Forces B/w Dislocations

• Consider two parallel (//) edge dislocations lying in the same slip plane.

• The two dislocations can be of same sign or different signs

(a) Same Sign (on same slip plane)

• When the two dislocations are separated by large distance: The total elastic energy per unit length of the dislocation is given by:

• When dislocations are very close together: The arrangement can be considered approximately a single dislocation with Burgers vector = 2b

In order to reduce the total elastic energy, same sign dislocations will repel each other (i.e., prefer large distance separation).

222 2 GbGbGbU sep

22 42)( GbbGUEnergyElastic couple

(14.34)

(14.35)

sepcouple UU

• Dislocations of opposite sign (on same slip plane)

• If the dislocations are separated by large distance:

• If dislocations are close together: Burgers vector = b - b = 0

Hence, in order to reduce their total energy, dislocations of opposite signs will prefer to come together and annihilate (cancel) each other.

222 2)( GbbGGbU sep

couplesep

couple

UUand

U

0

• The same conclusions are obtained for dislocation of mixed orientations

• (a) and (b) above can be summarized as:– Like dislocations repel and – unlike dislocations attract

Dislocations Not on the Same Slip Plane

• Consider two dislocations lying parallel to the z (x3) -axis:

• In order to solve this:

(a) We assume that dislocation “I” is at the origin

(b) We then find the interaction force on dislocation “II” due to dislocation “I”

x2

x1

x3

I

II

• Recall Eqn. 14.29

• Note that the dislocation at the origin (dislocation I) provides the stress field, while the Burgers vector and the dislocation length belongs to dislocation II

• Since is edge:

• Also bII is parallel to x1: Therefore,

This means that b2 = b3 = 0 and b1 = b

tH

tbF IIIII

)(

14.29

I 032233113

0

0

b

bII

• Since tII is parallel to x3, then

This means that t1 = t2 = 0 and t3 = 1

• From Eqn. 14.31, we can write:

Therefore

1

0

0

IIt

kHtHtjHtHtiHtHtF ˆˆ)(ˆ)(ˆ211213313223

jHtiHtF ˆˆˆ1323

But

Therefore, F along the x1 Direction is given as: 21b

This component of force is responsible for dislocation glide motion - i.e., for dislocation II to move along x1 axis.

3132121111 bbbH

3232221212 bbbH

3332321313 bbbH

22

221

22

21

1

2

)1(21

xx

xxx

GbFx

14.30

F along the x2 Direction is given as: - 11b

This component of force is responsible for climb (along x2).

• At ambient (low) temperature, Fx2 is not important (because, no climb).

• For edge dislocation, movement is by slip & slip occurs only in the plane contained by the dislocation line & its Burgers vector.

222

21

22

21

2

2

11)(

)3()1(22 xx

xxx

vGb

bFx

14.31

• Consider only component Fx1

For x1>0: Fx1 is negative (attractive)

when x1<x2 for same sign, or

x1>x2 for opposite sign.

For x1<0: Fx1 is positive (repulsive)

when x1>-x2 same sign disl. or x1<-x2 for opposite sign disl.

Fx1=0 when x1 = 0, x2,

)45( o)45( o

Usually for edge dislocationsof same sign

For edge dislocationsof opposite signs

• Hence

• Stable positions for two edge dislocations.

900 450

• Equations 14-30 and 14-31 can also be obtained by considering both the radial and tangential components. The force per unit length is given by:

• Because edge dislocations are mainly confined to the plane, the force component along the the x direction, which is the slip direction, is of most interest, and is given by:

rGb

F

rGb

Fr

2sin

12

112

2

2

14.32

14.33

Eqn. 14-34 is same as 14-30. Figure 14-5 is a plot of the variation of Fx with distance x, using equation 14-34. Where x is expressed in units of y. Curve A is for dislocations of the same sign; curve B is for dislocations of opposite sign. Note that dislocations of the same sign repel each other when x > y, and attract each other when x < y.

222

222

12

sincos

yx

yxxGb

FFF rx

14.34

Figure 14-5. Graphical representation of Eq. (14-21). Solid curve A is for two edge dislocations of same sign. Dashed curve B is for two unlike two dislocations.

• Example: A dislocations lies parallel to [100] with Burgers vector b<110>. Compute the force acting on the dislocation due to the stress field of a neighboring screw dislocation lying parallel to [001].

Assume that for the screw dislocations yzxz

x2

x1

x3

Solution:

S

Let the screw dislocation be dislocation I at the origin.

The stress field for screw dislocation is given by:

based on the assumption,

we have

011

100

100

0

00

00

yzxz

yz

xz

For the other dislocation

22

0

0

0

2/

2/

011

100

100

0

0

1

0

2/

2/

0

1

1

2

bbb

b

bH

t

b

bb

b

• (continued)

jb

bkji

ttt

HHH

kji

tHF

ˆ2

2

0012

200

321

321

(b)(a)

Figure 14-6. (a) Diffusion of vacancy to edge dislocation; (b) dislocation climbs up one lattice spacing