The strength of heat treated steel decreases with increasing tempering temperature
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Q: How a heat treatment process needs to be carried out to obtain a martensite structure in steel?
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Q: Figure 11.20 For cylindrical specimens of an oil-quenched 4140 steel, (a) tensile strength, (b)…
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Q: a) Explain the normalizing process in the heat treatment of metals. b) What are the final properties…
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- 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. 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 50A nitriding heat treatment of a BCC steel normally requires 2 h at 600 °C. What temperature would be required to reduce the heat treatment time to 1 h?(Q =76570 J/mol, Do =0.0047x10- 4 m2/s) and the last one. thanks for ur helping.A nitriding heat treatment of a BCC steel normally requires 2 h at 600 °C. What temperature would be required to reduce the heat treatment time to 1 h?(Q =76570 J/mol, Do =0.0047x10- 4 m2/s)
- 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 50(b) Figure Q4 (b) displays the time-temperature cooling paths for a 1080 steel on an isothermal transformation diagram. Start with the steels in the austenitic condition at time = 0 and 850°C. Interpret the final microstructures that will be obtained for paths (i, ii and iii) shown in the figure. 700 600 500 400 300 (iii) Ms 200 (i) (ii) Mf Figure Q4 (b)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 hardness
- Q6/ A. Show how the precipitation heat treatment of precipitation hardenable alloys differs from dispersion hardening process of composite material. B. What are the main constitutes of composite materials and what are the roles of each constitute?Some alloys use a combination of strain hardening and precipitation hardening to achieve particularly high strength levels. The usual order of strengthening is solution treatment, quenching, cold working, and finally precipitation heat treatment. Why not reverse the order of the cold working and precipitation heat treatment steps?Using the isothermal transformation curve for a Eutectoid steel, determine the final structure of alloy for the following cooling processes
- Use the following TTT Diagram for the following questions: 800 A 1400 -Eutectoid temperature 700 1200 600 1000 500 800 400 600 300 Mistart) 200 50% 400 M+A M(50%) M(90%) 100 200 10-1 10 102 103 10 105 Time (s) Temperature ("C) Termperature (°F)Q3: 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)jnd Retained austenite represents a large problem in hardened steel, because ....*. ..........