Lecture13-14 - Iron-Carbon (Fe-C) Phase Diagram T(°C) •...

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1 1 Lecture 13-14 Steel 2 Iron-Carbon (Fe-C) Phase Diagram • 2 important points -Eutectoid ( B ): γ⇒α +Fe 3 C -Eutectic ( A ): L ⇒γ +Fe 3 C Adapted from Fig. 9.24, Cal ister 7e . Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 123456 6 . 7 L γ (austenite) γ + L γ +Fe 3 C α +Fe 3 C α + γ L +Fe 3 C δ (Fe) C o , wt% C 1148°C T (°C) α 727°C = T eutectoid A S R 4.30 Result: Pearlite = alternating layers of α and Fe 3 C phases 120 μ m (Adapted from Fig. 9.27, Cal ister 7e .) γ γ γ γ R S 0.76 C eutectoid B Fe 3 C (cementite-hard) α (ferrite-soft) 3 Hypo eutectoid Steel Adapted from Figs. 9.24 and 9.29, Cal ister 7e . (Fig. 9.24 adapted from Binary Al oy Phase Diagrams , 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) 1600 1400 1200 1000 800 600 400 0 6 . 7 L γ (austenite) γ + L γ 3 C α 3 C L +Fe 3 C δ (Fe) C o , wt% C 1148°C T (°C) α 727°C (Fe-C System) C0 0.76 Adapted from Fig. 9.30, Cal ister 7e . proeutectoid ferrite pearlite 100 μ m Hypoeutectoid steel R S α w α = S /( R + S ) w Fe 3 C =(1- w α ) w pearlite = w γ pearlite r s w α = s /( r + s ) w γ =(1- w α ) γ γ γ γ α α α γ γ γ γ γ γ γ γ 4 Hyper eutectoid Steel 1600 1400 1200 1000 800 600 400 0 6 . 7 L γ (austenite) γ + L γ +Fe 3 C α +Fe 3 C L +Fe 3 C δ (Fe) C o , wt%C 1148°C T (°C) α Adapted from Figs. 9.24 and 9.32, Cal ister 7e . (Fig. 9.24 adapted from Binary Al oy Phase Diagrams , 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) (Fe-C System) C o Adapted from Fig. 9.33, Cal ister 7e . proeutectoid Fe 3 C 60 μ m Hypereutectoid steel pearlite R S w α = S /( R + S ) w Fe 3 C =(1- w α ) w pearlite = w γ pearlite s r w Fe 3 C = r /( r + s ) w γ =(1- w Fe 3 C ) Fe 3 C γ γ γ γ γ γ γ γ γ γ γ γ 5 Equilibrium Fe-C phase diagram 6 Example: Phase Equilibria For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following a) composition of Fe 3 C and ferrite ( α ) b) the amount of carbide (cementite) in grams that forms per 100 g of steel c) the amount of pearlite and proeutectoid ferrite ( α )
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2 7 Phase Equilibria Solution: g 3 . 94 g 5.7 C Fe g 7 . 5 100 022 . 0 7 . 6 022 . 0 4 . 0 100 x C Fe C Fe 3 C Fe 3 3 3 = α = = = = α + α α x C C C C o b) the amount of carbide (cementite) in grams that forms per 100 g of steel a) composition of Fe 3 C and ferrite ( α ) C O = 0.40 wt% C C α = 0.022 wt% C C Fe C = 6.70 wt% C 3 FeC (cementite) 1600 1400 1200 1000 800 600 400 0 123456 6 . 7 L γ (austenite) γ + L γ +Fe 3 C α 3 C L +Fe 3 C δ C o , wt% C 1148°C T (°C) 727°C C O R S C Fe C 3 C α 8 Phase Equilibria c. the amount of pearlite and proeutectoid ferrite ( α ) note: amount of pearlite = amount of γ just above T E C o = 0.40 wt% C C α = 0.022 wt% C C pearlite = C
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This note was uploaded on 09/15/2010 for the course MATSCIE 250 taught by Professor Yalisove during the Fall '08 term at University of Michigan.

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Lecture13-14 - Iron-Carbon (Fe-C) Phase Diagram T(°C) •...

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