PY3090 PY3090 –– 55

PY3090PY3090Preparation of MaterialsPreparation of MaterialsLecture 5Lecture 5

Colm StephensColm StephensSchool of PhysicsSchool of Physics

• Co < 2 wt% Sn• Result:

--at extreme ends--polycrystal of α grains

i.e., only one solid phase.

Microstructures Microstructures in Eutectic Systems: Iin Eutectic Systems: I

0

L+ α200

T(°C)

Co, wt% Sn10

2

20Co

300

100

L

α

30

α+β

400

(room T solubility limit)

TE(Pb-SnSystem)

αL

L: Co wt% Sn

α: Co wt% Sn

• 2 wt% Sn < Co < 18.3 wt% Sn• Result:

Initially liquid + αthen α alonefinally two phases

α polycrystalfine β-phase inclusions

Microstructures Microstructures in Eutectic Systems: IIin Eutectic Systems: II

Pb-Snsystem

L + α

200

T(°C)

Co , wt% Sn10

18.3

200Co

300

100

L

α

30

α+ β

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

Lα

L: Co wt% Sn

αβ

α: Co wt% Sn

• Co = CE• Result: Eutectic microstructure (lamellar structure)

--alternating layers (lamellae) of α and β crystals.

Microstructures Microstructures in Eutectic Systems: IIIin Eutectic Systems: III

Adapted from Fig. 9.14, Callister 7e.160μm

Micrograph of Pb-Sneutectic microstructure

Pb-Snsystem

L + β

α + β

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L

α βL+α

183°C

40

TE

18.3

α: 18.3 wt%Sn

97.8

β: 97.8 wt% Sn

CE61.9

L: Co wt% Sn

• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: α crystals and a eutectic microstructure

Microstructures Microstructures in Eutectic Systems: IVin Eutectic Systems: IV

18.3 61.9

SR

97.8

SR

primary αeutectic α

eutectic β

WL = (1-Wα) = 50 wt%

Cα = 18.3 wt% SnCL = 61.9 wt% Sn

SR + S

Wα= = 50 wt%

• Just above TE :

• Just below TE :Cα = 18.3 wt% SnCβ = 97.8 wt% Sn

SR + S

Wα= = 73 wt%

Wβ = 27 wt%

Pb-Snsystem

L+β200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

α βL+α

40

α+β

TE

L: Co wt% Sn LαLα

L+αL+β

α + β

200

Co, wt% Sn20 60 80 1000

300

100

L

α βTE

40

(Pb-SnSystem)

Hypoeutectic & HypereutecticHypoeutectic & Hypereutectic

160 μmeutectic micro-constituent

hypereutectic: (illustration only)

β

ββ

ββ

β

175 μm

α

α

α

αα

α

hypoeutectic: Co = 50 wt% Sn

T(°C)

61.9eutectic

eutectic: Co =61.9wt% Sn

Intermetallic CompoundsIntermetallic Compounds

Note: intermetallic compound forms a line - not an area -because stoichiometry (i.e. composition) is exact.

Mg2Pb

Eutectic Eutectic -- liquid in equilibrium with liquid in equilibrium with two solidstwo solids

LL αα + + ββ

EutecticEutectic

coolheat

PeritecticPeritectic -- liquid + solid 1 liquid + solid 1 solid 2solid 2SS1 1 + + LL SS22

δδ + + LL γγ (1493(1493ººC)C)coolheat

EutectoidEutectoid -- solid phase in equilibrium solid phase in equilibrium with two solid phaseswith two solid phasesSS22 SS11++SS33

γγ αα + Fe+ Fe33C C (727(727ººC)C)

intermetallic compound - cementite

coolheat

Eutectoid & Eutectoid & PeritecticPeritectic

Example: Eutectoid & Example: Eutectoid & PeritecticPeritectic

Cu-Zn Phase diagram

Eutectoid transition δ γ + ε

Peritectic transition γ + L δ

IronIron--Carbon Phase Diagram ExtractCarbon Phase Diagram Extract• 2 important

points

-Eutectoid (B):γ ⇒ α +Fe3C

-Eutectic (A):L ⇒ γ +Fe3C

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ+Fe3C

α+Fe3C

α+γ

L+Fe3C

δ

(Fe) Co, wt% C

1148°C

T(°C)

α 727°C = Teutectoid

ASR

4.30

γ γγγ

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard)α (ferrite-soft)

PearlitePearlite

Fe3C (cementite-hard)

α (ferrite-soft)

Result: Pearlite = alternating layers ofα and Fe3C phases

120 μm

HypoeutectoidHypoeutectoid SteelSteel

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α+ Fe3C

L+Fe3C

δ

(Fe) Co, wt% C

1148°C

T(°C)

α727°C

(Fe-C System)

C0

0.76

r s

wα =s/(r+s)wγ =(1- wα)

γγ γ

γαα

α

γγγ γ

γ γγγ

α

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγpearlite

Hypoeutectoid SteelHypoeutectoid Steel

Proeutectoidferrite

pearlite

100 μmα

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγpearlite

Hypereutectoid SteelHypereutectoid Steel

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ +Fe3C

α +Fe3C

L+Fe3C

δ

(Fe) Co, wt%C

1148°C

T(°C)

α

(Fe-C System)

0.76

Co

R S

wα =S/(R+S)wFe3C =(1-w α)

wpearlite = wγpearlite

sr

wFe3C =r/(r+s)wγ =(1-w Fe3C)

Fe3C

γγγ γ

γγγ γ

γγγ γ

Hypereutectoid SteelHypereutectoid Steel

proeutectoid Fe3C

60 μm

pearlite

wα =S/(R+S)wFe3C =(1-w α)

wpearlite = wγpearlite

ExampleExample

For a 99.6 wt% FeFor a 99.6 wt% Fe--0.40 wt% C at a 0.40 wt% C at a temperature just below the temperature just below the eutectoid, determine the followingeutectoid, determine the following

a)a) the amount of the amount of pearlitepearlite and and proeutectoidproeutectoid ferrite (ferrite (αα) per 100 g of ) per 100 g of steelsteel

b)b) composition of Fecomposition of Fe33C and ferrite (C and ferrite (αα))c)c) the amount of carbide (cementite) in the amount of carbide (cementite) in

grams that forms per 100 g of steelgrams that forms per 100 g of steel

SolutionSolutiona. the amount of pearlite and proeutectoid ferrite (α)

note: amount of pearlite = amount of γ just above TE

Co = 0.40 wt% CCα = 0.022 wt% CCpearlite = Cγ = 0.76 wt% C

γγ + α

=Co −CαCγ −Cα

x 100 = 51.2 g

pearlite = 51.2 gproeutectoid α = 48.8 g

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

δ

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CγCα

SolutionSolution

g 3.94g 5.7 CFe

g7.5100 022.07.6022.04.0

100xCFe

CFe

3

CFe3

3

3

=α

=

=−−

=

−−

=α+ α

α

x

CCCCo

c) the amount of carbide (cementite) in grams that forms per 100 g of steel

b) composition of Fe3C and ferrite (α)

CO = 0.40 wt% CCα = 0.022 wt% CCFe C = 6.70 wt% C3

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

δ

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CFe C3Cα

Alloying Steel with More ElementsAlloying Steel with More Elements

• Teutectoid changes:

TE

utec

toid

(°C

)

wt. % of alloying elements

Ti

Ni

MoSi

W

Cr

Mn

• Ceutectoid changes:

wt. % of alloying elements

Ceu

tect

oid

(wt%

C)

Ni

Ti

Cr

SiMn

WMo

Taxonomy of MetalsTaxonomy of MetalsMetal Alloys

Ferrous Nonferrous

Cu Al Mg TiSteels

SteelsSteels

increasing strength, cost, decreasing ductility

PY3090�Preparation of Materials�Lecture 5Microstructures �in Eutectic Systems: IMicrostructures �in Eutectic Systems: IIMicrostructures �in Eutectic Systems: IIIMicrostructures �in Eutectic Systems: IVHypoeutectic & HypereutecticIntermetallic CompoundsEutecticEutectoid & PeritecticExample: Eutectoid & PeritecticIron-Carbon Phase Diagram ExtractPearliteHypoeutectoid SteelHypoeutectoid SteelHypereutectoid SteelHypereutectoid SteelExampleSolutionSolutionAlloying Steel with More ElementsTaxonomy of MetalsSteels