The statistics of brittle fracture and case study

The chalk with which I write on the blackboard when I teach is a brittle solid. Some sticks of chalk are weaker than others. On average, I find (to my slight irritation), that about 3 out of 10 sticks break as soon as I start to write with them the other 7 survive. The failure probability, Pf, for this chalk, loaded in bending under my (standard) writing load is 3/10, that is 

When you write on a blackboard with chalk, you are not unduly inconvenienced if 3 pieces in 10 break while you are using it but if 1 in 2 broke, you might seek an alternative supplier. So the failure probability, Pf, of 0.3 is acceptable (just barely). If the component were a ceramic cutting tool, a failure probability of 1 in 100 (Pf= 10 ) might be acceptable, because a tool is easily replaced. But if it were the window of a vacuum system, the failure of which can cause injury, one might aim for a Pf of lO and for a ceramic protective tile on the re-entry vehicle of a space shuttle, when one failure in any one of 10,000 tiles could be fatal, you might calculate that a Pf of 10 was needed. 

The Swedish engineer, Weibull, invented the following way of handling the statistics of strength. Fie defined the survival probability PJ.Vg) as the fraction of identical samples, each of volume Vg, which survive loading to a tensile stress a. Fie then proposed that 

Thus a plot of In In (1/Ps(Vq)) against In ( 7/ 7q) is a straight line of slope m. Weibull-probability graph paper does the conversion for you (see Fig. 18.4). 

So much for the stress dependence of P. But what of its volume dependence We have already seen that the probability of one sample surviving a stress 7 is Ps(Vq). The probability that a batch of n such samples all survive the stress is just P fl/g) . If these n samples were stuck together to give a single sample of volume 1/ = nVo then its survival probability would still be (P fl/o) . So 

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