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Aviation History
1919
1919 - 0516.PDF
lESE im.uarr\ 1 has already been fixed by bearing pressure, hence "y" is determinable: Example : For -ft in. A.G.S. fork end F =* 5,700 lbs. = 2-53 tons; t = «22 in., then 1-5 x 20 x -22 x y = 2-53- 2-53 = '385 in., say -39 in. .-. v = — % I'5 X 20 X '22 " ." " ~* The next critical sections to be considered are x and x at the sides of the lug- These two areas added together must have a tensile strength equal to or greater than the specified strength of the fork joint. Let/, be the tensile strength of lug material = 26 tons per square inch for 26-ton mild steel plate, then 2 x x x t x ft must be equal to or greater than F. For example, for a &-in. A.G.S. fork end 2 x x x *22 2 *5S x 26 =2-53 tons. .-.x = ——-F =-22 in. JJ 2 x -22 x 26 Actua'ly, x should be made slightly greater than this to allow for eccentric loading when the eye of the lug is not exactly in the centre of the lug. The accompanying table (Fig. 5) has been calculated on the above assumptions. (b).—Strength of A.G.S. Standard Bolts in Direct Shear, Direct Tension, and Combined Shear and Tension. In designing many of the fittings of an aeroplane it becomes necessary, in order to provide the requisite factor of safety, to know the strength of the bolt or bolts holding down the particular fitting. In a few cases only are the stresses to which the bolts are subjected either direct tension stresses or direct shear stresses; more often a combination of the two is met with. To the average draughtsman it is sometimes a matter of some difficulty to determine how much to allow for one kind of stress and how much for the other, and in all cases a fair amount of calculations are necessary, which, it is hoped, the accompanying tables and charts may prove helpful in reducing to a minimum. Two of the accompanying tables (Figs. 6 and 7) give the A.G.S. numbers of bolts, nuts, screws and studs, and the dimensions of A.G.S. bolts, boltheads, and nuts respectively, while a third table, Fig. 8, gives the ultimate tensile and shearing values for all A.G.S. bolts, both on full area and on core area. In the cases of direct shear and direct tension the values are obtained by multiplying the ultimate shearing or tensile stress by the area considered. In Fig. 8, cases I, II and III give the strength values for the bolts when the pull on the bolt is inclined at 45 deg. CASE I.—It is assumed that a strut takes all the component of the load acting along the axis of the bolt; that is to say, that there is no tension on the bolt, but only pure shear. Assuming that the bolt is nutted up on the opposite side of the beam (as should always be done), the shear is taken by the full area of the bolt^ In this case the allowable inclined pull equals v/2 times the allowable shearing force acting at right angles to the axis of the bolt. CASE II.—It is assumed that the lug putting the inclined load on the bolt is carried beyond the bolt, and is anchored on several more bolts, so that the shear per bolt is negligible. In this case the stress on the bolt is purely tensional, and the allowable inclined pull is equal to Jz times the allowable direct tension measured on the core area. CASE III.—This case assumes that both components APRIL 17, 1919 of the inclined pull are taken on one belt, and this probably represents the most usual practical problem. The combined tension and shear occur at the headed end cf the bolt, that is, they are resisted by the full area of the shank cf the bolt. At the nutted end of the bolt there is no shear, and the allowable inclined pull, as fixed by the core area at this end, is the same as in Case II. It now becomes a question whether the allowable inclined pull as fixed by the combined stress in the shank is greater or less than the allowable inclined pull as fixed by the tension in the core area. Of course, the lesser of the inclined pulls is the one required. In the column under Case III, the inclined pulls, as fixed by the combined stress in the shanks, are given. For 6BA and 4BA bolts, these are greater than the pulls given under Case II, and the latter and lesser values must be taken. For the other sizes of bolts the ultimate inclined pulls are fixed by the stresses in the shank, which goes to show the efficiency of the bolts. The combined stresses in the shank of the bolt are obtained as follows :— Let F be inclined pull, Ft be tensile component, F,v be shearing component. F Since F is inclined at 45 deg., Ft = F« = —~ iSe^ bn ^ ^^ °f the Shank'then tensile ^ shearing stress on axis or at right angles to axis of bolt respectively = * \/2xA The principal tensile stress = 1 -=1 . / ~F>~ 2 /„ + V-v/2 x A v 4 x 2 x A; The principal shearing stress 7 + 2 X A* "A/: 4x2xA8 2xA "ch—C1Pal t6nSile StrCSS iS Umited t0 35 tons/sq uare 35 = L__+_^5xF F = A x 2 J2 A x 2 J2 35A x 2J2 1 + v/5 _ 70A J2 = j, = 30-6A. If principal shearing stress is limited to 25 tons/square inch— ' * 25= A/ZIL_ +~~F r~_ V5~*F v 4X2XA'_ 2xA* AX2V2 . ,? 25XAX2V2 50A./2 • • * " ~IT~ - ~~rr =3i-5A. *rom this it is seen that the principal tensile stress is critical. Example : |-in. B.S.F. Full area = -0491 sq. in. F = 30-6 x -0491 tons = 3>4°o lbs. On the accompanying chart, Fig. 9, the ultimate inclined pulls at any angle of pull are given. The strengths of the bolts are plotted on a system of polar co-ordinates. Distances along radial lines 516
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