WEBVTT 00:00:00.040 --> 00:00:02.470 The following content is provided under a Creative 00:00:02.470 --> 00:00:03.880 Commons license. 00:00:03.880 --> 00:00:06.920 Your support will help MIT OpenCourseWare continue to 00:00:06.920 --> 00:00:10.570 offer high-quality educational resources for free. 00:00:10.570 --> 00:00:13.470 To make a donation, or view additional materials from 00:00:13.470 --> 00:00:17.400 hundreds of MIT courses, visit MIT OpenCourseWare at 00:00:17.400 --> 00:00:18.650 ocw.mit.edu. 00:00:21.910 --> 00:00:23.210 PROFESSOR: Ladies and gentlemen, 00:00:23.210 --> 00:00:25.740 welcome to lecture four. 00:00:25.740 --> 00:00:28.240 In the last lecture, I presented to you a general 00:00:28.240 --> 00:00:32.000 formulation for finite element analysis. 00:00:32.000 --> 00:00:35.570 In this lecture, I would like to talk to you about the 00:00:35.570 --> 00:00:38.400 derivation of specific finite element matrices. 00:00:41.260 --> 00:00:44.870 The formulation that we will be using, or the derivation I 00:00:44.870 --> 00:00:48.240 should say, that we will be using, leads us to generalized 00:00:48.240 --> 00:00:50.770 finite element models. 00:00:50.770 --> 00:00:53.220 I will be talking later on about what we mean by 00:00:53.220 --> 00:00:54.870 generalized. 00:00:54.870 --> 00:00:58.450 Let's remember, let's recall what we arrived 00:00:58.450 --> 00:01:00.340 at in the last lecture. 00:01:00.340 --> 00:01:05.840 We had that the element stiffness matrix for element m 00:01:05.840 --> 00:01:11.040 is obtained by this integration here where we 00:01:11.040 --> 00:01:15.450 integrate the product of a strain-displacement matrix 00:01:15.450 --> 00:01:20.010 times a stress-strain matrix times the strain-displacement 00:01:20.010 --> 00:01:24.390 matrix over the volume of the element. 00:01:24.390 --> 00:01:28.270 We also had derived the expression for a body force 00:01:28.270 --> 00:01:32.200 vector, B. m, again, denoting element m. 00:01:32.200 --> 00:01:36.600 And here, we integrated over the volume of the element the 00:01:36.600 --> 00:01:41.560 product of the displacement interpolation matrix times the 00:01:41.560 --> 00:01:46.740 body forces in the element. 00:01:46.740 --> 00:01:52.050 We also had derived a surface force vector of the element in 00:01:52.050 --> 00:01:56.960 which we integrate the product of the surface interpolation 00:01:56.960 --> 00:02:02.100 matrix times the surface forces applied to the element. 00:02:02.100 --> 00:02:07.110 So in essence, we see then if we look at these expressions, 00:02:07.110 --> 00:02:11.750 we see that we need really a B matrix, a strain-displacement 00:02:11.750 --> 00:02:14.290 matrix for the element. 00:02:14.290 --> 00:02:15.230 Of course, we need the 00:02:15.230 --> 00:02:17.970 stress-strain law for the element. 00:02:17.970 --> 00:02:21.250 And that will really be given to us from the serial strength 00:02:21.250 --> 00:02:23.980 of materials. 00:02:23.980 --> 00:02:27.880 We also want the displacement interpolation 00:02:27.880 --> 00:02:31.010 matrix for the element. 00:02:31.010 --> 00:02:34.410 Once we have this displacement interpolation matrix, we can 00:02:34.410 --> 00:02:37.550 directly obtain the interpolation matrix for the 00:02:37.550 --> 00:02:42.590 surface displacement of the element by simply substituting 00:02:42.590 --> 00:02:47.740 into this matrix here, the surface coordinates. 00:02:47.740 --> 00:02:53.590 In summary then, we need the H matrix, the B matrix, and the 00:02:53.590 --> 00:02:57.620 stress-strain law for every element that we have in the 00:02:57.620 --> 00:02:59.790 assemblage. 00:02:59.790 --> 00:03:03.690 Notice that the stress-strain law, of course, must depend on 00:03:03.690 --> 00:03:06.330 what kind of element we are looking at. 00:03:06.330 --> 00:03:10.090 How we have mechanically idealized our system. 00:03:10.090 --> 00:03:13.490 In a truss structure, we will have simply the Young's 00:03:13.490 --> 00:03:15.120 modulus in here. 00:03:15.120 --> 00:03:17.740 In a plane stress idealization, we will have a 00:03:17.740 --> 00:03:19.310 plane stress law. 00:03:19.310 --> 00:03:21.420 In a plate bending idealization, we will have the 00:03:21.420 --> 00:03:28.960 plate bending stress-strain law in there, and so on. 00:03:28.960 --> 00:03:32.940 Well, I want to talk today about these matrices. 00:03:32.940 --> 00:03:36.240 How do we obtain these matrices? 00:03:36.240 --> 00:03:41.030 Once we have talked about that subject, I finally want to 00:03:41.030 --> 00:03:43.990 also spend a little bit of time talking about the 00:03:43.990 --> 00:03:46.170 convergence of the analysis results. 00:03:46.170 --> 00:03:49.540 Of course, in finite element analysis, we must be concerned 00:03:49.540 --> 00:03:52.960 about the accuracy of the analysis results that we are 00:03:52.960 --> 00:03:57.620 obtaining, and about the convergence of the results to 00:03:57.620 --> 00:04:03.610 the actual analytically, theoretically accurate result 00:04:03.610 --> 00:04:06.000 that we want to obtain. 00:04:06.000 --> 00:04:10.245 Well, let us then first talk about the derivation of the H, 00:04:10.245 --> 00:04:12.480 B, and C matrices. 00:04:12.480 --> 00:04:17.990 And the H, B, and C matrix, of course, depend on the 00:04:17.990 --> 00:04:21.711 particular problem that you're looking at. 00:04:21.711 --> 00:04:24.730 Let us therefore categorize the different types of 00:04:24.730 --> 00:04:28.320 problems that we encounter in structural analysis. 00:04:28.320 --> 00:04:33.060 Here we have a truss structure, and a truss 00:04:33.060 --> 00:04:38.430 structure is really an assemblage of truss elements. 00:04:38.430 --> 00:04:43.710 And in each truss element that lies between the nodes, we 00:04:43.710 --> 00:04:47.090 have one-dimensional stress conditions. 00:04:47.090 --> 00:04:51.470 In other words, the only stress that we see on the 00:04:51.470 --> 00:04:55.780 section AA would be the tau xx stress. 00:04:55.780 --> 00:04:59.230 And that is the only stress, no other stress is involved. 00:04:59.230 --> 00:05:01.640 Of course, just going a little bit ahead, we immediately 00:05:01.640 --> 00:05:05.480 would also say that the only strain we would be interested 00:05:05.480 --> 00:05:08.770 in is epsilon xx and the stress-strain law is simply 00:05:08.770 --> 00:05:11.650 Young's modulus. 00:05:11.650 --> 00:05:15.400 The second kind of analysis that we might be performing is 00:05:15.400 --> 00:05:18.290 a plane stress analysis. 00:05:18.290 --> 00:05:21.430 In plane stress conditions, the only stresses that are 00:05:21.430 --> 00:05:26.420 non-zero are tau xx, tau yy, and tau xy. 00:05:26.420 --> 00:05:29.600 Remember in lectures three, I had listed all six stress and 00:05:29.600 --> 00:05:30.680 strain components. 00:05:30.680 --> 00:05:34.100 And what I'm doing now is I take out just those stress and 00:05:34.100 --> 00:05:37.290 strain components that are really applicable for the kind 00:05:37.290 --> 00:05:40.260 of problem that I want to look at. 00:05:40.260 --> 00:05:43.460 Plane stress situations we will encounter in the analysis 00:05:43.460 --> 00:05:44.860 of a beam, for example. 00:05:44.860 --> 00:05:47.670 Here's a simple cantilever beam subjected to 00:05:47.670 --> 00:05:50.090 surface loading p. 00:05:50.090 --> 00:05:57.370 And if we look at an element here, it is redrawn here. 00:05:57.370 --> 00:05:59.790 We would find that tau xx would be a 00:05:59.790 --> 00:06:01.280 stress into this direction. 00:06:01.280 --> 00:06:03.610 Tau xy is the shear stress, and tau yy 00:06:03.610 --> 00:06:06.180 is that normal stress. 00:06:06.180 --> 00:06:08.760 The cantilever, of course, can be subjected to a surface 00:06:08.760 --> 00:06:12.360 loading, a bending moment, tip loading, whatever loading. 00:06:12.360 --> 00:06:15.310 The stress components, the only stress components that we 00:06:15.310 --> 00:06:20.720 assume of course, act in the structure would be these three 00:06:20.720 --> 00:06:22.480 stress components. 00:06:22.480 --> 00:06:26.180 In fact, of course, in the actual physical situation, we 00:06:26.180 --> 00:06:30.340 would find that around here, depending on the specific 00:06:30.340 --> 00:06:33.660 support conditions that we are using, we would have a much 00:06:33.660 --> 00:06:36.310 more complicated stress situation than the plane 00:06:36.310 --> 00:06:38.170 stress situation assumed. 00:06:38.170 --> 00:06:40.670 However, if we are only interested in the tip 00:06:40.670 --> 00:06:45.050 displacement here, or the displacements away from the 00:06:45.050 --> 00:06:48.250 support condition from the support. 00:06:48.250 --> 00:06:50.550 Similarly, if we are only interested in the stresses and 00:06:50.550 --> 00:06:54.580 strains away from the support, certainly it is accurate 00:06:54.580 --> 00:06:55.810 enough to-- 00:06:55.810 --> 00:06:58.790 it yields an accurate enough model to assume a 00:06:58.790 --> 00:07:01.070 plane stress situation. 00:07:01.070 --> 00:07:04.190 Another plane stress situation occurs here, for example, 00:07:04.190 --> 00:07:06.470 where we have a sheet with a hole. 00:07:06.470 --> 00:07:10.730 If the sheet is thin, the only stresses that we would see on 00:07:10.730 --> 00:07:14.440 an element, such as shown here, would be z stresses, the 00:07:14.440 --> 00:07:17.385 plane stress stresses. 00:07:17.385 --> 00:07:20.950 A next category of analysis that we would encounter in 00:07:20.950 --> 00:07:25.190 practice is a plane strain condition. 00:07:25.190 --> 00:07:31.970 Plane strain condition means that in the space xyz shown 00:07:31.970 --> 00:07:37.500 here with the displacements uvw, the displacement w is 0. 00:07:37.500 --> 00:07:42.450 It's identically 0 if our x, y-coordinate system lies in 00:07:42.450 --> 00:07:47.020 the plane of a typical section of the total problem. 00:07:47.020 --> 00:07:50.560 And let me show you one such typical-- 00:07:50.560 --> 00:07:53.000 or describe to you one such difficult problem. 00:07:53.000 --> 00:07:56.740 Here we have a dam that is very long. 00:07:56.740 --> 00:07:59.550 In fact, we might assume it infinitely long. 00:07:59.550 --> 00:08:05.020 And it is supported on this face here, so that w, the 00:08:05.020 --> 00:08:08.300 displacement, normal to that face is 0. 00:08:08.300 --> 00:08:10.320 And the same support conditions are 00:08:10.320 --> 00:08:12.480 at the other face. 00:08:12.480 --> 00:08:16.400 If the dam is subject to uniform water loading, water 00:08:16.400 --> 00:08:20.800 pressure loading all along its side, then by symmetry it 00:08:20.800 --> 00:08:24.980 follows that w, the displacement, at any section 00:08:24.980 --> 00:08:27.430 of the dam, the displacement normal to that 00:08:27.430 --> 00:08:29.490 section must be 0. 00:08:29.490 --> 00:08:35.030 Therefore, we have w 0 meaning directly epsilon zz 0. 00:08:35.030 --> 00:08:39.270 The strain into the z-direction is also 0. 00:08:39.270 --> 00:08:46.870 Similarly, the shear stresses, tau zx and tau zy are 0. 00:08:46.870 --> 00:08:49.140 And the only stresses that we are left with which are 00:08:49.140 --> 00:08:51.990 non-zero are those shown here. 00:08:51.990 --> 00:08:59.840 These are the stresses acting on a typical slice of the dam. 00:08:59.840 --> 00:09:05.040 A slice say, of unit length would be subjected to just 00:09:05.040 --> 00:09:07.070 these stress conditions. 00:09:07.070 --> 00:09:11.190 Therefore, in that array of six stresses and strains that 00:09:11.190 --> 00:09:14.410 I talked to you about in lecture three, we really only 00:09:14.410 --> 00:09:17.590 are concerned-- 00:09:17.590 --> 00:09:21.410 you only need to calculate these four components of those 00:09:21.410 --> 00:09:24.600 six that I talked about there. 00:09:24.600 --> 00:09:27.100 Another stress-strain condition that we are 00:09:27.100 --> 00:09:31.370 interested in frequently in practice is an axisymmetric 00:09:31.370 --> 00:09:32.830 analysis condition. 00:09:32.830 --> 00:09:37.100 This case we have an axis of revolution for the structure, 00:09:37.100 --> 00:09:40.490 and here we look at a cylinder, which is 00:09:40.490 --> 00:09:45.370 axisymmetric in geometry about that center line here, the 00:09:45.370 --> 00:09:46.750 axis of revolution. 00:09:46.750 --> 00:09:50.590 And the loading is also assumed in this particular 00:09:50.590 --> 00:09:52.470 case to be also axisymmetric. 00:09:52.470 --> 00:09:56.040 So typically, we are talking about a cylinder, circular, of 00:09:56.040 --> 00:09:59.010 course, when looking from a plane view subjected to 00:09:59.010 --> 00:10:03.220 internal pressure, which is the same all around. 00:10:03.220 --> 00:10:07.410 And of course, the geometry is also axisymmetric when we look 00:10:07.410 --> 00:10:11.810 at the axis of revolution, the center line of the cylinder. 00:10:11.810 --> 00:10:16.930 In this particular case, the only stress conditions that we 00:10:16.930 --> 00:10:19.710 need to calculate are shown here for an 00:10:19.710 --> 00:10:21.890 element in that cylinder. 00:10:21.890 --> 00:10:25.890 We have tau xx, tau yy, tau zz. 00:10:25.890 --> 00:10:29.510 This, of course, being the hoop stress. 00:10:29.510 --> 00:10:32.290 And the shear stress, tau xy. 00:10:32.290 --> 00:10:34.150 These are the stress conditions that we need to 00:10:34.150 --> 00:10:36.090 calculate for this particular case. 00:10:36.090 --> 00:10:42.080 Similarity, because we are talking about an axisymmetric 00:10:42.080 --> 00:10:46.260 condition, an axisymmetric condition where really, every 00:10:46.260 --> 00:10:51.080 one of these elements behaves in the same way, we can look 00:10:51.080 --> 00:10:57.205 at one radian of the total circumference in the analysis. 00:10:59.740 --> 00:11:04.420 So this is an axisymmetric analysis, which we also are 00:11:04.420 --> 00:11:08.450 concerned with frequently in practice. 00:11:08.450 --> 00:11:13.330 Finally, the very large number of analyses-- 00:11:13.330 --> 00:11:17.040 in a very large number of analyses, we consider plates 00:11:17.040 --> 00:11:19.080 and shells. 00:11:19.080 --> 00:11:23.930 If we look at a plate, an element in that plate, we 00:11:23.930 --> 00:11:30.240 would find that in general we would have the tau xx, tau yy. 00:11:30.240 --> 00:11:32.400 These are the normal stresses. 00:11:32.400 --> 00:11:37.530 And tau xy shearing stresses in the plane of the plate. 00:11:37.530 --> 00:11:40.060 We would also have transfer shearing stresses, 00:11:40.060 --> 00:11:43.230 tau xz and tau yz. 00:11:43.230 --> 00:11:47.290 Sorry, tau yz, shown here as the 00:11:47.290 --> 00:11:49.450 transfer shearing stresses. 00:11:49.450 --> 00:11:52.490 Of course, these transfers shearing stresses give us 00:11:52.490 --> 00:11:56.020 shear forces. 00:11:56.020 --> 00:12:03.890 And these stresses here give us membrane forces plus 00:12:03.890 --> 00:12:07.810 bending moments and twisting moments. 00:12:07.810 --> 00:12:11.330 Elementary theory of plates shows us, tells us, that these 00:12:11.330 --> 00:12:14.370 are the significant stresses in a plate. 00:12:14.370 --> 00:12:18.080 The important point is that in a plate, as well as in a 00:12:18.080 --> 00:12:23.790 shell, the stress normal to the plate is 0. 00:12:23.790 --> 00:12:25.320 The same holds for the shell. 00:12:25.320 --> 00:12:29.600 The stress normal to the shell is 0, and we have the stress 00:12:29.600 --> 00:12:30.330 components. 00:12:30.330 --> 00:12:35.426 However, we have to now remember that the xx and yy 00:12:35.426 --> 00:12:39.210 axes are aligned with the curvature of the shelf. 00:12:39.210 --> 00:12:41.420 Are aligned with the curvature of the shell. 00:12:41.420 --> 00:12:45.330 And then these are the stress components that are of 00:12:45.330 --> 00:12:48.010 interest in the analysis of shells. 00:12:48.010 --> 00:12:51.330 In many cases, when we look at the analysis of plates, the 00:12:51.330 --> 00:12:53.300 membrane forces-- 00:12:53.300 --> 00:12:56.690 the total membrane forces are 0. 00:12:56.690 --> 00:13:01.400 In which case, of course, we assume that tau yy has a 00:13:01.400 --> 00:13:04.480 linear variation through the thickness of the plate. 00:13:04.480 --> 00:13:10.230 As stipulated, for example, in the Kirchhoff plate theory. 00:13:10.230 --> 00:13:15.270 This much then for the static components, or rather the 00:13:15.270 --> 00:13:18.300 stress components of plates and shells when we look at the 00:13:18.300 --> 00:13:21.350 kinematics of plates and shells. 00:13:21.350 --> 00:13:25.910 Just to remind you, we assume that there's a normal to the 00:13:25.910 --> 00:13:27.800 mid-surface. 00:13:27.800 --> 00:13:31.520 And that normal in Kirchhoff plate theory remains during 00:13:31.520 --> 00:13:35.510 deformation normal to the plate surface. 00:13:35.510 --> 00:13:40.550 And all material particles on that normal AB remain on a 00:13:40.550 --> 00:13:41.480 straight line. 00:13:41.480 --> 00:13:42.460 These are the two things. 00:13:42.460 --> 00:13:46.720 I repeat, all material particles on the line AB 00:13:46.720 --> 00:13:51.450 remain during deformation on the line AB, and the line AB 00:13:51.450 --> 00:13:53.955 remains normal to the mid-surface in Kirchhoff plate 00:13:53.955 --> 00:13:57.750 theory, which does not exclude shear deformations. 00:13:57.750 --> 00:14:01.200 However, if we use a theory that does include shear 00:14:01.200 --> 00:14:04.970 deformations, we would still have that the material 00:14:04.970 --> 00:14:09.450 particles on line AB will still lie after deformations 00:14:09.450 --> 00:14:14.230 on the line A prime B prime say. 00:14:14.230 --> 00:14:16.850 But the line A prime B prime-- 00:14:16.850 --> 00:14:20.240 B is still a straight line-- is not anymore normal to the 00:14:20.240 --> 00:14:21.090 mid-surface. 00:14:21.090 --> 00:14:23.680 In this case, we have included shear deformations. 00:14:23.680 --> 00:14:26.830 We will see in the next lecture that, in fact, it is 00:14:26.830 --> 00:14:30.030 simpler to use a theory including shear deformations 00:14:30.030 --> 00:14:33.980 to derive effective finite elements. 00:14:33.980 --> 00:14:39.380 Well, this then is a review for the kinds of stress 00:14:39.380 --> 00:14:43.010 components and strain components that we are 00:14:43.010 --> 00:14:43.910 interested in. 00:14:43.910 --> 00:14:47.920 Of course, that information which I discussed so far is 00:14:47.920 --> 00:14:53.050 available from the strengths of material theory and 00:14:53.050 --> 00:14:54.760 continue mechanic theory. 00:14:54.760 --> 00:14:58.140 And as an engineer, however, we have to be able to decide 00:14:58.140 --> 00:15:00.470 what kind of stress and strain 00:15:00.470 --> 00:15:02.230 situations you want to analyze. 00:15:02.230 --> 00:15:04.520 And there, of course, is already one very important 00:15:04.520 --> 00:15:05.010 assumption. 00:15:05.010 --> 00:15:09.430 A very important assumption in laying out or starting with 00:15:09.430 --> 00:15:13.210 the right model for the actual physical structure. 00:15:13.210 --> 00:15:16.340 Whether we want to model the actual physical structure as a 00:15:16.340 --> 00:15:21.240 truss assemblage, as an assemblage of plates, beams, 00:15:21.240 --> 00:15:26.810 et cetera, all of those situations can be covered in 00:15:26.810 --> 00:15:28.090 the finite element theory. 00:15:28.090 --> 00:15:31.630 But we have to make the right assumptions right from the 00:15:31.630 --> 00:15:34.380 start in starting with the right model. 00:15:34.380 --> 00:15:36.460 So here I listed ones. 00:15:36.460 --> 00:15:38.850 In one table, the particular problems 00:15:38.850 --> 00:15:40.100 that we might encounter. 00:15:40.100 --> 00:15:42.640 We might have a bar. 00:15:42.640 --> 00:15:45.770 The displacement components then being simply u. 00:15:45.770 --> 00:15:51.020 In a beam, which I have not given earlier, but here we 00:15:51.020 --> 00:15:53.200 have a transverse displacement, w. 00:15:53.200 --> 00:15:57.350 In plane stress situations, we have two displacements, u and 00:15:57.350 --> 00:15:58.100 v. 00:15:58.100 --> 00:16:01.090 In a plane strain situation, u and v. 00:16:01.090 --> 00:16:03.230 In an axisymmetric situation, u and v. 00:16:03.230 --> 00:16:06.360 Three-dimensional situation, we have u, v, and w. 00:16:06.360 --> 00:16:08.760 In plate bending, simply the transverse displacements. 00:16:11.590 --> 00:16:15.600 In analysis of shells, in the analysis of shells we would, 00:16:15.600 --> 00:16:20.260 of course, have the transverse placements and also the 00:16:20.260 --> 00:16:21.770 membrane displacements included. 00:16:21.770 --> 00:16:25.710 We will talk about that in the next lecture. 00:16:25.710 --> 00:16:29.120 These are the displacement components for the particular 00:16:29.120 --> 00:16:31.270 problem that we are looking at. 00:16:31.270 --> 00:16:35.810 And correspondingly, we have particular strains. 00:16:35.810 --> 00:16:39.170 The strains that we are talking about are in the bar, 00:16:39.170 --> 00:16:40.480 simply the normals strains. 00:16:40.480 --> 00:16:44.640 In a beam, the second derivative of the transverse 00:16:44.640 --> 00:16:45.680 displacements. 00:16:45.680 --> 00:16:49.310 Here we have kappa xx defined. 00:16:49.310 --> 00:16:50.660 This is kappa xx defined. 00:16:50.660 --> 00:16:53.120 In plane stress, we have these three components. 00:16:53.120 --> 00:16:54.860 Plane strain, these three components. 00:16:54.860 --> 00:16:57.550 Axisymmetric these three components, and so on. 00:16:57.550 --> 00:17:00.500 And it is here, of course, where the engineer has to 00:17:00.500 --> 00:17:04.450 decide what kind of problem he has in hand when he actually-- 00:17:04.450 --> 00:17:07.880 when he looks at an actual physical situation. 00:17:07.880 --> 00:17:10.849 Corresponding to these strains, 00:17:10.849 --> 00:17:12.690 we have also stresses. 00:17:12.690 --> 00:17:15.010 And here we are listing the stresses. 00:17:15.010 --> 00:17:19.970 Notice that for the beam, we are talking about a bending 00:17:19.970 --> 00:17:21.760 moment, which we might consider to be 00:17:21.760 --> 00:17:23.150 a generalized stress. 00:17:23.150 --> 00:17:25.810 In plate bending, we talk similarly about bending 00:17:25.810 --> 00:17:28.380 moments, which you also might consider as 00:17:28.380 --> 00:17:30.440 being generalized stresses. 00:17:30.440 --> 00:17:34.680 Otherwise, we have the actual stresses that correspond to 00:17:34.680 --> 00:17:36.490 the strains. 00:17:36.490 --> 00:17:39.420 Having decided what stresses and strains we 00:17:39.420 --> 00:17:40.920 want to look at-- 00:17:40.920 --> 00:17:43.390 in other words, what kind of problem we have at hand, we 00:17:43.390 --> 00:17:45.630 also have to select the appropriate material matrix. 00:17:45.630 --> 00:17:48.840 Again, this one is given form strengths of material. 00:17:48.840 --> 00:17:52.140 Here, for a bar, a truss structure, we simply have 00:17:52.140 --> 00:17:52.820 Young's modulus. 00:17:52.820 --> 00:17:55.360 For a beam, we have the flexural rigidity. 00:17:55.360 --> 00:18:00.300 In plane stress analysis, this would be the material law that 00:18:00.300 --> 00:18:03.380 you would find in many textbooks on 00:18:03.380 --> 00:18:05.470 strengths of materials. 00:18:05.470 --> 00:18:09.640 Well, having therefore resolved the problem of how to 00:18:09.640 --> 00:18:12.420 establish the stress-strain law, or the stress-strain 00:18:12.420 --> 00:18:17.530 matrix c, we now go on to discuss the construction of 00:18:17.530 --> 00:18:20.860 the displacement interpolation matrix. 00:18:20.860 --> 00:18:22.770 For a one-dimensional bar element, 00:18:22.770 --> 00:18:25.570 we would have simply-- 00:18:25.570 --> 00:18:29.430 we can assume simply a polynomial with unknown 00:18:29.430 --> 00:18:34.980 coefficients, alpha 1, alpha 2, alpha 3. 00:18:34.980 --> 00:18:37.805 And this is simply a polynomial nx. 00:18:37.805 --> 00:18:41.620 x being, of course, the coordinate that is being used 00:18:41.620 --> 00:18:43.610 to describe the displacement for the element. 00:18:43.610 --> 00:18:46.230 For two-dimensional elements, here we're talking about plane 00:18:46.230 --> 00:18:47.890 stress, plane strain. 00:18:47.890 --> 00:18:51.360 Axisymmetric analysis, we would have, similarly, these 00:18:51.360 --> 00:18:52.390 assumptions here. 00:18:52.390 --> 00:18:58.060 alphas for the u displacements and betas for the v 00:18:58.060 --> 00:19:00.260 displacements. 00:19:00.260 --> 00:19:05.880 For the plate bending element in which we only discretize 00:19:05.880 --> 00:19:09.080 the w displacement, the transverse displacement, we 00:19:09.080 --> 00:19:12.150 would have this assumption here. 00:19:12.150 --> 00:19:15.130 For three-dimensional solid elements, similarly, we have 00:19:15.130 --> 00:19:16.540 these assumptions. 00:19:16.540 --> 00:19:22.660 Notice that these gammas, alphas, betas are unknowns for 00:19:22.660 --> 00:19:23.250 the element. 00:19:23.250 --> 00:19:26.990 And we will actually relate these coefficients to the 00:19:26.990 --> 00:19:29.040 nodal point displacements of an element. 00:19:31.930 --> 00:19:35.905 If we look, therefore, in general, at the description of 00:19:35.905 --> 00:19:40.330 an element, we really have on the left-hand side here, the 00:19:40.330 --> 00:19:44.350 actual displacements, the continuous displacements in 00:19:44.350 --> 00:19:53.260 the element being given by matrix phi, which contains the 00:19:53.260 --> 00:19:58.020 constant and x, x squared, x cubed terms, and so on. 00:19:58.020 --> 00:20:02.510 And this alpha vector here contains the generalized 00:20:02.510 --> 00:20:06.100 coordinates, the alphas, betas, gammas. 00:20:06.100 --> 00:20:08.910 Typically, therefore, just to focus our attention on a 00:20:08.910 --> 00:20:13.840 particular case, we might have if we say that u of x for a 00:20:13.840 --> 00:20:19.785 truss structure is equal to alpha 0 plus alpha 1x plus 00:20:19.785 --> 00:20:21.960 alpha 2x squared. 00:20:21.960 --> 00:20:27.110 In this particular case, we would have that we can write 00:20:27.110 --> 00:20:32.730 this, of course, as 1 x x squared in matrix form times 00:20:32.730 --> 00:20:36.610 alpha 0 alpha 1, alpha 2. 00:20:36.610 --> 00:20:42.150 And here, I'm talking about this being equal to the phi, 00:20:42.150 --> 00:20:45.090 and this being equal to the vector alpha. 00:20:45.090 --> 00:20:50.360 So just to give you an example for this particular case. 00:20:50.360 --> 00:20:55.970 If we now want to evaluate the alpha values. 00:20:55.970 --> 00:20:58.650 Of course, in two-dimension analysis as I shown earlier to 00:20:58.650 --> 00:21:00.780 you, we have also betas here. 00:21:00.780 --> 00:21:03.220 In three-dimensional analysis, we also would have gammas in 00:21:03.220 --> 00:21:04.640 this vector. 00:21:04.640 --> 00:21:08.580 If you want to evaluate these alpha values, then we can 00:21:08.580 --> 00:21:12.880 simply use this relation here, or that relation here, and 00:21:12.880 --> 00:21:15.880 apply it to the element nodal points. 00:21:15.880 --> 00:21:19.620 To the displacements at the nodal points of the elements. 00:21:19.620 --> 00:21:23.800 If we do that, in other words, we substitute into here, into 00:21:23.800 --> 00:21:27.770 the phi matrix, the actual coordinate of the nodal 00:21:27.770 --> 00:21:30.820 points, and on the left-hand side, of course, the 00:21:30.820 --> 00:21:34.160 displacement at that nodal point, we directly generate 00:21:34.160 --> 00:21:36.030 this relation. 00:21:36.030 --> 00:21:38.870 We can then invert this relation to obtain alpha. 00:21:38.870 --> 00:21:41.420 And here, we have now the generalized coordinates in 00:21:41.420 --> 00:21:46.030 terms of the nodal point displacements. 00:21:46.030 --> 00:21:49.650 This then would really complete the evaluation of the 00:21:49.650 --> 00:21:53.620 continuous displacements u in terms of the nodal points 00:21:53.620 --> 00:21:56.850 displacement u hat. 00:21:56.850 --> 00:22:00.620 If we have laid down this assumption, we can directly 00:22:00.620 --> 00:22:04.420 also obtain the relation for the strains of the elements. 00:22:04.420 --> 00:22:08.165 We have decided what kinds of strains we want to look at 00:22:08.165 --> 00:22:09.870 that we have to include. 00:22:09.870 --> 00:22:13.210 And by proper differentiation of the rows here, we directly 00:22:13.210 --> 00:22:14.620 generate this relation. 00:22:14.620 --> 00:22:16.820 Of course, this is a [UNINTELLIGIBLE] 00:22:16.820 --> 00:22:21.310 relation then for the material of the element. 00:22:21.310 --> 00:22:25.500 And we then directly have tau in terms of alpha. 00:22:25.500 --> 00:22:29.060 But alpha was already given in terms of u hat here. 00:22:29.060 --> 00:22:33.980 In summary, therefore, we have our displacement interpolation 00:22:33.980 --> 00:22:40.470 matrix here, given as phi times a inverse. 00:22:40.470 --> 00:22:44.130 This is obtained by simply substituting from here, 00:22:44.130 --> 00:22:47.790 substitute from here, directly into there. 00:22:47.790 --> 00:22:52.920 And you get that u is equal to H u hat. 00:22:52.920 --> 00:22:56.380 And our H being phi a inverse. 00:22:56.380 --> 00:23:00.280 Once again, the u is the continuous displacement in the 00:23:00.280 --> 00:23:02.240 elements. u hat has nodal point 00:23:02.240 --> 00:23:03.900 displacements in the elements. 00:23:03.900 --> 00:23:08.650 Similarly, if we substitute from here into there and 00:23:08.650 --> 00:23:13.990 recognize that our epsilon strain is equal to B times u 00:23:13.990 --> 00:23:18.570 hat, we directly obtain that B must be equal 00:23:18.570 --> 00:23:21.970 to E times A inverse. 00:23:21.970 --> 00:23:25.500 Well, let us now look at a particular case, a particular 00:23:25.500 --> 00:23:30.350 example to demonstrate how we proceed for a particular case. 00:23:30.350 --> 00:23:34.380 Here I'm looking at the analysis of a cantilever 00:23:34.380 --> 00:23:36.100 subjected to a load. 00:23:36.100 --> 00:23:38.510 And I've idealized that cantilever as an assemblage of 00:23:38.510 --> 00:23:40.560 four elements. 00:23:40.560 --> 00:23:44.040 Each element is a four nodal element. 00:23:44.040 --> 00:23:49.370 So for element 1, we have nodes 1, 2, 4, and 5 of the 00:23:49.370 --> 00:23:52.260 global assemblage. 00:23:52.260 --> 00:23:55.840 Notice that all together we have 9 nodes. 00:23:55.840 --> 00:23:58.700 And once again, 4 elements. 00:23:58.700 --> 00:24:05.160 The cantilever structure is SiN and we believe that it is 00:24:05.160 --> 00:24:09.600 appropriately modeled using a plane stress analysis. 00:24:09.600 --> 00:24:14.160 In which case, the only stresses that are of concern 00:24:14.160 --> 00:24:17.345 are the stresses in this plane, tau yy, 00:24:17.345 --> 00:24:19.280 tau xx, and tau xy. 00:24:19.280 --> 00:24:24.320 So these other stress components here are 0. 00:24:24.320 --> 00:24:28.660 If we look now at a particular element to focus our attention 00:24:28.660 --> 00:24:31.420 on one element, because all these elements 00:24:31.420 --> 00:24:32.530 are really the same. 00:24:32.530 --> 00:24:34.220 Basically, the same. 00:24:34.220 --> 00:24:36.190 Let's look at element 2. 00:24:36.190 --> 00:24:44.030 And we notice that element 2 being a four nodal element is 00:24:44.030 --> 00:24:49.980 shown once more here with the displacements U1, V1 at the 00:24:49.980 --> 00:24:54.110 local nodal point of the element 1. 00:24:54.110 --> 00:24:59.440 V2, U2 being the displacements of the local nodal point 2. 00:24:59.440 --> 00:25:02.920 And here, V3, U3 similarly for nodal point 3. 00:25:02.920 --> 00:25:06.200 And V4, U4 for nodal point 4. 00:25:06.200 --> 00:25:09.350 We have taken this element of the assemblage and what we 00:25:09.350 --> 00:25:15.330 want to really do is derive the displacement and strain 00:25:15.330 --> 00:25:19.790 interpolation matrices for this element corresponding to 00:25:19.790 --> 00:25:25.360 U1, V1; U2, V2; U3, V3; and U4, V4. 00:25:25.360 --> 00:25:30.080 If we have done that, we recognize, of course, that U1 00:25:30.080 --> 00:25:32.130 corresponds to a global displacement. 00:25:32.130 --> 00:25:35.410 Similarly, V1 corresponds to a global placement. 00:25:35.410 --> 00:25:40.015 Then we can directly construct the displacement and strain 00:25:40.015 --> 00:25:45.520 interpolations corresponding to the global displacements. 00:25:45.520 --> 00:25:48.800 But the finite element procedure is really to look 00:25:48.800 --> 00:25:50.480 locally at an element. 00:25:50.480 --> 00:25:51.900 Locally at an element. 00:25:51.900 --> 00:25:54.810 And work in terms of local displacements. 00:25:54.810 --> 00:25:57.400 Later on then, we relate the local displacements to the 00:25:57.400 --> 00:25:58.720 global displacements. 00:25:58.720 --> 00:26:02.270 The advantage of proceeding this way is that we really-- 00:26:02.270 --> 00:26:04.920 in looking at one element, we look at all four 00:26:04.920 --> 00:26:06.170 simultaneously. 00:26:09.510 --> 00:26:16.950 The procedure then is, once more in summary, we want this 00:26:16.950 --> 00:26:21.820 matrix here, the H2 matrix corresponding to element 2. 00:26:21.820 --> 00:26:25.970 The vector U lists all of the displacements, all of the 18 00:26:25.970 --> 00:26:27.170 displacements. 00:26:27.170 --> 00:26:37.365 However, we will look at the element 2 with local nodal 00:26:37.365 --> 00:26:39.980 point displacements, forgetting about the global 00:26:39.980 --> 00:26:41.340 nodal point displacement. 00:26:41.340 --> 00:26:47.600 And our objective is to derive a relationship of u equals H 00:26:47.600 --> 00:26:54.860 times u hat, where u hat lists U1, V1; U2, V2; 00:26:54.860 --> 00:26:58.800 U3, V3; U4 and V4. 00:26:58.800 --> 00:27:05.100 Once we have this H, which is, of course, a 2 by 8 matrix, 00:27:05.100 --> 00:27:08.856 because there are 2 displacements, u and v. The u 00:27:08.856 --> 00:27:12.960 and v displacement for the element. 00:27:12.960 --> 00:27:16.060 And there are 8 entries in u hat. 00:27:16.060 --> 00:27:18.970 Therefore, this H matrix must be a 2 by 8. 00:27:18.970 --> 00:27:23.930 Once we have obtained this H matrix, we can construct 00:27:23.930 --> 00:27:27.730 directly our H2 matrix that we talked about earlier. 00:27:30.470 --> 00:27:36.390 Well, let's proceed then in the way that I discussed in 00:27:36.390 --> 00:27:38.640 general form earlier. 00:27:38.640 --> 00:27:43.240 Since we have a four nodal element, let me just sketch it 00:27:43.240 --> 00:27:44.210 once more here. 00:27:44.210 --> 00:27:49.170 Since we have a four nodal element, and we are talking 00:27:49.170 --> 00:27:55.490 about u and v displacements, we really can only have four 00:27:55.490 --> 00:27:59.140 constants, four generalized coordinates. 00:27:59.140 --> 00:28:02.870 Generalized coordinates corresponding to the u and 00:28:02.870 --> 00:28:05.570 corresponding to the v displacements. 00:28:05.570 --> 00:28:11.450 Remember that the only unknown u displacements are these 4. 00:28:11.450 --> 00:28:18.850 And these 4 u displacements will give us alpha 1, alpha 2, 00:28:18.850 --> 00:28:20.620 alpha 3, and alpha 4. 00:28:20.620 --> 00:28:23.560 The same holds for the v displacements. 00:28:23.560 --> 00:28:28.290 This is a bilinear approximation of the u 00:28:28.290 --> 00:28:31.520 displacements in the element. 00:28:31.520 --> 00:28:35.510 We write this in matrix form as shown here. 00:28:35.510 --> 00:28:40.300 Noting that this capital Phi, notice this is a capital Phi 00:28:40.300 --> 00:28:42.190 because there are these cross bars. 00:28:42.190 --> 00:28:45.660 It's easy to confuse the capital Phi with the lowercase 00:28:45.660 --> 00:28:48.150 phi, which doesn't have the cross bars. 00:28:48.150 --> 00:28:50.730 So this capital Phi was the cross bars here 00:28:50.730 --> 00:28:52.580 is defined in here. 00:28:52.580 --> 00:28:56.290 Where the lowercase phi is simply an array 00:28:56.290 --> 00:29:00.020 of 1, x, y, x, y. 00:29:00.020 --> 00:29:03.190 This relationship here is nothing else than that 00:29:03.190 --> 00:29:06.340 relationship, but written in matrix form. 00:29:06.340 --> 00:29:11.020 The alpha lists all the generalized coordinates. 00:29:11.020 --> 00:29:12.050 The generalize coordinates. 00:29:12.050 --> 00:29:14.850 And this is the reason why we call this the generalized 00:29:14.850 --> 00:29:19.300 coordinate finite element formulation for this 00:29:19.300 --> 00:29:20.940 particular element. 00:29:20.940 --> 00:29:24.520 Let us now define the u hat vector, 00:29:24.520 --> 00:29:26.280 which I refer to earlier. 00:29:26.280 --> 00:29:31.140 This case, the u hat vector involves 4 u nodal point 00:29:31.140 --> 00:29:35.420 displacements and 4 v nodal point displacements. 00:29:35.420 --> 00:29:37.910 Now if we apply-- 00:29:37.910 --> 00:29:40.700 and this is really one important point. 00:29:40.700 --> 00:29:47.410 If we apply this relationship here, which of course must 00:29:47.410 --> 00:29:49.580 hold all over the element. 00:29:49.580 --> 00:29:52.470 In particular, it must also hold for the nodal points of 00:29:52.470 --> 00:29:53.770 the element. 00:29:53.770 --> 00:29:58.220 If we apply this relationship for the nodal points of the 00:29:58.220 --> 00:30:02.730 elements, we can apply it eight times because we have 8 00:30:02.730 --> 00:30:05.430 nodal point displacements. 00:30:05.430 --> 00:30:10.320 We directly generate this relationship here. 00:30:10.320 --> 00:30:14.470 u hat equals A times alpha, where A involve the 00:30:14.470 --> 00:30:17.600 coordinates of the nodal points. 00:30:17.600 --> 00:30:21.670 Hence, we have our H Phi times A inverse the way I have 00:30:21.670 --> 00:30:24.240 derived it earlier. 00:30:24.240 --> 00:30:28.920 So this is a very systematic approach of deriving the 00:30:28.920 --> 00:30:32.190 displacement interpolation matrix. 00:30:32.190 --> 00:30:35.530 It consists of evaluating the A matrix. 00:30:35.530 --> 00:30:39.250 And of course, we assume that that A matrix can be inverted. 00:30:39.250 --> 00:30:42.670 This is, for this element, certainly the case, as long as 00:30:42.670 --> 00:30:47.070 we are talking about the rectangular element or not 00:30:47.070 --> 00:30:48.890 very highly distorted elements. 00:30:48.890 --> 00:30:52.000 We can for the four nodal element, invert this matrix 00:30:52.000 --> 00:30:55.140 and directly get this relationship. 00:30:55.140 --> 00:31:00.130 Let's look now, once more back at the analysis that we wanted 00:31:00.130 --> 00:31:01.260 to perform. 00:31:01.260 --> 00:31:05.450 Remember that we looked at this element here, and we have 00:31:05.450 --> 00:31:10.910 now obtained the H matrix, the displacement interpolation 00:31:10.910 --> 00:31:13.200 matrix for this element. 00:31:13.200 --> 00:31:15.810 Once more, here is the H matrix. 00:31:15.810 --> 00:31:19.270 In fact, this H matrix here, since we have no superscript 00:31:19.270 --> 00:31:23.680 attached to it yet, really holds for every one of the 4 00:31:23.680 --> 00:31:25.290 elements that we are talking about. 00:31:25.290 --> 00:31:32.130 Provided we put into the u hat here the appropriate nodal 00:31:32.130 --> 00:31:34.350 point displacements of the elements. 00:31:34.350 --> 00:31:39.170 We can directly say that this matrix here holds for any one 00:31:39.170 --> 00:31:40.420 of the elements. 00:31:42.400 --> 00:31:45.230 If the elements, if the 4 elements are not completely 00:31:45.230 --> 00:31:48.630 identical, of course, the A inverse will change from 00:31:48.630 --> 00:31:49.880 element to element. 00:31:52.290 --> 00:31:59.620 Having gone through that step and having looked at element 2 00:31:59.620 --> 00:32:05.330 once more, we directly recognize how to establish now 00:32:05.330 --> 00:32:08.600 the H2 matrix. 00:32:08.600 --> 00:32:14.750 Remember our u1 will be related to the global U11. 00:32:14.750 --> 00:32:17.626 Our v1 will be related to the-- 00:32:17.626 --> 00:32:23.050 or will correspond I should rather say, to the global U12. 00:32:23.050 --> 00:32:25.760 Let's look at why this is the case. 00:32:25.760 --> 00:32:29.570 Why did I write down U11 here and U12 there? 00:32:29.570 --> 00:32:33.360 Well, let's look at this point there. 00:32:33.360 --> 00:32:36.710 Which is of course, the local nodal point 1. 00:32:36.710 --> 00:32:41.000 The local nodal point 1, right upperhand corner of element 1, 00:32:41.000 --> 00:32:45.100 corresponds here to nodal point 6. 00:32:45.100 --> 00:32:49.590 Well, the global nodal point displacements of 00:32:49.590 --> 00:32:53.870 nodal point 6 are U11. 00:32:53.870 --> 00:32:55.180 Let me put them in. 00:32:55.180 --> 00:32:56.760 And U12. 00:32:56.760 --> 00:32:58.220 Why is that the case? 00:32:58.220 --> 00:33:00.970 Well, if we start numbering here-- and I'm going to 00:33:00.970 --> 00:33:05.770 include now, I'm going to include now the displacement 00:33:05.770 --> 00:33:08.590 of these nodal points, although we know that they are 00:33:08.590 --> 00:33:11.200 actually 0, that boundary condition we can 00:33:11.200 --> 00:33:13.910 impose later on. 00:33:13.910 --> 00:33:20.035 So if I start numbering here, I would have U1, U2, 3, 4, 5, 00:33:20.035 --> 00:33:24.940 6, 7, 8, 9, 10, 11, 12. 00:33:24.940 --> 00:33:29.990 So the global nodal point displacements are U11, U12 00:33:29.990 --> 00:33:32.270 corresponding to nodal point 9. 00:33:32.270 --> 00:33:36.460 And those are the ones that appear right here 00:33:36.460 --> 00:33:41.400 corresponding to the local U1 and the local V1. 00:33:41.400 --> 00:33:46.290 Well, if we proceed in the same way for all of the other 00:33:46.290 --> 00:33:50.700 nodal points for element 2, and these are the results. 00:33:50.700 --> 00:33:56.810 You can look these up in the textbook figure 4.6, or in the 00:33:56.810 --> 00:33:58.970 accompanying notes. 00:33:58.970 --> 00:34:05.100 We can directly construct the H2 matrix for the element. 00:34:05.100 --> 00:34:13.239 And now the H2 matrix, of course, involves as the vector 00:34:13.239 --> 00:34:19.520 the global nodal point displacements, U1 to U18. 00:34:19.520 --> 00:34:22.690 The H matrix for element 2. 00:34:22.690 --> 00:34:26.239 And if the 4 elements are identical, we will have the 00:34:26.239 --> 00:34:30.300 same H matrix for all the elements. 00:34:30.300 --> 00:34:34.400 The H matrix here looks as shown here. 00:34:34.400 --> 00:34:36.800 Notice that we have a one quarter there. 00:34:36.800 --> 00:34:39.139 It's a little arrow here. 00:34:39.139 --> 00:34:41.750 The bracket should've been there. 00:34:41.750 --> 00:34:42.679 Bracket should have been there. 00:34:42.679 --> 00:34:45.780 So we have a one quarter outside of the bracket. 00:34:45.780 --> 00:34:49.679 And we have as the first entry, this one here. 00:34:49.679 --> 00:34:52.570 This corresponds to U1. 00:34:52.570 --> 00:34:54.600 This is the entry corresponding to U1. 00:34:54.600 --> 00:34:56.330 I left the entries corresponding 00:34:56.330 --> 00:34:59.350 to U2, U3, U4 out. 00:34:59.350 --> 00:35:02.030 This is the entry corresponding to V1. 00:35:02.030 --> 00:35:06.800 And I left out the entries corresponding to V2, V3, V4. 00:35:06.800 --> 00:35:09.450 This is our H matrix here. 00:35:09.450 --> 00:35:13.200 And now we basically want to blow it up to obtain the H2 00:35:13.200 --> 00:35:15.920 matrix corresponding to element 2. 00:35:15.920 --> 00:35:22.020 Well, if we recognize as we just did that U1, the local 00:35:22.020 --> 00:35:29.910 U1, the local V1, correspond to capital U11 and capital 00:35:29.910 --> 00:35:34.350 U12, and I just showed you that that is indeed the case. 00:35:34.350 --> 00:35:38.480 We would simply take the first element here, this element 00:35:38.480 --> 00:35:42.270 here, and put it right there. 00:35:42.270 --> 00:35:45.990 This element here would go right there. 00:35:45.990 --> 00:35:50.410 This element here would go right there. 00:35:50.410 --> 00:35:53.630 And this element here would go right there. 00:35:53.630 --> 00:35:57.410 In fact, H1 5 is 0 as you can see here. 00:35:57.410 --> 00:35:58.910 And H2 1 is 0. 00:35:58.910 --> 00:36:01.970 So H1 1 is this term here. 00:36:01.970 --> 00:36:04.960 And H2 5 is that term here. 00:36:04.960 --> 00:36:08.090 Those are two non-zero entries in the complete H matrix. 00:36:10.590 --> 00:36:13.840 Well, this is then the way how we actually can proceed to 00:36:13.840 --> 00:36:16.330 construct the H2 matrix. 00:36:16.330 --> 00:36:21.510 I pointed out earlier in lecture three already that we 00:36:21.510 --> 00:36:24.760 do not perform this step computationally. 00:36:24.760 --> 00:36:28.040 In fact, what we do computationally is we stay 00:36:28.040 --> 00:36:29.740 with this H matrix. 00:36:29.740 --> 00:36:33.220 We calculate a compacted stiffness matrix, a compacted 00:36:33.220 --> 00:36:36.950 load vector, which the stiffness matrix will be of 00:36:36.950 --> 00:36:38.680 order 8 by 8. 00:36:38.680 --> 00:36:44.390 And then we assemble these element matrices with 00:36:44.390 --> 00:36:48.040 identification arrays and connectivity arrays into the 00:36:48.040 --> 00:36:50.540 global structural stiffness matrix, into 00:36:50.540 --> 00:36:53.690 the global load vectors. 00:36:53.690 --> 00:36:57.650 So we really don't perform this step of 00:36:57.650 --> 00:37:00.370 blowing up the H matrix. 00:37:00.370 --> 00:37:03.660 And I will show you in the next lecture how we actually 00:37:03.660 --> 00:37:08.130 perform the computations of assembling element matrices. 00:37:08.130 --> 00:37:12.420 However, for theoretical purposes, it is neat if you 00:37:12.420 --> 00:37:17.030 understand how we can construct this H2 matrix from 00:37:17.030 --> 00:37:18.570 this H matrix. 00:37:18.570 --> 00:37:23.460 And of course, similarly, we could construct H1, H3, and H4 00:37:23.460 --> 00:37:26.480 from this basic H matrix. 00:37:26.480 --> 00:37:33.180 Once we have done that, we can directly write that k being 00:37:33.180 --> 00:37:36.060 equal to the sum of the km. 00:37:36.060 --> 00:37:39.670 In other words, the global structure stiffness matrix is 00:37:39.670 --> 00:37:43.400 equal to the sum of the element matrices, stiffness 00:37:43.400 --> 00:37:51.490 matrices, where each of these km's has the same order as the 00:37:51.490 --> 00:37:52.960 structure stiffness matrix k. 00:37:56.690 --> 00:38:02.810 Let us now look briefly at the development of the strain 00:38:02.810 --> 00:38:04.390 displacement matrix. 00:38:04.390 --> 00:38:08.430 In plane stress conditions, we have three entries. 00:38:08.430 --> 00:38:10.590 The strains are given as shown here. 00:38:10.590 --> 00:38:13.460 Elementary strengths materials tells us that 00:38:13.460 --> 00:38:15.210 these are the strains. 00:38:15.210 --> 00:38:21.480 And I showed earlier that the B matrix is constructed by 00:38:21.480 --> 00:38:23.950 multiplying the E times the A inverse. 00:38:23.950 --> 00:38:26.950 The A inverse was already used in the calculation of the 00:38:26.950 --> 00:38:30.060 displacement interpolation matrix. 00:38:30.060 --> 00:38:36.630 The E matrix is obtained by appropriately differentiating 00:38:36.630 --> 00:38:43.110 the entries in the displacement interpolations. 00:38:43.110 --> 00:38:50.970 And this differentiation is performed on the phi matrix 00:38:50.970 --> 00:38:52.240 right here. 00:38:52.240 --> 00:38:55.750 Or rather, on these entries here. 00:38:55.750 --> 00:38:59.650 By simply differentiating these entries here, we would 00:38:59.650 --> 00:39:06.390 obtain the strains and writing those in matrix form, we 00:39:06.390 --> 00:39:09.940 obtain the E matrix here. 00:39:09.940 --> 00:39:16.000 Once we have the B matrix, which in this particular case 00:39:16.000 --> 00:39:22.390 is 3 by 8 matrix for a typical element, we can, of course, 00:39:22.390 --> 00:39:29.080 construct the B2 matrix for element 2 just in the same way 00:39:29.080 --> 00:39:31.000 as the H2 matrix. 00:39:31.000 --> 00:39:33.190 Of course, we would again, blow up the B 00:39:33.190 --> 00:39:35.205 matrix from a 3 to 8. 00:39:35.205 --> 00:39:40.400 3 by 8 I should rather say, to a 3 by 18 now because we have 00:39:40.400 --> 00:39:42.360 actually 18 degrees of freedom in the 00:39:42.360 --> 00:39:44.910 complete structural model. 00:39:44.910 --> 00:39:47.190 This then completes what I wanted to say about the 00:39:47.190 --> 00:39:49.830 construction of the displacement and 00:39:49.830 --> 00:39:51.960 strain-displacement interpolation matrix for the 00:39:51.960 --> 00:39:55.470 element, the H2 and B2 matrices. 00:39:55.470 --> 00:39:59.380 However, let me just spend a few moments on the 00:39:59.380 --> 00:40:02.680 construction of the surface displacement 00:40:02.680 --> 00:40:03.880 interpolation matrix. 00:40:03.880 --> 00:40:06.980 In other words, the interpolation of the 00:40:06.980 --> 00:40:10.270 displacements along the surface of an element. 00:40:10.270 --> 00:40:14.860 Here, assume for example, that this is a surface of the 00:40:14.860 --> 00:40:16.890 elements that we are concerned with. 00:40:16.890 --> 00:40:19.890 And that there would be some loading applied to that 00:40:19.890 --> 00:40:21.510 surface of the element. 00:40:21.510 --> 00:40:26.080 So that we need, in other words, the HSM matrix the way 00:40:26.080 --> 00:40:28.650 I have been pointing it out earlier. 00:40:28.650 --> 00:40:33.300 What we would do to obtain this matrix here, is really 00:40:33.300 --> 00:40:38.330 simply use our earlier results for the HS-- 00:40:38.330 --> 00:40:41.890 sorry, for the HM matrix. 00:40:41.890 --> 00:40:46.830 And here we have the H matrix and the H2 00:40:46.830 --> 00:40:49.380 matrix for element 2. 00:40:49.380 --> 00:40:52.570 What we would do is we would use this matrix, or this one 00:40:52.570 --> 00:40:55.820 of course, this matrix is contained in this one. 00:40:55.820 --> 00:41:00.830 And simply, substitute the particular coordinates along 00:41:00.830 --> 00:41:01.790 the surface. 00:41:01.790 --> 00:41:04.440 Now, going back once more to the element under 00:41:04.440 --> 00:41:05.950 consideration. 00:41:05.950 --> 00:41:10.380 If we are looking at this surface, then y is constant 00:41:10.380 --> 00:41:13.700 and equal to this distance here. 00:41:13.700 --> 00:41:19.060 So knowing y, we would simply substitute that value of y 00:41:19.060 --> 00:41:27.480 into this function here, and thus obtain the HS matrix. 00:41:27.480 --> 00:41:31.480 So, in general, therefore the HS matrix is simply obtained 00:41:31.480 --> 00:41:36.700 from the H matrix, or the HS2 matrix here would simply be 00:41:36.700 --> 00:41:42.050 obtained from the H2 matrix by substituting the coordinates 00:41:42.050 --> 00:41:46.470 along the line, or along the surface that we are 00:41:46.470 --> 00:41:47.720 considering. 00:41:49.410 --> 00:41:53.860 Let us now to look at the overall process of the finite 00:41:53.860 --> 00:41:55.820 element solution. 00:41:55.820 --> 00:41:59.990 Here I have summarized what we all are concerned about. 00:41:59.990 --> 00:42:02.800 Of course we have the actual physical problem that we want 00:42:02.800 --> 00:42:04.350 to analyze. 00:42:04.350 --> 00:42:08.450 We have a geometric domain that we have to discretize. 00:42:08.450 --> 00:42:11.270 A material that has to be represented, a loading and the 00:42:11.270 --> 00:42:12.750 boundary conditions. 00:42:12.750 --> 00:42:16.000 All of those have to be represented. 00:42:16.000 --> 00:42:20.150 In the mechanic idealization, the idealization step that is 00:42:20.150 --> 00:42:23.870 a very important step that I referred to earlier, we would 00:42:23.870 --> 00:42:26.890 idealize the kinematics as being a truss, plane stress, 00:42:26.890 --> 00:42:29.930 three-dimensional, Kirchhoff plate, kinematics. 00:42:29.930 --> 00:42:34.000 The material isotopic elastic, the loading, and the boundary 00:42:34.000 --> 00:42:34.910 conditions. 00:42:34.910 --> 00:42:37.560 And this mechanical idealization really gives us a 00:42:37.560 --> 00:42:41.180 differential equation or equilibrium for 00:42:41.180 --> 00:42:45.160 one-dimensional stress situation, a bar. 00:42:45.160 --> 00:42:49.190 We will have an equation such as this one. 00:42:49.190 --> 00:42:52.420 In the finite element solution then, we operate on this 00:42:52.420 --> 00:42:53.810 equation basically. 00:42:53.810 --> 00:42:57.840 We want to solve the governing differential equation of the 00:42:57.840 --> 00:43:00.920 mechanical idealization. 00:43:00.920 --> 00:43:05.340 The errors that we are therefore, concerned with are 00:43:05.340 --> 00:43:08.580 discretization errors by the use of the finite element 00:43:08.580 --> 00:43:09.840 interpolations. 00:43:09.840 --> 00:43:13.070 And then, a number of other errors. 00:43:13.070 --> 00:43:14.550 Numerical integration in space. 00:43:14.550 --> 00:43:16.990 If we use numerical integration, there would be 00:43:16.990 --> 00:43:18.530 other errors introduced. 00:43:18.530 --> 00:43:21.270 In the evaluation of the constitutive relations, again, 00:43:21.270 --> 00:43:22.800 errors would be introduced. 00:43:22.800 --> 00:43:25.380 The solution of the dynamic equilibrium equations, by 00:43:25.380 --> 00:43:26.810 [UNINTELLIGIBLE] time integration, mode 00:43:26.810 --> 00:43:29.890 superposition further errors would be introduced. 00:43:29.890 --> 00:43:32.780 If you solved the governing equilibrium equations, ku 00:43:32.780 --> 00:43:36.180 equals r, by iteration there would be further errors 00:43:36.180 --> 00:43:37.120 introduced. 00:43:37.120 --> 00:43:40.130 And of course, finally there are round-off errors on the 00:43:40.130 --> 00:43:41.270 digital computer. 00:43:41.270 --> 00:43:46.720 What I want to do now in discussing convergence, I want 00:43:46.720 --> 00:43:50.260 to say that all of these errors are negligible. 00:43:50.260 --> 00:43:53.770 In other words, we are basically evaluating the 00:43:53.770 --> 00:43:56.380 integrations in space exactly. 00:43:56.380 --> 00:43:59.600 There is no round-off because we have an infinite precision 00:43:59.600 --> 00:44:01.200 machine, and so on. 00:44:01.200 --> 00:44:04.760 So all the errors that we are looking at now are the 00:44:04.760 --> 00:44:06.500 discretization errors, 00:44:06.500 --> 00:44:08.860 And if we look only at the discretization 00:44:08.860 --> 00:44:12.210 errors, we can say some-- 00:44:12.210 --> 00:44:15.540 make some very fundamental remarks about the convergence 00:44:15.540 --> 00:44:18.310 of the finite elements scheme. 00:44:18.310 --> 00:44:22.330 I've summarized here on the blackboard some of these very 00:44:22.330 --> 00:44:24.200 important concepts. 00:44:24.200 --> 00:44:26.300 When we talk about convergence, let us assume 00:44:26.300 --> 00:44:30.560 first now that we are having a compatible element layout. 00:44:30.560 --> 00:44:33.270 In other words, we're using a compatible element layout. 00:44:33.270 --> 00:44:37.480 And then, we know that we have monotonic convergence to the 00:44:37.480 --> 00:44:40.410 solution of the problem-governing differential 00:44:40.410 --> 00:44:45.460 equations provided the elements that we are using 00:44:45.460 --> 00:44:46.420 contain: 00:44:46.420 --> 00:44:49.800 Firstly, all required rigid body modes. 00:44:49.800 --> 00:44:52.950 This means, for example, that for in a plane stress 00:44:52.950 --> 00:44:56.850 analysis, the element must be able to undergo three rigid 00:44:56.850 --> 00:45:00.230 body modes, two translations, horizontally and vertically, 00:45:00.230 --> 00:45:03.040 and a rigid body rotation. 00:45:03.040 --> 00:45:07.100 And the element must be able to represent the required 00:45:07.100 --> 00:45:08.790 constant strain states. 00:45:08.790 --> 00:45:12.220 In a plane stress analysis, three constant strain states. 00:45:12.220 --> 00:45:17.170 Pulling this way, pulling that way, and the constant shear. 00:45:17.170 --> 00:45:20.600 If these two conditions are satisfied, we have monotonic 00:45:20.600 --> 00:45:23.790 convergence in a compatible element layout. 00:45:23.790 --> 00:45:26.380 What do we mean by a compatible element layout? 00:45:26.380 --> 00:45:30.930 Well, here I have show a small picture showing two elements, 00:45:30.930 --> 00:45:32.470 a white and an orange element. 00:45:32.470 --> 00:45:37.190 And these elements are compatible because we have 00:45:37.190 --> 00:45:41.050 three nodes along this line for each of the elements. 00:45:41.050 --> 00:45:44.440 So the white element displacements basically vary 00:45:44.440 --> 00:45:48.040 parabolically along this line, and so do the orange element 00:45:48.040 --> 00:45:49.820 displacements. 00:45:49.820 --> 00:45:53.460 No gap can develop between these two lines. 00:45:53.460 --> 00:45:55.950 Here we have an incompatible element layout. 00:45:55.950 --> 00:45:59.940 Two nodes that describe only a linear variation 00:45:59.940 --> 00:46:02.370 for the white element. 00:46:02.370 --> 00:46:06.760 But three nodes for the orange element describing a parabolic 00:46:06.760 --> 00:46:08.010 variation in displacement. 00:46:08.010 --> 00:46:11.660 Therefore a gap can open up and this is an incompatible 00:46:11.660 --> 00:46:13.740 element layout. 00:46:13.740 --> 00:46:17.090 Well, let us assume then that we have a compatible element 00:46:17.090 --> 00:46:18.940 layout fist of all. 00:46:18.940 --> 00:46:22.260 Then, typically for an analysis such as this one, a 00:46:22.260 --> 00:46:26.230 simple cantilever subjected to a tip load, the convergence 00:46:26.230 --> 00:46:29.140 that we would observe would be the following. 00:46:29.140 --> 00:46:32.240 As we increase the number of elements that idealize the 00:46:32.240 --> 00:46:36.490 cantilever, the displacement measured here would 00:46:36.490 --> 00:46:41.700 monotonically approach from below the exact displacement. 00:46:41.700 --> 00:46:44.130 What do we mean by exact displacement? 00:46:44.130 --> 00:46:46.220 Notice I put it into quotes. 00:46:46.220 --> 00:46:48.940 Well, the exact displacement that I'm talking about here is 00:46:48.940 --> 00:46:52.190 the tip displacements that we would calculate from the 00:46:52.190 --> 00:46:54.580 problem-governing differential equation. 00:46:57.670 --> 00:46:59.870 Of course, that problem-governing differential 00:46:59.870 --> 00:47:03.940 equation also corresponds to a mechanical idealization of the 00:47:03.940 --> 00:47:06.840 actual physical situation. 00:47:06.840 --> 00:47:09.050 And depending on which problem-governing differential 00:47:09.050 --> 00:47:13.230 equation we choose, we would have a different dashed line. 00:47:13.230 --> 00:47:16.850 But assuming now that we have made up our mind which one to 00:47:16.850 --> 00:47:19.945 choose, which mechanical idealization to choose, and we 00:47:19.945 --> 00:47:23.760 are choosing a corresponding finite element idealization, 00:47:23.760 --> 00:47:26.890 this would be the convergence that we would observe. 00:47:26.890 --> 00:47:29.630 And it would be monotonically from below. 00:47:29.630 --> 00:47:31.560 In other words, our finite element model, 00:47:31.560 --> 00:47:33.210 actually is too stiff. 00:47:33.210 --> 00:47:36.860 It is stiffer than the actual physical structure. 00:47:36.860 --> 00:47:41.300 Well, why I think we can intuitively understand that 00:47:41.300 --> 00:47:44.190 quite well if we just recognize that we are 00:47:44.190 --> 00:47:50.040 constraining the displacements within each element to only be 00:47:50.040 --> 00:47:57.130 able to basically, take patterns on that are contained 00:47:57.130 --> 00:47:57.840 in the element. 00:47:57.840 --> 00:48:01.155 In other words, for the four nodal element here, we really 00:48:01.155 --> 00:48:05.110 have for the U displacements and for the V displacements, 00:48:05.110 --> 00:48:08.560 only 4 displacement patterns each that 00:48:08.560 --> 00:48:09.840 the element can undergo. 00:48:09.840 --> 00:48:13.660 So we are constraining the exact displacements to be 00:48:13.660 --> 00:48:18.820 really, approximated by these displacement patterns. 00:48:18.820 --> 00:48:21.940 And that makes the actual finite element model stiffer 00:48:21.940 --> 00:48:25.350 than the actual physical structure. 00:48:25.350 --> 00:48:28.370 If we have an incompatible layout, and I showed earlier 00:48:28.370 --> 00:48:32.190 what we mean by that, then, in addition, 00:48:32.190 --> 00:48:34.830 every patch of elements-- 00:48:34.830 --> 00:48:38.490 by patch of elements we mean small assemblage of elements-- 00:48:38.490 --> 00:48:42.010 must be able to represent the constant strain states. 00:48:42.010 --> 00:48:44.150 It's not just that each element must be able to 00:48:44.150 --> 00:48:46.090 represent the constant strain state, but 00:48:46.090 --> 00:48:47.990 each patch of elements. 00:48:47.990 --> 00:48:53.250 Also, then if this holds true, then we have convergence. 00:48:53.250 --> 00:48:55.640 But non-monotonic convergence. 00:48:55.640 --> 00:48:58.940 By non-monotonic convergence, I mean the following. 00:48:58.940 --> 00:49:02.140 We might get this result for a certain set of elements. 00:49:02.140 --> 00:49:04.760 And as we increase the number of elements, we are 00:49:04.760 --> 00:49:09.450 oscillating about the correct result, the exact result in 00:49:09.450 --> 00:49:12.570 quotes again, and finally we converge. 00:49:12.570 --> 00:49:16.270 This is non-monotonic convergence because we can 00:49:16.270 --> 00:49:20.700 have solutions lying on either side of the exact result. 00:49:20.700 --> 00:49:31.490 Well, let me now show to you some of the facts that I just 00:49:31.490 --> 00:49:33.900 talked about on this view graph. 00:49:33.900 --> 00:49:35.480 What do these mean? 00:49:35.480 --> 00:49:39.770 Well, I said already that each element in every case must be 00:49:39.770 --> 00:49:43.480 able to represent the rigid body modes. 00:49:43.480 --> 00:49:46.910 Here for a plane stress analysis of a cantilever beam 00:49:46.910 --> 00:49:49.790 shown here, we have 3 rigid body modes. 00:49:49.790 --> 00:49:53.430 And as I mentioned earlier briefly already, the element 00:49:53.430 --> 00:49:57.260 must be able to go over, go up as a rigid body and rotate as 00:49:57.260 --> 00:49:59.370 a rigid body. 00:49:59.370 --> 00:50:00.610 Why is this the case? 00:50:00.610 --> 00:50:03.060 Well, if we look at an analysis of the cantilever 00:50:03.060 --> 00:50:06.280 where we have not a tip load, but a load say, at this end 00:50:06.280 --> 00:50:09.650 here, we know that the bending moment looks this way. 00:50:09.650 --> 00:50:16.700 Hence, these elements over here must be able to freely 00:50:16.700 --> 00:50:18.790 rotate and displace. 00:50:18.790 --> 00:50:22.940 In other words, we know that the exact condition, the exact 00:50:22.940 --> 00:50:25.630 physical condition, there should be no stress here. 00:50:25.630 --> 00:50:29.450 The stress should be 0 in this element. 00:50:29.450 --> 00:50:33.600 And this means that the tip element here must be able to 00:50:33.600 --> 00:50:35.780 translate freely as a rigid body and 00:50:35.780 --> 00:50:37.340 rotate as a rigid body. 00:50:37.340 --> 00:50:39.040 If this were not the case, then we would 00:50:39.040 --> 00:50:40.600 get stresses here. 00:50:40.600 --> 00:50:44.640 And of course, our analysis or finite element model can never 00:50:44.640 --> 00:50:48.250 predict the actual physical conditions accurately or 00:50:48.250 --> 00:50:52.450 closely, even if we take an infinite number of elements. 00:50:52.450 --> 00:50:56.860 So this somehow explains to you intuitively why the rigid 00:50:56.860 --> 00:50:59.330 body more criteria must be satisfied. 00:50:59.330 --> 00:51:02.260 How about the constant strain state criteria? 00:51:02.260 --> 00:51:06.940 Well, if we assume that we take more and more elements-- 00:51:06.940 --> 00:51:10.820 and let me draw in here such elements. 00:51:10.820 --> 00:51:14.020 For example, now we have gone from a certain number of 00:51:14.020 --> 00:51:17.940 elements to a larger number of elements. 00:51:17.940 --> 00:51:21.400 These elements, as we keep on refining the mesh, would 00:51:21.400 --> 00:51:23.290 become very small, and would go down. 00:51:23.290 --> 00:51:25.810 In fact, each element would go down to a point. 00:51:25.810 --> 00:51:29.860 But at a point we only have one constant stress basically. 00:51:29.860 --> 00:51:33.310 So the element must be able in the limit, to represent the 00:51:33.310 --> 00:51:36.240 constant stress at a point. 00:51:36.240 --> 00:51:40.270 And therefore, we have the constant stress criteria also. 00:51:40.270 --> 00:51:44.330 One way of finding out whether an element satisfies these 00:51:44.330 --> 00:51:48.180 criteria, the constant stress criteria and the rigid body 00:51:48.180 --> 00:51:50.580 mode criteria is to calculate its eigenvalues. 00:51:50.580 --> 00:51:52.260 Here, we have taken ones. 00:51:52.260 --> 00:51:54.770 A four node plane stress element and calculate the 00:51:54.770 --> 00:51:55.560 eigenvalues. 00:51:55.560 --> 00:52:00.790 Notice the lowest three eigenvalues are 0, 00:52:00.790 --> 00:52:03.710 representing the rigid body modes. 00:52:03.710 --> 00:52:07.600 And then, we have a flexural mode here, the fourth 00:52:07.600 --> 00:52:09.310 eigenvalue. 00:52:09.310 --> 00:52:12.170 The fifth eigenvalue is again the flexural mode. 00:52:12.170 --> 00:52:14.610 And here is the constant sharing mode. 00:52:14.610 --> 00:52:17.500 And here are the two constant, one 00:52:17.500 --> 00:52:19.350 compression and tension modes. 00:52:19.350 --> 00:52:23.310 So these are the constant strain states that the element 00:52:23.310 --> 00:52:24.800 can represent therefore. 00:52:24.800 --> 00:52:29.800 And earlier I showed you the three rigid body modes. 00:52:29.800 --> 00:52:34.660 So surely, this element therefore satisfies both of 00:52:34.660 --> 00:52:36.890 the criteria. 00:52:36.890 --> 00:52:43.770 Let's look now at what happens when we don't have a 00:52:43.770 --> 00:52:46.220 compatible mesh. 00:52:46.220 --> 00:52:49.600 We have an additional condition that every patch of 00:52:49.600 --> 00:52:52.810 elements must also be able to represent the 00:52:52.810 --> 00:52:55.030 constant strain states. 00:52:55.030 --> 00:52:58.000 Only then we will have convergence. 00:52:58.000 --> 00:53:00.820 But we will not necessarily have monotonic convergence. 00:53:00.820 --> 00:53:03.870 We can only talk about non-monotonic convergence. 00:53:03.870 --> 00:53:08.150 Well, here I've taken a patch of elements. 00:53:08.150 --> 00:53:12.640 Each element by itself satisfies the constant strain 00:53:12.640 --> 00:53:15.890 criterion and the rigid body more criterion. 00:53:15.890 --> 00:53:21.620 In this case here, we use a compatible element layout. 00:53:21.620 --> 00:53:26.440 Compatible because along each side, the adjacent elements 00:53:26.440 --> 00:53:29.210 shares the nodes. 00:53:29.210 --> 00:53:32.410 And we are subjecting this patch of elements to a total 00:53:32.410 --> 00:53:36.180 stress here of 100 Newton per square centimeters. 00:53:36.180 --> 00:53:38.880 Of course, the stress in each of these elements would be 00:53:38.880 --> 00:53:42.880 simply a stress this direction of 1,000. 00:53:42.880 --> 00:53:45.690 Excuse me, 1,000 not 100. 00:53:45.690 --> 00:53:49.570 The other stress, the vertical stress and the shear stress is 00:53:49.570 --> 00:53:52.240 0 for this patch of elements. 00:53:52.240 --> 00:53:54.840 Let us now perform the following experiment. 00:53:54.840 --> 00:54:00.530 Let us say that we are assigning two nodes, two 00:54:00.530 --> 00:54:05.570 different nodes to element 4 and 5 at this location. 00:54:05.570 --> 00:54:08.570 Here they share the same node. 00:54:08.570 --> 00:54:10.890 Let us do the same at this end. 00:54:10.890 --> 00:54:14.090 And let us say that node 17 only belongs to 00:54:14.090 --> 00:54:16.160 this element 4. 00:54:16.160 --> 00:54:18.730 Node 18 belongs only to element 6. 00:54:18.730 --> 00:54:22.170 Node 20 and 19 belong to element 5. 00:54:22.170 --> 00:54:28.130 In other words, node 17 can basically gives this 00:54:28.130 --> 00:54:29.080 displacement pattern. 00:54:29.080 --> 00:54:30.950 And 19 gives this displacement pattern. 00:54:30.950 --> 00:54:33.880 An incompatibility can reside here. 00:54:33.880 --> 00:54:35.940 And similarly, along this line. 00:54:35.940 --> 00:54:41.910 If we subject this patch of elements to the stress state 00:54:41.910 --> 00:54:45.920 of the external forces that we used here too, we would find 00:54:45.920 --> 00:54:50.230 that the stresses at the points A, B, C, D, E shown 00:54:50.230 --> 00:54:53.580 along here, which here were, of course, exactly equal to 00:54:53.580 --> 00:55:01.530 1,000 for this stress tau yy and 0 in this direction. 00:55:01.530 --> 00:55:02.820 And the sheer stress was 0. 00:55:02.820 --> 00:55:07.480 If you look at the tau yy at A, B, C, D, E now the 00:55:07.480 --> 00:55:11.540 incompatible element mesh layout, we would find the 00:55:11.540 --> 00:55:13.450 following stress result. 00:55:13.450 --> 00:55:16.380 Notice that the stresses now are not 1,000. 00:55:16.380 --> 00:55:18.930 In fact, they vary drastically from 1,000. 00:55:18.930 --> 00:55:22.300 359 instead of 1,000. 00:55:22.300 --> 00:55:27.480 This means that this patch of elements with the 00:55:27.480 --> 00:55:32.820 incompatibility here, does not represent exactly the constant 00:55:32.820 --> 00:55:36.560 stress state that it should represent because we are 00:55:36.560 --> 00:55:40.080 subjecting it to a constant stress state. 00:55:40.080 --> 00:55:43.980 Therefore, this patch of element does not satisfy the 00:55:43.980 --> 00:55:47.500 requirement for convergence. 00:55:47.500 --> 00:55:51.450 Not monotonic convergence and not non-monotonic convergence. 00:55:51.450 --> 00:55:57.350 It just should not be used in an actual practical analysis. 00:55:57.350 --> 00:56:01.610 This completes what I wanted to say then in this lecture. 00:56:01.610 --> 00:56:03.310 Thank you very much for your attention.