Aircraft structures for engineering students - part 10 pot

Problems 535 P.12.4 The symmetrical plane rigid jointed frame 1234567, shown in Fig. P.12.4, is fixed to rigid supports at 1 and 5 and supported by rollers inclined at 45" to the horizontal at nodes 3 and 7. It carries a vertical point load P at node 4 and a uniformly distributed load w per unit length on the span 26. Assuming the same flexural rigidity EI for all members, set up the stiffness equations which, when solved, give the nodal displacements of the frame. Explain how the member forces can be obtained. 2 z z Fig. P.12.4 P.12.5 The frame shown in Fig. P.12.5 has the planes xz and yz as planes of symmetry. The nodal coordinates of one quarter of the frame are given in Table P.12.5(i). In this structure the deformation of each member is due to a single effect, this being axial, bending or torsional. The mode of deformation of each member is given in Table P. 12.5(ii), together with the relevant rigidity. Fig. P.12.5 536 Matrix methods of structural analysis Table P.12.5(i) Node X Y z 2 0 0 0 3 L 0 0 I L 0.8L 0 9 L 0 L Table P.12.5(ii) Bending Torsional 23 - EI - 37 - - GJ = 0.8EI - EI L= EA = 6fi- - 29 Use the direct stzrness method to find all the displacements and hence calculate the forces in all the members. For member 123 plot the shear force and bending moment diagrams. Briefly outline the sequence of operations in a typical computer program suitable for linear frame analysis. Ans. S29 = S28 = AP/6 (tension) M3 = -MI = PL/9 (hogging), M2 = 2PL/9 (sagging) SF12 = -SF23 = P/3 Twisting moment in 37, PL/18 (anticlockwise). P.12.6 Given that the force-displacement (stiffness) relationship for the beam element shown in Fig. P.12.6(a) may be expressed in the following form: obtain the force-displacement (stiffness) relationship for the variable section beam (Fig. P.12.6(b)), composed of elements 12, 23 and 34. Such a beam is loaded and supported symmetrically as shown in Fig. P.12.6(c). Both ends are rigidly fixed and the ties FB, CH have a cross-section area al and the ties EB, CG a cross-section area q. Calculate the deflections under the loads, the forces in the ties and all other information necessary for sketching the bending moment and shear force diagrams for the beam. Neglect axial effects in the beam. The ties are made from the same material as the beam. Problems 537 L L L 4 b Y Fig. P.12.6 Am. Fy,l I "1 vg = vc = -5PL3/144EI, eB = -ec = PL~~EI Si =2P/3, S2 = &P/3 FJ,A = P/3: MA = -PL/4 P.12.7 The symmetrical rigid jointed grillage shown in Fig. P. 12.7 is encastrk at 6, 7,8 and 9 and rests on simple supports at 1,2,4 and 5. It is loaded with a vertical point load P at 3. Use the stiffness method to find the displacements of the structure and hence calcu- late the support reactions and the forces in all the members. Plot the bending moment diagram for 123. All members have the same section properties and GJ = 0.8EI. 538 Matrix methods of structural analysis Fig. P.12.7 Ans. Fy,l =Fy,5 = -PI16 Fy,z = Fy,4 = 9P/ 16 MZ1. =M45 = -P1/16 (hogging) MZ3 =M43 = -PI112 (hogging) Twisting moment in 62, 82, 74 and 94 is P1/96. P.12.8 It is required to formulate the stiffness of a triangular element 123 with coordinates (0, 0), (a, 0) and (0, a) respectively, to be used for 'plane stress' problems. (a) Form the [B] matrix. (b) Obtain the stiffness matrix [K']. Why, in general, is a finite element solution not an exact solution? P.12.9 It is required to form the stiffness matrix of a triangular element 123 for use in stress analysis problems. The coordinates of the element are (1, l), (2,l) and (2,2) respectively. (a) Assume a suitable displacement field explaining the reasons for your choice. (b) Form the [B] matrix. (c) Form the matrix which gives, when multiplied by the element nodal displace- ments, the stresses in the element. Assume a general [D] matrix. P.12.10 It is required to form the stiffness matrix for a rectangular element of side (a) Assume a suitable displacement field. (b) Form the [q matrix. 2a x 2b and thickness t for use in 'plane stress' problems. (c) Obtain Jvol [qT [Dl [q d v. Note that the stiffness matrix may be expressed as Problems 539 P.12.11 A square element 1234, whose corners have coordinates x, y (in metres) of (- 1, -l), (1 - l), (1,l) and (- 1 1) respectively, was used in a plane stress finite element analysis. The following nodal displacements (mm) were obtained: ~1=0.1, ~2=0.3, ~3~0.6, ~4~0.1 ~l=O.l, ~2~0.3, ~3=0.7, ~4~0.5 If Young’s modulus E = 200000N/mm2 and Poisson’s ratio v = 0.3, calculate the stresses at the centre of the element. Ans. ux = 51.65N/mm2, uy = 55.49N/mm2, rxJ = 13.46N/mm2. Elementary aeroelasticity Aircraft structures, being extremely flexible, are prone to distortion under load. When these loads are caused by aerodynamic forces, which themselves depend on the geo- metry of the structure and the orientation of the various structural components to the surrounding airflow, then structural distortion results in changes in aerodynamic load, leading to further distortion and so on. The interaction of aerodynamic and elastic forces is known as aeroelasticity. Two distinct types of aeroelastic problem occur. One involves the interaction of aerodynamic and elastic forces of the type described above. Such interactions may exhibit divergent tendencies in a too flexible structure, leading to failure, or, in an adequately stiff structure, converge until a condition of stable equilibrium is reached. In this type of problem static or steady state systems of aerodynamic and elastic forces produce such aeroelastic phenomena as divergence and control reversal. The second class of problem involves the inertia of the structure as well as aerodynamic and elastic forces. Dynamic loading systems, of which gusts are of primary importance, induce oscillations of structural components. If the natural or resonant frequency of the component is in the region of the frequency of the applied loads then the amplitude of the oscillations may diverge, causing failure. Also, as we observed in Chapter 8, the presence of fluctuating loads is a fatigue hazard. For obvious reasons we refer to these prob- lems as dynamic. Included in this group are flutter, buffeting and dynamic response. The various aeroelastic problems may be conveniently summarized in the form of a ‘tree’ as follows Aeroelasticity a Dynamic . . . Dynamic Static stability . Static stability 4-7 I Load Divergence Control Flutter Buffeting Dynamic distribution reversal response 13.1 load distribution and divergence 541 In this chapter we shall concentrate on the purely structural aspects of aeroelasticity; its effect on aircraft static and dynamic stability is treated in books devoted primarily to aircraft stability and control''2. rioad di stribution and divergence Redistribution of aerodynamic loads and divergence are closely related aeroelastic phenomena; we shall therefore consider them simultaneously. It is essential in the design of structural components that the aerodynamic load distribution on the com- ponent is known. Wing distortion, for example, may produce significant changes in lift distribution from that calculated on the assumption of a rigid wing, especially in instances of high wing loadings such as those experienced in manoeuvres and gusts. To estimate actual lift distributions the aerodynamicist requires to know the incidence of the wing at all stations along its span. Obviously this is affected by any twisting of the wing which may be present. Let us consider the case of a simple straight wing with the centre of twist (or flexural centre, see Chapters 9 and 10) behind the aerodynamic centre (see Fig. 13.1). The moment of the lift vector about the centre of twist causes an increase in wing incidence which produces a further increase in lift, leading to another increase in incidence and so on. At speeds below a critical value, called the divergence speed, the increments in lift converge to a condition of stable equilibrium in which the torsional moment of the aerodynamic forces about the centre of twist is balanced by the torsional rigidity of the wing. The calculation of lift distribution then proceeds from a knowledge of the distribution of twist along the wing. For a straight wing the redistribution of lift usually causes an outward spanwise movement of the centre of pressure, resulting in greater bending moments at the wing root. In the case of a swept wing a reduction in streamwise incidence of the outboard sections due to bending deflections causes a movement of the centre of pressure towards the wing root. All aerodynamic surfaces of the aircraft suffer similar load redistribution due to distortion. 13.1 .I Wing torsional divergence (two-dimensional case) The most common divergence problem is the torsional divergence of a wing. It is useful, initially, to consider the case of a wing of area S without ailerons and in a Lift A Wing twist, Centre of twist I Aerodynamic centre Fig. 13.1 Increase of wing incidence due to wing twist. 542 Elementary aeroelasticity L t AC Fig. 13.2 Determination of wing divergence speed (two-dimensional case). two-dimensional flow, as shown in Fig. 13.2. The torsional stiffness of the wing, which we shall represent by a spring of stiffness K, resists the moment of the lift vector, L, and the wing pitching moment Mo, acting at the aerodynamic centre of the wing section. For moment equilibrium of the wing section about the aerodynamic centre we have Mo + Lec = KO (13.1) where ec is the distance of the aerodynamic centre forward of the flexural centre expressed in terms of the wing chord, c, and 8 is the elastic twist of the wing. From aerodynamic theory MO = ~~v’sccM,~, L = 4pv2scL Substituting in Eq. (13.1) yields $~V~S(CCM,O + ecCL) = KO or, since in which (Y is the initial wing incidence or, in other words, the incidence corresponding to given flight conditions assuming that the wing is rigid and CL.o is the wing lift coefficient at zero incidence, then -pv2s CCM,~ + eCL:, + ec- (a + e) = Ke 2 l[ acL aa 1 where aCL/aa is the wing lift curve slope. Rearranging gives or (13.2) 13.1 Load distribution and divergence 543 Equation (13.2) shows that divergence occurs (Le. 6 becomes infinite) when 1 2 acL K = -pV Sec- 2 da The divergence speed vd is then (13.3) We see from Eq. (13.3) that vd may be increased either by stiffening the wing (increas- ing K) or by reducing the distance ec between the aerodynamic and flexural centres. The former approach involves weight and cost penalties so that designers usually prefer to design a wing structure with the flexural centre as far forward as possible. If the aerodynamic centre coincides with or is aft of the flexural centre then the wing is stable at all speeds. 13.1.2 Wing __U.r_ll.__IIm "-__U-~ ~~ torsional divergence (finite wing) __*.I____ 1 We shall consider the simple case of a straight wing having its flexural axis nearly perpendicular to the aircraft's plane of symmetry (Fig. 13.3(a)). We shall also assume that wing cross-sections remain undistorted under the loading. Applying strip theory in the usual manner, that is we regard a small element of chord c and spanwise width 6z as acting independently of the remainder of the wing and consider its equilibrium, we have from Fig. 13.3(b), neglecting wing weight (T +g&) - T + ALec + AM, = 0 (13.4) AY Line of ACs z Flexural axis dz (bl Fig. 13.3 Determination of wing divergence speed (three-dimensional case). 544 Elementary aeroelasticity where T is the applied torque at any spanwise section z and AL and AMo are the lift and pitching moment on the elemental strip acting at its aerodynamic centre. As Sz approaches zero, Eq. (13.4) becomes dT dL dMo -+ ec-+- = 0 dz dzdz (13.5) In Eq. (13.4) AL = -pV2cSz-(a 1 OCl + e) 2 Sa where dcl /acu is the local two-dimensional lift curve slope and in which c,,~ is the local pitching moment coefficient about the aerodynamic centre. Also from torsion theory (see Chapter 3) T = GJ dO/dz. Substituting for L, Mo and T in Eq. (13.5) gives (13.6) d28 4 pV2ec2 (acl /&)e - - $ p V2e? (del /da)a - i pV2c2c,,o -+ dz2 - GJ GJ GJ Equation (13.6) is a second-order differential equation in 0 having a solution of the standard form (13.7) where and A and B are unknown constants that are obtained from the boundary conditions; namely, 6 = 0 when z = 0 at the wing root and de/& = 0 at z = s since the torque is zero at the wing tip. From the first of these and from the second Hence +a (tanhsinAz+cosXz- 1) 6 = [ e(zTaa) I or rearranging (13.8) (13.9) [...]... or (1 -6 x 1 0-5 w2)v- 0.203 x 1 0-5 w28= 0 (4 (1 - 0.36 x 1 0-5 w2)8= 0 (4 -8 .8 x lO-'w%+ Solving Eqs (x) and (xi) as before gives w = 122 or 1300 from which the natural frequencies are 61 650 h= ,h =-7 r 7 r From Eq (x) v - - 0.203 x 8- 1 -6 x lOP5w2 1 0-5 w2 which is positive at the lowest natural frequency, corresponding to w = 122, and negative for w = 1300.The modes of vibration are therefore as shown in... 0 = -X2BsinXL- X2CcosXL+X2DsinhXL+X2FcoshAL (iii) + X3Csin XL + X3Dcosh XL + X3F sinh AL (iv) 0 = -X3Bcos X L From Eqs (i) and (ii), C = -F and B = -D Thus, replacing F and D in Eqs (iii) and (iv) we obtain B (- sin X - sinh XL) L + C (- cos XL - cosh XL) = 0 (v) and B (- cos X - cosh XL) + C(sin X - sinh XL) = 0 L L (vi) Eliminating B and C from Eqs (v) and (vi) gives (- sin X - sinh XL)(sinh XL - sin... and S12(= in Eqs (i) and (ii) and writing X = nzI3/(3 x 48EI), we obtain (1 - 1 6 X u 2 ) v l - 15Xw2v1 = 0 5XW2Vl - (1 - ~ X W ’ ) W ~ 0 = (4 (viii) 556 Elementary aeroelasticity For a non-trivial solution (1 - 1 6 d ) -1 5Xw’ -( 1 - 6 X J ) Expanding this determinant we have -( 1 - 16Xw2)( 1- 6 X J ) + 75(Xw2)’ = 0 or 21(AJ)’ - 22xw2 +1=0 Inspection of Eq (ix) shows that x w 2 = 1/21 or 1 Hence 3 x48EI... e =-& Substituting in Eqs (i) and (ii) gives (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 0.1522x 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 0.1522x 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 -6 x 1 0-5 w2)v-... motion, then the solution for the ith mode takes the form x = xf sin(wt E ) so that jii = - J x f sin(wt E ) = -w2xi Equation (13.40)may therefore be written as + + n - J C m j s i j x j + x i = o (i = 1,2, ,n) ( 13.41) j= 1 For a non-trivial solution, that is xi # 0, the determinant of Eqs (13.41)must be zero Hence (Jm1611 - 1) I w2misli w2mn s~,, w2mlS21 I dm2612 (w2m2S2 2- 1 ) w2miS2i w2mn % 6... frequencies of vibration are therefore 2 w = w 2-6 F h = g - ; m13 The system is therefore capable of vibrating at two distinct frequencies To determine the normal mode corresponding to each frequency we first take the lower frequencyfi and substitute it in either Eq (vii) or Eq (viii) From Eq (vii) -~ 2 15Xw2 15 x (1/21) 1 - 16Xw 2- 1 - 16 x (1/21) - which is a positive quantity Therefore, at the lowest natural... - S, - pA6zd2V = 0 at2 so that ( 13.44) From basic bending theory (see Eqs (9.20)) (13.45) It follows from Eqs (13.43), (13.44) and (13.45) that a2 (13.46) Equation (13.46) is applicable to both uniform and non-uniform beams In the latter case the flexural rigidity, EI, and the mass per unit length, pA, are functions of 3 For a beam of uniform section, Eq (13.46) reduces to 6 % + az4 a2v EI - P A -. .. Therefore, from Fig 13.9 O . 23 - EI - 37 - - GJ = 0.8EI - EI L= EA = 6fi- - 29 Use the direct stzrness method to find all the displacements and hence calculate the forces in all the members. For member. Substituting for L, Mo and T in Eq. (13.5) gives (13.6) d28 4 pV2ec2 (acl /&)e - - $ p V2e? (del /da)a - i pV2c2c,,o -+ dz2 - GJ GJ GJ Equation (13.6) is a second-order. 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.