Page 138 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Solution 1 (US Customary System):
Average pressure exerted on one side is pressure existent at C of the side times area A of the side. Since pressure varies linearly with
g
depth, as shown in Fig. 5.4, the C (center of pressure) is located below the C (center of gravity). The C is the location where a
p
single concentrated force P can be used to replace the triangle-shape distributed pressure forces for moment purposes. The pressure
f
existent at the area centroid times area equals force exerted normal to the area acting through the C .
h = 0.5 × d = 0.5 × 4ft = 2ft.
cg
C = 2 ft down from water surface.
g
3
P = ( h )A = I A = (62.4 lb/ft )(2 ft)(4 ft × 4ft) = 2,000 lb.
p
f
y = h = 2 ft in this particular case.
cg
b = 4ft. cg cg Water Hydraulics, Transmission, and Appurtenances g p p
h = 4ft.
3
4
3
I = bh ∕12 = 4 × 4 ∕12 = 21.33 ft .
cg
4
y = [I /y A] + y = [21.33 ft /(2 ft × 4ft × 4 ft)] + 2ft = 2.67 ft.
cg
cg
cp
cg
It is noted that 2.67∕4 = 2∕3.
C = 2.67 ft down from water surface.
p
Solution 2 (SI System):
Read the explanations in Solution 1.
h = 0.5 × d = 0.5 × 1.22 m = 0.61 m.
cg
C = 0.61 m downward from water surface.
g
3
P = ( h )A = I A = (9.8 kN/m )(0.5 × 1.22 m)(1.22 m × 1.22 m) = 8.9 kN.
f cg p
y = h = 0.5 × 1.22 m = 0.61 m in this particular case.
cg cg
b = 1.22 m.
h = 1.22 m.
3
4
3
I = bh ∕12 = 1.22 × 1.22 ∕12 = 0.185 m .
cg
4
y = [I /y A] + y = [0.185 m /(0.61 m × 1.22 m × 1.22 m)] + 0.61 m = 0.81 m.
cg
cg
cg
cp
C = 0.81 m downward from water surface.
p
EXAMPLE 5.10 DETERMINATION OF DEPTH OF CENTER OF PRESSURE
Determine the depth to the center of pressure y cp for a rectangular area (b × h) vertically submerged with the long side
(h = 8.2 ft = 2.5 m) at the water surface.
Solution 1 (US Customary System):
3
Using Eqs. (5.8) and (5.9), y = [I /y A] + y ,where y = h∕2, A = b × h,and I = bh ∕12.
cp cg cg cg cg cg
3
y = (bh ∕12)/[(h∕2)(bh)] + (h∕2) = 2h∕3.
cp
y = (2 × 8.2 ft)∕3 = 5.47 ft.
cp
Solution 2 (SI System):
3
Using Eqs. (5.8) and (5.9), y = [I /y A] + y ,where y = h∕2, A = b × h,and I = bh ∕12.
cg
cg
cg
cp
cg
cg
3
y = (bh ∕12)/[(h∕2)(bh)] + (h∕2) = 2h∕3.
cp
y = (2 × 2.5 m)∕3 = 1.67 m.
cp