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AAE 556 Aeroelasticity Lectures 22, 23 Typical dynamic instability problems and test review Purdue Aeroelasticity 22-1 How to recognize a flutter problem in the making Given: a DOF system with a parameter Q that creates loads on the system that are linear functions of the displacements M1 &&x1 K1 + M &&x2 0 x1 = Q K x2 p21 x1 x1 = ei ωt x 2 x p12 x1 x2 ( ) K = M1 ω 22 = ω − ω ω 12 + ω 22 + ω12ω 22 = Q is a real number If p12 and p21 have the same sign (both positive or both negative) can flutter occur? ω12 −ω + ω 12 Q − M p21 ( ( ) ∆ = −ω +ω 2 )( K2 M2 Q − p12 M1 x = 0 x 0 −ω + ω 22 ( ) ) Q=0 Q not zero Q2 −ω +ω − p12 p21 = M 1M 2 2 The modified determinant M1 −ω x1 K1 + M x2 ( ∆ = − ω + ω12 ω12 = 2 ω + ω 2 ωn = ± 2 )( x1 −Q K x2 p21 p12 x1 0 r = x2 0 ) Q − ω + ω22 − p12 p21 = M 1M K1 M1 ω 22 (ω −ω K2 = M2 ) 2 Q2 +4 p12 p21 M 1M If flutter occurs two frequencies must merge 2 ω + ω 2 ωn = ± 2 (ω −ω ) 2 Q2 +4 p12 p21 M 1M FLUTTER – Increasing Q must cause the term under the radical sign to become zero K1 ω1 = ω 22 = (ω M1 K2 M2 Q =− −ω ) Q2 =− p12 p21 M 1M ??? 2 ( M M ω 12 − ω 22 p12 p21 ) p12 p21 = − ( M M ω 12 − ω 22 ) 4Q For frequency merging flutter to occur, p12 and p21 must have opposite signs If one of the frequencies can be driven to zero then we have divergence ( ∆ = −ω +ω 2 ωn = )( ) Q2 −ω +ω − p12 p21 = M 1M 2 2 ( )( ) ω 12 ω 22 ( )( ) Q2 = p12 p21 M 1M KK Q = p12 p21 2 ∆ = = ω1 ω2 Q2 − p12 p21 M 1M M M 2ω 12ω 22 Q = p12 p21 p12 p21 = M M 2ω12ω 22 Q2 Divergence requires that the cross-coupling terms have the same sign Aero/structural interaction model TYPICAL SECTION What did we learn? L = qSCL α (αo + θ ) V lift e θ GJ KT ∝ span torsion spring KT qScCMAC αo + K T L = qSCLα − qSeC Lα K T Divergence-examination vs perturbation L= 1− qSCLα qSeCLα αo + KT Kh 1− qSCLα qSeC Lα KT qScC MAC K T h − L = KT θ MSC ∞ = + q + q + q + = + ∑ q n 1− q n=1 Perturbations & Euler’s Test V KT (∆θ ) > (∆L)e lift e θ torsion spring KT .result - stable - returns -no static equilibrium in perturbed state KT ( ∆θ ) < ( ∆L)e result - unstable -no static equilibrium - motion away from equilibrium state KT ( ∆θ ) = ( ∆L)e result - neutrally stable - system stays - new static equilibrium point Stability equation is original equilibrium equation with R.H.S.=0 ∆θ ≠ V θ lift e torsion spring KT (KT − qSeC Lα )= KT = The stability equation is an equilibrium equation that represents an equilibrium state with no "external loads" – Only loads that are deformation dependent are included The neutrally stable state is called self-equilibrating Multi-degree of freedom systems A 2KT 3KT panel panel e b/2 V 5 KT −2 A b/2 αο + θ2 αο + θ1 shear centers aero centers From linear algebra, we know that there is a solution to the homogeneous equation only if the determinant of the aeroelastic stiffness matrix is zero view A-A −2 θ1 −1 θ1 1 + qSeC Lα = qSeC Lα αo θ2 −1 θ 1 Three different definitions of roll effectiveness • Generation of lift – unusual but the only game in town for the typical section • Generation of rolling moment – • contrived for the typical section – reduces to lift generation • Multi-dof systems – this is the way to it • Generation of steady-state rolling rate or velocity-this is the information we really want for airplane performance • Reversal speed is the same no materr which way you it Control effectiveness q c CM δ 1+ qD e CLδ L = qSCL δ δ o =0 q 1− qD q c CM δ 1+ =0 qD e CLδ KT CL δ qR = − ScCLα CMδ Lift α0+ θ V MAC t orsion spring KT shear center e δ0 reversal is not an instability - large input produces small output opposite to divergence phenomenon Steady-state rolling motion qScCMδ v L = = qSCLα δo − + qSC Lα δ o KT V Lift α0+ θ V MAC t orsion spring KT shear center e δ0 Swept wings α structural= θ − φ tan Λ qn = qcos2 Λ K1 d f K2 αo Λ V C V cosΛ A b c B A Kφ −tb − Q Kθ −te b φ b Qα o θ = cosΛ e e B C Divergence bt ∆ = Kθ Kφ + Q Kθ − Kφe Kθ e c Kφ tan Λ crit = 2 c b Kθ 2.0 nondimensional divergence dynamic pressure Seao qD = b K tan Λ cos Λ 1− θ e K φ nondimensional divergence dynamic pressure vs wing sweep angle 1.5 sweep back sweep f orward 1.0 5.72 degrees 0 -0 -1.0 b/c=6 e/c=0.10 Kb/Kt=3 -1.5 -2.0 -90 -75 -60 -45 -30 -15 15 30 sweep angle (degrees) 45 60 75 90 Lift effectiveness lift eff ect iveness vs dynamic pressure 2.0 lif t ef f ect iveness unswept wing 1.5 unswept wing divergence 1.0 15 degrees sweep 30 degrees sweep 0 50 10 150 20 250 dynamic pressure (psf ) 30 350 Flexural axis x ref e r enc e ax is Λ θ E = θ − φ tanΛ β y Flexural axis - locus of points where a concentrated force creates no stream-wise twist (or chordwise aeroelastic angle of attack) θE = The closer we align the airloads with the flexural axis, the smaller will be aeroelastic effects How to recognize a flutter problem in the making Given: a DOF system with a parameter Q that creates loads on the system that are linear functions of the displacements M1 &&x1 K1 + M &&x2 0 x1 = Q K x2 p21 x1 x1 = ei ωt x 2 x p12 x1 x2 ( ) K = M1 ω 22 = ω − ω ω 12 + ω 22 + ω12ω 22 = Q is a real number If p12 and p21 have the same sign (both positive or both negative) can flutter occur? ω12 −ω + ω 12 Q − M p21 ( ( ) ∆ = −ω +ω 2 )( K2 M2 Q − p12 M1 x = 0 x 0 −ω + ω 22 ( ) ) Q=0 Q not zero Q2 −ω +ω − p12 p21 = M 1M 2 2 If flutter occurs two frequencies must merge 2 ω + ω 2 ωn = ± 2 (ω −ω ) 2 Q2 +4 p12 p21 M 1M FLUTTER – Increasing Q causes the term under the radical sign to be zero K1 ω1 = ω 22 = ( M1 ω 12 K2 M2 Q =− − ω 22 ) Q2 = −4 p12 p 21 M1 M ( M M ω 12 − ω 22 p12 p21 ) p12 p21 = − ( M M ω 12 − ω 22 ) 4Q For frequency merging flutter to occur, p12 and p21 must have opposite signs If one of the frequencies is driven to zero then we have divergence ωn = M1 &&x1 K1 + M &&x2 0 x1 = Q K x2 p21 ( )( ) 2 ∆ = = ω1 ω2 ( )( ) ω 12 ω 22 Q2 = p12 p21 M 1M KK Q = p12 p21 p12 x1 x2 Q2 − p12 p21 M 1M M M 2ω 12ω 22 Q = p12 p21 p12 p21 = M M 2ω12ω 22 Q2 Divergence requires that the cross-coupling terms are of the same sign Fuel line flutter A hollow, uniform-thickness, flexible tube has a mass per unit length of m slugs/ft and carries liquid fuel with density ρ to a rocket engine The fuel flow rate is U ft/sec through a pipe cross-section of A The tube is straight and has supports a distance L apart, the tube bending displacement is approximated to be πy 2π y w (y,t ) = φ1 sin + φ sin L L φ1 φ2 Unknown amplitudes of vibrational motion The free vibration frequencies when the fluid is not flowing are: π EI ω = L mo 2π EI ω 22 = L mo mo = m + ρA Fluid flow creates system coupling, but through the velocity, not the displacement ρ AU && φ1 − mo L & ρ AU π φ1 = ÷φ2 + ω1 − ÷÷ ÷ mo L ρ AU && φ2 − mo L & ρ AU 2π φ2 = ÷φ1 + ω2 − ÷÷ mo L ÷ Find the divergence speed Estimate the flow speed that flutter occurs, if it occurs Divergence is found by computing the determinant of the aeroelastic stiffness matrix ∆ aesm ρ AU φ&&1 − mo L & ρ AU π φ1 = ÷φ2 + ω1 − ÷÷ ÷ m L o ρ AU φ&&2 − mo L & ρ AU 2π φ2 = ÷φ1 + ω2 − ÷÷ ÷ m L o ρAU π ρAU 2π ω2 − =0 = ω1 − m L m L o o L U1 = ρA π mo ω12 L U2 = ρA 2π mo ω 22 π EI U div = L ρA Assume that coupling leads to flutter and find an estimate of the merging point ρ AU π φ&&1 + ω1 − φ1 = ÷÷ ÷ mo L ρ AU 2π 2 φ&&2 + ω2 − φ2 = ÷÷ ÷ mo L Harmonic motion? 2 2 2 ρ AU π ρ AU π 2 −ω + ω − m L −ω + ω − m L = o o The frequencies are approximated 2 2 2 π 2π 2 ρAU ρAU = − ω + ω − − ω + ω − mo L mo L ω = ω 12 2 ρAU π − mo L ρAU 2π 2 ω = ω2 − mo L 2 2 ρAU π ρAU 2π ω − ≅ ω2 − mo L mo L F 2 F 2 π EI mo L U ≅ ω2 − ω1 = 5 ρA π L ρA F ( )
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