AAE 556 Aeroelasticity Lecture 6 Multi-degree-of-freedom systems AAE556 – Spring 2008 - 6

Nonlinear models and finding the effects of aerodynamic stall on stability Math model is at zero angle of attack when a non-aero load P is applied as shown Lift is a nonlinear (cubic function of angle of attack) Structural stiffness reponse is linear Pd is an applied moment AAE556 – Spring 2008 - 6

Sum the moments about the shear center (the model pin) Define “Effective torsional stiffness” So - AAE556 – Spring 2008 - 6

Plot effective stiffness for 4 values of q bar AAE556 – Spring 2008 - 6

Also nondimensionalize the original static equilibrium equation by dividing by KT AAE556 – Spring 2008 - 6

Plotting displacement angle q shows that multiple equilibrium states are possible above divergence q and the use of nonlinear analysis reveals what the angles are AAE556 – Spring 2008 - 6

MDOF system study goals Identify similarities and differences between 1 DOF and MDOF models Define theoretical stability conditions for MDOF systems Reading - Multi-degree-of-freedom systems – TAW text p.63-76 AAE556 – Spring 2008 - 6

Torsional springs wing root Develop a 2 DOF segmented aeroelastic finite wing model that represents it as two discrete aerodynamic surfaces with flexible connections Torsional springs fuselage wing tip wing root Torsional degrees of freedom AAE556 – Spring 2008 - 6

Introduce “strip theory” aerodynamic modeling to represent twist dependent airloads Strip theory assumes that lift depends only on local angle of attack of the strip of aero surface why is this an assumption? Notice that q twist angles are measured from a common point AAE556 – Spring 2008 - 6

Double arrow vectors are torques The two twist angles are unknowns and we have to construct two free body diagrams Structural restoring torques depend on the difference between elastic twist angles Wing root Internal shear forces are present, but not drawn Wing tip Double arrow vectors are torques AAE556 – Spring 2008 - 6

Torsional static equilibrium is a special case of dynamic equilibrium Arrange these two simultaneous equations in matrix form AAE556 – Spring 2008 - 6

Summary Static equilibrium equations are necessary to solve aeroelastic problems Solution in terms of unknown displacements and known applied loads Matrix equation order, sign convention and listed ordering of unknowns is important AAE556 – Spring 2008 - 6

Problem solution Both equations on the left hand side multiply the same vector The aeroelastic stiffness matrix is Solve for q1 and q2 AAE556 – Spring 2008 - 6

The solution for the q’s requires inverting the aeroelastic stiffness matrix AAE556 – Spring 2008 - 6

The aeroelastic stiffness matrix determinant is a function of q The determinant is where When dynamic pressure increases, the determinant D tends to zero – what happens to the system then? AAE556 – Spring 2008 - 6

Solve for the twist angles created by an initial system input angle of attack ao AAE556 – Spring 2008 - 6

Plot the aeroelastic stiffness determinant D against dynamic pressure (parameter) The determinant of the stiffness matrix is always positive before the air is turned on Where do the original assumptions about small angles break down? AAE556 – Spring 2008 - 6

Twist deformation vs. dynamic pressure parameter Unstable q region panel twist, qi/ao divergence Outboard panel (2) determinant D is zero AAE556 – Spring 2008 - 6

Panel lift computation on each segment gives: Note that AAE556 – Spring 2008 - 6

More algebra - Flexible system lift Set the lift equal to half the airplane weight AAE556 – Spring 2008 - 6

Lift re-distribution due to aeroelasticity Observation - Outer wing panel carries more of the total load than the inner panel as q increases AAE556 – Spring 2008 - 6

Question – How do we determine divergence for an MDOF system? The general form of the aeroelastic static equilibrium equations is n degrees of freedom Perturb the system by an amount Euler question – “Is there an equilibrium solution to the following relationship?” AAE556 – Spring 2008 - 6

Static stability The static stability test reduces to the existence of a homogeneous matrix equation that must hold if the system is neutrally stable ? AAE556 – Spring 2008 - 6

Linear algebra says... only if … or This is an nth order determinant sometimes called the stability determinant AAE556 – Spring 2008 - 6

Summary System is stable if the aeroelastic stiffness matrix determinant is positive. When additional loads are applied, the system can absorb energy and go to a unique 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 - it must become dynamic and will move, looking for a new equilibrium state. AAE556 – Spring 2008 - 6