>> 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.