Hydrostatic Force on Submerged Surfaces: Calculations and Examples, Slides of Fluid Mechanics

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ENGR. BON RYAN ANIBAN

If the pressure over a plane area is uniform, as in the case of

horizontal surface of the bottom of the tank filled with

liquid, the total hydrostatic force (or the total pressure ) is

given by:

Where:

p = uniform, pressure

A = area of the horizontal surface subjected with the pressure

y axis

y axis

h

x

y

g

LOCATION OF TOTAL HYDROSTATIC FORCE,F

free surface

𝑐𝑔

𝑐𝑔

Where:

F = total hydrostatic force

acting normal to the plane,

N

A = area of the plane, m

2

ℎ = vertical distance of c g

from

the liquid surface

𝑔

= centroid of the plane

𝑝

𝑝

y axis

y axis

h

x

y

g

free surface

𝑝

𝑝

𝑝

𝑝

𝑝

2 𝑑𝐴

from statics

2 𝑑𝐴 =^ 𝐼

O

𝑜

Moment of inertia about O

𝑝

LOCATION OF TOTAL HYDROSTATIC FORCE, F

By trnasfer formula of moment of inertia:

𝑜

2

𝑝

2

𝑝

𝑔

2

𝑝

𝑝

𝑔

𝑔

𝑔

Where:

e = eccentricity , c g

to c p

I g

= centroidal moment of

Inertia

𝐴= Area of the plane surface

𝑦 ത= distance of cp from free

surface along the axis of

the plane surface

A vertical rectangular plate 1 m wide and 3 m. high is submerged in water with its top edge at the water surface. Find the total

pressure acting on of side of the plate and its location from the bottom.

Solution No. 2

𝑝 = 𝛾 h

3m

1m

Total Hydrostatic force by Pressure Diagram

𝑝 = 29. 430 kPa

= 29. 430 kPa

( 29. 43 kN/m

2 )( 3 m)( 1 m)

𝐹 = 44. 145 kN

Solving for the location of F

from bottom of the plate by centroid of the pressure diagram

x =

3 = 1 m

A vertical triangular surface of height d and horizontal base width b is submerged in a liquid with its vertex at the liquid

surface. Determine the total force F acting on one side and its location form the center of gravity of the gate.

Solution

1. 5 m

3 m

c g

2 m

𝐹 = ( 9. 81 x 0. 82 )( 3 ) (^) [

1. 5 3 ]

𝐹 = 54. 298 kN

𝑔

3

𝑒 = 0. 167 m.

c p

A vertical 10 m. ∅ circular gate is submerged half in oil (sp. gr. 0.8) and half in water such that its top edge is flushed with the

oil surface. What is the total force acting on the gate. Determine the distance of the total hydrostatic force from the oil surface

10 m

Solution

g

5 m

5 m

𝑔

4𝑟

3𝜋

=

4 ( 5 )

3𝜋

g

= 2. 122 m

2. 122 m

3. 878 m

𝑜𝑖𝑙

= 886. 951 kN 𝐹 𝑤𝑎𝑡𝑒𝑟

= 2358. 451 kN

𝐹 = 3254. 403 kN

p

water 𝑐 p

𝑔

Location of Foil

4

π 5

2

𝑒 = 0. 608 m.

𝑔

Location of 𝐹 water

4

π 5

2

Convert height of

oil to water

γℎ = γℎ

alternate solution to solve for 𝑦ത of water

π 5

2

ℎ = 6. 122 m.

𝑒 = 0. 286 m

oil

A vertical 10 m. ∅ circular gate is submerged half in oil (sp. gr. 0.8) and half in water such that its top edge is flushed with the

oil surface. What is the total force acting on the gate. Determine the distance of the total hydrostatic force from the oil surface

Solution

𝑜𝑖𝑙

= 886. 951 kN 𝐹 𝑤𝑎𝑡𝑒𝑟

= 2358. 451 kN

𝐹 = 3254. 403 kN

𝑒 = 0. 608 m. 𝑒 = 0. 286 m

10 m

g

p

g

5 m

5 m

𝑔

p

4𝑟

3𝜋

=

4 ( 5 )

3𝜋

= 2. 122 m

2. 122 m

3. 878 m

water

oil

The isosceles triangle gate shown in the figure is hinge at A and weights 1500 N. What is total hydrostatic force acting on one

side of the gate in kN , determine its vertical distance from point B.

3. 5 m

2 m

A

B

50 °

Oil (s= 0.83)

The isosceles triangle gate shown in the figure is hinge at A and weights 1500 N. What is total hydrostatic force acting on one

side of the gate in kN and determine its vertical distance from point B.

3. 5 m

2 m Oil (s= 0.83)

A

F 50 °

Solution

25

6

𝐹 = 44. 288 kN

𝑔

𝑦^ ത

= 5. 439

3

( 1 )( 2. 611 ) (^5.^439 )

𝑒 = 0. 070 m

x

𝑥 = ( 2. 611 − 0. 870 − e )sin(50)

𝑥 = 1. 28 m.

B

The gate shown in the figure is hinged at A and rest on the smooth floor at B. The gate is 3 m square and oil having sp.gr. Of

0.82 stands to a height of 1.5 m above the hinge A. The air above the oil surface is under a pressure of 7 kPa. If the gate weights

5 kN, determine the vertical force P required to open it.

B

A

c g

𝐏

Oil

s = 0. 82

Floor

Air, 𝑝 = 7 kPa

1. 5 m

Solution

𝐖

𝐅

c p

45 °

𝑐𝑔

𝑟

1 .5sin( 45 )

= 1. 061 m

𝑐𝑔

2. 561 m

𝑐𝑔

= 27. 598 kPa

2 )

𝐹 = 248. 386 kN

𝑔

3

2 )

ℎ = 3. 431 m.

sin( 45 )

sin( 45 )

45 °

𝑦 ത = 4. 852 m

𝑒 = 0. 155 m

The gate shown in the figure is hinged at A and rest on the smooth floor at B. The gate is 3 m square and oil having sp.gr. Of

0.82 stands to a height of 1.5 m above the hinge A. The air above the oil surface is under a pressure of 7 kPa. If the gate weights

5 kN, determine the vertical force P required to open it.

B

A

c g

𝐏

Oil

s = 0. 82

Solution

𝐖

𝐅 =

c p

45 °

45 °

𝐴

P𝑠𝑖𝑛 45 ( 3 ) −( 248. 386 )( 0. 155 + 1. 5 ) − 5 sin 45 ( 1. 5 ) = 0

P = 196. 235 kN