曝光臺 注意防騙 網(wǎng)曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

would be very useful for testing the accuracy of approximate methods of solution.
Flapwise stiffness EI
Mass per unit length
Spanwise co-ordinate Spanwise co-ordinate
Fig. 7.4 Mass and stiffness distributions of a typical uniform blade
Structural dynamics of elastic blades 243
The approximate methods usually used fall under two main headings:
(i) method of assumed modes,
(ii) method of lumped parameters.
7.2.2.1 Method of assumed modes
This method consists of choosing a finite sequence of functions which, preferably,
approximate to the expected mode shapes and satisfy the appropriate boundary
conditions. This latter requirement is not essential, however, since it is always possible
to include restraint conditions in the analysis.
We shall consider three methods employing assumed modes:
(a) Lagrange’s equations
(b) Rayleigh–Ritz procedure,
(c) Galerkin’s method.
(a) Lagrange’s equations
Let the sequence of functions which are to be used to approximate to the blade shape
be
γ1(x), γ2(x), …, γ
i(x), …, γn(x)
and let the displacement be
z x x
i
n
( , ) = i( )i( )
=1
ψ Σγ φψ
1
0.5
0
–0.5
S1(x)
0.2 0.4 0.6 0.8r/R 1 0.2 0.4 0.6 0.8 1
1
0.5
0
–0.5
–1
S3(x)
0.2 0.4 0.6 0.8 1
1
0.5
0
–0.5
–1
S2(x)
λ1 = 1
λ3 = 5.59
λ2 = 2.75
λ2 = 1.15
r/R 0.2 0.4 0.6 0.8 1
1
0.5
0
–0.5
–1
S4(x)
r/R
λ4 = 9.79
Rotating (k2 = 0.0055)
Non-rotating
r/R
Fig. 7.5 Mode shapes of rotating and non-rotating uniform blade
λ3 = 3.72–
λ4 = 7.77–
244 Bramwell’s Helicopter Dynamics
It will be assumed that the functions γ
i(x) satisfy the boundary conditions.
Then
∂z∂ψ γ xφ x
i
n
/ = i( )i( )
=1
Σ ′
and the kinetic energy T in term of blade axes (see section 7.3.2) is therefore
T R m x
i
n
j
n
i j i j = 12
2 3 d
=1 =1 0
1
Ω Σ Σ φ ′φ ′ ∫ γ γ (7.14)
=
=1 =1
Σ Σ ′′ ′′
i
n
j
n
Aijφiφj
where A R m x ij i j = 12
2 3 d
0
1
Ω ∫ γ γ
is a generalised mass or inertia coefficient.
It should be noted that the orthogonal properties do not apply to the integral of
eqn 7.14 since the functions γ
i(x) themselves are not exact solutions of the blade
bending equation, eqn 7.11.
The strain and potential energy U is, as will be seen from eqn 7.92.
U
R
El
d
x x
R G
x x
x
B
i
n
j
n
i j
i j i j
i
n
j
n
ij i j
= 1
2
d
d
d
+
d
d
d
d
d
=
=1 =1 0
1 2
2
2
2
2
=1 =1
Σ Σ
Σ Σ
∫ ⋅

 

 
φ φ γ γ γ γ
φ φ
where B
R
EI
x x
R G
x x
ij x
i j i j = 1
2
d
d
d
d
+
d
d
d
d
d
0
1 2
2
2
2
2 ∫ 
 

 
γ γ γ γ
is a generalised stiffness coefficient (including both structural and centrifugal stiffening
effects.
Then applying Lagrange’s equations eqn 7.95, with ∂W/∂φn = 0, gives
Σ ′′ Σ
j
n
ij j
j
n
A Bij j
=1 =1
φ + φ = 0 (7.15)
If φ
j = φ
j0 sin λ
jψ, eqn 7.15 becomes
Σ
j
n
Bij jAij j
=1
2
( – λ )φ0 = 0 (7.16)
As a simple example, let us calculate the first two mode shapes and frequencies of
a centrally hinged uniform blade and suppose that the deflection amplitude can be
expressed as
γ0 = φ10γ1 + φ20γ2
where
γ1(x) = x and γ2(x) = 10x3/3 – 10x4/3 + x5
Structural dynamics of elastic blades 245
　
中國航空網(wǎng) www.k6050.com
航空翻譯 www.aviation.cn
本文鏈接地址：Bramwell’s Helicopter Dynamics(122)