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Unformatted text preview: = h' 1 h c1 I xx A h c1 + = h c1 b L 1 3 12 b L 1 h c1 + = h c1 L 1 2 12 h c1 + = h' 1 6 ft 12 ft ( ) 2 12 6 ft + = h' 1 8 ft = Next find the force on OC: F 2 1.94 slug ft 3 32.2 ft s 2 12 ft 6 ft 6 ft lbf s 2 slug ft = F 2 27.0 kip = F 1 h 1 F 2 L 1 L 2 x 2 F AB L 1 Since the pressure is uniform over OC, the force acts at the centroid of OC, i.e., x' 2 3 ft = Summing moments about the hinge gives: F AB L 1 L 3 + ( ) F 1 L 1 h' 1 ( ) F 2 x' 2 + = Solving for the force in the bar: F AB F 1 L 1 h' 1 ( ) F 2 x' 2 L 1 L 3 + = Substituting in values: F AB 1 12 ft 3 ft + 27.0 10 3 lbf 12 ft 8 ft ( ) 27.0 10 3 lbf 3 ft = F AB 1800 lbf = Thus bar AB is in compression...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Statics, Fluid Mechanics, Gate

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