ENGR 376 - Lecture 4 - Phase Transformations

Phase Transformation Nucleation

Growth

APPLICATIONSAPPLICATIONS

Transformations in SteelPrecipitationSolidification & crystallizationGl i iGlass transition Recovery, Recrystallization & Grain growth

Dr. Ray Taheri Last revision: Sept. 01, 2009

4-Nov-09

PHASE TRANSFORMATIONSPHASE TRANSFORMATIONSBased on

Masst t

Diffusional Martensitictransport

PHASE TRANSFORMATIONSPHASE TRANSFORMATIONS

Based onorder

1nd ordernucleation & growth

2nd orderEntire volume transforms

order


4-Nov-09

Bulk Gibbs free energy ↓

Energies involved

Bulk Gibbs free energy ↓

Interfacial energy ↑

Strain energy ↑ Solid-solid transformation

New interface created

Volume of transforming materialf f g


4-Nov-09

1nd ordernucleation & growth

Nucleationof

β phaseTransformation

β+

Growthtill

α is =β phaseα → β exhausted


4-Nov-09

Liquid → Solid phase transformation

On cooling just below T solid becomes stableOn cooling just below Tm solid becomes stableBut solidification does not starte.g. liquid Ni can be undercooled 250 K below Tm

Liquid stableSolid stable

↑ T

→

ΔG

Solid (GS)

G →

Liquid (GL)ΔTΔG → +ve


4-Nov-09

Tm T →ΔT - Undercooling “For sufficientUndercooling”

ΔG ve

NucleationSolidification + Growth=

Nucleation Homogenous

Heterogenous

NucleationLiquid → solidwalls of container, inclusions

gSolid → solidinclusions, grain boundaries, dislocations, stacking faults

In Homogenous solidification the probability of nucleation occurring at points in the parent phase is same throughout the parent phase

, g f


4-Nov-09

In heterogeneous solidification nucleation there are some preferred sites in the parent phase where nucleation can occur

Homogenous nucleation Negligible in L → Stransformations

nucleationonchangeenergyFree

)((Surface))(Volume) (ΔG γ+Δ= Genergystrain in increase energy surfacein increase energy freebulk in Reduction

nucleationon changeenergy Free++

=

X)((Surface).)(Volume).( ΔG γ+Δ= vG

( ) )(4)(4ΔG 23 γππ rGr +Δ⎟⎞

⎜⎛ ( ) ).(4).(

3 ΔG γππ rGr v +Δ⎟

⎠⎜⎝

=

r3 )( TfGv Δ=Δr2

r


4-Nov-09

ro

4.5

3.5

4

2.5

3

ergy

1.5

2En

0.5

1

Volume energy

00 0.2 0.4 0.6 0.8 1 1.2

Radius

Surface energy


4-Nov-09

0.4

0

0.2

0 0 2 0 4 0 6 0 8 1 1 2y

-0.2

0 0.2 0.4 0.6 0.8 1 1.2

yste

m E

nerg

y

Critical Radius=RC

-0.6

-0.4

Tota

l Sy

-0.8

-1Radius


4-Nov-09


4-Nov-09

( ) ).(4 ).(34 ΔG 23 γππ rGr v +Δ⎟

⎠⎞

⎜⎝⎛=

3 ⎠⎝By setting dΔG/dr = 0 the critical values (corresponding to the maximum) are obtained (denoted by superscript *)Reduction in free energy is obtained only after r0 is obtained

0=ΔGd r −=

γ2*20* =r

As ΔGv is _ , r*is +0=

dr vGΔ201 =r

Trivialγ2*

0=Δdr

Gd

vGr

Δ−=

γ2*

3* 16 γ

→dr

0=ΔG

2*

316

vGG

Δ=Δ

γπ

ΔG

→

*r0r

3Supercritical nucleiEmbryos


4-Nov-09r →0=ΔG

vGr

Δ−=

γ30

Liquid stableSolid stable

S lid (G→ΔG

Solid (GS)

G →

Liquid (GL)ΔTΔG → +ve

TmΔT - Undercooling


4-Nov-09

)( TfGv Δ=Δ The bulk free energy reduction is a function of undercooling

G*

Tmas

ing

ΔG

Dec

rea

G →

Decreasing r*

r →

ΔG


4-Nov-09

T 2<T1


4-Nov-09

No. of critical sized particlesRate of nucleation x Frequency with which they

become supercritical=

dtdNI = ⎟

⎟⎠

⎞⎜⎜⎝

⎛ Δ−

= kTG

t eNN*

*⎟⎠⎞

⎜⎝⎛ Δ

−= kT

Hd

es' *ννdt t eNN es ννNo. of particles/volume in L

s* atoms of the liquid facing the nucleus

Critical nucleus

Jump taking particle to supercriticality→ nucleated (enthalpy of activation = ΔHd)


4-Nov-09

Critical sized nucleus

⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ+Δ−

= kTHG d

esNI

*

* νΔG* ↑ ⇒ I ↓

T ↑ ⇒ I ↑⎠⎝= t esNI ν

T

T ↑ ⇒ I ↑

T

Tm T = Tm → ΔG* = ∞ → I = 0

3

easi

ng Δ

T

2

3*

316

vGG

Δ=Δ

γπ

→In

cre

T (K

) →

0 T = 0 → I = 0


4-Nov-09

I →0 T = 0 → I = 0


4-Nov-09

Heterogenous nucleation

Consider the nucleation of β from α on a planar surface of inclusion δβ p

γαβα

A γC t d

Interfacial Energies

γ δ

β

γβδ

θAlens γαβ

A i l γβδ

Created

Createdγαδ

δ

γβδ Acircle γβδ

A

Created

γγδ Acircle γαδ

δβδβ γγθγ =+CosSurface tension force balance

αβ

βδαδ

γγγ

θ−

= Cos Lost

αδβδαβ γγγ )( )()(A )(V ΔG lenslens circlecirclev AAG −++Δ=

αδβδαβ γγθγ +CosSurface tension force balance


4-Nov-09

Vlens = πh2(3r-h)/3 Alens = 2πrh h = (1-Cosθ)r rcircle = r Sinθ

hetero Gr

Δ−= αβγ2* ( )θθ

γπ αβ 3

2

3* 32

34 CosCos

GGhetero +−

Δ=Δ0=

Δdr

Gd

vGΔ( )23 GvΔdr

( )θθ 3homo

* 3241 CosCosGG *

hetero +−Δ=Δ3* 16G =Δ

γπ

1

( )4

o→

ΔG*h t (0o) = 0

23 vGG

Δ=Δ π

0.75

/ ΔG

* hom

o ΔG hetero (0 ) 0no barrier to nucleation

ΔG*hetero (90o) = ΔG*

homo/2

ΔG*hetero (180o) = ΔG*

homo

0.5

ΔG

* hete

ro

hetero ( ) homo

0.25

Δ

Complete wetting No wettingPartial wetting


4-Nov-09

00 30 60 90 120 150 180

θ (degrees) →

Choice of heterogeneous nucleating agent

Small value of θ

Choosing a nucleating agent with a low value of γβδ (low energy βδ interface)Choosing a nucleating agent with a low value of γβδ (low energy βδ interface)

(Actually the value of (γαδ − γβδ) will determine the effectiveness of the heterogeneous nucleating agent → high γαδ or low γβδ)β

low value of γβδ → Crystal structure of β and δ are similar and lattice parameters are as close as

possiblepossible

Ni (FCC, a = 3.52 Å) is used a heterogeneous nucleating agent in the production of artificial diamonds (FCC, a = 3.57 Å) from graphitep ( , ) g p


4-Nov-09

Nucleationof

Trasformation+

Growthtill

α is=β phaseα → β α is

exhaustedGrowthAt transformation temperature the probability of jump of atom from α → β

(across the interface) is same as the reverse jump

G h d b l h f i h i h i iGrowth proceeds below the transformation temperature, wherein the activationbarrier for the reverse jump is higher

ΔHd – v t ΔGΔHd

ΔHd vatom ΔGv

α phase


4-Nov-09β phase


4-Nov-09

PHASE TRANSFORMATIONS

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

Fe

γ (Austenite)

Eutectoid transformation

C FCC

Fe3C (cementite)

α (ferrite)

+(BCC)

• How does the rate of transformation depend ontime and T?

FCC (ferrite) (BCC)

time and T?• How can we slow down the transformation so that

we can engineering non-equilibrium structures?• Are the mechanical properties of non-equilibrium


4-Nov-09 1

• Are the mechanical properties of non-equilibriumstructures better?

COOLING AUSTENITEAustenite

Pearlite

• Mainly interested in eutectoid cooling: γ α + Fe3C (pearlite), 0.78 wt% C• Cooling rate can result in a wide variety of phases and microstructures

Equilibrium phases: pearlite bainite


4-Nov-09

– Equilibrium phases: pearlite, bainite– Non-equilibrium phases: martensite

FRACTION OF TRANSFORMATION• Fraction transformed depends on time,

at constant temperature (e.g., γ pearlite)

nktey −−= 1

Transformation rate r = 1/t

Avrami equation(k, n are constants)


4-Nov-09

• Transformation rate , r = 1/t0.5

1

1.2

0.8

te

0.4

0.6

Y=R

at

0.2

00 10 20 30 40 50 60 70 80

Time (S)


4-Nov-09


4-Nov-09

EUTECTOID TRANSFORMATION RATE ~ ΔT

• Growth of pearlite from austenite:

• Reaction rate increases with ΔT.


4-Nov-09

EUTECTOID TRANSFORMATION RATE ~ ΔT

αAustenite (γ) grain

cementite (Fe3C)

ferrite (α)

Diffusive flow of C needed

α

• Growth of pearlite from austenite:ΔT

γαααα

αpearlite growth

gboundary

ferrite (α)

γ

α

αγ γ

αdirection α


4-Nov-09

TIME-TEMPERATURE TRANSFORMATION (TTT) DIAGRAMS(TTT) DIAGRAMS

• Fe-C system, Eutectoid composition (Co = 0.77wt%C)p ( )• Transformation at T = 675C.


4-Nov-09


4-Nov-09

PEARLITE MORPHOLOGY

Two cases:

• Ttransf just below TE--Larger T: diffusion is faster--Pearlite is coarser.

• Ttransf well below TE--Smaller T: diffusion is slower--Pearlite is finer.


4-Nov-09

2E-08

1.5E-08

/s)

1E-08

cien

t (M

2/

5E-09

on C

oeffi

c

D (A in B)D (A in C)

2.5E-22Diff

usio

-5E-09

0 1000 2000 3000 4000

T (K)


4-Nov-09

T (K)

Eutectoid & PeritecticCu-Zn Phase diagram

Peritectic transition γ + L δ

Adapted from Fig 9 21 Callister 7e


4-Nov-09

Fig. 9.21, Callister 7e.Eutectoid transition δ γ + ε

Peritectic Trans. L +δ ε

Figure 2. Microstructure of the Hyper Eutectoid.


4-Nov-09


4-Nov-09

FRACTION OF TRANSFORMATION• Fraction transformed depends on time,

at constant temperature (e.g., γ pearlite)

nktey −−= 1

Transformation rate r = 1/t

Avrami equation(k, n are constants)


4-Nov-09

• Transformation rate , r = 1/t0.5

EX: COOLING HISTORY Fe-C SYSTEM• Eutectoid composition, Co = 0.77wt%C• Begin at T > 727C• Rapidly cool to 625C and hold isothermally. C li t l t t lt i fi i t t• Cooling to lower temperatures results in finer microstructures


4-Nov-09

OTHER TRANSFORMATION PRODUCTS• Bainite:

--α strips with long, finerods of Fe3Crods of Fe3C

• Isothermal Transf. Diagram

Fe3C

(cementite)α (ferrite)

5 μm(Adapted from Fig. 10.8, Callister, 6e. (Fig. 10.8 from Metals Handbook, 8th ed.,Vol. 8, Metallography, Structures, and Phase Diagrams, American Society for Metals, Materials Park, OH, 1973.)

μ

Note: reaction rate increases with decreasing temperature first and then


4-Nov-09

temperature first, and then decreases

Bainite is a phase that exists in steel microstructures after certain heat treatments.First described by Davenport E. S. and Edgar Bain, it is one of the decompositionproducts that may form when austenite (the face centered cubic crystal structure of

) f °C ( ° )iron) is cooled past a critical temperature of 723 °C (about 1333 °F). Davenport andBain originally described the microstructure as being similar in appearance totempered martensite.

A fi l ll b i i l i f f i bid dA fine non-lamellar structure, bainite commonly consists of ferrite, carbide, andretained austenite. In these cases it is similar in constitution to pearlite, but with theferrite forming by a displacive mechanism similar to martensite formation, usuallyfollowed by precipitation of carbides from the

t t d f it t itsupersaturated ferrite or austenite.


4-Nov-09


4-Nov-09

NUCLEATION AND GROWTH• Reaction rate is a result of nucleation and growth

of crystals.

Nucleation rate increases with ΔTGrowth rate increases with T

• Examples:pearlite

γ γ γcolony

Nucleation rate high

T just below TE T moderately below TE T way below TENucleation rate low

Growth rate high

Nucleation rate med .Growth rate med. Growth rate low


4-Nov-09


4-Nov-09

TRANSFORMATIONS & UNDERCOOLING

• Can make it occur at: ...727ºC (cool it slowly)

• Eutectoid transf. (Fe-C System): γ ⇒ α +Fe3C0.77wt%C

0.022wt%C6.7wt%C

( y) ...below 727ºC (“undercool” it!)

1600T(°C)

Adapted from Fig. 9 21 C lli t 6 (Fi

1400

1200

L

γ γ+L

L+Fe3C

9.21,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International

α ferrite

1000

800

austenite

γ+Fe3C

Fe3C

α+γ

Eutectoid:

727°CEquil. cooling: Ttransf. = 727ºC

ASM International, Materials Park, OH, 1990.)

600

400 6.7

Fe3C

cementiteα+Fe3C

γ

0.7

7

727°CΔT

.02

2

Undercooling by ΔT: Ttransf. < 727ºC

q g transf.


4-Nov-09

4000 1 2 3 4 5 6 6

(Fe) Co, wt% C

00

NUCLEATION AND GROWTH• Reaction rate is a result of nucleation and growth

of crystals.100

% Pearlite

50

100

Nucleation

Growth regime

Nucleation rate increases w/ ΔT

Growth rate increases w/ T

• Examples:0

regime

log (time)t50

Adapted fromFig. 10.1, Callister 6e.

γ γ γ

pearlite colony

Nucleation rate high

T just below TE T moderately below TE T way below TENucleation rate low Nucleation rate med .


4-Nov-09

Growth rate high Growth rate med. Growth rate low

PEARLITE MORPHOLOGY

• Ttransf just below TE--Larger T: diffusion is faster

Two cases:• Ttransf well below TE

--Smaller T: diffusion is slower

μm

g--Pearlite is coarser. --Pearlite is finer.

10

μ

Adapted from Fig. 10.6 (a) and (b),Callister 6e. (Fig. 10.6 from R.M. Ralls et al., An Introduction to Materials Science and Engineering, p. 361, John Wiley and Sons, Inc., New York, 1976.)

- Smaller ΔT: colonies are larger

- Larger ΔT: colonies are smaller


4-Nov-09

g smaller

EX: COOLING HISTORY Fe-C SYSTEM• Eutectoid composition, Co = 0.77wt%C• Begin at T > 727C• Rapidly cool to 625C and hold isothermally. Rapidly cool to 625C and hold isothermally.

T(°C)Austenite (stable)

TE (727°C)

Adapted from Fig. 10.5,Callister 6e. (Fig 10 5 adapted from

700

Pearlite

E ( )

(Fig. 10.5 adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American

600

γ γ

γγγγ

γ

0% 100

50% Diagrams, American Society for Metals, 1997, p. 28.)

1 10 102 103 104 105 time (s)

500

%pearlite

00%

0%Dr. Ray Taheri Last revision: Sept. 01, 2009

4-Nov-09

1 10 102 103 104 105 time (s)

OTHER PRODUCTS: Fe-C SYSTEM (1)

α (ferrite)

• Spheroidite:--α crystals with spherical Fe3C--diffusion dependent. ( )

Fe3C

(cementite)

diffusion dependent.--heat bainite or pearlite for long times--reduces interfacial area (driving force)

• Isothermal Transf. Diagram (cementite)g800


PTEA

60 μm

Ad t d f Fi 10 9 C lli t 6

(Adapted from Fig. 10.10, Callister, 6e. (Fig. 10.10 copyright United States Steel Corporation, 1971.)

400

600P

BA

Spheroidite100% spheroidite

100% spheroidite

Adapted from Fig. 10.9,Callister 6e.(Fig. 10.9 adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1997, p. 28.)

400

200 0%

100%

50%

A


4-Nov-09 10

10 103 105time (s)10-1

%%


4-Nov-09

OTHER PRODUCTS: Fe-C SYSTEM (2)• Martensite:

--γ(FCC) to Martensite (BCT)(involves single atom jumps)

xx x

xx

xpotential C atom sites

Fe atom sites

(Adapted from Fig. 10 11 C lli t 6

• Isothermal Transf. Diagram800


TEA

10.11, Callister, 6e.

Adapted from Fig

(Adapted from Fig. 10.12, Callister, 6e. (Fig. 10.12 courtesy United States Steel Corporation.)

600

T( C)P

B

TEA

S

from Fig. 10.13, Callister 6e.

400

200

B

0%

100%50%

A

M + AM + A

0%50%

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


4-Nov-09 11time (s)10 103 10510-1

M + AM + A

50%90%

Stable

TTT diagram for eutectoid steelQUENCHING

H d R 65 austenite

'αγ →coolingrapid

Hardness RC 65

α’: martensite (M)

'αγ ⎯⎯⎯⎯ →⎯ gp

mart n t (M)

Extremely rapid, no C

unstable

C-curves

austenite

A+M

MsMs : Martensite start

temperature


4-Nov-09 M

MfMf : Martensite finish

temperature

Amount of martensite formed

Martensitic transformation

Amount of martensite formed does not depend upon time, only on temperature.

Atoms move only a fraction of atomic distance during the

f itransformation:

1. Diffusionless ( l diff i )(no long-range diffusion)

2. ShearBCT (one-to-one correspondence

between γ and α’ atoms)


4-Nov-09

3. No composition change


4-Nov-09

Martensitic transformation (contd.)

BCT unit cell

414.12 ==ac

BCT unit cell of α’ (martensite)

c 20.100.1 −=ac

Expand ~ 12%

0% Con. (BCC)

20 % Con. Contract ~ 20%


4-Nov-09

Martensitic transformation (contd.)

Hardness of martensite as a function of C content

60

40R C 40

dnes

s, R

20

Har

d

Wt % Carbon →0.2 0.4 0.6


4-Nov-09

Hardness of martensite depends mainly on C content and not on other alloying additions

Slow Cooling

Time in region indicates amount of microconstituent!


4-Nov-09

Medium Cooling

Cooling Rate, R, is Change in Temp / Time °C/s


4-Nov-09

Fast Cooling

This steel is very hardenable… 100% Martensite in ~ 1 minute of cooling!minute of cooling!


4-Nov-09

COOLING EX: Fe-C SYSTEM (1)• Co = Ceutectoid• Three histories...

Rapid cool to:

Hold for:

Rapid cool to:

Hold for:

Rapid cool to:

800

Case I

350°C

250°C

104s

102s

Troom

Troom

104s

102s

Troom

Troom

600

800


PA

S

650°C

20s

400°C

103s

Troom

Ad t d

400B

100

SA

100%A 100%BAdapted from Fig. 10.15, Callister 6e.

200

0%

100%50%

M + AM + AM + A

0%50%90%

100% Bainite

100%A


4-Nov-09 12

time (s)10 103 10510-1100% Bainite

COOLING EX: Fe-C SYSTEM (2)• Co = Ceutectoid• Three histories...

Rapid cool to:

Hold for:

Rapid cool to:

Hold for:

Rapid cool to:

Case II

350°C

250°C

104s

102s

Troom

Troom

104s

102s

Troom

Troom

600

800


PA

650°C

20s

400°C

103s

Troom

400

600

B

0 1

SA

Ad t d

200

0% 100%50%

M + AM + AM + A

0%50%90%

M + trace of A

100%AAdapted from Fig. 10.15, Callister 6e.


4-Nov-09

time (s)10 103 10510-1M + A M + trace of A

13

COOLING EX: Fe-C SYSTEM (3)Rapid cool to:

Hold for:

Rapid cool to:

Hold for:

Rapid cool to:

• Co = Ceutectoid• Three histories...

350°C

250°C

104s

102s

Troom

Troom

104s

102s

Troom

Troom 800

Case III

650°C

20s

400°C

103s

Troom

600

800T(°C)

Austenite (stable)

P

A

S

50%P, 50%A100%A

Ad t d

400

B

0

100%5

SA

50%P, 50%A50%P, 50%B

Adapted from Fig. 10.15, Callister 6e.

200

0%

00%50%

M + AM + AM + A

0%50%90%

50%


4-Nov-09 14

time (s)10 103 10510-150%P, 50%B


4-Nov-09

TEMPERING

Heating of quenched steel below the eutectoid t t h ldi f ifi d ti f ll d b

TEMPERING

temperature, holding for a specified time followed by ar cooling.

CFetempering3+⎯⎯⎯ →⎯′ αα

T<TE


4-Nov-09

Tempering (contd.)

α+Fe3C PEARLITE

A distribution of fine particles of Fe3C in α matrix known as TEMPERED MARTENSITEknown as TEMPERED MARTENSITE.

Hardness more than fine pearlite, ductility more than martensitethan martensite.

H d ss d d tilit t ll d b t m i Hardness and ductility controlled by tempering temperature and time.

Higher T or t > higher ductility lower strength


4-Nov-09

Higher T or t -> higher ductility, lower strength

Problems in Quenching Quench Cracks

High rate of cooling:

f l h i isurface cooler than interior

Surface forms martensite before the interior

Austenite martensite Volume expansionp

When interior transforms, the hard outer martensitic shell constrains this expansion martensitic shell constrains this expansion leading to residual stresses


4-Nov-09


4-Nov-09

Solution to Quench cracksSolution to Quench cracks

Shift the C-curve to the right (higher times)f g ( g )

More time at the noseMore time at the nose

Slower quenching (oil quench) can give Slower quenching (oil quench) can give martensite

But how to shift the C-curve to higher times?


4-Nov-09

By alloyingy y g

All alloying elements in steel (Cr, Mn, Mo, Ni, Ti, W, V) etc shift the C-curves to the right.

E i CException: Co

b l d ff f ll l Substitutional diffusion of alloying elements is slower than the interstitial diffusion of C


4-Nov-09

Plain C steel Alloy steel


4-Nov-09

Hardenabilityy

Ability or ease of hardening a steel by formation of martensite using as slow quenching as possible

Alloying elements in steels shift the C-curve to the right

Alloy steels have higher hardenability than plain C steelsC steels.


4-Nov-09

Hardnenability Hardnessy

Ability or ease of Resistance to plastic d f ti hardening a steel deformation as measured by indentation

Only applicable to steels Applicable to all materials

Alloying additions increase the hardenability of steels but not the hardness.

C increases both hardenability and hardness of steels.


4-Nov-09

High Speed steelHigh Speed steel

Alloy steels used for cutting tools operated at high speeds

Cutting at high speeds lead to excessive heating of Cutting at high speeds lead to excessive heating of cutting tools

h l d d f h This is equivalent to unintended tempering of the tools leading to loss of hardness and cutting edge

Alloying by W gives fine distribution of hard WC particles which counters this reduction in hardness:


4-Nov-09

particles which counters this reduction in hardness such steels are known as high speed steels.

holdingT

AT A

N

AT

TQ

Ntime

Annealing Furnace cooling RC 15g g

Normalizing Air cooling RC 30Quenching Water cooling RC 65Quenching Water cooling RC 65

Tempering Heating after quench RC 55


4-Nov-09

Austempering Quench to an inter- RC 45mediate temp and hold

MECHANICAL PROPERTIES

• Martensiteh ty• Tempered martensite• Bainite

Fine pearliteStre

ngt

Duc

tilit

• Fine pearlite• Coarse pearlite• Spheroidite pearliteSpheroidite pearlite

• Can control the formation of specific phases andmicrostructure so that desired properties result


4-Nov-09

microstructure so that desired properties result

MECHANICAL PROP: Fe-C SYSTEM (1)(1)

• Effect of wt%C

Pearlite (med)ferrite (soft)

Pearlite (med)Cementite

(hard)Adapted from Fig. 9.27,Callister6e. (Fig. 9.27 courtesy Republic Steel Corporation.)

Adapted from Fig. 9.30,Callister 6e. (Fig. 9.30 copyright 1971 by United States Steel Corporation.)

Co>0.77wt%C

Hypereutectoid

Co<0.77wt%C

Hypoeutectoid

ferrite (soft)

H HHypo Hyper

(hard)

Adapted from Fig. 10.20, Callister 6e. (Fig. 10.20 based on data from Metals

100%EL

zod

, ft-

lb)

80

900

1100YS(MPa)TS(MPa)

hardness

Hypo HyperHypo Hyper

data from Metals Handbook: Heat Treating, Vol. 4, 9th ed., V. Masseria (Managing Ed.), American Society for

50

ct

en

erg

y (I

z

0

40

500

700

hardness

Metals, 1981, p. 9.)

wt%C0 0.5 1

0 Imp

ac

300

wt%C0 0.5 1

0.7

7

0.7

7

Dr. Ray Taheri Last revision: Sept. 01, 2009 4-Nov-09

• More wt%C: TS and YS increase, %EL decreases.

MECHANICAL PROP: Fe-C SYSTEM (2)(2)

• Fine vs coarse pearlite vs spheroidite

90Hypo Hyper Hypo Hyper

240

320

dn

ess

fine pearlite

coarse 60

y (%

AR

)

spheroidite

160

Bri

ne

ll h

ard

coarse pearlitespheroidite

30

Du

cti

lity

fine

coarse pearlite

Adapted from Fig 10 21 Callister

80

wt%C0 0.5 1

B

0

wt%C0 0.5 1

fine pearlite

Adapted from Fig. 10.21, Callister 6e. (Fig. 10.21 based on data from Metals Handbook: Heat Treating, Vol. 4, 9th ed., V. Masseria (Managing Ed.), American Society for Metals, 1981, pp. 9 and 17.)

• Hardness: fine > coarse > spheroidite • %AR: fine < coarse < spheroidite


4-Nov-09

, , pp )

MECHANICAL PROP: Fe-C SYSTEM (3)

• Fine Pearlite vs Martensite:(3)

Hypo Hyper

600

ne

ss martensite

Hypo Hyper

Adapted from Fig. 10.23, Callister 6e. (Fig. 10.23 adapted from Edgar C. Bain, Functions of the Alloying Elements in Steel, American S i t f M t l 1939 36

400

Bri

ne

ll h

ard

n

Society for Metals, 1939, p. 36; and R.A. Grange, C.R. Hribal, and L.F. Porter, Metall. Trans. A, Vol. 8A, p. 1776.)

0

200

0 0 5 1

B

fine pearlite

• Hardness: fine pearlite << martensite.wt%C

0 0.5 1


4-Nov-09 17

TEMPERING MARTENSITE• reduces brittleness of martensite,• reduces internal stress caused by quenching.

YS(MPa)TS(MPa)

1800

TS

Adapted from Fig. 10.24, Callister 6e. (Fig. 10.24

Adapted from Fig. 10.25, Callister 6e. (Fig. 10.25 1200

1400

1600

60

TS

YS

9 μ

m

( gcopyright by United States Steel Corporation, 1971.)

( gadapted from Fig. furnished courtesy of Republic Steel Corporation.) 800

1000

1200

3040

50

60

%AR%AR

9

30200 400 600

Tempering T (°C)

• produces extremely small Fe3C particles surrounded by α


4-Nov-09 18

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


4-Nov-09

Austempering

B i itBainite

Short needles of Fe3C embedded in plates of ferrite


4-Nov-09


4-Nov-09

1

2

3

4

56


4-Nov-09

Cool down to 600 C and hold for 4 s



Q hQuench

Final structure consist of:50% pearlite25% upper Bainite12.5% lower Bainite12.5% Martensite12.5% Martensite


4-Nov-09


4-Nov-09

Stable austeniteStable austenite

Annealing:Annealing:coarse pearlite

Normalizing:gfine pearlite

unstable austenite


4-Nov-09

Design questionWhat will be a suitable tempering temperature andtime for a 1080 water quenched plain carbon steel toq phave maximum 230000 psi tensile strength?


4-Nov-09


4-Nov-09

At 425 C for 17 min


4-Nov-09

At 425 C for 17 min.


4-Nov-09

A heat treatment cycle is needed to produce a uniform

Example 1 Design of a Heat Treatment for an Axle

y pmicrostructure with minimum hardness of 200 BHN and minimum ductility of 35 in a 1050 steel axle.

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4-Nov-09


4-Nov-09

SOLUTION

1. Austenitize the steel at 770 + (30 to 55) = 805oC to 1. Austenitize the steel at 770 + (30 to 55) 805 C to

825oC, holding for 1 h and obtaining 100% γ.

2. Quench the steel to 600oC and hold for a minimum of 10 s. f b f h blPrimary ferrite begins to precipitate from the unstable

austenite after about 1.0 s. After 1.5 s, pearlite begins to grow, and the austenite is completely transformed to ferrite and pearlite after about 10 s After this treatment ferrite and pearlite after about 10 s. After this treatment, the microconstituents present are:

%36100)022.077.0(

0.5)(0.77αPrimary =×⎥⎦

⎤⎢⎣

⎡−−

=

%64100)022.077.0(

0.0218)(0.5Pearlite

)(

=×⎥⎦

⎤⎢⎣

⎡−

−=

⎦⎣

3. Cool in air-to-room temperature, preserving the equilibrium amounts of primary ferrite and pearlite. The microstructure and hardness are uniform because of the


4-Nov-09

isothermal anneal.


4-Nov-09

Example 2: Design of a Quench and Temper Treatmentp

A rotating shaft that delivers power from an electric motor is made f om a 4340 steel Its ield st ength sho ld be at least made from a 4340 steel. Its yield strength should be at least 200,000 psi, yet it should also have at least 40% AR. Design a heat treatment to heat treat this part.


4-Nov-09

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4-Nov-09


4-Nov-09

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4-Nov-09

SOLUTION

1. Austenitize above the A3 temperature of 760oC for 1 h. An appropriate temperature may be 760 + 55 = 815oC.

2 Oil q ench apidl to oom tempe at e Since the M is 2. Oil quench rapidly to room temperature. Since the Mf is about 250oC, martensite will form.

3. Temper by heating the steel to 420oC. Normally, 1 h will b ffi i t if th t l i t t thi kbe sufficient if the steel is not too thick.

4. Cool to room temperature.


4-Nov-09

©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.


4-Nov-09


4-Nov-09

Example : 3

Determination of Heat Treating Temperatures

Recommend temperatures for the process annealing, annealing, normalizing, and spheroidizing of 1020, 1077, and 10120 (1.2 %C) steels.0 , a d 0 0 ( %C) s ee s


4-Nov-09


4-Nov-09

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Figure 1 (a) The eutectoid portion Figure 1 (a) The eutectoid portion of the Fe-Fe3C phase diagram. (b) An expanded version of the Fe-C diagram, adapted from several sources.


4-Nov-09

SOLUTION

From Figure 1, we find the critical A1, A3, or Acm, temperatures for each steel. We can then specify the heat treatment based on these temperatures.treatment based on these temperatures.


4-Nov-09

Equivalent Carbon

Carbon, C 0.280 - 0.330 %

Chromium Cr 0 500 %

AISI 860Chromium, Cr 0.500 %

Iron, Fe 97.0 %

Manganese, Mn 0.800 %

Molybdenum, Mo 0.200 %

Nickel, Ni 0.550 %

Phosphorous, P <= 0.0350 %

Silicon, Si 0.230 %

Sulfur, S <= 0.0400 %

E =0 65


4-Nov-09

Ec 0.65

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Figure 3 Producing complicated structures

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by interrupting the isothermal heat treatment of a 1050

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4-Nov-09


4-Nov-09

Application of Hardenability

Jominy test - The test used to evaluate hardenability An

pp y

Jominy test - The test used to evaluate hardenability. An austenitized steel bar is quenched at one end only, thus producing a range of cooling rates along the bar.Hardenability curves - Graphs showing the effect of the Hardenability curves Graphs showing the effect of the cooling rate on the hardness of as-quenched steel.Jominy distance - The distance from the quenched end of a Jominy bar The Jominy distance is related to the a Jominy bar. The Jominy distance is related to the cooling rate.


4-Nov-09


Figure. The set-up for the Jominy test used for determining the hardenability of a steel.


4-Nov-09


4-Nov-09

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4-Nov-09


4-Nov-09

A gear made from 9310 steel which has an as quenched

Example Design of a Wear-Resistant GearA gear made from 9310 steel, which has an as-quenchedhardness at a critical location of HRC 40, wears at an excessiverate. Tests have shown that an as-quenched hardness of atleast HRC 50 is required at that critical location Select anleast HRC 50 is required at that critical location. Select anappropriate steel to satisfy hardness criterion.


4-Nov-09


4-Nov-09

SOLUTION

From the Figure, a hardness of HRC 40 in a 9310 steel corresponds to a Jominy distance of 10/16 in. (10oC/s). If we assume the same Jominy distance, the other steels shown in Fi h th f ll i h d t th iti l l tiFigure have the following hardnesses at the critical location:

1050 HRC 28 1080 HRC 36 4320 HRC 31

8640 HRC 52 4340 HRC 608640 HRC 52 4340 HRC 60

In Table 12-1, we find that the 86xx steels contain less alloying elements than the 43xx steels; thus the 8640 steel is probably l i th th 4340 t l d i ht b b t less expensive than the 4340 steel and might be our best choice. We must also consider other factors such as durability.


4-Nov-09

Example Design of a Quenching Process

Design a quenching process to produce a minimum hardness of HRC 40 at the center of a 1.75 in. diameter 4320 steel bar.HRC 40 at the center of a 1.75 in. diameter 4320 steel bar.


4-Nov-09

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Figure 2. The hardenability curves for several steels.

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4-Nov-09


Figure 1. The Grossman chart used to determine the hardenability at the center of a steel bar for different

h t


4-Nov-09

quenchants.

SOLUTIONSeveral quenching media are listed in Table 12-2 We can find Several quenching media are listed in Table 12 2. We can find an approximate H coefficient for each of the quenching media, then use Figure 1 to estimate the Jominy distance in a 1.75-in. diameter bar for each media. Finally, we can use the

fhardenability curve (Figure 2) to find the hardness in the 4320 steel. The results are listed below.

The last three media, brine or agitated water, are satisfactory. Using an unagitated brine quenchant might be least expensive, since no extra equipment is needed to agitate the quenching bath However H2O is less corrosive than the brine quenchantbath. However, H2O is less corrosive than the brine quenchant.


4-Nov-09


4-Nov-09

Surface Treatments

Selectively Heating the Surface - Rapidly heat the f f di b t l b th Asurface of a medium-carbon steel above the A3

temperature and then quench the steel.Case depth - The depth below the surface of a steel at

hi h h d i b f h d iwhich hardening occurs by surface hardening.Carburizing - A group of surface-hardening techniques by which carbon diffuses into steel.Cyaniding - Hardening the surface of steel with carbon and nitrogen obtained from a bath of liquid cyanide solution.Carbonitriding - Hardening the surface of steel with carbon and nitrogen obtained from a special gas atmosphere.


4-Nov-09


4-Nov-09

Nitriding is a surface-hardening heat treatment that introduces nitrogen into the surface of steel at a temperature range (500 to 550°C, or 930 to 1020°F), while it is in the ferrite condition. Thus, nitriding is similar to carburizing in that surface composition is altered, but different in that nitrogen is added into f it i t d f t it B it idi d t i l h ti i t th t it h fi ld dferrite instead of austenite. Because nitriding does not involve heating into the austenite phase field and a subsequent quench to form martensite, nitriding can be accomplished with a minimum of distortion and with excellent dimensional control.

The mechanism of nitriding is generally known, but the specific reactions that occur in different steels and e ec a s o d g s ge e a y o , bu e spec c eac o s a occu d e e s ee s a dwith different nitriding media are not always known. Nitrogen has partial solubility in iron. It can form a solid solution with ferrite at nitrogen contents up to about 6%. At about 6% N, a compound called gamma prime (γ’), with a composition of Fe4N is formed.

At nitrogen contents greater than 8% the equilibrium reaction product is ε compound Fe N NitridedAt nitrogen contents greater than 8%, the equilibrium reaction product is ε compound, Fe3N. Nitrided cases are stratified. The outermost surface can be all γ’ and if this is the case, it is referred to as the white layer. Such a surface layer is undesirable: it is very hard profiles but is so brittle that it may spall in use. Usually it is removed; special nitriding processes are used to reduce this layer or make it less brittle. The ε zone of the case is hardened by the formation of the Fe3N compound, and below this layer there is some 3solid solution strengthening from the nitrogen in solid solution.Principal reasons for nitriding are:

•To obtain high surface hardness•To increase wear resistance•To increase wear resistance•To improve fatigue life•To improve corrosion resistance (except for stainless steels)•To obtain a surface that is resistant to the softening effect of heat at temperatures up to the nitriding temperature


4-Nov-09


Figure 1. (a) Surface hardening by localized heating. (b) Only the surface heats above the A1 temperature and is quenched to martensite.


4-Nov-09

que c ed to a te s te


Figure 2. Carburizing of a low-carbon steel to produce a high-carbon, wear-resistant surface.


4-Nov-09

Example

Design of Surface-Hardening Treatments for a Drive Train

Design the materials and heat treatments for an automobile axle and drive gear.


4-Nov-09


SOLUTIONSOLUTION

The axle might be made from a forged 1050 steel containing a matrix of ferrite and pearlite. The axle could be surface-hardened, perhaps by moving the axle through an induction coil to selectively heat the surface of the steel above the A3temperature (about 770oC). After the coil passes any

ti l l ti f th l th ld i t i h th particular location of the axle, the cold interior quenches the surface to martensite. Tempering then softens the martensiteto improve ductility.

Carburize a 1010 steel for the gear. By performing a gas carburizing process above the A3 temperature (about 860oC), we introduce about 1.0% C in a very thin case at the surface of the gear teeth. This high-carbon case, which transforms to martensite during quenching, is tempered to control the hardness.


4-Nov-09


4-Nov-09

Example

Compare the structures in the heat-affected zones of

Structures of Heat-Affected Zones

pwelds in 1040 and 4340 steels if the cooling rate in the heat-affected zone is 9 oC/s.


4-Nov-09


4-Nov-09

SOLUTION

The cooling rate in the weld produces the following structures:

1040: 100% pearlite

4340: Bainite and martensite

The high hardenability of the alloy steel reduces the weldability, permitting martensite to form and embrittle the weld.


4-Nov-09

Weldability of Steele.

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a d cooling in a steel of

low hardenability, and (c) the structure after cooling in a steel of high hardenability.


4-Nov-09

Super question 1p qSeveral bar stocks in different diameters made of 8620 alloy areavailable. In order to increase the hardness and tensile strength of the bar

d t l hi ( ith i t il) D t iwe need to apply a quenching process (either in water or oil). Determinethe maximum diameter of the bar in order to have a minimum hardnessof 40 HRC at the center and minimum hardness of 55 HRC at 0.1 mmbelow the surface? The carburizing process should not take more 3below the surface? The carburizing process should not take more 3hours. Design a complete heat treatment process to accomplish requiredhardness criteria.

)20000exp(10688.6 3

RTPC −

×= −

P: Pressure in PaP: Pressure in Pa


4-Nov-09


4-Nov-09

18 mm


4-Nov-09


4-Nov-09

C b C 0 180 0 230 %

AISI 8620Carbon, C 0.180 - 0.230 %

Chromium, Cr 0.500 %

Iron, Fe 97.0 %

Manganese, Mn 0.800 % E =0 55Molybdenum, Mo 0.200 %

Nickel, Ni 0.550 %


Sili Si 0 230 %

Ec=0.55

Silicon, Si 0.230 %

Sulfur, S <= 0.0400 %


4-Nov-09


4-Nov-09

Step 1:Determine the equivalent diameter of the bar in order to get the requited hardness at the center.

Step 2: Realize the fact that the carbon content of the alloy is not enough to give the sufficient hardness at the surface or slightly belowsufficient hardness at the surface or slightly below.

Step 3:Determining which alloy gives the sufficient hardness at the surface.

St 4Step 4:Finding the suitable Austeniting temperature for the alloy, depending of the carbon content.

Step 5:pApplying the second Fick law and find the required surface carbon content.

Step 6:Finding the required pressure to accomplished the carburizing process.


4-Nov-09

Super question 2p qDetermine the minimum carbon content, maximum diameter, and quench medium for 86 series round bar alloy in order to get the minimum hardness of 45 HRC at the center. Required hardness at 0 5R is 50 HRC Available alloys are:Required hardness at 0.5R is 50 HRC. Available alloys are:8610, 8620, 8630, 8640, 8650 and 8660.

Is there a more realistic engineering approach to this problem?


4-Nov-09

86508650 (0.5 wt% C)


4-Nov-09


4-Nov-09

86508650 (0.5 wt% C)


4-Nov-09

Alloy Equivalent distance from quenched end (0.5R) mm

Equivalent distance from quenched end (C) mm

HRC Diameter

8620 - - - -

8630 4 6 44 29


4-Nov-09


4-Nov-09

86508650 (0.5 wt% C)


4-Nov-09



HRC Diameter

8620 - - - -

8630 4 6 44 29

8640 11 13 45 57


4-Nov-09


4-Nov-09

8650 (0.5 wt% C)


4-Nov-09



HRC Diameter

8620 - - - -

8630 4 6 44 29

8640 11 13 45 57

8650 20 27 47 100


86508650 (0.5 wt% C)


4-Nov-09



HRC Diameter

8620 - - - -

8630 4 6 44 29

8640 11 13 45 57

8650 20 27 47 100

8660 32 >32 >48 >100


8650 (0.5 wt% C)


4-Nov-09


4-Nov-09



HRC Diameter

8620 - - - -

8630 - - - -

8640 11 13 45 25


ExampleExampleThe following five alloys are available in form of round bar invariety of diameters We are asked to design a gear in whichvariety of diameters. We are asked to design a gear in whichthe minimum hardness at the center will be 40 HRC.Moreover, the hardness at approximately 5 mm below thesurface should be between 47-50 HRC. Also determine thequenching environment.

104051408640414043404340


4-Nov-09

5

15 20


4-Nov-09

15 2032


4-Nov-09

Therefore, 1040 is not a good candidate because the maximum allowable diameter is 25 mm and the hardness at 0.75 R, 3.125 mm below surface, isdiameter is 25 mm and the hardness at 0.75 R, 3.125 mm below surface, is about 47HRC, too low, when quenched in water.


4-Nov-09

5

14 20


4-Nov-09

14 2032

6565

7.52.5


4-Nov-09


4-Nov-09

Therefore, 5140 is not a good candidate because the maximum allowable diameter is 65 mm and the hardness at 0.75 R, 8.125 mm below surface, isdiameter is 65 mm and the hardness at 0.75 R, 8.125 mm below surface, is about 51 HRC, too high, when quenched in water.


4-Nov-09


4-Nov-09

51140 is a good candidate because the maximum allowable diameter , oil h d i 44 d th h d t 0 75 R 5 5 b l f iquenched, is 44 mm and the hardness at 0.75 R, 5.5 mm below surface, is

about 46 HRC . As a result with a great level of confidence, we can say at 5mm below the surface the hardness is around 47 HRC.


4-Nov-09

Example:pWhat would be the suitable carbon content of a 86 series alloy quenched in water, if the hardness of a cylindrical piece throughout the first 17mm from the center should be between 45‐50 HRC. The diameter range should be 50‐100 mm.


Wednesday, November 04, 2009

3



10

6

14 32 45

Alloy Distance from Jominyend, 45 HRC

Distance from Jominyend, 50 HRC

8620 6 38620 6 38630 14 108460 45 32



X

6565

30

3



12

Alloy Distance from the Jominy end

Required Distancefrom the Jominy ,

Diameter Distance from jominy ends y

(center) mmy ,

50 HRCj y

at 17 mm8620 6 4 30 38630 14 10 65 128630 14 10 65 128460 45 32 >100 -



Examples

The design criteria for a differential axle used in a heavy duty pickup truck are as follows:as follows:

1‐Minimum hardness throughout the component: 45 HRC2‐Maximum hardness at center: 50 HRC3‐ Range of diameter: 55‐70 mm4‐Minimum surface hardness 55 HRC 5‐ Quenching Medium: water



X

Too soft

15 466



411

15 23



Minimum hardness Criterion

Alloy Min distance from JominyAlloy Min distance from Jominyend (mm), 45 HRC

8620 Too soft8630 68630 68640 158660 461040 45140 11 4140 234140 234340 50


4-Nov-09

Minimum hardness on the surface

X


4-Nov-09

Alloy Min distance from JominyAlloy Min distance from Jominyend (center) mm

8620 Too soft8630 6 (too soft at the s rface)8630 6 (too soft at the surface)8640 158660 461040 45140 11 4140 234140 234340 50


4-Nov-09

X

Too soft

33



2 167



Alloy Max distance from Jominy end (mm) center

8640 108640 108660 331040 25140 75140 7 4140 164340 50


4-Nov-09

Alloy Max distance from Jominy end (mm) center

Min allowable diameter(mm)

8640 10 508640 10 508660 33 100<1040 2 Too small5140 7 305140 7 304140 16 714340 50 100<


4-Nov-09

3




4-Nov-09

Question Three:The design criteria for a gear used in an excavation drill gearbox are as follows:1‐Minimum hardness throughout the component: 45 HRC2‐Maximum hardness at center: 50 HRC3‐ Range of diameter: 55‐60 mm (find the maximum allowable diameter)4‐Maximum heat treatment time: 3.5 hours5‐Minimum hardness 0.1 mm below the surface: 64 HRC6 Quenching medium Agitated water6‐ Quenching medium: Agitated water7‐ Available alloys: 8610, 8620, 8630, 8640, and 86608‐ Relationship between pressure and carbon content given as follow:

)20000(106886 3 PC −−

P: in Pascal 9‐ Allowable range of pressure: 150 psi10‐Maximum achievable temperature: 940 C°

)exp(10688.6 3

RTPC ×=

pAssumption: Disregard the effect of alloying elements on carbon diffusion coefficient and process. Select the appropriate material and design a suitable process.



3 11 32



6



13



65

6 15



C b C 0 180 0 230 %

AISI 8640Carbon, C 0.180 - 0.230 %

Chromium, Cr 0.500 %

Iron, Fe 97.0 %

Manganese, Mn 0.800 % E =0 75Molybdenum, Mo 0.200 %

Nickel, Ni 0.550 %


Sili Si 0 230 %

Ec=0.75

Silicon, Si 0.230 %

Sulfur, S <= 0.0400 %



ENGR 376 - Lecture 4 - Phase Transformations

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Transcript of ENGR 376 - Lecture 4 - Phase Transformations