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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



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



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



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



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



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



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.



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



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



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



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:



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



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.



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.



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



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



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



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.



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.



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).



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



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



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



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



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 𝑡𝑕𝑖𝑠 𝑐𝑎𝑠𝑒

𝜎𝑖𝑛𝑑𝑢𝑐𝑒𝑑 ≤ 𝜎𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒

𝑁𝑜 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑡𝑕𝑖𝑠 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



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



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



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)



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



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



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.



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



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



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



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



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



Hence,

Bending Formula ,

M f E

I Y R

𝜎𝑚𝑎𝑥 =𝑃

𝐴+

𝑀𝑦𝑚𝑎𝑥

𝐼𝑥𝑥

𝜎𝑚𝑎𝑥 =𝑃

𝐴+

𝑃𝑒𝑦𝑚𝑎𝑥

𝐼𝑥𝑥

Similarly,

𝜎𝑚𝑖𝑛 =𝑃

𝐴−


𝐼𝑥𝑥

In case of compressive loads,

𝜎𝑚𝑎𝑥 = −𝑃

𝐴−


𝐼𝑥𝑥

𝜎𝑚𝑖𝑛 = −𝑃

𝐴+


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



Bending stress,

At distance „a‟ from free end


𝑍𝑥𝑥=

𝑃. 𝑎

𝑍𝑥𝑥

At a distance „l’ from free end


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



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)



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



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



True Strain (εt) :

0

0

0

0 0

0

0

0

0

.

ln ln 2ln

ln

ln ln(1 )

ln(1 ) .

L

t

L

t

t

t

LEngg strain

L

dL

l

L Final Length

L Initial Length

A dLor

L A d

L

L

as L L L

L L

L

Engg strain

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