AAE556 lecture 05 MDOF response

AAE 556 Aeroelasticity Lecture – 1) Compressibility; 2) Multi-DOF systems Reading: Sections 2-13 to 2-15 5-1 Purdue Aeroelasticity Homework for Monday? i Prob 2.1 – – i Uncambered (symmetrical sections) MAC = Lift acts at aero center (AC) a distance e ahead to the shear center Problem 2.3 – wait to hand in next Friday 5-2 Purdue Aeroelasticity Aeroelasticity matters Reflections on the feedback process 5-3 Purdue Aeroelasticity Topic - Flow compressibility (Mach number) has an effect on divergence because the liftcurve slope depends on Mach number Approximate the effect of Mach number by adding the Prandtl-Glauert correction factor for sub-sonic flow KT qD = SeC Lα qDo C Lα = KT = SeC Lα0 C Lα0 1− M KT − M qD = = qDo − M SeC Lα0 Plots as a curve vs M 5-4 Purdue Aeroelasticity But wait! – there’s more! Mach number depends on altitude and airspeed so two expressions must be reconciled qatmosphere = qa Physics 1 2 qa = ρV = ρ a M 2 M=V/a Speed of sound, “a," depends on temperature and temperature depends on altitude KT − M qD = = q Do − M SeC Lα 5-5 Purdue Aeroelasticity The divergence equation which contains Mach number must be consistent with the “physics” equation qDo − M D 2 2 = ρa M D = q1M D 2 q1 = ρa Choose an altitude Find the speed of sound Square both sides of the above equation and solve for MD ( )  qDo    − M D = M D  q1  5-6 Purdue Aeroelasticity Determining MD requires solving a quadratic equation  qDo   qDo   M D −   = +   q1   q1  qatmosphere = qa 1 2 qa = ρV = ρa M 2 qDo − M 250 dynamic pressure (lb/sq ft) MD 200 sea level q divergence q 150 20,000 f t 100 40,000 f t 50 0.00 0.25 0.50 0.75 Mach number Purdue Aeroelasticity 1.00 5-7 If we want to increase the divergence Mach number we must increase stiffness (and weight) to move the math line upward dynamic pressure (lb/sq ft) 400 350 300 sea level q divergence q 250 200 20,000 f t 150 100 40,000 ft 50 0.00 0.25 0.50 0.75 1.00 Mach number 5-8 Purdue Aeroelasticity Summary i Lift curve slope is one strong factor that determines divergence dynamic pressure – depends on Mach number i Critical Mach number solution for divergence dynamic pressure must be added to the solution process 5-9 Purdue Aeroelasticity Topic – Multi-degree-of-freedom (MDOF) systems i Develop process for analyzing MDOF systems i Define theoretical stability conditions for MDOF systems i Reading - Multi-degree-of-freedom systems – Section 2.14 5-10 Purdue Aeroelasticity Here is a DOF, segmented, aeroelastic finite wing model - two discrete aerodynamic surfaces with flexible connections used to represent a finite span wing (page 57) A 2KT 3KT Torsional springs fuselage panel panel e b/2 V shear centers A b/2 aero centers αο + θ2 αο + θ1 wing root view A-A wing tip Torsional degrees of freedom 5-11 Purdue Aeroelasticity Introduce “strip theory” aerodynamic modeling to represent twist dependent airloads i Strip theory assumes that lift depends only on local angle of attack of the strip of aero surface – why is this an assumption? L1 = qSCLα (α o + θ1 ) q twist angles are measured from a common reference L2 = qSCLα (α o + θ2 ) 5-12 Purdue Aeroelasticity The two twist angles are unknowns - we have to construct two free body diagrams to develop equations to find them Structural restoring torques depend on the difference between elastic twist angles Wing root Wing tip Internal shear forces are present, but not drawn Double arrow vectors are torques 5-13 Purdue Aeroelasticity This is the eventual lift re-distribution equation due to aeroelasticity – let’s see how we find it  L1  W 2 − q   =    L2  2(4 − q )   Observation - Outer wing panel carries more of the total load than the inner panel as q increases 5-14 Purdue Aeroelasticity Torsional static equilibrium is a special case of dynamic equilibrium   M = I θ ∑ 1 = = L1e − 3KTθ1 + KT (θ − θ1 )   M = I θ ∑ 2 = = L2e − KT (θ − θ1 ) Arrange these two simultaneous equations in matrix form  − 2 θ1  1 0 θ1  1 KT    − qSeC Lα    = qSeC Lα α o     − 2  θ  0 1 θ  1 5-15 Purdue Aeroelasticity Summary i The equilibrium equations are written in terms of unknown displacements and known applied loads due to initial angles of attack These lead to matrix equations i Matrix equation order, sign convention and ordering of unknown displacements (torsion angles) is important  − 2 θ1  −  θ1  1 KT    + qSeC Lα    = qSeC Lα α o     − 2  θ   − 1 θ  1 5-16 Purdue Aeroelasticity Problem solution outline  − 2 θ1  −  θ1  1 KT    + qSeCLα    = qSeC Lα α o     − 2  θ   − 1 θ  1 Combine structural and aero stiffness matrices on the left hand side θ1  1 [ KT ]θ  = qSCLα α o 1   2  KT  The aeroelastic stiffness matrix is Invert matrix and solve for θ1 and θ2 {θ i } = qSeCLα α o [ KT ] {1} −1 5-17 Purdue Aeroelasticity The solution for the θ’s requires inverting the aeroelastic stiffness matrix  K11  KT  =   K 21  K11   K 21 K12   K 22  −1 K12   K 22  =  ∆  − K 21 K 22  − K12   K11  ∆ = K11 K 22 − K12 K 21 5-18 Purdue Aeroelasticity The aeroelastic stiffness matrix determinant is a function of q i ∆ = K11 K 22 − K12 K 21 The determinant is ∆ = q − 7q + where q= qSeCLα ∆ = ( 1− q ) ( − q ) KT When dynamic pressure increases, the determinant ∆ tends to zero – what happens to the system then? 5-19 Purdue Aeroelasticity Plot the aeroelastic stiffness determinant D against dynamic pressure (parameter) determinant STABLE UNSTABLE -2 -4 -6 -8 Dynamic pressure parameter The determinant of the stiffness matrix is always positive until the air is turned on 5-20 Purdue Aeroelasticity Solve for the twist angles created by an input angle of attack αo θ1  qα o  = ∆ θ  4 − q    7 − q  q= ∆ = ( 1− q ) ( − q ) 5-21 Purdue Aeroelasticity qSeCLα KT Twist deformation vs dynamic pressure parameter A 2KT panel panel e b/2 V A b/2 αο + θ2 αο + θ1 shear centers view A-A divergence aero centers al angle panel twist/initi panel twist, θi/αo of attack 3KT outboard panel 10 outboard panel unstable Unstable q region region divergence -2 inboard panel -4 -6 inboard panel outboard Outboard panel panel (2) -8 -10 dynamic pressure parameter, q determinant ∆ is zero 5-22 Purdue Aeroelasticity Panel lift computation on each segment gives:  q ( − q )    1+  L1  ∆ = qSC α    Lα o  q − q )   L2  1 + (  ∆ L flex Lrigid L1 + L2 L1 + L2 = = q ( S ) CLα α o qStotal CLα α o Note that Stotal = S 5-23 Purdue Aeroelasticity More algebra - Flexible system lift L flex Lrigid q = + ( 5.5 − q ) ∆ L flex  q  = qStotal CLα α o 1 + ( 5.5 − q ) ÷  ∆  Set the wing lift equal to half the airplane weight L flex  q  Weight = qStotal CLα α o 1 + ( 5.5 − q ) ÷ =  ∆  5-24 Purdue Aeroelasticity Lift re-distribution due to aeroelasticity (originally presented on slide 13) W∆ αo = 2qStotal CLα ( ∆ + q ( 5.5 − q ) )  1 + q (4 − q )    L1    = qSCLα α o  q − q ∆  )   L2  1 + (  ∆  L1  W 2 − q   =    L2  2(4 − q )   Observation - Outer wing panel carries more of the total load than the inner panel as q increases 5-25 Purdue Aeroelasticity