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STRENGTH OF MATERIALS
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STRENGTH OF MATERIAL
0
INTRODUCTION
6
1 L0ADS
Introduction
9
On the basis of time of action of load
On the basis of direction of load
On the basis of area of acting the load
Couple
Pure Bending torsion
Free Body Diagram
2
STRESSES
Introduction
22
Strength
Classification of stresses
Normal stresses
Shear stress
Stress tensor
Effects of various loads acting on the body
3
STRAINS
Introduction
44
Classification of strain
Normal strain
Longitudinal and lateral strain
Volumetric strain
Shear strains
Sign conventions for shear strains
4
ELASTIC
CONSTANTS
Properties of materials
55
Young‟s modulus of elasticity
Modulus of rigidity
Bulk modulus
Poisson‟s ratio
State of simple shear
Relationships between various constants
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5
MECHANICAL
PROPERTIES OF
MATERIALS
Stress and strain diagram
73
Limit of proportionality
Elasticity
Plasticity
Ductility, Brittleness, Malleability
Yield strength, ultimate strength, rapture strength
6
STRAIN
ENERGY, RESIELENCE
AND
TOUGHNESS
Work done by load
81
Strain energy due to torsion
Strain energy due to bending
Resilience
Toughness
Effect of carbon percentage on properties
7
NORMAL
STRESSES AND STRAIN
Principle of superposition
98
Elongation of bar due to axial load
Bar of varying cross-section
Uniformly tapering circular bar
Uniformly tapering rectangular bar
Elongation of bar of uniform cross-section
due to self- weight
Compound bars or parallel bars
Statically indeterminate problems
8
THERMAL STRESSES
Thermal effects
108
Free expansion of bar
Temperature stresses in bar fixed at the both ends
Temperature stresses in composite bars
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9 PURE BENDING AND BENDING STRESSES
Pure bending
129
Theory of simple bending
Moment of resistance
Bending equation
Assumptions
Design criteria
Analysis of bending equation
10 SHEAR
STRESSES IN BEAMS
Distribution of shear stresses
144
Assumptions
Shear stress distribution –rectangular section
Circular section
I-section
11 TORSION
Pure torsion
151
Moment of resistance
Torsion equation
Assumptions
Shear stress distribution in shafts
Analysis of torque equation
Compound shafts
12 PRINCIPAL
STRESSES AND STRAINS
Introduction
167
Stresses on inclined section pq
State of stress at a point due to biaxial stress
State of stress due to simple shear
State of stress due to normal and shear stress
Normal and shear stress on a plane perpendicular to oblique plane
Mohr Circle
13 S.F.D AND B.M.D
Type of support
192
Types of beams
Sign conventions
SFD and BMD
Relationship b/w load, force and B.M.
Cantilever beam
Simply supported beams
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14 THIN CYLINDER
Thin cylindrical shell subjected to internal
pressure
244
Maximum shear stress in cylindrical shell
Volumetric strain of thin cylindrical shell
Design of thin cylinder
Thin spherical shells subjected to internal pressure
Volumetric strain in spherical shell
Cylindrical shell with hemispherical ends
15 THICK CYLINDER
Lame‟s theory
261
When only external pressure is zero
When only internal pressure is zero
When internal pressure is pr and external
pressure pR
For solid circular shaft, subjected only to
external pressure pr
Graphical representation of lame‟s theory
Compound cylinders
Shrinking another cylinder over the cylinder
Shrink fit allowance
Thick spherical shell
16 DEFLECTION OF BEAMS
Differential equation of the deflection curve of
beam
268
Sign conventions
Double integration method
Steps for solving the problems
Macaulay‟s method
Moment area method
Conjugate beam method
Strain energy method
17 SPRINGS
Introduction
307
Close coiled helical spring: axial pull
Closed – coiled helical springs: axial couple
or torque
Open – coiled helical spring: axial force
Open coiled helical spring: axial torque
Series and parallel arrangement of springs
Leaf or carriage springs Flat spiral springs
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18 COLUMNS
Introduction
314
Equilibrium of elastic body
Buckling stress
Slenderess ratio
Euler‟s theory
End conditions
Rankine theory
19 THEORIES OF ELASTIC FAILURE
Introduction
332
Maximum principal stress theory : Rankines‟s theory
Maximum principal strain theory: St.
Venant‟s theory
Maximum shear stress theory: guest‟s
theory
Maximum strain energy theory or haigh‟s
theory
Maximum shear strain energy (or distortion
energy) theory Mises- Henky theory
20 TESTS
Tests
352
Answer Key
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INTRODUCTION
There are three fundamental areas of engineering mechanics:
i. Statics
ii. Dynamics
iii. Strength of materials or mechanics of materials
Statics and dynamics deals with the effect of forces on rigid bodies
i.e. the bodies in which change in shape can be neglected.
Strength of material deals with the relation between externally
applied loads and their internal effects on sold bodies.
“Strength of material is a branch of applied mechanics that deals
with the behaviour of solid bodies subjected to various types of
loading.”
The principal objective of strength of materials is to determine the
stresses, strains, and displacements in structures and their
components due to the loads acting on them.
An understanding of mechanical behaviour is essential for the safe
design of all types of structures, whether airplanes and antennas,
buildings and bridges, machines and motors, or ships and spacecraft.
In designing, engineer must consider both dimensions and material
properties to satisfy the requirements of strength and rigidity.
Mechanics of Rigid Bodies
The mechanics of rigid bodies is primarily concerned with the static and
dynamic behavior under external forces of engineering components and
systems which are treated as infinitely strong and undeformable.
Primarily, we deal here with the forces and motions associated with
particles and rigid bodies.
A basic requirement for the study of the mechanics of deformable bodies
and the mechanics of fluids is essential for the design and analysis of
many types of structural members, mechanical components, electrical
devices, etc, encountered in engineering.
A rigid body does not deform under load
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Mechanics of deformable solids
Mechanics of Solids
The mechanics of deformable solids is more concerned with the internal
forces and associated changes in the geometry of the components
involved. Of particular importance are the properties of the materials
used, the strength of which will determine whether the components fail
by breaking in service, and the stiffness of which will determine whether
the amount of deformation they suffer is acceptable.
Therefore, the subject of mechanics of materials or strength of
materials is central to the whole activity of engineering design. Usually
the objectives in analysis here will be the determination of the stresses,
strains, and deflections produced by loads. Theoretical analyses and
experimental results have an equal role in this field.
In short, Mechanics of Solids deals with the relation between the loads applied
to a solid (non-rigid) body and the resulting internal forces and deformations
induced in the body.
Principle Objective = determine the stresses, strains, and displacements in
structures and their components due to loads acting on them.
Alternate Names = Strength of Materials or Mechanics of Deformable Bodies
These notes will provide a basis to determine:
The materials to be used in constructing a machine or structure to perform
a given function.
The optimal sizes and proportions of various elements of a machine or
structure.
If a given design is adequate and economical.
The actual load carrying capacity of a structure or machine. (structure may
have been design for a purpose other than one being considered).
Guru Gyan
Mass is a property of matter that does not change from one location to another.
Weight refers to the gravitational attraction of the earth on a body or quantity of mass.
Its magnitude depends upon the elevation at which the mass is located Weight of a
body is the gravitational force acting on it.
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CHAPTER 1
LOADS INTRODUCTION
Load may be defined as the external force or couple to which a
component is subjected during its functioning.
Load is a vector quantity.
All the external forces acting on bodies are SURFACE forces.
Externally applied forces may be due to
Working environment
Service conditions
Contact with other members
Fluid pressure
Gravity or inertia forces
The forces acting on the body due to volume of the body is called BODY
force.
Loads may be classified on following basis:
On the basis of time
On basis of direction of load
On the basis of area
ON THE BASIS OF TIME OF ACTION OF LOAD
On the basis of time of action of load, load may be classified as
Static load
Dynamic load
STATIC LOAD may be
Dead load
Gradually applied load
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DEAD LOAD
Dead load includes loads that are relatively constant over time,
including the weight of the structure itself.
GRADUALLY APPLIED LOAD
Gradually applied load may be defined as the load whose magnitude
increases gradually with the time.
Gradually applied loads become dead load after a certain period of time.
FIGURE 1.1
DYNAMIC LOAD may be
Impact
Fatigue
IMPACT LOAD
The load which are acting for short interval time are said to be impact
load.
If
t = time of application of load
T = time period of vibration.
Then
𝑡 <𝑇
2 𝑖𝑚𝑝𝑎𝑐𝑡 𝑙𝑜𝑎𝑑
𝑡 ≥ 3𝑇 𝑠𝑡𝑎𝑡𝑖𝑐 𝑙𝑜𝑎𝑑
𝑡 = 2𝑇 𝑎𝑠𝑠𝑢𝑚𝑒𝑑 𝑖𝑚𝑝𝑎𝑐𝑡 𝑙𝑜𝑎𝑑
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Since stress produced is greater than the static load.
𝜎𝑖𝑚𝑝𝑎𝑐𝑡 > 𝜎𝑠𝑡𝑎𝑡𝑖𝑐
𝜎𝑖𝑚𝑝𝑎𝑐𝑡 = 𝜎𝑠𝑡𝑎𝑡𝑖𝑐 × 𝑖𝑚𝑝𝑎𝑐𝑡 𝑓𝑎𝑐𝑡𝑜𝑟
FIGURE 1.2
𝒊𝒎𝒑𝒂𝒄𝒕 𝒇𝒂𝒄𝒕𝒐𝒓 = 𝟏 + 𝟏 +𝟐𝒉
𝜹𝒔𝒕
Where
𝜎𝑖𝑚𝑝𝑎𝑐𝑡 = Stress due to impact loads
𝜎𝑠𝑡𝑎𝑡𝑖𝑐 = Stress due to static loads
𝛿𝑠𝑡 = Static deflection
FATIGUE LOADS
Fatigue load are those load whose magnitude or direction or both
magnitude and direction change with respective to time.
FIGURE 1.3
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ON THE BASIS OF DIRECTION OF LOAD
On the basis of direction of load, load may be classified as
Normal loads
Shear loads
NORMAL LOADS
The loads which are acting perpendicular to the surface of the body.
Normal loads may be
Axial
Eccentric axial
AXIAL LOADS
The loads which are perpendicular to the surface and passes through
the longitudinal axis of the body are called axial loads.
Figure 1.4
ECCENTRIC AXIAL LOAD
The loads which are perpendicular to the surface and does not pass
through the axis of the body.
Note:
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Figure 1.5
Note: Eccentric Load creates → Axial Load + Bending Moment
SHEAR LOADS
The loads which acts parallel to the surface of body.
Shear loads may be
Transverse shear load
Eccentric transverse shear load
TRANSVERSE SHEAR LOAD
The loads which are parallel to the surface and passes through the
longitudinal axis of the body are called transverse shear loads.
Figure 1.6
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ECCENTRIC TRANSVERSE SHEAR LOAD
The loads which are parallel to the surface and do not pass through the
longitudinal axis of the body are called transverse shear loads.
Figure 1.7
ON THE BASIS OF AREA OF ACTING THE LOAD
On the basis of area of acting the load, load may be classified as
Concentrated point load
Uniformly distributed load
Uniform variable load
Uniformly distributed moment
Uniform variable moment
CONCENTRATED POINT LOAD
Figure 1.8
A point load is one which is concentrated at a point.
Self-weight of the body is assumed to be concentrated point load acting
through the centre of gravity.
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UNIFORMLY DISTRIBUTED LOAD
Figure 1.9
The load which is not acting through a point but is distributed
uniformly over some area.
UNIFORM VARIABLE LOAD
Figure 1.10
A uniformly varying load is one in which load intensity varies from one
end to other.
They are also called linearly varying load.
UNIFORMLY DISTRIBUTED MOMENT
It is the moment which is not acting at a point but uniformly
distributed over some area.
UNIFORM VARIABLE MOMENT
A uniformly varying moment is one in which moment intensity varies
from one end to other.
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COUPLE
A couple consists of two parallel forces that are equal in magnitude
and opposite in direction.
Figure 1.11
BENDING COUPLE
A couple is said to be bending couple if plane of couple is passing
through the longitudinal axis of the body and perpendicular to the
plane of cross-section.
Figure 1.12
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TWISTING COUPLE
A couple is said to be twisting couple when plane of the couple is
perpendicular to longitudinal axis of the member and parallel to the
cross-section of plane .
Figure 1.13
OBLIQUE COUPLE
When member is subjected to bending and twisting couple when plane
of couple acting in a plane inclined to the axis and plane is known as
oblique plane.
𝑀𝑥 = 𝑀𝑐𝑜𝑠𝛼 𝐵𝑒𝑛𝑑𝑖𝑛𝑔
𝑀𝑦 = 𝑀𝑐𝑜𝑠𝛽 𝐵𝑒𝑛𝑑𝑖𝑛𝑔
𝑀𝑧 = 𝑀𝑐𝑜𝑠𝛾 𝑡𝑤𝑖𝑠𝑡𝑖𝑛𝑔
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Figure 1.14
PURE BENDING
A member is said to be under pure bending when is subjected to two
equal and opposite couple in a plane passing through the longitudinal
axis of member and perpendicular to cross-section.
Figure 1.15
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PURE TORSION
A member is said to be in pure torsion couple in the plane
perpendicular long axis of member and parallel to cross section of plane
Figure 1.16
FREE BODY DIAGRAM
A diagram showing all the forces acting directly or indirectly on the
body that completely balance each other
Steps involved for drawing a free body diagram
Step I.
At the beginning, a clear decision is to be made by the analyst on the
choice of the body to be considered for free body diagram.
Step II.
Then that body is detached from all of its surrounding members including
ground and only their forces on the free body are represented.
Step III
The weight of the body and other external body forces like centrifugal,
inertia, etc., should also be included in the diagram and they are assumed
to act at the centre of gravity of the body.
Step IV
When a structure involving many elements is considered for free body
diagram, the forces acting in between the elements should not be brought
into the diagram.
Step V
The known forces acting on the body should be represented with proper
magnitude and direction.
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Step VI
If the direction of unknown forces like reactions can be decided, they
should be indicated clearly in the diagram.
Figure 1.17
SOLVED EXAMPLE 1.1
Consider the following figure. At point A, bar is rigidly fixed and at
point D, bar is loaded in negative x,y and z- direction with loads P,Q
and R respectively. Determine type and nature of load acting on each
bar.
SOLUTION:
On bar CD:
Load P is acting perpendicular to cross section hence is a direct
tensile load.
Load Q and R is acting parallel to surface, hence are transverse shear
loads.
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On bar BC:
Load P is acting parallel to surface, hence are transverse shear loads.
Load Q is acting perpendicular to surface but not passing through the
axis, hence is eccentric axial tensile load.
Load R is acting parallel to surface but not passing through the axis
hence is eccentric transverse shear loads.
On bar AB:
Load P is acting perpendicular to cross section but not passing
through the axis hence is a direct compressive load.
Load Q is acting parallel to surface and passing through the axis
hence is transverse shear load.
Load R is acting parallel to surface but not passing through the axis
hence is eccentric transverse shear loads
BAR AXIAL ECENTRIC
AXIAL
TRANSVERSE
SHEAR
ECCENTRIC
TRANS
V.
SHEAR
AB 0 -P Q R
BC 0 +Q P R
CD P 0 Q,R 0
Guru Gyan
Force: Magnitude (P), direction (arrow) and point of application (point A) is
important.
Change in any of the three specifications will alter the effect on any force.
In case of rigid bodies, line of action of force is important (not its point of
application if we are interested in only the resultant external effects of the
force).
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CHAPTER 2
STRESSES INTRODUCTION
Analysis of Stress and Strain
Concept of stress: Let us introduce the concept of stress as we know
that the main problem of engineering mechanics of material is the
investigation of the internal resistance of the body, i.e. the nature of
forces set up within a body to balance the effect of the externally applied
forces.
As we know that in mechanics of deformable solids, externally applied
forces acts on a body and body suffers a deformation. From equilibrium
point of view, this action should be opposed or reacted by internal forces
which are set up within the particles of material due to cohesion. These
internal forces give rise to a concept of stress.
Thus, we can say that when some external forces are applied on a body,
the body get deformed. For the equilibrium, this action must be
opposed or reacted by some internal forces which are set up in the body
due to cohesion. These internal forces give rise to the concept of
STRESS.
Thus, stress is the internal resistance offered by the body to
deformation when it is acted upon by the body some external force or
load.
A PRISMATIC BAR is a straight structural member having the same
cross section throughout its length.
Examples of prismatic bar are the members of a bridge truss,
connecting rod in automobile engines, spokes of bicycle wheels,
columns in buildings, and wing struts in small airplanes.
A section cut perpendicular to longitudinal axis of bar is called cross
section of the bar.
Let us consider a rectangular bar of some cross sectional area and
subjected to some load or force (in Newtons) Let us imagine that the
same rectangular bar is assumed to be cut into two halves at section XX.
The each portion of this rectangular bar is in equilibrium under the
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action of load P and the internal forces acting at the section XX has been
shown.
Now stress is defined as the force intensity or force per unit area. Here we use a symbol
to represent the stress.
𝜎 =𝑃
𝐴
Where 𝜎 represents the stress induced in the body.
This expression is valid only when the force is uniformly distributed
over the surface area.
Hence, Stress is also defined as the load applied per unit area of cross-
section.
If the force is not uniformly distributed, then we choose a small area δA
and a load 𝛿𝑃 acting on this area, then stress may be calculated as
𝜎 = lim𝛿𝐴→0
𝛿𝐹
𝛿𝐴
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Figure 2.1
𝑃 = 𝐹1 + 𝐹2 + 𝐹3 + − − − + 𝐹𝑛
If,
𝐹1 = 𝐹2 = 𝐹3 = − − −= 𝐹𝑛
𝑃 = 𝑛𝐹1
𝐹1 =𝑃
𝑛
𝜎1 =𝐹1
𝐴= 𝜎2 = 𝜎3 = − − − − −= 𝜎𝑛 = 𝜎𝑎𝑣𝑔
𝜎𝑎𝑣𝑔 =𝑃
𝐴
UNITS OF STRESS
Stress has the unit of force per unit area.
𝑆𝑡𝑟𝑒𝑠𝑠 = 𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎=
𝑁
𝑚2= 𝑃𝑎𝑠𝑐𝑎𝑙𝑠 (𝑃𝑎)
1 𝑁 𝑚𝑚2 = 106 𝑁 𝑚2 = 106𝑃𝑎 = 1𝑀𝑃𝑎
In MKS system, unit of force is kgf.
1 𝑘𝑔𝑓 = 9.807𝑁 ≈ 10 𝑁
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1 𝑡𝑜𝑛𝑛𝑒 = 9.807 𝑘𝑁 ≈ 10 𝑘𝑁
Hence unit of stress,
1𝑘𝑔
𝑐𝑚2=
105𝑁
𝑚2= 105𝑃𝑎 ≈ 0.1 𝑁/𝑚𝑚2
Saint Venant’s Principle – (used for gauge length in testing)
“It states that except in the region of extreme ends of a bar carrying
direct loading, the stress distribution over the cross section is uniform.”
This is also called Principal of Rapid distribution of localised stresses.
DIFFERENCE BETWEEN PRESSURE AND STRESS
Sr.
N
o
.
PRESSURE STRESS
1 It is always due to normal
force.
It may be due to normal and
shear force.
2 It is the force that has been
applied externally.
It is developed internally.
3 Pressure is a scalar
property.
Stress is a tensor or pointing
property.
4 It can be measured using
pressure gauge.
It cannot be measured.
5 Pressure can‟t be developed
due to stresses.
Stresses may be developed due
to pressure.
6. Example: Hydrostatic
Pressure
Example: Thermal Stresses
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Guru Gyan
Stress provides a measure of the intensity of an internal force.
Strain provides a measure of the intensity of a deformation.
STRENGTH
Strength is defined as the maximum or limiting value of stress at that a
material can with stand any failure or fracture.
It has the same unit that of stress.
Strength is a constant property for a given material.
𝜎𝑖𝑛𝑑𝑢𝑐𝑒𝑑 ≤ 𝜎𝑦𝑖𝑒𝑙𝑑𝑖𝑛𝑔
𝑁𝑜 𝑦𝑖𝑒𝑙𝑑𝑖𝑛𝑔 𝑜𝑐𝑐𝑢𝑟𝑠 𝑖𝑛 𝑡𝑖𝑠 𝑐𝑎𝑠𝑒
𝜎𝑖𝑛𝑑𝑢𝑐𝑒𝑑 ≤ 𝜎𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒
𝑁𝑜 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑡𝑖𝑠 𝑐𝑎𝑠𝑒
CLASSIFICATION OF STRESS
Stresses may be classified as
i. Normal stress or Direct stress
Tensile and compressive stresses
Axial and bending stresses
ii. Shear stresses
iii. Torsional stresses
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NORMAL STRESS
When a stress acts in a direction perpendicular to the cross-section, it
is called as normal or direct stress.
Normal stress occurs due to dimensional distortion.
Normal stress may be
Axial
Bending
AXIAL STRESS
The stress acting along the longitudinal axis of the bar which tend to
change the length of the body.
𝜎 =𝑃
𝐴
Stress
Normal
Axial Bending Tensile Compressive
Shear
Tensional Direct
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This is also known as uniaxial state of stress, because the stresses acts
only in one direction however, such a state rarely exists, therefore we
have biaxial and triaxial state of stresses where either the two mutually
perpendicular normal stresses acts or three mutually perpendicular
normal stresses acts as shown in the figures below :
Axial stress may be
Tensile
Compressive
Bending stress
When a sign convention for normal stresses is required, it is
customary to define tensile stresses as positive and compressive
stresses as negative.
TENSILE STRESS
When a body is stretched by a force, the resulting stress is known as
tensile stress.
Tensile stress exists between two parts when two parts draws each
other towards each other.
Figure 2.2
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COMPRESSIVE STRESS
When the forces are reversed and try to compress the body, the
resulting stress is known as compressive stress.
Compressive stress exists between two parts when they try to push
each other from it.
Figure 2.3
BENDING STRESS
Bending stresses are developed in beams due to bending of a member.
Loads perpendicular to the length of the beam causes bending moment.
The stresses produced at any section to resist the bending moment are
called bending stresses.
Figure 2.4
𝜎𝑏 =𝑀𝑥𝑥
𝑍𝑥𝑥
(Discussed in detail in chapter of bending stresses in beams)
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SHEAR STRESS
When the force is acting parallel or tangential to the surface, the
resulting stress is known as shear stress.
Shear stress exists between parts of a body when the two parts exerts
equal and opposite forces on each other laterally in a direction
tangential to their surface in contact.
Shear stress occurs due to shape deformation of body.
Shear stress may be
Direct Shear stress
Torsional shear stress
DIRECT SHEAR STRESS
Direct shear stress is developed in the body when the forces try to act
through the material directly.
Let us consider now the situation, where the cross sectional area of a block of material is subject to a distribution of forces which are parallel,
rather than normal, to the area concerned. Such forces are associated
with a shearing of the material, and are referred to as shear forces.
The resulting force intensities are known as shear stresses, the mean
shear stress being equal to
The Greek symbol τ (tau) (suggesting tangential) is used to denote shear stress.
Where P is the total force and A the area over which it acts.
As we know that the particular stress generally holds good only at a
point therefore we can define shear stress at a point as
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Figure 2.5
Shear stresses are also developed indirectly when a body is subjected to
bending and torsion.
TORSIONAL SHEAR STRESS
Torsional shear stress is developed in the body when a torque is applied
to body (due to twisting).
Figure 2.6
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Figure 2.7
(Discussed in detail in chapter of torsional stresses in beams)
However, it must be borne in mind that the stress (resultant stress) at any
point in a body is basically resolved into two components σ and τ one acts
perpendicular and other parallel to the area concerned, as it is clearly defined
in the following figure.
The single shear takes place on the single plane and the shear area is the cross
- sectional of the rivet, whereas the double shear takes place in the case of
Butt joints of rivets and the shear area is the twice of the X - sectional area of
the rivet.
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SIGN CONVENTION FOR SHEAR STRESS
Sign convention for shear stress a shear stress acing on positive face of
an element is positive if it acts in the positive directing of one of the co-
ordinate axes and is negative if it acts in the negative direction of an
axis.
Shear stress acting on a negative face of an element is positive if it acts
in the negative direction of an axis and negative if it acts in a positive
direction.
STRESS TENSOR
Stress tensor is used to define the state of stress (how many stress
components acting at a point.
Stress at a point in 3D is given by
𝜎 3𝐷 =
𝜎𝑥𝑥 𝜏𝑥𝑦 𝜏𝑥𝑧
𝜏𝑦𝑥 𝜎𝑦𝑦 𝜏𝑦𝑧
𝜏𝑧𝑥 𝜏𝑧𝑦 𝜎𝑧𝑧
The no. of stress component in a stress tensor for a point in 3D are
9 (3 normal and 6 shear).
Since complementary shear stress are equal and opposite
𝜏𝑥𝑦 = −𝜏𝑦𝑥
𝜏𝑥𝑧 = −𝜏𝑧𝑥
𝜏𝑦𝑧 = −𝜏𝑧𝑦
The number of stress components in a stress tensor to define the state
of stress at a point is 6 (3 normal and 3 shear )
PROB 1: How much quantity is sufficient to describe the stress at a point in a co-ordinate plane?
1] 3 2] 4 3] 6 4] 9
Correct Answer: 3
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The state of stress at a point
PROB 2: The steel and brass bars comprising the stepped shaft in figure are to suffer the same displacement under tensile force of 40 kN. The diameter of the brass bar is
(Take and )
1] 45.65 mm 2] 52.75 mm 3] 18.27 mm 4] 96.94 mm
Correct Answer: 4
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Figure 2.8
Plane Stress Problems
Stress tensor in 2D is given by
𝜎 2𝐷 = 𝜎𝑥𝑥 𝜏𝑥𝑦
𝜏𝑦𝑥 𝜎𝑦𝑦
The number of stress component in a stress tensor for a point in 2D are
4 (2 normal and 2 shear).
Since
𝜏𝑥𝑦 = −𝜏𝑦𝑥
The number of stress components in a stress tensor to define the state
of stress at a point is 3 (2 normal and 1 shear).
EFFECTS OF VARIOUS LOADS ACTING ON THE BODIES
EFFECT DUE TO AXIAL LOAD
Axial load causes elongation or compression of the member.
Tensile stress
Tensile stress causes elongation of the bar.
𝜎 =𝑃
𝐴
Change in length is given by
𝛿 =𝑃𝐿
𝐴𝐸
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Figure 2.9
Compressive Stress
Compressive stresses causes compression of the bar.
𝜎 = −𝑃
𝐴
Change in length is given by
𝛿 = −𝑃𝐿
𝐴𝐸
Figure 2.10
EFFECT OF ECCENTRIC AXIAL LOAD
Eccentric axial load is equivalent to axial load and constant bending
moment.
Shear force is zero in this case.
Hence resultant stress is given by
𝜎𝑟 = 𝜎𝑎𝑥𝑖𝑎𝑙 + 𝜎𝑏𝑒𝑛𝑑𝑖𝑛𝑔
𝜎𝑟 = 𝜎𝑎 + 𝜎𝑏
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𝜎𝑚𝑎𝑥 = 𝜎𝑎 + 𝜎𝑏 = 𝜎𝑡𝑜𝑝
𝜎𝑚𝑖𝑛 = 𝜎𝑎 − 𝜎𝑏 = 𝜎𝑏𝑜𝑡𝑡𝑜𝑚
Figure 2.11
Figure 2.12
In case of tensile loads,
𝜎𝑚𝑎𝑥 = 𝜎𝑎 + 𝜎𝑏
Bending stress will be maximum in the top fibre i.e. ymax
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Hence,
Bending Formula ,
M f E
I Y R
𝜎𝑚𝑎𝑥 =𝑃
𝐴+
𝑀𝑦𝑚𝑎𝑥
𝐼𝑥𝑥
𝜎𝑚𝑎𝑥 =𝑃
𝐴+
𝑃𝑒𝑦𝑚𝑎𝑥
𝐼𝑥𝑥
Similarly,
𝜎𝑚𝑖𝑛 =𝑃
𝐴−
𝑃𝑒𝑦𝑚𝑎𝑥
𝐼𝑥𝑥
In case of compressive loads,
𝜎𝑚𝑎𝑥 = −𝑃
𝐴−
𝑃𝑒𝑦𝑚𝑎𝑥
𝐼𝑥𝑥
𝜎𝑚𝑖𝑛 = −𝑃
𝐴+
𝑃𝑒𝑦𝑚𝑎𝑥
𝐼𝑥𝑥
EFFECT OF TRANSVERSE SHEAR LOAD
Transverse shear load is equivalent to a constant shear force and a
variable bending moment.
Direct shear stress,
𝜏 =𝑃
𝐴
Figure 2.13
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Bending stress,
At distance „a‟ from free end
𝜎𝑏 =𝑀𝑥𝑥
𝑍𝑥𝑥=
𝑃. 𝑎
𝑍𝑥𝑥
At a distance „l’ from free end
𝜎𝑏 =𝑀𝑥𝑥
𝑍𝑥𝑥=
𝑃. 𝑙
𝑍𝑥𝑥
In this case, bending moment will be varying along the length and shear
force will be constant throughout the length.
EFFECT OF ECCENTRIC TRANSVERSE SHEAR LOAD
Eccentric transverse shear load is equivalent to a direct shear, a
variable bending moment and a constant torque.
Figure 2.14
Direct shear stress,
𝜏 =𝑃
𝐴
Torsional Formula,
T f N
I Y L
Torsional stress,
𝜏𝑇 =𝑀𝑧𝑧
𝑍𝑧𝑧=
𝑃. 𝑒. 𝑦𝑚𝑎𝑥
𝐼𝑧𝑧
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Bending stresses,
𝜎𝑏 =𝑀𝑦𝑚𝑎𝑥
𝐼𝑥𝑥=
𝑃. 𝑎. 𝑦𝑚𝑎𝑥
𝐼𝑥𝑥
EXAMPLE 2.1
Determine the shear force, bending moment, torsional moment in the
following figure:-
Figure 2.15
On bar CD:
P is acting parallel to cross-section, hence a shear force P is acting.
A 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑃 × 𝐶𝐷 is induced at point C.
On bar BC:
Eccentric shear force P is acting on bar BC.
Due to eccentric P, a 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑃 × 𝐶𝐷 is induced.
Due to P, bending moment is induced at B, 𝐵𝑀 = 𝑃 × 𝐵𝐶.
On bar AB:
Shear force P is acting on bar AB at E.
Due to this P, 𝐵. 𝑀. = 𝑃 × (𝐴𝐵 − 𝐶𝐷) is induced at A.
𝑇𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑃 × 𝐵𝐶 is induced in bar.
BAR AXIAL
LOAD
SHEAR
FORCE
BENDING
MOMENT
TORSIONAL
MOMENT
CD 0 P P.(CD) 0
BC 0 P P.(BC) P.(CD)
AB 0 P P.(AB-CD) P.(BC)
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EXAMPLE 2.2
Determine the stresses acting at points A, B, C and D in the following
figure. The load P is acting out of plane and perpendicular to the plane.
Figure 2.16
SOLUTION:
At point A,
𝜎𝐴 =𝑃
𝐴−
𝑀𝑦𝑦
𝑍𝑦𝑦+
𝑀𝑥𝑥
𝑍𝑥𝑥
At point B,
𝜎𝐵 =𝑃
𝐴−
𝑀𝑦𝑦
𝑍𝑦𝑦−
𝑀𝑥𝑥
𝑍𝑥𝑥
At point C,
𝜎𝐶 =𝑃
𝐴+
𝑀𝑦𝑦
𝑍𝑦𝑦+
𝑀𝑥𝑥
𝑍𝑥𝑥
At point D,
𝜎𝐷 =𝑃
𝐴+
𝑀𝑦𝑦
𝑍𝑦𝑦−
𝑀𝑥𝑥
𝑍𝑥𝑥
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Concept of True Stress and strain:
True Stress (σt):
Original area (initial)
0
( )P
StressA
__________(1)
As with elongation, ceoss section area varies
Present area at that instant,
0
0
t
t
P
A
AP P
A A A
As volume remain same,
A0 L0 =A L
0
0
0 0 0
0
0
0
1
t
t
t
A L
A L
P L L
A L L
L L L
LL
L
Engg Stress