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Q: Steel Beam Sizing ( Answered,   1 Comment )
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 Subject: Steel Beam Sizing Category: Science > Physics Asked by: josephc68-ga List Price: \$10.00 Posted: 28 Oct 2006 12:57 PDT Expires: 27 Nov 2006 11:57 PST Question ID: 777843
 ```I am doing some reinforcement in my house to compensate for some poor building practices. I need to support a 2nd floor and attic load without exceeding the depth of my joists, 7 1/4". I used an engineered lumber program to compute my loads but do not know how to convert this load information to size a steel beam. I am pretty sure that a W6x25 would carry the load but would like this confirmed and to see the math behind this. Deflection Criteria: Specified(LL:L/360,TL:L/240). Load Group: Primary Load Group _________________ ^ 14' 11.50" ^ Max. Vertical Reaction Total (lbs) 6473 6473 Max. Vertical Reaction Live (lbs) 4550 4550 Required Bearing Length in 2.47(W) 2.47(S) Max. Unbraced Length (in) 35 Loading on all spans, LDF = 0.90 , 1.0 Dead Shear at Support (lbs) 1569 -1569 Max Shear at Support (lbs) 1896 -1896 Member Reaction (lbs) 1896 1896 Support Reaction (lbs) 1923 1923 Moment (Ft-Lbs) 7091 Loading on all spans, LDF = 1.00 , 1.0 Dead + 1.0 Floor Shear at Support (lbs) 5281 -5281 Max Shear at Support (lbs) 6384 -6384 Member Reaction (lbs) 6384 6384 Support Reaction (lbs) 6473 6473 Moment (Ft-Lbs) 23872 Live Deflection (in) 0.486 Total Deflection (in) 0.691``` Request for Question Clarification by redhoss-ga on 28 Oct 2006 16:03 PDT ```Just out of curiosity, what were the loads that you used to get these results. I can answer your question.```
 ```Okay josephc68, I guess you don't want to tell me how you got your numbers. We will just use your numbers and go ahead. The maximum bending moment is: M = Rx x L/2 = 6473 x 89.75 = 580,952 in-lbs The required section modulus is: S = M/allowable stress = 580,952/(36,000 x .55) = 29.34 in^3 The allowable deflection is L/360 = 179.5/360 = 0.5 in Using this deflection to calculate the required moment of inertia: Where P = 2 x Rx = 12,946 lb E is the modulus of elasticity of steel = 30,000,000 psi I = PL^3/48ED = 12,946 (179.5^3)/48 (30,000,000)(0.5) = 104 in^4 The properties for a W6x25 are: S = 16.7 in^3 I = 53.3 in^4 So, you see that the W6x25 is not adequate. It looks like your best choice would be a W8x40 with S = 35.5 and I = 146. Look this over and if you have something else you would like to try please ask for a clarification. Redhoss``` Request for Answer Clarification by josephc68-ga on 30 Oct 2006 11:10 PST ```Sorry Redhoss, I didn't get my email notification that there was an answer waiting for me. For simplicity my loads are a 10' tributary with 30 psf live load and 12 psf dead load. Do I have any other options in terms of a steel beam that would be less than a 2x8 (7.25") in height? In what sense is the W6x25 inadequate, deflection? The reason I ask is that I found a table online that shows the W6x25 can support 652 lb/ft over a 16' span. Thanks.``` Clarification of Answer by redhoss-ga on 30 Oct 2006 14:03 PST ```I wish that you had started with the info that you state in your clarification request. If you take your new numbers here is what we have: w (load in lbs. per foot) = 42 psf x 10 = 420 lb/ft So, the end reactions would be: Rx = 420 x 14.958 / 2 = 3,141 lbs. This is less that half the 6,473 number that your "engineered lumber program" gave you. I don't know what went wrong, but it yielded some bogus numbers. Here is a re-calculation using the correct numbers: The maximum bending moment is: M = Rx x L/2 = 3,141 x 89.75 = 281,905 in-lbs The required section modulus is: S = M/allowable stress = 281,905/(36,000 x .55) = 14.24 in^3 The allowable deflection is L/360 = 179.5/360 = 0.5 in Using this deflection to calculate the required moment of inertia: Where P = 2 x Rx = 6,282 lb E is the modulus of elasticity of steel = 30,000,000 psi I = PL^3/48ED = 6,282 (179.5^3)/48 (30,000,000)(0.5) = 50.46 in^4 Now the W6x25 looks just fine. Good choice.```
 ```Joe, I just wanted to gently point out that if it ever comes around again that you are restricted by height and so cannot get the load rating you need - just use two beams side-by-side. Even without joining plates they are fairly close to being double the strength of one beam. PHM ----------```