Thickness of steel required for a vacuum chamber

AI Thread Summary
To build a vacuum chamber for chemistry, the design involves a prism structure measuring 24" wide, 18" deep, and 18" high, with a Plexiglas window. The walls will be made of galvanized steel, needing to withstand significant pressure, with the front, back, top, and bottom requiring support for 6480 pounds each, and the sides for 4860 pounds. Calculating the necessary thickness involves treating each side as a flat plate under normal load, using formulas for 2D elastic deformation. It's crucial to ensure that the maximum stress does not exceed the yield stress of the materials used, incorporating a safety factor. Understanding these calculations will facilitate moving forward with the project.
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I'm trying to build a vacuum chamber for chemistry purposes. This isn't exactly for school, but it's something I'm doing for fun and it's close enough to a school problem. Anyway, my vacuum chamber will be a prism 24" wide, 18" deep, and 18" in height. It will also have a 8"x6" Plexiglas window on top. Obviously, a large amount of pressure will have to be dealt with. The walls of the chamber will be galvanized steel. Assuming about 15 PSI at sea level (I live in Florida), the front, back, top, and bottom (24"x18" each) must hold up to 6480 pounds each, and the left and right sides (18"x18" each) must hold up to 4860 pounds each. The Plexiglas window must also hold 720 pounds on its own.

Now, I've spent several hours trying to figure this out on my own, but it's just become too much of a headache and I have no idea where to start trying to solve this. Basically, I want to know how thick the steel/Plexiglas has to be to withstand the above forces. Or perhaps you could give me some formulas and whatnot to get me started. I'd really just like to get past this so I can start on the fun stuff.

Thanks,
υ
 
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Treat each of the 6 sides as a flat plate simply supported on the edge, with no tension normal to the edge (i.e., on a roller), and loaded normally. Look up the solution for the 2D elastic deformation of a flat plate, and make sure that the plate is thick enough such that the maximum stress does not exceed the yield stress of the material. Also apply a reasonable factor of safety to be sure.
 
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