HW14-Solution

HW14-Solution - HW14 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW14 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 6, Solution 61 Free body: Entire truss MA 0: K (8 m) (1.2 kN)(6 m) (1.2 kN)(12 m) 0 K 2.70 kN K 2.70 kN Free body: Lower portion MF 0: FEJ (12 m) (2.70 kN)(4 m) (1.2 kN)(6 m) (1.2 kN)(12 m) FEJ 0.900 kN 0 FEJ 0.900 kN T Fy 0 : FAF 0.9 kN 1.2 kN 1.2 kN FAF 0 1.500 kN FAF 1.500 kN T Chapter 6, Solution 67 Free body: Truss Fx 0: Fx 0 MH Fy Fy H 0 : 4.8(3a) 4.8(2a) 4.8a 2.4a Fy (2a) 13.20 kips 0 : H 13.20 kips 3(4.8 kips) 2(2.4 kips) 6.00 kips 0 F 13.20 kips 0 H 6.00 kips Free body: ABF We assume that counter BG is acting. Fy 0: 9.6 FBG 13.20 2(4.8) 14.6 FBG 0 5.475 FBG 5.48 kips T Since BG is in tension, our assumption was correct Free body: DEH We assume that counter DG is acting. Fy FDG 0: 9.6 FDG 14.6 6.00 2(2.4) 0 FDG 1.825 kips T 1.825 Since DG is in tension, O.K. ...
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This note was uploaded on 11/07/2010 for the course ME 211 taught by Professor Staff during the Fall '05 term at Cal Poly.

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