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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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