Excavation effect on vertical stress

Principle

Excavation removes soil weight, so it acts like a negative uniform load on the base of the excavation:

$$q = \gamma D_f = 2\times2.5 = 5\ \text{kN/m}^2 \quad (\text{unit weight as printed})$$

The reduction in vertical stress is obtained by the rectangular area (Fadum / Newmark) corner formula $I(m,n)$ with $m = B/z$, $n = L/z$, where $z$ is the depth below the excavation base. Depth required = 6 m from the ground, so $z = 6 - 2.5 = 3.5$ m.

(i) Below the centre

Divide the 3 m × 6 m area into four rectangles of 1.5 m × 3 m (corner at the centre).

$$m = \frac{1.5}{3.5} = 0.429, \quad n = \frac{3}{3.5} = 0.857 \;\Rightarrow\; I = 0.1013$$ $$\Delta\sigma_z = 4\,q\,I = 4\times5\times0.1013 = 2.03\ \text{kN/m}^2 \;(\text{reduction})$$

(ii) At a point 6 m from the centre (along the 6 m side direction)

The point is 3 m beyond the end of the excavation. Use a rectangle from 3 m to 9 m (length) and 1.5 m (half-width), doubled for the two halves:

$$I_1 = I\!\left(\frac{9}{3.5},\frac{1.5}{3.5}\right) = I(2.571, 0.429) = 0.1212, \qquad I_2 = I\!\left(\frac{3}{3.5},\frac{1.5}{3.5}\right) = 0.1013$$ $$\Delta\sigma_z = 2\,q\,(I_1 - I_2) = 2\times5\times(0.1212 - 0.1013) = 0.20\ \text{kN/m}^2 \;(\text{reduction})$$

(Along the 3 m side direction the same method gives 0.11 kN/m².)

Answer: The effective vertical stress at 6 m depth is reduced by about 2.0 kN/m² below the centre and about 0.2 kN/m² at 6 m from the centre.

If the intended unit weight is 20 kN/m³ ($q = 50$ kN/m²), all reductions are 10 times larger: 20.3 kN/m² and 2.0 kN/m². The reduction is small, so excavation has little effect on points far away.