Page 454 - Marine Structural Design

430 Pari IV Structural Reliability

To normalize the random variables in above equations, the followings can be
formed
R-4
XI =-
0.4
P-4
x2 =-
0.8
Based on the p index method described in section 23.3.3 and Example 23.2, the
followings can be obtained,
+
2, = 0.728~, - 0.686~~ 3.846
2, = 0.577~~ -0.816~~ +1.691
Accordingly, the reliability index of component 1 and 2 are estimated as

p 1=3.846
pz=I .691
The corresponding probabilities of failure are obtained as
P,,, = a(- p,) = 0.00006
P,,2 = a(- p,) = 0.04794
The failure probability of the system is approximated by
mm P,,i 5 p/.sys CP/,i

Hence,
0.04794 I Pf.r,,s 10.04800
Besides, the correlation coefficient of ZI and ZZ is obtained as

p = aibi = 0.728 x 0.577 + 0.686 x 0.861 = 0.98
For this system, the correlation coefficient is nearly equal to 1. Accordingly, the
failure probability of the system Pf, sys is equal to the lower bound approximately, i.e.
Pf, ,,=0.04794

23.1 1.4 Example 23.4: Reliability Calculation of Parallel System
Problem
Assume that a structure is composed of 4 parallel components, their corresponding
reliability index are, p1=3.57, p~3.41, p3=4.24 and p4=5.48 respectively. What are
the bounds of the failure probably of the parallel system?

Solution
The failure probability of each component can be estimated as