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1. Rank the following iron–carbon alloys and associated microstructures from the highest to the lowest tensile strength: (a) 0.25 wt%C with spheroidite (b) 0.25 wt%C with coarse pearlite (c) 0.60 wt%C with fine pearlite (d) 0.60 wt%C with coarse pearlite This ranking is as follows: 0.60 wt% C, fine pearlite 0.60 wt% C, coarse pearlite 0.25 wt% C, coarse pearlite 0.25 wt% C, spheroidite The 0.25 wt% C, coarse pearlite is stronger than the 0.25 wt% C, spheroidite since coarse pearlite is stronger than spheroidite; the composition of the alloys is the same. The 0.60 wt% C, coarse pearlite is stronger than the 0.25 wt% C, coarse pearlite, since increasing the carbon content increases the strength. Finally, the 0.60wt% C, fine pearlite is stronger than the 0.60 wt% C, coarse pearlite in as much as the strength of fine pearlite is greater than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine pearlite.

2. Briefly describe the microstructural difference between spheroidite and tempered martensite. Both tempered martensite and spheroidite have sphere-like cementite particles within a ferrite matrix; however, these particles are much larger for spheroidite.

3. For a eutectoid steel, describe isothermal heat treatments that would be required to yield specimens having the following Rockwell hardnesses: (Hint: Hardness of these materials are given in Figures 10.31, 10.33, and 10.36 in your book.) (a) 93 HRB (b) 40 HRC (c) 27 HRC For this problem we are asked to describe isothermal heat treatments required to yield specimens having several Brinell hardnesses. (a) From Figure 10.30a, in order for a 0.76 wt% C alloy to have a Rockwell hardness of 93 HRB, the microstructure must be coarse pearlite. Thus, utilizing the isothermal transformation diagram for this alloy, Figure10.22, we must rapidly cool to a temperature at which coarse pearlite forms (i.e., to about 675°C), allow the specimen to isothermally and completely transform to coarse pearlite. At this temperature an isothermal heat treatment for at least 200 s is required.(b) This portion of the problem asks for a hardness of 40 HRC the microstructure could consist of either (1)about 75% fine pearlite and 25% martensite (Figure 10.32), or (2) tempered martensite (Figure 10.35).For case (1), after austenitizing, rapidly cool to about 580°C (Figure 10.22), hold at this temperature for about 4 s (to obtain 75% fine pearlite), and then rapidly quench to room temperature. For case (2), after austenitizing, rapidly cool to room temperature in order to achieve 100% martensite. Then temper this martensite for about 2000 s at 535°C (Figure 10.35).(c) From Figure 10.30a, for a 0.76 wt% C alloy to have a Rockwell hardness of 27 HRC, the microstructure must be fine pearlite. Thus, utilizing the isothermal transformation diagram for this alloy, Figure10.22, we must rapidly cool to a temperature at

which fine pearlite forms (i.e., at about 580°C), allow the specimen to isothermally and completely transform to fine pearlite. At this temperature an isothermal heat treatment for at least 7 s is required.

4. A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 106 psi) and an original diameter of 3.8 mm (0.15 in.) will experience only elastic deformation when a tensile load of 2000 N (450 lbf) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.42 mm (0.0165 in.).

5. A cylindrical bar of steel 10 mm (0.4 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Using the data in Table 6.1, determine the force that will produce an elastic reduction of 3  10-3 mm (1.2  10-4 in.) in the diameter.

6. A cylindrical specimen of some alloy 8 mm (0.31 in.) in diameter is stressed elastically in tension. A force of 15,700 N (3530 lbf) produces a reduction in specimen diameter of 5  10-3 mm (2  10-4 in.). Compute Poisson's ratio for this material if its modulus of elasticity is 140 GPa (20.3  106 psi).

7. A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensile load of 50,000 N (11,240 lbf) is applied. Under these circumstances, what must be the radius of the specimen? Consider this brass alloy to have the stress-strain behavior shown in Figure 6.12.

8. 1 Figure 6.22 shows the tensile engineering stress–strain behavior for a steel alloy. (a) What is the modulus of elasticity? (b) What is the yield strength? (d) What is the tensile strength?

9.

A bar of a steel alloy that exhibits the stress-strain behavior shown in Figure 6.22 is subjected to a tensile load; the specimen is 300 mm (12 in.) long, and of square cross section 4.5 mm (0.175 in.) on a side. (a) Compute the magnitude of the load necessary to produce an elongation of 0.45 mm (0.018 in.). (b) What will be the deformation after the load has been released? Given Length=300mm Cross section length=4.5mm Steel alloy strain behavior

A0=0.45x0.45 =20.25mm2

Stress= 𝜎 =

𝐹 𝐴0

Magnitude of load, F= 𝜎 × 𝐴0 Then for magnitude of load need to calculate stress. To calculate stress, we need strain. 𝜀=

∆𝑙 0.45 = = 0.0015 𝑙 300

(a)From the stress strain curve, stress at 𝜀 = 0.0015 is approximately 300MPa The magnitude of load, 𝐹 = 300 × 20.25 F=6075 N (b)After the release of the load of 6075N there will not be any deformation because from the stress strain plot, it is evident that the sample has undergone elastic strain.

10. A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stress of 575 MPa (83,500 psi) is applied; for the same metal, the value of K in Equation 6.19 is 860 MPa (125,000 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87,000 psi).

11. A steel alloy specimen having a rectangular cross section of dimensions 12.7 mm × 6.4 mm (0.5 in. × 0.25 in.) has the stress–strain behavior shown in Figure 6.22. If this specimen is subjected to a tensile force of 38,000 N (8540 lbf) then (a) Determine the elastic and plastic strain values. (b) If its original length is 460 mm (18.0 in.), what will be its final length after the load in part (a) is applied and then released?

12. A cylindrical rod, to be constructed from a steel that has a yield strength of 310 MPa, is to withstand a load of 220,000 N without yielding. Assuming a value of 4 for N, specify a suitable bar diameter. Given σy = 310MPa N= 4 F=220,000N 𝜎 We have 𝜎𝑤 = 𝑁𝑦 𝑑 2

𝐹

𝐴0 = ( ) × 𝜋 = 𝜎 2 𝐹×𝑁

𝑑 = 2 × √𝜋×σy

𝑤

= 0.06m=60mm d = 60mm...