Water tower foundation on weak stratum

Given

Circular foundation $D = 10$ m ($R = 5$ m). Load $Q = 18000$ kN. Weak stratum at $z = 3$ m below the foundation level (assumed, as the depth is not printed), bearing capacity $q_a = 100$ kN/m².

Contact pressure:

$$q = \frac{Q}{\pi R^2} = \frac{18000}{\pi \times 5^2} = 229.2\ \text{kN/m}^2$$

Stress at the top of the weak stratum (below the centre)

Boussinesq's equation for a uniformly loaded circular area:

$$\Delta\sigma_z = q\left[1 - \frac{1}{\left(1 + (R/z)^2\right)^{3/2}}\right]$$ $$\frac{R}{z} = \frac{5}{3} = 1.667, \qquad 1 + (R/z)^2 = 3.778, \qquad (3.778)^{3/2} = 7.343$$ $$\Delta\sigma_z = 229.2\left[1 - \frac{1}{7.343}\right] = 229.2 \times 0.8638 = 198.0\ \text{kN/m}^2$$

(A quick check by the 2:1 method: $Q/[\frac{\pi}{4}(10+3)^2] = 135.6$ kN/m², also above 100.)

Check

$\Delta\sigma_z = 198\ \text{kN/m}^2 \approx 19.8\ \text{t/m}^2 > 100\ \text{kN/m}^2\ (10\ \text{t/m}^2)$

Answer: The stress at the top of the weak stratum is about 198 kN/m², which exceeds the bearing capacity of the stratum (100 kN/m²). It is not safe to construct the water tower with this foundation size.

The foundation must be enlarged (a diameter of about 15 m gives about 97 kN/m²) or the load shifted to a deeper stratum (piles).