Given the sigma 0 and ky for Mg, Al, and Fe plot yield strength vs d1/2 mm+1/2 f

Question

Given the sigma 0 and ky for Mg, Al, and Fe plot yield strength vs d1/2 mm+1/2 for both of these three systems. Watch your units! Calculate the strength for these three materials with a grain size of: 200 mu m 25 mu m Based upon the table given above, for which material(s) is strengthening through grain size reduction an effective strategy? Ky obviously depends upon the crystal structure. Explain why ky is larger for Mg w.r.t. Al. Explain why ky is larger for Mg w.r.t. Fe. Experimentally, it has been observed for single crystals of a number of metals that the yield strength sigma 0 is a function of the dislocation density PD as sigma ys = sigma 0 + where sigma 0 , G, b are given in the tabic in Problem 2, and A is a fitting constant for each material (assume 1.0 for this problem). For which material do you expect dislocation hardening (work hardening) to he most effective? Explain why using a mathematical argument. Calculate the yield strength of the three elements if the dislocation density is 108 cm-2 (Watch your units!). Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 degree and 35 degree, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa (900 psi), will an applied stress of 12 MPa (1750 psi) cause the single to yield? If not, what stress will be necessary?

Explanation / Answer

3)As shown inthe equation above, work hardening has a half root dependency on the number of dislocations. The material exhibits high strength if there are either high levels of dislocations (greater than 1014dislocations per m2) or no dislocations. A moderate number of dislocations (between 107and 109dislocations per m2) typically results in low strength.

The work-hardened steel bar has a large enough number of dislocations that the strain field interaction prevents all plastic deformation. Subsequent deformation requires a stress that varies linearly with thestrainobserved, the slope of the graph of stress vs. strain is the modulus of elasticity, as usual.

The work-hardened steel bar fractures when the applied stress exceeds the usual fracture stress and the strain exceeds usual fracture strain. This may be considered to be the elastic limit and theyield stressis now equal to thefracture toughness, which is of course, much higher than a non-work-hardened-steel yield stress.

Assuming a = 1.0

for Mg

yield strength = 542.355 GPa

for Al

yield strength = 743.61 GPa

for Fe

yield strength = 2033.67 GPa

2)posting in a few moments

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