Weibull Survival with Volume Scaling

This is the form Weibull actually derived, twelve years before the 1951 paper, in a 1939 monograph for the Swedish Royal Academy of Engineering Sciences. The argument is a chain and its weakest link. Imagine the body divided into a great many small elements, each with its own independent chance of containing a flaw severe enough to fail at the applied stress. The body survives only if every element survives. Multiply the survival probabilities of all the elements together, take the logarithm, and the number of elements — that is, the volume — comes out as a simple multiplier in the exponent. That is the whole derivation, and $V/V_0$ is what falls out of it.

The consequence is that $\sigma_0$ is meaningless without a volume attached. A characteristic strength quoted with no specimen size is half a sentence, and it is quoted that way constantly — on datasheets, in papers, in supplier literature. The number is real; it just belongs to the bar it was measured on. Move to a part of a different size and it moves with it, in a direction and by an amount this equation gives exactly.

Choosing what $V$ is takes more care than choosing its value. Two things are easy to get wrong here and both are silent.

The first is that only material in tension counts. Weakest-link failure is a tensile phenomenon: a crack opens under tension and is closed under compression, and a brittle material in compression is typically ten times stronger and fails by an entirely different mechanism. The compressive half of a beam contributes essentially nothing to this integral. Neither does a lightly stressed region, because the risk of failure is weighted by stress to the $m$-th power, and with $m = 10$ a region at half the peak stress contributes about one thousandth as much. This is what the effective-volume page exists to compute, and for a bend bar the answer is a couple of hundred times smaller than the bar.

The second is volume against area. The weakest-link argument does not care whether the elements are little cubes of bulk or little patches of surface — it only cares that they are independent and that there are many of them. If the flaws that decide strength are pores or inclusions inside the material, the population scales with volume and $V/V_0$ is right. If they are machining damage, grinding scratches, handling chips or a corroded skin, the population scales with surface area, and the ratio must be $A/A_0$ instead. For a part that is geometrically similar to its test bar the two give different answers, and the difference grows with the scale factor: volume goes as the cube of a linear dimension and area only as the square. Nothing in a Weibull plot distinguishes the two cases. The straight line looks identical. The only way to know is fractography — break the specimens, find the fracture origin on each broken surface under a microscope, and see whether the origins sit on the surface or inside. Quinn's NIST SP 960-16 is the public-domain manual for exactly this, and it is the single most useful free document in the subject.

A third possibility deserves naming: a bimodal population, where surface damage and bulk porosity compete. Then neither scaling is right on its own, the Weibull plot shows a visible kink, and a single $m$ fitted through it describes neither population. That case is discussed on the modulus page, because it is where fitting goes wrong most often.