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8/10/2019 Mg Slip Planes
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Slip by Dislocation Motion
*
DIETER: Ch. 4, p. 114-132; Ch. 5, p. 145-183; Ch. 8, p. 310-314
HERTZBERG: Ch. 2
. , . - HULL & BACON, Ch. 3, pp. 42-61
Reed-Hill & Abbaschian: Chs. 4 and 5
Prof. M.L. Weaver
, .
*This list does not mean that you need to read all of these chapters. It has been assembled toprovide you with suggested reading from that you may be using OR referring to in your course.Most of these chapters cover similar material. Any required reading will be noted separately.
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Types of Dislocation MotionTypes of Dislocation Motion
Glide (conservative motion): moves on a lane that contains both its line and Bur ers
vector. A that moves glides is called glissile. A that cant move is called sessile. glide plane and direction depend upon crystal structure.
- moves out of the glide plane perpendicular to the Burgers
vector.
Glide of many dislocations leads to slip which is themost common manifestation of plastic deformation.
Prof. M.L. Weaver
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Slip by Dislocation GlideSlip by Dislocation Glide
Glide requires relatively little atomic motion compared with
the process for slip that we outlined for perfect (i.e., defect
ree crys a s.
b
Af ter complete motion
through crystal
There is also no net chan e inMonotonic slip stepbonding. This makes the stress
to move a dislocation smaller than the theoretical stress to
Prof. M.L. Weaver
.
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Force is required to move individual dislocationsForce is required to move individual dislocations
We call it the Peierls-Nabarro force.
The concept was originally developed in 1940 by
Peierls1, but later refined by others2-4.
It is the frictional force that must be overcome to
move an individual dislocation.
It is a consequence of the distortion caused by the
presence o a s oca on n a crys a a ce.
1R. Peierls Proc. Ph s. Soc. v. 52 . 34 1940 .
Prof. M.L. Weaver
2F.R.N. Nabarro, Proc. Phys. Soc., v. 59, p. 256 (1947).3
G. Leibfried and K. Lcke, Z. Phys., v. 126, p. 450 (1949).4A.J. Foreman, M.A. Jawson, and J.K. Wood, Proc. Phys. Soc., v. 64, p. 156 (1951).
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A dislocation must pass through a higher energy configuration to move.
Energy EP-N
E
n-s ppe
state
ppe
state
Displacement,xb/20 b
The force required to move the dislocation (i.e., to overcome stressfield caused by lattice distortion) shear stress on a slip plane.
-
Prof. M.L. Weaver
relation between atoms.
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Force EP-N
or
Stress
P-N
Displacement,xb/20 b
The Peierls-Nabarro stress is the shear stress required to move anindividual dislocation on its slip plane.
Prof. M.L. Weaver
dislocation.
Amount of distortion described by dislocation width (w).
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u
Distortion described
in terms of
,
w
-b/2
>, ,
its maximum value (i.e., u/b > or -b/4 u b/4).
.
Core widths vary between b and 5b and depend upon:
Prof. M.L. Weaver
n era om c po en a ,
Crystal structure.The width, w, is
directly related tocrystal structure
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Fig. A. Close packedSchematicillustration of a
wide dislocation;
more typical of
close- acked
structures: w is
larger. Distortion
distributed through
b/4 b/4
metals. .
g. .
Schematic
illustration of a
narrow
dislocation; more
on-c ose pac e
structures: w is
smaller. Distortion
concentrated into a
b/4 b/4
typical ofceramics,
intermetallics,
and non-close-
smaller area.
Prof. M.L. Weaver
w .
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Peierls and Nabarro estimated the energy of the dislocation
2 2exp
(1 )P N
w
bE
bG
from which the shear stress required to move a dislocation
i.e. Peierls stress can be determined as:
22 2 2 2 2
exp exp(1 ) (1 ) (1 )P N P N b b b
G w dG
E
Values of P-Nvary with crystal structure. In general,
P-N
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b
d
A B C D
22 2 2 2 2
exp exp(1 ) (1 ) (1 )P N P N b b b
G w dG
E
Let G = 100 GPa and = 0.3.
-
Prof. M.L. Weaver
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Variation of Peierls Stress w/Variation of Peierls Stress w/ ddandand bb
12
14 b= 1.0b= 5.0
b = 10.0 For fixed b, P-Ndecreases
exp
(1 ) (1 )P N
b
8
10
s(x10
7)
d= 1 to 10
as ncreases.
An increase inb results in a
larger P-N.
4
6
ierlsStre
Slip via dislocation motion
occurs more readily in close
packed directions (lowest b)
0
2Pe and on widely spaced
planes (highest d). This is
because P-N values are
bb
Interplanar Spacing (arb. units)
.
The width, w, is
[d]
Prof. M.L. Weaver
d
A B C D
d
A B C D
directly related to
crystal structure
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Variation of Peierls Stress w/Variation of Peierls Stress w/ ddandand bb
12 d= 1.0d= 2.0
d= 5.0 For fixed d,P-N increases
exp
(1 ) (1 )P N
b
8 b= 1 to 10
s(x10
7) d= 10.0 as ncreases.
An increase indresults in a
reducedP-N.
4
ier
lsStre
Slip via dislocation is more
likely to occur more readily
in close packed directions
0
P (lowest b) and on widely
spaced planes (highest d).
This is becauseP-Nvalues
Atomic Separation (arb. units)
.
The width, w, isbb
[b]
Prof. M.L. Weaver
directly related to
crystal structured
A B C D
d
A B C D
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Close Packed PlanesClose Packed Planes Recall from your introductory materials courses:
Close acked lanes i.e. those with the smallest interatomic
separation, d) are the ones that are spaced farthest apart (i.e., those
with the largest b).
We can relate properties to atomic/ionic packing factors (APF/IPF) or
planar density.
Ionic Covalent
Diamond
APF
IPF0.74 0.68 0.52 0.725 0.67 0.68 0.627 0.34
1 3 7 2 5 3 6 8P N
Prof. M.L. Weaver
Rank 1 8 where 1 is lowest and 8 is highest.
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Slip vs. atomic densitySlip vs. atomic density
Close-packed planes/structuresSlipdistance, b
ma er
Larger d
spacing, d
P N p p ane(high atomic density)
Non close-packed planes/structuresSlip
distance, b
Planar
spacing, d
Larger b
Smaller d
Prof. M.L. Weaver
arger P N
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The Peierls-Nabarro stress is smaller than
the critical resolved shear stress or the
The CRSS and YS represent the
conditions to move lots of dislocations
Prof. M.L. Weaver
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Slip Systems in CrystalsSlip Systems in Crystals
A specific shear stress is required to
induce dislocations to move.
z
zz
Dislocations slip on specific slip
systems (i.e., specific crystal plane +
specific crystal direction on that plane).
y
The resolved shear stress on planeAs
x
s
coscos cos
y zz y zz
F F
o
Applies for single crystals and individual
z o
z z y za a xzF
Prof. M.L. Weaver
grains in polycrystals.90
is not necessarily 90
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Resolved Shear StressResolved Shear Stress
ons er an ar rary p ane or en e a ang e w respec o e app e oa .Lety = slip direction andz = slip plane normal.
Consider SLIP on the -plane.
Refer to
illustration on
NORMAL Force:FN=Fz =Fzcos
previous
viewgraph
s z .
Area of slip-plane: As =A/cos (check: As must have larger area thanAo.)
Resolved NORMAL Stress on the slip plane:
2/ cos / cos cosN N s oF A F A
Resolved SHEAR Stress on the slip plane in the slip direction:
/ cos / cos cos coss RSS s s oF A F A
Prof. M.L. Weaver
The slip direction is not necessarily in same direction as til t of the slip plane!
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Critical Resolved Shear StressCritical Resolved Shear Stress
coscos cos cosflow flowSchmid Factor
or
RSS
o
mA
flowRSS
m
.
stress as opposed to the flow stress, we get:
cos coscos cosy CRSS y ym or
Prof. M.L. Weaver
CRSS is theresolved shear stress requiredto initiate plastic
deformation via slip.
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Solution to Example Problem 1
plane (111) is:
2 2 2 2 2 2
(1)(1) ( 2)( 1) (0)( 1) 3 3
cos 5 3 15(1) ( 2) (0) (1) ( 1) ( 1)
Angle btw.
Tensile axis & slipplane normal
The angle between the tensile axis [120] and the slip direction [011] is:
2 2 2 2 2 2
(1)(0) ( 2)( 1) (0)(1) 2 2cos
5 2 10(1) ( 2) (0) (0) ( 1) ( 1)
Angle btw.
Tensile axis & slip
direction
Since CRSS = 10 MPa,
Prof. M.L. Weaver
10
cos cos 3/ 15 220
/.
11
04CRSS
P
A
MPa
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Solution to Example Problem 1
plane (111) is:
2 2 2 2 2 2
(1)(1) ( 2)( 1) (0)( 1) 3 3
cos 5 3 15(1) ( 2) (0) (1) ( 1) ( 1)
The angle between the tensile axis [120] and the slip direction [011] is:
2 2 2 2 2 2
(1)(0) ( 2)( 1) (0)(1) 2 2cos
5 2 10(1) ( 2) (0) (0) ( 1) ( 1)
Since CRSS = 10 MPa,
Prof. M.L. Weaver
10
cos cos 3/ 15 220
/.
11
04CRSS
P
A
MPaB*
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Room temperature slip systems and critical resolved shear stress formetal single crystals (from Dieter, 3rd Edition, p. 126).
Critical
Cr stal Sli Sli Shear stressMetal PurityStructure Plane Direction
99.999 (0001) [1120] 0.18 [1]
99.996 (0001) [1120] 0.77 [2]
99.996 (0001) [1120] 0.58 [3]
Zn HCP
Mg HCP
Cd HCP
(MPa) Reference
. .99.9 [1120] 90.1(1010)
99.999 (111)(11199.97
99.93
Ag FCC [110] 0.48 [5][110])0.73[110](111)
[110](111) 1.3
99.999 (111) [101] 0.65 [5]99.98 [101](111)
0.94
99.8 (111) [110] 5.7 [5]
99.96 (110) [111] 27.5 [6](112)
Cu FCC
Ni FCC
Fe BCC
(110) [111] 49.0 [7]
[1] D.C. Jillson,
Mo BCC
Tran
, v. 188, p. 1129 (1950).
[2] E.C. Burke and W.R. Hibbard, Jr., , v. 194, p. 295 (1952).
" "
s. AIME
Trans. AIME
. , , . ,
[4] A.T. Churchman, , v. 226A, p. 216 (1954)[5] F.D. Rosi, , v. 200, p. 1009 (1954)
[6] J.J. Cox, R.F. Mehl, and G.T. Horne, , v. 49, p. 118
Proc. R. Soc. London Ser. ATrans. AIME
Trans. Am. Soc. Met. (1957)
[7] R. Maddin and N.K. Chen, , v. 191, p. 937 (1951)Trans. AIME
Prof. M.L. Weaver
Note the differences in slip systems for different crystal structures. Slip occurs
when m is maximum. This means that we must determine which particular slipsystem has the maximum m to obtain the CRSS.
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Example Problem 2
Consider a c lindrical sin le cr stal of silver of 5 mmdiameter with its axis parallel to [321]. This crystal begins
to deform plastically in compression at a load of 39 N.
Determine the CRSS for this crystal.
Prof. M.L. Weaver
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Solution to Example Problem 2 .Silver has an FCC crystal structure. Thus the slip system is 111 110
There are 12 distinct slip systems for FCC. For each we must compute
and tabulate the corresponding angles and , as well as the Schmid
. .
1 2 2 1 1 2
1 1 1 2 2 22 2 2 2 2 2
1 1 1 2 2 2
cos h k l h k l
h k l h k l
Slip System
cos
cos
(a/2) [1 1 0] (1 1 1) 79.11 22.21 0.1749
(a/2) [1 0 1] (1 1 1) 67.79 22.21 0.3500
(a/2) [0 1 1] (1 1 1) 79.11 22.21 0.1749
. . .
(a/2) [1 0 1] (1 1 1) 40.89 51.89 0.4666
(a/2) [1 1 0] (1 1 1) 79.11 51.89 0.1166
(a/2) [1 1 0] (1 1 1) 19.11 72.02 0.2917
(a/2) [1 0 1] (1 1 1) 67.79 72.02 0.1167
a/2 0 1 1 1 1 1 55.46 72.02 0.1750
mmax
. . .
(a/2) [0 1 1] (1 1 1) 79.11 90.00 0.0000(a/2) [1 0 1] (1 1 1) 40.89 90.00 0.0000
(a/2) [1 1 0] (1 1 1) 19.11 90.00 0.0000
39N
Prof. M.L. Weaver
maxcos cos cos cos(5 ) / 4
2 0.466 0.935
CRSS
CRSS
mA
2mm
MPa MPa
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Solution to Example Problem 2 .Silver has an FCC crystal structure. Thus the slip system is 111 110
There are 12 distinct slip systems for FCC. For each we must compute
and tabulate the corresponding angles and , as well as the Schmid
. .
1 2 2 1 1 2
1 1 1 2 2 22 2 2 2 2 2
1 1 1 2 2 2
cos h k l h k l
h k l h k l
Slip System
cos
cos
(a/2) [1 1 0] (1 1 1) 79.11 22.21 0.1749
(a/2) [1 0 1] (1 1 1) 67.79 22.21 0.3500
(a/2) [0 1 1] (1 1 1) 79.11 22.21 0.1749
. . .
(a/2) [1 0 1] (1 1 1) 40.89 51.89 0.4666
(a/2) [1 1 0] (1 1 1) 79.11 51.89 0.1166
(a/2) [1 1 0] (1 1 1) 19.11 72.02 0.2917
(a/2) [1 0 1] (1 1 1) 67.79 72.02 0.1167
a/2 0 1 1 1 1 1 55.46 72.02 0.1750
mmax
. . .
(a/2) [0 1 1] (1 1 1) 79.11 90.00 0.0000(a/2) [1 0 1] (1 1 1) 40.89 90.00 0.0000
(a/2) [1 1 0] (1 1 1) 19.11 90.00 0.0000
39N
Prof. M.L. Weaver
maxcos cos cos cos(5 ) / 4
2 0.466 0.935
CRSS
CRSS
mA
2mm
MPa MPa
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Critical Resolved Shear StressCritical Resolved Shear Stress
Condition for dislocation motion: CRSSR
Crystal orientation can make
it easy or hard to move dislocation10-4 GPa to 10-2 GPa coscosR
R = 0=90
R = /2 =45
R = 0
=90
=45
Prof. M.L. Weaver
R = max.at = = 45
R = 0 at
= = 90
R = 0 at
= = 0
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Common Slip Planes in MetalsCommon Slip Planes in Metals
fcc112
111
y
110
011 101
Four 111 planes each with 3 110 slip directions
Prof. M.L. Weaver
[Adapted from Felbeck and Atkins, 2nd Ed., p. 117]
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Common Slip Planes in MetalsCommon Slip Planes in Metals1120
(b)(a)
1120 hcp
1210 1210
2110
2110
Figure. (a) Basal slip plane; (b) atomic arrangement on basal
One 0001 plane with three 1120 slip directions
p ane w poss e s p rec ons n ca e .
Three 1010 planes with one 1120 slip direction on each
Six 1011 planes with one 1120 slip direction on each
Prof. M.L. Weaver
12 slip systems possible. # active depends on c/a ratio
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Common Slip Planes in MetalsCommon Slip Planes in Metals1 11
oaz
111
bcc 122oa 101
y
010
10 1
1 11 z
x 101
123oa
1 10
111
2oax
z
120
243oa
213
111
Prof. M.L. Weaver[After Felbeck and Atkins, 2nd Ed., p. 118]
48 slip
systems5oax
y120
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Crystal Slip Slip Number of non- Slip directions Number ofstructure plane direction parallel planes per plane slip systems
.
fcc 111 110 4 3 4 3 12
bcc 110 111 6 2 6 2 12
112
111 12 1 12 1 12
123 111 24 1 24 1 24
hcp 0001 1120 1 3 1 3 3
1010 1120 3 1 3 1 3
1011 1120 6 1 6 1 6
Prof. M.L. Weaver
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Does bonding influence dislocation motion?Does bonding influence dislocation motion?
Metals: Disl. motion easier.++++++++
YES!
- -
-close-packed directionsfor slip. electron cloud ion cores
+
+
++ + + + + +
+++++++
Covalent Crystals
Si diamond : Motion hard.-directional (angular) bonding
on c rys a s a :Motion hard.-need to avoid ++ and - -
+ + + ++++
- - -----
Prof. M.L. Weaver
neighbors.- - -
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Limited #independent slipsystems
. .
Prof. M.L. Weaver
rom arter an orton, eram c ater a s c ence an ng neer ng, p.