Chapter 9: Phase Diagrams ISSUES to ADDRESS

Home , Eutectic system, Lever rule, Phase diagram, Phase rule

Chapter 9: Phase Diagrams ISSUES TO ADDRESS... • When we combine two elements... what equilibrium state do we get? • In particular, if we specify... --a composition (e.g., wt% Cu - wt% Ni), and --a temperature ( T) then... How many phases do we get? What is the composition of each phase? How much of each phase do we get?

Nickel atom Copper atom Chapter 9 - 1 Phase Equilibria: Solubility Limit Introduction – Solutions – solid solutions, single phase

– Mixtures – more than one phase Adapted from Fig. 9.1, Callister 7e. Sucrose/Water Phase Diagram • Solubility Limit : 100 Max concentration for Solubility which only a single phase 80 Limit L solution occurs. (liquid) 60 L + Question: What is the 40 (liquid solution S solubility limit at 20°C? i.e., syrup) (solid 20

Temperature (°C) Temperature sugar) Answer: 65 wt% sugar . If C < 65 wt% sugar: syrup o 0 20 40 60 65 80 100 If Co > 65 wt% sugar: syrup + sugar. Co =Composition (wt% sugar) Sugar Pure Pure Water Chapter 9 - 2 Components and Phases • Components : The elements or compounds which are present in the mixture (e.g., Al and Cu) • Phases : The physically and chemically distinct material regions that result (e.g., α and β).

Aluminum- β (lighter Copper phase) Alloy α (darker phase) Adapted from chapter-opening photograph, Chapter 9, Callister 3e. A phase maybe defined as a homogeneous portion of a system that has Chapter 9 - 3 uniform physical and chemical characteristics. Effect of T & Composition ( Co) • Changing T can change # of phases: path A to B. • Changing Co can change # of phases: path B to D. B (100°C,70) D (100°C,90) 1 phase 2 phases 100

80 L (liquid) water- 60 + sugar L S (liquid solution system 40 (solid i.e., syrup) sugar) 20 A (20°C,70) Temperature (°C) 2 phases

Adapted from 0 Fig. 9.1, 0 20 40 60 7080 100 Callister 7e. Co =Composition (wt% sugar) Chapter 9 - 4 Phase Equilibrium

Equilibrium: minimum energy state for a given T, P, and composition (i.e. equilibrium state will persist indefinitely for a fixed T, P and composition).

Phase Equilibrium: If there is more than 1 phase present, phase characteristics will stay constant over time.

Phase diagrams tell us about equilibrium phases as a function of T, P and composition (here, we’ll always keep P constant for simplicity).

Chapter 9 - 5 Unary Systems

Triple point Chapter 9 - 6 Phase Equilibria

Simple solution system (e.g., Ni-Cu solution)

Crystal electroneg r (nm) Structure Ni FCC 1.9 0.1246 Cu FCC 1.8 0.1278

• Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii ( W. Hume – Rothery rules ) suggesting high mutual solubility. • Ni and Cu are totally miscible in all proportions.

Chapter 9 - 7 Unary Systems

Single component system Consider 2 metals: Cu has melting T = 1085 oC Ni has melting T = 1453 oC (at standard P = 1 atm) T T

liquid

liquid 1453 oC

1085 oC solid solid

Cu Ni Chapter 9 - 8 What happens when Cu and Ni are mixed? Binary Isomorphous Systems

2 components Complete liquid and solid solubility

Expect T m of solution to lie in between T m of two pure components

T T For a pure liquid component, complete melting occurs before T liquid L 1453 oC increases (sharp phase transition). But for 1085 oC multicomponent solid systems, there is S solid usually a coexistence of L 0 100 and S. Cu wt% Ni Ni Chapter 9 - 9 Phase Diagrams

• Indicate phases as function of T, Co, and P. • For this course: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used). T(°C) 1600 • 2 phases: • Phase L (liquid) Diagram 1500 for Cu-Ni L (liquid) α (FCC solid solution) system 1400 • 3 phase fields: s du L ui α 1300 liq + us L + α L id ol 1200 s α α Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase 1100 (FCC solid Diagrams of Binary Nickel Alloys , P. Nash (Ed.), ASM International, Materials Park, solution) OH (1991). 1000 0 20 40 60 80 100 wt% Ni Chapter 9 - 10 Phase Diagrams : # and types of phases

• Rule 1: If we know T and Co, then we know: --the # and types of phases present. T(°C) • Examples: 1600 A(1100°C, 60): L (liquid) 1500 1 phase: α s Cu-Ni du ui iq phase B(1250°C, 35): 1400 l us lid 2 phases: L + α so diagram

(1250°C,35) α 1300 + α B L (FCC solid 1200 solution)

Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase 1100 A(1100°C,60) Diagrams of Binary Nickel Alloys , P. Nash (Ed.), ASM International, Materials Park, OH, 1991). 1000 0 20 40 60 80 100 wt% Ni

Chapter 9 - 11 Phase Diagrams : composition of phases • Rule 2: If we know T and Co, then we know: --the composition of each phase. Cu-Ni system • Examples: T(°C) A C = 35 wt% Ni T tie line s o A idu iqu At TA = 1320°C: 1300 L (liquid) l + α Only Liquid ( L) L B us T olid CL = C o ( = 35 wt% Ni) B s α At TD = 1190°C: + α 1200 L D (solid) Only Solid ( α) TD Cα = Co ( = 35 wt% Ni ) 20 3032 35 4043 50 At TB = 1250°C: CLCo Cα wt% Ni

Both α and L Adapted from Fig. 9.3(b), Callister 7e. (Fig. 9.3(b) is adapted from Phase Diagrams CL = Cliquidus ( = 32 wt% Ni here) of Binary Nickel Alloys , P. Nash (Ed.), ASM Cα = Csolidus ( = 43 wt% Ni here) International, Materials Park, OH, 1991.)

Chapter 9 - 12 Determining phase composition in 2-phase region: 1. Draw the tie line. 2. Note where the tie line intersects the liquidus and solidus lines (i.e. where the tie line crosses the phase boundaries). 3. Read off the composition at the boundaries:

Liquid is composed of CL amount of Ni (31.5 wt% Ni).

Solid is composed of Cααα amount of Ni (42.5 wt% Ni).

Chapter 9 - 13 Phase Diagrams : weight fractions of phases

• Rule 3: If we know T and Co, then we know: --the amount of each phase (given in wt%). Cu-Ni • Examples: T(°C) system Co = 35 wt% Ni T A A tie line dus qui At TA: Only Liquid (L) 1300 L (liquid) li α W = 100 wt%, W = 0 L + L α B s idu At TD: Only Solid ( α) TB R S sol α WL = 0, Wα = 100 wt% + α 1200 L D (solid) At TB: Both α and L TD S 43 − 35 20 30 40 50 W = = = 73 wt % 32 35 43 L R +S 43 − 32 CLCo Cα wt% Ni

Adapted from Fig. 9.3(b), Callister 7e. R (Fig. 9.3(b) is adapted from Phase Diagrams of Wα = = 27 wt% Binary Nickel Alloys , P. Nash (Ed.), ASM R +S International, Materials Park, OH, 1991.) Chapter 9 - 14 Lever Rule: Derivation

Since we have only 2 phases: W W (1) L + α =1 Conservation of mass requires that: Amount of Ni in α-phase + amount of Ni in liquid phase = total amount of Ni or W C W C C (2) α α + L L = o

From 1 st condition, we have: W W α =1− L W C W C C Sub-in to (2): 1( − L ) α + L L = o

W W Solving for L and α gives : C − C C − C W = α o W = o L L C C α C C α − L α − L Chapter 9 - 15 The Lever Rule

• Tie line – connects the phases in equilibrium with each other - essentially an isotherm T(°C) How much of each phase? tie line dus qui Think of it as a lever (teeter-totter) 1300 L (liquid) li + α M M L L α B s idu TB sol α + α 1200 L (solid) R S R S

20 30CL C 40C 50 o ααα M α ⋅S = M L ⋅R wt% Ni Adapted from Fig. 9.3(b), Callister 7e.

ML S Cα −C0 R C0 −CL WL = = = Wα = = ML + Mα R + S Cα −CL R + S Cα −CL

Chapter 9 - 16 Ex: Cooling in a Cu-Ni Binary

• Phase diagram: T(°C) L (liquid) L: 35wt%Ni Cu-Ni system. Cu-Ni system • System is: 1300 α A + -- binary L: 35 wt% Ni L α: 46 wt% Ni B i.e. , 2 components: 35 46 32 C Cu and Ni. 43 -- isomorphous D 24 36 L: 32 wt% Ni i.e., complete α α: 43 wt% Ni 1200 + solubility of one L E component in L: 24 wt% Ni another; α phase α α: 36 wt% Ni field extends from (solid) 0 to 100 wt% Ni.

• Consider 1100 Co = 35 wt%Ni . 20 3035 40 50 Adapted from Fig. 9.4, Co wt% Ni Callister 7e. Chapter 9 - 17 Microstructures in Isomorphous Alloys

Microstructures will vary on the cooling rate (i.e. processing conditions)

1. Equilibrium Cooling: Very slow cooling to allow phase equilibrium to be maintained during the cooling process. a (T>1260 oC): start as homogeneous liquid solution. b (T ~ 1260 oC): liquidus line reached. α phase begins to nucleate. Cα = 46 wt% Ni; C L = 35 wt% Ni c (T= 1250 oC): calculate composition and mass fraction of each phase.

Cα = 43 wt% Ni; CL = 32 wt% Ni d (T~ 1220 oC): solidus line reached. Nearly complete solidification.

Cα = 35 wt% Ni; C L = 24 wt% Ni e (T<1220 oC): homogeneous solid solution with 35 wt% Ni. Chapter 9 - 18 Non-equilibrium cooling

Since diffusion rate is especially low in solids, consider case where:

Cooling rate >> diffusion rate in solid Cooling rate << diffusion rate in liquid (equilibrium maintained in liquids phase)

Chapter 9 - 19 Non-equilibrium cooling a’ (T>1260 oC): start as homogeneous liquid solution. b’ (T ~ 1260 oC): liquidus line reached. α phase begins to nucleate. Cα = 46 wt% Ni; C L = 35 wt% Ni c’ (T= 1250 oC): solids that formed at pt b’ remain with same composition (46wt%) and new solids with 42 wt% Ni form around the existing solids (Why around them?). d’ (T~ 1220 oC): solidus line reached. Nearly complete solidification. •Previously solidified regions maintain original composition and further solidification occurs at 35 wt% Ni. e (T<1220 oC): Non-equilibrium solidification complete (with phase segregation).

Chapter 9 - 20 Cored vs Equilibrium Phases

• Cα changes as we solidify. • Cu-Ni case: First α to solidify has Cα = 46 wt% Ni. Last α to solidify has Cα = 35 wt% Ni. • Fast rate of cooling: • Slow rate of cooling: Cored structure Equilibrium structure Uniform C α: First α to solidify: 46 wt% Ni 35 wt% Ni Last α to solidify: < 35 wt% Ni

Cored structure can be eliminated by a homogenization heat treatment at a temperature below the solidus point for the particular alloy composition. Chapter 9 - 21 Mechanical Properties: Cu-Ni System • Effect of solid solution strengthening on: --Tensile strength ( TS ) --Ductility (% EL ,% AR ) At RT

) 60 %EL for pure Cu EL 400 50 %EL for TS for pure Ni pure Ni 40 300 TS for pure Cu 30

200 (% Elongation 20 0 20 40 60 80 100 0 20 40 60 80 100 Cu Tensile Strength (MPa) TensileStrength Ni Cu Ni Composition, wt% Ni Composition, wt% Ni Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e. --Peak as a function of Co --Min. as a function of Co

Chapter 9 - 22 Binary-Eutectic Systems has a special composition 2 components with a min. melting T. Cu-Ag T(°C) system Ex.: Cu-Ag system 1200 • 3 single phase regions L (liquid) (L, α, β ) 1000 • Limited solubility: α L + α 800 779°C L+β β α: mostly Cu TE 8.0 71.9 91.2 β: mostly Ag 600 • TE : No liquid below TE α + β 400 • CE : Min. melting TE composition 200 0 20 40 60 CE 80 100 • Eutectic transition Co , wt% Ag L(C ) α(C ) + β(C ) Adapted from Fig. 9.7, E αE βE Callister 7e.

Chapter 9 - 23 Eutectic Point

Eutectic point: Where 2 liquidus lines meet (pt. E). Sometimes also referred to as invariant point.

Eutectic Reaction: cool

L(C E) α(C αE) + β(C βE) heat similar to one component (pure) system except 2 solid phases.

Eutectic Isotherm: Horizontal solidus Cu-Ag phase diagram line at T = T E.

Single phase regions : α, β, L-phase 2-Phase coexistence regions: α+β, α+L and β+L Chapter 9 - 24 EX: Pb-Sn Eutectic System (1) • For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present: α + β Pb-Sn --compositions of phases: T(°C) system

CO = 40 wt% Sn 300 Cα = 11 wt% Sn L (liquid) Cβ = 99 wt% Sn --the relative amount L+α 200 α 183°C L+β β of each phase: 18.3 61.9 97.8 S Cβ - CO 150 Wα= = R S R+S C - C 100 β α α + β 99 - 40 59 = = = 67 wt% 99 - 11 88 0 11 2040 60 80 99 100 R CO -Cα Wβ = = Cα Co Cβ R+S Cβ -Cα C, wt% Sn Adapted from Fig. 9.8, 40 - 11 29 Callister 7e. = = = 33 wt% 99 - 11 88 Chapter 9 - 25 EX: Pb-Sn Eutectic System (2) • For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find... --the phases present: α + L Pb-Sn --compositions of phases: T(°C) system

CO = 40 wt% Sn 300 Cα = 17 wt% Sn L (liquid) CL = 46 wt% Sn L+α --the relative amount 220 200 α R S L+β β of each phase: 183°C CL -CO 46 - 40 W = = α C -C 46 - 17 100 L α α + β 6 = = 21 wt% 29 0 17 2040 46 60 80 100 Cα Co CL CO -Cα 23 C, wt% Sn WL = = = 79 wt% Adapted from Fig. 9.8, CL -Cα 29 Callister 7e.

Chapter 9 - 26 Microstructures in Eutectic Systems: I

Adapted from Fig. 9.11, L: C wt% Sn Callister 7e. • Co < 2 wt% Sn T(°C) o 400 L α 300 L 1. One component rich composition. L+ α a: start with homogeneous liquid. α 200 (Pb-Sn b: -phase solids with liquid. α: C wt% Sn α TE o Compositions and mass fractions System) can be found via tie lines and lever 100 rule. α+ β c: α-phase solid solution only.

Net result: polycrystalline α solid. 0 10 20 30 Co C , wt% Sn 2 o Cooling at this composition is similar (room T solubility limit) to binary isomorphous systems. Chapter 9 - 27 Microstructures

Adapted from Fig. 9.12, in Eutectic Systems: II Callister 7e.

• 2 wt% Sn < Co < 18.3 wt% Sn L: C wt% Sn T(°C) o 400 L 2. One-component rich but cooling to α + L βββ coexistence. 300 α d: homogeneous liquid. L + α α: C o wt% Sn e: α + L phase (same as previous but at α different compositions and mass fractions). 200 TE f: all α-phase solid solution. α g: α + β phase (passing through solvus line β leads to exceeding solubility limit and β 100 phase precipitates out). α+ β Pb-Sn Net result: polycrystalline α-solid with fine system β crystals. 0 10 20 30 C 2 o Co , wt% Sn (sol. limit at Troom ) 18.3 (sol. limit at TE) Chapter 9 - 28 Microstructures in Eutectic Systems: III • Co = CE • Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of α and β crystals. T(°C) Micrograph of Pb-Sn eutectic L: Co wt% Sn microstructure 300 L Pb-Sn L+α system α 200 183°C L + β β TE

100 160 µm β: 97.8 wt% Sn α + β Adapted from Fig. 9.14, Callister 7e. α: 18.3 wt%Sn

0 2040 60 80 100 97.8 18.3 CE Adapted from Fig. 9.13, 61.9 C, wt% Sn Callister 7e. Chapter 9 - 29 α Lamellar Eutectic Structure β

Sn α Pb rich L β Sn rich Pb Adapted from Figs. 9.14 & 9.15, Callister 7e. In order to achieve large Lamellar structure forms because homogeneous regions, long relatively short diffusion lengths Chapter 9 - 30 diffusion lengths are required. are required. Microstructures in Eutectic Systems: IV

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

• Just above TE : T(°C) L: Co wt% Sn α L C = 18.3 wt% Sn L α α C = 61.9 wt% Sn 300 L L Pb-Sn S L+α Wα= = 50 wt% system R + S α 200 R S L+β β WL = (1- Wα) = 50 wt% T E R S • Just below TE : 100 α+β Cα = 18.3 wt% Sn primary α Cβ = 97.8 wt% Sn eutectic α S eutectic β Wα= = 73 wt% 0 2040 60 80 100 R + S 18.3 61.9 97.8 Wβ = 27 wt% Adapted from Fig. 9.16, Callister 7e. Co, wt% Sn Chapter 9 - 31 Hypo eutectic & Hyper eutectic (Pb-Sn Microconstituent : an System) 300 element of a L microstructure with Adapted from Fig. 9.8, T(°C) Callister 7e. (Fig. 9.8 L+α identifiable and 200 α adapted from Binary Phase L+β β TE characteristic structure (at Diagrams , 2nd ed., Vol. 3, T.B. Massalski (Editor-in- α + β pt. m there are 2 Chief), ASM International, 100 microconstituents: primary Materials Park, OH, 1990.) α and eutectic structures)

0 2040 60 80 100 Co, wt% Sn eutectic hypoeutectic: Co = 50 wt% Sn 61.9 hypereutectic: (illustration only) (Figs. 9.14 and 9.17 from Metals Handbook , 9th ed., eutectic: Co =61.9 wt% Sn Vol. 9, α β Metallography and α β Microstructures , α α β β American Society for α β Metals, Materials Park, OH, 1985.) α β 175 µm 160 µm Adapted from eutectic micro-constituent Adapted from Fig. 9.17, Fig. 9.17, Callister 7e. Adapted from Fig. 9.14, Callister 7e. (Illustration only) Callister 7e. Chapter 9 - 32 Intermediate phases

Intermediate solid solutions (intermediate phases): Solid solutions that do not extend to pure components in the phase diagram. Cu-Zn

Terminal solid solutions: α and η.

Intermediate solid solutions: β, γ, δ and ε.

Tie lines and lever rule can be used to determine compositions and wt% of phases. e.g. at 800 oC with 70 wt% Zn

CL = 78 wt% Zn Cγ = 67 wt% Zn

Chapter 9 - 33 Intermetallic Compounds

Adapted from Fig. 9.20, Callister 7e.

Mg 2Pb

Note: intermetallic compound forms a discrete line - not an area - because stoichiometry (i.e. composition) is Chapterexact. 9 - 34 Eutectoid & Peritectic

• Eutectic - liquid in equilibrium with two solids L cool α + β heat • Eutectoid - solid phase in equation with two solid phases intermetallic compound S2 S1+S3 - cementite cool γ α + Fe 3C (727ºC) heat • Peritectic - liquid + solid 1 solid 2 (Fig 9.21)

S1 + LS2 δ + L cool γ (1493ºC) heat

Chapter 9 - 35 Eutectoid & Peritectic Peritectic transition γ + L δ Cu-Zn Phase diagram

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

Chapter 9 - 36 Congruent phase transformation

Congruent transformation: no change in composition upon phase transformation. (for example: allotropic transformation) Incongruent transformation: phase transformation where at least one of the phases go through composition change. (for example: isomorphous, eutectic and eutectoid system)

Chapter 9 - 37 Ceramic phase diagrams

Al 2O3-Cr 2O3 MgO-Al 2O3

Chapter 9 - 38 Gibbs Phase Rule A criterion for the number of phases that will coexist within a system at equilibrium Number of non- Number of compositional variables phases present P + F = C + N (Temperature & Pressure)

Degree of freedom (externally controllable Number of parameters: i.e. T, P, and C) components

e.g. Cu-Ag phase diagram

Cu and Ag are the only components -> C = 2

Temperature is the only non-compositional variable here (i.e. fixed pressure). -> N = 1 (but in general N = 2)

When 2 phases are present -> P = 2 which leads to F = C+N-P = 2+1-2 = 1

When only 1 phase is present. -> P = 1 which leads to F = 2

What does this mean? Why should you care? Chapter 9 - 39 Gibbs Phase Rule

In the previous example of Cu-Ag phase diagram, when F = 1, only one parameter (T or C) needs to be specified to completely define the system.

e.g. (for α+L region) If T is specified to be 1000 oC, compositions are already determined

(C α and C L).

If composition of the a phase is

specified to be C α then both T and C L are already determined.

The nature of the phases is important, not the relative phase amounts. C α CL Chapter 9 - 40 Gibbs Phase Rule

When F = 2, both T and C have to be specified to completely define the state of the system.

e.g.(for α region) o If T is specified to be 800 C, C α can be any where between 0 to ~8 wt% Ag)

If composition of the a phase is

specified to be C α = 3 wt%, then T and can be any where between ~600 to 1100 oC.

Cα

Chapter 9 - 41 Gibbs Phase Rule

Where in the Cu—Ag diagram, is there a 0 degree of freedom? (i.e. T, P, and C are all fixed)---eutectic isotherm F=3-P=0

Chapter 9 - 42 Concept Check

• Question: For a ternary system, three components are present; temperature is also a variable. What is the maximum number of phases that may be present for a ternary system, assuming that pressure is held constant?

Chapter 9 - 43 Iron-Carbon System structural material

Ferrite Magnets

TVR Tuscan Speed 6, high-performance sports The world's first bridge made of iron car with an austempered ductile iron crankshaft. in 1779. The entire structure is made of cast iron. (near Broseley, Chapter 9 - 44 UK) Iron-Carbon System Steel bridges

Millau Viaduct in France, the The Akashi Kaikyo bridge, a 3-span 2- highest bridge in the world. hinged truss-stiffened suspension bridge. completed in 1998. It connects Kobe with Awaji Island. It is the world's longest Golden Gate Bridge suspension bridge, with a span between the towers of 1.9 km.

Chapter 9 - 45 Iron-Carbon System

• Iron-carbon (Fe-C) system is one of the most important binary systems due to the versatile uses of the iron-based structural alloys. • This phase diagram is so important in understanding the equilibrium structure, and in the design of heat treatment process of iron alloys. • The most important part of the phase diagram is the region below6.7 w% carbon. All practical iron-carbon alloys contain C below 6.7 w%. This part is of the phase diagram is thus the most analyzed part of the iron carbon phase diagram.

Chapter 9 - 46 Iron-Carbon System

Typical metal (e.g. Cu) Iron T( oC) T Liquid Liquid 1538 Ferrite (BCC) 1394 Tm Austenite (FCC)

Solid 912 Ferrite (BCC)

Chapter 9 - 47 Iron-Carbon System Iron -Iron Carbide Diagram

Chapter 9 - 48 Remarks on Fe-C system

• C is an interstitial element in Fe matrix.

• C has limited solid solubility in the alpha BCC phase (narrow region close to pure iron). Max solid solubility of C in alpha iron is 0.022 w% at 727 °C.

• Alpha iron an be made magnetic below 768°C.

• Austenite phase is not stable below at 727 °C. Max solid solubility of C into austenite is 2.14 w% at 1147°C, much larger that that in alpha phase.

• Austenite is a non-magnetic phase, and heat treatment of Fe-C alloys involving austenite is so important.

• Cementite phase (intermetallic, Fe 3C) forms over a large region of the Fe-C phase diagram, but it is a metastable phase (heating above 650 °C for years decomposes this phase into alpha iron and graphite). Cementite is very hard and brittle.

• δ ferrite is stable only at relatively high T, it is of no technological importance and is not discussed further.

Chapter 9 - 49 Ferrite (90x) Austenite (x325) Relatively soft 3 Density: 7.88g/cm Chapter 9 - 50 Classification Scheme of Ferrous Alloys

• Iron Pure ion contains less than 0.008wt% C • Steel 0.008-2.14 wt% C, in practice,<1.0 wt% C • Cast Iron 2.14-6.70 wt% C, in practice, <4.5 wt% C

Chapter 9 - 51 Iron-Carbon (Fe-C) Phase Diagram Eutectoid cooling: cool γ (0.76 wt% C) α (0.022 wt% C) + Fe C (6.7 wt% C) • 2 important T(°C) heat 3 1600 points δ -Eutectic ( A): 1400 L L ⇒ γ +Fe C 3 γ γ+L 1200 A L+Fe C -Eutectoid ( B): (austenite) 1148°C 3 R S ⇒ γ α +Fe 3C 1000 γ γ γ+Fe 3C α γ γ + 800α γ B 727°C = T eutectoid

R S C(cementite) 3 600 α+Fe C

3 Fe 400 0 1 2 3 4 5 66.7 (Fe) 0.76 4.30 120 µm Co, wt% C Result: Pearlite = Fe 3C (cementite-hard) alternating layers of α (ferrite-soft) α and Fe 3C phases eutectoid (Adapted from Fig. 9.27, Callister 7e .) Adapted from Fig. 9.24, Callister 7e . C Chapter 9 - 52 Pearlite

colonies Pearl-microscope picture

(X500) Mechanically, pearlite has properties intermediate between the Chapter 9 - soft, ductile ferrite and the hard, brittle cementite. 53 Hypo eutectoid Steel T(°C) 1600 δ 1400 L (Fe-C γ γ γ γ+L γ γ 1200 1148°C L+Fe C System) (austenite) 3 γ γ 1000 γ + Fe C Adapted from Figs. 9.24 γ γ 3 and 9.29, Callister 7e . (Fig. 9.24 adapted from α 800 αγ γ r s 727°C Binary Alloy Phase α Diagrams , 2nd ed., Vol. γ γ R S C(cementite) α 3 1, T.B. Massalski (Ed.-in- wα =s/( r+s) 600 α+ Fe C Chief), ASM International, 3 Fe wγ =(1- wα) Materials Park, OH, 400 1990.) 0 1 2 3 4 5 66.7 α (Fe) C0 Co, wt% C

pearlite 0.76 wpearlite = wγ Hypoeutectoid 100 µm wα =S/( R+S) steel w =(1-w ) Fe 3C α pearlite proeutectoid ferrite Adapted from Fig. 9.30, Callister 7e . Chapter 9 - 54 Hyper eutectoid Steel T(°C) 1600 δ 1400 L (Fe-C System) γ γ γ γ+L 1200 1148°C L+Fe C γ γ (austenite) 3 γ γ 1000 γ +Fe C Adapted from Figs. 9.24 γ γ 3 and 9.32, Callister 7e . Fe C (Fig. 9.24 adapted from 3 800 r s γ γ Binary Alloy Phase Diagrams , 2nd ed., Vol. γ γ C(cementite) α R S 3 1, T.B. Massalski (Ed.-in- 600 Chief), ASM International, wFe C =r/( r +s) α +Fe C 3 3 Fe Materials Park, OH, wγ =(1-w Fe C) 3 400 1990.) 0 1Co 2 3 4 5 66.7 (Fe) C , wt%C

pearlite 0.76 o w pearlite = wγ wα =S/( R+S) Hypereutectoid w =(1-w α) 60 µm Fe 3C steel

pearlite proeutectoid Fe 3C Adapted from Fig. 9.33, Callister 7e . Chapter 9 - 55 Example: Phase Equilibria

For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following a) composition of Fe 3C and ferrite ( α) b) the amount of carbide (cementite) in grams that forms per 100 g of steel c) the amount of pearlite and proeutectoid ferrite ( α)

Chapter 9 - 56 Chapter 9 – Phase Equilibria

Solution: a) composition of Fe 3C and ferrite ( α)

b) the amount of carbide CO = 0.40 wt% C (cementite) in grams that Cα = 0.022 wt% C C = 6.70 wt% C forms per 100 g of steel Fe3 C 1600 δ Fe 3C Co −Cα 1400 L = x100 T(°C) Fe C + α C −C γ+L 3 Fe 3C α 1200 γ L+Fe C (austenite) 1148°C 3 4.0 − .0 022 = x 100 = 7.5 g 1000 7.6 − .0 022 γ + Fe 3C 800 727°C

R S C(cementite) Fe C = 5.7 g 3 3 600 α + Fe C

3 Fe α = 94 3. g 400 0 1 2 3 4 5 66.7 C C C ααα O Co, wt% C Fe3 C

Chapter 9 - 57 Chapter 9 – Phase Equilibria c. the amount of pearlite and proeutectoid ferrite ( α)

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

Co = 0.40 wt% C C = 0.022 wt% C α 1600 Cpearlite = Cγ = 0.76 wt% C δ 1400 L T(°C) γ Co −Cα γ γ+L = x 100 = 51.2 g 1200 1148°C L+Fe C γ + α C −C (austenite) 3 γ α 1000 γ + Fe 3C

800 727°C

R S C(cementite) 3 pearlite = 51.2 g 600 α + Fe C

3 Fe proeutectoid α = 48.8 g 400 C 0 C1 2 3 4 5 66.7 α C γγγ αα O Co, wt% C

Chapter 9 - 58 Effect of alloying elements on Fe-C phase diagram

Chapter 9 - 59 Alloying Steel with More Elements

• Teutectoid changes: • Ceutectoid changes:

Ti Si Mo

(wt%C) Ni

(°C) W Cr

Cr Si Mn W eutectoid Eutectoid Mn Ti Mo T Ni C

wt. % of alloying elements wt. % of alloying elements Adapted from Fig. 9.34, Callister 7e . (Fig. 9.34 Adapted from Fig. 9.35, Callister 7e . (Fig. 9.35 from Edgar C. Bain, Functions of the Alloying from Edgar C. Bain, Functions of the Alloying Elements in Steel , American Society for Metals, Elements in Steel , American Society for Metals, 1939, p. 127.) 1939, p. 127.)

Chapter 9 - 60 Summary

• Phase diagrams are useful tools to determine: --the number and types of phases, --the wt% of each phase, --and the composition of each phase for a given T and composition of the system. • Alloying to produce a solid solution usually --increases the tensile strength ( TS ) --decreases the ductility. • Binary eutectics and binary eutectoids allow for a range of microstructures.

Chapter 9 - 61