Forces on an underwater vessel

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SUMMARY

The discussion focuses on the stress analysis of a 16-inch diameter, 8-inch height fiberglass hemisphere used as an underwater vessel. It is established that the pressure differential between the compressed air inside and the surrounding water pressure is critical in determining the vessel's failure point. The vessel, positioned upside down, will not likely fail unless the height creates a significant pressure difference between the top and bottom. The air pressure inside the vessel will equalize, exerting uniform pressure on the walls, preventing catastrophic failure under normal conditions.

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  • Knowledge of material stress and strain concepts
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yellowcouch
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I have a question about a Fiberglass Hemisphere that is 16" in diameter and 8 " in height. It is being used as a vessel under the water so it has a closed top. I wanted to know how to figure out how much stress from water pressure it will handle before it fails.
thanks!
 
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Which way up is it? Are you saying it has a flat top, or a (possibly open) flat base?
 
I understand it this way that you have a kind of bucket upside down. The water pressure will compress the air inside, at the same time as the surrounding pressure (outside the vessel) is the same. The vessel will be filled with water as the air is compressed to a smaller volume. The vessel will probably not "blow apart" under such experiment unless the vessel is so tall that there is considerably less water pressure at the top compared to the bottom. The air inside will have approx the same pressure acting on the walls any place inside. Considerably higher on the top compared to the water pressure outside - too much difference and it will blow apart.

Vidar
 

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