UNIT
1
System of Measurement and
Units
Learning Objectives
The branch of science which deals with motion, forces and their effect on
bodies is called mechanics (2) The application of principles of mechanics to
common engineering problems is known as engineering mechanics.
Introduction
Fundamental Units
The physical quantities which do not depend up on other quantities are
known as “Fundamental Quantities” and units for such quantities are known as
“Fundamental Units” or “Base Units”. The internationally accepted fundamental
quantities are
(i) Length (ii) Mass
and
(iii) Time
Derived Units
If the units are expressed in other units which are derived from fundamental
units are known as “derived units’.
Ex: Units of Area, Velocity, Acceleration, Presure etc.
System of Units
There are four systems of units which are commonly used and universally
recognised. These are known as
(i) C.G..S. (ii) F.P.S.
(iii) M.K.S.
and
(iv) S.I. Units
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C.G.S. Units
In this system, the fundamental units of length, mass and time are centimeter,
gram and second.
F.P.S. Units
In this system, the fundamental units of length, mass and time are foot,
pound and second respectivley.
M.K.S. Units
In this system, the fundamental units of length, mass and time are metre,
kilogram and second respectively.
S.I. Units (International System of Units)
The eleventh general conference of weights and measures (GCWM) has
recommended a unified, systematically constituted system of fundamental and
derived units for international use. In this system the fundamental units are meter
(m) kilogram (kg) and second (s) respectively. But there is slight varaition in
their derived units. The following are the derived units.
Force
-
N (Newton)
Stress (or) Pressure
-
N / mm2 (or) N/m2
Workdone (in joules)
-
J =Nm
Power in watts
-
W
Temperature Degree Kelvin
-
K
Current (ampere)
-
A
Units of Physical Quantities
Engineering mechanics and strength of materials are essentially “quantitative
sciences”. They involve expressions of quantities. For example.
(i) Height of a building is 12m
(ii) Area of cross section of land is 220 mm2
(iii) Stress in bar is 150 Newtons per mm2.
(iv) Radius of gyration of the section is 12 mm.
In all the above expressions of quantities; we essentially state two items,
viz., a number and a known standard of measurement.
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In the statement “Height of building is 12m, the standard of length
measurement is 1m and the height of the building is 12 times the length of that
standard. The standard of measurement adopted is known as the “unit” of physical
quantity. Each physical quantity can be expressed in a number of units.
Eg.(1) Length can be measured interms of meters, millimeters, feet, yards
and so on.
(2) The unit of area is the area of a square of side 1m and is stated as 1m2.
(3) Unit of volume is the volume of a cube of side 1m and is stated as 1m3.
(4) Unit of velocity is unit displacement for unit time and stated as 1m/s
(5) Unit of mass density is unit mass per unit volume and is stated as 1 kg/
m3.
(6) Unit of acceleration is unit change of velocity per unit time and is stated
as 1m per second square (1m/sec2).
Multiples and Submultiples
Recommended prefixes for the formation of multiples and submultiples of
units are shown in table 1.
Table 1
S.No.
Multiplying factor
Prefire
1.
1012
Tera
T
2.
109
Giga
G
3.
106
Mega
M
4.
103
Kilo
K
5.
10-3
Milli
m
6.
10-6
Micro

7.
10-9
Nano
n
8.
10-12
Pico
p
Useful Units
1 KNm
=
106 N mm
1 M.Pa
=
N/mm2
Symbol
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1 G..Pa
=
109 Pa = 103 x 106 Pa
= 103 M Pa
= KN / mm2
 1 G.Pa
=
1 KN/mm2
1 MN/m2
=
1 N/mm2
1 KN/m
=
1 N/mm (for u.d.l)
1 tonne
=
103 kgs = 103 x 9.81 Newtons
Short Answer Type Questions
1. Define a) Base units &
b) Derived untis.
2. State the units for the following in S.I.System
a) length
b) Mass
c) Velocity
d) Workdone
3. Write the units in S.I. system for the following quantities
a) Area
b) Radius of Gyration
c) Sterss
d) Volume
4. Mention S.I. units for the following quantities
a) Density b) Acceleration
c) Current
d) Power
5. List the four systems of measurement commonly used.
6. Mention the base units and derived units used in S.I. Units.
UNIT
2
Forces and Moments
Learning Objectives
1. Mechanics is the science that deals with force, and the effect of force on
bodies. In short, it deals with the motion of bodies, state of not being
considered as a special case of motion.
2. In order to understand the principles of mechanics, it is essential that
force should be studied in detail and its effects understood.
3. Newton’s first law of motion helps us to define a force as an external
agency which tends to change the state of rest or of uniform motion of a
body.
Basic Concepts
Space: Space is the region which extends in all directions and contains
everything in it.
Ex: Sun and its Planets, Stars, etc.
Time: Time is a measure of duration between successive events. The unit
time is a second which is a fraction (1/86,400) of an average solar day.
Motion: A body is said to be in motion, when it changes its position with
respect to other bodies. The relative change in position is called motion.
Matter: Any substance which occupies space is called matter.
Body: Any matter that is bounded by a closed surface is called a body.
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Inertia: A resistance offered by matter to any change in its state of motion
is called inertia.
Mass: It is a quantitative measure. It is the resistance offered by any
substance to the change in its position.
Particle: A particle occupies no space, i.e. it has no size, but has a definite
mass concentrated at a point.
Rigid body: A Rigid body is that which does not undergo deformation on
the application of forces.
Definition of Force
According to Newton’s first law of motion, force is defined as follows.
“Force is an action that changes or tends to change the state of rest or
uniform motion of a body in a straight line”.
In simple words, the action of one body on any other body can be called a
force. These actions may be of various forms, pull or push on a body, gravitational
force known as weight of a body, force presented by an elastic spring, force
exerted by a locomotive on the train, resistance offered by the track etc.
“Force is specified with its magnitude, direction and the point of application”.
Unit of Force
The unit of force in the international system is the Newton.
The Newton is that force which when applied to a body having a mass of
1kg gives it an acceleration of 1m/s2
Characteristics of a Force
The folowing are the four essential characteristics of a force:
1. Point of application 2. Direction, 3. Magnitude and 4. Sense.
1. Point of Application : It indicates where the force is applied.
2. Direction : The direction of a force is stated with refernce to geographical
directions (i.e. North, East, South and West) or with any fixed reference line.
3. Magnitude : It is the quantity of force applied and it is expressed in
Newtons.
4. Sense: It is represented by placing an arrow head to force line denoting
the nature of force i.e. whether a push or pull.
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Fig 2.1 Characteristics of Force
Fig 2.1. Shown a force acting on a body. The point of application is A, and
acts in a direction which makes at 400 to the horizontal. It’s magnitude is 400N.
Reference Frame of Axes
We know that space is a region extending in all directions. Position of a
body in space can be defined only by referring measurement to a fixed frame of
axes. The measurement may be linear, angular (or) both.
Let us consider methods to define the position of a particle with respect to
reference frame. To determine the position of a particle in space, three
measurements are necessary to be made. For example the position “P” of a
particle with respect to the reference frame XYZ is defined by three co-ordinates
X, Y, Z as shown is Fig.2.2
Fig.2.2 defining position of “P” in space.
Incase the motion of a particle is taking place in a plane, its position can be
defined by only two measurements. This can be done in the following two ways:
1. Rectangular co-ordinate system.
2. polar co-ordinate system
1. Rectangular Co-ordinate system
In this system, the position of any particle “P” is determined by the
rectangular co-ordinates (x,y), where “x” is the linear distance of the particle
“P” from the origin “o” measured parallel to the X – axis and “Y” is the linear
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distance of the particle “p” from the origin “o” measured parallel to the “Y” axis.
This is shown is fig 2.3.
Fig: 2.3 Defining the position of a
point “P” in a plane through
Rectangular Co-ordinates.
2. Polar co-ordinate system
In polar co-ordinate system the position of a particle “P” is defined by
where “r” is the radial distance from of the particle “p” from the origin “o” and
““ is the angle made by the radial line OP with the base line OX as shown in
fig 2.4.
Fig: 2.4 position of a point “P” in a plane through polar co-ordinate system.
Quantities
A physical quantity is one which can be measured. All physical quantity can
be classified in to two main categories.
1. Scalar quantity
2. Vector quantity.
Scalar Quantity
Scalar quantities are those which can be completely defined by their
magnitude (or) numerical value only and no direction.
Ex: Mass, Length, Time, Speed etc.,
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Vector Quantity
Vector quantities are those which require not only magnitude but also
direction (line of action) and sense for completely defining them.
Ex: Displacement, veloncity Force, Momenturn etc.,
1. Representation of a vector:- A vector is represented by directed line
as shown in fig 2.5. It may be noted that the length OA represents the magnitude
of the vector OA
Fig 2.5 Vector OA
The direction of the vector OA is from’o’ (i.e. starting point) to A (i.e,end
point). It is know as vector P.
2. Unit Vector:- A vector, whose magnitude is unity, is know as unit vector.
3. Equal vectors:- The vectors, which are parallel to each other and have
same direction (i.e. same sense) and equal magnitude are known as equal vectors.
4. Like Vectors:- The vectors, which are parallel to each other and have
same sense but unequal magnitude, are known as like vectors.
5. Addition of vectors:- Consider two vectors PQ and RS, which are
required to be added as shown in Fig2.6(a)
S
C
A
R
P
B
Q
Fig 2.6 (a) Vector PQ and RS
(b) Sum of vectors
Take a point A, and draw line AB parallel and equal in magnitude to the
vector PQ to some convenient scale. Through B, draw BC parallel and equal to
vector RS to the same scale.
Join AC which will give the required sum of vectors PQ and RS as shown
is Fig 2.6 (b).This is the method of addition of vectors.
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6. Subtraction of Vectors:- Consider two vectors PQ and RS whose
difference is required to be found out as shown in fig2.7(a)
P
Q
S
A
B
R
C
Fig 2.7 (a) Vector PQ and RS
(b) Difference of vectors
Take a point A, and draw line AB parallel and equal in magnitude to the
vector PQ to some convenient scale. Through B draw BC parallel and equal to
the vector RS, but in opposite direction to that of the vector RS, to the same
scale. Join AC which will give the required difference of the vector PQ and RS
as shown is fig2.7(b)
System of Forces:- a combination of several forces acting on a body is
called a “system of forces” (or)”Force system”.
The system of forces can be classified according to the arrangement of the
lines of action of the forces of the system. The forces may be classified as
1. Coplanar & Non-Coplanar
2. Concurrent & Non-Concurrent
3. Parallel & Non-Parallel and
4. Collinear Forces.
1. Coplanar&Non Coplanar Forces:- The forces whose lines lie on the
same plane are known as “Coplanar Forces”.If the lines of action of the forces
do not lie in the same plane, then, the forces are called “Non-Coplanar Forces”.
2. Concurrent, Non-Concurrent Forces:- If the forces acting on a body
meet at a point, they are called “Concurrent Forces”.
Fig 2.8
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Forces a , b and c shown in 2.8 are concurrent forces as they are meeting
at point “0” whereas forces d , e and f are called “Non-Concurrent Forces”
because all the three forces are not meeting at a point.
3. Parallel, Non-Parallel Forces:- The forces whose lines of action are
parallel to one another are known as ‘Parallel Forces’. The parallel forces may
further be classified into two categories depending upon their directions.
a.
Like Parallel Forces
b.
Unlike Parallel Forces
a. Like Parallel Forces
The Forces whose lines of action are parallel
to each other and act in the same direction as
shown in fig 2.9 are known as “ Like Parallel Forces”
b. Unlike Parallel Forces
Fig 2.9like parallel
forces
The Forces whose lines of action are parallel
to each other and act is opposite directions as
shown in fig 2.10 are known as “Unlike Parallel Forces.”
Fig 2.10 Unlike
parallel forces
4. Collinear Forces:- The forces whose lines of action lie on the same
line are known as collinear forces (fig.2.11)
Fig 2.11 Collinear Forces
Non – parallel forces
The forces whose lines of action are not parallel
to one another are known as “Non – Parallel
forces”. As shown in fig 2.12 forces s , t , u are
Non-parallel forces.
Fig 2.11 Collinear Forces
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Resultant of forces at a point
The resultant of a force system can be defined as the simplest – single force
which can replace the original system without changing its external effect on a
rigid body.
Law of Parallelogram of forces
If the forces meet at a point, their resultant may be found by the law of
parallelogram of forces. It states that if two forces acting at a point are such that
they can be represented in magnitude and direction by the two adjacent sides of
parallelogram, the diagonal of the parallelogram passing through their point of
intersection gives the resultant in magnitude and direction.
Fig 2.13
Consider two forces “P” and “Q” acting a point “O” in the body as shown
in fig 2.13 (a). Their combined effect can be found out by constructing a
parallelogram using vector “P” and vector “Q” as two adjacent sides of the
parallelogram as shown in figure 2.13 (b). the diagonal passing through “O”
represents their resultant in magnitude and direction.
Mathematically, magnitude of resultant
R  P 2  Q  2PQ cos 
Where  = Angle between two forces P and Q
 = Angle between resultant R and force P
 = Angle between resultant R and force Q
Triangle law of Forces
It states “If two forces acting at a point be
represented is magnitude and direction by the two
sides of a triangle in order, their resultant – may be
represented in magnitude and direction by the third
side of a triangle taken in opposite order”.
Fig 2.14 Traingle
law of forces
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Lami’s Theorem
It states “If three forces acting at a point are in
equilibrium, then, each force is proportional to the sine of
the angle between the other two forces.”
Mathematically,
P
Q
R


Sin  Sin  Sin
Fig 2.15
Polygon law of Forces
It is an extension of triangle law of forces for more than two forces. It states
“ If a number of forces acting simultaneously on a particle, be represented in
magnitude and direction, by the sides of a polygon taken in order, then the
resultant may be represented, in magnitude and direction by the closing side of
the polygon taken in opposite order.
Fig 2.16
Resolution of a Force
The process of splitting up the given force into a number of components
without changing its effect on the body is called “resolution of forces”.A force is
generally resolved along two mutually perpenduclar directions.
Fig 2.17
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Analytical determination of the Resultant of a concurrent coplanar force
system
The process of resolution and composition can be very conveniently used
in determining the resultant of a concurrent, coplanar force system.
Fig 2.18
Fig 2.18 (a) shows a concurrent, coplanar force system consisting of four
forces F1, F2, F3 and F4 acting an a particle.
The point of concurrence of the force system is taken as the origin, and
rectangular cordinate-axis X and Y are selected as shown. The Forces F1, F2 F3
and F4 can now be reserved into rectangular components, which have directions
along the X and Y axes.
For example the components of force F1 or F1 cos  along the X-axis and
F1 sin  along the Y-axis as shown in fig 2.18(b). The components are taken
positive when they act along the positive directions of X and Y axes.
The calculations are conveniently done in a tabular form as shown in
table below.
Force
Angle with X-axis
X-component ( +) Y-component ( +)
F1
1
F1 cos 1
F2
2
F2 cos 2
F3
3
F3 cos 3
F4
4
F4 cos  4
Fx  F1 cos 1  F2 cos 2  F3 cos 3  F4 cos 4
F1 sin 1
F2 sin  2
F3 sin 3
F4 sin  4
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Fy  F1 sin 1  F2 sin  2  F3 sin 3  F4 sin 4
Resul tan t(R) 
  Fx 
2
   Fy 
2
The direction of ‘R’ can also be easily determined. If y is makes an angle
R with the x-axis, then
tan R 
 Fy
 Fx
(or)
 F 
R  tan 1  y 
  Fx 
This method is most general method. It determines the resultant of force
systems analytically. Using the graphical method of force polygon, the resultant
of force can also be determined.
Moment of Force
A force acting on a body not only produces linear motion but also produces
rotation when its effect is considered at any other point other than the point
contact. The effect of rotation is called “Moment of a force”.
Fig 2.19
The moment of a force is equal to the product of the force and the
perpendicular distance of the point, about which moment is required, and the
line of action of force as shown in fig. 2.19
Mathematically M = pxl
Where p = force acting on the body
l = perpendicular distance of the point and the line of action of the force.
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Building Construction and Maintenance Technician
Units of Moment
Moment is the product of force and distance, So if the force is in Newton’s
and distance, is in mm, then the unit of moment is Nm. in S.I units it is expressed
in KNm.
1KNm = 106 Nm
Types of Moements
The moments are classified in to following two types
1. Clock wise Moments
2. Anti clock wise Moments
1. Clockwise Moment: It is the moment of a force, whose effect is to
turn or rotate the body, in the same direction in which the hands of a clock
move.
2. Anticlockwise Moment: it is the moment of a force, whose effect is to
turn or rotate the body, in the opposite direction in which the hands of a clock
move.
Generally the sign convention for clockwise moment is taken as positive
and for anticlockwise moment – negative
Couple
Tow equal, opposite and parallel forces having different lines of action form
a “Couple”.
Moment of Couple
The moment of a couple is a product of force “P” and the perpendicular
distance (a) between the line of action of two equal and opposite forces as
shown in fig. 2.20.
Fig 2.20
Mathematically, moment of a couple,
M = Pxa
Where P = force and
a = perpendicular distance also know as arm of couple
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Equilibrium :- a body acted upon by a system of forces is said to be in
equilibrium, if it either continues in a state of rest or continues to move is a
straight line with uniform velocity.
Conditions of equilibrium:- A body is said to be in equilibrium if it satisfies
the following conditions.
i) The algebraic sum of all vertical forces is zero i.e V  0
ii) the algebraic sum of all Horizontal forces is zero i.e H  0
iii) the algebraic sum of moments about a point is zero i.e. M  0
V  0
H  0
M  0
Solved Examples
Example 2.1
A body is acted upon by an upward force of 200N and a horizontal force
of 400N. Find the magnitude and direction of resultant.
Solution
Upward Force
P = 200N
Horizontal Force Q = 400N
R  P 2  Q2
Resultant
 (200) 2  (400) 2
 447.21N
P 200
tan R  
 0.5
Q 400
Direction of Resultant
R  tan 1 (0.5)  26.560
 260331
Resultant is at an angle of 260331 with x-axis i.e. Horizontal axis.
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Building Construction and Maintenance Technician
Example 2.2
Two Co-planar forces act at a point with an angle of 600 between them. It
their resultant is 120N and one of the forces is 60KN. Calculate the other force.
Solution
Resultant
R = 120KN
Force
P = 60KN
Angle between two forces θ = 600
Let other force
=Q
R  P 2  Q 2  2PQCos
120  602  Q 2  2X60XQCos600
120  3600  Q 2  60Q
2
∴ 3600  Q  60q  14400
Q 2  60Q  14400  3600  10800
Q 2  60Q  10800  0
Resultant =
-60  602  4(1)(10800)
∴Q=
2X1
-60  3600  43200 -60  216.33


2
2
 78.165KN (taking  ve value)
∴ Other Force  78.165KN
Example 2.3
Find the Magnitude and Direction of a resultant of two forces 50N and
70N acting of a point with an included angle of 500 between them. The force
70N being horizontal.
Solution
Horizontal Force P = 70N
Other Force
Q = 50N
Angle between two forces θ = 500
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2
2
∴ Magnitudeof Re sul tan t R  P  Q  2PQCos
 (70)2  (50) 2  2X70X50XCos500
 109.08N
Let   Angle between resul tan t R and force P
∴ tan  
tan  
QSin
50 x Sin500

P  QCos 70  50 x Cos500
50(0.766)
 0.375
70  50(0.6428)
  tan 1 (0.375)  20.550  200331
Resultant acts at angle of 220331 with force ‘p’ i.e. 70N.
Example 2.4
Determine the magnitude and direction of the resultant of the two forces
260N and 180N acting at a point, at mutually perpendicular directions.
Solution
Horizontal Force P = 260N
Vertical Force
Q = 180N
Angle between two forces θ = 900
2
2
∴ Magnitudeof Resul tan t R  P  Q  2PQCos
 (260) 2  (180) 2  2X260X180XCos900
 (260) 2  (180) 2  316.23N
Let   Angle between resul tan t R and force P
∴ tan  
QSin
180 x Sin900
180


 0.6923
0
P  QCos 260  180 x Cos90
260
  tan 1 (0.6923)  34.690  340 411
Resultant 316.23N acts at an angle of 340411 with Horizontal Force ‘P’
i.e. 260N.
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Building Construction and Maintenance Technician
Example 2.5
Two forces of magnitude 20 KN and 30 KN act at a point angle of 600
find the magnitude and direction of resultant force.
Solution
P = 20 KN
Q = 30 KN
= 600
Magnitude of the resultant
R  P 2  Q 2  2PQ cos 

2
 20    30 
2
 2 2030 cos 600
= 43.59 KN
Resultant of the given forces = 43.59 KN
If  = angle between resultant ‘R’ and force ‘P’
tan  
30  0.866 
Qsin 
30sin 600


P  Q cos  20  30cos 600 20  30  0.5 

25.98
0.7423
35
  tan 1  0.7423  360351
 Direction of resultant – with respect to force ‘p’ = 360 350
Example 2.6
A particle ‘o’ is acted and by the following forces
i) 200 N inclined at 300 to north of east
ii) 250 N to words north
iii) 300 N to words North 450 west
Fig 2.21
iv) 350 N inclined at 400 to south of west as per the given data, the forces
are acting as shown in fig 2.21
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Solution:
As per the given data, the forces are acting as shown in fig 2.21
Resolving the forces horizontally
H  200 cos 30  300cos 450  350 cos 400  250cos900
173.21  212.13  268.11  0
  307.03N
Resolving the forces vertically
V  200 cos 300  250sin 900  300sin 450  350sin 400
100  250  212.13  224.97
 337.16N
Magnitude of the resultant
R

 H 
2
  V 
 307.03
2
2
  337.16 
2
Fig 2.22
 456 N
 V 337.16

H 307.03
  1.0981
tan R 
Direction of Resultant R  tan 1  1.0981
  47.680
 R  47 0 401 with x  axis asshown in fig 2.22
Example 2.7
Find the magnitude and direction of resultant force of following force acting
at a point.
a. 80 KN due North
b.20 KN due North – East
c. 40 KN due east
d.60 KN in a direction inclined 300 east to south
0
e. 70 KN in a direction inclined 60 south of west
Fig 2.23
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Building Construction and Maintenance Technician
Solution
As per the given data the forces are acting as shown in fig 2.23
Resolving the forces horizontally
H  40  20cos 450  60cos300  70cos 600  80 cos 900
 40  20  0.7071  60  0.866   70  0.5   0
 40  14.142  57.96  35
 71.102 KN
Resolving the force in vertically
V  80sin 900  20sin 450  40sin 00  60sin 30  70sin 600
 80  1  20  0.7071  40  0  60  0.5   70  0.866 
 80  14.142  30  60.62
 3.522 KN
Magnitude of resultant
R
2
 H     V 
2


 71.02    3.522 
2
 71.19KN
Direction of resultant tan R 
V 3.522

 0.0495
H 71.19
R  tan 1  0.0495   2.8340
 20 501
R  200 501 with x axis
Example 2.8
Determine the resultant magnitude direction of
the given system of coplanar concurrent.
Solution
The force diagram is drawn as shown in fig
2.24 and angles are needed with respect it horizontal
axis i.e., x – axis as shown in fig 2.25
Fig 2.24
Paper - III Engineering Mechanics
275
Fig 2.25
Resolving the forces horizontally
H  30cos 450  40cos300  60 cos 600  20cos300
 30  0.7071  40  0.866   60  0.5   20  0.866 
 21.21  34.64  30  17.32
 8.53KN
Resolving the force vertically
V  30sin 450  40sin 300  60sin 60  20sin 300
 30  0.7071  40  0.5   60  0.866   20  0.5 
 21.21  20  57.96  10
  40.75KN
Magnitude of Resultant
R

2
 H    V 
2
2
8.53   40.75
2
 41.63KN
V 40.75
tan R 

 4.727
H 8.53
Direction of Resulatant  R  tan 1  4.777 
  78.17 0
  780 101
 R  780101 with x axis.
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Building Construction and Maintenance Technician
Example 2.9
Four men pull a tree is the east, south east, south west and North West
directions with forces 200 N, 300N, 150 N and 350 N respectively. Find the
resultant for a ovel its direction.
Fig 2.26
Solution
Force diagram is drawn as shown in fig 2.26
Resolving forces horizontally
H  200cos 00  300 cos 450  150cos 450  350cos 450
 200  1 300  0.7071 150  0.7071  350  0.7071
 200  212.13 106.06  247.48
 58.59 N
Resolving the forces vertically
V  200sin 450 150sin 450  300sin 450  200sin 00
 350  0.7071 150  0.7071  300  0.7071  200  0
 247.48  106.06  212.13
 70.71N
Magnitude of Resultant
R  (H) 2   V 

2
2
 58.59    70.71
 91.83 N
2
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277
Direction of Resultant
tan R 
V 58.59

H 70.71
= -1.207
R  tan 1  1.207 
  50.360  500 21
R  500.211 with x  axis
Example 2.10
Two Forces of 100N and 80N act at a point making an angle of 600 between
them. Determine their resultant.
Solution
Let the two forces to be P and Q ; then P = 100N and Q = 80 N
Angle between the above two forces =600
The resultant force R of the tow forces
 P 2  Q 2  2PQ cos 

100   80   2 100  80cos 60
2
2
0
 10000  6400  8000
Resultant of the two given  P 2  Q 2  2PQ Cos
 (100) 2  (80) 2  2x100x80 Cos600
 10000  6400  8000
 24400
= 156.2N
Forces = 156.2 N
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Building Construction and Maintenance Technician
Short Answer Type Questions
1. Define scalar quality and vector quality with one example for each.
2. State the following with neat sketcure
a.
Parallelogram law of force
b.
Triangle law of force.
3. Define force and mention two characteristics of force.
4. Define the terms
a.
Resultant of a force
b.
Equilibrium of force
c.
Polygon law of forces
5. Define
a.
Resultant – of a system of forces
b.
Equilibrium of system of forces
6. State the conditions for equilibrium of rigid body
7. Define
a.
Co – planar forces
b.
Non- coplanar forces.
c.
Collinear forces.
Essay Answer Type Questions
1. Find the magnitude and direction of a resultant of two forces 60 N and
80 N acting at a point with an included angle 500 between them. The force 80 N
being horizontal
Ans: R = 127.164 r=28.810
2. Find the resultant of the force as shown in figure
Ans: R = 41.63 r =780.101
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279
3. Determine the magnitude and direction of the resultant of system of
forces as shown in figure.
Ans: R = 444.70 r =45.240W
4. Find the magnitude and direction of the resultant of the system of coplanar forces as shown in fig.
Ans:R=31.03KN r=470.401With X-axis
5. Calculate the magnitude and direction of the resultant of a system of co
–planar forces given below
i)
200 N inclined 300 to north of the east
ii)
250 N towards north
iii)
300 N north 450 west
iv)
350 N included 400 to south of west.
Ans: R= 456 r= 47.670 with x-axis in II quadrant
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Building Construction and Maintenance Technician
6. A telephone pole has five wires radiating from in IInd Floor Quadrant
the top of the pole, producing the following concurrent pulls.
i)
260 N due West
ii)
180 N due East
iii) 170 N at 300 East of North
iv) 240 N due North west
v) 200 N due South west
Find the resultant pull in magnitude and direction
Ans: R=352.87N r= N60.170W
7. Find the magnitude of two forces, such that if they act at right angles,
their resultant is  but if they act at 600, their resultant is 
Ans: 3N and 1N
3
UNIT
Centroid & Moment of
Inertia
Learning Objectives
After studying this unit, the student will be able to
• Know what is centre of gravity and centroid
• Calculate centroid of geometric sections
Centre of Gravity
Centre of Gravity (or) mass centre of a point in the body where entire mass
   weight – is assumed to be concentrated. In other words, it is a point in the
body, through which the resultant of the weights of different parts of the body is
assumed to be acting. It is generally written as C.G.
Centroid:
The plane figure like triangle, rectangle circle etc have only areas and mass
is negligible. The centre of area of such plane figures is called ‘Centroid’ (or)
“Centre of Area”. It is generally denoted by “G”
Centroidal Axis
The axis which passes through centre of gravity (or)
centroid is known as “Centroidal Axis” XX1, YY1, ZZ1
are called Centroidal Axis
Fig 3.1
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Building Construction and Maintenance Technician
Axis of Symmetry
Axis of Symmetry is the line dividing the figure into two equal parts like
mirror images the centroid always lies on the axis of symmetry.
Fig 3.2
A figure may contain one (or) more axis of symmetry. If there are more
axis of symmetry the cntroid lies at the intersection of axis of symmetry
Fig 3.3
Position of centroids for Standard Geometric Sections.
S. No
1
Name
Rectangle
Shape of figure
centroid
Position of
At int er sec tions of
Diagonals
L
x
2
B
y
2
ALXB
At int er sec tions of
Diagonals
2
Triangle
H
3
1
A  BH
2
y
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283
At int er sec tions of
Diagonals
3
Parallelogram
4
Circle
5
Semicircle
L
2
B
y
2
At int er sec tions of
x
Diagonals
D
x   Radius
2
D
y   Radius
2
D2
A
2
h 2a  b
(
)
3 a b
h
y1  ( 2aa bb )
3
h
A  (a  b)
2
y
6
7
Trapezium (sloping
on both sides)
Trapezium (One side
is vertical and
other side is sloping)
x
a 2  ab  b 2
3 (a  b)
A
h
(a  b)
2
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Building Construction and Maintenance Technician
Centroid of Composite sections
A composite section is a combination of simple regular shapes as rectangle,
triangle circle, semi circle etc. For determining the centroid of composite sections,
the entire area is divided into two (or) more regular simple shapes, Then the
principle of moments is applied to determine the centroid.
Centoid of plane figure having hollow Portion
The Centroid of plane figure having hallow portion is determined similar to
the composite sections by applying principle of moments, However the negative
sign is taken into consideration of hollow positions which are enclosed in a
regular shape.
Sections Symmetrical about both X and Y axes
Fig 3.4
Sections Symmetrical about – horizontal axis (XX)
Fig 3.5
Sections Symmetrical about the vertical axes (YY)
Fig 3.6
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285
Sections un symmetric about – the both axes (X-X,Y-Y)
Fig 3.7
Methods of determination of centroid
The following three methods are available to locate the cntroid of an area.
1. Analytical method
2. Graphical method
3. Experimental method
Analytical method for location of the centroid
Principle: The sum of the moments of a system of a coplanar forces about
any point in the plane is equal to the moment of their resultant about the same
point.
Fig 3.8
Consider a lamina in area “A” divided into number of elementary areas A1,
A2, A3, ….etc as shown in fig. 3.8.Let the centroids of these elementary areas
be at a distance of x1, x2, x3….. etc from vertical axis and y1 , y2, y3 from the
horizontal axis.
Let the centroids of the total area “A” is at a distance of x and y from
vertical and horizontal axis respectively. As per the principle of moments, the
sum of moments of all the elementary areas about horizontal axis OX is equal to
the moment of the total area about the same horizontal axis i.e OX.
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Building Construction and Maintenance Technician
y
  A1y1 
A
where A=A1+A2 +A3 +….
Similarly taking moments of areas about vertical axis i.e. OY
x 
  A1x1 
A
The terms  A1 y1 &  A1 x1 are know as First movement of area
about y-axis and x-axis respectively
First moment of area: The First moment of area about a line is the product
of area and the perpendicular distance of its centroid from the given line.
Important – Note
1. If the axis passer through the centroid, the moments of areas on one
side of the axis will be equal to the moments of areas on the other side of the
axis.
Example 3.2
Locate the centroids if the trapegezium as shown in figure 3.09
Fig 3.09
Solution
Dived the trapezium into rectangle of size a x h and triangle of base (b-a)
and height – “h”
Area of rectangle (1) A1 = a.h
1
(b-a)h
2h
Total area of trapezium A   a  b 
2
Area of rectangle (2) A2 =
Let the centroid of the trapezium be at a distance y above base and x
from last vertical side. Centroidal distance of rectangle from A i. e , x1  a
2
Paper - III Engineering Mechanics
287
ba
3
3a  b  a 2a  b


3
3
a 1
 2a  b 
ah     b  a  h 

A1x1  A 2 x 2
2
2
 3 
x 
  
h
A1  A 2
a  b
2
h  2  b  a  2a  b  
a 

2
2
2
2 
3
  3a  2ab  b  2a  ab

h
3a  b 
a  b
2
2
a  ab  b 2

3a  b 
Centroid distance of triangle from A i. e ; x 2  a 
Similarly
y
A1y1  A 2 y 2
A1  A 2
y1 
h
h
, y2 
2
3
h
h h
h h

ah     b  a   
ah   b  a  

2
2
3
 3  2

y  
h
h
a  b
a  b
2
2
3ah  bh  ah 2ah  bh h  2a  b 


 

3a  b 
3a  h  3  a  b 
centroid x 
a 2  ab  b 2
h  2a  b 
,y 
3a  b
3  a  b 
Example 3.3
Locate the position of centroid of lamina in fig 3.10
Solution
Y.Y Axis as symmetry centroid lies in this axis divide the section in to a
square and a triangle A1 = 100x100 = 10,000mm
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Building Construction and Maintenance Technician
y1 = 100  50mm
2
1
A 2   100  60  3000mm 2
2
0
60
Y2 100 
120 mm above base
3
Total area = A1 +A2 = 10,000+3000 = 13000 mm2
Let
Fig 3.10
be centroid distance from base
y = A1y1+A2y2 = 10000(50) + 3000(120)
13,000
A1+A2
=
500000 + 3,60,000
13,000
= 8,60,000
13,000
= 66.15 mm
The centroid of the lamina is 66.15 mm above the base.
Example 3.4
A trapezoidal lamina has uniform batter on both sides. Its top width is 200
mm. bottom width is 300mm and height is 600 mm. determine position of centroid
from base.
Solution
Top width a = 200 mm
Bottom width b = 30 mm
Height
h = 600 m
Fig 3.11
y = Position of centroid from the base
Paper - III Engineering Mechanics
y = h
3
289
=
(2a+b
a+b )
=
=
+ 300
( 2x200
200 + 300 )
200 (400 + 300 )
500
600
3
280 mm
Example 3.5
Find the centroid of the following T – section
Solution
Fig 3.12 shows T- Sections Y-Y axis as symmetrical axis. In this axis only.
Taking base X-X axis as reference line.
Fig 3.12
Dividing the “T” section in to two rectangles areas. (flange +web)
Area of rectangles flange A1 = 100x10 = 1000 mm2
y1 = 140 + 10 = 145 mm from the bottom of
2
the base
Area of rectangle web (2) A2 =140x10 = 1400 mm2
y2 = 140 = 70 mm from the bottom of the
2
web
X.X axis
distance of centroid from bottom of web XX i. e.,
y
A1y1  A 2 y 2 1000  145  1400  70

A1  A 2
1000  1400
98000  145000
 101.25mm
2400
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Building Construction and Maintenance Technician
Example 3.6(imp)
Find the position of the centroid of an I section given.
Top angle : 60 x 20 mm
Web
: 20 x 100 mm
Bottom angle : 100x20 mm
Solution
Fig 3.13 shows given I sectrim Y-Y axis as axis of symmetry so centroid
lien in this axis only.
we can find y
X-X axis is base of the bottom taken as reference line. Dividing I section in
to three rectangles.
A1 = 60 x 20 =1200 mm2
y1 = 20 + 100 + 20 = 130 mm from the base of the
2
bottom flange
Area of rectangle (1)
Area of rectangle (2) A2 = 100 x 20 = 2000 mm2
y2 = 20 + 100= 70 mm from the base of the
2
bottom flange
Area of rectangle (3)
A3 = 100 x 20 = 2000 mm2
20
Y3 =
= 10 mm from the base of the
2
botttom of the flange
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291
Example 3.7(imp)
A masonry dam of the trapezoidal section with one face is vertical. Top
width of dam is 3m, bottom width of dam is 6m and height is 6m. Find the
position of centroid.
Solution
(i) Applying for multa
Top width
a = 3 mt
Bottom width
b = 6 mt
Height of the dam
∴
and
= 6 mt
one face is vertical.
Fig 3.14
Let centroid of the dam be at a distance
turn use vertical face centroid.
x =
=
Y
about base
a2 + ab + b2
3(a+b)
32 + 3 x 6 + 6 2
3(3+6)
9+18+36
27
= 2.33 mt
=
h
3
(2aa+b+ b)
2x3+6
= 6 (
3+6 )
3
12
= 2( 9 )
y =
= 2.67 mt
IInd method
Trapezium OBCD is divided in to tow simple areas
1. Rectangle OLCD
2.
Triangle CLB
x
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Building Construction and Maintenance Technician
There is no axis as axis of symmetry line.
We can find both x and
y
∴
Finding x Vertical face OD as resume line.
For Rectangle OLCD area. A1 = 3x 6 = 18 m2
Xl = = 1.5. from vertical face OD.
For triangle CLB
Area
3
1
 3  6  9m 2
2
3
x 2  3   4m
3
A x  A 2 x 2 18  1.5  9  4
x 1 1

 2.33 m
A1  A 2
18  9
A2 
form vertical face.
Finding y Base of the dam OB taken as reference line.
Area of rectangle (1)
A1 = 18m2
y1 
Area of triangle (2)
y
6
 3mtfrom the base of the dam
2
A2 = 9m2
A1y1  A 2 y 2 18  3  9  2

 2.67m from the base
A1  A 2
18  9
of the dam
Example 3.8 (imp)
Determine the centroid of the channel section 200 x 100 x 10 mm as shown
is fig 3.15
Solution
Fig 3.15 shown the given channel section. X-X axis as axis of symmetry
line is this only centroid lies it.
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293
Fig 3.15
Finding only taking vertical outerface AB as reference line dividing the
section in to three rectangles
Rectangle (1) AreaA1 = 100  10 = 1000 mm2
x1 
Area of Rectangle(2)
100
 50mm from vertical face AB
2
Area A2 = 180 10 = 1800 mm2
x2 = 10 = 5mm from vertical face AB
2
A3 =100 10 = 100mm2
Area of rectangle (3)
x3 
x
100
 50mm 2 from vertical
2
face AB
A1x1  A 2 x 2  A 3 x 3
A1  A 2  A 3
1000  50  1800  5  1000  50
1000  1800  1000
50000  9000  50000

3800
 28.68mm From the vertical face AB

Example 3.9
Find the position of centroid for an angle of section from base as shown in
fig. 3.16
Solution
Fig 3.16 shows given angle section there is no X-X and Y-Y axis are axis
of symmetry.
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Building Construction and Maintenance Technician
We have to find both x and
Finding x
20
y
Vertical face AC as reference line.
120
20
Dividing the angle section as two rectangular
areas.
Areas of Rectangle 1
2400 mm2
x1 
120
Fi.g 3.16
A1 = 120 x 20 =
20
 10mm from vertical face AC
2
Area of Rectangle 2
A2 = 100 x 20 = 2000 mm2
100
 70mm from vertical from AC
2
A x  A 2 x 2 2400  10  2000  70 24000  140000
x 1 1


A1  A 2
2400  2000
4400
x 2  20 
y
= 37.27mm from vertical face AC.
Finding
Bottom AB as axis of reference.
120
 60 mm from bottom base AB
2
20
y 2  10 mm
2
from the bottom base AB
y1 
y
A1y1  A 2 y 2 2400  60  2000 10

A1  A 2
2400  2000
= 37.27 from the bottom base AB.
Example 3.10
Determine the centroids of the selection shown in figure 3.17
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295
Solution
100
20
Figure 3.17 shows selection has
80
y of
no X-X and Y-Y axisx as axis
symmetry. So both and
20
200
80
can be
20
100
determined.
x
Fig. 3.17
Finding
CD line vertical face taken as axis of reference line.
Dividing given Z selection in to three rectangular areas.
Area of rectangular
A1 = 80 x 20 = 1600mm2
x1 
Area of rectangle (2)
80
 40 mm (from vertical face CD)
2
A2 = 220x20 = 4400 mm2
x 2  80 
20
 90 mm from vertical face CD
2
Area of rectangle (3) A3 = 80 x 24 = 1920 mm2
x 3  100 
x
80
140 mm from the vertical face CD
2
A1x1  A 2 x 2  A3 x 3 1600  40  4400  90  1920 140

A1  A 2  A 3
1600  4400  1920

64000  396000  268800
7920
= 92.02 mm from the vertical face CD
Finding y
Bottom base AB as axis of reference
20
y1  200   210 mm from bottom base AB
200 2
y2 
110mm from bottom base AB
2 2
y3 
 12 mm
2
from bottom base of AB
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Building Construction and Maintenance Technician
y
A1y1  A 2  y 2  A 3 y3 1600  210  4400  110  1920  12

A1  A 2  A 3
1600  4400  1920
336000  484000  23040
7920
843040

7920

= 106.44 mm from bottom base AB
Example 3.11
Find the cenrodidal distance for the built up section shown in figure 3.18
Solution
Figure 3.18 shown Y-Y axis ofy symmetry
centroid lies in it we can find
y
Fig 3.18
Finding
Bottom most layer AB line a as axis of reference.
Built up section has divided in 5 rectangular areas.
Rectangular (1)
A1= 100 x 10 = 1000mm2
y1 10  20  150  20 
Rectangular (2)
A2 = 100 x 20 = 2000mm2
y 2 10  20  150 
Rectangular (3)
Rectangular (4)
150
 105mm
2
A4 = 20 x 200 = 4000mm2
y 4 10 
Rectangle (5)
20
 190 mm
2
A3 = 150 x 20 = 3000 mm2
y3 10  20 
20
 20 mm
2
A5 = 10 x 200 = 2000mm2
y5 
10
 205mm
2
10
 5mm
2
Paper - III Engineering Mechanics
297
= 82.5mm from bottom base AB
Review Questions
Short Answer Type Questions
1. Define centre of gravity
2. Determine the center of gravity and centroid
3. Locate the position of centroid of the following figures with a neat sketch
1) rectangle 2) triangle 3) circle 4) Semi circle
4. Find the centroid of triangle of base 80 mm and height 120 mm from the
base and the apex
Essay Answer Type Questions
1. A masonry dam is trapezoidal in section with one face vertical. Top
width is 3m and bottom width is 10 m height is 10 m. Find the position of
centroid axis
Ans. x 3.564m and y = 4.260 m
2. Determine the centre of gravity of I section having the following dimensions
Bottom flange = 300x100mm
Top flange
= 150x50mm
Web
= 50x400mm
Ans. 198.9mm from bottom flange
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Building Construction and Maintenance Technician
3. Find out the centroid of an un equal angle section 100mm x 80mm x
20mm
Ans x = 25 mm from left face
y = 35mm from bottom face
4. Find the centre of gravity of channel section 100 x 50 x 15 mm
Ans x= 17.8 mm from outer face of web
5. Find the centroid of the given “T” section
Top flange of
250mmx50mm
Web
50mmx200mm
Ans:
y = 169.44mm from bottom of the web.
6. Find the centroid of the section shown in figure
Ans:
x = 95.56mm from the left edge
y = 85.55mm from the bottom edge
Fig 3.19
Paper - III Engineering Mechanics
299
Moment of Inertia (M.I)
Definition
The product area (A) and perpendicular distance (x) between the point is
know as “the first moment of area (Ax) about the point. If this moment is again
multipled by the distance ‘x’ i.e. Ax.x = Ax2 is called moment of moment of area
(or) the second moment of area or simply moment of inertia. Its unit in SI system
is mm4.
Moment of inertia for some regular geometrical sections
Position of centroids for Standard Geometric Sections.
S. No
Name
1
Rectangle
2
Hollow
Rectangle
3
Solid Circular
section
4
Hollow Circular
section
5
Triangle
Shape of figure
MI about
XX(Ixx)
BD32
12
BD3 - bd 3
12
12
D 4
64
MI about
yy(Iyy)
DB32
12
DB3 - db3
12
12
D4
64


(D 4  d 4 )
(D 4  d 4 )
64
64
bh 3
36
bh 3
12
about cg about base
BC
300
Building Construction and Maintenance Technician
Parallel Axis Theorem
It states that if the moment of inertia of a plane area about an axis through
its centre of gravity (IGG), is shown in fig 3.20. Then the moment of inertia of the
area about an axis AB parallel to IGG at a distance of “h” from centre of gravity
is given by IAB = IGG + Ah2
Fig 3.20
Where
IAB = M.I of the area about an axis AB
IGG = M. I of the area about its C.G
A = Area of the section
h = distance between C.G of the section and the axis AB.
Radius of Gyration
Radius of gyration about a given axis is defined as the effective distance
from the given axis at which the whole are may be considered to be located with
respect to axis of rotation. It is denoted by “k” or “r”
I = Ak2 (or) Ar2
Where
I = moment of inertia
K(or) r = radius of gyration k (or) r =
I
A
A= area of cross section
Units for k or r in S.I system is mm
Perpendicular axis theorem
It states that if IXX and IYY be the moment of inertia of plane section about
two perpendicular axes meeting at “o” shown in figure 3.21 then, the moment of
inertia IZZ about the axis Z Z which is perpendicular to both XX and YY axises,
is given by
IZZ = IXX + IYY
Paper - III Engineering Mechanics
301
For symmetrical section like circular IXX = IYY
Fig 3.21
IZZ = IXX + Ixx
IZZ = 2IXX
J = 2 I where “J” is known as polar moment of Inertia
Polar moment of inertia.
Definition :- The moment of inertia of an area( IZZ ) about an axis
perpendicular to its plane is called “polar moment of Inertia”. It is denoted by
“J”
3
I XX
BD3 400   800 


1.706  1010 mm42
12
12
Solved Problems
Problem 3.12
Find the moment of inertia of a rectangular
section 400mm wide and 800mm deep about
its base.
Solution
Breadth of bearn B = 400mm
Depth
of beam D = 800 mm
Fig 3.22
M.I . about C.G i.e.
M. I about its base I AB  I  Ah 2
IAB = 1.706 x 1010 + (400x 800) (400)2
= 1.706 x 1010 + 5.12 x 1010
= 6.826 x 1010 mm4
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Building Construction and Maintenance Technician
Problem 3.13
Find the M.I of hollow circular sections whose external diameter is 60mm
and internal diameter is 50mm about Centroidal axis
Solutions
External dia D = 60 mm
Internal dia d = 50mm
IXX  I YY 

D4  d 4 

64

60 4  504 

64
 329.2 mm 4

Fig 3.23
Moment of inertia about Centroidal axis is = 329.2 mm4
Problem 3.14
Find the moment of inertia of a rectangle 60mm wide and 120mm deep
about Centroidal axis. Find also least radius of gyration.
Solutions
B = 60mm
D = 120mm
M. I about Centroidal axis
Fig 3.24
Area of rectangle A = BD = 60 x 120 = 7200mm2
Least radius of gyrations k (or) r =
IcG
=
A
8.64 x 106
7200
Paper - III Engineering Mechanics
∴
303
Least radius of gyration = 34.64mm
Problem 3.15
Find the radius of gyration of hollow circular sectors of external diameter
300mm and internal dia 200mm.
Solution
External dia D = 300mm
d = 200mm

Area
 2
D  d2 

4
Fig 3.25

=  300 2  2002   3.927  10 4 mm 2
4
Radius of gyration K 
I
3.191 108

 90.14mm
A
3.927  104
Alternate method K 

D4  d4 

D2  d 2
300 2  2002
64



4
4
D2  d2 

64
= 90.14mm
Problem 3.16
Find the radius of gyration of a triangle whose base is 40mm and height is
60mm about an axis passing through C.G and parallel to base.
Base
b = 40mm
H = 60mm
Area
=
1
bh
2
M.I of triangle about Centroidal axis
Fig 3.26
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Building Construction and Maintenance Technician
IXX 
bh 
Radius of gyration K =
36 ∴
Ixx
A
bh 3 2
h


36 bh
18
60
K 
14.14mm
18
K
Problem 3.17
Find the moment of inertia about Centroidal axis of hollow rectangular
sections shown in fig 3.27
Solution
B = 200mm
D = 400mm
b = 100mm
d = 200mm
Fig 3.27
M.I about XX axis for hollow rectangular sections.
Ixx =
1 [ 200 x 4003 - 100 x 2003]
12
= 1000x106 mm4
M. I about Y Y Axis for a hollow rectangular section
IYY
DB3 db 3 1


  400  2003  200  1003 
12
12 2
= 250 x 106mm4
Problem 3.18
Determine the position of centroid and calculate the moment of inertia about
its horizontal centroidal axis of a T – beam shown in figure 3.28
Solution
Paper - III Engineering Mechanics
305
Finding Centroid
YY axis is axis of symmetry centroid lies on it.
To Finding y
Fig 3.28
Take AB line axis of reference.
Dividing T section into two rectangular areas
A rea of rectangle (1) A 1 = 300 x 100 = 30000mm2
y1  200 
100
 250mm from bottom base AB
2
A2 = 200 x 100 = 2000mm2
Area of rectangle (2)
y2 
200
100mm from bottom base AB
2
Centroidal distance y from bottom 

A1y1  A 2 y 2
A1  A 2
30000  250  20000  100
30000  20000
= 190mm from bottom base AB.
M. I of a rectangle 1 about centroidal axis
IXX at (1) = IG + Ah12
300  1003
2

 300  100  y  y 
12
2
 2.5 107  300  100  250  190   h1  y1  y 
 2.5 107  300  100  602
= 2.5 x 107x108 x 106 = 25 x 106+108 x 106
= 133 x 106 mm4
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Building Construction and Maintenance Technician
M. I of a rectangle 2 of about Centroidal axis
I @ 2  IG  Ah 22
100  2003
 100  200  90 2 since h2 = y- y2
12
h2 = 190-100
 6.67  107  162  106

 66.7  106  162  106
= 90mm
 228.7 106 mm 4
Moment inertia of T – beam about its Centroidal axis
 I at (1)d  I at(2)
133  106  228.7  106
 361.70  106 mm 4
Problem 3.19
An un symmetrical I section has top flange 100x20mm web 100 x 120mm
and bottom flange 80x20 mm over all depth is 160mm.
Calculate centroid
Solution
Figure 3.30 Shows given I section.
YY-axis is axis of symmetric line
so centroid lies on it.
Finding y
Take line AB, passing through the bottom edge as axis of reference
Divide the section into three rectangular areas.
Area of rectangle (1)
A 1 = 100 x 20 = 2000 mm2
y1  20  120 
20
 150mm from base
2
Paper - III Engineering Mechanics
A2 = 10 60
x 120 = 1200mm2
Area of rectangle (2)
y 2  20 
Area of rectangle (3)
Ixx
y3 
307
120
2
 80mm from base
A3 = 80 x 20 = 1600mm2
20
 10mm from the base
2
A1 y1  A 2 y 2  A 3 y3 2000 150  1200  80  1600 10

A1  A 2  A 3
2000  1200  1600
300000  96000  16000

 85.83mm
4800
y
from the base AB.
Finding M.I of “I” section about X-X axis about centroid
M.I of rectangular (1) about X-X axis
I xx @ I  IG1  A1h12  h1  y1  y 
h1 = 150-85.83 = 64.17
100  203

 100  20  64.17 
12
 8.3106 mm 4
M. I of rectangle (2) about a X-axis
I xx at 2  IG 2  A 2 h 22
h2 = y - y2 = 85.83-50 = 5.83mm
10  1203
2
 10  120  5.83
12
1.48  106 mm 4

M. I of rectangle (3) about X-X axis
I xx at 3  I G3  A 3h 2
h3 = - y3 = 85.83 - 10 = 75.83 mm
80  203
2
 80  20  75.83 
12
 9.25  106 mm 4

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Building Construction and Maintenance Technician
Moment of inertia of given I section about X axis
 I xx at1  I xx at 2  I xx at 3
 8.3  106  1.48  106  9.25  106
19.03  106 mm 4
Problem 3.20
Determine the moment of inertia of the un equal angle section of size 150mm
x 100mm x 25mm about Centroidal axis.
Solution
Finding centroid
Finding
x
Vertical face CD has axis of reference, dividing L section has two
rectangular areas.
A1= 125 x 25 = 3125 mm2
Area of rectangle 1
x1 
Area of rectangle 2
25
12.5mm from vertical face CD
2
A2 = y100 x 25 = 2500mm2
100
 50mm from vertical face CD
2
A x  A 2 x 2 3125  12.5  2500  50
x 1 1

A1  A 2
3125  2500
x2 
39062.5  125000
5625
164062.5

5625

= 29.17mm from vertical face CD
Finding y
Bottom base AB has taken as axis of reference
125
y1  25 
 87.5mm from base
2
25
y 2   12.5 from base
2
Paper - III Engineering Mechanics
309
A1y1 + A2y2
3125 x 87.5 + 2500 x 12.5
A1 + A2 =
31.25 + 2500
y =
= 273437.5 + 31250
5625
304687.5
=
5625
= 54.17mm from the base AB.
Finding I xx
M. I of rectangle (1) about x-x axis
25  1253
2
 25  125  33.33
12
 4.07  106  3.47  106
= 7.54  106 mm4
I xx at1  IG1  A1h12 
y
h1 = y - y1
=54.17-37.5
= 33.33mm
M I of rectangle (2) about x-x axis
I xx @.2  I G2
h2 = y - y2
100  253
 A 2h 
 100  25  41.67  =54.17-12.5
12
2
2
= 41.67mm
= 0.13 x 106 + 4.34 x 106
= 4.47 x 106 mm4
Moment Inertia of given angular section about X-X axis
 I xx at I  I xx at2
 7.54  106  4.47  106
12.01  106 mm 4
Finding IYY
M I of rectangle (1) about Y-Y axis
IYY at 1 = IG1 + A1h12
h1 = x - x1
= 29.17 - 12.5
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Building Construction and Maintenance Technician
DB3
2

 A1  x  x1 
12
150  253
2

 3125  29.17  12.5 
12
 0.195  106  0.868 106
1.063  106 mm 4
M.I Rectangular (2) about Y-Y axis
I YY at 2  I G2  A 2 h 22
h2 = x2 - x = 50 - 29.17 = 20.83
25  1003
2
 2500  20.83
12
 2.08  106  1.08  106

 3.16 106 mm 4
M. I of a given angular section about Y-Y axis
 I YY at I  I YY at2
1.063  106  3.16  106
 4.223 106 mm 4
Review Questions
Short Answer Type Questions
1. Explain
a) Parallel axis Theorem
b) Perpendicular axis theorem
2. Define the termsa) Moment of inertia
b) Radius of gylation
3. Find the radius of gyration of circle having diameter “d”
Ans:
d
4
4. Find the radius of gyration of hollow circular plate of 60mm inner diameter
and 100mm outer diameter
(Ans:29.15mm)
Paper - III Engineering Mechanics
311
5. Find M.I of a rectangular section 200mm width and 400mm depth about
the base
(Ans. 4.267 x 109mm4)
Essay Answer Type Questions
1. Find the moment of Inertia of a T Section having flange150mm x 50mm
and web 50 x 150mm about xx and yy-axis through the C. G of the
section.
[ Ans: Ixx = 53.125 x 106 mm4
IYY =15.625 x 106mm4]
2. Determine the moment of Inertia of an unequal angle section of size
100mm x 80mm x 20mm about Centroidal axis
[ Ans: Ixx = 2.907 x 106 mm4
IYY =1.627 x 106mm4]
3. Determine the moment of inertia of an I section about XX axis given that
top flange 100mm x 10mm web = 200mm x 10mm different flange
160mm x 10mm
[Ans: Ixx = 34.38 x 106 mm4]
4. A built up section is formed by an I section and to flange plates of size
280 x 20mm are an each flange find the moment of inertia about
centrodial X-X axis as shown in below figure
[Ans: Ixx = 188.22 x 106 mm4]
Key Concepts
1. The C.G of a body is the fixed point at which its weight is assumed to be
concentrated.
2. The centroid of a surface is the fixed point at which the area of the
surface is assumed to be concentrated.
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Building Construction and Maintenance Technician
3. The centroid of a surface is determined from the equations:
x
A1x1
A1y1
and y 
A
A
4. The centroid of a composite area is treated by the principle of moments,
dividing it into regular simple figures.
5. The M.I of an area about a given axis is the sum of the values of “ax2”
where “a” is the area of each element and “x” is the distance of the
centroid of the element from the given axis
I  ax 2
6. Radius of gylation (Kxx) of an area about given axis is the distance from
the axis at which the area may be assumed concentrated to given the M.
I of the area about the given axis
K  
I
A
7. Parallel axis theorem :- if “XX” is an axis is parallel to the centrodal axis
C.G of surface of area A and if “d” is the distance between the two
parallel axis.
I  I CG  Ad 2
8. Perpendicular axis theorem: If XX and YY are two perpendicular axis
is the plane of the area and ZZ is an axis perpendicular to both of them
through their intersection.
Izz  I xx  I YY
9. The M.I about an axis perpendicular to its plane is known as its polar
M.I
10. M.I of a built up section = Sum of M.I of all elements of the section
about the same axis.
11.M.I of a rectangle bxd about axis through centroid parallel to
side b 
bd 3
12
12. M. I of
a triangle ‘bxh’ about axis through centroid parallel to base
3

bh
36
13. M. I of circle of dia ‘d’ about any diameter 
 d4
64
Paper - III Engineering Mechanics
313
14. M.I of hollow circular section of diameters “D” and ‘d’ about any

dia 
D4  d 4
64


15. Polar M.I of a solid shaft of dia ‘d’ about axis 
 4
d
32
16. Polar M.I of hollow shaft of dia of diameter ‘D’ and ‘d’
=

D4  d 4 

32
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Building Construction and Maintenance Technician
UNIT
4
Simple Stress and Strains
Learning Objectives
When an engineer under taken the design of a structure, it is essential that
he should have the concept of various forces acting on the structure. The main
objective of a civil engineer is to design a safe and feasible (most economical)
structure. Hence adequate knowledge of the properties of materials and their
behavior under various external loads is essential. When an external force acts
on a body, if body tends to under go some deformations. The effect of forces of
bodies are to be studied.
The properties of the materials and their behavior under load are explained
in this chapter a few tests assess the performance of the materials are also
explained.
When a body is subjected to a system of external loads, it undergoes deformation. At that time it offers resistance against the deformation. The internal
resistance exerted by the body to resist the applied load or force is termed as
stress. In other words it is defined as the force acting on unit area of crosssection.
External force
p
Stress 

Cross sectional area A
2
Units are N m or kN m 2 or N mm 2
According to Nature of Stress. It can be classified as
1) Tensile Stress 2) Compressive Stress 3) Shear Stress
Paper - III Engineering Mechanics
315
Tensile Stress: When an external force produces increase in length of the
body then that force is called as tensile force or the body is in tension or pulling
force.
Then the stress developed in the cross section of the body is called tensile
stress and is denoted by f t .
Compressive Stress: When an external force causes shortening of the
body then that force is called compressive force or the body is under compression or impushing force.
The stress developed in the body due to compressive force is called as
compressive stress. It is denoted by f c .
Shear Stress: The tangential force acting along the section of the body
then that force is called shear force.
The stress in the section due to shear force is called as shear stress. It is
denoted by fs .
Due to this force there is no increase or decrease in length. But there is
change in shape.
Strain: It is a measure of the deformation produced by the application of
the external forces.
Change in dimension
Strain ‘e’ = Original dimension
This strain is three types.
1) Tensile Strain
2) Compressive Strain
Increase in length
Tensile Strain et 
Original length
3) Shear Strain
Derease in length
Original length
Shear Strain es = It is a measure of the angle through which a body is
destorted by the applied force.
ds
Shear Strain 
 tan 
L
where  = Radian
Compressive Strain ec 
Volumetric Strain ev 
Change in volume v

Original volume
v
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Building Construction and Maintenance Technician
Mechanical Properties of Materials
1. Elasticity
2. Plasticity
3. Ductility
5. Malleability 6. Stiffness 7. Hardness
4. Brittleness
8. Toughness
9. Creep 10. Fatigue
1. Elasticity
The property of the material by which a body returns to its original shape
after the removal of external load is called Elasticity.
If the body regains completely its original shape, then it is said to be perfectly elastic.
Ex: Rubber, Steel and Mild Steel, Copper, Aluminium
2. Plasticity
It is the property of the material by which the material undergo permanent
deformation. That means it fails to regain its original shape after removal of load.
Ex: Gold, Lead, Copper
3. Ductility
It is the property of material by which the material can be drawn into thin
wires after under going a considerable deformation without rupture.
Ex: Mild Steel, For steel, Silver, Aluminium etc.
ex , ey , ez
4. Brittleness
It is the property of material by which it breaks without much deformation.
Ex: Glass, Chinaware, Concrete Rock Materials, Cast Iron etc.
Volumetric Strain
The volumetric strain is the algebric sum of all the linear (or) axial strains if
are the strains in three mutual perpendicular directions, then
Volumetric straing, e v  e x  e y  e z
When a solid cube is subjectred to equal normal forces of the same type
on all forces, wil have e x  e y  e z equal in value.
Then the volumetric strain equal to three times the linear strain e v  3.e .
Paper - III Engineering Mechanics
317
5. Malleability
It is the property of material by which it can be beaten or rolled into thin
sheets without rupture.
Ex: Copper, Ornamental gold, Ornamental silver, Wrought iron.
6. Stiffness
It is the property of material by which it offers resistance to bending action.
Stiffness is the load required to be applied on a body to produce unit
deflection and it is denoted by ‘S’ or ‘K’.
Load
p
Stiffness 'S' 

Deflection 
Ex: Springs
7. Hardness
It is the ability of material to resist impressing scratching or surface abrasion. It is the relative property of material. Every Material will have its own
hardness number.
Ex: Diamond, Graphtic, Talc, Mild Steel etc.
Diamond is the hardest substance and
Talc is the softest substance in nature.
8. Toughness
It is the property of material which enables it to absorb energy without
rupture.
Ex: Brass, Mild Steel
9. Creep
It is the property of material by which it develops the slow deformation
and strain with time due to constant stress.
Ex: Concrete
10. Fatigue
It is the property of material by which the material with stands to varying
and repeating loads.
Ex: Concrete and Prestressed Concrete.
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Building Construction and Maintenance Technician
Stress-Strain Curve of Mild Steel
Fig 4.1 stress-strain curve of Mild Steel
A = Limit of Proportionality
B= Elastic Limit
C= Upper Yield Point
D = Lower Yiled Point
E = Strain Hardenning Point
F = Ultimate Region
G = Breaking Point
Hooke’s Law: The stress is directly proportional to strain within a elastic
limit i.e. upto proportionality limit. In other words, the ratio of axial stress to the
corresponding axial strain is constant.
Proportionality Limit: This is the point upto which the stress is directly
proportional to strain. Hence upto this limit the stress-strain curve is a straight
line.
Elastic Limit: It is the limit upto which the strain produced will dissappear
completly on the removal of load. It means the body gets the original shape
after removal of load. But the stress is not proportional to strain between proportional limit and elastic limit.
Yiled Limit: i) When tensile load further increases stress reaches yield
stress and material starts yielding. Even for a small increase in stress the increase in strain is very large.
Paper - III Engineering Mechanics
319
Stress
Strain curve suddenly falls showing a decrease in stress. The point from
where sudden fall to curve occurs is known as upper yield point ‘C’.
The point upto which the fall of the curve occurs is known as lower yield
point ‘D’.
Sudden stretching of the material at constant stress from lower yiled point
‘D’ to the point ‘E’ is known as Strain Hardening.
The point where the stress is constant from lower yield point is known as
“Strain Hardening Point” ‘E’.
Beyond this, the stress increases with the increase of strain.
The portion of the curve beyond strain hardening represents the strain hardening range.
Ultimate Point: Ultimate load is defined as maximum load which can be
placed prior to the breaking of specimen. Stress corresponing to ultimate load is
known as ultimate stress.
Breaking load: After reaching ultimate stress, stress-strain curve suddenly
falls with rapid increase in strain and specimen breaks. The stress corresponding to breaking point “G” is known as “Breaking Stress”.
1) Linear Strain or Logitudinal Strain: The deformation or change in
length per unit length in longitudinal direction is known as linear strain or longitudinal strain.
Change in length 3l
Linear or Longitudinal Strain 

Original length
l
2) Lateral Strain : When a material is subjected to uni-axial stress within
elastic limit. It deforms not only longitudinally but also laterally. It tensile force is
applied the linear dimensions increase, whereas lateral dimensions decrease.
If compressive force is applied, the linear dimensions decrease where as
lateral dimensions increase.
Change in lateral dimension
Lateral Strain =
Original lateral dimension
For rectangular sections 
Change in width Change in depth

Original width
Original depth
320
Building Construction and Maintenance Technician
are lateral strains these two equal
b n
 .
b
n
For Circular sections
Poisson’s Ratio
1
or  or  .
The ratio of laternal strain to linear strain. It is denoted by
m
Laternal strain
Poissons Ratio 
Linear strain
1
The value of
for elastic materials is 0.25 to 0.33.
m
The value of
1
should not be greater than 0.5.
m
1
for steel lies between 0.25 to 0.30.
m
This poissons ratio is same both in tensile and in compression.
The value of
Volumetric Strain: The change in volume of an elastic body due to external forces per unit original volume is known on volumetric strain and is denoted
by ev .
Change in volume v
 Volumetric Strain 

Original in volume v
Bulk Modulus (K): a) When a body is subjected to uniform direct stress
in all the three mutually perpendicular directions, the ratio of volumetric stress
to the corresponding volumetric strain is found to be constant. This is called
Bulk Modullus ‘K’.
Volumetric Stress
Bulk Modulus ' K ' 
Volumetric Strain
The volumetric stress may be direct stress.
Relationship between Elastic Constants
1
Elastic Constants are , E, C, K .
m
1
where
= Poissons Ratio
m
E = Young’s Modulus or Elastic Modulus
C= Modulus of Rigidity
Paper - III Engineering Mechanics
321
K = Bulk Modulus
1
Case i): Relationship between E, C, and .
m
1

E  2C  1  
 m
1
Case ii): Relationship between E, K and .
m
2

E  3K 1  
 m
Case iii): Relationship between E, C, K.
9KC
E
3K  C
1
Case iv) : Relationship between C, K, .
m
1 3K  2C

m 6K  2C
Solved Problems
Short Answer Type Questions
1) A mild steel rod of 10mm diameter and 300 mm length elongates 0.18
mm under an axial pull of 10kN. Determine the Young’s Modulus of Material?
Ans: Axial load = 10kN = 10  103 N
= 10,000N
Diameter of rod = 10mm
d 2 22 1
   10  10
4
7 4
 78.54 mm 2

Load
10, 000
Stress f  A  C.S. Area  78.54
 127.32 N mm 2
Cross Sectional Area ‘A’ 
Change in length 3l 0.18
 
Original in length l 300
= 0.0006
Stress 127.32
=
Young’s Modulus ‘E’ =
Strain 0.0006
 212200 N mm 2
Strain e 
= 2.12 x 105 N/mm2
322
Building Construction and Maintenance Technician


or

Stress
A  pl

E 
l
l
Strain
A.3

l






2) A wooden size tie 50mm  100mm size is 2 mts long. It is subjected to
an axial pull of 20kN. Find out the elongation of tie. If the modulus of Elasticity
of wood is 1 104 N mm 2 ?
Ans: Axial pull =   20 kN  20 103 N
Cross Sectional Area A  50  100  5000 mm 2
Length of the tie = l  2m  2  1000  2000 mm
Young 's Modulus E  1104 N mm 2
l 20  103  2000

 0.8 mm
Elongation 3l 
AE
5000  1 104
3) A hollow cast iron column carries an axial load of 2000 kN. If the outer
diameter of the column is 30 cm and permissible stress = 8k N cm2 . Findout
the thickness of the column?
Ans: Outer diameter D = 30 cm
Inner diameter d =
Axial load = 2000kN
Permissible Stress = 8k N cm 2

Load
2000

C .S . Area C .S . Area

(D 2  d 2 )
4
D = External dia d = Internal dia
C.S. Area of hollow section =
But D  30 cm, Stress 
2000

 (30)2  d 2 
4
8
Paper - III Engineering Mechanics
323
22 1
2000
   302  d 2  
7 4
5
22
 900  d 2   250
28
 318.18
 900  d   250  22
28
2
d 2  900  318.18  581.82
d  581.82  24.12 cm
But
D  d  2t  2t  D  d t 
Dd
2
30.0  24.12 5.88

 2.94 cm
2
2
Thickness = 29.4 mm

4) A steel rod of 25mm diameter and 600 mm long is subjected to an axial
pull of 40,000N find 1) Intensity of stress 2) Elongation of the rod?
Diameter = 25mm
Length = 600mm
Load = 40,000N
Pull = 40, 000N
Young’s Modulus = 2  105 N mm2
 2 22 1
Cross Sectional Area   d    25  25
4
7 4
 491.07 mm 2
(a) Intensity of Stress
(b) Elongation l 
 40, 000

 81.45 N mm 2
A 491.07
l
40, 000  600

 0.224 mm
AE 491.07  2  105
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Building Construction and Maintenance Technician
E  Yong ' s Modulus 
Linear Stress
Linear Strain
Load

C.S . Area

 A
Change in length
l
Original length
l
 l
l
E  
A  l A l
 l 
l
AE
Direct for  l
(5) A short timber post of rectangular section has are side of cross section
equal to twice the other. When the post is axially loaded with 10kN (Compression). If contracts by 0.0521 mm per meta length. Calculate the cross section
dimensions of the post if ‘E’ for timber  1.5  104 N mm 2 ?
Ans: One side of cross section = t (breadth)
Other side of cross section = 2t (depth)
Cross sectional area A = breadth  depth
= t  2t  2t 2
Axial load pull = 10kn = 10,000N
Change in length  l  0.0521 mm
Original length l  1m  1000 mm
E  1.5 104 N mm 2
l 
l
 0.0521
AE
0.0521 
2d 2 
10, 000  1000
2d 2  1.5  104
1000
 6397.95
3.0  0.0521
Paper - III Engineering Mechanics
325
d  6397.95  79.98  80mm
b  2d  160 mm
Cross Sectional Dimensions are = 80  160 mm
6) A hollow steel column has to carry an axial load of 3.3MN. The yield
stress of steel is 282 N mm 2 . Assessing Factor of Safety 2 For steel. Deter-mine the external and internal diameters required for column section. If the ratio
of internal dia to external dia is to be 0.5?
Ans: Axial load    3.3 mn  3.3  106 N
Mega = 106
N = Newton
Gega = 109
m= meta
Kilo = 103
Paseal = 1 N m 2
Yield Stress of Steel = 282 N m 2
Factor of Safety = 2
Yield Stress
282
Permissible stress 

 141 N / mm 2
Factor of Safety
2
Load 3.3  106 N

N
Stress
141
m2
 23404.25 mm 2
Internal diameter
 0.5
Outer diameter
Cross Sectional Area A 
d
 0.5
D
d  0.5  D
 2
 D  d 2   23404.25
4
 2
 D  (0.5D) 2   23404.25
4
326
Building Construction and Maintenance Technician

 0.75  D 2  23404.25
4
D  199.32  200 mm
Internal dia d  0.5D  0.5  200
 100 mm
Outer dia D= 200 mm
Internal dia d = 100 mm
7) The following observations were made during a tension test on mild
steel bar of 20mm diameter Gange length = 200mm. Extension Gange at load of
31.4kN = 0.1mm, Yield load = 88kN, Ultimate load = 132kN, Breaking load
= 92kN, Total extension = 54mm, Diameter of rod at failure = 14.2mm, Determine a) Young’s modulus, b) Yield Stress, c) Ultimate Stress, d) Breakiung
Stress, e) % Elongation, f) % Reduction in Area?
 2 22 1
d 
  20  20
4
7 4
 314.3 mm 2
Ans: Cross sectional area A 
Load = 31.4 kN  31.4  1000  31400 N
Stress on this load 
Load
31400

 99.9 N mm 2
C.S. Area 314.3
Strain on this load 
Change in length l 0.1
 
 0.0005
Original lengh
l 200
1) Young 's Modulus ' E ' 
Stress
99.9

 1.998  105 N mm 2
Strain 0.0005
= 2.0  105 N mm 2
Yield load 88000
2) Yield Stress  C.S. Area  314.3  279.98  280
= 280.0 N mm 2
3)
Ultimate Stress 
Ultimate Load 132000

 420.0 N mm 2
C.S. Area
314.3
Paper - III Engineering Mechanics
4)
Breaking Stress 
327
Breaking load 92000

 292.7 N mm 2
C.S. Area
314.3
Total Extension
54
 100 
 100  27%
Original Length
200
6) Diameter at failure = 14.2 mm.
5)
% Elongation 
The area at failure i.e. at breaking stress changes i.e. going to be decrease.
Hence reduction in area = Original Area (i.e.) Area of (a1 ) Cross Section
at First - Area of Cross Section (a 2 ).

22

2
2
2
 a2  4  (14.2)  77  (14.2)  158.43 mm  at final or failure


2
 a1  a2  314.3  158.43  155.87 mm
% Reduction in Area =
a1  a2
100%
4

Re duction in area
 100
Original area
155.87
 100  49.6%
314.3
8) A copper bar of 20mm diameter and 300mm long registers an alongation
3
of 0.5mm and decrease in diameter
1 of 8.34  10 mm under a direct tensile
E
&
load of 47.1kN. Determine
of copper?
m
Ans: Direct tensile load = 47.1 kN = 47100 N

Diameter of the rod d = 20mm
  22 1
C.S. Area Sectional Area A  d    20  20
4
7 4
 314.3 N mm 2
Direct Stress
' f '
p
Load
47100


 149.85 N mm 2
C.S . Area C .S . Area 314.3
Elongation =  l  0.5 mm
328
Building Construction and Maintenance Technician
 l 0.5

 1.667  103 mm
 300
Decrease in dia meter  d  8.34  103 mm
Linear Strain =
Futeral Strain 

Change in lateral dimension
Original dimension
Change in dia
Original dia
8.34  103

 4.17  104
20
E
Linear stress
149.85

 89.9  103 N mm 2
3
Linear strain 1.667  10
Poisson ' s Ratio 
Linear stress 4.17 104

Linear strain 1.667 103
4.17  104

 2.50  101
3
1.667  10
2.5
 0.25
10
Young ' s Modulus  89.9 kN mm 2

Poissons Ratio = 0.25
9) A bar of 40  40 mm cross section is subjected to an axial load of 300
N. The contraction was found to be 0.42 mm over a guage length of 170mm
what will be the change in lateral dimension. If poisson’s ratio is 6.3? Find E= ?
Young’s Modulus
Ans:
Cross sectional area = 40  40  1600 mm 2
Axial load p = 300N
Original length l  170mm
Change in length  l  0.042mm
 l 0.042
 0.00247
Linear strain =  
l
170
Paper - III Engineering Mechanics
329
1 Lateral strain
Lateral strain


 0.3
m Linear strain
0.00247
Later Strain  0.3  0.00247 = 0.00072
pl
Change in length  l 
AE
Poisson ' s Ratio
pl
300  170

 75.89 N mm2
A. l 1600  0.42
10) The following results were obtained from tensil test on mild steel specimen.
E 
Diameter of Specimen = 50mm
Gange Length = 250mm
Length of Specimen of Failure = 300mm
Extension of Load of 42.5kN  444 104 mm
Load of Yield Point = 162.2 kN
Max Load = 250kN
Diameter of Neck = 36mm
Factor of Safety = 3
Calculate
a) E = Young Modulus b) Stress at Yield Point c) Ultimate Stress
d) Working Stress e) % Elongation f) % of Reduction in Area?
Ans:
Initial Diameter D = 50mm
Cross Sectional Area 
 2 22 1
d    50  50
4
7 4
 1964.3 N mm2
Original Length l  250mm
Extension of load (42.5kN) is 444  104 mm
 Axial load = 42.5  103  42500 N
Cross Sectional Area A  1964  3mm 2
330
Building Construction and Maintenance Technician
Stress 
Load
42500

 21.64 N m 2
C.S . Area 1964.3
Strain 
 l 444 10 4

 1.776  10 4
l
250
Young 's Modulus E 
Yield Stress 
Stress
21.64

Strain 1.776  10 4
 1.218 105 N mm 2
Load at Yield Po int 162.2 103

Original Area of C.S
1964.3
2
 82.57 N mm
Ultimate Stress 
Ultimate Load 250 103

C.S. Area
1964.3
 127.27 N mm 2
Factor of Safety 
Ultimate Stress
127.27

Working Stress Working Stress
127.27
;
Working Stress
 42.42 N mm 2
3
Percentage of Elongation 
Working Stress 
127.27
3
Change in Length
100
Original Length

(300  250)
 100
250
50
 100  20%
250
Length of specimen increase where as C.S. Area decreases. Due to tensile

test.
Paper - III Engineering Mechanics
331
The length at failure is less than original length.
A1  A 2
 100
A1
Diameter at failure is less than original diameter
% Reduction in Area 
 ( A 2 = Area at Failure) diameter at neck
= 36 mm
 A 2 = Area at Failure 
 2 22 1
d    36  36
4
7 4
22
 36  36
28
 1018.28 mm 2

% Re duction in Area 
A1  A 2
100
A1
1964.3  1018.28
 100
1964.3
 48.16%

11) A bar of 30mm diameter is subjected to a pull of 60kN. The measured
extension over a guage length of 200mm is 0.9mm and change in dia is found to
be 0.0039mm. Calculate a) E = Young’s Modulus b) Poisson’s Ratio
c) Modulus of Rigidity d) Bulk Modulus
Ans: Diameter of Bar = 30mm
Axial pull p  60  103 N  60, 000 N
Length of bar ‘l’ = 200mm; Change in length 3l=0.9mm
Change in dia 3d = 0.0039mm

22 1
Cross Sectional Area  d 2 
  30  30
4
7 4
 707.14 mm 2
Linear Stress 'f ' 
p 60, 000

 84.84 N mm 2
A 707.14
332
Building Construction and Maintenance Technician
Linear Strain 
l 0.9

 4.5  10 3
l 200
= 0.0045
Laterial Strain 
d 0.0039

 0.00013
d
30
Linear Stress
1) 1) Young 's Modulus E  Linear Strain
84.84

 18.853 103 N mm 2
0.0045
2) Poisson 's Ratio
1 LateralStrain 0.00013


m Linear Strain
0.0045
= 0.0288 = 0.029
1

3) E  2C 1  
 m
18.853 103  2  C (1  0.029)
C
18.853  103
 9.16  103 N mm 2
2 1.029
2

E  2K  1  
 m
E
18.853  103
K

2  3 1  2  0.029 

3 1  
 m
6.284  103
K
0.942
 6.67  103 N mm 2
12) A bar of 10mm 10mm size 400mm long is subjected to axial pull of
12kN. The elongation in length and contraction in lateral
1 dimension is found to
be 0.4mm & 0.0025mm respectively. Determine the
, Poissons Ratio, Young
m
Modullus ‘E’, Modulus of Rigidity ‘c’, Bulk Modulus K of the Material?
Ans: L.S. Area  10  10  10mm 2
Paper - III Engineering Mechanics
333
Length = 400mm
Side a = 10mm 
10
 0.01mm
1000
Axial pull  12kN  12 103  12, 000N
Change in Length = 0.4m
Change in dia = 0.0025mm
l 
pl
pl
 E 
AE
Al
E
12000  400
2
 1.2  105 N mm
100  0.4
Poisson 's Ratio 
Lateral Strain
Linear Strain
But Linear Strain 
l 0.4

 0.001
l
400
d 0.0025

 0.00025
d
10
Lateral Strain 
Lateral Strain 0.00025

 0.25
Linear Strain
0.001
C = Modulus of Rigidity
Poisson 's Ratio 
K = Bulk Modulus
1

E  2 1  
 m
C
E
1

2 1  
 m

5.2 105
2 1  0.25 
1.2  105
1.2

 105  0.48 105 N mm 2
2 1.25 
2.5
2
E
1.2 105

E  3K  1    K 

2  3 1  0.25 

 m
3 1  
 m

334
Building Construction and Maintenance Technician
1.2 105
1.2

 105  0.8  105 N mm 2
3 1  0.5  1.5
Poisson’s Ratio = 0.25
K
E  1.2  105 N mm 2
C  0.48  105 N mm 2
K  0.8  105 N mm 2
13) A steel bar of 400 mm length and 50mm  50mm cross sectional
dimensions is subjected to an axial pull of 300kN in the direction of length.
Calculate the volumetric strain change volume if = 0.25 .
Ans:
Length y steel bar l=400mm
Cross Sectional Area of the Bar = 50  50  2500 mm 2
Axial Pull = 300kN = 3,00,000N
Poisson’s Ratio = 0.25
E  2  105 N mm 2
Longitudinal Stress f 
p 300  100

 120 N mm 2
A
2500
If K = Bulk Modulus
2

Then E  3K  1  
 m
5
2  10  3K (1  2  0.25)
2 105
2 105 2 105
K


3(1  0.5) 3  0.5
1.5
5
2
 1.33 10 N mm
Bulk Modulus 
Stress
Volumetric Strain
Volumetric Strain e v 
Stress
K
Paper - III Engineering Mechanics

335
120
 90.22  105
5
1.33 10
Change in Volume
Original Volume
Change in Volume  e v  Original Volume
ev 
90.2 105  50  50  400
C.S. Area  Length
 90200000 105

 902 mm3
Volumetric Strain e v  90  10 5
Change in Volume  902 mm3
14) A steel bar of 240mm length and 20mm in dia meter was stretched by
1.2mm under an axial pull of 32kN. Determine Young’s Modulus and Shear
Modulus take Poisson’s Ratio as 0.25?
Ans: Length of Steel Bar l=2400mm
Diameter d = 20mm
Increase in Length  l  1.2mm
Axial Pull = 32kN = 32,000N
E = ? ; C=? ;
1
 0.25
m
22 1
  20  20
7 4
22  20  20

 314.3 N mm 2
28
Cross Sectioinal Area 
Linear Stress 
32, 000
 101.8 N mm 2
314.3
Linear Strain 
l 1.2

 0.0005
l 2400
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Building Construction and Maintenance Technician
Young 's Modulus 
Linear Stress 101 . 8

 203600
Linear Strain 0.0005
 2.04 105 N mm 2
Poisson’s Ratio = 0.25
Shear Modulus = Rigidity Modulus = ‘C’
1

E  2C 1  
 m
E
2.04 105 2.04 105
C


1  2(1  0.25) 2  (1.25)

2C  1  
 m
2.04 105
 0.816  105 N mm 2
2.50
 0.82 105 N mm 2

15) A rectangular steel bar 60 mm wide and 10 mm thick, 5m long is
subjected to an axial pull of 80kN. If the increase in length under the load is
1.5mm and decrease in thickness is 0.0014mm. Determine three elastic constants of the material and poisson’s ratio. Decrease in width under the load and
change in volume produced?
b = 60mm;
t = 10mm;
l=5m = 5000mm
Cross Sectional area  b  t  60 10  600 mm 2
Axial load = 80kN = 80,000N
Increase in Length = 1.5mm
Decrease in Thickness = 0.0014mm
E, C, K  ?
1
?
m
1) Linear Stress 
Load
80000

 133.33 N mm 2
C.S. Area
600
2) Linear Strain 
Increase in Length
1.5mm

 3  104  0.0003
Original Length
5000mm
Paper - III Engineering Mechanics
3) Young 's Modulus 
4) Lateral Strain 
337
Linear Stress 133.33

 4.44  105 N mm 2
Linear Strain 0.0003
d 0.0014

 0.00014
d
10
Lateral Strain 0.00014

 0.47
Linear Strain
0.0003
Poisson 's Ratio 
If C = Modulus of Rigidity
K = Bulk Modulus
1

E  2C 1  
 m
E
Modulus of Rigidity 'C ' 
1

2C 1  
 m
 1.52 105 N mm 2
Bulk Modulus K 

E
2

3 1  
 m


4.44 105
4.4 105

2(1  0.46) 2  (1.46)
4.44 105
3(0.08)
4.44  105
 18.5  105 N mm 2
0.24
Bulk Modulus K 
ev 
Volumetric Stress 133.33

Volumetric Strain
ev
133.33
 7.207  105
5
18.5  10
But e v 
Changein Volume
Original Volume
 v  e v  Original Volume
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Building Construction and Maintenance Technician
v 
7  207 60  10
5000


 105  216.2 mm3
ev
C.S.Area Length
t 0.0014
b

 0.00014 
t
10
b
b  0.00014  60  0.00084mm
Lateral Strain 
Essay Answer Type Questions
1) The following observations were made during a tension test on the mild
steel bar of 20mm diameter of gange length = 300mm.
Extension at a load of 30kN=0.1mm
Yield Point = 80kN
Ultimate Load = 130kN
Total Extension = 50mm
Diameter of Rod at Failure = 14.1mm
Calculate a) Young’s Modulus
d) % of Elongation
Ans:
b) Yield Stress c) Ultimate Stress
e) % Reduction in Area?
a) 2.56 105 N mm 2 b) 254.65 N mm 2
d) 16.67%
c) 413.8 N mm 2
e) 50.3%
2) A hallow pipe metal pupe 2.5m long is subjected to an axial pull of
300kN. The piple has an interval diameter 250mm assuming ‘E’ for the metal as
0.1 106 N mm 2 . Find the thickness of the pipe. If the elongation of pipe is
0.15mm.
Ans: 52.6mm
3) A rectangular bar 50mm wide and 10mm thick is 3000mm long. It is
subjected to an axial pull of 50N. If the change in length is 1.5mm and decrease
in thickness is 0.0014. Determine the for elastic constants?
Ans:
a) E  200 kN mm 2
c) K  151.52 kN mm 2
b) C1 78.125 N mm 2
d)  0.28
m
Paper - III Engineering Mechanics
339
4) A steel bar 2m long, 20mm wide and 10mm thick is subjected to a pull
of 20kN in the direction of its length. Find the change in length, breadth and
thickness Take E  1.0
2  105 N mm 2 and Poisson’s Ratio = 0.3?
Ans: l  1mm; b  0.003mm; t  0.0015mm
5) For a given Material E  1.0  105 N mm 2 and C  0.4  105 N mm 2 .
Find K and lateral contraction
1 of a round bar of 50mm diameter and 2-5m long
 0.25 .
when stretched 2-5mm
m
Ans: K  0.67  105 N mm 2 ; d  0.0125mm
6) A bar of 30mm diameter is subjected to a pull of 60kN. The measured
extension an gauge length of 200mm is 0.09 mm and change in diameter is
0.0039mm.
1
 Poissons ratio and the values of three elastic constants?
Calculate the
m
Ans:
a)
1
 0.288
m
b) E  188.67 kN mm 2
c) C  73.19 kN mm 2
d) K  149.85 kN mm 2
Key Concepts
1. Stress is internal resistance setup by material per unit area to resist
deformation
:
P
A
2. Strain is deformation per unit length.
3. Hookes law states that stress is proportional to strain within elastic
limits.
4. Young’s modulus is normal stress per unit normal strain within elastic
limits.
If young’s modulus is E, area of cross section A angle length L
load applied on the body “P” the deformation  L is the body then
E
PL
A L
 or   L 
PL
AE
5. Yield point is stress at which strain increases under a steady load.
6. Ultimate strength is stress corresponding to maximum load that can be
realised before failure.
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Building Construction and Maintenance Technician
7. Factor of Safety 
Ultimatestress
Working stress
8. Shear Stress is stress applied tangalially to a surface.
9. Shear strain is angular strain in radians produced under shear stress.
10. Modules of Rigidify is shear stress per unit shear strain within elastic
limits.
11. Ratio lateral strain to longitudinal strain within elastic limits is poisson’s
ratio.
12. Bulk modulus is normal stress per unit Volumetric strain within elastic
limits when the body is subjected to three equal mutually perpendicular normal
stresses of same kind.
13. The relationship between the three elastic moduli is given by
1
2


E  Z2 N  1    3k  1  
 m
 m
9KN 
1

By e liminations 

3K  N 
m
UNIT
5
Columns - Struts
Learning Objectives
• We come across several instances of members subjected to compressive
loads. Examples for such loading may be framed structures, rafters in buildings,
frames of presses etc.,
Columns
Columns are the members subjected to direct compressive force. A vertical
member subjected to direct compressive force is called a column or pillar.
Struts:- When a member of a structure is any position and carrying an
axial compressive load is called strut. Strut may be horizontal, inclined, or even
vertical.
Post:- Wooden member carrying compressive force called post.
Stanchion:- a built up rolled steel carrying compressive force is called
stanchion
Boom:- Main compressive member in a jib crane is called a boom.
Short Columns:- In this type of columns, the bucking stresses are very
small, compared to direct stress or bucking stresses. Fails due to direct stress.
b. Short columns is a column whose slenders ration is less than < 32 or
whose length is <8 times the least latered dimension
c. For mild steel columns slenderness ration is less then 80.
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Building Construction and Maintenance Technician
Medium columns or Intermediate Columns
Medium columns is a column which fairs either due to direct stress or
buckling stress.
a. For medium columns, the slenderness ration is more than 32 and less
than 120.
b. For medium columns, their length is more than 8 times but less than 30
times. Their least lateral dimmesion.
Long columns
Long Columns is column which fairs primary due to bending stress.
a. Long columns, The direct stress are very small compared to their backing
stresses.
b. Long columns is a column whose slenderness raton is greater than 120
or whose legth is more then 30 times the lest lateral dimension.
c. For mild Steel Column slender ness ration is >80.
Slenderness Ratio
Slenderness Ratio of a column is defined as the ration of column height to
its least radings of gyration
l
 
k
where
l = Equivalent length of the column
k = Least radius of gyration.
and x 
I
A
I = Least moment of Inertia
A= Cross sectional area
a. Load carrying capacity of long columns depends on slenderness ratio.
b. For good design the slenderness ration is as small as possible.
Effective length of a Column
The effective length of a given with given end conditions is the length of an
equivalent column of the same material and semicross section with hinged ends
having the value of the crippling load equal to that of a given column effective
length of column mainly depends upon end conditions.
Paper - III Engineering Mechanics
S. No
Types and
conditions
343
Endless
crippling
Actual Equivalent
length
length
 2 EI
L2
1
Both ends hinged
P
2
Both ends fixed
P
4 2 EI
hL2
3
One end fixed other
P
2 2 EI
L2
L
 2 EI
P
4L2
L
end free
4
One end fixed other
end minged.
L
l=L
L
l
L
2
l
L
2
l = 2L
P = Eulers crippling or buckling load.
E = Young’s modulus
I = Least moment of inertia
l = Equivalent length of columns for given end conditions.
Columns subjected axial load
Equals crippling load
or critical load
2
Pcr 
 EI
l2
Where EI = Flexural rigidity
l = effective length of columns
equals formula is used for lons columns or
l
 80 for mild steel columns,
k
Assumptions made in Euler’s formula
a. The column is initially perfectly straight
b. The Cross section of the column is uniform through out length.
c. We length of the column is very large when compared to its lateral
dimension.
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Building Construction and Maintenance Technician
d. The Shortening of the columns is neglected
e. the self weight of the columns is ignorable
f. The column fails due to bucking alone.
g. The column material is perfectly elastic, homogeneous is tropic & obeys
hook’s law.
h. The pin joints are friction less & fixed ends are rigid.
Rankines formula can be used for both short & long columns the following
values are for a columns both eudshinged.
S. NO
Material
Crushins Stress
Fc in N/mm2
1
Wrought Iron
250
2
Cast Iron
250
3
Mild steel 320
250
4
Timber
50
5
Medium carbon Steel
500
Rakine'scons tan t
fc
 2
 EI
1
9000
1
1600
1
7500
1
750
1
5000
The above values for a column both ends hinged.
Bucking loads:- the load at which the column just buckles is called buckling
load or critical load or crippling load and the column is said to have developed
an elastic instability.
The value of buckling load is low for long columns and relatively high for
short columns.
Factor of safety :- Factory Safety =
Column subjected to axial load :-
crippling load
Safe load
 2 EI
Euler’s is crippling load = Pcr  2
Ll
Paper - III Engineering Mechanics
345
EI =flexural rigidity
l = effective length of the column
Euler’s formula is only for long columns 
L
l
K
 80
11.Rankinl’s formula for crippling load :f c.A
crushing load
pcr 
L
1  
K
2

l
1   
k
2
fc = Allowable crushing stress
A = Area of cross section of the column
K= least radius of gyration
A = rankine’s constant.
Limitations of Euler’s Formula
1. Euler’s formula is used for long columns & negrech the stress due to
direct compressive loads.
2. If slenderness ration is less than 80, the Euler’s formula for mild steel
columns is not valid.
3. In Euler’s formula that the critical or allowable stress an a column
decreases with the increase in slenderness ratio.
4. Long axially loaded columns tends to defect about the axis of the least
moment of inertia, the least radius of gyration and it should be used
for determination the slenderness ration.
Solved Problems
1) A mild steel column 5m long and 50mm diameter which is fixed at one
end free at the other end determine the Euler’s crippling load take
E = 2 x 10 N/mm2
 2 EI
Euler’s crippling load = 2
l
This is condition for its both ends hinged
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Building Construction and Maintenance Technician
But one end fixed other end free
=> 1 = 2L, p =
 3.14 
p
2El
l2
2
 2  105  I
4  l2
But I for circular section
ITd 4 22
4
I
   50 
64
1
= 306919.6mm4
 3.14 
p
2
 2  105  306919.6
4   5000 
2
2
 2  306919.6  105
4  5000  5000
 6052.2 N
 3.14 

2. A hollow along tube of 5m long with external dia 40mm and internal dia
25mm was to extend 6mm under a tensile load of 60 KN. Find the bucking load
for the tube when used as strut in both ends pinned. Also find the safe load,
taking factor of safty as 4?
Length = 5m
Both ends pinned or hinged l  L  5m  5000mm
External = 40mm ; internal dia = 25mm
M.I hollow Cross section =


D4  d4 

64
22 1 
4
4

40    25  


7 64 
Paper - III Engineering Mechanics
347
22 1
  256000  335776
7 64
22 1
   256000  335776
7 64
 109421.7 mm 4

Area of hollow section 


 2 2
D  d 
4
22 1 
2
2
   40    25  


7 4
22
1600  625  766mm 2
28
l= 6mm; total load = 60KN = 60,000N
Stress 
60,000
60,000

 78.33N mm 2
c  s, area
766
Strain 
3l
6
stress 78.33

 0.0012 E 

l 5000
strain 0.0012
2
 2 EI  3.17   65275  109421.7
P 2 
 2816.9 N
2
L
 5000
Safe load =
crippling load 2816.9

Fs
4
= 704.2 N
3. A mild steel tube 4m long, 30mm internal dia and 5mm thick is used as
strut with both ends pinned find the Euler’s crippling load take E = 2x 105N/
mm2 ?
l = 4 m = 4000mm
Internal dia = 30mm
External dia = 5+30+5
= 40mm1
348
Building Construction and Maintenance Technician
I of hollow circular section 


D4  d 4 

64
22 1 
4
2
   40    30  

7 64 

22
 2560000  810000
7  64
22
 10,000  256  81
7  64
22

 1, 75,0,000  85937.5mm 4
7  64
 2 EI
Euller’s crippling load  2
L

Both ends pinned  l = L

 2 EI  2  2  105  85937.5

2
L2
 4000 
 2  2 105  85937.5  2  2  85937.5


4000  4000
160
P = 10610.6 N
Crippling load = 10610.6N
4. A cast graw hollow column, having 80mm extends dia and 60mm internal
dia, is used as column of 3m long using Rakine’s formula, determine crippling
load, when both ends are fixed? Take
fc =500N/mm2  
1
1600
D= external dia = 80mm internal dia d = 60mm
 2
22 1
2
2
D  d 2      80    60  



4
7 4
100 
 22
22
   6500  3600  
 2800 
28
 28

A
Paper - III Engineering Mechanics
349
= 2200mm2
 
22 1
4
4
4
4
 80    60     80    60  

64
7 64
 491.07  4096  1296
I
l
 491.07  2800 1374996mm 4
Least radius of gyration K 
I
1374996

A
2200
K = 25mm
L=3m

= 3000mm
Both ends fixed = L 
Fc A
P
L
1  
K
2
1
fe = 500N/mm2
1600
l 3000

1500mm
2
2

500  2200
1  1500 
1
1600  25 
2

1.1  50 6
1
1
1600
6
1.1  106

 338461.6 N
3.25
= 338.5KN
5. The cross section of mild steel column is hollow rectangular section with
the dimensions 300  200mm and vertal thickness 25mm. the length of column
is 3m and both ends are hinged. Farm the safe load is can carry using a Ramkine’s
formula:Take F-S = 3; Fc = 300N/mm2  
Feast M.I 

1
1500
1
 BD3  bd 3 
12
1 
3
3
300   200   250  150  

12 
l
350
Building Construction and Maintenance Technician
1
 10 4  2, 40, 000  84375 
12
1
  1,55625  12968  10 4 mm 4
12


L = 3000mm


fc ' A

 l  L,
2
Both ends hinged

L 
  
1 A
k 


2
I 12968  10 4
12968  10 4
K


 75.92
C
A
GS.area
225 00
C. S area = (BD-bd)= 300x200-250x150
= 60 000 -37 , 500 = 22,500mm2
P
300  22, 500
1 
1   
K
2
300 x 22500

1  3000 
1


7500  75.92 
2
300 x 22, 500
 5587749.3 N
1.208
= 5587.75 KN

Factory safety =3
Safe load =
5587.75
 1862.6KN
3
Short Answer Type Questions
1. What is mean of bucking load or criphists load? An what factors does it
depends?
2. Define the terms i) slenderness ratio ii) critical load iii) equivalent length
3. State the different Euler’s formulae for different end conditions of
columns?
4. Distinguish between long columns short column of mild steel?
5. Different between a)long columns b) medium columns c) short columns.
Paper - III Engineering Mechanics
351
6. State the assumptions made by Euler’s theory?
7. How do the Euler’s formula for crippling load to the following conditions?
i) both ends hinged ii) both ends fixed iii) one end fixed other end free iv) are end
fixed and other hinged?
8. What are the limitations of Euler’s buckling theory?
9. A mild steel tube 3m long 30mm internal dia and 4mm thick is used as
strut both ends hinged find the crippling load. Take E = 2105 N/mm2
Ans. Pcr = 13728.98 N
10. Calculate the safe compressive load on a hollow cast iron column with
one end fixed and other end hinged of 150 mm external dia and 100mm and 3m
length use angle’s formula with a factor of safety of 3 and E =0.95 x 106 N/mm2
(Safe load = 1384.89)
11. A column of timber section 100 x 200mm is 4.5m long end its both
ends are fixed calculate safe load for a column. It can take using Euler’s formula.
Take E = 17.5 x 103N/mm2 & factory safety = 3 ?
(Ans: 189.463 KN = safe load )
12. Determine the section of cast iron hollow cylindrical columns 5m long
with ends firmly build in (both ends field) if is carries axis load of 300 KN. The
ration of using factory of safety = 8, take
 1
Ramkin’s constant
1000
Ans D=171.1mm d = 128.3 mm
13. A hollow cast iron column with fixed ends supports axial load of
1000KN. If the column is 5m long and has and external dia of 250 mm find the
thickness of meta required. Use rankine’s formula taking a constant and assuming
the working stress of 80N/mm2
(Ans d = 215.6mm t = 17.19mm )
UNIT
7
Graphics Statics
Learning Objectives
• The graphical statics presents a less tediuos and practical solutions of a
problem in statics by graphical method.
• The accuracy of the graphical solution may not match with that of the
analytical one but is generally sufficient for all practical purposes.
Space Diagram, Bow’s Notation and Vector Diagram
The relative positions of the various vectors acting on a system are
represented, in a figure called the Space Diagram. It is drawn to a linear scale to
show the points of application and the directions of all the vectors. In naming the
vectors, a standard practice or notation is used. Bow’s notation is generally
followed.
In Bow’s notation, each space on either side of the line of action of each
vector is given a name.
The vector Diagram represents the magnitudes and directions of all the
vectors acting on the system. It is drawn to the scale of vectors.
Equilibrant and Resultant of Two Concurrent Forces
These are determined the help of the law of triangle of forces.
Example 7.1 : Determine the equilibrant and hence the resultant of two
forces of 150 N and 250 N acting at a point O if the angle between them is 600.
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Building Construction and Maintenance Technician
Fig 7.1
Fig 7.2
Force
Magnitude
Inclination with OX
Equilibrant
ca
350 N
-1420
Resultant
ac
320 N
380
1. Draw the space diagram to show the two given forces making 600 with
each other.
2. Name the two forces as per Bow’s notation, the 150 N by AB and the
250 N by BC as shown in Fig 7.1. The equilibrate will then be
represented by CA.
3. Select a convenient point (in Fig 7.2) to present the space A. Draw
through a, ab parallel to the direction of force AB (150 N). Mark the
point b on ab represents 150 N to the selected scale say 1 mm for 5 N.
4. From b, draw bc parallel to the 250 N force i.e., BC. The length bc is
selected such that the magnitude of BC is represented by it to the same
scale of 1 mm for 5 N.
5. Join ‘ca’ to get abc, the triangle of forces for the point O. Fig. 7.2 is
known as the Vector Diagram.
6. ‘ca’ represents the magnitude and direction of the equilibrant of the given
forces. Measure its magnitude, Draw a parallel to this direction in the
space diagram, tabulate results and measure the angle made by it with
OX.
7. ‘ac’ represents the magnitude and direction of the resultant of the two
given forces. Measures its magnitude and inclination with OX and
tabulate result.
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Note
1. To every space in the space diagram, there will be a corresponding point
in the vector diagram.
2. To every vector in the space diagram, there will be a straight line in the
vecot diagram.
3. The equilibrant and the resultant will be collinear, equal and opposite.
4. The vector diagram is a closed figure for a system of forces in equilibrium
Equilibrant and Resultant of more than two Concurent Forces
These are determined by the law of polygon of forces. This is only an
extension of the method of triangle of forces.
Example 7.2
Determine the equilibrant and resultant of 4 pulls of 300 N, 600N, 400 N
and 200 N making angles of 300, 1200, 2250 and 3300 respectively with a fixed
direction OX.
Procedure
1. In the space diagram (Fig ) draw the direction OX and the direction of
all the given forces making the stated angles with OX.
2. Name the given forces as AB, BC, CD and DE by using Bow’s notation
starting with 200 N and going clock wise about O. (See Fig ) Let EA
be the equilibrant to he system.
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Building Construction and Maintenance Technician
Results
Force
Magnitude
Inclination with OX
Equilibrant
ea
320 N
2980
Resultant
ae
320 N
1180
3. Draw ab (Fig ) parallel to the force AB=200 N and to represent its
magnitude to a scale of 1 mn for 10 N.
4. From ‘b’ draw be parallel to the next force in the order i.e., BC and
mark as such that be represents the forces of 300 N to the scale selected.
5. From ‘c’ draw cd parallel and proportional to force CD=600N and
form an draw be parallel and proportional to DE =400 N.
6. ‘abcde’ is the vector diagram. Hence by law of polygon of forces ‘ea’
represents the equilibrate of the given system of cocurrent forces.
Measure its magnitude to scale (=320 N) and draw a parallel to its direction
through O in the space diagram. Measure inclination of this line with OX (=2980)
and tabulate results.
7. ‘ae’ represents magnitude and direction of the resultant. Measure its
magnitude and inclination and tabulate results.
Note : The equilibrnat of a system of copanar concurrent forces is also a
coplanar force and is concurrent with the system. Hence the resultant passes
through O, the point of concurrency. Its direction will be parallel to the closing
sidce eea of the polygon of forces for the point O.
Reactions of Simply Supported Beams
To find the reactions at the supports of simply supported beams, proceed
as follows:
1. Draw space diagram, vector diagram and funicular polygon for all the
forces on the beam excepting the reactions.
2. Produce the first ray a/o and the last ray say d/o of the funicular polygon
to cut the lines of action of the reactions at the respective supports at p,
q respectively. The line pq will be the closing line ofthe funicular polygon.
3. Draw a ray parallel to the closing line ‘pq’ through the pole O of the
vector diagram to meet the load line at say e.
4. ‘ea’ will represent the reaction EA and ‘de’ will represent the reaction
DE.
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385
Graphical Method of Determing Centroid
As in the analytical method, the composite area is divided into elementary
figures whose area and centroid can be determined easily. The areas of all such
elements are considered as parallel forces acting in a convenient direction through
the respective centroid. The line of action of the resultants force is determined
graphically as per the method. The centoid of the given composite area lies on
the line of action of the resultant force. If there is no symmetry about any axis,
the centroid is then located at the intersection of the resultant forces in the two
assumed directions.
Example 4.9 : Determine graphically the centroid of a Tee section 180
mm/120mm / 20 mm.
The Tee section is symmetric about the axis of the web. Hence its centroid
lies on this axis. The areas of the flange and web will be treated as horizontal
forces through their centroids located by intersecting the diagnonals.
Example: 4.10 : Determine graphically the centroid of an unequal angle
100 mm x 80 mm x 10 mm.
Diagram is displayed in the next page
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Long Answer Type Questions
1. Determine graphically the euilibrant of the forces shown in Fig.
2. Two forces 200 N and 300 N act at an aggle of 1500. Find the magnitude
and direction of the resultnat by graphical method. The 200 N Force is horizontal.
3. Determine the distance of the centroid of the sections shown in Fig.
from the bottom most edge and the central vertical axis.