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Chapter 5. Dislocation
I. Critical shear stress of perfect crystal
- Consider displacement x of two atomic planes under a shear stress σ. For small
displacement, the shear strain is given as e=x/d, where d is the distance between the
layers.
- The force (and hence the shear stress) required is periodic because of the lattice
periodicity. The magnitude of force can be modeled as sinusoidal. So, σ = A sin (2πx/a)
where a is the lattice constant or distance between two atoms in the same plane.
- “Hooke’s law”: σ = Ce, but this linear relationship applies only to small strain e
(or small displacement x). Since σ = A sin (2πx/a) ⇒ σ = 2πAx/a
∴ Cx/d = 2πAx/a ⇒ C = 2πdA/a ⇒ A = Ca/2πd
∴σ = Ca/2πd sin (2πx/a)
a
Original equilibrium
position
x
d
d
x
x
Stress σ
d
- The amplitude σc = A = Ca/2πd is known as the critical shear stress at which the
lattice becomes unstable and the top layer will slide into a new equilibrium position. For
a ~ d, we have σc = C/2π. In other words, the critical shear stress is about 1/6 of the shear
modulus.
- In reality, the critical shear is much smaller and is only a few percent of the shear
modulus. Such a discrepancy can only be explained by dislocation.
II. Edge dislocation
- There are two types of dislocation: edge dislocation and screw dislocation (or a
mixture of both).
- To understand dislocation, we can imagine to start from a perfect, undeformed
lattice and cut the lattice along any of the planes as shown in the figure.
Stress
x
σ = Ca/2πd sin (2πx/a)
σ = Cx/d
- If one looks at the plane perpendicular to the dislocation line, the planes look like
a spiral ramp.
This is the reason why this is called screw dislocation. Again, there are two directions of
screw dislocation – right hand and left hand. Note that the “handness” of a screw is
independent of the direction at which it is looked at.
- The slip can occur in general in any direction, neither parallel nor perpendicular to
the dislocation line. This is the case of a mixed (of edge and screw) dislocation:
IV. Burgers vector
- In a lattice, if we draw a closed circuit loop by going right x lattice point, up y
lattice points, and then left the same x lattice points and down y lattice points. If the
lattice is perfect, we will come back to the same lattice points (bottom loop in the figure
below).
- If the lattice is not perfect (i.e. with dislocation), we will not come back to the
same lattice point (top loop in the figure below).
Similarly, if a screw dislocation moves completely across its slip plane:
Movement of dislocations results in plastic deformation.
- As the dislocations move, plastic deformation causes a very great increase in
dislocation density, typically from 10
8 to 10
11 eiwlocations/cm
2
. Dislocations multiply
during deformation!
- For the strength of a material, the dislocations have to be “pinned” so that they
cannot move freely to generate more dislocations. Examples: particles of iron carbide
are precipitated in the hardening of steel, and particles of Al 2 Cu are precipitated in the
hardening of aluminum.
VI. Stress fields of dislocations
- Young’s modulus (E) is defined as stress/strain for a tensile stress acting in one
direction, with the specimen sides left free. Poisson’s ratio (ν) is defined as (δw/w)(δl/l)
for this situation.
l
w (^) w-δw
l+δl
- For isotropic materials,
G is the rigidity modulus and it is related to E and ν as:
Reversing this matrix, we have:
λ and μ are related to the Young’s modulus and Poisson Ratio as
- In particular, let us look at the x-axis:
e
e
e
e
e
e
xy
zx
yz
zz
yy
xx
xy
zx
yz
zz
yy
xx
μ
μ
μ
λ λ λ μ
λ λ μ λ
λ μ λ λ
σ
σ
σ
σ
σ
σ
G
G
G
E
E E
E E
E
E E E
e
e
e
e
e
e
xy
zx
yz
zz
yy
xx
xy
zx
yz
zz
yy
xx
σ
σ
σ
σ
σ
σ
ν ν
ν ν
ν ν
E
G
and
E
z y x
2 z
y x z
2 y
x y z
2 x
xz xz
yx xy
xx xx
δ δ δ
σ σ
δ δ δ
σ σ
δ δ δ
σ σ
z y x
2 z
y x z
2 y
x yz
2 x
xz xz
yx xy
xx xx
δ δ δ
σ σ
δ δ δ
σ σ
δ δδ
σ σ
- For stationary dislocations:
- Consider the case of pure screw dislocation, with Burgers vector b = b k. For
screw dislocation, u=v=0.
Notes: If a circuit is made about the dislocation line, the arc tan function changed
by 2π and thus w changed by the amount –b.
- From this relation, we can calculate the strain of the screw dislocation:
And then stress follows:
- In cylindrical coordinates, r
2 =x
2 +y
2 , tan θ = y/x, z=z. Above results can be
transformed into simpler form:
y
w
x
w
x y
u ( ) y
v ( 2 ) x
v
x y
v ( ) y
u
x
u ( 2 )
2
2
2
2
2
2
2
2
2
2
2
2
2
2
x
y tan 2
b w -
y
w
x
w Guesssolutionof
2
2
2
2
e e e e 0
x y
x
b
y
w
z
v e
x y
y
b
x
w
z
u e
xx yy zz xy
yz 2 2
xz 2 2
x y
x
b
x y
y
b
xx yy zz xy
yz 2 2
xz 2 2
- Consider the case of pure edge dislocation, solution of the equations of motion are
given as:
2 r
b
, e e e e e 0 2 r
b e
,u u 0 2
b w -
z rr zz r rz
z rr zz r rz
r
σ σ σ σ σ π
σ
π
π
θ
θ θθ θ
θ θθ θ
θ
w 0
x y
y
C 2
x y log 2 2 ( 2 )
b v -
x y
x
x 2
y tan 2
b u -
2 2
2 2 2
2 2
2
λ μ
λ μ
λ μ
μ
π
λ μ
λ μ
π
