Boussinesq stress below circle center

Assumptions

Uniform pressure $q$ on a flexible circular area of radius $R$ on the ground surface; the soil is homogeneous, isotropic and elastic (Boussinesq). The stress is required at depth $z$ below the centre.

Derivation

Boussinesq's equation for a point load $dQ$ at horizontal distance $r$:

$$d\sigma_z = \frac{3\,dQ}{2\pi z^2}\left[\frac{1}{1+(r/z)^2}\right]^{5/2}$$

An elemental ring of radius $r$ and width $dr$ carries $dQ = q\,(2\pi r\,dr)$. All points of the ring are at the same distance $r$ from the axis, so:

$$\Delta\sigma_z = \int_0^R \frac{3q\,(2\pi r\,dr)}{2\pi z^2}\left[\frac{1}{1+(r/z)^2}\right]^{5/2} = 3q\int_0^R \frac{r\,dr}{z^2}\left[1+\frac{r^2}{z^2}\right]^{-5/2}$$

Let $t = 1 + r^2/z^2$, so $dt = 2r\,dr/z^2$:

$$\Delta\sigma_z = \frac{3q}{2}\int_1^{1+R^2/z^2} t^{-5/2}\,dt = q\left[1 - \frac{1}{\left(1+(R/z)^2\right)^{3/2}}\right]$$ $$\boxed{\Delta\sigma_z = q\left[1 - \frac{1}{\left(1+(R/z)^2\right)^{3/2}}\right]}$$

Example

Take $q = 100$ kPa and $R = 2$ m.

The stress decreases with depth, and is nearly $q$ close to the surface.