Material Properties and Phase Diagrams

Transcription

1 PY2M20 Material Properties and Phase Diagrams Lecture 6 P. Stamenov, PhD School of Physics, TCD PY2M20-6

2 Microstructures in Eutectic ti Systems: I C o < 2 wt% Sn Result: - at extreme ends - polycrystal of grains i.e., only one solid phase. T( C) 400 L: C o wt% Sn L 300 L 200 T E : C o wt% Sn L+ (Pb-Sn System) C o 2 (room T solubility limit) C o, wt% Sn

3 Microstructures in Eutectic ti Systems: II 2 wt% Sn < C o < 18.3 wt% Sn Result: Initially liquid + then alone finally two phases polycrystal fine -phase inclusions T( C) T E 100 L L + + L: C o wt% Sn L : C o wt% Sn Pb-Sn system C o (sol. limit at Troom) ) 18.3 (sol. limit at T E ) C o, wt% Sn

4 Microstructures in Eutectic ti Systems: III C o = C E Result: Eutectic microstructure (lamellar structure) - alternating layers (lamellae) of and crystals. Pb-Sn system T( C) 300 L L C L T E Micrograph of Pb-Sn eutectic L: C o wt% Sn microstructure t 100 : 97.8 wt% Sn : 18.3 wt%sn 160m Adapted from Fig. 9.14, Callister 7e C E C, wt% Sn

5 Lamellar Eutectic Structure Other possible eutectic structures are: rod-like, globular and acicular.

6 Microstructures in Eutectic ti Systems: IV 18.3 wt% Sn < C o < 61.9 wt% Sn Result: crystals and an eutectic microstructure Pb-Sn system T( C) L: C o wt% Sn L 300 L L+ L 200 R S L+ T E R S Just above T E : C = 18.3 wt% Sn C L = 61.9 wt% Sn S W= = 50 wt% R + S W L = (1- W) = 50 wt% Just below T E : C = 18.3 wt% Sn primary C = 97.8 wt% Sn eutectic eutectic S W= = 73 wt% R + S W = 27 wt% C o, wt% Sn 0 40

7 Hypoeutectic & Hypereutectic 300 L T( C) 200 L+ L+ TE + (Pb-Sn S t ) System) hypoeutectic: Co = 50 wt% Sn eutectic Co, wt% Sn hypereutectic: (illustration only) eutectic: Co = 61.9 wt% Sn 175 m m micro constituent eutectic micro-constituent

8 Intermetallic Compounds Mg 2 Pb Note: intermetallic compound forms a line - not an area - Note: intermetallic compound forms a line not an area because the stoichiometry (i.e. composition) is exact.

9 Eutectic Eutectic - liquid in equilibrium with two solids L cool + heat

10 Example: Eutectoid & Peritectic Cu-Zn Phase diagram Peritectic transition + L Eutectoid transition + Peritectoid solid state Peritectic

11 Eutectoid & Peritectic Eutectoid t -solid phase in equilibrium i with two solid phases S 2 S 1 +S 3 intermetallic compound intermetallic compound - cementite cool + Fe 3 C (727ºC) heat Peritectic - liquid + solid 1 solid 2 S 1 + L S 2 cool + L (1493ºC) περιτεκτικός περι - included heat

12 Iron-Carbon Phase Diagram Extract 2 important points -Eutectic (A): L + Fe 3 C -Eutectoid (B): +Fe 3 C 1600 T( C) 1400 L L (austenite) R B 1148 C R S A +Fe 3 C +Fe 3 C L+Fe 3 C S 727 C = Teutectoid α bcc (FM) β bcc (NM) obs. (Fe) C o, wt% C γ fcc (NM) Fe 3 C (cementite-hard) (ferrite-soft) δ bcc (NM) ε hcp(p >13 GPa) C eut tectoid e) Fe 3 C (ce ementit

13 Pearlite Fe 3 C (cementite-hard) (ferrite-soft) 120 m Result: Pearlite = alternating layers of and Fe 3 C phases 3 p

14 Hypoeutectoid Steel w =s/(r +s) w =(1- w) w =S/(R+S) w F C =(1-w) w Fe3 C 1600 T( C) 1400 L L (austenite) rs 727 C 1148 C + Fe 3 C +Fe 3 C L+Fe 3 C (Fe) C 0 pearlite w pearlite = w 0.76 Fe 3 C (ceme entite) C o, wt% C (Fe-C System)

15 Hypoeutectoid Steel pearlite w pearlite = w w =S/(R+S) w Fe3 C = (1-w) 100 m pearlite Proeutectoid ferrite proeutectoid phase the first phase that forms upon cooling the solid

16 Hypereutectoid Steel Fe 3 C 1600 T( C) 1400 L L (austenite) r s R S 1148 C +Fe 3 C (Fe-C System) L+Fe 3 C 600 w Fe3 C =r/(r +s) +Fe 3 C C w =(1-w o Fe3 C) pearlite (Fe) C o, wt%c w pearlite = w w =S/(R+S) w Fe3 C =(1-ww ) ) Fe 3 C (ce ementit te)

17 Hypereutectoid Steel pearlite w pearlite = w w =S/(R+S) w Fe 3 C =(1-w ) 60 mm pearlite proeutectoid Fe 3 C

18 Example For a 99.6 wt% Fe wt%Cata a temperature just below the eutectoid, determine the following a) the amount of pearlite and proeutectoid ferrite () per 100 g of steel b) composition of Fe 3 C and ferrite () c) the amount of carbide (cementite) in grams that forms per 100 g of steel

19 Solution a. the amount of pearlite and proeutectoid ferrite () note: amount of pearlite = amount of just above T E C o = 0.40 wt% C C = wt% C C pearlite = C = 0.76 wt% C 1600 C o C x g C C pearlite = 51.2 g proeutectoid = 48.8 g T( C) (austenite) R S +L 727 C L 1148 C +Fe 3 C Fe 3 C L+Fe 3 C C C C O C o, wt% C C O e) e 3 C (ce ementit F

20 Solution - continued b) composition of Fe 3 C and ferrite () c) the amount of carbide (cementite) in grams that forms per 100 g of steel Fe3C Fe C C Fe 3C 5.7 g 94.3 g C o Fe C C C x 100 x g T( C) (austenite) C O = 0.40 wt% C C = wt% C C Fe C = wt% t%c +L 3 L 1148 C +Fe 3 C C R S Fe 3 C L+Fe 3 C C C o, wt% C C C O e) e 3 C (ce ementit F C Fe 3 C

21 Alloying Steel with More Elements T eutectoid changes: C eutectoid changes: T Eu utectoid ( C ) Ti Mo Ni Si Cr Mn W (wt%c C) C eutect toid Ti Si Mo Cr Ni W Mn wt. % of alloying elements wt. % of alloying elements

22 Taxonomy of Metals Metal Alloys Ferrous Nonferrous Steels <1.4wt% C Cast Irons 3-4.5wt% C Cu Al Mg Ti T( C) L microstructure: ferrite, graphite cementite austenite +L 1148 C L+Fe 3 C Eutectic: 4.30 ferrite C Eutectoid: Fe 3 C +Fe 3 C Fe3C cementite (Fe) C o, wt% C

23 Steels increasing strength, cost, decreasing ductility

24 Steell St Types (( for for your information only Not required for the examination!)