HIDROSTATIKA

Fluids:Statics vs Dynamics

For a fluid at rest, pressure increases linearly with depth. As a consequence, large forces can develop on plane and curved surfaces. The water behind the Hoover dam, on the Colorado river, is approximately 715 feet deep and at this depth the pressure is 310 psi. To withstand the large pressure forces on the face of the dam, its thickness varies from 45 feet at the top to 660 feet at the base.

HydrostaticsHydrostaticsAnd the Hoover Dam

Macam Tekanan Hidrostatik

• Tekanan pada bidang datar• Tekanan pada bidang datar horisontal• Tekanan pada bidang miring• Tekanan pada bidang lengkung

• Tekanan yang diterima oleh suatu bidang arahnya selalu tegak lurus dengan bidang tersebut

bidang datar

bidang datarhorisontal

• Tekanan yang diterima oleh suatu bidang arahnya selalu tegak lurus dengan bidang tersebut

bidang datar

bidang miring

• Tekanan yang diterima oleh suatu bidang arahnya selalu tegak lurus dengan bidang tersebut

bidang datar

bidang lengkung

Tekanan pada bidang datar

Tekanan = Gaya per Unit Area

Atmospheric Pressure is the weight of the column of air above a unit area. For example, the atmospheric pressure felt by a man is the weight of the column of air above his body divided by the area the air is resting on

P = (Weight of column)/(Area of base)

Standard Atmospheric Pressure:

1 atmosphere (atm)

14.7 lbs/in2 (psi) 760 Torr (mm Hg) 1013.25 millibars = 101.3 kPascals

1kPa = 1N/m2

Konsep dasar

• Tekanan = Gaya per Unit Area• Gaya adalah berat liquid di atas titik tinjauan• Berat air = massa liquid dikalikan dengan g

(percepatan grafitasi)• Massa liquid = volume liquid x densitas liquid• Volume liquid = kedalaman titik x unit area• Jadi: hg

A

Ahg**

***(P)Tekanan

Fluid StaticsFluid Statics

Basic Principles:

Fluid is at rest : no shear forces

Pressure is the only force acting

What are the forces acting on the block?

Air pressure on the surface - neglect

Weight of the water above the block

Pressure only a function of depth

• Gambar berikut adalah empat bak berisi air. Bentuk bak dan volume air berbeda, namun luas dasar empat bak adalah SAMA. Apakah tekanan hidrostatis yang diterima oleh dasar bak berbeda?

TEKANAN PADA BIDANG DATAR HORISONTAL

Basic Concepts

Pressure = h = spec gravity of water

h = depth of water

C = Center of Mass of Gate

CP = Center of Pressure on Gate

Fr = Resultant Force acts at CP

γh

Penentuan Center of Pressure

Gaya hidrostatik pada bendungan

Tekanan pada tiap kedalaman

Di permukaan, P = 0Di dasar, P = ρ g H Pada h=H/2, P=ρ g H/2 FH=W= ρ g H/2 *bH

Momen=M=W.ee=H/2 – H/3 = H/6M=(1/12) ρ g bH3

C = Centroid or Center of Mass

CP = Center of Pressure

Fr = Resultant Force

I = Moment of Inertia

γh

For a Rectangular Gate:

Ixc = 1/12 bh3

Ixyc = 0

For a circle:

Ixc = r4 / 4

Ixyc = 0

Tekanan Hidrostatik pada bidang dinding

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

Hydrostatics Example Problem #1

Magnitude 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

TEKANAN PADA BIDANG MIRING

The 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

Tekanan Hidrostatik pada bidang miring

Dengan pendekatan rumus Phytagoras:

• WV = (1/2) ρ g HB B = L cos α

• WH = (1/2) ρ g H2

• P = W/(L * Lh)

22Hv WWW

W

α L

B

TEKANAN PADA BIDANG LENGKUNG

Komponen vertikal:

• FV=berat liquid di atas bidang lengkung = WAA’B’B = γVAA’B’B = γ (AAA’B’B . b)

Komponen horisontal:

• FH=gaya pada bidang hasil proyeksi bidang lengkung = γ.hOB.AOB

A

B

B’ A’

Fv

FH

F

O

22HV FFF

KESEIMBANGAN TEKANAN

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

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?

W = 62.4 lbs/ft3

FB

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

Will It Float?Ship Specifications:

Weight = 300 million pounds

Dimensions = 100’ wide by 150’ tall by 800’ long

Given Information: W = 62.4 lbs/ft3

LATIHAN SOAL-SOAL HIDROSTATIKA

ATTENTION

Pada Hari Kamis 7 Oktober 2010 dilaksanakan Quis 1.

Harap dipelajari materi kuliah pada pertemuan 1 – 6

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