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Strengthening mechanisms and dislocations

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?

  1. AIt occurs because cementite precipitated at grain boundaries hinders the generation of dislocations, so dissolving the cementite by quenching keeps it from appearing even after months at room temperature
  2. BIt occurs because carbon and nitrogen atoms gather at and pin dislocations; light cold rolling to add mobile dislocations removes it, but it can reappear if the steel is left at room temperature
  3. CThe upper yield point arises from dislocations piling up at grain boundaries, so coarsening the grains to reduce the boundaries prevents both the yield-point phenomenon and Lüders bands
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Answer: B. It occurs because carbon and nitrogen atoms gather at and pin dislocations; light cold rolling to add mobile dislocations removes it, but it can reappear if the steel is left at room temperature

In low-carbon steel, interstitial carbon and nitrogen atoms gather around dislocations (Cottrell atmospheres) and pin them. Yielding needs a large stress to tear dislocations away or create new ones (the upper yield point), and once they start moving deformation proceeds at a lower stress, so a deformed band travels along the specimen. Because this produces surface markings on pressed parts (stretcher strains), steel sheet is given about 1% temper rolling to add mobile dislocations before use. However, when dissolved carbon and nitrogen diffuse and pin the dislocations again, the yield point returns (strain aging); this does not happen in IF steel, where carbon and nitrogen are tied up as compounds by Ti or Nb. The options that explain it by cementite or by pile-ups at grain boundaries are wrong.

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More Strengthening mechanisms and dislocations questions

  1. 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?
  2. 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)?
  3. As precipitates grow during aging, the strength rises to a maximum and then falls (overaging). Which is the correct explanation in terms of the interaction between precipitates and dislocations? Take the volume fraction of precipitates as constant.
  4. In an aluminum alloy, the increase in shear stress due to the Orowan mechanism, in which dislocations bow out and pass between precipitates, is estimated as Δτ ≈ Gb/L (L is the precipitate spacing). With G = 26 GPa, b = 0.286 nm and L = 100 nm, and taking the increase in the yield stress of the polycrystal as Δσ = MΔτ (Taylor factor M = 3.06), which is closest to Δσ?
  5. Which statement about the characteristics of dislocations shearing through precipitates with an ordered structure, such as the γ' phase (an L1₂ ordered phase) in Ni-based superalloys, is correct?