Geology Reference
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The displacement components are then
x 1
x 1 ,
A
x 3
R 0
=
u 1
x 2
x 2 ,
A
x 3
R 0
u 2
=
(1.345)
.
x 3
R 0
A
1
R 0
λ +
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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