For pressure vessels in the shape of circular cylinders, we can use $\sigma_{hoop}=\frac{pr}{t}$ to find the minimum skin thickness by setting the hoop stress the maximum allowed value, and then solving for t. But, what about (the more complicated) elliptical cylinders?

Since ellipses have a semi-major axis and a semi-minor axis (as opposed to circles with just a single radius), it stands to reason that an elliptically cylindrical pressure vessel would have two minimum thickness (instead of just one). Is this correct? If so, how does one go about finding them?

  • $\begingroup$ Where do you find elliptical shaped pressure vessels? $\endgroup$
    – Solar Mike
    Aug 15, 2018 at 21:29
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    $\begingroup$ @SolarMike One application is tanks on vehicles where the lower center of mass compared with a cylindrical tank improves the dynamics of the vehicle as a whole. They are certainly a subject for academic research - e.g. pressurevesseltech.asmedigitalcollection.asme.org/… and slideplayer.com/slide/7986046. But note that "non-circular" pressure vessels are not necessarily "elliptical." The ends of circular pressure vessels are often non-spherical, and similar design considerations apply. $\endgroup$
    – alephzero
    Aug 15, 2018 at 22:05
  • $\begingroup$ @alephzero a non-circular shape was part of a pressure bulkhead that failed in an aircraft - consequences can be disastrous... but I was expecting the OP to fill out more of what they were looking for ie the field they are working in etc.... $\endgroup$
    – Solar Mike
    Aug 15, 2018 at 22:13
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    $\begingroup$ @SolarMike I'm not sure what point you are trying to make. I can't think of any aircraft that has a perfectly circular fuselage cross section! And the laws of physics (and continuum mechanics) are independent of what field the OP is working in, so why does the field matter? $\endgroup$
    – alephzero
    Aug 15, 2018 at 23:53
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    $\begingroup$ Nice old Air Force Space Systems document here. $\endgroup$
    – Phil Sweet
    Aug 16, 2018 at 1:15

3 Answers 3



While not a direct answer to your question, it should be quickly noted that ASME BPVC Section VIII, Mandatory Appendix 13, Section 10 has a section to design obround vessels (circular half sections with flat plate walls, see image below).

enter image description here

While not the same, these can typically optimize the available area compared to elliptical vessels (for a given dimension A and B). As such, these are almost always the preferred design choice compared to elliptical, and more commercially available. The equations, while handling a membrane stress discontinuity, are easier.

Thin wall assumption basis

We can easily derive a thin wall version of this. It can be noted that the ellipse shown above can be expressed as all points $$(\frac{b}{2}\cos(\phi),\frac{a}{2}\sin(\phi))$$ with a parameter $\phi$ that varies from $0$ to $2\pi$. Note this isn't the same as the angle from the center, $\theta$, but is related to this angle by the relationship $$\tan(\theta) = \frac{b}{a} \tan(\phi)$$ For a given $\phi$, the stress in a thin-wall vessel can be easily seen. Similar to a circular vessel, we take a cross section to a point at a specific $\phi$ through to the center. Using a similar thin wall approximation as cylindrical vessels, we presume the tangential (membrane) stress is uniform across this cut of a thickness $t$. We presume this is uniform regardless of $\phi$ as $t<<a ; t<<b$. The distance $D(\phi)$ is a function based upon $\phi$ from the edge, through the center to the other edge would be easily $$D(\phi) = \sqrt{b^2\cos^2(\phi) + a^2\sin^2(\phi)}$$. Since the wall is pushed uniformly across this cut with a pressure $P$, then it is clear that, similar to circles, for a load per unit length basis:

$$PD(\phi) = 2t\sigma(\phi)$$.
$$\sigma_t(\phi) = \frac{PD(\phi)}{2t}$$.

Presuming constant perimeter

Note that stress concentrations will definitely emerge, as the vessel will want to take on a cylindrical shape. Running a cut through both the x and y axis and running a statics analysis of this section, we note that a much smaller tangential stress comes from one axis than the other. The non-uniform strain will certainly cause bending moments in the thin wall material.

Without running a thick wall analysis, these stress concentrations cannot be determined exactly. For this thin wall presumption, we simply assume the perimeter will stay roughly the same before and after deflection. The perimeter of an ellipse is determined by using elliptical integrals. Defining: $$k^2 = 1-\frac{a^2}{b^2}$$

Then the perimeter for the ellipse is $2bE(k)$, where $E$ is the elliptical integral of the second kind. We note that $k$ is a function of $b$. Taking the derivative of this with respect to $b$, and a lot of math, we find the beautiful result that:

$$\frac{E(k)}{F(k)} = \frac{a^2}{b^2}$$

We approximate $E(k)$ and $F(k)$, the elliptical integral of the first kind, as a constant for small changes in $k$ due to changes in $a$ and $b$. As such, for a small contraction of $b$, $\Delta b$, we find an expansion of $a$, $\Delta a$ as follows:

$$\Delta a = - \frac{b^3}{a^3} \Delta b$$

The cubic relation implies that highly eccentric ellipses will quickly snap from the elliptical shape into a cylindrical shape! As such, the above approximations will quickly become invalid. The negative implying the expansion is in converse to the contraction. With this information, we can attempt to resolve the stress concentrations to a reasonable approximation. The "radial" stress from a statics view indicates it should have a linear radial growth, while the perimeter analysis shows a cubic radial growth. This deflection would present itself as a force, adding a bending concentration at the tips of edge b (worst case) of approximately:

$$M = \frac{Pb^2(b^2-a^2)}{4a^2}$$ $$\sigma_b = \frac{3Pb^2(b^2-a^2)}{2t^2a^2}$$

This stress would have $b$ swapped with $a$ for the other axis, to obtain the two maximum stresses in the vessel. Note the original tangential stress will need to be added to the bending stress. As such, the total stress in the long-axis side would be:

$$\sigma_b = \frac{3Pb^2(b^2-a^2)}{2t^2a^2} + \frac{Pb}{2t}$$

Adding together with the axial stress of your case-by-case analysis (complicated, but could be assumed to be reasonable by dividing the area of an elliptical cap by the strange perimeter of an ellipse), and you have the stresses for a Von-Mises stress. This, as a function of the thickness, would be set to maximum value, and the minimum thickness would be the result that allows that Von-Mises stress to reach the maximum value. This portion is commonly utilized in pressure vessel design, as environmental loading adds more stress than your equation presumes.

I would run some tests to verify, but this is my conclusion as a PE for 8 years. Others can disagree as they see fit, and I'd be happy with a better analysis that shows better work.

Notes on thick wall vessel

I did attempt a thick wall version, but it quickly became apparent that I needed to invest more time than I was able to dedicate to it. Note you did not ask for this - and since your question and equation related to the thin-wall case, I saw no reason to push forward. Some notes, if you are familiar with elliptical integrals:

  1. The practicalities are still an issue - obround works well even in thick-wall cases.

  2. Sources are elusive for other works - Roark's was empty, as was ASME code. As such, I again defer to obround as the better method of construction for all practical applications.


You can use a finite element code to simulate the stress distribution in the wall. Simulating elastic problems under pressure is straight forward to model and you get the advantage of getting the stress distributions in 2 or 3 dimensions in contrast with the other methods mentioned.


I would say that calculating for the thickness with the least stress would be the best thickness for the whole container. A variable thickness container is going to cost more to construct and be difficult to consistently manufacture to specifications.

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    $\begingroup$ Although that is a good practical consideration, it doesn't help the OP calculate what that thickness is. What's the equation for the minimum thickness at the semi-minor vertex, and what's the equation at the semi-major vertex? Which gives a smaller thickness? $\endgroup$
    – AndyT
    Aug 21, 2018 at 8:42
  • $\begingroup$ Pressure in a vessel is uniform. The one equation should be sufficient to calculate the thickness because material strength is only determined according to greater structural size. We're am egg internal pressure higher than surrounding atmospheric pressure, it would be equal on all surface points internally. Shell thickness is uniform in eggs for that reason but shape is to protect against external deformation from external forces. $\endgroup$
    – Rhodie
    Aug 21, 2018 at 9:53
  • $\begingroup$ Sounds like you've changed your answer. There's the word "edit" in grey beneath your answer, you can click it to edit your answer. You're still missing an equation though. $\endgroup$
    – AndyT
    Aug 21, 2018 at 9:56
  • $\begingroup$ "A variable thickness container is going to cost more to construct ..." I fail to see how this helps in any way. As I said in the comments to my question (in the comment right before yours, as a matter of fact). I never asked how to build a variable thickness pressure vessel. I asked how we calculate the required thicknesses. You can't build a pressure vessel with a single thickness if you don't know what the minimum thickness is for the part(s) of the vessel under the greatest stress. $\endgroup$ Aug 25, 2018 at 5:03
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    $\begingroup$ "Pressure in a vessel is uniform. The one equation should be sufficient to calculate the thickness because material strength is only determined according to greater structural size." This is factually incorrect. The internal pressure is uniform but the stresses in a non-circular pressure vessel are not uniform. The equation for circular cylinder absolutely isn't sufficient for an elliptical cylinder, as the internal pressure will conspire to make a circle out of that ellipse. If you don't know how to answer the question, you shouldn't pretend. $\endgroup$ Aug 25, 2018 at 5:09

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