# Stress in a pipe wall

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?

Related Classical Physics News on Phys.org
BvU
Homework Helper
2019 Award
yes

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...

Baluncore
2019 Award
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.

mfb
Mentor
Just like for every other material around us (with very few exceptions), yes.

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.

Baluncore
2019 Award
If you go to the calculator at the bottom of the referenced page, and enter zero internal and external pressure, you get zero stress.

Ebi
Baluncore
2019 Award
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))]

Ebi
mfb
Mentor
... 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.

Ebi