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NEEP 541 Hardening
Fall 2002
Jake Blanchard
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Outline Hardening
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Radiation Hardening Radiation tends to increase the strength
of metals Point defects Impurity atoms Depleted zones Dislocation loops Line dislocationsVoids
precipitates
negligible
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Two Mechanisms Increase stress needed to start
dislocation motion (source hardening)
Impede dislocation motion (frictionhardening)
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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
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Dislocation
Source
=pinning point
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Frank-Read Source
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Animation
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Frank-Read Source
Si
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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
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Force on a Dislocation
s
R
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Model for critical shear stress
R
Gb
bRT
sT
sT
LF
Rs
GbT
lengtharcs
tensionlineT
=
===
=
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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=
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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 +==
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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
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Dislocation Network Model
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Select a Unit Cell
Dislocation loop
Find force on loop from network of linedislocations
L determined by dislocation density
L
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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
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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
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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
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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
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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
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Obstacles
L
Area=A
Radius=r
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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
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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
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Picture of Model
R
LoLo
h
LoLo=distance betweenpinning points
L=distance betweenobstacles
Lo>L
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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
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Diagram
LoLa
If La
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Diagram
LoLa
If La>Lo, then
dislocation doesnot cut throughand La becomesthe pinning pointdistance
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Strain Rate Strain is determined by step size, which is b
Shear strain is b/a
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ModelingAssume N1 loops in a volume V
Assume each loop grows by amount dA
N1adA=dV
1/a=N1dA/dV
Dislocation density:
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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
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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
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Overcoming Obstacles
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Shearing Obstacles Slicing a sphere is easier off the diameter
Obstacle radius about 10 angstroms
Average radius is
r
rr3
2'=
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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
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Temperature Effects
=
=
=
rN
bvk
UT
T
T
T
T
c
co
co
2ln
2
1
1
2/33/2
3/23/2
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Temperature Dependence Plot
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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
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Saturation Modeling
VNdt
dN
densityclusterN
ss =
=
# zones percollision
Collision rate perunit volume
V=volume around depletedzone that is unavailable for
cascade production
Destruction rate
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Saturation Modeling
( )[ ]
( )Vt
N
VtV
N
N
ss
s
s
=
=
exp1
exp11
0)0(
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