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Steel Arching?

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bbartlet

Structural
Dec 20, 2007
25
Does anyone know where I could find information on the design of a single member arch?

We designed intermediate support for a fabric awning on an outdoor canopy structure with tubes at roughly 2' o.c. bent to a 10' radius. During construction flat steel straps which were bent to radius were provided instead. The structure is built, and is standing, but we want to check for snow loading that will undoubtedly occur this winter.

Any direction on where I can find information on how to check this?
 
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I forgot to mention that I've tried making a model of it in RISA 3D, but it simply goes into bending and doesn't appear to get any arching action. Which is really my only hope that this structure might pass.
 
Make sure that neither of the two supports are not on rollers, thus alowing a compression force to be generated. That could be your problem.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
 
Both ends are pinned, and it's most likely not working because it's a shallow radius, and it's slightly angled. This was sort of a last ditch effort and may be grasping at straws, but I thought I should at least test it.

Does anyone know of any formulas that I could do a hand calc with?
 
I use Castigliano's Theorem. It's much easier to make an influence line for the arch and then use that to get the forces.

I'll admit I have never done this by hand for work, but I had to a number of times for a grad class. Once you get the influence line, it's pretty straightforward, and the influence line isn't all that difficult.

I'll post a example - I think I have one with a circular arch that is less than a half-circle.
 
Even with a shallow arch, you should still be generating a compression force in the members. Are you sure that you have not released movement in the X direction?

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
 
I would double check the model, if you have it modeled with the approximate curvature and have both ends pinned, it should generate axial load. Even if you do something as simple as two straight members sloping up to meet - if you pin both ends and apply a vertical load you'll get compression. Here is an example, but it's not "quick and easy".

Just FYI, the answers are correct.
 
 http://files.engineering.com/getfile.aspx?folder=c6e740a7-2d9b-41c5-9b1c-8fdc38550d32&file=arch_example.pdf
How many members have you broken the arch up into? The greater the number of segments, the more precise the solution.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
 
I had the model segmented into 15 sections. I thought that it was just going into bending, but I think the straps are just grossly inadequate and are failing. I got the model to work and give me compression reactions at the ends. But now I'm getting an error message that "p-delta is diverging", so if I beef up the section to a wide flange I'm able to get a solution and everything works out fine.

I think there simply needs to be a more significant member than what was provided. I appreciate the help, all.
 
Unless the flats are substantial and/or span is short any unbalanced snow load is likely to kill it anyway, even if a constant UDL doesn't.
 
If the curve was a parabola, pinned at both ends, there would be no moment, just axial load under a UDL. Because this is not a parabola, there will be a mixture of moment and axial.

You could, perhaps, run it with one end pinned, using a section that is capable. Make note of the displacement of the roller and then run a case where you just push the roller back to its original position and superimpose the results.



Michael.
Timing has a lot to do with the outcome of a rain dance.
 
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