Question

For the loading conditions shown below, use LRFD and a) Select a W14 of A992 steel b) Select a steel pipe c) Select a rectangular HSS Note: After selection, be sure to perform all checks D = 90k L = 260k

Answer 1

- D=90 L=260k Giver, A 992 steel D2=90k, L2=260k 151-44 Ale to LRFO, Puz 1-2D2 +1622 - 12(90)+261260) 1. = 524 kips Unsupported hength of Columns 15!yy As bo the ends are fined fetive hemath factor k = 065 = K 2 = (0' 6.5) (15'4") =9",1" (approx) -KL het il y = 50 2 = 0661215 * For = 10.658 ) by = 41.64 ksi + Required Area of 418 = Ay = Pu De fr - 524 0.9141-64) 13.982 in2 =

select wloo 500 49 whose Ag = 14.4 inz r = 4.35 in ru=2.54 in ky Ly , 065* ((9x12) +11) = 30.452 2:54 alde = (30452 29000) = 0.4024215 For = 0:658 (004024)?(50) = 46.722 ksi o Be Pn= 10.9) (46.722) (144) = 605:Skips > Szylijs tenu safe & select a steel pipe select la 3" NPS 20 NPS 20 whose D=105" whose t = 0.308 Ad-0172434 718 I (1.052_0.434 ² ) = 0 . bus inz &A= Radius of gyration &= (0²+d² [105) + 10 792) 50.32142 in -=240-665 KL065(119) 0-3214 ode = (240.665 50 - = 3:1809 > 15 112129000)

& For = fy I de - 08779.7150) (13.1809) - 4.333 socPn=109)(0.718)241333) = 2.79 CNot ok) . D?d? select G" NPS 150 Ved whose D = 6.625" t = 0.8644 27 d = 4.89711 - 2.06 nchi (D2-d2] =15.632 in2 KL 10-657(119) — 37.5485. 2.06 de = 37.5485 / 50 V 52129000) = 0.496 <1,5 . for = 0.658 10496) (50) = 45.104 ksi (049612 & Po = (0.9)/45-104) 115.632) = 634.565 > 524 Kies Sto select xps 6" NPS iso (Hence ok).

ry is Rectangular Has select W 12*10* whose Ag = 14.1 in 2 - rox4.45 in r = 3.25 in . Ku Ly 0.65(119) = 23.8 - 3.25 de = (23.8) Dianos 0.314</05 > Fer= 0.658 (0.314% (50) = 47.971 ksi % Po - (0.9)147.971)(4.1) a 608.7579 524 kips Hence ok. :: Select H55 12x10x section
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