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ch7(6-16)

ch7(6-16) - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: ENGINEERING MECHANICS - 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—6. 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.ﬂﬂwnx From a free-body diagr:m' If for joint B: ‘ .I7E; V + T SE = -T sin 60° - 2800 = 0 Tan = ~3233 N 3 3.23 kN (C) Ans. From a free-body diagram for joint D: 0 2 0 + /” ZFX - TCD + TBD cos 30 s 2100 cos 30 = T + (—3233) cos 30° - 2100 cos 30° CD = 4619 N g 4.62 kN (T) Ans. . O . 0 + R\ BFy - TAD + TBD Sln 30 + 2100 Sln 30 = T + (—3233) sin 30° + 2100 sin 30° TAD = 566.6 N 3 0.567 kN (T) Ans. From a free-body diagram for joint C: 0 0 ' + —e 2Fx — -TBC — TCD cos 30 — -TBC - 4619 cos 30 — O [AW—“ﬂ TBC = -4000 N = 4.00 kN (C) us. From the free—body diagram for joint B: + —+ EFX = —T :7 : -T + (—4000) - (-3233) cos 60° = 0 TAB = -2383 N E 2.38 kN (C) Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-9* 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 _ Jnember forces can be D determined without solving for the support reactions. .70D From a free—body diagram i7¥ES for joint D: ' + ‘\ SE = - T sin 30° — 1000 sin 60° = 0 TBD = -1732.05 lb 8 1732 lb (C) Ans. 0 0 + /” ZFX ~ -TCD - TED cos 30 - 1000 cos 60 - — 1000 cos 60° = 0 = T - (-1732.05) cos 30° TCD = 1000 1b = 1000 lb (T) Ans. \ ‘2 ’.+ . \ ’/ From a free—body diagram \ TED for joint C: c + /' EFX = -TAC + TCD = -TAC + 1000 = O “The ‘ TBc TAC = 1000 lb = 1000 lb (T) Ans. 3 +'K\ 2F = -T = O 7 1 V 3° T I m Be . 1;... From a freevbody diagram . : LC? for joint B: + -+ 2Fx = -T - T cos 600 + T cos 60? AB BC BD — (0) + (—1732.05) cos 60° -T TAB = ~866.03 lb 3 866 1b (C) 'vrv .7: >7» v-—s—~———.-»“scissrwmwlw 7,,” E, ., W u U . m," MWWWWLMMWst‘ mmxmg: .r‘,~mr=,~;-Mm‘ e, w is w—_‘——*‘ ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-10* Use the method of joints to 33” 5*“ determine the force in each C E ‘- b‘ member of the truss shown in D. Fig. P7—10. State whether —1—3m each member is in tension or Fig. v7.10 i compression. SOLUTION From a free-body diagram for the complete truss: ,; + -9 25‘“ = A = O A = O Ans. X X + C 2MA = Dy(6) - 5(4.5) - 3(1.5) = o Dy=4.5 kN=4.5 kNT + (j EMD = -Ay(6) + 5(1.5) + 3(4.5) =\Q_ Ay=3.5kN=3.5kNT From a free—body diagram r 7 S T for joint D: CD +TZF=T sineo°+4.5=o ’-° y CD TCD = 45.196 kN s 5.20 M! (c; Ans. To: O —+ + EFX — —TDE — TCD cos 60 - 0 __ — ~TDE - (-3.196) cos 60 I- 0 TDE = 2.598 kN 3 2.60 kN (T) A From a free—body diagram for joint A: R T 2 = - ° . = . ' ""X + Fy TAB Sln 60 + 3 5 0 _ vThE = -4.o41 M] a 4.04 kN (C) Ans. 3.5 KN AB ~+ + SE = TA 0 x + TAB cos 60 E - __ o‘- — TAE + ( 4.041) cos 60 — 0 TAE = 2.021 kN g 2.02 kN (T) Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-10 (Continued) - From a free-body diagram for joint B: I _ o . 0 + T 2Fy -TAB Sln 60 — TBE s1n 60 3 in 60° — T - + TBE cos 60 - TAB cos 60 + (0.5769) cos 60° — (44.041) cos 60° = 0 -2.309 kN E 2.31 kN (C) From a free-body diagram for joint C: + —e 2Fx cos 60° - T cos 60o - T CE BC cos 60° — T cos 60° —(-2.309) = 0 —0.5774 kN g 0.577 kN (C) ENGINEERING MECHANICS ~ STATICS, 2nd. Ed; W. F. RILEY AND L. D. STURGES 7—13* Determine the forces in members AB, AE, and EF of the truss shown in Fig. P7-13. State whether each member is in tension or compression. 800 lb Fig. P7-13 SOLUTION For this simple truss, the required member forces can be determined without solving for the support reactions. From a free-body diagram bﬁjﬁ; for joint A: . o _ + T 2F — TAB Sln 30 - 600 - 0 TAB = 1200 lb = 1200 lb (T) 0 + TAB cos 30 T + 1200 cos 30° = 0 : -1039.2 lb 2 1039 lb (C) Ans. From a free—body diagram for joint B: + “\ ZFy - 750 = o TBE = —750 lb = 750 lb (C) From a free-body diagram for joint E: — . 0— . '0 . O- . 0 + R\ 2Fy — TEE Sln 60 TEF Sln 60 + TAE Sln 60 800 Bin 30 . 0 . 0 (-750) Sln 60 - TEF Sin 60 (—1039.2) sin 60° — 800 sin 30° = o ENGINEERING MECHANICS * STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES . 7-15 Determine the forces in 300010 200010 members AD, CD, anﬂ DE of the truss shown in Fig. P7—15. State whether each member is in tension or compression. Fig. P7-15 SOLUTION , ’ " 7 000 51:. 200015 From a free-body diagram for the complete truss: .. I + C EMA = By(6) — 3000(6) - 2000(12) .= 0 By = 7000 lb = 7000 lb T From a free-body diagram for joint B: + T 2F = T + 7000 = 0 4000 lb = 7000 lb (C) From a free-body diagram for joint F: + T 2F = «r» sin 45° - 2000 = 0 y DF Tm. : -2328 lb a: 2830 lb (C) 3" — - _ -° + —4 -Fx — TEF TDF cos 40 a ‘0 _ — --I‘EF - (-ZBZB) cos 4o — 0 . ‘ . 1' = 2000 1b = 2000 H: m EF m, mammm ‘p—gH‘ "an, 0 \$17.. W mem‘wmmmr ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES '7-15 (Continued) From a free—body diagram for joint E: T - T cos 45° EF CE 0 2000 — TCE cos 45 — 0 +->2F X 2828 1b = 2828 lb (T) + ——) M '13 ll : . o _ y --TCE Sln 45 - TDE - 3000 — 0 : -(2828) sin 4 — 3000 = O -—-_— —5000 1b = 5000 lb (0) From a free-body diagram for joint D: . o L o + T EFY — TDE + TDF Sln 45 — TAD Sln 45 — TED = (-5000) + (-2828) sin 45° - T sin 45° - (-7000) = 0 AD T = 0 Ans. ‘ AD cos 45° — T - T +—_)ZF:T CD AD x DF (-2828) cos 45° — T -2000 lb 2 2000 lb (C) 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 trussz, + 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 SE = T sin 30° + 11.00 = 0 V y D E TDE = -22.00 kN-= 22.0 kN (C) From a free—body diagram for joint D: O 0 — -TDF cos 30 — 8 cos 30 - 0 -8.00 kN = 8.00 kN (C) . O . O TDE ~ TCD + 8 Sin 30 — TDF Sin 30 (-22.00) — TCD + 3 sin 30° — (—3.00) sin 30° = o ~14.00 kN = 14.00 kN (C) Ans. From a free—body diagram for joint C: 0 0 + -+ EFX — —TBC cos 30 + TCD cos 30 -T cos 300 + (—14.00) cos 300 = 0 BC —14.00 kN = 14.00 kN (C) Ansj\\\\:> ...
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ch7(6-16) - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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