Materials Science and Engineering 8

Division of Mechanical Engineering, College of EngineeringDivision of Mechanical Engineering, College of Engineering

Page 1Edited by Byoung-Ho Choi

Chapter 8 Deformation and Strengthening Mechanisms

• Issues to address in this chapter:• 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?

Page 2


Edited by Byoung-Ho Choi

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

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

neighbors of like sign (- and +)

+ + + +

+++

+ + + +

- - -

----

- - -

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

for slip electron cloud ion cores

++

++

++++++++ + + + + +

+++++++

Dislocations & Materials Classes

Page 3



Dislocation 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.

• If dislocations can't move, plastic deformation doesn't occur! Adapted from Fig. 8.1, Callister & Rethwisch 3e.

Dislocation Motion

Page 4



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

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

Edge dislocation

Screw dislocation

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

Dislocation Motion

Page 5



Slip System• Slip plane - plane on which easiest slippage occurs

» Highest planar densities (and large interplanar spacings) • Slip directions - directions of movement

» Highest linear densitiesAdapted from Fig. 8.6, Callister & Rethwisch3e.

– 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.

Deformation Mechanisms

Page 6



• Resolved shear stress, tR– results from applied tensile stresses

slip planenormal, ns

Resolved shear stress: τR = Fs /A s

slipdirectio

n

AS

tR

tR

FS

slipdirectio

n

Relation between σ and τR

τR= FS /AS

Fcos λ A /cos φ

λF

FS

φnS

AS

A

Applied tensile stress: = F/Aσ

slipdirectio

n

FA

F

φλσ=τ coscosR

Stress and Dislocation Motion

Page 7



• Condition for dislocation motion: CRSS τ>τR• Ease of dislocation motion depends

on crystallographic orientation

10-4 GPa to 10-2 GPa

Typically

φλσ=τ coscosR

τ maximum at λ = φ = 45º

tR = 0λ=90°

σ

tR = σ/2λ=45°φ=45°

σ

tR = 0φ=90°

σ

Critical Resolved Shear Stress

Page 8



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


Single Crystal Slip

Page 9


Edited by Byoung-Ho ChoiSo the applied stress of 6500 psi will not cause the crystal to yield.

MPa 20.7 cos cos

=σφλσ=τ

λ = 35°φ = 60°

τcrss = 20.7 MPa

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

σ = 6500 psi


τ = (45 MPa) = (45 MPa) ( 0.41)τ = 18.5 MPa < τ crss =

(cos 35 o ) (cos 60 o )

20 .7 MPa

Ex: Deformation of single crystal

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psi 732541.0

psi 3000coscos

crss ==φλ

τ=σ∴ y

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

)41.0(cos cos psi 3000crss yy σ=φλσ==τ

psi 7325=σ≥σ y

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

Ex: Deformation of single crystal

Page 11



• Stronger - grain boundaries pin deformations

• Slip planes & directions (l, f) change from one crystal to another.

• tR will vary from one crystal to another.

• The crystal with the largest tR yields first.

• Other (less favorably oriented) crystals yield later.

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].)

σ

300 μm

Slip Motion in Polycrystals

Page 12



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

235 mm- isotropicsince grains are approx. spherical & randomly oriented.

- after rolling

- anisotropicsince rolling affects grainorientation and shape.

rolling direction

Adapted from Fig. 8.11, 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.)

Anisotropy in σy

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side view

1. Cylinder oftantalummachinedfrom arolled plate:

rollin

g dire

ction

2. Fire cylinderat a target.

• The noncircular end view showsanisotropic deformation of rolled material.

endview

3. Deformedcylinder

platethicknessdirection

Photos courtesyof G.T. Gray III,Los AlamosNational Labs. Used withpermission.

Anisotropy in Deformation

Page 14



• Grain boundaries arebarriers to slip.

• Barrier "strength"increases withIncreasing angle of misorientation.

• Smaller grain size:more barriers to slip.

• Hall-Petch Equation:21 /

yoyield dk −+σ=σ

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.)

4 Strategies for Strengthening Metals: 1: Reduce Grain Size

Page 15



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

impurity

Impurity generates local stress at A and Bthat opposes dislocation motion to the right.

A

B

• Larger substitutionalimpurity

Impurity generates local stress at C and Dthat opposes dislocation motion to the right.

C

D

4 Strategies for Strengthening Metals: 2: Solid Solutions

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Stress Concentration at Dislocations

Page 17



• small impurities tend to concentrate at dislocations• reduce mobility of dislocation ∴ increase strength


Strengthening by Alloying

Page 18



• large impurities concentrate at dislocations on low density side


Strengthening by Alloying

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

• Empirical relation:

• Alloying increases σy and TS.

21 /y C~σ

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

Tens

ile st

reng

th (M

Pa)

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yield

stre

ngth

(MPa

)wt.%Ni, (Concentration C)

60

120

180

0 10 20 30 40 50

Ex: Solid Solution Strengthening in Copper

Page 20



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

• Result:S

~y1 σ

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

S spacing

4 Strategies for Strengthening Metals: 3: Precipitation Strengthening

Page 21



• Internal wing structure on Boeing 767

• Aluminum is strengthened with precipitates formed by alloying.

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

1.5mm

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

Application: Precipitation Strengthening

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• Room temperature deformation.• Common forming operations change the cross sectional area:


-Forging

A o A d

force

dieblank

force-Drawing

tensile force

A oA ddie

die

-Extrusion

ram billet

container

containerforce die holder

die

A o

A dextrusion

100 x %o

doA

AACW −=

-Rolling

rollA o

A droll

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

Page 23



• Ti alloy after cold working:• Dislocations entangle

with one anotherduring cold work.

• Dislocation motionbecomes more difficult.

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

0.9 mm

Dislocations During Cold Work

Page 24



• Dislocation density =

• Carefully grown single crystal ca. 103 mm-2

• Deforming sample increases density 109-1010 mm-2

• Heat treatment reduces density 105-106 mm-2

• Yield stress increasesas rd increases:

total dislocation lengthunit volume

large hardeningsmall hardening

σ

ε

σy0σy1

Result of Cold Work

Page 25




Effects of Stress at Dislocations

Page 26




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

• As cold work is increased

Impact of Cold Work

Page 27



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

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.), American Society for Metals, 1979, p. 276 and 327.)

%6.35100 x % 2

22=

π

π−π=

o

dor

rrCW

% Cold Work

100

300

500

700

Cu

200 40 60

yield strength (MPa)

σy = 300MPa

300MPa

% Cold Work

tensile strength (MPa)

200Cu

0

400

600

800

20 40 60

ductility (%EL)

% Cold Work

20

40

60

20 40 6000

Cu

D o =15.2mm

Cold Work

Dd =12.2mm

Copper

340MPa

TS = 340MPa

7%

%EL = 7%

Cold Work Analysis

Page 28



• Results forpolycrystalline iron:

• σy and TS decrease with increasing test temperature.

• %EL increases with increasing test temperature.

• Why? Vacancieshelp dislocationsmove past obstacles.


2. vacancies replace atoms on the disl. half plane

3 . disl. glides past obstacle

-200°C

-100°C

25°C

800

600

400

200

0

Strain

Stre

ss (M

Pa)

0 0.1 0.2 0.3 0.4 0.5

1. disl. trapped by obstacle

obstacle

σ-ε Behavior vs. Temperature

Page 29



• 1 hour treatment at Tanneal: decreases TS and increases %EL.• Effects of cold work are reversed!

• 3 Annealing stages to 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.)

tensil

e stre

ngth

(MPa

)

ducti

lity (%

EL)

tensile strength

ductility

RecoveryRecrystallization

Grain Growth

600

300

400

500

60

50

40

30

20

annealing temperature (ºC)200100 300 400 500 600 700

Effect of Heating After %CW

Page 30



Annihilation reduces dislocation density.• Scenario 1

Results from diffusion

• Scenario 2

4. opposite dislocations meet and annihilate

Dislocations annihilate and form a perfect atomic plane.

extra half-plane of atoms

extra half-plane of atoms

atoms diffuse to regions of tension

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

Recovery

Page 31



• New grains are formed that:-- have a small dislocation density-- are small-- consume cold-worked grains.

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

33% coldworkedbrass

New crystalsnucleate after3 sec. at 580°C.

0.6 mm 0.6 mm

Recrystallization

Page 32



• All cold-worked grains are consumed.

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

After 4seconds

After 8seconds

0.6 mm0.6 mm

Further Recrystallization

Page 33



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

After 8 s,580ºC

After 15 min,580ºC

0.6 mm 0.6 mmAdapted from Fig. 8.21 (d),(e), Callister& Rethwisch 3e. (Fig. 8.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.)

• Empirical Relation:

Ktdd no

n =−elapsed time

coefficient dependenton material and T.

grain diam.at time t.

exponent typ. ~ 2

Ostwald Ripening

Grain Growth

Page 34



TR

Adapted from Fig. 8.22, Callister& Rethwisch 3e.

º

º

TR = recrystallizationtemperature

Effect of Heating

Page 35



TR = recrystallization temperature = point of highest rate of property change1. Tm => TR ≈ 0.3-0.6 Tm (K)2. Due to diffusion annealing time TR = f(time) shorter annealing time

=> higher TR3. Higher %CW => lower TR – strain hardening4. Pure metals lower TR due to dislocation movements

» Easier to move in pure metals => lower TR

Recrystallization Temperature, TR

Page 36



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

Page 37



If we directly draw to the final diameter what happens?

%CW =Ao − Af

Ao

⎛

⎝ ⎜

⎞

⎠ ⎟ x 100 = 1−

Af

Ao

⎛

⎝ ⎜

⎞

⎠ ⎟ x 100

= 1−πDf

2 4πDo

2 4

⎛

⎝ ⎜

⎞

⎠ ⎟ x 100 = 1−

0.300.40

⎛

⎝ ⎜

⎞

⎠ ⎟

2⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ x 100 = 43.8%

Do = 0.40 in

BrassCold Work

Df = 0.30 in

Coldwork Calculations Solution

Page 38



• For %CW = 43.8%– σy = 420 MPa– TS = 540 MPa > 380 MPa

540420

6

– %EL = 6 < 15• This doesn’t satisfy criteria…… what can we do?


Coldwork Calc Solution: Cont.

Page 39



For %EL > 15For TS > 380 MPa > 12 %CW

< 27 %CW

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

380

12

15

27


Coldwork Calc Solution: Cont.

Page 40



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

⎟⎠⎞

⎜⎝⎛ −=

502

02

100%1

.f

CWDD

⎟⎠⎞

⎜⎝⎛ −

=⇒

m 3350100201300

50

021 ..DD.

f =⎟⎠⎞

⎜⎝⎛ −==Intermediate diameter =

Coldwork Calc Soln: Recrystallization

Page 41



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

2. Anneal above D02 = Df13. Cold work D02= 0.335 in Df2 =0.30 m

Therefore, meets all requirements

20100 3350

301%2

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−= x

..CW

24%MPa 400MPa 340

==

=σ

ELTS

y

⇒

%CW1 = 1−0.335

0.4⎛

⎝ ⎜

⎞

⎠ ⎟ 2⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ x 100 = 30

Fig 7.19

Coldwork Calculations Solution

Page 42



• Hot work above TR

• Cold work below TR

• Smaller grains • stronger at low temperature• weaker at high temperature

t/RTBCt

kTERtR

1:note

log

logloglog 0

=

+=

−=−=

RT1

log t

start

finish

50%

Rate of Recrystallization

Page 43



• Fracture strengths of polymers ~ 10% of those for metals• Deformation strains for polymers > 1000%

– for most metals, deformation strains < 10%

brittle polymer

plastic

elastomerelastic moduli

– less than for metals Adapted from Fig. 7.22, Callister & Rethwisch 3e.

Mechanical Properties of Polymers - Stress-Strain Behavior

Page 44



brittle failure

plastic failure

σ(MPa)

ε

x

x

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

Rethwisch 3e.

InitialNear

Failure Initial

network polymer

NearFailure

Mechanisms of Deformation- Brittle Crosslinked and Network Polymers

Page 45



brittle failure

plastic failure

σ(MPa)

x

x

crystallineblock segments

separate

fibrillarstructure

near failure

crystallineregions align

onset of necking

undeformedstructure amorphous

regionselongate

unload/reload

Stress-strain curves adapted 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.)

ε

Mechanisms of Deformation - Semicrystalline (Plastic) Polymers

Page 46



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

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

Adapted from Fig. 8.28, Callister & Rethwisch 3e. (Fig. 8.28 is from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.)

Predeformation by Drawing

Page 47



• 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)

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

σ(MPa)

ε

initial: amorphous chains are kinked, cross-linked.

x

final: chainsare straighter,

stillcross-linked

elastomer

deformation is reversible (elastic)!

brittle failure

plastic failurex

x

Mechanisms of Deformation—Elastomers

Page 48



• Dislocations are observed primarily in metals and alloys.• Strength is increased by making dislocation motion difficult.• Particular ways to increase strength are to:

-- decrease grain size-- solid solution strengthening-- precipitate strengthening-- cold work

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

Summary

Documents

Materials Science and Engineering 8