pressure prism and curved plane

8/10/2019 Pressure Prism and Curved Plane

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Pressure Prism

- Another method in determining the magnitude of the FR andlocation of cp (graphical technique).- The use of pressure prisms is very convenient for the vertical planeproblem.

-

Consider the pressure distribution along a vertical wall of a tankof width b , which contains a liquid having a specific weight γ.

- Since the pressure must vary linearly with depth, we canrepresent the variation as is shown in Fig. 2.22 (a), where thepressure is equal to zero at the upper surface and equal to γhat the bottom and Fig. 2.22 (b) shows it 3D representation.

Fig. 2.22 : The pressure prism



Due to this method, the magnitude of the resultant force acting onthe surface is equal to the volume of the pressure prism or,

( )( )( )

21

pressure

2

2

2

1

bh

hbh

height widthbase

prismof volumeF R

γ

γ

=

=

×××=

=

Let’s CHECK!....

Agh AhF

hhce

Ah

bh

prism pressureof VolumeF

c c R

c

R

ρ γ

γ γ

==∴

=

×==

=

2sin

22

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,

h

hhbhh

bhh

Ah

I hh

c

xc c R

3

2

62

2

12

2

3

=

+=

×

+=

+=

/

/



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.

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

Specific values can be obtained by decomposing the pressureprism/trapezoidal into two parts, ABDE and BCD. Thus,

21 F F F R +=

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

2211 y F y F y F AR +=

Therefore the location where the FR acts measured from point A is



R

AF

y F y F y 2211 +=

Or hR (measured from free surface) = h1 + y A

# 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

( ) Ah pF s 11 +=

( ) ( )( )[ ][ ]( )6.06.0281.9100090.050000 ×××+=

kN 4.24=

Ahh

F

−=

212

2 γ



( ) ( )6.06.02

6.081.9100090.0 ×

××=

kN 954.0=

Total hydrostatic force,

kN F F F R 4.25954.04.2421 =+=+=

The location of FR measured from O, ho,

( ) ( )2030 21 .. F F hF oR +=

So,

( ) ( )

R

oF

F F h

2030 21 .. +=

( ) ( )

( ) m296.0

104.25

2.010954.03.0104.243

33

=×

×+×=

m296.0=

Therefore the location of FR measured from free surface, hR,

mhR 3.2296.06.2 =−=



2.8 Hydrostatic Forces Acting On Submerged CurvedSurfaces

The equations FR =ρghC A and hR =Ixc /hC.A + hC which obtained fromintegration are developed for the magnitude and location of theresultant force acting on a submerged surface only apply to planesurfaces. However,many surfaces of interest (such as thoseassociated with dams, pipes, and tank) are nonplanar.

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 FR acting on the curvedsurface by separating it into the horizontal and vertical components,FH and F V.

This is done by considering the free-body diagram of the fluid volume

enclosed by the curved surface of interest and the horizontal andvertical projections of this surface, as shown in Fig. 2.25 below.



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

x H F F = and W F F y V +=

The magnitude of resultant force, F R is then given by,

22

V H R F F F +=

And its direction,

= −

H

V

F

F 1tanθ

Summary of the procedure for computing the hydrostaticforce on submerged curved surface.

1.

Isolate the volume of fluid above/under the curved surface.2.

Sketch the free body diagram (FBD) of the fluid volume andshow all the forces involved with correct direction and location.

3.

Compute the F x =ρgh 2 .A vertical . (identify first the h 2 and vertical

projection surface area, A vertical for F x ).4. Compute F Y =ρgh 1 .A horizontal (identify first the h 1 and horizontal

projection surface area, A horizontal for F Y ).5.

Compute W = ρgV (identify first the fluid volume,V ).6.

Calculate F V and F H from the FBD.

7.

Calculate the resultant force, F R from22

V H R F F F += and its

direction from

= −

H

V

F

F 1tanθ .

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*.



# 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,

x H F F =

And W F F y V +=

where,



( )( ) ( )

kN

ghAF y x

1104

352

368191000

=

×

+=

=

.

ρ

( )( )( )( )

kN

ghAF x y

883

3568191000

=

×=

=

.

ρ

( )( ) ( )

kN

r g g W

347

54

38191000

4

2

2

=

×=

×=∀=

π

π ρ ρ

.

λ

Therefore, →== kN1104 x H F F

↓=+=+= kN1230347883W F F y V

Thus,

( ) ( )

kN

F F F V H R

1653

1230110422

22

=

+=

+=

and

ο

481104

123011 =

=

= −−

tantan H

V

F

F

θ



# 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 of

the cylinder and its line of action when the gate opens.

Fig. E2.20

Solution :

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

x H F F =

and W F F y V −=

where,

( )( ) ( )

kN

ghAF y x

136

1802

80248191000

.

..

..

=

×

+=

= ρ



( )( )( )( )

kN

ghAF x y

239

18058191000

.

..

=

×=

= ρ

( )

( )( ) ( ) ( )

( )

kN

R R g g W

31

14

80808191000

4

22

22

.

...

=

−=

−=∀=

π

π ρ ρ λ

Therefore, →== kN.136 x H F F

and ↓=−=−= kN.3.. 9731239W F F y V

Thus,

( ) ( )kN

F F F V H R

352937136

22

22

...

=+=

+=

and, ο446136

93711 ..

.tantan =

=

= −−

H

V

F

F θ

pressure prism and curved plane

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