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Calculating internal pressure

  1. Jan 8, 2014 #1
    1. The problem statement, all variables and given/known data
    What internal pressure (in the absence of an external pressure) can be sustained

    a)by a glass tube
    b)by a glass spherical flask

    if in both cases the wall thickness is ##\Delta r## and the radius of the tube and the flask equals ##r##?

    (##\sigma_m## is the glass strength)

    2. Relevant equations

    3. The attempt at a solution
    Honestly, I don't know how to begin here. From the definition,
    But I don't see how a strain is produced in both the cases to produce the corresponding stress. :confused:

    I understand that this is an easy problem (as it is the second problem of the exercise in the book) but I really have no idea to begin with. I need a few hints to begin with.

    Any help is appreciated. Thanks!
  2. jcsd
  3. Jan 8, 2014 #2


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    You need to know when glass breaks.
    Can you find that in your book?
  4. Jan 8, 2014 #3
    What should I exactly look for? :confused:
  5. Jan 8, 2014 #4


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    The definition of σm and eventually some numerical value.
    You then need to relate the deformations, the stresses and the pressure.
  6. Jan 8, 2014 #5


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    You need to think how they will break, i.e. where the boundaries of the (two) fragments will be.
  7. Jan 8, 2014 #6
    To be clear, I am using two books, one is a problem book and the other is a textbook. The problem book uses a different notation than the textbook. I am not sure what ##\sigma_m## represents, does it represent the young's modulus which is generally denoted as Y in the textbook?

    Why does it even break? We are not pushing or pulling it apart to break it. :confused:

    Sorry if these are stupid questions.
  8. Jan 8, 2014 #7
    Really? With zero external pressure and non-zero internal?
  9. Jan 8, 2014 #8
    I still don't get it. Do you mean that there is some internal force acting between the adjacent parts of tube or sphere which causes the objects to break? :confused:
  10. Jan 8, 2014 #9
    Take a shoe lace. Make it taut. Now press with your finger perpendicularly to it. Will there be any forces between the adjacent parts of the string? What happens if you keep pressing stronger (assuming your finger is ideal)?
  11. Jan 8, 2014 #10
    Gravitational force?
    At some point of time, the lace breaks.

    Are you hinting towards the tension being produced in the lace? :confused:
  12. Jan 8, 2014 #11
    Well, there will be some of that, but we usually ignore that for such tiny objects.

    This is really a problem of statics (before it breaks). The force exerted by your finger must be balanced by something. What is that something?
  13. Jan 8, 2014 #12
    Something=the maximum tension that the lace can bear?
  14. Jan 8, 2014 #13
    Yes. The correct term for that is ultimate tensile strength.
  15. Jan 8, 2014 #14
    I am still stumped. How do I write down the equations? Do I select a small part of the glass tube subtending some small angle at the centre as shown in the attachment?

    The blue arrows represent the tension forces tangent to the tube.

    Attached Files:

  16. Jan 8, 2014 #15
    It is probably easier to start with the spherical flask, because it is more symmetric.

    So let's say you have a spherical cap, radius ##r## and diameter ##d##. It is acted upon by pressure ##d##. What is the resultant force exerted on the cap?
  17. Jan 8, 2014 #16
    I think you mean pressure ##P##. Why do you give me both the radius and diameter? Do you mean that radius of spherical cap is ##r## and height of spherical cap is ##d##?
  18. Jan 8, 2014 #17
    Sorry, a typo. Read "pressure ##p##".
  19. Jan 8, 2014 #18
    I actually meant that the diameter of the circle at the bottom of the cap was ##d##. You could, of course, use its height ##h##.

    Note, however, that the next step would be to consider tensile forces acting at the bottom of the cap (so that it is held in place), and using the diameter might simplify formulae.
  20. Jan 8, 2014 #19


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    Looking ahead a bit, I suggest it will be more convenient to express things in terms of the angle which a radius of the cap subtends at the centre.
    Pranav, consider the net force pushing out that cap (P times something) and the net force retaining it (tension per unit length times something).
  21. Jan 8, 2014 #20
    Voko has been trying to get you pointed in the right direction. You are trying to determine the internal tensile stress within the wall of the tube or within the wall of the spherical shell. Voko suggested that you split the spherical shell into two halves, and do a force balance on half the shell (as a free body). There is a pressure force pushing upward over the cross sectional area, and there is a downward tensile force in the glass all around the rim. Let σ represent the tensile stress within the glass within the rim. The rim has a diameter of 2r and a thickness of Δr, so what is its area? What is the downward tensile force on the rim? The pressure is pushing upward over the entire cross section of the hemisphere. What is the magnitude of this force if the pressure of the gas is p? Set these two forces equal to one another. This will give you the tensile stress in terms of the pressure, the wall thickness, and the sphere diameter. Once you know this tensile stress, you can compare it with the critical stress for failure σm.

    This is a statically determinate problem, and you do not need to know the Young's modulus or the strain.

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