Weibull Modulus from Two Points

Also known as Weibull slope · two point Weibull fit · Weibull plot slope · graphical Weibull modulus · ln ln plot slope · estimate m from data

m=ln ⁣[ln(1/P1)ln(1/P2)]ln(σ1/σ2)m = \frac{\ln\!\left[ \dfrac{\ln(1/P_1)}{\ln(1/P_2)} \right]}{\ln(\sigma_1 / \sigma_2)}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

The two-parameter Weibull distribution straightens out beautifully. Take Ps=exp[(σ/σ0)m]P_s = \exp[-(\sigma/\sigma_0)^m], take the logarithm twice, and you get ln[ln(1/Ps)]=mlnσmlnσ0\ln[\ln(1/P_s)] = m\ln\sigma - m\ln\sigma_0 — a straight line of slope mm against lnσ\ln\sigma. That double-logarithm plot is the Weibull plot, it is how every fit in this field is presented, and this page is simply the slope taken between any two points that lie on it.

Two points is the cheapest possible estimator and the worst. It is here because it closes the algebra and because sometimes two published numbers are all you have. The proper method uses every specimen: break thirty or more bars, sort them by strength, assign each one a failure probability by median rank, plot the lot, and fit. ASTM C1239 is the authority, and it prefers maximum likelihood to least-squares regression on the plotted points, because regression treats the ranked probabilities as though they were independent measurements and they are not.

Here is the honesty that this whole subject turns on: mm from ten bars has roughly a ±30% confidence interval. A fitted modulus of 10 from a sample of ten is honestly reported as "somewhere between about 7 and 13". Quoting m=12.47m = 12.47 from ten specimens is theatre — the digits after the decimal point are noise wearing a lab coat, and they will not reproduce on the next batch of ten from the same drum of powder. Thirty specimens is the usual working minimum for a number worth designing to, and even thirty leaves an interval of roughly ±20%. The estimator is also biased at small sample sizes, not merely uncertain: maximum likelihood systematically overestimates mm when NN is small, which is the flattering direction, and C1239 publishes the unbiasing factors that correct it. If you take one thing from this page, take this one, because a mm sits in the exponent of the size effect and an optimistic modulus produces an optimistic scale-up.

And look at the plot before trusting any slope at all. Weakest-link theory assumes one flaw population. Real ceramics frequently have two: machining damage on the ground tensile surface, and porosity or inclusions in the bulk. Each population has its own distribution, and the specimen fails from whichever one happens to produce the worst flaw in that particular bar. The signature is a kink — a shallow slope over one range of stress meeting a steeper one over another, sometimes obvious and sometimes only visible once you know to look. A single mm fitted through a kinked plot is a fiction. It describes neither population, and it mispredicts worst in exactly the region that matters most: the weak tail, where the design allowable comes from.

The fix is not statistical. It is to look at the broken pieces. Fractography — reading the mirror, mist and hackle markings on a fracture surface back to the origin — will tell you which specimens failed from surface damage and which from the bulk, and the two groups can then be censored or fitted separately. Quinn's NIST SP 960-16 is the public-domain guide, and it is genuinely readable. The alternative, fitting one line through a bimodal set and reporting the number, is how a material gets qualified and then fails in service at a stress the report said was safe.

One small practical note on reading a plot: the vertical axis is sometimes ln[ln(1/(1Pf))]\ln[\ln(1/(1-P_f))] and sometimes ln[ln(1/Ps)]\ln[\ln(1/P_s)], which are the same thing written two ways, and sometimes log10\log_{10} is used instead of ln\ln on one axis or both. The slope only equals mm when the same base is used on both. Mixed bases are a quiet way to be wrong by a factor of 2.303.

Weibull Modulus from Two Points
m=ln ⁣[ln(1/P1)ln(1/P2)]ln(σ1/σ2)m = \frac{\ln\!\left[ \dfrac{\ln(1/P_1)}{\ln(1/P_2)} \right]}{\ln(\sigma_1 / \sigma_2)}
ln ln(1/Ps)ln σσ1σ2P1P2m
Where
  • mm= Weibull modulus (slope)
  • P1P_1= Survival probability at σ₁
  • P2P_2= Survival probability at σ₂
  • σ1\sigma_1= Stress at the first point (MPa)
  • σ2\sigma_2= Stress at the second point (MPa)
Missing one of these? Work it out first, then come back