**Question:**

The rule of thumb in beam design is length divide by 2 in inches as the beam depth. Use that number as starting point of the beam analysis. For a 30 ft span you are looking at W16, 16 inches deep I beam because there is no W15.

Suppose you have a 30ftx30ft garage that you want to be able to park your cars on top and use the bottom as game room or storage. For a concrete slab on metal deck, we will need two beams because the deck can only span 10 to 12 feet. Letâ€™s run the number and see what size beam we need.

Dead Load Calculations:

8 inch Slab = 8/12*150 = 100 psf

3 inch 20 Ga metal deck = 2.3 psf

Beam weight = 31*30/(30*10) = 3.1 psf

Duct work = 5 psf

Ceiling = 2 psf

Electrical light = 3 psf

Total DL = 115.4 psf

Live Load: = 40 psf

Total load = 155.4 psf

The two beams will be spaced evenly at 10 ft apart for tributary width 10ft.

Beam moment equation:

M = W*L^2/8; w = 155.4*10 = 1554 plf

M = 1554*30^2/8 = 174,825 lb-ft = 174.8kip-ft = 2098 kip-in

S = M/Fy = 2098/

A W16x40 with S=64.7 cubic inches will be sufficient for the load. Other beam options W14x48 S=70.3, W12x50 S=64.7, W18x40 S=68.4 will work as well.

The beam weight of 31 lb/ft used in the calculation is less than the 40lb/ft needed for the job, let's double check:

beam weight = 50x30/300 = 5 psf (50lb/ft)

M = 1573*900/8 = 2123.6 k-in

S = 2123.6/33 = 64.35 cubic inches which is less than 64.7 W16x40

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