Aircraft Structures 3E Episode 11 pptx

386 Stress analysis of aircraft components Choosing GREF = 27 600N/mm2 then, from Eq. (10.27) x 1.22 = 1.07mm 24 200 27 600 tip = - Similarly tf3 = r;4 = 1.07mm, t& = ti6 = t& = 0.69mm Hence Similarly 61zi = 250, 613 = 6% = 725, 634 233, 635 = 646 = 736, b56 = 368 Substituting the appropriate values of 6 in Eq. (10.24) for each cell in turn gives the following. For cell I For cell I1 [-250qI + qII(250 + 725 + 233 + 725) - 233q11~] (ii) de 1 dz - 2 x 355 OOOGREF _- For cell 111 (iii) 1 [-233q11+ q111(736 i- 233 + 736 + 368)] d6’ -= dz 2 x 161 OOOGREF In addition, from Eq. (10.22) 11.3 x lo6 = 2(258 OOOqI + 355 OOOqII + 161 OOO~III) (3 Solving Eqs (i) to (iv) simultaneously gives qr = 7.1N/mm, qIr = 8.9N/mm, qIII = 4.2N/mm The shear stress in any wall is obtained by dividing the shear flow by the actual wall thickness. Hence the shear stress distribution is as shown in Fig. 10.21. Fig. 10.21 Shear stress (Wrnm’) distribution in wing section of Example 10.7. 10.3 Wings 387 . I -I_I*iilPIUI_ Y_ ~~ ~ ~ 10.3.3 Shear Initially we shall consider the general case of an N-cell wing section comprising booms and skin panels, the latter being capable of resisting both direct and shear stresses. The wing section is subjected to shear loads S, and S, whose lines of action do not necessarily pass through the shear centre S (see Fig. 10.22); the resulting shear flow distribution is therefore due to the combined effects of shear and torsion. The method for determining the shear flow distribution and the rate of twist is based on a simple extension of the analysis of a single cell beam subjected to shear loads (Sections 9.4 and 9.9). Such a beam is statically indeterminate, the single redundancy being selected as the value of shear flow at an arbitrarily positioned ‘cut’. Thus, the N-cell wing section of Fig. 10.22 may be made statically determinate by ‘cutting’ a skin panel in each cell as shown. While the actual position of these ‘cuts’ is theoretically immaterial there are advantages to be gained from a numerical point of view if the ‘cuts’ are made near the centre of the top or bottom skin panel in each cell. Generally, at these points, the redundant shear flows are small so that the final shear flows differ only slightly from those of the determinate structure. The system of simultaneous equations from which the final shear flows are found will then be ‘well conditioned’ and will produce reliable results. The solution of an ‘ill conditioned’ system of equations would probably involve the subtraction of large numbers of a similar size which would therefore need to be expressed to a large number of significant figures for reasonable accuracy. Although this reasoning does not apply to a completely idealized wing section since the calculated values of shear flow are constant between the booms, it is again advantageous to ‘cut’ either the top or bottom skin panels for, in the special case of a wing section having a horizontal axis of symmetry, a ‘cut’ in, say, the top skin panels will result in the ‘open section’ shear flows (qb) being zero in the bottom skin panels. This decreases the arithmetical labour and simplifies the derivation of the moment equation, as will become obvious in Example 10.8. The above remarks regarding the ‘cutting’ of multicell wing sections are applicable only to this method of analysis. In the approximate analysis of multicell wing sections ES Moment centre \ I- = = :I -x 70 1 * A - e - I t I t s, Fig. 10.22 N-cell wing section subjected to shear loads. 388 Stress analysis of aircraft components & Fig. 10.23 Redundant shear flow in the Rth cell of an N-cell wing section subjected to shear. by the method of successive approximations ‘cuts’ are sometimes made in the spar webs although in some cases ‘cutting’ the top or bottom skin panels produces a more rapid convergence in the numerical iteration process. This approximate method is extremely useful when the number of cells is large since, in the above approach, it is clear that the greater the number of cells the greater the number of simultaneous equations requiring solution. The ‘open section’ shear flow qb in the wing section of Fig. 10.22 is given by Eq. (9.75), i.e. We are left with an unknown value of shear flow at each of the ‘cuts’, i.e. qs,o,I, qs,o,II, . . . , qs,O,N plus the unknown rate of twist de/& which, from the assumption of an undistorted cross-section, is the same for each cell. Therefore, as in the torsion case, there are N + 1 unknowns requiring N + 1 equations for a solution. Consider the Rth cell shown in Fig. 10.23. The complete distribution of shear flow around the cell is given by the summation of the ‘open section’ shear flow qb and the value of shear flow at the ‘cut’, qs,O,R. We may therefore regard qs,O,R as a constant shear flow acting around the cell. The rate of twist is again given by Eq. (9.42); thus By comparison with the pure torsion case we deduce that in which qb has previously been determined. There are N equations of the type (10.28) so that a further equation is required to solve for the N + 1 unknowns. This is obtained by considering the moment equilibrium of the Rth cell in Fig. 10.24. The moment Mq,R produced by the total shear flow about any convenient moment centre 0 is given by M~,R = f qRp0 ds (see Section 9.5) 10.3 Wings 389 U’ I I I‘ Fig. 10.24 Moment equilibrium of Rth cell. Substituting for qR in terms of the ‘open section’ shear flow qb and the redundant shear flow qs‘s;O,R, we have Mq,R = fR %PO dS + qs,O,R fRPO dS or Mq,R = fR qbP0 ds f 2ARqs,0:R The sum of the moments from the individual cells is equivalent to the moment of the externally applied loads about the same point. Thus, for the wing section of Fig. 10.22 (10.29) If the moment centre is chosen to coincide with the point of intersection of the lines of action of S, and S,, Eq. (10.29) becomes (10.30) Example 10.8 The wing section of Example 10.6 (Fig. 10.17) carries a vertically upward shear load of 86.8 kN in the plane of the web 572. The section has been idealized such that the booms resist all the direct stresses while the walls are effective only in shear. If the shear modulus of all walls is 27 600 N/m2 except for the wall 78 for which it is three times this value, calculate the shear flow distribution in the section and the rate of twist. Additional data are given below. Wall Length (mm) Thickness (mm) Cell area (mm’) ~~ 12, 56 1023 1.22 AI = 265000 23 1274 1.63 AI1 = 213 000 34 2200 2.03 A111 = 413 000 483 400 2.64 572 460 2.64 61 330 1.63 78 1270 1.22 390 Stress analysis of aircraft components Choosing GREF as 27 600 N/mm2 then, from Eq. (10.27) x 1.22 = 3.66mm 3 x 27600 27 600 t;, = Hence 1270 3.66 67, = - - - 347 Also 612 = 656 = 840, 623 = 783, 634 = 1083, 63, = 57, 684 = 95, 687 = 347, 627 = 68, 675 106, 616 = 202 We now ‘cut’ the top skin panels in each cell and calculate the ‘open section’ shear flows using Eq. (9.75) which, since the wing section is idealized, singly symmetrical (as far as the direct stress carrying area is concerned) and is subjected to a vertical shear load only, reduces to where, from Example 10.6, Ixx = 809 x 106mm4. Thus, from Eq. (i) 86.8 x IO3 809 x lo6 r=l n B,~, = - 1.07 x 10-~ B,~, r=l qb = - (ii) Since qb = 0 at each ‘cut’, then qb = 0 for the skin panels 12,23 and 34. The remaining qb shear flows are now calculated using Eq. (ii). Note that the order of the numerals in the subscript of qb indicates the direction of movement from boom to boom. qb:27 = -1.07 x qb,J6 = -1.07 x x 3880 x 230 = -95.5N/mm x 2580 x 165 = -45.5N/mm qb$5 -45.5 - 1.07 x io-4 x 2580 x (-165) = 0 qb.57 = 1.07 X qb;38 = -1.07 X qb.48 = - 1.07 X lop4 X 3230 x (-200) = 69.0 N/mm Therefore, as ,,83 = qb.48 (or qb,72 = qb,57), &7g = 0. The distribution of the shear flows is show n Fig. 10.25. The values of 6 and qb are now substituted in Eq. (10.28) for each cell in turn. For cell I X 3880 X (-230) = 95.5N/= X 3230 X 200 = -b9.0N/mm [qs,o:l( 1083 + 95 + 57) - 57qS,o,11 + 69 x 95 + 69 x 571 (iii) d0 1 dz - 2 x 265 OOOGREF _- 10.3 Wings 391 Fig. 10.25 qb distribution (N/mm). For cell I1 +95.5 x 68 - 69 x 571 For cell I11 +45.5 x 202 - 95.5 x 68 - 95.5 x 1061 (VI The solely numerical terms in Eqs (iii) to (v) represent fR qb(ds/t) for each cell. Care must be taken to ensure that the contribution of each qb value to this term is interpreted correctly. The path of the integration follows the positive direction of qS,o in each cell, i.e. anticlockwise. Thus, the positive contribution of qb,83 to fI qb(ds/t) becomes a negative contribution to fII qb(ds/r) and so on. The fourth equation required for a solution is obtained from Eq. (10.30) by taking moments about the intersection of the x axis and the web 572. Thus 0 = -69.0 x 250 x 1270 - 69.0 x 150 x 1270 + 45.5 x 330 x 1020 +2 x 265 OOOqS,o,I + 2 x 213 OOOqS,o,II + 2 x 413 OOOqS,o,III qs:o,r = 5.5 N/mm! 4S,O,II = 10.2 N/-, %,O:III = 16.5 N/= (vi> Simultaneous solution of Eqs (iii)-(vi) gives Superimposing these shear flows on the qb distribution of Fig. 10.25, we obtain the final shear flow distribution. Thus q34 5.5N/m~1. q23 = @7 10.2N/=, q12 = q56 16.5N/mm g61 = 62.0 N/mm, q57 = 79.0 N/mm, q72 = 89.2 N/mm q48 74.5N/1nm, q83 = 64.3N/mm Finally, from any of Eqs (iii)-(v) de - = 1.16 x 10-6rad/mm dz 392 Stress analysis of aircraft components 10.3.4 Shear centre The position of the shear centre of a wing section is found in an identical manner to that described in Section 9.4. Arbitrary shear loads Sx and Sy are applied in turn through the shear centre S, the corresponding shear flow distributions determined and moments taken about some convenient point. The shear flow distributions are obtained as described previously in the shear of multicell wing sections except that the N equations of the type (10.28) are sufficient for a solution since the rate of twist de/& is zero for shear loads applied through the shear centre. 10.3.5 Tapered wings Wings are generally tapered in both spanwise and chordwise directions. The effects on the analysis of taper in a single cell beam have been discussed in Section 10.1. In a multicell wing section the effects are dealt with in an identical manner except that the moment equation (10.16) becomes, for an N-cell wing section (see Figs 10.5 and 10.22) Example 10.9 A two-cell beam has singly symmetrical cross-sections 1.2 m apart and tapers symme- trically in the y direction about a longitudinal axis (Fig. 10.26). The beam supports loads which produce a shear force Sy = lOkN and a bending moment 1 1 Fig. 10.26 Tapered beam of Example 10.9. 10.3 Wings 393 M, = 1.65 kNm at the larger cross-section; the shear load is applied in the plane of the internal spar web. If booms 1 and 6 lie in a plane which is parallel to the yz plane calculate the forces in the booms and the shear flow distribution in the walls at the larger cross-section. The booms are assumed to resist all the direct stresses while the walls are effective only in shear. The shear modulus is constant throughout, the vertical webs are all l.0mm thick while the remaining walls are all 0.8 mm thick. ~oom areas: B~ = B~ = B~ = B6 = 600mm2, B2 = B~ = 9OOmm' At the larger cross-section I,, = 4 x 600 x 902 + 2 x 900 x 902 = 34.02 x 106mm4 The direct stress in a boom is given by Eq. (9.6) in which Ixv = 0 and MJ = 0, i.e. whence or 1.65 x 106y B 34.02 x lo6 = 0.08yrB,. p,,, = The value of P,,, is calculated from Eq. (i) in column 0 of Table 10.5; P.y,r and Py., follow from Eqs (10.10) and (10.9) respectively in columns @ and 0. The axial load P, is given by [a2 + O2 + a2I1l2 in column 0 and has the same sign as PZ,, (see Eq. (10.12)). The moments of P,,r and P,,,, columns @ and 0, are calculated for a moment centre at the mid-point of the internal web taking anticlockwise moments as positive. From column @ (as would be expected from symmetry). Table 10.5 1 2619.0 0 0.0417 0 109.2 2621.3 400 90 0 43680 2 3928.6 0.0833 0.0417 327.3 163.8 3945.6 0 90 -29457 0 3 2619.0 0.1250 0.0417 327.4 109.2 2641.6 200 90 -29466 21840 4 -2619.0 0.1250 -0.0417 -327.4 109.2 -2641.6 200 90 -29466 21840 5 -3928.6 0.0833 -0.0417 -327.3 163.8 -3945.6 0 90 -29457 0 6 -2619.0 0 -0.0417 0 109.2 -2621.3 400 90 0 -43680 394 Stress analysis of aircraft components From column @ 6 P,,r = 764.4 N r=l From column @ From column @ c P y:r ( r =-43680Nmm r= I From Eqs (10.15) Sx,w = 0, Sy,w = 10 x lo3 - 764.4 = 9235.6N Also, since Cx is an axis of symmetry, I,, = 0 and Eq. (9.75) for the ‘open section’ shear flow reduces to or 9235’6 cBryr = -2.715 x 10-42Bryr r=l r= 1 qb = - 34.02 x lo6 ‘Cutting’ the top walls of each cell and using Eq. (ii), we obtain the qb distribution shown in Fig. 10.27. Evaluating 6 for each wall and substituting in Eq. (10.28) gives for cell I 6 0 504 Fig. 10.27 qb (Wmrn) distribution in beam section of Example 10.9 (view along z axis towards C). 10.3 Wings 395 4.6 2 2.5 1 12.2 4 2.5 - 4.6 - 6 Fig. 10.28 Shear flow (Wmm) distribution in tapered beam of Example 10.9. Taking moments about the mid-point of web 25 we have, using Eq. (10.31) 0 = -14.7 x 180 x 400 + 14.7 x 180 x 200 + 2 x 36000q,:031 + 2 X 72oooq,,0.~1 -117846-43680 or 0 = -690 726 + 72000q,,o,~ + 144OOOq,,o,I~ Solving Eqs (iii)-(v) gives qs,o.r = 4.6 N/m7 4S,O,II = 2.5 N/mm and the resulting shear flow distribution is shown in Fig. 10.28. 10.3.6 Method of successive approximations - torsion It is clear from the torsion and shear loading of multicell wing sections that the greater the number of cells the greater the number of simultaneous equations requiring solution. Some modem aircraft have wings comprising a relatively large number of cells, for example, the Harrier wing shown in Fig. 7.8, so that the arithmetical labour involved becomes extremely tedious unless a computer is used; an approxi- mate but much more rapid method may therefore be preferable. The method of successive approximations provides a simple and rapid method for calculating the shear flow in many-celled wing sections and may be used with slight differences of treatment for both the pure torsion and shear loading cases. Initially we shall consider a wing section subjected to a pure torque. The mechanics of the method may be illustrated by considering the simple two-cell wing section shown in Fig. 10.29 and which carries a pure torque T. First we assume Fig. 10.29 Method of successive approximations applied to a two-cell wing section. [...]... 34 100 x lo3&= 2AIqI + 2AIIqIl+ 2A111qI11f 110 .1 x 460 x 1270 + 52 x 330 x 2290 which gives Es = 946.8 mm 404 Stress analysis of aircraft components Table 10.7 ~ ~ cell I Iqb w t 6 cs 6[=-(I4bW)/4 coq coq coq Corrective shear f o s lw Cell I1 12 166.0 112 8.7 0.08 -10.78 -0.50 6945.7 1870.0 0.134 0.086 -3.71 -0.86 0.29 -0.04 -0.03 -0.01 0 -0.08 -0.01 -11. 37 Cell I11 -4.36 -6218.9 2013.6 0.093 3.09... to cell I1 and CIIIqII the correction carry over factor from cell 111 to cell 11 As a first approximation in the solution we neglect the effect of the shear flows in adjacent cells so that 411 =4 1 Similarly 4 1 = rlf? qIII =4 h Substituting these values in Eq (10.38) we have (10.39) or 411 = 4 +4 1 1 (10.40) 402 Stress analysis of aircraft components Similarly and simultaneously, corrections qy and... weighted thicknesses t*; 6 is then Jds/t* 10.3 Wings 399 Table 10.6 Cell I Assumed q (N/mm) coq coq coq coq Final q (N/mm) 2Aq (Nmm) Total T (Nmm) Actual q (N/mm) ( T = 11. 3kNm) Cell I11 0.129 288 51.38 5.20 0.93 0.09 345.6 C S Cell 11 0.149 0 .112 367 37.15 18.76 6.63 4.97 0.67 0.25 0.12 0.07 0.121 155 41.10 2.10 0.56 0.03 435.6 3.09 x 10' 5.51 x IO* 8.9 1.78 x 10' 7.1 198.8 0.64 x 10' 4.1 10.3.7 Method... analysis of aircraft components The actual shear flows are obtained by factoring the final shear flows by the ratio of the applied torque to the torque corresponding to the final shear flows, Le Example 10.10 Solve Example 10.7 using the method of successive approximations From Example 10.7 61 = 1542 + 250 = 1792 + 725 + 233 + 725 = 1933 6 1 = 736 + 233 + 736 + 368 = 2073 11 6 1 = 250 1 61,~ 250 = 6 11 1... A I = 265 000 mm2, A11 = 580 000 m 2 , AIII = 410 000 mm2 Wall Length (mm) Thickness (mm) 12, 56 1025 1.25 23,45 1275 1.65 34" 2200 2.25 16 330 1.65 25 460 2.65 34' 400 2.65 Initially we calculate SI,SI,etc and the correction carry over factors CI,II, etc CII,I SI=-+-= 2200 2.25 112 8.7 400 2.65 Similarly n t I IOokN 7 L 6 1270 rnrn Fig 10.36 Three-cell wing section of Example 10 .11 1020 rnrn 10.3 Wings... flows 4, qfI and dII cells I, I1 and I11 respectively to reduce this twist to zero (Fig 10.34).On rejoining the cells it is clear that dIwill cause twist in cell I by its action on the web common to cells I and 11, that qi will cause twist in cell I1 and so on We therefore apply a second system dI1 of corrective shear flows 4, d1, to the separated cells I, I1 and I11 respectively However, since the cells... shown in 10.3 Wings 401 Fig 10.35 Final shear flow system in Cell II Fig 10.35 Since the cell does not twist then, from Eq (9.42) dB dz 1 ds q-=o 2AIIGf11 t -=- or Hence, from Fig 10.35 and taking anticlockwise torques as positive ds fII q b 7 - qI1 611 + qIbI,II f ~ I I I ~ I I I 1= 0 1 giving (10.37) The first term on the right-hand side of Eq (10.37) represents the proportion of the 'open section'... of a two-cell wing section where = J d s / t for the wall common to cells I and 1 Hence 1 4,II (10.32) 4: = 411 6 1 Similarly (10.33) Since 4 and qfI are expressed in terms of the shear flows in adjacent cells they are are referred to as correction carry over s h e a r f l o w The factors 61 .11/ 61 and 61,1~/6~~ known as correction carry over factors and may be written as 6I:II CIJI = -1 61 1 6I:Il...396 Stress analysis of aircraft components Fig 10.30 Shear flows giving G(dB/dz) = I for the separated cells of a two-cell wing section that each cell acts independently and that cell I is subjected to a constant shear flow qI such that G(dB/dz) for cell I is equal to unity From Eqs (9.49) and (9.52) where s - ds/t 1 -11 Hence 2 4 4 =1 61 Similarly, for G(dB/dz) to be unity... 2.65 qb shear flow = 12 166.0N/mm ds 7 = 6945.7 N/mm ds $11, qb 7 = -6218.9 N/mm The solution is now completed in Table 10.7 Note that in Table 10.7 a negative sign is used to indicate that 4 etc are in the opposite sense to qb The final shear flow distribution is shown in Fig 10.38 A check on the vertical equilibrium of the wing section gives (11. 4 x 400 + 73.6 x 400 + 2 x 4.4 x 30 + 103.0 x 460 + 2 . 233q11~] (ii) de 1 dz - 2 x 355 OOOGREF _- For cell 111 (iii) 1 [-233q11+ q 111( 736 i- 233 + 736 + 368)] d6’ -= dz 2 x 161 OOOGREF In addition, from Eq. (10.22) 11. 3. Example 10.7 61 = 1542 + 250 = 1792 611 = 250 + 725 + 233 + 725 = 1933 6111 = 736 + 233 + 736 + 368 = 2073 61,~ = 250 611, 111 = 233 Hence, from Eqs (10.34) 250. 2A111qI11f 110 .1 x 460 x 1270 + 52 x 330 x 2290 which gives Es = 946.8 mm. 404 Stress analysis of aircraft components Table 10.7 ~ ~ cell I I qb wt 12 166.0 6 112 8.7