Understanding Stress in Pipe Walls: Impact of Internal Air Pressure

AI Thread Summary
The discussion focuses on the effects of internal air pressure on pipe wall stress, noting that the air pressure inside the pipe does not crush it because it acts uniformly from both sides. It is established that the through-thickness stress in the pipe is equivalent to the internal air pressure, but this pressure is relatively insignificant at 101.325 kPa, often leading to it being ignored in practical applications. The wall thickness of the pipe is noted to be greatest in a vacuum and decreases with increased hydrostatic pressures. A referenced link suggests that stress through the thickness varies with the radius, but calculations indicate that zero pressure results in zero stress. Overall, the conversation highlights the relationship between pressure and pipe wall stress while addressing some nuances in stress calculations.
Ebi
Messages
13
Reaction score
3
Consider a pipe. The pipe is not crushed by the air pressure because the same air pressure is acting from inside the pipe. But this means that the material of the pipe is being compressed on both sides by some air pressure. So is it correct to say that the through-thickness stress in the pipe would be the same as the air pressure?
 
Physics news on Phys.org
yes
 
  • Like
Likes Chestermiller
Yes, but its insignificant. Most material strengths are rated in MPa. The measly 101.325 kPa we've got as air pressure is hardly a challenge. Thats why it's just ignored...
 
The wall will be thickest when the open pipe is in a vacuum.

The wall will become progressively thinner as the open pipe is subjected to greater hydrostatic pressures.
 
Just like for every other material around us (with very few exceptions), yes.
 
Baluncore said:
The wall will be thickest when the open pipe is in a vacuum.

The wall will become progressively thinner as the open pipe is subjected to greater hydrostatic pressures.

Thanks for your response. What you said is intuitive but if you look at link below, for Pi=Po, it seems stress through the thickness varies with the radius r.
https://www.engineeringtoolbox.com/stress-thick-walled-tube-d_949.html

I am not sure if I am missing something.
 
If you go to the calculator at the bottom of the referenced page, and enter zero internal and external pressure, you get zero stress.
 
  • Like
Likes Ebi
Ebi said:
Thanks for your response. What you said is intuitive but if you look at link below, for Pi=Po, it seems stress through the thickness varies with the radius r.
The r, or term disappears for zero pressure difference.
σr = [(Pi·Ri² - Po·Ro²) / (Ro² - Ri²)] + [Ri²·Ro²· (Po - Pi) / ( r²· (Ro² - Ri²))]
 
  • Like
Likes Ebi
  • #10
... and the first term simplifies to -Pi. Or -Po, same thing. Which means radial compression is simply the common pressure on both sides, as expected.
 
  • Like
Likes Ebi

Similar threads

Back
Top