Aircraft Structures 3E Episode 15 doc

546 Elementary aeroelasticity Reference \ axis ii"p Fig. 13.4 Effect of wing sweep on wing divergence speed. showed that wings with moderate or large sweepback cannot diverge. The opposite of course is true for swept-forward wings where bending deflections have a destabilizing effect and divergence speeds are extremely low. The determination of lift distributions and divergence speeds for swept-forward wings is presented in Ref. 3. The flexibility of the major aerodynamic surfaces (wings, vertical and horizontal tails) adversely affects the effectiveness of the corresponding control surfaces (ailerons, rudder and elevators). For example, the downward deflection of an aileron causes a nose down twisting of the wing which consequently reduces the aileron incidence. Thus, the wing twist tends to reduce the increase in lift produced by the aileron deflec- tion, and thereby the rolling moment to a value less than that for a rigid wing. The aerodynamic twisting moment on the wing due to aileron deflection increases as the square of the speed but the elastic restoring moment is constant since it depends on the torsional stiffness of the wing structure. Therefore, ailerons become markedly less effective as the speed increases until, at a particular speed, the aileron reversal speed, aileron deflection does not produce any rolling moment at all. At higher speeds reversed aileron movements are necessary in that a positive increment of wing lift requires an upward aileron deflection and vice versa. Similar, less critical, problems arise in the loss of effectiveness and reversal of the rudder and elevator controls. They are complicated by the additional deformations of the fuselage and tailplane-fuselage attachment points, which may be as important as the deformations of the tailplane itself. We shall concentrate in this section on the problem of aileron effectiveness and reversal. 13.2.1 Aileron effectiveness and reversal (two-dimensional case) We shall illustrate the problem by investigating, as in Section 13.1, the case of a wing- aileron combination in a two-dimensional flow. In Fig. 13.5 an aileron deflection < produces changes AL and AM, in the wing lift, L, and wing pitching moment M,; 13.2 Control effectiveness and reversal 547 L + AL Spring stiffness K V ____t t- Fig. Aileron effectiveness and reversal speed (two-dimensional tax,. these in turn cause an elastic twist, 8, of the wing. Thus (13.12) where aCL/aa has been previously defined and aCL/a( is the rate of change of lift coefficient with aileron angle. Also in which aCM,o/at is the rate of change of wing pitching moment coefficient with aileron deflection. The moment produced by these increments in lift and pitching moment is equilibrated by an increment of torque AT about the flexural axis. Hence Isolating e from Eq. (13.14) gives pv2sc[(acL/at)e f acM,O/ad< e= K - $pv2Sce(aCL/aa) Substituting for B in Eq. (13.12) we have ifv2sc{(acL/at)e f acM,O/%) acL I acL] E K - 3 p V2Sce ( dCL/da) aa at which simplifies to (13.15) 548 Elementary aeroelasticity The increment of wing lift is therefore a linear function of aileron deflection and becomes zero, that is aileron reversal occurs, when Hence the aileron reversal speed, Vr, is, from Eq. (13.17) (1 3.17) (13.18) We may define aileron effectiveness at speeds below the reversal speed in terms of aileron effectiveness = AL/ALR (13.19) the lift ALR produced by an aileron deflection on a rigid wing. Thus where (1 3.20) Hence, substituting in Eq. (13.19) for AL from Eq. (13.16) and ALR from Eq. (13.20), we have Equation (13.21) may be expressed in terms of the wing divergence speed Vd and aileron reversal speed V, , using Eqs (1 3.3) and (1 3.18) respectively; hence 1 - V’/V? 1 - V’/Vi aileron effectiveness = (13.22) We see that when V, = V,, which occurs when aCL/at = -(acM,o/at)/e, then the aileron is completely effective at all speeds. Such a situation arises because the nose-down wing twist caused by aileron deflection is cancelled by the nose-up twist produced by the increase in wing lift. Although the analysis described above is based on a two-dimensional case it is sometimes used in practice to give approximate answers for finite wings. The method is to apply the theory to a representative wing cross-section at an arbitrary spanwise station and use the local wing section properties in the formulae. 13.2.2 Aileron effectiveness and reversal (finite wing) We shall again apply strip theory to investigate the aeroelastic effects of aileron deflec- tion on a finite wing. In Fig. 13.6(a) the deflection of the aileron through an angle t produces a rolling velocity p radlsec, having the sense shown. The wing incidence at any section z is thus reduced due to p by an amount pz/ V. The downward aileron deflection shown here coincides with an upward deflection on the opposite wing, thereby contributing to the rolling velocity p. The incidence of the opposite wing is therefore increased by this direction of roll. Since we are concerned with aileron 13.2 Control effectiveness and reversal 549 t’ Lines z’ A’ X > T cc;y, +-!!I az dz ( b) Fig. 13.6 Aileron effectiveness and reversal speed (finite wing). effects we consider the antisymmetric lift and pitching moment produced by aileron deflection. Thus, in Fig. 13.6(b), the forces and moments are changes from the level flight condition. The lift AL on the strip shown in Fig. 13.6(b) is given by (13.23) where dcl /aa has been previously defined and ac, /a< is the rate of change of local wing lift coefficient with aileron angle. The function fa(.) represents aileron forces and moments along the span; for 0 < z < sl,fa(z) = 0 and for s1 < z < S,fa(Z) = 1. The pitching moment AMo on the elemental strip is given by ( 1 3.24) AM0 = 1 TpV2?6z-fa(z)E acm o a< in which acm,/a< is the rate of change of local pitching moment coefficient with aileron angle. Considering the moment equilibrium of the elemental strip of Fig. 13.6(b) we obtain, neglecting wing weight c~z + ALec + AMo = 0 dz or substituting for AL and AMo from Eqs (13.23) and (13.24) (13.25) 550 Elementary aeroelasticity Substituting for T in Eq. (13.26) from torsion theory (7‘ = GJd8/dz) and rearranging we have d28 ;pv2ec‘acl/aa ;pv2c2 [ acl PZ acl dz2 e=- e- - - e-fa(z)C - *f,cz)(] (13.27) GJ GJ aa v at aC -+ Writing we obtain It may be shown that the solution of Eq. (13.28), satisfying the boundary conditions 8=0 atz=O and dO/dz=O atz=s is ] (13.29) sinXz 5 sin X(s - sl) cos xr x [fa(z){ 1 - cos X(z - SI)} - where cos X(z - sl) = 0 when z < s1 The spanwise variation of total local wing lift coefficient is given by strip theory as (13.30) where 8 is known from Eq. (13.29) and a is the steady flight wing incidence. The aileron effectiveness is often measured in terms of the wing-tip helix angle (ps/ V) per unit aileron displacement during a steady roll. In this condition the rolling moments due to a given aileron deflection, [, wing twist and aerodynamic damping are in equilibrium so that from Fig. 13.6(a) and Eq. (13.23) and noting that ailerons on opposite wings both contribute to the rolling, we have from which (13.31) (13.32) 13.3 Structural vibration 551 Substituting for 8 from Eq. (13.29) into Eq. (13.32) gives Hence 1 sinX(s - sl) . sin Xz q{ (%+: 2) [r,e){l -cosx(z-s])) - cos As (13.33) Therefore, aileron effectiveness (ps/ V)/< is given by Integration of the right-hand side of the above equation gives 1 8cm.o - e(acl/aa) < (13.34) The aileron reversal speed occurs when the aileron effectiveness is zero. Thus, equat- ing the numerator of Eq. (13.34) to zero, we obtain the transcendental equation (%+: %) (COSXS - COSXS~) + cosXs = 0 (13.35) a€ Alternative methods of obtaining divergence and control reversal speeds employ matrix or energy procedures. Details of such treatments may be found in Ref. 3. The remainder of this chapter is concerned with dynamic problems of aeroelasticity, of whichflutter is of primary importance. Flutter has been defined as the dynamic instability of an elastic body in an airstream and is produced by aerodynamic forces which result from the deflection of the elastic body from its undeformed state. The determination of critical or Jlutter speeds for the continuous structure of 552 Elementary aeroelasticity I ‘-1 - L J ass -l m X Fig. 13.7 Oscillation of a masdspring system. an aircraft is a complex process since such a structure possesses an infinite number of natural or normal modes of vibration. Simplifying assumptions, such as breaking down the structure into a number of concentrated masses connected by weightless elastic beams (lumped mass concept) are made, but whatever method is employed the natural modes and frequencies of vibration of the structure must be known before flutter speeds and frequencies can be found. We shall discuss flutter and other dynamic aeroelastic phenomena later in the chapter; for the moment we shall consider methods of calculating normal modes and frequencies of vibration of a variety of beam and mass systems. Let us suppose that the simple mass/spring system shown in Fig. 13.7 is displaced by a small amount xo and suddenly released. The equation of the resulting motion in the absence of damping forces is mx+kx=O ( 13.36) where k is the spring stiffness. We see from Eq. (13.36) that the mass, m, oscillates with simple harmonic motion given by x = xo sin(wt + E) (13.37) in which 3 = k/m and E is a phase angle. The frequency of the oscillation is w/2~ cycles per second and its amplitude xo. Further, the periodic time of the motion, that is the time taken by one complete oscillation, is ~K/w. Both the frequency and periodic time are seen to depend upon the basic physical characteristics of the system, namely the spring stiffness and the magnitude of the mass. Therefore, although the amplitude of the oscillation may be changed by altering the size of the initial disturbance, its frequency is ked. This frequency is the normal or natural frequency of the system and the vertical simple harmonic motion of the mass is its normal mode of vibration. Consider now the system of n masses connected by (n - 1) springs, as shown in Fig. 13.8. If we specify that motion may only take place in the direction of the spring axes then the system has n degrees of freedom. It is therefore possible to set the system oscillating with simple harmonic motion in n different ways. In each of these n modes of vibration the masses oscillate in phase so that they all attain maximum amplitude at the same time and pass through their zero displacement 13.3 Structural vibration 553 Fig. 13.8 Oscillation of an n masdspring system. positions at the same time. The set of amplitudes and the corresponding frequency take up different values in each of the n modes. Again these modes are termed normal or natural modes of vibration and the corresponding frequencies are called normal or natural frequencies. The determination of normal modes and frequencies for a general spring/mass system involves the solution of a set of n simultaneous second-order differential equations of a type similar to Eq. (13.36). Associated with each solution are two arbitrary constants which determine the phase and amplitude of each mode of vibration. We can therefore relate the vibration of a system to a given set of initial conditions by assigning appropriate values to these constants. A useful property of the normal modes of a system is their orthogonality, which is demonstrated by the provable fact that the product of the inertia forces in one mode and the displacements in another results in zero work done. In other words displace- ments in one mode cannot be produced by inertia forces in another. It follows that the normal modes are independent of one another so that the response of each mode to an externally applied force may be found without reference to the other modes. Thus, by considering the response of each mode in turn and adding the resulting motions we can find the response of the complete system to the applied loading. Another useful characteristic of normal modes is their ‘stationary property’. It can be shown that if an elastic system is forced to vibrate in a mode that is slightly different from a true normal mode the frequency is only very slightly different to the corresponding natural frequency of the system. Reasonably accurate estimates of natural frequencies may therefore be made from ‘guessed’ modes of displacement. We shall proceed to illustrate the general method of solution by determining normal modes and frequencies of some simple beam/mass systems. Two approaches are possible: a stiyness or displacement method in which spring or elastic forces are expressed in terms of stiffness parameters such as k in Eq. (13.36); and aflexibility or force method in which elastic forces are expressed in terms of the flexibility 6 of the elastic system. In the latter approach 6 is defined as the deflection due to unit force; the equation of motion of the spring/mass system of Fig. 13.7 then becomes (13.38) mx+-=O Again the solution takes the form x = xo sin(wt + E) but in this case ~3 = l/rnS. Clearly by our definitions of k and 6 the product k6 = 1. In problems involving rotational oscillations m becomes the moment of inertia of the mass and S the rotation or displacement produced by unit moment. Let us consider a spring/mass system having a finite number, n, degrees of freedom. The term spring is used here in a general sense in that the n masses ml, X 6 554 Elementary aeroelasticity m2, mi, m, may be connected by any form of elastic weightless member. Thus, if mi is the mass at a point i where the displacement is xi and 6, is the displacement at the point i due to a unit load at a point j (note from the reciprocal theorem 6, = S,), the n equations of motion for the system are mlxlS11 + m2x2S12 + + mixiSli + + rnnjt,S1, + x1 = 0 mlxlS21 + m2x2622 + + mixi62i + + mnxnS2, + x2 = 0 mljilSil + m2x2Siz + + mixiSii + + mnxn,4 + xi = 0 mlxlS,l + m2x2Sn2 + + mixiSni + + mnxnS,, + x,, = 0 (1 3.39) 1 or n mjxjS, + xi = 0 (i = 1,2, n) j= 1 (13.40) Since each normal mode of the system oscillates with simple harmonic motion, then the solution for the ith mode takes the form x = xf sin(wt + E) so that jii = -Jxf sin(wt + E) = -w2xi. Equation (13.40) may therefore be written as n -JCmjsijxj +xi = o (i = 1,2,. ,n) (1 3.41) j= 1 For a non-trivial solution, that is xi # 0, the determinant of Eqs (13.41) must be zero. Hence (Jm1611 - 1) dm2612 w2mi sli w2mn s~,, w2mlS21 (w2m2S22 - 1) w2miS2i w2mn 6% I=O 2 w2ml ail w2m2si2 (w2miSii - 1) w m,S, I I 2 I w2mlsn1 Jm26n2 w miSni (w2wI,,~,,,, - 1) I (13.42) The solution of Eqs (13.42) gives the normal frequencies of vibration of the system. The corresponding modes may then be deduced as we shall see in the following examples. Example 13.1 Determine the normal modes and frequencies of vibration of a weightless cantilever supporting masses m/3 and m at points 1 and 2 as shown in Fig. 13.9. The flexural rigidity of the cantilever is EI. The equations of motion of the system are (m/3)ij1611 + mij2Sl2 + w1 = 0 (rn/3)ijlS2, + mij2S22 + wz = 0 (9 (ii) 13.3 Structural vibration 555 Fig. 13.9 Massheam system for Example 13.1. where wl and v2 are the vertical displacements of the masses 1 at any instant of time. In this example, displacements are-assumed to be caused by bending strains only; the flexibility coefficients Sll, S2, and S12(= 621) may therefore be found by the unit load method described in Section 4.8. From the first of Eqs (4.27) we deduce that (iii) where Mi is the bending moment at any section z due to a unit load at the point i and Mi is the bending moment at any section z produced by a unit load at the point j. Therefore, from Fig. 13.9 M1 = l(1-z) O<Z<l M2 = 1(1/2 - z) M2=0 112 < z < 1 0 < z < 112 Hence Integrating Eqs (iv), (v) and (vi) and substituting limits, we obtain 513 s -6 - 13 i3 611 =E 7 622 =- 24EI ’ l2 - 21 - 48EI Each mass describes simple harmonic motion in the normal modes of oscillation so that w1 = v’: sin(wt + E) and v2 = v! sin(wt + E). Hence iil = -w w1 and 32 = -&2. Substituting for ij,, w2, Sll, S2, and S12(= in Eqs (i) and (ii) and writing X = nzI3/(3 x 48EI), we obtain 2 (1 - 16Xu2)vl - 15Xw2v1 = 0 (4 5XW2Vl - (1 - ~XW’)W~ = 0 (viii) [...]... From Eq (vii) -~2 15Xw2 15 x (1/21) 1 - 16Xw2- 1 - 16 x (1/21) - which is a positive quantity Therefore, at the lowest natural frequency the cantilever oscillates in such a way that the displacement of both masses has the same sign at the same instant of time Such an oscillation would take the form shown in Fig 13.10 Substituting the second natural frequency in Eq (vii) we have 15Xw’ - 15 -1 - 1 6 h 2... the designer requires the information before the aircraft is built, although this type of test is carried out on the completed aircraft as a design check The alternative of building a scale model has found favour for many years Such models are usually designed to be as light as possible and to represent the stiffness characteristics of the full-scale aircraft The inertia properties are simulated by... accelerometers over a given frequency range Having obtained the resonant frequencies the aircraft is then excited at each of these frequencies in turn and all accelerometer records taken simultaneously Babister, A W., Aircraft Stability and Control, Pergamon Press, London, 1961 Duncan, W J., The Principles of the Control and Stability of Aircraft, Cambridge University Press, Cambridge, 1959 Bisplinghoff, R L., Ashley,... we obtain w1 = 1 .158 4dZ EI (iii) which value is only 0.02 per cent higher than the true value given above The estimation of higher natural frequencies requires the assumption of further, more complex, shapes for V ( z ) It is clear from the previous elementary examples of normal mode and natural frequency calculation that the estimation of such modes and frequencies for a complete aircraft is a complex... aircraft is a complex process However, the aircraft designer is not restricted to calculation for the solution of such problems, although the advent of the digital computer has widened the scope and accuracy of this approach Other possible methods are to obtain the natural frequencies and modes by direct measurement from the results of a resonance test on the actual aircraft or to carry out a similar test... (1 - u 2 r n G ) u - u-mr2 -e 7 312 = o 2EI (4 (vii) 560 Elementary aeroelasticity Inserting the values of m,r, I and EI we have 1435 x 4 x 0.763 1435 x 0 .152 2x 3 x 0.762 w2e = o (l -9.81 x 3 x 1.44 x lo6 w2)v9.81 x 2 x 1.44 x lo6 (viii) 1435 x 0 .152 2x 2 x 0.76 8 =0 (l 9.81 x 1.44 x lo6 w2) (ix) - 1435 x 3 x 0.762 9.81 x 2 x 1.44 x lo6 " + or (1 -6x 10-5w2)v- 0.203 x 10-5w28= 0 (4 (1 - 0.36 x 10-5w2)8=... modes and frequencies is also treated in Refs 3 and 4 13.4 to "wutter' We have previously defined flutter as the dynamic instability of an elastic body in an airstream It is found most frequently in aircraft structures subjected to large aerodynamic loads such as wings, tail units and control surfaces Flutter occurs at a critical or flutter speed Vf which in turn is defined as the lowest airspeed at which... the critical flutter speed for each mode may be found by calculating S for a range of speeds and determining the value of speed at which S = 0 A similar approach is used experimentally on actual aircraft The aircraft is flown at a given steady speed and caused to oscillate either by exploding a small detonator on the wing or control surface or by a sudden control jerk The resulting oscillations + Decay... pitching and translation of the complete aircraft It can be shown4 that control surface flutter can be prevented by eliminating the inertial coupling between the control rotation and the motion of the main surface This may be achieved by mass balancing the control surface whereby weights are attached to the control surface forward of the hinge line All newly designed aircraft are subjected early in the... such tests are to check the accuracy of the calculated normal modes on which the flutter predictions are based and to show up any unanticipated peculiarities in the vibrational behaviour of the aircraft Usually the aircraft rests on some low frequency support system or even on its deflated tyres Electrodynamic exciters are mounted in pairs on the wings and tail with accelerometers as the measuring devices . obtain 2 (1 - 16Xu2)vl - 15Xw2v1 = 0 (4 5XW2Vl - (1 - ~XW’)W~ = 0 (viii) 556 Elementary aeroelasticity For a non-trivial solution (1 - 16d) -15Xw’ -( 1 - 6XJ) Expanding. natural frequency in Eq. (vii) we have - - ~2 1 - 16Xw2 - 1 - 16 x (1/21) 15 - v1 15Xw’ -= ~2 1 - 16h2 1- 16 which is negative so that the masses have displacements. 0.763 1435 x 0 .152 2 x 3 x 0.762 w2e = o (viii) 8 = 0 (ix) (l -9.81 x 3 x 1.44 x lo6 w2)v- 9.81 x 2 x 1.44 x lo6 1435 x 3 x 0.762 1435 x 0 .152 2 x 2 x 0.76