Larson-Miller Parameter Worked Example

Problem

A component operates at 650 degrees Celsius under a stress of 100 MPa. The Larson-Miller master curve for the material gives LMP = 20.5 (in units of log(t) + C*T, with T in Kelvin and C = 20) at this stress. Estimate the rupture time.

Given

• T = 650 C = 923 K
• sigma = 100 MPa
• LMP = 20.5 (illustrative, at this stress)
• C = 20 (illustrative Larson-Miller constant)

Solution

Larson-Miller parameter:
  LMP = C * T * log10(t_r) + log10(t_r)
  (common form: LMP = T * (log10(t_r) + C))

Using the form LMP = T * (log10(t_r) + C):
  20.5 * 1000 = 923 * (log10(t_r) + 20)
  (Note: LMP is typically in units of K*log(hours))

  LMP = 20.5 * 1000 = 20500 (if LMP is in thousands)
  Actually, let us use the standard form:
  LMP = T * (log10(t_r) + C) where T is in Kelvin

  20500 = 923 * (log10(t_r) + 20)
  log10(t_r) + 20 = 20500 / 923 = 22.21
  log10(t_r) = 2.21
  t_r = 10^2.21 = 162 hours

The rupture time is approximately 162 hours at
650 C and 100 MPa for this illustrative LMP.

Interpretation

The rupture time of 162 hours is relatively short — this component would be at significant creep rupture risk if the required service life exceeds this value. The LMP and C value are illustrative — actual values must be determined from material-specific creep rupture data. The Larson-Miller parameter is an extrapolation tool and should be used with caution outside the data range.