1. Chapter 10 - 1 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? Chapter 10: Phase Diagrams Phase B Phase A Nickel atom Copper atom

2. Chapter 10 - 2 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 Diagram for Cu-Ni system Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). 2 phases: L (liquid)  (FCC solid solution) 3 different phase fields: L L +   wt% Ni 204060801000 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid solution) L +  liquidus solidus

3. Chapter 10 - 3 Cu-Ni phase diagram Isomorphous Binary Phase Diagram Phase diagram: Cu-Ni system. System is: Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). -- 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. wt% Ni 204060801000 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid solution) L +  liquidus solidus

4. Chapter 10 - wt% Ni 204060801000 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid solution) L +  liquidus solidus 4 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°C, 60 wt% Ni): 1 phase:  B(1250°C, 35 wt% Ni): 2 phases: L +  B (1250°C,35) A(1100°C,60) Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

5. Chapter 10 - 5 wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  Cu-Ni system 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: TATA A 35 C0C0 32 CLCL At T A = 1320°C: Only Liquid (L) present C L = C 0 ( = 35 wt% Ni) At T B = 1250°C: Both  and L present CLCL = C liquidus ( = 32 wt% Ni) CC = C solidus ( = 43 wt% Ni) At T D = 1190°C: Only Solid (  ) present C  = C 0 ( = 35 wt% Ni) Consider C 0 = 35 wt% Ni D TDTD tie line 4 CC 3 Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). B TBTB

6. Chapter 10 - 6 Rule 3: If we know T and C 0, then can determine: -- the weight fraction of each phase. Examples: At T A : Only Liquid (L) present W L = 1.00, W  = 0 At T D : Only Solid (  ) present WL WL = 0, W  = 1.00 Phase Diagrams : Determination of phase weight fractions wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  Cu-Ni system TATA A 35 C0C0 32 CLCL B TBTB D TDTD tie line 4 CC 3 R S At T B : Both  and L present = 0.27 WLWL  S R+S WW  R R+S Consider C 0 = 35 wt% Ni Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

7. Chapter 10 - 7 Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm The Lever Rule What fraction of each phase? Think of the tie line as a lever (teeter-totter) MLML MM RS wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  B T B tie line C0C0 CLCL CC S R Adapted from Fig. 10.3(b), Callister & Rethwisch 3e.

8. Chapter 10 - 8 wt% Ni 20 1200 1300 304050 1100 L (liquid)  (solid) L +  L +  T(°C) A 35 C0C0 L: 35wt%Ni Cu-Ni system Phase diagram: Cu-Ni system. Adapted from Fig. 10.4, Callister & Rethwisch 3e. Consider microstuctural changes that accompany the cooling of a C 0 = 35 wt% Ni alloy Ex: Cooling of a Cu-Ni Alloy 46 35 43 32  : 43 wt% Ni L: 32 wt% Ni B  : 46 wt% Ni L: 35 wt% Ni C E L: 24 wt% Ni  : 36 wt% Ni 24 36 D

9. Chapter 10 - Slow rate of cooling: Equilibrium structure Fast rate of cooling: Cored structure First  to solidify: 46 wt% Ni Last  to solidify: < 35 wt% Ni 9 C  changes as we solidify. Cu-Ni case: First  to solidify has C  = 46 wt% Ni. Last  to solidify has C  = 35 wt% Ni. Cored vs Equilibrium Structures Uniform C  : 35 wt% Ni

10. Chapter 10 - 10 Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: -- Tensile strength (TS)-- Ductility (%EL) Adapted from Fig. 10.6(a), Callister & Rethwisch 3e. Tensile Strength (MPa) Composition, wt% Ni Cu Ni 020406080100 200 300 400 TS for pure Ni TS for pure Cu Elongation (%EL) Composition, wt% Ni Cu Ni 020406080100 20 30 40 50 60 %EL for pure Ni %EL for pure Cu Adapted from Fig. 10.6(b), Callister & Rethwisch 3e.

11. Chapter 10 - 11 2 components has a special composition with a min. melting T. Adapted from Fig. 10.7, Callister & Rethwisch 3e. Binary-Eutectic Systems 3 single phase regions (L, ,  ) Limited solubility:  : mostly Cu  : mostly Ag T E : No liquid below T E : Composition at temperature T E C E Ex.: Cu-Ag system Cu-Ag system L (liquid)  L +  L+    C,wt% Ag 204060 80100 0 200 1200 T(°C) 400 600 800 1000 CECE TETE 8.0 71.9 91.2 779°C cooling heating Eutectic reaction L(C E )  (C  E ) +  (C  E )

12. Chapter 10 - 12 L+  L+   +  200 T(°C) 18.3 C, wt% Sn 206080 100 0 300 100 L (liquid)  183°C 61.997.8  For a 40 wt% Sn-60 wt% Pb alloy at 150°C, determine: -- the phases present Pb-Sn system EX 1: Pb-Sn Eutectic System Answer:  +  -- the phase compositions -- the relative amount of each phase 150 40 C0C0 11 CC 99 CC S R Answer: C  = 11 wt% Sn C  = 99 wt% Sn W  = C  - C 0 C  - C  = 99 - 40 99 - 11 = 59 88 = 0.67 S R+SR+S = W  = C 0 - C  C  - C  = R R+SR+S = 29 88 = 0.33 = 40 - 11 99 - 11 Answer: Adapted from Fig. 10.8, Callister & Rethwisch 3e.

13. Chapter 10 - 13 Answer: C  = 17 wt% Sn -- the phase compositions L+   +  200 T(°C) C, wt% Sn 206080 100 0 300 100 L (liquid)   L+  183°C For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine: -- the phases present: Pb-Sn system EX 2: Pb-Sn Eutectic System -- the relative amount of each phase W  = C L - C 0 C L - C  = 46 - 40 46 - 17 = 6 29 = 0.21 WLWL = C 0 - C  C L - C  = 23 29 = 0.79 40 C0C0 46 CLCL 17 CC 220 S R Answer:  + L C L = 46 wt% Sn Answer: Adapted from Fig. 10.8, Callister & Rethwisch 3e.

14. Chapter 10 - 14 For alloys for which C 0 < 2 wt% Sn Result: at room temperature -- polycrystalline with grains of  phase having composition C 0 Microstructural Developments in Eutectic Systems I 0 L +  200 T(°C) C,wt% Sn 10 2 20 C0C0 300 100 L  30  +  400 (room T solubility limit) TETE (Pb-Sn System)  L L: C 0 wt% Sn  : C 0 wt% Sn Adapted from Fig. 10.11, Callister & Rethwisch 3e.

15. Chapter 10 - 15 For alloys for which 2 wt% Sn < C 0 < 18.3 wt% Sn Result: at temperatures in  +  range -- polycrystalline with  grains and small  -phase particles Adapted from Fig. 10.12, Callister & Rethwisch 3e. Microstructural Developments in Eutectic Systems II Pb-Sn system L +  200 T(°C) C,wt% Sn 10 18.3 200 C0C0 300 100 L  30  +  400 (sol. limit at T E ) TETE 2 (sol. limit at T room ) L  L: C 0 wt% Sn    : C 0 wt% Sn

16. Chapter 10 - 16 For alloy of composition C 0 = C E Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of  and  phases. Adapted from Fig. 10.13, Callister & Rethwisch 3e. Microstructural Developments in Eutectic Systems III Adapted from Fig. 10.14, Callister & Rethwisch 3e. 160  m Micrograph of Pb-Sn eutectic microstructure Pb-Sn system LL  200 T(°C) C, wt% Sn 2060801000 300 100 L   L+  183°C 40 TETE 18.3  : 18.3 wt%Sn 97.8  : 97.8 wt% Sn CECE 61.9 L: C 0 wt% Sn

17. Chapter 10 - 17 Lamellar Eutectic Structure Adapted from Figs. 10.14 & 10.15, Callister & Rethwisch 3e.

18. Chapter 10 - 18 For alloys for which 18.3 wt% Sn < C 0 < 61.9 wt% Sn Result:  phase particles and a eutectic microconstituent Microstructural Developments in Eutectic Systems IV 18.361.9 SR 97.8 S R primary  eutectic   WLWL = (1-W  ) = 0.50 CC = 18.3 wt% Sn CLCL = 61.9 wt% Sn S R +S WW = = 0.50 Just above T E : Just below T E : C  = 18.3 wt% Sn C  = 97.8 wt% Sn S R +S W  = = 0.73 W  = 0.27 Adapted from Fig. 10.16, Callister & Rethwisch 3e. Pb-Sn system L+  200 T(°C) C, wt% Sn 2060801000 300 100 L   L+  40  +  TETE L: C 0 wt% Sn L  L 

19. Chapter 10 - 19 L+  L+   +  200 C, wt% Sn 2060801000 300 100 L   TETE 40 (Pb-Sn System) Hypoeutectic & Hypereutectic Adapted from Fig. 10.8, Callister & Rethwisch 3e. (Fig. 10.8 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.) 160  m eutectic micro-constituent Adapted from Fig. 10.14, Callister & Rethwisch 3e. hypereutectic: (illustration only)       Adapted from Fig. 10.17, Callister & Rethwisch 3e. (Illustration only) (Figs. 10.14 and 10.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) 175  m       hypoeutectic: C 0 = 50 wt% Sn Adapted from Fig. 10.17, Callister & Rethwisch 3e. T(°C) 61.9 eutectic eutectic: C 0 = 61.9 wt% Sn

20. Chapter 10 - 20 Intermetallic Compounds 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). Adapted from Fig. 10.20, Callister & Rethwisch 3e.

21. Chapter 10 - 21 Eutectoid – one solid phase transforms to two other solid phases S 2 S 1 +S 3   + Fe 3 C (For Fe-C, 727  C, 0.76 wt% C) intermetallic compound - cementite cool heat Eutectic, Eutectoid, & Peritectic Eutectic - liquid transforms to two solid phases L  +  (For Pb-Sn, 183  C, 61.9 wt% Sn) cool heat cool heat Peritectic - liquid and one solid phase transform to a second solid phase S 1 + L S 2  + L  (For Fe-C, 1493°C, 0.16 wt% C)

22. Chapter 10 - 22 Eutectoid & Peritectic Cu-Zn Phase diagram Adapted from Fig. 10.21, Callister & Rethwisch 3e. Eutectoid transformation   +  Peritectic transformation  + L 

23. Chapter 10 - 23 Ceramic Phase Diagrams MgO-Al 2 O 3 diagram: Adapted from Fig. 10.24, Callister & Rethwisch 3e. 

24. Chapter 10 - 24 Iron-Carbon (Fe-C) Phase Diagram 2 important points - Eutectoid (B):  + Fe 3 C - Eutectic (A): L  + Fe 3 C Adapted from Fig. 10.28, Callister & Rethwisch 3e. Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  +Fe 3 C   +   (Fe) C, wt% C 1148°C T(°C)  727°C = T eutectoid 4.30 Result: Pearlite = alternating layers of  and Fe 3 C phases 120  m (Adapted from Fig. 10.31, Callister & Rethwisch 3e.) 0.76 B    A L+Fe 3 C Fe 3 C (cementite-hard)  (ferrite-soft)

25. Chapter 10 - 25 Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  + Fe 3 C  L+Fe 3 C  (Fe) C, wt% C 1148°C T(°C)  727°C (Fe-C System) C0C0 0.76 Hypoeutectoid Steel Adapted from Figs. 10.28 and 10.33,Callister & Rethwisch 3e. (Fig. 10.28 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) Adapted from Fig. 10.34, Callister & Rethwisch 3e. proeutectoid ferrite pearlite 100  m Hypoeutectoid steel  pearlite              

26. Chapter 10 - 26 Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  + Fe 3 C  L+Fe 3 C  (Fe) C, wt% C 1148°C T(°C)  727°C (Fe-C System) C0C0 0.76 Hypoeutectoid Steel        sr W  = s/(r + s) W  =(1 - W  ) R S  pearlite W pearlite = W  W  ’ = S/(R + S) W =(1 – W  ’ ) Fe 3 C Adapted from Figs. 10.28 and 10.33,Callister & Rethwisch 3e. (Fig. 10.28 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) Adapted from Fig. 10.34, Callister & Rethwisch 3e. proeutectoid ferrite pearlite 100  m Hypoeutectoid steel

27. Chapter 10 - 27 Hypereutectoid Steel Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  +Fe 3 C  L+Fe 3 C  (Fe) C, wt%C 1148°C T(°C)  Adapted from Figs. 10.28 and 10.36,Callister & Rethwisch 3e. (Fig. 10.28 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) (Fe-C System) 0.76 C0C0 Fe 3 C             Adapted from Fig. 10.37, Callister & Rethwisch 3e. proeutectoid Fe 3 C 60  m Hypereutectoid steel pearlite

28. Chapter 10 - 28 Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  +Fe 3 C  L+Fe 3 C  (Fe) C, wt%C 1148°C T(°C)  Hypereutectoid Steel (Fe-C System) 0.76 C0C0 pearlite Fe 3 C     x v VX W pearlite = W  W  = X/(V + X) W =(1 - W  ) Fe 3 C’ W  =(1-W  ) W  =x/(v + x) Fe 3 C Adapted from Fig. 10.37, Callister & Rethwisch 3e. proeutectoid Fe 3 C 60  m Hypereutectoid steel pearlite Adapted from Figs. 10.28 and 10.36,Callister & Rethwisch 3e. (Fig. 10.28 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.)

29. Chapter 10 - 29 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.

30. Chapter 10 - 30 Solution to Example Problem b)Using the lever rule with the tie line shown a) Using the RS tie line just below the eutectoid C  = 0.022 wt% C C Fe 3 C = 6.70 wt% C Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  + Fe 3 C  L+Fe 3 C  C, wt% C 1148°C T(°C) 727°C C0C0 R S C Fe C 3 CC Amount of Fe 3 C in 100 g = (100 g)W Fe 3 C = (100 g)(0.057) = 5.7 g

31. Chapter 10 - 31 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  = 0.76 wt% C Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  + Fe 3 C  L+Fe 3 C  C, wt% C 1148°C T(°C) 727°C C0C0 V X CC CC Amount of pearlite in 100 g = (100 g)W pearlite = (100 g)(0.512) = 51.2 g

32. Chapter 10 - 32 Alloying with Other Elements T eutectoid changes: Adapted from Fig. 10.38,Callister & Rethwisch 3e. (Fig. 10.38 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) T Eutectoid (°C) wt. % of alloying elements Ti Ni Mo Si W Cr Mn C eutectoid changes: Adapted from Fig. 10.39,Callister & Rethwisch 3e. (Fig. 10.39 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) wt. % of alloying elements C eutectoid (wt% C) Ni Ti Cr Si Mn W Mo

33. Chapter 10 - 33 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. Summary

34. Chapter 10 - 34 Core Problems: Self-help Problems: ANNOUNCEMENTS Reading: