I won't argue with the calculator-- but I'd love for the OP to give it a go and get back to us next year.
Jim
I won't argue with the calculator-- but I'd love for the OP to give it a go and get back to us next year.
Jim
But that would allow perhaps someone to suggest a good alternative to accomplish the task.
Its always a good idea to state what you are trying to accomplish because quite often someone will say "did you consider x"
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So, buy a 14 and field fit...(saw it to length, that is)
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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.
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.
Metal but in practical terms, neither can I. Normally flitch plates aren't used for such shallow beams. I merely was making a suggestion if the OP must do it his way.
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That's the least I'd use for a 12' unsupported span-- And I'd specify Doug fir just to make sure.
Jim
My money is on it's a beam for a science, engineering, or shop class. Darro wants us to do his homework for him.
Hi Darro,
Sounds like an interesting situation, but there's not much information to go on.
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:
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!
Anthony Watson Mountain Software
On Wed, 13 Feb 2013 09:49:40 -0600, dpb wrote in Re Re: 12-foot wood beam - How to construct?:
Good advice there.
wants us to do his homework for him.
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
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... :)
Anthony Watson Mountain Software
Yup, clearly homework. I'd bet highschool, though, or maybe science for 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.
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)
wants us to do his homework for him.
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?
wants us to do his homework for him.
the relatively simple calculations to figure out what forces are involved and how to figure out how much of what material is necessary.
do that if we give him the answer?
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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