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By Hadjidakis, P. J.

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In respect of such materials (W EIBULL thinks of isotropic materials) we find that the risk of rupture dB for a small volume element dv is determined by the equation dB = − log(1 − S0 ) dv. (6) As has been mentioned in the above, log(1 − S0 ) is a function of σ only and is negative because 1 − S0 < 1. Hence we have dB = −n(σ) dv. (7) If the distribution of stresses in the body is arbitrary, the risk of rupture is B= n(σ) dv (8) and the probability of rupture S ≡ 1 − e−B = 1 − e− R n(σ) dv . 1c) leading to H(x) = [(x − a)/b]c , and (9) gives the well–known relation (see Table 2/1) F (x) = 1 − exp − H(x) .

We will present a subclass of the model described leading to some well–known lifetime distributions, the W EIBULL distribution being one of them. In this subclass the buildup of wear (or loss of strength) is assumed to be deterministic, and thus σ 2 (x) = 0. Under this assumption, the system state process X = {X(t), t ≥ 0} satisfies the following equation t X(t) = x + µ X(s) ds. 17b) with initial condition X(0) = x. The probability of system survival through time t is then given by   t   Pr(T > t) = exp − k X(s) ds I{τ >t} .

As with ROSIN /R AMMLER /S PERLINGS’s approach W EIBULL’s reasoning at the decisive part of his paper is empirical and heuristic and lacks theoretical argumentation. The relevant text in W EIBULL (1939a, p. ) for any length ℓ may be computed from the distribution curve S1 for the unit of length according to the formula 1 − Sℓ = (1 − S1 )ℓ or log(1 − S1 ) = ℓ · log(1 − S1 ). ) In this case the volume V of the stressed system is proportional to the length ℓ, and if S0 is the distribution curve for that length of rod which corresponds to the unit of volume, we have log(1 − S) = V · log(1 − S0 ).

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