1 00:00:00,090 --> 00:00:02,450 The following content is provided under a Creative 2 00:00:02,450 --> 00:00:03,830 Commons license. 3 00:00:03,830 --> 00:00:06,080 Your support will help MIT OpenCourseWare 4 00:00:06,080 --> 00:00:10,160 continue to offer high quality educational resources for free. 5 00:00:10,160 --> 00:00:12,710 To make a donation or to view additional materials 6 00:00:12,710 --> 00:00:16,630 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,630 --> 00:00:17,325 at ocw.mit.edu. 8 00:00:21,839 --> 00:00:23,880 PROFESSOR: We've got three lectures left-- today, 9 00:00:23,880 --> 00:00:27,620 Thursday, and next Tuesday. 10 00:00:27,620 --> 00:00:34,230 So today we're going to talk a little bit more 11 00:00:34,230 --> 00:00:46,710 about modal analysis, and we did initial conditions last time. 12 00:00:46,710 --> 00:00:53,350 This time we'll do harmonic excitation mostly 13 00:00:53,350 --> 00:00:55,390 with a little review there. 14 00:00:55,390 --> 00:01:01,350 Thursday-- let me say this slightly differently. 15 00:01:04,650 --> 00:01:07,710 Modal analysis looks at the vibration of a system 16 00:01:07,710 --> 00:01:12,000 with many degrees of freedom and looks at one mode at a time, 17 00:01:12,000 --> 00:01:15,664 but you can also just solve the whole thing at once. 18 00:01:15,664 --> 00:01:18,080 You don't have to break it down into the individual modes, 19 00:01:18,080 --> 00:01:20,870 so you can come up with a-- if you have a harmonic 20 00:01:20,870 --> 00:01:24,597 excitation on a system with many degrees of freedom, 21 00:01:24,597 --> 00:01:26,930 if you put in a harmonic force, the whole thing is going 22 00:01:26,930 --> 00:01:30,090 to shake, and you can solve it in one go, 23 00:01:30,090 --> 00:01:33,470 and that's essentially using a transfer function approach. 24 00:01:33,470 --> 00:01:36,980 So this is breaking it down one mode at a time. 25 00:01:36,980 --> 00:01:41,570 Here we're going to do this concept of transfer functions 26 00:01:41,570 --> 00:01:46,820 for Hij, so response at location i due to an excitation at j. 27 00:01:46,820 --> 00:01:48,780 In a multiple degree of freedom system, 28 00:01:48,780 --> 00:01:50,340 you get many different combinations. 29 00:01:50,340 --> 00:01:55,380 So this is steady state response, but done all at once. 30 00:01:55,380 --> 00:01:59,680 And Tuesday-- and also this has an application to something 31 00:01:59,680 --> 00:02:01,305 I called dynamic absorbers. 32 00:02:05,850 --> 00:02:08,240 I'll just be able to scratch the surface of that, 33 00:02:08,240 --> 00:02:12,020 but the Hancock Building across the way-- I told you a couple 34 00:02:12,020 --> 00:02:13,570 lectures back about it. 35 00:02:13,570 --> 00:02:16,100 In the 1970s, it was brand new, and the windows 36 00:02:16,100 --> 00:02:17,162 were falling out. 37 00:02:17,162 --> 00:02:18,870 And there was lots of windows falling out 38 00:02:18,870 --> 00:02:21,120 at the bottom and none at the top, 39 00:02:21,120 --> 00:02:22,920 and the thing was bending back and forth 40 00:02:22,920 --> 00:02:27,420 like just a cantilever, a reed in the wind, 41 00:02:27,420 --> 00:02:29,750 and one of the fixes for it is they 42 00:02:29,750 --> 00:02:34,000 put a thing called the dynamic absorber on the 58th floor. 43 00:02:34,000 --> 00:02:42,380 There are two 300 ton blocks of lead on a pressurized oil film 44 00:02:42,380 --> 00:02:45,230 so they can slide back and forth on the 58th floor 45 00:02:45,230 --> 00:02:46,660 of that building. 46 00:02:46,660 --> 00:02:49,210 They're called dynamic absorbers. 47 00:02:49,210 --> 00:02:52,590 And at the expense of letting these big blocks of lead slide 48 00:02:52,590 --> 00:02:55,110 back and forth, it keeps the building from vibrating. 49 00:02:55,110 --> 00:02:57,220 It keeps the windows from falling out. 50 00:02:57,220 --> 00:02:59,740 So I hope to get to talking a little bit 51 00:02:59,740 --> 00:03:02,380 about dynamic absorbers-- one other way 52 00:03:02,380 --> 00:03:04,860 of stopping problem vibration. 53 00:03:04,860 --> 00:03:13,630 And on Tuesday, we'll talk about strings and beams. 54 00:03:13,630 --> 00:03:15,845 Just a brief introduction to continuous systems. 55 00:03:19,140 --> 00:03:21,739 And that'll be our last lecture, and probably 56 00:03:21,739 --> 00:03:23,530 give you a very quick review then of what's 57 00:03:23,530 --> 00:03:25,080 going to be covered on the final, 58 00:03:25,080 --> 00:03:29,793 but mostly the final will be covering this last third 59 00:03:29,793 --> 00:03:32,890 of the course on vibration. 60 00:03:32,890 --> 00:03:38,460 So let's turn to modal analysis, and posted on Stellar 61 00:03:38,460 --> 00:03:43,260 is a little too page handout that gives you a just step 62 00:03:43,260 --> 00:03:47,375 by step cookbook approach to conducting a modal analysis. 63 00:03:47,375 --> 00:03:49,250 And we're going to hit the highlights of that 64 00:03:49,250 --> 00:03:51,900 this morning as a way of reviewing what we learned 65 00:03:51,900 --> 00:03:55,470 last time and moving on to calculating the response 66 00:03:55,470 --> 00:03:57,910 to harmonic forces. 67 00:03:57,910 --> 00:04:11,700 So we begin with some n degree of freedom system linearized. 68 00:04:11,700 --> 00:04:16,779 We're only dealing with linear equations of motion. 69 00:04:16,779 --> 00:04:20,300 And our first step is we need equations of motion. 70 00:04:20,300 --> 00:04:31,060 So you write them m-- in general, you'd 71 00:04:31,060 --> 00:04:33,990 write out your equations of motion for the system, 72 00:04:33,990 --> 00:04:37,100 and these x's here are just my generalized coordinates. 73 00:04:37,100 --> 00:04:39,640 They can be rotations, deflections, whatever 74 00:04:39,640 --> 00:04:42,694 makes sense in the problem. 75 00:04:42,694 --> 00:04:43,610 That's the first step. 76 00:04:43,610 --> 00:04:45,670 You need your equations of motion. 77 00:04:45,670 --> 00:04:55,690 Second step, find undamped natural frequency-- 78 00:04:55,690 --> 00:05:01,306 is the omega i's-- I'll just call it omega i's-- and mode 79 00:05:01,306 --> 00:05:01,805 shapes. 80 00:05:06,700 --> 00:05:09,380 And the mode shapes we put into a matrix 81 00:05:09,380 --> 00:05:11,070 that we call a mode shape matrix. 82 00:05:11,070 --> 00:05:13,340 I write it as u. 83 00:05:13,340 --> 00:05:15,010 So that's the next step in the problem. 84 00:05:18,260 --> 00:05:32,830 The third is basically to apply or invoke the modal expansion 85 00:05:32,830 --> 00:05:34,570 theorem. 86 00:05:34,570 --> 00:05:37,940 This is the key to the whole thing, 87 00:05:37,940 --> 00:05:42,560 and that is to say that you can write the generalized motions 88 00:05:42,560 --> 00:05:45,950 of the system-- the responses of the system-- 89 00:05:45,950 --> 00:05:56,800 as a weighted sum of the individual modes of the system. 90 00:05:56,800 --> 00:05:59,960 So the q's are the modal amplitudes. 91 00:05:59,960 --> 00:06:04,910 The u's are the mode shape-- each mode, like the first mode, 92 00:06:04,910 --> 00:06:07,640 has some amplitude and time dependence, 93 00:06:07,640 --> 00:06:11,370 and its motion is distributed to the whole system 94 00:06:11,370 --> 00:06:15,181 according to its mode shape, and that's what that statement is. 95 00:06:18,070 --> 00:06:21,190 And another way of writing it which is more intuitive 96 00:06:21,190 --> 00:06:24,350 is this is a summation then of i equals 97 00:06:24,350 --> 00:06:29,440 1 over the degrees of freedom of the system of the mode 98 00:06:29,440 --> 00:06:39,390 shapes times qi of t, the modal amplitudes of the response 99 00:06:39,390 --> 00:06:41,250 of that single degree of freedom system 100 00:06:41,250 --> 00:06:43,270 that describes each of the modes. 101 00:06:59,090 --> 00:07:09,200 Now, I should-- the qi's are the solution 102 00:07:09,200 --> 00:07:16,350 to n equations of the form mi qi double 103 00:07:16,350 --> 00:07:28,400 dot plus ci qi dot plus ki qi equals 104 00:07:28,400 --> 00:07:32,370 the excitation for that single degree of freedom system. 105 00:07:32,370 --> 00:07:34,150 So the modal expansion theorem says 106 00:07:34,150 --> 00:07:36,110 we're going to take each mode of the system, 107 00:07:36,110 --> 00:07:39,940 treat it like a single degree of freedom problem. 108 00:07:39,940 --> 00:07:42,700 And so for the first mode, I would be 1. 109 00:07:42,700 --> 00:07:46,120 You would have a modal mass, damping, stiffness, 110 00:07:46,120 --> 00:07:47,710 and a force for that mode. 111 00:07:47,710 --> 00:07:50,935 And you know how to solve for the response of a single degree 112 00:07:50,935 --> 00:07:53,820 of freedom system to initial conditions, 113 00:07:53,820 --> 00:07:56,900 or the steady state response to some harmonic input, 114 00:07:56,900 --> 00:07:59,080 and you would do that for each of the modes. 115 00:07:59,080 --> 00:08:01,130 And to get back to the final answer, 116 00:08:01,130 --> 00:08:03,540 you sum them back up again this way. 117 00:08:03,540 --> 00:08:09,100 So I've put up here on the board where 118 00:08:09,100 --> 00:08:12,340 that-- the description of that demo, 119 00:08:12,340 --> 00:08:15,110 and we went through all this information 120 00:08:15,110 --> 00:08:17,650 last time so you have it in your notes for last time. 121 00:08:17,650 --> 00:08:20,050 If you missed it last time, I put it up again. 122 00:08:20,050 --> 00:08:23,360 So here's the demo on the table. 123 00:08:23,360 --> 00:08:27,980 This is my diagram of it. 124 00:08:27,980 --> 00:08:31,640 Mass 1, mass 2, spring, lowercase k1, 125 00:08:31,640 --> 00:08:33,720 lowercase k2-- so the lower cases 126 00:08:33,720 --> 00:08:36,390 I'm going to use to describe system parameters. 127 00:08:36,390 --> 00:08:38,929 Uppercase K's and that sort of thing 128 00:08:38,929 --> 00:08:41,799 will be my modal parameters. 129 00:08:41,799 --> 00:08:45,440 So this has masses, stiffnesses, and damping, 130 00:08:45,440 --> 00:08:50,400 and this damper is connected to the non-moving frame, 131 00:08:50,400 --> 00:08:52,880 and so is this one to model the damping sliding 132 00:08:52,880 --> 00:08:55,320 up and down on that shaft. 133 00:08:55,320 --> 00:08:57,750 So the equations of motion of the system 134 00:08:57,750 --> 00:09:01,680 are here in your generalized coordinates. 135 00:09:01,680 --> 00:09:03,900 And I've included the possibility 136 00:09:03,900 --> 00:09:06,130 that I could have a force in the first mass, F1, 137 00:09:06,130 --> 00:09:07,360 a force in the second mass. 138 00:09:07,360 --> 00:09:11,900 So this is a completely general set up for this problem. 139 00:09:11,900 --> 00:09:14,910 Then you need to find the natural frequencies and mode 140 00:09:14,910 --> 00:09:20,440 shapes, so you assume a solution of the form e to i omega t. 141 00:09:20,440 --> 00:09:24,020 Plug in, and you get this algebraic equation. 142 00:09:24,020 --> 00:09:27,590 Minus omega squared m plus k times some vector, which will 143 00:09:27,590 --> 00:09:29,680 turn out to be the mode shapes. 144 00:09:29,680 --> 00:09:31,665 And either this is 0, which is trivial, 145 00:09:31,665 --> 00:09:34,870 or the determinant of this is 0, and this gives you 146 00:09:34,870 --> 00:09:36,840 the roots of this determinant, give you 147 00:09:36,840 --> 00:09:38,900 the natural frequencies of the system. 148 00:09:38,900 --> 00:09:42,680 If it's a four degree of freedom problem, you get four roots. 149 00:09:42,680 --> 00:09:45,280 So we found for this particular problem 150 00:09:45,280 --> 00:09:50,980 that the natural frequencies are 5.65 and 17.69, 151 00:09:50,980 --> 00:09:55,140 and the mass matrix, if you weren't here last time, 152 00:09:55,140 --> 00:10:06,840 in kilograms, 3.193, 0.63. 153 00:10:06,840 --> 00:10:11,750 And the stiffness matrix in newtons per meter 154 00:10:11,750 --> 00:10:14,780 for that system is this. 155 00:10:14,780 --> 00:10:22,250 Notice, stiffness matrices are always symmetric, 156 00:10:22,250 --> 00:10:26,150 and this one minus 36, minus 36. 157 00:10:26,150 --> 00:10:29,340 And many of you have been taking 2001. 158 00:10:29,340 --> 00:10:31,690 Did anybody invoke something called 159 00:10:31,690 --> 00:10:33,410 Maxwell's reciprocal theorem? 160 00:10:33,410 --> 00:10:36,760 Have you run into that? 161 00:10:36,760 --> 00:10:38,990 Well, this comes from structural analysis, 162 00:10:38,990 --> 00:10:41,660 this notion of the symmetry of the K matrix. 163 00:10:41,660 --> 00:10:44,710 The mass matrix is also going to be symmetric. 164 00:10:44,710 --> 00:10:47,910 So we need to find-- to do modal analysis, 165 00:10:47,910 --> 00:10:50,910 you make this calculation. 166 00:10:50,910 --> 00:10:52,550 You transpose mu. 167 00:10:52,550 --> 00:10:56,975 This is the transpose of the mode shape matrix mu, 168 00:10:56,975 --> 00:11:00,190 and if you do that calculation with these numbers 169 00:11:00,190 --> 00:11:05,640 and those mode shapes, you get a new mass-- 170 00:11:05,640 --> 00:11:07,420 you get the modal mass matrix, which 171 00:11:07,420 --> 00:11:09,520 is guaranteed to be diagonal, even 172 00:11:09,520 --> 00:11:10,850 if it wasn't to begin with. 173 00:11:10,850 --> 00:11:12,640 This is the numbers you get. 174 00:11:12,640 --> 00:11:15,630 The stiffness matrix certainly wasn't diagonal to begin with. 175 00:11:15,630 --> 00:11:18,950 It looked like that. 176 00:11:18,950 --> 00:11:23,580 And we multiply u transpose times k times u, 177 00:11:23,580 --> 00:11:24,890 and we get a diagonal matrix. 178 00:11:29,280 --> 00:11:31,890 This element here we'd call capital K1. 179 00:11:31,890 --> 00:11:34,600 It's the modal stiffness for mode 1. 180 00:11:34,600 --> 00:11:36,770 This is the modal mass for mode 1, 181 00:11:36,770 --> 00:11:41,370 and they represent a-- they're the numbers we 182 00:11:41,370 --> 00:11:45,050 need to write a single degree of freedom equation of motion 183 00:11:45,050 --> 00:11:47,850 of the kind at the bottom of that board up there. 184 00:11:47,850 --> 00:11:52,250 So we want to write M1q1 double dot 185 00:11:52,250 --> 00:12:03,570 plus some c1q1 dot plus k1q1 equals, in general, a-- this 186 00:12:03,570 --> 00:12:08,300 is a 1-- some excitation, but it's a single degree of freedom 187 00:12:08,300 --> 00:12:09,110 system. 188 00:12:09,110 --> 00:12:11,330 If it's a single degree of freedom system, 189 00:12:11,330 --> 00:12:13,830 what's its natural frequency using the parameters 190 00:12:13,830 --> 00:12:14,610 in this equation? 191 00:12:19,510 --> 00:12:20,490 AUDIENCE: k1/m1. 192 00:12:20,490 --> 00:12:23,660 PROFESSOR: Right, so omega 1 had better 193 00:12:23,660 --> 00:12:31,590 be the square root of k1/m1, and that's the square root of-- k1 194 00:12:31,590 --> 00:12:37,400 is 91-- nope. 195 00:12:37,400 --> 00:12:39,230 That's the real [INAUDIBLE]. 196 00:12:39,230 --> 00:12:41,030 I need my u transpose k here. 197 00:12:41,030 --> 00:12:44,640 113.71. 198 00:12:44,640 --> 00:12:46,490 113.71. 199 00:12:46,490 --> 00:12:56,420 And the modal mass for system one is 3.556. 200 00:12:56,420 --> 00:13:00,370 So that's 3 or so and 100 and something. 201 00:13:00,370 --> 00:13:03,160 That's neighborhood of 35, 36. 202 00:13:03,160 --> 00:13:06,890 Square root of 36 is about 6. 203 00:13:06,890 --> 00:13:09,450 So it needs to give you back exactly 204 00:13:09,450 --> 00:13:11,210 the correct natural frequency, which 205 00:13:11,210 --> 00:13:13,040 you found in the beginning when you 206 00:13:13,040 --> 00:13:14,900 solved for the natural frequencies and mode 207 00:13:14,900 --> 00:13:16,440 shapes of the system. 208 00:13:16,440 --> 00:13:19,420 So I repeated myself here. 209 00:13:19,420 --> 00:13:22,060 And if you write the same thing-- omega 2-- 210 00:13:22,060 --> 00:13:24,200 it better be equal to k2/m2. 211 00:13:24,200 --> 00:13:26,639 So these are just ways of verifying that you've 212 00:13:26,639 --> 00:13:27,930 done your arithmetic correctly. 213 00:13:32,370 --> 00:13:35,380 And I haven't done the damping matrix yet, 214 00:13:35,380 --> 00:13:37,730 because the damping matrices have 215 00:13:37,730 --> 00:13:39,500 to be treated rather carefully. 216 00:13:39,500 --> 00:13:42,550 They don't automatically uncouple, 217 00:13:42,550 --> 00:13:45,420 and so we'll address it in just a second. 218 00:13:48,100 --> 00:13:50,560 The damping matrix, however, would 219 00:13:50,560 --> 00:13:58,220 be written utcu would give me what 220 00:13:58,220 --> 00:14:06,225 I hope is a diagonalized modal damping matrix. 221 00:14:06,225 --> 00:14:08,440 I'm trying to make these look like caps or something 222 00:14:08,440 --> 00:14:11,610 to make them look a little different from the c's 223 00:14:11,610 --> 00:14:13,070 of the dashpots themselves. 224 00:14:13,070 --> 00:14:16,900 We hope to find a diagonalized damping matrix, 225 00:14:16,900 --> 00:14:21,890 but we have to make some effort to make that happen. 226 00:14:21,890 --> 00:14:25,900 Finally, one thing we haven't dealt with before. 227 00:14:25,900 --> 00:14:31,940 In order to-- when we derive the single degree of freedom modal 228 00:14:31,940 --> 00:14:34,680 systems, we had to multiply through 229 00:14:34,680 --> 00:14:40,980 by this u transpose, the original equations of motion, 230 00:14:40,980 --> 00:14:43,360 and so what we haven't worked with yet 231 00:14:43,360 --> 00:14:50,540 is how do you get the modal forces in the system? 232 00:14:50,540 --> 00:14:54,610 And they come from the calculation u transpose 233 00:14:54,610 --> 00:14:59,290 times the generalized forces in the system. 234 00:14:59,290 --> 00:15:01,520 So that's a vector times a matrix gives you 235 00:15:01,520 --> 00:15:04,650 a new vector, which are the generalized forces. 236 00:15:04,650 --> 00:15:06,430 And we're going to do that example in more 237 00:15:06,430 --> 00:15:07,760 detail in a few minutes. 238 00:15:18,190 --> 00:15:24,680 So last time we did the initial conditions problem, 239 00:15:24,680 --> 00:15:29,680 and we need the results of that to do-- 240 00:15:29,680 --> 00:15:31,990 we need the damping that we learned from that to do 241 00:15:31,990 --> 00:15:35,140 the force vibration problem. 242 00:15:35,140 --> 00:15:38,320 So let's go back and review a little bit about what we did. 243 00:15:38,320 --> 00:15:43,570 So let's let, for a moment, the generalized forces be 0. 244 00:15:43,570 --> 00:15:45,530 So no excitation. 245 00:15:45,530 --> 00:15:48,530 So we're only doing an initial conditions problem, 246 00:15:48,530 --> 00:15:55,470 and let's assume that we have some set of initial conditions 247 00:15:55,470 --> 00:15:59,430 on the generalized displacements at time 0. 248 00:15:59,430 --> 00:16:02,720 So those initial conditions on displacements-- 249 00:16:02,720 --> 00:16:07,290 they'll look something like x10 down to xn0, 250 00:16:07,290 --> 00:16:10,880 and we just have a two degree of freedom system in our example. 251 00:16:10,880 --> 00:16:14,875 And also you could have an x dot at t equals 0, 252 00:16:14,875 --> 00:16:21,270 and that would be some vector of v10's down to vn0's. 253 00:16:21,270 --> 00:16:23,300 And for today, we're going to let those be 0s. 254 00:16:23,300 --> 00:16:26,190 So we're just going to have an initial deflection 255 00:16:26,190 --> 00:16:27,650 in the system and see what happens. 256 00:16:33,300 --> 00:16:38,700 And we learned last time that, since the modal expansion 257 00:16:38,700 --> 00:16:43,160 theorem is what makes this whole thing work, 258 00:16:43,160 --> 00:16:52,490 the idea that x can be written as u times q-- 259 00:16:52,490 --> 00:16:56,360 then we ought to be able to say the same thing for-- if these 260 00:16:56,360 --> 00:16:58,240 are initial conditions, then we should 261 00:16:58,240 --> 00:17:01,870 be able to get the initial conditions in modal 262 00:17:01,870 --> 00:17:02,460 coordinates. 263 00:17:02,460 --> 00:17:11,319 So x at 0 here is also u at the q's at 0, 264 00:17:11,319 --> 00:17:14,140 but we normally would specify these. 265 00:17:14,140 --> 00:17:15,569 We need to know those. 266 00:17:15,569 --> 00:17:19,560 Well, to get these I just multiply through by u inverse 267 00:17:19,560 --> 00:17:24,940 so that the q, the vector of the initial displacements in modal 268 00:17:24,940 --> 00:17:31,980 coordinates, is just equal to u inverse times 269 00:17:31,980 --> 00:17:35,330 the initial conditions-- the initial deflections 270 00:17:35,330 --> 00:17:38,350 in the generalized coordinates. 271 00:17:38,350 --> 00:17:45,730 And q dot at 0, if you had non-zero initial velocities, 272 00:17:45,730 --> 00:17:50,940 would be u inverse times x dot at time 0. 273 00:18:02,820 --> 00:18:04,680 So let me tell you where we're going. 274 00:18:04,680 --> 00:18:07,955 I'm not going to go back all the way 275 00:18:07,955 --> 00:18:09,580 through the initial conditions problem. 276 00:18:09,580 --> 00:18:10,630 That was last lecture. 277 00:18:10,630 --> 00:18:12,690 I'm going to review the results of it, 278 00:18:12,690 --> 00:18:16,010 because what I want to get to is, how do you 279 00:18:16,010 --> 00:18:19,680 compute the response by modal analysis-- the response 280 00:18:19,680 --> 00:18:23,010 to a harmonic force? 281 00:18:23,010 --> 00:18:25,450 For example, if you get close to resonance 282 00:18:25,450 --> 00:18:27,120 for a single degree of freedom system, 283 00:18:27,120 --> 00:18:31,610 what controls the height of that peak? 284 00:18:31,610 --> 00:18:34,290 If you drive a single degree of freedom system at resonance, 285 00:18:34,290 --> 00:18:35,960 what's the most important parameter? 286 00:18:35,960 --> 00:18:36,870 AUDIENCE: Damping [INAUDIBLE]. 287 00:18:36,870 --> 00:18:37,661 PROFESSOR: Damping. 288 00:18:37,661 --> 00:18:39,570 So we really need to know damping. 289 00:18:39,570 --> 00:18:43,040 And the way we get damping is-- one of the ways is measure it. 290 00:18:43,040 --> 00:18:45,160 So I want to measure the right damping 291 00:18:45,160 --> 00:18:46,990 so that I can do-- what I'm trying 292 00:18:46,990 --> 00:18:50,940 to get to is to do the force vibration problem, 293 00:18:50,940 --> 00:18:54,830 but I need to get some estimates of damping to do it. 294 00:18:54,830 --> 00:18:56,560 Now, I said damping can sometimes 295 00:18:56,560 --> 00:19:01,330 be a problem, so I'm going to show you-- 296 00:19:01,330 --> 00:19:03,660 I alluded to this damp last time, 297 00:19:03,660 --> 00:19:07,280 but I wasn't able to kind of really clearly go through it. 298 00:19:07,280 --> 00:19:10,660 So I'm going to make a damping model, my damping matrix 299 00:19:10,660 --> 00:19:15,050 for my system such that it is proportional 300 00:19:15,050 --> 00:19:21,740 to the original mass and stiffness matrix of the system. 301 00:19:21,740 --> 00:19:24,482 So alpha is just a parameter that I get to choose. 302 00:19:24,482 --> 00:19:26,190 Beta is another one that I get to choose, 303 00:19:26,190 --> 00:19:29,010 and this is m and k, the original mass 304 00:19:29,010 --> 00:19:36,150 and stiffness matrices of this problem 305 00:19:36,150 --> 00:19:38,400 or any n degree of freedom problem. 306 00:19:38,400 --> 00:19:40,400 We're doing these two degree of freedom examples 307 00:19:40,400 --> 00:19:44,060 because they're tractable on the board. 308 00:19:44,060 --> 00:19:48,110 So if I make my damping matrix look like that, 309 00:19:48,110 --> 00:19:53,440 it is guaranteed to work when I do u transpose times that times 310 00:19:53,440 --> 00:19:56,950 u, because we know the mass and stiffness matrices give me 311 00:19:56,950 --> 00:20:00,710 diagonals, so I'm just essentially doing that again. 312 00:20:00,710 --> 00:20:03,015 So this is guaranteed to give me a diagonal matrix. 313 00:20:03,015 --> 00:20:03,780 It's what I want. 314 00:20:03,780 --> 00:20:04,400 Yes? 315 00:20:04,400 --> 00:20:07,330 AUDIENCE: So are those the original mass? 316 00:20:07,330 --> 00:20:09,690 PROFESSOR: Yes, these are the original ones. 317 00:20:09,690 --> 00:20:12,509 Usually, if I'm really trying to write the modal ones, 318 00:20:12,509 --> 00:20:15,050 I'll draw them with a diagonal through them or something just 319 00:20:15,050 --> 00:20:20,910 to-- so these are m and k, just like up here. 320 00:20:20,910 --> 00:20:21,650 Right there. 321 00:20:21,650 --> 00:20:23,600 They're right from the original system. 322 00:20:23,600 --> 00:20:25,620 They're the ones on the top of this board. 323 00:20:25,620 --> 00:20:29,940 There's the mass matrix, damping matrix, stiffness matrix 324 00:20:29,940 --> 00:20:33,020 for the original equations of motion. 325 00:20:33,020 --> 00:20:38,450 So I am saying that my unspecified damping matrix, 326 00:20:38,450 --> 00:20:42,010 which is written as c1, c2 there-- 327 00:20:42,010 --> 00:20:48,100 I'm going to represent it this way, 328 00:20:48,100 --> 00:20:53,010 so that u transpose cu, which is the calculation I need 329 00:20:53,010 --> 00:20:58,140 to be able to do-- what that gives me is an alpha, 330 00:20:58,140 --> 00:21:01,400 and here's my modal mass, diagonalized 331 00:21:01,400 --> 00:21:04,760 modal mass matrix that results, plus a beta 332 00:21:04,760 --> 00:21:08,280 times my diagonalized stiffness matrices. 333 00:21:08,280 --> 00:21:13,400 So Christina, these guys are the modal ones after doing 334 00:21:13,400 --> 00:21:15,550 u transpose mu, u transpose ku. 335 00:21:22,140 --> 00:21:25,150 So we have a two degree of freedom system, 336 00:21:25,150 --> 00:21:27,870 and when I do this calculation, I'll 337 00:21:27,870 --> 00:21:34,600 get modal mass m1, m2, k1, k2. 338 00:21:34,600 --> 00:21:39,350 So the final diagonalized stiffness matrix 339 00:21:39,350 --> 00:21:47,080 will end up looking like some capital C10, capital C2 here. 340 00:21:47,080 --> 00:21:50,150 That's my diagonalized damping matrix. 341 00:21:50,150 --> 00:21:54,890 It'll come from these, and I'll just write them out. 342 00:21:54,890 --> 00:22:02,390 So the C1 will be an alpha m1 plus beta k1, 343 00:22:02,390 --> 00:22:10,750 and C2 will be alpha m2 plus beta k2. 344 00:22:14,340 --> 00:22:16,130 So I have two free parameters with which 345 00:22:16,130 --> 00:22:19,810 I can fit-- I can fit those parameters to give me 346 00:22:19,810 --> 00:22:21,990 the damping that I want, and I'm going to measure 347 00:22:21,990 --> 00:22:24,000 the damping in the system. 348 00:22:24,000 --> 00:22:25,570 And then I'm going to find the two 349 00:22:25,570 --> 00:22:27,940 parameters that make that work. 350 00:22:27,940 --> 00:22:30,030 That's why I'm going through this. 351 00:22:43,510 --> 00:22:44,875 All right, running out of room. 352 00:22:50,550 --> 00:22:55,461 So zeta 1, the damping ratio for mode 1-- 353 00:22:55,461 --> 00:22:57,210 if it's a single degree of freedom system, 354 00:22:57,210 --> 00:23:01,240 you say, oh, well that's the damping constant over 2 355 00:23:01,240 --> 00:23:03,420 omega 1 m1. 356 00:23:03,420 --> 00:23:07,307 That's how we define damping ratio 357 00:23:07,307 --> 00:23:08,890 for a single degree of freedom system, 358 00:23:08,890 --> 00:23:11,160 but we know what these quantities are. 359 00:23:11,160 --> 00:23:17,760 This then is an alpha m1 over 2 omega 1 m1 360 00:23:17,760 --> 00:23:26,020 plus a beta k1 over 2 omega 1 m1. 361 00:23:26,020 --> 00:23:29,860 k1/m1 is omega 1 squared. 362 00:23:29,860 --> 00:23:31,970 So put omega 1 squared up here. 363 00:23:31,970 --> 00:23:33,750 Cancel with that. 364 00:23:33,750 --> 00:23:39,090 These two gives you-- this gives you alpha over 2 365 00:23:39,090 --> 00:23:47,130 omega 1 plus beta omega 1 over 2, 366 00:23:47,130 --> 00:23:49,010 and I can do the same thing for zeta 2. 367 00:23:51,710 --> 00:24:02,470 Be my c2 over 2 omega 2 m2, and that'll give me an alpha over 2 368 00:24:02,470 --> 00:24:08,650 omega 2 plus beta omega 2 over 2. 369 00:24:08,650 --> 00:24:13,350 So if I can measure a value for this damping and a value 370 00:24:13,350 --> 00:24:16,310 for that damping, I have. 371 00:24:16,310 --> 00:24:17,980 These are known then. 372 00:24:17,980 --> 00:24:21,580 I have two equations and two unknowns, alpha 373 00:24:21,580 --> 00:24:24,500 and beta-- just algebraic equations that I can solve. 374 00:24:28,800 --> 00:24:30,500 So now I need to conduct my experiment. 375 00:24:39,420 --> 00:24:42,020 I have this system. 376 00:24:42,020 --> 00:24:43,940 Without external excitation, it's 377 00:24:43,940 --> 00:24:53,080 just a free vibration system-- typical equations of motion. 378 00:24:53,080 --> 00:24:58,260 The first mode would look like m1q1 double dot 379 00:24:58,260 --> 00:25:08,040 plus c1q1 dot plus k1q1 equals 0. 380 00:25:08,040 --> 00:25:09,634 I'm looking for solutions to that, 381 00:25:09,634 --> 00:25:11,550 and there's a similar one for the second mode. 382 00:25:17,960 --> 00:25:23,169 And what I want to do is excite only one 383 00:25:23,169 --> 00:25:24,210 of these modes at a time. 384 00:25:24,210 --> 00:25:26,090 Now, you've seen this demo done before, 385 00:25:26,090 --> 00:25:28,330 but at a certain combination of deflections, 386 00:25:28,330 --> 00:25:31,560 it'll respond only in mode 1, and a different one 387 00:25:31,560 --> 00:25:33,620 will respond only in mode 2. 388 00:25:33,620 --> 00:25:36,400 And it's guaranteed that, if you deflect 389 00:25:36,400 --> 00:25:40,870 a system in the shape of one of its modes and let it go, 390 00:25:40,870 --> 00:25:42,660 it will only vibrate in that mode, 391 00:25:42,660 --> 00:25:44,900 but let's prove to ourselves that that actually 392 00:25:44,900 --> 00:25:47,190 is going to work. 393 00:25:47,190 --> 00:26:02,820 So I'm going to let, for example, initial conditions-- 394 00:26:02,820 --> 00:26:11,160 x10, x20-- be in the shape of mode 1. 395 00:26:11,160 --> 00:26:21,770 Well mode 1 is 1, 2.2667-- is the mode shape of mode 1. 396 00:26:21,770 --> 00:26:23,620 I'm just going to let that be. 397 00:26:23,620 --> 00:26:27,320 I'm going to deflect it in the shape of mode one, 398 00:26:27,320 --> 00:26:36,830 and I'm going to let x10 dot and x20 dot-- those are 0. 399 00:26:36,830 --> 00:26:42,090 And I need to know, if I do that, 400 00:26:42,090 --> 00:26:46,240 what are the resulting initial conditions 401 00:26:46,240 --> 00:26:47,890 in modal coordinates? 402 00:26:47,890 --> 00:26:58,010 Well, we know q10 and q20-- I can obtain them 403 00:26:58,010 --> 00:27:07,100 by doing u inverse times this. 404 00:27:07,100 --> 00:27:12,320 So this will be u inverse-- last time-- I'm going to write this. 405 00:27:12,320 --> 00:27:14,230 Last time I made a mistake. 406 00:27:14,230 --> 00:27:16,890 So u inverse-- I left out a 0. 407 00:27:16,890 --> 00:27:37,030 So 0.0898, 0.9102, 0.4016, minus 0.4016. 408 00:27:37,030 --> 00:27:40,430 So that's u inverse, and I'm going to multiply it 409 00:27:40,430 --> 00:27:43,130 by one of the mode shapes. 410 00:27:43,130 --> 00:27:49,310 1 and 2.2667. 411 00:27:49,310 --> 00:27:51,470 So I'm saying that my initial conditions 412 00:27:51,470 --> 00:27:54,190 are going to look exactly like one mode shape. 413 00:27:54,190 --> 00:27:58,390 To compute the equivalent modal initial conditions, 414 00:27:58,390 --> 00:28:04,050 I multiply the generalized the coordinate initial conditions 415 00:28:04,050 --> 00:28:05,370 by u inverse. 416 00:28:05,370 --> 00:28:09,940 Here's u inverse times that, and if I do that calculation, 417 00:28:09,940 --> 00:28:11,960 I get exactly 1, 0. 418 00:28:15,180 --> 00:28:17,230 And if instead of the first mode shape 419 00:28:17,230 --> 00:28:23,510 I put in the second mode shape, the 1 minus 0.2236-- 420 00:28:23,510 --> 00:28:28,510 if I did that, I would get exactly 0, 1. 421 00:28:31,200 --> 00:28:33,960 So I just wanted to go through that just so you'd see it-- 422 00:28:33,960 --> 00:28:38,560 that the math bears it out. 423 00:28:38,560 --> 00:28:40,670 If you put in a deflection that's exactly 424 00:28:40,670 --> 00:28:43,160 the shape of a mode, then you will get back 425 00:28:43,160 --> 00:28:48,889 an equivalent initial condition in the modal coordinates 426 00:28:48,889 --> 00:28:50,930 for only that mode and everything else will be 0. 427 00:28:50,930 --> 00:28:51,900 So that's q. 428 00:28:51,900 --> 00:28:56,220 This is q10-- is that guy. 429 00:28:56,220 --> 00:29:01,970 And this q20 in this case is 0. 430 00:29:01,970 --> 00:29:02,500 OK? 431 00:29:02,500 --> 00:29:05,040 So now we're ready to do the experiment. 432 00:29:05,040 --> 00:29:15,440 So if I deflected in the shape of mode 1-- come back here. 433 00:29:18,817 --> 00:29:21,400 And now would be a good time to lower the lights a little bit. 434 00:29:24,100 --> 00:29:27,040 And unfortunately we have that white chalk in the background 435 00:29:27,040 --> 00:29:29,670 to distract us, but it's now deflected only 436 00:29:29,670 --> 00:29:31,979 in the shape of mode 1, and now it's 437 00:29:31,979 --> 00:29:34,270 going to behave like a single degree of freedom system, 438 00:29:34,270 --> 00:29:35,470 right? 439 00:29:35,470 --> 00:29:40,760 So how would you estimate-- do a quick estimate 440 00:29:40,760 --> 00:29:41,980 of the damping of mode 1. 441 00:29:46,210 --> 00:29:48,970 I gave you a little quick, easy rule 442 00:29:48,970 --> 00:29:51,121 you could use a few times back. 443 00:29:51,121 --> 00:29:51,620 What was it? 444 00:29:58,242 --> 00:30:00,140 AUDIENCE: Do the 50% thing? 445 00:30:00,140 --> 00:30:02,660 PROFESSOR: Do the 50% thing she suggests. 446 00:30:02,660 --> 00:30:04,910 Now, can you be a little more specific. 447 00:30:04,910 --> 00:30:05,410 Pardon? 448 00:30:05,410 --> 00:30:06,910 AUDIENCE: The half life. 449 00:30:06,910 --> 00:30:09,440 PROFESSOR: Yeah, how many-- what am I looking for? 450 00:30:09,440 --> 00:30:10,180 How many-- 451 00:30:10,180 --> 00:30:11,010 AUDIENCE: Cycles. 452 00:30:11,010 --> 00:30:15,540 PROFESSOR: Cycles it takes for the thing to decay 50%. 453 00:30:15,540 --> 00:30:17,930 So the reference line is here. 454 00:30:17,930 --> 00:30:21,740 I've deflected it that far, and it'll start vibrating, 455 00:30:21,740 --> 00:30:23,660 and when the top of this on a vibration 456 00:30:23,660 --> 00:30:25,827 gets to the halfway point between here 457 00:30:25,827 --> 00:30:27,410 and the reference-- reference is where 458 00:30:27,410 --> 00:30:30,350 it starts-- that'll be my 50%. 459 00:30:30,350 --> 00:30:37,540 So one, two, three, four. 460 00:30:37,540 --> 00:30:41,740 Got about four cycles when it only went down halfway. 461 00:30:41,740 --> 00:30:43,755 Now, when I did it in my office the other day, 462 00:30:43,755 --> 00:30:46,920 I only got 2 and 1/2 cycles. 463 00:30:46,920 --> 00:30:49,070 So we're going to use 2 and 1/2 cycles because I 464 00:30:49,070 --> 00:30:51,380 ran the numbers for that. 465 00:30:51,380 --> 00:30:53,710 This thing is very sensitive if it's 466 00:30:53,710 --> 00:30:57,100 inclined a little bit, because that changes 467 00:30:57,100 --> 00:30:59,160 the friction on the shaft. 468 00:30:59,160 --> 00:31:01,660 So in my office the other day, there was a lot more damping. 469 00:31:01,660 --> 00:31:05,570 So anyway, pretend it's 2 and 1/2. 470 00:31:05,570 --> 00:31:09,320 Now, while we're at it, let's do the other case. 471 00:31:09,320 --> 00:31:12,970 So the other case we want to deflect it 472 00:31:12,970 --> 00:31:14,700 in the shape of mode 2. 473 00:31:14,700 --> 00:31:17,380 So I go down, say, a unit amount. 474 00:31:23,610 --> 00:31:29,170 It's now deflected downward some amount and upwards such 475 00:31:29,170 --> 00:31:31,890 that the ratio is in the mode shape of mode 2. 476 00:31:31,890 --> 00:31:38,080 So every unit I went down, I go up minus 22% of that. 477 00:31:38,080 --> 00:31:40,310 So that's what's been done here. 478 00:31:40,310 --> 00:31:45,450 So when I release this one, now the reference line 479 00:31:45,450 --> 00:31:49,580 for the second one is up here down to here. 480 00:31:49,580 --> 00:31:54,217 So when this upper one decays to halfway-- about here-- 481 00:31:54,217 --> 00:31:55,550 that'll be the number of cycles. 482 00:31:58,210 --> 00:32:02,950 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 483 00:32:02,950 --> 00:32:07,040 Anyway, in my office I got about 10. 484 00:32:07,040 --> 00:32:10,200 We're going to use my numbers, but that one is clearly 485 00:32:10,200 --> 00:32:11,380 a lot less damped. 486 00:32:11,380 --> 00:32:14,220 That thing just went on, an on, and on, and on, and on. 487 00:32:14,220 --> 00:32:31,680 So in our experiment then-- so we conducted our experiment. 488 00:32:37,580 --> 00:32:43,000 Zeta 1 is approximately equal to point 1.1 489 00:32:43,000 --> 00:32:47,460 over the number of cycles to decay 50%, 490 00:32:47,460 --> 00:32:49,760 and when I did the experiment in my office, 491 00:32:49,760 --> 00:32:54,300 that was 0.11 over 2.5. 492 00:32:54,300 --> 00:32:56,810 So it was really decaying fast, and so when you only 493 00:32:56,810 --> 00:32:58,830 have two or three cycles, it helps 494 00:32:58,830 --> 00:33:00,450 to use even fractional cycles. 495 00:33:00,450 --> 00:33:02,740 And it's OK to use fractional cycles. 496 00:33:02,740 --> 00:33:05,810 This is just an estimate. 497 00:33:05,810 --> 00:33:07,750 That's going to give us a number. 498 00:33:07,750 --> 00:33:12,850 Zeta 2 is approximately 0.11 over the number 499 00:33:12,850 --> 00:33:17,350 of cycles to decay 50%, and in this case, that's 500 00:33:17,350 --> 00:33:19,770 going to be 10. 501 00:33:19,770 --> 00:33:27,540 And so that gives me 0.011, or what's known as 1.1% damping. 502 00:33:27,540 --> 00:33:34,736 And the other one, 0.11/2.5, gives me 0.044, 503 00:33:34,736 --> 00:33:38,140 or 4.4% damping. 504 00:33:38,140 --> 00:33:39,030 Percent of what? 505 00:33:39,030 --> 00:33:43,210 Anybody remember what happens when you're at 1? 506 00:33:43,210 --> 00:33:44,975 Have a damping ratio of 1.0? 507 00:33:44,975 --> 00:33:46,100 AUDIENCE: Critical damping. 508 00:33:46,100 --> 00:33:46,820 PROFESSOR: That's critical damping. 509 00:33:46,820 --> 00:33:48,000 That's that crossover point. 510 00:33:48,000 --> 00:33:50,990 If you have 1 or greater, you get no oscillation. 511 00:33:50,990 --> 00:33:53,030 It just goes and stops. 512 00:33:53,030 --> 00:33:55,270 Less than 1, it will actually across 0 513 00:33:55,270 --> 00:33:56,640 and oscillate a little bit. 514 00:33:59,690 --> 00:34:07,810 Well, now I have values for 0.044 0.011 515 00:34:07,810 --> 00:34:10,699 that I can plug in to these two equations, 516 00:34:10,699 --> 00:34:13,610 and I now have two equations and two unknowns, alpha and beta, 517 00:34:13,610 --> 00:34:16,110 because I know omega 1 and omega 2. 518 00:34:16,110 --> 00:34:24,270 So solve for alpha, and for alpha I'm not 519 00:34:24,270 --> 00:34:25,680 going to do that on the board. 520 00:34:25,680 --> 00:34:29,510 It's kind of a waste of good lecture time. 521 00:34:29,510 --> 00:34:34,800 10.71. 522 00:34:34,800 --> 00:34:48,080 And beta, minus 0.033. 523 00:34:48,080 --> 00:34:51,650 So they can be positive or negative to make these work 524 00:34:51,650 --> 00:34:54,530 out, but those are the two values you need, 525 00:34:54,530 --> 00:35:00,210 and that says that we're going to model the damping matrix 526 00:35:00,210 --> 00:35:15,240 of the system as 10.71m plus minus 0.033k. 527 00:35:15,240 --> 00:35:21,460 And we'll compute the diagonalized damping matrix, 528 00:35:21,460 --> 00:35:27,700 which comes from utcu, and did I write that down? 529 00:35:34,250 --> 00:35:39,600 I didn't, but I don't need it, because I know the damping 530 00:35:39,600 --> 00:35:45,450 that I'm after 0.044 and 0.011, because in order 531 00:35:45,450 --> 00:35:49,250 to complete this problem, if I wanted to go to completion 532 00:35:49,250 --> 00:35:53,610 and have the transient decay of the system, 533 00:35:53,610 --> 00:36:01,860 I would have it in for each of the two modal systems. 534 00:36:01,860 --> 00:36:11,130 q of t for the first mode would be some e to the minus zeta 535 00:36:11,130 --> 00:36:22,600 1 omega 1t cosine omega 1-- I left out my q10. 536 00:36:25,410 --> 00:36:35,210 q10 cosine omega 1d times t-- that's the most 537 00:36:35,210 --> 00:36:47,400 of the initial displacement part-- plus q10 dot plus zeta 538 00:36:47,400 --> 00:36:53,770 1 omega 1 q10. 539 00:36:53,770 --> 00:36:55,890 All of that over omega 1d. 540 00:36:59,550 --> 00:37:07,130 All of this times sine omega 1d times time. 541 00:37:07,130 --> 00:37:10,880 So this is just the response to initial conditions for a single 542 00:37:10,880 --> 00:37:14,500 degree of freedom system, and you'd have a similar one for q2 543 00:37:14,500 --> 00:37:17,960 of t, except now all the ones would be replaced by 2's. 544 00:37:17,960 --> 00:37:26,020 So you have an e to the minus zeta 2 omega 2t times-- 545 00:37:26,020 --> 00:37:29,200 and those are your two-- that would 546 00:37:29,200 --> 00:37:33,960 be in general the response to initial conditions 547 00:37:33,960 --> 00:37:36,370 in modal terms. 548 00:37:36,370 --> 00:37:39,160 And if you want to get back to the final response 549 00:37:39,160 --> 00:37:44,950 in your generalized coordinates, then it's just summation 550 00:37:44,950 --> 00:37:52,100 i equals 1 to 2, in this case, of u mode shape, 551 00:37:52,100 --> 00:38:03,400 or mode 1, q1 plus mode shape for mode 2 q2 of t. 552 00:38:03,400 --> 00:38:07,772 So these two added together would give you total response. 553 00:38:07,772 --> 00:38:08,730 So that's the response. 554 00:38:08,730 --> 00:38:10,771 Kind of a quick review of what we learned before. 555 00:38:10,771 --> 00:38:12,960 Response to initial conditions, but now 556 00:38:12,960 --> 00:38:18,840 along the way we've learned how to experimentally fix 557 00:38:18,840 --> 00:38:21,140 the damping matrix, so it'll work, 558 00:38:21,140 --> 00:38:24,850 and actually get it to give us exact accurate results if we 559 00:38:24,850 --> 00:38:28,770 want to do a response to any initial conditions problem. 560 00:38:28,770 --> 00:38:31,710 But now we also have damping ratios for this system, 561 00:38:31,710 --> 00:38:35,425 and we can now go on to do the force vibration problem. 562 00:38:35,425 --> 00:38:37,050 And the force vibration problem will 563 00:38:37,050 --> 00:38:42,380 be pretty easy at this point. 564 00:38:42,380 --> 00:38:47,200 So that was a quick review of response to initial conditions 565 00:38:47,200 --> 00:38:52,970 by modal analysis-- how to find a damping matrix that works. 566 00:38:52,970 --> 00:38:56,070 We've also proven that, if you deflect 567 00:38:56,070 --> 00:39:00,220 a system in the shape of a mode, it responds only in that mode. 568 00:39:00,220 --> 00:39:01,660 That's all pretty use-- and that's 569 00:39:01,660 --> 00:39:04,510 generally true of vibration systems. 570 00:39:04,510 --> 00:39:06,357 So we know-- I banged on this before. 571 00:39:06,357 --> 00:39:08,190 We know these things like to shake a little. 572 00:39:08,190 --> 00:39:10,757 They're flexible, and this one has a mode shape. 573 00:39:10,757 --> 00:39:11,840 It's kind of a cantilever. 574 00:39:11,840 --> 00:39:12,940 It's first mode. 575 00:39:12,940 --> 00:39:15,540 It bends like that a little bit, and this thing 576 00:39:15,540 --> 00:39:16,826 has to bend with it. 577 00:39:16,826 --> 00:39:18,575 But if I were able to just give this thing 578 00:39:18,575 --> 00:39:21,350 an initial deflection just in the shape of that mode 579 00:39:21,350 --> 00:39:26,080 and let it go, it will vibrate in just that mode. 580 00:39:26,080 --> 00:39:28,550 And if I do it in some other contorted way 581 00:39:28,550 --> 00:39:32,251 that its initial shape isn't just one mode and let it go, 582 00:39:32,251 --> 00:39:34,000 it'll vibrate in a couple different modes. 583 00:39:34,000 --> 00:39:36,490 AUDIENCE: Is there a way to visualize 584 00:39:36,490 --> 00:39:37,980 the modal coordinates? 585 00:39:37,980 --> 00:39:41,326 Would they be the center of a mass of certain things, 586 00:39:41,326 --> 00:39:42,992 or would they would be the center masses 587 00:39:42,992 --> 00:39:45,420 of the springs or something? 588 00:39:45,420 --> 00:39:50,350 PROFESSOR: Yeah, I taught a vibration course 589 00:39:50,350 --> 00:39:53,690 for many, many years, and I scratched my head long and hard 590 00:39:53,690 --> 00:39:55,190 to try to come up with an example 591 00:39:55,190 --> 00:39:57,340 where you could do exactly what you asked. 592 00:39:57,340 --> 00:39:59,810 I know of one example that works. 593 00:39:59,810 --> 00:40:02,340 In general, the modal coordinates-- 594 00:40:02,340 --> 00:40:05,140 it's a coordinate transformation into some system 595 00:40:05,140 --> 00:40:07,810 that it's very hard to place yourself physically so you can 596 00:40:07,810 --> 00:40:10,280 see what's going on, but to answer your question 597 00:40:10,280 --> 00:40:15,630 I'll show you one and you still have enough time 598 00:40:15,630 --> 00:40:17,460 to get through the other part. 599 00:40:17,460 --> 00:40:19,240 Imagine this eraser. 600 00:40:19,240 --> 00:40:22,570 It's a car, a car automobile suspension system. 601 00:40:22,570 --> 00:40:25,650 So here's the car. 602 00:40:25,650 --> 00:40:30,340 Sits on springs, the tires and its suspension system. 603 00:40:30,340 --> 00:40:34,120 Got a center of mass that's here somewhere. 604 00:40:34,120 --> 00:40:38,070 And I'm only going to consider vertical motion of this thing. 605 00:40:38,070 --> 00:40:40,770 Now, you can imagine that it could go up and down, 606 00:40:40,770 --> 00:40:44,000 but you can imagine it can also pitch back and forth. 607 00:40:44,000 --> 00:40:49,860 So I'm going to have two generalized coordinates. 608 00:40:49,860 --> 00:40:53,330 One is the vertical deflection of its center of mass, 609 00:40:53,330 --> 00:40:57,140 and the other is the angular rotation 610 00:40:57,140 --> 00:40:58,660 of the center of mass with respect 611 00:40:58,660 --> 00:41:01,760 to my initial horizontal. 612 00:41:01,760 --> 00:41:04,400 Now, you get equations of motion of this. 613 00:41:04,400 --> 00:41:07,780 One, you'll get an mx double dot kind of equation 614 00:41:07,780 --> 00:41:10,280 by summing the forces on this and the forces 615 00:41:10,280 --> 00:41:11,100 from the springs. 616 00:41:11,100 --> 00:41:13,090 And of course you'll have dampers 617 00:41:13,090 --> 00:41:15,100 and those kinds of things. 618 00:41:15,100 --> 00:41:16,930 So you'll have one force equation 619 00:41:16,930 --> 00:41:18,230 by summing the forces on it. 620 00:41:18,230 --> 00:41:20,620 You'll have another equation that's sum of torques, 621 00:41:20,620 --> 00:41:24,330 and it's an i theta double dot kind of equation. 622 00:41:24,330 --> 00:41:26,737 So it'll be a two degree of freedom system. 623 00:41:26,737 --> 00:41:28,195 You'll get two natural frequencies, 624 00:41:28,195 --> 00:41:30,210 and two mode shapes. 625 00:41:30,210 --> 00:41:34,220 And the two natural frequencies look 626 00:41:34,220 --> 00:41:37,685 approximate-- the two mode shapes look-- here 627 00:41:37,685 --> 00:41:41,000 is the original undeflected system. 628 00:41:41,000 --> 00:41:46,190 One mode shape the system moves up and rotates up. 629 00:41:46,190 --> 00:41:50,730 It has both positive rotation and some positive deflection. 630 00:41:50,730 --> 00:41:53,760 And I need to draw this over here a little further. 631 00:41:53,760 --> 00:41:56,280 So here's the original system. 632 00:41:56,280 --> 00:41:58,940 It moves up in its mode shape like that, 633 00:41:58,940 --> 00:42:02,610 and when it goes down to its max negative, it's like that. 634 00:42:02,610 --> 00:42:05,880 So it goes through a motion. 635 00:42:05,880 --> 00:42:08,640 And if you think about it, if you extend these lines out 636 00:42:08,640 --> 00:42:14,150 here, you come to a point that this intersects. 637 00:42:14,150 --> 00:42:17,020 And so in fact, in the first mode 638 00:42:17,020 --> 00:42:23,060 of vibration of this thing, if you went and sat right here, 639 00:42:23,060 --> 00:42:26,440 you would just see this thing go through an angle. 640 00:42:26,440 --> 00:42:29,812 And that, in fact, is the modal coordinate for mode 1. 641 00:42:33,510 --> 00:42:37,310 And the second mode I've kind of forgotten. 642 00:42:37,310 --> 00:42:40,760 It goes has a sign change, and so when 643 00:42:40,760 --> 00:42:47,370 you're going-- deflecting upwards, 644 00:42:47,370 --> 00:42:53,580 it goes-- positive flexion upwards-- it tips downwards. 645 00:42:53,580 --> 00:43:01,210 And when it goes the other way, it goes like that. 646 00:43:01,210 --> 00:43:04,180 So this one, it goes up and rotates up. 647 00:43:04,180 --> 00:43:06,850 And the second mode, when it goes up, it rotates down. 648 00:43:09,470 --> 00:43:12,780 And out here is the point at which you could go sit there, 649 00:43:12,780 --> 00:43:15,937 and you would just see this thing rotate up, rotate down 650 00:43:15,937 --> 00:43:17,020 if you went and sat there. 651 00:43:17,020 --> 00:43:20,410 So you move yourself to a place where 652 00:43:20,410 --> 00:43:22,430 you can see with a single coordinate-- 653 00:43:22,430 --> 00:43:25,410 with only the angle measured by this point, 654 00:43:25,410 --> 00:43:28,556 you can completely describe that modal motion. 655 00:43:28,556 --> 00:43:30,680 And with only the angle measured around this point, 656 00:43:30,680 --> 00:43:33,380 you can describe the modal motion. 657 00:43:33,380 --> 00:43:37,320 But in general, you can't do it. 658 00:43:37,320 --> 00:43:40,600 So if I'm out here-- if my eye is at this point, 659 00:43:40,600 --> 00:43:42,111 I'll see this thing go like that. 660 00:43:46,530 --> 00:43:47,581 That help? 661 00:43:47,581 --> 00:43:48,523 AUDIENCE: Yeah, a lot. 662 00:43:55,130 --> 00:43:57,390 PROFESSOR: So one of the hardest things for me 663 00:43:57,390 --> 00:44:00,050 as a lecturer is, since I really like vibration 664 00:44:00,050 --> 00:44:03,840 and I've taught it for 35 years or so, 665 00:44:03,840 --> 00:44:07,800 to try to cram everything I know about vibration into the one 666 00:44:07,800 --> 00:44:08,950 third of this course. 667 00:44:08,950 --> 00:44:10,740 So I obviously don't do that. 668 00:44:10,740 --> 00:44:14,330 So I'm trying to give you-- my goal then becomes give you 669 00:44:14,330 --> 00:44:17,300 some basic insight about vibration 670 00:44:17,300 --> 00:44:20,240 so that, when you do get out there in the real world 671 00:44:20,240 --> 00:44:22,647 and you need to know something, need to solve a vibration 672 00:44:22,647 --> 00:44:24,230 problem, you'll know the fundamentals, 673 00:44:24,230 --> 00:44:26,835 and you'll know where to go look it up. 674 00:44:26,835 --> 00:44:28,960 And if you want to take another course or something 675 00:44:28,960 --> 00:44:32,170 in vibration, you can do that. 676 00:44:32,170 --> 00:44:35,670 All right, now we've got to do, in relatively short order, 677 00:44:35,670 --> 00:44:41,150 response of our initial system over here 678 00:44:41,150 --> 00:44:44,730 to some harmonic excitation. 679 00:44:44,730 --> 00:44:53,860 And the one I've chosen to do-- let's 680 00:44:53,860 --> 00:44:57,300 imagine that I just-- I'm just going 681 00:44:57,300 --> 00:45:02,900 to put a force, a harmonic force, on this mass, 682 00:45:02,900 --> 00:45:06,400 and I want the steady state response. 683 00:45:06,400 --> 00:45:10,300 And you can imagine, if I do it close to the natural frequency 684 00:45:10,300 --> 00:45:12,760 of one of the modes, you're going to get a lot of that, 685 00:45:12,760 --> 00:45:15,410 but if I do it close to the natural frequency 686 00:45:15,410 --> 00:45:18,630 of the other mode, I can do it too. 687 00:45:18,630 --> 00:45:23,485 So let's do the steady state harmonic excitation problem. 688 00:45:31,090 --> 00:45:32,690 So we're going to do this problem. 689 00:45:32,690 --> 00:45:43,810 F2 of t-- so my generalized forces are 0 some magnitude 690 00:45:43,810 --> 00:45:47,070 Ft e to the i omega t. 691 00:45:49,920 --> 00:45:58,210 And the modal forces are u transpose F, 692 00:45:58,210 --> 00:46:00,070 and that would be the calculation 693 00:46:00,070 --> 00:46:12,050 one and 2.2667, and 1, and minus 0.2236-- my two mode shapes 694 00:46:12,050 --> 00:46:16,490 transposed multiplied by 0 and F2. 695 00:46:20,450 --> 00:46:36,000 So that says that q1 is F2.2667 F2e to the i omega t. 696 00:46:36,000 --> 00:46:41,400 And q2, the modal force for mode 2, 697 00:46:41,400 --> 00:46:53,180 is this times that-- is minus 0.2236 F2 e to the i omega t. 698 00:46:53,180 --> 00:46:57,420 So I've put on just a force on one mass, 699 00:46:57,420 --> 00:47:01,560 but it gets distributed in a way that it'll excite both modes, 700 00:47:01,560 --> 00:47:06,470 and it'll excite mode one in an amount 2.2 times F2. 701 00:47:06,470 --> 00:47:10,710 And it'll excite mode 2 in an amount minus 0.22. 702 00:47:10,710 --> 00:47:13,420 So mode 1 is going to get more excitation 703 00:47:13,420 --> 00:47:19,090 in this particular case because of the shape of the modes. 704 00:47:19,090 --> 00:47:21,210 The bigger the modal deflection is 705 00:47:21,210 --> 00:47:23,282 at the point of application of the force, 706 00:47:23,282 --> 00:47:24,740 the more that mode is going to get. 707 00:47:33,340 --> 00:47:38,320 Well, now these gave you two equations of motion. 708 00:47:38,320 --> 00:47:48,100 m1q1 double dot plus c1q1 dot k1q1 709 00:47:48,100 --> 00:47:59,877 equals, in the case of mode 1, 2.2667 F2 e to the i omega t. 710 00:47:59,877 --> 00:48:01,710 And that's a single degree of freedom system 711 00:48:01,710 --> 00:48:03,990 excited by a harmonic force. 712 00:48:03,990 --> 00:48:06,080 We worked that problem. 713 00:48:06,080 --> 00:48:08,070 We know what the answer looks like. 714 00:48:08,070 --> 00:48:13,740 So for example, the magnitude of the response q1 715 00:48:13,740 --> 00:48:19,510 is given by the magnitude of the force q2 times 716 00:48:19,510 --> 00:48:23,900 the magnitude of a transfer function, which is the response 717 00:48:23,900 --> 00:48:31,380 q1 per unit input force q1 evaluated at whatever 718 00:48:31,380 --> 00:48:33,360 frequency I evaluate it at. 719 00:48:33,360 --> 00:48:35,319 So the magnitude of the response magnitude 720 00:48:35,319 --> 00:48:36,860 of the modal force times the transfer 721 00:48:36,860 --> 00:48:39,010 function, and this transfer function 722 00:48:39,010 --> 00:48:44,450 is this exactly the same form as when 723 00:48:44,450 --> 00:48:47,745 we did just the single degree of freedom system as x/F. 724 00:48:47,745 --> 00:48:51,730 That's what I called the response x over input force F. 725 00:48:51,730 --> 00:48:54,800 Same thing, where just now the response is q, 726 00:48:54,800 --> 00:48:57,510 and the input force is capital Q. 727 00:48:57,510 --> 00:49:04,290 Therefore, that looks like magnitude of Q-- 728 00:49:04,290 --> 00:49:06,220 and I made a mistake here. 729 00:49:06,220 --> 00:49:07,765 Q1. 730 00:49:07,765 --> 00:49:21,270 The magnitude of force Q1, which is this, times 1/k1-- 731 00:49:21,270 --> 00:49:27,230 this should look familiar-- 1 minus omega squared 732 00:49:27,230 --> 00:49:32,860 over omega 1 squared squared plus 2 zeta 733 00:49:32,860 --> 00:49:40,400 1 omega over omega 1 quantity squared square root. 734 00:49:40,400 --> 00:49:44,750 So there's that transfer function expression. 735 00:49:44,750 --> 00:49:48,680 At resonance, for example, all this goes to 0. 736 00:49:48,680 --> 00:49:50,860 Omega over omega 1 goes to 1. 737 00:49:50,860 --> 00:49:54,070 This whole denominator turns into 1/2 times 738 00:49:54,070 --> 00:49:55,450 the damping ratio, for example. 739 00:49:58,170 --> 00:50:01,540 And Q1 over k1 is the static deflection 740 00:50:01,540 --> 00:50:05,750 of the system under the load-- under a static load 741 00:50:05,750 --> 00:50:06,800 of that large. 742 00:50:12,440 --> 00:50:13,900 So I'll do the problem. 743 00:50:13,900 --> 00:50:18,580 If the excitation frequency happens 744 00:50:18,580 --> 00:50:21,990 to be right on the natural frequency for mode one, 745 00:50:21,990 --> 00:50:28,716 then I can evaluate this at 1 omega over omega 1 is 1, 746 00:50:28,716 --> 00:50:31,580 and I've worked that out. 747 00:50:31,580 --> 00:50:40,820 That says that magnitude of q1 here-- 748 00:50:40,820 --> 00:50:57,360 2.2667F2/k1 times 1 over 2 zeta one. 749 00:50:57,360 --> 00:50:59,270 That's what all this condenses to. 750 00:50:59,270 --> 00:51:01,560 We know what k1 is. 751 00:51:01,560 --> 00:51:03,940 We know what zeta 1 is. 752 00:51:03,940 --> 00:51:07,430 F2 is the thing we're specifying. 753 00:51:07,430 --> 00:51:13,910 And so zeta 1 for example is what-- it was 0.044. 754 00:51:13,910 --> 00:51:21,500 So you plug in your 0.044 here, and this whole thing works out 755 00:51:21,500 --> 00:51:26,660 to be-- actually I'll write down the numbers here-- 756 00:51:26,660 --> 00:51:40,620 2.2667F2/k1 is 10984, and not 109. 757 00:51:40,620 --> 00:51:41,120 That's k2. 758 00:51:44,010 --> 00:51:56,350 113.71 times 11.36, which is 1/2 times zeta. 759 00:51:56,350 --> 00:52:01,950 1/0.088 is 11.36. 760 00:52:01,950 --> 00:52:16,590 The final analysis-- q1 is 0.227F2. 761 00:52:16,590 --> 00:52:18,932 So it's just a single degree of freedom system excited 762 00:52:18,932 --> 00:52:20,640 by harmonic force, and you can figure out 763 00:52:20,640 --> 00:52:22,180 how big its response is. 764 00:52:22,180 --> 00:52:24,830 I could do the same thing for mode 2. 765 00:52:24,830 --> 00:52:29,234 Is mode 2 resonant? 766 00:52:29,234 --> 00:52:29,734 No. 767 00:52:35,460 --> 00:52:39,690 So this is instructive. 768 00:52:39,690 --> 00:52:46,820 The magnitude of the responsive q2 769 00:52:46,820 --> 00:52:50,250 is the magnitude of the force times the magnitude 770 00:52:50,250 --> 00:52:57,416 of the Hq2 per unit input 2q. 771 00:52:57,416 --> 00:53:05,730 It looks like that, but this time this is minus 772 00:53:05,730 --> 00:53:17,290 0.2236F2/109.84, which is k2. 773 00:53:17,290 --> 00:53:27,250 And then all of the denominator involving one minus-- and this 774 00:53:27,250 --> 00:53:34,060 will now be omega-- over omega 2 squared 775 00:53:34,060 --> 00:53:44,690 squared plus 2 zeta 2 omega over omega 2 squared square root. 776 00:53:44,690 --> 00:53:46,248 This number here is k2. 777 00:53:49,430 --> 00:53:52,435 The reason I'm going to this is this one 778 00:53:52,435 --> 00:53:58,740 is not resonant, and in fact, what's omega? 779 00:53:58,740 --> 00:54:00,792 Well, in fact, what is the excite-- 780 00:54:00,792 --> 00:54:01,750 I started this problem. 781 00:54:01,750 --> 00:54:06,350 I said let the excitation frequency be what? 782 00:54:06,350 --> 00:54:08,130 Natural frequency of mode 1. 783 00:54:08,130 --> 00:54:10,970 So omega in this case equals omega 1. 784 00:54:10,970 --> 00:54:15,090 Therefore, for this problem, omega over omega 2 785 00:54:15,090 --> 00:54:21,020 is omega 1 over omega 2, and that's 5.65 something 786 00:54:21,020 --> 00:54:23,310 over 17 point something. 787 00:54:23,310 --> 00:54:25,830 It's about 1/3. 788 00:54:25,830 --> 00:54:27,850 So we are exciting the second mode 789 00:54:27,850 --> 00:54:30,970 at about 1/3 its natural frequency. 790 00:54:30,970 --> 00:54:36,259 And so this is 1 minus 1/3 quantity squared and so forth. 791 00:54:36,259 --> 00:54:37,800 And if I run the numbers, I just want 792 00:54:37,800 --> 00:54:40,760 you to see what happens with the numbers. 793 00:54:40,760 --> 00:54:46,780 This turns out to be 0.806, and this turns out 794 00:54:46,780 --> 00:54:53,060 to be-- remember this is 0.01 something here. 795 00:54:53,060 --> 00:54:57,860 This number turns out to be 5 times 10 to the minus fifth. 796 00:54:57,860 --> 00:54:59,173 That's a really small number. 797 00:55:04,270 --> 00:55:07,500 And this whole square root of this stuff 798 00:55:07,500 --> 00:55:17,220 turns out to be 0.898. 799 00:55:17,220 --> 00:55:24,430 And in the end, q2 works out- the magnitude of q2 works out 800 00:55:24,430 --> 00:55:35,970 to be 0.0023F2. 801 00:55:35,970 --> 00:55:37,300 I did this on purpose. 802 00:55:37,300 --> 00:55:41,580 If, when you're at resonance, meaning the excitation 803 00:55:41,580 --> 00:55:43,410 frequency is close to the natural frequency 804 00:55:43,410 --> 00:55:51,140 for any linear vibration system, then this term is important, 805 00:55:51,140 --> 00:55:53,900 and it'll be the most important term in the denominator, 806 00:55:53,900 --> 00:55:55,190 because this term goes to 0. 807 00:55:55,190 --> 00:55:57,510 It's the only term in the denominator, 808 00:55:57,510 --> 00:55:58,820 and it's likely be quite small. 809 00:55:58,820 --> 00:56:03,230 That's why it gives you one over that big response, 810 00:56:03,230 --> 00:56:06,710 but when you're not-- when this denominator-- if you're not 811 00:56:06,710 --> 00:56:12,610 at resonance, this term almost always is negligible, 812 00:56:12,610 --> 00:56:14,290 and it's this term that governs it. 813 00:56:14,290 --> 00:56:17,380 So here you are at 0.8 versus 10 to the minus fifth. 814 00:56:17,380 --> 00:56:20,230 So away from resonance, this term is important, 815 00:56:20,230 --> 00:56:23,910 and on resonance, that one is important. 816 00:56:23,910 --> 00:56:27,130 And now, how would you get back to-- we've now 817 00:56:27,130 --> 00:56:31,800 got the two responses. 818 00:56:31,800 --> 00:56:35,990 How do we compute the total system response? 819 00:56:43,370 --> 00:56:44,846 How would you do it? 820 00:56:51,260 --> 00:56:53,220 The modal expansion theorem. 821 00:56:53,220 --> 00:56:55,020 AUDIENCE: Put it back into x [INAUDIBLE]. 822 00:56:55,020 --> 00:57:02,960 PROFESSOR: Yeah, so x equals u cubed. 823 00:57:02,960 --> 00:57:04,790 That's where we started. 824 00:57:04,790 --> 00:57:13,170 And q1, since this is a harmonic excitation problem, 825 00:57:13,170 --> 00:57:16,550 if the input had been e to the i omega t, 826 00:57:16,550 --> 00:57:20,430 then the output is some e to the i omega t minus a phase angle. 827 00:57:20,430 --> 00:57:24,160 So this is going to look like some q1 amplitude. 828 00:57:24,160 --> 00:57:29,710 Say, cosine omega 1t minus phi 1 but at resonance, 829 00:57:29,710 --> 00:57:32,920 we know the phase angle is pi over 2. 830 00:57:32,920 --> 00:57:39,280 And q2 of t is going to be the amplitude q2 831 00:57:39,280 --> 00:57:45,080 times some cosine omega 1t-- because that's the excitation 832 00:57:45,080 --> 00:57:48,660 frequency-- minus phi 2. 833 00:57:48,660 --> 00:57:58,260 And remember, each of these systems-- 834 00:57:58,260 --> 00:58:05,620 these transfer functions, Hq/Q for whichever one 835 00:58:05,620 --> 00:58:10,260 it happens to be looks like this. 836 00:58:10,260 --> 00:58:13,730 Different amounts of damping give you different heights 837 00:58:13,730 --> 00:58:15,350 of the peak at resonance. 838 00:58:15,350 --> 00:58:20,010 And this is omega over omega i. 839 00:58:20,010 --> 00:58:24,220 So when you're at resonance, you're at 1. 840 00:58:24,220 --> 00:58:28,220 So what we've done is we have a two degree of freedom system. 841 00:58:28,220 --> 00:58:32,420 We're exciting it at the natural frequency of mode 1. 842 00:58:32,420 --> 00:58:34,587 So it means for mode 1, we're right here. 843 00:58:34,587 --> 00:58:36,170 Let's say that's where our damping is, 844 00:58:36,170 --> 00:58:37,610 so that's going to be our transfer 845 00:58:37,610 --> 00:58:41,830 function for that mode, but drawn on the same figure, 846 00:58:41,830 --> 00:58:46,130 where are we for mode 2? 847 00:58:46,130 --> 00:58:49,370 We're at omega 1 over omega 2, which is somewhere 848 00:58:49,370 --> 00:58:52,262 in the neighborhood of 0.3. 849 00:58:52,262 --> 00:58:55,450 So we are in here. 850 00:58:55,450 --> 00:58:59,600 So this is omega over omega 2. 851 00:58:59,600 --> 00:59:02,500 This is omega over omega 1. 852 00:59:02,500 --> 00:59:05,100 So down here, we're at about 0.3. 853 00:59:05,100 --> 00:59:09,070 Here we're at 1.0, so we're at resonance for one of the modes, 854 00:59:09,070 --> 00:59:12,450 and we're here in what's called the stiffness controlled region 855 00:59:12,450 --> 00:59:14,050 for the other mode. 856 00:59:14,050 --> 00:59:17,010 This mode-- mode 2 basically acts like a spring. 857 00:59:19,670 --> 00:59:21,620 The dynamic amplification is about 1. 858 00:59:21,620 --> 00:59:25,140 It just gives you the static response for mode 2 859 00:59:25,140 --> 00:59:27,560 and the resonant response for mode 1. 860 00:59:36,380 --> 00:59:38,900 I know what I was going to draw to remind you. 861 00:59:41,550 --> 00:59:44,570 This figure has a phase diagram that goes with it, 862 00:59:44,570 --> 00:59:50,150 and for lightly damped systems, it goes from 0 to pi. 863 00:59:50,150 --> 00:59:55,780 And at resonance, all of them cross pi over 2. 864 00:59:55,780 --> 00:59:59,530 When it's a response to a force, a simple force, at resonance, 865 00:59:59,530 --> 01:00:03,610 the phase angle is pi over 2, so the response 866 01:00:03,610 --> 01:00:06,760 lags the input by 90 degrees. 867 01:00:06,760 --> 01:00:08,930 When you're down here in this region, 868 01:00:08,930 --> 01:00:11,950 the response moves with the input. 869 01:00:11,950 --> 01:00:14,730 The phase angle is basically 0, and up in here 870 01:00:14,730 --> 01:00:16,620 the phase angle is 180 degrees. 871 01:00:16,620 --> 01:00:18,544 It acts like a driving mass. 872 01:00:23,470 --> 01:00:25,860 So our two responses look like that, 873 01:00:25,860 --> 01:00:30,340 and to get back into modal coordinates, x1, 874 01:00:30,340 --> 01:00:46,010 x2 is going to look like u1 q1 of t plus u2 q2 of t. 875 01:00:46,010 --> 01:00:46,770 It's harmonic. 876 01:00:46,770 --> 01:00:48,490 Steady state response. 877 01:00:48,490 --> 01:00:50,575 You know the amplitudes, q1 and q2. 878 01:00:50,575 --> 01:00:52,540 We figured them out. 879 01:00:52,540 --> 01:01:06,860 One of them is 2-- q1 turned out to be 0.227F2, 880 01:01:06,860 --> 01:01:18,560 and q2 is 0.0023F2. 881 01:01:18,560 --> 01:01:31,470 And this one, point cosine cosine. 882 01:01:31,470 --> 01:01:36,145 So if we do that, if you put your excitation only on here, 883 01:01:36,145 --> 01:01:37,895 it's trying to tell you that mostly you'll 884 01:01:37,895 --> 01:01:44,380 get mode 1, and not much mode 2, but that's primarily 885 01:01:44,380 --> 01:01:47,130 caused because you chose to put the excitation 886 01:01:47,130 --> 01:01:52,940 frequency at where? 887 01:01:52,940 --> 01:01:56,970 The natural frequency of mode one. 888 01:01:56,970 --> 01:02:00,530 So if I had made my excitation here the same place, 889 01:02:00,530 --> 01:02:04,500 but made the frequency close to the natural frequency of mode 890 01:02:04,500 --> 01:02:08,440 2, which of the two modes would have dominated? 891 01:02:08,440 --> 01:02:10,720 Mode 2. 892 01:02:10,720 --> 01:02:16,608 So I've got another little demo. 893 01:02:16,608 --> 01:02:17,585 We're doing well. 894 01:02:23,760 --> 01:02:25,765 Out of necessity, to do it on the blackboard 895 01:02:25,765 --> 01:02:28,370 and in relatively short time, we've 896 01:02:28,370 --> 01:02:35,250 only talked about a two degree of freedom two rigid body 897 01:02:35,250 --> 01:02:37,190 system. 898 01:02:37,190 --> 01:02:38,800 This is a continuous system. 899 01:02:38,800 --> 01:02:39,780 It's a taut string. 900 01:02:39,780 --> 01:02:46,950 It's your violin string, and I've put some white tags on it 901 01:02:46,950 --> 01:02:49,075 so that you could see it against black backgrounds. 902 01:02:51,680 --> 01:02:55,030 Everything that we've learned about the behavior of this two 903 01:02:55,030 --> 01:02:58,170 degree of freedom system will apply to a three, 904 01:02:58,170 --> 01:03:00,100 or four, or five degree of freedom system, 905 01:03:00,100 --> 01:03:01,800 but actually the basic lessons apply 906 01:03:01,800 --> 01:03:03,890 to continuous systems, too. 907 01:03:03,890 --> 01:03:10,940 So the lesson we just learned is, if you excite a system, 908 01:03:10,940 --> 01:03:13,740 this has many natural frequencies. 909 01:03:13,740 --> 01:03:16,930 And in fact, if the first one is at 1 Hertz, 910 01:03:16,930 --> 01:03:19,430 which this just about is-- maybe a little more than that. 911 01:03:19,430 --> 01:03:21,280 First mode is like that. 912 01:03:21,280 --> 01:03:22,790 Maybe 2 Hertz. 913 01:03:22,790 --> 01:03:24,250 The second mode is twice that. 914 01:03:24,250 --> 01:03:25,650 Third mode is three times that. 915 01:03:25,650 --> 01:03:28,350 It happens to be really simple. 916 01:03:28,350 --> 01:03:31,360 So if I put a force-- the only force in this problem-- 917 01:03:31,360 --> 01:03:33,630 I'm going to do the analogous problem here. 918 01:03:33,630 --> 01:03:36,890 Harmonic excitation-- I'm going to do it in one little place 919 01:03:36,890 --> 01:03:41,980 right here, and if I drive this system at the natural frequency 920 01:03:41,980 --> 01:03:46,710 of mode 1, what do you see? 921 01:03:46,710 --> 01:03:49,000 What mode is dominating response? 922 01:03:49,000 --> 01:03:50,550 Mode 1. 923 01:03:50,550 --> 01:03:55,440 Very, very small responses of other modes. 924 01:03:55,440 --> 01:04:00,330 So now if I can get the system to stop shaking, if I drive it 925 01:04:00,330 --> 01:04:08,850 at exactly the same place at a different natural frequency, 926 01:04:08,850 --> 01:04:11,390 I see 3/2 sine waves there. 927 01:04:11,390 --> 01:04:14,246 So I was driving it at the natural frequency of? 928 01:04:14,246 --> 01:04:15,220 AUDIENCE: Mode 3. 929 01:04:15,220 --> 01:04:16,750 PROFESSOR: Mode 3. 930 01:04:16,750 --> 01:04:17,890 And there's no mode 1. 931 01:04:17,890 --> 01:04:20,520 No mode 2. 932 01:04:20,520 --> 01:04:22,085 Now I need a helper. 933 01:04:22,085 --> 01:04:24,000 Can somebody come hang onto this for me? 934 01:04:29,890 --> 01:04:32,160 You've got to keep the tension on it. 935 01:04:32,160 --> 01:04:35,190 So now I'm going to teach you a really important lesson 936 01:04:35,190 --> 01:04:40,300 about systems exciting systems. 937 01:04:40,300 --> 01:04:44,640 The second mode-- I might have a hard time driving it here. 938 01:04:44,640 --> 01:04:46,787 Let me see if I can get the second mode going. 939 01:04:54,130 --> 01:04:55,495 There's second mode. 940 01:04:55,495 --> 01:04:56,900 I has a node right here. 941 01:04:56,900 --> 01:04:58,650 There's a point right here with no motion. 942 01:05:01,470 --> 01:05:04,240 If I sit here, and I'm going to let the system stop-- 943 01:05:04,240 --> 01:05:07,580 if I drive this system at the second mode natural frequency 944 01:05:07,580 --> 01:05:09,090 right here, what will happen? 945 01:05:16,230 --> 01:05:16,922 Pardon? 946 01:05:16,922 --> 01:05:18,130 AUDIENCE: It would be mode 1. 947 01:05:18,130 --> 01:05:20,075 PROFESSOR: Maybe [INAUDIBLE] 1. 948 01:05:20,075 --> 01:05:22,926 But how much mode 2 will I get? 949 01:05:27,660 --> 01:05:33,700 So another lesson here is that continuous systems have nodes, 950 01:05:33,700 --> 01:05:36,474 points of no motion, and the second mode 951 01:05:36,474 --> 01:05:37,890 happens to have a node right here. 952 01:05:37,890 --> 01:05:39,580 This point doesn't move one when it's 953 01:05:39,580 --> 01:05:41,400 vibrating in the second mode. 954 01:05:41,400 --> 01:05:43,290 And the modal force looks something 955 01:05:43,290 --> 01:05:47,560 like the generalized force times a mode shape, 956 01:05:47,560 --> 01:05:50,880 u transpose F. That's how we got the-- you have 957 01:05:50,880 --> 01:05:56,210 to multiply the generalized external force times the mode 958 01:05:56,210 --> 01:05:59,340 shape to get the modal force. 959 01:05:59,340 --> 01:06:02,070 Well, here's my generalized force. 960 01:06:02,070 --> 01:06:05,670 What's the amplitude of the mode shape here? 961 01:06:05,670 --> 01:06:06,430 0. 962 01:06:06,430 --> 01:06:10,250 F times 0 is 0. 963 01:06:10,250 --> 01:06:13,100 There is just no way I can get this thing 964 01:06:13,100 --> 01:06:16,420 to vibrate in the second mode by driving it at a node, 965 01:06:16,420 --> 01:06:19,010 and that's just generally true. 966 01:06:19,010 --> 01:06:22,750 Take it to the bank and remember that. 967 01:06:22,750 --> 01:06:25,650 Thank you. 968 01:06:25,650 --> 01:06:28,430 All right. 969 01:06:28,430 --> 01:06:30,380 Something else I wanted to do. 970 01:06:34,244 --> 01:06:34,910 I need you back. 971 01:06:39,490 --> 01:06:48,380 So when you give a system an initial deflection, 972 01:06:48,380 --> 01:06:52,780 because when we want to get initial conditions, 973 01:06:52,780 --> 01:06:56,200 we said that the equivalent modal initial conditions were 974 01:06:56,200 --> 01:07:00,740 u inverse times the u inverse times to the initial conditions 975 01:07:00,740 --> 01:07:02,750 in generalized coordinates. 976 01:07:02,750 --> 01:07:04,990 So for a continuous system, this thing 977 01:07:04,990 --> 01:07:07,980 has mode shapes that look like sine and pi x over l. 978 01:07:07,980 --> 01:07:10,100 Those are the mode shapes. 979 01:07:10,100 --> 01:07:11,620 And if I take this thing and grab it 980 01:07:11,620 --> 01:07:14,634 about in the middle, wherever that node was, 981 01:07:14,634 --> 01:07:16,050 and give it an initial deflection, 982 01:07:16,050 --> 01:07:17,424 the shape that initial deflection 983 01:07:17,424 --> 01:07:18,940 is kind of a triangle. 984 01:07:18,940 --> 01:07:20,310 You've all had Fourier series. 985 01:07:23,100 --> 01:07:25,530 This a triangular shape. 986 01:07:25,530 --> 01:07:30,940 Could you express this shape as a Fourier sine series? 987 01:07:30,940 --> 01:07:32,367 Sure. 988 01:07:32,367 --> 01:07:34,700 And it just happens that sine waves are the mode shapes. 989 01:07:34,700 --> 01:07:36,970 So you would be coming up-- the Fourier 990 01:07:36,970 --> 01:07:40,660 coefficients are the modal amplitudes 991 01:07:40,660 --> 01:07:43,240 of n initial conditions. 992 01:07:43,240 --> 01:07:46,040 And so whatever the Fourier coefficients for this are 993 01:07:46,040 --> 01:07:49,070 are the modal amplitudes for each of the modes. 994 01:07:49,070 --> 01:07:50,550 So which mode do you think is going 995 01:07:50,550 --> 01:07:54,020 to-- which Fourier sign component is going 996 01:07:54,020 --> 01:07:55,654 to be largest in this one? 997 01:07:58,438 --> 01:07:59,370 AUDIENCE: Mode 1? 998 01:07:59,370 --> 01:08:00,230 PROFESSOR: Mode 1. 999 01:08:00,230 --> 01:08:02,890 By the way, all even numbered modes would be 0 1000 01:08:02,890 --> 01:08:04,560 because they're asymmetric. 1001 01:08:04,560 --> 01:08:08,450 This is a symmetrically shaped pulse. 1002 01:08:08,450 --> 01:08:12,920 The modal initial conditions for 2, 4, 6, 8, 10 are 0. 1003 01:08:12,920 --> 01:08:17,350 You will get non-zero Fourier coefficients for 1, 3, 5, 7. 1004 01:08:17,350 --> 01:08:18,950 The biggest one is mode 1. 1005 01:08:18,950 --> 01:08:22,640 So what do you expect to see if I let go of this? 1006 01:08:22,640 --> 01:08:23,939 Vibration primarily at? 1007 01:08:23,939 --> 01:08:24,855 AUDIENCE: [INAUDIBLE]. 1008 01:08:30,200 --> 01:08:32,020 PROFESSOR: And a little bit of some others. 1009 01:08:32,020 --> 01:08:34,145 Now, let's say they all had about the same damping. 1010 01:08:34,145 --> 01:08:36,927 Let's say they all had 10% damping. 1011 01:08:36,927 --> 01:08:37,510 No, excuse me. 1012 01:08:37,510 --> 01:08:38,779 All had 1% damping. 1013 01:08:38,779 --> 01:08:40,850 That means they'll go through about 10 cycles 1014 01:08:40,850 --> 01:08:44,010 to decay to halfway. 1015 01:08:44,010 --> 01:08:48,130 So if they all have the same damping, 1016 01:08:48,130 --> 01:08:50,180 even if there are several other modes present, 1017 01:08:50,180 --> 01:08:55,069 which ones are going to last longer in time? 1018 01:08:55,069 --> 01:08:58,540 Mode 1, because it takes just longer in time 1019 01:08:58,540 --> 01:09:00,979 to get to 10 cycles. 1020 01:09:00,979 --> 01:09:04,210 The other modes get there quicker. 1021 01:09:04,210 --> 01:09:07,979 If you take a guitar string and plunk it, or a violin, 1022 01:09:07,979 --> 01:09:10,069 or a piano, you'll hear the basic tone. 1023 01:09:10,069 --> 01:09:13,122 And what makes it sound nice, you have those nice overtones, 1024 01:09:13,122 --> 01:09:15,580 but if you listen carefully, the overtones die out usually, 1025 01:09:15,580 --> 01:09:17,580 and you're left with the fundamental at the end. 1026 01:09:17,580 --> 01:09:20,560 If you smack a piano key hard, you'll 1027 01:09:20,560 --> 01:09:22,460 get an interesting sound at the beginning, 1028 01:09:22,460 --> 01:09:23,740 and then it'll mellow out. 1029 01:09:23,740 --> 01:09:26,180 And you'll hear just a pure tone at the end, 1030 01:09:26,180 --> 01:09:30,569 and that's because the higher frequencies damp out quicker 1031 01:09:30,569 --> 01:09:34,100 because they get in more cycles per unit of time. 1032 01:09:34,100 --> 01:09:36,985 That's the other quick lesson. 1033 01:09:36,985 --> 01:09:38,990 I have one thing I want to explain to you 1034 01:09:38,990 --> 01:09:43,920 which will help maybe a little conceptual understanding 1035 01:09:43,920 --> 01:09:47,609 about-- this came up in a homework discussion-- 1036 01:09:47,609 --> 01:09:51,800 and that is just a note about stiffness matrices. 1037 01:10:05,790 --> 01:10:12,280 There's a really fast easy way to assemble stiffness matrices. 1038 01:10:12,280 --> 01:10:26,950 So here's a three mass system and a spring, spring, spring. 1039 01:10:26,950 --> 01:10:33,610 And I'm going to put a spring here and one here. 1040 01:10:36,730 --> 01:10:50,410 k1, k2, k3, k4, k5, and k6-- and I 1041 01:10:50,410 --> 01:10:52,345 want to get my stiffness matrix. 1042 01:10:56,620 --> 01:11:00,230 The stiffness matrix-- this is a three degree of freedom system. 1043 01:11:00,230 --> 01:11:04,130 It'll be a 3 by 3, and it'll have elements up here, 1044 01:11:04,130 --> 01:11:26,230 which I'll call k11, k12, k13, k21, k23, k31, k22, 1045 01:11:26,230 --> 01:11:28,260 and so forth. 1046 01:11:28,260 --> 01:11:32,148 So what's the new meaning of kij? 1047 01:11:35,850 --> 01:11:38,660 So ki11-- i as 1. 1048 01:11:38,660 --> 01:11:41,151 J is 1. 1049 01:11:41,151 --> 01:11:43,150 If you can understand the interpretation of what 1050 01:11:43,150 --> 01:11:45,620 a stiffness matrix is, it'll help 1051 01:11:45,620 --> 01:11:49,440 you make it much easier for you to find them. 1052 01:11:49,440 --> 01:12:07,060 So kij is the force required at i 1053 01:12:07,060 --> 01:12:21,070 due to a unit deflection at j due to sounds 1054 01:12:21,070 --> 01:12:22,830 kind of like a causal thing. 1055 01:12:22,830 --> 01:12:25,520 I don't quite mean that, but so let's think about this. 1056 01:12:25,520 --> 01:12:27,630 What's k11? 1057 01:12:27,630 --> 01:12:33,480 So k11 is the force required at 1 per unit deflection at 1. 1058 01:12:33,480 --> 01:12:39,520 So if I take this system and I make it move over one unit, 1059 01:12:39,520 --> 01:12:45,100 and the other-- this is now x1 here, by the way. x1, 1060 01:12:45,100 --> 01:12:49,160 x2, x300 so x2 and x3 are 0. 1061 01:12:49,160 --> 01:12:50,490 I do this one a time. 1062 01:12:50,490 --> 01:12:52,700 I move this over one unit. 1063 01:12:52,700 --> 01:12:56,140 How much force does it take to make that happen? 1064 01:12:56,140 --> 01:12:56,829 A real system. 1065 01:12:56,829 --> 01:12:57,620 You're grabbing it. 1066 01:12:57,620 --> 01:12:59,010 You're pulling it over one unit. 1067 01:12:59,010 --> 01:13:01,610 You're making some springs move. 1068 01:13:01,610 --> 01:13:04,140 How much force does it take to move that one one 1069 01:13:04,140 --> 01:13:05,339 unit holding these still? 1070 01:13:10,030 --> 01:13:14,150 Well, you're going to-- one, force on a spring is kx. 1071 01:13:14,150 --> 01:13:17,560 If x is one, the force for trying to stretch this spring 1072 01:13:17,560 --> 01:13:20,700 is k1 times 1. 1073 01:13:20,700 --> 01:13:24,350 The force required to push on that spring as k5 times 1074 01:13:24,350 --> 01:13:31,190 5-- k5 times 1. k2 times one. 1075 01:13:31,190 --> 01:13:35,800 The k11, this first element, is the sum 1076 01:13:35,800 --> 01:13:40,030 of all of the springs connected to it. 1077 01:13:40,030 --> 01:13:44,980 k1 plus k2 plus k5. 1078 01:13:44,980 --> 01:13:45,860 So how about k12? 1079 01:13:48,710 --> 01:13:57,620 k12 is the force required at 1-- actually, let me do k21. 1080 01:13:57,620 --> 01:14:00,110 Make a little more sense-- is the force 1081 01:14:00,110 --> 01:14:08,671 required at 2, because I've moved the system one unit at 1. 1082 01:14:08,671 --> 01:14:10,920 That's what the problem-- that's this thing I've done. 1083 01:14:10,920 --> 01:14:12,990 I've moved this one unit. 1084 01:14:12,990 --> 01:14:15,080 In order to keep this one from moving, 1085 01:14:15,080 --> 01:14:17,914 do I have to apply force to it? 1086 01:14:17,914 --> 01:14:20,349 How big? 1087 01:14:20,349 --> 01:14:22,300 AUDIENCE: k2. 1088 01:14:22,300 --> 01:14:25,890 PROFESSOR: k2 or minus k2? 1089 01:14:25,890 --> 01:14:28,090 This thing is pushing, going over one unit. 1090 01:14:28,090 --> 01:14:29,450 It compresses that spring. 1091 01:14:29,450 --> 01:14:31,310 It's pushing on this thing. 1092 01:14:31,310 --> 01:14:32,640 I say it cannot move. 1093 01:14:32,640 --> 01:14:34,150 What do I have to do? 1094 01:14:34,150 --> 01:14:42,000 Push back minus k2, and you can go through-- and then how 1095 01:14:42,000 --> 01:14:43,040 about number 3? 1096 01:14:43,040 --> 01:14:50,090 What's the force required at 3 because I've moved the one at 1 1097 01:14:50,090 --> 01:14:52,610 by 1 unit? 1098 01:14:52,610 --> 01:14:53,970 Move this over. 1099 01:14:53,970 --> 01:14:56,480 Are there any springs connected to mass 3 that 1100 01:14:56,480 --> 01:14:58,106 are affected by that motion? 1101 01:14:58,106 --> 01:14:58,980 AUDIENCE: k5. 1102 01:14:58,980 --> 01:15:01,905 PROFESSOR: k5, and it pushes on it through that spring, 1103 01:15:01,905 --> 01:15:15,300 so I have to resist by-- so k31 equals minus k5. 1104 01:15:15,300 --> 01:15:16,340 So now then you go on. 1105 01:15:16,340 --> 01:15:17,715 If you want to get the next ones, 1106 01:15:17,715 --> 01:15:19,110 OK, you go to the next system. 1107 01:15:19,110 --> 01:15:20,460 This is now can't move. 1108 01:15:20,460 --> 01:15:21,480 This can't move. 1109 01:15:21,480 --> 01:15:23,820 We're going to let this be unit deflection. 1110 01:15:23,820 --> 01:15:26,530 So unit deflection at two, add up the springs. 1111 01:15:26,530 --> 01:15:32,320 k2 plus k3 plus k6-- and that's all there is. 1112 01:15:32,320 --> 01:15:41,080 So k22, k2 plus k3 plus k6, and then you 1113 01:15:41,080 --> 01:15:43,000 go through all the ones that it affects, 1114 01:15:43,000 --> 01:15:45,870 and you'll get minus this and that. 1115 01:15:45,870 --> 01:15:48,290 So that's the meaning. 1116 01:15:48,290 --> 01:15:50,600 Each of the elements of that stiffness matrix 1117 01:15:50,600 --> 01:15:52,570 have that meaning to it. 1118 01:15:52,570 --> 01:16:10,310 And to give this closure, we have equation of motion, 1119 01:16:10,310 --> 01:16:14,950 but the stiffness matrices also applies to the statics problem. 1120 01:16:14,950 --> 01:16:22,120 So you have mx double dot plus cx dot plus kx equals 1121 01:16:22,120 --> 01:16:25,404 F. What if I only want to do a statics problem? 1122 01:16:25,404 --> 01:16:27,070 I'm going to put a static force on here, 1123 01:16:27,070 --> 01:16:29,780 and I want to know the deflections, 1124 01:16:29,780 --> 01:16:31,300 or I'm going to cause deflections, 1125 01:16:31,300 --> 01:16:33,750 and I want to know what force it takes to do it. 1126 01:16:33,750 --> 01:16:38,110 Well, the static problems-- let this be 0, this be 0. 1127 01:16:38,110 --> 01:16:44,840 And it says kx equals F. So in this three degree 1128 01:16:44,840 --> 01:16:49,750 of freedom system, what forces are required to cause 1129 01:16:49,750 --> 01:17:00,560 the deflection 0, 0, 1. 1130 01:17:00,560 --> 01:17:06,280 I want to deflect it one unit on the third mass only. 1131 01:17:06,280 --> 01:17:09,730 What forces do I apply it to the system to make it happen? 1132 01:17:09,730 --> 01:17:14,860 Well, you just multiply it out. 1133 01:17:14,860 --> 01:17:19,150 This is F1, F2, F3. 1134 01:17:19,150 --> 01:17:22,850 So it'll only be this one. 1135 01:17:22,850 --> 01:17:24,760 This times this, this, and this. 1136 01:17:24,760 --> 01:17:28,330 The only ones that matter are these three here, 1137 01:17:28,330 --> 01:17:37,150 and this will end up being k31 times 1. 1138 01:17:37,150 --> 01:17:44,940 This will be k-- I guess I've got to round-- 13. 1139 01:17:44,940 --> 01:17:51,275 And we get 21, 22, 23, k33. 1140 01:17:54,190 --> 01:17:56,710 Those are the three forces required. 1141 01:17:56,710 --> 01:18:01,230 You put on those forces. 1142 01:18:01,230 --> 01:18:02,910 Force is equal to these amounts. 1143 01:18:02,910 --> 01:18:06,262 You will get that deflection. 1144 01:18:06,262 --> 01:18:07,470 So that's just a little help. 1145 01:18:07,470 --> 01:18:10,400 That gives you a little insight as to what 1146 01:18:10,400 --> 01:18:14,000 stiffness matrices mean. 1147 01:18:14,000 --> 01:18:15,720 So you can do them by inspection. 1148 01:18:15,720 --> 01:18:18,192 Once you understand that, you can actually just fill them 1149 01:18:18,192 --> 01:18:21,890 in by inspection, just by doing unit displacements 1150 01:18:21,890 --> 01:18:25,360 at each place and adding up the forces. 1151 01:18:25,360 --> 01:18:28,112 See you on Thursday.