Therefore, TAdoes not change with H. The correct answers are obtained by reviewing
the above solution.
101
3.70: PROBLEM DEFINITION
Situation:
Water exerts a load on square panel.
d=1m,h=2m
Find:
(a) Depth of the centroid (m).
(b) Resultant force on the panel (kN).
(c) Distance from the centroid to the center of pressure (m).
Properties:
Water (15 ◦C), Table A.5: γ=9800N/m3.
PLAN
1. Locate the centroid by inspection (center of the panel).
3. Find the distance between the centroid and the CP using ycp −¯y=I/(¯yA)
SOLUTION
1. Depth of the centroid of area:
2. Hydrostatic equation:
4. Distance to CP:
102
•Find Iusing formula from Fig. A.1.
103
3.71: PROBLEM DEFINITION
Situation:
Seawaterexertsaloadonaroundviewingwindow.
h=1.5m,θ=60
0,D=0.5m
Find:
Hydrostatic force on the window (kN).
Locate the CP (center of pressure).
Properties:
Seawater: S=1.03,γ=1.03 ×9810 N/m3= 10100 N/m3.
PLAN
1. Find distances using trig.
3. Find the distance between the centroid and the CP using ycp −¯y=I/(¯yA)
SOLUTION
1. Distances:
•Slant height
3. Resultant force:
4. Distance to CP:
104
•Find ¯
Iusing formula from Fig. A.1.
105
3.72: PROBLEM DEFINITION
Situation:
Water exerts a load on a submerged gate.
Find:
Force of gate on block (kN).
SOLUTION
Hydrostatic force
Center of pressure
Equilibrium (sum moments about the pivot)
106
3.73: PROBLEM DEFINITION
Situation:
Wet concrete is held in place with forms.
Ties are spaced on 2 feet centers.
Find:
Hydrostatic force per foot on form (lbf/ft).
Force exerted on bottom tie (lbf).
Properties:
Concrete, γ=150lbf/ft3.
SOLUTION
Hydrostatic force
Center of pressure
Equilibrium (sum moments about the top tie)
108
3.74: PROBLEM DEFINITION
Situation:
A rectangular gate is hinged at the water line.
h=4ft,b=8ft.
Find:
Force to keep gate closed.
Properties:
From Table A.4, γWater =62.4lbf/ft3.
SOLUTION
Hydrostatic Force (magnitude):
Center of pressure. Since the gate extends from the free surface of the water, FG
acts at 2/3 depth or 8/3 ft. below the water surface.
Moment Equilibrium. (sum moments about the hinge)
109
3.75: PROBLEM DEFINITION
Situation:
A submerged gate sits at an angle.
h=6m,b=4m,θ=30◦.
Find:
Reaction at point A.
Assumptions:
Gate is weightless.
Properties:
Water, Table A.5: γ=9810N/m3.
PLAN
The reaction at A can be found by summing moments about the stop. The steps are
1. Find the hydrostatic force.
3. Sum moments about the stop.
SOLUTION
1. Hydrostatic force (magnitude)
2. Center of pressure:
110
3. Moment equilibrium about the stop:
111
3.76: PROBLEM DEFINITION
Situation:
A submerged gate holds back water.
b=2m
Find:
Force Prequired to begin to open gate (kN).
Assumptions:
Gate is weightless.
Properties:
Water, Table A.5: γ=9810N/m3.
SOLUTION
Hydrostatic force
Center of pressure
112
Equilibrium
113
3.77: PROBLEM DEFINITION
Situation:
A submerged gate opens when the water level reaches a certain value.
Find:
hin terms of to open gate.
PLAN
As depth of water increase, the center of pressure will move upward. The gate will
open when the center of pressure reaches the pivot.
SOLUTION
Center of pressure (when the gate opens)
Center of pressure (formula)
Combine Eqs. (1) and (2)
114
3.78: PROBLEM DEFINITION
Situation:
Abutterfly valve is described in the problem statement.
d=10ft,θ=30◦,¯y=30ft.
Find:
Torque required to hold valve in position (ft-lbf).
SOLUTION Hydrostatic force
Center of pressure
Torque
3.79: PROBLEM DEFINITION
Situation:
A submerged gate may fall due to its weight (or be held in place by pressure).
y1=1m,y2=4m,w=1m.
W= 150 kN,α=45
o.
Find:
Will the gate fall or stay in position?
Properties:
Water (10 ◦C), Table A.5, γ=9810N/m3.
SOLUTION
1. Geometry
•Slant height:
•Panel surface area
116
•Area moment of inertia from Fig. A.1:
5. Torques:
•Torque caused by hydrostatic force:
•Torque caused by the weight:
117
3.80: PROBLEM DEFINITION
Situation:
A submerged gate may fall due to its weight.
y1=3ft,y2=6ft,w=3ft.
W= 18000 lbf,α=45
o.
Find:
Will gate fall or stay in position?
Properties:
Water (50 ◦F), Table A.5, γ=62.4lbf/ft3.
SOLUTION
1. Hydrostatic Force:
•Area:
A=y2
sin α×w=6ft
sin 45o×3ft=25.46 ft2
2. Distance from CP to centroid:
•Area moment of inertia from Fig. A.1:
118
3. Torque due to weight:
4. Torque due hydrostatic pressure:
•Moment arm:
119
3.81: PROBLEM DEFINITION
Situation:
A submerged gate is described in the problem statement.
Find:
Hydrostatic force (F)on gate.
Ratio (RT/F )of the reaction force to the hydrostatic force.
SOLUTION
F=¯pA
120