neep541 hardening

8/3/2019 NEEP541 Hardening

1/41

NEEP 541 Hardening

Fall 2002

Jake Blanchard


2/41

Outline Hardening


3/41

Radiation Hardening Radiation tends to increase the strength

of metals Point defects Impurity atoms Depleted zones Dislocation loops Line dislocationsVoids

precipitates

negligible


4/41

Two Mechanisms Increase stress needed to start

dislocation motion (source hardening)

Impede dislocation motion (frictionhardening)


5/41

Source Hardening Stress to initiate dislocation motion is

associated with unpinning of Frank-

Read source This source increases dislocation

density as a result of deformation


6/41

Dislocation

Source

=pinning point


7/41

Frank-Read Source


8/41

Animation


9/41

Frank-Read Source

Si


10/41

What stress is required to

activate source? Shear stress acting on the dislocation,

which is pinned by defects, distorts

dislocation We can estimate the stress needed to

bend the dislocation beyond the critical

strain needed to activate the sourceand create a new loop


11/41

Force on a Dislocation

s

R


12/41

Model for critical shear stress

R

Gb

bRT

sT

sT

LF

Rs

GbT

lengtharcs

tensionlineT

=

===

=


13/41

Critical Stress Critical point is when radius is half the

distance between pinning points

(dislocation is semi-circular)

Decreasing distance between pinningpoints increases stress needed toinitiate motion

Gb

2=


14/41

Friction Hardening Defects impede dislocation motion

2 sources of resistive force Long range forces from interaction with

other dislocations Short range forces from obstacles

sLRi stressfriction +==


15/41

Long Range Stresses Dislocations repel each other because

of stress fields associated with

interruption of lattice structure Model dislocation as an ordered array of

defects


16/41

Dislocation Network Model


17/41

Select a Unit Cell

Dislocation loop

Find force on loop from network of linedislocations

L determined by dislocation density

L


18/41

Modeling Let =total length of dislocations in

cube/cube volume (dislocation density)

=(12/4)L/L3=3/L2 (each dislocation shared by 4 unit cells) L=(3/)1/2

Loop is only affected by paralleldislocations (4 top, 4 bottom)

Approximate force by force only on

parallel dislocations


19/41

Modeling( )

( )

( )

( ))(sin)(cos)(sin

)(sin)(cos)cos()sin(

12/

12/

222

22

2

2

=

=

=

=

y

x

yy

xx

f

f

y

fGblengthunitF

y

fGblengthunitF

Fy

Fx

y


20/41

Modeling Maximum force (Fx) is at angle where

fx is a maximum

Differentiate fx and set to 0 Maximum angle is 22.5 degrees

Maximum value of fx is 0.25

Let poissons ratio=1/2

Y=L/2


21/41

Modeling

( )

===

=

=

=

322

/

2

225.0

5.02/

22

d

LR

LRLR

LR

Gb

L

Gb

blengthunitF

L

Gb

L

GblengthunitF

Applied stress must overcome this force to movedislocation

Increasing dislocation density increases thisfriction stress


22/41

Short Range Forces Short range stresses are due to

obstacles lying in the slip plane

Force is exerted at point of contact Two types:

Athermal=bowing around obstacle

Thermal=climbing over or cutting throughbarrier (energy is supplied by thermalactivation)

Friction stress depends on distance

between obstacles


23/41

Obstacles

L

Area=A

Radius=r


24/41

Modeling N=particle density

Slab volume is 2rA

Number of particles in slab=2rAN

Average distance between particles=L

L2

*2rAN=A

rN

b

L

b

rNL

2

2

1

=

=

More defectsimplies higher

strength


25/41

Hardening by Depleted Zones Significant at low fluence and low

temperatures

Mechanism is thermally activatedfriction hardening

Thermal activation allows dislocation to

cut through or jump over obstacle Dislocation is moved by short range

stress


26/41

Picture of Model

R

LoLo

h

LoLo=distance betweenpinning points

L=distance betweenobstacles

Lo>L


27/41

Model

( )

3/12

2

2

42

22

222

2

2

2

2

=

=

+

=+

=

+=

=

=

GbL

L

Gb

L

L

L

L

L

Gb

h

hLR

hRLR

R

Gb

hLL

o

o

oo

o

o

o

So the dislocation line

adjusts its position until Losatisfies this equation


28/41

Diagram

LoLa

If La


29/41

Diagram

LoLa

If La>Lo, then

dislocation doesnot cut throughand La becomesthe pinning pointdistance


30/41

Strain Rate Strain is determined by step size, which is b

Shear strain is b/a


31/41

ModelingAssume N1 loops in a volume V

Assume each loop grows by amount dA

N1adA=dV

1/a=N1dA/dV

Dislocation density:


32/41

Modeling

bv

dt

dRb

dt

d

bdRd

RdRdA

RN

dAbN

a

bd

RNdV

dV

RN

dd

d

d

d

d

==

==

==

=

=

2

2

2

2

1

1

1

1

R=loop radius

V=dislocationglide velocity


33/41

Glide Velocity

=

=

=

kT

UbL

dt

d

kT

ULv

bvdt

d

d

d

dd

*

*

exp

exp

Velocity depends on T,activation energy, and

thermal vibrationfrequency

Increasing temperatureincreases strain rate

because it becomeseasier to overcomeobstacles


34/41

Overcoming Obstacles


35/41

Shearing Obstacles Slicing a sphere is easier off the diameter

Obstacle radius about 10 angstroms

Average radius is

r

rr3

2'=


36/41

Stress to penetrate obstacle The stress needed to cut a model can

be approximated as:

Grb

NU

rNL

GbLL

rbL

U

o

o

2

2/3

2

3/12

24

3

2

1

2

'4

>

=

=

>

R=obstacle size

N=obstacle density

B, G =material properties


37/41

Temperature Effects

=

=

=

rN

bvk

UT

T

T

T

T

c

co

co

2ln

2

1

1

2/33/2

3/23/2


38/41

Temperature Dependence Plot


39/41

Fluence DependenceAccording to the model, the strength is

proportional to the square root of the

fluence But saturation occurs

The theory is that as depleted zones

get too close, their hardening effect isdiminished


40/41

Saturation Modeling

VNdt

dN

densityclusterN

ss =

=

# zones percollision

Collision rate perunit volume

V=volume around depletedzone that is unavailable for

cascade production

Destruction rate


41/41

Saturation Modeling

( )[ ]

( )Vt

N

VtV

N

N

ss

s

s

=

=

exp1

exp11

0)0(

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