# Measure stress of gas acting on a spherical container

How can I measure stress in a thin- walled sphere containing a gas? Is there a way to predict what the stress will be without actually needing to fill a sphere with gas? I am talking about hoop stress/longitudinal stress.

The thickness of the walls of the container is unknown at this point; I know the radius and the material (EVOH); the internal pressure will be 10 atm. The question then is how will I know how much stress is too much for the material to handle? Would it be the material's yield stress value? Once that is found, how will I know how much of a safety factor is acceptable?

• stress is never measured directly. You can measure strain and calculate the stress if you know the elastic modulus. – agentp Jan 7 '17 at 2:04

## 2 Answers

Spherical vessels have a stress in the wall equal to,

$$\sigma = \frac{p r}{2 t}$$

where $p$ is the internal gage pressure, $r$ is the inner radius of the sphere, and $t$ is the thickness of the wall, where the only constraint is $r/t \geq 10$.

• I do know that this is a useful formula to use in this situation, however, the stress is unknown and thickness of the wall is also unknown. Is there a way to measure the stress in the container physically if a formula cannot be used to find the stress given the aforementioned constraints? – Hypnos Stratagem Jan 7 '17 at 1:59
• Are you doing this experimentally or are you trying to seek an analytical solution? – TRF Jan 7 '17 at 7:31
• I am hoping for an analytical solution, but currently, I am faced with an equation with two variables. I am also open to any suggestions for how to measure the stress experimentally as well. Any suggestions for a resolution to this using an analytical solution would be preferred. – Hypnos Stratagem Jan 7 '17 at 21:24
• Do you have the physical spherical vessel already made? – TRF Jan 7 '17 at 22:53
• @HypnosStratagem yes, that should provide the minimum thickness (note that the equation is only valid for $t < 10r$) as long as you don't need to hold the pressure for a long time or above room temperature, which will cause creep. – regdoug Jan 13 '17 at 23:46

Create a 3D stress tensor and calculate the known forces for a tube (easily googleable). You simply need to know the pressure inside the tube if it is in quasi-equilibrium. Given that there is no torsion it should be quite simple, but you must know the force exerted on the body. You can predict the pressure using PV=nRT assuming an ideal gas. If I'm missing something let me know and I'll modify my answer.

• Shouldn't the known forces for a sphere be calculated instead of a tube? Also, there is no torsion (meaning that the sphere is not twisted in any way). The pressure is known for the interior of the sphere as 10 ATM. Also, if I may ask, what exactly is a 3D stress tensor? (I am a chemistry guy.) – Hypnos Stratagem Jan 7 '17 at 21:29
• No engineer in his/her right mind would solve this problem in such a convoluted way. Just draw a free body diagram for half of the the sphere plus half the pressurized gas. Resultant force on the circular "cut" through the sphere = $\pi r^2 P$ (from the gas) - $2 \pi r t \sigma$ (from the stress in the sphere = 0. Therefore $\sigma = pr/2t$. – alephzero Jan 7 '17 at 22:23
• It should be noted that I do not know the thickness of the container wall or the stress on the container. Unless I am missing something about the post, this would leave me with an equation with two variables (please correct me if I am wrong about this conclusion). See the above conversation with TRF for more details. – Hypnos Stratagem Jan 7 '17 at 23:31
• @alephzero I suppose you should fill in the experts in the field of their incorrect approach. They can be contacted at Rensselaer polytechnic, the oldest technical school in the English speaking world. By the way the stresses described are encompassed in the 3D tensor. – ShuddaBeenCodin Jan 8 '17 at 7:43
• @Hypnos Stratagem a 3D stress tensor is just a matrix that expresses the stresses on an object in every direction with the diagonal stresses being principal and others being shear. Don't let the other comments confuse you, they are giving you equations for hoop stress etc, the tensor includes those stresses as it is the fundamental approach to analyzing a body in three dimensional space (real material behavior/understanding vs regurgitating equations). – ShuddaBeenCodin Jan 8 '17 at 8:41