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0
r
<
a
1
m
0
−
r
a
1
r
p
0
r
(
r
−
a
1
)
r
2
r
a
1
r
r
p
0
r
(
r
−
a
1
)
(
r
)
=
1
1
−
ν
dr
+
dr
r
≥
a
1
(
b
−
a
1
)
(
b
−
a
1
)
4
a
1
r
2
+
p
0
+
ν
45
3
+
ν
16
4
+
ν
36
1
−
ν
80
1
+
ν
12
r
a
1
0
r
3
a
1
r
2
a
1
−
a
1
ln
=−
r
−
a
1
−
+
(
b
−
a
1
)
(9)
Since the inner edge is fixed and the outer edge is free (cantilevered), the boundary condi-
tions are
w
r
=
a
=
w
0
=
0
,
θ
r
=
a
=
θ
0
=
0
,m
r
=
b
=
0
,
and
V
r
=
b
=
0
.
From (1), these conditions
lead to the two equations.
V
0
V
r
=
b
=
V
0
U
VV
(
b
)
+
(
b
)
=
0
(10)
m
0
m
r
=
b
=
V
0
U
mV
(
b
)
+
m
0
U
mm
(
b
)
+
(
b
)
=
0
which can be solved for
V
0
and
m
0
.
Equation (1) is now a fully defined response for this
plate.
For a third example of a plate with a center hole, place a concentrated line load
P
(force/length) at
r
a
1
as shown in Fig. 13.15d. O
n
e approach is to represent the con-
ce
ntrate
d
line load
P
in terms of the distributed load
p
z
using a delta function
=
δ
in the form
p
z
(
r
)
=
P
δ(
r, a
1
)
,
where the delta function is characterized by the sifting property
r
0
if
r
<
a
1
0
f
f
(
r
)δ(
r, a
1
)
dr
=
a
1
=
r
−
a
1
(
a
1
)
(11)
f
(
a
1
)
if
r
≥
a
0
is the unit step function of (5). The solution is given by (1), with the loading
terms (particular solutions)
−
where
r
a
1
r
r
1
r
p
z
r
K
r
r
1
r
Pr
K
δ(
1
r
1
r
0
w
(
r
)
=
dr
=
r, a
1
)
dr
a
a
0
r
<
a
1
=
K
r
r
r
a
1
r
r
P
1
1
r
a
1
dr
r
≥
a
1
a
1
a
1
r
r
r
r
a
1
P
K
1
r
1
r
a
1
dr
0
=
r
−
a
1
a
1
a
1
0
r
<
a
1
=
K
r
r
r
P
1
r
a
1
a
1
ra
1
ln
dr
r
≥
a
1
a
1
0
r
<
a
1
4
K
a
1
2
r
ln
r
dr
=
a
Pa
1
a
1
−
+
≥
r
r
a
1
0
r
<
a
1
=
4
K
r
2
a
1
ln
a
1
−
r
2
a
1
r
Pa
1
r
+
−
≥
a
1
4
K
r
2
a
1
ln
a
1
−
r
2
a
1
0
Pa
1
r
=
r
−
a
1
+
−
(12)
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