曝光臺 注意防騙 網曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

It can easily be seen that the functions satisfy the boundary conditions of a centrally
hinged blade. Then evaluating the integrals Aij and Bij we get
A A A A
B B B B k
11 12 21 22
11 12 21 22
2
= 1
3
, = = 16
63
, = 1304
6237
= 1
3
= = 16
63
, = 80
21
+ 1850
6237
,
the bars denoting that the integrals have been divided by 12
mΩ2R3, and where k2 =
EI/mΩ2R4. In this case the tension G is 12
mΩ2R3(1 – x2). substituting in eqn 7.16
gives, in matrix form,
1
3
16
63
16
63
+ 1850
6237
–
1
3
16
63
16
63
1304
6237
= 0
2
10
20
2
2
2
2
10
20
80
21
k
























φ
φ
λ
λ
λ
λ
φ
φ
or
1
3
( – 1)
16
63
( – 1)
16
63
( – 1)
– 80
21
– 1850
6237
= 0
2
2
2
2 2
10
20
λ
λ
λ
λ
φ
1304 φ
6237
k












The frequency equation is found by putting the determinant of the square matrix
to zero. If we take k2 = 0.004 as a typical value, the determinant gives
– 1 0.7619( – 1)
– 1 0.8232 – 1.2279
= 0
2 2
2 2
λ λ
λ λ
or (λ2 – 1)(λ2 – 7.5988) = 0
This equation gives λ = 1, a value we should expect since γ1 = x is known to be
an exact solution, and λ = 2.757. We cannot find the absolute amplitudes of the
modes but only the ratio of the amplitudes. Taking the second equation of the matrix
form of eqn 7.16 gives
(λ φ λ φ 1
2
10
2
– 1) + (0.8232 – 1.2279) 20 = 0
When λ = 1, we have φ20/φ10 = 0, i.e. if φ10 ≠ 0, φ20 = 0 indicating that in the
oscillation whose frequency is Ω the mode shape consists entirely of the function
γ = x. This is, again, just what we should expect, as γ = x is an exact solution of the
flapping equation. The higher frequency 2.757Ω gives φ10/φ20 = – 0.759 and the
corresponding mode shape can be written
S2(x) = –3.15x + 4.15x3(10/3 – 10x/3 + x2)
The two modes are shown in Fig. 7.6. The accuracy of the mode shapes and
frequencies improves with the number of functions chosen, the lower modes being
improved the most. In general, the stiffness and mass distribution will be such as to
require numerical integration of the integrals Aij and Bij.
246 Bramwell’s Helicopter Dynamics
(b) The Rayleigh–Ritz procedure
It can be shown1 from the calculus of variations that a function satisfying the fourth
order linear differential equation
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 RS

 

 


λ Ω (7.11)
in association with either set of boundary conditions for the hinged or hingeless
blade, is a function which gives a stationary value to the integral
L
R
m RS EI S
x
R G
S
x
= 1 x
2
– d
d
–
d
d
d
0
1
2 2 4 2
2
2
2
2
2 ∫ 
 

  


 

 
λ Ω (7.17)
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(123)