1 00:00:00,040 --> 00:00:02,470 The following content is provided under a Creative 2 00:00:02,470 --> 00:00:03,880 Commons license. 3 00:00:03,880 --> 00:00:06,920 Your support will help MIT OpenCourseWare continue to 4 00:00:06,920 --> 00:00:10,570 offer high quality educational resources for free. 5 00:00:10,570 --> 00:00:13,470 To make a donation, or view additional materials from 6 00:00:13,470 --> 00:00:17,400 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,400 --> 00:00:18,650 ocw.mit.edu. 8 00:00:21,290 --> 00:00:24,000 PROFESSOR: Ladies and gentlemen, welcome to 9 00:00:24,000 --> 00:00:25,840 lecture number 3. 10 00:00:25,840 --> 00:00:29,000 In the previous two lectures, we discussed some basic 11 00:00:29,000 --> 00:00:32,430 concepts related to finite element analysis. 12 00:00:32,430 --> 00:00:35,760 In this lecture, I would like to present to you a general 13 00:00:35,760 --> 00:00:38,640 formulation of the displacement-based finite 14 00:00:38,640 --> 00:00:40,500 element method. 15 00:00:40,500 --> 00:00:42,800 This is a very general formulation. 16 00:00:42,800 --> 00:00:47,510 We use it to analyze 1D, 2D, three-dimensional problems, 17 00:00:47,510 --> 00:00:49,770 plate and shell structures. 18 00:00:49,770 --> 00:00:53,340 And it provides the basis of almost all finite element 19 00:00:53,340 --> 00:00:57,470 analysis performed at present in practice. 20 00:00:57,470 --> 00:01:00,010 We will see that the formation is really a modern an 21 00:01:00,010 --> 00:01:03,410 application of the Ritz/Golerkin procedures that 22 00:01:03,410 --> 00:01:06,010 we discussed in the last lecture. 23 00:01:06,010 --> 00:01:09,400 We will consider in this lecture static and dynamic 24 00:01:09,400 --> 00:01:13,030 conditions, but as I pointed out earlier, we will only be 25 00:01:13,030 --> 00:01:16,750 concerned with linear analysis. 26 00:01:16,750 --> 00:01:21,990 On this first view graph, I've prepared schematically a 27 00:01:21,990 --> 00:01:25,220 sketch of a three-dimensional body. 28 00:01:25,220 --> 00:01:28,900 This three-dimensional body could represent, typically, a 29 00:01:28,900 --> 00:01:32,690 bridge, a shaft, a building-- 30 00:01:32,690 --> 00:01:35,050 whatever structure we want to analyze. 31 00:01:35,050 --> 00:01:38,710 And this three-dimensional body here is subject to the 32 00:01:38,710 --> 00:01:41,920 following forces-- it is subjected to concentrated 33 00:01:41,920 --> 00:01:51,190 forces, forces that have components, Fxi, Fyi, Fzi, at 34 00:01:51,190 --> 00:01:55,210 one point i, and there are many such points i. 35 00:01:55,210 --> 00:01:58,050 The body's also subjected to body force 36 00:01:58,050 --> 00:02:03,910 components, Fbx, Fby, Fbz. 37 00:02:03,910 --> 00:02:07,830 These are forces per unit volume. 38 00:02:07,830 --> 00:02:10,860 And we will see later on that we include in these forces the 39 00:02:10,860 --> 00:02:14,970 d'Alembert forces, when we consider dynamic analysis. 40 00:02:14,970 --> 00:02:20,200 The body is also subjected to distributed surface forces, 41 00:02:20,200 --> 00:02:26,690 with components, Fsx, Fsy, and Fsz. 42 00:02:26,690 --> 00:02:30,980 These surface forces would be, for example, distributed water 43 00:02:30,980 --> 00:02:35,570 pressure in a dam, frictional forces, et cetera. 44 00:02:35,570 --> 00:02:40,050 So we have, basically, concentrated surface forces, 45 00:02:40,050 --> 00:02:45,740 we have distributed surface forces, and volume forces-- 46 00:02:45,740 --> 00:02:50,150 externally applied volume forces, body forces. 47 00:02:50,150 --> 00:02:52,820 The body is, of course, also properly supported. 48 00:02:52,820 --> 00:02:55,840 We have here, typically, a support that prevents 49 00:02:55,840 --> 00:02:58,640 displacements in any direction. 50 00:02:58,640 --> 00:03:01,410 And here, we have another such support. 51 00:03:01,410 --> 00:03:03,910 Here, we have a roller support, which prevents 52 00:03:03,910 --> 00:03:07,650 displacements only in this direction. 53 00:03:07,650 --> 00:03:13,520 The body is defined in the coordinate system, XYZ, and 54 00:03:13,520 --> 00:03:16,476 notice that I'm using here capital XYZ's. 55 00:03:19,110 --> 00:03:24,240 And the displacements of the body are measured as U, V, and 56 00:03:24,240 --> 00:03:29,100 W into the capital X, Y, and Z directions. 57 00:03:29,100 --> 00:03:34,720 I'm using capital letters here to denote global displacements 58 00:03:34,720 --> 00:03:36,780 and global coordinates. 59 00:03:36,780 --> 00:03:40,580 We will later on in the finite element discretization also 60 00:03:40,580 --> 00:03:45,480 introduce small lowercase x, y and z's, and u, v, and w's to 61 00:03:45,480 --> 00:03:49,270 measure the displacement in the individual elements. 62 00:03:49,270 --> 00:03:53,300 So the problem is, other words, that we have this body, 63 00:03:53,300 --> 00:03:57,950 this general structure, subjected to certain forces, 64 00:03:57,950 --> 00:04:03,180 properly constrained, and we want to calculate the 65 00:04:03,180 --> 00:04:06,600 displacements of the body, the strains in the body, and the 66 00:04:06,600 --> 00:04:08,720 stresses, of course, in the body. 67 00:04:08,720 --> 00:04:12,200 Well, on this view graph, here, I listed the external 68 00:04:12,200 --> 00:04:18,010 forces once as vectors, here's our force FB, the body force 69 00:04:18,010 --> 00:04:21,860 per unit volume, with components into the x, y, and 70 00:04:21,860 --> 00:04:23,190 z directions. 71 00:04:23,190 --> 00:04:25,810 Here, we have the surface forces with components in the 72 00:04:25,810 --> 00:04:27,690 x, y, and z directions. 73 00:04:27,690 --> 00:04:30,430 And here, we have a typical concentrated force at that 74 00:04:30,430 --> 00:04:35,900 point i, with components FX, FY, and FZ again. 75 00:04:35,900 --> 00:04:39,250 The displacements of the body measured in the global 76 00:04:39,250 --> 00:04:43,130 coordinates are U, V and W, as shown here. 77 00:04:43,130 --> 00:04:46,680 Of course, notice that these U, V and W's are functions of 78 00:04:46,680 --> 00:04:50,570 the capital XYZ coordinates. 79 00:04:50,570 --> 00:04:54,180 The strains corresponding to these displacements, which of 80 00:04:54,180 --> 00:04:56,405 course, are known, are listed here. 81 00:04:56,405 --> 00:05:00,300 And the three-dimensional analysis, we have six such 82 00:05:00,300 --> 00:05:04,830 known strains, from epsilon XX to gamma ZX. 83 00:05:04,830 --> 00:05:07,100 Of course, the last three being the shearing strains, 84 00:05:07,100 --> 00:05:10,050 and the first three being the normal strains. 85 00:05:10,050 --> 00:05:13,720 The corresponding stresses are listed here. 86 00:05:13,720 --> 00:05:18,770 Again, six components from tau XX to tau ZX. 87 00:05:18,770 --> 00:05:21,290 Of course, if we were to actually analyze a 88 00:05:21,290 --> 00:05:24,120 two-dimensional problem, such as the plane stress problem, 89 00:05:24,120 --> 00:05:26,590 we would only use the appropriate quantities from 90 00:05:26,590 --> 00:05:30,880 here and from there, as we'll discuss later on. 91 00:05:30,880 --> 00:05:34,460 The starting point of our analysis, in which we want to 92 00:05:34,460 --> 00:05:37,700 calculate the stresses, the strains, and of course, the 93 00:05:37,700 --> 00:05:39,320 displacement also. 94 00:05:39,320 --> 00:05:42,560 The starting point is the principle of virtual 95 00:05:42,560 --> 00:05:43,630 displacements. 96 00:05:43,630 --> 00:05:46,910 Now, this is the principle which we already discussed 97 00:05:46,910 --> 00:05:49,050 very briefly in the last lecture. 98 00:05:49,050 --> 00:05:52,830 Remember, please that we can derive it by looking at the 99 00:05:52,830 --> 00:05:57,220 total potential of the system, which is given as the strain 100 00:05:57,220 --> 00:06:02,170 energy, minus the potential of the external loads, W, U being 101 00:06:02,170 --> 00:06:03,560 the strain energy. 102 00:06:03,560 --> 00:06:09,560 If we invoke the stationarity of pi, and we use the 103 00:06:09,560 --> 00:06:11,830 essential boundary conditions, which are the displacement 104 00:06:11,830 --> 00:06:16,340 boundary conditions, then we can derive the governing 105 00:06:16,340 --> 00:06:18,870 differential equations of equilibrium, and the force 106 00:06:18,870 --> 00:06:21,270 boundary conditions, the natural boundary conditions, 107 00:06:21,270 --> 00:06:23,520 as I have shown in the last lecture. 108 00:06:23,520 --> 00:06:27,570 Well, we will not derive these boundary conditions and the 109 00:06:27,570 --> 00:06:30,260 governing differential equations in this approach, 110 00:06:30,260 --> 00:06:33,770 but rather what we do is we invoke this principle, we set 111 00:06:33,770 --> 00:06:37,860 del pi equal to 0, and that gives us the principle of 112 00:06:37,860 --> 00:06:39,530 virtual displacements. 113 00:06:39,530 --> 00:06:43,040 And it is this principle which is a starting point all our 114 00:06:43,040 --> 00:06:45,190 finite element analysis. 115 00:06:45,190 --> 00:06:48,150 Let's recall once what does it mean. 116 00:06:48,150 --> 00:06:52,680 Well, here we have the body forces that I applied to the 117 00:06:52,680 --> 00:06:56,360 body, the surface forces that I applied to the body. 118 00:06:56,360 --> 00:07:00,180 These are externally applied loads and concentrated forces 119 00:07:00,180 --> 00:07:06,120 that are also applied to the body at the points, i. 120 00:07:06,120 --> 00:07:11,110 These forces are in equilibrium with 121 00:07:11,110 --> 00:07:13,690 the stresses, tau. 122 00:07:13,690 --> 00:07:16,930 Let's assume that we know the stresses, at this point. 123 00:07:16,930 --> 00:07:19,310 Then the principle states the following-- 124 00:07:19,310 --> 00:07:24,480 if we subject the body to any arbitrary virtual 125 00:07:24,480 --> 00:07:27,680 displacements, listed in here-- 126 00:07:27,680 --> 00:07:29,965 and I'm saying any arbitrary virtual displacements-- 127 00:07:35,570 --> 00:07:39,190 excuse me, that however satisfy the essential boundary 128 00:07:39,190 --> 00:07:41,710 conditions, and that means just the displacement boundary 129 00:07:41,710 --> 00:07:43,910 conditions. 130 00:07:43,910 --> 00:07:49,540 Then the work done by the loads, and that total work is 131 00:07:49,540 --> 00:07:50,260 given here. 132 00:07:50,260 --> 00:07:53,000 This is a virtual work because we are taking virtual 133 00:07:53,000 --> 00:07:58,350 displacements and subject the forces to these virtual 134 00:07:58,350 --> 00:07:59,560 displacements. 135 00:07:59,560 --> 00:08:04,840 Then the external virtual work done is equal to the internal 136 00:08:04,840 --> 00:08:11,150 virtual work done, which is obtained by multiplying the 137 00:08:11,150 --> 00:08:18,400 real stresses, which are in equilibrium with these 138 00:08:18,400 --> 00:08:22,660 extraordinary applied forces. 139 00:08:22,660 --> 00:08:27,190 Multiplying the real stresses by the virtual strains, which 140 00:08:27,190 --> 00:08:31,220 correspond to the virtual displacements. 141 00:08:31,220 --> 00:08:33,500 So let me use here a different color. 142 00:08:33,500 --> 00:08:39,159 These virtual strains correspond to these virtual 143 00:08:39,159 --> 00:08:42,780 displacements, and of course, these virtual displacements 144 00:08:42,780 --> 00:08:44,650 over the body-- 145 00:08:44,650 --> 00:08:48,600 these are, of course, a function of x, y and z. 146 00:08:48,600 --> 00:08:51,380 These virtual displacements over the body give us also 147 00:08:51,380 --> 00:08:54,590 virtual displacements on the surface of the body, which are 148 00:08:54,590 --> 00:08:55,910 listed in here. 149 00:08:55,910 --> 00:08:58,900 So let us put another arrow in there. 150 00:08:58,900 --> 00:09:04,000 And these virtual displacement also give us concentrated 151 00:09:04,000 --> 00:09:06,980 virtual displacements at those points where we have 152 00:09:06,980 --> 00:09:09,760 concentrated load supply. 153 00:09:09,760 --> 00:09:14,580 So once again, if we take the body and subject that body, 154 00:09:14,580 --> 00:09:20,820 who is in equilibrium under Fb, Fs, and Fi, with tau-- 155 00:09:20,820 --> 00:09:22,590 tau being the real stresses. 156 00:09:22,590 --> 00:09:26,440 If you take that body and subject it to any arbitrary 157 00:09:26,440 --> 00:09:29,460 virtual displacement that satisfy the displacement 158 00:09:29,460 --> 00:09:34,910 boundary conditions, then the external virtual work is equal 159 00:09:34,910 --> 00:09:37,750 to the internal virtual work. 160 00:09:37,750 --> 00:09:40,140 The internal virtual work being obtained by taking the 161 00:09:40,140 --> 00:09:43,830 real stresses, times the virtual strains, which 162 00:09:43,830 --> 00:09:47,200 correspond to the virtual displacements here, and 163 00:09:47,200 --> 00:09:51,120 integrating that product over the volume of the body. 164 00:09:51,120 --> 00:09:54,800 And the external pressure work is obtained by taking the real 165 00:09:54,800 --> 00:10:00,100 forces, multiplying these by the virtual displacements, and 166 00:10:00,100 --> 00:10:02,380 integrating these contributions over the 167 00:10:02,380 --> 00:10:04,770 complete body. 168 00:10:04,770 --> 00:10:06,720 Physically, what does this mean? 169 00:10:06,720 --> 00:10:11,180 Here, we have again, our general body. 170 00:10:11,180 --> 00:10:14,440 Let's see once, pictorially, what we're doing. 171 00:10:14,440 --> 00:10:18,630 Well, let's take a certain virtual displacement, which I 172 00:10:18,630 --> 00:10:20,210 depict here. 173 00:10:20,210 --> 00:10:23,650 Now, here, we have a boundary condition, so this point, P 174 00:10:23,650 --> 00:10:27,010 can only move over to there. 175 00:10:27,010 --> 00:10:30,080 It could not move this way because we have to satisfy, in 176 00:10:30,080 --> 00:10:34,150 the virtual displacements, the actual displacement boundary 177 00:10:34,150 --> 00:10:35,480 conditions. 178 00:10:35,480 --> 00:10:39,160 Here, the point cannot move at all, and here, this point can 179 00:10:39,160 --> 00:10:40,050 also not move. 180 00:10:40,050 --> 00:10:43,500 So a typical set of virtual displacement 181 00:10:43,500 --> 00:10:44,940 might look like that. 182 00:10:49,030 --> 00:10:51,960 Just sketched in here. 183 00:10:51,960 --> 00:10:55,230 This roller out has moved over there. 184 00:10:55,230 --> 00:11:00,040 So there's our new roller right there. 185 00:11:00,040 --> 00:11:04,300 What I satisfy are the order displacement conditions. 186 00:11:04,300 --> 00:11:06,600 Only horizontal movement was possible here. 187 00:11:06,600 --> 00:11:09,260 No movement here, no movement here. 188 00:11:09,260 --> 00:11:14,300 The virtual displacement vector here is that one. u 189 00:11:14,300 --> 00:11:16,740 bar, for a particular point. 190 00:11:16,740 --> 00:11:19,170 And that is the point that I'm looking at. 191 00:11:19,170 --> 00:11:22,230 Then what the principle says, once again, is that if I take 192 00:11:22,230 --> 00:11:25,750 these virtual displacements, multiply them by the real 193 00:11:25,750 --> 00:11:29,870 forces, integrate that product over the total body-- 194 00:11:29,870 --> 00:11:32,530 that is my external virtual work, and that external 195 00:11:32,530 --> 00:11:35,460 virtual work shall be equal to the internal virtual work, 196 00:11:35,460 --> 00:11:39,120 which is obtained by taking the real stresses, which are 197 00:11:39,120 --> 00:11:42,330 in equilibrium with these externally applied loads. 198 00:11:42,330 --> 00:11:45,230 And multiplying these real stresses by the virtual 199 00:11:45,230 --> 00:11:48,860 strains that correspond to these virtual displacements, 200 00:11:48,860 --> 00:11:52,640 and integrating that product, as the internal virtual work 201 00:11:52,640 --> 00:11:54,260 over the whole body. 202 00:11:54,260 --> 00:11:57,680 This is an extremely powerful principle and an extremely 203 00:11:57,680 --> 00:12:01,090 important principle, and provides a basis of our finite 204 00:12:01,090 --> 00:12:02,610 element formulation. 205 00:12:02,610 --> 00:12:07,560 In our finite element analysis, we are proceeding in 206 00:12:07,560 --> 00:12:11,320 the following way-- we say, well, let us idealize this 207 00:12:11,320 --> 00:12:15,130 complete body as an assemblage of elements, and what I've 208 00:12:15,130 --> 00:12:19,120 done here is to draw one typical element. 209 00:12:19,120 --> 00:12:23,130 This is an 8-node element, a brick element, a distorted 210 00:12:23,130 --> 00:12:27,590 brick element, to make it a little bit more general. 211 00:12:27,590 --> 00:12:32,460 It is an 8-node element because we have four nodes on 212 00:12:32,460 --> 00:12:35,700 the top surface, and four nodes at the bottom surface. 213 00:12:35,700 --> 00:12:38,870 There's another node here. 214 00:12:38,870 --> 00:12:43,050 This element here undergoes certain 215 00:12:43,050 --> 00:12:44,720 displacements, of course. 216 00:12:44,720 --> 00:12:48,100 And what I will be doing is I will express the displacements 217 00:12:48,100 --> 00:12:51,480 in that element as a function of the letter coordinates 218 00:12:51,480 --> 00:12:54,050 system, x, y, and z. 219 00:12:54,050 --> 00:12:55,760 The displacement in the element being 220 00:12:55,760 --> 00:12:59,140 lower u, v, and w. 221 00:12:59,140 --> 00:13:01,250 If we idealize the total body as an 222 00:13:01,250 --> 00:13:03,250 assemblage of such elements-- 223 00:13:03,250 --> 00:13:05,820 in other words, there's another element coming in from 224 00:13:05,820 --> 00:13:09,040 the top, and another element coming in from the sides, from 225 00:13:09,040 --> 00:13:10,650 the four sides, and another element 226 00:13:10,650 --> 00:13:12,930 coming in from the bottom. 227 00:13:12,930 --> 00:13:16,760 So if we idealize the total body as an assemblage of such 228 00:13:16,760 --> 00:13:21,830 brick elements that lie next to each other, et cetera. 229 00:13:21,830 --> 00:13:25,570 And we express the displacement in each of these 230 00:13:25,570 --> 00:13:29,170 brick elements, as a function of the nodal point 231 00:13:29,170 --> 00:13:32,120 displacements, of the displacements of the corners 232 00:13:32,120 --> 00:13:36,620 of the bricks, then we can, of course, express the total 233 00:13:36,620 --> 00:13:40,350 displacement in the body as a function of the nodal point 234 00:13:40,350 --> 00:13:41,500 displacement. 235 00:13:41,500 --> 00:13:43,480 And that is the important step in the 236 00:13:43,480 --> 00:13:44,820 finite element analysis. 237 00:13:44,820 --> 00:13:50,100 That the displacements in each of these sub-domains or 238 00:13:50,100 --> 00:13:53,460 elements are expressed in terms of nodal point 239 00:13:53,460 --> 00:13:54,930 displacements. 240 00:13:54,930 --> 00:13:59,690 These corner nodes, as shown here, for the brick element. 241 00:13:59,690 --> 00:14:05,320 And then since the total body is made up of an assemblage of 242 00:14:05,320 --> 00:14:09,480 such brick elements, we can express the total displacement 243 00:14:09,480 --> 00:14:12,600 in the body as a functional of these nodal point 244 00:14:12,600 --> 00:14:14,490 displacements. 245 00:14:14,490 --> 00:14:17,310 And invokes the principle of virtual displacements. 246 00:14:17,310 --> 00:14:20,350 Now let's go into the actual specifics. 247 00:14:20,350 --> 00:14:25,070 Well, for element m, this might be element 10, m in that 248 00:14:25,070 --> 00:14:26,870 case would be equal to 10. 249 00:14:26,870 --> 00:14:29,210 Then we have the following relationship-- 250 00:14:29,210 --> 00:14:32,140 and this is the important assumption of the finite 251 00:14:32,140 --> 00:14:33,350 element discretization. 252 00:14:33,350 --> 00:14:35,725 We say that the displacements-- 253 00:14:35,725 --> 00:14:39,190 there are three displacements, U, V, and W, of course, now. 254 00:14:39,190 --> 00:14:44,350 For element m, U, V and W are listed in this vector u, are 255 00:14:44,350 --> 00:14:48,000 equal to a displacement interpolation matrix, Hm, 256 00:14:48,000 --> 00:14:52,760 which is a function of x, y, and z, times the nodal point 257 00:14:52,760 --> 00:14:53,910 displacements. 258 00:14:53,910 --> 00:14:57,980 And what I'm listing in this u hat vector are all the nodal 259 00:14:57,980 --> 00:15:00,880 point displacements that I've called in the finite element 260 00:15:00,880 --> 00:15:02,310 discretization. 261 00:15:02,310 --> 00:15:09,120 For this brick element here, we have eight nodes, and 24 262 00:15:09,120 --> 00:15:10,440 nodal point displacements. 263 00:15:10,440 --> 00:15:16,260 At each node, we have U, V and W. But notice, once again, 264 00:15:16,260 --> 00:15:20,260 there are other brick elements on top of it, on the sides, 265 00:15:20,260 --> 00:15:21,890 and below of it. 266 00:15:21,890 --> 00:15:25,190 And each of these brick elements, of course, has a set 267 00:15:25,190 --> 00:15:28,810 of such nodal point displacements. 268 00:15:28,810 --> 00:15:31,870 We notice, however, that the element below it here-- 269 00:15:31,870 --> 00:15:35,630 if I take my pen here, and draw in another element, we 270 00:15:35,630 --> 00:15:40,670 notice that that element has the same 271 00:15:40,670 --> 00:15:42,350 node as the top element. 272 00:15:42,350 --> 00:15:45,310 In other words, this node here is common to this top element 273 00:15:45,310 --> 00:15:46,580 and the bottom element. 274 00:15:46,580 --> 00:15:50,400 And that's where we have the coupling between elements. 275 00:15:50,400 --> 00:15:53,540 We will see that more distinctly later. 276 00:15:53,540 --> 00:15:57,070 So what I'm doing here is I express the displacements of 277 00:15:57,070 --> 00:16:01,900 element m as a function of all the nodal point displacements, 278 00:16:01,900 --> 00:16:06,420 and I'm listing here in u hat these displacements for N, 279 00:16:06,420 --> 00:16:07,760 capital N nodal points. 280 00:16:07,760 --> 00:16:09,810 We have this vector. 281 00:16:09,810 --> 00:16:13,500 Now, in general, later on, we would simply call all of these 282 00:16:13,500 --> 00:16:17,370 components, Ui's, and so our u hat here, would be 283 00:16:17,370 --> 00:16:18,630 written in this way. 284 00:16:18,630 --> 00:16:22,122 Notice I use the transpose, the capital T here, to denote 285 00:16:22,122 --> 00:16:23,510 the transpose of a vector. 286 00:16:23,510 --> 00:16:31,170 So this UN is equal to that W, capital N. That's just for 287 00:16:31,170 --> 00:16:33,240 ease of notation. 288 00:16:33,240 --> 00:16:35,670 This is our major assumption. 289 00:16:35,670 --> 00:16:37,560 This is the major assumption in the 290 00:16:37,560 --> 00:16:38,910 finite element analysis. 291 00:16:38,910 --> 00:16:43,370 We will have to define for each element this displacement 292 00:16:43,370 --> 00:16:45,620 interpolation matrix. 293 00:16:45,620 --> 00:16:52,350 We notice that when we define it, there will be many columns 294 00:16:52,350 --> 00:16:57,870 that are simply 0's because only certain displacements in 295 00:16:57,870 --> 00:17:01,500 this vector listed here, in this vector, really affect the 296 00:17:01,500 --> 00:17:02,930 displacement in an element. 297 00:17:02,930 --> 00:17:08,940 In other words, typically, for this element here, if we look 298 00:17:08,940 --> 00:17:13,560 at this node, then the displacement at this node do 299 00:17:13,560 --> 00:17:17,160 not affect the displacement in this element because this node 300 00:17:17,160 --> 00:17:21,140 does not belong to the element. 301 00:17:21,140 --> 00:17:25,089 That is recognized by the fact that in this Hm matrix there 302 00:17:25,089 --> 00:17:28,720 will be many columns that are simply 0's. 303 00:17:28,720 --> 00:17:32,710 In fact, the only non-zero columns in this Hm matrix are 304 00:17:32,710 --> 00:17:37,850 those that correspond to nodal points in this vector, or 305 00:17:37,850 --> 00:17:40,130 nodal point displacements in this vector that 306 00:17:40,130 --> 00:17:43,170 belong to element m. 307 00:17:43,170 --> 00:17:46,170 Well, having laid down this assumption, and we will 308 00:17:46,170 --> 00:17:48,960 define, once again, later on the Hm matrix 309 00:17:48,960 --> 00:17:50,560 for specific elements-- 310 00:17:50,560 --> 00:17:53,380 1D, 2D, 3D elements and so on. 311 00:17:53,380 --> 00:17:56,540 Having laid down this assumption, we can derive the 312 00:17:56,540 --> 00:17:59,920 strains and the strains are simply given by this 313 00:17:59,920 --> 00:18:04,610 relationship, with the Bm matrix is the strain 314 00:18:04,610 --> 00:18:07,670 interpolation matrix, of element m. 315 00:18:07,670 --> 00:18:13,150 Notice that the rows in this matrix are obtained by using 316 00:18:13,150 --> 00:18:14,950 the rows in the Hm. 317 00:18:14,950 --> 00:18:19,810 And differentiating these rows and combining these rows in 318 00:18:19,810 --> 00:18:22,380 the appropriate way. 319 00:18:22,380 --> 00:18:24,210 I show examples later on. 320 00:18:24,210 --> 00:18:27,930 There is no more assumption in this step. 321 00:18:27,930 --> 00:18:33,320 This Bm matrix is simply obtained from the Hm matrix. 322 00:18:33,320 --> 00:18:37,540 By recognizing what strains we are talking about, and by 323 00:18:37,540 --> 00:18:40,840 recognizing that we can simply use the rows here, 324 00:18:40,840 --> 00:18:44,770 differentiate them, linearally combine them, if necessary, to 325 00:18:44,770 --> 00:18:46,860 obtain the Bm matrix. 326 00:18:46,860 --> 00:18:50,330 Of course, we also have to use our stress strain law to 327 00:18:50,330 --> 00:18:53,430 obtain stresses from the strains, and the stresses in 328 00:18:53,430 --> 00:18:55,730 element m are given as shown here. 329 00:18:55,730 --> 00:18:58,120 These are the strains that I talked about here already. 330 00:18:58,120 --> 00:19:00,660 This is our stress strain law, which can vary 331 00:19:00,660 --> 00:19:03,370 from element to element. 332 00:19:03,370 --> 00:19:07,670 I also introduce here an initial stress, which might 333 00:19:07,670 --> 00:19:09,260 already be in the body. 334 00:19:09,260 --> 00:19:12,590 This might be due to overburden pressure in an 335 00:19:12,590 --> 00:19:16,440 underground structure, as an example, and so on. 336 00:19:16,440 --> 00:19:20,030 So, this is our stress strain law, which we have to satisfy 337 00:19:20,030 --> 00:19:21,520 for the body, of course. 338 00:19:21,520 --> 00:19:24,570 Our compatibility conditions in the analysis 339 00:19:24,570 --> 00:19:26,090 will also be satisfied. 340 00:19:26,090 --> 00:19:28,680 The strain compatibility conditions are satisfied 341 00:19:28,680 --> 00:19:34,210 because we are deriving the strains from continuous 342 00:19:34,210 --> 00:19:38,190 displacements, within the element. 343 00:19:41,880 --> 00:19:46,930 We'll impose on to the different elements that they 344 00:19:46,930 --> 00:19:49,610 remain compatible under deformations. 345 00:19:49,610 --> 00:19:53,610 By that, I mean if we have an element coming in here and 346 00:19:53,610 --> 00:19:57,860 another element going out there that the elements 347 00:19:57,860 --> 00:20:01,400 underloading, when the top element here, and the bottom 348 00:20:01,400 --> 00:20:03,080 element, both of them are loaded. 349 00:20:03,080 --> 00:20:07,290 No gap is opening up here so that displacement 350 00:20:07,290 --> 00:20:10,490 compatibility between the elements is satisfied. 351 00:20:10,490 --> 00:20:13,320 So if we look at the three conditions that we have to 352 00:20:13,320 --> 00:20:16,580 satisfy in an analysis-- 353 00:20:16,580 --> 00:20:18,910 the first one being the stress strain law. 354 00:20:18,910 --> 00:20:21,410 That is satisfied because we are using this equation. 355 00:20:21,410 --> 00:20:24,650 The second one being compatibility. 356 00:20:24,650 --> 00:20:29,590 That is satisfied because we using this relationship here 357 00:20:29,590 --> 00:20:37,010 to calculate our strains from the Hm, via the Bm matrix. 358 00:20:37,010 --> 00:20:40,410 And we are satisfying, of course, that the elements 359 00:20:40,410 --> 00:20:43,680 remain together, so no gaps opening up. 360 00:20:43,680 --> 00:20:45,830 We will be talking about it later on when we talk about 361 00:20:45,830 --> 00:20:47,770 the convergence requirements also of 362 00:20:47,770 --> 00:20:49,360 finite element analysis. 363 00:20:49,360 --> 00:20:52,300 And finally, our equilibrium condition has to be satisfied. 364 00:20:52,300 --> 00:20:54,990 That is the third condition where that equilibrium 365 00:20:54,990 --> 00:20:59,240 condition is embodied in the principle of virtual work. 366 00:20:59,240 --> 00:21:01,910 Here, I have it once again written down. 367 00:21:01,910 --> 00:21:04,040 The equilibrium conditions are in this 368 00:21:04,040 --> 00:21:06,500 principle of virtual work. 369 00:21:06,500 --> 00:21:11,280 And as I stated earlier, that if this equation is satisfied 370 00:21:11,280 --> 00:21:15,440 for any and all the arbitrary virtual displacements that 371 00:21:15,440 --> 00:21:18,560 satisfy the displacement boundary conditions, the real 372 00:21:18,560 --> 00:21:23,050 displacement boundary conditions, then tau is in 373 00:21:23,050 --> 00:21:25,910 equilibrium with Fb, Fs, and Fi. 374 00:21:25,910 --> 00:21:30,200 Well, what we will be doing is we will be applying this 375 00:21:30,200 --> 00:21:32,900 principle of virtual displacements for our finite 376 00:21:32,900 --> 00:21:36,853 element discretization, which means that in an integral 377 00:21:36,853 --> 00:21:40,710 sense, we satisfy equilibrium. 378 00:21:40,710 --> 00:21:45,250 However, if we look into an element, then within the 379 00:21:45,250 --> 00:21:48,440 element, we will only satisfy the differential equations of 380 00:21:48,440 --> 00:21:50,670 equilibrium in an approximate way. 381 00:21:50,670 --> 00:21:52,160 We will not satisfy them exactly. 382 00:21:52,160 --> 00:21:58,680 However, if we have a proper finite element discretization, 383 00:21:58,680 --> 00:22:01,180 and by that, I mean if we satisfy all the convergence 384 00:22:01,180 --> 00:22:07,150 requirements that have to be satisfied in order to obtain a 385 00:22:07,150 --> 00:22:10,300 valid solution, or a reliable solution in a finite element 386 00:22:10,300 --> 00:22:14,420 analysis, then we know that as our elements become smaller 387 00:22:14,420 --> 00:22:18,040 and smaller and smaller, we will be finally-- 388 00:22:18,040 --> 00:22:19,780 and always, of course, applying the principle of 389 00:22:19,780 --> 00:22:21,050 virtual displacements-- 390 00:22:21,050 --> 00:22:23,980 we will be finally satisfying, also, the differential 391 00:22:23,980 --> 00:22:27,830 equations of equilibrium locally within each element. 392 00:22:27,830 --> 00:22:31,510 So stress strain law is satisfied, compatibility is 393 00:22:31,510 --> 00:22:33,960 satisfied, both of them exactly. 394 00:22:33,960 --> 00:22:36,735 The equilibrium requirements are only satisfied in an 395 00:22:36,735 --> 00:22:40,230 integral sense, if we have a coarse finite element mesh, 396 00:22:40,230 --> 00:22:43,750 but as the finite elements become more and more, as we 397 00:22:43,750 --> 00:22:46,870 refine our finite element mesh, we will be satisfying 398 00:22:46,870 --> 00:22:48,660 the equilibrium requirements. 399 00:22:48,660 --> 00:22:53,460 Also locally, within an element, always closer and 400 00:22:53,460 --> 00:22:56,790 closer, and we will be approximating, or we will be 401 00:22:56,790 --> 00:22:59,260 getting closer to the satisfaction of the 402 00:22:59,260 --> 00:23:01,060 differential equation of equilibrium. 403 00:23:03,870 --> 00:23:07,080 The first step now is to rewrite this principle of 404 00:23:07,080 --> 00:23:11,640 virtual displacement, in this form, namely as a sum of 405 00:23:11,640 --> 00:23:14,010 integrations over the elements. 406 00:23:14,010 --> 00:23:15,030 There's no assumption yet. 407 00:23:15,030 --> 00:23:18,510 All I've done is since our total body is idealized as a 408 00:23:18,510 --> 00:23:24,150 sum of volumes, namely the volumes over the elements, I 409 00:23:24,150 --> 00:23:27,960 can rewrite the total integral as a the sum 410 00:23:27,960 --> 00:23:30,030 over the element integrals. 411 00:23:30,030 --> 00:23:31,610 And that's what I have done here. 412 00:23:31,610 --> 00:23:35,380 Notice we have now here, m, denoting element m and we are 413 00:23:35,380 --> 00:23:37,110 summing over all of the elements. 414 00:23:37,110 --> 00:23:39,260 There's no assumption here yet. 415 00:23:39,260 --> 00:23:43,060 Now, however, I can substitute our assumption. 416 00:23:43,060 --> 00:23:47,050 Namely, that Um is given in this way, and epsilon m is 417 00:23:47,050 --> 00:23:47,850 given that way. 418 00:23:47,850 --> 00:23:49,750 Once again, this is actually not an assumption. 419 00:23:49,750 --> 00:23:51,900 This is the major assumption. 420 00:23:51,900 --> 00:23:55,710 That epsilon m follows from this assumption. 421 00:23:55,710 --> 00:23:59,640 This equation follows from that equation entirely. 422 00:23:59,640 --> 00:24:03,550 Substituting now from here, these equations into the 423 00:24:03,550 --> 00:24:07,710 principle of virtual displacements, we directly 424 00:24:07,710 --> 00:24:10,420 obtain the following equations. 425 00:24:10,420 --> 00:24:14,660 Here, I have on the left hand side-- 426 00:24:14,660 --> 00:24:17,660 let's go through this equation in detail. 427 00:24:17,660 --> 00:24:22,560 On the left hand side, I have the following part. 428 00:24:22,560 --> 00:24:25,980 I have an integral over all of the elements, that is the 429 00:24:25,980 --> 00:24:30,090 integral here, summing over element m and integrating over 430 00:24:30,090 --> 00:24:32,980 each of the element. 431 00:24:32,980 --> 00:24:39,310 This part here, Bm transposed, times U hat T, is equal to the 432 00:24:39,310 --> 00:24:42,100 epsilon bar, mT. 433 00:24:42,100 --> 00:24:47,930 So notice that our epsilon bar m, is a virtual strain, is 434 00:24:47,930 --> 00:24:49,220 given in this way. 435 00:24:52,200 --> 00:24:55,980 Before, I talked only about the real strains. 436 00:24:55,980 --> 00:24:57,920 In other words, the bar was not there. 437 00:24:57,920 --> 00:24:59,620 I put that now into brackets here. 438 00:24:59,620 --> 00:25:01,840 This bar was not there. 439 00:25:01,840 --> 00:25:04,500 Well, what we're doing in the finite element analysis is to 440 00:25:04,500 --> 00:25:08,220 use the same assumption for the virtual strains as we use 441 00:25:08,220 --> 00:25:09,650 for real strains. 442 00:25:09,650 --> 00:25:12,790 In other words, we use this equation without the bar and 443 00:25:12,790 --> 00:25:15,740 with the bar, on top of the epsilon and on 444 00:25:15,740 --> 00:25:18,850 top of the u hat. 445 00:25:18,850 --> 00:25:22,210 So here, we have our epsilon bar transposed. 446 00:25:22,210 --> 00:25:25,360 This part here is the stress. 447 00:25:25,360 --> 00:25:28,380 Tau m equals Cm epsilon m-- 448 00:25:28,380 --> 00:25:29,330 that is our cm. 449 00:25:29,330 --> 00:25:33,330 The Bm here comes in from the epsilon m. 450 00:25:33,330 --> 00:25:37,400 So Bm times U hat is equal to epsilon m, once written down 451 00:25:37,400 --> 00:25:38,700 again here. 452 00:25:38,700 --> 00:25:43,190 So this total part here is nothing else than-- 453 00:25:43,190 --> 00:25:47,250 let's go back once more to the previous view graph-- 454 00:25:47,250 --> 00:25:52,530 nothing else than this part here, than that part there. 455 00:25:52,530 --> 00:25:57,240 But with the initial stress, tau i, m being on 456 00:25:57,240 --> 00:25:58,000 the right hand side. 457 00:25:58,000 --> 00:26:01,180 Since we do know the initial stress, we put that one, of 458 00:26:01,180 --> 00:26:02,580 course, on the right hand side. 459 00:26:02,580 --> 00:26:04,080 It is a load contribution. 460 00:26:04,080 --> 00:26:08,600 And here, you see it as a minus sign because we put on 461 00:26:08,600 --> 00:26:12,360 the right hand side, and this part here. 462 00:26:12,360 --> 00:26:15,150 This part times this u hat, notice there's 463 00:26:15,150 --> 00:26:16,400 a big bracket here. 464 00:26:19,550 --> 00:26:26,120 That u hat bar here operating on that Bm transposed, there's 465 00:26:26,120 --> 00:26:29,220 a transposed here, too, gives us again the epsilon bar m 466 00:26:29,220 --> 00:26:31,310 transposed. 467 00:26:31,310 --> 00:26:34,700 So let us look now at what we have on the right hand side. 468 00:26:34,700 --> 00:26:36,840 On the right hand side, I want to have 469 00:26:36,840 --> 00:26:40,540 discretized this part here. 470 00:26:40,540 --> 00:26:43,910 Well, what we do is we substitute here from our 471 00:26:43,910 --> 00:26:46,930 displacement interpolation, here from our displacement 472 00:26:46,930 --> 00:26:51,230 interpolation, and each of these integrals can directly 473 00:26:51,230 --> 00:26:54,380 be expressed, in terms of the nodal point displacements. 474 00:26:54,380 --> 00:26:55,915 And that's what we have done here. 475 00:26:55,915 --> 00:27:02,400 The Hm times the u hat bar is the u bar m transposed. 476 00:27:02,400 --> 00:27:07,900 Here, we have the u bar s. 477 00:27:07,900 --> 00:27:14,110 That is the u hat bar times the Hsm. 478 00:27:14,110 --> 00:27:16,790 Notice that this part, you should not have been 479 00:27:16,790 --> 00:27:17,990 written that far. 480 00:27:17,990 --> 00:27:19,660 In other words, there is the end. 481 00:27:19,660 --> 00:27:25,350 The u bar s m transposed only goes up to there and it 482 00:27:25,350 --> 00:27:31,610 embodies this Hs m transposed and the u hat bar transposed. 483 00:27:31,610 --> 00:27:33,830 This part we talked about already. 484 00:27:33,830 --> 00:27:36,850 So what we have done then is to rewrite-- 485 00:27:36,850 --> 00:27:38,460 this is the important part-- 486 00:27:38,460 --> 00:27:43,360 is to rewrite this principle of virtual displacements, in 487 00:27:43,360 --> 00:27:46,670 which we had no assumption yet. 488 00:27:46,670 --> 00:27:50,520 We rewrote this in terms of the nodal point displacements 489 00:27:50,520 --> 00:27:55,260 and element interpolation matrices that we use for our 490 00:27:55,260 --> 00:27:57,330 finite element discretization. 491 00:27:57,330 --> 00:27:59,800 Now, we of course, have the assumption that the 492 00:27:59,800 --> 00:28:04,090 displacements within each element are given by the Hm 493 00:28:04,090 --> 00:28:08,230 matrix, the strains are given by the Bm matrix. 494 00:28:08,230 --> 00:28:12,120 So this is the result that we have obtained. 495 00:28:12,120 --> 00:28:16,150 And at this point, we now invoke the principle of 496 00:28:16,150 --> 00:28:17,560 virtual displacements. 497 00:28:17,560 --> 00:28:23,590 Since this principle here shall hold for any arbitrary 498 00:28:23,590 --> 00:28:28,540 virtual displacements that satisfy the displacement 499 00:28:28,540 --> 00:28:31,640 boundary conditions, we can now invoke 500 00:28:31,640 --> 00:28:35,700 this principle n times. 501 00:28:35,700 --> 00:28:41,600 And by that, I mean, once by imposing a unit displacement 502 00:28:41,600 --> 00:28:44,330 at the first displacement degree of freedom, and leaving 503 00:28:44,330 --> 00:28:46,440 all the others 0. 504 00:28:46,440 --> 00:28:49,810 Second time around, imposing a unit displacement at the 505 00:28:49,810 --> 00:28:53,160 second degree of freedom, all the others being 0. 506 00:28:53,160 --> 00:28:56,830 Third time around, imposing a unit displacement at the third 507 00:28:56,830 --> 00:28:59,670 degree of freedom, all the other displacements 508 00:28:59,670 --> 00:29:02,070 being 0, and so on. 509 00:29:02,070 --> 00:29:06,040 And that really amounts to then saying that this vector 510 00:29:06,040 --> 00:29:10,290 here becomes an identity matrix. 511 00:29:10,290 --> 00:29:14,420 And similarly, this one here becomes an identity matrix. 512 00:29:14,420 --> 00:29:20,430 And therefore, we can take those two out, and our 513 00:29:20,430 --> 00:29:26,410 resulting equation then is simply what we have left. 514 00:29:26,410 --> 00:29:29,670 Taking these two identity matrices out, and that is our 515 00:29:29,670 --> 00:29:32,610 finite element equilibrium equation, or I should rather 516 00:29:32,610 --> 00:29:36,370 say that these are n finite element equilibrium equations, 517 00:29:36,370 --> 00:29:40,110 namely corresponding to the n nodal point displacements that 518 00:29:40,110 --> 00:29:41,950 we are considering. 519 00:29:41,950 --> 00:29:46,810 In general, what one does most effectively is to really 520 00:29:46,810 --> 00:29:51,110 derive these corresponding to all displacements. 521 00:29:51,110 --> 00:29:57,290 Having removed the boundary conditions, and then later on, 522 00:29:57,290 --> 00:30:02,190 one imposes the known bounded displacements. 523 00:30:02,190 --> 00:30:06,440 And that's what I want to discuss a little bit later in 524 00:30:06,440 --> 00:30:07,290 more detail. 525 00:30:07,290 --> 00:30:11,020 So this is the result then that we obtained. 526 00:30:11,020 --> 00:30:15,030 And we obtained really in shorthand, Ku equals r. 527 00:30:15,030 --> 00:30:18,770 Where K is this matrix. 528 00:30:18,770 --> 00:30:20,890 It's the structural stiffness matrix. 529 00:30:20,890 --> 00:30:26,820 Notice that I'm summing here over the elements. 530 00:30:26,820 --> 00:30:30,580 This being here, the element stiffness matrix. 531 00:30:30,580 --> 00:30:35,350 Notice that this element stiffness matrix here is an n 532 00:30:35,350 --> 00:30:43,450 by n matrix, has the same order as this K matrix. 533 00:30:43,450 --> 00:30:47,530 However, we will recognize that a large number of columns 534 00:30:47,530 --> 00:30:50,240 and rows in this matrix are simply 0. 535 00:30:50,240 --> 00:30:57,030 In fact, all those columns and rows are filled with 0's that 536 00:30:57,030 --> 00:31:00,880 do not correspond to a nodal point displacement degree of 537 00:31:00,880 --> 00:31:04,640 freedom of element m. 538 00:31:04,640 --> 00:31:07,570 I show you later on some examples. 539 00:31:07,570 --> 00:31:14,740 However, by using this Bm here, and making the element 540 00:31:14,740 --> 00:31:18,260 stiffness matrix of the same order as this total structural 541 00:31:18,260 --> 00:31:23,100 stiffness matrix, we can directly sum over all of the 542 00:31:23,100 --> 00:31:24,800 elements stiffness matrices. 543 00:31:24,800 --> 00:31:28,130 And that, of course, is our direct stiffness procedure, 544 00:31:28,130 --> 00:31:33,280 which already I pointed out to you earlier. 545 00:31:33,280 --> 00:31:36,950 The direct stiffness procedure means that we are adding the 546 00:31:36,950 --> 00:31:43,040 element stiffness matrices into the total stiffness 547 00:31:43,040 --> 00:31:46,620 matrix via this summation here. 548 00:31:46,620 --> 00:31:50,310 So the Km here, being what I have here in the blue 549 00:31:50,310 --> 00:31:54,180 brackets, must be, of course, of the same order as K in 550 00:31:54,180 --> 00:31:56,420 order to be able to do that method 551 00:31:56,420 --> 00:31:57,730 theoretically, at least. 552 00:31:57,730 --> 00:32:01,190 Later on, we will see that we indeed only work with the 553 00:32:01,190 --> 00:32:06,120 non-zero rows and columns in the K matrix, and then use 554 00:32:06,120 --> 00:32:09,620 connectivity arrays to assemble Km effectively into 555 00:32:09,620 --> 00:32:11,590 the actual K matrix. 556 00:32:11,590 --> 00:32:14,220 For the RB vector, we have this part. 557 00:32:14,220 --> 00:32:17,490 Once again, we're using the direct stiffness procedure to 558 00:32:17,490 --> 00:32:21,460 add the contributions of all the elements in order to 559 00:32:21,460 --> 00:32:24,130 obtain the total RB vector. 560 00:32:24,130 --> 00:32:31,250 Once again, the rows now, or rather, the elements because 561 00:32:31,250 --> 00:32:36,190 this of course, is a vector here of n long now. 562 00:32:36,190 --> 00:32:39,580 Those elements that do not correspond to nodal point 563 00:32:39,580 --> 00:32:41,645 degrees of freedom will all be 0's. 564 00:32:44,660 --> 00:32:49,630 The RS vector, similarly, is obtained as shown here. 565 00:32:49,630 --> 00:32:53,130 Now, we of course, sum the element contributions as they 566 00:32:53,130 --> 00:32:58,390 arise from the surface forces and the RI vector is obtained 567 00:32:58,390 --> 00:33:00,850 as shown here, and the concentrated load vector is 568 00:33:00,850 --> 00:33:07,870 simply a vector listing all the concentrated forces in F. 569 00:33:07,870 --> 00:33:14,580 Notice that this HSM matrix here is directly obtained from 570 00:33:14,580 --> 00:33:19,670 this Hm matrix. 571 00:33:19,670 --> 00:33:22,930 Sometimes one has difficulties visualizing what this 572 00:33:22,930 --> 00:33:24,340 matrix really is. 573 00:33:24,340 --> 00:33:27,980 Well, we will see later on if this is an element here, and 574 00:33:27,980 --> 00:33:31,020 our coordinate system, say, lies in that element this way, 575 00:33:31,020 --> 00:33:34,710 then the Hm matrix, of course, gives us a displacement within 576 00:33:34,710 --> 00:33:35,910 the element. 577 00:33:35,910 --> 00:33:40,550 Whereas the HSM matrix gives us, say, the displacement on 578 00:33:40,550 --> 00:33:43,420 this surface element, if it is that surface 579 00:33:43,420 --> 00:33:45,430 that we want to consider. 580 00:33:45,430 --> 00:33:48,630 Now, to get the displacement on the surface of the element 581 00:33:48,630 --> 00:33:52,170 when we know the displacement within the total volume of the 582 00:33:52,170 --> 00:33:55,400 element, well, what we simply have to do is we have to 583 00:33:55,400 --> 00:33:59,690 substitute the coordinates of the surface in the Hm here to 584 00:33:59,690 --> 00:34:02,920 obtain the HSM. 585 00:34:02,920 --> 00:34:05,860 I will show you later on some examples. 586 00:34:05,860 --> 00:34:10,210 Now, in dynamic analysis, of course, the loads are time 587 00:34:10,210 --> 00:34:15,929 dependent and if we are considering a truly dynamic 588 00:34:15,929 --> 00:34:19,520 analysis, then we have to include inertia forces. 589 00:34:19,520 --> 00:34:23,659 And the inertia forces can directly be taken care of, or 590 00:34:23,659 --> 00:34:27,060 can directly be included in analysis if we use the 591 00:34:27,060 --> 00:34:28,800 d'Alembert principle. 592 00:34:28,800 --> 00:34:34,060 Here, we have the body loads, which are the externally 593 00:34:34,060 --> 00:34:37,210 applied forces per unit volume. 594 00:34:37,210 --> 00:34:40,690 And if we split these up into those forces that are 595 00:34:40,690 --> 00:34:44,679 externally applied, and those that are arising due to the 596 00:34:44,679 --> 00:34:47,070 d'Alembert forces, as shown here. 597 00:34:47,070 --> 00:34:48,630 Then we directly have the inertia 598 00:34:48,630 --> 00:34:50,219 effect in the analysis. 599 00:34:50,219 --> 00:34:53,739 We now can, of course, express our accelerations in the 600 00:34:53,739 --> 00:34:58,120 element in terms of nodal point accelerations again, and 601 00:34:58,120 --> 00:35:02,710 we are using here the same Hm matrix that we use already for 602 00:35:02,710 --> 00:35:04,540 the displacement interpolations. 603 00:35:04,540 --> 00:35:13,110 If we substitute from here and here into the RB which I had 604 00:35:13,110 --> 00:35:14,820 written down here. 605 00:35:14,820 --> 00:35:20,790 If we substitute into this RB here, this equation for fB, 606 00:35:20,790 --> 00:35:24,620 then we directly can write down this equation here, M U 607 00:35:24,620 --> 00:35:28,780 double dot plus Ku equals R, where the M matrix now is 608 00:35:28,780 --> 00:35:30,970 obtained as shown here. 609 00:35:30,970 --> 00:35:38,780 Notice that this R vector now only contains this RB part. 610 00:35:38,780 --> 00:35:44,320 In other words, not anymore the fB here, but 611 00:35:44,320 --> 00:35:46,280 rather an fB curl. 612 00:35:48,870 --> 00:35:52,670 Notice also that in this analysis now, or in this view 613 00:35:52,670 --> 00:35:56,130 graph, I've dropped the hat on the u. 614 00:35:56,130 --> 00:35:58,920 These are the nodal point displacements, these are the 615 00:35:58,920 --> 00:36:00,470 nodal point accelerations. 616 00:36:00,470 --> 00:36:02,920 I've dropped the hat just for convenience. 617 00:36:02,920 --> 00:36:06,430 I had already dropped it actually here also. 618 00:36:06,430 --> 00:36:09,940 We earlier had the hat there. 619 00:36:09,940 --> 00:36:13,830 Here, we had the hat still because I wanted to 620 00:36:13,830 --> 00:36:18,700 distinguish the actual nodal point displacements from the 621 00:36:18,700 --> 00:36:23,240 continuous displacements in the structure, or in the body. 622 00:36:23,240 --> 00:36:27,010 So here, the hat still being there. 623 00:36:27,010 --> 00:36:30,630 And here, I dropped the hat already, just for convenience 624 00:36:30,630 --> 00:36:35,540 of writing, and from now on, when we have this vector, U 625 00:36:35,540 --> 00:36:39,290 here, then that means that we're talking about the 626 00:36:39,290 --> 00:36:42,560 concentrated nodal point displacements, or the actual 627 00:36:42,560 --> 00:36:45,130 note point displacements of the finite element mesh. 628 00:36:48,010 --> 00:36:51,650 I mentioned earlier that it is most convenient to include in 629 00:36:51,650 --> 00:36:56,420 the formulation all of the nodal point displacements, 630 00:36:56,420 --> 00:37:00,750 including those that actually might be 0. 631 00:37:00,750 --> 00:37:03,630 In other words, for our three-dimensional body, to 632 00:37:03,630 --> 00:37:05,210 make a quick sketch here. 633 00:37:05,210 --> 00:37:11,430 If we have here our support in an actual analysis, it is most 634 00:37:11,430 --> 00:37:15,140 effective to say well, let us remove the support and assign 635 00:37:15,140 --> 00:37:19,260 a node there with three unknown displacements. 636 00:37:19,260 --> 00:37:23,540 Once we have derived these equations of equilibrium, of 637 00:37:23,540 --> 00:37:25,620 course, we now will have to impose the fact that the 638 00:37:25,620 --> 00:37:28,380 displacements are 0's there. 639 00:37:28,380 --> 00:37:32,180 And that is then done effectively, for example, as 640 00:37:32,180 --> 00:37:33,250 shown here. 641 00:37:33,250 --> 00:37:38,080 We have here the general equations, M U double dot plus 642 00:37:38,080 --> 00:37:43,130 Ku equals R. And what we are doing is we are listing the 643 00:37:43,130 --> 00:37:48,250 displacements and accelerations into vectors, U 644 00:37:48,250 --> 00:37:53,870 double dot A, U A, and U double dot b, and Ub, where 645 00:37:53,870 --> 00:37:59,430 the b components of the displacements and 646 00:37:59,430 --> 00:38:01,680 accelerations are known. 647 00:38:01,680 --> 00:38:04,870 And now they might be 0, as I showed here in this particular 648 00:38:04,870 --> 00:38:08,510 example, or they might be actual values 649 00:38:08,510 --> 00:38:10,480 that we want to impose. 650 00:38:10,480 --> 00:38:13,580 If they are known, well, we can look at the first 651 00:38:13,580 --> 00:38:17,980 equation, as shown here, and put all the known quantities 652 00:38:17,980 --> 00:38:20,750 on the right hand side, substitute for Ub and U double 653 00:38:20,750 --> 00:38:24,110 dot b, and we know then the right hand side load vector. 654 00:38:24,110 --> 00:38:27,280 Thus, we can calculate Ua, and U double dot a. 655 00:38:27,280 --> 00:38:30,530 Having now calculated the velocities, the accelerations, 656 00:38:30,530 --> 00:38:33,390 and displacements, we can go back and get the reactions. 657 00:38:33,390 --> 00:38:37,540 The reactions, of course, being Rb. 658 00:38:37,540 --> 00:38:40,600 Here, I assumed that the displacements which we are 659 00:38:40,600 --> 00:38:45,050 talking about in the vector here are actually the ones 660 00:38:45,050 --> 00:38:47,740 that we also might want to impose. 661 00:38:47,740 --> 00:38:50,590 Well, in some cases, of course, we might have defined 662 00:38:50,590 --> 00:38:54,220 in our finite element formulations the U and V 663 00:38:54,220 --> 00:38:56,830 displacement, as shown here. 664 00:38:56,830 --> 00:39:01,250 But a displacement that we want to impose is actually 665 00:39:01,250 --> 00:39:04,380 this one here, namely, that one might have to be 666 00:39:04,380 --> 00:39:07,470 restrained, and this one here might have to be free. 667 00:39:07,470 --> 00:39:11,110 In that case, if our finite element formulation has used 668 00:39:11,110 --> 00:39:15,230 the U and V displacement, we have to make a transformation 669 00:39:15,230 --> 00:39:16,460 as shown here. 670 00:39:16,460 --> 00:39:21,810 A well known transformation from the U to the U bar 671 00:39:21,810 --> 00:39:26,630 displacements, and this, in a more general sense, is written 672 00:39:26,630 --> 00:39:28,340 down here once again. 673 00:39:28,340 --> 00:39:33,350 When we have many more degrees of freedom, our T matrix would 674 00:39:33,350 --> 00:39:34,950 look as shown here. 675 00:39:34,950 --> 00:39:38,830 It's an identity matrix with the little transformation 676 00:39:38,830 --> 00:39:40,520 matrix that I've shown here. 677 00:39:40,520 --> 00:39:44,140 The cosine minus sine sine cosine matrix. 678 00:39:44,140 --> 00:39:47,920 Now, put into the appropriate rows and column. 679 00:39:47,920 --> 00:39:52,720 The i'th column, j'th column, i'th row, and j'th row would 680 00:39:52,720 --> 00:39:55,640 carry these 2x2 matrix. 681 00:39:55,640 --> 00:39:59,050 And otherwise, we just have 1's on the diagonal. 682 00:39:59,050 --> 00:40:02,430 So this is the more general transformation that we are 683 00:40:02,430 --> 00:40:06,420 using when we have many more degrees of freedom than just 684 00:40:06,420 --> 00:40:08,720 the two that we want to modify. 685 00:40:08,720 --> 00:40:13,660 Substituting from here into our equations of equilibrium M 686 00:40:13,660 --> 00:40:18,650 U double dot plus Ku equals R. We directly obtained this 687 00:40:18,650 --> 00:40:23,510 equation, where M bar now is shown here, K bar is shown 688 00:40:23,510 --> 00:40:25,710 here, and R bar is shown here. 689 00:40:25,710 --> 00:40:30,460 Let me mention here that this looks like a former matrix 690 00:40:30,460 --> 00:40:31,760 multiplication. 691 00:40:31,760 --> 00:40:34,870 In fact, two former matrix multiplications. 692 00:40:34,870 --> 00:40:38,870 M times T, And then the product should be taken times 693 00:40:38,870 --> 00:40:41,880 T transposed, pre-multiplied by T transpose. 694 00:40:41,880 --> 00:40:45,970 Well, in actuality, of course, all we need to do is combine 695 00:40:45,970 --> 00:40:47,980 rows and columns. 696 00:40:47,980 --> 00:40:51,230 The i'th and j'th rows and columns to obtain directly our 697 00:40:51,230 --> 00:40:53,830 M bar matrix, similarly for the K bar 698 00:40:53,830 --> 00:40:57,240 and the R bar matrices. 699 00:40:57,240 --> 00:41:00,860 Another procedure that is also used in practice-- 700 00:41:00,860 --> 00:41:01,950 can be very effective-- 701 00:41:01,950 --> 00:41:06,300 is an application of the penalty method. 702 00:41:06,300 --> 00:41:12,800 In this procedure, we impose, basically, physically a spring 703 00:41:12,800 --> 00:41:18,580 of very large stiffness, where K is much larger 704 00:41:18,580 --> 00:41:20,660 than K bar i i. 705 00:41:20,660 --> 00:41:27,820 And then we supplement our basic equations that are shown 706 00:41:27,820 --> 00:41:34,000 here by this equation here. 707 00:41:34,000 --> 00:41:41,730 So if this K is much larger than K bar i i, and if we 708 00:41:41,730 --> 00:41:46,790 supplement this equation or add this equation into this 709 00:41:46,790 --> 00:41:52,330 equation here, then we notice that the spring stiffness will 710 00:41:52,330 --> 00:41:55,570 wipe out basically the other stiffnesses that come into 711 00:41:55,570 --> 00:41:59,160 this degree of freedom, and our solution will simply be 712 00:41:59,160 --> 00:42:03,000 that U i is equal to b, which is the one that we want. 713 00:42:03,000 --> 00:42:07,850 So this penalty method can be used to impose displacement 714 00:42:07,850 --> 00:42:11,390 degrees of freedom, and it really physically amounts to 715 00:42:11,390 --> 00:42:14,440 adding a spring into the degree of freedom where we 716 00:42:14,440 --> 00:42:17,230 want to impose a certain displacement. 717 00:42:17,230 --> 00:42:20,860 And it's important, however, that if we use that method 718 00:42:20,860 --> 00:42:25,130 that we are always dealing with single degree of freedoms 719 00:42:25,130 --> 00:42:25,685 being imposed. 720 00:42:25,685 --> 00:42:27,730 And by that, I mean the following-- 721 00:42:27,730 --> 00:42:35,400 if we had a system, like this one here, and our original 722 00:42:35,400 --> 00:42:39,060 displacement degrees of freedom are these, then we 723 00:42:39,060 --> 00:42:41,990 first have to make the transformation onto that 724 00:42:41,990 --> 00:42:44,010 finite element element system to these 725 00:42:44,010 --> 00:42:45,840 degrees of freedom here. 726 00:42:45,840 --> 00:42:47,750 And now we add our spring in. 727 00:42:47,750 --> 00:42:52,690 In other words, we want to add our spring into this system of 728 00:42:52,690 --> 00:42:58,880 equations because now there's no coupling from this degree 729 00:42:58,880 --> 00:43:01,170 of freedom that we want to be impose into other degrees of 730 00:43:01,170 --> 00:43:03,180 freedom, through that spring. 731 00:43:03,180 --> 00:43:06,510 This spring only enters on the diagonal, and now it is a 732 00:43:06,510 --> 00:43:09,140 numerically stable process. 733 00:43:09,140 --> 00:43:14,060 However, if we were not to perform this transformation-- 734 00:43:14,060 --> 00:43:18,160 in other words, if we were still to deal with these 735 00:43:18,160 --> 00:43:22,880 degrees of freedom, and then add our spring in-- 736 00:43:22,880 --> 00:43:25,370 of course, that spring now would introduce coupling 737 00:43:25,370 --> 00:43:27,840 between these two degrees of freedom, and numerical 738 00:43:27,840 --> 00:43:32,050 difficulties may arise in the solution. 739 00:43:32,050 --> 00:43:36,220 So basically, then in summary, if we do have degrees of 740 00:43:36,220 --> 00:43:39,900 freedom to be imposed, we first go through this 741 00:43:39,900 --> 00:43:47,870 transformation to obtain the M bar, U double dot bar, K bar, 742 00:43:47,870 --> 00:43:51,360 U bar equals R bar system of equations where the 743 00:43:51,360 --> 00:43:55,530 displacement that we're talking about are containing 744 00:43:55,530 --> 00:43:59,070 those displacements that we actually want to impose. 745 00:43:59,070 --> 00:44:02,850 We then can impose these displacements using the 746 00:44:02,850 --> 00:44:06,670 penalty method, which is this one. 747 00:44:06,670 --> 00:44:13,790 Or we can impose these displacements using the more 748 00:44:13,790 --> 00:44:18,050 conventional procedure, using, in other words, simply this 749 00:44:18,050 --> 00:44:23,680 procedural of imposing Ub and rewriting the equations into 750 00:44:23,680 --> 00:44:24,710 two equations. 751 00:44:24,710 --> 00:44:28,570 The first equation we solved for the Ua now, of course, in 752 00:44:28,570 --> 00:44:32,290 this particular case, we would now have all bars on there. 753 00:44:32,290 --> 00:44:35,630 And if we have done a transformation, and in the 754 00:44:35,630 --> 00:44:37,830 second equation then, we obtain the reactions. 755 00:44:40,380 --> 00:44:44,350 Let me now go through a simple example to show you the 756 00:44:44,350 --> 00:44:48,550 application of what I have discussed. 757 00:44:48,550 --> 00:44:51,330 This is a very simple example, but a very 758 00:44:51,330 --> 00:44:53,310 illustrative example. 759 00:44:53,310 --> 00:44:56,880 In particular, it is also the example that we talked about 760 00:44:56,880 --> 00:45:01,540 already earlier in lecture 2, when we did a Ritz Analysis on 761 00:45:01,540 --> 00:45:03,390 this problem. 762 00:45:03,390 --> 00:45:06,620 In fact, our finite element analysis that we are now 763 00:45:06,620 --> 00:45:10,310 pursuing, using the general equation that I presented to 764 00:45:10,310 --> 00:45:13,530 you is really nothing else than a Ritz Analysis. 765 00:45:13,530 --> 00:45:17,180 And in fact, if you look at the earlier 766 00:45:17,180 --> 00:45:18,910 solutions that we obtained-- 767 00:45:18,910 --> 00:45:25,390 solution 2, in the Ritz analysis, corresponds to the 768 00:45:25,390 --> 00:45:26,790 finite element solution that I will be 769 00:45:26,790 --> 00:45:28,520 discussing with you now. 770 00:45:28,520 --> 00:45:30,350 So here is a problem once again. 771 00:45:30,350 --> 00:45:36,950 We have a bar of unit area, from here to there, and then 772 00:45:36,950 --> 00:45:40,670 of changing area, from here to there. 773 00:45:40,670 --> 00:45:44,790 The length here is 100, the length here is 80. 774 00:45:44,790 --> 00:45:50,560 The bar in actuality, is supported here, but as I 775 00:45:50,560 --> 00:45:54,360 mentioned earlier, we remove that support in our finite 776 00:45:54,360 --> 00:45:57,880 element formulation, and introduce, in fact, a 777 00:45:57,880 --> 00:46:00,030 displacement degree of freedom there. 778 00:46:00,030 --> 00:46:03,930 So here, I want to put down the first node. 779 00:46:03,930 --> 00:46:09,040 Now, the first step in any finite element analysis must, 780 00:46:09,040 --> 00:46:14,350 of course, be the step of idealizing the total structure 781 00:46:14,350 --> 00:46:16,060 as an assemblage of elements. 782 00:46:16,060 --> 00:46:17,710 And there are generally choices-- 783 00:46:17,710 --> 00:46:20,180 how many elements to take, what type of element to 784 00:46:20,180 --> 00:46:21,650 take, and so on. 785 00:46:21,650 --> 00:46:26,130 In this particular case, I know that there's a 786 00:46:26,130 --> 00:46:30,460 discontinuity in area here and for that reason, intuitively, 787 00:46:30,460 --> 00:46:35,560 I will put one element from here to there 788 00:46:35,560 --> 00:46:37,720 with a constant area. 789 00:46:37,720 --> 00:46:42,380 Also, let us consider for the moment, this also as being one 790 00:46:42,380 --> 00:46:45,240 element, and this then will correspond to the Ritz 791 00:46:45,240 --> 00:46:48,350 Analysis that we performed earlier. 792 00:46:48,350 --> 00:46:54,400 Notice that we have here a bar of unit area, a bar of 793 00:46:54,400 --> 00:46:56,070 changing area. 794 00:46:56,070 --> 00:47:01,950 This total bar assemblage is subjected to a load of 100, a 795 00:47:01,950 --> 00:47:05,690 concentrated load of 100, as shown here. 796 00:47:05,690 --> 00:47:08,620 The only strains that this bar can 797 00:47:08,620 --> 00:47:10,740 develop are normal strains. 798 00:47:10,740 --> 00:47:16,360 In other words, if a section originally is here, that 799 00:47:16,360 --> 00:47:20,980 section we move over a certain amount and by that amount. 800 00:47:20,980 --> 00:47:25,620 That is U. In the coordinate system that we are using, the 801 00:47:25,620 --> 00:47:28,850 y-coordinate being in this direction, this is the 802 00:47:28,850 --> 00:47:30,400 y-coordinate here. 803 00:47:30,400 --> 00:47:36,270 This would be our U of Y. However, since we are dealing 804 00:47:36,270 --> 00:47:40,780 with two elements to analyze this bar, what I will do is I 805 00:47:40,780 --> 00:47:44,170 will introduce a little coordinate system here. 806 00:47:44,170 --> 00:47:48,470 And I used little y in this particular case. 807 00:47:48,470 --> 00:47:50,850 So we put a little y here. 808 00:47:50,850 --> 00:47:54,120 There's also, for this element, a little y. 809 00:47:54,120 --> 00:48:00,720 And the area, in this particular element, is given 810 00:48:00,720 --> 00:48:05,150 as 1 plus y divided by 40 squared. 811 00:48:05,150 --> 00:48:09,310 So this is the changing area in this domain. 812 00:48:09,310 --> 00:48:14,050 However, also, remember please, if in our analysis, if 813 00:48:14,050 --> 00:48:18,830 an original section in this area was vertical like that, 814 00:48:18,830 --> 00:48:22,570 after deformations, due to the load here, it will still be 815 00:48:22,570 --> 00:48:26,140 vertical, and now our displacements in this element 816 00:48:26,140 --> 00:48:29,000 will also be given by a u. 817 00:48:29,000 --> 00:48:31,790 And that u is a function of-- 818 00:48:31,790 --> 00:48:34,730 if we look at this little y, if you use that little y of 819 00:48:34,730 --> 00:48:40,780 that y as this u here is a function of this y. 820 00:48:40,780 --> 00:48:44,350 This y corresponds to the y in this element, that y 821 00:48:44,350 --> 00:48:48,760 corresponds to the y in this element because we use 822 00:48:48,760 --> 00:48:51,430 different coordinate systems for each element. 823 00:48:51,430 --> 00:48:53,840 Now, this is actually an important point that we can 824 00:48:53,840 --> 00:48:56,870 use for each element, a different coordinate system. 825 00:48:56,870 --> 00:48:59,870 We could use Cartesian coordinate systems for each 826 00:48:59,870 --> 00:49:02,540 element, different ones. 827 00:49:02,540 --> 00:49:04,840 In fact, that is most generally done. 828 00:49:04,840 --> 00:49:10,030 If we have specific geometries, we might use 829 00:49:10,030 --> 00:49:12,860 cylindrical coordinates systems for certain elements, 830 00:49:12,860 --> 00:49:15,910 Cartesian coordinate systems for other elements, and so on. 831 00:49:15,910 --> 00:49:18,450 This is an extremely important point that we can have 832 00:49:18,450 --> 00:49:21,580 different coordinate systems for different elements because 833 00:49:21,580 --> 00:49:24,350 that eases the calculation of the 834 00:49:24,350 --> 00:49:26,890 element stiffness matrices. 835 00:49:26,890 --> 00:49:32,290 So here, our element 1, here our element 2. 836 00:49:32,290 --> 00:49:36,970 And the displacements that we are talking about are U 1 at 837 00:49:36,970 --> 00:49:40,840 this node, U 2 at this node, U 3 at that node. 838 00:49:40,840 --> 00:49:45,980 These three displacements shall give us the displacement 839 00:49:45,980 --> 00:49:46,960 distributions. 840 00:49:46,960 --> 00:49:50,220 Of course, only an approximate displacement distribution in 841 00:49:50,220 --> 00:49:52,490 the complete element mesh. 842 00:49:52,490 --> 00:49:55,210 Two elements make up our element-- 843 00:49:55,210 --> 00:49:59,370 complete element idealization or complete element mesh. 844 00:49:59,370 --> 00:50:03,410 Well, the equation that, of course I will now be operating 845 00:50:03,410 --> 00:50:08,290 on is this one, KU equals R. In this particular case, we 846 00:50:08,290 --> 00:50:11,000 recognize that we want to calculate our stiffness 847 00:50:11,000 --> 00:50:15,320 matrix, K. Here, we have two elements, so M, in this 848 00:50:15,320 --> 00:50:18,960 particular case, will be equal to 1 and 2. 849 00:50:18,960 --> 00:50:22,590 We don't have a body force vector, we don't have a 850 00:50:22,590 --> 00:50:26,570 surface force vector, we don't have an initial stress vector. 851 00:50:26,570 --> 00:50:29,560 However, we have a concentrated load vector. 852 00:50:29,560 --> 00:50:33,370 So what I will want to do then is calculate our K matrix, and 853 00:50:33,370 --> 00:50:37,110 establish our concentrated load vector. 854 00:50:37,110 --> 00:50:40,890 The K matrix embodies the strain displacement 855 00:50:40,890 --> 00:50:43,730 interpolations, which are obtained from the element 856 00:50:43,730 --> 00:50:45,310 displacement interpolations. 857 00:50:45,310 --> 00:50:48,300 Now, as I already pointed out, we have 858 00:50:48,300 --> 00:50:49,980 now here, two elements. 859 00:50:49,980 --> 00:50:54,250 The first element shown here, second element shown here. 860 00:50:54,250 --> 00:50:59,870 Notice that the U1, U2 correspond to these two 861 00:50:59,870 --> 00:51:02,550 displacements, U1, U2. 862 00:51:02,550 --> 00:51:10,030 The U2, U3 correspond to these displacements, U2, U3. 863 00:51:10,030 --> 00:51:14,370 Notice that there's a coupling between the elements because 864 00:51:14,370 --> 00:51:19,070 U2 is here a displacement of that element, and is here the 865 00:51:19,070 --> 00:51:21,470 displacement of element 2. 866 00:51:21,470 --> 00:51:23,760 The length of the element 1 is 100, length of 867 00:51:23,760 --> 00:51:25,900 element 2 is 80. 868 00:51:25,900 --> 00:51:30,080 Well, if we have two displacements to describe the 869 00:51:30,080 --> 00:51:34,020 displacement in an element, then we recognize immediately 870 00:51:34,020 --> 00:51:39,160 that all that we can have is a linear variation in 871 00:51:39,160 --> 00:51:41,920 displacement between the two end points, 872 00:51:41,920 --> 00:51:43,410 between these two nodes. 873 00:51:43,410 --> 00:51:48,100 And so the element displacement interpolations 874 00:51:48,100 --> 00:51:50,850 must involve these functions. 875 00:51:50,850 --> 00:51:55,510 For unit displacement at this end of an element, Y over L is 876 00:51:55,510 --> 00:51:58,500 the interpolation of the displacement. 877 00:51:58,500 --> 00:52:02,070 For unit displacement at this end of the element, this is 878 00:52:02,070 --> 00:52:03,420 the interpolation. 879 00:52:03,420 --> 00:52:06,460 Notice that the actual displacement, of course, goes 880 00:52:06,460 --> 00:52:09,860 into this direction, but I'm plotting it upwards to show 881 00:52:09,860 --> 00:52:14,230 you the magnitude of that displacement. 882 00:52:14,230 --> 00:52:19,180 If we have established these interpolation functions, 883 00:52:19,180 --> 00:52:22,390 recognizing, of course, that for element 1, L is 100, for 884 00:52:22,390 --> 00:52:26,070 element 2, L is equal to 80. 885 00:52:26,070 --> 00:52:30,270 Then we can directly write down our H1 and H2 matrices. 886 00:52:30,270 --> 00:52:31,720 They are given as shown here. 887 00:52:31,720 --> 00:52:40,810 Notice that Um is equal to Hmu, where U is equal to-- 888 00:52:40,810 --> 00:52:42,530 I write down the transpose-- 889 00:52:42,530 --> 00:52:46,310 U1, U2, U3. 890 00:52:46,310 --> 00:52:49,650 Now, we notice that element 1-- 891 00:52:49,650 --> 00:52:52,200 let's go back once more for element 1. 892 00:52:52,200 --> 00:52:58,380 Only U1 and U2 influence the displacements in that element. 893 00:52:58,380 --> 00:53:00,590 And that is shown here. 894 00:53:00,590 --> 00:53:06,670 U3 has 0 and does not influence the displacement in 895 00:53:06,670 --> 00:53:08,950 the element. 896 00:53:08,950 --> 00:53:13,610 Similarly for H2, U1 does not influence the displacement in 897 00:53:13,610 --> 00:53:15,190 that element. 898 00:53:15,190 --> 00:53:17,620 So we have these displacements. 899 00:53:17,620 --> 00:53:22,170 Notice also that I've written here, Um, of course, but that 900 00:53:22,170 --> 00:53:25,840 Um here, for our specific case is simply this 901 00:53:25,840 --> 00:53:27,290 displacement, Vm. 902 00:53:27,290 --> 00:53:30,420 We have only one displacement component. 903 00:53:30,420 --> 00:53:36,280 Taking the derivative of these relations here, we get 904 00:53:36,280 --> 00:53:39,380 directly the strains. 905 00:53:39,380 --> 00:53:43,550 Notice that here, we should have probably put an m there. 906 00:53:43,550 --> 00:53:49,420 This is the normal strain and we obtain these matrices by 907 00:53:49,420 --> 00:53:51,400 simply taking the derivatives. 908 00:53:51,400 --> 00:53:55,260 Well, now we have the components that we need to 909 00:53:55,260 --> 00:53:57,160 evaluate the K matrix. 910 00:53:57,160 --> 00:54:00,690 And as I mentioned earlier, that is the total K matrix 911 00:54:00,690 --> 00:54:01,800 obtained by summing the 912 00:54:01,800 --> 00:54:03,810 contributions over the elements. 913 00:54:03,810 --> 00:54:08,420 This is coming from element 1, this is coming from element 2. 914 00:54:08,420 --> 00:54:12,000 Notice that the area is 1, Young's modulus of stress 915 00:54:12,000 --> 00:54:13,080 strains law. 916 00:54:13,080 --> 00:54:15,080 We are integrating from 0 to 100. 917 00:54:15,080 --> 00:54:20,690 This is here, B1 transpose, that is B1. 918 00:54:20,690 --> 00:54:23,820 This is here, B2 transpose, that is B2. 919 00:54:23,820 --> 00:54:27,630 This is the area that I pointed out to you earlier. 920 00:54:27,630 --> 00:54:31,330 Evaluating these two matrices, we directly 921 00:54:31,330 --> 00:54:33,140 obtain these matrices. 922 00:54:33,140 --> 00:54:34,690 And notice the following-- 923 00:54:34,690 --> 00:54:39,270 that there's no coupling from the third degree of freedom 924 00:54:39,270 --> 00:54:40,660 into element 1. 925 00:54:40,660 --> 00:54:43,830 Similarly, there's no coupling from the first degree of 926 00:54:43,830 --> 00:54:45,970 freedom into element 2. 927 00:54:45,970 --> 00:54:50,400 In fact, what we will do later on is simply calculate the 928 00:54:50,400 --> 00:54:52,270 non-zero parts. 929 00:54:52,270 --> 00:54:57,160 We call these the compacted element stiffness matrices. 930 00:54:57,160 --> 00:55:01,290 And knowing these non-zero parts and knowing into which 931 00:55:01,290 --> 00:55:06,090 degrees of freedom they have to put in the assemblage phase 932 00:55:06,090 --> 00:55:09,070 to obtain the total stiffness matrix, we can directly 933 00:55:09,070 --> 00:55:10,700 assemble the stiffness matrix. 934 00:55:10,700 --> 00:55:14,300 In other words, if I know this part here and I know that the 935 00:55:14,300 --> 00:55:17,630 first column corresponds to the first column of the global 936 00:55:17,630 --> 00:55:21,050 stiffness matrix, the second column corresponds to the 937 00:55:21,050 --> 00:55:24,060 second column in the global stiffness matrix, then I can 938 00:55:24,060 --> 00:55:28,110 just add this contribution into this part here. 939 00:55:28,110 --> 00:55:32,150 Similarly, I can simply add this contribution here into 940 00:55:32,150 --> 00:55:34,800 that part there, without carrying 941 00:55:34,800 --> 00:55:36,690 always these 0's along. 942 00:55:36,690 --> 00:55:38,770 And that is, of course, a very important 943 00:55:38,770 --> 00:55:40,650 computational aspect. 944 00:55:40,650 --> 00:55:45,370 However, in theory, we are really still performing these 945 00:55:45,370 --> 00:55:49,240 additions as shown here, we really still perform the 946 00:55:49,240 --> 00:55:52,560 additions as we pointed them out in the 947 00:55:52,560 --> 00:55:54,150 direct stiffness procedure. 948 00:55:54,150 --> 00:55:57,730 In other words, we still perform this summation, as I 949 00:55:57,730 --> 00:55:59,470 pointed out to you earlier. 950 00:55:59,470 --> 00:56:04,280 The important point, however, is that we now have 951 00:56:04,280 --> 00:56:07,810 established the K matrix, corresponding to the system. 952 00:56:07,810 --> 00:56:10,720 Our R vector is simply, in this 953 00:56:10,720 --> 00:56:15,150 particular case, 0, 0, 100. 954 00:56:15,150 --> 00:56:17,440 Because we only have 100 applied at the 955 00:56:17,440 --> 00:56:18,670 third degree of freedom. 956 00:56:18,670 --> 00:56:22,870 We now have to impose that U1 is 0, we simply set U1 equal 957 00:56:22,870 --> 00:56:24,890 to 0 in the equilibrium equations, as 958 00:56:24,890 --> 00:56:26,020 I pointed out earlier. 959 00:56:26,020 --> 00:56:31,310 We solve for U2 and U3, and having obtained U2 and U3, we 960 00:56:31,310 --> 00:56:34,730 know the displacement in each of these parts, and we know, 961 00:56:34,730 --> 00:56:36,730 therefore, the strains and the stresses in 962 00:56:36,730 --> 00:56:38,410 each of these parts. 963 00:56:38,410 --> 00:56:42,460 The solution is plotted in the example that I discussed with 964 00:56:42,460 --> 00:56:43,710 you in lecture 2. 965 00:56:46,820 --> 00:56:50,750 This example, really, showed some off the basic points of 966 00:56:50,750 --> 00:56:52,050 finite element analysis. 967 00:56:52,050 --> 00:56:54,830 Of course, we have to discuss much more how we actually 968 00:56:54,830 --> 00:56:59,950 obtain the Hm matrices for more complex, more complicated 969 00:56:59,950 --> 00:57:03,160 elements that I use in actual practical analysis. 970 00:57:03,160 --> 00:57:05,480 However, this is all I wanted to mention in this lecture. 971 00:57:05,480 --> 00:57:06,730 Thank you for your attention.