Physical Metallurgy PrinciplesSI Version

Chapter Six:Elements of Grain Boundaries

Fourth Edition

What you have to learn from this chapter?♦ Dislocation Model for Grain Boundary;♦ Low Energy Dislocation Structure; ♦ Stress Field, Energy, and Surface Tension of a Grain

Boundary;♦ Special Boundary: Coincident site boundary;♦ Grain Boundary and Mechanical Properties

Chapter 6. Elements of Grain Boundaries

Most engineering objects are polycrystalline materials.Grain boundaries play an important part in determiningthe properties of a material; e.g. resistivity, diffusion,fracture mechanism, etc.

▲ Small-Angle Grain Boundary:Consists of an array of edgedislocation!The angle between two grains: θ;the spacing between dislocations: d;the Burgers vector: b;

θ /2

2d

b

θ

( ) dbdb / ; 1 when; 2/2/sin =<<= θθθA real idea case, difficult to find in realcase. Experimental evidence of smallangle boundary: see Fig. 6.3, 6.4 (not asimple tilt boundary, involving twisting).

The Five Degrees of Freedom of a Grain Boundary

▲ Stress Field of a Grain Boundary:The grain boundaries do not possess long-range stressfields. To demonstrate this claim, take the simple smallangle boundary as a case.

( )( )( )222

22

-12 yxyxx

xy′+

′−=

νπμ

τb

; y’ = -(nd-y)

( )( )[ ]( )[ ]222

22

-12 yndxyndxx

xy−+

−−=

νπμ

τb

x

a

nd

…

py

nd-y

d

Summation of all shear stressesthat act on point p; from n = -∞ to +∞

( ) ∑+∞=

−∞= +=

n

n ana 1cot ππΘ

( )( )[ ]( )[ ]∑

+∞=

−∞= −+

−−=

n

npxy

yndxyndxx

222

22

, -12 νπμ

τb

( )[ ] ( )[ ]{ } ∑∑+∞=

−∞=

+∞=

−∞= −++

++=−++∴

n

n

n

n iqpniqpniqpiqp 11 cotcot πππ

( )( ) ( ) ( )∑

+∞=

−∞= +++

=−

n

n pnqpn

pqp

222cos2cosh2sin

ππππ

=> ……… (1)

( )[ ] ( )[ ]{ } ∑∑+∞=

−∞=

+∞=

−∞= −+−

++=−−+∴

n

n

n

n iqpniqpniqpiqp 11 cotcot πππ

( )( ) ( ) ( )∑

+∞=

−∞= ++=

−

n

n pnqpqq

q 22

12cos2cosh

2sinhππ

ππ=> ……… (2)

Differentiating equation (1) with respect to p =>( ) ( )( ) ( )[ ]

( )( )[ ]∑

+∞=

−∞= ++

+−=

−− n

n pnqpnq

pqpq

222

22

22

2cos2cosh12cos2cosh2

πππππ=> (3)

Differentiating equation (2) with respect to p =>( ) ( )

( ) ( )[ ] ( )[ ]∑+∞=

−∞= ++

+=

−

n

n pnqpn

pqpq

q 2222

2

2cos2cosh2sin2sinhπππππ

=> (4)

Using eq. (3) : ( ) ( )[ ]( ) ( ) ( )[ ]22, /2cos/2cosh1

1/2cos/2coshdydxd

dydxxbpxy ππν

πππμτ−−

−=

=>For x >> d/2π ( ) ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ −

−≈

dy

dx

dxb

pxyππ

νππμτ 2cos2exp

12

2,

In the textbook, y = 0 ( ) ( )[ ]1/2cosh12, −−=

dxdxb

pxy πνπμτ=>

( ) ( ) 1sinh22cosh 2 += zzΘ

( ) ( )dxdxb

pxy /sinh12 22, πνπμτ

−==>

Using eq. (4) => pyypxx ,, and σσ

Shea

r stre

ss, τ

xy

Boundarystress

Stress of asingle edgedislocation

x

Number of b from boundary

CRSS

The boundary stressapproaches that of asingle dislocation as xbecome very small.See Fig. 6.7 (A).

CRSS: critical resolved shear stress.• Fig. 6.7 of the textbook assume d=22b => boundary

stress drop to negligible level at x about several tensof b.

▲ Grain Boundary Energy:Consider the work required to separate a set ofdislocation array (one positive and the other negativearray) form r0 to ∞

∞α/0 br = τxyb

Attractiveforce

r0: the distance wherethe core energy oftwo dislocationsshould cancel each other!

wgb: the energy per unit length per dislocation

( ) ( )∫∫∞∞

−===

00 /sinh1221

21

22

2

rr xygb dxdx

xd

bbdxL

Wwππ

νμτ

( ) ( )∫∞

−=⇒=≡≡

02

2

00 sinh14 // ; /

η ηηη

νμαππηπη dbwdbdrdx gb

γb: the energy per unit area per dislocation

=> ( ) ( )[ ]000

2

sinh2lncoth14

ηηην

μγ −−

==d

bd

wgbb

When η0 << 1, ( ) ( )db απηηη 2lnsinh2ln ; 1coth 000 ≈≈

Θθ =b/d; ( ) ( )[ ]; 1ln2/ln14

+−−

= θπαθν

μγ bb

α: a factor accountingfor the core energyShockley-Read equation

θ (degrees)

γ/γ

max

γmax1

030

See Fig.6.8!

θ (degrees)

γ

Large angle equation

Small angleequation

θ (degrees)

γ

α1

α2

α3

α1<α2<α3

▲ Low-Energy Dislocation Structure:

( )( )[ ]( )[ ]222

22

-12 ndxndxxn

xy+

−=

νπμ

τb

Follow previous calculation:

• Consider the case of simple tilt boundary again!

0

n τ+

-

when nd < x => τnxy >0;

when nd > x => τnxy <0

=> Only small n => large positivecontributions to shear stress

The overall effect of shears stress field is small atreasonable x. (already shown in previous section!!)

• The other two stress components (normal stresses) =>the stress equation is anti-symmetric =>σxx(x,y)=-σxx(x,-y) σyy(x,y)=-σyy(x,-y)!When n = 0, σ 0

xx(x,y)=σ 0yy(x,y) = 0,

when n ≠ 0 => σ nxx(x,y)= -σ- -n

xx(x,y);σn

yy(x,y)= -σ- -nyy(x,y)!

• Incorporation of a random dislocation into a tiltboundary:Energy per unit length of a boundary dislocation: wgb;Energy per unit length of a random dislocation: we.

d

wgb

/we

1 wgb/we =1: no energy difference fora dislocation to be in a grainboundary or to be a random one.wgb/we <1: a dislocation could lowerits energy by being part of a grainboundary.

Even with d = 500b => a dislocation could still lowerits energy significantly by being part of the tilt grainboundary.As d ↓ wgb/we ↓ => Tilt boundary attracts edgedislocations!

Dislocation spacing

• The tilt boundary is one of a very large number ofdislocation arrays which have the property of havinga low strain energy. => Kuhlmann-Wilsdorf proposedlow energy dislocation structures (LEDS).

• Grain and subgrain boundary structures normallybelong to LEDS!

TaylorLattice:

earliest formof LEDS

Slipplane

Kink band

• When the plastic deformation involves a number of different slip planes and Burgers vector => morecomplicated LEDS structure and difficult to perceive!Plastic deformation => forms cells with a low internaldislocation density and boundaries composed of

dislocation tangles.See Fig. 6.15. Plastic deformation ↑ cell size ↓ # ofcell ↑ ϑ = κρ -1/2; ϑ : average cell diameter; κ : aconstant; ρ: dislocation density

▲ Dynamic Recovery:• Recovery: after plastic deformation, a considerable

amount of energy is released by the localrearrangement of the dislocations in the tanglesand further release of energy occurs when LEDS areformed. => the process typically requires thermalenergy => Annealing is the typical process forrecovery (static) to happen!

• Recovery could also occur during plastic deformation(no matter the temperature of the sample is low orhigh) => dynamic recovery (a very important factor

in the deformation process) => affects the σ - εcurve (lower the effective rate of work hardening)

• Dynamic recovery occurs most strongly in metals of high stacking fault energy and shows limited effecton metals of low stacking fault energy!See Fig. 6.16. The major mechanism involvingdynamic recovery is thermally activated cross-slip.

▲ Surface Tension of the Grain Boundary:• The solid grain boundaries process a surface energy

=> surface tension (equivalent to that of a liquid).In solid, surface tension is a confusion term! Surfaceenergy is the energy required to create a unit surface.Surface tension is the work required to create a unitsurface. It costs less work to create a low energysurface!

• Typical value of surface tension of external metalsurface ~ 1.5 J/m2; of liquid surface ~ 0. 1 J/m2; of large angle grain boundary ~ 0. 4 J/m2.

Crystal 1

Crystal 2

Crystal 3

γa

γb

γc

ab

c Force balance requires:

cbacba

sinsinsinγγγ

==

γ : the surface energyof grain boundary

Grain boundary

• The kinetics of reaching force balance configurationis mainly controlled by the grain boundary motion.The grain boundary mobility is definitely a functionof temperature. The thermodynamics of the grainboundary movement is to reduce the total energy ofthe system: move through highly deformed region andleave behind strain free region; reducing the boundary

area (by straighten the boundaries, or grain growth), etc.• In pure metal, low angle boundaries are seldom

observed. => Fig. 6.8 => all grain boundaries havethe same energy! => 120o is the balanced anglebetween three grains

▲ Boundaries Between Crystals of Different Phases:• force balance

θγfs γfs

Grain 1 Grain 2γb

Force balance requires:bfs γθγ =)2/cos(2

Phase 1

Phase 1Phase 2 γ11

γ12

γ12

1112 )2/cos(2 γθγ =

θ θ

γ12 / γ110.5

180o

0

• When θ → 0, the second phase tends to form a thinfilm between crystals! If γ12/γ11 < 0.5 => the second

6-35© 2010. Cengage Learning,

Engineering. All Rights Reserved.

phase penetrate the single phase boundary => isolatedthe crystals of the first phase (occurs even at verysmall amount of second phase)

• E.g. Bi (a brittle metal) in Cu (a ductile metal) =>Θ very small γ12 => θ = 0 => when Bi could form acontinuous film (<0.05%) around Cu crystal => Culoses its ductility.

• Another important metallurgical case: second phaseimpurities remain in the liquid state until a temp.well below the freezing point of the major phase!

• A small amount of harmful impurities could do a lotof damage to the plastic properties of metals. Highinterfacial energy => the liquid tends to form discreteglobules; see Fig. 6.21 => less significant effect on theplastic properties of the metals!

▲ Grain Size:• An important parameter in a lot of properties, but

difficult to define precisely.• Most generally accepted method: linear intercept

method; mean grain intercept ; average distancebetween grain boundaries along a line laid on aphotograph (SEM, TEM).

• Quantitative metallography has shownwhere Sv is the surface area of grain boundaries perunit volume.

l

lS v 2=

▲ Effect of Grain Boundaries on Mechanical Properties:• Polycrystalline metals always show a strong effect of

grain size on hardness. => 2/10

−+= dkHH H

d-1/2

Har

dnes

s, H

d-1/2

Flow

-stre

ss, σ

ε1

ε2ε3

ε3 >ε2 > ε1H0

Slope = kH

H0: not necessarily the hardness of single crystal

2/10

−+= kdσσHall-Petch equation* Rationale: Considering the grain boundaries as a

barrier to the dislocation movement => a pile up atone grain could generate sufficiently large stressesto operate sources in an adjacent grain at a yieldstress! There are also other interpretations based ongrain boundary dislocations.

▲ Coincidence Site Boundaries:• A very important form of grain boundary ; first

observed in annealed cold rolled fcc metal (Cu)=> the two stages annealing; a lower temperatureanneal followed by a high temperature anneal(secondary recrystallization) => the secondaryrecrystallized crystals have specific orientationrelation to the old crystals. => could be characterizedby coincidence sites of atoms in two orientation.=> coincidence site boundaries!

* The above two cases are rotation of (100) plane! See Fig. 6.25.

22o38o

• Any two lattices with a specific rotation relation=> form a boundary with certain coincidence sites=> the more the coincidence sites=> the more important the boundary Θ higher

boundary mobility=> a rapid rate of grain growth.

♦ Density of Coincidence Sites: in previous two figures,one is 1/13 of the density of the original lattice,the other is 1/5 of the density of original lattice.Fig. 6.25 is 1/7! How to quickly count it?• The reciprocal of the density is commonly used as a

parameter to describe the boundary: ∑=13, ∑=5, ∑=7.

▲ The Ranganathan Relations:• Ground rules for rotation to generate a coincidence

sites in a 3-D lattice: 4 factors;(1) axis to rotate, [hkl];(2) rotation angle, θ, about axis [hkl];(3) the coordinates of a coincidence site in the

coincidence site (hkl) net, (x,y);(4) ∑ = A the reciprocal of the density of coincidence

sites in (hkl) net. • The relation between factors: the length ratio y/x=N1/2;

222

22

2/11

where

)/(tan2

lkhNNyx

xyN

++=

+=∑

= −θ

• Twist Boundaries:Examples 1, see Fig.6.26.Simple cubic lattice; rotation axis: <100> => N=1;* take x=2 and y=1 => θ =53.1o, ∑=5;* take x=3 and y=1 => θ =36.9.1o, ∑=10;see Fig. 6.27. 10/2 => 5!

• A could only be odd number; divided the A valueby smallest 2n to get odd number.

• Twist Boundaries: Examples 2, see Fig.6.25 and Fig. 6.28.FCC lattice; rotation axis: <111> => N=3;outline (111) as a tetragonal unit cell, see Fig. 6.28* take x=9 & y=1 ⇒ θ =21.8o, ∑=84; 84/12=7 ⇒∑=7* take x=5 & y=1 ⇒ θ =38.2o, ∑=28; 28/4=7 ⇒∑=7

• Tilt Boundaries:Examples 1, see Fig.6.29Simple cubic lattice; rotation axis <100> => N=1;* take surface steps: x=2 and y=1and join two surface

=> the tilt boundary => θ =53.1o, ∑=5; 51/2a = p;* take x=3 and y=1 => θ =36.9o, ∑=10; 10/2=5, ∑=5;

51/2a = p; see Fig. 6.30.

• Cases of overlapping atoms, see Fig. 6.31.=> Relieving by relative translation of the latticeabove and below the boundary, see Fig. 6.32.