Fe-Carbon Phase Diagram Prepared By: Behzad Heidarshenas Ph.D in Manufacturing Processes Mechanical Engineering Department

Fe-Carbon Diagram

IRON CARBON CONSTITUTIONAL DIAGRAM-II

The following phases are involved in the transformation, occurring with iron-carbon alloys: L - Liquid solution of carbon in iron; δ-ferrite – Solid solution of carbon in iron. Maximum concentration of carbon in δ-ferrite is 0.09% at 2719 ºF (1493ºC) – temperature of the peritectic transformation. The crystal structure of δ-ferrite is BCC (cubic body centered). Austenite – interstitial solid solution of carbon in γ-iron. Austenite has FCC (cubic face centered) crystal structure, permitting high solubility of carbon – up to 2.06% at 2097 ºF (1147 ºC). Austenite does not exist below 1333 ºF (723ºC) and maximum carbon concentration at this temperature is 0.83%. α-ferrite – solid solution of carbon in α-iron. α-ferrite has BCC crystal structure and low solubility of carbon – up to 0.25% at 1333 ºF (723ºC). α-ferrite exists at room temperature. Cementite – iron carbide, intermetallic compound, having fixed composition Fe3C.

CRITICAL TEMPERATURE Upper critical temperature (point) A3 is the temperature, below which ferrite starts to form as a result of ejection from austenite in the hypoeutectoid alloys. Upper critical temperature (point) ACM is the temperature, below which cementite starts to form as a result of ejection from austenite in the hypereutectoid alloys. Lower critical temperature (point) A1 is the temperature of the austenite-to-pearlite eutectoid transformation. Below this temperature austenite does not exist. Magnetic transformation temperature A2 is the temperature below which α-ferrite is ferromagnetic.

PHASE COMPOSITIONS OF THE IRON-CARBON ALLOYS AT ROOM TEMPERATURE Hypoeutectoid steels (carbon content from 0 to 0.83%) consist of primary proeutectoid) ferrite (according to the curve A3) and pearlite. Eutectoid steel (carbon content 0.83%) entirely consists of pearlite. Hypereutectoid steels (carbon content from 0.83 to 2.06%) consist of primary (proeutectoid) cementite (according to the curve ACM) and pearlite. Cast irons (carbon content from 2.06% to 4.3%) consist of proeutectoid cementite C2 ejected from austenite according to the curve ACM , pearlite and transformed ledeburite (ledeburite in which austenite transformed to pearlite.

PHASES OF IRON FCC (Austenite) BCC (Ferrite) HCP (Martensite)

Alpha “Ferrite”, BCC Iron Room Temperature Gamma “Austenite”, FCC Iron Elevated Temperatures These are PHASES of iron. Adding carbon changes the phase transformation temperature.

MICROSTRUCTURE OF AUSTENITE

MICROSTRUCTURE OF PEARLITE Photomicrographs of (a) coarse pearlite and (b) fine pearlite. 3000X

MICROSTRUCTURE OF MARTENSITE

IRON-CARBON (Fe-C) PHASE DIAGRAM (EXAMPLE 1) • 2 important Adapted from Fig. 9.24, Callister & Rethwisch 8e. Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L g (austenite) +L +Fe3C a + d (Fe) C, wt% C 1148ºC T(ºC) 727ºC = T eutectoid points - Eutectic (A): L Þ g + Fe3C A L+Fe3C - Eutectoid (B): g Þ a + Fe3C g Result: Pearlite = alternating layers of a and Fe3C phases 120 mm (Adapted from Fig. 9.27, Callister & Rethwisch 8e.) 0.76 B Fe3C (cementite-hard) a (ferrite-soft) 4.30

EXAMPLE 1 An alloy of eutectoid composition (0.76 wt% C) as it is cooled down from a temperature within the g-phase region (e.g., at 800 ºC). Initially the alloy is composed entirely of the austenitic phase having a composition of 0.76 wt% C As the alloy is cooled, no changes will occur until the eutectoid temperature (727 ºC). Upon crossing this temperature to point B, the austenite transforms according to: Eutectoid (B): g (0.76 wt% C) Þ a (0.022 wt% C) + Fe3C (6.7 wt% C)

EXAMPLE 1 (cont.) The microstructure for this eutectoid steel is slowly cooled through the eutectoid temperature consists of alternating layers or lamellae of the two phases (a and Fe3C) that form simultaneously during the transformation. Point B is called pearlite. Mechanically, pearlite has properties intermediate between the soft, ductile ferrite and the hard, brittle cementite.

EXAMPLE 1 (cont.) The alternating a and Fe3C layers in pearlite form as such for the same reason that the eutectic structure forms because the composition of austenite (0.76 %wt C) is different from either of ferrite (0.022 wt% C) and cementite (6.70 wt% C), and the phase transformation requires that there be a redistribution of the carbon by diffusion. Subsequent cooling of the pearlite from point B will produce relatively insignificant microstructural changes.

Hypoeutectoid Steel (EXAMPLE 2) 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) C, wt% C 1148ºC T(ºC) 727ºC (Fe-C System) C0 0.76 g g Adapted from Figs. 9.24 and 9.29,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) g a a pearlite Adapted from Fig. 9.30, Callister & Rethwisch 8e. proeutectoid ferrite pearlite 100 mm Hypoeutectoid steel

EXAMPLE 2 (cont.) Within the a + g region, most of the a particles will form along the original g grain boundaries. The particles will grow larger just above the eutectoid line. As the temperature is lowered below Te, all the g phase will transform to pearlite according to: There will be virtually no change in the a phase that existed just above the Te. This a that is formed above Te is called proeutectoid (pro=pre=before eutectoid) ferrite. g Þ a + Fe3C

EXAMPLE 2 (cont.) The ferrite that is present in the pearlite is called eutectoid ferrite. As a result, two microconstituents are present in the last micrograph (the one below Te): proeutectoid ferrite and pearlite

EXAMPLE 2 T(ºC) d L +L g (austenite) Fe3C (cementite) Wa = s/(r + s) 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) C, wt% C 1148ºC T(ºC) 727ºC (Fe-C System) C0 0.76 g a Adapted from Figs. 9.24 and 9.29,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) Wa = s/(r + s) Wg =(1 - Wa) s r a pearlite R S Wpearlite = Wg Wa’ = S/(R + S) W =(1 – Wa’) Fe3C Adapted from Fig. 9.30, Callister & Rethwisch 8e. proeutectoid ferrite pearlite 100 mm Hypoeutectoid steel

HYPEREUTECTOID STEEL (EXAMPLE 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 (Fe) C, wt%C 1148ºC T(ºC) (Fe-C System) g g Adapted from Figs. 9.24 and 9.32,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) Fe3C g pearlite C0 0.76 Adapted from Fig. 9.33, Callister & Rethwisch 8e. proeutectoid Fe3C 60 mm Hypereutectoid steel pearlite

EXAMPLE 3 (cont.) Fe3C (cementite) L g (austenite) +L a d T(ºC) g x v 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) C, wt%C 1148ºC T(ºC) (Fe-C System) Fe3C g Adapted from Figs. 9.24 and 9.32,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) W =(1-Wg) Wg =x/(v + x) Fe3C x v pearlite V X C0 0.76 Wpearlite = Wg Wa = X/(V + X) W =(1 - Wa) Fe3C’ Adapted from Fig. 9.33, Callister & Rethwisch 8e. proeutectoid Fe3C 60 mm Hypereutectoid steel pearlite

PROBLEM For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following: The compositions of Fe3C and ferrite (). The amount of cementite (in grams) that forms in 100 g of steel. The amounts of pearlite and proeutectoid ferrite () in the 100 gr.

SOLUTION TO PROBLEM a) Using the RS tie line just below the eutectoid Ca = 0.022 wt% C CFe3C = 6.70 wt% C Using the lever rule with the tie line shown 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 C , wt% C 1148ºC T(ºC) 727ºC C0 R S CFe C 3 C Amount of Fe3C in 100 g = (100 g)WFe3C = (100 g)(0.057) = 5.7 g

SOLUTION TO PROBLEM (CONT.) c) Using the VX tie line just above the eutectoid and realizing that C0 = 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 C, wt% C 1148ºC T(ºC) 727ºC C0 V X C C Amount of pearlite in 100 g = (100 g)Wpearlite = (100 g)(0.512) = 51.2 g

ALLOYING WITH OTHER ELEMENTS • Teutectoid changes: Adapted from Fig. 9.34,Callister & Rethwisch 8e. (Fig. 9.34 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) T Eutectoid (ºC) wt. % of alloying elements Ti Ni Mo Si W Cr Mn • Ceutectoid changes: Adapted from Fig. 9.35,Callister & Rethwisch 8e. (Fig. 9.35 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