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# Week 1-3 Notes MANE 4900

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This 80 page Bundle was uploaded by David Harris on Wednesday February 10, 2016. The Bundle belongs to MANE 4900 at Rensselaer Polytechnic Institute taught by Zvi Rusak in Spring 2016. Since its upload, it has received 72 views. For similar materials see Aeroelasticity & Structural Vibrations in Mechanical Engineering at Rensselaer Polytechnic Institute.

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Date Created: 02/10/16

Aeroelasticity is the science which studies the interaction among inertial (dynamics), elastic (structure), and aerodynamic forces. It was defined by Collar (1947) as "the study of the mutual interaction that takes place within the triangle of the inertial, elastic, and aerodynamic forces acting on structural members exposed to an airstream, and the influence of this study on design." The airplane structure is not completely rigid, and aeroelastic phenomena arise when structural deformations induce changes on aerodynamic forces. The additional aerodynamic forces cause an increase in the structural deformations, which leads to greater aerodynamic forces in a feedback process. These interactions may become smaller until a condition of equilibrium is reached, or may diverge catastrophically. 1. SteaAyeroelasticiy+E) Steady aeroelasticity studies the interaction between aerodynamic and elastic forces on an elastic structure. Mass properties are not significant in the calculations of this type of phenomena. Load Distribution (L) The influence of elastic deformations of the structure (of a wing) on the distribution of aerodynamic pressures over the structure (wing). Divergence (D) Divergence occurs when a lifting surface deflects under aerodynamic load so as to increase the applied load, or move the load so that the twisting effect on the structure is increased. The increased load deflects the structure further, which brings the structure to the limit loads (and to failure). Control surface effectiveness (C) and reversal (R) Control surface effectiveness (reversal) is the loss (or even reversal) of the expected response of a control surface, due to structural deformation of the main lifting surface. This phenomenon applies to the loss of effectiveness of the elevators, rudder and ailerons of an airplane at high speeds. 2. Flight Mechanics (A+I) Static Stability (SS) and Dynamic Stability (DS) Static Stability is the rigid-body response of the vehicle to static aerodynamic forces. Dynamic Stability is the vehicle dynamics in response to aerodynamic forces and the aerodynamic stability of a flying vehicle. This is taught in details in the coursMANE 4090 Flight Mechanics. 3. Structural Dynamics (E+I) Vibrations (V) The response of the vehicle structure to unsteady (oscillating) loads (mechanical vibrations). 4. Dynamic Aeroelasticity (A+E+I) Dynamic Aeroelasticity studies the interactions among aerodynamic, elastic, and inertial forces. Examples of dynamic aeroelastic phenomena are: Flutter (F) Flutter is a self-starting and potentially destructive vibration where aerodynamic forces on an object couple with a structure's natural mode of vibration to produce rapid periodic motion. Flutter can occur in any object within a strong fluid flow, under the conditions that a positive feedback occurs between the structure's natural vibration and the aerodynamic forces. That is, that the vibrational movement of the object increases an aerodynamic load which in turn drives the object to move further. If the energy during the period of aerodynamic excitation is larger than the natural damping of the system, the level of vibration will increase. The vibration levels can thus build up and are only limited when the aerodynamic or mechanical damping of the object match the energy input, this often results in large amplitudes and can lead to rapid failure. Because of this, structures exposed to aerodynamic forces - including wings, aerofoils, but also chimneys and bridges - are designed carefully within known parameters to avoid flutter. It is however not always a destructive force; recent progress has been made in small scale (table top) wind generators, for the third world designed specifically to take advantage of this effect. Dynamic response (Z) Dynamic response or forced response is the response of the vehicle (wing) to changes in inertia (drop a payload) or in the fluid flow such as aircraft to gusts and other external atmospheric disturbances. Forced response is a concern in axial compressor and gas turbine design, where one set of aerofoils pass through the wakes of the aerofoils upstream. Buffeting (B) Buffeting is a high-frequency instability, caused by airflow boundary-layer separation or by shock wave oscillations. It is a random forced vibration. 5. Other fields of study Other fields of physics may have an influence on aeroelastic phenomena. The dynamics of the control (auto-control by computers) of a vehicle may affect aeroelastic phenomena. This is called AeroServoElasticity (A+E+I+S). In aerospace vehicles, stress induced by high temperatures is important. This leads to the study of aerothermoelasticity (A+E+I+T). Prediction and cure Aeroelasticity involves not just the external aerodynamic loads and the way they change but also the structural, damping and mass characteristics of the aircraft. Prediction involves making a mathematical model of the aircraft as a series of masses connected by springs and dampers which are tuned to represent the dynamic characteristics of the aircraft structure. The model also includes details of applied aerodynamic forces and how they vary. The model can be used to predict the flutter margin and, if necessary, test fixes to potential problems. Small carefully-chosen changes to mass distribution and local structural stiffness can be very effective in solving aeroelastic problems. 6. Aeroelasticity Video Clips 1. The severe flutter of an A6 model 2. The light (non-explosive) flutter of a Boeing 747 model 3. The sever flutter of airplanes models 4. The flutter of the tail of the Comanche Civil Engineering - The Collapse of the Tacoma Narrows Bridge The Tacoma Narrows Bridge - the initiation of instability The Tacoma Narrows Bridge - Falling A mass–spring–damper system 1. Physical Model An ideal mass–spring–damper system with mass m, spring constant k, and viscous damper of damping coefficientc is subject to an oscillatory force and a damping force The values can be in any consistent system of units; for example, in SI units, m in kilograms, k in newtons per meter, and c in Newton-seconds per meter or kilograms per second. Treating the mass as afree body and applying Newton's second law, the total force F ontote body is where a is the acceleration of the mass and x is the displacement of the mass relative to a fixed point of reference. Figure 1: A mass–spring–damper system 2. Mathematical Model Since F tot + Fs, d This differential equation may be rearranged into The following parameters are then defined: The first parameter,ω , 0s called the (undamped) natural frequency of the system. The second parameter, ζ, is called the damping ratio. The natural frequency represents an angular frequency, expressed in radians per second. The damping ratio is a dimensionless quantity. The differential equation now becomes 3. Solution: Continuing, we can solve the equation by assuming a solution x such that: where the parameter (gamma) is, in general, a complex number. Substituting this assumed solution back into the differential equation gives which is the characteristic equation. Solving the characteristic equation will give two roots, and . The solution to the [1] differential equation is thus where A and B are determined by the initial conditions of the system: 4. System behavior The behavior of the system depends on the relative values of the two fundamental parameters, the natural frequencyω an0 the damping ratio ζ. In particular, the qualitative behavior of the system depends crucially on whether thequadratic equation for γ has one real solution, two real solutions, or two complex conjugate solutions. 4.1. Critical damping (ζ = 1) When ζ = 1, there is a double root γ (defined above), which is real. The system is said to be critically damped. A critically damped system converges to zero as fast as possible without oscillating (although overshoot can occur). An example of critical damping is the door closer seen on many hinged doors in public buildings. The recoil mechanisms in most guns are also critically damped so that they return to their original position, after the recoil due to firing, in the least possible time. γt γt In this case, with only one root γ, there is in addition to the solution x(t) = e a solution x(t) = te : where and are determined by the initial conditions of the system (usually the initial position and velocity of the mass): 4.2. Over-damping when ζ > 1 When ζ > 1, the system is over-damped and there are two different real roots. An over-damped door-closer takes longer to close than a critically damped door does. The solution to the motion equation is: where and are determined by the initial conditions of the system: 4.3. Under-damping when 0 ≤ ζ < 1 Finally, when 0 < ζ < 1, γ is complex, and the system is under-damped. In this situation, the system will oscillate at the natural damped frequency ω , dhich is a function of the natural frequency and the damping ratio. To continue the analogy, an underdamped door closer would close quickly, but would hit the door frame with significant velocity, or would oscillate in the case of a swinging door. In this case, the solution can be generally written as: where represents the damped frequency or ringing frequency of the system, and A and B are again determined by the initial conditions of the system: Comment: The "damped frequency" is not to be confused with the damped resonant frequency or peak frequency ω peak discussed in section 5, which is the frequency at which a moderately underdamped (ζ < 1/√2) simple 2nd-order harmonic oscillator has its maximum gain (or peak transmissibility) when driven by a sinusoidal input. For an under-damped system, the value of ζ can be found by examining the logarithm of the ratio of succeeding amplitudes of a system. This is called the logarithmic decrement. Time dependence of the system behavior on the value of the damping ratioζ, for undamped (blue), under-damped (green), critically damped (red), and over- damped (cyan) cases, for zero-velocity initial condition. 5. Driven harmonic oscillators Driven harmonic oscillators are damped oscillators further affected by an externally applied force F(t). Newton's second lawtakes the form It is usually rewritten into the form This equation can be solved exactly for any driving force, using the solutions z(t) which satisfy the unforced equation: and which can be expressed as damped sinusoidal oscillations, in the case where 0 < ζ ≤ 1. The amplitude A and phase φ determine the behavior needed to match the initial conditions. 5.1. Step input / Step response In the case 0 < ζ < 1 and a unit step input with x(0) = 0: the solution is: with phase φ given by The time an oscillator needs to adapt to changed external conditions is of the order τ = 1/(ζω ). In0 physics, the adaptation is calledrelaxation, and τ is called the relaxation time. A multiple of τ is called the settling time, i.e. the time necessary to ensure the signal is within a fixed departure from final value, typically within 10%. The term overshoot refers to the extent the maximum response exceeds final value, andundershoot refers to the extent the response falls below final value for times following the maximum response. 5.2. Sinusoidal driving force For a sinusoidal driving force: where is the driving amplitude and is the driving frequency for a sinusoidal driving mechanism. The general solution is a sum of a transient solution that depends on initial conditions, and a steady state that is independent of initial conditions and depends only on the driving amplitude , driving frequency, , undamped angular frequency , and the damping ratio . The steady-state solution is proportional to the driving force with an induced phase change of : where is the absolute value of the impedance or linear response function and is the phase of the oscillation relative to the driving force, if the arctan value is taken to be between -180 degrees and 0 (that is, it represents a phase lag, for both positive and negative values of the arctan's argument). For a particular driving frequency called theresonance, or resonant frequency , the amplitude (for a given ) is maximum. This resonance effect only occurswhen , i.e. for significantly underdamped systems. For strongly underdamped systems the value of the amplitude can become quite large near the resonance frequency. All of the aerostructures are characterized by small values of . The transient solutions are the same as the unforced ( ) damped harmonic oscillator and represent the systems response to other events that occurred previously. The transient solutions typically die out rapidly enough that they can be ignored. Steady state variation of amplitude with relative frequency and damping of a driven simple harmonic oscillator. 6. Special case When the force ???? ???? is given by ???? ???? ???????? ???? ???? = ???? + ???? + ???? ???? + ????(????) ???? ????????2 ???? ???????? ???? the problem becomes: 2 2 ???? ???? ???? + ???? ???????? + ???????? = ???? ???? ???? + ???? ???????? + ???? ???? + ????(????) ????????2 ???????? ???? ????????2 ???? ???????? ???? or 2 ???? ???? ???????? (???? − ???? ???? )???????? 2+ ???? − ???? ???? )???????? + (???? − ???? ???????? = ????(????) ( ) ( ) ( ) Here ???? − ???? ???? is the modified mass, ???? − ???? the modified rigidity, and ???? −???????? the modified damping. 2 For an aerodynamic force ???????? ,????????,????????are proportional to the square of the speed of flight, V . When under certain speeds of flight, the modified rigidity ???? − ????) ≤ 0 or the modified damping ???? − ???????? )≤ 0, the oscillations grow in time. The first situation describes divergence (loss of modified rigidity) while the second situation describes flutter (loss of modified damping). Structural Dynamics Addresses the dynamic deformation behavior in space and time of continuous structural elements such as strings, rods, beams in torsion and beams in bending under prescribed boundary conditions and unsteady external forces. We assume linearly elastic systems. We look to determine natural frequencies and mode shapes, response to initial conditions, forced response to unsteady forces, and frequency response. We will consider a physical problem for each element, formulate a physical model, develop a mathematical model of equations that describe the element motion (equations of motion = ODEs), solve them under certain boundary and initial conditions, and deduct conclusions of the physical behavior. String Dynamics Planar analysis: axial and vertical deformations are u(x,t) and v(x,t) y q(x,t) f(x,t) u(x,t) Fig. 1: Physical problem. q(x,t f(x,t Fig. 2: Physical model; Free body diagram. 1. From Newton’s 2 ndlaw of dynamics: The balance of forces acting on the differential element: Element mass is ρAdx. Let mA =ρ . In x-direction: ∂ u () dx 2= fxt) ∂t x+TTx ∂Td )c+s(θ − θ ) cosθ ∂x∂x ∂u −ηx dx. ∂t In y-direction: 2 () dx ∂ v =qxt,) ∂t 2 ∂T∂ θ x+Txd d )+inθ − ) siθ ∂x∂x −η ∂vdx . y∂t When dx tends to 0 we have: ∂ ∂uTu s)c θ m 2=+f(t,) −,x ∂t ∂x∂t ∂ v ∂v n)s θ m ∂t2 =q+xt) ∂x∂t −η.y 2. From geometry: Element length is: ds=++−+− [] ) dx] u 2 2v dx v ∂ ∂x ∂∂v22 =+ +1 ∂∂x) ( )dx . Element elongation is: e+=−−∂∂ (1 u )2 ( v ) 1. dx ∂∂x Then, ∂uu 1 ) 1 dx ++u ∂xxx +u + dx cosθ= = ds ds 1 e , x vxdvd∂vv sin = = ∂xxx . ds ds 1+e 3. From theory of linear elasticity: TE= . 4. Steady state case: no changes in time. Assume: EA=const. Exact solution is: f(t,(,,0= ux,)t,)x vxt,0 = θ x)0 = ex,)t)=x Tx(t T.x where Tx() c== (given pre-tension force), 0 e() = t s n oec , EA 0 u() = e0 . 5. Deformations from steady state: fxt0,qxt uxt ux uxt) ˆ vxt)0 ,)vxtˆ θ(xt0 (,+) xt ext ex ext ˆ ˆ Tx(,) () (.,) 6. Assume: small deformations from steady state (a simplification/idealization) i) All quantities |(ˆ)|<<1. .ii)01<e0 Then, cosθθ= = θθ~1, sin sin ~ ∂v, ∂ x e~e0+ ∂∂u, ˆ~ , ∂∂x A E A T cos ~Aee ( ++ )(1)~ ( )= 0 0 0 TTsiθ~ ∂vˆ. 0∂x The equations of motion become: ∂∂ ∂uˆ EA 22=+− ηx ∂t f( , ) rod dynamics eq. ∂∂t tTm x ∂=+− η ( , ) string dynamics eq. 0∂∂t2 y∂t When: fq== 0, = xy 0: 22 EA ∂∂uu=m rod dynamics eq. ∂xt2 2ˆˆ Tm0∂∂vv= string dynamics eq. ∂xt2 Both have the form of the wave equation. Boundary conditions: For a fixed string/rod at the two ends, for all time: utl, )=( , ) 0, vtv, )=( , ) 0. Solutions of the wave equation ∂∂uu EA 22 m rod dynamics eq. ∂xt Tm ∂∂v= string dynamics eq. 0∂xt22 Two types of solution: 1.Standing waves – use separation of variables method v(x,t) = X(x)T(t) 2.Traveling waves – wave form v(x,t) = f(x-at) +g(x+at) 3.Homogeneous solution of the string dynamics Standing waves: From: Then, From: Then, A=0 gives the trivial solution, i.e. non perturbed motion. For a non trivial solution: Then, Then, Mode i is defined by a frequency, shape function, and generalized coordinate Then Orthogonality of Mode Shapes 1. Two functions φ nxx() φ are defined i j orthogonal with respect to weighting function m(x) over the domain [0,l] if: 2. Mode i of string dynamics is: where Here T and m are constant. Since we find that or for mode i: Same for mode j: Multiplying first eq by φ jx and second eq by φ i)x and subtracting gives: Using the integration by parts formula: for the right hand side integrals where for first term and for second term we have Here we used the boundary conditions: φi i)=0=and= φ(φl) l 0,φj j) 0 and ( ) 0. We find that for every ij≠ : and for i = j: M is defined as mode i generalized mass. i Modes’ shape functions are orthogonal with respect to m over the domain [0,l]. 3. Analysis using orthogonality of modes’ shape functions Consider the initial conditions for the string deformation shape and speed: From the solution of the string dynamics eq: we have at t=0: Since we find that: Solution is: 4. Example Figure 2.4 Initial shape of plucked string. Initial shape is: Initial speed is g=0. Therefore, E=0. iolution is: where Solution is: ω i T0 fi== i ρ A . 2π l Traveling wave solutions of the wave eq. Solution of string dynamics eq with zero initial speed is: Using the identity: we find that: From the eq for f(x) above we find: Let Then, At t = 0: For t > 0: Here, we can identify the speed of the displacement: and 1. First property of f(x): 2. Second property of f(x): Generalized Eqs of Motion Lagrange’s dynamics eqs. (developed in App. A): Here: L = K - P where K is the system’s kinetic energy and P is the system’s potential energy. Then, Lagrange’s dynamics eqs. become String dynamics is described by: From: We have EA 2 + 2 e0. Using formula for the elongation e, l 0 ∂u The first term vanishes since∫ ∂xdx =.,)0(t −= t 0 The third term does not contribute to transverse motion and is neglected. The second term: 1 ~1, 1+e 0 Example of a quadratic polynomial: Then, Since (prove): Also, First term vanishes from boundary conditions. Since, φ '=− m ωφ , i ii T0 Kinetic energy increment: First term does not contribute to transverse dynamics. Then, Therefore, from Lagrange’s dynamic eqs we have for each mode i: Analogous to the 2 order ODE of a mass-spring system. Example: A concentrated force at x : c Define: such that Zero initial conditions Check: as Example 4: Calculation of Forced Response with Nonzero Initial Condition Multiply by sin( jx ) and (eq)sin( jx )dx : l 0 l Then: l Multiply by sin( jπ ) and ( )sin( jπ )x : l 0 l all A=0. Then, i Beam Dynamics in Torsion 1. The physical problem 2. The physical model. T =the internal twist moment 3. Rule of physics: balance of moments about x axis: where, the polar mass moment of inertia about the x axis is defined as: As dx tends to zero 4. From the theory of linear elasticity: Here, GJ is the Saint-Venant torsional rigidity. 5. We find the beam torsional dynamics eq: When GJ=constant, Same as string dynamics eq. 6. Suggested separation of variables solution: Assume: α > 0 Solutions of X(x) and Y(t) are: 7. Possible “simple” boundary conditions at the beam ends to determine α: i) Clamped end at x=l: no twist angle at x=l: ii) Free end x=l: no twist moment at x=l: iii) Elastic constraint at x=l: torsional spring at x=l: For a torsional spring at x=0: change sign of moment: iv) Inertial constraint at x=l: rigid body with mass moment of inertia I cat x=l: From: For a rigid body at x=0: change sign of moment: The α = 0 mode (the 0 mode): Use end (boundary) conditions to find a and b. Then, φ0xxB=+ 0 0,000 ξ tD= +=, . Beam Dynamics in Torsion: Solution procedure 1. Assign the end conditions an d write the mathematical boundary conditions. 2. Substitute suggested solution, derive the reduced boundary conditions for X(x), and construct the cha racteristic equation for αl. 3. Solve the characteristic eq a nd determine for each mode (i=1,2,..) the wave number αi, the natural frequency GJ ω i i ρ I α , the shape function p AB x⎢in⎥αx α)+ ⎜ ⎟/A cos( ) (note: ii⎣ i⎦ ⎝ ⎠ by the eqs in 2), and the generalized masi 4. Given an external moment, T(x,t), determine the generalized l force exerted on mode i,Ξ= ∫Tx()()φx . i i0 5. Solve the generalize d coordinate eq. ξ ()Ct t() ξ (,) =()ω n s+ ω os⎜⎟ i i i p i i i ih ih ⎝⎠ 6. Apply the given initial conditions: ∞ ∂θ ∞ x g x x x h φ θ) (x x = (∑ ∑(iiξ=0, 0(() () ii = 0 0 ∂t and sov(e nr i iii A A FD ∞ 7. Solution φ θ(ξx)t = (∑()tii x . i 0 A Clamped - Free Beam Represents half a wing clamped to a wall (a wind l=b/2 tunnel model where ). At x=0: the beam is clamped; θ(0,t)=0 or (0) =0. ∂θ At x=l: the beam is free; (o.'X= = ∂x Characteristic eq is: For mode i (i=1,2,…): M i= ρIp/2. In addition, boundary conditions give for the 0 mode: AB == x0φ(0,0= = ω 0. 00 0 (no zero mode since beam is clamped on one side) A Free - Free Beam Represents a wing in flight at slow speed where l=b . ∂θ At x=0: the beam is free; (0,t)=0 or '(0) 0. ∂x At x= l: the beam is free; ∂θ l)'X= = ∂x Characteristic eq is: For mode i: M i = ρIp/2. Boundary conditions give for the ω0 =0 mode: AB0 00, x 1; φ0).1= Pure rigid-body pitch motion. Clamped – Spring Restrained Beam Here At x=0: the beam is clamped; θ(0,t)=0 or (0) 0. At x= l: the beam is spring constrained; Characteristic eq is: For mode i (i=1,2,…): M i= ρIpl/2. Boundary conditions give for the 0 mode: AB00= x0φ(0,00 = ω 0.

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