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Q: Beam size to replace a bearing wall in first level down of a split level home ( Answered,   0 Comments )
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 Subject: Beam size to replace a bearing wall in first level down of a split level home Category: Science Asked by: kjelica-ga List Price: \$50.00 Posted: 07 Aug 2006 09:07 PDT Expires: 06 Sep 2006 09:07 PDT Question ID: 753426
 ```I am replacing a bearing wall that supports the second story of my home. The roof is supported by W trussed rafters mounted on the outside walls. The beam will need to support a 24 ft. wide upper floor for a 16 ft span with no center support. The supported floor is made from 2X10 joists with 1 and 1/2 inches of 3/4 in' plywood flooring. There is parguet flooring over half of the area. There are two main 2X4 and sheetrock walls partitioning the second floor, neither is bearing. There is foundation at both ends of the proposed span. What is the calculation? What size beam can I use? I would like to use a standard flange if possible, a conservative answer is fine by me.```
 ```Hello kjelica, I believe we can work out a solution for you. Floor live load is normally taken to be 40 psf. For the type construction you have I think a 10 psf dead load is probably a good number. The beam actually supports a 12' wide portion of the floor and the walls support the rest. The beam formulas for your 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 = 16 x 12 / 360 = 0.53 in w (load per foot) = (40 + 10)psf x 12' = 600 lb per ft l (beam span) = 16 ft Where E is a constant for steel = 30,000,000 psi And I is the moment of inertia Solving for M: M = 600 x 16^2 / 8 = 19,200 ft lb = 230,400 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 = 230,400/19,800 = 11.64 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 600 x 16^4 / 384 x 30,000,000 x 0.53) x 1728 NOTE: 1728 is a conversion factor to get the proper units for I I = 55.6 in^4 From my AISC manual it looks like good choices might be one of these wide flange beams: W8x17 Depth = 8" Flange width = 5.25" Weight = 17 lb/ft W10x15 Depth = 10" Flange width = 4.00" Weight = 15 lb/ft W12x14 Depth = 11.91" Flange width = 3.968" Weight = 14 lb/ft I just re-read your question and noticed you said "standard beam". Here is an "S beam": S8x18.4 Depth = 8" Flange width = 4.001" Weight = 18.4 lb/ft If there is some other type beam you would rather use, or you have a beam you would like me to check out just let me know. Please ask for a clarification if there is something that you don't understand. Good luck with your project, Redhoss```