mg slip planes

8/10/2019 Mg Slip Planes

1/34

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.


2/34

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


3/34

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

.


4/34

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).


5/34

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.


6/34

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).


7/34


8/34

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


9/34

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 .


10/34

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


11/34

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


12/34

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


13/34

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


14/34

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.


15/34

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


16/34

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


17/34

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


18/34

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!


19/34

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.


20/34


21/34

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


22/34

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*


23/34

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.


24/34

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


25/34

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


26/34

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


27/34

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


28/34

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]


29/34


30/34

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


31/34

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


32/34

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


33/34

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.- - -


34/34

Limited #independent slipsystems

. .

Prof. M.L. Weaver

rom arter an orton, eram c ater a s c ence an ng neer ng, p.

mg slip planes

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