Ch 3 Fluid Statics (Part B)

From Last Lecture

• Pressure is independent of direction• Pressure increases with depth

--constant density p = ρgh• Pressure scales

--units--datum

• Pressure measurement

Free surface

P = P0 +ρgh

Basic equation of fluid statics

Characters of pressure at a point

Pressure at any point is the same in all directions.

•absolute pressure - Pressure measured relative to absolute zero.• gage pressure - Pressure > Patm measured relative to Patm•vacuum pressure - Pressure < Patm measured relative to Patm•Patm - local absolute pressure due to the local atmosphere only.

standard Patm at sea level = 1atm = 101.3 kPa

Absolute, gage, and vacuum pressures

Outline

Hydrostatic Force on a Plane SurfaceHydrostatic Force on a Curved SurfaceBuoyancy and Stability

Lecture GoalsLecture Goals

1.force due to pressure of the fluid;

2. the direction of the resultant force; and

3.the position of the resultant.

Analysis and calculate

3.6 Force on a Plane Area

• Specifying the magnitude of the force.

• Specifying the direction of the force.

• Specifying the line of action of the force.

• To determine completely the resultant force acting on a submerged force.

3.6.1 Force on a Horizontal Plane Area

Simplest Case: Tank bottom with a uniform pressure distribution

hp γ=

Now, the resultant Force:

RF = pA

Acts through the Centroid

A = area of the Tank Bottom

3.6.2 Force on a Plane Area:General Case

General Shape: Planar View, in the x-y plane

θ is the angle the plane makeswith the free surface.

y is directed along the plane surface.

The origin O is at the Free Surface.

A is the area of the surface.

dA is a differential element of the surface.

dF is the force acting on the differential element.

C is the centroid.

CP is the center of Pressure

FR is the resultant force acting through CP

• Notation for hydrostatic force on an inclined plane surface of arbitrary shape.

3.6.2 Hydrostatic Force on a Plane Surface: General Case

Then the force acting on the differential element:

Then the resultant force acting on the entire surface:

With γ and θ taken as constant:

We note, the integral part is the first moment of area about the x-axis

Where yc is the y coordinate to the centroid of the object.

We note h = ysinθ

hc

Resultant Force

To obtain the net hydrostatic force F on a plane surface:

1. Determine depth of centroid hc for the area in contact with the fluid.2. Determine the (gage) pressure at the centroid Pc.3. Calculate F = PcA

1. Determine depth of centroid hc for the area in contact with the fluid.2. Determine the (gage) pressure at the centroid Pc.3. Calculate F = PcA

hc is the vertical distance from the fluid surface to the centroid of the area.

3.7 Center of Pressure: Location

Now, we must find the location of the center of Pressure where the Resultant Force Acts: “The Moments of the Resultant Force must Equal the Moment of the Distributed Pressure Force”

We note,

Moments about the x-axis:

Then,

Second moment of Intertia, Ix

Parallel Axis Thereom:

Ixc is the second moment of inertia through the centroid

Substituting the parallel Axis thereom, and rearranging:

We, note that for a submerged plane, the resultant force always acts below the centroid of the plane.

And, note h = ysinθ

Center of Pressure

Hydrostatic Force on a Plane Area: Geometric Properties

Centroid Coordinates

Areas

Moments of Inertia

CentroidsCentroids

Centroidal coordinates and momentsof inertia for some common areas

Hydrostatic Force: Vertical Wall

Find the Pressure on a Vertical Wall using Hydrostatic Force Method

Pressure varies linearly with depth by the hydrostatic equation:The magnitude of pressure at the bottom is p = γh

The width of the wall is “b” into the board

The depth of the fluid is “h” into the board

By inspection, the average pressure occurs at h/2, pav = γh/2

The resultant force act through the center of pressure, CP:

( )

hhhy

h

bhhbhy

R

R

32

26

22

12

3

=+=

+=

O

yR = 2/3h

y-coordinate: 3

121 bhI xc =

2hyc =

bhA =

Hydrostatic Force: Vertical Wall

x-coordinate:

( )

2

22

0

bx

b

bhhx

R

R

=

+=0=xycI

2byc =

bhA =

Center of Pressure:

⎟⎠⎞

⎜⎝⎛

32,

2hb

Now, we have both the resultant force and its location.

Summary

To find hydrostatic force on a plane surface:

1. Find area in contact with fluid.

2. Locate centroid of that area.

3. Find hydrostatic pressure Pc at centroid,

typically = (generally neglect Patm ).

4.Find force F = Pc A.

5.The location will not be at the centroid., but at a distance below the centroid.

1. Find area in contact with fluid.

2. Locate centroid of that area.

3. Find hydrostatic pressure Pc at centroid,

typically = (generally neglect Patm ).

4.Find force F = Pc A.

5.The location will not be at the centroid., but at a distance below the centroid.

chgρ

Example: Hydrostatic Force Acting on the Door of a Submerged Car

A heavy car plunges into a lake during an accident and lands at the bottom of the lake on its wheels. The door is 1.2 m high and 1 m wide,and the top edge of the door is 8 m below the free surface of the water.Determine the hydrostatic force on the door and the location of the pressure center, and discuss if the driver can open the door.

Pave = PC = ρghC = ρg(s + b/2)= 84.4 kN/m2

FR = PaveA = (84.4 kNm2) (1 m × 1.2 m) = 101.3 kN

yP = 8.61 m

Discussion A strong person can lift 100 kg, whose weight is 981 N or about 1 kN. Also, the person can apply the force at a point farthest from the hinges (1 m farther) for maximum effect and generate a moment of 1 kN · m. The resultant hydrostatic force acts under the midpoint of the door, and thus a distance of 0.5 m from the hinges. This generates a moment of 50.6 kN · m, which is about 50 times the moment the driver can possibly generate. Therefore, it is impossible for the driver to open the door of the car. The driver’s best bet is to let some water in (by rolling the window down a little, for example) and to keep his or her head close to the ceiling. The driver should be able to open the door shortly before the car is filled with water since at that point the pressures on both sides of the door are nearly the same and opening the door in water is almost as easy as opening it in air.

Pressure Diagram

Pressure Diagram

(1) According to a definite scale, static pressure can be expressed by a definite length line ；

( 2) The line expressing the direction of static pressure with arrow, is perpendicular to the acting wall.

The pressure diagram is a second way of analyzing the forces on a vertical wall.

Pressure diagram

Which of the following is wrong？A. B. C.

Pressure Diagram: Vertical Wall

bhFR2

21 γ=

Resultant Force:

Location of the Resultant Force, CP:

Center of Pressure:

⎟⎠⎞

⎜⎝⎛

32,

2hb

O

The force acts through the centroid of the pressure diagram

The area of this triangle represents the resultant force per unit width on the vertical wall

Atmospheric Pressure on a Vertical Wall

Gauge Pressure Analysis Absolute Pressure Analysis But,

So, in this case the resultant force is the same as the gag pressure analysis.It is not the case, if the container is closed with a vapor pressure above it.If the plane is submerged, there are multiple possibilities.

Le31

=

)(3)2(

21

21

hhhhLe

++

=

Example

A 6-m deep tank contains 4 m of water and 2-m of oil as shown in the diagram below. Determine the pressure at point A and at the bottom of the tank. Draw the pressure diagram.

Aoil

water

2 m

4 m

ρwater = 1000 kg/m3

ρ oil = 980 kg/m3

Pressure at oil water interface (PA) PA= ρoilghoil = 980 x 9.81 x 2

= 15696 Pa=15.7KPaPressure at the bottom of the tank;

PB = PA + ρwaterghwater=15.7x1000 + 1000 x 9.81 x 4 = 54940 Pa=54.9KPa

4 m

2 m

PA

PA=15.7 kPa

B

Aoil

water

PB = 54.9 kPaPressure Diagram

Solution:

Here is a vertical rectangular sluice gate as shown in figure,

h1=1m，h2=2m, width b=1.5m. What is the resultant force and the center of pressure?

Example

Here is a vertical rectangular sluice gate as shown in figure, h1=1m，h2=2m,width b=1.5m. What is the resultant force and the center of pressure?

Solution：

3.8 Force on a Curved Surface

Since this class of surface is curved, the direction of the force is different at each location on the surface.

Many surfaces of interest (such as those associated with dams, pipes, and tanks) are non-plannar.

Gates

Gates

Force on a Curved Surface

• FR on a curved surface is more involved since it requires integration of the pressure forces that change direction along the surface.

• Easiest approach: determine horizontal and vertical components FH and FVseparately.

3.8 Hydrostatic Force on a Curved Surface

• General theory of plane surfaces does not apply to curved surfaces• Many surfaces in dams, pumps, pipes or tanks are curved• No simple formulas by integration similar to those for plane surfaces• A new method must be used

Isolated VolumeBounded by AB an AC and BC

Then we mark a F.B.D. for the volume:

F1 and F2 is the hydrostatic force on each planar face

FH and FV is the component of the resultant force on the curved surface.

W is the weight of the fluid volume.

Hydrostatic Force on a Curved Surface

Now, balancing the forces for the Equilibrium condition:Horizontal Force:

Vertical Force:

Resultant Force:

The location of the Resultant Force is through O by sum of Moments:

HH

VVc

xFxFxFWxxF

==+

22

11Y-axis:

X-axis:

Horizontal Component (Fh)

Fh = Resultant force on the projection of the curved surface on a vertical plane

The horizontal force will act through the c.p. (not the centroid) of the projected area.

to determine the horizontal component of force on a curved surface in a hydrostatic fluid:

1. Project the curved surface into the appropriate vertical plane.

2. Perform all further calculations on the vertical plane.

3. Determine the location of the centroid - c.g. of the vertical plane.

4. Determine the depth of the centroid - hcg of the vertical plane.

5. Determine the pressure - Pcg = r g hcg at the centroid of the vertical plane.

6. Calculate Fh = Pcg A, where A is the area of the projection of the curved surface into the vertical plane, ie. the area of the vertical plane.

7. The location of Fh is through the center of pressure of the vertical plane, not the centroid.

1. Project the curved surface into the appropriate vertical plane.

2. Perform all further calculations on the vertical plane.

3. Determine the location of the centroid - c.g. of the vertical plane.

4. Determine the depth of the centroid - hcg of the vertical plane.

5. Determine the pressure - Pcg = r g hcg at the centroid of the vertical plane.

6. Calculate Fh = Pcg A, where A is the area of the projection of the curved surface into the vertical plane, ie. the area of the vertical plane.

7. The location of Fh is through the center of pressure of the vertical plane, not the centroid.

Vertical Component

The vertical component of force on a curved surface equals the weight of the pressure prism necessary to cause the pressure on the surface.

pressure prism

pressure prism

Pressure Prism

Components of Pressure Prism：

（1）Pressed curved surface；（2）Vertical surface drawn along the curved surface edge；（3）Free surface or extended surface of the free surface.

Real pressure prism virtual pressure prism

Classification of pressure prism

Real pressure prism and virtual pressure prism.

For real pressure prism Fh acts downwards.

For unreal pressure prism Fh acts upwards.

Example

To determine the vertical component of force on a curved surface in a hydrostatic fluid:

1 .Identify the Pressure Prism.

2.Determine the volume of the pressure prism.

3.Calculate the weight of the pressure prism

Fv = ρgV.

4.The location of Fv is through the centroid of V.

22VH FFF +=

H

V

FF

=αtan

H

V

FFarctg=α

Resultant force

The angle the resultant force makes to the horizontal is

Static Surface Forces SummaryStatic Surface Forces Summary

• Forces caused by gravity on submerged surfaces• horizontal surfaces• inclined surfaces (y coordinate has origin at

free surface)

• curved surfaces• Horizontal component • Vertical component(weight of fluid above surface )

RF hAg=

AhF cR γ= cc

xcR y

AyIy +=

AhF cR γ=

water= (3 m)(2 m)(1 m)γ +1/4π(2 m)2(1 m)γ

Example: Forces on Curved Surfaces

Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc.

FV =

FH = cp A

2 m

2 m

3 m W1

W2

W1 + W2

= 58.9 kN + 30.8 kN= 89.7 kN

= γ(4 m)(2 m)(1 m)= 78.5 kN y

cc hp xγ=

= 0.948 m (measured from A) with magnitude of 89.7 kN

Take moments about a vertical axis through A.

Example: Forces on Curved Surfaces

The vertical component line of action goes through the centroid of the volume of water above the surface.

water 2 m

2 m

3 m

A

W1

W2

21 3)m 2(4)m 1( WWFx VR π

+=

)kN 7.89(

)kN 8.30(3

)m 2(4)kN 9.58)(m 1(π

+=Rx

π34R

Example: Forces on Curved Surfaces

water 2 m

2 m

3 m

A

W1

W2

The location of the line of action of the horizontal component is given by

xcR c

c

Iy yy A

= +

3

12xcbaI =

b

a

xcI =xcI =

cy =cy =

( ) ( )( )[ ]( )

40.667 m 4 m 4.083 m4 m 2 m 1 mRy = + =

y

(1 m)(2 m)3/12 = 0.667 m4

4 m x

Example: Forces on Curved Surfaces

78.5 kN

89.7 kN4.083 m

0.94

8 m

119.2 kN

horizontal

vertical

resultant

Example

A C

B B

x

y 6 mHinge

Fy

Fx

kN

AyF CBcx

6.1761*6*3*9810

===γ

m

yy cR

41*6*312/6*1 3

=

+=

kN

VF ABCy

4.277

1*46*9810

2

=

=

=

π

γ

m

rx

55.2*3

6*434

==

=

π

π

An 1-m-diameter water pipe is made of a piece of steel plate. Its permission stress [σ] is 150MPa and the pressure expressed in the height of water column in pipe reaches 500m，what is the least thickness of the pipe? （Pressure difference between different points caused by elevation difference is negligible）

Example

Solution:

Assume cutting the pipe into two half as shown in figure, and analyze forces exerted on a half pipe as an isolated body.Horizontal force on inner surface of the half pipe is equal to the force on the projection upon a vertical plane .

The force is equal to the pulling forces,

pDlFT ==2Assume T distributes uniformly along the pipe thickness, thus：

Multiple Choice

Multiple Choice

A B C D

3.9 Buoyancy and Stability

• Buoyancy is due to the fluid displaced by a body. FB=ρfgV.

• Archimedes principal : The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.

3.9 Buoyancy and Stability

• Buoyancy force FB is equal only to the displaced volume ρfgVdisplaced.

• Three scenarios possible1. ρbody<ρfluid: Floating body2. ρbody=ρfluid: Neutrally buoyant3. ρbody>ρfluid: Sinking body

Stability of Immersed Bodies

• Rotational stability of immersed bodies depends upon relative location of center of gravity G( 重心)and center of buoyancy B(浮心).• G below B: stable• G above B: unstable • G coincides with B: neutrally stable.

Stability of Floating Bodies

• If body is bottom heavy (G lower than B), it is always stable.

• Floating bodies can be stable when G is higher than B due to shift in location of center buoyancy and creation of restoring moment.

• Measure of stability is the metacentric height (稳定高度)GM. If GM>1, ship is stable.

•Examples of Archimedes Principle

The Golden Crown of Hiero II, King of Syracuse

• Archimedes, 287-212 B.C.• Hiero, 306-215 B.C.• Hiero learned of a rumor where the

goldsmith replaced some of the gold in his crown with silver. Hiero asked Archimedes to determine whether the crown was pure gold.

• Archimedes had to develop a nondestructive testing method

The Golden Crown of Hiero II, King of Syracuse

• The weight of the crown and nugget are the same in air: Wc = ρcVc = Wn = ρnVn.

• If the crown is pure gold, ρc=ρnwhich means that the volumes must be the same, Vc=Vn.

• In water, the buoyancy force is B=ρH2OV.

• If the scale becomes unbalanced, this implies that the Vc ≠ Vn, which in turn means that the ρc ≠ρn

• Goldsmith was shown to be a fraud!