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 --> 00:00:17,590 hundreds of MIT courses, visit MIT OpenCourseWare at ocw dot 7 00:00:17,590 --> 00:00:18,840 MIT dot edu. 8 00:00:21,320 --> 00:00:23,780 PROFESSOR: Ladies and gentlemen, welcome to lecture 9 00:00:23,780 --> 00:00:25,420 number six. 10 00:00:25,420 --> 00:00:27,840 In this lecture I would like to discuss with you the 11 00:00:27,840 --> 00:00:33,160 formulation and calculation of isoparametric finite elements. 12 00:00:33,160 --> 00:00:36,860 We considered earlier already, in lecture four, the 13 00:00:36,860 --> 00:00:39,300 calculation all finite element matrices. 14 00:00:39,300 --> 00:00:42,050 But in that lecture we considered the generalized 15 00:00:42,050 --> 00:00:45,930 coordinate finite element models. 16 00:00:45,930 --> 00:00:49,020 The generalized coordinate finite element models were the 17 00:00:49,020 --> 00:00:52,330 first finite elements derived. 18 00:00:52,330 --> 00:00:56,110 However, I now want to discuss with you a more general 19 00:00:56,110 --> 00:01:00,210 approach of deriving the required interpolation 20 00:01:00,210 --> 00:01:02,450 matrices and element matrices. 21 00:01:02,450 --> 00:01:07,560 And this more general approach is the isoparametric finite 22 00:01:07,560 --> 00:01:09,462 element derivation. 23 00:01:09,462 --> 00:01:12,360 The isoparametric finite elements that I will be 24 00:01:12,360 --> 00:01:15,750 discussing in this and the next lecture are, in my 25 00:01:15,750 --> 00:01:19,920 opinion, the most effective elements currently available 26 00:01:19,920 --> 00:01:24,320 for plane stress, plane strain, axisymmetric analysis, 27 00:01:24,320 --> 00:01:28,750 three dimensional analysis, thick and thin shell analysis. 28 00:01:28,750 --> 00:01:32,160 These elements are being used, for example, in the computer 29 00:01:32,160 --> 00:01:33,810 program [? Aldena ?]. 30 00:01:33,810 --> 00:01:37,050 They are also used in other computer programs and 31 00:01:37,050 --> 00:01:39,790 represent a modern approach to the solution 32 00:01:39,790 --> 00:01:42,920 off structure problems. 33 00:01:42,920 --> 00:01:46,270 In this lecture, I would like to talk about the derivation 34 00:01:46,270 --> 00:01:49,470 of continuum elements. 35 00:01:49,470 --> 00:01:52,640 In the next lecture we will talk about the derivation of 36 00:01:52,640 --> 00:01:55,840 structural elements, [INAUDIBLE] and central 37 00:01:55,840 --> 00:01:58,190 elements, beam elements. 38 00:01:58,190 --> 00:02:01,510 The basic concept of isoparametric finite element 39 00:02:01,510 --> 00:02:05,830 analysis, is that we interpolate the geometry of an 40 00:02:05,830 --> 00:02:12,410 element, and the displacements of an element in 41 00:02:12,410 --> 00:02:15,360 exactly the same way. 42 00:02:15,360 --> 00:02:18,730 Let us look at the geometry interpolation. 43 00:02:18,730 --> 00:02:22,290 Here you see for three dimensional analysis, the 44 00:02:22,290 --> 00:02:26,640 interpolation of the x-coordinate within the 45 00:02:26,640 --> 00:02:32,560 element, where we use interpolation functions h i. 46 00:02:32,560 --> 00:02:35,600 Of course these interpolation functions are unknown and I 47 00:02:35,600 --> 00:02:37,840 will have to show you how we derive 48 00:02:37,840 --> 00:02:40,390 them for certain elements. 49 00:02:40,390 --> 00:02:43,306 The x i as a nodal point coordinate. 50 00:02:47,860 --> 00:02:52,870 The x i value for i equals 1, for example, is nothing else 51 00:02:52,870 --> 00:02:58,160 than the x-coordinate of nodal point one. 52 00:02:58,160 --> 00:02:58,780 And so on. 53 00:02:58,780 --> 00:03:03,330 So the x i, the nodal point coordinates for all nodes, 54 00:03:03,330 --> 00:03:06,560 they are given, they are input in the analysis the way we 55 00:03:06,560 --> 00:03:09,110 have been discussing it in the previous lecture. 56 00:03:09,110 --> 00:03:16,190 And if we know the h i, we have a direct relationship for 57 00:03:16,190 --> 00:03:21,580 the x-coordinates within the element, as a function of the 58 00:03:21,580 --> 00:03:24,460 nodal point coordinates. 59 00:03:24,460 --> 00:03:29,210 Again, I have to show you how we obtain the h i functions 60 00:03:29,210 --> 00:03:32,300 for the various elements that we are using. 61 00:03:32,300 --> 00:03:36,310 Similarly, for the y interpolation and similarly 62 00:03:36,310 --> 00:03:37,560 for the z interpolation. 63 00:03:40,550 --> 00:03:45,500 Having derived the h i functions we are using in the 64 00:03:45,500 --> 00:03:51,760 isoparametric finite element, the same functions also to 65 00:03:51,760 --> 00:03:54,300 interpolate the displacements. 66 00:03:54,300 --> 00:04:00,700 Notice that we're having here, the nodal point displacements 67 00:04:00,700 --> 00:04:07,980 u i, v i, and w i, and here we have the same interpolation 68 00:04:07,980 --> 00:04:11,560 functions that we already used in the coordinate 69 00:04:11,560 --> 00:04:13,810 interpolations. 70 00:04:13,810 --> 00:04:18,680 And if the number of nodes that are used in the 71 00:04:18,680 --> 00:04:20,420 description of the element-- 72 00:04:20,420 --> 00:04:24,550 and, in fact, we will see that N can be a variable, it can be 73 00:04:24,550 --> 00:04:30,870 equal to three, four, five, up to a large number of nodes. 74 00:04:30,870 --> 00:04:36,870 And in practice we generally don't go much further than 60. 75 00:04:36,870 --> 00:04:40,030 I mentioned that I want to discuss in this lecture 76 00:04:40,030 --> 00:04:41,180 continuum elements. 77 00:04:41,180 --> 00:04:46,730 Well, the continuum elements that we addressing ourselves 78 00:04:46,730 --> 00:04:51,190 to are the truss element, the two d elements-- plane stress, 79 00:04:51,190 --> 00:04:54,320 plane strain, and axisymmetric elements-- 80 00:04:54,320 --> 00:04:57,550 and then, the 3 d elements for three dimensional and thick 81 00:04:57,550 --> 00:04:59,640 shell analysis. 82 00:04:59,640 --> 00:05:01,700 These are continuum elements. 83 00:05:01,700 --> 00:05:06,720 We call them continuum elements, because we only use 84 00:05:06,720 --> 00:05:11,940 u, here u and v, and here, u v and w-- 85 00:05:11,940 --> 00:05:12,740 the displacements-- 86 00:05:12,740 --> 00:05:17,270 to describe the internal element displacements. 87 00:05:17,270 --> 00:05:21,440 We're only using the nodal point displacements u, u v, 88 00:05:21,440 --> 00:05:25,770 and u v w, for all of the nodal points to describe the 89 00:05:25,770 --> 00:05:29,080 internal element displacements. 90 00:05:29,080 --> 00:05:34,080 Structure elements, on the other hand, also use the 91 00:05:34,080 --> 00:05:39,360 rotations at the nodal point theta x, theta y, and theta z, 92 00:05:39,360 --> 00:05:40,940 in order to describe the 93 00:05:40,940 --> 00:05:43,050 deformations within the element. 94 00:05:43,050 --> 00:05:45,730 And I will be discussing structural elements in the 95 00:05:45,730 --> 00:05:49,620 next lecture, so, for the moment, we do not have any 96 00:05:49,620 --> 00:05:51,150 rotations at the nodes. 97 00:05:51,150 --> 00:05:57,730 Or what we will allow, u v and w displacements at the nodes. 98 00:05:57,730 --> 00:06:03,130 Typical examples are given on this Viewgraph. 99 00:06:03,130 --> 00:06:05,160 These are elements that are available in the computer 100 00:06:05,160 --> 00:06:06,980 program [? Aldena ?]. 101 00:06:06,980 --> 00:06:12,910 Here we have a truss element, a two-noded truss element. 102 00:06:12,910 --> 00:06:17,480 The only displacement of interest in the truss is the 103 00:06:17,480 --> 00:06:20,350 actual displacement here. 104 00:06:20,350 --> 00:06:24,390 Here we have a three-noded truss, or cable element. 105 00:06:24,390 --> 00:06:27,470 Here we have a set of two dimensional elements. 106 00:06:27,470 --> 00:06:31,120 Notice we have triangular elements here, a triangular 107 00:06:31,120 --> 00:06:32,100 element here. 108 00:06:32,100 --> 00:06:35,640 We have here an eight node element that is curved. 109 00:06:35,640 --> 00:06:37,920 We can construct curved elements in the 110 00:06:37,920 --> 00:06:39,420 isoparametric approach. 111 00:06:39,420 --> 00:06:41,980 And here we have a rectangular, eight node 112 00:06:41,980 --> 00:06:45,290 element, straight sides, in other words. 113 00:06:45,290 --> 00:06:47,950 These elements are used in plane stress, plane strain, 114 00:06:47,950 --> 00:06:49,295 and axisymmetric analysis. 115 00:06:52,100 --> 00:06:55,540 In three dimensional analysis we might be talking about-- 116 00:06:55,540 --> 00:06:57,050 we might be using-- 117 00:06:57,050 --> 00:07:00,550 such elements such as the eight node brick. 118 00:07:00,550 --> 00:07:06,940 Each node now has three displacements, u, v, and w. 119 00:07:06,940 --> 00:07:10,190 This is here a higher order element, where we have, in 120 00:07:10,190 --> 00:07:14,430 addition to the corner nodes, we also have mid-side nodes. 121 00:07:14,430 --> 00:07:16,250 You don't need to have a mid-side 122 00:07:16,250 --> 00:07:18,060 nodes along all sides. 123 00:07:18,060 --> 00:07:20,050 For example here, I did not put a 124 00:07:20,050 --> 00:07:21,960 mid-side node as an example. 125 00:07:21,960 --> 00:07:25,610 In fact we could simply have mid-side nodes here, and no 126 00:07:25,610 --> 00:07:28,030 mid-side nodes anywhere else. 127 00:07:28,030 --> 00:07:30,670 I will show you how we deconstruct these 128 00:07:30,670 --> 00:07:35,500 interpolation matrices in the next few minutes. 129 00:07:35,500 --> 00:07:41,220 Let us consider, as a very simple 130 00:07:41,220 --> 00:07:44,500 example, a special case. 131 00:07:44,500 --> 00:07:48,510 And the special case that I would like to look at is a 132 00:07:48,510 --> 00:07:52,940 truss element, which is two units long. 133 00:07:52,940 --> 00:07:57,170 In other words, the length from here to there, is equal 134 00:07:57,170 --> 00:08:01,720 to 1, and similarly the length from here to there, is also 135 00:08:01,720 --> 00:08:02,970 equal to one. 136 00:08:05,460 --> 00:08:11,700 I'm describing the element with a coordinate r that I set 137 00:08:11,700 --> 00:08:16,140 equal to zero at the midpoint of the element, plus one at 138 00:08:16,140 --> 00:08:19,570 the right end and minus one at the left end. 139 00:08:19,570 --> 00:08:24,350 This is special geometry and we will later on have to 140 00:08:24,350 --> 00:08:29,240 generalize our approach to more general geometries where 141 00:08:29,240 --> 00:08:31,750 the length is not equal to two and the element 142 00:08:31,750 --> 00:08:33,720 might even be curved. 143 00:08:33,720 --> 00:08:35,880 But for instructive purposes, let's look at 144 00:08:35,880 --> 00:08:38,250 this geometry first. 145 00:08:38,250 --> 00:08:43,330 Similarly, for two dimensional analysis, I want to look at an 146 00:08:43,330 --> 00:08:47,940 element that has length two this direction and length two 147 00:08:47,940 --> 00:08:50,050 that direction. 148 00:08:50,050 --> 00:08:53,390 In other words, a generalization of the concept 149 00:08:53,390 --> 00:08:55,920 that we just looked at for the truss. 150 00:08:55,920 --> 00:08:59,660 Now we have two dimensions and in each dimension we have a 151 00:08:59,660 --> 00:09:01,060 length of two. 152 00:09:01,060 --> 00:09:06,690 I embed into this element an rs-coordinate system. 153 00:09:06,690 --> 00:09:12,710 And this coordinate system now has its origin here at r 154 00:09:12,710 --> 00:09:17,130 equals zero and s equals zero. 155 00:09:17,130 --> 00:09:22,310 The r-axis bisects this side, and the s-axis 156 00:09:22,310 --> 00:09:25,560 bisects this side. 157 00:09:25,560 --> 00:09:29,280 Notice that this is a coordinate system that is 158 00:09:29,280 --> 00:09:31,670 embedded into the element. 159 00:09:31,670 --> 00:09:37,370 We also call that the natural coordinate system, or the 160 00:09:37,370 --> 00:09:40,300 Isoparametric coordinate system. 161 00:09:40,300 --> 00:09:45,700 This element, two by two, would lie in space in an x 162 00:09:45,700 --> 00:09:49,650 xy-coordinate system as indicated here. 163 00:09:49,650 --> 00:09:52,880 Now for three dimensional analysis we would proceed in 164 00:09:52,880 --> 00:09:56,750 the same way, then we would consider an element that is 165 00:09:56,750 --> 00:09:59,700 two by two by two units long into 166 00:09:59,700 --> 00:10:01,450 each coordinate direction. 167 00:10:01,450 --> 00:10:05,530 And then we would have an rs-axis and a t-axis coming 168 00:10:05,530 --> 00:10:10,530 out of the transparency in this particular case. 169 00:10:10,530 --> 00:10:15,260 Well, I want to look at the truss element, the special 170 00:10:15,260 --> 00:10:19,920 truss element, and the special two d element. 171 00:10:19,920 --> 00:10:23,010 And I want to show you how we can construct the 172 00:10:23,010 --> 00:10:26,260 interpolation matrices, the displacement interpolation 173 00:10:26,260 --> 00:10:29,210 matrices, h i. 174 00:10:29,210 --> 00:10:32,210 These are also the coordinate interpolation matrices. 175 00:10:32,210 --> 00:10:34,550 I want to show you how we construct them for the special 176 00:10:34,550 --> 00:10:39,220 elements, how we then can calculate the strain 177 00:10:39,220 --> 00:10:41,490 displacement interpolation matrices 178 00:10:41,490 --> 00:10:43,240 for the special elements. 179 00:10:43,240 --> 00:10:46,510 And then I want to go on and show you how we generalize to 180 00:10:46,510 --> 00:10:50,750 concepts to curved elements. 181 00:10:50,750 --> 00:10:54,790 And once we have discussed the two dimensional case, I think 182 00:10:54,790 --> 00:10:58,520 you can see yourself how the concepts are generalized to 183 00:10:58,520 --> 00:11:00,985 the three dimensional case. 184 00:11:00,985 --> 00:11:05,440 Let's look then at our two-noded truss first. 185 00:11:05,440 --> 00:11:12,000 Here we have once more, the truss, and we have node one on 186 00:11:12,000 --> 00:11:14,900 the right hand side, node two on the left hand side. 187 00:11:14,900 --> 00:11:18,820 Our r-coordinate system starts in the middle of the truss. 188 00:11:18,820 --> 00:11:22,900 Notice what we want to obtain is that we interpolate our u 189 00:11:22,900 --> 00:11:26,010 displacement via the interpolations. 190 00:11:26,010 --> 00:11:31,470 U being equal to h i times u i, the h i are the unknowns. 191 00:11:31,470 --> 00:11:34,430 I want to show you how we obtain the h i. 192 00:11:34,430 --> 00:11:37,600 The u i are the nodal point displacements. 193 00:11:37,600 --> 00:11:41,270 In this particular case it would be here, u two and here 194 00:11:41,270 --> 00:11:42,595 we would have u one. 195 00:11:46,120 --> 00:11:48,780 I in other words, goes from one to two for 196 00:11:48,780 --> 00:11:50,810 this particular case. 197 00:11:50,810 --> 00:11:55,140 The h i has to be a function of r. 198 00:11:55,140 --> 00:12:02,230 For a given r, however, we can evaluate the u displacement if 199 00:12:02,230 --> 00:12:04,530 we have u i is given. 200 00:12:04,530 --> 00:12:08,780 In other words, when u one and u two are given, then for a 201 00:12:08,780 --> 00:12:12,140 given r, we can evaluate directly from this relation 202 00:12:12,140 --> 00:12:15,810 the displacement at that point r. 203 00:12:15,810 --> 00:12:20,650 Well, with two nodes, we can only have a bi-- 204 00:12:20,650 --> 00:12:24,784 only a linear representation in the displacements. 205 00:12:24,784 --> 00:12:30,050 The h one, from this relation, must be this function. 206 00:12:30,050 --> 00:12:35,310 Because if we look at this, and we say let u two be equal 207 00:12:35,310 --> 00:12:38,390 to zero, we would simply have that u is 208 00:12:38,390 --> 00:12:41,270 equal to h one u one. 209 00:12:41,270 --> 00:12:46,100 Well, therefore, our h one must look this one, because u 210 00:12:46,100 --> 00:12:48,830 one has its full strength here. 211 00:12:48,830 --> 00:12:53,690 If we put u one equal to one, we would have u is equal to h 212 00:12:53,690 --> 00:12:57,700 one, and this would be the variation. 213 00:12:57,700 --> 00:13:03,240 Similarly, for h two, in this case, we would have u is equal 214 00:13:03,240 --> 00:13:07,290 to h two u two. 215 00:13:07,290 --> 00:13:10,160 H one is now equal to zero. 216 00:13:10,160 --> 00:13:13,990 If we put you two equal to one, as I have done in this 217 00:13:13,990 --> 00:13:19,170 particular case, then our linear variation is indicated 218 00:13:19,170 --> 00:13:22,680 as shown here , and the function h two 219 00:13:22,680 --> 00:13:24,340 is given right here. 220 00:13:24,340 --> 00:13:30,310 Notice that h two is equal to zero when r is equal to one. 221 00:13:30,310 --> 00:13:35,780 And h two is equal to one when r is equal to minus 1. 222 00:13:35,780 --> 00:13:38,570 So, this gives us h two. 223 00:13:38,570 --> 00:13:41,740 The two-noded truss, therefore, simply has this 224 00:13:41,740 --> 00:13:45,170 description here for the displacement and the 225 00:13:45,170 --> 00:13:46,660 coordinates. 226 00:13:46,660 --> 00:13:49,850 Remember that we're using the same h i for displacement and 227 00:13:49,850 --> 00:13:51,590 coordinates. 228 00:13:51,590 --> 00:13:54,720 It simply has this description where h one and h two are 229 00:13:54,720 --> 00:13:57,080 defined as shown. 230 00:13:57,080 --> 00:14:02,320 Let us now say that we want to add another node, that we want 231 00:14:02,320 --> 00:14:04,130 to put another node right there. 232 00:14:04,130 --> 00:14:07,040 In other words, we want to go from a two-noded description 233 00:14:07,040 --> 00:14:08,750 to a three-noded description. 234 00:14:08,750 --> 00:14:11,700 On one of the earlier Viewgraphs, you could see a 235 00:14:11,700 --> 00:14:13,680 three-noded cable element. 236 00:14:13,680 --> 00:14:19,540 Well, the way we proceed in the construction of the h i 237 00:14:19,540 --> 00:14:23,550 functions is as follows. 238 00:14:23,550 --> 00:14:29,556 The two-noded element simply had this description here. 239 00:14:29,556 --> 00:14:32,470 H one is given as shown here on the left 240 00:14:32,470 --> 00:14:35,100 side of the blue line. 241 00:14:35,100 --> 00:14:38,010 H two was simply this. 242 00:14:38,010 --> 00:14:42,680 And we knew that we could do no better than a linear 243 00:14:42,680 --> 00:14:46,300 description in displacements between two nodes. 244 00:14:46,300 --> 00:14:50,530 However, now we have a third node right here. 245 00:14:50,530 --> 00:14:55,890 And a third node means that we can use a parabolic 246 00:14:55,890 --> 00:14:57,960 description in displacements. 247 00:14:57,960 --> 00:15:04,230 Now h three must be equal to one at the third node and zero 248 00:15:04,230 --> 00:15:05,840 at both sides. 249 00:15:05,840 --> 00:15:06,540 Why? 250 00:15:06,540 --> 00:15:12,270 Well, because, remember we have u now equal to h i u I, 251 00:15:12,270 --> 00:15:15,220 where i equals one to three. 252 00:15:15,220 --> 00:15:21,180 And if we put u one equal to zero and u to equal to zero, 253 00:15:21,180 --> 00:15:26,690 we simply have u equals h three u three, and, therefore, 254 00:15:26,690 --> 00:15:29,060 h three is this function. 255 00:15:31,750 --> 00:15:35,640 I've written it down here when r is equal to zero its one 256 00:15:35,640 --> 00:15:39,750 when r is equal to minus 1 or plus h three is zero. 257 00:15:42,890 --> 00:15:44,810 So this is our h three. 258 00:15:44,810 --> 00:15:51,320 However, if we now look back to our earlier description of 259 00:15:51,320 --> 00:15:56,450 h one and h two for the two-noded element, we remember 260 00:15:56,450 --> 00:15:58,850 that we had a linear variation here. 261 00:15:58,850 --> 00:16:01,640 That we had a linear variation here and here. 262 00:16:01,640 --> 00:16:04,560 This one was a linear variation of h one-- 263 00:16:04,560 --> 00:16:08,220 let me take here the green color to show once more what 264 00:16:08,220 --> 00:16:09,370 we are talking about-- 265 00:16:09,370 --> 00:16:11,560 that was our linear variation. 266 00:16:11,560 --> 00:16:16,140 And here, this one here was our linear variation there. 267 00:16:16,140 --> 00:16:22,790 Well, now with the third node there, we recognize that, at 268 00:16:22,790 --> 00:16:25,635 this third node our h two-- 269 00:16:25,635 --> 00:16:28,650 our actual h two for the three-noded truss-- 270 00:16:28,650 --> 00:16:30,147 must be zero here. 271 00:16:32,730 --> 00:16:34,820 H one must be zero here. 272 00:16:34,820 --> 00:16:37,940 Well, how can we make it zero here? 273 00:16:37,940 --> 00:16:42,550 We can make it zero by subtracting from our two-noded 274 00:16:42,550 --> 00:16:47,690 truss h one, one half of this parabolic description. 275 00:16:47,690 --> 00:16:49,910 And that's what I have done here. 276 00:16:49,910 --> 00:16:55,525 Similarly, we have to subtract it here, because then, what we 277 00:16:55,525 --> 00:16:59,350 are doing is we are taking one half off this parabola, and we 278 00:16:59,350 --> 00:17:01,660 are putting it right on here. 279 00:17:01,660 --> 00:17:05,430 This is one half of the bottom parabola. 280 00:17:05,430 --> 00:17:06,780 We are putting it on there. 281 00:17:06,780 --> 00:17:10,280 And that brings this point back to that point. 282 00:17:10,280 --> 00:17:12,220 Remember this is equal to one. 283 00:17:12,220 --> 00:17:16,560 I want one half as a correction here, and that's 284 00:17:16,560 --> 00:17:21,920 why I take one half of one minus r squared, to bring this 285 00:17:21,920 --> 00:17:23,950 point back to there. 286 00:17:23,950 --> 00:17:30,100 This total description then, this total part-- 287 00:17:30,100 --> 00:17:33,060 this part here, all of that together-- 288 00:17:33,060 --> 00:17:38,500 is our h one for the three-noded element, for the 289 00:17:38,500 --> 00:17:40,030 three-noded element. 290 00:17:40,030 --> 00:17:46,170 And similarly, we would have four h two. 291 00:17:46,170 --> 00:17:51,510 This total part here is this function here for our 292 00:17:51,510 --> 00:17:53,050 three-noded element. 293 00:17:53,050 --> 00:17:56,470 Now the important point that I really would like you to 294 00:17:56,470 --> 00:18:02,940 understand is that, we have started off with a two-noded 295 00:18:02,940 --> 00:18:05,230 element description-- 296 00:18:05,230 --> 00:18:09,610 h one only the linear part, h two only the linear part. 297 00:18:09,610 --> 00:18:13,180 We have constructed the interpolation function for the 298 00:18:13,180 --> 00:18:18,470 third node, and then we have corrected the earlier 299 00:18:18,470 --> 00:18:24,200 two-noded interpolation functions via subtracting a 300 00:18:24,200 --> 00:18:27,990 certain part of the third interpolation function, in 301 00:18:27,990 --> 00:18:31,480 order to obtain the new h one and h two for 302 00:18:31,480 --> 00:18:33,950 the three-noded element. 303 00:18:33,950 --> 00:18:38,620 And this is indeed the actual procedure that we can use very 304 00:18:38,620 --> 00:18:42,520 effectively in constructing higher order elements. 305 00:18:42,520 --> 00:18:45,040 We are starting off with the lower order element 306 00:18:45,040 --> 00:18:46,810 descriptions-- 307 00:18:46,810 --> 00:18:51,070 the ones shown here dashed with a dashed line, the linear 308 00:18:51,070 --> 00:18:52,690 part only-- 309 00:18:52,690 --> 00:18:59,590 and we add in the higher order description, and subtract the 310 00:18:59,590 --> 00:19:03,620 correction from the lower order description. 311 00:19:03,620 --> 00:19:06,910 The correction has to be subtracted to bring this point 312 00:19:06,910 --> 00:19:11,130 back to zero, because h one total must now be equal to 313 00:19:11,130 --> 00:19:15,330 zero here, h two total must be equal to zero here, and this 314 00:19:15,330 --> 00:19:19,000 way we have constructed our new h one and h two for the 315 00:19:19,000 --> 00:19:22,500 three-noded truss element. 316 00:19:22,500 --> 00:19:28,950 Let's look at the four-noded element in two dimensions now. 317 00:19:28,950 --> 00:19:32,120 Here we use very similar concepts. 318 00:19:32,120 --> 00:19:35,500 These are the four nodes for the element. 319 00:19:35,500 --> 00:19:39,980 The description along a side is just like we have been 320 00:19:39,980 --> 00:19:44,450 discussing now for the truss, linear here and linear here. 321 00:19:44,450 --> 00:19:47,320 Notice h one can directly be written down. 322 00:19:47,320 --> 00:19:51,630 It has to be equal to one here and zero everywhere else. 323 00:19:51,630 --> 00:19:57,400 This is a function which is bilinear and which satisfies 324 00:19:57,400 --> 00:20:00,170 these conditions to be equal to one here and. 325 00:20:00,170 --> 00:20:02,770 Zero at the other nodes. 326 00:20:02,770 --> 00:20:07,980 It is a linear variation along this side, along this side, 327 00:20:07,980 --> 00:20:10,590 and also across the surface. 328 00:20:10,590 --> 00:20:18,040 In other words, for a given value of r, we have a linear 329 00:20:18,040 --> 00:20:21,140 variation across here too. 330 00:20:21,140 --> 00:20:22,960 How do we obtain h two? 331 00:20:22,960 --> 00:20:27,980 Well, we simply have to change the signs over here-- 332 00:20:27,980 --> 00:20:32,810 the plus signs here-- in an appropriate way. 333 00:20:32,810 --> 00:20:36,270 H two shall be equal to one here zero at all the other 334 00:20:36,270 --> 00:20:40,270 nodes, while we see that this function satisfies these 335 00:20:40,270 --> 00:20:44,030 conditions, let's put in r equal to minus 1. 336 00:20:44,030 --> 00:20:47,920 That makes this two, divided by four, gives us one half. 337 00:20:47,920 --> 00:20:50,790 S has to be equal to plus 1 at this point-- 338 00:20:50,790 --> 00:20:52,070 you get another two in here-- 339 00:20:52,070 --> 00:20:56,010 so h two is equal to one right there, and its 340 00:20:56,010 --> 00:20:57,480 zero everywhere else. 341 00:20:57,480 --> 00:21:00,010 Similarly we construct h three and h four. 342 00:21:00,010 --> 00:21:04,240 Indeed we can immediately observe that these signs-- 343 00:21:04,240 --> 00:21:09,720 plus in both cases here, a minus here, plus there, minus, 344 00:21:09,720 --> 00:21:12,620 minus, plus, minus-- these signs correspond to nothing 345 00:21:12,620 --> 00:21:17,570 else in the signs of the r- and s-coordinates of the nodal 346 00:21:17,570 --> 00:21:19,590 point under consideration. 347 00:21:19,590 --> 00:21:23,320 R and s is plus here, you have two plus signs here. 348 00:21:23,320 --> 00:21:26,460 R is negative here, you put a negative sign here. 349 00:21:26,460 --> 00:21:29,940 But s is positive here, you put a plus sign there. 350 00:21:29,940 --> 00:21:33,470 r and s both are negative here, so we have two negative 351 00:21:33,470 --> 00:21:36,030 signs here, et cetera. 352 00:21:36,030 --> 00:21:40,590 Let us now see how we construct from this basic four 353 00:21:40,590 --> 00:21:44,030 node element, which really corresponds to our basic two 354 00:21:44,030 --> 00:21:46,520 node element in the case of the truss, how we can 355 00:21:46,520 --> 00:21:48,620 construct higher order elements. 356 00:21:48,620 --> 00:21:52,425 Well, we proceed in much the same way as in the truss 357 00:21:52,425 --> 00:21:53,400 formulation. 358 00:21:53,400 --> 00:21:57,500 Here we have a four-noded basic element and we have 359 00:21:57,500 --> 00:22:00,100 added a fifth node to it. 360 00:22:00,100 --> 00:22:02,800 Now let's look at this in detail. 361 00:22:02,800 --> 00:22:06,950 Since we now have added a fifth node here, we know that 362 00:22:06,950 --> 00:22:09,820 we can allow a parabolic distribution. 363 00:22:09,820 --> 00:22:12,220 In fact, we should allow a parabolic distribution of 364 00:22:12,220 --> 00:22:14,950 displacements along the side. 365 00:22:14,950 --> 00:22:19,780 Well, if its a parabola along the side for s being plus 1, 366 00:22:19,780 --> 00:22:23,380 we immediately see that this function here-- 367 00:22:23,380 --> 00:22:27,500 this s equal to plus one, this is equal to two, knocks out 368 00:22:27,500 --> 00:22:29,420 that one half, we have a one minus r 369 00:22:29,420 --> 00:22:31,340 squared along the side-- 370 00:22:31,340 --> 00:22:34,910 just the same function that we already had in the truss. 371 00:22:34,910 --> 00:22:39,630 Well, along this direction we can only vary things linearly 372 00:22:39,630 --> 00:22:43,550 because we have two nodes only in along these directions. 373 00:22:43,550 --> 00:22:46,650 And this is the reason why we have to put in 374 00:22:46,650 --> 00:22:48,720 a one plus s here. 375 00:22:48,720 --> 00:22:51,400 The linear variation is given by one plus s. 376 00:22:51,400 --> 00:22:56,240 And we also notice that when s is equal to minus 1, in other 377 00:22:56,240 --> 00:22:59,790 words, we are looking at this side, this function is zero. 378 00:22:59,790 --> 00:23:02,960 When s is equal to plus one, as we pointed out earlier, 379 00:23:02,960 --> 00:23:04,180 this function here-- 380 00:23:04,180 --> 00:23:06,880 this part and that part-- 381 00:23:06,880 --> 00:23:10,330 gives us a one together, and we have simply a one minus r 382 00:23:10,330 --> 00:23:11,780 squared along here. 383 00:23:11,780 --> 00:23:15,990 So, here you can see these triangles that show the linear 384 00:23:15,990 --> 00:23:18,790 variation along this side. 385 00:23:18,790 --> 00:23:23,150 These parabolas here run really across here, but with a 386 00:23:23,150 --> 00:23:24,520 different intensity. 387 00:23:24,520 --> 00:23:29,250 The intensity off the parabola goes down linearly from one at 388 00:23:29,250 --> 00:23:33,160 this end, to is zero at that end. 389 00:23:33,160 --> 00:23:34,720 This is our h five. 390 00:23:34,720 --> 00:23:39,800 Well, if we have this as our h five, we remember that our 391 00:23:39,800 --> 00:23:45,520 original h one of the of the four-noded element had a 392 00:23:45,520 --> 00:23:49,310 linear variation along here, and also a linear variation 393 00:23:49,310 --> 00:23:51,210 along here. 394 00:23:51,210 --> 00:23:57,590 We will now have to take this h one and subtract some 395 00:23:57,590 --> 00:23:59,040 correction from it. 396 00:23:59,040 --> 00:24:05,610 In order to make this point here have zero displacement, 397 00:24:05,610 --> 00:24:06,860 four h one. 398 00:24:06,860 --> 00:24:10,890 Well, what we will do is we take this h five and subtract 399 00:24:10,890 --> 00:24:15,020 a multiple of h five from h one. 400 00:24:15,020 --> 00:24:18,620 In fact, you can see since the original h one is equal to one 401 00:24:18,620 --> 00:24:21,820 half here, we simply have to subtract one half of this 402 00:24:21,820 --> 00:24:25,050 function to obtain the new h one. 403 00:24:25,050 --> 00:24:28,740 And this is what I have done on this Viewgraph. 404 00:24:28,740 --> 00:24:33,340 The h one now is the original h one that we had. 405 00:24:33,340 --> 00:24:37,740 And we are subtracting one half of h five, which is one 406 00:24:37,740 --> 00:24:43,240 half of the parabolic distribution, to bring this 407 00:24:43,240 --> 00:24:46,490 point here down to zero in displacements. 408 00:24:46,490 --> 00:24:48,010 And that's what we have done here. 409 00:24:48,010 --> 00:24:52,760 The resulting function then is shown here. 410 00:24:52,760 --> 00:24:57,290 And our h one for the five-noded element is shown 411 00:24:57,290 --> 00:24:58,280 right here. 412 00:24:58,280 --> 00:25:01,070 Similarly, for h two-- 413 00:25:01,070 --> 00:25:05,180 this is the h two function for the five-noded element-- 414 00:25:05,180 --> 00:25:10,500 interesting to note that h three and h four are for the 415 00:25:10,500 --> 00:25:14,840 five-noded element, the same as for the four-noded element. 416 00:25:14,840 --> 00:25:20,160 Because this fifth node lies between nodes one and two. 417 00:25:20,160 --> 00:25:23,910 And there is no effect along this side and 418 00:25:23,910 --> 00:25:25,440 along this side here-- 419 00:25:25,440 --> 00:25:32,300 along that side four h three and h four-- so, we have our 420 00:25:32,300 --> 00:25:38,060 original functions also for the five-noded element. 421 00:25:38,060 --> 00:25:39,910 Let us look now at a 422 00:25:39,910 --> 00:25:42,210 generalization of this concept. 423 00:25:42,210 --> 00:25:46,320 Here we have a typical nine-noded element. 424 00:25:46,320 --> 00:25:49,940 A very effective element for many types of applications. 425 00:25:49,940 --> 00:25:53,970 I already show it here in its curved form, but think of it, 426 00:25:53,970 --> 00:25:56,270 please, as follows. 427 00:25:56,270 --> 00:25:57,970 This is the x-axis. 428 00:25:57,970 --> 00:25:59,630 This is the r-axis. 429 00:25:59,630 --> 00:26:04,160 S is equal to zero along this side. 430 00:26:04,160 --> 00:26:07,030 S is equal to plus one along this side. 431 00:26:07,030 --> 00:26:10,670 S is equal to minus 1 along this side. 432 00:26:10,670 --> 00:26:13,440 R is equal to plus one along this side. 433 00:26:13,440 --> 00:26:16,080 R is equal to minus one along the side. 434 00:26:16,080 --> 00:26:18,830 And r is zero along this axis. 435 00:26:18,830 --> 00:26:24,020 So, in the r s description, in the embedded coordinate 436 00:26:24,020 --> 00:26:30,480 system, this element is still a two by two squared element. 437 00:26:30,480 --> 00:26:37,180 Well, if we look then at the interpolation functions, for 438 00:26:37,180 --> 00:26:39,490 this element, they look as follows. 439 00:26:39,490 --> 00:26:43,020 Now maybe you have difficulty seeing all this information, 440 00:26:43,020 --> 00:26:47,590 so, please then refer to the study guide where you'll find 441 00:26:47,590 --> 00:26:49,270 this Viewgraph. 442 00:26:49,270 --> 00:26:53,510 For the four-noded element, we had these 443 00:26:53,510 --> 00:26:56,670 interpolation functions. 444 00:26:56,670 --> 00:27:02,300 If we want to deal with the five-noded element, what we 445 00:27:02,300 --> 00:27:06,920 have to do is, we add this interpolation function. 446 00:27:06,920 --> 00:27:07,620 And-- 447 00:27:07,620 --> 00:27:09,060 as I pointed out earlier-- 448 00:27:09,060 --> 00:27:14,880 we have to correct our h one and h two. 449 00:27:14,880 --> 00:27:17,830 But that is all of the correction that is required. 450 00:27:17,830 --> 00:27:20,900 H three and h four not corrected-- 451 00:27:20,900 --> 00:27:24,940 they are blank spots here, they are blank spots here. 452 00:27:24,940 --> 00:27:29,460 So, our five-noded element would have these interpolation 453 00:27:29,460 --> 00:27:32,800 functions now shown in red. 454 00:27:32,800 --> 00:27:35,410 For if we added a sixth node-- 455 00:27:35,410 --> 00:27:39,050 and I now should go back to our earlier picture-- 456 00:27:39,050 --> 00:27:42,630 if I wanted to add, in addition to the fifth node, 457 00:27:42,630 --> 00:27:44,250 also the sixth node. 458 00:27:44,250 --> 00:27:50,320 Well, then, I have to put another interpolation function 459 00:27:50,320 --> 00:27:53,430 down here, which now is parabolic in s 460 00:27:53,430 --> 00:27:54,900 and linear in r. 461 00:27:54,900 --> 00:27:58,900 And I have to correct h two and h three again, these are 462 00:27:58,900 --> 00:28:00,740 the corrections. 463 00:28:00,740 --> 00:28:05,360 So, now I have in green here shown to you the interpolation 464 00:28:05,360 --> 00:28:08,340 functions off a six-noded element. 465 00:28:08,340 --> 00:28:12,850 And like that we can proceed by adding interpolation 466 00:28:12,850 --> 00:28:17,230 functions and correcting the earlier constructed 467 00:28:17,230 --> 00:28:20,150 interpolation functions as shown. 468 00:28:20,150 --> 00:28:24,830 Like that we can proceed to directly obtain the 469 00:28:24,830 --> 00:28:30,610 interpolation functions for the five-noded, six-noded, 470 00:28:30,610 --> 00:28:33,370 seven-, eight-, and nine-noded element. 471 00:28:33,370 --> 00:28:38,280 In fact, its also important to notice that we could have this 472 00:28:38,280 --> 00:28:43,190 node, that node, this node, and that node, and just h nine 473 00:28:43,190 --> 00:28:44,330 also added. 474 00:28:44,330 --> 00:28:47,690 We could have, in other words, a five-noded element which 475 00:28:47,690 --> 00:28:48,520 looks like this. 476 00:28:48,520 --> 00:28:52,030 It has this node, that one, that one, and that one, and 477 00:28:52,030 --> 00:28:53,390 that one in the middle. 478 00:28:53,390 --> 00:28:59,230 Another five-noded element would be this one, this one, 479 00:28:59,230 --> 00:29:02,250 this one, this one with that node in there. 480 00:29:02,250 --> 00:29:06,960 So there is no necessity in having all the nodes below a 481 00:29:06,960 --> 00:29:07,910 certain number. 482 00:29:07,910 --> 00:29:12,220 But we can simply use four nodes, and then add whichever 483 00:29:12,220 --> 00:29:17,480 nodes we want to have into the element. 484 00:29:17,480 --> 00:29:21,690 This is, then, how we construct 485 00:29:21,690 --> 00:29:23,690 interpolation functions. 486 00:29:23,690 --> 00:29:29,020 And once we had the h i, we directly can obtain the h 487 00:29:29,020 --> 00:29:33,840 matrix, the matrix that gives the displacements in terms of 488 00:29:33,840 --> 00:29:37,070 the nodal point displacements. 489 00:29:37,070 --> 00:29:40,170 Notice that the h matrix really is 490 00:29:40,170 --> 00:29:43,410 constructed from the hi's. 491 00:29:43,410 --> 00:29:45,910 And the elements of the b matrix-- 492 00:29:45,910 --> 00:29:48,490 the strain displacement interpolation matrix-- 493 00:29:48,490 --> 00:29:52,330 are the derivatives of the h i or zero. 494 00:29:52,330 --> 00:29:54,810 And I will show you an example right now. 495 00:29:54,810 --> 00:29:58,160 Because we are using four, we are still looking at the 496 00:29:58,160 --> 00:30:02,150 special case of a two by two by two element in a truss 497 00:30:02,150 --> 00:30:03,500 case, only this two. 498 00:30:03,500 --> 00:30:06,000 Plane stress, plane strain, axisymmetry, two dimensional 499 00:30:06,000 --> 00:30:08,770 analysis, we have these two two's. 500 00:30:08,770 --> 00:30:10,960 In other words, we're talking about a two by two element. 501 00:30:10,960 --> 00:30:13,510 And in three dimensional analysis we would talk about a 502 00:30:13,510 --> 00:30:15,120 two by two by two element. 503 00:30:15,120 --> 00:30:18,270 In these cases, we have x equal to r, y equal to s, z 504 00:30:18,270 --> 00:30:19,270 equal to t. 505 00:30:19,270 --> 00:30:20,930 So the strains-- 506 00:30:20,930 --> 00:30:24,440 which are derivatives with respect to x the actual 507 00:30:24,440 --> 00:30:26,100 physical coordinates-- 508 00:30:26,100 --> 00:30:28,790 can also directly be obtained by simply taking the 509 00:30:28,790 --> 00:30:31,710 derivative with respect to r, and then 510 00:30:31,710 --> 00:30:35,230 similarly for s and t. 511 00:30:35,230 --> 00:30:39,920 Let us look at a four node, two dimensional element. 512 00:30:39,920 --> 00:30:42,760 This is really the element that we have used in our 513 00:30:42,760 --> 00:30:46,860 earlier example of the cantilever analysis. 514 00:30:46,860 --> 00:30:50,730 Here we would simply have that u r s, v r s, are 515 00:30:50,730 --> 00:30:53,820 described as shown. 516 00:30:53,820 --> 00:30:58,100 Notice that in the first row we are really saying nothing 517 00:30:58,100 --> 00:31:02,890 else than u is a summation h i u i. 518 00:31:02,890 --> 00:31:06,150 Where i equals one to four because we have 519 00:31:06,150 --> 00:31:07,725 a four-noded element. 520 00:31:11,510 --> 00:31:12,830 H i u i. 521 00:31:12,830 --> 00:31:15,930 In the second row we are really saying nothing else 522 00:31:15,930 --> 00:31:22,450 than v being the summation of i equals one to four h i v i. 523 00:31:22,450 --> 00:31:26,670 And all I have done is I have taken these hi's and assembled 524 00:31:26,670 --> 00:31:30,545 them into a matrix form to obtain our h matrix. 525 00:31:33,180 --> 00:31:35,290 This h matrix here-- 526 00:31:35,290 --> 00:31:39,120 the entries in that h matrix-- are dependent on the ordering 527 00:31:39,120 --> 00:31:43,660 that you're using here for u one, v one, u two, et cetera. 528 00:31:43,660 --> 00:31:46,100 With this ordering, these are the entries. 529 00:31:46,100 --> 00:31:49,650 Notice there are zeroes here, because the v degrees of 530 00:31:49,650 --> 00:31:54,860 freedom at the nodes have no contribution to the u 531 00:31:54,860 --> 00:31:57,110 displacement in the element. 532 00:31:57,110 --> 00:32:01,470 Well, if we look at the plane stress case and want to 533 00:32:01,470 --> 00:32:06,260 construct our b matrix, remember the b matrix gives 534 00:32:06,260 --> 00:32:11,970 the strains in terms of the nodal point displacements. 535 00:32:11,970 --> 00:32:18,190 And we remember also that our epsilon xx is equal to a dy 536 00:32:18,190 --> 00:32:26,800 dx, our epsilon yy is equal to dv dy and our gamma xy is 537 00:32:26,800 --> 00:32:32,370 equal to a du dy plus dv dx. 538 00:32:32,370 --> 00:32:36,260 Well, if we recognize these facts. 539 00:32:36,260 --> 00:32:40,560 And we also note again that for the two by two element, r 540 00:32:40,560 --> 00:32:44,330 is identical to x, s is identical to y. 541 00:32:44,330 --> 00:32:53,630 Then, we can obtain the epsilon rr or epsilon xx by 542 00:32:53,630 --> 00:32:58,750 simply taking the derivatives of the hi's with respect to r. 543 00:32:58,750 --> 00:33:04,940 Notice here, since u is equal to the summation of h i u i, 544 00:33:04,940 --> 00:33:10,060 dy dr, which is equal to du dx, is nothing else then the 545 00:33:10,060 --> 00:33:16,730 summation of partial h i dr u i. 546 00:33:16,730 --> 00:33:21,110 And this part, which runs from one to four-- 547 00:33:21,110 --> 00:33:22,800 i going from one to four-- 548 00:33:22,800 --> 00:33:26,170 I simply put right in there. 549 00:33:26,170 --> 00:33:29,510 I proceed similarly for epsilon yy. 550 00:33:29,510 --> 00:33:34,540 Now I'm talking about the v displacements, which are 551 00:33:34,540 --> 00:33:37,170 stored after the u displacements. 552 00:33:37,170 --> 00:33:38,620 That's why I'm looking here at the second 553 00:33:38,620 --> 00:33:40,640 column, the last column. 554 00:33:40,640 --> 00:33:44,670 And here we have dh one ds, because we are talking about 555 00:33:44,670 --> 00:33:47,870 the derivative with respect to y or with respect to s, which 556 00:33:47,870 --> 00:33:49,050 is the same thing. 557 00:33:49,050 --> 00:33:54,540 So here we have the entries for the epsilon ss part. 558 00:33:54,540 --> 00:34:02,610 Now for the strain part, we are talking du dy or du ds, dv 559 00:34:02,610 --> 00:34:08,510 dx, dv dr. And all we have to do in now is take this term 560 00:34:08,510 --> 00:34:12,530 and put it right in there, and take this term and put it 561 00:34:12,530 --> 00:34:15,219 right in there. 562 00:34:15,219 --> 00:34:17,760 And then the last row here gives us 563 00:34:17,760 --> 00:34:20,719 the shearing strength. 564 00:34:20,719 --> 00:34:23,080 So this is our b matrix for the 565 00:34:23,080 --> 00:34:26,639 special two by two element. 566 00:34:26,639 --> 00:34:30,040 It is constructed in a very simple way. 567 00:34:30,040 --> 00:34:34,320 The h i are known, and if we had five or six or seven 568 00:34:34,320 --> 00:34:36,889 nodes, we would proceed in exactly the same way. 569 00:34:36,889 --> 00:34:40,550 All we would have to do is include additional columns in 570 00:34:40,550 --> 00:34:45,300 the b matrix that would give us the appropriate entries for 571 00:34:45,300 --> 00:34:50,000 strains generated by the additional nodal point 572 00:34:50,000 --> 00:34:51,690 displacements. 573 00:34:51,690 --> 00:34:56,285 Let us now look how we can generalize these concepts to 574 00:34:56,285 --> 00:35:01,710 the element that is not, anymore, in the physical x and 575 00:35:01,710 --> 00:35:05,090 y space, two by two element. 576 00:35:05,090 --> 00:35:08,210 In the physical x and y space, this element-- 577 00:35:08,210 --> 00:35:10,330 four-noded element now-- 578 00:35:10,330 --> 00:35:12,540 might look as shown here. 579 00:35:12,540 --> 00:35:15,870 However, what we do is we still deal with the r- and 580 00:35:15,870 --> 00:35:18,930 s-coordinate system embedded on the element. 581 00:35:18,930 --> 00:35:24,340 We are on s one here, minus one s plus one here, r and s 582 00:35:24,340 --> 00:35:28,920 both minus one here, plus one here s minus one. 583 00:35:28,920 --> 00:35:33,030 So in the natural coordinate space, the rs space, we still 584 00:35:33,030 --> 00:35:34,715 have a two by two element. 585 00:35:38,409 --> 00:35:42,820 The interpolation of the displacement therefore-- 586 00:35:42,820 --> 00:35:45,610 the interpolation of the displacements-- 587 00:35:45,610 --> 00:35:49,360 even for the distorted element, is exactly still as 588 00:35:49,360 --> 00:35:50,820 shown here. 589 00:35:50,820 --> 00:35:53,260 Provided we are entering always with the 590 00:35:53,260 --> 00:35:55,176 appropriate r and s. 591 00:35:55,176 --> 00:35:57,460 H one is a function of r and s. 592 00:35:57,460 --> 00:36:02,050 So if we look at a point in the general element, if we 593 00:36:02,050 --> 00:36:05,420 look at a point in the general element, here, for example, at 594 00:36:05,420 --> 00:36:06,380 such a point. 595 00:36:06,380 --> 00:36:09,960 Because that point has a specific r and s value. 596 00:36:09,960 --> 00:36:12,760 And if we want to find the displacement at that point, 597 00:36:12,760 --> 00:36:17,960 well, we would have to put the r and s value of that point 598 00:36:17,960 --> 00:36:21,350 into h one, h two, h three, h four, and that gives us then 599 00:36:21,350 --> 00:36:25,290 the displacement at that point, in terms of the nodal 600 00:36:25,290 --> 00:36:26,440 point displacement. 601 00:36:26,440 --> 00:36:31,010 So, the displacement interpolation for this element 602 00:36:31,010 --> 00:36:33,820 can still be done in the r and s space. 603 00:36:33,820 --> 00:36:36,590 However, difficulties arise-- 604 00:36:36,590 --> 00:36:39,480 or additional considerations I should say rather-- 605 00:36:39,480 --> 00:36:43,120 arise when we talk about strains, because the physical 606 00:36:43,120 --> 00:36:47,150 strains that we have to deal with are derivatives with 607 00:36:47,150 --> 00:36:51,850 respect to x and y, and not r and s anymore. 608 00:36:51,850 --> 00:36:55,000 Well, so what we have to do is, use a Jacobian 609 00:36:55,000 --> 00:36:56,960 Transformation. 610 00:36:56,960 --> 00:36:58,850 What we want are the derivatives with 611 00:36:58,850 --> 00:37:01,710 respect to x and y. 612 00:37:01,710 --> 00:37:05,400 What we can find easily are derivatives with respect to r 613 00:37:05,400 --> 00:37:08,570 and s on the displacements, because the displacements are 614 00:37:08,570 --> 00:37:12,090 given in terms of r and s values. 615 00:37:12,090 --> 00:37:22,480 So, remember u is equal to some of h i u i and the h i is 616 00:37:22,480 --> 00:37:27,330 a function of r and s, so, we can directly find derivatives 617 00:37:27,330 --> 00:37:29,950 with respect to r and s of u. 618 00:37:29,950 --> 00:37:35,240 What we cannot find easily are derivatives with respect to x. 619 00:37:35,240 --> 00:37:37,960 This is the relationship that we use. 620 00:37:37,960 --> 00:37:43,090 It gives us a transformation from derivatives of x and y to 621 00:37:43,090 --> 00:37:45,560 derivatives r and s. 622 00:37:45,560 --> 00:37:49,240 A question must immediately be in your mind, why do we not 623 00:37:49,240 --> 00:37:52,720 write down directly this relationship, which is given 624 00:37:52,720 --> 00:37:53,860 by the [INAUDIBLE]. 625 00:37:53,860 --> 00:37:58,440 If we want d dx of displacements, why not just 626 00:37:58,440 --> 00:38:03,130 use d dx being equal to d dr, dr dx, and so on. 627 00:38:03,130 --> 00:38:05,260 Well, the difficultly is that we cannot 628 00:38:05,260 --> 00:38:08,510 find dr dx very easily. 629 00:38:08,510 --> 00:38:13,230 We have x being this function of h i x i. 630 00:38:13,230 --> 00:38:17,960 This is the interpolation which we have to use now. 631 00:38:17,960 --> 00:38:20,130 I mentioned earlier on the first slide that we 632 00:38:20,130 --> 00:38:24,410 interpolate displacement and coordinates in the same way. 633 00:38:24,410 --> 00:38:27,210 So, here we have a linear interpolation of the 634 00:38:27,210 --> 00:38:30,960 coordinates, from this node to that node, and in 635 00:38:30,960 --> 00:38:32,540 between here too. 636 00:38:32,540 --> 00:38:35,210 So we are using this interpolation here on the 637 00:38:35,210 --> 00:38:36,580 coordinates. 638 00:38:36,580 --> 00:38:47,540 And we can easily dx dr, but we cannot easily find dr dx. 639 00:38:47,540 --> 00:38:50,880 You would have to invert this relationship somehow so that 640 00:38:50,880 --> 00:38:53,360 we have r in terms of x. 641 00:38:53,360 --> 00:38:56,820 Well, it is easier, therefore, to write this relationship 642 00:38:56,820 --> 00:38:59,510 down, which is really the chain rule. 643 00:38:59,510 --> 00:39:01,910 This is also the chain rule, its a chain rule the other way 644 00:39:01,910 --> 00:39:10,180 around, which gives d dr being equal to d dx, dx dr plus d dy 645 00:39:10,180 --> 00:39:13,030 dy dr, if we multiply this out. 646 00:39:13,030 --> 00:39:16,950 And its this relationship that we can use effectively. 647 00:39:16,950 --> 00:39:21,430 Well, this relationship in three dimensional analysis 648 00:39:21,430 --> 00:39:24,780 would involve a third row and third column. 649 00:39:24,780 --> 00:39:29,360 In one dimensional analysis we only talk about one by one. 650 00:39:29,360 --> 00:39:31,620 In other words, in one dimension analysis for a truss 651 00:39:31,620 --> 00:39:33,770 we just have that entry. 652 00:39:33,770 --> 00:39:36,700 In general we can write it in this way, where j is the 653 00:39:36,700 --> 00:39:42,620 Jacobian transformation from the xyz-coordinate system to 654 00:39:42,620 --> 00:39:46,030 the rst-coordinate system. 655 00:39:46,030 --> 00:39:48,890 And since we want these derivative-- 656 00:39:48,890 --> 00:39:53,180 because these derivatives give us actual strains, we have to 657 00:39:53,180 --> 00:39:56,660 invert this relationship. 658 00:39:56,660 --> 00:40:02,320 Having constructed then, these derivatives in terms of these 659 00:40:02,320 --> 00:40:06,050 derivatives, which we can find just as we have done before, 660 00:40:06,050 --> 00:40:10,070 we can now establish the b matrix in much 661 00:40:10,070 --> 00:40:11,320 the same way as earlier. 662 00:40:13,960 --> 00:40:19,020 And since we now have h and b matrices for an element-- 663 00:40:19,020 --> 00:40:22,890 these are a function r, s, and t-- 664 00:40:22,890 --> 00:40:25,970 we would perform the integration. 665 00:40:25,970 --> 00:40:29,220 And now I'm referring to the integration of 666 00:40:29,220 --> 00:40:30,290 the stiffness matrix. 667 00:40:30,290 --> 00:40:37,390 Remember k is equal to b, transposed cb over the volume. 668 00:40:37,390 --> 00:40:41,880 Now notice that since the b matrix that we are using in 669 00:40:41,880 --> 00:40:48,260 here is a function of r and s in a two dimensional analysis. 670 00:40:48,260 --> 00:40:52,140 And r runs from minus one to plus one, and s runs from 671 00:40:52,140 --> 00:40:54,940 minus one to plus one. 672 00:40:54,940 --> 00:41:02,090 We now have to use a transformation also on dv to 673 00:41:02,090 --> 00:41:06,920 integrate over the r s volume. 674 00:41:06,920 --> 00:41:08,300 And that integration-- 675 00:41:08,300 --> 00:41:11,900 and that dv element is expressed as shown here-- 676 00:41:11,900 --> 00:41:18,830 and that is given to us from mathematical analysis. 677 00:41:18,830 --> 00:41:21,680 So basically what we are saying here is that we are 678 00:41:21,680 --> 00:41:25,610 replacing this integral over the physical volume by an 679 00:41:25,610 --> 00:41:31,160 integral minus one to plus one, minus one to plus one. 680 00:41:31,160 --> 00:41:36,010 And that signifies from minus one to plus one over r and 681 00:41:36,010 --> 00:41:38,270 over s, if we had a three dimensional analysis we would 682 00:41:38,270 --> 00:41:41,780 have another integral sign here. 683 00:41:41,780 --> 00:41:47,570 B transposed now r and s function of r and s, cb, 684 00:41:47,570 --> 00:41:50,250 function of r and s. 685 00:41:50,250 --> 00:41:55,860 And then our dv, now in terms of r and s. 686 00:41:55,860 --> 00:41:59,230 So this is how we really do things. 687 00:41:59,230 --> 00:42:03,540 And remember this dv here, this dv, is that one there. 688 00:42:06,750 --> 00:42:17,020 This integration is effectively performed using 689 00:42:17,020 --> 00:42:22,540 numerical integration and I will discuss later on. 690 00:42:22,540 --> 00:42:25,760 Let's look once at the Jacobian transformation for 691 00:42:25,760 --> 00:42:29,250 some very simple examples-- 692 00:42:29,250 --> 00:42:30,940 the Jacobian transformation for 693 00:42:30,940 --> 00:42:33,590 some very simple examples-- 694 00:42:33,590 --> 00:42:36,780 just to make things a little clearer. 695 00:42:36,780 --> 00:42:41,540 In this case we really have taken our two by two element 696 00:42:41,540 --> 00:42:45,690 and we have stretched it into the x- and y-axes. 697 00:42:45,690 --> 00:42:50,040 That stretching is giving us a three here and a two there 698 00:42:50,040 --> 00:42:54,840 because our two by two element has a length of two here, and 699 00:42:54,840 --> 00:42:58,490 six divided by two gives us three. 700 00:42:58,490 --> 00:43:01,110 Similarly here we have stretched the element by a 701 00:43:01,110 --> 00:43:03,220 factor of two. 702 00:43:03,220 --> 00:43:07,990 This relationship here is, in general, calculated-- 703 00:43:07,990 --> 00:43:11,040 the j is, in general, calculated-- 704 00:43:11,040 --> 00:43:12,260 as shown here. 705 00:43:12,260 --> 00:43:15,410 And the result of that, by putting these interpolations 706 00:43:15,410 --> 00:43:20,130 into there are these values here. 707 00:43:20,130 --> 00:43:23,760 Physically what these mean is a stretching, in this 708 00:43:23,760 --> 00:43:25,990 particular case, into the x- and y-axis. 709 00:43:29,050 --> 00:43:30,300 Somewhat-- 710 00:43:32,360 --> 00:43:34,910 the case where we cannot directly-- 711 00:43:34,910 --> 00:43:39,910 not easily directly write down to j matrix is this one. 712 00:43:39,910 --> 00:43:45,870 Here we would go through the actual evaluation the way I 713 00:43:45,870 --> 00:43:47,250 have indicated it. 714 00:43:47,250 --> 00:43:49,770 In other words, we would go through this actual 715 00:43:49,770 --> 00:43:54,850 evaluation, substituting from here and of course for y also. 716 00:43:54,850 --> 00:43:57,640 And this would be the result Notice that we now have a 717 00:43:57,640 --> 00:44:03,740 stretching here of three, compression from a two length 718 00:44:03,740 --> 00:44:04,920 to a one length-- 719 00:44:04,920 --> 00:44:07,300 therefore we have a one half here-- and there's also an 720 00:44:07,300 --> 00:44:11,940 angle change that gives us the off-diagonal element here. 721 00:44:11,940 --> 00:44:17,170 Another interesting case here, as an example, we have the 722 00:44:17,170 --> 00:44:19,100 same lengths here-- 723 00:44:19,100 --> 00:44:23,540 two same lengths here-- but a distortion in the element 724 00:44:23,540 --> 00:44:27,920 because this node two has come down from 725 00:44:27,920 --> 00:44:30,490 there to its midpoint. 726 00:44:30,490 --> 00:44:35,700 And the resulting Jacobian is given here. 727 00:44:35,700 --> 00:44:40,060 Now notice that that Jacobian is a function r and s, so the 728 00:44:40,060 --> 00:44:43,250 inverse, which is used in the construction of the b matrix 729 00:44:43,250 --> 00:44:47,660 would also be a function of r and s. 730 00:44:47,660 --> 00:44:53,650 A particularly interesting case is the one where we shift 731 00:44:53,650 --> 00:44:56,300 nodes to advantage. 732 00:44:56,300 --> 00:44:59,430 See here we have our original-- 733 00:44:59,430 --> 00:45:02,070 our three-noded element that we talked about already-- in 734 00:45:02,070 --> 00:45:03,800 the r space now. 735 00:45:03,800 --> 00:45:06,230 Its a truss element. 736 00:45:06,230 --> 00:45:11,020 And let's say that in our actual physical space, we have 737 00:45:11,020 --> 00:45:14,490 this node there, the three node there, and 738 00:45:14,490 --> 00:45:16,530 the two node there. 739 00:45:16,530 --> 00:45:19,580 The element in the actual physical space 740 00:45:19,580 --> 00:45:22,100 has a length of l. 741 00:45:22,100 --> 00:45:25,650 We have taken this node and shifted it to the quarter 742 00:45:25,650 --> 00:45:29,370 point of the element, to the quarter point of the element. 743 00:45:29,370 --> 00:45:33,820 What I will show you right now is that by having done so-- by 744 00:45:33,820 --> 00:45:37,620 having taken this node from its midpoint and shift it over 745 00:45:37,620 --> 00:45:40,520 in the actual physical space to the quarter point-- 746 00:45:40,520 --> 00:45:46,850 we will find that the strain has a singularity here of one 747 00:45:46,850 --> 00:45:49,540 over square root x. 748 00:45:49,540 --> 00:45:56,430 This is a very important point which can be used in the 749 00:45:56,430 --> 00:46:00,360 analysis of fracture problems, because we know that in the 750 00:46:00,360 --> 00:46:05,040 analysis of fracture problems, we have a one over square root 751 00:46:05,040 --> 00:46:11,460 x singularity at a crack tip. 752 00:46:11,460 --> 00:46:15,400 And, if we want to predict the actual stress there or the 753 00:46:15,400 --> 00:46:20,010 displacement around the crack tip, it can be of advantage to 754 00:46:20,010 --> 00:46:25,460 use this fact, shifting nodes to quarter points in order to 755 00:46:25,460 --> 00:46:28,920 capture the stress singularity more accurately. 756 00:46:28,920 --> 00:46:32,450 Well let me show you then how this strain or stress 757 00:46:32,450 --> 00:46:35,950 singularity comes about. 758 00:46:35,950 --> 00:46:41,310 If we look at this element here and we use our 759 00:46:41,310 --> 00:46:45,450 interpolation on the coordinates, this 760 00:46:45,450 --> 00:46:47,530 would be the result. 761 00:46:47,530 --> 00:46:51,910 Now, notice that we have substituted 762 00:46:51,910 --> 00:46:54,050 the x i values here. 763 00:46:54,050 --> 00:47:00,205 X one is zero, x two is l, x three is l over four. 764 00:47:00,205 --> 00:47:03,610 We have substituted those values and directly come up 765 00:47:03,610 --> 00:47:05,770 with this result. 766 00:47:05,770 --> 00:47:09,300 Well, we can see that this indeed is true. 767 00:47:09,300 --> 00:47:13,150 Let's put r equal to plus one n. 768 00:47:13,150 --> 00:47:16,780 In other words, the right hand side node-- 769 00:47:16,780 --> 00:47:19,260 for the right hand side node-- and we would have a two here 770 00:47:19,260 --> 00:47:22,630 squared, gives us four, goes out with that four. 771 00:47:22,630 --> 00:47:26,450 So at i equals plus 1 we have x equal to l, which 772 00:47:26,450 --> 00:47:27,810 is correct of course. 773 00:47:27,810 --> 00:47:32,650 Let's put i equal to minus one n, we find x is equal to zero. 774 00:47:32,650 --> 00:47:36,510 Let's put i equal to zero n we find x is 775 00:47:36,510 --> 00:47:38,600 equal to l over four. 776 00:47:38,600 --> 00:47:43,030 In other words, this has been a simple check in that we have 777 00:47:43,030 --> 00:47:44,350 the right interpolation-- 778 00:47:44,350 --> 00:47:45,790 geometry interpolation-- 779 00:47:45,790 --> 00:47:48,360 for this element. 780 00:47:48,360 --> 00:47:53,940 Our j now is simply, dx dr, X is given here. 781 00:47:53,940 --> 00:47:55,960 If you take the differentiation of that you 782 00:47:55,960 --> 00:47:59,000 get the two in front that gives l over two times one 783 00:47:59,000 --> 00:48:04,390 plus r, this value here, in other words. 784 00:48:04,390 --> 00:48:07,980 Then our b matrix is constructed by the 785 00:48:07,980 --> 00:48:11,530 inverse of the j-- 786 00:48:11,530 --> 00:48:13,160 that is this one-- 787 00:48:13,160 --> 00:48:19,160 two times the r derivative of the interpolation functions. 788 00:48:19,160 --> 00:48:22,100 Of course here we talk only about one strain. 789 00:48:22,100 --> 00:48:26,420 Remember, in the truss, the only displacement of concern 790 00:48:26,420 --> 00:48:32,020 is the u displacement and the strain is simply epsilon xx, a 791 00:48:32,020 --> 00:48:34,120 strain into this direction also. 792 00:48:36,790 --> 00:48:40,000 Well, this is our b matrix then. 793 00:48:40,000 --> 00:48:43,560 And if we take our h one, two, and three, and we 794 00:48:43,560 --> 00:48:45,680 differentiate these-- 795 00:48:45,680 --> 00:48:49,500 as indicated here with respect to r-- we directly obtain 796 00:48:49,500 --> 00:48:51,790 these functions here. 797 00:48:51,790 --> 00:49:00,260 If we now recognize that since we have x related to r here, 798 00:49:00,260 --> 00:49:03,490 we can also invert this relationship. 799 00:49:03,490 --> 00:49:06,830 We can write r in terms of x. 800 00:49:06,830 --> 00:49:10,810 And if we have done so we can take that relationship and put 801 00:49:10,810 --> 00:49:12,320 it right in here. 802 00:49:12,320 --> 00:49:15,690 Then we would get b in terms of x. 803 00:49:15,690 --> 00:49:17,770 And that's what I have done here. 804 00:49:17,770 --> 00:49:23,260 The first line shows simply r in terms of x now, and I have 805 00:49:23,260 --> 00:49:28,470 substituted that r value into the b matrix and this is there 806 00:49:28,470 --> 00:49:29,090 is the result. 807 00:49:29,090 --> 00:49:34,940 Notice that we have a strain singularity. 808 00:49:34,940 --> 00:49:38,730 This is the first element here. 809 00:49:38,730 --> 00:49:40,830 The next element-- 810 00:49:40,830 --> 00:49:41,600 this is here-- 811 00:49:41,600 --> 00:49:46,470 the next element in b matrix, and that is the third element 812 00:49:46,470 --> 00:49:47,660 in the b matrix. 813 00:49:47,660 --> 00:49:52,820 Notice that we have in this element, that one, and that 814 00:49:52,820 --> 00:49:56,990 one the one over square root x, which means that we have 815 00:49:56,990 --> 00:49:59,006 one over square root x singularity 816 00:49:59,006 --> 00:50:01,780 at x equal to zero. 817 00:50:01,780 --> 00:50:07,450 Well, this fact is used very effectively in fracture 818 00:50:07,450 --> 00:50:09,370 mechanics analysis. 819 00:50:09,370 --> 00:50:16,035 Assume that we have crack here and that we want to analyze 820 00:50:16,035 --> 00:50:20,150 the stress conditions around that crack. 821 00:50:20,150 --> 00:50:25,680 What we can do is we use now two dimension elements, as its 822 00:50:25,680 --> 00:50:28,230 a plane stress situation. 823 00:50:28,230 --> 00:50:32,540 We would put a two dimensional triangular element there and 824 00:50:32,540 --> 00:50:36,050 we shift the midpoint nodes. 825 00:50:36,050 --> 00:50:38,420 This is now very small here, but I hope 826 00:50:38,420 --> 00:50:39,760 you can still follow. 827 00:50:39,760 --> 00:50:44,090 You shift these midpoint nodes to the quarter point, just the 828 00:50:44,090 --> 00:50:48,810 way we have been doing it here. 829 00:50:48,810 --> 00:50:51,815 You are putting the third node the quarter point and the 830 00:50:51,815 --> 00:50:57,105 result is that at this crack tip we have a one over square 831 00:50:57,105 --> 00:51:05,290 root x singularity, using this element layout. 832 00:51:05,290 --> 00:51:10,020 And we know that in fact, there is a one over square 833 00:51:10,020 --> 00:51:14,570 root x singularity in linear fracture mechanics analysis. 834 00:51:14,570 --> 00:51:18,160 And so this is an effective way of capturing this 835 00:51:18,160 --> 00:51:22,990 singularity, and has been used or is currently being used 836 00:51:22,990 --> 00:51:26,060 very abundantly in practice. 837 00:51:26,060 --> 00:51:30,810 The important point that I wanted to make really is that 838 00:51:30,810 --> 00:51:37,930 we can shift nodes in the element to our advantage. 839 00:51:37,930 --> 00:51:42,470 But, we really do that in specific applications such as 840 00:51:42,470 --> 00:51:43,720 fracture mechanics. 841 00:51:45,840 --> 00:51:50,170 In general we will see later when I talk about modeling of 842 00:51:50,170 --> 00:51:53,730 finite element systems, in general, it is most effective 843 00:51:53,730 --> 00:52:00,880 to leave the mid-side nodes at their physical midpoints. 844 00:52:00,880 --> 00:52:03,510 In other words for an eight-noded element, two 845 00:52:03,510 --> 00:52:07,380 dimensional analysis, we would put this mid-node in the 846 00:52:07,380 --> 00:52:11,670 physical space also, actually at the midpoint. 847 00:52:11,670 --> 00:52:13,150 We would not shift it. 848 00:52:13,150 --> 00:52:17,010 Then the element has good convergence characteristics 849 00:52:17,010 --> 00:52:20,230 into all directions and this is really how the element is 850 00:52:20,230 --> 00:52:23,860 used most effectively for general applications. 851 00:52:23,860 --> 00:52:27,090 However, in specific applications, such as fracture 852 00:52:27,090 --> 00:52:31,760 mechanics analysis, it can be of advantage to shift these 853 00:52:31,760 --> 00:52:35,970 mid-side nodes to pick up certain strain or stress 854 00:52:35,970 --> 00:52:40,590 singularities that we know do exist. 855 00:52:40,590 --> 00:52:45,720 Now on the last transparency that I wanted to show you, I 856 00:52:45,720 --> 00:52:49,850 wanted to indicate something to you that I will be talking 857 00:52:49,850 --> 00:52:54,020 about in later lectures more abundantly, namely the fact 858 00:52:54,020 --> 00:52:56,960 that we're using numerical integration. 859 00:52:56,960 --> 00:53:01,860 B, for the k matrix as an example, is once again now a 860 00:53:01,860 --> 00:53:05,220 function of r and s. 861 00:53:05,220 --> 00:53:09,210 This part here is also a function of r and s. 862 00:53:09,210 --> 00:53:13,870 So we have here function of r and s. 863 00:53:13,870 --> 00:53:17,130 and we have here also a function of r and s. 864 00:53:17,130 --> 00:53:21,930 So this f matrix here is a function of r and s. 865 00:53:21,930 --> 00:53:27,400 Notice that the b also includes the inversion of the 866 00:53:27,400 --> 00:53:30,550 j, the Jacobian matrix. 867 00:53:30,550 --> 00:53:34,060 It includes the inversion of the j, because we had to 868 00:53:34,060 --> 00:53:39,785 construct the x and y and z derivatives from the r, s, and 869 00:53:39,785 --> 00:53:41,180 t derivatives. 870 00:53:41,180 --> 00:53:47,940 So, what we do in practical analysis is that we use 871 00:53:47,940 --> 00:53:51,540 numerical integration to evaluate the k matrix. 872 00:53:51,540 --> 00:53:54,690 I have indicated that here schematically, if you look at 873 00:53:54,690 --> 00:53:56,670 this element here. 874 00:53:56,670 --> 00:54:02,900 What we do is, we evaluate the f matrix-- this is a matrix. 875 00:54:02,900 --> 00:54:07,340 In two dimensional analysis we would only run over i and j. 876 00:54:07,340 --> 00:54:08,630 I is this direction. 877 00:54:08,630 --> 00:54:09,610 J is that direction. 878 00:54:09,610 --> 00:54:12,640 In three dimensional analysis, which is in general analysis, 879 00:54:12,640 --> 00:54:16,630 we run i, j, and k this way. 880 00:54:16,630 --> 00:54:21,130 We evaluate the f matrix here at specific 881 00:54:21,130 --> 00:54:24,410 points, r, s, and t. 882 00:54:24,410 --> 00:54:26,850 T now being this axis. 883 00:54:26,850 --> 00:54:31,560 And then multiply that f matrix by certain weight 884 00:54:31,560 --> 00:54:36,040 constants and sum these contributions over all i, j, 885 00:54:36,040 --> 00:54:42,010 and k, in order to obtain an accurate enough approximation 886 00:54:42,010 --> 00:54:43,815 to the actual stiffness matrix. 887 00:54:46,370 --> 00:54:53,360 The order of approximation with which we obtain the 888 00:54:53,360 --> 00:54:55,330 actual stiffness matrix-- 889 00:54:55,330 --> 00:54:59,100 or rather how closely the numerically calculated 890 00:54:59,100 --> 00:55:02,610 stiffness matrix approximates the actual stiffness matrix-- 891 00:55:02,610 --> 00:55:06,440 depends on, number one, how many integration points we are 892 00:55:06,440 --> 00:55:09,260 using and what kind of integration 893 00:55:09,260 --> 00:55:11,880 scheme we are using. 894 00:55:11,880 --> 00:55:15,460 These points here correspond to the Gauss numerical 895 00:55:15,460 --> 00:55:16,370 integration. 896 00:55:16,370 --> 00:55:19,310 In this case for two dimensional analysis, we would 897 00:55:19,310 --> 00:55:21,820 use a two by two integration. 898 00:55:21,820 --> 00:55:26,550 In other words, i and j would both run from one to two. 899 00:55:26,550 --> 00:55:29,840 K is not applicable and we would have altogether four 900 00:55:29,840 --> 00:55:34,190 evaluations off the Fs here. 901 00:55:34,190 --> 00:55:37,870 Multiply each of them by weighting factors, which has 902 00:55:37,870 --> 00:55:42,870 been derived for us by Gauss some long time ago. 903 00:55:42,870 --> 00:55:47,860 And summing up these contributions gives us a close 904 00:55:47,860 --> 00:55:50,310 enough approximation to the k matrix. 905 00:55:50,310 --> 00:55:53,650 Of course, the question of how many points we have to use, 906 00:55:53,650 --> 00:55:56,960 what integration scheme we should use, is a very 907 00:55:56,960 --> 00:55:58,360 important one. 908 00:55:58,360 --> 00:56:01,790 We must use enough integration points to get a close enough 909 00:56:01,790 --> 00:56:05,240 approximation to the actual stiffness matrix and I will be 910 00:56:05,240 --> 00:56:09,300 addressing those questions in a later lecture. 911 00:56:09,300 --> 00:56:11,520 This is all I wanted to say in this lecture. 912 00:56:11,520 --> 00:56:12,920 Thank you very much for your attention.