Equilibrium about the Lead-Lag Hinge

The equilibrium of the blade about the lead-lag hinge is also determined by a balance of centrifugal and aerodynamic moments. In this case, the aerodynamic moments are generated by the aerodynamic drag of the blade as it rotates. The lag angle, £, is defined as positive in the lagging direction (Fig. 4.4). The centrifugal force on the blade element is

Подпись: (4.15)d(FcF) = mQ2y dy

(4.17)

Equilibrium about the Lead-Lag HingeПодпись: (4.16)

Equilibrium about the Lead-Lag Hinge

The aerodynamic forces acting on the blade in the plane of rotation include both induced drag and profile drag components. To keep the following analysis as simple as possible, it will be assumed that the resultant of all these drag forces is denoted by Fq and acts at a distance yD from the lead-lag hinge. The equation of equilibrium is therefore

Like the coning angle, this equation indicates that, for a constant aerodynamic loading, the mean lag angle is inversely proportional to both blade mass and to S32.

Equilibrium about the Lead-Lag Hinge Подпись: (4.20)

Again, in practice, like the flapping hinge, the lead-lag hinge is also offset from the rotational axis. Notice that the lag hinge offset may be different from the flap hinge offset. In this case, the centrifugal force as a result of lag about the offset hinge is

Equilibrium about the Lead-Lag Hinge(4.21)

Equilibrium about the Lead-Lag Hinge

Therefore, the centrifugal moment about the hinge is

Подпись: (4.22)MQ2teR( l+e)
2

where the blade mass is mR(l — e) and this shows that the centrifugal force acts at a distance R( 1 + e)/2 from the rotational axis (i. e., at the center of gravity of the blade, ycg).

The resultant of the aerodynamic forces because of blade drag is denoted by Fq and acts at a distance у о from the hinge axis. The shear force on the lead-lag hinge because of the aerodynamic and centrifugal forces must be equal to the shaft torque, Q, divided by the hinge offset, eR. The equation of force equilibrium is, therefore,

CR О

Fd cosf — / mO,2y sin £ dy = —. (4.23)

JeR eR

This equation indicates that, for a given rotor, the mean drag angle of the blades will essentially be a function of Q/&2.