Geology Reference
In-Depth Information
As in the Boussinesq problem, we begin by specialising Kelvin's problem to the
case where the concentrated force is applied at x 1 , x 2 ,0 ,onthesurface x 3
=
0,
and now entirely in the x 1 -direction. Then, the Galerkin vector is G
=
AR 0 e 1 , with
A again an arbitrary constant. The displacement components are then
,
x 1
x 1 2
λ +
A
1
R 0
λ + μ +
u 1
=
(1.380)
R 0
x 1
x 1 x 2
x 2
A
=
u 2
,
(1.381)
R 0
A
R 0 x 1
x 3
x 1 .
u 3
=
(1.382)
The associated normal stresses are
,
3 x 1
x 1 2
x 1
τ 11 =− A x 1
R 0 + μ
(1.383)
R 0
λ + μ
,
3 x 2
x 2 2
x 1
τ 22 =− A x 1
R 0 μ
(1.384)
R 0
λ + μ
,
A x 1 x 1
R 0
3 x 3
μ
λ + μ
τ 33 =−
R 0
(1.385)
while the shear stresses are
,
3 x 1
x 1 2
A x 3
R 0
μ
λ + μ
τ 13
=−
R 0 +
(1.386)
3 A x 1
x 1 x 2
x 2 x 3
τ 23
=−
,
(1.387)
R 0
.
3 x 1
x 1
2
R 0 +
x 2
A x 2
μ
λ + μ
τ 12
=−
(1.388)
R 0
On the plane x 3 =
0, the two shear stresses,τ 13 andτ 23 , vanish but the normal stress,
τ 33 , does not.
In contrast to the Boussinesq problem, we need two additional solutions for the
plane x 3
0 to be a free surface.
The first extra displacement field is given by the gradient of the scalar potential,
=
x 1
L = B x 1
x 3 ,
(1.389)
R 0
+
 
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