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What is Taylor factor?

The ratio M of the yield stress of a polycrystal to the critical resolved shear stress of a single crystal. It is set by the distribution of grain orientations and is about 3.06 for randomly oriented FCC metals.

Japanese name: テイラー因子

Taylor factor appears in 3 practice questions across 1 topic, including Strengthening mechanisms and dislocations.

Field: Strengthening mechanisms and dislocations

Try it: questions about Taylor factor

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
Show answer and explanation

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.

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?

  1. AAbout 74 MPa
  2. BAbout 228 MPa
  3. CAbout 24 MPa
Show answer and explanation

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.

Taylor factor in the explanations

Questions that mention Taylor factor (3)

  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?
  2. 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?
  3. 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?

Related terms

Dislocation Critical resolved shear stress Schmid factor Hall–Petch relationship Orowan mechanism Work hardening Bauschinger effect Dispersion strengthening Peierls stress

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