Problem 5-c The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A d 1.8 ft B 30o

Solving Problems on Your Own Problem 5-c Solving Problems on Your Own The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A d 1.8 ft B 30o Assuming the submerged body has a width b, the load per unit length is w = brgh, where h is the distance below the surface of the fluid. 1. First, determine the pressure distribution acting perpendicular the surface of the submerged body. The pressure distribution will be either triangular or trapezoidal.

Solving Problems on Your Own Problem 5-c Solving Problems on Your Own The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A d 1.8 ft B 30o 2. Replace the pressure distribution with a resultant force, and construct the free-body diagram. 3. Write the equations of static equilibrium for the problem, and solve them.

Problem 5-c Solution Determine the pressure distribution acting perpendicular the surface of the submerged body. 1.7 ft A PA PA = 1.7 rg PB = (1.7 + 1.8 cos 30o)rg (1.8 ft) cos 30o B PB

P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb Ay Problem 5-c Solution A Ax Replace the pressure distribution with a resultant force, and construct the free-body diagram. 1.7 rg (1.8 ft) cos 30o LAB/3 P1 FB LAB/3 The force of the water on the gate is P2 B LAB/3 1 2 1 2 (1.7 + 1.8 cos 30o)rg P = Ap = A(rgh) 1 2 P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb 1 2 P2 = (1.8 ft)2(62.4 lb/ft3)(1.7 + 1.8 cos 30o)ft = 329.43 lb

S MA = 0: ( LAB)P1 + ( LAB)P2 - LABFB = 0 Ay Problem 5-c Solution A Ax Write the equations of static equilibrium for the problem, and solve them. 1.7 rg (1.8 ft) cos 30o LAB/3 P1 FB S MA = 0: LAB/3 + P2 B 1 3 2 3 ( LAB)P1 + ( LAB)P2 LAB/3 (1.7 + 1.8 cos 30o)rg - LABFB = 0 P1 = 171.85 lb P2 = 329.43 lb 1 3 2 3 (171.85 lb) + (329.43 lb) - FB = 0 FB = 276.90 lb 30o FB = 277 lb