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panel on front of building - wind loads?

panel on front of building - wind loads?

panel on front of building - wind loads?

(OP)
See the attachment below - that's a random example I picked from google images.

It's a HDG frame with a commercial banner in, mounted on the front of a building.

Non-seismic zone.
Apart from it's own weight, what else is there? Only wind? I'll also take the weight of two guys hanging from it, as the bottom bar can be reached without a ladder.

How do I get the wind loading (I guess negative pressure will be present) ?

Any other pointers, I suppose someone has done this before?

Thanks in advance.

RE: panel on front of building - wind loads?

USe 30 psf outward load.. I take it you're looking for attachment loads..

And for your shear, use someone hanging on it as you have suggested.

RE: panel on front of building - wind loads?

The model alone has to be 150#.

Mike McCann
MMC Engineering

RE: panel on front of building - wind loads?

For your direct shear, include self-weight plus an obese american hanging from it. Or Canuck if Canadian.

rednose

RE: panel on front of building - wind loads?

I would take self weight plus an outward wind load per ASCE 7-10 (components and cladding), plus the live load as you described. Tension/pullout (wind) and shear (live load+self weight) on the connecting elements to the masonry, tension (wind) in the ropes, bending (wind, or live load+self weight) in the framing. Wind load towards it won't do anything for you calculations-wise.

No other loads (seismic, vibrations, etc.) present. Seismic won't govern unless the thing is really heavy, which it's not.

RE: panel on front of building - wind loads?

kingnero, you should also check signage tesnioning force exerted on the steel frame.

RE: panel on front of building - wind loads?

(OP)
Thanks for al your replies.
I was indeed talking about the attachment loads.

@ hetgen, I do not understand what you are saying - do you mean that the rubber ropes pull on the frame (like an inward pressure)?
If so, OK. When looking at it as a uniform divided pressure, it's low enough that it can be neglected. If that's not what you mean, can you please explain?

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