DISLOCATION MOVEMENT

• Because a dislocation represents the boundary b/w the slipped an unslipped region of a crystal, therefore it must:– be a closed loop, or

– end at the free surface of a crystal or

– end at a grain boundary.

• As shown in Fig. 12-1, dislocations will ordinarily take the form of curves or loops, which in three dimensions form an interlocking dislocation network.

• In considering a dislocation line, it can be:– edge,

– screw or

– mixed type, which can be resolved into edge and screw components.

• For example, in Figure 12-1, the dislocation loop is:– pure screw at point A

– pure edge at point B,

– along most of its length it has mixed edge and screw components.

Figure 12-1. Dislocation loop lying in a slip plane

• The process of cross slip illustrated in Fig 12-2, will serve as an example of dislocation loops.

• In Fig. 12-2(a), a small loop of dislocation line with

is moving on a (111) plane in an fcc crystal.

• The dislocation loop is pure positive edge at “w” and pure negative edge at “y”.

• At “x” the dislocation is a right-handed screw while at “z” the dislocation loop is a pure left-handed screw dislocation.

]011[2/oab

• At some stage (Fig. 12-2b), the shear causing expansion of the loop tends to move the dislocation on the intersecting plane.

• Since the dislocation is pure screw at “z”, it is free to move on this second plane.

• In Fig. 11-2(d), double cross slip has taken place as the loop glides back onto the original (111) plane.

)111(

)111(

Figure 12-2. Cross slip in a face-centered cubic crystal.

Cross-Slip Mechanism

Figure 12-3. Movement (glide) of an edge dislocation.

Figure 12-4. (a) Macroscopic deformation of a cube produced by glideof an edge dislocation. (b) Macroscopic deformation of a cube by glideof a screw dislocation. Note that the end result is identical for both situations.

• The macroscopic slip produced by the motion of an edge dislocation is shown in Fig. 12-4a and by a screw dislocation in Fig.12-4b.

• Both the shear stress and final deformation are identical for both situations, but for an edge dislocation line moves parallel to the slip direction while the screw dislocation moves at right angles to it.

• Recall the geometric properties of dislocations. This is summarized in Table 10-1.

Table 10-1 Geometric properties of dislocations Type of dislocation

Dislocation property Edge Screw Relationship b/w dislocation perpendicular parallelline l and b

Slip direction parallel to b parallel to b

Direction of dislocation line l parallel perpendicularmovement relative to b (slip direction)

Process by which dislocation climb cross-slip may leave slip plane

• Aside from the shear stress associated with the glide of edge, screw, and mixed dislocations, it is possible for edge dislocations to produce tensile strains by moving in a direction perpendicular to the Burgers vector [Fig. 12-5(a)].

• This type of motion is called dislocation climb and involves the transport of atoms (or vacancies) away from or to the dislocation (edge of the extra half plane) by diffusion.

• Because dislocation climb is diffusion-controlled, it is important only at relatively high temperatures where the diffusion rates are fairly rapid.

• Motion of an edge dislocation showing both glide and climb motion is illustrated in Fig. 12-5(b).

Figure 12-5(a) . Illustration of climb of edge dislocation

Figure 12-5(b) . Illustration of glide and climb of edge dislocation.

Relation Between Dislocation Movement and Plastic Flow

• To demonstrate how we can relate the motion of individual dislocations to macroscopic plastic flow, let us consider the strain produced by the motion of a single dislocation.

• Suppose we have a crystal in the shape of a cube , where the edge length is L.

• Under the application of an appropriate shear stress, a dislocation is somehow produced on the left-hand side of the crystal and is allowed to move all the way through the crystal as shown in Fig 12-6.

• The plastic shear strain associated with the passage of this dislocation had not passed all the way through the crystal but had gone only a distance x , where x<L.

• Then the overall shear strain would somewhere between 0 and b/L, the exact value being proportional to the fraction of the slip plane area that the dislocation has traversed. In moving a distance x, the dislocation has swept put x/L of its slip plane and the resulting shear strain is given by

• If instead of one dislocation we had considered N dislocation, all moving an average distance x [ Fig 12-7], the resulting strain would be

Lb

Lx

p (12.1)

2L

bxNp (12.2)

Figure 12-6. Relation between shear strain and dislocationmovement in glide.

Figure 12-7 . Relation between shear strain and dislocation movement in glide.

• Which can be rewritten

• where NL is the total length of the dislocation line and is the dislocation density, NL/L-3. In units of length per unit volume. The shear strain is then the product of the dislocation density (the number of dislocations that have moved), the Burgers vector, and the average distance each dislocation has moved. If the shear strain b occurs over a time t, then we can write

or

xbL

bxNLb

3(12.3)

txb

tb

(12.4)

vbp (12.5)

• Where is the shear strain rate and is the average dislocation velocity. Equation (12.5) is a type of transport equation that often occurs in physics.

• It simply states that the strain rate is equal to the density of defects producing strain (dislocations) times the strength of each defect (the amount of strain associated with each defect), times the average velocity of the defects.

• Later we shall show that an identical equation exists for describing electrical conduction. For electrical conduction we write

p v

vnqJ (12.6)

Stress Required for Slip

• The flow stresses of crystals are highly anistropic.

• For instance, the yield stress of zinc under uniaxial tension varies by a factor of at least 6.

• Consequently, it is very important to specify the orientation of the load.

• In shear or torsion tests, the shear plane and directions are precisely known.

• Because dislocations can glide only under the effect of shear stresses, these shear stresses have to be determined.

• In uniaxial tensile and compressive test ( the most common tests), one has to determine mathematically the shear component of the applied stress acting on the plane in which slip is taking place.

• Figure 12-8 shows a crystal with a normal cross-sectional area A upon which a tensile load acts, generating a uniaxial stress P/A.

Figure 12-8. Relationship between stress axis and slipplane and direction.

• The slip plane and direction are indicated, respectively, by the angles and that they make with the tensile axis.

• We take the normal n of the plane that makes an angle

with the loading direction .

• The areas A1 and A are related by the angle . Area A is the projection of A1 onto the horizontal plane; thus, we can write

cos1AA

(12.7)

• The shear stress acting on the slip plane and along the slip direction S is obtained by dividing the resolved load along the slip direction ( ) by A1:

• But is the normal stress applied to the specimen. Hence,

• Note that

coscoscos

1 AP

AP

cosP

AP /

coscos

sincos

(12.8)

(12.9)

• This equation shows that will be zero when either or is equal to 0o.

• When both and are equal to 45o, the shear component is maximum and we have, in this case,

• The angle between any two directions a and b can be obtained from the scalar product of these vectors:

or

245cos45cosmax

oo

cosbaba

babacos

(12.10)

(12.11)

(12.12)

• For cubic crystals, planes and directions with the same indices are perpendicular, and the angle is determined from the coefficients, h, k, and l. For two vectors

and

the angle is given by

2/122

22

22

2/121

21

21

212121

)()(cos

lkhlkh

llkkhh

kljkiha 111

kljkihb 222

(12.13a)

(12.13b)

(12.14)

• If the two directions are perpendicular, their dot product is zero; and the same is true for a direction that is contained in a plane.

• From Eq. 12.11, it is possible to obtain the and terms for all desired crystallographic directions of a crystal.

• For instance, if the loading directions is [123] for an FCC crystal, then the Schmid factors of the various slip systems are found by obtaining the angles of [123] with <111> (perpendicular to slip planes) and [123] with (110) (in slip directions).

cos cos

• Note that each slip plane contains three slip directions and that 12 values (4 x 3) have to be obtained.

• Schmid and coworkers used the variation in the resolved shear stress to explain the great differences in the yield stresses of monocrystals of certain metals.

• They proposed the following rationalization, known as the Schmid law: Metals flows plastically when the resolved shear stress acting in the plane and along the direction of slip reaches the critical value

• where the factor M is usually known as the Schmid factor

ooc M cossin

coscoscossin M

(11.15)

(11.16)

• A single crystal subjected to a shear stress can deform extensively with slip on only a single slip system. However, this is not the case for polycrystals.

• Since each grain in a polycrystal has a different crystallographic orientation, each will respond differently (slip in a different direction) when subjected to a shear stress.

• Thus, unless the region around grain boundaries can be undergo arbitrary shape changes, voids at grain boundaries will be opened up and the material will fracture at low strains.

• Example: An NaCl single crystal is loaded in uniaxial compression with the [001] direction parallel to the compression axis. Dislocation motion is first observed on the slip system when the applied tensile stress is 500 psi. Calculate the inherent lattice resistance to dislocation motion.

• Solution: The resolved shear stress on the slip system is given by Eq.(6.7), where

• or for slip since . The shear stress necessary to overcome the inherent lattice resistance to dislocation motion is therefore 250 psi.

]110)[101(

coscos

]110)[101(

2/ ]110)[101( o45