AAELecture 22 Typical dynamic instability problems and test review

AAE 556 Aeroelasticity Lecture 22 Typical dynamic instability problems and test review ARMS 3326 6:00-8:00 PM 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    & x&  K1 1 +    M   & x& 2 0   x1   = Q    K   x2   p21  x   x    =  ei ωt  x  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 ω12 sign (both positive or both negative) can flutter occur?   −ω + ω 12     Q  − M p21    ( ( ) ∆ = −ω +ω 2 )( Q=0 K2 M2  Q   − p12       M1   x  = 0   x  0  − ω + ω 22      ( Q not zero ) ) Q2 −ω +ω − p12 p21 = M 1M 2 2 23-2 Purdue Aeroelasticity If flutter occurs two frequencies must merge 2 ω + ω 2 ωn = ± 2 (ω −ω ) 2 Q2 +4 p12 p21 M 1M For Flutter – Increasing Q must cause the term under the radical sign to become zero and then go negative The zero condition is: ( K1 M1 K ω 22 = M2 ω12 = ω 12 Q =− − ω 22 ) Q2 = −4 p12 p 21 M1 M ( M M ω 12 − ω 22 p12 p21 ) p12 p21 = − ( M M ω 12 4Q For frequency merging flutter to occur, p12 and p21 must have opposite signs 23-3 Purdue Aeroelasticity − ω 22 ) If one of the frequencies is driven to zero then we have divergence M −ω   0   x1   K1  +   M   x2   p12   x1      x2  ( )( ) ωn = ∆ = = ω 12 ω 22 ( )( ) ω 12   x1   = Q    K   x2   p21 ω 22 Q2 = p12 p21 M 1M KK Q = p12 p21 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 23-4 Purdue Aeroelasticity Aero/structural interaction model TYPICAL SECTION What did we learn? L = qSCL α (α o + θ ) V lift torsion spring KT e θ GJ KT ∝ span  qScCMAC   αo +  K T  L = qSCLα   − qSeC Lα    K   T 23-5 Purdue Aeroelasticity Divergence-examination vs perturbation L= 1− qSCLα qSeCLα αo + KT  Kh  1− qSCLα qSeC Lα KT  h   − L   =   KT θ  MSC  ∞ = + q + q + q + = + ∑ q n 1− q n=1 23-6 Purdue Aeroelasticity  qScC  MAC    K   T Perturbations & Euler’s Test V KT ( ∆θ ) > ( ∆L )e lift torsion spring KT e θ .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 23-7 Purdue Aeroelasticity 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 23-8 Purdue Aeroelasticity Multi-degree of freedom systems A 2KT 3KT panel panel e b/2 V A b/2 aero centers there is a solution to the homogeneous equation only if the determinant of the aeroelastic stiffness matrix is zero αο + θ2 αο + θ1 5 KT  −2 shear centers From linear algebra, we know that view A-A −2  θ1   −1  θ1  1   + qSeC Lα    = qSeC Lα α o     θ2  −1   θ  1 23-9 Purdue Aeroelasticity MDOF stability Mode shapes? Eigenvectors and eigenvalues [KT ]{∆θ i }= {0} KT = Kij − qAij = System is stable if the aeroelastic stiffness matrix determinant is positive Then the system can absorb energy in a static deformation mode If the stability determinant is negative then the static system, when perturbed, cannot absorb all of the energy due to work done by aeroelastic forces and must become dynamic 23-10 Purdue Aeroelasticity 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 23-11 Purdue Aeroelasticity Control effectiveness  q  c  CMδ   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 cent er reversal is not an instability large input produces small output opposite to divergence δ0 e phenomenon 23-12 Purdue Aeroelasticity Steady-state rolling motion  qScCMδ v   L = = qSCLα δo − + qSC Lα δ o  KT V Lif t α0+ θ V MAC t orsion spring KT shear center δ0 e 23-13 Purdue Aeroelasticity Swept wings α structural= θ − φ tan Λ qn = qcos Λ K1 f K2 αo V C V cosΛ C b  Kφ    −tb  − Q Kθ   −te b  φ b   Qαo    θ  = cosΛ   e   e 23-14 Purdue Aeroelasticity 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) 23-15 Purdue Aeroelasticity 45 60 75 90 Lift effectiveness lif t ef f 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 30 350 dynamic pressure (psf) 23-16 Purdue Aeroelasticity Flexural axis Λ θ E = θ − φ tanΛ β x 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 23-17 Purdue Aeroelasticity