Chapter 11: Phase 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 temperature (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 Copper atom AMSE 205 Spring 2016 Chapter 11-1
Phase Equilibria: Solubility imit Solution solid, liquid, or gas solutions, single phase Mixture more than one phase Solubility imit: Maximum concentration for which only a single phase solution exists. Question: What is the solubility limit for sugar in water at 20 o C? Temperature ( o C) Answer: 65 wt% sugar. At 20 C, if C < 65 wt% sugar: syrup At 20 C, if C > 65 wt% sugar: syrup + sugar 100 80 60 40 20 Sugar/Water Phase Diagram Solubility imit (liquid solution i.e., syrup) (liquid) + S (solid sugar) 0 20 40 6065 80 100 C = Composition (wt% sugar) Water Adapted from Fig. 11.1, Callister & Rethwisch 9e. Sugar AMSE 205 Spring 2016 Chapter 11-2
Components and Phases Components: The elements or compounds which are present in the alloy (e.g., Al and Cu) Phases: The physically and chemically distinct material regions that form (e.g., and β). Aluminum- Copper Alloy Adapted from chapteropening photograph, Chapter 9, Callister, Materials Science & Engineering: An Introduction, 3e. β (lighter phase) (darker phase) 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. watersugar system Fig. 11.1, Callister & Rethwisch 9e. Temperature ( o C) 100 80 60 40 20 0 0 (liquid solution i.e., syrup) B (100 o C, C = 70) 1 phase (liquid) + S (solid sugar) A (20 o C, C = 70) 2 phases 20 40 60 70 80 100 C = Composition (wt% sugar) D (100 o C, C = 90) 2 phases 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 crystal structure (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( o C) Phase Diagram for Cu-Ni system 1600 1500 1400 1300 1200 1100 1000 0 (liquid) (FCC solid solution) 20 40 60 80 100 2 phases: (liquid) (FCC solid solution) 3 different phase fields: + 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.) wt% Ni AMSE 205 Spring 2016 Chapter 11-6
Isomorphous Binary Phase Diagram Phase diagram: Cu-Ni system. System is: -- binary i.e., 2 components: Cu and Ni. -- isomorphous i.e., complete solubility of one component in another; phase field extends from 0 to 100 wt% Ni. T( o C) 1600 1500 1400 1300 1200 1100 1000 0 (liquid) (FCC solid solution) 20 40 60 80 100 Cu-Ni phase diagram 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 C o, then we know: -- which phase(s) is (are) present. Examples: A(1100 o C, 60 wt% Ni): 1 phase: B(1250 o C, 35 wt% Ni): 2 phases: + 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.) T( o C) 1600 1500 1400 1300 1200 1100 1000 0 (liquid) B (1250ºC,35) (FCC solid solution) A(1100ºC,60) 20 40 60 80 100 Cu-Ni phase diagram wt% Ni AMSE 205 Spring 2016 Chapter 11-8
Phase Diagrams: Determination of phase compositions Rule 2: If we know T and C 0, then we can determine: -- the composition of each phase. Examples: Consider C 0 = 35 wt% Ni At T A = 1320 o C : Only iquid () present C = C 0 ( = 35 wt% Ni) At T D = 1190 o C : Only Solid () present C = C 0 ( = 35 wt% Ni) At T B = 1250 o C : Both and present C = Cliquidus ( = 32 wt% Ni) C = Csolidus ( = 43 wt% Ni) T( o C) T A 1300 T B 1200 T D 20 (liquid) Cu-Ni system A B D tie line (solid) 303235 4043 50 C C0 C Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) wt% Ni AMSE 205 Spring 2016 Chapter 11-9
Phase Diagrams: Determination of phase weight fractions Rule 3: If we know T and C 0, then can determine: -- the weight fraction of each phase. Examples: Consider C 0 = 35 wt% Ni At T A : Only iquid () present W = 1.00, W = 0 At T D : Only Solid ( ) present W = 0, W = 1.00 At T B : W W Both and present S R +S R R +S 43 43 = 0.27 35 32 0.73 T( o C) T A 1300 T B 1200 T D 20 (liquid) Cu-Ni system (solid) 30 40 50 CC0 wt% Ni AMSE 205 Spring 2016 Chapter 11-10 A B R S D 3235 tie line 4 C 3 Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)
The ever Rule Tie line connects the phases in equilibrium with each other also sometimes called an isotherm T( o C) 1300 TB 1200 (liquid) R B tie line S (solid) What fraction of each phase? Think of the tie line as a lever (teeter-totter) M M 20 30C 40 50 C 0 C wt% Ni Adapted from Fig. 11.3(b), Callister & Rethwisch 9e. R S AMSE 205 Spring 2016 Chapter 11-11
Ex: Cooling of a Cu-Ni Alloy Phase diagram: Cu-Ni system. Consider microstuctural changes that accompany the cooling of a C 0 = 35 wt% Ni alloy T( o C) 1300 : 35 wt% Ni : 46 wt% Ni 1200 (liquid) 35 32 (solid) A B C 24 D 36 E : 35 wt%ni 43 46 Cu-Ni system : 32 wt% Ni : 43 wt% Ni : 24 wt% Ni : 36 wt% Ni : 35 wt% Ni 110 0 20 Adapted from Fig. 11.4, Callister & Rethwisch 9e. 35 30 40 50 C 0 wt% Ni AMSE 205 Spring 2016 Chapter 11-12
Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: -- Tensile strength (TS) -- Ductility (%E) Tensile Strength (MPa) 400 300 TS for pure Cu 200 0 20 40 60 80 100 Cu Ni Composition, wt% Ni Adapted from Fig. 11.5(a), Callister & Rethwisch 9e. TS for pure Ni Elongation (%E) 60 50 40 30 %E for pure Cu 20 0 20 40 60 80 100 Cu Ni %E for pure Ni Composition, wt% Ni Adapted from Fig. 11.5(b), Callister & Rethwisch 9e. AMSE 205 Spring 2016 Chapter 11-13
Binary-Eutectic Systems 2 components Ex.: Cu-Ag system 3 single phase regions (,, β) imited solubility: : mostly Cu β: mostly Ag T E : No liquid below T E C E : Composition at temperature T E Eutectic reaction T( o C) 1200 1000 600 400 200 0 (C E ) (C E ) + β(c βe ) cooling has a special composition with a min. melting T. + (liquid) β Cu-Ag system +β β T 800 779 C E 8.0 71.9 91.2 20 40 60 C E 80 100 C, wt% Ag Fig. 11.6, Callister & Rethwisch 9e [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]. heating AMSE 205 Spring 2016 Chapter 11-14
EX 1: Pb-Sn Eutectic System For a 40 wt% Sn-60 wt% Pb alloy at 150 o C, determine: -- the phases present Answer: + β T( o C) Pb-Sn -- the phase compositions Answer: C = 11 wt% Sn C β = 99 wt% Sn -- the relative amount of each phase Answer: W = = W β = S R+S = C β - C 0 C β - C 99-40 99-11 = 59 88 = 0.67 R = R+S 40-11 = 99-11 C 0 -C C β -C = 29 88 = 0.33 300 200 150 100 0 + (liquid) 183 C +β 18.3 61.9 97.8 R + β system 11 20 40 60 80 99100 C C β C 0 C, wt% Sn Fig. 11.7, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] AMSE 205 Spring 2016 Chapter 11-15 S β
EX 2: Pb-Sn Eutectic System For a 40 wt% Sn-60 wt% Pb alloy at 220 o C, determine: -- the phases present: Answer: + T ( o C) Pb-Sn -- the phase compositions Answer: C = 17 wt% Sn C = 46 wt% Sn -- the relative amount of each phase Answer: W = C -C 0 C -C = = 6 29 = 0.21 46-40 46-17 W = C 0 -C C -C = 23 29 = 0.79 300 220 200 100 0 + R S (liquid) 183 C + β system +β 1720 40 46 60 80 100 C C 0 C C, wt% Sn Fig. 11.7, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] AMSE 205 Spring 2016 Chapter 11-16 β
Microstructural Developments in Eutectic Systems I For alloys for which C 0 < 2 wt% Sn Result: at room temperature -- polycrystalline with grains of phase having composition C 0 T( o C) 400 300 200 T E : C 0 wt% Sn : C 0 wt% Sn + (Pb-Sn System) 100 + β Fig. 11.10, Callister & Rethwisch 9e. 0 C 0 10 20 2 (room T solubility limit) 30 C, wt% Sn AMSE 205 Spring 2016 Chapter 11-17
Microstructural Developments in Eutectic Systems II For alloys for which 2 wt% Sn < C 0 < 18.3 wt% Sn Result: at temperatures in + β range -- polycrystalline with grains and small β-phase particles 400 300 200 T( o C) T E + : C 0 wt% Sn : C 0 wt% Sn β 100 + β Pb-Sn system Fig. 11.11, Callister & Rethwisch 9e. 0 10 20 2 C 0 (sol. limit at Troom) 18.3 (sol. limit at T E ) 30 C, wt% Sn AMSE 205 Spring 2016 Chapter 11-18
Microstructural Developments in Eutectic Systems III For alloy of composition C 0 = C E Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of and β phases. 300 Pb-Sn system 200 T E T( o C) + 183 C : C 0 wt% Sn β β Micrograph of Pb-Sn eutectic microstructure 100 0 Fig. 11.12, Callister & Rethwisch 9e. β β: 97.8 wt% Sn : 18.3 wt%sn 20 40 60 80 100 18.3 C E 97.8 61.9 C, wt% Sn 160μm Fig. 11.13, Callister & Rethwisch 9e. (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 11-19
amellar 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 < C 0 < 61.9 wt% Sn Result: phase particles and a eutectic microconstituent Just above T E : T( o C) 300 Pb-Sn system 200 T E 100 0 + Fig. 11.15, Callister & Rethwisch 9e. R R + β : C 0 wt% Sn S 20 40 60 80 100 18.3 61.9 97.8 S +β C, wt% Sn β primary eutectic eutectic β C = 18.3 wt% Sn C = 61.9 wt% Sn W S = = 0.50 R + S W = (1- W) = 0.50 Just below T E : C = 18.3 wt% Sn C β = 97.8 wt% Sn W S = = 0.73 R + S W β = 0.27 AMSE 205 Spring 2016 Chapter 11-21
Hypoeutectic & Hypereutectic Fig. 11.7, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 300 T( o C) 200 T E 100 + + β +β β (Pb-Sn System) (Figs. 11.13 and 11.16 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.) C, wt% Sn 0 20 40 60 80 100 eutectic hypoeutectic: C 0 = 50 wt% Sn 61.9 hypereutectic: (illustration only) 175 μm Fig. 11.16, Callister & Rethwisch 9e. eutectic: C 0 =61.9wt% Sn 160 μm eutectic micro-constituent Fig. 11.13, Callister & Rethwisch 9e. β β β β β Adapted from Fig. 11.16, Callister & Rethwisch 9e. (Illustration only) β AMSE 205 Spring 2016 Chapter 11-22
Intermediate Phases (Intermediate Solid Solutions) AMSE 205 Spring 2016 Chapter 11-23
Intermetallic Compounds Fig. 11.19, Callister & Rethwisch 9e. [Adapted from Phase Diagrams of Binary Magnesium Alloys, A. A. Nayeb-Hashemi and J. B. Clark (Editors), 1988. Reprinted by permission of ASM International, Materials Park, OH.] Mg 2 Pb Note: intermetallic compound exists as a line on the diagram - not an area - because of stoichiometry (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 cool + β (For Pb-Sn, 183 o C, 61.9 wt% Sn) heat Eutectoid one solid phase transforms to two other solid phases intermetallic compound S 2 S 1 +S - cementite 3 cool γ + Fe 3 C (For Fe-C, 727 o C, 0.76 wt% C) heat Peritectic - liquid and one solid phase transform to a second solid phase S 1 + S 2 cool heat δ + γ (For Fe-C, 1493 o 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 cool + β (For Co-Cr, 976 o C, 37.6 wt% Cr) heat Monotectic liquid 1 decomposes to liquid 2 and a solid cool 1 2 + (For Cu-Pb, 955 o C, 36 wt% Pb) heat AMSE 205 Spring 2016 Chapter 11-27
Eutectoid & Peritectic Cu-Zn Phase diagram Peritectic transformations + + γ + δ + Eutectoid transformat δ γ + 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
P+F = C+2 Gibbs Phase Rule 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 T( o C) 1200 1000 + (liquid) Cu-Ag system β+ β T 800 779 C E 8.0 71.9 91.2 600 400 200 0 β 20 40 60 C E 80 100 C, wt% Ag AMSE 205 Spring 2016 Chapter 11-32
Iron-Carbon (Fe-C) Phase Diagram 2 important points - Eutectic (A): γ + Fe 3 C - Eutectoid (B): γ +Fe 3 C 1600 δ 1400 1200 1000 800 600 T( C) γ γ + (austenite) B γ γ γ γ 1148 C A γ+fe 3 C 727 C = T eutectoid +Fe 3 C +Fe 3 C Fe 3 C (cementite) 120 μm Result: Pearlite = alternating layers of and Fe 3 C phases Fig. 11.26, Callister & Rethwisch 9e. (From Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.) 400 0 1 2 3 4 5 6 6.7 (Fe) 0.76 4.30 Fig. 11.23, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] C, wt% C Fe 3 C (cementite-hard) (ferrite-soft) AMSE 205 Spring 2016 Chapter 11-33
γ γ γ γ γ γ γ γ γ γ γ γ 1600 δ 1400 1200 1000 800 600 T( C) γ γ+ (austenite) 400 0 1 2 3 4 5 6 6.7 (Fe) pearlite C 0 Hypoeutectoid Steel 0.76 727 C 1148 C γ +Fe 3 C +Fe 3 C +Fe 3 C C, wt% C Fe 3 C (cementite) (Fe-C System) Adapted from Figs. 11.23 and 11.28, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] pearlite 100 μm Hypoeutectoid steel proeutectoid ferrite Adapted from Fig. 11.29, Callister & Rethwisch 9e. (Photomicrograph courtesy of Republic Steel Corporation.) AMSE 205 Spring 2016 Chapter 11-34
γ γ γ γ W = s/(r + s) W γ =(1 - W ) 1600 δ 1400 1200 1000 800 600 T( C) γ γ+ (austenite) r pearlite 400 0 1 2 3 4 5 6 6.7 (Fe) C 0 Hypoeutectoid Steel s R S 0.76 W pearlite = W γ W ʼ = S/(R + S) W Fe 3 C =(1 W ) pearlite 727 C 1148 C γ +Fe 3 C +Fe 3 C +Fe 3 C C, wt% C Fe 3 C (cementite) (Fe-C System) 100 μm Hypoeutectoid steel Adapted from Figs. 11.23 and 11.28, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] proeutectoid ferrite Adapted from Fig. 11.29, Callister & Rethwisch 9e. (Photomicrograph courtesy of Republic Steel Corporation.) AMSE 205 Spring 2016 Chapter 11-35
Fe 3 C γ γ γ γ γ γ γ γ γ γ γ γ 1600 δ 1400 1200 1000 800 600 T( o C) γ γ+ (austenite) 400 0 1 2 3 4 5 6 6.7 (Fe) pearlite Hypereutectoid Steel C 0 0.76 C 0 727 C 1148 C γ +Fe 3 C +Fe 3 C +Fe 3 C C, wt% C Fe 3 C (cementite) (Fe-C System) Adapted from Figs. 11.23 and 11.31, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] pearlite 60 μm Hypereutectoid steel proeutectoid Fe 3 C Adapted from Fig. 11.32, Callister & Rethwisch 9e. (Copyright 1971 by United States Steel Corporation.) AMSE 205 Spring 2016 Chapter 11-36
Fe 3 C γ γ γ γ W γ =x/(v + x) W =(1-W γ ) Fe 3 C W pearlite = W γ W = X/(V + X) 1600 δ 1400 1200 1000 W =(1 - W ) Fe 3 Cʼ 800 T( o C) γ γ+ (austenite) 600 pearlite +Fe 3 C 400 0 1 2 3 4 5 6 6.7 (Fe) Hypereutectoid Steel C 0 V 0.76 v C 0 x pearlite X 727 C 1148 C γ +Fe 3 C +Fe 3 C C, wt% C Fe 3 C (cementite) (Fe-C System) 60 μm Hypereutectoid steel proeutectoid Fe 3 C Adapted from Figs. 11.23 and 11.31, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 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 Fe 3 C and ferrite (). b) The amount of cementite (in grams) that forms in 100 g of steel. c) The amounts of pearlite and proeutectoid ferrite () 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 C Fe3 C = 6.70 wt% C b) Using the lever rule with the tie line shown Amount of Fe 3 C in 100 g 1600 δ 1400 T( C) 1200 1000 800 600 γ γ+ (austenite) R Fig. 11.23, Callister & Rethwisch 9e. [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 727 C 1148 C γ +Fe 3 C S +Fe 3 C +Fe 3 C Fe 3 C (cementite) = (100 g)w Fe3 C = (100 g)(0.057) = 5.7 g 400 0 1 2 3 4 5 6 6.7 C, wt% C C C 0 C Fe C 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 C 0 = 0.40 wt% C C = 0.022 wt% C C pearlite = C γ = 0.76 wt% C Amount of pearlite in 100 g 1600 δ 1400 T( C) 1200 1000 800 600 γ γ+ (austenite) V X Fig. 11.23, Callister & Rethwisch 9e. [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 727 C 1148 C γ +Fe 3 C +Fe 3 C +Fe 3 C Fe 3 C (cementite) = (100 g)w pearlite = (100 g)(0.512) = 51.2 g 400 0 1 2 3 4 5 6 6.7 C C γ C, wt% C C 0 AMSE 205 Spring 2016 Chapter 11-40
Alloying with Other Elements T eutectoid changes: C eutectoid changes: TEutectoid (ºC) Ti Mo Ni Si W Cr Mn Ceutectoid (wt% C) Ti Si Mo Ni Cr W Mn wt. % of alloying elements Fig. 11.33, Callister & Rethwisch 9e. (From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.) wt. % of alloying elements Fig. 11.34,Callister & Rethwisch 9e. (From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM 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