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# qz 36 - Determine the reactions at supports A and E and the...

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Unformatted text preview: Determine the reactions at supports A and E and the force exerted by the pin at B on member ABC of the frame shown in Fig. 97—97. ' 16in. 16in. 16in. Fig. P7-97 SOLUTION From a free-body diagram for the complete frame: + C 2M = Ey(50) - 750(16) — 1600(50) = 0 A Ey = 1840 lb = 1840 lb T + C‘ZME = —Ay(50) — 750(16) = o Ay = -240.0 lb = 240 lb ¢ From a free-body diagram for bar ABC: + C 2MB = Ax(32) + 240(18) - 1600(32) = 0 = 1465 lb = 1465 lb -+ = By - 240 - 1600 = 0 = 1840 lb = 1840 lb T = 1465 + B = 0 X -1465 lb = 1465 lb +- 'ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-97 (Continued) FrOm the free-body diagram for the complete frame: +——>EF A+E +750 x x x 1465 + Ex + 750 = 0 —2215 1b = 2215 lb +- /(A )2 + (A )2 = /(l465)2 + (-240)2 = 1484.5 lb s 1845 lb x y . -1 -2 0 _ _ - o A tan 1465 — 9.304 K = 1845 lb 8 8.3o° Ans. (Bx)2 + (By)2 = /(-1465)2 + (1840)2 = 2352 lb s 2350 lb -1 1840 _ 0 GB tan -1465 — 128.53 E = 2350 lb 5 51 5° Ans. /(Ex)2 + (Ey)2 = /(-2215)2 + (1840)2 = 2879.6 lb a 2880 lb _ —1 1840 _ o 6E tan :EEIE — 140.28 E = 2880 lb 5 39.7° Ans. <ﬁ‘1l ...
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