Geology Reference
In-Depth Information
Substituting the values of the constants A , B , C , D , and multiplying by 2μ,wefind
the components of the displacement field, given by the Galerkin vector (1.439) for
Mindlin problem I for a point force of magnitude 8πμ(λ +
2μ)/(λ + μ) to be
ξ 1 ) x 3 ξ 3
R 3
(3
4σ)( x 3 ξ 3 )
Q 3
u 1
=
( x 1
+
3 x 3 ( x 3
+ ξ 3 )
4 (1
σ)(1
2σ)
+
,
Q ( Q
+
+ ξ 3 )
Q 5
x 3
( x 2 ξ 2 ) x 3
ξ 3
R 3
(3
4σ)( x 3
ξ 3 )
u 2 =
+
Q 3
6 ξ 3 x 3 ( x 3 + ξ 3 )
Q 5
4 (1
σ)(1
2σ)
+
,
Q ( Q
+
x 3 + ξ 3 )
σ) 2
ξ 3 ) 2
R 3
3
4 R +
8 (1
(3
4σ)
( x 3
u 3 =
+
Q
4σ)( x 3 + ξ 3 ) 2
6 ξ 3 x 3 ( x 3 + ξ 3 ) 2
Q 5
(3
3 x 3
+
+
.
(1.527)
Q 3
In this form of the displacement field, the elastic constants appear exclusively
through the dimensionless Poisson's ratio (1.255).
For Mindlin problem II, the Galerkin vector, 2μ HPR e 1 , for Kelvin's problem
with a horizontal point force in the lower half-space is the starting point. For a
point force with magnitude 8πμ(λ +
2μ)/(λ + μ) in the x 1 direction, Mindlin and
Cheng (1950) give the Galerkin vector for Mindlin problem II as
e 1
Q
2
3
2σ) ( x 3 + ξ 3 ) log( Q + x 3 + ξ 3 )
R + Q
Q +
4 (1
σ)(1
e 3 2 ξ 3 ( x 1 ξ 1 )
Q
+ ξ 3 )
+
+
2 (1
2σ)( x 1
ξ 1 ) log( Q
+
x 3
.
(1.528)
This Galerkin vector is a linear combination of the Galerkin vector for the Kelvin
problem of a horizontal point force in the lower half-space with the additional
Galerkin vectors, e 1 Q , e 1 ( x 3
+ ξ 3 ) log( Q
+
x 3
+ ξ 3 ), e 3 ( x 1
ξ 1 ) log( Q
+
x 3
+ ξ 3 ),
e 1 / Q and e 3 ( x 1
ξ 1 )/ Q . All of the additional Galerkin vectors represent solutions
singular at the image point. Once again, the solution can be found, as in Mindlin
problem I, through the determination of the constants of the linear combination
by imposing the conditions of total force balance and that the surface, x 3
0, be
stress free. The details of the solution are left to the reader. The components of the
resulting displacement are,
=
 
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