# hw 6 - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY AND L...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-35* A beam is loaded and supported as shown in Fig. P6—35. The beam has a uniform cross 500 lb 250 lb section and weighs 425 lb. Determine the reactions at supports A and E. SOLUTION A free-body diagram of the beam is shown at the right. The reaction at support A is represented by force components IX and Ky. The reaction at E is a force E normal to the support surface. The weight w of the beam acts through the center of gravity G of the beam and is directed toward the center of the earth. + C EMA = E(9) — 500(1) + 250(4) - 425(4.5) - 400(7) = 0 E = 468.1 lb 8 468 lb lb = 468 lb T +—->2F :0 X X + C ZME = -Ay(9) + 500(8) - 250(5) + 425(4.5) + 400(2) = 0 A = 606.9 lb 3 607 lb 607 3 lb = 607 lb T ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-37 A structural member is loaded and supported as shown in Fig. P6-37. The member has a uniform cross section and weighs 208 lb. Determine the reactions at supports A and B. SOLUTION A free-body diagram of the beam is shown at the right. The reaction at support A is represented by force components xx and Ky. The reaction at B is a vertical force 5. The weight w of the beam acts through the center of gravity G of the beam and is directed toward the center of the earth. Lid1 + dez + Lads 6(3) + 4(6) + 3(7.5) G = 4.962 ft Ax + 200 - 300 = o 100 lb = -Ay(2) - 208(2.962) - 200(2) - 100(7) - 300(2) = 0 = -1158.0 lb = -1158 lb {(100)2 + (-1153)2 = 1162.3 1b a 1162 lb _ tan—1 -1162 100 X = —35.03° a -35.1° A = 100 T - 1158 J lb = 1162 lb E 85.10 + C EN = 3(2) - 208(4.962) - 200(2) - 100(9) - 300(2) = 0 A B = 1466.0 lb = 1466 lb 3 = 1466 3 lb = 1466 lb 1 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-43* Determine the reactions at supports A and B of the curved bar shown in Fig. P6—43- SOLUTION From a free-body diagram for the curved bar: + c 2MB = A sin 45°(6 + 6 sin 30°) + A cos 45°(6 cos 30°) 50(14 + 6 sin 30°) = o A = 84.68 1b a 84.7 lb K = 84.7 lb 1 45.o° Bx + A cos 45° Bx + 84.68 cos 45° = 0 -59.88 lb 8 —59.9 1b = By + A sin 45° — 50 = By + 84.68 sin 45° - 50 = o -9.878 1b 8 -9.88 lb B: + B: = /(59.88)2 + (9.88)2 = 60.69 1b a 60.7 lb ‘1 _9l88 an . 0 O _59.88 — a -170I6 — 9.88 3 N = 60.7 N r 9.37° ENGINEERING MECHANICS - STATICS, 2nd. Ed. 6-60* The coal wagon shown in Fig. P6-60 is used to haul coal from a mine. If the mass of the coal and wagon is 2000 kg, determine the force P required to move the wagon at a constant velocity and the forces exerted on the front and rear wheels by the inclined surface. SOLUTION W = 2000(9.807) = 19,614 N From a free-body diagram of the coal wagon: + /” ZFX P — w sin 30° P - 19,614 sin 30° = 0 9807 N a 9.81 kN 9.31 kN a 30° Ans.‘ W. F. w cos 30° (1) + w sin 30° (1) s FR (3) RILEY AND L. 19,614 cos 30° (1) + 19,614 sin 300 (1) - FR(3) = 0 8931 N a 8.93 kN 3.93 kN a 60° 0 FF + FR - W cos 30 F + 3931 - 19,614 cos 30° = 0 F 8055 N 3 8.06 kN = 8.06 kN 5 60° D. STURGES an“.-.___._... . ..._~._...i‘._.c..—. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-62* A frame is loaded and l—-ﬁJ'W—-T--03m--i I mmNm. supported as shown in 4uonm, Fig. P6—62. Determine the reactions at supports A and E. SOLUTION The distributed load can be replaced on the free—body diagram by a resultant force R at a distance dx from joint C. R 400(0.6) + %(400)(0.6) 240 + 120 = 360 N = A‘Id‘1 + Azd2 400(0.6)(0.3) + %(400)(0.6)(0.4) a = 0.3333 m From a free-body diagram for the complete frame: + c EMA = E cos 30° (0.6) + E sin 30° (0.6 + 0.6 tan 15°) - 360(0.3333 + 0.6 tan 15°) = 0 E 197.63 N a 197.6 N E = 197.6 N L 150.o° Ans. + —+ EFX A - 197.63 cos 30°: 0 A = 171.15 N a 171.2 N X X Ay + 197.63 sin 30° - 360 = 0 Av = 261.19 N z 261 N /(Ax)2 + (A )2 = {(171.15)2 + (261.19)2 = 312.3 N z 312 N 261.1 0 171.1 K s 312 N A 56.8 Ans. = 56.76° a 56.3° ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-86 Beam CD of Fig. P6-86 is supported at the left end C by a smooth pin and bracket and at the right end D by a continuous cable that passes around a frictionless pulley. The lines of action of the force in the cable pass through point D. Determine the components of the reaction at support C and the force in the cable when a 5-kN load W is being supported by the beam. SOLUTION’ Since the cable is continuous: I TAD=TB =l TAD: 8/0 v---. [ ’i- “1+3E 2 8 /( )2 + (-8)2 + (3)2 9117T 3 + 0.3419T E -2 8 3L 2 = 0.2279T i - 0. r = r = T n T A TBD = Tee ’1‘— “14-32 ] ID /(-2)2 + (-8)2 + (3)2 = -U.2279T i - 0.9117T 3 + 0.3419T 2 From a free-body diagram for the beam: For moment equilibrium: 2“c = Cc + [(rD/C) x (TAD + TBD) + (rs/c x a) A = J + MCZ E + [(3 j) x (-1.3234T j + 0.6838T 2)] + [(4 j) x (—5 E)! = (5.4704T - 20) 1 + Cy) J + (Mcz) E = 0 Solving yields: J+MCZIE=U T = 3.656 kN E 3.66 kN m r .‘l )- “aw—.1 »:_wu-ngym—Hé "ma-um ENGINEERING MECHANICS ~ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6—86 (Continued) TAD = 3.656(0.2279 i - 0.9117 3 + 0.3419 2) = 0.8332 3 - 3.3332 3 + 1.2500 E kN TED = 3.656(-0.2279 t - 0.9117 3 + 0.3419 2) = -0.8332 3 — 3.3332 3 + 1.2500 E kN For force equilibrium: SE = RC + TAD + TED + W = RC)‘ 1 + RCy 3 + RCz E + 0.8332 1 - 3.3332 3 + 1.2500 R - 0.8332 3 - 3.3332 3 + 1.2500 R — 2.5 E 1 3 + (Rey — 6.6664) 3 + (RC2 - 2.500) E = U 1 + Rev J + RCz E = 6.666 3 + 2.500 2 RN 3 6.67 3 + 2.50 2 kN RC = /(6.666)2 + (2.500)2 = 7.119 RN 2 7.12 kN ‘1’??? ...
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## This note was uploaded on 07/16/2009 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

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hw 6 - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY AND L...

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