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?

Answer: B. About 228 MPa

Δτ = 26×10⁹ × 0.286×10⁻⁹/(100×10⁻⁹) ≈ 7.4×10⁷ Pa, about 74 MPa. Converting to the increase in the yield stress of a polycrystal means multiplying by the Taylor factor, Δσ = 3.06 × 74 ≈ 228 MPa. About 74 MPa leaves the value as a shear stress and forgets M, and about 24 MPa divides by M. The expression comes from the line tension of a dislocation, about Gb²/2, and the fact that a dislocation can pass once it has bowed out into a semicircle between precipitates. At the same volume fraction, finer particles have a smaller spacing L, so the Orowan stress is larger. More precise expressions include a factor for the dislocation character and particle size and a logarithmic term, which give values smaller than this estimate.