Question

3.25. A structural engineer believes the yield stress S in mild steel reinforcing bars to be lognormally distributed with parameters is - 40.0 ksi and Ons - 0.1, while the bar areas A of nominal %-in.-diameter bars ("no. 9" bars) are lognormally dis- tributed with mean 1.0 in.. and standard deviation 0.05 in. Find the distribution of the yield force F-SA, assuming independence of 8 and A. Sketch this density function and find PF S 35 kips) and the value follow such that PIF S Sallow) - 0.01

show your work, please.

Answer 1

- Clearly, here we are dealing with two independently dentrocluided log normal nandom variables s and A and to find the distribution of the produst of e parameters maa son lognormal distribution with moon 40.0 Khi we want to find the distribe Sand A, where 1 and 6 Is = 0.1 and An lognormal cseribution with mean to . and how standard deviation o.oo Now, we know a continuous random variable Xn lognormal distribution =) legg a normal distribution. Now, il xn lognormal (M, 0p) = 3Y = 109 xn nlm, 0 ?). YNN (V,0?) • F (4) 4 , 5 Vào(Y): an(y): MGF of y: pult 12220? - E Plexy) se mi to ROV...lis but it als we get Ele"). eM+102: . inol (pjere p butting -2 --> F(x) = 8M+ 2 H + (x) = 0 Mt 102 Qy Agoun, E(XP); Elepy): een 24 +202. Var (x): F(XP) -E2(x) RM+2.02 - - QP + 2 = 2448? leo? ).

has mean 1.0 and a.d. 0.05 Wow, A has mean i. no e Meteor a) Wt 02.1093.0 -) Mt 102=0. 20+020 now, 2M +02 lepel)=10.0512 => Boreri) = 0.0029 is eu?: 1.0025 =).02 = ghosts) en 11.0025 ) A 0.0025 -0.00125 -0.0005 no A vognormal (HE40.0, 0, 2, (0.1)? son lognormal I Mo.-0.00125,822 = 0.0025). we have to find distribullon of F = AS. wow & Toge say Uz logan normal (M1,02).. :) indefondently v=log su normal (Mg, 122) indepenas Aceu sevda . We have to find distribution of AS - e utv. vow, UNN(Misrie) VnNCAMA, ODP). Then, Utv NNCHANA, 2002) [ Proof: I Using characteristic functions UN NO Mo, Oja)folt)= Fleila), Demi larky, Qult). Mat + que o .... (1) Mta270,2 Matu Ovld)

. So, Eur (it) Eleit (u+vs) Eleitu) Eleito) as U and v are independently distributed neendom Larichen Plondise) using cii) and (ii) il (Mi+ha) – which is the most characteristic function of N(Mitha, O2102?). So, we know characterestin fundion uniquely determines the distribution function no U+V NN ( HitMQ, 0,84092). --) e lutv) vlog normal (MyMp; 0724092). =JF= AS n10g normal (Motha, 0,240,22). here, in this case, Hit Ma = (40.07 0.00195) = 40.00125 8,2 0QP = (0.138 + 0.0025 = 0.0125 AFN lognormal (40.00125, 0.0126) P[FI 35] ( 109.3" fly is an increasing ] [as function [as 10g x pllonge ?109,35] ] 160.00125,0.012 1 Entognormal 190.00125,0.0128) 10ger 3.55] iegef-40.00125 - 109NN 140.0012 530-01 p AN(Oil) 10.0195 3 0 .0125 ? 3.55–40.001951togef - 40.00125 Ve o.0795 10.0125 - 2~N(011) P[ $-326,6297 $(-326.029) i Find value using table standard normal eDF )