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PREPARED BY, ENROLLMENT NO :- 150490106033 150490106034 150490106035 150490106036 150490106037 TOPIC - STATICS FORCES ON SURFACE SUBJECT :- FLUID MECHANICS SNPIT&RC,UMRAKH

# Fm hydrostatics

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PREPARED BY,ENROLLMENT NO :-

150490106033150490106034150490106035150490106036150490106037

TOPIC - STATICS FORCES ON SURFACE

SUBJECT :- FLUID MECHANICS

SNPIT&RC,UMRAKH

FLUID STATICS

Deals with problems associated with fluid at rest.

No relative motion between adjacent layers. Thus, no shear stress (tangential stress) to deform the fluid.

Liquid – hydrostatic; Gas – aerostatic.

FLUID STATICS

The only stress in fluid statics is normal stress (perpendicular to surface)

Normal stress is due to pressure (Pressure: gravity field-weight of fluid)

Variation of pressure is due only to the weight of the fluid → fluid statics is only relevant in presence of gravity fields.

HYDROSTATIC FORCES

F=PAOnly when the pressure distribution is uniform over the entire area of interest.

HYDROSTATIC FORCES ON PLANE SURFACES

Problem To find the resultant force FR (magnitude) and its

point of application (center of pressure) for non-uniformly distributed pressure.

**Atmospheric pressure Patm can be neglected when it acts on both sides of the surface.

WHAT IS A RESULTANT FORCE AND CENTRE OF PRESSURE ?

Engineering Mechanics

Resultant force: A force that results from

the combination of two or more forces.

COP:A point where the entire

forces (Resultant force) were concentrated at a single point.

HOW TO DETERMINE THE RESULTANT FORCE

Consider a flat plate completely submerged in a liquid.

Plane of the top surface intersects with horizontal surface with an angle Ө, the line of intersection will be the x axis.

HOW TO DETERMINE THE RESULTANT FORCE, FR

Absolute pressure at any point of the fluid,P = Po + ρgh, h = y sin Ө = Po + ρgy sin Ө,FR = ∫PdA =∫ (Po + ρgy sin Ө)dA = PoA + ρg sin Ө ∫ydA**∫ydA is the first moment of area is related to the y coordinate of the centroid (or centre) of the surface by , yc= 1/A ∫ydA.

FR = (Po + ρgyc sin Ө) A = (Po + ρghc) A

= PcAThe magnitude of the resultant force, FR acting on a plane surface of a completely submerged plate in a homogeneous (constant density) fluid is = the product of the pressure Pc at the centroid of the surface and the area A of the surface.

CENTROID AND CENTROIDAL MOMENTS OF INERTIA FOR SOME COMMON GEOMETRIES

SUBMERGED RECTANGULAR PLATE : HOW TO DETERMINE THE RESULTANCE FORCE, FR.

HOW TO DETERMINE THE LOCATION OF THE COP?

Line of action of resultant force FR=PCA does not pass through the centroid of the surface. In general, it lies below the centroid because pressure increase with depth.

HOW TO DETERMINE THE LOCATION OF THE RESULTANT FORCE (CENTER OF PRESSURE) ?

The vertical location of the line of action is determined by equating the moment of the resultant force, FR to the moment of the distributed pressure force about the x-axis.

Then, ypFR = ∫yPdA =∫ y(Po + ρgy sin Ө)dA

= Po ∫ ydA + ρgsin Ө ∫y2dA= PoycA + ρgsin Ө Ixx,o

Moment of the resultance force = Moment of the distributed pressure force about the x axis.

xx,oco

A

2

Ao

Ao

ARp

ρgsinθIAyP

dAyρgsinθydAP

ρgysinθ)dAy(P

yPdAFy

Ixx,o = ∫y2dA is actually the second moment of area about the x axis passing through point O.

Normally, the second moment of area is given about the axes passing through the centroid of the area. Therefore, we need parallel axis theorem to relate the Ixx,o and Ixx,c

AyII ccxxoxx2

,,

AgPyI

yyoc

cxxcp )sin/(

,

Therefore,

If Po = 0

AyI

yyc

cxxcp

,

SUBMERGED RECTANGULAR PLATE : HOW TO DETERMINE THE C.O.P.

AgPyI

yyoc

cxxcp )sin/(

,

3

, 12;;

2abIabAbsy cxxc

abgPbs

abbsyo

p

)sin/(

2

122

3

IF Ө=90

abbs

abbsy p

2

122

3

Example 11-1; Calculate FR, C.O.P and discuss whether the driver can open the door or not.

APF CR

]m1*m2.1A[;

101.3kNAPF cR

2cc

84.4kN/m

b/2)ρg(sρghP

my

abbs

abbsy

p

p

61.82

122

3

The moment acting on the mid point of the submerged door is 101.3kN x 0.5m = 50.6kNm, which is 50 times of the moment the driver can possibly generate. CAN’T OPEN.

HYDROSTATIC FORCES ON CURVED SURFACES

Complicated: FR on a curved surface requires integration of the pressure forces that change direction along the surface.

Easiest approach: Determine horizontal and vertical components FH and FV separately.

HYDROSTATIC FORCES ON CURVED SURFACES

1. Vertical surface of the liquid block, BC = projection of the curved surface on a vertical plane (vertical projection).

2. Horizontal surface of the liquid block, AB = projection of the curved surface on a horizontal plane (horizontal projection).

3. Newton’s 3rd Law – Action and reaction. The resultant force acting on the curve liquid surface = the force on curved solid surface.

• Assume that the direction to the right and up is positive,Horizontal force component on curved surface: FH – FX = 0. The horizontal component of the hydrostatic force (FH) acting on a curved surface is (both magnitude and line of action) equal to the hydrostatic force (FX) acting on the vertical projection of the curved surface.

• Vertical force component on curved surface: FV - FY – W = 0, where W is the weight of the liquid in the enclosed block W=gV.The vertical component of the hydrostatic force (FV) acting on a curved surface is (both magnitude and line of action) equal to the hydrostatic force (FY) acting on the horizontal projection of the curved surface, plus or minus (depend on direction) the weight of the fluid block.

HYDROSTATIC FORCES ON CURVED SURFACES

Magnitude of force FR=(FH2+FV

2)1/2

Angle of force is a = tan-1(FV/FH)

A GRAVITY CONTROLLED CYLINDRICAL GATE

Example 11-2 A long solid cylinder, R=0.8, hinged at point A.When water level at 5m, gate open.Determine: 1. The hydrostatic force acting on the cylinder and its line of action when the gates open

2. The weight of cylinder per m length of the cylinder

Friction at hinge is negligible The other side of the gate is

exposed to the atmosphere, therefore, the Patm is cancelled out.

A) Determine the net horizontal and vertical force, FH and FV respectively.

FH = FX = PCA = ρghcA; hc = 4.2 + 0.8/2

FY = ρghcA; hc = 5m ; W=ρg(R2 – (πR2/4))

FY - FV - W = 0;

Solve for resultant force and angle.

B) Think about the moment acting at point A due to the cylinder weight and also resultant force

THANK YOU