HW16-Solution

HW16-Solution - HW16 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW16 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 6, Solution 112 Member FBDs: I FBD I: MB MD MG 0: aC y a 1 2 1 2 1 2 FAF FEH FAF 0 FAF 0 FEH a 1 2 FEH II 2C y 2C y 0P P 2 FBD II: 0: aC y 0: a FBDs combined: aP a 1 2 2C y 1 2 2C y 2 PC 2 2 PT 2 Cy so FAF FEH FBD I: Fy 0: 1 2 FAF 1 2 FBG P Cy 0 P 2 1 2 FBG P P 2 0 FBG 0 FBD II: Fy 0: Cy 1 2 FDG 1 2 FEH 0 P 2 1 2 FDG P 2 0 FDG 2P C Chapter 6, Solution 132 Free body: Member ABC MC 0: (25 lb)(8 in.) B(5.196 in.) 0 B Fx 0 : 38.49 lb C x Cx 0 38.49 lb 38.49 lb Fy 0: 25 lb C y Cy 0 25 lb Free body: Member CD sin 1 3 ; 8 22.024 CD cos MD 0: M (8 in.) cos 22.024 7.416 in. 0 (25 lb)(3 in.) (38.49 lb)(7.416 in.) M 360.4 lb in. M 360 lb in. Chapter 6, Solution 141 FBD AB: By symmetry: and Ay By Ax 400 lb Bx 6 (400 lb) 5 480 lb Note: so FBD DEF: MF D Dx Dy B 480 lb 400 lb 0 (10.5 in.)(400 lb) (15.5 in.)(480 lb) (12 in.) E x Ex 970 lb Fx Fx 0 490 lb E 970 lb Fx 0: 480 lb 970 lb Fy 0 : 400 lb Fy Fy 0 400 lb F 633 lb 39.2 Chapter 6, Solution 147 FBD jaw AB: Fx MB 0: Bx 0 0 : (0.5 in.)Q (1.5 in.) A 0 A Q 3 Fy 0 : A Q By By 0 AQ 4Q 3 FBD handle ACE: By symmetry and FBD jaw DE: D Ex Ey MC A Bx By Q 3 0 4Q 3 Q 3 (0.75 in.) 4Q 3 0 0: (5.25 in.)(50 lb) (0.75 in.) Q 350 lb ...
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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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