Mechanics of Solids
Mechanics of Solids
Mechanics of Solids
OF SOLIDS
This page
intentionally left
blank
MECHANICS
OF SOLIDS
S.S. Bhavikatti
Emeritus Fellow (AICTE)
BVB College of Engineering and Technology, Hubli
(Formerly Principal, RYMEC, Bellary
S.S. Bhavikatti
Professor & Dean
SDMCET, Dharwad and NITK, Surathkal)
Copyright © 2010, New Age International (P) Ltd., Publishers
Published by New Age International (P) Ltd., Publishers
—Author
(v)
This page
intentionally left
blank
Contents
Preface v
3 TRUSSES 65–93
3.1 Perfect, Deficient and Redundant Trusses ................................................................ 65
3.2 Assumptions ............................................................................................................... 66
3.3 Nature of Forces in Members ..................................................................................... 67
3.4 Methods of Analysis .................................................................................................... 68
3.5 Method of Joints ......................................................................................................... 68
3.6 Method of Section ........................................................................................................ 81
Important Formula .................................................................................................... 87
Theory Questions ........................................................................................................ 87
Problems for Exercise ................................................................................................. 88
5 FRICTION 161–190
5.1 Coefficient of Friction ............................................................................................... 161
5.2 Laws of Friction ....................................................................................................... 162
5.3 Angle of Friction, Angle of Repose and Cone of Friction .......................................... 162
5.4 Problems on Blocks Resting on Horizontal and Inclined Planes ............................. 164
5.5 Application to Wedge Problems ................................................................................ 174
5.6 Application to Ladder Problems ............................................................................... 177
5.7 Belt Friction ............................................................................................................. 180
Important Formulae ................................................................................................. 187
Theory Questions ...................................................................................................... 187
Problems for Exercise ............................................................................................... 187
9 BEAMS 283–312
9.1 Introduction .............................................................................................................. 283
9.2 Types of Supports ..................................................................................................... 283
9.3 Types of Beams ......................................................................................................... 284
9.4 Types of Loading ....................................................................................................... 285
9.5 Reactions from Supports of Beams ........................................................................... 286
9.6 Shear Force and Bending Moment ........................................................................... 291
9.7 Sign Convention ....................................................................................................... 293
9.8 Relationship between Load Intensity, Shear Force and Bending Moment .............. 293
9.9 Shear Force and Bending Moment Diagrams .......................................................... 294
9.10 SFD and BMD for a few Standard Cases ................................................................. 295
9.11 Short-cut Procedure .................................................................................................. 307
Important Formulae ................................................................................................. 310
Theory Questions ...................................................................................................... 310
Problems for Exercise ............................................................................................... 310
1
2 MECHANICS OF SOLIDS
The bodies which do not change their shape or size appreciably when the forces are applied
are termed as Solids while the bodies which change their shape or size appreciably even when small
forces are applied are termed as Fluids. Stone, steel, concrete etc. are the example of solids while
water, gases are the examples of fluids.
In this book application of Newtonian mechanics to solids is dealt with.
Rigid Body
A body is said to be rigid, if the relative positions of any two particles do not change under the
action of the forces acting on it. In Fig. 1.1 (a), point A and B are the original positions in a body.
After the application of forces F1, F2, F3, the body takes the position as shown in Fig. 1.1(b). A′
and B′ are the new positions of A and B. If the body is treated as rigid, the relative position of A′B′
and AB are the same i.e.
A′B′ = AB
Many engineering problems can be solved by assuming bodies rigid
F2
B
B′
A′
A F1
F3
(a) (b)
Fig. 1.1
Particle
A particle may be defined as an object which has only mass and no size. Theoretically speaking
such a body cannot exist. However in dealing with problems involving distances considerably larger
compared to the size of the body, the body may be treated as a particle, without sacrificing
accuracy.
For example:
— A bomber aeroplane is a particle for a gunner operating from the ground.
— A ship in mid sea is a particle in the study of its relative motion from a control tower.
— In the study of movement of the earth in celestial sphere, earth is treated as a particle.
Force
Force is an important term used in solid mechanics. Newton’s first law states that everybody
continues in its state of rest or of uniform motion in a straight line unless it is compelled by an
external agency acting on it. This leads to the definition of force as ‘force is an external agency
which changes or tends to change the state of rest or uniform linear motion of the body’.
Magnitude of force is defined by Newton’s second law. It states that the rate of change of
momentum of a body is directly proportional to the impressed force and it takes place in the
direction of the force acting on it. Noting that rate of change of velocity is acceleration, and the
product of mass and velocity is momentum we can derive expression for the force as given below:
From Newton’s second law of motion
Force ∝ rate of change of momentum
∝ rate of change of (mass × velocity)
4 MECHANICS OF SOLIDS
Characteristics of a Force
— Direction.
In Fig. 1.2, AB is a ladder kept against a wall. At point C, a person
weighing 600 N is standing. The force applied by the person on the
ladder has the following characters:
A
— magnitude is 600 N Fig. 1.2
— the point of application is C which is at 2 m from A along the
ladder
— the line of action is vertical
— the direction is downward.
It may be noted that in the figure
— magnitude is written near the arrow
— the line of arrow shows the line of application
— the arrow head shows the point of application
— the direction of arrow represents the direction of the force.
INTRODUCTION TO MECHANICS OF SOLIDS 5
1.2 UNITS
Length (L), mass (M) and time (S) are the fundamental units used in mechanics. The units of all
other quantities may be expressed in terms of these basic units. The three commonly used systems
are
— Metre, Kilogram, Second (MKS)
— Centimetre, Gram, Second (CGS)
— Foot, Pound, Second (FPS).
The systems are named after the units used to define the fundamental quantities length, mass
and time. Using these basic units, the units of other quantities can be found. For example in MKS
the units for various quantities are
Quantity Unit
Area m2
Volume m3
Velocity m/sec
Acceleration m/sec2
Momentum kg-m/sec [Since it is = mass × velocity]
2
Force kg-m/sec [Since it is = mass × acceleration]
S.I. Units
Presently the whole world is in the process of switching over to SI-system of units. SI units stands
for the System International d′ units or International System of units. As in MKS units in SI also
the fundamental units are metre for length, kilogram for mass and second for time. The difference
between MKS and SI system arises mainly in selecting the unit of force. In MKS unit of force is
kg-wt while in SI units it is newton. As we have already seen one kg-wt is equal to 9.81 newtons.
The prefixes used in SI when quantities are too big or too small are shown in Table 1.1.
Table 1.1. Prefixes in SI Units
1012 tera T
10 9 giga G
6
10 mega M
3
10 kilo k
10 0 — —
–3
10 milli m
10–6 micro m
–9
10 nano n
–12
10 pico p
10–15 femto f
–18
10 atto a
6 MECHANICS OF SOLIDS
The parallelogram law of vectors enables us to determine the single vector called resultant vector
which can replace the two vectors acting at a point with the same effect as that of the two vectors.
This law was formulated based on exprimental results on a body subjected to two forces. This law
can be applied not only to the forces but to any two vectors like velocities, acceleration, momentum
etc. Though stevinces employed it in 1586, the credit of presenting it as a law goes to Varignon
and Newton (1687). This law states that if two forcer (vectors) acting simultaneously on a body
at a point are represented in magnitude and directions by the two adjacent sides of a parallelogram,
their resultant is represented in magnitude and direction by the diagonal of the parallelogram which
passes thorough the point of intersection of the two sides representing the forces (vectors).
In the Fig. 1.3, the force F1 = 4 units and the force F2 = 3 unit are acting on a body at a
point A. To get the resultant of these forces, according to this law, construct the parallelogram
ABCD such that AB is equal to 4 units to the linear scale and AC is equal to 3 units. Then according
to this law, the diagonal AD represents the resultant in magnitude and direction. Thus the resultant
of the forces F1 and F2 is equal to the units corresponding to AD in the direction α to F1.
C D
F2 R
R
3 θ
θ
F1 α
A B
4
Fig. 1.3
INTRODUCTION TO MECHANICS OF SOLIDS 7
Referring to Fig. 1.3 (b), it can be observed that the resultant AD may be obtained by constructing
the triangle ABD. Line AB is drawn to represent F1 and BD to represent F2. Then AD should
represent the resultant of F1 and F2. Thus we have derived the triangle law of forces from the
fundamental law of parallelogram. The Triangle Law of Forces (vectors) may be stated as if two
forces (vectors) acting on a body are represented one after another by the sides of a triangle, their
resultant is represented by the closing side of the triangle taken from the first point to the last point.
If more than two forces (vectors) are acting on a body, two forces (vectors) at a line can be
combined by the triangle law, and finally resultant of all forces (vectors) acting on the body may
be obtained.
A system of four concurrent forces acting on a body are shown in Fig. 1.4. AB represents F1
and BC represent F2. Hence according to triangle law of forces AC represents the resultant of F1
and F2, say R1.
E
F4
F3 = 30 kN
D
F4 = 30 kN
R
F2 = 25 kN F3
R2
C
F1 = 20 kN R1
F2
B
F1
O A
Fig. 1.4
If CD is drawn to represent F3, then from the triangle law of forces AD represents the resultant
of R1 and F3. In other words, AD represents the resultant of F1, F2 and F3. Let it be called as R2.
Similarly the logic can be extended to conclude that AE represents the resultant of F1, F2, F3
and F4. The resultant R is represented by the closing line of the polygon ABCDE in the direction
form A to E. Thus we have derived the polygon law of the forces (vectors) and it may be stated
as if a number of concurrent forces (vectors) acting simultaneously on a body are represented in
magnitude and direction by the sides of a polygon, taken in a order, then the resultant is represented
in magnitude and direction by the closing side of the polygon, taken from the first point to the last
point.
Parallelogram law, triangle law and polygonal law of vectors can be used to find the resultant
graphically. This method gives a clear picture of the work being carried out. However the main
disadvantage is that it needs drawing aids like pencil, scale, drawing sheets. Hence there is need for
analytical method.
8 MECHANICS OF SOLIDS
Consider the two forces F1 and F2 acting on a particle as shown in Fig 1.5(a). Let the angle
between the two forces be θ. If parallelogram ABCD is drawn as shown in Fig. 1.5(b) with AB
respresenting F1 and AD representing F2 to some scale, according to parallelogram law of forces
AC represents the resultant R. Drop perpendicular CE to AB.
D C
F2
F2
R
F2
O F1 A F1 B E
(a) (b)
Fig. 1.5
∴ R = ( F1 + F2 cos θ) 2 + ( F2 sin θ) 2
= F1 2 + 2 F1 F2 cos θ + F2 2
Since, sin2 θ + cos2 θ = 1.
The inclination of resultant to the direction of F1 is given by α, where
CE CE F2 sin θ
tan α = = =
AE AB + BE F1 + F2 cos θ
F2 sin θ
Hence α = tan–1
F1 + F2 cos θ
Particular cases:
F2
F1 F2
F1 F1 F2
(a) (b) (c)
Fig. 1.6
Resolution of Vectors
Since the resolution of vectors is exactly opposite process of composition of vectors, exactly the
opposite process of composition can be employed to get the resolved components of a given force.
F1
F F
β F2
β
α α θ=α+β
F2 F1
(a)
Fy
F
F
Fy
F2 Fx
(b)
F4
F3
F F F4
F2 F1 F3
F2
F1
(c)
Fig. 1.7
In Fig. 1.7(a), the given force F is resolved into two components making angles α and β with F.
In Fig. 1.7(b) the force F is resolved into its rectangular components Fx and Fy.
In Fig. 1.7(c), the force F is resolved into its four components F1, F2, F3 and F4.
It may be noted that all component forces act at the same point as the given force. Resolution
of forces into its rectangular components is more useful in solving the problems in mechanics. In
this case, if the force F makes angle θ with x-axis, from Fig. 1.7(a), it is clear that
Fx = F cos θ and Fy = F sin θ.
10 MECHANICS OF SOLIDS
Example 1.1. A boat is rowed at a velocity of 20 km/hour across a river. The velocity of stream
is 8 km/hour. Determine the resultant velocity of the boat.
Solution: Taking downstream direction as x and direction across the river as y, it is given that
Vx = 8 km/hour
Vy = 20 km/hour
∴ The resultant velocity
V = 82 + 20 2 = 21.54 km/hour
Vy 20
α = tan–1 = tan–1 = 68.20°, as shown in Fig. 1.8
Vx 8
Vy =20 km/hour
V Downstream
Vx = 8 km/hour
Fig. 1.8
Example. 1.2. The guy wire of the electrical pole shown in Fig. 1.9(a) makes 60° to the horizontal
and is carrying a force of 60 kN. Find the horizontal and vertical components of the force.
kN
20
F
Fy
60° 60°
Fx
(a) (b)
Fig. 1.9
Solution: Figure 1.9(b) shows the resolution of force F = 20 kN into its components in horizontal
and vertical components. From the figure it is clear that
Fx = F cos 60° = 20 cos 60° = 10 kN (to the left)
Fy = F sin 60° = 20 sin 60° = 17.32 kN (downward)
INTRODUCTION TO MECHANICS OF SOLIDS 11
20°
W
C
B
20° 20°
(a) (b)
Fig. 1.10
Solution: The plane makes an angle of 20° to the horizontal. Hence the normal to the plane makes
an angles of 70° to the horizontal i.e., 20° to the vertical [Ref. Fig. 1.10(b)]. If AB represents the
given force W to some scale, AC represents its component normal to the plane and CB represents
its component parallel to the plane.
Thus from ∆ ABC,
Component normal to the plane = AC
= W cos 20°
= 10 cos 20°
= 9.4 kN as shown in Fig. 1.10(b)
Component parallel to the plane = W sin 20° = 10 sin 20°
= 3.42 kN, down the plane
From the above example, the following points may be noted:
1. Imagine that the arrow drawn represents the given force to some scale.
2. Travel from the tail to head of arrow in the direction of the coordinates selected.
3. Then the direction of travel gives the direction of the component of vector.
4. From the triangle of vector, the magnitudes of components can be calculated.
Example 1.4. The resultant of two forces, one of which is double the other is 260 N. If the direction
of the larger force is reversed and the other remain unaltered, the magnitude of the resultant
reduces to 180 N. Determine the magnitude of the forces and the angle between the forces.
Solution: Let the magnitude of the smaller force be F. Hence the magnitude of the larger force is
2F.
Thus F1 = F and F2 = 2F
Let θ be the angle between the two forces.
∴ From the condition 1, we get
R = F1 2 + 2 F1 F2 cos θ + F2 2 = 260
i.e., F2 + 2F (2F) cos θ + (2F)2 = 2602
5F2 + 4F2 cos θ = 67600 ...(i)
12 MECHANICS OF SOLIDS
∴ cos θ = 0.20577
∴ θ = 78.13°
Noting that
R sin α = F2 sin θ
260 sin 78.13°
we get sin α = = 0.489
520
∴ α = 29.29°
Example 1.7. Fig. 1.13 shows a particular position of 200 mm connecting rod AB and 80 mm long
crank BC. At this position, the connecting rod of the engine experience a force of 3000 N on the
crank pin at B. Find its
(a) horizontal and vertical component
(b) component along BC and normal to it.
Vertical
Connecting rod
B Crank
80
3000 N
0 mm 60°
mm
20
60° Horizontal
C
A
(a) (b)
Fig. 1.13
Solution: The force of 3000 N acts along line AB. Let AB make angle α with horizontal. Then,
obviously 200 sin α = 80 sin 60°
∴ α = 20.268°
Referring to Fig. 1.13(b), we get
Horizontal component = 3000 cos 20.268° = 2814.2 N
Vertical component = 3000 sin 20.268° = 1039.2 N
Components along and normal to crank:
The force makes angle α + 60° = 20.268 + 60 = 80.268° with crank.
∴ Component along crank = 3000 cos 80.268° = 507.1 N
Component normal to crank = 3000 sin 80.268° = 2956.8 N
IMPORTANT FORMULAE
THEORY QUESTIONS
1. The resultant of two forces one of which is 3 times the other is 300 N. When the direction of
smaller force is reversed, the resultant is 200 N. Determine the two forces and the angle between
them. [Ans. F1 = 80.6 N, F2 = 241.8 N, θ = 50.13°]
2. A rocket is released from a fighter plane at an angle upward 20° to the vertical with an acceleration
of 8 m/sec2. The gravitational acceleration is 9.1 m/sec2 downward. Determine the instantaneous
acceleration of the rocket when it was fired. [Ans. 9.849 m/sec2, θ = 49.75° to vertical]
2
Fundamentals of Statics
In this chapter principles of statics is explained and their applications to concurrent and non-concurrent
force system in plane is illustrated by solving several engineering problems.
As already discussed in first chapter, Newton’s first law gave definition of the force and second law
gave basis for quantifying the force. There are two more Newton’s laws:
a. Newton’s Third Law
b. Newton’s Law of Gravitation
These laws are explained in this article.
(a) Newton’s Third Law
It states that for every action there is an equal and opposite reaction. Consider the two bodies in
contact with each other. Let one body apply a force F on another. According to this law the second
body develops a reactive force R which is equal in magnitude to force F and acts in the line same
as F but in the opposite direction. Figure 2.1 shows the action of a ball on the floor and the reaction
of floor to this action. In Fig. 2.2 the action of a ladder on the wall and the floor and the reactions
from the wall and the floor are shown.
R-reaction
F-action
Fig. 2.1
15
16 MECHANICS OF SOLIDS
R1
F1
R2
F2
Fig. 2.2
1 2
F F
m1 m2
Fig. 2.3
N × m2
Hence unit of G = = Nm2/kg2
kg × kg
It has been proved by experiments that the value of G = 6.673 × 10–11 Nm2/kg2. Thus if two
bodies one of mass 10 kg and the other of 5 kg are at a distance of 1 m, they exert a force
6.673 × 10 −11 × 10 × 5
F= 2 = 33.365 × 10–10 N
1
on each other.
FUNDAMENTALS OF STATICS 17
According to this law the state of rest or motion of the rigid body is unaltered, if a force acting on
the body is replaced by another force of the same magnitude and direction but acting anywhere on
the body along the line of action of the replaced force.
Let F be the force acting on a rigid body at point A as shown in Fig. 2.4. According to this law,
this force has the same effect on the state of body as the force F applied at point B, where AB is
in the line of force F.
F
B
F A
Fig. 2.4
In using law of transmissibility it should be carefully noted that it is applicable only if the body
can be treated as rigid. Hence if we are interested in the study of internal forces developed in a body,
the deformation of body is to be considered and hence this law cannot be applied in such studies.
This has been already explained in chapter 1 along with the derived laws i.e., triangle and polygonal
law.
It states that the action of a force on a body is not affected by the action of any other force on the
body.
18 MECHANICS OF SOLIDS
It states that the net effect of a system of forces on a body is same as the combined of individual
forces acting on the body. Since a system of forces in equilibrium do not have any effect on a rigid
body this principle is stated in the following form also: ‘The effect of a given system of forces on
a rigid body is not changed by adding or subtracting another system of forces in equilibrium.’
Collinear forces Line of action of all the forces act Forces on a rope in a tug of
along the same line. war
Coplanar All forces are parallel to each other System of forces acting on a
parallel forces and lie in a single plane. beam subjected to vertical
loads (including reactions)
Coplanar All forces are parallel to each other, Weight of a stationary train
like parallel forces lie in a single plane and are acting in on a rail when the track is
the same direction. straight
Coplanar Line of action of all forces pass Forces on a rod resting against
concurrent forces through a single point and forces a wall
lie in the same plane.
Coplanar All forces do not meet at a point, Forces on a ladder resting
non-concurrent forces but lie in a single plane. against a wall when a person
stands on a rung which is not
at its centre of gravity
Non-coplanar All the forces are parallel to each The weight of benches in a
parallel forces other, but not in the same plane. class room
Non-coplanar All forces do not lie in the same A tripod carrying a camera
concurrent forces plane, but their lines of action pass
through a single point.
Non-coplanar All forces do not lie in the same Forces acting on a moving bus
non-concurrent forces plane and their lines of action do
not pass through a single point.
action of the force. The point about which the moment is considered is called moment centre and
the perpendicular distance of the point from the line of action of the force is called moment arm.
Referring to Fig. 2.5, if d1 is the perpendicular distance of point 1 from the line of action of force
F, the moment of F about point 1 is given by
M 1 = F d1 ...(2.2)
Similarly, moment about point 2 is given by
M 2 = F d2 ...(2.3)
If the moment centre 3 lies on the line of action of the force F, the moment arm is zero and
hence,
M3 = F × 0 = 0 ...(2.4)
Thus, it may be noted that if a point lie on the line of action of a force, the moment of the force
about that point is zero.
The moment of a force has got direction also. In Fig. 2.5 it may be noted
that M1 is clockwise and M2 is anticlockwise. To find the direction of the
F
moment, imagine that the line of action of the force is connected to the point d
2x d 1
by a rigid rod pinned at the point and is free to move around the point. The 2 1x
B d2 y
d1 R B
d R
d
F2 F2
B1
F1
A
A F1 x
(a) (b)
Fig. 2.6
Similarly, if F1x and F2x are the components of F1 and F2, in x direction, respectively, then
F1 d1 = AB F1x ...(b)
and F2 d2 = AB F2x ...(c)
From Eqns. (b) and (c)
F1 d1 + F2 d2 = AB (F1x + F2x)
= AB × Rx ...(d)
From equation (a) and (d), we get
Rd = F1 d1 + F2 d2
If a system of forces consists of more than two forces, the above result can be extended as given
below:
Let F1, F2, F3 and F4 be four concurrent forces and R be their resultant. Let d1, d2, d3, d4 and
a be the distances of line of action of forces F1, F2, F3, F4 and R, respectively from the moment
centre O, [Ref. Fig 2.7].
If R1 is the resultant of F1 and F2 and its distance from O is a1, then applying Varignon’s
theorem:
R 1 a 1 = F 1 d1 + F 2 d 2
If R2 is the resultant of R1 and F3 (and hence of F1, F2 and F3) and its distance from O is a2,
then applying Varignon’s theorem:
R2a2 = R1a1 + F3d3
= F 1 d1 + F 2 d2 + F 3 d3
d
4 O
d2
F3 d
3
F2
a
d1
R
F1
F4
Fig. 2.7
FUNDAMENTALS OF STATICS 21
100N
B 60°
500 mm
400 mm
Fig. 2.8
3
C
Fy 4
B Fx
y
A x
Fig. 2.9
22 MECHANICS OF SOLIDS
Solution: 5000 N force is shifted to a point B along its line of action (law of transmissibility) and
it is resolved into its x and y components (Fx and Fy as shown in Fig. 2.9).
4
Fx = 5000 cos θ = 5000 × = 4000 N
5
3
and Fy = 5000 sin θ = 5000 ×= 3000 N.
5
By Varignon's theorem, moment of 5000 N force about A is equal to moment of its component
forces about the same point.
8000 = 4000 × y + 3000 × 0
∴ y = 2 m.
2.5 COUPLE
Two parallel forces equal in magnitude and opposite in direction and separated by a definite
distance are said to form a couple. The sum of the forces forming a couple is zero, since they are
equal and opposite, which means the translatory effect of the couple is zero.
An interesting property can be observed if we consider rotational effect of a couple about any
point. Let the magnitude of the forces forming the couple be F and the perpendicular distance
between the two forces be d. Consider the moment of the two forces constituting a couple about point
1 as shown in Fig. 2.10(a). Let the moment be M1 then,
M1 = Fd1 + Fd2
= F (d1 + d2) = Fd
Now, consider the moment of the forces about point 2 which is outside the two forces as shown
in Fig. 2.10(b). Let M2 be the moment.
Then,
M2 = Fd3 – Fd4
= F (d3 – d4) = Fd
Similarly it can be seen that M3 = Fd
Thus at any point M = Fd ...(2.6)
F F
3 F
d1
1d d3 d5
d 2
d d4 d6
2 d
F F
(a) (b) (c)
F
Fig. 2.10
Thus, moment of a couple about any point is the same. Now we can list the following characteristics
of a couple:
– A couple consists of a pair of equal and opposite parallel forces which are separated by a
definite distance;
FUNDAMENTALS OF STATICS 23
F A F
A A
d = d = M = Pd
B B
B
F
(a) (b) (c)
Fig. 2.11
Now it can be shown that F at A may be resolved into force F at B and a couple of magnitude
M = F × d, where d is the perpendicular distance of B from the line of action of F through A.
By applying equal and opposite forces F at B the system of forces is not disturbed. Hence the
system of forces in Fig. 2.11(b) is the same as the system given in Fig. 2.11(a). Now the original
force F at A and the opposite force F at B form a couple of magnitude Fd. The system in Fig. 2.11(b)
can be replaced by the system shown in Fig. 2.11(c). Thus, the given force F at A is replaced by
a force F at B and a moment Fd.
(iii) The direction of the travel gives the direction of component forces
(iv) From the triangle law of forces, the magnitude of the components can be calculated.
After finding the components of all the forces in the system in the two mutually perpendicular
directions, the component in each direction are algebraically added to get the two components. These
two components, which are mutually perpendicular, are combined to get the resultant.
Let F1, F2, F3 and F4 shown in Fig. 2.12(a) be the system of four forces the resultant of which
is required.
Y
F2
R
F1
F1y
F3x F2y F2x O F4x F1x Fy
x
F3y
F4y
F3 Fx
F4 (b)
(a)
Fig. 2.12
( )
2
R= (Σ Fx )2 + Σ Fy ...(2.7)
Example 2.3. Determine the resultant of the three forces acting on a hook as shown in Fig. 2.13(a).
y
70 N
80 N
52.07
25° R
25°
o
x
152.86
45°
(b)
50 N
(a)
Fig. 2.13
120 N
200 N
4
65.54
1 R
3 2
x
60º
40° 146.16
(b)
100 N
50 N
(a)
Fig. 2.14
26 MECHANICS OF SOLIDS
N
F θ°
H o rizon ta l
Fig. 2.15
Solution: In this problem, note that selecting X and Y axes parallel to the plane and perpendicular
to the plane is convenient.
Rx = ΣFx = T – F – W sin θ
= 1200 – 100 – 1000 sin 60° = 233.97 N
Ry = ΣFy = N – W cos 60° = 500 – 1000 cos 60° = 0.
∴ Resultant is force of 233.97 N directed up the plane.
Example 2.6. Two forces acting on a body are 500 N and 1000 N as shown in Fig. 2.16(a).
Determine the third force F such that the resultant of all the three forces is 1000 N directed at 45°
to x axis.
Solution: Let the third force F make an angle θ with x axis.
FUNDAMENTALS OF STATICS 27
Y
1000 N
R=1000 N
30°
500 N 225.9 N
45°
30°
X
408.91 N
(a) (b)
Fig. 2.16
F 408.91I
θ = tan −1
H 255.9 K = 61.08° as shown in Fig. 2.16.
Example 2.7. Three forces acting at a point are shown in Fig. 2.17. The direction of the 300 N forces
may vary, but the angle between them is always 40°. Determine the value of θ for which the resultant
of the three forces is directed parallel to b-b.
Solution: Let the x and y axes be as shown in Fig. 2.17. If the resultant is directed along the x axis,
its component in y direction is zero.
i.e., 0 = ΣFy = 300 sin θ + 300 sin (40 + θ) – 500 sin 30°
= 0.8333 Y
40° x
∴ sin θ + sin(40 + θ) = 0.8333 b
FG 40 + θ + θ IJ × cos FG 40 + θ − θ IJ = 0.8333
2 sin
H 2 K H 2 K 500 N
30
°
∴ θ = 6.35°
Fig. 2.17
28 MECHANICS OF SOLIDS
F3 F3
F1
0
F1
Mo
0
F2 F2
(a) (b)
R
R
0
Mo d
0
(c) (d)
Fig. 2.18
The force R and moment ΣMO shown in Fig. 2.18(c) can be replaced by a single force R acting
at a distance d from O such that the moment produced by this force R is equal to ΣMO [Ref. 2.18(d)].
Thus, we get a single force R acting at a distance d from the point O which gives the same effect
as the constituent forces of the systems. Thus, the resultant of the given forces may be reduced to
a single force.
FUNDAMENTALS OF STATICS 29
Mathematically,
R = ( Σ Fx ) 2 + ( Σ Fy ) 2 U|
tan α =
Σ Fy V| ...(2.11)
Σ Fx W
Σ Mo
d=
R
where, ΣFx – algebraic sum of the components of all forces in x direction
ΣFy – algebraic sum of the components of all forces in y direction
α – inclination of the resultant R to x direction
ΣMO – algebraic sum of the moments of all the forces about point O
d – is distance of the resultant R from the point O.
Note: R is marked at distance d such that it produces the same direction of moment about point O as ΣMO.
Sometimes the values of ΣFx and ΣFy may come out to be zero, but ΣMO may exist. This means
that the resultant of the system gets reduced to a pure couple.
A F1
y F2
R
d
F3
X
O x B
Fig. 2.19
Rd = ΣMO
Rx × 0 + Ry x = ΣMO
Σ Mo Σ Mo
∴ x = = ...(2.13)
Ry Σ Fy
R
x
B Rx
O x
Ry
Fig. 2.20
Similarly, resolving the resultant into its components at A, it can be shown that:
Σ Mo Σ Mo
y = = ...(2.14)
Rx Σ Fx
Example 2.8. A system of loads acting on a beam is shown in Fig. 2.21(a). Determine the resultant
of the loads.
Solution: Taking horizontal direction towards left as x axis and the vertical downward direction as
y axis.
ΣFx = 20 cos 60° = 10 kN
ΣFy = 20 + 30 + 20 sin 60° = 67.3205
2 0 kN 3 0 kN R 2 0 kN
R Σy
α 6 0°
B α
A
1 .5 m 1 .5 m 3 .0 m 2 .0 m
d Σx
x
(a) (b)
Fig. 2.21
∴ R = b Σxg + b Σy g
2 2
= 102 + ( 67. 3205) 2
i.e., R = 68.0592 kN.
Σ Fy
tan α = = 6.7321
Σ Fx
∴ α = 81.55°.
FUNDAMENTALS OF STATICS 31
y B 80 N
100 N
R Fy
R
60°
C
x A
120 N Fx
30°
(b)
100 mm
x
(a)
Fig. 2.22
←
Rx = 73.92 N
ΣFy = 80 + 120 sin 30° – 100 sin 60°
Ry = 53.40 N
Example 2.10. Four forces having magnitudes of 200 N, 400 N, 600 N and 800 N respectively, are
acting along the four sides (1 m each) of a square ABCD taken in order, as shown in Fig. 2.23.
Determine the magnitude and direction of the resultant force.
400 N
600 N D 400 N
C
q
R
400 N
q
200 N
R A B (b)
x 800 N
(a)
Fig. 2.23
←−−−
Solution: ΣFx = 200 – 600 = –400 N = 400 N
ΣFy = 400 – 800 = –400 N = 400↓ N
∴ R = (Σx ) 2 + (Σy )2 = 4002 + 4002 = 400 2
= 565.68 N
400
θ = tan −1 = 45° , as shown in Fig. 2.23(b).
400
ΣMA = 400 × 1 + 600 × 1 = 1000 N-m.
Let x be the distance from A along x axis, where resultant cuts AB. Then
ΣM A 1000
x = = = 2.5 m as shown in the Fig. 2.23(a).
ΣFy 400
Example 2.11. Forces 2, 3 , 5, 3 and 2 kN respectively act at one of the angular points of a
regular hexagon towards five other angular points. Determine the magnitude and direction of the
resultant force.
Solution: Let the system of forces be as shown in Fig. 2.24 shown below:
Let O be the centre of the encircling circle A, B, C, D, E and F. In regular hexagon each side
is equal to the radius AO. Hence OAB is equilateral triangle.
∴ ∠OAB = 60°
In ∆ABC, BA = BC
∴ ∠CAB = ∠BCA
But ∠CAB + ∠BCA = interior angle at B
= 180° – 120° = 60°
FUNDAMENTALS OF STATICS 33
2 kN
F 3 kN
E
θ4
θ3
A O 5 kN
θ2 D
θ1
B C
2 kN 3 kN
Fig. 2.24
60
∴ ∠CAB = ∠BCA = = 30°
2
θ1 = 30°
∴ θ2 = 60° – 30° = 30°
Similarly θ3 = 30°
and θ4 = 30°
∴ ΣFx = 2 cos 60° + 3 cos 30° + 5 + 3 cos 30° + 2 cos 60°
= 10 kN
ΣFy = –2 sin 60° – 3 sin 30° + 0 + 3 sin 30° + 2 sin 60°
= 0
∴ R = 102 + 02 = 10 kN
θ = 0° i.e., the resultant is in the direction x.
Example 2.12. Find the resultant of a set of coplanar forces acting on a lamina as shown in
Fig. 2.25(a). Each square has side of 10 mm.
y 2 kN
1 R
R
Fy
d
2
Fx
3 (b)
x
O
(a)
1.5 kN 5 kN
Fig. 2.25
34 MECHANICS OF SOLIDS
Solution: If θ1, θ2 and θ3 are the slopes of the forces 2 kN, 5 kN and 1.5 kN forces with respect
to x axis, then
10
tan θ1 = = 1 ∴ θ1 = 45°
10
30
tan θ2 = ∴ θ2 = 36.87°
40
10
tan θ3 = ∴ θ3 = 26.565
20
Rx = ΣFx = 2 cos 45° + 5 cos 36.87° – 1.5 cos 26.565° = 4.072 kN
R y= ΣFy = 2 sin 45° – 5 sin 36.87° – 1.5 sin 26.565° = 2.26 kN
R = ( Σ Fx ) 2 + ( Σ Fy ) 2 = 4.66 kN.
2.26
tan α =
4.072
∴ α = 28.99°.
Distance d of the resultant from O is given by
Rd = ΣMO
4.66 d = 2 × cos 45° × 30 + 5 × sin 36.87° × 50 + 1.5 × sin 26.565° × 10
= 199.13
d = 42.77 mm as shown in Fig. 2.25(a).
Note: To find moment of forces about O, 2 kN force is resolved at it’s intersection with y axis and
5 kN and 1.5 kN forces are resolved at their intersection with x axis, and then Varignon theorem is used.
Example 2.13 Determine the resultant of four parallel forces acting on the axle of a vehicle as shown
in Fig. 2.26.
60 kN 20 kN 30 kN 40 kN
x
A B C D
1m 2m 1m
x
y R
Fig. 2.26
∴ R= 0 2 + 150 2 = 150 kN
Taking clockwise moment as +ve,
ΣMA = 60 × 0 + 20 × 1 + 30 × 3 + 40 × 4
= 270 kN-m
∴ Distance of resultant from A
270
x=
= 1.8 m as shown in the figure.
150
Example 2.14. Determine the resultant of system of parallel forces acting on a beam as shown in
Fig. 2.27.
80 kN 30 kN 40 kN 50 kN 60 kN
x
A B C D E
2m 2m 4m 2m
x
y R
10 m
Fig. 2.27
Example 2.15. The system of forces acting on a bell crank is shown in Fig. 2.28(a). Determine the
magnitude, direction and the point of application of the resultant.
700 N
0
15
500 N
R
1200 N R
0
Fy
15
60°
60°
x O
150
Fx
150
1000 N
Solution: Rx = ΣFx = 500 cos 60° – 700 = – 450 N = 450 N (from right to left)
Ry = ΣFy = – 1200 – 1000 – 500 sin 60° = – 2633.01 N
= 2633.01 N (downward)
R = 4502 + ( 2633. 01) 2
R = 2671.19 N.
2633. 01
tan α =
450
α = 80.30°, as shown in Fig. 2.28(b).
Let the point of application of the resultant be at a distance x from the point O along the
horizontal arm. Then,
x × 2633.01 = 500 sin 60° × 300 + 1000 × 150
– 1200 × 150 cos 60° + 700 × 300 sin 60°
x = 141.195 mm, as shown in Fig. 2.28(a).
Example 2.16. Various forces to be considered for the stability analysis of a dam are shown in the
Fig. 2.29. The dam is safe if the resultant force passes through middle third of the base. Verify
whether the dam is safe.
Solution: Rx = ΣFx = 500 kN
Ry = ΣFy = + 1120 – 120 + 420 = 1420 kN
= 1420 kN (downward)
Let x be the distance from O where the resultant cuts the base.
120 kN
500 kN 2m
1120 kN
4m
4m 420 kN
5m
x
O
7m
y
Fig. 2.29
Example 2.17. A building frame is subjected to wind loads as shown in Fig. 2.30. Determine the
resultant of the loads.
5 kN 5 kN
1 0 kN 1 0 kN
d 1 .5 m
1 0 kN 1 0 kN 3m
R
5 kN 5 kN
1m 1m 1m 1m 1m 1m C
6m
Fig. 2.30
Solution: The roof is inclined at 45° to horizontal and loads are at 90° to the roof. Hence, the loads
are also inclined at 45° to vertical/horizontal.
Now,
Rx = ΣFx = (5 + 10 + 10 + 5 + 5 +10 + 10 + 5) cos 45°
1
= 60 ×= 42. 426 kN
2
ΣFy = –(5 + 10 + 10 + 5) sin 45° + (5 + 10 + 10 + 5) sin 45°
= 0
∴ R = Σ Fx = 42.426 kN
and its direction is horizontal.
Let R be at a distance d from the ridge A.
Then, Rd = ΣMA
1 3 2 1 1 2 3
60 × × d = 5× + 10 × + 10 × + 10 × + 10 × + 5×
2 2 2 2 2 2 2
d = 1.5 m
∴ Resultant is a horizontal force of magnitude 42.426 kN at 1.5 m below A.
Example 2.18. Determine the magnitude, direction and line of action of the equilibriant of the given
set of coplanar forces acting on a planar structure shown in Fig. 2.31.
Solution: The two 40 kN forces acting on the smooth pulley may be replaced by a pair of 40 kN
forces acting at centre of pulley C and parallel to the given forces, since the sum of moments of the
two given forces about C is zero.
Now, Rx = ΣFx = 20 cos 45° – 30 cos 60° – 50 cos 30° + 40 cos 20° – 40 sin 30°
ΣFx = – 26.5714 kN
= 26.5714 kN (from right to left)
38 MECHANICS OF SOLIDS
20°
C
40 kN
30° 50 kN
40 kN 30° 20 kN
3m
R
113.4447
30 kN
E
x 60°
A 45° B
20 kN 26.5741
20 kN
2m 2m 2m
(a) (b)
Fig. 2.31
113. 4447
tan α =
26.5714
α = 76.82° as shown in Fig. 2.31(b)
Let the resultant intersect AB at a distance x from A. Then,
x ΣFy = ΣMA
x × 113.4447 = 20 × 4 – 20 × 4 + (30 sin 60°) × 6 + (50 sin 30°)
× 2 – (50 cos 30°) × 2 + (40 cos 20°) × 3
– (40 sin 30°) × 3
x = 1.516 m
The equilibriant is equal and opposite to the resultant. Hence, it is as shown in Fig. 2.31(a) in
which E = 116.515 kN, α = 76.82° and x = 1.516 m.
If the surface of contact is smooth, the direction of the reaction is normal to the surface of
contact. If the surface of contact is not smooth, apart from normal reaction, there will be frictional
reaction also. Hence the resultant reaction will not be normal to the surface of contact.
B a ll
T
S m oo th
B a ll R
6 00 N
S m o o th
6 00 N
W R1
L ad de r G
P
P
S m o oth
R2
B lock w eigh in g 60 0 N
4 00 N
6 00 N
6 00 N
R
(a) The algebraic sum of the component of forces along each of the two mutually perpendicular
directions is zero (translatory motion is zero).
(b) The algebraic sum of moment of all the forces about any point in the plane is zero
(rotational moment is zero).
The above conditions for coplanar concurrent and non-concurrent forces is discussed and illustrated
in this article.
b
F3
F1
a
F3
(a) (b)
Fig. 2.32
42 MECHANICS OF SOLIDS
Fig. 2.33
Notes:
1. The string can have only tension in it (it can pull a body), but cannot have compression in it (cannot
push a body).
2. The wall reaction is a push, but cannot be a pull on the body.
3. If the magnitude of reaction comes out to be negative, then assumed direction of reaction is wrong. It
is acting exactly in the opposite to the assumed direction. However, the magnitude will be the same. Hence
no further analysis is required. This advantage is not there in using Lami's equation. Hence, it is advisable for
beginners to use equations of equilibrium, instead of Lami's theorem even if the body is in equilibrium under
the action of only three forces.
Example 2.20. Determine the horizontal force P to be applied to a block of weight 1500 N to hold
it in position on a smooth inclined plane AB which makes an angle of 30° with the horizontal [Fig.
2.34(a)].
Solution: The body is in equilibrium under the action of applied force P, self-weight 1500 N and
normal reaction R from the plane. Since R, which is normal to the plane, makes 30° with the vertical
(or 60° with the horizontal),
ΣFy = 0, gives
R cos 30°– 1500 = 0
R = 1732.06 N.
ΣFx = 0, gives
P – R sin 30° = 0
P = R sin 30°
P = 866.03 N.
1 50 0 N
B Y
P P
X
3 0°
3 0°
A R
(a ) (b )
Fig. 2.34
Note: Since the body is in equilibrium under the action of only three forces the above problem can be solved
using Lami’s theorem as given below:
R P 1500
= =
sin 90° sin(180 − 30) sin( 90 + 30)
R = 1732.06 and P = 866.03.
Example 2.21. A roller of weight 10 kN rests on a smooth horizontal floor and is connected to the
floor by the bar AC as shown in Fig. 2.35 (a). Determine the force in the bar AC and reaction from
floor, if the roller is subjected to a horizontal force of 5 kN and an inclined force of 7 kN as shown
in the figure.
44 MECHANICS OF SOLIDS
Solution: A bar can develop a tensile force or a compressive force. Let the force developed be a
compressive force S (push on the cylinder). Free Body Diagram of the cylinder is as shown in
Fig. 2.35(b).
10 kN
7 kN 7 kN
45° 45°
5 kN 5 kN
C 30°
A 30° S
R
(a) (b)
Fig. 2.35
Since there are more than three forces in the system, Lami’s equations cannot be applied.
Consider the components in horizontal and vertical directions.
ΣH = 0
S cos 30° + 5 – 7 cos 45° = 0
7 cos 45° − 5
S= = – 0.058 kN
cos 30°
Since the value of S is negative the force exerted by the bar is not a push, but it is pull
(tensile force in bar) of magnitude 0.058 kN.
ΣV = 0
R – 10 – 7 sin 45° + S sin 30° = 0
R = 10 + 7 sin 45° – S sin 30°
= 10 + 7 sin 45° – (– 0.058) sin 30°
R = 14.979 kN.
Example 2.22. A cord ACB 5 m long is attached at points A and B to two vertical walls 3 m apart
as shown in Fig. 2.36(a). A pully C of negligible radius carries a suspended load of 200 N and is
free to roll without friction along the cord. Determine the position of equilibrium, as defined by the
distance X, that the pulley will assume and also the tensile force in the cord.
Solution: The pulley C is in equilibrium under the action of tensile forces in CA and CB and vertical
downward load 200 N. The tensile forces in segment CA and CB are the same since the pulley is
frictionless. Now consider the equilibrium of pulley C shown in Fig. 2.36(b).
ΣH = 0
T cos θ1 – T cos θ2 = 0
∴ θ 1 = θ2, say, equal to θ
Now, let BC be extended to D.
∆CFD = ∆CFA
∴ CD = AC
FUNDAMENTALS OF STATICS 45
BD = BC + CD = BC + AC = length of chord = 5 m
DE = 3 m
∴ BE = 4 m
B
3m
1m
x T T
H
A I
T T
θ1 θ2
θ θ
F G
θ C
2 00 N
D E 2 00 N
(a ) (b )
Fig. 2.36
B
P P
O O
r 3 00 α
A C
150
R
(a )
2 00 0 N
(b)
Fig. 2.37
P A
θ 2 00 0 N
C
θ
R B
(a) (b)
Fig. 2.38
Fig. 2.39
Solution: Free body diagrams of points B and D are shown in Fig. 2.39(b). Let the forces in the
members be as shown in the figure.
Applying Lami’s theorem to the system of forces at point D,
T1 T2 250
= =
sin 120° sin 135° sin 105°
∴ T1 = 224.14 N.
T2 = 183.01 N.
Consider the system of forces acting at B.
ΣV = 0
T3 cos 30° – 200 – T2 cos 60° = 0
200 + 183. 01 cos 60°
T3 =
cos 30°
T3 = 336.60 N.
ΣH = 0
T4 – T2 sin 60° – T3 sin 30° = 0
T4 = 183.01 × sin 60° + 336.60 sin 30°
T4 = 326.79 N.
Example 2.25. A rope AB, 4.5 m long is connected at two points A and B at the same level 4 m apart.
A load of 1500 N is suspended from a point C on the rope 1.5m from A as shown in Fig. 2.40(a).
What load connected at a point D on the rope, 1 m from B will be necessary to keep the position
CD level ?
48 MECHANICS OF SOLIDS
x
A E F B
α β
T1 T3 T1 T3
y 1.
5m T2 T2
0m
T2 T2
α β
1.
C
2 .0 m D
1 50 0 N W=?
1 50 0 N W
(a ) (b ) (c)
Fig. 2.40
(2 – x)2 + y2 = 1 ...(iii)
From (1) and (3)
x2 – (2 – x)2 = 1.25
i.e., x2 – 4 + 4x – x2 = 1.25
x = 1.3125 m
FG 1.3125IJ = 28.955°
∴ α = cos−1
H 1.5 K
FG 2 − 1. 3125IJ = 46.567°
β = cos−1
H 1 K
Applying Lami’s theorem to the system of forces acting at point C [Fig. 2.40(b)], we get
T1 T2 1500
= =
sin 90° sin 118.955° sin (180 − 28.955)°
T1 = 3098.39 N
T2 = 2711.09 N
FUNDAMENTALS OF STATICS 49
A D
30º T1 θ
B
T2 T3
50º T1
2 0 kN T2 T2 T3
T3 30º
C 50º θ
T2
50º
3 0 kN
2 0 kN 3 0 kN
(a )
(b ) (c )
Fig. 2.41
Solution: Writing equations of equilibrium for the system of forces at joints B and C [Figs. 2.42(b)
and (c)], we have
T2 sin θ = T1 sin 30° ...(i)
T2 cos θ = T1 cos 30° – 20 ...(ii)
A D
6 0°
3 0° T1
B T2 3 0°
T3 T3
T2 T1
θ C T2 T2 6 0°
θ
2 0 kN θ
2 5 kN
2 0 kN 2 5 kN
(a )
(b ) (c)
Fig. 2.42
RD
D D
3 00 m m
6 00 m m
m
0
m RC RA RC α
15
O 1 50 m m
r≡
C O A
RB
6 00 N
B
RB
(a ) (b ) (c)
Fig. 2.43
As the frame is in equilibrium under the action of three forces only, they must be concurrent
forces. In other words, reaction at D has line of action alone OD. Hence, its inclination to horizontal
is given by:
450
tan α = =3
150
α = 71.5650°
ΣV = 0, gives
RD sin α = RB = 600
∴ RD = 632.456 N.
ΣH = 0, gives
RC = RD cos α
∴ RC = 200 N.
From (2), ∴ RA = 200 N.
Example 2.29. Two smooth spheres each of radius 100 mm and weight 100 N, rest in a horizontal
channel having vertical walls, the distance between which is 360 mm. Find the reactions at the points
of contacts A, B, C and D shown in Fig. 2.44(a).
Solution: Let O1 and O2 be the centres of the first and second spheres. Drop perpendicular O1P to
the horizontal line through O2. Figures 2.44(b) and 2.44(c) show free body diagram of the sphere
1 and 2, respectively. Since the surface of contact are smooth, reaction of B is in the radial direction,
i.e., in the direction O1O2. Let it make angle a with the horizontal. Then,
O2 P 360 − O1 A − O2 D 360 − 100 − 100
cos α = = = = 0.8
O1O2 O1B + BO2 100 + 100
∴ sin α = 0.6.
Consider sphere No. 1.
ΣV = 0, gives
RB × 0.6 = 100
RB = 166.67 N.
ΣH = 0, gives
RA = RB × 0.8
∴ RA = 133.33 N.
52 MECHANICS OF SOLIDS
3 60 m m 1 00 N RA
O1
B A RB
α RD (b )
D
O2 P
1 00 N
C
RC
(a ) (c)
Fig. 2.44
A
15° P
60°
15°
B
60° 45°
(a)
15° P
C 60°
15°
15° 45°
60°
4000 N 2000 N
60° C
45°
60°
(b) (c)
Fig. 2.45
FUNDAMENTALS OF STATICS 53
Solution: Figures 2.45(b) and 2.45(c) show the free body diagram of the two cylinders. Applying
Lami’s theorem to the system of forces on cylinder A, we get
C 4000
=
sin 60° sin(60 + 90 − 15)
C = 4898.98 N
Consider cylinder B.
Summation of the forces parallel to the inclined plane (45° to horizontal) = 0, gives:
P cos 15° + 2000 cos 45° – C cos 60° = 0
4898.98 cos 60° − 2000 cos 45°
∴ P =
cos 15°
P = 1071.8 N.
ΣF y R
A
x
d1 ΣF x
d2 d
B
Fig. 2.46
54 MECHANICS OF SOLIDS
If the condition ΣMA = 0 is satisfied, we can conclude that the resultant R should be a force
passing through A. If ΣMB = 0 is satisfied, the line of action of the resultant passes through B, i.e.,
AB is the line of action. Now, if ΣMC = 0 is also satisfied, the resultant should be zero since the line
of action of the resultant (line AB) cannot pass through C also. Thus if A, B, C are not collinear
points, the following conditions are necessary and sufficient conditions of equilibrium.
ΣM A = 0 U|
and ΣM B = 0 V| ...(2.17)
ΣMC = 0 W
The equilibrium Equations 2.16 are not independent of equilibrium Equations 2.15. Two of them
are common to the two sets. Referring to Fig. 2.46, R can be resolved into its components perpendicular
to and parallel to AC.
Then, ΣMC = (ΣFx)d
∴ if ΣMC = 0, (ΣFx)d = 0 or ΣFx = 0
Now, ΣMB = 0
i.e., (ΣFx)d1 + (ΣFx)d2 = 0
But ΣFx = 0
∴ (ΣFy)d2 = 0
Since ABC are not colinear, d2 ≠ 0
∴ ΣFy = 0
Thus, ΣMC = 0 is identical to ΣFx = 0 and ΣMB = 0 is identical to ΣFy = 0. Hence any one of
the following sets may be used as equations of equilibrium:
(1) ΣFx = 0; ΣFy = 0; ΣM A = 0
(2) If line AB is not in y direction,
ΣFy = 0; ΣM A = 0; ΣM B = 0
(3) If line AB is not in x direction ...(2.18)
ΣFx = 0; ΣM A = 0; ΣM B = 0
(4) If A, B and C are non-collinear
ΣM A = 0; ΣM B = 0; ΣM C = 0
It can be proved that if a system is in equilibrium under the action of three forces, those three
forces must be concurrent.
Let P1, P2 and P3 be the forces acting on a body as shown in Fig. 2.47 and let P1 and P2 intersect
at A. Then applying moment equilibrium condition about A, we get
ΣMA = 0 i.e. P3d = 0
where d is distance of line of action of P3 from A.
Since P3 is not zero,
d = 0, i.e. P3 also must pass through A.
Hence, the proposition is proved.
FUNDAMENTALS OF STATICS 55
P3
P1
A
d
P2
Fig. 2.47
Three problems are solved in this article. The problem of finding the reactions at supports of
beams and trusses also fall under this categories. But these problems are solved in the chapters
analysis of beams and trusses.
Example 2.31. The 12 m boom AB weighs 1 kN, the distance of the centre of gravity G being 6 m
from A. For the position shown, determine the tension T in the cable and the reaction at B [Ref.
Fig. 2.48(a)].
Solution: The free body diagram of the boom is shown in Fig. 2.48(b).
ΣMA = 0, gives
T sin 15° × 12 – 2.5 × 12 cos 30° – 1 × 6 cos 30° = 0
T = 10.0382 kN.
∑ H = 0, gives
HA – T cos 15° = 0
HA = 9.6962 kN
B
1 5°
1 5° T B
2 .5 kN 6m 1 5°
C
6m
G HA 3 0° 1 kN 2 .5 kN
3 0°
A
A
VA
(a ) (b )
VA RA
α
HA
(c)
Fig. 2.48
∑V = 0, gives
VA = 1 + 2.5 + T sin 15° = 6.0981 kN
RA = V A2 + H A2
RA = 11.4544 kN.
56 MECHANICS OF SOLIDS
−16. 0981
α = tan
9. 6962
= 32.17° as shown in Fig. 2.48(c).
Example 2.32. A cable car used for carrying materials in a hydroelectric project is at rest on a track
formed at an angle of 30° with the vertical. The gross weight of the car and its load is 60 kN and
its centroid is at a point 800 mm from the track half way between the axles. The car is held by a
cable as shown in Fig. 2.49. The axles of the car are at a distance 1.2 m. Find the tension in the
cables and reaction at each of the axles neglecting friction of the track.
T
Cable
600
30°
800
G
R2
0
60
60 kN
0
60
R1
60°
Fig. 2.49
Solution: Let T be the tension in the cable and the reaction at the pair of wheels be R1 and R2 as
shown in Fig. 2.49.
Now, ∑ of forces parallel to the track = 0, gives
T – 60 sin 60° = 0
T = 51.9615 kN.
Taking moment equilibrium condition about upper axle point on track, we get
R1 × 1200 + T × 600 – 60 sin 60° × 800 – 60 cos 60° × 600 = 0
R1 = 23.6603 kN.
∑ of forces normal to the plane = 0, gives
R1 + R2 – 60 cos 60° = 0
R2 = 30 – 23.6603
R2 = 6.3397 kN.
Example 2.33. A hollow right circular cylinder of radius 800 mm is open at both ends and rests on
a smooth horizontal plane as shown in Fig. 2.50(a). Inside the cylinder there are two spheres having
weights 1 kN and 3 kN and radii 400 mm and 600 mm, respectively. The lower sphere also rests on
the horizontal plane. Neglecting friction find the minimum weight W of the cylinder for which it will
not tip over.
Solution: Join the centres of spheres, O1 and O2 and drop O1D perpendicular to horizontal through
O2 .
Now, O1O2 = 400 + 600 = 1000 mm
O2D = 1600 – 400 – 600 = 600 mm
FUNDAMENTALS OF STATICS 57
1600
W 1 kN O
O1 R1 1
R1
h1
α O2 R2 α O2
D D R2
h2 3 kN
A B
(a ) (b ) (c)
R3
Fig. 2.50
Free body diagrams of cylinder and spheres are shown in Fig. 2.50(b) and (c). Considering the
equilibrium of the spheres,
∑ M O2 = 0, gives
R1 × O1O2 sin α – 1 × O2D = 0
R1 × 1000 × 0.8 = 1 × 600
R1 = 0.75 kN
∑ H = 0, gives
R2 = R1 = 0.75 kN
∑V = 0, gives
R3 = 1 + 3 = 4 kN
Now consider the equilibrium of cylinder. When it is about to tip over A, there is no reaction
from ground at B. The reaction will be only at A. ∑ M A = 0 gives,
R1 h1 – R2 h2 – W × 800 = 0
R1(h1 – h2) – W × 800 = 0
Since R1 = R2
0.75 × O1D = W × 800
0.75 × 1000 × 0.8 = W × 800
∴ W = 0.75 kN.
IMPORTANT FORMULAE
5. Varignon’s theorem
Ra = P1d1 + P2d2 + P3d3 + ....
6. Magnitude of a couple
M = Pd.
7. P at any point A may be resolved into force P at B plus a moment P × d, where ‘d’ is the
perpendicular distance of B from the line of action of P through A.
8. Rx = R cos α = ΣFx
Ry = R sin α = ΣFy
ΣFy
tan α =
ΣFx
d ΣM o
and y= = .
cos α ΣFx
10. Lami’s theorem:
F1 F F
= 2 = 3 .
sin α sin β sin γ
THEORY QUESTIONS
1. A body is subjected to the three forces as shown in Fig. 2.51. If possible determine the direction
of the force F so that the resultant is in x-direction, when
(a) F = 5000 N
(b) F = 3000 N
3 00 0 N
2 00 0 N
6 0°
x
θ
Fig. 2.51
B
A
30°
D
°
G
30
C 20 kN
Fig. 2.52
W 10 kN
Fig. 2.53
60 MECHANICS OF SOLIDS
4. Three bars, hinged at A and D and pinned at B and C as shown in Fig. 2.54 form a four-linked
mechanism. Determine the value of P that will prevent movement of bars.
B C
45° 60°
75° 45°
2000 N
P
A D
Fig. 2.54
[Ans. P = 3047.2 N]
5. Two identical rollers, each of weights 100 N are supported by an inclined plane and a vertical
wall as shown in Fig. 2.55. Assuming smooth surfaces, find the reactions induced at the points
of supports A, B, C and D.
1
B
D 2 A
C
30°
Fig. 2.55
4
45°
Fig. 2.56
7. Two smooth spheres each of weight W and radius ‘r’ are in equilibrium in a horizontal channel
of width ‘b’ (b < 4r) and vertical sides as shown in Fig. 2.57. Find the three reactions from the
sides of the channel which are all smooth. Also find the force exerted by each spheres on the
other.
FUNDAMENTALS OF STATICS 61
Fig. 2.57
W
[Ans. RA = RD = W cot α, RC = 2W and RB =
sin α
b − 2r
where cos α = ]
2r
8. Determine the resultant of parallel force system shown in Fig. 2.58.
50 kN 40 kN 40 kN 40 kN
A D
B C
2m 3m 3m
Fig. 2.58
600 N
1000 N
60° O
30°
x
10°
40°
2000 N
400 N
d
D = 600 m m
R
Fig. 2.59
50 kN
B
60 kN
A C
30 kN
R d
80 kN
Fig. 2.60
11. Determine the magnitude, direction and the line of application of the equilibriant of the set of
forces shown in Fig. 2.61, which will keep the plane body ABCDEFGH in equilibrium.
[Ans. E = 23.6517 kN; α = 24.37°; x from A = 1.041 m]
20 kN 10 kN
D E 20 kN
1.0 m
C F
0.5 m
B G 20 kN
30 kN A H
0.5 m 0.5 m
10 kN α
0.5 m 1m
45°
E 10 kN
Fig. 2.61
FUNDAMENTALS OF STATICS 63
12. Determine the resultant of the four forces acting on a body as shown in Fig. 2.62.
[Ans. R = 200; α = 60°; at a distance y = 8.7679 m below O]
0.5 m 1m
60°
3 m 200 N
3
4
500 N
3m
O 400 N
300 N
Fig. 2.62
13. A bracket is subjected to the system of forces and couples as shown in Fig. 2.63. Find the
resultant of the system and the point of intersection of its line of action with (a) line AB,
(b) line BC and (c) line CD.
[Ans. R = 485.4 N; α = 34.50°; YBA = 112.5 mm; XBC = 163.6 mm; YCD = 93.75 mm]
150 N A
125 N
Fig. 2.63
14. Determine the resultant of the three forces acting on the dam section shown in Fig. 2.64 and
locate its intersection with the base AB. For a safe design this intersection should occur
within the middle third. Is it a safe design?
[Ans. Resultant intersects AB 3.333 m from A. It is a safe design]
64 MECHANICS OF SOLIDS
50 kN 2m
3m 120 kN 30 kN
1m
60°
A B
6m
Fig. 2.64
15. A 1000 N cylinder is supported by a horizontal rod AB and a smooth uniform rod CD which
weighs 500 N as shown in Fig. 2.65. Assuming the pins at A, B, C and D, to be frictionless
and weight of AB negligible, find the reactions at C and D.
[Ans. VC = 937.1 N (upward); HC = 577.3 N (towards left) RD = 562.9 N (upward)]
C
A B
m
1.5
m
3 .5
3 0°
D
Fig. 2.65
16. The frame shown in Fig. 2.66 is supported by a hinge at E and by a roller at D. Determine
the horizontal and vertical components of the reactions at hinge C as it acts upon member
BD. [Ans. HC = 140 N (towards right); VC = 35 N (upward)]
Fig. 2.66
3
Trusses
A truss is a structure made up of slender members pin-connected at ends and is capable of taking
loads at joints. They are used as roof trusses to support sloping roofs and as bridge trusses to support
deck. In many machines steel trusses are used. Transmission towers are also the examples of trusses.
In the case of wooden trusses, the ends are connected by making suitable joints or by nailing and
bolting whereas in steel trusses ends are connected by bolting or welding. The trusses are also known
as ‘pinjointed frames’.
A truss in which all the members lie in a single plane is called as a plane truss. In such trusses
loads act in the plane of the truss only. Roof trusses and bridge trusses can be considered as plane
trusses. If all the members of a truss do not lie in a single plane, then it is called a space truss. Tripod
and transmission towers are the examples of space trusses. In this chapter, the analysis of only plane
trusses is considered.
65
66 MECHANICS OF SOLIDS
4
2 4
1 6
2 5
1 5
3 3 7
Fig. 3.3
P
2 4 6 2 4
6
1 5 1 5
3 3
(a) (b)
Fig. 3.4
3.2 ASSUMPTIONS
In the theory that is going to be developed in this chapter, the following assumptions are made:
(1) The ends of the members are pin-connected (hinged);
(2) The loads act only at the joints;
(3) Self-weights of the members are negligible;
(4) Cross-section of the members is uniform.
If at all the cross-section varies, the centre of gravity of the section is assumed to be located
along the same longitudinal line.
In reality the members are connected by bolting, riveting or by welding. No special care is taken
to ensure perfect pin-connections. However, experiments have shown that assuming pin-connected
ends is quite satisfactory since the members used are slender.
TRUSSES 67
In most of the frames the loads act at the joints. Even if a load is not acting at a joint, it can
be replaced by its reaction at the joint and a local bending effect on the member. The frame may
be analysed for the joint loads and the local bending effect on the member superposed in the design
of that member.
In most of the trusses, the self-weight is really small compared to the loads they carry. Hence
self-weight of the members may be neglected.
It is the duty of construction engineer to see that the centroid of all cross-sections lie along a
single axis so that the member is held in equilibrium by the two forces acting at its ends.
Because of the assumption of pin-connected ends, it is more appropriate to call the theory that
is going to be developed in this chapter as analysis of pin-connected plane trusses. Analysis of rigid
frames is not covered in this book.
A D
Fig. 3.7(a)
Fig. 3.7(b)
The member AE is subjected to tensile force T. Its effect on the joints A and E are as shown
in Fig. 3.7(b). In the analysis of frame we mark the forces on the joints, instead of the forces in the
members as shown in Fig. 3.7(c). It may be noted that compressive force in a member is represented
in a figure by two arrows going away from each other and a tensile force by two arrows coming
towards each other. This is quite logical considering the fact that the markings on the members
represent the internal reactive forces developed which are opposite in direction to the applied forces.
68 MECHANICS OF SOLIDS
C O M PR ESSIO N
TEN SIO N
Fig. 3.7(c)
EXAMPLES A B
Solution: Step 1: Determine the inclinations of all inclined members. In this case,
3
tan θ = =1
3
∴ θ = 45°
Step 2: Look for a joint at which there are only two unknowns. If such a joint is not available,
determine the reactions at the supports, and then at the supports these unknowns may reduce to only
two.
Now at joints C, there are only two unknowns, i.e., forces in members CB and CD, say FCB and
FCD .
Note: Usually in cantilever type frames, we find such joints without the need to find reactions.
Step 3: Now there are two equations of equilibrium for the forces meeting at the joint and two
unknown forces. Hence, the unknown forces can be determined.
At joint C [Ref. Fig. 3.8(b)] ∑V = 0 condition shows that the force FCB should act away from
the joint C so that its vertical component balances the vertical downward load at C.
FCB sin 45° = 40 FCB
∴ FCB = 40 2 kN
Now ∑ H = 0 indicates that FCD should act towards C. 45°
C
FCD – FCB cos 45° = 0 FCD
1
FCD = FCB cos 45° = 40 2 × = 40 kN 40 kN
2
Fig. 3.8(b)
Note: If the assumed direction of unknown force is opposite, the value will be negative. Then reverse the
direction and proceed.
Step 4: On the diagram of the truss, mark arrows on the members near the joint analysed to indicate
the forces on the joint. At the other end, mark the arrows in the reverse direction.
In the present case, near the joint C, the arrows are marked on the members CB and CD to
indicate forces FCB and FCD directions as found in the analysis of joint C. Then reversed directions
are marked in the members CB and CD near joints B and D, respectively. FDB
Step 5: Look for the next joint where there are only two unknown
forces and analyse that joint.
In this case, there are only two unknown forces at the joint FDE D 40 kN
D as shown in Fig. 3.8(c).
∑V = 0
40 kN
FDB = 40 kN
Fig. 3.8(c)
∑H = 0 F BA B
FDE = 40 kN 45° 45°
Step 6: Repeat steps 4 and 5 till forces in all the members are found.
In the present case, after marking the forces in the members F BE
40 2
DB and DE, we find that analysis of joint B can be taken up. 40 kN
Referring to Fig. 3.8(d). Fig. 3.8(d)
70 MECHANICS OF SOLIDS
∑V = 0, gives
FBE sin 45° – 40 – 40 2 × sin 45° = 0
∴ FBE = 80 2 kN
∑H = 0
FBA – FBE cos 45° – 40 2 × cos 45° = 0
1 1
FBA = 80 2 × + 40 2 ×
2 2
∴ FBA = 120 kN
The directions of these forces are marked on the diagram. Now the analysis is complete since
the forces in all the members are determined.
Step 7: Determine the nature of forces in each member and TE N S IO N
tabulate the results. Note that if the arrow marks on a member
C O M P R E S S IO N
are towards each other, then the member is in tension and if the
arrow marks are away from each other, the member is in Fig. 3.8(e)
compression [Ref. Fig. 3.8(e)]. In this case,
AB 120 Tension
BC 40 2 Tension
CD 40 Compression
DE 40 Compression
BE 80 2 Compression
BD 40 Tension
Example 3.2. Determine the forces in all the members of the truss shown in Fig. 3.9(a) and indicate
the magnitude and nature of forces on the diagram of the truss. All inclined members are at 60° to
horizontal and length of each member is 2 m.
40 kN 50 kN
B C
Solution: Now, we cannot find a joint with only two unknown forces without finding reactions.
Consider the equilibrium of the entire frame.
∑ M A = 0, gives
TRUSSES 71
RD × 4 – 40 × 1 – 60 × 2 – 50 × 3 = 0
∴ RD = 77.5 kN
∑ H = 0, gives
∴ HA = 0
∴ Reaction at A is vertical only
∑V = 0, gives
RA + 77.5 = 40 + 60 + 50
FAB
∴ RA = 72.5 kN
Joint A: ∑V = 0, gives 60°
A
FAB sin 60° = RA = 72.5 FAE
60° D
FDE
Joint D: ∑V = 0, gives
FDC sin 60° = RD = 77.5 RD
∑ H = 0, gives
FDE – 87.4893 cos 60° = 0
∴ FDE = 44.7446 kN (Tension)
Joint B: ∑V = 0, gives
40 kN
FBE sin 60° – FAB sin 60° + 40 = 0
72.5 − 40
∴ FBE = = 37.5278 (Tension) B F BC
sin 60°
60°
60°
F BE
∑ H = 0, gives
F AB
FBC – FAB cos 60° – FBE cos 60° = 0
Fig. 3.9(d)
FBC = (83.7158 + 37.5274) × 0.5
FBC = 60.6218 kN (Comp.)
72 MECHANICS OF SOLIDS
Joint C: ∑V = 0, gives
FCE sin 60° + 50 – FDC sin 60° = 0
77.5 − 50
FCE = = 31.7543 kN (Tension)
sin 60°
40 kN
F BC
B
60°
60°
F BE
F AB
Fig. 3.9(e)
Now the forces in all the members are known. If joint E is analysed it will give the check for
the analysis. The results are shown on the diagram of the truss in Fig. 3.9(f).
40 kN 50 kN
6 0 .6 2 1 8 C O M P R E S S IO N
B C
TE N S IO N
37
89
8
43
715
.5 2
.4 8
.7 5
83.
78
93
31
A D
4 1 .8 5 7 9 E 4 4 .7 4 4 6
60 kN
Fig. 3.9(f)
Example 3.3. Determine the forces in all the members of the truss shown in Fig. 3.10 (a).
20 kN
B
60° 30°
A D
C
3m 3m
RA 10 kN RD
Fig. 3.10(a)
TRUSSES 73
Tension
.09
20
11
23
.55
A D
11.55 C 17.32
Fig. 3.10(e)
74 MECHANICS OF SOLIDS
Example 3.4. Determine the forces in the members of truss shown in Fig. 3.11(a).
30 kN 50 kN 40 kN
B C D
20 kN
4m
HA E
A F
30 kN
4m 4m
VA
Fig. 3.11(a)
Joint E:
ΣV = 0, gives, FED = 90 kN (Comp.)
ΣH = 0, gives, FEF = 0
FEF = 0 E
60
Fig. 3.11(d)
TRUSSES 75
ΣH = 0, gives 30 – FCD = 0
or FCD = 30 kN [Comp.] 30 C FCD
FCF
Joint D: Noting that DF is at 45° as shown in Fig. 3.11(f)
Fig. 3.11(e)
ΣV = 0 →
40 kN
– FDF cos 45° + 90 – 40 = 0
50
or FDF = = 70.71 kN [Tensile]
cos 45° 30 kN D 20 kN
45°
Check ΣH = 0, gives 45°
30 50 40
30 30
20
1
42
60 50 .7 90
70
.4
3
20
20 0
30
60 90
Fig. 3.11(g)
4
∴ sin θ =
5
A θ θ E
3 HA
cos θ = 3m F 3m
5
VA 20 kN
Fig. 3.12(a)
76 MECHANICS OF SOLIDS
As soon as a joint is analysed the forces on the joint are marked on members [Fig. 3.12(b)]
C
RC
kN
2 0 kN
25
θ1 5 kN
B D
25
O
kN
O
2 0 kN
A θ θ
E
HA 1 5 kN F 1 5 kN
VA 20 kN
Fig. 3.12(b)
Joint E: ∑V = 0, gives F ED
4
FED × – 20 = 0
5 θ
E
∴ FED = 25 kN (Tension) F EF
∑ H = 0, gives
FEF – FED cos θ = 0 20 kN
3
∴ FEF = 25 × = 15 kN (Comp.) Fig. 3.12(c)
5
At this stage as no other joint is having only two unknowns, no further progress is possible. Let
us find the reactions at the supports considering the whole structure. Let the reaction be as shown
in Fig. 3.12(b).
∑ M A = 0, gives
RC × 8 – 20 × 6 = 0
RC = 15 kN
∑V = 0, gives
VA = 20 kN
∑ H = 0, gives
H A = RC = 15 kN
Joint A: ∑V = 0, gives
F AB
FAB – VA = 0
A
FAB = 20 kN (Comp.) HA F AF
∑ H = 0, gives
VA
FAF – HA = 0
FAF = 15 kN (Comp.) Fig. 3.12(d)
TRUSSES 77
Joint C: ∑ H = 0, gives
3
FCB × – RC = 0 RC
5
5
FCB = 15 × = 25 kN (Comp.) FCB FCD
3
∑V = 0, gives Fig. 3.12(e)
FCD = FCB sin θ
4
= 25 × = 20 kN (Tension)
5
Joint B: ∑V = 0, gives F BC
4 4
FBF × – FBC × + FAB = 0
5 5 θ
B
4 4 θ F BD
FBF × = 25 × – 20 = 0
5 5
∴ FBF = 0 FBF
F AB
∑ H = 0, gives
Fig. 3.12(f)
3
FBD – 25 × = 0
5
FBD = 15 kN (Tension)
Joint F:
∑V = 0
FFD = 0 (since FBF = 0)
Fig. 3.12(g)
Note: When three members are meeting at an unloaded joint and out of them two are collinear, then the force
in third member will be zero. Such situations are illustrated in Fig. 3.12(h) and (i).
A A
C B D C B D
Fig. 3.12(h) and (i)
78 MECHANICS OF SOLIDS
Example 3.6. Find the forces in all the members of the truss shown in Fig. 3.13(a).
4
Solution: tan θ1 = = 33.69°
6
FG 8 × 1 IJ = 53.13°
θ2 = tan −1
H 3 2K
FG 4 IJ = 53.13°
H 3K
−1
θ3 = tan
θ2= θ3 = θ
4 3
sin θ = and cos θ =
5 5
2 0 kN
1 2 kN
2m 2m 2m 3m
A B D F H
θ1 θ1 θ2 θ3
2m
E
Fig. 3.13(a)
Joint-by-joint analysis is carried out as given below and the joint forces are marked in Fig. 3.13(b).
Then nature of the force in the members is determined.
12 k N 20 k N
A 15 k N B 15 k N D 15 k N F 15 k N H
C O M P R E S S IO N
18
R A k .0 2 7
N 8
C 25 k N
18 T E N S IO N
.0
kN 27 8
E
18
.0
kN 278
G
RG
Fig. 3.13(b)
Joint H: ∑V = 0, gives
FHG sin θ3 = 20
TRUSSES 79
5
∴ FHG = 20 × = 25 kN (Comp.)
4
Σ H = 0, gives
FHF – FHG cos θ2 = 0
3
FHF = 25 × = 15 kN (Tension)
5
Now ∑ M G = 0, gives
RA × 6 – 20 × 3 = 0
RA = 10 kN (Downword)
∑V = 0, gives
RG – 10 – 12 – 20 = 0
RG = 42 kN
Joint A: ∑V = 0, gives
FAC sin θ1 – 10 = 0
FAC = 18.0278 kN (Comp.)
∑ H = 0, gives
FAB – FAC cos θ1 = 0
FAB = 15 kN (Tension)
Joint B: ∑V = 0, gives
FBC = 0
∑ H = 0, gives
FBD = FBA = 15 kN (Tension)
Joint C: ∑ Forces normal to AC = 0, gives
FCD = 0 since FBC = 0
∑ Forces parallel to CE = 0
FCE = FCA = 18.0278 (Comp.)
Joint D: ∑V = 0, gives
FDE = 0
∑ H = 0, gives
FDF = FDB = 15 kN (Tension)
Joint E: ∑ Forces normal to CG = 0, gives
FEF = 0 and
∑ Forces in the direction of CG = 0, gives
FEG = FCE = 18.0278 kN (Comp.)
Joint F: ∑V = 0, gives
FFG – 12 = 0
FFG = 12 kN (Tension)
80 MECHANICS OF SOLIDS
Example 3.7. Analyse the truss shown in Fig. 3.14(a). All the members are of 3 m length.
40 kN 30 kN
B D F
10 kN
60 ° 60 ° 60 ° 60 ° 60 ° 60 °
A G
C E
3 × 3 × 3 = 9 cm
20 kN
Fig. 3.14(a)
Solution: Since all members are 3 m long, all triangles are equilateral and hence all inclined members
are at 60° to horizontal. Joint-by-joint analysis is carried out and the forces are represented in
Fig. 3.14(b). Then nature of the force is determined.
Joint G: ∑V = 0, gives
FGF sin 60° = 20
FGF = 23.0940 kN (Tension)
∑ H = 0, gives
FGE – FGE cos 60° = 0
FGE = 11.5470 kN (Comp.)
Joint F: ∑V = 0, gives
FFG sin 60° – FGF sin 60° = 0
FFG = FGF = 23.0940 kN (Comp.)
∑ H = 0, gives
FFD + 10 – FGF cos 60° – FFE cos 60° = 0
FFD = 13.0940 kN (Tension)
Now, without finding reaction we cannot proceed. Hence, consider equilibrium of the entire truss.
40 kN 30 kN
B 13 .6 60 3 D 13 .0 94 F
10 kN
44
23
43
9 .4
094
.0 7
.0 9
.7 5
337
33
47
4
38
23.
7
9 .4
VA RE 20 kN
C O M P R E S S IO N T E N SIO N
Fig. 3.14(b)
∑ MA = 0
RE × 6 + 10 × 3 × sin 60° – 40 × 1.5 – 30 × 4.5 – 20 × 9 = 0
∴ RE = 58.1699 kN
TRUSSES 81
∑V = 0, gives
VA = 40 + 30 + 20 – RE = 31.8301 kN
∑ H = 0, gives
HA = 10 kN
Joint A: ∑V = 0, gives
FAB sin 60° – 31.8301 = 0
FAB = 36.7543 kN (Comp.)
∑ H = 0, gives
FAC – FAB cos 60° + 10 = 0
FAC = 8.3771 kN (Tension)
Joint B: ∑V = 0, gives
FBC sin 60° + FAB sin 60° – 40 = 0
FBC = 9.4337 kN (Comp.)
∑ H = 0, gives
FBD + FBC cos 60° – FBA cos 60° = 0
FBD = 13.6603 kN (Comp.)
Joint C: ∑V = 0, gives
FCD sin 60° – FBC sin 60° = 0
FCD = FBC = 9.4337 kN (Tension)
∑ H = 0, gives
FCE + FAC – FCD cos 60° – FBC cos 60° = 0
1
FCE = 2 × 9.4337 × – 8.3771 = 1.0566 kN (Comp.)
2
Joint D: ∑V = 0, gives
FDE sin 60° – FCD sin 60° – 30 = 0
FDE = 44.0747 kN (Comp.)
Thus, the method of section is the application of nonconcurrent force system analysis whereas
the method of joints, described in previous article was the application of analysis of concurrent force
system.
Under the following two conditions the method of section is preferred over the method of joints:
(1) In a large truss in which forces in only few members are required;
(2) In the situation where the method of joints fails to start/proceed with analysis.
The method of section is illustrated with the examples 3.8 to 3.11. Examples 3.8 and 3.9 are the
cases in which method of section is advantageous since forces in only few members are required.
Examples 3.10 and 3.11 are the cases in which method of joints fails to start/proceed to get the
solution. In practice the frames may be analysed partly by method of section and partly by method
of joints as illustrated in example 3.11.
Example 3.8. Determine the forces in the members FH, HG and GI in the truss shown in Fig. 3.15(a).
Each load is 10 kN and all triangles are equilateral with sides 4 m.
10 kN 10 kN 10 kN 10 kN 10 kN 10 kN 10 kN
B D F A H J L N
A O
C E G AI K M
7 × 4 = 28 m
Fig. 3.15(a)
Fig. 3.15(b)
ΣMG = 0, gives
FFH × 4 sin 60° – 35 × 12 + 10 × 10 + 10 × 6 + 10 × 2 = 0
FFH = 69.2820 kN (Comp.)
∑V = 0, gives
FGH sin 60° + 10 + 10 + 10 – 35 = 0
FGH = 5.7735 kN (Comp.)
∑ H = 0, gives
TRUSSES 83
1
U3
U2 U4
U1 U5
9m
8m
6m
L6
L0 L1 L2 L3 L4 L5
200 kN 200 kN 150 kN 100 kN 100 kN
6 × 6 = 36
R1 1 R2
Fig. 3.16(a)
Now, ∑ M LO = 0, gives
R2 × 36 – 200 × 6 – 200 × 12 – 150 × 18 – 100 × 24 – 100 × 30 = 0
R2 = 325 kN
∑V = 0, gives
R1 = 200 + 200 + 150 + 100 + 100 – 325 = 425 kN
Take the section (1)–(1) and consider the right hand side part.
F U 3U 4
θ1 U
4
U5
θ2
4 L3
FU
L 3 FL L L6
3 4 L4 L5
1 00 kN 1 00 kN R2
Fig. 3.16(b)
L3U4 = 62 + 82 = 10
sin θ2 = 0.6 cos θ2 = 0.8
Σ M U 4 = 0, gives
FL × 8 – 325 × 12 + 100 × 6 = 0
3L4
FL = 412.5 kN (Tension)
3 L4
Σ ML = 0, gives
3
FU × cos θ1 × 9 + 100 × 6 + 100 × 12 – 325 × 18 = 0
4U3
FU = 456.2072 kN (Comp.)
4U3
∑H = 0
FU sin θ2 – FU cos θ1 + FL = 0
4U3 4U3 4L3
456. 2072 × 0. 9864 − 412. 5
FU =
4U3 0. 6
= 62.5 kN (Tension)
Example 3.10. Find the forces in the members (1), (2) and (3) of French truss shown in Fig. 3.17(a).
20 kN
A
20 kN C 20 kN
a
20 kN 20 kN
a 1
20 kN 20 kN
a 2
a
A 30° 3 B
E
3m 3m 6m 3m 3m
18 m
A
Fig. 3.17(a)
20 kN C
20 kN F1
20 kN F2
θ
D F3 E
R A = 70 kN
Fig. 3.17(b)
Take Section (A)–(A) and consider the equilibrium of left hand side part of the French Truss
shown in Fig. 3.17(b).
Drop perpendicular CE on AB.
Now CE = 9 tan 30° and DE = 3 m
9 tan 30° 1
∴ tan θ = = 3× = 3
3 3
∴ θ = 60°
∑ M A = 0, gives
F2 sin 60° × 6 – 20 × 2.5981 cos 30° – 20 × 2 × 2.5981 cos 30° – 20 × 3 × 2.5981 cos 30° = 0
(1 + 2 + 3)
F2 = 20 × 2.5981 since sin 60° = cos 30°
6
F2 = 51.9615 kN (Tension)
∑V = 0, gives
F1 sin 30° – 70 + 20 + 20 + 20 – 51.9615 sin 60° = 0
F1 = 110 kN (Comp.)
∑ H = 0, gives
F3 + F2 cos 60° – F1 cos 30° = 0
F3 = 69.2820 (Tension)
Note: In this problem, the method of joints cannot give complete solution.
Example 3.11. Find the forces in all the members of the truss shown in Fig. 3.18(a).
30 kN
1
C
30 kN a 30 kN
15°
a B
15 kN F G D 15 kN
45°
A 30° E
1H
10 m
Fig. 3.18(a)
86 MECHANICS OF SOLIDS
30 kN C
FBC
15°
B F FC
15 kN F
45°
A 30° E
FAE
60 kN
Fig. 3.18(b)
∴ FFC = 71.4042 × 2 − 60
= 40.98 kN (Tension)
Assumed directions of FBC and FFC are correct.
Therefore, FBC is in compression and FFC is in tension.
Now we can proceed with method of joints to find the forces in other members. Since it is a
symmetric truss, analysis of half the truss is sufficient. Other values may be written down by making
use of symmetry.
Joint B:
∑ forces normal to AC = 0, gives
FBF – 30 cos 45° = 0
FBF = 21.21 kN (Comp.)
∑ forces parallel to AC = 0, gives
FAB – FBC – 30 sin 45° = 0
FAB = 71.40 + 21.21
= 92.61 kN (Comp)
Joint A: ∑V = 0, gives
FAF sin 30° – FAB sin 45° – 15 + 60 = 0
FAF = 40.98 kN (Tension)
The results are tabulated below:
Members Force in kN
AB and ED – 92.61
BC and DC – 71.40
BF and DG – 21.21
AF and EG + 40.98
FC and GC + 40.98
AE + 30.00
+ means tension and – means compression
IMPORTANT FORMULA
1. In a perfect truss
m = 2j – 3
where m = number of members
j = number of joints.
THEORY QUESTIONS
1. Bring out the differences among perfect, deficient and redundant trusses.
2. State the assumptions made in the analysis of pin jointed trusses.
3. How method of joint differs from the method of section in the analysis of pin jointed trusses?
88 MECHANICS OF SOLIDS
1 to 17: Determine the forces in all the members of the trusses shown in Fig. 3.19 to 3.35. Indicate
the nature of forces using the convention tension as +ve and compression as –ve.
1.
6m 6m 20 kN
A B
4m
F
E D
3m 6m
[Ans. FAB = +67.5 kN; FBC = +15 kN; FCD = –25 kN; FDE = –30 kN;
FEF = –105 kN; FAE = +62.5 kN; FBE = –62.5 kN; FBD = +25 kN]
2.
A
B 20 kN
4m
C
D
G F E
3m 3m 3m
20 kN
[Ans. FAB = +82.0738 kN; FBC = +73.866 kN; FCD = 49.2443 kN; FDE = –45 kN;
FEF = –45 kN; FFG = –67.5 kN; FBG = – 10.0 kN; FFC = +24.622 kN; FCE = 0; FBF = 10 kN]
3.
A
C
E
G
30°
F
100 kN
D
B
3m 3m 3m
Fig. 3.21 (Prob. 3)
TRUSSES 89
[Ans. FAC = FCE = FEG = +193.1852 kN; FBD = FDE = FFG = –193.1852 kN; all others
are zero members]
4.
E
200 kN
2m
C
A
2m
D B
4m 4m
Fig. 3.22 (Prob. 4)
[Ans. F EC = +447.2136 kN; FCA = +400 kN; FAB = –447.2136 kN; FBD = –400 kN;
FCD = 0; FCB = –200 kN] F
D 3m B
5. [Ans. FDB = FBA = +5.7735 kN; FBC = FDE = –5.7738 kN; 90°
FAC = –2.8868 kN; FCE = –14.4338 kN; FDC = +17.3205 kN; 3m 3m
FDF = +20.0 kN] 3m 3m
6. [Ans. FAB = –30 kN; FAC = –160 kN; FBC = +50 kN; 3m 3m A
F BD = –200 kN; F CD = –50 kN; F CE = –120 kN; E C
FDF = –266.67 kN; FDE = +83.33 kN] 10 kN 5 kN
Fig. 3.23 (Prob. 5)
160 kN 160 kN
30 kN
B
A
4m
G H
8. [Ans. FBD = −2 2 kN; FBA = +3 kN; FAC = +3 2 kN; FAD = –3 kN; FDC = –2 kN;
FDF = –5 kN; FCF = − 2 kN; FCE = +6 kN; FFE = + 1 kN; FFH = –4kN; FEH = − 2 kN;
FEG = +5 kN; FGH = +1 kN]
A B
1 kN
9. [Ans. FAC = –100 5 kN; FAB = +200 kN; FBD = +200 kN;
2 kN
FBC = –100 kN; FCD = + 50 5 kN; FCE = −150 5 kN; FDE 2m
C
= +35.0 kN; FDF = 300 2 kN; FEF = –300 kN] D
10. [Ans. FAB = + 5 2 kN; FAC = –5 kN; FBC = –5 kN; 2m
FBD = +5 kN; FCD = +15 2 kN; FCE = –20 kN;
E F
FDE = –15 kN; FDF = +20 2 kN; FEH = –15 kN;
2m
FEF = –20 kN; FFG = + 30 2 kN; FFH = +10 2 kN]
G H
2m 2m
5m
100 kN C
F
E
5m 5m 5m
D B
2m
F
A
E C
2m 10 kN 5 kN
G H
2m 2m 2m 2m
40 kN 30 kN
45° B D 45°
2m
A E
C
2m 2m
11. [Ans. FAB = –15 kN; FAC = +12 2 kN; FBD = −27. 5 2 2 0 kN
2m 2 0 kN
kN; F BC = –12.5 2 kN; FCE = 0; FCD = +25 kN; D
2m 2 0 kN
B
FED = −27. 5 2 kN] 60 °
E 30 ° A
12. [Ans. FAB = –17.32 kN; FAC = +5 kN; FBC = –20 kN; C
F BD = –17.32 kN; F CD = +20 kN; FCE = –15 kN; Fig. 3.30 (Prob. 12)
FDE = –30 kN]
13. [Ans. FAB = 60 kN; FAC = +51.96 kN; FBC = –20 kN; FBD = –40 kN; FCD = +40 kN; symmetry]
20 kN
20 kN D 20 kN
10 kN B E 10 kN
14. [Ans. F AC = −4.5 13 kN; F AB = +13.5 kN; F BC = +6 kN; FBD = +13.5 kN; F CD
= −0. 5 13 kN; FCE = −4 10 kN; FDE = 8 kN]
E
C F 1m
2m
A H
B D G
6 kN 6 kN 6 kN
3 × 4 = 12 m
15. [Ans. FAB = +10 13 kN; FAC = –20 kN; FCB = –48.75 kN; FCE = –20 kN; FCD = –7.5 kN;
FBE = +6. 25 13 kN; FDE = 18.75 kN; FDF = −3. 75 13 kN; FFE = –7.5 kN]
3 0 kN
B D
1 5 kN
3m
A C E F
2m 2m 2m
3 0 kN
16. [Ans. FAB = 16.91 kN; FAF = +31.55 kN; FBF = +23.91 kN; FBD = –23.91 kN; FBC = +40 kN;
FCD = –40 kN; FDE = –63.1 kN; FDF = +23.91 kN; FEF = +31.55 kN]
C 4 0 kN
B D
6 0°
6 0° 6 0° 6 0° 6 0°
A E
3m F 3m
4 0 kN
Fig. 3.34 (Prob. 16)
17. [Ans. FAC = –67.48 kN; FAB = +53.99 kN; FBC = +10 kN; FCD = –8.33 kN; FCE = –59.15 kN;
F EF = –24.5 kN; F ED = +52.81 kN; F FD = +47.21 kN; F FG = –34.64 kN;
FDG = +47.32 kN]
E
3 0 kN
C
3m
1.5 m
F
3 0° D B A
G 1 0 kN
3m 2m 3m
18. Find the force in the member FG of the triangular Howe truss shown in Fig. 3.36.
4 kN
1
F
8 kN
H
D
8 kN 4.5 m
J
B
4 kN
A L
C E G H K
2m 2m 2m 2m 2m
6 kN 6 kN 6 kN 6 kN 6 kN
1
Fig. 3.36
(Hint: Take section (1)–(1) and find force in FD. Then analyse joint F) [Ans. +28 kN]
TRUSSES 93
19. Determine the forces in the members AB, AC, DF and CE of the scissors truss shown in
Fig. 3.37.
W
1 F
W W
D G
W W
E
B C H I
1
A J
3m 3m 3m 3m 3m 3m
Fig. 3.37
(Hint: Find reaction RA and analyse joint A. Take section (1)–(1) and find force in DF and CE)
[Ans. FAB = –6.25 W; FAC = 4.51 W; FDF = –3.75 W; FCE = +2.75 W]
20. Find the force in member KL of the French truss shown in Fig. 3.38.
2 0 kN
2 0 kN 1 5 kN
A E
2 0 kN 1 5 kN
D F
2 0 kN C G 1 5 kN
L M
B H
A 30° 30° I
J K AN O
3m 3m 6m 3m 3m
3 × 6 = 18 m
Fig. 3.38
(Hint: Take section (A)–(A) and find FLE and FDE. From joint D find FDL. Then analyse joint
L to get FKL) [Ans. +41.96 kN]
4
Distributed Forces, Centre of
Gravity and Moment of Inertia
The term ‘distributed force’ has been explained in Art 2.11 and this applied force has been classified
into linear surface and the body force. The number of such forces acting on a body is infinite.
However, these forces can be replaced by their resultant which acts through a point, known as the
centre of gravity of the body. In this chapter the method of finding areas of given figures and
volumes is explained. Then the terms centroid and centre of gravity and second moment of area
(moment of inertia of areas) are explained and method of finding them is illustrated with examples.
Theorem of Pappus-Guldinus is introduced which is very useful for finding surface areas and volumes
of solids then the method of finding centre of gravity and moment of inertia (mass moment of inertia)
of solids is illustrated.
= bd. O x
d/2
If we take element as shown in Fig. 4.2,
b/2 b/2
Fig. 4.1
94
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 95
d/2 d/2
A = ò dA = ò b ⋅ dy d/2 dy
y
-d / 2 -d / 2
= b [y] d- d/ 2/ 2 d/2
= bd
b
Fig. 4.2
(ii) Area of a triangle of base width ‘b’ height ‘h’: Referring to
Fig. 4.3, let the element be selected as shown by hatched
y
lines
h
dy
y b
Then dA = b′dy = b dy
h
b
h h
y
A = ò dA = ò b dy Fig. 4.3
0 0 h
h
b é y2 ù bh
= ê ú =
h ë 2 û0 2
2ð R
∴ A= ò ò r dθ dr
0 0
dr
rd
r d
2ð é 2 ùR
r R
= ò ê ú dθ O
x
0 ë 2 û0
2ð
R2
= ò dθ
0 2
R2 Fig. 4.4
= [θ ] 02ð
2
R2
= ⋅ 2π = πR2
2
In the above derivation, if we take variation of θ from 0 to π, we get the area of semicircle as
ðR2
ðR 2
and if the limit is from 0 to π/2 the area of quarter of a circle is obtained as .
2 4
96 MECHANICS OF SOLIDS
=
é R2 ù R2
= ê èú = (2α) = R2α. Fig. 4.5
ë 2 û -= 2
= kx2 dx
a a 2
∴ A = ò dA = ò kx2 dx y = kx
h
0 0
x
x x=a
éx 3 ùa 3
ka
=k ê ú = dx
ë 3 û0 3 Fig. 4.6
We know, when x = a, y = h
h
i.e., h = ka2 or k =
a2
ka3 h a3 1 1
∴ A= = 2 = ha = rd the area of rectangle of size a × h
3 a 3 3 3
2
Case b: In this case y = kx
Referring to Fig. 4.7
dA = y dx = kx dx
a a
A = ò y dx = ò kx dx
0 0
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 97
a y
= k éê x3/ 2 ùú = k a3/2
2 2 2
y = kx
ë 3 û0 3
We know that, when x = a, y = h
h
2
h
∴ h2 = ka or k= x
a x=a
dx
h 2
Hence A= × . a3/2 Fig. 4.7
a 3
2 2
i.e., A= ha = rd the area of rectangle of size a × h.
3 3
(vi) Surface area of a cone
Consider the cone shown in Fig. 4.8. Now,
x
y= R
h dl
Surface area of the element, R
y
x x dx
dA = 2π y dl = 2π R dl
h
x dx h
= 2π R
h sin α Fig. 4.8
∴
2 πR x 2 LM OP h
N Q
A=
h sin α 2 0
πRh
= = π Rl
sin α
(vii) Surface area of a sphere
Consider the sphere of radius R shown in Fig. 4.9. The
element considered is the parallel circle at distance y
from the diametral axis of sphere.
dy x
dS = 2π x Rdθ Rd
= 2π R cos θ Rdθ, since x = R cos θ y
d
ð/2
∴ S = 2π R 2
ò cos θ dθ
-ð / 2
= 2π R2 [sin θ] ð- ð/ 2/ 2
= 4πR2
Fig. 4.9
98 MECHANICS OF SOLIDS
x2 x
dV = πy2 . dx = π 2 R2dx, since y = R
h h
z LM OP h
π h π x3
V = 2 R2 x dx = 2 R 2
2
h 0 h 3 N Q 0
π 2h πR h
3 2
= R =
h2 3 3
(iii) Volume of a sphere
Referring to Fig. 4.9
dV = π x2 . dy
But x2 + y2 = R2
i.e., x2 = R2 – y 2
∴ dV = π (R2 – y2)dy
R
V = ò π (R2 – y2)dy
-R
R
é y3 ù
= π ê R2 y - ú
ë 3û
-R
é R3 ìï ïù
( - R )3 ü
= π ê R2 × R - - í- R3 - ýú
ëê 3 ï î 3 þ ï ûú
é 1 1ù 4
= π R3 ê1 - + 1 - ú = πR3
ë 3 3û 3
The surface areas and volumes of solids of revolutions like cone, spheres may be easily found using
theorems of Pappus and Guldinus. This will be taken up latter in this chapter, since it needs the term
centroid of generating lines.
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 99
T T
1 1
2
w1
G
W
1
2
1
W = Sw1 W = Sw1
(a) (b)
Fig. 4.10
xc Wi
z dL
yc (xi, yc)
Wi = A dL
x
Fig. 4.12 Fig. 4.13
If the body is a wire of uniform cross-section in plane x, y (Ref. Fig. 4.13) the equation 4.1 reduces
to
Lxc = Σ Lixi = Ñò x dL ü
Lyc = Σ Liyi = Ñò y dL ý ...(4.5)
þ
The term centre of gravity is used only when the gravitational forces (weights) are considered. This
term is applicable to solids. Equations 4.2 in which only masses are used the point obtained is termed
as centre of mass. The central points obtained for volumes, surfaces and line segments (obtained by
eqn. 4.3, 4.4 and 4.5) are termed as centroids.
L
L é x2 ù L2
Lxc = ò x dx = ê ú =
0 ë2û 2
0
L
∴ xc =
2
Thus the centroid lies at midpoint of a straight line, whatever be the orientation of line
(Ref. Fig. 4.15).
y y
G
G
O x G
L L cos
L 2
2 L
2 sin
L 2
x
O
Fig. 4.15
z
x
α d
x cL = xdL
−α O x
z
α
i.e., xc R 2α = R cos θ . Rdθ
−α
2 LMsin θOP α
N Q
=R ...(i)
−α
and yc L z α
−α
y dL = z α
−α
R sin θ . Rdθ
LMcos θOP α
N Q
2
=R ...(ii)
−α
=0
∴ yc = 0
From equation (i) and (ii) we can get the centroid of semicircle shown in Fig. 4.17 by putting
α = π/2 and for quarter of a circle shown in Fig. 4.18 by putting α varying from zero to π/2.
102 MECHANICS OF SOLIDS
G
R R
G
2R
For semicircle xc =
π
yc = 0
For quarter of a circle,
2R
xc =
π
2R
yc =
π
(iii) Centroid of composite line segments
The results obtained for standard cases may be used for various segments and then the equations
4.5 in the form
xcL = ΣLixi
ycL = ΣLiyi
may be used to get centroid xc and yc. If the line segments is in space the expression zcL = ΣLizi
may also be used. The method is illustrated with few examples below:
Example 4.1. Determine the centroid of the wire shown in Fig. 4.19.
D
y
G3
30
0
m
m
45° C
G2 200 mm
G1
A B k
600 mm
Fig. 4.19
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 103
Solution: The wire is divided into three segments AB, BC and CD. Taking A as origin the coordinates
of the centroids of AB, BC and CD are
G1(300, 0); G2(600, 100) and G3 (600 – 150 cos 45°, 200 + 150 sin 45°)
i.e., G3 (493.93, 306.07)
L1 = 600 mm, L2 = 200 mm, L3 = 300 mm
∴ Total length L = 600 + 200 + 300 = 1100 mm
∴ From the eqn. Lxc = ΣLixi, we get
1100 xc = L1x1 + L2x2 + L3x3
= 600 × 300 + 200 × 600 + 300 × 493.93
∴ xc = 407.44 mm
Now, Lyc = ΣLiyi
1100 yc = 600 × 0 + 200 × 100 + 300 × 306.07
∴ yc = 101.66 mm
Example 4.2. Locate the centroid of the uniform wire bent as shown in Fig. 4.20.
m D
G2 0m
25 G3
G1 150 mm
A B 30°
C
400 mm
All dimensions in mm
Fig. 4.20
Solution: The composite figure is divided into three simple figures and taking A as origin coordinates
of their centroids noted down as shown below:
AB—a straight line
L1 = 400 mm, G1 (200, 0)
BC—a semicircle
F 2 × 150 I
L2 = 150 π = 471.24,
H
G2 475,
π K
i.e. G2 (475, 95.49)
CD—a straight line
250
L3 = 250; x3 = 400 + 300 + cos 30° = 808.25 mm
2
y3 = 125 sin 30° = 62.5 mm
∴ Total length L = L1 + L2 + L3 = 1121.24 mm
∴ Lxc = ΣLixi gives
1121.24 xc = 400 × 200 + 471.24 × 475 + 250 × 808.25
xc = 451.20 mm
Lyc = ΣLiyi gives
104 MECHANICS OF SOLIDS
r = 140
y
B C 45°
0
10
r=
A
x
D
Fig. 4.21
Solution: The length and the centroid of portions AB, BC and CD are as shown in table below:
Table 4.1
Portion Li xi yi zi
2 × 100
AB 100π 100 0
π
2 × 140
BC 140π 0 140
π
CD 300 300 sin 45° 280 + 300 cos 45°
= 492.13 0
From the above equation we can make the statement that distance of centre of gravity of a body
from an axis is obtained by dividing moment of the gravitational forces acting on the body, about the
axis, by the total weight of the body. Similarly from equation 4.4, we have,
SAi xi SAi yi
xc = , yc =
A A
By terming ΣAix: as the moment of area about the axis, we can say centroid of plane area from any
axis is equal to moment of area about the axis divided by the total area. The moment of area ΣAix: is
termed as first moment of area also just to differentiate this from the term ΣAix i2 , which will be dealt
latter. It may be noted that since the moment of area about an axis divided by total area gives the
distance of the centroid from that axis, the moment of area is zero about any centroidal axis.
SAi xi O X
A
Consider the two elemental areas shown in Fig. 4.22, which are
equal in size and are equidistant from the axis, but on either side. Now
the sum of moments of these areas cancel each other since the areas and
distances are the same, but signs of distances are opposite. Similarly, Fig. 4.22
we can go on considering an area on one side of symmetric axis and
corresponding image area on the other side, and prove that total moments of area (ΣAixi) about the
symmetric axis is zero. Hence the distance of centroid from the symmetric axis is zero, i.e., centroid
always lies on symmetric axis.
Making use of the symmetry we can conclude that:
(1) Centroid of a circle is its centre (Fig. 4.23);
b d
(2) Centroid of a rectangle of sides b and d is at distance and from the corner as shown in
Fig. 4.24. 2 2
106 MECHANICS OF SOLIDS
b
b/2
G G
d
d/2
y =
ò ydA
A
x = ò
xdA
A
The location of the centroid using the above equations may be considered as finding centroid from
first principle. Now, let us find centroid of some standard figures from first principle.
Centroid of a Triangle
Consider the triangle ABC of base width b and height h as shown in Fig. 4.25. Let us locate the
distance of centroid from the base. Let b1 be the width of elemental strip of thickness dy at a distance
y from the base. Since DAEF and DABC are similar triangles, we can write:
b1 h- y
=
b h A
æ h - yö æ yö dy
b1 = ç b = ç1 - ÷ b
è h ÷ø è hø E F
h
b1
∴ Area of the element y
= dA = b1dy
B
C
æ yö b
= ç1 - ÷ b dy
è hø Fig. 4.25
1
Area of the triangle A= bh
2
∴ From eqn. 4.4
Moment of area ò ydA
y = =
Total area A
h
æ yö
Now, ò ydA = ò y çè1 - ÷ø b dy
0 h
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 107
h æ y2 ö
= ò ç y - ÷ b dy
0 è hø
h
é y 2 y3 ù
= bê - ú
ë 2 3h û0
bh 2
=
6
∴ y =
ò ydA =
bh2
´1
1
A 6 2
bh
h
∴ y =
3
h 2h
Thus the centroid of a triangle is at a distance from the base (or from the apex) of the
3 3
triangle, where h is the height of the triangle.
Centroid of a Semicircle
Consider the semicircle of radius R as shown in Fig. 4.26. Due to symmetry centroid must lie on y
axis. Let its distance from diametral axis be y . To find y , consider an element at a distance r from
the centre O of the semicircle, radial width being dr and bound by radii at θ and θ + dθ.
Area of element = r dθdr.
Y
Its moment about diametral axis x is given by:
rdθ × dr × r sin θ = r2 sin θ dr dθ
∴ Total moment of area about diametral axis,
zz z
LM r OP sin θ dθ
π R π R
3
d
r2 sin θ dr dθ =
0 0 N3Q
0 0
r
dr
R L O π O X
M − cos θP
3 R
3 N Q
=
0 Fig. 4.26
R3 2R3
= [1 + 1] =
3 3
1 2
Area of semicircle A= πR
2
2 R3
Moment of area
∴ y = = 3
Total area 1 2
ðR
2
4R
=
3π
4R
Thus, the centroid of the circle is at a distance from the diametral axis.
3ð
108 MECHANICS OF SOLIDS
Consider the sector of a circle of angle 2α as shown in Fig. 4.27. Due to symmetry, centroid lies on x
axis. To find its distance from the centre O, consider the elemental area shown.
Area of the element =rdθ dr
Its moment about y axis
Y
= rdθ × dr × r cos θ
dr
= r2 cos θ drdθ
∴ Total moment of area about y axis r d
zz
α R
O
2
= r2 cos θ drdθ
G X
−α 0
é r3 ù
= ê ú sin θ
R
LM OP α R
ë 3 û0 N Q −α
R3 Fig. 4.27
= 2 sin α
3
Total area of the sector
zz
α R
= rdrdθ
−α 0
z LMN OP
α R
r2
dθ
=
−α
2 Q 0
R2 LMθOP α
=
2 NQ −α
=R α 2
2 R3
sin α
2R
= 3 2 = sin α
R α 3α
Consider the parabolic spandrel shown in Fig. 4.28. Height of the element at a distance x from O is
y = kx2
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 109
Width of element = dx
∴ Area of the element = kx2dx
a
a é kx 3 ù
∴ Total area of spandrel = ò kx2dx = ê ú
0 ë 3 û0
ka 3 Y y = kx
2 h
=
3
– –
G(x, y)
Moment of area about y axis
O
a x X
2 dx
= ò kx dx × x a
0
Fig. 4.28
a
3
= ò kx dx
0
a
é k x4 ù
= ê ú
ë 4 û0
ka 4
=
4
k 2 a5
=
10
ka 4 ka 3 3a
∴ x = ¸ =
4 3 4
k 2 a 5 ka 3 3
y = ¸ = ka2
10 3 10
From the Fig. 4.28, at x = a, y = h
h
∴ h = ka2 or k =
a2
3 h 3h
∴ y = ´ 2 a2 =
10 a 10
æ 3a 3h ö
Thus, centroid of spandrel is ç , ÷
è 4 10 ø
Centroids of some common figures are shown in Table 4.2.
110 MECHANICS OF SOLIDS
h bh
Triangle h G —
3 2
x
b
4R πR 2
Semicircle G r 0
3π 2
x
y
G 4R 4R πR 2
Quarter circle
x 3π 3π 4
R
y
2R
Sector of a circle G x sin a 0 αR 2
2 3α
3h 4ah
Parabola h G 0
5 3
x
2a
y
3 3h 2ah
Semi parabola h a
8 5 3
x
a
y
3a 3h ah
Parabolic spandrel h
G 4 10 3
x
a
A1 x1 + A2 x2
=
A
1800 ´ 75 + 2256 ´ 6 12
= = 36.62 mm
4056 Y
All dimensions in mm
Moment about x axis
y = Fig. 4.30
Total area
A y + A2 y2
= 1 1
A
1800 ´ 6 + 2256 ´ 106
= = 61.62 mm
4056
Thus, the centroid is at x = 36.62 mm and y = 61.62 mm as shown in the figure.
112 MECHANICS OF SOLIDS
Example 4.6. Locate the centroid of the I-section shown in Fig. 4.31.
Y
100
20 A1 g1
20
A2
100 g2
G
–
y
30 A3 g3
X
O
150
All dimensions in mm
Fig. 4.31
Solution: Selecting the coordinate system as shown in Fig. 4.31, due to symmetry centroid must
lie on y axis,
i.e., x =0
Now, the composite section may be split into three rectangles
A1 = 100 × 20 = 2000 mm2
Centroid of A1 from the origin is:
20
y1 = 30 + 100 + = 140 mm
2
Similarly A2 = 100 × 20 = 2000 mm2
100
y2 = 30 + = 80 mm
2
A3 = 150 × 30 = 4500 mm2,
30
and y3 = = 15 mm
2
A1 y1 + A2 y2 + A3 y3
∴ y =
A
2000 ´ 140 + 2000 ´ 80 + 4500 ´ 15
=
2000 + 2000 + 4500
= 59.71 mm
Thus, the centroid is on the symmetric axis at a distance 59.71 mm from the bottom as
shown in Fig. 4.31.
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 113
Example 4.7. Determine the centroid of the section of the concrete dam shown in Fig. 4.32.
2m 2m 3m
1m
1.5 m
A2
g2 A3 5.0 m
6.0 m A1 G
y g3
–
y
g1
A4 g4 1.0 m
O – x
x
8m
Fig. 4.32
Solution: Let the axis be selected as shown in Fig. 4.32. Note that it is convenient to take axis in such
a way that the centroids of all simple figures are having positive coordinates. If coordinate of any
simple figure comes out to be negative, one should be careful in assigning the sign of moment of area
of that figure.
The composite figure can be conveniently divided into two triangles and two rectangles, as
shown in Fig. 4.32.
1
Now, A1 = × 2 × 6 = 6 m2
2
A2 = 2 × 7.5 = 15 m2
1
A3 = × 3 × 5 = 7.5 m2
2
A4 = 1 × 4 = 4 m 2
A = total area = 32.5 m2
Centroids of simple figures are:
2 4
x1 = ×2= m
3 3
1
y1 = ×6=2m
3
x2 = 2 + 1 = 3 m
7.5
y2 = = 3.75 m
2
1
x3 = 2 + 2 + ×3=5m
3
114 MECHANICS OF SOLIDS
1 8
y3 = 1 + ×5= m
3 3
4
x4 = 4 + =6m
2
y4 = 0.5 m
A1 x1 + A2 x2 + A3 x3 + A4 x4
x =
A
6´ 4
3
+ 15 ´ 3 + 7.5 ´ 5 + 4 ´ 6
=
32.5
= 3.523 m
A1 y1 + A2 y2 + A3 y3 + A4 y4
y =
A
g3
m g2 4m
=2
R g1
O x
6m 3m
Fig. 4.33
Solution: The composite section is divided into three simple figures, a triangle, a rectangle and a
semicircle
1
Now, area of triangle A1 = × 3 × 4 = 6 m2
2
Area of rectangle A2 = 6 × 4 = 24 m2
1
Area of semicircle A3 = × π × 22 = 6.2832 m2
2
∴ Total area A = 36.2832 m2
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 115
6
5
4 200
130
160
1 2 3 140
120
50
70 70 70 70 X
Fig. 4.34
Solution: The composite area is equal to a rectangle of size 160 × 280 mm plus a triangle of size 280
mm base width and 40 mm height and minus areas of six holes. In this case also the Eqn. 4.4 can be
used for locating centroid by treating area of holes as negative. The area of simple figures and their
centroids are as shown in Table 4.3.
116 MECHANICS OF SOLIDS
Table 4.3
Figure Area in mm 2 xi in mm yi in mm
∴ A = ΣAi = 48221.70
560
∴ ΣAixi = 44800 × 140 + 5600 × – 363.05 (70 + 140 + 210 + 70 + 140 + 210)
3
= 7012371.3 mm3
SAi xi
x = = 145.42 mm
A
ΣAiyi = 44800 × 80 + 5600 × 173.33 – 363.05 (50 × 3 + 120 + 130 + 140)
= 4358601 mm3
SAi yi 4358601
y = =
A 48221.70
= 90.39 mm
Thus, the coordinates of centroid of composite figure is given by:
x = 145.42 mm
y = 90.39 mm
Example 4.10. Determine the coordinates xc and yc of the Y
100
centre of a 100 mm diameter circular hole cut in a thin
plate so that this point will be the centroid of the remain-
ing shaded area shown in Fig. 4.35 (All dimensions are
in mm).
100
Solution: If xc and yc are the coordinates of the centre of 150
1 é æ 100 ö ù F
200 ´ 150 ´ 100 - ´ 100 ´ 75 ´ ê 200 - ç ÷ø ú - ´ 100 ´ xc
2
2 ë è 3 û 4
xc = x =
18396
1 F
∴ xc(18396) = 200 × 150 × 100 – × 100 × 75 × 166.67 – × 1002 xc
2 4
2375000
xc = = 90.48 mm
26250
Similarly,
1 F
18396 yc = 200 × 150 × 75 – × 100 × 75 × (150 – 25) – × 1002 yc
2 4
1781250.0
∴ yc = = 67.86 mm
26250
Centre of the circle should be located at (90.48, 67.86) so that this point will be the centroid of the
remaining shaded area as shown in Fig. 4.35.
Note: The centroid of the given figure will coincide with the centroid of the figure without circular hole.
Hence, the centroid of the given figure may be obtained by determining the centroid of the figure without the
circular hole also.
Example 4.11. Determine the coordinates of the centroid of the plane area shown in Fig. 4.36 with
reference to the axis shown. Take x = 40 mm.
4x 6x 4x
4x
=
4x Y R 4x
4x
8x
4x
4x
R=
O X
2x 8x 4x 6x
Fig. 4.36
Solution: The composite figure is divided into the following simple figures:
(1) A rectangle A1 = (14x) × (12x) = 168x2
x1 = 7x; y1 = 6x
118 MECHANICS OF SOLIDS
1
(2) A triangle A2 = (6x) × (4x) = 12x2
2
x2 = 14x + 2x = 16x
4x
y2 =
3
(3) A rectangle to be subtracted
A3 = (–4x) × (4x) = –16x2
x3 = 2x; y3 = 8x + 2x = 10x
(4) A semicircle to be subtracted
1
A4 = – π (4x)2 = –8πx2
2
x4 = 6x
4R 4( x ) 16 x
y4 = =4× =
3π 3π 3π
(5) A quarter of a circle to be subtracted
1
A5 = - × π (4x)2 = –4πx2
4
F I = 12.3023x
4x
H K
4R
x5 = 14x – = 14x – (4)
3π 3π
F 4 x I = 10.3023x
y5 = 12x – 4 × H 3π K
Total area A = 168x2 + 12x2 – 16x2 – 8πx2 – 4πx2
= 126.3009x2
SAi xi
x =
A
ΣAixi = 168x2 × 7x + 12x2 × 16x – 16x2 × 2x – 8πx2 × 6x – 4πx2 × 12.3023x
= 1030.6083x3
1030.6083x 3
∴ x =
126.3009 x 2
= 8.1599x = 8.1599 × 40 (since x = 40 mm)
= 326.40 mm
SAi yi
y =
A
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 119
4x
ΣAi yi = 168x2 × 6x + 12x2 × – 16x2 × 10x
3
16 x
– 8πx2 × – 4πx2 × 10.3023x
3π
= 691.8708x3
691.8708 x 3
∴ y =
126.3009 x 2
= 5.4780x
= 219.12 mm (since x = 40 mm)
Centroid is at (326.40, 219.12).
Y
dA
dA B
x
r
y
X
A
(a) (b)
Fig. 4.37
The term rdA may be called as moment of area, similar to moment of a force, and hence r2 dA
may be called as moment of area or the second moment of area. Thus, the moment of inertia of area
is nothing but second moment of area. In fact, the term ‘second moment of area’ appears to correctly
signify the meaning of the expression Σr2 dA. The term ‘moment of inertia’ is rather a misnomer.
However, the term moment of inertia has come to stay for long time and hence it will be used in this
book also.
Though moment of inertia of plane area is a purely mathematical term, it is one of the important
properties of areas. The strength of members subject to bending depends on the moment of inertia of
its cross-sectional area. Students will find this property of area very useful when they study subjects
like strength of materials, structural design and machine design.
The moment of inertia is a fourth dimensional term since it is a term obtained by multiplying
area by the square of the distance. Hence, in SI units, if metre (m) is the unit for linear measurements
120 MECHANICS OF SOLIDS
used then m4 is the unit of moment of inertia. If millimetre (mm) is the unit used for linear measure-
ments, then mm4 is the unit of moment of inertia. In MKS system m4 or cm4 and in FPS system ft4 or
in4 are commonly used as units for moment of inertia.
O x
I
k= ...(4.8)
A
where k = radius of gyration,
I = moment of inertia,
and A = the cross-sectional area
Suffixes with moment of inertia I also accompany the term radius of gyration k. Thus, we can
have,
I xx
kxx =
A
I yy
kyy =
A
I AB
kAB =
A
and so on.
The relation between radius of gyration and moment of A
inertia can be put in the form:
I = Ak2 ...(4.9)
k
From the above relation a geometric meaning can be
assigned to the term ‘radius of gyration.’ We can consider k as Axis
the distance at which the complete area is squeezed and kept as
a strip of negligible width (Fig. 4.39) such that there is no Fig. 4.39
change in the moment of inertia.
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 121
Σ2yyc dA = 2yc Σy dA
SydA
= 2yc A
A
SydA
In the above term 2yc A is constant and is the distance of centroid from the reference axis
A
ydA
GG. Since GG is passing through the centroid itself is zero and hence the term Σ2yycdA is zero.
A
Now, the third term,
Σyc2dA = yc2ΣdA
= Ayc2
∴ IAB = IGG + Ayc2
Note: The above equation cannot be applied to any two parallel axis. One of the axis (GG) must be centroidal
axis only.
-d / 2 d/2
éy 3 ùd / 2 y
= bê ú
ë 3 û- d / 2 x x
éd3 d3 ù
=b ê + ú d/2
ë 24 24 û
bd 3
Ixx =
12 b
Fig. 4.42
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 123
é h3 h 4 ù
= bê - ú
ë 3 4h û
bh3
IAB =
12
(3) Moment of Inertia of a Circle about its Diametral d
Axis: Moment of inertia of a circle of radius R is required
about it’s diametral axis as shown in Fig. 4.44
rd
Consider an element of sides rdθ and dr as shown in the
figure. It’s moment of inertia about the diametral axis x-x: R r d
= y2dA
dr
= (r sin θ)2 r dθ dr X X
= r3 sin2 θ dθ dr
∴ Moment of inertia of the circle about x-x is given by
zz
R 2π
Ixx = r3 sin2 θ dθ dr
0 0
Fig. 4.44
124 MECHANICS OF SOLIDS
zz
R 2π
(1 − cos 2θ)
= dθ dr
2
z LMN
0 0
R
r3 sin 2θ OP 2π
=
0
2
θ−
2 Q 0
dr
R
é r4 ù 2π 4
= ê ú [2π – 0 + 0 – 0] = R
8
ë 8 û0
πR 4
Ixx =
4
If d is the diameter of the circle, then
d
R=
2
π d F I 4
∴ Ixx =
4 2 H K
πd 4
Ixx =
64
y
Moment of Inertia of Standard Sections
Rectangle: Referring to Fig. 4.45.
bd 3 d/2
(a) Ixx = as derived from first principle.
12
x x
db 3
(b) Iyy = can be derived on the same lines.
12 d/2
(c) About the base AB, from parallel axis theorem,
IAB = Ixx + Ayc2 A B
2 b
bd 3 ædö d y
= + bd ç ÷ , since yc =
12 è 2ø 2 Fig. 4.45
bd 3 bd 3
= +
12 4
bd 3
=
3
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 125
BD 3 bd 3
= -
12 12
x x
1 d
= (BD3 – bd3)
12 D/2
Fig. 4.46
Triangle—Referring to Fig. 4.47.
(a) About the base:
As found from first principle
G h
3
bh
IAB = h/3
12 A
b B
(b) About centroidal axis, x-x parallel to base:
From parallel axis theorem, Fig. 4.47
IAB = Ixx + Ayc2
h
Now, yc, the distance between the non-centroidal axis AB and centroidal axis x-x, is equal to .
3
2
bh3 1 æ hö
∴ = Ixx + bh ç ÷
12 2 è 3ø
bh3
= Ixx +
18
bh3 bh3
∴ Ixx = –
12 18
bh3
=
36
Moment of Inertia of a Circle about any diametral axis
πd 4
= (as found from first principle)
64
126 MECHANICS OF SOLIDS
F D4 Fd4 A B
= –
64 64
F
= (D4 – d4)
64
d
D
Fig. 4.48
Moment of Inertia of a Semicircle: (a) About Diametral Axis:
If the limit of integration is put as 0 to π instead of 0 to 2π in
the derivation for the moment of inertia of a circle about
diametral axis the moment of inertia of a semicircle is
obtained. It can be observed that the moment of inertia of a G
semicircle (Fig. 4.49) about the diametral axis AB: x
yc
x
1 πd 4 πd 4 A B
= × =
2 64 128 d
πd 4 πd 2 F I
2d 2
128
= Ixx +
8
×
H K
3π
πd 4 d4
Ixx = −
128 18π
= 0.0068598 d 4
Moment of Inertia of a Quarter of a Circle: (a) About the Base: If the limit of integration is put as 0
π
to instead of 0 to 2π in the derivation for moment of inertia of a circle the moment of inertia of a
2
quarter of a circle is obtained. It can be observed that moment of inertia of the quarter of a circle
about the base AB.
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 127
1 πd 4 πd 4
×= =
4 64π 256
(b) About Centroidal Axis x-x:
Now, the distance of centroidal axis yc from the base is given by: G
X X 4R
4 R 2d 3
yc = = A B
3p 3p
R
1 pd 2 pd 2 Fig. 4.50
and the area A= ´ =
4 4 16
From parallel axis theorem,
IAB = Ixx + Ay c2
2
pd 4 pd 2 æ 2 d ö
= Ixx + ç ÷
256 16 è 3p ø
pd 4 d4
-
Ixx =
256 36 p
= 0.00343 d 4
The moment of inertia of common standard sections are presented in Table 4.4.
bd 3
Rectangle (a) Centroidal axis x-x Ixx =
12
Y
3
(b) Centroidal axis y-y Iyy = db
d/2 12
X X bd 3
(c) A – B IAB =
d/2 3
A B
b/2 b/2
Y
BD 3 - bd 3
Hollow Rectangle Centroidal axis x-x Ixx =
12
X b DX
(Contd.)
128 MECHANICS OF SOLIDS
bh 3
Triangle (a) Centroidal axis x-x Ixx =
36
h bh 3
G (b) Base AB IAB =
X X 12
h/3
A
b B
4
Circle Diametral axis I = pd
64
A B
pd 4
Semicircle (a) A – B IAB =
128
(b) Centroidal axis Ixx = 0.0068598 d 4
G
X X
A B
d
pd 4
Quarter of a circle (a) A – B IAB =
256
d/2 (b) Centroidal axis x-x Ixx = 0.00343 d 4
G
X X
A B
d/2
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 129
A3 g3
A1
y3
g1
g2 y1
A2 y2
A
Fig. 4.51
(2) The centroids of the rectangle (g1), triangle (g2) and semicircle (g3) are located. The distances
y1, y2 and y3 are found from the axis AB.
(3) The moment of inertia of the rectangle about it’s centroid (Ig1) is calculated using standard
expression. To this, the term A1 y 12 is added to get the moment of inertia about the axis AB as:
I1 = Ig1 + A1 y12
Similarly, the moment of inertia of the triangle (I2 = Ig2 + A2 y22) and of semicircle (I3 = Ig3
+ A3 y32) about axis AB are calculated.
(4) Moment of inertia of the composite section about AB is given by:
IAB = I1 + I2 + I3
= Ig1 + A1 y 12 + Ig2 + A2 y 22 + Ig3 + A3 y32 ...(4.12)
In most engineering problems, moment of inertia about the centroidal axis is required. In such
cases, first locate the centroidal axis as discussed in 4.4 and then find the moment of inertia about this
axis.
Referring to Fig. 4.52, first the moment of area about any reference axis, say AB is taken and is
divided by the total area of section to locate centroidal axis x-x. Then the distances of centroid of
130 MECHANICS OF SOLIDS
individual figures yc1, yc2 and yc3 from the axis x-x are determined. The moment of inertia of the
composite section about the centroidal axis x-x is calculated using the expression:
Ixx = Ig1 + A12 yc1 + Ig2 + A22 yc2 + I g3 + A23 yc3 ...(4.13)
g1
yc3
g
yc1
g2 yc2
–
y
g3
A B
Fig. 4.52
Sometimes the moment of inertia is found about a conveninet axis and then using parallel axis
theorem, the moment of inertia about centroidal axis is found.
In the above example, the moment of inertia IAB is found and yc, the distance of CG from axis
AB is calculated. Then from parallel axis theorem,
IAB = Ixx + Ay2c
Ixx = IAB – Ay2c
where A is the area of composite section.
Example 4.12. Determine the moment of inertia of the section y
150
shown in Fig. 4.53 about an axis passing through the centroid
10
and parallel to the top most fibre of the section. Also determine
–
y g2
moment of inertia about the axis of symmetry. Hence find radii A1
of gyration.
X G X
Solution: The given composite section can be divided into two
rectangles as follows: 140
g1
Area A1 = 150 × 10 = 1500 mm2 A2
Area A2 = 140 × 10 = 1400 mm2
Total Area A = A1 + A2 = 2900 mm2
10
Due to symmetry, centroid lies on the symmetric axis y-y. y
The distance of the centroid from the top most fibre is Fig. 4.53
given by:
Sum of moment of the areas about the top most fibre
yc =
Total area
1500 ´ 5 + 1400(10 + 70)
=
2900
= 41.21 mm
Referring to the centroidal axis x-x and y-y, the centroid of A1 is g1 (0.0, 36.21) and that of A2 is
g2 (0.0, 38.79).
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 131
I
k=
A
I xx
∴ kxx =
A
6372442.5
=
2900
kxx = 46.88 mm
2824166.66
Similarly, kyy =
2900
kyy = 31.21 mm
Example 4.13. Determine the moment of inertia of the L-section 1
shown in the Fig. 4.54 about its centroidal axis parallel to the
legs. Also find out the polar moment of inertia. 10 Y
Solution: The given section is divided into two rectangles A1
and A2. A1
æ 75 ö
120 ´ 5 + 750 ç10 + ÷
è 2ø
i.e., x =
2000
= 20.94 mm
Similarly,
the distance of the centroid from the axis (2)–(2)
125
1250 ´ + 750 ´ 5
= y= 2 = 40.94 mm
2000
With respect to the centroidal axis x-x and y-y, the centroid of A1 is g1 (15.94, 21.56) and that of
A2 is g2 (26.56, 35.94).
∴ Ixx = Moment of inertia of A1 about x-x axis + Moment of inertia of A2 about x-x axis
10 ´ 1253 75 ´ 103
∴ Ixx = + 1250 × 21.562 + + 750 × 39.942
12 12
i.e., Ixx = 3411298.9 mm4
Similarly,
Similarly,
13.5 ´ 100 3
× 21.682 + + 1350 × 24.272
12
Iyy = 52,72557.6 mm4
Example 4.16. Determine the polar moment of inertia of the I-section shown in the Fig. 4.57. Also
determine the radii of gyration with respect to x-x axis and y-y axis.
y
80
12
A1 g1
12
A2
150
g2
x x
A3
g3
10
1 1
120
y
Fig. 4.57
Solution: The section is divided into three rectangles as shown in Fig. 4.57
Area A1 = 80 × 12 = 960 mm2
Area A2 = (150 – 22) × 12 = 1536 mm2
Area A3 = 120 × 10 = 1200 mm2
Total area A = 3696 mm2
Due to symmetry, centroid lies on axis y-y. The bottom fibre (1)–(1) is chosen as reference axis
to locate the centroid.
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 135
æ 128 ö
960 ´ (150 – 6) + 1536 ´ ç + 10÷ + 1200 ´ 5
è 2 ø
=
3696
= 69.78 mm
With reference to the centroidal axis x-x and y-y, the centroid of the rectangles A1 is g1 (0.0,
74.22), that of A2 is g2 (0.0, 4.22) and that of A3 is g3 (0.0, 64.78).
I xx 1, 24,70,027
∴ kxx = =
A 3696
= 58.09 mm
I yy 19,70, 432
kyy = =
A 3696
= 23.09 mm.
Example 4.17. Determine the moment of inertia of the built-up section shown in Fig. 4.58 about its
centroidal axis x-x and y-y.
Solution: The given composite section may be divided into simple rectangles and triangles as shown
in the Fig. 4.58
Area A1 = 100 × 30 = 3000 mm2
Area A2 = 100 × 25 = 2500 mm2
Area A3 = 200 × 20 = 4000 mm2
1
Area A4 = × 87.5 × 20 = 875 mm2
2
1
Area A5 = × 87.5 × 20 = 875 mm2
2
Total area A = 11250 mm2
136 MECHANICS OF SOLIDS
y
100
30 g1 A1
25
80 A2
g2
x x
20 g4 g5 A5
A4
20 g3 A3
1 1
200
y
Fig. 4.58
Due to symmetry, centroid lies on the axis y-y.
A reference axis (1)–(1) is choosen as shown in the figure.
The distance of the centroidal axis from (1)–(1)
sum of moment of areas about (1)–(1)
y =
Total area
æ1 ö
3000 ´ 135 + 2500 ´ 70 + 4000 ´ 10 + 875 ç ´ 20 + 20÷ ´ 2
è3 ø
=
11250
= 59.26 mm
With reference to the centroidal axis x-x and y-y, the centroid of the rectangle A1 is g1 (0.0, 75.74),
that of A2 is g2(0.0, 10.74), that of A3 is g3 (0.0, 49.26), the centroid of triangle A4 is g4 (41.66, 32.59)
and that of A5 is g5 (41.66, 32.59).
100 ´ 303 25 ´ 1003 200 ´ 203
Ixx = + 3000 × 75.742 + + 2500 × 10.742 + + 4000
12 12 12
87.5 ´ 203 87.5 ´ 203
× 49.262 + + 875 × 32.592 + + 875 × 32.592
36 36
Ixx = 3,15,43,447 mm4
30 ´ 1003 100 ´ 253 20 ´ 2003 20 ´ 87.53
Iyy = + + + + 875 × 41.662
12 12 12 36
20 ´ 87.53
+ + 875 × 41.662
36
Iyy = 1,97,45,122 mm4.
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 137
Example 4.18. Determine the moment of inertia of the built-up section shown in the Fig. 4.59 about
an axis AB passing through the top most fibre of the section as shown.
400
A B
20
10
10
380
10
100 150 100
Fig. 4.59
Solution: In this problem, it is required to find out the moment of inertia of the section about an axis
AB. So there is no need to find out the position of the centroid.
The given section is split up into simple rectangles as shown in Fig. 4.59.
Now,
Moment of inertia about AB = Sum of moments of inertia of the rectangle about AB
400 ´ 203
+ 400 × 20 × 102 +
LM
100 × 10 3
+ 100 × 10 × ( 20 + 5) 2
OP × 2
N Q
=
12 12
LM100 × 380 + 10 × 380 × (30 + 190) OP × 2
3
2
+
N 12 Q
+ M
L100 × 10 + 100 × 10 × (20 + 10 + 380 + 5) OP × 2
3
2
N 12 Q
IAB = 8.06093 × 108 mm4.
Example 4.19. Calculate the moment of inertia of the 250
A B
built-up section shown in Fig. 4.60 about a centroidal axis 50
parallel to AB. All members are 10 mm thick.
Solution: The built-up section is divided into six simple X X
10
rectangles as shown in the figure.
The distance of centroidal axis from AB 250
250 ´ 103
= + 250 × 10 × (73.03 – 5)2
12
LM10 × 40 3
OP
+ 40 × 10 ( 73.03 − 30) 2 × 2
N 12 Q
+
40 ´ 103 10 ´ 2503
+ + 40 × 10 (73.03 – 15)2 + + 250
12 12
40 ´ 103
× 10 (73.03 – 135)2 + + 40 × 10 (73.03 – 255)2
12
Ixx = 5,03,99,395 mm4.
Example 4.20. A built-up section of structural steel consists of a flange plate 400 mm × 20 mm, a web
plate 600 mm × 15 mm and two angles 150 mm × 150 mm × 10 mm assembled to form a section as
shown in Fig. 4.61. Determine the moment of inertia of the section about the horizontal centroidal
axis.
Solution: Each angle is divided into two rectangles as shown Web 600 × 15
in Fig. 4.61.
The distance of the centroidal axis from the bottom fibres
of section
Sum of the moment of the areas about bottom fibres 600
=
Total area of the section
å Ai yi
=
A
Now, ΣAiyi = 600 × 15 × ç
æ 600 ö X X
+ 20÷ + 140 × 10 –
y
è 2 ø
20
× (70 + 30) × 2 + 150 × 10 × (5 + 20) 400
× 2 + 400 × 20 × 10 Flange plate 400 × 20
= 33,15,000 mm3 Fig. 4.61
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 139
å Ai yi 3315000
∴ y = =
A 22800
= 145.39 mm
Moment of inertia of the UV RS
Sum of the moments of inertia of the
W T
section about centroidal axis = all simple figures about centroidal axis
15 ´ 6003
= + 600 × 15(145.39 – 320)2
12
LM10 × 140 3
+ 1400(145.39 − 100) 2 × 2
OP
N 12 Q
+
+M
L150 × 10 3
+ 1500 × (145.39 − 15) 2 × 2
OP
N 12 Q
400 ´ 203
+ + 400 × 20 × (145.39 – 10)2
12
Ixx = 7.45156 × 108 mm4.
Example 4.21. Compute the moment of inertia of the 100 mm × 150 mm rectangle shown in Fig. 4.62
about x-x axis to which it is inclined at an angle
D
æ 4ö
θ = sin –1
çè ÷ø .
5 A1
N
15
60 60
∴ FD = = = 75 mm
sin θ sin 5313
.
140 MECHANICS OF SOLIDS
∴ AF = 150 – FD = 75 mm
Hence FL = ME = 75 sin θ = 60 mm
AB 100
AE = FC = = = 125 mm
cos( 90° − θ ) 0.8
UV RS
Moment of inertia of the Sum of the moments of inertia of indivi-
section about x-x axis W T
= dual triangular areas about x-x axis
125 ´ 603 1
2
1
= + × 125 × 60 × 60 + × 60
36 2 3
125 ´ 603 1
125 ´ 603 1
2
2
+ + × 125 × 60 × × 60 + + × 125
36 2 3 36 2
125 ´ 603 1
2 2
1 1
× 60 × × 60 + + × 125 × 60 × × 60
3 36 2 3
Ixx = 3,60,00,000 mm4.
Example 4.22. Find moment of inertia of the shaded area shown in the Fig. 4.63 about the axis AB.
80
P Q
A 40 B
40
S
40 40
Fig. 4.63
Solution: The section is divided into a triangle PQR, a semicircle PSQ having base on axis AB and a
circle having its centre on axis AB.
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 141
Now,
R|Moment of inertia of triangle PQR about
Moment of inertia of the U S AB + Moment of inertia of semicircle
section about axis AB VW = | PSQ about AB − moment of inertia of
Tcircle about AB
80 ´ 803 p p
= + × 804 – × 404
12 128 64
IAB = 42,92,979 mm4.
Example 4.23. Find the second moment of the shaded portion shown in the Fig. 4.64 about its
centroidal axis.
Y
30 50
B
20
A C
B
G
X X
40 –
y
20
=
R
E D
20 40 20
Y
Fig. 4.64
Solution: The section is divided into three simple figures viz., a triangle ABC, a rectangle ACDE and
a semicircle.
Total Area = Area of triangle ABC + Area of rectangle ACDE – Area of semicircle
1 1
A= × 80 × 20 + 40 × 80 – × π × 202
2 2
= 3371.68
1 æ1 ö 1 4 ´ 20
Ay = × 80 × 20 è ´ 20 + 40ø + 40 × 80 × 20 – × π × 202 ×
2 3 2 3p
= 95991.77
∴ y = 95991.77 = 28.47 mm
3371.6
1 2 1 æ1 ö
Ax = × 30 × 20 × × 30 + × 50 × 20 × è ´ 50 ´ 30ø
2 3 2 3
1
+ 40 × 80 × 40 – × π × 202 × 40
2
142 MECHANICS OF SOLIDS
= 132203.6
Ax 132203.6
∴ x = = = 37.21 mm
A 3371.68
80 ´ 203
2
1 2
∴ Ixx = + × 80 × 20 60 – × 20 – 28.47
36 2 3
80 ´ 403
+ + 80 × 40 × (28.47 – 20)2
12
LM 1 F 4 × 20 I OP2
– 0.0068598 × 20 4 +
N 2 H
π × 20 2 28.47 −
3π KQ
Ixx = 6,86,944 mm4.
20 ´ 303 1
2
2 20 ´ 503
Similarly, Iyy = + × 20 × 30 39.21 – × 30 +
36 2 3 36
2
1 é 1 ù 40 ´ 80 3
+ × 20 × 50 × ê 39.21 – 30 + × 50 ú +
2
ëê 3 ûú 12
1 p 1 p
+ 40 × 80(39.21 – 40)2 – ´ × 404 – ×
2 64 2 4
× 402 (40 – 39.21)2
= 1868392 mm4.
y y
A B
x x
(a) Cylinder Solid cylinder
y y
B
A
x x
(b) Cone Solid cone
y y
2
y = kx
B
A
x x
(c) General curve General solid
Semicircle
A B
(d) Sphere Solid sphere
y
x
(e) Torus Solid Torus
Generating surface of revolution Generating solid of revolution
Fig. 4.65
Theorem I
The area of surface generated by revolving a plane curve about a non-intersecting axis in the plane
of the curve is equal to the length of the generating curve times the distance travelled by the centroid
of the curve in the rotation.
144 MECHANICS OF SOLIDS
Proof: Figure 4.66 shows the isometric view of the plane curve rotated about x-axis by angle θ. We
are interested in finding the surface area generated by rotating the curve AB. Let dL be the elemental
length on the curve at D. Its coordinate be y. Then the elemental surface area generated by this
element at D
dA = dL(y θ)
∴ A = ò dL(y θ)
= θ ò y dL
= θ Lyc
= L (yc θ)
dL
D
B
y
A
x
Fig. 4.66
Thus we get area of the surface generated as length of the generating curve times the distance
travelled by the centroid.
Theorem II
The volume of the solid generated by revolving a plane area about a non-intersecting axis in the
plane is equal to the area of the generating plane times the distance travelled by the centroid of the
plane area during the rotation.
Proof: Consider the plane area ABC, which is rotated through an angle θ about x-axis as shown in
Fig. 4.67.
B
y
B dA
y
A
C x
Fig. 4.67
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 145
Let dA be the elemental area of distance y from x-axis. Then the volume generated by this area
during rotation is given by
dV = dA/yθ
∴ V = ò dA/yθ
= θ ò y dA
= θ A yc
= A(yc θ)
Thus the volume of the solid generated is area times the distance travelled by its centroid during
the rotation. Using Pappus-Guldinus theorems surface area and volumes of cones and spheres can be
calculated as shown below:
(i) Surface area of a cone: Referring to Fig. 4.68(a),
Length of the line generating cone = L
R
Distance of centroid of the line from the axis of rotation = y =
2
In one revolution centroid moves by distance = 2πy = πR
∴ Surface area = L × (πR) = πRL
L/2
G R R
L/2
y = R/2 G
y1 = R/3
h x
(a) (b)
Fig. 4.68
R
∴ The distance moved by the centroid in one revolution = 2πy = 2π
3
1 2pR
∴ Volume of solid cone = hR ×
2 3
pR2 h
=
3
146 MECHANICS OF SOLIDS
(iii) Surface area of sphere: Sphere of radius R is obtained by rotating a semi circular arc of
radius R about its diametral axis. Referring to Fig. 4.69(a),
Length of the arc = π R
2R
Centroid of the arc is at y = from the diametral axis (i.e. axis of rotation)
p
∴ Distance travelled by centroid of the arc in one revolution
2R
= 2π y = 2π = 4R
p
∴ Surface area of sphere = π R × 4R
= 4π R2
(iv) Volume of sphere: Solid sphere of radius R is obtained by rotating a semicircular area about
its diametral axis. Referring to Fig. 4.69(b).
pR 2
Area of semicircle =
2
Distance of centroid of semicircular area from its centroidal axis
4R
=y=
3p
∴ The distance travelled by the centroid in one revolution
4 R 8R
= 2πy = 2π =
3p 3
ðR 2 8 R
∴ Volume of sphere = ´
2 3
4pR 3
=
3
G G
y = 2R y = 4R
3
x x x x
(a) (b)
Fig. 4.69
centre of gravity may be found from first principle and the results obtained for standard solids may
be used to find centre of gravity of composite solids. The procedure is illustrated with examples 4.24
to 4.27.
Example 4.24. Locate the centre of gravity of the right circular cone of base radius r and height h
shown in Fig. 4.70.
dx Axis of
rotation
O C A
X
Z
r
x D
B
h
Y z
Fig. 4.70
Solution: Taking origin at the vertex of the cone and selecting the axis as shown in Fig. 4.70, it can
be observed that due to symmetry the coordinates of centre of gravity y and z are equal to zero, i.e.
the centre of gravity lies on the axis of rotation of the cone. To find its distance x from the vertex,
consider an elemental plate at a distance x. Let the thickness of the elemental plate be dx. From the
similar triangles OAB and OCD, the radius of elemental plate z is given by
x
z= r
h
∴ Volume of the elemental plate dv
r2
dv = π z2 dx = π x2 dx
h2
If γ is the unit weight of the material of the cone, then weight of the elemental plate is given by:
r2
dW = γπx2 dx ...(i)
h2
h pr 2
W= ò g x2 dx
0 h2
h
pr 2 é x3 ù
=γ 2 ê ú
h ë 3 û0
r 2h
= γπ ...(ii)
3
LMNote: πr h is volume of coneOP
2
N 3 Q
Now, substituting the value of dW in (i), above, we get:
h
pr 2
ò x . dW = ò γ x2 ⋅ x ⋅ dx
0 h2
148 MECHANICS OF SOLIDS
h
pr 2 é x 4 ù
=γ 2 ê ú
h ë 4 û0
pr 2 h 2
=γ ...(iii)
4
From eqn. 4.1,
W x = ò x dW
pr 2 h gpr 2 h 2
i.e., x =
3 4
3
∴ x = h
4
3
Thus, in a right circular cone, centre of gravity lies at a distance h from vertex along the axis
4
h
of rotation i.e., at a distance from the base.
4
Example 4.25. Determine the centre of gravity of a solid hemisphere of radius r from its diametral
axis.
Solution: Due to symmetry, centre of gravity lies on the axis of rotation. To find its distance x from
the base along the axis of rotation, consider an elemental plate at a distance x as shown in Fig. 4.71.
Now, x2 + z2 = r2
z2 = r2 – x2 ...(i)
Volume of elemental plate
dv = πz2 dx = π(r2 – x2)dx ...(ii)
x
∴ Weight of elemental plate O x
2 gpr 3
= ...(iv)
3
Moment of weight about z axis
r
= ò xdW
0
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 149
r
= ò x π(r2 – x2)dx
0
r
é x2 x4 ù
= π êr 2 - ú
ë 2 4 û0
pr 4
= ...(v)
4
∴ x , the distance of centre of gravity from base is given by:
r
W x = ò x dw
0
i.e., From (iv) and (v) above, we get
2 gpr 3 gpr 4 3
x = x = r
3 4 8
3
Thus, the centre of gravity of a solid hemisphere of radius r is at a distance r from its
8
diametral axis.
Example 4.26. Determine the maximum height h of the cylindrical portion of the body with hemi-
spherical base shown in Fig. 4.72 so that it is in stable equilibrium on its base.
Axis of rotation
Solution: The body will be stable on its base as long as its centre of
gravity is in hemispherical base. The limiting case is when it is on the
plane x-x shown in the figure. m1
g1
Centroid lies on the axis of rotation. h
Mass of cylindrical portion h/2
X X
m1 = πr2hρ, where ρ is unit mass of material. r 3/8 r
g2 m2
Its centre of gravity g1 is at a height
h
z1 = from x axis.
2 Z
Mass of hemispherical portion Fig. 4.72
2 pr 3
m2 = ρ
3
and its CG is at a distance
3r
z2 = from x-x plane.
8
Since centroid is to be on x-x plane z = 0
i.e., Σmizi = 0
m1h 3
∴ – m2 r = 0
2 8
h 2 pr 3 3
πr2hρ =ρ r
2 3 8
150 MECHANICS OF SOLIDS
1 2
∴ h2 = r
2
r
or h= = 0.707 r
2
Example 4.27. A concrete block of size 0.60 m × 0.75 m × 0.5 m is cast with a hole of diameter 0.2
m and depth 0.3 m as shown in Fig. 4.73. The hole is completely filled with steel balls weighing 2500
N. Locate the centre of gravity of the body. Take the weight of concrete = 25000 N/m3.
0.75 m
0.5 m
X
0.2 m
0. 4m
m
0.6 Z
0.3 m
Y
0.5 m
Fig. 4.73
x = å i i =å i i
Wx Wx 3241.57
∴ x = = 0.411 m
W åWi 7889.38
2593.25
Similarly, y = = 0.329 m
7887.38
1745.91
z = = 0.221 m
7889.38
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 151
IMPORTANT FORMULAE
ðR 2 h
5. Volume of a cone =
3
4
6. Volume of a sphere = πR3
3
R sin α
7. Centroid of a arc of a circle is at xc = from the centre of circle on the symmetric axis.
α
8. Centroid of a composite figure is given by
ΣAi x i ΣAi yi
xc = , yc = .
A A
9. Centroid of simple figure from the reference axis
y=
A
. z y dA
z
10. For centroid of standard figures refer Table 4.2.
11. Iyy = Σxi2 dAi and Ixx = Σyi2 dAi, Izz = Σri2dAi = r 2 dA .
I
12. Radius of gyration k = i.e. I = Ak2.
A
13. Izz = Ixx+ Iyy.
14. IAB = IGG + Ayc2.
15. Moment of inertia of standard sections are as shown in Table 4.4.
16. Pappus-Guldinus Theorems:
(i) The area of surface generated by revolving a plane curve about a non-intersecting axis in
the plane of the curve is equal to the length of the generating curve times the distance
travelled by the centroid of the curve in the rotation.
(ii) The volume of the solid generated by a plane area about a non-intersecting axis in the
plane is equal to the area of the generating plane times the distance travelled by the centroid
of the plane area during the rotation.
152 MECHANICS OF SOLIDS
3
17. In a right circular cone, the centre of gravity lies at a distance × height from the vertex along
4
the axis of rotation.
3
18. The centre of gravity of a solid hemisphere of radius r is at a distance × r from its diametral
8
axis.
THEORY QUESTIONS
1. Determine the centroid of the built-up section in Fig. 4.74. Express the coordinates of centroid
with respect to x and y axes shown. [Ans. x = 48.91 mm; y = 61.30 mm]
100
10
20
80
10
y
40 10 20
20
O x
120
Fig. 4.74
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 153
2. Determine the centroid of the reinforced concrete retaining wall section shown in Fig. 4.75.
[Ans. x = 1.848 m; y = 1.825 m]
0.3 m
6.0 m
0.6 m
O x
1.4 m 0.5 2.1 m
Fig. 4.75
3. Find the coordinates of the centroid of the shaded area with respect to the axes shown in
Fig. 4.76. [Ans. x = 43.98 mm; y = 70.15 mm]
60 60
60 y
R
80 =
80
X
40 80
Fig. 4.76
4. A circular plate of uniform thickness and of diameter 500 mm as shown in Fig. 4.77 has two
circular holes of 40 mm diameter each. Where should a 80 mm diameter hole be drilled so that
the centre of gravity of the plate will be at the geometric centre.
[Ans. x = 50 mm; y = 37.5 mm]
154 MECHANICS OF SOLIDS
200
X
40 5 00
150
D=
40
Fig. 4.77
5. With respect to the coordinate axes x and y locate the centriod of the shaded area shown in
Fig. 4.78. [Ans. x = 97.47 mm; y = 70.77 mm]
Y 50
30
30
100
70
50
R=
X
100 100
Fig. 4.78
80
45
0
R=5
30
R=
y 80
225 X
Fig. 4.79
7. Determine the coordinates of the centroid of shaded as shown in Fig. 4.80 with respect to the
corner point O. Take x = 40 mm. [Ans. x = 260.07 mm; y = 113.95 mm]
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 155
2x 2x 6x 2x
X
2x
2x 2x
r=
3x 3x 3x
R=
2x 2x
2x
Y
Fig. 4.80
8. ABCD is a square section of sides 100 mm. Determine the ratio of moment of inertia of the
section about centroidal axis parallel to a side to that about diagonal AC. [Ans. 1]
9. The cross-section of a rectangular hollow beam is as shown in Fig. 4.81. Determine the polar
moment of inertia of the section about centroidal axes.
[Ans. Ixx = 1,05,38,667 mm4; Iyy = 49,06,667 mm4; Izz = 1,54,45,334 mm4]
20 40 20
20
40
120
60
80
Fig. 4.81
10. The cross-section of a prestressed concrete beam is shown in Fig. 4.82. Calculate the moment
of inertia of this section about the centroidal axes parallel to and perpendicular to top edge.
Also determine the radii of gyration. [Ans. Ixx = 1.15668 × 1010 mm4; kxx = 231.95 mm;
Iyy = 8.75729 × 109 mm4; kyy = 201.82 mm]
1000
100
150 350
250
250
Fig. 4.82
156 MECHANICS OF SOLIDS
11. The strength of a 400 mm deep and 200 mm wide I-beam of uniform thickness 10 mm, is
increased by welding a 250 mm wide and 20 mm thick plate to its upper flanges as shown in
Fig. 4.83. Determine the moment of inertia and the radii of gyration of the composite section
with respect to cetroidal axes parallel to and perpendicular to the bottom edge AB.
[Ans. Ixx = 3.32393 × 108 mm4; kxx = 161.15 mm;
Iyy = 3,94,06,667 mm4; kyy = 55.49 mm]
250
20
10
10
400
10
A B
200
Fig. 4.83
12. The cross-section of a gantry girder is as shown in Fig. 4.84. It is made up of an I-section of
depth 450 mm, flange width 200 mm and a channel of size 400 mm × 150 mm. Thickness of
all members is 10 mm. Find the moment of inertia of the section about the horizontal centroid
axis. [Ans. Ixx = 4.2198 × 108 mm4]
400
150
10
10 450
200
Fig. 4.84
13. A plate girder is made up of a web plate of size 400 mm × 10 mm, four angles of size 100
mm × 100 mm × 10 mm and cover plates of size 300 mm × 10 mm as shown in Fig. 4.85.
Determine the moment of inertia about horizontal and vertical centroidal axes.
[Ans. Ixx = 5.35786 × 108 mm4; Iyy = 6,08,50,667 mm4]
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 157
300
10
4 equal
angles of size 10
100 × 100 × 10 mm 400
Fig. 4.85
14. Determine the moment of inertia and radii of gyration of the area shown in Fig. 4.86 about
the base A-B and the centroidal axis parallel to AB.
[Ans. IAB = 48,15,000 mm4; Ixx = 18,24,231 mm4
kAB = 35.14 mm; kxx = 21.62 mm]
30
30
20
30
A 100 B
Fig. 4.86
15. Determine the moment of inertia of the section shown in Fig. 4.87 about the vertical centroidal
axis. [Ans. Iyy = 5,03,82,857 mm4]
20 80
100
60 120 60
Fig. 4.87
158 MECHANICS OF SOLIDS
16. A semi-circular cut is made in rectangular wooden beam as shown in Fig. 4.88. Determine the
polar moment of inertia of the section about the centroidal axis.
[Ans. Ixx = 3,35,81,456 mm4; Iyy = 1,00,45,631 mm4; Izz = 2,20,98,980 mm4]
150
100
Fig. 4.88
17. Determine the moment of inertia of the section shown in the Fig. 4.89 about the horizontal
centroidal axis. Also find the moment of inertia of the section about the symmetrical axis.
Hence find the polar moment of inertia.
[Ans. Ixx = 54,09,046 mm4; Iyy = 14,55,310 mm4; Izz = 68,64,356 mm4]
Semicircle
40
r=
30 30
100
Fig. 4.89
18. The cross-section of a machine part is as shown in Fig. 4.90. Determine its moment of inertia
and radius of gyration about the horizontal centroidal axis.
[Ans. Ixx = 5249090.85 mm4; kxx = 27.05 mm]
DISTRIBUTED FORCES, CENTRE OF GRAVITY AND MOMENT OF INERTIA 159
O
60
r=
100
100
Fig. 4.90
19. The cross-section of a plain concrete culvert is as shown in Fig. 4.91. Determine the moment
of inertia about the horizontal centroidal axes. [Ans. Ixx = 5.45865 × 1010 mm4]
400 1000 400
1000
0
65
r=
Fig. 4.91
20. Determine the centroid of the built-up section shown in Fig. 4.92 and find the moment of
inertia and radius of gyration about the horizontal centroidal axis.
[Ans. Ixx = 1267942 mm4; kxx = 18.55 mm]
20
20 20
20
40
20
r=
20 40 20
Fig. 4.92
LMAns. x = 3 hOP
21. Determine the centre of gravity of the pyramid shown in Fig. 4.93.
N 4 Q
160 MECHANICS OF SOLIDS
O x
a
y
2y
h
Fig. 4.93
22. A steel ball of diameter 150 mm rests centrally over a concrete cube of size 150 mm. Determine
the centre of gravity of the system, taking weight of concrete = 25000 N/m3 and that of steel
80000 N/m3. [Ans. 168.94 mm from base]
23. Locate the centre of gravity of the wire shown in Fig. 4.94. Portion BC is in x-y plane and
semicircle CD is parallel to x-z plane.
[Ans. x = 124.02 mm ; y = 110.41 mm ; z = 11.28 mm]
B
x
0 25
20 0
30°
A C 60 D
z
Fig. 4.94
5
Friction
When a body moves or tends to move over another body, a force opposing the motion develops at the
contact surfaces. The force which opposes the movement or the tendency of movement is called the
frictional force or simply friction. So far, in earlier chapters, we had ignored this force and considered
contacting surfaces are smooth. Actually in almost all cases the contacting surfaces are not smooth.
There are minutely projecting particles which develop frictional force to oppose the tendency to
movement of one surface over the other surface. In this chapter, the additional terminology used in
connection with frictional forces are explained and laws of dry friction (wet friction excluded) are
presented. Applications of these laws to many engineering problems are illustrated.
161
162 MECHANICS OF SOLIDS
F Fig. 5.1
µ= ...(5.1)
N
F F q
tan θ = R
N N
F
As frictional force increases the angle θ increases and it can Fig. 5.2
reach maximum value a when limiting value of friction is
reached. Thus, when motion is impending
FRICTION 163
F
tan α = =µ ...(5.2)
N
and this value of α is called angle of limiting friction. Hence, the angle of limiting friction can be
defined as the angle between the resultant reaction and the normal to the plane on which the motion of
the body is impending.
Angle of Repose
It is very well-known that when grains (food grains, sand, cement, soil etc.) are heaped, there exists a
limit for the inclination of the heap. Beyond that the grains start rolling down. The limiting angle up to
which the grains repose (sleep) is called angle of repose.
Now consider the block of weight W shown in Fig. 5.3 which is resting on an inclined plane that
makes angle θ with the horizontal. When θ is a small, block rests on the plane. If θ is increased
gradually a stage is reached at which the block starts sliding. The angle made by the plane with the
horizontal is called angle of friction for the contacting surfaces. Thus, the maximum inclination of the
plane on which the body, free from external forces, can repose is called angle of repose.
Consider the equilibrium of the block shown in Fig. 5.3. Since the surface of contact is not smooth,
not only normal reaction but frictional force also develops. As the body tends to slide down, the
frictional resistance will be up the plane.
Σ Forces normal to plane = 0, gives
N = W cos θ ...(i)
Σ Forces parallel to plane = 0, gives
F = W sin θ ...(ii)
Dividing eqn. (ii) by eqn. (i) we get,
F
= tan θ Fig. 5.3
N
If φ is the value of θ when motion is impending, frictional force will be limiting friction and hence
F
tan φ = ...(5.3)
N
i.e., tan φ = µ = tan α
or φ = α.
Thus, the value of angle of repose is the same as the value of limiting angle of friction.
Cone of Friction
When a body is having impending motion in the direction of P, the frictional force will be the limiting
friction and the resultant reaction R will make limiting frictional angle a with the normal as shown in
164 MECHANICS OF SOLIDS
(a) (b)
Fig. 5.5
FRICTION 165
∑ FV = 0 →
N1 – 1000 = 0 or N1 = 1000 newton.
Since F1 is limiting friction,
F1 1
= µ1 =
N1 4
1
∴ F1 = ´ 1000 = 250 newton.
4
∑ FH = 0 →
F1 – T = 0 or T = F1, i.e. T = 250 newton.
Consider the equilibrium of block B.
∑ FV = 0 →
N2 – N1 – 2000 = 0.
∴ N2 = N1 + 2000 = 1000 + 2000 = 3000 newton.
Since F2 is limiting friction,
1
F2 = µ2 N2 = ´ 3000 = 1000 newton.
3
∑ FH = 0 →
P – F1 – F2 = 0
∴ P = F1 + F2 = 250 + 1000 = 1250 newton.
(b) When P is inclined: Free body diagrams for this case are shown in Fig. 5.5(c).
Considering equilibrium of block A, we get
∑ FV = 0 → N1 = 1000 newton.
1
∴ F1 = ´ 1000 = 250 newton.
4
∑ FH = 0 → T = F1 = 250 newton.
∑ FV = 0 →
166 MECHANICS OF SOLIDS
N2 – 2000 – N1 + P sin 30 = 0
or N2 + 0.5P = 3000, since N1 = 1000 newton.
From law of friction
1
F2 = µ2N2 = ´ (3000 – 0.5P)
3
0.5
= 1000 – P.
3
∑ FH = 0 →
P cos 30 – F1 – F2 = 0 Fig. 5.5(c)
0.5
∴ P cos 30 – 250 – 1000 − P = 0
3
0.5
∴ P cos 30 + = 1250
3
∴ P = 1210.4 newton
Example 5.2. What should be the value of θ in Fig. 5.6(a) which will make the motion of 900 N block
down the plane to impend? The coefficient of friction for all contact surfaces is 1/3.
N
300
N
900
Fig. 5.6(a)
Solution: 900 N block is on the verge of moving downward. Hence frictional forces F1 and F2 [Ref.
Fig. 5.6(b)] act up the plane on 900 N block. Free body diagrams of the blocks are as shown in Fig.
5.6(b).
Consider the equilibrium of 300 N block.
Σ Forces normal to plane = 0 →
N1 – 300 cos θ = 0 or N1 = 300 cos θ ...(i)
FRICTION 167
(a) (b)
Fig. 5.7
168 MECHANICS OF SOLIDS
When block starts moving down the plane [Ref. Fig. 5.7(a)]
Frictional forces oppose the direction of the movement. Hence F1 is up the plane and F2 down the
plane. Since it is limiting case
F
= µ.
N
S forces perpendicular to the plane = 0 →
N – 500 cos θ = 0 or N = 500 cos θ ...(i)
From law of friction,
F1 = µN = 500 µ cos θ ...(ii)
Σ Forces parallel to the plane = 0 →
F1 + 200 – 500 sin θ = 0
Substituting the value of F1 from eqn. (ii), we get
500 sin θ – 500 µ cos θ = 200 ...(iii)
When the block starts moving up the plane [Fig. 5.7(b)]
50
∴ µ= = 0.115
500 cos 30
Example 5.4. Block A weighing 1000 N and block B weighing 500 N are connected by flexible wire.
The coefficient of friction between block A and the plane is 0.5 while that for block B and the plane is
0.2. Determine what value of inclination of the plane the system will have impending motion down the
plane? [Ref. Fig. 5.8].
FRICTION 169
(a) (b)
Fig. 5.8
Solution: Let θ be the inclination of the plane for which motion is impending. Free body diagrams of
blocks A and B are as shown in Fig. 5.8(b). Considering equilibrium of block A,
Σ Forces normal to plane = 0 →
N1 – 1000 cos θ = 0 or N1 = 1000 cos θ ...(i)
∴ From law of friction
F1 = µ1 N1 = 0.5 ´ 1000 cos θ = 500 cos θ ...(ii)
Σ Forces parallel to plane = 0 →
F1 – T – 1000 sin θ = 0
or T = 500 cos θ – 1000 sin θ ...(iii)
Consider the equilibrium of block B,
Example 5.5. What is the value of P in the system shown in Fig. 5.9(a) to cause the motion to impend?
Assume the pulley is smooth and coefficient of friction between the other contact surfaces is 0.2.
(a) (b)
Fig. 5.9
Solution: Free body diagrams of the blocks are as shown in Fig. 5.9(b). Consider the equilibrium of
750 N block.
∑ FV = 0 →
N2 – 500 + P sin 30 = 0
i.e., N2 + 0.5 P = 500 ...(iv)
From law of friction,
F2 = µN2 = 0.2 (500 – 0.5P) = 100 – 0.1P ...(v)
∑ FH = 0 →
P cos 30 – T – F2 = 0
i.e., P cos 30 – 724.5 – 100 + 0.1P = 0
∴ P = 853.5 N
FRICTION 171
Example 5.6. Two identical planes AC and BC, inclined at 60° and 30° to the horizontal meet at C as
shown in Fig. 5.10. A load of 1000 N rests on the inclined plane BC and is tied by a rope passing over
a pulley to a block weighing W newtons and resting on the plane AC. If the coefficient of friction
between the load and the plane BC is 0.28 and that between the block and the plane AC is 0.20, find
the least and greatest values of W for the equilibrium of the system.
(a) (b)
(c)
Fig. 5.10
Solution:
(a) Least value of W:
In this case motion of 1000 N block is impending down the plane and block W has impending motion
up the plane. Hence free body diagrams for the blocks are as shown in Fig. 5.10(b). Considering the
equilibrium of 1000 N block,
Σ Forces normal to the plane = 0 →
N1 – 1000 cos 30 = 0 ∴ N1 = 866.0 newton ...(i)
From the law of friction
F1 = µ1 N1 = 0.28 ´ 866.0 = 242.5 newton ...(ii)
257.5
∴ W= = 266.6 N.
0.1 + sin 60
(b) For the greatest value of W:
In such case 1000 N block is on the verge of moving up the plane and W is on the verge of moving
down the plane. For this case free body diagrams of the blocks are as shown in Fig. 5.10(c).
Considering the block of 1000 N,
Σ Forces normal to plane = 0 →
N1 – 1000 cos 30 = 0 ∴ N1 = 866.0 newton ...(vi)
From law of friction,
F1 = µ1 N1 = 0.28 ´ 866.0 = 242.5 N ...(vii)
Σ Forces parallel to the plane = 0 →
T – 1000 sin 30 – F1 = 0
∴ T = 500 + 242.5 = 742.5 newton ...(viii)
Considering the equilibrium of block weighing W,
Σ Forces normal to plane = 0 →
N2 – W cos 60 = 0 or N2 = 0.5 W ...(ix)
∴ F2 = µ2 N2 = 0.2 ´ 0.5 W = 0.1 W ...(x)
Σ Forces parallel to plane = 0 →
T – W sin 60 + F2 = 0 ...(xi)
Substituting the values of T and F2 from eqns. (viii) and (x), we get,
742.5 – W sin 60 + 0.1 W = 0
742.5
or W= = 969.3 newton
sin 60 − 0.1
FRICTION 173
20°
(a) (b)
Fig. 5.11
Solution: Free body diagrams for blocks A and B are as shown in Fig. 5.11(b). Consider block B.
µ = tan 20°, given.
∴ F1 = N1 tan 20°
∑ FV = 0 →
N1 sin 30 + F1 sin 60 – 5 = 0
0.5 N1 + N1 tan 20 sin 60 = 5
N1 = 6.133 kN
Hence, F1 = 6.13 tan 20 = 2.232 kN
∑ FH = 0 →
C + F1 cos 60 – N1 cos 30 = 0
C + 2.232 cos 60 – 6.133 cos 30 = 0
∴ C = 4.196 kN
Now consider the equilibrium of block A.
∑ FH = 0 →
174 MECHANICS OF SOLIDS
F2 – C = 0 or F2 = C = 4.196 kN
From law of friction,
F2 = mN2
i.e., 4.196 = 0.4 N2
∴ N2 = 10.49 kN
Then, ∑ FV = 0 →
N2 – W = 0
or W = N2 = 10.49 kN
(a) (b)
(c) (d)
Fig. 5.12
Solution: As wedge is driven, it moves towards left and the block moves upwards. When motion is
impending limiting friction develops. Hence resultant force makes limiting angle of 15° with normal.
The care is taken to mark 15° inclination such that the tangential component of the resultant opposes
the impending motion.
The free body diagrams of the block and wedge are shown in Fig. 5.12(b). The forces on block and
wedge are redrawn in Figs. 5.12(c) and (d) so that Lami’s theorem can be applied conveniently.
Applying Lami’s theorem to the system of forces on block
R1 R2 20
= =
sin (180 − 15 − 20) sin (90 − 15) sin (15 + 20 + 90 + 15)
R1 R2 20
i.e., = =
sin 145 sin 75 sin 140
∴ R1 = 17.85 kN
and R2 = 30.05 kN
Applying Lami’s theorem to system of forces on the wedge, we get
P R2
=
sin 130 sin 105
∴ P = 23.84 kN
Example 5.9. A block weighing 160 kN is to be raised by means of the wedges A and B as shown in
Fig. 5.13(a). Find the value of force P for impending motion of block C upwards, if coefficient of
friction is 0.25 for all contact surfaces. The self weight of wedges may be neglected.
Solution: Let φ be the angle of limiting friction.
on wedge A are shown in the form convenient for applying Lami’s theorem [Ref. Figs. 5.13(c) and
(d)].
(a) (b)
(c) (d)
Fig. 5.13
R1 160
i.e., = , since φ = 14.036°.
sin 149.96 sin 60.072
∴ R1 = 92.42 kN
Consider the equilibrium of the wedge A. Applying Lami’s theorem, we get
P R1
=
sin (180 − φ − φ − 16) sin (90 + φ)
600 N
NB
B B
600 N
4m
3m
2m 200 N 200 N
60°
P 60° P
A FA
A
NA
(a) (b)
Fig. 5.14
Solution: The free body diagram of the ladder is as shown in Fig. 5.14(b).
∑MA = 0 →
200 ´ 2 cos 60 + 600 ´ 3 cos 60 – FB ´ 4 cos 60 – NB ´ 4 sin 60 = 0
Dividing throughout by 4 and rearranging the terms, we get
or NB = 284.7 newton.
∴ FB = 56.94 newton.
178 MECHANICS OF SOLIDS
∑ FV = 0 →
NA – 200 – 600 + FB = 0
NA = 743.06 newton, since FB = 56.94
∴ FA = µA NA
= 0.3 ´ 743.06 = 222.9 newton
∑ FH = 0 →
P + FA – NB = 0
∴ P = NB – FA = 284.7 – 222.9
i.e., P = 61.8 newton
Example 5.11. The ladder shown in Fig. 5.15(a) is 6 m long and is supported by a horizontal floor
and a vertical wall. The coefficient of friction between the floor and the ladder is 0.25 and between the
wall and the ladder is 0.4. The weight of the ladder is 200 N and may be considered as a concentrated
load at G. The ladder supports a vertical load of 900 N at C which is at a distance of 1 m from B.
Determine the least value of = at which the ladder may be placed without slipping. Determine the
reaction at that stage.
FB
= 0.4
900 N B
NB
1m B
900 N
C
C
6m
G
200 N
200 N
3m
FA
A = 0.25
A
NA
(a) (b)
Fig. 5.15
Solution: Figure 5.15(b) shows the free body diagram of the ladder.
From law of friction,
FRICTION 179
FA = 0.25 NA ...(i)
FB = 0.40 NB ...(ii)
Σ FV = 0 →
NA – 200 – 900 + FB = 0
i.e., NA + 0.4 NB = 900 + 200 = 1100 ...(iii)
åF H =0→
FA – NB = 0 i.e., FA = NB
i.e., 0.25 NA = NB ...(iv)
From eqns. (iii) and (iv), we get
NA (1 + 0.4 ´ 0.25) = 1100
or NA = 1000 newton
∴ FA = 0.25 ´ NA = 0.25 ´ 1000 = 250 N
From eqn. (iv) NB = 0.25 NA = 250 N
∴ FB = 0.4 ´ NB = 0.4 ´ 250 = 100 N
∑MA = 0 →
200 ´ 3 cos α + 900 ´ 5 cos α – FB ´ 6 cos α – NB ´ 6 sin α = 0
∴ Substituting the values of FB and NB, we get
200 ´ 3 cos α + 900 ´ 5 cos α – 100 ´ 6 cos α – 250 ´ 6 sin α = 0.
or 4500 cos α = 1500 sin α
or tan α = 3
∴ α = 71.57°
Example 5.12. A ladder of length L rests against a wall, the angle of inclination being 45°. If the
coefficient of friction between the ladder and the ground and that between ground and the wall is
0.5 each, what will be the maximum distance on ladder to which a man whose weight is 1.5 times
the weight of ladder may ascend before the ladder begins to slip?
Solution: Figure 5.16(a) shows the ladder when it is about to slip when the man weighing 1.5 W
is at a distance, ‘aL’ from the end A. Its free body diagram is shown in Fig. 5.16(b).
Since ladder is on the verge of slipping, from law of friction,
FA = µNB = 0.5 NB ...(1)
and FB = µNB = 0.5 NB ...(2)
ΣH = 0 →
180 MECHANICS OF SOLIDS
FB
1.5 W B 1.5 W B
NB
L L
L L
a a
L L
5 5
0. 0.
W W
45°
FA A
A
NA
Fig. 5.16
FA – NB = 0 or NB = FA = 0.5 NA ...(3)
∴ FB = 0.5 NB = 0.25 NA ...(4)
ΣV = 0 →
NA + FB = W + 1.5 W
i.e., NA + 0.25 NA = 2.5 W
2.5
or NA = W = 1.667 W ...(5)
1.5
ΣMA = 0 →
–FB L cos 45° – NB L sin 45° + 1.5 W aL cos 45º + W 0.5 L cos 45º = 0
Since sin 45° = cos 45°, we get
FB + NB = 1.5 aW + 0.5 W
0.25 NA + 0.5 NA = 1.5 aW + 0.5 W
i.e., 0.75 × 1.667 W = 1.5 aW + 0.5 W
i.e., 1.25 = 1.5 a + 0.5
0.75
or a = = 0.5
1.5
Thus in this case the man can ascend up to ‘0.5 L’ of ladder.
T2 T2 dθ
P W T1 T1
(a) (b) (c) (d)
Fig. 5.17
∴ N −T
dθ
2
− T+dT b dθ
2
g
= 0 ...(1)
FG T + d T IJ d θ
i.e., N =
H 2K
From the law of friction,
FG dT IJ
F =µN = µ T+
H 2
dθ
K ...(2)
dθ dθ
Since is small, cos = 1
2 2
∴ T + dT = F + T
or dT = F ...(3)
FG dT IJ
From (2) and (3), dT = µ T +
H 2
dθ
K
182 MECHANICS OF SOLIDS
T2 θ
∴ log T T1
= µθ 0
T2 T
log = µθ ; since log T2 – log T1 = log 2
T1 T1
T2
= eµθ
T1
i.e., T2 = T1eµθ ...(5.4)
Note: θ should be in radians.
1
Example 5.13. A rope making 1 turns around a stationary horizontal drum is used to support a
4
weight W (Fig. 5.18). If the coefficient of friction is 0.3 what range of weight can be supported by
exerting a 600 N force at the other end of the rope?
6 00 N
Solution: Angle of contact = 1.25 × 2π = 2.5π
(1) Let the impending motion of the weight be downward.
Then,
T1 = 600 N; T2 = W
W W
= eµ 2.5π = e 0.3 × 2.5π = e0.75π
600
Fig. 5.18
W = 6330.43 N
(2) Let the impending motion of weight be upwards. Then
T1 = W; T2 = 600 N
µθ
T2 = T1 e
600 = W e0.75π
W = 56.87 N
Thus, a 600 N force can support a range of loads between 56.87 N to 6330.43 N weight on the
other side of drum.
Example 5.14. In Fig. 5.19 (a) The coefficient of friction is 0.20 between the rope and the fixed
drum, and between other surface of contact µ = 0.3. Determine the minimum weight W to prevent
downward motion of the 1000 N body.
FRICTION 183
T1
W T2 1 00 0 N
T1 T2
3 3
4 N1 3
W N 4 4
0 0
10 F1 F2
F1 N1
α N2
(a) (b)
Fig. 5.19
Solution: Since 1000 N weight is on the verge of sliding downwards the rope connecting it is the
tight side and the rope connecting W is the slack side. Free body diagrams for W and 1000 N body
are shown in Fig. 5.19(b).
Now, cos α = 0.8
sin α = 0.6
Consider the equilibrium of weight W,
Σ Forces perpendicular to the plane = 0, gives
N1 = W cos α
N1 = 0.8 W ...(1)
∴ F1 = µN1 = 0.3 × 0.8 W
F1 = 0.24 W ...(2)
Σ Forces parallel to the plane = 0, gives
T1 = F1 + W sin α = 0.24 W + 0.6 W
= 0.84 W ...(3)
Angle of contact of rope with the pulley = 180° = π radians
Applying friction equation, we get
T2 = T1eµθ = T1 e0.3π
T2 = 2.566 T1
Substituting the value of T1 from (3)
T2 = 2.156 W ...(4)
Now, consider 1000 N body,
Σ forces perpendicular to the plane = 0, gives
N2 – N1 – 1000 cos α = 0
Substituting the value of N1 from (1),
N2 = 0.8 W + 1000 × 0.8 = 0.8 W + 800
∴ F2 = 0.3 N2 = 0.24 W + 240 ...(5)
Σ forces parallel to the plane = 0, gives
F1 + F2 – 1000 sin α + T2 = 0
184 MECHANICS OF SOLIDS
9 0° P
T2 T1
3 00 m m 9 0°
50 m m
R
(a) (b)
Fig. 5.20
Solution: Figure 5.20 (b) shows free body diagrams of brake drum and the lever arm.
Now T2 = T1 eµθ
250π
θ = radians and µ = 0.3
180
∴ µ θ = 1.309
∴ T2 = T1 e1.309 = 3.7025 T1
Now, (T2 – T1) r = M
(3.7025 – 1) T1 × 250 = 300 × 103
∴ T1 = 444.04 N
∴ T2 = 1644.058 N
Consider the lever arm. Taking moment about the hinge, we get
T2 × 50 = P × 300
P = 274.0 N.
Example 5.16. Two parallel shafts 3 m apart are to be connected by a belt running over the pulleys
of diamter 500 mm and 100 mm respectively. Determine the length of belt required and the angle
of contacts between belt and each pulley if it is crossed. What power can be transferred if the larger
pulley rotates at 220 revolutions per minute. Given: Maximum permissible tension in the belt = 1 kN
and coefficient of friction between the belt and the pulley is 0.25.
Solution: The crossed belt drive system t is shown in Fig. 5.21.
From the geometry of the system,
250 + 50
sin α = = 0.1
3000
∴ α = 5.739°
∴ θ 2 = θ 1 = 180 + 2α = 191.478°
= 3.342 radians.
FRICTION 185
G
C
α E
α
A α
5 00 θ θ2 1 00
B
α
α F
D
3 00 0
Fig. 5.21
α
G E
α
α θ2
5 00 θ1 1 00
A B
α
α F
Fig. 5.22
T2
= eµθ2 = e0.25 × 3.008 = 2.1212
T1
T2 1000
∴ T1 = = = 471.4 Newton
2.1212 2.1212
FRICTION 187
2π × 220 2π × 220
Velocity of belt = r × = 250 ×
60 60
= 5759.6 mm/sec.
∴ Maximum power that can be transferred without slippage
= (T2 – T1) × Velocity
= (1000 – 471.4) × 5759.6 = 3044517 N mm/sec
= 3.044517 kW.
IMPORTANT FORMULAE
Limiting F
1. µ = = tan α
N
2. T2 = T1 eµθ
THEORY QUESTIONS
1. A pull of 180 N applied upward at 30° to a rough horizontal plane was required to just move
a body resting on the plane while a push of 220 N applied along the same line of action was
required to just move the same body downwards. Determine the weight of the body and the
coefficient of friction. [Ans. W = 990 N; µ = 0.1732]
2. The block A shown in Fig. 5.23 weighs 2000 N. The cord attached to if passes over a
frictionless pulley and supports a weight equal to 800 N. The value of coefficient friction
between A and the horizontal plane is 0.35. Determine the horizontal force P : (i) If the motion
is impending towards the left. (ii) if the motion is impending towards the right.
[Ans. (i) 1252.82 N (ii) 132.82 N]
800 N
30°
P 2000 N
A
Fig. 5.23
188 MECHANICS OF SOLIDS
Fig. 5.24
Fig. 5.27
1 5°
10. The level of precast beam weighing 20,000 N is to W = 2 0 ,00 0 N
P
be adjusted by driving a wedge as shown in Fig.
5.31. If coefficient of friction between the wedge P ier
and pier is 0.35 and that between beam and the wedge
is 0.25, determine the minimum force P required on Fig. 5.31
the wedge to make adjustment of the beam. Angle of the wedge is 15°. (Hint: Vertical component
of reaction on wedge at contact with beam = 1/2 vertical load on beam = 10,000 kN)
[Ans. 9057.4 N]
11. A ladder 5 m long rests on a horizontal ground and leans against a smooth vertical wall at an
angle of 70° with the horizontal. The weight of the ladder is 300 N. The ladder is on the verge
of sliding when a man weighing 750 N stands on a rung 1.5 m high. Calculate the coefficient
of friction between the ladder and the floor.
[Ans. µ = 0.1837] B
8 00 N
determine the coefficient of friction between the ladder and the
floor. [Ans. 0.3548] 2 00 N
13. A uniform ladder of length ‘L’ rests against a vertical wall making 6 0°
an angle of 60° with the horizontal. Coefficient of friction between
A
wall and ladder and ground and ladder are 0.3 and 0.25 Fig. 5.32
respectively. A man weighing 650 N ascends the ladder. How
high he will be able to go before the ladder slips? Find the magnitude of weight to be put at the
bottom of the ladder so as to make it just sufficient to permit the man to go to the top. Take
ladder’s weight = 900 N.
P
(Hint. Find P as found in Example 11. Then W = )
µ
[Ans. (i) To a length of 0.4345 L i.e., to a height of 0.435 L sin 60° = 0.367 L metres.
(ii) W = 777.68 N]
190 MECHANICS OF SOLIDS
A W
3 0°
Fig. 5.34
16. Block A shown in Fig. 5.35 weighs 2000 N. The
cord attached to A passes over a fixed drum and
supports a weight equal to 800 N. The value of 800 N
coefficient of friction between A and the horizontal
30°
plane is 0.25 and between the rope and the fixed P A
drum is 0.1. Solve for P: (1) if motion is impending
towards the left, (2) if the motion is impending
towards the right.
Fig. 5.35
[Ans. (1) 1230.94 N; (2) 143.0 N]
r = 250
°
P
Fig. 5.36. Determine the torque M exerted on the 5 00 m m
drum if the load P = 50 N. Assume coefficient of
mm
50
kinetic friction between rope and drum to be 0.15.
[Ans. 747.685 N-m] 30 m
m
Fig. 5.36
18. A belt drive is required to transmit 12 kW power. The velocity of the belt is 9.6 m/sec. If
coefficient of friction is 0.25 and the angle of contact is 150º, determine the maximum tension
is the belt. [Ans. T1 = 2.602 kN]
19. Two parallel shafts are having pulleys of diameters 300 mm and 500 mm. The distance between
the shafts is 2.5 m. They are connected by crossed belt drive system. The speed of the larger
pulley is 500 rpm and coefficient of friction µ = 0.3, determine the maximum power that can
be transmitted, if maximum permissible tension in the belt is 2.4 kN. [Ans. P = 20.3 kN]
6
Simple Machines
A simple machine is a device with the help of which heavy loads are lifted by applying small effects
in a convenient direction. Pulley used to lift water from a well and screw jacks used to lift motor
car are some of the common examples of simple machines. In this chapter some of the terms
connected with simple machines are explained first followed by the description of the characteristic
features of levers, systems of pulleys, wheel and axle, Weston differential pulley block, inclined
plane, simple screw jack, differential screw jack and winch crab.
6.1 DEFINITIONS
The terms commonly used while dealing with simple machines are defined below:
Load: This is the resistance to be overcome by the machine.
Effort: This is the force required to overcome the resistance to get the work done by the
machine.
Mechanical Advantage: This is the ratio of load lifted to effort applied. Thus, if W is the load
and P is the corresponding effort, then
W
Mechanical Advantage = ...(6.1)
P
Velocity Ratio: This is the ratio of the distance moved by the effort to the distance moved
by the load in the same interval of time. Thus,
D
Velocity Ratio = ...(6.2)
d
where, D – distance moved by effort
d – distance moved by the load.
Input: The work done by the effort is known as input to the machine. Since work done by
a force is defined as the product of the force and the distance moved in the direction of the force,
Input = P × D ...(6.3)
If force P is in newton and distance D is in metre, the unit of input will be N-m. One
N-m work is also known as one Joule (J).
Output: It is defined as useful work got out of the machine, i.e., the work done by the load.
Thus,
Output = W × d ...(6.4)
191
192 MECHANICS OF SOLIDS
Efficiency: This is defined as the ratio of output to the input. Thus, if we use notation η for
efficiency,
Output W × d W d
η= = = ×
Input P×D P D
1
= Mechanical Advantage (MA) ×
Velocity Ratio (VR)
MA
= ...(6.5)
VR
Mechanical Advantage
i.e., Efficiency =
Velocity Ratio
Ideal Machine: A machine whose efficiency is 1 (i.e., 100%) is called an ideal machine. In
other words, in an ideal machine, the output is equal to the input. From eqn. (6.5), in an ideal
machine,
Velocity Ratio = Mechanical Advantage
Ideal Effort: Ideal effort is the effort required to lift the given load by the machine assuming
the machine to be ideal.
For ideal machine,
VR = MA
If Pi is the ideal effort, then
W
VR =
Pi
W
∴ Pi = ...(6.6)
VR
Ideal Load: Ideal load is the load that can be lifted using the given effort by the machine,
assuming it to be ideal.
For the ideal machine,
VR = MA
If Wi is the ideal load, then
Wi
VR =
P
∴ Wi = VR × P ...(6.7)
Then,
P – Pi is called effort lost in friction and W – Wi is called frictional resistance.
MA W 1
Now, η = = ×
VR P VR
W
From eqn. (6.6), Pi =
VR
Pi
∴ η =
P
Similarly from eqn. (6.7), Wi = VR × P
W
∴ η=
Wi
Pi W
Thus, η = = ...(6.8)
P Wi
Example 6.1. In a lifting machine, an effort of 500 N is to be moved by a distance of
20 m to raise a load of 10,000 N by a distance of 0.8 m. Determine the velocity ratio, mechanical
advantage and efficiency of the machine. Determine also ideal effort, effort lost in friction, ideal load
and frictional resistance.
Solution: Load, W = 10,000 N
Effort P = 500 N
Distance moved by the effort D = 20 m
Distance moved by the load d = 0.8 m
W 10,000
Mechanical advantage, MA = =
P 500
= 20
D 20
Velocity ratio, VR = =
d 0.8
= 25
MA 20
Efficiency, η= =
VR 25
= 0.8 = 80 per cent
W 10,000
Ideal effort, Pi = = = 400 N
VR 25
Effort lost in friction = P – Pi
= 500 – 400
= 100 N
Ideal load, Wi = P × VR = 500 × 25
= 12,500 N
194 MECHANICS OF SOLIDS
Frictional resistance = Wi – W
= 12,500 – 10,000
= 2500 N
ac h in e
al m
a c tu
For B
C
E ffort (P )
θ F
A m
1
θ
E Fo r ide al m ach ine
O D L oa d (W )
Fig. 6.1
1
m
M.A.
O Load (W)
Fig. 6.2
MA
From eqn. (6.5), the efficiency of the machine is given by η = . Using the eqn. (6.10),
VR
1 1
η = × ...(6.11)
VR m + C
W
Since the velocity ratio (VR) is constant for a machine, variation of efficiency with load is
similar to the variation of mechanical advantage with the load.
The maximum efficiency is approached as the load approaches infinity (W → ∞) and its value
1 1
is equal to × . The variation of the efficiency with load is shown in Fig. 6.3.
VR m
196 MECHANICS OF SOLIDS
1 1
×
VR m
Load (W)
Fig. 6.3
Example 6.2. In a simple machine, whose velocity ratio is 30, a load of 2400 N is lifted by an effort
of 150 N and a load of 3000 N is lifted by an effort of 180 N. Find the law of machine and calculate
the load that could be lifted by a force of 200 N. Calculate also:
(1) The amount of effort wasted in overcoming the friction,
(2) Mechanical advantage, and
(3) The efficiency.
Solution: Let the law of machine be
P = mW + C
In the first case, P = 150 N, and W = 2400 N
In the second case, P = 180 N and W = 3000 N
∴ 150 = 2400 m + C ...(1)
180 = 3000 m + C ...(2)
Subtracting eqn. (1) from eqn. (2), we get
30 = 600 m
m = 0.05
Substituting this value in eqn. (1), we get
150 = 120 + C
∴ C = 30
Hence, the law of machine is
P = 0.05 W + 30 ...(3)
When a force of 200 N is applied:
From the law of machine (3),
200 = 0.05 W + 30
∴ W = 3400 N
Ideal effort is given by:
W 3400
Pi = = = 113.33 N
VR 30
SIMPLE MACHINES 197
∴ m = 0.01818
∴ Maximum mechanical advantage
1 1
= =
m 0.01818
= 55
1 1 1 1
Maximum efficiency × = = ×
m VR 0.01818 85.55
= 0.6429
= 64.29%
Example 6.4. The efforts required for lifting various loads by a lifting machine are tabulated below:
Load lifted in N 100 200 300 400 500 600
Effort required in N 16.0 22.5 28.0 34.0 40.5 46.5
Determine the law of machine. If the velocity ratio is 25, calculate efficiency at each load and plot
efficiency versus load curve. From this curve, determine the maximum efficiency.
Solution: Figure 6.4 shows the graph of effort versus load. From this figure, C = 10 N and slope
30
m= = 0.06
500
∴ The law of machine is
P = 0.06 W + 10
50
40
30
E ffort (P )
P = 30
20
10 W = 500
C
0
0 1 00 2 00 3 00 4 00 5 00 6 00
L oa d (W )
Fig. 6.4
MA W 1 W 1 W
η= = × = × =
VR P VR P 25 25 P
SIMPLE MACHINES 199
60
50
40
% Efficiency
30
20
10
0
0 100 200 300 400 500 600 700 800 900
Load (W)
Fig. 6.5
From the graph (Fig. 6.5) maximum efficiency is seen as 57%. Actually if it is plotted for
infinitely large load, maximum efficiency will be equal to
1 1 1 1
× = × = 0.6667 = 66.67%
m VR 0.06 25
P
A
W
C
B
b
Fig. 6.6
Pa = Wb
b
or P =W
a
W a
∴ Mechanical advantage = = ...(6.12)
P b
6.8 PULLEYS
A systematic arrangement of one or more pulleys may provide a simple and convenient lifting
machine. In its simplest form, it consists only one pulley over which a rope or chain passes as shown
in the Fig. 6.7. In this case, velocity ratio is equal to one since distance moved by effort is equal
to the distance moved by the load. It just performs changing the direction of the applied force.
Depending on the arrangement, pulleys are classified as:
(1) First order pulley system
(2) Second order pulley system
(3) Third order pulley system.
At times, it may be difficult or may be detour to find velocity ratio directly. In such cases ideal
conditions may be assumed (neglecting friction) and mechanical advantage may be found first. Then
applying VR = MA for ideal machine, the velocity ratio is found. This method of determining
velocity ratio is used for various pulley systems considered here.
P
Effort
W
Load
Fig. 6.7
202 MECHANICS OF SOLIDS
A first order pulley system is shown in the Fig. 6.8. Pulley No. 1 is fixed in position to a support
at top. A rope passes over this pulley and one end of this rope is tied to the support at the top, making
a loop, in which pulley No. 2 is suspended and effort is applied at the other end. One end of another
rope is tied to pulley No. 2 and the other end to the top support which makes a loop in which pulley
No. 3 is suspended. Similarly, a number of pulleys can be arranged as shown, when an effort is
applied to lift the load except first pulley all other pulleys move vertically. Therefore, first pulley is
termed as fixed pulley and the others as movable pulleys.
Let an effort P be applied to lift a load W. In an ideal pulley system (friction = 0), the rope
which passes over pulley No. 4 is subjected to a tension W . Then tension in rope which passes over
2
W W
pulley No. 3 is and tension in the rope which passes over pulley No. 2 is . Hence, an effort
4 8
W
equal to is required to lift a load W.
8
W W W
∴ P = and hence MA = = = 8 = 23
8 P W
8
But in an ideal machine, VR = MA ∴ VR = 23
1
W W
8 8
W W
2 8 8
E ffort
W
W
4
4
W
3 4
W W
2 2
W
4
2
W L oa d
Fig. 6.8
It is to be noted that in the system considered, there are three movable pulleys and the velocity
ratio is 23. If there are only two movable pulleys, then velocity ratio would be 4 (i.e., 22). In general,
in the first order pulley system, velocity ratio (VR) is given by 2n, where, n is the number of movable
pulleys present in the system. Thus, in first order pulley system
VR = 2n ...(6.13)
SIMPLE MACHINES 203
P
E ffort
1 1
W
L oa d
Fig. 6.9
Figure 6.9 shows a second order pulley system. This system consists of a top pulley block and a
bottom pulley block. In a pulley block pulleys may be arranged side by side or may be one below
the other as shown in Fig. 6.9. The top pulley block is fixed in position to the top support whereas
bottom pulley block can move vertically along with the load which is attached to it. One end of the
rope is attached to the hook provided at the bottom of the top pulley block and the effort is applied
at the other end. A single rope goes round all the pulleys. Let an effort P be applied to lift a load W.
Neglecting frictional losses, the tension in the rope all along the length is P. Take the section
along (1)–(1) and consider the equilibrium of the bottom pulley block.
The load W is lifted using six ropes having equal tension P
W W W
∴ W = 6P, ∴ P = , ∴ MA = = = 6.
6 P W
6
But VR = MA in ideal condition
∴ VR = 6.
In general, in the second order pulley system velocity ratio is equal to twice the number of
movable pulleys in the system.
204 MECHANICS OF SOLIDS
The arrangement of the pulleys in the third order system is shown in the Fig. 6.10. In this system
a pulley (No. 1) is fixed to the top support, over which a rope passes. One end of the rope is attached
to a rigid base at the bottom. The other end is attached to a second pulley. Over this pulley another
rope passes, whose one end is attached to the same rigid base and the other end to a third pulley as
shown. Likewise a series of pulleys can be arranged. The load to be lifted will be attached to the
rigid base.
1
4P
4P
4P
2
2P 2P
2P
P
3
P
P P
4P 2P Effort
Rigid base
Fig. 6.10
Referring to the Fig. 6.10, let the effort required be P to lift a load W. Then neglecting friction,
Tension in the rope which passes over pulley No. 3 = P
Tension in the rope which passes over pulley No. 2 = 2P
Tension in the rope which passes over pulley No. 1 = 4P
∴ A total force of 7P is acting on the base.
∴ Lifting force produced on the base = 7P
Considering the equilibrium of rigid base,
W
7P = W ∴ MA = =7
P
But in an ideal machine, VR = MA, and hence VR = 7.
It can be easily seen that, if there are only two pulleys, VR = 3 and if there is only one pulley,
VR = 1. Therefore, in general, for the third order pulley system:
VR = 2n – 1 ...(6.15)
where, n = number of pulleys.
SIMPLE MACHINES 205
A
B
d
W
P P
Effort Load
Effort
(a)
Load W
(b)
Fig. 6.11
One end of a rope is tied to the pin provided on the wheel and the rope is wound around the
wheel. The other end of the rope provides the means for the application of the effort. One end of
another rope is tied to the pin provided on the axle and wound around the axle in the opposite
direction to that of rope wound to the wheel. To the other end of this rope the load is attached. If
the whole assembly is rotated, one rope gets wound up and the other gets unwound.
Suppose the assembly is moved by one complete revolution, then the distance moved by the
effort = π D and distance moved by the load = π d
distance moved by effort
∴ VR =
distance moved by load
πD
= ,
πd
D
or VR = ...(6.16)
d
d1
P
E ffort
W
L oa d
Fig. 6.12
206 MECHANICS OF SOLIDS
This consists of a differential axle of diameter d1 and d2 (d1 < d2) and a wheel of diameter
D, fixed uniaxially as shown in Fig. 6.12. One end of the rope is tied to the pin provided on the axle
portion having diameter d1 and a part of the rope is wound around it. The other end of the rope is
wound around axle on the portion having diameter d2 in the opposite direction. This pattern of
winding forms a loop and a simple pulley is installed in this loop as shown in the figure. The load
is attached to this pulley. The second rope is wound to the wheel in such a direction that if it is
unwound, the rope around the bigger diameter axle gets wound up and the rope around smaller
diameter axle gets unwound. One end of this second rope provides means for application of the
effort.
Suppose the whole system makes one complete revolution due to the applied effort, then
Total distance moved by the effort at the differential axle = πD
Length of winding of the rope = πd2
Length of unwinding of rope = πd1
∴ Net wound length = πd2 – πd1
= π(d2 – d1)
But, the rope is continuous and the load is to be lifted by the pulley block in the loop.
π( d 2 − d1 )
∴ Total height over which pulley is lifted =
2
π( d 2 − d1 )
and hence the distance moved by the load =
2
πD
∴ VR =
π( d2 − d1 )
2
2D
VR =
d2 − d1
Hence, velocity ratio in wheel and differential axle is given by:
2D
VR = ...(6.17)
d2 − d1
W
Load
Fig. 6.13
SIMPLE MACHINES 207
This system consists of two pulley blocks, one at the top attached to the support and the other
at the bottom hanging in the chain loop. The top block consists of two wheels of different diameters,
but fixed co-axially. The bottom block is a simple pulley to which the load W is attached. An endless
chain is wound around the pulley system as shown in the figure. All the wheels are made with teeth
so as to accommodate the links of the chain. The chain is essentially used to avoid slipping.
To determine velocity ratio of the system, let us consider pulley block as an ideal machine
and determine its mechanical advantage first. In ideal machine VR = MA. Let the diameter of the
larger wheel of the top block be D and the diameter of the smaller wheel of the top block be d [Fig.
6.14(a)]. Let the effort required to lift the load W be P.
D
d
D
d
W
2 P
W
2
W
W
2 W
2 W
2 2
(b)
W
(a)
Fig. 6.14
W
Then the tension in the chain loop in which pulley is hanging is .
2
Now, taking moment about the axis of top block [Fig. 6.14(b)],
W D W d D
× = × + P×
2 2 2 2 2
W PD
∴ (D – d) =
4 2
W 2D
=
P (D − d)
2D
i.e., MA =
(D − d)
In an ideal machine,
2D
VR = MA = ...(6.18)
(D − d)
208 MECHANICS OF SOLIDS
P
E ffort W
Fig. 6.15
Figure 6.15 shows a typical inclined plane. Here the load is a roller which is to be lifted to
a higher elevation. One end of a rope is tied to the roller and the rope is passed over a pulley attached
at the top of the inclined plane. At the other end of the rope the effort is applied.
Let the angle of inclination of the plane be θ and the length of the inclined plane be L. Then,
if the roller is made to roll from bottom to top, applying an effort P, the load is lifted through a height
of L sin θ. In this process the effort P moves through a distance L vertically downwards.
distance moved by the effort
∴ VR =
distance through which load is lifted
L 1
= =
L sin θ sin θ
1
∴ VR = ...(6.19)
sin θ
Example 6.6. In a first order system of pulleys there are three movable pulleys. What is the effort
required to raise a load of 6000 N ? Assume efficiency of the system to be 80%.
If the same load is to be raised using 520 N, find the number of movable pulleys that are
necessary.
Assume a reduction of efficiency of 5% for each additional pulley used in the system.
Solution: VR = 2n, where n is the number of movable pulleys.
VR = 23 = 8
Now, MA = η × VR
= 0.8 × 8
= 6.4
W
i.e., = 6.4
P
W 6000
∴ P= =
6.4 6.4
SIMPLE MACHINES 209
i.e., P = 937.5 N
In the second case,
Effort = 520 N
Efficiency η = 0.80 – n1 × 0.05
where n1 = number of additional pulleys required and equal to (n – 3).
MA = η × VR
W
i.e., = η × VR
P
∴ W = P × η × 2n
= P(0.8 – n1 × 0.05) × 2n
= P[0.8 – (n – 3) × 0.05] 2n
By going for a trial and error solution, starting with one additional pulley i.e., totally with four
pulleys,
W = 520 [0.8 – (4 – 3) × 0.05] 24 = 6240 N
i.e., if four pulleys are used, a load of 6240 N can be raised with the help of 520 N effort.
∴ Number of movable pulleys required = 4
Example 6.7. What force is required to raise the load W shown in Fig. 6.16 ? Assume efficiency of
the system to be 85%.
Solution: The pulley system shown in the Fig. 6.16 is a variation of the second order pulley system.
W = 1 2 00 0 N
Fig. 6.16
Example 6.8. Find the pull required to lift the load W shown in Fig. 6.17(a) assuming the efficiency
of the system to be 78%.
P P
W = 12,000 N
Fig. 6.17
Solution: The pulley system shown in Fig. 6.17(a) is a combination of a first order system and a
second order system as shown in Figs. 6.17(b) and (c).
Let load W be lifted by a distance x. Consider the first order system portion [Fig. 6.17(b)].
Here there are two movable pulleys. Hence
VR = 22 = 4
In this portion P moves by 4x.
Now, consider the second order pulley system portion [Fig. 6.17(c)]. Here there are two
movable pulleys. Hence VR = 2 × 2 = 4.
∴ Distance moved by the effort in this system = 4x.
Hence, the total distance moved by the effort in the given system = 4x + 4x = 8x
8x
∴ VR = =8
x
Now, MA = η × VR = 0.78 × 8
= 6.24
W
i.e., = 6.24
P
12,000
∴ P= = 1923.08 N
6.24
Example 6.9. A lifting machine consists of pulleys arranged in the third order system. There are
three pulleys in the system. A load of 1000 N is lifted by an effort of 180 N. Find the efficiency of
the machine and the effort lost in friction.
Solution: For the third order system of pulleys,
VR = 2n – 1
where, n is the number of pulleys in the system.
VR = 23 – 1 = 7
SIMPLE MACHINES 211
Now, MA = η × VR
W
i.e., = η × VR
P
W 1
∴ η= ×
P VR
1000 1
= × = 0.7937
180 7
i.e., η = 79.37%
Now,
W
Ideal effort, Pi =
VR
1000
=
7
= 142.86 N
∴ Effort lost in friction = P – Pi
= 180 – 142.86
= 37.14 N
Example 6.10. What force P is required to raise a load of 2500 N in the system of pulleys shown
in Fig. 6.18(a). Assume efficiency of the system to be equal to 70%.
Solution: Figure 6.18(a) can be split into two simple systems as shown in Figs. 6.18(b) and 6.18(c).
What is shown in Fig. 6.18(b) is a third order pulley system having two pulleys.
∴ VR = 2n – 1
= 22 – 1 = 3
P
W
W (c)
P
(b)
(a)
Fig. 6.18
∴ VR = 22 – 1 = 3
∴ VR of the whole system = 3 + 3 = 6
Now, MA = η × VR
W
i.e., = η × VR
P
W
∴ P=
η × VR
2500
=
0.7 × 6
i.e., P = 595.24 N
Example 6.11. In a wheel and axle, diameter of the wheel is 500 mm and that of the axle is 200 mm.
The thickness of the cord on the wheel is 6 mm and that of the axle is 20 mm. Find the velocity ratio
of the machine. If the efficiency when lifting a load of 1200 N with a velocity of 10 metres per minute
is 70%, find the effort necessary.
6 6
Solution: Effective wheel diameter = + 500 +
2 2
= 506 mm
20 20
Effective axle diameter = + 200 +
2 2
= 220 mm.
D
For a wheel and axle, the velocity ratio is given by
d
506
∴ VR = = 2.30
220
Mechanical advantage = Efficiency × velocity ratio
= 0.7 × 2.30 = 1.61
W
MA =
P
1200
∴ P=
1.61
i.e., P = 745.34 N
Example 6.12. A load of 20 kN is to be lifted by a differential wheel and axle. It consists of
differential axle of 250 mm and 300 mm diameter and the wheel diameter is 800 mm. Find the effort
required if the efficiency of the machine is 55%.
SIMPLE MACHINES 213
2 × 500
= = 3.33
500 − 200
S crew W
h ea d
L ever
Nut
d
S ta n d
Fig. 6.19
The device consists of a nut and a screw. Monolithically cast nut and stand form the body of
the jack. The load is carried by the screw head fitted onto the screw as shown in the figure. The body
(consisting of nut) is fixed and the screw is rotated by means of a lever.
The axial distance moved by the nut (or by the screw, relative to each other) when it makes
one complete revolution is known as lead of the screw head. The distance between consecutive
threads is called pitch (of a screw thread). If the screw is single threaded, then lead of the screw is
equal to the pitch. If the screw is double threaded then lead of the screw is twice the pitch.
Let R be the length of the lever and d be the mean diameter of the screw.
Let a load W be lifted using an effort P.
If an effort P is applied at the lever end, it is equivalent to an effort P1 at the screw [Fig. 6.20(a)]
and P1 is given by the condition:
R1
P1
Fig. 6.20(a)
d
P × R = P1 ×
2
2PR
∴ P1 =
d
Now, consider one complete revolution of the lever. The load W is lifted up by a distance p
equal to the lead of the screw.
SIMPLE MACHINES 215
p
θ
πd
Fig. 6.20(b)
p
This can be compared with that of inclined plane having inclination = tan–1
πd
where, p – lead of the screw
d – mean diameter of screw.
Applying an effort P at the end of the lever is as good as applying an effort P1 (at the screw)
on this inclined plane. [Fig. 6.20(c)].
W
P1
F θ φ R1
θ
Fig. 6.20(c)
p
where, tan θ =
πd
If the load is descending, then the friction will be acting in the reverse direction so that the
resultant reaction R shifts as shown in Fig. 6.20(d).
W
P1 F
φ
θ
θ
R1
Fig. 6.20(d)
We have
d µ + tan θ
P= W
2R 1 − µ tan θ
p
and tan θ =
πD
6.667
= = 0.05305
π × 40
∴
40 LM
0.12 + 0.05305 OP
P=
2 × 400
× 40,000
N
1 − (0.12 × 0.05305) Q
i.e., P = 348.32 kN
Example 6.15. A screw jack has square threads 50 mm mean diameter and 10 mm pitch. The load
on the jack revolves with the screw. The coefficient of friction at the screw thread is 0.05.
(1) Find the tangential force required at the end of 300 mm lever to lift a load of 6000 N.
(2) State whether the jack is self-locking. If not, find the torque which must be applied to keep the
load from descending.
p 10
Solution: tan θ = = = 0.0637
πd π × 50
∴ θ = 3.6426°
tan φ = 0.05
∴ φ = 2.8624°
d
(1) P= × W tan (θ + φ)
2R
50
= × 6000 tan (3.6426° + 2.8624°)
2 × 300
P = 57.01 N
(2) We have
2 πR 2 π × 300
VR = =
p 10
= 188.496
W 6000
MA = =
P 57.01
= 105.245
MA 105.245
Efficiency = =
VR 188.496
= 0.5583
218 MECHANICS OF SOLIDS
Fig. 6.21
Let R be the radial distance (from the centre line of A and B) at which an effort P is applied
Distance moved by the effort 2πR
Now, VR = = ...(6.23)
Distance moved by the load pA − p B
It can be seen from eqn. (6.23) that the velocity ratio in the differential screw jack is increased
as compared to that of simple screw jack (eqn. 6.22).
Example 6.16. The following are the specifications for a differential screw jack:
(1) Pitch of smaller screw, 5.0 mm
(2) Pitch of larger screw, 10.0 mm
(3) Lever arm length from centre of screw = 500 mm.
The screw jack raises a load of 15 kN with an effort of 185 N. Determine the efficiency of
the differential screw jack at this load.
If the above jack can raise a load of 40 kN with an effort of 585 N, determine the law of
machine.
Solution: Now, pA = 10.0 mm
pB = 5.0 mm
Lever arm length, R = 500 mm
2 πR 2 π × 500
VR = = = 628.32
pA − pB 10 − 5.0
W 15,000
MA = = = 81.08
P 185
MA 81.08
∴ η = =
VR 628.32
= 0.129 = 12.9 per cent.
To find law of machine:
Let law of machine be P = mW + C
From first case: 185 = m × 15,000 + C ...(1)
From second case: 585 = m × 50,000 + C ...(2)
(2) minus (1) gives, 400 = 35,000 m.
4
or m=
350
Substituting this value in eqn. (1), we get
185 = 171.43 + C
∴ C = 13.57 N
4
∴ Law of machine is P = W + 13.57
350
Line diagram of a single purchase winch crab is shown in Fig. 6.22. It consists of a load drum of
radius r connected to an axle by gears. The toothed wheel on load drum is called spur wheel and
the toothed wheel on axle is called pinion. Pinion is always smaller in size and it contains less
number of teeth as compared to that on the spur wheel. The axle is provided with a handle of arm
length R. Let the number of teeth on pinion and spur wheel be T1 and T2, respectively. Let the effort
be applied at the end of the handle. When one revolution is made, the distance moved by the effort
is given by:
D = 2πR
When axle makes one revolution, due to gear arrangement load drum moves by T1 number
T
of teeth, which means that it makes a revolution of 1 .
T2
∴ The distance over which the load moves:
T1
d = 2πr ×
T2
∴ Velocity ratio,
D 2πR
VR = =
d 2 πr × T1
T2
R T
i.e., VR = × 2 ...(6.24)
r T1
H a nd le
P inion
A xle T1
L oa d d ru m
T2
S p ur w he el
Fig. 6.22
Velocity ratio of a winch crab can be increased by providing another axle with a pair of pinion and
gear as shown in Fig. 6.23. Since two pairs of pinion and gear are used it is called a double purchase
winch crab. This is used for lifting heavier loads.
SIMPLE MACHINES 221
H a nd le
A xle -A T1
A xle -B T3 T2
L oa d d ru m
T4
Fig. 6.23
Let the number of teeth on various wheels be T1, T2, T3 and T4 as shown in Fig. 6.23. Let
the handle makes one revolution.
Distance moved by effort P is given by:
D = 2πR ...(6.25)
T1
When axle A makes one revolution, axle B is moved by T1 teeth, i.e., it makes revolutions.
T2
T1
The number of teeth by which spur wheel is moved is × T3 and hence load drum makes
T2
FG T IJ × FG T IJ
HT K HT K
1 3
revolutions.
2 4
FG T IJ × F T I
H T K GH T JK
1 3
d = 2πr ×
2 4
D 2 πR
VR =
d
=
FG IJ FG IJ
T T
H K H K
2πr × 1 × 3
T2 T4
R FT I FT I
×G J ×G J
r HT K HT K
2 4
i.e., VR = ...(6.26)
1 3
MA 56.25
∴ η= =
VR 120
i.e., η = 0.4688 = 46.88%
Example 6.18. In a double purchase crab, the pinions have 15 and 20 teeth, while the spur wheels
have 45 and 40 teeth. The effort handle is 400 mm while the effective diameter of the drum is 150
mm. If the efficiency of the winch is 40%, what load will be lifted by an effort of 250 N applied at
the end of the handle?
Solution: T1 = 15; T2 = 45; T3 = 20; T4 = 40
Length of handle, R = 400 mm
150
Radius of the load drum, r = = 75 mm
2
R T T
∴ VR = × 2 × 4
r T1 T3
400 45 40
= × ×
75 15 20
= 32
MA
Now, η=
VR
MA
∴ 0.40 =
32
or MA = 12.8
W
i.e., = 12.8, but P = 250 N
P
∴ W = 12.8 × 250
i.e., W = 3200 N
Applied effort lifts a load of 3200 N
IMPORTANT FORMULAE
Load
1. Mechanical advantage = .
Effort
Distance moved by effort
2. Velocity ratio = .
Distance moved by load
3. Input = Effort × Distance moved by effort.
4. Output = Load × Distance moved by the load.
Output Mechanical advantage
5. Efficiency = = .
Input Velocity ratio
6. Law of machine:
P = mW + C.
7. A machine is reversible, if η > 50%.
224 MECHANICS OF SOLIDS
THEORY QUESTIONS
4. With neat sketches explain the different types of pulley systems and derive the expressions for
their efficiency.
5. With neat sketches explain the following simple machines:
(i) Wheel and axle
(ii) Wheel and differential axle
(iii) Weston differential pulley block.
d µ + tan θ
6. Show that in a screw jack effort P required to lift a load W is given by P = W
2 R 1 − µ tan θ
p
where φ is angle of friction and tan θ = , p being the pitch of screw.
πd
7. Give a neat sketch of
(i) Single purchase winch crab
(ii) Double purchase winch crab
1. In a lifting machine an effort of 1.5 kN is to be moved through a distance of 7.2 m to move a load
of 24 kN through a distance of 300 mm. Determine: (1) mechanical advantage, (2) velocity ratio,
(3) efficiency, (4) ideal effort, (5) effort lost in friction, (6) ideal load and (7) frictional resistance.
[Ans. (1) 16; (2) 24; (3) 66.67%; (4) 1.0 kN; (5) 0.5 kN; (6) 36 kN and (7) 12 kN]
2. In a lifting machine an effort of 400 N is required to raise a load of 3000 N and an effort of 640 N,
to raise a load of 5000 N. How much load can be lifted with an effort of 760 N? If the velocity
ratio is 16, determine the efficiency of the machine when an effort of 760 N is applied. Is it a
reversible machine? [Ans. W = 6000 N; η = 49.34; It is not reversible]
3. The following observations were made in an experiment on a lifting machine:
Load in N 500 1000 1500 2000 2500 3000
Effort in N 26 47 76 95 105 127
Draw the load versus effort graph and determine the law of machine. If the velocity ratio is 30
determine the efficiency while lifting a load of 1800 N.
What is the maximum efficiency of the machine ?
[Ans. P = 0.04W + 6; η = 51.28%; maximum η = 83.33%]
4. A lifting machine having velocity ratio 28 starts raising a load of 6420 N on applying an effort of
450 N to it. If suddenly the effort is removed find whether the load comes down or not?
[Ans. η = 50.95%, hence the load comes down]
5. In the first order pulley system having three movable pulleys, how much effort is required to
raise a load of 5780 N if the same system raises a load of 1200 N with an effort of 200 N? Assume
the efficiency to be constant for the pulley system. [Ans. P = 963.33 N]
6. For the arrangement of pulleys shown in Fig. 6.24, find the effort required to raise the given load
7280 N. Assume efficiency of the system as 75%. [Ans. P = 2436.7 N]
226 MECHANICS OF SOLIDS
1280 N
Fig. 6.24
7. For the combination of first order and second order pulley system shown in Fig. 6.25 what will be
velocity ratio?
8200 N
Fig. 6.25
Assuming efficiency to be 80%, calculate what effort is required to raise the load of 8200 N.
[Ans. VR = 8; P = 1281.25 N]
8. For a third order pulley system having six movable pulleys, an effort of 720 N is required to raise
a load of 30 kN. Calculate the efficiency of the system. [Ans. 66.14%]
9. For a wheel and axle, the following details are available:
Diameter of the wheel = 540 mm
Diameter of the axle = 270 mm
Thickness of the cord on the wheel = 6 mm
Thickness of the cord on the axle = 20 mm
Calculate the efficiency of the device if an effort of 725 N is required to lift a load of 1000 N.
[Ans. η = 73.26%]
10. A differential axle and wheel consists of a differential axle of 240 mm and 320 mm diameter; and
a wheel of diameter 750 mm. Assuming efficiency to be 57%, find the effort required to raise a
load of 24 kN. [Ans. P = 2.2456 kN]
11. A Weston differential pulley block of diameters 400 mm and 800 mm is used to lift a load of
40,000 N. Find the effort required if efficiency of the system is 60%. [Ans. 26,667 N]
SIMPLE MACHINES 227
10. Thermal conductivity: It is the ability of the material to transfer heat. Metals have higher
thermal conductivity. Moist materials have higher heat conductivity compared to dry porous
materials. Rubber is having very low thermal conductivity.
11. Thermal resistivity: It is the reciprocal of thermal conductivity and is defined as time taken
for the flow of unit heat.
12. Fire resistance: It is the property by virtue of which a material resists the action of high
temperature without undergoing substantial changes in shape and loss of strength. Steel has
poor fire resistance. Concrete is better than steel in resisting fire. Bricks are having very
good fire resistance.
13. Sound absorption: It is the property of reducing the reflection of sound waves. Porous
materials have better sound absorption property. In auditoriums and cinema halls we have
to look for materials with good sound absorption properties.
14. Chemical resistance: It is the ability of the material to withstand the action of acids,
alkalies, gases etc. Materials are subjected to the action of these chemicals, which are present
in air or water or land. Stone and wood have poor resistance to chemical actions while bricks
have good chemical resistance.
15. Corrosion: It is the destruction of the material due to slow oxidation. Steel is susceptible
to corrosion. The corrosion rate is high in marine environment.
16. Soundness: A material is said to be sound, if it has good resistance to heat, alternate
freezing and thawing and the other destructive actions of the atmosphere.
17. Durability: It is the ability to resist the combined effects of atmosphere, rain and other
effects and maintain the original strength characters for a long period.
P P
(a)
P pA
(b)
pA pA
(c)
Fig. 7.1
230 MECHANICS OF SOLIDS
shows a bar under tension and stresses at a cross-section and also an element of bar under
tensile stress. In this P is the tensile force applied, p is the tensile stress (resistance per unit
area) and A is the cross-sectional area.
∴ P = pA ...(7.1)
(b) Compressive stress: Instead of pull, if push ‘P’ acts on the element, it tries to shorten the
bar and the internal resistance developed per unit area is called compressive stress
P
(Ref. Fig. 7.2). Here also p = .
A
P P
P pA
pA
Fig. 7.2
2. Properties related to shear loading: If the applied force is trying to shear off a particular
section of the element, the resistance developed for unit area in such case along that section is called
shearing stress. The maximum stress that material can resist is called shearing strength of the
material. Figure 7.3 shows a section of bolt under shearing stress.
Q
Q
Q
Q
Fig. 7.3
If Q is the shear force, q is the shear stress, then the equilibrium condition of any one part
considered shows,
Q = qA, where A is the cross-sectional area of the bolt.
Q
∴ q= ...(7.2)
A
3. Properties related to torsional moment: A member is said to be in torsion when it is
subjected to a moment about its axis, Fig. 7.5 shows a shaft in torsion. The effect of a torsional
member is to twist it and hence a torsional moment is also called as a twisting moment. In engineering
PHYSICAL AND MECHANICAL PROPERTIES OF STRUCTURAL MATERIALS 231
problems many members are subjected to torsion. Shafts transmitting power from engine to rear axle
of automobile, from a motor to machine tool and from a turbine to electric motor are the common
examples of members in torsion. Ring beams of circular water tanks and beams of grid floors are
also the examples of members in torsion. The applied torsion is resisted by the material of the
member by developing shear stresses. This shear stress varies linearly from zero at centre of the shaft
to the maximum value at the extreme fibre.
da
T
r
T
qda
T = Σqrda
Fig. 7.4
4. Properties related to bending: When a member is supported at two or more points and
subjected to transverse load it bends and develops resistance to the load. The cross-sections of the
members are subjected to bending moment and shear force. Finally the load gets transferred to the
support by end shear. The shear force introduces shear stresses in the material while bending moment
introduces tension in some parts and compression in other parts as shown in Fig. 7.5.
P1 P2 P3
(a) Beam
M
F
(c) Resistance to shear
M M
Fig. 7.5
232 MECHANICS OF SOLIDS
5. Fatigue: Many structural memberes are subjected to fluctuation and reversal of stresses as
shown in Fig. 7.6 (a) and (b). A bridge deck is a common example of fluctuation of stresses and
the shaft is the mechanical component subjected to reversal of stresses. This type of stress when
applied repeatedly reduces the strength of material and this effect is called fatigue. This property is
shown in Fig. 7.6 (c). The maximum stress at which even a billion reversal of stresses cannot cause
failure of the material is called endurance limit.
Stress
O
No. of cycles
(a) Fluctuation of stress
O
Stress
No. of cycles
A
A - Static breaking stress
Breaking stress
B - Endurance limit
No. of cycles
Fig. 7.6
6. Abrasive resistance: This is the property by virtue of which a material resists the forces
acting at contact surfaces when one material rubs/moves over the other. This is an important property
to be studied for the materials used for road surfaces and flooring.
7. Impact strength: This property refers to the ability of the material to resist shock loads due
to heavy loads falling on the surface.
8. Hardness: It is the ability of the material to resist penetration from another material. A
number called hardness number is used to measure hardness of various materials. It is based on
hardnesses of ten minerals arranged in the increasing hardness. Diamond has hardness number 10.
Sometimes hardness is measured by size of indentation of steel balls under standard pressure using
hardness testing machine.
PHYSICAL AND MECHANICAL PROPERTIES OF STRUCTURAL MATERIALS 233
9. Elasticity: It is the property of the material by virtue of which it regains its original shape
and size after the removal of external load. The maximum stress level before which if the load is
removed the material regains its shape and size fully is called its elastic limit.
10. Plasticity: It is the property of the material to retain its changed shape and size after the
loads are removed. It is a required property when a material is to be moulded into different shape.
11. Creep: It is the property of the material by virtue of which it undergoes changes in size with
time under the action of constant load. Concrete possesses this property.
12. Toughness: It is the property of a material whereby it absorbs energy due to straining
actions by undergoing plastic deformation.
The other terms proof stress, factor of safety, working stress and load factors are explained in
the next chapter after explaining the stress strain curves. There are standard tests specified by codes
to quantity various properties of the materials. Knowing the requirements of materials properties for
different purposes, the engineer has to select or reject the materials to be used in construction.
THEORY QUESTIONS
When a member is subjected to loads it develops resisting forces. To find the resisting forces
developed a section plane may be passed through the member and equilibrium of any one part may
be considered. Each part is in equilibrium under the action of applied forces and internal resisting
forces. The resisting forces may be conveniently split into normal and parallel to the section plane.
The resisting force parallel to the plane is called shearing resistance. The intensity of resisting force
normal to the sectional plane is called intensity of Normal Stress (Ref. Fig. 8.1).
Resisting Force
Normal to Plane
p
Shearing
Force q
Section
Plane
Fig. 8.1
234
SIMPLE STRESSES AND STRAINS 235
P P
(a)
PA
Fig. 8.2
Similarly stress near the hole or at fillets will not be uniform as shown in Figs. 8.3 and 8.4. It
is very common that at some points in such regions maximum stress will be as high as 2 to 4 times
the average stresses.
236 MECHANICS OF SOLIDS
P P
When Newton is taken as unit of force and millimetre as unit of area, unit of stress will be
N/mm2. The other derived units used in practice are kN/mm2, N/m2, kN/m2 or MN/m2. A stress of
one N/m2 is known as Pascal and is represented by Pa.
Hence, 1 MPa = 1 MN/m2 = 1 × 106 N/(1000 mm)2 = 1 N/mm2.
Thus one Mega Pascal is equal to 1 N/mm2. In most of the standard codes published unit of stress
has been used as Mega Pascal (MPa or N/mm2).
Consider a bar subjected to force P as shown in Fig. 8.5. To maintain the equilibrium the end forces
applied must be the same, say P.
P P
A Sectional Plane
R
P
P
R
P P
P R
8.4 STRAIN
No material is perfectly rigid. Under the action of forces a rubber undergoes changes in shape and
size. This phenomenon is very well known to all since in case of rubber, even for small forces
deformations are quite large. Actually all materials including steel, cast iron, brass, concrete, etc.
undergo similar deformation when loaded. But the deformations are very small and hence we cannot
see them with naked eye. There are instruments like extensometer, electric strain gauges which can
measure extension of magnitude 1/100th, 1/1000th of a millimetre. There are machines like universal
testing machines in which bars of different materials can be subjected to accurately known forces of
magnitude as high as 1000 kN. The studies have shown that the bars extend under tensile force and
shorten under compressive forces as shown in Fig. 8.7. The change in length per unit length is known
as linear strain. Thus,
Change in Length
Linear Strain =
Original Length
238 MECHANICS OF SOLIDS
∆
e = ...(8.7)
L
b¢ b
L
(Original Length) (Extension)
b¢ b
(Shortening) b
L
(Original Length)
Fig. 8.7
When changes in longitudinal direction is taking place changes in lateral direction also take
place. The nature of these changes in lateral direction are exactly opposite to that of changes in
longitudinal direction i.e., if extension is taking place in longitudinal direction, the shortening of
lateral dimension takes place and if shortening is taking place in longitudinal direction extension
takes place in lateral directions (See Fig. 8.7). The lateral strain may be defined as changes in the
lateral dimension per unit lateral dimension. Thus,
Change in Lateral Dimension
Lateral Strain =
Original Lateral Dimension
b ′ − b δb
= = ...(8.8)
b b
The stress-strain relation of any material is obtained by conducting tension test in the laboratories
on standard specimen. Different materials behave differently and their behaviour in tension and in
compression differ slightly.
8.5.1 Behaviour in Tension
Mild steel. Figure 8.8 shows a typical tensile test specimen of mild steel. Its ends are gripped into
universal testing machine. Extensometer is fitted to test specimen which measures extension over the
length L1, shown in Fig. 8.8. The length over which extension is mesured is called gauge length.
The load is applied gradually and at regular interval of loads extension is measured. After certain
load, extension increases at faster rate and the capacity of extensometer to measure extension comes
to an end and, hence, it is removed before this stage is reached and extension is measured from scale
on the universal testing machine. Load is increased gradually till the specimen breaks.
SIMPLE STRESSES AND STRAINS 239
L1 L2 Cup
Cone
Fig. 8.8. Tension Test Specimen Fig. 8.9. Tension Test Specimen after Breaking
follows down the loading curve shown in Fig. 8.6. If unloading is made after loading the specimen
beyond elastic limit, it follows a straight line parallel to the original straight portion as shown by line
FF′ in Fig. 8.10. Thus if it is loaded beyond elastic limit and then unloaded a permanent strain (OF)
is left in the specimen. This is called permanent set.
Stress-strain relation in aluminium and high strength steel. In these elastic materials there is
no clear cut yield point. The necking takes place at ultimate stress and eventually the breaking point
is lower than the ultimate point. The typical stress-strain diagram is shown in Fig. 8.11. The stress
p at which if unloading is made there will be 0.2 per cent permanent set is known as 0.2 per cent
proof stress and this point is treated as yield point for all practical purposes.
F
py
Stress
Stress
F
0.2
Strain Strain
Stress-strain relation in brittle material. The typical stress-strain relation in a brittle material
like cast iron, is shown in Fig. 8.12.
In these material, there is no appreciable change in rate of strain. There is no yield point and
no necking takes place. Ultimate point and breaking point are one and the same. The strain at failure
is very small.
Percentage elongation and percentage reduction in area. Percentage elongation and percentage
reduction in area are the two terms used to measure the ductility of material.
(a) Percentage Elongation: It is defined as the ratio of the final extension at rupture to original
length expressed, as percentage. Thus,
L′ − L
Percentage Elongation = × 100 ...(8.9)
L
where L – original length, L′– length at rupture.
The code specify that original length is to be five times the diameter and the portion
considered must include neck (whenever it occurs). Usually marking are made on tension
rod at every ‘2.5 d’ distance and after failure the portion in which necking takes place is
considered. In case of ductile material percentage elongation is 20 to 25.
(b) Percentage Reduction in Area: It is defined as the ratio of maximum changes in the cross-
sectional area to original cross-sectional area, expressed as percentage. Thus,
SIMPLE STRESSES AND STRAINS 241
A − A′
Percentage Reduction in Area = × 100 ...(8.10)
A
where A–original cross-sectional area, A′–minimum cross-sectional area. In case of ductile
material, A′ is calculated after measuring the diameter at the neck. For this, the two broken
pieces of the specimen are to be kept joining each other properly. For steel, the percentage
reduction in area is 60 to 70.
8.5.2 Behaviour of Materials under Compression
As there is chance to bucking (laterally bending) of long specimen, for compression tests short
specimens are used. Hence, this test involves measurement of smaller changes in length. It results
into lesser accuracy. However precise measurements have shown the following results:
(a) In case of ductile materials stress-strain curve follows exactly same path as in tensile test
up to and even slightly beyond yield point. For larger values the curves diverge. There will
not be necking in case of compression tests.
(b) For most brittle materials ultimate compresive stress in compression is much larger than in
tension. It is because of flows and cracks present in brittle materials which weaken the
material in tension but will not affect the strength in compression.
So far our discussion on direct stress is based on the value obtained by dividing the load by original
cross-sectional area. That is the reason why the value of stress started dropping after neck is formed
in mild steel (or any ductile material) as seen in Fig. 8.10. But actually as material is stressed its
cross-sectional area changes. We should divide load by the actual cross-sectional area to get true
stress in the material. To distinguish between the two values we introduce the terms nominal stress
and true stress and define them as given below:
Load
Nominal Stress = ...(8.11a)
Original Cross-sectional Area
Load
True Stress = ...(8.11b)
Actual Cross-sectional Area
True Stress-Strain Curve
Strain
Robert Hooke, an English mathematician conducted several experiments and concluded that stress
is proportional to strain up to elastic limit. This is called Hooke’s law. Thus Hooke’s law is, up to
elastic limit
p ∝ e ...(8.13a)
where p is stress and e is strain
Hence, p = Ee ...(8.13b)
where E is the constant of proportionality of the material, known as modulus of elasticity or Young’s
modulus, named after the English scientist Thomas Young (1773–1829).
However, present day sophisticated experiments have shown that for mild steel the Hooke’s law
holds good up to the proportionality limit which is very close to the elastic limit. For other materials,
as seen in art. 1.5, Hooke’s law does not hold good. However, in the range of working stresses,
assuming Hooke’s law to hold good, the relationship does not deviate considerably from actual
behaviour. Accepting Hooke’s law to hold good, simplifies the analysis and design procedure
considerably. Hence Hooke’s law is widely accepted. The analysis procedure accepting Hooke’s law
is known as Linear Analysis and the design procedure is known as the working stress method.
SIMPLE STRESSES AND STRAINS 243
P P
L
P P
L
Fig. 8.14
P
From equation (8.6), Stress p =
A
∆
From equation (8.7), Strain, e =
L
From Hooke’s Law we have,
Stress p P / A PL
E= = = =
Strain e ∆ / L A∆
PL
or ∆ = . ...(8.14)
AE
Example 8.1. A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile force 40
kN. The modulus of elasticity for steel may be taken as 200 kN/mm2. Find stress, strain and elongation
of the bar due to applied load.
Solution: Load P = 40 kN = 40 × 1000 N
E = 200 kN/mm2 = 200 × 103 N/mm2
L = 500 mm
Diameter of the rod d = 16 mm
A = πd = π × 162
2
Therefore, sectional area
4 4
= 201.06 mm2
P 40 × 1000
Stress p = = = 198.94 N/mm2
A 201.06
p 198.94
Strain e = = = 0.0009947
E 200 × 10 3
PL 4.0 × 1000 × 500
Elongation ∆ = = = 0.497 mm
AE 201.06 × 200 × 10 3
244 MECHANICS OF SOLIDS
Example 8.2. A Surveyor’s steel tape 30 m long has a cross-section of 15 mm × 0.75 mm. With this,
line AB is measure as 150 m. If the force applied during measurement is 120 N more than the force
applied at the time of calibration, what is the actual length of the line?
Take modulus of elasticity for steel as 200 kN/mm2.
Solution: A = 15 × 0.75 = 11.25 mm2
P = 120 N, L = 30 m = 30 × 1000 mm
E = 200 kN/mm2 = 200 × 103 N/mm2
PL 120 × 30 × 1000
Elongation ∆ = = = 1.600 mm
AE 11.25 × 200 × 10 3
Hence, if measured length is 30 m.
Actual length is 30 m + 1.600 mm = 30.001600 m
150
∴ Actual length of line AB = × 30.001600 = 150.008 m
30
Example 8.3. A hollow steel tube is to be used to carry an axial compressive load of
160 kN. The yield stress for steel is 250 N/mm2. A factor of safety of 1.75 is to be used in the design.
The following three class of tubes of external diameter 101.6 mm are available.
Class Thickness
Light 3.65 mm
Medium 4.05 mm
Heavy 4.85 mm
Which section do you recommend?
Solution: Yield stress = 250 N/mm2
Factor of safety = 1.75
Therefore, permissible stress
250
p = = 142.857 N/mm2
1.75
Load P = 160 kN = 160 × 103 N
P
but p =
A
160 × 10 3
i.e. 142.857 =
A
160 × 103
∴ A = = 1120 mm2
142.857
For hollow section of outer diameter ‘D’ and inner diameter ‘d’
π
A = (D 2 – d2) = 1120
4
π
(101.62 – d2) = 1120
4
SIMPLE STRESSES AND STRAINS 245
d2 = 8896.53 ∴ d = 94.32 mm
D − d 101.6 − 94.32
∴ t= = = 3.63 mm
2 2
Hence, use of light section is recommended.
Example 8.4. A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to
destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is
102 kN. The maximum load is 130 kN.
The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find
(i) The stress at elastic limit.
(ii) Young’s modulus.
(iii) Percentage elongation.
(iv) Percentage reduction in area.
(v) Ultimate tensile stress.
Solution: Diameter d = 20 mm
πd 2
Area A = = 314.16 mm2
4
Load at elastic limit
(i) Stress at elastic limit =
Area
102 × 10 3
= = 324.675 N/mm2
314.16
Stress
(ii) Young’s modulus E = within elastic limit
Strain
P / A 80 × 10 3 /314.16
= =
∆/ L 0.25/200
= 203718 N/mm2
Final extension
(iii) Percentage elongation =
Original length
56
= × 100 = 28
200
(iv) Percentage reduction in area
Initial area − Final area
= × 100
Initial area
π π
× 20 2 − × 152
= 4 4 × 100 = 43.75
π
× 20 2
4
246 MECHANICS OF SOLIDS
Ultimate Load
(v) Ultimate Tensile Stress =
Area
130 × 10 3
= = 413.80 N/mm2.
314.16
A typical bar with cross-sections varying in steps and subjected to axial load is as shown in Fig.
8.15(a). Let the length of three portions be L1, L2 and L3 and the respective cross-sectional areas of
the portion be A1, A2, A3 and E be the Young’s modulus of the material and P be the applied axial
load.
Figure 8.15(b) shows the forces acting on the cross-sections of the three portions. It is obvious
that to maintain equilibrium the load acting on each portion is P only. Hence stress, strain and
extension of each of these portions are as listed below:
P 1 2 3 A3 P
A1 A2
L1 L2 L3
(a)
P P P P P
(b)
P p1 P PL1
1 p1 = e1 = = ∆1 =
A1 E A1E A1E
P p2 P PL2
2 p2 = e2 = = ∆2 =
A2 E A2E A2E
P p3 P PL3
3 p3 = e3 = = ∆3 =
A3 E A3E A3E
Example 8.5. The bar shown in Fig. 8.16 is tested in universal testing machine. It is observed that
at a load of 40 kN the total extension of the bar is 0.280 mm. Determine the Young’s modulus of
the material.
P d1 = 25 mm d2 = 20 mm d3 = 25 mm P
Fig. 8.16
PL1 40 × 10 3 × 150
Solution: Extension of portion 1, =
A1 E π
× 252 E
4
PL2 40 × 10 3 × 250
Extension of portion 2, =
A2 E π
× 20 2 E
4
PL3 40 × 10 3 × 150
Extension of portion 3, =
A3 E π
× 252 E
4
RS
40 × 10 3 4 150 250 150 UV
Total extension =
E
×
T + +
π 625 400 625 W
40 × 10 3 4 1.112
0.280 = × ×
E π E
2
E = 200990 N/mm
Example 8.6. The stepped bar shown in Fig. 8.17 is made up of two different materials. The material
1 has Young’s modulus = 2 × 105 N/mm, while that of material 2 is 1 × 105 N/mm2. Find the
extension of the bar under a pull of 30 kN if both the portions are 20 mm in thickness.
b = 40 Material 2
30 kN b = 30 P = 30 kN
Material 1
600 800
Fig. 8.17
PL1 30 × 10 3 × 600
Extension of portion 1, = = 0.1125 mm.
A1 E1 800 × 2 × 10 5
PL2 30 × 10 3 × 800
Extension of portion 2, = = 0.4000 mm.
A2 E2 600 × 1 × 10 5
∴ Total extension of the bar = 0.1125 + 0.4000 = 0.5125 mm.
Example 8.7. A bar of length 1000 mm
and diameter 30 mm is centrally bored for
400 mm, the bore diameter being 10 mm 30 kN D = 30 d = 30 30 kN
as shown in Fig. 8.18. Under a load of 30
kN, if the extension of the bar is 0.222 mm, 400
what is the modulus of elasticity of the bar? 1000
π
A2 = × (302 – 102) = 200 π
4
PL1
∆1 =
A1 E
PL2
∆2 =
A2 E
FG
P L1 L2 IJ
∴ ∆ = ∆1 + ∆2 =
H +
E A1 A2 K
FG
30 × 10 3 600 400 IJ
i.e., 0.222 =
E H +
225 π 200 π K
∴ E = 200736 N/mm2.
When the cross-section varies continuously, an elemental length of the bar should be considered and
general expression for elongation of the elemental length derived. Then the general expression should
be integrated over entire length to get total extension. The procedure is illustrated with Examples 8.8
and 8.9.
Example 8.8. A bar of uniform thickness ‘t’ tapers uniformly from a width of b1 at one end to b2
at other end in a length ‘L’ as shown in Fig. 8.18. Find the expression for the change in length of
the bar when subjected to an axial force P.
SIMPLE STRESSES AND STRAINS 249
b1
P b P b
b2
t
x dx
Cross-section
L
Fig. 8.19
Solution: Consider an elemental length dx at a distance x from larger end. Rate of change of breadth
b − b2
is 1 .
L
b1 − b2
Hence, width at section x is b = b1 – x = b1 – kx
L
b1 − b2
where k =
L
∴ Cross-section area of the element = A = t(b1 – kx)
Since force acting at all sections is P only,
Pdx
Extension of element = [where length = dx]
AE
Pdx
=
( b1 − kx )tE
FG IJ LMlog (b − kx)OP
P 1
L
=
H KN
tE − k Q
1
0
P L b − b IO
− log F b −
L
tEk MN
x P
H KQ
1 2
= 1
L 0
P P b
= [– log b2 + log b1] = log 1
tEk tEk b2
PL b
= log 1 . ...(8.16)
tE(b 1 − b 2 ) b2
250 MECHANICS OF SOLIDS
Example 8.9. A tapering rod has diameter d1 at one end and it tapers uniformly to a diameter d2
at the other end in a length L as shown in Fig. 8.20. If modulus of elasticity of the material is E,
find its change in length when subjected to an axial force P.
d1
P b P
d2 d
x dx
Cross-section
L
Fig. 8.20
πd 2 π
Cross-sectional area A = = (d1 – kx)2
4 4
P dx
∴ Extension of the element =
π
( d1 − kx ) 2 E
4
π
4
P dx
( d1 − kx ) 2 E
=
4P
πE z L
0
dx
( d1 − kx )2
4P 1 FG
IJ L
=
πEk d1 − kx H
K 0
F 1 − 1 IJ , since d
πE( d − d ) GH d
4P
d K
= 1 – kL = d2
1 2 2 1
L
4 PL ( d − d2 ) 4PL
∴ ∆ = × 1 = . ...(8.17)
πE( d1 − d2 ) d1 d2 πEd 1 d 2
SIMPLE STRESSES AND STRAINS 251
PL PL
Note: For bar of uniform diameter extension is and for tapering rod it is .
πd 2 π
E d d
1 2 E
4 4
Thus the change is, instead of d2 for uniform bar, d1 d2 term appears if the bar tapers uniformly.
Example 8.10. A steel flat of thickness 10 mm tapers uniformly from 60 mm at one end to 40 mm
at other end in a length of 600 mm. If the bar is subjected to a load of 80 kN, find its extension.
Take E = 2 × 105 MPa. What is the percentage error if average area is used for calculating
extension?
Solution: Now, t = 10 mm b1 = 60 mm b2 = 40 mm
L = 600 mm P = 80 kN = 80000 N
2
Now, 1 MPa = 1 N/mm
Hence E = 2 × 105 N/mm2
Extension of the tapering bar of rectangular section
PL b
∆ = log 1
tE( b1 − b2 ) b2
80000 × 600 60
= log
10 × 2 × 10 (60 − 40)
5
40
= 0.4865 mm
If averages cross-section is considered instead of tapering cross-section, extension is given by
PL
∆ =
Aav E
60 × 10 + 40 × 10
Now Aav = = 500 mm2
2
80000 × 600
∆= = 0.480 mm
500 × 2 × 10 5
0.4865 − 0.48
∴ Percentage error = × 100
0.4865
= 1.348
Example 8.11. A 2.0 m long steel bar is having uniform diameter of 40 mm for a length of 1 m and
in the next 0.5 m its diameter gradually reduces from 40 mm to 20 mm as shown in Fig. 8.21.
Determine the elongation of this bar when subjected to an axial tensile load of 200 kN. Given
E = 200 GN/m2.
20
200 kN
40 200 kN
1500 500
Fig. 8.21
252 MECHANICS OF SOLIDS
PL 200 × 10 3 × 1500
∆1 = = = 1.194 mm.
AE π
× 40 2 × 2 × 10 5
4
Extension of tapering portion
4 PL 4 × 200 × 10 3 × 500
∆2 = =
Eπd1d2 2 × 10 5 × π × 60 × 40
= 0.265 mm
Total extension = ∆1 + ∆2 = 1.194 + 0.265 = 1.459 mm
Example 8.12. The extension of a bar uniformly tapering from a diameter of d + a to
d – a in a length L is calculated by treating it as a bar of uniform cross-section of average diameter
d. What is the percentage error?
Solution: Actual extension under load
4 PL 4PL
P= =
πd1d2 E π( d + a)( d − a) E
4 PL
=
πE ( d 2 − a 2 )
If it is treated as a bar of uniform diameter ‘d’, erroneous extension calculated
PL 4 PL
= =
π πd 2 E
×d E
2
4
4 PL 4 PL
−
πE( d 2 − a 2 ) πEd 2
∴ Percentage error = × 100
4 PL
πE(d 2 − a 2 )
1 1
−
(d 2 − a 2 ) d 2
= × 100
1
(d 2 − a 2 )
SIMPLE STRESSES AND STRAINS 253
FG d 2 − a2 IJ × 100
= 1−
H d2 K
a2
= 100
d2
Q Q
Q R
Q
Q
R = ∫ q dA = q ∫ dA = qA
For equilibrium Q = R = qA
Q
i.e., q = ...(8.18)
A
Thus, the direct stress is equal to shearing force per unit area.
qab × AB × t = qcd × CD × t
qab = qcd = q(say)
qbc
Since qab and qcd are giving rise to equal and opposite forces B C
q × AB × t, with moment arm = AD, they form a couple of qab
magnitude equal to
q × AB × t × AD ...(1) qcd
This can be balanced by another couple only. i.e. qbc and qcd A D
should form an equal and opposite couple. Hence their direction qad
will be as shown in Fig. 8.24b. (b) Element under Shear
q a C C
B p
a a
a Dq A D q
A
(a) (b)
Fig. 8.25
Since it is square element AC = a 2 . Consider section along AC and let ‘p’ be the stress on this
section. From equilibrium condition of system of forces in the direction normal to AC, we get
p × AC × t = q CD t cos 45° + q AD t sin 45°
pa 2t = qat 1/ 2 + qat 1/ 2
= qat 2
i.e., p = q
Thus in case of simple shear tensile stress of the same magnitude as C C
B
shearing stress develops at 45° to shearing plane. By taking stresses on B
diagonal BD it can be shown that compressive stress of magnitude q acts
on this plane. Thus simple shear gives rise to tensile and compressive
stresses across planes inclined at 45° to the shearing planes, the intensity
A D
of direct stresses being of same magnitude as shearing stresss.
Fig. 8.26
SIMPLE STRESSES AND STRAINS 255
Shearing strain. Shearing stress has a tendency to distort the element to position AB′C′D from
the original position ABCD as shown in Fig. 8.26.
This deformation is expressed in terms of angular displacement and is called shear strain. Thus
BB′
Shear strain = = tan φ
AB
= φ, since angle φ is small
When a material undergoes changes in length, it undergoes changes of opposite nature in lateral
directions. For example, if a bar is subjected to direct tension in its axial direction it elongates and
at the same time its sides contract (Fig. 8.27).
L
If we define the ratio of change in axial direction to original length as linear strain and change
in lateral direction to the original lateral dimension as lateral strain, it is found that within elastic limit
there is a constant ratio between lateral strain and linear strain. This constant ratio is called
Poisson’s ratio. Thus,
Lateral strain
Poisson’s ratio = ...(8.19)
Linear strain
1
It is denoted by , or µ. For most of metals its value is between 0.25 to 0.33. Its value for steel
m
is 0.3 and for concrete 0.15.
When a member is subjected to stresses, it undergoes deformation in all directions. Hence, there will
be change in volume. The ratio of the change in volume to original volume is called volumetric
strain.
δV
Thus ev = ...(8.20)
V
where eV = Volumetric strain
δV = Change in volume
V = Original volume
It can be shown that volumetric strain is sum of strains in three mutually perpendicular directions.
i.e., ev = ex + ey + ez
256 MECHANICS OF SOLIDS
For example consider a bar of length L, breadth b and depth d as shown in Fig. 8.28.
z
y
d
x
b
L
Fig. 8.28
Now, V = Lbd
Since volume is function of L, b and d.
δV = δL bd + L δb d + Lb δd
δV δv
=
V Lbd
δL δb δd
eV = + +
L b d
eV = ex + ey + ez
Now, consider a circular rod of length L and diameter ‘d’ as shown in Fig. 8.29.
z
y
d
x
Fig. 8.29
π 2
Volume of the bar V= dL
4
π π 2
∴ δV = 2dδd L + d δL (since v is function of d and L).
4 4
δV δd δL
∴ = 2 +
π 2 d L
d L
4
δd
eV = ex + ey + ez; since ey = ez =
d
In general for any shape volumetric strain may be taken as sum of strains in three mutually
perpendicular directions.
Modulus of elasticity, modulus of rigidity and bulk modulus are the three elastic constants. Modulus
of elasticity (Young’s Modulus) ‘E’ has been already defined as the ratio of linear stress to linear
strain within elastic limit. Rigidity modulus and Bulk modulus are defined in this article.
SIMPLE STRESSES AND STRAINS 257
Modulus of Rigidity: It is defined as the ratio of shearing stress to shearing strain within elastic
limit and is usually denoted by letter G or N. Thus
q
G= ...(8.21)
φ
where G = Modulus of rigidity
q = Shearing stress
and φ = Shearing strain
Bulk Modulus: When a body is subjected to identical stresses p in three mutually perpendicular
directions, (Fig. 8.30), the body undergoes uniform changes in three directions without undergoing
distortion of shape. The ratio of change in volume to original volume has been defined as volumetric
strain (ev). Then the bulk modulus, K is defined as
p
K =
ev
where p = identical pressure in three mutually perpendicular directions
∆v
ev = , Volumetric strain
v
∆v = Change in volume
v = Original volume
Thus bulk modulus may be defined as the ratio of identical pressure ‘p’ acting in three mutually
perpendicular directions to corresponding volumetric strain.
p p
p
p
p p p p
p
p
p p
(a) (b)
Fig. 8.30
Figure 8.30 shows a body subjected to identical compressive pressure ‘p’ in three mutually
perpendicular directions. Since hydrostatic pressure, the pressure exerted by a liquid on a body within
it, has this nature of stress, such a pressure ‘p’ is called as hydrostatic pressure.
q
8.17 RELATIONSHIP BETWEEN MODULUS OF ELASTICITY E B C¢ C
AND MODULUS OF RIGIDITY F¢
q
Consider a square element ABCD of sides ‘a’ subjected to pure shear ‘q’ a
q
as shown in Fig. 8.31. AEC′D shown is the deformed shape due to shear
j
q. Drop perpendicular BF to diagonal DE. Let φ be the shear strain and A D
G modulus of rigidity. a
Fig. 8.31
258 MECHANICS OF SOLIDS
DE − DF
Now, strain in diagonal BD =
DF
EF
=
DB
EF
=
AB 2
Since angle of deformation is very small we can assume ∠BEF = 45°, hence EF = BE cos 45°
EF BE cos 45°
∴ Strain in diagonal BD = =
BD AB 2
a tan φ cos 45°
=
a 2
1 1
= tan φ = φ (Since φ is very small)
2 2
1 q q
= × , since φ = ...(1)
2 G G
Now, we know that the above pure shear gives rise to axial tensile stress q in the diagonal
direction of DB and axial compression q at right angles to it. These two stresses cause tensile strain
along the diagonal DB.
q q q
Tensile strain along the diagonal DB = + µ = (1 + µ ) ...(2)
E E E
From equations (1) and (2), we get
1 q q
× = (1 + µ )
2 G E
E = 2G(1 + µ) ...(8.22)
p
ez = (1 − 2 µ ) ...(1)
E
3p
∴ Volumetric strain ev = ex + ey + ez = (1 − 2 µ)
E
From definition, bulk modulus K is given by
p p
K= =
ev 3 p (1 − 2µ )
E
or E = 3K(1 – µ) ...(2)
Relationship between EGK:
We know E = 2G(1 + µ) ...(a)
and E = 3K(1 – 2µ) ...(b)
By eliminating µ between the above two equations we can get the relationship between E, G,
K, free from the term µ.
E
From equation (a) µ = −1
2G
Substituting it in equation (b), we get
LM F I OPE
N H KQ
E = 3K 1 − 2
2G
−1
F I F
E E I
H K H
= 3K 1 − + 2 = 3K 3 −
G G K
3KE
= 9K –
G
F I
3K
∴
HE 1+
K
G
= 9K
F I
G + 3K
or
HE
G K = 9K ...(c)
9 KG
or E= ...(8.23a)
G + 3K
Equation (c) may be expressed as
9 G + 3K
=
E KG
9 3 1
i.e., = + ...(8.23b)
E G K
Example 8.13. A bar of 25 mm diameter is tested in tension. It is observed that when a load of 60
kN is applied, the extension measured over a guage length of 200 mm is 0.12 mm and contraction
in diameter is 0.0045 mm. Find Poisson’s ratio and elastic constants E, G, K.
Solution: Now, P = 60 kN = 60000 N
π
Area A = × 252 = 156.25π mm2
4
Guage length L = 200 mm
260 MECHANICS OF SOLIDS
∆ = 0.12 mm
∆d = 0.0045 mm
∆ 0.12
Linear strain = = = 0.0006
L 200
∆d 0.0045
Lateral strain = = = 0.00018
d 25
Lateral strain 0.00018
∴ Poisson’s ratio = =
Linear strain 0.0006
µ = 0.3
PL
Now, ∆=
AE
60000 × 200
0.12 =
156.25π × E
or E = 203718.3 N/mm2
Using the relation E = 2G(1 + µ)
E 203718.3
We get G= = = 78353.2 N/mm2
2(1 + µ ) 2(1 + 0.3)
From the relation, E = 3K(1 – 2µ), we get
E 203718.3
K= = = 169765.25 N/mm2
3(1 − 2µ) 3(1 − 2 × 0.3)
Example 8.14. A circular rod of 25 mm diameter and 500 mm long is subjected to a tensile force
of 60 kN. Determine modulus of rigidity, bulk modulus and change in volume if Poisson’s ratio =
0.3 and Young’s modulus E = 2 × 105 N/mm2.
Solution: From the relationship
E = 2G(1 + µ) = 3k(1 – 2µ)
E 2 × 10 5
We get, G= = = 0.7692 × 105 N/mm2
2(1 + µ ) 2(1 + 0.3)
E 2 × 10 5
and K= = = 1.667 × 105 N/mm2
3(1 + 2µ) 3(1 − 2 × 0.3)
P 60 × 10 3
Longitudinal stress = = = 122.23 N/mm2
A π × 252
4
Stress 122.23
Linear strain = = = 61.115 × 10–5
E 2 × 10 5
Lateral strain = ey = – µex and ez = – µex
Volumetric strain ev = ex + ey + ez
= ex(1 – 2µ)
= 61.115 × 10–5 (1 – 2 × 0.3)
= 24.446 × 10–5
Change in volume
but = ev
v
SIMPLE STRESSES AND STRAINS 261
∴ Change in volume = ev × v
π
= 24.446 × 10–5 × × (252) × 500
4
= 60 mm3
Example 8.15. A 400 mm long bar has rectangular cross-section 10 mm × 30 mm. This bar is
subjected to
(i) 15 kN tensile force on 10 mm × 30 mm faces,
(ii) 80 kN compressive force on 10 mm × 400 mm faces, and
(iii) 180 kN tensile force on 30 mm × 400 mm faces.
Find the change in volume if E = 2 × 105 N/mm2 and µ = 0.3.
180 kN 30
80 15 kN
10
80 kN
z
y x 400
15 kN
180 kN
Fig. 8.33
Solution: The forces on the bar are as shown in Fig. 8.33. Let the x, y, z be the mutually perpendicular
directions as shown in the figure.
15 × 10 3
Now, px = = 50 N/mm2 (Tensile)
10 × 30
80 × 10 3
py = = 20 N/mm2 (Comp.)
10 × 400
180 × 10 3
pz = = 15 N/mm2 (Tensile)
30 × 400
Noting that a stress produces a strain of p/E in its own direction, the nature being same as that
p
of stress and µ in lateral direction of opposite nature, and taking tensile stress as +ve, we can write
E
expression for strains ex, ey, ez as
50 20 15
ex = +µ −µ
E E E
1 1
= (50 + 0.3 × 20 – 0.3 × 15) = (51.5)
E E
50 20 15
ey = – µ − −µ
E E E
1 39.5
= (– 0.3 × 50 – 20 – 0.3 × 15) = –
E E
262 MECHANICS OF SOLIDS
50 20 15
ez = – µ +µ +
E E E
1 6
= (– 0.3 × 50 + 20 × 0.3 + 15) =
E E
Volumetric strain: ev = ex + ey + ez
1 18
= (51.5 – 39.5 + 6) =
E E
Change in volume
But = ev
Volume
∴ Change in volume = ev × Volume
18
= × 10 × 30 × 400
2 × 10 5
= 10.8 mm3
Example 8.16. A bar of rectangular section shown in Fig. 8.34 is subjected to stresses px, py and
pz in x, y and z directions respectively. Show that if sum of these stresses is zero, there is no change
in volume of the bar.
pz
py px
py
px
pz
z
y x
Fig. 8.34
p
Solution: A stress p produces strain of magnitude p/E in its direction and a strain µ of opposite
E
nature at right angles to its direction.
p py p
Hence ex = x − µ −µ z
E E E
px p y p
ey = – µ + −µ z
E E E
p p y pz
ez = – µ x − µ +
E E E
SIMPLE STRESSES AND STRAINS 263
Now, ev = ex + ey + ez
p py p
= x (1 − 2µ ) + (1 − 2µ ) + z (1 − 2µ )
E E E
dV ( px + py + pz )
i.e., = (1 – 2µ)
V E
Hence if px + py + pz = 0
dV
ev = =0
V
∴ dV = 0
i.e., there is no volumetric change, if px + py + pz = 0.
Example 8.17. In a laboratory, tensile test is conducted and Young’s modulus of the material is
found to be 2.1 × 105 N/mm2. On the same material torsion test is conducted and modulus of rigidity
is found to be 0.78 × 105 N/mm2. Determine Poisson’s Ratio and bulk modulus of the material.
[Note: This is usual way of finding material properties in the laboratory].
Solution: E = 2.1 × 105 N/mm2
G = 0.78 × 105 N/mm2
Using relation E = 2G(1 + µ)
we get 2.1 × 105 = 2 × 0.78 × 105 (1 + µ)
1.346 = 1 + µ
or µ = 0.346
From relation E = 3K(1 – 2µ)
we get 2.1 × 105 = 3 × K(1 – 2 × 0.346)
K = 2.275 × 105 N/mm2
Example 8.18. A material has modulus of rigidity equal to 0.4 × 105 N/mm2 and bulk modulus equal
to 0.8 × 105 N/mm2. Find its Young’s Modulus and Poisson’s Ratio.
Solution: G = 0.4 × 105 N/mm2
K = 0.8 × 105 N/mm2
9GK
Using the relation E=
3K + G
9 × 0.4 × 10 5 × 0.8 × 10 5
E=
3 × 0.8 × 10 5 + 0.4 × 10 5
E = 1.0286 × 105 N
From the relation E = 2G(1 + µ)
we get 1.0286 × 105 = 2 × 0.4 × 105(1 + µ)
1.2857 = 1 + µ
or µ = 0.2857
264 MECHANICS OF SOLIDS
P
Rigid support
Fig. 8.35
Consider a member with two materials. Let the load shared by material 1 be P1 and that by
material 2 be P2. Then
(i) From equation of equilibrium of the forces, we get
P = P1 + P2 ...(8.24a)
(ii) Since the ends are held securely, we get
∆l1 = ∆l2
where ∆l1 and ∆l2 are the extension of the bars of material 1 and 2 respectively
P1 L1 PL
i.e. = 2 2 ...(8.24b)
A1 E1 A2 E2
Using equations 8.24(a) and (b), P1 and P2 can be found uniquely. Then extension of the system
P1 L1 P2 L2
can be found using the relation ∆l = or ∆l = since ∆l = ∆l1 = ∆l2.
A1 E1 A2 E2
The procedure of the analysis of compound bars is illustrated with the examples below:
Example 8.19. A compound bar of length 600 mm consists of a strip of aluminium 40 mm wide and
20 mm thick and a strip of steel 60 mm wide × 15 mm thick rigidly joined at the ends. If elastic
modulus of aluminium and steel are 1 × 105 N/mm2 and 2 × 105 N/mm2, determine the stresses
developed in each material and the extension of the compound bar when axial tensile force of 60
kN acts.
SIMPLE STRESSES AND STRAINS 265
Pa L 18462 × 600
Extension of the compound bar = =
Aa Ea 800 × 1 × 10 5
∆l = 0.138 mm.
Copper tube
Example 8.20. A compound bar consists of a circular rod of steel of 25
mm diameter rigidly fixed into a copper tube of internal diameter 25 mm
and external diameter 40 mm as shown in Fig. 8.37. If the compound bar
is subjected to a load of 120 kN, find the stresses developed in the two
materials. Steel rod
Take Es = 2 ×105 N/mm2 Fig. 8.37
and Ec = 1.2 × 105 N/mm2.
266 MECHANICS OF SOLIDS
π
Solution: Area of steel rod As = × 252 = 490.87 mm2
4
π
Area of copper tube Ac = (402 – 252) = 765.76 mm2
4
From equation of equilibrium,
Ps + Pc = 120 × 1000 ...(1)
where Ps is the load shared by steel rod and Pc is the load shared by the copper tube.
From compatibility condition, we have
∆s = ∆c
Ps L PL
= c
As Es Ac Ec
Ps Pc
=
490.87 × 2 × 10 5 765.76 × 1.2 × 10 5
∴ Ps = 1.068 Pc ...(2)
From eqns. (1) and (2), we get
1.068 Pc + Pc = 120 × 1000
120 × 1000
∴ Pc = = 58027 N
2.068
∴ Ps = 1.068 Pc = 61973 N
58027
∴ Stress in copper = = 75.78 N/mm2
9765.76
61973
Stress in steel = = 126.25 N/mm2
490.87
Example 8.21. A reinforced concrete column of size 300 mm × 500 16 dia. steel
mm has 8 steel bars of 16 mm diameter as shown in Fig. 8.38. If the
column is subjected to an axial compressive force of 800 kN, find the
Es 300
stresses developed in steel and concrete. Take = 18.
Ec
π
Solution: Area of steel = 8 × × 162 = 1608.5 mm2 600
4
∴ Area of concrete = 300 × 500 – 1608.5 = 148391.5 mm2 Fig. 8.38
From equilibrium condition,
Pc + Ps = 800 × 1000 ...(1)
From compatibility condition,
∆s = ∆c
Ps L PL
= c
As Es Ac Ec
Ps Pc
=
1608.5 Es 148391.5 Ec
SIMPLE STRESSES AND STRAINS 267
148391.5 Ec
Pc = × Ps
1608.5 Es
148391.5 1
= × Ps
1608.5 18
= 5.125 Ps ...(2)
From eqns. (1) and (2), we get
Ps + 5.125 Ps = 800 × 1000
∴ Ps = 130612 N.
Hence Pc = 5.125 Ps = 669388 N
Pc 669388
Hence stress in concrete = = = 4.51 N/mm2
Ac 148391.5
Ps 130612
and stress in steel = = = 81.2 N/mm2
As 1608.5
Example 8.22. Three pillars, two of aluminium and one of steel support a rigid platform of 250 kN
as shown in Fig. 8.39. If area of each aluminium pillar is 1200 mm2 and that of steel pillar is 1000
mm2, find the stresses developed in each pillar.
250 kN
Take Es = 2 × 105 N/mm2 and Ea = 1 × 106 N/mm2.
Aluminium
Aluminium
Solution: Let force shared by each aluminium pillar be Pa and that
160 mm
Steel
240 mm
∴ The forces in vertical direction = 0 →
Pa + Ps + Pa = 250
2Pa + Ps = 250 ...(1)
From compatibility condition, we get
∆s = ∆a Fig. 8.39
Ps Ls PL
= a a
As Es Aa Ea
Ps × 240 Pa × 160
=
1000 × 2 × 10 5
1200 × 1 × 10 5
∴ Ps = 1.111 Pa ...(2)
From eqns. (1) and (2), we get
Pa (2 + 1.111) = 250
∴ Pa = 80.36 kN
Hence from eqn. (1),
Ps = 250 – 2 × 80.36 = 89.28 kN
268 MECHANICS OF SOLIDS
Steel bolt
Steel bolt
600 mm (b)
(a)
Fig. 8.40
Solution: Figure 8.40 shows the assembly. Let the force shared by bolt be Ps and that by tube be
Pc. Since there is no external force, static equilibrium condition gives
Ps + Pc = 0 or Ps = – Pc
i.e., the two forces are equal in magnitude but opposite in nature. Obviously bolt is in tension and
tube is in compression.
1
Let the magnitude of force be P. Due to quarter turn of the nut, the nut advances by × pitch
4
1
= × 2 = 0.5 mm.
4
[Note. Pitch means advancement of nut in one full turn]
During this process bolt is extended and copper tube is shortened due to force P developed. Let
∆s be extension of bolt and ∆c shortening of copper tube. Final position of assembly be ∆, then
∆s + ∆ c = ∆
Ps Ls PL
i.e. + c c = 0.5
As Es Ac Ec
P × 600 P × 600
+ = 0.5
( π / 4) × 20 × 2 × 10
2 5
( π / 4) ( 40 − 282 ) × 1.2 × 10 5
2
P × 600 LM 1
+
1 OP = 0.5
N
( π / 4 ) × 10 5 20 2 × 2 (40 2 − 28 2 ) × 1.2 Q
∴ P = 28816.8 N
SIMPLE STRESSES AND STRAINS 269
Ps 28816.8
∴ ps = = = 91.72 N/mm2
As ( π / 4) × 20 2
Pc 28816.8
pc = = = 44.96 N/mm2
Ac ( π / 4 ) ( 40 2 − 28 2 )
Steel 12 × 10–6/°C
Copper 17.5 × 10–6/°C
Stainless steel 18 × 10 –6/°C
Brass, Bronze 19 × 10 –6/°C
Aluminium 23 × 10–6/°C
If the expansion of the member is freely permitted, as shown in Fig. 8.41, no temperature
stresses are induced in the material.
αtL
If the free expansion is prevented fully or partially the stresses are induced in the bar, by the
support forces. Referring to Fig. 8.42,
270 MECHANICS OF SOLIDS
L
(a)
αtL
(b)
R P
(c)
Fig. 8.42
L δ
αtL
R=P P
Fig. 8.43
P 1.668 × 2 × 10 5
or p = = = 27.8 N/mm2
A 12 × 1000
P
(iii) If the stress developed is 20 N/mm2, then p = = 20
A
If δ is the gap, ∆ = α tL – δ
PL
∴ = 3.168 – δ
AE
12 × 1000
i.e. 20 × = 3.168 – δ
2 × 10 5
∴ δ = 3.168 – 1.20 = 1.968 mm
272 MECHANICS OF SOLIDS
Example 8.25. The temperature of a steel ring is raised through 150°C in order to fit it on a wooden
wheel of 1.2 m diameter. Find the original diameter of the steel ring and also the stresses developed
in the ring, when it cools back to normal temperature. Assume Es = 2 × 105 N/mm2 and αs = 12
× 10–6/°C.
Solution: Let D be the diameter of ring after heating and ‘d’ be its diameter before heating
∴ D = 1.2 m = 1200 mm.
Circumference of ring after heating = πD
Circumference of ring before heating (L) = πd
∴ πD = πd + αs t (πd)
D = d + αs t d = (1 + αs t)d
i.e., 1200 = [1 + 12 × 10–6 × 150] d
∴ d = 1197.84 mm.
when it cools expansion prevented
∆ = π(D – d) = αstπd = 12 × 10–6 × 150 πd
= 1800 × 10–6 πd
PL
∴ = 1800 × 10–6 πd
AE
F PI × π × d
H AK E = 1800 × 10–6 πd
P
Stress p = = 1800 × 10–6 × 2 × 105 = 360 N/mm2
A
Example 8.26. The composite bar shown in Fig. 8.44 is rigidly fixed at the ends A and B. Determine
the reaction developed at ends when the temperature is raised by 18°C. Given
Ea = 70 kN/mm2
Es = 200 kN/mm2
αa = 11 × 10–6/°C
αs = 12 × 10–6/°C
2 2
Aa = 600 mm As = 400 mm
Aluminium Steel
1.5 m 3.0 m
(a)
(b)
Fig. 8.44
SIMPLE STRESSES AND STRAINS 273
LM 1500 + 3000 OP
i.e. 0.945 = P
N 600 × 70000 400 × 200 × 1000 Q
0.945 = 73.214 × 10–6 P
or P = 12907.3 N
Example 8.27. The steel bar AB shown in Fig. 8.45 varies linearly in diameter from 25 mm to 50
mm in a length 500 mm. It is held between two unyielding supports at room temperature. What is
the stress induced in the bar, if temperature rises by 25°C? Take Es = 2 × 105 N/mm2 and
αs = 12 × 10–6/°C.
25 mm
A B
50 mm
500 mm
Fig. 8.45
∴ P = 58905 N
P 58905
∴ Corresponding maximum stress = =
Amin π × 25 2
4
= 120 N/mm2.
Bar-1 Bar-1 P1
A
1tL
C
Fig. 8.46
Since the two bars are rigidly connected at the ends, the final position of the end will be
somewhere between AA and BB, say at CC. It means Bar–1 will experience compressive force P1
which contracts it by ∆1 and Bar–2 experience tensile force P2 which will expand it by ∆2.
The equilibrium of horizontal forces gives,
P1 = Pc, say P
From the Fig. 8.46 (b), it is clear,
α1 tL – ∆1 = α2 tL + ∆2
∴ ∆1 + ∆2 = α1 tL – α2 tL = (α1 – α2) tL.
If the cross-sectional areas of the bars are A1 and A2, we get
PL PL
+ = (α1 – α2) t L ...(8.28)
A1 E1 A2 E2
From the above equation force P can be found and hence the stresses P1 and P2 can be determined.
SIMPLE STRESSES AND STRAINS 275
Example 8.28. A bar of brass 20 mm is enclosed in a steel tube of 40 mm external diameter and
20 mm internal diameter. The bar and the tubes are initially 1.2 m long and are rigidly fastened at
both ends using 20 mm diameter pins. If the temperature is raised by 60°C, find the stresses induced
in the bar, tube and pins.
Given: Es = 2 × 105 N/mm2
Eb = 1 × 105 N/mm2
αs = 11.6 × 10–6/°C
αb = 18.7 × 10–6/°C.
Solution:
stL
Pin C Steel tube
B
s
A
20 40
b A
B
1200 mm C
btb Brass rod
Fig. 8.47
LM 1
+
1 OP = 7.1 × 10
N 942.48 × 2 × 10 Q
–6
P × 1200 × 60 × 1200
5
314.16 × 1 × 10 5
∴ P = 11471.3 N
P 11471.3
∴ Stress in steel = = = 12.17 N/mm2
As 942.48
P 11471.3
and Stress in brass = = = 36.51 N/mm2
Ab 314.16
276 MECHANICS OF SOLIDS
The pin resist the force P at the two cross-sections at junction of two bars.
P
∴ Shear stress in pin =
2 × Area of pin
11471.3
= = 18.26 N/mm2
2 × π / 4 × 20 2
Example 8.29. A compound bar is made of a steel plate 50 mm wide and 10 mm thick to which
copper plates of size 40 mm wide and 5 mm thick are connected rigidly on each side as shown in
Fig. 8.48. The length of the bar at normal temperature is 1 m. If the temperature is raised by 80°,
determine the stresses in each metal and the change in length. Given αs =12 × 10–6/°C, αc = 17 ×
10–6/°C, Es = 2 × 105 N/mm2, Ec = 1 × 105– N/mm2.
Copper stL C
A
B PC
PS
B
A PC
Copper L ctL
Steel
C
Fig. 8.48
Solution: Now, L = 1000 mm, As = 50 × 10 = 500 mm2, Ac = Free expansion of copper αctL is
greater than free expansion of steel αstL . To bring them to the same position, tensile force Ps acts
on steel plate and compressive force Pc acts on each copper plate.
∴ Ps = 2Pc
If ∆s and ∆c are changes in the length of steel and copper plates.
∆s + ∆c = (αc – αs) tL
Ps L PL
∴ + c = (αc – αs) tL
As Es Ac Ec
But Ps = 2Pc
2 Pc Pc
∴ + = (17 – 12) × 10–6 × 80
500 × 2 × 10 5
200 × 1 × 10 5
∴ Pc = 5714.28 N
∴ Ps = 11428.56 N
Pc 5714.28
∴ Stress in copper = = = 28.57 N/mm2
Ac 200
Ps 5714.28 × 2
Stress in steel = = = 22.86 N/mm2
As 500
SIMPLE STRESSES AND STRAINS 277
D/2 θ
dθ p
p D
(b)
(a)
Fig. 8.49
Consider half the section shown in Fig. 8.49(b). The normal pressure on the element of unit
length as shown in Fig. 8.49(b)
D
=p× dθ
2
pD
Its vertical component = dθ cos θ
2
∴ Bursting force normal to the horizontal section
=2
z
0
π/ 2
p
d
2
cos θ dθ = pd sin θ
π/2
0
= pd.
This bursting force is resisted by the hoop stresses f. Let ‘t’ be the thickness of cylinder. Then
resisting force per unit length of cylinder.
= 2 × ft
Equating resisting force to bursting force, we get
2ft = pd
pd
or f= ...(8.29)
2t
278 MECHANICS OF SOLIDS
IMPORTANT FORMULAE
1. If stress is uniform
P
p=
A
Change in length
2. (i) Linear strain =
Original length
Change in lateral dimension
(ii) Lateral strain =
Original lateral dimension
Lateral strain
3. Poisson’s ratio = , within elastic limit.
Linear strain
L′ − L
4. Percentage elongation = × 100.
L
A − A′
5. Percentage reduction in area = × 100.
A
Load
6. Nominal stress = .
Original cross-sectional area
Load
7. True stress = .
Actual cross-sectional area
Ultimate stress
8. Factor of safety =
Working stress
Yield stress
However in case of steel, = .
Working stress
9. Hooke’s Law, p = Ee.
PL
10. Extension/shortening of bar = .
AE
PL b
11. Extension of flat bar with linearly varying width and constant thickness = log 1 .
tE( b1 − b2 ) b2
4 PL PL
12. Extension of linearly tapering rod = = .
πE d1d2 (π / 4 d1d2 ) E
SIMPLE STRESSES AND STRAINS 279
Q
13. Direct shear stress = .
A
δV
14. Volumetric strain ev = = ex + ey + ez.
V
15. E = 2G (1 + µ) = 3K (1 – 2µ)
9 3 1
or = + .
E G K
16. Extension due to rise in temperature:
∆ = α tL
17. Thermal force, P is given by
PL
= extension prevented.
AE
THEORY QUESTIONS
PL PL
+ = (α1 – α2) tL
A1 E1 A2 E2
where A1, A2–Cross-sectional areas of bar 1 and bar 2 respectively
E1, E2–Young’s moduli of bar 1 and bar 2 respectively and
α1, α2–are coefficient of thermal expansion of bars 1 and 2 respectively.
12. Show that in a thin cylinder subject to internal pressure ‘p’ is given by
pd
f =
2t
where d–diameter of the cylinder
t–thickness of cylinder.
1. Tension test was conducted on a specimen and the following readings were recorded.
Diameter = 25 mm
Gauge length of extensometer = 200 mm
Least count of extensometer = 0.001 mm
At a load of 30 kN, extensometer reading = 60
At a load of 50 kN, extensometer reading = 100
Yield load = 160 kN
Maximum load = 205 kN
Diameter neck = 17 mm
Final extension over 125 mm original length = 150 mm
Find Young’s Modulus, yield stress, ultimate stress, percentage elongation and percentage
reduction in area.
[Ans. E = 2.0372 × 105 N/mm2, py = 325.95 N/mm2. Ultimate stress = 417.62 N/mm2,
% elongation = 20% reduction in area = 53.76]
2. The composite bar shown in Fig. 8.50 is subjected to a tensile force of 30 kN. The extension
observed is 0.44. Find the Young’s modulus of brass, if Young’s modulus of steel is 2 × 105
N/mm2. [Ans. 99745 N/mm2]
25 mm
20 mm
Steel Brass P = 30 kN
500 mm 300 mm
Fig. 8.50
3. The steel flat shown in Fig. 8.51 has uniform thickness of 20 mm. Under an axial load of 100 kN,
its extension is found to be 0.21 mm. Determine the Young’s modulus of the material.
[Ans. E = 2.06293 × 105 N/mm2]
SIMPLE STRESSES AND STRAINS 281
b2 = 40 mm
b1 = 80
100 kN P = 100 kN
500 mm
Fig. 8.51
4. Find the extension of the bar shown in Fig. 8.52 under an axial load of 25 kN.
[Ans. ∆ = 0.555 mm]
1 = 25 mm
2 = 15 mm
P P = 25 kN
Fig. 8.52
5. The compound bar shown in Fig. 8.54 consists of three materials and supports a rigid horizontal
platform. Find the stresses developed in each bar, if the platform remains horizontal even after
the loading with 100 kN. Given
Bar 1 Bar 2 Bar 3
6. The composite bar shown in Fig. 8.53 is 0.20 mm short of distance between the rigid supports at
room temperature. What is the maximum temperature rise which will not produce stresses in the
bar ? Find the stresses induced in the bar when temperature rise is 60°. Given:
αs = 12 × 10–6/°C αc = 17.5 × 10–6/°C
Es = 2 × 105 N/mm2 Ec = 1.2 × 105 N/mm2
As = 400 mm2 Ac = 300 mm2
[Ans. (a) 42.10 °C, (b) ps = 36 N/mm2, pc = 48 N/mm2]
282 MECHANICS OF SOLIDS
0.2 mm
Copper 100 mm
Steel 250 mm
Fig. 8.53
7. At room temperature the gap between bar A and bar B shown in Fig. 8.54 is 0.2 mm. What are the
stresses developed in the bars, if temperature rise is 30°C ? Given:
Aa = 800 mm2 Ab = 600 mm2
5 2
Ea = 2 × 10 N/mm Eb = 1 × 105 N/mm2
αa = 12 × 10–6/°C αb = 23 × 10–6 /°C
La = 250 mm Lb = 200 mm.
[Ans. pa = 7.15 N/mm2 ; pb = 9.53 N/mm2]
Bar A Bar B
250 mm 200 mm
Fig. 8.54
9
Beams
9.1 INTRODUCTION
A beam may be defined as a structural element which has one dimension considerably larger than
the other two dimensions, namely breadth and depth, and is supported at few points. The distance
between two adjacent supports is called span. It is usually loaded normal to its axis. The applied
loads make every cross-section to face bending and shearing. The load finally get transferred to
supports. The system of forces consisting of applied loads and reactions keep the beam in equilibrium.
The reactions depend upon the type of supports and type of loading. In this chapter type of supports,
types of beams and types of loading are first explained and then the methods of finding reactions,
bending moment and shear forces are illustrated for the following determinate beams:
(a) Simply supported beams
(b) Cantilever beams and
(c) Overhanging beams.
283
284 MECHANICS OF SOLIDS
HA = R cos θ
Fixed Support: At such supports, the beam end is not free to translate or rotate. Translation
is prevented by developing support reaction in any required direction.
Referring to Fig. 9.4 the support reaction R which is at an angle θ to x axis may be
represented by HA and VA, where
VA = R sin θ
HA = R cos θ
Rotation is prevented by developing support moment MA as shown in Fig. 9.4. Thus at fixed
support A, there are three reactions HA, VA and MA.
HA A HA A MA
G G
R R
VA VA
In the case of simply supported beams, beams with one end hinged and the other on rollers,
cantilever and over-hanging beams, it is possible to determine the reactions for given loadings by
using the equations of equilibrium only. In the other cases, the number of independent equilibrium
equations are less than the number of unknown reactions and hence it is not possible to analyse them
by using equilibrium equations alone. The beams which can be analysed using only equilibrium
equations are known as Statically Determinate beams and those which cannot be analysed are known
as Statically Indeterminate beams. The latter beams can be analysed using the conditions of continuity
in deformations in addition to equilibrium equations. Such cases will not be treated in this book.
Example 9.1. The beam AB of span 12 m shown in Fig. 9.17 (a) is hinged at A and is on rollers
at B. Determine the reactions at A and B for the loading shown in the Figure.
Solution: At A the reaction can be in any direction. Let this reaction be represented by its components
VA and HA as shown in Fig. 9.17 (b). At B the reaction is in vertical direction only. The beam is in
equilibrium under the action of system of forces shown in Fig. 9.17 (b).
10 kN 15 kN 20 kN 10 kN 15 kN 20 kN
30° 45° 30° 45°
A B HA
4m 2m 4 m 2m
VA RB
12 m (b)
(a)
HA
RA VA
(c)
Fig. 9.17
Now, ∑ H = 0, gives
HA – 15 cos 30° – 20 cos 45° = 0
HA = 27.1325 kN
∑ M A = 0, gives
RB × 12 – 10 × 4 – 15 sin 30° × 6 – 20 sin 45° × 10 = 0
RB = 18.8684 kN.
∑V = 0, gives
VA + 18.8684 – 10 – 15 sin 30° – 20 sin 45° = 0
∴ VA = 12.7737 kN
∴ RA = H A2 + VA2 = 27.13252 + 12. 77372
i.e., RA = 29.989 kN.
(12. 7737)
and α = tan −1 = 25.21°.
( 27.1325)
[Note: For finding moments, inclined loads are split into their vertical and horizontal components.
Horizontal components do not produce moment about A.]
BEAMS 287
Example 9.2. Find the reactions at supports A and B in the beam AB shown in Fig. 9.18 (a).
80 kN
60 kN 50 kN HA
HA α
6 0° 7 5° 6 0°
A 2 m 2 .5 m B 3 0° R A VA
1 m
6 m
6 0° RB
VA
(a)
(b)
Fig. 9.18
Solution: The reaction at B will be at right angles to the support, i.e., at 60° to horizontal as shown
in the figure. Let the components of the reactions at A be HA and VA. Then
∑ M A = 0 gives
RB sin 60° × 6 – 60 sin 60° × 1 – 80 × sin 75° × 3 – 50 × sin 60° × 5.5 = 0
∴ RB = 100.4475 kN.
∑ H = 0, gives
HA + 60 cos 60° – 80 cos 75° + 50 cos 60° – RB cos 60° = 0
HA = –60 cos 60° + 80 cos 75° – 50 cos 60° + 100.4475 cos 60°
= 15.9293 kN
∑V = 0, gives
VA + RB sin 60° – 60 sin 60° – 80 sin 75° – 50 sin 60° = 0
VA = –100.4475 sin 60° + 60 sin 60° + 80 sin 75° + 50 sin 60°
= 85.5468 kN
∴ RA = 15. 92932 + 85.54682
i.e., RA = 87.0172 kN.
85.5468 −1
α = tan
15.9293
i.e., α = 79.45°, as shown in Fig. 9.18(b).
Example 9.3. Find the reactions at supports A and B of the loaded beam shown in Fig. 9.19(a).
2 0 kN 6 0 kN HB
3 0 kN /m
4 5° B α
A
HB
RB
VB
2m 4m 1m 2m
9m
RA VB
(a) (b)
Fig. 9.19
288 MECHANICS OF SOLIDS
Solution: The reaction at A is vertical. Let HB and VB be the components of the reaction at B.
∑ M B = 0, gives
RA × 9 – 20 × 7 – 30 × 4 × 5 – 60 sin 45° × 2 = 0
∴ RA = 91.6503 kN.
∑ H A = 0, gives
HB – 60 cos 45° = 0
∴ HB = 42.4264 kN.
∑VA = 0
91.6503 + VB – 20 – 30 × 4 – 60 sin 45° = 0
VB = 90.7761 kN.
∴ RB = 42. 42642 + 90. 77612
∴ RB = 100.2013 kN.
90. 7761
−1
α = tan
42. 4264
α = 64.95°, as shown in Fig. 9.19(b).
Example 9.4. The cantilever shown in Fig. 9.20 is fixed at A and is free at B. Determine the reactions
when it is loaded as shown in the Figure.
2 0 kN 1 2 kN 1 0 kN
1 6 kN /m
HA
A B
MA 2m 1m 1m
VA
Fig. 9.20
20kN/m
15 kN 10 kN
B
MA
A 30 kN-m
2m 1m 1m 1m
Fig. 9.21
∴ VA = 65 kN.
ΣM = 0, gives
MA – 20 × 2 × 1 – 15 × 3 – 30 – 10 × 5 = 0
MA = 165 kN-m
Example 9.6. Determine the reactions at supports A and B of the overhanging beam shown in
Fig. 9.22.
30 kN 20 kN/m 40 kN
A B
1m 1m 3m 1.5 m
RA RB
Fig. 9.22
Solution: As supports A and B are simple supports and loading is only in vertical direction, the
reactions RA and RB are in vertical directions only.
ΣMA = 0, gives
RB × 5 – 30 × 1 – 20 × 3 × (2 + 1.5) – 40 × 6.5 = 0
∴ RB = 100 kN.
ΣV = 0, gives
RA + RB – 30 – 20 × 3 – 40 = 0
∴ RA = 130 – RB = 130 – 100 = 30 kN.
Example 9.7. Find the reactions at supports A and B of the beam shown in Fig. 9.23.
20 kN/m 60 kN 20 kN
A B
30 kN
2m 4m 3m 2m 2m
Fig. 9.23
Solution: Let VA and HA be the vertical and the horizontal reactions at A and VB be vertical reaction
at B.
ΣH = 0, gives
HA = 0.
290 MECHANICS OF SOLIDS
ΣMA = 0, gives
– 20 × 2 × 1 + 60 × 4 + 30 + 20 × 11 – VB × 9 = 0
∴ VB = 50 kN.
ΣV = 0, gives
– 20 × 2 + VA – 60 + VB – 20 = 0
VA = 120 – VB = 120 – 50
VA = 70 kN.
Example 9.8. Determine the reactions at A and B of the overhanging beam shown in Fig. 9.24(a).
4 0 kN -m 3 0 kN
2 0 kN /m A HA
HA A 4 5°
B α
VA RA
3m 2m 1m 2m
VA (a ) RB (b )
Fig. 9.24
Solution: ∑ MA = 0
RB × 6 – 40 – 30 sin 45° × 5 – 20 × 2 × 7 = 0
RB = 71.0110 kN.
∑H = 0
HA = 30 cos 45° = 21.2132 kN
∑V = 0
VA – 30 sin 45° + RB – 20 × 2 = 0
VA = 30 sin 45° – RB + 40
VA = –9.7978
(Negative sign show that the assumed direction of VA is wrong. In other words, VA is acting
vertically downwards).
RA = VA2 + H A2
RA = 23.3666 kN.
−1 V A
α = tan
HA
α = 24.79°, as shown in Fig. 9.24(b).
Example 9.9. A beam AB 20 m long supported on two intermediate supports 12 m apart, carries
a uniformly distributed load of 6 kN/m and two concentrated loads of 30 kN at left end A and 50
kN at the right end B as shown in Fig. 9.25. How far away should the first support C be located
from the end A so that the reactions at both the supports are equal ?
Solution: Let the left support C be at a distance x metres from A.
Now, RC = RD (given)
∑V = 0, gives
RC + RD – 30 – 6 × 20 – 50 = 0
BEAMS 291
Fig. 9.25
A B
2m 2m 2m 1m
Fig. 9.26
Now to find what is happening at a section, consider the section at C which is at a distance of
3 m from A. Imagining a cut at this section, left hand side portion and right hand side portions are
drawn separately in Fig. 9.27.
2 0 kN 2 0 kN
1 00 kN -m
A 4 0 kN 6 0 kN
3m
1 00 kN -m
4 0 kN C B
4m
2 0 kN 8 0 kN
Fig. 9.27
292 MECHANICS OF SOLIDS
Considering the algebraic sum of vertical forces acting on left hand side portion, it is found that
a net vertical force of 40 – 20 = 20 kN is experienced at the section. This effect is shown in Fig.
9.27 by dotted line. Again considering the portion on left hand side, the moment experienced at C
is given by
MC = 40 × 3 – 20 × 1
= 100 kN-m clockwise.
This moment is also shown on left hand side portion of the beam at C by dotted line.
Now, considering the right hand side portion:
Force experienced at C
= 80 – 60 – 40 = –20 kN
= 20 kN. downward
and the moment experienced is M = 80 × 4 – 60 × 3 – 40 × 1
= 100 kN-m. anticlockwise
These forces and moments are also shown in Fig. 9.27 on right hand side portion of the beam
at C.
Thus the section C is subjected to a force of 20 kN, which is trying to shear off the beam as
shown in Fig. 9.28(a), and is also subjected to a moment of 100 kN-m which is trying to bend the
beam as shown in Fig. 9.28(b). Since this force of 20 kN is trying to shear off the section, it is called
as shear force at section C. The moment is trying to bend the beam at C and hence it is called as
bending moment at that section. The shear force and bending moment at a section in a beam may
be defined as follows:
F
A B
C
F
(a ) S h ea r fo rce a t C
M
A B
C
M
C
(b ) B e nd in g m o m en t a t C
Fig. 9.28
“Shear Force at a section in a beam (or any structural member) is the force that is trying to shear
off the section and is obtained as algebraic sum of all the forces acting normal to the axis of beam
either to the left or to the right of the section”.
“Bending Moment at a section in a beam is the moment that is trying to bend the beam and is
obtained as algebraic sum of moment of all the forces about the section, acting either to the left or
to the right of the section”.
BEAMS 293
Hence to find shear force or bending moment at a section, a cut at the section is to be imagined
and any one portion with all the forces acting on that portion, is to be considered. It may be noted
that for finding bending moment at a section, the moment of the forces are to be found about the
section considered.
Fig. 9.29
(b) The bending moment that is trying to sag the beam shall be taken as positive bending
moment. If left portion is considered positive bending moment comes out to be clockwise moment
(Fig. 9.30).
Fig. 9.30
To decide the sign of moment due to a force about a section, assume the beam is held tightly
at that section and observe the deflected shape. Then looking at the shape sign can be assigned. Thus
in the problem shown in Fig. 9.26 and 9.27, 40 kN reaction at A produces positive moment at C and
20 kN load produces negative moment.
∑V = 0 leads to
– F + F + δF – wδx = 0
δF
or =w
δx
In the limiting case as δx → 0, ...(9.1)
dF
=w
dx
(a ) P o sition of E lem en t
w /U n it Le n gth
M M +d M
F + dF
F
dx
Fig. 9.31
of change of bending moment (i.e., shape of bending moment diagram curve) is equal to the shear
force at that section. From equation 2, it can also be concluded that the bending moment will be
maximum/minimum where shear force (dM/dx) is zero.
At any section, if moment changes its sign the point representing that section is called the point
of contraflexure. Obviously, the moment at that section is zero.
Consider the section X–X at a distance x from free end in a cantilever beam shown in
Fig. 9.32(a).
From left hand side segment of beam,
F = –W
Thus shear force is constant i.e., it will not vary with x. Hence the SFD is as shown in
Fig. 9.32(b).
M = –Wx, linear variation.
At x = 0, MA = 0
W
x x
A x B
(a ) Lo a d D iag ram
– ve W
(b ) S F D
– ve Wl
(c) B M D
Fig. 9.32
At x = l, MB = –Wl
Hence BMD is as shown in Fig. 9.32(c).
Considering the left hand side portion of the beam from the section X–X which is at a distance
x from the free end A,
F = –Wx, linear variation
At x = 0, FA = 0
At x = l, FB = –wl
x
w /u n it len gth
A x B
x
l
(a ) L oa d D ia gra m
– ve
wl
(b ) S F D
– ve w l2
2
(c) B M D
Fig. 9.33
Let W be the concentrated load acting on beam AB at a distance ‘a’ from the end A as shown in Fig.
9.34(a).
Wb Wa
Now RA = and RB =
l l
Consider the portion AC. At any distance x from A,
Wb
F = RA = , constant
l
Wb
M = RA x = x, linear variation.
l
At x = 0, MA = 0
Wab
At x = a, MC =
l
BEAMS 297
A a C B
b
l
(a) L oad D iag ra m
Wb
+ ve
l
– ve Wa
l
(b) S F D
W ab
l
(c) B M D
Fig. 9.34
SF and BM diagrams for this portion can now be drawn. Hence SFD and BMD for the beam is
as shown in Fig. 9.34(b) and 9.34(c) respectively.
Particular case:
When a = b = l/2
Wb W ( l / 2 ) W
F = = =
l l 2
and moment under the load (centre of span since a = b = l/2)
Wab W l / 2 × l / 2 Wl
M = = =
l l 4
Let the beam AB of span l be subjected to uniformly distributed load of intensity w/unit length as
shown in Fig. 9.35(a).
wl. l / 2 wl
RA = =
l 2
298 MECHANICS OF SOLIDS
wl
RB =
2
At a section X–X which is at a distance x from A,
wl
F = RA – wx = − wx, linear variation
2
wl
At x = 0, FA =
2
wl
At x = l, FB = −
2
∴ SFD is as shown in Fig. 9.35(b).
w /u nit len gth
x
A x B
x
l
(a ) L o a d D iag ra m
wl
2 + ve
– ve wl
(b ) S F D 2
w l2
8
(c) B M D
Fig. 9.35
Consider the overhanging beam ABC of span AB = l and overhang BC = a, subjected to a concentrated
load W at free end as shown in Fig. 9.36(a).
W
A C
l B a
RA (a ) L oa d D iag ra m
W
+ ve
Wa – ve
l
(b ) S F D
– ve
(c) B M D Wa
Fig. 9.36
Wa
RA = , downward
l
Wa FG
a IJ
∴ RB = W +
l
=W l +
H
l K
Portion AB:
Measuring x from A and considering left side of the section.
Wa
F = –RA = − , constant
l
Wax
M = –RAx = − , linear variation.
l
At x = 0, MA = 0
At x = l, MB = –Wa
Portion BC:
Measuring x from C, shear force and bending moments at that section are given by
F = W, constant
M = –Wx, linear variation.
At x = 0, MC = 0
At x = a, MB = –Wa.
SFD and BMD for the entire beam is shown in Fig. 9.36(b) and 9.36(c) respectively.
Example 9.10. Draw shear force and bending moment diagram for the cantilever beam shown in
Fig. 9.37(a).
300 MECHANICS OF SOLIDS
A 1m B 2m C
(a ) L oa d D iag ra m
20
40
– ve
80
1 20
(b ) S F D
30 – ve
(c) B M D
2 30
Fig. 9.37
BEAMS 301
Example 9.11. Draw the SF and BM diagrams for the beam shown in Fig. 9.38(a) and find out the
position and the magnitude of maximum moment.
Solution: ∑ MA = 0 →
RB × 10 = 20 × 5 × 2.5 + 20 × 5 + 40 × 7.5 + 20 × 8.5
∴ RB = 82 kN.
∑V = 0 →
∴ RA = 20 × 5 + 20 + 40 + 20 – 82 = 98 kN.
Portion AC:
Measuring x from A,
F = 98 – 20x, linear variation
At x = 0, FA = 98 kN
At x = 5 m, FB = 98 – 100 = –2 kN
Points where shear force is zero is given by,
0 = 98 – 20x
or x = 4.9 m
2 0 kN /m 2 0 kN /m 2 0 kN 4 0 kN 2 0 kN
A B
C D E
5m 2 .5 m 1 m 1 .5 m
(a ) L oa d D iag ra m
98
+ ve 2
22 – ve
62 82
82
(b ) S F D
2 40
1 85
1 23
2 40 .1
4 .9 m
(c) B M D
Fig. 9.38
Moment is given by
M = 98x – 20x x , parabolic variation.
2
At x = 0, MA = 0
20 × 52
At x = 5 m, MB = 98 × 5 – = 240 kN-m
2
302 MECHANICS OF SOLIDS
dM
Maximum moment occurs at x = 4.9 m where F = =0
dx
4. 92
Mmax = 98 × 4.9 – 20 × = 240.1 kN-m
2
Portion CD:
Measuring x from B and considering right hand side segment of the beam,
F = –82 + 20 + 40 = –22 kN, constant
M = 82x – 20(x – 1.5) – 40(x – 2.5)
= 22x + 130, linear variation.
At x = 2.5 m, MD = 22 × 2.5 + 130 = 185 kN-m
At x = 5 m, MC = 22 × 5 + 130 = 240 kN-m.
Portion DE:
Measuring x from B and considering the portion of the beam on the right side of the section,
F = –82 + 20 = –62 kN, constant
M = 82x – 20(x – 1.5) linear variation
At x = 1.5 m,
ME = 82 × 1.5 = 123 kN-m.
At x = 2.5, M = 82 × 2.5 – 20 × 1 = 185 kN-m.
Portion EB:
Measuring x from B and considering the right side segment,
F = –82 kN, constant
M = 82x, linear variation.
At x = 0, MB = 0
At x = 1.5 m, ME = 82 × 1.5 = 123 kN-m.
SFD and BMD are shown in Fig. 9.38(b) and 9.38(c) respectively, for the entire beam.
Example 9.12. A beam of span 8 m has roller support at A and hinge support at B as shown in
Fig. 9.39(a). Draw SF and BM diagrams when the beam is subjected to udl, a concentrated load and
an externally applied moment as shown in the Figure.
Solution: ∑ MA = 0 →
RB × 8 – 10 × 4 × 2 – 20 × 4 + 240 = 0
RB = –10 kN (upwards)
= 10 kN. (downwards)
∑V = 0 →
RA = 10 × 4 + 20 + 10 = 70 kN.
Portion AC:
Measuring x from A and considering left hand side segment of the beam,
F = 70 – 10x, linear variation
At x = 0, FA = 70 kN
At x = 4 m, FC = 70 – 40 = 30 kN
BEAMS 303
x
M = 70x – 10x , parabolic variation
2
At x = 0, MA = 0
4
At x = 4 m, MC = 70 × 4 – 10 × 4 × = 200 kN-m.
2
Portion CD: Measuring x from B,
F = 10 kN, constant
M = –10x + 240, linear variation
x = 4 m, MC = 200 kN-m
At x = 2 m, MD = –10 × 2 + 240 = 220 kN-m.
Portion DB:
Measuring x from B,
F = 10 kN, constant
M = –10x, linear variation
At x = 0, MB = 0
At x = 2 m, MD = –10 × 2 = –20 kN-m.
1 0 kN /m 2 0 kN
2 40 kN -m
A B
C D
RA
4 m 2 m 2 m RB
(a ) L oa ding D ia gra m
70
30
+ ve 10 10
(b ) SF D
2 20
2 00
+ ve
– ve
20
(c) BM D
Fig. 9.39
Solution: Σ ΜΑ = 0 →
RB × 6 = 20 × 8 + 30 × 3 × 1.5
∴ RB = 49.167 kN.
ΣV = 0 →
RA = 30 × 3 + 20 – 49.167 = 60.833 kN.
Portion AD:
Measuring x from A and considering left side segment,
F = 60.833 – 30x, linear variation
At x = 0, FA = 60.833 kN.
At x = 3 m,
FD = 60.833 – 30 × 3 = –29.167 kN.
The point of zero shear is given by
0 = 60.833 – 30x
or, x = 2.028 m.
At section X–X, the moment is given by
x2
M = 60.833 x – 30 , parabolic variation
2
At x = 0, MA = 0
At x = 3 m,
9
MD = 60.833 × 3 – 30 × = 47.5 kN-m.
2
dM
Maximum moment occurs at x = 2.028 m. since here F = =0
dx
2. 0282
Mmax = 60.833 × 2.028 – 30 × = 61.678 kN-m.
2
2 0 kN
3 0 kN /m
A C
D B
3m 3m 2m
(a ) L oa ding D ia gram
6 0.8 33
+ ve 20 20
+ ve
– ve
2 9.1 67
2 9.1 67
(b ) S F D
4 7.5
6 1.6 78 + ve 3 .37 1 m
2 .02 8 – ve
(c) B M D 40
Fig. 9.40
BEAMS 305
Portion DB:
Measuring x from the free end C and considering right hand side segment,
F = 20.0 – 49.167
= 29.167 kN, constant.
M = –20x + 49.167 (x – 2), linear variation
At x = 5 m, MD = –20 × 5 + 49.167 × 3 = 47.5 kN-m.
At x = 2 m, MB = –20 × 2 = – 40 kN-m.
In this portion the bending moment changes the sign. The point of contraflexure is given by the
expression
0 = –20x + 49.167 (x – 2)
i.e., x = 3.371 m from free end.
Portion BC:
Measuring x from free end,
F = 20 kN, constant
M = –20x, linear variation
At x = 0, MC = 0
At x = 2 m, MB = – 40 kN-m.
Hence SF and BM diagrams are as shown in Fig. 9.40(b) and 9.40(c) respectively.
Example 9.14. Draw BM and SF diagrams for the beam shown in Fig. 9.41(a), indicating the values
at all salient points.
Solution: ∑ MB = 0 →
RE × 4 + 20 × 1 – 30 × 2 × 1 – 40 × 3 – 25 × 1 × 4.5 = 0
2 0 kN 4 0 kN
3 0 kN /m
B 2 5 kN /m
A C D E
F
1m 2m 1m 1m 1m
RA (a ) L oa d D iag ra m RE
5 6.8 75
25
+ ve + ve
– ve
3 .12 5 – ve
20
4 3.1 25 4 3.1 25
(b ) S F D
3 3.7 5 3 0.6 25
3 3 .9 1 5 1 .29 m
1 .39 2 m
– ve – ve
20 1 2.5
1 .89 6 m
(c) B M D
Fig. 9.41
306 MECHANICS OF SOLIDS
RE = 68.125 kN.
∑V = 0 →
RB = 20 + 30 × 2 + 40 + 25 × 1 – 68.125
= 76.875 kN.
Portion AB:
Measuring x from A,
F = –20 kN, constant
M = –20x, linear variation
At x = 0, MA = 0
At x = 1 m, MB = –20 kN-m
Portion BC:
Measuring x from B,
F = –20 + 76.875 – 30x, linear variation
At x = 0, F = 56.875 kN.
At x = 2 m, F = –3.125 kN-m.
The point of zero shear force is given by
0 = –20 + 76.875 – 30x
x = 1.896 m from B.
At distance x from B the moment is given by
x
M = –20(x + 1) + 76.875x – 30x
2
= –20 + 56.875x – 15x2, parabolic variation
At x = 0, MB = –20 kN-m.
At x = 2 m, MC = –20 + 56.875 × 2 – 15 × 4
= 33.75 kN-m.
Maximum moment occurs where SF = 0. i.e., at x = 1.896 m.
∴ Mmax = –20 + 56.875 × 1.896 – 15 × 1.8962
= 33.913 kN-m.
The bending moment is changing its sign in this portion. Hence the point of contraflexure exists
in this portion. It is given by
0 = –20 + 56.875x – 15x2
∴ x = 0.392 m.
i.e., the point of contraflexure is at 1.392 m from the free end A.
Portion CD:
Measuring x from F,
Shear force = 25 × 1 – 68.125 – 40
= –3.125 kN, constant
M = –25 × 1 × (x – 0.5) + 68.125 (x – 1) – 40 ( x – 2), linear variation
At x = 3 m, MC = 33.75 kN-m.
At x = 2 m, MD = –25 × 1.5 + 68.125
= 30.625 kN-m.
BEAMS 307
Portion DE:
Measuring x from free end F,
Shear force = 25 – 68.125 = – 43.125 kN, constant
M = –25 (x – 0.5) + 68.125(x – 1), linear variation
At x = 2 m, MD = 30.625 kN-m.
At x = 1 m, ME = –12.5 kN-m.
Portion EF: M = 0, at x = 1.29 m
Measuring x from free end F,
Shear force F = 25x, linear constant
At x = 0, FF = 0
At x = 1 m, FE = 25 kN
x
M = –25x. , parabolic variation
2
At x = 0, MF = 0
At x = 1 m, ME = –12.5 kN-m.
SF and BM diagrams are as shown in Fig. 9.41(b) and 9.41(c) respectively.
Load SF BM
The following points also may be noted while drawing shear force and bending moment
diagrams:
308 MECHANICS OF SOLIDS
(a) The bending moment is maximum when shear force is zero. The location and the value of
maximum moment should always be indicated in bending moment diagram.
(b) The point of contraflexure is an important point in a BMD, hence if it exists, its location
should be indicated.
This method of drawing SFD and BMD is illustrated with two problems below:
Example 9.15. Determine the reactions and construct the shear force and bending moment diagrams
for the beam shown in Fig. 9.42(a). Mark the salient points and the values at those points.
Solution: ∑ MA = 0 →
RB × 6 + 120 – 60 × 4 – 60 × 7 = 0
∴ RB = 90 kN.
∑V = 0 → RA = 60 + 60 – 90 = 30 kN.
SFD:
In portion ADE, F = 30 kN.
In portion EB, F = 30 – 60 = –30 kN.
In portion BC, F = 60 kN.
SFD is as shown in Fig. 3.45(b).
BMD:
At A, MA = 0
Just to the left of D, M = 30 × 2 = 60 kN-m.
Just to the right of D, M = 30 × 2 – 120 = –60 kN-m.
At E, M = –60 × 3 + 90 × 2 = 0
At B, M = –60 × 1 = –60 kN-m.
At C, M=0
BMD is as shown in Fig. 9.42(c).
6 0 kN 6 0 kN
1 20 kN -m
B
A C
D E
2m 2m 2m 1m
RB
RA (a ) L oa d D iag ra m
30 30 60 60
+ ve
+ ve
– ve
30 30
(b ) S F D
60
+ ve 0
– ve – ve
60
(c) B M D 60
Fig. 9.42
BEAMS 309
Example 9.16. Draw the bending moment and shear force diagram for the beam loaded as shown
in Fig. 9.43. (a). Mark the values at the salient points Determine the point of contraflexure also.
10kN/m 15 kN 20 kN
A F
B C D
50 kN-m
RA RE E
4m 2m 2m 2m 2m
37 (a)
+
0
(b) SFD 3 –
18
38 38
68 62
+ 26
–
(b) BMD
50
Fig. 9.43
IMPORTANT FORMULAE
1. Shear force at a section of the beam = Σ all forces to the left or right of the section.
2. Bending moment at a section of the beam = Σ moments of all forces to the left or to the right of the
section, moment centre being the section.
dF dM
3. = w and =–F
dx dx
THEORY QUESTIONS
1. Determine the reaction at A and B on the overhanging beam shown in Fig. 9.44.
RA 30 kN
20 kN /m
α 45 °
A B
40 kN -m RB
3m 2m 1m 2m
Fig. 9.44
1m 4m
Fig. 9.45
[Ans. RA = 45 kN; RB = 15 kN]
BEAMS 311
3. In all the problems given below draw SFD and BMD indicating values at salient point. (See Fig.
9.46 to 9.50).
2 kN 3 kN
1 k N /m
A B C D E
2 k N -m
1m 1m 1 .5 m 1 .5 m
Fig. 9.46
A B C D E
3m 2m 2m 2m
Fig. 9.47
A B C D E F
2m 2m 2m 2m 2m
Fig. 9.48
A B C D
2m 4m 2m
Fig. 9.49
50 kN 20 kN
1 0 k N /m
2 0 k N -m
A
B C D E
1m 1m 3m 1 .5 m
Fig. 9.50
A
D B C
4m 2m 1m
Fig. 9.51
5. A bar of length ‘l’ is supported at A and B which are at distances ‘a’ from the ends as shown in
Fig. 9.52. Find the distance ‘a’ such that maximum moment is least. Using the above results find
the most economical length of railway sleeper if the rails are 1.6 m apart.
w /u n it len g th
A B E C D
a b a
l
Fig. 9.52
Under compression
Neutral
layer Neutral
axis
Under tension
(a) Sagging moment case
Under
tension Neutral
layer
Neutral
axis
Under
compression
Due to pure bending, beams sag or hog depending upon the nature of bending moment as shown
in Fig. 10.1. It can be easily observed that when beams sag, fibres in the bottom side get stretched
while fibres on the top side are compressed. In other words, the material of the beam is subjected
to tensile stresses in the bottom side and to compressive stresses in the upper side. In case of hogging
the nature of bending stress is exactly opposite, i.e., tension at top and compression at bottom. Thus
bending stress varies from compression at one edge to tension at the other edge. Hence somewhere
in between the two edges the bending stress should be zero. The layer of zero stress due to bending
is called neutral layer and the trace of neutral layer in the cross-section is called neutral axis [Refer
Fig. 10.1].
313
314 MECHANICS OF SOLIDS
10.1 ASSUMPTIONS
Theory of simple bending is developed with the following assumptions which are reasonably acceptable:
(i) The material is homogeneous and isotropic.
(ii) Modulus of elasticity is the same in tension and in compression.
(iii) Stresses are within the elastic limit.
(iv) Plane section remains plane even after deformations.
(v) The beam is initially straight and every layer of it is free to expand or contract.
(vi) The radius of curvature of bent beam is very large compared to depth of the beam.
A¢ B¢
A B
E¢ F¢
E F
y
G H H¢
G¢
C D C¢ D¢
(a) (b)
Fig. 10.2
Let after bending A, B, C, D, E, F, G and H take positions A′, B′, C′, D′, E′, F′, G′ and H′
respectively as shown in Fig. 10.2(b). Let R be the radius of curvature and φ be the angle subtended
by C′A′ and D′B′ at centre of radius of curvature. Then,
EF = E′F′, since EF is neutral axis
= Rφ ...(i)
Final length – Initial length
Strain in GH =
Initial length
STRESSES IN BEAMS 315
G′ H ′ − GH
=
GH
But GH = EF (The initial length)
= Rφ
and G′H′ = (R + y) φ
( R + y ) φ − Rφ
∴ Strain in layer GH =
Rφ
y
= ...(ii)
R
Since strain in GH is due to tensile forces, strain in GH = f/E ...(iii)
where f is tensile stress and E is modulus of elasticity.
From eqns. (ii) and (iii), we get
f y
=
E R
f E
or = ...(10.1)
y R
(ii) Relationship between bending moment and radius of curvature: Consider an elemental area
δa at distance y from neutral axis as shown in Fig. 10.3.
a
Fig. 10.3
∑
E 2
M′ = y δa
R
∑
E
= y 2 δa
R
From the definition of moment of inertia (second moment of area) about centroidal axis, we
know
I = Σy2 δa
E
∴ M′ = I
R
From equilibrium condition, M = M′ where M is applied moment.
E
∴ M = I
R
M E
or = ...(10.2)
I R
From eqns. (10.1) and (10.2), we get
M f E
= = ...(10.3)
I y R
where M = bending moment at the section
I = moment of inertia about centroid axis
f = bending stress
y = distance of the fibre from neutral axis
E = modulus of elasticity and
R = radius of curvature of bent section.
Equation (10.3) is known as bending equation.
∑
E
= y δa
R
STRESSES IN BEAMS 317
E
= Σy δa
R
Since there is no other horizontal force, equilibrium condition of horizontal forces gives
E
Σy δa = 0
R
E
As is not zero,
R
Σy δa = 0 ...(i)
If A is total area of cross-section, from eqn. (i), we get
y δa
∑ A
= 0 ...(ii)
Σyδa
Noting that Σyδa is the moment of area about neutral axis, should be the distance of
A
Σyδa
centroid of the area from the neutral axis. Hence = 0 means the neutral axis coincides with
A
the centroid of the cross-section.
I
Z = ...(10.4)
ymax
and M = fper Z ...(10.5)
Note : If moment of inertia has unit mm4 and ymax has mm, Z has the unit mm3.
The eqn. (10.5) gives permissible maximum moment on the section and is known as moment
carrying capacity of the section. Since there is definite relation between bending moment and the
loading given for a beam it is possible to find the load carrying capacity of the beam by equating
maximum moment in the beam to moment carrying capacity of the section. Thus
Mmax = fper Z ...(10.6)
If permissible stresses in tension and compressions are different for a material, moment carrying
capacity in tension and compression should be found separately and equated to maximum values of
moment creating tension and compression separately to find the load carrying capacity. The lower
of the two values obtained should be reported as the load carrying capacity.
ymax = d/2
1
I= bd 3 ymax d/2
12
G
I 1/12 bd 3 N A
∴ Z = =
ymax d /2
d/2
i.e., Z = 1/6 bd2 ...(10.7)
Fig. 10.4
ymax = D/2
D/2
I 1 ( BD3 − bd 3 )
∴ Z= =
ymax 12 D/2
1 BD 3 − bd 3 Fig. 10.5
i.e. Z= ...(10.8)
6 D
STRESSES IN BEAMS 319
(iii) Circular section of diameter ‘d’. Typical section is shown in Fig. 10.6. For this,
π 4
I= d ymax = d/2
64 d
ymax = d/2 N A
G
I π / 64 d 4
∴ Z= =
ymax d /2
Fig. 10.6
π 3
i.e., Z= d
32
(iv) Hollow circular tube of uniform section. Referring to Fig. 10.7,
π 4 π 4
I = D − d ymax
64 64
G
π N d D
= (D4 – d4) A
64
ymax = D/2
I π ( D4 − d 4 )
∴ Z = = Fig. 10.7
ymax 64 D/ 2
π D4 − d 4
i.e., Z = ...(10.9)
32 D
(v) Triangular section of base width b and height ‘h’. Referring to Fig. 10.8,
bh 3
I =
36
ymax = 2h/3
2
ymax = h
3 h
G A
I bh 3 / 36 N
∴ Z = =
ymax 2/3 h
bh 2
i.e., Z = ...(10.10) b
24
Fig. 10.8
Example 10.1. A simply supported beam of span 3.0 m has a cross-section 120 mm × 180 mm. If
the permissible stress in the material of the beam is 10 N/mm2, determine
(i) maximum udl it can carry
(ii) maximum concentrated load at a point 1 m from support it can carry.
Neglect moment due to self weight.
320 MECHANICS OF SOLIDS
Solution:
1 d
Here b = 120 mm, d = 180 mm, I = bd3, ymax =
12 2
1 2
∴ Z = bd
6
1
= × 120 × 180 2 = 648000 mm3
6
fper = 10 N/mm2
∴ Moment carrying capacity of the section
= fper × Z
= 10 × 648000 N-mm
(i) Let maximum udl beam can carry be w/metre length as shown in Fig. 10.9.
wL2
In this case, we know that maximum moment occurs at mid span and is equal to Mmax = .
8
Equating it to moment carrying capacity, we get,
w/m
w × 32
× 106 = 10 × 648000
8
∴ w = 5.76 kN/m.
3m
(ii) Concentrated load at distance 1 m from the sup-
port be P kN. Referring to Fig. 10.10. Fig. 10.9
P × a × b P ×1× 2
Mmax = = P
L 3
2P
= kN-m
3 a = 1m b=2m
2P L=3m
= × 10 6 N-mm
3 Fig. 10.10
Equating it to moment carrying capacity, we get
2P
× 10 6 = 10 × 648000
3
∴ P = 9.72 kN-m.
Example 10.2. A circular steel pipe of external diameter 60 mm and thickness 8 mm is used as a
simply supported beam over an effective span of 2 m. If permissible stress in steel is 150 N/mm2,
determine the maximum concentrated load that can be carried by it at mid span.
Solution:
External diameter D = 60 mm
Thickness = 8 mm
STRESSES IN BEAMS 321
8 mm
P=?
60 mm
2m
(a) (b)
Fig. 10.11
I 452188
∴ Z = = = 15073 mm3.
y max 30
P×2
= kN-m
4
= 0.5 P ×106 N-mm.
Equating maximum bending moment to moment carrying capacity, we get
0.5P × 106 = 150 × 15073
∴ P = 4.52 kN.
Example 10.3: Figure 10.12 (a) shows the cross-section of a cantilever beam of 2.5 m span. Material
used is steel for which maximum permissible stress is 150 N/mm2. What is the maximum uniformly
distributed load this beam can carry?
Solution: Since it is a symmetric section, centroid is at mid depth.
I = MI of 3 rectangles about centroid
322 MECHANICS OF SOLIDS
180 mm 10 mm
400 mm
10 mm
w/m = ?
2m
10 mm
(a) (b)
Fig. 10.12
1
= × 180 × 10 3 + 180 × 10 (200 – 5)2
12
1
+ × 10 × ( 400 − 20) 3 + 10 × (400 – 20) × 02
12
1
+ × 180 × 103 + 180 × 10 (200 – 5)2
12
= 182.6467 × 106 mm4
[Note: Moment of above section may be calculated as difference between MI of rectangle of size 180 × 400
and 170 × 380. i.e.,
1 1
I = × 180 × 4003 – × 170 × 380 3
12 12
ymax = 200 mm.
I 182.6467 × 10 6
∴ Z= = = 913233 mm3.
y max 200
∴ Moment carrying capacity
= fper × Z
= 180 × 913233
= 136985000 N-mm.
If udl is w kN/m, maximum moment in cantilever
= wL = 2w kN-mm
= 2w × 106 N-mm
Equating maximum moment to movement carrying capacity of the section, we get
2w × 106 = 136985000
∴ w = 68.49 kN/m
STRESSES IN BEAMS 323
Example 10.4. Compare the moment carrying capacity of the section given in example 10.3 with
equivalent section of the same area but
(i) square section
(ii) rectangular section with depth twice the width and
(iii) a circular section.
Solution:
Area of the section = 180 × 10 + 380 × 10 + 180 × 10
= 7400 mm2
(i) Square section
If ‘a’ is the size of the equivalent square section,
a2 = 7400 ∴ a = 86.023 mm.
Moment of inertia of this section
1
= × 86.023 × 86.0233
12
= 4563333 mm4
I 4563333
Z= = = 106095.6 mm3
y max 86.023 / 2
Moment carrying capacity = f Z = 150 × 106095.6
= 15.914 × 106 N-mm
Moment carrying capacity of I section 136985000
∴ =
Moment carrying capacity of equivalent square section 15.914 × 10 6
= 8.607.
(ii) Equivalent rectangular section of depth twice the width.
Let b be the width
∴ Depth d = 2b.
Equating its area to area of I-section, we get
b × 2b = 7400
b = 60.8276 mm
ymax = d/2 = b = 60.8276
I 1 b × (2 b ) 3
M=f = 150 × ×
y max 12 b
8 3 8
= 150 × b = 150 × × 60.82763
12 12
= 22506193 N-mm.
Moment carrying capacity of I section 136985000
∴ = = 6.086.
Moment carrying capacity of this section 22506193
324 MECHANICS OF SOLIDS
240 mm
10 mm
8 mm thick
400 mm
180 mm
Fig. 10.13
Let centroid of the section be at a distance y from the bottom most fibre. Then
A y = 240 × 10 × 405 + 180 × 8 × (400 – 4) + 384 × 8 × 200 + 180 × 8 × 4
STRESSES IN BEAMS 325
75 mm fc
50 mm
25
100 mm
–
y
50 mm
ft
150 mm
(a) (b)
Fig. 10.14
1
∴ I = × 75 × 503 + 75 × 50 (175 – 79.54)2
12
1
+ × 25 × 1003 + 25 × 100 (100 – 79.54)2
12
1
+ × 150 × 503 + 150 × 50 (25 – 79.54)2
12
= 61.955493 × 106 mm4.
Extreme fibre distances are
ybottom = y = 79.54 mm.
ytop = 200 – y = 200 – 79.54 = 120.46 mm.
Top fibres are in compression. Hence from consideration of compression strength, moment
carrying capacity of the beam is given by
I
M1 = fper in compression ×
ytop
61.955493 × 10 6
= 90 ×
120.46
= 46.289178 × 106 N-mm
= 46.289178 kN-m.
Bottom fibres are in tension. Hence from consideration of tension, moment carrying capacity of
the section is given by
I
M2 = fper in tension ×
ybottom
STRESSES IN BEAMS 327
30 × 61.955493 × 10 6
=
79.54
= 21.367674 × 106 N-mm
= 21.367674 kN-m.
Actual moment carrying capacity is the lower value of the above two values. Hence moment
carrying capacity of the section is
= 21.367674 kN-m.
Maximum moment in a simply supported beam subjected to udl of w/unit length and span L is
wL2
=
8
Equating maximum moment to moment carrying capacity of the section, we get maximum load
carrying capacity of the beam as
52
w× = 21.367674
8
∴ w = 6.838 kN/m.
Example 10.7. The diameter of a concrete flag post varies from 240 mm at base to 120 mm at top
as shown in Fig. 10.15. The height of the post is 10 m. If the post is subjected to a horizontal force
of 600 N at top, find the section at which stress is maximum. Find its value also.
Solution: Consider a section y metres from top. Diameter at this section is
y 120 mm
d = 120 + (240 – 120)
10
= 120 + 12y mm
ym
π 4
∴ I= d
64
I π 3
Z = = d
d /2 32 10 m
π
= [120 + 12y]3
32
At this section, moment is given by
M = 600 y N-m
= 600000 y N-mm.
Equating moment of resistance to moment at the section, we get 240 mm
f Z = M Fig. 10.15
where f is extreme fibre stress
df
For ‘f ’ to be maximum, = 0
dy
600000 × 32 [(120 + 12y)–3 + y(– 3) (120 + 12y)–4 × 12] = 0
i.e., (120 + 12y)–3 = 36 (120 + 12y)–4 y
i.e., 1 = 36 (120 + 12y)–1 y
i.e., 120 + 12y = 36y
∴ y = 5 m.
Stress at this section f is given by
5
f = 600000 × 32 ×
π (120 + 12 × 5)3
f = 5.24 N/mm2.
Example 10.8. A circular log of timber has diameter D. Find the dimensions of the strongest
rectangular section one can cut from this.
Solution: Let the width and depth of strongest section that can be cut from
the log be ‘b’ and ‘d’ respectively. Then,
D 2 = b2 + d2
or d2 = D2 – b2
For rectangular section
d
D
1
I = bd 3
12
ymax = d/2.
b
I 1
∴ z = = bd2
ymax 6 Fig. 10.16
1 1
= b (D2 – b2) = (bD2 – b3)
6 6
The beam is strongest if section modulus is maximum. Hence the condition is
dz
= 0
db
1
[D2 – 3b2] = 0
6
i.e., D2 = 3b2
D
or b = .
3
D2
∴ d = (D2 − b2 ) = D2 − = D 2 /3
3
Thus the dimensions of strongest beam
D
= wide × 2/3 D deep.
3
STRESSES IN BEAMS 329
Consider an elemental length ‘δx’ of beam shown in Fig. 10.17 (a). Let bending moment at section
A-A be M and that at section B-B be M + δM. Let CD be an elemental fibre at distance y from neutral
axis and its thickness be δy. Then,
Bending stress on left side of elemental fibre
M
= y
I
M M + M
A B b
C D y yt y
y y
A B
x x
(a) (b)
C D
M yb dy q M + M yb dy
I I
C D
(c)
Fig. 10.17
= z y
yt dM
I
y b δy
=
y
dM
I z yt
y b dy =
δM yt
I y
yb dy z
Let intensity of shearing stress on element CD be q. [Refer Fig. 10.17 (c)]. Then equating
resisting shearing force to unbalanced horizontal force, we get
q b δx =
δM
I zy
yt
yb dy
∴ q =
δM 1
δx bI z yt
y
yb dy
dM 1
As δx → 0, q = (a y )
dx bI
where a y = Moment of area above the section under consideration about neutral axis.
dM
But we know = F
dx
F
∴ (a y )
q = ...(10.11)
bI
The above expression gives shear stress at any fibre y distance above neutral axis.
Variation of shear stresses across the following three cases are discussed below:
(i) Rectangular
(ii) Circular and
(iii) Isosceles triangle.
(i) Rectangular section. Consider the rectangular section of width ‘b’ and depth shown in Fig.
10.18(a). Let A-A be the fibre at a distance y from neutral axis. Let the shear force on the
section be F.
Parabolic
A variation
d/2 A
y
b
(a) (b)
Fig. 10.18
STRESSES IN BEAMS 333
F b 2
∴ q = (d /4 – y2)
1 3 2
b bd
12
6F
= (d2/4 – y2)
bd 3
This shows shear stress varies parabolically.
When y = ± d/2, q = 0
6F d 2 F
At y = 0, qmax = 3 = 1.5
bd 4 bd
= 1.5 qav
F
where qav = is average shear stress.
bd
Thus in rectangular section maximum shear stress is at neutral axis and it is 1.5 times average
shear stress. It varies parabolically from zero at extreme fibres to 1.5 qav at mid depth as shown in
Fig. 10.18(b).
(ii) Circular section. Consider a circular section of diameter ‘d’ as shown in Fig. 10.19 (a) on
which a shear force F is acting. Let A-A be the section at distance ‘y’ from neutral axis at
which shear stress is to be found. To find moment of area of the portion above A-A about
neutral axis, let us consider an element at distance ‘z’ from neutral axis. Let its thickness be
dz. Let it be at an angular distance φ and A-A be at angular distance θ as shown in figure.
334 MECHANICS OF SOLIDS
b/2 b/2
dz Parabolic
variation
d/2
A A
Z y
d
(a) (b)
Fig. 10.19
d
Width of element b = 2. cos φ
2
= d cos φ
d
z = sin φ
2
d
∴ dz = cos φ dφ
2
∴ Area of the element
d
a = bdz = d cos φ . cos φ dφ
2
d2
= cos2 φ dφ
2
Moment of this area about neutral axis
= area × z
d2 d
= cos2 φ dφ sin φ
2 2
d3
=cos2 φ sin φ dφ
4
∴ Moment of area about section A-A about neutral axis
(a y ) =
z
θ
π/ 2 d2
4
cos2 φ sin φ dφ
d 3 − cos 3 φLM OP π/ 2
N Q
=
4 3 θ
d3 LM− cos π OP
∴ (a y ) =
N + cos 3 θ
Q
2
4×3 2
d
= cos3 θ
12
πd 4
Now I=
64
F
∴ q = (a y )
bI
F d3
= × cos3 θ
π 4 12
d cos θ d
64
= 64 F 2 cos2 θ
12 πd
16 F
= [1 – sin2 θ]
3 πd 2
LM1 − F y I 2
OP
MN GH d / 2 JK
16 F
PQ
=
3 πd 2
16 F LM1 − 4 y OP2
=
3 πd 2 N d Q 2
(iii) Isosceles triangular section. Consider the isosceles triangular section of width ‘b’ and
2h
height ‘h’ as shown in Fig. 10.20(a). Its centroid and hence neutral axis is at from top
3
fibre. Now shear stress is to be found at section A-A which is at a depth ‘y’ from top fibre.
2y/3
y h/2
2h/3 2h/3
g
h
qmax = 1.5qav
G
qcentroid = 4/3qav
h/3
(a) (b)
Fig. 10.20
y
At A-A width b′ = b
h
1
Area above A-A a= b′y
2
1 b 2
= y
2 h
2y
Its centroid from top fibre is at .
3
2h 2 y
∴ Distance of shaded area above the section A-A from neutral axis y = − .
3 3
1 b 2 F 2h − 2 y I
∴ ay =
2 h
y
H 3 3K
1b 2
= y (h – y)
3h
Moment of inertia of the section
bh 3
I = .
36
STRESSES IN BEAMS 337
Example 10.12. Draw the shear stress variation diagram for the I-section shown in Fig. 10.21(a)
if it is subjected to a shear force of 100 kN.
180 mm
19.217
10 mm 1.068
400 mm 29.10
10 mm
10 mm
80 mm
(a) (b)
Fig. 10.21
100 × 1000
= × (180 × 10 × 195)
180 × 182.646666 × 10 6
= 1.068 N/mm2
Shear stress in the web at the junction with flange
100 × 1000
= (180 × 10 × 195)
10 × 182.646666 × 10 6
= 19.217 N/mm2
Shear stress at N-A
120 mm
2
2.9 N/mm
2
29 N/mm
12 mm
2
31.17 N/mm
120 mm
12 mm
(a) (b)
Fig. 10.22
Solution: Let y be the distance of centroid of the section from its top fibre. Then
Moment of area about top fibre
yt =
Total area
F 120 − 12 I
=
H
120 × 12 × 6 + (120 − 12) × 12 × 12 +
2 K
120 × 12 + (120 − 12) × 12
= 34.42 mm
340 MECHANICS OF SOLIDS
F 108 I 2
H K
1
+ × 12 × 1083 + 12 × 108 34.42 −
12 2
= 2936930 mm4
Shear stresses are zero at extreme fibres.
Shear stress at bottom of flange:
Area above this level, a = 120 × 12 = 1440 mm2
Centroid of this area above N-A
y = 34.42 – 6 = 28.42 mm
= 2.90 N/mm2
Shear stress at the same level but in web, where width b = 12 mm
25 × 1000
= × 1440 × 28.42
12 × 2936930
= 29.0 N/mm2
Shear stress at neutral axis:
For this we can consider a y term above this section or below this section. It is convenient to
consider the term below this level.
a = 12 × (120 – 34.42) = 1026.96 mm2
The distance of its centroid from N-A
120 − 34.42
= = 42.79 mm.
2
Width at this section b = 12 mm.
25 × 1000
∴ q = × 1026.96 × 42.79
12 × 2936930
= 31.17 N/mm2
Hence variation of shear stress across the depth is as shown in Fig. 10.22(b).
STRESSES IN BEAMS 341
Example 10.14. The unsymmetric I-section shown in Fig. 10.23(a) is the cross-section of a beam,
which is subjected to a shear force of 60 kN. Draw the shear stress variation diagram across the
depth.
100 mm 2
2.61 N/mm 2
20 13.03 N/mm
yt 20 mm
2
160 18.37 N/mm
200 mm
2
20 2 15.24 N/mm
2.04 N/mm
150 mm
(a) (b)
Fig. 10.23
Solution: Distance of neutral axis (centroid) of the section from top fibre be yt. Then
F
160 I
100 × 20 × 10 + (200 − 20 − 20) × 20 × 20 +
2H K
+ 150 × 20 × (200 − 10)
yt =
100 × 20 + 160 × 20 + 150 × 20
= 111 mm
I = 1 × 100 × 203 + 100 × 20 (111 – 10)2
12
+ 1 × 20 × 1603 + 160 × 20 (111 – 100)2
12
+ 1 × 150 × 203 + 150 × 20 (111 – 190)2
12
= 46505533 mm4
Shear stress at bottom of top flange
F
= ay
bI
60 × 1000
= × 100 × 20 × (111 – 10)
100 × 46505533
= 2.61 N/mm2
∴ Shear stress at the same level, but in web
60 × 1000
= × 100 × 20 (111 – 10)
20 × 46505533
= 13.03 N/mm2
342 MECHANICS OF SOLIDS
IMPORTANT FORMULAE
M f E
1. Bending equation: = = .
I y R
I
2. Modulus of section Z= .
ymax
3. Moment carrying capacity of section = fper Z.
4. Section modulus of various sections:
1 1 BD3 − bd 3
(i) Rectangular: bd2 (ii) Hollow rectangular:
6 6 D
π 3
(iv) Hollow circular section: π D − d
4 4
(iii) Solid circular section: d
32 32 D
bh 2
(v) Solid triangular section:
24
STRESSES IN BEAMS 343
F
5. Shear stress in a beam q = ( ay )
bI
6. In rectangular sections,
qmax = 1.5 qav, at y = d/2
4
In circular sections qmax = q , at centre
3 av
h
In triangular section, qmax = 1.5 qav, at y = .
2
THEORY QUESTIONS
1. A I-section has flanges of size 200 × 12 mm and its overall depth is 360 mm. Thickness of web
is also 12 mm. It is used as a simply supported beam over a span of 4 m to carry a load of
60 kN/m over its entire span. Draw the variation of bending and shearing stresses across the depth.
2
117.8 N/mm 1.37 N/mm
2
2
22.78 N/mm
2
32.02 N/mm
2
117.8 N/mm
(a) Variation of bending stress (b) Variation of shear stress
Fig. 10.24
344 MECHANICS OF SOLIDS
2. Unsymmetric I-section shown in Fig. 10.25 is used as a cantilever of span 2 m to carry uni-
formly distributed load of 6 kN/m over entire span. Draw the variation of bending stress across
the depth marking the values at salient point.
120 mm 2
33.56 N/mm
10 mm 2
31.04 N/mm
133.27
240 mm
10 mm
2
23.10 N/mm
10 mm
2
180 mm 26.88 N/mm
(a) (b)
Fig. 10.25
3. Calculate the variation of shear stresses at various salient level near fixed support in the above
problem and give the sketch. [Ans. See Fig. 10.26]
2
3.87 N/mm
2
0.32 N/mm
133.27
mm
2
5.79 N/mm
2
0.26 N/mm
2
4.61 N/mm
Fig. 10.26
4. A cantilever beam of 1.2 m span is having cross-section as shown in Fig. 10.27. The permissible
stresses in tension and compressions are 20 N/mm2 and 80 N/mm2 respectively. Determine the
maximum concentrated load W it can carry at the free end. [Ans. W = 44.576 kN]
[Note: In cantilever tension is at top.]
STRESSES IN BEAMS 345
60 60 60
120
120
60
180
All dimensions in mm
Fig. 10.27
5. A timber beam is to be designed to carry a load of 6 kN/m over a simply supported span of 5 m.
Permissible stress is 10 N/mm2. Keeping the depth twice the width, design the beam. If the
permissible stress in shear is 1 N/mm2, check for shear.
[Ans. Required b = 141.18, d = 282.31. Hence select 150 × 300 mm section]
11
Principal Stresses and Strains
A structural member need not be always under simple (only one type) of stress. It may be subjected
to direct stresses in different directions and may be subjected to shear stresses also. A beam is usually
subjected to axial stresses due to bending and also for shear stresses. A shaft is subjected to shear
stresses due to torsion and axial stresses due to bending/direct thrust. The stresses may vary from
point to point in the member.
In a three-dimensional system, stresses acting at a point may be represented as shown in Fig. 11.1.
py py
q
q
qyx
qyz qxy
qzy px px px
qzx qxz
pz q
q
py
In many engineering problems, two dimensional idealization is made as shown in Fig. 11.2. In
this chapter discussion is limited to two dimensional stress system. First general expression for
stresses on a plane inclined at a selected axis is discussed. Then the terms principal stresses and
planes are explained and the expressions to get them are presented. A number of problems are solved
to make the concept clear. At the end analysis is given for principal strains also.
346
PRINCIPAL STRESSES AND STRAINS 347
py
py
E q
B q q E C
C pt
px px px
O θ pn θ
A
q D D
py
(a) (b)
Fig. 11.3
Now we are interested in finding state of stress on plane DE which makes anticlockwise angle
θ with the plane of stress px, in other words with y-axis.
For simplicity let us consider thickness of the element as unity. We are interested in finding
normal and tangential stresses acting on the plane DE. Let normal stress be pn and tangential stress
pt as shown in Fig. 11.3(b). Since the system is in equilibrium,
Σ Forces normal to DE = 0 gives
pn × DE × 1 = px × CD × 1 × cos θ + q × CD × 1 × sin θ
+ py × CE × 1 × sin θ + q × CE × 1 × cos θ
CD CE CD CE
∴ pn = p x cos θ + py sin θ + q sin θ + q cos θ
DE DE DE DE
CD CE
Since = cos θ and = sin θ, we get
DE DE
pn = px cos2 θ + py sin2 θ + q cos θ sin θ + q sin θ cos θ
F 1 + cos 2θ I + p F 1 − cos 2θ I + 2q sin θ cos θ ,
= px
H 2 K H 2 K y
1 + cos 2θ 1 − cos 2θ
since cos2 θ = and sin2 θ =
2 2
px + py px − py
Thus pn = +cos 2 θ + q sin 2 θ ...(11.1)
2 2
Similarly from equilibrium condition of forces tangential to plane DE, we get
pt × DE × 1 = px × CD × 1 × sin θ – q × CD × 1 × cos θ
– py × CE × 1 × cos θ + q × CE × 1 × sin θ
CD CE CD CE
∴ pt = p x sin θ − p y cos θ − q cos θ + q sin θ
DE DE DE DE
CD CE
But = cos θ and = sin θ.
DE DE
∴ pt = px sin θ cos θ – py sin θ cos θ – q cos2 θ + q sin2 θ
348 MECHANICS OF SOLIDS
R= pn 2 + pt 2
which is inclined at ‘α’ to the plane (Ref. Fig. 11.3c).
pn
tan α =
pt
In other words, the resultant is inclined at θ + α to the plane of px .
py
pt
E Cq
R α px
pn
Fig. 11.3(c)
q 2q
∴ tan 2θ = = ...(11.3)
( px − py )/2 px − py
There are two values for 2θ which differ by 180° for which
2
eqn. (11.3) can be satisfied. Let 2θ1 and 2θ2 be the solution. 2 +4
q
Referring to Fig. 11.4, we find – p y)
(p x 2q
2q
sin 2θ1 = ...(11.4a)
( px − p y ) 2 + 4q 2 2q
px – py
px − py
cos 2θ1 = ...(11.4b)
Fig. 11.4
( px − p y ) 2 + 4q 2
Similarly,
− 2q
sin 2θ2 = ...(11.5a)
( px − p y ) 2 + 4q 2
− ( px − p y )
and cos 2θ2 = ...(11.5b)
( px − p y ) 2 + 4q 2
2θ1 and 2θ2 differ by 180°. Hence we can say θ1 and θ2 differ by 90°. Thus direction of principal
planes to the plane of pn are given by eqn. (11.4). Another principal plane is at right angles to it.
Principal Stresses
Principal stresses are the normal stresses on principal planes. Hence the values of principal stresses
may be obtained by substitutes θ1 and θ2 values for θ in the expression for pn. Denoting the values
as p1 and p2, we get
p1 = pn at θ = θ1
px + p y px − p y
= + cos 2θ1 + q sin 2θ1
2 2
px + py p x − py px − py 2q
= + +q
2 2 ( px − p y ) 2 + 4 q 2 ( p x − py ) 2 + 4q 2
px + py 1 ( p x − py ) + 4q
2 2
= +
2 2 ( px − p y ) 2 + 4q 2
px + py 1
= + ( px − py ) 2 + 4 q 2
2 2
px + p y FG p − py IJ 2
H K
x
= + + q2
2 2
and p2 = pn at θ = θ2
px + p y p x − py
= + cos 2θ2 + q sin 2θ2
2 2
px + py p x − py [ − ( p x − p y )] q(− 2 q)
= + +
2 2 ( px − p y ) + 4 q
2 2
( p x − py )2 + 4q 2
350 MECHANICS OF SOLIDS
px + py 1 ( px − p y ) + 4q
2 2
= −
2 2 ( px − py ) 2 + 4q 2
px + py 1
= − ( p x − p y ) 2 + 4q 2
2 2
px + p y FG p − py IJ 2
H K
x
= − + q2
2 2
Thus the principal stresses are given by
px + p y
+ G
F p − p IJ + q 2
H 2 K
x y 2
p1 = ...(11.6a)
2
p +p
− G
F p − p IJ + q 2
H 2 K
x y x y 2
and p2 = ...(11.6b)
2
It can be proved that principal stresses are maximum and minimum stresses also. To find
extreme value of normal stress pn,
dpn
= 0
dθ
px − py
i.e., 2(– sin 2θ) + q 2 cos 2θ = 0
2
px − py
i.e., sin 2θ – q cos 2θ = 0
2
i.e., pt = 0
Thus the principal planes are the planes of maximum/minimum normal stresses also. Plane
corresponding to θ1 gives the maximum value while plane corresponding to θ2 gives minimum
normal stress.
Plane of Maximum Shear Stress
For maximum shear
dpt
=0
dθ θ = θ′
px − py
i.e., . 2 cos 2θ′ – q 2 (– sin 2θ′) = 0
2
− ( px − py )
or tan 2θ′ = ...(11.7)
2q
From eqn. (11.3) and eqn. (11.7), we get
tan 2θ × tan 2θ′ = –1
Hence 2θ′ and 2θ values differ by 90°. In other word planes of extreme shearing stresses are
90
at = 45° to the principal planes.
2
PRINCIPAL STRESSES AND STRAINS 351
Value of maximum shearing stress may be obtained by substituting θ′ for θ in equation for pt
(eqn. 11.2)
− ( px − py )
Now tan 2θ′ = q
2
2q 2 +4
– p y)
From Fig. 11.5, we get (p x (px – py)
( px − py ) 2q¢
sin 2θ′ =
( px − p y ) 2 + 4q 2 – 2q
Fig. 11.5
− 2q
cos 2θ′ =
( px − p y ) 2 + 4q 2
px − py
∴ pt (max) = sin 2θ′ – q cos 2θ′
2
px − py px − py q( − 2 q )
= −
2 ( p x − p y ) + 4q
2 2
( px − py ) 2 + 4q 2
1 ( p x − py ) + 4q
2 2
=
2 ( p x − py ) 2 + 4q 2
1
= ( px − py )2 + 4 q 2
2
FG p − py IJ 2
H K
x
= + q2 ...(11.8a)
2
1
=
the difference between p1 and p2
2
p − p2
= 1 ...(11.8b)
2
The direction of principal planes and the planes of maximum shearing stresses may be indicated
as shown in Fig. 11.6(a) or 11.6(b).
p1 = plane y y
p2
p1
q
ma
x pla
ne
45° q q
q
45° x x
45°
p2-plane
ne
pla
x
ma
q
(a) (b)
Fig. 11.6
352 MECHANICS OF SOLIDS
Example 11.1. Show that sum of normal stresses in any two mutually perpendicular directions is
constant in case of a general two dimensional stress.
Solution: Let px and py as shown in Fig. 11.3(a) be normal stresses and q be the shearing stresses.
Let the thickness of element be unity.
Taking element as shown in Fig. 11.3(b) and considering the equation of equilibrium in the
direction normal to the plane DE, we get
pn × 1 × DE = px × 1 × CD cos θ + py × 1 × CE sin θ
+ q × 1 × CD sin θ + q × 1 × CE cos θ
CD CE CD CE
∴ pn = p x cos θ + py sin θ + q sin θ + q cos θ
DE DE DE DE
CD CE
Noting that = cos θ and = sin θ, we get
DE DE
pn = px cos2 θ + py sin2 θ + q cos θ sin θ + q sin θ cos θ
F 1 + cos 2θ I + p F 1 − cos 2θ I
i.e., pn = p x
H 2 K H 2 K y + 2q sin θ cos θ
px + p y px − p y
= + cos 2θ + q sin 2θ ...(1)
2 2
If pn′ is the stress on a plane at right angles to CD, θ′ = θ + 90°. Hence from the above general
expression for stress on an inclined plane, we get
px + p y px − p y
pn ′ = + cos 2(θ + 90) + q sin 2(θ + 90)
2 2
px + p y p x − py
= − cos 2θ – q sin 2θ ...(2)
2 2
Adding eqns. (1) and (2), we get
pn + pn′ = px + py
Thus sum of normal stresses in any two mutually perpendicular planes is constant and is equal
to px + py .
Example 11.2. A material has strength in tension, compression and shear as 30N/mm2, 90 N/mm2
and 25 N/mm2, respectively. If a specimen of diameter 25 mm is tested in tension and compression
identity the failure surfaces and loads.
Solution:
(a) In tension: Let axial direction be x direction.
Since it is uniaxial loading,
py = 0, q = 0 and only px exists.
∴ when the material is subjected to full tensile stress, px = 30 N/mm2.
30 + 0 F 30 − 0 I 2
p1 =
2
+
H 2 K + 0 2 = 30 N/mm2
PRINCIPAL STRESSES AND STRAINS 353
P
30 + 0 F 30 − 0 I 2
p2 =
2
−
H 2 K + 02 = 0
px − p y30 − 0
qmax = =
= 15 N/mm2 < 25 N/mm2.
2 2
Hence failure criteria is normal stress p1 = 30 N/mm2.
Plane of p1
Corresponding load P is obtained by
P
p =
A
P
30 = or P = 14726 N.
( π / 4) × 252 P
Failure surface is given by plane of p1 which is as shown in Fig. 11.7. Fig. 11.7
(b) In case of compression test
px = – 90 N/mm2, py = q = 0
∴ At failure when px = – 90 N/mm2, load is
P
– 90 =
( π 4) × 252
or P = – 44178 N i.e., 44178 N compressive.
FG p − p IJ + q 2
H 2 K
x y 2
At this stage qmax =
F − 90 − 0 I + 0 2
H 2 K
2
=
FG p − py IJ 2
F p − 0I 2
H K H 2 K
x
qmax = + q2 = x
+ 02
2 Failure plane
45°
px
=
2
∴ px = 2qmax = 2 × 25 = 50 N/mm2.
∴ Corresponding axial load is given by
P
50 = px
( π 4) × 252
Fig. 11.8
or P = 24543 N
The plane of qmax is at 45° to the plane of px . This failure surface is shown in Fig. 11.8.
354 MECHANICS OF SOLIDS
Example 11.3. The direct stresses at a point in the strained material are 120 N/mm2 compressive
and 80 N/mm2 tensile as shown in Fig. 11.9. There is no shear stress. Find the normal and tangential
stresses on the plane AC. Also find the resultant stress on AC.
2
120 N/mm
D C R
pn pt C
a
2 2 2
80 N/mm 80 N/mm 80 N/mm
30°
A B A B
2 2
120 N/mm 120 N/mm
(a) (b)
Fig. 11.9
Solution: The plane AC makes 30° (anticlockwise) to the plane of px (y-axis). Hence θ = 30°.
px = 80 N/mm2 py = – 120 N/mm2 q = 0
px + p y px − p y
∴ pn = + cos 2θ + q sin 2θ
2 2
80 − 120 80 − ( − 120 )
= + cos (2 × 30) + 0
2 2
= – 20 + 100 cos 60
Thus pn = 30 N/mm2
px − py
pt = sin 2θ – q cos 2θ
2
80 − ( − 120)
= sin (2 × 30 ) − 0
2
Thus pt = 86.6 N/mm2.
The resultant of pn and pt is given by
p= pn 2 + pt 2 = 30 2 + 86.6 2
p = 91.65 N/mm2
pn 30
Angle made by the resultant stress with pt is given by tan α = =
pt 86.6
∴ α = 19.1° as shown in Fig. 11.9(b).
PRINCIPAL STRESSES AND STRAINS 355
Example 11.4. The state of stress at a point in a strained material is as shown in Fig. 11.10. Determine
(i) the direction of principal planes
(ii) the magnitude of principal stresses and
(iii) the magnitude of maximum shear stress.
Indicate the direction of all the above by a sketch.
2
150 N/mm
2
100 N/mm
2
200 N/mm
Fig. 11.10
px + p y FG p − p IJ + q 2
H 2 K
x y
p =
1
+ 2
2
+ F
200 + 150 200 − 150 I 2
H 2 K + 100
2
=
2
= 175 + 103.08
p1 = 278.08 N/mm2
px + p y FG p − py IJ 2
H K
x
and p2 = − + q2
2 2
∴ p2 = 175 – 103.08 = 71.92 N/mm2.
FG p − py IJ 2
H K
x
qmax = + q2
2
i.e., qmax = 103.08 N/mm2
356 MECHANICS OF SOLIDS
The planes of maximum shear are at 45° to the principal planes. Principal planes and planes of
maximum shear are shown in Fig. 11.11.
p1-plane y
qmax plane
45° q
x
45°
45°
p2 - plane
qmax plane
Fig. 11.11
Example 11.5. The state of stress in a material stressed to two-dimensional state of stress is as
shown in Fig. 11.12. Determine principal stresses and maximum shear stress and the planes on
which they act.
2
60 N/mm
2
20 N/mm
2
80 N/mm
Fig. 11.12
Solution:
px + p y FG p − py IJ 2
H K
x
p1, 2 = ± + q2
2 2
In this problem,
px = 80 N/mm2 py = – 60 N/mm2 q = 20 N/mm2.
80 + ( − 60) F 80 − (− 60)I 2
∴ p1, 2 =
2
±
H 2 K + 20 2
= 10 ± 70 2 + 20 2
= 10 ± 72.8
∴ p1 = 82.8 N/mm2
PRINCIPAL STRESSES AND STRAINS 357
FG p − py IJ 2
H K
x
qmax = + q2
2
= 72.8 N/mm2
Let θ be the inclination of principal stress to the plane of px. Then,
2q 2 × 20
tan 2θ = = =2
px − py 80 − 60
∴ 2θ = 63.44° or 63.44 + 180
∴ θ = 31.72° or 121.72°
Planes of maximum shear make 45° to the above planes
∴ θ′ = 15.86° and 60.86°
Example 11.6. The state of stress in two-dimensionally stressed body at a point is as shown in
Fig. 11.13(a). Determine the principal planes, principal stresses, maximum shear stress and their
planes
2
75 N/mm
2
100 N/mm
2
50 N/mm
Fig. 11.13(a)
px + p y
+ G
F p − p IJ + q 2
H 2 K
x y 2
∴ p =
1
2
+ F
− 100 − 75 − 100 + 75 I 2
H K + (− 50)
2
=
2 2
= – 87.5 + 51.54
i.e., p1 = – 35.96 N/mm2
px + p y FG p − py IJ 2
H K
x
p2 = – − + q2
2 2
358 MECHANICS OF SOLIDS
= – 87.5 – 51.54
i.e., p2 = – 139.04 N/mm2
FG p − py IJ 2
H K
x
qmax = + q2
2
i.e., qmax = 51.51.
Let principal plane of p1 make angle θ with x-axis. Then
2q 2 ( − 50)
tan 2θ = = =4
px − py − 100 + 75
∴ 2θ = 75.96 and 75.96 + 180
or θ = 37.98° and 127.98°.
The planes of maximum shear stresses are at 45° to the principal planes. These planes are shown
in Fig. 11.13(b).
p1 - plane y qmax plane
qmax plane
45° θ
x
45°
45°
p2 - plane
Fig. 11.13(b)
Example 11.7. State of stress at a point in a material is as shown in the Fig. 11.14(a). Determine
(i) principal stresses
(ii) maximum shear stress
(iii) plane of maximum shear stress and
(iv) the resultant stress on the plane of maximum shear stress.
2
100 N/mm
2
75 N/mm
y
2
50 N/mm
x
Fig. 11.14(a)
PRINCIPAL STRESSES AND STRAINS 359
px + p y FG p − p IJ + q 2
H 2 K
x y
∴ p1 = + 2
2
+ F
− 50 − 100 I
2
− 50 + 100
=
2 H 2 K + 752
= 25 + 106.07
= 131.07 N/mm2.
px + p y FG p − py IJ 2
H K
x
p2 = − + q2
2 2
= 25 – 106.07
= – 81.07 N/mm2.
FG p − py IJ 2
H K
x
qmax = + q2
2
= 106.07 N/mm2.
The principal plane makes an angle θ to y-axis in anticlockwise direction. Then
2q 2 × 75
tan 2θ =
px − py
=
− 50 − 100F I =–2
2 H K
∴ 2θ = – 63.43°
or θ = – 31.72° = 31.72° clockwise.
Plane of maximum shear makes 45° to it
θ = – 31.72 + 45.00 = 13.28°.
Normal stress on this plane is given by
px + p y px − p y
px = + cos 2θ + q sin 2θ
2 2
− 50 + 100 − 50 − 100
= + cos 2(13.28) + 75 sin ( 2 × 13.28)
2 2
= 25 – 67.08 + 33.54
= – 8.54 N/mm2
pt = qmax = 106.07 N/mm2.
∴ Resultant stress p = ( − 8.54) 2 + 106.07 2
= 106.41 N/mm2
360 MECHANICS OF SOLIDS
H 2 K
y
p1 = + + q2
2
px Fp I 2
+
H 2K + q2
x
=
2
px Fp I 2
−
H2K + q2
x
and p2 =
2
Since the second term is larger than the first term, naturally p1 is +ve and p2 is –ve. Thus major
and minor principal stresses are having opposite nature.
M F px q
A
y
N A N A
Fig. 11.15
px px
q
Stresses at point A
Fig. 11.16
Example 11.9. A shear force of 100 kN and a sagging moment of 80 kN-m act at a certain cross-section
of rectangular beam 100 mm wide and 200 mm deep. Compute the principal stresses at a point 30
mm below the top surface.
Solution: Referring to Fig. 11.17,
1
× 100 × 2003 = 66.667 × 106 mm4
I=
12
At point A, which is at 30 mm below top fibre
y = 100 – 30 = 70 mm
M 80 × 10 6
∴ fx = y= × 70 = 84 N/mm2 (compressive)
I 66.667 × 10 6
∴ px = – 84 N/mm2
100 mm
30 mm
100 mm
A
q
y
2
N A 84 N/mm
100 mm
Fig. 11.17
362 MECHANICS OF SOLIDS
F
Shearing stress q= ( ay )
bI
100 × 10 3
= [100 × 30 × (100 – 15)]
100 × 66.667 × 10 6
= 3.82 N/mm2 as shown in Fig.
Thus, px = – 84 N/mm2, py = 0, q = 3.82 N/mm2
− 84 + 0 F − 84 + 0 I 2
∴ p1,2 =
2
±
H 2 K + (3.82) 2
= – 42 ± 42.17
∴ p1 = 0.17 N/mm2
p2 = – 84.17 N/mm2
Example 11.10. A simply supported beam of 4 m span carries loads as shown in Fig. 11.18. The
cross-section of the beam is 100 mm wide and 180 mm deep. At a section 1.5 m from left support,
calculate the bending and shearing stresses at distances 0, 45 mm and 90 mm above the neutral axis.
Find the principal planes and principal stresses at these points.
20 kN 20 kN 20 kN
A B
1m 1m 1m 1m
Fig. 11.18
0.833 0.625
Fig. 11.19
p1 = 0.833 N/mm2
p2 = – 0.833 N/mm2.
2 × 0.833
Inclination to the plane of px , tan 2θ = = ∝
0
∴ 2θ = 90° and 270°
∴ θ = 45° and 135°
(b) At y = 45 mm
px = – 32.4 N/mm2 py = 0, q = 0.625 N/mm2
H 2 K
2
∴ p1, 2 = ±
2
= – 16.2 ± 16.212
∴ p1 = 0.012 N/mm2
364 MECHANICS OF SOLIDS
p2 = – 32.412 N/mm2
2 × 0.625
tan 2θ = = 0.09876
− 32.4 − 0
2θ = 5.64° and 185.64°
∴ θ = 2.82° and 92.82°
(c) At y = 90 mm,
px = – 64.8 N/mm2, py = 0, q = 0
− 64.8 + 0 F − 64.8 − 0 I 2
∴ p1, 2 =
2
±
H 2 K + 02
= – 32.4 ± 32.4
∴ p1 = 0 N/mm2
p2 = – 64.8 N/mm2
tan 2θ = 0 ∴ 2θ = 0° and 180° or θ = 0° and 90°
Example 11.11. A simply supported beam of span 6 m has I-section as shown in Fig. 11.20(a). It
carries uniformly distributed load (inclusive self weight) of 60 kN/m over entire span. Calculate the
principal stresses and the maximum shearing stress at 100 mm above neutral axis of the beam at
a section 1.5 m from support.
200
10 mm
400
10 mm
10 mm
Fig. 11.20
Solution: L = 6 m, w = 60 kN/m
6 × 60
∴ Reaction at support = = 180 kN
2
∴ Moment at 1.5 m from support
1.5 2
M = 180 × 1.5 – 60 × = 202.5 kN-m
2
Shear force at 1.5 m from support
F = 180 – 1.5 × 60 = 90 kN
PRINCIPAL STRESSES AND STRAINS 365
− 102.35 + 0 F − 102.35 − 0 I 2
∴ p1, 2 =
2
±
H 2 K + (23.68) 2
= – 51.175 ± 56.388
∴ p1 = 5.21 N/mm2 p2 = – 107.56 N/mm2
F − 102.35 − 0 I 2
H 2 K
2
qmax = + (23.68) 2 = 56.39 N/mm
py
a (b)
(a)
Fig. 11.21
366 MECHANICS OF SOLIDS
Referring to Fig. 11.21(a), which represent state of stress at A with their positive senses it may
be noted that direct strain ex and ey are tensile strains and shearing strain γxy reduces the angle θ.
Figure 11.22 shows the element at A with its de-formed shape (shown with dotted lines) with
ex,ey and γxy positive. Now we are interested in finding strains en, et and γnt on a plane at ‘θ’ to the
plane of px forces. The size of element (a × b) is so selected that the diagonal AE is normal to the
plane DE i.e. ∠CAD is also θ, for the convenience. n
(a) To find en B1
be y C1
C
Let AC = l, B Q
E1
∴ AD = l cos θ = a
xy
AB = l sin θ = b
Drop ⊥ C1, P to AD A
D D1 P
⊥ BQ to C1P.
t
Now,
AP = AD + DD1 + D1P aex bxy
= a + aex + bγxy Fig. 11.22
= l cos θ + l cos θ × ex + l sin θ × γxy
= l [cos θ + ex cos θ + γxy sin θ]
PC = PQ + QC
= b + bey
= l sin θ + l sin θ × ey = l sin θ (1 + ey)
pt
∴ AC1 = AP 2 + PC 2
aex bxy
Fig. 11.23
After straining let point E1 move to E2. Draw E2R parallel to E1 D [Ref. Fig. 11.23]
∴ Total shearing strain
= ∠CAC1 + ∠D1E2R = φ1 + φ2
To find φ1 :
Now : CC3 = CC2 sin θ
= (CQ – C2Q) sin θ
LM
= ( ae x + bγ xy ) − be y
1
sin θ
OP
N tan θ Q
= (aex + bγxy) sin θ – bey cos θ
Noting that l cos θ = a and l sin θ = b, we get
CC3 = lex cos θ sin θ + l sin2 θ γxy – ley sin θ cos θ
= l (ex – ey) sin θ cos θ + l γxy sin2 θ
CC3
∴ φ1 = = (ex – ey) sin θ cos θ + γxy sin2 θ ...(1)
l
To find φ2 :
E3E2 = Extension of AE1
= AE1 en = a cos θ ex = l cos2 θ × ex ...(2)
RD2 DD2 − RD2 DD2 − E3 E2
φ2 = = =
E1 D E1 D E1 D
ae x cos θ − l cos 2 θ en
=
a sin θ
F1γ I
tan 2θ =
2
H2 K =xy γ xy
...(11.12)
ex − ey ex − e y
The magnitude of principal planes may be obtained exactly on the same line as the principal
stresses were obtained. The final result will be
FG e − ey IJ + F 1 γ I
2 2
K H2 K
1
H
x
e1, 2 = (ex + ey) ± xy
2 2
1 1
(ex + ey) ±
= ( e x − e y ) 2 + γ 2xy ...(11.13)
2 2
The maximum shearing strain occurs at 45° to the plane of principal plane and its magnitude is
given by
1
γmax = (e x − e y ) 2 + γ 2xy ...(11.14)
2
Commonly used strain rosettes are 45° rosette and 120° rosette which are as shown in Fig. 11.24.
45° rosette is also known as rectangular rosette.
B
120° 60°
C C B
B 120°
A
45°
120°
45° 120° 60° 60° 120°
C
O A A
(a) Rectangular rosette (b) 120° rosette (c) Another form of 120° rosette
Fig. 11.24
Treating one of the strain gauge direction, strains eθ1 and eθ2 of the other two gauges is known
writing eθ1 and eθ2 in terms of ex, ey and γxy, it is possible to find ex,,ey and γxy.
After finding ex, ey and γxy it is possible to find the corresponding stresses. We know,
Ee1 = p1 – µp2 ...(1)
Ee2 = p2 – µp1 ...(2)
Multiplying Eqn. (2) with µ and adding it to eqn.1, we get,
E(e1 + µe2) = p1 – µ2p1 = p1(1 – µ2)
E (e1 + µe2 )
∴ p1 = ...(11.15a)
1 − µ2
E (e2 + µe1 )
Similarly p2 = ...(11.15b)
1 − µ2
Example 11.12. At a point strains measured with rectangular rosettes are eA = 600 microns, eB =
300 microns and ec = – 200 microns. Determine the principal strains. Find principal stresses also
taking E = 2 × 105 and µ = 0.3.
Solution: Taking the direction of strain gauge as x-axis,
ex = 600 microns
eB = e45 = 300 microns and ec = e90 = – 200 microns = ey
1 1 1
∴ 300 = (ex + ey) + (e – ey) cos 2 × 45 + γ sin 2 × 45
2 2 x 2 xy
1 1 1
= (600 – 200) + (600 + 200) × 0 + γ
2 2 2 xy
γxy = 200 microns.
∴ Principal strains are
1 FG e − ey IJ + F 1 γ I
2 2
(ex + ey) ±
H K H2 K
e1, 2 = x
2 xy
2
370 MECHANICS OF SOLIDS
ex + ey FG e IJ + F 1 γ I
− ey 2 2
H 2 K H2 K
x
∴ e1, 2 = ± xy
2
IMPORTANT FORMULAE
px + py px − p y
1. pn = + cos 2θ + q sin 2θ.
2 2
px − py
2. pt = sin 2θ – q cos 2θ.
2
3. Principal planes are given by
2q
tan 2θ = .
px − py
px + p y FG p − py IJ 2
H K
x
4. p1, 2 = ± + q2 .
2 2
p1 − p2 FG p − py IJ 2
H K
x
5. Maximum shear stress = = + q2 .
2 2
6. Plane of maximum shear is at 45° to the principal planes.
7. Obliquity of resultant stress on a plane at θ to the plane of px
= θ + α to the plane of px
pn
where α = tan–1 .
pt
372 MECHANICS OF SOLIDS
ex + ey ex − ey 1
8. en = + cos 2θ + γ sin 2θ
2 2 2 xy
1 (ex − ey ) 1
rnt = sin 2θ – cos 2θ
2 2 2γ xy
1 FG e IJ + F 1 γ I
− ey
2 2
H2 K H2 K
e1, 2 = (ex + ey) ± x
2 xy
e1 − e2 FG e − e IJ + F 1 γ I .
2 2
H 2 K H2 K
x y
γmax = = xy
2
THEORY QUESTIONS
1. Derive the expressions for normal and tangential stresses on a plane inclined at θ to the plane of
px force. Take a general two dimensional state of stress.
2. State the equations for normal and tangential stresses on an inclined plane, in an element under
general two dimensional stress system. Derive the expressions for principal planes, principal
stresses and maximum shear stress.
3. Explain the terms principal stresses and principal strains.
1. A point in a strained material is subjected to tensile stresses px = 180 N/mm2 and py = 120 N/mm2.
Determine the intensities of normal, tangential and resultant stresses on a plane inclined at 30°
anticlockwise to the axis of minor stress.
[Note : Axis of minor stress means the plane of major stress]
[Ans. pn = 165 N/mm2, pt = 25.98 N/mm2, p = 167.03 N/mm2, α = 111.05° anticlockwise to the
axis of minor principal plane]
2. The state of stress at a point in a strained material is as shown in Fig. 11.25. Determine the
normal, tangential and the resultant stress on plane DE. Determine the direction of resultant also.
B E C
2
120 N/mm
60°
A D
2
40 N/mm
Fig. 11.25
3. The state of stress at a point is as shown in Fig. 11.26. Determine the principal stresses and
maximum shear stress. Indicate their planes on a separate sketch.
[Ans. p1 = 133.13 N/mm2, p2 = – 93.13 N/mm2, qmax = 113.13 N/mm2, θ = – 22.5° and 67.5°,
θ′ = 22.5 and 112.5°]
y-axis
qmax - plane p1 - plane
2
100 N/mm
.5°
2
80 N/mm p2 - plane
22
45°
2 45°
60 N/mm x-axis
45°
qmax plane
(a)
(b)
Fig. 11.26
4. The state of stress in a two dimensionally stressed material is as shown in Fig. 11.27. Determine
the principal stresses, principal planes and the maximum shear stress. Determine normal and
tangential stresses on plane AC also.
2
60 N/mm
60° 2
80 N/mm
2
C 60 N/mm
Fig. 11.27
[Ans. p1 = – 9.18 p2 = – 130.82 N/mm2, qmax = 60.82 N/mm2, θ = – 40.27° and 49.73°,
N/mm2,
pn = – 116.96 N/mm2, pt = 21.34 N/mm2]