in mind that the procedure really 613 00:37:38,840 --> 00:37:43,120 predicts only physically realistic buckling loads when 614 00:37:43,120 --> 00:37:46,910 we have structures that behave close to the 615 00:37:46,910 --> 00:37:48,460 Euler column type. 616 00:37:48,460 --> 00:37:53,150 In other words, that buckle with very small initial 617 00:37:53,150 --> 00:37:57,640 displacements, or pre-buckling displacements, I should say. 618 00:37:57,640 --> 00:37:59,930 Let's look at an example. 619 00:37:59,930 --> 00:38:03,930 Here we have the example of an arch 620 00:38:03,930 --> 00:38:07,540 subjected to uniform pressure. 621 00:38:07,540 --> 00:38:11,050 The geometric data of the arch are given here. 622 00:38:11,050 --> 00:38:12,890 The material data are given as well. 623 00:38:12,890 --> 00:38:14,890 Notice it's an elastic arch. 624 00:38:14,890 --> 00:38:21,250 And it's an arch with cross section b and h here, being 625 00:38:21,250 --> 00:38:23,680 both equal to 1. 626 00:38:23,680 --> 00:38:25,970 We also want to consider only the two-dimensional 627 00:38:25,970 --> 00:38:27,200 motion of the arch. 628 00:38:27,200 --> 00:38:28,980 In other words, two-dimensional action. 629 00:38:28,980 --> 00:38:33,700 We do not allow out-of-plane bucking for this arch. 630 00:38:33,700 --> 00:38:40,240 Now the finite element model that we select for the arch is 631 00:38:40,240 --> 00:38:43,800 a model of 10 2-node isoparametric beam elements. 632 00:38:43,800 --> 00:38:46,780 We will talk about these isoparametric beam elements-- 633 00:38:46,780 --> 00:38:49,190 we call them also isobeam elements-- 634 00:38:49,190 --> 00:38:50,840 in a later lecture. 635 00:38:50,840 --> 00:38:55,780 And for the two-dimensional motion of the arch, we model 636 00:38:55,780 --> 00:38:58,200 the complete arch. 637 00:38:58,200 --> 00:39:01,810 The purpose of the analysis is to determine the collapse 638 00:39:01,810 --> 00:39:05,580 mechanism, the collapse load level of the structure. 639 00:39:05,580 --> 00:39:11,250 And not only this particular part, but we also want to 640 00:39:11,250 --> 00:39:16,150 calculate the post-collapse response of the structure. 641 00:39:16,150 --> 00:39:22,750 Let us now go through the solution that was performed 642 00:39:22,750 --> 00:39:24,140 for this arch. 643 00:39:24,140 --> 00:39:29,180 As a first step we calculated the linearized buckling loads 644 00:39:29,180 --> 00:39:30,390 and corresponding mode shapes. 645 00:39:30,390 --> 00:39:32,480 And we calculated two. 646 00:39:32,480 --> 00:39:35,620 Interesting to note that the first mode corresponds to this 647 00:39:35,620 --> 00:39:42,000 pressure, and it's an anti-symmetric mode. 648 00:39:42,000 --> 00:39:48,010 The second buckling mode shape looks like this. 649 00:39:48,010 --> 00:39:50,370 It's a symmetric buckling mode shape. 650 00:39:50,370 --> 00:39:54,430 Notice that, in other words, the anti-symmetric buckling 651 00:39:54,430 --> 00:39:59,660 mode corresponds to a lower load than the symmetry 652 00:39:59,660 --> 00:40:02,320 buckling mode shape. 653 00:40:02,320 --> 00:40:05,350 Having performed the linearized buckling analysis, 654 00:40:05,350 --> 00:40:08,580 we next use our automatic load stepping algorithm to 655 00:40:08,580 --> 00:40:11,630 calculate the displacement response of the arch as the 656 00:40:11,630 --> 00:40:13,000 load increases. 657 00:40:13,000 --> 00:40:16,630 Here on this view graph, we have plotted the pressure 658 00:40:16,630 --> 00:40:19,680 vertically up here, and the displacement of the center of 659 00:40:19,680 --> 00:40:21,410 the arch horizontally. 660 00:40:21,410 --> 00:40:26,790 Notice this black curve here shows a computed response 661 00:40:26,790 --> 00:40:30,210 using about 60 steps in the analysis. 662 00:40:30,210 --> 00:40:34,060 Notice it is a collapse load, of course, here, and this is a 663 00:40:34,060 --> 00:40:36,950 post-collapse response. 664 00:40:36,950 --> 00:40:40,480 It's interesting to note that the collapse load predicted 665 00:40:40,480 --> 00:40:42,950 using the buckling analysis and linearized buckling 666 00:40:42,950 --> 00:40:48,240 analysis given by this blue line lies below this actual 667 00:40:48,240 --> 00:40:52,600 ultimate load predicted for using the automatic load 668 00:40:52,600 --> 00:40:54,870 stepping algorithm. 669 00:40:54,870 --> 00:40:58,470 And we have to ask ourselves why is that the case? 670 00:40:58,470 --> 00:41:05,740 Well, the computed response of the the perfect symmetry arch 671 00:41:05,740 --> 00:41:09,900 does not allow the anti-symmetric displacements 672 00:41:09,900 --> 00:41:15,130 to take place, and it is the anti-symmetric displacements 673 00:41:15,130 --> 00:41:20,550 that, of course, initialize, so to say, anti-symmetric 674 00:41:20,550 --> 00:41:22,880 buckling response. 675 00:41:22,880 --> 00:41:25,910 We have computed the response of a perfect symmetric arch 676 00:41:25,910 --> 00:41:31,580 that is subjected to a perfectly symmetric load, and 677 00:41:31,580 --> 00:41:36,810 this does not allow the anti-symmetric buckling mode 678 00:41:36,810 --> 00:41:39,800 of the arch to come into effect. 679 00:41:39,800 --> 00:41:45,020 However, a real structure will contain imperfections, and 680 00:41:45,020 --> 00:41:51,990 hence it will go into the anti-symmetric behavior, and 681 00:41:51,990 --> 00:41:56,450 the actual collapse load will be below the one that we have 682 00:41:56,450 --> 00:41:59,630 predicted in our first analysis using the automatic 683 00:41:59,630 --> 00:42:02,180 load stepping algorithm, and the results of which 684 00:42:02,180 --> 00:42:05,470 I've just shown you. 685 00:42:05,470 --> 00:42:09,640 Therefore, to really obtain a realistic collapse load, we 686 00:42:09,640 --> 00:42:16,410 have to now impose onto the perfect symmetry arch a 687 00:42:16,410 --> 00:42:22,550 geometric imperfection which allows the anti-symmetric 688 00:42:22,550 --> 00:42:25,030 behavior to take place. 689 00:42:25,030 --> 00:42:31,470 And we do so by adding to the geometry of the arch to the 690 00:42:31,470 --> 00:42:36,040 nodal point coordinates of the arch a multiple of the 691 00:42:36,040 --> 00:42:39,580 anti-symmetric mode shape. 692 00:42:39,580 --> 00:42:43,400 This collapse mode is scaled so that the magnitude of 693 00:42:43,400 --> 00:42:47,600 imperfection is less than 100. 694 00:42:47,600 --> 00:42:51,240 This resulting imperfect arch is, of course, no longer 695 00:42:51,240 --> 00:42:55,830 symmetric, and we now solve for the response of that arch 696 00:42:55,830 --> 00:43:00,210 using our automatic load stepping algorithm again. 697 00:43:00,210 --> 00:43:03,700 This is the response that you'll see. 698 00:43:03,700 --> 00:43:07,280 Pressure and displacement of the arch. 699 00:43:07,280 --> 00:43:12,850 Notice this is well below the linearized buckling analysis 700 00:43:12,850 --> 00:43:16,420 solution, the blue solution, that I showed you on the 701 00:43:16,420 --> 00:43:19,380 earlier response graph. 702 00:43:19,380 --> 00:43:24,520 And this is here a much more realistic estimate for the 703 00:43:24,520 --> 00:43:26,060 collapse load of the arch. 704 00:43:26,060 --> 00:43:29,270 This is the realistic estimate that you would be using, for 705 00:43:29,270 --> 00:43:32,860 example, in the design of the actual structure. 706 00:43:32,860 --> 00:43:35,540 It's also interesting to look at the deflective shape of the 707 00:43:35,540 --> 00:43:40,390 structure, and here we have the deflective shape of the 708 00:43:40,390 --> 00:43:45,400 perfect arch, of being symmetric at this 709 00:43:45,400 --> 00:43:47,350 displacement value. 710 00:43:47,350 --> 00:43:52,430 And here we have a deflective shape of the imperfect arch at 711 00:43:52,430 --> 00:43:54,500 this particular displacement value. 712 00:43:54,500 --> 00:43:56,750 Of course, this one is not a totally 713 00:43:56,750 --> 00:43:59,230 symmetric deflective shape. 714 00:43:59,230 --> 00:44:02,640 Well, we have discussed now quite a number of solution 715 00:44:02,640 --> 00:44:07,290 schemes that we use to solve the finite element equations 716 00:44:07,290 --> 00:44:09,160 in static non-linear analysis. 717 00:44:09,160 --> 00:44:13,100 We have looked in the previous lecture at some schemes-- the 718 00:44:13,100 --> 00:44:15,880 full Newton-Raphson method, the modified Newton-Raphson 719 00:44:15,880 --> 00:44:19,320 method, the BFGS method, the initial stress method-- 720 00:44:19,320 --> 00:44:21,535 in which we have to prescribe the load levels. 721 00:44:21,535 --> 00:44:26,780 For each load step we have to prescribe the load level prior 722 00:44:26,780 --> 00:44:30,570 to the analysis, by the input of the computer program, and 723 00:44:30,570 --> 00:44:33,390 in this lecture I have shared with you some experiences 724 00:44:33,390 --> 00:44:35,940 regarding an automatic load stepping scheme regarding 725 00:44:35,940 --> 00:44:37,960 linearized buckling analysis. 726 00:44:37,960 --> 00:44:40,560 What we have not done yet, really, in my opinion is to 727 00:44:40,560 --> 00:44:43,510 look at enough examples, and that's what I would like to do 728 00:44:43,510 --> 00:44:44,630 in the next lecture. 729 00:44:44,630 --> 00:44:46,300 Thank you very much for your attention.