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qz 35 - 6ch q d‘sMw/fleaas moan“ ENGINEERING MECHANICS...

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Unformatted text preview: 6ch \% q d‘sMw/fleaas moan“ ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-73* The awning structure of Fig. P7-73 is supported by ball- and—socket joints at A and C and by a short link at B. The 50-lb force is parallel to the y-axis and the 150-lb force is parallel to the z- axis. Determine the support reactions and the force in each member of this space 150m1,2n truss. Fig. P7-73 SOLUTION From a free-body diagram for joint D: 2F=TAD+T +T +FD BD CD TAD-2 1 - 4 1 — 1 E + T + 2 E + TCD-z 1 + 4 J — 1 E _ 150 E /§T 21 (—0.4364TAD - 0.7071TBD - 0.4364TCD) 1 + (-0.8729TAD + 0.8729TCD) j + (—0.21321AD + 0.7071TBD — 0.2182TCD— 150) E = 0 Solving yields: TAD = TCD = —114.59 lb 3 114.6 lb (C) TBD 141.44 lb E 141.4 lb (T) From a free-body diagram for joint B: )Z‘F‘=TM+TBC+TBD+B+FB A TAB:£_1_:_§_§ + Twill—31-g + (141.44)§—l—1—§—- y’fi y’fi {$3 (100.00 - By) 1 + (-0.8000TAB + 0.8000TBc - 50) J + (—0.6000TAB - 0.6000TBC - 100.00) E = 0 Solving yields: By = -100.00 lb 3 = -100.0 1 lb Ans. -114.58 lb 3 114.6 lb (C) Ans. -52.08 lb 3 52.1 lb (C) Ans. 7—73 (Continued) From a free-body diagram for joint A: .1 (-50.01 + A‘) i + (—91.66 - 100.02 + Ay) 3 + (-68.75 — 25.00 + A2) E = 0 yields: 50.01 lb 191.68 lb 93.75 lb A z 50 i + 191.7 3 + 93.3 E lb From a free-body diagram for joint C: In -7...“ I I I y--- 1 / l a.“ n I I I I ’§ ‘1‘, \1 P I 51' C, 2? = TBC + TCD + C = (-52.03133—1—1-3—g + (-114.59)2 1 ‘ 4 + 1 E + Cx i + c flfi fifi ‘ (-50.01 + Ox) 1 + (41.66 + 100.02 + cy) 3 + (—31.25 - 25.00 + 02) E = 5 Solving yields: = 50.01 lb -14l.68 lb 56.25 1b 1 - 141.7 3 + 56.3 E 1b ...
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