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Q: i beam calcualtion for a bridge. ( Answered,   0 Comments )
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.
Subject: Re: i beam calcualtion for a bridge.
Answered By: redhoss-ga on 29 Aug 2006 07:43 PDT
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")

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
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