# How do I calculate the maximum pressure a container can withhold?

Say I have a spherical container with a certain gas inside, heated to a very high temperature, and thus a very high pressure. The gas will exert a force radially outwards onto the container, and at some critical point the container will no longer be able to contain the gas and deform or break. At first, to solve for the maximum pressure a given container could hold, I thought that maybe I could use Young's Modulus, which related stress on an object to the strain felt by an object. The change in 'length' of the sphere I would get by considering changing the radius of the sphere a small amount $dr$ and seeing what happened to a small element (that was my plan anyway). Then I realized that since the force applied onto the object wasn't really in the direction of stretching, but rather perpendicular to it (as the metal should stretch in the direction of the blue arrows(I think)) I thought maybe a better substitute would be Shear Modulus. But that didn't really make sense in my head.

Finally I seemed to have a breakthrough, remember that Ultimate tensile strength exists! Ultimate tensile strength is given in units of pressure, so I thought that perhaps the answer to my question was simply the number listed on the Wikipedia article. Not only did that feel supremely unsatisfactory, it also seems wrong, because Ultimate tensile strength seems to imply that the force of stretching be in the same direction of stretching. No more good ideas have come to me in a while, so I thought I should ask, How do I calculate the maximum pressure a given container can contain?

Edit: I realize ultimate tensile strength can't be the answer (or the full answer anyway) because the thickness of the container must come into the equation at some point! (I think). So far none of my methods seem to have included this crucial aspect...

• In an engineering context, one would consult the ASME Boiler Code (provided the sphere has a diameter greater than 6 inches). Now for physics... Ultimate tensile strength (or most elasticity failure criteria) depend on the resolved forces. So, does the steel really care, locally, exactly how the force is applied to the body? Or does it only worry about the forces that it sees right where it is? – Jon Custer Jul 30 '15 at 13:54
• This question may be much more difficult to answer than you suspect. While you can look up ASME boiler code data, I am sure that such data is also associated with temperature. Boilers will not operate at temperatures higher than 705 deg F (the critical temperature of water), so if your hypothetical sphere is operated at substantially higher temperatures, you will probably have a difficult time finding design data that is appropriate. Note that temperature is important in your answer because metals lose strength as they get hotter. – David White Jul 30 '15 at 16:12

For now, let's ignore the expansion of the container due to the heating and just focus on the stress in the wall of the pressure vessel. I will also examine the case of a thin-walled, spherical vessel, but the same procedure may be applied for other geometries.

## Compute Stress State

First, you must compute the pressure in the walls. To do this, imagine a cutting plane through any diameter of the vessel. Now, balance forces on one half of the vessel. In one direction, you have the force of the fluid pressure, which totals to $F = \pi R^2 P$. In the other direction, the only balancing force is the stress in the walls acting over the exposed cross-section. The force here equals $F = 2\pi R t \sigma_{wall}$, where $t$ is the thickness of the wall. By setting these forces equal, we can compute that the stress in the wall is

$$\sigma_{wall} = \frac{P R}{2t}.$$

Now, due to the symmetry of the problem, and making the (valid) simplification that of zero through-thickness stress, we can write the full stress state of any point in the wall:

$$\mathbf{\sigma} = \sigma_{wall} \left( \mathbf{e}_{\theta} \otimes \mathbf{e}_{\theta} + \mathbf{e}_{\phi} \otimes \mathbf{e}_{\phi} \right) + 0\; (\mathbf{e}_{r} \otimes \mathbf{e}_{r})$$

## Check Yield Criterion

To compute the point where yield happens, you must compute the Mises equivalent tensile stress. There are a variety of equivalent definitions, but all will lead you to the conclusion that in order to prevent plastic deformation, $\sigma_{wall} < \sigma_y$, where $\sigma_y$ is the material's uniaxial tensile yield stress. Thus, the maximum pressure the vessel can contain before yield is $$P < \frac{2\,t\,\sigma_{y}}{R}$$

## Other Considerations

You still have to consider fracture, but that's a separate discussion that is more suited to an engineering class. (If this is what you're looking for, I can go there...) With this comes the "leak before break" criterion, which puts an upper bound (counter-intuitive, but it checks out) on the safe thickness of the vessel walls.

Another point to keep in mind is that the pressure in the tank will change with temperature. Be sure to account for this in your analysis with $PV = nRT$ or some other appropriate state equation.

As others have mentioned in the comments on your question, there are very strict codes that put factors of safety on essentially every aspect of the tank design. This would be the place to start if you're actually going to build something.