Homework set #2 Due February 4, 2011 A-1 AAE556 Three identical uncambered wing segments are connected to each other by torsional springs and to the wind tunnel walls by additional torsion springs, as indicated in the figure The four Problem 2.1 torsion springs have the same spring constant K T The wing segments are mounted on bearings on a spindle attached to the tunnel walls 2.1 (a) Develop the equations of static equilibrium at neutral stability The three degrees of freedom are θ1, θ2 and (b) θ3 Write the expression for the strain energy stored in this configuration as a function of the torsional displacements Use the energy method to derive the system stiffness matrix in terms of the three torsional deflections Compare this result to that found in part (a) They should be identical (c) Solve for the three dynamic pressures that create neutral stability (d) Find the mode shapes at all neutral stability points Describe the mode shapes (how the surfaces deflect relative to each other?) (e) Which of the three dynamic pressures in part (c) is the divergence dynamic pressure? A-2 Problem 2.2 Two wing sections are mounted on shafts attached to each otherProblem and to2.2 a wind tunnel wall, as indicated Note that the two torsion spring stiffnesses are equal, but are offset an amount d This configuration is similar to, but not identical to, the example in Section 2.18 When the sections are placed in the wind tunnel at an angle of attack αo, the two springs deform, as indicated in the diagram Lift on the two, identical, uncambered wing sections is L1 = qSCLα (α o + θ1 ) L2 = qSCLα (αo + θ ) (a) Place an aileron on the outer (right-hand) section This aileron has aerodynamic coefficients CLδ and CMACδ Develop the static equilibrium equations for the system when there is no initial angle of (b) (c) attack, but the control surface is deflected downward an amount δo Specialize the result in part (a) by making the aileron flap-to-chord ratio E = 0.15 with d/e = and e/c = 0.10 Solve for the rolling moment generated by the aileron as a function of dynamic pressure, q Use Section 2.18 as an example of how this is done Solve for the reversal speed Part (a): The static equilibrium equations for the system when there is no initial angle of attack, but the control surface is deflected downward an amount δo 2K − K d CL c CMAC δ d 1 + ÷ δ ÷+ ÷ − e CLα ÷ − K θ1 e CLα e θ1 = qSeCLα δ o − qSeCLα K θ + d θ d CLδ ÷ ÷ e ÷ e C L α CLδ CLα = 0.4805 and The aileron coefficients are computed to be: get: CMACδ CLα = −0.0966 With d/e = and c/e = 10 I −1 θ1 1 −1 θ1 −0.0051 − q = q δ o −1 θ 0 θ 2 2 0.4805 A-3 ÷ ÷ ( 0.4754 − 0.4703q ) ( 0.956 − 0.4805q ) θ1 qδ o = θ ( 2q − 4q + 1) The lift on each section is: 0.4754 − 0.4703q L1 1 2.081q = qSC δ + qSC δ Lδ o Lδ o 2q − 4q + ( 0.956 − 0.4805q ) 1 L2 ( ) 1.021q − 3.011q + ÷ L1 = qSCLδ δ o ÷ q − 4q + 1.00q − 2.011q + ÷ L2 = qSCLδ δ o ÷ q − 4q + b M Roll = 4 ( ) ( ) L1 3b L1 b = = ( L1b + 3L2b ) L2 L2 M Roll = ( ( 2q ) qSCLδ δ o 1.005q − 2.261q + A-4 ) − 4q + Problem 2.3 The wing idealization is shown in the figure will be tested in the wind tunnel at several different airspeeds and in several different configurations All testing is to be done liftat sea level conditions and at such low speed that the flow field is incompressible Changes in the position of the wing box change the wing shear center position and the offset distance between the shear center and wingthe boxaerodynamic center For instance, when the shear center is at the mid-chord, when the shear center is at the wing mid-chord the offset distance e is equal to c/4 Preliminary wind tunnel testing shows that the wing divergence dynamic 100 lb / ft pressure is shear center range The team aerodynamicist predicts that the ratio CMAC CLα tunnel wing segment model cross-section Wind / is -0.075 and the lift curve slope is 4.0; the S =10 ft.2 reference planform area is αo The section angle of attack for all tests will be pre-set to 3o Problem a) Develop the expression for the lift force when the shear center is located at the mid-chord Find the twist angle θ as a function of wind tunnel airspeed b) Consider the same wing with different wing shear center locations The offset distance eo = c/4 when ei the shear center is located at the mid-chord (part (a)) is to be used as the reference and when the shear center is moved Develop the equations for wing lift ei ratio L is the offset and twist θ as functions of the eo Write these expressions in nondimensional form, but use the divergence q for the reference ei wing as the reference in all expressions, because the divergence speed changes with ei (c) Plot wing lift and twist angle vs airspeed for four airspeed range between zero and 150 ft/sec A-5 eo eo values of 0, 0.25, 0.50 and 0.75 Use an Hints for Problem 2.3 in the text A rigid wing is attached to a wind-tunnel wall by a linear torsion spring This wing also has a small wing attached to LS its tip The idealized tip surface produces lift V according to the assumed simple relationship St torsional spring KT LS = qStipCL αS α S A es aerodynamic center αS CLα S A Stip shear center where is the streamwise angle of attack of this small surface; is the tip surface lift curve slope, while gap accentuated is the tip surface area for illustration β αo + θ The tip device is connected to the wing tip by a torsion spring with stiffness k in-lb/radian When the tip device rotates an angle β with respect to the wing tip a restoring torque kβ is generatedview to oppose A-A this rotation V L = qSCL α α The idealized wing lift is enlarged view of cross-sectional geometry α = αo + θ with Problem 2.4 - Wing with flexible tip support Problem statement a) Write the equations of torsional equilibrium for this system when the entire system is given an initial angle q= qSeCLα of attack Express these equations in terms of the following non-dimensional parameters; SR = Stip CLαδ es kR = SCLα e ; k KT KT ; qD Derive the characteristic equation for aeroelastic divergence these parameters k R = 10 qD b) Plot the divergence dynamic pressure parameter as a function of SR when in terms of kR = and Partial answers qSeCLα 1 − ÷ KT qS e C − tip s Lαδ ÷ k KT KT k − ÷ KT qStip es CLαδ − KT A-6 qSeC θ Lα α o Stip esCLαδ = KT β SeC Lα ÷ kR kR kR + + k ± + + k R÷ R ÷ −4 SR SR SR qD = qSeCLα q= KT SR = Stip CLαδ es kR = SCLα e with k KT and SCLα e kR k SCLα e k = = ÷ S R KT Stip CLαδ es KT Stip CLαδ es qDtip ÷= ÷ qD wing The ratio is the ratio of divergence q’s for the tip device alone and the wing alone q= qSeCLα KT = 0.9083 When kr = 10 and SR = 0.1 then q= qSeCLα KT = 0.6592 When kr = 10 and SR = 0.5 then The larger the size of the tip device, the lower the value of divergence dynamic pressure R The divergence dynamic pressure is plotted against the ratio of tip device area beginning with S = 0.1 and ending R with S = System divergence q vs SR when kR = 0.1 A-7 A-8