1. Consider the steel rod provided in the following. VS Figure 1: steel rod = - The rod carries a normal distributed load, S with us 31kN and os = 3.0kN. The resistance, R, of the rod is given by the following product: R = A· fy , where A is the area of the rod, equal to 105mm2 and fy is the yield stress modelled as a normal distributed random variable with mean li fy = 458 · 10-3kN/mm2 and standard deviation 20 · 10-3kN/mm². (a) Calculate the mean uz and the standard deviation oz of the safety margin, Z assuming Z= R – S. O fy =
(b) Estimate the rod's probability of failure Pf. Note: You can estimate the standard normal distribution value either using standard nor- mal distribution table form the appendix or using Excel's direct function: NORM.S.DIST(z)
(c) What new mean value of the stress ūs is needed, to reach a reliability index of ß = 3.2? (Assume that the standard deviation stays the same)
1. Consider the steel rod provided in the following. VS Figure 1: steel rod = - The rod carries a normal distributed loa
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1. Consider the steel rod provided in the following. VS Figure 1: steel rod = - The rod carries a normal distributed loa
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