Question

QUESTION 4 (10 points) a. Using the isothermal transformation diagram for iron carbon alloy of eutectoid composition. Specify and sketch the resulting microstructure from the following heat treatments: (Assume that, in each case, the specimen begins at 760 C and that it attains a homogeneous austenite structure) (N.B. ATTACH THE T-T-T DIAGRAM SHEET WITH THE ANSWER SHEET) 11) Cool rapidly to 370 C, hold for 100 s, and then quench to room temperature. [2] Cool rapidly to 250 C, hold for 10 s, then quench to room temperature. 13) Reheat the resulting microstructure obtained in part (2) to 400 for 1 hour and slowly cool to room temperature. [4] Cool rapidly to450 C, hold for 10 s, rapidly cool to 250 C hold for 1000 s then quench to room temperature. [5] Reheat the resulting microstructure obtained in part (4) at 700 C for 24 hours and slowly cool to room temeprarture. -Eutectold temperature 1400 Temperature (°C) Temperature start) 200- 50%) M90%) 100 200 10-4 1 10 102 Times) 103 104 109

Answer 1

1200 Austenite ( 1100- 1000 - 7 + Fec 900 - 800 - T(°C) 723°C 0.80 700 – Ferrite (a) 600 - a + Fe :C 500 tos to is zo 3.0 Fe Wt%C

solution : - toward Labelling the of T-T-T diagram &e then moving questions step by step :- labelling of T-T-T diagram for eutectoid steel step 1:- eutectoid reaction:- Iron carbon diagram is shown in pic below the entectoid reaction happens to occur at 0.8%. Wt , 723°c where upon cooling 723°C Y = x + Fezc part 1 starting point KE " - 760°C 11A= y iron Y 0.8.1.wt eutedoid T - - - - - - - - T-723 - - - - - - - - - - - - - - - CP - corseser Pearlite A Pearlite start line 50%. Atpun..... FP- Fine Peaclite - - - - - - - - -. - - - - + -- T°C Bainite Staut line 400 + B = Bainite B (1=504., BE Apoo & structure) 801) 300 t Mstart (Martensite) 200 € -.Moso7. --.: 1 10 10 10 t (sec) → the 50% curve for (B=80%. , so for M= 80%) pevet 1, it will be (A + B)

1 hour = 60X60 - 3600 sec part 3:- Į A = e-iron - - - - - - - - . - - - - - ___723°C - - 700 - Pearlite start line Sot -- Ati -..-..- Go + CP To → .- 1 7 -FP 600 Bainite start 400 final Structure 300 M start - - - - - DO part 2 100 1000 tell you (sec) 10 1 A = yiron - - - - - - - - - - 723°c 700 + Pearlite start line A+P------- > CD ou 600 -90%... Atp line 500 + A+B B = Bainete 400 y - 100% 300 Bainite start/ line + - M = 100% L M start line : 11 HO 200 10 Too Noo → + (sec)

part :- line - - - - - ---7230 - - - start line - - - - 700 + pearlite A line - ÁP CP = course peaulite 50% line 600 ---7. Fp - Fine peaelite 500 t B = Bainite A +B 400 + lune Bainite stat ... 300- - Mstart ~ -- 200 (Martensite_(80%) Į 22) - 10-'sec 1sec losec 1000 sec stage stage stage stage stage oso 18 LOOD 00000000000 000 02 x= 100/ y=100% x=soy. 7 = 504 B-80% x = 25% B = 15% B-007.

part 5: 723c 700 + - - - - - stucture 08 heB - - - - - B = Bainite CP CP= 25%. B - 75-). 1000) 100 300 t 2000 OOOR - -. -- y = 25.4 8=759/ - M start 2000 - - - - - - - - - - - - - 10-1 1 10 100 sec 1000 sec → Time (sec)