Stress Fields and Energies of Dislocation

Stress Field Around Dislocations

• Dislocations are defects; hence, they introduce stresses and strains in the surrounding lattice of a material.

• The mathematical treatment of these stresses and strains can be substantially simplified if the medium is considered to be isotropic and continuous.

• Under conditions of isotropy, a dislocation is completely described by the line and Burgers vectors.

• With this in mind, and considering the simplest situation, dislocations are assumed to be straight, infinitely long lines.

• Figure 14-1 shows a hollow cylinder sectioned along the longitudinal direction. This is an idealization of the strains around an edge dislocation.

Figure 14-1. Simple model for edge dislocation.

• The deformation fields can be obtained by cutting a slit longitudinally along a thick-walled cylinder and displacing the surface by b perpendicular to the dislocation line.

Edge Dislocation

Figure 14-2b. Deformation of a circle containing an edge dislocation. The unstrained circle is shown by a dashed line. The solid line represents the circle after the dislocation has been introduced.

• The cylinder, with external radius R, was longitudinally and transversally displaced by the Burgers vector b, which is perpendicular to the cylinder axis in the representation of an edge dislocation.

• An internal hole with radius ro is made through the center.

• This is done to simplify the mathematical treatment.

• In a continuous medium, the stresses on the center would build up and become infinite in the absence of a hole; in real dislocations the crystalline lattice is periodic, and this does not occur.

• In mechanics terminology, this is called singularity. A “singularity” is a spike, or a single event. For instance, the Kilimanjaro is a singularity in the African plans.

• Therefore, we “drill out” the central core, which is a way of reconciling the continuous-medium hypothesis with the periodic nature of the structure.

• To analyze the stresses around a dislocation, we use the formal theory of elasticity.

• For that, one has to use the relationships between stresses and strains (constitutive relationships), the equilibrium equations, the compatibility equations, and the boundary conditions.

• Hence, the problem is somewhat elaborate.

22

22 )3(

yx

yxbyox

222

22

)(

)(

yx

yxbyoy

222

)(yx

vyv o

yxz

(14.1)

(14.2)

(14.3)

Stress Field Due to Edge Dislocations

where

)1(2 vG

o

(14.4)

222

22

)(

)(

yx

yxbxoxy

0 yzxy

(14.5)

(14.6)

z

yxy

xyx

edge

bygivenisndislocatioedgeforfieldstressThe

00

0

0

:

(14.7)

• The largest normal stress is along the x-axis. – This is compressive--- above slip plane.

– tensile---------- below slip plane.

xy shear stress is maximum in the slip plane, i.e. when y=0

• The stress field can also be written in Polar Coordinates, and this is given as:

r

bor

sin

rb

orr

cos

(14.8)

(14.9)

x

Figure 14-2a. Simple model for screw dislocation.

• The deformation field can be obtained by cutting a slit longitudinally along a thick-walled cylinder and displacing a surface by b parallel to the dislocation line.

Screw Dislocation

Stress Field Due to Screw Dislocations

• This has complete cylindrical symmetry

• The non zero components are:

• In Cartesian coordinate, the stress field matrix is given as:

222 yx

yGbxz

222

yxxGb

yz

(14.10)

(14.11)

0

00

00

yzxz

yz

xz

screw

(14.12)

• There are no extra half plane of atoms.

Therefore, there are no compressive or tensile normal stresses.

• The stress field of the screw dislocation can also be expressed in Polar-coordinate system as:

rGb

2 (14.13)

Strain Energy

• The elastic deformation energy of a dislocation can be found by integrating the elastic deformation energy over the whole volume of the deformed crystal. The deformation energy is given for

(a) Edge Dislocation

1 1 cos21

21 2rr

rr oro o r

drbbdrU

orr

vGb

U 12

ln)1(4

(14.14)

(14.15)

(b) Screw Dislocation

• Note that:

for both edge and screw dislocations

• If we add the core energy (ro ~ b), the total Energy will be given by:

o

rr z r

rGbbdrU

o

12

ln42

11

(14.16)

brGb

Ut1

2

ln4

(14.18)

2GbU (14.17)

• For an annealed crystal: r1~ 10-5cm, b ~ 2 x10-8 cm

Therefore,

• Strain energy of dislocation ~ 8eV for each atom plane threaded by the dislocation.

• Core energy ~0.5eV per atom plane

• Free energy of crystal increases by introducing a dislocation.

2ln 1

br

2

2GbU (14.19)

Forces on Dislocations

• When a sufficiently high stress is applied to a crystal:

Dislocation move

Produce plastic deformation

Slip (glide) Climb (high Temperatures)

• When dislocations move it responds as though it experiences a force equal to the work done divided by the distance it moves

• The force is regarded as a glide force if no climb is involved.

b

dl

ds

Figure 14-3. Force acting on a dislocation line.

• The crystal planes above & below the slip plane will be displaced relative to each other by b

• Average shear displacement =

where, A is the area of the slip plane

• The external force on the area is

Therefore, work done when the elements of slip occur is:

bAdlds

A

bAdsdl

AdW

(14-20)

(14-21)

• The glide force F on a unit length of dislocation is defined as the work done when unit length of dislocation moves unit distance.

Therefore,

• Shear stress in the glide plane resolved in the direction of b

bF

bdAdW

dldsdW

F

(14.22)

(14.23)

Line Tension:

• In addition to the force due to an externally applied stress, a dislocation has a line tension, T which is defined as the energy per unit length = force tending to straighten the line

• The is analogous to the surface tension of a soap bubble or a liquid.

• Consider the curved dislocation. The line tension will produce forces tending to straighten the line & so reduce the total energy of the line.

2GbUel (Energy)

Figure 14-4. Forces on a curved dislocation line.

• The direction of the net force is perpendicular to the dislocation and towards the center of curvature

• For small , F ~ 2T

• But

dislocation segment Radius of curvature

Rds 1

*2

sin

sin2TF (14.24)

• The line will only remain curved if there is a shear stress which produces a force on the dislocation line in the opposite sense. [recall equation 14.23]

• equation 14.25 & 14.26 gives

RTds

Rds

TF

1*

2*2

(14.25)

bdsF (14.26)

RTds

bds

Recall

Stress required to bend a dislocation to a radius R

bRT

2GbT

RGb

bRGb

2

(14.27)

(14.28)

• A more general form of eqn 14.23 is given as

where, t is the dislocation line vector

• Expanding equation 14.29 gives:

Gt

or

btF

)(

bG

321

321

GGG

ttt

kji

F

(14.29)

(14.30)

• Note eqns. 14.26, 14.29 & 14.31 are the same

• A particular direct applicant of these is in the understanding of the Frank-Read dislocation multiplication source

kGtGtjGtGtiGtGtF ))()( 122131132332

kGtGtjGtGtiGtGtF ))()( 211213313223 (14.31)

• Forces on dislocations can be due to other dislocations, precipitates, point defects, thermal gradients, second-phases, etc.

3132121111 bbbG

3232221212 bbbG

3332321313 bbbG

3

2

1

333231

232221

131211

b

b

b

bG

(14.32)

(14.33)