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High-temperature materials and creep

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?

  1. AAbout 920 K
  2. BAbout 854 K
  3. CAbout 600 K
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Answer: A. About 920 K

From the test condition, P = 1000 × (20 + log₁₀1000) = 1000 × 23 = 23000. For 10⁵ h, log₁₀t_r = 5, so T = 23000/(20 + 5) = 920 K. Lowering the temperature by only 80 K extends the life a hundredfold, and estimating long lives at lower temperatures from short tests at higher temperatures is what this parameter is used for. About 854 K uses the natural logarithm instead of the common logarithm, and about 600 K forgets the constant C and takes P = T log₁₀t_r. However, if the microstructure changes or the deformation mechanism shifts over long times, the extrapolation fails, so an extrapolation as large as a hundredfold, as in this example, must be treated with care, and the extrapolation range should not be stretched needlessly.

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