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Aviation History
1929
1929 - 0635.PDF
MARCH 28, 1929 19 THE AIRCRAFT ENGINEER SUPPLEMENT TOFLIGHT By changing the vertical scale, Fig. 4 may be made to represent the loading curve. The expression for the shear force is found by integrating the load curve. 3Shear = L k | x>dx = - hkz' 4 (4) The shape of the shear force curve is shown on Pig. 5. The expression for the bending moment is found by integ- rating the shear load curve. o r o o Bending moment, M = - Lfcix'ndx = - \Jc - xl 4 I 4 7 = — Lfe5 28 (5) The shape of the bending-moment curve is shown on Pig. 6- The spar flange sizes, also the size of web, may now be calculated if a definite relationship between x and the spar depth be determined. The ratio of wing thickness to chord is made constant, so we may express the spar depth in terms of the chord by multiplying the chord expression, y — ten*, by a factor c; then, if D represents the spar depth, D = c . fcei (6^ In practice c may vary between the limits of 0-1 to 0-18. The convention will be adopted that, as the spar is deep in comparison with the thickness of the flange, the moment of resistance may be expressed as aDf, where a is the area of the flange and / is the maximum permissible stress in the material. Fig. 7 illustrates the spar proportions that will be con- sidered in the following investigation. In practice of course, the flange " a '' and the web " b " would be spread in some such manner as is shown by Fig. 8, but at the present stage we are not so much concerned with details of construction as with a conventional method of finding what amount of material is required to be put into the spar to resist the bending moment. '-kjc* PLAIN FORM AND LOADING FIG 4 jy-4 MAX- SHEAR FIG. 6. FLANGE. AREA'a WEB AREA b" FIG 7 FIG. 8 FIG. 9 To find the flange area we have :— Moment of resistance = Bending moment. 9 a.c.kxff = 28 a = (7) . for both flanges, area = —14 c./. ' The total volume of material required for the flanges from wing tip to wing root may now be found by integrating the expression for the flange area between the limits of x = o to x = I. Flange volume = — —r- x2 dx14 C-J Jo 3 LI3 = 14 ~e7f The weight of the flanges will then be :— (8)14 rc./. where p is the density of the material used. To find the web size and weight we can adopt the usual approximations applied to deep " I" girders, i.e. that the shear load is all taken by the web and that the shear stress is nearly uniform. Shear load Shear stress. /„ = where Dt t = the thickness of the web shear load t = fsB 3L 4/sc X.The cross sectional area of the web, b is i X D, substituting for the values of t and D we have:— • 4 /, The volume of the web is found by integrating the expres- sion for the web area, between the limits of a: = o to a; = I 3 Lfc ,Web volume = — — I x'< 4 /. jo no f ^° Js258e
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