

How to calculate deflection of plate under pressure
I have a large steel plate for a lid on what I plan to use as a vacuum chamber for vacuum casting and degassing. The door is about 38" x 45" and made from 1/4" electroless nickel plated steel. How can I figure the deflection of the plate under a vacuum?

Not an answer but this was a similar problem posed on an engineering site. However the question was not asked correctly.
http://www.engineering.com/Ask/tabid...8/Default.aspx
Have you already converted your proposed inches of vacuum to PSI?
Walter A.

If you're going to pull vacuum on a plate the load is equally distributed and is simply the differential pressure (atmospheric pressure  internal pressure) per unit of area. The total load is the differential pressure times the whole area, you'll find the total load is a very substantial number.
The edge conditions of the plate are critical: is it simply supported or restrained? In other words if the edges of the plate are clamped such that they can neither translate or rotate then the edges are restrained. If they are merely resting on a support structure then the edges are simply supported. It can't be over stated how important the edge support conditions are for the analysis you want to do.
In the real world there is most often some combination of these conditions and the analysis can get very cumbersome and uncertain. Often in such cases the most cost effective approach is to use analysis to make sure the design is safe and deflections are found by test.
Be aware that a flat plate is absolutely the worst possible configuration to resist such a load regardless of the edge conditons. It will take the most material to do the job safely.
Vacuum systems can fall prey to an implosion failure mode of collapse caused by buckling instability which isn't predicted by simple analysis. Be aware of what you're dealing with and be careful.
I'll part with an demonstration sometimes given to vacuum system workers. A pressure vessel will be internally pressurized to an number of atmospheres with no ill effects. The same pressure vessel will then be evacuated and it catastrophically collapses long before it reaches even a low vacuum.

A domed lid would be best. If you are stuck with the flat plate, firmly attach a hefty xbrace on the outside. The brace pieces could be Ibeams or rectangular steel tubing. If you can't weld them to the plate (distortion issue), use lots of strong screws and seal with Loctite.
Larry

Watch out  under full vacuum you will get a bigger deflection than usual formulas can handle accurately. You need a copy of Roark's Formulas for Stress & Strain. For a simply supported rectangular plate under uniform pressure, it gives for the maximum deflection (at the center, of course),
ymax = (c p b^4)/(E t^3)
where, for your dimensions,
c ~ 0.060
b = 38 in
E = 30 x 10^6 psi (Young's modulus)
t = 0.25 in
p = 14.7 psi
This formula is only suitable for small deflections, i.e., on the order of the half thickness of the plate or less, and the formula gives something like 4 inches deflection under full vacuum. The formula is therefore close to useless as far as accuracy is concerned.
The minus sign is because the deflection is in the inward direction for positive pressure on the outside. It's just the chose sign convention.
A simply supported plate is one whose edges are free to rotate along the line of the edge but cannot move out of plane. If you clamp the plate with fasteners, you won't have simple support, and your deflection will be smaller than given by the formula. It will be about one third as big for a plate perfectly rigidly clamped along all four edges. But that answer is only valid for small deflections, i.e., small pressure differentials, such that the maximum deflection remains within one half of a plate thickness.

This has been a very interesting search well outside my level of knowledge. Like the others I would be able to offer opinions of how to stiffen the lid but that is not what you asked. I did not find an answer but did follow several paths that while interesting did not lead to the formula. Although I did find a close one last night but could not retrace my path there.
Some of the ways I approached this was to first look at the relationship to vacuum measurements converted to PSI. Since drawing a vacuum would create pressure applied to the outside.
I felt a good starting place once the amount of pressure was known is to calculate the deflection of a 1/4" plate 45" long supported only at the ends. Along those lines of reasoning I thought that the deflection of the same 1/4" plate but 38" long should somehow be inserted into the formula.
Finally I did find some very little information on the effect of a bolted connection would change the formula.
Searching for this data while interesting took me away from a couple of projects I now need to get back to. My last path was looking for deflection rates of steel trench covers as I felt this was a very similar situation.
Sorry I could not help with an answer. I too have an inquiring mind but just not enough skill to follow through.
The good thing about this forum is someone can always learn more about almost any subject.
Walter A.

There is a simple free software program that will help. It's called beamboy.
http://www.softlookup.com/display.asp?id=27851
Although the software analyzes simply supported beams, it should be a good approximation at the center (max deflection).
You can do point loads, distributed loads, cantilevered beams etc.

My initial thinkthrough suggests that the plate is far too thin to stabley resist 15 psi over tha enourmous area. !/4 inch plate? 1 1/2" would be more suitable. The total pressure load may seem large but that's only numbers however the effect of load on a flat plate may scallop the plate enough and stress the fasteners so that no means of fastening it will achieve a gas tight seal short of welding.
I srongly suggest you weld a grid of 4" T beams (split along the web axis from 8" Jr beams) on 6 centers. Full penetration weld the flanges at the intersections. Is should finish looking like a ship's watertight door.
A domed hatch fastened to a circular flange is by far the preferred pressure boundry for an access.

Can't remember, but the book Building Scientific Apparatus may have something on this. FWIW, look at commercial vacuum components. Even in 812" sizes, flanges and covers are about 1" thick. Whatever you calculate, use a generous safety factor, and make a comparison with commercial vacuum equipment of the same diameter.

Have played with vacuum a bit and imploded a few tanks my gut feeling is that 1/4" is way too thin over that area; but I have a question. What is the construction of the rest of your chamber? Are the walls of the chamber flat or arched/domed, are they the same thickness as your intended door or thicker? Are you sure your chamber can handle vacuum, never mind the plate you use for a door.

Thanks for the info. I have worked with a few vacuum systems but nothing resembling a box like this is. The lid will be secured with a locking pin toggle clamps.
I do plan on reinforcing the top as well as the other sides. I justy wanted to get some sort of worst cast scenario. The lid is such that and excessive deformation would cause the seal to open and vent to atmosphere.
I dont plan on going to a full vacuum. There is no need for this kind of application. Again, worst cast here.

i went had a look at a manhole on a vacuum degasseer [in a steelplant hooked up to 4 steam ejectors and quite happy to pull 35 tons of molten steel up its nozzle a couple of meters] the door was 24" round and 1/2" thick suffering a fast cycle of atmospheric up to very hard vacuum in a minuite or less, couldent see any deflection. it seemed that all the platework/pipework was 1/2 inch, there was a bit of 3/8 but only a small access plate bout 6"dia, hope this helps [vacmetal vacuum degasser]http://www.vacmetal.com/products/
mark
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