Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30• When we combine two elements... what equilibrium state do we get?• In particular, if we specify... --a composition (e.g., wt%Cu - wt%Ag), and --a temperature (T)PHASE DIAGRAMSPhase BPhase ASilver atomCopper atom• Solubility Limit:Max concentration for which only a solution occurs.(No precipitate)• Ex: Phase Diagram: Water-Sugar SystemQuestion: What is the solubility limit at 20C?• Solubility limit increases with T: e.g., if T = 100C, solubility limit = 80wt% sugar.THE SOLUBILITY LIMITAnswer: 65wt% sugar. If Comp < 65wt% sugar: syrup If Comp > 65wt% sugar: syrup + sugar coexist3Solubility Limit3• Changing T can change # of phases: path A to B EFFECT OF T & COMPOSITION (Co)ABC• Changing Co can change # of phases: path B to C• Each point on this phase diagram represents equilibriumWATER-SALT PHASE DIAGRAMSolubility limitReduction in freezing point• Components: The elements or compounds which are mixed initially (e.g., Al and Cu, or water and sugar)Aluminum-CopperAlloyAdapted from Fig. 9.0, Callister 3e.COMPONENTS AND PHASES• Phases: The physically and chemically distinct material regions that result (e.g., α and β, or syrup and sugar)• Tell us about phases as function of T, Co, P• Phase Diagram for Cu-Ni system• Isomorphous system: i.e., complete solubility of onecomponent in anotherAdapted from Fig. 9.2(a), Callister 6e.(Fig. 9.2(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).PHASE DIAGRAMS• For this course: --binary systems: just 2 components. --independent variables: T and Co (P = 1 atm is always used)Note change in melting point• Rule 1: If we know T and Co, then we know: --the # and types of phases present.• Examples:PHASE DIAGRAMS: # and types of phasesCu-NiphasediagramA: 1 phase (α)B: 2 phases (L + α)• Rule 2: If we know T and Co, then we know: --the composition of each phase.Examples:Cu-Ni systemPHASE DIAGRAMS: composition of phases•C0 = 35 wt% Ni•At 1300 C:–Only liquid (L)–CL = C0 (= 35 wt% Ni)•At 1150 C:–Only solid (α)–Cα = C0 (= 35 wt% Ni)•At TB:–Both α and L–CL = Cliquidus (= 32 wt% Ni)–Cα = Csolidus (=43 wt% Ni)10• Rule 3: If we know T and Co, then we know: --the amount of each phase (given in wt%).Cu-NisystemPHASE DIAGRAMS: weight fractions of phases•C0 = 35 wt% Ni•At 1300 C:–Only liquid (L)–WL = 100 wt%, Wα = 0 wt%•At 1150 C:–Only solid (α)–WL = 0 wt%, Wα = 100 wt%•At TB:–Both α and L–WL = S/(R+S) = (43-35)/(43-32) = 73 wt%–Wα = R/(R+S) = (35-32)/(43-32) = 27 wt%The lever rule• Sum of weight fractions:• Conservation of mass (Ni):• Combine above equations:• A geometric interpretation:THE LEVER RULE: A PROOF12• System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; α phase field extends from 0 to 100wt% Ni.• Consider Co = 35wt%Ni.COOLING A Cu-Ni BINARY• Equilibrium cooling• Cα changes as we solidify.• Cu-Ni case:• Fast rate of cooling: Cored structure• Slow rate of cooling: Equilibrium structureFirst α to solidify has Cα = 46wt%Ni.Last α to solidify has Cα = 35wt%Ni.NON-EQUILIBRIUM PHASES• Effect of solid solution strengthening on:--Tensile strength (TS) --Ductility (%EL,%AR)Adapted from Fig. 9.5(a), Callister 6e. Adapted from Fig. 9.5(b), Callister 6e.MECHANICAL PROPERTIES: Cu-Ni System2 componentshas a special compositionwith a min. melting T.Adapted from Fig. 9.6, Callister 6e. (Fig. 9.6 adaptedfrom Binary Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)Cu-AgsystemBINARY-EUTECTIC SYSTEMS• 3 single phase regions (L, αβ) • Limited solubility: α: FCC, mostly Cu β: FCC, mostly Ag • TE: No liquid below TE • CE: Min. melting T composition Ex.: Cu-Ag system • 3 two phase regions• Cooling along dotted line: L (71.9%)  α (8%) + β (91.2%)• For a 40wt%Sn-60wt%Pb alloy at 150C, find... --the phases present --the compositions of the phases --the relative amounts of each phasePb-SnsystemAdapted from Fig. 9.7, Callister 6e. (Fig. 9.7 adaptedfrom Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)EX: Pb-Sn EUTECTIC SYSTEM (1)• For a 40wt%Sn-60wt%Pb alloy at 150C, find... --the phases present: α + β --the compositions of the phases: Cα = 11wt%Sn Cβ = 99wt%Sn --the relative amounts of each phase: (lever rule)Pb-SnsystemAdapted from Fig. 9.7, Callister 6e. (Fig. 9.7 adaptedfrom Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)EX: Pb-Sn EUTECTIC SYSTEM (2)• Co < 2wt%SnAdapted from Fig. 9.9, Callister 6e.MICROSTRUCTURESIN EUTECTIC SYSTEMS-I• Result: --polycrystal of α grains.• 2wt%Sn < Co < 18.3wt%SnPb-SnsystemAdapted from Fig. 9.10, Callister 6e.MICROSTRUCTURESIN EUTECTIC SYSTEMS-II• Result: --α polycrystal with fine β crystals.• Co = CE (Eutectic composition)Pb-SnsystemAdapted from Fig. 9.11, Callister 6e.Adapted from Fig. 9.12, Callister 6e. (Fig. 9.12 from Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)MICROSTRUCTURESIN EUTECTIC SYSTEMS-III• Result: Eutectic microstructure --alternating layers of α and β crystals.Pb-Snsystem• 18.3wt%Sn < Co < 61.9wt%SnAdapted from Fig. 9.14, Callister 6e.MICROSTRUCTURESIN EUTECTIC SYSTEMS-IV• Result: α crystals and a eutectic microstructureAdapted from Fig. 9.7, Callister 6e. (Fig. 9.7 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)(Figs. 9.12 and 9.15 from Metals Handbook, 9th ed.,Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)Adapted from Fig. 9.15, Callister 6e.Adapted from Fig. 9.12, Callister 6e. Adapted from Fig. 9.15, Callister 6e. (Illustration