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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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