Information Technology Reference
In-Depth Information
9.3.4
Points on the Boundary
In order to calculate the unknown values of
p
i
and
u
i
in Eq. (9.80), it is necessary to move
the point
to the boundary, so that all the unknowns in Eq. (9.80) are on the boundary, and a
system of linear equations can be formed to solve for the unknowns. When the point
ξ
is on
the boundary, however, singularities develop, and they must be given special consideration.
Assume that the boundary can be represented as shown in Fig. 9.1, where
S
ξ
=
(
S
−
S
)
+
S
,
in which
S
.
The first integral on the right-hand side of Eq. (9.80b) can be written as
is a hemispherical surface of radius
u
ij
u
ij
u
ij
p
j
dS
=
lim
→
p
j
dS
+
lim
→
p
j
dS
0
0
S
S
−
S
S
S
,u
ij
does not involve any singularity, the first
Since on the part of the boundary
S
−
integral will not be altered when
→
0
,
i.e.,
u
ij
u
ij
lim
→
p
j
dS
=
p
j
dS
(9.81)
0
S
−
S
S
Assume that
S
is very small, so that
p
j
can be treated as being constant on
S
. Then
u
ij
u
ij
dS
lim
→
p
j
dS
=
p
j
lim
→
(9.82)
0
0
S
S
Although the two-dimensional case will be used to study this integral, similar arguments
hold for three-dimensional cases. As shown in Fig. 9.3 and from the relationships given in
Eq. (9.71),
,
1
=
,
1
=
1
,
2
=
,
2
=
2
dS
=
d
θ
,
=
cos
ϕ
,
=
sin
ϕ
(9.83)
Substitute Eqs. (9.74) and (9.83) into Eq. (9.82) to obtain
δ
ij
+
,i
,j
dS
1
ln
1
u
ij
dS
lim
→
=
lim
→
(
3
−
4
ν)
8
π(
1
−
ν)
G
0
0
S
S
1
ln
1
=
(
−
ν)
δ
+
ϕ
ϕ
θ
lim
→
3
4
cos
sin
d
ij
8
π(
1
−
ν)
G
0
S
S
1
ln
1
=
(
3
−
4
ν)
lim
→
d
θδ
ij
+
lim
→
cos
ϕ
sin
ϕ
d
θ
8
π(
1
−
ν)
G
0
0
S
For the second integral here,
θ
=
γ
+
ϕ
(Fig. 9.3), so that
d
θ
=
d
ϕ
,
and when
θ
varies from
0to
π
,
ϕ
varies from
−
γ
to
π
−
γ
. Then
π
−
γ
lim
→
cos
ϕ
sin
ϕ
d
θ
=
lim
→
0
cos
ϕ
sin
ϕ
d
ϕ
=
0
0
S
−
γ
For the first integral,
d
d
ln
1
ln
(
ln
)
lim
→
S
d
θ
=−
lim
→
d
θ
=−
lim
→
1
d
θ
=
lim
→
0
()
d
θ
=
0
1
d
d
0
0
0
S
S
S
Search WWH ::
Custom Search