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
2μ
1
R
0
3μ
λ
+
μ
+
u
1
=
(1.380)
R
0
x
1
x
1
x
2
x
2
−
−
A
2μ
=
u
2
,
(1.381)
R
0
A
2μ
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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