I have a question about the pressure placed on a plate of material X, how the force is distributed and what would be the material property that would determine its failure.

To simplify things "suppose" I wanted to build a vacuum chamber with a viewing window made of material X, that could handle a external pressure of 0.990Bar, and has a dimension of 500mm x 500mm x thickness. How could I calculate a for safety reasons the minimum thickness needed for the window so that it doesn't implode, ignoring safety factors for now. Also could this calculation be then used on a box of some arbitrary dimensions to calculate the minimum thickness of the material?



Thickness, $t$, for a window made of a brittle material (like most glasses) is: $$ t=\sqrt{\frac{pr^2}{\sigma_{MOR}}} $$ Where $p$ is the pressure difference and $\sigma_{MOR}$ the Modulus of Rupture, which is roughly equal to the tensile strength, it's listed for most materials in data books or http://www.sgpinc.com/materials.htm Be careful that the pressure and modulus are in the same units .

In real life you want a decent safety factor of 2-4 and you also need to consider what optical effects the window bending will have.

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  • $\begingroup$ What if Hypothetically the window was not round? I'm assuming the r^2 is coming from the area of the window. $\endgroup$ – Lpaulson Nov 28 '13 at 18:37
  • $\begingroup$ r is the "unsupported radius", so for a square I would use 1/2 the diagonal or just (width*height)/4 Since you then add a factor of 4 'safety' the extra sqrt(2) doesn't really matter. $\endgroup$ – Martin Beckett Nov 28 '13 at 20:34

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