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small for low values of κξ, indicating that the motion consists almost entirely of lag,
agreeing with the corresponding uncoupled motion, but that for large values of κξ we
find that β0 /ξ0 is also large, indicating that the motion now consists mainly of flapping,
in agreement with the uncoupled frequency curve which lies close to it. Thus, along
either of the frequency curves of coupled motion the mode changes its character in
the region of the uncoupled cross-over point.
A similar situation arises when flap and lag stiffness are kept constant but the
frequencies change with rotor speed such that there is an intersection of a flap and a
lag mode. For example, the second lag frequency might intersect the frequency curve
of the fourth flap mode, as indicated in Fig. 7.23. Again, we would find that the
lagging motion at low speeds would become mainly flapping motion after passing
the uncoupled cross-over point, and vice versa.
It appears, therefore, that the coupling is important in the region where the frequency
curves of two uncoupled modes would intersect one another.
Fig. 7.22 Effect of stiffness coupling on flap and lag frequencies
Uncoupled lag frequency
Uncoupled flap frequency
Mainly flap
Mainly lag
Mainly lag
Mainly flap
ωβ, ωξ
√(κξ/IΩ2)
Structural dynamics of elastic blades 267
50
40
30
20
10
Natural frequency, Hz
1st lag
1st flap
1st torsion
2nd flap
3rd flap
2nd lag
4th flap
0 50 100 150 200 250 300
10Ω
9Ω
8Ω
4th flap
7Ω
6Ω
1st torsion
5Ω
3Ω
2Ω
3rd flap
2nd lag
4Ω
2nd flap
1Ω
1st flap
1st lag
Rotor speed, rev/min
Fig. 7.23 Variation of blade bending frequencies with rotor speed
7.3 Forced response of rotor blades
7.3.1 Orthogonal property of the normal modes
The mode shapes possess a very important orthogonal property which can be deduced
from the flapwise bending equation, eqn 7.11. Let Sm(x) and Sn(x) be two different
solutions of eqn 7.11, with their associated frequency ratios λm and λn.
Thus, we have
d
d
d
d
– d
d
d
d
– = 0
2
2
2
2
2 22 4
x
EI
S
x
R
x
G
S
x
m m m RS
m m

 

 

λ
Ω
(
7.77)
and d
d
d
d
– d
d
d
d
– = 0
2
2
2
2
2 22 4
x
EI
S
x
R
x
G
S
x
n n m RS
n n

 

 

λ Ω (7.78)
Now multiply eqn 7.77 by Sn and eqn 7.78 by Sm. Consider the first equation:
S
x
EI
S
x
R S
x
G
S
x
n m RSS
m
n
m
d m m n
d
d
d
– d
d
d
d
– = 0
2
2
2
2
2 22 4 
 

 


λ Ω (7.79)
Integrating the first term by parts from x = 0 to x = 1 gives
0
1 2
2
2
2
2
2
0
1
d
d
d
d
d = d
d
d
∫ d 
 

 

 

 

 

 
S
x
EI
S
x
x S
x
EI
S
x n
m
n
m
–
d
d
d
d
d
d
d
0
1 2
∫ 2 
 

 
S
x x
EI
S
x
n m x
268 Bramwell’s Helicopter Dynamics
The term in the square brackets vanishes on account of the boundary conditions
for both the hinged and hingeless blade. Further,
0
1 2
2
2
2
0
1
0
1 2
2
2
2
0
1 2
2
2
2
d
d
d
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(134)