Annular ring, stress at five depths

Given

Ring diameters 4 m and 6 m (as printed: taken as outer $D = 6$ m, inner $d = 4$ m), so $R = 3$ m and $r = 2$ m. Pressure $q = 100$ kN/m².

Method

The stress under the centre of a ring equals that of the full circle of radius $R$ minus that of the inner circle (radius $r$) at the same pressure:

$$\Delta\sigma_z = q\left[\left(1 - \frac{1}{\left(1+(R/z)^2\right)^{3/2}}\right) - \left(1 - \frac{1}{\left(1+(r/z)^2\right)^{3/2}}\right)\right]$$

Calculation

Distribution curve

 depth z (m)   stress (kPa)
   0 |*   0
 0.5 |*  1
   1 |* 5.8
   2 |---* 18.3
   3 |----* 22.2   (maximum, z ≈ 3 m)
   4 |----* 20.4
   8 |--* 9.2

The stress along the axis is near zero at the surface (the centre is unloaded), increases to a maximum of about 22 kPa at $z \approx 3$ m, and then decreases.

Answer: $\Delta\sigma_z = 0.98,\ 5.78,\ 18.29,\ 20.35,\ 9.22$ kN/m² at 0.5, 1, 2, 4 and 8 m.