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 00:00:22,070 --> 00:00:23,830 PROFESSOR: Ladies and gentlemen, welcome to this 9 00:00:23,830 --> 00:00:26,660 lecture on nonlinear finite element analysis of solids and 10 00:00:26,660 --> 00:00:28,210 structures. 11 00:00:28,210 --> 00:00:31,250 In this lecture I would like to discuss with you the total 12 00:00:31,250 --> 00:00:34,880 Lagrangian formulation of the two-noded truss element. 13 00:00:34,880 --> 00:00:37,210 In the previous lecture, we considered the updated 14 00:00:37,210 --> 00:00:40,390 Lagrangian formulation of the same element. 15 00:00:40,390 --> 00:00:43,360 We started from the general continuum mechanics equation, 16 00:00:43,360 --> 00:00:46,180 and obtained the governing finite element matrices. 17 00:00:46,180 --> 00:00:48,440 I would like to do the same now for the total Lagrangian 18 00:00:48,440 --> 00:00:49,460 formulation. 19 00:00:49,460 --> 00:00:52,690 And we will see that to find element matrices are indeed 20 00:00:52,690 --> 00:00:55,690 the same that we obtained in the updated Lagrangian 21 00:00:55,690 --> 00:00:57,110 formulation. 22 00:00:57,110 --> 00:01:01,210 Here we have the governing equation of the principle of 23 00:01:01,210 --> 00:01:04,519 virtual work that we derived in an earlier lecture. 24 00:01:04,519 --> 00:01:07,660 This, of course, is a governing equation that we 25 00:01:07,660 --> 00:01:11,100 apply to obtain the governing equation, finite element 26 00:01:11,100 --> 00:01:15,450 equations, for two or three dimensional analysis, and also 27 00:01:15,450 --> 00:01:16,580 for the truss element. 28 00:01:16,580 --> 00:01:21,280 It's a very general equation that we discussed earlier in 29 00:01:21,280 --> 00:01:23,290 great detail. 30 00:01:23,290 --> 00:01:26,870 We use this equation now for the truss element. 31 00:01:26,870 --> 00:01:31,620 And the first point that we have to look at is what stress 32 00:01:31,620 --> 00:01:34,040 measure and corresponding strain measure do we 33 00:01:34,040 --> 00:01:35,290 need to deal with? 34 00:01:35,290 --> 00:01:39,340 Now for the truss, it is important to realize that the 35 00:01:39,340 --> 00:01:44,670 only non-zero stress is this stress here, t0 S11. 36 00:01:44,670 --> 00:01:47,990 In other words, only the 11 component is non-zero. 37 00:01:47,990 --> 00:01:51,460 Let us look at this fact in detail. 38 00:01:51,460 --> 00:01:54,250 There's first of all, what we might call a mathematical 39 00:01:54,250 --> 00:01:55,820 explanation. 40 00:01:55,820 --> 00:01:59,050 Let us assume, of course, that the cross-section here is 41 00:01:59,050 --> 00:02:02,020 constant, the same way as we did in the discussion of the 42 00:02:02,020 --> 00:02:03,790 updated Lagrangian formulation. 43 00:02:03,790 --> 00:02:08,430 And here we have the element in its original configuration, 44 00:02:08,430 --> 00:02:11,790 and the element has undergone large displacements and large 45 00:02:11,790 --> 00:02:16,700 rotation to get into this configuration at time t. 46 00:02:16,700 --> 00:02:21,180 Notice the length is almost constant. 47 00:02:21,180 --> 00:02:23,190 We have an epsilon there. 48 00:02:23,190 --> 00:02:26,140 That, of course, stands for the strain that the truss has 49 00:02:26,140 --> 00:02:28,670 undergone, but that epsilon is assumed to 50 00:02:28,670 --> 00:02:31,590 be very, very small. 51 00:02:31,590 --> 00:02:36,430 We will later on also look at a similar picture, but this 52 00:02:36,430 --> 00:02:40,410 red truss element at time t will have to be moved into 53 00:02:40,410 --> 00:02:45,870 this configuration here, starting, in other words, 54 00:02:45,870 --> 00:02:50,130 starting on the left-hand side here, we simply move this 55 00:02:50,130 --> 00:02:54,150 truss element here straight back so that this node lies on 56 00:02:54,150 --> 00:02:56,290 top of that node. 57 00:02:56,290 --> 00:02:59,240 In other words, if I just show it to you like that, it will 58 00:02:59,240 --> 00:03:00,810 have rotated like that. 59 00:03:00,810 --> 00:03:04,170 Notice that as far as the derivation of the stiffness 60 00:03:04,170 --> 00:03:07,770 matrix of the truss element in this configuration, and as far 61 00:03:07,770 --> 00:03:11,150 as the element force vector derivation are concerned, it 62 00:03:11,150 --> 00:03:15,210 does not matter whether we look at this particular truss 63 00:03:15,210 --> 00:03:19,750 element in this configuration, or whether we move it back to 64 00:03:19,750 --> 00:03:23,600 make this point coincide with that point. 65 00:03:23,600 --> 00:03:29,670 Well if we drive using what we have discussed earlier the 66 00:03:29,670 --> 00:03:32,650 deformation gradient for this particular element, we 67 00:03:32,650 --> 00:03:35,330 recognize that this is the general expression. 68 00:03:35,330 --> 00:03:41,960 Notice that X, of course, is equal to R times U, R being 69 00:03:41,960 --> 00:03:45,190 the orthogonal matrix, and U being the stretch matrix. 70 00:03:45,190 --> 00:03:50,340 Notice that in this X matrix now, this here really is the R 71 00:03:50,340 --> 00:03:55,970 component, and the U matrix is nothing else but the identity 72 00:03:55,970 --> 00:04:00,510 matrix with an epsilon added to the 11 component. 73 00:04:00,510 --> 00:04:04,060 And the product of R times U gives you then, of course, 74 00:04:04,060 --> 00:04:06,860 this right-hand side. 75 00:04:06,860 --> 00:04:12,410 Since the truss carries only the Cauchy stress, tau11, 76 00:04:12,410 --> 00:04:15,890 corresponding to the tensile force in the truss, we can 77 00:04:15,890 --> 00:04:19,209 directly write down the Cauchy stress in the stationary 78 00:04:19,209 --> 00:04:22,280 coordinate frame as given here. 79 00:04:22,280 --> 00:04:24,490 This, of course, is nothing else but a Mohr's circle 80 00:04:24,490 --> 00:04:25,710 transformation. 81 00:04:25,710 --> 00:04:30,670 Notice that our truss element moves through the stationary 82 00:04:30,670 --> 00:04:31,500 coordinate frame. 83 00:04:31,500 --> 00:04:34,870 I pointed out earlier very strongly that we always deal 84 00:04:34,870 --> 00:04:37,290 with stationary coordinate frames. 85 00:04:37,290 --> 00:04:44,670 So this is the expression of tau, the stress in the truss, 86 00:04:44,670 --> 00:04:47,920 but written now in the stationary coordinate frame. 87 00:04:47,920 --> 00:04:51,140 So we have sheer components and normal components. 88 00:04:51,140 --> 00:04:56,220 Of course, if theta's is equal to 0, then we have only here a 89 00:04:56,220 --> 00:04:59,110 1, and all of these are the components drop to 0. 90 00:05:01,790 --> 00:05:05,380 If we now use our general formula, that once again we 91 00:05:05,380 --> 00:05:09,140 discussed earlier, to evaluate from Cauchy stresses, second 92 00:05:09,140 --> 00:05:11,820 Piola-Kirchhoff stresses, we directly 93 00:05:11,820 --> 00:05:14,160 obtain this result here. 94 00:05:14,160 --> 00:05:17,400 Since epsilon is small, we set this equal to 1. 95 00:05:17,400 --> 00:05:20,950 And we have, indeed, that's the only non-zero stress 96 00:05:20,950 --> 00:05:27,720 component in this stress tensor is the S11 component. 97 00:05:27,720 --> 00:05:31,370 Well there's also a physical explanation. 98 00:05:31,370 --> 00:05:35,880 And that physical explanation goes as follows. 99 00:05:35,880 --> 00:05:39,730 We have initially, say, this situation here in the 100 00:05:39,730 --> 00:05:45,170 configuration at time zero where 0S refer to the 101 00:05:45,170 --> 00:05:49,700 configuration at time 0 is equal to tau0, and that is 102 00:05:49,700 --> 00:05:52,340 equal to nothing else than the initial force in the truss 103 00:05:52,340 --> 00:05:56,470 divided by the area, 0 components, of course, here. 104 00:05:56,470 --> 00:06:04,180 If we now think of first extending that truss element 105 00:06:04,180 --> 00:06:10,570 by a very small amount as shown here, notice that blue 106 00:06:10,570 --> 00:06:15,130 truss element here corresponds to the extended truss element. 107 00:06:15,130 --> 00:06:18,460 It lies really on top of the black, but in separating it 108 00:06:18,460 --> 00:06:22,150 slightly out so that you can see the different coloration. 109 00:06:22,150 --> 00:06:26,730 Then this blue truss element can be sort of corresponding 110 00:06:26,730 --> 00:06:32,320 to, and I'm showing it now up here, equal to a time T star, 111 00:06:32,320 --> 00:06:35,000 which is really a conceptual time. 112 00:06:35,000 --> 00:06:40,220 We are pulling out the truss first, and this means of 113 00:06:40,220 --> 00:06:44,830 course, corresponding to that T star conceptual time, we can 114 00:06:44,830 --> 00:06:47,650 write down this equation here. 115 00:06:47,650 --> 00:06:52,020 Surely there are 0 components here. 116 00:06:52,020 --> 00:06:58,780 Now, we start rotating the truss about the left node, and 117 00:06:58,780 --> 00:07:02,820 the result is shown on this U graph. 118 00:07:02,820 --> 00:07:07,670 Now we have the RAD configuration, theta is here, 119 00:07:07,670 --> 00:07:11,180 and the stress tensor that we are talking 120 00:07:11,180 --> 00:07:13,760 about has not changed. 121 00:07:13,760 --> 00:07:15,480 Why has it not changed? 122 00:07:15,480 --> 00:07:20,290 Well because of this statement down here, the components of 123 00:07:20,290 --> 00:07:23,830 the second Piola-Kirchhoff stress tensor do not change 124 00:07:23,830 --> 00:07:26,725 doing a rigid body motion, in particular doing 125 00:07:26,725 --> 00:07:29,170 a rigid body rotation. 126 00:07:29,170 --> 00:07:32,600 We discussed this statement earlier in an earlier lecture. 127 00:07:32,600 --> 00:07:35,300 It's a very important property of the second Piola-Kirchhoff 128 00:07:35,300 --> 00:07:36,230 stress tensor. 129 00:07:36,230 --> 00:07:40,870 And this means, of course, that the 0s which we had here 130 00:07:40,870 --> 00:07:44,470 in blue remains 0's here in red. 131 00:07:44,470 --> 00:07:46,650 Very important point. 132 00:07:46,650 --> 00:07:52,980 Well having identified then that S11 is the only non-zero 133 00:07:52,980 --> 00:07:55,810 stress component that we really need to deal with in 134 00:07:55,810 --> 00:07:59,090 the general principle of virtual work, we can directly 135 00:07:59,090 --> 00:08:02,500 write down this relationship here. 136 00:08:02,500 --> 00:08:05,070 It's nothing else than the application of the general 137 00:08:05,070 --> 00:08:08,860 equation, which we looked at earlier, reduced to the 138 00:08:08,860 --> 00:08:13,360 special form of the truss element taking only the S11 139 00:08:13,360 --> 00:08:15,070 component into account. 140 00:08:15,070 --> 00:08:18,680 See here, S11, of course, corresponding strains-- 141 00:08:18,680 --> 00:08:24,790 eta11, and here we have only the 11 components as well. 142 00:08:24,790 --> 00:08:30,560 The next step is to look at what are the individual terms 143 00:08:30,560 --> 00:08:33,210 in that equation. 144 00:08:33,210 --> 00:08:43,340 Recognize that S11 is nothing else than tP over A. C1111 is 145 00:08:43,340 --> 00:08:45,940 equal to E. The volume, of course, is 146 00:08:45,940 --> 00:08:47,700 given as shown here. 147 00:08:47,700 --> 00:08:51,770 And of course, we also recognize that the stress and 148 00:08:51,770 --> 00:08:54,310 strain states are constant along the truss. 149 00:08:54,310 --> 00:08:57,350 Once again, because we're dealing with two nodes only, 150 00:08:57,350 --> 00:09:00,090 and we can only have a lean interpolation of displacement 151 00:09:00,090 --> 00:09:02,790 between these two nodes, meaning that the strains must 152 00:09:02,790 --> 00:09:05,660 be constant, and the corresponding stresses must be 153 00:09:05,660 --> 00:09:08,000 constant as well. 154 00:09:08,000 --> 00:09:12,540 We substitute these things into, these relationships up 155 00:09:12,540 --> 00:09:15,740 here, into the equation that I showed you earlier and we 156 00:09:15,740 --> 00:09:17,900 directly obtain this equation. 157 00:09:17,900 --> 00:09:21,160 Notice that in this equation now, we want to, of course, 158 00:09:21,160 --> 00:09:28,570 express the strain terms e11, eta11 in terms of 159 00:09:28,570 --> 00:09:32,330 displacements, nodal point displacements. 160 00:09:32,330 --> 00:09:37,950 And that is achieved as usual via a strain 161 00:09:37,950 --> 00:09:39,470 displacement matrix. 162 00:09:39,470 --> 00:09:43,120 This is the relationship here corresponding to the linear 163 00:09:43,120 --> 00:09:46,910 strain increment, and that is the relationship corresponding 164 00:09:46,910 --> 00:09:50,350 to the variation on the nonlinear strain increment. 165 00:09:50,350 --> 00:09:53,010 Notice again, and I've pointed it out a number of times 166 00:09:53,010 --> 00:09:57,500 already earlier, that we write this right-hand side in such 167 00:09:57,500 --> 00:10:00,760 way that the right-hand side is equal to 168 00:10:00,760 --> 00:10:02,020 the left-hand side. 169 00:10:02,020 --> 00:10:06,480 Therefore, these BNLs are defined such that the 170 00:10:06,480 --> 00:10:10,070 right-hand side is equal to the left-hand side. 171 00:10:10,070 --> 00:10:14,740 Also remember, u hat here, this hat up there, the nodes, 172 00:10:14,740 --> 00:10:15,710 the [? discrete ?] 173 00:10:15,710 --> 00:10:19,090 nodal point displacements are listed in this vector. 174 00:10:19,090 --> 00:10:20,700 And here they are. 175 00:10:20,700 --> 00:10:26,410 Notice upper superscript refers to the nodal point. 176 00:10:26,410 --> 00:10:31,340 Lower subscript refers to the directions, the coordinate 177 00:10:31,340 --> 00:10:33,250 directions that we are looking at. 178 00:10:33,250 --> 00:10:37,910 Similarly, of course, for node two, 2 up there, and 179 00:10:37,910 --> 00:10:41,460 1 and 2 down here. 180 00:10:41,460 --> 00:10:45,060 We go back to our general continuum mechanics relations 181 00:10:45,060 --> 00:10:49,300 that we discussed earlier, and identify that the total 182 00:10:49,300 --> 00:10:53,010 increment in the Green-Lagrangian strain is 183 00:10:53,010 --> 00:10:56,150 given via this relationship here. 184 00:10:56,150 --> 00:11:00,285 Notice that this relationship is obtained from the general 185 00:11:00,285 --> 00:11:06,210 equation of this epsilon 0 IJ. 186 00:11:06,210 --> 00:11:09,500 Of course, I and J are now equal to 11 for the truss 187 00:11:09,500 --> 00:11:13,700 because S11 is the only non-zero stress that we just 188 00:11:13,700 --> 00:11:15,960 have been considering earlier. 189 00:11:15,960 --> 00:11:21,170 So we'll go into our earlier view graphs, the information 190 00:11:21,170 --> 00:11:23,320 that we discussed earlier. 191 00:11:23,320 --> 00:11:29,030 To look up the equation on 0 epsilon IJ, we set I and J 192 00:11:29,030 --> 00:11:31,080 equal to 1. 193 00:11:31,080 --> 00:11:34,190 There is a k also in that equation if you look back. 194 00:11:34,190 --> 00:11:38,010 That k goes from 1 to 2 now because we're looking at the 195 00:11:38,010 --> 00:11:40,460 two dimensional motion of the truss. 196 00:11:40,460 --> 00:11:42,700 If you have a three dimensional motion, then, of 197 00:11:42,700 --> 00:11:47,210 course, we would have to let k go from 1 to 2 to 3. 198 00:11:47,210 --> 00:11:50,960 Here we have k only going from 1 to 2, that's why you see the 199 00:11:50,960 --> 00:11:53,580 2 at the highest subscript here. 200 00:11:53,580 --> 00:11:57,780 And automatically, you would get this equation. 201 00:11:57,780 --> 00:12:01,850 Notice that the linear part in here gives us the linear 202 00:12:01,850 --> 00:12:03,100 strain increment. 203 00:12:05,200 --> 00:12:09,910 And that the nonlinear part in here gives us the nonlinear 204 00:12:09,910 --> 00:12:10,720 strain increment. 205 00:12:10,720 --> 00:12:14,360 And if we take the variation on that part, we directly 206 00:12:14,360 --> 00:12:17,690 obtain this relationship here. 207 00:12:17,690 --> 00:12:21,430 We like to write this now in matrix form, and that is 208 00:12:21,430 --> 00:12:24,800 achieved as shown down here. 209 00:12:24,800 --> 00:12:28,130 Notice this part, of course, is nothing else than what we 210 00:12:28,130 --> 00:12:31,180 have up here written in matrix form. 211 00:12:33,920 --> 00:12:36,585 There is, of course, in the total Lagrangian formulation 212 00:12:36,585 --> 00:12:39,410 the initial displacement effect. 213 00:12:39,410 --> 00:12:43,040 And that initial displacement fact lies in 214 00:12:43,040 --> 00:12:44,960 these derivatives here. 215 00:12:44,960 --> 00:12:47,070 This one and that one. 216 00:12:47,070 --> 00:12:48,800 And we need to evaluate them. 217 00:12:48,800 --> 00:12:53,350 We can evaluate them from kinematics, as shown in this 218 00:12:53,350 --> 00:12:53,910 picture here. 219 00:12:53,910 --> 00:12:56,180 Notice that we have moved the truss now, as 220 00:12:56,180 --> 00:12:57,090 I pointed out earlier. 221 00:12:57,090 --> 00:13:02,910 We would do to the nodal point to coincide such that, I 222 00:13:02,910 --> 00:13:06,940 should say such that the nodal points or nodal 223 00:13:06,940 --> 00:13:09,510 point one has not moved. 224 00:13:09,510 --> 00:13:12,030 Once again, that is quite OK. 225 00:13:12,030 --> 00:13:15,230 If we want to derive the stiffness matrix and force 226 00:13:15,230 --> 00:13:18,540 vector of the element, because those quantities are not 227 00:13:18,540 --> 00:13:23,860 affected by this rigid body translation. 228 00:13:23,860 --> 00:13:31,150 Here we have theta, and notice these are displacements that 229 00:13:31,150 --> 00:13:36,460 we call delta tU2, delta tU1 with a negative sign there 230 00:13:36,460 --> 00:13:41,150 because this displacement would be measured positive in 231 00:13:41,150 --> 00:13:43,560 the opposite direction. 232 00:13:43,560 --> 00:13:48,180 And we can directly see that this differentiation here is 233 00:13:48,180 --> 00:13:51,730 nothing else than this value right here. 234 00:13:51,730 --> 00:13:52,710 From the kinematics. 235 00:13:52,710 --> 00:13:56,540 Similar, this differentiation is nothing else than sin 236 00:13:56,540 --> 00:14:00,870 theta, once again, from the kinematics. 237 00:14:00,870 --> 00:14:05,270 So we have the initial displacement effect in these 238 00:14:05,270 --> 00:14:11,060 derivatives, and can evaluate those as shown by these 239 00:14:11,060 --> 00:14:12,820 functions here. 240 00:14:12,820 --> 00:14:17,480 We also want to, of course, evaluate these derivatives 241 00:14:17,480 --> 00:14:23,910 here, 0U11, and in fact, 0U2 comma 1. 242 00:14:23,910 --> 00:14:27,640 And we recognize that, as I pointed out earlier, the 243 00:14:27,640 --> 00:14:30,310 strains are constant. 244 00:14:30,310 --> 00:14:32,360 Therefore, this is a constant value, and in this 245 00:14:32,360 --> 00:14:36,370 differentiation it's really nothing else than this 246 00:14:36,370 --> 00:14:37,780 relationship here. 247 00:14:37,780 --> 00:14:41,290 We take the difference in the nodal point displacement, and 248 00:14:41,290 --> 00:14:46,470 divide by L. Quite the same way as we did it for the 249 00:14:46,470 --> 00:14:48,540 updated Lagrangian formulation. 250 00:14:48,540 --> 00:14:51,840 You can, therefore, write down this differentiation, these 251 00:14:51,840 --> 00:14:55,450 two derivatives here, as shown on the right-hand side. 252 00:14:59,130 --> 00:15:03,750 Now, we are ready to take these expressions and 253 00:15:03,750 --> 00:15:09,300 substitute into the right-hand side of the 254 00:15:09,300 --> 00:15:10,580 linear strain term. 255 00:15:10,580 --> 00:15:15,000 Notice that this gives us one term here, and 256 00:15:15,000 --> 00:15:17,750 another term there. 257 00:15:17,750 --> 00:15:19,160 Let's look at these terms. 258 00:15:19,160 --> 00:15:22,530 This term here comes from that part there. 259 00:15:25,840 --> 00:15:33,670 And this term here, down here, comes from that part there. 260 00:15:33,670 --> 00:15:37,580 Let's look at this second term a little closer. 261 00:15:37,580 --> 00:15:41,440 This here is, of course, the initial displacement effect. 262 00:15:41,440 --> 00:15:43,210 All of that I should say is really the initial 263 00:15:43,210 --> 00:15:44,240 displacement effect. 264 00:15:44,240 --> 00:15:48,000 But if these terms are 0 here, then, of course, initial 265 00:15:48,000 --> 00:15:49,820 displacement effects drop out. 266 00:15:49,820 --> 00:15:55,180 So therefore, these two terms here are the important ones, 267 00:15:55,180 --> 00:16:00,720 and we can see that they are written in this vector here. 268 00:16:00,720 --> 00:16:02,980 We just derived them earlier. 269 00:16:02,980 --> 00:16:08,650 Notice that this matrix, this vector here, is given by this 270 00:16:08,650 --> 00:16:11,820 matrix there multiplied by the nodal point 271 00:16:11,820 --> 00:16:14,390 displacement vector. 272 00:16:14,390 --> 00:16:17,760 And of course, this total here gives us the initial 273 00:16:17,760 --> 00:16:21,620 displacement effect, which is captured in t0BL1. 274 00:16:25,000 --> 00:16:30,320 The sum of t0BL1 and t0BL0, of course, gives us a linear 275 00:16:30,320 --> 00:16:31,585 strain displacement matrix. 276 00:16:34,670 --> 00:16:37,360 And that summation is performed on 277 00:16:37,360 --> 00:16:39,250 this view graph here. 278 00:16:39,250 --> 00:16:43,430 Notice we have written these terms once again. 279 00:16:43,430 --> 00:16:46,630 Of course, if you had to multiply out in order to get 280 00:16:46,630 --> 00:16:51,640 t0BL1, the sum of these two gives us this strain matrix, 281 00:16:51,640 --> 00:16:56,190 the linear strain displacement matrix. 282 00:16:56,190 --> 00:16:59,370 And we now notice that in the linear strain displacement 283 00:16:59,370 --> 00:17:02,840 matrix, cosines and sines of thetas of 284 00:17:02,840 --> 00:17:05,200 the angle theta appear. 285 00:17:05,200 --> 00:17:08,310 That is something to keep in mind. 286 00:17:08,310 --> 00:17:13,670 For the nonlinear strain displacement matrix, we look 287 00:17:13,670 --> 00:17:17,930 at our nonlinear variation on the nonlinear strain term. 288 00:17:17,930 --> 00:17:19,730 Here it is on the left-hand side. 289 00:17:19,730 --> 00:17:24,140 And we substitute for the differentations the way we 290 00:17:24,140 --> 00:17:27,030 have been doing it for the updated Lagrangian formulation 291 00:17:27,030 --> 00:17:28,040 of the truss. 292 00:17:28,040 --> 00:17:35,010 And we recognize, of course, that this part here is given 293 00:17:35,010 --> 00:17:39,610 by what is underlined as blue. 294 00:17:39,610 --> 00:17:42,620 And this part here is given by, again what's 295 00:17:42,620 --> 00:17:44,330 underlined in blue. 296 00:17:44,330 --> 00:17:49,420 And then this part here is the nonlinear strain displacement 297 00:17:49,420 --> 00:17:52,040 matrix, the BNL. 298 00:17:52,040 --> 00:17:55,080 That part right there. 299 00:17:55,080 --> 00:18:01,400 Well if we now use all this information, we directly 300 00:18:01,400 --> 00:18:06,712 obtain the stiffness matrix and stiffness matrices, and 301 00:18:06,712 --> 00:18:10,050 force vector off the truss element. 302 00:18:10,050 --> 00:18:15,740 We substitute into here from what we derived earlier, and 303 00:18:15,740 --> 00:18:17,490 directly obtained this expression. 304 00:18:17,490 --> 00:18:21,790 And what's underlined here in blue is, of course, the linear 305 00:18:21,790 --> 00:18:24,390 strain stiffness matrix of the truss element. 306 00:18:26,980 --> 00:18:29,910 And notice that this linear strain stiffness matrix is 307 00:18:29,910 --> 00:18:34,570 identically equal to the linear strain stiffness matrix 308 00:18:34,570 --> 00:18:38,990 of the updated Lagrangian formulation truss element. 309 00:18:38,990 --> 00:18:43,750 We use this term to obtain the nonlinear 310 00:18:43,750 --> 00:18:45,490 strain stiffness matrix. 311 00:18:45,490 --> 00:18:48,380 And we notice that this nonlinear strain stiffness 312 00:18:48,380 --> 00:18:53,340 matrix is identically equal to what we derived earlier in the 313 00:18:53,340 --> 00:18:55,410 updated Lagrangian formulation as well. 314 00:18:58,290 --> 00:19:03,740 The force vector, the F vector for the truss element is given 315 00:19:03,740 --> 00:19:06,930 on this view graph and is obtained from this 316 00:19:06,930 --> 00:19:08,650 relationship here. 317 00:19:08,650 --> 00:19:12,250 Notice it's, once again, identically equal to the force 318 00:19:12,250 --> 00:19:15,570 vector of the updated Lagrangian formulated truss 319 00:19:15,570 --> 00:19:17,390 element as well. 320 00:19:17,390 --> 00:19:20,290 And so we notice that to find all the matrices that we 321 00:19:20,290 --> 00:19:23,020 obtain in this total Lagrangian formulation are the 322 00:19:23,020 --> 00:19:27,130 same matrices that we already had derived in the updated 323 00:19:27,130 --> 00:19:28,380 Lagrangian formulation. 324 00:19:31,010 --> 00:19:34,630 Important point to recognize is that the coordinate 325 00:19:34,630 --> 00:19:38,540 transformation used in the U.L. formulation is contained 326 00:19:38,540 --> 00:19:43,430 in the initial displacement effect of the T.L. formation. 327 00:19:43,430 --> 00:19:46,090 Let's look at that a little bit closer. 328 00:19:46,090 --> 00:19:51,400 What this means is that in the updated Lagrangian 329 00:19:51,400 --> 00:19:55,120 formulation, we did the evaluation corresponding to 330 00:19:55,120 --> 00:19:58,740 the configuration at time t with a 331 00:19:58,740 --> 00:20:01,300 local coordinate system. 332 00:20:01,300 --> 00:20:06,030 And then after that derivation was performed, we transformed 333 00:20:06,030 --> 00:20:10,050 the nodal point variables to the global coordinate system. 334 00:20:10,050 --> 00:20:14,360 In the total Lagrangian formulation, we obtain the 335 00:20:14,360 --> 00:20:16,980 matrices directly corresponding to the global 336 00:20:16,980 --> 00:20:21,240 coordinate system through the initial displacement effect. 337 00:20:21,240 --> 00:20:25,090 Now, the same can also be shown, of course, for other 338 00:20:25,090 --> 00:20:28,450 elements, and we have looked here, particularly at a 339 00:20:28,450 --> 00:20:32,030 two-noded beam element. 340 00:20:32,030 --> 00:20:37,060 And we have asked ourself the question then which method or 341 00:20:37,060 --> 00:20:40,920 which formulation is always more effective? 342 00:20:40,920 --> 00:20:45,800 If you use numerical integration, as we might want 343 00:20:45,800 --> 00:20:49,620 to use for the beam element, particular we have to use if 344 00:20:49,620 --> 00:20:53,690 we want to capture elastoplasticity effects. 345 00:20:53,690 --> 00:20:57,230 Then clearly, the updated Lagrangian formulation must be 346 00:20:57,230 --> 00:21:00,680 more effective because what you are doing in the total 347 00:21:00,680 --> 00:21:06,370 Lagrangian formulation is to carry these cosines and sines 348 00:21:06,370 --> 00:21:11,170 theta terms along for every integration that you're 349 00:21:11,170 --> 00:21:12,420 considering. 350 00:21:14,390 --> 00:21:20,970 In the numerical integration, the product of B transpose CB, 351 00:21:20,970 --> 00:21:24,380 with the B containing the initial displacement effect 352 00:21:24,380 --> 00:21:29,990 now, is much more expensive than the product for the 353 00:21:29,990 --> 00:21:34,250 updated Lagrangian formulation where the B does not contain 354 00:21:34,250 --> 00:21:36,410 this initial displacement effect. 355 00:21:36,410 --> 00:21:39,830 So in the updated Lagrangian formulation, the B matrix is 356 00:21:39,830 --> 00:21:40,760 much simpler. 357 00:21:40,760 --> 00:21:44,010 We do the numerical integration with a simpler B 358 00:21:44,010 --> 00:21:48,960 matrix along the beam element, and the details are given in 359 00:21:48,960 --> 00:21:50,080 this paper. 360 00:21:50,080 --> 00:21:53,240 And once we have done the numerical integration, we 361 00:21:53,240 --> 00:21:57,220 simply transform the nodal point displacements from the 362 00:21:57,220 --> 00:21:59,580 local coordinate system. 363 00:21:59,580 --> 00:22:02,100 The [? coord ?] system, we call it a [? coord ?] system 364 00:22:02,100 --> 00:22:04,220 when we discuss the updated Lagrangian formulation of the 365 00:22:04,220 --> 00:22:06,970 truss element, and we transform these nodal point 366 00:22:06,970 --> 00:22:12,570 displacements to the global system only once by this t 367 00:22:12,570 --> 00:22:15,550 transpose kt product. 368 00:22:15,550 --> 00:22:20,060 So this means then that certainly for a two-noded beam 369 00:22:20,060 --> 00:22:25,750 element, the total Lagrangian formulation is not as 370 00:22:25,750 --> 00:22:28,620 effective as the updated Lagrangian formulation, and it 371 00:22:28,620 --> 00:22:30,930 is very natural to use the updated Lagrangian formulation 372 00:22:30,930 --> 00:22:35,830 because you get the same result in the end. 373 00:22:35,830 --> 00:22:39,720 For the two-noded truss, of course, the numerical 374 00:22:39,720 --> 00:22:44,280 operations involved are so close, so close, that one can 375 00:22:44,280 --> 00:22:47,490 really say the updated and total Lagrangian formulations 376 00:22:47,490 --> 00:22:50,510 are basically the same. 377 00:22:50,510 --> 00:22:55,370 Let's now look at an example. 378 00:22:55,370 --> 00:22:59,730 Here we have a truss, a simple truss structure. 379 00:22:59,730 --> 00:23:05,350 And we perform the analysis of this truss structure using the 380 00:23:05,350 --> 00:23:09,430 U.L. formulation, the T.L., total Lagrangian formulation 381 00:23:09,430 --> 00:23:13,380 would give the same result, as I just discussed. 382 00:23:13,380 --> 00:23:17,670 And we want to particularly calculate the collapse load of 383 00:23:17,670 --> 00:23:19,580 this truss. 384 00:23:19,580 --> 00:23:23,520 We also want to test the MNO, the Materially Nonlinear Only 385 00:23:23,520 --> 00:23:24,470 formulation. 386 00:23:24,470 --> 00:23:29,880 The reason for testing this formulation is to identify 387 00:23:29,880 --> 00:23:35,630 what differences do occur, and to understand more clearly the 388 00:23:35,630 --> 00:23:38,950 U.L. Formulation, as well as the MNO formulation. 389 00:23:38,950 --> 00:23:42,420 Notice that for this truss, the [? geo ?] matrix, the 390 00:23:42,420 --> 00:23:44,910 geometric data are given here. 391 00:23:44,910 --> 00:23:50,620 The material data are given here in this table. 392 00:23:50,620 --> 00:23:54,110 Notice that, of course, these nodes dawn here are fixed. 393 00:23:54,110 --> 00:23:58,260 Notice that we are putting a load that pulls on the truss, 394 00:23:58,260 --> 00:24:00,720 P, as shown. 395 00:24:00,720 --> 00:24:05,530 And that we have two degrees of freedom, U and V. So it's a 396 00:24:05,530 --> 00:24:09,430 fairly simple problem, yet there is quite a bit of meat 397 00:24:09,430 --> 00:24:13,370 in it when one looks closely at the solution. 398 00:24:13,370 --> 00:24:16,060 And that's what we like to do now. 399 00:24:16,060 --> 00:24:19,250 We calculate first analytically, the limit load, 400 00:24:19,250 --> 00:24:20,880 the elastic limit load. 401 00:24:20,880 --> 00:24:25,630 In other words, the load for which the yield point is 402 00:24:25,630 --> 00:24:28,680 reached in one of the members. 403 00:24:28,680 --> 00:24:31,300 And in fact, for this particular load, the side 404 00:24:31,300 --> 00:24:34,720 trusses just become plastic, and this is the 405 00:24:34,720 --> 00:24:36,100 value of that load. 406 00:24:36,100 --> 00:24:41,160 The ultimate limit load, when the side trusses are plastic 407 00:24:41,160 --> 00:24:45,990 and the center truss is plastic as well, is reached at 408 00:24:45,990 --> 00:24:47,240 this level. 409 00:24:49,150 --> 00:24:53,940 If we use an automatic load stepping incrementation, we 410 00:24:53,940 --> 00:24:58,550 will discuss that in one of the next lectures, we obtain 411 00:24:58,550 --> 00:25:00,200 this response. 412 00:25:00,200 --> 00:25:04,730 Notice P plotted vertically here, and V, the displacement 413 00:25:04,730 --> 00:25:08,490 corresponding to P, plotted horizontally. 414 00:25:08,490 --> 00:25:15,340 Notice that we come up here in the step-by-step solution and 415 00:25:15,340 --> 00:25:18,860 march along here. 416 00:25:18,860 --> 00:25:22,940 There are two solutions given, actually. 417 00:25:22,940 --> 00:25:29,220 One with v1, very small, and one with v1 larger. 418 00:25:29,220 --> 00:25:32,490 This is here the parameters that we use to start a 419 00:25:32,490 --> 00:25:34,620 solution algorithm. 420 00:25:34,620 --> 00:25:39,210 We have to start the solution algorithm by imposing a small 421 00:25:39,210 --> 00:25:43,730 value of displacement to a node, and then the solution 422 00:25:43,730 --> 00:25:46,610 algorithm can perform automatically the total 423 00:25:46,610 --> 00:25:48,230 step-by-step solution. 424 00:25:48,230 --> 00:25:50,910 We will discuss that in one of the next lectures. 425 00:25:50,910 --> 00:25:54,430 And as you see, with the larger value here, we get 426 00:25:54,430 --> 00:25:55,780 these discrete points. 427 00:25:55,780 --> 00:25:59,260 No difficulty solving for this response. 428 00:25:59,260 --> 00:26:03,290 Notice also that there's no difficulty going beyond this 429 00:26:03,290 --> 00:26:08,480 analytic elastic limit load, shown here by the blue line. 430 00:26:08,480 --> 00:26:12,190 If we were to look very closely at this slope here, we 431 00:26:12,190 --> 00:26:14,780 would see that it is slightly positive. 432 00:26:14,780 --> 00:26:16,760 You can hardly see it from this graph. 433 00:26:19,760 --> 00:26:23,360 This is the solution obtained with the U.L. formulation, 434 00:26:23,360 --> 00:26:24,580 with the U.L. formulation. 435 00:26:24,580 --> 00:26:26,930 And once again, the T.L. formulation would give 436 00:26:26,930 --> 00:26:30,000 identically the same result. 437 00:26:30,000 --> 00:26:33,190 We now consider an MNO analysis. 438 00:26:33,190 --> 00:26:35,640 MNO means Materially Nonlinear Only. 439 00:26:35,640 --> 00:26:38,490 No large displacement effect included. 440 00:26:38,490 --> 00:26:41,940 No geometric stiffening effect included. 441 00:26:41,940 --> 00:26:44,830 Geometric stiffening, meaning the effect of the nonlinear 442 00:26:44,830 --> 00:26:47,450 strain stiffness matrix being included. 443 00:26:47,450 --> 00:26:51,210 So we want to use this formulation, and once again, 444 00:26:51,210 --> 00:26:54,150 apply the automatic load stepping incrementation. 445 00:26:54,150 --> 00:26:57,560 If the stiffness matrix in that automatic load step 446 00:26:57,560 --> 00:27:00,460 incrementation is not reformed, in other words, we 447 00:27:00,460 --> 00:27:05,030 keep the original stiffness matrix, then almost identical 448 00:27:05,030 --> 00:27:08,760 results are obtained to the U.L. or T.L. formulation 449 00:27:08,760 --> 00:27:10,890 solution results. 450 00:27:10,890 --> 00:27:17,160 However, if the stiffness matrix is reformed in the MNO 451 00:27:17,160 --> 00:27:21,140 formulation, and for a load level larger than the elastic 452 00:27:21,140 --> 00:27:22,460 limit load-- 453 00:27:22,460 --> 00:27:25,320 remember, the elastic limit load is quite a bit lower than 454 00:27:25,320 --> 00:27:27,510 the ultimate limit load-- 455 00:27:27,510 --> 00:27:32,990 then the structure is found to be unstable, as 0 pivot is 456 00:27:32,990 --> 00:27:35,790 obtained in the stiffness matrix. 457 00:27:35,790 --> 00:27:41,250 I like to ask the question why is that so to you now? 458 00:27:41,250 --> 00:27:45,100 And ask you to just think a little bit about it. 459 00:27:45,100 --> 00:27:46,400 A few minutes. 460 00:27:46,400 --> 00:27:49,680 And then when I come back, I give you the explanation by 461 00:27:49,680 --> 00:27:53,580 looking at the problem once again. 462 00:28:30,960 --> 00:28:34,020 I'd like to now address the question that I asked you 463 00:28:34,020 --> 00:28:35,270 before the break. 464 00:28:35,270 --> 00:28:39,510 Namely, why, if we use the MNO formulation do we get 465 00:28:39,510 --> 00:28:41,910 virtually identically the same results as in the U.L. 466 00:28:41,910 --> 00:28:47,370 formulation if we only set up a stiffness matrix that 467 00:28:47,370 --> 00:28:51,050 corresponds to a configuration below the elastic limit load. 468 00:28:51,050 --> 00:28:53,840 If on the other hand, we setting up a difference matrix 469 00:28:53,840 --> 00:28:57,210 above the elastic limit load in the MNO formulation, then 470 00:28:57,210 --> 00:28:59,120 we cannot, so it's a problem. 471 00:28:59,120 --> 00:29:06,740 Well the answer is that as soon as the two trusses on the 472 00:29:06,740 --> 00:29:11,100 side have become plastic, the MNO formulation does not have 473 00:29:11,100 --> 00:29:13,760 any more stiffness corresponding to this degree 474 00:29:13,760 --> 00:29:18,240 of freedom, meaning the matrix is singular, and we cannot 475 00:29:18,240 --> 00:29:20,540 solve the system of equations. 476 00:29:20,540 --> 00:29:26,510 Whereas if we look at the U.L. formulation, then we see that 477 00:29:26,510 --> 00:29:30,530 the geometric stiffening effect, or the KNL effect, the 478 00:29:30,530 --> 00:29:36,830 nonlinear strain stiffness effect, introduces the forces 479 00:29:36,830 --> 00:29:40,460 that are acting in these two elements into the structure 480 00:29:40,460 --> 00:29:44,040 into the structure assemblage as the stiffening part. 481 00:29:44,040 --> 00:29:47,500 And of course, is a KNL matrix corresponding to 482 00:29:47,500 --> 00:29:49,170 this element as well. 483 00:29:49,170 --> 00:29:53,450 So in other words, these three KNL matrices corresponding to 484 00:29:53,450 --> 00:29:56,940 these three elements, these three truss elements, 485 00:29:56,940 --> 00:30:02,930 introduce a stiffness into the total structure assemblage, 486 00:30:02,930 --> 00:30:06,500 meaning that we can still solve the system of equations. 487 00:30:06,500 --> 00:30:11,320 We also notice that these KNL effect that we are carrying 488 00:30:11,320 --> 00:30:16,615 along in the U.L. formulation, when measured on the KL effect 489 00:30:16,615 --> 00:30:18,520 are rather small. 490 00:30:18,520 --> 00:30:22,020 And this means that the response of the U.L. 491 00:30:22,020 --> 00:30:25,700 formulation calculated with the U.L. formulation is 492 00:30:25,700 --> 00:30:29,430 basically the same as the response calculated with the 493 00:30:29,430 --> 00:30:30,770 MNO formulation. 494 00:30:30,770 --> 00:30:33,730 As long, of course, we use in the MNO formulation a 495 00:30:33,730 --> 00:30:37,040 stiffness matrix that corresponds to a configuration 496 00:30:37,040 --> 00:30:40,240 below the elastic limit load. 497 00:30:40,240 --> 00:30:43,860 Well this completes this example. 498 00:30:43,860 --> 00:30:46,000 Let us look at another example. 499 00:30:46,000 --> 00:30:50,900 And in this example, I like to consider with you a pre-stress 500 00:30:50,900 --> 00:30:59,980 cable, as shown up here, that has initially a span of S. And 501 00:30:59,980 --> 00:31:04,230 we then prescribe on the right-hand side a displacement 502 00:31:04,230 --> 00:31:08,780 to this node to bring this node towards the left, meaning 503 00:31:08,780 --> 00:31:11,350 that the cable will sag through. 504 00:31:11,350 --> 00:31:15,520 We want to calculate the deformations of the cable, and 505 00:31:15,520 --> 00:31:19,320 in particular, we want to have the final deformations when S 506 00:31:19,320 --> 00:31:21,420 is equal to 30 meters. 507 00:31:21,420 --> 00:31:23,820 Notice that the cable carries an initial 508 00:31:23,820 --> 00:31:26,580 tension of 500 Newton. 509 00:31:26,580 --> 00:31:32,130 And here we see the other data corresponding to the cable. 510 00:31:32,130 --> 00:31:32,940 This, of course, is a 511 00:31:32,940 --> 00:31:35,350 geometrically nonlinear problem. 512 00:31:35,350 --> 00:31:38,950 We want to use here the U.L. formulation. 513 00:31:38,950 --> 00:31:45,320 And the interesting part is that to solve this problem we 514 00:31:45,320 --> 00:31:49,820 have to use relatively small time steps, or load steps, 515 00:31:49,820 --> 00:31:51,830 because they're considering it to be, of 516 00:31:51,830 --> 00:31:54,680 course, a static solution. 517 00:31:54,680 --> 00:31:58,180 We need to only allow small perturbations in the nodal 518 00:31:58,180 --> 00:32:04,170 point coordinates, because if we do will allow too large 519 00:32:04,170 --> 00:32:06,610 perturbations in the nodal point coordinates, we find 520 00:32:06,610 --> 00:32:10,520 that the out of balance loads at the nodes become very large 521 00:32:10,520 --> 00:32:13,140 and we have difficulties solving the equation. 522 00:32:13,140 --> 00:32:17,700 So we use many load steps with equilibrium iterations so that 523 00:32:17,700 --> 00:32:21,980 the configuration of the cable is never far from an 524 00:32:21,980 --> 00:32:24,720 equilibrium position. 525 00:32:24,720 --> 00:32:30,150 In this particular problem, we used full Newton iterations 526 00:32:30,150 --> 00:32:31,150 without line searches. 527 00:32:31,150 --> 00:32:36,750 We will talk about this scheme in detail in the next lecture. 528 00:32:36,750 --> 00:32:40,560 We use also convergence criteria because we're 529 00:32:40,560 --> 00:32:43,850 iterating, we have to break up that iteration or stop the 530 00:32:43,850 --> 00:32:45,500 iteration at some point. 531 00:32:45,500 --> 00:32:49,270 And we will use these two convergence criteria. 532 00:32:49,270 --> 00:32:52,490 We consider these convergence criteria in quite some detail 533 00:32:52,490 --> 00:32:54,090 in the next lecture. 534 00:32:54,090 --> 00:32:56,230 Notice that basically we are saying is that the out of 535 00:32:56,230 --> 00:33:01,010 balance loads shall be small enough, and here we have an 536 00:33:01,010 --> 00:33:04,805 energy criterion that also has to be-- 537 00:33:04,805 --> 00:33:08,050 this measure has to be small enough for convergence. 538 00:33:08,050 --> 00:33:11,480 Once again, we will talk about these in detail when we 539 00:33:11,480 --> 00:33:15,550 consider the solution of the equation. 540 00:33:15,550 --> 00:33:18,665 Here we have a table that summarizes the results. 541 00:33:21,490 --> 00:33:24,470 In the first time step, puts the gravity loading onto the 542 00:33:24,470 --> 00:33:28,820 cable with a mass density that was given, of course, on the 543 00:33:28,820 --> 00:33:30,030 earlier view graph. 544 00:33:30,030 --> 00:33:33,050 And we notice that with this iterative scheme, we need to 545 00:33:33,050 --> 00:33:35,400 14 iterations. 546 00:33:35,400 --> 00:33:41,860 We then from time or load step 2 to 1,001 prescribed the 547 00:33:41,860 --> 00:33:46,230 displacements in 1,000 equal steps on the right-hand side 548 00:33:46,230 --> 00:33:47,320 of the cable. 549 00:33:47,320 --> 00:33:51,790 And we notice that we need less than 5 or equal to 5 550 00:33:51,790 --> 00:33:54,480 iterations per load step. 551 00:33:54,480 --> 00:33:56,990 This might appear to be a fairly large 552 00:33:56,990 --> 00:33:57,970 number of load steps. 553 00:33:57,970 --> 00:33:59,640 Of course, one could now experiment 554 00:33:59,640 --> 00:34:01,000 and bring this down. 555 00:34:01,000 --> 00:34:02,980 If you try to bring this down, this number, of 556 00:34:02,980 --> 00:34:05,870 course, would go up. 557 00:34:05,870 --> 00:34:07,610 And one could experiment around. 558 00:34:07,610 --> 00:34:11,770 But it shows what kind of considerations are necessary 559 00:34:11,770 --> 00:34:15,090 when you solve this problem, namely that you don't want to 560 00:34:15,090 --> 00:34:17,870 be too far away from the equilibrium configuration 561 00:34:17,870 --> 00:34:21,489 force of the cable in any one of the load steps, and if you 562 00:34:21,489 --> 00:34:24,199 are, then, of course, you will need a large number of 563 00:34:24,199 --> 00:34:27,010 iterations here, and your iterative scheme might not 564 00:34:27,010 --> 00:34:28,540 even converge. 565 00:34:28,540 --> 00:34:32,530 Pictorally, the results are shown on this view graph here. 566 00:34:32,530 --> 00:34:37,780 Notice we have the undeformed cable here. 567 00:34:37,780 --> 00:34:42,600 And then with S equal to 55 meters, the cable takes on 568 00:34:42,600 --> 00:34:45,929 this configuration with S equal to 30 meters. 569 00:34:45,929 --> 00:34:47,909 We see this configuration. 570 00:34:47,909 --> 00:34:51,530 And if you were to try to bring this point in further, 571 00:34:51,530 --> 00:34:54,230 then to model the curvature down here, of course, 572 00:34:54,230 --> 00:34:57,200 properly, you would have to have a finite discretization. 573 00:34:57,200 --> 00:35:02,660 So this is probably the largest amount that you want 574 00:35:02,660 --> 00:35:05,950 to bring this point over with this particular 575 00:35:05,950 --> 00:35:07,200 discretization. 576 00:35:09,080 --> 00:35:14,000 If you were to use a finite element discretization, this 577 00:35:14,000 --> 00:35:16,400 curve, of course, would hardly change. 578 00:35:16,400 --> 00:35:20,040 And this one here will change a bit down here. 579 00:35:20,040 --> 00:35:23,740 But these results are quite acceptable up to S equal to 30 580 00:35:23,740 --> 00:35:26,140 meters with this coarse discretization. 581 00:35:26,140 --> 00:35:29,230 And we have learned a bit from this problem. 582 00:35:29,230 --> 00:35:30,810 Thank you very much for your attention.