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Q: I-Beam size calaculation for residential construction ( Answered ,   0 Comments )
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 Subject: I-Beam size calaculation for residential construction Category: Science > Physics Asked by: bosarge66-ga List Price: \$50.00 Posted: 05 Sep 2006 05:36 PDT Expires: 05 Oct 2006 05:36 PDT Question ID: 762357
 ```I would like to calculate the size of I-Beams needed to carry the load of a two story house. Outside Dimensions of house: 33'2" X 34'0". We'll be using (4) I-beams equally spaced running parallel to the 33'2" dimension. The longest span of any beam will be 9'9.3125". Exterior walls will be 2X6 studs on 16" centers. First and second floor joists will be 2X12 on 16" centers. 3/4 plywood and hardwood floors through out.``` Request for Question Clarification by redhoss-ga on 05 Sep 2006 06:07 PDT ```Is the roof load transfered to the exterior walls or supported by the interior walls. Since you don't mention anything about the roof in your question, I would guess that it would be excluded from the beam loads.``` Clarification of Question by bosarge66-ga on 05 Sep 2006 10:32 PDT ```Sorry....Yes the load would be transferred to the exterior walls. Pitch of roof (main roof) is 12 12. Roof will be constructed of 3/4 decking with 26ga. 5V-Crimp galvalum. Thanks```
 ```Okay bosarge66, not having to deal with the roof loads is good. We can use 10 psf dead load and 40 psf live load for the two floors. So, we have a total of 100 psf on the beams. The beam formulas for this loading are: M (maximum bending moment) = wl^2/8 D (deflection @ center of span) = 5wl^4/384 EI w = (34'/5 spaces) x 100 psf = 6.8' x 100 psf = 680 lb per ft l = 9.75' D = l/360 = (9.75 x 12) / 360 = .325" Where E is a constant for steel = 30,000,000 psi And I is the moment of inertia Solving for M: M = 680 x 9.75^2 / 8 = 8,080 ft lb = 96,960 in lb The allowable bending stress for structural steel (s) = 0.55 x 36,000 psi = 19,800 psi The section modulus of the required beam (S) = M/s = 96,960/19,800 = 4.89 in^3 Now we must calculate the required I (moment of inertia): Solving for I in the above formula for deflection we get: I = 5wl^4/384 ED = (5 x 680 x 9.75^4 / 384 x 30,000,000 x 0.325) x 1728 NOTE: 1728 is a conversion factor to get the proper units for I I = 14.2 in^4 Now we can look for a beam with these minimum properties. A good choice would be a 6 inch deep Wide Flange beam weighing 8.5 # per ft (W6x8.5) S = 5.08 in^3 I = 14.8 in^4 If you have some other beam you would like to use, or there is anything that you don't understand, please ask for a clarification and I will try and help. Good luck with your project, Redhoss```
 bosarge66-ga rated this answer: `Thanks for the prompt answers.......Very Helpful!`