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I was watching a YouTube video to understand the concepts of hoop/circumferential stress and axial/longitudinal stress in thin-walled cylinders and their formula derivation.

So the formula of the Hoop Stress: $$\sigma_{hoop}=\frac{(p * r )}{t}$$

where $p$ is the internal pressure of the cylinder, $r$ is the internal radius of the cylinder, and $t$ is the thickness of the cylinder,

The formula of Axial Stress: $$\sigma_{axial}=\frac{(p*r)}{2t}$$

My question is: They have mentioned in the video that since the axial stress is less than the hoop stress, then the thin walled cylinder is more likely to experience failure along its axis than along its circumference.

I still don't agree that just because the stress along its axis is smaller than that along its hoop, then it is more likely to undergo failure along its axis.

The stress is meant to resist deformation, and to resist the applied force. The axial stress is smaller just because the internal pressure applied along the axis is smaller than the pressure applied along the hoop.

So yes the axial stress is smaller than the hoop stress, but on the other hand the applied internal pressure is less along the axis than it is along the hoop of the cylinder.

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Actually, the pressure inside the vessel is uniform and constant everywhere. What happens is that geometry results in reduced stresses in one direction and not in the other.

There may be some confusion here with respect to stress and pressure being the same thing. Although they share a few similarities ($\frac{F}{A}$ and units $[Pa] or [psi]$, they have a few differences, like stress being a vector, while pressure being a scalar quantity.

It is nice to go through the derivation of the axial and the hoop stress to understand why this happens. The video appears to go quickly through the steps of derivation.

What you should understand is that independent of the pressure the stresses in the circuferential direction are always greater than the axial stresses.

Practical application (tongue in cheek): Try putting a straight hot dog wiener in a saucepan or on a grill.

enter image description here

It is in its own right a thin waled pressure vessel. As you heat it up, pressure rises and so do internal stresses (hoop and circumferential). At some point, one of the two increases over the maximum allowable stress of the material. What you get, (almost invariably) is the following.

enter image description here

Bonus to avoid the meat getting out of the hot dog what you can do is use a knife to make slits, like so.

enter image description here

What this do is: a) They relieve internal pressure b) if crack start to develop, the slits arrest the development of the crack early on.

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  • $\begingroup$ Thank you for your answer and information, especially for the nice examples of everyday life. $\endgroup$ – Unknown Sep 27 at 12:51
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In the old days we used the formula- Stress = PD/2t , and the hoop stress was twice the axial /longitudinal stress . Pressure vessel heads were usually half the thickness of the walls because of the lower stress relative to the walls.

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  • $\begingroup$ Thank you. So, since the stress at the hoop of the cylinder is twice the stress at the hemispherical ends, then the cylinder is made in such a way that the thickness at the ends is half the thickness at the hoop? And according to the formula, when we increase the thickness, the stress decreases, so to compensate the decrease of stress at the ends, we decrease its thickness. Am I thinking right? $\endgroup$ – Unknown Sep 27 at 18:24
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    $\begingroup$ Yes , but more a matter of cost than adjusting stress. The vessel heads may be thinner . But there are other factors , such as shell penetrations may all be in the heads to keep the shell "clean". Then the heads may be thicker to meet ASME requirements for reinforcements at shell penetrations . $\endgroup$ – blacksmith37 Sep 27 at 20:26
  • $\begingroup$ Thank you. What do you mean by shell penetrations? Like making small holes in the heads, that could decrease the internal pressure, so when the thickness is small then its easier to make penetrations? $\endgroup$ – Unknown Oct 2 at 10:54
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    $\begingroup$ By penetrations I mean a few 8 " and 4 " holes in an 8 ft, diameter head. The attached pipe would have at least a one inch wall thickness. For process flow, hydrogen injection and instrumentation. $\endgroup$ – blacksmith37 Oct 2 at 16:24
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    $\begingroup$ By not having shell penetrations ,the stress calculations are much simpler. $\endgroup$ – blacksmith37 Oct 2 at 16:26

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