1 00:00:07,280 --> 00:00:11,220 PROFESSOR: Resume by going back to our one-dimensional 2 00:00:11,220 --> 00:00:15,630 body that has undergone some elastic deformation. 3 00:00:15,630 --> 00:00:24,530 And what I would like to do now is to distinguish between 4 00:00:24,530 --> 00:00:28,510 displacement of an object and fractional change of length, 5 00:00:28,510 --> 00:00:32,540 which turned out to be measured by the same thing, 6 00:00:32,540 --> 00:00:34,710 that thing that we're going to name strain 7 00:00:34,710 --> 00:00:37,430 when properly defined. 8 00:00:37,430 --> 00:00:41,130 OK, here is our one-dimensional case. 9 00:00:41,130 --> 00:00:48,190 And we said that originally some point, P, at a location x 10 00:00:48,190 --> 00:00:54,260 gets mapped to a point P prime that is at x plus some 11 00:00:54,260 --> 00:01:02,510 displacement U. So this is the displacement vector U. 12 00:01:02,510 --> 00:01:11,110 Our point Q, which is originally at some location x 13 00:01:11,110 --> 00:01:14,770 plus delta x, where delta x is the original separation 14 00:01:14,770 --> 00:01:20,990 between P and Q, gets mapped to a point Q prime, which is 15 00:01:20,990 --> 00:01:27,910 going to be equal to a whole collection of terms. 16 00:01:27,910 --> 00:01:32,850 It's going to be equal to x plus delta x, the original 17 00:01:32,850 --> 00:01:39,708 location, plus the linear variation of U with x that has 18 00:01:39,708 --> 00:01:44,870 to go U times x plus delta x. 19 00:01:48,060 --> 00:01:51,770 And if we simplify this a little bit, Q prime is going 20 00:01:51,770 --> 00:01:59,360 to be at a location, factoring out x plus delta x, x plus 21 00:01:59,360 --> 00:02:02,150 delta x times 1 plus e. 22 00:02:06,254 --> 00:02:07,515 AUDIENCE: Is e represented here? 23 00:02:07,515 --> 00:02:12,570 PROFESSOR: e is the linear relation between displacement 24 00:02:12,570 --> 00:02:14,670 U and position along the body. 25 00:02:20,510 --> 00:02:23,390 OK, so what has happened here? 26 00:02:23,390 --> 00:02:37,280 The relative change of length is going to be P prime 27 00:02:37,280 --> 00:02:41,108 Q prime minus PQ. 28 00:02:43,706 --> 00:02:48,360 And if we just substitute in our two expressions, the 29 00:02:48,360 --> 00:02:56,420 relative change of length is going to be equal to delta x 30 00:02:56,420 --> 00:03:00,795 times 1 plus e minus delta x. 31 00:03:05,330 --> 00:03:07,960 This is the relative change of length. 32 00:03:07,960 --> 00:03:10,500 And I should label this such. 33 00:03:15,590 --> 00:03:19,600 That's going to be equal to delta x times 1 plus e minus 34 00:03:19,600 --> 00:03:23,250 delta x divided by delta x. 35 00:03:23,250 --> 00:03:28,410 And that simply going to be e, which is equal to delta 36 00:03:28,410 --> 00:03:30,720 U over delta x. 37 00:03:30,720 --> 00:03:34,610 And that is what we define as strain. 38 00:03:34,610 --> 00:03:38,170 And that is a dimensionless quantity. 39 00:03:38,170 --> 00:03:41,630 Because it has units of length over length. 40 00:03:47,935 --> 00:03:52,710 All right, how can we patch this one-dimensional sort of 41 00:03:52,710 --> 00:03:58,406 behavior up to a three-dimensional situation? 42 00:04:05,680 --> 00:04:09,850 What we are saying in this one-dimensional case is that 43 00:04:09,850 --> 00:04:14,530 the displacement of a particular point P varies 44 00:04:14,530 --> 00:04:18,680 linearly with position in the body. 45 00:04:18,680 --> 00:04:22,610 So let's generalize that into three dimensions by saying 46 00:04:22,610 --> 00:04:26,610 that the component of displacement U1-- and this is 47 00:04:26,610 --> 00:04:31,100 going to look like an algebraic identity. 48 00:04:31,100 --> 00:04:36,100 It's going to be the way in which U changes with x1 times 49 00:04:36,100 --> 00:04:39,090 the position x1. 50 00:04:39,090 --> 00:04:41,890 So this is saying that the displacement will vary 51 00:04:41,890 --> 00:04:43,800 linearly with x1. 52 00:04:43,800 --> 00:04:48,500 And the rate will be du1 dx1. 53 00:04:48,500 --> 00:04:53,250 But also U1 is going to change with the coordinate x2. 54 00:04:53,250 --> 00:05:02,650 And the coefficient there will be du1 dx2. 55 00:05:02,650 --> 00:05:06,690 And there's going to be a third term that will be the 56 00:05:06,690 --> 00:05:09,900 way in which U1, the x1 component of displacement, 57 00:05:09,900 --> 00:05:15,110 changes with x3 times the position in the body x3. 58 00:05:15,110 --> 00:05:17,780 So this is saying exactly the same thing that we did in one 59 00:05:17,780 --> 00:05:22,010 dimension, that displacement depends linearly with 60 00:05:22,010 --> 00:05:25,550 position, except that there's now three 61 00:05:25,550 --> 00:05:27,340 coordinates for position. 62 00:05:27,340 --> 00:05:33,560 And the displacement will change with an increment of a 63 00:05:33,560 --> 00:05:36,550 change in position along each of those three axes. 64 00:05:36,550 --> 00:05:40,190 Similarly, we can say that U2 is going to be equal to the 65 00:05:40,190 --> 00:05:45,890 way in which U2 changes with x1 times the position x1, the 66 00:05:45,890 --> 00:05:50,420 way in which U2 changes with x2 times the coordinate within 67 00:05:50,420 --> 00:05:56,930 the body x2, and the same for U2 dx3 and the actual 68 00:05:56,930 --> 00:05:59,340 positional coordinate x3. 69 00:05:59,340 --> 00:06:03,000 And so, in general, we're going to propose that the i-th 70 00:06:03,000 --> 00:06:07,520 component of displacement is given by the way in which 71 00:06:07,520 --> 00:06:15,730 displacement changes with x sub j times x sub j. 72 00:06:22,050 --> 00:06:26,270 Or you could do the same thing, not in terms of actual 73 00:06:26,270 --> 00:06:43,170 displacement, but differences in displacement, the same way 74 00:06:43,170 --> 00:06:48,070 we could define strain as the fractional change of length of 75 00:06:48,070 --> 00:06:51,310 our line segment on the elastic band or, 76 00:06:51,310 --> 00:06:54,870 alternatively, as the shift in position. 77 00:06:54,870 --> 00:07:01,000 So we could define it also as the difference in distances or 78 00:07:01,000 --> 00:07:02,720 displacements let me call them. 79 00:07:10,370 --> 00:07:12,920 And that really is a trivial change. 80 00:07:12,920 --> 00:07:18,700 We'll say the change of a length along x1 delta U1 is 81 00:07:18,700 --> 00:07:23,140 going to be the way in which U1 changes with x1 times delta 82 00:07:23,140 --> 00:07:30,580 x1 plus the way in which U2 changes with x1 times the way 83 00:07:30,580 --> 00:07:35,290 in which U1 changes with x2 times delta x2 plus the way in 84 00:07:35,290 --> 00:07:43,550 which U1 changes with x3 times delta x3 or, in general, that 85 00:07:43,550 --> 00:07:49,720 the fractional change in length is going to be equal to 86 00:07:49,720 --> 00:07:55,590 dui dxj times delta xj. 87 00:07:58,375 --> 00:08:03,480 All right, so what we're going to define now is dui. 88 00:08:03,480 --> 00:08:08,510 Dxj will be defined as something like strain. 89 00:08:08,510 --> 00:08:12,360 It's not exactly strain yet. 90 00:08:12,360 --> 00:08:16,540 And this is going to be a measure of deformation. 91 00:08:16,540 --> 00:08:24,350 I'm hedging my words because of something that we'll want 92 00:08:24,350 --> 00:08:28,960 to impose upon the proper strain tensor. 93 00:08:28,960 --> 00:08:32,010 But first, let's see what these 94 00:08:32,010 --> 00:08:35,100 three-dimensional terms mean. 95 00:08:35,100 --> 00:08:43,505 And to see that, let me look at a place within the body. 96 00:08:43,505 --> 00:08:48,520 And I'm going to look at a line segment that goes between 97 00:08:48,520 --> 00:08:52,430 a P and Q that is oriented along x1. 98 00:08:52,430 --> 00:08:55,370 And I'm picking that deliberately. 99 00:08:55,370 --> 00:08:57,720 Because I want to keep things simple and isolate one of 100 00:08:57,720 --> 00:09:00,890 these terms so we can identify its meaning. 101 00:09:00,890 --> 00:09:05,810 So let's say here we have two points P and Q separated by a 102 00:09:05,810 --> 00:09:08,530 distance delta x1 prior to deformation. 103 00:09:11,140 --> 00:09:13,850 And now we squish the body. 104 00:09:13,850 --> 00:09:17,590 And what happens is that P will move to 105 00:09:17,590 --> 00:09:20,930 a position P prime. 106 00:09:20,930 --> 00:09:24,950 Q will move to a position Q prime. 107 00:09:24,950 --> 00:09:29,450 And this will be the new line segment, P prime Q prime. 108 00:09:34,840 --> 00:09:38,580 This will be the original delta x1. 109 00:09:38,580 --> 00:09:41,150 To that we're going to tack on an instrument 110 00:09:41,150 --> 00:09:43,310 which is delta U1. 111 00:09:43,310 --> 00:09:46,155 But then there will also be a delta U2. 112 00:09:49,370 --> 00:09:52,770 And clearly what has happened is that the length of the line 113 00:09:52,770 --> 00:09:55,410 segment P prime Q prime has changed. 114 00:09:55,410 --> 00:10:00,860 But also the orientation of the line segment has changed 115 00:10:00,860 --> 00:10:03,595 by an angle, which I'll define as phi. 116 00:10:06,320 --> 00:10:07,890 So again, we have a line segment. 117 00:10:07,890 --> 00:10:09,930 We deform the body. 118 00:10:09,930 --> 00:10:11,270 P is displaced. 119 00:10:11,270 --> 00:10:13,650 Q is displaced. 120 00:10:13,650 --> 00:10:17,960 The component of that line segment along x1 has changed 121 00:10:17,960 --> 00:10:21,350 in length by an amount delta U1. 122 00:10:21,350 --> 00:10:28,220 The coordinate along x2 will change by an amount delta U2. 123 00:10:28,220 --> 00:10:29,820 So we not only change the length. 124 00:10:29,820 --> 00:10:31,580 But we rotate the line segment. 125 00:10:52,750 --> 00:10:56,570 We can express these in terms of our general relation that I 126 00:10:56,570 --> 00:10:58,470 proposed a moment ago. 127 00:10:58,470 --> 00:11:08,810 Delta U1 is going to be equal to du1 dx1 times delta x1. 128 00:11:08,810 --> 00:11:15,100 And that is going to be given by the element epsilon 1, 1 129 00:11:15,100 --> 00:11:17,450 times delta x1. 130 00:11:17,450 --> 00:11:22,280 The value of x2 was going to be equal to du2 131 00:11:22,280 --> 00:11:25,970 dx1 times delta x1. 132 00:11:25,970 --> 00:11:28,415 And that's going to be by definition e2, 133 00:11:28,415 --> 00:11:30,220 1 times delta x1. 134 00:11:32,720 --> 00:11:35,520 And the reason this is so simple is that I initially 135 00:11:35,520 --> 00:11:38,450 picked the line segment which was parallel to x1. 136 00:11:38,450 --> 00:11:44,200 So not all of the four terms that would be present in the 137 00:11:44,200 --> 00:11:47,280 x1, x2 system have come in. 138 00:11:47,280 --> 00:11:52,320 So there's no contribution of delta x2 because of the fact 139 00:11:52,320 --> 00:11:53,905 that I've picked this special orientation. 140 00:12:01,360 --> 00:12:26,970 OK, the fractional change of length resolved on x1 is going 141 00:12:26,970 --> 00:12:35,600 to be, well, it's going to be exactly delta x plus delta U1 142 00:12:35,600 --> 00:12:40,190 quantity squared, it's going to be this horizontal line 143 00:12:40,190 --> 00:12:48,675 segment, plus delta U2 quantity squared all 144 00:12:48,675 --> 00:12:52,160 to the power 1/2. 145 00:12:52,160 --> 00:12:55,030 And now I'm going to say that delta U2 is negligible 146 00:12:55,030 --> 00:12:58,680 compared to this big delta x plus delta U1. 147 00:12:58,680 --> 00:13:03,480 And I'll say that this is approximately equal to delta x 148 00:13:03,480 --> 00:13:15,150 plus delta U1, OK, just taking the whole works outside of the 149 00:13:15,150 --> 00:13:16,400 square root sign. 150 00:13:19,120 --> 00:13:29,450 So delta U1 over delta x1 is going to be equal to the term 151 00:13:29,450 --> 00:13:35,150 e1, 1 from this expression here. 152 00:13:35,150 --> 00:13:41,760 So the term e1, 1 represents the tensile strain along x1. 153 00:13:55,680 --> 00:13:59,470 So we can see how these derivatives are going to enter 154 00:13:59,470 --> 00:14:02,620 into changes of length. 155 00:14:02,620 --> 00:14:07,940 The P prime Q prime has also been rotated by phi. 156 00:14:14,410 --> 00:14:18,530 And we can say exactly what that is that the tangent of 157 00:14:18,530 --> 00:14:28,860 phi is going to be given exactly by delta U2 divided by 158 00:14:28,860 --> 00:14:34,330 the original length of the line segment delta x1 plus the 159 00:14:34,330 --> 00:14:40,930 little increment of displacement along x1. 160 00:14:40,930 --> 00:14:42,850 And this is-- 161 00:14:42,850 --> 00:14:44,700 I'll walk down here so I can see it. 162 00:14:44,700 --> 00:14:46,560 This is delta U1. 163 00:14:46,560 --> 00:14:51,830 This is delta U2 over times delta x1-- 164 00:14:56,310 --> 00:14:59,430 sorry, I can't see what I've got down here-- delta 165 00:14:59,430 --> 00:15:00,870 U1 plus delta x1. 166 00:15:03,430 --> 00:15:07,130 OK, so it's the amount of displacement along x2 over the 167 00:15:07,130 --> 00:15:10,680 original line segment delta x1 plus the change in 168 00:15:10,680 --> 00:15:12,800 displacement delta U1. 169 00:15:12,800 --> 00:15:18,900 And clearly delta U1 can be claimed to be small with 170 00:15:18,900 --> 00:15:20,460 respect to delta x1. 171 00:15:20,460 --> 00:15:26,130 So this is approximately equal to delta U2 over delta x1. 172 00:15:26,130 --> 00:15:29,670 And tangent of phi is going to be tangent of 173 00:15:29,670 --> 00:15:31,310 a very small angle. 174 00:15:31,310 --> 00:15:35,930 So this will be delta U2 over delta x1. 175 00:15:35,930 --> 00:15:44,100 And so this angle phi is, for small strains, going to be 176 00:15:44,100 --> 00:15:50,180 equal to delta U2 over delta x1. 177 00:15:50,180 --> 00:16:00,800 And that is the definition of our element of strain e1, 2. 178 00:16:00,800 --> 00:16:19,780 So e1, 2 corresponds to a rotation of a line segment 179 00:16:19,780 --> 00:16:29,125 that was originally parallel to x1 in the direction of x2-- 180 00:16:35,474 --> 00:16:36,270 AUDIENCE: [INAUDIBLE]. 181 00:16:36,270 --> 00:16:40,790 PROFESSOR: e2, 1, I'm sorry, e2, 1, yeah, along x1 in the 182 00:16:40,790 --> 00:16:43,440 direction of x2. 183 00:16:43,440 --> 00:16:49,120 And that is counterintuitive as I have just demonstrated. 184 00:16:49,120 --> 00:16:54,480 e2, 1 is a rotation of a line segment along x1 in the 185 00:16:54,480 --> 00:16:58,600 direction of x2, which is just the reverse of the subscript. 186 00:16:58,600 --> 00:17:05,339 So eij, a general off-diagonal element of the array eij, is 187 00:17:05,339 --> 00:17:20,880 going to be a rotation of a line segment initially along x 188 00:17:20,880 --> 00:17:27,590 sub j in the direction of x sub i. 189 00:17:33,770 --> 00:17:37,910 And for very small strains, numerically that term eij will 190 00:17:37,910 --> 00:17:40,596 give an angle in radiants. 191 00:17:40,596 --> 00:17:43,520 AUDIENCE: For the equation you have over there, should it be 192 00:17:43,520 --> 00:17:47,492 delta x plus delta x2? 193 00:17:47,492 --> 00:17:49,550 PROFESSOR: No, I would say this-- 194 00:17:49,550 --> 00:17:52,880 AUDIENCE: Because aren't you saying delta U1? 195 00:17:52,880 --> 00:17:55,380 PROFESSOR: Oh, OK, you're right. 196 00:17:55,380 --> 00:17:57,380 Yeah, I didn't take the difference here. 197 00:17:57,380 --> 00:17:59,380 AUDIENCE: Delta U1 is basically negligible? 198 00:17:59,380 --> 00:18:03,150 PROFESSOR: Yep, yep, so I'm saying that that [INAUDIBLE] 199 00:18:03,150 --> 00:18:05,420 should be delta x1-- 200 00:18:05,420 --> 00:18:06,900 AUDIENCE: Plus delta U2? 201 00:18:06,900 --> 00:18:07,930 PROFESSOR: --plus delta U2. 202 00:18:07,930 --> 00:18:09,540 You're right. 203 00:18:09,540 --> 00:18:10,790 No, I'm throwing this out. 204 00:18:13,180 --> 00:18:14,190 And that's right. 205 00:18:14,190 --> 00:18:16,730 So I'm saying that this is essentially delta 206 00:18:16,730 --> 00:18:20,590 x1 plus delta U1. 207 00:18:20,590 --> 00:18:23,170 I'm saying that this is negligible. 208 00:18:23,170 --> 00:18:27,200 So strictly speaking the distance between P prime and Q 209 00:18:27,200 --> 00:18:30,050 prime is the square root of this 210 00:18:30,050 --> 00:18:31,290 squared plus this squared. 211 00:18:31,290 --> 00:18:35,750 But if this thing is tiny, P prime Q prime is essentially 212 00:18:35,750 --> 00:18:38,860 going to be this distance, delta x1 plus delta U1. 213 00:18:38,860 --> 00:18:40,110 So that's right. 214 00:18:49,510 --> 00:18:54,390 OK, so we have something that looks like it measures 215 00:18:54,390 --> 00:18:56,730 deformation. 216 00:18:56,730 --> 00:19:02,640 But I would like to ask if this is a suitable measure of 217 00:19:02,640 --> 00:19:05,580 deformation. 218 00:19:05,580 --> 00:19:19,100 And I would like to show that, unless the tensor eij is 219 00:19:19,100 --> 00:19:23,420 symmetric, that we have included in our definition of 220 00:19:23,420 --> 00:19:30,650 strain rigid body rotation as well as true deformation. 221 00:19:30,650 --> 00:19:33,770 So in order to demonstrate that, what I'm going to look 222 00:19:33,770 --> 00:19:38,790 at is a case where we have actually, by the way in which 223 00:19:38,790 --> 00:19:42,300 we apply a stress to the material, actually done 224 00:19:42,300 --> 00:19:47,440 nothing more than rotate x1 to x1 prime and 225 00:19:47,440 --> 00:19:53,750 rotate x2 to x2 prime. 226 00:19:53,750 --> 00:19:57,060 And in general, for a real amount of deformation, that is 227 00:19:57,060 --> 00:19:58,900 something that is going to happen. 228 00:19:58,900 --> 00:20:05,420 Let's say I decide to deform this eraser in shear before 229 00:20:05,420 --> 00:20:06,210 your very eyes. 230 00:20:06,210 --> 00:20:07,260 And I try shearing it. 231 00:20:07,260 --> 00:20:08,500 Nothing much has happened. 232 00:20:08,500 --> 00:20:09,720 And I squeeze a little harder. 233 00:20:09,720 --> 00:20:12,240 And finally, I've got it wrestled down into an 234 00:20:12,240 --> 00:20:14,270 orientation like this. 235 00:20:14,270 --> 00:20:20,170 And when I finished, this was the original 236 00:20:20,170 --> 00:20:22,300 position of the eraser. 237 00:20:22,300 --> 00:20:23,170 Here's x1. 238 00:20:23,170 --> 00:20:24,690 Here's x2. 239 00:20:24,690 --> 00:20:30,060 And by the time I've wrestled it around to a deformed state, 240 00:20:30,060 --> 00:20:33,205 it sits down like this, maybe deformed a little bit. 241 00:20:36,310 --> 00:20:42,180 But would you say that this angle here is a measure of 242 00:20:42,180 --> 00:20:44,100 deformation? 243 00:20:44,100 --> 00:20:47,300 No, that's clearly a rigid body rotation. 244 00:20:47,300 --> 00:20:50,310 And what I would intuitively do, if I wanted to measure 245 00:20:50,310 --> 00:20:55,950 true deformation, if this were the original body and after 246 00:20:55,950 --> 00:21:01,130 deformation it went to a location with some deformation 247 00:21:01,130 --> 00:21:06,950 to be sure but this has been rotated to here. 248 00:21:06,950 --> 00:21:09,760 This has been rotated down. 249 00:21:09,760 --> 00:21:12,230 And I wouldn't want to say that this is a measure of 250 00:21:12,230 --> 00:21:13,460 deformation. 251 00:21:13,460 --> 00:21:15,160 This would be e1, 2. 252 00:21:15,160 --> 00:21:18,290 What I would want to do intuitively would be to 253 00:21:18,290 --> 00:21:23,370 position the body symmetrically between the axes 254 00:21:23,370 --> 00:21:27,860 and then say that this is a measure of shear strain. 255 00:21:27,860 --> 00:21:31,850 So let me show you now in a more rigorous treatment that, 256 00:21:31,850 --> 00:21:35,460 if there is a component of rigid body rotation, let me 257 00:21:35,460 --> 00:21:40,980 show you what a rotation of the coordinate system to a new 258 00:21:40,980 --> 00:21:44,250 orientation x1, x prime would do. 259 00:21:44,250 --> 00:21:47,950 So let's say that this is an original point Q1. 260 00:21:47,950 --> 00:21:53,030 And after deformation, it moves by a rotation phi that's 261 00:21:53,030 --> 00:21:59,110 equal to e2, 1 to a new location Q1 prime. 262 00:21:59,110 --> 00:22:02,220 Let's suppose that this is a position Q2. 263 00:22:02,220 --> 00:22:08,080 And after what we think is the deformation, this moves to a 264 00:22:08,080 --> 00:22:11,000 location Q2 prime. 265 00:22:11,000 --> 00:22:14,050 And this would be e1, 2. 266 00:22:14,050 --> 00:22:16,000 That's equal to minus phi. 267 00:22:16,000 --> 00:22:17,280 This is e2, 1. 268 00:22:17,280 --> 00:22:18,790 And that's equal to plus phi. 269 00:22:27,490 --> 00:22:29,670 So what we would do is like to get rid of 270 00:22:29,670 --> 00:22:31,450 that rigid body rotation. 271 00:22:36,810 --> 00:22:44,670 And what I'll do is to show that, if we have a point 272 00:22:44,670 --> 00:22:48,310 that's out at the end of a radial vector, R, which has 273 00:22:48,310 --> 00:22:54,520 coordinates xi, x1, x2, x3, and if we have a displacement, 274 00:22:54,520 --> 00:22:58,890 which is absolutely perpendicular to that radial 275 00:22:58,890 --> 00:23:04,410 vector, that this would be characteristic 276 00:23:04,410 --> 00:23:08,080 of rigid body rotation. 277 00:23:08,080 --> 00:23:18,900 And I will say that, if Ui is perpendicular to the radial 278 00:23:18,900 --> 00:23:27,750 vector xi, then U.R should be equal to 0 for 279 00:23:27,750 --> 00:23:31,361 every position xi. 280 00:23:31,361 --> 00:23:36,780 So if we just multiply this out, this is saying that Ui Ri 281 00:23:36,780 --> 00:23:39,700 should be equal to 0 for rigid body rotation. 282 00:23:47,280 --> 00:23:53,110 And let's simply carry out this expansion. 283 00:23:53,110 --> 00:24:04,340 And I will have then eij xi xj forming this dot product. 284 00:24:04,340 --> 00:24:08,980 And that is going to be 0 for a rigid body rotation. 285 00:24:08,980 --> 00:24:13,790 And if I expand this, this will contain terms like e1, 1 286 00:24:13,790 --> 00:24:21,950 times x1 x1 plus e2, 2 times x2 x2 plus 287 00:24:21,950 --> 00:24:25,310 e3, 3 times x3 squared. 288 00:24:25,310 --> 00:24:30,690 And then they'll be cross-terms e1, 3 plus e3, 1 289 00:24:30,690 --> 00:24:47,510 times x1 x3 plus e1, 2 plus e2, 1 times x1 x2 plus e2, 3 290 00:24:47,510 --> 00:24:52,770 plus e3, 2 times x2, x3. 291 00:24:52,770 --> 00:24:56,530 And my claim now is that, if this represents rigid body 292 00:24:56,530 --> 00:25:01,995 rotation, then that should be 0. 293 00:25:05,960 --> 00:25:09,810 That is to say the displacement is absolutely 294 00:25:09,810 --> 00:25:12,420 perpendicular to the radius vector for small 295 00:25:12,420 --> 00:25:13,370 displacements. 296 00:25:13,370 --> 00:25:21,120 This must be 0 for all xi. 297 00:25:21,120 --> 00:25:24,770 And the only way that's going to be possible for any value 298 00:25:24,770 --> 00:25:30,290 of x1, x2, x3 is that each of these six terms vanish 299 00:25:30,290 --> 00:25:31,782 individual. 300 00:25:31,782 --> 00:25:34,960 It's the only way I'm going to be able to get the whole thing 301 00:25:34,960 --> 00:25:40,570 to disappear for any value of coordinate x1, x2, x3. 302 00:25:40,570 --> 00:25:48,220 So we're going to have to then have e1, 1 equals e2, 2 equals 303 00:25:48,220 --> 00:25:51,080 e3, 3 equals 0. 304 00:25:51,080 --> 00:25:55,650 All the diagonal terms are going to have to be 0. 305 00:25:55,650 --> 00:25:58,040 In order to get the fourth term to vanish, I'm going to 306 00:25:58,040 --> 00:26:04,930 have to have e1, 3 equal to the negative of e3, 1 and e1, 307 00:26:04,930 --> 00:26:13,140 2 the negative of e2, 1 and e2, 3 the negative of e3, 2. 308 00:26:16,000 --> 00:26:21,700 So what is this going to look like for the part of the e 309 00:26:21,700 --> 00:26:25,960 tensor that corresponds to rigid body rotation? 310 00:26:25,960 --> 00:26:29,000 The diagonal terms are all going to be 0. 311 00:26:32,670 --> 00:26:36,260 And the off-diagonal terms are going to be the negative of 312 00:26:36,260 --> 00:26:36,800 one another. 313 00:26:36,800 --> 00:26:38,480 So this is e1, 2. 314 00:26:38,480 --> 00:26:42,447 e2, 1 would have to be equal to minus e1, 2. 315 00:26:42,447 --> 00:26:49,290 e3, 1 would have to be equal to the negative of e1, 3. 316 00:26:49,290 --> 00:26:51,390 And so we will have something like this. 317 00:26:56,870 --> 00:27:06,140 So this suggests that our definition of the true 318 00:27:06,140 --> 00:27:20,240 deformation, which will be a tensor epsilon ij-- 319 00:27:20,240 --> 00:27:23,550 I bet you kind of guessed I was going to call it epsilon 320 00:27:23,550 --> 00:27:30,435 and not E. I can write this as a sum of two parts. 321 00:27:34,680 --> 00:27:39,440 I'll turn that around and say I'll write epsilon ij plus 322 00:27:39,440 --> 00:27:45,640 another three by three array omega ij is equal to the 323 00:27:45,640 --> 00:27:49,190 tensor eij. 324 00:27:49,190 --> 00:27:50,500 That's a novel idea. 325 00:27:50,500 --> 00:27:54,590 This is addition of two tensors element by element. 326 00:27:57,800 --> 00:28:07,390 And if I do that and define epsilon ij equal to 1/2 of eij 327 00:28:07,390 --> 00:28:14,800 plus eji, so for the diagonal elements that is going to take 328 00:28:14,800 --> 00:28:17,910 1/2 of e1, 1 plus e1, 1. 329 00:28:17,910 --> 00:28:20,960 And that's just going to give me e 1, 1 back again. 330 00:28:23,590 --> 00:28:33,450 And I'm going to define the terms omega ij as 1/2 331 00:28:33,450 --> 00:28:39,505 of eij minus eji. 332 00:28:39,505 --> 00:28:43,130 And if I do that, the resulting 333 00:28:43,130 --> 00:28:45,275 tensor will be symmetric. 334 00:28:51,900 --> 00:28:58,410 OK, so from tensor eij we can split it up into two parts and 335 00:28:58,410 --> 00:29:00,240 a part omega ij. 336 00:29:00,240 --> 00:29:01,820 And I won't bother to write it out. 337 00:29:01,820 --> 00:29:07,340 But you can see that omega ij plus eij is going to be equal 338 00:29:07,340 --> 00:29:18,440 to eij minus 1/2 of 2eij minus 1/2 of 0. 339 00:29:18,440 --> 00:29:20,280 And that's just going to be eij. 340 00:29:20,280 --> 00:29:22,850 So this plus this is indeed going to give 341 00:29:22,850 --> 00:29:24,590 me the tensor eij. 342 00:29:29,570 --> 00:29:34,690 So given a tensor eij, which is not symmetric, I will 343 00:29:34,690 --> 00:29:41,950 define eij as the sum of two parts, a tensor epsilon ij, 344 00:29:41,950 --> 00:29:49,520 which is true strain, and a part omega ij, which is rigid 345 00:29:49,520 --> 00:29:50,770 body rotation. 346 00:29:55,879 --> 00:30:02,950 And now at last we have a satisfactory measure of true 347 00:30:02,950 --> 00:30:07,530 deformation in terms of the displacement of points within 348 00:30:07,530 --> 00:30:11,110 a deformed body where that displacement can arise from 349 00:30:11,110 --> 00:30:14,280 either rigid body rotation and or deformation. 350 00:30:20,140 --> 00:30:27,850 So finally, we have a tensor epsilon ij, which is the 351 00:30:27,850 --> 00:30:29,100 strain tensor. 352 00:30:36,095 --> 00:30:42,910 And now I can finally make my claim that, if this really is 353 00:30:42,910 --> 00:30:45,985 true deformation, that for cubic crystals-- 354 00:30:50,410 --> 00:30:51,760 you can see the same [INAUDIBLE] 355 00:30:51,760 --> 00:30:53,210 window coming again-- 356 00:30:53,210 --> 00:30:56,520 since second ranked tensors have to be symmetric for cubic 357 00:30:56,520 --> 00:30:59,790 crystals, the form of strain for a cubic 358 00:30:59,790 --> 00:31:02,722 crystal can only be this. 359 00:31:02,722 --> 00:31:08,600 It has to be diagonal if the tensor is to 360 00:31:08,600 --> 00:31:12,140 conform to cubic symmetry. 361 00:31:12,140 --> 00:31:14,260 And you say, ah, ah, you tried to pull that on us with 362 00:31:14,260 --> 00:31:15,020 stress, too. 363 00:31:15,020 --> 00:31:18,250 And I can squish a crystal anyway I want. 364 00:31:18,250 --> 00:31:21,650 But you can only develop a strain if 365 00:31:21,650 --> 00:31:22,980 you deform a crystal. 366 00:31:22,980 --> 00:31:26,020 And the symmetry of the crystal is going to determine 367 00:31:26,020 --> 00:31:28,710 how the crystal deforms. 368 00:31:28,710 --> 00:31:32,380 So therefore, a cubic crystal should be able to deform only 369 00:31:32,380 --> 00:31:36,510 in a way that stays invariant to the transformations of a 370 00:31:36,510 --> 00:31:37,780 cubic symmetry, right? 371 00:31:37,780 --> 00:31:43,420 So cubic crystals can only deform isotropically. 372 00:31:43,420 --> 00:31:48,020 So if I wanted to design springs for an automobile so 373 00:31:48,020 --> 00:31:50,750 that elastic energy could be stored most efficiently in a 374 00:31:50,750 --> 00:31:54,060 solid, I would want to make these automobile springs out 375 00:31:54,060 --> 00:31:55,880 of a triclinic metal. 376 00:31:55,880 --> 00:31:59,950 Because then the tensor could be one of general deformation. 377 00:32:04,000 --> 00:32:07,490 OK, well, this, obviously, is a swindle like my assertion 378 00:32:07,490 --> 00:32:15,070 that stress could only be an isotropic compressional stress 379 00:32:15,070 --> 00:32:16,140 for a cubic crystal. 380 00:32:16,140 --> 00:32:19,480 But the argument is a little more convoluted. 381 00:32:19,480 --> 00:32:24,090 And strain, indeed, is also a field tensor. 382 00:32:24,090 --> 00:32:27,820 Because, even though I can only create the deformation by 383 00:32:27,820 --> 00:32:31,040 application of a stress or perhaps an electric field or 384 00:32:31,040 --> 00:32:34,870 something of that sort in that the link between the 385 00:32:34,870 --> 00:32:40,690 deformation and what I do to the crystal is coupled by a 386 00:32:40,690 --> 00:32:44,970 property of the crystal, which has to conform to the symmetry 387 00:32:44,970 --> 00:32:51,300 of the crystal, nevertheless, given a tensor that describes 388 00:32:51,300 --> 00:32:55,720 deformation, I can always, independent of the symmetry of 389 00:32:55,720 --> 00:33:00,840 the crystal, devise a particular set of stresses 390 00:33:00,840 --> 00:33:03,890 that would produce any state of strain I want. 391 00:33:03,890 --> 00:33:06,500 I would have to design, though, a particular state of 392 00:33:06,500 --> 00:33:10,130 stress to produce a desired state of strain. 393 00:33:10,130 --> 00:33:12,270 And that coupling has to conform to the 394 00:33:12,270 --> 00:33:13,560 symmetry of the crystal. 395 00:33:13,560 --> 00:33:17,980 But there's no reason why strain itself has to conform 396 00:33:17,980 --> 00:33:20,220 to the symmetry of the crystal. 397 00:33:20,220 --> 00:33:23,050 I can create any state of strain I like by choosing the 398 00:33:23,050 --> 00:33:25,600 appropriate stress. 399 00:33:25,600 --> 00:33:28,120 OK, so that is a swindle. 400 00:33:28,120 --> 00:33:37,390 But everything now that I can say about the behavior of a 401 00:33:37,390 --> 00:33:40,990 second ranked tensor applies to the strain tensor. 402 00:33:45,190 --> 00:33:48,420 I can take a strain tensor epsilon ij. 403 00:33:48,420 --> 00:33:54,840 And I can perform the surface epsilon ij xi xj equals 1. 404 00:33:54,840 --> 00:33:57,180 And that will be the strain quadric. 405 00:34:06,980 --> 00:34:14,000 The strain quadric will have the property like any quadric 406 00:34:14,000 --> 00:34:21,500 constructed from a tensor that the value of the radius in a 407 00:34:21,500 --> 00:34:29,320 particular direction is going to be such that the strain in 408 00:34:29,320 --> 00:34:34,249 that direction is equal to 1 over the radius squared. 409 00:34:36,900 --> 00:34:37,860 So what does that mean? 410 00:34:37,860 --> 00:34:44,900 Well, we have to look at what the strain tensor is relating. 411 00:34:44,900 --> 00:34:53,750 The direction, remember now that the definition of the 412 00:34:53,750 --> 00:34:58,750 strain tensor is that U sub i equals epsilon 413 00:34:58,750 --> 00:35:01,920 ij times x sub j. 414 00:35:01,920 --> 00:35:12,590 So the quote "applied vector" is the direction in a solid. 415 00:35:12,590 --> 00:35:19,640 And we will get, in general, a displacement, U, which is not 416 00:35:19,640 --> 00:35:23,890 parallel to the direction that we're considering. 417 00:35:23,890 --> 00:35:29,330 And the epsilon in a particular direction is going 418 00:35:29,330 --> 00:35:35,790 to be equal to the part of U that's parallel to the 419 00:35:35,790 --> 00:35:41,650 direction of interest divided by distance in that direction. 420 00:35:41,650 --> 00:35:47,430 And what this is going to give us is the tensile component of 421 00:35:47,430 --> 00:36:00,815 deformation in the direction that we're examining. 422 00:36:13,520 --> 00:36:17,475 Now, the radius normal property works in this case. 423 00:36:30,540 --> 00:36:33,980 And what the radius normal property says, if you look at 424 00:36:33,980 --> 00:36:39,290 a certain direction in the solid and then look at the 425 00:36:39,290 --> 00:36:44,880 normal to the surface at that particular point, this will be 426 00:36:44,880 --> 00:36:46,830 the direction of what happened. 427 00:36:46,830 --> 00:36:48,180 So I've not drawn this correctly. 428 00:36:48,180 --> 00:36:50,930 This should be the direction of U. And this 429 00:36:50,930 --> 00:36:53,080 is the radio vector. 430 00:36:53,080 --> 00:36:57,130 So the normal to the surface gives us the direction of the 431 00:36:57,130 --> 00:36:59,000 displacement that's going to occur. 432 00:36:59,000 --> 00:37:02,830 And the reason I can say that is by definition that that 433 00:37:02,830 --> 00:37:05,760 property holds only for a symmetric tensor. 434 00:37:05,760 --> 00:37:09,330 And by definition, we have defined the strained tensor 435 00:37:09,330 --> 00:37:10,790 such that it is symmetric. 436 00:37:10,790 --> 00:37:12,535 So the radius normal property holds. 437 00:37:20,480 --> 00:37:26,850 If we want to change from one coordinate system to another, 438 00:37:26,850 --> 00:37:32,900 we do that by the all for transformation of a second 439 00:37:32,900 --> 00:37:36,230 ranked tensor that, if we change coordinate system, the 440 00:37:36,230 --> 00:37:40,160 elements of strain change to new values epsilon ij prime 441 00:37:40,160 --> 00:37:49,940 that are given by cil cjm times epsilon lm, the same law 442 00:37:49,940 --> 00:37:53,600 for transformation of a second ranked tensor that we have 443 00:37:53,600 --> 00:37:54,850 seen earlier. 444 00:38:03,220 --> 00:38:09,240 Finally, we can, because there is a quadric, we can change 445 00:38:09,240 --> 00:38:14,030 that by finding eigenvectors and eigenvalues, take a 446 00:38:14,030 --> 00:38:19,600 general form of the tensor epsilon 1, 1, epsilon 1, 2, 447 00:38:19,600 --> 00:38:27,660 epsilon 1, 3, and so on, and by solving the eigenvalue 448 00:38:27,660 --> 00:38:32,298 problem, convert this to a value epsilon 1, 1 prime 0, 0, 449 00:38:32,298 --> 00:38:38,420 0, epsilon 2, 2 prime 0, 0, 0, epsilon 3, 3 prime to find out 450 00:38:38,420 --> 00:38:43,150 what the extreme values of tensile deformation are and 451 00:38:43,150 --> 00:38:46,310 the orientations within the original description of the 452 00:38:46,310 --> 00:38:49,060 body in which these extreme values occur. 453 00:39:00,540 --> 00:39:03,410 OK, I'm just about out of time. 454 00:39:03,410 --> 00:39:06,550 I'll save until next time another neat feature of the 455 00:39:06,550 --> 00:39:08,940 strain tensor. 456 00:39:08,940 --> 00:39:16,230 And that is contained within these elements is information 457 00:39:16,230 --> 00:39:19,260 on the volume change that the body undergoes. 458 00:39:19,260 --> 00:39:21,150 And this is something called the dilation. 459 00:39:24,730 --> 00:39:27,520 And we'll see that this is related in a very simple way 460 00:39:27,520 --> 00:39:30,040 to the elements of the strain tensor. 461 00:39:33,280 --> 00:39:37,880 OK, we've spent a lot of time developing the formalism of 462 00:39:37,880 --> 00:39:39,770 stress and strain. 463 00:39:39,770 --> 00:39:43,850 And next thing we'll do is to move on to some interesting 464 00:39:43,850 --> 00:39:48,350 properties of crystals, which are defined in terms of 465 00:39:48,350 --> 00:39:51,420 tensors of rank higher than two. 466 00:39:51,420 --> 00:39:55,430 So we'll look at some third ranked tensor properties. 467 00:39:55,430 --> 00:40:00,590 That will include things like piezoelectricity, the stress 468 00:40:00,590 --> 00:40:04,570 optical effect, and things of that sort. 469 00:40:04,570 --> 00:40:08,020 These are going to be really anisotropic. 470 00:40:08,020 --> 00:40:10,540 Second ranked tensors were anisotropic enough. 471 00:40:10,540 --> 00:40:13,750 But the variation of property with direction was a 472 00:40:13,750 --> 00:40:18,150 quasi-ellipsoidal variation When as 1 over the square of 473 00:40:18,150 --> 00:40:20,310 the radius of the quadric. 474 00:40:20,310 --> 00:40:24,340 For higher ranked tensor properties, we will encounter 475 00:40:24,340 --> 00:40:28,520 some absolutely weird surfaces with dimples and lumps and 476 00:40:28,520 --> 00:40:34,200 lobes, not this smooth quasi-ellipsoidal variation, a 477 00:40:34,200 --> 00:40:36,800 property with direction, which was the case for second ranked 478 00:40:36,800 --> 00:40:38,070 tensor properties. 479 00:40:38,070 --> 00:40:41,280 So we will get into some exotic anisotropies very 480 00:40:41,280 --> 00:40:46,070 shortly to finish up the semester. 481 00:40:46,070 --> 00:40:49,800 So I will probably see some of you at the MRS 482 00:40:49,800 --> 00:40:52,330 meeting next week. 483 00:40:52,330 --> 00:40:55,880 I won't see you on Thursday for reasons that I need not 484 00:40:55,880 --> 00:40:56,780 elaborate upon. 485 00:40:56,780 --> 00:40:59,510 So I hope you have a happy holiday. 486 00:40:59,510 --> 00:41:05,960 Relax, suck in air, and come back refreshed next Tuesday 487 00:41:05,960 --> 00:41:09,950 for some really wild anisotropy of physical 488 00:41:09,950 --> 00:41:11,200 properties.