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Chapter 8: Deformation & Strengthening Mechanisms

School of Mechanical Engineering

Choi Hae Jin

Chapter 8 -

Choi, Hae-Jin

Materials Science - Prof. Choi, Hae-Jin 1

ISSUES TO ADDRESS…

• Why are the number of dislocations present greatest in metals ?

• How are strength and dislocation motion related?

• Why does heating alter strength and other properties?

Chapter 8 - 2

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Dislocations & Materials Classes

• Metals (Cu, Al): Dislocation motion easiest- non-directional bonding

+

+

+

+

+++++++

+ + + + + +

+++++++

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

g- close-packed directions

for slipelectron cloud ion cores

++++++++

Chapter 8 - 3

• Ionic Ceramics (NaCl):Motion difficult- need to avoid nearest

neighbors of like sign (- and +)

+ + + +

+++

+ + + +

- - -

----

- - -

Dislocation MotionDislocation motion & plastic deformation

• Metals - plastic deformation occurs by slip – an edge dislocation (extra half-plane of atoms) slides over adjacent plane half-planes of atoms.

Chapter 8 - 4

• If dislocations can't move, plastic deformation doesn't occur!

Adapted from Fig. 8.1, Callister & Rethwisch 3e.

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Dislocation Motion

• A dislocation moves along a slip plane in a slip directionperpendicular to the dislocation line

• The slip direction is the same as the Burgers vector directionp g

Edge dislocation

Adapted from Fig. 8.2, Callister & Rethwisch 3e.

Chapter 8 - 5

Screw dislocation

Deformation MechanismsSlip System

– Slip plane - plane on which easiest slippage occurs• Highest planar densities (and large interplanar spacings)

Slip directions directions of movement– Slip directions - directions of movement • Highest linear densities

Adapted from Fig. 8.6, Callister & Rethwisch 3e.

Chapter 8 - 6

– FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed)

=> total of 12 slip systems in FCC– For BCC & HCP there are other slip systems.

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Stress and Dislocation Motion• Resolved shear stress, R

– results from applied tensile stresses

Resolved shear t F /A

Relation between d

Applied tensile t F/A

slip plane

normal, ns

stress: R =Fs/As

AS

R

FS

and R

R=FS/AS

Fcos A/cos

F

F

nS

A

stress: = F/A

FA

Chapter 8 -

coscosR

7

RFS ASF

Critical Resolved Shear Stress

• Condition for dislocation motion: CRSS R

• Ease of dislocation motion depends on crystallographic orientation

typicallyon crystallographic orientation

10-4 GPa to 10-2 GPa coscosR

R = 0

= /2

R = 0

Chapter 8 - 8

maximum at = = 45º

R = 0

=90°

R = /2=45°=45°

R 0

=90°

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Single Crystal Slip

Adapted from Fig. 8 9 Callister &8.9, Callister & Rethwisch 3e.

Chapter 8 - 9

Adapted from Fig. 8.8, Callister & Rethwisch 3e.

Ex: Deformation of single crystal

= 35°

= 60°crss = 20.7 MPa

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

MPa 20.7

cos cos

= 35

Adapted from Fig. 8.7, Callister &

(45 MPa)

(45 MP ) (0 41) (cos35) (cos60)

Chapter 8 -

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

10

= 6500 psi

Callister & Rethwisch 3e.

(45 MPa) (0.41)

18.5 MPa crss 20.7 MPa

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Ex: Deformation of single crystal

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

psi 732541.0

psi 3000

coscoscrss

y

)41.0(cos cos psi 3000crss yy

S f d f ti t th li d t t

Chapter 8 - 11

psi 7325 y

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

Slip Motion in Polycrystals• Stronger - grain boundaries

pin deformations

• Slip planes & directions

• Slip planes & directions(, ) change from onecrystal to another.

• R will vary from onecrystal to another.

• The crystal with the

Adapted from Fig. 8.10, Callister & Rethwisch 3e.(Fig. 8.10 is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)

Chapter 8 - 12

The crystal with thelargest R yields first.

• Other (less favorablyoriented) crystalsyield later.

g, ] )

300 m

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Anisotropy in y

• Can be induced by rolling a polycrystalline metal

- before rolling - after rollingAdapted from Fig. 8.11, Callister & Rethwisch 3e

235 m

rolling direction

Callister & Rethwisch 3e.(Fig. 8.11 is from W.G. Moffatt, G.W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 140, John Wiley and Sons, New York, 1964.)

Chapter 8 - 13

- isotropicsince grains areapprox. spherical& randomlyoriented.

- anisotropicsince rolling affects grainorientation and shape.

Strategies for Strengthening Metals

• Grain Size Reduction

• Solid SolutionsSolid Solutions

• Precipitation Strengthening

• Cold Work

Chapter 8 - 14

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4 Strategies for Strengthening Metals: 1: Reduce Grain Size

• Grain boundaries areb i t li

21 /yoyield dk

barriers to slip.• Barrier "strength"

increases withIncreasing angle of misorientation.

• Smaller grain size:more barriers to slip

Adapted from Fig. 8.14, Callister & Rethwisch 3e. (Fig. 8.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

Chapter 8 - 15

more barriers to slip.

• Hall-Petch Equation:

4 Strategies for Strengthening Metals: 2: Solid Solutions

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

• Smaller substitutionalimpurity

A

B

• Larger substitutionalimpurity

C

D

Chapter 8 - 16

Impurity generates local stress at Aand B that opposes dislocation motion to the right.

B

Impurity generates local stress at Cand D that opposes dislocation motion to the right.

D

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Stress Concentration at Dislocations

Chapter 8 - 17

Adapted from Fig. 8.4, Callister & Rethwisch 3e.

Strengthening by Alloying

• small impurities tend to concentrate at dislocations

• reduce mobility of dislocation increase strength

Chapter 8 - 18

Adapted from Fig. 8.17, Callister & Rethwisch 3e.

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Strengthening by Alloying

• large impurities concentrate at dislocations on low density side

Chapter 8 - 19

Adapted from Fig. 8.18, Callister & Rethwisch 3e.

Ex: Solid SolutionStrengthening in Copper

• Tensile strength & yield strength increase with wt% Ni.

Pa) a) 180

Adapted from Fig. 8.16 (a) and (b), Callister & Rethwisch 3e.

Tens

ile s

tren

gth

(MP

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yie

ld s

tren

gth

(MP

a

wt.%Ni, (Concentration C)

60

120

0 10 20 30 40 50

Chapter 8 - 20

• Empirical relation:

• Alloying increases y and TS.

21 /y C~

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4 Strategies for Strengthening Metals: 3: Precipitation Strengthening

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

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

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

Side View

precipitate

Top ViewUnslipped part of slip plane

S

Chapter 8 - 21

• Result:S

~y

1

pinning sites with S.spacing

Slipped part of slip plane

Sspacing

Application:Precipitation Strengthening

• Internal wing structure on Boeing 767Adapted from chapter-opening photograph,

• Aluminum is strengthened with precipitates formedby alloying.

Adapted from chapter-

p g p g p ,Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

Chapter 8 - 22

Adapted from chapteropening photograph, Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

1.5m

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4 Strategies for Strengthening Metals: 4: Cold Work (%CW)

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

sectional area:

Adapted from Fig. 14.2, Callister & Rethwisch 3e.

-Forging

Ao Ad

force

dieblank

force-Drawing -ExtrusionA

-Rolling

roll

AoAd

roll

Chapter 8 - 23

tensile force

AoAddie

dieram billet

container

containerforce

die holder

die

Ao

Adextrusion

100 x %o

do

A

AACW

Dislocations During Cold Work• Ti alloy after cold working:

• Dislocations entanglewith one anotherwith one anotherduring cold work.

• Dislocation motionbecomes more difficult.

Chapter 8 - 24

Adapted from Fig. 5.11, Callister & Rethwisch 3e. (Fig. 5.11 is courtesy of M.R. Plichta, Michigan Technological University.)

0.9 m

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Result of Cold Work

Dislocation density =

– Carefully grown single crystal

total dislocation lengthunit volume

y g g y

ca. 103 mm-2

– Deforming sample increases density

109-1010 mm-2

– Heat treatment reduces density

105-106 mm-2

Chapter 8 - 25

• Yield stress increasesas d increases:

large hardening

small hardening

y0y1

Effects of Stress at Dislocations

Ad d f FiAdapted from Fig. 8.5, Callister & Rethwisch 3e.

Chapter 8 - 26

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Impact of Cold Work

As cold work is i d

Adapted from Fig. 8.20, Callister & Rethwisch 3e.

• Yield strength (y) increases.

• Tensile strength (TS) increases.

increased

Chapter 8 - 27

• Ductility (%EL or %AR) decreases.

Cold Work Analysis• What is the tensile strength &

ductility after cold working?

%6.35100x%2

22

do rr

CW

Cold Work

Copper

%6.35100 x %2 or

CW

300

500

700

Cu

yield strength (MPa)

300MPa

tensile strength (MPa)

400

600

800

ductility (%EL)

20

40

60

Cu

Do =15.2mm Dd =12.2mm

340MPa

Chapter 8 - 28

Adapted from Fig. 8.19, Callister & Rethwisch 3e. (Fig. 8.19 is adapted from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 226; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th ed., H. Baker (Managing Ed.),

% Cold Work

100

Cu

200 40 60

y = 300MPa% Cold Work

200Cu

0 20 40 60% Cold Work20 40 6000

340MPa

TS = 340MPa

7%

%EL = 7%

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Effect of Heating After %CW• 1 hour treatment at Tanneal...

decreases TS and increases %EL.• Effects of cold work are reversed!

• 3 Annealingstages todiscuss s

tren

gth

(MP

a)

uctil

ity (

%E

L)

tensile strength600

400

500

60

50

40

30

annealing temperature (ºC)200100 300 400 500 600 700

Chapter 8 - 29

discuss...Adapted from Fig. 8.22, Callister & Rethwisch 3e. (Fig. 8.22 is adapted from G. Sachs and K.R. van Horn, Practical Metallurgy, Applied Metallurgy, and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.)

tens

ile duductility

300

400 30

20

Recrystallization• New grains are formed that:

-- have a small dislocation density-- are small

consume cold worked grains-- consume cold-worked grains.

Adapted from Fig. 8.21 (a),(b), Callister & Rethwisch 3e. (Fig. 8.21 (a),(b) are courtesy of J E B k

0.6 mm 0.6 mm

Chapter 8 - 30

J.E. Burke, General Electric Company.)

33% coldworkedbrass

New crystalsnucleate after3 sec. at 580C.

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Further Recrystallization• All cold-worked grains are consumed.

0.6 mm0.6 mm

Adapted from Fig. 8.21 (c),(d), Callister & Rethwisch 3e. (Fig. 8.21 (c),(d) are courtesy of J.E. Burke, General Electric

Chapter 8 - 31

General Electric Company.)

After 4seconds

After 8seconds

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

is reduced.0 6 0 6

After 8 s,580ºC

After 15 min,580ºC

0.6 mm 0.6 mm

Adapted from Fig. 8.21 (d),(e), Callister & Rethwisch 3e. (Fig. 8.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.)

Chapter 8 - 32

580ºC 580ºC

• Empirical Relation:

Ktdd no

n elapsed time

coefficient dependenton material and T.

grain diam.at time t.

exponent typ. ~ 2

Ostwald Ripening

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º

TR = recrystallization temperature

TR

Adapted from Fig. 8.22, Callister & Rethwisch 3e

Chapter 8 - 33

Callister & Rethwisch 3e.

º

Recrystallization Temperature, TR

TR = recrystallization temperature = point of highest rate of property changeg p p y g1. Tm => TR 0.3-0.6 Tm (K)

2. Due to diffusion annealing time TR = f(time) shorter annealing time => higher TR

3. Higher %CW => lower TR – strain hardening

4. Pure metals lower TR due to dislocation movements

Chapter 8 -

• Easier to move in pure metals => lower TR

34

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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 Across 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.

Chapter 8 - 35

Coldwork Calculations Solution

If we directly draw to the final diameter what happens?

%CWAo Af

x 100 1

Af

x 100

Do = 0.40 in

BrassCold Work

Df = 0.30 in

Chapter 8 - 36

%CW o f

Ao

x 100 1 f

Ao

x 100

1 Df2 4

Do2 4

x 100 1 0.30

0.40

2

x 100 43.8%

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Coldwork Calc Solution: Cont.

F %CW 43 8%

540420

6

Adapted from Fig 8 19

Chapter 8 -

• For %CW = 43.8%

37

– y = 420 MPa– TS = 540 MPa > 380 MPa– %EL = 6 < 15

• This doesn’t satisfy criteria…… what can we do?

Adapted from Fig. 8.19, Callister & Rethwisch 3e.

Coldwork Calc Solution: Cont.

380

12

15

27

Chapter 8 - 38

For %EL > 15

For TS > 380 MPa > 12 %CW

< 27 %CW

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

Adapted from Fig. 8.19, Callister & Rethwisch 3e.

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Coldwork Calc Soln: RecrystallizationCold 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

D

Dx

D

DCW ff

50%

.CWD 2fD

D

Chapter 8 - 39

02

2

100

%1f CW

D

D

50

202

100%

1.

f

CWD

m 335010020

130050

021 ..DD.

f

Intermediate diameter =

Coldwork Calculations Solution

Summary:

1. Cold work D01= 0.40 in Df1 = 0.335 m

2. Anneal above D02 = Df1

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

302

MPa 340y

%CW1 10.335

0.4

2

x 100 30

Fig 7.19

Chapter 8 -

Therefore, meets all requirements

40

20100 3350

301% 2

x

.

.CW

24%

MPa 400

EL

TSy

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Mechanical Properties of Polymers –Stress-Strain Behaviorbrittle polymer

plasticelastomer

elastic moduli – less than for metals Adapted from Fig. 7.22,

Callister & Rethwisch 3e.

Chapter 8 - 4141

• Fracture strengths of polymers ~ 10% of those for metals

• Deformation strains for polymers > 1000%

– for most metals, deformation strains < 10%

Mechanisms of Deformation—Brittle Crosslinked and Network Polymers

(MPa)Near Nearbrittle failure

plastic failure

( )

x

xInitial Failure Initial Failure

Chapter 8 - 4242

aligned, crosslinkedpolymer Stress-strain curves adapted from Fig. 7.22,

Callister & Rethwisch 3e.

network polymer

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Mechanisms of Deformation —Semicrystalline (Plastic) Polymers

brittle failure(MPa)

x

fibrillar structure

nearStress-strain curves adapted from Fig 7 22 Callister &

plastic failure

x

near failureonset of

necking

unload/reload

from Fig. 7.22, Callister & Rethwisch 3e. Inset figures along plastic response curve adapted from Figs. 8.27 & 8.28, Callister & Rethwisch 3e. (Figs. 8.27 & 8.28 are from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.)

Chapter 8 - 4343

crystallineblock segments

separate

crystallineregions align

undeformedstructure amorphous

regionselongate

Predeformation by Drawing• Drawing…(ex: monofilament fishline)

-- stretches the polymer prior to use-- aligns chains in the stretching direction

• Results of drawing:-- increases the elastic modulus (E) in the

stretching direction-- increases the tensile strength (TS) in the

stretching direction-- decreases ductility (%EL)

• Annealing after drawing...

Adapted from Fig. 8.28, Callister & Rethwisch 3e. (Fig. 8.28 is from J.M. Schultz, Polymer Materials Science, Prentice-Hall,

Chapter 8 - 44

g g-- decreases chain alignment-- reverses effects of drawing (reduces E and

TS, enhances %EL)• Contrast to effects of cold working in metals!

Inc., 1974, pp. 500-501.)

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Mechanisms of Deformation—Elastomers

Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 3e.Inset figures along

(MPa)brittle failurex

elastomer curve (green) adapted from Fig. 8.30, Callister & Rethwisch 3e. (Fig. 8.30 is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.)

x

final: chainsare straighter,

stillcross-linked

elastomer

plastic failurex

Chapter 8 - 45

• Compare elastic behavior of elastomers with the:-- brittle behavior (of aligned, crosslinked & network polymers), and-- plastic behavior (of semicrystalline polymers)

(as shown on previous slides)

initial: amorphous chains are kinked, cross-linked.

deformation is reversible (elastic)!

Summary

• Dislocations are observed primarily in metalsand alloys.

St th i i d b ki di l ti• Strength is increased by making dislocationmotion difficult.

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

Chapter 8 - 46

cold work

• Heating (annealing) can reduce dislocation densityand increase grain size. This decreases the strength.