chapter 2 hydro static forces

8/2/2019 Chapter 2 Hydro Static Forces

1/28

CHAPTER 2 : FLUI D STATI CS

Hydrostati c Forces

The design of many engineering systems such as water dams andliquid storage tanks requires the determination of the forces actingon the surfaces using fluid statics. The complete description of theresultant hydrostatic force acting on a submerged surface requiresthe determination of its magnitude, its point of application or itsdirection, and the line of action of the force.


2/28

2.7 Hydrostati c Forces On Subm erged Plane Surfaces

When a surface such as a gate valve in a dam, the wall of a liquidstorage tank, or the hull of a ship is submerged in a fluid, forces

develope on the surface due to the fluid pressure. For fluids at restwe know that the force must be perpendicular to the surface sincethere are no shearing stresses present. We also know that thepressure will vary linearly with depth if the fluid is incompressible.Fig. 2.18 below shows the hydrostatics pressure distribution on thesubmerged tilted, vertical and horizontal plates.

Fig. 2.18 : Hydrostatic force distribution on the submerged plates.

In most cases, the other side of the plate is open to the atmosphere(such as the dry side of a gate), and thus atmospheric pressure actson both sides of the plate, yielding a zero resultant. In such cases, itis convenient to subtract atmospheric pressure and work with thegauge pressureonly.

On a plane surface, we often need to determine the magnitude of theresultant force and its point of application, which is called the centerof pressure.


3/28

2.7.1 Hydrostat ic Forces Act ing On Horizont al Subm ergedPlane Sur faces

For the horizontal surface, such as the bottom surface of a tank as

shown in Fig. 2.19, the magnitude of the resultant force is simply

FR =pA =gh.A

where ;p = uniform pressure on the bottomA = area of the bottom

Fig. 2.19 : Pressure distribution on the bottom horizontal surface oftank

Since the pressure is constant and uniformly distributed over thebottom, the resultant force acts through the centroid of the area.


4/28

# Example 2.13 : Hydrostatic Force on the Horizontal Plane

The rigid L-shaped gate OAB is 3 m width and hinged at O and rests against a rigidsupport at B. Find the hydrostatic force acting on the plate AB.

Fig E2.13

Solution :

( )( )( )( )

ABplateofcentroidthethroughacting----kN

.

412

3278191000

=

=

== AghApFR

To be continued at Example 2.14


5/28

2.7.2 Hydrostati c Forces Act ing On Vert ical and I nclinedSubm erged Plane Sur faces

For the more general cases in which a submerged plane surface isinclined, as is illustrated in Fig. 2.20, the determination of theresultant force acting on the surface is more involved.

Fig. 2.20 : Arbitrary shape plane submerged in liquid

For the present we will assume that the fluid surface is open tothe atmosphere. Let the plane in which the surface lies intersect thefree surface at O and make an angle with this surface. The xycoordinate system is defined so that 0 is the origin and y is directedalong the surface as shown. The area can have an arbitrary shape asshown. Now, we wish to determine the direction, location, and

magnitude of the resultant force acting on one side of this area dueto the liquid in contact with the area.

At any given depth, h, the force acting on dA is dAhdF .= andis acting perpendicular to the surface. Thus, the magnitude of the


6/28

resultant force can be found by summing these differential forcesover the entire surface. In equation form,

with

=

AR

dAhF . sinyh= = A dAy sin

= A ydAsinThe integral of ydAis the first moment of the area with respect tothe x axis or ydA = ycA, so we can write,

AyAyF ccR .sin.sin == Or more simply as,

Agh

AhF

c

cR

= =

where ;hc= vertical distance from free surface to the centroid of the area.Note that the magnitude of the force is independent of the angle anddepends only on the specific weight of the fluid, the total area, andthe depth of the centroid of the area below the surface.

In effect, the equation indicates that the magnitude of the resultantforce is equal to the pressure at the centroid of the area multiplied bythe total area. Since all the differential forces that were summed toobtain are perpendicular to the surface, the resultant must alsobe perpendicular to the surface.

Besides the magnitude of resultant force, the location where this FR isacting on also has to be determined. This point or location is calledas center of pressure, CP. The location of CP normally is described interms of vertical distance from free surface, hRor hcp or inclineddistance from free surface, yR (or sometimes also known as ycp).


7/28

Note : From the equation FR= ghc.A, our intuition might suggestthat the resultant force should pass through the centroid of the area,this is not actually the case.

The yRcan be determined by summing the moments aroundthe x axis. That is, the moment of the resultant force must equal themoment of the distributed pressure force, or

=

=

ARR

ARR

dAyyF

ydFyF

2sin

Thus, Ay

dAy

Ay

dAy

F

dAy

yc

A

c

A

R

AR

===

222

sin

sinsin

But, Ay2dAis the second moment of the area (moment of inertia,

Ix), with respect to an axis formed by the intersection of the planecontaining the surface and the free surface (x axis). Thus, we canwrite,

2cxc

c

xR AyI

Ay

Iy +== xIwith

where Ixc is the second moment of the area with respect to an axispassing through its centroid and parallel to the x axis. Thus,

c

c

xcR y

Ay

Iy +=

Or, in vertical distance, hcp @ hR

c

c

xcR h

Ah

Ih +=

2sinsince yR= hR/sin and yc= hc/sin


8/28

Both result shows that the resultant force does not pass through thecentroid but is always belowthe centroid, since Ixc/ycA> 0.

Fig. 2.21 shows the Ixcproperties of some common shape.

Fig. 2.21 : Geometric properties of some common shapes.


9/28

Procedure for comput ing t he hydrostat ic force on asubmerged plane area

1. Identify the point where the angle of inclination of the area ofinterest intersects the level of free surface of the fluid. This mayrequire the extension of the angled surface or the fluid surfaceline. Call this point as origin, 0.

2. Locate the centroid of the area from its geometry.3. Determine hc as the vertical distance from the level of the free

surface down to the centroid of the area.4. Determine yc as the inclined distance from the level of the free

surface down to the centroid of the area. This is the distance from0 to the centroid. Note that hc and yc are related by hc = yc sin .

5. Calculate the resultant force from AghAhF ccR == 6. Then calculate the Ixc, the moment of inertia of the area about itscentroidal-x axis.

7. Determine the location of cp by calculating the yR from(i) c

c

xcRcp y

Ay

Iyoy +=r

* Or hR for vertical plane cases which is using the equation of

(ii) cc

xcRcp h

Ah

Ihoh +=

2sinr

8. Sketch the FRacting on the cp, perpendicular to the area (or thewhole free body diagram if necessary).

9. Calculate other force magnitude if required using M=0 forequilibrium.

Note : For inclined plane case, follow all steps 1 to 7. For vertical plane case, follow steps 2, 3, 5, 6 and 7 (ii). Step 8 and 9 required if problem involved solving using M=0.


10/28

# Example 2.14 : Hydrostatic Force on the Vertical Plane

The rigid L-shaped gate, OAB, of Fig. E2.14 is hinged at O and rests against a rigidsupport at B. What minimum horizontal force, P, is required to hold the gate closed ifits width is 3 m? Neglect the weight of the gate and friction in the hinge. The back ofthe gate is exposed to the atmosphere.

Fig E2.14

In this case, we should decompose the force analysis of L gate into two parts:i. determination the resultant force FR(AB) which acting on plate AB.ii. determination of the resultant force FR(OA) which acting on the plate OA.

Solution :

iii.

From Eg. E2.13, FR(OB) acting on plate AB has been calculating which isequalto 412 Kn.

iv. For plate OA,


11/28

( )

( )( )( )(

kN6.588

34581.91000

=

=

== AghApF ccOAR

)

( ) 433 m161243

12=== bdIxc

( )

( )( )m

Ah

Ihh

c

xccOAR

27.5

345

165

.

=

+=

+=

FREE BODY DIAGRAM

Refer to free body diagram above, for equilibrium,

( )( ) ( ) ( )

( ) ( )

kNP

P

PFF

M

ABROAR

hinge

437

4141200027.2588600

)4(1327.5

0

=

=+

=+

=


12/28


An open tank has a vertical partition and on one side contains gasoline with adensity 700 kg/m3 at a depth of 4 m, as shown in Fig. E2.15. A rectangular gate thatis 4 m high and 2 m wide and hinged at one end is located in the partition. Water isslowly added to the empty side of the tank. At what depth, h, will the gate start toopen?

Fig E2.15

Solution :

( )( )( )( )kN9.109

24281.9700

=

=

== AghApF ccgasolineR

( ) 433

m67.1012

42

12===

bdIxc

( )( )m

Ah

Ihh

c

xccgasolineR

67.2

242

67.102

.

=

+=

+=


13/28

( )( )( )( )

/.

29810

228191000

h

hh

AghApF ccwaterR

=

=

==

( ) 4333

m167.012

2

12h

hbdIxc ===

( )( )

h

hh

hh

Ah

Ihh

c

xccwaterR

6670

22

16702

3

.

/

./

.

=

+=

+=

For equilibrium,

( ) ( )

( ) ( )

mh

hhhh

hFhhF

M

gasolineRgasolineRwaterRwaterR

hinge

563

146167733266672410990066709810

4

0

3

2

.

...

=

=

=

=

=


14/28


A pressurized tank contains oil (SG = 0.9) and has a square, 0.6-m by 0.6-m platebolted to its side, as is illustrated in Fig. E2.16. When the pressure gauge on the topof the tank reads 50 kPa, what is the magnitude and location of the resultant force onthe attached plate? The outside of the tank is at atmospheric pressure.

Fig E2.16

Solution :

( )

( ) ( )( )( )[ ]( )kN.

....

.

325

6060328199001050 3

=

+=

+== AghpApF caircR

( ) 433

m0108.012

6.06.0

12===

bdIxx

( )( )m

Ah

I

hhc

xc

cR

31.2

6.06.03.2

0108.03.2

.

=

+=

+=


15/28

# Example 2.17 : Hydrostatic Force on the Inclined Plane

The 4 m diameter circular gate of Fig. E2.17 is located in the inclined wall of a largereservoir containing water. The gate is mounted on a shaft along its horizontaldiameter. For a water depth of 10 m above the shaft, determine:

(a) the magnitude and location of the resultant force exerted on the gate by thewater,

(b) and the moment that would have to be applied to the shaft to open the gate.

Fig E2.17

Solution :

a. The magnitude and location of resultant force

AghF cR .=

( )( )( ) ( ) kN. 12302108191000 2 == RF ( ) 4m44

2

4

44

===R

Ixx


16/28

Ay

Iyy

c

xxcR

.+=

( ) (( ))2260104

60

10

sin/sin+=

m611.= b. The moment that would have to be applied to the shaft to open the gate.

( ) ( ) .1007.10866.0101230 53 NmyyFM cRR ===

Pressure Prism

The pressure prism is a graphical representation of thehydrostatic force on a plain surface. The magnitude of the resultantfluid force is equal to the volume of the pressure prismand passesthrough the centroid.

Consider the pressure distribution along a vertical wall of a tankof width b, which contains a liquid having a specific weight . Sincethe pressure must vary linearly with depth, we can represent thevariation as is shown in Fig. 2.22 (a), where the pressure is equal tozero at the upper surface and equal to h at the bottom.

Fig. 2.22 : The pressure prism


17/28

The pressure distribution shown in Fig. 2.22 (a)applies across

the vertical surface so we can draw the three-dimensionalrepresentation of the pressure distribution as shown in Fig. 2.22 (b).

The base of this volume in pressure-area space is the plane surfaceof interest, and its altitude at each point is the pressure. This volumeis called thepressure prism, and it is clear that the magnitude oftheresultant force acting on the surface is equal tothevolume of thepressure prism or,

( )( )

2

2

22

1

bh

bhh

bhh

prismofvolumeFR

=

==

=

.

pressure

JUST CHECK!....

AghAhFh

hce

Ah

bh

prismpressureofVolumeF

ccR

c

R

===

==

=

2sin

22

1 2

The resultant force must pass through the centroid of the pressureprism which is located along the vertical axis of symmetry of thesurface, and at a distance h/3 above the base (since the centroid of atriangle is located at h/3 above its base) or 2/3h from the upper end.Or it can be proved as follows,


18/28

h

hh

bhh

bhh

Ah

Ihh

c

xccR

3

2

62

2

12

2

3

=

+=

+=

+=

/

/

Note :

The use of pressure prisms for determining the force on submerged

plane areasis convenient if the area is rectangularso the volumeand centroid can be easily determined. However, for othernonrectangular shapes, integration would generally be needed todetermine the volume and centroid. In these circumstances it is moreconvenient to use the equations developed in the previous section.

This same graphical approach can be used for plane surfaces that donot extend up to the fluid surface as illustrated in Fig. 2.23. In thisinstance, the cross section of the pressure prism is trapezoidal.

However, the resultant force is still equal in magnitude to the volumeof the pressure trapezoidal, and it passes through the centroid of thevolume.


19/28

Fig. 2.23 : The pressure distribution on the vertical plates located farbelow from the free surface.

Specific values can be obtained by decomposing the pressure

prism/trapezoidal into two parts, ABDE and BCD. Thus,

21 FFFR += where the components can readily be determined by inspection forrectangular surfaces. The location of FRcan be determined bysumming moments about some convenient axis, such as one passingthrough A. In this instance,

2211 yFyFyF AR += Therefore the location where the FRacts measured from point A is

R

AF

yFyFy 2211

+=


20/28

# Example 2.18 : Determination of Hydrostatic Force on the Vertical PlaneUsing Pressure Prism Technique

Solve Example 2.16 using pressure prism technique.

Solution :

Fig E2.18

( )AhpF s 11 += ( ) ( )( )[ ][ ]( )6.06.0281.9100090.050000 +=

kN4.24=

Ahh

F

=

212

2

( ) ( )6.06.026.081.9100090.0

=

kN954.0=


21/28

Total hydrostatic force,

kNFFFR 4.25954.04.2421 =+=+=

The location of FR measured from O, yo,

( ) ( )2.03.0 21 FFyF oR +=So,

( ) ( )

R

oF

FFy

2.03.0 21 +=

( ) ( )( ) m296.0104.25 2.010954.03.0104.24 333

=

+= m296.0=

Therefore the location of FR measured from free surface, hR,mhR 3.2296.06.2 ==


22/28

2.8 Hydrostati c Forces Act ing On Subm erged Curved

Surfaces

The equations FR=ghCA and hR=Ixc/hC.A + hC are developed for themagnitude and location of the resultant force acting on a submergedsurface only apply to plane surfaces. However,many surfaces ofinterest (such as those associated with dams, pipes, and tank) arenonplanar.

Fig. 2.24 : Examples of curved or nonplanar surfaces.

For submerged curved surface, the determination of the resultantforce (FR) typically requires the integration of the pressure force thatchange along the curve surface.

However, the easiest way to determine the FRacting on the curvedsurface by separating it into the horizontal and vertical components,FH and FV.

This is done by considering the free-body diagram of the fluid volumeenclosed by the curved surface of interest and the horizontal andvertical projections of this surface, as shown in Fig. 2.25 below.


23/28

Fig. 2.25 : Hydrostatic force on the curved surface

The forces acting on this enclosed volume include :

1. W is the weight of the enclosed fluid volume and actsdownwardthrough the centroid of this volume which is simply given by,

W=gV

2. Fx is the hydrostatic force acting on the vertical projection surfacearea, through the centroid of this vertical surface where,Fx=gh2.Avertical

3. Fy is the hydrostatic force acting on the horizontal projectionsurface area, through the centroid of this horizontal surfacewhere,

FY=gh1.Ahorizontal

Note that :vertical surface = the projection of the curved surface on a

vertical plane,horizontal surface= the projection of the curved surface on a

horizontal plane.


24/28

From the Fig. 2 25 (b), for the equilibrium, the force balances in thehorizontal and vertical directions give;

andxH FF = WFF yV +=

The magnitude of resultant force, FR is then given by,

22

VHR FFF +=

And its direction,

=

H

V

F

F1tan

Summary of the procedure for computing the hydrostatic

force on submerged curved sur face.

1. Isolate the volume of fluid above/under the curved surface.2. Sketch the free body diagram (FBD) of the fluid volume and

show all the forces involved with correct direction and location.3. Compute the Fx=gh2.Avertical. (identify first the h2and vertical

projection surface area,Averticalfor Fx).4. Compute FY=gh1.Ahorizontal (identify first the h1and horizontal

projection surface area,Ahorizontalfor FY).5. Compute W=gV (identify first the fluid volume,V).6. Calculate FV and FH from the FBD.7. Calculate the resultant force, FR from 22 VHR FFF += and its

direction from

=

H

V

F

F1tan .

8. Show the resultant force acting on the curved surface in such adirection that its line of action passes through the center ofcurvature of the surface.

9. Sketch the FBD and solve problem using M=0 if required*.


25/28

# Example 2.19: Hydrostatic Force on the Curve Surface

A 5 m width curved gate is located in the side of a reservoir containing water asshown in Fig. E2.19. Determine the magnitude of the resultant force and its location.

Fig. E2.19

Solution :

Fig. E2.19 (b)From the free body diagram of the fluid on the curve surface,

xH FF =

And WFF yV +=

where,


26/28

( )( ) ( )

kN

ghAF yx

1104

352

368191000

=

+=

=

.

( )( )( )(

kN

)

ghAF xy

883

3568191000

=

=

=

.

( )( )( )

kN

rggW

347

54

38191000

4

2

2

=

=

==

.

l

Therefore, == kN1104xH FF

=+=+=kN1230347883WFF yV

Thus,

( ) ( )

kN

FFF VHR

1653

1230110422

22

=

+=

+=

and

o

481104

123011 =

=

=

tantan H

V

F

F


27/28

# Example 2.20: Hydrostatic Force on the Curve Surface

A long solid cylinder of radius 0.8 m hinged at point A is used as an automaticgate, as shown in Fig. E2.20. When the water level reaches 5 m, the gate opens byturning about the hinge at point A. Determine the hydrostatic force per m length ofthe cylinder and its line of action when the gate opens.

Fig. E2.20Solution :

From the free body diagram of the fluid under curve surface,

xH FF =

and WFF yV =

where,

( )( ) ( )

kN

ghAF yx

136

1802

80248191000

.

..

..

=

+=

=


28/28

chapter 2 hydro static forces

Documents