Pertemuan 05 - 07Hydrostatics 2

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Outline• Pressure Forces on Plane Surface• Pressure Forces on Curved Surface• Pressure on Spillway Sections of Dam• Stability of Dam• Buoyancy

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Pressure Forces on Plane Surfacehttp://www.ce.utexas.edu/prof/kinnas/319LAB/Applets/PresPla

ne4.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• 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

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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

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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

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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

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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

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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

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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

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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