1 00:00:00,000 --> 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 --> 00:00:17,872 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,872 --> 00:00:19,122 ocw.mit.edu. 8 00:00:21,290 --> 00:00:23,770 PROFESSOR: Ladies and gentlemen, welcome to lecture 9 00:00:23,770 --> 00:00:25,060 number seven. 10 00:00:25,060 --> 00:00:27,260 In this lecture, I would like to present to you the 11 00:00:27,260 --> 00:00:29,740 formulation of structural elements. 12 00:00:29,740 --> 00:00:34,040 We will be discussing beam, plate, and shell elements, and 13 00:00:34,040 --> 00:00:38,450 I would like to introduce to you the isoparametric approach 14 00:00:38,450 --> 00:00:40,570 for interpolations. 15 00:00:40,570 --> 00:00:42,380 There are two approaches in the 16 00:00:42,380 --> 00:00:44,170 formulation that we can follow. 17 00:00:44,170 --> 00:00:47,870 The first one is a strength of materials approach, in which 18 00:00:47,870 --> 00:00:51,560 case we look at a straight beam element, we use a beam 19 00:00:51,560 --> 00:00:53,990 theory including shear effects. 20 00:00:53,990 --> 00:00:57,120 If you look at a plate element, we use a plate theory 21 00:00:57,120 --> 00:00:59,040 including shear effects, also. 22 00:00:59,040 --> 00:01:01,990 The names associated with these theories that we're 23 00:01:01,990 --> 00:01:06,900 using are the names of Reissner and Mindlin. 24 00:01:06,900 --> 00:01:09,850 In the second approach, we have the continuum mechanics 25 00:01:09,850 --> 00:01:13,360 approach, in which we use the general principle of virtual 26 00:01:13,360 --> 00:01:18,210 displacement, but we exclude the stress components not 27 00:01:18,210 --> 00:01:19,390 applicable. 28 00:01:19,390 --> 00:01:22,460 For example, in a plate, we set the stress through the 29 00:01:22,460 --> 00:01:25,480 thickness of the plate equal to zero. 30 00:01:25,480 --> 00:01:29,560 In addition, also, we have to impose in the use of the 31 00:01:29,560 --> 00:01:32,240 principle of virtual displacement the kinematic 32 00:01:32,240 --> 00:01:37,730 constraints for particles on sections or originally normal 33 00:01:37,730 --> 00:01:39,320 to the mid-surface. 34 00:01:39,320 --> 00:01:43,730 Namely, we have to put the constraint into the structure 35 00:01:43,730 --> 00:01:47,160 that the particles remain on a straight line during 36 00:01:47,160 --> 00:01:48,410 deformation. 37 00:01:48,410 --> 00:01:51,650 Well, as examples, I've plotted here, I've shown here 38 00:01:51,650 --> 00:01:57,020 two structures, a beam and a shell. 39 00:01:57,020 --> 00:01:59,120 Let's look first at a beam. 40 00:01:59,120 --> 00:02:05,020 In this case, we have that the original particles normal to 41 00:02:05,020 --> 00:02:08,210 the mid-surface, or the neutral axis of the beam, are 42 00:02:08,210 --> 00:02:09,630 on this orange line. 43 00:02:09,630 --> 00:02:12,880 I've shown here a large number of particles. 44 00:02:12,880 --> 00:02:16,630 The kinematic constraint that we're talking about is that 45 00:02:16,630 --> 00:02:20,380 during deformations, these particles remain 46 00:02:20,380 --> 00:02:21,420 on a straight line. 47 00:02:21,420 --> 00:02:24,710 They move over to the yellow line here. 48 00:02:24,710 --> 00:02:28,040 In other words, point A goes to point a prime. 49 00:02:28,040 --> 00:02:31,400 Another particle here goes over to this particle here. 50 00:02:31,400 --> 00:02:34,620 This particles goes over to that particle, and so on, and 51 00:02:34,620 --> 00:02:37,210 these particles remain on a straight line. 52 00:02:37,210 --> 00:02:39,970 That is the basic kinematic assumption. 53 00:02:39,970 --> 00:02:40,420 However, 54 00:02:40,420 --> 00:02:44,660 We should also notice that there's a right angle between 55 00:02:44,660 --> 00:02:49,080 the mid-surface, or neutral axis off the beam, and this 56 00:02:49,080 --> 00:02:52,970 line of particles initially. 57 00:02:52,970 --> 00:02:56,140 This right angle is not preserved during deformation. 58 00:02:56,140 --> 00:03:01,700 In other words, this angle here is not a right angle 59 00:03:01,700 --> 00:03:04,310 after deformations anymore. 60 00:03:04,310 --> 00:03:06,400 In the case of the shell, the kinematic 61 00:03:06,400 --> 00:03:08,470 constraint is quite similar. 62 00:03:08,470 --> 00:03:11,230 Here, we have now the mid-surface of the shell shown 63 00:03:11,230 --> 00:03:12,860 as a dashed line. 64 00:03:12,860 --> 00:03:17,700 The particles on a line normal to that mid-surface are shown 65 00:03:17,700 --> 00:03:20,670 here again in orange. 66 00:03:20,670 --> 00:03:22,600 This is the initial line. 67 00:03:22,600 --> 00:03:26,000 And during deformation, these particles remain 68 00:03:26,000 --> 00:03:27,100 on a straight line. 69 00:03:27,100 --> 00:03:32,110 Now they have come to be the yellow line here, and we 70 00:03:32,110 --> 00:03:36,280 notice that, again, there's a right angle initially here, 71 00:03:36,280 --> 00:03:40,350 but that right angle is not preserved during deformations 72 00:03:40,350 --> 00:03:44,530 because, in each case, we are including shear effects. 73 00:03:44,530 --> 00:03:47,270 Here, we include shear effects, and similarly here, 74 00:03:47,270 --> 00:03:50,010 we also include shear effects. 75 00:03:50,010 --> 00:03:57,340 Well, I've prepared some view graphs to show these facts a 76 00:03:57,340 --> 00:03:59,010 little bit more distinct. 77 00:03:59,010 --> 00:04:03,590 Here, we have a first view graph on which I show the 78 00:04:03,590 --> 00:04:07,880 assumptions of the basic Bernoulli-Euler beam theory 79 00:04:07,880 --> 00:04:10,290 that is used in the development of conventional 80 00:04:10,290 --> 00:04:11,620 beam elements. 81 00:04:11,620 --> 00:04:19,160 We have here the original beam element with its neutral axis 82 00:04:19,160 --> 00:04:20,990 in a dashed line. 83 00:04:20,990 --> 00:04:26,220 And that beam element, during deformation, becomes this. 84 00:04:26,220 --> 00:04:28,740 It goes into this shape here. 85 00:04:28,740 --> 00:04:30,840 We notice-- and this is important-- 86 00:04:30,840 --> 00:04:36,320 that this section here, which I now mark in blue, goes over 87 00:04:36,320 --> 00:04:42,840 into that section, and that the displacement is w at the 88 00:04:42,840 --> 00:04:49,600 mid-surface, and that the slope here at right angles to 89 00:04:49,600 --> 00:04:54,520 that section is nothing else than dw dx. 90 00:04:54,520 --> 00:04:57,690 In other words, this angle here is really nothing else 91 00:04:57,690 --> 00:05:01,030 than that angle, dw dx. 92 00:05:01,030 --> 00:05:04,550 This is the Bernoulli-Euler beam theory excluding shear 93 00:05:04,550 --> 00:05:06,130 deformations. 94 00:05:06,130 --> 00:05:11,040 The important point is that when we use this beam theory, 95 00:05:11,040 --> 00:05:16,410 we have to match between two elements, w, in other words, 96 00:05:16,410 --> 00:05:19,475 the displacement at the mid-surface has to be the same 97 00:05:19,475 --> 00:05:22,780 for element one and element two. 98 00:05:22,780 --> 00:05:26,620 And in addition, this slope has to be the same for both 99 00:05:26,620 --> 00:05:31,190 elements, dw dx for element one on the left-hand side must 100 00:05:31,190 --> 00:05:34,750 be equal to dw dx on the right-hand side. 101 00:05:34,750 --> 00:05:39,750 This is the conventional beam theory that is used to develop 102 00:05:39,750 --> 00:05:42,770 the Hermitian beam elements. 103 00:05:42,770 --> 00:05:47,620 And what I'd like to introduce to you now is the beam theory 104 00:05:47,620 --> 00:05:52,980 that we're using in the isoparametric formulation, 105 00:05:52,980 --> 00:05:58,290 namely in the development of modern beam elements, pipe 106 00:05:58,290 --> 00:06:01,400 elements, shell elements, and plate elements. 107 00:06:01,400 --> 00:06:05,130 I should say at this point that the conventional 108 00:06:05,130 --> 00:06:09,090 Hermitian beam element that you're probably familiar with 109 00:06:09,090 --> 00:06:12,520 is more effective in engineering analysis than the 110 00:06:12,520 --> 00:06:15,670 beam element that I'm talking about here when we just look 111 00:06:15,670 --> 00:06:17,660 at a straight beam. 112 00:06:17,660 --> 00:06:22,240 However, I want to look at the straight beam here as an 113 00:06:22,240 --> 00:06:26,890 example to introduce to you the formulation of the 114 00:06:26,890 --> 00:06:29,560 structural elements that I'm talking about. 115 00:06:29,560 --> 00:06:32,000 The formulation is very well displayed, very well 116 00:06:32,000 --> 00:06:32,980 demonstrated. 117 00:06:32,980 --> 00:06:36,040 The basic features are very well demonstrated looking at 118 00:06:36,040 --> 00:06:39,920 the beam element, although the application of this 119 00:06:39,920 --> 00:06:44,000 formulation to a straight beam element is not as effective as 120 00:06:44,000 --> 00:06:47,230 just the usage of a Hermitian beam element. 121 00:06:47,230 --> 00:06:50,310 However, if we talk about curved beam elements, pipe 122 00:06:50,310 --> 00:06:52,140 elements, then this formulation 123 00:06:52,140 --> 00:06:55,230 is indeed very effective. 124 00:06:55,230 --> 00:06:57,540 And, of course, for plates and shells, it 125 00:06:57,540 --> 00:06:59,320 is really very effective. 126 00:06:59,320 --> 00:07:02,080 Let me show to you, then, the basic points-- 127 00:07:02,080 --> 00:07:04,360 the basic important points-- 128 00:07:04,360 --> 00:07:07,980 that are being used in the formulation. 129 00:07:07,980 --> 00:07:12,250 Here we have, again, our original beam element. 130 00:07:12,250 --> 00:07:16,020 The neutral access is shown here, and this beam element 131 00:07:16,020 --> 00:07:22,120 now moves over into this piece, into that shape, during 132 00:07:22,120 --> 00:07:23,630 deformations. 133 00:07:23,630 --> 00:07:29,250 We have a section here, and as I pointed out earlier, that 134 00:07:29,250 --> 00:07:33,420 section has to remain straight during deformations. 135 00:07:33,420 --> 00:07:40,770 In fact, it moves right to that section here. 136 00:07:40,770 --> 00:07:45,970 Now notice that what we're talking about is a slope, dw 137 00:07:45,970 --> 00:07:51,690 dx here, which is this angle here, plus a shear deformation 138 00:07:51,690 --> 00:07:58,260 angle, gamma, and dw dx minus gamma is this angle. 139 00:07:58,260 --> 00:08:01,390 And that is the angle beta. 140 00:08:01,390 --> 00:08:09,080 In fact, this is the rotation of this line of particles. 141 00:08:09,080 --> 00:08:12,050 In other words, we are not talking just about one state 142 00:08:12,050 --> 00:08:16,810 variable, w, as we do in the Hermitian formulation, but we 143 00:08:16,810 --> 00:08:21,620 talk about two state variables now, beta and w. 144 00:08:21,620 --> 00:08:25,950 Beta and w both are independent, and we will see 145 00:08:25,950 --> 00:08:28,370 that, later on, we will interpolate them as 146 00:08:28,370 --> 00:08:30,560 independent quantities. 147 00:08:30,560 --> 00:08:33,330 The important point, then, that if you look at two 148 00:08:33,330 --> 00:08:37,350 elements, if you do interpolate w and beta 149 00:08:37,350 --> 00:08:41,900 independently, then we need, between two elements, 150 00:08:41,900 --> 00:08:48,620 continuity in w and continuity in beta. 151 00:08:48,620 --> 00:08:53,290 We do not talk about continuity in dw dx. 152 00:08:53,290 --> 00:08:57,560 We do not talk about that, and that is the important point, 153 00:08:57,560 --> 00:09:00,480 particularly when we talk about the formulation of plate 154 00:09:00,480 --> 00:09:02,750 elements and shell elements. 155 00:09:02,750 --> 00:09:06,430 The fact that we'll talk about independent interpolations of 156 00:09:06,430 --> 00:09:11,160 w and beta, including shear effects, of course, in an 157 00:09:11,160 --> 00:09:15,990 approximate way, but including shear affects, that fact 158 00:09:15,990 --> 00:09:18,745 alleviates us of many difficulties that 159 00:09:18,745 --> 00:09:21,700 we encounter otherwise. 160 00:09:21,700 --> 00:09:24,080 In other words, that we encounter if we use a 161 00:09:24,080 --> 00:09:29,600 classical plate theory excluding shear deformations. 162 00:09:29,600 --> 00:09:35,220 A good starting point for the development of the elements is 163 00:09:35,220 --> 00:09:40,550 the use of the total potential, pi, of an element. 164 00:09:40,550 --> 00:09:45,000 That total potential I've written down here, and we have 165 00:09:45,000 --> 00:09:49,640 set pi is equal to a contribution from the bending 166 00:09:49,640 --> 00:09:56,250 part plus a contribution from the shearing part, or the 167 00:09:56,250 --> 00:10:01,680 shearing deformations, and, of course, there is the external 168 00:10:01,680 --> 00:10:05,530 work due to distributed pressure, p, on a beam 169 00:10:05,530 --> 00:10:09,330 element, and moments, externally applied moments, m, 170 00:10:09,330 --> 00:10:11,100 onto the beam element. 171 00:10:11,100 --> 00:10:14,070 Now, notice the quantities that I'm using here. 172 00:10:14,070 --> 00:10:19,890 The bending part is given by dw dx squared, of course with 173 00:10:19,890 --> 00:10:26,030 the flex or rigidity in front, and this part here is given 174 00:10:26,030 --> 00:10:29,720 only in terms of the section rotation beta, which is 175 00:10:29,720 --> 00:10:36,370 independent of the translation of the neutral axes, w. 176 00:10:36,370 --> 00:10:41,000 Here, we have the shearing part, dw dx, minus beta As, 177 00:10:41,000 --> 00:10:43,140 the shear strength. 178 00:10:43,140 --> 00:10:46,720 And they have been written down here once more. 179 00:10:46,720 --> 00:10:50,840 If we look at this equation, here we get dw dx minus beta 180 00:10:50,840 --> 00:10:53,880 is equal to gamma. 181 00:10:53,880 --> 00:10:56,690 The other quantities, of course, that I used in the 182 00:10:56,690 --> 00:10:59,030 derivation of this-- 183 00:10:59,030 --> 00:11:05,770 pi, the stress being equal to v over as, v being the shear 184 00:11:05,770 --> 00:11:10,410 force on the section, As is the shear area. 185 00:11:10,410 --> 00:11:14,760 Notice that we are assuming the shear strength through the 186 00:11:14,760 --> 00:11:18,130 thickness of the beam element to be constant. 187 00:11:18,130 --> 00:11:23,140 They are constant because of this equation here, basically. 188 00:11:23,140 --> 00:11:27,010 w, of course, varies along the length of the beam. 189 00:11:27,010 --> 00:11:31,220 Beta varies along the length of the beam, but that means 190 00:11:31,220 --> 00:11:34,350 gamma is constant through to the thickness of the beam. 191 00:11:34,350 --> 00:11:37,035 Now, since gamma is constant through the thickness of the 192 00:11:37,035 --> 00:11:42,330 beam, we have to say, of course, that also our shear 193 00:11:42,330 --> 00:11:46,330 stress is constant through the thickness of the beam, and we 194 00:11:46,330 --> 00:11:50,800 have to introduce a shear correction factor, k, which is 195 00:11:50,800 --> 00:11:54,770 equal to As over A, where As is an equivalent shear area. 196 00:11:57,360 --> 00:12:03,570 Using these quantities we obtain this pi functional, 197 00:12:03,570 --> 00:12:08,150 where, once again, we simply add the bending contribution, 198 00:12:08,150 --> 00:12:12,480 bending strain energy to the shear strain energy, and we 199 00:12:12,480 --> 00:12:17,330 subtract the total potential of the external loads. 200 00:12:17,330 --> 00:12:21,150 If we invoke the stationarity of this functional, in other 201 00:12:21,150 --> 00:12:25,650 words, we invoke that delta pi is equal to zero, we obtain 202 00:12:25,650 --> 00:12:29,640 the principle of virtual work, or principle of virtual 203 00:12:29,640 --> 00:12:33,690 displacement, which I have discussed with you in an 204 00:12:33,690 --> 00:12:35,330 earlier lecture. 205 00:12:35,330 --> 00:12:40,340 The result of invoking that del pi is equal to zero, or 206 00:12:40,340 --> 00:12:45,250 invoking the stationarity of pi is this equation here. 207 00:12:45,250 --> 00:12:49,090 Now notice that in this equation we have, if we want 208 00:12:49,090 --> 00:12:52,400 to interpret it once physically here, basically the 209 00:12:52,400 --> 00:12:59,970 real stress part, here, the virtual strain part. 210 00:12:59,970 --> 00:13:04,690 Similarly here, the real stress part and the virtual 211 00:13:04,690 --> 00:13:07,360 strain part, and, of course, the virtual work of the 212 00:13:07,360 --> 00:13:10,020 external loads. 213 00:13:10,020 --> 00:13:12,900 The important point is that we only integrate along the 214 00:13:12,900 --> 00:13:15,270 length of the beam, and not through the thickness anymore, 215 00:13:15,270 --> 00:13:20,780 because we talk about quantities stress resultants 216 00:13:20,780 --> 00:13:22,660 over the thickness. 217 00:13:22,660 --> 00:13:28,330 Well, once we have arrived at this equation, we can proceed 218 00:13:28,330 --> 00:13:31,370 in much the same way as we have been proceeding in the 219 00:13:31,370 --> 00:13:35,160 development of continuum isoparametric elements. 220 00:13:35,160 --> 00:13:38,230 Here, we look at a particular case. 221 00:13:38,230 --> 00:13:41,340 Let us say we have a beam elements such as shown here. 222 00:13:41,340 --> 00:13:45,040 Of course, this is the loading applied, P. The bending moment 223 00:13:45,040 --> 00:13:47,650 that they're talking about here is shown here. 224 00:13:47,650 --> 00:13:49,980 It is a distributed bending moment over 225 00:13:49,980 --> 00:13:51,260 any part of the beam. 226 00:13:51,260 --> 00:13:53,010 Similarly, P is only applied over a 227 00:13:53,010 --> 00:13:54,610 certain part of the beam. 228 00:13:54,610 --> 00:14:00,030 The depth of the beam is b, the width of the beam is a. 229 00:14:00,030 --> 00:14:04,040 As a shear factor for a rectangular beam, the sheer 230 00:14:04,040 --> 00:14:08,160 fact k that I introduced to you briefly is 5/6. 231 00:14:08,160 --> 00:14:10,930 Of course, I, the moment of inertia, is ab 232 00:14:10,930 --> 00:14:14,230 cubed divided by 12. 233 00:14:14,230 --> 00:14:17,300 The interpolations that we would use for such a beam 234 00:14:17,300 --> 00:14:19,860 element are one dimensional interpolations. 235 00:14:19,860 --> 00:14:23,050 We only integrate along the length, x. 236 00:14:23,050 --> 00:14:26,220 And the one dimension interpolations we discussed 237 00:14:26,220 --> 00:14:27,860 already earlier. 238 00:14:27,860 --> 00:14:31,800 We simply use the same that we have been using already in the 239 00:14:31,800 --> 00:14:34,090 formulation of truss elements. 240 00:14:34,090 --> 00:14:38,460 Two point interpolation would be this element here. 241 00:14:38,460 --> 00:14:41,010 Here, we use a three point interpolation. 242 00:14:41,010 --> 00:14:41,800 Notice-- 243 00:14:41,800 --> 00:14:42,970 and this is important-- 244 00:14:42,970 --> 00:14:47,610 that we're talking about w and beta, the section rotations, 245 00:14:47,610 --> 00:14:49,550 as independent quantities. 246 00:14:49,550 --> 00:14:54,820 So for a three point, or three nodal point beam, we would 247 00:14:54,820 --> 00:15:00,710 have, in just a plane analysis, we would have six 248 00:15:00,710 --> 00:15:02,240 degrees of freedom. 249 00:15:02,240 --> 00:15:06,960 For a cubic element, we have eight degrees of freedom. 250 00:15:06,960 --> 00:15:10,420 Of course, for a Hermitian beam element, we would have 251 00:15:10,420 --> 00:15:14,150 only these degrees of freedom and those degrees of freedom, 252 00:15:14,150 --> 00:15:17,710 w and dw dx here, and w, dw dx here. 253 00:15:17,710 --> 00:15:20,200 So, that is the reason why the Hermitian beam 254 00:15:20,200 --> 00:15:22,050 element is more effective. 255 00:15:22,050 --> 00:15:25,650 However, we can use these interpolations directly to 256 00:15:25,650 --> 00:15:30,460 develop curved beam elements, pipe elements, and then, of 257 00:15:30,460 --> 00:15:32,990 course, this approach is-- even for beam elements-- 258 00:15:32,990 --> 00:15:35,060 more effective, as I mentioned earlier. 259 00:15:35,060 --> 00:15:38,680 Well, let us then write down the basic interpolations that 260 00:15:38,680 --> 00:15:39,360 we are using. 261 00:15:39,360 --> 00:15:43,870 Here, we have w being the sum of hi wi. 262 00:15:43,870 --> 00:15:46,350 hi, we discussed earlier already. 263 00:15:46,350 --> 00:15:49,980 And these, of course, are nodal point transverse 264 00:15:49,980 --> 00:15:51,800 displacement. 265 00:15:51,800 --> 00:15:55,390 These are the nodal point rotations of 266 00:15:55,390 --> 00:15:57,830 the sections, beta. 267 00:15:57,830 --> 00:16:00,330 In other words, I could have used here, as a notation, beta 268 00:16:00,330 --> 00:16:04,390 i, but I chose to use theta i. 269 00:16:04,390 --> 00:16:07,860 If you write these equations in matrix form, we directly 270 00:16:07,860 --> 00:16:14,900 obtain this relation here, where hw simply list hi, u 271 00:16:14,900 --> 00:16:19,550 lists the displacements and section rotations, and similar 272 00:16:19,550 --> 00:16:23,300 here, beta is given in terms of an h beta matrix. 273 00:16:23,300 --> 00:16:27,570 We take the differentiations of hw and h beta, and we get 274 00:16:27,570 --> 00:16:35,490 dw dx equal to Bw times u d beta dx equal beta times u. 275 00:16:35,490 --> 00:16:40,110 Once we have the principle virtual work established for 276 00:16:40,110 --> 00:16:44,170 the element that we're considering and have chosen 277 00:16:44,170 --> 00:16:47,160 our interpolations, the approach of developing the 278 00:16:47,160 --> 00:16:51,680 stiffness matrices, the load vectors, is exactly the same 279 00:16:51,680 --> 00:16:56,740 as in the development of continuum elements. 280 00:16:56,740 --> 00:17:01,790 Well here, I have written down the various quantities. 281 00:17:01,790 --> 00:17:06,560 U transposed lists, as I said earlier, the displacement 282 00:17:06,560 --> 00:17:09,900 vector nodal points and the rotation vector nodal points. 283 00:17:09,900 --> 00:17:13,450 Hw simply gives the interpolation functions, q, of 284 00:17:13,450 --> 00:17:15,170 course, being equal to the number of nodal 285 00:17:15,170 --> 00:17:17,200 points we are using. 286 00:17:17,200 --> 00:17:19,930 And here, we have H beta. 287 00:17:19,930 --> 00:17:22,380 The Bw is given right here. 288 00:17:22,380 --> 00:17:25,640 Notice our J inverse comes in there because we have to 289 00:17:25,640 --> 00:17:29,950 transform from r to x, x being the actual physical coordinate 290 00:17:29,950 --> 00:17:31,160 along the length. 291 00:17:31,160 --> 00:17:33,390 Similarly, B beta being given here. 292 00:17:33,390 --> 00:17:37,450 Again, the J inverse here to transform from r to x 293 00:17:37,450 --> 00:17:39,450 coordinates. 294 00:17:39,450 --> 00:17:43,020 Once we have written down these, we can directly 295 00:17:43,020 --> 00:17:49,020 substitute into the principal of virtual work, and we come 296 00:17:49,020 --> 00:17:53,610 up with the stiffness matrix, given here, and the load 297 00:17:53,610 --> 00:17:55,120 vector given here. 298 00:17:55,120 --> 00:17:58,370 Let me point out a few important things here. 299 00:17:58,370 --> 00:18:02,870 Of course, this B beta matrix here-- just to remind you-- 300 00:18:02,870 --> 00:18:07,100 comes from d beta dx. 301 00:18:07,100 --> 00:18:12,800 It's in fact, really, d beta dr, but because we're 302 00:18:12,800 --> 00:18:16,660 integrating from -1 to plus 1, however, we have our 303 00:18:16,660 --> 00:18:19,720 determinant J there to take into account the volume 304 00:18:19,720 --> 00:18:22,190 transformation. 305 00:18:22,190 --> 00:18:26,870 Here, of course, we have d beta dx transposed. 306 00:18:26,870 --> 00:18:29,730 This comes from the virtual strains, and that comes from 307 00:18:29,730 --> 00:18:32,120 the real strains. 308 00:18:32,120 --> 00:18:33,460 Of course, we have the stresses. 309 00:18:33,460 --> 00:18:37,870 Here, so these strains times the stress strain law, E, 310 00:18:37,870 --> 00:18:41,980 being the Young's modulus, gives us the stresses. 311 00:18:41,980 --> 00:18:44,700 Here, we talk about shearing deformations. 312 00:18:44,700 --> 00:18:52,261 Notice that we have here dw dx, that is, the 313 00:18:52,261 --> 00:18:55,510 Bw minus the beta. 314 00:18:55,510 --> 00:19:00,320 So we have the derivative in here of w, but no derivative 315 00:19:00,320 --> 00:19:04,780 in H beta because we are simply interpolating here the 316 00:19:04,780 --> 00:19:06,890 beta values. 317 00:19:06,890 --> 00:19:12,270 Again, of course, a transformation to x, and 318 00:19:12,270 --> 00:19:15,970 therefore we have a determine J in there. 319 00:19:15,970 --> 00:19:21,660 The load vector looks just the same way as in the development 320 00:19:21,660 --> 00:19:23,260 of continuum elements. 321 00:19:23,260 --> 00:19:27,600 This is the transverse loading applied P. This is, of course, 322 00:19:27,600 --> 00:19:28,730 interpolating-- 323 00:19:28,730 --> 00:19:29,230 this [UNINTELLIGIBLE] 324 00:19:29,230 --> 00:19:33,270 interpolates the virtual transverse displacements. 325 00:19:33,270 --> 00:19:39,230 Here, we have the moment loads, the real moment loads, 326 00:19:39,230 --> 00:19:43,610 and this interpolates here the section rotations, beta, along 327 00:19:43,610 --> 00:19:45,700 the lengths of the beam. 328 00:19:45,700 --> 00:19:51,800 Of course, again, the volume transformation from r to x. 329 00:19:51,800 --> 00:19:54,960 This is really a straightforward application of 330 00:19:54,960 --> 00:19:56,840 what we discussed earlier. 331 00:19:56,840 --> 00:19:59,990 There is one important point, however, now, that I have to 332 00:19:59,990 --> 00:20:01,910 point out to you. 333 00:20:01,910 --> 00:20:08,390 If we consider the functional pi, as I mentioned earlier, 334 00:20:08,390 --> 00:20:10,940 there's a bending part here and there's a 335 00:20:10,940 --> 00:20:13,970 shearing part here. 336 00:20:13,970 --> 00:20:18,680 Notice that in this development I have divided 337 00:20:18,680 --> 00:20:24,010 through by EI over 2, so that I introduce a value alpha 338 00:20:24,010 --> 00:20:29,770 there, and that alpha is really GAK divided by EI. 339 00:20:29,770 --> 00:20:37,090 Now, if we look at that alpha value, and let's look at the 340 00:20:37,090 --> 00:20:44,430 value for a rectangular section, we would see that the 341 00:20:44,430 --> 00:20:54,680 a value, of course, is a times b, and the I value gives us 342 00:20:54,680 --> 00:21:00,896 really an ab cubed, if this is, here, b, and that is a. 343 00:21:00,896 --> 00:21:04,190 An ab cubed over 12, of course. 344 00:21:04,190 --> 00:21:10,120 And we have, also, the GK that gives us these, and of course 345 00:21:10,120 --> 00:21:11,540 an E in front here. 346 00:21:11,540 --> 00:21:14,950 But the important point that I want to now concentrate on is 347 00:21:14,950 --> 00:21:18,580 really this b over b cubed. 348 00:21:18,580 --> 00:21:23,840 We can see that as the element gets thinner and thinner, as 349 00:21:23,840 --> 00:21:28,380 the element gets thinner and thinner, that alpha gets 350 00:21:28,380 --> 00:21:29,660 larger and larger. 351 00:21:32,500 --> 00:21:36,790 If alpha gets larger and larger, this term here will be 352 00:21:36,790 --> 00:21:39,240 predominant. 353 00:21:39,240 --> 00:21:41,060 This term will be predominant. 354 00:21:41,060 --> 00:21:47,940 Now this means, however, that if we want to finally converge 355 00:21:47,940 --> 00:21:52,540 to a beam in which the shear strengths are negligible, in 356 00:21:52,540 --> 00:21:54,800 other words, in which the shear strengths are extremely 357 00:21:54,800 --> 00:21:58,650 small, what we would have to be able to represent in the 358 00:21:58,650 --> 00:22:03,550 formulation is that this value here goes to zero. 359 00:22:03,550 --> 00:22:06,150 And what happens in the formulation, really, is that 360 00:22:06,150 --> 00:22:11,430 as this value gets larger and larger, any error introduced 361 00:22:11,430 --> 00:22:15,520 in the formulation, due to the fact that this is not exactly 362 00:22:15,520 --> 00:22:20,080 zero in the finite element interpolation, that error is 363 00:22:20,080 --> 00:22:21,340 largely magnified. 364 00:22:21,340 --> 00:22:26,570 Is magnified and, in fact, can introduce a very large error 365 00:22:26,570 --> 00:22:33,020 if this value is not zero due to the fact that alpha becomes 366 00:22:33,020 --> 00:22:35,600 larger and larger if the element 367 00:22:35,600 --> 00:22:38,410 becomes thinner and thinner. 368 00:22:38,410 --> 00:22:42,500 In other words, in summary once more, if we are talking 369 00:22:42,500 --> 00:22:47,270 about a beam element that gets thinner and thinner for which 370 00:22:47,270 --> 00:22:51,340 we know the shear strengths should becomes smaller and 371 00:22:51,340 --> 00:22:56,260 smaller, our finite element interpolation must be able to 372 00:22:56,260 --> 00:22:58,950 represent this fact. 373 00:22:58,950 --> 00:23:01,600 Now, if we look at the shear strengths, and this is, of 374 00:23:01,600 --> 00:23:05,100 course here, nothing else then basically the shear strength 375 00:23:05,100 --> 00:23:13,650 squared, we now identify that dw dx minus beta, when 376 00:23:13,650 --> 00:23:19,300 interpolated using our interpolation functions, must 377 00:23:19,300 --> 00:23:23,570 be able to be very, very, very small. 378 00:23:23,570 --> 00:23:27,270 And that is a restriction on the formulation. 379 00:23:27,270 --> 00:23:31,390 So what we have to do, really, is use high enough order 380 00:23:31,390 --> 00:23:36,850 interpolations so that dw dx minus beta can be smaller for 381 00:23:36,850 --> 00:23:38,230 thin elements. 382 00:23:38,230 --> 00:23:41,630 Of course for thick beam elements, that is not really a 383 00:23:41,630 --> 00:23:44,360 constraint because we know that there are shearing 384 00:23:44,360 --> 00:23:46,690 deformations, and the shear deformations can be quite 385 00:23:46,690 --> 00:23:47,800 significant. 386 00:23:47,800 --> 00:23:52,550 However, for thin elements, we must be able to represent the 387 00:23:52,550 --> 00:23:57,220 fact that gamma is small, and therefore, we have to use high 388 00:23:57,220 --> 00:23:58,580 order interpolations. 389 00:23:58,580 --> 00:24:02,810 In fact, the parabolic interpolation is really the 390 00:24:02,810 --> 00:24:05,460 lowest interpolation that one can recommend. 391 00:24:05,460 --> 00:24:08,650 It would be better to use cubic interpolation. 392 00:24:08,650 --> 00:24:11,990 In fact, we use the cubic interpolation in practice. 393 00:24:11,990 --> 00:24:18,520 In that case, this gamma value can be small, and we run into 394 00:24:18,520 --> 00:24:22,480 no difficulties and the element can be 395 00:24:22,480 --> 00:24:24,520 very, very, very thin. 396 00:24:24,520 --> 00:24:26,880 Another approach would be to use a discrete Kirchhoff 397 00:24:26,880 --> 00:24:30,270 theory or reduced numerical integration. 398 00:24:30,270 --> 00:24:32,860 These approaches have been developed 399 00:24:32,860 --> 00:24:34,970 for low order elements. 400 00:24:34,970 --> 00:24:38,710 The discrete Kirchhoff theory approach is very effective. 401 00:24:38,710 --> 00:24:41,240 The reduced numerical integration can also be 402 00:24:41,240 --> 00:24:43,300 effective, but has to be used with care. 403 00:24:43,300 --> 00:24:47,250 In particular, as I will point out in the next lecture, we 404 00:24:47,250 --> 00:24:51,710 have to be careful that we do not introduce spurious rigid 405 00:24:51,710 --> 00:24:53,050 body modes into the system. 406 00:24:55,990 --> 00:25:03,540 The development that I just talked about really is 407 00:25:03,540 --> 00:25:10,350 applicable to an element that has a rectangular section and 408 00:25:10,350 --> 00:25:13,110 the element was also lying in a plane. 409 00:25:13,110 --> 00:25:14,570 We looked at a straight element. 410 00:25:14,570 --> 00:25:19,450 Let us now see how we can generalize these concepts 411 00:25:19,450 --> 00:25:23,640 directly to the formulation of general curved beam elements. 412 00:25:23,640 --> 00:25:28,630 And for that purpose I've shown here-- 413 00:25:28,630 --> 00:25:30,310 I'm showing here-- 414 00:25:30,310 --> 00:25:33,300 a more general beam element that lies in a 415 00:25:33,300 --> 00:25:34,920 three-dimensional space. 416 00:25:34,920 --> 00:25:38,160 It's still rectangular, however, we could also have a 417 00:25:38,160 --> 00:25:40,410 circular section instead. 418 00:25:40,410 --> 00:25:43,350 In fact, when we look at a pipe element, we do talk 419 00:25:43,350 --> 00:25:45,630 about-- we have, of course, a circular section. 420 00:25:48,470 --> 00:25:53,190 Well, in this particular beam element, notice I'm looking at 421 00:25:53,190 --> 00:25:56,660 node one here, node two there, and generally node three here 422 00:25:56,660 --> 00:25:59,810 and node four here, because we want to pick up the curvature 423 00:25:59,810 --> 00:26:01,740 of the element. 424 00:26:01,740 --> 00:26:05,010 I have this element lying in a three dimensional 425 00:26:05,010 --> 00:26:07,990 space, x, y, z. 426 00:26:07,990 --> 00:26:13,810 Notice that that element has local coordinates r, s, t. 427 00:26:13,810 --> 00:26:15,980 These are the isoparametric coordinates. 428 00:26:15,980 --> 00:26:23,260 And psi, eta, zeta, these are the actual continuous physical 429 00:26:23,260 --> 00:26:26,010 coordinates in the beam element. 430 00:26:26,010 --> 00:26:32,390 We define normals at each nodal point. 431 00:26:32,390 --> 00:26:38,430 The normal in the t direction here is 0 Vt 1. 432 00:26:38,430 --> 00:26:42,770 The normal into the s direction is 0 Vs 1. 433 00:26:42,770 --> 00:26:47,720 And similarly, we have two normals at each nodal point. 434 00:26:47,720 --> 00:26:50,330 Notice that for this particular rectangular beam 435 00:26:50,330 --> 00:26:54,760 element, I have the thickness, a1, here and b1, there, and a2 436 00:26:54,760 --> 00:26:56,020 here, b2 there. 437 00:26:56,020 --> 00:26:58,360 These thicknesses can be different. 438 00:26:58,360 --> 00:27:05,370 Now what I want to use are the same basic assumptions that we 439 00:27:05,370 --> 00:27:09,440 have familiarize ourselves already with when we looked at 440 00:27:09,440 --> 00:27:13,050 the special case of a straight beam element in planar 441 00:27:13,050 --> 00:27:13,930 deformations. 442 00:27:13,930 --> 00:27:15,990 I want to use those basic assumption now in the 443 00:27:15,990 --> 00:27:19,450 development of this more general beam element. 444 00:27:19,450 --> 00:27:22,870 And I want to use a continuum approach. 445 00:27:22,870 --> 00:27:26,080 Well, what we are doing, then, is the following. 446 00:27:26,080 --> 00:27:34,160 We interpolate the coordinates, x, y, and z, 447 00:27:34,160 --> 00:27:37,750 along the beam element in terms the nodal point 448 00:27:37,750 --> 00:27:45,095 coordinates of the nodes that lie on the neutral axes of the 449 00:27:45,095 --> 00:27:52,950 beam plus an effect that comes in due to the 450 00:27:52,950 --> 00:27:55,120 thickness of the beam. 451 00:27:55,120 --> 00:27:58,460 Now let us go and look in detail at the x 452 00:27:58,460 --> 00:27:59,590 interpolations. 453 00:27:59,590 --> 00:28:05,020 The L denotes 0 or 1, 0 being the initial configuration, 1 454 00:28:05,020 --> 00:28:07,750 being the final configuration. 455 00:28:07,750 --> 00:28:12,240 So let's put simply a 0 in there, think in terms of a 0 456 00:28:12,240 --> 00:28:16,410 there, and let's look at the initial configuration first. 457 00:28:16,410 --> 00:28:21,980 Well, here we have the initial x-coordinates of the nodal 458 00:28:21,980 --> 00:28:25,760 points, and there are q of them. 459 00:28:25,760 --> 00:28:30,160 These are the one dimensional interpolation functions, just 460 00:28:30,160 --> 00:28:34,240 the same that we use for a truss element, for example. 461 00:28:34,240 --> 00:28:38,030 Here, we have the t-axis. 462 00:28:38,030 --> 00:28:46,440 This is the t-axis into the direction of the normal into 463 00:28:46,440 --> 00:28:48,340 the t direction. 464 00:28:48,340 --> 00:28:50,560 In other words, this is a t-axis here. 465 00:28:50,560 --> 00:28:52,980 Notice that that t-axis here corresponds 466 00:28:52,980 --> 00:28:54,760 to this normal here. 467 00:28:54,760 --> 00:28:59,430 That s axis here corresponds to this normal here. 468 00:28:59,430 --> 00:29:07,350 So here, we have t/2, and ak being the total thickness of 469 00:29:07,350 --> 00:29:13,125 the beam corresponding to that t direction, hk being the one 470 00:29:13,125 --> 00:29:16,350 dimensional interpolation functions again, and these are 471 00:29:16,350 --> 00:29:26,020 the direction cosines of the normal in the t direction. 472 00:29:26,020 --> 00:29:28,250 Here, I should really say, this is a direction cosine 473 00:29:28,250 --> 00:29:32,180 corresponding to the x-axis, corresponding to the x-axis. 474 00:29:32,180 --> 00:29:38,180 When we talk later about the y and z-axis, then we use the y 475 00:29:38,180 --> 00:29:40,960 direction cosine and the z direction cosine. 476 00:29:43,820 --> 00:29:48,770 This part comes in because the beam basically has a thickness 477 00:29:48,770 --> 00:29:51,670 into the t direction. 478 00:29:51,670 --> 00:29:57,490 Now, we have also to introduce the s direction part, and here 479 00:29:57,490 --> 00:30:02,160 we s/2, the thickness into s direction, the one dimensional 480 00:30:02,160 --> 00:30:06,920 interpolation functions, and the direction cosine in the x 481 00:30:06,920 --> 00:30:12,950 direction of the normal in the s direction. 482 00:30:12,950 --> 00:30:17,130 Well, if we want to find, in other words, the coordinates 483 00:30:17,130 --> 00:30:21,010 of any point in the beam element, and let us look, once 484 00:30:21,010 --> 00:30:23,390 more, back at the beam element. 485 00:30:23,390 --> 00:30:26,440 If I want to find the coordinate of a point, p, 486 00:30:26,440 --> 00:30:30,490 lying in that beam element here-- there's, say, point p-- 487 00:30:30,490 --> 00:30:35,000 what I have to do is I have to identify the r, s, and t 488 00:30:35,000 --> 00:30:42,000 coordinates of that point, p, and then substitute these r, 489 00:30:42,000 --> 00:30:47,320 s, and t-coordinates into this part here. 490 00:30:47,320 --> 00:30:51,720 And I would get the corresponding x-coordinate. 491 00:30:51,720 --> 00:30:56,430 I proceed similarly with the y and z-coordinates. 492 00:30:56,430 --> 00:30:59,940 Notice, as I pointed out earlier, we are talking still 493 00:30:59,940 --> 00:31:07,300 about the thickness ak hk here, here, and here, but 494 00:31:07,300 --> 00:31:10,950 we're using the x direction cosines, the y direction 495 00:31:10,950 --> 00:31:14,680 cosines, and the z direction cosines here. 496 00:31:14,680 --> 00:31:20,110 And similarly for the s direction, we are using 497 00:31:20,110 --> 00:31:22,020 similar quantities. 498 00:31:22,020 --> 00:31:30,780 So this is the interpolation of the beam element, and using 499 00:31:30,780 --> 00:31:34,270 these three formerly, we can directly obtain-- 500 00:31:34,270 --> 00:31:36,300 we can directly obtain-- 501 00:31:36,300 --> 00:31:42,010 the x, y, and z-coordinates in this system of axes of any 502 00:31:42,010 --> 00:31:45,030 point in the beam element. 503 00:31:45,030 --> 00:31:49,150 This is the most important fact. 504 00:31:49,150 --> 00:31:53,240 The interpolations that I've listed here are the starting 505 00:31:53,240 --> 00:31:58,870 point of the development of the strain displacement 506 00:31:58,870 --> 00:32:02,970 matrices and displacement interpolation matrices. 507 00:32:02,970 --> 00:32:09,800 Now, let us identify that if these are the original 508 00:32:09,800 --> 00:32:15,930 coordinates for L being 0, then we can also apply, of 509 00:32:15,930 --> 00:32:19,860 course, after the deformation, the same interpolation, and we 510 00:32:19,860 --> 00:32:22,610 put L equal to 1. 511 00:32:22,610 --> 00:32:29,690 If we subtract the x1 minus zero x, we should get the 512 00:32:29,690 --> 00:32:31,400 displacements, u. 513 00:32:31,400 --> 00:32:35,830 Notice that the displacements, u, are in the directions of 514 00:32:35,830 --> 00:32:37,020 the x-axis. 515 00:32:37,020 --> 00:32:39,560 Well, this is exactly how we proceed. 516 00:32:39,560 --> 00:32:45,200 We use these interpolations for before deformation and 517 00:32:45,200 --> 00:32:46,670 after deformation. 518 00:32:46,670 --> 00:32:50,500 And we subtract these interpolations as shown here 519 00:32:50,500 --> 00:32:54,890 and directly obtain the u, v, and w displacements, of 520 00:32:54,890 --> 00:32:59,360 course, as a functional of r, s, and t. 521 00:32:59,360 --> 00:33:03,320 Notice that if I proceed this way-- 522 00:33:03,320 --> 00:33:06,820 notice that if I proceed this way, I have used the basic 523 00:33:06,820 --> 00:33:11,750 assumption that plane sections remain plane during 524 00:33:11,750 --> 00:33:12,820 deformation. 525 00:33:12,820 --> 00:33:17,210 This was the assumption that I pointed out to your earlier. 526 00:33:17,210 --> 00:33:20,280 And we are using it in this general information just in 527 00:33:20,280 --> 00:33:22,610 the same way as in the special application that 528 00:33:22,610 --> 00:33:24,190 I showed you earlier. 529 00:33:24,190 --> 00:33:28,140 Well, having then-- 530 00:33:28,140 --> 00:33:32,120 just to refresh your memory, the u, v, and w, in terms of 531 00:33:32,120 --> 00:33:38,950 r, s, and t, of course, via these subtractions, we obtain 532 00:33:38,950 --> 00:33:41,970 directly these equations here. 533 00:33:41,970 --> 00:33:48,090 Notice that here we have, now, the nodal point displacements, 534 00:33:48,090 --> 00:33:53,790 uk, vk, wk, and we have the change in 535 00:33:53,790 --> 00:33:55,370 the direction cosines. 536 00:33:55,370 --> 00:33:57,520 These are the changes in the direction cosines. 537 00:34:00,460 --> 00:34:03,690 Well, these changes in the direction cosines we want to 538 00:34:03,690 --> 00:34:07,810 express in terms nodal points rotations. 539 00:34:07,810 --> 00:34:13,020 And that is achieved as shown on this view graph. 540 00:34:13,020 --> 00:34:16,860 We can express these changes in the direction cosines 541 00:34:16,860 --> 00:34:21,550 directly by taking the cross product of a vector of nodal 542 00:34:21,550 --> 00:34:25,889 points rotations, and I've listed here this vector, this 543 00:34:25,889 --> 00:34:30,340 is a nodal point rotation about the x, y, and z-axis at 544 00:34:30,340 --> 00:34:32,780 nodal point, k. 545 00:34:32,780 --> 00:34:35,940 You're taking the cross product of this vector times 546 00:34:35,940 --> 00:34:38,880 the original normal. 547 00:34:38,880 --> 00:34:40,590 Time the original normal. 548 00:34:40,590 --> 00:34:43,560 And we get, then, the change in the normal, of course, for 549 00:34:43,560 --> 00:34:45,870 the t direction and for the s direction. 550 00:34:48,469 --> 00:34:55,080 If we substitute this relation here, of course, remember that 551 00:34:55,080 --> 00:34:56,449 these quantities are known. 552 00:34:56,449 --> 00:34:57,910 They are given. 553 00:34:57,910 --> 00:34:59,920 So the unknowns now are theta k. 554 00:35:03,100 --> 00:35:08,560 If you substitute from here into our relation here that I 555 00:35:08,560 --> 00:35:13,740 developed earlier, we directly obtain the displacements of 556 00:35:13,740 --> 00:35:18,770 any point, p, in the beam in terms of nodal point 557 00:35:18,770 --> 00:35:21,630 displacements and nodal points rotations. 558 00:35:21,630 --> 00:35:25,470 Because these quantities here now have been eliminated and 559 00:35:25,470 --> 00:35:28,710 have been expressed in terms of nodal point rotations. 560 00:35:28,710 --> 00:35:32,530 Well, now we have all the quantities that we need to 561 00:35:32,530 --> 00:35:36,060 develop our strain displacement matrix for the 562 00:35:36,060 --> 00:35:37,110 beam element. 563 00:35:37,110 --> 00:35:41,130 Remember, all we need are the coordinate interpolations, 564 00:35:41,130 --> 00:35:46,290 which I have developed already, and the displacement 565 00:35:46,290 --> 00:35:47,270 interpolations. 566 00:35:47,270 --> 00:35:50,820 With those two quantities, we can immediately calculate, via 567 00:35:50,820 --> 00:35:53,110 the procedures that I discussed with you earlier, 568 00:35:53,110 --> 00:35:55,740 the strain displacement transformation matrix. 569 00:35:55,740 --> 00:35:57,570 And here it is given. 570 00:35:57,570 --> 00:36:02,180 Notice that we are now talking about strain into the eta 571 00:36:02,180 --> 00:36:03,000 directions. 572 00:36:03,000 --> 00:36:06,240 In other words, the eta, psy, and zeta directions. 573 00:36:06,240 --> 00:36:10,940 These are the directions that I pointed out to you earlier, 574 00:36:10,940 --> 00:36:12,510 which are the physical coordinate 575 00:36:12,510 --> 00:36:14,650 directions along the beam. 576 00:36:14,650 --> 00:36:17,740 Let me show it to you once more, the picture. 577 00:36:17,740 --> 00:36:21,590 The r, s, and t-coordinates are the isoparametric 578 00:36:21,590 --> 00:36:22,590 coordinates. 579 00:36:22,590 --> 00:36:25,910 The eta, psy, and zeta coordinates are the physical 580 00:36:25,910 --> 00:36:27,670 coordinates along the beam. 581 00:36:27,670 --> 00:36:29,820 In other words, for the one dimensional beam that we 582 00:36:29,820 --> 00:36:33,980 looked at, this eta axis was, in fact, the x-axis. 583 00:36:33,980 --> 00:36:37,330 Of course, our x, y, and z-axes are now global 584 00:36:37,330 --> 00:36:40,110 Cartesian axes. 585 00:36:40,110 --> 00:36:46,950 Well then, with that information, we can directly 586 00:36:46,950 --> 00:36:51,270 calculate the strain displacement matrix here. 587 00:36:51,270 --> 00:36:55,760 This is done effectively using numerical integration, as I 588 00:36:55,760 --> 00:36:57,790 will be discussing in the next lecture. 589 00:36:57,790 --> 00:37:02,820 The transformations that are necessary are also done on the 590 00:37:02,820 --> 00:37:05,400 integration point level. 591 00:37:05,400 --> 00:37:10,660 Notice that the uk here lists the nodal point displacements 592 00:37:10,660 --> 00:37:14,410 and the nodal point rotations, and, of course, we have to 593 00:37:14,410 --> 00:37:18,070 also remember one important fact, that for the beam we are 594 00:37:18,070 --> 00:37:20,980 talking about stresses into the eta, psy, and zeta 595 00:37:20,980 --> 00:37:25,670 directions, normal stresses, shear stresses here, that are 596 00:37:25,670 --> 00:37:29,640 related via this stress strain law to the normal strains and 597 00:37:29,640 --> 00:37:31,160 shear strains. 598 00:37:31,160 --> 00:37:35,030 Notice that there's again the shear correction factor, k, 599 00:37:35,030 --> 00:37:38,690 which we want to also include in the formulation because we 600 00:37:38,690 --> 00:37:43,480 have assumed constant shearing strains through the thickness 601 00:37:43,480 --> 00:37:46,230 of the beam, whereas we know that for a rectangular beam, 602 00:37:46,230 --> 00:37:51,810 for example, we have parabolic shear strain distributions if 603 00:37:51,810 --> 00:37:53,730 the beam is straight. 604 00:37:56,360 --> 00:38:01,800 I'd like to now go on with the development of plate elements. 605 00:38:01,800 --> 00:38:06,440 Here, we are talking basically about the same approach. 606 00:38:06,440 --> 00:38:11,220 As I mentioned already once, the beam element that I'm 607 00:38:11,220 --> 00:38:12,650 talking about here-- 608 00:38:12,650 --> 00:38:15,440 that I have been talking about-- is really only an 609 00:38:15,440 --> 00:38:18,100 effective formulation when we talk about, and we want to 610 00:38:18,100 --> 00:38:21,780 develop, a curved beam element. 611 00:38:21,780 --> 00:38:25,280 For pipe elements also, in the case of pipe elements, I 612 00:38:25,280 --> 00:38:28,950 should briefly mention that, of course, we have to 613 00:38:28,950 --> 00:38:32,250 introduce, also, an ovalization degree of freedom. 614 00:38:32,250 --> 00:38:38,580 That ovalization degree of freedom interpolates basically 615 00:38:38,580 --> 00:38:44,300 the ovalization along the curved pipe. 616 00:38:44,300 --> 00:38:46,320 That is an additional degree of freedom that has to be 617 00:38:46,320 --> 00:38:50,710 introduced in the curved beam element formulation. 618 00:38:50,710 --> 00:38:54,450 So, the beam element formulation is effective for 619 00:38:54,450 --> 00:38:57,000 curved beams, pipes. 620 00:38:57,000 --> 00:39:01,260 However, it does show the basic procedure that we are 621 00:39:01,260 --> 00:39:03,240 following also in the development of 622 00:39:03,240 --> 00:39:04,880 plate and shell elements. 623 00:39:04,880 --> 00:39:09,390 And here, we have a typical plate element. 624 00:39:09,390 --> 00:39:11,110 In other words, a flat shell. 625 00:39:11,110 --> 00:39:14,940 The u, v, and w displacements are now 626 00:39:14,940 --> 00:39:16,480 interpolated in this way. 627 00:39:16,480 --> 00:39:20,170 Notice u be the displacement into the x direction, v the 628 00:39:20,170 --> 00:39:22,410 displacement into the y direction, w being the 629 00:39:22,410 --> 00:39:26,100 transverse displacement, and again, we're talking about 630 00:39:26,100 --> 00:39:29,580 section rotations. 631 00:39:29,580 --> 00:39:34,580 Beta x being the section rotation about the y-axis. 632 00:39:34,580 --> 00:39:37,550 That is, the beta x section rotation. 633 00:39:37,550 --> 00:39:42,090 Beta y is a section rotation about the x-axis. 634 00:39:42,090 --> 00:39:46,290 So that our v, measuring z positive upwards. 635 00:39:46,290 --> 00:39:50,540 Notice, our [? v4 ?] point here is negative, and that's 636 00:39:50,540 --> 00:39:52,790 why we have a negative sign there. 637 00:39:52,790 --> 00:39:56,950 Well, with that then given, we can immediately develop our 638 00:39:56,950 --> 00:40:02,830 strains by using the strength of material equations that 639 00:40:02,830 --> 00:40:07,785 tell us epsilon xx is del u del x, epsilon yy is del v del 640 00:40:07,785 --> 00:40:08,850 y, et cetera. 641 00:40:08,850 --> 00:40:11,960 And, of course, we're getting also our shear strengths. 642 00:40:11,960 --> 00:40:17,040 Having developed the strains, we also recognize that our 643 00:40:17,040 --> 00:40:22,080 stresses are given in terms of these formulae, where we have 644 00:40:22,080 --> 00:40:27,870 now the stress strain law for planed stress analysis, 645 00:40:27,870 --> 00:40:34,020 because we are looking at the plate as an assemblage of thin 646 00:40:34,020 --> 00:40:36,540 elements, plane stress elements, lying 647 00:40:36,540 --> 00:40:37,760 on top of each other. 648 00:40:37,760 --> 00:40:41,570 The stress through the plate, of course, is 0, and this is, 649 00:40:41,570 --> 00:40:45,650 therefore, the plane stress [? material ?] 650 00:40:45,650 --> 00:40:47,690 that we have been putting in here. 651 00:40:47,690 --> 00:40:52,650 And the z times this vector here gives us the strains. 652 00:40:52,650 --> 00:40:57,790 The shearing stresses are given here, and our functional 653 00:40:57,790 --> 00:41:01,370 pi that I used also for the beam element 654 00:41:01,370 --> 00:41:03,370 already is given here. 655 00:41:03,370 --> 00:41:06,150 Notice that we now have to integrate through this 656 00:41:06,150 --> 00:41:08,970 thickness of the plate element. 657 00:41:08,970 --> 00:41:11,650 Here's our shear correction factor again, which is 658 00:41:11,650 --> 00:41:16,190 introduced just the same way as in the beam element. 659 00:41:16,190 --> 00:41:20,980 Notice here we have the work, or the total potential of the 660 00:41:20,980 --> 00:41:24,580 external loads, I should say. 661 00:41:24,580 --> 00:41:27,450 It is convenient now to integrate through the 662 00:41:27,450 --> 00:41:30,220 thickness because we can integrate prior to 663 00:41:30,220 --> 00:41:35,820 interpolating the quantities, and that then yields this 664 00:41:35,820 --> 00:41:41,580 value for pi, where our Cb parts and Cs part here, these 665 00:41:41,580 --> 00:41:44,800 two matrices, embody the fact that we have integrated 666 00:41:44,800 --> 00:41:49,350 through the thickness, so we have the 667 00:41:49,350 --> 00:41:51,800 following definitions here. 668 00:41:51,800 --> 00:41:58,580 Kappa simply lists basically the bending strains, or I 669 00:41:58,580 --> 00:42:04,350 should say the rotations of the sections. 670 00:42:04,350 --> 00:42:07,390 Of course here, we have the shearing strains. 671 00:42:07,390 --> 00:42:12,290 Gamma lists the shearing strains. 672 00:42:12,290 --> 00:42:17,440 Cb now is a function of h cubed, just like in classic 673 00:42:17,440 --> 00:42:18,000 plate theory. 674 00:42:18,000 --> 00:42:21,190 Of course, we also have an h cubed entering into the 675 00:42:21,190 --> 00:42:22,250 formulation. 676 00:42:22,250 --> 00:42:26,560 And this Cb matrix embodies the fact that we have 677 00:42:26,560 --> 00:42:28,660 integrated through the thickness. 678 00:42:28,660 --> 00:42:30,400 Here is our Cs. 679 00:42:30,400 --> 00:42:34,440 Of course, in the Cs, we also have the k part. 680 00:42:34,440 --> 00:42:39,030 Notice that, again, we have an h cubed here and an h there. 681 00:42:39,030 --> 00:42:43,340 Therefore, to use our interpolation for plate and 682 00:42:43,340 --> 00:42:46,350 shell elements, we will have to use high enough shell 683 00:42:46,350 --> 00:42:49,460 interpolations to be able to represent the fact that the 684 00:42:49,460 --> 00:42:54,070 shear strains go to 0 for thin plates if we want to use this 685 00:42:54,070 --> 00:42:56,780 formulation for thin plates and shells. 686 00:42:56,780 --> 00:43:02,470 Well, invoking now the fact that pi shall be stationary, 687 00:43:02,470 --> 00:43:05,850 we directly obtain this equation here. 688 00:43:05,850 --> 00:43:07,770 And this, of course, is nothing else than the 689 00:43:07,770 --> 00:43:09,610 principal of virtual displacement 690 00:43:09,610 --> 00:43:11,880 for the plate element. 691 00:43:11,880 --> 00:43:16,050 Notice that from this point onwards, we simply need to 692 00:43:16,050 --> 00:43:18,810 substitute only our interpolations. 693 00:43:18,810 --> 00:43:22,740 The interpolations that we are using are now interpolations 694 00:43:22,740 --> 00:43:27,870 for w, beta x, and beta y. 695 00:43:27,870 --> 00:43:30,860 And, of course, we also interpolate x and y. 696 00:43:34,250 --> 00:43:37,870 These interpolations, beta x and beta y, are independent 697 00:43:37,870 --> 00:43:41,080 from the interpolations of w, and that is the important 698 00:43:41,080 --> 00:43:43,510 points, as I mentioned in the 699 00:43:43,510 --> 00:43:44,960 development of the beam element. 700 00:43:47,590 --> 00:43:51,480 The fact that we are dealing, here, with three 701 00:43:51,480 --> 00:43:56,130 interpolations of course means that an each nodal point, we 702 00:43:56,130 --> 00:44:02,330 have three unknowns, w's, and section rotations. 703 00:44:02,330 --> 00:44:06,440 Let us now look very briefly at shell elements. 704 00:44:06,440 --> 00:44:10,730 The same concept that we used to develop the general beam 705 00:44:10,730 --> 00:44:15,690 element after having discussed the special beam element is 706 00:44:15,690 --> 00:44:18,620 also employed now in the development of what you might 707 00:44:18,620 --> 00:44:22,410 call a general shell element, versus the special plate 708 00:44:22,410 --> 00:44:25,790 element that I just discussed, or the special shell element 709 00:44:25,790 --> 00:44:31,150 that I just discussed, because a flat shell is of course 710 00:44:31,150 --> 00:44:33,450 nothing else than a plate, if we also don't 711 00:44:33,450 --> 00:44:37,950 have membrane forces. 712 00:44:37,950 --> 00:44:42,280 Well here, I'm showing a shell element, a 713 00:44:42,280 --> 00:44:44,000 nine-noded shell element. 714 00:44:44,000 --> 00:44:48,190 And notice that in this case, now, we are talking about this 715 00:44:48,190 --> 00:44:49,190 normal only. 716 00:44:49,190 --> 00:44:54,660 In the beam, we had two normals, vt and vs. Now we 717 00:44:54,660 --> 00:44:56,900 only have one normal. 718 00:44:56,900 --> 00:44:59,730 At a nodal point, we are defining the membrane 719 00:44:59,730 --> 00:45:04,250 displacements, uk, vk, wk, being a transverse 720 00:45:04,250 --> 00:45:11,070 displacement, and the rotations, alpha k and beta k. 721 00:45:11,070 --> 00:45:18,550 These rotations are defined about the axis v1 k and v2 k. 722 00:45:18,550 --> 00:45:23,830 Now notice that these two rotations, alpha k and beta k, 723 00:45:23,830 --> 00:45:32,560 will give us the change in vn during deformations. 724 00:45:32,560 --> 00:45:35,460 And this is really how we use alpha k and beta k. 725 00:45:35,460 --> 00:45:41,110 We express a change in vn in terms of alpha k and beta k. 726 00:45:41,110 --> 00:45:43,290 The procedure is the same as in the 727 00:45:43,290 --> 00:45:45,490 case of the beam element. 728 00:45:45,490 --> 00:45:50,660 First, we express our x, y, and z-coordinates. 729 00:45:50,660 --> 00:45:57,530 And we're using here the original normal, vn, the x, y 730 00:45:57,530 --> 00:46:00,660 and z direction cosines. 731 00:46:03,180 --> 00:46:08,820 These are, here, the x, y and z-coordinates of the 732 00:46:08,820 --> 00:46:10,300 nodal point, k. 733 00:46:10,300 --> 00:46:13,770 We have our two dimensional interpolation functions, hk, 734 00:46:13,770 --> 00:46:16,160 now here, because we talk about a two dimensional 735 00:46:16,160 --> 00:46:19,490 surface, the mid-surface of the shell. 736 00:46:19,490 --> 00:46:22,330 Of course, at each nodal point, the shell can have a 737 00:46:22,330 --> 00:46:25,340 different thickness, and that is denoted by using a 738 00:46:25,340 --> 00:46:30,250 different ak value at each nodal point. 739 00:46:30,250 --> 00:46:35,440 Applying this interpolation here to the initial 740 00:46:35,440 --> 00:46:41,520 configuration and the final configuration and subtracting 741 00:46:41,520 --> 00:46:48,220 0x from 1x, and similar for y and z, we directly obtain the 742 00:46:48,220 --> 00:46:51,050 displacements u, v, and w. 743 00:46:51,050 --> 00:46:54,740 Notice that the displacements now are involving the nodal 744 00:46:54,740 --> 00:46:59,910 point displacements and the change in the direction 745 00:46:59,910 --> 00:47:04,410 cosines of the normal, denoted here. 746 00:47:04,410 --> 00:47:08,890 These changes in the direction cosines of the normal can 747 00:47:08,890 --> 00:47:13,840 directly be expressed in terms of the rotations, 748 00:47:13,840 --> 00:47:17,100 alpha k and beta k. 749 00:47:17,100 --> 00:47:24,440 Now, notice here that once we have done this, of course here 750 00:47:24,440 --> 00:47:32,550 we involve now, as I mentioned earlier, the v1 and v2 751 00:47:32,550 --> 00:47:33,600 directions. 752 00:47:33,600 --> 00:47:37,530 And these v1 and v2 directions are arbitrarily selected. 753 00:47:37,530 --> 00:47:38,640 In fact, they are-- 754 00:47:38,640 --> 00:47:43,610 for our shell element here-- selected as shown here. 755 00:47:43,610 --> 00:47:48,110 But once we have selected v1 and v2 at each nodal point, 756 00:47:48,110 --> 00:47:52,670 and they can vary from nodal point to nodal point, then we 757 00:47:52,670 --> 00:47:56,770 can use this relation to attain directly the change in 758 00:47:56,770 --> 00:48:03,930 the direction cosines of the normal during deformations, 759 00:48:03,930 --> 00:48:08,330 when the deformations are alpha k and beta k. 760 00:48:08,330 --> 00:48:12,530 So with this equation, then, and the earlier equation that 761 00:48:12,530 --> 00:48:18,850 I've given to you, we can directly obtain the 762 00:48:18,850 --> 00:48:21,580 displacement interpolation matrix and the strain 763 00:48:21,580 --> 00:48:22,800 interpolation matrix. 764 00:48:22,800 --> 00:48:25,740 One important point that I should briefly mention is 765 00:48:25,740 --> 00:48:28,980 that, of course, for the shell, we have 0 stresses 766 00:48:28,980 --> 00:48:31,390 through the thickness. 767 00:48:31,390 --> 00:48:37,510 So, we have to use this stress strain law here. 768 00:48:37,510 --> 00:48:41,600 Notice there are 0's in this row and column. 769 00:48:41,600 --> 00:48:45,450 And this is, here, the plane stress part for the bending, 770 00:48:45,450 --> 00:48:49,800 and that is the shear part here. 771 00:48:49,800 --> 00:48:54,890 This is the stress strain law defined in a local convected 772 00:48:54,890 --> 00:48:58,970 coordinate system where we are talking about the stresses 773 00:48:58,970 --> 00:49:03,570 through the thickness being this direction here. 774 00:49:03,570 --> 00:49:05,965 And these other stresses are aligned with 775 00:49:05,965 --> 00:49:07,490 the coordinate system. 776 00:49:07,490 --> 00:49:10,810 We have to transform this one here to the global coordinate 777 00:49:10,810 --> 00:49:14,900 system in order to be able to use it directly in our 778 00:49:14,900 --> 00:49:16,330 formulation. 779 00:49:16,330 --> 00:49:18,820 And that transformation is achieved via these 780 00:49:18,820 --> 00:49:21,160 transformation matrices. 781 00:49:21,160 --> 00:49:25,160 Now, this element has been effectively implemented in the 782 00:49:25,160 --> 00:49:29,270 ADINA computer program, and we want to use it using high 783 00:49:29,270 --> 00:49:32,860 order interpolation, as I mentioned earlier, as the 784 00:49:32,860 --> 00:49:36,160 basic element that therefore is very useful, which can be 785 00:49:36,160 --> 00:49:39,890 used as a flat element, a curved element, all curved 786 00:49:39,890 --> 00:49:47,010 element, or it can have curvature in both directions. 787 00:49:47,010 --> 00:49:50,000 Also, this is the basic element that is being used. 788 00:49:50,000 --> 00:49:54,290 We can also collapse nodes and derive other elements. 789 00:49:54,290 --> 00:49:59,010 As I pointed out earlier, the low order elements should only 790 00:49:59,010 --> 00:50:02,950 be used in very special cases. 791 00:50:02,950 --> 00:50:05,020 I would not recommend these elements. 792 00:50:05,020 --> 00:50:10,570 Although they can be used in principle, the element that is 793 00:50:10,570 --> 00:50:13,960 really useful is this one and that one. 794 00:50:13,960 --> 00:50:17,650 Both, of course, can be used as curved elements as I 795 00:50:17,650 --> 00:50:19,190 pointed out. 796 00:50:19,190 --> 00:50:22,650 Another feature, and this is the final view graph that I 797 00:50:22,650 --> 00:50:26,960 wanted to show in this lecture, is that we can use 798 00:50:26,960 --> 00:50:31,830 these elements also in transition regions. 799 00:50:31,830 --> 00:50:36,460 Namely, here we have the shell element, now being flat-- 800 00:50:36,460 --> 00:50:37,550 that I discussed-- 801 00:50:37,550 --> 00:50:40,320 and we can directly couple this element into another 802 00:50:40,320 --> 00:50:44,320 element, which we call a transition element, which has 803 00:50:44,320 --> 00:50:47,450 the shell degrees of freedom at these nodes, but 804 00:50:47,450 --> 00:50:50,090 translational degrees of freedom only here. 805 00:50:50,090 --> 00:50:53,200 In other words, a continuum element degrees of freedom 806 00:50:53,200 --> 00:50:53,890 right here. 807 00:50:53,890 --> 00:50:57,110 Notice, three degrees of freedom at this node, only 808 00:50:57,110 --> 00:51:00,490 translations, whereas here, we would have five degrees of 809 00:51:00,490 --> 00:51:03,600 freedom-- three translations, two rotations. 810 00:51:03,600 --> 00:51:07,620 Similarly, here we have a curved shell going into a 811 00:51:07,620 --> 00:51:11,200 solid, and again here, we have a transition element. 812 00:51:11,200 --> 00:51:14,700 Here, we show the five degrees of freedom at a shell node and 813 00:51:14,700 --> 00:51:18,670 the three degrees of freedom at a continuum element node. 814 00:51:18,670 --> 00:51:21,870 This is an effective approach to be able to couple director 815 00:51:21,870 --> 00:51:24,930 shell elements into solid elements. 816 00:51:24,930 --> 00:51:29,150 I have not talked, of course, about the actual derivation of 817 00:51:29,150 --> 00:51:31,870 the matrices used in the formulation of 818 00:51:31,870 --> 00:51:33,510 the transition element. 819 00:51:33,510 --> 00:51:36,210 That is a little bit beyond what I wanted to present in 820 00:51:36,210 --> 00:51:37,110 this lecture. 821 00:51:37,110 --> 00:51:40,280 But the basic concepts are those that we discussed 822 00:51:40,280 --> 00:51:44,530 already in the earlier lecture for continuum element and in 823 00:51:44,530 --> 00:51:46,510 this lecture for structural elements. 824 00:51:46,510 --> 00:51:48,200 Thank you very much for your attention.