HW13-Solution

# HW13-Solution - HW13 Solution Terry Cooke Cal Poly San Luis...

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Unformatted text preview: HW13 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 6, Solution 44 See solution of Problem 6.43 for free-body diagram of truss and determination of reactions: A 8400 lb k 3600 lb We pass a section through members EG, FG, and FH, and use the free body shown. MF 0: (3600 lb)(31.25 ft) FEG FEG (15 ft) 7500 lb 0 FEG 7500 lb T Fy 0: 15 FFG 16.25 3600 lb FFG 0 3900 lb 0 FFH 6000 lb C FFG 3900 lb C MG 0 : FFH (15 ft) (3600 lb)(25 ft) FFH 6000 lb Chapter 6, Solution 49 Reactions at supports: Because of the symmetry of the loading, Ax Ay 0 I 1 (Total load) 2 1 (8 kN) 2 A I 4 kN We pass a section through members CD, DE, and DF, and use the free body shown. (We moved FDE to E and FDF to F ) Slope Slope BJ 2.16 m 9.6 m 9 40 DE a MD 1m 5 2.4 m 12 0.46 m 0.46 m 9 Slope BJ 40 2.0444 m 0 FCE 7.20 kN T 0 : FCE (1 m) (1 kN)(2.4 m) (4 kN)(2.4 m) FCE 7.20 kN MK 0 : (4 kN)(2.0444 m) (1 kN)(2.0444 m) (2 kN)(4.4444 m) FDE 5 FDE (6.8444 m) 13 1.047 kN 0 FDE 1.047 kN C ME 0 : (1 kN)(4.8 m) (2kN)(2.4 m) (4 kN)(4.8 m) 40 FDF (1.54 m) 41 FDF 0 6.39 kN FDF 6.39 kN C Chapter 6, Solution 55 Reactions: Mk Fx 0: 36(2.4) B(13.5) 20(9) 20(4.5) 0: 36 K x 0 Kx 36 kN 0 B 26.4 kN Fy 0: 26.4 20 20 K y 0 K y 13.6 kN MC 0: 36(1.2) 26.4(2.25) FAD (1.2) FAD 0 13.5 kN FAD 13.5 kN C FCD 0 MA MD 0: 8 FCD (4.5) 17 0 0: 15 FCE (2.4) 26.4(4.5) 17 FCE 0 56.1 kN FCE 56.1 kN T ...
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