Problem 3.64 - = h' 1 h c1 I xx A h c1 + = h c1 b L 1 3 12...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 3.64 [Difficulty: 3] Given: Gate AOC, hinged along O, has known width; Weight of gate may be neglected. Gate is sealed at C. b6 f t = Find: Force in bar AB Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g = (Hydrostatic Pressure - h is positive downwards) F R p c A = (Hydrostatic Force) y' y c I xx Ay c + = (Location of line of action) Σ M z 0 = (Rotational equilibrium) F 1 h 1 F 2 L 1 L 2 x 2 F AB L 1 Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate (4) No resisting moment in hinge at O (5) No vertical resisting force at C Integrating the hydrostatic pressure equation: p ρ g h = The free body diagram of the gate is shown here: F 1 is the resultant of the distributed force on AO F 2 is the resultant of the distributed force on OC F AB is the force of the bar C x is the sealing force at C First find the force on AO: F 1 p c A 1 = ρ g h c1 b L 1 = F 1 1.94 slug ft 3 32.2 × ft s 2 6 × ft 6 × ft 12 × ft lbf s 2 slugft × = F 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Page1 / 2

Problem 3.64 - = h' 1 h c1 I xx A h c1 + = h c1 b L 1 3 12...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online