Plastic Deformation Elementary Dislocation Theorytra-erf.org/attachments/Dislocation/disloc-3.pdf · Plastic Deformation & Elementary Dislocation Theory Lecture Course for the Students

Plastic Deformation &

Elementary Dislocation Theory

Lecture Course for the

Students of Metallurgical Engineering

V V Kutumbarao

LECTURE - 3

2

Brief Review of

Part – 1: The Etchpit

Enter the Dislocation

3

Modes of Plastic Deformation - Slip

4

• Rough approximation for Theoretical Shear Sress

required to cause slip = G/2p

• G for most metals = 20 to 150 GPa

• Therefore, theoretical shear strength should be

3 to 30 GPa

• Experiments show that the critical shear stress for plastic

deformation in single crystals is mostly 0.5 to 50 MPa

• Large discrepancy of over a 1000 times!

• Therefore, real materials are a 1000 times weaker than

they should theoretically be.

SO EXPLAIN THIS TERRIBLE WEAKNESS

OF REAL MATERIALS

5

Caterpillar analogy applied to deformation

• Slip occurs in small steps and progresses by the

movement of the boundary between the deformed and

undeformed regions inside the crystal.

☺We call it a dislocation!

Boundary

Slipped Not yet deformed

t

t

6

Brief Review of

Part – 2: The Prism Loop

Going in Deeper

7

Geometrical properties of dislocations

Dislocation is coplanar with the slip plane

Dislocation is a line that can have any shape including

a loop

Dislocation can only move in its own slip plane

Dislocation always moves in such a way as to increase

the slipped area

Displacement produced when a dislocation moves

completely out of the crystal is called the

“Burgers Vector” of the dislocation

Dislocation line perpendicular to b : Edge

Dislocation line parallel to b : Screw

Dislocation line inclined to b : Mixed

8

• Burgers Circuit Rules– Circuit traversed in the same manner as a rotating R-H screw

advancing in the direction of the dislocation.

– Circuit must close completely in a perfect crystal and must go completely around the dislocation in the real crystal.

• Vector that closes the circuit in the imperfect crystal is the Burgers Vector

9

Burgers Vectors of Dislocations in

different Crystal Structures

Crystal Slip Plane Slip Direction Burgers Vector

sc {100} <110> a[110]

fcc {111} <110> a/2 [110]

bcc {110} <111> a/2 [111]

{112} {123}

hcp (0001) <1120> a[1120]

Atomic Arrangement around a Screw Dislocation

10

How dislocations overcome obstacles

on the slip plane

11

Cross slip of Screw

dislocation

12

Stress Fields around Dislocations

Edge Dislocation

• sx is the largest normal stress

compressive for y > 0 tensile for y < 0

• txy is maximum at y=0

222

22

)(

)(

)1(2 yx

yxyGby

ps

)(

2

12 22 yx

yGbz

ps

222

22

)(

)(

)1(2 yx

yxxGbyx

pt

222

22

)(

)3(

)1(2 yx

yxyGbx

ps

Screw Dislocation

Polar: tqz = Gb/2p r

i.e.,independent of q

• No normal stresses

• Only a shear stress field which has complete radial symmetry

)(2

)(2

0

0

22

22

yx

xGbyx

yGb

yz

xz

xy

zyx

pt

pt

tsss

Polar coordinatesEdge:

Screw: tqz = Gb/2p r

• Stress becomes infinite at r = 0

• So for a small cylindrical region r = r0

around dislocation (called the core), the equations are not valid

• r0 is of the order of 0.5 to1nm (b to 4b)

r

SinGbr

q

pss q

)1(2

r

CosGbrr

q

ptt qq

)1(2

13

• Escrew ½ Gb² L

• Energy per unit length ½ Gb²

• Eedge ½ G b² L/(1-)

• If = ⅓, Eedge 3/2 Escrew for same

length

• Energy length, so dislocations tend to have minimum length

• Preferred shape is a straight line or a circular loop

• Thus dislocation may be thought of as having a line tension T = dE/dl ½Gb²

• Analogous to surface tension of a liquid.

Strain Energy of a Dislocation

14

So far so good.

We postulated the existence of a dislocation

and derived some of its properties.

OK, but what about our basic problem?

….. an explanation for the observation that

strength of real materials is

far less than theoretical shear strength.

Can our dislocation theory explain it?

15

Shear stress required to move a dislocation in a periodic lattice

where w = ‘width’ of the dislocation dhkl/b

• For d b and ½

tp 2 x10⁻⁵ G, close to observed

shear strength

• Metals: w 10b

tp = 4G exp (-2p x10) ~ 2x10⁻²⁷G

i.e., almost negligible

• Real metals are much stronger than predicted!

b

dGb

wG

hkl

p

)1(

)2(exp

1

2

)2(exp

1

2

p

p

t

• To minimize the energy of interface

between slipped and unslipped

regions, interface thickness w should

be low.

• w = width of the dislocation

• Smaller the width, lower the

interfacial energy

• Wider the dislocation, lower the

elastic energy

• So an equilibrium width exists.

Force required to move a dislocation: Peierls – Nabarro force

16

Some other useful properties of dislocations

• Force on a dislocation F due to

an applied shear stress t is

F = t b• F is the same at all locations

and is always perpendicular to dislocation line

• Shear stress required to bend a dislocation of length l to a radius R

t = Gb/2R = Gb/ l

• Shear strain g due to dislocation movement by x

g = r b x

where r is disloc density = no. of dislocs per unit area, usually

around 10⁵/cm²

• Consider a 1cm³ crystal of well annealed copper

• b = 0.256 nm; r = 10⁵/cm² ; x = 0.5 cm

• g = 10⁵ x 0.256 x 10⁻⁹x 0.5 1.28

x10⁻³ 0.128%

• Even if we consider r =10⁶, strain produced when all dislocations move out of the crystal is < 2.6%.

• Very low compared to observed values: 20 to 50%.

• Mechanism required for generating additional dislocations during plastic deformation.

• One such is the Frank-Read source

17

Dislocation Multiplication

The Frank-Read Source

18

Operation of a Frank – Read Source

• Consider a dislocation

line DD′ lying in its slip

plane and moving along a

direction perpendicular to

b (screw dislocation).

• It is pinned at D and D′,

so immobile at these two

points. (D and D′ are

nodes or impurity atoms).

• So the dislocation starts

to bend and bulges out to

a radius of curvature

R = G b / 2 t

• Shear stress required to

bend a dislocation of

length l to a radius R

t = Gb/2R = Gb/ l

19

• R decrease as t increases

• Eventually when R = L / 2

(semi circle)

R = L/2 = Rmin

(R cannot be reduced further) and

t= tmax

• The stress required to

produce this configuration

tmax = Gb / L

Operation of a Frank – Read Source

20

• Beyond this, R will increase.

• The dislocation loop expands under a decreasing stress

• Eventually starts bending on itself producing points m&n - edge components of opposite sign very close to each other.

• When m&n meet they annihilate each other producing a perfect lattice.

Operation of a Frank – Read Source

21

• The dislocation regains its original shape and is ready to reproduce again. Eventually the dislocation loops emitted by this source pile up at a grain boundary and cause a back stress.

• When this back stress

equals tmax, the source stops emitting dislocations

Operation of a Frank – Read Source

22

Dislocation Multiplication: The Frank-Read Source

• Pinned at D and D’, screw disloc

bulges out to a radius of curvature

R = G b/2 t

•R decreases as t increases R = L/2 = Rmin at t = t max = Gb/L

•Then R increases, disloc loop expands, starts bending on itself producing points m&n - edge components of opposite sign.

•When m& n meet they annihilate each other producing a perfect lattice

•Disloc regains its original shape and is ready to reproduce again.

•Emitted dislocs pile up at a grain boundary and cause a back stress. When back stress =tmax, the source stops emitting.

23

Part 3: The Partials

Getting Real

Some General Vector Operations

Test for coplanarity of (hkl) plane & [uvw] direction:

hu+kv+lw=0

Thus slip directions in a (111) plane are

Or their negatives

Rule for vector addition:

c₁ a [h₁ k₁ l₁] + c₂ a [h₁k₁l₁] = a [c₁h₁+c₂h₂, c₁k₁+c₂k₂, c₁l₁+c₂l₂]

Or

c₁c₂ a [h₁k₁l₁] + c₁c₂ a [h₂k₂l₂] = c₁c₂ a [h₁+h₂, k₁+k₂, l₁+l₂]

For example

] 111[ 2

] 333 [ 6

] 1 11[ 6

] 224 [6

] 1 11[ 6

] 112 [ 3

000000aaaaaa

24

] 1 01 [], 1 0 1 [], 0 1 1 [

] 1 1 0 [], 1 0 1 [], 10 1 [

Partial Dislocations• Dislocation with b = one

lattice spacing: unit dislocation or perfect dislocation

• Possible to lower energy by dissociation

• Energy Criterion for dislocation dissociation:

• If

• Then

• Dislocations with b < 1 atomic spacing possible in close packed lattices

• called Partial Dislocations

b3b2b1

b3 b2b1²

25

Partial Dislocations in fcc: Shockley Partial

Perfect dislocation (shortest vector)

Slip on {111} planes in <110> directions

It is easier for slip to occur by following the path b2 b3 than b1If b1 = Then b2 =

b3 =

Energetically favorable since b1² > b2 +b3

]110[2

0a

26

] 1 0 1 [ 2

a

] 1 1 2 [ 6

a

] 2 1 1 [ 6

a

Shockley Partial (contd..)

• This dissociation results inStacking Fault

• Combination of the two partials :

Extended Dislocation • b2 and b3 at 60° angle to

each other− Repel each other but surface tension tends to

pull them together • So equilibrium separation

- determined byStacking Fault Energy

27

• Vacancy aggregation on one close packed plane

• Collapse of adjoining parts of neighbouring planes

• Dislocation bounding the collapsed sheet

• Encloses an area of stacking fault which can not move with the dislocation

• b = a/3 [111]

− Normal to the (111) plane

− Not a slip plane

− So, immovable (sessile)28

Partial Dislocations in fcc: Frank Partial

Partial Dislocations in fccLomer – Cotterell dislocation

• Dislocation on the ( 1 1 1) plane

Plus

• Dislocation on the plane

• dissociate and move towards each other

29

]101[2

0a

)111(

]101[2

0

a

]112[6

]121[6

]101[2

000 aaa

]211[6

]211[6

]101[2

000 aaa

Lomer – Cotterell Stair Rod dislocation (contd..)

• The two leading partials interact to produce a third partial:

• lies parallel to the line of intersection of the slip planes

− Has a pure edge character in the (100) plane.

− its Burgers vector does not lie in either of the planes of the stacking fault. So, Sessile

− Called a Stair Rod dislocation

30

]011[6

]211[6

]121[6

000aaa

]110[6

0a

31

Dislocations of like sign on same slip plane:

• Large separation: act independent of each other

− Energy ∝ 2 X

• Small separation: act like one dislocation:

− Energy ∝

= 2 X above

∴ Dislocations of like sign on the same slip plane tend to repel each other

p2

2Gb

p2

)2( 2bG

Dislocation –Dislocation Interactions

Dislocation –Dislocation Interactions (contd..)

32

• Dislocations of like sign on parallel slip planes:

Form an array

• Line up above one other cancelling out each other’s stress fields

• Lattice bending around the array: Small angle boundaries

Tilt , if array is of edges

Twist, if array is of screws

33

• Unlike dislocations on same slip plane:

− Attract and annihilate each other forming a perfect lattice

Dislocation –Dislocation Interactions (contd..)

• Unlike dislocations on parallel slip planes: − Attract each other and form a row of

vacancies or interstitials

Dislocation Interactions (contd..)Dislocations on intersecting slip planes

• Interaction produces a sharp break of

two types

Jog: break out of slip plane

Kink: break in slip plane

Case 1: Two edges of b

perpendicular to each other

• Produces a jog PP’ in AB:

• Length = b1, b = b2

• Edge orientation can move readily

with AB

• No jog produced in XY, only an

increase in length by b2, b remains

= b1

• (In general: jog produced if Burgers

vector of intersecting dislocation is

normal to the other dislocation line)

34

A

B

P

P’

b1

X Y

b1X

Y

A

B

b2

Case2: Orthogonal edges with parallel b • Both acquire jogs on crossing

• Jog in AB is PP’ = b1 with b = b2

• Jog in XY is QQ’ = b2 with b = b1

• Both are screw in character and lie in the original slip plane

• So they are only kinks (in-plane offsets)

• Unstable, since during glide they line up and annihilate the offset.

35

X

Y

BA b2

b1

P

P’A

X

Y

Q

Q’

b1

Dislocation Interactions (contd..)Dislocations on intersecting slip planes

b1

b2

Case 3: Intersection of edge and screw

• Jog with edge orientation in edge

• Kink with edge orientation in screw

Case 4: Intersection of two screws

• Jogs of edge orientation in both screw dislocations

• Important type of interaction

36

Dislocation Interactions (contd..)Dislocations on intersecting slip planes

X

Y

b1

A

B

A

B

X

Y

b1

A

B

X

Y

b1

A

B

X

Y

b1

• In general jogs in edge dislocation do not impede their motion

• Jogs in screw dislocations all have edge orientation

• Can only move by climb of the edge segment

37

Jogs in Dislocations

AP , P’B : screw

segments

PP’ : edge Jog

PP’BC: its slip plane

PP’Q’ Q : extra half

plane

b

A

B

P

P’Q

A’

B’

Q’C

Jogs in Dislocations (contd..)

• Stable jog = Increase in length of dislocation line

= increase in energy

• Energy of a dislocation = ∝Gb² (∝ = 0.5 to 1)

• Energy of jog of length b2 in dislocation of b = b1 is

Uj= ∝Gb₁²b₂ (∝ ~ 0.2)

• less for the jog than the dislocation

• Energy of a jog is ~0.5 to 1.0 ev in metals

• Intersection of screw dislocations produces non conservative jogs.

• Jogs are of edge character, so cannot move with the dislocation

• If they move along with the rest of the dislocation, they either create vacancies on interstitials.

• Vacancy jogs more likely than interstitial jogs as they have less energy.

ed 2

38

39

Stress Fields Around Dislocations

Edge Dislocation

– Elastic stress fields surround dislocation lines

– Interact with other dislocations and solute atoms.

– Expressions using theory of elasticity

– Consider the X- section of a cylindrical piece of material,

assumed to be continuous and elastically isotropic

– Edge dislocation can be made by a radial cut along OA

(y=0) and sliding cut surfaces by AA1.

– Edge dislocation running through point O normal to

paper, parallel to Z axis: so strains along Z are zero

– Plane strain situation, so stresses are:

40

Stress Fields around Dislocations : Edge

•Consider the X- section of

a cylindrical piece of

material, assumed to be

continuous and elastically

isotropic

•Edge dislocation can be

made by a radial cut along

OA (y=0) and sliding cut

surfaces by AA1.

•Edge dislocation running

through point O normal to

paper, parallel to Z axis: so

strains along Z are zero

back

41

Stress Fields around Dislocations: Edge

Plane strain situation:

Expressing the stresses

in polar co-ordinates

sr = sq = - Gb sin q2p(1-) r

(sr along r ; sq along plane

perpendicular to r)

sz = tqz = 0trq = tqr = Gb cos q

2p(1- ) r

back

42

Stress Fields around Dislocations: Edge

G b y (x² –y²)sy =

2p(1-) (x² + y²)²

G b y (3x²+y²)sx = -

2p(1-) (x² + y²)²

sz = (sx + sy)

G b 2 y= -

2p(1-) (x² + y²)

G b x (x²-y²) txy =

2p(1-) (x² + y²)

txz = tyz= 0

gxz = gyz = 0

gxz = 1/G txz and

gyz = 1/G tyz

Stresses in Rectilinear coordinates

2

. .

.

sx is the largest normal stress

compressive for y > 0

tensile for y < 0

txy is maximum at y=0back

43

sx = sy = sz = 0

txy = 0

txz = - Gb y

2p (x²+y²)

tyz = Gb x

2p (x²+y²)

Polar: tqz = Gb/2pr, i.e., independent of q

eqz = ezq b/4pr

g = b/2pr

Screw Dislocation

– Shear displacement along

Z direction only.

– No strain in the X and Y

directions.

– So no stress in the X and

Y direction either

– Only a shear stress field

which has complete radial

symmetry

Stress Fields around Dislocations: Screw

back

44

Stress Fields around Dislocations (contd..)

Edge: sr = sq = - Gb sin q

2p(1-) r

and trq = tqr = Gb cos q

2p(1- ) r

Screw: tqz = Gb/2pr

Note: From the above expressions it is seen that

Stress becomes at r = 0

So for a small cylindrical region r = r0 around

dislocation (called the core), the equations are

not valid.

r0 is of the order of ~ 0.5 to1nm (b to 4b usually)

back

45

Strain Energy of a Dislocation

• Consider a cylindrical

crystal of length L

with a screw

dislocation of Burgers

vector b along its axis

• Elastic shear strain in

a thin annular section

of radius r and

thickness dr is

g = b / 2pr

46

Strain Energy of a Dislocation

• Energy per unit volume of the annulus,

dE/ dv = ½ tg = ½ Gg²

= G/2 b /2pr}²

dv = 2p r L dr

dE = G b² L/4p . dr/r

• Total strain energy E = Gb²L/4p ln R /r0

• Total including core = Gb²L/4p ln R /r0 + Eo

47

• For r0 ~ b, E0 is negligible

• Also approximate ln R/ r0 ~ 2p

Thus E ~ Gb² L/2

or Energy per unit length ~ Gb² /2

• Energy of an edge dislocation:

E ~ G b² L/4p(1-) ln (R/r0) + E0

~ G b² L/2(1)

• If = 1/3, Eedge ~ 3/2 Escrew for same length

Strain Energy of a Dislocation

back

48

Strain Energy of a Dislocation (contd..)

• Energy b dislocations tend to have minimum b slip direction is the closest packed direction in the crystal

• Energy l dislocations tend to have minimum l preferred shape is a straight line or a circular loop

• Thus dislocation may be thought of as having a line tension T - dE ~ Gb²

dl 2

• Analogous to surface tension of a liquid.

2

49

Force required to move a dislocation

• Perfect crystal: Same force

opposing the movement of

atoms acts on all the atoms.

• If a dislocation is present:

Only a small movement of

atoms near the dislocation.

Distant atoms remain in

minimum energy positions.

• Atoms around dislocation symmetrically placed on opposite

sides of the extra half plane: equal and opposite forces

oppose and aid this motion.

Atom movements around an edge dislocation

50

Force required to move a dislocation (contd..)

• To a first approximation no net force on the

dislocation, so stress required to move it is zero!

• To minimize the energy of interface between

slipped and unslipped regions, interface

thickness w should be low.

• w = width of the dislocation

• Smaller the width, lower the interfacial energy

• Wider the dislocation, lower the elastic energy

• So an equilibrium width exists.

51

Peierls – Nabarro force

tp = 2G exp - 2pw

1- b

~ 2G exp (-2p dhkl)

1- (1-)b

w = ~ dhkl

1-

where dhkl = Interplanar spacing of slip planes

• Note w appears in the exponential term

A net force is required to move the dislocation in a

periodic lattice : the Peierls – Nabarro force.

52

• Say d ~ b and ~ ½

tp = 2G exp (-2p .1) ~ 4G exp (-4p)

1-½ 1-½

• Thus tp ~ 2 x10 G, close to observed shear strength.

• Highly ductile metals: w ~ 10b

tp = 4G exp (-2p x10) ~ 2x10 G

i.e., almost negligible

So, now explain why real metals are much stronger than predicted!

- 27

- 5

Peierls – Nabarro force

53

Peierls – Nabarro force

• Ceramics, w ~ 1b

tp = 4G exp (-2p) ~ 7.5x10 G

i.e, ~10 times larger than in metals!

• Also in ceramics, restricted slip systems due to the

requirements of electrostatic forces. Results in the low

degree of plasticity observed in ceramics.

• At high temperatures, ceramics become more ductile

because thermal activation helps the dislocations to

overcome the energy barrier.

- 3

24

back

54

Force on a Dislocation due to an Applied Stress

Dislocations move under external

force to produce slip.

Equivalent to saying that there is

force acting on a dislocation line

tending to drive it forward.

Consider dislocation segment ds

moving along b under uniform

external stress t

ds is displaced by dl

Area swept by dislocation ~ ds dl

Let total area of slip plane = A

55

Force on a Dislocation (contd..)

This occurs under the external force

= stress x area = t A

Work done on the dislocation

dw = t A (dsdl) b/A

Let F be the force per unit length on dislocation line

Work done by dislocation = F ds .dl (dl is the distance over which it moves)

F ds dl = t(ds dl) b

Or F = t bF is the same at all locations and is always perpendicular to dislocation line

Displacement produced when dislocation sweeps whole

area = b (definition of b)

Displacement produced when dislocation sweeps area

ds dl = ds dl b/A

back

56

dq/2

T

T

dq/2

T Sin dq/2

T Sin dq/2

F = t b dS

Force required to bend a dislocation

back

57

Force required to bend a

dislocation

• Line tension T will tend to straighten the

dislocation line.

• Dislocation can only remain curved if the

shear stresses produce a force sufficient

to resist the line tension.

• Let the dislocation be bent to a radius of

curvature R

• Shear stress required to maintain it at

this radius can be estimated as follows

58

Angle subtended by an element of the dislocation line ds

dq =ds / R

Outward force on dislocation line F= tb ds

Inward force due to line tension on ds = 2T sin dq/2~ 2T dq/2 for small dq

Force required to bend a dislocation (contd..)

For balance

T dq = tb ds

t = T / bR = Gb² / 2bR = Gb / 2R

Thus shear stress required to bend a dislocation

= Gb/2R = Gb/ l

back

59

Shear Strain due to Dislocation Movement

Dislocation produces offset of length b when it completely goes out of a crystal of length say L and height h

Shear strain g = b / h

60

For partial movement by a distance x, displacement di

would be proportional to the fractional distance x / L

Thus di = (xi / L) b

Total displacement of top of crystal relative to bottom due to many dislocations on many slip planes is

N

Δ = Σdi = (b/L) Σ xi1

where N is the total number of dislocations that moved.

Shear Strain due to Dislocation Movement (contd..)

61

Shear Strain due to Dislocation Movement (contd..)

Macroscopic shear strain

N

g = Δ / h = (b/hL) Σ xi1

Average distance

dislocations have moved

N

x = Σ xi / N1

Then

g = b N x / h L

62

Dislocation density r is defined as

r = no. of dislocations per unit area (no. / cm2 )

= total dislocation line length per unit volume (cm / cm3)

r = N / h L

g = r b x

Expressed in terms of rates:

Shear strain rate g = r b v

where

v = average dislocation velocity

= dx / dt

Shear Strain due to Dislocation Movement (contd..)

back

63

• Strain due to dislocation movement g = r b x

• Consider a 1cm³ crystal of copper (b = 0.256 nm)

Well annealed: r = 10 / cm²

x = 0.5 cm

g = 10 x 0.256 x 10 x 0.5

~ 1.28 x10 ~ 0.128%

• Even if we consider 10 dislocations / cm², the total strain produced when all the dislocations move out of the crystal is < 2.6%.

• Very low compared to observed values: 20 to 50%.

• So a mechanism is required for generating additional dislocations during plastic deformation.

One such is the Frank-Read source

- 9

5

6

5

- 3

Shear Strain due to Dislocation Movement (contd..)

Point defects in Crystals

64back

Elementary Dislocation Theory

End of Lecture 3