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