WEBVTT 00:00:00.090 --> 00:00:02.450 The following content is provided under a Creative 00:00:02.450 --> 00:00:03.830 Commons license. 00:00:03.830 --> 00:00:06.080 Your support will help MIT OpenCourseWare 00:00:06.080 --> 00:00:10.160 continue to offer high quality educational resources for free. 00:00:10.160 --> 00:00:12.710 To make a donation or to view additional materials 00:00:12.710 --> 00:00:16.630 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.630 --> 00:00:17.325 at ocw.mit.edu. 00:00:21.839 --> 00:00:23.880 PROFESSOR: We've got three lectures left-- today, 00:00:23.880 --> 00:00:27.620 Thursday, and next Tuesday. 00:00:27.620 --> 00:00:34.230 So today we're going to talk a little bit more 00:00:34.230 --> 00:00:46.710 about modal analysis, and we did initial conditions last time. 00:00:46.710 --> 00:00:53.350 This time we'll do harmonic excitation mostly 00:00:53.350 --> 00:00:55.390 with a little review there. 00:00:55.390 --> 00:01:01.350 Thursday-- let me say this slightly differently. 00:01:04.650 --> 00:01:07.710 Modal analysis looks at the vibration of a system 00:01:07.710 --> 00:01:12.000 with many degrees of freedom and looks at one mode at a time, 00:01:12.000 --> 00:01:15.664 but you can also just solve the whole thing at once. 00:01:15.664 --> 00:01:18.080 You don't have to break it down into the individual modes, 00:01:18.080 --> 00:01:20.870 so you can come up with a-- if you have a harmonic 00:01:20.870 --> 00:01:24.597 excitation on a system with many degrees of freedom, 00:01:24.597 --> 00:01:26.930 if you put in a harmonic force, the whole thing is going 00:01:26.930 --> 00:01:30.090 to shake, and you can solve it in one go, 00:01:30.090 --> 00:01:33.470 and that's essentially using a transfer function approach. 00:01:33.470 --> 00:01:36.980 So this is breaking it down one mode at a time. 00:01:36.980 --> 00:01:41.570 Here we're going to do this concept of transfer functions 00:01:41.570 --> 00:01:46.820 for Hij, so response at location i due to an excitation at j. 00:01:46.820 --> 00:01:48.780 In a multiple degree of freedom system, 00:01:48.780 --> 00:01:50.340 you get many different combinations. 00:01:50.340 --> 00:01:55.380 So this is steady state response, but done all at once. 00:01:55.380 --> 00:01:59.680 And Tuesday-- and also this has an application to something 00:01:59.680 --> 00:02:01.305 I called dynamic absorbers. 00:02:05.850 --> 00:02:08.240 I'll just be able to scratch the surface of that, 00:02:08.240 --> 00:02:12.020 but the Hancock Building across the way-- I told you a couple 00:02:12.020 --> 00:02:13.570 lectures back about it. 00:02:13.570 --> 00:02:16.100 In the 1970s, it was brand new, and the windows 00:02:16.100 --> 00:02:17.162 were falling out. 00:02:17.162 --> 00:02:18.870 And there was lots of windows falling out 00:02:18.870 --> 00:02:21.120 at the bottom and none at the top, 00:02:21.120 --> 00:02:22.920 and the thing was bending back and forth 00:02:22.920 --> 00:02:27.420 like just a cantilever, a reed in the wind, 00:02:27.420 --> 00:02:29.750 and one of the fixes for it is they 00:02:29.750 --> 00:02:34.000 put a thing called the dynamic absorber on the 58th floor. 00:02:34.000 --> 00:02:42.380 There are two 300 ton blocks of lead on a pressurized oil film 00:02:42.380 --> 00:02:45.230 so they can slide back and forth on the 58th floor 00:02:45.230 --> 00:02:46.660 of that building. 00:02:46.660 --> 00:02:49.210 They're called dynamic absorbers. 00:02:49.210 --> 00:02:52.590 And at the expense of letting these big blocks of lead slide 00:02:52.590 --> 00:02:55.110 back and forth, it keeps the building from vibrating. 00:02:55.110 --> 00:02:57.220 It keeps the windows from falling out. 00:02:57.220 --> 00:02:59.740 So I hope to get to talking a little bit 00:02:59.740 --> 00:03:02.380 about dynamic absorbers-- one other way 00:03:02.380 --> 00:03:04.860 of stopping problem vibration. 00:03:04.860 --> 00:03:13.630 And on Tuesday, we'll talk about strings and beams. 00:03:13.630 --> 00:03:15.845 Just a brief introduction to continuous systems. 00:03:19.140 --> 00:03:21.739 And that'll be our last lecture, and probably 00:03:21.739 --> 00:03:23.530 give you a very quick review then of what's 00:03:23.530 --> 00:03:25.080 going to be covered on the final, 00:03:25.080 --> 00:03:29.793 but mostly the final will be covering this last third 00:03:29.793 --> 00:03:32.890 of the course on vibration. 00:03:32.890 --> 00:03:38.460 So let's turn to modal analysis, and posted on Stellar 00:03:38.460 --> 00:03:43.260 is a little too page handout that gives you a just step 00:03:43.260 --> 00:03:47.375 by step cookbook approach to conducting a modal analysis. 00:03:47.375 --> 00:03:49.250 And we're going to hit the highlights of that 00:03:49.250 --> 00:03:51.900 this morning as a way of reviewing what we learned 00:03:51.900 --> 00:03:55.470 last time and moving on to calculating the response 00:03:55.470 --> 00:03:57.910 to harmonic forces. 00:03:57.910 --> 00:04:11.700 So we begin with some n degree of freedom system linearized. 00:04:11.700 --> 00:04:16.779 We're only dealing with linear equations of motion. 00:04:16.779 --> 00:04:20.300 And our first step is we need equations of motion. 00:04:20.300 --> 00:04:31.060 So you write them m-- in general, you'd 00:04:31.060 --> 00:04:33.990 write out your equations of motion for the system, 00:04:33.990 --> 00:04:37.100 and these x's here are just my generalized coordinates. 00:04:37.100 --> 00:04:39.640 They can be rotations, deflections, whatever 00:04:39.640 --> 00:04:42.694 makes sense in the problem. 00:04:42.694 --> 00:04:43.610 That's the first step. 00:04:43.610 --> 00:04:45.670 You need your equations of motion. 00:04:45.670 --> 00:04:55.690 Second step, find undamped natural frequency-- 00:04:55.690 --> 00:05:01.306 is the omega i's-- I'll just call it omega i's-- and mode 00:05:01.306 --> 00:05:01.805 shapes. 00:05:06.700 --> 00:05:09.380 And the mode shapes we put into a matrix 00:05:09.380 --> 00:05:11.070 that we call a mode shape matrix. 00:05:11.070 --> 00:05:13.340 I write it as u. 00:05:13.340 --> 00:05:15.010 So that's the next step in the problem. 00:05:18.260 --> 00:05:32.830 The third is basically to apply or invoke the modal expansion 00:05:32.830 --> 00:05:34.570 theorem. 00:05:34.570 --> 00:05:37.940 This is the key to the whole thing, 00:05:37.940 --> 00:05:42.560 and that is to say that you can write the generalized motions 00:05:42.560 --> 00:05:45.950 of the system-- the responses of the system-- 00:05:45.950 --> 00:05:56.800 as a weighted sum of the individual modes of the system. 00:05:56.800 --> 00:05:59.960 So the q's are the modal amplitudes. 00:05:59.960 --> 00:06:04.910 The u's are the mode shape-- each mode, like the first mode, 00:06:04.910 --> 00:06:07.640 has some amplitude and time dependence, 00:06:07.640 --> 00:06:11.370 and its motion is distributed to the whole system 00:06:11.370 --> 00:06:15.181 according to its mode shape, and that's what that statement is. 00:06:18.070 --> 00:06:21.190 And another way of writing it which is more intuitive 00:06:21.190 --> 00:06:24.350 is this is a summation then of i equals 00:06:24.350 --> 00:06:29.440 1 over the degrees of freedom of the system of the mode 00:06:29.440 --> 00:06:39.390 shapes times qi of t, the modal amplitudes of the response 00:06:39.390 --> 00:06:41.250 of that single degree of freedom system 00:06:41.250 --> 00:06:43.270 that describes each of the modes. 00:06:59.090 --> 00:07:09.200 Now, I should-- the qi's are the solution 00:07:09.200 --> 00:07:16.350 to n equations of the form mi qi double 00:07:16.350 --> 00:07:28.400 dot plus ci qi dot plus ki qi equals 00:07:28.400 --> 00:07:32.370 the excitation for that single degree of freedom system. 00:07:32.370 --> 00:07:34.150 So the modal expansion theorem says 00:07:34.150 --> 00:07:36.110 we're going to take each mode of the system, 00:07:36.110 --> 00:07:39.940 treat it like a single degree of freedom problem. 00:07:39.940 --> 00:07:42.700 And so for the first mode, I would be 1. 00:07:42.700 --> 00:07:46.120 You would have a modal mass, damping, stiffness, 00:07:46.120 --> 00:07:47.710 and a force for that mode. 00:07:47.710 --> 00:07:50.935 And you know how to solve for the response of a single degree 00:07:50.935 --> 00:07:53.820 of freedom system to initial conditions, 00:07:53.820 --> 00:07:56.900 or the steady state response to some harmonic input, 00:07:56.900 --> 00:07:59.080 and you would do that for each of the modes. 00:07:59.080 --> 00:08:01.130 And to get back to the final answer, 00:08:01.130 --> 00:08:03.540 you sum them back up again this way. 00:08:03.540 --> 00:08:09.100 So I've put up here on the board where 00:08:09.100 --> 00:08:12.340 that-- the description of that demo, 00:08:12.340 --> 00:08:15.110 and we went through all this information 00:08:15.110 --> 00:08:17.650 last time so you have it in your notes for last time. 00:08:17.650 --> 00:08:20.050 If you missed it last time, I put it up again. 00:08:20.050 --> 00:08:23.360 So here's the demo on the table. 00:08:23.360 --> 00:08:27.980 This is my diagram of it. 00:08:27.980 --> 00:08:31.640 Mass 1, mass 2, spring, lowercase k1, 00:08:31.640 --> 00:08:33.720 lowercase k2-- so the lower cases 00:08:33.720 --> 00:08:36.390 I'm going to use to describe system parameters. 00:08:36.390 --> 00:08:38.929 Uppercase K's and that sort of thing 00:08:38.929 --> 00:08:41.799 will be my modal parameters. 00:08:41.799 --> 00:08:45.440 So this has masses, stiffnesses, and damping, 00:08:45.440 --> 00:08:50.400 and this damper is connected to the non-moving frame, 00:08:50.400 --> 00:08:52.880 and so is this one to model the damping sliding 00:08:52.880 --> 00:08:55.320 up and down on that shaft. 00:08:55.320 --> 00:08:57.750 So the equations of motion of the system 00:08:57.750 --> 00:09:01.680 are here in your generalized coordinates. 00:09:01.680 --> 00:09:03.900 And I've included the possibility 00:09:03.900 --> 00:09:06.130 that I could have a force in the first mass, F1, 00:09:06.130 --> 00:09:07.360 a force in the second mass. 00:09:07.360 --> 00:09:11.900 So this is a completely general set up for this problem. 00:09:11.900 --> 00:09:14.910 Then you need to find the natural frequencies and mode 00:09:14.910 --> 00:09:20.440 shapes, so you assume a solution of the form e to i omega t. 00:09:20.440 --> 00:09:24.020 Plug in, and you get this algebraic equation. 00:09:24.020 --> 00:09:27.590 Minus omega squared m plus k times some vector, which will 00:09:27.590 --> 00:09:29.680 turn out to be the mode shapes. 00:09:29.680 --> 00:09:31.665 And either this is 0, which is trivial, 00:09:31.665 --> 00:09:34.870 or the determinant of this is 0, and this gives you 00:09:34.870 --> 00:09:36.840 the roots of this determinant, give you 00:09:36.840 --> 00:09:38.900 the natural frequencies of the system. 00:09:38.900 --> 00:09:42.680 If it's a four degree of freedom problem, you get four roots. 00:09:42.680 --> 00:09:45.280 So we found for this particular problem 00:09:45.280 --> 00:09:50.980 that the natural frequencies are 5.65 and 17.69, 00:09:50.980 --> 00:09:55.140 and the mass matrix, if you weren't here last time, 00:09:55.140 --> 00:10:06.840 in kilograms, 3.193, 0.63. 00:10:06.840 --> 00:10:11.750 And the stiffness matrix in newtons per meter 00:10:11.750 --> 00:10:14.780 for that system is this. 00:10:14.780 --> 00:10:22.250 Notice, stiffness matrices are always symmetric, 00:10:22.250 --> 00:10:26.150 and this one minus 36, minus 36. 00:10:26.150 --> 00:10:29.340 And many of you have been taking 2001. 00:10:29.340 --> 00:10:31.690 Did anybody invoke something called 00:10:31.690 --> 00:10:33.410 Maxwell's reciprocal theorem? 00:10:33.410 --> 00:10:36.760 Have you run into that? 00:10:36.760 --> 00:10:38.990 Well, this comes from structural analysis, 00:10:38.990 --> 00:10:41.660 this notion of the symmetry of the K matrix. 00:10:41.660 --> 00:10:44.710 The mass matrix is also going to be symmetric. 00:10:44.710 --> 00:10:47.910 So we need to find-- to do modal analysis, 00:10:47.910 --> 00:10:50.910 you make this calculation. 00:10:50.910 --> 00:10:52.550 You transpose mu. 00:10:52.550 --> 00:10:56.975 This is the transpose of the mode shape matrix mu, 00:10:56.975 --> 00:11:00.190 and if you do that calculation with these numbers 00:11:00.190 --> 00:11:05.640 and those mode shapes, you get a new mass-- 00:11:05.640 --> 00:11:07.420 you get the modal mass matrix, which 00:11:07.420 --> 00:11:09.520 is guaranteed to be diagonal, even 00:11:09.520 --> 00:11:10.850 if it wasn't to begin with. 00:11:10.850 --> 00:11:12.640 This is the numbers you get. 00:11:12.640 --> 00:11:15.630 The stiffness matrix certainly wasn't diagonal to begin with. 00:11:15.630 --> 00:11:18.950 It looked like that. 00:11:18.950 --> 00:11:23.580 And we multiply u transpose times k times u, 00:11:23.580 --> 00:11:24.890 and we get a diagonal matrix. 00:11:29.280 --> 00:11:31.890 This element here we'd call capital K1. 00:11:31.890 --> 00:11:34.600 It's the modal stiffness for mode 1. 00:11:34.600 --> 00:11:36.770 This is the modal mass for mode 1, 00:11:36.770 --> 00:11:41.370 and they represent a-- they're the numbers we 00:11:41.370 --> 00:11:45.050 need to write a single degree of freedom equation of motion 00:11:45.050 --> 00:11:47.850 of the kind at the bottom of that board up there. 00:11:47.850 --> 00:11:52.250 So we want to write M1q1 double dot 00:11:52.250 --> 00:12:03.570 plus some c1q1 dot plus k1q1 equals, in general, a-- this 00:12:03.570 --> 00:12:08.300 is a 1-- some excitation, but it's a single degree of freedom 00:12:08.300 --> 00:12:09.110 system. 00:12:09.110 --> 00:12:11.330 If it's a single degree of freedom system, 00:12:11.330 --> 00:12:13.830 what's its natural frequency using the parameters 00:12:13.830 --> 00:12:14.610 in this equation? 00:12:19.510 --> 00:12:20.490 AUDIENCE: k1/m1. 00:12:20.490 --> 00:12:23.660 PROFESSOR: Right, so omega 1 had better 00:12:23.660 --> 00:12:31.590 be the square root of k1/m1, and that's the square root of-- k1 00:12:31.590 --> 00:12:37.400 is 91-- nope. 00:12:37.400 --> 00:12:39.230 That's the real [INAUDIBLE]. 00:12:39.230 --> 00:12:41.030 I need my u transpose k here. 00:12:41.030 --> 00:12:44.640 113.71. 00:12:44.640 --> 00:12:46.490 113.71. 00:12:46.490 --> 00:12:56.420 And the modal mass for system one is 3.556. 00:12:56.420 --> 00:13:00.370 So that's 3 or so and 100 and something. 00:13:00.370 --> 00:13:03.160 That's neighborhood of 35, 36. 00:13:03.160 --> 00:13:06.890 Square root of 36 is about 6. 00:13:06.890 --> 00:13:09.450 So it needs to give you back exactly 00:13:09.450 --> 00:13:11.210 the correct natural frequency, which 00:13:11.210 --> 00:13:13.040 you found in the beginning when you 00:13:13.040 --> 00:13:14.900 solved for the natural frequencies and mode 00:13:14.900 --> 00:13:16.440 shapes of the system. 00:13:16.440 --> 00:13:19.420 So I repeated myself here. 00:13:19.420 --> 00:13:22.060 And if you write the same thing-- omega 2-- 00:13:22.060 --> 00:13:24.200 it better be equal to k2/m2. 00:13:24.200 --> 00:13:26.639 So these are just ways of verifying that you've 00:13:26.639 --> 00:13:27.930 done your arithmetic correctly. 00:13:32.370 --> 00:13:35.380 And I haven't done the damping matrix yet, 00:13:35.380 --> 00:13:37.730 because the damping matrices have 00:13:37.730 --> 00:13:39.500 to be treated rather carefully. 00:13:39.500 --> 00:13:42.550 They don't automatically uncouple, 00:13:42.550 --> 00:13:45.420 and so we'll address it in just a second. 00:13:48.100 --> 00:13:50.560 The damping matrix, however, would 00:13:50.560 --> 00:13:58.220 be written utcu would give me what 00:13:58.220 --> 00:14:06.225 I hope is a diagonalized modal damping matrix. 00:14:06.225 --> 00:14:08.440 I'm trying to make these look like caps or something 00:14:08.440 --> 00:14:11.610 to make them look a little different from the c's 00:14:11.610 --> 00:14:13.070 of the dashpots themselves. 00:14:13.070 --> 00:14:16.900 We hope to find a diagonalized damping matrix, 00:14:16.900 --> 00:14:21.890 but we have to make some effort to make that happen. 00:14:21.890 --> 00:14:25.900 Finally, one thing we haven't dealt with before. 00:14:25.900 --> 00:14:31.940 In order to-- when we derive the single degree of freedom modal 00:14:31.940 --> 00:14:34.680 systems, we had to multiply through 00:14:34.680 --> 00:14:40.980 by this u transpose, the original equations of motion, 00:14:40.980 --> 00:14:43.360 and so what we haven't worked with yet 00:14:43.360 --> 00:14:50.540 is how do you get the modal forces in the system? 00:14:50.540 --> 00:14:54.610 And they come from the calculation u transpose 00:14:54.610 --> 00:14:59.290 times the generalized forces in the system. 00:14:59.290 --> 00:15:01.520 So that's a vector times a matrix gives you 00:15:01.520 --> 00:15:04.650 a new vector, which are the generalized forces. 00:15:04.650 --> 00:15:06.430 And we're going to do that example in more 00:15:06.430 --> 00:15:07.760 detail in a few minutes. 00:15:18.190 --> 00:15:24.680 So last time we did the initial conditions problem, 00:15:24.680 --> 00:15:29.680 and we need the results of that to do-- 00:15:29.680 --> 00:15:31.990 we need the damping that we learned from that to do 00:15:31.990 --> 00:15:35.140 the force vibration problem. 00:15:35.140 --> 00:15:38.320 So let's go back and review a little bit about what we did. 00:15:38.320 --> 00:15:43.570 So let's let, for a moment, the generalized forces be 0. 00:15:43.570 --> 00:15:45.530 So no excitation. 00:15:45.530 --> 00:15:48.530 So we're only doing an initial conditions problem, 00:15:48.530 --> 00:15:55.470 and let's assume that we have some set of initial conditions 00:15:55.470 --> 00:15:59.430 on the generalized displacements at time 0. 00:15:59.430 --> 00:16:02.720 So those initial conditions on displacements-- 00:16:02.720 --> 00:16:07.290 they'll look something like x10 down to xn0, 00:16:07.290 --> 00:16:10.880 and we just have a two degree of freedom system in our example. 00:16:10.880 --> 00:16:14.875 And also you could have an x dot at t equals 0, 00:16:14.875 --> 00:16:21.270 and that would be some vector of v10's down to vn0's. 00:16:21.270 --> 00:16:23.300 And for today, we're going to let those be 0s. 00:16:23.300 --> 00:16:26.190 So we're just going to have an initial deflection 00:16:26.190 --> 00:16:27.650 in the system and see what happens. 00:16:33.300 --> 00:16:38.700 And we learned last time that, since the modal expansion 00:16:38.700 --> 00:16:43.160 theorem is what makes this whole thing work, 00:16:43.160 --> 00:16:52.490 the idea that x can be written as u times q-- 00:16:52.490 --> 00:16:56.360 then we ought to be able to say the same thing for-- if these 00:16:56.360 --> 00:16:58.240 are initial conditions, then we should 00:16:58.240 --> 00:17:01.870 be able to get the initial conditions in modal 00:17:01.870 --> 00:17:02.460 coordinates. 00:17:02.460 --> 00:17:11.319 So x at 0 here is also u at the q's at 0, 00:17:11.319 --> 00:17:14.140 but we normally would specify these. 00:17:14.140 --> 00:17:15.569 We need to know those. 00:17:15.569 --> 00:17:19.560 Well, to get these I just multiply through by u inverse 00:17:19.560 --> 00:17:24.940 so that the q, the vector of the initial displacements in modal 00:17:24.940 --> 00:17:31.980 coordinates, is just equal to u inverse times 00:17:31.980 --> 00:17:35.330 the initial conditions-- the initial deflections 00:17:35.330 --> 00:17:38.350 in the generalized coordinates. 00:17:38.350 --> 00:17:45.730 And q dot at 0, if you had non-zero initial velocities, 00:17:45.730 --> 00:17:50.940 would be u inverse times x dot at time 0. 00:18:02.820 --> 00:18:04.680 So let me tell you where we're going. 00:18:04.680 --> 00:18:07.955 I'm not going to go back all the way 00:18:07.955 --> 00:18:09.580 through the initial conditions problem. 00:18:09.580 --> 00:18:10.630 That was last lecture. 00:18:10.630 --> 00:18:12.690 I'm going to review the results of it, 00:18:12.690 --> 00:18:16.010 because what I want to get to is, how do you 00:18:16.010 --> 00:18:19.680 compute the response by modal analysis-- the response 00:18:19.680 --> 00:18:23.010 to a harmonic force? 00:18:23.010 --> 00:18:25.450 For example, if you get close to resonance 00:18:25.450 --> 00:18:27.120 for a single degree of freedom system, 00:18:27.120 --> 00:18:31.610 what controls the height of that peak? 00:18:31.610 --> 00:18:34.290 If you drive a single degree of freedom system at resonance, 00:18:34.290 --> 00:18:35.960 what's the most important parameter? 00:18:35.960 --> 00:18:36.870 AUDIENCE: Damping [INAUDIBLE]. 00:18:36.870 --> 00:18:37.661 PROFESSOR: Damping. 00:18:37.661 --> 00:18:39.570 So we really need to know damping. 00:18:39.570 --> 00:18:43.040 And the way we get damping is-- one of the ways is measure it. 00:18:43.040 --> 00:18:45.160 So I want to measure the right damping 00:18:45.160 --> 00:18:46.990 so that I can do-- what I'm trying 00:18:46.990 --> 00:18:50.940 to get to is to do the force vibration problem, 00:18:50.940 --> 00:18:54.830 but I need to get some estimates of damping to do it. 00:18:54.830 --> 00:18:56.560 Now, I said damping can sometimes 00:18:56.560 --> 00:19:01.330 be a problem, so I'm going to show you-- 00:19:01.330 --> 00:19:03.660 I alluded to this damp last time, 00:19:03.660 --> 00:19:07.280 but I wasn't able to kind of really clearly go through it. 00:19:07.280 --> 00:19:10.660 So I'm going to make a damping model, my damping matrix 00:19:10.660 --> 00:19:15.050 for my system such that it is proportional 00:19:15.050 --> 00:19:21.740 to the original mass and stiffness matrix of the system. 00:19:21.740 --> 00:19:24.482 So alpha is just a parameter that I get to choose. 00:19:24.482 --> 00:19:26.190 Beta is another one that I get to choose, 00:19:26.190 --> 00:19:29.010 and this is m and k, the original mass 00:19:29.010 --> 00:19:36.150 and stiffness matrices of this problem 00:19:36.150 --> 00:19:38.400 or any n degree of freedom problem. 00:19:38.400 --> 00:19:40.400 We're doing these two degree of freedom examples 00:19:40.400 --> 00:19:44.060 because they're tractable on the board. 00:19:44.060 --> 00:19:48.110 So if I make my damping matrix look like that, 00:19:48.110 --> 00:19:53.440 it is guaranteed to work when I do u transpose times that times 00:19:53.440 --> 00:19:56.950 u, because we know the mass and stiffness matrices give me 00:19:56.950 --> 00:20:00.710 diagonals, so I'm just essentially doing that again. 00:20:00.710 --> 00:20:03.015 So this is guaranteed to give me a diagonal matrix. 00:20:03.015 --> 00:20:03.780 It's what I want. 00:20:03.780 --> 00:20:04.400 Yes? 00:20:04.400 --> 00:20:07.330 AUDIENCE: So are those the original mass? 00:20:07.330 --> 00:20:09.690 PROFESSOR: Yes, these are the original ones. 00:20:09.690 --> 00:20:12.509 Usually, if I'm really trying to write the modal ones, 00:20:12.509 --> 00:20:15.050 I'll draw them with a diagonal through them or something just 00:20:15.050 --> 00:20:20.910 to-- so these are m and k, just like up here. 00:20:20.910 --> 00:20:21.650 Right there. 00:20:21.650 --> 00:20:23.600 They're right from the original system. 00:20:23.600 --> 00:20:25.620 They're the ones on the top of this board. 00:20:25.620 --> 00:20:29.940 There's the mass matrix, damping matrix, stiffness matrix 00:20:29.940 --> 00:20:33.020 for the original equations of motion. 00:20:33.020 --> 00:20:38.450 So I am saying that my unspecified damping matrix, 00:20:38.450 --> 00:20:42.010 which is written as c1, c2 there-- 00:20:42.010 --> 00:20:48.100 I'm going to represent it this way, 00:20:48.100 --> 00:20:53.010 so that u transpose cu, which is the calculation I need 00:20:53.010 --> 00:20:58.140 to be able to do-- what that gives me is an alpha, 00:20:58.140 --> 00:21:01.400 and here's my modal mass, diagonalized 00:21:01.400 --> 00:21:04.760 modal mass matrix that results, plus a beta 00:21:04.760 --> 00:21:08.280 times my diagonalized stiffness matrices. 00:21:08.280 --> 00:21:13.400 So Christina, these guys are the modal ones after doing 00:21:13.400 --> 00:21:15.550 u transpose mu, u transpose ku. 00:21:22.140 --> 00:21:25.150 So we have a two degree of freedom system, 00:21:25.150 --> 00:21:27.870 and when I do this calculation, I'll 00:21:27.870 --> 00:21:34.600 get modal mass m1, m2, k1, k2. 00:21:34.600 --> 00:21:39.350 So the final diagonalized stiffness matrix 00:21:39.350 --> 00:21:47.080 will end up looking like some capital C10, capital C2 here. 00:21:47.080 --> 00:21:50.150 That's my diagonalized damping matrix. 00:21:50.150 --> 00:21:54.890 It'll come from these, and I'll just write them out. 00:21:54.890 --> 00:22:02.390 So the C1 will be an alpha m1 plus beta k1, 00:22:02.390 --> 00:22:10.750 and C2 will be alpha m2 plus beta k2. 00:22:14.340 --> 00:22:16.130 So I have two free parameters with which 00:22:16.130 --> 00:22:19.810 I can fit-- I can fit those parameters to give me 00:22:19.810 --> 00:22:21.990 the damping that I want, and I'm going to measure 00:22:21.990 --> 00:22:24.000 the damping in the system. 00:22:24.000 --> 00:22:25.570 And then I'm going to find the two 00:22:25.570 --> 00:22:27.940 parameters that make that work. 00:22:27.940 --> 00:22:30.030 That's why I'm going through this. 00:22:43.510 --> 00:22:44.875 All right, running out of room. 00:22:50.550 --> 00:22:55.461 So zeta 1, the damping ratio for mode 1-- 00:22:55.461 --> 00:22:57.210 if it's a single degree of freedom system, 00:22:57.210 --> 00:23:01.240 you say, oh, well that's the damping constant over 2 00:23:01.240 --> 00:23:03.420 omega 1 m1. 00:23:03.420 --> 00:23:07.307 That's how we define damping ratio 00:23:07.307 --> 00:23:08.890 for a single degree of freedom system, 00:23:08.890 --> 00:23:11.160 but we know what these quantities are. 00:23:11.160 --> 00:23:17.760 This then is an alpha m1 over 2 omega 1 m1 00:23:17.760 --> 00:23:26.020 plus a beta k1 over 2 omega 1 m1. 00:23:26.020 --> 00:23:29.860 k1/m1 is omega 1 squared. 00:23:29.860 --> 00:23:31.970 So put omega 1 squared up here. 00:23:31.970 --> 00:23:33.750 Cancel with that. 00:23:33.750 --> 00:23:39.090 These two gives you-- this gives you alpha over 2 00:23:39.090 --> 00:23:47.130 omega 1 plus beta omega 1 over 2, 00:23:47.130 --> 00:23:49.010 and I can do the same thing for zeta 2. 00:23:51.710 --> 00:24:02.470 Be my c2 over 2 omega 2 m2, and that'll give me an alpha over 2 00:24:02.470 --> 00:24:08.650 omega 2 plus beta omega 2 over 2. 00:24:08.650 --> 00:24:13.350 So if I can measure a value for this damping and a value 00:24:13.350 --> 00:24:16.310 for that damping, I have. 00:24:16.310 --> 00:24:17.980 These are known then. 00:24:17.980 --> 00:24:21.580 I have two equations and two unknowns, alpha 00:24:21.580 --> 00:24:24.500 and beta-- just algebraic equations that I can solve. 00:24:28.800 --> 00:24:30.500 So now I need to conduct my experiment. 00:24:39.420 --> 00:24:42.020 I have this system. 00:24:42.020 --> 00:24:43.940 Without external excitation, it's 00:24:43.940 --> 00:24:53.080 just a free vibration system-- typical equations of motion. 00:24:53.080 --> 00:24:58.260 The first mode would look like m1q1 double dot 00:24:58.260 --> 00:25:08.040 plus c1q1 dot plus k1q1 equals 0. 00:25:08.040 --> 00:25:09.634 I'm looking for solutions to that, 00:25:09.634 --> 00:25:11.550 and there's a similar one for the second mode. 00:25:17.960 --> 00:25:23.169 And what I want to do is excite only one 00:25:23.169 --> 00:25:24.210 of these modes at a time. 00:25:24.210 --> 00:25:26.090 Now, you've seen this demo done before, 00:25:26.090 --> 00:25:28.330 but at a certain combination of deflections, 00:25:28.330 --> 00:25:31.560 it'll respond only in mode 1, and a different one 00:25:31.560 --> 00:25:33.620 will respond only in mode 2. 00:25:33.620 --> 00:25:36.400 And it's guaranteed that, if you deflect 00:25:36.400 --> 00:25:40.870 a system in the shape of one of its modes and let it go, 00:25:40.870 --> 00:25:42.660 it will only vibrate in that mode, 00:25:42.660 --> 00:25:44.900 but let's prove to ourselves that that actually 00:25:44.900 --> 00:25:47.190 is going to work. 00:25:47.190 --> 00:26:02.820 So I'm going to let, for example, initial conditions-- 00:26:02.820 --> 00:26:11.160 x10, x20-- be in the shape of mode 1. 00:26:11.160 --> 00:26:21.770 Well mode 1 is 1, 2.2667-- is the mode shape of mode 1. 00:26:21.770 --> 00:26:23.620 I'm just going to let that be. 00:26:23.620 --> 00:26:27.320 I'm going to deflect it in the shape of mode one, 00:26:27.320 --> 00:26:36.830 and I'm going to let x10 dot and x20 dot-- those are 0. 00:26:36.830 --> 00:26:42.090 And I need to know, if I do that, 00:26:42.090 --> 00:26:46.240 what are the resulting initial conditions 00:26:46.240 --> 00:26:47.890 in modal coordinates? 00:26:47.890 --> 00:26:58.010 Well, we know q10 and q20-- I can obtain them 00:26:58.010 --> 00:27:07.100 by doing u inverse times this. 00:27:07.100 --> 00:27:12.320 So this will be u inverse-- last time-- I'm going to write this. 00:27:12.320 --> 00:27:14.230 Last time I made a mistake. 00:27:14.230 --> 00:27:16.890 So u inverse-- I left out a 0. 00:27:16.890 --> 00:27:37.030 So 0.0898, 0.9102, 0.4016, minus 0.4016. 00:27:37.030 --> 00:27:40.430 So that's u inverse, and I'm going to multiply it 00:27:40.430 --> 00:27:43.130 by one of the mode shapes. 00:27:43.130 --> 00:27:49.310 1 and 2.2667. 00:27:49.310 --> 00:27:51.470 So I'm saying that my initial conditions 00:27:51.470 --> 00:27:54.190 are going to look exactly like one mode shape. 00:27:54.190 --> 00:27:58.390 To compute the equivalent modal initial conditions, 00:27:58.390 --> 00:28:04.050 I multiply the generalized the coordinate initial conditions 00:28:04.050 --> 00:28:05.370 by u inverse. 00:28:05.370 --> 00:28:09.940 Here's u inverse times that, and if I do that calculation, 00:28:09.940 --> 00:28:11.960 I get exactly 1, 0. 00:28:15.180 --> 00:28:17.230 And if instead of the first mode shape 00:28:17.230 --> 00:28:23.510 I put in the second mode shape, the 1 minus 0.2236-- 00:28:23.510 --> 00:28:28.510 if I did that, I would get exactly 0, 1. 00:28:31.200 --> 00:28:33.960 So I just wanted to go through that just so you'd see it-- 00:28:33.960 --> 00:28:38.560 that the math bears it out. 00:28:38.560 --> 00:28:40.670 If you put in a deflection that's exactly 00:28:40.670 --> 00:28:43.160 the shape of a mode, then you will get back 00:28:43.160 --> 00:28:48.889 an equivalent initial condition in the modal coordinates 00:28:48.889 --> 00:28:50.930 for only that mode and everything else will be 0. 00:28:50.930 --> 00:28:51.900 So that's q. 00:28:51.900 --> 00:28:56.220 This is q10-- is that guy. 00:28:56.220 --> 00:29:01.970 And this q20 in this case is 0. 00:29:01.970 --> 00:29:02.500 OK? 00:29:02.500 --> 00:29:05.040 So now we're ready to do the experiment. 00:29:05.040 --> 00:29:15.440 So if I deflected in the shape of mode 1-- come back here. 00:29:18.817 --> 00:29:21.400 And now would be a good time to lower the lights a little bit. 00:29:24.100 --> 00:29:27.040 And unfortunately we have that white chalk in the background 00:29:27.040 --> 00:29:29.670 to distract us, but it's now deflected only 00:29:29.670 --> 00:29:31.979 in the shape of mode 1, and now it's 00:29:31.979 --> 00:29:34.270 going to behave like a single degree of freedom system, 00:29:34.270 --> 00:29:35.470 right? 00:29:35.470 --> 00:29:40.760 So how would you estimate-- do a quick estimate 00:29:40.760 --> 00:29:41.980 of the damping of mode 1. 00:29:46.210 --> 00:29:48.970 I gave you a little quick, easy rule 00:29:48.970 --> 00:29:51.121 you could use a few times back. 00:29:51.121 --> 00:29:51.620 What was it? 00:29:58.242 --> 00:30:00.140 AUDIENCE: Do the 50% thing? 00:30:00.140 --> 00:30:02.660 PROFESSOR: Do the 50% thing she suggests. 00:30:02.660 --> 00:30:04.910 Now, can you be a little more specific. 00:30:04.910 --> 00:30:05.410 Pardon? 00:30:05.410 --> 00:30:06.910 AUDIENCE: The half life. 00:30:06.910 --> 00:30:09.440 PROFESSOR: Yeah, how many-- what am I looking for? 00:30:09.440 --> 00:30:10.180 How many-- 00:30:10.180 --> 00:30:11.010 AUDIENCE: Cycles. 00:30:11.010 --> 00:30:15.540 PROFESSOR: Cycles it takes for the thing to decay 50%. 00:30:15.540 --> 00:30:17.930 So the reference line is here. 00:30:17.930 --> 00:30:21.740 I've deflected it that far, and it'll start vibrating, 00:30:21.740 --> 00:30:23.660 and when the top of this on a vibration 00:30:23.660 --> 00:30:25.827 gets to the halfway point between here 00:30:25.827 --> 00:30:27.410 and the reference-- reference is where 00:30:27.410 --> 00:30:30.350 it starts-- that'll be my 50%. 00:30:30.350 --> 00:30:37.540 So one, two, three, four. 00:30:37.540 --> 00:30:41.740 Got about four cycles when it only went down halfway. 00:30:41.740 --> 00:30:43.755 Now, when I did it in my office the other day, 00:30:43.755 --> 00:30:46.920 I only got 2 and 1/2 cycles. 00:30:46.920 --> 00:30:49.070 So we're going to use 2 and 1/2 cycles because I 00:30:49.070 --> 00:30:51.380 ran the numbers for that. 00:30:51.380 --> 00:30:53.710 This thing is very sensitive if it's 00:30:53.710 --> 00:30:57.100 inclined a little bit, because that changes 00:30:57.100 --> 00:30:59.160 the friction on the shaft. 00:30:59.160 --> 00:31:01.660 So in my office the other day, there was a lot more damping. 00:31:01.660 --> 00:31:05.570 So anyway, pretend it's 2 and 1/2. 00:31:05.570 --> 00:31:09.320 Now, while we're at it, let's do the other case. 00:31:09.320 --> 00:31:12.970 So the other case we want to deflect it 00:31:12.970 --> 00:31:14.700 in the shape of mode 2. 00:31:14.700 --> 00:31:17.380 So I go down, say, a unit amount. 00:31:23.610 --> 00:31:29.170 It's now deflected downward some amount and upwards such 00:31:29.170 --> 00:31:31.890 that the ratio is in the mode shape of mode 2. 00:31:31.890 --> 00:31:38.080 So every unit I went down, I go up minus 22% of that. 00:31:38.080 --> 00:31:40.310 So that's what's been done here. 00:31:40.310 --> 00:31:45.450 So when I release this one, now the reference line 00:31:45.450 --> 00:31:49.580 for the second one is up here down to here. 00:31:49.580 --> 00:31:54.217 So when this upper one decays to halfway-- about here-- 00:31:54.217 --> 00:31:55.550 that'll be the number of cycles. 00:31:58.210 --> 00:32:02.950 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 00:32:02.950 --> 00:32:07.040 Anyway, in my office I got about 10. 00:32:07.040 --> 00:32:10.200 We're going to use my numbers, but that one is clearly 00:32:10.200 --> 00:32:11.380 a lot less damped. 00:32:11.380 --> 00:32:14.220 That thing just went on, an on, and on, and on, and on. 00:32:14.220 --> 00:32:31.680 So in our experiment then-- so we conducted our experiment. 00:32:37.580 --> 00:32:43.000 Zeta 1 is approximately equal to point 1.1 00:32:43.000 --> 00:32:47.460 over the number of cycles to decay 50%, 00:32:47.460 --> 00:32:49.760 and when I did the experiment in my office, 00:32:49.760 --> 00:32:54.300 that was 0.11 over 2.5. 00:32:54.300 --> 00:32:56.810 So it was really decaying fast, and so when you only 00:32:56.810 --> 00:32:58.830 have two or three cycles, it helps 00:32:58.830 --> 00:33:00.450 to use even fractional cycles. 00:33:00.450 --> 00:33:02.740 And it's OK to use fractional cycles. 00:33:02.740 --> 00:33:05.810 This is just an estimate. 00:33:05.810 --> 00:33:07.750 That's going to give us a number. 00:33:07.750 --> 00:33:12.850 Zeta 2 is approximately 0.11 over the number 00:33:12.850 --> 00:33:17.350 of cycles to decay 50%, and in this case, that's 00:33:17.350 --> 00:33:19.770 going to be 10. 00:33:19.770 --> 00:33:27.540 And so that gives me 0.011, or what's known as 1.1% damping. 00:33:27.540 --> 00:33:34.736 And the other one, 0.11/2.5, gives me 0.044, 00:33:34.736 --> 00:33:38.140 or 4.4% damping. 00:33:38.140 --> 00:33:39.030 Percent of what? 00:33:39.030 --> 00:33:43.210 Anybody remember what happens when you're at 1? 00:33:43.210 --> 00:33:44.975 Have a damping ratio of 1.0? 00:33:44.975 --> 00:33:46.100 AUDIENCE: Critical damping. 00:33:46.100 --> 00:33:46.820 PROFESSOR: That's critical damping. 00:33:46.820 --> 00:33:48.000 That's that crossover point. 00:33:48.000 --> 00:33:50.990 If you have 1 or greater, you get no oscillation. 00:33:50.990 --> 00:33:53.030 It just goes and stops. 00:33:53.030 --> 00:33:55.270 Less than 1, it will actually across 0 00:33:55.270 --> 00:33:56.640 and oscillate a little bit. 00:33:59.690 --> 00:34:07.810 Well, now I have values for 0.044 0.011 00:34:07.810 --> 00:34:10.699 that I can plug in to these two equations, 00:34:10.699 --> 00:34:13.610 and I now have two equations and two unknowns, alpha and beta, 00:34:13.610 --> 00:34:16.110 because I know omega 1 and omega 2. 00:34:16.110 --> 00:34:24.270 So solve for alpha, and for alpha I'm not 00:34:24.270 --> 00:34:25.680 going to do that on the board. 00:34:25.680 --> 00:34:29.510 It's kind of a waste of good lecture time. 00:34:29.510 --> 00:34:34.800 10.71. 00:34:34.800 --> 00:34:48.080 And beta, minus 0.033. 00:34:48.080 --> 00:34:51.650 So they can be positive or negative to make these work 00:34:51.650 --> 00:34:54.530 out, but those are the two values you need, 00:34:54.530 --> 00:35:00.210 and that says that we're going to model the damping matrix 00:35:00.210 --> 00:35:15.240 of the system as 10.71m plus minus 0.033k. 00:35:15.240 --> 00:35:21.460 And we'll compute the diagonalized damping matrix, 00:35:21.460 --> 00:35:27.700 which comes from utcu, and did I write that down? 00:35:34.250 --> 00:35:39.600 I didn't, but I don't need it, because I know the damping 00:35:39.600 --> 00:35:45.450 that I'm after 0.044 and 0.011, because in order 00:35:45.450 --> 00:35:49.250 to complete this problem, if I wanted to go to completion 00:35:49.250 --> 00:35:53.610 and have the transient decay of the system, 00:35:53.610 --> 00:36:01.860 I would have it in for each of the two modal systems. 00:36:01.860 --> 00:36:11.130 q of t for the first mode would be some e to the minus zeta 00:36:11.130 --> 00:36:22.600 1 omega 1t cosine omega 1-- I left out my q10. 00:36:25.410 --> 00:36:35.210 q10 cosine omega 1d times t-- that's the most 00:36:35.210 --> 00:36:47.400 of the initial displacement part-- plus q10 dot plus zeta 00:36:47.400 --> 00:36:53.770 1 omega 1 q10. 00:36:53.770 --> 00:36:55.890 All of that over omega 1d. 00:36:59.550 --> 00:37:07.130 All of this times sine omega 1d times time. 00:37:07.130 --> 00:37:10.880 So this is just the response to initial conditions for a single 00:37:10.880 --> 00:37:14.500 degree of freedom system, and you'd have a similar one for q2 00:37:14.500 --> 00:37:17.960 of t, except now all the ones would be replaced by 2's. 00:37:17.960 --> 00:37:26.020 So you have an e to the minus zeta 2 omega 2t times-- 00:37:26.020 --> 00:37:29.200 and those are your two-- that would 00:37:29.200 --> 00:37:33.960 be in general the response to initial conditions 00:37:33.960 --> 00:37:36.370 in modal terms. 00:37:36.370 --> 00:37:39.160 And if you want to get back to the final response 00:37:39.160 --> 00:37:44.950 in your generalized coordinates, then it's just summation 00:37:44.950 --> 00:37:52.100 i equals 1 to 2, in this case, of u mode shape, 00:37:52.100 --> 00:38:03.400 or mode 1, q1 plus mode shape for mode 2 q2 of t. 00:38:03.400 --> 00:38:07.772 So these two added together would give you total response. 00:38:07.772 --> 00:38:08.730 So that's the response. 00:38:08.730 --> 00:38:10.771 Kind of a quick review of what we learned before. 00:38:10.771 --> 00:38:12.960 Response to initial conditions, but now 00:38:12.960 --> 00:38:18.840 along the way we've learned how to experimentally fix 00:38:18.840 --> 00:38:21.140 the damping matrix, so it'll work, 00:38:21.140 --> 00:38:24.850 and actually get it to give us exact accurate results if we 00:38:24.850 --> 00:38:28.770 want to do a response to any initial conditions problem. 00:38:28.770 --> 00:38:31.710 But now we also have damping ratios for this system, 00:38:31.710 --> 00:38:35.425 and we can now go on to do the force vibration problem. 00:38:35.425 --> 00:38:37.050 And the force vibration problem will 00:38:37.050 --> 00:38:42.380 be pretty easy at this point. 00:38:42.380 --> 00:38:47.200 So that was a quick review of response to initial conditions 00:38:47.200 --> 00:38:52.970 by modal analysis-- how to find a damping matrix that works. 00:38:52.970 --> 00:38:56.070 We've also proven that, if you deflect 00:38:56.070 --> 00:39:00.220 a system in the shape of a mode, it responds only in that mode. 00:39:00.220 --> 00:39:01.660 That's all pretty use-- and that's 00:39:01.660 --> 00:39:04.510 generally true of vibration systems. 00:39:04.510 --> 00:39:06.357 So we know-- I banged on this before. 00:39:06.357 --> 00:39:08.190 We know these things like to shake a little. 00:39:08.190 --> 00:39:10.757 They're flexible, and this one has a mode shape. 00:39:10.757 --> 00:39:11.840 It's kind of a cantilever. 00:39:11.840 --> 00:39:12.940 It's first mode. 00:39:12.940 --> 00:39:15.540 It bends like that a little bit, and this thing 00:39:15.540 --> 00:39:16.826 has to bend with it. 00:39:16.826 --> 00:39:18.575 But if I were able to just give this thing 00:39:18.575 --> 00:39:21.350 an initial deflection just in the shape of that mode 00:39:21.350 --> 00:39:26.080 and let it go, it will vibrate in just that mode. 00:39:26.080 --> 00:39:28.550 And if I do it in some other contorted way 00:39:28.550 --> 00:39:32.251 that its initial shape isn't just one mode and let it go, 00:39:32.251 --> 00:39:34.000 it'll vibrate in a couple different modes. 00:39:34.000 --> 00:39:36.490 AUDIENCE: Is there a way to visualize 00:39:36.490 --> 00:39:37.980 the modal coordinates? 00:39:37.980 --> 00:39:41.326 Would they be the center of a mass of certain things, 00:39:41.326 --> 00:39:42.992 or would they would be the center masses 00:39:42.992 --> 00:39:45.420 of the springs or something? 00:39:45.420 --> 00:39:50.350 PROFESSOR: Yeah, I taught a vibration course 00:39:50.350 --> 00:39:53.690 for many, many years, and I scratched my head long and hard 00:39:53.690 --> 00:39:55.190 to try to come up with an example 00:39:55.190 --> 00:39:57.340 where you could do exactly what you asked. 00:39:57.340 --> 00:39:59.810 I know of one example that works. 00:39:59.810 --> 00:40:02.340 In general, the modal coordinates-- 00:40:02.340 --> 00:40:05.140 it's a coordinate transformation into some system 00:40:05.140 --> 00:40:07.810 that it's very hard to place yourself physically so you can 00:40:07.810 --> 00:40:10.280 see what's going on, but to answer your question 00:40:10.280 --> 00:40:15.630 I'll show you one and you still have enough time 00:40:15.630 --> 00:40:17.460 to get through the other part. 00:40:17.460 --> 00:40:19.240 Imagine this eraser. 00:40:19.240 --> 00:40:22.570 It's a car, a car automobile suspension system. 00:40:22.570 --> 00:40:25.650 So here's the car. 00:40:25.650 --> 00:40:30.340 Sits on springs, the tires and its suspension system. 00:40:30.340 --> 00:40:34.120 Got a center of mass that's here somewhere. 00:40:34.120 --> 00:40:38.070 And I'm only going to consider vertical motion of this thing. 00:40:38.070 --> 00:40:40.770 Now, you can imagine that it could go up and down, 00:40:40.770 --> 00:40:44.000 but you can imagine it can also pitch back and forth. 00:40:44.000 --> 00:40:49.860 So I'm going to have two generalized coordinates. 00:40:49.860 --> 00:40:53.330 One is the vertical deflection of its center of mass, 00:40:53.330 --> 00:40:57.140 and the other is the angular rotation 00:40:57.140 --> 00:40:58.660 of the center of mass with respect 00:40:58.660 --> 00:41:01.760 to my initial horizontal. 00:41:01.760 --> 00:41:04.400 Now, you get equations of motion of this. 00:41:04.400 --> 00:41:07.780 One, you'll get an mx double dot kind of equation 00:41:07.780 --> 00:41:10.280 by summing the forces on this and the forces 00:41:10.280 --> 00:41:11.100 from the springs. 00:41:11.100 --> 00:41:13.090 And of course you'll have dampers 00:41:13.090 --> 00:41:15.100 and those kinds of things. 00:41:15.100 --> 00:41:16.930 So you'll have one force equation 00:41:16.930 --> 00:41:18.230 by summing the forces on it. 00:41:18.230 --> 00:41:20.620 You'll have another equation that's sum of torques, 00:41:20.620 --> 00:41:24.330 and it's an i theta double dot kind of equation. 00:41:24.330 --> 00:41:26.737 So it'll be a two degree of freedom system. 00:41:26.737 --> 00:41:28.195 You'll get two natural frequencies, 00:41:28.195 --> 00:41:30.210 and two mode shapes. 00:41:30.210 --> 00:41:34.220 And the two natural frequencies look 00:41:34.220 --> 00:41:37.685 approximate-- the two mode shapes look-- here 00:41:37.685 --> 00:41:41.000 is the original undeflected system. 00:41:41.000 --> 00:41:46.190 One mode shape the system moves up and rotates up. 00:41:46.190 --> 00:41:50.730 It has both positive rotation and some positive deflection. 00:41:50.730 --> 00:41:53.760 And I need to draw this over here a little further. 00:41:53.760 --> 00:41:56.280 So here's the original system. 00:41:56.280 --> 00:41:58.940 It moves up in its mode shape like that, 00:41:58.940 --> 00:42:02.610 and when it goes down to its max negative, it's like that. 00:42:02.610 --> 00:42:05.880 So it goes through a motion. 00:42:05.880 --> 00:42:08.640 And if you think about it, if you extend these lines out 00:42:08.640 --> 00:42:14.150 here, you come to a point that this intersects. 00:42:14.150 --> 00:42:17.020 And so in fact, in the first mode 00:42:17.020 --> 00:42:23.060 of vibration of this thing, if you went and sat right here, 00:42:23.060 --> 00:42:26.440 you would just see this thing go through an angle. 00:42:26.440 --> 00:42:29.812 And that, in fact, is the modal coordinate for mode 1. 00:42:33.510 --> 00:42:37.310 And the second mode I've kind of forgotten. 00:42:37.310 --> 00:42:40.760 It goes has a sign change, and so when 00:42:40.760 --> 00:42:47.370 you're going-- deflecting upwards, 00:42:47.370 --> 00:42:53.580 it goes-- positive flexion upwards-- it tips downwards. 00:42:53.580 --> 00:43:01.210 And when it goes the other way, it goes like that. 00:43:01.210 --> 00:43:04.180 So this one, it goes up and rotates up. 00:43:04.180 --> 00:43:06.850 And the second mode, when it goes up, it rotates down. 00:43:09.470 --> 00:43:12.780 And out here is the point at which you could go sit there, 00:43:12.780 --> 00:43:15.937 and you would just see this thing rotate up, rotate down 00:43:15.937 --> 00:43:17.020 if you went and sat there. 00:43:17.020 --> 00:43:20.410 So you move yourself to a place where 00:43:20.410 --> 00:43:22.430 you can see with a single coordinate-- 00:43:22.430 --> 00:43:25.410 with only the angle measured by this point, 00:43:25.410 --> 00:43:28.556 you can completely describe that modal motion. 00:43:28.556 --> 00:43:30.680 And with only the angle measured around this point, 00:43:30.680 --> 00:43:33.380 you can describe the modal motion. 00:43:33.380 --> 00:43:37.320 But in general, you can't do it. 00:43:37.320 --> 00:43:40.600 So if I'm out here-- if my eye is at this point, 00:43:40.600 --> 00:43:42.111 I'll see this thing go like that. 00:43:46.530 --> 00:43:47.581 That help? 00:43:47.581 --> 00:43:48.523 AUDIENCE: Yeah, a lot. 00:43:55.130 --> 00:43:57.390 PROFESSOR: So one of the hardest things for me 00:43:57.390 --> 00:44:00.050 as a lecturer is, since I really like vibration 00:44:00.050 --> 00:44:03.840 and I've taught it for 35 years or so, 00:44:03.840 --> 00:44:07.800 to try to cram everything I know about vibration into the one 00:44:07.800 --> 00:44:08.950 third of this course. 00:44:08.950 --> 00:44:10.740 So I obviously don't do that. 00:44:10.740 --> 00:44:14.330 So I'm trying to give you-- my goal then becomes give you 00:44:14.330 --> 00:44:17.300 some basic insight about vibration 00:44:17.300 --> 00:44:20.240 so that, when you do get out there in the real world 00:44:20.240 --> 00:44:22.647 and you need to know something, need to solve a vibration 00:44:22.647 --> 00:44:24.230 problem, you'll know the fundamentals, 00:44:24.230 --> 00:44:26.835 and you'll know where to go look it up. 00:44:26.835 --> 00:44:28.960 And if you want to take another course or something 00:44:28.960 --> 00:44:32.170 in vibration, you can do that. 00:44:32.170 --> 00:44:35.670 All right, now we've got to do, in relatively short order, 00:44:35.670 --> 00:44:41.150 response of our initial system over here 00:44:41.150 --> 00:44:44.730 to some harmonic excitation. 00:44:44.730 --> 00:44:53.860 And the one I've chosen to do-- let's 00:44:53.860 --> 00:44:57.300 imagine that I just-- I'm just going 00:44:57.300 --> 00:45:02.900 to put a force, a harmonic force, on this mass, 00:45:02.900 --> 00:45:06.400 and I want the steady state response. 00:45:06.400 --> 00:45:10.300 And you can imagine, if I do it close to the natural frequency 00:45:10.300 --> 00:45:12.760 of one of the modes, you're going to get a lot of that, 00:45:12.760 --> 00:45:15.410 but if I do it close to the natural frequency 00:45:15.410 --> 00:45:18.630 of the other mode, I can do it too. 00:45:18.630 --> 00:45:23.485 So let's do the steady state harmonic excitation problem. 00:45:31.090 --> 00:45:32.690 So we're going to do this problem. 00:45:32.690 --> 00:45:43.810 F2 of t-- so my generalized forces are 0 some magnitude 00:45:43.810 --> 00:45:47.070 Ft e to the i omega t. 00:45:49.920 --> 00:45:58.210 And the modal forces are u transpose F, 00:45:58.210 --> 00:46:00.070 and that would be the calculation 00:46:00.070 --> 00:46:12.050 one and 2.2667, and 1, and minus 0.2236-- my two mode shapes 00:46:12.050 --> 00:46:16.490 transposed multiplied by 0 and F2. 00:46:20.450 --> 00:46:36.000 So that says that q1 is F2.2667 F2e to the i omega t. 00:46:36.000 --> 00:46:41.400 And q2, the modal force for mode 2, 00:46:41.400 --> 00:46:53.180 is this times that-- is minus 0.2236 F2 e to the i omega t. 00:46:53.180 --> 00:46:57.420 So I've put on just a force on one mass, 00:46:57.420 --> 00:47:01.560 but it gets distributed in a way that it'll excite both modes, 00:47:01.560 --> 00:47:06.470 and it'll excite mode one in an amount 2.2 times F2. 00:47:06.470 --> 00:47:10.710 And it'll excite mode 2 in an amount minus 0.22. 00:47:10.710 --> 00:47:13.420 So mode 1 is going to get more excitation 00:47:13.420 --> 00:47:19.090 in this particular case because of the shape of the modes. 00:47:19.090 --> 00:47:21.210 The bigger the modal deflection is 00:47:21.210 --> 00:47:23.282 at the point of application of the force, 00:47:23.282 --> 00:47:24.740 the more that mode is going to get. 00:47:33.340 --> 00:47:38.320 Well, now these gave you two equations of motion. 00:47:38.320 --> 00:47:48.100 m1q1 double dot plus c1q1 dot k1q1 00:47:48.100 --> 00:47:59.877 equals, in the case of mode 1, 2.2667 F2 e to the i omega t. 00:47:59.877 --> 00:48:01.710 And that's a single degree of freedom system 00:48:01.710 --> 00:48:03.990 excited by a harmonic force. 00:48:03.990 --> 00:48:06.080 We worked that problem. 00:48:06.080 --> 00:48:08.070 We know what the answer looks like. 00:48:08.070 --> 00:48:13.740 So for example, the magnitude of the response q1 00:48:13.740 --> 00:48:19.510 is given by the magnitude of the force q2 times 00:48:19.510 --> 00:48:23.900 the magnitude of a transfer function, which is the response 00:48:23.900 --> 00:48:31.380 q1 per unit input force q1 evaluated at whatever 00:48:31.380 --> 00:48:33.360 frequency I evaluate it at. 00:48:33.360 --> 00:48:35.319 So the magnitude of the response magnitude 00:48:35.319 --> 00:48:36.860 of the modal force times the transfer 00:48:36.860 --> 00:48:39.010 function, and this transfer function 00:48:39.010 --> 00:48:44.450 is this exactly the same form as when 00:48:44.450 --> 00:48:47.745 we did just the single degree of freedom system as x/F. 00:48:47.745 --> 00:48:51.730 That's what I called the response x over input force F. 00:48:51.730 --> 00:48:54.800 Same thing, where just now the response is q, 00:48:54.800 --> 00:48:57.510 and the input force is capital Q. 00:48:57.510 --> 00:49:04.290 Therefore, that looks like magnitude of Q-- 00:49:04.290 --> 00:49:06.220 and I made a mistake here. 00:49:06.220 --> 00:49:07.765 Q1. 00:49:07.765 --> 00:49:21.270 The magnitude of force Q1, which is this, times 1/k1-- 00:49:21.270 --> 00:49:27.230 this should look familiar-- 1 minus omega squared 00:49:27.230 --> 00:49:32.860 over omega 1 squared squared plus 2 zeta 00:49:32.860 --> 00:49:40.400 1 omega over omega 1 quantity squared square root. 00:49:40.400 --> 00:49:44.750 So there's that transfer function expression. 00:49:44.750 --> 00:49:48.680 At resonance, for example, all this goes to 0. 00:49:48.680 --> 00:49:50.860 Omega over omega 1 goes to 1. 00:49:50.860 --> 00:49:54.070 This whole denominator turns into 1/2 times 00:49:54.070 --> 00:49:55.450 the damping ratio, for example. 00:49:58.170 --> 00:50:01.540 And Q1 over k1 is the static deflection 00:50:01.540 --> 00:50:05.750 of the system under the load-- under a static load 00:50:05.750 --> 00:50:06.800 of that large. 00:50:12.440 --> 00:50:13.900 So I'll do the problem. 00:50:13.900 --> 00:50:18.580 If the excitation frequency happens 00:50:18.580 --> 00:50:21.990 to be right on the natural frequency for mode one, 00:50:21.990 --> 00:50:28.716 then I can evaluate this at 1 omega over omega 1 is 1, 00:50:28.716 --> 00:50:31.580 and I've worked that out. 00:50:31.580 --> 00:50:40.820 That says that magnitude of q1 here-- 00:50:40.820 --> 00:50:57.360 2.2667F2/k1 times 1 over 2 zeta one. 00:50:57.360 --> 00:50:59.270 That's what all this condenses to. 00:50:59.270 --> 00:51:01.560 We know what k1 is. 00:51:01.560 --> 00:51:03.940 We know what zeta 1 is. 00:51:03.940 --> 00:51:07.430 F2 is the thing we're specifying. 00:51:07.430 --> 00:51:13.910 And so zeta 1 for example is what-- it was 0.044. 00:51:13.910 --> 00:51:21.500 So you plug in your 0.044 here, and this whole thing works out 00:51:21.500 --> 00:51:26.660 to be-- actually I'll write down the numbers here-- 00:51:26.660 --> 00:51:40.620 2.2667F2/k1 is 10984, and not 109. 00:51:40.620 --> 00:51:41.120 That's k2. 00:51:44.010 --> 00:51:56.350 113.71 times 11.36, which is 1/2 times zeta. 00:51:56.350 --> 00:52:01.950 1/0.088 is 11.36. 00:52:01.950 --> 00:52:16.590 The final analysis-- q1 is 0.227F2. 00:52:16.590 --> 00:52:18.932 So it's just a single degree of freedom system excited 00:52:18.932 --> 00:52:20.640 by harmonic force, and you can figure out 00:52:20.640 --> 00:52:22.180 how big its response is. 00:52:22.180 --> 00:52:24.830 I could do the same thing for mode 2. 00:52:24.830 --> 00:52:29.234 Is mode 2 resonant? 00:52:29.234 --> 00:52:29.734 No. 00:52:35.460 --> 00:52:39.690 So this is instructive. 00:52:39.690 --> 00:52:46.820 The magnitude of the responsive q2 00:52:46.820 --> 00:52:50.250 is the magnitude of the force times the magnitude 00:52:50.250 --> 00:52:57.416 of the Hq2 per unit input 2q. 00:52:57.416 --> 00:53:05.730 It looks like that, but this time this is minus 00:53:05.730 --> 00:53:17.290 0.2236F2/109.84, which is k2. 00:53:17.290 --> 00:53:27.250 And then all of the denominator involving one minus-- and this 00:53:27.250 --> 00:53:34.060 will now be omega-- over omega 2 squared 00:53:34.060 --> 00:53:44.690 squared plus 2 zeta 2 omega over omega 2 squared square root. 00:53:44.690 --> 00:53:46.248 This number here is k2. 00:53:49.430 --> 00:53:52.435 The reason I'm going to this is this one 00:53:52.435 --> 00:53:58.740 is not resonant, and in fact, what's omega? 00:53:58.740 --> 00:54:00.792 Well, in fact, what is the excite-- 00:54:00.792 --> 00:54:01.750 I started this problem. 00:54:01.750 --> 00:54:06.350 I said let the excitation frequency be what? 00:54:06.350 --> 00:54:08.130 Natural frequency of mode 1. 00:54:08.130 --> 00:54:10.970 So omega in this case equals omega 1. 00:54:10.970 --> 00:54:15.090 Therefore, for this problem, omega over omega 2 00:54:15.090 --> 00:54:21.020 is omega 1 over omega 2, and that's 5.65 something 00:54:21.020 --> 00:54:23.310 over 17 point something. 00:54:23.310 --> 00:54:25.830 It's about 1/3. 00:54:25.830 --> 00:54:27.850 So we are exciting the second mode 00:54:27.850 --> 00:54:30.970 at about 1/3 its natural frequency. 00:54:30.970 --> 00:54:36.259 And so this is 1 minus 1/3 quantity squared and so forth. 00:54:36.259 --> 00:54:37.800 And if I run the numbers, I just want 00:54:37.800 --> 00:54:40.760 you to see what happens with the numbers. 00:54:40.760 --> 00:54:46.780 This turns out to be 0.806, and this turns out 00:54:46.780 --> 00:54:53.060 to be-- remember this is 0.01 something here. 00:54:53.060 --> 00:54:57.860 This number turns out to be 5 times 10 to the minus fifth. 00:54:57.860 --> 00:54:59.173 That's a really small number. 00:55:04.270 --> 00:55:07.500 And this whole square root of this stuff 00:55:07.500 --> 00:55:17.220 turns out to be 0.898. 00:55:17.220 --> 00:55:24.430 And in the end, q2 works out- the magnitude of q2 works out 00:55:24.430 --> 00:55:35.970 to be 0.0023F2. 00:55:35.970 --> 00:55:37.300 I did this on purpose. 00:55:37.300 --> 00:55:41.580 If, when you're at resonance, meaning the excitation 00:55:41.580 --> 00:55:43.410 frequency is close to the natural frequency 00:55:43.410 --> 00:55:51.140 for any linear vibration system, then this term is important, 00:55:51.140 --> 00:55:53.900 and it'll be the most important term in the denominator, 00:55:53.900 --> 00:55:55.190 because this term goes to 0. 00:55:55.190 --> 00:55:57.510 It's the only term in the denominator, 00:55:57.510 --> 00:55:58.820 and it's likely be quite small. 00:55:58.820 --> 00:56:03.230 That's why it gives you one over that big response, 00:56:03.230 --> 00:56:06.710 but when you're not-- when this denominator-- if you're not 00:56:06.710 --> 00:56:12.610 at resonance, this term almost always is negligible, 00:56:12.610 --> 00:56:14.290 and it's this term that governs it. 00:56:14.290 --> 00:56:17.380 So here you are at 0.8 versus 10 to the minus fifth. 00:56:17.380 --> 00:56:20.230 So away from resonance, this term is important, 00:56:20.230 --> 00:56:23.910 and on resonance, that one is important. 00:56:23.910 --> 00:56:27.130 And now, how would you get back to-- we've now 00:56:27.130 --> 00:56:31.800 got the two responses. 00:56:31.800 --> 00:56:35.990 How do we compute the total system response? 00:56:43.370 --> 00:56:44.846 How would you do it? 00:56:51.260 --> 00:56:53.220 The modal expansion theorem. 00:56:53.220 --> 00:56:55.020 AUDIENCE: Put it back into x [INAUDIBLE]. 00:56:55.020 --> 00:57:02.960 PROFESSOR: Yeah, so x equals u cubed. 00:57:02.960 --> 00:57:04.790 That's where we started. 00:57:04.790 --> 00:57:13.170 And q1, since this is a harmonic excitation problem, 00:57:13.170 --> 00:57:16.550 if the input had been e to the i omega t, 00:57:16.550 --> 00:57:20.430 then the output is some e to the i omega t minus a phase angle. 00:57:20.430 --> 00:57:24.160 So this is going to look like some q1 amplitude. 00:57:24.160 --> 00:57:29.710 Say, cosine omega 1t minus phi 1 but at resonance, 00:57:29.710 --> 00:57:32.920 we know the phase angle is pi over 2. 00:57:32.920 --> 00:57:39.280 And q2 of t is going to be the amplitude q2 00:57:39.280 --> 00:57:45.080 times some cosine omega 1t-- because that's the excitation 00:57:45.080 --> 00:57:48.660 frequency-- minus phi 2. 00:57:48.660 --> 00:57:58.260 And remember, each of these systems-- 00:57:58.260 --> 00:58:05.620 these transfer functions, Hq/Q for whichever one 00:58:05.620 --> 00:58:10.260 it happens to be looks like this. 00:58:10.260 --> 00:58:13.730 Different amounts of damping give you different heights 00:58:13.730 --> 00:58:15.350 of the peak at resonance. 00:58:15.350 --> 00:58:20.010 And this is omega over omega i. 00:58:20.010 --> 00:58:24.220 So when you're at resonance, you're at 1. 00:58:24.220 --> 00:58:28.220 So what we've done is we have a two degree of freedom system. 00:58:28.220 --> 00:58:32.420 We're exciting it at the natural frequency of mode 1. 00:58:32.420 --> 00:58:34.587 So it means for mode 1, we're right here. 00:58:34.587 --> 00:58:36.170 Let's say that's where our damping is, 00:58:36.170 --> 00:58:37.610 so that's going to be our transfer 00:58:37.610 --> 00:58:41.830 function for that mode, but drawn on the same figure, 00:58:41.830 --> 00:58:46.130 where are we for mode 2? 00:58:46.130 --> 00:58:49.370 We're at omega 1 over omega 2, which is somewhere 00:58:49.370 --> 00:58:52.262 in the neighborhood of 0.3. 00:58:52.262 --> 00:58:55.450 So we are in here. 00:58:55.450 --> 00:58:59.600 So this is omega over omega 2. 00:58:59.600 --> 00:59:02.500 This is omega over omega 1. 00:59:02.500 --> 00:59:05.100 So down here, we're at about 0.3. 00:59:05.100 --> 00:59:09.070 Here we're at 1.0, so we're at resonance for one of the modes, 00:59:09.070 --> 00:59:12.450 and we're here in what's called the stiffness controlled region 00:59:12.450 --> 00:59:14.050 for the other mode. 00:59:14.050 --> 00:59:17.010 This mode-- mode 2 basically acts like a spring. 00:59:19.670 --> 00:59:21.620 The dynamic amplification is about 1. 00:59:21.620 --> 00:59:25.140 It just gives you the static response for mode 2 00:59:25.140 --> 00:59:27.560 and the resonant response for mode 1. 00:59:36.380 --> 00:59:38.900 I know what I was going to draw to remind you. 00:59:41.550 --> 00:59:44.570 This figure has a phase diagram that goes with it, 00:59:44.570 --> 00:59:50.150 and for lightly damped systems, it goes from 0 to pi. 00:59:50.150 --> 00:59:55.780 And at resonance, all of them cross pi over 2. 00:59:55.780 --> 00:59:59.530 When it's a response to a force, a simple force, at resonance, 00:59:59.530 --> 01:00:03.610 the phase angle is pi over 2, so the response 01:00:03.610 --> 01:00:06.760 lags the input by 90 degrees. 01:00:06.760 --> 01:00:08.930 When you're down here in this region, 01:00:08.930 --> 01:00:11.950 the response moves with the input. 01:00:11.950 --> 01:00:14.730 The phase angle is basically 0, and up in here 01:00:14.730 --> 01:00:16.620 the phase angle is 180 degrees. 01:00:16.620 --> 01:00:18.544 It acts like a driving mass. 01:00:23.470 --> 01:00:25.860 So our two responses look like that, 01:00:25.860 --> 01:00:30.340 and to get back into modal coordinates, x1, 01:00:30.340 --> 01:00:46.010 x2 is going to look like u1 q1 of t plus u2 q2 of t. 01:00:46.010 --> 01:00:46.770 It's harmonic. 01:00:46.770 --> 01:00:48.490 Steady state response. 01:00:48.490 --> 01:00:50.575 You know the amplitudes, q1 and q2. 01:00:50.575 --> 01:00:52.540 We figured them out. 01:00:52.540 --> 01:01:06.860 One of them is 2-- q1 turned out to be 0.227F2, 01:01:06.860 --> 01:01:18.560 and q2 is 0.0023F2. 01:01:18.560 --> 01:01:31.470 And this one, point cosine cosine. 01:01:31.470 --> 01:01:36.145 So if we do that, if you put your excitation only on here, 01:01:36.145 --> 01:01:37.895 it's trying to tell you that mostly you'll 01:01:37.895 --> 01:01:44.380 get mode 1, and not much mode 2, but that's primarily 01:01:44.380 --> 01:01:47.130 caused because you chose to put the excitation 01:01:47.130 --> 01:01:52.940 frequency at where? 01:01:52.940 --> 01:01:56.970 The natural frequency of mode one. 01:01:56.970 --> 01:02:00.530 So if I had made my excitation here the same place, 01:02:00.530 --> 01:02:04.500 but made the frequency close to the natural frequency of mode 01:02:04.500 --> 01:02:08.440 2, which of the two modes would have dominated? 01:02:08.440 --> 01:02:10.720 Mode 2. 01:02:10.720 --> 01:02:16.608 So I've got another little demo. 01:02:16.608 --> 01:02:17.585 We're doing well. 01:02:23.760 --> 01:02:25.765 Out of necessity, to do it on the blackboard 01:02:25.765 --> 01:02:28.370 and in relatively short time, we've 01:02:28.370 --> 01:02:35.250 only talked about a two degree of freedom two rigid body 01:02:35.250 --> 01:02:37.190 system. 01:02:37.190 --> 01:02:38.800 This is a continuous system. 01:02:38.800 --> 01:02:39.780 It's a taut string. 01:02:39.780 --> 01:02:46.950 It's your violin string, and I've put some white tags on it 01:02:46.950 --> 01:02:49.075 so that you could see it against black backgrounds. 01:02:51.680 --> 01:02:55.030 Everything that we've learned about the behavior of this two 01:02:55.030 --> 01:02:58.170 degree of freedom system will apply to a three, 01:02:58.170 --> 01:03:00.100 or four, or five degree of freedom system, 01:03:00.100 --> 01:03:01.800 but actually the basic lessons apply 01:03:01.800 --> 01:03:03.890 to continuous systems, too. 01:03:03.890 --> 01:03:10.940 So the lesson we just learned is, if you excite a system, 01:03:10.940 --> 01:03:13.740 this has many natural frequencies. 01:03:13.740 --> 01:03:16.930 And in fact, if the first one is at 1 Hertz, 01:03:16.930 --> 01:03:19.430 which this just about is-- maybe a little more than that. 01:03:19.430 --> 01:03:21.280 First mode is like that. 01:03:21.280 --> 01:03:22.790 Maybe 2 Hertz. 01:03:22.790 --> 01:03:24.250 The second mode is twice that. 01:03:24.250 --> 01:03:25.650 Third mode is three times that. 01:03:25.650 --> 01:03:28.350 It happens to be really simple. 01:03:28.350 --> 01:03:31.360 So if I put a force-- the only force in this problem-- 01:03:31.360 --> 01:03:33.630 I'm going to do the analogous problem here. 01:03:33.630 --> 01:03:36.890 Harmonic excitation-- I'm going to do it in one little place 01:03:36.890 --> 01:03:41.980 right here, and if I drive this system at the natural frequency 01:03:41.980 --> 01:03:46.710 of mode 1, what do you see? 01:03:46.710 --> 01:03:49.000 What mode is dominating response? 01:03:49.000 --> 01:03:50.550 Mode 1. 01:03:50.550 --> 01:03:55.440 Very, very small responses of other modes. 01:03:55.440 --> 01:04:00.330 So now if I can get the system to stop shaking, if I drive it 01:04:00.330 --> 01:04:08.850 at exactly the same place at a different natural frequency, 01:04:08.850 --> 01:04:11.390 I see 3/2 sine waves there. 01:04:11.390 --> 01:04:14.246 So I was driving it at the natural frequency of? 01:04:14.246 --> 01:04:15.220 AUDIENCE: Mode 3. 01:04:15.220 --> 01:04:16.750 PROFESSOR: Mode 3. 01:04:16.750 --> 01:04:17.890 And there's no mode 1. 01:04:17.890 --> 01:04:20.520 No mode 2. 01:04:20.520 --> 01:04:22.085 Now I need a helper. 01:04:22.085 --> 01:04:24.000 Can somebody come hang onto this for me? 01:04:29.890 --> 01:04:32.160 You've got to keep the tension on it. 01:04:32.160 --> 01:04:35.190 So now I'm going to teach you a really important lesson 01:04:35.190 --> 01:04:40.300 about systems exciting systems. 01:04:40.300 --> 01:04:44.640 The second mode-- I might have a hard time driving it here. 01:04:44.640 --> 01:04:46.787 Let me see if I can get the second mode going. 01:04:54.130 --> 01:04:55.495 There's second mode. 01:04:55.495 --> 01:04:56.900 I has a node right here. 01:04:56.900 --> 01:04:58.650 There's a point right here with no motion. 01:05:01.470 --> 01:05:04.240 If I sit here, and I'm going to let the system stop-- 01:05:04.240 --> 01:05:07.580 if I drive this system at the second mode natural frequency 01:05:07.580 --> 01:05:09.090 right here, what will happen? 01:05:16.230 --> 01:05:16.922 Pardon? 01:05:16.922 --> 01:05:18.130 AUDIENCE: It would be mode 1. 01:05:18.130 --> 01:05:20.075 PROFESSOR: Maybe [INAUDIBLE] 1. 01:05:20.075 --> 01:05:22.926 But how much mode 2 will I get? 01:05:27.660 --> 01:05:33.700 So another lesson here is that continuous systems have nodes, 01:05:33.700 --> 01:05:36.474 points of no motion, and the second mode 01:05:36.474 --> 01:05:37.890 happens to have a node right here. 01:05:37.890 --> 01:05:39.580 This point doesn't move one when it's 01:05:39.580 --> 01:05:41.400 vibrating in the second mode. 01:05:41.400 --> 01:05:43.290 And the modal force looks something 01:05:43.290 --> 01:05:47.560 like the generalized force times a mode shape, 01:05:47.560 --> 01:05:50.880 u transpose F. That's how we got the-- you have 01:05:50.880 --> 01:05:56.210 to multiply the generalized external force times the mode 01:05:56.210 --> 01:05:59.340 shape to get the modal force. 01:05:59.340 --> 01:06:02.070 Well, here's my generalized force. 01:06:02.070 --> 01:06:05.670 What's the amplitude of the mode shape here? 01:06:05.670 --> 01:06:06.430 0. 01:06:06.430 --> 01:06:10.250 F times 0 is 0. 01:06:10.250 --> 01:06:13.100 There is just no way I can get this thing 01:06:13.100 --> 01:06:16.420 to vibrate in the second mode by driving it at a node, 01:06:16.420 --> 01:06:19.010 and that's just generally true. 01:06:19.010 --> 01:06:22.750 Take it to the bank and remember that. 01:06:22.750 --> 01:06:25.650 Thank you. 01:06:25.650 --> 01:06:28.430 All right. 01:06:28.430 --> 01:06:30.380 Something else I wanted to do. 01:06:34.244 --> 01:06:34.910 I need you back. 01:06:39.490 --> 01:06:48.380 So when you give a system an initial deflection, 01:06:48.380 --> 01:06:52.780 because when we want to get initial conditions, 01:06:52.780 --> 01:06:56.200 we said that the equivalent modal initial conditions were 01:06:56.200 --> 01:07:00.740 u inverse times the u inverse times to the initial conditions 01:07:00.740 --> 01:07:02.750 in generalized coordinates. 01:07:02.750 --> 01:07:04.990 So for a continuous system, this thing 01:07:04.990 --> 01:07:07.980 has mode shapes that look like sine and pi x over l. 01:07:07.980 --> 01:07:10.100 Those are the mode shapes. 01:07:10.100 --> 01:07:11.620 And if I take this thing and grab it 01:07:11.620 --> 01:07:14.634 about in the middle, wherever that node was, 01:07:14.634 --> 01:07:16.050 and give it an initial deflection, 01:07:16.050 --> 01:07:17.424 the shape that initial deflection 01:07:17.424 --> 01:07:18.940 is kind of a triangle. 01:07:18.940 --> 01:07:20.310 You've all had Fourier series. 01:07:23.100 --> 01:07:25.530 This a triangular shape. 01:07:25.530 --> 01:07:30.940 Could you express this shape as a Fourier sine series? 01:07:30.940 --> 01:07:32.367 Sure. 01:07:32.367 --> 01:07:34.700 And it just happens that sine waves are the mode shapes. 01:07:34.700 --> 01:07:36.970 So you would be coming up-- the Fourier 01:07:36.970 --> 01:07:40.660 coefficients are the modal amplitudes 01:07:40.660 --> 01:07:43.240 of n initial conditions. 01:07:43.240 --> 01:07:46.040 And so whatever the Fourier coefficients for this are 01:07:46.040 --> 01:07:49.070 are the modal amplitudes for each of the modes. 01:07:49.070 --> 01:07:50.550 So which mode do you think is going 01:07:50.550 --> 01:07:54.020 to-- which Fourier sign component is going 01:07:54.020 --> 01:07:55.654 to be largest in this one? 01:07:58.438 --> 01:07:59.370 AUDIENCE: Mode 1? 01:07:59.370 --> 01:08:00.230 PROFESSOR: Mode 1. 01:08:00.230 --> 01:08:02.890 By the way, all even numbered modes would be 0 01:08:02.890 --> 01:08:04.560 because they're asymmetric. 01:08:04.560 --> 01:08:08.450 This is a symmetrically shaped pulse. 01:08:08.450 --> 01:08:12.920 The modal initial conditions for 2, 4, 6, 8, 10 are 0. 01:08:12.920 --> 01:08:17.350 You will get non-zero Fourier coefficients for 1, 3, 5, 7. 01:08:17.350 --> 01:08:18.950 The biggest one is mode 1. 01:08:18.950 --> 01:08:22.640 So what do you expect to see if I let go of this? 01:08:22.640 --> 01:08:23.939 Vibration primarily at? 01:08:23.939 --> 01:08:24.855 AUDIENCE: [INAUDIBLE]. 01:08:30.200 --> 01:08:32.020 PROFESSOR: And a little bit of some others. 01:08:32.020 --> 01:08:34.145 Now, let's say they all had about the same damping. 01:08:34.145 --> 01:08:36.927 Let's say they all had 10% damping. 01:08:36.927 --> 01:08:37.510 No, excuse me. 01:08:37.510 --> 01:08:38.779 All had 1% damping. 01:08:38.779 --> 01:08:40.850 That means they'll go through about 10 cycles 01:08:40.850 --> 01:08:44.010 to decay to halfway. 01:08:44.010 --> 01:08:48.130 So if they all have the same damping, 01:08:48.130 --> 01:08:50.180 even if there are several other modes present, 01:08:50.180 --> 01:08:55.069 which ones are going to last longer in time? 01:08:55.069 --> 01:08:58.540 Mode 1, because it takes just longer in time 01:08:58.540 --> 01:09:00.979 to get to 10 cycles. 01:09:00.979 --> 01:09:04.210 The other modes get there quicker. 01:09:04.210 --> 01:09:07.979 If you take a guitar string and plunk it, or a violin, 01:09:07.979 --> 01:09:10.069 or a piano, you'll hear the basic tone. 01:09:10.069 --> 01:09:13.122 And what makes it sound nice, you have those nice overtones, 01:09:13.122 --> 01:09:15.580 but if you listen carefully, the overtones die out usually, 01:09:15.580 --> 01:09:17.580 and you're left with the fundamental at the end. 01:09:17.580 --> 01:09:20.560 If you smack a piano key hard, you'll 01:09:20.560 --> 01:09:22.460 get an interesting sound at the beginning, 01:09:22.460 --> 01:09:23.740 and then it'll mellow out. 01:09:23.740 --> 01:09:26.180 And you'll hear just a pure tone at the end, 01:09:26.180 --> 01:09:30.569 and that's because the higher frequencies damp out quicker 01:09:30.569 --> 01:09:34.100 because they get in more cycles per unit of time. 01:09:34.100 --> 01:09:36.985 That's the other quick lesson. 01:09:36.985 --> 01:09:38.990 I have one thing I want to explain to you 01:09:38.990 --> 01:09:43.920 which will help maybe a little conceptual understanding 01:09:43.920 --> 01:09:47.609 about-- this came up in a homework discussion-- 01:09:47.609 --> 01:09:51.800 and that is just a note about stiffness matrices. 01:10:05.790 --> 01:10:12.280 There's a really fast easy way to assemble stiffness matrices. 01:10:12.280 --> 01:10:26.950 So here's a three mass system and a spring, spring, spring. 01:10:26.950 --> 01:10:33.610 And I'm going to put a spring here and one here. 01:10:36.730 --> 01:10:50.410 k1, k2, k3, k4, k5, and k6-- and I 01:10:50.410 --> 01:10:52.345 want to get my stiffness matrix. 01:10:56.620 --> 01:11:00.230 The stiffness matrix-- this is a three degree of freedom system. 01:11:00.230 --> 01:11:04.130 It'll be a 3 by 3, and it'll have elements up here, 01:11:04.130 --> 01:11:26.230 which I'll call k11, k12, k13, k21, k23, k31, k22, 01:11:26.230 --> 01:11:28.260 and so forth. 01:11:28.260 --> 01:11:32.148 So what's the new meaning of kij? 01:11:35.850 --> 01:11:38.660 So ki11-- i as 1. 01:11:38.660 --> 01:11:41.151 J is 1. 01:11:41.151 --> 01:11:43.150 If you can understand the interpretation of what 01:11:43.150 --> 01:11:45.620 a stiffness matrix is, it'll help 01:11:45.620 --> 01:11:49.440 you make it much easier for you to find them. 01:11:49.440 --> 01:12:07.060 So kij is the force required at i 01:12:07.060 --> 01:12:21.070 due to a unit deflection at j due to sounds 01:12:21.070 --> 01:12:22.830 kind of like a causal thing. 01:12:22.830 --> 01:12:25.520 I don't quite mean that, but so let's think about this. 01:12:25.520 --> 01:12:27.630 What's k11? 01:12:27.630 --> 01:12:33.480 So k11 is the force required at 1 per unit deflection at 1. 01:12:33.480 --> 01:12:39.520 So if I take this system and I make it move over one unit, 01:12:39.520 --> 01:12:45.100 and the other-- this is now x1 here, by the way. x1, 01:12:45.100 --> 01:12:49.160 x2, x300 so x2 and x3 are 0. 01:12:49.160 --> 01:12:50.490 I do this one a time. 01:12:50.490 --> 01:12:52.700 I move this over one unit. 01:12:52.700 --> 01:12:56.140 How much force does it take to make that happen? 01:12:56.140 --> 01:12:56.829 A real system. 01:12:56.829 --> 01:12:57.620 You're grabbing it. 01:12:57.620 --> 01:12:59.010 You're pulling it over one unit. 01:12:59.010 --> 01:13:01.610 You're making some springs move. 01:13:01.610 --> 01:13:04.140 How much force does it take to move that one one 01:13:04.140 --> 01:13:05.339 unit holding these still? 01:13:10.030 --> 01:13:14.150 Well, you're going to-- one, force on a spring is kx. 01:13:14.150 --> 01:13:17.560 If x is one, the force for trying to stretch this spring 01:13:17.560 --> 01:13:20.700 is k1 times 1. 01:13:20.700 --> 01:13:24.350 The force required to push on that spring as k5 times 01:13:24.350 --> 01:13:31.190 5-- k5 times 1. k2 times one. 01:13:31.190 --> 01:13:35.800 The k11, this first element, is the sum 01:13:35.800 --> 01:13:40.030 of all of the springs connected to it. 01:13:40.030 --> 01:13:44.980 k1 plus k2 plus k5. 01:13:44.980 --> 01:13:45.860 So how about k12? 01:13:48.710 --> 01:13:57.620 k12 is the force required at 1-- actually, let me do k21. 01:13:57.620 --> 01:14:00.110 Make a little more sense-- is the force 01:14:00.110 --> 01:14:08.671 required at 2, because I've moved the system one unit at 1. 01:14:08.671 --> 01:14:10.920 That's what the problem-- that's this thing I've done. 01:14:10.920 --> 01:14:12.990 I've moved this one unit. 01:14:12.990 --> 01:14:15.080 In order to keep this one from moving, 01:14:15.080 --> 01:14:17.914 do I have to apply force to it? 01:14:17.914 --> 01:14:20.349 How big? 01:14:20.349 --> 01:14:22.300 AUDIENCE: k2. 01:14:22.300 --> 01:14:25.890 PROFESSOR: k2 or minus k2? 01:14:25.890 --> 01:14:28.090 This thing is pushing, going over one unit. 01:14:28.090 --> 01:14:29.450 It compresses that spring. 01:14:29.450 --> 01:14:31.310 It's pushing on this thing. 01:14:31.310 --> 01:14:32.640 I say it cannot move. 01:14:32.640 --> 01:14:34.150 What do I have to do? 01:14:34.150 --> 01:14:42.000 Push back minus k2, and you can go through-- and then how 01:14:42.000 --> 01:14:43.040 about number 3? 01:14:43.040 --> 01:14:50.090 What's the force required at 3 because I've moved the one at 1 01:14:50.090 --> 01:14:52.610 by 1 unit? 01:14:52.610 --> 01:14:53.970 Move this over. 01:14:53.970 --> 01:14:56.480 Are there any springs connected to mass 3 that 01:14:56.480 --> 01:14:58.106 are affected by that motion? 01:14:58.106 --> 01:14:58.980 AUDIENCE: k5. 01:14:58.980 --> 01:15:01.905 PROFESSOR: k5, and it pushes on it through that spring, 01:15:01.905 --> 01:15:15.300 so I have to resist by-- so k31 equals minus k5. 01:15:15.300 --> 01:15:16.340 So now then you go on. 01:15:16.340 --> 01:15:17.715 If you want to get the next ones, 01:15:17.715 --> 01:15:19.110 OK, you go to the next system. 01:15:19.110 --> 01:15:20.460 This is now can't move. 01:15:20.460 --> 01:15:21.480 This can't move. 01:15:21.480 --> 01:15:23.820 We're going to let this be unit deflection. 01:15:23.820 --> 01:15:26.530 So unit deflection at two, add up the springs. 01:15:26.530 --> 01:15:32.320 k2 plus k3 plus k6-- and that's all there is. 01:15:32.320 --> 01:15:41.080 So k22, k2 plus k3 plus k6, and then you 01:15:41.080 --> 01:15:43.000 go through all the ones that it affects, 01:15:43.000 --> 01:15:45.870 and you'll get minus this and that. 01:15:45.870 --> 01:15:48.290 So that's the meaning. 01:15:48.290 --> 01:15:50.600 Each of the elements of that stiffness matrix 01:15:50.600 --> 01:15:52.570 have that meaning to it. 01:15:52.570 --> 01:16:10.310 And to give this closure, we have equation of motion, 01:16:10.310 --> 01:16:14.950 but the stiffness matrices also applies to the statics problem. 01:16:14.950 --> 01:16:22.120 So you have mx double dot plus cx dot plus kx equals 01:16:22.120 --> 01:16:25.404 F. What if I only want to do a statics problem? 01:16:25.404 --> 01:16:27.070 I'm going to put a static force on here, 01:16:27.070 --> 01:16:29.780 and I want to know the deflections, 01:16:29.780 --> 01:16:31.300 or I'm going to cause deflections, 01:16:31.300 --> 01:16:33.750 and I want to know what force it takes to do it. 01:16:33.750 --> 01:16:38.110 Well, the static problems-- let this be 0, this be 0. 01:16:38.110 --> 01:16:44.840 And it says kx equals F. So in this three degree 01:16:44.840 --> 01:16:49.750 of freedom system, what forces are required to cause 01:16:49.750 --> 01:17:00.560 the deflection 0, 0, 1. 01:17:00.560 --> 01:17:06.280 I want to deflect it one unit on the third mass only. 01:17:06.280 --> 01:17:09.730 What forces do I apply it to the system to make it happen? 01:17:09.730 --> 01:17:14.860 Well, you just multiply it out. 01:17:14.860 --> 01:17:19.150 This is F1, F2, F3. 01:17:19.150 --> 01:17:22.850 So it'll only be this one. 01:17:22.850 --> 01:17:24.760 This times this, this, and this. 01:17:24.760 --> 01:17:28.330 The only ones that matter are these three here, 01:17:28.330 --> 01:17:37.150 and this will end up being k31 times 1. 01:17:37.150 --> 01:17:44.940 This will be k-- I guess I've got to round-- 13. 01:17:44.940 --> 01:17:51.275 And we get 21, 22, 23, k33. 01:17:54.190 --> 01:17:56.710 Those are the three forces required. 01:17:56.710 --> 01:18:01.230 You put on those forces. 01:18:01.230 --> 01:18:02.910 Force is equal to these amounts. 01:18:02.910 --> 01:18:06.262 You will get that deflection. 01:18:06.262 --> 01:18:07.470 So that's just a little help. 01:18:07.470 --> 01:18:10.400 That gives you a little insight as to what 01:18:10.400 --> 01:18:14.000 stiffness matrices mean. 01:18:14.000 --> 01:18:15.720 So you can do them by inspection. 01:18:15.720 --> 01:18:18.192 Once you understand that, you can actually just fill them 01:18:18.192 --> 01:18:21.890 in by inspection, just by doing unit displacements 01:18:21.890 --> 01:18:25.360 at each place and adding up the forces. 01:18:25.360 --> 01:18:28.112 See you on Thursday.