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Aviation History
1934
1934 - 1454.PDF
SUPPLEMENT TO FLIGHT 1O0 2& THE AIRCRAFT ENGINEER SEPTEMBER 27, 1934 Y / -kxi \ FIG.4. ^ -~-—"" i y 5' Equation to shear force diagram: y = $bifxzdx = | x Eq uation to bending moment : y = ^6fx^dx = f x 346*3 = 148.2*3 The plan form, loading, shear force, and B.M. curves are shown by Figs. 5, 6 and 7. Relief loads due to engines, etc. : lb. Weight of engine unit .. .. 1,400 Weight of half undercarriage .. 240 1,640 Factored load = 7.5 X 1,640 = 12,300 lb. On the starboard side the engine torque is subtracting from the relief B.M. due to the above weights. Engines give 640 h.p. at 1,200 r.p.m. : Engine torque, T = —'•—' = 2,810 1b. ft. ° ^ 2iX I,2OO 2 8lO Reactions at A and B = = 1,405 lb. 2 T J Relief B.M. at root due to petrol and oil weight = 7.5 X 16.1 x 392 2 • . ; - = 92,000 lb. in.The relief bending moments are shown plotted on Fig. 7. Method of Estimating the Moment of Inertia and Bending Modulus of the Spar Section The following method enables a rapid and accurate estimate to be made of the moment of inertia of the spar section about its neutral axis. LOAD FT/LBSjCHOKD 0 ((•^•685x4 PIAN FORM & LOADING FIG.5. 15 20 25 30 35 SEMI SPAN FT 10000 LBS . • SHEAR FOt I C£ / End ronjue < A FIG.6. 10" 15 TO 25 30 40 SEMI SRftN Ft 10 15 20 25 50 35 SEMI SPAN FT. For the purpose of calculation, each corrugated arc may be replaced by a plain arc of radius equal to the mean radius of the corrugated arc and of a thickness such that the cross sections of the corrugated and plain arcs are equal. 1.—To find the height of the neutral axis above the spar centre line we have :— Distance of the e.g. of each arc from x x, H chord = r x arc . a sin- = 2f . 2 A - - 1405 —^—— <•* 24"- B 1405 • ^ 39' - /BODY FRAME \ FIG.86150 6150 FIG 10 . >v r\ /// 9 CROSS SeCTiONAL AREAS OF ARCS 2 -i 5 AI'.C IN RADIANS H 13 12 II 10 in UJ la z t « •4 6 5 4 HEIGHTS OF CGs OF \ ARCS ABOVE THEIR s^ ^-> ^^ \ \ s \ \\ \\ s\ \\ s CIRCULAR CENTRES 1 H \ V \ N \ \ FIG 12 1-5 2-5 3 ARC IN RADIANS
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