Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

P=F/A problem

  1. Mar 21, 2008 #1
    I have a tube 1 1/2 diam. wall thickness .120 We need to crush each end about 1 inch in from the end. We are in need of the requirements of a press that will do the job???
    your help is needed thank you
  2. jcsd
  3. Mar 21, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It makes a difference what the material is and what its strength is. Also, it's length must be sufficiently short to prevent buckling before it crushes, or at least it should be sideways supported against buckling, and also to prevent local buckling of the tube wall. What material you talking? Steel, fiberglass, cardboard??
  4. Mar 21, 2008 #3
    steel 3 feet long it will be supported at the other end by a stop bracket the press will have two dies
  5. Mar 21, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, I misuderstood, i think. I was assuming from your title that you wanted to place an axial force on the pipe. But do you want to instead 'squeeze' the ends to flatten the pipe, kind of close it off? If so, I don't think I can help. But please clarify.
  6. Mar 21, 2008 #5
    yes it is to be squeezed
  7. Mar 21, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, sorry. I don't know for sure. These presses are often used to make electrical splices between 2 wires by using an alumimum or steel tube installed over each end, and compressed (squeezed) with a 2 piece circular or hex shaped die. They require usually a minimum 15 ton hydraulic press, sometimes a 60 or even a 100 ton press for diameters over 2". I'm not sure of the wall thickness, though, in this application. You might want to check the tubing manufacturer's recommendations (Burndy and Anderson-Fargo and Alcoa come to mind).
    Last edited: Mar 21, 2008
  8. Mar 21, 2008 #7


    User Avatar
    Science Advisor

    For your particular pipe, collapse is governed by yield strength collapse.

    Yield strength collapse pressure is given by:

    [tex]P_{yp} = 2Y_p \cdot \frac{(\frac{D}{t} - 1)}{(\frac{D}{t})^2}[/tex]


    [tex] P_{yp} [/tex] is the external pressure required to generate minimum yield stress in the pipe
    [tex] Y_p [/tex] is the yield strength of the pipe
    D is the diameter of the pipe
    t is the wall thickness of the pipe

    Hope this helps.

  9. Mar 22, 2008 #8
    Stewartcs formula worked well. I came up with 3,827.2 Pyb

    Thank you all for your in put

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: P=F/A problem
  1. JSF F-35B Tests (Replies: 21)

  2. A/f Ratio and BSFC (Replies: 5)