Pertemuan 05 - 07
Hydrostatics 2
Outline
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Pressure Forces on Plane Surface
Pressure Forces on Curved Surface
Pressure on Spillway Sections of Dam
Stability of Dam
Buoyancy
Pressure Forces on Plane Surface
http://www.ce.utexas.edu/prof/kinnas/319LAB/Applets/PresPlane4.html
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Hydrostatics Force on Immersed Surfaces
 Immersed Surfaces are Subject to hydrostatics pressure
 May generally be horizontal, vertical or inclined
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Horizontal Surface
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The total weight of the
fluid above the surface
is equal to the volume
of the liquid above the
surface multiplied by
the specific weight of
the liquid
h
• F = gA h
 g: specific weight of
the liquid
 A : the area of the
surface
 F : the force acting on
the immersed surface
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A horizontal surface immersed in a liquid
Hydrostatic Force on a Plane Surface
The Center of Pressure YR lies below the centroid - since pressure
increases with depth
FR = g A YC sinq
or FR
= g A Hc
YR = (Ixc / YcA) + Yc
XR = (Ixyc / YcA) + Xc
but for a rectangle or circle:
XR = Xc
For 90 degree walls:
FR = g A Hc
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Hydrostatics Example Problem # 1
What is the Magnitude and Location of the
Resultant force of water on the door?
gW = 62.4 lbs/ft3
Water Depth = 6 feet
Door Height = 4 feet
Door Width = 3 feet
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Hydrostatics Example Problem #1
Important variables:
HC and Yc = 4’
Xc = 1.5’
A = 4’ x 3’ = 12’
Ixc = (1/12)bh3
Magnitude of Resultant
Force:
FR = gW A HC
FR = 62.4 x 12 x 4 = 2995.2 lbs
= (1/12)x3x43 = 16 ft4
Location of Force:
YR = (Ixc / YcA) + Yc
YR = (16 / 4x12) + 4 = 4.333 ft down
XR = Xc (symmetry) = 1.5 ft from the
corner of the door
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Buoyancy
Archimedes Principle: Will it Float?
The upward vertical force felt by a submerged, or partially submerged,
body is known as the buoyancy force. It is equal to the weight of the
fluid displaced by the submerged portion of the body. The buoyancy
force acts through the centroid of the displaced volume, known as
the center of buoyancy. A body will sink until the buoyancy force is
equal to the weight of the body.
FB = g x Vdisplaced
FB = gW x Vdisp
= Vdisp
FB
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W = FB
FB
Buoyancy Example Problem # 1
A 500 lb buoy, with a 2 ft radius is tethered to the bed of
a lake. What is the tensile force T in the cable?
FB
gW = 62.4
lbs/ft3
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Buoyancy Example Problem # 1
Buoyancy Force:
Displaced Volume of Water:
FB = gW x Vdisp-w
Vdisp-W = 4/3 x p x R3
FB = 62.4 x 33.51
Vdisp-W = 33.51 ft3
FB = 2091.024 lbs up
Sum of the Forces:
SFy = 0 = 500 - 2091.024 + T
T = 1591.024 lbs down
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Will It Float?
Ship Specifications:
Weight = 300 million pounds
Dimensions = 100’ wide by 150’ tall by 800’ long
Given Information: gW = 62.4 lbs/ft3
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Assume Full Submersion:
FB = Vol x gW
FB = (100’ x 150’ x 800’) x 62.4 lbs/ft3
FB = 748,800,000 lbs
Weight of Boat = 300,000,000 lbs
The Force of Buoyancy is greater than the Weight of the Boat
meaning the Boat will float!
How much of the boat will be submerged?
Assume weight = Displaced Volume
WB = F B
300,000,000 = (100’ x H’ x 800’) x 62.4 lbs/ft3
H = Submersion depth = 60.1 feet
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