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Aviation History
1938
1938 - 0271.PDF
JANUARY 27, 1938 THE AIRCRAFT ENGINEER SUPPLEMENT TO FLIGHT 9<>g The term outside the integral has a constant numerical value for all points along the semi-span in any particular case. The evaluation of the integral gives :— sin 2 0,,\ n — 2 Substituting n = 2, 3, 4, we get : hi 2 sin 4 0O sin 2 (w —i)80 sin 2 n 80 n - 1) « o\ , r. r • /> sin 4^oi - \ — I A3I sin2 0o-——— A, A, sin 60O 2 sin 80O 2 3 J ' L 3 4 Multiplying this expression by the constant terrn:- P •]) -(5) s.C,.- .V«=.S.C,.g gives S.F. = s.CB.q A,«, + A3a3 *.] where = I sin «,--*( I sin 2 0O — sin 4 0O 2 sin 60O 3 2 sin 40O 2 sin 60, 3 sin 80, -(7) These coefficients (7) are the shear force coefficients (S.F.)c and are seen to be solely functions of 80, i.e., of the fraction of the semi-span at which the shear force is to be calculated, obviously. Thus, it is unnecessary to proceed further with the loading calculation than the evaluation of the- coefficients Aj* — 1, in order to evaluate the shear ; the same will be shown to be true of the bending moment. Since the total lift on a wing depends only on the co- efficient Aj (the other coefficients merely modifying the shape of the curve), the maximum shear, which occurs at the axis of symmetry is also solely dependent on A,, and is therefore given by :— S.F. = C.s.f.Aj.a, (8) The coefficients a,, a,, o5, a7, are plotted in Figures 2 and 3, and the numerical values are given in Table 1. .- : Elliptic Loading •';'..'",•' In the particular case of elliptic loading (7) is true at all points along the semi-span, since A-j, A6 and A, are each zero. y/s 0 •156 •3°y •454 .588 •707 .nog •OCII •951 .988 1030 al •7853 .6296 .4813 •3475 •2333 •1425 .0763 •°33^ .00995 .00105 0 TABLE «3 O •1507 .2658 .3211 .3112 .25OO .1643 .0834 .O28l .OO378 O I «5 O - .I409 - .1981 - I446— 0245 •O833.1224 .O93I .0396 .OO6O5 O <*7 O .1269 .Il6O — .OIIO — .IO84 - -0833 .OIO5 •OO25 .O425 .OO798 O file coefficients as and a7 are seen, from Fig. 3, to have erojalue at several points along the semi-span (excluding • ~ ° and 1). The exact values of r/s are found by equating «5 and a, in (7) to zero. anH °5 iUs occurs at yl* = -OI3. and for «7 atJ/5 = -443^ -799- (9) Thus, we have :— S.F. at y/s = .443 :— s.Cp.?[.357 A,. + .319 A3 - .152 A5j S.F. at _v/s = .613 s.C0.?[.2i4 A, -I- .302 A3 — .114 A,] S.F. at yls — .799 s.C0.c/[.o83 At 4- .10S A3 + .122 At] These expressions (9), combined with equation (8), give a slightly simplified method of calculating the shear at four points along the semi-span. Other values may be quickly obtained by plotting these against yjs. (b) Aerodynamic Bending Moment The aerodynamic bending moment at a point distant as from the axis of symmetry is given by :— B.M. = 5» [ l(%^)(c«-;-Vf)0' - «0 <v = s«.co.!:.v*| /CLCNI 1 v.dv — >» •1 —1 IO) \cB/ a •/ a Putting a — — cos.0o, and substituting lor (3), we get:— r* »=4 B.M. = s2.Cn.- V7t I —I ^ Ain.,. sin(2« — 1) J gan=i 6. sin 2 B.dB + cos 8, •20 •15 •10 •05 -05 -10 , j ^ A2w-1.sin (2» — 1) 8. sin fl.tf0. : *....;. ^ •-'•^ .. .. (11) / \ \ \^ \ \ y \ \ / Jx/ / ~7_/ FIG.3. kA •I 2 •9 1-0 The term outside the bracket has a constant numerical value in any particular case, as before, and the evaluation of the integrals gives :— n = 1 A,.cos 0O sin (in — 3)0O sin (in + i)0o (2M - 3) sin 2 0, ,2n cos 8, « = 40 V - 2n — 2 'sin 2 n — 1) 9a sin 2 (n - 1) Substituting » = 1, 2, 3, 4, we get :— sin 3 0r\ cos 0O (12) A.,|^(sin 0o - j( sin 0O - — ^0 H sin 2 6, sin sin 3 80 sin 7 sin 2 cos fl0/ cos 0O/Sin 4^o sm 60. 4 \ 2 3 cos 0O sin 60O sin 80O (13) Multiplying the expression by the constant term *c
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