Forces Due to Static Fluids

Introduction

• Pressure is force divided by the area on which it acts, p=F/A

• When the pressure varies over the surface of interest, the location of the resultant force (center of pressure) must be located to analyse the effect of that force

• It is important especially in designing retaining wall, reservoir, aquarium observation

Gases Under Pressure

• A pneumatic cylinder has an internal diameter of 2 in and operates at a pressure of 300psig, calculate the forced on the ends of the cylinder

𝐹 = 𝑝𝐴 =300𝑙𝑏

𝑖𝑛2𝜋22

4= 942𝑙𝑏

Gage pressure was used in this calculation instead of absolute pressure.The additional atmospheric pressure acts on both side of the area hence balanced out.

Horizontal Flat Surfaces Under Liquids• If the drum is open to the

atmospheric pressure at the top, calculate the force on the bottom

𝑝𝐵 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑏𝑜𝑡𝑡𝑜𝑚

𝑝𝐵 = 𝑝𝑎𝑡𝑚 + 𝛾𝑜𝑖𝑙 2.4𝑚 + 𝛾𝑤 1.5𝑚

𝑝𝐵 = 0𝑃𝑎(𝑔𝑎𝑔𝑒) + 𝛾𝑜𝑖𝑙 2.4𝑚 + 𝛾𝑤 1.5𝑚

𝑝𝐵 =35.9kPa(gage)

𝐴 =𝜋 3𝑚 2

4= 7.07𝑚2 ∴ 𝐹 = 𝑝𝐵𝐴 = 253.8𝑘𝑁

Force would be the same for both container – pressure at bottom dependant on depth and sp.weight only

Rectangular walls• The walls exposed to a pressure varying from zero on the surface of

the fluid to a maximum at the bottom of the wall.

• The force due to the fluid pressure tends to overturn the wall or break it at the place where it is fixed to the bottom

• For purpose of analysis, it is desirable to determine the center of pressure

𝐹𝑅 = 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 = 𝛾ℎ

2𝐴

The location of the center of pressure is at a vertical distance of h/3 from the bottom of the wall

The resultant force is calculated:

Where 𝛾 is the sp weight of the fluidℎ is the total depth of the fluid𝐴 is the total area of the wall

Example• Figure shows a dam 30.5m long that retains 8m of fresh

water and is inclined at an angle θ of 60o. Calculate the magnitude of the resultant force on the dam and the location of the center of pressure.

Calculate the area of the wall

𝐿 =ℎ

𝑠𝑖𝑛𝜃=

8

𝑠𝑖𝑛60= 9.24𝑚

𝐴 = 9.24 30.5 = 281.8𝑚2

Calculate the resultant force

𝐹𝑅 = 𝛾ℎ

2𝐴 = 9.81

8

2281.8 = 11060𝑘𝑁 = 11.06𝑀𝑁

Calculate the location of center of pressure

• Center of pressure is at a vertical distance of ℎ

3=

8𝑚

3= 2.67𝑚

• Measured from the bottom of dam along the surface,

=𝐿

3=

9.24𝑚

3= 3.08𝑚

• Measured along the face of the dam, we define𝐿𝑝 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑟𝑒𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝐿𝑝 = 𝐿 − 𝐿/3𝐿𝑝 = 9.24 − 3.08 = 6.16𝑚

=6.16m

=3.08m=2.67m

= 11.06MN

= 8m

= 60o

=9.24m

Submerged Plane Areas - General

• This procedure applies to problem dealing with plane areas that are completely submerged in the fluid

• This procedure will enable us to calculate the magnitude of the resultant force on an area and the location of the center of pressure where we can assume the resultant force to act

Term:

• 𝐹𝑅 resultant force on the area due to the fluid pressure

• The center of pressure of the area is the point at which the resultant force can be considered to act

• The centroid of the area is the point at which the area would be balanced if suspended from that point;it is equivalent to the center of gravity of solid body

Term:Angle of inclination of the area

Depth of fluid from the free surface to the centroid of the area

Distance from the level of the free surface of the fluid to the centroid of the area, measured along the angle of inclination of the area

Distance from the level of the free surface of the fluid to the centre of pressure of the area, measured along the angle of inclination of the area

Vertical distance from the free surface to the center of pressure of the area

Dimensions of the area

𝜃

ℎ𝑐

𝐿𝑐

𝐿𝑝

ℎ𝑝

B, H

Procedure for computing the force on a submerged plane area1. Identify the point where the angle of inclination of the area of

interest intersects the level of the free surface of the fluid. This may require the extension of the angled surface of the fluid surface line. Call this point S.

2. Locate the centroid of the area from its geometry.

3. Determine ℎ𝑐4. Determine 𝐿𝑐, ℎ𝑐 = 𝐿𝑐𝑠𝑖𝑛𝜃

5. Calculate the total area A on which the force is to be determined

6. Calculate the resultant force

Resultant Force on a Submerged Plane Area

𝐹𝑅 = 𝛾ℎ𝑐𝐴

where 𝛾 is the specific weight of the fluid.

Resultant force is the product of the pressure at the centroid of the area and the total area.

7. Calculate 𝐼𝑐, the moment of inertia of the area about its centroidalaxis

8. Calculate the location of the centre of pressure, 𝐿𝑝 = 𝐿𝑐 +𝐼𝑐

𝐿𝑐𝐴

9. Sketch the resultant force 𝐹𝑅 acting at the center of pressure, perpendicular to the area.

10. Show the dimension 𝐿𝑝 on the sketch

11. Draw the dimension lines for 𝐿𝑐 and 𝐿𝑝 from a reference line drawn through point S and perpendicular to the angle of inclination of the area.

12. If it is desired to compute the vertical depth to the center of pressure, ℎ𝑝, either of two methods can be used.

ℎ𝑝 = 𝐿𝑝𝑠𝑖𝑛𝜃 if ℎ𝑝 has been computed

ℎ𝑝 = ℎ𝑐 +𝐼𝑐𝑠𝑖𝑛

2𝜃

ℎ𝑐𝐴

Example

• The tank shown in figure contains a lubricating oil with a specific gravity of 0.91. A rectangular window with the dimensions B = 1m and H = 0.5m is placed in the inclined wall of the tank (θ=60o). The centroid of the window is at a depth of 1.3m from the surface of the oil. Calculate (a) the magnitude of the resultant force FR on the window and (b) the location of the center of pressure.

Example

s.g = 0.91

60o

1.3m

Step 4:

ℎ𝑐 = 𝐿𝑐𝑠𝑖𝑛𝜃 ∴ 𝐿𝑐 =ℎ𝑐𝑠𝑖𝑛𝜃

=1.3

𝑠𝑖𝑛60= 1.5𝑚

Step 5:𝐴 = 𝐵𝐻 = 1 0.5 = 0.5𝑚2

Step 6:

𝐹𝑅 = 𝛾ℎ𝑐𝐴

𝛾 = 9.81 𝑠. 𝑔 = 9.81 0.91 = 8.927𝑘𝑁/𝑚3

𝐹𝑅 = 8.927 1.3 0.5 = 5.8𝑘𝑁

Step 7:

𝐼𝑐 = 𝐵𝐻3/12 =0.5 13

12= 0.042𝑚4

Step 8:

𝐿𝑝 = 𝐿𝑐 +𝐼𝑐𝐿𝑐𝐴

= 1.5 +0.042

1.5(0.5)= 1.5065𝑚

Step 9:

60o

1.3m

5.8kN

Piezometric Head

• A change is required in the procedure if the pressure above the free surface of the fluid is different from the ambient pressure outside the area.

• Piezometric head is a concept in which the actual pressure above the fluid, pa is converted into an equivalent depth of the fluid ha that would create the same pressure

ℎ𝑎 = 𝑝𝑎/𝛾

• This depth is added to any depth h below the free surface to obtain an equivalent depth ℎ𝑒 that is

ℎ𝑒 = ℎ + ℎ𝑎

ℎ𝑎 = 𝑝𝑎/𝛾ℎ𝑒 = ℎ + ℎ𝑎

Then ℎ𝑒 can be used in any calculation requiring a depth to compute pressure.

For previous example, the equivalent depth to the centroid is ℎ𝑐𝑒 = ℎ𝑐 + ℎ𝑎

Example60o

• No vent

• Pressure of 12.4kPa gage above the oil

• Convert the pressure into piezometric head using density of oil

𝛾 =8.927𝑘𝑁

𝑚3

ℎ𝑎 =𝑝𝑎𝛾=

12.4

8.927= 1.389𝑚

• The equivalent depth to the centroid is ℎ𝑐𝑒 = ℎ𝑐 + ℎ𝑎 = 1.3 + 1.389 = 2.689𝑚

• The resultant force 𝐹𝑅 = 𝛾ℎ𝑐𝑒𝐴 = 8.927 2.689 0.5 = 12.002kN

• Centre of pressure

𝐿𝑐𝑒 =ℎ𝑐𝑒𝑠𝑖𝑛𝜃

= 3.104

𝐿𝑝 = 𝐿𝑐 +𝐼𝑐𝐿𝑐𝐴

= 3.104 +0.01

3.104(0.5)= 3.111𝑚

1.389m

1.3m

Distribution of Force on a Submerged Curved Surface• Figure shows a tank holding a liquid

with its top surface open to the atmosphere

• Lower portion is a segment of a cylinder

• We want to analyse the force acting on the curved surface due to fluid pressure

• Visualize the total force system involved by isolating the volume of the fluid directly above the surface of interest as afree body diagram

• Show all the forces acting on it as shown in figure

• Determine the horizontal forces FH and the vertical forces FV exerted on the fluid by the curved surface

• Determine the resultant force FR

Steps:

Horizontal Component

• F1 acts at distance h/3 from bottom of vertical wall

• F2a on the upper part to a depth of = F1

• F2b magnitude and location can be found using the procedure developed for plane surface

𝐹2𝑏 = 𝛾ℎ𝑐𝐴

Horizontal component

𝐹2𝑏 = 𝛾ℎ𝑐𝐴

where ℎ𝑐 is the depth to the centroid of the projected area.

• For the type of surface shown in the figure, the projected area is a rectangle

Hence: ℎ𝑐 = ℎ +𝑠

2and 𝐴𝑟𝑒𝑎 = 𝑠𝑤

where w is the width of the curved surface

Horizontal component

• Hence 𝐹2𝑏 = 𝐹𝐻 = 𝛾𝑠𝑤 ℎ +𝑠

2

• The location of F2b is the center of pressure of the projected area.

ℎ𝑝−ℎ𝑐=𝐼𝑐ℎ𝑐𝐴

=𝑤𝑠3/12

ℎ𝑐𝑠𝑤=

𝑠2

12ℎ𝑐

Vertical component

• Only weight of the fluid acts downward and only vertical component Fv acts upward.

• Then Fv = weight

𝐹𝑉 = 𝛾 × 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝛾𝐴𝑤

Resultant Force

• Total resultant force FR is

𝐹𝑅 = 𝐹𝐻2 + 𝐹𝑉

2

• The resultant force acts at an angle relative from horizontal:∅ = tan−1(𝐹𝑉/𝐹𝐻)

Example

• Compute the horizontal and vertical components of the resultant force on the curved surface and the resultant force itself. Show these vectors on a sketch

=3m

=4.5m

=2.5m

Fluid=water (9.81kN/m3)

• s = h2-h1 = 1.5m = R

• Area = A1+A2

• Area = (3)(1.5)+(1

4)π(1.5)2

=6.267m2

• Volume = Aw = 6.267(2.5)

= 15.67m3

• Weight = 𝛾V = 9.81(15.67)

= 153.7kN

Hence weight = FV = 153.7kN

Vertical Component

Location of vertical component

• The location of the centroid is found using composite-area technique

Centroid of quadrant 𝑥2= 0.424R (appendix L)

=0.424(1.5)=0.636m

Centroid of rectangle 𝑥1= b/2 = 1.5/2 = 0.75m

∴ 𝑥 =𝐴1𝑥1 + 𝐴2𝑥2𝐴1 + 𝐴2

=4.5(0.75) + 1.767(0.636)

4.5 + 1.767= 0.718𝑚

Horizontal Component and Resultant Force

• ℎ𝑐 = ℎ1 +𝑠

2= 3.75𝑚

• 𝐹2𝑏 = 𝐹𝐻 = 𝛾𝑠𝑤 ℎ +𝑠

2= 9.81 1.5 (2.5) 3 +

1.5

2=138kN

• The depth of the line of action of the horizontal component:

ℎ𝑝 = ℎ𝑐 +𝑠2

12ℎ𝑐= 3.75 +

1.52

12(3.75)= 3.8𝑚

• 𝐹𝑅 = 𝐹𝐻2 + 𝐹𝑉

2= 1382 + 153.72 = 206.5𝑘𝑁

• ∅ = tan−1(𝐹𝑉/𝐹𝐻) = tan−1153.7

138= 48.1𝑜

Effect Of a Pressure Above The Fluid Surface

• In the previous calculation of force on a submerged curve surface, the magnitude of force was directly dependent on the depth of the static fluid above the surface of interest

• If additional pressure exist above the fluid or if the fluid itself is pressurized:

• the effect is to add a depth of fluid ha = p/ϒ to the actual depth of fluid (piezometric head)

ExamplePa=7.5kPa

• s = R = 0.75

• Area = A1+A2

• Area =

(1.85)(0.75)+(1

4)π(0.75)2

=1.829m2

• Volume = Aw = 1.829(2)

= 3.689m3

• Weight = 𝛾V = 9.81(3.689)

= 35.892kN

Hence weight = FV = 35.892kN

Vertical Component (WITH VENT0.75m

1.85m

0.75m0.75m

2m

Location of vertical component

• The location of the centroid is found using composite-area technique

Centroid of quadrant 𝑥2= 0.424R (appendix L)

=0.424(0.75)=0.318m

Centroid of rectangle 𝑥1= b/2 = 0.75/2 = 0.375m

∴ 𝑥 =𝐴1𝑥1 + 𝐴2𝑥2𝐴1 + 𝐴2

=1.3875(0.375) + 0.442(0.318)

1.829= 0.361𝑚

• ha=pa/ 𝛾=7.5/9.81=0.765m• s = R = 0.75• Area = A1+A2

• Area = (1.85+0.765)(0.75)+(

1

4)π(0.75)

2

=2.403m2

• Volume = Aw = 2.403(2)= 4.806m3

• Weight = 𝛾V = 9.81(4.806)= 47.147kN

Hence weight = FV = 47.147kN

Vertical Component (WITH 7.5kPa)0.75m

1.85+0.765=2.615m

0.75m0.75m

2m

Location of vertical component (with 7.5kPa)

Centroid of quadrant 𝑥2= 0.424R (appendix L)

=0.424(0.75)=0.318m

Centroid of rectangle 𝑥1= b/2 = 0.75/2 = 0.375m

∴ 𝑥 =𝐴1𝑥1 + 𝐴2𝑥2𝐴1 + 𝐴2

=1.961(0.375) + 0.442(0.318)

2.403= 0.365𝑚

Horizontal Component and Resultant Force (with vent)

• ℎ𝑐 = ℎ1 +𝑠

2= 1.85 +

0.75

2= 2.225𝑚

• 𝐹2𝑏 = 𝐹𝐻 = 𝛾𝑠𝑤 ℎ𝑐 = 9.81 0.75 2 2.225 = 𝟑𝟐. 𝟕𝟒𝟏𝐤𝐍

• The depth of the line of action of the horizontal component:

ℎ𝑝 = ℎ𝑐 +𝑠2

12ℎ𝑐= 2.225 +

0.752

12(2.225)= 2.246𝑚

• 𝐹𝑅 = 𝐹𝐻2 + 𝐹𝑉

2= 32.7412 + 35.8922 = 48.582𝑘𝑁

• ∅ = tan−1(𝐹𝑉/𝐹𝐻) = tan−135.892

32.741= 47.628𝑜

Horizontal Component and Resultant Force (with 7.5kPa)

• ℎ𝑐𝑒 = ℎ1 +𝑠

2= 1.85 + 0.765 +

0.75

2= 2.99𝑚

• 𝐹2𝑏 = 𝐹𝐻 = 𝛾𝑠𝑤 ℎ𝑐𝑒 = 9.81 0.75 2 2.99 = 𝟒𝟑. 𝟗𝟗𝟖𝐤𝐍

• The depth of the line of action of the horizontal component:

ℎ𝑝 = ℎ𝑐𝑒 +𝑠2

12ℎ𝑐𝑒= 2.99 +

0.752

12(2.99)= 3.006𝑚

• 𝐹𝑅 = 𝐹𝐻2 + 𝐹𝑉

2= 43.9982 + 47.1472 = 64.488𝑘𝑁

• ∅ = tan−1(𝐹𝑉/𝐹𝐻) = tan−147.147

43.998= 46.979𝑜

Forces On A Curved Surface With Fluid Below It

• Fluid pressure causes forces that push upward and to the right

• The surface and its connection would have to exert reaction forces downward and to the left on the contained fluid

Force to the right

Forc

e u

pw

ard

Forces to contained the fluid

• The pressure in the fluid at any point is dependent on the depth of fluid to that point from the level of the free surface

• This is equivalent to having the curved surface supporting a volume of liquid above it, except for the direction of the force vectors

• The volume of the imaginary fluid can be calculated and piezometric height may be added if the fluid is under additional pressure

• The horizontal component of the force exerted by the curved surface on the fluid is the force on the projection of the curved surface on a vertical plane.

• The vertical component is equal to the weight of the imaginary volume of fluid above the surface

Example

• Area = A1+A2

• Area = (2.8)(1.2)+(1.2 x 1.2)(1

4)π(1.2)2

=3.669m2

• Volume = Aw = 3.669(1.5)

= 5.504m3

• Weight = 𝛾V = 9.81(5.504)

= 54kN

Hence weight = FV = 54kN

Vertical Component0.75m

2m

A1

imaginary

A2

Horizontal Component and Resultant Force

• ℎ𝑐 = ℎ1 +𝑠

2= 2.8 +

1.2

2= 3.4𝑚

• 𝐹2𝑏 = 𝐹𝐻 = 𝛾𝑠𝑤 ℎ𝑐 = 9.81 1.2 1.5 3.4 = 𝟔𝟎. 𝟎𝟑𝟕𝐤𝐍

• The depth of the line of action of the horizontal component:

ℎ𝑝 = ℎ𝑐 +𝑠2

12ℎ𝑐= 2.99 +

1.22

12(3.4)= 3.025𝑚

• 𝐹𝑅 = 𝐹𝐻2 + 𝐹𝑉

2= 60.0372 + 542 = 80749𝑘𝑁

• ∅ = tan−1(𝐹𝑉/𝐹𝐻) = tan−154

60.037= 41.969𝑜

Forces on curved surfaces with fluid above and below• The horizontal component force acts on the projection of the surface

on a vertical plane

• The vertical force – 2 forces. Top of curved has vertical force acting downward while bottom curve has vertical force acting upward

• The net vertical forces is the difference between the two forces which is equal to the weight of semicylindrical volume of fluid displaced by the curve surface.

Example

• As figure 4.19 with surface 2m long

• Area = semicircle

• Area = (1

2)π(0.7)2

=0.77m2

• Volume = Aw = 0.77(2)

= 1.54m3

• Weight = 𝛾V = 9.81(0.9)(1.54)

= 13.597kN

Hence weight = FV = 13.597kN

Vertical Component

• ℎ𝑐 = ℎ1 +𝑠

2= 1.75 +

1.4

2= 2.45𝑚

• 𝐹𝐻 = 𝛾𝑠𝑤 ℎ𝑐 = 9.81 × 0.9 0.7 2 2.45

= 𝟑𝟑. 𝟔𝟒𝟖𝐤𝐍

ℎ𝑝 = ℎ𝑐 +𝑠2

12ℎ𝑐= 2.45 +

0.72

12(2.45)= 2.433𝑚

• 𝐹𝑅 = 𝐹𝐻2 + 𝐹𝑉

2= 33.6482 + 13.5972

= 36.291𝑘𝑁

• ∅ = tan−1(𝐹𝑉/𝐹𝐻)