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 00:00:22,440 --> 00:00:24,880 PROFESSOR: Ladies and gentlemen, welcome to this set 9 00:00:24,880 --> 00:00:28,760 of lectures on the finite element method. 10 00:00:28,760 --> 00:00:32,640 In these lectures I would like to give you an introduction to 11 00:00:32,640 --> 00:00:37,190 the linear analysis of solids and structures. 12 00:00:37,190 --> 00:00:39,950 You are probably well aware that the finite element method 13 00:00:39,950 --> 00:00:42,980 is now widely used for analysis of structural 14 00:00:42,980 --> 00:00:45,090 engineering problems. 15 00:00:45,090 --> 00:00:48,700 The method is used in civil, aeronautical, mechanical, 16 00:00:48,700 --> 00:00:52,970 ocean, mining, nuclear, biomechanical, and other 17 00:00:52,970 --> 00:00:55,470 engineering disciplines. 18 00:00:55,470 --> 00:00:59,230 Since the first applications two decades ago of the finite 19 00:00:59,230 --> 00:01:03,370 element method we now see applications in linear, 20 00:01:03,370 --> 00:01:07,020 nonlinear, static, and dynamic analysis. 21 00:01:07,020 --> 00:01:09,580 However, in this set of lectures, I would like to 22 00:01:09,580 --> 00:01:13,220 discuss with you only the linear, static, and dynamic 23 00:01:13,220 --> 00:01:16,000 analysis of problems. 24 00:01:16,000 --> 00:01:18,700 The finite element method is used today in 25 00:01:18,700 --> 00:01:21,025 various computer programs. 26 00:01:21,025 --> 00:01:22,960 And its use is very significant. 27 00:01:26,700 --> 00:01:29,780 My objective in this set of lectures is to introduce to 28 00:01:29,780 --> 00:01:34,800 you the finite element methods or some of the finite element 29 00:01:34,800 --> 00:01:38,220 methods that are used for linear analysis of solids and 30 00:01:38,220 --> 00:01:39,340 structures. 31 00:01:39,340 --> 00:01:43,120 And here we understand linear to mean that we're talking 32 00:01:43,120 --> 00:01:48,300 about infinitesimally small displacements and that we are 33 00:01:48,300 --> 00:01:51,740 using a linear elastic material law. 34 00:01:51,740 --> 00:01:54,660 In other words, Hooke's law applies. 35 00:01:54,660 --> 00:01:58,070 We will consider, in this set of lectures, the formulation 36 00:01:58,070 --> 00:02:01,200 of the finite element equilibrium equations, the 37 00:02:01,200 --> 00:02:06,230 calculation of finite element matrices of the matrices that 38 00:02:06,230 --> 00:02:09,090 arise in the equilibrium equations. 39 00:02:09,090 --> 00:02:12,170 We will be talking about the methods for solution of the 40 00:02:12,170 --> 00:02:16,340 governing equations in static and dynamic analysis. 41 00:02:16,340 --> 00:02:21,280 And we will talk about actual computer implementations. 42 00:02:21,280 --> 00:02:26,000 I will emphasize modern and effective techniques and their 43 00:02:26,000 --> 00:02:27,250 practical usage. 44 00:02:29,760 --> 00:02:34,820 The emphasis, in this set of lectures, is given to physical 45 00:02:34,820 --> 00:02:38,460 explanations of the methods, techniques that we are using 46 00:02:38,460 --> 00:02:42,200 rather than mathematical derivations. 47 00:02:42,200 --> 00:02:46,720 The techniques that we will be discussing are those employed 48 00:02:46,720 --> 00:02:51,320 largely in the computer programs SAP and ADINA. 49 00:02:51,320 --> 00:02:56,000 SAP stands for Structural Analysis Program and you might 50 00:02:56,000 --> 00:03:00,990 very well be aware that there is a series of such programs, 51 00:03:00,990 --> 00:03:04,430 SAP I to SAP VI now. 52 00:03:04,430 --> 00:03:08,540 And ADINA stands for Automatic Dynamic 53 00:03:08,540 --> 00:03:11,200 Incremental Nonlinear Analysis. 54 00:03:11,200 --> 00:03:16,285 However, this program is also very effectively employed for 55 00:03:16,285 --> 00:03:17,700 linear analysis. 56 00:03:17,700 --> 00:03:21,560 The nonlinear analysis being then a next step in the usage 57 00:03:21,560 --> 00:03:24,250 of the program. 58 00:03:24,250 --> 00:03:27,290 In fact, the elements in ADINA, the numerical methods 59 00:03:27,290 --> 00:03:30,840 that are used in ADINA, I consider to be the most 60 00:03:30,840 --> 00:03:34,020 effective, the most modern state of the art techniques 61 00:03:34,020 --> 00:03:36,500 that are currently available. 62 00:03:36,500 --> 00:03:40,980 These few lectures really represent a very brief and 63 00:03:40,980 --> 00:03:43,480 compact introduction to the field of 64 00:03:43,480 --> 00:03:45,210 finite element analysis. 65 00:03:45,210 --> 00:03:48,330 We will go very rapidly through some or the basic 66 00:03:48,330 --> 00:03:52,450 concepts, practical applications, and so on. 67 00:03:52,450 --> 00:03:56,230 We shall follow quite closely, however, certain sections in 68 00:03:56,230 --> 00:03:59,160 my book entitled Finite Element Procedures in 69 00:03:59,160 --> 00:04:03,960 Engineering Analysis to be published by Prentice Hall. 70 00:04:03,960 --> 00:04:07,970 And I will be referring in the study guide of this set of 71 00:04:07,970 --> 00:04:12,190 lectures extensively to this book to the specific sections 72 00:04:12,190 --> 00:04:17,370 that we're considering in the lectures in this book. 73 00:04:17,370 --> 00:04:22,700 The finite element solution process can be described as 74 00:04:22,700 --> 00:04:24,620 given on this viewgraph. 75 00:04:24,620 --> 00:04:29,880 You can see here that we talk about a physical problem. 76 00:04:29,880 --> 00:04:33,220 We want to analyze an actual physical problem. 77 00:04:33,220 --> 00:04:36,440 And our first step, of course, is to establish a finite 78 00:04:36,440 --> 00:04:38,950 element model of that physical problem. 79 00:04:41,710 --> 00:04:45,410 Then, in the next step, we solve that model. 80 00:04:45,410 --> 00:04:48,550 And then we have to interpret the results. 81 00:04:48,550 --> 00:04:51,600 Because the interpretation of the results depends very much 82 00:04:51,600 --> 00:04:54,830 on how we established the finite element model, what 83 00:04:54,830 --> 00:04:56,900 kind of model we used, and so on. 84 00:04:56,900 --> 00:04:59,650 And in establishing the finite element model, we have to be 85 00:04:59,650 --> 00:05:03,640 aware of what kinds of elements, techniques, and so 86 00:05:03,640 --> 00:05:05,610 on are available to us. 87 00:05:05,610 --> 00:05:11,250 Well, therefore, I will be talking, in the set of 88 00:05:11,250 --> 00:05:15,990 lectures, about these three steps basically here for 89 00:05:15,990 --> 00:05:19,800 different kinds of physical problems. 90 00:05:19,800 --> 00:05:23,560 Once we have interpreted the results we might go back from 91 00:05:23,560 --> 00:05:28,350 down here to there to revise or refine our model and go 92 00:05:28,350 --> 00:05:32,510 through this process again until we feel that our model 93 00:05:32,510 --> 00:05:35,520 has been an adequate one for the solution of the physical 94 00:05:35,520 --> 00:05:37,610 problem of interest. 95 00:05:37,610 --> 00:05:41,690 Let me give you or show you some models that have been 96 00:05:41,690 --> 00:05:45,100 used in actual structural analysis. 97 00:05:45,100 --> 00:05:47,890 You might have seen similar models in textbooks, in 98 00:05:47,890 --> 00:05:49,580 publications already. 99 00:05:49,580 --> 00:05:53,190 This, for example, is a model that was used for the analysis 100 00:05:53,190 --> 00:05:54,650 of a cooling tower. 101 00:05:54,650 --> 00:05:57,930 The basic process of the finite element method is that 102 00:05:57,930 --> 00:06:01,570 we are taking the continuous system, and we are idealizing 103 00:06:01,570 --> 00:06:03,660 it as an assemblage of elements. 104 00:06:03,660 --> 00:06:08,110 I'm drawing here a typical three-noded triangular shell 105 00:06:08,110 --> 00:06:10,710 element that was used in the analysis of 106 00:06:10,710 --> 00:06:13,010 this cooling tower. 107 00:06:13,010 --> 00:06:16,680 We talk about very many elements in order to obtain an 108 00:06:16,680 --> 00:06:18,630 accurate response prediction. 109 00:06:18,630 --> 00:06:21,240 And, of course, that means that we will be dealing with a 110 00:06:21,240 --> 00:06:24,100 large set of equations to be solved. 111 00:06:24,100 --> 00:06:26,980 And there's a significant computer effort required. 112 00:06:26,980 --> 00:06:28,630 I will be addressing all of these 113 00:06:28,630 --> 00:06:32,360 questions in these lectures. 114 00:06:32,360 --> 00:06:35,400 Here you see the finite element model of a dam. 115 00:06:35,400 --> 00:06:39,380 The earth below the dam was idealized as an assemblage all 116 00:06:39,380 --> 00:06:42,240 such elements here, triangular elements now. 117 00:06:42,240 --> 00:06:45,280 And the dam itself was also idealized as an 118 00:06:45,280 --> 00:06:47,400 assemblage of elements. 119 00:06:47,400 --> 00:06:51,530 We will be talking about how such assemblages are best 120 00:06:51,530 --> 00:06:54,320 created, what kinds of elements to select, what 121 00:06:54,320 --> 00:06:56,910 assumptions are in the selection of these elements, 122 00:06:56,910 --> 00:07:00,120 and then how do we solve the resulting finite element 123 00:07:00,120 --> 00:07:01,820 equilibrium equations. 124 00:07:01,820 --> 00:07:05,390 Here you see the finite element analysis, or the mesh 125 00:07:05,390 --> 00:07:08,770 that was used in the finite element analysis of a tire. 126 00:07:08,770 --> 00:07:12,170 This wall is half of the tire, as you can see, and this was 127 00:07:12,170 --> 00:07:13,760 the finite element mesh used. 128 00:07:13,760 --> 00:07:17,140 Again, we have to judiciously choose the kinds of finite 129 00:07:17,140 --> 00:07:18,790 elements to be employed. 130 00:07:18,790 --> 00:07:22,080 And we will be talking about that in this set of lectures. 131 00:07:22,080 --> 00:07:26,750 Here you see the finite element model employed in the 132 00:07:26,750 --> 00:07:29,240 analysis of a spherical cover of a laser 133 00:07:29,240 --> 00:07:32,010 vacuum target chamber. 134 00:07:32,010 --> 00:07:33,390 This is the finite element mesh used. 135 00:07:33,390 --> 00:07:35,940 Again, specific elements were employed here. 136 00:07:35,940 --> 00:07:39,050 And we will be talking about the characteristics of these 137 00:07:39,050 --> 00:07:41,030 elements in this set of lectures. 138 00:07:41,030 --> 00:07:47,270 Here you see the model of the shell structure subjected to a 139 00:07:47,270 --> 00:07:48,120 pinching load. 140 00:07:48,120 --> 00:07:50,790 There's a load up here and a load down there. 141 00:07:50,790 --> 00:07:53,670 These are the triangular elements that were used in the 142 00:07:53,670 --> 00:07:58,670 idealization of that shell and the resulting bending moments 143 00:07:58,670 --> 00:08:02,110 and displacements along the line DC are plotted here that 144 00:08:02,110 --> 00:08:06,570 have been predicted by the finite element analysis. 145 00:08:06,570 --> 00:08:09,920 Finally here you see the finite element idealization of 146 00:08:09,920 --> 00:08:14,140 a wind tunnel that was used for the dynamic analysis of 147 00:08:14,140 --> 00:08:15,270 this tunnel. 148 00:08:15,270 --> 00:08:18,740 You can see a large number of shell elements were employed 149 00:08:18,740 --> 00:08:21,860 in the idealization of the shelf of the tunnel. 150 00:08:21,860 --> 00:08:25,640 Then, of course, supports were provided here for that shell. 151 00:08:25,640 --> 00:08:29,680 And this was a very large system that was solved. 152 00:08:29,680 --> 00:08:33,010 And the eigenvalues of this system were calculated using 153 00:08:33,010 --> 00:08:35,600 the subspace iteration method that we would be also talking 154 00:08:35,600 --> 00:08:37,900 about in this set of lectures. 155 00:08:37,900 --> 00:08:41,190 Well, with this short introduction then, I would 156 00:08:41,190 --> 00:08:45,850 like to go now and discuss with you some basic concepts 157 00:08:45,850 --> 00:08:47,460 of engineering analysis. 158 00:08:47,460 --> 00:08:50,340 There's a lot of work ahead in this set of lectures. 159 00:08:50,340 --> 00:08:54,310 So let me take off my jacket with your permission, and let 160 00:08:54,310 --> 00:08:58,270 us just go right on with the actual discussion of the 161 00:08:58,270 --> 00:09:01,650 theory of the finite element method. 162 00:09:01,650 --> 00:09:07,390 The basic concepts that I address here, in this first 163 00:09:07,390 --> 00:09:13,210 lecture, is summarized basically here once more. 164 00:09:13,210 --> 00:09:16,430 We are talking about the idealization of a system. 165 00:09:16,430 --> 00:09:18,800 We are talking about the formulation of the equilibrium 166 00:09:18,800 --> 00:09:23,110 equations, then the solution of the equations, and then, as 167 00:09:23,110 --> 00:09:24,290 I mentioned earlier already, the 168 00:09:24,290 --> 00:09:26,110 interpretation of the results. 169 00:09:26,110 --> 00:09:29,180 These are really the four steps that have to be 170 00:09:29,180 --> 00:09:35,090 performed in the analysis of an engineering system or of a 171 00:09:35,090 --> 00:09:38,550 physical system that we want to analyze. 172 00:09:38,550 --> 00:09:41,280 Now when we talk about systems, we are really talking 173 00:09:41,280 --> 00:09:45,380 about discrete and continuous systems where, however, in 174 00:09:45,380 --> 00:09:51,160 reality, we recognize that all systems are really continuous. 175 00:09:51,160 --> 00:09:57,110 However, if the system consists of a set of springs, 176 00:09:57,110 --> 00:10:01,680 dashpots, beam elements, then we might refer to this 177 00:10:01,680 --> 00:10:04,650 continuous system as a discrete system because we can 178 00:10:04,650 --> 00:10:11,210 see already, it is obvious, so to say, how to idealize a 179 00:10:11,210 --> 00:10:17,050 system into a set of elements, discrete elements. 180 00:10:17,050 --> 00:10:19,900 In that case, the response is described by variables at a 181 00:10:19,900 --> 00:10:22,090 finite number of points. 182 00:10:22,090 --> 00:10:25,730 And this means that we have to set up a set of algebraic 183 00:10:25,730 --> 00:10:29,690 questions to solve that system. 184 00:10:29,690 --> 00:10:32,840 So here I'm talking about elementary systems ofl 185 00:10:32,840 --> 00:10:37,720 springs, dashpots, discrete beam elements, and so on. 186 00:10:37,720 --> 00:10:41,950 In the analysis of a continuous system the response 187 00:10:41,950 --> 00:10:44,360 is described really by variables at an infinite 188 00:10:44,360 --> 00:10:46,220 number of points. 189 00:10:46,220 --> 00:10:49,370 And, in this case, we really come up with a differential 190 00:10:49,370 --> 00:10:52,100 equation, obviously a set of differential equations, that 191 00:10:52,100 --> 00:10:54,170 we have to solve. 192 00:10:54,170 --> 00:11:00,030 The analysis of a complex continuous system requires a 193 00:11:00,030 --> 00:11:03,630 dissolution of the differential equations using 194 00:11:03,630 --> 00:11:05,990 numerical procedures. 195 00:11:05,990 --> 00:11:08,880 And this solution via numerical procedures-- 196 00:11:08,880 --> 00:11:10,940 and, of course, in this set of lectures we will be talking 197 00:11:10,940 --> 00:11:14,240 about the finite element method numerical procedures-- 198 00:11:14,240 --> 00:11:19,130 really reduces a continuous system to a discrete form. 199 00:11:19,130 --> 00:11:23,490 The powerful mechanism that we talk about here is the finite 200 00:11:23,490 --> 00:11:26,970 element method implemented on a digital computer. 201 00:11:26,970 --> 00:11:29,620 The problem types that I will be talking about are 202 00:11:29,620 --> 00:11:33,940 steady-state problems, or static analysis, propagation 203 00:11:33,940 --> 00:11:37,810 problems, dynamic analysis, and eigenvalue problems. 204 00:11:37,810 --> 00:11:41,660 And these three types of problems, of course, arise for 205 00:11:41,660 --> 00:11:45,240 discrete and continuous systems. 206 00:11:45,240 --> 00:11:48,520 Now let us talk first about the analysis of discrete 207 00:11:48,520 --> 00:11:50,430 systems in this first lecture. 208 00:11:50,430 --> 00:11:53,120 Because many of the characteristics that we are 209 00:11:53,120 --> 00:11:56,080 using in the analysis of discrete systems, discrete 210 00:11:56,080 --> 00:12:01,580 meaning, springs, dashpots, et cetera, we can directly see 211 00:12:01,580 --> 00:12:04,090 the discrete elements of the system. 212 00:12:04,090 --> 00:12:07,270 The steps involved in the analysis of such discrete 213 00:12:07,270 --> 00:12:13,150 systems are very similar to the analysis of complex finite 214 00:12:13,150 --> 00:12:14,990 element systems. 215 00:12:14,990 --> 00:12:16,550 The steps involved are the system 216 00:12:16,550 --> 00:12:17,860 idealization into elements. 217 00:12:17,860 --> 00:12:21,300 And that idealization is somewhat obvious because we 218 00:12:21,300 --> 00:12:24,120 have the discrete elements already. 219 00:12:24,120 --> 00:12:27,780 The evaluation of the element equilibrium requirements, the 220 00:12:27,780 --> 00:12:30,920 element assemblage, and the solution of the response. 221 00:12:30,920 --> 00:12:33,080 Notice when we later on talk about the analysis of 222 00:12:33,080 --> 00:12:37,810 continuous systems instead of discrete systems, then the 223 00:12:37,810 --> 00:12:42,490 system idealization into finite elements here is not an 224 00:12:42,490 --> 00:12:45,690 obvious step and needs much attention. 225 00:12:45,690 --> 00:12:50,430 But these three steps here are the same in the finite element 226 00:12:50,430 --> 00:12:52,460 analysis of a continuous system. 227 00:12:52,460 --> 00:12:56,960 And I would like to now discuss all of these steps 228 00:12:56,960 --> 00:13:00,460 here just to show you, basically, some of the basic 229 00:13:00,460 --> 00:13:03,660 concepts that we're using in the finite element analysis. 230 00:13:03,660 --> 00:13:08,730 Let us look at this discrete system here as an example. 231 00:13:08,730 --> 00:13:13,500 And let us display the basic ideas in the analysis of this 232 00:13:13,500 --> 00:13:14,620 discrete system. 233 00:13:14,620 --> 00:13:18,830 Here we have a set of rigid carts, three rigid carts, 234 00:13:18,830 --> 00:13:21,930 vertical carts that are supported on 235 00:13:21,930 --> 00:13:23,850 rollers down here. 236 00:13:23,850 --> 00:13:27,340 This means that each of these carts can just roll 237 00:13:27,340 --> 00:13:29,580 horizontally. 238 00:13:29,580 --> 00:13:37,120 The carts are connected via springs, k2, k3, k4, k5. 239 00:13:37,120 --> 00:13:42,640 And the first cart here is connected via k1 to a rigid 240 00:13:42,640 --> 00:13:45,440 support that does not move. 241 00:13:45,440 --> 00:13:48,430 The displacement all of this cart here is u1. 242 00:13:48,430 --> 00:13:50,970 The load applied is R1. 243 00:13:50,970 --> 00:13:55,110 Notice that u1 is the displacement of each of these 244 00:13:55,110 --> 00:13:59,060 springs since this cart is rigid. 245 00:13:59,060 --> 00:14:02,200 The displacement of this cart here is u2. 246 00:14:02,200 --> 00:14:05,450 And R2 is the load applied. 247 00:14:05,450 --> 00:14:07,760 The displacement of this cart is u3. 248 00:14:07,760 --> 00:14:10,390 And R3 is the load applied. 249 00:14:10,390 --> 00:14:17,080 We now want to analyze this system when R1, R2, looking at 250 00:14:17,080 --> 00:14:22,270 it, we can directly see the elements of the system k1 to 251 00:14:22,270 --> 00:14:28,000 k5, and we can see directly, of course, how these elements 252 00:14:28,000 --> 00:14:29,760 are interconnected. 253 00:14:29,760 --> 00:14:32,930 The steps that we will be talking about in the analysis 254 00:14:32,930 --> 00:14:37,570 of this discrete system are really very similar to the 255 00:14:37,570 --> 00:14:39,930 steps that we're using in the finite element analysis of 256 00:14:39,930 --> 00:14:41,330 continuous systems. 257 00:14:41,330 --> 00:14:44,340 What we will be doing is that we look at the equilibrium 258 00:14:44,340 --> 00:14:51,390 requirements for each spring as a first step. 259 00:14:51,390 --> 00:14:54,660 Then we look at the interconnection requirements 260 00:14:54,660 --> 00:15:01,440 between these springs that, in other words, the force on 261 00:15:01,440 --> 00:15:05,900 these springs here at this cart, and that spring, must be 262 00:15:05,900 --> 00:15:07,830 balanced by R1. 263 00:15:07,830 --> 00:15:11,480 And then, of course, we have a compatibility requirement that 264 00:15:11,480 --> 00:15:15,940 u1 is a displacement of each of these springs here. 265 00:15:15,940 --> 00:15:19,410 So we are talking about the constitutive relations, the 266 00:15:19,410 --> 00:15:21,880 equilibrium requirements, and the compatibility 267 00:15:21,880 --> 00:15:22,730 requirements. 268 00:15:22,730 --> 00:15:24,860 These are, of course, the three requirements that we 269 00:15:24,860 --> 00:15:28,110 also have to satisfy in the analysis of a continuous 270 00:15:28,110 --> 00:15:31,200 system using, later on, finite element methods. 271 00:15:31,200 --> 00:15:35,070 Notice that these springs here are our finite elements, if 272 00:15:35,070 --> 00:15:37,890 you want to think of it that way, a very 273 00:15:37,890 --> 00:15:40,400 simple set of elements. 274 00:15:40,400 --> 00:15:43,430 In a more complex analysis, these springs here would be 275 00:15:43,430 --> 00:15:46,060 plane stress elements, plane strain elements, three 276 00:15:46,060 --> 00:15:47,740 dimension elements, shell elements. 277 00:15:47,740 --> 00:15:50,320 And we will be talking about how we derive the 278 00:15:50,320 --> 00:15:52,880 characteristics of these elements. 279 00:15:52,880 --> 00:15:57,090 And we will, however, interconnect these elements, 280 00:15:57,090 --> 00:15:59,860 these more complex elements, later in exactly the same way 281 00:15:59,860 --> 00:16:02,850 as we connect these simple elements. 282 00:16:02,850 --> 00:16:06,370 So the connections between the elements are established in 283 00:16:06,370 --> 00:16:09,760 the same way, and the solution of the equilibrium equations 284 00:16:09,760 --> 00:16:12,830 is also performed in the same way. 285 00:16:12,830 --> 00:16:17,060 But in this simple analysis, we are given directly the 286 00:16:17,060 --> 00:16:18,800 spring stiffnesses. 287 00:16:18,800 --> 00:16:23,470 And one other important point is that the spring stiffnesses 288 00:16:23,470 --> 00:16:25,610 here are exact stiffnesses. 289 00:16:25,610 --> 00:16:28,530 In a finite element analysis of a continuous system, we 290 00:16:28,530 --> 00:16:31,585 have a choice on what kind of interpolations we can use for 291 00:16:31,585 --> 00:16:32,980 an element. 292 00:16:32,980 --> 00:16:35,820 We have a choice on what assumptions we want to lay 293 00:16:35,820 --> 00:16:37,010 down for an element. 294 00:16:37,010 --> 00:16:39,940 And then using different assumptions we are coming up 295 00:16:39,940 --> 00:16:43,880 with different stiffnesses of the element domain that we 296 00:16:43,880 --> 00:16:45,270 will be talking about. 297 00:16:45,270 --> 00:16:48,750 And we will also find that the equilibrium in that element 298 00:16:48,750 --> 00:16:50,980 domain is not satisfied. 299 00:16:50,980 --> 00:16:53,800 It will only be satisfied in the limit as the elements 300 00:16:53,800 --> 00:16:55,710 become smaller, and smaller, and smaller. 301 00:16:55,710 --> 00:16:59,230 Whereas in the analysis of this discrete system, the 302 00:16:59,230 --> 00:17:03,880 equilibrium in each spring is always satisfied. 303 00:17:03,880 --> 00:17:08,020 So this is a very simple finite element analysis if you 304 00:17:08,020 --> 00:17:10,490 want to think of it that way. 305 00:17:10,490 --> 00:17:14,150 The elements here than are k1. 306 00:17:14,150 --> 00:17:17,290 And notice that the equilibrium requirement for 307 00:17:17,290 --> 00:17:21,390 this element says simply that k1u1 is equal to the force 308 00:17:21,390 --> 00:17:22,849 applied to this node. 309 00:17:22,849 --> 00:17:25,839 It's a force, the external force, applied to this node. 310 00:17:25,839 --> 00:17:29,360 The equilibrium requirement of this element, k2, is written 311 00:17:29,360 --> 00:17:32,730 down here in matrix form. 312 00:17:32,730 --> 00:17:36,060 k2 is the physical stiffness of the spring. 313 00:17:36,060 --> 00:17:41,740 And F1, F2 are the forces applied at these two ends. 314 00:17:41,740 --> 00:17:46,210 Notice, please, that the superscript here refers to the 315 00:17:46,210 --> 00:17:50,490 element number, superscript 1 here for element 1, 316 00:17:50,490 --> 00:17:54,190 superscript 2 here for element 2. 317 00:17:54,190 --> 00:18:02,820 And notice that we would find that F1(2) is minus F2(2) 318 00:18:02,820 --> 00:18:04,490 given u1 and u2. 319 00:18:04,490 --> 00:18:08,590 Of course, that means the element is in equilibrium. 320 00:18:08,590 --> 00:18:13,450 Notice also, if you look at this matrix closer, that if u1 321 00:18:13,450 --> 00:18:19,560 is greater than u2, then we would find that, in other 322 00:18:19,560 --> 00:18:23,260 words, u1 greater than u2 means that the spring is in 323 00:18:23,260 --> 00:18:24,260 compression. 324 00:18:24,260 --> 00:18:28,680 We would find that F1(2) is positive by simply multiplying 325 00:18:28,680 --> 00:18:30,060 this out here. 326 00:18:30,060 --> 00:18:33,720 And F2(2) is negative, which corresponds to the physical 327 00:18:33,720 --> 00:18:35,820 situation that we actually have. 328 00:18:35,820 --> 00:18:40,710 If u1 is greater than u2, this force here is positive, and 329 00:18:40,710 --> 00:18:46,230 that force is negative because the spring is compressed. 330 00:18:46,230 --> 00:18:48,230 Well, similarly, we can write down the equilibrium 331 00:18:48,230 --> 00:18:51,450 requirement for the spring 3. 332 00:18:51,450 --> 00:18:53,020 And I've written down the matrix here. 333 00:18:53,020 --> 00:18:55,250 The only difference to the equilibrium requirements for 334 00:18:55,250 --> 00:18:58,720 spring 2 are that we're using now k3 here. 335 00:18:58,720 --> 00:19:02,980 And, of course, the superscript now is 3. 336 00:19:02,980 --> 00:19:06,420 We can then proceed to write down the equilibrium equation 337 00:19:06,420 --> 00:19:11,700 for spring 4, which is the same form as before, now k4 338 00:19:11,700 --> 00:19:16,010 here and the 4 superscript denoting element 4. 339 00:19:16,010 --> 00:19:21,080 And, finally for k5, we have k5 here and superscripts 5 340 00:19:21,080 --> 00:19:24,250 here to denote element 5. 341 00:19:24,250 --> 00:19:30,280 Now we should also point out one other important point. 342 00:19:30,280 --> 00:19:36,980 Namely, if we look at this cart systems here, notice that 343 00:19:36,980 --> 00:19:42,370 this k1 spring is connected to u1. 344 00:19:42,370 --> 00:19:44,440 It's connected to u1. 345 00:19:44,440 --> 00:19:48,450 k4 is connected to u1 and u3. 346 00:19:48,450 --> 00:19:54,460 So if we look at the equilibrium requirements here, 347 00:19:54,460 --> 00:19:59,960 you will notice that I have F1 here for k1 348 00:19:59,960 --> 00:20:02,850 because this is u1 here. 349 00:20:02,850 --> 00:20:04,990 That is the global displacement u1. 350 00:20:04,990 --> 00:20:09,020 And looking now at k4, a more complicated case which is 351 00:20:09,020 --> 00:20:21,080 connected to u1 and u3, I have for that spring the u1 and u3 352 00:20:21,080 --> 00:20:22,080 denoted here. 353 00:20:22,080 --> 00:20:26,210 And we have F1 and F3 here, F1 and F3. 354 00:20:26,210 --> 00:20:30,620 So these are the forces that go directly into the degrees 355 00:20:30,620 --> 00:20:34,840 of freedom 1 and 3 respectively, and similarly 356 00:20:34,840 --> 00:20:36,460 for the other springs. 357 00:20:36,460 --> 00:20:40,810 Now if we want to assemble the global equilibrium equations 358 00:20:40,810 --> 00:20:47,460 for this structure with the unknowns u1, u2, u3, the loads 359 00:20:47,460 --> 00:20:52,660 R1, R2, and R3 are known, then we have to use now the 360 00:20:52,660 --> 00:20:57,730 equilibrium requirement at these degrees of freedom u1, 361 00:20:57,730 --> 00:21:02,740 u2, and u3, or rather at the cart 1, 2, and 3. 362 00:21:02,740 --> 00:21:05,830 And that equilibrium requirement then means that 363 00:21:05,830 --> 00:21:11,010 the sum of the forces acting onto the individual springs 1, 364 00:21:11,010 --> 00:21:17,720 2, 3, and 4 at degree of freedom 1 must be equal to R1. 365 00:21:17,720 --> 00:21:20,030 Now let us look once at this first 366 00:21:20,030 --> 00:21:22,430 equation back here again. 367 00:21:22,430 --> 00:21:29,110 Notice u1 couples into this spring 1, spring 2, spring 3, 368 00:21:29,110 --> 00:21:30,670 and spring 4. 369 00:21:30,670 --> 00:21:34,320 And that coupling is seen right here in spring 370 00:21:34,320 --> 00:21:36,830 1, 2, 3, and 4. 371 00:21:36,830 --> 00:21:40,490 And summing all these forces that are acting individually 372 00:21:40,490 --> 00:21:43,860 onto the springs, the sum of these forces must be equal to 373 00:21:43,860 --> 00:21:45,250 the external load. 374 00:21:45,250 --> 00:21:47,420 That is the interconnection 375 00:21:47,420 --> 00:21:50,730 requirement between the springs. 376 00:21:50,730 --> 00:21:54,010 The equilibrium requirements within the springs are 377 00:21:54,010 --> 00:21:59,300 expressed by these individual matrices here that we looked 378 00:21:59,300 --> 00:21:59,820 at already. 379 00:21:59,820 --> 00:22:01,610 These are the equilibrium requirements for the 380 00:22:01,610 --> 00:22:02,820 individual springs. 381 00:22:02,820 --> 00:22:06,270 Now I'm talking about the equilibrium requirement at the 382 00:22:06,270 --> 00:22:07,730 carts, the interconnection 383 00:22:07,730 --> 00:22:10,310 requirements between the springs. 384 00:22:10,310 --> 00:22:14,450 Similarly, we can sum the forces that have to be equal 385 00:22:14,450 --> 00:22:18,890 to R2 and sum the forces that have to be equal to R3. 386 00:22:18,890 --> 00:22:22,960 And these three equations then set up in matrix form by 387 00:22:22,960 --> 00:22:28,140 substituting for F1(1), F1(2), and so on from the equilibrium 388 00:22:28,140 --> 00:22:32,030 requirements of the springs, we directly obtain this set 389 00:22:32,030 --> 00:22:35,820 off equations, KU equals R. Notice that K 390 00:22:35,820 --> 00:22:38,630 now is a 3 by 3 matrix. 391 00:22:38,630 --> 00:22:41,540 U is a 3 by 1 vector. 392 00:22:41,540 --> 00:22:43,490 R is a 3 by 1 vector. 393 00:22:43,490 --> 00:22:47,950 I denote matrices and vectors by bars under the symbols. 394 00:22:47,950 --> 00:22:51,782 As you can see here there are bars under these symbols. 395 00:22:51,782 --> 00:22:54,890 Well, if we look at these equilibrium equations, we 396 00:22:54,890 --> 00:23:04,480 notice that our U vector, this vector U here contains u1, u2, 397 00:23:04,480 --> 00:23:06,400 and u3 as the unknowns. 398 00:23:06,400 --> 00:23:11,610 Notice this T here, this superscript T means transpose. 399 00:23:11,610 --> 00:23:17,670 The actual vector U actually looks this way u1, u2, u3. 400 00:23:17,670 --> 00:23:20,240 It lists the displacements vertically downwards. 401 00:23:20,240 --> 00:23:23,110 But it is easier to write it this way by 402 00:23:23,110 --> 00:23:24,350 transposing as a vector. 403 00:23:24,350 --> 00:23:27,340 So UT, capital T there, means transpose. 404 00:23:27,340 --> 00:23:31,790 Similarly for R we have R1, R2, and R3 as the components. 405 00:23:31,790 --> 00:23:37,180 And the K matrix that we have obtained by substituting into 406 00:23:37,180 --> 00:23:39,680 these equations from the element equilibrium 407 00:23:39,680 --> 00:23:44,160 requirements, the K matrix is this one here. 408 00:23:44,160 --> 00:23:48,400 Now let us look a little closer at how do we construct 409 00:23:48,400 --> 00:23:49,520 this K matrix. 410 00:23:49,520 --> 00:23:55,760 Well, we note that the total K matrix can be constructed by 411 00:23:55,760 --> 00:24:01,820 summing all of the individual element matrices from 1 to 5. 412 00:24:01,820 --> 00:24:05,820 And these individual element matrices are, for two 413 00:24:05,820 --> 00:24:07,150 extremes, written down here. 414 00:24:07,150 --> 00:24:09,790 K1 is a 3 by 3 matrix now. 415 00:24:09,790 --> 00:24:12,200 Not anymore the 1 by 1 or 2 by 2. 416 00:24:12,200 --> 00:24:17,500 It's a 3 by 3 matrix with just k1 in the 1,1 position. 417 00:24:17,500 --> 00:24:19,690 All the other elements are 0. 418 00:24:19,690 --> 00:24:22,470 K2 is this matrix. 419 00:24:22,470 --> 00:24:26,300 So what I have done then is I have taken the 2 by 2 matrix 420 00:24:26,300 --> 00:24:29,180 which appeared in the element equilibrium requirement and 421 00:24:29,180 --> 00:24:32,640 has blown this matrix up filling zeroes for the third 422 00:24:32,640 --> 00:24:34,240 degree of freedom. 423 00:24:34,240 --> 00:24:37,120 Similarly we would obtain K3 and so on. 424 00:24:37,120 --> 00:24:41,530 The zeroes always appear in those rows and columns into 425 00:24:41,530 --> 00:24:45,320 which the element does not couple, in other words, into 426 00:24:45,320 --> 00:24:47,500 those degrees of freedom that the element does not couple. 427 00:24:47,500 --> 00:24:53,130 For example, k1, this element 1 here, couples only into the 428 00:24:53,130 --> 00:24:54,480 degree of freedom 1. 429 00:24:54,480 --> 00:24:58,380 So, therefore, we have the second and third rows be 0. 430 00:24:58,380 --> 00:25:03,800 Element 2 couples only into degree of freedom 1 and 2. 431 00:25:03,800 --> 00:25:07,080 Therefore, the third degree of freedom contains all 432 00:25:07,080 --> 00:25:09,360 zeroes, and so on. 433 00:25:09,360 --> 00:25:13,550 This assemblage process is called the direct stiffness 434 00:25:13,550 --> 00:25:19,330 method, an extremely important concept that is very well 435 00:25:19,330 --> 00:25:21,200 implemented in a computer program. 436 00:25:21,200 --> 00:25:25,410 It represents the basis of the implementation of the finite 437 00:25:25,410 --> 00:25:30,120 element method in almost every code that is currently in use. 438 00:25:30,120 --> 00:25:34,900 The direct stiffness method has also a very nice physical 439 00:25:34,900 --> 00:25:35,740 explanation. 440 00:25:35,740 --> 00:25:39,030 And this is what I really want to talk to you about now for 441 00:25:39,030 --> 00:25:40,180 the next five minutes. 442 00:25:40,180 --> 00:25:44,260 The steady-state analysis, of course, then is completed. 443 00:25:44,260 --> 00:25:46,440 The steady-state analysis of this system, of course, is 444 00:25:46,440 --> 00:25:50,660 completed by solving this system of equations here, 445 00:25:50,660 --> 00:25:51,810 equations a. 446 00:25:51,810 --> 00:25:57,000 Once we know U we can go back to the elements and calculate 447 00:25:57,000 --> 00:25:59,820 the forces in the elements themselves by going to the 448 00:25:59,820 --> 00:26:02,070 element equilibrium requirements. 449 00:26:02,070 --> 00:26:08,100 Well let us look then at what we are doing when we perform 450 00:26:08,100 --> 00:26:14,810 this process here by summing, in other words, the K element, 451 00:26:14,810 --> 00:26:17,510 the stiffnesses of the elements into a global 452 00:26:17,510 --> 00:26:18,520 stiffness matrix. 453 00:26:18,520 --> 00:26:21,670 And let us look at what we're doing physically. 454 00:26:21,670 --> 00:26:24,040 Because that really, of course, is the direct 455 00:26:24,040 --> 00:26:26,210 stiffness method that we are using here. 456 00:26:26,210 --> 00:26:31,040 And it is, I think, very nice if you can clearly see what is 457 00:26:31,040 --> 00:26:32,590 happening in that method. 458 00:26:32,590 --> 00:26:36,120 Well, the basic process is the following. 459 00:26:36,120 --> 00:26:39,600 Here I have drawn the carts without any strings. 460 00:26:39,600 --> 00:26:41,220 Of course, our degrees of freedom are 461 00:26:41,220 --> 00:26:45,160 here, u1, u2, and u3. 462 00:26:45,160 --> 00:26:47,280 And the loads are R1, R2, R3. 463 00:26:47,280 --> 00:26:49,160 I don't need to put them in again. 464 00:26:49,160 --> 00:26:55,700 This system here corresponds to a K matrix with zeroes 465 00:26:55,700 --> 00:26:57,410 everywhere. 466 00:26:57,410 --> 00:27:02,000 Blanks in these positions here denote zeroes. 467 00:27:02,000 --> 00:27:06,670 So this is a system that we're starting off with in this 468 00:27:06,670 --> 00:27:11,200 direct stiffness method, a system without any elements, a 469 00:27:11,200 --> 00:27:14,470 matrix without any elements also. 470 00:27:14,470 --> 00:27:17,370 The process, then, is the following. 471 00:27:17,370 --> 00:27:21,860 We are using this cart system. 472 00:27:21,860 --> 00:27:23,810 And we're adding one spring on. 473 00:27:23,810 --> 00:27:25,510 That is the first edition. 474 00:27:25,510 --> 00:27:28,680 That is spring k1. 475 00:27:28,680 --> 00:27:33,390 Mathematically this means that we're going through the 476 00:27:33,390 --> 00:27:34,760 following process. 477 00:27:34,760 --> 00:27:39,110 We are taking our K matrix with blanks everywhere, and 478 00:27:39,110 --> 00:27:43,370 we're adding into it this one element, k1. 479 00:27:43,370 --> 00:27:47,380 Now this is a K matrix, this stiffness matrix governing-- 480 00:27:47,380 --> 00:27:48,980 and this is very important-- 481 00:27:48,980 --> 00:27:53,650 governing this system. 482 00:27:53,650 --> 00:27:58,220 Once again, this is a K matrix governing this system. 483 00:27:58,220 --> 00:28:00,460 Of course, this is not a stable system yet because 484 00:28:00,460 --> 00:28:04,320 there are no connections between these carts here. 485 00:28:04,320 --> 00:28:08,220 Well, with a second edition we're adding 486 00:28:08,220 --> 00:28:11,350 in the second spring. 487 00:28:11,350 --> 00:28:13,970 And that means we are putting this spring there. 488 00:28:13,970 --> 00:28:16,110 That is k2. 489 00:28:16,110 --> 00:28:20,520 Well in our matrix formulation then, what that means in our 490 00:28:20,520 --> 00:28:25,750 direct stiffness method is that we are going from this 491 00:28:25,750 --> 00:28:30,170 system over on this K matrix to that K matrix. 492 00:28:30,170 --> 00:28:32,800 We are adding this second spring 493 00:28:32,800 --> 00:28:35,070 stiffness into the K matrix. 494 00:28:35,070 --> 00:28:39,970 So this is the stiffness matrix that governs the 495 00:28:39,970 --> 00:28:43,940 equilibrium of this physical system. 496 00:28:47,070 --> 00:28:50,390 Notice that this spring here, the second spring, couples 497 00:28:50,390 --> 00:28:52,070 into u1 and u2. 498 00:28:52,070 --> 00:28:56,130 And therefore we have added these blue elements 499 00:28:56,130 --> 00:28:58,730 corresponding to the second spring into degrees of 500 00:28:58,730 --> 00:29:01,500 freedom 1 and 2. 501 00:29:01,500 --> 00:29:04,750 Next in the direct stiffness method we're adding the next 502 00:29:04,750 --> 00:29:09,820 spring element, and that is spring element number 3. 503 00:29:09,820 --> 00:29:12,900 Again, it couples into u1 and u2. 504 00:29:12,900 --> 00:29:17,930 And the stiffness matrix that we are now talking about is 505 00:29:17,930 --> 00:29:18,640 the following. 506 00:29:18,640 --> 00:29:23,200 We're going from this stiffness matrix to that 507 00:29:23,200 --> 00:29:28,260 stiffness matrix here, adding the green k3 in there. 508 00:29:28,260 --> 00:29:37,230 Now next we go from this system to add into the system 509 00:29:37,230 --> 00:29:42,620 the spring 4, spring 4 now. 510 00:29:42,620 --> 00:29:45,400 Please notice that this is now a stable system. 511 00:29:45,400 --> 00:29:49,950 It is a stable system because if I want to put u3 over here, 512 00:29:49,950 --> 00:29:51,690 then I have to do work on this spring. 513 00:29:51,690 --> 00:29:53,690 So this is now a stable system. 514 00:29:53,690 --> 00:29:56,910 In our mathematical formulation, or in our direct 515 00:29:56,910 --> 00:30:00,050 stiffness method rather, what this then corresponds to is 516 00:30:00,050 --> 00:30:04,050 that we are going from this system here, or this stiffness 517 00:30:04,050 --> 00:30:06,220 matrix, to that stiffness matrix. 518 00:30:06,220 --> 00:30:09,080 Notice we have added a k4 into here. 519 00:30:09,080 --> 00:30:13,730 And that, in fact, allows us now to solve at this level. 520 00:30:13,730 --> 00:30:18,360 We could solve the equations KU equals R. Of course, this 521 00:30:18,360 --> 00:30:23,880 now is a stiffness matrix corresponding to this system. 522 00:30:23,880 --> 00:30:26,230 We have not quite yet reached the system 523 00:30:26,230 --> 00:30:28,070 that we want to analyze. 524 00:30:28,070 --> 00:30:33,230 But we reach it by adding the final spring in k5. 525 00:30:33,230 --> 00:30:34,870 k5 now is here. 526 00:30:34,870 --> 00:30:41,210 And that corresponds in our direct stiffness method to 527 00:30:41,210 --> 00:30:45,210 adding this spring in there. 528 00:30:45,210 --> 00:30:47,970 These elements here, k5. 529 00:30:47,970 --> 00:30:52,120 Notice that this spring now here couples into degrees of 530 00:30:52,120 --> 00:30:55,500 freedom 4 and five 5, and that's why it appears in 531 00:30:55,500 --> 00:30:56,775 quadrant column 4 and 5 here. 532 00:31:00,480 --> 00:31:07,020 And this spring. 533 00:31:07,020 --> 00:31:10,130 I should've have said, this spring here couples into 534 00:31:10,130 --> 00:31:17,190 column 2 and 3, column 2 and 3 meaning u2 and u3. 535 00:31:17,190 --> 00:31:21,270 And here we see, of course, that the spring indeed goes 536 00:31:21,270 --> 00:31:24,890 into degree of freedom u2 and u3, into degree of 537 00:31:24,890 --> 00:31:26,490 freedom u2 and u3. 538 00:31:26,490 --> 00:31:29,470 So this then is to final system that we want to 539 00:31:29,470 --> 00:31:33,440 analyze, and this is the final stiffness matrix 540 00:31:33,440 --> 00:31:35,710 that we had to obtain. 541 00:31:35,710 --> 00:31:38,600 Notice, once again, this matrix has been obtained by 542 00:31:38,600 --> 00:31:44,235 taking the sum over all the element stiffness matrices. 543 00:31:44,235 --> 00:31:47,520 We are summing from i equals 1 to 5. 544 00:31:47,520 --> 00:31:50,950 And this mathematical process, once again, which we call the 545 00:31:50,950 --> 00:31:56,570 direct stiffness method has a physical analog. 546 00:31:56,570 --> 00:31:59,030 You can understand it physically in the way I've 547 00:31:59,030 --> 00:32:00,450 shown here. 548 00:32:00,450 --> 00:32:05,010 Namely you're starting off with a blank K matrix, no 549 00:32:05,010 --> 00:32:09,660 elements in it at all, and you simply add one element after 550 00:32:09,660 --> 00:32:12,930 the other into that K matrix filling up the 551 00:32:12,930 --> 00:32:14,290 K matrix that way. 552 00:32:14,290 --> 00:32:17,590 And the additions are carried out-- 553 00:32:17,590 --> 00:32:18,670 this is important-- 554 00:32:18,670 --> 00:32:24,560 by taking the element matrices and adding them into the 555 00:32:24,560 --> 00:32:28,800 appropriate columns and rows of the K matrix. 556 00:32:28,800 --> 00:32:32,280 For example, this element here couples into degree of 557 00:32:32,280 --> 00:32:34,850 freedom 1 and 3. 558 00:32:34,850 --> 00:32:39,850 And if we go once more back to the process that we have been 559 00:32:39,850 --> 00:32:46,530 carrying out here, notice our k4 here corresponds to degree 560 00:32:46,530 --> 00:32:50,680 of freedom 1 and degree of freedom 3, the first row and 561 00:32:50,680 --> 00:32:52,750 column and third row and column. 562 00:32:52,750 --> 00:32:56,030 That's where these elements appear. 563 00:32:56,030 --> 00:33:00,310 So there's a neat physical explanation for the direct 564 00:33:00,310 --> 00:33:03,900 stiffness method which I wanted to 565 00:33:03,900 --> 00:33:05,240 discuss with you here. 566 00:33:05,240 --> 00:33:09,750 Now as another approach, instead of using the direct 567 00:33:09,750 --> 00:33:15,380 formulation of the equations KU equals R, the equilibrium 568 00:33:15,380 --> 00:33:19,410 equations of the system, we can also use 569 00:33:19,410 --> 00:33:20,950 a variational approach. 570 00:33:20,950 --> 00:33:23,940 We will be talking about that variational approach in the 571 00:33:23,940 --> 00:33:24,790 second lecture. 572 00:33:24,790 --> 00:33:28,030 And I would like to discuss it, or introduce it to you, 573 00:33:28,030 --> 00:33:31,880 now very briefly for the analysis of this discrete 574 00:33:31,880 --> 00:33:33,980 system that we just looked at. 575 00:33:33,980 --> 00:33:37,010 The basic process here is that we are constructing a 576 00:33:37,010 --> 00:33:42,320 functional pi which is equal to u minus w where u is the 577 00:33:42,320 --> 00:33:46,510 strain energy of the system and w is the total potential 578 00:33:46,510 --> 00:33:48,390 of the loads. 579 00:33:48,390 --> 00:33:53,110 The equilibrium equations that we just looked at, KU equals R 580 00:33:53,110 --> 00:33:59,685 in other words, are obtained by invoking that del-pi shall 581 00:33:59,685 --> 00:34:03,740 be 0, the stationality condition on pi. 582 00:34:03,740 --> 00:34:10,770 And this means that del-pi, del-ui, shall be 0 for all ui. 583 00:34:10,770 --> 00:34:15,360 This then gives three equations, And these three 584 00:34:15,360 --> 00:34:18,805 questions are obtained as follows. 585 00:34:18,805 --> 00:34:25,770 If we use u, the strain energy of the system is given right 586 00:34:25,770 --> 00:34:29,340 here, 1/2 U transpose KU. 587 00:34:29,340 --> 00:34:33,270 If you were to multiply this out substituting for U and for 588 00:34:33,270 --> 00:34:36,360 K with the values that I've given to you, you would find 589 00:34:36,360 --> 00:34:39,449 that this indeed is the strain energy in the system. 590 00:34:39,449 --> 00:34:43,260 The potential of the total loads is given by U transposed 591 00:34:43,260 --> 00:34:49,250 R. Notice please that there is no 1/2 here in front. 592 00:34:49,250 --> 00:34:52,270 Simply U transpose R is the potential of the loads. 593 00:34:52,270 --> 00:34:57,110 Now if we invoke this condition that del-pi, del-ui 594 00:34:57,110 --> 00:35:03,790 shall be 0, we directly obtain KU equals R. 595 00:35:03,790 --> 00:35:05,830 Now there's one important point. 596 00:35:05,830 --> 00:35:11,510 To obtain u and w, u and w here, we again, can add up the 597 00:35:11,510 --> 00:35:14,810 contributions from all the elements using the direct 598 00:35:14,810 --> 00:35:15,550 stiffness method. 599 00:35:15,550 --> 00:35:21,410 In other words, this K here can be constructed as we have 600 00:35:21,410 --> 00:35:26,990 shown by summing over the elements, by summing the 601 00:35:26,990 --> 00:35:28,890 contributions over all of the elements. 602 00:35:28,890 --> 00:35:36,130 And since this is true, we can also write this total u as 603 00:35:36,130 --> 00:35:40,380 being the sum of the ui's, if you want to, the strain 604 00:35:40,380 --> 00:35:44,220 energies of all of the individual elements. 605 00:35:44,220 --> 00:35:48,400 So here too we could use the direct stiffness method. 606 00:35:48,400 --> 00:35:51,640 Of course, in actuality, in actual practical analysis, we 607 00:35:51,640 --> 00:35:56,180 never form this u, we never form that w when we want to 608 00:35:56,180 --> 00:36:00,470 calculate KU equals R. This is simply a theoretical concept 609 00:36:00,470 --> 00:36:02,810 that I wanted to introduce to you, a theoretical concept 610 00:36:02,810 --> 00:36:06,630 that we will be using later on in the construction of KU 611 00:36:06,630 --> 00:36:12,000 equals R. We never really calculate these measures if we 612 00:36:12,000 --> 00:36:14,460 only want to calculate KU equals R. 613 00:36:14,460 --> 00:36:17,240 It might be of interest to us to calculate this in order to 614 00:36:17,240 --> 00:36:20,750 find out how much strain energy is put into individual 615 00:36:20,750 --> 00:36:23,730 elements in finite element analysis. 616 00:36:23,730 --> 00:36:28,480 But this is only done if you want to evaluate error bounds 617 00:36:28,480 --> 00:36:30,710 on the finite element solution and so on. 618 00:36:30,710 --> 00:36:34,840 If we only want the calculate KU equals R and obtain the use 619 00:36:34,840 --> 00:36:40,185 in other words, to be able to predict the displacements and 620 00:36:40,185 --> 00:36:43,440 the stresses in the elements, then we would not calculate 621 00:36:43,440 --> 00:36:45,610 these two quantities. 622 00:36:45,610 --> 00:36:50,090 Now this then were the essence of the analysis of a 623 00:36:50,090 --> 00:36:55,650 steady-state problem for discrete systems. 624 00:36:55,650 --> 00:36:58,390 I pointed out already that if we have an extra finite 625 00:36:58,390 --> 00:37:00,560 element system there, of course, many additional 626 00:37:00,560 --> 00:37:02,750 concepts that we have to talk about, a selection of 627 00:37:02,750 --> 00:37:05,520 elements, the kinds of interpolations to be used, 628 00:37:05,520 --> 00:37:08,070 and, of course, we then have to also talk about how do we 629 00:37:08,070 --> 00:37:10,330 solve these equations, and so on. 630 00:37:10,330 --> 00:37:11,720 We will address these questions 631 00:37:11,720 --> 00:37:13,340 in the later lectures. 632 00:37:13,340 --> 00:37:16,660 However, another class of problems that we will be 633 00:37:16,660 --> 00:37:19,660 talking about are propagation problems. 634 00:37:19,660 --> 00:37:23,160 The main characteristics of propagation problems are that 635 00:37:23,160 --> 00:37:25,580 the response changes with time. 636 00:37:25,580 --> 00:37:28,690 Therefore, we need to include the d'Alembert forces. 637 00:37:28,690 --> 00:37:31,280 Now basically what we are saying then is that we're 638 00:37:31,280 --> 00:37:37,490 looking at static equilibrium as a function of time but also 639 00:37:37,490 --> 00:37:39,780 taking into account the d'Alembert forces. 640 00:37:39,780 --> 00:37:43,560 And that, together then, makes it a dynamic problem. 641 00:37:43,560 --> 00:37:49,150 Of course, if the displacement varies very slow, in other 642 00:37:49,150 --> 00:37:56,400 words, the load varies very slow, then the inertia forces 643 00:37:56,400 --> 00:37:59,950 can be neglected, and we would simply have this set of 644 00:37:59,950 --> 00:38:03,586 equations where R of t is a function of time and U of t 645 00:38:03,586 --> 00:38:04,760 would be a function of time. 646 00:38:04,760 --> 00:38:09,440 However, when R of t acts rapidly or suddenly inertia 647 00:38:09,440 --> 00:38:12,430 conditions are applied to the system, then the inertia 648 00:38:12,430 --> 00:38:14,650 forces can be very important. 649 00:38:14,650 --> 00:38:15,920 We have to include their effect. 650 00:38:15,920 --> 00:38:18,820 And then we have a true propagation problem, a truly 651 00:38:18,820 --> 00:38:20,930 dynamic problem that has to be solved. 652 00:38:20,930 --> 00:38:26,690 For our example, the M matrix here would be this 3 by 3 653 00:38:26,690 --> 00:38:33,750 matrix where m1 is simply the mass of the cart 1, m2 is the 654 00:38:33,750 --> 00:38:37,560 mass of the cart 2, m3 is the mass of the cart 3. 655 00:38:37,560 --> 00:38:39,680 Of course these masses would have to be given. 656 00:38:39,680 --> 00:38:46,780 And notice that we would evaluate them by basically 657 00:38:46,780 --> 00:38:51,420 saying that this total mass here can be evaluated by 658 00:38:51,420 --> 00:38:54,170 taking the mass per unit volume times the volume. 659 00:38:54,170 --> 00:38:56,750 And that would be the mass that we're talking about when 660 00:38:56,750 --> 00:39:02,040 we accelerate that cart into this direction. 661 00:39:02,040 --> 00:39:04,970 So these masses here are very simply evaluated. 662 00:39:04,970 --> 00:39:08,200 When we talk later on about actually finite elements, we 663 00:39:08,200 --> 00:39:12,180 will be talking about similar mass matrices where we simply 664 00:39:12,180 --> 00:39:16,600 take the total volume of an element and lump that volume 665 00:39:16,600 --> 00:39:18,440 of the element to its nodes. 666 00:39:18,440 --> 00:39:22,050 We will also talk about consistent mass matrices where 667 00:39:22,050 --> 00:39:25,260 this mass matrix is a little bit more complicated. 668 00:39:25,260 --> 00:39:27,170 In other words, some of these off-diagonal 669 00:39:27,170 --> 00:39:29,820 elements are not 0. 670 00:39:29,820 --> 00:39:33,350 Finally, we will also talk about eigenvalue problems. 671 00:39:33,350 --> 00:39:36,920 In the solution of eigenvalue problems, we will be talking 672 00:39:36,920 --> 00:39:41,140 about generalized eigenvalue problems, in particular, which 673 00:39:41,140 --> 00:39:46,650 are Av equals lambda Bv, which can be written down in this 674 00:39:46,650 --> 00:39:51,920 form where A and B are symmetric matrices of order n, 675 00:39:51,920 --> 00:39:56,850 v is a vector of order n, and lambda is a scalar. 676 00:39:56,850 --> 00:40:00,530 As an example, for example here in dynamic analysis, what 677 00:40:00,530 --> 00:40:07,150 we will see there is K phi equals omega squared M phi 678 00:40:07,150 --> 00:40:09,850 where K is the stiffness matrix that I 679 00:40:09,850 --> 00:40:11,280 talked about already. 680 00:40:11,280 --> 00:40:14,350 This which would be for the cart system here simply as 3 681 00:40:14,350 --> 00:40:17,030 by 3, this 3 by 3 stiffness matrix that I 682 00:40:17,030 --> 00:40:18,160 introduced to you. 683 00:40:18,160 --> 00:40:20,430 And it's a mass matrix that we just had 684 00:40:20,430 --> 00:40:22,870 here on this viewgraph. 685 00:40:22,870 --> 00:40:24,020 That is the mass matrix. 686 00:40:24,020 --> 00:40:28,790 And phi is the vector. 687 00:40:28,790 --> 00:40:31,510 If we find a solution, in other words, if this equation 688 00:40:31,510 --> 00:40:38,280 is satisfied, we put an i on there and satisfy for phi i 689 00:40:38,280 --> 00:40:41,200 and omega i squared. 690 00:40:41,200 --> 00:40:44,320 Omega i squared will be a frequency. 691 00:40:44,320 --> 00:40:46,880 I will be discussing it just now a little more. 692 00:40:46,880 --> 00:40:49,470 And then we're talking about an eigenpair. 693 00:40:49,470 --> 00:40:52,210 But notice that is a typical problem that we will be 694 00:40:52,210 --> 00:40:54,340 discussing which arises, in other 695 00:40:54,340 --> 00:40:56,580 words, in dynamic analysis. 696 00:40:56,580 --> 00:41:00,050 Notice also that what we're really saying here is that the 697 00:41:00,050 --> 00:41:01,860 right-hand side is a load vector. 698 00:41:04,830 --> 00:41:09,470 And if we know v, if we know lambda, then we 699 00:41:09,470 --> 00:41:11,130 know the load vector. 700 00:41:11,130 --> 00:41:16,540 What we would calculate then is the same v that we have 701 00:41:16,540 --> 00:41:17,430 substituted here. 702 00:41:17,430 --> 00:41:21,420 In other words, if we consider this to be a set of loads 703 00:41:21,420 --> 00:41:25,430 where v is now known, lambda is known, then we could 704 00:41:25,430 --> 00:41:30,170 evaluate R. In solving Av equals R, we would get back 705 00:41:30,170 --> 00:41:32,320 our v that we substituted into here. 706 00:41:32,320 --> 00:41:34,670 And that is the main characteristic of an 707 00:41:34,670 --> 00:41:36,630 eigenvalue problem. 708 00:41:36,630 --> 00:41:39,610 Well, they arise in dynamic and buckling analysis, and let 709 00:41:39,610 --> 00:41:43,770 us look at one example where we actually obtain this 710 00:41:43,770 --> 00:41:45,030 eigenvalue problem. 711 00:41:45,030 --> 00:41:47,850 And the example is simply the system of rigid carts that we 712 00:41:47,850 --> 00:41:50,120 considered already earlier. 713 00:41:50,120 --> 00:41:53,670 We obtain the eigenvalue problem by looking at the 714 00:41:53,670 --> 00:41:56,980 equilibrium equations when no loads are applied. 715 00:41:56,980 --> 00:42:00,860 And we call these the free vibration conditions, free 716 00:42:00,860 --> 00:42:03,990 because there are no loads applied, free of loads. 717 00:42:03,990 --> 00:42:09,540 If we let U be equal to phi times sine omega t minus tau 718 00:42:09,540 --> 00:42:13,090 where the time dependency now in the response is in this 719 00:42:13,090 --> 00:42:17,980 function here, in the sine function only, and if we take 720 00:42:17,980 --> 00:42:22,700 the second derivative of U, meaning that we get a cosine 721 00:42:22,700 --> 00:42:26,820 and then a minus sign in here, and, of course, this omega 722 00:42:26,820 --> 00:42:30,580 twice outside, so we have a sign change here. 723 00:42:30,580 --> 00:42:35,160 We have a minus omega squared M phi sine omega t minus tau 724 00:42:35,160 --> 00:42:37,270 for this part here. 725 00:42:37,270 --> 00:42:42,940 And for this part KU we obtain K phi sine omega t minus tau 726 00:42:42,940 --> 00:42:46,130 by simply substituting from here into there. 727 00:42:46,130 --> 00:42:51,470 And, of course, the sum of these two must be equal to 0. 728 00:42:51,470 --> 00:42:56,260 Now this equation must hold for any time, t. 729 00:42:56,260 --> 00:43:00,420 So we can simply cancel out this part and that part. 730 00:43:00,420 --> 00:43:04,790 And the resulting set of equations that we are 731 00:43:04,790 --> 00:43:08,590 obtaining then are given on the last viewgraph, namely 732 00:43:08,590 --> 00:43:12,520 those equations being K phi equals omega squared M phi. 733 00:43:12,520 --> 00:43:15,000 So that is the generalized eigenvalue problem which we 734 00:43:15,000 --> 00:43:17,130 obtain in dynamic analysis. 735 00:43:17,130 --> 00:43:21,540 We will be later on talking about how we solve this 736 00:43:21,540 --> 00:43:24,690 generalized eigenvalue problem for the eigenvalues and 737 00:43:24,690 --> 00:43:25,400 eigenvectors. 738 00:43:25,400 --> 00:43:28,500 In the case of of the 3 by 3 system that we are considering 739 00:43:28,500 --> 00:43:31,780 here, in other words, the analysis of the cart system, 740 00:43:31,780 --> 00:43:38,030 we only have three solutions, omega 1 phi 1, omega2 phi2, 741 00:43:38,030 --> 00:43:39,770 omega3 phi3. 742 00:43:39,770 --> 00:43:43,850 And we call each of the solutions an eigenpair. 743 00:43:43,850 --> 00:43:46,910 So there are three eigenpairs that satisfy 744 00:43:46,910 --> 00:43:48,410 this particular equation. 745 00:43:48,410 --> 00:43:51,320 Notice that this is, in other words, the equation that I 746 00:43:51,320 --> 00:43:53,310 talked about here earlier. 747 00:43:53,310 --> 00:43:59,820 And the eigenpairs, phi i, omega i squared are the 748 00:43:59,820 --> 00:44:02,110 solutions to this equation. 749 00:44:02,110 --> 00:44:06,220 We are really interested in omega i because that is the 750 00:44:06,220 --> 00:44:10,260 frequency in radians per second, and the eigenvalue, 751 00:44:10,260 --> 00:44:12,660 however, being omega squared. 752 00:44:12,660 --> 00:44:15,430 In general when we have an n by n system-- 753 00:44:15,430 --> 00:44:19,090 and I have already written down here the n by n, let me 754 00:44:19,090 --> 00:44:21,130 put it bigger once more here-- 755 00:44:21,130 --> 00:44:23,480 and we have a general n by n system, in other words, and 756 00:44:23,480 --> 00:44:28,000 not being equal to 3 just as we have in our cart system, 757 00:44:28,000 --> 00:44:30,410 then we have n solutions. 758 00:44:30,410 --> 00:44:34,080 And, however, we will find that in finite element 759 00:44:34,080 --> 00:44:36,670 analysis we do not necessarily need to 760 00:44:36,670 --> 00:44:38,600 calculate all n solutions. 761 00:44:38,600 --> 00:44:42,080 In fact, when we consider large eigensystems where n is 762 00:44:42,080 --> 00:44:46,320 equal to 1,000 or even more, then certainly we do not want 763 00:44:46,320 --> 00:44:47,790 to calculate all eigenvalues. 764 00:44:47,790 --> 00:44:51,720 It would be exorbitantly expensive, much too expensive 765 00:44:51,720 --> 00:44:53,810 to calculate all of the eigenvalues and eigenvectors. 766 00:44:53,810 --> 00:44:57,250 We don't need to have them all in analysis. 767 00:44:57,250 --> 00:45:00,260 And, therefore, we will talk about eigenvalue solution 768 00:45:00,260 --> 00:45:03,440 methods that only calculate the eigenvalues and 769 00:45:03,440 --> 00:45:07,210 eigenvectors that we are actually interested in. 770 00:45:07,210 --> 00:45:10,450 We also, of course, have to, before we actually get to that 771 00:45:10,450 --> 00:45:13,410 topic which is the topic of the last lecture, we will talk 772 00:45:13,410 --> 00:45:15,740 about how we actually construct these K matrices, 773 00:45:15,740 --> 00:45:18,840 how we calculate them, construct them for different 774 00:45:18,840 --> 00:45:20,500 finite element systems. 775 00:45:20,500 --> 00:45:24,640 Well, this then does complete what I wanted 776 00:45:24,640 --> 00:45:26,000 to say in this lecture. 777 00:45:26,000 --> 00:45:27,850 Thank you very much for your attention.