Mechanics of Solids [3 1 0 4]
CIE 101 / 102
First Year B.E. Degree
Mechanics of Solids
PART- I
PART- II
Mechanics of Rigid
Bodies
Mechanics of Deformable
Bodies
COURSE CONTENT IN BRIEF
PART I Mechanics of Rigid Bodies
1. Resultant of concurrent and non-concurrent coplanar forces.
2. Equilibrium of concurrent and non-concurrent coplanar forces.
3. Centroid of plane areas
4. Moment of Inertia of plane areas
5. Kinetics: Newton’s second law, D’Alembert’s principle, Work- Energy,
and Impulse- Momentum principle.
PART II
Mechanics of Deformable bodies
6. Simple stresses and strains
7. Statically indeterminate problems and thermal stresses
8. Stresses on inclined planes
9. Stresses due to fluid pressure in thin cylinders
Books for Reference
1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.
2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India.
3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition
4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.
5. Machanics of Materials, by E.P.Popov
6. Machanics of Materials, by E J Hearn
7. Strength of materials, by Beer and Johnston
8. Strength of materials, by F L Singer & Andrew Pytel
9. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa
10. Strength of Materials, by Ramamruthum
11. Strength of Materials, by S S Bhavikatti
PART - I
MECHANICS OF RIGID BODIES
PART - I
Mechanics of Rigid Bodies
INTRODUCTION
Definition of Mechanics :
In its broadest sense the term ‘Mechanics’ may be defined as
the ‘Science which describes and predicts the conditions of rest
or motion of bodies under the action of forces’.
This Course on Engineering Mechanics comprises of
Mechanics of Rigid bodies and the sub-divisions that come
under it.
Branches of Mechanics
Engineering Mechanics
Mechanics of Solids
Rigid Bodies
Deformable
Bodies
Statics Dynamics
Kinematics
Mechanics of Fluids
Ideal
Fluids
Viscous
Fluids
Compres
Fluids
Theory of Theory of Plasticity
Strength of Elasticity
Materials
Kinetics
Concept of Rigid Body :
It is defined as a definite amount of matter the parts of which
are fixed in position relative to one another under the
application of load.
Actually solid bodies are never rigid; they deform under the
action of applied forces. In those cases where this deformation
is negligible compared to the size of the body, the body may be
considered to be rigid.
Particle
A body whose dimensions are negligible when compared to the
distances involved in the discussion of its motion is called a
‘Particle’.
For example, while studying the motion of sun and earth, they
are considered as particles since their dimensions are small
when compared with the distance between them.
Force
It is that agent which causes or tends to cause, changes or
tends to change the state of rest or of motion of a mass.
A force is fully defined only when the following four
characteristics are known:
(i) Magnitude
(ii) Direction
(iii) Point of application
(iv) Sense.
Force:
characteristics of the force 100 kN are :
(i) Magnitude = 100 kN
(ii) Direction = at an inclination of 300 to the x-axis
(iii) Point of application = at point A shown
100 kN
(iv) Sense = towards point A
A
300
Scalars and Vectors
A quantity is said to be a ‘scalar’ if it is completely defined by
its magnitude alone.
Example : Length, Area, and Time.
A quantity is said to be a ‘vector’ if it is completely defined only
when its magnitude and direction are specified.
Example : Force, Velocity, and Acceleration.
Principle of Transmissibility : It is stated as follows :
‘The
external effect of a force on a rigid body is the same for all points
of application along its line of action’.
P
A
B
P
For example, consider the above figure. The motion of the block will be
the same if a force of magnitude P is applied as a push at A or as a pull
at B.
P
P
O
The same is true when the force is applied at a point O.
1. RESULTANT OF COPLANAR FORCES
Resultant, R : It is defined as that single force which can
replace a set of forces, in a force system, and cause the
same external effect.
R
F2
F1
A
F3
=
θ
A
R = F1 + F2 + F3
external effect on particle, A is same
Resultant of two forces acting at a point
Parallelogram law of forces : ‘If two forces acting at a point are
represented in magnitude and direction by the two adjacent
sides of a parallelogram, then the resultant of these two forces
is represented in magnitude and direction by the diagonal of
the parallelogram passing through the same point.’
B
P2
O
α
C
R
θ
P1
A
Contd..
B
P2
O
α
C
R
θ
P1
A
In the above figure, P1 and P2, represented by the sides OA and OB have
R as their resultant represented by the diagonal OC of the parallelogram
OACB.
It can be shown that the magnitude of the resultant is given by:
R = √P12 + P22 + 2P1P2Cos α
Inclination of the resultant w.r.t. the force P1 is given by:
θ = tan-1 [( P2 Sin α) / ( P1 + P2 Cos α )]
Resultant of two forces acting at a point at right angle
B
P2
O
α
C
R
θ
A
P1
If α = 900 , (two forces acting at a point are at right angle)
B
C
R
P2
O
R = P1 + P 2 2
2
θ
P1
A
P2
tan θ =
P1
Triangle law of forces
‘If two forces acting at a point can be represented both in
magnitude and direction, by the two sides of a triangle taken in
tip to tail order, the third side of the triangle represents both in
magnitude and direction the resultant force F, the sense of the same is
defined by its tail at the tail of the first force and its tip at the tip of
the second force’.
Triangle law of forces
Let F1 and F2 be the two forces acting at a point A and θ is the
included angle.
F1
F1
R
θ
A
θ
=
F2
F2
‘Arrange the two forces as two sides of a triangle taken in tip to
tail order, the third side of the triangle represents both in magnitude
and direction the resultant force R.
the sense of the resultant force is defined by its tail at the tail of the
first force and its tip at the tip of the second force’.
Triangle law of forces
F1
θ
A
R
α
F1
=
F2
β F2
F1
R
θ
F2
F1
F2
R
=
=
sin β sin α sin(180 − α − β )
(180 - α - β) = θ
where α and β are the angles made by the resultant force
with the force F1 and F2 respectively.
Component of a force :
Component of a force, in simple terms, is the effect of a
force in a certain direction. A force can be split into infinite number
of components along infinite directions.
Usually, a force is split into two mutually perpendicular
components, one along the x-direction and the other along ydirection (generally horizontal and vertical, respectively).
Such components that are mutually perpendicular are called
‘Rectangular Components’.
The process of obtaining the components of a force is called
‘Resolution of a force’.
Rectangular component of a force
F
F
θx
θx
Fx
Fy
=
F
θx
Fx
Fy
Consider a force F making an angle θx with x-axis.
Then the resolved part of the force F along x-axis is given by
Fx = F cos θx
The resolved part of the force F along y-axis is given by
Fy = F sin θx
Oblique component of a force
Let F1 and F2 be the oblique components of a force F. The
components F1 and F2 can be found using the ‘triangle law of
forces’.
N
F2
β
α
M
F
F
F1
β F2
α
F1
O
The resolved part of the force F along OM and ON can
obtained by using the equation of a triangle.
F1 / Sin β = F2 / Sin α = F / Sin(180 - α - β)
Sign Convention for force components:
y
y
x
+ve
+ve
x
The adjacent diagram gives the sign convention for
force components, i.e., force components that are directed
along positive x-direction are taken +ve for summation along
the x-direction.
Also force components that are directed along +ve y-direction are
taken +ve for summation along the y-direction.
Classification of force system
Force system
Coplanar Forces
Concurrent
Non-concurrent
Like parallel
Non-Coplanar
Forces
Concurrent
Non-concurrent
Unlike parallel
Like parallel
Unlike parallel
A force that can replace a set of forces, in a force system,
and cause the same ‘external effect’ is called the Resultant.
RESULTANT OF COPLANAR NON CONCURRENT
FORCE SYSTEM
Coplanar Non-concurrent Force System:
This is the force system in which lines of action of
individual forces lie in the same plane but act at different points
of applications.
F1
F3
Fig. 1
F2
F2
F1
F5
F3
F4
Fig. 2
1. Parallel Force System – Lines of action of individual
forces are parallel to each other.
2. Non-Parallel Force System – Lines of action of the forces
are not parallel to each other.
MOMENT OF A FORCE ABOUT AN AXIS
The applied force can also tend to rotate the body about
an axis in addition to motion. This rotational tendency is
known as moment.
Definition: Moment is the
tendency of a force to make a
rigid body to rotate about an
axis.
This is a vector quantity
having both magnitude
and direction.
MOMENT OF A FORCE ABOUT AN AXIS
Moment Axis: This is the axis about which rotational
tendency is determined. It is perpendicular to the plane
comprising moment arm and line of action of the force (axis
0-0 in the figure)
Moment Center: This is
the position of axis on coplanar system. (A).
Moment Arm:
Perpendicular distance
from the line of action of
the force to moment
center. Distance AB = d.
Magnitude of moment:
It is computed as the product of the of the force and
the perpendicular distance from the line of action
to the point about which moment is computed.
(Moment center).
MA = F×d
= Rotation effect because of
the force F, about the point A
(about an axis 0-0)
Unit – kN-m, N-mm etc.
Sense of moment:
The sense is obtained by ‘Right Hand Thumb’ rule.
‘If the fingers of the right hand are curled in the direction
of rotational tendency of the body, the extended thumb
represents the sense of moment vector’.
For the purpose of additions,
the moment direction may be
considered by using a suitable
sign convention such as +ve
for counterclockwise and –ve
for clockwise rotations or viceversa.
M
M
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Statement: The moment of a force about a moment center or
axis is equal to the algebraic sum of the moments of its
component forces about the same moment center (axis).
P
P sinθ
P
θ
P cosθ
θ d
d1
A
d2
A
Moment of Force P about the
Algebraic sum of Moments of
= components of the Force P
point A,
about the point A,
Pxd
P cosθ x d1 + P sinθ x d2
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Proof (by Scalar Formulation):
Let ‘R’ be the given force.
‘P’ & ‘Q’ are component forces of ‘R’.
‘O’ is the moment center.
Y
p, r and q are moment arms from ‘O’
of P, R and Q respectively.
R
α, β and γ are the inclinations of ‘P’,
‘R’ and ‘Q’ respectively w.r.to X –
axis.
Q
q
A
α
r P
βγ p
O
X
Y
We have,
Ry = Py + Qy
R Sinβ = P Sinα + Q Sin γ ----(1)
From ∆le AOB, p/AO = Sin α
From ∆le AOC, r/AO = Sin β
From ∆le AOD, q/AO = Sin γ
From (1),
∴ R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)
i.e., R × r = P × p + Q × q
Moment of resultant R about O = algebraic
sum of moments of component forces P &
Q about same moment center ‘O’.
Ry
Q
R
D
Qy
Py
γ
α
A
C
β
q r
P
B
p
O
X
COUPLE
Two parallel, non collinear (separated by certain
distance) forces that are equal in magnitude and opposite
in direction form ‘couple’.
F
The algebraic summation of the
d
Hence, couple does not produce any
translation and produces only rotation.
=
two forces forming couple is zero.
F
M=Fxd
RESOLUTION OF A FORCE INTO A
FORCE-COUPLE SYSTEM
Replace the force F acting at the point A to the point B
F
B
A
Apply two equal and opposite forces of same magnitude &
direction as Force F at point B, so that external effect is
unchanged
F
B
F
F
A
d
F
B
F
F
A
d
=
F
B
A
M=Fxd
Of these three forces, two forces i.e., one at A and the other
oppositely directed at B form a couple.
Moment of this couple, M = F × d.
Third force at B is acting in the same direction as that at P.
Thus, the force F acting at a point such as A in a rigid body can
be moved to any other given point B, by adding a couple M. The
moment of the couple is equal to moment of the force in its
original position about B.
TYPES OF LOADS ON BEAMS
W kN
1. Concentrated Loads – This is the load
acting for very small length of the beam.
(also known as point load, Total load W is
acting at one point )
w kN/m
2. Uniformly distributed load – This is
the load acting for a considerable
length of the beam with same intensity
of w kN/m throughout its spread.
Total intensity, W = w × L
L
W = (w x L) kN
L/2
(acts at L/2 from one end of the spread)
L
3. Uniformly varying load – This load acts
for a considerable length of the beam with
intensity varying linearly from ‘0’ at one end
to w kN/m to the other representing a
triangular distribution.
Total intensity of load = area of triangular
spread of the load
W = 1/2× w × L.
w kN/m
L
W=½×L×w
1/3 ×L
(acts at 2×L/3 from ‘Zero’ load end) 2/3 ×L
L
EXERCISE PROBLEMS
1. Resultant of force system
Q1
A body of negligible weight, subjected to two forces
F1= 1200N, and F2=400N acting along the vertical, and the
horizontal respectively, is shown in figure. Find the
component of each force parallel, and perpendicular to the
plane.
Y
F2 = 400 N
4
F1 = 1200 N
X
3
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N
EXERCISE PROBLEMS
1. Resultant of force system
Q2. Determine the X and Y components of each of the forces
shown in the figure.
F2 = 390 N
12
Y
X
5
40º
F3 =400 N
30º
F1 = 300 N
(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,
F3X = -306.42 N, F3Y= -257.12N )
EXERCISE PROBLEMS
1. Resultant of force system
Q3. Obtain the resultant of the concurrent coplanar forces
shown in the figure
600N
800N
40º
20º
30º
200N
(Ans: R = 522.67 N, θ = 68.43º)
EXERCISE PROBLEMS
1. Resultant of force system
Q4. A disabled ship is pulled by means of two tug boats as shown in FIG.
4. If the resultant of the two forces T1 and T2 exerted by the ropes is a
300 N force acting parallel to the X – direction, find :
(a) Force exerted by each of the tug boats knowing α = 30º.
(b) The value of α such that the force of tugboat 2 is minimum,
while that of 1 acts in the same direction.
Find the corresponding force to be exerted by tugboat 2.
T2
α
20º
R = 300 N
X - direction
FIG. 4
T1
( Ans: a. T1= 195.81 N, T2 = 133.94 N
b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )
EXERCISE PROBLEMS
1. Resultant of force system
Q5. An automobile which is disabled is pulled by two ropes
as shown in the figure. Find the force P and resultant R,
such that R is directed as shown in the figure.
P
20º
40º
R
Q = 5 kN
(Ans: P = 9.4 kN , R = 12.66 kN)
EXERCISE PROBLEMS
1. Resultant of force system
Q6. A collar, which may slide on a vertical rod, is subjected to three
forces as shown in figure. The direction of the force F may be varied
Determine the direction of the force F, so that resultant of the three
forces is horizontal, knowing that the magnitude of F is equal to
(a) 2400 N, (b)1400N
1200 N
800 N
60º
θ
F
COLLAR
ROD
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)
EXERCISE PROBLEMS
1. Resultant of force system
Q7. Determine the angle α and the magnitude of the force Q
such that the resultant of the three forces on the pole is
vertically downwards and of magnitude 12 kN. Refer figure
5kN
8kN
α
30º
Q
Fig. 7
(Ans: α = 10.7 º, Q = 9.479 kN )
EXERCISE PROBLEMS
1. Resultant of force system
Q8. Determine the resultant of the parallel coplanar
force system shown in figure.
600 N
2000 N
60º
10º
o
30º
60º
1000 N
400 N
(Ans. R=800N towards left, d=627.5mm)
EXERCISE PROBLEMS
1. Resultant of force system
Q9. Four forces of magnitudes 10N, 20N, 30N and 40N
acting respectively along the four sides of a square
ABCD as shown in the figure. Determine the
magnitude, direction and position of resultant w.r.t. A.
20N
D
C
30N
a
A
a
B
10N
40N
(Ans:R=28.28N, θ=45º, x=1.77a)
EXERCISE PROBLEMS
1. Resultant of force system
Q10. Four parallel forces of magnitudes 100N, 150N, 25N
and 200N acting at left end, 0.9m, 2.1m and
2.85m respectively from the left end of a horizontal
bar of 2.85m. Determine the magnitude of resultant
and also the distance of the resultant from the left
end.
(Ans: R = 125 N, x = 3.06 m)
EXERCISE PROBLEMS
1. Resultant of force system
Q11. Reduce the given forces into a single force and a
couple at A.
70.7 kN
200 kN
45º
30º
1.5m
A
1m
30º
100 N
80 N
(Ans:F=320kN, θ=14.48º, M=284.8kNm)
EXERCISE PROBLEMS
1. Resultant of force system
Q12. Determine the resultant w.r.t. point A.
150 Nm
150 N
1.5m
3m
1.5m
A
100 N 500 N
(Ans: R = 450 kN, X = 7.5 kNm)
2. EQUILIBRIUM OF FORCE SYSTEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Definition:If a system of forces acting on a body, keeps the body in a
state of rest or in a state of uniform motion along a straight
line, then the system of forces is said to be in equilibrium.
ALTERNATIVELY, if the resultant of the force system is zero,
then, the force system is said to be in equilibrium.
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Conditions for Equilibrium :
A coplanar concurrent force system will be in equilibrium if it
satisfies the following two conditions:
i)
∑ Fx = 0; and ii) ∑ Fy = 0
i.e. Algebraic sum of components of all the forces of the system,
along two mutually perpendicular directions, is ZERO.
Y
X
Graphical conditions for Equilibrium
Triangle Law: If three forces are in equilibrium, then, they form a
closed triangle when represented in a Tip to Tail arrangement, as
shown in Fig 2.1
F
3
F2
F2
F1
Fig 2.1
F3
F1
Polygonal Law: If more than three forces are in equilibrium,
then, they form a closed polygon when represented in a Tip to
Tail arrangement, as shown in Fig. 2.2.
F
F4
3
F2
F4
F5
2
F1
F
F5
F3
Fig 2.2
F1
LAMI’S THEOREM
If a system of three forces is in equilibrium, then, each force of the
system is proportional to sine of the angle between the other two
forces (and constant of proportionality is the same for all the
forces). Thus, with reference to Fig.2.3, we have,
F3
F3
F1
F2
=
=
Sin α
Sin β
Sin γ
F2
α
γ
β
F1
Fig. 2.3
Note: While using Lami’s theorem, all the three forces
should be either directed away or all directed towards the
point of concurrence.
EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM
When a body is in equilibrium, it has neither translatory nor
rotatory motion in any direction.
Thus the resultant force R and the resultant couple M are both
zero, and we have the equilibrium equations for two
dimensional force system
∑ Fx = 0;
∑ Fy = 0
∑M = 0
These requirements are both necessary and sufficient
conditions for equilibrium.
SPACE DIAGRAMS & FREE BODY DIAGRAMS
Space Diagram (SPD) : The sketch showing the physical
conditions of the problem, like, the nature of supports
provided; size, shape and location of various bodies; forces
applied on the bodies, etc., is known as space diagram.
eg, Fig 2.4 is a space diagram
Weight of sphere = 0.5 kN,
Radius = 1m
3m θ
Cable
P = 2kN
30°
Sphere
wall
Fig. 2.4 SPD
Free Body Diagram (FBD) :
It is an isolated diagram of the body being analyzed (called
free body), in which, the body is shown freed from all its
supports and contacting bodies/surfaces. Instead of the
supports and contacting bodies/surfaces, the reactive
forces exerted by them on the free body is shown, along
with all other applied forces.
A Few Guidelines for Drawing FBD
1) Tensile Force: It is a force trying to pull or extend the body. It
is represented by a vector directed away from the body.
2) Compressive Force: It is force trying to push or contract the
body. It is represented by a vector directed towards the body.
3) Reactions at smooth surfaces: The reactions of smooth
surfaces, like walls, floors, Inclined planes, etc. will be normal to
the surface and pointing towards the body.
4)Forces in Link rods/connecting rods: These forces will be acting
along the axis of the rod, either towards or away from the body.
(They are either compressive or tensile in nature).
5) Forces in Cables (Strings or Chords): These can only be
tensile forces. Thus, these forces will be along the cable and
directed away from the body.
Free Body Diagrams of the sphere shown in Fig. 2.4
T = Tension in the cable
θ
T
P = 2kN
30°
Rw
Sphere
W=0.5kN
Fig. 2.5 F B D of Sphere
Rw = Reaction of the wall
W = self weight of the sphere
P = external load acting on
the sphere
Detach the sphere from all contacts
and replace that with forces like:
Cable contact is replaced by the
force tension = T
Contact with the smooth wall is
replaced by the reaction Rw.
Supports: A structure is subjected to external forces and
transfers these forces through the supports on to the
foundation. Therefore the support reactions and the external
forces together keep the structure in equilibrium.
There are different types of supports.
a) Roller Support
b) Hinged or pinned support
c) Fixed or built in support
Supports
Types of Supports
Action on body
(a) Flexible cable ,belt ,chain, rope
BODY
BODY
T
Force exerted by cable is
always a tension away from
the body in the direction of
cable
(b) Smooth surfaces
900
F
900
F
Contact forces are normal to
the surfaces
Supports
A
(c) Roller
support
A
Contact force is normal to the
surface on which the roller moves.
The reaction will always be
perpendicular to the plane of the
roller . Roller support will offer only
one independent reaction
component. (Whose direction is
known.)
Supports
(d) pinned Support / hinged support
A
Rh
A
θ
R
Rv
This support does not allow any translatory movement of the
rigid body. There will be two independent reaction
components at the support. The resultant reaction can be
resolved into two mutually perpendicular components. Or it
can be shown as resultant reaction inclined at an angle with
respect to a reference direction.
Supports
(e) Fixed or Built-in Support
A
RAH
M
A
RAV
This type of support not only prevents the translatory
movement of the rigid body, but also the rotation of the rigid
body. Hence there will be 3 independent reaction
components of forces. Hence there will be 3 unknown
components of forces, two mutually perpendicular reactive
force component and a reactive moment as shown in the
figure.
TYPES OF BEAMS
A member which is subjected to predominantly transverse loads
and supported in such a way that rigid body motion is prevented
is known as beam. It is classified based on the support
conditions. A beam generally supported by a hinge or roller at
the ends having one span (distance between the support) is
called as simply supported beam. A beam which is fixed at one
end and free at another end is called as a cantilever beam.
A
B
A
B
span
span
(a) Simply supported beam
TYPES OF BEAMS
A
B
span
RH M
A
B
Rv
(b) Cantilever beam
TYPES OF BEAMS
If one end or both ends of the beam project beyond
the support it is known as overhanging beam.
A
A
B
B
(c) Overhanging beam
(right overhang)
Statically determinate beam
Using the equations of equilibrium given below, if all
the reaction components can be found out, then the
beam is a statically determinate beam
the equations of equilibrium
∑ Fx = 0;
∑ Fy = 0 ∑M = 0
FRICTION
Friction is defined as the contact resistance exerted by
one body upon another body when one body moves or tends
to move past another body. This force which opposes the
movement or tendency of movement is known as frictional
resistance or friction. Friction is due to the resistance offered by
minute projections at the contact surfaces. Hence friction is the
retarding force, always opposite to the direction of motion.
Friction has both advantages & disadvantages.
Disadvantages ----Æ Power loss, wear and tear etc.
Advantages ----Æ Brakes, traction for vehicles etc.
FRICTION
W
P
F (Friction)
N
Hills & Vales
Magnified Surface
Frictional resistance is dependent on the amount of wedging
action between the hills and vales of contact surfaces. The
wedging action is dependent on the normal reaction N.
FRICTION
Frictional resistance has the remarkable property of
adjusting itself in magnitude of force producing or tending
to produce the motion so that the motion is prevented.
When P = 0, F = 0 Æ block under equilibrium
When P increases, F also increases proportionately to
maintain equilibrium. However there is a limit beyond
which the magnitude of this friction cannot increase.
FRICTION
When the block is on the verge of motion(motion of the
block is impending) F attains maximum possible value,
which is termed as Limiting Friction. When the applied force
is less than the limiting friction, the body remains at rest and
such frictional resistance is called the static friction.
Further if P is increased, the value of F decreases rapidly
and then remains fairly a constant thereafter. However at
high speeds it tends to decrease. This frictional resistance
experienced by the body while in motion is known as
Dynamic friction OR Kinetic Friction.
FRICTION
Sliding frictionÆ friction experienced
when a body slides over another surface.
Dynamic Friction
Rolling friction Æ friction experienced by
a body when it rolls over a surface.
FRICTION
FαN
W
P
Fmax = µN
Where Fmax = Limiting Friction
Fmax
µ =Coefficient of friction
φ
N
µ=
N= Normal Reaction between the
contact surfaces
R
Fmax
N
Note : Static friction varies from zero to a maximum value. Dynamic
friction is fairly a constant.
FRICTION
Angle of Friction
W
The angle between N & R depends
on the value of F.
This angle θ, between the resultant
R and the normal reaction N is
termed as angle of friction.
As F increases, θ also increases and will
reach to a maximum value of φ when F is
Fmax (limiting friction)
i.e. tanφ = (Fmax )/N = µ
Angle φ is known as Angle of limiting Friction.
P
φ
N
Fmax
R
FRICTION
Angle of limiting friction is defined as the angle between the
resultant reaction (of limiting friction and normal reaction) and
the normal to the plane on which the motion of the body is
impending.
Angle of repose
When granular material is heaped, there exists a limit for the
inclination of the surface. Beyond that angle, the grains start
rolling down. This limiting angle upto which the grains repose
(sleep) is called the angle of repose of the granular material.
FRICTION
Significance of Angle of repose:
The angle that an inclined plane makes with the
horizontal, when the body supported on the plane is on
the verge of motion due to its self -weight is equal to the
angle of repose.
Angle of repose is numerically equal to Angle of
limiting friction
Laws of dry friction
FRICTION
1. The magnitude of limiting friction bears a constant ratio
to the normal reaction between the two surfaces.
(Experimentally proved)
2. The force of friction is independent of the area of contact
between the two surfaces.
3. For low velocities the total amount of friction that can
be developed is practically independent of velocity.
It is less than the frictional force corresponding
to impending motion.
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q1. A 10kN roller rests on a smooth horizontal floor and is
held by the bar AC as shown in Fig(1). Determine the
magnitude and nature of the force in the bar AC and reaction
from the floor under the action of the forces applied on the
roller. [Ans:FAC=0.058 kN(T),R=14.98 kN]
7kN
C
A
300
Fig(1)
450
5kN
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q2.
A 10 kN weight is suspended from a rope as shown in
figure. Determine the magnitude and direction of the least force P
required to pull the rope, so that, the weight is shifted horizontally
by 0.5m. Also, determine, tension in the rope in its new position.
[Ans: P= 2.43 kN, θ = 14.480 ; T= 9.7kN.]
2m
θ
10kN
P
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q3. Determine the value of P and the nature of the forces in the
bars for equilibrium of the system shown in figure.
[Ans: P = 3.04 kN, Forces in bars are Compressive.]
45
75
2kN
60
45
P
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q4. A cable fixed as shown in Fig. supports three loads.
Determine the value of the load W and the inclination of the
segment BC. [Ans: W=25kN, θ = 54.780]
A
D
30
B
60
θ
20
Loads are in kN
C
W
22.5
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q5. Find the reactions at A,B,C and D for the beam loaded
as shown in the figure. (Ans.RA=RB =34kN;RC=28.84kN;
MC=-140kNm ; θC=-33.69 ˚ )
12kN/m
20 kN
4kN/m
12kN/m
4kN/m
30kN
A
4
B
3
C
40kNm
1m
2m
1m
1m
2m
1m
1m
2m
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q6. A uniform bar AB of weight 50N shown in the
figure supports a load of 200N at its end. Determine the
tension developed in the string and the force supported
by the pin at B. (Ans. T=529.12N;RB=807.15N, θB=64.6˚)
2.5m
string
B
60˚
A
200N
2.5m
2.5m
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q7. Find the position of the hinged support (x),such that the
reactions developed at the supports of the beam are equal..
(Ans.x=2m.)
10kN/m
2.0m
15kN
1.0m 0.6 1.4m
18kN/m
3.0m
x
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q8. A right angled bar ABC hinged at A as shown in fig carries
two loads W and 2W applied at B &C .Neglecting self weight of
the bar find the angle made by AB with vertical (Ans:θ =18.44˚)
Lm
A
θ
B
W
0.5L
C
2W
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q9. For the block shown in fig., determine the smallest
force P required
a) to start the block up the plane
b) to prevent the block moving down the plane.
Take µ = 0.20
[Ans.:
(a) Pmin
=
59.2N
(b) Pmin = 23.7N
(b) θ = 11.3o]
P
100N
θ
25°
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q10. A block of weight 2000 N is attached to a cord
passing over a frictionless pulley and supporting a weight
of 800N as shown in fig. If µ between the block and the
plane is 0.35, determine the unknown force P for
impending motion
(a) to the right
(b) to the left
[Ans.: (a) P = 132.8N (b) P = 1252N]
2000N
30°
P
800N
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q11. Determine value of angle θ to cause the motion of
500N block to impend down the plane, if µ for all contact
surfaces is 0.30.
200N
500N
θ=?
[Ans.: θ = 28.4°]
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q12. A horizontal bar 10m long and of negligible weight
rests on rough inclines as shown in fig. If angle of friction
is 15o, how close to B may the 200N force be applied
before the motion impends.
100N
A
2m
30°
200N
X=?
B
60°
[Ans.: x = 3.5m]
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q13. Determine the vertical force P required to drive the
wedge B downwards in the arrangements shown in fig.
Angle of friction for all contact surfaces is 12o.Weight of
block A= 1600 N.
P
A
[Ans.: P = 328.42N]
B
20°
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q14.
Determine the force P which is necessary to start
the wedge to raise the block A weighing 1000N. Self
weight of the wedge may be ignored. Take angle of friction,
φ = 15o for all contact surfaces.
A
P
[Ans.: P = 1192N]
20
°
wedge
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q15.
A ladder of weight 200N, 6m long is supported as
shown in fig. If µ between the floor and the ladder is 0.5 &
between the wall and the ladder is 0.25 and it supports a
vertical load of 1000N, determine
a) the least value of α at which the ladder may be placed
without slipping
b) the reactions at A & B
1000N
5m
α
A
B
[Ans.: (a) α = 56.3o
(b) RA = 1193 N,
RB = 550N]
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q16. An uniform ladder of weight 250N is placed against a
smooth vertical wall with its lower end 5m from the wall. µ
between the ladder and the floor is 0.3. Show that the ladder
remains in equilibrium in this position. What is the frictional
resistance on the ladder at the point of contact between the
ladder and the floor?
Smooth wall
B
12m
[Ans.: FA = 52 N]
A
5m
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q17. A ladder of length 5m weighing 500N is placed at 45o
against a vertical wall. µ between the ladder and the wall
is 0.20 & between ladder and ground is 0.50. If a man
weighing 600N ascends the ladder, how high will he be
when the ladder just slips. If a boy now stands on the
bottom rung of the ladder, what must be his least weight
so that the man can go to the top of the ladder.
[Ans.: (a) x = 2.92m (b) Wboy = 458N]
3.
CENTROID OF PLANE AREA
3.
CENTROID
Centre of gravity : of a body is the point at which the whole
weight of the body may be assumed to be concentrated.
A body is having only one center of gravity for all
positions of the body.
It is represented by CG. or simply G or C.
Contd.
CENTRE OF GRAVITY
Consider a block of uniform thickness and
having a uniform mass m.
W
R
It is possible to support (hold) the
block in stable position by a rod as
shown in the figure provided rod must
be positioned exactly at the point of
intersection of the diagonals.
Or the rod must be supported
exactly below where the total weight of
the block act.
Contd.
W
CENTRE OF GRAVITY
The block can be supported from any position
provided the support rod and the line of action
of weight are in same line.
This indicates that the whole weight of the
block act through one point. This point is
called as centre of gravity.
R
Centre of gravity is that point about which
the summation of the first moments of the
weights of the elements of the body is zero.
Contd.
W
CENTRE OF GRAVITY
x
W3
W4
To determine mathematically the
W2
W1
location of the “centre of gravity”
of any body, we apply the
X1
“principle of moments” to the
X2
parallel system of gravitational
forces
The moment of the
resultant gravitational
force W, about any axis
=
the algebraic sum of the
moments about the same axis
of the gravitational forces dW
acting on all infinitesimal
elements of the body. Contd.
W
CENTRE OF GRAVITY
x
dW3
dW4
dW2
dW1
X1
x ⋅ W = ∫ x × dW
X2
x ⋅ W = dW1 × x1 + dW2 × x2 + dW3 × x3 + ................dWn × xn
Where
W =
∫ dW
Contd.
x
CENTRE OF GRAVITY
where
x = x- coordinate of centre of gravity
x ⋅ dW
∫
x=
W
x
Similarly, y and z coordinates of the centre of gravity are
y ⋅ dW
∫
y=
W
and
z ⋅ dW
∫
z=
W
----(1)
x
CENTRE OF MASS
With the substitution of W= m g
and dW = g dm
(if ‘g’ is assumed constant for all particles, then )
x ⋅ dm
∫
x=
m
,
y ⋅ dm
∫
y=
m
,
z ⋅ dm
∫
z=
m
----(2)
Equation 2 is independent of g and therefore define a unique point in
the body which is a function solely of the distribution of mass. This point
is called the centre of mass and clearly coincides with the centre of
gravity as long as the gravity field is treated as uniform and parallel
Contd.
When speaking of an actual physical body, we use the term
“centre of mass”.
The term centroid is used when the calculation concerns a
geometrical shape only.
Calculation of centroid falls within three distinct categories,
depending on whether we can model the shape of the body
involved as a line, an area or a volume.
Contd.
The centroid “C” of the Volume segment,
x ⋅ dV
∫
x=
,
V
y ⋅ dV
∫
y=
,
V
z ⋅ dV
∫
z=
V
The centroid “C” of the line segment,
x ⋅ dL
∫
x=
L
,
y ⋅ dL
∫
y=
L
,
z ⋅ dL
∫
z=
L
The centroid “C” of the Area segment,
AREA: when the density ρ, is constant and the body
has a small constant thickness t, the body can be
modeled as a surface area.
The mass of an element becomes dm = ρ t dA.
If ρ and t are constant over entire area, the
coordinates of the ‘centre of mass’ also becomes the
coordinates of the centroid, C of the surface area and
which may be written as
x ⋅ dA
∫
x=
,
A
y =
∫ y ⋅ dA ,
A
z ⋅ dA
∫
z=
A
Contd.
Centroid of Simple figures: using method of moment ( First
moment of area)
¾ Centroid of an area may or may not lie on the area in
question.
¾ It is a unique point for a given area regardless of the
choice of the origin and the orientation of the axes about
which we take the moment.
The coordinates of the centroid of the surface area
about any axis can be calculated by using the equn.
(A) x = (a1) x1 + (a2) x2 + (a3) x3 +
……….+(an) xn
= First moment of area
Moment of Total
area ‘A’ about yaxis
=
Algebraic Sum of
moment of elemental
‘dA’ about the same
axis
where (A = a1 + a2 + a3 + a4 + ……..+ an)
AXIS of SYMMETRY:
It is an axis w.r.t. which for an elementary area on one side
of the axis , there is a corresponding elementary area on
the other side of the axis (the first moment of these
elementary areas about the axis balance each other)
™If an area has an axis of symmetry, then the centroid
must lie on that axis.
™If an area has two axes of symmetry, then the centroid
must lie at the point of intersection of these axes.
Contd.
For example:
The rectangular shown in
the figure has two axis of
symmetry, X-X and Y-Y.
Therefore intersection of
these two axes gives the
centroid of the rectangle.
Y
B/2 B/2
da
D/2
X
da × x = da × x
Moment of areas,da
about y-axis cancel
each other
da × x + da × x = 0
D/2
x
x
da
D
X
B
Y
Contd.
AXIS of SYMMETYRY
‘C’ must lie
on the axis
of symmetry
‘C’ must lie on
the axis of
symmetry
‘C’ must lie at the intersection
of the axes of symmetry
To locate the centroid of simple rectangular
area from first principles
D
B
To locate the centroid w.r.t. the base line x-x
Let the distance of centroid
from the base line x-x be y
D
Then from the Principle of
Moments
y
X
X
A ⋅ y = ∫ y ⋅ da
Moment of
Total area A
about x-axis
=
Sum of moment of
elemental area dA
about the same axis
Contd.
Consider a elemental area dA at a distance y from the
base line (x-x)
Let the thickness of the element be ‘dy’
B
dy
dA
y
x
Area of small element
= dA = B .dy
Moment of this elemental area
about x-x axis
y
x
= (area) x (distance)
= (B.dy) . (y)
Contd.
Sum of Moment of all such elemental areas
comprising the total area =
= ∫ y ⋅ da
= ∫ B ⋅ dy ⋅ y
D
⎡ BD 2 ⎤
=⎢
⎥
2
⎣
⎦
⎡ By ⎤
=⎢
⎥
⎣ 2 ⎦0
2
Then from the Principle of Moments
BD 2
y=
2A
,
BD 2
y=
2 BD
,
BD 2
Ay =
2
D
y=
2
Contd.
Similarly we can show
B
Y
x
Y
x
dx
A ⋅ x = ∫ x ⋅ da
x = B/2
= ∫ x ⋅ (D ⋅ dx )
B
⎡ Dx ⎤
=⎢
⎥
2
⎣
⎦0
2
To locate the centroid of simple right angle triangular
area from first principles
H
B
To locate the centroid w.r.t. the base line x-x.
Let the distance of
centroid from the base
line x-x be y
H
y
x
x
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
Moment of Total area
A about
x-axis
=
Algebraic Sum of moment of
elementary area ‘dA’ about the same
axis.
Contd.
Consider a small elemental area ‘dA’ at a distance ‘y’
from the base line (x-x)
Let the thickness of the element be ‘dy’
Area of small element
= dA = b .dy
b
dy
dA
x
y
y
x
Moment of this small
elemental area about xx axis
= (area) x (distance)
= (b.dy) . (y)
Contd.
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
da = b × dy =
b=
A ⋅ y = ∫ b ⋅ dy ⋅ y
A⋅ y = ∫
B (H − y )
× dy
H
B (H − y )
H
B (H − y )
⋅ dy ⋅ y
H
H/3
y = H/3
H
y
X
B
X
Contd.
Similarly we can prove that x = B/3
da = h ⋅ dx =
A ⋅ x = ∫ x ⋅ da
A ⋅ x = ∫ b ⋅ dx ⋅ x
A⋅ x = ∫
H (B − x )
⋅ dx
B
Y
H (B − x )
⋅ dx ⋅ x
B
b=
H (B − x )
B
dx
H
x = B/3
x
b
x
B/3
Y
B
Contd.
The centroid of simple right angled triangle
area from the base
B/3
Centroid
H
B
H/3
To locate the centroid of Semi Circular Area w.r.t. the
diameter AB from first principles
Consider a semicircle of
radius R,
2
π
D
A = area =
D = 2R
8
Let ‘G’ be the centroid of the Semicircle, and y is its
distance from the diameter AB.
Contd.
dA = r × dθ × dr
Consider a small elemental
area da, located at distance y
from the diameter AB,
Let ‘r’ =radial distance of area ‘da’ from centre of the semi
circle.
da = r × dθ × dr
y = r × sin θ
Moment of this elemental area about the diameter AB =
= r 2 sin θ ⋅ dr ⋅ dθ
Contd.
Moment of all such elemental
area ‘da’ about the diameter
AB
π R
= ∫ ∫ r 2 ⋅ sin θ ⋅ dr.dθ
OO
π
R
⎡r ⎤
= ∫ ⎢ ⎥ ⋅ sin θ ⋅ .dθ
3 ⎦O
O⎣
3
R3
π
=
(−[cos θ ]O )
3
2.R 3
=
3
Contd.
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
A⋅ y =
y
π R
2
r
∫ ∫ ⋅ sin θ ⋅ dr.dθ
Centroid
OO
A⋅ y =
2.R 3
3
R
4R
y=
3π
Because of symmetry
x
x
y
x=0
R
4R
y=
3π
To locate the centroid of Quarter Circular Area w.r.t. the
boundary radial line AB from first principle
A ⋅ y = ∫ y ⋅ da
π
A⋅ y =
dr
2 R
2
r
∫ ∫ ⋅ sin θ ⋅ dr.dθ
OO
r
y
dθ
A
θ
B
4R
y=
3π
R
Contd.
Centroid of Quarter Circular Area w.r.t. the boundary radial
lines
x=
4R
3π
Centroid
y=
4R
3π
4R
y=
3π
To locate the centroid of Circular Sector w.r.t. the y-axis
shown from first principle
y
Xc=(2/3)Rcosθ
da
Consider a triangle of
differential area = da =
=
1
× base × height
2
1
= × R × dθ × R
2
Distance of the differential area
‘da’, from y-axis = xc =
=
2
× R × cos θ
3
Contd.
To locate the centroid of Circular Sector w.r.t. the y-axis
shown from first principle
Consider a triangle of
differential area = da =
=
1
× base × height
2
1
= × R × dθ × R
2
Distance of the differential area
‘da’, from y-axis = xc =
=
2
× R × cos θ
3
Contd.
Then from the Principle of Moments
α
A=∫
1
−α 2
A × x = ∫ da × xc
R × dθ × R = R × α
2
(
α
)
2
1 2
R × α x = ∫ ⋅ R cos θ × R dθ
3
2
−α
2
(
)
2
R × α x = × R 3 × sin α
3
2
2 R × sin α
x=
3
α
2 R × sin α
x=
α
3
For a semicircular area 2α = π, if we use
this value in the above formula we get
y =
4 R
3π
To locate the centroid of area under the curve x = k y3
from x = 0 to x = a from first principle
Consider a vertical element of
area da = y dx at a distance x
from the y-axis.
da
To find x- coordinate,
x
x
X
A × x = ∫ da × x
a
At x = a, y = b,
0
i.e. a = k b3,
A = ∫ y ⋅ dx
k = a/b3
Contd.
Substituting y = (x/k)1/3 and k = a/b3
a
a
0
0
x ∫ ydx = ∫ xydx
a
1
3
a
1
3
⎛ x⎞
⎛ x⎞
x ∫ ⎜ ⎟ dx = ∫ x⎜ ⎟ dx
k⎠
k⎠
0⎝
0 ⎝
3ab
3a 2b
⋅x=
4
7
4
x= a
7
Contd.
To find y, Coordinate of centroid of the rectangular
element is
yc = y/2
A × y = ∫ yc × dA
a
a
⎛ y⎞
y ∫ ydx = ∫ ⎜ ⎟( ydx )
2⎠
0
0⎝
Substituting,
y = b( x/a)1/3
3ab
3ab 2
⋅y=
4
10
2
y= b
5
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.1:
Locate the centroid of the shaded area shown
50
10
10
40
Ans: x=12.5, y=17.5
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.2:
Locate the centroid of the shaded area shown
D=600
r=600
300
300
500
1000 mm
500
1000 mm
Ans: x=474mm, y=474mm
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.3:
Locate the centroid of the shaded area w.r.t. to the axes
shown
y-axis
r=40
x-axis
60
20
120
20
90
Ans: x=34.4, y=40.3
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.4:
Locate the centroid of the shaded area w.r.t. to the axes
shown
y-axis
250 mm
10
200 mm
20
10
380
10
x-axis
Ans: x= -5mm, y=282mm
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.5
Locate the centroid of the shaded area w.r.t. to the axes
shown
y 30
50
40
40
30
20
20
x
r=20
Ans:x =38.94, y=31.46
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.6
Locate the centroid of the shaded area w.r.t. to the axes
shown
1.0
2.4 m
y
r=0.6
1.5
1.0
1.0
x
1.5
Ans: x=0.817, y=0.24
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.7
Locate the centroid of the shaded area w.r.t. to the axes
shown
Ans: x= -30.43, y= +9.58
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.8
Locate the centroid of the shaded area.
20
Ans: x= 0, y= 67.22(about base)
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.9
Locate the centroid of the shaded area w.r.t. to the base
line.
2
Ans: x=5.9, y= 8.17
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.10
Locate the centroid of the shaded area w.r.t. to the axes
shown
Ans: x=21.11, y= 21.11
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.11
Locate the centroid of the shaded area w.r.t. to the axes
shown
Ans: x= y= 22.22
EXERCISE PROBLEMS
3. Centroid of plane area
Problem No.12
Locate the centroid of the shaded area w.r.t. to the axes
shown
Y
R=25
R=25
80
X
50
50
Ans: x= 24.33 y= 4.723
75
4. MOMENT OF INERTIA OF
PLANE AREA
4. MOMENT OF INERTIA
Moment of Inertia( Second moment area)
The product of the elemental area and square of the
perpendicular distance between the centroid of area and the
axis of reference is the “Moment of Inertia” about the reference
axis
y
Iox= da1 y12 + da2 y22+ da3 y32+ -= ∑ da
y2
dA
x
Ioy = da1 x12 + da2 x22 + da3 x32+ ---= ∑ da x2
y
o
Radius of Gyration
Elemental area
A
r1
A
k
r2
k
r3
k
B
B
Radius of gyration is defined as a constant distance of all
elemental areas which have been rearranged with out
altering the total moment of inertia.
IAB= da k2 + da k2 + -----------
IAB= A k2
IAB = ∑ da k2
k=√IAB/A
Polar moment of Inertia
(Perpendicular Axes theorem)
The moment of inertia of an area about an axis perpendicular
to the plane of the area is called “Polar Moment of Inertia” and
it is denoted by symbol Izz or J or Ip. The moment of inertia of
an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 +
y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy
Y
x
r
y
O
x
z
Polar moment of Inertia
(Perpendicular Axes theorem)
Hence polar M.I. for an area w.r.t. an axis perpendicular to its
plane of area is equal to the sum of the M.I. about any two
mutually perpendicular axes in its plane, passing through the
point of intersection of the polar axis and the area.
Parallel Axis Theorem
dA
x
*G
y
x
d
A
B
Moment of inertia of any area about an axis AB is equal to the
M.I. about parallel centroidal axis plus the product of the total
area and square of the distance between the two axes.
IAB =∑dA (d +y)2
= ∑dA (d2 + y2 + 2 × d × y)
=∑dA. d2 +∑dA y2 + ∑ 2×d×dA y
= ∑dA. d2 +∑dA y2 + 2×d. ∑ y dA
In the above term (2×d) is constant & ∑ y dA = 0
IAB = Ixx + A.d2
MOMENT OF INERTIA BY DIRECT INTEGRATION
Moment of inertia of rectangular area about centroidal
horizontal axis by direct integration
M.I. about its horizontal centroidal axis :
+D / 2
I x x = ∫ dA × y 2
−D / 2
+D / 2
dy
D/2
= ∫ ( B × dy ) × y
x
−D / 2
BD 3
=
12
2
y
D
G
B
x
Moment of Inertia of rectangular area about its base(about the line AB) using Parallel Axis Theorem
IAB = IXX + A(d)2
.
Where d = D/2, the distance
between axes xx and AB
dy
D/2
3
=BD /12+(BD)(D/2)
2
x
y
D
G
x
=BD3/12+BD3/4
=BD3/3
A
B
B
Moment of inertia of Triangular area about the base
by direct integration
dy
h
(h-y)
x
x
x
y
h/3
A
B
b
From similar triangles
b/h = x/(h-y)
∴ x = b . (h-y)/h
dy
h
x
x
A
B
b
Ixx = ∫
h
(b . (h-y) y2.dy) /h
0
= b[ h (y3/3) – y4/4 ]/h
= bh3/12
x
y
h/3
IAB = ∫ dA.y2 = ∫ (x.dy)y2
(h-y)
Moment of inertia of Triangular area about the
centroidal horizontal axis
Using Parallel axis theorem .
MI about any line(AB) = MI about cenroidal parallel axis + Ad2
IAB
= Ixx + Ad2
Ixx = MI about centroidal axis x x
IAB= MI about the Base line AB
Ixx = IAB – Ad2
dy
h
x
(h-y)
x
h/3
A
x
y
B
= bh3/12 – bh/2 . (h/3)2
b
= bh3/36
Centroidal horizontal axis
Moment of inertia of Circular area about the centroidal
horizontal axis
Ixx = ∫ dA . y2
dθ
R 2π
= ∫ ∫ (x.dθ.dr) r2Sin2θ
0 0
R 2π
=∫ ∫
r3.dr
Sin2θ
r
dθ
0 0
R
θ
x
R
2π
=∫ r3 dr ∫ {(1- Cos2θ)/2} dθ
0
=[r4/4]
R
0
[θ/2 – Sin2θ/4]
0
2π
0
= R4/4[π - 0] = πR4/4
IXX = π R4/4 = πD4/64
y=rSinθ
A
B
x
Moment of inertia of Semi-circular area about the
Base & centroidal horizontal axis
IAB = ∫ dA . y2
R π
= ∫ ∫ (r.dθ.dr) r2Sin2θ
0 0
R
π
=∫ r3.dr ∫ Sin2θ
0
R π
=∫ ∫ r3
y0
dθ
0
dr (1- Cos2θ)/2) dθ
0 0
=[R4/4]
[θ/2 – Sin2θ/4]
π
0
= R4/4[π/2 - 0]
= πR4/8 = (πD4/32)
x
4R/3π
A
R
y0
x
B
Moment of inertia of Semi-circular area about the
centriodal horizontal axis
using parallel axis theorem:
2
IAB = Ixx + A(d)
Ixx = IAB – A(d)2
= π R4/8
Ixx = 0.11R4
πR2/2 . (4R/3π)2
Moment of inertia of Quarter-circular area about the
base & centroidal horizontal axis
D
IAB = ICD
y
R π/2
IAB = ∫ ∫ (r.dθ.dr). r2Sin2θ
0
R
0
π/2
x
=∫ r3.dr ∫ Sin2θ dθ
0
4R/3π
0
R
π/2
=∫ r3 dr ∫ (1- Cos2θ)/2) dθ
0
x
0
C
π/2
=[R4/4] [θ/2 – (Sin2 θ)/4]
0
= R4 (π/16 – 0) = πR4/16
A
4R/3π
y
B
Moment of inertia about Centroidal axis,
Ixx = IAB - Ad2
= πR4/16 - πR2. (0. 424R)2
= 0.055R4
Sl.No
Y
1
x0
x
2
Figure
b
Y
x0
x0
x
b
3
O
y0
y0
x
5
y
y0
y0
I xx
I yy
bd3/12
-
bd3/3
-
bh3/36
-
bh3/12
-
πR4/4
πR4/4
-
-
0.11R4
πR4/8
πR4/8
-
0.055R4
0.055R4
πR4/16
πR4/16
h/3
x
x0
4R/3π
x0
x
x0 4R/3π
4R/3π
-y
0 0
y0
R
x0
Iy
x
Xo
h
4
-x
0 0
d
x0
d/2
x0
Ix
EXERCISE PROBLEMS
Q1. Determine the moment of inertia about the
centroidal axes.
30mm
30mm
20
30mm
100mm
[Ans: Y = 27.69mm Ixx = 1.801 x 106mm4
Iyy = 1.855 x 106mm4]
EXERCISE PROBLEMS
Q2. Determine second moment of area about the centroidal
horizontal and vertical axes.
300mm
300mm
200
200mm
900mm
[Ans: X = 99.7mm from A, Y = 265 mm
Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]
EXERCISE PROBLEMS
Q3. Determine M.I. Of the built up section about the horizontal
and vertical centroidal axes and the radii of gyration.
200mm
20
60
140mm
20
100mm
[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4
rxx = 62.66mm, ryy = 45.63mm]
EXERCISE PROBLEMS
Q4. Determine the horizontal and vertical centroidal M.I. Of the
shaded portion of the figure.
60
X
60
20
20
X
60
[Ans: X = 83.1mm
Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]
EXERCISE PROBLEMS
Q5. Determine the spacing of the symmetrically placed vertical
blocks such that Ixx = Iyy for the shaded area.
200mm
400mm
200mm
d
200mm
200mm
600mm
[Ans: d/2 = 223.9mm d=447.8mm]
EXERCISE PROBLEMS
Q6. Find the horizontal and vertical centroidal moment of inertia
of the section shown in Fig. built up with R.S.J. (I-Section) 250 x
250 and two plates 400 x 16 mm each attached one to each.
Properties of I section are
Ixx = 7983.9 x 104mm4
Iyy = 2011.7 x
104mm4
160mm
2500mm
Cross sectional area=6971mm2
160mm
4000mm
[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]
EXERCISE PROBLEMS
Q7. Find the horizontal and vertical centroidal moment of inertia
of built up section shown in Figure. The section consists of 4
symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.
Properties of ISA
Cross sectional area = 4400mm2
Ixx = Iyy ;Cxx = Cyy =18.5mm
200mm
18.5mm
18.5mm
20mm
300mm
[Ans: Ixx = 111.078 x 107mm4, Iyy = 39.574 x 107mm4]
EXERCISE PROBLEMS
Q8. The R.S. Channel section ISAIC 300 are placed back to
back with required
to keep them in place. Determine the
clear distance d between them so that Ixx = Iyy for the composite
section.
Properties of ISMC300
C/S Area =
Y
4564mm2
Ixx = 6362.6 x 104mm4
Iyy = 310.8 x
23.6mm
104mm4
Cyy = 23.6mm
X 380mm
X
d
[Ans: d = 183.1mm]
Lacing
Y
EXERCISE PROBLEMS
Q9. Determine horizontal and vertical centroidal M.I. for the
section shown in figure.
40mm
160mm
40mm
40mm
90mm
[Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]
5. Kinetics of rectilinear motion
5. Kinetics of rectilinear motion
In this chapter we will be studying the
relationship between forces on a body/particle
and the accompanying motion
Newton’s Second law of motion:
Newton’s first and third law of motion
were used extensively in the study of statics
(the bodies at rest) whereas Newton’s second
law of motion is used extensively in the study
of the kinetics.
Work done by force:
Work done by a force is the product of the force and the
distance moved by the point of application in the direction of
the force. It is a scalar quantity.
F sinθ
F
F sinθ
F
F cosθ
A
α
B
F cosθ
α
s
Work done = (F cosα ) × s
X- component of force F moves through distance a S,
S = displacement of force from A to B
Unit: Nm ( Joule )
POWER:It is defined as the time rate of doing work.
Power = work done /Time= (force × distance) /Time
= force × velocity
Unit: (Nm)/s = [watt]
(kN m)/s = [kilo watt]
1 metric H.P=735.75 watts.
Energy:It is defined as the capacity to do work. It is a scalar
quantity.
Unit :- N m (Joule)
Momentum:Quantity of motion possessed by a body is called momentum. It
is the product of mass and velocity. It is a vector quantity.
Unit:- N s.
Impulse of a Force:It is defined as the product of force and the time over which it
acts. It is a vector quantity.
Unit:- N s.
Newton’s second law of motion.
“If the resultant force acting on a particle is not zero , the
particle will have an acceleration proportional to the
magnitude of the resultant force and its direction is along
that of the resultant force.”
Fαa
F =Resultant of forces
a = Acceleration of the particle.
F = ma
m= mass of the particle.
The constant value obtained for the ratio of the
magnitude of the force and acceleration is
characteristic of the particle and is denoted by ‘m’.
Where ‘m’ is mass of the particle
Since ‘m’ is a +ve scalar, the vectors of force ‘F’and
acceleration ‘a’ have the same direction.
Units
Force in Newtons (N)
Acceleration in m/s2
N = 1 Kgm/s2
e
h
t t
n
i an
e
a
v ult
o
m
s
m
e
=
l
r
l f
i
R
w o
y
d tion
o
B rec
di
F2
F1
F3
R = Resultant of forces F1,F2 and F3
Using the rectangular coordinate system we have
components along axes as,
ΣFx = max
ΣFy = may
ΣFz = maz
where Fx ,Fy Fz and ax , ay ,az are rectangular
components of resultant forces and accelerations
respectively.
Newton’s second law may also be expressed by
considering a force vector of magnitude ‘ma’ but of sense
opposite to that of the acceleration. This vector is denoted
by (ma)rev. The subscript indicates that the sense of
acceleration has been reversed and is called the inertia
force vector.
o
m
ill
w t
y
d ltan
o
B su
re
in
e
v
t
d
e
h
R
t io
c
e
ir
=
m
n
of
e
c
r
o
F
a
a
i
t
m
r
e
=
In
R
a
F2
F2
F1
F3
R = Resultant of forces F1,F2 and F3
F1
F3
a
e
c
m or
=
F
R tia
=
F Iner
F2
F1
F3
If the inertia force vector is
added to the forces acting
on the particle we obtain a
system of forces whose
resultant is zero.
Resultant of forces F1,F2, F3 and Inertia force = 0
F1 + F2 + F3 + ma = 0
The particle may thus be considered to be in
equilibrium. (THIS IS DYNAMIC EQUILIBRIUM)
It was pointed out by D’Alembert (Alembert, Jean le
Rond d’ (1717-1783), French mathematician and
philosopher) that problems of kinetics can be solved by
using the principles of statics only (the equations of
equilibrium) by considering an inertia force in a direction
directly opposite to the acceleration in addition to the real
forces acting on the system
D’Alembert’s principle states that
When different forces act on a system such that it is in
motion with an acceleration in a particular direction, the
vectorial sum of all the forces acting on the system
including the inertia force (‘ma’ taken in the opposite
direction to the direction of the acceleration) is zero.
The problem under consideration may be solved by using
the method developed earlier in statics. The particle is said
to be in dynamic equilibrium.
If
ΣFx = 0
ΣFy= 0
including inertia force vector
ΣFz = 0
This principle is known as D’Alembert’s principle
Work-Energy relation for translation
From Newton’s second law of motion
∑F=m×a =
Also
-----------(1)
a = dv/dt =( dv/dt) ×( ds/ds) = v × dv/ds
substituting in (1)
∑ F = m × v × dv/ds
ΣF × ds = m × v × dv ------------------(2)
Let the initial velocity be u and the final velocity after it
moves through a distance ‘ s’ be v
Integrating both sides, we get
s
v
0
u
∑ F ∫ ds = m ∫ v dv
v2
v
F
s
m
(
)
×
=
∑
2
u
1
2
2
∑ F × s = m(v − u )
2
Therefore work done by a system of forces acting on a body
while causing a displacement is equal to the change in kinetic
energy of the body during the displacement.
Impulse-momentum relationship
F=m×a
F = m × (v - u)/t = (mv – mu)/t
Force = Rate of change of momentum
F × t = mv – mu
Impulse = final momentum – Initial momentum
The component of the resultant linear impulse along any
direction is equal to change in the component of
momentum in that direction.
EXERCISE PROBLEMS
5. Kinetics
Q1. Blocks A and B of mass 10 kg and 30 kg respectively
are connected by an inextensible cord passing over a
smooth pulley as shown in Fig. Determine the velocity of
the system 4 sec. after starting from rest. Assume
coefficient of friction =0.3 for all surfaces in contact.
B
60o
Ans: v=13.6m/s
A
30o
EXERCISE PROBLEMS
5. Kinetics
Q2.
A tram car weighs 150kN. The tractive
resistance being 1% of the weight of car. What
power will be required to move the car at uniform
speed of 20 kmph
(i) Up an incline 1 in 300
(ii) (ii) Down an incline 1 in 250. Take efficiency
75%.
Ans:
Pull = 2 kN
a) Output power=11.12kW
EXERCISE PROBLEMS
5. Kinetics
Q3. Two masses of 5 kg and 3 kg rest on two smooth inclined
plane, each of inclination 30º and are connected by a
string passing over a common apex. Find the velocity of 3 kg
mass after 2 sec when released from rest. Find the distance
it will cover before changing direction of motion, if 5kg mass
is cut off after two sec of its release from rest.
5kg
30º
V = 4.45 m/s
s = 0.61 m
3kg
30º
EXERCISE PROBLEMS
5. Kinetics
Q4. A locomotive weighing 900 kN pulls a train of 10 coaches each
weighing 300 kN at 72 Kmph on a level track against a resistance of 7
N/kN. If the rear 4 coaches get snapped from the train, find the speed of
the engine and the remaining coaches after 120 secs. Assume no
change in resistance and draw bar pull. Find also distance traveled by
detached coaches before coming to rest.
4x300=1200kN
6x300=1800kN
900kN
P
V = 23.66 m/s
s = 2.9 km
EXERCISE PROBLEMS
5. Kinetics
Q5. Find the tension in the cord supporting body C in Fig.
below. The pulleys are frictionless and of negligible
weight.
A
C
150 kN
B
450 kN
Ans : T=211.72 kN
300 kN
Assume all blocks
moving either
downward or
upward and
accordingly draw
FBD
0=aA +2aB +cC
EXERCISE PROBLEMS
5. Kinetics
Q6. Two blocks A and B are released from rest on a 30o
inclined plane with horizontal, when they are 20m apart.
The coefficient of friction under the upper block is 0.2
and that under lower block is 0.4. compute the time
elapsed until the block touch. After they touch and move
as a unit what will be the constant forces between them.
(Ans : t = 4.85 s, contact force=8.65 N)
EXERCISE PROBLEMS
5. Kinetics
Q7. An elevator cage of a mine shaft weighing 8kN when
empty is lifted or lowered by means of rope. Once a man
weighing 600N entered it and lowered at uniform
acceleratin such that when a distance of 187.5 m was
covered, the velocity of the cage was 25m/s. Determine
the tension in the cable and force exerted by man on the
floor of the cage.
(Ans: T=7139 N and R=498 N)
EXERCISE PROBLEMS
5. Kinetics
Q8. A small block starts from rest at point a and slides down the
inclined plane. At what distance along the horizontal will it travel
before coming to rest . Take µk=0.3 [Ans :s=6m ]
5m
A
3
4
C
B
s
EXERCISE PROBLEMS
5. Kinetics
Q9. The system starts from rest in the position shown . How
much further will block ‘A’ move up the incline after block B hits
the ground . assume the pulley to be frictionless and massless
and µ is 0.2 .WA=1000N, WB=2000N. [ Answer s =1.27m]
A
4
3
B
3m
EXERCISE PROBLEMS
5. Kinetics
Q10. A 1500Kg automobile travels at a uniform rate of
50kmph to 75kmph . During the entire motion, the
automobile is traveling on a level horizontal road and
rolling resistance is 2 % of weight of automobile . Find
(i) maximum power developed (ii) power required to
maintain a constant speed of 75kmph.
[ ANSWER: power developed = 6.131KN]
EXERCISE PROBLEMS
5. Kinetics
Q11. Two bodies A and B weighing 2000N and 5200N are
connected as shown in the figure . find the further distance
moved by block a after the block B hits Wall. µ=0.2 .[ Answer
s=1.34m]
A
5
12
B
3m
EXERCISE PROBLEMS
5. Kinetics
Q12. A spring is used to stop 60kg package which is sliding
on a horizontal surface . the spring has a constant k = 20kN/m
and is held by cable such that it is initially compressed at
120mm. knowing that the package has a velocity of 2.5m/s in
position shown and maximum additional displacement of
spring is 40mm . Determine the coefficient of kinetic friction
between package and surface. (Answer µk=0.2)
2.5 m/s
60kg
600m
EXERCISE PROBLEMS
5. Kinetics
Q13. The system shown in figure has a rightward velocity of
4m/s, just before force P is applied. Determine the value of P
that will give a leftward velocity of 6m/s in a time interval of
20sec. Take µ = 0.2 & assume ideal pulley. [Answer
P=645.41N]
P
1000N
400N
EXERCISE PROBLEMS
5. Kinetics
Q14. A locomotive of weight 500kN pulls a train of weight of
2500kN. The tractive resistance, due to friction is 10N/kN. The
train can go with a maximum speed of 27kmph on a grade of
1in100. Determine (a) Power of the locomotive. (b) Maximum
speed it can attain on a straight level track with the tractive
resistance remaining same. [Answer (a) Power= 450kN (b) v=15m/s]
Q15. A wagon weighing 400kN starts from rest, runs 30m down
a 1% grade & strikes a post. If the rolling resistance of the track
is 5N/kN, find the velocity of the wagon when it strikes the post.
If the impact is to be cushioned by means of one bumper string,
which compresses 1mm per 20 kN weight, determine how much
the bumper spring will be compressed. [Answer v=1.716m/s, x=77.5mm]
EXERCISE PROBLEMS
5. Kinetics
Q16. A train whose weight is 20kN moves at the rate of
60kmph. After brakes are applied, it is brought to rest in
500m. Find the force exerted, assuming it to be uniform
a) Use work-energy relation
b) Use D’Alemberts equation.
Ans: F = 5.663 kN
EXERCISE PROBLEMS
5. Kinetics
Q17. The blocks A and B having weights 100 N and 300
N start from rest. The horizontal plane and the pulleys
are frictionless. Determine the acceleration and the
tension in the string.
string
Frictionless pulley
A
string
Frictionless pulley
Ans: aA=8.403m/s2
aB=4.201 m/s2
T= 85.71 N
B
EXERCISE PROBLEMS
5. Kinetics
Q18. The blocks A and B having weights 100 N and 300
N start from rest. The horizontal plane and the pulleys
are frictionless. Determine the velocity of block B after
0.5 seconds and the tension in the string. Use impulsemomentum relation.
string
Frictionless pulley
A
string
Frictionless pulley
B
EXERCISE PROBLEMS
5. Kinetics
Q19. The blocks A and B having weights 100 N and 300
N start from rest when a load of 100 N is applied on the
block A as shown in the figure. The horizontal plane and
the pulleys are frictionless. Determine the acceleration
and the tension in the string.
string
100 N
Frictionless pulley
A
string
Frictionless pulley
B
EXERCISE PROBLEMS
5. Kinetics
Q20. Two blocks A and B are connected as shown in the
figure. At the instant of their release if the block A, which
is on smooth horizontal plane has a left ward velocity of
2 m/s, what would be its velocity 5 seconds after their
release. The blocks A and B weigh 100 N and 300 N
respectively.
string
Frictionless pulley
A
string
Frictionless pulley
B
EXERCISE PROBLEMS
5. Kinetics
Q21. An engine weighing 500 kN drags carriages weighing
1500kN up an incline of 1 in 100 against a resistance of
5N/kN starting from rest. It attains a velocity of 36 kmph
(10m/s) in 1 km distance with a constant draw bar pull
supplied by the engine. What is the power required for the
same ? What is the tension developed in the link
connecting the engine and carriages?
1500N
500N
P
100
1
Ans:
Pull=40.19kN,
power=401kW,
T=30.15kN
EXERCISE PROBLEMS
5. Kinetics
Q22. what velocity the block A will attain after 2
seconds starting from rest? Take µ = 0.2. WA =
1500N, WB = 2000N. Use impulse-momentum
relation.
A
4
3
3
4
B
PART - II
Mechanics of Deformable Bodies
COURSE CONTENT IN BRIEF
6. Simple stresses and strains
7. Statically indeterminate problems and thermal
stresses
8. Stresses on inclined planes
9. Stresses due to fluid pressure in thin cylinders
6. Simple stresses and strains
The subject strength of materials deals with the relations
between externally applied loads and their internal effects on
bodies. The bodies are no longer assumed to be rigid and the
deformations, however small, are of major interest
The subject, strength of materials or mechanics of materials
involves analytical methods for determining the strength ,
stiffness (deformation characteristics), and stability of various
load carrying members.
Alternatively the subject may be called the mechanics of solids.
GENERAL CONCEPTS
STRESS
No engineering material is perfectly rigid and hence,
when a material is subjected to external load, it
undergoes deformation.
While undergoing deformation, the particles of the
material offer a resisting force (internal force). When
this resisting force equals applied load the equilibrium
condition exists and hence the deformation stops.
These internal forces maintain the externally applied
forces in equilibrium.
STRESS
The internal force resisting the deformation per unit area is
called as stress or intensity of stress.
Stress = internal resisting force / resisting cross sectional
area
R
=
A
STRESS
SI unit for stress
N/m2 also designated as a pascal (Pa)
Pa = N/m2
kilopascal, 1kPa = 1000 N/m2
megapascal, 1 MPa = 1×106 N/m2
= 1×106 N/(106mm2) = 1N/mm2
1 MPa = 1 N/mm2
gigapascal, 1GPa = 1×109 N/m2
= 1×103 MPa
= 1×103 N/mm2
AXIAL LOADING – NORMAL STRESS
P
P
R
Consider a uniform bar of cross
sectional area A, subjected to a
tensile force P.
Consider a section AB normal to
the direction of force P
Let R is the total resisting force
acting on the cross section AB.
B
A
STRESS
Then for equilibrium condition,
R
P
R=P
P
Symbol:
Then from the definition of stress,
normal stress = σ = R/A = P/A
σ = Normal Stress
AXIAL LOADING – NORMAL STRESS
STRESS
Direct or Normal
Stress:
Intensity of resisting force perpendicular to or normal
to the section is called the normal stress.
Normal stress may be tensile or compressive
Tensile stress:
stresses that cause pulling on the surface of
the section, (particles of the materials tend to pull apart
causing extension in the direction of force)
Compressive stress: stresses that cause pushing on the
surface of the section, (particles of the materials tend to push
together causing shortening in the direction of force)
STRESS
• The resultant of the internal forces for
an axially loaded member is normal
to a section cut perpendicular to the
member axis.
• The force intensity on that section is
defined as the normal stress.
∆F
σ = lim
∆A→0 ∆A
P
σ ave =
A
STRAIN
STRAIN :
when a load acts on the material it will undergo
deformation. Strain is a measure of deformation produced by
the application of external forces.
If a bar is subjected to a direct load, and hence a stress, the
bar will changes in length. If the bar has an original length L
and change in length by an amount δL, the linear strain
produced is defined as,
δL
Change in length
ε
=
=
Linear strain,
Original length
L
Strain is a dimensionless quantity.
Linear Strain
P
σ = = stress
A
2P P
=
σ=
2A A
ε=
ε=
δ
L
= normal strain
δ
L
P
A
2δ δ
ε=
=
2L L
σ=
STRESS-STRAIN DIAGRAM
In order to compare the strength of various materials it is
necessary to carry out some standard form of test to establish
their relative properties.
One such test is the standard tensile test in which a circular
bar of uniform cross section is subjected to a gradually
increasing tensile load until failure occurs.
Measurement of change in length over a selected gauge
length of the bar are recorded throughout the loading
operation by means of extensometers.
A graph of load verses extension or stress against strain is
drawn as shown in figure.
STRESS-STRAIN DIAGRAM
Proportionality limit
Typical tensile test curve for mild steel
STRESS-STRAIN DIAGRAM
Typical tensile test curve for mild steel showing upper yield point
and lower yield point and also the elastic range and plastic range
Stress-strain Diagram
Limit of Proportionality :
From the origin O to a point called proportionality limit the
stress strain diagram is a straight line. That is stress is
proportional to strain. Hence proportional limit is the maximum
stress up to which the stress – strain relationship is a straight
line and material behaves elastically.
From this we deduce the well known relation, first postulated
by Robert Hooke, that stress is proportional to strain.
Beyond this point, the stress is no longer proportional to strain
PP
σP =
= Load at proportionality limit
A
Original cross sectional area
Stress-strain Diagram
Elastic limit:
It is the stress beyond which the material will not return to its
original shape when unloaded but will retain a permanent
deformation called permanent set. For most practical purposes
it can often be assumed that points corresponding proportional
limit and elastic limit coincide.
Beyond the elastic limit plastic deformation occurs and strains
are not totally recoverable. There will be thus some permanent
deformation when load is removed.
PE Load at proportional limit
σE =
=
A Original cross sectional area
Stress-strain Diagram
Yield point:
It is the point at which there is an appreciable elongation or
yielding of the material without any corresponding increase of
load.
PY
σY =
=
A
Load at yield point
Original cross sectional area
Ultimate strength:
It is the stress corresponding to
maximum load recorded during
the test. It is stress corresponding
to maximum ordinate in the
stress-strain graph.
PU
σU =
=
A
Maximum load taken by the material
Original cross sectional area
Stress-strain Diagram
Rupture strength (Nominal Breaking stress):
It is the stress at failure.
For most ductile material including structural steel breaking
stress is somewhat lower than ultimate strength because the
rupture strength is computed by dividing the rupture load
(Breaking load) by the original cross sectional area.
PB
σB =
=
A
load at breaking (failure)
Original cross sectional area
True breaking stress =
load at breaking (failure)
Actual cross sectional
area
Stress-strain Diagram
After yield point the graph becomes much more shallow and
covers a much greater portion of the strain axis than the
elastic range.
The capacity of a material to allow these large plastic
deformations is a measure of ductility of the material
Ductile Materials:
The capacity of a material to allow large extension i.e. the
ability to be drawn out plastically is termed as its ductility.
Material with high ductility are termed ductile material.
Example: Low carbon steel, mild steel, gold, silver, aluminum
Stress-strain Diagram
A measure of ductility is obtained by measurements of the
percentage elongation or percentage reduction in area,
defined as,
increase in gauge length (up to fracture)
×100
=
original gauge length
Percentage elongation
Reduction in cross sectional area
of necked portion (at fracture)
Percentage reduction in =
area
original area
Cup and cone fracture for a Ductile
Material
×100
Stress-strain Diagram
Brittle Materials :
A brittle material is one which exhibits relatively small
extensions before fracture so that plastic region of the tensile
test graph is much reduced.
Example: steel with higher carbon content, cast iron,
concrete, brick
Stress-strain diagram for a typical brittle material
HOOKE’S LAW
Hooke’s Law
For all practical purposes, up to certain limit the relationship
between normal stress and linear strain may be said to be
linear for all materials
stress (σ) α strain (ε)
stress (σ) constant
strain (ε) =
Thomas Young introduced a constant of proportionality that
came to be known as Young’s modulus.
stress (σ) E
strain (ε) =
=
Young’s Modulus
or
Modulus of Elasticity
HOOKE’S LAW
Young’s Modulus is defined as the ratio of normal stress to
linear strain within the proportionality limit.
stress (σ) =
E = strain (ε)
P δL PL
÷
=
A L AδL
The value of the Young’s modulus is a definite property of a
material
From the experiments, it is known that strain is always a very
small quantity, hence E must be large.
For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2
Deformations Under Axial Loading
• From Hooke’s Law:
σ = Eε
ε=
σ
E
=
P
AE
• From the definition of strain:
ε=
δ
L
• Equating and solving for the
deformation,
PL
δ =
AE
• With variations in loading, crosssection or material properties,
Pi Li
δ =∑
i Ai Ei
Consider an element of length, δx at a distance x from A
W
W
A
d1
B
x
dx
Diameter at x, = d1
(
d 2 − d1 )
+
×x
L
= d1 + k × x
Change in length over a
length dx is
Change in length over a
length L is
d2
c/s area at x,
2
d1
π
=
⎛
⎞
⎜
⎟
Wdx
⎛ PL ⎞
⎟
=⎜
⎟ =⎜
⎝ AE ⎠ dx ⎜⎜ π (d + kx )2 × E ⎟⎟
1
⎝4
⎠
⎛
⎞
⎟
L⎜
Wdx
⎟
=∫ ⎜
0
⎜ π (d + kx )2 × E ⎟
⎜
⎟
1
⎝4
⎠
4
=
π
4
(d1 + kx )2
Consider an element of length, δx at a distance x from A
⎛
⎞
⎜
⎟
L
Wdx
⎟
=∫ ⎜
0
⎜ π (d + kx )2 × E ⎟
⎜
⎟
1
4
⎝
⎠
Change in length over a
length L is
Put d1+kx = t,
dt ⎞
⎛
W
⎜
⎟
L
k ⎟
=∫ ⎜
0
⎜ π (t )2 × E ⎟
⎜
⎟
4
⎝
⎠
L
⎤
4W ⎡ t
4W
=
⎢
⎥ =
πEk ⎣ − 1 ⎦ 0 πEk
− 2 +1
4WL
WL
=
=
πEd1d 2 πd1d 2 × E
4
Then k dx = dt
L
− 4W
⎡ − 1⎤
⎢⎣ t ⎥⎦ = πEk
0
L
⎡
⎤
1
⎢
⎥
(
d
+
kx
)
⎣ 1
⎦0
Derive an expression for the total extension of the tapered bar
AB of rectangular cross section and uniform thickness, as
shown in the figure, when subjected to an axial tensile load ,W.
W
W
d1
b
d2
B
A
L
b
W
d1
b
W
d2
B
A
x
b
dx
Consider an element of length, δx at a distance x from A
depth at x,
= d1 +
(d 2 − d1 ) × x
L
= d1 + k × x
Change in length over a
length dx is
c/s area at x, = (d1 + kx )b
⎛
⎞
Wdx
⎛ PL ⎞
⎜
⎟⎟
=⎜
⎟ =⎜
⎝ AE ⎠ dx ⎝ (d1 + kx )b × E ⎠
Change in length over a
length L is
⎛
⎞
Wdx
⎟⎟
= ∫ ⎜⎜
0
⎝ (d1 + kx )b × E ⎠
L
P
(log e d 2 − log e d1 )
=
b× E ×k
2.302 × P × L
(log d 2 − log d1 )
=
b × E × (d 2 − d1 )
Derive an expression for the total extension produced by self
weight of a uniform bar, when the bar is suspended vertically.
L
Diameter
d
dx
x
Diameter
d
P1 = weight of the bar below
element
the section,
= volume × specific weight
dx
= (π d2/4)× x × γ
P1
= A× x ×γ
Extension of
the element
due to weight
of the bar
below that,
P1 dx ( A × x × ρ ) dx
⎡ PL ⎤
=
=
=⎢
⎥
AE
AE
⎣ AE ⎦ dx
Hence the total extension
entire bar
=∫
L
0
The above expression
can also be written as
L
( A × x × γ )dx ⎡ γx ⎤
γL2
=⎢ ⎥ =
AE
⎣ 2E ⎦ 0 2E
2
=
γL2
2E
×
A ( ALγ ) × L 1 PL
=
= ×
A
2 AE
2 AE
Where, P = (AL)×γ
= total weight of the bar
SHEAR STRESS
Consider a block or portion of a material shown in Fig.(a)
subjected to a set of equal and opposite forces P. then there is
a tendency for one layer of the material to slide over another to
produce the form failure as shown in Fig.(b)
P
P
P
Fig. a
R
Fig. b
P
R
Fig. c
The resisting force developed by any plane ( or section) of the
block will be parallel to the surface as shown in Fig.(c).
The resisting forces acting parallel to the surface per unit area is
called as shear stress.
Shear stress (τ)
Shear resistance
Area resisting shear
=
=
P
A
This shear stress will always be tangential to the area on which
it acts
Shear strain
If block ABCD subjected to shearing stress as shown in
Fig.(d), then it undergoes deformation. The shape will not
remain rectangular, it changes into the form shown in Fig.(e),
as AB'C'D.
τ
τ
B'
C'
C
C
B
B
A
τ
Fig. d
D
A
τ
Fig. e
D
τ
B'
B
C
φ
A
Shear strain is defined as
C'
the change in angle
between two line element
which are originally right
angles to one another.
τ
D
Fig. e
BB′
shear strain =
= tan φ ≈ φ
AB
The angle of deformation
φ
is then termed as shear strain
The angle of deformation is measured in radians and hence
is non-dimensional.
SHEAR MODULUS
For materials within the proportionality limit the shear strain is
proportional to the shear stress. Hence the ratio of shear stress
to shear strain is a constant within the proportionality limit.
Shear Modulus
Shear stress (τ)
or
Shear strain (φ) = constant = G =
Modulus of Rigidity
The value of the modulus of rigidity is a definite property
of a material
For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2
example: Shearing Stress
• Forces P and P‘ are applied
transversely to the member AB.
• Corresponding internal forces act in
the plane of section C and are
called shearing forces.
• The resultant of the internal shear
force distribution is defined as the
shear of the section and is equal to
the load P.
• The corresponding average shear stress is,
τ ave =
P
A
• The shear stress distribution cannot be
assumed to be uniform.
State of simple shear
Consider an element ABCD in a strained material
subjected to shear stress, τ as shown in the figure
A
D
τ
B
C
τ
Force on the face AB = P = τ × AB × t
Where, t is the thickness of the
element.
Force on the face DC is also equal to
P
State of simple shear
Now consider the equilibrium of the element.
(i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.)
For the force diagram shown,
P
A
B
D
C
ΣFx = 0, & ΣFy = 0,
But ΣM = 0
The element is subjected
force
to a clockwise moment
P × AD = (τ × AB × t) × AD
P
But, as the element is actually in equilibrium, there must be
another pair of forces say P' acting on faces AD and BC,
such that they produce a anticlockwise moment equal to ( P
× AD )
State of simple shear
A
P ' × AB = P × AD
= (τ × AB × t)× AD ----- (1)
If τ1 is the intensity of the shear
stress on the faces AD and BC,
then P ' can be written as,
Equn.(1) can be written as
A
P'
P
τ
τ'
C
B
τ'
D
(τ ' × AD× t ) × AB = (τ × AB × t) × AD ----- (1)
τ' =τ
B
P'
D
P ' = τ ' × AD × t
P
τ
C
State of simple shear
Thus in a strained material a shear stress is always
accompanied by a balancing shear of same intensity at
right angles to itself. This balancing shear is called
“complementary shear”.
τ
A
The shear and the
complementary shear together
constitute a state of simple
shear
τ'= τ
D
B
τ'= τ
τ
C
Direct stress due to pure shear
Consider a square element of side ‘a’ subjected to
shear stress as shown in the Fig.(a). Let the thickness
of the square be unity.
τ
τ
A
a
τ
τ
a
D
B
B
A
τ
Fig.(a).
C
a
τ
a
D
τ
τ
C
Fig.(b).
Fig.(b) shows the deformed shape of the element. The length of
diagonal DB increases, indicating that it is subjected to tensile
stress. Similarly the length of diagonal AC decreases indicating
that compressive stress.
Direct stress due to pure shear
Now consider the section, ADC of the element, Fig.(c).
X
σn
A
τ
A
a
a
a
D
τ
( 2 )a
a
D
C
C
Fig.(c).
Resolving the forces in σn direction, i.e., in the X-direction
shown
For equilibrium
∑ Fx = 0
=σn ×
σn =τ
(
)
2 × a ×1 − 2(τ × a × cos 45)
Direct stress due to pure shear
Therefore the intensity of normal tensile stress
developed on plane BD is numerically equal to the
intensity of shear stress.
Similarly it can be proved that the intensity of compressive
stress developed on plane AC is numerically equal to the
intensity of shear stress.
POISSON’S RATIO
Poisson’s Ratio:
Consider the rectangular bar shown in Fig.(a) subjected to a
tensile load. Under the action of this load the bar will increase
in length by an amount δL giving a longitudinal strain in the
bar of
δl
εl =
l
Fig.(a)
POISSON’S RATIO
The bar will also exhibit, reduction in dimension laterally, i.e.
its breadth and depth will both reduce. These change in
lateral dimension is measured as strains in the lateral
direction as given below.
δb
δd
ε lat = − = −
b
d
The associated lateral strains will be equal and are of
opposite sense to the longitudinal strain.
Provided the load on the material is retained within the elastic
range the ratio of the lateral and longitudinal strains will
always be constant. This ratio is termed Poisson’s ratio (µ)
(− δd )
d
=
POISSON’S RATIO =
δl
Longitudinal strain
l
Lateral strain
(− δb )
b
OR δl
l
Poisson’s Ratio = µ
For most engineering metals the value of µ lies between 0.25 and
0.33
In general
y
Lz
Ly
P
P
x
Lx
z
Poisson’s
Ratio
=
Lateral strain
Strain in the direction of
load applied
− δl y
=
δl x
ly
lx
− δl z
OR = δl
x
lz
lx
Poisson’s Ratio = µ
In general
y
Lz
Ly
Px
Px
x
Lx
z
Strain in Y-direction =
Strain in X-direction = εx
=
δl x
εy
=
lx
Strain in Z-direction = εz
=
δl z
lz
=µ
δl x
lx
δl y
ly
=µ
δl x
lx
Load applied in Y-direction
y
Py
Lz
Ly
x
Lx
z
Poisson’s
Ratio
Py
=
− δl x
Lateral strain
Strain in the direction of
load applied
Strain in X-direction = εx
=
δl x
lx
=µ
=
δl y
ly
δl y
lx
ly
− δl z
OR = δl
y
lz
ly
Load applied in Z-direction
y
Pz
Lz
Ly
x
Pz
z
Poisson’s
Ratio
=
Lx
− δl x
Lateral strain
Strain in the direction of
load applied
Strain in X-direction = εx
=
δl x
lx
=µ
=
δl z
lz
δl z
lx
lz
− δl y
OR
=
δl z
ly
lz
Load applied in X & Y direction
y
Py
Strain in X-direction = εx
Ly
Lz
Px
Px
Lx
z
=
x
Py
Strain in Y-direction = εy
Strain in Z-direction = εz
=
σy
E
−µ
= −µ
σy
E
σx
E
−µ
σx
E
σx
E
−µ
σy
E
Py
General
case:
Pz
y
Strain in X-direction = εx
Px
Px
z
x
εx =
Py
Pz
Strain in Y-direction = εy
εy =
σy
E
−µ
σx
E
σy
σz
E
Strain in Z-direction = εz
σy
σx
σz
εz =
−µ
−µ
E
E
−µ
σx
E
E
−µ
σy
E
−µ
σz
σx
σx
σz
σy
σz
E
Bulk Modulus
Bulk Modulus
A body subjected to three mutually perpendicular equal direct
stresses undergoes volumetric change without distortion of
shape.
If V is the original volume and dV is the change in volume,
then dV/V is called volumetric strain.
A body subjected to three mutually perpendicular equal direct
stresses then the ratio of stress to volumetric strain is called
Bulk Modulus.
σ
=
Bulk modulus, K
⎛ dV ⎞
⎜
⎟
⎝ V ⎠
Relationship between volumetric strain and linear strain
Consider a cube of side 1unit, subjected to
three mutually perpendicular direct
stresses as shown in the figure.
Relative to the unstressed state, the change
in volume per unit volume is
[
]
[
dV
= 1 − (1 + ε x )(1 + ε y )(1 + ε z ) = 1 − 1 + ε x + ε y + ε z
1
= εx +εy +εz
= change in volume per unit volume
]
Relationship between volumetric strain and linear strain
Volumetric strain
dV
= εx +εy +εz
V
σy
⎛σx
σ
= ⎜⎜
−µ
−µ z
E
E
⎝ E
⎞ ⎛σ y
σ
σ
⎟⎟ + ⎜⎜
−µ x −µ z
E
E
⎠ ⎝ E
1 − 2µ
(σ x + σ y + σ z )
=
E
⎞ ⎛σ
⎟⎟ + ⎜ z − µ σ y − µ σ x ⎞⎟
E
E ⎟⎠
⎠ ⎜⎝ E
For element subjected to uniform hydrostatic pressure,
σx =σy =σz =σ
dV 1 − 2 µ
(σ x + σ y + σ z )
=
V
E
dV 1 − 2 µ
(3σ )
=
V
E
K=
E
K=
= bulk modulus
3(1 − 2 µ )
or
E = 3K (1 - 2 µ )
σ
⎛ dV ⎞
⎜
⎟
V
⎝
⎠
Relationship between young’s modulus of elasticity (E)
and modulus of rigidity (G) :A1 A
H
a
45˚
a
φτ
φ
D
B1 B
τ
C
Consider a square element ABCD of side ‘a’ subjected to pure shear
‘τ’. DA'B'C is the deformed shape due to shear τ. Drop a perpendicular
AH to diagonal A'C.
Strain in the diagonal AC = τ /E – µ (- τ /E)
[ σn= τ ]
= τ /E [ 1 + µ ] -----------(1)
Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC
In ∆le AA'H
Cos 45˚ = A'H/AA'
A'H= AA' × 1/√2
AC = √2 × AD
( AC = √ AD2 +AD2)
Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=φ/2 ----(2)
Modulus of rigidity = G = τ /φ
φ = τ /G
Substituting in (2)
Strain along the diagonal AC = τ /2G -----------(3)
Equating (1) & (3)
τ /2G = τ /E[1+µ]
E=2G(1+ µ)
Relationship between E, G, and K:We have
E = 2G( 1+ µ) -----------(1)
E = 3K( 1-2µ) -----------(2)
Equating (1) & (2)
2G( 1+ µ) =3K( 1- 2µ)
2G + 2Gµ=3K- 6Kµ
µ= (3K- 2G) /(2G +6K)
Substituting in (1)
E = 2G[ 1+(3K – 2G)/ (2G+6K)]
E = 18GK/( 2G+6K)
E = 9GK/(G+3K)
Working stress: It is obvious that one cannot take risk of
loading a member to its ultimate strength, in practice. The
maximum stress to which the material of a member is
subjected to in practice is called working stress.
This value should be well within the elastic limit in elastic
design method.
Factor of safety: Because of uncertainty of loading
conditions, design procedure, production methods, etc.,
designers generally introduce a factor of safety into their
design, defined as follows
Factor of safety = Maximum stress or Yield stress (or proof stress)
Allowable working
Allowable working
stress
stress
Malleability: A property closely related to ductility, which
defines a material’s ability to be hammered out in to thin
sheets
Homogeneous: A material which has a uniform structure
throughout without any flaws or discontinuities.
Isotropic: If a material exhibits uniform properties throughout
in all directions ,it is said to be isotropic.
Anisotropic: If a material does not exhibit uniform properties
throughout in all directions ,it is said to be anisotropic or
nonisotropic.
Exercise Problems
Q1.
An aluminum tube is rigidly fastened between a brass
rod and steel rod. Axial loads are applied as indicated in
the figure. Determine the stresses in each material and total
deformation. Take Ea=70GPa, Eb=100GPa, Es=200GPa
20kN
Ab=700mm2
brass
500mm
Aa=1000mm2
15kN
15kN
aluminum
600mm
As=800mm2
steel
10kN
700mm
Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm
Q2.
A 2.4m long steel bar has uniform diameter of 40mm for
a length of 1.2m and in the next 0.6m of its length its
diameter gradually reduces to ‘D’ mm and for remaining
0.6m of its length diameter remains the same as shown in
the figure. When a load of 200kN is applied to this bar
extension observed is equal to 2.59mm. Determine the
diameter ‘D’ of the bar. Take E =200GPa
200kN
200kN
1000mm
500mm 500mm
Ф = 40mm
Ф = D mm
Q3. The diameter of a specimen is found to reduce by
0.004mm when it is subjected to a tensile force of 19kN.
The initial diameter of the specimen was 20mm. Taking
modulus of rigidity as 40GPa determine the value of E and
µ
Ans: E=110GPa, µ=0.36
Q.4 A circular bar of brass is to be loaded by a shear load of
30kN. Determine the necessary diameter of the bars (a) in
single shear (b) in double shear, if the shear stress in
material must not exceed 50MPa.
Ans: 27.6, 19.5mm
Q.5 Determine the largest weight W that can be supported
by the two wires shown. Stresses in wires AB and AC are
not to exceed 100MPa and 150MPa respectively. The cross
sectional areas of the two wires are 400mm2 for AB and
200mm2 for AC.
Ans: 33.4kN
C
B
300
450
A
W
Q.6 A homogeneous rigid bar of weight 1500N carries a
2000N load as shown. The bar is supported by a pin at B
and a 10mm diameter cable CD. Determine the stress in
the cable
D
C
A
3m
2000 N
B
3m
Ans: 87.53MPa
Q.7. A stepped bar with three different cross-sectional
areas, is fixed at one end and loaded as shown in the
figure. Determine the stress and deformation in each
portions. Also find the net change in the length of the
bar. Take E = 200GPa
300mm2
20kN
250mm
450mm2
250mm2
320mm
40kN
10kN
270mm
Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm
Q.8
a)
b)
c)
The coupling shown in figure is constructed from steel
of rectangular cross-section and is designed to transmit a
tensile force of 50kN. If the bolt is of 15mm diameter
calculate:
The shear stress in the bolt;
The direct stress in the plate;
The direct stress in the forked end of the coupling.
Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa
Q.9 The maximum safe compressive stress in a hardened
steel punch is limited to 1000MPa, and the punch is used to
pierce circular holes in mild steel plate 20mm thick. If the
ultimate shearing stress is 312.5MPa, calculate the
smallest diameter of hole that can be pierced.
Ans: 25mm
Q.10 A rectangular bar of 250mm long is 75mm wide and
25mm thick. It is loaded with an axial tensile load of 200kN,
together with a normal compressive force of 2000kN on
face 75mm×250mm and a tensile force 400kN on face
25mm×250mm. Calculate the change in length, breadth,
thickness and volume. Take E = 200GPa & µ=0.3
Ans: 0.15,0.024,0.0197mm, 60mm3
Q.11 A piece of 180mm long by 30mm square is in
compression under a load of 90kN as shown in the figure. If
the modulus of elasticity of the material is 120GPa and
Poisson’s ratio is 0.25, find the change in the length if all
lateral strain is prevented by the application of uniform
lateral external pressure of suitable intensity.
90kN
30
30
180
Ans: 0.125mm
Q.12 Define the terms: stress, strain, elastic limit,
proportionality limit, yield stress, ultimate stress, proof
stress, true stress, factor of safety, Young’s modulus,
modulus of rigidity, bulk modulus, Poisson's ratio,
Q.13 Draw a typical stress-strain diagram for mild steel rod
under tension and mark the salient points.
Q.14 Diameter of a bar of length ‘L’ varies from D1 at one end
to D2 at the other end. Find the extension of the bar under
the axial load P
Q.15 Derive the relationship between Young’s modulus and
modulus of rigidity.
Q.16 Derive the relationship between Young’s modulus and
Bulk modulus.
Q.17 A flat plate of thickness ‘t’ tapers uniformly from a width
b1at one end to b2 at the other end, in a length of L units.
Determine the extension of the plate due to a pull P.
Q.18 Find the extension of a conical rod due to its own weight
when suspended vertically with its base at the top.
Q.19 Prove that a material subjected to pure shear in two
perpendicular planes has a diagonal tension and
compression of same magnitude at 45o to the planes of
shear.
Q.20
For a given material E=1.1×105N/mm2&
G=0.43×105N/mm2 .Find bulk modulus & lateral
contraction of round bar of 40mm diameter & 2.5m
length when stretched by 2.5mm.
ANS: K=83.33Gpa, Lateral contraction=0.011mm
Q.21
The modulus of rigidity of a material is 0.8×105N/mm2 ,
when 6mm×6mm bar of this material subjected to an axial
pull of 3600N.It was found that the lateral dimension of the
bar is changed to 5.9991mm×5.9991mm. Find µ & E.
ANS: µ=0.31, E= 210Gpa.
7. STATICALLY INDETERMINATE MEMBERS
&
THERMAL STRESSES
STATICALLY INDETERMINATE MEMBERS
Structure for which equilibrium equations are sufficient to
obtain the solution are classified as statically determinate.
But for some combination of members subjected to axial
loads, the solution cannot be obtained by merely using
equilibrium equations. The structural problems with
number of unknowns greater than the number independent
equilibrium equations are called statically indeterminate.
The following equations are required to solve the problems
on statically indeterminate structure.
1) Equilibrium equations based on free body diagram of the
structure or part of the structure.
2) Equations based on geometric relations regarding elastic
deformations, produced by the loads.
COMPOUND BAR
Material(2)
Material(1)
L1
L2
W
A compound bar is one which is made of two or more than
two materials rigidly fixed, so that they sustain together an
externally applied load. In such cases
(i) Change in length in all the materials are same.
(ii) Applied load is equal to sum of the loads carried by
hb
(dL)1 = (dL)2
(σ1/ E1)L1 = (σ2 /E2)L2
σ1 = σ2 ×( E1/E2)(L1/L2)
(1)
E1/E2 is called modular ratio
Total load = load carried by material (1) + load carried by
material(2)
W = σ 1 A1 + σ 2 A2
(2)
From Equation (1) & (2) σ1 and σ2 can be calculated
Temperature Stress
B
A
L
A
A
B
B´
P
L
B
αTL
L
Any material is capable of expanding or contracting freely
due to rise or fall in temperature. If it is subjected to rise in
temperature of T˚C, it expands freely by an amount ‘αTL’ as
shown in figure. Where α is the coefficient of linear
expansion, T˚C = rise in temperature and L = original length.
From the above figure it is seen that ‘B’ shifts to B' by an
amount ‘αTL’. If this expansion is to be prevented a
compressive force is required at B'.
Temperature strain = αTL/(L + αTL) ≈ αTL/L= αT
Temperature stress = αTE
Hence the temperature strain is the ratio of expansion or
contraction prevented to its original length.
If a gap δ is provided for expansion then
Temperature strain = (αTL – δ) / L
Temperature stress = [(αTL – δ)/L] E
Temperature stress in compound
bars:-
α1TL
x
P1
Material(1)
Material(2)
(dL)2 (dL)1
P2
α2TL
x
∆
When a compound bar is subjected to change in temperature,
both the materials will experience stresses of opposite nature.
Compressive force on material (1) = tensile force on material (2)
load)
σ1A1 = σ2A2 (there is no external
σ1=( σ2A2)/A1
(1)
As the two bars are connected together, the actual position of the
bars will be at XX.
Actual expansion in material (1) = actual expansion in material (2)
α1TL – (dL)1 = α2TL + (dL)2
α1TL – (σ1 / E1) L =α2TL + (σ2 / E2) L
αT – (σ1 / E1) = α2T + σ2 / E2 --------------------------(2)
From (1) and (2) magnitude of σ1 and σ2 can be found out.
Exercise problems
Q.1 A circular concrete pillar consists of six steel rods of
22mm diameter each reinforced into it. Determine the
diameter of pillar required when it has to carry a load of
1000kN. Take allowable stresses for steel & concrete as
140Mpa & 8Mpa respectively. The modular ratio is 15
ANS: D=344.3mm
Q.2 Determine the stresses & deformation induced in
Bronze & steel as shown in figure. Given As=1000mm2,
Ab=600mm2, Es= 200Gpa, Eb= 83Gpa ANS: ( σb=55Mpa,
σs=93.5Mpa, dLs=dLb=0.093mm)
Bronze
Bronze
160kN
Steel
Q..3
A cart wheel of 1.2m diameter is to be provided with
steel tyre. Assume the wheel to be rigid. If the stress in steel
does not exceed 140MPa, calculate minimum diameter of steel
tyre & minimum temperature to which it should be heated
before on to the wheel.
ANS: d=1199.16mm T=58.330C
Q.4 A brass rod 20mm diameter enclosed in a steel tube of
25mm internal diameter & 10mm thick. The bar & the tube are
initially 2m long & rigidly fastened at both the ends. The
temperature is raised from 200C to 800C. Find the stresses in
both the materials.
If the composite bar is then subjected to an axial pull of 50kN, find
the total stress. Es=200GPa, Eb=80GPa, αs=12×10-6/0C,
αb=19×10-6/0C.
ANS: σb=8.81N/mm2 ( C ) , σs=47.99N/mm2( T )
8. STRESSES ON INCLINED PLANES
INTRODUCTION
The state of stress on any plane in a strained body is said to be
‘Compound Stress’, if, both Normal and Shear stresses are acting on
that plane.
For, example, the state of stress on any vertical plane of a beam
subjected to transverse loads will, in general, be a Compound
Stress.
In actual practice the state of Compound Stress is of more common
occurrence than Simple state of stress.
In a compound state of stress, the normal and shear stress may
have a greater magnitude on some planes which are inclined (or,
Oblique) to the given stress plane.
Hence in compound state of stresses it is necessary to find the following
(i) The normal and shear stress on a plane which is inclined
(Oblique) to the given stress plane;
(ii) The inclination of max. and min. normal stress planes and
values of the normal stress (max. / min.) on them;
(iii) The inclination of max. shear stress planes and
the values of the shear stress (max.) on them.
EQUATIONS, NOTATIONS & SIGN CONVENTIONS :
Consider an element ABCD subjected to a
state of compound stress as shown in the Fig.
Let:
C
σx Æ Normal Stress in x- direction
σx
σy Æ Normal Stress in y- direction
D
τ ÆShear Stresses in x & y – directions
θ Æ Angle made by inclined plane AE wrt vertical τ
σθ ÆNormal Stress on inclined plane
τθ ÆShear Stress on inclined plane
σy
τ
τθ E B
σθ
θ
σx
A
σy
(i) Normal & Shear stress on plane inclined (Oblique) to given stress plane:
Normal Stress, σθ, and Shear stress, τθ, on inclined plane are given by:
⎛σ x +σ y
σ θ = ⎜⎜
2
⎝
τθ
⎛σ
= ⎜⎜
⎝
x
⎞ ⎛σ x −σ y
⎟⎟ + ⎜⎜
2
⎠ ⎝
−σ
2
y
⎞
⎟⎟ cos 2θ + τ sin 2θ − − − (1)
⎠
⎞
⎟⎟ sin 2 θ − τ cos 2 θ − − − ( 2 )
⎠
(ii) The inclination of max. and min. normal stress planes and
the values of the normal stress (max. / min.) on them
Let, θP be the inclination of the plane of max. or min. normal stress
and σP be the value of the max. or min. normal stress on that plane,
then, from Eqn. (1):
⎛σ x +σ y ⎞ ⎛σ x −σ y ⎞
⎟⎟ + ⎜⎜
⎟⎟ cos 2θ P + τ sin 2θ P - - - (1)
σ P = ⎜⎜
⎝
2
⎠ ⎝
2
⎠
Thus, the condition for max. or
min. normal stress to occur on a
dσ P
=0
For σ P to be max. or min.,
plane is, shear stress on that
dθ
plane should be zero.
⎛ σ x −σ y ⎞
⎟⎟(− 2 sin 2θ P ) + τ (2 cos 2θ P ) = 0 These planes on which shear
⇒ ⎜⎜
⎝ 2 ⎠
stress is zero and the normal
⎛ σ x −σ y ⎞
stress on them being either the
⎟⎟(sin 2θ P ) − τ (cos 2θ P ) = 0
⇒ ⎜⎜
⎝ 2 ⎠
max. or the min. are called
⇒ τθ = 0
‘PRINCIPAL PLANES’.
P
From,τ θ P = 0, we have,
⎛σ x −σ y ⎞
τ
⎜⎜
⎟⎟(sin 2θ P ) − τ (cos 2θ P ) = 0 ⇒ tan 2θ P =
- - - (3)
−
σ
σ
⎛ x
y ⎞
⎝ 2 ⎠
⎜⎜
⎟⎟
⎝ 2 ⎠
ƒ The above Eqn. (3), gives two values for θP, which differ by 900.
Thus, there are two mutually perpendicular Principal planes, on
which there are only normal stresses, shear stress being zero on
them.
ƒ On one of them, the value of the normal stress is the max.; it is
called the ‘Major Principal plane’, the max. normal stress on it is
called the ‘Major Principal Stress’.
ƒ On the other principal plane, the value of the normal stress is the
min.; it is called the ‘Minor Principal plane’, the min. normal stress on
it is called the ‘Minor Principal Stress’.
τ2
From tan 2θ P =
y /2
] 2+
2θP
sin 2θ P = ±
±√
[(σ
x
-σ
(σx-σy)/ 2
cos 2θ P = ±
τ
τ
⎛σ x −σ y
⎜⎜
2
⎝
⎞
⎟⎟
⎠
, we get,
τ
2
⎡ (σ X − σ Y ) ⎤
+τ
⎢⎣
⎥
2
⎦
(σ X − σ Y ) / 2
2
2
⎡ (σ X − σ Y ) ⎤
⎢⎣
⎥⎦ + τ
2
2
Substituting for sin 2θP and cos 2θP in Eqn. (1), and simplifying, we
get the equation for principal stresses as:
σP =
(σ
x
+σ
2
y
)±
⎛σ x −σ
⎜⎜
2
⎝
2
y
⎞
⎟⎟ + τ
⎠
2
− − − (4)
The above equation (4) gives two values for principal stresses.
The numerically max. of the two values (+ ve or − ve) is the Major
Principal Stress, (σMajor or σMax);
The numerically min. (+ ve or − ve) is the Minor Principal Stress (σMinor
or σMin).
(iii) Inclination of max. shear stress planes, Max. shear stress Equation.
Let, θS be the inclination of the plane of max. or min. shear stress
and τS be the value of the max. or min. shear stress on that plane,
then, from Eqn. (2):
τS
⎛σ
= ⎜⎜
⎝
x
−σ
2
y
⎞
⎟⎟ sin 2 θ S − τ cos 2 θ S
⎠
dτ
For τ S to be max. or min., S = 0
dθS
⎛ σ x −σ y ⎞
⎟⎟(2 cos 2θ S ) − τ (− 2 sin 2θ S ) = 0
⇒ ⎜⎜
⎝ 2 ⎠
⎛ σ x −σ y ⎞
⎜⎜
⎟⎟
2 ⎠
⇒ tan 2θ S = − ⎝
- - - (5)
τ
NOTE : We have tan2θ P × tan 2θ S = −1
− − − (2)
Eqn. (5) gives two values for θS,
which differ by 900. Thus, there
are two mutually perpendicular
planes, on which shear stress
are max.; numerically equal but
opposite in sense.
The planes of Max. Shear
stresses are inclined at 450 to
the Principal planes.
±√
[(σ
x
−σ
y )/
2] 2
+
τ2
⎛σ x −σ y
⎜⎜
2
From tan 2θ S = − ⎝
cos 2θ s = ±
(σx-σy)/ 2
2θS
sin 2θ S = ±
τ
τ
τ
⎞
⎟⎟
⎠ , we get,
2
⎡ (σ x − σ y ) ⎤
⎥ +τ
⎢
2
⎦
⎣
(σ x − σ y )/ 2
2
2
⎡ (σ x − σ y ) ⎤
+τ
⎢
⎥
2
⎣
⎦
2
Substituting for sin 2θS and cos 2θS in Eqn. (2), we get the
equation for Max. shear stresses as:
τ max .
⎛σ x −σ
= ± ⎜⎜
2
⎝
2
y
⎞
⎟⎟ + τ
⎠
2
⎛ σ Major − σ Minor
⇒ τ max . = ± ⎜⎜
2
⎝
⎞
⎟⎟
⎠
⎤
⎥
⎥
⎥ − − − (6 )
⎥
⎥
⎦
The above equation (6) gives two values for Max. shear stresses,
which are numerically equal but opposite in sense.
NOTATIONS & SIGN CONVENTIONS
σy
σx Æ Normal Stress in x- direction
σy Æ Normal Stress in y- direction
τ ÆShear Stresses in x & y – directions
θ Æ Angle made by inclined plane wrt vertical
σθ ÆNormal Stress on inclined plane AE
τθ ÆShear Stress on inclined plane AE
τ
τθ E B
C
σx
σθ
θ
D
σx
A
τ
σy
θP Æ Inclination of Principal planes
σP Æ Principal stresses
θS Æ Inclination of Max. shear stress planes [θS = θP + 450].
All the parameters are shown in their +ve sense in the Fig.
Normal stresses, σ Æ Tensile stresses +ve.
Shear Stresses, τ, in x – direction & Inclined Plane Æ Clockwise +ve.
Shear Stresses, τ, in y – direction Æ Anti-Clockwise +ve.
Angle, θ Æ measured w r t vertical, Anti-Clockwise +ve.
Exercise Problems:
Q.1
The principal stresses at a point in a strained material are 80
MPa(C) and 40 MPa(T). Find the normal, tangential and resultant
stress on a plane inclined at 50o to the major principal plane.
[Ans: - 9.58MPa, -59MPa].
Q.2 The stresses in a strained material is as shown in Fig. Find the
normal and shear stresses on plane inclined at 30o to the horizontal.
Also determine the intensity and position of the plane upon which
there is only shear stress. Sketch the plane.
60MPa
90
MPa
30o
Ans:
σ = − 22.5MPa, τ = − 64.95MPa.
Pure shear plane σ = 0,
Ө = 39.23o w.r.t. horizontal.
τpure = − 73.48MPa.
Q.3 A plane element is subjected to the system of stresses as shown
in Fig. Determine (i) the principal stresses and inclination of their planes
(ii) maximum shearing stresses and inclination of their planes.
Represent your answers in neat sketches.
160 MPa
80 MPa
200 MPa
Ans:
σ1 = 262.46MPa, σ2 = 97.54MPa.
Өp = 37.98 or 127.98 with horizontal
τtmax = 82.46, Өs = 82.98 or 172.98 with horizontal
Q.4
At a point in a structural member subjected to stresses as
shown in fig. determine the principal stresses and the maximum shear
stress. Also determine and sketch planes on which these stresses
act.
80
[Ans:131.23, 48.77, − 37.98o, − 127.98o
41.23, 7o, 970 , Angles w.r.t. horizontal]
100
4
0
Q.5 At a point in a material under stress, the intensity of resultant
stress on a certain plane is 60 MPa, directed outwards and inclined
at 30o to the normal of that plane. The stress on the plane at right
angles to this has a normal stress component of 40 Mpa (T). Find (i)
the principal stresses and inclination of their planes, (ii) the maximum
shear stresses and inclination of their planes .
[Ans: 76.57MPa, 15.39MPa, 39.36o , 129.36o and 30.59 MPa,
84.36o , 174.36o, Angles w.r.t. horizontal]
9. Stresses due to fluid
pressure in thin cylinders
9 -THIN CYLINDERS
INTRODUCTION:
In many engineering applications, cylinders are
frequently used for transporting or storing of liquids, gases
or fluids.
Eg: Pipes, Boilers, storage tanks etc.
These cylinders are subjected to fluid pressures. When
a cylinder is subjected to internal pressure, at any point on
the cylinder wall, three types of stresses are induced on
three
mutually perpendicular planes.
They are,
1. Hoop or Circumferential Stress (σC) – This is directed
along the tangent to the circumference and tensile in
nature. Thus, there will be increase in diameter.
2. Longitudinal Stress (σL) – This stress is directed along
the length of the cylinder. This is also tensile in nature
and tends to increase the length.
3. Radial pressure (σ r) – It is compressive in nature. Its
magnitude is equal to fluid pressure on the inside wall
and zero on the outer wall if it is open to atmosphere.
σC
σC
p
σC
σL
σL
p
σC
1. Hoop
/Circumferential
Stress (σC)
σr
p
σL
σL
2. Longitudinal Stress (σL)
Element on the cylinder
wall subjected to these
three stresses
σr
3. Radial Stress (σr)
σC
σL
σL
σC
σr
A cylinder or spherical shell is considered to be thin
when the metal thickness is small compared to internal
diameter.
i. e., when the wall thickness, ‘t’ is equal to or less than
‘d/20’, where ‘d’ is the internal diameter of the cylinder or
shell, we consider the cylinder or shell to be thin, otherwise
thick.
Magnitude of radial pressure is very small compared to
other two stresses in case of thin cylinders and hence
neglected.
p×d
Circumferential stress, σ c =
........................(1)
2× t
p×d
Longitudinal stress, σL =
...................( 2)
4× t
Maximum Shear stress, τmax = (σc- σL) / 2
τmax = ( p x d) / (8 x t)
Where
p = internal fluid pressure
d= internal diameter,
t = thickness of the wall
EVALUATION OF STRAINS
σ L=(pd)/(4t)
σ C=(pd)/(2t)
σ C=(pd)/(2t)
σ L=(pd)/(4t)
A point on the surface of thin cylinder is subjected to
biaxial stress system, (Hoop stress and Longitudinal stress)
mutually perpendicular to each other, as shown in the figure.
The strains due to these stresses i.e., circumferential and
longitudinal are obtained by applying Hooke’s law and
Poisson’s theory for elastic materials.
ŒL=(pd)/(4t)
Circumferential strain, ε C :
σC
σL
σL
σL
−µ×
= 2×
−µ×
εC =
E
E
E
E
σL
=
× (2 − µ)
E
i.e.,
ŒC=(pd)/(2t)
ŒC=(pd)/(2t)
ŒL=(pd)/(4t)
δd
p×d
εC =
=
× (2 − µ)................................(3)
d 4× t × E
Longitudinal strain, ε L :
σC σL
(2 × σ L ) σ L
σL
−µ×
=
−µ×
=
× (1 − 2 × µ)
εL =
E
E
E
E
E
i.e.,
δl
p×d
εL = =
× (1 − 2 × µ)................................(4)
L 4× t × E
dv
VOLUMETRIC STRAIN,
V
σL=(pd)/(4t)
π 2
σC=(pd)/(2t)
σC=(pd)/(2t)
We have volume, V = × d × L.
4
π 2
π
dV = × d × dL + L × × 2 × d × dd σL=(pd)/(4t)
4
4
π 2
π
× d × dL + L × × 2 × d × dd
dv 4
dL
dd
4
=
=
+ 2×
π 2
V
L
d
×d ×L
4
p×d
p×d
dV
=
(1 − 2 × µ) + 2 ×
(2 − µ)
= ε +2× εC
V L
4× t × E
4× t × E
dv
p×d
i.e.,
=
(5 − 4 × µ).................(5)
V 4× t × E
JOINT EFFICIENCY
The cylindrical shells like boilers are having two types of
joints namely Longitudinal and Circumferential joints. Due to
the holes for rivets, the net area of cross section decreases
and hence the stresses increase. If the efficiencies of these
joints are known, the stresses can be calculated as follows.
Let ηL=Efficiency of Longitudinal joint
and ηC=Efficiency of Circumferential joint.
Circumferential stress is given by,
p×d
σC =
2 × t × ηL
.............(1)
Longitudinal stress is given by,
p×d
σL =
..........
...(2)
4×t×ηC
Note: In longitudinal joint, the circumferential stress is
developed and in circumferential joint, longitudinal stress is
developed.
Exercise Problems
Q.1
Calculate the circumferential and longitudinal strains for a boiler of
1000mm diameter when it is subjected to an internal pressure of 1MPa. The
wall thickness is such that the safe maximum tensile stress in the boiler
material is 35 MPa. Take E=200GPa and µ= 0.25.
(Ans: ε C=0.0001531, ε L=0.00004375)
Q.2
A water main 1m in diameter contains water at a pressure head of
120m. Find the thickness of the metal if the working stress in the pipe
metal is 30 MPa. Take unit weight of water = 10 kN/m3.
(Ans: t=20mm)
Q.3
A gravity main 2m in diameter and 15mm in thickness. It is subjected to an
internal fluid pressure of 1.5 MPa. Calculate the hoop and longitudinal
stresses induced in the pipe material. If a factor of safety 4 was used in the
design, what is the ultimate tensile stress in the pipe material?
(Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa)
Q.4
At a point in a thin cylinder subjected to internal fluid pressure, the value of
hoop strain is 600×10-4 (tensile). Compute hoop and longitudinal stresses.
How much is the percentage change in the volume of the cylinder? Take
E=200GPa and µ= 0.2857.
(Ans: σC=140 MPa,
σL=70 MPa, %age change=0.135%.)
Q.5
A cylindrical tank of 750mm internal diameter and 1.5m long is to be filled
with an oil of specific weight 7.85 kN/m3 under a pressure head of 365 m. If
the longitudinal joint efficiency is 75% and circumferential joint efficiency is
40%, find the thickness of the tank required. Also calculate the error of
calculation in the quantity of oil in the tank if the volumetric strain of the tank
is neglected. Take permissible tensile stress as 120 MPa, E=200GPa and
µ= 0.3 for the tank material.
(Ans: t=12 mm, error=0.085%.)