Mechanics of Solids Fundamentals.pdf

Introduction
Strength
The Strength is the most important property of material from design point of view. It
enables material to resist fracture under load. The load required to cause fracture
divided by the cross sectional area of the test specimen is termed as ultimate strength
of the material & expressed in the unit of stress. In short ,it is the ability of a structure
to resist loads is called “Strength”
Mechanics OR Strength Of Materials
Mechanics is the the branch of physical science which deals with the state of
rest or motion of bodies that are subjected to the action of Forces
Mechanics
Mechanics of rigid bodies Mechanics of deformable bodies Fluid Mechanics
Statics Dynamics OR
Mechanics Of Materials
OR
Strength Of Materials
Statics- Study of body at rest
Dynamics- Study of body in motion
Statics & Dynamics are devoted primarily to the study of the extended effects of
forces on rigid bodies i.e. the bodies for which the change in shape ( or deformation
be neglected)
Mechanics Of Materials OR Strength Of Materials-
It deals with the relation between externally applied loads & their internal effects on
solid bodies.
Note :- The purpose of studying strength of the materials is to ensure that the
structure used will be safe against maximum internal effects that may be produced by
any combination of loading.
Definitions-
Matter- Any substance that occupies space.
Body- Matter that is bounded by a closed surface.
Body
Rigid body Deformable body
Inertia- Is an inherent property of matter by which it resists any change in its state
Mass- It is a quantitative measure of mass ( inertia). Bodies undergo different degrees
of deformations under the action of forces. Thus steel has more ( inertia) mass than
aluminium, as under the action of same force on two identical bodies of steel &
aluminium, the aluminium body will undergo more change in motion than steel.
Force- It is a physical quantity that changes or tries to change the state of rest or of
uniform motion of an object.
Mechanical Properties Of Materials-
The properties which are related with the behavior of materials under the
action of forces under different conditions are called as Mechanical Properties Of
Materials. Following are the basic properties of materials:-
Strength-

The strength of materials enables materials to resist fracture under loading ; i.e.
it is the ability of a structure to resist loads with undue distortion, collapse or rupture.
The load required to cause fracture divided by the cross sectional area of the
test specimen is termed as ultimate strength of the material or tenacity.
Note :- For strength purpose, the actual strength of a structure must exceed the
required strength.
Elasticity -
The property of material by virtue of which it regains its original shape & size
(original dimension) after deformation when loads causing deformation are removed.
The materials exhibiting this property are known as elastic materials. e.g. Rubber,
Spring etc.
If a body regains completely its original shape & size, it is said to be perfectly
elastic & if the body does not return back completely its original shape & size, after
the removal of the external force it is said to be partially elastic.
The property of elasticity is only possible, if the deformation caused by
external force is within a certain limit, such a limit is called elastic limit. e.g. steel,
aluminium, copper stone, concrete, etc may be considered to be perfectly elastic
within certain limit. If a material exhibit the same elastic properties at all points then
that material is said to be homogeneous material. It has same composition throughout
the body.
Plasticity-
Plasticity is the converse of Elasticity, i.e. a material in plastic state is
permanently deformed by the application of load & it has no tendency to recover. The
characteristics of the material by which it undergoes inelastic strains beyond those at
the elastic limit is known as plasticity.
Plasticity is important when a substance has to be moulded into components.
Many materials become plastic at large values of stress or at high temperature.
Ductility-
Ductility is the characteristic which permits a material to be drawn out
longitudinally to a reduced section, under the action of a tensile force, i.e. it possible
to draw thin wires of a metal. Therefore in a ductile material large plastic deformation
is possible before absolute failure or rupture takes place.
Ductility is a tensile quality of material. It is measured in the tensile test of
specimen of the material, either in terms of percentage elongation or in terms of
percentage reduction in the cross sectional area of the test specimen. Higher the
percentage elongation more ductile is material. Example :- Gold, Platinum, copper,
aluminium, mild steel nickel,lead etc.
Note :- As temperature increases ductility increases.
Definition :- Under the action of a tensile force on a material it possible to draw out
material longitudinally to a reduced section i.e. it possible to draw thin wires of a
metal.
Brittleness-
Brittleness implies lack of ductility. The material does not have the capacity to
undergo large deformation before failure. Brittle fractures take place without warning.
Example ;- cast iron, high carbon steel, concrete, glass, stone etc. Brittle materials are
suitable for resisting compressive loads but usually less suitable for resisting tensile &
impact loads. Therefore the compression test is generally performed for testing the
brittleness of a material.
Note :-Materials having less than 5% elongation are considered as brittle.

Fig.shows the strain – stress curve for a typical brittle material which fails
with only little elongation after proportional limit (point A) is exceeded & the fracture
stress is the same as ultimate stress.
Malleability-
Malleability is a property of a material which permits the materials to be
extended in all directions without rupture i.e. it is the ability of a material to be rolled
or beaten up into thin sheets without cracking by rolling & hammering. A malleable
material possesses a high degree of plasticity i.e. it gets permanently deformed by
compression but not necessarily great strength. This property is utilized in many
operations such as forging hot rolling etc. Example:- Gold, Silver, Copper Tin etc.
Note :- Malleability is a compressive quality of a material.
Resilience-
Resilience is the ability of a material to recover its shape & size after
deformation.
OR
The strain energy stored by the body within elastic limit, when loaded
externally is called resilience.
Proof Resilience-
The maximum strain energy which a body stores up to a elastic limit is called
proof resilience. It is the capacity of a material to be on shocks.
Modulus Of Resilience-
The Proof Resilience per unit volume of piece is called Modulus Of
Resilience.
Strain Energy -
When an elastic body is loaded it undergoes deformation i.e. its dimensions
change & when it is released of the load it regains its original shape. For the time
loaded energy is stored in it, the same is given up or released by the loading when the
load is removed. This energy is called Strain Energy.
Hardness-
Hardness is the resistance of a material to indentation, scratching, cutting or
wear by surface abrasion. Tests such as Brinell, Rockwell are generally performed to
measure the hardness of a material.
Fatigue-
Fatigue is the phenomenon of a material failing under very little stress due to
repeated cycles of loading.
OR

Fatigue is deterioration of a material subjected to repeated cycles of stress & strain
resulting in progressive cracking eventually producing fracture.
This is important in case a component is likely to be subjected to cyclic or reversal
loading, as in machine foundations or members subjected repeated dynamic loads,
Example :- Earthquake
Reversal loading-
The member may be subjected to repeated tensile & compressive stresses.
Note - The members subjected to repeated stresses fail at stresses lower than ultimate
stresses.
Toughness-
Toughness is the property which enables a material to absorb large amount of
energy by undergoing large plastic deformation, particularly due to shock loading.
The toughness of a material is measured by using Charpy test or Izod test. Toughness
property is important where impact loads are applied. Example :-Brass, mild steel,
manganese, wrought iron.
Stiffness-
It relates to the deformation of the member. Mathematically it is defined as the
force/moment required to cause unit deformation in the member.
Creep-
Creep is the property by which a material undergoes deformations at constant
stress over a period of time. The stress remaining constant, it is found that
deformations increase over a period of time due to creep.

Unit -I
Simple Stress and Strain
Simple Stresses –
When a body is acted upon by external force or load, internal resisting force is
set up in the body. Such a body is then said to be in the state of stress. Therefore the
stress is the internal resistance offered by the body to deformation against the
externally applied force is called as stress.
Types of Stresses –
1)Normal Stress OR Direct Stress
1. Tensile Stress
2. Compressive Stress
2)Shear Stress OR Tangential Stress.
Normal Stress-
It is the stress acting perpendicular to the cross section of the member.
Cross section –
The section which is normal or perpendicular to the longitudinal axis of the bar, is
known as cross section.
. ˙ . Normal Stress ꞊ σn ꞊ P/A ꞊ R/A ꞊ N/m2
꞊ Pascal (Pa)
1.Tensile Stress :-
When the resistance offered by a section of a member is against an increase in
length, the section is said to offer a Tensile Stress. Fig. shows a body subjected to
tensile forces, are a pair of forces acting away from each other. They tend to elongate
or stretch the body.
If P is the force and A is the area of cross section of the bar, then the tensile stress (σt )
is given by
σt ꞊ P/A ꞊ Force/Area = N/m2
=R/A - - - - For equilibrium R = P
Note :- It is considered as positive stress
2.Compressive Stress :-
When the resistance offered by a section of a member is against an decrease in
length, the section is said to offer a Compressive Stress. Fig. shows a body subjected
to compressive forces, are a pair of forces acting towards each other. They tend to
shorten the body.

If P is the force and A is the area of cross section of the bar, then the compressive
stress (σc ) is given by
σc ꞊ P/A ꞊ Force/Area = N/m2
=R/A - - - - For equilibrium R = P
Note :- It is considered as negative stress
3.Shear Stress OR Tangential Stress:-
If the external forces tend to shear the member along the cross section, internal forces
developed act parallel or tangential to the cross section. Thus, shear stress exists
between the parts of a body when the two parts exert equal and opposite forces on
each other laterally in a direction tangential to their surface in contact.
Mathematically, Shear Stress τ = P/A ꞊ Shearing Force/Area of c/s = N/m2
Bearing Stress:
When one object presses against another, it is referred to a bearing stress ( They are in
fact the compressive stresses ).
Strain-
1.Normal Strain-
If a bar is subjected to a direct load, and hence a stress the bar will change in
length. If the bar has an original length L and changes by an amount dL, the strain
produce is defined as follows. Strain is thus, a measure of the deformation of the
material and is a nondimensional Quantity i.e. it has no units. It is simply a ratio of
two quantities with the same unit.
It is denoted by е or є (Epsilon)
change in length L
strain
original length L


  

Sign convention for strain-
Tensile strains are positive whereas compressive strains are negative. The
strain defined earlier was known as linear strain or normal strain or the longitudinal
strain.
2.Shear Strain-(Tangential Strain)
The measure of the small distortion of the block caused by the shear force is
called as shear strain.
From fig. consider a rectangular block ABCD fixed at one face and subjected to shear
stress. After application of shear stress it distorts through an angle ‘ ’ and it occupies
new position ABC1D 1. Therefore the shear strain is given by
Shear strain ‘ ‘ = tan  = 1
CC
AD


Since  is very small, tan = 
Linear Strain: l
l
L

 
Lateral Strain : d
d
OR
b
b
Lt


 
Volumetric Strain:
 







2
1





L
v
v
d
dd
b
b
l
l
v
v
Hooks Law
It states that within elastic limit stress is proportional to strain. Mathematically
E= Stress /Strain
Where E = Young’s Modulus
Hooks law holds good equally for tension and compression.
Poisson’s Ratio;
The ratio lateral strain to longitudinal strain produced by a single stress is known as
Poisson’s ratio. Symbol used for Poisson's ratio is µ or 1/ m .
Modulus of Elasticity (or Young’s Modulus)
Young’s modulus is defined as the ratio of stress to strain within elastic limit.
Deformation of a body due to load acting on it

Shear Strain
The distortion produced by shear stress on an element or rectangular block is shown
in the figure. The shear strain or ‘slide’ is expressed by angle ϕ and it can be defined
as the change in the right angle. It is measured in radians and is dimensionless in
nature.
Modulus of Rigidity
For elastic materials it is found that shear stress is proportional to the shear strain
within elastic limit. The ratio is called modulus rigidity. It is denoted by the symbol
‘G’ or ‘C’.
G=shear stress /shear strain
G =
τ
∅
N/mm2
Relation between elastic constants:
Elastic constants: These are the relations which determine the deformations
produced by a given stress system acting on a particular material. These factors are
constant within elastic limit, and known as modulus of elasticity E, modulus of
rigidity G, Bulk modulus K and Poisson’s ratio μ.
E=3K (1-2µ)
E=2G (1+µ)
E=9KG/(G+3K)
Generalized Hooke’s Law:
Thermal Stresses:
Stress is the force acting per unit area. The force can be of any form. When the
applied force is in the form of temperature the resultant stress is called Thermal stress.
It is observed when an object expands or contracts due to a change in temperature.
Thus, we can define Thermal stress as: “Stress caused due to the change in
temperature”
Thermal stresses can have a significant effect on the structural strength and stability if
we do not consider it correctly. The potential to creak cracks and breaks in various
regions of major weakness often occurs due to negligence in understanding Thermal
stress. One of the most common examples of thermal stress is the fracturing of glass

that occurs when we heat the glass to a high temperature and immediately dip into
cold water. The crack and fracture in the glass are different than the one that occurs as
a result of the impact.
Free Expansion Fully Restricted Bar Partially restriction by
allowing yielding
Change in Temperature = T= T2 - T1
Change in length=
support
at
is
yielding
f
.......
..........
support
at
not
is
yielding
f
.......
..........
i
T
L
L
i
T
L
L










Thermal Strain=
support
at
yielding
no
if
........
support
at
yielding
if
........
T
L
T
L
L
L
L
T
L
L
L
L
L

















Thermal Stress=
support
at
yielding
no
if
........
support
at
yielding
if
........
T
E
L
T
L
E
E
L
T
L
E
E
L
L







 







 












Unit -II
Bending and Shear Stress
Introduction
Beam is a structural member which has negligible cross-section compared to
its length. It carries load perpendicular to the axis in the plane of the beam. Due to the
loading on the beam, the beam deforms and is called as deflection in the direction of
loading. This deflection is due to bending moment and shear force generated as
resistance to the bending. Bending Moment is defined as the internal resistance
moment to counteract the external moment due to the loads and mathematically it is
equal to algebraic sum of moments of the loads acting on one side of the section. It
can also be defined as the unbalanced moment on the beam at that section.
Shear force is the internal resistance developed to counteract the shearing
action due to external load and mathematically it is equal to algebraic sum of vertical
loads on one side of the section and this act tangential to cross section.
Moment of force
If forces act on a finite size body they may produce a turning effect. This is
measured by the moment of the force. The moment of a force F (applied at a point P )
about a point A is equal to ± Fd
where:
• F is the magnitude of the force.
•d is the perpendicular distance from
Radius of gyration
The distance from an axis at which the mass or entire area of a body may be
assumed to be concentrated and at which the moment of inertia will be equal to the
moment of inertia of the actual mass about the axis. It is equal to the square root of
the quotient of the moment of inertia and the mass or area.
Mathematically
rxx= √Ixx /A and ryy= √Iyy /A
The unit of radius of gyration is mm, cm etc.

Second Moment of Area :(Moment of inertia)
Moment of Inertia for Other Shapes-
Rectangle: y
x
y`
x`
b
h C
b/
h/2
3
1
12
x
I bh
  3
1
3
x
I bh

3
1
12
y
I b h
  3
1
3
y
I b h

0
xy
I   Area = bh
Triangle:
b
h
x
x`
C
h/3
s
3
1
36
x
I bh
  3
1
12
x
I bh

2
( 2 )
72
xy
b b s h
I


1
2
Area bh

Circle:
C
r
y
x
4
1
4
x y
I I r

 
0
xy
I  
2
Area r


Semi-circle:
C
r
y
x
4r/3
x`
4
1
8
x y
I I r

 
4
'
8
8 9
x
I r


 
 
 
 
0
xy
I  
2
2
r
Area


Ellipse:
C
r
y
x
a
b
3
1
4
x
I ab

 3
1
4
y
I a b


0
xy
I  
Area ab


Moment of inertia for other than standard sections is determined using parallel axis
theorem by dividing the fig. Into parts as,
IXX1=Ixx + Ah2
Where, h is the distance between CG of part to the CG of whole section

Theory of simple bending or pure bending
If at every cross section of a certain portion of beam there is only bending moment
and there is no other forces (either shear or axial) then that portion of beam is said to
be under the action of pure bending. In such case value of BM is same at every cross
section of that portion. Such portion bends in the form of arc of circle.
In reality a state of pure bending does not theoretically exists, because such
state needs a weightless member.
Assumptions in Simple Bending
The following assumptions are made in the theory of simple bending:
1 The beam is initially straight and unstressed.
2 The material of the beam is perfectly homogeneous and isotropic, i.e. of the same
density and elastic properties throughout.
3 The elastic limit is nowhere exceeded.
4 Young's modulus for the material is the same in tension and compression.
5 Plane cross-sections remain plane before and after bending.
6 Every cross-section of the beam is symmetrical about the plane of bending, i.e.
about an axis perpendicular to the N.A.
7 There is no resultant force perpendicular to any cross-section.
8 The radius of curvature is large compared to depth of beam.
Therefore the bending stresses are determined using the Flexure formula
expressed as,
y
R
E
I
M 


Where,
M = Bending Moment at a section (N-mm).
I = Moment of Inertia of the cross section of the beam about Neutral axis (mm4).
σ = Bending stress in a fibre located at distance y from neutral axis (N/mm2). This
stress could be compressive or tensile stress depending on the location of the fibre.
y = Distance of the fibre under consideration from neutral axis (mm).
E = Young's Modulus of the material of the beam (N/mm2).
R = Radius of curvature of the bent beam (mm).

Section Modulus: it is the ratio of moment of inertia of c/s to the distance of
maximum fiber from neutral axis.
z=I/y
Modulus of Resistance: it is defined as maximum bending moment resist by the beam
material.

 Z
y
I
R
M 

.
.
For symmetric section: (MR)c=(MR)t t
c 
 
For unsymmetric section: (MR)c ≠ (MR)t t
c 
 
c = bending stress in compression
t= bending stress in tension
Shear Stress in beams:
 Although we are considering pure bending while deriving bending stress formula,
this case is rarely encountered in engineering problems.
 Therefore practically shear force is not zero.
 The resistance of the beam against the maximum shear force is called as shear
stress.
 Shear formula:
Ib
y
FA

 where, =shear stress, F= shear force at a point, A=Area of above or
below portion of c/s at a point, y =distance between CG of c/s
and CG of area considered, I=moment of inertia, b= width of c/s
Bending stress distribution diagram Shear Stress distribution Diagram
 Bending stresses are maximum at top
and bottom face.
 Bending stresses are zero at top and
bottom face
 Bending stresses between two point
is joined by linear line.
 Bending stresses not depend on
depth.
 Shear stresses are zero at top and
bottom faces.
 Shear stresses are maximum at
neutral axis
 Shear stresses are vary parabolically
from one point to other.
 Shear stress is inversely proportional
to width of c/s

Unit -III
Shear force and Bending Moment Diagram
SHEAR FORCE is defined as the algebraic sum of all the vertical forces either to
left or to the right hand side of the section.
BENDING MOMENT is defined as the algebraic sum of the moments of all the
forces either to the left or to the right of a section.
SIGN CONVENTIONS:
 A clockwise shear will be taken as positive and an anti-clockwise shear will
be taken as negative.
 A bending moment causing concavity upward will be taken as positive and
will be called as Sagging Bending Moment.
 A bending moment causing convexity upward will be taken as negative and
will be called a Hogging Bending Moment.
SHEAR FORCE DIAGRAM:
Shear Force Diagram is a graphical representation of the variation of shear force
along the length of the beam. The ordinate of the SFD at any section gives the value
of the Shear Force at that section, due to the fixed load positions on the beam
 Shear Force Diagram will be Rectangular between POINT LOAD.
 Shear Force Diagram will be Triangular between UNIFORMLY
DISTRIBUTED LOAD.
 Shear Force Diagram will be Parabolic between UNIFORMLY VARYING
LOAD.
BENDING MOMENT DIAGRAM:
Bending Moment Diagram is a graphical representation of the variation of bending
moment along the length of the beam. The ordinate of the BMD at any section gives
the value of Bending Moment at that section, due to the fixed load positions on the
beam.
 Bending Moment Diagram will be Triangular between POINT LOAD.
 Bending Moment Diagram will be Parabolic between UNIFORMLY
DISTRIBUTED LOAD.
 Bending Moment Diagram will be Hyperbolic between UNIFORMLY
VARYING LOAD.
PROPERTIES OF SHEAR FORCE DIAGRAM:
 Shear Force Diagram consists of a rectangle, if the beam is loaded with point
loads.

 Shear Force Diagram consists of inclined line for the portion on
which uniformly distributed load is acting.
 Shear Force Diagram consists of parabolic curve for the portion over
which triangular or trapezoidal load distribution is acting.
 Shear Force Diagram consists of cubic curve for the portion over
which parabolic load distribution is acting.
 Shear Force does not change at the point of application of a couple.
PROPERTIES OF BENDING MOMENT DIAGRAM:
 Bending Moment Diagram consists of inclined lines for the beam loaded
with point loads.
 Bending Moment Diagram consists of parabolic curve for the portion over
which uniformly distributed load is acting.
 Bending Moment Diagram of cubic or third degree curve for the portion
over which uniformly varying load is acting.
 Bending Moment Diagram consists of fourth degree curve if the load
distribution is parabolic.
 Bending Moment is maximum where shear force is zero or changes sign.
 Bending Moment abruptly at the point of application of couple.
POINT OF CONTRAFLEXURE:
The bending moments of opposite nature always produce curvature of beams in
opposite directions. In a beam if the bending moment changes sign at at point, the
point itself having zero bending moment, the beam changes curvature at this point of
zero bending moment and this point is called the point of contraflexure. So at a point
of contraflexure the beam flexes in opposite direction. The point of contraflexure is
called the point of inflexion or a virtual hinge.
USE OF SHEAR FORCE DIAGRAM AND BENDING MOMENT DIAGRAM:
Every structure will try to resist the deformation by developing the internal resistance
(Shear, Moment and/or Bending Moment). However, the properties of the body will
decide how much resistance would be provided against loading.
For example,
 A beam of concrete material has very high resistance to shear. It is however very
weak in tension.
 A steel beam has high resistance in both shear as well as bending. However, it is
very expensive when compared to concrete.
 Hence, while designing a beam of concrete we find the points of maximum SFD
and BMD values. These values in turn give us the idea as to the kind of material
and geometry we need to provide to resist deformation (or failure in general).

Since, concrete is very cheap we try to provide as much of it as possible.
However, it is weak in tension.
 Hence, for zones of maximum Bending Moment (which we find from BMD) we
provide Steel reinforcements. We also find zones of maximum Shear Force (from
SFD). If the shear force values surpass the concrete’s capacity (resistance) we
provide steel shear stirrups in order to resist the additional shear.
 The same concept is applicable for columns as well.
Shear Diagram Moment Diagram
 Point loads cause a vertical jump in
the shear diagram. The direction of
the jump is the same as the sign of
the point load.
 Uniform distributed loads result in
a straight, sloped line on the shear
diagram. The slope of the line is
equal to the value of the distributed
load.
 The shear diagram is horizontal for
distances along the beam with no
applied load.
 The shear at any point along the
beam is equal to the slope of the
moment at that same point

 The moment diagram is a straight,
sloped line for distances along the
beam with no applied load. The
slope of the line is equal to the
value of the shear.
 Uniform distributed loads result in
a parabolic curve on the moment
diagram.
 The maximum/minimum values of
moment occur where the shear
line crosses zero.
 The moment at any point along
the beam is equal to the area under
the shear diagram up to that point:
 M = ∫ V dx

Unit -IV
Torsion
Torsional shear stress is indirect shear stress produced inside the body subjected to
torsion/twisting. In an engineering context, the elements or members that are
generally subjected to twisting have a circular cross-section, for example, shafts to
transfer power from engines or motors, bolted connections, etc. The moment MZ will
cause rotation about the longitudinal axis (z-axis). Hence, MZ is a twisting moment.
The twisting moment is also referred to as torque. Torque will tend to cause angular
displacement in the body. Therefore, shear stress is generated in the body, referred to
as torsional shear stress. Basically, torsional shear stress is the internal resistance of
the body to resist twisting.
Assumptions for the Theory of Pure Torsion
When a circular shaft is subjected to torque, it will only generate torsional shear stress
in the shaft (pure torsion). Our analysis makes certain assumptions to determine the
torsional shear stress that will be produced. These assumptions are listed below -
 The material should be linearly elastic, isotropic and homogeneous.
 Hooke’s law is valid.
 The cross-section remains plane and circular before and after twisting.
 The shaft should be of uniform cross-section
 The shaft should be perfectly straight
Torsion Formula:
J
T
L
G
R



max
Where, T=twisting moment (Nmm); G= rigid modulus (N/mm2
) ; R= Radius of
Shaft (mm) ; = angle of twist (radians); max=maximum shear stress (N/mm2
);
J= polar moment of inertia (mm4
)= Izz= Ixx+Iyy
The torsional shear stress in a solid shaft generated by the torque will be zero at the
centroidal axis and it will be maximum at the outermost fibre. It is shown in the figure
below.
Torsional Rigidity: Product of shear modulus and polar moment of inertia,
Torsional rigidity = G J
Torsional Stiffness: it is the torque required for unit twist.
L
GJ
T  ………(=1)

Torsional Flexibility: Angle/twist required for unit torque = (L/GJ)
Torsional section modulus: ratio of polar moment of inertia to radius.= J/R
Power transmitted by shaft:
60
2 NT
P


Where, T=torque (KNm or Nmm)
N= no. Of revolution per minute
Solid Shaft Hollow Shaft
Polar moment of inertia 4
32
D
J

  
4
4
32
d
D
J 


Section modulus 3
16
D
R
J 
  
D
d
D
R
J 4
4
16



Tmax 3
max
16
D
T 

  
D
d
D
T
4
4
max
16




 In case of composite shafts two shafts are behaves like a single shaft.
 Total torque transmitted by the composite shaft is shared by two shafts.
 Angle of twist remains same for the composite shaft. (Total angle of
twist=1+θ2+……)
Replacement of a shaft / % saving in material:
% saving in material =
  100
2
2
1
2
1
2
x
D
d
D
D 


Where, D1 and d1 are external and internal diameter of hollow shaft
D is the dia. of solid shaft

Unit -V
Principal Stresses and Strains
Principal stress:is defined as when any load is applied to a body, normal and shear
stresses are developed. Due to this, applied loading concentrated on a point where all
three planes X, Y, and Z, are perpendicular to that point. The resultant stress on these
planes is called principal stress.
The sustainability of principal stress is to check the allowable maximum force on a
body, and this force is designed according to the normal stress. All normal stress
centred at a location on a plane is defined as resultant principal stress. For any
structural design, we always consider the design stress. Hence to design the structure,
we need to know the maximum or minimum stress this is the sustainability of
principal stresses.
 Principal stress is the maximum or minimum normal stress which may be
developed on a loaded body. It is classified as major principal stress and minor
principal stress. On the plane of principal stress shear stress value is termed as
zero.
 When principal stress gets on the major principal plane, it is called major
principal stress and when it is found on the minor principal plane, it is known as
minor principal stress. These are denoted as σ1 and σ2, respectively.
Principal Stress Theory
Shear stress is a sloping applied force that causes deformation due to lateral load. In
any stress block surface, there is applied shear stress over the plane, but to stabilize
the body there, we need to apply complementary shear stress. This is known as
complementary shear stress. Mock Test
The principle of complementary shear stress is when the same intensity but opposite
direction shear stress is applied over the surface of the body, it creates a couple and
this couple stabilizes the body. This balancing couple mechanism is known as the
principle of complementary shear stress.
Maximum and Minimum Principal Stress
Principal stress is calculated on the principal plane as maximum stress is called major
principal stress, and minimum stress on the principal plane is called minor principal
stress.
Here;
 σx = Stress in x direction
 σY = Stress in the Y direction
 τn = Normal shear stress
 Ө = inclination angle of stress to the principal axis
 X and Y are the axes of the plane.
2
2
2
2
2
1
2
2
2
2
xy
xy
y
x
y
x
p
y
x
y
x
p




















 











 



Where
 σp1 is the maximum principal stress, and
 σp2 is the minimum principal stress. it is also termed as σMax and σMin.

Maximum shear stress (τmax) = (Maximum principal stress- Minor principal
stress)/2
2
2
2
1
2
2
max
p
p
y
x
xy
















 

The maximum shear stress is the magnitude of that point with an equal radius in a
mohr circle, and principal stress is the end point of the circle's diameter.
Here we take only two-dimensional theory in which we consider σ1 ( the maximum
principal stress ) and σ2 (the minimum principal stress). But in the case of three-
dimensional theory, we take σ1, σ2 and σ3. In this case, we decide on the major or
minor principal stress by using principal stress theory or Rankine, lames or maximum
principal stress theory.
Principle Plane: 90
2
2
tan 1
2
1 


 




 and
y
x
xy
Plane carry maximum shear/tangential stress: 45
45 3
4
1
3 


 


 and
Stresses along the Oblique Plane:
Other than the principal plane On the arbitrarily chosen oblique plane, there will
be normal stress n and shear or tangential stress . The stresses change with the
inclination of the planes passing through that point i.e. the stress on the faces of the
element vary as the angular position of the element changes. These stresses along the
oblique plane are determined as,














2
cos
2
sin
2
2
sin
2
cos
2
2
xy
y
x
xy
y
x
y
x
n








Where, n= normal stress
=shear/tangential stress
θ is w.r.t. horizontal axis
Normal stresses are tensile (positive) and shear stresses are clockwise positive

Combined Bending and Torsion:
Such a Cases arise for ex. in propeller shafts of ships where a shaft is subjected to
direct thrust in addition to bending moment and torsion. In such cases the direct
stresses due to bending moment and the axial thrust have to be combined into a single
resultant.
For Solid Shaft-
Shear stress due to torsion: 3
16
D
T

 
Bending stress due to bending: 3
32
D
M

 
Maximum Principal stress: 2
2
1
2
2



 








p
Minimum Principal stress: 2
2
2
2
2



 








p

Unit -VI
Axially Loaded Column
Introduction: A column is a vertical member subjected to either axial loading or ecce
ntric loading.
An axially loaded column may undergo two different kinds of deformations
depending upon the height of the column; it’s cross sectional area and the load to
which it is subjected. These deformations are as follows
The column may get compressed, eventually leading to crushing of the material as
the load increases beyond a certain point.
The column may bend outwardly leading to a condition called buckling and finally
may fail due to this buckling.
The failure of an axial compression member is dependent of three variables
1. Axial Compressive load applied
2. Cross section of the member
3. Height of the member/ (Effective length – end conditions, to be discussed later)
Accordingly a column is classified as follows
1. Short column – one which fails by crushing.
2. Long column – one which fails by buckling.
Euler’s Theory
Assumptions in Euler’s Theory:
· 1. The material of the column is Isotropic and Homogenous
· 2. The cross section is uniform throughout the length of the column.
· 3. The load is placed axially and the column is straight when the load is placed.
· 4. The column is long and is going to fail by buckling only
· 5. The stresses are within the elastic limit.
2
2
Le
EI
PE

 for both ends of column hinged
Where, Pe - is the Euler’s crippling load
E - is the modulus of elasticity
I - is the minimum (Ixx or Iyy) moment of inertia
Le -is the effective length

Limitations Of Euler’s Theory
If K is radius of Gyration (the distance at which the area of a plane lamina can be
safely assumed to be concentrated so that the Moment of Inertia of the area about its
own c.g. is the same as the double moment of area about that point. So =
Euler’s theory is limited to long columns. This Limitation of Euler’s Theory is
overcome by another theory known as Rankine’s Theory and is discussed later on in
this chapter.Greater the Slenderness Ratio Lesser the Buckling Load and Lesser the
Slenderness Ratio Greater the Buckling Load
Effective Length of a Column Member
1. Both ends Fixed Le=L/2
2. One end Fixed and one end Hinged Le=L/√2
3. Both ends Hinged Le=L
4. One end Fixed and One end Free Le=2L
These conditions and their implications can be seen in the diagrams given below.
Slenderness Ratio and Material:
As explained earlier the ratio of the effective length of a column to the least radius of
gyration of its cross section is called the slenderness ratio
1. A short steel column is one whose slenderness ratio does not exceed 50; an interme
diate length steel column has a slenderness ratio ranging from about 50 to 200, and ar
e dominated by the strength limit of the material, while a long steel column may be as
sumed to have a slenderness ratio greater than 200 and its behavior is dominated by th
e modulus of elasticity of the material.
2. A short concrete column is one having a ratio of unsupported length to least dimens
ion of the cross section equal to or less than 12. If the ratio is greater than 12, it is con
sidered a long column (sometimes referred to as a slender column).
3. Timber columns may be classified as short columns if the ratio of the length to least
dimension of the cross section is equal to or less than 10. The dividing line between i
ntermediate and long timber columns cannot be readily evaluated

Rankine’s Theory
It has been shown that Euler’s formula is valid for long column having l/k ratio
greater than a certain value for a particular material (97 in case of steel). Euler’s
formula does not give a reliable result for short column and length of column
intermediate between very long to short. An empirical formula has been proposed by
Rankine for columns of all lengths. The proposed formula is
C
E
R P
P
P
1
1
1


Where PR =Rankine’s load, PC = Crushing load = σc × A,
PE = Euler’s load = ( 2
2
Le
EI
PE

 )
 For very Short columns, PE is very large compared to PC so 1/ PE is very small,
thus P = PC,
 For very Long Columns PC is very large compared to PE, SO 1/PC is very small,
thus P = PE
2
1 







K
Le
A
P c
R


Eccentrically Loaded Column:
Eccentrically loaded columns are subjected to moment, in addition to axial force. The
moment can be converted to a load P and eccentricity e. The moment can be uniaxial,
as in the case when two adjacent panels are not similarly loaded, such as columns
along the edge and middle of floor plan. A column is considered biaxially loaded
when the bending occurs about the X and Y axes, such as in the case of the corner
column.
In case of eccentric loading, there will be produced direct stress as well as bending
stress in the column.
Direct Stress:
A
P

0

Bending Stress: y
I
M
b 

P -is the Load
M- Bending Moment =M X e
e - is eccentricity
b


 
 0
If b

 
0 stresses throughout the section will be same i. e compressive
b

 
0 stresses throughout the section will be partly tensile and partly
compressive
b

 
0 this is the ideal condition which designer tries to achieve.

Core or Kernal of Section
The centrally located portion of a within which the load must act so as to produce
only compressive stress and no tensile stress is called a core of the section.

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