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ln
(
1
/
r
)
=−
ln
r,
to obtain
r
i
a
i
r
2
2
π ω(ξ )
+
q
ln
rdS
−
S
ω
dS
=
0
(13)
S
ξ
where
is a point inside the cross-section and the quantities in the integrands of the line
integrals are referred to point
x
on the boundary.
To process the boundary integrals, move the point
to the boundary, and utilize the
geometry and notation of Fig. 9.3. Use the procedure of Section 9.2.1 to evaluate the contri-
butions to the integral equation of the integration along the semicircle of Fig. 9.3. Then (13)
can be written as
ξ
dS
r
i
a
i
r
2
c
ω(ξ)
+
q
ln
rdS
−
S
ω
=
0
(14)
S
with
2
π
inside the cross-section
c
=
π
on the boundary, if smooth at
ξ
Constant
on the boundary, if not smooth at
ξ
This boundary integral equation can be solved using the methodology of Section 9.2.2.
The torsional constant is found from [Chapter 1, Eq. (1.154)]
x
2
+
dA
x
2
∂ω
∂
x
3
∂ω
∂
x
3
−
J
=
x
3
+
x
2
A
∂
∂
dA
∂
∂
=
I
p
+
x
3
(
−
x
2
ω)
+
x
2
(
x
3
ω)
A
where
I
p
is the polar moment of inertia. Use Gauss' theorem [Eq. (II.8) of Appendix II],
where, for a two-dimensional problem, the volume
V
in Eq. (II.8) becomes the boundary
S
of the cross-section.
∂
∂
dA
∂
∂
x
3
(
−
x
2
ω)
+
x
2
(
x
3
ω)
=
S
(
−
x
2
ω
a
3
+
x
3
ω
a
2
)
dS
A
Thus,
S
ω(
−
=
+
+
)
J
I
p
x
2
a
3
x
3
a
2
dS
(15)
The values of
ω
and its derivatives at point
ξ
inside the cross-section can be calculated
using (13)
r
2
dS
1
2
r
i
a
i
ω(ξ)
=
−
q
ln
r
+
ω
(16)
π
S
r
i
a
i
r
2
dS
∂ω(ξ )
∂ξ
1
2
∂
∂ξ
+
ω
∂
∂ξ
2
=
−
q
ln
r
π
S
2
2
r
4
r
2
r
2
1
2
q
1
r
r
∂ξ
∂
1
∂
∂ξ
r
i
a
i
∂
∂ξ
=
−
2
+
ω
r
i
a
i
−
π
S
2
2
q
r
2
dS
r
2
a
2
r
3
a
2
1
2
+
2
r
2
r
3
a
3
−
=
r
2
+
ω
(17)
π
r
4
S
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