Question

4. A two story office building has interior columns spaced 25 ft apart in two perpendicular directions. The columns are fixed at the roof, the second floor and the ground floor in both directions. Consider the dead load (4 inch-thick reinforced concrete slab at the roof and 6 inch thick reinforced concrete slab at the floor) as well as appropriate live load. Ignore other dead weight other than shown. Determine the following using A992 steel: (20 points) 1) Dead load and live load supported by interior column at ground level 2) Select the most economical W12 section for the interior column at ground level. I II - 서 I I 2@25' = 50 C Interior Column 내 I 2@25' = 50

Answer 1

ROOF LOADS Thickness of slab Live Load 4 in L 20 psf CALCULATIONS Tributary area of structural member At 25 x 25 625 ft? Dead Load D 4 /12 150 pcf x 50 psf Axial load on column Dead Lolad X 625 /1000 = 50 psf 31.25 kips 20 psf 12.5 kips Live Load х 625 /1000 FLOOR LOADS Thickness of slab Live Load 6 in L 50 psf Number of floors 1 CALCULATIONS Tributary area of structural member AT 625 ft? х 1 floors 625 ft? Dead Load D 6/12 150 pcf x 75 psf Axial load on column Dead load 625 /1000 75 psf x 46.88 kips 50 psf x 31.25 kips Live Load 625 /1000 TOTAL LOADS Dead Load D 31.25 + 46.9 78.13 kips 12.5 + 31.25 43.75 kips Live Load, = Therefore, factored design load, P 1.2 D + 1.6 L 164 kips

GIVEN DATA Grade of steel : A 992 Factored Load, Pu= 164 kips Restraints Along x direction Unsupported Length of column, Bottom End Top end Lx = 24 ft : fixed : fixed Along y direction Unsupported Length of column, Bottom End Top end Ly = 24 ft : fixed : fixed CALCULATIONS Yield stress of steel Modulus of Elasticity of steel Fy = 50 ksi E 29000 ksi Effective length factors, 0.65 Ky = Ky = 0.65 Assuming Kl/ry 100 KL F r y 7 E 1.322 < 1.5 Note: For q, <1.5, Fc = (0.65842 )F, ac > 1.5, For = 0.877 2 Therefore, critical stress, Fc = 24.06 ksi Therefore, required area of cross-section, Ag = Pu/ocFc = 163.75 / 0.9 X 24.06 7.56 in? Select W 12x40 CHECK CAPACITY Properties of Steel section Column Section W 12x40

Ag = 11.7 in 2 Gross area of cross section, Radii of gyration, rx = 5.13 in 1.94 in Since, KxLx/rx Kyby/ry, Kyly/ry controls Kyly/ry 24.0 x 12 in) / 1.94 0.65 x 96.495 F K,L, ry πΕ 1.275 < 1.5 Note: For « <1.5, F, = (0.6581c%)F, 2 > 1.5, Fc = 0.877 2² y Therefore, critical stress, Fas 25.32 ksi Hence, Design axial strength of column, O.PL 0.F, Ag 25.32 0.9 x 267 kips SAFE X 11.7 164 kips = Pu Hence, Thus, the member size is OK