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 is possible to derive some valuable conclusions by simpler approaches. We shall
describe first an approach given by Mil13.
Let us suppose that the chassis mode consists of lateral oscillations of amplitude
y0 and frequency p in the form
y = y0 sin pt (9.42)
Then, according to eqn 9.34 the motion of the blade is given by
ξ˙˙k κ δξ˙k κ ξk ψk
y
l
+ 2 + = 0 p sin pt cos
b
Ω 2Ω2 2
= 12
y0 sin ( + ) + 2 + sin ( – ) – 2
l
p p t k
b
p t k
b Ω Ω π π { } { } 


(9.43)
Equation 9.43 represents a second order system excited by a pair of simpleharmonic
forcing functions. The solution must therefore be of the form
ξk = ξ1 sin [(p + Ω)t + ψ1] + ξ2 sin [(p – Ω)t + ψ2] (9.44)
where ξ1, ξ2, ψ1, ψ2 are amplitudes and phase angles which can be determined from
the known response of such systems, Appendix A.4. Now, the blade motion would be
expected to reach large amplitudes if one of the forcing frequencies lies close to the
natural frequency of the lagging motion of the blade. Let us first consider resonance
in the case
κΩ = |p – Ω|
The amplitude of the blade oscillations will then be determined almost entirely by
the second of the forcing terms on the right-hand side of eqn 9.43 and the response
can be written approximately as
Aeroelastic and aeromechanical behaviour 349
ξk = ξ0 cos [(p – Ω)t – 2πk/b] (9.45)
where
ξ0 = – y0p2/4δκΩ(p – Ω)
Now, from eqn 9.30 the centre of gravity of the blades when the chassis mode
consists of oscillations in the form given by eqn 9.41 is
y y r b t kb
k
b
rg g =0 k
–1
= + ( / ) Σ ξ cos (Ω + 2π/ )
Substituting for ξk from eqn 9.45 and expressing the products of the cosine terms
as sums gives
y y r b pt pt kb
k
b
rg g 0 =0
–1
= + (ξ/2 ) Σ [cos + cos {(2Ω – ) + 4π/ }]
From an analysis similar to that of Appendix A.3, we can easily show that
Σ k
b
p t k b
=0
–1
cos [(2Ω – ) + 4π / ] = 0
so that
yrg r pt
12
= gξ0 cos
The inertia force Pin acting on the chassis due to the oscillating centre of gravity
of the blades is given by
Pin = – bMb ˙y˙rg
= – –
8 ( – )
b cos
b 0
4
g
b
bM y
bM y p r
l p
˙˙ pt
δ κΩ Ω
and from eqn 9.35 the equation of motion of the chassis is
(M + bMb)˙y˙ + 2 c c (M + bMb )y˙ + (M + bMb ) c
δ κΩ κ2Ω2
= –
8 ( – )
cos b 0
4
b
bM y p r
l p
pt g
δ κΩ Ω
or
˙y˙ y˙ y
y p r
l p
+ 2 + = – pt
8 ( – )
c c c cos
2 2 0
4
g
b
δ κ κ
μ
Ω Ω δ κΩ Ω
But the solution to this equation must correspond to the motion assumed originally,
eqn 9.41 and this requires that
p = κcΩ
and
δδ μκ
c κ κ
c
2
c
g
b
=
8 (1 – )
⋅
r
l
(9.46)
350 Bramwell’s Helicopter Dynamics
remembering also that
κΩ = |κ c – 1 | Ω
From these three relationships the critical rotor speed may be calculated. To ensure
that the oscillations at resonance will be no worse than neutrally damped, the product
of the dampings of the chassis and of the blade should satisfy the relationship eqn 9.46.
If we now consider the other possible resonant state represented by
κ Ω = | p + Ω |
we find that the above formulae remain the same but that Ω is everywhere replaced
by – Ω. Then, according to eqn 9.46 it appears that for undamped oscillations δδc
would have to be negative, i.e. that either δ and δc would have to be negative. Since
this is impossible, we conclude that ground resonance does not occur for κ Ω =
| p + Ω |.
Referring to eqn 9.46 again, and since δδc cannot be negative, undamped vibrations
can only occur if Ω > p. In Figs 9.19 and 9.20 this is satisfied only by point C, i.e.
the intersection of the branch of negative (regressive) values of κ – 1 with the chassis
　
中國航空網 www.k6050.com
航空翻譯 www.aviation.cn
本文鏈接地址：Bramwell’s Helicopter Dynamics(172)