Analysis of some simple examples. Accuracy11
Analysis of some simple examples. Accuracy 11
Simplified example, buckling of Euler beam:Eulerbucklingload:Pcr=?El/lk2Elis bending stiffness,is buckling lengthConsiderfirst the case where only X, =lk is a random variablewhile the bending stiffness isassumed tobe given.Thebucklinglength which is applied for design is Ikc*The limit state function then becomes: g(x) = [(lkc)2/(lk)2] - 1= [(X1c)2/(X1)3] - 1The"exact"bucklingload is next approximated bya responsesurfacewhichisthe combinationof afirstanda secondorderpolynomial12
Simplified example, buckling of Euler beam: • Euler buckling load: Pcr = 2EI/lk 2 – EI is bending stiffness – l k is buckling length • Consider first the case where only X1 = lk is a random variable, while the bending stiffness is assumed to be given. The buckling length which is applied for design is lkc. • The limit state function then becomes: g(x) = [(lkc) 2 /(lk ) 2 ] – 1 = [(X1c) 2 /(X1 ) 2 ] - 1 • The ”exact” buckling load is next approximated by a response surface which is the combination of a first and a second order polynomial 12
Simplified example, buckling of Euler beam:The buckling length is taken to be Gaussian distributed with amean value equal to lk,mean and a Coefficient of variation equal to0.20The characteristic value of the buckling length, lkc (i.e. the valuewhich is applied for design) is assumed to be equal to the meanvalue plus two standard deviations (i.e.two timesjk).The"true"failure function then becomes:g(x) = [(1+2*C.0.V.)2/(lk/ lk,mean)] - 1 = [(1+2*0.2)2/(X1/X1,mean)}] - 113
Simplified example, buckling of Euler beam: • The buckling length is taken to be Gaussian distributed with a mean value equal to lk,mean and a Coefficient of variation equal to 0.20 • The characteristic value of the buckling length, lkc (i.e. the value which is applied for design) is assumed to be equal to the mean value plus two standard deviations (i.e. two times lk). • The ”true” failure function then becomes: g(x) = [(1+2*C.o.V.)2 /(lk / lk,mean) 2 ] – 1 = [(1+2*0.2)2 /(X1 /X1,mean) 2 ] - 1 13
Simplified example, buckling of Euler beam, cont.:Thethreeapproximationpointsfora secondorder"responsesurface"aretaken as the mean value plus/minus two times the standard deviationBasedonthisresponsesurface,theapproximateexpressionforthelimitstate function becomes: (Y= (Xi/X1,mean)g(x) = [(1+2*0.2)2*(4.02*2-10.9*Y+7.9)] - 1The"true reliabilityindex"based onthe"true limitstate function"nowbecomes 2.00 (i.e. which corresponds to a failure probability of 0.0228)Thereliabilityindexbasedonthesecondorderapproximationbecomes1.52 (i.e.a failure probability equal to 0.064)Thisresultisverysensitivetowheretheapproximationisperformed14
Simplified example, buckling of Euler beam, cont.: • The three approximation points for a second order ”response surface” are taken as the mean value plus/minus two times the standard deviation • Based on this response surface, the approximate expression for the limit state function becomes: (Y = (X1/X1,mean))) g(x) = [(1+2*0.2)2*(4.02*Y2 -10.9*Y+7.9)] – 1 • The ”true reliability index” based on the ”true limit state function” now becomes 2.00 (i.e. which corresponds to a failure probability of 0.0228) • The reliability index based on the second order approximation becomes 1.52 (i.e. a failure probability equal to 0.064) • This result is very sensitive to where the approximation is performed 14
Simplified example, buckling of Euler beam, cont.:"Exact"limitstatefunctionaftertransformationtothestandardnormalvariable,u:(Thereliabilityindexcorrespondstothepointg(u)=o)Exact limit state function4030-Limit state function2010--2-12-33Normalized variable
Simplified example, buckling of Euler beam, cont.: • ”Exact” limit state function after transformation to the standard normal variable, u: (The reliability index corresponds to the point g(u)=0)