Problem Set #5 Problem 5.1 - Solution Kφ The purpose of this problem is to principal bending stiffness axis Kθ principal torsional stiffness axis θ Λ aile principal axes V γ φ illustrate how tailoring of the wing K Vco orientation affects aileron effectiveness The idealized model to used is identical to that used sΛ in Section 3.11: an aileron is added to line of aero centers the semi-span as shown in the figure ron Preliminaries reference axis wing-fuselage junction The load-deflection relationship is: φ [Kij ]θ = Mφθ M with Kφ cos γ + sin γ ⌢ Kθ K ij = Kθ K ij = Kθ Kφ − 1 sin γ cos γ K θ Kφ − 1 sin γ cos γ Kθ Kφ 2 sin γ + cos γ Kθ Without bending and twisting, the lift per unit length along the line of aerodynamic centers due to control surface deflection, δ is: l ( y ) = ( q cos Λ ) cclδ δ o The pitching moment per unit length (positive nose-up or leading edge up) about the line of aerodynamic centers is mac = ( q cos Λ ) c cmδ δ o With bending and twisting deformation included, the lift per unit span length, constant along the span is ( l ( y ) = ( qc cos Λ ) clδ δ o + ao (θ − φ tan Λ ) ) where a0 = clα The twisting moment per unit span length along the wing reference axis is: ( mac = ( qn c ) ccmδ o + eclδ δ o + eao (θ − φ tan Λ ) ) Integrating along the wing span, the total wing lift due to the aileron deflection is given by L flex = qn Sao (θ − φ tan Λ ) + qn Sclδ δ o cl L flex = Q (θ − φ tan Λ ) + Q δ ao Q = ( q cos Λ ) Sao = qn Sao δo Summing moments about the wing root and the spanwise reference axis by integrating Eqns and along the wing span, the following matrix equation for static equilibrium results: K11 K21 K12 φ = qnSea0δ c l K22 θ δ a0 b a 2e c1 = Qe δ c c mδ c2 1 + e cl δ cl c1 = δ ao c cmδ 1 + e clδ cl δ where b 2e cl c2 = δ ao bQ tan Λ K11 = Kθ Kˆ 11 + Kθ Q K12 = Kθ Kˆ12 − b Kθ bQ e K21 = Kθ Kˆ 21 + tan Λ Kθ b bQ e K22 = Kθ Kˆ 22 − Kθ b Solving for θ and φ Qeδ ( K 22c1 − K12c2 ) ∆ Qeδ θ= ( − K 21c1 + K11c2 ) ∆ Qb ˆ e Kˆ 21 Kˆ 22 ˆ e ∆ = Kθ2 Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 − − − K12 tan Λ K11 − Kθ b b φ= ( ) (The determinant is computed in part(a) below) (a) Show that the divergence dynamic pressure parameter is ⌢ ⌢ ⌢ ⌢ Kθ e K11 K 22 − K 21 K12 qD = ⌢ ⌢ K 21 K 22 e ⌢ Sao e b cos Λ e ⌢ K − − − K tan Λ b 11 b 12 The divergence dynamic pressure is obtained from setting the determinant to zero: ∆ = K11 K 22 − K 21 K12 = Qb tan Λ ˆ Qb e Qb ˆ Qb e 2 ∆ = Kθ2 Kˆ 11 + tan Λ = K 22 − − Kθ Kˆ 12 − K 21 + Kθ Kθ b Kθ Kθ b Qb ˆ Qb tan Λ ˆ Qb e ˆ Qb e tan Λ = Kˆ 11 + K 22 − − K12 − K 21 + Kθ Kθ b Kθ Kθ b Q b e ˆ QD b tan Λ QD b tan Λ e + K 22 − Kˆ 11 Kˆ 22 − Kˆ 11 D Kθ b Kθ Kθ b Q be QDb QD b e tan Λ − Kˆ 12 Kˆ 21 − Kˆ 12 D + =0 tan Λ + Kˆ 21 Kθ b Kθ K θ b ( Q b e Kˆ Q b Kˆ e Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 − D Kˆ 11 − 21 + D 22 − Kˆ 12 tan Λ = Kθ b Kθ b QD b = Kθ ˆ e K11 − b ) ( Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 ) Kˆ 21 Kˆ 22 ˆ e − K12 tan Λ − b which yields: Kφ Kθ e c Kθ qD = ⌢ ⌢ Sao e c b cos Λ ⌢ e c K 21 ⌢ e c K 22 + K − K − t an Λ 11 c b 12 c b The parameter qo = Kθ/Seao is the divergence dynamic pressure for the unswept wing If the wing were rigid, the lift produced by the aileron deflection would be ( ) Lrigid = qScl cos Λ δ = Q δ cl δ a0 δ0 The flexible wing lift is: cl L flex = Q (θ − φ tan Λ ) + Q δ ao δo Substituting the relationships for the displacements: L flex = ∆ c cm c cm b b δ − Qe K 22 − K12 + δ tan Λ + ∆ Qe − K 21 + K11 1 + 2e 2e e clδ e clδ Qδ clδ 1 e c cmδ 1 e c cm = − Qb K 22 − K12 1 + δ tan Λ + ∆ Qb − K 21 + K11 1 + ∆ ao 2 b e clδ 2 b e clδ L flex = L flex clδ Qδ Qe ( − K 21c1 + K11c2 ) − Qe ( K 22c1 − K12c2 ) tan Λ + ∆ ao Qδ clδ ∆ ao Qb ˆ e Kˆ 21 Kˆ 22 ˆ e with ∆ = Kθ2 Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 − − − K12 tan Λ K11 − ∆ L flex cl Q δ δ ao ( ) Kθ b b = Qb − K + K e 1 + c cmδ − Qb K − K e + c cmδ 21 11 22 12 2 b e clδ 2e b e clδ Qb ˆ e Kˆ 21 Qb Kˆ 22 ˆ e + Kθ2 Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 − K11 − + − K12 tan Λ Kθ b Kθ b ( ) tan Λ e c cm = qn Sao b Kθ δ b e cl qn Saoδ clδ δ ∆ ao L flex Kθ2 Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 Kˆ 11 + Kˆ12 tan Λ + q Sa b n o ( ) ( ) qn Sao b e c cmδ ˆ K11 + Kˆ12 tan Λ + Kˆ11 Kˆ 22 − Kˆ 12 Kˆ 21 Kθ b e clδ L flex = qn Sclδ δ Qb ˆ e Kˆ 21 Kˆ 22 ˆ e − − K12 tan Λ Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 − K11 − Kθ b b ( ( L flex qn Sclδ δ = L flex Lrigid = ) ( ) ) qn Scao cmδ Kθ clδ ˆ ˆ K11 K 22 − Kˆ 12 Kˆ 21 ( ) Kˆ 11 + Kˆ 12 tan Λ + Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 q Sa e b ˆ − n o Kˆ11 + Kˆ12 tan Λ − K 21 + Kˆ 22 tan Λ 2e Kθ ( ) ( ( ) ) ( ) The answer is qn Scao cmδ ˆ K + Kˆ12 tan Λ Kθ clδ 11 1+ Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 ( L flex qn Sclδ δ = L flex Lrigid = ( ( ) ) ) ( ) b ˆ ˆ ˆ ˆ q Sa e K11 + K12 tan Λ − 2e K 21 + K 22 tan Λ 1 − n o ˆ ˆ ˆ ˆ Kθ K11 K 22 − K12 K 21 ( ) or qn Scao cmδ Kθ clδ 1+ L flex qn Sclδ δ = L flex Lrigid = q Sa e 1 − n o Kθ ( Kˆ 11 + Kˆ12 tan Λ Kφ Kθ b ˆ Kˆ 11 + Kˆ12 tan Λ − K 21 + Kˆ 22 tan Λ 2e Kφ Kθ ( ) ) Reversal occurs when ( ) Kˆ 11 Kˆ 22 − Kˆ 12 Kˆ 21 Kθ e qR = − Seao c cos Λ cmδ ˆ K + Kˆ 12 tan Λ cl 11 δ ( ) ( ) Kθ e qR = − Seao c cos Λ cmδ cl δ Kθ ˆ ˆ K11 + K12 tan Λ Kθ e qR = − Seao c cos Λ cmδ cl δ Kθ Kφ Kφ 2 cos γ + sin γ + − 1 sin γ cos γ tan Λ K K θ θ Kφ ( ) Kφ This last answer is due to the fact that the determinant of the stiffness matrix does not depend on the structural angle We can see this by doing the following calculations Kφ cos γ + sin γ Kθ K ij = Kθ Kφ − 1 sin γ cos γ K θ Kφ − 1 sin γ cos γ Kθ Kφ 2 sin γ + cos γ Kθ K Kφ Kφ φ 2 2 ∆ = Kθ cos γ + sin γ sin γ + cos γ − − 1 sin γ cos γ Kθ Kθ K θ Kφ 2 Kφ Kφ 2 4 2 = Kθ cos γ sin γ + cos γ + sin γ + c o s γ sin γ K Kθ Kθ θ Kφ 2 Kφ 2 2 2 γ γ γ − Kθ s i n γ co s γ + sin cos − sin cos γ Kθ Kθ K = Kθ2 φ ( cos γ + sin γ cos γ + sin γ ) = Kθ Kφ Kθ (e)Plot aileron reversal dynamic pressure for wing sweep angles 0o, 30o and -30o as a function of structural sweep angle, γ, with: Kφ Kθ = 2; b e = 6; = 0.10; c c flap − to − chord − ratio E = 0.15; qDo = Kθ = 250 lb / ft Seao (f)Plot aileron reversal dynamic pressure as a function of sweep angle for the range -30o