1 00:00:00,040 --> 00:00:02,470 The following content is provided under a Creative 2 00:00:02,470 --> 00:00:03,880 Commons license. 3 00:00:03,880 --> 00:00:06,920 Your support will help MIT OpenCourseWare continue to 4 00:00:06,920 --> 00:00:10,570 offer high quality educational resources for free. 5 00:00:10,570 --> 00:00:13,470 To make a donation, or view additional materials from 6 00:00:13,470 --> 00:00:17,400 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,400 --> 00:00:18,650 ocw.mit.edu. 8 00:00:20,896 --> 00:00:23,250 PROFESSOR: Ladies and gentlemen, welcome to 9 00:00:23,250 --> 00:00:25,410 lecture number 8. 10 00:00:25,410 --> 00:00:27,780 In this lecture, I would like to discuss with you the 11 00:00:27,780 --> 00:00:32,950 numerical integration and some modeling considerations. 12 00:00:32,950 --> 00:00:36,010 You might recall that in the isoparametric formulation that 13 00:00:36,010 --> 00:00:39,220 we discussed in the early lectures, we used numerical 14 00:00:39,220 --> 00:00:40,910 integration. 15 00:00:40,910 --> 00:00:44,920 And this is an important fact, and I want to discuss with you 16 00:00:44,920 --> 00:00:47,820 in this lecture the Newton-Cotes formulas that 17 00:00:47,820 --> 00:00:50,420 we're using, and in particular, the Gauss 18 00:00:50,420 --> 00:00:51,440 integration-- 19 00:00:51,440 --> 00:00:54,010 the Gauss numerical integration that we are using 20 00:00:54,010 --> 00:00:57,550 in finite element analysis. 21 00:00:57,550 --> 00:00:59,880 There are various considerations. 22 00:00:59,880 --> 00:01:02,030 First of all, I'd like to present to you a little bit of 23 00:01:02,030 --> 00:01:05,230 the theory that is being used and in particular, however, I 24 00:01:05,230 --> 00:01:08,770 also like to present to you some practical considerations. 25 00:01:08,770 --> 00:01:14,380 And finally, together with these aspects, I would like to 26 00:01:14,380 --> 00:01:18,380 also present to you some considerations regarding the 27 00:01:18,380 --> 00:01:21,190 choice of elements, and, of course, regarding the choice 28 00:01:21,190 --> 00:01:23,810 of the order of numerical integration that has to be 29 00:01:23,810 --> 00:01:27,120 used in the finite element formulation. 30 00:01:27,120 --> 00:01:31,740 Well, just to refresh your memory, on this view graph, I 31 00:01:31,740 --> 00:01:34,570 put together once the various matrices 32 00:01:34,570 --> 00:01:36,870 that we have to evaluate. 33 00:01:36,870 --> 00:01:39,280 Here, we have to stiffness matrix, which of course, is 34 00:01:39,280 --> 00:01:45,130 obtained by the integral of B transpose C, B. B is the 35 00:01:45,130 --> 00:01:48,260 strain displacement transformation matrix. 36 00:01:48,260 --> 00:01:49,870 C is the stress strain law. 37 00:01:49,870 --> 00:01:53,290 Of course, this B-matrix is a function of x and y in the 38 00:01:53,290 --> 00:01:57,680 physical space, but if you use isoparametric interpolation in 39 00:01:57,680 --> 00:02:00,320 two-dimensional analysis, this B-matrix would be a function 40 00:02:00,320 --> 00:02:03,500 of r and s, as we discussed earlier. 41 00:02:03,500 --> 00:02:07,970 The m matrix is given here, the RB load vector is given 42 00:02:07,970 --> 00:02:12,780 here, the RS load vector is given here, and the initial 43 00:02:12,780 --> 00:02:14,990 stress load vector is given here. 44 00:02:14,990 --> 00:02:20,040 We derived all of these quantities in an earlier 45 00:02:20,040 --> 00:02:24,060 lecture and all I like to now discuss with you is how do we 46 00:02:24,060 --> 00:02:28,700 evaluate these quantities using numerical integration. 47 00:02:28,700 --> 00:02:32,510 In isoparametric finite element analysis, I mentioned 48 00:02:32,510 --> 00:02:36,260 already the displacement transformation matrix is a 49 00:02:36,260 --> 00:02:38,210 function of r, s, and t. 50 00:02:38,210 --> 00:02:40,880 The strain-displacement interpolation matrix is also a 51 00:02:40,880 --> 00:02:43,860 functional of r, s, and t in three-dimensional analysis. 52 00:02:43,860 --> 00:02:46,440 Of course, in one- or two-dimensional analysis, we 53 00:02:46,440 --> 00:02:50,110 only talk about r and s, respectively. 54 00:02:50,110 --> 00:02:54,230 Remember please, that r, s, and t vary from minus 1 to 55 00:02:54,230 --> 00:02:57,980 plus 1, and that we, therefore, need the volume 56 00:02:57,980 --> 00:03:02,720 transformation here, dV being the determinant of J, where J 57 00:03:02,720 --> 00:03:05,730 is a Jacobian transformation between the x- and y-, and r- 58 00:03:05,730 --> 00:03:09,490 and s-axis, or x-, y-, and z- and r-, s-, and t-axis. 59 00:03:09,490 --> 00:03:13,190 And of course, dr, dr, dt are the respective differentiates 60 00:03:13,190 --> 00:03:14,940 that have to be used here, too. 61 00:03:14,940 --> 00:03:19,170 We discussed that earlier, and just to refresh your memory 62 00:03:19,170 --> 00:03:25,490 again, what we came up with is that the stiffness matrix K 63 00:03:25,490 --> 00:03:28,470 two-dimensional analysis is given in terms of of B 64 00:03:28,470 --> 00:03:32,830 transpose C, B, times determinant of J, dr, ds, and 65 00:03:32,830 --> 00:03:35,930 we are integrating from minus 1 to plus 1. 66 00:03:35,930 --> 00:03:40,330 The mass matrix is given, as shown here, where H is, of 67 00:03:40,330 --> 00:03:42,630 course, once again, the displacement 68 00:03:42,630 --> 00:03:44,370 transformation matrix. 69 00:03:44,370 --> 00:03:47,080 [? Row ?] is the mass density. 70 00:03:47,080 --> 00:03:52,920 Just a preliminary thought, please recognize that B 71 00:03:52,920 --> 00:03:57,050 involves derivatives of displacements, We have here 72 00:03:57,050 --> 00:04:00,510 derivatives of displacements, here we have the actual 73 00:04:00,510 --> 00:04:01,540 displacements. 74 00:04:01,540 --> 00:04:06,030 So the product of B transpose B will, in general, be of 75 00:04:06,030 --> 00:04:11,530 lower order than the product of H transpose H. Therefore, 76 00:04:11,530 --> 00:04:14,690 when we talk about numerical integration, we will find that 77 00:04:14,690 --> 00:04:18,209 to evaluate the M matrix exactly, we will have to use a 78 00:04:18,209 --> 00:04:23,090 higher order integration than in the exact evaluation of the 79 00:04:23,090 --> 00:04:24,540 stiffness matrix. 80 00:04:24,540 --> 00:04:28,180 I will refer to that again later. 81 00:04:28,180 --> 00:04:32,780 We also mentioned already that in the numerical integration, 82 00:04:32,780 --> 00:04:38,700 the basic process is that we are evaluating the function 83 00:04:38,700 --> 00:04:47,120 Fij, as shown here, at particular points, i, j, 84 00:04:47,120 --> 00:04:48,720 denoted by i, j. 85 00:04:48,720 --> 00:04:52,460 And the alpha i, j, are weight coefficients. 86 00:04:52,460 --> 00:04:57,310 Therefore, what we do is, once again, to evaluate this 87 00:04:57,310 --> 00:05:02,630 product at a particular point within the element, multiply 88 00:05:02,630 --> 00:05:06,870 that product by a weight coefficient, and then do the 89 00:05:06,870 --> 00:05:12,180 same for a certain selected number of points, and sum all 90 00:05:12,180 --> 00:05:17,080 these evaluations up, as shown in this equation. 91 00:05:17,080 --> 00:05:19,960 Now, what this means, of course, is that we really have 92 00:05:19,960 --> 00:05:21,890 to use specific points-- 93 00:05:21,890 --> 00:05:26,060 and I have to discuss with you what points we are using-- 94 00:05:26,060 --> 00:05:29,100 and how many points is another question that we 95 00:05:29,100 --> 00:05:30,490 have to also discuss. 96 00:05:30,490 --> 00:05:36,440 Well, in 2x2 integration, in isoparametric finite element 97 00:05:36,440 --> 00:05:42,150 analysis, we would use four points altogether. 98 00:05:42,150 --> 00:05:45,390 2x2 means two points this direction, and 99 00:05:45,390 --> 00:05:47,550 two points that direction. 100 00:05:47,550 --> 00:05:49,610 In three- dimensional analysis, we would, of course, 101 00:05:49,610 --> 00:05:51,970 talk about 2x2x2. 102 00:05:51,970 --> 00:05:55,030 You would have another two points in this direction, in 103 00:05:55,030 --> 00:05:56,660 the t direction. 104 00:05:56,660 --> 00:06:01,540 These points in Gauss integration, are given by 105 00:06:01,540 --> 00:06:04,390 these values here. 106 00:06:04,390 --> 00:06:07,320 In other words, point 4, for example, would have the 107 00:06:07,320 --> 00:06:16,150 values, r equals plus 0.577 and s equal to plus 0.577. 108 00:06:16,150 --> 00:06:20,800 Point 1 would have the same values, but both negative. 109 00:06:20,800 --> 00:06:24,370 These values here are the Gauss values, and I will 110 00:06:24,370 --> 00:06:27,340 discuss briefly with you a little later how these have 111 00:06:27,340 --> 00:06:28,240 been obtained. 112 00:06:28,240 --> 00:06:32,910 Of course, Gauss derived them for us some 100 years ago, and 113 00:06:32,910 --> 00:06:34,930 we are now simply using them. 114 00:06:34,930 --> 00:06:39,140 In 3x3 integration, we talk about nine integration points 115 00:06:39,140 --> 00:06:40,580 altogether. 116 00:06:40,580 --> 00:06:44,830 Once again, three integration points lay, as so to say, in 117 00:06:44,830 --> 00:06:47,680 this direction, and into that direction. 118 00:06:47,680 --> 00:06:49,790 If we use three-point integration in 119 00:06:49,790 --> 00:06:53,510 three-dimensional analysis, we would have altogether 27 120 00:06:53,510 --> 00:06:56,880 integration points because each of these layers, of 121 00:06:56,880 --> 00:07:01,290 course, applies then three times, and because we also 122 00:07:01,290 --> 00:07:04,220 have three layers into the t direction, and therefore 123 00:07:04,220 --> 00:07:09,270 altogether, 9x3, being 27 integration points. 124 00:07:09,270 --> 00:07:15,280 The basic process of numerical integration is that if we want 125 00:07:15,280 --> 00:07:17,180 to integrate a function. 126 00:07:17,180 --> 00:07:21,490 And I've shown here an actual function-- this heavy line is 127 00:07:21,490 --> 00:07:25,130 the actual function f-- if we want to integrate that actual 128 00:07:25,130 --> 00:07:30,640 function, with respect to x here. 129 00:07:30,640 --> 00:07:35,420 What we do is we are putting an interpolating polynomial 130 00:07:35,420 --> 00:07:38,060 down as an approximation to that function. 131 00:07:38,060 --> 00:07:43,140 Now, if we want to integrate this function from a to b, the 132 00:07:43,140 --> 00:07:46,480 simplest interpolating polynomial would be to put a 133 00:07:46,480 --> 00:07:50,920 straight line from this point to that point. 134 00:07:50,920 --> 00:07:56,740 And then the area under the straight line is assumed to be 135 00:07:56,740 --> 00:08:03,450 approximately, at least, really sufficiently close, to 136 00:08:03,450 --> 00:08:07,790 the area under the actual function f. 137 00:08:07,790 --> 00:08:12,410 So what we are doing then is we are replacing the integral 138 00:08:12,410 --> 00:08:19,140 of the actual function f, over this integral, by the integral 139 00:08:19,140 --> 00:08:26,250 of the interpolating polynomial over that integral. 140 00:08:26,250 --> 00:08:30,620 If we do that with just two points, and the two points are 141 00:08:30,620 --> 00:08:37,130 these two, we will derive the trapezoidal rule. 142 00:08:37,130 --> 00:08:42,169 If we use those three points, and I have here a view graph 143 00:08:42,169 --> 00:08:45,370 that shows how we would use three points. 144 00:08:45,370 --> 00:08:48,665 One point here, second point there, and now we have also a 145 00:08:48,665 --> 00:08:49,930 third point there. 146 00:08:49,930 --> 00:08:53,290 Then we would obtain the Simpson's rule. 147 00:08:53,290 --> 00:08:56,260 Once again, instead of integrating the actual 148 00:08:56,260 --> 00:09:04,820 function, f, which is shown by this line here, we are, in 149 00:09:04,820 --> 00:09:09,880 fact, integrating the area under this red line. 150 00:09:09,880 --> 00:09:11,760 This is a little the weak. 151 00:09:11,760 --> 00:09:14,371 Well, now there it comes. 152 00:09:14,371 --> 00:09:18,190 We are integrating the area under the red line. 153 00:09:18,190 --> 00:09:23,830 In other words, the assumption is then that this area is 154 00:09:23,830 --> 00:09:31,220 close enough to the area under blue line, and what we need to 155 00:09:31,220 --> 00:09:38,260 do is to evaluate, of course, the actual function, f, at 156 00:09:38,260 --> 00:09:41,260 these three points. 157 00:09:41,260 --> 00:09:45,950 And we will have to somehow develop certain constants, the 158 00:09:45,950 --> 00:09:51,180 alpha i, j's that I talked about earlier that multiply 159 00:09:51,180 --> 00:09:58,120 this function, f, in order to evaluate the red area here 160 00:09:58,120 --> 00:09:59,990 accurately. 161 00:09:59,990 --> 00:10:03,590 Well, if we do that-- 162 00:10:03,590 --> 00:10:09,890 in other words, if we integrate the red area using 163 00:10:09,890 --> 00:10:15,200 these three function points then we obtain the Simpson's 164 00:10:15,200 --> 00:10:20,470 rule, and this is just one rule of the more general 165 00:10:20,470 --> 00:10:23,930 Newton-Cotes integration. 166 00:10:23,930 --> 00:10:27,360 In the Newton-Cotes integration, we use 167 00:10:27,360 --> 00:10:29,860 equally-spaced sampling points. 168 00:10:29,860 --> 00:10:35,680 In other words, if we want to integrate from a to b, the 169 00:10:35,680 --> 00:10:39,790 simplest way of proceeding is to use a sampling point here, 170 00:10:39,790 --> 00:10:41,860 and a sampling point there. 171 00:10:41,860 --> 00:10:44,540 And then, of course, as I pointed out earlier, we are 172 00:10:44,540 --> 00:10:46,820 putting, really, a straight line down and we're assuming 173 00:10:46,820 --> 00:10:51,610 that the area under this straight line is equal to the 174 00:10:51,610 --> 00:10:57,650 area to under the actual function, f, that might be 175 00:10:57,650 --> 00:11:00,280 looking like that. 176 00:11:00,280 --> 00:11:06,310 Well, that is the trapezoidal rule and Newton-Cotes 177 00:11:06,310 --> 00:11:13,080 integration with n equal to 1. 178 00:11:13,080 --> 00:11:17,470 In other words, with two constants here, is equal to 179 00:11:17,470 --> 00:11:20,180 this trapezoidal rule. 180 00:11:20,180 --> 00:11:23,970 What we are saying is that the area from a to b-- now I used 181 00:11:23,970 --> 00:11:29,150 here r, this is the r-axis. say, from a to b-- 182 00:11:29,150 --> 00:11:30,590 this is the actual integration-- 183 00:11:30,590 --> 00:11:38,370 is replaced by the summation of the value at this point, 184 00:11:38,370 --> 00:11:43,640 the F0 value, the F1 value, and each of these values is 185 00:11:43,640 --> 00:11:47,030 multiplied by a certain constant. 186 00:11:47,030 --> 00:11:51,840 So in this particular case, for the trapezoidal rules that 187 00:11:51,840 --> 00:11:55,080 I'm talking about, the right hand side looks as follows-- b 188 00:11:55,080 --> 00:12:01,200 minus a, that is this length here, times two values, the 189 00:12:01,200 --> 00:12:08,050 C01, n is equal to 1 in this particular case, times F0. 190 00:12:08,050 --> 00:12:16,200 Plus C1 1 1 times F1. 191 00:12:16,200 --> 00:12:21,800 And of course then, plus an error, R in this particular 192 00:12:21,800 --> 00:12:25,870 case, n, once again, is equal to 1, plus R1. 193 00:12:25,870 --> 00:12:29,670 This error of course, is assumed to be very small in 194 00:12:29,670 --> 00:12:32,850 the analysis, and we would actually neglect it. 195 00:12:32,850 --> 00:12:35,880 So once again, what we are saying is that we can replace 196 00:12:35,880 --> 00:12:42,050 the integration of F of R dr, from a to b, via this 197 00:12:42,050 --> 00:12:43,240 summation here. 198 00:12:43,240 --> 00:12:47,250 Notice that b minus a times that constant, that is our 199 00:12:47,250 --> 00:12:50,710 alpha 1, that I referred to earlier. 200 00:12:50,710 --> 00:12:53,960 And b minus a times that constant would be another 201 00:12:53,960 --> 00:12:55,830 alpha value. 202 00:12:55,830 --> 00:12:59,280 And, of course, these alpha values here multiply the 203 00:12:59,280 --> 00:13:04,770 actual function values at this station, and at that station. 204 00:13:04,770 --> 00:13:10,560 Well in the Simpson rule, n is equal to 2, b minus a is still 205 00:13:10,560 --> 00:13:17,707 there, but now we're talking about C0 2 0 times F0 plus C1 206 00:13:17,707 --> 00:13:26,640 2 times F1 plus C2 2 times F2, and of course, an 207 00:13:26,640 --> 00:13:29,880 error of 2's there. 208 00:13:29,880 --> 00:13:37,000 Well, the constants, the C0 1, C1 1 and so on. 209 00:13:37,000 --> 00:13:43,440 The constant Cin have been evaluated for us, and they are 210 00:13:43,440 --> 00:13:47,190 tabulated in this table here. 211 00:13:47,190 --> 00:13:52,340 Notice, that for the trapezoidal rule, we have the 212 00:13:52,340 --> 00:13:55,740 simple constants, 1/2, 1/2. 213 00:13:55,740 --> 00:14:01,740 The error here is shown here, 10 to the minus 1 times the 214 00:14:01,740 --> 00:14:05,370 second derivative of F. So if the second derivative of the 215 00:14:05,370 --> 00:14:11,220 actual function is 0, well, then our integration is exact. 216 00:14:11,220 --> 00:14:15,670 Of course, what that means is that our function is indeed a 217 00:14:15,670 --> 00:14:16,950 straight line. 218 00:14:16,950 --> 00:14:20,640 Because only then the second derivative is 0. 219 00:14:20,640 --> 00:14:23,590 So our integration is exact when we 220 00:14:23,590 --> 00:14:25,040 integrate a straight line. 221 00:14:25,040 --> 00:14:29,910 Well, if you look at our earlier graph here, surely, if 222 00:14:29,910 --> 00:14:32,630 our actual red line is a straight line between this 223 00:14:32,630 --> 00:14:36,190 point and that point, then our integration, using the 224 00:14:36,190 --> 00:14:38,350 trapezoidal rule, must be exact. 225 00:14:38,350 --> 00:14:42,570 And that is verified by looking at the error bound, 226 00:14:42,570 --> 00:14:44,460 given in this table. 227 00:14:44,460 --> 00:14:50,700 The Simpson's rule has constants, 1/6, 4/6, 1/6. 228 00:14:50,700 --> 00:14:53,340 n is equal to 2 in this particular case, as I 229 00:14:53,340 --> 00:14:55,040 mentioned earlier. 230 00:14:55,040 --> 00:14:59,720 The error here is now given to the force derivative. 231 00:14:59,720 --> 00:15:01,280 And so on. 232 00:15:01,280 --> 00:15:03,510 Here, we have integrals 1 to 6. 233 00:15:03,510 --> 00:15:08,220 In practice, however, we would only use this the trapezoidal 234 00:15:08,220 --> 00:15:11,840 rule, the Simpson rule, and we might then still 235 00:15:11,840 --> 00:15:15,940 use this rule here. 236 00:15:15,940 --> 00:15:20,450 Notice that the error of this rule here is of the same order 237 00:15:20,450 --> 00:15:23,850 as the error of this rule, involving five points. 238 00:15:23,850 --> 00:15:27,920 So the four-point integration rule, of course, is preferable 239 00:15:27,920 --> 00:15:30,880 when you compare to the five-point integration rule. 240 00:15:30,880 --> 00:15:35,230 The other important point is that if we use the 241 00:15:35,230 --> 00:15:41,010 Newton-Cotes formally, we are really always involving the 242 00:15:41,010 --> 00:15:42,750 endpoints of the integration. 243 00:15:42,750 --> 00:15:50,090 In other words, if we want to integrate from a to b then we 244 00:15:50,090 --> 00:15:53,625 will involve these two points in the actual integration. 245 00:15:56,220 --> 00:15:59,020 So if we talk about the integration through the 246 00:15:59,020 --> 00:16:02,760 thickness off the shell, or through a beam element 247 00:16:02,760 --> 00:16:07,250 thickness, then we would have here the top and bottom 248 00:16:07,250 --> 00:16:08,500 surfaces involved. 249 00:16:15,380 --> 00:16:16,200 Bad point-- 250 00:16:16,200 --> 00:16:19,570 well, I shouldn't say bad, but certainly it's not a very 251 00:16:19,570 --> 00:16:24,070 advantageous point, of the Newton-Cotes integration is 252 00:16:24,070 --> 00:16:29,910 that we need to involve a fairly large number of 253 00:16:29,910 --> 00:16:33,130 integration points, or integration point stations to 254 00:16:33,130 --> 00:16:37,240 obtain an acceptable accuracy when we talk about finite 255 00:16:37,240 --> 00:16:38,790 element analysis. 256 00:16:38,790 --> 00:16:43,250 And therefore, the Gauss numerical integration is a 257 00:16:43,250 --> 00:16:45,350 more attractive scheme. 258 00:16:45,350 --> 00:16:48,230 And, in fact, we will find that in finite element 259 00:16:48,230 --> 00:16:53,730 analysis, we almost always use Gauss numerical integration. 260 00:16:53,730 --> 00:16:58,010 They're only certain special cases in analysis of beams, 261 00:16:58,010 --> 00:17:01,250 plates, and shells where we, in fact, use the Newton-Cotes 262 00:17:01,250 --> 00:17:02,030 integration. 263 00:17:02,030 --> 00:17:04,290 And the reason then, of course, is that we want to 264 00:17:04,290 --> 00:17:08,560 obtain also surface stresses and strains. 265 00:17:08,560 --> 00:17:10,310 That would be, for example, one reason. 266 00:17:10,310 --> 00:17:13,710 Now, in Gauss numerical integration, the basic concept 267 00:17:13,710 --> 00:17:17,069 is quite similar to the concept 268 00:17:17,069 --> 00:17:18,130 that we just discussed. 269 00:17:18,130 --> 00:17:22,170 That we again use an interpolating polynomial to 270 00:17:22,170 --> 00:17:27,630 lay through the function over the interval, a to b. 271 00:17:27,630 --> 00:17:32,520 The important point, however, now is that we use-- 272 00:17:32,520 --> 00:17:35,200 if we look at our function here. 273 00:17:35,200 --> 00:17:38,280 Here is the interval, a to b. 274 00:17:38,280 --> 00:17:39,980 Our rx is here. 275 00:17:39,980 --> 00:17:43,470 If you look at that function, it's that we, in Gauss 276 00:17:43,470 --> 00:17:49,220 numerical integration, do not use equal-spaced intervals. 277 00:17:49,220 --> 00:17:52,770 In Newton-Cotes integration, we use equal-spaced intervals, 278 00:17:52,770 --> 00:17:55,520 and involves the end points. 279 00:17:55,520 --> 00:17:59,900 In Gauss numerical integration, we are using 280 00:17:59,900 --> 00:18:07,960 points, stations, 1 to 2, and our stations that can be 281 00:18:07,960 --> 00:18:10,400 anywhere in the interval. 282 00:18:10,400 --> 00:18:16,550 And Gauss has developed the optimum stations for us. 283 00:18:16,550 --> 00:18:22,110 in other words, what he has said, at that time is let us 284 00:18:22,110 --> 00:18:26,630 optimize the particular locations of the integration 285 00:18:26,630 --> 00:18:31,990 points stations, and in addition, let us optimize the 286 00:18:31,990 --> 00:18:35,620 weights, alpha 1 to alpha n. 287 00:18:35,620 --> 00:18:40,000 In the Newton-Cotes integration, we only optimize, 288 00:18:40,000 --> 00:18:42,540 so to say, the alpha values. 289 00:18:42,540 --> 00:18:46,210 The stations were fixed because we involved, for 290 00:18:46,210 --> 00:18:49,340 example, in the trapezoidal rule, we involved this end 291 00:18:49,340 --> 00:18:50,880 point, and that point. 292 00:18:50,880 --> 00:18:54,070 In the Simpson's rule, we used the midpoint. 293 00:18:54,070 --> 00:18:55,070 And so on. 294 00:18:55,070 --> 00:18:57,820 In Gauss integration, we say, well, let us does not 295 00:18:57,820 --> 00:19:00,680 necessarily fix those stations but rather let us just 296 00:19:00,680 --> 00:19:03,370 optimize the location of these stations. 297 00:19:03,370 --> 00:19:08,460 Well, since we then can optimize the values of the 298 00:19:08,460 --> 00:19:14,410 alpha 1, to alpha n, this of course are n, components, and 299 00:19:14,410 --> 00:19:18,530 the stations R1 to Rn, which are also n components. 300 00:19:18,530 --> 00:19:22,530 We really can optimize two n components. 301 00:19:22,530 --> 00:19:26,350 Now, if we look at the fact that we're dealing with two n 302 00:19:26,350 --> 00:19:30,990 components that we optimize now, and the fact that the 303 00:19:30,990 --> 00:19:34,400 interpolating polynomial of course, also involves one 304 00:19:34,400 --> 00:19:39,670 constant, then we find that the interpolating polynomial 305 00:19:39,670 --> 00:19:43,230 can now be of order, 2n minus 1. 306 00:19:43,230 --> 00:19:45,650 2n minus 1. 307 00:19:45,650 --> 00:19:51,340 Whereas in the Simpson's rule, the interpolating polynomial 308 00:19:51,340 --> 00:19:55,940 with n stations could only be of order, n minus 1. 309 00:19:55,940 --> 00:20:00,200 You see, in the Simpsons rule, for example, with three 310 00:20:00,200 --> 00:20:06,220 stations, the n was, in the Simpson's rule equal to 3, but 311 00:20:06,220 --> 00:20:11,280 the order of the interpolating polynomial was 2. 312 00:20:11,280 --> 00:20:17,630 In other words, a parabola could be integrated exactly. 313 00:20:17,630 --> 00:20:21,860 Now in the Gauss integration, we are talking about the 314 00:20:21,860 --> 00:20:26,640 interpolating polynomial that can be of order, 2n minus 1, a 315 00:20:26,640 --> 00:20:31,880 much higher order for n stations. 316 00:20:31,880 --> 00:20:35,660 Here, with the Newton-Cotes rule for n stations, we only 317 00:20:35,660 --> 00:20:39,100 get an interpreting polynomial of n minus 1. 318 00:20:39,100 --> 00:20:48,180 Well, Gauss has developed the specific stations, R1 to Rn, 319 00:20:48,180 --> 00:20:50,270 to be used [UNINTELLIGIBLE] 320 00:20:50,270 --> 00:20:54,630 in order to fit the interpolating polynomial 321 00:20:54,630 --> 00:21:01,030 through the actual function that we want to integrate. 322 00:21:01,030 --> 00:21:05,680 And of course, we also know now the error that is 323 00:21:05,680 --> 00:21:07,860 involved, Rn. 324 00:21:07,860 --> 00:21:11,190 The error of course, being much smaller than in the 325 00:21:11,190 --> 00:21:14,110 Newton-Cotes formula because we are dealing with a 326 00:21:14,110 --> 00:21:17,340 higher-order interpolating polynomial. 327 00:21:17,340 --> 00:21:21,100 The Gauss values are given here. 328 00:21:21,100 --> 00:21:25,655 When n is equal to 1, well, we only use the midpoint and the 329 00:21:25,655 --> 00:21:27,820 alpha value is 2. 330 00:21:27,820 --> 00:21:30,600 When n is equal to 2, in other words, we are talking about 331 00:21:30,600 --> 00:21:35,670 two-point integration, then ri is equal to plus 332 00:21:35,670 --> 00:21:38,750 minus 0.577, and 1. 333 00:21:38,750 --> 00:21:44,330 In this particular case, we are using the fact that we 334 00:21:44,330 --> 00:21:48,660 want to integrate 4 minus 1, to plus 1. 335 00:21:48,660 --> 00:21:52,230 In other words, our r here, is this. 336 00:21:52,230 --> 00:21:54,880 We coordinate here, and we're integrating from 337 00:21:54,880 --> 00:21:57,450 minus 1 to plus 1. 338 00:21:57,450 --> 00:22:03,090 So in two-point integration, we using a point here, which 339 00:22:03,090 --> 00:22:07,290 is minus 0.577, and another point here, 340 00:22:07,290 --> 00:22:11,090 which is plus 0.577. 341 00:22:11,090 --> 00:22:14,740 Notice that we are feeding in these points to a higher-order 342 00:22:14,740 --> 00:22:19,760 accuracy in a computer program, typically, in a 343 00:22:19,760 --> 00:22:21,990 computer program that runs on a CDC machine 344 00:22:21,990 --> 00:22:23,800 with 14-digit precision. 345 00:22:23,800 --> 00:22:25,350 We would actually put these points 346 00:22:25,350 --> 00:22:28,440 in to 14-digit precision. 347 00:22:28,440 --> 00:22:33,660 Of course, they are actually put into this [? outputing. ?] 348 00:22:33,660 --> 00:22:35,880 They are there, and the computer program just picks up 349 00:22:35,880 --> 00:22:38,850 this value and evaluates then, of course, the function that 350 00:22:38,850 --> 00:22:41,740 we want to integrate at that station. 351 00:22:41,740 --> 00:22:45,420 The factor that the function has to be multiplied by is 352 00:22:45,420 --> 00:22:47,690 given here. 353 00:22:47,690 --> 00:22:51,960 In three-point integration, we use three points. 354 00:22:51,960 --> 00:22:53,720 This one is the easiest to look at-- 355 00:22:53,720 --> 00:22:56,590 it's right to center point, and then there are two more 356 00:22:56,590 --> 00:23:03,050 points, one at minus 0.77 and one at plus 0.77. 357 00:23:03,050 --> 00:23:07,945 In finite element analysis, we know, of course, in 358 00:23:07,945 --> 00:23:10,410 isoparametric finite element analysis, I should really say, 359 00:23:10,410 --> 00:23:12,530 we know, of course, that our integration runs from 360 00:23:12,530 --> 00:23:13,910 minus 1 to plus 1. 361 00:23:13,910 --> 00:23:17,510 So these are the actual values to be used. 362 00:23:17,510 --> 00:23:19,390 And these are the actual values that you would find, 363 00:23:19,390 --> 00:23:23,220 for example, in the computer program such as ADINA. 364 00:23:23,220 --> 00:23:32,720 If we want to integrate an interval from a to b-- 365 00:23:32,720 --> 00:23:36,130 I'd like to simply mention that, of course, the same 366 00:23:36,130 --> 00:23:39,530 scheme can directly be used. 367 00:23:39,530 --> 00:23:43,540 If we have the ri and alpha i for the interval from minus 1 368 00:23:43,540 --> 00:23:49,370 to plus 1, then these would be the value to be used for the 369 00:23:49,370 --> 00:23:51,330 integration point stations. 370 00:23:51,330 --> 00:23:54,690 If our interval is not from minus 1 to plus 1. but from a 371 00:23:54,690 --> 00:23:58,680 to b, and this would be the weight that would have to be 372 00:23:58,680 --> 00:24:03,030 used for this particular case. 373 00:24:03,030 --> 00:24:07,440 The i in alpha i, once again, are the same ri and alpha 1 374 00:24:07,440 --> 00:24:11,640 that I listed here in this table. 375 00:24:11,640 --> 00:24:17,290 Well, so far we really talked about one-dimensional 376 00:24:17,290 --> 00:24:19,180 integration. 377 00:24:19,180 --> 00:24:22,130 However, the same concept can directly be used for multiple 378 00:24:22,130 --> 00:24:23,790 dimension integration. 379 00:24:23,790 --> 00:24:29,590 In other words, if we talk about the integration, 380 00:24:29,590 --> 00:24:32,460 minus 1 to plus 1. 381 00:24:32,460 --> 00:24:35,150 For example, as we would integrate, of course, in the 382 00:24:35,150 --> 00:24:37,810 formation of a tress element. 383 00:24:37,810 --> 00:24:40,250 If we now want to go through the integration of a 384 00:24:40,250 --> 00:24:43,750 two-dimensional element, then we would have to integrate a 385 00:24:43,750 --> 00:24:47,630 two-dimensional function in r and s, and that means 386 00:24:47,630 --> 00:24:50,495 integrating twice, once with respect to r, once with 387 00:24:50,495 --> 00:24:54,030 respect to s, and we can use the same concept that we used 388 00:24:54,030 --> 00:24:54,620 in [? analytical ?] 389 00:24:54,620 --> 00:24:58,880 integrations, and we first integrate with respect to r. 390 00:24:58,880 --> 00:25:01,110 That gives us one alpha i here, and then 391 00:25:01,110 --> 00:25:03,040 with respect to s. 392 00:25:03,040 --> 00:25:07,440 And the final result is shown here, that we have to bring in 393 00:25:07,440 --> 00:25:11,020 a weight factor, alpha i, and a weight factor, alpha j. 394 00:25:11,020 --> 00:25:13,312 This one for the i integration, that one for the 395 00:25:13,312 --> 00:25:14,340 a integration. 396 00:25:14,340 --> 00:25:18,750 And of course, stations ri and sj are the stations at which 397 00:25:18,750 --> 00:25:20,840 we would evaluate the actual function. 398 00:25:20,840 --> 00:25:25,440 Notice one important point that we could use a different 399 00:25:25,440 --> 00:25:27,900 order of integration into the two directions. 400 00:25:27,900 --> 00:25:31,320 For example, if we have an element such as this one, 401 00:25:31,320 --> 00:25:33,250 two-dimension element, two nodes 402 00:25:33,250 --> 00:25:36,020 here, in the s direction. 403 00:25:36,020 --> 00:25:38,440 But say three nodes in the r direction. 404 00:25:38,440 --> 00:25:45,665 So our r-axis is this one, s-axis is that one. 405 00:25:45,665 --> 00:25:48,420 Then in this particular case, since we're using a higher 406 00:25:48,420 --> 00:25:52,290 order interpolation into the r direction, in this particular 407 00:25:52,290 --> 00:25:55,340 case, we might say, well, let us use three-point integration 408 00:25:55,340 --> 00:26:00,610 this way, but two-point integration into the s-axis. 409 00:26:00,610 --> 00:26:03,350 So here, two-point integration into the s-axis, but 410 00:26:03,350 --> 00:26:06,150 three-point integration into the r-axis. 411 00:26:06,150 --> 00:26:08,990 And of course, the same concept applies also in 412 00:26:08,990 --> 00:26:10,430 three-dimensional analysis. 413 00:26:10,430 --> 00:26:14,060 In three-dimensional analysis as an example here, we would 414 00:26:14,060 --> 00:26:19,370 have the F now being a function of r, s, and t. 415 00:26:19,370 --> 00:26:22,470 Three times integration from minus 1 to plus 1. 416 00:26:22,470 --> 00:26:26,440 And the final result just derived the same way as in 417 00:26:26,440 --> 00:26:28,270 two-dimensional analysis as given here. 418 00:26:28,270 --> 00:26:31,980 Notice once again, we can use different orders of 419 00:26:31,980 --> 00:26:35,330 integration into the three directions. 420 00:26:35,330 --> 00:26:38,830 I should also mention at this point that when we talk about 421 00:26:38,830 --> 00:26:42,210 an actual stiffness matrix, of course, we talk about an 422 00:26:42,210 --> 00:26:48,920 array, n by n, for three-dimensional element, say 423 00:26:48,920 --> 00:26:50,460 a 20-node brick. 424 00:26:50,460 --> 00:26:55,250 This would be a 60 by 60 stiffness matrix, and every 425 00:26:55,250 --> 00:26:59,700 one of these elements here, in that stiffness matrix would be 426 00:26:59,700 --> 00:27:03,490 an F, such an F. Every one of these elements would be 427 00:27:03,490 --> 00:27:05,310 integrated as shown here. 428 00:27:08,140 --> 00:27:11,150 Let us now look at the practical use of numerical 429 00:27:11,150 --> 00:27:13,140 integration. 430 00:27:13,140 --> 00:27:20,070 Well, the first important point is that the order of 431 00:27:20,070 --> 00:27:24,060 integration that is required to evaluate a specific element 432 00:27:24,060 --> 00:27:28,950 matrix can, of course, be evaluated by studying the 433 00:27:28,950 --> 00:27:31,300 function, f, to be integrated. 434 00:27:31,300 --> 00:27:36,990 In other words, if I want to evaluate a certain stiffness 435 00:27:36,990 --> 00:27:40,880 matrix, K. Here is the stiffness matrix given. 436 00:27:40,880 --> 00:27:45,140 If I want to evaluate that stiffness matrix exactly, I 437 00:27:45,140 --> 00:27:47,990 could look at each all of these elements, and there are 438 00:27:47,990 --> 00:27:51,550 now many of these F functions in there. 439 00:27:51,550 --> 00:27:53,970 Let us look at that one. 440 00:27:53,970 --> 00:27:59,880 I would look at that F value and identify its dependency on 441 00:27:59,880 --> 00:28:02,270 i, s, and t. 442 00:28:02,270 --> 00:28:05,280 Linear, cubic, parabolic and so on. 443 00:28:05,280 --> 00:28:09,500 And then I could, by entering into-- 444 00:28:09,500 --> 00:28:12,980 by the use of the Gauss numeric integration, or the 445 00:28:12,980 --> 00:28:17,780 Newton-Cotes integration, and by looking up the errors 446 00:28:17,780 --> 00:28:22,000 involved, I could identify what integration order I need 447 00:28:22,000 --> 00:28:27,500 in order to evaluate this integral exactly. 448 00:28:27,500 --> 00:28:34,080 That can be done and for simple elements, relatively 449 00:28:34,080 --> 00:28:34,920 simple elements-- 450 00:28:34,920 --> 00:28:39,770 in other words, 4-noded elements, that are square, or 451 00:28:39,770 --> 00:28:42,020 8-noded elements that are square. 452 00:28:42,020 --> 00:28:46,050 We can fairly simply evaluate the order of integration that 453 00:28:46,050 --> 00:28:47,090 is required. 454 00:28:47,090 --> 00:28:49,500 However, when the element is curved. 455 00:28:49,500 --> 00:28:52,040 In other words, when we talk about a general 456 00:28:52,040 --> 00:28:54,030 elements like that-- 457 00:28:54,030 --> 00:28:58,930 the required integration order is not that easily assessed, 458 00:28:58,930 --> 00:29:02,290 in order to evaluate the stiffness matrix exactly. 459 00:29:02,290 --> 00:29:04,840 And the reason for it is that we have the Jacobian 460 00:29:04,840 --> 00:29:07,300 transformation in there. 461 00:29:07,300 --> 00:29:11,400 Therefore, in practice, the integration is frequently not 462 00:29:11,400 --> 00:29:15,530 performed exactly, but we have to remember that the 463 00:29:15,530 --> 00:29:19,060 integration order must be high enough. 464 00:29:19,060 --> 00:29:23,020 Well, considering then the evaluation of the element 465 00:29:23,020 --> 00:29:28,560 matrices, we should notice the following requirements. 466 00:29:28,560 --> 00:29:31,190 For the stiffness matrix evaluation, we note that the 467 00:29:31,190 --> 00:29:35,650 element matrix should not contain any spurious zero 468 00:29:35,650 --> 00:29:37,400 energy modes. 469 00:29:37,400 --> 00:29:38,570 What do I mean by that? 470 00:29:38,570 --> 00:29:44,550 Well, I mean that if I have an element such as 471 00:29:44,550 --> 00:29:46,420 that element here. 472 00:29:49,750 --> 00:29:54,540 We know that if you're talking about a plane stress element, 473 00:29:54,540 --> 00:29:57,370 there only three rigid body modes. 474 00:29:57,370 --> 00:30:01,390 One rigid translation this way, one rigid translation 475 00:30:01,390 --> 00:30:03,810 that way, and one rigid rotation of the 476 00:30:03,810 --> 00:30:04,950 elements that way. 477 00:30:04,950 --> 00:30:09,250 These are the only three rigid body modes that the element 478 00:30:09,250 --> 00:30:12,000 stiffness matrix should contain. 479 00:30:12,000 --> 00:30:17,870 It should, of course, contain these in order to satisfy the 480 00:30:17,870 --> 00:30:20,360 convergence requirements that we discussed earlier. 481 00:30:20,360 --> 00:30:24,230 What I'm saying now is that my integration order should be 482 00:30:24,230 --> 00:30:29,800 high enough so that there is no additional 0 igon value. 483 00:30:32,300 --> 00:30:36,160 If there is an additional 0 igon value, I might run into 484 00:30:36,160 --> 00:30:40,230 serious numerical difficulties in the solution of the overall 485 00:30:40,230 --> 00:30:42,030 governing equilibrium equations. 486 00:30:42,030 --> 00:30:45,500 And I will give you just now a very simple example. 487 00:30:45,500 --> 00:30:50,010 The second requirement is that the element stiffness matrix 488 00:30:50,010 --> 00:30:53,600 should contain the required constant strain states. 489 00:30:53,600 --> 00:31:01,200 So basically, what I'm saying here is that the numerically 490 00:31:01,200 --> 00:31:05,100 integrated stiffness matrix should still contain the 491 00:31:05,100 --> 00:31:09,720 requirements that we discussed, which the element 492 00:31:09,720 --> 00:31:14,170 must satisfy for convergence. 493 00:31:14,170 --> 00:31:18,170 For the mass matrix evaluation, here we only need 494 00:31:18,170 --> 00:31:21,150 to satisfy that the total mass must be included. 495 00:31:21,150 --> 00:31:24,140 In other words, the numerical integration should pick up the 496 00:31:24,140 --> 00:31:26,150 actual mass of the element. 497 00:31:26,150 --> 00:31:30,850 If that were not the case, then considering the total 498 00:31:30,850 --> 00:31:33,750 analysis, we would have lost some mass. 499 00:31:33,750 --> 00:31:36,570 The force vector evaluation valuation here, also is a 500 00:31:36,570 --> 00:31:37,350 total loads. 501 00:31:37,350 --> 00:31:40,550 Must be included, must be picked up, so to say, by the 502 00:31:40,550 --> 00:31:43,270 numerical integration. 503 00:31:43,270 --> 00:31:47,890 Let us look at this requirement here, and then I 504 00:31:47,890 --> 00:31:50,436 have a very simple example. 505 00:31:50,436 --> 00:31:55,820 Here we have the 8-node element, plane stress element, 506 00:31:55,820 --> 00:31:58,360 which is supported here on a hinge, and 507 00:31:58,360 --> 00:31:59,990 here on a stiff string. 508 00:31:59,990 --> 00:32:02,870 I apply a load, p, there. 509 00:32:02,870 --> 00:32:06,680 If I use 2x2 Gauss integration, 510 00:32:06,680 --> 00:32:09,200 I get absurd results. 511 00:32:09,200 --> 00:32:12,190 If I use 3x3 Gauss integration, I 512 00:32:12,190 --> 00:32:13,540 get the correct results. 513 00:32:13,540 --> 00:32:16,510 The correct results, of course, being that the 514 00:32:16,510 --> 00:32:19,780 elements simply rotates about this point, a. 515 00:32:19,780 --> 00:32:23,340 It simply rotates about the point, a, as a rigid body. 516 00:32:26,570 --> 00:32:29,090 And, of course, extending in that rotation's a 517 00:32:29,090 --> 00:32:31,610 spring at point b. 518 00:32:31,610 --> 00:32:34,940 The absurd results here come about-- 519 00:32:34,940 --> 00:32:37,290 using 2x2 integration-- 520 00:32:37,290 --> 00:32:43,710 because the stiffness matrix k does not contain just three 521 00:32:43,710 --> 00:32:50,150 rigid body modes, but four rigid body modes, with 2x2 522 00:32:50,150 --> 00:32:51,560 Gauss integration. 523 00:32:51,560 --> 00:32:56,270 Now, three rigid body modes are suppressed by the hinge 524 00:32:56,270 --> 00:32:59,880 here and the spring there, they are suppressed. 525 00:32:59,880 --> 00:33:03,350 But if I look at the fact that we are now having four in the 526 00:33:03,350 --> 00:33:07,250 stiffness matrix, if I suppress three, I still am 527 00:33:07,250 --> 00:33:11,990 left with one rigid body mode, and that 528 00:33:11,990 --> 00:33:13,540 means one 0 igon value. 529 00:33:13,540 --> 00:33:19,920 So this is still an unstable structural stiffness matrix, 530 00:33:19,920 --> 00:33:22,840 and when I subject that stiffness matrix two a load p, 531 00:33:22,840 --> 00:33:25,020 I get absurd results. 532 00:33:25,020 --> 00:33:27,990 Therefore, this is a very simple example that 533 00:33:27,990 --> 00:33:32,000 demonstrates that the Gauss integration has to be high 534 00:33:32,000 --> 00:33:36,520 enough in order to prevent instabilities occurring in the 535 00:33:36,520 --> 00:33:38,990 solution of the equations. 536 00:33:38,990 --> 00:33:41,970 In practice, of course, what happens frequently is that we 537 00:33:41,970 --> 00:33:46,790 are using these elements in a mesh, so we have an 8-node 538 00:33:46,790 --> 00:33:51,290 element here, and another 8-node element here. 539 00:33:51,290 --> 00:33:54,900 And if we use 2x2 integration for this element, 2x2 540 00:33:54,900 --> 00:33:57,720 integration for that element, we might very well be able to 541 00:33:57,720 --> 00:34:01,380 solve the equations very nicely, and we might, in fact 542 00:34:01,380 --> 00:34:05,830 get good results in a analysis of a more complicated problem. 543 00:34:05,830 --> 00:34:10,219 Because the stiffness of this element and that element, 544 00:34:10,219 --> 00:34:17,280 they, each other, compensate for the loss for the rigid 545 00:34:17,280 --> 00:34:22,320 body mode that is still available in 546 00:34:22,320 --> 00:34:24,530 this one element here. 547 00:34:24,530 --> 00:34:29,350 In other words, although this one element is unstable, when 548 00:34:29,350 --> 00:34:31,989 you look at it numerically, when we solve the equations 549 00:34:31,989 --> 00:34:34,199 for a single element, as shown here. 550 00:34:34,199 --> 00:34:38,449 When the element is surrounded by other elements, they 551 00:34:38,449 --> 00:34:41,980 provide enough stiffness into this element so that the 552 00:34:41,980 --> 00:34:47,030 overall solution of the mesh can still be obtained, and we 553 00:34:47,030 --> 00:34:49,320 obtain, in fact, good results. 554 00:34:49,320 --> 00:34:52,610 However, if you do use reduced integration, we call this 555 00:34:52,610 --> 00:34:56,340 reduced integration because we are not evaluating the 556 00:34:56,340 --> 00:34:58,670 stiffness matrix exactly. 557 00:34:58,670 --> 00:35:01,320 If you do use reduced integration, you should be 558 00:35:01,320 --> 00:35:06,190 aware of this fact that for a single element, or for a 559 00:35:06,190 --> 00:35:10,600 certain element mesh layouts, we can run into numerical 560 00:35:10,600 --> 00:35:12,720 difficulties. 561 00:35:12,720 --> 00:35:17,370 Let's look briefly at the stress calculations. 562 00:35:17,370 --> 00:35:19,890 Surely, stresses can be calculated at any 563 00:35:19,890 --> 00:35:22,550 point of the element. 564 00:35:22,550 --> 00:35:25,910 And they would be evaluated as shown here. 565 00:35:25,910 --> 00:35:27,270 The stress strain law is here. 566 00:35:27,270 --> 00:35:29,260 Here we have to the strain-displacement 567 00:35:29,260 --> 00:35:30,330 interpolation matrix. 568 00:35:30,330 --> 00:35:33,140 Here, we have seen the nodal point displacement vector. 569 00:35:33,140 --> 00:35:36,340 Of course, these nodal points displacements are now known, 570 00:35:36,340 --> 00:35:37,380 they are now given. 571 00:35:37,380 --> 00:35:41,400 This is an initial stress vector which is also given. 572 00:35:41,400 --> 00:35:43,450 You should notice that the stresses are, in general, 573 00:35:43,450 --> 00:35:45,625 discontinuous across element boundaries. 574 00:35:48,900 --> 00:35:51,090 If we look at a [? practical ?] analysis, we 575 00:35:51,090 --> 00:35:54,040 might actually not evaluate the stresses at the boundary 576 00:35:54,040 --> 00:35:57,190 of an element, but we might evaluate the stresses at the 577 00:35:57,190 --> 00:35:58,140 integration points. 578 00:35:58,140 --> 00:36:02,260 So if you look at a mesh of elements, as shown here. 579 00:36:02,260 --> 00:36:06,860 We might evaluate the stresses at the 2x2 integration points, 580 00:36:06,860 --> 00:36:09,880 shown here for that element. 581 00:36:09,880 --> 00:36:13,390 If we were to evaluate the stresses of this element at 582 00:36:13,390 --> 00:36:16,410 this boundary, and of this element at that boundary, we 583 00:36:16,410 --> 00:36:18,900 would see a stress jump. 584 00:36:18,900 --> 00:36:22,360 I should mention here, just as a side note, that in fact, one 585 00:36:22,360 --> 00:36:26,470 can show that the stresses at the Gauss integration points 586 00:36:26,470 --> 00:36:29,150 are somewhat better predicted in a finite element analysis 587 00:36:29,150 --> 00:36:32,630 than along the boundary of the element. 588 00:36:32,630 --> 00:36:36,850 So that is another reason why we use, rather, why we predict 589 00:36:36,850 --> 00:36:40,660 stresses rather, at the Gauss integration points, and not on 590 00:36:40,660 --> 00:36:45,460 the boundaries of the element, using directly this approach. 591 00:36:45,460 --> 00:36:50,150 As an example here, let us look at a very simple problem. 592 00:36:50,150 --> 00:36:52,980 Here we have a cantilever modeled 593 00:36:52,980 --> 00:36:54,780 with two 8-node elements. 594 00:36:54,780 --> 00:36:59,100 Now remember, that these two 8-node elements each contain 595 00:36:59,100 --> 00:37:01,560 the parabolic displacement variation. 596 00:37:01,560 --> 00:37:05,950 So since this is a bending moment that we apply to that 597 00:37:05,950 --> 00:37:09,550 cantilever, and since this bending moment, of course, 598 00:37:09,550 --> 00:37:14,780 gives parabolic displacement variationn The finite element 599 00:37:14,780 --> 00:37:17,790 solution would be correct, will be exact in this 600 00:37:17,790 --> 00:37:18,810 particular case. 601 00:37:18,810 --> 00:37:22,880 And therefore, there would be no stress jump here. 602 00:37:22,880 --> 00:37:26,590 Since the elements contain the displacements that shall be 603 00:37:26,590 --> 00:37:31,920 predicted, that are the analytically-correct results. 604 00:37:31,920 --> 00:37:35,770 The finite element mesh will predict these displacements, 605 00:37:35,770 --> 00:37:38,800 and, of course, we have stress continuity. 606 00:37:38,800 --> 00:37:41,416 This is shown here by-- 607 00:37:41,416 --> 00:37:43,090 we have dissected these elements here. 608 00:37:43,090 --> 00:37:47,160 This is element 2, this is here, element 1. 609 00:37:47,160 --> 00:37:50,790 And we have plotted the stress distribution along 610 00:37:50,790 --> 00:37:52,400 the line a, b, c. 611 00:37:52,400 --> 00:37:53,590 Here's the line, a, b, c. 612 00:37:53,590 --> 00:37:56,200 And as you can see, there's 900 here, 613 00:37:56,200 --> 00:37:58,150 and there's 900 there. 614 00:37:58,150 --> 00:38:00,510 And of course, a linear variation stresses, so no 615 00:38:00,510 --> 00:38:02,152 stress jumps there. 616 00:38:02,152 --> 00:38:07,580 However, if you look at the same problem, the same 617 00:38:07,580 --> 00:38:11,950 cantilever, I should say, with now an edge share load, as 618 00:38:11,950 --> 00:38:13,340 shown here. 619 00:38:13,340 --> 00:38:18,310 Then these elements cannot represent the exact 620 00:38:18,310 --> 00:38:22,190 displacement distribution in the cantilever because they 621 00:38:22,190 --> 00:38:25,830 only contain a parabolic variation in displacement. 622 00:38:25,830 --> 00:38:30,050 And in this case, we have the stress discontinuity. 623 00:38:30,050 --> 00:38:33,250 tau x x, in other words, the stress into this direction 624 00:38:33,250 --> 00:38:35,240 here, so normal stress. 625 00:38:35,240 --> 00:38:39,140 For this element along, for element 2, along a,b,c, you 626 00:38:39,140 --> 00:38:42,080 can see 1744 up there. 627 00:38:42,080 --> 00:38:42,850 And then [UNINTELLIGIBLE] 628 00:38:42,850 --> 00:38:44,440 a neutral axis because of symmetry 629 00:38:44,440 --> 00:38:46,010 conditions, of course. 630 00:38:46,010 --> 00:38:53,000 And for element 1, along a, b, c, we have 1502 right there. 631 00:38:53,000 --> 00:38:56,470 So there is, in fact, this stressed discontinuity that I 632 00:38:56,470 --> 00:38:58,740 was talking about earlier. 633 00:38:58,740 --> 00:39:02,320 The share stresses here have also a slight discontinuity, 634 00:39:02,320 --> 00:39:09,000 as you can see, 291.38, 296.48. 635 00:39:09,000 --> 00:39:14,220 The reason that we have this stress discontinuity here is, 636 00:39:14,220 --> 00:39:16,930 of course, that the elements cannot 637 00:39:16,930 --> 00:39:18,505 represent the exact solution. 638 00:39:21,550 --> 00:39:27,710 And therefore, they each try to, in the integral sense of 639 00:39:27,710 --> 00:39:31,850 the finite element formulation, try to predict 640 00:39:31,850 --> 00:39:37,060 the solution accurately within its domain, and the result is 641 00:39:37,060 --> 00:39:38,140 that we have a stress 642 00:39:38,140 --> 00:39:41,210 discontinuity between elements. 643 00:39:41,210 --> 00:39:46,430 Finally, let us look then at some modeling considerations. 644 00:39:46,430 --> 00:39:51,300 In order to solve a problem, we should really have a 645 00:39:51,300 --> 00:39:55,330 qualitative knowledge of the response to be predicted. 646 00:39:55,330 --> 00:39:58,000 We should also have a thorough knowledge of the principles of 647 00:39:58,000 --> 00:40:02,470 mechanics and the finite element procedures available. 648 00:40:02,470 --> 00:40:05,820 And this, of coure, is the subject of my 649 00:40:05,820 --> 00:40:08,880 lectures here to you. 650 00:40:08,880 --> 00:40:12,530 If we have these two knowledges, then we should 651 00:40:12,530 --> 00:40:16,320 still know a little bit more about what kind of elements we 652 00:40:16,320 --> 00:40:23,410 should use, how one element performs versus another. 653 00:40:23,410 --> 00:40:27,200 And here, the first remark that I like to make is that 654 00:40:27,200 --> 00:40:31,270 parabolic undistorted elements are usually most effective. 655 00:40:31,270 --> 00:40:35,910 Here, please see that I'm saying usually most effective 656 00:40:35,910 --> 00:40:39,190 because in some earlier lecture, I talked about the 657 00:40:39,190 --> 00:40:41,510 analysis of fracture mechanics problems, where we, in fact, 658 00:40:41,510 --> 00:40:44,550 distort elements to pick up stress singularities. 659 00:40:44,550 --> 00:40:47,380 But if we don't talk about stress singularities or 660 00:40:47,380 --> 00:40:51,410 special situations, usually the parabolic undistorted 661 00:40:51,410 --> 00:40:54,820 elements are most effective. 662 00:40:54,820 --> 00:41:00,630 Here, I have put together in a table the kinds of elements 663 00:41:00,630 --> 00:41:03,680 that I would recommend for usage. 664 00:41:03,680 --> 00:41:06,550 These are typically also the elements that I used in 665 00:41:06,550 --> 00:41:09,020 general purpose code such ADINA. 666 00:41:09,020 --> 00:41:11,480 Tress or cable element, the 2-node 667 00:41:11,480 --> 00:41:13,690 element is quite effective. 668 00:41:13,690 --> 00:41:16,430 In some cases, we want to use a three-node element or 669 00:41:16,430 --> 00:41:19,100 four-node element, but the 2-node element is cheap and 670 00:41:19,100 --> 00:41:20,970 very effective in modeling. 671 00:41:20,970 --> 00:41:22,840 For two-dimensional plane stress, plane strain, 672 00:41:22,840 --> 00:41:26,300 axisymetric analysis, the 8-node or 8-node element. 673 00:41:26,300 --> 00:41:29,070 Those are the elements most effective, and as I just 674 00:41:29,070 --> 00:41:32,520 mentioned, we should use them undistorted, in other words, 675 00:41:32,520 --> 00:41:35,080 rectangular. 676 00:41:35,080 --> 00:41:38,600 The three-dimensional analysis then. 677 00:41:42,530 --> 00:41:45,360 In three-dimensional analysis, we would effectively use the 678 00:41:45,360 --> 00:41:48,550 20-node brick element, which is really the counterpart of 679 00:41:48,550 --> 00:41:51,060 the 8- or 9-node element. 680 00:41:51,060 --> 00:41:57,560 For the 3D that is curved, we would use three-moded elements 681 00:41:57,560 --> 00:41:58,920 or four-noded elements-- better 682 00:41:58,920 --> 00:42:00,450 even four-noded elements. 683 00:42:00,450 --> 00:42:03,230 Of course, if we have a straight beam, then we would 684 00:42:03,230 --> 00:42:06,420 use simply the engineering [? animation, ?] 685 00:42:06,420 --> 00:42:08,190 two-noded beam. 686 00:42:08,190 --> 00:42:11,420 But if you talk about a curved beam, then the isoparametric 687 00:42:11,420 --> 00:42:13,630 beam that I presented to you in 688 00:42:13,630 --> 00:42:17,060 lecture 7 is very effective. 689 00:42:17,060 --> 00:42:21,880 The plate and shell elements that I would recommend are 690 00:42:21,880 --> 00:42:24,520 this element here, for plate analysis-- 691 00:42:24,520 --> 00:42:26,600 I mentioned that already earlier. 692 00:42:26,600 --> 00:42:31,370 And for the shell analysis, the nine-noded element and the 693 00:42:31,370 --> 00:42:33,790 16-node element. 694 00:42:33,790 --> 00:42:37,530 Even for plates, it might pay sometimes to use the 16-node 695 00:42:37,530 --> 00:42:40,110 element, which is very effective, and particularly 696 00:42:40,110 --> 00:42:43,720 works for very, very thin plates and shells. 697 00:42:43,720 --> 00:42:48,000 It does not have the deteriorating behavior of the 698 00:42:48,000 --> 00:42:53,790 locking phenomenon that I referred to in lecture 7. 699 00:42:53,790 --> 00:43:00,390 Let us look also at some mesh considerations. 700 00:43:00,390 --> 00:43:06,350 If we perform an analysis, we might find that in some area 701 00:43:06,350 --> 00:43:08,430 of the element idealization. 702 00:43:08,430 --> 00:43:12,440 We want to use 4-node elements lower order elements, and in 703 00:43:12,440 --> 00:43:15,320 another area, we want to use higher order elements. 704 00:43:15,320 --> 00:43:19,460 Well, here, I've show a transition region going from a 705 00:43:19,460 --> 00:43:22,530 4- to a 5-node to take an 8-node element. 706 00:43:22,530 --> 00:43:28,130 Notice that we have full compatibility along this side. 707 00:43:28,130 --> 00:43:32,250 So this is a comparable element transition from 4- to 708 00:43:32,250 --> 00:43:32,940 8-node elements. 709 00:43:32,940 --> 00:43:35,850 And similarly, of course, in three-dimensional analysis, 710 00:43:35,850 --> 00:43:39,720 you could go from 8-node bricks to 20-node bricks by 711 00:43:39,720 --> 00:43:43,960 using such transition elements. 712 00:43:43,960 --> 00:43:49,960 An alternative approach would be to use a 4-node element 713 00:43:49,960 --> 00:43:53,730 here, and say, two 4-node elements there, but notice 714 00:43:53,730 --> 00:43:59,370 that if you do so, the displacements at node a, have 715 00:43:59,370 --> 00:44:02,980 to be constraint, if you want to preserve 716 00:44:02,980 --> 00:44:04,580 compatibility here. 717 00:44:04,580 --> 00:44:10,310 Notice that along BC, for this 4-node element, we have a 718 00:44:10,310 --> 00:44:12,150 linear variation and displacements because we have 719 00:44:12,150 --> 00:44:15,340 only UB, VB, UC, VC there. 720 00:44:15,340 --> 00:44:18,620 And of course, here, for these two 4-noded elements, we have 721 00:44:18,620 --> 00:44:21,370 a linear varition from here to there, and another linear 722 00:44:21,370 --> 00:44:23,940 variatoin from here to there in displacements. 723 00:44:23,940 --> 00:44:27,500 To preserve full compatibility along this line, these two 724 00:44:27,500 --> 00:44:30,240 constraint equations have to be used. 725 00:44:30,240 --> 00:44:34,810 And that means, really, that VA and UA are set to the mean 726 00:44:34,810 --> 00:44:39,720 of UB, UC, and VB, BC, as shown here. 727 00:44:39,720 --> 00:44:45,020 Another transition approach is shown here on the last view 728 00:44:45,020 --> 00:44:47,290 graphs that I wanted to present to you. 729 00:44:47,290 --> 00:44:52,020 And here we're going from 8-node elements, or 7-node 730 00:44:52,020 --> 00:44:55,880 elements, if you wanted to use 8-noded elements here, then of 731 00:44:55,880 --> 00:45:00,050 course, we would have to put another node in there each. 732 00:45:00,050 --> 00:45:05,290 So here we're going from two layers of 8-node elements, 733 00:45:05,290 --> 00:45:08,100 over into one layer of 8-node not elements. 734 00:45:08,100 --> 00:45:12,570 Notice that in this particular transition region here, we are 735 00:45:12,570 --> 00:45:16,190 using a distorted element. 736 00:45:16,190 --> 00:45:21,400 Of course, that means the order of accuracy that we can 737 00:45:21,400 --> 00:45:25,000 expect here in the stress predictions and so on, is not 738 00:45:25,000 --> 00:45:27,260 quite as good as we would like to see it. 739 00:45:27,260 --> 00:45:32,110 So this transition region should be away from the area 740 00:45:32,110 --> 00:45:34,540 of interest. the area of interest being somewhere over 741 00:45:34,540 --> 00:45:37,690 here, far way from the transition reason. 742 00:45:37,690 --> 00:45:41,460 However, we do have a compatible element layout, and 743 00:45:41,460 --> 00:45:45,390 that, of course, is an important point, which I 744 00:45:45,390 --> 00:45:48,570 mentioned to you earlier, that we would like to preserve 745 00:45:48,570 --> 00:45:50,930 compatibility as much as possible. 746 00:45:50,930 --> 00:45:53,330 And for that reason, of course, we have developed 747 00:45:53,330 --> 00:45:56,820 these variable number node elements. 748 00:45:56,820 --> 00:45:59,650 This concludes what I wanted to say in this lecture. 749 00:45:59,650 --> 00:46:00,900 Thank you for your attention.