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PRESSURE in Inflatable Material Objects

  1. Aug 3, 2011 #1
    If you wanted to calculate the pressure on the shell of an airtight enclosed concentric or asymmetrical/symetrical object, do you simply need the surface area of the object and then divide it into the applied force?

    1. So if you apply 50Newtons onto a grounded football for example,it would be.... (Force)/(4.pie.radius)>>>>surface area of the football.....??? And what if the football had an inch thick shell, would I need the surface area on the inner or outer of the shell? Correct me, but I assume its all about surface area and that shape wouldn't make a difference...

    ...but does thickness make a difference...?
    2. I am trying to figure out the pressure on the walls of a exhaust powered car jack, it's basically an inflatable cylinder(although when inflated it asymmetrical as it bulges). The one I am trying to design will have plates on top and bottom, so how would the variation in surface thicknesses affect the spread of pressure?

    May help:


    Pressure = Force/Area
    surface area of cylinder = 2.pie.r(r + h)


    Thanks,
    James

    Oh and have a look on google images for exhaust powered car jack if it will help.
     
  2. jcsd
  3. Aug 3, 2011 #2
    You should look at the "hoop stress" as follows:

    hoop stress = Internal Pressure X Cylinder Radius / wall thickness

    The hoop stress is the pressure (technically stress) inside the walls of the cylinder.

    For this formula to be accurate, it must be "thin walled."

    wall thickness / diameter < 0.1.
     
  4. Aug 4, 2011 #3

    SteamKing

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    If it is inflatable, then the gas laws will be part of determining the internal pressure.
     
  5. Aug 7, 2011 #4
    Thanks edgepflow about the hoop stress. One of my teachers gave me a handout on it without me mentioning it i have started to use it even though i can't figure out why a bigger diameter = more stress, i thought the bigger the diameter the more the stress would be divided into it?? I am missing something, i know that at least.

    SteamKing I bet if I was better in thermodynamics this would probably be quite an easy question, but I have too much to learn about that so i wanted to keep it as close to
    P = F/A as possible im afraid :( .
     
  6. Aug 7, 2011 #5
    Oh and I used the analogy of a car tyre to help me. (Tyre Pressure = weight of car/ Tyre to ground contact point). still not sure how to divide that pressure i have found which is = 1.38 bar or 138.74Kpa or 20.12 PSI evenly amonst differing thickness of walls.
     
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