164
Chapter 3
(c)
Substituting
f = ~7 " -- 11.6436 x 10
-3
Now the torsional frequency is
f = 3.69714[ ]G~3g 1/2
1/2
The value is to be greater than 200 Hertz so constraint 1,
torsional frequency is
200Hz[
L ]
3.69714
~ < 1
Constraint 1 torsional frequency
F L -]1/2
54.0958 l~-~-~] _<1
(3.50)
Bending frequency constraint [3.5]
CO 1 [g]
1/2
fa -- 2g -- 2~ L37J
where 6st is
WL 3
fiST -- 3EI
Substituting
L F 3(386"4)1 l/= FE/I 1/2
fa = 2~ L 2 Ib J LL3j
[e’]
fa = 3.83164
The I is half of J
fe = 3.83264[~33q
1/2
LL J
fe is to be greater than 200 hertz
200Hz r L3 ]1/2<
3.83164,/TIE----~I - 1
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