# Engineering Materials Week 3

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5.1 Briefly explain the difference between self-diffusion and interdiffusion.

5.7 A sheet of steel 2.5mm thick has nitrogen atmospheres on both sides at 900C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient of nitrogen in steel at 900 C is 1.2 10^-10 m^2/s, and the diffusion flux is found to be 1.0 10^-7 kg/m^2-s. The concentration of nitrogen in the steel at the high-pressure side is 2kg/m^3. How far into the sheet from this high pressure side will the concentration be 0.5kg/m^3? Assume a linear concentration profile.

5.17 Using the data in Table 5.2, compute the value of D for the diffusion of magnesium in aluminum at 400C.

5.30 The outer surface of a steel gear is to be hardened by increasing its carbon content; the carbon is to be supplied from an external carbon-rich atmosphere that is maintained at an elevated temperature. A diffusion heat treatment at 600C (873K) for 100 min increases the carbon concentration to 0.75 wt% at a position 0.5mm below the surface. Estimate the diffusion time required at 900C (1173K) to achieve this same concentration also at 0.5mm position.

6.5 An aluminum bar 125mm (5.0 in.) long and having a square cross section 16.5 mm (0.65 in.) on an edge is pulled in tension with a load of 66,700 N (15,000 lb) and experiences an elongation of 0.43 mm. Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

6.24 Figure 6.21 shows the tensile engineering stress-strain behavior for a steel alloy. a. What is the modulus of elasticity?

1. What is the proportional limit?
2. What is the yield strength at a strain off-set of 0.002?
3. What is the tensile strength?

6.38 For some metal alloy, a true stress of 345 MPa produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 415 MPa is applied if the original length is 500mm. Assume a value of 0.22 for the strain-hardening exponent, n.

6.46

1. A 10 mm diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this material?
2. What will the diameter of an indentation to yield a hardness of 300HB when a 500 kg load is used?
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