Static Aeroelasticity: Control Effectiveness

Question 1

Hoa can the stability derivatives CLβ and CMβ be found?

Answer 1

There are graphs which show the relation between these derivatives and CLα, which depend on E = flap chord/total chord. Typically Cmβ < 0.

Question 2

Make some remarks about the twist due to flap rotation.

Answer 2

*Theta depends on the flap deflection
* The denominator is the effective stiffness k = Ks - qKa
* The derivatives of the numerator are physically opposite: Rotating the flap downward, there is an increase of lift generated by the flap and a consequent negative pitching moment.

Question 3

What is the control elastic efficiency?

Answer 3

The control elastic efficiency is: (CLβ)e/CLβ = (1+qSc CLaCmβ/KsCLβ)/(1-q/qD). The higher is q, the smaller is the numerator which means that the control efficiency is reduced.

Question 4

What is aileron control reversal?

Answer 4

1) The rotation of the flap produces an increase of lift (CLβ>0) and consequently a negative pitching moment (Cmβ<0).
2) The wing submitted to Cmβ <0, twists with θ<0.
3) θ<0 causes a reduction in the AoA, and consequently a reduction of lift that balances the increase of lift.

When the control elastic efficiency is equal to 0, control reversal occurs. The dynamic pressure at which this occurs qR, depends on the structural stiffness, the stability derivatives and ScCLa.

This condition is critical when flying close to qR, becasue the control surface becomes ineffective, compromising aircraft controllability.

Question 5

Discuss about the control surface load distribution.

Answer 5

When the aileron is activates, the center of pressure is moved backwards (towards the trailing edge) causing a nose-down pitch moment. The elastic twist reduces the incidence (AoA) and so loads and pitch moments. So, the effect in terms of loading is positive becasue there is a favorable redistribution of loads.

Question 6

How can the control reversal problem be formulated?

Answer 6

It is possible to formulate the control reversal problem in a matrix form, by taking the moment equilibrium about the elastic axis and the equation of lift. As L and theta are the only variations with respect to the trim values, these are the unknowns.

Two approaches exist to then find qR.
1. Lift increment. The equations are solved with respect to L and theta. From these equations, either the lift and twist increment can be found for an assigned value of dynamic pressure q, either the control reversal condition qR can be found for L=0.
2. Eigenvalue problem. Given a target variation of lift L, the required rotation β and the resulting twist can be obtained. When L = 0, the problem is transformed to an eigenvalue problem with two solutions, qR = 0 and qR the already calculated control reversal dynamic pressure.

Question 7

What if the control surface stiffness is added?

Answer 7

Then there is 1 more equation, which is the equlibrium around the flap hinge. H is the aerodynamic hinge moment that depends on the flap surface and flap chord, and has two stability derivatives, CHα and CHβ.

Then a system can be built, which has the form: Ax = F, where x = {θ, β}, and F is equal to {0 kβ)β0 and acts as an forcing term.

Question 8

What is the effect of the control surface stiffness addition on the control reversal and divergence speed?

Answer 8

1. Reversal is possible also in this case and qR is the same as for infinite flap hinge stiffness.
2. There is a change in the divergence speed, which depends on all the stability derivatives.

Question 9

Describe the model for the rolling of a straight wing (typical section model).

Answer 9

The simple typical section rotates freely around the roll axis with an roll angular speed p. The hypothesis of the model are:

1. The motion is considered the sum of a rigid movement plus the deformation of the wing structure
2. It is supposed that the structure deformation behaves statically, meaning that the characteristic time required by the deformable structure to adapt to a change of loading is much smaller than the characteristic time of the rigid movement. Also, the structure is deformed as if the load is applied statically, meaning that the different loading conditions during the maneuver are equivalent to a sequence of static deformations.
3. Aerodynamics also behave statically.

Question 10

Is the reversal dynamic pressure different in the rolling model and the non rolling model?

Answer 10

Yes, it is the same. In the typical section, the control reversal is the speed at which no lift is generated. In this model, control reversal is the point at which no roll is generated, meaning that no lift is generated.

Question 11

What consistent perfomance problems can be identified using this model?

Answer 11

1. Compute the angular speed generated by a rotation of the movable surface β at regime p, i.e. when p_dot = 0.
2. Compute the initial acceleration po_dot generated by the rotation of the movable surface β, i.e. when p=0