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

Let us suppose that the solution of eqn 7.83 to a known blade loading can be
expressed as
Z R S x
n
= n( )n( )
=1
Σ ∞
φ ψ (7.84)
where the terms Sn(x) are the blade mode shapes calculated from solving eqn 7.11 but
the terms φn(ψ) have yet to be determined. Substituting eqn 7.84 into eqn 7.83 gives
Σ ∞

 

 



 

 
n
n
n n
x
EI
S
x
R
x
G
S
=1 x
2
2
2
2
( ) 2
d
d
d
d
– d
d
d
d φ ψ
+
d
d
2 4 =
=1
2
2
m R S R2 F
n x
n Ω Σ
∞ φ
ψ
∂
∂
(a) (b)
Fig. 7.24 Possible deflections of the teetering rotor
270 Bramwell’s Helicopter Dynamics
But from eqn 7.77 the terms in the large bracket can be replaced by
m R S
n
n n Ω Σ
∞
2 4
=1
λ 2
giving
Σ ∞

 

 
n Ω
n
n n Sn x
m R
F
=1 x
2
2
2
2 4
d
d
+ ( ) ( ) = 1
φ
ψ
λ φ ψ ∂
∂ (7.85)
Multiplying eqn 7.85 through by mSm(x), integrating from 0 to 1, and applying the
orthogonal properties as defined by eqn 7.82, gives
d
d
+ = 1
( )
( )d
2
2
2
2 2
0
φ 1
ψ
λ φ ∂
∂
n
n n n R f n
F
x
S x x
Ω ∫ (7.86)
where f(n) =
0
1
∫ mSn2(x)dx is the generalised mass or inertia term for the nth mode.
Thus, the terms φn of eqn 7.84 are determined from the solutions of eqn 7.86.
Unfortunately, although the orthogonal properties ‘decouple’ the inertia and elastic
terms, represented by the left-hand side of eqn 7.83, the blade loading ∂F/∂r depends
on the blade deflections which will, in general, contain all the blade modes of motion.
Otherwise, each azimuth co-ordinate φn could be calculated individually from
eqn 7.86. We shall discuss means of solving eqn 7.86 later in the chapter.
We may also derive the forced response equation from energy considerations.
It is convenient to express the energy forms in terms of the blade axes, which are
non-Newtonian due to the rotation present. However, energy forms relative to these
axes may be used in the same way as those for absolute energy forms, provided the
inertia forces resulting from the rotation are treated as normal external forces14,15 in
developing energy forms. In the present case, the inertia forces acting on a mass
element in the blade are the centrifugal force and the Coriolis force. The former is
normally considered to provide a potential energy term arising from the centrifugal
effect, which can then be added to the potential or strain energy from elastic forces;
the latter is usually removed to the right-hand side of the equations of motion,
together with the other external, i.e. aerodynamic, forces.
The kinetic energy T of the mass elements of the blade relative to axes rotating
with the blade is
Τ ∂
∂ = 12
d
0
R 2
m
Z
t
∫ r 
 
 
= 12
2 3 d
0
1 2
Ω 
 
 
R ∫m
z
x
∂
∂ψ (7.87)
m being, as usual, the mass per unit length of the blade.
Assuming as before
Structural dynamics of elastic blades 271
z S x
n
= n( )n( )
=1
Σ ∞
φ ψ
we have
∂
∂ψ
φ
ψ
z
S x
n
n
n = ( )
d
=1 d
Σ ∞
so that substitution in eqn 7.87 we find
Τ φ
ψ
φ
ψ = 12
d
d
d
d
2 3 d
=1 =1 0
1
Ω Σ Σ
∞ ∞ R ∫mS S x
m n
m n
m n
= 12
d
d
2 3 ( )
=1
2
Ω 


Σ ∞
R fn
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(136)