Stress in a thick-walled pressure vessel 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.
 A: Generalities
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:
$$\epsilon_{rr} = \frac{dr(1 + u'(r)) - dr}{dr} = u'(r)$$
$$\epsilon_{\theta\theta} = \frac{(r + u(r))\,d\theta - r\,d\theta}{r\,d\theta} = \frac{u(r)}{r}$$
$$\epsilon_{\phi\phi} = \frac{(r + u(r))\cos\theta\,d\phi - r\cos\theta\,d\phi}{r\cos\theta\,d\phi} = \frac{u(r)}{r}$$
Symmetry
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.
Math
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:
$$\sigma_{rr} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 - \frac{R_o^3}{r^3}\right)$$
$$\sigma_{tt} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 + \frac{1}{2}\frac{R_o^3}{r^3}\right)$$
A: 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.

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.
