Structural analysis and design of airplanes (1920)
United States. Army. Air Corps · [Washington] · leaf 39
X300 + m.,x50 = — 2,500,000 10 (4) m.,X50 + 2m, (50+110) -2,000X110= (50:;+1103) 4 (5) m2X50 + m,X320 = —3,480,000 Solve equations (3) and (5) simultaneously—m2 = —6,695 in. lbs; m3 = —9,835 in. lbs. SHEARS AND REACTIONS (6) SV = —25X10 + s_1 = 0; s^ — +250 lbs. 1002X10 (7) —6695 =—3,125 + 100Xs+1… … <u PQ X 3 C 3 C C in O C { ) O o 73 £ed -o s~ e 03 03 cn Q -o o 03 J _c -fl cn .ti T3 < C w £ c cd a» cuPQ £o 3 s O oc 3C C •rt ■a G c 0 a> u PQ ^ <u a £ a o D J -— —■ r\i ro (>i <\J 20 Art. 22 Determination of Signs (—9835+6695) 502XlO (12) s+2 = 1 =+187.2 lbs. 50 2X50 (13) r2 =…
Airplanes
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