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 00:00:22,040 --> 00:00:24,720 PROFESSOR: Ladies and gentlemen, welcome to lecture 9 00:00:24,720 --> 00:00:26,040 number two. 10 00:00:26,040 --> 00:00:29,400 In this lecture, I would like to discuss some basic concepts 11 00:00:29,400 --> 00:00:33,040 of finite element analysis with regard to the analysis of 12 00:00:33,040 --> 00:00:35,590 continuous systems. 13 00:00:35,590 --> 00:00:38,710 We discussed in the first lecture already some basic 14 00:00:38,710 --> 00:00:42,965 concepts of analysis of discrete systems. 15 00:00:42,965 --> 00:00:47,080 However, in actuality, in the analysis of a complex system, 16 00:00:47,080 --> 00:00:50,600 we are dealing with a continuous system, and there 17 00:00:50,600 --> 00:00:54,660 are some additional basic concepts that are used in 18 00:00:54,660 --> 00:00:57,310 analysis of continuous systems, using 19 00:00:57,310 --> 00:00:58,870 finite element methods. 20 00:00:58,870 --> 00:01:02,150 And those additional concept that are used, I want to 21 00:01:02,150 --> 00:01:04,069 discuss in this lecture. 22 00:01:04,069 --> 00:01:09,370 Well, when we talk about the analysis of a continuous 23 00:01:09,370 --> 00:01:15,010 system, we can analyze that system via a differential 24 00:01:15,010 --> 00:01:19,320 formulation or a variational formulation. 25 00:01:19,320 --> 00:01:23,430 If we use a differential formulation or variational 26 00:01:23,430 --> 00:01:27,920 formulation, of course we obtain continuous variables, 27 00:01:27,920 --> 00:01:34,380 and we have an infinite number of state variables, or rather, 28 00:01:34,380 --> 00:01:37,180 if we talk about displacements, a U 29 00:01:37,180 --> 00:01:41,300 displacement, for example, of a rod, as I will be discussing 30 00:01:41,300 --> 00:01:46,220 just now, we will have infinite values of that 31 00:01:46,220 --> 00:01:48,520 displacement along the rod. 32 00:01:48,520 --> 00:01:51,100 In the differential formulation or the variational 33 00:01:51,100 --> 00:01:54,330 formulation, we would have to solve for that continuous 34 00:01:54,330 --> 00:01:55,970 variable along the rod. 35 00:01:55,970 --> 00:02:00,470 Well, we will also notice that in the analysis of a complex 36 00:02:00,470 --> 00:02:04,940 system, we cannot solve the differential equations that we 37 00:02:04,940 --> 00:02:09,539 are arriving at directly, and we have to resort 38 00:02:09,539 --> 00:02:11,020 to numerical methods. 39 00:02:11,020 --> 00:02:14,840 Now, some concepts that have been used for a long time are 40 00:02:14,840 --> 00:02:17,310 the weighted residual methods. 41 00:02:17,310 --> 00:02:19,980 These have been used by Galerkin least squares 42 00:02:19,980 --> 00:02:24,410 approaches to solve the differential equations that 43 00:02:24,410 --> 00:02:29,100 govern the equilibrium motion of the system approximately. 44 00:02:29,100 --> 00:02:31,420 In the variational formulation, the Ritz method 45 00:02:31,420 --> 00:02:35,580 has been used for quite a long time. 46 00:02:35,580 --> 00:02:37,800 These are classical techniques, therefore, 47 00:02:37,800 --> 00:02:40,280 weighted residual method and the Ritz method. 48 00:02:40,280 --> 00:02:43,580 And what I want to show to you in this lecture is how the 49 00:02:43,580 --> 00:02:47,780 finite element method is really an extension off these 50 00:02:47,780 --> 00:02:51,110 methods, or how this method, the finite element method, is 51 00:02:51,110 --> 00:02:55,050 related to these classical techniques. 52 00:02:55,050 --> 00:03:00,180 Well, when we talk about the differential formulation, we 53 00:03:00,180 --> 00:03:05,060 are looking at the differential equilibrium, or 54 00:03:05,060 --> 00:03:07,020 the equilibrium of a differential 55 00:03:07,020 --> 00:03:10,020 element of the system. 56 00:03:10,020 --> 00:03:15,350 Now, I want to show to you the basic ideas of a differential 57 00:03:15,350 --> 00:03:20,240 formulation by looking at, or analyzing, this rod. 58 00:03:20,240 --> 00:03:25,090 Here we have a rod that is fixed at the left-hand side, 59 00:03:25,090 --> 00:03:27,120 on rollers. 60 00:03:27,120 --> 00:03:29,640 x is a variable along the rod. 61 00:03:29,640 --> 00:03:34,530 u is the displacement of that rod into this direction. 62 00:03:34,530 --> 00:03:37,730 The rod is subjected, as shown here, to a load , R0 63 00:03:37,730 --> 00:03:39,400 at its right end. 64 00:03:39,400 --> 00:03:44,420 Notice that in this analysis, we assume that plane sections 65 00:03:44,420 --> 00:03:45,950 remain plane. 66 00:03:45,950 --> 00:03:52,300 In other words, a section that was originally there at time t 67 00:03:52,300 --> 00:03:56,150 greater than 0 has moved to here. 68 00:03:56,150 --> 00:03:59,830 And this movement here is the u displacement. 69 00:03:59,830 --> 00:04:05,070 But notice that the vertical section here remains vertical 70 00:04:05,070 --> 00:04:06,170 during the motion. 71 00:04:06,170 --> 00:04:12,280 So at every section, we have only 1 degree of freedom. 72 00:04:12,280 --> 00:04:14,010 There's no rotation of that section. 73 00:04:14,010 --> 00:04:17,810 However, this one degree of freedom, u, varies 74 00:04:17,810 --> 00:04:20,320 continuously along the rod. 75 00:04:20,320 --> 00:04:23,900 Therefore, the rod itself has really an infinite number of 76 00:04:23,900 --> 00:04:26,510 degrees of freedom. 77 00:04:26,510 --> 00:04:30,020 For this very simple system, we could 78 00:04:30,020 --> 00:04:32,160 obtain an exact solution. 79 00:04:32,160 --> 00:04:38,080 However, I want to show you how we proceed in analyzing 80 00:04:38,080 --> 00:04:41,580 this rod via differential formulation, a variational 81 00:04:41,580 --> 00:04:45,700 formulation, and so on, simply as an example. 82 00:04:45,700 --> 00:04:49,630 Therefore, the basic ideas, really, that I will be putting 83 00:04:49,630 --> 00:04:52,090 forth to you, that I will be discussing with you, are 84 00:04:52,090 --> 00:04:56,710 really the important things that I want to expose to you. 85 00:04:56,710 --> 00:05:00,260 It's not the analysis of this very specific problem. 86 00:05:00,260 --> 00:05:03,270 It's really the basic idea that I want to clarify to you 87 00:05:03,270 --> 00:05:05,970 by looking at this one problem. 88 00:05:05,970 --> 00:05:10,010 Well, for this problem here, the governing differential 89 00:05:10,010 --> 00:05:13,210 equation of motion is shown here. 90 00:05:13,210 --> 00:05:17,500 Notice u once again is the displacement of a section. 91 00:05:17,500 --> 00:05:22,410 That is, this displacement here at that coordinate x. 92 00:05:22,410 --> 00:05:26,540 c is given here as square root E over rho, where E is Young's 93 00:05:26,540 --> 00:05:28,960 modulus of the material, rho is the mass 94 00:05:28,960 --> 00:05:30,910 density of the material. 95 00:05:30,910 --> 00:05:32,320 t, of course, is the time variable. 96 00:05:34,950 --> 00:05:37,850 Notice the cross-sectional area a here, I have also 97 00:05:37,850 --> 00:05:39,150 written down here. 98 00:05:39,150 --> 00:05:43,320 This cross-sectional area cancels out on both sides, as 99 00:05:43,320 --> 00:05:44,880 you will see just now. 100 00:05:44,880 --> 00:05:48,550 Now this question here is obtained in the differential 101 00:05:48,550 --> 00:05:52,980 formulation by looking at the equilibrium of an element. 102 00:05:52,980 --> 00:05:58,070 And we might consider this to be the element that I will now 103 00:05:58,070 --> 00:06:00,590 focus our attention on. 104 00:06:00,590 --> 00:06:03,540 Here we have that element drawn again. 105 00:06:03,540 --> 00:06:10,450 This element here of length dx is subjected at its left side, 106 00:06:10,450 --> 00:06:17,420 because x comes from here as a variable and brings us up to 107 00:06:17,420 --> 00:06:18,950 this station. 108 00:06:18,950 --> 00:06:22,360 At the left side here, the element is subjected to sigma, 109 00:06:22,360 --> 00:06:24,110 the stress sigma. 110 00:06:24,110 --> 00:06:27,140 At the right side, we have the stress sigma plus 111 00:06:27,140 --> 00:06:29,070 partial d sigma dx. 112 00:06:29,070 --> 00:06:33,370 This here really means nothing else than a d sigma, an 113 00:06:33,370 --> 00:06:35,450 increment in the sigma. 114 00:06:35,450 --> 00:06:41,020 Well, the equilibrium requirement for this element 115 00:06:41,020 --> 00:06:48,560 here is now that the force on this side here, that is, sigma 116 00:06:48,560 --> 00:06:54,780 times a on this side, which is actually this one here, I 117 00:06:54,780 --> 00:06:57,900 should have pointed to this one-- 118 00:06:57,900 --> 00:07:03,730 and this force here, which is the force on the right side. 119 00:07:03,730 --> 00:07:07,430 If we subtract these two forces, that must be equal to 120 00:07:07,430 --> 00:07:09,970 the force applied. 121 00:07:09,970 --> 00:07:14,030 Or rather, the d'Alembert force. 122 00:07:14,030 --> 00:07:18,020 Notice that if we look at this element, there's a force on 123 00:07:18,020 --> 00:07:20,540 the right, there's a force on the left. 124 00:07:20,540 --> 00:07:24,890 And this being the force on the right, let's call that, 125 00:07:24,890 --> 00:07:27,580 say, R1, and let's call that R2. 126 00:07:27,580 --> 00:07:31,770 So I put here R1, I put here R2. 127 00:07:31,770 --> 00:07:38,860 And R1 minus R2 must be equal to the d'Alembert force, which 128 00:07:38,860 --> 00:07:41,930 is due to the inertia of the material. 129 00:07:41,930 --> 00:07:45,750 This is the basic Newton's law applied to this 130 00:07:45,750 --> 00:07:47,060 differential element. 131 00:07:47,060 --> 00:07:49,870 Now, if we referred back to how we proceeded in the 132 00:07:49,870 --> 00:07:53,680 analysis of a discrete system, we really proceeded in exactly 133 00:07:53,680 --> 00:07:54,650 the same way. 134 00:07:54,650 --> 00:07:58,470 But our element then was a discrete element, a discrete 135 00:07:58,470 --> 00:07:59,380 spring element. 136 00:07:59,380 --> 00:08:03,200 We now use the same concept, but apply those concepts to a 137 00:08:03,200 --> 00:08:05,340 differential element. 138 00:08:05,340 --> 00:08:10,120 So this is the equilibrium equation of the element. 139 00:08:10,120 --> 00:08:12,660 The constitutive relation is given here-- 140 00:08:12,660 --> 00:08:15,670 that the stress is equal to e, the Young's 141 00:08:15,670 --> 00:08:17,470 Modulus, times the strain. 142 00:08:17,470 --> 00:08:18,780 This is the strain. 143 00:08:18,780 --> 00:08:21,770 And notice once again, since we are considering sections to 144 00:08:21,770 --> 00:08:26,720 be remaining plane, and simply move horizontally, the only 145 00:08:26,720 --> 00:08:28,960 strain that we're talking about is this one. 146 00:08:28,960 --> 00:08:32,370 If we combine these two equations here, we directly 147 00:08:32,370 --> 00:08:34,110 obtain that equation here. 148 00:08:34,110 --> 00:08:37,270 Notice as I mentioned earlier, a cancels out. 149 00:08:37,270 --> 00:08:41,100 If a is constant, that's why a does not enter into this 150 00:08:41,100 --> 00:08:45,770 equation, and this substitution for sigma into 151 00:08:45,770 --> 00:08:50,420 here gives us the second derivative here. 152 00:08:50,420 --> 00:08:54,260 Of course, this part here cancels out that part there, 153 00:08:54,260 --> 00:08:57,910 and that second derivative is this one here. 154 00:08:57,910 --> 00:09:01,710 The E brought over to this side gives us a 1 over c 155 00:09:01,710 --> 00:09:07,240 squared, where c, once again, is defined as shown here. 156 00:09:07,240 --> 00:09:11,210 The important point, really, is that we are looking here in 157 00:09:11,210 --> 00:09:15,035 the differential formulation at a differential element of 158 00:09:15,035 --> 00:09:20,130 length dx at a particular station x. 159 00:09:20,130 --> 00:09:22,670 That we're looking at this element and we establish the 160 00:09:22,670 --> 00:09:26,620 equilibrium requirement of that element. 161 00:09:26,620 --> 00:09:34,820 R1 minus R2 shall be equal to the mass of the element times 162 00:09:34,820 --> 00:09:37,760 the acceleration. 163 00:09:37,760 --> 00:09:42,450 We also introduced a constitutive relation. 164 00:09:42,450 --> 00:09:45,640 So far, clearly, we have used two conditions for the 165 00:09:45,640 --> 00:09:46,830 solution of the problem. 166 00:09:46,830 --> 00:09:49,030 The first one is the equilibrium condition. 167 00:09:49,030 --> 00:09:51,990 The second is the constitutive condition, or constitutive 168 00:09:51,990 --> 00:09:52,810 requirement. 169 00:09:52,810 --> 00:09:55,960 We have to ask ourselves, where do we satisfy the 170 00:09:55,960 --> 00:09:57,420 compatibility condition? 171 00:09:57,420 --> 00:09:59,750 Because there are always these three conditions that we have 172 00:09:59,750 --> 00:10:00,840 to satisfy. 173 00:10:00,840 --> 00:10:03,940 Well, the compatibility condition is satisfied by 174 00:10:03,940 --> 00:10:10,940 solving this differential equation for this rod, and 175 00:10:10,940 --> 00:10:13,820 obtaining a u that is continuous. 176 00:10:13,820 --> 00:10:19,080 In other words, a u that tells that all the sections have 177 00:10:19,080 --> 00:10:19,930 remained together. 178 00:10:19,930 --> 00:10:22,730 We did not cut that bar apart. 179 00:10:22,730 --> 00:10:26,280 In the discrete system analysis of the spring system 180 00:10:26,280 --> 00:10:30,330 of lecture 1, if you were to think back to it, we had to 181 00:10:30,330 --> 00:10:34,510 satisfy the compatibility condition explicitly, in 182 00:10:34,510 --> 00:10:36,950 establishing the equilibrium equations, because we had to 183 00:10:36,950 --> 00:10:41,340 make sure that all the springs remain attached to the carts. 184 00:10:41,340 --> 00:10:43,800 Here we satisfy the compatibility condition by 185 00:10:43,800 --> 00:10:46,900 solving this equation for a continuous u. 186 00:10:46,900 --> 00:10:49,070 Well, the boundary conditions, of course, 187 00:10:49,070 --> 00:10:51,130 also have to be stated. 188 00:10:51,130 --> 00:10:54,610 And here we have a boundary condition on the 189 00:10:54,610 --> 00:10:56,380 left end of the rod. 190 00:10:56,380 --> 00:11:00,930 Remember, please, that the rod is fixed at its left end, so 191 00:11:00,930 --> 00:11:05,090 we have this condition here, and clearly u must be 0 for 192 00:11:05,090 --> 00:11:08,240 all times t at x equals 0. 193 00:11:08,240 --> 00:11:11,930 At the right end, we apply a load R0. 194 00:11:11,930 --> 00:11:17,160 And there we have, this being here the area times the stress 195 00:11:17,160 --> 00:11:20,820 at x equals L, E times a du dx. 196 00:11:20,820 --> 00:11:25,300 Notice that E du dx is, of course, the stress, and so we 197 00:11:25,300 --> 00:11:29,160 have here this total force at the right end 198 00:11:29,160 --> 00:11:31,600 being equal to R0. 199 00:11:31,600 --> 00:11:35,070 We also have initial conditions for the solution of 200 00:11:35,070 --> 00:11:40,310 this equation that I showed to you, this equation here. 201 00:11:40,310 --> 00:11:43,320 We have to have two spatial conditions, two boundary 202 00:11:43,320 --> 00:11:46,640 conditions, one at the left and one at the right end. 203 00:11:46,640 --> 00:11:48,150 The ones that I just showed to you. 204 00:11:48,150 --> 00:11:50,870 We also have to have to have two initial conditions, one on 205 00:11:50,870 --> 00:11:53,580 the displacement and one on the velocity. 206 00:11:53,580 --> 00:11:56,570 Well, their initial conditions, in this particular 207 00:11:56,570 --> 00:11:58,680 example, might be as shown here. 208 00:11:58,680 --> 00:12:02,190 At time 0, all of the displacement are 0. 209 00:12:02,190 --> 00:12:04,400 And at time 0, all the velocities 210 00:12:04,400 --> 00:12:06,470 along the rod are 0. 211 00:12:06,470 --> 00:12:11,160 So the basic differential equation given here, once 212 00:12:11,160 --> 00:12:16,740 again, plus these boundary conditions, plus these initial 213 00:12:16,740 --> 00:12:20,700 conditions, define the complete problem. 214 00:12:20,700 --> 00:12:24,540 I also like to point out here that this boundary condition 215 00:12:24,540 --> 00:12:28,820 here, which does not involve any derivative, is called an 216 00:12:28,820 --> 00:12:32,380 essential, or displacement boundary condition. 217 00:12:32,380 --> 00:12:35,690 An essential boundary condition because it does not 218 00:12:35,690 --> 00:12:40,090 involve any derivative when, and this is important, the 219 00:12:40,090 --> 00:12:45,060 highest derivative in this differential equation is 2. 220 00:12:45,060 --> 00:12:49,390 The right-hand side boundary condition is called a natural 221 00:12:49,390 --> 00:12:51,110 force boundary condition. 222 00:12:51,110 --> 00:12:53,150 It's really involving forces. 223 00:12:53,150 --> 00:12:57,000 And it involves, as a highest derivative, a derivative of 224 00:12:57,000 --> 00:13:02,280 order 1 when the differential equation here involves as the 225 00:13:02,280 --> 00:13:06,740 highest derivative a derivative of order 2. 226 00:13:06,740 --> 00:13:08,480 So in general-- 227 00:13:08,480 --> 00:13:11,830 and this is a very important point-- 228 00:13:11,830 --> 00:13:14,360 we can say the following. 229 00:13:14,360 --> 00:13:17,970 If the highest order of the spatial derivative in the 230 00:13:17,970 --> 00:13:24,650 problem governing differential equation is 2m, in our case, m 231 00:13:24,650 --> 00:13:28,350 is equal to 1 for our problem. 232 00:13:28,350 --> 00:13:30,730 The highest order of the spacial derivative in the 233 00:13:30,730 --> 00:13:33,840 essential boundary condition is m minus 1. 234 00:13:33,840 --> 00:13:36,380 In other words, in this case, of order 0. 235 00:13:36,380 --> 00:13:39,060 The highest order the spatial derivative in the natural 236 00:13:39,060 --> 00:13:42,380 boundary conditions that I just discussed is 2m minus 1, 237 00:13:42,380 --> 00:13:45,590 which is 1 in our particular case. 238 00:13:45,590 --> 00:13:51,030 Then if we talk about this problem, we talk about a C m 239 00:13:51,030 --> 00:13:54,190 minus 1 variational problem. 240 00:13:54,190 --> 00:13:56,440 It will become apparent to you later on why we 241 00:13:56,440 --> 00:13:57,560 call it this way. 242 00:13:57,560 --> 00:14:02,980 C m minus 1 means continuity of order m minus 1. 243 00:14:02,980 --> 00:14:05,790 In fact, in the Ritz analysis that we will be performing 244 00:14:05,790 --> 00:14:08,570 later on, that I will discuss with you later on, we find 245 00:14:08,570 --> 00:14:14,420 that we need, in the solution of that kind of problem, only 246 00:14:14,420 --> 00:14:20,080 continuity in the Ritz functions of order m minus 1. 247 00:14:20,080 --> 00:14:23,610 Well, let us now look at the variational formulation. 248 00:14:23,610 --> 00:14:26,510 I mentioned earlier that we have two different approaches. 249 00:14:26,510 --> 00:14:28,670 The first approach is a differential formulation, the 250 00:14:28,670 --> 00:14:31,080 second approach is the variational formulation. 251 00:14:31,080 --> 00:14:35,690 The variational formulation operates in much the same way 252 00:14:35,690 --> 00:14:37,760 as I introduced it to you for the 253 00:14:37,760 --> 00:14:39,320 analysis of discrete systems. 254 00:14:39,320 --> 00:14:43,620 We talk about pi a functional, being equal to the strain 255 00:14:43,620 --> 00:14:47,150 energy minus the potential of the loads. 256 00:14:47,150 --> 00:14:50,390 Now for the rod, the strain energy is given here. 257 00:14:50,390 --> 00:14:55,260 Notice this is 1/2 times the stress times the strain, and 258 00:14:55,260 --> 00:14:59,290 integrated over the total volume of the element, or of 259 00:14:59,290 --> 00:15:01,320 the rod, I should rather say. 260 00:15:01,320 --> 00:15:06,240 The total potential of the loads is given here. 261 00:15:06,240 --> 00:15:09,060 I could have written it this way, with a minus out there 262 00:15:09,060 --> 00:15:10,550 and a plus in there-- 263 00:15:10,550 --> 00:15:11,440 same thing. 264 00:15:11,440 --> 00:15:15,860 This, then, is really nothing other than the loads 265 00:15:15,860 --> 00:15:18,620 multiplied by the total displacement. 266 00:15:18,620 --> 00:15:20,590 And of course, there's an integration involved here, 267 00:15:20,590 --> 00:15:23,480 because the body forces, the body loads that I introduced 268 00:15:23,480 --> 00:15:28,560 here, fB, are varying along the length of the rod. 269 00:15:28,560 --> 00:15:31,740 I introduce these fB body forces because I want to use, 270 00:15:31,740 --> 00:15:35,560 later on, the d'Alembert principle, put these, in other 271 00:15:35,560 --> 00:15:40,230 words, equal to minus the acceleration forces, and can 272 00:15:40,230 --> 00:15:42,620 directly apply what I discussed now, also to the 273 00:15:42,620 --> 00:15:45,860 dynamic analysis of this rod, which we considered in the 274 00:15:45,860 --> 00:15:47,790 differential formulations. 275 00:15:47,790 --> 00:15:51,670 Together with stating pi as shown here, we also have to 276 00:15:51,670 --> 00:15:54,620 state the left-hand boundary condition, which is an 277 00:15:54,620 --> 00:15:56,550 essential boundary condition. 278 00:15:56,550 --> 00:15:59,570 We have to list all the essential boundary conditions 279 00:15:59,570 --> 00:16:01,860 here, or the displacement boundary conditions. 280 00:16:01,860 --> 00:16:04,020 Essential and displacement mean the same 281 00:16:04,020 --> 00:16:05,220 thing, in that sense. 282 00:16:05,220 --> 00:16:10,410 Well, then we invoke the stationality of pi. 283 00:16:10,410 --> 00:16:14,890 We are saying that del pi shall be equal to 0 for any 284 00:16:14,890 --> 00:16:20,500 arbitrary variations of u that satisfy, however, the 285 00:16:20,500 --> 00:16:22,590 essential boundary conditions. 286 00:16:22,590 --> 00:16:24,340 This boundary condition here. 287 00:16:24,340 --> 00:16:28,380 So this has to hold, this statement shall hold, for any 288 00:16:28,380 --> 00:16:30,930 arbitrary variations in u. 289 00:16:30,930 --> 00:16:36,460 However, del u 0 shall be 0, that satisfy, in other words, 290 00:16:36,460 --> 00:16:39,980 the essential boundary condition. 291 00:16:39,980 --> 00:16:46,330 Well, if we apply this variation on pi, we obtain 292 00:16:46,330 --> 00:16:49,920 directly this result here. 293 00:16:49,920 --> 00:16:56,600 Notice that this part here is obtained by applying the 294 00:16:56,600 --> 00:17:00,750 variation on this part here. 295 00:17:00,750 --> 00:17:04,430 The variation operator operates, del operates, much 296 00:17:04,430 --> 00:17:06,730 in the same way as a differential operator. 297 00:17:06,730 --> 00:17:11,990 So this tool cancels this at 1/2, and we are left with EA 298 00:17:11,990 --> 00:17:15,849 del u del x times the variation on del u del x. 299 00:17:15,849 --> 00:17:18,819 And that is given right there. 300 00:17:18,819 --> 00:17:26,119 The variation on this part here gives us simply a del u 301 00:17:26,119 --> 00:17:32,370 times fB, and a del uL times R. Of course, uL is equal to 302 00:17:32,370 --> 00:17:37,980 is the displacement at x equal to L. And this is, therefore, 303 00:17:37,980 --> 00:17:39,290 the final result. 304 00:17:39,290 --> 00:17:45,360 Now if we look at this, we recognize, really, that this 305 00:17:45,360 --> 00:17:48,360 is the principle of virtual displacement. 306 00:17:48,360 --> 00:17:50,180 It's a principle of virtual displacement 307 00:17:50,180 --> 00:17:51,420 governing the problem. 308 00:17:51,420 --> 00:17:55,760 In general, we can write this principle as follows. 309 00:17:55,760 --> 00:17:59,400 Notice that here, we have variations in strains, which 310 00:17:59,400 --> 00:18:02,270 is that part there. 311 00:18:02,270 --> 00:18:08,610 The real stresses are given there, which are those. 312 00:18:08,610 --> 00:18:14,310 And here we have variations in displacement, those operating 313 00:18:14,310 --> 00:18:17,120 on the body forces. 314 00:18:17,120 --> 00:18:19,030 There, there. 315 00:18:19,030 --> 00:18:22,640 Notice, I sum here over all body forces in general. 316 00:18:22,640 --> 00:18:24,360 We have three components in fB-- 317 00:18:24,360 --> 00:18:27,030 the x, y, and z component. 318 00:18:27,030 --> 00:18:31,460 So I list these components in a vector that I call fB. 319 00:18:31,460 --> 00:18:34,560 Similarly, of course, we have three displacement components 320 00:18:34,560 --> 00:18:40,520 that appear here in this vector U. Tau in general has 6 321 00:18:40,520 --> 00:18:41,870 components. 322 00:18:41,870 --> 00:18:44,750 Del epsilon also has 6 components. 323 00:18:44,750 --> 00:18:47,830 Putting their transpose on the del epsilon vector means that 324 00:18:47,830 --> 00:18:49,380 we're summing the product of the 325 00:18:49,380 --> 00:18:51,230 strains times these stresses. 326 00:18:51,230 --> 00:18:54,330 Similarly, we are summing here the product of the 327 00:18:54,330 --> 00:18:57,620 displacement components times the force components. 328 00:18:57,620 --> 00:19:00,445 We also have here a contribution 329 00:19:00,445 --> 00:19:02,270 due to surface forces. 330 00:19:02,270 --> 00:19:04,420 fS are the surface forces. 331 00:19:04,420 --> 00:19:07,540 3 components, again, these are the variations in the surface 332 00:19:07,540 --> 00:19:08,610 displacement. 333 00:19:08,610 --> 00:19:13,510 So this part here, del uLR, really corresponds to this 334 00:19:13,510 --> 00:19:16,200 part here, involving surface forces. 335 00:19:16,200 --> 00:19:19,450 So the surface forces read again, and the variations in 336 00:19:19,450 --> 00:19:22,430 the surface displacements, are those here. 337 00:19:22,430 --> 00:19:26,240 Now notice that this principle here, or this equation, I 338 00:19:26,240 --> 00:19:30,540 should rather say, once again has to be satisfied for any 339 00:19:30,540 --> 00:19:35,810 arbitrary variations in displacements that satisfy the 340 00:19:35,810 --> 00:19:37,610 essential boundary conditions. 341 00:19:37,610 --> 00:19:42,100 For the problem of the rod, these del u's have to satisfy 342 00:19:42,100 --> 00:19:45,840 the condition that the variation at the left-hand 343 00:19:45,840 --> 00:19:48,670 boundary on u is 0, because that is the 344 00:19:48,670 --> 00:19:50,390 essential boundary condition. 345 00:19:50,390 --> 00:19:54,880 Also, of course, notice that this variation in the u's 346 00:19:54,880 --> 00:19:59,750 corresponds to these variations in the strains. 347 00:19:59,750 --> 00:20:05,230 In other words, these strains here are obtained from the 348 00:20:05,230 --> 00:20:08,650 variations in the displacement. 349 00:20:08,650 --> 00:20:09,460 That's important. 350 00:20:09,460 --> 00:20:10,630 They are linked together. 351 00:20:10,630 --> 00:20:13,980 So if we impose certain variations in the 352 00:20:13,980 --> 00:20:17,520 displacement, we have to impose here the corresponding 353 00:20:17,520 --> 00:20:19,981 variations in strains. 354 00:20:19,981 --> 00:20:22,770 Of course, these variations in the displacements give us also 355 00:20:22,770 --> 00:20:24,590 variations in surface displacement. 356 00:20:24,590 --> 00:20:28,580 So these here are again linked up with that. 357 00:20:28,580 --> 00:20:32,090 Later on in our formulation of the finite element method, we 358 00:20:32,090 --> 00:20:35,990 will write the variations in the strains here as virtual 359 00:20:35,990 --> 00:20:40,210 strains, arbitrary virtual strains. 360 00:20:40,210 --> 00:20:42,490 And we are talking about the strain vector with a 361 00:20:42,490 --> 00:20:43,830 bar on top of it. 362 00:20:43,830 --> 00:20:46,650 Similarly, a bar here, a bar here, instead of 363 00:20:46,650 --> 00:20:48,050 the variation sign. 364 00:20:48,050 --> 00:20:50,900 However, the meaning is quite identical-- 365 00:20:50,900 --> 00:20:53,030 what we are saying here, and this is the principle of 366 00:20:53,030 --> 00:20:57,030 virtual displacement that we will be talking about later 367 00:20:57,030 --> 00:21:01,050 on, when we formulate the finite element equations, what 368 00:21:01,050 --> 00:21:04,190 we are saying here really is that this equation has to be 369 00:21:04,190 --> 00:21:09,860 satisfied for any arbitrary virtual displacements and 370 00:21:09,860 --> 00:21:12,550 corresponding virtual strains. 371 00:21:12,550 --> 00:21:15,260 However, the virtual displacements have to satisfy 372 00:21:15,260 --> 00:21:17,590 the displacement boundary conditions, the essential 373 00:21:17,590 --> 00:21:19,280 boundary conditions. 374 00:21:19,280 --> 00:21:23,880 Well, from this, or rather that, which of course these 375 00:21:23,880 --> 00:21:25,810 equations are completely equivalent, we 376 00:21:25,810 --> 00:21:29,120 can go one step further. 377 00:21:29,120 --> 00:21:34,600 And if we go one step further, by applying integration by 378 00:21:34,600 --> 00:21:42,050 parts and recognizing that this part here is completely 379 00:21:42,050 --> 00:21:46,780 equal to that, by that I mean taking the variation on the 380 00:21:46,780 --> 00:21:52,518 derivative of x is the same as taking the sorry... by taking 381 00:21:52,518 --> 00:21:58,290 variation on the derivative of u with respect to x, that is 382 00:21:58,290 --> 00:22:03,020 completely identical to first taking the variation on u, and 383 00:22:03,020 --> 00:22:05,760 then taking the differentiation of that 384 00:22:05,760 --> 00:22:08,530 variation on u with respect to x. 385 00:22:08,530 --> 00:22:10,770 Well, if we recognize that these two things are 386 00:22:10,770 --> 00:22:19,200 identical, then using integration by parts on this 387 00:22:19,200 --> 00:22:24,210 equation here, which means, really, on that equation here, 388 00:22:24,210 --> 00:22:26,560 for the special case of the rod that we're now 389 00:22:26,560 --> 00:22:32,680 considering, we directly obtain this equation here. 390 00:22:32,680 --> 00:22:38,060 The integration by parts is performed by integrating this 391 00:22:38,060 --> 00:22:41,830 relation here first, and notice that if we do integrate 392 00:22:41,830 --> 00:22:45,910 this, we want to lower the differentiation of the virtual 393 00:22:45,910 --> 00:22:50,230 part here and increase the order of differentiation on 394 00:22:50,230 --> 00:22:52,210 this part here. 395 00:22:52,210 --> 00:22:58,620 And that then directly gives us these two terms 396 00:22:58,620 --> 00:23:01,700 and that term here. 397 00:23:01,700 --> 00:23:08,830 If we then also list this part here together with what we 398 00:23:08,830 --> 00:23:13,810 obtain from that part, we directly obtain this equation 399 00:23:13,810 --> 00:23:18,680 here, and we proceed similarly for the coefficients of del 400 00:23:18,680 --> 00:23:21,500 uL, and we obtain this part here. 401 00:23:21,500 --> 00:23:26,230 So this equation here is obtained by simply using 402 00:23:26,230 --> 00:23:30,970 integration by parts on, I repeat, this equation here. 403 00:23:30,970 --> 00:23:34,420 We have not used any assumption. 404 00:23:34,420 --> 00:23:37,980 All we did is a mathematical manipulation of this equation 405 00:23:37,980 --> 00:23:39,980 in a different form. 406 00:23:39,980 --> 00:23:42,500 And this is the new form that we obtained. 407 00:23:42,500 --> 00:23:49,260 Del u0, of course, is the variation of u at x equals 0. 408 00:23:49,260 --> 00:23:53,500 Now when we look at this relation, we can extract now 409 00:23:53,500 --> 00:23:57,310 the differential equation of equilibrium and the natural or 410 00:23:57,310 --> 00:23:59,230 force boundary conditions. 411 00:23:59,230 --> 00:24:00,350 How do we extract them? 412 00:24:00,350 --> 00:24:03,970 Well, the first point is that I mentioned already earlier, 413 00:24:03,970 --> 00:24:09,690 this part here is 0, because del u0 is imposed to be 0. 414 00:24:09,690 --> 00:24:11,310 So that part is 0. 415 00:24:11,310 --> 00:24:13,850 We can strike it out directly. 416 00:24:13,850 --> 00:24:17,770 Now when we look at this part here and that part here and 417 00:24:17,770 --> 00:24:23,570 recognize that del u is now arbitrary, we can directly 418 00:24:23,570 --> 00:24:28,420 extract the relation that this must be 0 and that must be 0. 419 00:24:28,420 --> 00:24:29,430 How this is done? 420 00:24:29,430 --> 00:24:31,890 Well, first of all, we recognize that this 421 00:24:31,890 --> 00:24:37,370 integration here really goes from 0 plus to L minus, if you 422 00:24:37,370 --> 00:24:39,320 want to be really exact. 423 00:24:39,320 --> 00:24:43,020 Because at the boundary, we have the boundary conditions. 424 00:24:43,020 --> 00:24:48,140 Of course, this 0 plus means it is infinitesimally close to 425 00:24:48,140 --> 00:24:53,350 0, and L minus, we are integrating up to a distance 426 00:24:53,350 --> 00:24:59,680 infinitesimally small to actually L. So putting the 0 427 00:24:59,680 --> 00:25:03,010 plus and L minus here is just for conceptual understanding 428 00:25:03,010 --> 00:25:04,260 really necessary. 429 00:25:04,260 --> 00:25:07,950 Well, if we then impose the following variations, say that 430 00:25:07,950 --> 00:25:14,680 del uL is 0, is exactly 0, then this part is out. 431 00:25:14,680 --> 00:25:18,190 And then we only have to look at this part. 432 00:25:18,190 --> 00:25:25,810 Now we can apply any arbitrary variation on u from 0 to L 433 00:25:25,810 --> 00:25:30,590 minus, and this, then, this total integration, must be 434 00:25:30,590 --> 00:25:31,490 equal to 0. 435 00:25:31,490 --> 00:25:33,920 It must satisfy the 0 condition. 436 00:25:33,920 --> 00:25:40,060 And that can only be true when this part here, what is in the 437 00:25:40,060 --> 00:25:41,340 brackets, is 0. 438 00:25:41,340 --> 00:25:45,470 Because if this is not 0, I can always select a certain 439 00:25:45,470 --> 00:25:50,310 delta u which, when multiplied by that and integrated as 440 00:25:50,310 --> 00:25:51,920 shown here-- 441 00:25:51,920 --> 00:25:54,080 remember, this part is not there-- 442 00:25:54,080 --> 00:25:55,840 will not give us 0. 443 00:25:55,840 --> 00:25:58,810 So therefore, this part must be 0. 444 00:25:58,810 --> 00:26:01,950 And that is our first condition, which is the 445 00:26:01,950 --> 00:26:05,960 equilibrium condition of a differential element. 446 00:26:05,960 --> 00:26:09,340 It's the equilibrium condition of a differential element. 447 00:26:09,340 --> 00:26:13,710 Now, if we say, let us look at the right-hand side boundary, 448 00:26:13,710 --> 00:26:17,520 and put del u0 everywhere along the length of the rod, 449 00:26:17,520 --> 00:26:28,310 except at x equals exactly L. Then this part would be 0, and 450 00:26:28,310 --> 00:26:33,780 this part here is non-zero, provided-- 451 00:26:36,350 --> 00:26:38,460 or rather, this part would be present. 452 00:26:38,460 --> 00:26:42,850 And the only way that this part can be 0, as it must be, 453 00:26:42,850 --> 00:26:46,850 is that this part here, the coefficient on del uL, is 454 00:26:46,850 --> 00:26:48,020 actually 0. 455 00:26:48,020 --> 00:26:52,810 So this way, we have extracted two points, two conditions. 456 00:26:52,810 --> 00:26:55,920 This part must be 0 and that part must be 0. 457 00:26:55,920 --> 00:26:58,830 We extract these conditions by looking at specific 458 00:26:58,830 --> 00:27:00,340 variations on u. 459 00:27:00,340 --> 00:27:05,610 First we look at the variation on u where del uL is 0, and 460 00:27:05,610 --> 00:27:11,000 otherwise arbitrary from x equals 0 to L minus. 461 00:27:11,000 --> 00:27:14,200 And then we can directly conclude this must be 0, and 462 00:27:14,200 --> 00:27:19,730 second time around, we say, let del u equal 0 from x 463 00:27:19,730 --> 00:27:23,380 equals 0 to L minus, which makes all of this integral 0, 464 00:27:23,380 --> 00:27:27,910 and we can focus all our attention on this part here, 465 00:27:27,910 --> 00:27:30,630 and we directly can conclude that this part now 466 00:27:30,630 --> 00:27:32,070 must also be 0. 467 00:27:32,070 --> 00:27:36,000 So this way, then, we extracted the differential 468 00:27:36,000 --> 00:27:39,490 equation of equilibrium and the natural boundary 469 00:27:39,490 --> 00:27:40,370 conditions. 470 00:27:40,370 --> 00:27:41,580 Very important. 471 00:27:41,580 --> 00:27:44,700 Differential equation of equilibrium and the natural 472 00:27:44,700 --> 00:27:47,540 boundary condition. 473 00:27:47,540 --> 00:27:49,780 And now we really recognize. 474 00:27:49,780 --> 00:27:54,280 of course, that if we put fB equal to the d'Alembert force, 475 00:27:54,280 --> 00:27:59,920 or minus the d'Alembert force, equal to this right-hand side, 476 00:27:59,920 --> 00:28:03,910 substitute from here back into there, cancel out A, we 477 00:28:03,910 --> 00:28:06,840 directly obtain this differential equation. 478 00:28:06,840 --> 00:28:09,565 Notice that c is E divided by rho square root 479 00:28:09,565 --> 00:28:10,860 E divided by rho. 480 00:28:10,860 --> 00:28:15,070 Now, the important point really is that by having 481 00:28:15,070 --> 00:28:25,400 started off with this pi functional and that condition, 482 00:28:25,400 --> 00:28:30,150 this condition also, and that the variations on u at x 483 00:28:30,150 --> 00:28:35,620 equals 0 shall be identically 0, we directly can extract 484 00:28:35,620 --> 00:28:38,820 from this pi functional the differential equation of 485 00:28:38,820 --> 00:28:41,850 equilibrium and the natural boundary conditions. 486 00:28:41,850 --> 00:28:44,070 The natural boundary conditions, in fact, are 487 00:28:44,070 --> 00:28:46,200 contained right in there. 488 00:28:46,200 --> 00:28:48,450 That's where the natural boundary conditions appear. 489 00:28:48,450 --> 00:28:51,250 The essential boundary conditions are there, and have 490 00:28:51,250 --> 00:28:53,940 to be satisfied by the variations. 491 00:28:53,940 --> 00:29:02,350 So in general, then, we find the following points. 492 00:29:02,350 --> 00:29:06,860 The important points are that by invoking del pi equals 0 493 00:29:06,860 --> 00:29:13,730 and using the essential boundary conditions only, we 494 00:29:13,730 --> 00:29:16,790 generate the principle of virtual displacement, an 495 00:29:16,790 --> 00:29:18,880 extremely important fact. 496 00:29:18,880 --> 00:29:23,680 And this will be the starting equation that we will be using 497 00:29:23,680 --> 00:29:27,210 to generate our finite element equations later on. 498 00:29:27,210 --> 00:29:29,570 We also can extract the problem governing 499 00:29:29,570 --> 00:29:32,030 differential equation. 500 00:29:32,030 --> 00:29:34,200 Therefore, the problem governing differential 501 00:29:34,200 --> 00:29:36,700 equation is contained in the principle of virtual 502 00:29:36,700 --> 00:29:40,990 displacement, and this one is contained in del pi equals 0, 503 00:29:40,990 --> 00:29:42,880 in the del pi equals 0 condition. 504 00:29:42,880 --> 00:29:47,610 Of course also satisfying the essential boundary condition. 505 00:29:47,610 --> 00:29:50,460 We also can extract the natural boundary conditions. 506 00:29:50,460 --> 00:29:54,770 So these are also contained, in essence, in pi, and as I 507 00:29:54,770 --> 00:29:58,270 showed to you, they're contained really in w. 508 00:29:58,270 --> 00:29:59,640 That's where they appear. 509 00:29:59,640 --> 00:30:03,310 Now in the derivation of the problem governing differential 510 00:30:03,310 --> 00:30:06,630 equation, we used integration by parts. 511 00:30:06,630 --> 00:30:09,050 And the highest spatial derivative in 512 00:30:09,050 --> 00:30:11,700 pi is of order m. 513 00:30:11,700 --> 00:30:17,430 We used integration by parts m times, and what we're finding 514 00:30:17,430 --> 00:30:20,970 is that the highest spatial derivative in the problem 515 00:30:20,970 --> 00:30:24,350 governing differential equation is then 2m. 516 00:30:24,350 --> 00:30:27,840 And this then puts together the complete structure of the 517 00:30:27,840 --> 00:30:29,970 equations that we're talking about. 518 00:30:29,970 --> 00:30:32,730 Here I have another view graph that summarizes the 519 00:30:32,730 --> 00:30:34,880 process once more. 520 00:30:34,880 --> 00:30:38,060 We are starting with the total potential pi of the system 521 00:30:38,060 --> 00:30:41,350 being equal to the strain energy minus the total 522 00:30:41,350 --> 00:30:42,910 potential of the loads. 523 00:30:42,910 --> 00:30:46,070 We're using this condition and the essential boundary 524 00:30:46,070 --> 00:30:47,250 condition-- 525 00:30:47,250 --> 00:30:48,610 very important-- 526 00:30:48,610 --> 00:30:51,030 to generate the principle of virtual displacement. 527 00:30:51,030 --> 00:30:54,920 At this stage, we can solve the problem, and we will do so 528 00:30:54,920 --> 00:30:57,870 by using finite element methods. 529 00:30:57,870 --> 00:31:02,190 We can, however, also go on from the principle of virtual 530 00:31:02,190 --> 00:31:06,920 displacement using integration by parts, and then we would 531 00:31:06,920 --> 00:31:09,665 derive the differential equation of equilibrium and 532 00:31:09,665 --> 00:31:12,050 the natural boundary conditions. 533 00:31:12,050 --> 00:31:13,820 We can solve the problem. 534 00:31:13,820 --> 00:31:17,250 Well, we can solve the problem at this level really only when 535 00:31:17,250 --> 00:31:21,010 we can solve the differential equation of equilibrium, 536 00:31:21,010 --> 00:31:23,150 subject to the natural boundary conditions. 537 00:31:23,150 --> 00:31:27,290 And that we can really only do for very simple systems. 538 00:31:27,290 --> 00:31:32,550 Therefore, this process here, from here onward, can really 539 00:31:32,550 --> 00:31:36,000 be followed only for relatively simple systems. 540 00:31:36,000 --> 00:31:40,760 For complex systems, shell structures, complicated shell 541 00:31:40,760 --> 00:31:45,640 beam structures, plane stress systems, real engineering 542 00:31:45,640 --> 00:31:49,820 structural analysis systems, this is not possible, and we 543 00:31:49,820 --> 00:31:53,760 stop right there, and we solve our problem this way, using 544 00:31:53,760 --> 00:31:55,200 finite element methods. 545 00:31:55,200 --> 00:32:00,320 Now, the one important point, however, I want to make once 546 00:32:00,320 --> 00:32:01,600 more clear-- 547 00:32:01,600 --> 00:32:05,910 when we proceed this way, we are deriving the differential 548 00:32:05,910 --> 00:32:10,470 equation of equilibrium for each differential element. 549 00:32:10,470 --> 00:32:14,410 And this means that when we go this route for the simple 550 00:32:14,410 --> 00:32:19,045 systems that we can go this route, we satisfy the 551 00:32:19,045 --> 00:32:23,200 equilibrium condition on each differential element. 552 00:32:23,200 --> 00:32:26,910 However, when we go this route, we will find that we 553 00:32:26,910 --> 00:32:31,430 only satisfy the equilibrium conditions in a global sense, 554 00:32:31,430 --> 00:32:36,150 in an integrated sense, using finite element methods, and 555 00:32:36,150 --> 00:32:42,300 that to actually satisfy the equilibrium conditions in a 556 00:32:42,300 --> 00:32:47,130 local sense, meaning for each differential element, we have 557 00:32:47,130 --> 00:32:52,670 to use many elements, and only then, of course, our finite 558 00:32:52,670 --> 00:32:56,040 element solution will converge to the solution that we would 559 00:32:56,040 --> 00:32:59,660 have obtained solving the differential equation of 560 00:32:59,660 --> 00:33:00,950 equilibrium. 561 00:33:00,950 --> 00:33:05,270 Therefore, when we start from the principle of virtual 562 00:33:05,270 --> 00:33:09,880 displacement, if we, as I will show you, use many elements, 563 00:33:09,880 --> 00:33:13,440 we really obtain the same solution as here. 564 00:33:13,440 --> 00:33:16,040 However, if we do not use many elements, if we 565 00:33:16,040 --> 00:33:18,140 use a coarse mesh-- 566 00:33:18,140 --> 00:33:20,470 we will talk about a coarse mesh later on-- 567 00:33:20,470 --> 00:33:23,790 then we will see that we only satisfy the equilibrium 568 00:33:23,790 --> 00:33:26,605 conditions in a global sense for the complete structure, in 569 00:33:26,605 --> 00:33:29,410 an integrated sense for the complete structure. 570 00:33:29,410 --> 00:33:32,390 For each finite element, we will satisfy the equilibrium 571 00:33:32,390 --> 00:33:35,180 conditions, but we will not satisfy the equilibrium 572 00:33:35,180 --> 00:33:40,610 conditions accurately for each differential element-- 573 00:33:40,610 --> 00:33:46,760 dx, dy, dz being arbitrarily small-- 574 00:33:46,760 --> 00:33:50,230 in the continuous body. 575 00:33:50,230 --> 00:33:52,650 That will become clearer, then, when we actually go 576 00:33:52,650 --> 00:33:54,120 through an example. 577 00:33:54,120 --> 00:34:00,120 Now I mentioned earlier that a whole class of classical 578 00:34:00,120 --> 00:34:02,850 methods are the weighted residual methods. 579 00:34:02,850 --> 00:34:05,730 In the weighted residual methods, we proceed in the 580 00:34:05,730 --> 00:34:06,910 following way. 581 00:34:06,910 --> 00:34:10,139 We consider the steady state problem, which is given here, 582 00:34:10,139 --> 00:34:11,860 with these boundary conditions. 583 00:34:11,860 --> 00:34:15,840 Now this operator L2m phi equals R-- 584 00:34:15,840 --> 00:34:20,830 this L2m is the operator that governs the problem. 585 00:34:20,830 --> 00:34:24,920 Like in our particular rod problem, L2m would be equal to 586 00:34:24,920 --> 00:34:28,460 delta u delta x squared. 587 00:34:28,460 --> 00:34:32,500 So the highest derivative in this spacial operator being 588 00:34:32,500 --> 00:34:36,830 2m, in this case, 2, for our example. 589 00:34:36,830 --> 00:34:39,560 The boundary conditions can be written this way. 590 00:34:39,560 --> 00:34:43,380 And the basic step, then, in the weighted residual and the 591 00:34:43,380 --> 00:34:46,730 Ritz analysis is to assume a solution all of this form, 592 00:34:46,730 --> 00:34:51,550 where phi bar gives us the assumed solution ai are 593 00:34:51,550 --> 00:34:56,560 parameters that are unknown, and fi are bases functions. 594 00:34:56,560 --> 00:34:59,800 These are functions that have to be assumed. 595 00:34:59,800 --> 00:35:05,000 Well, there is, of course, considerable concern on what 596 00:35:05,000 --> 00:35:06,750 kind of functions to choose. 597 00:35:06,750 --> 00:35:10,220 In the weighted residual method, these functions here, 598 00:35:10,220 --> 00:35:14,070 if you directly operate on this equation here, should 599 00:35:14,070 --> 00:35:16,730 satisfy all boundary conditions. 600 00:35:16,730 --> 00:35:19,830 Notice that in the weighted residual method, here I'm 601 00:35:19,830 --> 00:35:24,940 really talking about going on this view graph through the 602 00:35:24,940 --> 00:35:27,480 differential equations of equilibrium and natural 603 00:35:27,480 --> 00:35:30,870 boundary conditions, deriving them, if you want, via this 604 00:35:30,870 --> 00:35:34,090 route, and then trying to solve the problem here 605 00:35:34,090 --> 00:35:36,370 numerically. 606 00:35:36,370 --> 00:35:38,340 We will actually, as I said earlier, in the finite element 607 00:35:38,340 --> 00:35:43,050 process, go this route, but we can also go that route with 608 00:35:43,050 --> 00:35:44,710 the weighted residual methods. 609 00:35:44,710 --> 00:35:49,930 In fact, there is a close relationship between using 610 00:35:49,930 --> 00:35:52,860 weighted residual methods at this level and the Ritz method 611 00:35:52,860 --> 00:35:56,230 at that level, the way I will be describing it later on. 612 00:35:56,230 --> 00:35:59,440 But in the weighted residual method, this is the 613 00:35:59,440 --> 00:36:01,340 assumption. 614 00:36:01,340 --> 00:36:06,000 And then if we look at this equation here, we can 615 00:36:06,000 --> 00:36:08,990 construct an error R, substituting 616 00:36:08,990 --> 00:36:10,780 from here into there. 617 00:36:10,780 --> 00:36:16,270 And that capital R error is given by this equation. 618 00:36:16,270 --> 00:36:18,540 Notice that this is, of course, our trial function 619 00:36:18,540 --> 00:36:19,550 that we have. 620 00:36:19,550 --> 00:36:24,290 And if the right-hand side is 0, everywhere over the domain, 621 00:36:24,290 --> 00:36:28,130 then of course our error would be 0, and we would have solved 622 00:36:28,130 --> 00:36:32,570 our equation that we're looking at here. 623 00:36:32,570 --> 00:36:34,470 That is the equation we want to solve. 624 00:36:34,470 --> 00:36:38,400 Since if all of these functions fi satisfy all of 625 00:36:38,400 --> 00:36:42,490 the boundary conditions, then we are satisfying these 626 00:36:42,490 --> 00:36:45,170 equations, and all we have to worry about further is to 627 00:36:45,170 --> 00:36:47,630 satisfy this equation. 628 00:36:47,630 --> 00:36:50,390 If R is 0, we would also satisfy the differential 629 00:36:50,390 --> 00:36:51,880 equation of equilibrium, and we would have, 630 00:36:51,880 --> 00:36:53,510 in fact, the solution. 631 00:36:53,510 --> 00:36:57,560 However, that, of course, would be a very lucky choice 632 00:36:57,560 --> 00:36:59,150 on the fi functions. 633 00:36:59,150 --> 00:37:02,350 In general, R will not be identically 0 634 00:37:02,350 --> 00:37:03,890 all over the domain. 635 00:37:03,890 --> 00:37:09,910 In fact, how much R, or how close R will be to 0 will of 636 00:37:09,910 --> 00:37:11,890 course depend on the ai. 637 00:37:11,890 --> 00:37:17,270 And this is then basically our objective, namely, to 638 00:37:17,270 --> 00:37:21,310 calculate ai values that are making this R, this left-hand 639 00:37:21,310 --> 00:37:24,350 side capital R, as close as possible to 0. 640 00:37:24,350 --> 00:37:25,660 And that can be achieved via the 641 00:37:25,660 --> 00:37:28,170 Galerkin method, for example. 642 00:37:28,170 --> 00:37:29,380 This is the basic process. 643 00:37:29,380 --> 00:37:33,150 Here we're substituting the R. These are the trial functions, 644 00:37:33,150 --> 00:37:36,290 and we're integrating the product of these over the 645 00:37:36,290 --> 00:37:37,830 total domain. 646 00:37:37,830 --> 00:37:41,340 The domain here, in the case of our rod, would simply be 647 00:37:41,340 --> 00:37:43,880 the volume of the rod. 648 00:37:43,880 --> 00:37:48,780 This is the mechanism that generates to us n equations in 649 00:37:48,780 --> 00:37:52,550 the trial parameters ai. 650 00:37:52,550 --> 00:37:55,770 In another approach, the least squares method, we would 651 00:37:55,770 --> 00:37:59,900 operate on the square of the error, and minimize the square 652 00:37:59,900 --> 00:38:02,750 of the error when integrated over the total domain with 653 00:38:02,750 --> 00:38:06,910 respect to the trial parameters ai. 654 00:38:06,910 --> 00:38:10,720 That again gives us n equations, and we set up these 655 00:38:10,720 --> 00:38:13,410 n equations just as we're doing here, to 656 00:38:13,410 --> 00:38:14,730 solve for the ai. 657 00:38:14,730 --> 00:38:19,820 Knowing, then, the ai, we can back substitute into this 658 00:38:19,820 --> 00:38:22,510 assumption here, and now we have our approximate 659 00:38:22,510 --> 00:38:24,320 solution, phi bar. 660 00:38:24,320 --> 00:38:27,590 If the trial functions have been selected to satisfy the 661 00:38:27,590 --> 00:38:30,770 boundary conditions, then of course phi bar will satisfy 662 00:38:30,770 --> 00:38:33,500 the boundary conditions. 663 00:38:33,500 --> 00:38:36,650 However, what phi bar will not satisfy exactly is this 664 00:38:36,650 --> 00:38:37,836 equation here. 665 00:38:37,836 --> 00:38:41,110 However, we have minimized the error in the satisfaction of 666 00:38:41,110 --> 00:38:45,790 this equation in some sense using the Galerkin method or 667 00:38:45,790 --> 00:38:47,280 the least squares method. 668 00:38:47,280 --> 00:38:52,950 These methods can also be extended when the fi trial 669 00:38:52,950 --> 00:38:57,460 functions do not satisfy all of the boundary conditions, 670 00:38:57,460 --> 00:38:59,620 namely, not the natural boundary conditions. 671 00:38:59,620 --> 00:39:02,410 They can be extended to their places, but classically, they 672 00:39:02,410 --> 00:39:06,640 have been used with trial functions that satisfy all 673 00:39:06,640 --> 00:39:08,140 boundary conditions. 674 00:39:08,140 --> 00:39:14,360 When we were to extend the Galerkin method for the case 675 00:39:14,360 --> 00:39:18,510 where the trial functions do not satisfy the natural 676 00:39:18,510 --> 00:39:22,430 boundary conditions, then we really talk basically about 677 00:39:22,430 --> 00:39:27,050 already a Ritz analysis, and that is the next procedure 678 00:39:27,050 --> 00:39:28,860 that I want to introduce to you. 679 00:39:28,860 --> 00:39:35,670 Now of course, if we wanted to really start deriving the Ritz 680 00:39:35,670 --> 00:39:39,310 method from the Galerkin approach-- 681 00:39:39,310 --> 00:39:42,020 in other words, if we wanted to derive this Ritz analysis 682 00:39:42,020 --> 00:39:44,580 method from the Galerkin approach, we would have to 683 00:39:44,580 --> 00:39:47,460 extend, first of all, this Galerkin approach to include 684 00:39:47,460 --> 00:39:49,440 the natural boundary conditions, and then we would 685 00:39:49,440 --> 00:39:54,340 have to perform integrations on this equation, and we would 686 00:39:54,340 --> 00:39:58,510 obtain the Ritz analysis method. 687 00:39:58,510 --> 00:40:03,240 The actual way of starting, of introducing the Ritz analysis 688 00:40:03,240 --> 00:40:07,790 method, is to introduce it as a separate tool. 689 00:40:07,790 --> 00:40:10,750 And it is introduced in the following way. 690 00:40:10,750 --> 00:40:15,370 Let pi be the functional of the C m minus 1 variational 691 00:40:15,370 --> 00:40:18,690 problem that is equivalent to the differential formulation 692 00:40:18,690 --> 00:40:20,360 that we talked about earlier. 693 00:40:20,360 --> 00:40:23,960 Now, the differential formulation that I talked 694 00:40:23,960 --> 00:40:26,120 about here is this one. 695 00:40:26,120 --> 00:40:27,380 That's a differential formulation. 696 00:40:30,440 --> 00:40:34,380 And as an example once again, the operator L2m is del 2 u 697 00:40:34,380 --> 00:40:37,400 del x squared. 698 00:40:37,400 --> 00:40:38,430 There's a constant here. 699 00:40:38,430 --> 00:40:40,610 We could also put an EA in front here, but 700 00:40:40,610 --> 00:40:42,080 that only is a constant. 701 00:40:42,080 --> 00:40:45,280 It doesn't change the character of the operator. 702 00:40:45,280 --> 00:40:47,690 But this is basically the operator for the problem that 703 00:40:47,690 --> 00:40:50,660 we considered, and of course, our boundary conditions are 704 00:40:50,660 --> 00:40:51,380 also there. 705 00:40:51,380 --> 00:41:00,270 So if we have a pi functional that is equivalent to the 706 00:41:00,270 --> 00:41:03,120 differential formulation given in those two equations that I 707 00:41:03,120 --> 00:41:06,350 just pointed out to you again, then in the Ritz method, we 708 00:41:06,350 --> 00:41:10,320 substitute the trial functions, phi bar-- 709 00:41:10,320 --> 00:41:12,780 let us look at them again, these are the trial 710 00:41:12,780 --> 00:41:14,230 functions-- 711 00:41:14,230 --> 00:41:21,870 into pi, and we generate n simultaneous equations for the 712 00:41:21,870 --> 00:41:27,660 parameters that appear in this assumption here by invoking 713 00:41:27,660 --> 00:41:29,770 the stationality of pi. 714 00:41:29,770 --> 00:41:36,890 Notice that by invoking that del pi is 0, we really say 715 00:41:36,890 --> 00:41:44,130 that del pi ai is 0 for all i. 716 00:41:44,130 --> 00:41:47,820 And that gives us the condition which we used to set 717 00:41:47,820 --> 00:41:51,740 up the individual equations that we need to set up to 718 00:41:51,740 --> 00:41:55,900 solve for the trial parameters ai. 719 00:41:55,900 --> 00:41:58,980 Well, let us look at some of the properties. 720 00:41:58,980 --> 00:42:02,080 The trial functions used in the Ritz analysis need only 721 00:42:02,080 --> 00:42:05,550 satisfy the essential boundary conditions, an extremely 722 00:42:05,550 --> 00:42:06,840 important fact. 723 00:42:06,840 --> 00:42:10,260 In the classical weighted residual method, as I pointed 724 00:42:10,260 --> 00:42:13,020 out, the trial function should satisfy all boundary 725 00:42:13,020 --> 00:42:13,730 conditions. 726 00:42:13,730 --> 00:42:16,540 Therefore, they can be very difficult to choose. 727 00:42:16,540 --> 00:42:19,750 In the Ritz analysis method, we only need to satisfy the 728 00:42:19,750 --> 00:42:22,080 essential boundary conditions. 729 00:42:22,080 --> 00:42:25,780 The application of del pi equals 0 generates the 730 00:42:25,780 --> 00:42:28,010 principle of virtual displacement. 731 00:42:28,010 --> 00:42:30,640 I mentioned that to you earlier already. 732 00:42:30,640 --> 00:42:34,460 And therefore, in effect, we use in the Ritz analysis this 733 00:42:34,460 --> 00:42:37,270 principle of virtual displacement. 734 00:42:37,270 --> 00:42:41,380 By invoking del pi equals 0, we minimize basically the 735 00:42:41,380 --> 00:42:45,670 violation of the internal equilibrium requirements and 736 00:42:45,670 --> 00:42:49,070 the violation of the natural boundary conditions. 737 00:42:49,070 --> 00:42:55,050 Well, remember that invoking del pi equal to 0 and then 738 00:42:55,050 --> 00:42:58,400 using integration by parts, we actually generate, we could 739 00:42:58,400 --> 00:43:02,020 generate the differential equations of equilibrium and 740 00:43:02,020 --> 00:43:04,160 the natural boundary conditions the way I've shown 741 00:43:04,160 --> 00:43:07,270 it to you, for the simple bar structure. 742 00:43:07,270 --> 00:43:09,910 Now, what I'm saying here is that we do not want to 743 00:43:09,910 --> 00:43:12,230 generate these differential equations of equilibrium and 744 00:43:12,230 --> 00:43:13,180 natural boundary conditions. 745 00:43:13,180 --> 00:43:17,140 However, please recognize that they are contained in the 746 00:43:17,140 --> 00:43:19,980 equation del pi equals 0. 747 00:43:19,980 --> 00:43:23,500 Therefore, by substituting our trial functions, we violate 748 00:43:23,500 --> 00:43:26,150 the internal equilibrium requirements and the natural 749 00:43:26,150 --> 00:43:29,410 boundary conditions, but we will see that we are 750 00:43:29,410 --> 00:43:34,630 minimizing that violation in these conditions here. 751 00:43:34,630 --> 00:43:36,930 Also, we will see that we generate a symmetric 752 00:43:36,930 --> 00:43:40,960 coefficient matrix, and K and the governing equations then, 753 00:43:40,960 --> 00:43:44,370 our KU equals R, that we want to solve. 754 00:43:44,370 --> 00:43:49,670 And that is really the basis of the finite element method 755 00:43:49,670 --> 00:43:52,370 for the analysis of continuous systems. 756 00:43:52,370 --> 00:43:56,180 Let me now go in detail through an example. 757 00:43:56,180 --> 00:44:01,330 Here we have is simple bar structure which has an area 1 758 00:44:01,330 --> 00:44:06,860 square centimeter from A to B, and from B to C, B being this 759 00:44:06,860 --> 00:44:08,995 point here where the area changes and C 760 00:44:08,995 --> 00:44:10,500 being that point there. 761 00:44:10,500 --> 00:44:13,850 From B to C, we have a varying area. 762 00:44:13,850 --> 00:44:16,420 This variation in the area is shown here. 763 00:44:16,420 --> 00:44:21,750 It's 1 plus y/40 squared is the area at any station y, y 764 00:44:21,750 --> 00:44:25,290 being measured from point B, as you can see here. 765 00:44:25,290 --> 00:44:27,200 The length here is 100 centimeters. 766 00:44:27,200 --> 00:44:29,340 This length is 80 centimeters. 767 00:44:29,340 --> 00:44:33,340 The structure is subjected to a load of 100 newtons here. 768 00:44:33,340 --> 00:44:36,930 Notice that this arrow really lies on top of that dashed 769 00:44:36,930 --> 00:44:40,390 line we just separated out for you to understand that there 770 00:44:40,390 --> 00:44:41,290 is this arrow. 771 00:44:41,290 --> 00:44:43,810 So this is the load applied at the 772 00:44:43,810 --> 00:44:46,820 mid-line of this structure. 773 00:44:46,820 --> 00:44:51,690 We assume once again for the structure also that there is 774 00:44:51,690 --> 00:44:54,840 only the following displacement mechanism. 775 00:44:54,840 --> 00:45:00,670 If a section was originally there, and it is a vertical 776 00:45:00,670 --> 00:45:04,260 section to the midline, then it has moved over, and I 777 00:45:04,260 --> 00:45:08,730 grossly exaggerate now, to this position, where this is 778 00:45:08,730 --> 00:45:11,530 the displacement that we're talking about u. 779 00:45:11,530 --> 00:45:15,050 Grossly exaggerated, of course. 780 00:45:15,050 --> 00:45:18,400 So we're having a bar structure subjected to a 781 00:45:18,400 --> 00:45:21,550 concentrated load fixed at the left end. 782 00:45:21,550 --> 00:45:27,360 And our objective now is to solve this structure, to solve 783 00:45:27,360 --> 00:45:32,300 for the unknown displacement u, being 0 here, of course as 784 00:45:32,300 --> 00:45:35,180 a function of x, when this structure is 785 00:45:35,180 --> 00:45:37,130 subjected to that load. 786 00:45:37,130 --> 00:45:41,690 Well, in the calculation of this example, I want to 787 00:45:41,690 --> 00:45:44,590 display to you as many of the concepts 788 00:45:44,590 --> 00:45:47,090 that we just discussed. 789 00:45:47,090 --> 00:45:50,580 Here we have pi being equal to this value here. 790 00:45:50,580 --> 00:45:53,050 The strain energy is given here. 791 00:45:53,050 --> 00:45:57,090 Notice this is 1/2 times the stress times the strain 792 00:45:57,090 --> 00:46:00,190 integrated over the volume of the structure. 793 00:46:00,190 --> 00:46:04,800 The integration goes from 0 to 180, because that is the 794 00:46:04,800 --> 00:46:06,750 length of that structure. 795 00:46:06,750 --> 00:46:11,860 The total potential of the external load is 100, which is 796 00:46:11,860 --> 00:46:15,270 the intensity of the load times the displacement at x 797 00:46:15,270 --> 00:46:18,910 equal to 180... at x equal to 180. 798 00:46:18,910 --> 00:46:22,120 Well, the essential boundary condition is that u is 799 00:46:22,120 --> 00:46:25,180 0 at x equals 0. 800 00:46:25,180 --> 00:46:28,260 I'd like to now consider two different cases 801 00:46:28,260 --> 00:46:29,880 for the Ritz analysis. 802 00:46:29,880 --> 00:46:34,580 In the case one I want to use a function that spans u, which 803 00:46:34,580 --> 00:46:36,550 spans continuously-- 804 00:46:36,550 --> 00:46:38,430 and let me draw it out here-- 805 00:46:38,430 --> 00:46:44,760 from u, from x equals 0, to x equals 180. 806 00:46:44,760 --> 00:46:46,530 That is the endpoint. 807 00:46:46,530 --> 00:46:51,510 So here we have this function, this part here and that part 808 00:46:51,510 --> 00:46:53,800 there, these are the two trial parameters that we 809 00:46:53,800 --> 00:46:55,010 want to solve for. 810 00:46:55,010 --> 00:46:58,200 And we will select them, we will calculate them, rather, 811 00:46:58,200 --> 00:47:00,820 using the Ritz analysis. 812 00:47:00,820 --> 00:47:09,860 Case two, I also use trial functions, but notice now that 813 00:47:09,860 --> 00:47:11,740 I'm performing the following. 814 00:47:11,740 --> 00:47:14,190 We have a domain AB-- 815 00:47:14,190 --> 00:47:15,680 let me go back once more-- 816 00:47:15,680 --> 00:47:19,510 a domain AB, and a domain BC. 817 00:47:19,510 --> 00:47:24,280 And I want to now use one function for AB and one 818 00:47:24,280 --> 00:47:26,820 function for BC. 819 00:47:26,820 --> 00:47:35,020 The AB function is simply this one here, a linear variation 820 00:47:35,020 --> 00:47:36,280 up to this point. 821 00:47:36,280 --> 00:47:40,220 Now notice that UB is our trial parameter-- 822 00:47:40,220 --> 00:47:43,700 that's the one we don't know, our trial function parameter. 823 00:47:43,700 --> 00:47:46,150 x/100 is simply is the function 824 00:47:46,150 --> 00:47:47,210 that I'm talking about. 825 00:47:47,210 --> 00:47:50,100 And notice that this function only is applicable for this 826 00:47:50,100 --> 00:47:54,580 domain where, let me put down here the length that is x 827 00:47:54,580 --> 00:47:58,270 equal to 100, and this is here x equal to 180. 828 00:47:58,270 --> 00:48:04,650 Now for this part here, I use this function here. 829 00:48:04,650 --> 00:48:07,680 Now notice what this function does. 830 00:48:07,680 --> 00:48:13,130 Well, if we look at this part here in front, it involves uB, 831 00:48:13,130 --> 00:48:15,980 which is also there, and it involves uC. 832 00:48:15,980 --> 00:48:19,770 uB, by the way, is the physical displacement right 833 00:48:19,770 --> 00:48:21,460 here into this direction. 834 00:48:21,460 --> 00:48:24,390 Of course, I'm plotting u upwards here to be able to 835 00:48:24,390 --> 00:48:25,420 show it to you. 836 00:48:25,420 --> 00:48:31,610 But the displacement, uB, is the displacement of this point 837 00:48:31,610 --> 00:48:33,340 B to the right. 838 00:48:33,340 --> 00:48:38,460 uC is the displacement of this point C to the right. 839 00:48:38,460 --> 00:48:42,730 Then we recognize that this part here corresponds really 840 00:48:42,730 --> 00:48:45,490 to a variation such as that. 841 00:48:45,490 --> 00:48:51,590 Notice when x is equal to 100, which is that point, this 842 00:48:51,590 --> 00:48:53,560 function here is 1. 843 00:48:53,560 --> 00:49:00,380 When x is equal to 180, which is that point there, this part 844 00:49:00,380 --> 00:49:04,870 is equal to 0, because 180 minus 100 is 80, divided by 80 845 00:49:04,870 --> 00:49:07,480 is 1, and 1 minus 1 is 0. 846 00:49:07,480 --> 00:49:10,580 So this dashed line corresponds to 847 00:49:10,580 --> 00:49:12,400 this function here. 848 00:49:12,400 --> 00:49:16,080 Let me put a dashed line underneath there. 849 00:49:16,080 --> 00:49:21,470 Well, if we now look at this part here, we notice that this 850 00:49:21,470 --> 00:49:26,890 part is 0, or this trial function here is 0 at this 851 00:49:26,890 --> 00:49:31,540 point B. And it varies linearly like that across, 852 00:49:31,540 --> 00:49:36,660 where this part here, of course, denotes uC. 853 00:49:36,660 --> 00:49:41,270 That is this one here, solid black line, and here also, 854 00:49:41,270 --> 00:49:42,530 solid black line. 855 00:49:42,530 --> 00:49:49,750 Now the superposition of both these functions, the dashed 856 00:49:49,750 --> 00:49:55,760 blue and the solid black line, give us this function here. 857 00:49:55,760 --> 00:50:01,420 So the actual function that I'm talking about is a linear 858 00:50:01,420 --> 00:50:04,950 variation along here, and a linear variation along here, 859 00:50:04,950 --> 00:50:07,610 where I plot it vertically up here-- 860 00:50:07,610 --> 00:50:12,300 uB and uC here. 861 00:50:12,300 --> 00:50:16,070 Now, this is a specific case that I want to draw your 862 00:50:16,070 --> 00:50:19,330 attention on, because this really corresponds, as we 863 00:50:19,330 --> 00:50:22,720 shall see, to a true finite element analysis. 864 00:50:22,720 --> 00:50:25,730 And the reason for it is that we're talking about one domain 865 00:50:25,730 --> 00:50:27,960 here and another domain there. 866 00:50:27,960 --> 00:50:32,890 And both of these domains are identified as finite elements. 867 00:50:32,890 --> 00:50:39,200 Well, the first step now is to use pi, invoke the 868 00:50:39,200 --> 00:50:43,250 stationality condition as we did earlier, and this gives us 869 00:50:43,250 --> 00:50:45,910 the principle of virtual displacement. 870 00:50:45,910 --> 00:50:47,620 I mentioned it earlier already. 871 00:50:47,620 --> 00:50:49,710 Our virtual strains are here. 872 00:50:49,710 --> 00:50:51,750 The stresses are here. 873 00:50:51,750 --> 00:50:55,290 The virtual work is on this side. 874 00:50:55,290 --> 00:50:57,380 I discussed it earlier already. 875 00:50:57,380 --> 00:51:01,650 We do not want to go now via this route. 876 00:51:01,650 --> 00:51:05,300 We first of all want to now obtain the exact solution. 877 00:51:05,300 --> 00:51:09,860 The exact solution is obtained by using integration by parts 878 00:51:09,860 --> 00:51:16,150 on delta pi, being, of course, equal to 0, and extracting the 879 00:51:16,150 --> 00:51:18,850 differential equation of equilibrium for each 880 00:51:18,850 --> 00:51:22,450 differential element in this structure. 881 00:51:22,450 --> 00:51:24,560 This means that if we are talking here about the 882 00:51:24,560 --> 00:51:27,050 differential element equilibrium of each 883 00:51:27,050 --> 00:51:30,880 differential element dx long anywhere along the structure, 884 00:51:30,880 --> 00:51:33,170 in other words, the equilibrium of typically an 885 00:51:33,170 --> 00:51:36,860 element like that. 886 00:51:36,860 --> 00:51:38,980 That is a differential equation of equilibrium. 887 00:51:38,980 --> 00:51:41,980 And we also, of course, have the natural boundary 888 00:51:41,980 --> 00:51:42,520 conditions. 889 00:51:42,520 --> 00:51:44,970 We can also derive the natural boundary conditions. 890 00:51:44,970 --> 00:51:47,910 The solution to this is obtained by integration, and 891 00:51:47,910 --> 00:51:50,380 this is the solution given. 892 00:51:50,380 --> 00:51:55,110 Well, the stresses, then, of course are obtained by 893 00:51:55,110 --> 00:51:58,430 differentiation of the u's to get strains, and multiplying 894 00:51:58,430 --> 00:52:01,100 those by E, and these are the stresses in the bar. 895 00:52:01,100 --> 00:52:04,490 These are the exact stresses in the bar that satisfy the 896 00:52:04,490 --> 00:52:07,190 differential equations of equilibrium and the natural 897 00:52:07,190 --> 00:52:08,260 boundary conditions. 898 00:52:08,260 --> 00:52:12,290 This is the exact solution of this bar problem, the way I 899 00:52:12,290 --> 00:52:13,750 have formulated it. 900 00:52:13,750 --> 00:52:15,970 Now will perform our Ritz analysis. 901 00:52:15,970 --> 00:52:19,840 In case one, we use pi equal to this. 902 00:52:19,840 --> 00:52:22,730 Notice that I have substituted now our trial functions 903 00:52:22,730 --> 00:52:27,350 corresponding to case one into the functional pi. 904 00:52:27,350 --> 00:52:32,910 That gives us this term, that term here, and that term here. 905 00:52:32,910 --> 00:52:36,930 Notice that I have broken up the integration from 0 to 100, 906 00:52:36,930 --> 00:52:44,010 and 100 to 180, because the area changes from over this 907 00:52:44,010 --> 00:52:47,200 length here, and only for that reason, really, I have broken 908 00:52:47,200 --> 00:52:49,520 up the integrations. 909 00:52:49,520 --> 00:52:50,740 Now this is pi. 910 00:52:50,740 --> 00:52:56,260 And if we now invoke, we can integrate this out, and then 911 00:52:56,260 --> 00:52:59,600 invoke that del pi shall be 0, we 912 00:52:59,600 --> 00:53:01,870 obtain this set of equations. 913 00:53:01,870 --> 00:53:05,850 We solve for a1 and a2, substitute back into our 914 00:53:05,850 --> 00:53:09,700 assumption that we had earlier, and we got this u. 915 00:53:09,700 --> 00:53:16,330 Notice that of course this u displacement does satisfy the 916 00:53:16,330 --> 00:53:17,800 essential boundary conditions. 917 00:53:17,800 --> 00:53:19,940 It does satisfy the essential boundary 918 00:53:19,940 --> 00:53:22,040 conditions at x equals 0. 919 00:53:22,040 --> 00:53:24,660 You can just substitute x equals 0, and you would see 920 00:53:24,660 --> 00:53:25,890 that u is 0. 921 00:53:25,890 --> 00:53:30,560 It does not satisfy, however, the natural boundary condition 922 00:53:30,560 --> 00:53:33,190 at x equal to 180. 923 00:53:33,190 --> 00:53:37,300 Sigma is given here, obtained by calculating the strains 924 00:53:37,300 --> 00:53:39,260 from here and multiplying by E-- 925 00:53:39,260 --> 00:53:43,240 this is our approximate solution to the problem. 926 00:53:43,240 --> 00:53:46,880 We are satisfying the compatibility conditions, 927 00:53:46,880 --> 00:53:50,280 because the bar has remained together. 928 00:53:50,280 --> 00:53:54,340 No material has been cut away from it. 929 00:53:54,340 --> 00:53:59,380 Also, we are satisfying the constitutive relations, but we 930 00:53:59,380 --> 00:54:03,920 do not satisfy the internal equilibrium on a differential 931 00:54:03,920 --> 00:54:06,190 local elements sense. 932 00:54:06,190 --> 00:54:09,700 We do not satisfy the differential equilibrium, and 933 00:54:09,700 --> 00:54:11,880 we do not satisfy the natural boundary conditions. 934 00:54:11,880 --> 00:54:15,330 But we satisfy them in an approximate sense. 935 00:54:15,330 --> 00:54:16,370 Case two. 936 00:54:16,370 --> 00:54:21,450 Here now we're talking about our two linear functions. 937 00:54:21,450 --> 00:54:24,620 And here we naturally integrate from 0 to 100 for 938 00:54:24,620 --> 00:54:29,040 the first linear function, and from 100 to 180 for the second 939 00:54:29,040 --> 00:54:30,760 linear function. 940 00:54:30,760 --> 00:54:36,040 Notice this is, again, the area, and notice that this is 941 00:54:36,040 --> 00:54:37,435 here the strain squared. 942 00:54:40,190 --> 00:54:42,980 It's strain squared here because our E is out there, 943 00:54:42,980 --> 00:54:45,510 which would give us the stress. 944 00:54:45,510 --> 00:54:51,310 And the area here, of course, is equal to 1, which we did 945 00:54:51,310 --> 00:54:52,280 not write down. 946 00:54:52,280 --> 00:54:55,040 The important point is that this is now our pi for these 947 00:54:55,040 --> 00:54:55,650 two functions. 948 00:54:55,650 --> 00:54:59,690 We again invoke del pi equal to 0. 949 00:54:59,690 --> 00:55:01,820 We obtain now this set of equations. 950 00:55:01,820 --> 00:55:04,010 We are solving from this set of equations uB 951 00:55:04,010 --> 00:55:07,000 and uC, given here. 952 00:55:07,000 --> 00:55:10,320 Having got uB and uC, of course we now have the 953 00:55:10,320 --> 00:55:14,240 complete displacements along the bar, because we only need 954 00:55:14,240 --> 00:55:19,500 to substitute back into our original approximations that 955 00:55:19,500 --> 00:55:22,280 we looked at earlier. 956 00:55:22,280 --> 00:55:25,320 Let me just get them once more here. 957 00:55:25,320 --> 00:55:27,370 We had them here. 958 00:55:27,370 --> 00:55:31,840 If we now substitute from uB and uC into these two 959 00:55:31,840 --> 00:55:33,360 equations, we have the complete 960 00:55:33,360 --> 00:55:35,170 displacement solution. 961 00:55:35,170 --> 00:55:36,510 Of course, this is an approximate 962 00:55:36,510 --> 00:55:37,720 displacement solution. 963 00:55:37,720 --> 00:55:40,650 And similarly, our stresses are approximate. 964 00:55:40,650 --> 00:55:45,800 Now on these last few graphs, I have plotted the solution. 965 00:55:45,800 --> 00:55:50,650 And notice that this is here the direction x. 966 00:55:50,650 --> 00:55:55,200 Here we have the point B, here we have the point C, here we 967 00:55:55,200 --> 00:56:00,400 have the point A. Our exact solution, which satisfies the 968 00:56:00,400 --> 00:56:04,300 constitutive relations, compatibility relations, and 969 00:56:04,300 --> 00:56:07,270 the differential equations of equilibrium, and all bounded 970 00:56:07,270 --> 00:56:09,780 conditions, is the solid line here. 971 00:56:09,780 --> 00:56:13,150 Our solution one, case one, Ritz analysis, is the dashed 972 00:56:13,150 --> 00:56:17,790 line here, and the solution two is this dashed dotted 973 00:56:17,790 --> 00:56:19,350 line, down there. 974 00:56:19,350 --> 00:56:22,610 Notice that we are quite close in our Ritz analysis to the 975 00:56:22,610 --> 00:56:25,370 exact solution in the displacement. 976 00:56:25,370 --> 00:56:29,100 However, the strains and stresses are obtained by the 977 00:56:29,100 --> 00:56:32,530 differentiation of these displacement solutions, and 978 00:56:32,530 --> 00:56:36,240 here I show to you the calculated stresses. 979 00:56:36,240 --> 00:56:40,620 Again, point A here, point B here, point C there. 980 00:56:40,620 --> 00:56:43,430 The important point is the following. 981 00:56:43,430 --> 00:56:49,430 In the exact solution, we have the stress of 100 in domain 982 00:56:49,430 --> 00:56:52,810 AB, and then we have this curve here, a 983 00:56:52,810 --> 00:56:54,670 very high slope there. 984 00:56:54,670 --> 00:56:59,460 And in our solution one, we had this variation in stress. 985 00:56:59,460 --> 00:57:02,340 Notice that it goes continuously over the complete 986 00:57:02,340 --> 00:57:08,620 domain, because our assumed displacement function was 987 00:57:08,620 --> 00:57:11,640 continuous also over this domain, and its first 988 00:57:11,640 --> 00:57:15,100 derivative was continuous over this complete domain. 989 00:57:15,100 --> 00:57:18,860 So that's why our solution one is continuous there, and in 990 00:57:18,860 --> 00:57:21,220 fact, we're seeing just the straight line there, because 991 00:57:21,220 --> 00:57:23,660 our displacement approximation was parabolic. 992 00:57:23,660 --> 00:57:29,300 Our solution two is exact here, 100, and very 993 00:57:29,300 --> 00:57:32,350 approximate here for the displacement. 994 00:57:32,350 --> 00:57:35,830 But notice that at the midpoint between B and C, we 995 00:57:35,830 --> 00:57:39,300 get very good results. 996 00:57:39,300 --> 00:57:43,180 Now the important point really is shown here on 997 00:57:43,180 --> 00:57:44,820 the last view graph. 998 00:57:44,820 --> 00:57:48,520 We note that in this last analysis, we use trial 999 00:57:48,520 --> 00:57:50,840 functions that do not satisfy the natural boundary 1000 00:57:50,840 --> 00:57:54,680 condition, and I'm talking now about the piecewise linear 1001 00:57:54,680 --> 00:57:57,660 functions, in other words, from A to B and B to C each, 1002 00:57:57,660 --> 00:57:59,030 just a straight line. 1003 00:57:59,030 --> 00:58:01,440 We use trial functions that do not satisfy the natural 1004 00:58:01,440 --> 00:58:02,490 boundary conditions. 1005 00:58:02,490 --> 00:58:05,560 The trial functions themselves are continuous, but the 1006 00:58:05,560 --> 00:58:10,720 derivatives are discontinuous at point B. Notice our 1007 00:58:10,720 --> 00:58:15,080 stresses here are discontinuous at point B. 1008 00:58:15,080 --> 00:58:18,700 For a cm minus 1 variational problem, the way I've defined 1009 00:58:18,700 --> 00:58:22,150 it, we only need continuity in the m minus first derivatives 1010 00:58:22,150 --> 00:58:23,830 of the functions. 1011 00:58:23,830 --> 00:58:27,610 In this problem, m is 1, and therefore we only need 1012 00:58:27,610 --> 00:58:35,300 continuity in the functions themselves, and not in any 1013 00:58:35,300 --> 00:58:39,330 derivatives, because we only need continuity in the m minus 1014 00:58:39,330 --> 00:58:41,050 first derivative. 1015 00:58:41,050 --> 00:58:46,390 The domains A and B and B and C are finite elements, and in 1016 00:58:46,390 --> 00:58:49,850 actuality, we've performed a finite element analysis. 1017 00:58:49,850 --> 00:58:51,990 This is all I wanted to say in this lecture. 1018 00:58:51,990 --> 00:58:53,240 Thank you for your attention.