CIVE 303 - HW #1_Solutions(2)

CIVE 303 HW #1 Solutions: Problem 2-3 (a) Determine the factored axial load or the required axial strength, P u of a column in an office building with a regular roof configuration. The service axial loads on the column are as follows P D = 200 kips (dead load) P L = 300 kips (floor live load) P S = 150 kips (snow load) P W = ±60 kips (wind load) P E = ±40 kips (seismic load) (b) Calculate the required nominal axial compression strength, P n of the column. 1: P u = 1.4 P D = 1.4 (200k) = 280 kips 2: P u = 1.2 P D + 1.6 P L + 0.5 P S = 1.2 (200) + 1.6 (300) + 0.5 (150) = 795 kip s (governs) 3 (a): P u = 1.2 P D + 1.6 P S + 0.5P L = 1.2 (200) + 1.6 (150) + 0.5(300) = 630 kips 3 (b): P u = 1.2 P D + 1.6 P S + 0.5 P W = 1.2 (200) + 1.6 (150) + 0.5 (60) = 510 kips 4: P u = 1.2 P D + 1.0 P W + 0.5 P L + 0.5 P S = 1.2 (200) + 1.0 (60) + 0.5(300) + 0.5 (150) = 525 kips 5: P u = 1.2 P D + 1.0 P E + 0.5 P L + 0.2 P S = 1.2 (200) + 1.0 (40) + 0.5 (300) + 0.2 (150) = 460 kips Note that P D must always oppose P W and P E in load combination 6 6: P u = 0.9 P D + 1.0 P W = 0.9 (200) +1.0 (-60) = 120 kips (no net uplift) 7: P u = 0.9 P D + 1.0 P E = 0.9 (200) + 1.0 (-40) = 140 kips (no net uplift) φ P n > P u φ c = 0.9 (0.9)(P n ) = (795 kips) P n = 884 kips

Problem 2-4 (a) Determine the ultimate or factored load for a roof beam subjected to the following service loads: Dead Load = 29 psf (dead load) Snow Load = 35 psf (snow load) Roof live load = 20 psf Wind Load = 25 psf upwards / 15 psf downwards (b) Assuming the roof beam span is 30 ft and tributary width of 6 ft, determine the factored moment and shear. Since, S = 35psf > L r = 20psf, use S in equations and ignore L r . 1: p u = 1.4D = 1.4 (29) = 40.6 psf 2: p u = 1.2 D + 1.6 L + 0.5 S = 1.2 (29) + 1.6 (0) + 0.5 (35) = 52.3 psf 3 (a): p u = 1.2D + 1.6S + 0.5W = 1.2 (29) + 1.6 (35) + 0.5 (15) = 98.3 psf (governs) 3 (b): p u = 1.2D + 1.6S + 0.5L = 1.2 (29) + 1.6 (35) + (0) = 90.8 psf 4: p u = 1.2 D + 1.0 W + L + 0.5S = 1.2 (29) + 1.0 (15) + (0) + 0.5 (35) = 67.3 psf 5: p u = 1.2 D + 1.0 E + 0.5L + 0.2S = 1.2 (29) + 1.0 (0) + 0.5(0) + 0.2 (35) = 41.8 psf 6: p u = 0.9D + 1.0W ( D must always oppose W in load combinations 6 and 7) = 0.9 (29) + 1.0( - 25) ( upward wind load is taken as negative ) = 1.1 psf (no net uplift) 7: p u = 0.9D + 1.0E ( D must always oppose E in load combinations 6 and 7) = 0.9 (29) + 1.6(0) ( upward wind load is taken as negative ) = 26.1 psf (no net uplift)` w u = (98.3psf)(6ft) = 590 plf (downward) downward No net uplift (590)(30) 2 2 u u w L V = = = 8850 lb. . 2 2 (590)(30) 8 8 u u w L M = = = 66375 ft-Ib = 66.4 ft-kips Problem 2-5