Stress Fields Around Dislocations -...
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Stress Fields Around Dislocations The crystal lattice in the vicinity of a dislocation is distorted (or strained). The stresses that accompanied the strains can be calculated by elasticity theory beginning from a radial distance about 5b, or ~ 15 Å from the axis of the dislocation. The dislocation core is universally ignored in calculating the consequences of the stresses around dislocations. The stress field around a dislocation is responsible for several important interactions with the environment. These include: 1. An applied shear stress on the slip plane exerts a force on the dislocation line, which responds by moving or changing shape. 2. Interaction of the stress fields of dislocations in close proximity to one another results in forces on both which are either repulsive or attractive. 3. Edge dislocations attract and collect interstitial impurity atoms dispersed in the lattice. This phenomenon is especially important for carbon in iron alloys.
Transcript of Stress Fields Around Dislocations -...
Stress Fields Around DislocationsThe crystal lattice in the vicinity of a dislocation is distorted (or strained). The stresses that accompanied the strains can be calculated by elasticity theory beginning from a radial distance about 5b, or ~ 15 Åfrom the axis of the dislocation. The dislocation core is universally ignored in calculating the consequences of the stresses around dislocations. The stress field around a dislocation is responsible for severalimportant interactions with the environment. These include:1. An applied shear stress on the slip plane exerts a force on the dislocation line, which responds by moving or changing shape.2. Interaction of the stress fields of dislocations in close proximity to one another results in forces on both which are either repulsive or attractive.3. Edge dislocations attract and collect interstitial impurity atoms dispersed in the lattice. This phenomenon is especially important for carbon in iron alloys.
Screw DislocationAssume that the material is an elastic continuous and a perfect crystal of cylindrical shape of length L and radius r. Now, introduce a screw dislocation along AB. The Burger’s vector is parallel to the dislocation line ζ . Now let us, unwrap the surface of the cylinder into the plane of the paper
A
B
L
b
2πr
rGbG
rb
πγτ
θπ
γ
2
tan2
==
==
Then, the strain energy per unit volume is:22
2
82 rπGbγτrgyStrain ene =
×=
We have identified the strain at any point with cylindrical coordinates (r,θ,z)
r
A
B
Slip planez
θ r
A
B
Slip planez
θ
τθZτZθ
rGbGZ π
γτθ 2== The elastic energy associated with an element is its
energy per unit volume times its volume.The volume of a pipe is rrδπ2
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛== ∫
0
121
0 22 ln
41
22
21
rrGb
πδr
πrπrGbth unit LengEnergy per
r
r
To obtain the total energy locked in the crystal due to the screw dislocation we need to integrate the above equation for all values of rwith a ro (minimum) of 5b.
Shear plane
Escrew =
Gb2
4πln
rr0
⎛
⎝ ⎜
⎞
⎠ ⎟ ≈
Gb2
2
Energy per unit length of screw dislocation (integrating from r0 to r):
Elasticity theory breaks down for r0~5b so core energy is ignored here.
14π
lnrr0
⎛
⎝ ⎜
⎞
⎠ ⎟ ≈
12
Roughly, due to r dependence
The total strain energy of a dislocation is the sum of the elastic strain energy plus the energy of the core of the dislocation (about 1/15th of the total energy – quantum mechanical calculations).
We have shown the distortion of a cylindrical element by a screwdislocation and the equivalent to a simple shear type of distortion. When translated to a coordinate system, the only shear possible are those with a z‐component.
The strains given in cartesian and cylindrical coordinates are:
( )
( ) rθ
πb
yxx
πbγγ
rθ
πb
yxy
πbγγ
cos22
sin22
223223
223113
=+
==
−=
+−
==
All the other strains should be zero in an isotropic material. The associated stresses are given by: ( )
( ) rθ
πGb
yxx
πGb
rθ
πGb
yxy
πGb
cos22
sin22
0
223223
223113
2112332211
=+
==
−=
+−
==
=====
σσ
σσ
σσσσσ
The strain field surrounding the core of a screw dislocation can be represented as:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
00000
3231
23
13
εεεε
ε
Where each value εij depends on the x‐yposition for dislocation lying along the z‐axis (or 3‐axis).
Energy of Edge Dislocations
Eedge =
Gb2
4π(1−ν)ln
rr0
⎛
⎝ ⎜
⎞
⎠ ⎟ ≈
Gb2
2(1−ν)
For idealized edge component, one entire plane has been pushed into the other planes above the glide plane but not below (tensile + compressive stresses). Hence, there is Poisson Effect along length of line, which yields a (1‐v) in denominator for strain.
compression
Idealized Edge
For many metals, ν~ 1/3, so
Eedge =
32
Escrew
Elasticity theory breaks down for r0~5b so core energy is ignored here.
Slides presentations taken from http://web.mse.uiuc.edu/courses/mse406/Handouts/index.html
Dislocation Stress Fields: Edges
See book, Hirth and Loath.
zxryxrxx
=====
3
2
1
sincos
θθ
Edges with z‐axis line direction best describe in x‐y plane in cartesian coordinates. Use:
General trend:• Above the edge (x=0, y>0), pure compression.• Below the edge (x=0, y<0), pure tension.• Along the slip plane (y=0), pure shear.• All other locations, compression + tension + shear.
+Edge at center with u=(001) and b=(100) interacting with +edge somewhere nearby
y=x
y=–xOther stresses are zero!
compression
)()1(2)1(
32)(
31
)(–
)1(2
)(1
)1(–)(
)(–
)1(2
)(3
)1(2–
22
222
22
22
222
22
222
22
yxyGbp
yxyxGbx
yxGby
yxyxGbyyxyxGby
zzyyxx
yxxy
yyxxzz
yy
xx
+−+=++−=
+−+
==
+−=+=
+−+
=
++
−=
νπνσσσ
νπττ
νπνσσνσ
νπσ
νπσ
The elastic displacements around edge dislocations in isotropic materials include all three normal strains εxx, εyy and εzz, and the shear strains in the x‐y plane γxy.
For an edge dislocation with a core along the z‐axis and the Burger’s vector in the positive x‐direction ⎥
⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
33
2221
1211
0000
εεεεε
ε
Energy and Forces between Edge dislocations
Eedge ≈
Gb2
2(1−ν)Idealized
Roughly, your expectation should be (as found from intuition):
b
–bb=0
Energy before: 2Gb2
Energy after: Gbtot2 = 0
b b
b=2
Energy before: 2Gb2
Energy after: Gbtot2 = G(2b)2 = 4Gb2
Should attract
Should repel
Mixed DislocationsMixed dislocations are dislocation segments wherein the angle between the Burgers vector and the line direction is neither 90o (edge) or 0o
(screw).
Each mixed dislocation can be resolved into edge and screw components.
Emixed =
G(b⊥)2
4π(1−ν)ln
rr0
⎛
⎝ ⎜
⎞
⎠ ⎟ +
G(b||)2
4πln
rr0
⎛
⎝ ⎜
⎞
⎠ ⎟
Energy has component from both types:
Edge Screw
b
b
bu
b|| = bsinθ
b⊥ = bcosθ
θ
Edge
Screwmixed
Emixed =
Gb2
4π(1−ν)ln
rr0
⎛
⎝ ⎜
⎞
⎠ ⎟ (1−ν cos2θ)
14π
lnrr0
⎛
⎝ ⎜
⎞
⎠ ⎟ ~
12
(core energy < 10% of E)
Combining (Screw, Mixed, Edge):
π2
−θ
Etotal = Ecore + Emixed (θ)
Where θ is the angle between the Burger’s vector and the line direction.
The elastic energy of a dislocation can be generalized as:
lGbE ndislocatio2α=
Where α is a dimensionless factor (0.5‐1.0) and l is the dislocation length. It can be noted that smaller values of b lead to smallest energy.
(111) fcc plane
Partial Dislocations b = b1 + b2
a2
1 01[ ]=a6
2 11[ ]+a6
1 1 2[ ]
b1yb1
b1x b2x
b2b2y
b1y and b2y are attractive screw segments b1x and b2x are repulsive edge segments
b
FCC Partial Dislocations and Stacking Faults
b
Here partials form, edge repulsion wins out, which creates stacking faulted region in between.
Energies of Full and Partials are E1 = Gb2 E1+2 = Gb1
2 + Gb22
E1+2 = 2
Ga2
36(4 + 1+ 1) =
Ga2
32)101(
4
222 GaGabGbGbE =++=•== >
Favorable for partials to form, i.e. dislocation disassociate.
Dislocations may be sessile if not on the correct slip plane.
If energy is favorable, Gb2 > Gb12 + Gb22, then partial dislocation form. (We need to show: Ga2/2 > Ga2/3)
due to …ABC… stacking
A AB BC
FCC
HCP
C
partial
b
Motion of partials
Separation of partials
Partial dislocations move apart. As they move apart leave hcp SF ribbon.ABC = 3 layersAB = 2 layersABCABC converts to ABABAB
Partial Dislocations b = b1 + b2
a2
1 01[ ]=a6
2 11[ ]+a6
1 1 2[ ]In FCC, due to …ABC… stacking, if partials form, edge repulsion wins out, which creates stacking faulted region in between. Green Partials Separate.
Stacking Faults are defects that cost energyEnergy balance between separating partials to lower elastic energy and creation of more SF.
Non‐conservative Motion for Edge Dislocation: Vacancy‐assisted Climb
Edge climbs up
Climb is non‐conservative in work.
Only a part of the dislocation line climbs up, hence it will generate jogs. Edge can climb down too!
⊥
Vacancy:Missing atom
Swapped with atom at bottom of edge
b, τ
bxu2
n
bxn
Slip plane
Not a slip plane
With jogs, an edge dislocation will have sections that are sessile! Why? Because segments are not laying in possible slip plane.
Moves = Glissile Does not move = Sessile
b, τ
n
bxn
Slip plane
• n is the slip plane normal.• b is the Burger’s vector of edge dislocation.• τ is shear stress (could be applied or from other dislocation lines).
Vacancy‐assisted Climb creates jogs!
τ
bxn
b
Jogs will create sessile edge dislocation segments
Conservative Motion for Screws: Cross‐Slip
b u
The Burger’s vector is : This is a ScrewdislocationIt moves in the direction of
b u
Cross‐Slip of Screw Component
For a FCC
( )1 1 11 =n( )1 1 12 =n
( )1 1 11 =n
( ) ( ) [ ]1 0 11 1 11 1 121 =×=×= nnb
( )1 1 11 =n( )1 1 12 =n
( )1 1 11 =n
( ) ( ) [ ]1 2 11 0 11 1 11 =×=×= bnd
Here partials are favorable, Gb2 > Gb12 + Gb22, since Ga2/2 > Ga2/3.
Partial Dislocations b = b1 + b2a2
1 01[ ]=a6
2 11[ ]+a6
1 1 2[ ]
b1 yb2x
b1 b2b2y
b1x
a4
1 01[ ]
screw
edge edgea4
1 01[ ]
u for edge
Separation of Partials: Stacking Fault
In FCC crystals, the magnitude of the Burgers’ vector is:
2ab =
As the edge components have the same direction, the b1x and b2x components of the partials will repel.As the screw components of the partial dislocations b1yand b2y are in opposite direction these will attract.
[ ][ ]121
12
12112
2
1
ab
ab
y
y
=
=
dGbbGbF s
sScrew πγτ
2
2
±=±=±=
The attractive force between the parallel screw dislocations separated by a distance d is:
As the partials separate, the energy increase by d*SFE, where SFE is the stacking fault energy and d is the fault separation distance. Thus, the “chemical” force resisting separation is SFE (dimensions of joules per square meter or force per unit length).
The repulsive force between the edge components of the dislocations is:
222
222
)(–
)1(2 yxyxxGbFedge +−
+=
νπ
Since the dislocations are in the same plane and separated by a distance d, then y=0 and x=d d
GbFedge )1(2
2
νπ −=
Since the units of SFE are in force per unit length, and so are those of the forces above, we can just equate them directly.
dGb
dGbSFE 1
)1(21
2
22
νππ −=+
EdgeScrew FFSFE =+
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−
−⋅= 1
)1(1
2
2
νπ SFEGbd
For a perfect dislocation 0 1 12a
6 112
6abab ==For a partial dislocation
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−
−⋅= 1
)1(1
12
2
νπ SFEGad
High SFE low separation, low SFE large separation in TEM.
Stacking Faults and Energy Partial dislocation repel and leave stacking faults
)(221
SFEbGbdsf π
•=
For ideal case:
Stacking Faults are defects that cost energyEnergy balance between separating partials to lower elastic energy and creation of more SF. Partial Dislocations b = b1 + b2
b
b1
b2dSF
hcp
fccfcc
TEM image
In TEM, you see contrast between faulted and unfaulted regions, hence, you can measure dSFand get SFE.
D
A B
C
αβ
γ
δ
Full (Mixed) Dislocation Recombination: Lomer‐Cottrell “Locks”
Consider two slip planes in FCC crystals, δ and γ,namely (111) and (11‐1) planes .
[ ] [ ] [ ] 1 1 02
1 0 12
0 1 12 321
ababab ===
The three perfect dislocations in the (111) plane are
[ ] [ ] [ ] 1 1 02
1 0 12
0 1 12 654
ababab ===
The three perfect dislocations in the (11‐1)plane are
The direction b1 and b4 are in opposite direction and they will cancel each other. The combination b2 and b5 will result in a higher energy (not possible). The sole combination that result in a energy reduction is the b3 and b5. [ ] [ ] [ ]
222
0112
1012
1102
222
53
aaa
a a abb
>+
=+=+ This dislocation is not mobile in either (111) or (11‐1) planes.
Full (Mixed) Dislocation Recombination: Cottrell “Locks”
(MIXED) full dislocation reaction: b = b1 + b2
b1 =
a2
101[ ]
b2 =
a2
01 1 [ ] n2 = 111( )
n1 = 11 1 ( ) b1 • n1 = 0
b2 • n2 = 0
Check slip planes?
b
Mixed dislocations: u’s are all parallel to intersection, and b’s are not ⊥ to u’s.
n=(001)
motion
motion
11 1 ( )
111( )
u1 = 1 1 0[ ]
u2 = 1 1 0[ ]
u = 1 1 0[ ]
b1 =
a2
101[ ]
b2 =
a2
01 1 [ ] b =
a2
110[ ]
Burger’s vector, b? (See figure above in cube)
Favorable to recombine? Yes, Gb12 + Gb22 > Gb2
Slip Plane? does not lie in either of the two slip planes, but does lie in n = b x u= (001).
Glissile or Sessile? Sessile, not {111} fcc slip plane
b1 + b2 =
a2
101[ ]+a2
01 1 [ ]= b =a2
110[ ]
b1
2+ b2
2= a2 > b2 =
a2
2
b =
a2
110[ ]
• Unless lock (sessile dislocation) is removed, dislocation on same plane cannot move past. • Going back b => b1 + b2 would allow other dislocation to glide again.
It impedes slip and therefore is called a lock.
Schockley Partial Dislocations Recombination: “Locks”Partial dislocations reaction: b = bp1 + bp2
b1 =
a2
101[ ]→ b1p1 + b1
p2 =a6
112[ ]+a6
2 1 1[ ]
Lormer‐Cottrell lock.But if full b’s combine, it is Cottrell lock.
b1 •n1 = 0 n1= 111 ( )
b2 • n2 = 0 n2 = 111( )
b2 =
a2
01 1 [ ]→ b2p1 + b2
p2 =a6
1 2 1 [ ]+a6
112 [ ]
n
motion
motion
11 1 ( )
111( )
u1 = 1 1 0[ ]
u = 1 1 0[ ]
b1p 1⊥u
b2p 2⊥u
b1p 2
b2p 1
Burger’s vector? Leading partials combine
Favorable to recombine? Check Gb12 + Gb22 > Gb2
Line Direction?
Slip Plane?
Glissile or Sessile? Sessile, not {111} fcc slip plane
b1
p2 + b2p1 =
a6
2 1 1[ ]+a6
1 2 1 [ ]= b =a6
110[ ]
b1
2+ b2
2=
a2
3> b2 =
a2
18
u = n1 ×n2 =i j k1 1 1
1 1 1
= i2 − j2 + k0 → [11 0]
n = b × u = [00 1 ]
Unless lock (sessile dislocation) is removed, dislocation on same plane cannot move past. Other possible combinations give:
b =a3
110[ ]
Edge‐Edge Interactions: creates edge jogs
Dislocation 1 got a “jog” in direction of b2e of the other dislocation; thus, it got longer.Extra atoms in half‐plane increases length.
This dislocation got a jog in direction of b1e.
after
b1e
b2e
before
b1e
b2e
**Dislocations each acquire a jog equal to the component of the other dislocation’s Burger’s vector that is normal to its own slip plane.
Energy cost of jog: Gb2/2 (Energy/length) x b (length of jog) = Gb3/2
What happens when dislocations are extended, i.e. composed of two trailing partials?
Screw‐Edge Dislocation Interactions: creates edge jogs
Time snap shots
Energy cost of jog is Gb3/2
Edge jog is in direction of bs.
Jogs slow motion of dislocation.
Why does screw also have jog?
be
bs
be
edge
edge (later)
B of screw
Direction of screw dislocation motion
bsline
beThis end of edge goes underneath creates jog .
Screw‐Edge Dislocation Interactions: creates edge jogs
edge (early)
edge (later)
B of screw
Direction of screw dislocation motion
bsline
be
Edge jog is in direction of bs.
Jogs slow motion of dislocation.
be
bs
Screw jog is in direction of be.
Why is there a jog in screw?
Screw‐Screw Dislocation Interactions: creates edge jogs
screw
screw (later)
bsline
bs
Energy cost of jog Gb3/2
Jogs slow motion of dislocation.
In screw‐screw case, jog has to move via CLIMB, or generate a row VACANCIES or INTERSTITIALS.
Climb is non‐conservative, and point‐defects costs more energy.
Time snap shots
be
bs
bs
Multiplication of DislocationsTo account for large plastic strain that can be produced in crystals, it is necessary to have regenerative multiplication of dislocations. Of course, there are many variants that lead to many effects.
Two important mechanisms for this are:
• Frank‐Read sources and multiple cross glide
From Hall and Bacon 4th Ed
Fig. 3.10 Cross slip of single crystal of Fe‐3.25%Si
Marked pts could be from cross slip
Dislocations Generation: single‐ended Frank‐Read source
Single‐ended Frank‐Read source leads to regenerative multiplication.This mechanisms can be attained from a “superjog”, where an extended line is out of the slip plane and thus sessile.
For super‐jogs, see book and Gilman and Johnston, Solid State Physics 13, 147 (1962).
CEF
Sessile segment
• Segment BC is edge anchored at one end. (a)• Moves by rotating. (b)• Each revolution around B displaces the crystal above slip plane by b, so nrevolutions gives nb slip.• Spiraling around B increases line length.
Dislocation Generation: Frank‐Read SourceAfter effects of dislocation‐dislocation interaction
rTθ/2
rθ
θ/2
Line Tension (E/L) = 2T sinθ/2 ~ Tθ = θGb2/2
opposes bowing via shear τ:F/L * bowing arc = τb rθ
So, τb rθ = θGb2/2τ = Gb/2r
Radius of curvature r smallest for semicircular arc of r = L/2. Larger L easier to deform.
τmax = Gb/L
Small jog τ applied shear stress
L
Generated a dislocation in place of old one, which is now a loop.
(Shaded area has 1 unit of slip.)
Larger density. Back stresses hinder motions.
Tsinθ/2Tsinθ/2
τ
What type of dislocations?What can happen?
Shear bowing of lineUnstable position: loop expands
Screws annihilate
Shape due to Sidirectional bonding
A more likely mechanism for dislocation loop formation is the Frank Read Source – dislocation pinned at both ends.Radius of curvature depends on applied shear stress.Critical bow out for R = L/2 (L = AB)Further steps are the formation of a kidney‐shaped loop and the annihilation of dislocation segments with the same b vector but opposite line sense.
Dislocation Generation: Frank‐Read Source via Cross Slip
What type of dislocation is in (a)?
Bowing of cross slipped dislocation line is similar to jogged dislocations.
τ applied shear stress can be parallel and perpendicular to b.
jogτ applied shear stress
L
Looks like two pints on (111) plane, as in Si case
Summary
• Dislocations (line defects) give rise to complicated interactions in a crystal.
• Dislocation multiplication is responsible for the very large increases in YS.
• Dislocation‐dislocation interactions, or dislocations interacting with other defects, lead to higher stresses required to move the dislocations further (work‐hardening). For example, dislocation pile‐up, jogs, trasnfer across grain boundaries, etc., all contribute to YS increases.
• Dislocations interacting with anything lead to other defects (point, planar, volumetric).
• Consequences are found in the allowed slip and strengthening of materials.
• Be familiar/conversant with how dislocations interact and the consequences.
Are these consequences able to be mathematically described?
