GMR Institute of Technology

Rajam, Andhra Pradesh

(An Autonomous Institution Affiliated to JNTUK, AP)

Cohesive Teaching – Learning Practices (CTLP)

Class 3rdSem. – B. Tech. (CIVIL) Department: CIVIL
Course Solid Mechanics I Course Code 20CE305
Prepared by Mr. Bimalendu Dash, Mr.B.P.R.V.S Priyatham
Lecture Topic Types of stresses and strains - Stress – strain diagram for mild steel, Hooke’s law
Course Outcome (s) CCE305.4 Program Outcome (s) PO1, PO 12
Duration 50 Min Lecture 23 of 45 Unit III
Pre-requisite (s) ENGINEERING MECHANICS

At the end of this session the students will able to:

A. Classify the Normal and Shear stress

B. Classify the Normal and Shear strain
C. Show the salient features in stress strain diagram of mild steel
D. Identify the sitautions in daily life related to normal and shear stress

1. 2D Mapping of ILOs with Knowledge Dimension and Cognitive Learning Levels of

RBT

Cognitive Learning Levels

Knowledge
Remember Understand Apply Analyse Evaluate Create
Dimension
Factual
Conceptual A, B, C D
Procedural
Meta Cognitive

2. Teaching Methodology

❖ Chalk & Board, Visual Presentation

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
3. Evocation

6.Deliverables

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. The
externally applied forces are termed as loads. These externally applied forces may be due to
any one of the reason.

(i) Due to service conditions

(ii) Due to the environment in which the component works
(iii) Through contact with other members
(iv) Due to fluid pressures
(v) Due to gravity or inertia 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.
Therefore, let us define a term stress

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
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.
𝑃
𝜎=
𝐴

Where A is the area of the X – section

Here we are using an assumption that the total force or total load carried by the rectangular
bar is uniformly distributed over its cross – section.
But the stress distributions may be for from uniform, with local regions of high stress known
as stress concentrations.

As a particular stress generally holds true only at a point, therefore it is defined

mathematically as

If the force carried by a component is not uniformly distributed over its cross – sectional area,
A, we must consider a small area, ‘dA' which carries a small load ‘dP’, of the total force ‘P',
Then definition of stress is
𝛿𝑃
𝜎=
𝛿𝐴
As a particular stress generally holds true only at a point, therefore it is defined
mathematically as

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
 𝛿𝑃
𝜎 = lim
𝛿𝐴→0 𝛿𝐴
Units :
The basic units of stress in S.I units i.e. (International system) are N / m2 (or Pa)
MPa = 106 Pa
GPa = 109 Pa
KPa = 103 Pa
Some times N / mm2 units are also used, because this is an equivalent to MPa. While US
customary unit is pound per square inch psi.

TYPES OF STRESSES :

only two basic stresses exists : (1) normal stress and (2) shear shear stress. Other stresses either are
similar to these basic stresses or are a combination of these e.g. bending stress is a combination
tensile, compressive and shear stresses. Torsional stress, as encountered in twisting of a shaft is a
shearing stress.
Let us define the normal stresses and shear stresses in the following sections.
Normal stresses : We have defined stress as force per unit area. If the stresses are normal to the areas
concerned, then these are termed as normal stresses. The normal stresses are generally denoted by a
Greek letter ( )

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 :

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
Tensile or compressive stresses :
The normal stresses can be either tensile or compressive whether the stresses acts out of the area or
into the area

Bearing Stress : When one object presses against another, it is referred to a bearing stress ( They are in
fact the compressive stresses).

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
Shear stresses :
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 resulting force intensities are known as shear stresses, the mean shear stress being equal to

𝛿𝑃
𝜏=
𝛿𝐴

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

𝛿𝑃
𝜏 = lim
𝛿𝐴→0 𝛿𝐴

The greek symbol τ ( tau ) ( suggesting tangential ) is used to denote shear stress.
However, it must be borne in mind that the stress ( resultant stress ) at any point in a body is basically
resolved into two components, one acts perpendicular and the other parallel to the area concerned, as
it is clearly defined in the following figure.11

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
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.

CONCEPT OF STRAIN
Concept of 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 L, the strain produce is
defined as follows:

𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝛿𝑙
𝑆𝑡𝑎𝑟𝑖𝑛 = =
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑙

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
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
Since in practice, the extensions of materials under load are very very small, it is often
convenient to measure the strain in the form of strain x 10-6 i.e. micro strain, when the symbol
used becomes .
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 now let us define
the shear strain.
Definition: An element which is subjected to a shear stress experiences a deformation as
shown in the figure below. The tangent of the angle through which two adjacent sides rotate
relative to their initial position is termed shear strain. In many cases the angle is very small
and the angle it self is used, ( in radians), instead of tangent, so that  = ∠AOB - ∠A'OB' = 
Shear strain: As we know that the shear stresses acts along the surface. The action of the
stresses is to produce or being about the deformation in the body consider the distortion
produced on an element or rectangular block

This shear strain or slide is ‘’ and can be defined as the change in right angle. or The angle of
deformation ‘’ is then termed as the shear strain. Shear strain is measured in radians & hence is non –
dimensional, i.e. it has no unit.So we have two types of strain i.e. normal stress & shear stresses.
Uniaxial Tension Test: This test is of static type, i.e. the load is increased comparatively slowly from
zero to a certain value. Standard specimens are used for the tension test.

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
There are two types of standard specimen's which are generally used for this purpose, which have
been shown below
Specimen I:
This specimen utilizes a circular X-section.

Specimen II:
This specimen utilizes a rectangular X-section.

lg = gauge length i.e. length of the specimen on which we want to determine the mechanical
properties.The uniaxial tension test is carried out on tensile testing machine and the following steps
are performed to conduct this test.
(i) The ends of the specimen's are secured in the grips of the testing machine.
(ii) There is a unit for applying a load to the specimen with a hydraulic or mechanical drive.
(iii) There must be a some recording device by which you should be able to measure the final output in
the form of Load or stress. So the testing machines are often equipped with the pendulum type lever,
pressure gauge and hydraulic capsule and the stress Vs strain diagram is plotted which has the
following shape.
A typical tensile test curve for the mild steel has been shown below.
Nominal stress – Strain OR Conventional Stress – Strain diagrams:
Stresses are usually computed on the basis of the original area of the specimen; such stresses are often
referred to as conventional or nominal stresses.

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
True stress – Strain Diagram:
Since when a material is subjected to a uniaxial load, some contraction or expansion always takes
place. Thus, dividing the applied force by the corresponding actual area of the specimen at the same
instant gives the so called true stress.

SALIENT POINTS OF THE GRAPH:

(A) So it is evident from the graph that the stress is proportional to strain (or) elongation is
proportional to the load giving a st. line relationship. This law of proportionality is valid upto a point A.
Or we can say that point A is some ultimate point when the linear nature of the graph ceases or there is
a deviation from the linear nature. This point is known as the limit of proportionality or the
proportionality limit.

Initial part of the loading indicates a linear relationship between stress and strain, and the
deformation is completely recoverable in this region for both ductile and brittle materials.

This linear relationship, i.e., stress is directly proportional to strain, is popularly known as
Hooke's law.

Eσ =ε
(B) For a short period beyond the point A, the material may still be elastic in the sense that the
deformations are completely recovered when the load is removed. The limiting point B is termed as
Elastic Limit.
(C) and (D) - Beyond the elastic limit plastic deformation occurs and strains are not totally
recoverable. There will be thus permanent deformation or permanent set when load is removed.

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
These two points are termed as upper and lower yield points respectively. The stress at the yield point
is called the yield strength.
A study a stress – strain diagrams shows that the yield point is so near the proportional limit that for
most purpose the two may be taken as one. However, it is much easier to locate the former. For
material which do not possess a well define yield points, In order to find the yield point or yield
strength, an offset method is applied.
In this method a line is drawn parallel to the straight line portion of initial stress diagram by off setting
this by an amount equal to 0.2% of the strain as shown as below and this happens especially for the
low carbon steel.

(E) A further increase in the load will cause marked deformation in the whole volume of the metal. The
maximum load which the specimen can withstand without failure is called the load at the ultimate
strength.
The highest point ‘E' of the diagram corresponds to the ultimate strength of a material.
σu = Stress which the specimen can withstand without failure & is known as Ultimate Strength or
Tensile Strength.
σu is equal to load at E divided by the original cross-sectional area of the bar.
(F) Beyond point E, the bar begins to formthe neck. The load falling from the maximum until fracture
occurs at F.
[Beyond point E, the cross-sectional area of the specimen begins to reduce rapidly over a relatively
small length of the bar and the bar is said to form a neck. This necking takes place whilst the load
reduces, and fracture of the bar finally occurs at point F]
Note: Owing to large reduction in area produced by the necking process the actual stress at fracture is
often greater than the above value. Since the designers are interested in maximum loads which can be
carried by the complete cross section, hence the stress at fracture is seldom of any practical value.

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
Consider a rod under an axial tensile load P as shown in figure 1.6 such that the material is
within the elastic limit. The normal stress on x plane is xx PA σ = and the associated
longitudinal strain in the x direction can be found out from xx x E σ ε = . As the material
elongates in the x direction due to the load P, it also contracts in the other two mutually
perpendicular directions, i.e., y and z directions.

Hence, despite the absence of normal stresses in y and z directions, strains do exist in those
directions and they are called lateral strains.

The ratio between the lateral strain and the axial/longitudinal strain for a given material is
always a constant within the elastic limit and this constant is referred to as Poisson's ratio. It is
denoted by . ν
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑎𝑖𝑛
ν=
𝑎𝑥𝑖𝑎𝑙 / 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
Since the axial and lateral strains are opposite in sign, a negative sign is introduced in the above
equation to make ν positive.

Using equation, the lateral strain in the material can be obtained by

𝜎
y = z = - ν.x = - ν 𝐸𝑥

Poisson's ratio can be as low as 0.1 for concrete and as high as 0.5 for rubber.
In general, it varies from 0.25 to 0.35 and for steel it is about 0.3.

7.Keywords

❖ Stress
❖ Strain
❖ Normal stress
❖ Shear stress
8.Sample Questions

Remember
1. Define normal and shear stress
2. Define normal and shear strain
3. What are the units of staess and strain

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
 Understand

1. Differentiate between tensile and compressive stress.

2. Contrast between normal and shear stress.
3. Illustrate the salient features of stress- strain diagram
Apply
1. Identify the examples of normal stress and shear stress in daily life?
2. Identify the examples of normal stress and shear strain in daily life?

9. Stimulating Question (s)

1. A spring balance of weight W= 40 N is hanged from ceiling having an area of 4 m 2 find the
stress applied to the ceiling? Which kind of stress is that? Find the magnitude of it.

10. Mind Map

Fig.Mind map of types of stresses

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh
 Fig.Mind map of types of strains

11. Student Summary

At the end of this session, the facilitator (Teacher) shall randomly pic-up few students to
summarize the deliverables

12. Reading Materials

❖ https://www.scribd.com/document/299256556/Chapter-1-Stress-and-strain-pdf
❖ http://www.nprcet.org/mech/document/SM.pdf
13.Scope for Mini Project

❖ NA

Form No. AC 04. 01.2016 (R: 06.10.2016)– GMRIT, Rajam, Andhra Pradesh