What is Larson–Miller parameter?

The value P = T(C + log₁₀t_r) that combines the temperature and time of creep rupture. Taken as constant at a given stress, it is used to estimate long lives from short tests.

Japanese name: ラーソン・ミラー・パラメータ

Larson–Miller parameter appears in 1 practice question across 1 topic, including High-temperature materials and creep.

Field: High-temperature materials and creep

The idea behind the parameter

The time to creep rupture, t_r, shortens rapidly with rising temperature in an Arrhenius manner. Larson and Miller showed that when rupture time and temperature are combined into a single value, P = T(C + log₁₀t_r), P is nearly constant at a given stress. T is the absolute temperature, t_r is in hours (h), and C = 20 is commonly used for many steels and heat-resistant alloys. When data are plotted with P on the horizontal axis and stress on the vertical axis, results from tests at different temperatures collapse onto a single curve, and reading P from the stress lets you calculate the rupture time at any temperature.

Estimating life

High-temperature components may be used for 100,000 hours (about 11 years) or more, and tests of that length cannot all be performed. Instead, tests of several hundred to several thousand hours are run at temperatures above the service temperature, and the life at the service temperature is estimated from the value of P. For example, with C = 20, rupture in 1000 h at 1000 K corresponds to P = 23000, which at the same stress corresponds to a life of 10⁵ h at 920 K. That lowering the temperature by only 80 K extends the life a hundredfold also shows how important temperature control is in high-temperature components.

Limits and cautions

The method assumes that the same creep mechanism operates at the test and service temperatures and that the microstructure changes in the same way. If precipitates coarsen, other phases precipitate, or the deformation mechanism shifts from dislocation creep to diffusion creep over long times, extrapolation from short tests can overestimate the life. The value of C also varies between materials, so it is preferably determined from measured data, and it is important not to stretch the extrapolation range needlessly. Other parameters, such as the Orr–Sherby–Dorn parameter, are used for the same purpose.

Try it: questions about Larson–Miller parameter

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
Show answer and explanation

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.

FAQ

Why is the constant C equal to 20?

Fitting data for many steels and heat-resistant alloys gives C in the range of roughly 15 to 25, and 20 is commonly used as a representative value. It is not fixed by theory, so when accuracy is needed, C is determined from test data at several temperatures for the material in question.

What does a large P mean?

P expresses the severity of the combination of temperature and time; a larger P corresponds to a higher temperature or longer time. On a curve of P against rupture stress, the rupture stress falls as P increases. A material with a higher rupture stress at the same P is more resistant to creep rupture.

Can the rupture time be expressed in seconds?

The form of the equation is the same, but the value of log₁₀t_r changes, so the value of C that fits also changes. C = 20 is the value for rupture time expressed in hours (h), so units must match when comparing with other data.

Questions that mention Larson–Miller parameter (1)

  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?

Related terms

Creep Deformation mechanism map Diffusion creep Nickel-based superalloy Parabolic rate law Pilling–Bedworth ratio Power-law creep Thermal barrier coating

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