>> Now as we move on into
Chapter 7, we're going to look

at what's going on when
materials are deformed.

The title of the chapter
is about dislocations

and strengthening mechanisms.

In Chapter 6 we learned about
the basic mechanical behavior

of materials; what happens
when you stress the material

and it starts to strain.

We learned about elastic
strain and plastic strain.

What we need to do now is,
if we want to understand how

to strengthen the material,
how to make it stronger,

we need to know what's
happening inside of it

when it actually does
plastically deform.

So we're going to look

at how atoms rearrange
themselves under a stress.

And once we understand that,

then we can understand how
we can do different things

to make one material stronger
than another material.

The basic concept behind
plastic deformation has to do

with dislocations, and we first
saw dislocations in Chapter 4

when we talked about
defects in crystals.

A dislocation is a
line defect where are,

we have basically planes of the
atoms are not stacking properly.

And it's basically the
movement of these dislocations

which is what causes atoms
to rearrange themselves and,

thus, plastically deform.

We call the motion
of dislocations slip.

As a dislocation slides through
a material, we call it slip.

As a dislocation moves
through a material,

basically we have a shearing
of atoms, of one plane

of atoms relative to another.

Here's a little diagram where
we see a dislocation, this,

it starts off here
in column or plane A,

where we have a half-plane
of atoms.

This is an edge dislocation.

We have an applied stress
where at the top portion

of the crystal we have
forces pushing to the right,

and on the bottom side we have
forces pushing to the left.

And what happens is, essentially
what happens is the bonds

between the atoms here
along plane B break,

but just temporarily,
to be replaced

by a bond along plane A.

So basically all these
atoms along the bottom half

of plane B break off from the
rest of plane B and then form,

connect with the half,
extra half-plane over here

at plane A. Now the
dislocation's moved.

Now the dislocation is here
on plane B. That continues.

Now we, the bonds here on
plane C could break and form

with plane B and so on.

Eventually that dislocation
keeps moving across the crystal

until it's moved all the way
across and we have one unit step

of slip where all the atoms
above the dislocation line,

plane, have moved to
the right relative

to the atoms on the bottom.

This dislocation motion slips
along a certain slip plane,

that's the plane that
it's sliding around in,

and it's happening in a
certain slip direction.

So in this case, it'd
be in the direction

that the shear occurred.

The motion of a dislocation
to cause shear

in a crystal is compared a
lot of times to the motion,

or the movement,
of the caterpillar.

The caterpillar does
not move by moving all

of its legs all at once.

Instead it bunches up its
back legs and has kind

of a little hump in it,
and then that little hump

in the caterpillar
kind of moves forward

until it moves all the
way across the caterpillar

and the caterpillar has
moved one step forward.

In the same way, a dislocation
basically is the hump

of the material where one
little extra half-plane

of atoms is squished together.

As that squished-together
extra half-plane of atoms moves

across the crystal,
the atoms are moving

until it gets all the way across
that crystal and the entire set

of atoms above the
slip plane have moved.

Now, to cause a dislocation
to move,

we have to have a shear stress.

We have to have a stress such
that we have a force applying

in one direction
above the slip plane

and another direction
below the slip plane.

It's easiest always to
look at edge dislocations.

Edge dislocations are these
extra half-planes of atoms.

And basically they
move in a direct --

we have this one
unit step of slip.

Screw dislocations actually
would move perpendicularly

relative to the applied stress,

the stresses being
applied this way,

you're essentially
tearing the crystal apart.

And so the tear is
actually going

into the crystal
perpendicularly.

However, notice that the unit
step of slip is still the same.

The atoms are still shearing
to the right; however,

the screw dislocation
is moving into the page

or perpendicular to
the shear stress.

So although the dislocations
move in different directions,

the actual deformation
is the same

for both types of dislocations.

I brought up the idea

of dislocation density
also in Chapter 4.

This is a measure of the
amount of dislocations

in a volume of material.

And this is going
to be important,

as we look at strengthening
of metals,

that we understand
what this all means.

The dislocation density is
given in units of millimeters

to the negative 2 power,

which is a very strange
unit at first to look at.

But it makes a bit
more sense if you think

that basically that's
just a simplified form

of saying millimeters
per cubic millimeter.

Obviously if you simplified
that, canceled out one

of the units of millimeters,
you'd have 1

over millimeters squared,

which is just millimeters
to the negative 2.

However, this form makes
some sense in what it means.

It means that for a given
cubic millimeter of material,

this is the total
millimeters, the total length

of dislocations present
in that material.

So if I said there's a
dislocation density of 10

to the 7th millimeters to the
negative 2, that's the same

as saying there's 10 to the
7th millimeters of dislocations

in every 1 cubic
millimeter of material.

So the 10 to the 7th gives
you the length of dislocations

in millimeters for 1 cubic
millimeter of volume.

Now, this dislocation
density will change.

It will increase as
you deform a material.

Not only are the dislocations
moving, but they're multiplying;

more of them are being created.

So the dislocation
density increases

as the materials are deformed.

They can go up to about
10 to the 9th or 10

to the 10th millimeters to the
negative 2 for deformed metals.

You can actually heal a material

or decrease its dislocation
density

by heat treating the material.

You can get it down to around 10

to the 5th millimeters
to the negative 2.

They all sound kind of
big, but that's, you know,

by orders of magnitude smaller
than the dislocation density

of the deformed material.

The lower the dislocation
density,

basically the more
perfect the crystal is,

the less defects there
are in that crystal.

Now, remember, about
dislocations,

is that they cause a
strain, they create a strain

in the lattice or
in the crystal.

For an edge dislocation,
above that edge dislocation

where we have this extra
half-plane of atoms,

we have a very crowded region

where all these extra
atoms are crowded together.

This is an area of compression

where the atoms are
compressed into each other.

Below that edge dislocation,
below that half-plane of atoms,

the atoms are actually
stretched apart

because they are missing an
extra half-plane of atoms,

so they're being pulled apart by
that extra half-plane of atoms.

So that's an area of tension.

That's an edge dislocation.

For screw dislocations,
there's also distortions,

but they're shear distortions,

so it's a little
harder to picture.

But instead of being just
straight up pushed together

or stretched apart,
they're kind of twisted

around in the crystal.

What's the effect of
these lattice strains?

Well, since there's regions
of crowding and regions

of being stretched apart,

different dislocations can
actually either repel each other

or attract each other.

Repulsion is shown here
on the upper right.

When we have two
edge dislocations

that are both aligned
in the same way,

where they both have the
extra half-plane of atoms

above this same slip
plane, they're going

to repel each other, because
basically you have this crowded

region here, and you have
another crowded region here.

And for those, if those
dislocations tried to get close

to each other, they'd
be even more crowded.

So they're going to kind of
push away from each other.

Effectively there's
a force between,

from those two dislocations
pushing each other away.

However, if one dislocation was
upside down from the other one

so that the extra
half-plane in this one was

above the slip plane,
the extra half-plane

in this one is below the
slip plane, now I have,

as I bring them together,

this area of being crowded
is getting close to an area

where it's being
stretched apart.

And so they actually get
attracted to each other

because basically
the crowded nature

of this dislocation can
be eased by getting close

to the stretched out
nature of this dislocation

and vice versa on the bottom.

And actually what can happen,
if they completely line

up with each and are
on the same slip plane,

the two extra half-planes
of atoms can line up

and the dislocation would
be annihilated or disappear

because now you have one nice,
instead of an extra half-plane

of atoms here and an extra
half-plane of atoms over here,

the extra half-planes
are together

and you will have one
complete solid plane.

Okay, as we analyzed the
movement of dislocations,

we talked about their
slip systems.

A slip system is
defined by the slip plane

that the dislocation is moving
on, and its slip direction,

the direction in
which it's moving.

So, now, different
crystals are going

to have different slip systems

on which the dislocations
are going to move.

Basically dislocations
are going to move

so that the actual distortion

on the crystal is
kept to a minimum.

The implication of this is that
the preferred slip systems,

the way that the
dislocations would want to move,

are on slip planes
that are highly packed

and slip directions that
are also closely packed.

So if we looked at some of our
common crystals, FCC crystals,

the close packed planes of those
FCC crystals are the 111 planes.

Remember the 111 planes?

If I look down here, that
means it intersects the y

and z axis at 1, 1, and 1.

So it's, this plane right
here is the 111 plane,

they're shaded.

That's a close-packed plane.

It means the atoms are
completely touching each other,

just like in this drawing here.

The closest directions
would be the directions

where the nearest neighbors
are, right along the diagonals.

So going from A to B to
D or from A to C to F,

as shown in this figure, are
the nearest neighbor directions.

Those would be in the family
of 110, need the combination

of 110, 101, or negatives
of those are fine.

If we looked at this more
deeply, we'd say, well,

of the close-packed
planes, the 111 planes,

we've got four different planes.

You've got 111, you've got
bar 111, you've got 1-bar 11,

and you have 11-bar 1.

If you did two negatives
or three negatives,

they would just be the
same plane as these

because they would
just be the opposites.

So there's four unique slip
planes in an FCC crystal.

Each one of those has three
unique slip directions.

So if I look at this 111 plane
in the drawing, I have --

basically the three sides

of this triangle are three
unique slip directions.

So if there's three on each
one -- there's three here,

three here, three
here, and three here --

that's a total of 12
different slip systems.

And that's a lot.

Basically what that means is,
is each dislocation has lots

of different paths
that it can take.

So since there's lots
of different paths

for dislocation motion, that
means that the material ends

up being very ductile,

because it's very easy for
the dislocations to move,

which means it's easy for
plastic deformation to occur,

which means that the
material is going

to deform a lot before
it finally breaks.

Remember, ductile means a lot of
strain before a final fracture.

So by having lots of pathways
for dislocations to move,

you end up with a
very ductile material.

And they're moving on
these defined slip systems.

If you look at BCC crystals,

they do not have the
close-packed planes

that FCC crystals have.

They have some that are
closer packed than others.

The closest-packed plane is not
a perfectly close-packed plane,

but it would be of
the 110 family,

would be the closest-packed
plane,

and its most preferred slip
system is the 110 family

of directions.

There's some other
ones also available.

If you look down
at this table here,

these are common slip systems

for different crystal
structures.

You'll see in body
centered cubic

for several different
metals that are BCC,

the slip plane would
be in the 110 family.

The slip direction would be
one of the 111 directions.

But there are other possible
slip planes for BCC crystals;

211, 321 are also possible and
these are their slip directions.

Up here notice this is
the phase centered cubic

that we already talked about.

The 111 slip plane and
the 110 slip direction.

What you can see
from this table is

that the BCC crystals
have anywhere from 12

to 24 slip systems, depending
on which ones they have,

so once again, there's
lots of options,

lots of different pathways
that dislocations can travel,

so BCC metals are also
generally very ductile.

Now, in contrast to
this are the HCP.

Remember, that's the
hexagonal closest packed.

A hexagonal crystal does not
give very many slip systems.

If we look at these numbering
systems are these hexagonal

coding systems that we
didn't really get into.

But you can see over here
very few slip systems

for the hexagonal
close-packed crystals.

So that means if there's a
small number of slip systems,

then these crystals
tend to be very brittle

because there's just
not a lot of pathways

that dislocations can take.

So if one pathway gets blocked,

and we'll see why they
might get blocked later on,

they've got no other
pathways to go, possibly.

And so dislocation
motion doesn't occur,

so at some point the material's
just going to snap in two;

it's just going to fracture,
instead of ductilely deforming.

The Burgers vector we
talked about when we talked

about dislocations is important

when it comes to
dislocation slip.

The Burgers vector indicates,

in this context it
indicates the direction

that the slip is occurring

and the magnitude
of the deformation.

So, remember, it's a vector

so it has both direction
and magnitude.

The direction tells
you the direction

that deformation is occurring,
that the slip's occurring,

and then the magnitude

of the vector tells you how
much you've actually sheared

the atoms.

Here's some formulas.

You don't have to worry
about the HCP so much,

since we're not really
doing that, but you do want

to be able to use these two.

But the Burgers vector for
either FCC or BCC is equal

to the unit cell edge length --

now remember, Chapter 3 we
learned formulas for both FCC

and BCC on calculating
the length,

edge length of a unit cell --

divided by 2, multiplied
by basically this vector.

For FCC they're all
in the 110 direction,

so it'd be that vector.

For BCC it's in the
111 direction.

So real quick example.

Let's say we had an FCC atom,
and let's say we were given

that its atomic radius
was .124 nanometers.

Well, first thing we'd
want to do is figure

out the unit cell length.

So for FCC that would be
2R times square root 2,

which in this case would
give us .351 nanometers.

Once I see that, then I'd say,
okay, the Burgers vector then

for an FCC crystal, for
this specific FCC crystal,

would be .351 divided by
2 times the 110 vector.

So in our case it would
be .175 times 110.

If you want to put
that in vector form,

that means we'd have .175i
plus .175j plus 0k where i, j,

and k are indicating
the magnitudes

in the x, y, and z directions.

So what does this all mean?

This indicates the
components of deformation

in the x, y, and z directions.

So for this particular
Burgers vector,

we're not getting any
deformation in the z direction.

All the deformation's in
the x and y directions.

Now, it's simplest to
see how this is all going

to occur in a single crystal.

So let's first assume we have
one big solid chunk that's all

one single crystal, all
aligned in the same direction.

Then we'll talk about
polycrystals.

We need to have a
shear stress in order

to cause the dislocations
to move.

We typically, though, are
pulling the material in tension.

Well, a tensile stress can
create a shear stress along a

direction that's at some angle
to that applied tensile stress.

What we can do is we can
calculate what's called the

resolved shear stress.

That's the amount of
shear stress we get

in a certain direction,

so tau sub r represents
the shear stress,

the resolved shear stress
in a certain direction,

and it's equal to the tensile
stress times cosine phi times

cosine lambda.

You can see in the diagram
here that phi is the angle

between the tensile direction

and the normal to the
slip plane right here.

So here's the slip plane.

We think of a normal
vector coming off of that,

and we think, what's the angle

between the tensile
direction and that normal?

Lambda, notice, is the angle
between the tensile direction

and the actual slip
direction, the direction

that the actual dislocation's
slipping in.

Now, so we pull with a certain
tensile stress that's equivalent

to getting a certain
amount of shear stress

in a certain direction.

In order to get the
dislocation to move,

if that resolved shear
stress is greater

than a critical resolved
shear stress,

tau with the subscript crss,
then you get dislocation motion.

So what you can do
is you can say, okay,

here's how much shear
stress I need

to get the dislocation to move.

Based on my tensile
stress and my geometry,

here's how much shear
stress I get.

If this is at least
equal to this,

then the dislocation
is going to move.

Some slip systems are
going to be oriented

so that they're more
favorable than others.

The ones that are most favorable
are the ones that are going

to get the highest
resolved shear stress.

And so for a given tensile
stress, that means the ones

that have the highest value of
the product of these cosines.

That actually has a term;
it's called the Schmid factor.

Now, if you pull actual
single crystals, you do notice

that dislocation occurs
on certain slip planes.

You'll have a certain
preferred slip plane

where dislocations
start to move.

And here's an actual
picture of a single crystal.

And it looks like, you know,
all these are little slips

that have occurred
all on the same plane

or parallel to that same plane.

And so we get all these steps

because of all the different
dislocations that have slipped

through that material.

Now, to be able to use this
formula, you need to be able

to find these cosines
and those angles.

So one more formula here.

Define the angle
between two different

crystallographic directions.

We have this big ugly-looking
formula here and basically what

that means is if I have one
direction, which is u1, v1, w1,

and a second direction which
is u2, v2, w2, then I plug them

into this thing to
figure out the cosine

of the angle between them.

Let's try an example.

Let's say our two
directions are 110 and 200.

Plugging this formula, I'd say,
okay, then cosine of theta,

the angle between
them, on the top,

I'd simply multiply
the matching components

and then add those
different products together.

So 1 times 2 would be
2; 1 times 0 would be 0;

and 0 times 0 would be 0.

So notice how I just matched
up the matching indices

and multiplied them together
and then added up the products.

On the bottom I've got this
square root of a product.

The first is the sum of the
squares of the first set

of indices, so it'd be 1 squared
plus 1 squared plus 0 squared.

And the second product
would be the same thing

for the second one, 2 squared
plus 0 squared plus 0 squared.

So our cosine theta for this
example is going to be 2

over the square root
of 2 times 4.

So we're going to get
cosine theta equals 2

over the square root of 8.

2 over the square
root of 8 works

out to be approximately .7071.

A lot of times you can just
get that far and just use

that in the formula because
you don't really need

to know the angle itself.

You just need to know
the cosine of the angle.

But if you wanted to take
this all the way and figure

out what this angle is,
you'd take the inverse cosine

of that number and find that,

for this example,
it's 45 degrees.

So that's the angle
between the 110 direction

and the 200 direction.

We'll do more with this
as you do an example

at the end of this section.

So that's talking
about single crystals.

What about polycrystals?

Those are much more common,
when you have a material

with many different grains or
differently oriented crystals.

Well, the random orientation

of different grains
complicates things.

Each individual grain is going

to have slip along its
favorable directions.

Here's a micrograph
where you can see this,

where you can see here's the
slip direction for one grain

and the adjacent grain
has slip direction going

in this direction.

This one, so you can kind
of see the different lines.

These are the different ways
each grain wanted to slip.

They're not all in
the same direction.

So they're kind of, what
happens is because these are,

have borders between them and
these borders kind of have

to move together,

each individual grain
is then constrained.

This one's constrained
by the deformation

of the neighboring grain.

This one's constrained
by that one

and its next neighboring grain.

So what happens is polycrystal
materials have higher yield

strengths than single
crystals because it's harder

for dislocations to move.

They start moving, but then
they're inhibited by the fact

that they have these
grain boundaries.

So polycrystals are stronger

or they have a higher yield
strength than single crystals.

And then the more
different grains you have,

the stronger it's going to be.

So just think of that.

It means that smaller
grain sizes,

which leads to also more
grains, is going to lead

to higher yield strengths.

Here's a couple of other
micrographs of this.

When you do deform a
polycrystalline material,

basically the grains
go from being equiaxed,

meaning that they're all
kind of about the same shape,

they're generally, you know,

there's no one preferred
direction

where they're longer
than the others.

After deformation,
they become elongated.

These have been stretched
in, lengthwise,

they've been stretched
out this way

and so they've been elongated.

They get stretched out.

One final comment
here about mechanisms

of deformation is there
is another mechanism,

but it's not as common.

I should note that
right from the start.

It's not as common
as dislocation slip.

There's a different mechanism.

It doesn't involve dislocations.

It's called twinning, where if
we have applied shear stress,

we have a number of atoms
that all shift together.

So these atoms kind of shift
positions, and they end

up forming, they form a
mirror plane or a twin plane.

Here is, here is the twin
plane they show right here,

and notice that the
black ones are showing

where they've deformed into.

Each of these now, basically we
have a mirror image on this side

of the plane as we have
over here after deformation.

Likewise, basically there's
another twin plane over here.

So we have this one region of
deformation right in between.

That's the region
where we've had atoms

that have kind of shifted.

Instead of having atoms
that are all perfectly

in the same crystal orientation,

between these two lines we
basically have this twin portion

that's formed.

And all the atoms
have shifted over,

kind of created a mirror
image of the adjacent atom

and we have deformation
in that region.

Over here shows you
kind of the difference

between dislocation
slip, which causes kind

of dislocation slip
along specific planes;

and twin deformation, where you
get kind of this uniform kind

of shifting of atoms across to
form this kind of mirror image.

Not as common, but
worth mentioning

since we're talking about this.

Okay, we're going to have
one next section that goes

over a calculation on how
to do these calculations

of resolved shear stress.