Chapter 4
物理冶金
魏茂國
Introduction to Dislocations
 The discrepancy between the theoretical and observed yield stresses of crystals
 Dislocations
 The Burgers vector
 Vector notation for dislocations
 Dislocations in the face-centered cubic lattice
 Intrinsic and extrinsic stacking faults in face-centered cubic metals
 Extended dislocations in hexagonal metals
 Climb of edge dislocations
 Dislocation interactions
 The stress field of a screw dislocation
 The stress field of an edge dislocation
 The force on a dislocation
 The strain energy of a screw dislocation
 The strain energy of an edge dislocation
1
Discrepancy between Theoretical & Observed Yield Stresses魏茂國
物理冶金
 Discrepancy between the theoretical and observed yield stresses of crystals
- The stress-strain curve of a typical magnesium single crystal, oriented with the
basal plane inclined at 45 to the stress axis and strained in tension, is shown in
Fig. 4.1.
At the low tensile stress of 0.7 MPa, the crystal yields plastically and then easily
stretches out to a narrow ribbon which may be 4 or 5 times longer than the original
crystal.
- If one examines the surface of the deformed crystal, markings can be
seen which run more or less continuously around the specimen in the
form of ellipses (Fig. 4.2)
Fig. 4.1 Tensile stress-strain curve for a
Mg single crystal.
Fig. 4.2 Slip 2lines
on Mg crystal.
Discrepancy between Theoretical & Observed Yield Stresses魏茂國
物理冶金
- These markings (in Fig. 4.2) recognized as a series of fine steps that have formed on
the surface. The nature of these steps is shown schematically in Fig. 4.3. Evidently,
as a result of the applied force, the crystal has been sheared on a number of parallel
planes. Crystallographic analyses of the markings, furthermore, show that these are
basal (0002) planes and, therefore, the closest packed plane of the crystal.
- When this type of deformation occurs, the crystal is said to have undergone “slip,”
the visible markings on the surface are called slip lines, or slip traces, and the
crystallographic plane on which the shear has occurred is
(A)
(B)
called the slip plane (滑移面).
- The shear stress at which plastic flow begins in a single
crystal is amazingly small when compared to the theoretical
shear strength of a perfect crystal.
Fig. 4.3 (A) Magnified schematic view of slip lines (side view). (B) Magnified
schematic view of slip lines (front view)
3
Discrepancy between Theoretical & Observed Yield Stresses魏茂國
物理冶金
- An estimate of the strength can be obtained in the following manner.
Fig. 4.4A shows 2 adjacent planes of a hypothetic crystal. A shearing stress, acting as
indicated by the vectors marked ,ends to move the atoms of the upper plane to the
left.
Each atom of the upper plane rises to a maximum position (Fig. 4.4B) as it slides over
its neighbor in the plane below. This maximum position represents a saddle point.
A shear of one atomic distance requires that the atoms of the upper plane in Fig. 4.4A
be brought to a position equivalent to that in Fig. 4.4B, after which they move on
their own accord into the next equilibrium position, Fig. 4.4C.
Since the separation of the 2 planes is of the order of 2 atomic radii, the shear strain at
the saddle point is approximately equal to one half.

a 1

2a 2
(4.1)
where  is shear strain.
4
Discrepancy between Theoretical & Observed Yield Stresses魏茂國
物理冶金
(A)
(B)
(c)
Fig. 4.4 (A) Initial position of the atoms on a slip plane. (B) The saddle point for the shear of one plane of
5
atoms over another. (C) Final position of the atoms after shear by one atomic distance.
Discrepancy between Theoretical & Observed Yield Stresses魏茂國
物理冶金
- In a perfectly elastic crystal, the ratio of shear stress to shear strain is equal to the
shear modulus:

(4.2)

where  is shear strain,  is shear stress, and  is shear modulus.
- Substituting the value ½ for  the shear strain, and the value 17.2 GPa for , which is
of the order of magnitude of the shear modulus for magnesium, we obtain for  the
stress at the saddle point,

1
2
     17,200 MPa  8.6  103 MPa
- Real crystals deform at small fractions of their theoretical strengths (1/1000 to
1/100,000).
0.7 MPa
5

8
.
1

10
8.6 103 MPa
6
物理冶金
Dislocations
魏茂國
 Dislocations (差排，錯位)
- The discrepancy between the computed and real yield stresses is because real
crystals contain defects.
- If the transmission foil has been prepared properly and contains a section of a slip
plane, where it is examined in the microscope one may obtain a photograph of the
type shown schematically in Fig. 4.5A. The lines (a-a and b-b) have been drawn on
the figure to indicate the positions where the slip plane intersects the foil surfaces.
- It should be noted that the drawing in Fig. 4.5A is a 2dimensional projection of a 3-dimensional specimen.
- Fig. 4.5 B demonstrates that the dark lines in the
photograph run across the slip plane from the top to the
bottom surfaces of the foil.
- In a crystal which has undergone slip, lattice defects tend
to accumulate along the slip planes. These defects are
called dislocations.
7
Fig. 4.5(A)
Dislocations
物理冶金
魏茂國
- The points where dislocations intersect a specimen surface can often be made visible
by etching the surface with a suitable etching solution. As a result, etch pits may form
(Fig. 4.5C).
(A)
(B)
(C)
Fig. 4.5 (A) Schematic representation of an electron microscope photograph showing a section of a slip plane.
(B) A 3-dimensional view of the same slip plane section. (C) Termination of dislocations can also be revealed
by etch pits.
8
物理冶金
Dislocations
魏茂國
- Fig. 4.6 shows a portion of a foil of an aluminum specimen with a grain containing a
slip plane with dislocations.
The specimen was polycrystalline. The dark region at the upper right-hand corner
represents a second grain.
- The best evidence now indicates that
dislocations are boundaries on the slip
planes where a shearing operation has
ended.
Fig. 4.6 An electron micrograph of a foil removed from
an aluminum specimen. Note the dislocations lying
along a slip plane, in agreement with Fig. 4.5.
9
物理冶金
Dislocations
魏茂國
 Edge dislocation (刃差排)
- Fig. 4.7A represents a simple cubic crystal that is assumed to be subjected to
shearing stresses, , on its upper and lower surfaces. The line SP represents a
possible slip plane in the crystal. As a result of the applied shear stress, the righthand half of the crystal is displaced along SP so that the part above the slip plane is
moved to the left with respect to the part below the slip plane. The amount of this
shear is assumed to equal one interatomic spacing in a direction parallel to the slip
plane.
- As may be seen in Figs. 4.7B and C, this will leave an extra half-plane cd below the
slip plane at the right and outside the crystal. It will also form an extra vertical halfplane ab above the slip plane and in the center of the crystal.
- Fig. 4.7B clearly shows that the crystal is badly distorted where this half-plane
terminates at the slip plane. It can also be deduced that this distortion decreases in
intensity as one moves away from the edge of this half-plane. This is because at
large distances from this lower edge of the extra plane, the atoms tend to be
arranged as they would be in a perfect crystal. The distortion in the crystal is
centered around the edge of the extra plane. This boundary of the additional plane
is
10
called an edge dislocation.
物理冶金
Dislocations
(A)
魏茂國
(B)
(C)
Fig. 4.7 An edge dislocation. (A) A perfect crystal.
(B) When the crystal is sheared one atomic distance
over part of the distance S-P, an edge dislocation is
formed. (C) 3 dimensional view of slip.
11
物理冶金
Dislocations
魏茂國
- Fig. 4.8 represents a 3-dimensional sketch of the edge dislocations of Fig. 4.7. The
figure clearly shows that the dislocation has the dimensions of a line.
Another important fact is that the dislocation line marks the boundary between the
sheared and unsheared parts of the slip plane.
- In fact, a dislocation may be defined as a line that forms a boundary on a slip plane
between a region that has slipped and one that has not.
Fig. 4.8 This 3-dimensional view of a
crystal containing an edge dislocation
shows that the dislocation forms the
boundary on the slip plane between a
region that has been sheared and a
region that has not been sheared.
12
Dislocations
物理冶金
魏茂國
- As a result of the applied stress, atom c in Fig. 4.9A may move to the position marked
c’ in Fig. 4.9B. The final result is that the crystal is sheared across the slip plane by
one atomic distance, as shown in Fig. 4.9C.
- Each step in the motion of the dislocation (Fig. 4.9) requires only a slight
rearrangement of the atoms in the neighborhood of the extra plane. As a result, a very
small force will move a dislocation.
- In 1934, Orowan, Polyani, and Taylor presented papers which are said to have laid
the foundation for the modern theory of slip due to dislocations.
(A)
(B)
(C)
13
Fig. 4.9 Three stages in the movement of an edge dislocation through a crystal.
物理冶金
Dislocations
魏茂國
 Screw dislocation (螺旋差排)
- The movement of a single dislocation completely through a crystal produces a step
on the surface, the depth of which is one atomic distance. Many hundreds or
thousands of dislocations must move across a slip plane in order to produce a
visible slip line.
- Fig. 4.10A shows a screw dislocation, where each small cube can be considered to
represent an atom. Fig. 4.10B represents the same crystal with the position of the
dislocation line marked by the line DC.
(A)
(B)
Fig. 4.10 Two representations of a screw dislocation. Notice that the planes in this dislocation spiral
around the dislocation like a left-hand screw.
14
物理冶金
Dislocations
魏茂國
- The designation “screw” for this lattice defect
is derived from the fact that the lattice planes
of the crystal spiral the dislocation line DC.
This statement can be proved by starting an
point x in Fig. 4.10A and then proceeding
upward and around the crystal in the direction
Fig. 4.10(A)
of the arrows. One circuit of the crystal ends at point y; continued circuits will finally
end at point z.
- Fig. 4.10B plainly shows that a dislocation in
a screw orientation also represents the boundary
between a slipped and an unslipped area. Here
the dislocation, centered along line DC,
separates the slipped area ABCD from the
remainder of the slip plane in back of the
dislocation.
Fig. 4.10(B)
15
Materials Science and Engineering
Dislocations-Linear Defects
魏茂國
Dislocation line could be seen as a line to separate the deformed and undeformed
regions.
Figure 4.4 (a) A screw dislocation within a crystal. (b) The screw dislocation in (a) as
viewed from above. The dislocation line extends along line AB. Atom positions above
16
the slip plane are designated by open circles, those below by solid circles.
物理冶金
Dislocations
魏茂國
 Dislocations
- The edge dislocation shown in Fig. 4.7B has an
incomplete plane which lies above the slip plane.
It is also possible to have the incomplete plane
below the slip plane. The 2 cases are differentiated
by calling the former a positive edge dislocation,
and the latter a negative edge dislocation.
Fig. 4.7(B)
Symbols representing these 2 forms are ┴ and ┬, respectively, where the
horizontal line represents the slip plane and the vertical line the incomplete plane.
- The screw dislocation shown in Fig. 4.10 has lattice planes that spiral the line DC
like a left-hand screw.
- Both forms of the edge and the screw
dislocations, respectively, are shown in
Fig. 4.12.
17
Fig. 4.10(B)
Materials Science and Engineering
Dislocations-Linear Defects
Positive edge dislocation
魏茂國
Negative edge dislocation
18
Dislocations
物理冶金
魏茂國
(A)
(B)
(C)
(D)
Fig. 4.12 The ways that the 4 basic orientations of a dislocation move under the same applied stress:
19
(A) Positive edge, (B) Negative edge, (C) Left-hand screw, and (D) Right-hand screw.
物理冶金
Dislocations
魏茂國
- Fig. 4.12 shows that a positive edge dislocation moves to the left when the upper half
of the lattice is sheared to the left. On the other hand, a negative edge dislocation
moves to the right, but produces the identical shear of the crystal. The right-hand
screw moves forward and the left-hand screw moves to the rear, again producing the
same shear of the lattice.
- Dislocations cannot end inside a crystal.
- In Fig. 4.13, the 2 dislocation segments a and b form a continuous path through the
crystal from front to top surfaces.
Fig. 4.13 Dislocations can vary in direction. This shaded
extra plane forms a dislocation with edge components a
and b.
20
物理冶金
Dislocations
魏茂國
- It is also possible for all 4 edges of an incomplete plane to lie inside a crystal,
forming a 4-sided closed edge dislocation at the boundaries of the plane.
- Furthermore, a dislocation that is an edge in one orientation can change to a screw in
another orientation, as is illustrated in Figs. 4.14 and 4.15.
Fig. 4.14 A 2-component dislocation composed
of an edge and a screw component.
Fig. 4.15 Atomic configuration corresponding to the
dislocation of Fig. 4.14 viewed from above. Opencircle atoms are above the slip plane, dot atoms are
below the slip plane.
21
物理冶金
Dislocations
魏茂國
- A dislocation does not need to be either pure screw or pure edge, but may have
orientations intermediate to both. Fig. 4.16 shows a change in orientation from edge to
screw, but here the change is not abrupt.
- Fig. 4.17 consists of the 4 elementary types of dislocations. Sides a and c are positive
and negative edge dislocations, respectively, while b and d are right- and left-hand
screws, respectively.
Fig. 4.17 A closed dislocation loop consisting of (a) positive edge,
(b) right-hand screw, (c) negative edge, and (d) left-hand screw.
Fig. 4.16 A dislocation that changes its orientation from a screw to an edge as viewed from above 22
looking down on its slip plane.
物理冶金
Dislocations
魏茂國
- A dislocation cannot inside a crystal. This is because a dislocation represents the
boundary between a slipped and an unslipped area.
- If the slipped area on the slip plane does not touch the specimen surface, as in
Fig. 4.18, then its boundary is continuous and the dislocation has to be a closed loop.
Fig. 4.18 A curved dislocation loop lying in a slip plane.
23
物理冶金
Burgers Vector
魏茂國
 The Burgers vector
- In Fig. 4.17, although the dislocation varies in orientation in the slip plane ABCD,
the variation in shear across the dislocation is everywhere the same, and the slip
vector b is therefore a characteristic property of the dislocation. By definition, this
vector is called the Burgers vector of the dislocation.
- The Burgers vector of a dislocation is an important property of a dislocation
because, if the Burgers vector and the orientation of the dislocation line are known,
the dislocation is completely described.
Fig. 4.17
24
物理冶金
Burgers Vector
魏茂國
- In Fig. 4.19A a counterclockwise circuit of atom-to-atom steps in a perfect crystal
closes, but when the same step-by-step circuit is made around a dislocation in an
imperfect crystal (Fig. 4.19B), the end point of the circuit fails to coincide with the
starting point. The vector b connecting the end point with the starting point is the
Burgers vector of the dislocation.
(A)
(B)
25
Fig. 4.19 The Burgers circuit for an edge dislocation: (A) Perfect crystal and (B) crystal with dislocation.
物理冶金
Burgers Vector
魏茂國
- The procedure can be used to find the Burgers vectors of any dislocation if the
following rules are observed:
1. The circuit is traversed in the same manner as a rotating right-hand screw
advancing in the positive direction of the dislocation.
2. The circuit must close in a perfect crystal and must go completely around the
dislocation in the real crystal.
3. The vector that closes the circuit in the imperfect crystal (by connecting the end
point to the starting point) is the Burgers vector.
- The above convention involving a right-hand (RH) circuit around the dislocation line
yields a Burgers vector pointing from the finish to the start (FS) of the circuit, and
because the closure failure is measured in an imperfect crystal, it is called a local
Burger vector or more completely a RHFS local Burgers vector.
26
Burgers Vector
物理冶金
魏茂國
- Fig. 4.20 shows a Burgers circuit around a left-hand screw dislocation. In Fig. 4.20A,
the circuit is indicated for the perfect crystal. Fig. 4.20B shows the same circuit
transferred to a crystal containing a screw dislocation.
(A)
(B)
Fig. 4.20 The Burgers circuit for a dislocation in a screw orientation. (A) Perfect crystal and
(B) crystal with dislocation.
27
Burgers Vector
物理冶金
魏茂國
 Characteristics of both edge and screw dislocations
- Edge dislocation
1. An edge dislocation lies perpendicular to its Burgers vector.
2. An edge dislocation moves (in its slip plane) in the direction of the Burgers
vector (slip direction). Under a shear-stress sense 
 a positive dislocation 

moves to the right, a negative one
to the left.
- Screw dislocation
1. A screw dislocation lies parallel to its Burgers vector.
2. A screw dislocation moves (in the slip plane) in a direction perpendicular to the
Burgers vector (slip direction).
28
Materials Science and Engineering
Dislocations-Linear Defects
魏茂國
Edge dislocation
Dislocation line
Burgers vector
Slip direction
Screw dislocation
Dislocation line
Burgers vector
Slip direction
29
物理冶金
Burgers Vector
魏茂國
 Slip plane (滑移面)
- The slip plane is the plane containing both the Burgers vector and the dislocation.
- The slip plane of an edge dislocation is thus uniquely defined because the Burgers
vector and the dislocation are perpendicular.
- The slip plane of a screw dislocation can be any plane containing the dislocation
because the Burgers vector and dislocation have the same direction.
- Edge dislocations are confined to move or glide in a unique plane, but screw
dislocations can glide in any direction as long as they move parallel to their original
orientation.
30
物理冶金
Vector Notation for Dislocations
魏茂國
 Vector notation for dislocations
- In any crystal form, the distance between atoms in a close-packed direction
corresponds to the smallest shear distance that will preserve the crystal structure
during a slip movement. Dislocations with Burgers vectors equal to this shear are
energetically the most favored in a given crystal structure.
- With regard to the vector notation, the direction of a Burgers vector can be
represented by the Miller indices of its direction, and the length of the vector can
be expressed by a suitable numerical factor placed in front of the Miller indices.
- In a simple cubic lattice, the distance between atoms in a
close-packed direction equals the length of one edge of
a unit cell. A dislocation with a Burgers vector in a simple
cubic lattice is represented by [100].
31
物理冶金
Vector Notation for Dislocations
魏茂國
- In Fig. 4.21, the close-packed direction in a face-centered cubic lattice is a face
diagonal, and the distance between atoms in this direction is equal to one-half the
length of the face diagonal. As a result, a dislocation in a face-centered cubic lattice
having a Burgers vector lying in the [101] direction should be written ½[101].
- In the body-centered cubic lattice, the close-packed direction is a cube diagonal, or a
direction of the form <111>. The distance between atoms in these directions is onehalf the length of the diagonal, so that a dislocation having a Burgers vector parallel
to [111] is written ½[111].
001
011
101
111
000
Fig. 4.21 The spacing between atoms in the close-packed
directions of the different cubic systems: face-centered
cubic, body-centered cubic, and simple cubic.
010
100
110
32
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
 Dislocations in the face-centered cubic lattice
- The primary slip plane in the face-centered cubic lattice is the octahedral plane
{111}. Fig. 4.22 shows a plane of this type looking down on the extra plane of an
edge dislocation. Notice that in the latter plane a zigzag row of atoms is missing.
This corresponds to the missing plane of the edge dislocation.
Fig. 4.22 A total dislocation (edge orientation) in a face-centered cubic lattice as viewed
when looking down on the slip plane.
33
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
- The vector b represents the Burgers vector of the dislocation, which is designed as
½[110] (Fig. 4.24). As is to be expected, this dislocation movement shears the upper
half of the crystal (above the plane of the paper) one unit b to the right relative to the
bottom half (below the plane of the paper).
- In Fig. 4.22, the movement of a zigzag plane of atoms, such as aa, through the
horizontal distance b would involve a very large lattice strain, because each white
atom at the slip plane would be forced to climb over the dark atom below it and to its
right.
- What actually is believed to happen is that the indicated plane of atoms makes the
move indicated by the vectors marked c in Fig. 4.23. This
movement can occur with a much smaller strain
of the lattice. A second movement of the same
type, indicated by the vectors marked d, brings
the atoms to the same final positions as the
single displacement b of Fig. 4.22.
34
Fig. 4.22
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
- The atom arrangement of Fig. 4.23 is particularly significant because it shows how a
single-unit dislocation can break down into a pair of partial dislocations.
- In Fig. 4.24, the Burgers vector of the total dislocation equals the distance B1B2, while
the Burgers vectors of the 2 partial dislocations c and d of Fig. 4.23 are the same as
the distance B1C and CB2. The Burgers vector of the total dislocation is ½[110]. The
B1C lies in the [121] direction. Since B1C is just one-third of line mn, the Burgers
vector for this partial dislocation, c, is 1/6[121]. In the same manner,
the partial dislocation, d, can be represented as 1/6[211].
101
001
[110]
11/21/2
011
1/211/2
100
010
110
Fig. 4.23 Partial dislocation in a face-centered cubic lattice.
Fig. 4.24 The orientation relationship
between the Burgers vectors of a35
total
dislocation and its partial dislocations.
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
- The total face-centered cubic dislocation ½[110] is thus able to dissociate into 2
partial dislocations according to the relation
     
1
1
1
110  121  211
6
6
2
(4.4)
- When a total dislocation breaks down into a pair of partials, the strain energy of the
lattice is decreased. This results because the energy of a dislocation is proportional to
the square of its Burgers vector and because the square of the Burgers vector of the
total dislocation is more than twice as large as the square of the Burgers vector of a
partial dislocation.
 
2
 
2
 
2
1
1
1







1
10
1
1
0


2
4
2


1
1
1







1
2
1
1
4
1


6
36
6


1
1
1







2
11
4
1
1


6
36
6


1 1 1 1
  
2 6 6 3
36
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
- A total dislocation that has dissociated into a pair of separated partials like those in
Fig. 4.25 is known as an extended dislocation (擴展差排).
- If we assume that the dark-colored atoms occupy A positions in a stacking sequence
and the white atoms at either end of the figure, B positions, then the white atoms
between the 2 partial dislocations lie on C positions.
In this region, the ABCABCABC… stacking sequence of the face-centered cubic
lattice suffers a discontinuity and becomes ABCACABCA…. The arrows indicate the
discontinuity.
- Discontinuity in the stacking
order of the {111}, or closepacked planes, are called
stacking faults (疊差).
37
Fig. 4.25 An extended dislocation.
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
- The stacking fault occurs on the slip plane (between the dark and white atoms) and is
bounded at its end by what are known as Shockley partial dislocations.
- In all cases, if a stacking fault terminates inside a crystal, its boundaries will form a
partial dislocation.
- The partial dislocations of stacking faults, in general, may be either of the Shockley
type, with the Burgers vector of the dislocation lying in the plane of the fault, or of
the Frank type, with the Burgers vector normal to the stacking fault.
38
Fig. 4.25
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
- The defect marked by A in Fig. 4.26 shows stacking faults bordered by 2 dislocations
in the lower-left- and upper-right-hand side.
- The defect marked by B shows a narrow region of stacking fault surrounded by
Shockley partial dislocations.
Fig. 4.26 An electron micrograph of a thin foil of lightly
deformed Cu + 15.6 at.% Al alloy. The beam direction
is close to [101] and g is indicated. 44,000.
39
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
 Stacking fault (疊差)
- Since the atoms on either side of a stacking fault are not at the positions they would
normally occupy in a perfect lattice, a stacking fault possesses a surface energy
which, in general, is small compared with that of an ordinary grain boundary, but
nevertheless finite.
- The stacking-fault energy plays an important part in determining the size of an
extended dislocation. The larger the separation between the partial dislocations, the
smaller is the repulsive force between them. On the other hand, the total surface
energy associated with the stacking fault increases with the distance between partial
dislocations. The separation between the 2 partials thus represents an equilibrium
between the repulsive energy of the dislocations and the surface energy of the fault.
- Seeger and Schoeck have shown that the separation of the partial dislocations in an
extended dislocation depends on a dimensionless parameter Ic/Gb2, where I is the
specific surface energy of the stacking fault, c is the separation between adjoining
slip planes, G is the shear modulus in the slip plane, and b is the magnitude of the
Burgers vector.
40
物理冶金
Dislocations in the Face-Centered Cubic Lattice
魏茂國
In certain face-centered cubic metals typified by aluminum, this parameter is larger
than 10-2 and the separation between dislocations is only of the order of a single
atomic distance. These metals are said to have high stacking-fault energies.
When the parameter is less than 10-2, a metal is said to have a low stacking-fault
energy.
 Movement of an extended dislocation
- If the moving dislocation meets obstacles, such as other dislocations, or even
second-phase particles, the width of the stacking fault should vary.
- Thermal vibrations may also cause the width of the stacking fault to vary locally
along the dislocation, the vibration being a function of time.
- An extend dislocation can be pictured as a pair of partial dislocations, separated by
a finite distance, which move in consort through the crystal.
41
物理冶金
Intrinsic and Extrinsic Stacking Faults in FCC Metals
魏茂國
 Intrinsic and extrinsic stacking faults in face-centered cubic metals
- The movement of a Shockley partial across the slip plane of a fcc metal has been
shown to produce a stacking sequence ABCA  CABC…. A fault of this type is called
an intrinsic stacking fault. An intrinsic stacking fault may also be developed in a fcc
crystal by removing part of a close-packed plane, as shown in Fig. 4.27A. Its
Burgers vector is equal to one-third of a total dislocation and therefore may be
written 1/3<111>.
Fig. 4.27A An intrinsic stacking fault can also be formed in a face-centered cubic crystal by removing
42
part of a close-packed plane.
物理冶金
Intrinsic and Extrinsic Stacking Faults in FCC Metals
魏茂國
- The addition of a portion of an octahedral plane produces a different type of stacking
sequence which is ABCA  C  BCABC…. In this fault (Fig. 4.27B) a plane has been
inserted that is not correctly stacked with respect to the planes on either side of the
fault. This type of fault is called an extrinsic or double stacking fault.
The Burgers vector for the extrinsic fault, shown in Fig. 4.27B, is 1/3<111>.
An extrinsic stacking fault could be formed by the precipitation of interstitial atoms
on an octahedral plane.
Fig. 4.27B The addition of a portion of an extra close-packed plane to a face-centered cubic crystal
produces an extrinsic stacking fault.
43
物理冶金
Extended Dislocations in Hexagonal Metals
魏茂國
 Extended dislocations in hexagonal metals
- In the hexagonal system, the dissociation of a total dislocation into a pair of partials
on the basal plane is expressed in the following fashion:
     
1
1
1
1210  0110  1100
3
3
3
(4.5)
- For the face-centered cubic system, the dissociation of a total dislocation into a pair
of partials is expressed in the following fashion:
     
1
1
1
110  121  211
2
6
6
(4.6)
44
Materials Science and Engineering
Crystallographic Directions
魏茂國
[2110]
a2
1
2u'v'
3
HCP: [uvtw]  t = -(u+v)
a2 = ⅓[1210]
a2
a3
a1
a3
a1
a1 = ⅓[2110]
a2
a3
[1120]
a1
a3 = ⅓[1120]
[1210]
45
Climb of Edge Dislocations
物理冶金
魏茂國
 Climb of edge dislocations
- The slip plane of a dislocation is defined as the plane that contains both the
dislocation and its Burgers vector.
- Since the Burgers vector is parallel to a screw dislocation, any plane containing the
dislocation is a possible slip plane (Fig. 4.28A).
- The Burgers vector of an edge dislocation is perpendicular to the dislocation, and
there is only one possible slip plane (Fig. 4.28B).
(A)
(B)
Fig. 4.28 (A) Any plane containing the dislocation is a slip plane for a screw dislocation. (B) There is
46only
one slip plane for an edge dislocation. It contains both the Burgers vector and the dislocation.
物理冶金
Climb of Edge Dislocations
魏茂國
- A screw dislocation may move by slip or glide in any direction perpendicular to its
self, but an edge dislocation can only glide in its single slip plane.
- Fig. 4.29 demonstrates a positive climb of an edge dislocation and results in a
decrease in size of the extra plane.
- Negative climb corresponds to the opposite of the above in that the extra plane grows
in size instead of shrinking. A mechanism for negative climb is illustrated in
Fig. 4.30.
(A)
(B)
(C)
Fig. 4.29 Positive climb of an edge dislocation.
47
物理冶金
Climb of Edge Dislocations
魏茂國
- Because we are removing material from inside the crystal as the extra plane itself
grows smaller, the effect of positive climb on the crystal is to cause it to shrink in a
direction parallel to the slip plane (perpendicular to the extra plane).
Positive climb is therefore associated with a compressive strain and will be promoted
by a compressive stress component perpendicular to the extra plane.
- A tensile stress applied perpendicular to the extra plane of an edge dislocation
promotes the growth of the plane and thus negative climb.
- Slip occurs as the result of shear stress; climb as the result of a normal stress (tensile
or compressive).
(A)
Fig. 4.30 Negative climb of an edge
dislocation.
(B)
48
物理冶金
Climb of Edge Dislocations
魏茂國
- Both positive and negative climb require that vacancies move through the lattice,
toward the dislocation in the first case and away form it in the second case.
- If the concentration of vacancies and their jump rate is very low, then it is not
expected that edge dislocation will climb.
- Climb is a phenomenon that becomes increasingly important as the temperature rises.
Slip, on the other hand, is only slightly influenced by temperature.
49
物理冶金
Dislocation Intersections
魏茂國
 Dislocation intersections
- In Fig. 4.31 it is assumed that a dislocation has moved across the slip plane ABCD,
thereby shearing the top half of the rectangular crystal relative to the bottom half by
the length of its Burgers vector b. A second (vertical) dislocation, having a loop that
intersects the slip plane at 2 points, is shown in Fig. 4.31. It is assumed that this
loop is in the edge orientation where it intersects the slip plane.
- The indicated displacement of the crystal (Fig. 4.31) also shears the top half of the
vertical-dislocation loop relative to its bottom
half by the amount of the Burgers vector
b. The displacement lengthens the
vertical-dislocation loop by an
amount equal to the 2 horizontal
steps (Fig. 4.31).
Fig. 4.31 In the figure a dislocation is assumed to
have moved across the horizontal plane ABCD
and, in cutting through the vertical-dislocation
loop, it forms a pair of jogs in the latter.
50
Dislocation Intersections
物理冶金
魏茂國
This result is characteristic of the intersection of dislocations, for whenever
a dislocation cuts another dislocation, both dislocations acquire steps of a size equal
to the other’s Burgers vector.
- The first case, where the step lies in the slip plane of a dislocation, is called a kink
(Fig. 4.32). The second case, where the step is normal to the slip plane of a
dislocation, is called a jog (Fig. 4.33).
- The kink in the edge dislocation has a screw orientation (Burgers vector parallel to
line on), while the step in the screw dislocation has an edge orientation (Burgers
vector normal to the line on) (Fig. 4.32).
(A)
(B)
Fig. 4.32 Dislocations with kinks that lie in the slip plane of the dislocations.
51
物理冶金
Dislocation Intersections
魏茂國
Both of these steps can easily be eliminated by moving line mn over to the position of
the dashed line. This movement in both cases can occur by simple slip. Since the
elimination of a step lowers the energy of the crystal by the amount of the strain
energy associated with a step, it can be assumed that steps of this type may tend to
disappear.
- An edge and a screw dislocation, with steps normal to the primary slip plane, are
shown in Fig. 4.33. This type of discontinuity is called a jog.
(A)
(B)
Fig. 4.33 Dislocations with jogs normal to their slip planes.
52
物理冶金
Dislocation Intersections
魏茂國
- The edge dislocation with a jog (Fig. 4.33A) is free to move on the stepped surface,
for all 3 segments of the dislocation, mn, no, and op, are in a simple edge orientation
with their respective Burgers vectors lying in the crystal planes that contain the
dislocation segments.
- The screw dislocation with a jog (Fig. 4.33B) represents quite a different case. Here
the jog is an edge dislocation with an incomplete plane lying in the stepped surface.
Here the jog (line no), which is an edge orientation, is not capable of gliding along
the vertical surface because its Burgers vector is not in the surface of the step but is
normal to it. The only way that the jog can move across the surface of the step is for
it to move by dislocation climb.
Fig. 4.33
53
物理冶金
Stress Field of a Screw Dislocation
魏茂國
 Stress field of a screw dislocation
- The elastic strain of a screw dislocation is shown in Fig. 4.35. Consider the circular
Burgers circuit shown in Fig. 4.35. Such a path results in an advance (parallel to the
dislocation line) equal to the Burgers vector b. The strain in the lattice is the
advance divided by the distance around the dislocation.
 
b
2r
(4.7)
where r is the radius of the Burgers circuit.
This strain is accompanied by a corresponding state of stress in the crystal.
54
Fig. 4.35 Shear strain associated with a screw dislocation.
物理冶金
Stress Field of a Screw Dislocation
魏茂國
- Assuming the crystals to be homogeneous isotropic bodies, the elastic stress field
surrounding a screw dislocation is written:
   
b
2r
(4.8)
where  id the shear modulus of the material of the crystal.
- The analysis of the stress close to the center of the dislocation is extremely difficult,
and no completely satisfactory theory has yet been developed.
55
物理冶金
Stress Field of an Edge Dislocation
魏茂國
 Stress field of an edge dislocation
- It will be assumed that an edge dislocation lies in an infinitely large and elastically
isotropic material and that the dislocation line coincides with the z axis of a
Cartesian coordinate system. Under these conditions the stress can be considered to
be independent of position along the z direction. In Fig. 4.36A, with the aid of
elasticity theory it may be shown that the stress at some point, with coordinates x
and y, has the following components:
(A)
(B)
Fig. 4.36 (A) An edge dislocation aligned along
the z-axis. (B) The stress components at point
x, y.
56
物理冶金
Stress Field of an Edge Dislocation
 xx
 yy
 xy




y x  y 
b


2 1  v  x  y 
xx  y 
b


2 1  v  x  y 
魏茂國
y 3x 2  y 2
 b


2 1  v  x 2  y 2 2
2
2
2
2
2
2 2
(4.9)
2
2 2
where xx and yy are tensile stress components in the x and y directions, respectively,
and xy is the shear stress as shown in Fig. 4.36B.
- In the general case, that is, for an arbitrary position with coordinates x and y, the
stress will contain both normal and shear stress components. However, for points
along the x-axis the normal stress components both vanish and the state of stress is
pure shear. Also note that the sense of the shear stress along the x-axis, which
corresponds to the slip plane, is reversed if one moves from the right of the
dislocation to its left, as may be seen in Fig. 4.37.
57
物理冶金
Stress Field of an Edge Dislocation
魏茂國
- Also, as may be seen in Fig. 4.37, above and below the dislocation, that is, along the
y axis, there is no shear stress component. Here the stress is a biaxial normal stress
with xx = yy.
- Above the dislocation the lattice is under a compressive stress, while below it the
lattice is in a state of tensile stress. Note that the magnitude of the stress depends only
on the distance from the dislocation, r, and varies as 1/r.
(A)
(B)
58
Fig. 4.37 Stress and strain associated with an edge dislocation.
Stress Field of an Edge Dislocation
物理冶金
魏茂國
- A significant fact is that the stress field around a dislocation can be described in a
somewhat simpler fashion if one uses polar coordinates as defined in Fig. 4.38.
 b sin 

2 1  v  r
b
cos 


2 1  v  r
 rr    
 r
(4.10)
(B)
(A)
Fig. 4.38 (A) An edge dislocation in polar coordinates. (B) The corresponding stresses.
59
Force on a Dislocation
物理冶金
魏茂國
 The force on a dislocation
- Fig. 4.39 shows a left-hand screw dislocation, which is assumed to lie far enough
away from the crystal ends that surface end effects may be ignored. The length of
this dislocation, L, equals the crystal width. Now imagine that the dislocation moves
along the slip plane through a distance ∆x. This causes a section of the top half of
the crystal of width L and length ∆x to be displaced to the left by a Burgers vector,
b, relative to the bottom half.
- The external work W done by the applied
stress in this movement of the dislocation
is equivalent to that of a force L∆x moving
through a distance b or
W  Lxb
(4.11)
where  is the applied shear stress, L the
crystal width, and ∆x the distance the
dislocation moves.
60
Fig. 4.39 A left-hand screw dislocation in a long crystal.
物理冶金
Force on a Dislocation
魏茂國
- The internal work performed, as the dislocation moves, can be expressed as fL∆x,
where f is the virtual force per unit length on the dislocation, L the dislocation length,
and ∆x the distance through which the total force on the dislocation, fL, moves.
fLx  Lxb
f  b
(4.12)
The force per unit length on the dislocation, f, lies in the slip plane and is normal to
the dislocation line, that is, it is toward the front of the crystal.
- In Fig. 4.40 a positive edge dislocation is assumed to move through a distance ∆y
under an applied shear stress . The dislocation line length and crystal width are
assumed to equal L. It may be easily shown that the force per unit length on the edge
component of a dislocation line is also
f  b
(4.13)
This force also lies in the slip plane and is normal to the dislocation line.
61
物理冶金
Force on a Dislocation
魏茂國
Fig. 4.40 A positive edge dislocation.
- It can also be shown that if a part of a dislocation has a mixed-part edge and part
screw-character, the corresponding force per unit length, f, on the segment is also
normal to it.
Thus, a dislocation loop will be subjected to a force normal to the dislocation f = b
everywhere around the loop.
62
Force on a Dislocation
物理冶金
魏茂國
- Consider Fig. 4.41, where a tensile stress  is shown applied to a crystal containing a
positive edge dislocation. The given tensile stress will act to make the extra plane
increase in size or undergo negative climb. As a result, the dislocation line moves
downward. The force per unit length on the dislocation is given by
f  b
(4.14)
where f is the force on the dislocation line,
 the tensile stress, and b the Burgers vector
of the dislocation.
- The climb force on a dislocation is not only
normal to the dislocation line but it is also
normal to the slip plane of the dislocation.
- If the applied stress is purely compressive,
the climb force will point toward the positive
z direction and the dislocation should climb
upward.
63
Fig. 4.41 Climb force on an edge dislocation.
物理冶金
Force on a Dislocation
魏茂國
- It is common practice to define the orientation of the dislocation line, at a given point
along it, by the unit vector tangent to the dislocation line at the point. This vector
designated  and has the components x, y, and z.
- The Burgers vector can be expressed in the form
b  bxi  by j  bz k
(4.15)
where bx, by, and bz are 3 components of the Burgers vector and i, j, and k are unit
vectors in the x, y, and z directions, respectively.
64
物理冶金
Force on a Dislocation
魏茂國
- When a general stress , which may have both normal and shear components in all 3
directions, is applied to a crystal containing a dislocation, the force on the dislocation
is given by the cross-product of  and , which can be written as
i
j
k
f      x
y
z
x
y
z
(4.16)
where f is the force per unit length on the dislocation, and x, y, and z are given by
the following summations:
 x   xx bx   xy by   xz bz
 y   yx bx   yy by   yz bz
(4.17)
 z   zx bx   zy by   xx bz
Because of the cross-product between  and , the force on the dislocation is normal
to the dislocation.
- Peach-Kohler equation: The relation between the dislocation, its Burgers vector, and
the stress can be written
F  b    
(4.18)
65
Strain Energy of a Screw Dislocation
物理冶金
魏茂國
 Strain energy of a screw dislocation
- An important property of a dislocation is its strain energy, normally expressed as its
energy per unit length.
- According to linear elasticity theory, the strain energy density in the stress field of a
screw dislocation is 2/2, and by Eq. 4.8 the stress, , is b/2r. This suggests that
the strain energy per unit length of this screw dislocation might be estimated with
the following integration.
2
2
r ' b
b 2  r ' 
 b   1 
  2rdr  r
 ln  (4.19)
dr 
ws  r 
 
0
0
4r
4
 2r   2  
 r0 
where ws is the energy per unit length of the screw dislocation,  the shear modulus,
r'
b the Burgers vector, r0 an inner radius that excludes the dislocation core, and r’ an
outer limiting radius for the integration.
- Eq. 4.19 clearly indicates that as r0  0, ws ∞, implying that a lower limit for r0
is necessary. It has been suggested that one take r0 = b/, where  is a constant.
The value of  has been variously suggested to be between 2 and 4. For the present
let  = 4.
66
物理冶金
Strain Energy of a Screw Dislocation
魏茂國
- Eq. 4.19 also indicates that ws   as r’  . The stress field of a dislocation may
be assumed to be neutralized by those of its neighbors at a distance r’ equal to
one-half the average spacing between dislocations.
67
物理冶金
Strain Energy of an Edge Dislocation
魏茂國
 Strain energy of an edge dislocation
- An equation for the strain energy per unit length may also be derived for an
infinitely long edge dislocation using an approach similar to that used to obtain
Eq. 4.19. The result is
we 
b2
 r' 
 ln 
4 1  v   b 
(4.20)
where we is the strain energy per unit length of an edge dislocation,  the shear
modulus, v Poisson’s ratio, b the Burgers vector, and r’ the outer radius of the
volume over which the integration is carried out.
- Note that the strain energy for the edge dislocation differs from that of the screw
dislocation by a factor of 1/(1-v). Since v for most metals is near 1/3, the strain
energy for the edge dislocation is thus about 50% larger than that for the screw
dislocation.
68