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Approximate methods for calculating probability of failureMonte Carlo simulationFirst-order second-moment method (FOSM)Working with normal distributions is appealingThe reliability indexMost probable pointFirst order reliability method (FORM)The Rosenblatt transformation

Monte Carlo SimulationGiven a random variable X and a function h(X): sample X: [x1,x2,,xn]; Calculate [h(x1),h(x2),,h(xn)]; use to approximate statistics of h.

Example: X is U[0,1]. Use MCS to find mean of X2x=rand(10); y=x.^2; %generates 10x10 random matrixsumy=sum(y)/10sumy =0.4017 0.5279 0.1367 0.3501 0.3072 0.3362 0.3855 0.3646 0.5033 0.2666sum(sumy)/10 ans =0.3580

What is the true mean

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Evaluating probabilities of failureFailure 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, NThe accuracy of the estimate is poor unless N is much larger than 1/PfFor small Pf

ExampleEstimate the probability that x=N(0,1)>1x=randn(1,1000); x1=0.5*(sign(x-1)+1); pf=sum(x1)/1000.; pf =0.1550Repeating 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.

The normal distribution is attractiveIt 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.

Approximation about meanPredecessor of FORM called first-order second-moment method (FOSM)

Beam under central load exampleProbability of exceeding plastic moment capacity

PLW (plastic section modulus)T (yield stress)mean10kN8m0.0001m^3600,000 kN/m^2Standard deviation2kN0.1m0.00002m^3100,000kN/m^2

Reliability index for exampleUsing 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 Pf=0.0069)

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.

Response minus capacity illustrationr=randn(1000,1)*1.25+10;c=randn(1000,1)*1.5+13;f=@(x) x;fplot(f,[5,20])hold onplot(r,c,'ro') xlabel('r')ylabel('c')

Recipe for finding MPP with independent normal variablesTransform 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.

Visual

Response minus capacityOriginal and transformed variables

Distance in standard normal space

MPP and reliability index

First order reliability method (FORM)Limit state g(X). Failure when g