Chapter 1 Hydrostatic Force

CHAPTER 1 : HYDROSTATIC FORCE CC501-HYDRAULIC 2

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Hydrostatic Forces

F = P * A

CHAPTER 1

HYDROSTATIC FORCE

1.1 Introduction

This chapter focuses on how to determine hydrostatic forces exerted on a plane or

curved surface submerged in a static fluid. It covers the force produced by the pressure

in a fluid that acts on the wall and the location of the resultant force, called the centre of

pressure.

At the end of this chapter, student should be able to:

a) Define the terms of ‘Pressure’ and ‘Hydrostatic Force’.

b) Understand the terms centroid and center of pressure.

c) Understand the basic knowledge of hydrostatic forces acting on plane surface.

d) Understand the basic knowledge of hydrostatic forces action on curve surface.

e) Understand the basic knowledge of hydrostatic forces action on inclined

surface.

1.2 Hydrostatic Force

Hydrostatic is the branch of fluid mechanics that related to the fluids at rest. In

other word, it's deal with pressures and forces resulting from the weight of fluids at

rest. By referring Figure 1.1, the fluid exerts force and pressure against the walls of its

container, whether it is stored in a tank or flowing in a pipe. But there is a difference

between force and pressure, although they are closely related.

Figure 1.1: Hydrostatic Forces in Tank

In summarize, the formula of hydrostatic forces (F) is:


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Where, P = pressure & A = area over which the force is distributed.

Unit is in Newton (N).

1.2.1 Hydrostatic Pressure

The pressure water exerts is called hydrostatic pressure. These principles always

apply to hydrostatic pressure:

a) Pressure depends only on the depth of water above the point in question (not

on the water surface area).

b) Pressure increases in direct proportion to the depth of water.

c) Pressure in a continuous volume of water is the same at all points that are at

the same depth or elevation.

d) Pressure at any point in the water acts in all directions at the same magnitude.

Specifically, pressure is defined as force per unit area. In S.I units, pressure is

usually expressed in Newton per square meter (N/m2). For convenience, the unit N/m2

is called a Pascal (Pa). In this equation, pressure can be expressed as:

P = F / A

Where, P = pressure,

F = hydrostatic force

A = area over which the force is distributed

Depending on the benchmark used (with/without atmospheric pressure), pressure

can be described as absolute pressure or relative pressure.

a) Atmospheric pressure (ρa) is defined as the pressure at any given point in

the earth atmosphere caused by the weight of air above the measurement

point. Atmosphere pressure at sea level (standard) is approximately

101.325 kPa or 760 mmHg.


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ρr = ρabs - ρa

b) Absolute pressure (ρabs) is the pressure with its zero point set at the

vacuum pressure.

c) Relative pressure (ρr) is the terms of pressure with its zero set at the

atmospheric pressure. This pressure is more widely use in engineering

than absolute pressure.

There is the relationship between them is:

1.2.2 Head Pressure

It is often convenient to express pressure in terms of the height of a column of

water in meters instead of terms of kPa. This is called pressure head, h. The water

that filled into a tank will produce pressure at the sides and bottom of the tank. If the

h is high, the hydrostatic pressure is calculated from the bottom of the tank:

P = ρgh

Where is ρ fluid density, g is gravity acceleration and h is head pressure.

Problem 1.1

What is the pressure and force at the bottom of the cylindrical containers?

2.4 m

1.5 m

OIL(s.g = 0.9)

WATER(s.g = 1.0)

3.0 m


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Figure 1.2: Centroid Location

Solution:

The pressure for cylindrical tank:

P = ρgh(oil) + ρgh(water)

= (0.9 x 1000 x 9.81 x 2.4) + (1.0 x 1000 x 9.81 x 1.5)

= 21189.6 + 14715

= 35904.6 Pa

= 359.046 kPa

The force for cylindrical tank:

F = P x A

= 359.046 x (π x 32 / 4)

= 253.795 kN

1.2.3 Definition of Centroid, C

The centroid gives a definition of the mean

position of an area (volume). It is closely related

to the center of mass a body. One adds up

position of x for all the little pieces dAi of the

Area, A to get average x position, xc. The x and y

coordinates of the centroid are evaluated

mathematically as:


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1.2.4 The First Moment of Area

By referring Figure 1.2 above, the 1st moments of areas are the average

displacement of an area about an axis of rotation. They are closely related to the

centroid.

The first moment of area about the y -axis is:

So;

The first moment of area about x –axis is:

The first moment of area have units of m3.

1.2.5 The Second Moment of Area, Ix

The 2nd moments of areas are the average (displacement) of an area about an axis

of rotation. Have units of m4. The second moment of area about the x -axis is:

It is sometimes called the moment of inertia of the area. The second moment of

inertia is always positive since y2 > 0. The second moment of area about the y -axis is:

The products of inertia about a xy coordinate axes:


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Working out the second moments

would be troublesome as the axes of

rotations moved but for the parallel

axes theorem. The moments of many

objects through their centroid are

known.

The 2nd Moment of Area is:

One writes down second moment through centroid, then determines distance of

centroid to axis of rotation and finally applies the parallel axis theorem.

1.3 Hydrostatic Force on Plane Area

There are several steps to be followed for determine the hydrostatic force on

plane area:

a) Specify the magnitude of the force.

b) Specifying the direction of the force.

c) Specifying the line of action of the force.

d) To determine completely the resultant force acting on submerged force.

1.3.1 Hydrostatic Force on Horizontal Plane Area

This is the simplest cases to find forces on horizontal plane area. By referring

figure above, the tank bottom showed a uniform pressure distribution on the entire

plane. So, the pressure at the bottom is:


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Figure 1.3: Tank Bottom

P = γhNow the Resultant force, FR:

FR = PA = γhAWhere, A is bottom area of the tank.

1.3.2 Hydrostatic Force on Iclined Plane Surface

Consider a plane shown in Figure below:Where:

The origin O is at the free surface.

θ is the angle that plane makes

with the free surface.

y is directed along the plane

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.

Then, Differential force acting on differential area dA of plane:

dF = (Pressure) . (Area) = (γh). (dA) (Perpendicular to plane)

Then, the magnitude of total resulting force, FR acting on the entire surface:

Where, h = y sin θ


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*Note that dF = γh dA and h = y sinθ

With γ and θ take as constant:

Note that the integral part is the first moment area about the x-axis:

Then,

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

Now, we must find the location of the center of pressure where the resultant force acts

as “The moment of resultant force must equal to the moment of the

distribution force”.

Moment about the x-axis is:

We note that, F = y Ayc sin θ

Then,

Where, yc = y coordinate of the centroidfor the object.

Equation 1.1


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Where, Ix = ∫ y2 dA is 2nd Moment of Inertia

Where, Ixc = 2nd Moment of Inertia through centroid

By using Parallel Axis Theorem:

Substitute parallel axis theorem and rearranging:

hp = Ixc sin2 + hshc A

where,

hp = center of pressure height from fluid surface

Ixc= second moment of area

= angle of the inclined object

A= area of the object

hc= centroid height from fluid surface

Note that for a submerged plane, the resultant force always acts below the centroid of

the plane.

Summary from above formula to determine Hydrostatic Force on Plane Surface:

a) Find area in contact with fluid.

b) Locate centroid of that area.

c) Find hydrostatic pressure Pc at centroid, typically = (generally neglect

Patm).

d) Find force F = Pc A.

e) The location will not be at the centroid, but at a distance below the

centroid.

Equation 1.2


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1.3.3 Geometric Properties for locating Centroid Coordinates, Areas & Moment of

Inertia

Shape Area, A CentroidLocation, C

Moment of 2nd

Inertia, Ixcx y

ba b2

a2

ba3

12

ba2

2b3

a3

ba3

36

πR2 R RπR2

4

πR2

2R 4R

3π0.1098R4

πR2

44R3π

4R3π

0.05488R4


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1.3.3.1 Calculation of centroid, C or hc

To determine the hydrostatic force, first we must locate centroid of the area.

Centroid means the geometric center of the object’s shape, or in other word center of

gravity for the object/mass. Below are the some examples to calculate the centroid:

Problem 1.2

An object is immersed in the water that has specific weight, γ = 9.81kN/m3. Determine

the centroid of object, hc for the following cases below:

i. If object is immersed horizontally.

ii. If object is immersed vertically.

iii. If object is immersed at an inclination of θ˚.

Case 1:

Solution:

Height of rectangular is 0.6m. Centroid of the rectangular,

C = 0.6/2

= 0.3m

Locate centroid of the object from the surface, hc is:

hc = 1.1 + 0.3

= 1.4m

Case 2:

1.1m hc60cm

90cm

1.1m hc

d = 1.0m

hc

1.1m


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Solution:

Diameter of circle is 1.0m. Centroid of the circle,

C = 1.0/2

= 0.5m


hc = 1.1 + 0.5

= 1.6m

Case 3

Solution:

First, define θ:

sin θ = 1.0/1.3

= 0.769

θ = sin-1 0.769

= 50.3˚

Then; define centroid of triangle, C:

C = 2/3 (1.3)

= 0.8666m


Sin 50.3˚ = y/0.8666

y = sin 50.3˚ (0.8666)

= 0.668m

So;

hc

0.5m

1.3m

1.5m

1.5m

θ1.0m

1.3m

1/3h

2/3hh

50.3˚y

0.8666m

1.168m0.5m


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hc = 0.5 + 0.668

= 1.168m

Problem 1.3

By referring the Figure below, determine the hydrostatic force on the plate and location

of the center pressure.

Solution:

The Hydrostatic Force, F = ρgAhc

= 1000 (9.81) [(1.2 x 1.0)/2] [3 + (1.0/3)]

= 9810 (0.6) (3.333)

= 19618.038 N

= 19.62 kN

The Center of Pressure, hp = [Ixc*sin2θ / Ahc] + hc

= [0.0333(1) / (0.6 x 3.333)] + 3.333

= 3.35 m

Problem 1.4

A pool has a water gate AB with 60˚diagonal to the water surface and has been fixed to

pool wall. The water gate has a rectangular shape and 2m width shown in Figure below.

Calculate the hydrostatic force and determine the center of the pressure gate.

Where;

Ixc = 1.2 x (1.0)3

36= 0.0333 m4

Sin 900= 1Sin2900= 1

Pool Base

A

B

60˚

3m


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Solution:

Find Hydrostatic Force, F = ρgAhc

Where, hc = distance from water surface to the gravity center

hc = 3/2 = 1.5m

Where, h = 3 / (sin 60˚)

= 3.464m

F = 1000 (9.81) (3.464 x 2) (1.5)

= 101945.52N = 101.95kN

The Center of Pressure, hp = [Ixc sin2/ Ahc] + hc

= [6.928 (0.75)/ (6.928 x 1.5)] + 1.5

= 2.0m

1.4 Hydrostatic Force on Curved Surface

Curved surface occurs in many hydraulic structures for example dams, tanks and

cross sections of circular pipes. Since this class of surface is curved, the direction of

the force is different at each location on the surface. The

pressure forces are divided into horizontal and vertical

component.

Look at forces acting on wedge of water ABC by referring

the right figure. Weight force W due to weight of volume of

water. F1 and F2 is the hydrostatic force on each planar face.

Reaction Forces of FH and FV due to wall of tank. The weight

force W passes through the center of gravity of the wedge.

To determine the horizontal force on static equilibrium:

FH = F2

= ρgAhc

60˚

h3m

Where;

A = 2 x 3.464= 6.928m2

Ixc = bh3 /12= 2 x 3.4643

12= 6.928mm4

Sin 60=

Sin260= 0.75m


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The vertical component of the force on a curved surface

may be determined by considering the fluid enclosed by the

BC curved surface and AC vertical projection lines extending

to the free surface. Thus:

FV = F1 + W

= ρgV1 + ρgVABC

= ρgV

Where, V is the volume on the curve BC.

The Resultant or Magnitude Forces, FR is a triangular

combination of the horizontal and vertical parts. So:

FR = (FV2 + FH

2)

The direction of Resultant Force is determined by using following formula:

= tan-1 FVFH

Problem 1.5

Determine the resultant force on the curved part of the base by referring Figure below.

Solution: A

B5m

12m


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1 m long

We know that resultant force, FR = (FV2 + FH

2)

So, let’s find horizontal force first:

FH = ρgAhc

= 1000 x 9.81 x (12 x 1) x (5 + 12/2)

= 1294920 N = 1294 kN

Then, calculate vertical force:

FV = ρgV

But, the problem is how to find volumes, V?

Separate the Figure above to two blocks known as V1 volume for full blocks & V2 volume for

quadrant (quarter blocks).

V = V1 - V2

= (12 x 17 x 1) - [(π x 122 / 4) x 1]

= 204 – 113.112 = 90.888 m3

FV = 1000 x 9.81 x 90.888

= 891611N = 891.6 kN

So, the Resultant Force, FR = (FV2 + FH

2)

= [(1294)2 + (891.6)2]

= 2469386.56 = 1571.4 kN

Therefore, Direction of Resultant Force,

= tan-1 (FV / FH)

= tan-1 (891.6/ 1294)

= 34.56˚

17 m

12 m

FH

FV

R


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Problem 1.6

Determine the resultant force on the curved part of the base by referring Figure below.

V1

V2

Solution:

Area, A = 2 4 = 8 m 2

Centroid height, hc= 3 +22 = 4m

Total volume, V = v 1 + v 2

v 1 = 4 23 = 24m 3

v 2 =

4

)2(142.3 2

4 = 12.57 m 3

Total volume,V = 12.57 + 24 = 36.57 m 3

Horizontal Force, F H = cgAh = 10 3 (9.81)(8)(4)

= 313920 N = 313.92 kN

Vertical Force, F v = gV = 1000 (9.81)(36.57)

= 358751 N = 358.75 kN

3m

FV

FH

j = 2m

4m

R

RFV


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Resultant Force, R = (FV2 + FH

2)

R = [(313.92)2 + (358.75)2] = 476.7 kN

Direction of Resultant Force, R = tan 1

H

VF

F

= tan 1

92.31375.358 = 48.8 0

1.5 Hydrostatic Force on A Vertical And Inclined Rectangular Wall

The dam is a structure that is built to hold and retain water. The dam surface contact

with the water experiences a hydrostatic force. This hydrostatic forces acting perpendicular to

the dam wall surface area that contact with water, either square-shaped dam or trapezium-

shaped. Dam to withstand the force of gravity, weight or mass of the dam structure and act on

the center of gravity (centroid).

1.5.1 Hydrostatic Force Exerted On A Vertical Rectangular Wall

Problem 1.7

A concrete wall with 7m high and 5m long has been used to hold water up to 4.5m asshown Figure below. Determine: -

i. Hydrostatic Force (F) over the dam wall.ii. Center of pressure height from surface water (hp).

3m

4.5m

7m

FH


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Solution:

i. Hydrostatic Force, F = ghcA

= 1000(9.81)(4.5/2)(4.5*5)

= 496631.25 N = 496.63 kN

ii. Center of pressure, hp = Ixc sin2 + hs

hcA

= 37.96875(1) + 2.25

2.25(22.5)

= 3 meter

1.5.2 Hydrostatic Force Exerted On An Inclined Trapezoidal Wall

Problem 1.8A concrete dam with trapezium-shaped are 10 m high and 350 m long has been hold

water depth of 8m. Determine hydrostatic force over an inclined dam surfaced and

center of pressure.

10m

5m

7m

8m

2m

where; = 1000 kg/m3

g = 9.81 m/s2

A= 4.5 * 5 = 22.5 m2

hs= 4.5/2 = 2.25 m

Ixc = bd3 = 5(4.53)

12 12

= 37.96875 m4

= 900

sin2 = 1


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Solution :

Determine the values of , A, hc and Ixc.

How to define ………

How to define area of section, A ………

W

F

Center of Pressure

FH

FV

10m

2m

10/2 = tan

@ tan = 5

= tan-1 5= 78.690

= 78.690

8mx

Sin = 8/x

x = 8/sin 78.690

x = 8.16 m

Area, A = Ldam. * x= 350 * 8.16

= 2856 m2


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How to define hc ………

How to define Ixc ………

Ixc = bd3

12

= 350(8.163)

12

= 196244.7448

12

= 15847.37 m4

i.Hydrostatic Force, F = ghcA

= 1000(9.81)(8/2)(8.16*350)

= 112069 kN

ii.Center of pressure, hp = Ixc sin2 + hc

hcA

= 15847.37(0.96) + 4

4(2856)= 15213.48 + 4

8m

hc

centroid

hc = d/2

= 8/2= 4 m

Front elevation of dam

L @ b = 350m

x @ d = 8.16 m

where; = 1000 kg/m3

g = 9.81 m/s2

A = 2886.1 m2

hc = 4 m

Ixc = 16353.73m4

= 78.690

sin2 = 0.96


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11424

= 5.33 m

Chapter 1 Hydrostatic Force

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