3. A copper-nickel diffusion couple similar to that shown in Figure 5.1a is fashioned. After a 700-h heat treatment at 1100 C (1373 K) the

Transcription

1 ENT 145 Tutorial 3 1. A sheet of steel 1.8 mm thick has nitrogen atmospheres on both sides at 1200 C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is m 2 /s, and the diffusion flux is found to be kg/m 2 s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m 3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m 3? Assume a linear concentration profile. 2. Determine the carburizing time necessary to achieve a carbon concentration of 0.45 wt% at a position 2 mm into an iron carbon alloy that initially contains 0.20 wt% C. The surface concentration is to be maintained at 1.30 wt% C, and the treatment is to be conducted at 1000 C. Use the diffusion data for γ-fe in Table 5.2.

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3 3. A copper-nickel diffusion couple similar to that shown in Figure 5.1a is fashioned. After a 700-h heat treatment at 1100 C (1373 K) the concentration of Cu is 2.5 wt% at the 3.0-mm position within the nickel. At what temperature must the diffusion couples need to be heated to produce this same concentration (i.e., 2.5 wt% Cu) at a 2.0-mm position after 700 h? The preexponential and activation energy for the diffusion of Cu in Ni are given in Table 5.2.

4 4. An FCC iron carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.0 wt%. If after 49.5 h the concentration of carbon is 0.35 wt% at a position 3.5 mm below the surface, determine the temperature at which the treatment was carried out.

5 5. Compute the mass fractions of α-ferrite and cementite in pearlite.

6 6. (a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? Answer: (a) A hypoeutectoid steel has a carbon concentration less than the eutectoid; on the other hand, a hypereutectoid steel has a carbon content greater than the eutectoid. (b) For a hypoeutectoid steel, the proeutectoid ferrite is a microconstituent that formed above the eutectoid temperature. The eutectoid ferrite is one of the constituents of pearlite that formed at a temperature below the eutectoid. The carbon concentration for both ferrites is wt% C. 7. Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 725 C (998 K). (a) What is the proeutectoid phase? (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form?

7 8. Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron carbon alloy containing 0.30 wt% C. 9. Using the isothermal transformation diagram for an iron carbon alloy of eutectoid composition (Figure 2), specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time temperature treatments. In each case assume that the specimen begins at 760 C (1033 K) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Cool rapidly to 700 C (973 K), hold for 10 4 s, then quench to room temperature. (b) Reheat the specimen in part (a) to 700 C (973 K) for 20 h. (c) Rapidly cool to 600 C (873 K), hold for 4 s, rapidly cool to 448 C (721 K), hold for 10 s, then quench to room temperature. (d) Cool rapidly to 398 C (671 K), hold for 2 s, then quench to room temperature. (e) Cool rapidly to 398 C (671 K), hold for 20 s, then quench to room temperature. (f) Cool rapidly to 398 C (671 K), hold for 200 s, then quench to room temperature. (g) Rapidly cool to 575 C (848 K), hold for 20 s, rapidly cool to 350 C (623 K), hold for 100 s, then quench to room temperature. (h) Rapidly cool to 250 C (523 K), hold for 100 s, then quench to room temperature in water. Reheat to 315 C (588 K) for 1 h and slowly cool to room temperature

8 ANSWER ; (a) 50% pearlite 50% martensite (b) Spherodite (c) 50% pearlite, 25% bainite, 25% martensite

9 (d) 100% martensite (e) 50% bainite 50% martensite (f) 100% bainite

10 (g) 100% pearlite (h) 100% tempered martensite

11 10. What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of mm and a crack length of mm when a tensile stress of 170 MPa is applied? 11. Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.28 mm and having a tip radius of curvature of mm when a stress of 1200 MPa is applied.

12 12. A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPa m is exposed to a stress of 1000 MPa. Will this specimen experience fracture if it is known that the largest surface crack is 0.76 mm long? Why or why not? Assume that the parameter Y has a value of Figure 1 shows S-N diagram for a brass alloy. (a) Determine the fatigue strength at cycles. (b) Determine the fatigue life for 200 MPa. (c) Suppose that the fatigue data for the brass alloy were taken from torsional tests, and that a shaft of this alloy is to be used for a coupling that is attached to an electric motor operating at 1500 rpm. Give the maximum torsional stress amplitude possible for each of the following lifetimes of the coupling: (a) 1 year, (b) 1 month, (c) 1 day, and (d) 2 hours. (a) (b)

13 (C) For each lifetime, first compute the number of cycles, and then read the corresponding fatigue strength from the above plot. (a) Fatigue lifetime = (1 yr)(365 days/yr)(24 h/day)(60 min/h)(1500 cycles/min) = cycles. The stress amplitude corresponding to this lifetime is about 130 MPa. (b) Fatigue lifetime = (30 days)(24 h/day)(60 min/h)(1500 cycles/min) = cycles. The stress amplitude corresponding to this lifetime is about 145 MPa. (c) Fatigue lifetime = (24 h)(60 min/h)(1500 cycles/min) = cycles. The stress amplitude corresponding to this lifetime is about 195 MPa. (d) Fatigue lifetime = (2 h)(60 min/h)(1500 cycles/min) = cycles. The stress amplitude corresponding to this lifetime is about 315 MPa. 15. List two measures that may be taken to increase the resistance to fatigue of a metal alloy. ANSWER: Measures that may be taken to increase the fatigue resistance of a metal alloy (1) Polish the surface to remove stress amplification sites. (2) Reduce the number of internal defects (pores, etc.) by means of altering processing and fabrication techniques. (3) Modify the design to eliminate notches and sudden contour changes. (4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot peening. 16. For a cylindrical S-590 alloy specimen (Figure 4) originally 10 mm in diameter and 505 mm long, what tensile load is necessary to produce a total elongation of 145 mm after 2,000 h at 730 C (1003 K)? Assume that the sum of instantaneous and primary creep elongations is 8.6 mm.

14 17. A cylindrical component constructed from an S-590 alloy (Figure 3) has a diameter of 12.5 mm. Determine the maximum load that may be applied for it to survive 500 h at 925 C (1198 K). Figure 1 S-N curves

15 Figure 2 TTT Diagrams

16 Figure 3 Stress vs rupture lifetime Figure 4 Stress vs steady-state creep rate