Page 139 - Water Engineering Hydraulics, Distribution and Treatment

EXAMPLE 5.11 DETERMINATION OF HYDROSTATIC FORCE AND LOCATION OF CENTER OF PRESSURE
Determine the hydrostatic force and the location of the center of pressure on the 92-ft (28-m)-long dam shown in Fig. 5.5. The face
◦
◦
of the dam is at an angle of 60 . The water temperature is 77 F(25 C) and the depth is 26.25 ft (8 m).
Dam
P f
60°
Figure 5.5 Hydrostatic force on a dam. h Water surface ◦ 5.1 Fluid Mechanics, Hydraulics, and Water Transmission 117
Solution 1 (US Customary System):
◦
3
From Appendix 3, = 62.28 lb/ft at 77 F.
3 ◦
P = ( h )A = (62.28 lb∕ft )(26.25 ft∕2)(92 ft)(26.25 ft∕sin 60 ) = 2,279.54 lb.
f
cg
The center of pressure is at two-thirds of the total water depth, (2∕3)(26.25 ft) = 17.5 ft.
Solution 2 (SI System):
◦
3
From Appendix 3, = 9.779 kN/m at 25 C.
3 ◦
P = ( h )A = (9.779 kN∕m )(8 m∕2)(28 m)(8m∕sin 60 ) = 10,118 kN.
f
cg
The center of pressure is at two-thirds of the total water depth, (2∕3)(8 m) = 5.33 m.
EXAMPLE 5.12 DETERMINATION OF RESULTANT FORCE DUE TO WATER PRESSURE ACTING ON
VERTICAL GATE
Determine the resultant force and its location due to the water acting on a 4 ft by 8 ft (1.22 m by 2.44 m) rectangular gate AB shown
in Fig. 5.6. The top of the gate is 16 ft (4.88 m) below the water surface. The dimensions in Fig. 5.6 are a = 16 ft = 4.88 m; h = 8ft =
2.44 m; and b = 4ft = 1.22 m.

Water surface
O

a
h cp = y cp h cg = y cg
A
C g
h
C p
B
b

Figure 5.6 Vertical rectangular gate in a water tank.

Solution 1 (US Customary System):
3
P = ( h )A = (62.4lb∕ft )(16 ft + 8ft × 0.5)(4 ft × 8ft) = 39,936 lb.
f
cg
The above resultant force acts at the center of pressure, which is at a distance y from water surface O.
cp
At vertical position, h = y .
cg
cg
y = 16 ft + 8ft × 0.5 = 20 ft.
cg
2
A = 4ft × 8ft = 32 ft .