A 0.65 percent C hypoeutectoid plain-carbon steel is slowly cooled from about 950°C to a temperature just slightly above 723°C. Calculate the weight percent austenite and weight percent proeutectoid ferrite in this steel.
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- Question-6 A 0.25% C hypoeutectoid plain-carbon steel is slowly cooled from 950°C to a temperature just slightly below 723°C. (a) Calculate the weight percent proeutectoid ferrite in the steel. (b) Calculate the weight percent eutectoid ferrite and weight percent eutectoid cementite in the steel.Describe the final microstructure obtained in a 1050 steel after heat at 8200C, cooled to 7000C and hold for 5 s, cooled to room temperature. 900- a+y 800 A, 700 A1 F, Pr a+ pearlite 23 600 -y+a+pearlite 30 500 B1 y+ bainite Bainite 39 F=Ferrite B, M, 400 49 P=Pearlite 300 Y+ martensite B=Bainite 200 Martensite M=Martensite 62 100- 102 10 10 10 10 0.1 1 10 Time (s) TTT curve for carbon steel AISI 1050 Temperature (°C) Rockwell C hardnessThis equilibrium phase of steel only occurs at temperatures above 723° C: martensite ferrite austenite O cementite
- 3.The TTT diagram of a plain carbon steel is given in Fig. 1. Identify whether this steel is a hypoeutectoid, eutectoid or hypereutectoid steel. Explain why the transformation happens slowly at 850°C as well as at 300°C. Determine the microstructures expected in this type of steel after the following heat treatment processes. a) Austenize at 900°C, quench to 400°C and hold for 1000 s and quench to 25°C. b) Austenize at 900°C, quench to 25°C. c) Austenize at 900°C, quench to 675°C and hold for 1 s, quench to 400°C and hold for 900 s and slowly cool to 25°C. 900 Acm Cs 800 A1 FezC + Y 33 Fe3C + pearlite 45 700 Ps 600 ++ Fe3C + pearlite Y+ bainite Bs 500 46 Bainite 400 Yu Bf 300 57 Ms 200 60 Mf Y+ martensite 100 62 Martensite 102 65 103 104 105 106 0.1 1 10 Time (s) Fig. 1 Temperature (°C) Rockwell C hardnessQ3: With a moderately agitated water, a cylindrical piece of steel with 80 mm diameter is to be quenched. The hardnesses of the surface and center must be at least 55 and 40 HRC, respectively. Which of these alloys will satisfy the following requirements: 1040, 5140, 4340, 4140, 8620, 8630, 8640, and 8660? Cooling rate at 700c Cooling rate at 700°c 170 70 31 18 5.6 3.9 "C 270 170 70 31 18 9 60 2 "Os 5.6 3.9 2.8 100 100 50 4340 80 75 3 Surface 40 4140 Center 8640 30 5140 25 1040 20 10 20 30 40 50 mm 10 20 30 mm Distance from quenched end Equivalent distance from quenched end Figure 3 Figure 4 Hardness, HRC Percent martensite Diameter of bar (mm)Excellent combinations of hardness, strength, and toughness are obtained from bainite. One heat treater austenitized a eutectoid steel at 750°C, quenched and held the steel at 250°C for 15 min, and finally permitted the steel to cool to room temperature. Did he produce the required bainitic structure? Use the diagram below in your answer 900 600 700 14 00 Y. 400 12 52 M. 00 57 M. 100 0.3 10 10 10 Answer: Time Temperature ("C) Rockwell Chardnens
- Q1: Austenitized 40 mm diameter 5140 alloy steel bar is quenched in agitated oil. Predict what is the Rockwell hardness of this bar will be at (a) its surface and (b) its center (c) What do you think about the difference in hardness number between the center and surface (d) Differentiate between hardness and hardenability (e) Rank the steels in the figure below from lowest to highest hardenability and explain why. 600- Bar diameter (mm) 100 80 60 40 20 0 300 0 Cooling rate at 700°C (°C/sec). -150 55 0 تنا 25 ------- 5 S 10 12.5 8 M-R L 1/2 34-R Agitated oil 15 20 ¼ ¾ Distance from quenched end. De (Jominy distance) 5,5 54 Car Bar diameter (in.) 0 25 mm. 1 in. Hardness (Rockwell C) Where (C = center, S = surface, M-R = mid-radius) 2828 292 65 60- 55- 50 45 40 35 30 25 20 15 10 0 J 10 5140 30 20 Distance from quenched end (mm) 4340 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 Distance from quenched end (sixteenths of an inch) 40 9840 4140 8640 50Q1: Austenitized 40 mm diameter 5140 alloy steel bar is quenched in agitated oil. Predict what is the Rockwell hardness of this bar will be at (a) its surface and (b) its center (c) What do you think about the difference in hardness number between the center and surface (d) Differentiate between hardness and hardenability (e) Rank the steels in the figure below from lowest to highest hardenability and explain why. Cooling rate at 700°C (°C/sec) 300 150 011/3 55 100 600 Bar diameter (mm) 80 60 40 20 0 OLL 0 0 ww 25 12.5 8 5 S -------------- M-R 10 3/4-R Agitated oil 15 20 1/4 3/4 Distance from quenched end. De (Jominy distance) 5.5 4 3 Bar diameter (in.) 0 25 mm 1 in. Hardness (Rockwell C) Where (C = center, S = surface, M-R mid-radius) 65 60 55 50 45 40 35 30 25 20 15 10 0 10 1 20 Distance from quenched end (mm) 5140 1 30 L 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 Distance from quenched end (sixteenths of an inch) 4340 40 9840 4140 8640 50The TTT diagram for a 1076 eutectoid steel is given: What is the heat treatment process to form 50% Coarse Pearlite + 50% Martensite. 800 A 1400 Eutectoid temperature 700 A 1200 600 1000 500 800 400 A 600 300 M(start) 200 50% 400 M + A M(50%) M(90%) 100 200 10-1 1 10 102 103 104 105 Time (s) Temperature (°C) Temperature (°F)
- Determine the tensile strength, yield strength and percentage elongation (% ductility) for an iron- carbon alloy with 0.76 wt % carbon composition. O Yield strength - 60000 psi : Tensile strength - 110000 psi: Ductility - 15% O Yield strength - 130000 psi : Tensile strength - 68000 psi;: Ductility - 10% O Yield strength - 68000 psi : Tensile strength - 130000 psi; Ductility - 10% O Yield strength 50000 psi: Tensile strength - 100000 psi: Ductility - 20% O Yield strength - 86000 psi : Tensile strength 150000 psi: Ductility 25%A steel contains 18% pearlite and 82%primary ferrite at room temperature. Esti-mate the carbon content of the steel. Is thesteel hypoeutectoid or hypereutectoid?The chemical composition of X38CrMoV5-3 steel is given in the table below. b) This steel is austenitized at 1050 ° C and hardened by giving water in oil (completely martensite structure has been formed and has a hardness of 57 HRC) and tempered at 510 ° C to reduce its hardness (54 HRC) and gain toughness. In addition, it is desired to harden the surface by nitration. In order to achieve the desired surface hardness, liquid nitration should be done at 600 ° C, gas nitration at 510 ° C and plasma nitration at 450 ° C. Which nitration method do you prefer? Why is that?