pertemuan 05 - 07 hydrostatics 2. bina nusantara outline pressure forces on plane surface pressure...
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Bina Nusantara
Outline• Pressure Forces on Plane Surface• Pressure Forces on Curved Surface• Pressure on Spillway Sections of Dam• Stability of Dam• Buoyancy
Bina Nusantara
Pressure Forces on Plane Surfacehttp://www.ce.utexas.edu/prof/kinnas/319LAB/Applets/PresPla
ne4.html
Bina Nusantara
Hydrostatics Force on Immersed Surfaces Immersed Surfaces are Subject to hydrostatics
pressure May generally be horizontal, vertical or inclined
Bina Nusantara
Horizontal Surface• 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
• F = gA h
g: specific weight of the liquid
A : the area of the surface
F : the force acting on the immersed surface
h
A horizontal surface immersed in a liquid
Bina Nusantara
Hydrostatic Force on a Plane SurfaceThe Center of Pressure YR lies below the centroid - since pressure
increases with depth
FR = A YC sin
or FR = A Hc
YR = (Ixc / YcA) + Yc
XR = (Ixyc / YcA) + Xc
but for a rectangle or circle: XR = Xc
For 90 degree walls:
FR = A Hc
Bina Nusantara
Hydrostatics Example Problem # 1
What is the Magnitude and Location of the
Resultant force of water on the door?
W = 62.4 lbs/ft3
Water Depth = 6 feet
Door Height = 4 feet
Door Width = 3 feet
Bina Nusantara
Hydrostatics Example Problem #1Magnitude of Resultant
Force:
FR = W A HC
FR = 62.4 x 12 x 4 = 2995.2 lbs
Important variables:
HC and Yc = 4’
Xc = 1.5’
A = 4’ x 3’ = 12’
Ixc = (1/12)bh3
= (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
Bina Nusantara
BuoyancyArchimedes 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 = x Vdisplaced
= Vdisp
FB
FB
W = FB
FB = W x Vdisp
Bina Nusantara
Buoyancy Example Problem # 1A 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?
W = 62.4 lbs/ft3
FB
Bina Nusantara
Buoyancy Example Problem # 1
Displaced Volume of Water:
Vdisp-W = 4/3 x x R3
Vdisp-W = 33.51 ft3
Buoyancy Force:
FB = W x Vdisp-w
FB = 62.4 x 33.51
FB = 2091.024 lbs up
Sum of the Forces:
Fy = 0 = 500 - 2091.024 + T
T = 1591.024 lbs down
Bina Nusantara
Will It Float?Ship Specifications:
Weight = 300 million pounds
Dimensions = 100’ wide by 150’ tall by 800’ longGiven Information: W = 62.4 lbs/ft3
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Assume Full Submersion:
FB = Vol x W 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 = FB
300,000,000 = (100’ x H’ x 800’) x 62.4 lbs/ft3
H = Submersion depth = 60.1 feet