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Strengthening mechanisms and dislocations

The yield stress σy of an FCC polycrystal is about three times the critical resolved shear stress τc of a single crystal (σy ≈ Mτc, M ≈ 3.06). Which is the correct reason why this factor M (the Taylor factor) is larger than 2, the reciprocal of the maximum Schmid factor of 0.5?

  1. AIn a polycrystal only the most favorably oriented grains slip first while the others stay elastic, so M becomes larger to make up for the lower average strength
  2. BFor neighboring grains to deform without opening gaps or overlapping at the grain boundaries, at least five independent slip systems must operate in each grain, including poorly oriented ones
  3. CA polycrystal has far more grain-boundary area than a single crystal, and the grain-boundary energy adds to the work of deformation, so M is about 3 regardless of grain size
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Answer: B. For neighboring grains to deform without opening gaps or overlapping at the grain boundaries, at least five independent slip systems must operate in each grain, including poorly oriented ones

In a polycrystal, if each grain deformed into any shape it liked, gaps and overlaps would open at the grain boundaries. To accommodate an arbitrary change of shape at constant volume, five independent slip systems are needed (the von Mises criterion). As a result, poorly oriented systems must operate along with those of high Schmid factor, and averaging over a randomly oriented FCC polycrystal gives a Taylor factor of about 3.06. If yielding were set only by the most favorably oriented grains, a factor larger than 2 could not be explained. The Taylor factor is set by the distribution of grain orientations and does not depend on grain-boundary area or grain size; the effect of grain size is added separately through the Hall–Petch relationship. The same criterion explains why polycrystals of HCP metals, which lack enough independent slip systems, have poor ductility.

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More Strengthening mechanisms and dislocations questions

  1. The yield stress of BCC metals such as iron rises sharply as the temperature is lowered, whereas in FCC metals such as copper and aluminum it changes little with temperature. Which is the main cause of this difference?
  2. In a substitutional solid solution, assume that the increase in yield stress Δσ is proportional to the square root of the solute concentration c (the Fleischer model). If Δσ = 30 MPa at 1 at% solute, which is closest to Δσ at 4 at% solute? Ignore interactions between solute atoms and precipitation.
  3. When annealed low-carbon steel is pulled in tension, the stress drops after an upper yield point and a yield-point phenomenon appears in which Lüders bands travel along the specimen. Which statement about this phenomenon and its remedy is correct?
  4. The flow stress of a work-hardened metal is estimated with the Taylor equation σ = MαGb√ρ (ρ is the dislocation density). For cold-worked copper with M = 3.06, α = 0.3, G = 48 GPa, b = 0.256 nm and ρ = 1.0×10¹⁴ m⁻², which is closest to σ?
  5. When a metal is plastically deformed in tension, unloaded and then compressed in the reverse direction, the yield stress in compression is lower than the flow stress it would have had if tension had continued. Which is the correct explanation of this phenomenon (the Bauschinger effect)?