Geology Reference
In-Depth Information
The displacement components are then
x
1
x
1
,
A
2μ
x
3
R
0
=
−
u
1
x
2
x
2
,
A
2μ
x
3
R
0
u
2
=
−
(1.345)
⎣
⎦
.
x
3
R
0
A
2μ
1
R
0
λ
+
3μ
λ
+
μ
+
u
3
=
The associated normal stresses are
⎣
⎦
,
3
x
1
x
1
2
−
A
x
3
R
0
μ
λ
+
μ
τ
11
=−
R
0
−
(1.346)
⎣
⎦
,
3
x
2
x
2
2
−
A
x
3
R
0
μ
λ
+
μ
τ
22
=−
R
0
−
(1.347)
⎣
⎦
,
3
x
3
A
x
3
R
0
μ
λ
+
μ
=−
R
0
+
τ
33
(1.348)
while the shear stresses are
⎣
⎦
,
x
1
3
x
3
A
x
1
−
μ
λ
+
μ
τ
13
=−
R
0
+
(1.349)
R
0
⎣
⎦
,
x
2
3
x
3
A
x
2
−
μ
λ
+
μ
=−
R
0
+
τ
23
(1.350)
R
0
3
A
x
1
x
1
x
2
x
2
x
3
−
−
τ
12
=−
.
(1.351)
R
0
On the plane
x
3
=
0, the normal stresses vanish but the two shear stresses, τ
13
given
by (1.349) and τ
23
given by (1.350), do not.
For the plane
x
3
0 to be a free surface, we need an additional, lamellar solution
of the Navier equation, represented by the scalar potential in the Papkovich-Neuber
formulation, or by the scalar potential,
L
, in the Helmholtz separation (1.272). Take
L
=
x
3
), with
B
a constant to be determined. The extra displacement
field, given by the gradient of
L
,is
=
B
log(
R
0
+
x
1
x
2
x
1
−
x
2
−
B
R
0
.
u
1
=
B
x
3
)
,
u
2
=
B
x
3
)
,
u
3
=
(1.352)
R
0
(
R
0
+
R
0
(
R
0
+
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