# footbridge question

I would like to construct a very simple footbridge over our creek. The distance is around 12-14 feet from bank to bank. I figured on using two 16' pieces of lumber as the support (about 18-20 inches apart), with 1x6 decking boards along the top. About 2.5' to 3' wide.
My question is what should I use for the support beams. I don't want it to get too high off the ground, because we would like to easily ski over it in the winter.
I was conteplating either using 4x6 lumber or 2x8 lumber. I'm not sure which will provide a stiffer bridge - which is what I want. How much weight can I expect such a bridge to hold? Should I be using a third support beam?
Thanks.
-Jonathan
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a 4x4 will hold a truck, a 4x4 will hold a house if it aint got no wind, Blow a bit, and call back.
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A 4x4 spanning 12-14 feet won't hold a truck.
To the OP, the stiffness of a rectangular beam of depth d and width varies as bd^3 (that's b*d*d*d). For a 4x6, that's 3.5 * 5.5^3 = 582, and for a 2x8 that's 1.5 * 7.25^3 = 571. So they are almost equivalent in stiffness.
Cheers, Wayne
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Thanks for the info Wayne,
A friend has a bridge of similar length (but just over a marshy depression) that is built using 4x4x16s, and that thing would b a bit too bouncy for comfort when the drop is closer to 2.5'.
I'm not sure what the units are (distance^4 ?) or how the length of the span figures in, but from the equation, it looks like using 4x6s should provide almost four times the stiffness of 4x4s (3.5 * 3.5^2 = 150).
Anyone know how to calculate the weight that can be supported if the weight is centered over a 14' span?
Thanks.
-Jonathan
Wayne Whitney wrote:

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Yes, the units are distance^4. The relevant formula for the moment of inertia is actually I = 1/12 * b * d^3, but for comparison the factor of 1/12 is unimportant.

For a point load of fixed weight, deflection at midspan will vary as L^3. For a distributed load, where the total load increases as the length increases, deflection will vary as L^4.

Yes, the exact ratio is (5.5/3.5)^3 = 3.88.

I'm afraid answering that question would take a bit longer than space permits. If you know what you are doing, the span calculators at www.awc.org are very useful, although they are for distributed loads, not point loads. You should understand that allowable load may be limited by deflection (where the deflection criteria is specified as e.g. L/360, that midspan deflection should be less that 1/360 times the span) or may be limited by strength.
Cheers, Wayne
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A 4x4 with S = bd^2/6 = 7.15 in^4 and M = fs = 7146 in-lb will support a W = 8M/L = 340 pound distributed load (including the decking boards) over an L = 14x12" span.
Nick
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At 14' wide, a 16' beam only gives you a 1' overhang on each shore.
Can you safely dig down to the frost line in your neck of the woods for your posts at a distance of only 1' from the water?
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You may want to check with local authorities. I would not be surprised if there are codes that need to be addressed. Do that before you start buying materials. If the local authorities don't have any requirements, and maybe if they do, also check with your insurance company.

--
Joseph Meehan

Dia \'s Muire duit
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I had a neighbor once whose wife wanted about a .25 acre "pond", with an island, with @ a 25' bridge.
He dug it out, and built the bridge (steel), and I helped set it in place.
"Looks like a million bucks", I said.
"Thanks, I only got about half that in it". -----
- gpsman
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If you are thinking of spanning the gap with lumber laid straight across, I'd forget that right away.
You need to construct a triangle shaped support then span the gap with lumber hung from the triangle:
* * * * * ==*===========*=* *
4x4s (*) and 2x6s (=) would probably do the job. You need to plant the 4x4s in cement.
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Thanks for all the suggestions.
I think I'm going to try the straight across route. It's hard to measure the actual distance from bank to bank, because the banks are not straight lines and they slope down to the creek (not a sheer drop). It could even be as short as 10' in some places - but I'm trying to be conservative in my estimates so I don't end up with a bridge that's too short. The actual width of water in the creek is typically about two to four feet, depending on how much water is flowing. If it's been dry, there will be just a trickle or no water in it at all for much of the summer.
I won't be pouring any concrete or digging down to the frostline. I'm planning to just rest the bridge on top of a couple of 4x4s at each end - maybe dug half-way into the ground.
Thanks.
-Jonathan
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re: I won't be pouring any concrete or digging down to the frostline. I'm planning to just rest the bridge on top of a couple of 4x4s at each end - maybe dug half-way into the ground.
How often are you planning to rebuild/reset the bridge?
What's going to happen when the ground gets soft from rain or snow melt? I'd think that things would start to sink and/or shift due to the weight of both the structure itself and the live-load of the skiiers/walkers.
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I guess I'll post back in a few years and let you know. The ground is pretty rocky.
When we moved into our current house (7 years ago), I built a wood shed. It rests directly on the ground on two adjacent sides, and on a couple of old railroad ties that were laying around on the other two sides. The railroad ties are pretty much resting on ground at one corner and almost fully buried in the ground at the adjacent corners, which created a level surface for the wood shed on the slightly sloping ground. This structure has not budged since.
If the bridge shifts a little, I'll adjust it. In fact, if get a few strong helpers, I could even move it to a different location without too much trouble.
-JJ
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