Lec16_Structural_Reliability_Example

# Lec16_Structural_Reliability_Example - R – 0.5 Var[ln(R =...

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Lecture 16; page 12 Structural Reliability - Revisited S = load effect (lognormal) lognormal) R = resistance of structure to load lognormal) V = R/S = safety factor for structure < 1 failure > 1 no failure P[ V < 1 ] = P[ R S < 1]

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Lecture 16; page 13 SOLUTION STRATEGY: Resistance Load Safety Factor Real Space variables R S R/S = exp[ U ] | | /\ \/ \/ | Log Space variables lnR lnS —> U = lnR – lnS ~ N
Lecture 16; page 14 Let R and S be independent and lognormal where E[R] = 120 E[S] = 75 Var[R] = (10) 2 Var[S] = (15) 2 Find mean and variance of ln(R) Var[ ln(R) ] = ln[ 1 + σ R 2 R 2 ] = ln[ 1 + (10) 2 /120 2 ] = 0.00692 = (0.0832) 2 E[ ln(R) ] = ln[μ

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Unformatted text preview: R ] – 0.5 Var[ln(R)] = ln[120] – 0.5 (0.0832) 2 = 4.78 Doing the same for ln[S] Var[ ln(S) ] = ln[ 1 + σ S 2 /μ S 2 ] = ln[ 1 + (15) 2 /75 2 ] = 0.0392 = (0.2) 2 E[ ln(S) ] = ln[μ S ] – 0.5 Var[ln(S)] = ln[75] – 0.5 (0.2) 2 = 4.30 Lecture 16; page 15 The variable of interest is U = ln(R) – ln(S) ~ N(μ lnR – μ lnS , σ lnR 2 + σ lnS 2 ) = N( 4.78 – 4.30, (0.0832) 2 + (0.2) 2 ) = N( 0.486, 0.046 ) = N( 0.486, (0.21) 2 ) P[ R/S < 1] = P[ ln(R) – ln(S) < 0 ] = P[ (U-μ U )/ σ U < –0.486/0.21 ] = P[ Z < –0.486/0.21 ] = P[ Z < – 2.26 ] = 0.012 = 1.2% (using tables) Using bivariate normal model obtained failure probability of 0.6% —> model choice matters....
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## This note was uploaded on 12/05/2011 for the course CEE 3040 at Cornell.

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Lec16_Structural_Reliability_Example - R – 0.5 Var[ln(R =...

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