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Strengthening mechanisms and dislocations

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.

  1. AAbout 0.31 MPa
  2. BAbout 1.6 MPa
  3. CAbout 2.1 MPa
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Answer: C. About 2.1 MPa

The resolved shear stress on a slip system is τ = σ cosφ cosλ (Schmid's law), and cosφ cosλ is called the Schmid factor. Here cos60° × cos40° = 0.500 × 0.766 ≈ 0.383, so slip begins at σ = 0.80/0.383 ≈ 2.1 MPa. About 0.31 MPa multiplies the critical resolved shear stress by the Schmid factor instead of dividing, and about 1.6 MPa divides only by cosφ and forgets cosλ. The Schmid factor reaches its maximum of 0.5 when the slip-plane normal and the slip direction both make 45° with the tensile axis, and it is 0 when the slip plane is parallel or perpendicular to the axis, in which case that system does not slip. In a single crystal, the system with the largest factor slips first.

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More Strengthening mechanisms and dislocations questions

  1. 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?
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