The thin-wall formula rests on one quiet assumption: that the hoop stress is the same all the way through the wall. Equilibrium only fixes the AVERAGE, and for a thin wall the average is all there is. The working rule is that the wall thickness $t$ must be less than one twentieth of the internal diameter $d$, that is, less than $\dfrac{d}{20}$. It is the same as saying $d/t$ is above 20. Past that line the stress at the bore climbs well above the average, and the bore is where cracks start. Hydraulic cylinders, gun barrels and high-pressure tube all live on the thick side of it.

Gabriel Lamé solved the thick cylinder in 1833. For internal pressure only, the hoop stress at any radius is $\sigma_{\theta} = \dfrac{p_i r_i^{2} \left(r^{2} + r_o^{2}\right)}{r^{2} \left(r_o^{2} - r_i^{2}\right)}$, read aloud sigma-theta equals p-i r-i squared, times r squared plus r-o squared, over r squared times r-o squared minus r-i squared. The subscripts are the whole convention: i is inside and o is outside. $p_i$ is the internal gauge pressure in MPa, $r_i$ the inside radius and $r_o$ the outside radius, both in mm. The plain $r$, with no subscript, is the radius at which you are asking, anywhere from $r_i$ to $r_o$. $\sigma_{\theta}$ is the hoop stress there, in MPa; theta is the Greek letter for the direction around the circle.

Its companion runs through the wall: $\sigma_{r} = \dfrac{p_i r_i^{2} \left(r^{2} - r_o^{2}\right)}{r^{2} \left(r_o^{2} - r_i^{2}\right)}$, sigma-r equals the same thing with a minus in the bracket. $\sigma_r$ is the radial stress in MPa. Because $r$ never exceeds $r_o$ it is negative everywhere, which means compressive: the pressure squeezes the wall through its thickness. It equals $-p_i$ exactly at the bore and exactly zero at the free outside surface.

Three facts to carry. The hoop stress is largest at the bore, where it is $p_i \dfrac{r_o^{2} + r_i^{2}}{r_o^{2} - r_i^{2}}$. It falls steadily to the outside, and the drop from bore to outside is exactly $p_i$, whatever the geometry. And at every radius $\sigma_{\theta} + \sigma_{r}$ comes to the same constant, which is the cheapest check on both answers. One honest consequence: adding wall to a thick cylinder pays less and less, because the new metal sits where the stress is lowest.