Problem 5.158 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,

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Presentation transcript:

Problem 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 B d 1.8 ft 30 o

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 B d 1.8 ft 30 o Assuming the submerged body has a width b, the load per unit length is w = b  gh, 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. Problem 5.158

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 B d 1.8 ft 30 o 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 5.158

Problem Solution A B Determine the pressure distribution acting perpendicular the surface of the submerged body. 1.7 ft (1.8 ft) cos 30 o PAPA PBPB P A = 1.7  g P B = ( cos 30 o )  g

Problem Solution A B (1.8 ft) cos 30 o Replace the pressure distribution with a resultant force, and construct the free-body diagram. AyAy AxAx FBFB 1.7  g ( cos 30 o )  g L AB /3 P1P1 P2P2 The force of the water on the gate is P = Ap = A(  gh) P 1 = (1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft) = lb 1212 P 2 = (1.8 ft) 2 (62.4 lb/ft 3 )( cos 30 o )ft = lb 1212

Problem Solution A B (1.8 ft) cos 30 o AyAy AxAx FBFB 1.7  g ( cos 30 o )  g L AB /3 P1P1 P2P2 P 1 = lbP 2 = lb Write the equations of static equilibrium for the problem, and solve them.  M A = 0: ( L AB )P 1 + ( L AB )P L AB F B = 0 + ( lb) + ( lb) - F B = F B = lb 30 o F B = 277 lb