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Aviation History
1937
1937 - 1604.PDF
SUPPLEMENT TO FLIGHT 6086 36 THE AIRCRAFT ENGINEER JUNE 17, 1937 -x- H I t~ t~~ FIG.I -X- 1 s^ FIG r 2. f L 1 t— I f 1 1 1 Y f 1 y re i,3 -J T rf t FIG.4. FIG.5. FIG.6. t 1 — V y 6 I, - 2 FIG. 7. - t if J . J -t— 1 tru 1 -M— FIG 8. — I I, --* — 1 H* FIG.9. Gain of area due to fillet or loss of area due to rounded corner per fillet or rounded corner = -r2(2 cot 77 + 6J = -r2 x B 2 B and C are plotted for values of 6 between 400 and 1800 on Graph 2. Correction to A for Slope. Increase of A = b(cosec a — 1) — b X Da for channel (Fig. 2) and Z- section (Fig. 4) = 6(cosec a — 1) + 2<2(cosec /8 — 1) = 6 X Da + 2d X Da for H-section (Fig. 6) b = - X D^ for T-section (Fig. 9) = 2d X Da for V-section (Fig. 10) D is plotted for values of a or 0 between 400 and 90° on Graph 3. Correction to I for thickness and slope. bt3 Increase of I-. = — for channel (Fig. 1) 12 bt3 — — for angle (Fig. 7) and T-section (Fig. 8) 24 ^ (^sin^ + cos^ bt3 24 X E6 for T-section (Fig. 9) bt3 td3 ((A2 cos2a 1 = h -r 11 3 1- cosec a — 1 V 12 6 [\dj sin a j — f- — X Fd for V-section (Fig. 10) 12 6 \ b / <«3 Increase of I„„ = — for channel (Fig. 1) 6 dt3 tb3 = i X F 6 for channel (Fig. 2) 6 12 = —- for Z-section (Fig. 3) dt3 tb3 = —- -\ X F,, for Z-section (Fig. 4) dt3 = — for H-section (Fig. 5) 6 dt3 tb3 = —- X E „ -j x F 6 for H-section (Fig. 6) 6 12 E is plotted on a logarithmic scale for values of /3 between t t o and 500 and values of -or — between o and .3 on Graph 4. F is plotted on a linear scale for values of a between t t 400 and 900 and values of - or — between o and .3 on b d Graph 5. Table 2 shows a complete summary of the method of obtaining constants and corrections to be applied. The approximations for the constants are usually suffi ciently accurate for the right angle sections (Figs. 1, 3, 5, 7 and 8) depending on the ratio of t or r to b and d. This also applies to the other sections (Figs. 2, 4, 6, 9 and 10) except that the corrections to A and I for slope are usually necessary, depending on the value of a or 0. EXAMPLE :— Find A, I„ and Z„ for T-section shown in Fig. 12. d = 1.25 = 1.175m. 2 b = 2(1.5 — —J = 2.85U1. .*. A == - X .15 (2.85 + 2 X 1.175) n .390 m.2
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