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High-temperature materials and creep

At room temperature, refining the grains raises strength (the Hall–Petch relation), but components loaded for long times at high temperature often use coarse grains or single crystals. Which is the correct reason?

  1. ACoarser grains give a higher Young's modulus and so a smaller elastic strain at the same stress, which keeps the creep strain small even under long loading at high temperature
  2. BAt high temperature grain boundaries become stronger than the grain interiors, so reducing the boundaries keeps dislocations from being stopped too much at them and instead lets dislocation motion in the grains relax the stress
  3. CAt high temperature grain boundaries become fast diffusion paths and sliding sites, so grain boundary diffusion creep, grain boundary sliding and void formation on boundaries all become easier the more boundaries there are
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Answer: C. At high temperature grain boundaries become fast diffusion paths and sliding sites, so grain boundary diffusion creep, grain boundary sliding and void formation on boundaries all become easier the more boundaries there are

At high temperature, diffusion of atoms along grain boundaries is fast and grain boundary sliding also occurs, so grain boundaries become weak spots. The temperature at which grain boundaries and grain interiors are equally strong is called the equicohesive temperature, and above it the boundaries are the weaker. Diffusion creep (especially Coble creep, inversely proportional to the cube of grain size), grain boundary sliding, and rupture by the nucleation and coalescence of voids on boundaries therefore occur more readily in fine-grained material with many boundaries. That is why coarse grains, columnar grains parallel to the stress, and even single crystals are used in high-temperature components. Young's modulus is set by interatomic bonding and hardly depends on grain size, so it is no advantage of coarse grains. The statement that boundaries become stronger than the interiors at high temperature reverses the relationship.

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More High-temperature materials and creep questions

  1. A heat-resistant alloy ruptured by creep in 1000 h at 1000 K under a certain stress. Assuming the Larson–Miller parameter P = T(C + log₁₀t_r) (T is the absolute temperature, t_r is the rupture time in h, and C = 20) is constant at the same stress, which temperature is closest to that giving a rupture life of 10⁵ h at the same stress?
  2. Which statement about the γ' phase that provides the high-temperature strength of nickel-based superalloys is correct?
  3. Why are directionally solidified columnar-grained or single-crystal nickel-based superalloys used for jet engine turbine blades?
  4. What is the main purpose of adding a few percent of heavy elements such as rhenium (Re) to single-crystal nickel-based superalloys?
  5. When an alloy was oxidized in air at high temperature, it followed the parabolic law x² = k_p t, in which diffusion through the dense oxide scale controls the rate, and the scale was 2.0 µm thick after 100 h. Which is closest to the scale thickness after 900 h of oxidation at the same temperature?