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Q: i beam calcualtion for a bridge. ( Answered,   0 Comments )
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 Subject: i beam calcualtion for a bridge. Category: Miscellaneous Asked by: raymond111-ga List Price: \$50.00 Posted: 28 Aug 2006 10:42 PDT Expires: 27 Sep 2006 10:42 PDT Question ID: 760211
 ```I am building a walking bridge over a creek. the span from support points is 36' The width is to be 5 feet and the deck will be 2 x 8 lumber. the maximum weight to be used for calcs is 2000# live load in case i drive my tractor over it. can i use 8" H beams? and if so what weight / foot. I have two ways to go. one is to use two beams 3 feet apart with one foot over hang on each side to give me my 5 feet width . second I could put two beams 5 feet apart and a third onein the middle. (this is to minimze the stress on the decking boards. can you help me?``` Request for Question Clarification by redhoss-ga on 28 Aug 2006 16:13 PDT ```Can you describe your tractor. What are the front and rear wheel spacings side to side.```
 ```Okay raymond111, I read your question again and it is not necessary for you to answer my request for clarification. We will do exactly what you have asked for. I will use the 2000# tractor weight as a point load at the bridge center. This will give us a "worst case" solution, but we don't have enough information (such as axle loads) to do anything else. We will not consider the beam weight or decking. Case 1 (two beams 3 feet apart): M (maximum bending moment) = Pl/4 D (deflection at center of span) = Pl^3/48EI (D is limited to l/360 = 1.2") Where: P = point load (per beam) = 2000/2 = 1000# l = 36' E (modulus of elasticity) = 30,000,000 psi (constant for steel) I (moment of inertia of beam) M = Pl/4 = (1000 x 36)/4 = 9,000 ft lb = 108,000 in lb The required section modulus of the beam (S) can then be found: S = M/19,800 psi (allowable bending stress) = 5.45 in^3 Solving the deflection formula for I we have: NOTE: 1728 is a conversion factor to get the answer in proper units I = Pl^3/48ED = [(1000 x 36^3) / (48 x 30,000,000 x 1.2)] x 1728 = 46.7 in^4 Looking in the AISC handbook for 8' wide flange beams (I assume that when you say "H beam" you actually mean wide flange): W8x15 I = 48.1 S = 11.8 Depth = 8.12" Width = 4.015' Weight = 15 lb per ft Case 2 (two beams 5 feet apart and a third one in the middle): I don't know exactly how to handle this case. If the 2 x 8 lumber transfered the 2000# load uniformly across the 3 beams, it would be an easy solution. We could just divide the 2000# load by 3 and calculate a smaller beam. However, that is not what would happen. The two outside beams would still carry most of the load. The 2 x 8 deck spanning 3' between the beams will certainly support foot traffic. Putting two beams at 5' and one in the center would cut the deck span down to 2 1/2' and would certainly make the bridge stronger. I would suggest running 3 of the 8W15 beams in this case. I hope this makes sense to you and all is understood. Please ask for a clarification if there is anything else I can help you with. Good luck with your bridge, Redhoss```