Mechanical resistence with vacuum

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The discussion focuses on calculating the number and dimensions of "plugs" needed to hold two plates (aluminum and carbon fiber) flat while maintaining a vacuum between them. Participants clarify that the vacuum does not exert force, but atmospheric pressure acts on the outer surfaces, creating a compressive force that the plugs must counteract. Suggestions include performing stress analysis by applying external pressure to the plates and calculating the required surface area for the plugs based on material properties. The importance of considering buckling strength and axial stress in the calculations is emphasized, along with the need for consistent units in the analysis. Ultimately, the key takeaway is to subtract the internal vacuum pressure from the external atmospheric pressure to determine the necessary support structure.
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Hello!

I have to calculate some pieces (brown "plugs" and green ring) which hold two plates of 2 mm each. They are made of an insulation material (different the green and the brown one) with known properties.

One plate is made of aluminium and the other of carbon fibre.

Inside the two plates there is vacuum.

The point is I want to hold the two plates keeping them flat (no dent) and calculate the number and dimension of the "plugs" (they have to be placed in a homogeneous way, the paint drawing is not very good...)

Can anyone help me?

The problem I have is the vacuum, I made some calculations with forces but never with vacuum.

Thank you very much.

Regards
 

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davidgruty said:
Hello!

I have to calculate some pieces (brown "plugs" and green ring) which hold two plates of 2 mm each. They are made of an insulation material (different the green and the brown one) with known properties.

One plate is made of aluminium and the other of carbon fibre.

Inside the two plates there is vacuum.

The point is I want to hold the two plates keeping them flat (no dent) and calculate the number and dimension of the "plugs" (they have to be placed in a homogeneous way, the paint drawing is not very good...)

Can anyone help me?

The problem I have is the vacuum, I made some calculations with forces but never with vacuum.

Thank you very much.

Regards

Well I'm quite certain I do not understand your drawing or question, however, I may be able to offer one piece of advice that may help: Vacuums do not exert a force on anything.

CS
 
So basically you are saying that you have 14.7 psia acting on the outer surfaces creating a compressive force through the plates to the insulation pieces which appear to be rods?
 
davidgruty: To perform your stress analysis (finite element analysis), apply an external pressure of p = FSu*(101 325 Pa) to the outer surfaces of your aluminum and carbon fibre circular plates (and also to the outer surface of your black cylinder, if applicable), where FSu = ultimate factor of safety. After you finish this stress analysis, also check the buckling strength of the brown rod having the highest axial load, and check the buckling strength of a longitudinal strip of your green cylinder.
 
Hello all,

Thank you for your responses.

- stewartcs:

Sorry for my drawing; it's actually very bad. The idea is we have two metal layers separated by some plugs which I have to calculate because between the two layer there is vacuum (0,03 Torr = 0.000580103 psi). And out of the layers there is atmosphere pressure which try to crush the layers (that's why we place the plugs).

- FredGarvin:

Yes. you are right.

Maybe the solution is to subtract the in-pressure from the atmosphere pressure and then since I know the properties of the material I can calculate the surface needed. The problem is the number of plugs and the separation.

- nvn:

I would like to do this "by hand" not using any FEA software.

Thank you all!
 
davidgruty: If you do not want to use FEA, divide the circular plate into tributary areas for the brown rods and the green cylinder. If the green cylinder is relatively much more flexible than the brown rod, then make the tributary area for the outer brown rods relatively bigger, or even extending to the edge of the circular plate. If the brown rod is relatively much more flexible than the green cylinder, then make the tributary area for the green cylinder relatively bigger. Because your internal pressure is 0.03 torr = 4 Pa (absolute), apply an external pressure of p = FSu*(101 321 Pa) to the outer surface of your aluminum and carbon fibre circular plates, where FSu = ultimate factor of safety. Compute the axial stress on a brown rod, sigma = p*Apr/Ar, where Apr = tributary area for brown rod, and Ar = brown rod cross-sectional area. Ensure sigma/Scur < 100 %, where Scur = brown rod material compressive ultimate strength. (But if the brown rod or green cylinder material has a yield strength, let us know.) Also ensure sigma/sigma_crr < 100 %, where sigma_crr = brown rod buckling strength.

Compute the axial stress on a longitudinal strip (of unit width) of the green cylinder, sigma = p*Apc/Ac, where Apc = tributary area for green cylinder unit width, and Ac = green cylinder cross-sectional area per unit width. Ensure sigma/Scuc < 100 %, where Scuc = green cylinder material compressive ultimate strength. Also ensure sigma/sigma_crc < 100 %, where sigma_crc = green cylinder buckling strength per unit width. If the black cylinder is applying significant pressure to the green cylinder, then the analysis of the green cylinder becomes more complicated than described above.

Use consistent units, N, m, Pa; or N, mm, MPa. If you use N, mm, MPa, apply an external pressure of p = FSu*(0.101 321 MPa) to the outer surface of your aluminum and carbon fibre circular plates.

Also compute the axial contraction (shortening), delta, of the brown rod when the external pressure you apply to the circular plate tributary area is p = 101 321 Pa, ignoring the circular plate deflection; and ensure delta does not exceed the maximum amount of circular plate displacement you want. Likewise, check the axial shortening of the green cylinder per unit width.
 
nvn,

Thank you very much for your answer.

Sorry for the delay (I had to draw and postpone the calculations)

The most important point was the pressure around the two plates. It's just subtract the external from the internal.

Thank you; you helped to the science go forward!
 
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