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

where e is a column matrix giving the distances of the ends of the blade elements
from the flapping hinge. Then substituting eqn 7.45 into eqn 7.44 gives
Ω2bh(I + d)eα = – ω2(ah + Ω2bhc)(ZE + ZR)
or
α = – ω2[Ω2bh(I + d)e]–1(ah + Ω2bhc)(ZE + ZR)
so that eqn 7.45 can be written
ZR = – ω2e[Ω2bh(I + d)e]–1(ah + Ω2bhc)(ZE + ZR) (7.46)
Substituting for ZR in eqn 7.41 gives
ZE = ω2[c – de{Ω2bh(I + d)e}–1(ah + Ω2bhc)](ZE + ZR) (7.47)
and on adding eqns 7.46 and 7.47 we get
(ZE + ZR) = ω2[c – (I + d)e{Ω2bh(I + d)}–1(ah + Ω2bhc)](ZE + ZR) (7.48)
Defining c* to be the matrix in the square brackets, we have
ZE + ZR = ω2c*(ZE + ZR)
or
[c* – I/ω2](ZE + ZR) = 0 (7.49)
The eigen values and eigen vectors of c* yield the required frequencies and mode
shapes of the hinged blade.
The above method need not be confined to straight elements. For a given number
of elements, the accuracy can be greatly improved by taking curved elements – for
example, a cubic variation – the curvature being determined by the deformation of
neighbouring elements. The relationships eqns 7.37 and 7.39 have to be modified
accordingly, but the net result is a considerable increase of accuracy for only a slight
increase of computer time.
7.2.2 Lagwise bending
When considering motion of the blade in the lagging plane*, i.e. in the plane of
rotation, we have to note that the centrifugal force on a blade element is directed
radially outwards from the hub and therefore, unlike the flapping case, the direction,
as well as the magnitude of the force, varies along the blade. Consider the forces,
inertial and aerodynamic, acting on a given element of the blade and their moment
about another point P of the blade, Fig. 7.16. If r1 is the radial distance of the element
and r the radial distance of the point P from the hub, the moment of all the forces
about P is
*Lagging and chordwise motion are the same only if the blade chord is parallel to the plane of
rotation, i.e. when the pitch angle is zero.
258 Bramwell’s Helicopter Dynamics
M EI Y
r
m r r r r m r Y Y r
r
R
r
R
= = ( – ) sin d – ( – ) cos d
2
2
2
1 1 1 1
2
1 1 1 1
∂
∂
∫Ω α ∫Ω α
– ( – )d
2
1
2 1 1
r
R
m
Y
t
∫ r r r ∂
∂
(7.50)
Differentiating with respect to r we have, after cancelling certain terms,
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
M
r
m Y r
Y
r
m r r m
Y
t
r
m Y r
Y
r
G r m
Y
t
r
r
R
r
R
r
R
r
R
r
R
= – d + d + d
= – d + ( ) + d
2
1 1
2
1 1
2
1
2 1
2
1 1
2
1
2 1
∫ ∫ ∫
∫ ∫
Ω Ω
Ω
Differentiating again gives
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
2
2
2
2
2
2
2
2
M = = – – 2
r r
EI
Y
r r
G
Y
r
m
Y
r
Y

 

 

 

  

 
Ω
or
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
2
2
2
2
2
2
– + – 2 = 0
r
EI
Y
r r
G
Y
r
m
Y
r
Y

 

 

   

 

 
Ω (7.51)
where it is understood that the value of EI refers to lagwise bending.
To discuss the free lagging motion of the blade, assume a solution
Y = RT(x)χ(t)
where T(x) is a function of x alone and χ(t) is a function of t alone. By the same
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(129)