#### Transcript First order reliability method (FORM)

Approximate methods for calculating probability of failure • Working with normal distributions is appealing • First-order second-moment method (FOSM) • Most probable point • First order reliability method (FORM) • Section 4.1 in Choi, Grandhi, & Canfield 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. 1 ( Pf ) Pf ( ) is the normal CDF Approximation about mean • Predecessor of FORM called first-order second-moment method (FOSM) g ( X ) g ( X ) g ( X )T X X Then, easy to show 1/2 n g 2 g g ( X ) g xi i 1 xi (How can you use that to get rid of unimportant variables?) The reliability index and probability of failure are 2 g Pf ( ) g is the normal CDF Beam under central load example Example 4.2 • Probability of exceeding plastic moment capacity g (P, L,W , T ) WT PL / 4 P L W (plastic section modulus) T (yield stress) mean 10kN 8m 0.0001m^3 600,000 kN/m^2 Standard deviation 2kN 0.1m 0.00002m^3 100,000kN/m^2 • All normal Reliability index for example • Using the linear approximation get g ( P, L,W , T ) WT PL / 4 g g ( X ) 0.0001 600, 000 10 8 / 4 60 20 40kNm (safety factor of 3!) g W 0.0001 T g L / 4 2 P g T 600, 000 W g P / 4 2.5 L 1/2 n g 2 g x2i i 1 xi (0.0001100, 000) 2 (600, 000 0.00002) 2 2 2 (2.5 0.1) 2 16.13kNm 2 g 2.48, g Pf (2.48) 0.0066 • Example 4.2 of CGC shows that if we change to g=T0.25PL/W we get 3.48 (0.00025) • MCS with 10 million samples gave 2.85 (0.00218) Top Hat question • In the beam example, the error in estimating the reliability index was due to – Linear approximation – g is not normal – both 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 illustration r=randn(1000,1)*1.25+10; c=randn(1000,1)*1.5+13; [email protected](x) x; fplot(f,[5,20]) hold on plot(r,c,'ro') xlabel('r') ylabel('c') 20 c 15 10 5 5 10 15 r 20 Recipe for finding MPP with independent normal variables • Transform into standard normal variables (zero mean and unity standard deviation) ui xi xi x i • 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. min U TU 1/2 U g (U ) 0 First order reliability method (FORM) • Limit state g(X). Failure when g<0. • Linear approximation of limit state together with assumption that random variables are normal. • Approximate around most probable point. • Then limit state is also normal variable. • Reliability index is the distance of the mean of g from zero measured in standard deviations. Visual Linear Example • • • • For linear example g R C, R N (10,1.25 ) 10 c 13 Then u r1.25 u g 1.25u 1.5u 3 1.5 The failure boundary g 0 u (5 / 6)u 2 Distance from origin 2 u12 5u1 / 6 2 C N(13,1.52 ) 2 1 2 1 2 2 1 3 2 2 1 0 61 u2 1.1528(r c 11.27) 60 Pf ( ) 0.0621 u1 u2 d 2 25 10 2u1 u1 0 du1 18 3 -1 -2 1.537 -3 -4 -5 -3 -2 -1 0 1 u1 2 3 4 Check by MCS • • • • r=randn(1000,1)*1.25+10; c=randn(1000,1)*1.5+13; >> s=0.5*(sign(r-c)+1);pf=sum(s)/1000=0.0550 Six repetitions gave: 0.0660, 0.0640, 0.0670,0.0450, 0.0550, 0.0640 • With million samples got 0.06258 • So even with a million samples, accurate only to two digits. Problems FORM • Repeat the linear example of Slide 12 with a slightly different limit state g=R2-C2 with both FORM and MCS. • Do the beam problem of slides 4 and 5 with FORM. Note that you are now in 4 dimensional random space. General case • If random variables are normal but correlated, a linear transformation will transform them to independent variables. • If random variables are not normal, can be transformed to normal with similar probability of failure. See Section 4.1.5 of CGC (It is called the Rosenblatt transformation) • Murray Rosenblatt, Remarks on a Multivariate Transformation, Ann. Math. Statist. Volume 23, Number 3 (1952), 470-472. Approximate transformation • If we want the transformed variable u to have the same CDF as the original x at a certain value of x we would require (u ) F ( x) • Unfortunately we cannot enforce that everywhere. • If we focus on MPP we get the following transformation (Eqs. 4.38, 4.39), which must be used iteratively, starting from some guess. u X* replaced by xMPP in equations. x x' x' x' 1 F ( x MPP ) f ( x MPP ) x ' x MPP 1 F ( x MPP ) x '