HYDROSTATIC FORCE ON PLANE SURFACES I need help analyzing my experiment. For a Fully Submerged surface: - what is the comparison between the results of W x r with the calculated values of F x (xcp + hs)? What is the reason for their discrepancies? What is the theories behind these results?
HYDROSTATIC FORCE ON PLANE SURFACES I need help analyzing my experiment. For a Fully Submerged surface: - what is the comparison between the results of W x r with the calculated values of F x (xcp + hs)? What is the reason for their discrepancies? What is the theories behind these results?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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HYDROSTATIC FORCE ON PLANE SURFACES
I need help analyzing my experiment.
For a Fully Submerged surface:
- what is the comparison between the results of W x r with the calculated values of F x (xcp + hs)? What is the reason for their discrepancies? What is the theories behind these results?
Transcribed Image Text:Trial No.
1 Base trial
1 partially submerged
2 partially submerged
3 partially submerged
1 fully submerged
2 fully submerged
3 fully submerged
d (mm) h (mm) hs = 200- h-d (F (N)
69
84
54
62
101
121
148
0
0
0
0
1
21
48
xcp (mm)
131 174.3578657
116 258.4056082
146 106.7900728
138 140.7753908 41.33333333
98 380.9798103 710973.6667
58 722.2978755 2067793.667
4 1322.497636 5748010.667
46
56
36
F × (xcp + hs)(N- mm) W (N)
30861.34223
44445.7646
19435.79324
25245.72009
270903948.7
1493604866
7601735810
50
100
10
30
170
250
360
Wxr (N mm)
13970
27940
2794
8382
47498
69850
100584
Temperature (C)
Density (g/mL)
Temperature Coefficient
Width (m)
gravity
r (mm)
27
0.99652
0.99831
0.0075
9.8
279.4
![for a fully submerged surface:
F = = √yb[(h+d)² + h²]
2
Xep - (3h² +3hd+d²) (2h+d)
=
3
for a partially submerged surface:
here:
F
Xcp
Y
h
d
do
where:
W
r
F
Xcp
hs
F-Vbd²
=
Taking moments about the fulcrum,
Xcp
=
2
3d
total force on the end face of the quadrant, [kN]
depth of the center of pressure, [m or mm]
unit weight of the fluid, [kN/m²]
height of the submerged portion of the end face of the quadrant, [m or mm]
d = do, if the end face of the quadrant is fully submerged
height of the end face of the quadrant, [m or mm]
eq. 2
vertical height from the water surface to the top of the quadrant plane surface, [m or mm
h = 0, if the end face of the quadrant is partially submerged only
Wr = F(xcp +hs)
weight applied at the end of the lever arm, [kN]
perpendicular distance from the hanger to the fulcrum, [m or mm]
total force on the end face of the quadrant, [kN]
depth of the center of pressure, [m or mm]
vertical height from the water surface to the fulcrum, [m or mm]
eq. 3
eq. 4
eq. 5
eq. 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc91beade-9729-4bed-93b5-a690a61abffc%2F3442f787-b408-43ce-8c49-338e64577092%2Fhfsz12q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:for a fully submerged surface:
F = = √yb[(h+d)² + h²]
2
Xep - (3h² +3hd+d²) (2h+d)
=
3
for a partially submerged surface:
here:
F
Xcp
Y
h
d
do
where:
W
r
F
Xcp
hs
F-Vbd²
=
Taking moments about the fulcrum,
Xcp
=
2
3d
total force on the end face of the quadrant, [kN]
depth of the center of pressure, [m or mm]
unit weight of the fluid, [kN/m²]
height of the submerged portion of the end face of the quadrant, [m or mm]
d = do, if the end face of the quadrant is fully submerged
height of the end face of the quadrant, [m or mm]
eq. 2
vertical height from the water surface to the top of the quadrant plane surface, [m or mm
h = 0, if the end face of the quadrant is partially submerged only
Wr = F(xcp +hs)
weight applied at the end of the lever arm, [kN]
perpendicular distance from the hanger to the fulcrum, [m or mm]
total force on the end face of the quadrant, [kN]
depth of the center of pressure, [m or mm]
vertical height from the water surface to the fulcrum, [m or mm]
eq. 3
eq. 4
eq. 5
eq. 6
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