English
Menu

Strengthening mechanisms and dislocations

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).

  1. AMartensitic transformation or deformation twinning during deformation keeps the work-hardening rate dσ/dε high as strain proceeds, delaying the onset of necking
  2. BDislocations recover and disappear during deformation so that the work-hardening rate dσ/dε approaches zero, keeping the flow stress constant so that necking does not occur
  3. CRefining the grains to a few tens of nanometers raises the yield stress while keeping the work-hardening rate high, increasing strength and uniform elongation together
Show answer and explanation

Answer: A. Martensitic transformation or deformation twinning during deformation keeps the work-hardening rate dσ/dε high as strain proceeds, delaying the onset of necking

By Considère's criterion, necking begins when the work-hardening rate dσ/dε has fallen to the flow stress σ. Raising σ by strengthening without raising the work-hardening rate reaches this condition sooner and reduces the uniform elongation. In TRIP steels retained austenite gradually transforms to hard martensite during deformation, and in TWIP steels deformation twins form one after another in high-Mn austenite with a low stacking-fault energy, with the twin boundaries finely dividing the paths of dislocations. In both, the work-hardening rate stays high as strain proceeds, so high strength and large uniform elongation are achieved together. If recovery brings the work-hardening rate close to zero, necking begins almost at once. In grains a few tens of nanometers in size, dislocations hardly accumulate within the grains, so the work-hardening rate falls and the uniform elongation becomes small.

Report an error

Practice this topic Back to topic

More Strengthening mechanisms and dislocations questions

  1. 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?
  2. 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.
  3. 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?
  4. The yield stress of BCC metals such as iron rises sharply as the temperature is lowered, whereas in FCC metals such as copper and aluminum it changes little with temperature. Which is the main cause of this difference?
  5. In a substitutional solid solution, assume that the increase in yield stress Δσ is proportional to the square root of the solute concentration c (the Fleischer model). If Δσ = 30 MPa at 1 at% solute, which is closest to Δσ at 4 at% solute? Ignore interactions between solute atoms and precipitation.