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 00:00:21,520 --> 00:00:23,260 PROFESSOR: Ladies and gentlemen, welcome to 9 00:00:23,260 --> 00:00:25,220 Lecture Number 10. 10 00:00:25,220 --> 00:00:27,600 In this lecture, I would like to discuss a solution of 11 00:00:27,600 --> 00:00:29,520 dynamic equilibrium equations. 12 00:00:29,520 --> 00:00:32,240 And in particular, I would like to talk about the direct 13 00:00:32,240 --> 00:00:35,860 integration solution of the equilibrium equations in 14 00:00:35,860 --> 00:00:37,790 dynamic analysis. 15 00:00:37,790 --> 00:00:39,980 These, of course, are the equilibrium equations that we 16 00:00:39,980 --> 00:00:42,100 derived already earlier. 17 00:00:42,100 --> 00:00:45,270 M as a mass matrix, C as a damping matrix, K as a 18 00:00:45,270 --> 00:00:48,490 stiffness matrix, U as a displacement vector, and here 19 00:00:48,490 --> 00:00:50,490 we have the velocities and the 20 00:00:50,490 --> 00:00:52,535 accelerations at the null points. 21 00:00:52,535 --> 00:00:56,440 R, of course, being the load vector and this load vector is 22 00:00:56,440 --> 00:00:59,320 now time dependent. 23 00:00:59,320 --> 00:01:02,200 I also should draw your attention on the fact that I 24 00:01:02,200 --> 00:01:04,170 would like to talk about direct 25 00:01:04,170 --> 00:01:06,500 integration of these equations. 26 00:01:06,500 --> 00:01:10,140 We will talk about the modes of opposition procedures in 27 00:01:10,140 --> 00:01:12,100 the next lecture. 28 00:01:12,100 --> 00:01:16,010 By direct integration, we mean that we integrate these 29 00:01:16,010 --> 00:01:18,970 equations directly without-- 30 00:01:18,970 --> 00:01:20,120 that is, without-- 31 00:01:20,120 --> 00:01:23,280 a transformation of these equations into a different 32 00:01:23,280 --> 00:01:25,740 form prior to the integration. 33 00:01:25,740 --> 00:01:28,090 In modes of opposition analysis, now you will see in 34 00:01:28,090 --> 00:01:30,620 the next lecture, we're actually transforming these 35 00:01:30,620 --> 00:01:35,040 equations first, and then we integrate the response. 36 00:01:35,040 --> 00:01:38,150 So the procedures that I'd like to talk about are 37 00:01:38,150 --> 00:01:42,410 explicit, implicit integration procedures that directly 38 00:01:42,410 --> 00:01:45,980 operate on these equations without a transformation and 39 00:01:45,980 --> 00:01:48,670 integrate, solve these equations. 40 00:01:48,670 --> 00:01:51,670 I also would like to talk about some computational 41 00:01:51,670 --> 00:01:55,530 considerations with regard to these explicit and implicit 42 00:01:55,530 --> 00:01:58,750 integration procedures, the number of operations involved, 43 00:01:58,750 --> 00:02:00,810 the costs involved in performing these 44 00:02:00,810 --> 00:02:02,820 integrations and so on. 45 00:02:02,820 --> 00:02:06,450 And then there is an important consideration, namely the 46 00:02:06,450 --> 00:02:09,900 selection of the solution time step, delta-t. 47 00:02:09,900 --> 00:02:12,320 We will see that when we integrate the dynamic 48 00:02:12,320 --> 00:02:15,050 equilibrium equations, we have to select the time step 49 00:02:15,050 --> 00:02:19,000 delta-t with which we perform the integration. 50 00:02:19,000 --> 00:02:23,560 And this time step has to be selected in order to preserve 51 00:02:23,560 --> 00:02:26,990 stability of the integration and to preserve of course or 52 00:02:26,990 --> 00:02:29,960 to obtain accuracy in the integration. 53 00:02:29,960 --> 00:02:32,870 Finally, I'd like to talk about some modeling 54 00:02:32,870 --> 00:02:36,640 considerations pertaining to the solution of the dynamic 55 00:02:36,640 --> 00:02:38,160 equilibrium equations. 56 00:02:38,160 --> 00:02:44,600 Well, here on this view graph, I'm showing once again the 57 00:02:44,600 --> 00:02:47,570 dynamically equilibrium equations here, noticing once 58 00:02:47,570 --> 00:02:51,820 more that R, the load vector, is now a function of time. 59 00:02:51,820 --> 00:02:55,080 The same equations here have been rewritten once here, 60 00:02:55,080 --> 00:02:59,750 where FI denotes inertia forces, now time 61 00:02:59,750 --> 00:03:00,930 dependent, of course. 62 00:03:00,930 --> 00:03:03,740 The mass matrix is constant, but the accelerations of 63 00:03:03,740 --> 00:03:05,750 course depend on time. 64 00:03:05,750 --> 00:03:10,770 Then this part here, which are the damping forces, has been 65 00:03:10,770 --> 00:03:13,820 expressed here as FD a function of time again. 66 00:03:13,820 --> 00:03:17,480 The C matrix is constant with time, but the velocities, of 67 00:03:17,480 --> 00:03:18,580 course, vary with time. 68 00:03:18,580 --> 00:03:20,340 So here we have damping forces. 69 00:03:20,340 --> 00:03:22,700 It's a nodes of the finite element system. 70 00:03:22,700 --> 00:03:27,020 And these are the elastic forces, KU, which are really 71 00:03:27,020 --> 00:03:30,110 due to the internal element stresses. 72 00:03:30,110 --> 00:03:34,370 And on the right hand side, we have the external forces. 73 00:03:34,370 --> 00:03:39,070 So what we're saying is that we want to solve this force 74 00:03:39,070 --> 00:03:42,810 equilibrium over all times t. 75 00:03:42,810 --> 00:03:45,910 The inertia forces plus the damping forces plus the 76 00:03:45,910 --> 00:03:51,180 elastic forces, it's the nodes of the finite element system, 77 00:03:51,180 --> 00:03:55,340 of course, calculated in the virtual work sense, the way 78 00:03:55,340 --> 00:03:58,040 we've been talking about in the earlier lectures. 79 00:03:58,040 --> 00:04:00,750 The sum of these forces here must be equal to the 80 00:04:00,750 --> 00:04:03,890 externally applied nodal point logs. 81 00:04:03,890 --> 00:04:07,940 Well, the procedure that we will be 82 00:04:07,940 --> 00:04:10,550 following is the following. 83 00:04:10,550 --> 00:04:16,019 We will consider time steps delta-t a part at which we 84 00:04:16,019 --> 00:04:19,180 want to calculate the response of the system. 85 00:04:19,180 --> 00:04:20,810 In order to do so, of course, we have to 86 00:04:20,810 --> 00:04:22,710 define our loads first. 87 00:04:22,710 --> 00:04:26,970 And here, I have schematic shown once the description of 88 00:04:26,970 --> 00:04:31,770 the load Ri, that is, the load at degree of freedom i as a 89 00:04:31,770 --> 00:04:33,510 function of time t. 90 00:04:33,510 --> 00:04:39,490 Notice that we are having here this distribution of that load 91 00:04:39,490 --> 00:04:43,990 component as a function of time. 92 00:04:43,990 --> 00:04:46,120 The distribution is quite simple. 93 00:04:46,120 --> 00:04:48,690 In a computer program, of course, all we would have to 94 00:04:48,690 --> 00:04:52,670 put in is this point, this intensity here, that load 95 00:04:52,670 --> 00:04:54,930 intensity, at a specified time. 96 00:04:54,930 --> 00:04:58,620 That load intensity, at a specified time, and that load 97 00:04:58,620 --> 00:05:00,780 intensity at a specified time. 98 00:05:00,780 --> 00:05:04,570 In between, these intensities, the computer program would 99 00:05:04,570 --> 00:05:08,880 simply interpolate linearly. 100 00:05:08,880 --> 00:05:12,370 This is the load at the component i, displacement 101 00:05:12,370 --> 00:05:13,370 component i. 102 00:05:13,370 --> 00:05:16,160 This is the load at displacement component j. 103 00:05:16,160 --> 00:05:18,860 Notice here, of course, we would have in general a 104 00:05:18,860 --> 00:05:22,470 different curve at degree of freedom j, then the curve is 105 00:05:22,470 --> 00:05:25,570 used at degree of freedom i. 106 00:05:25,570 --> 00:05:31,140 I show here time steps, discrete time steps, delta-t1, 107 00:05:31,140 --> 00:05:33,880 delta-t2, delta-t3. 108 00:05:33,880 --> 00:05:37,260 These of course apply for the complete system, for the i 109 00:05:37,260 --> 00:05:41,250 component and the j component of displacements or loads, and 110 00:05:41,250 --> 00:05:45,160 for all other displacements and loads also. 111 00:05:45,160 --> 00:05:47,970 Notice that they can vary with time. 112 00:05:47,970 --> 00:05:51,760 In many analysis, however, we simply pick one time step and 113 00:05:51,760 --> 00:05:54,500 keep it constant all along. 114 00:05:54,500 --> 00:05:56,740 The objective is the following. 115 00:05:56,740 --> 00:06:01,200 In the direct integration of the equations of equilibrium, 116 00:06:01,200 --> 00:06:04,740 the objective is knowing the initial conditions. 117 00:06:04,740 --> 00:06:07,980 In other words, knowing the displacement at time 0, the 118 00:06:07,980 --> 00:06:11,190 velocities at time 0. 119 00:06:11,190 --> 00:06:15,540 We want to calculate the displacements, velocities, and 120 00:06:15,540 --> 00:06:21,120 accelerations at these discrete time points. 121 00:06:21,120 --> 00:06:24,920 We are matching here with the time interval delta-t1. 122 00:06:24,920 --> 00:06:29,780 And here we are changing the time interval to delta-t2. 123 00:06:29,780 --> 00:06:32,970 Knowing all the velocities, accelerations, and 124 00:06:32,970 --> 00:06:36,120 displacement at this time, we want to calculate the same 125 00:06:36,120 --> 00:06:40,200 quantities at the discrete time increments of 126 00:06:40,200 --> 00:06:42,180 delta-t2, and so on. 127 00:06:42,180 --> 00:06:45,100 So basically what we are saying is that knowing the 128 00:06:45,100 --> 00:06:51,010 initial conditions, we want to calculate in a step-by-step 129 00:06:51,010 --> 00:06:54,860 procedure, the displacements, velocities, and accelerations 130 00:06:54,860 --> 00:06:57,850 at these discrete time points. 131 00:06:57,850 --> 00:07:02,790 Of course, the real problem then reduces to the following. 132 00:07:02,790 --> 00:07:08,980 Given the initial conditions at the particular time t, we 133 00:07:08,980 --> 00:07:13,220 might call this time t, and by initial conditions, I mean we 134 00:07:13,220 --> 00:07:15,640 know the displacements, velocities, accelerations, we 135 00:07:15,640 --> 00:07:19,580 know all the quantities at time t. 136 00:07:19,580 --> 00:07:24,740 Our objective is to calculate the unknown quantities at time 137 00:07:24,740 --> 00:07:29,080 t plus delta-t1 in this particular case. 138 00:07:29,080 --> 00:07:35,230 So this point is t plus delta-t1. 139 00:07:35,230 --> 00:07:41,230 In general, or in many cases, we keep a constant time step, 140 00:07:41,230 --> 00:07:44,070 and so we will simply talk about t plus 141 00:07:44,070 --> 00:07:46,810 delta-t from now onwards. 142 00:07:46,810 --> 00:07:50,610 We should remember, however, the delta-t might change doing 143 00:07:50,610 --> 00:07:51,910 the response solution. 144 00:07:51,910 --> 00:07:54,980 I repeat once more, this is an important point, given the 145 00:07:54,980 --> 00:07:58,290 initial conditions or given the conditions at time t, our 146 00:07:58,290 --> 00:08:01,900 objective is to calculate the conditions, displacements, 147 00:08:01,900 --> 00:08:07,100 velocities, accelerations, at time t plus delta-t. 148 00:08:07,100 --> 00:08:13,380 In the explicit integration, we proceed in 149 00:08:13,380 --> 00:08:14,980 the solution as follows. 150 00:08:14,980 --> 00:08:22,590 We use the equations of dynamic equilibrium at time t 151 00:08:22,590 --> 00:08:26,500 to obtain the solution at time t plus delta-t. 152 00:08:26,500 --> 00:08:27,820 This is important. 153 00:08:27,820 --> 00:08:31,620 In explicit integration, we use the condition, the 154 00:08:31,620 --> 00:08:34,590 equilibrium equations, the equilibrium equation, the 155 00:08:34,590 --> 00:08:39,200 dynamic equilibrium equation, and by that I mean, just to 156 00:08:39,200 --> 00:08:41,830 refresh your memory, this equation-- we use that 157 00:08:41,830 --> 00:08:46,760 equation at time t in explicit integration to obtain the 158 00:08:46,760 --> 00:08:49,980 solution at time t plus delta-t. 159 00:08:49,980 --> 00:08:53,210 Once we have the solution at time t plus delta-t, of course 160 00:08:53,210 --> 00:08:55,550 we can march ahead in the same way again. 161 00:08:55,550 --> 00:08:58,980 In implicit integration, however, we are looking at the 162 00:08:58,980 --> 00:09:03,430 equilibrium equations at time t plus delta-t t to obtain the 163 00:09:03,430 --> 00:09:05,610 solution at time t plus delta-t. 164 00:09:05,610 --> 00:09:09,345 In other words, if you look at once more at the equations 165 00:09:09,345 --> 00:09:12,410 here, at the equilibrium equations, these equations 166 00:09:12,410 --> 00:09:15,280 would be applied at time t plus delta-t to obtain the 167 00:09:15,280 --> 00:09:18,190 solution at time t plus delta-t. 168 00:09:18,190 --> 00:09:21,500 This is indeed the fundamental difference between explicit 169 00:09:21,500 --> 00:09:22,980 and implicit integration. 170 00:09:22,980 --> 00:09:26,790 In explicit integration, we are looking at equilibrium 171 00:09:26,790 --> 00:09:29,270 equations at time t to obtain the solution 172 00:09:29,270 --> 00:09:31,150 at time t plus delta-t. 173 00:09:31,150 --> 00:09:33,640 And in implicit integration, we are looking at the 174 00:09:33,640 --> 00:09:37,090 equilibrium equations at time t plus delta-t to attain the 175 00:09:37,090 --> 00:09:39,970 solution at time t plus delta-t. 176 00:09:39,970 --> 00:09:42,830 In the following now, I would like to show to you an 177 00:09:42,830 --> 00:09:45,740 explicit and an implicit integration scheme. 178 00:09:45,740 --> 00:09:48,720 The explicit integration scheme that is abundantly used 179 00:09:48,720 --> 00:09:51,640 in practice is the central difference method. 180 00:09:51,640 --> 00:09:53,950 I will later on then discuss how this method 181 00:09:53,950 --> 00:09:55,550 is effectively used. 182 00:09:55,550 --> 00:09:58,710 In the central difference method, we are applying, as I 183 00:09:58,710 --> 00:10:02,980 said earlier, the equilibrium equations at time t, because 184 00:10:02,980 --> 00:10:05,780 it's an explicit integration scheme. 185 00:10:05,780 --> 00:10:09,460 The two additional questions that we're using to solve for 186 00:10:09,460 --> 00:10:13,580 the unknown displacements at time t plus delta-t, is 187 00:10:13,580 --> 00:10:17,180 firstly an equation for the accelerations at time t. 188 00:10:17,180 --> 00:10:22,030 And notice that this equation involves as the unknown, as 189 00:10:22,030 --> 00:10:24,340 the only unknown, the displacement 190 00:10:24,340 --> 00:10:27,110 at time t plus delta-t. 191 00:10:27,110 --> 00:10:29,960 In addition, of course, we are using an equation for the 192 00:10:29,960 --> 00:10:34,690 velocities at time t and again this equation only involves as 193 00:10:34,690 --> 00:10:36,500 the only unknown the displacement 194 00:10:36,500 --> 00:10:38,300 at time t plus delta-t. 195 00:10:38,300 --> 00:10:42,000 Substituting these two equations into the equilibrium 196 00:10:42,000 --> 00:10:48,060 equation at time t gives us one equation in the unknown t, 197 00:10:48,060 --> 00:10:50,620 displacement at time t plus delta-t. 198 00:10:50,620 --> 00:10:54,280 We can simply solve for it, as shown, now on 199 00:10:54,280 --> 00:10:56,240 the next view graph. 200 00:10:56,240 --> 00:10:59,530 You can see here the equation that is obtained by 201 00:10:59,530 --> 00:11:03,420 substituting the relations for the acceleration and for the 202 00:11:03,420 --> 00:11:07,350 velocities in the equilibrium equation at time t. 203 00:11:07,350 --> 00:11:08,990 The unknown appears here. 204 00:11:08,990 --> 00:11:11,430 This is the displacement at time t plus 205 00:11:11,430 --> 00:11:14,270 delta-t, which is unknown. 206 00:11:14,270 --> 00:11:17,310 And there is a coefficient matrix here, that involves the 207 00:11:17,310 --> 00:11:19,830 mass matrix and the damping matrix. 208 00:11:19,830 --> 00:11:22,730 And on the right hand side, of course, we are calculating a 209 00:11:22,730 --> 00:11:23,840 load vector. 210 00:11:23,840 --> 00:11:27,270 And that load vector, notice once again, corresponds to the 211 00:11:27,270 --> 00:11:29,650 conditions at time t. 212 00:11:29,650 --> 00:11:32,420 That's why it's an explicit integration scheme. 213 00:11:32,420 --> 00:11:35,780 The difference operator, the essential difference operator 214 00:11:35,780 --> 00:11:40,045 that I've been used here, simple operators which you 215 00:11:40,045 --> 00:11:44,060 have seen on the previous slide, that involve only the 216 00:11:44,060 --> 00:11:48,000 displacement at time t minus delta-t and the 217 00:11:48,000 --> 00:11:49,500 displacement at time t. 218 00:11:49,500 --> 00:11:52,660 And notice that here are these two 219 00:11:52,660 --> 00:11:55,990 displacement vectors appearing. 220 00:11:55,990 --> 00:12:00,803 Notice that this equation is cheaply-- 221 00:12:00,803 --> 00:12:04,930 or can be used in a very cheap way to calculate the 222 00:12:04,930 --> 00:12:07,450 displacement, the unknown displacement at time t plus 223 00:12:07,450 --> 00:12:12,370 delta-t, if the mass matrix is a diagonal matrix. 224 00:12:12,370 --> 00:12:14,920 And, say, the damping is neglected. 225 00:12:14,920 --> 00:12:17,520 In other words, this term not being there, this being a 226 00:12:17,520 --> 00:12:22,390 diagonal mass matrix, then the solution is very trivial. 227 00:12:22,390 --> 00:12:25,620 In fact, all we need to do is calculate the right hand side 228 00:12:25,620 --> 00:12:30,470 load vector and divide each component of that load vector 229 00:12:30,470 --> 00:12:35,390 by this coefficient here which, if m once again is a 230 00:12:35,390 --> 00:12:38,080 diagonal mass matrix, is simply Mii 231 00:12:38,080 --> 00:12:40,540 divided by delta-t squared. 232 00:12:40,540 --> 00:12:44,310 And in fact, this is really how was the procedure is 233 00:12:44,310 --> 00:12:47,580 usually applied, namely, with a diagonal mass 234 00:12:47,580 --> 00:12:49,370 matrix and no damping. 235 00:12:49,370 --> 00:12:54,050 The damping usually is not taken into account, or at 236 00:12:54,050 --> 00:12:57,450 most, we have diagonal dampers also, and the mass matrix 237 00:12:57,450 --> 00:13:01,450 being a diagonal matrix, then the solution is 238 00:13:01,450 --> 00:13:02,760 very cheaply performed. 239 00:13:02,760 --> 00:13:05,530 There's one further important point. 240 00:13:05,530 --> 00:13:11,120 This load vector of course is constructed in the usual way. 241 00:13:11,120 --> 00:13:16,630 This multiplication here is quite cheap if the mass matrix 242 00:13:16,630 --> 00:13:18,410 is a diagonal mass matrix. 243 00:13:18,410 --> 00:13:20,010 K times tU. 244 00:13:20,010 --> 00:13:23,470 Well, K here is of course a banded matrix. 245 00:13:23,470 --> 00:13:27,770 It's the stiffness matrix of the system, constant. 246 00:13:27,770 --> 00:13:31,270 Well, there are some operations involved here. 247 00:13:31,270 --> 00:13:35,280 And if we look at this operation here in detail, we 248 00:13:35,280 --> 00:13:38,590 see here how it can be effectively performed. 249 00:13:38,590 --> 00:13:40,270 We have a stiffness matrix times the 250 00:13:40,270 --> 00:13:41,780 displacement vector here. 251 00:13:41,780 --> 00:13:44,920 And of course, the stiffness matrix remember is obtained by 252 00:13:44,920 --> 00:13:47,350 summing the element contributions. 253 00:13:47,350 --> 00:13:51,500 So you can write this K matrix in this way. 254 00:13:51,500 --> 00:13:55,540 Since tU is independent of the elements, we can also take it 255 00:13:55,540 --> 00:13:57,230 into the summation sign. 256 00:13:57,230 --> 00:14:02,270 And if we look at this product closely, we find that is 257 00:14:02,270 --> 00:14:04,970 nothing else than tFm. 258 00:14:04,970 --> 00:14:13,900 It's a nodal point vector of loads or elements forces that 259 00:14:13,900 --> 00:14:15,260 correspond-- 260 00:14:15,260 --> 00:14:18,940 element elastic forces-- that correspond to the current 261 00:14:18,940 --> 00:14:20,420 displacements. 262 00:14:20,420 --> 00:14:25,280 And this tF vector is much more cheaply calculated than 263 00:14:25,280 --> 00:14:27,390 the Km times tU. 264 00:14:27,390 --> 00:14:30,740 And in fact, this is the way how we actually perform the 265 00:14:30,740 --> 00:14:33,370 calculation of this product here. 266 00:14:33,370 --> 00:14:37,130 Now notice the following important further point. 267 00:14:37,130 --> 00:14:41,980 If we were to leave this K times tU the way it stands 268 00:14:41,980 --> 00:14:44,800 here, we would have to assemble a global stiffness 269 00:14:44,800 --> 00:14:46,710 matrix of the system. 270 00:14:46,710 --> 00:14:49,700 However, if we write that product this way, we never 271 00:14:49,700 --> 00:14:52,060 need to assemble a global stiffness matrix 272 00:14:52,060 --> 00:14:53,380 of the total system. 273 00:14:53,380 --> 00:14:58,010 In fact, all we need to do is calculate element forces and 274 00:14:58,010 --> 00:15:01,860 add these elements forces via the direct stiffness procedure 275 00:15:01,860 --> 00:15:03,940 the way I've been discussing earlier. 276 00:15:03,940 --> 00:15:09,410 So this is a very effective way of calculating K times tU. 277 00:15:09,410 --> 00:15:12,560 We never need to assemble a global stiffness matrix. 278 00:15:12,560 --> 00:15:17,060 In fact, this is one of the main reasons why the central 279 00:15:17,060 --> 00:15:20,040 difference method is effective. 280 00:15:20,040 --> 00:15:25,250 It's this reason that we can calculate the KtU part, this 281 00:15:25,250 --> 00:15:29,530 times that, very effectively via this procedure. 282 00:15:29,530 --> 00:15:30,500 That is the first reason. 283 00:15:30,500 --> 00:15:34,980 And the second reason is that if M is diagonal, we really 284 00:15:34,980 --> 00:15:37,160 never solve equations. 285 00:15:37,160 --> 00:15:41,550 The equation solving here is so trivial, that we might just 286 00:15:41,550 --> 00:15:43,270 say that we do not solve equations. 287 00:15:43,270 --> 00:15:45,530 There is no factorization involved. 288 00:15:45,530 --> 00:15:49,910 There's no factorization of a stiffness matrix or another 289 00:15:49,910 --> 00:15:52,900 coefficient matrix involved, because if M is diagonal, and 290 00:15:52,900 --> 00:15:57,300 as I pointed out, we simply divide the right hand side by 291 00:15:57,300 --> 00:16:01,310 Mii divided by delta-t squared. 292 00:16:01,310 --> 00:16:05,550 Notice that if we start the integration process, we know 293 00:16:05,550 --> 00:16:07,040 of course the initial conditions. 294 00:16:07,040 --> 00:16:08,540 We know tU. 295 00:16:08,540 --> 00:16:11,930 And we do not know-- 296 00:16:11,930 --> 00:16:14,830 usually directly the t minus delta-tU. 297 00:16:14,830 --> 00:16:17,570 That is the displacement prior to the solution. 298 00:16:17,570 --> 00:16:22,160 And to obtain those, we can use this formula, which is 299 00:16:22,160 --> 00:16:27,420 really the central difference method applied at time t, 300 00:16:27,420 --> 00:16:29,790 where we however now involve the accelerations 301 00:16:29,790 --> 00:16:31,940 corresponding to time 0. 302 00:16:31,940 --> 00:16:35,270 We involve the accelerations corresponding to time zero 303 00:16:35,270 --> 00:16:38,290 right here, and the velocities corresponding to times 0, and 304 00:16:38,290 --> 00:16:40,580 the displacements corresponding to time zero. 305 00:16:40,580 --> 00:16:42,670 And of course, these quantities must all be given 306 00:16:42,670 --> 00:16:46,070 as the initial conditions to the solution. 307 00:16:46,070 --> 00:16:48,840 I pointed out already that we're using this technique 308 00:16:48,840 --> 00:16:51,130 mostly with the lump mass matrix. 309 00:16:51,130 --> 00:16:53,880 In addition I should also point out that we're using it 310 00:16:53,880 --> 00:16:56,250 primarily with lower order elements, 311 00:16:56,250 --> 00:16:57,890 with lower order elements. 312 00:16:57,890 --> 00:17:04,430 The reason for that is the following, namely, one of 313 00:17:04,430 --> 00:17:08,190 stability and accuracy of the central difference method. 314 00:17:08,190 --> 00:17:15,460 The method can only be applied if delta-t, the time step that 315 00:17:15,460 --> 00:17:18,200 I'm talking about here, is smaller than a 316 00:17:18,200 --> 00:17:20,119 critical time step. 317 00:17:20,119 --> 00:17:25,240 And that critical time step is given by the smallest period 318 00:17:25,240 --> 00:17:30,180 in the system divided by pi, 3.14 et cetera. 319 00:17:30,180 --> 00:17:32,730 The smallest period in the system can be 320 00:17:32,730 --> 00:17:34,930 of course very small. 321 00:17:34,930 --> 00:17:39,140 And therefore the critical time step that we're talking 322 00:17:39,140 --> 00:17:42,410 about here can be very small. 323 00:17:42,410 --> 00:17:46,060 The smallest period in the system, just to give you an 324 00:17:46,060 --> 00:17:50,210 example, would be extremely small if you have a bar 325 00:17:50,210 --> 00:17:52,990 element, and you have some very short 326 00:17:52,990 --> 00:17:54,470 truss element in there. 327 00:17:54,470 --> 00:17:58,000 The shorter the truss element, the stiffer of course it is. 328 00:17:58,000 --> 00:18:02,380 AU over L is the stiffness of a truss element, and 329 00:18:02,380 --> 00:18:06,860 therefore, a very small period would be in that 330 00:18:06,860 --> 00:18:10,470 assemblage of elements. 331 00:18:10,470 --> 00:18:14,240 Therefore, what is required really is to have a mesh that 332 00:18:14,240 --> 00:18:19,760 is more or less uniform, so that there is no artificially 333 00:18:19,760 --> 00:18:23,320 small time step required in the integration of the 334 00:18:23,320 --> 00:18:24,770 equilibrium equations. 335 00:18:24,770 --> 00:18:27,060 Because there is this condition that 336 00:18:27,060 --> 00:18:28,650 delta-t must be smaller-- 337 00:18:28,650 --> 00:18:30,980 we should really say must be smaller or equal-- 338 00:18:30,980 --> 00:18:36,380 to delta-t critical, but to be on the conservative side, we 339 00:18:36,380 --> 00:18:40,200 are selecting delta-t smaller than delta-t critical in 340 00:18:40,200 --> 00:18:43,310 practical analysis of complex meshes. 341 00:18:43,310 --> 00:18:46,130 Because this condition has to be satisfied, we're talking 342 00:18:46,130 --> 00:18:50,320 about a conditionally stable scheme. 343 00:18:50,320 --> 00:18:55,380 It is only stable when delta-t is smaller than that value. 344 00:18:55,380 --> 00:18:59,920 If we were to select delta-t larger than delta-t critical, 345 00:18:59,920 --> 00:19:05,140 we would see that after just a few time steps, the solution-- 346 00:19:05,140 --> 00:19:06,770 it gets out of bounds. 347 00:19:06,770 --> 00:19:10,860 And after 10 or 20 or 30 time step, in a computer program, 348 00:19:10,860 --> 00:19:12,150 you would all of a sudden notice 349 00:19:12,150 --> 00:19:13,560 overflow of the numbers. 350 00:19:13,560 --> 00:19:16,380 So the solution wouldn't make any sense whatsoever anymore. 351 00:19:16,380 --> 00:19:18,490 And this is an absolute condition that 352 00:19:18,490 --> 00:19:20,380 does have to be satisfied. 353 00:19:20,380 --> 00:19:23,370 There's no way you can use a time step larger than delta-t 354 00:19:23,370 --> 00:19:26,570 critical, and still get an acceptable, reasonable 355 00:19:26,570 --> 00:19:31,380 solution to your equations of motion. 356 00:19:31,380 --> 00:19:35,010 In practice, of course, we have to estimate, in a 357 00:19:35,010 --> 00:19:37,210 conservative way, a time step. 358 00:19:37,210 --> 00:19:39,380 And that means we have to have an estimate 359 00:19:39,380 --> 00:19:41,680 also of delta-t critical. 360 00:19:41,680 --> 00:19:43,440 For continuum elements-- 361 00:19:43,440 --> 00:19:46,350 by continuum elements, I mean plain stress, plain strain, 362 00:19:46,350 --> 00:19:48,560 axis a matrix, three-dimensional element, the 363 00:19:48,560 --> 00:19:50,670 way we have been discussing it earlier. 364 00:19:50,670 --> 00:19:57,430 A good way of obtaining time step delta-t is to use this 365 00:19:57,430 --> 00:20:03,650 formula here, delta-t being delta-L over C. C is the wave 366 00:20:03,650 --> 00:20:05,800 velocity through the system. 367 00:20:05,800 --> 00:20:06,390 [UNINTELLIGIBLE] 368 00:20:06,390 --> 00:20:10,890 is divided by rho, square root out of it gives us C. And 369 00:20:10,890 --> 00:20:19,210 delta-L is an element length that has to be defined. 370 00:20:19,210 --> 00:20:24,290 In fact, it's the smallest distance between nodes. 371 00:20:24,290 --> 00:20:27,990 The smallest distance between nodes if we talk about low 372 00:20:27,990 --> 00:20:29,040 order elements. 373 00:20:29,040 --> 00:20:30,590 What I mean by that is the following. 374 00:20:30,590 --> 00:20:34,250 If we have a mesh here that looks like that. 375 00:20:34,250 --> 00:20:38,125 And I deliberately show elements that are not equal in 376 00:20:38,125 --> 00:20:42,960 size, then delta-E in this particular case would be this 377 00:20:42,960 --> 00:20:43,860 distance here. 378 00:20:43,860 --> 00:20:48,050 That is our delta-L, sorry, our delta-L is 379 00:20:48,050 --> 00:20:51,610 that distance there. 380 00:20:51,610 --> 00:20:55,320 This, of course, as I have here in the heading, assumes 381 00:20:55,320 --> 00:20:57,040 that we have low-order elements. 382 00:20:57,040 --> 00:21:00,460 That means only corner nodes in the elements. 383 00:21:00,460 --> 00:21:02,240 For higher-order elements-- 384 00:21:02,240 --> 00:21:05,650 let me put down here one higher-order element, where we 385 00:21:05,650 --> 00:21:11,420 have these nodes now, say an 8 node element. 386 00:21:11,420 --> 00:21:14,310 In this particular case, we take the smallest distance 387 00:21:14,310 --> 00:21:19,710 between nodes, again, just as in this case. 388 00:21:19,710 --> 00:21:21,940 And in this particular case, it would be, say, this 389 00:21:21,940 --> 00:21:23,140 distance here. 390 00:21:23,140 --> 00:21:26,200 That is our smallest distance between nodes. 391 00:21:26,200 --> 00:21:29,460 However, we have to also divide this distance by a 392 00:21:29,460 --> 00:21:31,800 relative stiffness factor. 393 00:21:31,800 --> 00:21:34,430 And why does that relative stiffness factor 394 00:21:34,430 --> 00:21:35,860 come into the picture? 395 00:21:35,860 --> 00:21:39,320 Well, it does come into the picture, because remember that 396 00:21:39,320 --> 00:21:44,190 the stiffness at an interior node is larger than the 397 00:21:44,190 --> 00:21:47,570 stiffness at a corner for the element. 398 00:21:47,570 --> 00:21:50,870 The amount it is larger depends on the particular 399 00:21:50,870 --> 00:21:53,600 element used. 400 00:21:53,600 --> 00:21:57,490 But since there's a lot of stiffness here, we have to put 401 00:21:57,490 --> 00:22:01,430 here a relative stiffness factor into the calculation of 402 00:22:01,430 --> 00:22:06,970 delta-L, the effective lengths that we're using in this 403 00:22:06,970 --> 00:22:10,230 particular formula. 404 00:22:10,230 --> 00:22:14,450 Now, for this element here, for example, the relative 405 00:22:14,450 --> 00:22:16,910 stiffness factor would be conservatively 406 00:22:16,910 --> 00:22:20,410 just equal to 4. 407 00:22:20,410 --> 00:22:23,690 But for other elements, and particular also when there's 408 00:22:23,690 --> 00:22:27,260 node shifting involved and so on, the relative stiffness 409 00:22:27,260 --> 00:22:30,330 factor has to be calculated, has to be estimated. 410 00:22:30,330 --> 00:22:33,950 And we should always remember that it should be estimated 411 00:22:33,950 --> 00:22:41,320 here in a conservative way, because a smaller delta-L gave 412 00:22:41,320 --> 00:22:44,570 us a smaller delta-t. 413 00:22:44,570 --> 00:22:47,680 And then we are then conservatively applying the 414 00:22:47,680 --> 00:22:50,900 fact that delta-t shall be smaller than the delta-t 415 00:22:50,900 --> 00:22:55,220 critical, value that I have been referring hereto. 416 00:22:55,220 --> 00:23:00,120 The method is used mainly for wave propagation analysis. 417 00:23:00,120 --> 00:23:03,650 The reason being that we have to use a fairly large number 418 00:23:03,650 --> 00:23:07,260 of time steps to predict the solution. 419 00:23:07,260 --> 00:23:11,220 Large, because a time step delta-t is usually very small. 420 00:23:11,220 --> 00:23:13,940 And in wave propagation analysis, many modes of the 421 00:23:13,940 --> 00:23:16,590 system are being excited. 422 00:23:16,590 --> 00:23:20,340 So we really want to pick up a large number of 423 00:23:20,340 --> 00:23:22,170 modes in the mesh. 424 00:23:22,170 --> 00:23:27,990 And the small time step integrating accurately many 425 00:23:27,990 --> 00:23:31,540 modes in the mesh goes together with the fact that we 426 00:23:31,540 --> 00:23:35,180 want to predict many modes in the mesh or a large number of 427 00:23:35,180 --> 00:23:38,960 modes in the mesh, and therefore the method is 428 00:23:38,960 --> 00:23:42,060 effectively used for wave propagation analysis. 429 00:23:42,060 --> 00:23:46,500 The number of operations are proportional to the number of 430 00:23:46,500 --> 00:23:52,042 elements that we are using and the number of time steps. 431 00:23:52,042 --> 00:23:55,830 Of course, this is an approximate proportionality 432 00:23:55,830 --> 00:23:58,550 here that I'm talking about. 433 00:23:58,550 --> 00:24:06,340 This fact derives directly from the consideration that we 434 00:24:06,340 --> 00:24:07,440 talked about here. 435 00:24:07,440 --> 00:24:10,640 Notice that as we include more and more elements in the 436 00:24:10,640 --> 00:24:13,880 solution process, of course, we have to sum over more 437 00:24:13,880 --> 00:24:15,560 elements here. 438 00:24:15,560 --> 00:24:18,800 And as we go more and more time steps, we have to apply 439 00:24:18,800 --> 00:24:23,720 this equation as many times also more and more. 440 00:24:23,720 --> 00:24:27,440 The important point is that we don't assemble a K matrix, so 441 00:24:27,440 --> 00:24:30,850 the bandwidth off the system, the bandwidth of the system 442 00:24:30,850 --> 00:24:33,300 does not appear in an operation count. 443 00:24:33,300 --> 00:24:36,130 It's the number of elements that do appear in the 444 00:24:36,130 --> 00:24:37,380 operation count. 445 00:24:40,010 --> 00:24:48,330 The next procedure that I like to talk about 446 00:24:48,330 --> 00:24:50,100 is the Newmark method. 447 00:24:50,100 --> 00:24:55,600 And in the Newmark method, this being an implicit 448 00:24:55,600 --> 00:24:59,710 integration scheme now, we proceed as follows. 449 00:24:59,710 --> 00:25:00,610 We apply-- 450 00:25:00,610 --> 00:25:02,730 because it's an implicit integration scheme-- 451 00:25:02,730 --> 00:25:05,450 the equilibrium equation, the dynamic equilibrium equation 452 00:25:05,450 --> 00:25:08,150 at time t plus delta-t. 453 00:25:08,150 --> 00:25:12,450 Here it is given the additional formula that we're 454 00:25:12,450 --> 00:25:16,560 using is this one here for the velocity at time t plus 455 00:25:16,560 --> 00:25:19,140 delta-t, and this one here for the displacement of time t 456 00:25:19,140 --> 00:25:20,530 plus delta-t. 457 00:25:20,530 --> 00:25:25,380 Now notice that with this set of equations, with these three 458 00:25:25,380 --> 00:25:31,070 equations, knowing the displacements and velocities 459 00:25:31,070 --> 00:25:37,620 at time t and accelerations at time t, we have with these 460 00:25:37,620 --> 00:25:40,410 three equations, three unknowns. 461 00:25:40,410 --> 00:25:44,440 The velocity at time t plus delta-t, the acceleration at 462 00:25:44,440 --> 00:25:47,415 time t plus delta-t, and the displacement 463 00:25:47,415 --> 00:25:49,720 at time t plus delta-t. 464 00:25:49,720 --> 00:25:52,165 These are the only three unknowns that appear. 465 00:25:52,165 --> 00:25:55,540 There are three linearly independent equations. 466 00:25:55,540 --> 00:25:57,550 And we can simply use them to 467 00:25:57,550 --> 00:25:59,515 calculate these three unknowns. 468 00:25:59,515 --> 00:26:02,030 Of course, we want to do so in a very effective way. 469 00:26:02,030 --> 00:26:04,320 And I will show you how we proceed. 470 00:26:04,320 --> 00:26:11,640 The important point is that with this scheme, we are using 471 00:26:11,640 --> 00:26:16,170 an implicit approach, as I defined it earlier. 472 00:26:16,170 --> 00:26:21,070 And when we do substitute from here into there, we will find 473 00:26:21,070 --> 00:26:25,540 that on the left hand side, we have a coefficient matrix that 474 00:26:25,540 --> 00:26:30,500 involves a bandwidth even if the mass matrix is diagonal. 475 00:26:30,500 --> 00:26:33,415 And this coefficient matrix here is written 476 00:26:33,415 --> 00:26:35,730 here down as K hat. 477 00:26:35,730 --> 00:26:40,640 This coefficient matrix invoice this K matrix. 478 00:26:40,640 --> 00:26:43,350 I should say maybe a few words to these formulae. 479 00:26:43,350 --> 00:26:46,640 These formulae can be obtained by Taylor Series expansion, 480 00:26:46,640 --> 00:26:49,970 around the conditions at times t plus delta-t. 481 00:26:49,970 --> 00:26:53,070 They have been proposed originally by Newmark. 482 00:26:53,070 --> 00:26:56,330 That's why it's called now the Newmark method. 483 00:26:56,330 --> 00:27:00,420 Newmark in particular used the values of delta equals 1/2 and 484 00:27:00,420 --> 00:27:02,430 alpha equal to 1/4. 485 00:27:02,430 --> 00:27:06,800 And then it's a constant average acceleration scheme. 486 00:27:06,800 --> 00:27:14,960 The constant average acceleration schemes use these 487 00:27:14,960 --> 00:27:17,700 values as I pointed out already earlier. 488 00:27:17,700 --> 00:27:20,310 The method is unconditionally stable. 489 00:27:20,310 --> 00:27:25,550 This means that the time step delta-t does not to have to be 490 00:27:25,550 --> 00:27:28,480 smaller than a certain value, as it was the case in the 491 00:27:28,480 --> 00:27:30,490 central difference method. 492 00:27:30,490 --> 00:27:33,970 The method is primarily was for analysis of structural 493 00:27:33,970 --> 00:27:36,410 dynamic problems. 494 00:27:36,410 --> 00:27:40,030 I will talk about why this is so just now. 495 00:27:40,030 --> 00:27:42,560 And the number of operations that we're talking about here 496 00:27:42,560 --> 00:27:48,110 are an initial factorization of the K hat matrix. 497 00:27:48,110 --> 00:27:51,610 This is the K hat matrix, And that initial factorization 498 00:27:51,610 --> 00:27:55,900 involves this K matrix, so the bandwidth of the K hat matrix 499 00:27:55,900 --> 00:27:59,130 is the same as the bandwidth of the K matrix. 500 00:27:59,130 --> 00:28:04,170 And therefore the operations here are one half NM squared. 501 00:28:04,170 --> 00:28:11,160 And for each solution, in time, because this equation 502 00:28:11,160 --> 00:28:16,380 here has to be solved for each time step, you're starting 503 00:28:16,380 --> 00:28:19,870 with delta-t, and then we go to 2 delta-t, 3 delta-t, 4 504 00:28:19,870 --> 00:28:22,090 delta-t, and so on. 505 00:28:22,090 --> 00:28:25,820 And since this solution is to be formed for each time step, 506 00:28:25,820 --> 00:28:28,490 however, remember the factorization is only done 507 00:28:28,490 --> 00:28:36,520 once because this K hat matrix is independent of the time. 508 00:28:36,520 --> 00:28:39,410 Therefore, we have a forward reduction and back 509 00:28:39,410 --> 00:28:41,710 substitution in each time step. 510 00:28:41,710 --> 00:28:45,500 And the number of operations in each time step are 2nm. t 511 00:28:45,500 --> 00:28:47,600 is the number of time steps here. 512 00:28:47,600 --> 00:28:51,650 t is the number of time steps here. 513 00:28:51,650 --> 00:28:54,320 Well, let us look then at the accuracy 514 00:28:54,320 --> 00:28:56,600 considerations that we have. 515 00:28:56,600 --> 00:28:59,770 I mentioned already the method is unconditionally stable, 516 00:28:59,770 --> 00:29:04,470 which means that the delta-t time step that we're using can 517 00:29:04,470 --> 00:29:08,360 be any amount, and we will never have a blow up of the 518 00:29:08,360 --> 00:29:11,460 solution, a blow up of the solution due to round-off 519 00:29:11,460 --> 00:29:13,440 errors that I involved. 520 00:29:13,440 --> 00:29:15,350 In the central difference method, of course, the 521 00:29:15,350 --> 00:29:17,650 explicit time integration scheme, I mentioned that 522 00:29:17,650 --> 00:29:20,730 delta-t has to be smaller than a critical time step. 523 00:29:20,730 --> 00:29:24,760 So in the implicit Newmark integration scheme, what we 524 00:29:24,760 --> 00:29:29,600 can do is we select our of time step based solely on 525 00:29:29,600 --> 00:29:32,290 accuracy considerations. 526 00:29:32,290 --> 00:29:35,270 Let us consider then these equations that we want to 527 00:29:35,270 --> 00:29:39,220 solve, and I'm now leaving out the damping term in order to 528 00:29:39,220 --> 00:29:42,410 obtain some understanding of how we would select the time 529 00:29:42,410 --> 00:29:43,700 step, delta-t. 530 00:29:43,700 --> 00:29:47,200 Of course, remember the larger the time step, the less normal 531 00:29:47,200 --> 00:29:50,440 operations I involved in the solution. 532 00:29:50,440 --> 00:29:52,410 So we want to select the large time step. 533 00:29:52,410 --> 00:29:55,500 However, if we make the time step extremely large, then 534 00:29:55,500 --> 00:29:58,610 surely we cannot expect any accuracy in the solution. 535 00:29:58,610 --> 00:30:02,070 In fact, if delta-t is very large, we simply predict the 536 00:30:02,070 --> 00:30:03,670 static solution. 537 00:30:03,670 --> 00:30:05,430 This is one property of the Newmark method. 538 00:30:05,430 --> 00:30:07,950 If you were to use a computer program with the Newmark 539 00:30:07,950 --> 00:30:10,160 method, if you're interested in it, why don't you try, say, 540 00:30:10,160 --> 00:30:12,860 for example, using ADINA, in which we have the Newmark 541 00:30:12,860 --> 00:30:15,520 method, and just put delta-t very large, and all you would 542 00:30:15,520 --> 00:30:17,660 be getting out is a static solution. 543 00:30:17,660 --> 00:30:20,360 Well, since we're interested in the dynamic response, we 544 00:30:20,360 --> 00:30:22,840 have to therefore select delta-t small enough to 545 00:30:22,840 --> 00:30:28,550 integrate accurately the solution response. 546 00:30:28,550 --> 00:30:31,740 And these are the equations that I want to now consider, 547 00:30:31,740 --> 00:30:36,500 and I want to show you how can we select delta-t rationally 548 00:30:36,500 --> 00:30:39,340 and to obtain the required accuracy. 549 00:30:39,340 --> 00:30:44,040 The displacement vector here can be expressed in terms of 550 00:30:44,040 --> 00:30:48,630 the mode shape vectors times xi values. 551 00:30:48,630 --> 00:30:53,360 I will be talking about how these phi i vectors are 552 00:30:53,360 --> 00:30:57,780 calculated in the next lectures. 553 00:30:57,780 --> 00:31:00,350 This involves the solution of the eigenvalue problem 554 00:31:00,350 --> 00:31:01,890 written down here. 555 00:31:01,890 --> 00:31:04,460 The free vibration eigenvalue a problem corresponding to 556 00:31:04,460 --> 00:31:06,380 this equation. 557 00:31:06,380 --> 00:31:11,090 I like to defer the discussion at this point, because we will 558 00:31:11,090 --> 00:31:13,390 be talking about it later on in detail. 559 00:31:13,390 --> 00:31:17,520 All I like to say is that this transformation here, phi i 560 00:31:17,520 --> 00:31:21,550 being a vector time independent and xi being a 561 00:31:21,550 --> 00:31:23,830 scalar, dependent on time. 562 00:31:23,830 --> 00:31:29,560 This transformation can be used in this solution here, or 563 00:31:29,560 --> 00:31:32,100 can be used rather to transform this system of 564 00:31:32,100 --> 00:31:36,330 equations into a different form. 565 00:31:36,330 --> 00:31:42,390 And in that transformation, we use the fact that when we 566 00:31:42,390 --> 00:31:45,480 calculate phi transpose K times 5-- 567 00:31:45,480 --> 00:31:47,510 notice these are capital phis. 568 00:31:47,510 --> 00:31:50,280 I've crossed bars here, capital phis. 569 00:31:50,280 --> 00:31:54,090 This is a lower phi, no crossed bars top and bottom. 570 00:31:54,090 --> 00:31:58,770 This capital phi here stores all of the lowercase phi 571 00:31:58,770 --> 00:32:02,000 vectors, and their n of such vectors. 572 00:32:02,000 --> 00:32:07,520 If I use this capital phi and calculate this product here, I 573 00:32:07,520 --> 00:32:08,900 get a diagonal matrix. 574 00:32:08,900 --> 00:32:12,620 In fact, the diagonal matrix here stores on its diagonal 575 00:32:12,620 --> 00:32:15,030 the frequency squared, the free 576 00:32:15,030 --> 00:32:16,920 vibrational frequency squared. 577 00:32:16,920 --> 00:32:21,870 Also phi transposed M phi gives us an identity matrix. 578 00:32:21,870 --> 00:32:27,000 These, once again, the vectors in this matrix, capital phi, 579 00:32:27,000 --> 00:32:31,170 are these phi i vectors which are the 580 00:32:31,170 --> 00:32:33,590 eigenvectors of this problem. 581 00:32:33,590 --> 00:32:37,450 And I will talk later on about how we calculate them. 582 00:32:37,450 --> 00:32:41,280 Well, if we apply this transformation-- 583 00:32:41,280 --> 00:32:44,000 whoops, let me put it right once more-- if we apply this 584 00:32:44,000 --> 00:32:52,090 transformation to this system of equations, and we use this 585 00:32:52,090 --> 00:32:58,100 property right here, we obtain n decoupled 586 00:32:58,100 --> 00:33:01,670 equations of this form. 587 00:33:01,670 --> 00:33:05,600 X double ought i plus omega i squared, xi equals phi i 588 00:33:05,600 --> 00:33:09,150 transposed times R. And they are in such equations, in 589 00:33:09,150 --> 00:33:13,340 other words I run c from 1 to n. 590 00:33:13,340 --> 00:33:16,900 In fact, if we use multiple position analysis, we actually 591 00:33:16,900 --> 00:33:19,780 perform this transformation. 592 00:33:19,780 --> 00:33:22,570 What I now want to do is simply refer to it 593 00:33:22,570 --> 00:33:23,270 theoretically. 594 00:33:23,270 --> 00:33:25,150 I do not want to transform it. 595 00:33:25,150 --> 00:33:27,600 You're talking about direct integration of the solution of 596 00:33:27,600 --> 00:33:30,470 equations, in which there is no transformation involved. 597 00:33:30,470 --> 00:33:33,000 So I do not want to go through this transformation. 598 00:33:33,000 --> 00:33:38,300 However, I want to refer to it theoretically. 599 00:33:38,300 --> 00:33:43,750 And it gives me a tremendous insight into what happens. 600 00:33:43,750 --> 00:33:48,250 So basically, what I can say theoretically is that the 601 00:33:48,250 --> 00:33:51,890 direct step-by-step solution of this system of equations, 602 00:33:51,890 --> 00:33:54,590 using the explicit integration scheme, the central difference 603 00:33:54,590 --> 00:33:58,150 method, or the Newmark implicit integration scheme, 604 00:33:58,150 --> 00:34:01,980 corresponds completely to the direct step-by-step solution 605 00:34:01,980 --> 00:34:07,630 of n equations, of this form, where the displacements here, 606 00:34:07,630 --> 00:34:11,940 U, are given in this way. 607 00:34:11,940 --> 00:34:15,980 x being now a function of time, so if I differentiate 608 00:34:15,980 --> 00:34:21,179 here, your double dot would be obtained by differentiating xi 609 00:34:21,179 --> 00:34:24,710 also twice, with respect to time. 610 00:34:24,710 --> 00:34:26,620 Once again, this is an important, 611 00:34:26,620 --> 00:34:28,350 very important, fact. 612 00:34:28,350 --> 00:34:32,969 I look at these equations, and I say that's a solution of 613 00:34:32,969 --> 00:34:37,280 these set of coupled differential equations. 614 00:34:37,280 --> 00:34:42,310 It's completely equivalent to the solution of n equations of 615 00:34:42,310 --> 00:34:48,478 this form with this transformation used. 616 00:34:48,478 --> 00:34:50,950 No dots here or dots there. 617 00:34:50,950 --> 00:34:53,820 In other words, this equation holds for displacements 618 00:34:53,820 --> 00:34:55,820 without the red dots and for the accelerations 619 00:34:55,820 --> 00:34:58,280 with the red dots. 620 00:34:58,280 --> 00:34:59,820 It is equivalent. 621 00:34:59,820 --> 00:35:03,270 This solution is equivalent to that solution. 622 00:35:03,270 --> 00:35:06,990 However, in direct integration, I never perform 623 00:35:06,990 --> 00:35:08,420 actually the transformation. 624 00:35:08,420 --> 00:35:11,820 All I know is that when I operate on this system of 625 00:35:11,820 --> 00:35:14,570 equations, using the Newmark method or the central 626 00:35:14,570 --> 00:35:18,180 difference method, I, in fact, operate on n 627 00:35:18,180 --> 00:35:20,130 such decoupled equations. 628 00:35:20,130 --> 00:35:21,590 That's important. 629 00:35:21,590 --> 00:35:25,270 And this means that for an analysis on our direct 630 00:35:25,270 --> 00:35:29,420 integration scheme, we can in fact analyze our direct 631 00:35:29,420 --> 00:35:33,310 integration scheme by looking at these equations, instead of 632 00:35:33,310 --> 00:35:34,700 these equations. 633 00:35:34,700 --> 00:35:37,190 You see if I were to analyze my direct integration scheme 634 00:35:37,190 --> 00:35:40,450 using these equations, I would have to deal with this huge 635 00:35:40,450 --> 00:35:45,720 matrix K, the huge matrix M. All I now need to deal with, 636 00:35:45,720 --> 00:35:50,150 in the analysis of the direct integration scheme, is with n 637 00:35:50,150 --> 00:35:53,900 such simple one degree of freedom equations. 638 00:35:53,900 --> 00:35:57,810 And in fact, I don't even need to deal with n such equations, 639 00:35:57,810 --> 00:36:02,020 for the analysis purpose, I only need to deal with one 640 00:36:02,020 --> 00:36:04,210 equation, where I make now omega a 641 00:36:04,210 --> 00:36:06,710 variable and r a variable. 642 00:36:06,710 --> 00:36:13,070 So my analysis on this equation, using this 643 00:36:13,070 --> 00:36:17,000 particular time integration scheme that I want to analyze, 644 00:36:17,000 --> 00:36:22,140 is completely giving me all the information that I need to 645 00:36:22,140 --> 00:36:25,780 know when I use my direct integration scheme in the 646 00:36:25,780 --> 00:36:28,200 solution of this equation. 647 00:36:28,200 --> 00:36:32,020 For analysis purposes of my direct integration scheme, I 648 00:36:32,020 --> 00:36:35,330 can simply analyze my direct integration scheme using this 649 00:36:35,330 --> 00:36:38,030 equation and get all the relevant 650 00:36:38,030 --> 00:36:40,840 information from that analysis. 651 00:36:40,840 --> 00:36:44,140 This is an extremely important fact. 652 00:36:44,140 --> 00:36:48,060 Once again, I do not actually do perform the transformation. 653 00:36:48,060 --> 00:36:50,190 We don't do the transformation. 654 00:36:50,190 --> 00:36:53,780 Only in theory for analysis purposes. 655 00:36:53,780 --> 00:36:57,610 I'm looking at the solution of this equation using the 656 00:36:57,610 --> 00:36:59,100 Newmark integration scheme. 657 00:36:59,100 --> 00:37:02,290 And I get all the relevant information on the accuracy, 658 00:37:02,290 --> 00:37:07,650 stability, et cetera of the integration scheme by looking 659 00:37:07,650 --> 00:37:10,520 at the solution of this equation with that direct 660 00:37:10,520 --> 00:37:12,030 integration scheme. 661 00:37:12,030 --> 00:37:15,030 Now, even in the analysis of this 662 00:37:15,030 --> 00:37:17,000 equation, of course, remember-- 663 00:37:17,000 --> 00:37:20,710 or in the use of this equation, remember that r is 664 00:37:20,710 --> 00:37:22,170 available now. 665 00:37:22,170 --> 00:37:25,660 And I have some choice what to put in there and so on. 666 00:37:25,660 --> 00:37:30,220 But the important point is that what I want to identify 667 00:37:30,220 --> 00:37:34,650 is the characteristics of the integration scheme. 668 00:37:34,650 --> 00:37:39,620 And the characteristics of the integration scheme are really 669 00:37:39,620 --> 00:37:44,150 quite well seen already if we look at a very simple case, 670 00:37:44,150 --> 00:37:45,640 namely this case. 671 00:37:45,640 --> 00:37:48,260 I set r deliberately equal to zero. 672 00:37:48,260 --> 00:37:52,090 And I look at the case where I have initial condition x, the 673 00:37:52,090 --> 00:37:54,950 displacements being one, the velocities being zero. 674 00:37:54,950 --> 00:37:57,420 this means, of course, that the accelerations are minus 675 00:37:57,420 --> 00:37:59,050 omega squared. 676 00:37:59,050 --> 00:38:04,680 If I use the Newmark method, the one that I just talked to 677 00:38:04,680 --> 00:38:10,870 you about, on that system of equations, then I can plot 678 00:38:10,870 --> 00:38:13,090 errors that reside in that integration. 679 00:38:13,090 --> 00:38:17,400 And what I've plotted here the percentage period elongation, 680 00:38:17,400 --> 00:38:23,810 pe divided by t times 100, which is this amount here that 681 00:38:23,810 --> 00:38:27,530 we see when we integrate the differential equation that I 682 00:38:27,530 --> 00:38:32,260 showed you via as the Newmark method with different delta-t 683 00:38:32,260 --> 00:38:35,300 over t values. 684 00:38:35,300 --> 00:38:41,820 On the bottom here, we are plotting delta-t divided by t, 685 00:38:41,820 --> 00:38:49,510 the period t, of course, being given for the particular 686 00:38:49,510 --> 00:38:51,250 oscillator that we're looking at. 687 00:38:51,250 --> 00:38:52,870 We are starting off with one. 688 00:38:52,870 --> 00:38:56,570 The exact solution, I should take better now another color, 689 00:38:56,570 --> 00:39:01,760 the exact solution would look something like that, 690 00:39:01,760 --> 00:39:03,410 finishing off here. 691 00:39:03,410 --> 00:39:04,710 That would be the exact solution. 692 00:39:04,710 --> 00:39:10,030 And what we're seeing now is an amplitude decay. 693 00:39:10,030 --> 00:39:10,865 It's already there. 694 00:39:10,865 --> 00:39:13,260 It would come in right there. 695 00:39:13,260 --> 00:39:15,090 That is the exact solution. 696 00:39:15,090 --> 00:39:16,140 We'll stop here. 697 00:39:16,140 --> 00:39:19,190 The period is of course this t here, as shown. 698 00:39:19,190 --> 00:39:23,280 This would be the exact solution for one period. 699 00:39:23,280 --> 00:39:27,130 What we are seeing as an amplitude decay, a drop in the 700 00:39:27,130 --> 00:39:30,650 numerical solution, and period elongation in 701 00:39:30,650 --> 00:39:32,150 the numerical solution. 702 00:39:32,150 --> 00:39:35,470 And the percentage period elongation is plotted here. 703 00:39:35,470 --> 00:39:39,470 We notice that the Newmark method corresponds to this 704 00:39:39,470 --> 00:39:40,480 curve here. 705 00:39:40,480 --> 00:39:45,650 So with delta-t over t being 0.10. 706 00:39:45,650 --> 00:39:47,020 We have about-- 707 00:39:47,020 --> 00:39:49,830 let us just look it up here-- 708 00:39:49,830 --> 00:39:54,550 we have about 3%, 3% period elongation 709 00:39:54,550 --> 00:39:56,050 for the Newmark method. 710 00:39:56,050 --> 00:39:58,110 I'm also showing other techniques here, which are 711 00:39:58,110 --> 00:40:01,320 also implicit techniques, the Wilson Theta method with Theta 712 00:40:01,320 --> 00:40:07,970 1.4 has about 5% period elongation for delta-t over t 713 00:40:07,970 --> 00:40:11,050 being 0.10. 714 00:40:11,050 --> 00:40:14,380 Another point of interest is of course also is the 715 00:40:14,380 --> 00:40:15,990 amplitude decay. 716 00:40:15,990 --> 00:40:20,520 And that amplitude decay is plotted on this graph here, 717 00:40:20,520 --> 00:40:23,190 percentage amplitude decay as a function of 718 00:40:23,190 --> 00:40:25,370 delta-t over t again. 719 00:40:25,370 --> 00:40:26,540 Delta-t over t. 720 00:40:26,540 --> 00:40:30,020 And notice that the Newmark method does not appear. 721 00:40:30,020 --> 00:40:34,080 In fact, the Newmark method has no amplitude decay. 722 00:40:34,080 --> 00:40:36,860 The solution is right down here. 723 00:40:36,860 --> 00:40:41,230 No amplitude decay for any delta-t over t, whereas the 724 00:40:41,230 --> 00:40:45,690 Wilson methods and the Houbolt method is also shown here, 725 00:40:45,690 --> 00:40:49,170 have the amplitude decay shown by these curves. 726 00:40:49,170 --> 00:40:52,660 Notice that the Wilson method has less amplitude decay then 727 00:40:52,660 --> 00:40:54,040 the Houbolt method. 728 00:40:54,040 --> 00:40:58,990 The period elongation for the Houbolt method was also shown 729 00:40:58,990 --> 00:41:01,370 on this view graph here. 730 00:41:01,370 --> 00:41:06,840 So the Newmark method does have period elongation, but it 731 00:41:06,840 --> 00:41:14,410 does not have any amplitude decay in this solution. 732 00:41:14,410 --> 00:41:17,150 I should also mention this, of course, as one important 733 00:41:17,150 --> 00:41:20,000 point, I forgot to mention that I'm looking here at the 734 00:41:20,000 --> 00:41:24,530 Newmark method with delta 1/2 alpha equal to 1/4. 735 00:41:24,530 --> 00:41:29,900 This is the constant average acceleration method. 736 00:41:29,900 --> 00:41:32,540 Well, let us look then at how can we 737 00:41:32,540 --> 00:41:35,460 actually perform the solution? 738 00:41:35,460 --> 00:41:40,020 How can we select an appropriate time step delta-t? 739 00:41:40,020 --> 00:41:47,770 Here we notice that the dynamic load factor that we 740 00:41:47,770 --> 00:41:53,890 are knowing that we can derive for the solution of this 741 00:41:53,890 --> 00:41:56,660 differential equation here-- 742 00:41:56,660 --> 00:41:58,510 here I included damping-- 743 00:41:58,510 --> 00:42:01,620 the dynamic load factor plotted along here as a 744 00:42:01,620 --> 00:42:03,590 functional of p over omega. 745 00:42:03,590 --> 00:42:07,760 p being the exciting free frequency, omega being the 746 00:42:07,760 --> 00:42:10,290 frequency of the system. 747 00:42:10,290 --> 00:42:13,280 The dynamic load factor looks as shown here. 748 00:42:13,280 --> 00:42:17,800 If we have no damping, you're talking about this curve. 749 00:42:17,800 --> 00:42:19,800 Now notice the important point. 750 00:42:19,800 --> 00:42:23,140 The dynamic load factor, of course, is 1 for static 751 00:42:23,140 --> 00:42:26,900 response, and it really gives us a magnification of the 752 00:42:26,900 --> 00:42:28,680 static response. 753 00:42:28,680 --> 00:42:32,290 In a static response, we would have this solution. 754 00:42:32,290 --> 00:42:40,670 And if the p over omega value is say 2, we would have that 755 00:42:40,670 --> 00:42:44,990 maximum solution in the dynamic response analysis. 756 00:42:44,990 --> 00:42:48,500 If p over omega is equal to 0.5. 757 00:42:48,500 --> 00:42:52,920 we would have, for zero damping, this maximum solution 758 00:42:52,920 --> 00:42:55,430 in the dynamic response history. 759 00:42:55,430 --> 00:42:58,130 Or rather this would be the maximum dynamic response that 760 00:42:58,130 --> 00:43:00,620 you would see in the dynamic solution. 761 00:43:00,620 --> 00:43:04,370 This is how the dynamic load factor is defined. 762 00:43:04,370 --> 00:43:08,140 Of course, at p over omega equal to 1, we have resonance. 763 00:43:08,140 --> 00:43:11,090 In practice, we have always some damping, and if you look 764 00:43:11,090 --> 00:43:15,430 at this amount of damping, this would be the maximum with 765 00:43:15,430 --> 00:43:20,470 p over omega equal to 1, we would have this as the maximum 766 00:43:20,470 --> 00:43:23,520 response that is ever measured in the 767 00:43:23,520 --> 00:43:26,900 dynamic response solution. 768 00:43:26,900 --> 00:43:33,610 Well, if we look at this graph, we recognize of course 769 00:43:33,610 --> 00:43:38,920 that as p over omega goes against 0, in other words as 770 00:43:38,920 --> 00:43:45,020 the exciting frequency, as over the frequency of the 771 00:43:45,020 --> 00:43:49,060 system goes against 0, we really talk about static 772 00:43:49,060 --> 00:43:49,980 conditions. 773 00:43:49,980 --> 00:43:54,580 As p over omega is very large, we are talking about no 774 00:43:54,580 --> 00:43:56,050 response whatsoever. 775 00:43:56,050 --> 00:44:00,390 And the important dynamic response really lies somewhat 776 00:44:00,390 --> 00:44:04,640 in this region here with that being the tail end of it. 777 00:44:04,640 --> 00:44:08,350 But the important dynamic response really lies in here, 778 00:44:08,350 --> 00:44:11,250 that we have to include in the solution. 779 00:44:11,250 --> 00:44:14,030 Out here we have basically a static response. 780 00:44:14,030 --> 00:44:18,150 Let us look at two simple cases. 781 00:44:18,150 --> 00:44:25,420 For p over omega 0.05, we can see that the static response 782 00:44:25,420 --> 00:44:28,640 here shown in a dashed line, the dynamic response showing 783 00:44:28,640 --> 00:44:29,890 as a solid line. 784 00:44:29,890 --> 00:44:31,100 They're basically the same. 785 00:44:31,100 --> 00:44:33,710 In other words, a dynamic response being equal to the 786 00:44:33,710 --> 00:44:37,360 static response, practically equal to the static response. 787 00:44:37,360 --> 00:44:40,960 We are down at this range here. 788 00:44:40,960 --> 00:44:44,815 As another example, if we have p over omega equal to 3, we 789 00:44:44,815 --> 00:44:47,570 are down over here. 790 00:44:47,570 --> 00:44:53,070 Then the response, the static response is shown here in the 791 00:44:53,070 --> 00:44:57,030 dashed line, and the dynamic response is shown 792 00:44:57,030 --> 00:44:58,380 as the solid line. 793 00:44:58,380 --> 00:45:01,110 Very little response here at all. 794 00:45:01,110 --> 00:45:05,280 Because as p over omega goes to infinity, there is no 795 00:45:05,280 --> 00:45:07,870 response for the system at all anymore. 796 00:45:07,870 --> 00:45:10,780 Basically, what we're saying then is that the frequency of 797 00:45:10,780 --> 00:45:13,230 the applied loading is so fast, that the 798 00:45:13,230 --> 00:45:14,950 structure can;t just move. 799 00:45:14,950 --> 00:45:16,330 It just cannot move at all. 800 00:45:16,330 --> 00:45:18,960 And we don't have any response in the structure. 801 00:45:18,960 --> 00:45:27,240 Well, the important point is that if we know the frequency 802 00:45:27,240 --> 00:45:33,720 content of the loading, then what we have to do in the use 803 00:45:33,720 --> 00:45:39,220 of the Newmark method is to calculate a delta-t small 804 00:45:39,220 --> 00:45:42,700 enough to pick up the loading, of course, because that is 805 00:45:42,700 --> 00:45:48,960 given in terms of this great steps or this great input 806 00:45:48,960 --> 00:45:52,240 points to the computer program, and the time step has 807 00:45:52,240 --> 00:45:56,360 to be small enough to integrate the dynamic response 808 00:45:56,360 --> 00:45:58,260 of importance accurately. 809 00:45:58,260 --> 00:46:02,090 The static response, the static response is contained 810 00:46:02,090 --> 00:46:09,120 in the solution anyway, because the K matrix is on the 811 00:46:09,120 --> 00:46:11,270 left hand side of the solution. 812 00:46:11,270 --> 00:46:15,390 So the complete solution process therefore can be 813 00:46:15,390 --> 00:46:17,060 summarized as follows. 814 00:46:17,060 --> 00:46:21,850 Identify the frequencies contained in the loading, and 815 00:46:21,850 --> 00:46:24,910 using a Fourier analysis, if necessary. 816 00:46:24,910 --> 00:46:28,220 Choose a finite element mesh that accurately represents all 817 00:46:28,220 --> 00:46:32,350 frequencies up to about four times the highest frequency 818 00:46:32,350 --> 00:46:35,210 omega u contained in the loading. 819 00:46:35,210 --> 00:46:37,670 The highest frequency contained in the loading is 820 00:46:37,670 --> 00:46:42,920 really the cutoff frequency that will give us a time step. 821 00:46:42,920 --> 00:46:46,630 And that then, the time step is in fact obtained via these 822 00:46:46,630 --> 00:46:47,340 consideration. 823 00:46:47,340 --> 00:46:48,690 Perform a direct integration analysis. 824 00:46:48,690 --> 00:46:52,640 The time step delta-t for this solutions should equal about 1 825 00:46:52,640 --> 00:46:56,930 over 20 Tu, where Tu is 2 pi over omega u, which is the 826 00:46:56,930 --> 00:46:59,320 highest frequency contained in the loading. 827 00:47:02,340 --> 00:47:05,260 What we are saying basically is that we want to select a 828 00:47:05,260 --> 00:47:09,160 finite element mesh to represent the highest 829 00:47:09,160 --> 00:47:13,600 frequency in the loading correctly, accurately, and 830 00:47:13,600 --> 00:47:17,650 then we want to select a time step in the data integration 831 00:47:17,650 --> 00:47:22,250 solution using the Newmark method that integrates the 832 00:47:22,250 --> 00:47:25,900 highest frequency contained in the loading also accurately, 833 00:47:25,900 --> 00:47:28,360 or the response in that highest frequency contained in 834 00:47:28,360 --> 00:47:29,610 the loading also accurately. 835 00:47:32,000 --> 00:47:36,760 And this then gives us basically a step-by-step 836 00:47:36,760 --> 00:47:41,090 procedure for the selection of the finite element mesh and 837 00:47:41,090 --> 00:47:42,070 the times step. 838 00:47:42,070 --> 00:47:46,310 Notice that the mesh selection and the time step selection 839 00:47:46,310 --> 00:47:51,360 both of course hinge on the structure and the particular 840 00:47:51,360 --> 00:47:53,290 frequency content of the loading. 841 00:47:53,290 --> 00:47:55,070 That's important. 842 00:47:55,070 --> 00:47:58,906 This is the modeling procedure that we would use for a 843 00:47:58,906 --> 00:48:01,340 structural vibration problem. 844 00:48:01,340 --> 00:48:04,890 We would use this time step here in the implicit 845 00:48:04,890 --> 00:48:08,850 integration, because we get all the accuracy we want. 846 00:48:08,850 --> 00:48:11,990 In an explicit integration, we would have to use a 847 00:48:11,990 --> 00:48:13,320 smaller time step. 848 00:48:13,320 --> 00:48:16,390 The sentence down here "or be smaller for stability reasons" 849 00:48:16,390 --> 00:48:19,290 really refers to the explicit time integration. 850 00:48:19,290 --> 00:48:22,780 In the Newmark method, we would not use that condition. 851 00:48:22,780 --> 00:48:28,010 We would simply use Tu, being 2 pi omega u and delta-t being 852 00:48:28,010 --> 00:48:30,280 1 over 20 times Tu. 853 00:48:30,280 --> 00:48:34,400 So in the Newmark method, I blank this out, and this is 854 00:48:34,400 --> 00:48:37,760 here the delta-t that we would be using. 855 00:48:37,760 --> 00:48:43,610 And just to summarize once the modeling of wave propagation 856 00:48:43,610 --> 00:48:46,520 problem here, here what we would do is we assume that the 857 00:48:46,520 --> 00:48:48,870 wavelength is Lw. 858 00:48:48,870 --> 00:48:53,180 The total time for the wave to travel past a point is then tw 859 00:48:53,180 --> 00:48:57,500 equal Lw over c, where c is the wave speed. 860 00:48:57,500 --> 00:49:00,500 Assuming that n time steps are necessary to represent the 861 00:49:00,500 --> 00:49:05,750 wave, we get our delta-t and the effective length, which I 862 00:49:05,750 --> 00:49:08,800 talked about earlier, of the finite element would be then 863 00:49:08,800 --> 00:49:09,990 given by this formula. 864 00:49:09,990 --> 00:49:14,260 So the procedure here is somewhat different in that if 865 00:49:14,260 --> 00:49:18,870 we have a wave that looks like this say, we now select a 866 00:49:18,870 --> 00:49:23,370 certain number of times steps, delta-t, to pick up the wave 867 00:49:23,370 --> 00:49:24,730 accurately. 868 00:49:24,730 --> 00:49:31,280 And this being the length Lw, the time for the wave to 869 00:49:31,280 --> 00:49:33,500 travel past the point is tw. 870 00:49:33,500 --> 00:49:38,400 And delta-t then is the number of time points, or time steps, 871 00:49:38,400 --> 00:49:40,880 that are used to represent the wave accurately. 872 00:49:43,870 --> 00:49:47,420 And once we have the time step, we select the effective 873 00:49:47,420 --> 00:49:48,530 length of an element. 874 00:49:48,530 --> 00:49:52,030 And then, of course it depends on whether we have a low-order 875 00:49:52,030 --> 00:49:56,120 element or a high-order element, that will give us 876 00:49:56,120 --> 00:50:00,990 then the procedure of how to apply the 877 00:50:00,990 --> 00:50:03,620 effective lengths criteria. 878 00:50:03,620 --> 00:50:06,440 I like to finally now summarize the complete 879 00:50:06,440 --> 00:50:07,690 solution process. 880 00:50:09,680 --> 00:50:15,560 This is a table that I put together for you to show how 881 00:50:15,560 --> 00:50:19,740 the complete solution process for explicit and implicit time 882 00:50:19,740 --> 00:50:22,110 integration can be very effectively implemented in a 883 00:50:22,110 --> 00:50:23,860 computer program. 884 00:50:23,860 --> 00:50:27,340 We form a linear stiffness matrix K, the mass matrix M, 885 00:50:27,340 --> 00:50:30,230 the damping matrix C, whichever applicable. 886 00:50:30,230 --> 00:50:31,870 We calculate the following constants. 887 00:50:31,870 --> 00:50:34,510 If we select the Newmark method, these are the 888 00:50:34,510 --> 00:50:37,040 constants that you would use. 889 00:50:37,040 --> 00:50:38,610 Let's not go into details. 890 00:50:38,610 --> 00:50:43,270 These constants are given in the textbook. 891 00:50:43,270 --> 00:50:46,270 They are given in the ADINA manual, if you are using the 892 00:50:46,270 --> 00:50:47,490 ADINA computer program. 893 00:50:47,490 --> 00:50:52,000 They are widely published. 894 00:50:52,000 --> 00:50:55,000 We simply cast them in this form. 895 00:50:55,000 --> 00:50:56,520 In the central difference method, we 896 00:50:56,520 --> 00:50:59,810 would use these constants. 897 00:50:59,810 --> 00:51:02,420 So there is some initialization involved. 898 00:51:02,420 --> 00:51:05,710 Then in the central difference method, we also want to 899 00:51:05,710 --> 00:51:13,150 calculate our first displacement, delta-t U, as 900 00:51:13,150 --> 00:51:15,330 shown here. 901 00:51:15,330 --> 00:51:18,910 And in the Newmark method, we don't need to do that, because 902 00:51:18,910 --> 00:51:22,340 we simply march ahead from time t to time t plus delta-t. 903 00:51:22,340 --> 00:51:25,820 You only need the condition at the conditions at time t to 904 00:51:25,820 --> 00:51:29,090 obtain the solution at times t plus delta-t. 905 00:51:29,090 --> 00:51:32,070 In the central difference method, remember we need two 906 00:51:32,070 --> 00:51:33,420 previous values. 907 00:51:33,420 --> 00:51:37,150 We then form an effective linear coefficient matrix. 908 00:51:37,150 --> 00:51:39,570 In implicit time integration, this is 909 00:51:39,570 --> 00:51:40,780 the coefficient matrix. 910 00:51:40,780 --> 00:51:42,590 The stiffness matrix being there, as I 911 00:51:42,590 --> 00:51:43,840 showed to you earlier. 912 00:51:43,840 --> 00:51:47,890 The mass matrix and damping matrix being there. 913 00:51:47,890 --> 00:51:56,340 If we look at this a0, we notice that a0 is 1 over 914 00:51:56,340 --> 00:51:59,700 delta-t squared and there's a constant alpha here. 915 00:51:59,700 --> 00:52:02,660 The Newmark constant. 916 00:52:02,660 --> 00:52:05,080 But it's 1 over delta-t squared. 917 00:52:05,080 --> 00:52:06,640 That is the important point. 918 00:52:06,640 --> 00:52:12,280 So as delta-t gets larger and larger, larger and larger, 919 00:52:12,280 --> 00:52:15,680 this effect goes against 0. 920 00:52:15,680 --> 00:52:19,030 The same holds for the a1 constant. 921 00:52:19,030 --> 00:52:21,060 As delta-t gets larger and larger, this effect 922 00:52:21,060 --> 00:52:22,150 goes against 0. 923 00:52:22,150 --> 00:52:25,560 And this is the reason why the Newmark method really goes 924 00:52:25,560 --> 00:52:31,550 down or collapses to the static analysis, if we have 925 00:52:31,550 --> 00:52:32,970 very large time steps. 926 00:52:32,970 --> 00:52:38,320 In explicit integration, we use this coefficient matrix. 927 00:52:38,320 --> 00:52:43,740 And now for each integration step, or for each time step, 928 00:52:43,740 --> 00:52:46,320 we perform the following calculations. 929 00:52:46,320 --> 00:52:51,780 We calculate the t plus delta-t R hat. 930 00:52:51,780 --> 00:52:55,780 And implicit time integration, the Newmark method, in 931 00:52:55,780 --> 00:52:59,050 explicit time integration, we calculate this vector. 932 00:52:59,050 --> 00:53:02,450 Notice at tF here, no k times tU. 933 00:53:02,450 --> 00:53:06,170 k times 2u is equal to tF. 934 00:53:06,170 --> 00:53:08,890 In implicit time integration, having calculated for each 935 00:53:08,890 --> 00:53:15,530 time step this load step, we now use the equations K hat 936 00:53:15,530 --> 00:53:20,320 times t plus delta-t u, equals t plus delta-t R hat to 937 00:53:20,320 --> 00:53:22,850 calculate t plus delta-t u. 938 00:53:22,850 --> 00:53:27,010 Of course, we would triangulize the K hat matrix 939 00:53:27,010 --> 00:53:29,310 prior to going into this loop. 940 00:53:29,310 --> 00:53:31,750 We do it once and forward. 941 00:53:31,750 --> 00:53:36,790 In explicit time integration, we use this matrix or this 942 00:53:36,790 --> 00:53:39,850 vector here, I should say, this vector now, and we can 943 00:53:39,850 --> 00:53:44,850 calculate directly for the t plus delta-t U also. 944 00:53:44,850 --> 00:53:48,300 The two equations that I used are summarized here. 945 00:53:48,300 --> 00:53:50,120 In implicit time integration, this is the 946 00:53:50,120 --> 00:53:53,410 equation that we use. 947 00:53:53,410 --> 00:53:55,750 Notice we only need to factorize this 948 00:53:55,750 --> 00:53:57,080 K hat matrix once. 949 00:53:57,080 --> 00:53:59,910 We do it outside the step-by-step integration, as I 950 00:53:59,910 --> 00:54:01,470 pointed out earlier. 951 00:54:01,470 --> 00:54:05,330 And then this gives us the increment in displacements. 952 00:54:05,330 --> 00:54:09,230 In explicit time integration, this is the equation we use. 953 00:54:09,230 --> 00:54:13,580 If M and C are diagonal, it's a trivial operation as I 954 00:54:13,580 --> 00:54:15,410 pointed out earlier. 955 00:54:15,410 --> 00:54:18,070 With this then, we have the new displacements. 956 00:54:18,070 --> 00:54:21,990 And on the last view graph, I show you the formulae that we 957 00:54:21,990 --> 00:54:25,490 are then using to calculate the accelerations in the 958 00:54:25,490 --> 00:54:29,050 Newmark method from the incremental displacements and 959 00:54:29,050 --> 00:54:30,880 variables that we knew already. 960 00:54:30,880 --> 00:54:33,780 Having now calculated the accelerations, we can get the 961 00:54:33,780 --> 00:54:36,370 velocities at time t plus delta-t. 962 00:54:36,370 --> 00:54:38,990 And of course, we have already our displacement at time t 963 00:54:38,990 --> 00:54:39,980 plus delta-t. 964 00:54:39,980 --> 00:54:42,520 In the central difference method, we use our basic 965 00:54:42,520 --> 00:54:46,010 equations that we already applied earlier in the 966 00:54:46,010 --> 00:54:48,780 development of the method to calculate the velocities and 967 00:54:48,780 --> 00:54:51,510 the accelerations at time t. 968 00:54:51,510 --> 00:54:56,750 This completes the procedure that is effectively used in a 969 00:54:56,750 --> 00:55:00,400 computer program for the integration of the dynamic 970 00:55:00,400 --> 00:55:01,680 equilibrium equations. 971 00:55:01,680 --> 00:55:06,460 I like to just now mention that I've placed heavy 972 00:55:06,460 --> 00:55:09,910 emphasis on the Newmark method in an implicit integration. 973 00:55:09,910 --> 00:55:11,900 Of course, there are other implicit integration formulae 974 00:55:11,900 --> 00:55:16,570 available, the Wilson method, the Houbolt method, and so on. 975 00:55:16,570 --> 00:55:18,760 The Newmark method has shown, has 976 00:55:18,760 --> 00:55:20,750 proven to be quite effective. 977 00:55:20,750 --> 00:55:24,030 And it's certainly a good example to demonstrate to you 978 00:55:24,030 --> 00:55:28,150 how the actual procedure is implemented and how an 979 00:55:28,150 --> 00:55:30,370 implicit integration scheme is being used. 980 00:55:30,370 --> 00:55:33,560 Other implicit integration methods would be used in much 981 00:55:33,560 --> 00:55:34,790 the same way. 982 00:55:34,790 --> 00:55:37,990 This is all I wanted to mention to you, discuss with 983 00:55:37,990 --> 00:55:39,180 you in this lecture. 984 00:55:39,180 --> 00:55:40,530 Thank you very much for your attention.