Geology Reference
In-Depth Information
Subtracting equation (1.519) from equation (1.517) gives
=−
C
2ξ
3
,
(1.522)
and then equation (1.517) yields
4ξ
3
μ
A
=−
λ
+
μ
=−
4ξ
3
(1
−
2σ).
(1.523)
These values of the linear combination coe
cients
A
,
B
,
C
,
D
are found to sat-
isfy equation (1.516). Once again including the factor 2μ, they also produce the
Galerkin vector (1.439), as quoted for Mindlin problem I.
The linear combination of the Galerkin vector
R
e
3
, for Kelvin's problem of a
downward point force in the lower half-space, with the preceding four Galerkin
vectors produces a displacement field with components
(
x
1
−
ξ
1
)(
x
3
−
ξ
3
)
1
2μ
A
2μ
x
1
−
ξ
1
Q
3
u
1
=
+
R
3
−
ξ
1
)
x
3
+
ξ
3
Q
3
B
2μ
1
+
(
x
1
−
Q
(
Q
+
x
3
+
ξ
3
)
1
C
2μ
x
1
−
ξ
1
Q
3
3
x
3
(
x
3
+
ξ
3
)
Q
2
D
2μ
(
x
1
−
ξ
1
)(
x
3
+
ξ
3
)
+
−
+
,
(1.524)
Q
3
1
2μ
(
x
2
−
ξ
2
)(
x
3
−
ξ
3
)
A
2μ
x
2
−
ξ
2
Q
3
u
2
=
+
R
3
−
ξ
2
)
x
3
+
ξ
3
Q
3
B
2μ
1
+
(
x
2
−
Q
(
Q
+
x
3
+
ξ
3
)
1
C
2μ
x
2
−
ξ
2
Q
3
3
x
3
(
x
3
+
ξ
3
)
Q
2
D
2μ
(
x
2
−
ξ
2
)(
x
3
+
ξ
3
)
+
−
+
,
(1.525)
Q
3
3
−
ξ
3
)
2
R
2
1
2μ
1
R
(
x
3
A
2μ
x
3
+
ξ
3
Q
3
u
3
=
−
4σ
+
+
2
+
ξ
3
)
2
Q
2
B
2μ
1
Q
(
x
3
C
2μ
4σ)
x
3
+
ξ
3
Q
3
+
−
4σ
+
−
(2
−
Q
3
1
3
+
ξ
3
)
2
Q
2
+
ξ
3
)
2
Q
2
C
2μ
x
3
3
(
x
3
D
2μ
1
Q
(
x
3
+
−
+
−
4σ
+
.
(1.526)
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