I can find many references that give the stress in the walls of a pressure vessel for spheres and tubes, but they all seem to be limited to a thin-wall approximation. I'll limit my writing here to spherical vessels (I don't think it should be very different to find the answer for any structure with symmetry). My main trouble is in setting up the integral.

$$ \sigma = \frac{P R }{2 t}$$

So my question is: how do you find material stress for a pressure vessel with a wall that goes from $R_i$ to $R_o$, and you want a solution that is still perfectly valid when $R_i \ll R_o$? My guess is that you would take the equation for the thin wall case, and recast it into an integral.

$$ P = \frac{2 \sigma t }{ R }$$

In the more general case, would we write something like this?

$$ P = \int_{R_i}^{R_o} \frac{2 \sigma }{r} dr$$

This would only be valid if the stress was constant for all differential shells. I'm wondering if that's a bad assumption, but if the material was highly elastic I think it would be decent assumption. My intuition is that the material would have to be much more elastic for it to hold with the inner radius many times smaller than the outer radius.

But regardless of that detail, is the above integral correct? Is there some good intuitive logic to justify it? And is there any other reference that gives the expression for stress in this case? It should be a logarithm if the above is correct, but I haven't seen this before.

  • 1
    $\begingroup$ Check section 4.1.4 of this reference. (If elasticity cannot be assumed, the problem is much more complex.) $\endgroup$
    – mmc
    Jul 10, 2013 at 22:48
  • $\begingroup$ @mmc Please correct me if I'm wrong, but I believe that in your reference, $\nu$ is the Poisson's ratio. So it's not the elastic constant? Given that, it's doing something a bit different than what I was expecting. Still, it does appear to be a case even more general than the "general case" I was asking about. It is also a clearly polynomial result - indicating that my formulation is incorrect. This is useful information. Perhaps if I set $\nu=0$ and $p_B=0$ I can get an expression for $P=$ in the above nomenclature. Then hopefully work backward to see the right process. $\endgroup$ Jul 11, 2013 at 0:36
  • 1
    $\begingroup$ Yes, the Poisson's ration appears in the radial stress ($\sigma_{RR}$) because it "connects" tangential and radial stresses. The tangential stress distribution doesn't depend on $E$ because the deformations are linear on $E^{-1}$ and the stress distribution only depends on relative deformations (remember that these are "infinitesimal" deformations). $\endgroup$
    – mmc
    Jul 11, 2013 at 3:19
  • $\begingroup$ @mmc well I guess I'm at the limit of my understanding at this point. I'm not sure how the compressive radial stress makes any sense. IMO, there should only be tangential tensile stress. I can make sense of one of their tangential stress terms, but not the other. I was expecting there to be tangential stress and the isotropic pressure only. $\endgroup$ Jul 11, 2013 at 21:28
  • $\begingroup$ I have written an answer summarizing the solution process and trying to avoid "tricks". Please ask me if you are unsure about any of the steps. $\endgroup$
    – mmc
    Jul 13, 2013 at 23:46

2 Answers 2



The problem has spherical symmetry, so it makes sense to use spherical coordinates ($r$, $\theta$, $\phi$). We can divide the vessel into differential elements like the one shown in this post.

Deformation and strain

Only radial deformations are allowed by the spherical symmetry, so let's parametrize the deformation by

$$r \rightarrow r + u(r)$$

Assuming small deformations:

$$dr \rightarrow dr(1 + u'(r))$$

$$r\,d\theta \rightarrow (r + u(r))\,d\theta$$

$$r\cos\theta\,d\phi \rightarrow (r + u(r))\cos\theta\,d\phi$$

The associated strains:

$$\epsilon_{rr} = \frac{dr(1 + u'(r)) - dr}{dr} = u'(r)$$

$$\epsilon_{\theta\theta} = \frac{(r + u(r))\,d\theta - r\,d\theta}{r\,d\theta} = \frac{u(r)}{r}$$

$$\epsilon_{\phi\phi} = \frac{(r + u(r))\cos\theta\,d\phi - r\cos\theta\,d\phi}{r\cos\theta\,d\phi} = \frac{u(r)}{r}$$


By spherical symmetry we have:

$$\epsilon_{\theta\theta} = \epsilon_{\phi\phi} = \epsilon_{tt}$$

$$\sigma_{r\theta} = \sigma_{r\phi} = \sigma_{\theta\phi} = 0$$

$$\sigma_{\theta\theta} = \sigma_{\phi\phi} = \sigma_{tt}$$

$$\frac{\partial\,\sigma_{rr}}{\partial\,\theta} = \frac{\partial\,\sigma_{\theta\theta}}{\partial\,\theta} = \frac{\partial\,\sigma_{\phi\phi}}{\partial\,\theta} = \frac{\partial\,\sigma_{rr}}{\partial\,\phi} = \frac{\partial\,\sigma_{\theta\theta}}{\partial\,\phi} = \frac{\partial\,\sigma_{\phi\phi}}{\partial\,\phi} = 0$$

Equilibrium condition

The equilibrium condition can be expressed in spherical coordinates as

$$2\sigma_{rr} + r\frac{\partial\,\sigma_{rr}}{\partial\,r}-\sigma_{\theta\theta}-\sigma_{\phi\phi} = 0$$

or, using the symmetry conditions,

$$2\sigma_{rr}(r) + r\sigma_{rr}'(r) - 2\sigma_{tt} = 0$$

Hooke's law

Applying Hooke's law for isotropic materials we get

$$\epsilon_{rr} = \frac{1}{E}(\sigma_{rr} - 2 \nu \sigma_{tt})$$

$$\epsilon_{tt} = \frac{1}{E}(\sigma_{tt} - \nu (\sigma_{rr} + \sigma_{tt}))$$

where $E$ is the elastic modulus and $\nu$ is the Poisson's ratio.


Combining the previous results we get the following equations:

$$u'(r) = \frac{1}{E}(\sigma_{rr} - 2 \nu \sigma_{tt})$$

$$\frac{u(r)}{r} = \frac{1}{E}(\sigma_{tt} - \nu (\sigma_{rr} + \sigma_{tt}))$$

$$2\sigma_{rr}(r) + r\sigma_{rr}'(r) - 2\sigma_{tt} = 0$$

From there we can get the following differential equation for the deformation (by a tedious but quite straightforward path):

$$2 u'(r) - 2 \frac{u(r)}{r} + r\,u''(r) = 0$$

From there, it's easy to get

$$u(r) = A\,r + \frac{B}{r^2}$$

and, as a consequence,

$$\sigma_{tt} = E \frac{A}{1 - 2\nu} + E\frac{B}{1 + \nu}\frac{1}{r^3}$$

$$\sigma_{rr} = E \frac{A}{1 - 2\nu} - 2 E\frac{B}{1 + \nu}\frac{1}{r^3}$$

where $A$ and $B$ are integration constants determined to be consistent with the boundary conditions.

Solving the original problem

The original problem has zero external pressure, internal pressure $P$, interior radius $R_i$ and exterior radius $R_o$. For continuity, we must have

$$\sigma_{rr}(R_i) = -P$$

$$\sigma_{rr}(R_o) = 0$$

Getting the constants from these boundary conditions:

$$A = \frac{P(1 - 2\nu)}{E}\frac{R_i^3}{R_o^3 - R_i^3}$$

$$B = \frac{P(1 + \nu)}{2 E}\frac{R_o^3 R_i^3}{R_o^3 - R_i^3}$$

Getting the stresses:

$$\sigma_{rr} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 - \frac{R_o^3}{r^3}\right)$$

$$\sigma_{tt} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 + \frac{1}{2}\frac{R_o^3}{r^3}\right)$$

  • $\begingroup$ If I were looking for the pressure at which the material breaks, then I think I would equate your result to the listed material limit as $\sigma=\sigma_{tt}-\sigma_{rr}$, that being the only non-isotropic element from the stress tensor diagonals. Is my understanding correct? btw thanks for this detailed solution. $\endgroup$ Jul 14, 2013 at 23:45
  • $\begingroup$ Still trudging through the nomenclature... so $\sigma_{rr}$ is basically the pressure. The amount that $\sigma_{tt}$ differs from that is basically the tangential tension. I can see how this limits to the thin-wall case referenced in my question and also fits the boundary conditions. The same method used for the thin-wall case won't give the right answer, and I don't know why. In all, I have 3 original methods that all give different wrong answers. Your final answer doesn't have $E$, making me think I can replicate it, but I'm still struggling. $\endgroup$ Jul 15, 2013 at 3:08
  • 1
    $\begingroup$ @AlanSE The right solution will depend on the yield criteria you adopt. For metals, a good one is the von Mises yield criterion: $\sigma_y = \sqrt{\tfrac{1}{2}[(\sigma_{rr} - \sigma_{tt})^2 + (\sigma_{rr} - \sigma_{tt})^2 + (\sigma_{tt} - \sigma_{tt})^2]} = |\sigma_{rr} - \sigma_{tt}|$ $\endgroup$
    – mmc
    Jul 15, 2013 at 16:40
  • $\begingroup$ @AlanSE At the risk of being a bit inaccurate by ignoring the compression, the constant stress assumption fails in the thick wall case "because" the inner part of the wall needs to deform more for a given radial deformation. If the inner radius of the vessel is 20 cm and the outer radius is 40 cm, both circumferences will grow ~6.3 cm if the radius of the vessel increases by 1 cm. But 6.3 cm is a much bigger relative deformation (strain) for the inner section of the vessel. $\endgroup$
    – mmc
    Jul 15, 2013 at 16:55

Well I had 3 different answers, but now I've eliminated a flaw in reasoning of 2. Thereby, I believe I've come up with another valid solution to the problem rooted in a different assumption about the material.

To recap, I believe the prior answer is telling us that for setting engineering limits, you would have the following equation for a thick-walled pressure vessel.

$$ \sigma = \frac{3}{2} P \frac{a^3}{b^3 - a^3} \frac{b^3}{r^3} $$

Here a is the inner radius and b is the outer radius. The maximum stress is obviously at $r=a$ and is:

$$ \sigma = \frac{3}{2} P \frac{b^3}{b^3 - a^3} $$

Just to have another form to use later I'll rewrite this.

$$ P = \frac{2}{3} \sigma \frac{ b^3 - a^3 }{ b^3 } = \frac{2}{3} \sigma \left( 1 - (a/b)^3 \right) $$

I worked to find a solution without elasticity included. That is describing a physical situation where every unit of material is stressed equally. Then, I wanted to apply the relation for a thin walled pressure vessel of $2 t \sigma = p r$. That is, two times the thickness times the stress is equal to the pressure times the radius.

To use this, we see a differential form. I believe the following will do for this.

$$ dp = \frac{2 \sigma}{r} dr $$

Here, the thickness has been replaced by dr. That makes sense to me, because I am changing a measurably thick vessel into a infinitely thin slice. The above differential equation can be solved by simple integration as $dP/dr$ is isolated. Do this from b to a.

$$ P = 2 \sigma \ln \frac{b}{a} $$

I now believe this is a meaningful result of sorts. To recap, the condition for a thin walled vessel can be written as follows. From there, you can substitute in either a or b for the value of r. Since the entire point of the approximation is that the wall is thin, the model is ignorant of the difference.

$$ P = \frac{ 2 \sigma (b - a) }{ r } $$

For comparison, let's take my result, the prior answer, and the two forms of the thin wall approach and put things in terms of b/a. This is the natural choice since $b>a$, which is a mathematical requirement we will enforce.

Here is a graph of these 4 functions. Firstly, the fact that they start out with the same slope adds credibility to my expression and that of the prior answer.

different approximations

These are expressions that give the pressure your vessel can achieve (as a fraction of the material maximum stress), as a function of the inner to outer radius ratio. Going from top to bottom, these represent the most optimistic to most realistic. They are:

  • Thin wall approximation using $r \approx a$
  • Constant stress approximation
  • Thin wall approximation using $r \approx b$
  • Elastic constant nonzero, Poisson ratio equal to zero

The order of these also fits expectations well enough.


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.