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then the required deflection δ is given by
Sn+1
Mn+1
Gn+1 Zn+1 ln Zn
Sn
Gn
Mn
Fig. 7.14 Forces on blade element
Structural dynamics of elastic blades 255
δ = ( / )d
0
R
∫ MM1 EI r
Let us consider the contribution to this deflection due to one of the blade elements.
Let MA and MB be the moments at the ends of the element due to the applied loading,
and let MA1 and MB1 be the moments due to the unit load, Fig. 7.15, at which it is
required to know the deflection. If we suppose that these bending moments vary
linearly across the element, the contribution δAB to the deflection can easily be
calculated and expressed conveniently in matrix form as
δ AB
AB
= ∫ (MM1/EI )dr
= [ , ]
3 6
6 3
M1 M1
l
EI
l
EI
l
EI
l
EI
M
M
A B
A
B












where l is the length of the element.
If a number of successive elements is now considered, with the corresponding
moment distributions MA1, MB1, …, MN1, … and MA, MB, … , MN, … , the deflection
due to all these elements is
δ = [MA1, MB1; … , MN1, … .]
·
3 6
0 0 0
6 3
+
3 6
6 3
+
3 6
:
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
M
M
M
A
B
N
0 0
0 0
0 0 0





 





M (7.38)
To define the deformation of the complete blade, we need to know the deflections
MA1
MA
MB1
MB
1
Fig. 7.15 Moment due to application of unit load
.............................................................................................................
256 Bramwell’s Helicopter Dynamics
at the ends of each of the elements. This will require repeating the above calculations
using the bending moment distributions corresponding to unit loads applied at each
station. When this is done, the row matrix [MA1, MB1], …] then becomes a square
matrix M1, say, the columns of which give the bending moment distributions due to
each of the applied unit loads. The centre matrix of eqn 7.38 is called a flexibility
matrix and will be denoted by f. The total set of deflections is then given by
Z MTfM
E 1 = (7.39)
where ZE is the matrix of the values of δ. Eliminating M between eqns 7.37 and 7.39
gives
ZE = ω2a*(ZE + ZR) + Ω2b*(ZE + ZR) (7.40)
where a* and b* are matrices resulting from the addition and multiplication of
previously defined matrices. Rearranging eqn 7.40 gives
(I – Ω2b*)ZE = ω2a*(ZE + ZR) + Ω2b*ZR
giving ZE = ω2[I – Ω2b*]–1a*(ZE + ZR) + Ω2[I – Ω2b*]–1b*ZR
or ZE = ω2c(ZE + ZR) + Ω2dZR (7.41)
where c and d are square matrices.
The hingeless rotor
If no flapping hinge is present, ZR = 0, giving
ZE = ω2cZE
or (c – I/ω2)ZE = 0 (7.42)
The eigenvalues and eigenvectors of eqn 7.42 give the required frequencies and
mode shapes.
The hinged rotor
When there is a flapping hinge, the moment at the hinge line must be zero and from
eqn 7.37 we have
Mh = ω2ah(ZE + ZR) + Ω2bh(ZE + ZR) = 0 (7.43)
the suffix h indicating that the row of the matrix corresponding to the hinge line is
used. Substituting the value of ZE given by eqn 7.41 into the second bracket of eqn
7.43 gives
ω2ah(ZE + ZR) + Ω2bh[ω2c(ZE + ZR) + dZR + ZR] = 0
or, on rearranging,
ω2(ah + Ω2bhc)(ZE + ZR) + Ω2bh(I + d)ZR = 0 (7.44)
Now the unknowns in ZR are linearly related through an unknown rotation α about
the flapping hinge, i.e.,
Structural dynamics of elastic blades 257
ZR = eα (7.45)
　
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