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Aviation History
1932
1932 - 1082.PDF
74 SUPPLEMENT TO FLIGHT THE AIRCRAFT ENGINEER OCTOBER 27, 1932 TABLE I course, be considered as a separate unit. The object of the wire bracing is to take the torsion and drag loads, lift loads being taken by the spar in the normal manner. In the case of wing bracing it is usual to use the leading edge as the member DE, but if a wire is used it is necessary to have a wire on the opposite side to cater for the anti-drag. It is convenient to stress the structure under torsion and drag loads separately. Fig. 2 and Table I give the dimensions of a typical symmetrical pyramid. A 0 MS (P \ 0 a 1 ^^ &OA- T FIG.3 B bh \\ 5 6.-C » Member AB ab Aa Bb CD AC AD aC aD BC BD bC &D DE X + 18 + 18 0 0 0 + 9 + 9 + 9 + 9 - 9 - 9 — 9 - 9 + 21 y 0 0 0 0 + 20 — 10 + 10 - 10 + 10 - 10 + 10 - 10 + io + 3 z 0 0 - 16 - 16 0 - 8 — 8 + 8 + 8 - 8 - 8 + 8 + 8 0 Length 18-0 18-0 160 16-0 20-0 15-65 15-65 15-65 15-65 15-65 15-65 15-65 15-65 21-21 (a) Drag.—Drag loads of 500 lb. each are applied at joints A and a, as shown in Fig. 3. The equations of equilibrium for each joint are then set down in the manner described above ; the values of the tension coefficients found by their solution are also given. Joint A xj + 18 «AB + 9 tAT> + 9 <AC yj + 10 /AD - 10 (AC 8 (AC - 8(AD tAB = - 63-61 (AC -500 16 (Aa + 38-61 tAa = - 63-63 Joint a xj + 18 tab + 9(aD + 9 toC yj + 10 taD - 10 toC 2/ + 8 taT> + 8 toC + 16 taA tab = - 63-61 toD = + 88-61 toC 500 38-61 Joint C xj A- 9(CB + 9«C6- 9<CA - 9<Ca y\ + 10 tCB + 10 iCb + 10 (CA + 10 tCa + 20(CD 2/ + 8 (CB - 8 tCb + 8 (CA - 8 tCa tCB=+ 38-61 tCb = + 38-61 (CD =- 77-20 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 Joint D xj + 9 (DB + 9(D& - 9 (DA - yj - 10 (DB - 10«D6 - 10 tT>A - zj + 8 (DB — 8(D& + 8 (DA - (DA = + 88-61 (DB = 0 9(Da + 2KDE = 0 10(Da + 3(DE -20(DC = 0 8(Da = 0 (Db =0 For this case the members DB and Db are inoperative. At each joint there are, however, at least four unknowns, despite the fact that (DB = 0 and (Db = 0. If, by the " method of sections," moments are taken about the z axis through Bb, the value of (DE can be found immediately, thus:— (2KDE x 10) + (3(DE X 9) = (1,000 237 tDE = 18,000 *DE = + 75-95 18) We then have three equations at joint D with the unknowns tDA, fDC and tDa, which can therefore he found at once. The substitution of their values in the equations of joints A and a enable these joints to be solved in a similar way and thence to joint 0. It will be observed that there is no necessity for the member DC to be held to the spar. The loads in the members are shown on the diagram. (b) Torsion. Loads of 1,000 lb. each but of opposite sense are applied at joints A and a as shown in Fig. 4. For this case the members AC, aD, DB, C6 and DE are inoperative, the solution of all the equations follows immediately without the slightest difficulty. By adding the loads shown in Figs. 3 and 4 we obtain the net 1008
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