264
8 Elastic Solutions in Geomechanics
Now, the total resultant force on the surface z = h is given by,
Resultant upward force =
∞
0
σ z (2πrdr)
=
∞
0
P
4π(1 − v)
(1 − 2v)h
R 3
+
3h
3
R 5
(2πrdr)
=
∞
0
P
2(1 − v)
(1 − 2v)h
R 3
+
3h
3
R 5
r dr
To simplify the integration, introduce the angle ψ as shown in Fig. 8.2.
Here, r = h tanψ and dr = h sec
2
ψ dψ
Therefore, resultant upward force =
π/2
0
P
2(1 − v)
(1 − 2v) sin ψ + 3 cos
2
ψ sin ψ
dψ
(8.3)
Solving, we get resultant upward force on the lower plane = P which is exactly
one-half the applied load.
Further, if we consider a similar surface z =−h, shown in Fig. 8.3, we will find
tensile stresses of the same magnitude as the compressive stresses on the lower plane.
Hence, resultant force on the upper plane = −P (tensile force). Combining the
two resultant forces, we get 2P which exactly equilibrate the applied load.
8.3 Boussinesq’s Problem
The problem of a point load acting normal to the surface of an elastic half-space
was solved by the French mathematician Joseph Boussinesq (1878). The problem
geometry is illustrated in the following Fig. 8.4. The half-space is assumed to be
homogeneous, isotropic and elastic. The point load is applied at the origin of coordinates on the half-space surface. Let P be the magnitude of the point load. Now,
consider the stress function
φ = B
r
2
+ z
2
1
2
(8.4)
where B is a constant.
The stress components are given by