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MINISTRY OF SCIENCE AND TECHNOLOGY


DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION










CE-02016
FLUID MECHANICS






A.G.T.I (Second Year)





Civil Engineering



PART ONE
CONTENTS



Page



Chapter 1 Properties of Fluid 1

2 Fluid Pressure And Its Measurement 10

3 Hydrostatics Forces On Submerged Plane Areas 25

4 Buoyancy And Stability Of Floating Bodies 46

5 Fluid Flow Concepts And Basic Equations 56

6 Flow Through Pipes 70

1

CHAPTER 1

PROPERTIES OF FLUID



1.1 Introduction
Fluid mechanics is that branch of science which deals with the behaviour of the fluids
at rest as well as in motion. The problems, man encountered in the fields of water supply,
irrigation, navigation and water power, resulted in the development of the fluid mechanics.
It deals with the statics, kinematics and dynamics of fluids. Available methods of
analysis stem from the application of the following principles, concepts, and laws:
- Newton's law of motion
- The first and second laws of thermodynamics
- The principle of conservation of mass, and
- Newton's law of viscosity.
In the development of the principles of fluid mechanics, some fluid properties play
principal roles. In fluid statics, specific weight (or unit weight) is important property, whereas
in fluid flow, density and viscosity are predominant properties.

1.2 Definition of a Fluid
A fluid may be defined as a substance which is capable of flowing. It has no definite
shape of its own, but conforms to the shape of the containing vessel. Fluids can be classified
as liquids or gasses.
A 'liquid' is a fluid, which possesses a definite volume, which varies only slightly with
temperature and pressure. Since under ordinary conditions liquids are difficult to compress,
they may be for all practical purposes regarded as incompressible.
A 'gas' is a fluid, which is compressible and possesses no definite volume but it
always expands until its volume is equal to that of the container. Even a slight change in the
temperature of a gas has a significant effect on its volume and pressure.
The fluids are also classified as ideal fluids and real fluids. 'Ideal fluids' are those
fluids which have no viscosity and surface tension and they are incompressible. However, in
nature the ideal fluids do not exist and therefore, these are only imaginary fluids. 'Real fluids'

2

are those fluids which are actually available in nature. These fluids possess the properties
such as viscosity, surface tension and compressibility.

1.3 Units of Measurement
There are in general four systems of units, two in metric system and two in the
English system. Of the two, one is known as the absolute system and the other as the
gravitational system. Table below lists the various units of measurement for some of the basic
or fundamental quantities.
Metric Units English Units
Quantity
Gravitational Absolute Gravitational Absolute
Length

Mass

Force
m

metric slug(msl)

kg(f)
m

gm

dyne
ft

slug

lb(f)
ft

lb

poundal(pdl)

International System of Units ( SI )
Mass kg
Force N
Pressure N/m
2
(Pa)
Mass density kg/m
3

Weight density (Specific weight) N/m
3

Work J
Power Watt
Dynamic viscosity N.s/m
2

Kinematic viscosity m
2
/s



1.4 Mass density, Specific weight, Specific volume, Specific gravity
Mass Density (


C in different systems of units is 102 msl/m
3
(or) 1 gm/cc (or)
1000 kg/m
3
(or) 1.94 slug/ft
3
(or) 62.4 lb/ft
3
.

3

 = mass
volume

Specific Weight / Unit Weight ( or r): Specific weight of a fluid is the weight it possesses
per unit volume. The specific weight of water at 4 C is 1000 kg(f)/m
3
(or) 981 dynes/cm
3

(or) 9810 N/m
3
(or) 62.4 lb(f)/ft
3
(or) 62.4x32.2 pdl/ft
3
.

 = g = weight
volume

Specific volume (V
s
): of a fluid is the volume of the fluid per unit (mass).

1

s
V


Specific Gravity (S or G): is the ratio of specific weight (or mass density) of a fluid to the
specific weight (or mass density) of a standard fluid. For liquids, the standard fluid chosen for
comparison is pure water at 4 C. For gases, the standard fluid chosen is either hydrogen or
air at some specified temperature and pressure.

waterofvolumeequalofweight
cesubsofweight
S
tan


waterofweightspecific
cesubsofweightspecific
S
tan


waterofdensitymass
cesubsofdensitymass
S
tan

Note

-Specific gravity of water at 4 C is equal to 1
-Specific gravity of mercury  13.6




4


Example 1.1 If 6 m
3
of oil weighs 47 KN, calculate its specific weight, mass density and
specific gravity.

Specific weight  = weight = 47 = 7.833 KN/m
3

vol 6
Mass density  = r = 7833 = 798 kg/m
3

g 9.81
Specific gravity S = Specific weight of oil
Specific weight of water

= 7.833/9.81
 0.8



Example 1.2 Carbon-tetra chloride has a mass density of 162.5 msl/m
3
. Calculate its mass
density, specific weight and specific volume in the English system of units. Also calculate its
specific gravity.
Mass density of carbon tetra chloride = 162.5 msl/m
3

Specific weight of carbon tetra chloride in metric system = 162.5 x 9.81 = 1594.125 kg(f)/m
3

Specific weight of carbon tetra chloride in English system
=1594.125 x 2.205/(3.281)
3
[ 1 kg(f) = 2.205 lb(f)]
= 99.54 lb(f)/ft
3

Mass density of carbon tetra chloride in English system =99.54/32.2 = 3.09 slug/ft
3

Specific volume = 1/specific weight
= 1/99.54
= 0.0105 ft
3
/lb(f)
Specific gravity = Specific weight of carbon tetra chloride
Specific weight of water
= 99.54/62.4
= 1.595


5

1.5 Viscosity
The viscosity of a fluid is that property which determines the amount of its resistance
to a shearing force.
V
Moving Plate

F
dv
dy Y


Fixed Plate
Fig.1.1 Fluid motion between two parallel plates

Experiments show that shear force varies with the area of the plate A, with velocity V
and inversely with distance Y.
Y
AV
F 
Since by similar triangles

dy
dv
Y
V



dy
dv
AF 

)()( stressshear
dy
dv
A
F
 

If a proportionality constant  , called dynamic viscosity, is introduced

dy
dv
 
Newton Law of Viscosity
strainshearofrate
stressshear
dy
dv
ityvisDynamic 

cos
The dynamic viscosity may be defined as the shear stress required to produce unit rate
of angular deformation.

6

Units of  are:

FPS lb(f) sec
ft
2

CGS poise / dyne.sec
cm
2

SI N.S
m
2


Another viscosity coefficient, the coefficient of kinematic viscosity, defined as



 
g
r
densitymass
ityvisabsolute
ityviskinematic



 
cos
cos
r
g




Kinematic viscosity is the ratio of viscosity to mass density.

Units of  are:

FPS ft
2
/s
CGS stokes ( cm
2
/sec)
SI m
2
/sec






Example 1.3 Refer to figure, a fluid has absolute viscosity 0.001 lb-sec/ft
2
and specific
gravity 0.913. Calculate the velocity gradient and the intensity of shear stress at the boundary
and points 1 in, 2 in and 3 in from boundary, assuming a straight line velocity distribution.


7


45 in/sec




3"




For a straight line assumption, the relation between velocity and distance is V =15 Y
dv = 15 dy
velocity gradient dv/dy = 15

For y =0 ,V=0
dy
dv
 

= 0.001 x 15
= 0.015 lb/ft
2

Similarly for the other values of y, we also obtain  = 0.015 lb/ft
2


Example 1.4 At a certain point in castor oil the shear stress is 0.216 N/m
2
and the velocity
gradient is 0.216 s
-1
. If the mass density of castor oil is 959.42 kg/m
3
, find the kinematic
viscosity.

dy
dv
 

0.216 =  (0.216)
 = 1 N.s/m
2


 =  /
= 1 /959.42
= 1.04 x 10
-3
m
2
/s

8

1.6 Surface Tension ( )
The surface tension of a liquid is the work that must be done to bring enough
molecules from inside the liquid to the surface to form one new unit area of that surface.
 = 0.073 N/m for air-water interface
 = 0.48 N/m for air-mercury interface




1.7 Capillarity
Rise or fall of liquid in a capillary tube is caused by surface tension and depends on
the relative magnitude of the cohesion of the liquid and the adhesion of the liquid to the walls
of the containing vessel.
Liquids, such as water, which wet a surface cause capillary rise. In nonwetting liquids
(e.g mercury) capillary depression is caused.
Rr
Cos
h


2


where h = height of capillary rise ( or depression)
 = wetting angle
R = radius of tube
R
R

   

h
h
 
 
Capillary Rise Capillary Depression
Fig.1.2 Capillarity in circular glass tubes
Example 1.5 A clean tube of internal diameter 3 mm is immersed in a liquid with a
coefficient of surface tension 0.48 N/m. The angle of contact of the liquid with the glass can

9

be assumed to be 130. The density of the liquid is 13600 kg/m
3
. What would be the level of
the liquid in the tube relative to the free surface of the liquid outside the tube.

Rr
Cos
h


2

)1015()81.913600(
13048.02
3



xx
Cos
h
= - 3.08 x 10
-3
m
There is a capillary depression of 3.08 mm

1.8 Bulk Modulus of Elasticity
The bulk modulus of elasticity expresses the compressibility of a fluid. It is the ratio
of the change in unit pressure to the corresponding volume change per unit of volume.
)/( Vdv
dP
E


The units of E are Pa (or) lb/in
2


Example 1.6 At a great depth in the ocean, the pressure is 80 MPa. Assume that specific
weight at the surface is 10 KN/m
3
and the average bulk modulus of elasticity is 2.34 GPa .
Find change in specific volume at that depth.

r
g
V
s


1

=
3
1010
81.9


= 9.81 x 10
-4
m
3
/kg
s
s
V
dv
dp
E



2.34 x10
9
= (80 x10
6
) -0
- dv
s

9.81x10
-4

dv
s
= -0.335 x10
-4
m
3
/kg

10
CHAPTER 2

FLUID PRESSURE AND ITS MEASUREMENT




2.1 Fluid Pressure at a Point
Pressure or 'intensity of pressure' may be defined as the force exerted on a unit area. If
F represents the total force uniformly distributed over an area, the pressure at any point
p=F/A. However, if the force is not uniformly distributed, the expression will give the
average value only. When the pressure varies from point to point on an area, the magnitude of
pressure at any point can be obtained by the following expression
p = dF / dA
where dF represents the force acting on an infinitesimal area dA.
The forces so exerted always acts in the direction normal to the surface in contact.
The normal force exerted by a fluid per unit area of the surface is called the fluid pressure.

2.2 Variation of Pressure in a Fluid
The pressure intensity p at any point in a static mass of fluid does not vary in x and y
directions and it varies only in z direction.

dz
dp

The above equation is the basic differential equation representing the variation of
pressure in a fluid at rest, which holds for both compressible and incompressible fluids. It
indicates that within a body of fluid at rest the pressure increases in the downward direction
at the rate equivalent to the specific weight  of the liquid.
A liquid may be considered as incompressible fluid for which  is constant and
hence integration of above equation gives
p = -z + C
in which p is the pressure at any point at an elevation z in the static mass of liquid and C is
the constant of integration. Liquids have a free surface at which the pressure of atmosphere
acts. Thus as shown in Fig.2.1 for a point lying in the free surface of the liquid z = (H+z
o
)

11
and if p
a
is the atmospheric pressure at the liquid surface then from above equation the
constant of integration C=[p
a
+ (H+z
o
)]. Substituting this value of C in equation, it becomes
p = - z + [p
a
+  (H+z
o
)]
Now if a point is lying in the liquid mass at a vertical depth h below the free surface
of the liquid then as shown in Fig.2.1 for this point z = (H + z
o
- h) and from above equation
p = p
a
+  h

Free Liquid Surface


H h
Liquid of specific
weight 

z
o
z = (H+z
o
-h)
Datum

Fig.2.1 Pressure at a point in liquid
It is evident from the above equation the pressure at any point in a static mass of
liquid depends only upon the vertical depth of the point below the free surface and the
specific weight of the liquid, and it does not depend the shape and size of the bounding
containers. Since the atmospheric pressure at a place is constant, at any point in a static mass
of liquid, often only the pressure in excess of the atmospheric pressure is considered, in
which case the above equation becomes
hp



The vertical height of the free surface above any point in a liquid at rest is known as
pressure head.

p
h 

If h
1
and h
2
are the heights of the columns of liquids of specific weights 
1
and 
2

required to develop the same pressure p, at any point
2
2
1
1
hhp  

12
If S
1
and S
2
are the specific gravities of the two liquids and  is the specific weight
of water then since 
1
= S
1
 and 
2
= S
2
 , equation may also be written as

S
1
h
1
= S
2
h
2





Example 2.1 Convert a pressure head of 100 m of water to (a) Kerosene of specific
gravity 0.81 (b) Carbon tetrachloride of specific gravity 1.6.

(a) hS
1
= hS
2

100 x 1 = 0.81 x h
2

h
2
= 123.4 m of kerosene
(b) 100 x 1 = 1.6 x h
2

h
2
= 62.5 m of carbon tetra chloride

2.3 Pressure the Same in All Directions - Pascal's Law
The pressure at any point in a fluid at rest has the same magnitude in all directions. In
other words when a certain pressure is applied at any point in a fluid at rest, the pressure is
equally transmitted in all directions and to every other point in the fluid.
p
s
= p
x
= p
z


2.4 Atmospheric, Absolute, Gage and Vacuum Pressure
The atmospheric air exerts a normal pressure upon all surfaces with which it is in
contact, and it is known as atmospheric pressure. The atmospheric pressure varies with the
altitude and it can be measured by means of a barometer. As such it is also called the
barometric pressure. At sea level under normal conditions the equivalent values of the
atmospheric pressure are 1.03 kg/cm
2
; 101.3 kPa or 10.3 m of water; or 76 cm of mercury.
Fluid pressure may be measured with respect to any arbitrary datum. The two most
common datum used are (i) absolute zero pressure and (ii) local atmospheric pressure. When
pressure is measured above zero (or complete vacuum), it is called an absolute pressure.
When it is measured either above or below atmospheric pressure as datum, it is called gage

13
pressure. If the pressure of a fluid is below atmospheric it is designated as vacuum pressure;
and its gage value is the amount by which it is below that of the atmospheric pressure. A gage
which measures vacuum pressure is known as vacuum gage. Fig.2.2 illustrates the relation
between absolute, gage and vacuum pressures.

A 
Gage Pressure at A

Local Atmospheric Pressure (or Gage Zero)

Vacuum Pressure or Negative Gage Pressure at B

 B Absolute Pressure at A

Local Barometric Pressure
Absolute Pressure at B

Absolute Zero (or Complete Vacuum)

Fig.2.2 Relationship between absolute, gage and vacuum pressure

Absolute Pressure = Atmospheric Pressure + Gage Pressure
Absolute Pressure = Atmospheric Pressure - Vacuum Pressure


2.5 Measurement of Pressure
The various devices adopted for measuring fluid pressure may be broadly classified
under the following two heads:
(1) Manometers
(2) Mechanical Gages
Manometers are those pressure measuring devices which are based on the principle of
balancing the columns of liquid whose pressure is to be found by the same or another
column of liquid. The manometers are classified as simple manometers and differential
manometers.

14
Simple manometers are those which measure pressure at a point in a fluid contained
in a pipe or vessel. On the other hand differential manometers measure the difference of
pressure between any two points in a fluid contained in a pipe or a vessel.

Simple Manometers

In general a simple manometer consists of a glass tube having one of its ends
connected to the gage point where the pressure is to be measured and the other remains open
to atmosphere. Some of the common types of simple manometers are: (i) Piezometer (ii) U
tube manometer and (iii) single column manometers.

(i) Piezometer
A piezometer is the simplest form of manometer which can be used for measuring
moderate pressures of liquids. It consists of a glass tube inserted in the wall of a pipe or a
vessel, containing a liquid whose pressure is to be measured. Piezometers measure gage
pressure only since the surface of the liquid in the tube is subjected to atmospheric pressure.
The pressure at any point in the liquid is indicated by the height of the liquid in the
tube above that point, which can be read on the scale attached to it. Thus, if w is the specific
weight of the liquid, then the pressure at any point A in Fig.2.3(a) is

P
A
=  g h
A
=  h
A


h
A

h




(a) (b)
Fig.2.3 Piezometers

Negative gage pressures can be measured by means of the piezometer shown in Fig.
2.3(b). It is evident that if the pressure in the container is less than the atmospheric no column
A


15
of liquid will rise in the ordinary piezometer. Neglecting the weight of the air caught in the
portion of the tube, the pressure on the free surface in the container is the same as that at free
surface in the tube which may be expressed as p = - h, where  is the specific weight of the
liquid used in the vessel.



(ii) U- tube Manometer
Piezometers cannot be used when large pressures in the lighter liquids are to be
measured, since this would require very long tubes, which cannot be handled conveniently.
Furthermore gas pressures cannot be measured by means of piezometers because a gas forms
no free atmospheric surface. U tube manometer consists of a glass tube bent in U-shape, one
end of which is connected to the gage point and the other end remains open to the
atmosphere. The tube contains a liquid of specific gravity greater than that of the fluid of
which the pressure is to be measured. For the measurement shown in Fig.2.4 (a) the gage
equation may be written as indicated below.

V D
Manometric Liquid (Sp.gr S
2
)

y

B' B C
Fluid of Sp.gr S
1
z

A'

(a)
Pressure head at A , h =P
A
/ 
1
g = P
A
/(S
1
 ) g
Pressure head at A' = Pressure head at A
Pressure head at B' = Pressure head at A' -z = (P
A
/ S
1
 g )- z
Pressure head at B = Pressure head at B'
Pressure head at C =Pressure head at B =(P
A
/ S
1
 g )- z
Pressure head at D = Pressure head at at C - y x S
2
/S
1
( in terms of liquid at A)
A


16
At D, there being atmospheric pressure, the pressure head =0, in terms of gage pressure.

p
A
- z - y. S
2
= 0
S
1
 g S
1
p
A
= z + y S
2
(represents the pressure heads in terms of the liquid at A)
S
1
 g S
1

p
A
= z S
1
+ y S
2
(represents the pressure heads in terms water)
g


Fluid of Sp.gr S
1
D
V
A' h
1


h
2

B C
Manometric Liquid (Sp.gr S
2
)

(b)
Fig.2.4 U Tube Simple Manometer

Fig.2.4(b) shows another arrangement for measuring pressure at A by means of a U-
tube manometer. By following the same procedure as indicated above the gage equation for
this arrangement can also be written,

P
A
= h
1
S
2
- h
2
( in terms of liquid at A)
S
1
S
1
P
A
= h
1
S
2
- h
2
S
1
( in terms of water)

A U tube manometer can also be used to measure negative or vacuum pressure. For
measurement of small negative pressure, a U tube manometer without any manometric liquid
may be used, which is as shown in Fig.2.5(a).

A


17

Fluid of Sp.gr S
1


A'
h

B C


(a)
P
A
+ h = 0
S
1
 g
P
A
= - h ( m of liquid at A)
S g
P
A
= -S
1
h ( m of water)

For measuring greater negative pressures a manometric liquid of greater specific
gravity is employed, for which the arrangement shown in Fig.2.5(b) may be employed.

Fluid of Sp.gr S
1


A'
z
B y
Manometric liquid C
(Sp.gr S
2
)
( b)
Fig.2.5 Measurement of negative pressure by U-tube simple manometer

P
A
= - z - y S
2
( in terms of liquid at A)
 S
1
S
1
P
A
= -z S
1
-y S
2
( in terms of water)

A
A


18
Differential Manometers

For measuring the difference of pressure between any two points in a pipe line or in
two pipes or containers, a differential manometer is employed. In general a differential
manometer consists of a bent glass tube, the two ends of which are connected to each of the
two gage points between which the pressure difference is required. Some of the common
types of differential manometers are :
(i) Two-Piezometer Manometer
(ii) Inverted U Tube Manometer
(iii) U Tube Differential Manometer
(iv) Micromanometer

(i) Two Piezometer Manometer
The difference in the levels of the liquid raised in the two tubes will denote the
pressure difference between the two points. Evidently this method is useful only if the
pressure at each of the two points is small. Moreover it cannot be used to measure the
pressure difference in gases, for which the other types of differential manometers described
below may be employed.


P
1
h P
2

S
1
g S
1
g


 
1 2

Fig.2.6 Two Piezometer Manometer
P
1
-P
2
= h ( m of liquid in the pipe)
S
1
 g
P
1
-P
2
= S
1
h ( m of water)
g
(ii) U -tube Differential Manometer
It consists of glass tube bent in U-shape, the two ends of which are connected to the
two gage points between which the pressure difference is required to be measured. Fig.2.7

19
shows such an arrangement for measuring the pressure difference between any two points A
and B. The lower part of the manometer contains a manometric liquid which is heavier than
the liquid for which the pressure difference is to be measured and is immiscible with it.
Fluid of Sp.gr S
1


A'
y
D
C h C'
Manometric liquid Sp.gr S
2


Fig.2.7 U Tube Differential Manometer

P
A
+ y + h - h.S
2
- y = P
B

S
1
 g S
1
S
1
 g
P
A
-P
B
= h. S
2
- h = h (S
2
- 1) ( m of fluid of sp.gr S
1
)
S
1
 g S
1
S
1

P
A
-P
B
= h ( S
2
-S
1
) ( m of water)
g

(iii) Inverted U -tube Manometer
It consists of a glass tube bent in U-shape and held inverted as shown in Fig.2.8.
When the two ends of the manometer are connected to the points between which the pressure
difference is required to be measured, the liquid under pressure will enter the two limbs of the
manometer, thereby causing the air within the manometer to get compressed. The presence of
the compressed air results in restricting the heights of the columns of liquids raised in the two
limbs of the manometer. An air cook as shown in Fig.2.8, is usually provided at the top of the
inverted U tube which facilities the raising of the liquid columns to suitable level in both the
limbs by driving out a portion of the compressed air. Inverted U tube manometers are suitable
for the measurement of small pressure difference in liquids.



B

A


20


Air Cock
Air
C C'
h
D

y




Liquid Sp.gr S
1

Fig.2.8 Inverted U Tube Manometer
Since the specific weight of air is negligible as compared with that of liquid, between C' and
D may be neglected.

P
A
- y + ( y-h) = P
B

S
1
 g S
1
 g
P
A
-P
B
= h ( m of liquid of sp.gr S
1
)
S
1
 g
P
A
- P
B
= h S
1
( m of water)
 g



Example 2.2 The left leg of a U-tube mercury manometer is connected to a pipe line
conveying water, the level of mercury in the leg being 60 cm below the centre of pipe line
and the right leg is open to atmosphere. The level of mercury in the right leg is 45 cm above
that in the left leg and the space above mercury in the right leg contains Benzene (specific
gravity 0.88) to a height of 30 cm. Find the pressure in the pipe.
B

A


21

E


D

60 cm
C C'


P
A
/ + 0.6 = 0.45 x 13.6 + 0.3 x 0.88
P
A
/ = 5.784 m of water
5.784 x 1000
P
A
=
10
4

= 0.578 kg/cm
2


Example 2.3 A U tube manometer is used to measure the pressure of oil (sp.gr 0.8)
flowing in a pipeline. Its right limb is open to the atmosphere and the left limb is connected to
the pipe. The centre of the pipe is 9 cm below the level of mercury (sp.gr 13.6) in the right
limb. If the difference of mercury level in the two limbs is 15 cm, determine the absolute
pressure of the oil in the pipe in KPa.


9 cm
15 cm

Oil (Sp.gr 0.8) Mercury (Sp.gr 13.6)




P/s
1
g + 0.06 - 0.15 x S
2
/S
1
=0
P/s
1
g = 2.49 m of oil
45 cm

30 cm

A

Benzene

Mercury

Water


22
P = 2.49 x 0.8 x1000 x 9.81
= 19.541 KPa ( Gage Pressure)
Absolute Pressure = 19.541 + 101.3 = 120.841 KPa



Example 2.4 For a gage pressure at A of -0.15 kg/cm
2
, determine the specific gravity of
the gage liquid B in the figure given below.



Air E 10.25 m


9. 5 m B F G 9.6 m

Lquid A (Sp.gr 1.6) C D Liquid B ( Sp.gr S)
9.0 m


Pressure at C = Pressure at D
-(0.15) x 10
4
+ (1000 x 1.6 x 0.5) = P
D

P
D
= -0.7 x 10
3
kg/m
2


Between point D and E, since there is an air column which can be neglected.
P
D
= P
E
P
F
= P
G

P
G
= 0 =P
F
( point G being at atmospheric pressure)
Thus P
F
= P
E
+ S x 1000 (10.25-9.60) = 0
S = 1.077

Example 2.5 As shown in the accompanying figure, pipe M contains carbon-tetra
chloride of specific gravity 1.594 under a pressure of 1.05 kg/cm
2
and pipe N contains oil of
specific gravity 0.8. If the pressure in the pipe N is 1.75 kg/cm
2
and the manometric fluid is
mercury, find the difference x between the levels of mercury.

A


23
Carbontetra chloride
M

2.5 m
Oil
1.5 m
Z x Z'
Mercury

Equate the pressure heads at Z an Z' as shown in above figure
Pressure head at Z in terms of water
= [ 1.05 x 10
4
+ (2.5+1.5) 1.594 + x (13.6)]
1000
Similarly pressure head at Z' in terms of water
= [ 1.75 x 10
4
+ (1.5 x 0.8) + x (0.8) ]
1000
Equating the above two
10.5 + 6.376 + 13.6 x = 17.5 + 1.2 + 0.8x
x = 0.142 m =14.2 cm



Example 2.6 The tank in figure is closed at top and contains air at a pressure p
A
.
Calculate the value of p
A
for the manometer readings shown.
A

Open Tube Air

200 cm Oil (sp.gr 0.75) 150 cm
Water
10 cm
X X Mercury


M

N



24
p
A
+ 1.5 x 0.75 x 9810 + 0.5 x 9810 + 0.1 x 9810 - 0.1 x 13.6 x 9810 = 0
p
A
= -3580.65 Pa

Exampe 2.7 Petrol of specific gravity 0.8 flows upwards through a vertical pipe. A and
B are two points in the pipe, B being 30 cm higher than A. Connections are led from A and B
to a U-tube containing mercury. If the difference of pressure between A and B is 0.18
kg(f)/cm
2
, find the reading shown by the differential mercury gage.


B

30 cm Petrol
A
y
x Mercury


p
A
+ ( x + y) x 0.8 = p
B
+ (0.3 + y) 0.8 + ( x x 13.6)
 
( p
A
- p
B
) = 12.8 x + 0.24
 
x = 0.122 m

25

CHAPTER 3

HYDROSTATIC FORCES ON SUBMERGED PLANE AREAS


3.1 Total Pressure on a Plane Surface
(a) Total Pressure on a Horizontal Plane Surface

Fig.3.1 Total Pressure on a Horizontal Plane Surface
Since every point on the surface is at the same depth below the free surface of the
liquid, the pressure intensity is constant over the entire plane surface.
Pressure intensity p = wh
If A is the total area of the surface,
Total pressure on the horizontal surface P = pA =(h)A = Ah
(b) Total Pressure on a Vertical Plane Surface

Fig.3.2 Total Pressure on a Vertical Plane Surface

26

Since the depth of liquid varies from point to point on the surface, the pressure
intensity is not constant over the entire surface. Consider on the plane surface a horizontal
strip of thickness dx and width b lying at a vertical depth x below the free surface of the
liquid. Since the thickness of the strip is very small, for this strip the pressure intensity may
be assumed to be constant.
Pressure intensity of strip p = x
Area of strip dA = b dx
Total pressure on the strip dp = p dA = x . (bdx)
Total pressure on the entire plane surface P =

dP = 

x (bdx)


x bdx = A.x
P =  A x ---------------- (3.1)

( c ) Centre of Pressure for Vertical Plane Surface
For a plane surface immersed horizontally, since the pressure intensity is uniform,
the total pressure would pass through the centroid of the area, i.e, in this case the centroid of
the area and the centre of pressure coincide with each other. However, for a plane surface
immersed vertically the centre of pressure does not coincide with the centroid of the area.
As shown in Fig.3.2 let h be the vertical depth of the centre of pressure for the plane
surface immersed vertically. Then the moment of the total pressure P about axis OO is equal
to (Ph ).
Total pressure of the strip, dP =  x ( bdx ) and
Moment about axis OO , (dP) x = x
2
(bdx)
The sum of the moments of the total pressure on all strips =
 
bdxxxdP



2
)( 
By using principle of moments,
The moment of total pressure about axis OO,
 
bdxxhP



2
 --------- (3.2)
 

bdxx
2
represents the sum of the second moment of the areas of strips about axis OO,
which is equal to moment of inertia I
o
of the plane surface about axis OO. That is
 

 bdxxI
o
2
------------ (3.3)
Introducing equation 3.3 in equation 3.2 and solving for h,
P
I
h
o





27

Substituting for the total pressure from equation 3.1, we obtain



xA
I
h
O

From the 'Parallel axes theorem' for the moment of inertia,
2


 xAI
G
Io
-----------------(3.4)
where I
G
is the moment of inertia of the area about on axis passing through the centroid of
the area and parallel to axis 00
Introducing equation 3.4 in equation 3.3, it becomes

x
A
I
xh
G
 -----------------(3.5)
Equation 3.5 gives the position of centre of pressure on a plane surface immersed
vertically in a static mass of liquid. The position at which the resultant pressure P may be
taken as acting is called the centre of pressure.

( d ) Total Pressure on Inclined Plane Surface


Fig.3.3 Total Pressure on Inclined Plane Surface

Consider a plane surface of arbitrary shape and total area A , wholly submerged in a
static mass of liquid of specific weight  . The surface is held inclined such that the plane of
surface makes an angle  with the horizontal as shown in Fig.3.3. The intersection of this

28

plane with the free surface of the liquid is represented by axis OO, which is normal to the
plane of the paper. Let x be the vertical depth of the centroid of the plane surface below
the free surface of the liquid, and the inclined distance of the centroid from axis OO
measured along the inclined plane be y .
Total pressure on the strip = dP = x (dA)
Since x = y Sin 
dP =  ( y Sin  ) (dA)
Total pressure on the entire surface P = ( Sin  )

y (dA)
Again

y (dA) represents the sum of the first moments of the area of the strips
about axis OO, which is equal to the product of the area A and the inclined distance of the
centroid of the surface area y from axis OO.

y (dA) = A y
P = A ( y Sin )
x =y Sin

P =  Ax ----------- (3.6)

Equation 3.6 is the same as equation 3.1, thereby indicating that for a plane surface
wholly submerged in a static mass of liquid and held either vertical or inclined, the total
pressure is equal to the product of the pressure intensity at the centroid of the area and the
area of the plane surface.

(e) Centre of Pressure for Inclined Surface





xA
SinI
xh
G

2
------------- (3.7)
The equation 3.7 gives the vertical depth of centre of pressure below free surface of
liquid, for an inclined plane surface, wholly immersed in a static mass of liquid.
Table 3.1 gives the moments of inertia and other geometric properties of different
plane surfaces which are commonly met in actual practice.




29

Table.3.1 Moment of Inertia and other Geometric Properties of Plane Surface
















30

3.2 Pressure Diagram
Total pressure as well as centre of pressure for a plane surface wholly submerged in a
static mass of liquid, either vertically or inclined, may also be determined by drawing a
pressure diagram. A pressure diagram is a graphical representation of the variation of the
pressure intensity over a surface. Such a diagram may be prepared by plotting to some
convenient scale the pressure intensities at various points on the surface. Since pressure at
any point acts in the direction normal to the surface, the pressure intensities at various points
on the surface are plotted normal to the surface. Fig.3.4 shows typical pressure diagrams for
horizontal, vertical and inclined plane surfaces.

Fig.3.4 Pressure diagrams for horizontal, vertical and inclined plane surfaces

As an example consider a rectangular plane surface of width l and depth b, held
vertically submerged in a static mass of liquid of specific weight w, as shown in Fig.3.5. Let
the top and bottom edges of the surface area be at vertical depths of h
1
and h
2
respectively
below the free surface of the liquid. Thus for every point near the top edge of the surface area
the pressure intensity is
p
1
=  h
1

Similarly for every point near the bottom edge of the surface area the pressure
intensity is
p
2
=  h
2


31




Fig.3.5 Pressure diagram for a vertical rectangular plane surface

Since the pressure intensity at any point varies linearly with the depth of the point
below the free surface of the liquid, the pressure diagram may be drawn as shown in Fig 3.5,
which will be trapezium with the length of the top edge equal to h
1
, the length of the bottom
edge equal to wh
2
and its height equal to b, the depth of the rectangular plane surface. In the
same manners if the pressure diagrams are drawn for all the vertical sections of the surface
area, a trapezoidal prism will be developed as shown in Fig.3.5 The volume of the prism
gives the total pressure on the plane surface, which in the present case is

lb
hh
P 








2
21

---------------- (3.8)
The result obtained by equation 3.8 may also be obtained by using equation 3.1.






Example 3.1 A 3.6 m by 1.5 m wide rectangular gate MN is vertical and is hinged at point
15 cm below the centre of gravity of the gate. The total depth of water is 6 m. What
horizontal force must be applied at the bottom of the gate to keep the gate closed?


32





x M
h Gate
6 m
 3.6 m
P
F
N

Total pressure acting on the plane surface of the gate
P =  g Ax
= 1000 x 9.81 x (3.6 x 1.5) x 4.2
= 222491 N
The depth of centre of pressure



xA
xh
G
I


 
 
2.46.35.1
6.35.1
12
1
2.4
3



= 4.457 m
Let F be the force required to be applied at the bottom of the gate to keep it closed.
Taking moment about the hinge,
F (1.8-0.15) - 222491 (0.257 - 0.15) = 0
F = 14428 N
Example 3.2 A triangular gate which has a base of 1.5 m and an altitude of 2 m lies in a
vertical plane. The vertex of the gate is 1 m below the surface of a tank which contains oil of
specific gravity 0.8. Find the force exerted by the oil on the gate and the position of the centre
of pressure.
 = 0.8 x 9810 = 7848 N/m
3


15 cm

33

A =1/2 x 1.5 x 2 = 1.5 m
2

x = (1+ 2/3 x 2) = 2.33 m
The force exerted on the gate
P =  Ax
= 7848 x 1.5 x 2.33
= 27428 N
The position of the centre of pressure




xA
I
xh
G

= 2.33 + 0.33
1.5 x 2.33
= 2.43 m




Example 3.3 A rectangular door 2 m high and 1 m wide closes an opening in the vertical
side of a bulkhead which retains water on one side of it to a depth of 2 m above the top of the
door. The door is supported by two hinges placed 10 cm, from the top and bottom of one of
the vertical sides, and it is fastened by a bolt fixed at the centre of the opposite vertical side.
Determine the forces on each hinge and the force exerted on the bolt.




2 m x
h 1 m
F
T
 

F

bolt 



2 m


2 m P
F
B






34

Total pressure P =  Ax
= 9810 x 2 x1 x 3
= 58.86 KN




xA
I
xh
G

=
3
2
12
21
3
3
x
x

= 3.1 m
One half of P is taken by the hinges and the other half by the bolt.
Force on the bolt F = P/2 = 29.43 KN
Taking moment about at the bottom hinge
F
T
(1.8) = P x 0.8 - F x 0.9
F
T
= 11.445 KN
F
B
= 29.43-11.445 = 17.985 KN

Example 3.4 Gate PQ shown in the figure below is 1.25 m wide and 2 m high and it is
hinged at P. Gage G reads 1.5 x 10
4
N/m
2
. The left hand tank contains water and the right
hand tank oil of specific gravity 0.75 up to the heights shown in the figure. What horizontal
force must be applied at Q to keep the gate closed?


G
Air

IWS 1.5 m
O O'

6 m Water
P Hinge Oil  P
2 m Gate 2 m P
water
P
Oil
F
Q
Q

35

P
oil
=  g Ax = 1000 x 0.75 x 9.81 x (2 x1.25) x 1 = 18394 N




xA
I
xh
G

=
1)225.1(
)2)(25.1(
12
1
1
3
xx


= 1.34 m
For LHS of gate, it is necessary to convert the negative pressure due to the air to its
equivalent in meters of water
h = P / g = -1.5 x 10
4
= -1.5 m
10
3
x 9.81
This negative pressure head is equivalent to having the water level in the tank reduced by 1.5
m.
P
water
= 10
3
x 9.81 x ( 2 x 1.25) x (4.5-1) = 85838 N
OOsurfacewaterimaginarythebelowm
xx
h 595.3
5.3)225.1(
)2()25.1(
12
1
5.3
3



Taking the moments of all the forces about the hinge and equating the sum of all the
moments to zero for equilibrium of the gate.
F x 2 + P
oil
x 1.34 - P
water
(3.595 -2.5) = 0
F = 34672.5 N acting at Q to the left



Example 3.5 A trapezoidal plate 3 m wide at the base and 6 m at the top is 3 m high.
Determine the total pressure exerted on the plate and the depth to the centre of pressure when
the plate is immersed normally in water up to its upper edge.
6m

x
3m b
dx


3 m
By Integration


36

Intensity of pressure on strip, dp = g x
Total pressure on strip, dP = gx . bdx
From figure
b/2-1.5 = 1.5
3-x 3
b = 6-x
dP = g x (6-x) dx
Total Pressure P =

dP = g

3
0
x (6-x) dx
= 176.6 kN
Moment of total pressure on strip about OO
dM = gx (6-x) dx . x = gx
2
(6-x) dx
P.h =

dM = g

3
0
x
2
(6-x) dx
176.6 x 10
3
x h = g [ 6x
3
/3 -x
4
/4]
h = 1.875 m

[or]

P =  Ax
= 9810 x 13.5 x 1.333
= 176. 5 kN




xA
I
xh
G

=1.333 +
333
.
1
5
.
13
75.9
x

= 1.875 m

Example 3.6 A circular plate 2.5 m diameter is immersed in water, its greatest and least
depth below the free surface being 3 m and 1 m respectively. Find (a) the total pressure on
one face of the plate and (b) the position of the centre of pressure.


37

Free liquid surface

x 1 m
3 m h
P

Edge View of circular plate


View normal to circular plate

x = 2 m
A = /4 (2.5)
2
= 4.906 m
2

P =  g A x
= 1000 x 9.81 x 4.906 x 2
= 96256 N
The depth of the centre of pressure




xA
SinI
xh
G

2

I
G
=  /64 (2.5)
4
= 1.917 m
4

h = 2 + (1.917) (0.8)
2

4.906 x 2
= 2.125 m



Example 3.7 An opening in a reservoir is closed by a plate 1m square which is hinged at
the upper horizontal edge as shown in the figure. The plate is inclined at an angle 60 to the
horizontal and its top edge is 2 m below the surface of the water. If this plate is opened by
means of a chain attached to the centre of the lower edge, find the necessary pull T in the
chain. The line of action of the chain makes an angle of 45 with the plate. Weight of the
plate is 200 kg(f).

38

Water surface in reservoir Dam

2 m
T
Chain Hinge
1 m Opening
Plate
60


Area of plate A = 1m
2


Depth of CG below the free surface of water = 2 + 1/2 Sin 60 = 2.433 m

Total pressure acting on the plate P = Ax
= 1000x 1 x 2.433
= 2433 kg(f)




xA
SinI
xh
G

2

= 2.46 m

Distance of total pressure P from hinge along the plate = (2.46 -2) 1/Sin 60 = 0.53 m

Taking moment about the hinge,

T Sin 45 x1 = 2433 x 0.53 + 200 Cos 60 x 1/2
T = 1894 kg(f)







39

3.3 Total Pressure on Curved Surfaces
The horizontal component of the resultant fluid pressure acting on a curved surface is
the pressure exerted on the projected area, and this will act at the centre of pressure of the
vertical projection.

P
H
= . g. A
x
.x
The vertical component of the hydrostatic force on curved surface is equal to the
weight of the volume of liquid extending above the surface of the object to the level of free
surface. This vertical component passes through the centre of gravity of the volume
considered.

P
V
=  g V

Resultant force P
R
= P
H
2
+P
V
2

The direction of the resultant force P is given by

H
V
P
P
tan

where  is the angle made by the resultant force P with the horizontal.


R
x
3
4


R
x
3
4



G P
V
h H G
P
H
 P
H




P
R
P
R



W P
V
= W
(a) (b)
Fig.3.6 Total Pressure on Curved Surface

40

Hh
H
x
3
2
2





If the fluid pressure acts on the opposite side of the curve surface as shown in
Fig.3.6(a), the same approach may be used but the forces act in the opposite direction.

Example 3.8 Determine and locate the components of the force due to the water acting on
the curved surface AB as shown in figure, per meter of its length.
x
A C
Hinge
6 m
 G
P
H

B

P
V

m
R
x
mHh
m
H
x
546.2
6
3
4
3
4
46
3
2
3
2
3
2
6
2








P
H
= Force on vertical projection CB
=  g A
x 
x
= 1000 x 9.81 x (6x1) x 3
= 176.58 KN acting 4 m from C
P
V
= Weight of the water above surface AB
=  g V
=1000 x 9.81 x (/4 R
2
x 1)
= 277.37 KN
22
VHR
PPP  =328.81 KN;
H
V
P
P
1
tan

 = 57 32'

41

Example 3.9 A cylinder 2.4m diameter weighs 200 kg and rests on the bottom of a tank
which is 1m long. As shown in figure below water and oil are poured into the left and right
hand portions of the tank to depths 60 cm and 1.2 m respectively. Find the magnitude of the
horizontal and vertical components of the force which will keep the cylinder touching the
tank at B.
2.4 m


A Oil
Water C D D Sp.gr 0.75
60 cm 1.2 m
B
Net P
H
= Component on AB to left -Component on CB to right
={ 0.75 x 1000 x (1.2x1)x1.2/2}-{1000 x (0.6x1) x 0.6/2}
= 360 kg(f) to left
Net P
V
= Component upward on AB + Component upward on CB
= {0.75 x 1000 x (1/4 x  /4 x 2.4
2
) x 1} + [1000{ /6 x 1.2
2
-1/2 x 0.6 x 1.08} x 1]
= 1289.7 kg(f)
Net downward force to hold the cylinder in place
= 1289.7 -200
= 1089.7 kg(f)

Example 3.10 The face of a dam retaining water is shaped according to the relationship y =
x
2
/4 as shown in figure. The height of the water surface above the bottom of dam is 12 m.
Determine the magnitude and direction of the resultant water pressure per meter breadth.


x
12 m

dy
P
H



P
V


D

O


42

P
H
= Total pressure on the projected area
=  g A
x
x
= 706.32 KN acting at 4 m from the base
P
V
= Weight of the water above the curve OA
=  g

12
0
x dy x 1
=  g

12
0
2 y
1/2
dy
= 543. 47 KN
22
VHR
PPP  = 891.2 KN
H
V
P
P
1
tan

 = 37 34'

3.4 Practical Applications of Total Pressure and Centre of Pressure

In practice there exist several hydraulic structures which are subjected to hydrostatic
pressure forces. In the design of these structures it is therefore necessary to compute the
magnitude of these forces and to locate their points of application on the structures. Some of
the common types of such structures are (i) Dams (ii) Gates and (iii) Tanks.
In several hydraulic structures, openings are required to be provided in order to carry
water from the place of its storage to place of its utilisation for various purposes. The flow of
water though such openings, called sluices, is controlled by means of gates which are known
as sluice gates. Another type of gates which are used to change the water level in a canal or a
river are known as lock gates. The water level is required to be raised or lowered in a canal or
a river used for navigation, at a section where the bed of the canal or the river has a vertical
fall. As such a section of a canal or a river, in order to facilitate the transfer of a boat from
the upper water level to the lower one or vice versa a chamber known as lock is constructed
by providing two pairs of lock gates. If a boat is to be transferred from the upper water level
to the lower water level, the lock is filled up through the openings provided in the upstream
pair of lock gates and keeping the similar openings in the downstream pair of lock gates
closed. When the level of water in the lock becomes equal to the upper water level, the
upstream gates are opened and the boat is transferred to the lock.


43

Example 3.11 A masonry weir is of trapezoidal cross section with a top width of 2.0 m and
of height 5m. If the weir has water stored up to its crest on the u/s side and has a tail water of
2m depth on the d/s, calculate the resultant force on the base of the weir per unit length.
Assume specific weight of masonry as 22 KN/m
3
and neglect uplift forces.

2 m
P
V1
C D


W
2

5 m W
1
1 P
V2

P
H1
W
3
0.75


P
H2
2 m
A B

0.5m
2 m

3.75 m


6.25 m


W
1
= (1/2 x 0.5 x 5) x 1 x 22 = 27.5 KN
W
2
= (2 x 5) x 1 x 22 = 220 KN
W
3
= (1/2 x 3.75 x 5 ) x 1 x 22 = 206.25 KN
P
V1
= (1/2 x 0.5 x 5) x 1 x 9.81 = 12.26 KN
P
V2
= (1/2 x 1.5 x 2.0) x1 x 9.81 = 14.71 KN
P
H1
= (5 x 1) x 2.5 x 9.81 = 122.62 KN
P
H2
= (2 x 1) x 1 x 9.81 = 19.62 KN
KNHVR 6.491
22

 




H
V
P
P
tan
 = 7754'




44


Example 3.12 The end gates of a lock are 5 m high and include an angle of 120 in the
closed position. The width of the lock is 6.25 m. Each gate is carried on two hinges on the top
and the bottom of the gate. If the water levels are 4m and 2m on u/s and d/s sides
respectively, determine the magnitude of the forces on the hinges due to the water pressure.



5 m

4 m
P
u
2 m
P
d


Hinge

ELEVATION



u/s side
6.25 m
120 Lock
R d/s side
P 30
F
PLAN
Width of the gate = 6.25 = 3.61 m
2 Cos 30
Total pressure on the u/s face of gate is
P
u
=  A x
= 28880 kg(f)
Depth of the centre of pressure on the u/s face

45





xA
I
xh
G
u

= 2.67 m
Total Pressure on the d/s face of the gate is
P
d
=  A x
= 7220 kg(f)
Depth of the centre of pressure on the d/s side




xA
I
xh
G
d

= 1.33 m
Resultant pressure on each gate is P = P
u
- P
d
= 21,660 kg(f)
If x is the height of the point of application of the resultant water pressure on the gate,
P.x = P
u
(4-h
u
) - P
d
(2-h
d
)
x = 1.56m
Resolving parallel to the gate,
F Cos  = R Cos 
F = R

Resolving normal to the gate,
R = P = P

2 Sin 30
R = F = P = 21660 kg(f)
R
T
+ R
B
= R = 21660
Resultant hinge reaction is assumed to act at the same height as the resultant pressure. Taking
the moments of the hinge reactions about the bottom hinge,
R
T
x 5 = R x 1.56
R
T
= 6758 kg(f)
R
B
= 21660 -6758 = 14902 kg(f)




46
CHAPTER 4

BUOYANCY AND STABILITY OF FLOATING BODIES


4.1 Buoyant Force
The basic principle of buoyancy and floatation was first discovered and stated by
Archimedes. Archimedes' principle may be stated as follows; When a body is immersed in a
fluid either wholly or partially, it is buoyed or lifted up by a force which is equal to the
weight of the fluid displaced by the body. This force is known as the "buoyant force".
The point of application of the force of buoyancy on the body is known as "centre of
buoyancy".


W

G 
F
B


F
B


Fig.4.1 Buoyant force on floating and submerged bodies

For a body immersed [either wholly or partially] in the fluid, the self weight of the
body always acts in the vertical downward direction. As such if a body floating in a fluid
is to be in equilibrium the buoyant force must be equal to the weight of the body

F
B
=  g V = W

in which F
B
is the buoyant force and V is the volume of fluid displaced. Equation
represents the principle of floatation which states that the weight of a body floating in a
fluid is equal to the buoyant force which in turn is equal to the weight of the fluid
displaced by the body.

47
A hydrometer uses the principle of buoyant force to determine specific
gravities of liquids. Fig.4.2 shows a hydrometer in two liquids. It has a stem of
prismatic cross section a. Considering the liquid on the left to be distilled water
(S=1), the hydrometer floats in equilibrium when
W =  g V …………..(1)
in which V is the volume submerged and W is the weight of hydrometer.

h
1.0



Fig.4.2 Hydrometer in water and in liquid of specific gravity S
The position of the liquid surface is marked 1 on the stem to indicate unit specific
gravity S. When the hydrometer is floated in another liquid, the equation of equilibrium
becomes
(V-V) S  g =W …………………(2)
in which V =a. h
Solving eqtn (1) and (2)
S
S
a
V
h
1




Example 4.1 A hydrometer weighs 0.0216 N and has a stem at the upper end that is
cylindrical and 2.8 mm in diameter. How much deeper will it float in oil of sp.gr 0.78 than in
alcohol of sp.gr 0.821.

h




Alcohol Oil

48
In position 1, in alcohol
Weight of hydrometer = Weight of displaced liquid
0.0216 = 0.821 x 9810 x V
1

V
1
= 2.68 x 10
-6
m3 ( in alcohol)
In position 2, in oil
0.0216 = 0.78 x 9810 (V
1
+A.h)
h = 0.023 m = 23 mm

4.2 Metacentre and Metacentric Height
Consider a body floating in a liquid. If it is statically in equilibrium, it is acted upon
by two forces viz., the weight of the body W acting at the centre of gravity G of the body and
the buoyant force F
B
acting at the centre of buoyancy B. The forces F
B
and W are equal and
opposite and as shown in Fig.4.3, the points G and B lie along the same vertical line which is
the vertical axis of the body.
Let this body be tilted slightly or it undergoes a small angular displacement . It is
assumed that the portion of the centre of gravity G remains unchanged relative to the body.
The centre of buoyancy B, however, does not remain fixed relative to the body.

M

G G
W
B B B1

F
B

W F
B

Fig.4.3 Metacentre for a floating body
'Metacentre' may be defined as the point of intersection between the axis of the
floating body passing through the points B and G and a vertical line passing through the new
centre of buoyancy B
1
. The distance between the centre of gravity G and the metacentre M of
a floating body is known as 'metacentric height'.(GM)




49
4.3 Stability of Submerged and Floating Bodies
When a submerged or a floating body is given a slight angular displacement, it may
have either of the following three conditions of equilibrium developed
(i) Stable equilibrium
(ii) Unstable equilibrium
(iii) Neutral equilibrium

(i) Stable equilibrium- A body is said to be in a state of stable equilibrium if a small angular
displacement of the body sets up a couple that tends to oppose the angular displacement of
the body, thereby tending to bring back to its original position.

(ii) Unstable equilibrium - A body is said to be in a state of unstable equilibrium if a small
angular displacement of the body sets up a couple that tends to further increase in the
angular displacement of the body, thereby not allowing the body to restore its original
position.

(iii) Neutral equilibrium-A body is said to be in a state of neutral equilibrium if a small
angular displacement of the body does not set up a couple of any kind, and therefore the body
adopts the new position given to it by the angular displacement, without either returning to its
original position, or increasing the angular displacement.

Stability of a Wholly Submerged Body

In general a wholly submerged body is considered to be in a stable state of
equilibrium if its centre of gravity is below the centre of buoyancy. On the other hand a
wholly submerged body will be in an unstable state of equilibrium if its centre of buoyancy is
below its centre of gravity. If centre of gravity and the centre of buoyancy of a wholly
submerged body coincide with each other, it is rendered in a neutral state of equilibrium.

Stability of a Partially Immersed (or floating) Body

A body floating in a liquid (or a partially immersed body) which is initially in
equilibrium when undergoes a small angular displacement, the centre of buoyancy moves
relative to the body. Consider a floating body which has undergone a small angular
displacement in the clockwise direction as shown in Fig.4.4.

50
If the new centre of buoyancy B
1
is such that the metacentre M lies above the centre
of gravity G of the body, as shown in Fig.4.4(a), the buoyant force F
B
and the weight W
produce a couple acting on the body in the anticlockwise direction, which is thus a restoring
couple, tending to restore the body to its original position. Hence it may be stated that for a
floating body if the metacentre lies above its centre of gravity, then the body is in a stable

state of equilibrium
.
As shown in Fig.4 4 (b), if for a floating body slightly tilted in clockwise direction,
the metacentre M lies below the centre of gravity, G of the body, then the buoyant force and
the weight produce a couple acting on the body in the clockwise direction, which is thus an
overturning couple, tending to increase the angular displacement of the body still further. The
body is then considered to be in unstable equilibrium. Thus it may be stated that for a floating
body if the metacentre lies below its centre of gravity, then the body is said to be in an
unstable equilibrium
.
However, if for a floating body the metacentre coincides with the centre of gravity of
the body, then the body will be in a neutral state of equilibrium
. This is because there will be
neither a restoring couple nor an overturning couple developed when the body is tilted
slightly.
In the design of the floating objects such as boats, ships etc; care has to be taken to
keep the metacentre well above the centre of gravity of the object.
Overturning couple


M

G W G
B B B
1


F
B
= W
W F
B


Restoring couple

(a) Floating body in stable equilibrium


51
Overturning couple



G G
W M

B B