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substitute these into Eq. (13.60) for harmonic motion (of frequency
ω
) to obtain
d
z
dx
=
Az
+
P
(13.66)
with
0
−
1
0
0
m
νβ
ψ
m
0
0
1
/
K
K
n
m
2
m
1
2
m
A
=
(
n
y
−
ν
n
)β
ψ
+
k
0
0
−
+
ν
K
β
ψ
m
2
4
m
2
+
K
(
1
−
ν
)β
−
ρω
2
m
0
2
K
(
1
−
ν)β
ψ
1
0
m
0
0
L
y
L
y
α
L
y
β
p
m
=
L
y
φ
m
φ
m
dy,
P
=
,
ψ
m
=
p
z
(
x, y, t
)φ
m
dy
−
p
m
0
m
0
0
2 Simply supported-simply supported
α
=
1 Fixed-fixed or fixed-simply supported
Boundary Conditions
ψ
m
m
=
1
,
2
,
...
Case
y
=
0
y
=
L
y
ψ
1
ψ
2
ψ
3
ψ
m
m
large
1
Simply Supported
Simply Supported
1
1
1
1
η
m
−
1
2.9317
η
1
6.0686
η
2
9.2095
η
3
2
Fixed
Simply Supported
η
m
η
m
−
2
2.6009
η
1
5.8634
η
2
8.9984
η
3
3
Fixed
Fixed
η
m
The solution of Eq. (13.66) in transfer matrix and stiffness matrix form is given in Pilkey
(1994). This reference contains matrices for very general plates, including plates with arbi-
trary loading. Also tabulated are mass matrices for rectangular plates.
EXAMPLE 13.3 Simply Supported Rectangular Plate under Uniform Loading
Consider a rectangular plate simply supported on all edges. The dimensions of the plate
are
L
x
=
L
in the
x
d
irection and
L
y
in the
y
direction. The loading on the plate is uniform
with the intensity
p
z
=
L
y
are simply supported, the
state variables can be expressed as in Eq. (13.61), and the loading can also be expanded into
a sine series of Eq. (13.62), with Eq. (13.63) giving
p
0
.
Since the edges at
y
=
0 and
y
=
4
p
0
m
m
=
1
,
3
,
5
,
...
π
p
m
=
(1)
0
m
=
2
,
4
,
6
,
...
These lead to the first order differential relation of Eq. (13.64). Follow the procedure in
Chapter 4 to obtain the transfer matrix solution
U
i
z
0
+
z
i
z
x
=
(2)
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