4. Solve this problem again and this time find the force acting on hinge A, if water depth is increased at 0.5m increments starting from 1m up to 5m meters high from B. Plot the force at hinge with water depth. Challenge! A rectangular plate hinged about a horizontal axis along its upper edge blocks a fresh water channel. The plate is restrained from opening by a fixed ridge at a point B. The force exerted to the plate by the ridge is to be determined. Solution Assumptions Atmospheric pressure acts on both sides of the plate, and thus it can be ignored in calculations for convenience. Properties We take the density of water to be 1000 kg/m throughout. The average pressure on a surface is the pressure at the centroid Analysis (midpoint) of the surface, and is determined to be s-Im Pre = Pe = pghc = pg(h/2) 1kN - (1000 kg/m X9.81 m/s (5/2 m 24.53 kN/m 1000 kg m/s Then the resultant hydrostatic force on each wall becomes F h-4 m F= PA=(24.53 kN/m (6 m × 5 m) = 735.9 m The line of action of the force passes through the pressure center, which is 2h/3 from the free surface, Fadge B 2h 2x(5 m) - 3.333 m 3 3 Taking the moment about point A and setting it equal to zero gives F (s+yp)= Fndp AB Solving for Fridge and substituting, the reaction force is determined to be (1+3.333) m F. ridge S+yp (735.9 kN) = 638 kN %3D AB 5m
4. Solve this problem again and this time find the force acting on hinge A, if water depth is increased at 0.5m increments starting from 1m up to 5m meters high from B. Plot the force at hinge with water depth. Challenge! A rectangular plate hinged about a horizontal axis along its upper edge blocks a fresh water channel. The plate is restrained from opening by a fixed ridge at a point B. The force exerted to the plate by the ridge is to be determined. Solution Assumptions Atmospheric pressure acts on both sides of the plate, and thus it can be ignored in calculations for convenience. Properties We take the density of water to be 1000 kg/m throughout. The average pressure on a surface is the pressure at the centroid Analysis (midpoint) of the surface, and is determined to be s-Im Pre = Pe = pghc = pg(h/2) 1kN - (1000 kg/m X9.81 m/s (5/2 m 24.53 kN/m 1000 kg m/s Then the resultant hydrostatic force on each wall becomes F h-4 m F= PA=(24.53 kN/m (6 m × 5 m) = 735.9 m The line of action of the force passes through the pressure center, which is 2h/3 from the free surface, Fadge B 2h 2x(5 m) - 3.333 m 3 3 Taking the moment about point A and setting it equal to zero gives F (s+yp)= Fndp AB Solving for Fridge and substituting, the reaction force is determined to be (1+3.333) m F. ridge S+yp (735.9 kN) = 638 kN %3D AB 5m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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