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Q: I-Beam size calaculation for residential construction ( Answered ,   0 Comments ) Question
 Subject: I-Beam size calaculation for residential construction Category: Science Asked by: skierbob-ga List Price: \$50.00 Posted: 07 Aug 2006 05:42 PDT Expires: 06 Sep 2006 05:42 PDT Question ID: 753367
 ```I would like to calculate the size of two I-Beams with a span of 20 feet for a two story residential house addition. I would like to keep the height of both beams to a minimum by using heavier beams if possible. The first beam will carry the second floor 40 psf live and 20 psf dead. The second beam will be directly above and in the same plane so the lower beam will not be carrying the load of the ceiling and roof. The first beam could have a supported at its center so there would be two 10 foot spans. I would add this center support if it would reduce the height of the beam. The second beam will carry the ceiling at 10 psf live and 5 psf dead and the roof at 20 psf live load and 10 psf dead load. The second beam will be the full 20? clear span.``` Request for Question Clarification by redhoss-ga on 08 Aug 2006 07:15 PDT ```Your description of the loadings is very good. However, you have left out one important thing. I need to know the size of the area supported so that we can determine how much tributary area the beams bear. If you can give me this, we can size the beams.``` Request for Question Clarification by redhoss-ga on 08 Aug 2006 12:37 PDT ```When I asked for the size of the area supported, what I need is the other dimension of the house. You give me the 20' dimension, I just need the dimension in the other direction.``` Clarification of Question by skierbob-ga on 11 Aug 2006 05:42 PDT `da-- the other dimension is 28'-- it is a 20' x 28' garage with second floor.` Subject: Re: I-Beam size calaculation for residential construction Answered By: redhoss-ga on 12 Aug 2006 06:38 PDT Rated: ```Okay skierbob, we have enough info to proceed. In each case the beam supports 14' of the load. The rest being supported by the walls. FIRST BEAM 20' SPAN: The beam formulas for this beam loadings are: M (maximum bending moment) = wl^2/8 D (deflection @ center of span) = 5wl^4/384 EI NOTE: Maximum deflection is limited to D = l/360 = 20 x 12 / 360 = 0.66 in w (load per foot) = (40 + 20)psf x 14' = 840 lb per ft l (beam span) = 20 ft Where E is a constant for steel = 30,000,000 psi And I is the moment of inertia Solving for M: M = 840 x 20^2 / 8 = 42,000 ft lb = 504,000 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 = 504,000/19,800 = 25.45 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 840 x 20^4 / 384 x 30,000,000 x 0.66) x 1728 NOTE: 1728 is a conversion factor to get the proper units for I I = 152.73 in^4 It looks like a good choice for this beam would be: W8x48 Depth = 8.5" Flange width = 8.117 Weight = 48 lb per ft FIRST BEAM WITH CENTER SUPPORT (10 FT. SPANS): The beam formulas for this beam loadings are: M (maximum bending moment) = 49 wl^2/512 D (deflection @ center of span) = 0.0092 wl^4/EI NOTE: Maximum deflection is limited to D = l/360 = 10 x 12 / 360 = 0.33 in w (load per foot) = (40 + 20)psf x 14' = 840 lb per ft l (beam span) = 10 ft Where E is a constant for steel = 30,000,000 psi And I is the moment of inertia Solving for M: M = 49 x 840 x 10^2 / 512 = 8,039 ft lb = 96,469 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 = 96,469/19,800 = 4.87 in^3 Now we must calculate the required I (moment of inertia): Solving for I in the above formula for deflection we get: I = 0.0092 wl^4/ED = (0.0092 x 840 x 10^4 / 30,000,000 x 0.33) x 1728 NOTE: 1728 is a conversion factor to get the proper units for I I = 13.49 in^4 It looks like a good choice for this beam would be: W5x16 Depth = 5.0" Flange width = 5.0 Weight = 16 lb per ft SECOND BEAM: The beam formulas for this beam loadings are: M (maximum bending moment) = wl^2/8 D (deflection @ center of span) = 5wl^4/384 EI NOTE: Maximum deflection is limited to D = l/360 = 20 x 12 / 360 = 0.66 in w (load per foot) = (10 + 5 + 20 + 10)psf x 14' = 630 lb per ft l (beam span) = 20 ft Where E is a constant for steel = 30,000,000 psi And I is the moment of inertia Solving for M: M = 630 x 20^2 / 8 = 31,500 ft lb = 378,000 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 = 378,000/19,800 = 19.09 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 630 x 20^4 / 384 x 30,000,000 x 0.66) x 1728 NOTE: 1728 is a conversion factor to get the proper units for I I = 114.55 in^4 It looks like a good choice for this beam would be: W8x35 Depth = 8.12" Flange width = 8.027 Weight = 35 lb per ft If there is any of this you don't understand or you don't like the beam choices, please ask for a clarification and I will try and answer. Good luck with your project, Redhoss```
 skierbob-ga rated this answer: ```Excellent – this is exactly what I was looking for and the turn around was a lot quicker than I expected – Thanks```  