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r
r
1
r
p
z
r
K
1
r
0
θ
(
r
)
=−
dr
a
0
r
<
a
1
2
K
r
ln
a
1
r
2
r
r
2
=
(13)
Pa
1
r
1
−
a
1
−
−
≥
a
1
r
1
r
0
r
<
a
1
V
0
(
)
=−
=
r
p
z
rdr
P
a
r
−
≥
r
a
1
a
r
p
z
rdr
r
2
r
a
r
1
r
p
z
rdr
1
r
1
−
ν
m
0
(
r
)
=−
+
(14)
a
<
0
r
a
1
Pa
2
r
2
r
−
2
1
=
(15)
a
1
a
1
−
1
−
(
+
ν)
−
≥
1
ln
a
1
The inner and outer rims are fixed and the boundary conditions
w
r
=
a
=
w
0
=
0
,
θ
r
=
a
=
θ
0
=
0
,
w
r
=
b
=
θ
r
=
b
=
0 imposed on (1) lead to
0
w
=
V
0
U
(
b
)
+
m
0
U
(
b
)
+
w
(
b
)
=
0
r
=
b
w
V
w
m
(16)
0
θ
=
V
0
U
θ
V
(
b
)
+
m
0
U
θ
m
(
b
)
+
θ
(
b
)
=
0
r
=
b
Substitution into (1) of
V
0
and
m
0
,
which are obtained from (16), provides a fully defined
response.
An alternative approach for handling concentrated applied line loads (applied force or
moment) is to take advantage of the similarity between an in-span li
ne
load and the initial
stress resultants
V
0
and
m
0
of (1). For example, the ap
pli
ed line load
P
has the same effect
on the response as the initial shear force
V
0
.
O
f course
P
of Fig. 13.15d is applied at
r
=
a
1
,
while
V
0
occurs at
r
Also, whereas
P
acts downwards,
V
0
is (Fig. 13.14) directed
upwards. Thus, from (1), the influence of
V
0
on the deflection
=
a
.
w
is
a
4
K
r
a
+
(
a
2
r
2
r
2
a
2
+
−
(
+
)
−
)
V
0
ln
(17)
while the contribution of the applied load
P
to the deflection would be
a
1
−
a
1
+
r
2
ln
a
1
+
r
2
a
1
4
K
r
0
−
P
−
=
w
(18)
It is evident from (12) that this is the same result found by integration. Similar reasoning
leads to the expressions for
0
,V
0
,
and
m
0
θ
.
13.3.3 Variational (Integral) Relationships
The internal virtual work is given by
A
δ
T
s
dA
δ
W
i
=−
with
s
D
u
u
taken from Eq. (13.34). Supplement this
with the external virtual work to find the displacement form (Eq. 2.58)
=
E
taken from Eq. (13.35) and
=
u
T
k
D
u
u
T
p
ds
δ
W
=
δ
W
i
+
δ
=−
A
δ
(
−
p
V
)
+
S
p
δ
=
W
e
dA
0
(13.82)
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