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– 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)
and
d
d
+ = 0
2
2
2 φ
ψ
λ φ (7.12)
The boundary conditions to be satisfied are as follows.
(a) Hinged blade (including flapping hinge offset)
At x = e (at flapping hinge)
S = 0
d2S/dx2 = 0 (zero bending moment)
At x = 1 (blade tip)
d2S/dx2 = 0 (zero bending moment)
d3S/dx3 = 0 (zero shear force)
(b) Hingeless blade
At x = 0
S = 0
dS/dx = 0 (zero slope)
Structural dynamics of elastic blades 241
At x = 1
d2S/dx2 = 0
d3S/dx3 = 0
Equation 7.7 will be a solution of eqn 7.6 provided S(x) and φ (ψ) satisfy eqns 7.11
and 7.12 and the appropriate boundary conditions.
There is an infinite number of solutions of eqn 7.11 since λ is not a fixed number
but can be adjusted to satisfy the equation in association with the appropriate boundary
conditions. These solutions represent the blade shape and are called normal modes on
account of the orthogonal property to be proved in section 7.3.
Equation 7.12 is the equation of simple harmonic motion, and the infinite number
of discrete values of λ determine the frequencies ω = λΩ of the corresponding mode
shape. λn is therefore the ratio of the blade natural frequency ωn to the shaft rotational
frequency Ω for the nth mode shape.
It can easily be verified that when there is no flapping offset, i.e. e = 0, a solution
of eqn 7.11 is S(x) = x, with λ = 1, that is, the first normal mode shape of the centrally
hinged blade is a straight line whose flapwise frequency is exactly equal to the
rotational frequency of the shaft. This is the ‘rigid’ blade shape already assumed in
the previous chapters. If e ≠ 0, the mode shape is not exactly a straight line, although
for typical values of e it is very close to it.
Methods for calculating the blade mode shapes and frequencies are dealt with
below. Figure 7.3 shows the first four shapes for a typical helicopter blade.
Fig. 7.3 Mode shapes of typical helicopter blade
1
0
–1
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1
0
–1
S3(x), λ3 = 4.60
S4(x), λ4 = 7.18
S1(x), λ1 = 1
r/R
r/R
S2(x), λ2 = 2.58
242 Bramwell’s Helicopter Dynamics
7.2.2 Calculation of blade mode shapes and frequencies
We must now attempt to solve eqn 7.11 in order to obtain the blade mode shapes Sn(x)
and associated frequencies λnΩ. In general, both the blade elastic stiffness EI and
mass distribution m will be complicated functions of the radial station x, and it is
obvious that a simple analytical solution of the blade bending equation is out of the
question. Typical spanwise variations of stiffness and mass for a uniform blade are
shown in Fig. 7.4.
It can be seen that over the greater part of the blade the stiffness and mass distributions
are practically constant, but that near the blade root large and often discontinuous
changes occur because of the root attachment.
Now it is well known that for a uniform non-rotating beam there is an exact closed
solution of the mode equation and that other exact solutions exist for some particular
mass and stiffness distributions. However, there is no known closed solution for the
rotating beam, even for the apparently simple case of constant mass and stiffness, the
equation of which is given below, eqn 7.13:
k z
x x
x zx
n z
2
4
4
d 2 2
d
– 12
d
d
(1 – ) d
d
[ ] – λ = 0 (7.13)
where k2 = EI/mΩ4 is constant.
The mode shapes and frequencies of a rotating uniform blade for k2 = 0.0055 have
been calculated and are shown in Fig. 7.5. The broken lines show the mode shapes
and frequencies for a non-rotating blade of the same thickness (standard results for
a uniform vibrating beam). It is perhaps a little surprising to note that the effect of
rotation is quite small, which suggests that the non-rotating mode shapes might serve
as useful approximations in the calculation of the rotating modes.
The lack of an exact solution of eqn 7.13 is unfortunate because, although the
mass and stiffness distributions might not be representative of a practical blade, it
　
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