hw 8 - ENGINEERING MECHANICS STATICS 2nd Ed Use the method...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. Use the method of joints to determine the force in each member of the truss shown in Fig. P7-7. State whether each member is in tension or compression. SOLUTION For this simple truss, the member forces can be determined without solving for the support reactions. From a free-body diagram for joint C: + T 2Fy = TBC = 0 T = 0 BC Ans. From a free-body diagram for joint B: 0 AB cos 45 + TBD , 0 AB Sln 45 - TBD + —+ 2F = —T X -T + T 2F = v Solving yields: TAB = TBD = -141.42 lb 8 141.4 lb (C) -989.95 lb 3 990 lb (C) From a free—body diagram for joint D: ' + —9 XFX = -T - T CD BD CD T = CD 700 lb = 700 lb (T) From the free—body diagram for joint C: + T = _T c0 AC 700 lb = 700 lb (T) -T + 700 = 0 +—)2Fx= Ac TAC = Ans. Ans. Ans. Ans. w. F‘ cos 45° + 600 = 0 sin 45° - 300 = o RILEY AND L. D. cos 45° = —T - (—939.95) cos 45° = o STURGES ENGINEERING MECHANlCS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Use the method of joints to determine the force in each member of the truss shown in Fig. P7-9. State whether each member is in tension or compression. SOLUTION For this simple truss, the member forces can be determined without solving for the support reactions. From a free-body diagram for joint D: + “\ 2F - T sin 30° — 1000 sin 60° = 0 y 30 TBD -1732.05 lb a 1732 lb (C) ' Ans. 0 ° + /” 2Fx -TCD - TBD cos 30 - 1000 cos 60 -TCD — (—1732.05) cos 30° — 1000 cos 60° = 0 TCD 1000 lb = 1000 lb (T) Ans. From a free—body diagram for joint C: + /” EFX = -TAC + TCD = -TAC + 1000 = 0 TAc {1000 1b = 1000 lb (T) + “\ EFy -TBC = o TBC = 0 From a free-body diagram for joint B: 0 0 + -+ 2Fx - -TAB - TBe cos 60 + TBD cos 60 0 -TAB - (0) + (-1732.05) cos 60 -866.03 lb 3 866 lb (C) Ans. 5)7 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Determine the forces in members BC, CD, and DF of the truss shown in Fig. P7-16. State whether each member is in tension or compression. SOLUTION From a free-body diagram for the complete truss: + C EMA = Ey(8) - 10(4) - 8(6) = o Ey = 11.00 kN = 11.00 kN T From a free—body diagram for joint E: + T 2F T sin 30° + 11.00 = o y DE. TDE = -22.00 kN = 22.0 kN (C) BKN ll! for joint D: T g l c 6 ’ = -T cos 30° - 8 cos 300 = 0 D ’ DF I -8.00 kN = 8.00 kN (C) Ans. To; ,’D \ ,go/ 706‘” From a free-body diagram " - ‘ ’X . 0 . O - TDE — TCD + 8 sin 30 - TDF sin 30 (—22.00) - TCD + 8 sin 30° — (-8.00) sin 30° = 0 -14.00 kN = 14.00 kN (C) Ans. From a free-body diagram for joint C: 0 0 + —+ ZFX — -TBC cos 30 + TCD cos 30 -TBC cos 30° + (:14.00) cos 30° = 0 -14.00 kN = 14.00 kN (C) Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES w A A r r mass-1:1"; m 7—43* The length of each truss member in Fig. P7-43 is 5 ft. Use the method of :l ' sections to determine the itf‘""" forces in members CD, DE, ' lOOOlb 8001b 6001b 8001b I a) and EF' Fm.PL43 SOLUTION From a free—body diagram for the complete truss: o goo 1.00 300 ")1 n, )1. )L A = 0 A = 0 X X C D = -Ay(25) + 1000(20) ' + 800(15) + 600(10) + 800(5) = 0 = 1680.0 lb = 1680 lb T From a free-body diagram of the part of the truss to the left of member DF: “£34? “an A'a,».-;,.,,sm',.r.-.-:,.«.J;n «em “a ,5 , t t1 . z + c 2MB = 1000(5) - 1680(10) - TCD(5 sin 60°) = 0 TCD = -2725 lb a 2730 lb (C) .14.,ng Lyn-.1) y + C 2MB = 800(2.5) + 1000(7 5) — 1680(12.5) + TEF(5 sin 60°) = 0 TEF = 2656 lb 2 2660 lb (T) T + EFy = TDE sin 60° — 300 — 1000 + 1680 = 0 "DE = 138.56 lb 3 138.6 (T) 545 7-59* Use the method of sections to determine the forces in members CD, GK and LK of the Howe roof truss shown in Fig. P7—59. SOLUTION From a free—body diagram for the complete truss: + —e ZFX A = 0 A = 0 X X + C EMG = -Ay(36) + 1000(30) + 600(18) + 800(12) = 0 AV = 1400.0 lb = 1400 lb T From a free-body diagram of the part of the truss to the left of member DK: 29.05° 48.01° r ‘ 'n i n V -1400(18) + 1000(12) . O O _ TCD s1n 29.05 (6) - TCD cos 29.05 (6.667) - 0 -1510.0 lb = 1510 lb (C) -1000(6) - TCK cos 48.01° (6.667) - TcK sin 48.01° (12) = 0 -448.45 lb E 448 1b (C) Ans. TLK(6.667) + 1000(6) - 1400(12) = 0 1619.9 lb 2 1620 lb (T) m ENGINEERING MECHANICS - STATICS, 2nd. Ed. 7-68 Use the method of sections to determine the forces in members CD, DF, EF, and FG of the transmission line truss shown in Fig. SOLUTION From a free-body diagram for the part of the truss above member CG. + C EMD = 5 cos 30° (6) - TFG(3) = o TFG = 8.660 kN = 8.66 kN (T) Ans. O + C ZMG = TCD(3) + 5 cos 30 (9) — 5 sin 30° (4) = o TCD = -9.657 RN 3 9.66 kN (C) From a free-body diagram for joint F: 1 4 e = tan' 5 = 53.13° sin 51.13 - 8.660 = 0 + T XFY : TEF TEF = 10.825 kN E 10.83 kN (T) Ans. -T - T cos 9 BF EF 0 - TDF - 10.825 cos 53.13 — 0 -6.495 kN z 6.50 kN (C) Ans. WI F. RILEY AND L. DI STURGES ...
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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 8 - ENGINEERING MECHANICS STATICS 2nd Ed Use the method...

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