Topic Outline
Types
Slip
What is Slip?
Where does Slip Occur?
Slip Systems
Example
Burger’s Vector
Schmid’s Law
Slip Bands and Critical Stress
Your Responsibilities
Sample Calculations
CHE 294 1
Types
We recognize two types of line defects.
Looks as though we are
Screw tearing apart the material.
A dislocation line marks
the boundary of the tear.
As we walk around
the line, we spiral
through the crystal
like a cork screw. Edge
Looks like it could contain an additional
plane of atoms ... this is not the case
because one plane is
missing.
Line defects are commonly called dislocations.
CHE 294 2
What is Slip?
Slip is atom motion due to stress.
unit cell under
no stress
apply a stress
bonds stretch to obtain a strain
their limit before apply more
breaking stress
atoms slip
broken bond release the
leads to a stress, and the
defect inside atoms relax
the material the defect caused by slip remains back
CHE 294 3
Where does Slip Occur?
Compare these two planes.
So slip occurs on
lattice planes close packed
can easily slip planes in close
packed
now apply directions.
a stress
lattice planes do not slip as easily
a (110) plane
versus
BCC Crystal
a (100) plane
CHE 294 4
Slip Systems
Slip Planes + Slip Directions gives Slip Systems.
Lattice Slip Planes Slip Directions # of Slip Systems
FCC {111} 4 x 3 = 12
BCC {110} 6 x 2 = 12
{112} 12
{123} 24
HCP {0001} 3
You are responsible for learning
• placements of all slip systems in FCC
• placements of all {110} slip systems in BCC
• the total number of slip systems in FCC, BCC, and HCP
CHE 294 5
Example
What are the Miller indices for each of the following directions?
[001]
[111] z
[110]
y
x
[111]
(1 1 0) plane of BCC crystal
CHE 294 6
Defining the Burger’s Vector
The Burger’s vector completes a loop around
the dislocation line.
start here to
Edge Dislocation make a loop
+1 in x (this is arbitrary)
+4 in y +3 in x The length of the
Burgers vector |b| is
the distance between
the two atoms that
Burger’s complete the loop.
Vector You should learn
-4 in x -4 in y how to determine
|b| for cubic metals
Dislocation
and ceramics.
Line
CHE 294 7
Slip in Ceramics
Slipping ions is difficult!
+ +
+
+ +
+ + Consider slip along
+ +
+ in NaCl
+ +
We cannot move a cation into the slot
of an anion during this slip.
We have to slip the cation past other cations into
the slot of the next closest cation.
Slip in ceramics is limited by both ionic repulsion (like charges
repel each other) and the different sizes of the ions (cations
are larger, making the slip plane less smooth).
CHE 294 8
Line and Atom Motion
How do the lines and atoms move during slip?
b
Atoms move parallel to b
(in the slip direction
on the slip plane)
Line moves parallel to b
(on the slip plane)
How does this picture change for a screw dislocation?
CHE 294 9
General Characteristics
These are the general characteristics of dislocations.
b Vector Line Atoms
Type Points Moves Move
Edge to line || to b || to b
Screw || to line to b || to b
means perpendicular || means parallel
Atoms are always moving parallel to the Burgers vector in the
slip direction on the slip plane.
Convince yourself that you understand this table.
CHE 294 10
Force Required for Slip
How much applied force is needed to cause slip?
Consider a
single crystal Fresolved 3. Resolve that force on
to the slip plane ... in the slip direction.
rod. slip
1. Apply a
direction τresolved = Fresolved / As
tensile force
4. The resolved
λ
along its
axis. stress is a shear
F As stress.
Ao φ slip plane 5. Express the
resolved shear
σ = F /Ao
stress in terms of
the applied
normal to tensile stress.
2. Calculate the
τrss = σ cos φ cos λ
slip plane
applied tensile stress.
This formulation is known as Schmid’s law.
CHE 294 11
Vector Notation
Schmid’s law can
be expressed in
R
vector notation
with Miller cos λ = T • R / | T | | R|
indices.
λ
You already know how T
to calculate angles
between vectors using φ
Miller indices and dot
products.
n
An example using cos φ = T • n / | T | | n |
Miller indices is ...
(hT hR + kT kR + lT lR)
cos λ =
(hT2 + kT2 + lT2 ) (hR2 + kR2 + lR2 )
CHE 294 12
Example
What is τrss in terms of σ for this example in a BCC crystal?
slip τ along [1 1 1]
σ
σ along [1 1 0]
φ = 90o (by inspection)
λ = 35.3o (by calculation)
(110) plane τrss = 0! Why?
CHE 294 13
Slip Bands and Critical Stress
Consider stress applied to a single crystal rod or grain.
σ
n
Slip causes bands to
appear in a single crystal or τ
within a grain.
The stress needed to start slip is the
critical resolved shear stress τcrss
CHE 294 14
Your Responsibilities
You should learn ....
• the slip systems given previously
• where to locate the Burgers vector in screw
and edge dislocations
• how to calculate the length of the Burgers
vector in metals and ceramics
• the direction of motion of dislocation lines
and atoms relative to the Burgers vector in
edge and screw dislocations
• how to use Schmid’s law backwards and
forwards
CHE 294 15