Phase Diagram Fe3C

100X

500X

Ferrite and Austenite Ferrite Austenite

Hypoeutectoid Steel T(°C) d L +L g (austenite) Fe3C (cementite) a 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L g (austenite) +L + Fe3C a L+Fe3C d (Fe) Co , wt% C 1148°C T(°C) 727°C (Fe-C System) C0 0.76 g g r s w a = /( + ) g (1- R S a w = S /( R + ) Fe3C (1- pearlite g proeutectoid ferrite pearlite 100 mm Hypoeutectoid steel

Hypereutectoid Steel Fe3C (cementite) L g (austenite) +L a d T(°C) g g 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L g (austenite) +L +Fe3C a L+Fe3C d (Fe) Co , wt%C 1148°C T(°C) (Fe-C System) g g s r w Fe3C = /( + ) g =(1- proeutectoid Fe3C 60 mm Hypereutectoid steel pearlite R S w a = /( + ) Fe3C (1- g Co 0.76

Iron-Carbon (Fe-C) Phase Diagram • 2 important Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L g (austenite) +L +Fe3C a + L+Fe3C d (Fe) Co, wt% C 1148°C T(°C) 727°C = T eutectoid points -Eutectic (A): L Þ g + Fe3C A S R 4.30 -Eutectoid (B): g Þ a + Fe3C g 120 mm 0.76 C eutectoid B R S Fe3C (cementite-hard) a (ferrite-soft) Result: Pearlite = alternating layers of a and Fe3C phases

Example: Phase Equilibria For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following composition of Fe3C and ferrite () the amount of carbide (cementite) in grams that forms per 100 g of steel the amount of pearlite and proeutectoid ferrite ()

Solution: a) composition of Fe3C and ferrite () 3 the amount of carbide (cementite) in grams that forms per 100 g of steel CO = 0.40 wt% C Ca = 0.022 wt% C CFe C = 6.70 wt% C 3 Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L g (austenite) +L + Fe3C a L+Fe3C d Co , wt% C 1148°C T(°C) 727°C CO R S CFe C 3 C

Co = 0.40 wt% C Ca = 0.022 wt% C Cpearlite = C = 0.76 wt% C the amount of pearlite and proeutectoid ferrite () note: amount of pearlite = amount of g just above TE Co = 0.40 wt% C Ca = 0.022 wt% C Cpearlite = C = 0.76 wt% C Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L g (austenite) +L + Fe3C a L+Fe3C d Co , wt% C 1148°C T(°C) 727°C CO R S C C pearlite = 51.2 g proeutectoid  = 48.8 g

Alloying Steel with More Elements • Teutectoid changes: • Ceutectoid changes: Ti (wt%C) Si (°C) Mo W Ni Cr Cr Eutectoid eutectoid Si Mn Mn W Ti Mo T C Ni wt. % of alloying elements wt. % of alloying elements

Summary • Phase diagrams are useful tools to determine: --the number and types of phases, --the wt% of each phase, --and the composition of each phase for a given T and composition of the system. • Alloying to produce a solid solution usually --increases the tensile strength (TS) --decreases the ductility. • Binary eutectics and binary eutectoids allow for a range of microstructures.

End