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External pressure on a Torus

  1. Aug 7, 2008 #1
    Right now I'm trying to design a floatation device for putting 1000m under the sea (~1500 psi external pressure). The nicest looking and most convenient shape would be a torus but I'm having trouble estimating stress on the thing. I'm looking for around 20lb of buoyancy or more and dimension are roughly:

    OD: 16"
    ID: 5"
    Cross section d~5"

    At the company I work for they've done testing before with plastic floats of this kind with not so good results. As soon as they deform a little it becomes a positive feedback type situation creating a greater stress concentration and it crushed at relatively low pressure.

    I guess this means it's a hoop stress type buckling situation but I'm a little fuzzy on how to approach the calculations.
     
  2. jcsd
  3. Aug 7, 2008 #2

    Q_Goest

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    Are there laws/regulations that dictate what to use here?

    If not, I'd suggest using ASME BPV code first, or possibly Roark's.
     
  4. Aug 7, 2008 #3
    There are no laws or regulations at all as far as I know. I'm looking to use 6061 aluminum to avoid galvanic corrosion.

    Most pressure vessel calculations have the higher pressure on the inside so there's no buckling threshold.

    I'm not familiar with ASME BPV codes and I don't think my company has a copy.
     
  5. Aug 7, 2008 #4

    stewartcs

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    Since the stress field is isotropic, the stress will be equal to the hydrostatic pressure of 1500 psi.

    Are you asking at what point the torus will collapse? If so, that will depend on the diameter, wall thickness, and the material used for the tube.

    Like Q pointed out, Roark has some information on this (look under buckling due to external pressure) or API 5C3 for collapse resistance calculations (I prefer API since it considers the yield strength of the tube).

    Hope this helps.

    CS
     
  6. Aug 7, 2008 #5
    Yes I suppose I am asking when the torus will collapse. I guess the real information I'm trying to find is what wall thickness the torus would need to be to prevent crushing.

    I'll see if I can look up those resources you mentioned.

    Thanks for the help.
     
  7. Aug 7, 2008 #6

    Q_Goest

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    ASME Code provides for shells and tubes under external pressure in para UG-28 and for the use of stiffening rings to reinforce the shell (so it doesn't need to be as thick/heavy) in para UG-29. How to attach stiffening rings is also covered by para UG-30.

    Para UG-32 and UG-33 cover formed heads under pressurized load (ie: external or internal pressure).

    Material considerations (ie: material allowable stress) is covered in Subpart 3 of section II, Part D.
     
  8. Aug 7, 2008 #7
    I've found some charts from URI that tell collapse depth for long cylindrical tubes made from 6061-T6.

    I think a long cylindrical tube would be more prone to collapse than a torus correct? If I go by these design standards do you think it will be a fairly conservative design?
     
  9. Aug 7, 2008 #8

    stewartcs

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    The collapse resistance of a tube may occur in different regions (e.g. elastic buckling, yield, plastic, transitional). So you'll need to ensure you are in the correct region. For example, for thin-walled tubes, elastic buckling would be the correct region if the r/t ratio is greater than 10 (r is radius, and t is wall thickness) according to Roark. The length plays a part in this as well in the sense that the end caps of the tube help support or reinforce the tube which would help resist buckling in short tubes (not in long tubes though since the caps are far apart). There is usually a statement to that effect that restricts the use of certain book formulas to certain length tubes. That equates to shorter tubes having a slight increase in collapse resistance if using elastic buckling equations.

    As far as long cylindrical tubes being more prone to collapse as opposed to a torus, I would imagine they would be about the same.

    Just use a factor of safety and then pressure test your design to ensure it will work.

    BTW what is the application?

    CS
     
  10. Aug 11, 2008 #9

    Mech_Engineer

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    Regardless of the geometry of the flotation device, if it's going to stay at depth you could just fill it with 1500psi of air and the resultant force on the float will be zero. If it has to be filled at sea level and then sunk, or it has to be able to rise to the surface from maximum depth, your problem will be significantly more difficult.

    The buoyancy of the object is basically determined by the volume of water it displaces, not necessarily the internal pressure (unless the pressurized fluid's weight cannot be approximated as zero.) So to have a buoyancy of 20 lb, you would be displacing about 9 liters of water (not including the weight of the buoy itself).
     
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