Basics of Fluid Mechanics, 2014a

4.5. FLUID FORCES ON SURFACES 113
All the other geometrical values are obtained from
Tables 3.1 and 3.2. and substituting the proper values
results in
y cr =
A arc
{}}{
θr 2
2
y c
{ }} {
4 r sin ( y c
(
θ
2)
cos
θ
)
2
−
3 θ
θr 2
− r2 sin θ cos θ
}{{}
2
} {{
2
}
A arc
A triangle
{ }} {
2 r cos θ
3
This value is the reverse value and it is
y cr =1.65174[m]
A triangle
{ }} {
sin θr 2
2
4 r sin ( (
θ
2)
cos
θ
2
3 θ
θ
)
4 r sin ( )
θ
2
3 θ
Fig. -4.30. Area above the dam arc
calculation for the center.
The result of the arc center from point “O” (above
calculation area) is
The moment is
y c = r − y cr =2− 1.65174 ∼ 0.348[m]
M v = y c F y ∼ 0.348 × 22375.2 ∼ 7792.31759[N × m]
The center pressure for x area is
x p = x c + I xx
x c A = r cosθ 0
+
2
The moment due to hydrostatic pressure is
I xx
{ }} {
✁b (r cos θ 0 ) 3
36
r cosθ 0
✁b (r cos θ 0 )
} {{
2
}
x c
M h = x p F x = 5 r cosθ 0
F x ∼ 15399.21[N × m]
9
The total moment is the combination of the two and it is
M total = 23191.5[N × m]
= 5 r cos θ 0
9
For direct integration of the moment it
is done as following
dF = PdA=
∫ θ0
0
ρg sin θbrdθ
θ θ/2
θ/2
( ) π − θ
2
O
l =2r sin
( π
2
)
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
dF θ/2
Fig. -4.31. Moment on arc element around
Point “O.”