I'm currently researching the strength of domes underwater and I went down the rabithole of material science. The basics of this research (or though experiment) is to bring a dome filled with air to the seafloor and see if it implodes or not. The specifics are as follows.

 The dome is 3D-printed
 Outer Diameter: 10 cm
 Inner Diameter: 8 cm (so the thickness of the dome is 1 cm) (this can vary)
 Material: PLA

QUESTION: What is the maximum amount of pressure this dome can take before it breaks?

I've already found the following data from this source

Properties PLA:
 - Impact strength:
   - STD: 1.8 ft-lb/in
   - MAX: 4.1 ft-lb/in
 - Compressive Strength
   - STD: 2600 PSI
   - MAX: 13600 PSI
 - Tensile Strength
   - STD: 6783 PSI
   - MAX: 9531 PSI
 - Flexural Strength
   - STD: 8970 PSI
   - MAX: 13731 PSI

It seems likely that I'll be needing the compressive strength here, because the theoretical dome is being put at the bottom of the theoretical sea. It's going to be compressed. But these numbers have very little context (at least for me, I don't have an engineering degree). I mainly worry about thickness. In my mind, when a wall is thicker, it means it can hold more weight, can withstand heavier punches and generally breaks less easily. I don't see this reflected in this number. It also doesn't take the shape of the object in mind. A sphere is supposedly stronger then any other shape (I found this cool site selling dome bunkers). But this too doesn't seem reflected in the compressive strength value.

It might also have to do with the size effect, but that seems to weaken - not strengthen - structures and would only be applicable if I made the outer diameter variable.

If anyone can help me any further it would be very much appreciated. I really want to make clear that I'm not asking you to do my homework (this isn't really homework, but still). Pointing me vaguely in the right direction is also extremely helpful. Thanks in advance!


2 Answers 2


A shell structure under external pressure is likely to fail in buckling before it reaches the compressive strength of the material.

Buckling of curved shells is a complex problem in general, but for a complete sphere a simple formula (source here) is $$P = \frac{2E}{\sqrt{3(1-\nu^2)}}\frac{t^2}{r^2}$$

Where $E$ and $\nu$ are Young's modulus and Poisson's ratio, $t$ is the shell thickness and $r$ is the mean radius.

Taking $E$ = 510,000 psi and $\nu$ = 0.33 (sources, here and here) This gives about $31,000$ psi for your dimensions.

These buckling formulas are very imprecise and I would take a safety factor of at least 10 times, which would give a safe buckling load of about 3000 psi.

Since the maximum pressure using the "STD" compressive stress is about 1200 psi (using the formula in mg4w's answer), in this case buckling is not the limiting factor, but the buckling load is proportional to $t^2$ while the compressive load is proportional to $t$, so for a thinner shell the buckling load decreases faster than the compressive load.

So taking the pressure gradient in water as about 0.5 psi per foot of depth, (there's no point using more than one significant figure here!) a sphere with your dimensions should be good for a depth of about 2,500 feet.

But a dome with a flat base would fail before that, since the base would break at the edges where it joined the sphere. Assuming the base is rigidly supported by the dome around its edges, the maximum stress (source here) is $$\frac{3Pr^2}{4t^2}$$.

Taking the maximum tensile strength as 6783 psi, that gives a pressure of about 450 psi at failure - much smaller than the other failure modes considered above, and limiting the depth to about 900 feet (with no safety factor applied!)

  • $\begingroup$ I'm filling in the following thing into my calculator and get something completely different: 510000*2/sqrt(3*(1-0.33^2))*0.01^2/0.1^2 = 6238.4. Which is about 5 times less then your solution. How did you arrive at your answer? $\endgroup$
    – Trashtalk
    Dec 17, 2018 at 19:36

For simplicity let's first consider the case of a sphere with thickness much smaller than the radius ('Thin-Walled' pressure vessel). Say $\frac{r}{t} < 10$, with $r$ the radius of the sphere and $t$ the thickness. It is straightforward to show (see here) that the stress in two orthogonal circumferential directions is:

$$ \sigma = \frac{Pr}{2t} $$

where P is the difference between the external and internal pressures. If $P$ is compressive (exterior pressure is larger than internal) then $\sigma$ is compressive. You will then need to apply a failure model appropriate to your material (not sure about PLA). But a simple criteria could be if this stress is greater than the maximum stress your material can handle then it will fail.

If your pressure vessel is not 'thin' then you need to consider variation of stress through the thickness which complicates matters significantly ('Thick-Walled' pressure vessel). A discussion of this case can be found here.

A dome is much more complicated since the flat face on the bottom will also be under the same pressure and will therefore be subject to bending. The interface between the hemisphere and the base also complicates the stress state. This is why you virtually never see pressure vessels with flat faces.

In reality you will need to consider imperfections in the fabrication, variation in material properties, buckling, etc.

  • $\begingroup$ Thanks a lot for this answer, if I could mark two answers as 'solved' I would! $\endgroup$
    – Trashtalk
    Dec 17, 2018 at 18:33
  • $\begingroup$ I believe that if you clicked the check-mark, you can click it again to remove it. $\endgroup$ Nov 2, 2023 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.