1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum pressure a hull can stand

  1. Dec 10, 2014 #1

    oxi

    User Avatar

    Hello,

    First, please understand that I'm lacking fundamentals in physics and engineering.

    I'm trying to find the maximum pressure a submarine hull can resist. For the sake of simplicity I'm just considering a tube closed with magical caps that keep a different pressure inside and outside.

    After much research and a few calculations, I think I've gone wrong somewhere, because I'm getting that a 2m diameter, 2m long steel tube 1cm thick will plastically deform under the pressure at 114m depth only.

    From here (http://www.calctool.org/CALC/other/games/depth_press) I'm getting that the pressure at 114m depth is 1.24723 MPa = 1,247,230 Pascals or Newtons/square meter

    My magical hull is a tube 2.02m outer diameter and 2m long, again without caps, and keeping the pressure difference, and I guess the inner pressure would be zero. So its surface area should be: D x pi x height = 2.02 x 3.14 x 2 = 12.69m2

    That'd mean it's getting a total force exertion of 1,247,230 x 12.69 = 15,827,349 Newtons / 9.8 = 1,615,035 Kg = 1,615 metric tons. (Which seems ridiculously high)

    Then I'm calculating the yield moment using this (http://en.wikipedia.org/wiki/Section_modulus#Elastic_section_modulus):

    S =
    pi x (oD^4 - iD^4)
    -----------------------
    32 x oD

    pi x (2.02^4 - 2^4)
    -----------------------
    32 x 2.02

    3.14 x (16.64 - 16)
    ------------------------
    64.64

    S= 0.031

    If my hull is made of structural steel its got a yield strength of about 250MPa = 250,000,000 Newtons/square meter

    Yield moment = S x yield strength = 0.031 x 250,000,000 = 7,893,642 Newtons/meter / 9.8 = 805,474 Kg/meter

    Since my hull is 2m long, it should be able to withstand 1,610,947 Kg before plastic deformation occurs.

    Where did I go wrong? And please remember that I have little physics/engineering background, so explain as you would to a child.

    Thanks!

    P.S.: Sorry for the format, but I can't bother to learn Latex or whatever it's used for writing equations just for one question.
     
  2. jcsd
  3. Dec 10, 2014 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You didn't.

    I haven't checked every little bit of your calculation for arithmetic errors, or your figures for steel strength vs. mine, but you've come up with the correct order of magnitude (power of ten) and are easily within 20 to 30 percent of crush depth.
     
  4. Dec 10, 2014 #3

    russ_watters

    User Avatar

    Staff: Mentor

    Well, there might be one error: you are assuming a submarine with no internal support structure.
     
  5. Dec 10, 2014 #4

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yeah, one pistol shrimp and it's done for --- but it's reasonable for a beginning look at hoop stress.
     
  6. Dec 11, 2014 #5

    rollingstein

    User Avatar
    Gold Member

    Should you worry about yielding or buckling? I assumed that buckling would be the mode you'd worry about.

    Besides, are you imagining a failure under tension or compression.
     
  7. Dec 11, 2014 #6

    oxi

    User Avatar

    Thank you!

    So a simple tube wouldn't stand that much...

    How is this support like? Is it pillars crossing the hull diametrically?

    Should I be using these equations instead? http://en.wikipedia.org/wiki/Cylinder_stress

    (Seriously, a shrimp? o_O)

    I guess I'm looking at compression. I don't really know, but I'll have a read at this http://en.wikipedia.org/wiki/Buckling

    Thanks!

    P.S.: Excuse my English
     
  8. Dec 11, 2014 #7

    rollingstein

    User Avatar
    Gold Member

    For most hollow vessels under external pressure that I've come across, the ruling mode of failure has been bucking.

    It's take a very short vessel for compression to be the relevant failure mode. I could be wrong.
     
  9. Dec 11, 2014 #8

    Danger

    User Avatar
    Gold Member

    :oldlaugh:
     
  10. Dec 12, 2014 #9
  11. Dec 14, 2014 #10

    Baluncore

    User Avatar
    Science Advisor

    The hoop forces acting around a high pressure cylinder work out to be twice that of the axial forces due to the pressure on the ends of the pipe. That is why pipes tend to split along their length. A tube with a longitudinal flush welded seam will usually fail along the heat effected zone next to the seam weld. Some pipes are helically formed from a long narrow parallel strip of sheet material. The single folded seam provides increased hoop strength, while avoiding the weakness of an axial seam.

    Open ended pipes can be made with a corrugated wall, see; http://www.atlanticcivil.com.au/hel-cor/
    The properties of flat wall with a welded or folded helical seam. http://www.spiralmfg.com/downloads/physical_properties.pdf

    Buckling is compressive failure. The cylindrical shape of a pressure vessel, with hemispherical ends, is designed to balance the forces. As pressure increases with depth a submarine becomes smaller. It is important that there are no members such as struts that cross from one side of the pressure hull to the other. Any rigid “point” or “longitudinal” attachments to the pressure hull would change the force distribution and result in the formation of a wrinkle and early failure. Any structural reinforcement would need to be in the form of closely spaced internal circular hoops. They would be designed to keep the hull cylindrical by preventing the formation of longitudinal wrinkles. Internal bulkheads would need to attach to the pressure hull in a way that does not behave differently to any internal hoop structure fitted.
     
  12. Dec 15, 2014 #11

    rollingstein

    User Avatar
    Gold Member

    A typical way to increase buckling resistance of cylindrical hulls is to reduce unstiffened length by adding stiffening rings I think. To some limit, the closer your stiffening rings the higher the external pressure your cylindrical hull can resist without buckling.
     
  13. Dec 15, 2014 #12

    Baluncore

    User Avatar
    Science Advisor

    The same is true of surface vessel hulls. Vertical frames (and bulkheads) control the curvature of the external skin. Frames are spaced regularly along the axis of the vessel. Thinner plates require closer frames.

    Here is an example of the structural skeleton showing through the stretched plates on the stern of an old ship. They have fixed the worst examples which were at the bow.https://www.pearlharbortours.us/home_rotor/8.jpg [Broken]
     
    Last edited by a moderator: May 7, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Maximum pressure a hull can stand
Loading...