Problem #1 The 1 DOF idealized wing whose cross-section is shown in Figure 1.1 has leading edge and trailing edge control surfaces. There is no initial angle of attack when the two control surfaces are undeflected. Three lift components act on the idealized wing. Each force has its own location with respect to the shear center pin shown in the figure. Twisting is resisted by the torsional spring with torsional stiffness KT. The entire assembly is mounted at the end of a long bar a distance r from the center of rotation. This assembly will move upward at a terminal speed v = pr when the surfaces are deflected. The leading edge surface and the trailing edge flap are geared together so that the three lift components are as follows:

L1 = qSCLα θ + qSCLα −v

V( ) (where v = pr )

The rolling moment is Mroll = Lr = (L1 + L2+ L3)r.

Problem

a) Solve for the twist θ in terms of the aerodynamic derivatives and v/V.

b) Solve for the equation for the steady state

roll rate p .

c) Solve for the reversal dynamic pressure Partial Solution Assume that we are rolling at a constant roll rate so that the rolling moment due to the aileron is balanced by a damping in roll moment (see the notes) due to an upward velocity. In this case, the total lift on the surface is L = L1 + L2 + L3

L2 = qSCLδ δ0

L3 = 0.25qSCLδδ0

Figure 1.2 – Top view of wing assembly showing

rotational rate, p.

Figure 1.1

or 1

4L L L o L o

vL qS C C C C

Vα α δ δθ δ δ = − + +

The FBD is shown below.

Figure 1.3-Free-body diagram for θθθθ computation

Sum moments using +θ as the positive direction (clockwise). Let 11

L

T

qSe Cq

Kα= . I get

( )

3 21 1

1 1

1

14

1

Lo

L

Ce evq q

V e e C

q

δ

α

δ

θ

− + − =−

The rolling moment is 1

4roll L L L o L o

vM Lr qSr C C C C

Vα α δ δθ δ δ = = − + +

I get

( )

1 2 1

1

1

15 4 5

4 41

01

Lo

L

roll L

Ce evV C

M L

eq

r C Srq

eq

δ

α

α

δ − − + +

− + = = = −

1 2 11

1

5 4 5

4 4L

oL

e e ev prq

V V e

C

Cδ

α

δ − − + = = +

v = pr

For reversal

1 2 11

1

5 4 50

4 4

e e eprq

V e

− − + = = +

The answer is ( )321 45

5

eeeSC

Kq

L

Treversal −+

=α

The aileron reversal problem has two meanings and two approaches. In the first case we restrict

the upward velocity, v, to be zero and solve for the lift (or rolling moment) generated by an

aileron deflection. Reversal is defined as the airspeed (or dynamic pressure) at which the lift is

zero. By definition, at reversal in this case, both lift and upward speed are zero, but only because

the upward speed, v, is restrained from the beginning and we solve for the value of q to make lift

(or moment) zero.

In the second case, the upward airspeed (or steady-state roll moment) is constant, unrestrained and non-zero. To have a constant upward speed we must have the lift (or rolling moment) equal to zero. This is a constraint that allows us to solve for v. At reversal, both v and lift (or moment) are zero. We use the “v equation” to find the value of q at reversal. Because, in the end, no matter which of the two approaches we use for reversal, the reversal dynamic pressure must be the same since both conditions result in a zero upward velocity and zero lift due to aileron deflection. Only the path to reversal is different.

Problem 2 A mechanism with a rigid lifting surface (with negligible weight) with the ability to “bend” and rotate is shown in the figure. Two degrees of rotational freedom describe the motion of this device and are shown as θ and φ on the figure. The mechanism is restrained in twist by a torsion spring. Bending rotation is restrained by a bending spring located as shown. The lift per unit length along the swept y-axis is ( ) ( )tann ol y q ca θ φ= − Λ . The

mechanism is loaded by a small weight, W lb., placed as indicated.

Problem statement (a) Develop the two equations of static

equilibrium for this device. Note that the bending spring reference axis is

not at the wing surface mid-chord.

Free body diagram in the θθθθ direction

Free body diagram in the φφφφ direction

Problem 2.2

I get ( ) ( ) tan

0.75

tan2 2

K Q d e Q d ed c e

WQb QbbK

θ

φ

θφ

+ − − − Λ + − = −− + Λ

where 2coslQ qcbcα

= Λ .

(b) Solve for the divergence dynamic pressure in terms of the parameters, appearing in the

equations developed in part (a) (c)

I get

qD =

KθSeCLα

cos2 Λ 1− de

− KθKφ

btanΛ2e

Problem 3 Flight speed and aircraft weight change during flight. These changes mean that the wing design is non-optimal for major parts of the mission. Design features such as wing camber and twist distribution can be changed in flight by “morphing” devices that alter the wing shape to bring it back into optimal performance. Wings with active twist control using internal or external active control devices have been proposed and many patents exist. The drawing of the active twist wing shown in Figure 4.1.1 is typical of these types of devices. This homework problem will examine the interactions created by active twist wings by using a simple idealization of a swept wing.

Patent drawing of mechanically controlled wing with internal mechanisms

The idealized swept wing model shown in the figure below simulates an active twist wing; it consists of the familiar semi-rigid surface restrained by bending and torsion springs to resist rotations and , respectively, along the swept axis of the idealization. Wing planform dimensions are: swept semi-span, b; and, chord, c. A mechanism contained inside a wing has the effect of applying a torque To to the end of the wing, as shown in the diagram. This torques can be changed in flight. A feedback system will produce measure the wing bending slope φ and produce a torque according to the control law T kθ φ= .

Swept wing model planform and geometry showing wing detached from fuselage. The double arrows on the fuselage joint represent spring reactions. The swept semi-span dimension is b. The wing chord is c, just like

on the notes. Problem

φ θ

a) The swept wing/aircraft combination is given an angle of attack αo with respect to the free stream. The free body diagrams required to derive the two wing static equilibrium equations is

shown below. Find the equations of static equilibrium in matrix form in terms of the vector φθ

.

Part (a) solution

Answer

( ) ( )0

1tan2 2 2

2 costan

n l n l

n l

n l n l

b bK q S c q S c b

q S c eq Sec k K q Sec b

α α

α

α α

φ

θ

φ αθ

+ Λ − = Λ Λ − −

(b) Derive the characteristic equation for neutral static stability. Solve for the divergence dynamic

pressure. Answer

2cos 1 tan2 2

lD

KSec

qKb k b

e K K e

α

θ

θ

φ φ

=

Λ − Λ +

(c) Solve for the critical value of the gain k above which (or below which) wing divergence cannot occur. This answer will be a function of sweep angle and other model parameters

Figure 1 - Oblique wing aircraft and planform view of the idealized oblique wing. The wing pivot is at the ¼ chord. A drawing of this model seen from the perspective A-A is presented in Figure 3.

Problem 4 An oblique wing aircraft has one wing sweptforward and the other wing sweptback, as indicated. The idealized oblique wing model consists of two semi-rigid wing sections attached by two bending springs at the fuselage centerline. There is no torsional degree of freedom for either wing section. An oblique wing has a tendency to develop unsymmetrical lift when clamped at the center.

When the wing is rigid, this tendency is not severe. When aeroelasticity is included, the wing deformation can produce a severe lift distribution distortion. In flight this cannot be allowed to occur because the wing would roll. The problem is related to the tendency of the forward swept wing to diverge and the tendency of the aft swept wing to become lift ineffective. The purpose of this problem is to assess the severity of the problem at different dynamic pressures. The figure on the next page shows another view of the model idealization with the perspective labeled A-A in Figure 1. This view shows the two wing bending degrees of freedom, φ1 and φ2, the bending springs and the downward weight. Note that in Figure 1 the aircraft roll angle is measured as φo parallel to the freestream direction. In Figure 3, the component of this roll angle in the wing chordwise direction is Φο = φo cosΛ.

Figure 2-Oblique wing lift distribution showing rig id wing lift distribution and flexible lift distributi on at high speed. Note that this wing has a sweep opposite

to that shown in the figure above

Figure 3 - Oblique wing bending freedoms showing the two bending springs. The view is in the chordwise direction, as

indicated by the view A-A in Figure 1.

Problem statement

a) The oblique wing is given an angle of attack, measured with respect to the streamwise axis. Let K1 = K2. The wing pivot is at the ¼ chord. Compute the lift on each wing and the roll moment about the centerline of the aircraft as a function of angle of attack αo and dynamic pressure.

Partial Solution-part (a) Draw the free body diagrams and write the equilibrium equations. The sweptforward aileron rotates down (+) while the sweptback aileron rotates upward (-).

Sweptforward wing FBD

1

1

2 cos

tan2

n L Loo

L

n L

q SbC C

C

q SbCK

α δ

α

α

α δ

φ

+

Λ =− Λ

Sweptback wing FBD

2

2

2 cos

tan2

n L Loo

L

n L

q SbC C

C

q SbCK

α δ

α

α

α δ

φ

−

Λ =+ Λ

The lift expressions are:

1

1

1

cos1 tan

2

Lon L o

n LL

CL q SC

q SbCC

K

δ

ααα

α δ

= +

Λ − Λ

2

2

1

cos1 tan

2

Lon L o

n LL

CL q SC

q SbCC

K

δ

ααα

α δ

= −

Λ + Λ

The moment about the centerline in the direction of positive φo is:

( )

1 2

1 2

1 2

cos cos

2 2

tan tan2 2

21 tan 1 tan

2 2

roll

n L n L

n Lo

n L n L

b bM L L

q SbC q SbC

q SbC K K

q SbC q SbC

K K

α α

α

α α

α

Λ Λ = −

Λ + Λ = − Λ + Λ

When the spring constants are equal I get:

( )tan

21 tan 1 tan

2 2

n L

n Lroll o

n L n L

q SbCq SbC KM

q SbC q SbC

K K

α

α

α α

α

Λ = − Λ + Λ

b) Full span ailerons are added to each wing to counteract the roll moment due to aeroelasticity. The

aileron deflections are opposite in direction and equal to δo. Compute the aileron deflection δo required to make the oblique wing roll moment equal to zero. This deflection will be a function of angle of attack.

Answer 1

sin2

L Lo o

L

C qSbC

C Kα α

δ

δ α

= − Λ

Note that the aileron deflection expression is negative, telling us to deflect the sweptforward aileron up and the sweptback aileron down. Note also that the dynamic pressure is not multiplied by the cosine of the sweep angle.

c) Derive expressions for the angle of attack and the aileron deflection required for an airplane with weight, W.

Solution-part (c) The equation for aileron deflection in terms of wing angle of attack is a constraint equation for the aircraft. When we equate the total lift to the aircraft weight we will get a second equation that will enable us to solve for both aileron deflection and required angle of attack. The second equation is:

1 2L L W+ =

1 2

1 2

1 1

cos1 tan 1 tan

2 2

1 1

1 tan 1 tan2 2

on L

n L n L

Ln L o

n L n LL

W q SCq SbC q SbC

K K

Cq SC

q SbC q SbCC

K K

αα α

δ

αα αα

α

δ

= + Λ − Λ + Λ

+ −

− Λ + Λ

If the spring constants are equal I get the following:

1sin

2L L

o oL

C qSbC

C Kα α

δ

δ α

= − Λ

and tan2

n Lq SbCQ

Kα= Λ

Answer cos

2 2 coson L L

W W

q SC qSCα α

α Λ= =Λ

Answer tan4

Lo

L

CWb

K Cα

δ

δ

= − Λ

d) Derive the characteristic equation for fixed wing divergence and find the divergence dynamic

pressure. Solution-part (d)

1 20 tan tan2 2

n L n Lq SbC q SbCK Kα α

∆ = = − Λ + Λ

The sweptforward wing will give a positive value of dynamic pressure.

1

sin cos2

D

L

Kq

bS C

α

=Λ Λ