Blade Loading in Hover

There are two significant forces on a blade in hover: aerodynamic force and centrifugal force. We have already assumed that the blades are perpendicular to the axis of rotation. The centrifugal force thus acts directly along the blade, i.e., it is a pure tension force. There are three aerodynamic forces: lift, drag, and pitching moment. The pitching moment disappears when the airfoil is symmetric, and so we make that assumption. (The only stipulation is that the spar must be located at the point about which the moment is zero, typically about 1/4 of the chord back from the leading edge.) Drag force is usually very small compared to lift force; while important for performance issues, it is not so important in structural considerations. It is also somewhat more difficult to calculate than lift. For this first-order preliminary design, drag is considered negligible.

In a panel, the constant tension force is approximately the average centrifugal force in the panel. Distributed centrifugal force is given by

$$dF_c = \rho A \Omega^2 x \ dx,$$

where $F_c$ is centrifugal force, $\rho$ is the density of the blade material, $A$ is the cross-sectional area of the blade (not including the empty space in the cells), and $\Omega$ is the rotor rotational speed. For a panel, this integrates to

$$F_c = \frac12\rho\Omega^2A(x_o^2 - x^2)+F_{c_o},$$

where $x_o$ is the $x$-coordinate of the outer edge of the panel, and $F_{c_o}$ is the resultant centrifugal force at the outer edge of the panel. $F_{c_o}$ for the outer panel is zero; for other panels, $F_{c_o}$ matches the $F_c$ at the inside edge of the panel to its outside. The average $F_c$ over the panel (which is used for the beam axial force) is

$$N_x = \overline{F_c} = \frac{\int_{x_i}^{x_o} F_c \ dF_c}{x_o-x_i} =\frac16\rho\Omega^2A(2x_o^2-x_ox_i+x_i^2) + F_{c_o},$$

where $x_i$ is the $x$-coordinate of the inner edge of the panel.

The shear force in a panel is due to the lift force. For simplicity, we assume the lift distribution is constant. (This is a somewhat dubious assumption. It is possible to calculate the lift force on a panel from the current twist angle; however, this requires an iterative solution at two levels for it to be accurate.) The shear force due to lift, as a function of $x$, is

$$F_s = \frac T{N_b}(1-x),$$

where $T$ is the rotor thrust (which for the V-22, is half the gross weight or drag, because there are two rotors), and $N_b$ is the number of blades per rotor. The average shear force over a panel is

$$\overline{F_s} = \frac T{N_b}\left(1-\frac{x_o-x_i}2\right).$$

By definition, lift always acts perpendicular to the relative wind. Thus, for a blade at an angle-of-attack, the lift has components in both the $y$- and $z$-directions. To determine the component of lift in each direction, we need the rotor downwash, given by: [4]

$$v=\sqrt{\frac T{2\pi\rho_a R^2}},$$

where $\rho_a$ is the atmospheric density. Then, the absolute angle of attack is

$$\alpha=\theta - \frac v{\Omega x},$$

where $\theta$ is the local pitch angle (i.e., the angle of twist at that point, relative to horizontal). Then, the $y$- and $z$-components of lift (which are the beam shear forces) are

$$Q_y = -\overline{F_s}\cos\alpha\qquad Q_z = -\overline{F_s}\sin\alpha.$$

In an ordinary beam with a free end, a bending moment accompanies the shear force. However, in a helicopter blade, the bending moment is balanced by the component of centrifugal force normal to the blade. Both lift and centrifugal force are distributed loads, meaning that the blade need not carry the bending moments very far. For this first-order design, the bending moment is considered negligible. Although we assumed that centrifugal force acts along the blade, we acknowledge its effect on the bending moment.

The final two generalized forces, the twisting moment and the generalized warping force, are zero, as neither lift nor centrifugal force contribute to them.