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Similarly, multiply both sides of Eq. (13.76) by sin
n
φ
and integrate and note the relationship
2
π
π
m
=
n, m
>
0
sin
m
φ
sin
n
φ
d
φ
=
0
m
=
n
0
Then
2
π
1
π
p
s
m
(
r
)
=
p
z
(
r,
φ)
sin
m
φ
d
φ
m
>
0
(13.77c)
0
Several properties of these series are of interest. For symmetrical response of a plate,
m
is equal to zero, and the series expansions reduce to
0
0
w(
r,
φ)
=
w(
r
)
=
w
(
r
)
,
θ(
r,
φ)
=
θ(
r
)
=
θ
(
r
)
,
m
0
(
V
0
(
m
r
(
r,
φ)
=
m
r
(
r
)
=
r
)
,
V
(
r,
φ)
=
V
(
r
)
=
r
)
p
c
m
(
p
z
(
φ)
=
p
z
(
)
=
)
r,
r
r
If
p
z
(
r,
φ)
is an odd function of
φ
, i.e., if
p
z
(
r,
φ)
=−
p
z
(
r,
−
φ)
,
then
p
c
m
(
r
)
=
0
for
m
=
0
,
1
,
2
,
...
,
and Eq. (13.75) reduces to a sine series. For even functions of
φ
, i.e.,
p
z
(
r,
φ)
=
p
z
(
r,
−
φ).
p
s
m
(
r
)
=
0
for
m
=
0
,
1
,
2
,
...
,
and Eq. (13.75) contains only cosine terms.
The Fourier series expansions for
w
and
p
z
placed in Eq. (13.69b) lead to
d
2
dr
2
+
r
2
d
2
r
2
w
m
m
2
m
2
1
r
d
dr
−
w
m
dr
2
1
r
d
w
m
dr
p
j
m
+
−
=
(13.78a)
=
=
...
j
c
or
s,
m
0
,
1
,
2
,
which is readily shown to have the complementary functions
C
3
r
2
C
4
r
2
ln
r
m
=
0:
w
m
=
C
1
+
C
2
ln
r
+
+
C
3
r
3
m
=
1:
w
=
C
1
r
+
C
2
/
r
+
+
C
4
r
ln
r
(13.78b)
m
C
1
r
m
C
2
r
−
m
C
3
r
2
+
m
C
4
r
2
−
m
m
≥
2:
w
=
+
+
+
m
where the arbitrary constants are different for each
m
. With the addition of particular so-
lutions, these functions can be used to find the response of a circular plate with arbitrary
loading.
Insertion of the series expansions for
z
(Eq. 13.75) and
p
z
(Eq. 13.76) in
∂
r
z
=
Az
+
P
of
Eq. (13.74) leads to ordinary governing differential equations
d
z
dr
=
Az
+
P
(13.79a)
or
j
m
dr
=−
θ
d
w
m
j
=
c
or
s,
m
=
0
,
1
,
2
,
3
,
...
m
dr
m
j
m
m
2
r
2
d
θ
K
−
r
θ
−
ν
j
m
j
m
=
w
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