ABSTRACT
Example 6.12 A single concentrated load of 1000 kN acts at the ground surface. Construct an isobar for C1
z = 40 kN/m2 by making use of the Boussinesq equation.
Solution From Eq. (6.1a) we have
3Q [ 1 ]% CYz = 2m 2 1+ (r/z)2
We may now write by rearranging an equation for the radial distance r as
Now for Q = 1000 kN, C1z = 40 kN/m2, we obtain the values of rl' r2, r3, etc. for different depths zl' Z2' Z3' etc. The values so obtained are
z (m) r (m) 0.25 1.34 0.50 1.36 1.0 1.30 2.0 1.04 3.0 0.60 3.455 0.00
y y
Figure Ex. 6.12
The isobar for {jz = 40 kN/m2 may be obtained by plotting z against r as shown in Fig. Ex. 6.12.
