Physics/Pressure Vessel Wall Thickness
QUESTION: This may not be the sort of thing you answer. If not, I apologize for the inconvenience. I would like to build a pressure vessel, cubicle in design, comprised of six equal sides with inside dimensions of 12" X 12" X 12", one side, the top, being bolted on by means of a flange welded to the top perimeter of the box and the lid being over-sized with corresponding bolt holes to fit the flange. The vessel should be sufficiently strong to hold 150 p.s.i. of compressed air. I understand cylinders and spheres are preferred, but I must make mine with flat walls for two reasons: first, the contents to be pressurized fit much better in a cube, and secondly, the only materials I have at my disposal are flat sheets of .250" thick 304L stainless steel. I know the pressure exerted on the interior walls and lid would be 144 sq. in. X 150 p.s.i., or, 21,600 lbs. My question is how can I calculate how much, if any, the centers of my walls and lid will bulge and how big the weld beads needs to be to hold that kind of pressure. As for the bulging, I can weld stiffeners (.250" thick X ?" wide, on edge) to the edges of the lid and across the centers of the lid and exterior walls to limit it if needed. I just want to make sure my materials are up to the task. As the lid will be bolted on, I know how to figure the strength of each bolt needed given "x" number of bolts. I'm not sure how many bolts I need to begin with though. I know this would depend on how stiff the lid's edges are. I also don't know how to figure out how thick the flange should be. I would like to make the flange 1" wide with the bolt hole's centered in its width and probably 12 bolts equally spaced around the perimeter (1 at each corner with 2 others centered in between on each side). Again, this may not be the kind of question you answer, if not, my apologize. Any help or insight you can offer would be greatly appreciated.
ANSWER: This is a little more engineering, but I have some experience here and I'll try to help. The basic bulging can be calculated, assuming you're not in the failure zone, with the shear modulus of 6" (midpoint) of stainless steel and the thickness of 12"x0.5". Divide by at least 2, since it's gotta flex all sides in order to bulge. It's approximate, since bulging isn't quite the simple physics application it would appear. However, it should be ballpark and you want to be conservative in applications where things could potentially explode.
I'm better with vacuum than pressure, but it sounds like a lot (roughly 10 atmospheres). It's a fairly large chamber you're talking about, the energy density is pretty high, so be careful. I'd say that based on my off-the-cuff vacuum thinking you need a bolt for every four bolt widths for structural integrity (with washers and such, it'll be pretty packed). Much more and you'll have too many bolt holes. Much less and the there won't be enough bolt thread to hold it, since that's what ultimately holds the whole thing together. The seal will be critical, I hope you have good plans for that part.
Flanges you can get pretty good ideas on from some company like Lesker: http://www.lesker.com/newweb/menu_flanges.cfm
They make all the best vacuum stuff. Keep in mind that you're talking about 10x the pressure inside over vacuum, but they're talking about 1/trillionth or so of an atmosphere...so you don't need to be quite that high-end but you need to account for the bolts holding that high pressure system in place. Good luck, and do post a follow-up to let us all know how it turned out.
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QUESTION: I need clarification. I don't understand the first paragraph of your answer: "thickness 12" X 0.5"". The thickness of the stainless steel is .250". I'm not an engineer and never took physics classes so I am ignorant in this regard to say the least. I'm not sure what I am supposed to "divide by 2" either. Is there a simple formula for figuring the deflection of a square plate secured on all four sides given the thickness of the stainless steel (.250"), the size (12" X 12"), and its tensile strength (42,000 psi) or its Young's modulus of elasticity (28,000 ksi)?
OK, looking back over it, the 0.5" was a typo missing the 2. I did use 0.25" later. The "divide by two" an estimate, being conservative (if I was considering just the flexing of 4 independent bars I'd say "divide by 4" because you have 4 bars resisting the strain) because you're not flexing all along every bar. It's a guesstimate. The plates will flex less than if there was just one 6"x12"x0.25" plate going to the center with all that force on it, but calculating exactly how much is a good lengthy problem for an engineer. Either go pay one or estimate as above. You need some finite element analysis or more detailed calculation than belongs in this forum if you want an exact number. If you don't like it, don't use it at all and be reeeally conservative with you numbers, it'll give you more flex.
Young's modulus is related, but the shear modulus is really what you want to use. You're not stretching a bar, you're flexing a plate. It's different. An answer with Young's modulus will be way too low at the sizes you're talking about.