Question 6.12: A single concentrated load of 1000 kN acts at the ground sur...

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.

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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.

6.12

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