Question

The interior floor system shown in the figure has W24 x 55 sections spaced 8 ft. 6 in. = _ on center and is supporting a floor dead load of 50 + 2 psf. = -- and a live floor load of 80 + 2 psf. = . Determine the governing load in lb./ft. that each beam must support. (If your PID is “5684659”, then A = 5, B = 6, C = 8, D = 4, E = 6, F = 5 and G = 9) - - -- -- -- - Area of slab supported per foot of length by W24 x 55 - - - - -- - ------------- - - Area supported by one beam (shaded) -- -- - - - - W24 x 55 wzx55 w22x55 W24 x 55 8 ft. 6 in. 8 ft. 6 in. 8 ft. 6 in.

Answer 1

W24 X 55 section = 4f3 in Spaced 8ft. bin Floor dead lood - 50 + 2psf = 50+ 2 pst = (16.67 + 2)psf · 18.67 pet. { Total floor load - 50psf. No. of slaks - 3 (Floor dead load of Islab – 18.67 psf) Live floor load - 80 + 2pet 8 + 2 psf = 26.67 + apst = 28.67 psf. s Total floor load - 80 + 2pot. No. of slats - 3 Floor bv land of Islab - 28.67 psf live

TT. eft bin w PH bin &Hong 8tfon Area = of slab supported by one bean (4 f1. 3in) x2 xil 8846 in 8.5 ft. Load supported by each bean & = 8.5 ft x Total load pee of sq. ft dead lood Total load per sq. ft = Dead load the - 18.67 + 28.67 = 47.34 psf - 47.34 lbfft Load suppated by each beam Vokal / 8.5 x 47.34 E 402.39 lbst

Note: The assumption made in the above calculation is that the weight provided is taken as total weight of three slabs as in the diagram , if the data given are for individual weight of the slabs then do not divide the total weight by 3 , where it is done so .