A single concentrated load of 1000 kN acts at the ground surface. Construct an isobar for \sigma_{z}=40 kN / m ^{2} by making use of the Boussinesq equation.
A single concentrated load of 1000 kN acts at the ground surface. Construct an isobar for \sigma_{z}=40 kN / m ^{2} by making use of the Boussinesq equation.
From Eq. (6.la) we have
I_{B}=\text { Boussinesq stress coefficient }=\frac{3}{2 \pi} \frac{1}{\left[1+(r / z)^{2}\right]^{5 / 2}} (6.1a)
\sigma_{z}=\frac{3 Q}{2 \pi z^{2}}\left[\frac{1}{1+(r / z)^{2}}\right]^{5 / 2}
We may now write by rearranging an equation for the radial distance r as
r=\sqrt{z} \sqrt{\frac{3 Q}{2 \pi z^{2} \sigma_{z}}^{2 / 5}-1}
Now for Q=1000 kN , \sigma_{z}=40 kN / m ^{2}, \text { we obtain the values of } r_{1}, r_{2}, r_{3}, etc. for different depths z_{1}, z_{2}, z_{3}, etc. The values so obtained are
z(m) | r(m) |
0.25 | 1.34 |
0.5 | 1.36 |
1.0 | 1.30 |
2.0 | 1.04 |
3.0 | 0.60 |
3.455 | 0.00 |
The isobar for \sigma_{z}=40 kN / m ^{2} may be obtained by plotting z against r as shown in Fig. Ex. 6.12.