Draw isobar diagram of 0.1Q

The isobar of $0.1Q$ is the contour of points in the soil below a point load $Q$ where the vertical stress equals $0.1Q$ (per m², with $Q$ in kN and distances in m). Use Boussinesq's equation:

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

with $I_B = 0.4775\,[1+(r/z)^2]^{-5/2}$.

Computation

Setting $\Delta\sigma_z = 0.1\,Q$: $z^2 = I_B/0.1$, so $z = \sqrt{10\,I_B}$ and $r = (r/z)\,z$.

Diagram

Plot the points $(\pm r, z)$ with $z$ downward and join them by a smooth curve.

        Q (load)
  ------●-------------  surface
     .  |  .       r max = 0.94 m at z = 1.18 m
    .   |   .
    .   |   .
     .  |  .
      . | .
        ●    z max = 2.19 m on the axis (r = 0)

The isobar is a closed bulb that begins and ends at the load point on the surface and extends to $z = 2.19$ m on the axis. Inside the bulb the stress is greater than $0.1Q$.