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PROFESSOR: We had begun to
discuss, really what is,

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although you would not gather
it from last time, a fairly

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straightforward matter.

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Set up our usual Cartesian
coordinate system.

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And then look at the surface of
the solid, and applied to

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this solid a force per unit
area, which we'll represent by

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a vector k.

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And then put on some
labels here.

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Let this be O. Let this be A,
let this be B, and this be

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point C. Then we
proceeded to--

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I will say, attempt to--

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set up a balance of forces
between the external force per

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unit area and the internal
forces per unit area, which

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have to push back on the
external force if the solid is

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to remain motionless.

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The thing that I did not carry
through is that there are two

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directions involved.

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One is the direction to the
normal, to the particular

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surface that we're looking at.

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We'll call that N. And there is
the direction that k, the

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vector k, makes with respect to
the three reference x axes,

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x1, x2, x3.

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Let's take k and immediately
split it up into the three

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components, k1, k2, k3, because
throughout, we're

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going to be dealing with the
balance of those three forces

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per unit area in the directions
of x1, x2, and x3.

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We want to take this surface
area and resolve it into three

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separate areas that are
normal to x1, x2, x3.

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And what I did at the end of
last hour was to look at k and

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try to express these internal
areas in terms of the

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direction cosines of k.

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Forget that.

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We've split up k into its three
components, and I didn't

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draw in the direction
of the normal to the

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surface N last time.

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So I kept saying, direction
cosines of k, and people in

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the audience said,
no, no, no, no.

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And I said, yeah, because you
see, that's the ratio

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of these two areas.

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And everybody said, no, that's
wrong, and I said, ha.

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And that's where we ended
my most inglorious hour.

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And they've got this
all down on tape.

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My reputation's going
to be ruined.

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So, anyway, let's now
do it properly.

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And what I will say is that in
the x1 direction, we have to

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have internal forces that push
back, and we have to balance

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those forces in all
three directions.

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So I'll say is that in the x1
direction, and this as you saw

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last time, gets very cumbersome
because we have to

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deal with all these clumsy
areas, that the force per unit

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area on ABC has to be balanced
for each of the three

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components of k by internal
forces acting

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along x1, x2, and x3.

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So we can say is that the first
in the x1 direction, the

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component of k that acts on
the surface ABC in the x1

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direction has to be balanced
by internal forces.

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And let's look first at
the x1 direction.

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So there are three internal
surfaces.

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They are the area BOC.

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That is this back surface here,
and the normal to that

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surface is x1.

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So let me define some forces
per unit area that are

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internal, and I'll use the
term sigma for that.

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And I will use a first subscript
that gives the

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direction in which that force
per unit area acts, and I'll

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use a second subscript that
indicates the normal to the

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internal surface.

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So area BOC obviously, as we
observed a moment ago, has x1

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as its normal.

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But there are three
internal surfaces.

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There will also be an internal
area, AOC, on which a force

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per unit area acts in
the x1 direction.

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So I'll call this one
by analogy to what I

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did a moment ago.

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I'll call this sigma 1, because
it acts in the x1

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direction, and it acts on the
surface AOC, which has x2 as

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its normal.

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Then finally, there will be a
third force per unit area,

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sigma 1 3, acting in the x1
direction, on a surface whose

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normal is perpendicular
to-- along x3.

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The surface is perpendicular
to x3.

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And that's area AOB.

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OK, so this is just a simple
balance of forces, external

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and internal balancing forces,
which act in the x1 direction.

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Then we'll write a similar thing
for the x2 direction.

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And we'll say that the x2
component of force per unit

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area external to the volume
element acting on area ABC is

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going to be balanced by, again,
forces now acting in

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the x2 direction.

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So first, there will be a
component of 4 sigma 2 acting

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on the area that's normal to x1,
and that's the same area

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BOC, plus another force per
unit area acting in the x2

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direction on the surface, whose
normal is a long x2.

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And that is, once
more, area AOC.

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And that would be a force per
unit area that I'll define,

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because it's my ball game
and I own the ball.

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So again, that would be sigma 2
2, and then finally a sigma

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2 3 times the area whose
normal is x3.

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And I think I need not write
the third term that follows

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automatically from what
I've written so far.

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So now comes the geometry part,
where I'm going to get

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rid of these messy areas, and
I'm going to divide through by

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the external area on the
surface of the solid.

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I will then write accordingly
that k1 is equal to sigma 11

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times the ratio of area BOC
divided by area ABC plus sigma

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12 times area AOC divided
by area AOB--

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ABC, excuse me.

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Then finally sigma 1 3 times
area AOB divided by area ABC.

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So now I'll introduce a little
bit of geometry that caused

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such difficulties last time.

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If I have an area and project
it onto another area that

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makes an angle C with respect
to the first, this area A

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prime is equal to the area A
times the cosine of Phi and

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you could define that Phi in a
different way, namely take the

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normal to one of the two areas
and let M prime be the other

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normal, so C then is exactly
the same thing as the angle

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between the normals to
these two surfaces.

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So let's now look at what the
ratio of these two areas that

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we have is.

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What we're dealing with here is
an angle between the normal

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to the surface, not k, but the
normal to the surface, and

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either x1, x2, or x3.

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So the ratio of--

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let's look at one of these
specific terms.

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The ratio of area BOC through
the ratio of the area ABC

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should be the cosine of the
angle between the normal to

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the surface and the normal
to area BOC.

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Well, the normal to area BOC is
x1, so that first ratio of

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areas is just going to be the
cosine of this angle.

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And that cosine is just the
direction cosine of N1 along

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the x1 of M along the
x1 direction.

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So this is just the direction
cosine between N, the normal

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to the surface, and x1.

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And let me say then that, by
definition, I'll let the

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direction to the normal be its
three direction cosines, L 1

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plus L 2, plus J plus
L3 times k.

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These are the direction cosines
of the normal to the

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surface, and not k.

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So this then simply comes down
to sigma 11 times L1, the

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ratio of the area AOC to ABC
is going to be the angle

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between N and x2.

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And that's the angle whose
direction cosine is L2.

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And similarly, the ratio of
these two areas will be cosine

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minus 1 of L3.

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So this cumbersome construction
with ratios of

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areas just reduces to sigma 11
L1 plus sigma 12 times L2 plus

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sigma 13 times L3.

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And the other two equations will
follow a similar form.

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K2 will be sigma 21 times L1
plus sigma 22 times L2 plus

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sigma 23 times L3.

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Or in general, case of I will be
given by sigma I J times L

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to J. Easy.

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Nice and compact.

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Nice and simple.

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What can we say about this?

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K is a vector which has
components k1, k2, and k3.

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The direction cosines can be,
as we said a moment ago, can

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be regarded as the components
of a unit vector pointing

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along the normal
to the surface.

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So k is a vector.

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These three direction cosines
are the direction cosine to a

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unit vector normal
to the surface.

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If we were to be so foolish to
try to change the coordinate

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system., having just gotten it
right for the first time in

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one coordinate system, we know
exactly how the two vectors

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will change because
we have a law for

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transformation of a vector.

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If we were to change the
coordinate system, the

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components of k would wink on
end of and take new values and

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we know exactly how the
components of a vector will

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change as we change
coordinate system.

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OK, if this transforms like a
vector and this transforms

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like the vector, then the 3 by
3 array of coefficients which

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relate those to quantities
must be a tensor.

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So we've proven rigorously that
sigma IJ is a tensor, in

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particular, it's a second
rank tensor.

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And were we to change coordinate
system, we know

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from what we've done in the
past, exactly how the numbers

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00:13:31,730 --> 00:13:35,520
that make up this 3 by 3
array will transform.

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00:13:40,730 --> 00:13:47,940
These quantities, sigma IJ,
are something that you've

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probably encountered, introduced
to be sure,

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probably in a little
different way.

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00:13:53,140 --> 00:13:59,060
These are the elements of
stress, and sigma IJ is the

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stress tensor.

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00:14:13,530 --> 00:14:17,680
We can immediately, on physical
grounds, established

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00:14:17,680 --> 00:14:24,720
that sigma IJ must be
a symmetric tensor.

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So let's set up two axes, x1,
and x2, and x3 obviously is

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normal to the board.

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00:14:35,300 --> 00:14:40,270
And let us look at some
off-diagonal terms like sigma

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00:14:40,270 --> 00:14:43,360
12 and sigma 21.

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00:14:45,900 --> 00:14:52,080
Sigma 1 2 would be a force
acting on the x1 direction on

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00:14:52,080 --> 00:14:54,940
a surface whose normal is x2.

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00:14:54,940 --> 00:15:00,510
So this is going to be a sheer
stress and that, to prevent

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00:15:00,510 --> 00:15:05,290
any translation of the volume
element, must be balanced by a

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00:15:05,290 --> 00:15:09,270
sigma 12 on the other
face that acts in

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00:15:09,270 --> 00:15:11,240
the opposite direction.

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00:15:11,240 --> 00:15:15,320
We're going to need a convention
for defining sign,

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S-I-G-N, of what constitutes a
positive element of stress and

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00:15:20,450 --> 00:15:22,730
a negative element of stress.

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00:15:22,730 --> 00:15:38,365
And I'll define a plus sigma
IJ as a force per unit area

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00:15:38,365 --> 00:15:52,490
acting in the plus xi direction
on a surface whose

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00:15:52,490 --> 00:15:54,636
normal is plus xj.

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So if this is a force per unit
area acting on this surface,

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00:16:07,960 --> 00:16:13,380
and it acts in the direction
of x1 on a surface whose

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00:16:13,380 --> 00:16:19,130
normal is x2, what I've shown
here is a positive sigma 12.

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00:16:19,130 --> 00:16:22,560
A sheer force going in the
opposite direction would be a

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00:16:22,560 --> 00:16:25,560
negative sigma 12.

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00:16:25,560 --> 00:16:31,640
A positive sigma 11 would be
a stress that's tensile, a

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00:16:31,640 --> 00:16:36,950
negative sigma 11 would be a
stress that's compressive.

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00:16:36,950 --> 00:16:41,140
So here's a sigma 12, a sigma
12 on the opposite face.

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00:16:41,140 --> 00:16:44,640
If we went away and turned our
back, this volume element

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00:16:44,640 --> 00:16:47,740
would start accelerating,
getting an angular

213
00:16:47,740 --> 00:16:51,020
acceleration, would start
furiously spinning about the

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00:16:51,020 --> 00:16:53,120
x3 axis like a top.

215
00:16:53,120 --> 00:16:57,510
So something, if this volume
element is, by definition, an

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00:16:57,510 --> 00:17:01,930
equilibrium, something has
to balance this couple.

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00:17:01,930 --> 00:17:05,720
And if it's to stay put, we must
have a sheer force like

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00:17:05,720 --> 00:17:08,210
this and a sheer force
like this.

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00:17:08,210 --> 00:17:10,380
And that's the only thing
available from the nine

220
00:17:10,380 --> 00:17:15,359
elements of stress that will
give a net torque on the

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00:17:15,359 --> 00:17:16,540
volume element.

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00:17:16,540 --> 00:17:18,319
So what are these?

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00:17:18,319 --> 00:17:23,630
This is a stress that
acts on the surface

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00:17:23,630 --> 00:17:27,280
whose normal is x2--

225
00:17:27,280 --> 00:17:30,010
whose normal as x1 in
the x2 direction.

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00:17:30,010 --> 00:17:34,950
So this is sigma 21.

227
00:17:34,950 --> 00:17:47,820
It acts in x2 direction,
and its normal is x1.

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00:17:53,090 --> 00:17:54,760
An x, and then [? plus ?]

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00:17:54,760 --> 00:17:56,170
x2 direction on [? I ?]

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00:17:56,170 --> 00:18:03,660
surface, which has a normal
that is along x1.

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00:18:09,350 --> 00:18:14,720
So the only way we can avoid an
angular acceleration is to

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00:18:14,720 --> 00:18:18,960
say that sigma 12 has to be
identical to sigma 21 if

233
00:18:18,960 --> 00:18:23,220
there's no rotation about x3.

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00:18:23,220 --> 00:18:27,440
And looking at the couple's that
act about x2, and x3, we

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00:18:27,440 --> 00:18:32,350
could say in general that sigma
IJ is going to have to

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00:18:32,350 --> 00:18:38,425
be equal to sigma JI for a
volume element in equilibrium.

237
00:18:49,670 --> 00:18:55,260
So stress then is equilibrium
is in place, has to be a

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00:18:55,260 --> 00:18:56,510
symmetric tensor.

239
00:19:00,690 --> 00:19:06,140
It is, however, a plain old
second-rank tensor like all of

240
00:19:06,140 --> 00:19:11,090
the other second-rank tensors
that we've spent the last

241
00:19:11,090 --> 00:19:12,340
month talking about.

242
00:19:19,450 --> 00:19:23,970
And I think I tried to swindle
you on this one before.

243
00:19:23,970 --> 00:19:29,580
If this is a second-rank tensor,
the stress tensor must

244
00:19:29,580 --> 00:19:43,220
be symmetric and diagonal
for a cubic crystal.

245
00:19:48,330 --> 00:19:51,900
In other words, the stress
tensor sigma IJ must have the

246
00:19:51,900 --> 00:19:59,270
form sigma 0 0, 0 sigma 0, 0 0
sigma because we've shown that

247
00:19:59,270 --> 00:20:02,310
that is what is required of a
tensor if the crystal has

248
00:20:02,310 --> 00:20:05,010
cubic symmetry.

249
00:20:05,010 --> 00:20:10,570
So surprisingly, you cannot
subject a single crystal with

250
00:20:10,570 --> 00:20:13,160
cubic material to
sheer stress.

251
00:20:21,350 --> 00:20:22,470
Any quarrel with that?

252
00:20:22,470 --> 00:20:25,480
Should we move on?

253
00:20:25,480 --> 00:20:30,000
We better not, but somebody
better raise an objection.

254
00:20:30,000 --> 00:20:31,620
That's not true, obviously.

255
00:20:31,620 --> 00:20:36,820
I can take a cubic crystal and
shear the stuffings out of it.

256
00:20:36,820 --> 00:20:37,710
So what's going on here?

257
00:20:37,710 --> 00:20:39,670
AUDIENCE: The tensor property
that you are applying to

258
00:20:39,670 --> 00:20:41,140
[INAUDIBLE]

259
00:20:41,140 --> 00:20:43,100
property of the [? crystal. ?]

260
00:20:43,100 --> 00:20:49,470
But if it is an outside force,
it doesn't [INAUDIBLE].

261
00:20:49,470 --> 00:20:51,450
PROFESSOR: This is one
of your good days.

262
00:20:51,450 --> 00:20:52,920
AUDIENCE: [INAUDIBLE]

263
00:20:52,920 --> 00:20:56,350
society [INAUDIBLE].

264
00:20:56,350 --> 00:20:58,310
PROFESSOR: That makes it an
especially good day because

265
00:20:58,310 --> 00:21:02,230
you're paying attention to
what I said for a change.

266
00:21:02,230 --> 00:21:04,510
The distinction here, and I did
make it last time-- it was

267
00:21:04,510 --> 00:21:06,090
one of the things that
I did correctly--

268
00:21:06,090 --> 00:21:09,600
that there are really two kinds
of things that have to

269
00:21:09,600 --> 00:21:11,840
be described by second-rank
tensors.

270
00:21:11,840 --> 00:21:15,660
There are the things that we
have been spending most of our

271
00:21:15,660 --> 00:21:23,540
time on, and these are property
tensors and--

272
00:21:23,540 --> 00:21:27,420
I mentioned last time, too, so
I know I said this before--

273
00:21:27,420 --> 00:21:29,350
Nye calls them field tensors.

274
00:21:29,350 --> 00:21:30,590
I'm sorry.

275
00:21:30,590 --> 00:21:33,820
Nye calls them property
tensors, also.

276
00:21:33,820 --> 00:21:37,460
And these are something that
depends on the structure and

277
00:21:37,460 --> 00:21:38,680
bonding and other

278
00:21:38,680 --> 00:21:41,010
characteristics of the material.

279
00:21:41,010 --> 00:21:43,920
The stress is something that we
we're taking and imposing

280
00:21:43,920 --> 00:21:47,400
on an innocent, unsuspecting
crystal, and that is

281
00:21:47,400 --> 00:21:48,450
something we do.

282
00:21:48,450 --> 00:21:53,860
And we can, by design, twist it,
shear it, stretch it, as

283
00:21:53,860 --> 00:21:56,330
hard as we like, provided
we don't break it.

284
00:21:56,330 --> 00:21:59,640
And so this is something that
is externally imposed on the

285
00:21:59,640 --> 00:22:05,860
crystal, and this is something
that's called a field tensor.

286
00:22:05,860 --> 00:22:07,740
We mentioned last time
the meaning of field.

287
00:22:07,740 --> 00:22:12,980
The field is a volume of space
and a field tensor is a space

288
00:22:12,980 --> 00:22:18,240
in which a value of the tensor
is defined as every point, x1,

289
00:22:18,240 --> 00:22:20,200
x2, x3, in the space.

290
00:22:20,200 --> 00:22:24,850
Just as an electric field is a
vector field and you define

291
00:22:24,850 --> 00:22:28,690
the electric field we as a
function of every point x1,

292
00:22:28,690 --> 00:22:30,820
x2, x3, in the space.

293
00:22:30,820 --> 00:22:33,600
So there's no reason whatsoever
that there need be

294
00:22:33,600 --> 00:22:37,860
restrictions on the tensor
that we apply.

295
00:22:37,860 --> 00:22:42,750
And when we can cross out that
interesting, but incorrect,

296
00:22:42,750 --> 00:22:44,000
supposition.

297
00:22:46,110 --> 00:22:51,430
But they can be a lot of
different forms, special

298
00:22:51,430 --> 00:22:58,880
forms, of stress tensors which
have special forms simply

299
00:22:58,880 --> 00:23:01,900
because of the nature
of what we do.

300
00:23:01,900 --> 00:23:05,190
And these have restrictions and
equalities that have no

301
00:23:05,190 --> 00:23:07,080
counterpart in property
tensors.

302
00:23:07,080 --> 00:23:13,960
For example, we can have the
cancer sigma 00000000, which

303
00:23:13,960 --> 00:23:17,170
is something we could never
have identically so for a

304
00:23:17,170 --> 00:23:19,170
property tensor.

305
00:23:19,170 --> 00:23:21,850
And what does this represent?

306
00:23:21,850 --> 00:23:34,340
This is a tensor for which sigma
IJ is 0, unless I equals

307
00:23:34,340 --> 00:23:37,140
J equals 1.

308
00:23:37,140 --> 00:23:40,010
So this is the only thing that's
going on, is a force

309
00:23:40,010 --> 00:23:42,220
per unit area along
the x1 direction.

310
00:23:48,830 --> 00:23:50,655
So this is a uniaxial stress.

311
00:23:57,420 --> 00:24:00,870
And that is a stress field
which we could apply on a

312
00:24:00,870 --> 00:24:03,850
piece of material very easily.

313
00:24:03,850 --> 00:24:08,860
Just screw a hook into the
material and hang a weight on

314
00:24:08,860 --> 00:24:13,330
the hook and you've got
uniaxial stress.

315
00:24:13,330 --> 00:24:15,990
We can also have something
that's a little harder to

316
00:24:15,990 --> 00:24:18,295
visualize, a biaxial stress.

317
00:24:22,890 --> 00:24:27,430
And this would be a stress that
had elements sigma 11,

318
00:24:27,430 --> 00:24:28,420
register 0--

319
00:24:28,420 --> 00:24:30,690
at least when we refer
to principal axes--

320
00:24:30,690 --> 00:24:35,970
0 sigma 2 2 0 0 0 0.

321
00:24:35,970 --> 00:24:38,190
When it's in diagonal
form, two of the

322
00:24:38,190 --> 00:24:39,715
diagonal terms are 0.

323
00:24:42,430 --> 00:24:46,850
That doesn't seem to make sense
because, if this is a

324
00:24:46,850 --> 00:24:50,030
force per unit area along x1,
and this is a force per unit

325
00:24:50,030 --> 00:24:53,700
area along x2, can't we add up
those vectors and say that

326
00:24:53,700 --> 00:24:57,250
that's a uniaxial stress?

327
00:24:57,250 --> 00:24:58,750
Depends on your coordinate
system.

328
00:24:58,750 --> 00:25:05,260
If you want an example of a
biaxial stress, imagine some

329
00:25:05,260 --> 00:25:08,360
firemen underneath a burning
building, and they're holding

330
00:25:08,360 --> 00:25:10,870
out a blanket for somebody
to jump into.

331
00:25:10,870 --> 00:25:14,310
You have one pair of firemen
pulling in this direction to

332
00:25:14,310 --> 00:25:17,300
keep the blanket taut, and
another pair of firemen are

333
00:25:17,300 --> 00:25:19,560
not quite so vigorous, and
they're pulling in an

334
00:25:19,560 --> 00:25:21,820
orthogonal direction.

335
00:25:21,820 --> 00:25:29,350
And those two forces, sigma 11
and sigma 22, create a biaxial

336
00:25:29,350 --> 00:25:31,310
stress event material.

337
00:25:31,310 --> 00:25:34,820
You could change the form of
the tensor by rotating the

338
00:25:34,820 --> 00:25:37,670
coordinate system, but
this is one example

339
00:25:37,670 --> 00:25:40,280
of a biaxial stress.

340
00:25:40,280 --> 00:25:43,650
The final general form, a
triaxial stress, obviously by

341
00:25:43,650 --> 00:25:48,380
extension, is going to be where
of all three diagonals,

342
00:25:48,380 --> 00:25:56,980
sigma 11, sigma 22, and sigma
33, are non-zero when you put

343
00:25:56,980 --> 00:25:59,690
the tensor in diagonal form.

344
00:25:59,690 --> 00:26:02,080
This would be a general
triaxial stress.

345
00:26:07,660 --> 00:26:12,130
And an example of that would
be when the jumping person

346
00:26:12,130 --> 00:26:16,560
lands on the blanket, there's
going to be a force that his

347
00:26:16,560 --> 00:26:20,230
or her mass exerts on the
blanket, and that would be a

348
00:26:20,230 --> 00:26:22,350
third sigma 3 3.

349
00:26:22,350 --> 00:26:24,532
So that would be a
triaxial stress.

350
00:26:27,870 --> 00:26:34,410
Let me give you a couple of
special cases of specialized

351
00:26:34,410 --> 00:26:35,660
stress tensors.

352
00:26:38,950 --> 00:26:46,250
And suppose I had a triaxial
stress, but all the diagonal

353
00:26:46,250 --> 00:26:48,215
terms were equal.

354
00:26:53,800 --> 00:26:56,980
That's something that I tried
a moment ago to hoodwink you

355
00:26:56,980 --> 00:27:00,380
into believing was required by
a cubic crystal, but suppose

356
00:27:00,380 --> 00:27:02,890
that was the form of
the stress tensor.

357
00:27:02,890 --> 00:27:03,570
What would that represent?

358
00:27:03,570 --> 00:27:05,331
AUDIENCE: [INAUDIBLE]

359
00:27:05,331 --> 00:27:08,860
PROFESSOR: That is something
that's going to be--

360
00:27:08,860 --> 00:27:10,270
yeah, very good.

361
00:27:10,270 --> 00:27:11,240
That's not a pressure.

362
00:27:11,240 --> 00:27:12,680
It's a dilation.

363
00:27:12,680 --> 00:27:16,990
But if I make it into a more
familiar form, put it here,

364
00:27:16,990 --> 00:27:22,900
minus P, minus P, minus P. It's
a force prevented area.

365
00:27:22,900 --> 00:27:24,530
That's where the stress is.

366
00:27:24,530 --> 00:27:27,730
If it's a hydrostatic pressure,
it's squeezing the

367
00:27:27,730 --> 00:27:32,070
volume element so it has to be,
by our definition, acting

368
00:27:32,070 --> 00:27:35,590
in a negative x1 direction
on a surface whose

369
00:27:35,590 --> 00:27:37,250
normal is plus x1.

370
00:27:37,250 --> 00:27:40,300
So minus P, minus P,
minus P, represents

371
00:27:40,300 --> 00:27:41,550
a hydrostatic pressure.

372
00:27:53,340 --> 00:27:56,150
And now, unless you've seen it
before, or you're reading the

373
00:27:56,150 --> 00:27:58,950
notes instead of listening to
me, let me give you a really

374
00:27:58,950 --> 00:28:00,460
strange-looking one.

375
00:28:00,460 --> 00:28:05,950
What is this stress tensor?

376
00:28:09,920 --> 00:28:15,740
Where sigma 22 is opposite
in sign but equal in

377
00:28:15,740 --> 00:28:18,192
magnitude to sigma 11.

378
00:28:18,192 --> 00:28:19,485
AUDIENCE: [INAUDIBLE]

379
00:28:19,485 --> 00:28:20,625
PROFESSOR: Yeah.

380
00:28:20,625 --> 00:28:21,890
Good.

381
00:28:21,890 --> 00:28:23,340
The way to see that

382
00:28:23,340 --> 00:28:25,210
geometrically is the following.

383
00:28:25,210 --> 00:28:32,970
What we're doing is, we're
pushing with a force per unit

384
00:28:32,970 --> 00:28:34,160
area in this direction.

385
00:28:34,160 --> 00:28:38,680
That would be sigma 11, and
then we squeeze in the

386
00:28:38,680 --> 00:28:40,310
orthogonal direction.

387
00:28:40,310 --> 00:28:44,290
And that is a sigma 22,
which is negative.

388
00:28:44,290 --> 00:28:53,400
If we rotate the solid by 90
degrees, then we've got this,

389
00:28:53,400 --> 00:28:57,040
plus this, acting on this face,
same forces per unit

390
00:28:57,040 --> 00:28:58,460
area as we have here.

391
00:28:58,460 --> 00:29:02,840
And that is a net force per
unit area like this.

392
00:29:02,840 --> 00:29:06,990
On the opposite surface, we've
got this force per unit area

393
00:29:06,990 --> 00:29:11,740
and this force per unit
area, component

394
00:29:11,740 --> 00:29:13,150
going in this direction.

395
00:29:13,150 --> 00:29:19,170
And that is going to be a force
that goes like this.

396
00:29:19,170 --> 00:29:22,680
That's going to be a force per
unit area like this, and

397
00:29:22,680 --> 00:29:26,740
similarly, this combined with
this is going to be a force

398
00:29:26,740 --> 00:29:29,540
acting in the reverse
direction.

399
00:29:29,540 --> 00:29:35,432
That would be, again, this with
this, and this would be

400
00:29:35,432 --> 00:29:36,920
this with that.

401
00:29:36,920 --> 00:29:41,730
So clearly, that is a body
that's being subjected to pure

402
00:29:41,730 --> 00:29:51,250
sheer, and if we rotated the
axes x1 and x2 by 45 degrees

403
00:29:51,250 --> 00:29:58,510
to an x1 prime and x2 prime
like this, and then

404
00:29:58,510 --> 00:30:05,020
transformed the tensor, it
would take the form 0

405
00:30:05,020 --> 00:30:12,850
sigma sigma 0 0.

406
00:30:12,850 --> 00:30:16,300
And then, what do we want now?

407
00:30:16,300 --> 00:30:17,740
This would be the only
thing we get.

408
00:30:20,580 --> 00:30:24,000
Except I should be careful to
put a sigma prime on this

409
00:30:24,000 --> 00:30:27,480
because, when we change axes,
these sigmas will not be of

410
00:30:27,480 --> 00:30:29,520
the same magnitude as
the original ones.

411
00:30:29,520 --> 00:30:32,620
There's going to be a square
root of 2 coming from cosine

412
00:30:32,620 --> 00:30:35,330
of 45 degrees in there.

413
00:30:35,330 --> 00:30:38,370
But, anyway, just rotating the
axes 45 degrees of as I've

414
00:30:38,370 --> 00:30:41,600
indicated schematically here
with vector sums is going to

415
00:30:41,600 --> 00:30:49,650
give me two off diagonal terms
equal in magnitude that would

416
00:30:49,650 --> 00:30:53,320
correspond to sigma
12 and sigma 21.

417
00:31:01,770 --> 00:31:06,540
So those are some special forms
of a stress tensor when

418
00:31:06,540 --> 00:31:11,150
you place it in a diagonalized
form.

419
00:31:11,150 --> 00:31:14,550
But let's conclude this part of
our discussion by observing

420
00:31:14,550 --> 00:31:20,720
that everything that we've said
about second-rank tensors

421
00:31:20,720 --> 00:31:23,410
holds for the stress tensor.

422
00:31:23,410 --> 00:31:38,660
For example, you can always
select new axes that put the

423
00:31:38,660 --> 00:31:40,530
tensor in diagonal form.

424
00:31:56,160 --> 00:31:59,266
So that's one general
property.

425
00:32:01,870 --> 00:32:15,720
You can define a stress quadric,
and we'll do that

426
00:32:15,720 --> 00:32:23,340
with the equation sigma
ij xi xj equals 1.

427
00:32:23,340 --> 00:32:23,830
Why not?

428
00:32:23,830 --> 00:32:25,080
It's a second-ranked tensor.

429
00:32:25,080 --> 00:32:28,250
We can take the elements and
construct a surface from it.

430
00:32:32,070 --> 00:32:38,410
This quadric will have the
properties of any other

431
00:32:38,410 --> 00:32:40,820
quadric for a second-rank
tensor.

432
00:32:40,820 --> 00:32:46,790
In particular, if we construct
the quadric, it's now for sure

433
00:32:46,790 --> 00:32:50,430
going to be able to be either
an ellipsoid, an hyperboloid

434
00:32:50,430 --> 00:32:53,950
of one or two sheets, or in this
case, even an imaginary

435
00:32:53,950 --> 00:32:58,340
ellipsoid for a stress that's
everywhere compressive.

436
00:32:58,340 --> 00:33:00,600
The quadric could have
its diagonal

437
00:33:00,600 --> 00:33:04,110
elements all negative.

438
00:33:04,110 --> 00:33:07,690
The radius of the stress
quadric in a particular

439
00:33:07,690 --> 00:33:22,540
direction is going to be the 1
over the square root of the

440
00:33:22,540 --> 00:33:25,550
property in that direction.

441
00:33:25,550 --> 00:33:29,560
What does that mean, stress
in a given direction?

442
00:33:29,560 --> 00:33:31,990
Well, let's again remember
how we have

443
00:33:31,990 --> 00:33:34,470
defined the stress tensor.

444
00:33:34,470 --> 00:33:40,130
A force per unit area has three
components, k1, k2, k3,

445
00:33:40,130 --> 00:33:45,060
and they're equal to
sigma ij times lj.

446
00:33:47,830 --> 00:33:51,270
So the value of the property in
this direction is going to

447
00:33:51,270 --> 00:33:54,890
be the value of the property
in the direction defined by

448
00:33:54,890 --> 00:34:01,880
[? elsa ?] j and the values of
[? elsa ?] j are components of

449
00:34:01,880 --> 00:34:06,010
the unit vector that point
in that direction.

450
00:34:06,010 --> 00:34:11,389
So if this is the direction of
k, [? off ?] like this, the

451
00:34:11,389 --> 00:34:20,400
value of this quadric in this
direction is going to be the

452
00:34:20,400 --> 00:34:24,810
value of the part of k that's
parallel to the radius vector

453
00:34:24,810 --> 00:34:27,489
over the magnitude of
the radius vector.

454
00:34:27,489 --> 00:34:31,350
But the [? radius ?] vector is
a unit vector, so it's simply

455
00:34:31,350 --> 00:34:36,650
going to be equal
to k parallel.

456
00:34:36,650 --> 00:34:41,500
And if k parallel is the part
of the vector that is along

457
00:34:41,500 --> 00:34:46,330
the surface normal, that is
going to be the component of k

458
00:34:46,330 --> 00:34:47,885
which is purely tensile.

459
00:34:50,770 --> 00:34:55,110
This is a force per unit area,
has parts that are, components

460
00:34:55,110 --> 00:34:59,030
that are normal to the
surface normal.

461
00:34:59,030 --> 00:35:00,730
Those are shear components.

462
00:35:00,730 --> 00:35:03,890
The component that's directly
along the normal to the

463
00:35:03,890 --> 00:35:07,330
surface is going to be a purely
tensile component.

464
00:35:07,330 --> 00:35:11,940
And, therefore, the radius
of the quadric in a given

465
00:35:11,940 --> 00:35:25,220
direction gives us the tensile
component of the force density

466
00:35:25,220 --> 00:35:28,575
vector k that is transmitted
across the face--

467
00:35:46,850 --> 00:35:48,605
so let's not say face,
but interface--

468
00:35:52,340 --> 00:36:08,400
whose normal is in the
direction of R.

469
00:36:08,400 --> 00:36:10,020
So this is the radius vector.

470
00:36:10,020 --> 00:36:15,310
That's the direction of the
normal to a surface and 1 over

471
00:36:15,310 --> 00:36:18,190
the radius of--

472
00:36:18,190 --> 00:36:20,830
The radius of the quadric in
that direction is going to be

473
00:36:20,830 --> 00:36:24,740
1 over the square root of the
tensile component of stress

474
00:36:24,740 --> 00:36:28,110
that's transmitted across an
interface in this direction.

475
00:36:49,170 --> 00:36:51,000
How are we doing on time?

476
00:36:51,000 --> 00:36:52,750
Any questions at this point?

477
00:37:06,061 --> 00:37:12,470
AUDIENCE: Can you explain how
you know that's [INAUDIBLE]?

478
00:37:12,470 --> 00:37:16,945
PROFESSOR: Remember what our
radius normal property said.

479
00:37:19,820 --> 00:37:29,650
If we have a tensor like good
conductivity, where a current

480
00:37:29,650 --> 00:37:33,835
density is related to an
applied electric field,

481
00:37:33,835 --> 00:37:36,500
[? essa ?] j and we've--

482
00:37:36,500 --> 00:37:38,060
unfortunately, that's
segment 2.

483
00:37:38,060 --> 00:37:41,563
This is the conductivity
tensor.

484
00:37:41,563 --> 00:37:47,440
What we said is that, if we
take the components of the

485
00:37:47,440 --> 00:37:52,830
tensor, sigma ij, and construct
the surface sigma ij

486
00:37:52,830 --> 00:37:58,690
xij equals unity, that will be
some sort of quadratic form

487
00:37:58,690 --> 00:38:03,490
that has the property that the
radius in any given direction

488
00:38:03,490 --> 00:38:12,740
is 1 over the square root of
the value of the property,

489
00:38:12,740 --> 00:38:14,820
sigma conductivity
in this case.

490
00:38:14,820 --> 00:38:15,760
Bad case.

491
00:38:15,760 --> 00:38:17,890
I shouldn't have picked
things where the

492
00:38:17,890 --> 00:38:19,620
sigma is also involved.

493
00:38:19,620 --> 00:38:22,640
And the value of the property
in the given direction is

494
00:38:22,640 --> 00:38:26,560
going to be the parallel
component of what happens over

495
00:38:26,560 --> 00:38:30,770
the magnitude of what you do.

496
00:38:30,770 --> 00:38:34,030
So in the case of electrical
conductivity, if we apply a

497
00:38:34,030 --> 00:38:38,160
field in this direction, the
value of the property is going

498
00:38:38,160 --> 00:38:44,570
to be that part of the current
flow, which is parallel to the

499
00:38:44,570 --> 00:38:47,590
electric field, divided
by the magnitude of

500
00:38:47,590 --> 00:38:50,600
the electric field.

501
00:38:50,600 --> 00:38:56,140
So that is a relation that you
know by heart and have come to

502
00:38:56,140 --> 00:38:59,360
not only know, but to love.

503
00:38:59,360 --> 00:39:01,070
So what are we saying here?

504
00:39:01,070 --> 00:39:05,730
For the stress tensor, we can
take the elements of the

505
00:39:05,730 --> 00:39:08,940
stress tensor and construct
a quadric.

506
00:39:08,940 --> 00:39:12,990
And the thing that's related
here is a force per unit area

507
00:39:12,990 --> 00:39:19,530
that we're applying to the
surface of the crystal, and

508
00:39:19,530 --> 00:39:23,470
that's transmitted through the
volume of the crystal by the

509
00:39:23,470 --> 00:39:27,520
relation that we have defined
as the stress tensor.

510
00:39:27,520 --> 00:39:29,850
So, unfortunately,
again, sigma Ij.

511
00:39:29,850 --> 00:39:33,560
And now what is involved are the
direction cosines of the

512
00:39:33,560 --> 00:39:35,580
normal to the surface.

513
00:39:35,580 --> 00:39:38,770
So what is the applied vector?

514
00:39:38,770 --> 00:39:43,170
The applied vector is a unit
vector, and this is the normal

515
00:39:43,170 --> 00:39:44,420
to the surface.

516
00:39:48,800 --> 00:39:52,780
And the surface in question
would be a surface like this.

517
00:39:52,780 --> 00:40:00,490
The thing that results is
a force per unit area k.

518
00:40:00,490 --> 00:40:06,350
So what we're doing is taking
k, the value of

519
00:40:06,350 --> 00:40:07,770
sigma, in that direction.

520
00:40:07,770 --> 00:40:11,060
A scalar quantity will be the
part of k that's parallel to

521
00:40:11,060 --> 00:40:13,060
the normal vector,
divided by the

522
00:40:13,060 --> 00:40:18,700
magnitude of the unit vector.

523
00:40:18,700 --> 00:40:21,840
And that's, a magnitude
of n is 1, so

524
00:40:21,840 --> 00:40:23,910
that's just k parallel.

525
00:40:23,910 --> 00:40:25,710
So this is the surface.

526
00:40:25,710 --> 00:40:27,260
This is its normal.

527
00:40:27,260 --> 00:40:31,250
And what the radius of the
quadric is going to give us is

528
00:40:31,250 --> 00:40:37,010
that part of the force density,
which is the force

529
00:40:37,010 --> 00:40:40,150
density vector resolved onto
the normal vector.

530
00:40:40,150 --> 00:40:42,230
And this is a tensile.

531
00:40:42,230 --> 00:40:46,750
This is the part of the
tensile stress that's

532
00:40:46,750 --> 00:40:47,310
transmitted.

533
00:40:47,310 --> 00:40:49,770
This is the total stress
that has a sheer part

534
00:40:49,770 --> 00:40:50,790
and a tensile part.

535
00:40:50,790 --> 00:40:53,420
The radius of the quadric is
going to give you the part

536
00:40:53,420 --> 00:40:56,930
that's transmitted across the
surface in this orientation.

537
00:40:56,930 --> 00:40:59,460
That is purely a
tensile force.

538
00:41:04,590 --> 00:41:05,760
That make sense?

539
00:41:05,760 --> 00:41:07,460
I should have done it
without sigmas being

540
00:41:07,460 --> 00:41:08,710
present in both tensors.

541
00:41:11,670 --> 00:41:16,900
So everything you want to know
about stress that's applied to

542
00:41:16,900 --> 00:41:22,510
a solid is contained in a
second-rank tensor that is

543
00:41:22,510 --> 00:41:26,940
symmetric and obeys all of the
characteristics that we

544
00:41:26,940 --> 00:41:29,940
derived earlier for property
tensors, even though this is a

545
00:41:29,940 --> 00:41:31,760
field tensor, and not
a property tensor.

546
00:41:42,690 --> 00:41:45,870
Let us then turn to the next
thing that I want to discuss,

547
00:41:45,870 --> 00:41:48,930
again something you've
heard before.

548
00:41:48,930 --> 00:41:52,080
But we'll do it within the
context of the tensor algebra

549
00:41:52,080 --> 00:41:53,850
that we've developed before.

550
00:41:53,850 --> 00:41:56,905
And this is the concept
of strength.

551
00:42:03,930 --> 00:42:11,020
And let me introduce
the subject by a

552
00:42:11,020 --> 00:42:13,650
one-dimensional analog.

553
00:42:13,650 --> 00:42:18,250
Suppose we have a very thin
elastic band fastened to some

554
00:42:18,250 --> 00:42:22,535
immovable surface, and this
will be the direction x.

555
00:42:26,090 --> 00:42:30,950
And then we pull on it.

556
00:42:30,950 --> 00:42:34,540
And here again, we have to be
complete and note that there

557
00:42:34,540 --> 00:42:37,910
are several different
kinds of behavior.

558
00:42:37,910 --> 00:42:44,020
One type of behavior is a case
where the elastic band

559
00:42:44,020 --> 00:42:51,820
stretches, and if there's some
point P on the initial state

560
00:42:51,820 --> 00:42:57,000
that point P will move to
a point P prime that is

561
00:42:57,000 --> 00:43:05,930
displaced from the original
location by a vector U.

562
00:43:05,930 --> 00:43:12,790
There's another point Q part
way along the elastic band.

563
00:43:12,790 --> 00:43:16,670
If the thing that we have
fastened to the elastic band

564
00:43:16,670 --> 00:43:20,100
slipped, and there's no
deformation at all, Q would be

565
00:43:20,100 --> 00:43:24,080
displaced by the same amount U
if there was just translation

566
00:43:24,080 --> 00:43:27,400
of the elastic because it wasn't
fastened down solidly.

567
00:43:27,400 --> 00:43:31,160
But if it's stretching here
between this point and this

568
00:43:31,160 --> 00:43:33,810
point, it's going to stretch
up here, too.

569
00:43:33,810 --> 00:43:38,320
And, in general, it will move
over to some point Q prime,

570
00:43:38,320 --> 00:43:42,490
and this displacement vector is
going to be the original U

571
00:43:42,490 --> 00:43:52,620
plus delta U. And what we care
about in defining strength is

572
00:43:52,620 --> 00:43:55,370
not the absolute coordinates,
but the interval

573
00:43:55,370 --> 00:43:57,770
between these points.

574
00:43:57,770 --> 00:44:07,530
This separation is U, and this
separation is U plus delta U.

575
00:44:07,530 --> 00:44:11,610
Now this is a common, but
not a necessarily

576
00:44:11,610 --> 00:44:13,360
unique, sort of behavior.

577
00:44:13,360 --> 00:44:17,280
This is something where the
displacement does not change

578
00:44:17,280 --> 00:44:19,465
along the length of
the elastic bands.

579
00:44:19,465 --> 00:44:21,660
This is something that's
called homogeneous

580
00:44:21,660 --> 00:44:22,910
deformation.

581
00:44:30,840 --> 00:44:35,260
And this is characterized by a
displacement view that varies

582
00:44:35,260 --> 00:44:39,680
linearly with position along
the elastic band.

583
00:44:39,680 --> 00:44:40,930
Does something like this.

584
00:44:46,110 --> 00:44:50,410
It's possible, in a [? grade ?]
of material, if

585
00:44:50,410 --> 00:44:53,960
you'd like a real example,
that U, as a function of

586
00:44:53,960 --> 00:44:56,230
distance along this
one-dimensional object, does

587
00:44:56,230 --> 00:44:57,480
something like this.

588
00:45:00,410 --> 00:45:01,660
Why not?

589
00:45:04,550 --> 00:45:08,540
That's what might happen
for example in--

590
00:45:08,540 --> 00:45:11,230
that's almost an oxymoron, a
one-dimensional elastic band

591
00:45:11,230 --> 00:45:15,020
whose cross-sectional area
changes with distance in that

592
00:45:15,020 --> 00:45:18,110
one dimension.

593
00:45:18,110 --> 00:45:20,840
So this is homogeneous
deformation, as opposed to

594
00:45:20,840 --> 00:45:23,590
inhomogeneous deformation.

595
00:45:23,590 --> 00:45:30,490
There's another type of
behavior, and this is apparent

596
00:45:30,490 --> 00:45:37,870
if you place a thin piece of
metal or plastic, let's say,

597
00:45:37,870 --> 00:45:43,770
between the pair of grips in the
Instron machine, and after

598
00:45:43,770 --> 00:45:46,090
deformation, it does something
like this.

599
00:45:48,720 --> 00:45:52,580
There's an inhomogeneous
deformation, which is to

600
00:45:52,580 --> 00:45:56,120
referred to colloquially
as necking.

601
00:45:56,120 --> 00:45:59,740
A very good example of that is
what happens when you grab

602
00:45:59,740 --> 00:46:05,180
hold of some cheap plastic
shrink wrap that has been

603
00:46:05,180 --> 00:46:07,880
shrunk around a package that
has been mailed to you.

604
00:46:07,880 --> 00:46:10,220
And you want to get it
off, so you grab it.

605
00:46:10,220 --> 00:46:12,220
And it doesn't deform
elastically.

606
00:46:12,220 --> 00:46:14,780
What happens is that it necks
in places where there's a

607
00:46:14,780 --> 00:46:16,220
stress concentration.

608
00:46:16,220 --> 00:46:18,550
And you get something
exactly like this.

609
00:46:18,550 --> 00:46:22,500
And if we were to indicate that
sort of behavior, U as a

610
00:46:22,500 --> 00:46:26,180
function of position would be
one U, and then it would

611
00:46:26,180 --> 00:46:27,720
increase very rapidly.

612
00:46:27,720 --> 00:46:29,440
The [? UDX ?]

613
00:46:29,440 --> 00:46:33,970
would be very much larger than
the places which had uniform

614
00:46:33,970 --> 00:46:37,550
elastic deformation, so it would
be a nonlinear behavior

615
00:46:37,550 --> 00:46:38,880
of this sort.

616
00:46:38,880 --> 00:46:40,130
And that would be necking.

617
00:46:44,310 --> 00:46:46,730
If you've never noticed that
sort of behavior before, I

618
00:46:46,730 --> 00:46:49,510
give you an assignment tonight
of going home and getting a

619
00:46:49,510 --> 00:46:53,960
piece of Saran wrap from your
kitchen drawer and grabbing

620
00:46:53,960 --> 00:46:55,210
hold of it and pulling it.

621
00:46:55,210 --> 00:46:59,150
And it deforms non-uniformly in
a fashion that's indicative

622
00:46:59,150 --> 00:47:00,400
of necking.

623
00:47:07,690 --> 00:47:09,530
OK, it's just about five of.

624
00:47:09,530 --> 00:47:11,300
Why don't I stop there.

625
00:47:11,300 --> 00:47:16,440
And what I would like to do next
is to cast this relation,

626
00:47:16,440 --> 00:47:19,600
which is a one-dimensional
stream, into a

627
00:47:19,600 --> 00:47:21,550
three-dimensional situation.

628
00:47:21,550 --> 00:47:26,460
And we'll look at some of the
properties of the description

629
00:47:26,460 --> 00:47:28,620
of a body that's been deformed
in three dimensions.