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 00:00:21,910 --> 00:00:23,210 PROFESSOR: Ladies and gentlemen, 9 00:00:23,210 --> 00:00:25,740 welcome to lecture four. 10 00:00:25,740 --> 00:00:28,240 In the last lecture, I presented to you a general 11 00:00:28,240 --> 00:00:32,000 formulation for finite element analysis. 12 00:00:32,000 --> 00:00:35,570 In this lecture, I would like to talk to you about the 13 00:00:35,570 --> 00:00:38,400 derivation of specific finite element matrices. 14 00:00:41,260 --> 00:00:44,870 The formulation that we will be using, or the derivation I 15 00:00:44,870 --> 00:00:48,240 should say, that we will be using, leads us to generalized 16 00:00:48,240 --> 00:00:50,770 finite element models. 17 00:00:50,770 --> 00:00:53,220 I will be talking later on about what we mean by 18 00:00:53,220 --> 00:00:54,870 generalized. 19 00:00:54,870 --> 00:00:58,450 Let's remember, let's recall what we arrived 20 00:00:58,450 --> 00:01:00,340 at in the last lecture. 21 00:01:00,340 --> 00:01:05,840 We had that the element stiffness matrix for element m 22 00:01:05,840 --> 00:01:11,040 is obtained by this integration here where we 23 00:01:11,040 --> 00:01:15,450 integrate the product of a strain-displacement matrix 24 00:01:15,450 --> 00:01:20,010 times a stress-strain matrix times the strain-displacement 25 00:01:20,010 --> 00:01:24,390 matrix over the volume of the element. 26 00:01:24,390 --> 00:01:28,270 We also had derived the expression for a body force 27 00:01:28,270 --> 00:01:32,200 vector, B. m, again, denoting element m. 28 00:01:32,200 --> 00:01:36,600 And here, we integrated over the volume of the element the 29 00:01:36,600 --> 00:01:41,560 product of the displacement interpolation matrix times the 30 00:01:41,560 --> 00:01:46,740 body forces in the element. 31 00:01:46,740 --> 00:01:52,050 We also had derived a surface force vector of the element in 32 00:01:52,050 --> 00:01:56,960 which we integrate the product of the surface interpolation 33 00:01:56,960 --> 00:02:02,100 matrix times the surface forces applied to the element. 34 00:02:02,100 --> 00:02:07,110 So in essence, we see then if we look at these expressions, 35 00:02:07,110 --> 00:02:11,750 we see that we need really a B matrix, a strain-displacement 36 00:02:11,750 --> 00:02:14,290 matrix for the element. 37 00:02:14,290 --> 00:02:15,230 Of course, we need the 38 00:02:15,230 --> 00:02:17,970 stress-strain law for the element. 39 00:02:17,970 --> 00:02:21,250 And that will really be given to us from the serial strength 40 00:02:21,250 --> 00:02:23,980 of materials. 41 00:02:23,980 --> 00:02:27,880 We also want the displacement interpolation 42 00:02:27,880 --> 00:02:31,010 matrix for the element. 43 00:02:31,010 --> 00:02:34,410 Once we have this displacement interpolation matrix, we can 44 00:02:34,410 --> 00:02:37,550 directly obtain the interpolation matrix for the 45 00:02:37,550 --> 00:02:42,590 surface displacement of the element by simply substituting 46 00:02:42,590 --> 00:02:47,740 into this matrix here, the surface coordinates. 47 00:02:47,740 --> 00:02:53,590 In summary then, we need the H matrix, the B matrix, and the 48 00:02:53,590 --> 00:02:57,620 stress-strain law for every element that we have in the 49 00:02:57,620 --> 00:02:59,790 assemblage. 50 00:02:59,790 --> 00:03:03,690 Notice that the stress-strain law, of course, must depend on 51 00:03:03,690 --> 00:03:06,330 what kind of element we are looking at. 52 00:03:06,330 --> 00:03:10,090 How we have mechanically idealized our system. 53 00:03:10,090 --> 00:03:13,490 In a truss structure, we will have simply the Young's 54 00:03:13,490 --> 00:03:15,120 modulus in here. 55 00:03:15,120 --> 00:03:17,740 In a plane stress idealization, we will have a 56 00:03:17,740 --> 00:03:19,310 plane stress law. 57 00:03:19,310 --> 00:03:21,420 In a plate bending idealization, we will have the 58 00:03:21,420 --> 00:03:28,960 plate bending stress-strain law in there, and so on. 59 00:03:28,960 --> 00:03:32,940 Well, I want to talk today about these matrices. 60 00:03:32,940 --> 00:03:36,240 How do we obtain these matrices? 61 00:03:36,240 --> 00:03:41,030 Once we have talked about that subject, I finally want to 62 00:03:41,030 --> 00:03:43,990 also spend a little bit of time talking about the 63 00:03:43,990 --> 00:03:46,170 convergence of the analysis results. 64 00:03:46,170 --> 00:03:49,540 Of course, in finite element analysis, we must be concerned 65 00:03:49,540 --> 00:03:52,960 about the accuracy of the analysis results that we are 66 00:03:52,960 --> 00:03:57,620 obtaining, and about the convergence of the results to 67 00:03:57,620 --> 00:04:03,610 the actual analytically, theoretically accurate result 68 00:04:03,610 --> 00:04:06,000 that we want to obtain. 69 00:04:06,000 --> 00:04:10,245 Well, let us then first talk about the derivation of the H, 70 00:04:10,245 --> 00:04:12,480 B, and C matrices. 71 00:04:12,480 --> 00:04:17,990 And the H, B, and C matrix, of course, depend on the 72 00:04:17,990 --> 00:04:21,711 particular problem that you're looking at. 73 00:04:21,711 --> 00:04:24,730 Let us therefore categorize the different types of 74 00:04:24,730 --> 00:04:28,320 problems that we encounter in structural analysis. 75 00:04:28,320 --> 00:04:33,060 Here we have a truss structure, and a truss 76 00:04:33,060 --> 00:04:38,430 structure is really an assemblage of truss elements. 77 00:04:38,430 --> 00:04:43,710 And in each truss element that lies between the nodes, we 78 00:04:43,710 --> 00:04:47,090 have one-dimensional stress conditions. 79 00:04:47,090 --> 00:04:51,470 In other words, the only stress that we see on the 80 00:04:51,470 --> 00:04:55,780 section AA would be the tau xx stress. 81 00:04:55,780 --> 00:04:59,230 And that is the only stress, no other stress is involved. 82 00:04:59,230 --> 00:05:01,640 Of course, just going a little bit ahead, we immediately 83 00:05:01,640 --> 00:05:05,480 would also say that the only strain we would be interested 84 00:05:05,480 --> 00:05:08,770 in is epsilon xx and the stress-strain law is simply 85 00:05:08,770 --> 00:05:11,650 Young's modulus. 86 00:05:11,650 --> 00:05:15,400 The second kind of analysis that we might be performing is 87 00:05:15,400 --> 00:05:18,290 a plane stress analysis. 88 00:05:18,290 --> 00:05:21,430 In plane stress conditions, the only stresses that are 89 00:05:21,430 --> 00:05:26,420 non-zero are tau xx, tau yy, and tau xy. 90 00:05:26,420 --> 00:05:29,600 Remember in lectures three, I had listed all six stress and 91 00:05:29,600 --> 00:05:30,680 strain components. 92 00:05:30,680 --> 00:05:34,100 And what I'm doing now is I take out just those stress and 93 00:05:34,100 --> 00:05:37,290 strain components that are really applicable for the kind 94 00:05:37,290 --> 00:05:40,260 of problem that I want to look at. 95 00:05:40,260 --> 00:05:43,460 Plane stress situations we will encounter in the analysis 96 00:05:43,460 --> 00:05:44,860 of a beam, for example. 97 00:05:44,860 --> 00:05:47,670 Here's a simple cantilever beam subjected to 98 00:05:47,670 --> 00:05:50,090 surface loading p. 99 00:05:50,090 --> 00:05:57,370 And if we look at an element here, it is redrawn here. 100 00:05:57,370 --> 00:05:59,790 We would find that tau xx would be a 101 00:05:59,790 --> 00:06:01,280 stress into this direction. 102 00:06:01,280 --> 00:06:03,610 Tau xy is the shear stress, and tau yy 103 00:06:03,610 --> 00:06:06,180 is that normal stress. 104 00:06:06,180 --> 00:06:08,760 The cantilever, of course, can be subjected to a surface 105 00:06:08,760 --> 00:06:12,360 loading, a bending moment, tip loading, whatever loading. 106 00:06:12,360 --> 00:06:15,310 The stress components, the only stress components that we 107 00:06:15,310 --> 00:06:20,720 assume of course, act in the structure would be these three 108 00:06:20,720 --> 00:06:22,480 stress components. 109 00:06:22,480 --> 00:06:26,180 In fact, of course, in the actual physical situation, we 110 00:06:26,180 --> 00:06:30,340 would find that around here, depending on the specific 111 00:06:30,340 --> 00:06:33,660 support conditions that we are using, we would have a much 112 00:06:33,660 --> 00:06:36,310 more complicated stress situation than the plane 113 00:06:36,310 --> 00:06:38,170 stress situation assumed. 114 00:06:38,170 --> 00:06:40,670 However, if we are only interested in the tip 115 00:06:40,670 --> 00:06:45,050 displacement here, or the displacements away from the 116 00:06:45,050 --> 00:06:48,250 support condition from the support. 117 00:06:48,250 --> 00:06:50,550 Similarly, if we are only interested in the stresses and 118 00:06:50,550 --> 00:06:54,580 strains away from the support, certainly it is accurate 119 00:06:54,580 --> 00:06:55,810 enough to-- 120 00:06:55,810 --> 00:06:58,790 it yields an accurate enough model to assume a 121 00:06:58,790 --> 00:07:01,070 plane stress situation. 122 00:07:01,070 --> 00:07:04,190 Another plane stress situation occurs here, for example, 123 00:07:04,190 --> 00:07:06,470 where we have a sheet with a hole. 124 00:07:06,470 --> 00:07:10,730 If the sheet is thin, the only stresses that we would see on 125 00:07:10,730 --> 00:07:14,440 an element, such as shown here, would be z stresses, the 126 00:07:14,440 --> 00:07:17,385 plane stress stresses. 127 00:07:17,385 --> 00:07:20,950 A next category of analysis that we would encounter in 128 00:07:20,950 --> 00:07:25,190 practice is a plane strain condition. 129 00:07:25,190 --> 00:07:31,970 Plane strain condition means that in the space xyz shown 130 00:07:31,970 --> 00:07:37,500 here with the displacements uvw, the displacement w is 0. 131 00:07:37,500 --> 00:07:42,450 It's identically 0 if our x, y-coordinate system lies in 132 00:07:42,450 --> 00:07:47,020 the plane of a typical section of the total problem. 133 00:07:47,020 --> 00:07:50,560 And let me show you one such typical-- 134 00:07:50,560 --> 00:07:53,000 or describe to you one such difficult problem. 135 00:07:53,000 --> 00:07:56,740 Here we have a dam that is very long. 136 00:07:56,740 --> 00:07:59,550 In fact, we might assume it infinitely long. 137 00:07:59,550 --> 00:08:05,020 And it is supported on this face here, so that w, the 138 00:08:05,020 --> 00:08:08,300 displacement, normal to that face is 0. 139 00:08:08,300 --> 00:08:10,320 And the same support conditions are 140 00:08:10,320 --> 00:08:12,480 at the other face. 141 00:08:12,480 --> 00:08:16,400 If the dam is subject to uniform water loading, water 142 00:08:16,400 --> 00:08:20,800 pressure loading all along its side, then by symmetry it 143 00:08:20,800 --> 00:08:24,980 follows that w, the displacement, at any section 144 00:08:24,980 --> 00:08:27,430 of the dam, the displacement normal to that 145 00:08:27,430 --> 00:08:29,490 section must be 0. 146 00:08:29,490 --> 00:08:35,030 Therefore, we have w 0 meaning directly epsilon zz 0. 147 00:08:35,030 --> 00:08:39,270 The strain into the z-direction is also 0. 148 00:08:39,270 --> 00:08:46,870 Similarly, the shear stresses, tau zx and tau zy are 0. 149 00:08:46,870 --> 00:08:49,140 And the only stresses that we are left with which are 150 00:08:49,140 --> 00:08:51,990 non-zero are those shown here. 151 00:08:51,990 --> 00:08:59,840 These are the stresses acting on a typical slice of the dam. 152 00:08:59,840 --> 00:09:05,040 A slice say, of unit length would be subjected to just 153 00:09:05,040 --> 00:09:07,070 these stress conditions. 154 00:09:07,070 --> 00:09:11,190 Therefore, in that array of six stresses and strains that 155 00:09:11,190 --> 00:09:14,410 I talked to you about in lecture three, we really only 156 00:09:14,410 --> 00:09:17,590 are concerned-- 157 00:09:17,590 --> 00:09:21,410 you only need to calculate these four components of those 158 00:09:21,410 --> 00:09:24,600 six that I talked about there. 159 00:09:24,600 --> 00:09:27,100 Another stress-strain condition that we are 160 00:09:27,100 --> 00:09:31,370 interested in frequently in practice is an axisymmetric 161 00:09:31,370 --> 00:09:32,830 analysis condition. 162 00:09:32,830 --> 00:09:37,100 This case we have an axis of revolution for the structure, 163 00:09:37,100 --> 00:09:40,490 and here we look at a cylinder, which is 164 00:09:40,490 --> 00:09:45,370 axisymmetric in geometry about that center line here, the 165 00:09:45,370 --> 00:09:46,750 axis of revolution. 166 00:09:46,750 --> 00:09:50,590 And the loading is also assumed in this particular 167 00:09:50,590 --> 00:09:52,470 case to be also axisymmetric. 168 00:09:52,470 --> 00:09:56,040 So typically, we are talking about a cylinder, circular, of 169 00:09:56,040 --> 00:09:59,010 course, when looking from a plane view subjected to 170 00:09:59,010 --> 00:10:03,220 internal pressure, which is the same all around. 171 00:10:03,220 --> 00:10:07,410 And of course, the geometry is also axisymmetric when we look 172 00:10:07,410 --> 00:10:11,810 at the axis of revolution, the center line of the cylinder. 173 00:10:11,810 --> 00:10:16,930 In this particular case, the only stress conditions that we 174 00:10:16,930 --> 00:10:19,710 need to calculate are shown here for an 175 00:10:19,710 --> 00:10:21,890 element in that cylinder. 176 00:10:21,890 --> 00:10:25,890 We have tau xx, tau yy, tau zz. 177 00:10:25,890 --> 00:10:29,510 This, of course, being the hoop stress. 178 00:10:29,510 --> 00:10:32,290 And the shear stress, tau xy. 179 00:10:32,290 --> 00:10:34,150 These are the stress conditions that we need to 180 00:10:34,150 --> 00:10:36,090 calculate for this particular case. 181 00:10:36,090 --> 00:10:42,080 Similarity, because we are talking about an axisymmetric 182 00:10:42,080 --> 00:10:46,260 condition, an axisymmetric condition where really, every 183 00:10:46,260 --> 00:10:51,080 one of these elements behaves in the same way, we can look 184 00:10:51,080 --> 00:10:57,205 at one radian of the total circumference in the analysis. 185 00:10:59,740 --> 00:11:04,420 So this is an axisymmetric analysis, which we also are 186 00:11:04,420 --> 00:11:08,450 concerned with frequently in practice. 187 00:11:08,450 --> 00:11:13,330 Finally, the very large number of analyses-- 188 00:11:13,330 --> 00:11:17,040 in a very large number of analyses, we consider plates 189 00:11:17,040 --> 00:11:19,080 and shells. 190 00:11:19,080 --> 00:11:23,930 If we look at a plate, an element in that plate, we 191 00:11:23,930 --> 00:11:30,240 would find that in general we would have the tau xx, tau yy. 192 00:11:30,240 --> 00:11:32,400 These are the normal stresses. 193 00:11:32,400 --> 00:11:37,530 And tau xy shearing stresses in the plane of the plate. 194 00:11:37,530 --> 00:11:40,060 We would also have transfer shearing stresses, 195 00:11:40,060 --> 00:11:43,230 tau xz and tau yz. 196 00:11:43,230 --> 00:11:47,290 Sorry, tau yz, shown here as the 197 00:11:47,290 --> 00:11:49,450 transfer shearing stresses. 198 00:11:49,450 --> 00:11:52,490 Of course, these transfers shearing stresses give us 199 00:11:52,490 --> 00:11:56,020 shear forces. 200 00:11:56,020 --> 00:12:03,890 And these stresses here give us membrane forces plus 201 00:12:03,890 --> 00:12:07,810 bending moments and twisting moments. 202 00:12:07,810 --> 00:12:11,330 Elementary theory of plates shows us, tells us, that these 203 00:12:11,330 --> 00:12:14,370 are the significant stresses in a plate. 204 00:12:14,370 --> 00:12:18,080 The important point is that in a plate, as well as in a 205 00:12:18,080 --> 00:12:23,790 shell, the stress normal to the plate is 0. 206 00:12:23,790 --> 00:12:25,320 The same holds for the shell. 207 00:12:25,320 --> 00:12:29,600 The stress normal to the shell is 0, and we have the stress 208 00:12:29,600 --> 00:12:30,330 components. 209 00:12:30,330 --> 00:12:35,426 However, we have to now remember that the xx and yy 210 00:12:35,426 --> 00:12:39,210 axes are aligned with the curvature of the shelf. 211 00:12:39,210 --> 00:12:41,420 Are aligned with the curvature of the shell. 212 00:12:41,420 --> 00:12:45,330 And then these are the stress components that are of 213 00:12:45,330 --> 00:12:48,010 interest in the analysis of shells. 214 00:12:48,010 --> 00:12:51,330 In many cases, when we look at the analysis of plates, the 215 00:12:51,330 --> 00:12:53,300 membrane forces-- 216 00:12:53,300 --> 00:12:56,690 the total membrane forces are 0. 217 00:12:56,690 --> 00:13:01,400 In which case, of course, we assume that tau yy has a 218 00:13:01,400 --> 00:13:04,480 linear variation through the thickness of the plate. 219 00:13:04,480 --> 00:13:10,230 As stipulated, for example, in the Kirchhoff plate theory. 220 00:13:10,230 --> 00:13:15,270 This much then for the static components, or rather the 221 00:13:15,270 --> 00:13:18,300 stress components of plates and shells when we look at the 222 00:13:18,300 --> 00:13:21,350 kinematics of plates and shells. 223 00:13:21,350 --> 00:13:25,910 Just to remind you, we assume that there's a normal to the 224 00:13:25,910 --> 00:13:27,800 mid-surface. 225 00:13:27,800 --> 00:13:31,520 And that normal in Kirchhoff plate theory remains during 226 00:13:31,520 --> 00:13:35,510 deformation normal to the plate surface. 227 00:13:35,510 --> 00:13:40,550 And all material particles on that normal AB remain on a 228 00:13:40,550 --> 00:13:41,480 straight line. 229 00:13:41,480 --> 00:13:42,460 These are the two things. 230 00:13:42,460 --> 00:13:46,720 I repeat, all material particles on the line AB 231 00:13:46,720 --> 00:13:51,450 remain during deformation on the line AB, and the line AB 232 00:13:51,450 --> 00:13:53,955 remains normal to the mid-surface in Kirchhoff plate 233 00:13:53,955 --> 00:13:57,750 theory, which does not exclude shear deformations. 234 00:13:57,750 --> 00:14:01,200 However, if we use a theory that does include shear 235 00:14:01,200 --> 00:14:04,970 deformations, we would still have that the material 236 00:14:04,970 --> 00:14:09,450 particles on line AB will still lie after deformations 237 00:14:09,450 --> 00:14:14,230 on the line A prime B prime say. 238 00:14:14,230 --> 00:14:16,850 But the line A prime B prime-- 239 00:14:16,850 --> 00:14:20,240 B is still a straight line-- is not anymore normal to the 240 00:14:20,240 --> 00:14:21,090 mid-surface. 241 00:14:21,090 --> 00:14:23,680 In this case, we have included shear deformations. 242 00:14:23,680 --> 00:14:26,830 We will see in the next lecture that, in fact, it is 243 00:14:26,830 --> 00:14:30,030 simpler to use a theory including shear deformations 244 00:14:30,030 --> 00:14:33,980 to derive effective finite elements. 245 00:14:33,980 --> 00:14:39,380 Well, this then is a review for the kinds of stress 246 00:14:39,380 --> 00:14:43,010 components and strain components that we are 247 00:14:43,010 --> 00:14:43,910 interested in. 248 00:14:43,910 --> 00:14:47,920 Of course, that information which I discussed so far is 249 00:14:47,920 --> 00:14:53,050 available from the strengths of material theory and 250 00:14:53,050 --> 00:14:54,760 continue mechanic theory. 251 00:14:54,760 --> 00:14:58,140 And as an engineer, however, we have to be able to decide 252 00:14:58,140 --> 00:15:00,470 what kind of stress and strain 253 00:15:00,470 --> 00:15:02,230 situations you want to analyze. 254 00:15:02,230 --> 00:15:04,520 And there, of course, is already one very important 255 00:15:04,520 --> 00:15:05,010 assumption. 256 00:15:05,010 --> 00:15:09,430 A very important assumption in laying out or starting with 257 00:15:09,430 --> 00:15:13,210 the right model for the actual physical structure. 258 00:15:13,210 --> 00:15:16,340 Whether we want to model the actual physical structure as a 259 00:15:16,340 --> 00:15:21,240 truss assemblage, as an assemblage of plates, beams, 260 00:15:21,240 --> 00:15:26,810 et cetera, all of those situations can be covered in 261 00:15:26,810 --> 00:15:28,090 the finite element theory. 262 00:15:28,090 --> 00:15:31,630 But we have to make the right assumptions right from the 263 00:15:31,630 --> 00:15:34,380 start in starting with the right model. 264 00:15:34,380 --> 00:15:36,460 So here I listed ones. 265 00:15:36,460 --> 00:15:38,850 In one table, the particular problems 266 00:15:38,850 --> 00:15:40,100 that we might encounter. 267 00:15:40,100 --> 00:15:42,640 We might have a bar. 268 00:15:42,640 --> 00:15:45,770 The displacement components then being simply u. 269 00:15:45,770 --> 00:15:51,020 In a beam, which I have not given earlier, but here we 270 00:15:51,020 --> 00:15:53,200 have a transverse displacement, w. 271 00:15:53,200 --> 00:15:57,350 In plane stress situations, we have two displacements, u and 272 00:15:57,350 --> 00:15:58,100 v. 273 00:15:58,100 --> 00:16:01,090 In a plane strain situation, u and v. 274 00:16:01,090 --> 00:16:03,230 In an axisymmetric situation, u and v. 275 00:16:03,230 --> 00:16:06,360 Three-dimensional situation, we have u, v, and w. 276 00:16:06,360 --> 00:16:08,760 In plate bending, simply the transverse displacements. 277 00:16:11,590 --> 00:16:15,600 In analysis of shells, in the analysis of shells we would, 278 00:16:15,600 --> 00:16:20,260 of course, have the transverse placements and also the 279 00:16:20,260 --> 00:16:21,770 membrane displacements included. 280 00:16:21,770 --> 00:16:25,710 We will talk about that in the next lecture. 281 00:16:25,710 --> 00:16:29,120 These are the displacement components for the particular 282 00:16:29,120 --> 00:16:31,270 problem that we are looking at. 283 00:16:31,270 --> 00:16:35,810 And correspondingly, we have particular strains. 284 00:16:35,810 --> 00:16:39,170 The strains that we are talking about are in the bar, 285 00:16:39,170 --> 00:16:40,480 simply the normals strains. 286 00:16:40,480 --> 00:16:44,640 In a beam, the second derivative of the transverse 287 00:16:44,640 --> 00:16:45,680 displacements. 288 00:16:45,680 --> 00:16:49,310 Here we have kappa xx defined. 289 00:16:49,310 --> 00:16:50,660 This is kappa xx defined. 290 00:16:50,660 --> 00:16:53,120 In plane stress, we have these three components. 291 00:16:53,120 --> 00:16:54,860 Plane strain, these three components. 292 00:16:54,860 --> 00:16:57,550 Axisymmetric these three components, and so on. 293 00:16:57,550 --> 00:17:00,500 And it is here, of course, where the engineer has to 294 00:17:00,500 --> 00:17:04,450 decide what kind of problem he has in hand when he actually-- 295 00:17:04,450 --> 00:17:07,880 when he looks at an actual physical situation. 296 00:17:07,880 --> 00:17:10,849 Corresponding to these strains, 297 00:17:10,849 --> 00:17:12,690 we have also stresses. 298 00:17:12,690 --> 00:17:15,010 And here we are listing the stresses. 299 00:17:15,010 --> 00:17:19,970 Notice that for the beam, we are talking about a bending 300 00:17:19,970 --> 00:17:21,760 moment, which we might consider to be 301 00:17:21,760 --> 00:17:23,150 a generalized stress. 302 00:17:23,150 --> 00:17:25,810 In plate bending, we talk similarly about bending 303 00:17:25,810 --> 00:17:28,380 moments, which you also might consider as 304 00:17:28,380 --> 00:17:30,440 being generalized stresses. 305 00:17:30,440 --> 00:17:34,680 Otherwise, we have the actual stresses that correspond to 306 00:17:34,680 --> 00:17:36,490 the strains. 307 00:17:36,490 --> 00:17:39,420 Having decided what stresses and strains we 308 00:17:39,420 --> 00:17:40,920 want to look at-- 309 00:17:40,920 --> 00:17:43,390 in other words, what kind of problem we have at hand, we 310 00:17:43,390 --> 00:17:45,630 also have to select the appropriate material matrix. 311 00:17:45,630 --> 00:17:48,840 Again, this one is given form strengths of material. 312 00:17:48,840 --> 00:17:52,140 Here, for a bar, a truss structure, we simply have 313 00:17:52,140 --> 00:17:52,820 Young's modulus. 314 00:17:52,820 --> 00:17:55,360 For a beam, we have the flexural rigidity. 315 00:17:55,360 --> 00:18:00,300 In plane stress analysis, this would be the material law that 316 00:18:00,300 --> 00:18:03,380 you would find in many textbooks on 317 00:18:03,380 --> 00:18:05,470 strengths of materials. 318 00:18:05,470 --> 00:18:09,640 Well, having therefore resolved the problem of how to 319 00:18:09,640 --> 00:18:12,420 establish the stress-strain law, or the stress-strain 320 00:18:12,420 --> 00:18:17,530 matrix c, we now go on to discuss the construction of 321 00:18:17,530 --> 00:18:20,860 the displacement interpolation matrix. 322 00:18:20,860 --> 00:18:22,770 For a one-dimensional bar element, 323 00:18:22,770 --> 00:18:25,570 we would have simply-- 324 00:18:25,570 --> 00:18:29,430 we can assume simply a polynomial with unknown 325 00:18:29,430 --> 00:18:34,980 coefficients, alpha 1, alpha 2, alpha 3. 326 00:18:34,980 --> 00:18:37,805 And this is simply a polynomial nx. 327 00:18:37,805 --> 00:18:41,620 x being, of course, the coordinate that is being used 328 00:18:41,620 --> 00:18:43,610 to describe the displacement for the element. 329 00:18:43,610 --> 00:18:46,230 For two-dimensional elements, here we're talking about plane 330 00:18:46,230 --> 00:18:47,890 stress, plane strain. 331 00:18:47,890 --> 00:18:51,360 Axisymmetric analysis, we would have, similarly, these 332 00:18:51,360 --> 00:18:52,390 assumptions here. 333 00:18:52,390 --> 00:18:58,060 alphas for the u displacements and betas for the v 334 00:18:58,060 --> 00:19:00,260 displacements. 335 00:19:00,260 --> 00:19:05,880 For the plate bending element in which we only discretize 336 00:19:05,880 --> 00:19:09,080 the w displacement, the transverse displacement, we 337 00:19:09,080 --> 00:19:12,150 would have this assumption here. 338 00:19:12,150 --> 00:19:15,130 For three-dimensional solid elements, similarly, we have 339 00:19:15,130 --> 00:19:16,540 these assumptions. 340 00:19:16,540 --> 00:19:22,660 Notice that these gammas, alphas, betas are unknowns for 341 00:19:22,660 --> 00:19:23,250 the element. 342 00:19:23,250 --> 00:19:26,990 And we will actually relate these coefficients to the 343 00:19:26,990 --> 00:19:29,040 nodal point displacements of an element. 344 00:19:31,930 --> 00:19:35,905 If we look, therefore, in general, at the description of 345 00:19:35,905 --> 00:19:40,330 an element, we really have on the left-hand side here, the 346 00:19:40,330 --> 00:19:44,350 actual displacements, the continuous displacements in 347 00:19:44,350 --> 00:19:53,260 the element being given by matrix phi, which contains the 348 00:19:53,260 --> 00:19:58,020 constant and x, x squared, x cubed terms, and so on. 349 00:19:58,020 --> 00:20:02,510 And this alpha vector here contains the generalized 350 00:20:02,510 --> 00:20:06,100 coordinates, the alphas, betas, gammas. 351 00:20:06,100 --> 00:20:08,910 Typically, therefore, just to focus our attention on a 352 00:20:08,910 --> 00:20:13,840 particular case, we might have if we say that u of x for a 353 00:20:13,840 --> 00:20:19,785 truss structure is equal to alpha 0 plus alpha 1x plus 354 00:20:19,785 --> 00:20:21,960 alpha 2x squared. 355 00:20:21,960 --> 00:20:27,110 In this particular case, we would have that we can write 356 00:20:27,110 --> 00:20:32,730 this, of course, as 1 x x squared in matrix form times 357 00:20:32,730 --> 00:20:36,610 alpha 0 alpha 1, alpha 2. 358 00:20:36,610 --> 00:20:42,150 And here, I'm talking about this being equal to the phi, 359 00:20:42,150 --> 00:20:45,090 and this being equal to the vector alpha. 360 00:20:45,090 --> 00:20:50,360 So just to give you an example for this particular case. 361 00:20:50,360 --> 00:20:55,970 If we now want to evaluate the alpha values. 362 00:20:55,970 --> 00:20:58,650 Of course, in two-dimension analysis as I shown earlier to 363 00:20:58,650 --> 00:21:00,780 you, we have also betas here. 364 00:21:00,780 --> 00:21:03,220 In three-dimensional analysis, we also would have gammas in 365 00:21:03,220 --> 00:21:04,640 this vector. 366 00:21:04,640 --> 00:21:08,580 If you want to evaluate these alpha values, then we can 367 00:21:08,580 --> 00:21:12,880 simply use this relation here, or that relation here, and 368 00:21:12,880 --> 00:21:15,880 apply it to the element nodal points. 369 00:21:15,880 --> 00:21:19,620 To the displacements at the nodal points of the elements. 370 00:21:19,620 --> 00:21:23,800 If we do that, in other words, we substitute into here, into 371 00:21:23,800 --> 00:21:27,770 the phi matrix, the actual coordinate of the nodal 372 00:21:27,770 --> 00:21:30,820 points, and on the left-hand side, of course, the 373 00:21:30,820 --> 00:21:34,160 displacement at that nodal point, we directly generate 374 00:21:34,160 --> 00:21:36,030 this relation. 375 00:21:36,030 --> 00:21:38,870 We can then invert this relation to obtain alpha. 376 00:21:38,870 --> 00:21:41,420 And here, we have now the generalized coordinates in 377 00:21:41,420 --> 00:21:46,030 terms of the nodal point displacements. 378 00:21:46,030 --> 00:21:49,650 This then would really complete the evaluation of the 379 00:21:49,650 --> 00:21:53,620 continuous displacements u in terms of the nodal points 380 00:21:53,620 --> 00:21:56,850 displacement u hat. 381 00:21:56,850 --> 00:22:00,620 If we have laid down this assumption, we can directly 382 00:22:00,620 --> 00:22:04,420 also obtain the relation for the strains of the elements. 383 00:22:04,420 --> 00:22:08,165 We have decided what kinds of strains we want to look at 384 00:22:08,165 --> 00:22:09,870 that we have to include. 385 00:22:09,870 --> 00:22:13,210 And by proper differentiation of the rows here, we directly 386 00:22:13,210 --> 00:22:14,620 generate this relation. 387 00:22:14,620 --> 00:22:16,820 Of course, this is a [UNINTELLIGIBLE] 388 00:22:16,820 --> 00:22:21,310 relation then for the material of the element. 389 00:22:21,310 --> 00:22:25,500 And we then directly have tau in terms of alpha. 390 00:22:25,500 --> 00:22:29,060 But alpha was already given in terms of u hat here. 391 00:22:29,060 --> 00:22:33,980 In summary, therefore, we have our displacement interpolation 392 00:22:33,980 --> 00:22:40,470 matrix here, given as phi times a inverse. 393 00:22:40,470 --> 00:22:44,130 This is obtained by simply substituting from here, 394 00:22:44,130 --> 00:22:47,790 substitute from here, directly into there. 395 00:22:47,790 --> 00:22:52,920 And you get that u is equal to H u hat. 396 00:22:52,920 --> 00:22:56,380 And our H being phi a inverse. 397 00:22:56,380 --> 00:23:00,280 Once again, the u is the continuous displacement in the 398 00:23:00,280 --> 00:23:02,240 elements. u hat has nodal point 399 00:23:02,240 --> 00:23:03,900 displacements in the elements. 400 00:23:03,900 --> 00:23:08,650 Similarly, if we substitute from here into there and 401 00:23:08,650 --> 00:23:13,990 recognize that our epsilon strain is equal to B times u 402 00:23:13,990 --> 00:23:18,570 hat, we directly obtain that B must be equal 403 00:23:18,570 --> 00:23:21,970 to E times A inverse. 404 00:23:21,970 --> 00:23:25,500 Well, let us now look at a particular case, a particular 405 00:23:25,500 --> 00:23:30,350 example to demonstrate how we proceed for a particular case. 406 00:23:30,350 --> 00:23:34,380 Here I'm looking at the analysis of a cantilever 407 00:23:34,380 --> 00:23:36,100 subjected to a load. 408 00:23:36,100 --> 00:23:38,510 And I've idealized that cantilever as an assemblage of 409 00:23:38,510 --> 00:23:40,560 four elements. 410 00:23:40,560 --> 00:23:44,040 Each element is a four nodal element. 411 00:23:44,040 --> 00:23:49,370 So for element 1, we have nodes 1, 2, 4, and 5 of the 412 00:23:49,370 --> 00:23:52,260 global assemblage. 413 00:23:52,260 --> 00:23:55,840 Notice that all together we have 9 nodes. 414 00:23:55,840 --> 00:23:58,700 And once again, 4 elements. 415 00:23:58,700 --> 00:24:05,160 The cantilever structure is SiN and we believe that it is 416 00:24:05,160 --> 00:24:09,600 appropriately modeled using a plane stress analysis. 417 00:24:09,600 --> 00:24:14,160 In which case, the only stresses that are of concern 418 00:24:14,160 --> 00:24:17,345 are the stresses in this plane, tau yy, 419 00:24:17,345 --> 00:24:19,280 tau xx, and tau xy. 420 00:24:19,280 --> 00:24:24,320 So these other stress components here are 0. 421 00:24:24,320 --> 00:24:28,660 If we look now at a particular element to focus our attention 422 00:24:28,660 --> 00:24:31,420 on one element, because all these elements 423 00:24:31,420 --> 00:24:32,530 are really the same. 424 00:24:32,530 --> 00:24:34,220 Basically, the same. 425 00:24:34,220 --> 00:24:36,190 Let's look at element 2. 426 00:24:36,190 --> 00:24:44,030 And we notice that element 2 being a four nodal element is 427 00:24:44,030 --> 00:24:49,980 shown once more here with the displacements U1, V1 at the 428 00:24:49,980 --> 00:24:54,110 local nodal point of the element 1. 429 00:24:54,110 --> 00:24:59,440 V2, U2 being the displacements of the local nodal point 2. 430 00:24:59,440 --> 00:25:02,920 And here, V3, U3 similarly for nodal point 3. 431 00:25:02,920 --> 00:25:06,200 And V4, U4 for nodal point 4. 432 00:25:06,200 --> 00:25:09,350 We have taken this element of the assemblage and what we 433 00:25:09,350 --> 00:25:15,330 want to really do is derive the displacement and strain 434 00:25:15,330 --> 00:25:19,790 interpolation matrices for this element corresponding to 435 00:25:19,790 --> 00:25:25,360 U1, V1; U2, V2; U3, V3; and U4, V4. 436 00:25:25,360 --> 00:25:30,080 If we have done that, we recognize, of course, that U1 437 00:25:30,080 --> 00:25:32,130 corresponds to a global displacement. 438 00:25:32,130 --> 00:25:35,410 Similarly, V1 corresponds to a global placement. 439 00:25:35,410 --> 00:25:40,015 Then we can directly construct the displacement and strain 440 00:25:40,015 --> 00:25:45,520 interpolations corresponding to the global displacements. 441 00:25:45,520 --> 00:25:48,800 But the finite element procedure is really to look 442 00:25:48,800 --> 00:25:50,480 locally at an element. 443 00:25:50,480 --> 00:25:51,900 Locally at an element. 444 00:25:51,900 --> 00:25:54,810 And work in terms of local displacements. 445 00:25:54,810 --> 00:25:57,400 Later on then, we relate the local displacements to the 446 00:25:57,400 --> 00:25:58,720 global displacements. 447 00:25:58,720 --> 00:26:02,270 The advantage of proceeding this way is that we really-- 448 00:26:02,270 --> 00:26:04,920 in looking at one element, we look at all four 449 00:26:04,920 --> 00:26:06,170 simultaneously. 450 00:26:09,510 --> 00:26:16,950 The procedure then is, once more in summary, we want this 451 00:26:16,950 --> 00:26:21,820 matrix here, the H2 matrix corresponding to element 2. 452 00:26:21,820 --> 00:26:25,970 The vector U lists all of the displacements, all of the 18 453 00:26:25,970 --> 00:26:27,170 displacements. 454 00:26:27,170 --> 00:26:37,365 However, we will look at the element 2 with local nodal 455 00:26:37,365 --> 00:26:39,980 point displacements, forgetting about the global 456 00:26:39,980 --> 00:26:41,340 nodal point displacement. 457 00:26:41,340 --> 00:26:47,600 And our objective is to derive a relationship of u equals H 458 00:26:47,600 --> 00:26:54,860 times u hat, where u hat lists U1, V1; U2, V2; 459 00:26:54,860 --> 00:26:58,800 U3, V3; U4 and V4. 460 00:26:58,800 --> 00:27:05,100 Once we have this H, which is, of course, a 2 by 8 matrix, 461 00:27:05,100 --> 00:27:08,856 because there are 2 displacements, u and v. The u 462 00:27:08,856 --> 00:27:12,960 and v displacement for the element. 463 00:27:12,960 --> 00:27:16,060 And there are 8 entries in u hat. 464 00:27:16,060 --> 00:27:18,970 Therefore, this H matrix must be a 2 by 8. 465 00:27:18,970 --> 00:27:23,930 Once we have obtained this H matrix, we can construct 466 00:27:23,930 --> 00:27:27,730 directly our H2 matrix that we talked about earlier. 467 00:27:30,470 --> 00:27:36,390 Well, let's proceed then in the way that I discussed in 468 00:27:36,390 --> 00:27:38,640 general form earlier. 469 00:27:38,640 --> 00:27:43,240 Since we have a four nodal element, let me just sketch it 470 00:27:43,240 --> 00:27:44,210 once more here. 471 00:27:44,210 --> 00:27:49,170 Since we have a four nodal element, and we are talking 472 00:27:49,170 --> 00:27:55,490 about u and v displacements, we really can only have four 473 00:27:55,490 --> 00:27:59,140 constants, four generalized coordinates. 474 00:27:59,140 --> 00:28:02,870 Generalized coordinates corresponding to the u and 475 00:28:02,870 --> 00:28:05,570 corresponding to the v displacements. 476 00:28:05,570 --> 00:28:11,450 Remember that the only unknown u displacements are these 4. 477 00:28:11,450 --> 00:28:18,850 And these 4 u displacements will give us alpha 1, alpha 2, 478 00:28:18,850 --> 00:28:20,620 alpha 3, and alpha 4. 479 00:28:20,620 --> 00:28:23,560 The same holds for the v displacements. 480 00:28:23,560 --> 00:28:28,290 This is a bilinear approximation of the u 481 00:28:28,290 --> 00:28:31,520 displacements in the element. 482 00:28:31,520 --> 00:28:35,510 We write this in matrix form as shown here. 483 00:28:35,510 --> 00:28:40,300 Noting that this capital Phi, notice this is a capital Phi 484 00:28:40,300 --> 00:28:42,190 because there are these cross bars. 485 00:28:42,190 --> 00:28:45,660 It's easy to confuse the capital Phi with the lowercase 486 00:28:45,660 --> 00:28:48,150 phi, which doesn't have the cross bars. 487 00:28:48,150 --> 00:28:50,730 So this capital Phi was the cross bars here 488 00:28:50,730 --> 00:28:52,580 is defined in here. 489 00:28:52,580 --> 00:28:56,290 Where the lowercase phi is simply an array 490 00:28:56,290 --> 00:29:00,020 of 1, x, y, x, y. 491 00:29:00,020 --> 00:29:03,190 This relationship here is nothing else than that 492 00:29:03,190 --> 00:29:06,340 relationship, but written in matrix form. 493 00:29:06,340 --> 00:29:11,020 The alpha lists all the generalized coordinates. 494 00:29:11,020 --> 00:29:12,050 The generalize coordinates. 495 00:29:12,050 --> 00:29:14,850 And this is the reason why we call this the generalized 496 00:29:14,850 --> 00:29:19,300 coordinate finite element formulation for this 497 00:29:19,300 --> 00:29:20,940 particular element. 498 00:29:20,940 --> 00:29:24,520 Let us now define the u hat vector, 499 00:29:24,520 --> 00:29:26,280 which I refer to earlier. 500 00:29:26,280 --> 00:29:31,140 This case, the u hat vector involves 4 u nodal point 501 00:29:31,140 --> 00:29:35,420 displacements and 4 v nodal point displacements. 502 00:29:35,420 --> 00:29:37,910 Now if we apply-- 503 00:29:37,910 --> 00:29:40,700 and this is really one important point. 504 00:29:40,700 --> 00:29:47,410 If we apply this relationship here, which of course must 505 00:29:47,410 --> 00:29:49,580 hold all over the element. 506 00:29:49,580 --> 00:29:52,470 In particular, it must also hold for the nodal points of 507 00:29:52,470 --> 00:29:53,770 the element. 508 00:29:53,770 --> 00:29:58,220 If we apply this relationship for the nodal points of the 509 00:29:58,220 --> 00:30:02,730 elements, we can apply it eight times because we have 8 510 00:30:02,730 --> 00:30:05,430 nodal point displacements. 511 00:30:05,430 --> 00:30:10,320 We directly generate this relationship here. 512 00:30:10,320 --> 00:30:14,470 u hat equals A times alpha, where A involve the 513 00:30:14,470 --> 00:30:17,600 coordinates of the nodal points. 514 00:30:17,600 --> 00:30:21,670 Hence, we have our H Phi times A inverse the way I have 515 00:30:21,670 --> 00:30:24,240 derived it earlier. 516 00:30:24,240 --> 00:30:28,920 So this is a very systematic approach of deriving the 517 00:30:28,920 --> 00:30:32,190 displacement interpolation matrix. 518 00:30:32,190 --> 00:30:35,530 It consists of evaluating the A matrix. 519 00:30:35,530 --> 00:30:39,250 And of course, we assume that that A matrix can be inverted. 520 00:30:39,250 --> 00:30:42,670 This is, for this element, certainly the case, as long as 521 00:30:42,670 --> 00:30:47,070 we are talking about the rectangular element or not 522 00:30:47,070 --> 00:30:48,890 very highly distorted elements. 523 00:30:48,890 --> 00:30:52,000 We can for the four nodal element, invert this matrix 524 00:30:52,000 --> 00:30:55,140 and directly get this relationship. 525 00:30:55,140 --> 00:31:00,130 Let's look now, once more back at the analysis that we wanted 526 00:31:00,130 --> 00:31:01,260 to perform. 527 00:31:01,260 --> 00:31:05,450 Remember that we looked at this element here, and we have 528 00:31:05,450 --> 00:31:10,910 now obtained the H matrix, the displacement interpolation 529 00:31:10,910 --> 00:31:13,200 matrix for this element. 530 00:31:13,200 --> 00:31:15,810 Once more, here is the H matrix. 531 00:31:15,810 --> 00:31:19,270 In fact, this H matrix here, since we have no superscript 532 00:31:19,270 --> 00:31:23,680 attached to it yet, really holds for every one of the 4 533 00:31:23,680 --> 00:31:25,290 elements that we are talking about. 534 00:31:25,290 --> 00:31:32,130 Provided we put into the u hat here the appropriate nodal 535 00:31:32,130 --> 00:31:34,350 point displacements of the elements. 536 00:31:34,350 --> 00:31:39,170 We can directly say that this matrix here holds for any one 537 00:31:39,170 --> 00:31:40,420 of the elements. 538 00:31:42,400 --> 00:31:45,230 If the elements, if the 4 elements are not completely 539 00:31:45,230 --> 00:31:48,630 identical, of course, the A inverse will change from 540 00:31:48,630 --> 00:31:49,880 element to element. 541 00:31:52,290 --> 00:31:59,620 Having gone through that step and having looked at element 2 542 00:31:59,620 --> 00:32:05,330 once more, we directly recognize how to establish now 543 00:32:05,330 --> 00:32:08,600 the H2 matrix. 544 00:32:08,600 --> 00:32:14,750 Remember our u1 will be related to the global U11. 545 00:32:14,750 --> 00:32:17,626 Our v1 will be related to the-- 546 00:32:17,626 --> 00:32:23,050 or will correspond I should rather say, to the global U12. 547 00:32:23,050 --> 00:32:25,760 Let's look at why this is the case. 548 00:32:25,760 --> 00:32:29,570 Why did I write down U11 here and U12 there? 549 00:32:29,570 --> 00:32:33,360 Well, let's look at this point there. 550 00:32:33,360 --> 00:32:36,710 Which is of course, the local nodal point 1. 551 00:32:36,710 --> 00:32:41,000 The local nodal point 1, right upperhand corner of element 1, 552 00:32:41,000 --> 00:32:45,100 corresponds here to nodal point 6. 553 00:32:45,100 --> 00:32:49,590 Well, the global nodal point displacements of 554 00:32:49,590 --> 00:32:53,870 nodal point 6 are U11. 555 00:32:53,870 --> 00:32:55,180 Let me put them in. 556 00:32:55,180 --> 00:32:56,760 And U12. 557 00:32:56,760 --> 00:32:58,220 Why is that the case? 558 00:32:58,220 --> 00:33:00,970 Well, if we start numbering here-- and I'm going to 559 00:33:00,970 --> 00:33:05,770 include now, I'm going to include now the displacement 560 00:33:05,770 --> 00:33:08,590 of these nodal points, although we know that they are 561 00:33:08,590 --> 00:33:11,200 actually 0, that boundary condition we can 562 00:33:11,200 --> 00:33:13,910 impose later on. 563 00:33:13,910 --> 00:33:20,035 So if I start numbering here, I would have U1, U2, 3, 4, 5, 564 00:33:20,035 --> 00:33:24,940 6, 7, 8, 9, 10, 11, 12. 565 00:33:24,940 --> 00:33:29,990 So the global nodal point displacements are U11, U12 566 00:33:29,990 --> 00:33:32,270 corresponding to nodal point 9. 567 00:33:32,270 --> 00:33:36,460 And those are the ones that appear right here 568 00:33:36,460 --> 00:33:41,400 corresponding to the local U1 and the local V1. 569 00:33:41,400 --> 00:33:46,290 Well, if we proceed in the same way for all of the other 570 00:33:46,290 --> 00:33:50,700 nodal points for element 2, and these are the results. 571 00:33:50,700 --> 00:33:56,810 You can look these up in the textbook figure 4.6, or in the 572 00:33:56,810 --> 00:33:58,970 accompanying notes. 573 00:33:58,970 --> 00:34:05,100 We can directly construct the H2 matrix for the element. 574 00:34:05,100 --> 00:34:13,239 And now the H2 matrix, of course, involves as the vector 575 00:34:13,239 --> 00:34:19,520 the global nodal point displacements, U1 to U18. 576 00:34:19,520 --> 00:34:22,690 The H matrix for element 2. 577 00:34:22,690 --> 00:34:26,239 And if the 4 elements are identical, we will have the 578 00:34:26,239 --> 00:34:30,300 same H matrix for all the elements. 579 00:34:30,300 --> 00:34:34,400 The H matrix here looks as shown here. 580 00:34:34,400 --> 00:34:36,800 Notice that we have a one quarter there. 581 00:34:36,800 --> 00:34:39,139 It's a little arrow here. 582 00:34:39,139 --> 00:34:41,750 The bracket should've been there. 583 00:34:41,750 --> 00:34:42,679 Bracket should have been there. 584 00:34:42,679 --> 00:34:45,780 So we have a one quarter outside of the bracket. 585 00:34:45,780 --> 00:34:49,679 And we have as the first entry, this one here. 586 00:34:49,679 --> 00:34:52,570 This corresponds to U1. 587 00:34:52,570 --> 00:34:54,600 This is the entry corresponding to U1. 588 00:34:54,600 --> 00:34:56,330 I left the entries corresponding 589 00:34:56,330 --> 00:34:59,350 to U2, U3, U4 out. 590 00:34:59,350 --> 00:35:02,030 This is the entry corresponding to V1. 591 00:35:02,030 --> 00:35:06,800 And I left out the entries corresponding to V2, V3, V4. 592 00:35:06,800 --> 00:35:09,450 This is our H matrix here. 593 00:35:09,450 --> 00:35:13,200 And now we basically want to blow it up to obtain the H2 594 00:35:13,200 --> 00:35:15,920 matrix corresponding to element 2. 595 00:35:15,920 --> 00:35:22,020 Well, if we recognize as we just did that U1, the local 596 00:35:22,020 --> 00:35:29,910 U1, the local V1, correspond to capital U11 and capital 597 00:35:29,910 --> 00:35:34,350 U12, and I just showed you that that is indeed the case. 598 00:35:34,350 --> 00:35:38,480 We would simply take the first element here, this element 599 00:35:38,480 --> 00:35:42,270 here, and put it right there. 600 00:35:42,270 --> 00:35:45,990 This element here would go right there. 601 00:35:45,990 --> 00:35:50,410 This element here would go right there. 602 00:35:50,410 --> 00:35:53,630 And this element here would go right there. 603 00:35:53,630 --> 00:35:57,410 In fact, H1 5 is 0 as you can see here. 604 00:35:57,410 --> 00:35:58,910 And H2 1 is 0. 605 00:35:58,910 --> 00:36:01,970 So H1 1 is this term here. 606 00:36:01,970 --> 00:36:04,960 And H2 5 is that term here. 607 00:36:04,960 --> 00:36:08,090 Those are two non-zero entries in the complete H matrix. 608 00:36:10,590 --> 00:36:13,840 Well, this is then the way how we actually can proceed to 609 00:36:13,840 --> 00:36:16,330 construct the H2 matrix. 610 00:36:16,330 --> 00:36:21,510 I pointed out earlier in lecture three already that we 611 00:36:21,510 --> 00:36:24,760 do not perform this step computationally. 612 00:36:24,760 --> 00:36:28,040 In fact, what we do computationally is we stay 613 00:36:28,040 --> 00:36:29,740 with this H matrix. 614 00:36:29,740 --> 00:36:33,220 We calculate a compacted stiffness matrix, a compacted 615 00:36:33,220 --> 00:36:36,950 load vector, which the stiffness matrix will be of 616 00:36:36,950 --> 00:36:38,680 order 8 by 8. 617 00:36:38,680 --> 00:36:44,390 And then we assemble these element matrices with 618 00:36:44,390 --> 00:36:48,040 identification arrays and connectivity arrays into the 619 00:36:48,040 --> 00:36:50,540 global structural stiffness matrix, into 620 00:36:50,540 --> 00:36:53,690 the global load vectors. 621 00:36:53,690 --> 00:36:57,650 So we really don't perform this step of 622 00:36:57,650 --> 00:37:00,370 blowing up the H matrix. 623 00:37:00,370 --> 00:37:03,660 And I will show you in the next lecture how we actually 624 00:37:03,660 --> 00:37:08,130 perform the computations of assembling element matrices. 625 00:37:08,130 --> 00:37:12,420 However, for theoretical purposes, it is neat if you 626 00:37:12,420 --> 00:37:17,030 understand how we can construct this H2 matrix from 627 00:37:17,030 --> 00:37:18,570 this H matrix. 628 00:37:18,570 --> 00:37:23,460 And of course, similarly, we could construct H1, H3, and H4 629 00:37:23,460 --> 00:37:26,480 from this basic H matrix. 630 00:37:26,480 --> 00:37:33,180 Once we have done that, we can directly write that k being 631 00:37:33,180 --> 00:37:36,060 equal to the sum of the km. 632 00:37:36,060 --> 00:37:39,670 In other words, the global structure stiffness matrix is 633 00:37:39,670 --> 00:37:43,400 equal to the sum of the element matrices, stiffness 634 00:37:43,400 --> 00:37:51,490 matrices, where each of these km's has the same order as the 635 00:37:51,490 --> 00:37:52,960 structure stiffness matrix k. 636 00:37:56,690 --> 00:38:02,810 Let us now look briefly at the development of the strain 637 00:38:02,810 --> 00:38:04,390 displacement matrix. 638 00:38:04,390 --> 00:38:08,430 In plane stress conditions, we have three entries. 639 00:38:08,430 --> 00:38:10,590 The strains are given as shown here. 640 00:38:10,590 --> 00:38:13,460 Elementary strengths materials tells us that 641 00:38:13,460 --> 00:38:15,210 these are the strains. 642 00:38:15,210 --> 00:38:21,480 And I showed earlier that the B matrix is constructed by 643 00:38:21,480 --> 00:38:23,950 multiplying the E times the A inverse. 644 00:38:23,950 --> 00:38:26,950 The A inverse was already used in the calculation of the 645 00:38:26,950 --> 00:38:30,060 displacement interpolation matrix. 646 00:38:30,060 --> 00:38:36,630 The E matrix is obtained by appropriately differentiating 647 00:38:36,630 --> 00:38:43,110 the entries in the displacement interpolations. 648 00:38:43,110 --> 00:38:50,970 And this differentiation is performed on the phi matrix 649 00:38:50,970 --> 00:38:52,240 right here. 650 00:38:52,240 --> 00:38:55,750 Or rather, on these entries here. 651 00:38:55,750 --> 00:38:59,650 By simply differentiating these entries here, we would 652 00:38:59,650 --> 00:39:06,390 obtain the strains and writing those in matrix form, we 653 00:39:06,390 --> 00:39:09,940 obtain the E matrix here. 654 00:39:09,940 --> 00:39:16,000 Once we have the B matrix, which in this particular case 655 00:39:16,000 --> 00:39:22,390 is 3 by 8 matrix for a typical element, we can, of course, 656 00:39:22,390 --> 00:39:29,080 construct the B2 matrix for element 2 just in the same way 657 00:39:29,080 --> 00:39:31,000 as the H2 matrix. 658 00:39:31,000 --> 00:39:33,190 Of course, we would again, blow up the B 659 00:39:33,190 --> 00:39:35,205 matrix from a 3 to 8. 660 00:39:35,205 --> 00:39:40,400 3 by 8 I should rather say, to a 3 by 18 now because we have 661 00:39:40,400 --> 00:39:42,360 actually 18 degrees of freedom in the 662 00:39:42,360 --> 00:39:44,910 complete structural model. 663 00:39:44,910 --> 00:39:47,190 This then completes what I wanted to say about the 664 00:39:47,190 --> 00:39:49,830 construction of the displacement and 665 00:39:49,830 --> 00:39:51,960 strain-displacement interpolation matrix for the 666 00:39:51,960 --> 00:39:55,470 element, the H2 and B2 matrices. 667 00:39:55,470 --> 00:39:59,380 However, let me just spend a few moments on the 668 00:39:59,380 --> 00:40:02,680 construction of the surface displacement 669 00:40:02,680 --> 00:40:03,880 interpolation matrix. 670 00:40:03,880 --> 00:40:06,980 In other words, the interpolation of the 671 00:40:06,980 --> 00:40:10,270 displacements along the surface of an element. 672 00:40:10,270 --> 00:40:14,860 Here, assume for example, that this is a surface of the 673 00:40:14,860 --> 00:40:16,890 elements that we are concerned with. 674 00:40:16,890 --> 00:40:19,890 And that there would be some loading applied to that 675 00:40:19,890 --> 00:40:21,510 surface of the element. 676 00:40:21,510 --> 00:40:26,080 So that we need, in other words, the HSM matrix the way 677 00:40:26,080 --> 00:40:28,650 I have been pointing it out earlier. 678 00:40:28,650 --> 00:40:33,300 What we would do to obtain this matrix here, is really 679 00:40:33,300 --> 00:40:38,330 simply use our earlier results for the HS-- 680 00:40:38,330 --> 00:40:41,890 sorry, for the HM matrix. 681 00:40:41,890 --> 00:40:46,830 And here we have the H matrix and the H2 682 00:40:46,830 --> 00:40:49,380 matrix for element 2. 683 00:40:49,380 --> 00:40:52,570 What we would do is we would use this matrix, or this one 684 00:40:52,570 --> 00:40:55,820 of course, this matrix is contained in this one. 685 00:40:55,820 --> 00:41:00,830 And simply, substitute the particular coordinates along 686 00:41:00,830 --> 00:41:01,790 the surface. 687 00:41:01,790 --> 00:41:04,440 Now, going back once more to the element under 688 00:41:04,440 --> 00:41:05,950 consideration. 689 00:41:05,950 --> 00:41:10,380 If we are looking at this surface, then y is constant 690 00:41:10,380 --> 00:41:13,700 and equal to this distance here. 691 00:41:13,700 --> 00:41:19,060 So knowing y, we would simply substitute that value of y 692 00:41:19,060 --> 00:41:27,480 into this function here, and thus obtain the HS matrix. 693 00:41:27,480 --> 00:41:31,480 So, in general, therefore the HS matrix is simply obtained 694 00:41:31,480 --> 00:41:36,700 from the H matrix, or the HS2 matrix here would simply be 695 00:41:36,700 --> 00:41:42,050 obtained from the H2 matrix by substituting the coordinates 696 00:41:42,050 --> 00:41:46,470 along the line, or along the surface that we are 697 00:41:46,470 --> 00:41:47,720 considering. 698 00:41:49,410 --> 00:41:53,860 Let us now to look at the overall process of the finite 699 00:41:53,860 --> 00:41:55,820 element solution. 700 00:41:55,820 --> 00:41:59,990 Here I have summarized what we all are concerned about. 701 00:41:59,990 --> 00:42:02,800 Of course we have the actual physical problem that we want 702 00:42:02,800 --> 00:42:04,350 to analyze. 703 00:42:04,350 --> 00:42:08,450 We have a geometric domain that we have to discretize. 704 00:42:08,450 --> 00:42:11,270 A material that has to be represented, a loading and the 705 00:42:11,270 --> 00:42:12,750 boundary conditions. 706 00:42:12,750 --> 00:42:16,000 All of those have to be represented. 707 00:42:16,000 --> 00:42:20,150 In the mechanic idealization, the idealization step that is 708 00:42:20,150 --> 00:42:23,870 a very important step that I referred to earlier, we would 709 00:42:23,870 --> 00:42:26,890 idealize the kinematics as being a truss, plane stress, 710 00:42:26,890 --> 00:42:29,930 three-dimensional, Kirchhoff plate, kinematics. 711 00:42:29,930 --> 00:42:34,000 The material isotopic elastic, the loading, and the boundary 712 00:42:34,000 --> 00:42:34,910 conditions. 713 00:42:34,910 --> 00:42:37,560 And this mechanical idealization really gives us a 714 00:42:37,560 --> 00:42:41,180 differential equation or equilibrium for 715 00:42:41,180 --> 00:42:45,160 one-dimensional stress situation, a bar. 716 00:42:45,160 --> 00:42:49,190 We will have an equation such as this one. 717 00:42:49,190 --> 00:42:52,420 In the finite element solution then, we operate on this 718 00:42:52,420 --> 00:42:53,810 equation basically. 719 00:42:53,810 --> 00:42:57,840 We want to solve the governing differential equation of the 720 00:42:57,840 --> 00:43:00,920 mechanical idealization. 721 00:43:00,920 --> 00:43:05,340 The errors that we are therefore, concerned with are 722 00:43:05,340 --> 00:43:08,580 discretization errors by the use of the finite element 723 00:43:08,580 --> 00:43:09,840 interpolations. 724 00:43:09,840 --> 00:43:13,070 And then, a number of other errors. 725 00:43:13,070 --> 00:43:14,550 Numerical integration in space. 726 00:43:14,550 --> 00:43:16,990 If we use numerical integration, there would be 727 00:43:16,990 --> 00:43:18,530 other errors introduced. 728 00:43:18,530 --> 00:43:21,270 In the evaluation of the constitutive relations, again, 729 00:43:21,270 --> 00:43:22,800 errors would be introduced. 730 00:43:22,800 --> 00:43:25,380 The solution of the dynamic equilibrium equations, by 731 00:43:25,380 --> 00:43:26,810 [UNINTELLIGIBLE] time integration, mode 732 00:43:26,810 --> 00:43:29,890 superposition further errors would be introduced. 733 00:43:29,890 --> 00:43:32,780 If you solved the governing equilibrium equations, ku 734 00:43:32,780 --> 00:43:36,180 equals r, by iteration there would be further errors 735 00:43:36,180 --> 00:43:37,120 introduced. 736 00:43:37,120 --> 00:43:40,130 And of course, finally there are round-off errors on the 737 00:43:40,130 --> 00:43:41,270 digital computer. 738 00:43:41,270 --> 00:43:46,720 What I want to do now in discussing convergence, I want 739 00:43:46,720 --> 00:43:50,260 to say that all of these errors are negligible. 740 00:43:50,260 --> 00:43:53,770 In other words, we are basically evaluating the 741 00:43:53,770 --> 00:43:56,380 integrations in space exactly. 742 00:43:56,380 --> 00:43:59,600 There is no round-off because we have an infinite precision 743 00:43:59,600 --> 00:44:01,200 machine, and so on. 744 00:44:01,200 --> 00:44:04,760 So all the errors that we are looking at now are the 745 00:44:04,760 --> 00:44:06,500 discretization errors, 746 00:44:06,500 --> 00:44:08,860 And if we look only at the discretization 747 00:44:08,860 --> 00:44:12,210 errors, we can say some-- 748 00:44:12,210 --> 00:44:15,540 make some very fundamental remarks about the convergence 749 00:44:15,540 --> 00:44:18,310 of the finite elements scheme. 750 00:44:18,310 --> 00:44:22,330 I've summarized here on the blackboard some of these very 751 00:44:22,330 --> 00:44:24,200 important concepts. 752 00:44:24,200 --> 00:44:26,300 When we talk about convergence, let us assume 753 00:44:26,300 --> 00:44:30,560 first now that we are having a compatible element layout. 754 00:44:30,560 --> 00:44:33,270 In other words, we're using a compatible element layout. 755 00:44:33,270 --> 00:44:37,480 And then, we know that we have monotonic convergence to the 756 00:44:37,480 --> 00:44:40,410 solution of the problem-governing differential 757 00:44:40,410 --> 00:44:45,460 equations provided the elements that we are using 758 00:44:45,460 --> 00:44:46,420 contain: 759 00:44:46,420 --> 00:44:49,800 Firstly, all required rigid body modes. 760 00:44:49,800 --> 00:44:52,950 This means, for example, that for in a plane stress 761 00:44:52,950 --> 00:44:56,850 analysis, the element must be able to undergo three rigid 762 00:44:56,850 --> 00:45:00,230 body modes, two translations, horizontally and vertically, 763 00:45:00,230 --> 00:45:03,040 and a rigid body rotation. 764 00:45:03,040 --> 00:45:07,100 And the element must be able to represent the required 765 00:45:07,100 --> 00:45:08,790 constant strain states. 766 00:45:08,790 --> 00:45:12,220 In a plane stress analysis, three constant strain states. 767 00:45:12,220 --> 00:45:17,170 Pulling this way, pulling that way, and the constant shear. 768 00:45:17,170 --> 00:45:20,600 If these two conditions are satisfied, we have monotonic 769 00:45:20,600 --> 00:45:23,790 convergence in a compatible element layout. 770 00:45:23,790 --> 00:45:26,380 What do we mean by a compatible element layout? 771 00:45:26,380 --> 00:45:30,930 Well, here I have show a small picture showing two elements, 772 00:45:30,930 --> 00:45:32,470 a white and an orange element. 773 00:45:32,470 --> 00:45:37,190 And these elements are compatible because we have 774 00:45:37,190 --> 00:45:41,050 three nodes along this line for each of the elements. 775 00:45:41,050 --> 00:45:44,440 So the white element displacements basically vary 776 00:45:44,440 --> 00:45:48,040 parabolically along this line, and so do the orange element 777 00:45:48,040 --> 00:45:49,820 displacements. 778 00:45:49,820 --> 00:45:53,460 No gap can develop between these two lines. 779 00:45:53,460 --> 00:45:55,950 Here we have an incompatible element layout. 780 00:45:55,950 --> 00:45:59,940 Two nodes that describe only a linear variation 781 00:45:59,940 --> 00:46:02,370 for the white element. 782 00:46:02,370 --> 00:46:06,760 But three nodes for the orange element describing a parabolic 783 00:46:06,760 --> 00:46:08,010 variation in displacement. 784 00:46:08,010 --> 00:46:11,660 Therefore a gap can open up and this is an incompatible 785 00:46:11,660 --> 00:46:13,740 element layout. 786 00:46:13,740 --> 00:46:17,090 Well, let us assume then that we have a compatible element 787 00:46:17,090 --> 00:46:18,940 layout fist of all. 788 00:46:18,940 --> 00:46:22,260 Then, typically for an analysis such as this one, a 789 00:46:22,260 --> 00:46:26,230 simple cantilever subjected to a tip load, the convergence 790 00:46:26,230 --> 00:46:29,140 that we would observe would be the following. 791 00:46:29,140 --> 00:46:32,240 As we increase the number of elements that idealize the 792 00:46:32,240 --> 00:46:36,490 cantilever, the displacement measured here would 793 00:46:36,490 --> 00:46:41,700 monotonically approach from below the exact displacement. 794 00:46:41,700 --> 00:46:44,130 What do we mean by exact displacement? 795 00:46:44,130 --> 00:46:46,220 Notice I put it into quotes. 796 00:46:46,220 --> 00:46:48,940 Well, the exact displacement that I'm talking about here is 797 00:46:48,940 --> 00:46:52,190 the tip displacements that we would calculate from the 798 00:46:52,190 --> 00:46:54,580 problem-governing differential equation. 799 00:46:57,670 --> 00:46:59,870 Of course, that problem-governing differential 800 00:46:59,870 --> 00:47:03,940 equation also corresponds to a mechanical idealization of the 801 00:47:03,940 --> 00:47:06,840 actual physical situation. 802 00:47:06,840 --> 00:47:09,050 And depending on which problem-governing differential 803 00:47:09,050 --> 00:47:13,230 equation we choose, we would have a different dashed line. 804 00:47:13,230 --> 00:47:16,850 But assuming now that we have made up our mind which one to 805 00:47:16,850 --> 00:47:19,945 choose, which mechanical idealization to choose, and we 806 00:47:19,945 --> 00:47:23,760 are choosing a corresponding finite element idealization, 807 00:47:23,760 --> 00:47:26,890 this would be the convergence that we would observe. 808 00:47:26,890 --> 00:47:29,630 And it would be monotonically from below. 809 00:47:29,630 --> 00:47:31,560 In other words, our finite element model, 810 00:47:31,560 --> 00:47:33,210 actually is too stiff. 811 00:47:33,210 --> 00:47:36,860 It is stiffer than the actual physical structure. 812 00:47:36,860 --> 00:47:41,300 Well, why I think we can intuitively understand that 813 00:47:41,300 --> 00:47:44,190 quite well if we just recognize that we are 814 00:47:44,190 --> 00:47:50,040 constraining the displacements within each element to only be 815 00:47:50,040 --> 00:47:57,130 able to basically, take patterns on that are contained 816 00:47:57,130 --> 00:47:57,840 in the element. 817 00:47:57,840 --> 00:48:01,155 In other words, for the four nodal element here, we really 818 00:48:01,155 --> 00:48:05,110 have for the U displacements and for the V displacements, 819 00:48:05,110 --> 00:48:08,560 only 4 displacement patterns each that 820 00:48:08,560 --> 00:48:09,840 the element can undergo. 821 00:48:09,840 --> 00:48:13,660 So we are constraining the exact displacements to be 822 00:48:13,660 --> 00:48:18,820 really, approximated by these displacement patterns. 823 00:48:18,820 --> 00:48:21,940 And that makes the actual finite element model stiffer 824 00:48:21,940 --> 00:48:25,350 than the actual physical structure. 825 00:48:25,350 --> 00:48:28,370 If we have an incompatible layout, and I showed earlier 826 00:48:28,370 --> 00:48:32,190 what we mean by that, then, in addition, 827 00:48:32,190 --> 00:48:34,830 every patch of elements-- 828 00:48:34,830 --> 00:48:38,490 by patch of elements we mean small assemblage of elements-- 829 00:48:38,490 --> 00:48:42,010 must be able to represent the constant strain states. 830 00:48:42,010 --> 00:48:44,150 It's not just that each element must be able to 831 00:48:44,150 --> 00:48:46,090 represent the constant strain state, but 832 00:48:46,090 --> 00:48:47,990 each patch of elements. 833 00:48:47,990 --> 00:48:53,250 Also, then if this holds true, then we have convergence. 834 00:48:53,250 --> 00:48:55,640 But non-monotonic convergence. 835 00:48:55,640 --> 00:48:58,940 By non-monotonic convergence, I mean the following. 836 00:48:58,940 --> 00:49:02,140 We might get this result for a certain set of elements. 837 00:49:02,140 --> 00:49:04,760 And as we increase the number of elements, we are 838 00:49:04,760 --> 00:49:09,450 oscillating about the correct result, the exact result in 839 00:49:09,450 --> 00:49:12,570 quotes again, and finally we converge. 840 00:49:12,570 --> 00:49:16,270 This is non-monotonic convergence because we can 841 00:49:16,270 --> 00:49:20,700 have solutions lying on either side of the exact result. 842 00:49:20,700 --> 00:49:31,490 Well, let me now show to you some of the facts that I just 843 00:49:31,490 --> 00:49:33,900 talked about on this view graph. 844 00:49:33,900 --> 00:49:35,480 What do these mean? 845 00:49:35,480 --> 00:49:39,770 Well, I said already that each element in every case must be 846 00:49:39,770 --> 00:49:43,480 able to represent the rigid body modes. 847 00:49:43,480 --> 00:49:46,910 Here for a plane stress analysis of a cantilever beam 848 00:49:46,910 --> 00:49:49,790 shown here, we have 3 rigid body modes. 849 00:49:49,790 --> 00:49:53,430 And as I mentioned earlier briefly already, the element 850 00:49:53,430 --> 00:49:57,260 must be able to go over, go up as a rigid body and rotate as 851 00:49:57,260 --> 00:49:59,370 a rigid body. 852 00:49:59,370 --> 00:50:00,610 Why is this the case? 853 00:50:00,610 --> 00:50:03,060 Well, if we look at an analysis of the cantilever 854 00:50:03,060 --> 00:50:06,280 where we have not a tip load, but a load say, at this end 855 00:50:06,280 --> 00:50:09,650 here, we know that the bending moment looks this way. 856 00:50:09,650 --> 00:50:16,700 Hence, these elements over here must be able to freely 857 00:50:16,700 --> 00:50:18,790 rotate and displace. 858 00:50:18,790 --> 00:50:22,940 In other words, we know that the exact condition, the exact 859 00:50:22,940 --> 00:50:25,630 physical condition, there should be no stress here. 860 00:50:25,630 --> 00:50:29,450 The stress should be 0 in this element. 861 00:50:29,450 --> 00:50:33,600 And this means that the tip element here must be able to 862 00:50:33,600 --> 00:50:35,780 translate freely as a rigid body and 863 00:50:35,780 --> 00:50:37,340 rotate as a rigid body. 864 00:50:37,340 --> 00:50:39,040 If this were not the case, then we would 865 00:50:39,040 --> 00:50:40,600 get stresses here. 866 00:50:40,600 --> 00:50:44,640 And of course, our analysis or finite element model can never 867 00:50:44,640 --> 00:50:48,250 predict the actual physical conditions accurately or 868 00:50:48,250 --> 00:50:52,450 closely, even if we take an infinite number of elements. 869 00:50:52,450 --> 00:50:56,860 So this somehow explains to you intuitively why the rigid 870 00:50:56,860 --> 00:50:59,330 body more criteria must be satisfied. 871 00:50:59,330 --> 00:51:02,260 How about the constant strain state criteria? 872 00:51:02,260 --> 00:51:06,940 Well, if we assume that we take more and more elements-- 873 00:51:06,940 --> 00:51:10,820 and let me draw in here such elements. 874 00:51:10,820 --> 00:51:14,020 For example, now we have gone from a certain number of 875 00:51:14,020 --> 00:51:17,940 elements to a larger number of elements. 876 00:51:17,940 --> 00:51:21,400 These elements, as we keep on refining the mesh, would 877 00:51:21,400 --> 00:51:23,290 become very small, and would go down. 878 00:51:23,290 --> 00:51:25,810 In fact, each element would go down to a point. 879 00:51:25,810 --> 00:51:29,860 But at a point we only have one constant stress basically. 880 00:51:29,860 --> 00:51:33,310 So the element must be able in the limit, to represent the 881 00:51:33,310 --> 00:51:36,240 constant stress at a point. 882 00:51:36,240 --> 00:51:40,270 And therefore, we have the constant stress criteria also. 883 00:51:40,270 --> 00:51:44,330 One way of finding out whether an element satisfies these 884 00:51:44,330 --> 00:51:48,180 criteria, the constant stress criteria and the rigid body 885 00:51:48,180 --> 00:51:50,580 mode criteria is to calculate its eigenvalues. 886 00:51:50,580 --> 00:51:52,260 Here, we have taken ones. 887 00:51:52,260 --> 00:51:54,770 A four node plane stress element and calculate the 888 00:51:54,770 --> 00:51:55,560 eigenvalues. 889 00:51:55,560 --> 00:52:00,790 Notice the lowest three eigenvalues are 0, 890 00:52:00,790 --> 00:52:03,710 representing the rigid body modes. 891 00:52:03,710 --> 00:52:07,600 And then, we have a flexural mode here, the fourth 892 00:52:07,600 --> 00:52:09,310 eigenvalue. 893 00:52:09,310 --> 00:52:12,170 The fifth eigenvalue is again the flexural mode. 894 00:52:12,170 --> 00:52:14,610 And here is the constant sharing mode. 895 00:52:14,610 --> 00:52:17,500 And here are the two constant, one 896 00:52:17,500 --> 00:52:19,350 compression and tension modes. 897 00:52:19,350 --> 00:52:23,310 So these are the constant strain states that the element 898 00:52:23,310 --> 00:52:24,800 can represent therefore. 899 00:52:24,800 --> 00:52:29,800 And earlier I showed you the three rigid body modes. 900 00:52:29,800 --> 00:52:34,660 So surely, this element therefore satisfies both of 901 00:52:34,660 --> 00:52:36,890 the criteria. 902 00:52:36,890 --> 00:52:43,770 Let's look now at what happens when we don't have a 903 00:52:43,770 --> 00:52:46,220 compatible mesh. 904 00:52:46,220 --> 00:52:49,600 We have an additional condition that every patch of 905 00:52:49,600 --> 00:52:52,810 elements must also be able to represent the 906 00:52:52,810 --> 00:52:55,030 constant strain states. 907 00:52:55,030 --> 00:52:58,000 Only then we will have convergence. 908 00:52:58,000 --> 00:53:00,820 But we will not necessarily have monotonic convergence. 909 00:53:00,820 --> 00:53:03,870 We can only talk about non-monotonic convergence. 910 00:53:03,870 --> 00:53:08,150 Well, here I've taken a patch of elements. 911 00:53:08,150 --> 00:53:12,640 Each element by itself satisfies the constant strain 912 00:53:12,640 --> 00:53:15,890 criterion and the rigid body more criterion. 913 00:53:15,890 --> 00:53:21,620 In this case here, we use a compatible element layout. 914 00:53:21,620 --> 00:53:26,440 Compatible because along each side, the adjacent elements 915 00:53:26,440 --> 00:53:29,210 shares the nodes. 916 00:53:29,210 --> 00:53:32,410 And we are subjecting this patch of elements to a total 917 00:53:32,410 --> 00:53:36,180 stress here of 100 Newton per square centimeters. 918 00:53:36,180 --> 00:53:38,880 Of course, the stress in each of these elements would be 919 00:53:38,880 --> 00:53:42,880 simply a stress this direction of 1,000. 920 00:53:42,880 --> 00:53:45,690 Excuse me, 1,000 not 100. 921 00:53:45,690 --> 00:53:49,570 The other stress, the vertical stress and the shear stress is 922 00:53:49,570 --> 00:53:52,240 0 for this patch of elements. 923 00:53:52,240 --> 00:53:54,840 Let us now perform the following experiment. 924 00:53:54,840 --> 00:54:00,530 Let us say that we are assigning two nodes, two 925 00:54:00,530 --> 00:54:05,570 different nodes to element 4 and 5 at this location. 926 00:54:05,570 --> 00:54:08,570 Here they share the same node. 927 00:54:08,570 --> 00:54:10,890 Let us do the same at this end. 928 00:54:10,890 --> 00:54:14,090 And let us say that node 17 only belongs to 929 00:54:14,090 --> 00:54:16,160 this element 4. 930 00:54:16,160 --> 00:54:18,730 Node 18 belongs only to element 6. 931 00:54:18,730 --> 00:54:22,170 Node 20 and 19 belong to element 5. 932 00:54:22,170 --> 00:54:28,130 In other words, node 17 can basically gives this 933 00:54:28,130 --> 00:54:29,080 displacement pattern. 934 00:54:29,080 --> 00:54:30,950 And 19 gives this displacement pattern. 935 00:54:30,950 --> 00:54:33,880 An incompatibility can reside here. 936 00:54:33,880 --> 00:54:35,940 And similarly, along this line. 937 00:54:35,940 --> 00:54:41,910 If we subject this patch of elements to the stress state 938 00:54:41,910 --> 00:54:45,920 of the external forces that we used here too, we would find 939 00:54:45,920 --> 00:54:50,230 that the stresses at the points A, B, C, D, E shown 940 00:54:50,230 --> 00:54:53,580 along here, which here were, of course, exactly equal to 941 00:54:53,580 --> 00:55:01,530 1,000 for this stress tau yy and 0 in this direction. 942 00:55:01,530 --> 00:55:02,820 And the sheer stress was 0. 943 00:55:02,820 --> 00:55:07,480 If you look at the tau yy at A, B, C, D, E now the 944 00:55:07,480 --> 00:55:11,540 incompatible element mesh layout, we would find the 945 00:55:11,540 --> 00:55:13,450 following stress result. 946 00:55:13,450 --> 00:55:16,380 Notice that the stresses now are not 1,000. 947 00:55:16,380 --> 00:55:18,930 In fact, they vary drastically from 1,000. 948 00:55:18,930 --> 00:55:22,300 359 instead of 1,000. 949 00:55:22,300 --> 00:55:27,480 This means that this patch of elements with the 950 00:55:27,480 --> 00:55:32,820 incompatibility here, does not represent exactly the constant 951 00:55:32,820 --> 00:55:36,560 stress state that it should represent because we are 952 00:55:36,560 --> 00:55:40,080 subjecting it to a constant stress state. 953 00:55:40,080 --> 00:55:43,980 Therefore, this patch of element does not satisfy the 954 00:55:43,980 --> 00:55:47,500 requirement for convergence. 955 00:55:47,500 --> 00:55:51,450 Not monotonic convergence and not non-monotonic convergence. 956 00:55:51,450 --> 00:55:57,350 It just should not be used in an actual practical analysis. 957 00:55:57,350 --> 00:56:01,610 This completes what I wanted to say then in this lecture. 958 00:56:01,610 --> 00:56:03,310 Thank you very much for your attention.