English
Menu

Strengthening mechanisms and dislocations

Estimating the shear stress needed to displace the atomic planes above and below the slip plane all at once in a perfect crystal (the theoretical shear strength) as τ ≈ G/2π gives about 7.6 GPa for copper (shear modulus G = 48 GPa). Which is the correct reason why the shear stress at which a real copper single crystal begins to slip is orders of magnitude smaller, about 1 MPa?

  1. AWhen a dislocation in the crystal moves, bonds across the slip plane only need to be rearranged one row at a time, so the whole plane need not be displaced at once
  2. BVacancies in the crystal diffuse and gather on the slip plane, and atomic planes separate all at once where vacancies are concentrated, reducing the number of bonds to break
  3. CThe oxide film on the surface of the single crystal acts as a lubricant between atomic planes, so displacement begins at a force weaker than the internal bonds
Show answer and explanation

Answer: A. When a dislocation in the crystal moves, bonds across the slip plane only need to be rearranged one row at a time, so the whole plane need not be displaced at once

The theoretical shear strength assumes that all the atoms above and below the slip plane shift to the neighboring positions simultaneously; it is G/2π ≈ 7.6 GPa, and even with a more realistic form of the interatomic force it is about G/30, or about 1.6 GPa. Real yielding occurs at less than a thousandth of this because the crystal contains dislocations, and when a dislocation moves only the bonds at the dislocation line are rearranged in turn, so the displacement spreads a little at a time. Vacancy diffusion does not carry slip at room temperature; it is a mechanism that matters in high-temperature creep and similar processes. A surface oxide film, if anything, can hinder dislocations from leaving through the surface. Strengths close to the theoretical value have been measured in whiskers, which contain almost no dislocations.

Report an error

Practice this topic Back to topic

More Strengthening mechanisms and dislocations questions

  1. A single-crystal rod is pulled in tension. For a slip system whose slip-plane normal makes 60° and whose slip direction makes 40° with the tensile axis, the critical resolved shear stress is 0.80 MPa. Which is closest to the tensile stress at which this slip system begins to slip? Ignore other slip systems.
  2. 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?
  3. 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?
  4. 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.
  5. 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?