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Strengthening mechanisms and dislocations

The yield stress of an alloy is estimated from contributions of 50 MPa friction stress, 60 MPa solid-solution strengthening, 200 MPa precipitation strengthening and 100 MPa dislocation strengthening. If the friction stress and solid-solution strengthening are simply added, while the contributions of precipitates and dislocations, which are obstacles of similar strength, are combined as the square root of the sum of squares (σ = σ₀ + σss + √(σp² + σd²)), which is closest to the yield stress?

  1. AAbout 334 MPa
  2. BAbout 410 MPa
  3. CAbout 237 MPa
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Answer: A. About 334 MPa

The precipitate and dislocation contributions combine to √(200² + 100²) = √50000 ≈ 224 MPa, and adding 50 + 60 = 110 MPa gives about 334 MPa. About 410 MPa is the simple sum of everything, and about 237 MPa combines everything as the root sum of squares. When point obstacles of similar strength are mixed on the same slip plane, a dislocation is caught on them without distinguishing between them, so the strength approaches the root sum of squares. Obstacles of very different strengths, or resistances of a different nature such as lattice friction, are closer to simple addition. Intermediate exponents are also used, and the choice of combination changes the estimate by tens of percent, so comparisons should state which expression was used.

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More Strengthening mechanisms and dislocations questions

  1. Which method of strengthening steel raises the yield strength while also lowering the ductile-to-brittle transition temperature, improving toughness at low temperatures?
  2. Strengthening a metal usually reduces its uniform elongation. Which statement correctly describes the idea behind TRIP and TWIP steels, which show large uniform elongation even at high strength? Assume that necking begins when dσ/dε = σ (σ and ε are true stress and true strain).
  3. 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?
  4. 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.
  5. 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?