Question

HM 5 Compression Members (50pts) Problem 1 Determine iſ the built-up column described below can resist the axial loads. The design loads are Pu = 40 kips and P1 = 120 kips. The column's unbraced length is 15 ft and the ends are pinned in both axes. Use LRFD. The due date is Monday, December 2, 2019 at noon. Problem Data: ASTM A572 Grade 50 Overall depth d= 8.00 in. Flange width bf = 10.5 in. Flange thickness t[= 3/8 in. Web depth h = 7.25 in. Web thickness lw = 1/4 in P. - 40 kips P - 120 kips . W 1079" h-714" L = 15-0" 1, %" + .%

Answer 1

= |A= 10.5x332 + 7.25× 主 - 9.6875 in . 是xx 7.25 +2 x 10.5 * (3) + 10.5% 3122.5の = x 7.25 (4) 2 ×(9%) xbo.s]. A巧 ( 2) = - 72.36 in - 3.56 . : 2.73 in .

GIVEN DATA Properties of Steel section Grade of steel Column Section Gross area of cross section, Radii of gyration, A 572 Gr 50 Built-up column A, = 9.69 in? rx = 3.56 in ry= 2.73 in Restraints Along x direction Unsupported Length of column, Bottom End Top end L = 15 ft : pinned : pinned Along y direction Unsupported Length of column, Bottom End Top end Ly = 15 ft pinned : pinned Loads Dead load, Live load, = PL = 40 kips 120 kips CALCULATIONS Yield stress of steel Modulus of Elasticity of steel F, = E = 50 ksi 29000 ksi Factored axial load, P = 1.2 Po + 1.6 P. = 240 kips Effective length factors, 1 1 KL. 1x ( 15 x 12 in )/ 3.56 = 50.562 - 1x 15 x 12 in)/ 2.73 = 65.934 Since KxLx/g < K, Ly/ty, KyLy/ry controls. À = K,L, F, = 0.871 < 1.5

Note: For À. <1.5, F, = (0.6584)F, *>15, F, =(0,5377) Therefore, critical stress, For = 36.397 ksi Hence, Design axial strength of column, 6 P = = = SAFE 0F,A, 0.9 x 36.4 317 kips > X 9.69 240 kips [=Pu] Hence,