the probability distribution function of the error

where xo = (x,.0,x2.0,..,x„.o )7. is the point in which the linearization is performed normally chosen as the mean value point and of (x) = 1,2,..n are the first order partial derivatives ax,
of f (x) taken in x = . From Equation (11.12) and Equation (11.6) it is seen that the expected value of the error E[e] can be assessed by: EH= f (p.) (11.17) and its variance Var[e] can be determined by: yam = tiaM 10.2 (a./(11 ax, x ‘ ax, „, ax, ‘
Provided that the distribution functions for the random variables are known, e.g. normal distributed the probability distribution function of the error is easily assessed. It is, however, important to notice that the variance of the error as given by Equation (11.14) depends on the linearizationpoint, i.e. x. = )T .z,o
Example 1— Linear Safety Margin Consider a steel rod under pure tension loading. The rod will fail if the applied stresses on the rod cross-sectional area exceed the steel yield stress. The yield stress R of the rod and the loading stress on the rod S are assumed to be uncertain modelled by uncorrelated normal distributed variables. The mean values and the standard deviations of the yield strength and the loading are given as g„ = 350,6„ = 35 MPa and US = 200,a, = 40 MPa respectively. The limit state function describing the event of failure may be written as: g(x) = r — s whereby the safety margin M may be written as: M = R — S The mean value and standard deviation of the safety margin M arc thus: Pa = 350 — 200 =150 QM = 4352 +402 = 53.15 whereby we may calculate the reliability index as: = 5150. 2.84 53.15 Finally we have that the failure probability is determined as: Pr = 4,(-2.84) = 2.4.10′




Hello! Need help with your assignments? We are here
Don`t copy text!