7. A continuous cooling curve is shown below for AISI 15B41 steel containing nom

Question

7. A continuous cooling curve is shown below for AISI 15B41 steel containing nominally 0.41 wt. % C and 1.6 % Mn. a) According to this curve, how many seconds do you have in order to comfortably form 100 % martensite? b) What is the minimum cooling rate in °C/s in order to miss forming any proeutectoid ferrite and pearlite? How much martensite would form at this cooling rate and what other phase would there be? 1,600 0.42 C. 161 Mn. 0.29 Si 0.006 P.0.019 S.0.004 B ??.-1.333 F AISI 15B41 800 1,400 Grain size. ASTM No. 7-8 F-Ferrite P- Pearlite B- Bainite M- Martensite 35% Ferrite 1,200 65% Pearlite 600 15 458 1,000 E 800 40 400F 600 70% Bainite 30 1.200 50 250 0.50 400 200 Hardness Dph Rockwell 689 695 698 C60 C60 C60C 275. C31 ? C24 C B99 B96 200 10 104 Cooling time. sec

Explanation / Answer

(a) The cooling rate should be high enough so that 100% martensite is formed. For this should not touch the C portion of curve. Hence from graph we can see that for 100 %martensite to be formed, cooling time from austenite phase must be around 36 s.

(b) As there is no peoeutectoid and pearlite phase formations, cooling time less than 120 s is needed. Thus will completely avoid formation of the above two phases( 740 C/120= 6.17C/s). As this cooling rate increases the amount of martensite formed increases. More over formation of martensite and Bainite us competitive to each other. The fractions formed depends on cooling rate.

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