Ring foundation water tank max stress

Given

Outer $D = 10$ m ($R = 5$ m), inner $d = 7.5$ m ($r = 3.75$ m), $q = 160$ kN/m², $z = 4$ m below the foundation.

Stress on the axis (below the centre)

The usual textbook method takes the point on the axis, where Boussinesq's equation for a uniformly loaded circle can be used directly (full circle minus inner circle):

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

• Outer: $R/z = 1.25$, $(1+1.5625)^{3/2} = 4.102$, factor $= 1 - 0.2438 = 0.7562$
• Inner: $r/z = 0.9375$, $(1+0.8789)^{3/2} = 2.575$, factor $= 1 - 0.3883 = 0.6117$

$$\Delta\sigma_z = 160\,(0.7562 - 0.6117) = 160\times0.1445 = 23.1\ \text{kN/m}^2$$

Note on the true maximum

The centre of a ring is unloaded, so at a shallow depth like 4 m (less than the ring diameter) the stress on the axis is not the largest at that depth. Integrating Boussinesq's point-load equation over the ring for points off the axis gives about 34 kN/m² below the ring itself, at a radius of about 3.5 to 4 m from the centre.

Answer: Vertical stress at 4 m depth on the axis $\approx 23.1$ kN/m² (the standard textbook answer). Strictly, the maximum at this depth occurs below the ring, at about 34 kN/m².