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Suppose there is a rod, on the wall or at the centre, in a pressurized vessel, the pressure inside the vessel being P. The rod has a shear strength of S. Is it true that the rod will break, in shear, if P>S, otherwise it will not? To me it seems so, but I am not able to get my head around it.

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    $\begingroup$ where is the other end? $\endgroup$ – Solar Mike May 20 '17 at 22:42
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    $\begingroup$ @SolarMike, I don't think it matters, really. A rod in the middle of a pressurized vessel will suffer uniform pressure around its entire perimeter, so it will only suffer radial compression. $\endgroup$ – Wasabi May 21 '17 at 2:55
  • $\begingroup$ What I was thinking, if the rod is across the diameter, then it is in tension but the vessel would probably be weakest... $\endgroup$ – Solar Mike May 21 '17 at 5:26
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It won't. Solid objects don't break in hydrostatic compression. The pressure is pushing all parts of the rod together, so there's nowhere for them to go except stay together. You can imagine an object on the ocean floor. Pressure could easily exceed shear strength but even soft animals remain intact, as do rocks.

The von Mises failure criterion says a material that follows it can sustain infinite hydrostatic pressure or tension without reaching the failure surface. See the diagram here. The yield surface is a tube that extends out to infinity in the hydrostatic compression direction.

https://en.wikipedia.org/wiki/Von_Mises_yield_criterion

There can be other failure modes, like phase change. Or it could collapse if it contains air pockets. But that's not determined by the pressure exceeding the shear strength.

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