Question

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

please solve it manually and using ETABS.

Answer 1

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