Chapter 8

Deformation and Strengthening

Mechanisms

Chapter 8

Heat Treatment

Issues to Address...

• Why are dislocations observed primarily in metals and alloys?

• How are strength and dislocation motion related?

• How do we increase strength?

Deformation and Strengthening

• How can heating change strength and other properties?

Deformation and Strengthening

Dislocations and Plastic Deformation

In crystalline solids,

the onset of plastic deformation is caused

by the movement of dislocations.

• Metals: Disl. motion easier-non-directional bonding-close-packed directions

for slip. electron cloud

• Covalent Ceramics(Si, diamond): Motion hard.-directional (angular) bonding

• Ionic Ceramics (NaCl): Motion hard-need to avoid ++ and -- neighbors.

+ + + +

+ + +

+ + + +

- - -

- - - -

- - -

+ +

+

+ + + + + + + +

+ + + + + +

+ + + + + + + ion cores

Dislocations & Materials Classes

Dislocation Motion_1

Dislocations & plastic deformation• Cubic & hexagonal metals - plastic deformation by

plastic shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations).

• If dislocations don't move, deformation doesn't occur!

Dislocation Motion_2

• Dislocation moves along slip plane in slip directionperpendicular to edge dislocation line

• Slip direction same direction as Burgers vector

Edge dislocation

Screw dislocation

Dislocation Motion_3

Stress Concentration at Dislocation_1Stress Concentration at Dislocation_1

Stress Concentration at Dislocation_2Stress Concentration at Dislocations_2

Deformation Mechanisms

Plastic flow of crystalline materials usually occurs by slipalong certain crystal planes and in certain crystal directions.

A slip plane and direction is called a slip system.

Rule for slip

1. Slip always occurs in closed packed direction - in direction of shortest lattice translation vector.

2. Slip usually occurs on the smoothest and most widely spaced (or densely packed) plane, but this rule is not always obeyed.

Deformation Mechanisms

Slip SystemSlip System

Slip Planes and Directions

b(FCC) = a/2<110>b(BCC) = a/2<111>b(HCP) = a/2<1120>

Slip Planes and Directions

Dislocations & Crystal Structure

• Structure: close-packedplanes & directionsare preferred.

• Comparison among crystal structures:FCC: many close-packed planes/directions;HCP: only one plane, 3 directions;BCC: none

close-packed plane (bottom) close-packed plane (top)

close-packed directions

Mg (HCP)

Al (FCC)tensile direction

• Results of tensiletesting.

view onto twoclose-packedplanes.

Dislocations & Crystal Structure

Dislocation Density_1

ρ ≡ line length per unit Volume (cm/cm3)

L

Dislocation Density_1

Dislocation Density_2

Typical Dislocation Densitiesheavily coldworked metalsdeformed metals 108 ∼ 109 /cm2

annealed metals 107 ∼ 108 /cm2

Semiconductors 100 ∼ 102 /cm2

whiskers 0

n

L

L

n : # of dislocations

1011 ∼ 1012 /cm2

Plastic Flow and Dislocation Motion

: average velocity of dislocation.

shear strain increment

during the time increment

∴ Average shear strain rate is,

but n/(Lh) is the dislocations density

: mobile dislocation density

Taylor - Orowan Relation

※ for electrical conductivity

v

ρm

Slip in Single Crystals

1) Dislocations move in response to shear.

2) Applied tensile stress in one direction acts like a shear stress on any plane angled to that direction.

Compression

Dislocations can move by a shear stress.In σ - ε curve, we learned strength of the material characterizedby tensile yield strength

Shear or tensile?

Slip in Single Crystals

Schmid Law

F: Applied load

Single crystal yields when

s

ss

s

Schmid Law

Ex: Deformation of Single Crystal_1

So the applied stress of 6500 psi will not cause the crystal to yield.

cos cos 6500 psi

=35°=60°

(6500 psi) (cos35)(cos60) (6500 psi) (0.41) 2662 psi crss 3000 psi

crss = 3000 psi

a) Will the single crystal yield? b) If not, what stress is needed?

= 6500 psi

ex : Deformation of Single Crystal_1

Ex: Deformation of Single Crystal_2

psi 732541.0

psi 3000coscos

crss

y

What stress is necessary (i.e., what is the yield stress, sy)?

)41.0(cos cos psi 3000crss yy

psi 7325 y

So for deformation to occur the applied stress must be greater than or equal to the yield stress

ex : Deformation of Single Crystal_2

Critical Resolved Shear Stress

• Condition for dislocation motion:

• Crystal orientation can makeit easy or hard to move disl.

10-4G to 10-2G

typically

R cos cos

R maximum at = = 45º

R = 0l =90°

R = s/2l =45°f =45°

R = 0f =90°

σ

Critical Resolved Shear Stress

σσ

Single Crystal SlipSingle Crystal Slip

300 m

Disl. Motion in Polycrystals

Twin Deformation

• Grain Size Hardening

• Solid Solution Hardening

• Precipitation Hardening

• Strain Hardening

Four Strengthening Mechanisms

• Grain boundaries arebarriers to slip.

• Barrier "strength"increases withmisorientation.

• Smaller grain size:more barriers to slip.

• Hall-Petch Equation:

grain boundary

slip plane

grain Agra

in B

yield o kyd1/2

Strengthening Strategy 1 : Grain Size

• 70wt%Cu-30wt%Zn brass alloy

yield o kyd1/2

• Data:

[grain size (mm)]-0.5

yie

ld(M

Pa

)

50

100

150

200

04 8 12 16

10-1 10-2 5x10-3grain size, d (mm)

1

ky

0

0.75mm

Grain Size Hardening

Grain Size Hardening (Kinetics)

more specifically

parametercomplex : K

grain singlea of σ average : σ

relationPetch - Hall d+K= σσ

y

y0

5.0y0y-

Grain Size Hardening (Kinetics)

Strengthening Strategy 2 : Solid Solutions

• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.

• Smaller substitutionalimpurity

• Larger substitutionalimpurity

Impurity generates local shear at A and B that opposes disl motion to the right.

Impurity generates local shear at C and D that opposes disl motion to the right.

C

D

A

B

Strengthening by Alloying_1

• small impurities tend to concentrate at dislocations

• reduce mobility of dislocation increase strength

Strengthening by Alloying_2

• large impurities concentrate at dislocations on low density side

Ex : Solid Solution Strengthening in Copperex : Solid Solution Strengthening in Copper

• Tensile strength & yield strength increase w/wt% Ni.

• Empirical relation:• Alloying increases σy and TS.

y ~ C1/2

Adapted from Fig. 7.14 (a) and (b), Callister 6e.

Yie

ld s

tre

ng

th (

MP

a)

wt. %Ni, (Concentration C)

60

120

180

0 10 20 30 40 50

Te

nsi

le s

tre

ng

th (

MP

a)

wt. %Ni, (Concentration C)

200

300

400

0 10 20 30 40 50

Strengthening Strategy 3 : Precipitation

• Result:S

~y1

Large shear stress needed to move dislocation toward precipitate and shear it.

Dislocation “advances” but precipitates act as “pinning” sites with

S.spacing

Side View

precipitate

Top View

Slipped part of slip plane

Unslipped part of slip plane

Sspacing

• Hard precipitates are difficult to shear.Ex: Ceramics in metals (SiC in Iron or Aluminum).

Application : Precipitation Strengthening

• Internal wing structure on Boeing 767

• Aluminum is strengthened with precipitates formedby alloying.

1.5mm

Strengthening Strategy 4 : Cold Work (%CW)

• Room temperature deformation.• Common forming operations change the cross

sectional area:

%CW

Ao AdAo

x100

Ao Ad

force

dieblank

force

-Forging -Rolling

-Extrusion-Drawing

Adapted from Fig. 11.7, Callister 6e.

tensile force

AoAddie

dieram billet

container

containerforce

die holder

die

Ao

Adextrusion

roll

AoAd

roll

Strain Hardening (Work Hardening)

Increase the density of dislocation

→ increase yield strength

HW .

Dislocations during Cold Work

• Ti alloy after cold working:

• Dislocations entanglewith one anotherduring cold work.

• Dislocation motionbecomes more difficult.

0.9 m

Anisotropy in Deformation

1. Cylinder ofTantalummachinedfrom arolled plate: side view

endview

• The noncircular end view shows:anisotropic deformation of rolled material.

rolli

ng d

irect

ion

2. Fire cylinderat a target.

3. Deformedcylinder

platethicknessdirection

Anisotropy in σyield

• Can be induced by rolling a polycrystalline metal -before rolling -after rolling

235 mm-isotropicsince grains areapprox. spherical& randomlyoriented.

-anisotropicsince rolling affects grainorientation and shape.

rolling direction

Anisotropy in σyield

Result of Cold Work_1

• Dislocation density (ρd) goes up:Carefully prepared sample: ρd ~ 103 mm/mm3

Heavily deformed sample: ρd ~ 1010 mm/mm3

• Ways of measuring dislocation density:

OR

length, l1

length, l2length, l3

Volume, V

l1 l2 l3V

d d N

A

Area, A

N dislocation pits (revealed by etching)

dislocation pit

40mm

Result of Cold Work_2

• Dislocation density

– Carefully grown single crystal 103 mm-2

– Deforming sample increases density 109-1010 mm-2

– Heat treatment reduces density 105-106 mm-2

• Yield stress increasesas ρd increases:

large hardeningsmall hardeningσy0

σy1

Impact of Cold Work

• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.

As cold work is increased

• Yield strength (σy) increases.

Dislocation – Dislocation Trapping

• Dislocation generate stress.• This traps other dislocations.

Red dislocation generates shear at pts A and B that opposes motion of green disl. from left to right.

A

B

Impact of Cold Work

Str

ess

% cold work Strain

• Yield strength (σy) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.

Impact of Cold Work

Cold Work Analysis

• What is the tensile strength &ductility after cold working? Cold

work ----->

Do=15.2mm Dd=12.2mm

Copper

ductility (%EL)

7%

%EL=7%% Cold Work

20

40

60

20 40 6000

Cu

% Cold Worky=300MPa

100

300

500

700

Cu

200 40 60

yield strength (MPa)

300MPa

% Cold Work

tensile strength (MPa)

340MPa

TS=340MPa

200

Cu

0

400

600

800

20 40 60

Cold Work Analysis

%6.35=100xr

rr=CW% 2

o

2d

2o

-

Concepts for Microstructural Control

Soft while fabrication, Strong under service !

casting

forging

annealing - heating to a suitable T followed by

cooling at a moderate rate

in order to reduce hardness, improve machinability

cold working

hot working : materials worked above recrystallization T

recrystallization : 0.3Tm for pure metal

0.5Tm for alloys ( ~ up to 0.7 Tm )

Concepts for Microstructural Control

Effect of Heating After %CW

• 1 hour treatment at Tanneal...decreases TS and increases %EL.

• Effects of cold work are reversed!

• 3 annealingstages todiscuss...

tens

ile s

tren

gth

(MPa

)

duct

ility

(%EL

)tensile strength

ductility

600

300

400

500

60

50

40

30

20

annealing temperature (ºC)200100 300 400 500 600 700

Effect of Heating after %CW

cold worked and recovered

strain E by dislocation VS surface E

The Relation between Mechanical Properties and Microstructural Control

Recovery

Annihilation reduces dislocation density.• Scenario 1

atoms diffuse to regions of tension

extra half-plane of atoms

extra half-plane of atoms

Disl. annhilate and form a perfect atomic plane.

• Scenario 2 annihilation

4. opposite dislocations meet and annihilate

2. grey atoms leave by vacancy diffusion allowing disl. to “climb”

tR

1. dislocation blocked; can’t move to the right

Obstacle dislocation

3. “Climbed” disl. can now move on new slip plane

σ-ε Behavior vs Temperature

• Results forpolycrystalline iron:

• sy and TS decrease with increasing test temperature.• %EL increases with increasing test temperature.• Why? Vacancies

help dislocationspast obstacles.

1. disl. trapped by obstacle

2. vacancies replace atoms on the disl. half plane

3. disl. glides past obstacle

obstacle

-200C

-100C

25C

800

600

400

200

0

Strain

Stre

ss (M

Pa)

0 0.1 0.2 0.3 0.4 0.5

Recrystalization Temperature, TR

• TR = recrystallization temperaturepoint of highest rate of property change

or temperature at which the crystallization reaches completion in 1hr

– Tm => TR 0.3-0.6 Tm (K)– Due to diffusion annealing time TR = f(t)

shorter annealing time => higher TR

– Higher %CW => lower TR – strain hardening– Pure metals => lower TR due to easier movement of grain

boundary

Recrystalization

• New crystals are formed that:--have a small disl. density--are small--consume cold-worked crystals.

33% coldworkedbrass

New crystalsnucleate after3 sec. at 580C.

0.6 mm 0.6 mm

Further Recrystalization

• All cold-worked crystals are consumed.

After 4seconds

After 8seconds

0.6 mm0.6 mm

TR

º

TR = recrystallization temperature

Grain Growth_1

• At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy)

is reduced.

• Empirical Relation:

After 8 s,580C

After 15 min,580C

dn do

n Ktelapsed time

coefficient dependenton material and T.

grain diam.at time t.

exponent typ. ~ 2

0.6 mm 0.6 mm

Grain Growth_2

Coldwork Calculations

A cylindrical rod of brass originally 0.40 in (10.2 mm) in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 55,000 psi (380 MPa) and a ductility of at least 15 %EL are desired. Further more, the final diameter must be 0.30 in (7.6 mm). Explain how this may be accomplished.

Coldwork Calculations

Coldwork Calculations Solution

If we directly draw to the final diameter what happens?

%843100 x 4003001100 x

441

100 1100 x %

2

2

2

...

DD

xAA

AAACW

o

f

o

f

o

fo

D o = 0.40 in

BrassColdWork

Df = 0.30 in

Coldwork Calculations Solution

Coldwork Calculation Solution : Cont._1

• For %CW = 43.8%

540420

– y = 420 MPa– TS = 540 MPa > 380 MPa

6

– %EL = 6 < 15

Coldwork Calculation Solution : Cont._2

380

12

15

27

For %EL < 15For TS > 380 Mpa > 12 %CW

< 27 %CW

our working range is limited to %CW = 12-27

Coldwork Calculation Solution : Recyrstalization

• Cold draw-anneal-cold draw again• For objective we need a cold work of %CW 12-27

– We’ll use %CW = 20• Diameter after first cold draw (before 2nd cold draw)?

– must be calculated as follows:

100%1 100 1% 2

02

22

202

22 CW

DDx

DDCW ff

50

02

2

100%1

.f CW

DD

50

202

100%1

.f

CWDD

in 335.010020130.0

5.0

021

DDfIntermediate diameter =

Coldwork Calculations Solution

Summary:1.Cold work D01= 0.40 in Df1 = 0.335 in

2.Anneal above D02 = Df1

3.Cold work D02= 0.335 in Df2 =0.30 in

Therefore, meets all requirements24%EL=

MPa400TS=

MPa340=σy

%CW1 1 0.3350.4

2

x 100 30

20=100x335.0

3.01=CW

2

2 %

Summary

• Dislocations are observed primarily in metalsand alloys.

• Here, strength is increased by making dislocationmotion difficult.

• Particular ways to increase strength are to:--decrease grain size--solid solution strengthening--precipitate strengthening--cold work

• Heating (annealing) can reduce dislocation densityand increase grain size.