Question

based on problem 2.49 in text A semicircular gate is installed in the side of water storage reservoir that is at angle of 0=60 reservoir. The depth of water in the tank is H =7.5m. At the bottom of the gate, a force Fp normal to the gate as shown is applied at P to keep the gate shut against the hydrostatic force. The water in the tank may be assumed to be at room temperature ( 20° C) and at standard atmospheric pressure. O with respect to the horizontal. The gate has a radius of R =3.6m, and is supported at its top (Q) by a horizontal shaft attached to the Gate Shaft Water H Gate R P

What is the resultant hydrostatic force on the gate?

What is the distance of the line of action (or the center of pressure) of the resultant hydrostatic from the shaft?

If the weight of the gate is assumed negligible, what is the magnitude of the force  

Answer 1

7.5m Jl. W9 € 3-6m Centre of Gravity (CG) of Semicircle from diameter BR PR=R-4B = 3.6-4X3: 6 = 2.072 m > X = 7.5- PRSim 60° = 7.5-2.072 x Sm 600 = 5.705 m → Hydrostatic force, F = y. A 8 = 9.81*(TX3:04 x 5.705 F - 1139.43 KN

surface, distance of centre of pressure from free h = x+ IG Sim²0 A.X Io = moment of Imerta of Semicircle about its Centroid = 0.1097R4 ñ = 5.705 + 0.1097 x 3.67 x Sm?60° Tx 3.6 x 5.705 5 = 5.824 m so, QS=1.442 = 1.665m Sm60 - 1.665me 4.382 m :5-824 Distance of centre of pressure from shaft = 1.665m e 1.442 m 3.118m 7.5m 360° Now PS = 3.6 -1.665 = 1.935m Pa 1.8m LW999.12 Taking moment of forces about B=0 Fx1.665 - Fpx 3.6=0 F 1139.43* 1.665 Fp - 3.6m 3.6 Fp = 526.99 RN lic