ENS 167 Materials Science & Engineering Lecture Notes PDF

ENS 167 Fundamentals of Materials Science and Engineering Source (book and ppt): Callister, Materials Science and Engineering: An Introduction, 8th Edition  Chapter 9:Phase Diagrams ISSUES TO ADDRESS... When we combine two elements... what is the resulting e...

ENS 167 Fundamentals of Materials Science and Engineering Source (book and ppt): Callister, Materials Science and Engineering: An Introduction, 8th Edition  Chapter 9: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 Phase B A Nickel atom Copper atom  Phase Equilibria: Solubility Limit Solution – solid, liquid, or gas solutions, single phase Mixture – more than one phase Adapted from Fig. 9.1, Callister & Rethwisch 8e. Sugar/Water Phase Diagram Solubility Limit: 100 Maximum concentration for Solubility Temperature (ºC) which only a single phase 80 Limit L solution exists. (liquid) 60 L + Question: What is the 40 (liquid solution solubility limit for sugar in S i.e., syrup) water at 20ºC? 20 (solid sugar) Answer: 65 wt% sugar. 0 20 80 100 At 20ºC, if C < 65 wt% sugar: syrup 40 60 Sugar Water 65(wt% sugar) C = Composition At 20ºC, if C > 65 wt% sugar: syrup + sugar  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., α and β). Aluminum- β (lighter Copper Alloy phase) α (darker Adapted from chapter- opening photograph, phase) Chapter 9, Callister, Materials Science & Engineering: An Introduction, 3e.  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 (100ºC,C = 70) D (100ºC,C = 1 phase 90) 100 2 phases 80 L Temperature (ºC) (liquid) water- 60 + sugar L S syste 40 (liquid solution (solid m i.e., syrup) sugar) 20 A (20ºC,C = 70) 2 phases Adapted from Fig. 9.1, 0 2040 60 70 80 100 Callister & Rethwisch 8e. 0 C = Composition (wt% sugar)  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.  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). Phase T(ºC) 1600 2 phases: Diagram 1500 L (liquid) for Cu-Ni L (liquid) α (FCC solid solution) system 1400 3 different phase fields: 1300 L α 1200 α L+ (FCC solid α 1100 Adapted from Fig. 9.3(a), Callister & solution) Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, 1000 20 40 60 80 100 0 wt% Ni P. Nash (Ed.), ASM International, Materials Park, OH (1991).  Isomorphous Binary Phase Diagram Phase diagram: T(ºC) Cu-Ni system. 1600 System is: 1500 L (liquid) Cu-Ni -- binary phase i.e., 2 components: 1400 diagram Cu and Ni. 1300 -- isomorphous i.e., complete 1200 α solubility of one (FCC solid component in 1100 another; α phase solution) field extends from 1000 20 40 60 80 100 wt% Ni 0 0 to 100 wt% Ni. Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).  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(1100ºC, 60 wt% Ni): L (liquid) 1500 B (1250ºC,35) 1 phase: α Cu-Ni B(1250ºC, 35 wt% Ni): 1400 phase diagram 2 phases: L + α 1300 α 1200 (FCC solid Adapted from Fig. 9.3(a), Callister & solution) Rethwisch 8e. (Fig. 9.3(a) is adapted 1100 A(1100ºC,60) from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, 1000 20 40 60 80 100 Materials Park, OH (1991). 0 wt% Ni  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 T(ºC) system Examples: A Consider C0 = 35 wt% Ni TA tie line At TA = 1320ºC: 1300 L (liquid) Only Liquid (L) present B TB CL = C0 ( = 35 wt% Ni) α At TD = (solid) 1200 D 1190ºC: Only Solid (α) present TD Cα = C0 ( = 35 wt% Ni) 20 30 50 32 40 At TB = 1250ºC: 35 43 Cα Both and L C 9.3(a), Callister wt% Adapted from Fig.C & Ni α 0 L 9.3(a) Rethwisch 8e. (Fig. is adapted from Phase Diagrams of Binary Nickel Alloys, C =C present ( = 32 wt% Ni) L liquidus ( = 43 wt% Ni) P. Nash (Ed.), ASM International, Cα = C Materials Park, OH (1991). solidus  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 C0 = 35 wt% Ni TA A tie line At TA : Only Liquid (L) 1300 L (liquid) present W = 1.00, W = 0 B At TD : Only Solid ( α) present L TB R S α α 1200 D (solid) At TB : WL = 0, W = 1.00 α TD Both and L present 30 S 43 − 35 0.73 α 20 32 40 50 W = R+S L = = 35 43 Cα 43 − 32 wt% Ni Adapted fromCFig. 9.3(a), Callister & C Rethwisch 8e. (Fig. L 09.3(a) is adapted from R = 0.27 Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, = Materials Park, OH (1991). Wα R + S  The Lever Rule 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 ML Mα B α 1200 (solid) R S 20 40 50 R S 30 CL C0 Cα wt% Ni Adapted from Fig. 9.3(b), Callister & Rethwisch 8e. Mα x S = ML x R ML S Cα − C0 R C − WL = = = C −C Wα = = 0 M L + Mα R+S α L CL R + S Cα − CL  Ex: Cooling of a Cu-Ni Alloy Phase diagram: T(ºC) L: 35wt%Ni Cu-Ni system. L (liquid) Cu-Ni system Consider 130 A L: 350 wt% Ni microstuctural α: 46 wt% Ni 35 B changes that 32 C 46 43 accompany the D 24 36 L: 32 wt% Ni cooling of a α: 43 wt% Ni 120 E C0 = 35 wt% Ni alloy 0 L: 24 wt% Ni α: 36 wt% Ni α (solid) 110 0 20 30 35 40 50 Adapted from Fig. 9.4, C0 wt% Ni Callister & Rethwisch 8e.  Cored vs Equilibrium Structures Cα changes as we solidify. Cu-Ni case: First α to solidify has C = 46 wt% Ni. α Last α to solidify has C = 35 wt% Ni. Slow rate of cooling: α Fast rate of cooling: Equilibrium structure Cored structure Uniform C : α 35 wt% Ni First α to solidify: 46 wt% Ni Last α to solidify: < 35 wt% Ni  Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: -- Tensile strength (TS) -- Ductility (%EL) 60 Tensile Strength (MPa) Elongation (%EL) %EL for pure Cu 400 50 %EL for TS pure Ni pure Ni for 40 300 TS for pure Cu 30 200 20 0 20 40 60 80 0 20 40 60 80 100 Cu 100 Ni Cu Ni Composition, wt% Ni Composition, wt% Ni Adapted from Fig. 9.6(a), Adapted from Fig. 9.6(b), Callister & Rethwisch 8e. Callister & Rethwisch 8e.  Binary-Eutectic Systems has a special composition 2 components with a min. melting T. Cu-Ag T(ºC) system Ex.: Cu-Ag system 1200 3 single phase regions L (liquid) 1000 (L, α, β) α α L+ β Limited solubility: TE 800 779ºC L+ 8.0 91.2 β α: mostly Cu 71.9 600 β: mostly Ag α+β TE : No liquid below TE 400 CE : Composition at 200 temperature TE 0 20 40 80 100 60 CE C, wt% Ag Eutectic L(CEreaction ) α(CαE) + β(CβE) Adapted from Fig. 9.7, Callister & Rethwisch 8e. cooling L(71.9 wt% α(8.0 wt% Ag) + β(91.2 wt% Chapter heating Ag) Ag)  EX 1: Pb-Sn Eutectic System For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, determine: -- the phases present Pb-Sn Answer: α + β T(ºC) syste -- the phase compositions m 300 Answer: Cα = 11 wt% Sn L (liquid) Cβ = 99 wt% Sn α α 200 L+ 183ºC β -- the relative amount L+ 18.3 61.9 97.8 of each phase 150 R S β Answer: W S Cβ - 100 = = α R+S C Cβ - Cα α+ 0 β 99 - = 59 = 99 - 11 88 = 0 11 40 60 80 40 100 0.67 20 99 Wβ R C0 - Cα C0 C, wt% Sn = R+S = Cβ - Cα Cα from Fig. 9.8, Adapted Cβ Callister & Rethwisch 8e. 40 - 11 29 = 99 - = 88 = 0.33  EX 2: Pb-Sn Eutectic System For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine: -- the phases present: Pb-Sn Answer: α + L T(ºC) syste -- the phase compositions m 300 Answer: Cα = 17 wt% Sn L (liquid) α CL = 46 wt% Sn 220 L+ α R S β -- the relative amount 200 L+ β 183ºC of each phase Answer: CL - C0 46 - 40 100 α+ Wα = L α = β C- C 46 - 6 17 0 17 46 100 = 20 40 60 80 29= 0.21 Cα C0 CL C, wt% Sn Adapted from Fig. 9.8, C - 23 Callister & Rethwisch 8e. WL = C0 α = = 0.79 CL - Cα 29  Microstructural Developments in Eutectic Systems I For alloys for which T(ºC) L: C0 wt% Sn 400 C0 < 2 wt% Sn L Result: at room temperature α 300 L -- polycrystalline with grains of α phase having L α α + 200 (Pb-Sn composition C0 TE α: C0 wt% Sn System) 100 β α+ 0 10 20 30 C0 Adapted from Fig. 9.11, 2 C , wt% Callister & Rethwisch 8e. (room T solubility limit) Sn  Microstructural Developments in Eutectic Systems II L: C0 wt% Sn For alloys for which T(ºC) 400 2 wt% Sn < C0 < 18.3 wt% Sn L Result: L at temperatures in α + β range 300 α L+ -- polycrystalline with α grains α α: C0 wt% Sn and small β-phase particles 200 α TE α β 100 β Pb-Sn α+ system Adapted from Fig. 9.12, 0 10 20 30 Callister & Rethwisch 8e. 2 C0 C, wt% Sn (sol. limit at T ) 18.3 room (sol. limit at TE)  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. T(ºC) Micrograph of Pb-Sn eutectic L: C0 wt% Sn microstructure 300 L Pb-Sn α system L+ α 183ºC L +β 200 TE β 100 160 μm α+β β: 97.8 wt% Sn Adapted from Fig. 9.14, α: 18.3 wt%Sn Callister & Rethwisch 8e. 0 20 40 60 80 100 18.3 CE 97.8 Adapted from Fig. 9.13, 61.9 C, wt% Sn Callister & Rethwisch 8e.  Lamellar Eutectic Structure Adapted from Figs. 9.14 & 9.15, Callister & Rethwisch 8e.  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 Just above TE : T(ºC) wt% L L: C0 α Cα = 18.3 wt% Sn Sn L 300 L CL = 61.9 wt% α Pb-Sn Sn S = 0.50 L+ = R +S system α α Wα 200 R S L+ β WL = (1- W ) = 0.50 R S β α TE Just below T E: 100 α+β C = 18.3 wt% Sn α α primary α eutectic C = 97.8 wt% Sn eutectic β Wβα S 0 20 40 60 80 100 = R + = 0.73 18.3 61.9 97.8 S W = 0.27 Adapted from Fig. 9.16, Callister & Rethwisch 8e. C, wt% Sn β  Hypoeutectic & Hypereutectic 300 L Adapted from Fig. 9.8, T(ºC) α α Callister & Rethwisch 8e. 200 L+ ββ (Pb-Sn (Fig. 10.8 adapted from TE L+ Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. α+ System) Massalski (Editor-in-Chief), 100 ASM International, β Materials Park, OH, 1990.) 020 40 60 80 100 C, wt% Sn eutectic hypoeutectic: C0 = 50 wt% Sn 61.9 hypereutectic: (illustration only) (Figs. 9.14 and 9.17 from Metals Handbook, 9th ed., eutectic: C0 = 61.9 wt% Vol. 9, α Sn 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 & Adapted from Fig. 9.14, Callister & Rethwisch 8e. Rethwisch 8e. Callister & Rethwisch 8e. (Illustration only)  Intermetallic Compounds Adapted from Fig. 9.20, Callister & Rethwisch 8e. Mg2Pb 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).  Eutectic, Eutectoid, & Peritectic Eutectic - liquid transforms to two solid phases L cool α + β (For Pb-Sn, 183ºC, 61.9 wt% Sn) heat Eutectoid – one solid phase transforms to two other solid phases intermetallic compound S2 S1+S3 - cementite γ cool α + Fe3C (For Fe-C, 727ºC, 0.76 wt% C) heat Peritectic - liquid and one solid phase transform to a second solid phase S1 + L S2 cool δ + heat γ (For Fe-C, 1493ºC, 0.16 wt% C) L  Eutectoid & Peritectic Peritectic transformation γ δ Cu-Zn Phase diagram+ L Adapted from Fig. 9.21, Callister & Rethwisch 8e. Eutectoid transformation γ+ε δ  Iron-Carbon (Fe-C) Phase Diagram T(ºC) 2 important 1600 points δ 1400 L - Eutectic (A): ⇒γ+ γ γ A L Fe3C 1200 1148ºC L+Fe3C Fe3C (cementite) (austenite) +L - Eutectoid (B): 1000 γ γ⇒α+ γ +Fe 3C Fe3C 80 γ B γ 727ºC = T eutectoid 0 α γ α +Fe3C 600 400 0 2 3 4 6 120 μm 1 0.76 4.30 5 6.7 (Fe) Result: Pearlite = C, wt% C Fe3C (cementite-hard) alternating layers of α α and Fe3C phases (ferrite-soft) (Adapted from Fig. 9.27, Callister & Rethwisch 8e.) Adapted from Callister & Fig. 9.24, Rethwisch 8e.  Hypoeutectoid Steel T(ºC) 1600 δ 1400 L γ γ γ (Fe-C γ +L γ γ 1148ºC L+Fe3C System) Fe3C (cementite) 1200 (austenite) γ 1000 γγ Adapted from Figs. 9.24 + Fe3C and 9.29,Callister & γ Rethwisch 8e. α 800 αγ γ 727ºC (Fig. 9.24 adapted from Binary Alloy Phase γγ γ α Diagrams, 2nd ed., Vol. 600 1, T.B. Massalski (Ed.-in- α + Fe3C Chief), ASM International, α Materials Park, OH, 400 α 0 1 2 3 4 5 6 1990.) (Fe)C0 C, wt% C 0.76 6.7 pearlite Hypoeutectoid 100 μm steel pearlite proeutectoid ferrite Adapted from Fig. 9.30, Callister & Rethwisch 8e.  Hypoeutectoid Steel T(ºC) 1600 δ 1400 L γ (Fe-C 1200 γ +L γ γα 1148ºC L+Fe3C System) Fe3C (cementite) α (austenite) α γ γ 1000 Adapted from Figs. 9.24 Wα = s/(r + s) + Fe3C and 9.29,Callister & γ Rethwisch 8e. rs 727ºC (Fig. 9.24 adapted from Wγ =(1α - W ) 800 Binary Alloy Phase αR Diagrams, 2nd ed., Vol. α 600 S 1, T.B. Massalski (Ed.-in- + Fe 3C Chief), ASM International, pearlite α Materials Park, OH, 4000 1 2 3 4 5 6 1990.) (Fe)C0 C, wt% C 0.76 6.7 Wpearlite =W γ Wα’ = S/(R + S) 100 μm Hypoeutectoid steel WFe=(1 C – Wα’) 3 pearlite proeutectoid ferrite Adapted from Fig. 9.30, Callister & Rethwisch 8e.  Hypereutectoid Steel T(ºC) 1600 δ 1400 L (Fe-C γ γ γ γ System) 1200 +L 1148ºC L+Fe3C γ Fe3C (cementite) (austenite) γ γ 1000 γ Adapted from Figs. 9.24 +Fe3C and 9.32,Callister & Fe3C γ γ Rethwisch 8e. (Fig. 9.24 800 γ adapted from Binary Alloy Phase Diagrams, γ α 2nd ed., Vol. 1, T.B. γ 600 α Massalski (Ed.-in-Chief), +Fe 3C ASM International, γ Materials Park, OH, γ 400 0 1 2 3 4 5 66.7 1990.) 0.76 (Fe) C0 C, wt%C pearlite Hypereutectoid 60 μm steel pearlite proeutectoid Fe3C Adapted from Fig. 9.33, Callister & Rethwisch 8e.  Hypereutectoid Steel T(ºC) 1600 δ 1400 L (Fe-C γ γ System) Fe3C 1200 +L 1148ºC L+Fe3C γ Fe3C (cementite) (austenite) 1000 γ Adapted from Figs. 9.24 γ Wγ =x/(v +Fe3C and 9.32,Callister & γ + x) 800 v x γ Rethwisch 8e. (Fig. 9.24 adapted from Binary WFe3C =(1-Wγ) Alloy Phase Diagrams, α V X 2nd ed., Vol. 1, T.B. 600 α Massalski (Ed.-in-Chief), pearlite +Fe 3C ASM International, Materials Park, OH, 400 0 1 2 3 4 5 66.7 1990.) 0.76 (Fe) C0 Wpearlite =W C, wt%C γ Wα = X/(V + X) Hypereutectoid 60 μm steel WFe3C’ =(1 - Wα) pearlite proeutectoid Fe3C Adapted from Fig. 9.33, Callister & Rethwisch 8e.  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.  Solution to Example Problem a) Using the RS tie line just below the eutectoid Cα = 0.022 wt% C CFe3C = 6.70 wt% C b) Using the lever rule with 1600 the tie line shown δ L R C −C 1400 WFe 3=C T(ºC) γ α R + S CFe C− C α 0 γ +L Fe C (cementite) = 1200 1148ºC L+Fe3C 3 (austenite) 0.40 − 1000 = = 0.057 γ 0.022 + Fe3C 6.70 − 800 727ºC R S Amount0.022 of Fe3C in 100 g 600 + Fe3C α = (100 g)WFe 400 0 1 2 3 4 5 6 3 C Cα C0 C, wt% C 6.7 Fe C = (100 g)(0.057) = 5.7 g C 3  Solution to Example Problem c) (cont.) Using the VX tie line just above the eutectoid and realizing that C0 = 0.40 wt% C Cα = 0.022 wt% C Cpearlite = Cγ = 0.76 wt% C 1600 δ 1400 L V C − T(ºC) γ Wpearlite = 0 α γ +L 1148ºC Fe C (cementite) V + X =CCγ − α 1200 (austenite) L+Fe3C C − 0.40 1000 = = 0.512 γ 0.022 + Fe3C 800 0.76 − VX 727ºC Amount of pearlite in 100 g 0.022 600 + Fe3C α = (100 g)Wpearlite 400 2 3 4 5 6 0 Cα C0 Cγ C, wt% C 6.7 1 = (100 g)(0.512) = 51.2 g  Alloying with Other Elements Teutectoid changes: Ceutectoid changes: (wt% C) T Si i Mo W Ni Cr Eutectoid eutectoid Cr Si Mn (ºC) Mn W T Mo C T Ni i wt. % of alloying elements wt. % of alloying elements Adapted from Fig. 9.34,Callister & Rethwisch 8e. Adapted from Fig. 9.35,Callister & Rethwisch 8e. (Fig. 9.34 from Edgar C. Bain, Functions of the (Fig. 9.35 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Metals, 1939, p. 127.)  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.  Quiz What are the phase compositions of L and α at point B?