7-3, 7-12, 7-24 - Use the method of joints to determine the...

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Unformatted text preview: Use the method of joints to determine the force in each member of the truss shown in Fig. P7—3. 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: + T 2F? 2 Tan — 3000 = 0 Tan = 3000 lb = 3000 lb (Tl From a free-body diagram for joint B: I . o_ . ‘o_ + —a 2Fx — TBC Bin 30 TAB Sln 00 — 0 0 O + T 2Fy - —TBC cos 30 - TAB cos 60 — TBB — 0 0 0 -{1.732 T83) cos 30 - TAB cos 60 - 3000 - 0 -1500.0 lb = 1500 lb (C) T = 1.?32 TAB: 1.732(-1500.01 = -2598 lb 3 2600 lb {C} From a free-body diagram for joint C: . O + —9 SF3 - —TCD - TBC 51D 30 . 0 _ —TCD - {-2598} sin 30 — 0 TCD = 1299 lb = 1299 lb (T) Ans. From the free-body diagram for joint D: + —+ 2Fx = -TAD + TCD = “TAD + 1299 = 0 TM) 1299 lb = 1299 lb (T) Ans. M Use the method of joints to determine the force in each member of the truss shown in Fig. Pf-12. State whether each member is in tension or compression. SOLUTION From a free—body diagram for the complete truss: + C ZMD = 8(2113 cos 45°) + 10:3 sin 15°) — Ayifl cos 450} = 0 8.66? kN = 8.66? kN T .',|_' II 5' + T 2F = A — 8 + D 'x' Y Y = 8.66? — 8 + D = 0 Y D = —u.66? kN = 0.667 kN i + -4 2F = 10 + Dy = 0 D = -10 kN = 10 kN «— From a free—body diagram for joint A: + —4 EFX = THE cos 45° = 0 TAE = 0 Ans. + T 2F? = T53 + TAE sin 45° + 3.667 : TAB + (0) + 8.667 = O -8.66T kN a 8.6? kN {C} H II Ans. IIiEIWContinued) From a free—body diagram for joint B: . 0 + T EFy — TBC 51n 45 — TAB . 0 __ TBC Sln 45 - I-8.66?} — 0 —12.257 RN 3 12.26 kN {C} 0 TBE + TBC cos 45 + 10 THE + (—12.2571 cos 45° + 10 = 0 -l.3330 kN = 1.333 kN {C} Ans. From a free—body diagram for joint C: + T 29y —T sin 45° — T sin 45° - 3 BC CE —(—12.257} sin 45° — TCE sin 45° — 3 = 0.9433 kN 8 0.943 kN (T) Ans. . o _ 0 Ten + TCE cos 45 TBC cos 45 TCD + (0.9433) cos 45o - (-12.26?) cos 450 = 0 -9.334 RN 3 9.33 kN C} Ans. ‘ a» \ From a free-body diagram for joint E: 0 + /” EFK, TDE — TBE cos 45 - TAE 0 TDE - (—1.3330} cos 45 - O — 0 -0.9426 kN 3 0.943 kN {C} Ans. Determine the forces in members DE, EI, and HI of the truss shown in Fig. P7‘24. SOLUTION From a free—body diagram for the complete truss: + C EMA = 6(16} - 3(4) — 3(a) — 5(12} = o G = 8.50 kN = 8.50 kN T From a free—body diagram of joint G: ¢ = tan-1 20.550 6 tan" 56.310 0 0 + —+ 2Fx -TFG cos 56.31 ~ TGH cos 20.56 - 0 1 O . D _ = TFG Sin 56.31 + TGH Sin 20.56 + 8.50 — 0 -l3.622 kN H 13.62 kN (C) 8.070 kN = 8.07 kN IT) From a free—body diagram of joint F: + /; EFx’ = -TFH = 0 + E\ SE T + 13.622 = 0 3:" EF TEF -13.622 kN a 13.62 kN (cl 7-24 (Continued) From a free—body diagram of joint H: + /” ZFXJ TEH + E\ SE , y TR] 0 TEH cos 20.56 — 0 0 THI - 8.070 = 0 8.070 kN = 8.07 kN (T) From a free—body diagram of joint E: + T 2F y + -% 2F 3! E1 DE *6 “'3 '-'.| [-11 ‘ TDE UIIH'— gum m h-q 10.557 kN 3 10.56 kN {T} -16.002 kN E 16.00 kN (C) Ans. ' + 13.622 sin 56.310 = 0 - 13.622 cos 56.310 = 0 Ans. Ans. )LOOZKN E .3 TE! kawi + C3 13.1.2210: ...
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This homework help was uploaded on 04/16/2008 for the course ENGR 1100 taught by Professor Anderson during the Fall '06 term at Rensselaer Polytechnic Institute.

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7-3, 7-12, 7-24 - Use the method of joints to determine the...

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