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 Subject: Roof support Category: Reference, Education and News > Teaching and Research Asked by: hankconti-ga List Price: \$50.00 Posted: 13 Sep 2005 07:43 PDT Expires: 13 Oct 2005 07:43 PDT Question ID: 567547
 ```I'm bulding a barn 50' x 120' the 120 section have an opening of 30' for an 30' x 30' adition. Roof trusses sits on 2' on center. Reaction on the trusses is 3300 lbs. Can you help me with sizing an I-beam for this opening?``` Request for Question Clarification by redhoss-ga on 14 Sep 2005 16:36 PDT ```I can help you, but I need to know something first. The 3,300 lb. number sounds a little high if it is per end. What is your live load and dead load. Could the 3,300 lbs. be a total number, say 1,650 lbs. per end. Are you in a heavy snow area. You would have to be in a 60 PSF area to get 3,300 lbs. per side. Once we get past this, the rest is easy.``` Clarification of Question by hankconti-ga on 15 Sep 2005 04:32 PDT ```Loading (psf) TCLL 30.0 TCDL 15.0 BCLL 0.0 BCDL 10.0```
 ```Hello hankconti, sorry it took so long to get back with you. We will use your truss reaction of 3,300 lbs. First we will calculate the maximum bending moment: Using the formula for a simple beam-uniformly distributed load from the AISC (American Institute of Steel Construction): M = (w x L^2)/8 Where "w" in your case is the total load on the beam divided by the length or w = 3,300 lb x 14 (trusses on beam)/30 ft = 1540 lb/ft M = (1540 x 30^2)/8 = 173,250 ft-lb Section modulus = M / allowable stress For a standard structural beam (yield stress = 36 ksi) the allowable stress is usually taken as 0.6 x 36 ksi of 22,000 psi Section modulus = 173,250/22,000 = 7.85 in.^3 This was an interesting exercise, but in your case the beam design is controlled by deflection. The amount of deflection should be limited to D = L/360 or (30 ft x 12 in/ft)/360 = 1 in. The formula for maximum deflection is: D = 5wL^4/384 EI Where E (Modulus of Elasticity) is a constant for steel = 30,000,000 psi and I is the "Moment of Inertia" required. Solving the above equation for I we get: I = 5wL^4/384 ED I = ((5 x 1540 x 30^4)/(384 x 30,000,000 x 1)) x 1728 in^3/Ft^3 Note: the 1728 factor is required to get the units correct I = 935.55 in^4 In a wide flange beam a good choice would be a W21x49 with I = 971 in^4. This is a beam that is 21 inches deep and weighs 49 lb. per ft. I hope you have been able to follow this and I will be glad to answer any questions you might have. Please ask for a clarification if you need additional help. Good luck with your project, Redhoss```