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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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