DISLOCATION STRESS FIELDS

Dislocation stress fields → infinite body

Dislocation stress fields → finite body

Image forces

Interaction between dislocations

Theory of Dislocations J. P. Hirth and J. Lothe

McGraw-Hill, New York (1968)

Advanced reading (comprehensive)

MATERIALS SCIENCEMATERIALS SCIENCE&&

ENGINEERING ENGINEERING

Anandh Subramaniam & Kantesh Balani

Materials Science and Engineering (MSE)

Indian Institute of Technology, Kanpur- 208016

Email: anandh@iitk.ac.in, URL: home.iitk.ac.in/~anandh

AN INTRODUCTORY E-BOOKAN INTRODUCTORY E-BOOK

Part of

http://home.iitk.ac.in/~anandh/E-book.htmhttp://home.iitk.ac.in/~anandh/E-book.htm

A Learner’s GuideA Learner’s GuideA Learner’s GuideA Learner’s Guide

We start with the dislocation elastic stress fields in an infinite body The core region is ignored in these equations (which hence have a singularity at x = 0, y = 0)

(Core being the region where the linear theory of elasticity fails) Obviously a real material cannot bear such ‘singular’ stresses

Stress fields of dislocations

2 2

2 2 2

(3 )

2 (1 ) ( )

xx

Gb y x y

x y

2 2

yy 2 2 2

( )

2 (1 ) ( )

Gb y x y

x y

2 2

xy 2 2 2

( ) =

2 (1 ) ( )xy

Gb x x y

x y

2 2 2 2

2 2 2

[(3 ) 2 ( )] =

4 (1 ) ( )

xx

b y x y x y

x y

2 2 2 2

2 2 2

[( ) 2 ( )] =

4 (1 ) ( )

yy

b y x y x y

x y

2 2

2 2 2

( ) =

4 (1 ) ( )

xy

b x x y

x y

12 2

1tan

2 2(1 ) ( )x

b y xyu

x x y

2 22 2

2 2

( )(1 2 ) ln( )

8 (1 ) ( )

y

b x yu x y

x y

stress fields

Strain fields

Displacement fields

Plots in the coming slides

The material is considered isotropic (two elastic constants only- E & or G & ) → in reality crystals are anisotropic w.r.t to the elastic properties

Edge dislocation

2 2

2 2 2

(3 )

2 (1 ) ( )

xx

Gb y x y

x y

2 2

yy 2 2 2

( )

2 (1 ) ( )

Gb y x y

x y

Note that the region near the dislocation has stresses of the order of GPa

Material properties used in the plots are in the last slide

286 Å

286

Å

Stress values in GPa

yyxx

2 2

2 2 2

(3 )

2 (1 ) ( )

xx

Gb y x y

x y

2 2

yy 2 2 2

( )

2 (1 ) ( )

Gb y x y

x y

Position of the Dislocation line into the plane

More about this in the next slide

xx

Left-right mirror symmetry

Up down ‘inversion’ symmetry(i.e. compression goes to tension)

Tensile

Compressive

In an infinite body the xx stresses in one half-space maintain a constant sign (remain tensile or compressive) → in a finite body this situation is altered.

We consider here stresses in a finite cylindrical body. The core region is again ignored in the equations. The material is considered isotropic (two elastic constants only).

Stress fields in a finite cylindrical body

Plots in the coming slides

Finite cylindrical body

The results of edge dislocation in infinite homogeneous media are obtained by letting r2 → ∞

2

2 22

Eb sin r = 1

4 (1 ) rr r

2

2 22

Eb sin r = 1 3

4 (1 ) rr

2

2 22

Eb cos r = 1

4 (1 ) rr r

2 22 2

2 2 2 2 2 22 2

Eb r 3r = 3 1 + 1 y

4 (1 ) ( + y ) r rx

yx

x

2 22 2

2 2 2 2 2 22 2

Eb r r = 1 + 1 y

4 (1 ) ( + y ) r ry

yx

x

2 22 2

2 2 2 2 2 22 2

Eb r r = 1 + 1 y

4 (1 ) ( + y ) r rxy

xx

x

Cartesian coordinatesPolar coordinates

Str

ess

fiel

ds in

a f

init

e cy

lind

rica

l bod

y

286 Å

286

Å

Stress values in GPa

xx yy

xx

Left-right mirror symmetry

Up down ‘inversion’ symmetry(i.e. compression goes to tension)

Not fully tensile

Compressive stress

Like the infinite body the symmetries are maintained.But, half-space does not remain fully compressive or tensile

Tensile stress

The screw dislocation is associated with shear stresses only

Stress fields of dislocations Screw dislocation

xz zx 2 2

Gby = =

2 x y

yz zy 2 2

Gbx = =

2 x y

xx yy zz xy yz 0

xz zx

Gb Sin( ) = =

2 r

yz zy

Gb Cos( ) = =

2 r

Cartesian coordinates

Polar coordinates

Plots in the next slide

xz yz

572 Å57

2 Å

Stress values in GPa

For a mixed dislocation how to draw an effective “fraction” of an ‘extra half-plane’? For a mixed dislcation how to visualize the edge and screw component?

This is an important question as often the edge component is written as bCos →does this imply that the Burgers vector can be resolved (is it not a crystallographically determined constant?)

Understanding stress fields of mixed dislocations: an analogy

STRESS FIELD OF A EDGE DISLOCATIONSTRESS FIELD OF A EDGE DISLOCATIONX – FEM SIMULATED CONTOURS

(MPa) (x & y original grid size = b/2 = 1.92 Å)

27 Å

28 Å

FILM

SUBSTRATE

b

CONCEPT OF IMAGE FORCES & STRESS FIELDS IN THE PRESENCE OF A FREE SURFACE

A hypothetical negative dislocation is assumed to exist across the free-surface for the calculation of the force (attractive) experienced by the dislocation in the proximal presence of

a free-surface

A dislocation near a free surface (in a semi-infinite body) experiences a force towards the free surface, which is called the image force.

The force is called an ‘image force’ as the force can be calculated assuming an negative hypothetical dislocation on the other side of the surface (figure below).

Image force can be thought of as a ‘configurational force’ → the force tending to take one configuration of a body to another configuration.

The origin of the force can be understood as follows:◘ The surface is free of tractions and the dislocation can lower its energy by positioning itself closer to the surface. ◘ The slope of the energy of the system between two adjacent positions of the dislocation gives us the image force (Fimage = Eposition 1→2 /b)

In a finite crystal each surface will contribute to an ‘image dislocation’ and the net force experienced by the dislocation will be a superposition of these ‘image forces’.

d

GbFimage )1(4

2

An approximate formula derived using ‘image construction’

Importance of image stresses:If the image stresses exceed the Peierls stress then the dislocation can spontaneously move in the absence of externally applied forces and can even become dislocation free!

In a finite crystal each surface will contribute to an ‘image dislocation’ and the net force experienced by the dislocation will be a superposition of these ‘image forces’.

The image force shown below is the glide component of the image force (i.e. along the slip plane, originating from the vertical surfaces)

It must be clear that no image force is experienced by a dislocation which is positioned symmetrically in the domain.

2 2

2 2

1 1 2

4 (1 ) (1 ) 4image

Gb Gb xF

d L d L x

Superposition of two images

Glide

Climb

Superposition of two images

Similarly the climb component of the image force can be calculated (originating from the horizontal surfaces)

150b

= 4

29 Å

200b = 572 Å

Plot of x stress contours

Stress contour values in GPa

Highly Asymmetric Stress ContoursHighly Asymmetric Stress Contours

Edge Dislocation Near a Free Surface

Deformation of the free surfaceDeformation of the free surface

+7.0

+0.3

+0.18

+0.06

0.06

0.18

0.3

7.1

Simulation domain size (Lh): 500b 200b Scale for deformed shape: 3

F ree surface

30b

150b

= 4

29 Å

150b

= 4

29 Å

200b = 572 Å200b = 572 Å

Plot of x stress contours

Stress contour values in GPa

Highly Asymmetric Stress ContoursHighly Asymmetric Stress Contours

Edge Dislocation Near a Free Surface

Deformation of the free surfaceDeformation of the free surface

+7.0

+0.3

+0.18

+0.06

0.06

0.18

0.3

7.1

+7.0

+0.3

+0.18

+0.06

0.06

0.18

0.3

7.1

Simulation domain size (Lh): 500b 200b Scale for deformed shape: 3

F ree surface

30b

Deformation of the free surface in the proximity of a dislocation (edge here) leads to a breakdown of the formulae for image

forces seen before!

Left-right mirror symmetry of the stress fields broken due to the presence of free surfaces

Stress fields in the presence of an edge dislocation

Material Properties

Aluminum Silicon

Crystal structure Cubic Closed Packed Diamond cubic

Lattice type FCC FCC

Young’s Modulus 70.576 GPa 165.86 GPa

Shear Modulus 26.18 GPa 68.12 GPa

Poisson’s ratio 0.348 0.2174

C11 108 GPa 165.7 GPa

C12 62 GPa 63.9 GPa

Elastic

constants

C44 28.3 GPa 79.6 GPa

Anisotropic factor, H 10.6 GPa 57.4 GPa

Lattice parameter, a 4.04 Å 5.431 Å

bx[110] 2.86 Å 3.84 Å

by[110] 0 0

bx[100] --- 5.431 Å

Burgers

vector

by[100] --- 0

Material properties of Aluminium and Silicon used in the analysis