sign in sign up
<<
Home , Alloy, Eutectic system

Chapter 11: Diagrams ISSUES TO ADDRESS... • When we combine two elements... what is the resulting equilibrium state? • In particular, if we specify... -- the composition (e.g., wt% Cu - wt% Ni), and -- the (T) then... How many phases form? What is the composition of each phase? What is the amount of each phase?

Phase A Phase B

Nickel atom AMSE 205 Spring ‘2016 Chapter 11 - 1 Phase Equilibria: Limit

• – , , or gas , single phase • – more than one phase Adapted from Fig. 11.1, Callister & Rethwisch 9e. Sugar/ • Solubility Limit: 100 Maximum for

) Solubility C

which only a single phase o 80 Limit L solution exists. (liquid) 60 L + Question: What is the 40 (liquid solution S solubility limit for sugar in i.e., syrup) (solid o 20 water at 20 C ? ( Temperature sugar) Answer: 65 wt% sugar. 0 20 At 20°C, if C < 65 wt% sugar: syrup 40 6065 80 100 C = Composition (wt% sugar) At 20°C, if C > 65 wt% sugar: Sugar syrup + sugar Water AMSE 205 Spring ‘2016 Chapter 11 - 2 Components and Phases • Components: The elements or compounds which are present in the (e.g., Al and Cu) • Phases: The physically and chemically distinct regions that form (e.g., α and β).

Aluminum- β (lighter Copper phase) Alloy α (darker Adapted from chapter- phase) opening photograph, Chapter 9, Callister, Science & : An Introduction, 3e.

AMSE 205 Spring ‘2016 Chapter 11 - 3 Effect of Temperature & Composition • Altering T can change # of phases: path A to B. • Altering C can change # of phases: path B to D. B (100 oC, C = 70) D (100 oC, C = 90) 1 phase 2 phases 100 L ) 80

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

Fig. 11.1, Callister & 0 Rethwisch 9e. 0 20 40 60 70 80 100 C = Composition (wt% sugar)

AMSE 205 Spring ‘2016 Chapter 11 - 4 Criteria for Solid Solubility

Simple 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 (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility. • Ni and Cu are totally soluble in one another for all proportions.

AMSE 205 Spring ‘2016 Chapter 11 - 5 Phase Diagrams • Indicate phases as a function of T, C, and P. • For this course: - binary systems: just 2 components. - independent variables: T and C (P = 1 atm is almost always used). T(oC) Phase 1600 • 2 phases: Diagram L (liquid) 1500 for Cu-Ni L (liquid) α (FCC ) system 1400 • 3 different phase fields: L 1300 L + α 1200 α α Fig. 11.3(a), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary (FCC solid Alloys, P. Nash, Editor, 1991. Reprinted 1100 by permission of ASM International, Materials solution) Park, OH.) 1000 0 20 40 60 80 100 wt% Ni AMSE 205 Spring ‘2016 Chapter 11 - 6 Isomorphous Binary Phase Diagram

• Phase diagram: T(oC) Cu-Ni system. 1600 • System is: 1500 L (liquid) Cu-Ni -- binary 1400 phase i.e., 2 components: diagram Cu and Ni. -- isomorphous 1300 i.e., complete 1200 α solubility of one component in 1100 (FCC solid another; α phase solution) field extends from 1000 0 20 40 60 80 100 wt% Ni 0 to 100 wt% Ni.

Fig. 11.3(a), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)

AMSE 205 Spring ‘2016 Chapter 11 - 7 Phase Diagrams: Determination of phase(s) present

• Rule 1: If we know T and Co, then we know: -- which phase(s) is (are) present. T(oC) • Examples: 1600 A(1100 oC, 60 wt% Ni): L (liquid) 1 phase: α 1500 Cu-Ni B(1250 oC, 35 wt% Ni): 1400 phase 2 phases: L + α diagram 1300 (1250ºC,35) α B (FCC solid 1200 Fig. 11.3(a), Callister & Rethwisch 9e. solution) (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted 1100 A(1100ºC,60) by permission of ASM International, Materials Park, OH.) 1000 0 20 40 60 80 100 wt% Ni

AMSE 205 Spring ‘2016 Chapter 11 - 8 Phase Diagrams: Determination of phase compositions

• Rule 2: If we know T and C0, then we can determine: -- the composition of each phase. Cu-Ni system • Examples: T(oC) A Consider C0 = 35 wt% Ni TA tie line o 1300 L (liquid) At TA = 1320 C : Only Liquid (L) present B T CL = C0 ( = 35 wt% Ni) B o α At TD = 1190 C : 1200 D (solid) Only Solid (α) present TD Cα = C0 ( = 35 wt% Ni) 20 303235 4043 50 o At TB = 1250 C : CL C0 Cα wt% Ni Both α and L present Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary CL = Cliquidus ( = 32 wt% Ni) Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Cα = Csolidus ( = 43 wt% Ni) Park, OH.)

AMSE 205 Spring ‘2016 Chapter 11 - 9 Phase Diagrams: Determination of phase weight fractions

• Rule 3: If we know T and C0, then can determine: -- the weight fraction of each phase. Cu-Ni • Examples: T(oC) system Consider C = 35 wt% Ni A 0 TA tie line At TA : Only Liquid (L) present 1300 L (liquid) WL = 1.00, W = 0 B T At TD : Only Solid (α) present B R S α WL = 0, Wα = 1.00 1200 (solid) At T : Both α and L present D B TD S 43 35 20 303235 4043 50 WL    0.73 R +S 43  32 CL C0 Cα wt% Ni Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary R Nickel Alloys, P. Nash, Editor, 1991. Reprinted W  = 0.27 by permission of ASM International, Materials α R +S Park, OH.)

AMSE 205 Spring ‘2016 Chapter 11 - 10 The

• Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm T(oC) What fraction of each phase? tie line Think of the tie line as a lever 1300 L (liquid) (teeter-totter) B T B ML Mα α 1200 (solid) R S

20 30C 40 50 R S L C0 C αAdapted from Fig. 11.3(b), wt% Ni Callister & Rethwisch 9e.

AMSE 205 Spring ‘2016 Chapter 11 - 11 Ex: Cooling of a Cu-Ni Alloy

o • Phase diagram: T( C) L (liquid) L: 35 wt%Ni Cu-Ni system. Cu-Ni system • Consider 1300 A L: 35 wt% Ni microstuctural α: 46 wt% Ni B 35 46 changes that 32 C 43 accompany the D L: 32 wt% Ni cooling of a 24 36 α: 43 wt% Ni C = 35 wt% Ni alloy 1200 E 0 L: 24 wt% Ni α α: 36 wt% Ni (solid) α: 35 wt% Ni

110 0 20 3035 40 50 wt% Ni Adapted from Fig. 11.4, C0 Callister & Rethwisch 9e. AMSE 205 Spring ‘2016 Chapter 11 - 12 Mechanical Properties: Cu-Ni System • Effect of solid solution strengthening on: -- Tensile strength (TS) -- (%EL)

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

200 Elongation (% 20 0 20406080100 020406080100

Tensile (MPa) Strength Tensile Cu Ni Cu Ni Composition, wt% Ni Composition, wt% Ni Adapted from Fig. 11.5(a), Adapted from Fig. 11.5(b), Callister & Rethwisch 9e. Callister & Rethwisch 9e.

AMSE 205 Spring ‘2016 Chapter 11 - 13 Binary-Eutectic Systems has a special composition 2 components with a min. T. Cu-Ag T(oC) 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+ββ α: mostly Cu TE 8.0 71.9 91.2 β: mostly Ag 600 • TE : No liquid below TE α β 400 • CE : Composition at temperature TE 200 0 20 40 60 CE 80 100 • Eutectic reaction C, wt% Ag Fig. 11.6, Callister & Rethwisch 9e [Adapted from Binary Alloy Phase Diagrams, 2nd edition, L(CE) α(CαE) + β(CβE) Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]. cooling

heating AMSE 205 Spring ‘2016 Chapter 11 - 14 EX 1: Pb-Sn • For a 40 wt% Sn-60 wt% Pb alloy at 150 oC, determine: -- the phases present Pb-Sn Answer: α + β T(oC) system -- the phase compositions 300 Answer: 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 Answer: 150 R S S Cβ - C0 100 Wα = = R+S Cβ - Cα α + β 99 - 40 59 = = = 0.67 99 - 11 88 0 11 2040 60 80 99100 C C C R C0 -Cα α 0 β W = = C, wt% Sn β C -C Fig. 11.7, Callister & Rethwisch 9e. R+S β α [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- 40 - 11 29 Chief), 1990. Reprinted by permission of ASM = = = 0.33 International, Materials Park, OH.] 99 - 11 88 AMSE 205 Spring ‘2016 Chapter 11 - 15 EX 2: Pb-Sn Eutectic System • For a 40 wt% Sn-60 wt% Pb alloy at 220 oC, determine: -- the phases present: Pb-Sn Answer: α + L T (oC) system -- the phase compositions 300 Answer: 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 Answer: 100 CL -C0 46 - 40 α + β Wα = = CL -Cα 46 - 17 6 17 100 = = 0.21 0 2040 46 60 80 29 Cα C0 CLC, wt% Sn Fig. 11.7, Callister & Rethwisch 9e. C0 -Cα 23 [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- WL = = = 0.79 Chief), 1990. Reprinted by permission of ASM CL -Cα 29 International, Materials Park, OH.] AMSE 205 Spring ‘2016 Chapter 11 - 16 Microstructural Developments in Eutectic Systems I

o • For alloys for which T( C) L: C0 wt% Sn 400 C0 < 2 wt% Sn L • Result: at room temperature α -- polycrystalline with grains of 300 L α phase having L+ α α composition C0 200 (Pb-Sn T α: C0 wt% Sn E System)

100 α +β

0 10 20 30 Fig. 11.10, Callister & C0 C, wt% Sn Rethwisch 9e. 2 (room T solubility limit)

AMSE 205 Spring ‘2016 Chapter 11 - 17 Microstructural Developments in Eutectic Systems II

L: C wt% Sn • For alloys for which T(oC) 0 400 2 wt% Sn < C < 18.3 wt% Sn 0 L • Result: L at temperatures in α + β range 300 α L + α -- polycrystalline with α grains α: C0 wt% Sn α and small β-phase particles 200 TE α β 100 α + β Pb-Sn system

Fig. 11.11, Callister & 0 10 20 30 Rethwisch 9e. 2 C0 C, wt% Sn (sol. limit at Troom) 18.3 (sol. limit at TE) AMSE 205 Spring ‘2016 Chapter 11 - 18 Microstructural Developments in Eutectic Systems III

• For alloy of composition C0 = CE • Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of α and β phases.

o Micrograph of Pb-Sn T( C) eutectic L: C0 wt% Sn microstructure 300 L Pb-Sn L+ α system α 200 183°C Lβ β TE

100 160μm β: 97.8 wt% Sn αβ Fig. 11.13, Callister & Rethwisch 9e. α: 18.3 wt%Sn (From Handbook, 9th edition, Vol. 9, and Microstructures, 1985. Reproduced by permission of ASM 0 2040 60 80 100 International, Materials Park, OH.) 97.8 18.3 CE Fig. 11.12, Callister & 61.9 C, wt% Sn Rethwisch 9e. AMSE 205 Spring ‘2016 Chapter 11 - 19 Lamellar Eutectic Structure

Figs. 11.13 & 11.14, Callister & Rethwisch 9e. (Fig. 11.13 from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.)

AMSE 205 Spring ‘2016 Chapter 11 - 20 Microstructural Developments in Eutectic Systems IV

• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn • Result: α phase particles and a eutectic microconstituent

o • Just above TE : T( C) L: C0 wt% Sn L α C = 18.3 wt% Sn L α 300 L α CL = 61.9 wt% Sn Pb-Sn S L+α W = = 0.50 system α R + S α 200 R S L+β β WL = (1- Wα)= 0.50 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 = = 0.73 0 2040 60 80 100 R + S 18.3 61.9 97.8 Wβ = 0.27 Fig. 11.15, Callister & Rethwisch 9e. C, wt% Sn AMSE 205 Spring ‘2016 Chapter 11 - 21 Hypoeutectic & Hypereutectic

300 L T(oC) Fig. 11.7, Callister & Rethwisch α L+α 200 L+β β 9e. [Adapted from Binary Alloy Phase (Pb-Sn TE Diagrams, 2nd edition, Vol. 3, T. B. System) Massalski (Editor-in-Chief), 1990. α + β Reprinted by permission of ASM International, Materials Park, OH.] 100

0 2040 60 80 100 C, wt% Sn eutectic hypoeutectic: C0 = 50 wt% Sn 61.9 hypereutectic: (illustration only) (Figs. 11.13 and 11.16 from Metals Handbook, 9th ed., Vol. 9, Metallography and eutectic: C0 =61.9wt% Sn Microstructures, 1985. α β Reproduced by permission of ASM International, α β Materials Park, OH.) α α β β α β α β 175 μm 160 μm Fig. 11.16, Callister & eutectic micro-constituent Adapted from Fig. 11.16, Rethwisch 9e. Fig. 11.13, Callister & Callister & Rethwisch 9e. Rethwisch 9e. (Illustration only)

AMSE 205 Spring ‘2016 Chapter 11 - 22 Intermediate Phases (Intermediate Solid Solutions)

AMSE 205 Spring ‘2016 Chapter 11 - 23 Compounds

Fig. 11.19, Callister & Rethwisch 9e. [Adapted from Phase Diagrams of Binary Alloys, A. A. Nayeb-Hashemi and J. B. Clark (Editors), 1988. Reprinted by permission of ASM International, Materials Park, OH.]

Mg2Pb

Note: intermetallic compound exists as a line on the diagram - not an area - because of (i.e. composition of a compound is a fixed value). AMSE 205 Spring ‘2016 Chapter 11 - 24 Congruent melting vs. Incongruent melting

AMSE 205 Spring ‘2016 Chapter 11 - 25 Eutectic, Eutectoid, & Peritectic

• Eutectic - liquid transforms to two solid phases L cool α + β (For Pb-Sn, 183 oC, 61.9 wt% Sn) heat • Eutectoid – one solid phase transforms to two other solid phases intermetallic compound - S2 S1+S3 γ cool α + Fe C (For Fe-C, 727 oC, 0.76 wt% C) heat 3 • Peritectic - liquid and one solid phase transform to a second solid phase

S1 + LS2 cool o δ + L heat γ (For Fe-C, 1493 C, 0.16 wt% C)

AMSE 205 Spring ‘2016 Chapter 11 - 26 Peritectoid & Monotectic

• Peritectoid – one solid and another solid react to produce a new solid α + β cool  (For Co-Cr, 976 oC, 37.6 wt% Cr) heat

• Monotectic – liquid 1 decomposes to liquid 2 and a solid cool o L1 heat L2 +  (For Cu-Pb, 955 C, 36 wt% Pb)

AMSE 205 Spring ‘2016 Chapter 11 - 27 Eutectoid & Peritectic Cu-Zn Phase diagram

Peritectic transformations + L  + L  γ + L δ  + L 

Eutectoid transformat δγ+ e

AMSE 205 Spring ‘2016 Chapter 11 - 28 Peritectic reaction

AMSE 205 Spring ‘2016 Chapter 11 - 29 Peritectoid

AMSE 205 Spring ‘2016 Chapter 11 - 30 Monotectic

AMSE 205 Spring ‘2016 Chapter 11 - 31 Gibbs Phase Rule • P+F = C+2 P: # of phase; F: # of degree of freedom; C: # of components when Pressure is fixed at 1 atm • P+F = C+1 F = 3 – P for binary system Cu-Ag o T( C) system 1200 L (liquid) 1000 α α +L 800 779°C β+L β TE 8.0 71.9 91.2 600 α β 400

200 0 20 40 60 CE 80 100 AMSE 205 Spring ‘2016 C, wt% Ag Chapter 11 - 32 - (Fe-C) Phase Diagram • 2 important T(°C) 1600 points δ - Eutectic (A): 1400 L L  γ + Fe C 3 γ γ +L 1200 1148°C A L+Fe C - Eutectoid (B): () 3 γ  α +Fe C 3 1000 γ γ γ+Fe C γ γ 3 800α B 727°C = Teutectoid C (cementite) 600 3 α+Fe3C Fe 400 0 1234566.7 120 μm (Fe) 0.76 4.30 C, wt% C Result: = alternating layers of α and Fe C phases Fe3C (cementite-hard) 3 Fig. 11.23, Callister α (-soft) Fig. 11.26, Callister & Rethwisch 9e. & Rethwisch 9e. (From Metals Handbook, Vol. 9, 9th ed., [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Metallography and Microstructures, 1985. Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted Reproduced by permission of ASM by permission of ASM International, Materials Park, OH.] International, Materials Park, OH.) AMSE 205 Spring ‘2016 Chapter 11 - 33 Hypoeutectoid 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 γ γ +Fe C Adapted from Figs. 11.23 3 and 11.28, Callister & Rethwisch 9e. α 800 αγ γ 727°C [Figure 9.24 adapted from α Binary Alloy Phase Diagrams, γ γ 2nd edition, Vol. 1, T. B.  C (cementite) 600 3 Massalski (Editor-in-Chief), α +Fe3C 1990. Reprinted by permission Fe of ASM International, Materials Park, OH.] 400 α 0 1234566.7 (Fe)C0 C, wt% C

pearlite 0.76

100 μm Hypoeutectoid steel pearlite proeutectoid ferrite

Adapted from Fig. 11.29, Callister & Rethwisch 9e. (Photomicrograph courtesy of Republic Steel Corporation.)

AMSE 205 Spring ‘2016 Chapter 11 - 34 Hypoeutectoid Steel T(°C) 1600 δ 1400 L (Fe-C γ γ+L γ α γ 1200 1148°C L+Fe C System) α (austenite) 3 γ α γ 1000 γ +Fe C Adapted from Figs. 11.23 W = s/(r + s) 3 and 11.28, Callister & α Rethwisch 9e. 800 r s 727°C [Figure 9.24 adapted from Wγ =(1 - Wα) Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. α R S C (cementite) α 600 3 Massalski (Editor-in-Chief), α +Fe3C 1990. Reprinted by permission Fe of ASM International, Materials pearlite Park, OH.] 400 0 1234566.7 (Fe)C0 C, wt% C

Wpearlite = Wγ 0.76 Hypoeutectoid Wαʼ = S/(R + S) 100 μm steel WFe C =(1 – Wα’) 3 pearlite proeutectoid ferrite

Adapted from Fig. 11.29, Callister & Rethwisch 9e. (Photomicrograph courtesy of Republic Steel Corporation.) AMSE 205 Spring ‘2016 Chapter 11 - 35 Hypereutectoid Steel T(oC) 1600 δ 1400 L (Fe-C γ γ γ γ+L 1200 1148°C L+Fe C System) γ γ (austenite) 3 γ γ 1000 γ γ +Fe C Adapted from Figs. 11.23 γ 3 and 11.31, Callister & Fe3C Rethwisch 9e. γ 800 727°C [Figure 9.24 adapted from γ Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. γ γ α C (cementite) 600 3 Massalski (Editor-in-Chief), α +Fe3C 1990. Reprinted by permission Fe of ASM International, Materials Park, OH.] 400 0 1234566.7C0 (Fe)C0 C, wt% C pearlite 0.76

60 μmHypereutectoid steel

pearlite proeutectoid Fe3C

Adapted from Fig. 11.32, Callister & Rethwisch 9e. (Copyright 1971 by United States Steel Corporation.)

AMSE 205 Spring ‘2016 Chapter 11 - 36 Hypereutectoid Steel T(oC) 1600 δ 1400 L (Fe-C Fe C γ γ+L 3 1200 1148°C L+Fe C System) γ γ (austenite) 3 γ γ 1000 γ +Fe C Adapted from Figs. 11.23 W =x/(v + x) 3 and 11.31, Callister & γ Rethwisch 9e. 800 v x W =(1-W ) 727°C [Figure 9.24 adapted from Fe3C γ Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. α C (cementite)

V X 3 600 Massalski (Editor-in-Chief), α +Fe3C 1990. Reprinted by permission pearlite Fe of ASM International, Materials Park, OH.] 400 0 1234566.7C0 (Fe)C0 C, wt% C 0.76 Wpearlite = Wγ

Wα = X/(V + X) 60 μmHypereutectoid W =(1 - W ) steel Fe3Cʼ α pearlite proeutectoid Fe3C

Adapted from Fig. 11.32, Callister & Rethwisch 9e. (Copyright 1971 by United States Steel Corporation.)

AMSE 205 Spring ‘2016 Chapter 11 - 37 Example Problem

For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following: a) The compositions of Fe3C and ferrite (α). b) The amount of cementite (in grams) that forms in 100 g of steel. c) The amounts of pearlite and proeutectoid ferrite (α) in the 100 g.

AMSE 205 Spring ‘2016 Chapter 11 - 38 Solution to Example Problem a) Using the RS tie line just below the eutectoid

C = 0.022 wt% C α Fig. 11.23, Callister & Rethwisch 9e. C = 6.70 wt% C [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. Fe3C B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

b) Using the lever rule with 1600 the tie line shown δ 1400 L T(°C) γ γ+L 1200 1148°C L+Fe C (austenite) 3 1000 γ +Fe3C

800 727°C

R S C (cementite) Amount of Fe C in 100 g 600 3 3 α +Fe3C Fe 400 = (100 g)WFe3C 0 1234566.7 C C0 C, wt% C CFe C = (100 g)(0.057) = 5.7 g α 3 AMSE 205 Spring ‘2016 Chapter 11 - 39 Solution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and realizing that Fig. 11.23, Callister & Rethwisch 9e. C0 = 0.40 wt% C [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by Cα = 0.022 wt% C permission of ASM International, Materials Park, OH.] C = C = 0.76 wt% C pearlite γ 1600 δ 1400 L T(°C) γ γ+L 1200 1148°C L+Fe C (austenite) 3 1000 γ +Fe3C

800 727°°C

V X C (cementite) Amount of pearlite in 100 g 600 3 α +Fe3C Fe 400 = (100 g)Wpearlite 0 1234566.7 C Cα C γ C, wt% C = (100 g)(0.512) = 51.2 g 0 AMSE 205 Spring ‘2016 Chapter 11 - 40 Alloying with Other Elements

• Teutectoid changes: • Ceutectoid changes:

Ti Si Mo Ni (ºC)

W (wt% C) Cr

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

wt. % of alloying elements wt. % of alloying elements

Fig. 11.33, Callister & Rethwisch 9e. Fig. 11.34,Callister & Rethwisch 9e. (From Edgar C. Bain, Functions of the Alloying Elements (From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.) International, Materials Park, OH.)

AMSE 205 Spring ‘2016 Chapter 11 - 41 Summary

• Phase diagrams are useful tools to determine: -- the number and types of phases present, -- the composition of each phase, -- and the weight fraction of each phase given the temperature and composition of the system. • The microstructure of an alloy depends on -- its composition, and -- whether or not cooling rate allows for maintenance of equilibrium. • Important phase diagram phase transformations include eutectic, eutectoid, and peritectic.

AMSE 205 Spring ‘2016 Chapter 11 - 42