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Aviation History
1926
1926 - 0122.PDF
SUPPLEMENT TO FLIGHT 16 THE AIRCRAFT ENGINEER FEBRUARY 25, 1926 FIG. FIG. 2 (b) (c) (Q) (b) (a) (b) FIG.4 \ (2) FIG.5 Length of flange = 0 , (50 X 0 • 15) + (150 X 0 -4) •+ (100 X 0 • 1) 1 \ ^— — U • 18 (a -r b cos 6)- bd 6 =3-064 in. Area of flanges = 3-064 x 0-022 X 2 = 0 • 1348 sq. in. Length of Web = o /(270 X 0-1) + (70 X 0-2) + (70 X 0-5) \ OLA I =3-41 ins. Area of Webs = 3 -41 X 0-018 X 2 = 0 • 123 sq. in. Total area = 0 • 1348 + 0 • 123 = 0 • 2578 sq. in. The Moment of Inertia is best found by use of calculus. If we take any rectangle (Fig. 4 (a)) of length I, breadth b, and distance d from a line xx = tb = tb (a- -f 2ab cos 0 -f 6- cos- 6) d6 I b1 a~ + 2aZ> cos — cos 26 ) &6. Ixx = B ^ + _ j if — is very small it will be sufficiently accurate to say Ixx —d Now take the arc AB (Fig. 4 (6)) of radius b and thickness t, t being very small in relation to the distance from xx. The inertia is required about xx. CD is an infinitesimally small portion of the arc and its distance from xx will be a + b cos 6 and its inertia about xx, t (a + b cos 0)- bd 6, i.e., distance squared x length X thickness. The inertia of the whole arc AB will therefore be : = tb I ( a--t ' )e + 2ab sin 6 + - sin 26 Using this method for our spar we have :— f (1.518 _ 0-15 cos «)* 0-15 Jo 50 plOO + (1-08 -0-1 cos0) 2O-ld<9 + 0-18 X 0-9812 J n (1-167 -r 0-4 cos ^)- 0-4r/(9 2 /57-3 -2 X 1-518 x 0-15 X0-766 + 0'152 X 0-98ol4 J \0d
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