Probability of failure

The estimated probabilities of each of these events occurring are multiplied together to estimate the POS, since they must a//occur simultaneously if a hydrocarbon accumulation is to be formed. If the POS is estimated at say 30%, then the probability of failure must be 70%, and the expectation curve for an exploration prospect may look as shown in figure 6.9. 

Risk sensitivity results are also very useful in identifying key elements in your existing loss prevention program. For example, suppose your fire protection system was assumed to have a very low probability of failure because you test it weekly. Fire protection failures may not show up as an important contributor to your total risk (because failure is so unlikely), but your total risk estimate may be extremely sensitive to any change in the probability of fire protection failures. Flence you should not divert resources away from testing the fire protection system unless the alternate use of funds will decrease risk more than the reduced testing will increase risk. 

A logic model that graphically portrays the combinations of failures that can lead to a particular main failure (TOP event) or accident of interest. Given appropriate data, fault tree models can be quantitatively solved for an array of system performance characteristics (mean time between failures, probability of failure on demand, etc.)... 

We can demonstrate the notions of risk and risk assessment using Figure 1.18. For a given probability of failure occurrence and severity of consequence, it is possible to map the general relationship of risk and what this means in terms of the action required to eliminate the risk. 

An understanding of statisties is required Knowledge seattered througliout engineering texts Output as a probability of failure or reliability Better understanding of the effeets of variability... 

Equation 2.11 recognizes that for a failure to occur, there must be fault, and that other events may need to combine with the fault to bring about a failure. The equation states that the probabilities of each of these factors occurring must be multiplied together to calculate the probability of failure. 

In summary, for a eomponent/eharaeteristie it is possible to define an area of aeeeptable design on a graph of oeeurrenee versus severity. The aeeeptability of the design ean be enhaneed somewhat by the addition of inspeetion and test operations. The requirements of proeess eapability may be relaxed to a degree, as the eonditional probability of failure reduees, but this should be subjeet to a generous safety margin. 

As ean be seen from the above equation, for brittle materials like glass and eeramies, we ean seale the strength for a proposed design from a test speeimen analysis. In a more useful form for the 2-parameter Weibull distribution, the probability of failure is a funetion of the applied stress, L. 

The theory is concerned with the problem of determining the probability of failure of a part which is subjected to a loading stress, L, and which has a strength, S. It is assumed that both L and S are random variables with known PDFs, represented by f S) and f L) (Disney et al., 1968). The probability of failure, and hence the reliability, can then be estimated as the area of interference between these stress and strength functions (Murty and Naikan, 1997). 

Using the SND theory from Appendix I, the probability of failure, P, ean be determined from the Standard Normal variate, z, by ... 

We already know SM = 2.34 because it is the positive value of the Standard Normal variate, z, calculated above. The probability of failure per application of load... 

The calculated loading stress, L, on a component is not only a function of applied load, but also the stress analysis technique used to find the stress, the geometry, and the failure theory used (Ullman, 1992). Using the variance equation, the parameters for the dimensional variation estimates and the applied load distribution, a statistical failure theory can then be formulated to determine the stress distribution, f L). This is then used in the SSI analysis to determine the probability of failure together with material strength distribution f S). 

From Table 1 in Appendix I, the probability of failure P = 0.135666. This suggests that when the holding torque level is reaehed in serviee, due to a malfunetion stopping the hub from rotating, there is about a 1 in 7 ehanee that the shaft will yield before the hub slips. Clearly this is not adequate if we want to proteet the shaft and its transmission from damage. 

The material selected for the pin was 070M20 normalized mild steel. The pin was to be manufactured by machining from bar and was assumed to have non-critical dimensional variation in terms of the stress distribution, and therefore the overload stress could be represented by a unique value. The pin size would be determined based on the —3 standard deviation limit of the material s endurance strength in shear. This infers that the probability of failure of the con-rod system due to fatigue would be very low, around 1350 ppm assuming a Normal distribution for the endurance strength in shear. This relates to a reliability R a 0.999 which is adequate for the... 

Disney, R. L., Sneth, N. J. and Lipson, C. 1968 The DeteiTnination of Probability of Failure by Stress/Strength Interference Theory. In Proceedings of Annual Symposium on Reliability, 417-422. 

The total survival probability of the volume V under stress a, containing 2NV crack tips, is thus (Ps), and the total probability of failure of the specimen or stressed volume may be written as ... 

Benefit of Success) x (Probability of Success) - (Cost of Failure) x (Probability of Failure), or... 

WASH-1400 treated the probability of failure with time as being exponentially distributed with constant X. It treated X itself as being lognormally distributed. There are better a priori reasons... 

The cumulative binomial distribution is given by equation 2.5-33, where M is the number of f ailures out of items each having a probability of failure p. This can be worked backH tirLh lo find tlic implied value of p for a specified P(M, p,... 

Example Given there have been 200,000 scram tests and scram actuations with no failure. With 90% confidence, what is the probability of failure on demand ... 

A lrLL[uently encountered problem requires estimating a failure probability based on the number of failures, M, in N tests. These updates are assumed to be binomially distributed (equation 2.4-10) as p r N). Conjugate to the binomial distribution is the beta prior (equation 2.6-20), where / IS the probability of failure. 

In the introduction to this section, two differences between "classical" and Bayes statistics were mentioned. One of these was the Bayes treatment of failure rate and demand probttbility as random variables. This subsection provides a simple illustration of a Bayes treatment for calculating the confidence interval for demand probability. The direct approach taken here uses the binomial distribution (equation 2.4-7) for the probability density function (pdf). If p is the probability of failure on demand, then the confidence nr that p is less than p is given by equation 2.6-30. 

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