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Strengthening mechanisms and dislocations

Which statement about dispersion strengthening with oxide particles, as in copper containing fine Al₂O₃ particles or ODS steels with Y₂O₃ particles dispersed in iron, is correct?

  1. AIt uses coherent oxides precipitated from the matrix by aging, so the strength can be adjusted just as in precipitation hardening by choosing the solution-treatment and aging temperatures
  2. BHard oxides that do not dissolve in the matrix and resist coarsening force dislocations to bypass them, so strength tends to remain even at high temperatures where precipitates dissolve or coarsen
  3. CIt strengthens through the antiphase boundary created when dislocations shear the oxide particles, so larger particles give more strength, and at high temperature the particles dissolve and ductility returns
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Answer: B. Hard oxides that do not dissolve in the matrix and resist coarsening force dislocations to bypass them, so strength tends to remain even at high temperatures where precipitates dissolve or coarsen

In dispersion strengthening, oxide particles that hardly dissolve in or react with the matrix are produced and dispersed by internal oxidation or by mechanical alloying, in which powders are mixed intensely. Hard, incoherent particles are not sheared by dislocations and strengthen through the Orowan mechanism. Precipitates lose their strength at high temperature by dissolving or coarsening through Ostwald ripening, but oxide particles change little up to near the melting point, so strength and creep resistance at high temperature are retained. Al₂O₃-dispersed copper is used for welding electrodes and other parts that must resist softening at high temperature, and ODS steels are candidate materials for nuclear fuel cladding. The option about precipitating by aging describes precipitation hardening; oxides do not dissolve in the matrix, so they cannot be solution treated. Antiphase boundaries concern the shearing of ordered precipitates.

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More Strengthening mechanisms and dislocations questions

  1. 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?
  2. Which method of strengthening steel raises the yield strength while also lowering the ductile-to-brittle transition temperature, improving toughness at low temperatures?
  3. 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).
  4. 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?
  5. 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.