approximate methods for calculating probability of failure monte carlo simulation first-order...

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Approximate methods for calculating probability of failure • Monte Carlo simulation • First-order second-moment method (FOSM) • Working with normal distributions is appealing – The reliability index • Most probable point • First order reliability method (FORM) • The Rosenblatt transformation

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Slide 2 Approximate methods for calculating probability of failure Monte Carlo simulation First-order second-moment method (FOSM) Working with normal distributions is appealing The reliability index Most probable point First order reliability method (FORM) The Rosenblatt transformation Slide 3 Monte Carlo Simulation Given a random variable X and a function h(X): sample X: [x 1,x 2,,x n ]; Calculate [h(x 1 ),h(x 2 ),,h(x n )]; use to approximate statistics of h. Example: X is U[0,1]. Use MCS to find mean of X 2 x=rand(10); y=x.^2; %generates 10x10 random matrix sumy=sum(y)/10 sumy =0.4017 0.5279 0.1367 0.3501 0.3072 0.3362 0.3855 0.3646 0.5033 0.2666 sum(sumy)/10 ans =0.3580 What is the true mean SOURCE: http://schools.sd68.bc.ca/ed611/akerley/question.jpg SOURCE: http://www.sz-wholesale.com/uploadFiles/041022104413s.jpg Slide 4 Evaluating probabilities of failure Failure is defined in terms of a limit state function where failure occurs when g(r)>0, where r is a vector of random variables. Probability of failure is estimated as the ratio of number of positive gs, m, to total MC sample size, N The accuracy of the estimate is poor unless N is much larger than 1/P f For small P f Slide 5 Example Estimate the probability that x=N(0,1)>1 x=randn(1,1000); x1=0.5*(sign(x-1)+1); pf=sum(x1)/1000.; pf =0.1550 Repeating the process obtained: 0.136, 0.159, 0.160, 0.172, 0.154, 0.166. Exact value 0.1587. In general, for 10% accuracy of probability you need 100 failed samples. Slide 6 The normal distribution is attractive It has the nice property that linear functions of normal variables are normally distributed. Also, the sum of many random variables tends to be normally distributed. Probability of failure varies over many orders of magnitude. Reliability index, which is the number of standard deviations away from the mean solves this problem. Slide 7 Approximation about mean Predecessor of FORM called first-order second-moment method (FOSM) Slide 8 Beam under central load example Probability of exceeding plastic moment capacity PLW (plastic section modulus) T (yield stress) mean10kN8m0.0001m^3600,000 kN/m^2 Standard deviation2kN0.1m0.00002m^3100,000kN/m^2 Slide 9 Reliability index for example Using the linear approximation get Example 4.2 of CGC shows that if we change to g=T- 0.25PL/W we get 3.48 (0.00025, exact is 2.46 or P f =0.0069) Slide 10 Most probable point (MPP) The error due to the linear approximation is exacerbated due to the fact that the expansion may be about a point that is far from the failure region (due to the safety margin). Hasofer and Lind suggested remedying this problem by finding the most probable point and linearizing about it. The joint distribution of all the random variables assigns a probability density to every point in the random space. The point with the highest density on the line g=0 is the MPP. Slide 11 Response minus capacity illustration r=randn(1000,1)*1.25+10; c=randn(1000,1)*1.5+13; f=@(x) x; fplot(f,[5,20]) hold on plot(r,c,'ro') xlabel('r') ylabel('c') Slide 12 Recipe for finding MPP with independent normal variables Transform into standard normal variables (zero mean and unity standard deviation) Find the point on g=0 of minimum distance to origin. The point will be the MPP and the distance to the origin will be the reliability index based on linear approximation there. Slide 13 Visual Slide 14 Response minus capacity Original and transformed variables Distance in standard normal space MPP and reliability index Slide 15 First order reliability method (FORM) Limit state g(X). Failure when g