ISSUES TO ADDRESS... Transforming one phase into another takes time. How does the rate of transformation depend on time and T ? How can we slow down the

ISSUES TO ADDRESS...• Transforming one phase into another takes time.

• How does the rate of transformation depend on time and T ?• How can we slow down the transformation so that we can engineer non-equilibrium structures?• Are there other means to improve mechanical behavior?

Feg

(Austenite)

Eutectoid transformation

C FCC

Fe3C(cementite)

a (ferrite)

+

(BCC)

Chapter 10: Phase Transformations – Considering Kinetic and Heat Treatment

Nucleation – nuclei (like biological seeds) act as

template to grow crystals– for nuclei to form, the rate of addition of

atoms to any nucleus must be faster than rate of loss

– once nucleated, the “seed” must grow until they reach the predicted equilibrium

Phase Transformations

Phase Transformations

Driving force to nucleate increases as we increase T– supercooling (eutectic, eutectoid)– superheating (peritectic)

• With a Small amount of supercooling few nuclei - large crystals

• With a Large amount of supercooling rapid nucleation - many nuclei, small crystals

Solidification: Nucleation Processes

• Homogeneous nucleation – nuclei form in the bulk of liquid metal (as

“native chemistry”)– requires sufficient supercooling (typically 80-

300°C max)

• Heterogeneous nucleation– much easier since stable “nucleus” is already

present (they are non-native chemically)• Could be wall of mold or impurities in the liquid

phase– allows solidification with only minimal

supercooling (0.1-10ºC)

r* = critical nucleus: nuclei < r* shrink; nuclei>r* grow (to reduce energy)

Homogeneous Nucleation & Energy Effects

DGT = Total Free Energy = DGS + DGV

Surface Free Energy- destabilizes the nuclei (it takes energy to make an interface)

24 rGS

g = surface tension

Volume (Bulk) Free Energy – stabilizes the nuclei (releases energy)

GrGV3

3

4

volume unit

energy free volume G

Solidification

r* decreases as T increases

For typical T r* is around 100Å (10 nm)

TH

Tr

S

m

2

*

Note: HS is a strong function of T is a weak function of T

HS = latent heat of solidification

Tm = melting temperature

g = surface free energy

DT = Tm - T = supercooling

r* = critical radius

Rate of Phase Transformations

Kinetics - measures approach to equilibrium vs. time

• Hold temperature constant & measure conversion vs. time– How is the amount of conversion measured?

• X-ray diffraction – have to do many samples• electrical conductivity – follows a single sample• sound waves (insitu ultrasonic) – follows a single

sample

Thus, the rate of nucleation is a product of two curves that represent two opposing factors (instability and diffusivity).

Rate of Phase Transformation

– Modeled by the Avrami Rate Equation:

t0.5

All out of material – “done!”

Fixed Temp.

maximum rate reached – now amount unconverted decreases so rate slows

rate increases as surface area increases & nuclei grow

Log Time

1nkty e

Avrami EquationAvrami rate equation → y = 1- exp (-ktn)

k & n are fit for any specific sample

• By convention we define: r = 1 / t0.5 as “the rate of transformation” – it is simply the inverse of the time to complete half of the transformation

• The initial slow rate can be attributed to the time required for a significant number of nuclei of the new phase to form and begin growing.

• During the intermediate period the transformation is rapid as the nuclei grow into particles and consume the old phase while nuclei continue to form in the remaining parent phase.

• Once the transformation begins to near completion there is little untransformed material for nuclei to form in and the production of new particles begins to slow. Further, the particles already existing begin to touch one another, forming a boundary where growth stops.

http://en.wikipedia.org/w/index.php?title=Parent_phase&action=edit&redlink=1

Rate of Phase Transformations

• In general, rate increases as T

r = 1/t0.5 = A e -Q/RT

– R = gas constant– T = temperature (K) (higher causes higher rate too)– A = ‘preexponential’ rate factor– Q = activation energy

• r is often small so equilibrium is not possible!

Arrhenius expression

adapted from B.F. Decker and D. Harker, "Recrystallization in Rolled Copper", Trans AIME, 188, 1950, p. 888.

135C 119C 113C 102C 88C 43C

1 10 102 104

Transformations & Undercooling

• M. Eng. Can make it occur at: ...727ºC (cool it slowly)...below 727ºC (“supercool” or “Undercool” it!)

• Eutectoid transf. (Fe-C System): g Þ a + Fe3C0.76 wt% C

0.022 wt% C6.7 wt% C

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g +Fe3C

a +Fe3C

L+Fe3C

d

(Fe) Co , wt%C

1148°C

T(°C)

aferrite

727°C

Eutectoid:Equil. Cooling: Ttransf. = 727ºC

DT

Undercooling by DTtransf. < 727C

0.7

6

0.0

22

adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990

Eutectoid Transformation Rate

Coarse pearlite formed at higher T - softer

Fine pearlite formed at low T - harder

Diffusive flow of C needed

a

ag g

a

• Growth of pearlite from austenite:

Adapted from Fig. 9.15, Callister 7e.

gaaaa

a

a

pearlite growth direction

Austenite (g)grain boundary

cementite (Fe3C)

Ferrite (a)

g

• Recrystallization rate increases with DT.

52°C (675˚C)

0

50

y (%

pea

rlite

)127°C

(600 ˚C)77°C

100

• Reaction rate is a result of nucleation and growth of crystals.

• Examples from previous slide:

Nucleation and Growth

T: just below TE

Nucleation rate low

Growth rate high

g

pearlite colony

T: moderately below TE

g

Nucleation rate med

Growth rate med.

Nucleation rate high

T: way below TE

g

Growth rate low

The ideas of “reality” and the “ideal” meet in the Material Engineering’s Transformation

Curves

adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 369.

Isothermal Transformation (TTT) Diagrams

• Fe-C system, Co = 0.76 wt% C• Transformation at T = 675°C.

100

50

01 102 104

T = 675°C

y,

% tr

ansf

orm

ed

time (s)

400

500

600

700

1 10 102 103 104 105

0%pearlite

100%

50%

Austenite (stable) TE (727C)Austenite (unstable)

Pearlite

T(°C)

time (s)

isothermal transformation at 675°C

• Eutectoid composition, Co = 0.76 wt% C• Begin at T > 727°C• Rapidly cool to 625°C and hold isothermally.


Effect of Cooling History in Fe-C System

400

500

600

700

0%pearlite

100%

50%

Austenite (stable)TE (727C)

Austenite (unstable)

Pearlite

T(°C)

1 10 102 103 104 105

time (s)

g g

g

g g

g

Transformations with Proeutectoid Materials

Hypereutectoid composition – proeutectoid cementite

a

CO = 1.13 wt% C

TE (727°C)

T(°C)

time (s)

A

A

A+

C

P

1 10 102 103 104

500

700

900

600

800

A+

P



Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g(austenite)

g+L

g +Fe3C

a +Fe3C

L+Fe3C

d

(Fe) Co , wt%C

T(°C)

727°CDT

0.7

6

0.0

22

1.13

Non-Equilibrium Transformation Products in Fe-C

• Bainite: --a lathes (strips) with long

rods of Fe3C --diffusion controlled.• Isothermal Transf. Diagram


from Metals Handbook, 8th ed., Vol. 8, Metallography, Structures, and Phase Diagrams, American Society for Metals, Materials Park, OH, 1973.)

Fe3C

(cementite)

5 mm

a (ferrite)

10 103 105

time (s)10-1

400

600

800

T(°C)Austenite (stable)

200

P

B

TE

0%

100%

50%

pearlite/bainite boundary

A

A

100% bainite

100% pearlite

From: George Krauss, Steels: Processing, Structure, and Performance, ASM International, 2006.

TTT Curves showing the Bainite Transformation (a) Plain Carbon Steels; (b) Alloy Steel w/ distinct

Bainite “Nose”

• Martensite: --g(FCC) to Martensite (BCT)

courtesy United States Steel Corporation.

• Isothermal Transf. Diagram

• g to M transformation.. -- is rapid! -- % transf. depends on T only.

Martensite: Fe-C System

Martensite needlesAustenite

60 m

10 103 105 time (s)10-1

400

600

800

T(°C)Austenite (stable)

200

P

B

TE

0%

100%50%

A

A

M + AM + A

M + A

0%50%90%

xx x

xx

xpotential C atom sites

Fe atom sites

(involves single atom jumps)

Transformation to Martensite

Martensite formation requires that the steel be subject to a minimum – Critical – Cooling Rate (this value is ‘TTT’ or ‘CCT’ chart dependent for alloy of interest)

For most alloys it indicates a quench into a RT oil or water bath

(FCC) (BCC) + Fe3C

Martensite Formation

slow cooling

tempering

quench

M (BCT)

M = martensite is body centered tetragonal (BCT)

Diffusionless transformation BCT if C > 0.15 wt%

BCT few slip planes hard, brittle

Martensite Transformation Crystallography: FCC Austenite to BCT Martensite


Austenite to Martensite: Size Issues and Material Response


• Spheroidite: --a grains with spherical Fe3C --diffusion dependent. --heat bainite or pearlite for long times (below the AC1

critical temperature) --driven by a reduction in interfacial area of Carbide

Spheroidite: Fe-C System

(Fig. 10.19 copyright United States Steel Corporation, 1971.)

60 m

a(ferrite)

(cementite)

Fe3C

Phase Transformations of Alloys

Effect of adding other elementsChange transition temp.

Cr, Ni, Mo, Si, Mn

retard + Fe3C

transformation delaying the time to entering the diffusion controlled transformation reactions – thus promoting “Hardenability’ or Martensite development

Continuous Cooling Transformations (CCT)

• Isothermal Transformations are “Costly” requiring careful “gymnastics” with heated (and cooling) products

• CC Transformations change the observed behavior concerning transformation – With Plain Carbon Steels when cooled

“continuously” we find that the Bainite Transformation is suppress(see Figure 10.26)

Cooling CurvePlot:

temp vs. time

CCT for Eutectoid Steel

Figure: 10-26

CCT for Eutectoid Steel

Alloy Steel CCT Curve – again note distinct Bainite Nose


Dynamic Phase Transformations

On the isothermal transformation diagram for 0.45 wt% C Fe-C alloy, sketch and label the time-temperature paths to produce the following microstructures:a) 42% proeutectoid ferrite and 58% coarse

pearlite

b) 50% fine pearlite and 50% bainite

c) 100% martensite

d) 50% martensite and 50% austenite

A + B

A + P

A + aA

BP

A50%

0

200

400

600

800

0.1 10 103 105

time (s)

M (start)M (50%)M (90%)

Example Problem for Co = 0.45 wt%

a) 42% proeutectoid ferrite and 58% coarse pearlite

first make ferrite

then pearlite

coarse pearlite

higher T

Adapted from Fig. 10.29,Callister 5e.

T (°C)

b) 50% fine pearlite and 50% bainite

first make pearlite

then bainite

fine pearlite

lower T

T (°C)

A + B

A + P

A + aA

BP

A50%

0

200

400

600

800

0.1 10 103 105

time (s)

M (start)M (50%)M (90%)



NOTE: This “2nd step” is sometimes referred to as an “Austempering” step, quenching into a heated salt bath held at the temperature of need


c) 100 % martensite: quench @ 380C/s {(850-600)/.7s}


d)

A + B

A + P

A + aA

BP

A50%

0

200

400

600

800

0.1 10 103 105

time (s)

M (start)M (50%)M (90%)

c)

T (°C)

d) 50%martensite/50%(retained) austenite

CT for CrMo Med-carbon steel

Hardness of cooled samples at various cooling rates in bubbles -- Dashed lines are IT solid lines are CT regions

Tempering Martensite• reduces brittleness of martensite,• reduces internal stress caused by quenching.

copyright by United States Steel Corporation, 1971.

• decreases TS, YS but increases %RA• produces extremely small Fe3C particles surrounded by a.

from Fig. furnished courtesy of Republic Steel Corporation.) 9

mm

YS(MPa)TS(MPa)

800

1000

1200

1400

1600

1800

30

40

50

60

200 400 600Tempering T (°C)

%RA

TS

YS

%RA

The microstructure of tempered martensite, although an equilibrium mixture of α-Fe and Fe3C, differs from those for pearlite and bainite. This micrograph produced in a scanning electron microscope (SEM) shows carbide clusters in relief above an etched ferrite. (From ASM Handbook, Vol. 9: Metallography and Microstructures, ASM International, Materials Park, OH, 2004.)

Temper Martensite Embrittlement – an issue in Certain Steels

Suspected to be due to the deposition of very fine carbides during 2nd and 3rd phase tempering along original austenite G. B. from the transformation of retained austenite,


Increasing Strength and Hardness of Alloy involves some Heat Treatment

• The effect of quenching in steels is determined by the Jominey End Quench Test

• Precipitation Hardness is a method used for may alloy systems (mostly Non-Ferrous ones)

• Grain Size control is also an important consideration – which can be controlled by annealing processes

• Recovery after cold work (cold work can also increase strength of alloys)

• Recrystalization• Grain Growth

Schematic illustration of the Jominy end-quench test for hardenability. (After W. T. Lankford et al., Eds., The Making, Shaping, and Treating of Steel, 10th ed., United States Steel, Pittsburgh, PA, 1985. Copyright 1985 by United States Steel Corporation.)

Hardenability--Steels• Ability to form martensite• Jominy end quench test to measure hardenability.

• Hardness versus distance from the quenched end.

(adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill Book Company, New York, 1978.)

24°C water

specimen (heated to g phase field)

flat ground

Rockwell Chardness tests

Har

dnes

s, H

RC

Distance from quenched end

Figure 10.22 The cooling rate for the Jominy bar (see Figure 10.21) varies along its length. This curve applies to virtually all carbon and low-alloy steels. (After L. H. Van Vlack, Elements of Materials Science and Engineering, 4th ed., Addison-Wesley Publishing Co., Inc., Reading, MA, 1980.)

Figure 10.23 Variation in hardness along a typical Jominy bar. (From W. T. Lankford et al., Eds., The Making, Shaping, and Treating of Steel, 10th ed., United States Steel, Pittsburgh, PA, 1985. Copyright 1985 by United States Steel Corporation.)

• The cooling rate varies with position.

(adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 376.)

Why Hardness Changes W/Position

distance from quenched end (in)Ha

rdn

ess

, H

RC

20

40

60

0 1 2 3

600

400

200A ® M

A ®

P

0.1 1 10 100 1000

T(°C)

M(start)

Time (s)

0

0%100%

M(finish) Martensite

Martensite + Pearlite

Fine Pearlite

Pearlite

Hardenability vs Alloy Composition• Jominy end quench results, C = 0.4 wt% C

• "Alloy Steels" (4140, 4340, 5140, 8640) --contain Ni, Cr, Mo (0.2 to 2wt%) --these elements shift the "nose". --martensite is easier to form.

(adapted from figure furnished courtesy Republic Steel Corporation.)

Cooling rate (°C/s)

Har

dne

ss, H

RC

20

40

60

100 20 30 40 50Distance from quenched end (mm)

210100 3

4140

8640

5140

1040

50

80

100

%M4340

T(°C)

10-1 10 103 1050

200

400

600

800

Time (s)

M(start)M(90%)

shift from A to B due to alloying

BA

TE

Figure 10.24 Hardenability curves for various steels with the same carbon content (0.40 wt %) and various alloy contents. The codes designating the alloy compositions are defined in Table 11.1. (From W. T. Lankford et al., Eds., The Making, Shaping, and Treating of Steel, 10th ed., United States Steel, Pittsburgh, PA, 1985. Copyright 1985 by United States Steel Corporation.)

• Effect of quenching medium:

Mediumairoil

water

Severity of Quenchlow

moderatehigh

Hardnesslow

moderatehigh

• Effect of geometry: When surface-to-volume ratio increases: --cooling rate increases --hardness increases

Positioncentersurface

Cooling ratelowhigh

Hardnesslowhigh

Quenching Medium & Geometry

0 10 20 30 40 50wt% Cu

L+La

+a q q

+L

300

400

500

600

700

(Al)

T(°C)

composition range needed for precipitation hardening

CuAl2

A

Adapted from Fig. 11.24, Callister 7e. (Fig. 11.24 adapted from J.L. Murray, International Metals Review 30, p.5, 1985.)

Precipitation Hardening• Particles impede dislocations.• Ex: Al-Cu system• Procedure:

--Pt B: quench to room temp.--Pt C: reheat to nucleate small q crystals within a crystals.

• Other precipitation systems: • Cu-Be • Cu-Sn • Mg-Al

Temp.

Time

--Pt A: solution heat treat (get a solid solution)

Pt A (sol’n heat treat)

B

Pt B

C

Pt C (precipitate ) Consider: 17-4 PH St. Steel and Ni-Superalloys too!

Figure 10.25 Coarse precipitates form at grain boundaries in an Al–Cu (4.5 wt %) alloy when slowly cooled from the single-phase (κ) region of the phase diagram to the two-phase (θ + κ) region. These isolated precipitates do little to affect alloy hardness.

Figure 10.26 By quenching and then reheating an Al–Cu (4.5 wt %) alloy, a fine dispersion of precipitates forms within the κ grains. These precipitates are effective in hindering dislocation motion and, consequently, increasing alloy hardness (and strength). This process is known as precipitation hardening, or age hardening.

Figure 10.27 (a) By extending the reheat step, precipitates coalesce and become less effective in hardening the alloy. The result is referred to as overaging. (b) The variation in hardness with the length of the reheat step (aging time).

Figure 10.28 (a) Schematic illustration of the crystalline geometry of a Guinier–Preston (G.P.) zone. This structure is most effective for precipitation hardening and is the structure developed at the hardness maximum shown in Figure 10.27b. Note the coherent interfaces lengthwise along the precipitate. The precipitate is approximately 15 nm × 150 nm.

(From H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. 3: Mechanical Behavior, John Wiley & Sons, Inc., NY, 1965.) (b) Transmission electron micrograph of G.P. zones at 720,000×. (From ASM Handbook, Vol. 9: Metallography and Microstructures, ASM International, Materials Park, OH, 2004.)

Figure 10.30 Annealing can involve the complete recrystallization and subsequent grain growth of a cold-worked microstructure. (a) A cold-worked brass (deformed through rollers such that the cross-sectional area of the part was reduced by one-third). (b) After 3 s at 580°C, new grains appear. (c) After 4 s at 580°C, many more new grains are present.

(d) After 8 s at 580°C, complete recrystallization has occurred. (e) After 1 h at 580°C, substantial grain growth has occurred. The driving force for this growth is the reduction of high-energy grain boundaries. The predominant reduction in hardness for this overall process had occurred by step (d)

All micrographs have a magnification of 75×. (Courtesy of J. E. Burke, General Electric Company, Schenectady, NY.)

Figure 10.31 The sharp drop in hardness identifies the recrystallization temperature as ~290°C for the alloy C26000, “cartridge brass.” (From Metals Handbook, 9th ed., Vol. 4, American Society for Metals, Metals Park, OH, 1981.)

Recrystallization Temperature, TR

TR = recrystallization temperature = point of highest rate of property change1. TR 0.3-0.6 Tm (K)

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

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

4. Pure metals lower TR due to easier dislocation movements

Figure 10.32 Recrystallization temperature versus melting points for various metals. This plot is a graphic demonstration of the rule of thumb that atomic mobility is sufficient to affect mechanical properties above approximately 1/3 to 1/2 Tm on an absolute temperature scale.

(From L. H. Van Vlack, Elements of Materials Science and Engineering, 3rd ed., Addison-Wesley Publishing Co., Inc., Reading, MA, 1975.)

Figure 10.33 For this cold-worked brass alloy, the recrystallization temperature drops slightly with increasing degrees of cold work.

(From L. H. Van Vlack, Elements of Materials Science and Engineering, 4th ed., Addison-Wesley Publishing Co., Inc., Reading, MA, 1980.)

• 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 mm

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

Grain Growth

• Empirical Relation:

Ktdd no

n

coefficient dependent on material & Temp.

grain dia. At time t.

elapsed timeExponent is typ. 2

This is: Ostwald Ripening

Figure 10.34 Schematic illustration of the effect of annealing temperature on the strength and ductility of a brass alloy shows that most of the softening of the alloy occurs during the recrystallization stage.

(After G. Sachs and K. R. Van Horn, Practical Metallurgy: Applied Physical Metallurgy and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, Cleveland, OH, 1940.)

Figure 10.29 Examples of cold-working operations: (a) cold-rolling a bar or sheet and (b) cold-drawing a wire. Note in these schematic illustrations that the reduction in area caused by the cold-working operation is associated with a preferred orientation of the grain structure.

We can then find that the “cold working of an alloy” is an effect tool for improving performance –

• if done properly!– so as not to cause the material to exceed its

% EL which was a fracture deformation limit as we saw earlier

– If we impose appropriate intermediate recrystalization (and maybe even grain growth steps)

– Finish with a cold working step to achieve the desired hardness and finished size

Coldwork Hardening Example

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 Solution

If we directly draw to the final diameter what happens?

%843100 x 400

3001100 x

4

41

100 1100 x %

2

2

2

..

.

D

D

xA

A

A

AACW

o

f

o

f

o

fo

Do = 0.40 in

BrassCold Work

Df = 0.30 in

Coldwork Calc Solution: Cont.

• For %CW = 43.8%

540420

– y = 420 MPa– TS = 540 MPa > 380 MPa

6

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

Coldwork Calc Solution: Cont.


380

12

15

27

For %EL > 15

For TS > 380 MPa > 12 %CW

< 27 %CW

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

This process Needs an Intermediate Recrystallization

i.e.: 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:2 2

2 2

2 22 2

%% 1 100 1

100f f

s s

D D CWCW x

D D

0.52

2

%1

100f

s

D CW

D

22 0.5

%1

100

fs

DD

CW

0.5

1 2

200.30 1 0.335 in

100f sD D

Intermediate diameter =

Summary:

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

2. Anneal above TR Ds2 = Df1

3. Secondary Cold work Ds2= 0.335 in Df 2 =0.30 in

Therefore, we have met all requirements

Coldwork Calculations Solution

20100 3350

301%

2

2

x

.

.CW

24%

MPa 400

MPa 340

EL

TSy

%CW1 10.335

0.4

2

x 100 30

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ISSUES TO ADDRESS... Transforming one phase into another takes time. How does the rate of transformation depend on time and T ? How can we slow down the