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Then Eq. (9.70) can be written as
1
1
u
ji
=
[
−
r
,i j
]
=
r
r
,i
r
,j
(9.72a)
16
π
G
(
1
−
ν)
16
π
G
(
1
−
ν)
for
i
=
j
and
1
u
ji
=
[2
(
1
−
ν)
r
,kk
−
r
,i j
]
16
π
G
(
1
−
ν)
4
1
1
r
−
1
r
(
=
(
1
−
ν)
1
−
r
,i
r
,i
)
16
π
G
(
1
−
ν)
1
=
r
[
(
3
−
4
ν)
+
r
,i
r
,i
]
(9.72b)
16
π
G
(
1
−
ν)
for
i
=
j
. Combine Eqs. (9.72a) and (9.72b) to obtain
1
u
ji
=
Gr
[
(
3
−
4
ν)δ
+
r
,i
r
,j
]
(9.73)
ij
16
π(
1
−
ν)
For two-dimensional plane strain problems, a similar procedure leads to
ln
1
r
r
,i
r
,j
1
u
ji
=
(
3
−
4
ν)
δ
+
(9.74)
ij
8
π(
1
−
ν)
G
The tractions
p
ji
can be obtained by substituting the displacements into the strain-
displacement relations and the constitutive equations of Chapter 1, Eqs. (1.19) and (1.34),
to obtain
σ
ij
=
S
ki j
a
k
(9.75)
with
1
r
α
S
ki j
=−
(
−
ν)(δ
+
δ
−
δ
)
+
β
[
1
2
ki
r
,j
kj
r
,i
ij
r
,k
r
,i
r
,j
r
,k
]
where
3
,
2 for three-dimensional problems and two-dimensional plane
strain problems, respectively. The tractions are [Chapter 1, Eq. (1.58)]
α
=
2
,
1
,
and
β
=
p
i
=
σ
ij
a
j
on the boundary
(9.76)
where here (and for the following two equations)
a
j
is the direction cosine of the outer
normal to the boundary. Then from
p
i
=
p
ji
a
j
(9.77)
it follows that
[
−
1
r
,i
r
,j
]
∂
r
p
ji
=
(
1
−
2
ν)δ
+
β
n
−
(
1
−
2
ν)(
r
,i
a
j
−
r
,j
a
i
)
(9.78)
ij
4
απ(
1
−
ν)
r
α
∂
where
∂
r
/∂
n
is the derivative of
r
with respect to the outer normal of the boundary.
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