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2
2
n
∂
w
n
y
∂
y
2
−
∂
w
n
x
∂w
x
−
∂
n
y
∂
y
∂w
−
x
2
−
y
=
0
∂
∂
∂
∂
∂
and
∂
n
y
∂
If
∂
n
∂
are very small, then
x
y
2
2
n
∂
w
n
y
∂
w
−
x
2
−
y
2
=
0
∂
∂
Together with the condition of equilibrium of Eq. (13.30a), the equilibrium of forces in the
z
direction becomes
2
2
∂
q
x
∂
x
+
∂
q
y
∂
n
∂
w
n
y
∂
w
y
−
x
2
−
y
2
=−
p
z
(13.59)
∂
∂
Furthermore, if the plate is on an elastic foundation with the elastic constant
k
, and the
dynamic response is of interest, two terms,
k
2
t
2
, should be added to the right
w
and
ρ∂
w/∂
side of Eq. (13.59). The quantity
in the inertia term is the mass per unit area and
k
is the
modulus of the elasti
c
fou
n
dation in force/length
3
. In the dynamic case, the applied load
should be written as
p
z
=
ρ
p
z
(
x, y, t
).
Then the governing equations of motion become
∂w
∂
x
=−
θ
∂θ
∂
2
x
=
ν
∂
w
m
x
K
y
2
+
∂
n
y
2
m
x
4
2
2
∂
V
∂
)
∂
w
)
∂
w
1
K
K
∂
2
x
=
K
(
1
−
ν
y
4
+
(
n
y
−
ν
n
y
2
−
+
ν
(13.60)
∂
∂
∂
2
w
+
ρ
∂
w
+
k
t
2
−
p
z
(
x, y, t
)
∂
2
∂
m
x
∂
K
∂
θ
x
=−
2
(
1
−
ν)
y
2
+
V
∂
These governing equations can be solved to establish the transfer matrix as well as the
stiffness matrix, of an element, or for special cases, these relationships will provide the
complete solution, as illustrated in the following section.
Responses
Consider a rectangular plate with dimension
L
x
along the
x
direction, and
L
y
along the
other edge. The
y
derivatives of the governing relationships of Eq. (13.58) can be eliminated
by expanding the state variables
z
in a sine series.
w(
x, y
)
w
(
x
)
m
=
∞
θ(
x, y
)
θ
(
x
)
sin
m
π
y
m
z
=
(13.61)
(
)
(
)
L
y
V
x, y
V
m
x
m
=
1
m
x
(
x, y
)
m
m
(
x
)
These expansions correspond to boundary conditions of
w
=
0 and
m
y
=
0 along
y
=
0
and
y
=
L
y
.
Hence, the plate being considered here is simply supported along
y
=
0
and
y
Normally, use of a few terms in the series expansion will suffice to achieve
acceptable accuracy.
=
L
y
.
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