Weight of floor=(4/12)*150=50 psf
Additional dead load=40 psf
Total dead load=50+40=90 psf
Live load=100 psf
Factored load on floor=1.2*90+1.6*100=268 psf
Tributtary width of beam=beam spacing=15 ft
Uniform load on beam=268*15/1000=4.02 kip/ft
Span=35 ft
Maximimum moment in beam=4.02*35²/8=615.6 kip-ft
From table 3-2 of aisc, lighter w section that has flexural capacity greater than 615.6 is W24x68
1.) Select the lightest W beam to support the following loads 4 in. concrete slab 15 ft 40 psf dead load 15 ft 100...
please solve it manually and using ETABS. please solve it manually and using ETABS. 1.) Select the lightest W beam to support the following loads 4 in. concrete slab 15 ft 40 psf dead load 15 ft 100 psf live load Fy 50 ksi 35 ft 15 ft Wu 35 ft 1.) Select the lightest W beam to support the following loads 4 in. concrete slab 15 ft 40 psf dead load 15 ft 100 psf live load Fy 50...
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A one-way solid concrete slab is to be used for a simple span of 16 ft. In addition to its own weight, the slab carries a superimposed dead load of 20 psf and a live load of 90 psf. Design the slab for minimum thickness if ƒy is 40 ksi, and ƒ’c is 3 ksi 9 in 8 in 6 in 7.6 in
Problem 1 (100 pts) Select the lightest W24 beam section with Fy = 50 ksi using LRFD for the following span and loading. The unbraced length of the compression flange is 30 ft (Lb = 30'). Consider Cb > 1. The given dead load does not include beam weight. Verify that the selected beam has adequate shear strength. Maximum allowable deflection due to live load is L720. Maximum allowable deflection due to total load is L/360. WD = 1.2 k/ft...
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