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