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Forces to crack a pouch after a fall

  1. Aug 30, 2009 #1

    rah

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    Hi

    I have a question I need help with.

    Im trying to find out how much forces that a pouch (balloon) creates on the inside walls when it hits the floor after a fall.

    A pouch (balloon) with one litre of water falls from 1 meter down to the floor.

    How much force must the "walls" stand before they crack?

    I have tryed to put a known force on top of the pouch and it stands quite a lot.
    But when I drop it on the floor it cracks "a lot easier".


    R
     
  2. jcsd
  3. Aug 30, 2009 #2

    Astronuc

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    Staff: Mentor

    1 l of water = 1 kg and 1 m, mgh = 1 kg * 9.81 m/s2 * 1 m = 9.81 J.

    Does the static force on top of the balloon apply a for equivalent to 9.8 J?


    There are dynamic effects and stress concentrations that play a role. When the balloon drops there is a strain rate effect as well as a non-uniformity in the stress field that can cause the balloon material to tear. Some of the stress non-uniformity will arise from friction between the balloon and impact surface, as well as non-uniformity in the wall of the balloon.
     
  4. Aug 30, 2009 #3

    rah

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    Thank you for your quick reply.

    If I understood you right the force that the balloon hits the floor with is 9,8J or Newton.

    The force I use when I pressure test the pouch is 2000N and the pouch doesent have any sign of cracking.

    Is it possible to calculate the pressure the water creates on the walls. Is that hydraulic pressure?

    R
     
  5. Aug 30, 2009 #4

    Astronuc

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    Staff: Mentor

    When the balloon drops, the gavitational potential energy converts to kinetic energy until the balloon hits the floor (9.8 J). When the balloon hits the floor, the force is determined by the impulse relationship, but it's complicated since the balloon is not rigid. The water deforms as it collapses and the balloon is pushed out sideways (radially) - and there is probably an acoustic/shock wave reverberating through the water.
     
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