Materials Science - AAST

Materials Science

ME 274

Dr Yehia M. Youssef

1Copyright © YM Youssef, 4-Oct-10 Materials Science

Chapter 7: Dislocations & Strengthening M h iMechanisms

ISSUES TO ADDRESS...• Why are dislocations observed primarily in metals

and alloys?

• How are strength and dislocation motion related?

• How do we increase strength?

H h ti h t th d th ti ?• How can heating change strength and other properties?


Dislocations & Materials Classes

• Metals: Disl. motion easier.di ti l b di

++++++++-non-directional bonding-close-packed directions

for slip. electron cloud ion cores

++

++ + + + + +

+++++++

• Covalent Ceramics

p electron cloud ion cores

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

• Ionic Ceramics (NaCl):Motion hard + + + +- - -Motion hard.

-need to avoid ++ and - -neighbors.

+++

+ + + +

----

- - -


Dislocation MotionDi l ti & l ti d f tiDislocations & plastic deformation• Cubic & hexagonal metals - plastic deformation by

plastic shear or slip where one plane of atoms slidesplastic 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!

Adapted from Fig. 7.1, Callister 7e.

4

deformation doesn't occur!Copyright © YM Youssef, 4-Oct-10 Materials Science

Dislocation Motion

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

• Slip direction same direction as Burgers vector

Edge dislocation


Screw dislocation


Sli S tDeformation Mechanisms

Slip System– Slip plane - plane allowing easiest slippage

• Wide interplanar spacings - highest planar densities

– Slip direction - direction of movement - Highest linear densitiesdensities


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

=> total of 12 slip systems in FCCin BCC & HCP other slip systems occur

6

– in BCC & HCP other slip systems occurCopyright © YM Youssef, 4-Oct-10 Materials Science

Stress and Dislocation Motion• Crystals slip due to a resolved shear stress, τR. • Applied tension can produce such a stress.

Resolved shear stress: τR =Fs/As

Relation between σ and τR

Applied tensile stress: = F/Aσ

F slip planenormal, ns

AS

τRτR=FS/AS

Fcos λ A/cos φ

FA

ASFS

φ

λF φnS

AτR FS AS

AF

φλσ=τ coscosR

7

R

Copyright © YM Youssef, 4-Oct-10 Materials Science

Critical Resolved Shear Stress

• Condition for dislocation motion: CRSSτ>τR

• Crystal orientation can make i ll• Crystal orientation can makeit easy or hard to move dislocation

10-4 GPa to 10-2 GPatypically

φλσ=τ coscos φλσ=τ coscosR σ σ σ

τR = 0 τR = σ/2 τR = 0R

λ=90°τR σ/2λ =45°φ =45°

R

φ=90°

τ maximum at λ = φ = 45º8

τ maximum at λ = φ = 45Copyright © YM Youssef, 4-Oct-10 Materials Science

Single Crystal Slip




Ex: Deformation of single crystal) Will th i l t l i ld?

φ=60°a) Will the single crystal yield? b) If not, what stress is needed?

τ = σ cos λ cos φλ=35° τcrss = 3000 psi

φ σ = 6500 psi

τ = (6500 psi) (cos35 )(cos60 )(6500 i) (0 41)Adapted from

= (6500 psi) (0.41)τ = 2662 psi < τcrss = 3000 psi

6500 i

pFig. 7.7, Callister 7e.

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

σ = 6500 psi

10

crystal to yield.Copyright © YM Youssef, 4-Oct-10 Materials Science

Ex: Deformation of single crystal

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

)41.0(coscospsi3000crss yy σ=φλσ==τ

i7325psi 3000crssτ

)(pcrss yy φ

psi732541.0

pcoscos

crss ==φλ

=σ∴ y

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

psi 7325=σ≥σ y


St i b d i

Slip Motion in Polycrystalsσ• Stronger - grain boundaries

pin deformations

σ

• Slip planes & directions(λ, φ) change from onecr stal to another

Adapted from Fig. 7.10, Callister 7e.(Fig. 7.10 is crystal to another.

• τR will vary from one

( gcourtesy of C. Brady, National Bureau of Standards [now the National Institute ofcrystal to another.

• The crystal with the

National Institute of Standards and Technology, Gaithersburg, MD].)

ylargest τR yields first.

• Other (less favorably Other (less favorablyoriented) crystalsyield later.

300 μm


Anisotropy in σy

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

Adapted from Fig. 7.11, Callister 7e. (Fig. 7.11 is from W.G. Moffatt, G.W. Pearsall, and J. Wulff, The Structure and Properties of Materialsand Properties of Materials, Vol. I, Structure, p. 140, John Wiley and Sons, New York, 1964.)

235 μm

rolling direction

- isotropicsince grains areapprox. spherical

- anisotropicsince rolling affects grainorientation and shape.approx. spherical

& randomlyoriented.

orientation and shape.


Anisotropy in Deformation

1. Cylinder ofTantalummachined

2. Fire cylinderat a target.

3. Deformedcylinder

side view

machinedfrom arolled plate:

Photos courtesyof G.T. Gray III,Los AlamosNational Labsside view

ctio

n

National Labs. Used withpermission.

lling

dire

cro

endview

platethickness

• The noncircular end view showsanisotropic deformation of rolled material.

view thicknessdirection

14

anisotropic deformation of rolled material.Copyright © YM Youssef, 4-Oct-10 Materials Science

4 Strategies for Strengthening: 1: Reduce Grain Size1: Reduce Grain Size

• Grain boundaries arebarriers to slip.

• Barrier "strength"increases withIncreasing angle ofIncreasing angle of misorientation.

• Smaller grain size:Adapted from Fig. 7.14, Callister 7e.(Fig. 7.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Smaller grain size:

more barriers to slip.Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

• Hall-Petch Equation: 21 /yoyield dk −+σ=σ


4 Strategies for Strengthening: 2: Solid Solutions

• Impurity atoms distort the lattice & generate stress

2: Solid Solutions

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

impurity Larger substitutional

impurity

A C

B D

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

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


Stress Concentration at Dislocations



Strengthening by Alloying

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



Strengthening by Alloying

• large impurities concentrate at dislocations on low d it iddensity side



Ex: Solid SolutionStrengthening in CopperStrengthening in Copper

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

Adapted from Fig. (MP

a)

400 (MP

a) 180

7.16 (a) and (b), Callister 7e.

stre

ngth

300

stre

ngth

(

120

Tens

ile

wt.% Ni, (Concentration C)

2000 10 20 30 40 50 Yi

eld

s

wt.%Ni, (Concentration C)

600 10 20 30 40 50

• Empirical relation: 21 /y C~σ

, ( ) , ( )

• Alloying increases σy and TS.


4 Strategies for Strengthening: 3: Precipitation Strengthening

• Hard precipitates are difficult to shear.

3: Precipitation Strengthening

Ex: Ceramics in metals (SiC in Iron or Aluminum).precipitate

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

Side View

Dislocation “advances” but precipitates act as “ i i ” it ith

Top View Unslipped part of slip plane

“pinning” sites with S.spacing

Slipped part of slip plane

S

• Result:S

~y1 σ


Application:Precipitation Strengthening

• Internal wing structure on Boeing 767Precipitation Strengthening

Ad t d f h tAdapted from chapter-opening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G.

• Aluminum is strengthened with precipitates formed

Miller, Boeing Commercial Airplane Company.)

Aluminum is strengthened with precipitates formedby alloying.

Adapted from Fig.Adapted from Fig. 11.26, Callister 7e.(Fig. 11.26 is courtesy of G.H. Narayanan and A.G. Miller, Boeing CommercialBoeing Commercial Airplane Company.)

1.5μm

22

1.5μm


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

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

sectional area:-Forging force -Rolling

roll

Adapted from Fig.

g g

Ao Addie

blankroll

AoAd

roll

11.8, Callister 7e.

force-Drawing

Addie

-Extrusioncontainer

Ao

roll

tensile force

AoAddie

dieram billet

container

containerforce die holder

dieAdextrusion

100 x %o

doA

AACW −=

23

oCopyright © YM Youssef, 4-Oct-10 Materials Science

Dislocations During Cold Work

• Ti alloy after cold working:

Di l ti t l• Dislocations entanglewith one anotherduring cold workduring cold work.

• Dislocation motionbecomes more difficult.

Adapted from Fig. 4.6, Callister 7e.(Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)0.9 μm


Result of Cold Work

Dislocation density = total dislocation length

unit volume

– Carefully grown single crystal ca. 103 mm-2

– Deforming sample increases density 109-1010 mm-2

– Heat treatment reduces density 105-106 mm-210 10 mm

• Yield stress increasesas ρd increases: large hardening

σ

σy1as ρd increases: g gsmall hardeningσy0

σy1

25

εCopyright © YM Youssef, 4-Oct-10 Materials Science

Effects of Stress at Dislocations

Adapted from Fig. 7.5, Callister 7e.7.5, Callister 7e.


Impact of Cold WorkAs cold work is increased• Yield strength (σy) increases.• Tensile strength (TS) increases.

As cold work is increased

g ( )• Ductility (%EL or %AR) decreases.



• What is the tensile strength &Cold Work Analysis

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

22 π−π do rrCold Work

Copper

%6.35100x % 2 =π

ππ=

o

dor

rrCW

yield strength (MPa) tensile strength (MPa) ductility (%EL)

Do =15.2mm Dd =12.2mm

500

700

y g ( )

600

800ductility (%EL)

40

60

300

500

Cu300MPa

C400

600

20

40

Cu340MPa

% Cold Work

100200 40 60

300MP% Cold Work

200Cu

0 20 40 60% Cold Work20 40 6000

7%

Adapted from Fig. 7.19, Callister 7e. (Fig. 7.19 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

σy = 300MPa TS = 340MPa %EL = 7%

28

y(Managing Ed.), American Society for Metals, 1979, p. 276 and 327.)


σ- ε Behavior vs. Temperature• Results for

polycrystalline iron:-200°C

100°C

800

600

MP

a)

-100°C

25°C

400

200

Stre

ss (M

Adapted from Fig. 6.14, Callister 7e. 0

Strain

S

0 0.1 0.2 0.3 0.4 0.5

• σy and TS decrease with increasing test temperature.• %EL increases with increasing test temperature.• Why? Vacancies 3. disl. glides past obstacle Why? Vacancies

help dislocationsmove past obstacles.

2. vacancies replace atoms on the disl half obstacledisl. half plane 1. disl. trapped

by obstacle


1 ho r treatment at TEffect of Heating After %CW

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

• Effects of cold work are reversed! Effects of cold work are reversed!a) 600 60

annealing temperature (ºC)200100 300 400 500 600 700

• 3 Annealinggth

(MP

a

%E

L)tensile strength600

500

60

50

• 3 Annealingstages todiscuss...e

stre

ng

uctil

ity (

400

40

30

Adapted from Fig. 7.22, Callister 7e. (Fig.7.22 is adapted from G. Sachs and K.R. van Horn, Practical Metallurgy, Applied Metallurgy, and the Industrial Processing of Ferrous and

tens

ile d

ductility300

20

and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.)


RecoveryAnnihilation reduces dislocation density.• Scenario 1

Dislocationsextra half-plane

of atomsResults from diffusion

Dislocations annihilate and form a perfect

of atoms

atoms diffuse t i a perfect

atomic plane.extra half-plane

of atoms

to regions of tension

• Scenario 2of atoms

τR3. “Climbed” disl. can now

4. opposite dislocations meet and annihilate

2. grey atoms leave by vacancy diffusion

ll i di l t “ li b”

Rmove on new slip plane

meet and annihilateallowing disl. to “climb”1. dislocation blocked; can’t move to the right

Obstacle dislocation


N i f d th tRecrystallization

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

0.6 mm 0.6 mm

Adapted from Fig 7 21 (a) (b)Fig. 7.21 (a),(b), Callister 7e.(Fig. 7.21 (a),(b) are courtesy of J.E. Burke, G l El t iGeneral Electric Company.)

33% cold New crystals33% coldworkedbrass

New crystalsnucleate after3 sec. at 580°C.


Further Recrystallization

• All cold-worked grains are consumed.

0.6 mm0.6 mm

Adapted from Fig. 7.21 (c),(d), Callister 7e.(Fig. 7.21 (c),(d) are courtesy of J.E. Burke, General ElectricGeneral Electric Company.)

After 4 After 8After 4seconds

After 8seconds


At l ti l i ll

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

is reducedis reduced.0.6 mm 0.6 mm

Adapted from Fig 7 21 (d) (e)Fig. 7.21 (d),(e), Callister 7e.(Fig. 7.21 (d),(e) are courtesy of J.E. Burke,

After 8 s,580ºC

After 15 min,580ºC

General Electric Company.)

580 C 580 C

• Empirical Relation: coefficient dependenton material and T.exponent typ ~ 2

Ktdd no

n =−elapsed timegrain diam.

at time t.

exponent typ. ~ 2

Ostwald Ripening

34

Ostwald Ripening


º

T = recrystallizationTR = recrystallization temperature

TR


35ºCopyright © YM Youssef, 4-Oct-10 Materials Science

Recrystallization Temperature, TR

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

shorter annealing time => higher TR

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

movements• Easier to move in pure metals => lower TR


Coldwork Calculations

A cylindrical rod of brass originally 0.40 in (10.2 mm)in diameter is to be cold worked by drawing. Thein diameter is to be cold worked by drawing. Thecircular cross section will be maintained duringdeformation. A cold-worked tensile strength in excessof 55,000 psi (380 MPa) and a ductility of at least 15%EL are desired. Further more, the final diametermust be 0 30 in (7 6 mm) Explain how this may bemust be 0.30 in (7.6 mm). Explain how this may beaccomplished.


Coldwork Calculations Solution

If we directly draw to the final diameter what happens?what happens?

BrassColdCold Work

AAA ⎟⎞

⎜⎛

⎟⎞

⎜⎛ −

Do = 0.40 in Df = 0.30 in

3004

1001100x %

22D

xAA

AAACW

o

f

o

fo

⎞⎛ ⎞⎛⎞⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

%843100 x 4003001100 x

441

2

2

2

...

DD

o

f =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ππ

−=


Coldwork Calc Solution: Cont.

540420 540420

6

• For %CW = 43.8% Adapted from Fig. 7.19, Callister 7e.– σy = 420 MPa

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

• This doesn’t satisfy criteria what can we do?39

• This doesn t satisfy criteria…… what can we do?Copyright © YM Youssef, 4-Oct-10 Materials Science

Coldwork Calc Solution: Cont.

380 15

12 27


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

< 27 %CWFor %EL < 15 < 27 %CW

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

40

g gCopyright © YM Youssef, 4-Oct-10 Materials Science

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:%11001%

22

22 CWDxDCW ff =⇒⎟⎟

⎞⎜⎜⎛

=100

1 1001% 202

202 D

xD

CW =−⇒⎟⎟⎠

⎜⎜⎝

−=

50% .CWD ⎞⎛ 2fDD02

2

100%1f CW

DD

⎟⎠⎞

⎜⎝⎛ −= 50

202

100%1

.f

CWDD

⎟⎠⎞

⎜⎝⎛ −

=⇒

m 3350100201300

50

021 ..DD.

f =⎟⎠⎞

⎜⎝⎛ −==Intermediate diameter =

41

⎠⎝Copyright © YM Youssef, 4-Oct-10 Materials Science

Coldwork Calculations Solution

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

%CW1 = 1−0.335⎛

⎜ ⎞ ⎟ 2⎛

⎜ ⎜

⎞ ⎟ ⎟ x 100 = 30

2 Anneal above D = D

%CW1 = 1−0.4⎝

⎜ ⎠ ⎟

⎝ ⎜

⎠ ⎟ x 100 = 30

2. Anneal above D02 = Df1

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

MP34020100

3350301%

2

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−= x

..CW

MPa 400MPa340

=

=σ

TSy

⇒Fig 7.19

Therefore, meets all requirements24% =EL


Rate of Recrystallization

ERtR logloglog 0 −=−= start50%

BCt

kTRtR

log

logloglog 0

+= T1

start

finish

t/RT

Ct

1:note

log

=

+RT

• Hot work above TRlog t

Hot work above TR

• Cold work below TR

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

43

– weaker at high temperatureCopyright © YM Youssef, 4-Oct-10 Materials Science

Summary

• Dislocations are observed primarily in metals Dislocations are observed primarily in metalsand alloys.

• Strength is increased by making dislocationmotion difficult.

• Particular ways to increase strength are to:d i i--decrease grain size

--solid solution strengthening--precipitate strengtheningp p g g--cold work

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


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