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. 00:00:21.520 --> 00:00:23.260 PROFESSOR: Ladies and gentlemen, welcome to 00:00:23.260 --> 00:00:25.220 Lecture Number 10. 00:00:25.220 --> 00:00:27.600 In this lecture, I would like to discuss a solution of 00:00:27.600 --> 00:00:29.520 dynamic equilibrium equations. 00:00:29.520 --> 00:00:32.240 And in particular, I would like to talk about the direct 00:00:32.240 --> 00:00:35.860 integration solution of the equilibrium equations in 00:00:35.860 --> 00:00:37.790 dynamic analysis. 00:00:37.790 --> 00:00:39.980 These, of course, are the equilibrium equations that we 00:00:39.980 --> 00:00:42.100 derived already earlier. 00:00:42.100 --> 00:00:45.270 M as a mass matrix, C as a damping matrix, K as a 00:00:45.270 --> 00:00:48.490 stiffness matrix, U as a displacement vector, and here 00:00:48.490 --> 00:00:50.490 we have the velocities and the 00:00:50.490 --> 00:00:52.535 accelerations at the null points. 00:00:52.535 --> 00:00:56.440 R, of course, being the load vector and this load vector is 00:00:56.440 --> 00:00:59.320 now time dependent. 00:00:59.320 --> 00:01:02.200 I also should draw your attention on the fact that I 00:01:02.200 --> 00:01:04.170 would like to talk about direct 00:01:04.170 --> 00:01:06.500 integration of these equations. 00:01:06.500 --> 00:01:10.140 We will talk about the modes of opposition procedures in 00:01:10.140 --> 00:01:12.100 the next lecture. 00:01:12.100 --> 00:01:16.010 By direct integration, we mean that we integrate these 00:01:16.010 --> 00:01:18.970 equations directly without-- 00:01:18.970 --> 00:01:20.120 that is, without-- 00:01:20.120 --> 00:01:23.280 a transformation of these equations into a different 00:01:23.280 --> 00:01:25.740 form prior to the integration. 00:01:25.740 --> 00:01:28.090 In modes of opposition analysis, now you will see in 00:01:28.090 --> 00:01:30.620 the next lecture, we're actually transforming these 00:01:30.620 --> 00:01:35.040 equations first, and then we integrate the response. 00:01:35.040 --> 00:01:38.150 So the procedures that I'd like to talk about are 00:01:38.150 --> 00:01:42.410 explicit, implicit integration procedures that directly 00:01:42.410 --> 00:01:45.980 operate on these equations without a transformation and 00:01:45.980 --> 00:01:48.670 integrate, solve these equations. 00:01:48.670 --> 00:01:51.670 I also would like to talk about some computational 00:01:51.670 --> 00:01:55.530 considerations with regard to these explicit and implicit 00:01:55.530 --> 00:01:58.750 integration procedures, the number of operations involved, 00:01:58.750 --> 00:02:00.810 the costs involved in performing these 00:02:00.810 --> 00:02:02.820 integrations and so on. 00:02:02.820 --> 00:02:06.450 And then there is an important consideration, namely the 00:02:06.450 --> 00:02:09.900 selection of the solution time step, delta-t. 00:02:09.900 --> 00:02:12.320 We will see that when we integrate the dynamic 00:02:12.320 --> 00:02:15.050 equilibrium equations, we have to select the time step 00:02:15.050 --> 00:02:19.000 delta-t with which we perform the integration. 00:02:19.000 --> 00:02:23.560 And this time step has to be selected in order to preserve 00:02:23.560 --> 00:02:26.990 stability of the integration and to preserve of course or 00:02:26.990 --> 00:02:29.960 to obtain accuracy in the integration. 00:02:29.960 --> 00:02:32.870 Finally, I'd like to talk about some modeling 00:02:32.870 --> 00:02:36.640 considerations pertaining to the solution of the dynamic 00:02:36.640 --> 00:02:38.160 equilibrium equations. 00:02:38.160 --> 00:02:44.600 Well, here on this view graph, I'm showing once again the 00:02:44.600 --> 00:02:47.570 dynamically equilibrium equations here, noticing once 00:02:47.570 --> 00:02:51.820 more that R, the load vector, is now a function of time. 00:02:51.820 --> 00:02:55.080 The same equations here have been rewritten once here, 00:02:55.080 --> 00:02:59.750 where FI denotes inertia forces, now time 00:02:59.750 --> 00:03:00.930 dependent, of course. 00:03:00.930 --> 00:03:03.740 The mass matrix is constant, but the accelerations of 00:03:03.740 --> 00:03:05.750 course depend on time. 00:03:05.750 --> 00:03:10.770 Then this part here, which are the damping forces, has been 00:03:10.770 --> 00:03:13.820 expressed here as FD a function of time again. 00:03:13.820 --> 00:03:17.480 The C matrix is constant with time, but the velocities, of 00:03:17.480 --> 00:03:18.580 course, vary with time. 00:03:18.580 --> 00:03:20.340 So here we have damping forces. 00:03:20.340 --> 00:03:22.700 It's a nodes of the finite element system. 00:03:22.700 --> 00:03:27.020 And these are the elastic forces, KU, which are really 00:03:27.020 --> 00:03:30.110 due to the internal element stresses. 00:03:30.110 --> 00:03:34.370 And on the right hand side, we have the external forces. 00:03:34.370 --> 00:03:39.070 So what we're saying is that we want to solve this force 00:03:39.070 --> 00:03:42.810 equilibrium over all times t. 00:03:42.810 --> 00:03:45.910 The inertia forces plus the damping forces plus the 00:03:45.910 --> 00:03:51.180 elastic forces, it's the nodes of the finite element system, 00:03:51.180 --> 00:03:55.340 of course, calculated in the virtual work sense, the way 00:03:55.340 --> 00:03:58.040 we've been talking about in the earlier lectures. 00:03:58.040 --> 00:04:00.750 The sum of these forces here must be equal to the 00:04:00.750 --> 00:04:03.890 externally applied nodal point logs. 00:04:03.890 --> 00:04:07.940 Well, the procedure that we will be 00:04:07.940 --> 00:04:10.550 following is the following. 00:04:10.550 --> 00:04:16.019 We will consider time steps delta-t a part at which we 00:04:16.019 --> 00:04:19.180 want to calculate the response of the system. 00:04:19.180 --> 00:04:20.810 In order to do so, of course, we have to 00:04:20.810 --> 00:04:22.710 define our loads first. 00:04:22.710 --> 00:04:26.970 And here, I have schematic shown once the description of 00:04:26.970 --> 00:04:31.770 the load Ri, that is, the load at degree of freedom i as a 00:04:31.770 --> 00:04:33.510 function of time t. 00:04:33.510 --> 00:04:39.490 Notice that we are having here this distribution of that load 00:04:39.490 --> 00:04:43.990 component as a function of time. 00:04:43.990 --> 00:04:46.120 The distribution is quite simple. 00:04:46.120 --> 00:04:48.690 In a computer program, of course, all we would have to 00:04:48.690 --> 00:04:52.670 put in is this point, this intensity here, that load 00:04:52.670 --> 00:04:54.930 intensity, at a specified time. 00:04:54.930 --> 00:04:58.620 That load intensity, at a specified time, and that load 00:04:58.620 --> 00:05:00.780 intensity at a specified time. 00:05:00.780 --> 00:05:04.570 In between, these intensities, the computer program would 00:05:04.570 --> 00:05:08.880 simply interpolate linearly. 00:05:08.880 --> 00:05:12.370 This is the load at the component i, displacement 00:05:12.370 --> 00:05:13.370 component i. 00:05:13.370 --> 00:05:16.160 This is the load at displacement component j. 00:05:16.160 --> 00:05:18.860 Notice here, of course, we would have in general a 00:05:18.860 --> 00:05:22.470 different curve at degree of freedom j, then the curve is 00:05:22.470 --> 00:05:25.570 used at degree of freedom i. 00:05:25.570 --> 00:05:31.140 I show here time steps, discrete time steps, delta-t1, 00:05:31.140 --> 00:05:33.880 delta-t2, delta-t3. 00:05:33.880 --> 00:05:37.260 These of course apply for the complete system, for the i 00:05:37.260 --> 00:05:41.250 component and the j component of displacements or loads, and 00:05:41.250 --> 00:05:45.160 for all other displacements and loads also. 00:05:45.160 --> 00:05:47.970 Notice that they can vary with time. 00:05:47.970 --> 00:05:51.760 In many analysis, however, we simply pick one time step and 00:05:51.760 --> 00:05:54.500 keep it constant all along. 00:05:54.500 --> 00:05:56.740 The objective is the following. 00:05:56.740 --> 00:06:01.200 In the direct integration of the equations of equilibrium, 00:06:01.200 --> 00:06:04.740 the objective is knowing the initial conditions. 00:06:04.740 --> 00:06:07.980 In other words, knowing the displacement at time 0, the 00:06:07.980 --> 00:06:11.190 velocities at time 0. 00:06:11.190 --> 00:06:15.540 We want to calculate the displacements, velocities, and 00:06:15.540 --> 00:06:21.120 accelerations at these discrete time points. 00:06:21.120 --> 00:06:24.920 We are matching here with the time interval delta-t1. 00:06:24.920 --> 00:06:29.780 And here we are changing the time interval to delta-t2. 00:06:29.780 --> 00:06:32.970 Knowing all the velocities, accelerations, and 00:06:32.970 --> 00:06:36.120 displacement at this time, we want to calculate the same 00:06:36.120 --> 00:06:40.200 quantities at the discrete time increments of 00:06:40.200 --> 00:06:42.180 delta-t2, and so on. 00:06:42.180 --> 00:06:45.100 So basically what we are saying is that knowing the 00:06:45.100 --> 00:06:51.010 initial conditions, we want to calculate in a step-by-step 00:06:51.010 --> 00:06:54.860 procedure, the displacements, velocities, and accelerations 00:06:54.860 --> 00:06:57.850 at these discrete time points. 00:06:57.850 --> 00:07:02.790 Of course, the real problem then reduces to the following. 00:07:02.790 --> 00:07:08.980 Given the initial conditions at the particular time t, we 00:07:08.980 --> 00:07:13.220 might call this time t, and by initial conditions, I mean we 00:07:13.220 --> 00:07:15.640 know the displacements, velocities, accelerations, we 00:07:15.640 --> 00:07:19.580 know all the quantities at time t. 00:07:19.580 --> 00:07:24.740 Our objective is to calculate the unknown quantities at time 00:07:24.740 --> 00:07:29.080 t plus delta-t1 in this particular case. 00:07:29.080 --> 00:07:35.230 So this point is t plus delta-t1. 00:07:35.230 --> 00:07:41.230 In general, or in many cases, we keep a constant time step, 00:07:41.230 --> 00:07:44.070 and so we will simply talk about t plus 00:07:44.070 --> 00:07:46.810 delta-t from now onwards. 00:07:46.810 --> 00:07:50.610 We should remember, however, the delta-t might change doing 00:07:50.610 --> 00:07:51.910 the response solution. 00:07:51.910 --> 00:07:54.980 I repeat once more, this is an important point, given the 00:07:54.980 --> 00:07:58.290 initial conditions or given the conditions at time t, our 00:07:58.290 --> 00:08:01.900 objective is to calculate the conditions, displacements, 00:08:01.900 --> 00:08:07.100 velocities, accelerations, at time t plus delta-t. 00:08:07.100 --> 00:08:13.380 In the explicit integration, we proceed in 00:08:13.380 --> 00:08:14.980 the solution as follows. 00:08:14.980 --> 00:08:22.590 We use the equations of dynamic equilibrium at time t 00:08:22.590 --> 00:08:26.500 to obtain the solution at time t plus delta-t. 00:08:26.500 --> 00:08:27.820 This is important. 00:08:27.820 --> 00:08:31.620 In explicit integration, we use the condition, the 00:08:31.620 --> 00:08:34.590 equilibrium equations, the equilibrium equation, the 00:08:34.590 --> 00:08:39.200 dynamic equilibrium equation, and by that I mean, just to 00:08:39.200 --> 00:08:41.830 refresh your memory, this equation-- we use that 00:08:41.830 --> 00:08:46.760 equation at time t in explicit integration to obtain the 00:08:46.760 --> 00:08:49.980 solution at time t plus delta-t. 00:08:49.980 --> 00:08:53.210 Once we have the solution at time t plus delta-t, of course 00:08:53.210 --> 00:08:55.550 we can march ahead in the same way again. 00:08:55.550 --> 00:08:58.980 In implicit integration, however, we are looking at the 00:08:58.980 --> 00:09:03.430 equilibrium equations at time t plus delta-t t to obtain the 00:09:03.430 --> 00:09:05.610 solution at time t plus delta-t. 00:09:05.610 --> 00:09:09.345 In other words, if you look at once more at the equations 00:09:09.345 --> 00:09:12.410 here, at the equilibrium equations, these equations 00:09:12.410 --> 00:09:15.280 would be applied at time t plus delta-t to obtain the 00:09:15.280 --> 00:09:18.190 solution at time t plus delta-t. 00:09:18.190 --> 00:09:21.500 This is indeed the fundamental difference between explicit 00:09:21.500 --> 00:09:22.980 and implicit integration. 00:09:22.980 --> 00:09:26.790 In explicit integration, we are looking at equilibrium 00:09:26.790 --> 00:09:29.270 equations at time t to obtain the solution 00:09:29.270 --> 00:09:31.150 at time t plus delta-t. 00:09:31.150 --> 00:09:33.640 And in implicit integration, we are looking at the 00:09:33.640 --> 00:09:37.090 equilibrium equations at time t plus delta-t to attain the 00:09:37.090 --> 00:09:39.970 solution at time t plus delta-t. 00:09:39.970 --> 00:09:42.830 In the following now, I would like to show to you an 00:09:42.830 --> 00:09:45.740 explicit and an implicit integration scheme. 00:09:45.740 --> 00:09:48.720 The explicit integration scheme that is abundantly used 00:09:48.720 --> 00:09:51.640 in practice is the central difference method. 00:09:51.640 --> 00:09:53.950 I will later on then discuss how this method 00:09:53.950 --> 00:09:55.550 is effectively used. 00:09:55.550 --> 00:09:58.710 In the central difference method, we are applying, as I 00:09:58.710 --> 00:10:02.980 said earlier, the equilibrium equations at time t, because 00:10:02.980 --> 00:10:05.780 it's an explicit integration scheme. 00:10:05.780 --> 00:10:09.460 The two additional questions that we're using to solve for 00:10:09.460 --> 00:10:13.580 the unknown displacements at time t plus delta-t, is 00:10:13.580 --> 00:10:17.180 firstly an equation for the accelerations at time t. 00:10:17.180 --> 00:10:22.030 And notice that this equation involves as the unknown, as 00:10:22.030 --> 00:10:24.340 the only unknown, the displacement 00:10:24.340 --> 00:10:27.110 at time t plus delta-t. 00:10:27.110 --> 00:10:29.960 In addition, of course, we are using an equation for the 00:10:29.960 --> 00:10:34.690 velocities at time t and again this equation only involves as 00:10:34.690 --> 00:10:36.500 the only unknown the displacement 00:10:36.500 --> 00:10:38.300 at time t plus delta-t. 00:10:38.300 --> 00:10:42.000 Substituting these two equations into the equilibrium 00:10:42.000 --> 00:10:48.060 equation at time t gives us one equation in the unknown t, 00:10:48.060 --> 00:10:50.620 displacement at time t plus delta-t. 00:10:50.620 --> 00:10:54.280 We can simply solve for it, as shown, now on 00:10:54.280 --> 00:10:56.240 the next view graph. 00:10:56.240 --> 00:10:59.530 You can see here the equation that is obtained by 00:10:59.530 --> 00:11:03.420 substituting the relations for the acceleration and for the 00:11:03.420 --> 00:11:07.350 velocities in the equilibrium equation at time t. 00:11:07.350 --> 00:11:08.990 The unknown appears here. 00:11:08.990 --> 00:11:11.430 This is the displacement at time t plus 00:11:11.430 --> 00:11:14.270 delta-t, which is unknown. 00:11:14.270 --> 00:11:17.310 And there is a coefficient matrix here, that involves the 00:11:17.310 --> 00:11:19.830 mass matrix and the damping matrix. 00:11:19.830 --> 00:11:22.730 And on the right hand side, of course, we are calculating a 00:11:22.730 --> 00:11:23.840 load vector. 00:11:23.840 --> 00:11:27.270 And that load vector, notice once again, corresponds to the 00:11:27.270 --> 00:11:29.650 conditions at time t. 00:11:29.650 --> 00:11:32.420 That's why it's an explicit integration scheme. 00:11:32.420 --> 00:11:35.780 The difference operator, the essential difference operator 00:11:35.780 --> 00:11:40.045 that I've been used here, simple operators which you 00:11:40.045 --> 00:11:44.060 have seen on the previous slide, that involve only the 00:11:44.060 --> 00:11:48.000 displacement at time t minus delta-t and the 00:11:48.000 --> 00:11:49.500 displacement at time t. 00:11:49.500 --> 00:11:52.660 And notice that here are these two 00:11:52.660 --> 00:11:55.990 displacement vectors appearing. 00:11:55.990 --> 00:12:00.803 Notice that this equation is cheaply-- 00:12:00.803 --> 00:12:04.930 or can be used in a very cheap way to calculate the 00:12:04.930 --> 00:12:07.450 displacement, the unknown displacement at time t plus 00:12:07.450 --> 00:12:12.370 delta-t, if the mass matrix is a diagonal matrix. 00:12:12.370 --> 00:12:14.920 And, say, the damping is neglected. 00:12:14.920 --> 00:12:17.520 In other words, this term not being there, this being a 00:12:17.520 --> 00:12:22.390 diagonal mass matrix, then the solution is very trivial. 00:12:22.390 --> 00:12:25.620 In fact, all we need to do is calculate the right hand side 00:12:25.620 --> 00:12:30.470 load vector and divide each component of that load vector 00:12:30.470 --> 00:12:35.390 by this coefficient here which, if m once again is a 00:12:35.390 --> 00:12:38.080 diagonal mass matrix, is simply Mii 00:12:38.080 --> 00:12:40.540 divided by delta-t squared. 00:12:40.540 --> 00:12:44.310 And in fact, this is really how was the procedure is 00:12:44.310 --> 00:12:47.580 usually applied, namely, with a diagonal mass 00:12:47.580 --> 00:12:49.370 matrix and no damping. 00:12:49.370 --> 00:12:54.050 The damping usually is not taken into account, or at 00:12:54.050 --> 00:12:57.450 most, we have diagonal dampers also, and the mass matrix 00:12:57.450 --> 00:13:01.450 being a diagonal matrix, then the solution is 00:13:01.450 --> 00:13:02.760 very cheaply performed. 00:13:02.760 --> 00:13:05.530 There's one further important point. 00:13:05.530 --> 00:13:11.120 This load vector of course is constructed in the usual way. 00:13:11.120 --> 00:13:16.630 This multiplication here is quite cheap if the mass matrix 00:13:16.630 --> 00:13:18.410 is a diagonal mass matrix. 00:13:18.410 --> 00:13:20.010 K times tU. 00:13:20.010 --> 00:13:23.470 Well, K here is of course a banded matrix. 00:13:23.470 --> 00:13:27.770 It's the stiffness matrix of the system, constant. 00:13:27.770 --> 00:13:31.270 Well, there are some operations involved here. 00:13:31.270 --> 00:13:35.280 And if we look at this operation here in detail, we 00:13:35.280 --> 00:13:38.590 see here how it can be effectively performed. 00:13:38.590 --> 00:13:40.270 We have a stiffness matrix times the 00:13:40.270 --> 00:13:41.780 displacement vector here. 00:13:41.780 --> 00:13:44.920 And of course, the stiffness matrix remember is obtained by 00:13:44.920 --> 00:13:47.350 summing the element contributions. 00:13:47.350 --> 00:13:51.500 So you can write this K matrix in this way. 00:13:51.500 --> 00:13:55.540 Since tU is independent of the elements, we can also take it 00:13:55.540 --> 00:13:57.230 into the summation sign. 00:13:57.230 --> 00:14:02.270 And if we look at this product closely, we find that is 00:14:02.270 --> 00:14:04.970 nothing else than tFm. 00:14:04.970 --> 00:14:13.900 It's a nodal point vector of loads or elements forces that 00:14:13.900 --> 00:14:15.260 correspond-- 00:14:15.260 --> 00:14:18.940 element elastic forces-- that correspond to the current 00:14:18.940 --> 00:14:20.420 displacements. 00:14:20.420 --> 00:14:25.280 And this tF vector is much more cheaply calculated than 00:14:25.280 --> 00:14:27.390 the Km times tU. 00:14:27.390 --> 00:14:30.740 And in fact, this is the way how we actually perform the 00:14:30.740 --> 00:14:33.370 calculation of this product here. 00:14:33.370 --> 00:14:37.130 Now notice the following important further point. 00:14:37.130 --> 00:14:41.980 If we were to leave this K times tU the way it stands 00:14:41.980 --> 00:14:44.800 here, we would have to assemble a global stiffness 00:14:44.800 --> 00:14:46.710 matrix of the system. 00:14:46.710 --> 00:14:49.700 However, if we write that product this way, we never 00:14:49.700 --> 00:14:52.060 need to assemble a global stiffness matrix 00:14:52.060 --> 00:14:53.380 of the total system. 00:14:53.380 --> 00:14:58.010 In fact, all we need to do is calculate element forces and 00:14:58.010 --> 00:15:01.860 add these elements forces via the direct stiffness procedure 00:15:01.860 --> 00:15:03.940 the way I've been discussing earlier. 00:15:03.940 --> 00:15:09.410 So this is a very effective way of calculating K times tU. 00:15:09.410 --> 00:15:12.560 We never need to assemble a global stiffness matrix. 00:15:12.560 --> 00:15:17.060 In fact, this is one of the main reasons why the central 00:15:17.060 --> 00:15:20.040 difference method is effective. 00:15:20.040 --> 00:15:25.250 It's this reason that we can calculate the KtU part, this 00:15:25.250 --> 00:15:29.530 times that, very effectively via this procedure. 00:15:29.530 --> 00:15:30.500 That is the first reason. 00:15:30.500 --> 00:15:34.980 And the second reason is that if M is diagonal, we really 00:15:34.980 --> 00:15:37.160 never solve equations. 00:15:37.160 --> 00:15:41.550 The equation solving here is so trivial, that we might just 00:15:41.550 --> 00:15:43.270 say that we do not solve equations. 00:15:43.270 --> 00:15:45.530 There is no factorization involved. 00:15:45.530 --> 00:15:49.910 There's no factorization of a stiffness matrix or another 00:15:49.910 --> 00:15:52.900 coefficient matrix involved, because if M is diagonal, and 00:15:52.900 --> 00:15:57.300 as I pointed out, we simply divide the right hand side by 00:15:57.300 --> 00:16:01.310 Mii divided by delta-t squared. 00:16:01.310 --> 00:16:05.550 Notice that if we start the integration process, we know 00:16:05.550 --> 00:16:07.040 of course the initial conditions. 00:16:07.040 --> 00:16:08.540 We know tU. 00:16:08.540 --> 00:16:11.930 And we do not know-- 00:16:11.930 --> 00:16:14.830 usually directly the t minus delta-tU. 00:16:14.830 --> 00:16:17.570 That is the displacement prior to the solution. 00:16:17.570 --> 00:16:22.160 And to obtain those, we can use this formula, which is 00:16:22.160 --> 00:16:27.420 really the central difference method applied at time t, 00:16:27.420 --> 00:16:29.790 where we however now involve the accelerations 00:16:29.790 --> 00:16:31.940 corresponding to time 0. 00:16:31.940 --> 00:16:35.270 We involve the accelerations corresponding to time zero 00:16:35.270 --> 00:16:38.290 right here, and the velocities corresponding to times 0, and 00:16:38.290 --> 00:16:40.580 the displacements corresponding to time zero. 00:16:40.580 --> 00:16:42.670 And of course, these quantities must all be given 00:16:42.670 --> 00:16:46.070 as the initial conditions to the solution. 00:16:46.070 --> 00:16:48.840 I pointed out already that we're using this technique 00:16:48.840 --> 00:16:51.130 mostly with the lump mass matrix. 00:16:51.130 --> 00:16:53.880 In addition I should also point out that we're using it 00:16:53.880 --> 00:16:56.250 primarily with lower order elements, 00:16:56.250 --> 00:16:57.890 with lower order elements. 00:16:57.890 --> 00:17:04.430 The reason for that is the following, namely, one of 00:17:04.430 --> 00:17:08.190 stability and accuracy of the central difference method. 00:17:08.190 --> 00:17:15.460 The method can only be applied if delta-t, the time step that 00:17:15.460 --> 00:17:18.200 I'm talking about here, is smaller than a 00:17:18.200 --> 00:17:20.119 critical time step. 00:17:20.119 --> 00:17:25.240 And that critical time step is given by the smallest period 00:17:25.240 --> 00:17:30.180 in the system divided by pi, 3.14 et cetera. 00:17:30.180 --> 00:17:32.730 The smallest period in the system can be 00:17:32.730 --> 00:17:34.930 of course very small. 00:17:34.930 --> 00:17:39.140 And therefore the critical time step that we're talking 00:17:39.140 --> 00:17:42.410 about here can be very small. 00:17:42.410 --> 00:17:46.060 The smallest period in the system, just to give you an 00:17:46.060 --> 00:17:50.210 example, would be extremely small if you have a bar 00:17:50.210 --> 00:17:52.990 element, and you have some very short 00:17:52.990 --> 00:17:54.470 truss element in there. 00:17:54.470 --> 00:17:58.000 The shorter the truss element, the stiffer of course it is. 00:17:58.000 --> 00:18:02.380 AU over L is the stiffness of a truss element, and 00:18:02.380 --> 00:18:06.860 therefore, a very small period would be in that 00:18:06.860 --> 00:18:10.470 assemblage of elements. 00:18:10.470 --> 00:18:14.240 Therefore, what is required really is to have a mesh that 00:18:14.240 --> 00:18:19.760 is more or less uniform, so that there is no artificially 00:18:19.760 --> 00:18:23.320 small time step required in the integration of the 00:18:23.320 --> 00:18:24.770 equilibrium equations. 00:18:24.770 --> 00:18:27.060 Because there is this condition that 00:18:27.060 --> 00:18:28.650 delta-t must be smaller-- 00:18:28.650 --> 00:18:30.980 we should really say must be smaller or equal-- 00:18:30.980 --> 00:18:36.380 to delta-t critical, but to be on the conservative side, we 00:18:36.380 --> 00:18:40.200 are selecting delta-t smaller than delta-t critical in 00:18:40.200 --> 00:18:43.310 practical analysis of complex meshes. 00:18:43.310 --> 00:18:46.130 Because this condition has to be satisfied, we're talking 00:18:46.130 --> 00:18:50.320 about a conditionally stable scheme. 00:18:50.320 --> 00:18:55.380 It is only stable when delta-t is smaller than that value. 00:18:55.380 --> 00:18:59.920 If we were to select delta-t larger than delta-t critical, 00:18:59.920 --> 00:19:05.140 we would see that after just a few time steps, the solution-- 00:19:05.140 --> 00:19:06.770 it gets out of bounds. 00:19:06.770 --> 00:19:10.860 And after 10 or 20 or 30 time step, in a computer program, 00:19:10.860 --> 00:19:12.150 you would all of a sudden notice 00:19:12.150 --> 00:19:13.560 overflow of the numbers. 00:19:13.560 --> 00:19:16.380 So the solution wouldn't make any sense whatsoever anymore. 00:19:16.380 --> 00:19:18.490 And this is an absolute condition that 00:19:18.490 --> 00:19:20.380 does have to be satisfied. 00:19:20.380 --> 00:19:23.370 There's no way you can use a time step larger than delta-t 00:19:23.370 --> 00:19:26.570 critical, and still get an acceptable, reasonable 00:19:26.570 --> 00:19:31.380 solution to your equations of motion. 00:19:31.380 --> 00:19:35.010 In practice, of course, we have to estimate, in a 00:19:35.010 --> 00:19:37.210 conservative way, a time step. 00:19:37.210 --> 00:19:39.380 And that means we have to have an estimate 00:19:39.380 --> 00:19:41.680 also of delta-t critical. 00:19:41.680 --> 00:19:43.440 For continuum elements-- 00:19:43.440 --> 00:19:46.350 by continuum elements, I mean plain stress, plain strain, 00:19:46.350 --> 00:19:48.560 axis a matrix, three-dimensional element, the 00:19:48.560 --> 00:19:50.670 way we have been discussing it earlier. 00:19:50.670 --> 00:19:57.430 A good way of obtaining time step delta-t is to use this 00:19:57.430 --> 00:20:03.650 formula here, delta-t being delta-L over C. C is the wave 00:20:03.650 --> 00:20:05.800 velocity through the system. 00:20:05.800 --> 00:20:06.390 [UNINTELLIGIBLE] 00:20:06.390 --> 00:20:10.890 is divided by rho, square root out of it gives us C. And 00:20:10.890 --> 00:20:19.210 delta-L is an element length that has to be defined. 00:20:19.210 --> 00:20:24.290 In fact, it's the smallest distance between nodes. 00:20:24.290 --> 00:20:27.990 The smallest distance between nodes if we talk about low 00:20:27.990 --> 00:20:29.040 order elements. 00:20:29.040 --> 00:20:30.590 What I mean by that is the following. 00:20:30.590 --> 00:20:34.250 If we have a mesh here that looks like that. 00:20:34.250 --> 00:20:38.125 And I deliberately show elements that are not equal in 00:20:38.125 --> 00:20:42.960 size, then delta-E in this particular case would be this 00:20:42.960 --> 00:20:43.860 distance here. 00:20:43.860 --> 00:20:48.050 That is our delta-L, sorry, our delta-L is 00:20:48.050 --> 00:20:51.610 that distance there. 00:20:51.610 --> 00:20:55.320 This, of course, as I have here in the heading, assumes 00:20:55.320 --> 00:20:57.040 that we have low-order elements. 00:20:57.040 --> 00:21:00.460 That means only corner nodes in the elements. 00:21:00.460 --> 00:21:02.240 For higher-order elements-- 00:21:02.240 --> 00:21:05.650 let me put down here one higher-order element, where we 00:21:05.650 --> 00:21:11.420 have these nodes now, say an 8 node element. 00:21:11.420 --> 00:21:14.310 In this particular case, we take the smallest distance 00:21:14.310 --> 00:21:19.710 between nodes, again, just as in this case. 00:21:19.710 --> 00:21:21.940 And in this particular case, it would be, say, this 00:21:21.940 --> 00:21:23.140 distance here. 00:21:23.140 --> 00:21:26.200 That is our smallest distance between nodes. 00:21:26.200 --> 00:21:29.460 However, we have to also divide this distance by a 00:21:29.460 --> 00:21:31.800 relative stiffness factor. 00:21:31.800 --> 00:21:34.430 And why does that relative stiffness factor 00:21:34.430 --> 00:21:35.860 come into the picture? 00:21:35.860 --> 00:21:39.320 Well, it does come into the picture, because remember that 00:21:39.320 --> 00:21:44.190 the stiffness at an interior node is larger than the 00:21:44.190 --> 00:21:47.570 stiffness at a corner for the element. 00:21:47.570 --> 00:21:50.870 The amount it is larger depends on the particular 00:21:50.870 --> 00:21:53.600 element used. 00:21:53.600 --> 00:21:57.490 But since there's a lot of stiffness here, we have to put 00:21:57.490 --> 00:22:01.430 here a relative stiffness factor into the calculation of 00:22:01.430 --> 00:22:06.970 delta-L, the effective lengths that we're using in this 00:22:06.970 --> 00:22:10.230 particular formula. 00:22:10.230 --> 00:22:14.450 Now, for this element here, for example, the relative 00:22:14.450 --> 00:22:16.910 stiffness factor would be conservatively 00:22:16.910 --> 00:22:20.410 just equal to 4. 00:22:20.410 --> 00:22:23.690 But for other elements, and particular also when there's 00:22:23.690 --> 00:22:27.260 node shifting involved and so on, the relative stiffness 00:22:27.260 --> 00:22:30.330 factor has to be calculated, has to be estimated. 00:22:30.330 --> 00:22:33.950 And we should always remember that it should be estimated 00:22:33.950 --> 00:22:41.320 here in a conservative way, because a smaller delta-L gave 00:22:41.320 --> 00:22:44.570 us a smaller delta-t. 00:22:44.570 --> 00:22:47.680 And then we are then conservatively applying the 00:22:47.680 --> 00:22:50.900 fact that delta-t shall be smaller than the delta-t 00:22:50.900 --> 00:22:55.220 critical, value that I have been referring hereto. 00:22:55.220 --> 00:23:00.120 The method is used mainly for wave propagation analysis. 00:23:00.120 --> 00:23:03.650 The reason being that we have to use a fairly large number 00:23:03.650 --> 00:23:07.260 of time steps to predict the solution. 00:23:07.260 --> 00:23:11.220 Large, because a time step delta-t is usually very small. 00:23:11.220 --> 00:23:13.940 And in wave propagation analysis, many modes of the 00:23:13.940 --> 00:23:16.590 system are being excited. 00:23:16.590 --> 00:23:20.340 So we really want to pick up a large number of 00:23:20.340 --> 00:23:22.170 modes in the mesh. 00:23:22.170 --> 00:23:27.990 And the small time step integrating accurately many 00:23:27.990 --> 00:23:31.540 modes in the mesh goes together with the fact that we 00:23:31.540 --> 00:23:35.180 want to predict many modes in the mesh or a large number of 00:23:35.180 --> 00:23:38.960 modes in the mesh, and therefore the method is 00:23:38.960 --> 00:23:42.060 effectively used for wave propagation analysis. 00:23:42.060 --> 00:23:46.500 The number of operations are proportional to the number of 00:23:46.500 --> 00:23:52.042 elements that we are using and the number of time steps. 00:23:52.042 --> 00:23:55.830 Of course, this is an approximate proportionality 00:23:55.830 --> 00:23:58.550 here that I'm talking about. 00:23:58.550 --> 00:24:06.340 This fact derives directly from the consideration that we 00:24:06.340 --> 00:24:07.440 talked about here. 00:24:07.440 --> 00:24:10.640 Notice that as we include more and more elements in the 00:24:10.640 --> 00:24:13.880 solution process, of course, we have to sum over more 00:24:13.880 --> 00:24:15.560 elements here. 00:24:15.560 --> 00:24:18.800 And as we go more and more time steps, we have to apply 00:24:18.800 --> 00:24:23.720 this equation as many times also more and more. 00:24:23.720 --> 00:24:27.440 The important point is that we don't assemble a K matrix, so 00:24:27.440 --> 00:24:30.850 the bandwidth off the system, the bandwidth of the system 00:24:30.850 --> 00:24:33.300 does not appear in an operation count. 00:24:33.300 --> 00:24:36.130 It's the number of elements that do appear in the 00:24:36.130 --> 00:24:37.380 operation count. 00:24:40.010 --> 00:24:48.330 The next procedure that I like to talk about 00:24:48.330 --> 00:24:50.100 is the Newmark method. 00:24:50.100 --> 00:24:55.600 And in the Newmark method, this being an implicit 00:24:55.600 --> 00:24:59.710 integration scheme now, we proceed as follows. 00:24:59.710 --> 00:25:00.610 We apply-- 00:25:00.610 --> 00:25:02.730 because it's an implicit integration scheme-- 00:25:02.730 --> 00:25:05.450 the equilibrium equation, the dynamic equilibrium equation 00:25:05.450 --> 00:25:08.150 at time t plus delta-t. 00:25:08.150 --> 00:25:12.450 Here it is given the additional formula that we're 00:25:12.450 --> 00:25:16.560 using is this one here for the velocity at time t plus 00:25:16.560 --> 00:25:19.140 delta-t, and this one here for the displacement of time t 00:25:19.140 --> 00:25:20.530 plus delta-t. 00:25:20.530 --> 00:25:25.380 Now notice that with this set of equations, with these three 00:25:25.380 --> 00:25:31.070 equations, knowing the displacements and velocities 00:25:31.070 --> 00:25:37.620 at time t and accelerations at time t, we have with these 00:25:37.620 --> 00:25:40.410 three equations, three unknowns. 00:25:40.410 --> 00:25:44.440 The velocity at time t plus delta-t, the acceleration at 00:25:44.440 --> 00:25:47.415 time t plus delta-t, and the displacement 00:25:47.415 --> 00:25:49.720 at time t plus delta-t. 00:25:49.720 --> 00:25:52.165 These are the only three unknowns that appear. 00:25:52.165 --> 00:25:55.540 There are three linearly independent equations. 00:25:55.540 --> 00:25:57.550 And we can simply use them to 00:25:57.550 --> 00:25:59.515 calculate these three unknowns. 00:25:59.515 --> 00:26:02.030 Of course, we want to do so in a very effective way. 00:26:02.030 --> 00:26:04.320 And I will show you how we proceed. 00:26:04.320 --> 00:26:11.640 The important point is that with this scheme, we are using 00:26:11.640 --> 00:26:16.170 an implicit approach, as I defined it earlier. 00:26:16.170 --> 00:26:21.070 And when we do substitute from here into there, we will find 00:26:21.070 --> 00:26:25.540 that on the left hand side, we have a coefficient matrix that 00:26:25.540 --> 00:26:30.500 involves a bandwidth even if the mass matrix is diagonal. 00:26:30.500 --> 00:26:33.415 And this coefficient matrix here is written 00:26:33.415 --> 00:26:35.730 here down as K hat. 00:26:35.730 --> 00:26:40.640 This coefficient matrix invoice this K matrix. 00:26:40.640 --> 00:26:43.350 I should say maybe a few words to these formulae. 00:26:43.350 --> 00:26:46.640 These formulae can be obtained by Taylor Series expansion, 00:26:46.640 --> 00:26:49.970 around the conditions at times t plus delta-t. 00:26:49.970 --> 00:26:53.070 They have been proposed originally by Newmark. 00:26:53.070 --> 00:26:56.330 That's why it's called now the Newmark method. 00:26:56.330 --> 00:27:00.420 Newmark in particular used the values of delta equals 1/2 and 00:27:00.420 --> 00:27:02.430 alpha equal to 1/4. 00:27:02.430 --> 00:27:06.800 And then it's a constant average acceleration scheme. 00:27:06.800 --> 00:27:14.960 The constant average acceleration schemes use these 00:27:14.960 --> 00:27:17.700 values as I pointed out already earlier. 00:27:17.700 --> 00:27:20.310 The method is unconditionally stable. 00:27:20.310 --> 00:27:25.550 This means that the time step delta-t does not to have to be 00:27:25.550 --> 00:27:28.480 smaller than a certain value, as it was the case in the 00:27:28.480 --> 00:27:30.490 central difference method. 00:27:30.490 --> 00:27:33.970 The method is primarily was for analysis of structural 00:27:33.970 --> 00:27:36.410 dynamic problems. 00:27:36.410 --> 00:27:40.030 I will talk about why this is so just now. 00:27:40.030 --> 00:27:42.560 And the number of operations that we're talking about here 00:27:42.560 --> 00:27:48.110 are an initial factorization of the K hat matrix. 00:27:48.110 --> 00:27:51.610 This is the K hat matrix, And that initial factorization 00:27:51.610 --> 00:27:55.900 involves this K matrix, so the bandwidth of the K hat matrix 00:27:55.900 --> 00:27:59.130 is the same as the bandwidth of the K matrix. 00:27:59.130 --> 00:28:04.170 And therefore the operations here are one half NM squared. 00:28:04.170 --> 00:28:11.160 And for each solution, in time, because this equation 00:28:11.160 --> 00:28:16.380 here has to be solved for each time step, you're starting 00:28:16.380 --> 00:28:19.870 with delta-t, and then we go to 2 delta-t, 3 delta-t, 4 00:28:19.870 --> 00:28:22.090 delta-t, and so on. 00:28:22.090 --> 00:28:25.820 And since this solution is to be formed for each time step, 00:28:25.820 --> 00:28:28.490 however, remember the factorization is only done 00:28:28.490 --> 00:28:36.520 once because this K hat matrix is independent of the time. 00:28:36.520 --> 00:28:39.410 Therefore, we have a forward reduction and back 00:28:39.410 --> 00:28:41.710 substitution in each time step. 00:28:41.710 --> 00:28:45.500 And the number of operations in each time step are 2nm. t 00:28:45.500 --> 00:28:47.600 is the number of time steps here. 00:28:47.600 --> 00:28:51.650 t is the number of time steps here. 00:28:51.650 --> 00:28:54.320 Well, let us look then at the accuracy 00:28:54.320 --> 00:28:56.600 considerations that we have. 00:28:56.600 --> 00:28:59.770 I mentioned already the method is unconditionally stable, 00:28:59.770 --> 00:29:04.470 which means that the delta-t time step that we're using can 00:29:04.470 --> 00:29:08.360 be any amount, and we will never have a blow up of the 00:29:08.360 --> 00:29:11.460 solution, a blow up of the solution due to round-off 00:29:11.460 --> 00:29:13.440 errors that I involved. 00:29:13.440 --> 00:29:15.350 In the central difference method, of course, the 00:29:15.350 --> 00:29:17.650 explicit time integration scheme, I mentioned that 00:29:17.650 --> 00:29:20.730 delta-t has to be smaller than a critical time step. 00:29:20.730 --> 00:29:24.760 So in the implicit Newmark integration scheme, what we 00:29:24.760 --> 00:29:29.600 can do is we select our of time step based solely on 00:29:29.600 --> 00:29:32.290 accuracy considerations. 00:29:32.290 --> 00:29:35.270 Let us consider then these equations that we want to 00:29:35.270 --> 00:29:39.220 solve, and I'm now leaving out the damping term in order to 00:29:39.220 --> 00:29:42.410 obtain some understanding of how we would select the time 00:29:42.410 --> 00:29:43.700 step, delta-t. 00:29:43.700 --> 00:29:47.200 Of course, remember the larger the time step, the less normal 00:29:47.200 --> 00:29:50.440 operations I involved in the solution. 00:29:50.440 --> 00:29:52.410 So we want to select the large time step. 00:29:52.410 --> 00:29:55.500 However, if we make the time step extremely large, then 00:29:55.500 --> 00:29:58.610 surely we cannot expect any accuracy in the solution. 00:29:58.610 --> 00:30:02.070 In fact, if delta-t is very large, we simply predict the 00:30:02.070 --> 00:30:03.670 static solution. 00:30:03.670 --> 00:30:05.430 This is one property of the Newmark method. 00:30:05.430 --> 00:30:07.950 If you were to use a computer program with the Newmark 00:30:07.950 --> 00:30:10.160 method, if you're interested in it, why don't you try, say, 00:30:10.160 --> 00:30:12.860 for example, using ADINA, in which we have the Newmark 00:30:12.860 --> 00:30:15.520 method, and just put delta-t very large, and all you would 00:30:15.520 --> 00:30:17.660 be getting out is a static solution. 00:30:17.660 --> 00:30:20.360 Well, since we're interested in the dynamic response, we 00:30:20.360 --> 00:30:22.840 have to therefore select delta-t small enough to 00:30:22.840 --> 00:30:28.550 integrate accurately the solution response. 00:30:28.550 --> 00:30:31.740 And these are the equations that I want to now consider, 00:30:31.740 --> 00:30:36.500 and I want to show you how can we select delta-t rationally 00:30:36.500 --> 00:30:39.340 and to obtain the required accuracy. 00:30:39.340 --> 00:30:44.040 The displacement vector here can be expressed in terms of 00:30:44.040 --> 00:30:48.630 the mode shape vectors times xi values. 00:30:48.630 --> 00:30:53.360 I will be talking about how these phi i vectors are 00:30:53.360 --> 00:30:57.780 calculated in the next lectures. 00:30:57.780 --> 00:31:00.350 This involves the solution of the eigenvalue problem 00:31:00.350 --> 00:31:01.890 written down here. 00:31:01.890 --> 00:31:04.460 The free vibration eigenvalue a problem corresponding to 00:31:04.460 --> 00:31:06.380 this equation. 00:31:06.380 --> 00:31:11.090 I like to defer the discussion at this point, because we will 00:31:11.090 --> 00:31:13.390 be talking about it later on in detail. 00:31:13.390 --> 00:31:17.520 All I like to say is that this transformation here, phi i 00:31:17.520 --> 00:31:21.550 being a vector time independent and xi being a 00:31:21.550 --> 00:31:23.830 scalar, dependent on time. 00:31:23.830 --> 00:31:29.560 This transformation can be used in this solution here, or 00:31:29.560 --> 00:31:32.100 can be used rather to transform this system of 00:31:32.100 --> 00:31:36.330 equations into a different form. 00:31:36.330 --> 00:31:42.390 And in that transformation, we use the fact that when we 00:31:42.390 --> 00:31:45.480 calculate phi transpose K times 5-- 00:31:45.480 --> 00:31:47.510 notice these are capital phis. 00:31:47.510 --> 00:31:50.280 I've crossed bars here, capital phis. 00:31:50.280 --> 00:31:54.090 This is a lower phi, no crossed bars top and bottom. 00:31:54.090 --> 00:31:58.770 This capital phi here stores all of the lowercase phi 00:31:58.770 --> 00:32:02.000 vectors, and their n of such vectors. 00:32:02.000 --> 00:32:07.520 If I use this capital phi and calculate this product here, I 00:32:07.520 --> 00:32:08.900 get a diagonal matrix. 00:32:08.900 --> 00:32:12.620 In fact, the diagonal matrix here stores on its diagonal 00:32:12.620 --> 00:32:15.030 the frequency squared, the free 00:32:15.030 --> 00:32:16.920 vibrational frequency squared. 00:32:16.920 --> 00:32:21.870 Also phi transposed M phi gives us an identity matrix. 00:32:21.870 --> 00:32:27.000 These, once again, the vectors in this matrix, capital phi, 00:32:27.000 --> 00:32:31.170 are these phi i vectors which are the 00:32:31.170 --> 00:32:33.590 eigenvectors of this problem. 00:32:33.590 --> 00:32:37.450 And I will talk later on about how we calculate them. 00:32:37.450 --> 00:32:41.280 Well, if we apply this transformation-- 00:32:41.280 --> 00:32:44.000 whoops, let me put it right once more-- if we apply this 00:32:44.000 --> 00:32:52.090 transformation to this system of equations, and we use this 00:32:52.090 --> 00:32:58.100 property right here, we obtain n decoupled 00:32:58.100 --> 00:33:01.670 equations of this form. 00:33:01.670 --> 00:33:05.600 X double ought i plus omega i squared, xi equals phi i 00:33:05.600 --> 00:33:09.150 transposed times R. And they are in such equations, in 00:33:09.150 --> 00:33:13.340 other words I run c from 1 to n. 00:33:13.340 --> 00:33:16.900 In fact, if we use multiple position analysis, we actually 00:33:16.900 --> 00:33:19.780 perform this transformation. 00:33:19.780 --> 00:33:22.570 What I now want to do is simply refer to it 00:33:22.570 --> 00:33:23.270 theoretically. 00:33:23.270 --> 00:33:25.150 I do not want to transform it. 00:33:25.150 --> 00:33:27.600 You're talking about direct integration of the solution of 00:33:27.600 --> 00:33:30.470 equations, in which there is no transformation involved. 00:33:30.470 --> 00:33:33.000 So I do not want to go through this transformation. 00:33:33.000 --> 00:33:38.300 However, I want to refer to it theoretically. 00:33:38.300 --> 00:33:43.750 And it gives me a tremendous insight into what happens. 00:33:43.750 --> 00:33:48.250 So basically, what I can say theoretically is that the 00:33:48.250 --> 00:33:51.890 direct step-by-step solution of this system of equations, 00:33:51.890 --> 00:33:54.590 using the explicit integration scheme, the central difference 00:33:54.590 --> 00:33:58.150 method, or the Newmark implicit integration scheme, 00:33:58.150 --> 00:34:01.980 corresponds completely to the direct step-by-step solution 00:34:01.980 --> 00:34:07.630 of n equations, of this form, where the displacements here, 00:34:07.630 --> 00:34:11.940 U, are given in this way. 00:34:11.940 --> 00:34:15.980 x being now a function of time, so if I differentiate 00:34:15.980 --> 00:34:21.179 here, your double dot would be obtained by differentiating xi 00:34:21.179 --> 00:34:24.710 also twice, with respect to time. 00:34:24.710 --> 00:34:26.620 Once again, this is an important, 00:34:26.620 --> 00:34:28.350 very important, fact. 00:34:28.350 --> 00:34:32.969 I look at these equations, and I say that's a solution of 00:34:32.969 --> 00:34:37.280 these set of coupled differential equations. 00:34:37.280 --> 00:34:42.310 It's completely equivalent to the solution of n equations of 00:34:42.310 --> 00:34:48.478 this form with this transformation used. 00:34:48.478 --> 00:34:50.950 No dots here or dots there. 00:34:50.950 --> 00:34:53.820 In other words, this equation holds for displacements 00:34:53.820 --> 00:34:55.820 without the red dots and for the accelerations 00:34:55.820 --> 00:34:58.280 with the red dots. 00:34:58.280 --> 00:34:59.820 It is equivalent. 00:34:59.820 --> 00:35:03.270 This solution is equivalent to that solution. 00:35:03.270 --> 00:35:06.990 However, in direct integration, I never perform 00:35:06.990 --> 00:35:08.420 actually the transformation. 00:35:08.420 --> 00:35:11.820 All I know is that when I operate on this system of 00:35:11.820 --> 00:35:14.570 equations, using the Newmark method or the central 00:35:14.570 --> 00:35:18.180 difference method, I, in fact, operate on n 00:35:18.180 --> 00:35:20.130 such decoupled equations. 00:35:20.130 --> 00:35:21.590 That's important. 00:35:21.590 --> 00:35:25.270 And this means that for an analysis on our direct 00:35:25.270 --> 00:35:29.420 integration scheme, we can in fact analyze our direct 00:35:29.420 --> 00:35:33.310 integration scheme by looking at these equations, instead of 00:35:33.310 --> 00:35:34.700 these equations. 00:35:34.700 --> 00:35:37.190 You see if I were to analyze my direct integration scheme 00:35:37.190 --> 00:35:40.450 using these equations, I would have to deal with this huge 00:35:40.450 --> 00:35:45.720 matrix K, the huge matrix M. All I now need to deal with, 00:35:45.720 --> 00:35:50.150 in the analysis of the direct integration scheme, is with n 00:35:50.150 --> 00:35:53.900 such simple one degree of freedom equations. 00:35:53.900 --> 00:35:57.810 And in fact, I don't even need to deal with n such equations, 00:35:57.810 --> 00:36:02.020 for the analysis purpose, I only need to deal with one 00:36:02.020 --> 00:36:04.210 equation, where I make now omega a 00:36:04.210 --> 00:36:06.710 variable and r a variable. 00:36:06.710 --> 00:36:13.070 So my analysis on this equation, using this 00:36:13.070 --> 00:36:17.000 particular time integration scheme that I want to analyze, 00:36:17.000 --> 00:36:22.140 is completely giving me all the information that I need to 00:36:22.140 --> 00:36:25.780 know when I use my direct integration scheme in the 00:36:25.780 --> 00:36:28.200 solution of this equation. 00:36:28.200 --> 00:36:32.020 For analysis purposes of my direct integration scheme, I 00:36:32.020 --> 00:36:35.330 can simply analyze my direct integration scheme using this 00:36:35.330 --> 00:36:38.030 equation and get all the relevant 00:36:38.030 --> 00:36:40.840 information from that analysis. 00:36:40.840 --> 00:36:44.140 This is an extremely important fact. 00:36:44.140 --> 00:36:48.060 Once again, I do not actually do perform the transformation. 00:36:48.060 --> 00:36:50.190 We don't do the transformation. 00:36:50.190 --> 00:36:53.780 Only in theory for analysis purposes. 00:36:53.780 --> 00:36:57.610 I'm looking at the solution of this equation using the 00:36:57.610 --> 00:36:59.100 Newmark integration scheme. 00:36:59.100 --> 00:37:02.290 And I get all the relevant information on the accuracy, 00:37:02.290 --> 00:37:07.650 stability, et cetera of the integration scheme by looking 00:37:07.650 --> 00:37:10.520 at the solution of this equation with that direct 00:37:10.520 --> 00:37:12.030 integration scheme. 00:37:12.030 --> 00:37:15.030 Now, even in the analysis of this 00:37:15.030 --> 00:37:17.000 equation, of course, remember-- 00:37:17.000 --> 00:37:20.710 or in the use of this equation, remember that r is 00:37:20.710 --> 00:37:22.170 available now. 00:37:22.170 --> 00:37:25.660 And I have some choice what to put in there and so on. 00:37:25.660 --> 00:37:30.220 But the important point is that what I want to identify 00:37:30.220 --> 00:37:34.650 is the characteristics of the integration scheme. 00:37:34.650 --> 00:37:39.620 And the characteristics of the integration scheme are really 00:37:39.620 --> 00:37:44.150 quite well seen already if we look at a very simple case, 00:37:44.150 --> 00:37:45.640 namely this case. 00:37:45.640 --> 00:37:48.260 I set r deliberately equal to zero. 00:37:48.260 --> 00:37:52.090 And I look at the case where I have initial condition x, the 00:37:52.090 --> 00:37:54.950 displacements being one, the velocities being zero. 00:37:54.950 --> 00:37:57.420 this means, of course, that the accelerations are minus 00:37:57.420 --> 00:37:59.050 omega squared. 00:37:59.050 --> 00:38:04.680 If I use the Newmark method, the one that I just talked to 00:38:04.680 --> 00:38:10.870 you about, on that system of equations, then I can plot 00:38:10.870 --> 00:38:13.090 errors that reside in that integration. 00:38:13.090 --> 00:38:17.400 And what I've plotted here the percentage period elongation, 00:38:17.400 --> 00:38:23.810 pe divided by t times 100, which is this amount here that 00:38:23.810 --> 00:38:27.530 we see when we integrate the differential equation that I 00:38:27.530 --> 00:38:32.260 showed you via as the Newmark method with different delta-t 00:38:32.260 --> 00:38:35.300 over t values. 00:38:35.300 --> 00:38:41.820 On the bottom here, we are plotting delta-t divided by t, 00:38:41.820 --> 00:38:49.510 the period t, of course, being given for the particular 00:38:49.510 --> 00:38:51.250 oscillator that we're looking at. 00:38:51.250 --> 00:38:52.870 We are starting off with one. 00:38:52.870 --> 00:38:56.570 The exact solution, I should take better now another color, 00:38:56.570 --> 00:39:01.760 the exact solution would look something like that, 00:39:01.760 --> 00:39:03.410 finishing off here. 00:39:03.410 --> 00:39:04.710 That would be the exact solution. 00:39:04.710 --> 00:39:10.030 And what we're seeing now is an amplitude decay. 00:39:10.030 --> 00:39:10.865 It's already there. 00:39:10.865 --> 00:39:13.260 It would come in right there. 00:39:13.260 --> 00:39:15.090 That is the exact solution. 00:39:15.090 --> 00:39:16.140 We'll stop here. 00:39:16.140 --> 00:39:19.190 The period is of course this t here, as shown. 00:39:19.190 --> 00:39:23.280 This would be the exact solution for one period. 00:39:23.280 --> 00:39:27.130 What we are seeing as an amplitude decay, a drop in the 00:39:27.130 --> 00:39:30.650 numerical solution, and period elongation in 00:39:30.650 --> 00:39:32.150 the numerical solution. 00:39:32.150 --> 00:39:35.470 And the percentage period elongation is plotted here. 00:39:35.470 --> 00:39:39.470 We notice that the Newmark method corresponds to this 00:39:39.470 --> 00:39:40.480 curve here. 00:39:40.480 --> 00:39:45.650 So with delta-t over t being 0.10. 00:39:45.650 --> 00:39:47.020 We have about-- 00:39:47.020 --> 00:39:49.830 let us just look it up here-- 00:39:49.830 --> 00:39:54.550 we have about 3%, 3% period elongation 00:39:54.550 --> 00:39:56.050 for the Newmark method. 00:39:56.050 --> 00:39:58.110 I'm also showing other techniques here, which are 00:39:58.110 --> 00:40:01.320 also implicit techniques, the Wilson Theta method with Theta 00:40:01.320 --> 00:40:07.970 1.4 has about 5% period elongation for delta-t over t 00:40:07.970 --> 00:40:11.050 being 0.10. 00:40:11.050 --> 00:40:14.380 Another point of interest is of course also is the 00:40:14.380 --> 00:40:15.990 amplitude decay. 00:40:15.990 --> 00:40:20.520 And that amplitude decay is plotted on this graph here, 00:40:20.520 --> 00:40:23.190 percentage amplitude decay as a function of 00:40:23.190 --> 00:40:25.370 delta-t over t again. 00:40:25.370 --> 00:40:26.540 Delta-t over t. 00:40:26.540 --> 00:40:30.020 And notice that the Newmark method does not appear. 00:40:30.020 --> 00:40:34.080 In fact, the Newmark method has no amplitude decay. 00:40:34.080 --> 00:40:36.860 The solution is right down here. 00:40:36.860 --> 00:40:41.230 No amplitude decay for any delta-t over t, whereas the 00:40:41.230 --> 00:40:45.690 Wilson methods and the Houbolt method is also shown here, 00:40:45.690 --> 00:40:49.170 have the amplitude decay shown by these curves. 00:40:49.170 --> 00:40:52.660 Notice that the Wilson method has less amplitude decay then 00:40:52.660 --> 00:40:54.040 the Houbolt method. 00:40:54.040 --> 00:40:58.990 The period elongation for the Houbolt method was also shown 00:40:58.990 --> 00:41:01.370 on this view graph here. 00:41:01.370 --> 00:41:06.840 So the Newmark method does have period elongation, but it 00:41:06.840 --> 00:41:14.410 does not have any amplitude decay in this solution. 00:41:14.410 --> 00:41:17.150 I should also mention this, of course, as one important 00:41:17.150 --> 00:41:20.000 point, I forgot to mention that I'm looking here at the 00:41:20.000 --> 00:41:24.530 Newmark method with delta 1/2 alpha equal to 1/4. 00:41:24.530 --> 00:41:29.900 This is the constant average acceleration method. 00:41:29.900 --> 00:41:32.540 Well, let us look then at how can we 00:41:32.540 --> 00:41:35.460 actually perform the solution? 00:41:35.460 --> 00:41:40.020 How can we select an appropriate time step delta-t? 00:41:40.020 --> 00:41:47.770 Here we notice that the dynamic load factor that we 00:41:47.770 --> 00:41:53.890 are knowing that we can derive for the solution of this 00:41:53.890 --> 00:41:56.660 differential equation here-- 00:41:56.660 --> 00:41:58.510 here I included damping-- 00:41:58.510 --> 00:42:01.620 the dynamic load factor plotted along here as a 00:42:01.620 --> 00:42:03.590 functional of p over omega. 00:42:03.590 --> 00:42:07.760 p being the exciting free frequency, omega being the 00:42:07.760 --> 00:42:10.290 frequency of the system. 00:42:10.290 --> 00:42:13.280 The dynamic load factor looks as shown here. 00:42:13.280 --> 00:42:17.800 If we have no damping, you're talking about this curve. 00:42:17.800 --> 00:42:19.800 Now notice the important point. 00:42:19.800 --> 00:42:23.140 The dynamic load factor, of course, is 1 for static 00:42:23.140 --> 00:42:26.900 response, and it really gives us a magnification of the 00:42:26.900 --> 00:42:28.680 static response. 00:42:28.680 --> 00:42:32.290 In a static response, we would have this solution. 00:42:32.290 --> 00:42:40.670 And if the p over omega value is say 2, we would have that 00:42:40.670 --> 00:42:44.990 maximum solution in the dynamic response analysis. 00:42:44.990 --> 00:42:48.500 If p over omega is equal to 0.5. 00:42:48.500 --> 00:42:52.920 we would have, for zero damping, this maximum solution 00:42:52.920 --> 00:42:55.430 in the dynamic response history. 00:42:55.430 --> 00:42:58.130 Or rather this would be the maximum dynamic response that 00:42:58.130 --> 00:43:00.620 you would see in the dynamic solution. 00:43:00.620 --> 00:43:04.370 This is how the dynamic load factor is defined. 00:43:04.370 --> 00:43:08.140 Of course, at p over omega equal to 1, we have resonance. 00:43:08.140 --> 00:43:11.090 In practice, we have always some damping, and if you look 00:43:11.090 --> 00:43:15.430 at this amount of damping, this would be the maximum with 00:43:15.430 --> 00:43:20.470 p over omega equal to 1, we would have this as the maximum 00:43:20.470 --> 00:43:23.520 response that is ever measured in the 00:43:23.520 --> 00:43:26.900 dynamic response solution. 00:43:26.900 --> 00:43:33.610 Well, if we look at this graph, we recognize of course 00:43:33.610 --> 00:43:38.920 that as p over omega goes against 0, in other words as 00:43:38.920 --> 00:43:45.020 the exciting frequency, as over the frequency of the 00:43:45.020 --> 00:43:49.060 system goes against 0, we really talk about static 00:43:49.060 --> 00:43:49.980 conditions. 00:43:49.980 --> 00:43:54.580 As p over omega is very large, we are talking about no 00:43:54.580 --> 00:43:56.050 response whatsoever. 00:43:56.050 --> 00:44:00.390 And the important dynamic response really lies somewhat 00:44:00.390 --> 00:44:04.640 in this region here with that being the tail end of it. 00:44:04.640 --> 00:44:08.350 But the important dynamic response really lies in here, 00:44:08.350 --> 00:44:11.250 that we have to include in the solution. 00:44:11.250 --> 00:44:14.030 Out here we have basically a static response. 00:44:14.030 --> 00:44:18.150 Let us look at two simple cases. 00:44:18.150 --> 00:44:25.420 For p over omega 0.05, we can see that the static response 00:44:25.420 --> 00:44:28.640 here shown in a dashed line, the dynamic response showing 00:44:28.640 --> 00:44:29.890 as a solid line. 00:44:29.890 --> 00:44:31.100 They're basically the same. 00:44:31.100 --> 00:44:33.710 In other words, a dynamic response being equal to the 00:44:33.710 --> 00:44:37.360 static response, practically equal to the static response. 00:44:37.360 --> 00:44:40.960 We are down at this range here. 00:44:40.960 --> 00:44:44.815 As another example, if we have p over omega equal to 3, we 00:44:44.815 --> 00:44:47.570 are down over here. 00:44:47.570 --> 00:44:53.070 Then the response, the static response is shown here in the 00:44:53.070 --> 00:44:57.030 dashed line, and the dynamic response is shown 00:44:57.030 --> 00:44:58.380 as the solid line. 00:44:58.380 --> 00:45:01.110 Very little response here at all. 00:45:01.110 --> 00:45:05.280 Because as p over omega goes to infinity, there is no 00:45:05.280 --> 00:45:07.870 response for the system at all anymore. 00:45:07.870 --> 00:45:10.780 Basically, what we're saying then is that the frequency of 00:45:10.780 --> 00:45:13.230 the applied loading is so fast, that the 00:45:13.230 --> 00:45:14.950 structure can;t just move. 00:45:14.950 --> 00:45:16.330 It just cannot move at all. 00:45:16.330 --> 00:45:18.960 And we don't have any response in the structure. 00:45:18.960 --> 00:45:27.240 Well, the important point is that if we know the frequency 00:45:27.240 --> 00:45:33.720 content of the loading, then what we have to do in the use 00:45:33.720 --> 00:45:39.220 of the Newmark method is to calculate a delta-t small 00:45:39.220 --> 00:45:42.700 enough to pick up the loading, of course, because that is 00:45:42.700 --> 00:45:48.960 given in terms of this great steps or this great input 00:45:48.960 --> 00:45:52.240 points to the computer program, and the time step has 00:45:52.240 --> 00:45:56.360 to be small enough to integrate the dynamic response 00:45:56.360 --> 00:45:58.260 of importance accurately. 00:45:58.260 --> 00:46:02.090 The static response, the static response is contained 00:46:02.090 --> 00:46:09.120 in the solution anyway, because the K matrix is on the 00:46:09.120 --> 00:46:11.270 left hand side of the solution. 00:46:11.270 --> 00:46:15.390 So the complete solution process therefore can be 00:46:15.390 --> 00:46:17.060 summarized as follows. 00:46:17.060 --> 00:46:21.850 Identify the frequencies contained in the loading, and 00:46:21.850 --> 00:46:24.910 using a Fourier analysis, if necessary. 00:46:24.910 --> 00:46:28.220 Choose a finite element mesh that accurately represents all 00:46:28.220 --> 00:46:32.350 frequencies up to about four times the highest frequency 00:46:32.350 --> 00:46:35.210 omega u contained in the loading. 00:46:35.210 --> 00:46:37.670 The highest frequency contained in the loading is 00:46:37.670 --> 00:46:42.920 really the cutoff frequency that will give us a time step. 00:46:42.920 --> 00:46:46.630 And that then, the time step is in fact obtained via these 00:46:46.630 --> 00:46:47.340 consideration. 00:46:47.340 --> 00:46:48.690 Perform a direct integration analysis. 00:46:48.690 --> 00:46:52.640 The time step delta-t for this solutions should equal about 1 00:46:52.640 --> 00:46:56.930 over 20 Tu, where Tu is 2 pi over omega u, which is the 00:46:56.930 --> 00:46:59.320 highest frequency contained in the loading. 00:47:02.340 --> 00:47:05.260 What we are saying basically is that we want to select a 00:47:05.260 --> 00:47:09.160 finite element mesh to represent the highest 00:47:09.160 --> 00:47:13.600 frequency in the loading correctly, accurately, and 00:47:13.600 --> 00:47:17.650 then we want to select a time step in the data integration 00:47:17.650 --> 00:47:22.250 solution using the Newmark method that integrates the 00:47:22.250 --> 00:47:25.900 highest frequency contained in the loading also accurately, 00:47:25.900 --> 00:47:28.360 or the response in that highest frequency contained in 00:47:28.360 --> 00:47:29.610 the loading also accurately. 00:47:32.000 --> 00:47:36.760 And this then gives us basically a step-by-step 00:47:36.760 --> 00:47:41.090 procedure for the selection of the finite element mesh and 00:47:41.090 --> 00:47:42.070 the times step. 00:47:42.070 --> 00:47:46.310 Notice that the mesh selection and the time step selection 00:47:46.310 --> 00:47:51.360 both of course hinge on the structure and the particular 00:47:51.360 --> 00:47:53.290 frequency content of the loading. 00:47:53.290 --> 00:47:55.070 That's important. 00:47:55.070 --> 00:47:58.906 This is the modeling procedure that we would use for a 00:47:58.906 --> 00:48:01.340 structural vibration problem. 00:48:01.340 --> 00:48:04.890 We would use this time step here in the implicit 00:48:04.890 --> 00:48:08.850 integration, because we get all the accuracy we want. 00:48:08.850 --> 00:48:11.990 In an explicit integration, we would have to use a 00:48:11.990 --> 00:48:13.320 smaller time step. 00:48:13.320 --> 00:48:16.390 The sentence down here "or be smaller for stability reasons" 00:48:16.390 --> 00:48:19.290 really refers to the explicit time integration. 00:48:19.290 --> 00:48:22.780 In the Newmark method, we would not use that condition. 00:48:22.780 --> 00:48:28.010 We would simply use Tu, being 2 pi omega u and delta-t being 00:48:28.010 --> 00:48:30.280 1 over 20 times Tu. 00:48:30.280 --> 00:48:34.400 So in the Newmark method, I blank this out, and this is 00:48:34.400 --> 00:48:37.760 here the delta-t that we would be using. 00:48:37.760 --> 00:48:43.610 And just to summarize once the modeling of wave propagation 00:48:43.610 --> 00:48:46.520 problem here, here what we would do is we assume that the 00:48:46.520 --> 00:48:48.870 wavelength is Lw. 00:48:48.870 --> 00:48:53.180 The total time for the wave to travel past a point is then tw 00:48:53.180 --> 00:48:57.500 equal Lw over c, where c is the wave speed. 00:48:57.500 --> 00:49:00.500 Assuming that n time steps are necessary to represent the 00:49:00.500 --> 00:49:05.750 wave, we get our delta-t and the effective length, which I 00:49:05.750 --> 00:49:08.800 talked about earlier, of the finite element would be then 00:49:08.800 --> 00:49:09.990 given by this formula. 00:49:09.990 --> 00:49:14.260 So the procedure here is somewhat different in that if 00:49:14.260 --> 00:49:18.870 we have a wave that looks like this say, we now select a 00:49:18.870 --> 00:49:23.370 certain number of times steps, delta-t, to pick up the wave 00:49:23.370 --> 00:49:24.730 accurately. 00:49:24.730 --> 00:49:31.280 And this being the length Lw, the time for the wave to 00:49:31.280 --> 00:49:33.500 travel past the point is tw. 00:49:33.500 --> 00:49:38.400 And delta-t then is the number of time points, or time steps, 00:49:38.400 --> 00:49:40.880 that are used to represent the wave accurately. 00:49:43.870 --> 00:49:47.420 And once we have the time step, we select the effective 00:49:47.420 --> 00:49:48.530 length of an element. 00:49:48.530 --> 00:49:52.030 And then, of course it depends on whether we have a low-order 00:49:52.030 --> 00:49:56.120 element or a high-order element, that will give us 00:49:56.120 --> 00:50:00.990 then the procedure of how to apply the 00:50:00.990 --> 00:50:03.620 effective lengths criteria. 00:50:03.620 --> 00:50:06.440 I like to finally now summarize the complete 00:50:06.440 --> 00:50:07.690 solution process. 00:50:09.680 --> 00:50:15.560 This is a table that I put together for you to show how 00:50:15.560 --> 00:50:19.740 the complete solution process for explicit and implicit time 00:50:19.740 --> 00:50:22.110 integration can be very effectively implemented in a 00:50:22.110 --> 00:50:23.860 computer program. 00:50:23.860 --> 00:50:27.340 We form a linear stiffness matrix K, the mass matrix M, 00:50:27.340 --> 00:50:30.230 the damping matrix C, whichever applicable. 00:50:30.230 --> 00:50:31.870 We calculate the following constants. 00:50:31.870 --> 00:50:34.510 If we select the Newmark method, these are the 00:50:34.510 --> 00:50:37.040 constants that you would use. 00:50:37.040 --> 00:50:38.610 Let's not go into details. 00:50:38.610 --> 00:50:43.270 These constants are given in the textbook. 00:50:43.270 --> 00:50:46.270 They are given in the ADINA manual, if you are using the 00:50:46.270 --> 00:50:47.490 ADINA computer program. 00:50:47.490 --> 00:50:52.000 They are widely published. 00:50:52.000 --> 00:50:55.000 We simply cast them in this form. 00:50:55.000 --> 00:50:56.520 In the central difference method, we 00:50:56.520 --> 00:50:59.810 would use these constants. 00:50:59.810 --> 00:51:02.420 So there is some initialization involved. 00:51:02.420 --> 00:51:05.710 Then in the central difference method, we also want to 00:51:05.710 --> 00:51:13.150 calculate our first displacement, delta-t U, as 00:51:13.150 --> 00:51:15.330 shown here. 00:51:15.330 --> 00:51:18.910 And in the Newmark method, we don't need to do that, because 00:51:18.910 --> 00:51:22.340 we simply march ahead from time t to time t plus delta-t. 00:51:22.340 --> 00:51:25.820 You only need the condition at the conditions at time t to 00:51:25.820 --> 00:51:29.090 obtain the solution at times t plus delta-t. 00:51:29.090 --> 00:51:32.070 In the central difference method, remember we need two 00:51:32.070 --> 00:51:33.420 previous values. 00:51:33.420 --> 00:51:37.150 We then form an effective linear coefficient matrix. 00:51:37.150 --> 00:51:39.570 In implicit time integration, this is 00:51:39.570 --> 00:51:40.780 the coefficient matrix. 00:51:40.780 --> 00:51:42.590 The stiffness matrix being there, as I 00:51:42.590 --> 00:51:43.840 showed to you earlier. 00:51:43.840 --> 00:51:47.890 The mass matrix and damping matrix being there. 00:51:47.890 --> 00:51:56.340 If we look at this a0, we notice that a0 is 1 over 00:51:56.340 --> 00:51:59.700 delta-t squared and there's a constant alpha here. 00:51:59.700 --> 00:52:02.660 The Newmark constant. 00:52:02.660 --> 00:52:05.080 But it's 1 over delta-t squared. 00:52:05.080 --> 00:52:06.640 That is the important point. 00:52:06.640 --> 00:52:12.280 So as delta-t gets larger and larger, larger and larger, 00:52:12.280 --> 00:52:15.680 this effect goes against 0. 00:52:15.680 --> 00:52:19.030 The same holds for the a1 constant. 00:52:19.030 --> 00:52:21.060 As delta-t gets larger and larger, this effect 00:52:21.060 --> 00:52:22.150 goes against 0. 00:52:22.150 --> 00:52:25.560 And this is the reason why the Newmark method really goes 00:52:25.560 --> 00:52:31.550 down or collapses to the static analysis, if we have 00:52:31.550 --> 00:52:32.970 very large time steps. 00:52:32.970 --> 00:52:38.320 In explicit integration, we use this coefficient matrix. 00:52:38.320 --> 00:52:43.740 And now for each integration step, or for each time step, 00:52:43.740 --> 00:52:46.320 we perform the following calculations. 00:52:46.320 --> 00:52:51.780 We calculate the t plus delta-t R hat. 00:52:51.780 --> 00:52:55.780 And implicit time integration, the Newmark method, in 00:52:55.780 --> 00:52:59.050 explicit time integration, we calculate this vector. 00:52:59.050 --> 00:53:02.450 Notice at tF here, no k times tU. 00:53:02.450 --> 00:53:06.170 k times 2u is equal to tF. 00:53:06.170 --> 00:53:08.890 In implicit time integration, having calculated for each 00:53:08.890 --> 00:53:15.530 time step this load step, we now use the equations K hat 00:53:15.530 --> 00:53:20.320 times t plus delta-t u, equals t plus delta-t R hat to 00:53:20.320 --> 00:53:22.850 calculate t plus delta-t u. 00:53:22.850 --> 00:53:27.010 Of course, we would triangulize the K hat matrix 00:53:27.010 --> 00:53:29.310 prior to going into this loop. 00:53:29.310 --> 00:53:31.750 We do it once and forward. 00:53:31.750 --> 00:53:36.790 In explicit time integration, we use this matrix or this 00:53:36.790 --> 00:53:39.850 vector here, I should say, this vector now, and we can 00:53:39.850 --> 00:53:44.850 calculate directly for the t plus delta-t U also. 00:53:44.850 --> 00:53:48.300 The two equations that I used are summarized here. 00:53:48.300 --> 00:53:50.120 In implicit time integration, this is the 00:53:50.120 --> 00:53:53.410 equation that we use. 00:53:53.410 --> 00:53:55.750 Notice we only need to factorize this 00:53:55.750 --> 00:53:57.080 K hat matrix once. 00:53:57.080 --> 00:53:59.910 We do it outside the step-by-step integration, as I 00:53:59.910 --> 00:54:01.470 pointed out earlier. 00:54:01.470 --> 00:54:05.330 And then this gives us the increment in displacements. 00:54:05.330 --> 00:54:09.230 In explicit time integration, this is the equation we use. 00:54:09.230 --> 00:54:13.580 If M and C are diagonal, it's a trivial operation as I 00:54:13.580 --> 00:54:15.410 pointed out earlier. 00:54:15.410 --> 00:54:18.070 With this then, we have the new displacements. 00:54:18.070 --> 00:54:21.990 And on the last view graph, I show you the formulae that we 00:54:21.990 --> 00:54:25.490 are then using to calculate the accelerations in the 00:54:25.490 --> 00:54:29.050 Newmark method from the incremental displacements and 00:54:29.050 --> 00:54:30.880 variables that we knew already. 00:54:30.880 --> 00:54:33.780 Having now calculated the accelerations, we can get the 00:54:33.780 --> 00:54:36.370 velocities at time t plus delta-t. 00:54:36.370 --> 00:54:38.990 And of course, we have already our displacement at time t 00:54:38.990 --> 00:54:39.980 plus delta-t. 00:54:39.980 --> 00:54:42.520 In the central difference method, we use our basic 00:54:42.520 --> 00:54:46.010 equations that we already applied earlier in the 00:54:46.010 --> 00:54:48.780 development of the method to calculate the velocities and 00:54:48.780 --> 00:54:51.510 the accelerations at time t. 00:54:51.510 --> 00:54:56.750 This completes the procedure that is effectively used in a 00:54:56.750 --> 00:55:00.400 computer program for the integration of the dynamic 00:55:00.400 --> 00:55:01.680 equilibrium equations. 00:55:01.680 --> 00:55:06.460 I like to just now mention that I've placed heavy 00:55:06.460 --> 00:55:09.910 emphasis on the Newmark method in an implicit integration. 00:55:09.910 --> 00:55:11.900 Of course, there are other implicit integration formulae 00:55:11.900 --> 00:55:16.570 available, the Wilson method, the Houbolt method, and so on. 00:55:16.570 --> 00:55:18.760 The Newmark method has shown, has 00:55:18.760 --> 00:55:20.750 proven to be quite effective. 00:55:20.750 --> 00:55:24.030 And it's certainly a good example to demonstrate to you 00:55:24.030 --> 00:55:28.150 how the actual procedure is implemented and how an 00:55:28.150 --> 00:55:30.370 implicit integration scheme is being used. 00:55:30.370 --> 00:55:33.560 Other implicit integration methods would be used in much 00:55:33.560 --> 00:55:34.790 the same way. 00:55:34.790 --> 00:55:37.990 This is all I wanted to mention to you, discuss with 00:55:37.990 --> 00:55:39.180 you in this lecture. 00:55:39.180 --> 00:55:40.530 Thank you very much for your attention.