Maximum pressure a hull can stand

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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.
 

Answers and Replies

  • #2
Bystander
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Where did I go wrong?
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.
 
  • #3
russ_watters
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Well, there might be one error: you are assuming a submarine with no internal support structure.
 
  • #4
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Yeah, one pistol shrimp and it's done for --- but it's reasonable for a beginning look at hoop stress.
 
  • #5
rollingstein
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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.
 
  • #6
oxi
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Thank you!

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

Well, there might be one error: you are assuming a submarine with no internal support structure.
How is this support like? Is it pillars crossing the hull diametrically?

Yeah, one pistol shrimp and it's done for --- but it's reasonable for a beginning look at hoop stress.
Should I be using these equations instead? http://en.wikipedia.org/wiki/Cylinder_stress

(Seriously, a shrimp? o_O)

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.
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
 
  • #7
rollingstein
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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.
 
  • #8
Danger
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  • #10
Baluncore
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Should I be using these equations instead? http://en.wikipedia.org/wiki/Cylinder_stress
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.
 
  • #11
rollingstein
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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. Th
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.
 
  • #12
Baluncore
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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]
 
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