AAE 556 Aeroelasticity Lecture – Multi-DOF systems Reading: Sections 2-13 to 2-15 Homework for Friday? i Problem 2.3 Multi-degree-of-freedom (MDOF) systems i i i Develop process for analyzing MDOF systems Define theoretical stability conditions for MDOF systems Reading - Multi-degree-of-freedom systems – Section 2.14 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) 2KT 3KT fuselage Torsional springs A panel panel e b/2 V wing root shear centers A b/2 aero centers αο + θ2 αο + θ1 view A-A Torsional degrees of freedom wing tip 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 Double arrow vectors are torques Internal shear forces are present, but not drawn Torsional static equilibrium is a special case of dynamic equilibrium ∑ M = I1θ1 = = L1e − 3KTθ1 + KT (θ − θ1 ) L1 = qSCLα ( α o L2 = qSCLα ( α o + θ ) M = I θ ∑ 2 = = L2e − KT (θ − θ1 ) + θ ) 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 Strain energy 1 2 U = (3KT )θ1 + (2 KT ) ( θ − θ1 ) 2 ∂U = (3KT )θ1 + (2 KT ) ( θ − θ1 ) ( −1) ∂θ1 ∂ 2U = (3KT ) + (2 KT ) ( −1) ( −1) = KT = K11 ∂θ1 ∂ 2U = (2 KT ) ( 11 ) ( −1) = −2 KT = K12 ∂θ1∂θ Summary i 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 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 Problem solution outline − 2 θ1 − θ1 1 KT + qSeC Lα = 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 The aeroelastic stiffness matrix is KT Invert matrix and solve for θ1 and θ2 {θ i } = qSeCLα α o [ KT ] {1} −1 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 ∆ = K11 K 22 − K12 K 21 − K12 K11 The aeroelastic stiffness matrix determinant is a function of q i The determinant is ∆ = K11 K 22 − K12 K 21 ∆ = q − 7q + where ∆ = ( 1− q ) ( − q ) q= qSeCLα KT When dynamic pressure increases, the determinant ∆ tends to zero – what happens to the system then? Plot the aeroelastic stiffness determinant ∆ against dynamic pressure (parameter) determinant STABLE UNSTABLE -2 -4 -6 -8 Dynamic pressure parameter The determinant of the stiffness matrix is always positive turning the air on reduces its size Solve for the twist angles created by an input angle of attack αo θ1 qα o = ∆ θ 4 − q 7 − q ∆ = ( 1− q ) ( − q ) q= qSeCLα KT Twist deformation vs dynamic pressure parameter A 2KT shear centers e b/2 V A b/2 aero centers αο + θ2 αο + θ1 view A-A divergence panel twist, θi/αo panel panel panel twist/initial angle 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 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 ) ÷ = ∆ Lift re-distribution due to aeroelasticity (originally presented on slide 13) W∆ αo = 2qStotal CLα ( ∆ + q ( 5.5 − q ) ) q ( − q ) 1+ L1 ∆ = qSCLα α o L2 + q(7 − q) ∆ L1 W − q = L2 2( − q ) Observation - Outer wing panel carries more of the total load than the inner panel as q increases MDOF Divergence θ1 Q1 [ K S ] − q [ K A ] = i In general we have θ Q2 i (S ) matrix relationships θ1 θ1 ∆θ1 = (S ) + developed from θ ∆ θ θ EOM’s ∆θ1 0 [ K S ] − q [ K A ] These can be = ∆θ 0 converted into [ KS ] − q [ K A ] = perturbation relationships KT { ∆θ i } = { 0} The aeroelastic stiffness matrix determinant is a function of q i The determinant is ∆ = K11 K 22 − K12 K 21 ∆ = q − 7q + where ∆ = ( 1− q ) ( − q ) q= qSeCLα KT When dynamic pressure increases, the determinant ∆ tends to zero – divergence occurs Determinant ∆ plotted against dynamic pressure parameter ∆ = q − 7q + determinant ∆ = ( − q D ) ( − qD ) = STABLE UNSTABLE -2 qD = 1, -4 -6 -8 Dynamic pressure parameter This nth order determinant is called the stability determinant or the characteristic equation Eigenvectors −2 θ1 qSeCLα −1 θ1 0 −2 θ + K −1 θ = 0 2 2 T −2 θ1 −1 θ1 −2 θ1 0 + (1) −2 θ −1 θ = −2 θ = 0 2 2 2 with qD = θ2 4θ1 − 2θ = ⇒ =2 θ1 θ −2θ1 + θ = ⇒ = θ1 with qD = θ2 =− θ1 θ −2θ1 − 4θ = ⇒ = − θ1 −θ1 − 2θ = ⇒ Twist deformation vs dynamic pressure parameter A 2KT shear centers e b/2 V A b/2 aero centers αο + θ2 αο + θ1 view A-A divergence ∆ = q − 7q + panel twist, θi/αo panel panel panel twist/initial angle of attack 3KT outboard panel 10 -2 inboard panel -4 -6 inboard panel outboard Outboard panel panel (2) -8 -10 ∆ = ( 1− q ) ( − q ) outboard panel unstable Unstable q region region divergence dynamic pressure parameter, q determinant ∆ is zero