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.