Web Class 13a Final

8/12/2019 Web Class 13a Final

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TwinningDislocation Reactions


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Deformation by Twinning

The second important mechanism by which metalsdeform is the process known as twinning.

Twinning results when a portion of the crystal takes upan orientation that is related to the orientation of therest of the untwinned lattice in a definite, symmetricalway.

The twinned portion of the crystal is a mirror image ofthe parent crystal.

The plane of symmetry between the two portions iscalled the twinning plane.


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Note that the twin is visible on the polished surface because of the change in elevation produced by thedeformation and because of the difference incrystallographic orientation between the deformed andunreformed regions.

It should be noted that twinning differs from slip inseveral specific respects.

In slip, the orientation of the crystal above and below

the slip plane is the same after deformation as before,while twinning results in an orientation differenceacross the twin plane.


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Slip is usually considered to occur in discretemultiples of the atomic spacing, while in twinning theatom movements are much less than an atomicdistance.

Slip occurs on relatively widely spread planes, but inthe twinned region of a crystal every atomic plane isinvolved in the deformation.


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Figure 12-2. Classic picture of twinning.


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In general, a dislocation line cannot end inside of acrystal. The exception is at a node, where three or fourdislocation lines meet.

At a node two dislocations with Burgers vectors b 1 and b 2 combine to produce a resultant dislocation b 3.

The vector b 3 is given by the vector sum of b 1 and b 2 .

Dissociation or Combination of Dislocations


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A dislocation with a Burgers vector equal to one latticespacing is said to be a dislocation of unit strength .

Because of energy considerations, dislocations withstrengths larger than unity are generally unstable anddissociate into two or more dislocations of lowerstrength.


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The criterion for deciding whether or not thedissociation will occur is based on two

conditions:(a) The strain energy of a dislocation, which is

proportional to the square of its Burgers vector.The dissociation will occur if

The reaction will not occur if

(b) The vector addition (or subtraction) of the Burgersvector b must agree.

321 bbb

23

22

21 bbb

23

22

21 bbb

(13.1a)

(13.2a)


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A Burgers vector is specified by giving its componentsalong the axes of the crystallographic structure cell.

Thus, the Burgers vector for slip in a cubic lattice from acube corner to the center of one face has the components

(see Figure 11-7)

The Burgers vector is , or generallywritten as .

The magnitude of a dislocation with Burgers vectoris given as:

0,2/,2/ oaoa

]02/2/[ oaoa

]110)[2/( oab

][uvwa o

2/1]222[ wvuo

ab (13.3)


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Therefore, the magnitude of the Burgers vector givenabove is .2/oab


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Example Determine whether the dislocation dissociationreaction is feasible.

Since this vector equation the x, y, and z components of

the right-hand side of the equation must equal the x, y, andz components of the left side (original dislocation).

]211[6

]121[6

]110[2

321aaa

bbb

21

62

61

21

:

21

61

62

21

:

61

61

0:

components z

components y

components x


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For the dissociation to be energetically favorable 2322

21 bbb

666

])2()1()1[(6

666

])1()2()1[(6

222])1()1(0[

2

223

2/12223

222

2/12222

221

2/121

ab

aab

ab

aab

abaab

feasibleisreactionndislocatiotheand bbb 23

2

2

2

1


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Dislocations in the Face-centered cubic lattice

Slip occurs in the fcc lattice on the {111} plane in the direction.

The shortest lattice vector is (a o/2)[110], which

connects an atom at a cube corner with a neighboringatom at the center of a cube face.

The Burgers vector is therefore (a o/2)[110].


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However, consideration of the atomic arrangement onthe {111} slip plane shows that slip will not take place

so simply.

Figure 13-3 represents the atomic packing on a close- packed (111) plane.

It has already been shown (see Fig 11-4a) that the{111} planes are stacked on a sequence ABCABC.


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Figure 13-3. Slip in a close-packed (111) plane in anfcc lattice


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The vector defines one of the observed slipdirections.

The same shear displacement as produced by b 1 can beaccomplished by the two-step path b 2 + b 3 .

The latter displacement is more energetically favorable butit causes the perfect dislocation to decompose into two

partial dislocations.

]110[2

oab

]211[6

]112[6

]110[2

321

ooo aaabbb


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The above reaction is energetically favorable since there is adecrease in strain energy proportional to the change

_______________________________________________Original dislocation Product of reaction

2/1

4

10

4

11

oab

2/1

36

1

36

1

36

42 oab

oab 22

1 62oab

63oab

2

221

oab 6

223

oab 6

222

oab

2

1b

2

3

2

2 bb


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Slip by this two-stage process creates a stacking faultABCAC ABC in the stacking sequence.

As Fig. 13-4 shows, the dislocation with Burgers vectorhas been dissociated into two partial dislocations and . This dislocation reaction was suggested by Heidenreich and

Shockley.

Therefore this dislocation arrangement is often known asShockley partials , since the dislocations are imperfect ones which do not produce complete lattice translations.

Figure 13-4 represents the situation looking down on (111)

along .]111[

1b2b 3b


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Figure 13-4. Dissociation of a dislocation into two Shockleypartial dislocations.


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With the sequence ABC AC ABC, we have four planes inwhich the stacking is CA CA, which is exactly the

stacking of HCP structure.

This structure has a higher Gibbs free energy than theequilibrium FCC structure.

This specific array of planes is called the stacking fault ,and the energy associated with it determines the separation

between the two partial dislocations


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The repulsive force between the two partials is balanced bythe attraction trying to minimize the region with the stackingfault.

The equation for the calculation of the equilibrium separation between the partial dislocations d is given as:

where is the stacking-fault free energy (SFE) per unit area,

22cos21

12

8

2

d bG pSF

2 sinsincoscos2212121d

bGbSF

(13.4)

(13.5)


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b p is the Burgers vector of the partial dislocation, and isthe angle of the Burgers vector with the dislocation line.Table 13.1 presents the SFEs for some materials

Table 13-1. SFEs and Shockley Separations of MaterialsMetal

(mJ/m2

)

ao (nm) b (nm) G

(GPa)

d (nm)

Al 166 0.41 0.286 26.1 1

Cu 78 0.367 2.55 48.3 3.2

Au 45 0.408 0.288 27.0

Ni 128 0.352 0.249 76.0 2.9

Ag 22 0.409 0.289 30.3 9


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From the preceding equations, it can be seen that d isinversely proportional to .

Aluminum, which has a high SFE (166 mJ/m2

), does exhibita very small separation between partials: 1 nm. SFE is very sensitive to composition Addition of alloying elements in a material helps to

decrease the SFE. Examples: Brasses have SFE lower than that of Cu, Al alloys have SFE lower than that of Al, the addition of Al to Cu drops the SFE from 78 to 6 mJ/m 2.

Figure 13-5 shows stacking faults in 302 stainless steelviewed by TEM - seen as characteristic fringe (////) pattern.


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Low-SFE metals tend to exhibit a deformation substructurecharacterized by banded, linear arrays of dislocations.Cross-slip is more difficult, because the dislocations haveto constrict in order to change slip planes.

High-SFE metals tend to exhibit dislocations arranged intangles or cells.


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Figure 13-5. Group of stacking faults in 302 stainless steel stoppedat boundary on left-hand side.

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