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named. The original solutions were generalised and expanded to consideration of
nuclei of strain by Mindlin and Cheng (1950).
In contrast to the Boussinesq and Cerruti problems, where the conditions of equi-
librium were established by integrating tractions over a hemisphere centred on the
point of application of the force, for the Mindlin problems we use the method of
images and combine additional solutions, which are singular at the image point
in the upper half-space, to ensure equilibrium. The additional solutions are also
chosen to make the common bounding surface of the two half-spaces stress free,
allowing removal of the upper half-space.
A point force of strength P is located at (ξ 1 2 3 ) in the lower half-space. The
radius to the field point at ( x 1 , x 2 , x 3 ) is once again
( x 1
ξ 1 ) 2
ξ 2 ) 2
ξ 3 ) 2
R
=
+
( x 2
+
( x 3
.
(1.437)
The additional solutions combine at the image point to produce a point force of
strength
P , producing overall force equilibrium. The image point is located at
1 2 , ξ 3 ), in the upper half-space. The radius from there to the field point is then
( x 1
ξ 1 ) 2
ξ 2 ) 2
+ ξ 3 ) 2
=
+
( x 2
+
( x 3
Q
.
(1.438)
For Mindlin problem I, the Galerkin vector, 2μ HPR e 3 , for Kelvin's problem
with a downward 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 3 direction, Mindlin and
Cheng (1950) give the Galerkin vector for Mindlin problem I as
e 3 R
8σ(1
1 Q
+
σ)
) ( 1
σ ) x 3 σξ 3 log( Q + x 3 + ξ 3 )
2 ξ 3 x 3
Q
+
4 ( 1
, (1.439)
with the elastic constants (excepting μ) expressed in terms of the dimensionless
Poisson's ratio (1.255). It is thus seen to be a linear combination of the Galerkin
vector for the Kelvin problem of a downward point force in the lower half-space,
proportional to e 3 R (omitting the factor 2μ HP ), with the Galerkin vectors,
G 1
=
A e 3 log( Q
+
x 3
+ ξ 3 ),
(1.440)
G 2
=
B e 3 ( x 3
+ ξ 3 ) log( Q
+
x 3
+ ξ 3 ),
(1.441)
C e 3 x 3
G 3 =
Q ,
(1.442)
G 4 = D e 3 Q .
(1.443)
 
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