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2
h
3
h
3
y
FIGURE P1.65
Equilateral triangle.
z
y
b
a
FIGURE P1.66
Bar of rectangular cross-section.
z
Hint:
The boundary conditions lead to
c
n
sin
n
z
sinh
n
y
∞
2
b
2
b
ω
=
yz
+
n
=
1
,
3
,
5
...
)
(
n
+
1
)/
2
32
b
2
n
3
1
c
n
=
(
−
1
π
3
cosh
(
n
π
a
/
2
b
)
1.67 Show that for a bar of rectangular cross-section, Fig. P1.66
4
π
5
φ
b
4
∞
16
3
1
n
5
tanh
n
a
2
b
π
φ
b
3
a
M
t
=
G
−
G
n
=
1
,
3
,
5
...
∞
φ
b
16
G
1
n
2
cosh
(
n
π
y
/
2
b
)
sin
n
z
2
b
π
φ
z
)
(
n
+
1
)/
2
τ
xy
=
2
G
+
(
−
1
π
2
cosh
(
n
π
a
/
2
b
)
n
=
1
,
3
,
5
...
∞
φ
b
16
G
1
n
2
sinh
(
n
π
y
/
2
b
)
cos
n
z
2
b
π
)
(
n
+
1
)/
2
τ
=
(
−
1
xz
π
2
cosh
(
n
π
a
/
2
b
)
n
=
1
,
3
,
5
...
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