10.4 Fuselage frames and wing ribs 413 300 rnrn 300 mrn Fig. 10.48 Wing rib of Example 10.14. Taking moments about flange 3 2(50000+95000)q23 +2 x 95000q12 = -15000 x 300Nmm (iii) Solution of Eqs (i), (ii) and (iii) gives q12 = 13.0N/~, q23 = -7.ON/m, 431 = 43.0N/m Consider now the nose portion of the rib shown in Fig. 10.49 and suppose that the shear flow in the web immediately to the left of the stiffener 24 is ql. The total vertical shear force Sy,l at this section is given by Sy,l = 7.0 x 300 = 2100N The horizontal components of the rib flange loads resist the bending moment at this section. Thus 2 x 50 000 x 7.0 = 2333.3 300 px,4 = px,2 = The corresponding vertical components are then Py,2 = Py,4 = 2333.3 tan 15" = 625.2N Thus the shear force carried by the web is 2100 - 2 x 625.2 = 849.6N. Hence 849.6 300 41 =-=2.8N/m py.4 p4 - 4' Fig. 10.49 Equilibrium of nose portion of the rib. 414 Stress analysis of aircraft components - 7.0 N/rnrn 320 rnm I Fig. 10.50 Equilibrium of rib forward of intermediate stiffener 56. The axial loads in the rib flanges at this section are given by Pz = P4 = (2333.32 + 625.22)''2 = 2415.6N The rib flange loads and web panel shear flows, at a vertical section immediately to the left of the intermediate web stiffener 56, are found by considering the free body diagram shown in Fig. 10.50. At this section the rib flanges have zero slope so that the flange loads P5 and P6 are obtained directly from the value of bending moment at this section. Thus P5=P6=2[(50000+46000) ~7.0-49000~ 13.0]/320=218.8N The shear force at this section is resisted solely by the web. Hence 32092 = 7.0 x 300 + 7.0 x 10 - 13.0 x 10 = 2040N so that 42 6.4N/- The shear flow in the rib immediately to the right of stiffener 56 is found most simply by considering the vertical equilibrium of stiffener 56 as shown in Fig. 10.51. Thus 320q3 = 6.4 x 320 + 15000 which gives q3 = 53.3N/mm Fig. 10.51 Equilibrium of stiffener 56. 10.5 Cut-outs in wings and fuselages 41 5 15000N I Fig. 10.52 Equilibrium of the rib forward of stiffener 31. Finally, we shall consider the rib flange loads and the web shear flow at a section immediately forward of stiffener 31. From Fig. 10.52, in which we take moments about the point 3 M3=2[(50000+95000) x7.0-95000~ 13.0]+15000x300=4.06x 106Nmm The horizontal components of the flange loads at this section are then 4.06 x lo6 300 = 13 533.3N PX,l = px,3 = and the vertical components are Py,l = Py,3 = 3626.2N Hence PI = P3 = J13 533.32 + 3626.22 = 14010.7N The total shear force at this section is 15 000 + 300 x 7.0 = 17 100 N. Therefore, the shear force resisted by the web is 17 100 - 2 x 3626.2 = 9847.6N so that the shear flow g3 in the web at this section is 9847.6 93 =- = 32.8 N/mm So far we have considered wings and fuselages to be closed boxes stiffened by transverse ribs or frames and longitudinal stringers. In practice it is necessary to provide openings in these closed stiffened shells. Thus, wings may have openings on their undersurfaces to accommodate retractable undercarriages; other openings might be required for fuel tanks, engine nacelles and weapon installations. Fuselage structures have openings for doors, cockpits, bomb bays, windows in passenger cabins etc. Other openings provide means of access for inspection and maintenance. These openings or 'cut-outs' produce discontinuities in the otherwise continuous shell 416 Stress analysis of aircraft components structure so that loads are redistributed in the vicinity of the cut-out affecting loads in the skin, stringers, ribs and frames of the wing and fuselage. Frequently these regions must be heavily reinforced resulting in unavoidable weight increases. 10.5.1 Cutats in wings Initially we shall consider the case of a wing subjected to a pure torque in which one bay of the wing has the skin on its undersurface removed. The method is best illustrated by a numerical example. Example 10.15 The structural portion of a wing consists of a three-bay rectangular section box which may be assumed to be hly attached at all points around its periphery to the aircraft fuselage at its inboard end. The skin on the undersurface of the central bay has been removed and the wing is subjected to a torque of lOkNm at its tip (Fig. 10.53). Calculate the shear flows in the skin panels and spar webs, the loads in the corner hges and the forces in the ribs on each side of the cut-out assuming that the spar hges carry all the-direct loads while the skin panels and spar webs are effective only in shear. If the wing structure were continuous and the effects of restrained warping at the built-in end ignored, the shear flows in the skin panels would be given by Eq. (9.49), i.e. = 31.3N/m~l T 10 x lo6 q= 2A - 2 x 200 x 800 Statio I Fig. 10.53 Three-bay wing structure with cut-out of Example 10.1 5. 10.5 Cut-outs in wings and fuselages 417 200 mm I Station 3000 Is 1500 rnm _I Fm;t FIRear war 10 kN m w (a) ( b) Fig. 10.54 Differential bending of front spar. and the flanges would be unloaded. However, the removal of the lower skin panel in bay @ results in a torsionally weak channel section for the length of bay @ which must in any case still transmit the applied torque to bay (iJ and subsequently to the wing support points. Although open section beams are inherently weak in torsion (see Section 9.6), the channel section in this case is attached at its inboard and outboard ends to torsionally stiff closed boxes so that, in effect, it is built-in at both ends. We shall examine the effect of axial constraint on open section beams subjected to torsion in Chapter 11. An alternative approach is to assume that the torque is transmitted across bay @ by the differential bending of the front and rear spars. The bending moment in each spar is resisted by the flange loads P as shown, for the front spar, in Fig. 10.54(a). The shear loads in the front and rear spars form a couple at any station in bay @ which is equivalent to the applied torque. Thus, from Fig. 10.54(b) 800s = 10 x 106Nmm i.e. S = 12500N The shear flow q1 in Fig. 10.54(a) is given by 12 500 200 41 = - = 62.5 N/mm Midway between stations 1500 and 3000 a point of contraflexure occurs in the front and rear spars so that at this point the bending moment is zero. Hence 200P = 12 500 x 750 N lzl~ll so that P = 46 875N 418 Stress analysis of aircraft components 4500 Fig. 10.55 Loads on bay @ of the wing of Example 10.15. Alternatively, P may be found by considering the equilibrium of either of the spar flanges. Thus 2P = 1500ql = 1500 x 62.5N whence P = 46875N The flange loads P are reacted by loads in the flanges of bays (iJ and 0. These flange loads are transmitted to the adjacent spar webs and skin panels as shown in Fig. 10.55 for bay 0 and modify the shear flow distribution given by Eq. (9.49). For equilibrium of flange 1 1500q2 - 150093 = P 46 875 N or 42 - q3 = 31.3 6) The resultant of the shear flows q2 and q3 must be equivalent to the applied torque. Hence, for moments about the centre of symmetry at any section in bay 0 and using Eq. (9.79) 200 x 800q2 + 200 x 800q3 = 10 x lo6 N mm or q2 + q3 = 62.5 Solving Eqs (i) and (ii) we obtain 92 = 46.9N/mm, 93 = 15.6N/n~n Comparison with the results of Eq. (9.49) shows that the shear flows are increased by a factor of 1.5 in the upper and lower skin panels and decreased by a factor of 0.5 in the spar webs. 10.5 Cut-outs in wings and fuselages 419 46.9 (9, -93) 46875N 1 46875 N Fig. 10.56 Distribution of load in the top flange of the front spar of the wing of Example 10.1 5. 46.9 I The flange loads are in equilibrium with the resultants of the shear flows in the adjacent skin panels and spar webs. Thus, for example, in the top flange of the front spar P(st.4500) = 0 P(st.3000) = 1500q2 - 1500q3 = 46 875 N (Compression) P(~t.2250) = 15ooq2 - 1500q3 - 750ql = 0 The loads along the remainder of the flange follow from antisymmetry giving the distribution shown in Fig. 10.56. The load distribution in the bottom flange of the rear spar will be identical to that shown in Fig. 10.56 while the distributions in the bottom flange of the front spar and the top flange of the rear spar will be reversed. We note that the flange loads are zero at the built-in end of the wing (station 0). Generally, however, additional stresses are induced by the warping restraint at the built-in end; these are investigated in Chapter 11. The loads on the wing ribs on either the inboard or outboard end of the cut-out are found by considering the shear flows in the skin panels and spar webs immediately inboard and outboard of the rib. Thus, for the rib at station 3000 we obtain the shear flow distribution shown in Fig. 10.57. The shear flows in the wing rib panels and the loads in the flanges and stiffeners are found as previously described in Section 10.4. In Example 10.15 we implicitly assumed in the analysis that the local effects of the cut-out were completely dissipated within the length of the adjoining bays which were equal in length to the cut-out bay. The validity of this assumption relies on St. Venant’s principle (Section 2.4). It may generally be assumed therefore that the effects of a cut-out are restricted to spanwise lengths of the wing equal to the length of the cut-out on both inboard and outboard ends of the cut-out bay. - 46.9 Fig. 10.57 Shear flows (Wmm) on wing rib at station 3000 in the wing of Example 10.1 5. 420 Stress analysis of aircraft components We shall now consider the more complex case of a wing having a cut-out and subjected to shear loads which produce both bending and torsion. Again the method is illustrated by a numerical example. Example IO. 16 A wing box has the skin panel on its undersurface removed between stations 2000 and 3000 and carries lift and drag loads which are constant between stations 1000 and 4000 as shown in Fig. 10.58(a). Determine the shear flows in the skin panels and spar webs and also the loads in the wing ribs at the inboard and outboard ends of the cut-out bay. Assume that all bending moments are resisted by the spar flanges while the skin panels and spar webs are effective only in shear. The simplest approach is first to determine the shear flows in the skin panels and spar webs as though the wing box were continuous and then to apply an equal and opposite shear flow to that calculated around the edges of the cut-out. The shear flows in the wing box without the cut-out will be the same in each bay and are calculated using the method described in Section 9.9 and illustrated in Example 9.14. This gives the shear flow distribution shown in Fig. 10.59. We now consider bay @ and apply a shear flow of 75.9 N/mm in the wall 34 in the opposite sense to that shown in Fig. 10.59. This reduces the shear flow in the wall 34 to zero and, in effect, restores the cut-out to bay 0. The shear flows in the remaining walls of the cut-out bay will no longer be equivalent to the externally applied shear 120 kN WO mm2 20 kN 4 - I_ mmm I mmm Station 4 4Ooo la1 lbl Fig. 10.58 Wing box of Example 10.16. Fig. 10.59 Shear flow (Wmm) distribution at any station in the wing box of Example 10.1 6 without cut-out. 10.5 Cut-outs in wings and fuselages 421 Fig. 10.60 Correction shear flows in the cut-out bay of the wing box of Example 10.16. loads so that corrections are required. Consider the cut-out bay (Fig. 10.60) with the shear flow of 75.9 N/mm applied in the opposite sense to that shown in Fig. 10.59. The correction shear flows d2, d2 and d4 may be found using statics. Thus, resolving forces horizontally we have 800d2 = 800 x 75.9N whence 42 = 75.9N/m Resolving forces vertically 200632 = 504'12 - 50 X 75.9 - 30044 = O and taking moments about 0 in Fig. 10.58(b) we obtain 2 x 52000& - 2 x 40000q'32 + 2 x 52000 x 75.9 - 2 x 60000qi4 = 0 Solving Eqs (i) and (ii) gives qi2 = 117.6N/mm, d4 = 53.1 N/mm (i) (ii) The final shear flows in bay @ are found by superimposing d2, qi2 and 44 on the shear flows in Fig. 10.59, giving the distribution shown in Fig. 10.61. Alternatively, these shear flows could have been found directly by considering the equilibrium of the cut-out bay under the action of the applied shear loads. The correction shear flows in bay @ (Fig. 10.60) will also modify the shear flow distributions in bays (iJ and 0. The correction shear flows to be applied to those shown in Fig. 10.59 for bay 0 (those in bay @ will be identical) may be found by determining the flange loads corresponding to the correction shear flows in bay @ . 4 Fig. 10.61 Final shear flows (Wmm) in the cut-out bay of the wing box of Example 10.16. 422 Stress analysis of aircraft components It can be seen from the magnitudes and directions of these correction shear flows (Fig. 10.60) that at any section in bay @ the loads in the upper and lower flanges of the front spar are equal in magnitude but opposite in direction; similarly for the rear spar. Thus, the correction shear flows in bay @ produce an identical system of flange loads to that shown in Fig. 10.54 for the cut-out bays in the wing structure of Example 10.15. It follows that these correction shear flows produce differential bending of the front and rear spars in bay 0 and that the spar bending moments and hence the flange loads are zero at the mid-bay points. Thus, at station 3000 the flange loads are PI = (75.9 + 53.1) x 500 = 64 500N (Compression) P4 = 64 500 N (Tension) Pz = (75.9 + 117.6) x 500 = 96 750 N (Tension) P3 = 96750N (Tension) These flange loads produce correction shear flows &, d3, d3 and dl in the skin panels and spar webs of bay @) as shown in Fig. 10.62. Thus for equilibrium of flange 1 1OOOd1+ 1000& = 64 500 N (iii) and for equilibrium of flange 2 1000& + lOOO& = 96750N 64 For equilibrium in the chordwise direction at any section in bay @) 80041 = 800d3 96 750 N 64 500 N 2 64 500 N 3 Station 4 4000 Fig. 10.62 Correction shear flows in bay 0 of the wing box of Example 10.16. [...]... q23 = i o ~ ( m 0 .88 68 - 0 .88 6~’ 0.5~’) x q34 = i 2 ~ p o o 0 .86 6e’ - ~r) x (i) (E) (iii) (4 q41 = 1.OGu’ For horizontal equilibrium 500 x 0 .88 6q4 1- 500 x O .86 6q23 = 0 giving q41 = q23 For vertical equilibrium 375q12 - 125q34 - 25oq23 = 22 000 For moment equilibrium about point 1 500 x 375 x 0 .88 6q23 + 125 x 500 x 0 .88 6q34= 22000 x 100 or 3q23 + q34 = 40.6 (vii) Substituting for qlz etc from Eqs... (10.45), the longitudinal strain in the bar is 25.0 &I = - 5. 68 = 44000 x 1 0-~ The lengthening, A,, of the bar is then AI = 5. 68 x x 500 i.e A = 0. 284 1 The major Poisson's ratio for the bar is found from Eq (10.50) Thus 80 x 40 vlt =m x 0.2 x 10 + '80 x 50 x 0.3 = 0.22 80 Hence the strain in the bar across its thickness is & , = -0 .22 5. 68 1 0 - ~ -1 .25 = 1 0-~ The reduction in thickness, A,, of the bar is then... become (10.57) From Eqs (1.27), (1. 28) and the derivation of Eq (5.13) we see that E, a'w = -z-, ax2 a2W EY = -z-, aY2 rxy -2 z= a'W axay 10.6 laminated composite structures 431 so that Eqs (10.57) may be rewritten as (10. 58) From Section 5.3 rtl2 M , = J ’ axzdz, M~ = 412 I, 1-1 12 tl2 ayz dz, Mxy = - J ‘ -t/2 T,~Z dz Substituting for a, cyand rxy , from Eqs (10. 58) and integrating, we obtain Writing... -1 20000 N m and -3 0000Nm respectively at the section shown, calculate the direct stresses in the booms Neglect axial constraint effects and assume that the lift and drag vectors are in vertical and horizontal planes Boom areas: B1 = B4 = B5 = B8 = 1OOOmm2 B2 = B3 = B6 = B = 600mm2 7 AFZS.~1 = -1 90.7N/m2, U4 = -1 63 .8 U7 N/lIUll2, = 189 .6N/m2, ~ 02 = -1 81 .7N/mm2, 65 = Us 14ON/mI'Il2, 63 66 = = -1 72.8N/mm2,... that dotted line 45 is not a wall Wall Length (mm) Thickness (mm) G (N/mm2) Boom Area (mrn') 34, 56 12.23,61, 18 36 ,81 45 380 356 306 610 0.915 0.915 1.220 1.220 20 700 24 200 24 80 0 24 80 0 1, 3, 6, 8 2,4, 5: 1 1290 645 Nose area N,= 51 500mm' 4 153 X\WN +- I53 3 3 +_ _ _x - - ' - mm + I-mm Fig P.10.11 P.10.12 A singly symmetric wing section consists of two closed cells and one The webs 25, 34 and... epoxy resin and the carbon filament 80 mm f Fig 10.69 Cross-section of the bar of Example 10.17 * 10.6 Laminated composite structures 429 From Eq (10. 48) the modulus of the bar is given by 80 x 40 80 x 10 E = 200000 x - 5000 x 1 80 x 50 80 x 50 + i.e El = 44000N/m111' The direct stress, q,in the longitudinal direction is given by a = 1 100 x io3 = 25.0 N / m 2 80 x 50 Therefore, from Eq (10.45), the longitudinal... into Eqs (v), (vi) and (vii), and solving for 8 , u‘ alid v’, gives 8 = 0.122/G, u‘ = 9.71/G, v‘ = 42.9/G The values of#, u’ and v’ are now inserted in Eqs (i), (ii), (iii) and (iv), giving qI2= 68. 5N/mm, q23 = 9 .8 N/mm, q34 = 11.9N/mm, q41 = 9 .8 N/mm from which q2= 42 .8 N/mm2, ~ 2 = 7-4 1 3 = 9 .8 N/mm2: r34 9.9 N/mm2 = 4 48 Structural constraint Fig 11.3 Built-in end of a beam section having a curved... direction produced by the transverse stress at is given by ut a t ut ufl- = urn- = qE t E t Ef (10.51) From the last two of Eqs (10.51) E q=-uf Em m Substituting in Eq (10.50) or, from Eq (10. 48) Now substituting for urnin the first two of Eqs (10.51) u1 -t= - Yt Et E t or (10.52) 4 28 Stress analysis of aircraft components Fig 10. 68 Determination of Glt Finally, the shear modulus qt(= is determined by... displacement of the beam cross-section is completely defined by the displacements u, v,w and the rotation 8 referred to an arbitrary system of axes Oxy The shear flow q at any section of the beam is then given by Eq (9.40), that is d0 dz - du dv + - cos$ + - sin$ + dz dz Fig 11.1 Cross-section of a thin-walled beam at the built-in end 446 Structural constraint At the built-in end, aw/as is zero and hence... D12 (10.59) Similarly (10.60) and (10.61) For a lamina subjected to a distributed load of intensity q per unit area we see, by substituting for M,, M yand Mxy from Eqs (10.59 )-( 10.61) into Eq (5.19), that (10.62) Further, for a lamina subjected to in-plane loads in addition to q we obtain, by a comparison of Eq (10.62) with Eq (5.33) - a2W -q+Nx-+2N ax2 a2W -+ N ’ay’ a2W xya~ay (10.63) Problems involving . 10.6 Laminated composite structures 429 From Eq. (10. 48) the modulus of the bar is given by 80 x 40 80 x 50 + 5000 x - E1 = 200000 x - 80 x 10 80 x 50 i.e. El = 44000N/m111'. Eqs (1.27), (1. 28) and the derivation of Eq. (5.13) we see that a'w . a2W a'W E, = -z-, EY = -z-, rxy = -2 z- ax2 aY2 axay 10.6 laminated composite structures 431. 500 i.e. A1 = 0. 28 4- The major Poisson's ratio for the bar is found from Eq. (10.50). Thus x 0.3 = 0.22 &apos ;80 x 10 x 0.2 + - vlt =m 80 x 50 80 x 40 Hence the