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4.5 Rackwitz-Fiessler Procedure If we know the distribution of the random variables, which are not normal distributed then the Rackwitz-Fiessler Procedure should be introduced. The basic idea behind the procedure begins with the calculation of “equivalent normal” values of the mean and standard deviation for each non-normal random variable. Suppose that a particular random variable X with mean μX and standard deviation σX is described by a Cumulative Distributed Function (CDF) FX(x) and a Probability Density Function (PDF) fX(x). To obtain the equivalent normal mean μ Xe and standard deviation σXe , we require that the CDF and the PDF of the actual function be equal to the normal CDF and normal PDF at the value of the variable x* on the failure boundary described by g=0. Mathematically, these requirements are expressed as 1 e ⎛ x* − μ X FX ( x ) = Φ⎜ ⎜ σe X ⎝ * ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ (1) e ⎛ x* − μ X f X (x ) = e φ⎜ e σX ⎜ σX ⎝ * 1 (2) where Φ is the CDF for the standard normal distribution and φ is the PDF for the standard normal distribution. Equation (1) simply requires the cumulative probabilities to be equal at x*. Eq.(2) is obtained by differentiating both sides of Eq.(1) with respect to x* . By manipulating these equations , we can obtain expressions forμXe andσXe as follows: e e μ X = x * − σ X [Φ −1 ( FX ( x * ))] e ⎛ x* − μ X 1 σ = φ⎜ * e fX (x ) ⎜ σ X ⎝ e X (3) (4) ⎞ 1 ⎟= φ [Φ −1 ( FX ( x* ))] ⎟ f ( x* ) X ⎠ The basic steps in the iteration procedure are the same as those in the matrix procedure given previously except that we must add a step to calculate the equivalent normal 2 parameters. Example Calculate the reliability index for a steel compact beam with the limit state function g ( Z , Fy , M ) = ZF y − M The distributions and distribution parameters are listed in following Table. Table Variable Distribution Mean Coefficient (check of unit) variation,% Z Normal 100m3 4 Fy Lognormal 40kN/m2 10 M Extreme 2000kN-m 10 Type I Solution 1. Formulate the limit state function and probability distributions. This has been done. Here we will refer to Z as variable 3 X1 , Fy as X2 , and M as X3. 2. Guess an initial design point. We assume values for x1* and x2* and use g=0 to find * * x3*. Try x1 = 100 and x2 = 40 . Then, using * g=0, we find that x3 = 4000 . 3. Determine equivalent normal parameters. Since X 1 = Z is normal, we do not have to find equivalent normal parameters. For X 2 = Fy , since it is lognormal, we know that σ 2 ln Fy 2 ⎛ σ Fy = ln⎜1 + 2 ⎜ μF y ⎝ ⎞ ⎟ = 9.95 ×10−3 ⎟ ⎠ y ⇒ σ ln F y = 0.0998 2 μln F = ln(μ F ) − 0.5σ ln F = 3.68 y y Therefore, FFy ( x ) = Φ( * 2 * ln x2 − μ ln Fy σ ln F ) = Φ (0.0499) = 0.520 y * f Fy ( x 2 ) = * 1 ln x 2 − μ ln Fy 2 ) ] = 0.0999 exp[− ( * σ ln Fy 2 2π σ ln Fy x 2 1 1 4 The equivalent normal parameters for X2 are therefore e σF = y 1 1 * φ[Φ −1 ( FFy ( x2 ))] = φ (0.0499) = 3.99 * f Fy ( x2 ) 0.0999 y y e e * * μ F = x 2 − σ F [Φ −1 ( FF ( x 2 ))] = 39.8 y Now we need to go through similar calculation for M. Since M follows an extreme Type I distribution, we know FM (m) = exp[− exp(−a(m − u ))] f M (m) = a{exp(−a(m − u ))}exp[− exp(−a(m − u ))] where a and u are distribution parameters that are related to the mean and standard deviation of M by u = μM 0.5772 ; − a π2 a= 2 6σ M Plugging in the values of μ M , σ M , we find a = 0.00641, u = 1910. The values of CDF and PDF at x*3 are * FM ( x3 ) = 9.99 × 10 −1 , * f M ( x3 ) = 9.69 × 10 −9 Thus the equivalent normal parameters for M are 5 e σM = 1 1 * φ [Φ −1 ( FM ( x3 ))] = φ [Φ −1 (0.999)] = 759 * −9 f M ( x3 ) 9.69 × 1`0 * * e e μ M = x3 − σ M [Φ −1 ( FM ( x3 ))] = 456 4. Determine the values of the reduced variables: z = * 1 * x1 − μ Z σZ = 0; z = * 2 * e x 2 − μ Fy e σF y = 0.0501; z = * 3 * e x3 − μ M e σM = 4.69 5. Determine the {G} vector: G1 = − G2 = − G3 = − ∂g ∂g * =− * * σ Z = − x 2σ Z ∂Z 1 {zi } ∂X 1 {xi } ∂g ∂g e * e = − x1σ Fy * = − * σF ∂Z 2 {z i } ∂X 2 {xi } y ∂g ∂g e 1 e * = − * σ M = − ( )σ M ∂Z 3 {z i } ∂X 3 {xi } 6. Calculate an estimate of β {G}T {z * } β= = 4.04 T {G} {G} 7. Calculate {α}: ⎧− 0.183⎫ {α } = {G} = ⎪− 0.457⎪ ⎨ ⎬ T {G} {G} ⎪ 0.870 ⎪ ⎩ ⎭ 6 z i* for n-1 of 8. Determine new values of the variables. We will find * z1 = α 1 β = (−0.183)(4.04) = −0.739 * z2 = α 2 β = (−0.457)(4.04) = −1.85 * * 9. Determine the value of x1 and x2 using * * the updated z1 and z2 * * x1 = μ Z + z1 σ Z = 97.0 * * e x2 = μ Fy + z2σ Fy = 32.4 * x3 using the 10. Determine the value of limit state equation g=0. For this case * x3 = 3143 . 11. Iteration until the value of β and the design point converge. The results of subsequent iterations are listed in the following Table. The final value of β is 4.03. 7 Table Iteration number 1 2 3 100 97.0 96.8 40 32.4 32.9 4000 3143 3186 4 96.8 32.9 3189 5 6 96.8 32.9 3189 7 x1* x2* x3* β 4.04 4.02 4.02 4.03 4.03 97.0 96.8 96.8 96.8 96.8 32.4 32.9 32.9 32.9 32.9 3143 3186 3189 3189 3189 x1* x2* x3* 8 4.6 Correlated Random Variables Thus far, we have considered limit state equations in which the random variables are all uncorrelated. However, in many practical applications, some of the random variables may be correlated, and the correlation can have a significant impact on the calculated reliability index. We could take the following approaches: To modify the procedure previously by introducing a correlation matrix [ ρ ]. The modified equations are as follows: β= {G}T {z * } {G}T {z* } T changes to β = {G}T [ ρ ]{G} {G} {G} {G} {α } = [ ρT]{G} {G} [ ρ ]{G} {G}T {G} changes to {α } = 9 10