

Subject:
Steel beam specifications
Category: Science Asked by: waltallga List Price: $50.00 
Posted:
15 Apr 2006 19:00 PDT
Expires: 15 May 2006 19:00 PDT Question ID: 719327 
I need the specifications of a steel Ibeam to replace a wooden timber beam. This room has a 23 foot 4"x6" white oak beam supported on each end by a 2"x4" stud and headered wall with an 8"x8" wooden post at its center. Perpendicular to the wooden beam are 4"x6" rough cut unplained full dimension wooden beams spaced 24" on center and 23 feet long with the ends resting on 2"x4" wood studded walls and the centers supported by the main center beam described above. I would like to replace the center beam with a steel beam and eliminate the center post in order to open up the room. There is also an option of eliminating the center post and positioning a new steel beam perpendicular to and supporting the main center beam with supporting poles at each end. The area supported by the beam structures is a conventionally constructed wood frame floor with 1 and 1/2 inches plywood floors, 4 foot knee walls, and a 2 x 8 wood framed attic ceiling, attic, and roof system with 7 1/8 plywood roof sheathing and fiberglass shingles. What size steel beam is required for this project?  
 
 
 


Subject:
Re: Steel beam specifications
Answered By: redhossga on 17 Apr 2006 07:45 PDT Rated: 
Okay waltall, I think we have enough info to get started. First we need to calculate loading: Attic live load = 20 PSF 4 x 6 timber = 8.25 #/ft (based on 50# per cubic ft) 3/4 plywood = 2.2 PSF 3/8 plywood = 1.1 PSF 1" styrofoam = 0.25 PSF (based on 3# per cubic ft) The center beam supports 1/2 (11.5') of the total load. The remainder is supported by the walls. Calculating the beam loading in lbs. per ft: 4 x 6 beams @ 24' o/c (say 11 beams) x 11.5' x 8.25 #/ft / 23 ft = 46 ppf plywood = (2.2 + 1.1) 3.3 PSF x 11.5' = 38 ppf styrofoam = .25 x 11.5' = 3 ppf Live load = 230 ppf Total = 317 ppf (NOTE: weight of steel beam not included) The formula for maximum bending moment is: M = wl^2/8 Where: w = 317 ppf l = unsupported length of beam = 23' M = (317 x 23^2) / 8 = 20,962 ftlbs = 251,544 inlbs S (section modulus) = M / s (allowable bending stress = .55 x 36 ksi) S = 12.7 in^3 The formula for deflection is: D = 5wl^4/384EI Where: E (modulus of elasticity for steel) = 30,000,000 psi I = moment of inertia required D = l/360 = (23 x 12) / 360 = 0.76 inches Solving for I: I = 5wl^4/384ED I = [(5 x 317 x 23^4) / (384 x 30,000,000 x 0.76)] x 1728 (conversion factor) I = 88 in^4 Now we are ready to choose a beam. Looking in the AISC tables, a good beam would be (keeping in mind that we want to reduce headroom as little as practical): A wide flange beam 8 inches deep weighing 28 lbs. per ft. looks good: W8x28 Depth = 8.06 inches Weight = 28 ppf Flange width = 6.54 inches S = 24.3 in^3 I = 97.8 in^4 You could save weight/money by going with a W10x19 if you don't mind sacrificing another 2 inches in headroom. The other option (as you suggested) would be to use an additional steel beam to support the existing wood beam. I feel that this would not be a good plan as it would not look very good and severely reduce your headroom. Look this over and see what you think. If you have any questions, please ask for a clarification. Redhoss 
waltallga
rated this answer:
and gave an additional tip of:
$10.00
This is perfect. I have located a beam and a contractor to install it. I appreciate your thouroughness and the formulas add a great deal of confidence to the answer. Thank you redhossga. 

Subject:
Re: Steel beam specifications
From: redfoxjumpsga on 15 Apr 2006 20:25 PDT 
Building codes may guide or even prevent your project, While a beam of the same size s the existing timber will be stronger you are asking it to take a heavier load. Do you have an idea of how much load the center post was taking? (Be a shame if the beam flexed enough to crack the plaster.) Earthquake prone area? 
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