# 12-foot wood beam - How to construct?

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• posted on February 13, 2013, 3:53 pm
wrote:

Using Eastern Hemlock, I confirm about 1/2 " total compared to my earlier manual calculation. Remember to use floating ends (conservative) and don't forget to include dead load with live load.
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• posted on February 13, 2013, 4:21 pm
On Tuesday, February 12, 2013 6:29:22 PM UTC-5, elbrecht wrote:

My money is on it's a beam for a science, engineering, or shop class. Darro wants us to do his homework for him.
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• posted on February 13, 2013, 10:13 pm
On Wed, 13 Feb 2013 08:21:54 -0800 (PST), snipped-for-privacy@gmail.com wrote:

That would explain why it has to be easily dismantled-- and he hasn't seemed concerned if it lasts more than an hour.
I wouldn't fault him [her?] for asking here. We *are* a resource-- sometimes a real good one.
Jim
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• posted on February 14, 2013, 9:39 pm
On Wednesday, February 13, 2013 5:13:07 PM UTC-5, elbrecht wrote:

Yeah, except if he'd been paying attention in class they taught him how to do the relatively simple calculations to figure out what forces are involved and how to figure out how much of what material is necessary.
He's going to have to prove his design on paper to get a grade. How's he gonna do that if we give him the answer?
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• posted on February 16, 2013, 2:11 pm
On Thu, 14 Feb 2013 13:39:40 -0800 (PST), snipped-for-privacy@gmail.com wrote:

the relatively simple calculations to figure out what forces are involved and how to figure out how much of what material is necessary.

Yep. I once saw a guy in college cheat and I was amazed he cheated as long as he did according to some others. I remember saying to myself that you are only cheating yourself because once you get into the office, they'll know within a week or less, if you know your stuff. I have no idea if he graduated or what became of him thereafter.
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• posted on February 14, 2013, 5:48 pm

If that were the case, I would think you could make a couple of torsion boxes (thin plywood skins over internal webbing, like interior door slabs). Then bolt them together with long brackets on each side.
Light weight, easy to dismantle, and should be able to support 30 pounds easy enough (if it is evenly distributed as originally described).
Never tried it though. Just a guess for class... :)
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• posted on February 14, 2013, 6:46 pm
On Thursday, February 14, 2013 12:48:13 PM UTC-5, HerHusband wrote:

nonmajors.
The problem is trivial as stated, but in the real world there are always constraints - must weigh less than X, must cost less than Y, must resist dynamic loads, etc.
The safest solution is the double 2x6, on edge. If I were the teacher I'd fail that one, so it's safe only in the shop or backyard, not the classroom.
You can make your beam with much less material if you make what is commonly called an I-beam (or W-beam). But if you need to resist torsion, make it a box instead.
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• posted on February 14, 2013, 6:51 pm
He didn't say how much sag was acceptable, did he?
Some sag is inevitable, even if the beam is a solid steel bar.
But it has to handle hardly any load. 30 pounds spread over 12 feet is nothing. and it has to be light, and assemblable.
So, why not an inflatable tube instead? Really impress your teacher by calculating how much pressure you need to inflate it to resist the sag. (build it slightly convex up, obviously)
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• posted on February 13, 2013, 12:36 am

Change your plans to steel if you need that small a cross section. Or increase the size if you are going to use wood.
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• posted on February 13, 2013, 1:00 am

Since I hadn't considered the sagging that will inevitably occur as the wood fibres stretch over time, I'll go back to the drawing board and take a close look at steel alternatives.
My thanks to everyone who took the time to help me out!
Dennis
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• posted on February 13, 2013, 1:17 pm
Darro wrote:

Your problem wouldn't be stretching wood fibers, it would be warping. Keep it painted and wood should be no problem. Personally, I wouldn't hesitate using wood to do what you want to do. BTW, if you turn the thing 90 degrees it will be even stronger.
Also BTW, you still haven't said what the thing is for...inquiring minds want to know :)
--

____________________________
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• posted on February 13, 2013, 3:49 pm
On 2/12/2013 7:00 PM, Darro wrote:

Since softwoods are generally about 35 lb/cu-ft, your 3x3 (roughly) will weigh about 25 lb or so while you're talking of only 30 lb load uniform total load which is only 2.5 lb/ft.
An unsupported 12-ft 2x4 will sag some w/ time on edge and definitely will laid flat--as somebody else said, you want the long edge down.
Your best plan would be to buy two 2x6 14-s and cut to fit for simplicity.
If you really want the smaller cross section use 2 tubafors and stiffen 'em w/ a 1/2" plywood gusset between them--nail and construction adhesive or use one of the waterproof glues.
Alternatively, use 1x material and make it a box beam--cut the beam weight by roughly a third.
But, w/o knowing what it is you're actually trying to do, my initial response is still to just go get a couple 2x6's, nail 'em together and cut 'em to length and be done w/ it.
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• posted on February 13, 2013, 3:56 pm
-snip-

That's the least I'd use for a 12' unsupported span-- And I'd specify Doug fir just to make sure.
Jim
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• posted on February 13, 2013, 5:36 pm

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• posted on February 13, 2013, 6:51 am
wrote:

I did a quick calculation between yawns <g> and included the weight of the beams in addition to your 30 pounds. I calculated almost 1/2 " max deflection at mid span for a uniformly loaded beam length of 12' 3" .
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• posted on February 13, 2013, 4:32 pm
Hi Darro,

Sounds like an interesting situation, but there's not much information to go on.
1. What is the load being supported?
2. Why is the height limited to 2-1/2 inches?
3. Why can't the lengths be longer than 8 feet?
4. Why can't you use a single 12' board instead?
5. Does it need to be a square beam, or can a round beam work?
6. Why does it need to be disassembled?
A 2-1/2" tall wood beam of any width isn't going to carry much weight, and will probably sag under it's own weight over a 12' span.
Without knowing the intended purpose it is difficult to offer an appropriate solution. Based solely on the limited information you provided, there are a few solutions I can think of:
1. A single 12' long, 3" round, steel pipe. You could probably gain the extra 3 inch length with appropriate brackets at each end. And you could take the pipe down when needed. It's going to be heavy though. You might look for thin wall steel conduit, but I don't recall seeing lengths over 10' unless you want to cut down a 20 footer.
2. A steel cable. If you have solid anchors at each end you could stretch a cable between them, probably with turnbuckles to get it nice and tight. You may get some sag under load, but it's easy to install and take down, and can easily support 30 pounds over that distance. Of course, this is really only an option for loads that "hang". Clothes, hanging flower pots, etc.
3. A cable supported beam. Similar to a suspension bridge, you would have a cable anchored at each end with the "beam" essentially hanging from the cable. Of course, the anchor points on each end would have to be higher than the beam and you would need midpoint supports between the beam and suspension cable.
4. The obvious. A temporary support post at mid-span. Grab a couple of 6' boards, anchor at each end and the other ends meet over the temporary center post. You would probably have to build some kind of bracket or have a hole you could drop the post into so the whole assembly doesn't fall over sideways, but it's doable.
Clearly, we can come up with solutions all day to theoretical problems, but they won't be of much help if we don't know the intended goal.
Good luck!